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The mathematician John is having trouble remembering his girlfriend Alicia's 7-digit phone number. He remembers that the first four digits consist of one 1 , one 2 , and two 3 s . He also remembers that the fifth digit is either a 4 or 5 . While he has no memory of the sixth digit, he remembers that the seventh digit is 9 minus the sixth digit. If this is all the information he has, how many phone numbers does he have to try if he is to make sure he dials the correct number?
|
There are $\frac{4!}{2!}=12$ possibilities for the first four digits. There are two possibilities for the fifth digit. There are 10 possibilities for the sixth digit, and this uniquely determines the seventh digit. So he has to dial $12 \cdot 2 \cdot 10=240$ numbers.
|
240
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
The mathematician John is having trouble remembering his girlfriend Alicia's 7-digit phone number. He remembers that the first four digits consist of one 1 , one 2 , and two 3 s . He also remembers that the fifth digit is either a 4 or 5 . While he has no memory of the sixth digit, he remembers that the seventh digit is 9 minus the sixth digit. If this is all the information he has, how many phone numbers does he have to try if he is to make sure he dials the correct number?
|
There are $\frac{4!}{2!}=12$ possibilities for the first four digits. There are two possibilities for the fifth digit. There are 10 possibilities for the sixth digit, and this uniquely determines the seventh digit. So he has to dial $12 \cdot 2 \cdot 10=240$ numbers.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nSolution: "
}
|
ae62fa82-5370-570b-a1a9-3251ace071fe
| 611,018
|
How many real solutions are there to the equation
$$
||||x|-2|-2|-2|=||||x|-3|-3|-3| ?
$$
|
6. The graphs of the two sides of the equation can be graphed on the same plot to reveal six intersection points.
|
6
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
How many real solutions are there to the equation
$$
||||x|-2|-2|-2|=||||x|-3|-3|-3| ?
$$
|
6. The graphs of the two sides of the equation can be graphed on the same plot to reveal six intersection points.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n11. ",
"solution_match": "\nSolution: "
}
|
6d8f1093-7400-5bce-9325-e21ef28a0d3f
| 611,019
|
An omino is a 1-by-1 square or a 1-by-2 horizontal rectangle. An omino tiling of a region of the plane is a way of covering it (and only it) by ominoes. How many omino tilings are there of a 2 -by- 10 horizontal rectangle?
|
There are exactly as many omino tilings of a 1 -by- $n$ rectangle as there are domino tilings of a 2 -by- $n$ rectangle. Since the rows don't interact at all, the number of omino tilings of an $m$-by- $n$ rectangle is the number of omino tilings of a 1-by- $n$ rectangle raised to the $m$ th power, $F_{n}^{m}$. The answer is $89^{2}=7921$.
|
7921
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
An omino is a 1-by-1 square or a 1-by-2 horizontal rectangle. An omino tiling of a region of the plane is a way of covering it (and only it) by ominoes. How many omino tilings are there of a 2 -by- 10 horizontal rectangle?
|
There are exactly as many omino tilings of a 1 -by- $n$ rectangle as there are domino tilings of a 2 -by- $n$ rectangle. Since the rows don't interact at all, the number of omino tilings of an $m$-by- $n$ rectangle is the number of omino tilings of a 1-by- $n$ rectangle raised to the $m$ th power, $F_{n}^{m}$. The answer is $89^{2}=7921$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n14. ",
"solution_match": "\nSolution: "
}
|
c611047e-f623-55c9-9765-59b381221f29
| 611,022
|
How many sequences of 0 s and 1 s are there of length 10 such that there are no three 0 s or 1 s consecutively anywhere in the sequence?
|
We can have blocks of either 1 or 20 s and 1 s , and these blocks must be alternating between 0 s and 1 s . The number of ways of arranging blocks to form a sequence of length $n$ is the same as the number of omino tilings of a $1-b y-n$ rectangle, and we may start each sequence with a 0 or a 1 , making $2 F_{n}$ or, in this case, 178 sequences.
|
178
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many sequences of 0 s and 1 s are there of length 10 such that there are no three 0 s or 1 s consecutively anywhere in the sequence?
|
We can have blocks of either 1 or 20 s and 1 s , and these blocks must be alternating between 0 s and 1 s . The number of ways of arranging blocks to form a sequence of length $n$ is the same as the number of omino tilings of a $1-b y-n$ rectangle, and we may start each sequence with a 0 or a 1 , making $2 F_{n}$ or, in this case, 178 sequences.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n15. ",
"solution_match": "\nSolution: "
}
|
f65b2fb3-3daa-5017-b560-27ca227305bc
| 611,023
|
Divide an $m$-by- $n$ rectangle into $m n$ nonoverlapping 1-by-1 squares. A polyomino of this rectangle is a subset of these unit squares such that for any two unit squares $S, T$ in the polyomino, either
(1) $S$ and $T$ share an edge or
(2) there exists a positive integer $n$ such that the polyomino contains unit squares $S_{1}, S_{2}, S_{3}, \ldots, S_{n}$ such that $S$ and $S_{1}$ share an edge, $S_{n}$ and $T$ share an edge, and for all positive integers $k<n, S_{k}$ and $S_{k+1}$ share an edge.
We say a polyomino of a given rectangle spans the rectangle if for each of the four edges of the rectangle the polyomino contains a square whose edge lies on it.
What is the minimum number of unit squares a polyomino can have if it spans a 128 -by343 rectangle?
|
To span an $a \times b$ rectangle, we need at least $a+b-1$ squares. Indeed, consider a square of the polyomino bordering the left edge of the rectangle and one bordering the right edge. There exists a path connecting these squares; suppose it runs through $c$ different rows. Then the path requires at least $b-1$ horizontal and $c-1$ vertical steps, so it uses at least $b+c-1$ different squares. However, since the polyomino also hits the top and bottom edges of the rectangle, it must run into the remaining $a-c$ rows as well, so altogether we need at least $a+b-1$ squares. On the other hand, this many squares suffice - just consider all the squares bordering the lower or right edges of the rectangle. So, in our case, the answer is $128+343-1=470$.
|
470
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Divide an $m$-by- $n$ rectangle into $m n$ nonoverlapping 1-by-1 squares. A polyomino of this rectangle is a subset of these unit squares such that for any two unit squares $S, T$ in the polyomino, either
(1) $S$ and $T$ share an edge or
(2) there exists a positive integer $n$ such that the polyomino contains unit squares $S_{1}, S_{2}, S_{3}, \ldots, S_{n}$ such that $S$ and $S_{1}$ share an edge, $S_{n}$ and $T$ share an edge, and for all positive integers $k<n, S_{k}$ and $S_{k+1}$ share an edge.
We say a polyomino of a given rectangle spans the rectangle if for each of the four edges of the rectangle the polyomino contains a square whose edge lies on it.
What is the minimum number of unit squares a polyomino can have if it spans a 128 -by343 rectangle?
|
To span an $a \times b$ rectangle, we need at least $a+b-1$ squares. Indeed, consider a square of the polyomino bordering the left edge of the rectangle and one bordering the right edge. There exists a path connecting these squares; suppose it runs through $c$ different rows. Then the path requires at least $b-1$ horizontal and $c-1$ vertical steps, so it uses at least $b+c-1$ different squares. However, since the polyomino also hits the top and bottom edges of the rectangle, it must run into the remaining $a-c$ rows as well, so altogether we need at least $a+b-1$ squares. On the other hand, this many squares suffice - just consider all the squares bordering the lower or right edges of the rectangle. So, in our case, the answer is $128+343-1=470$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n16. ",
"solution_match": "\nSolution: "
}
|
8155aaf4-17c5-5e04-9cb8-5fd0408fceac
| 611,024
|
Find the number of pentominoes (5-square polyominoes) that span a 3 -by- 3 rectangle, where polyominoes that are flips or rotations of each other are considered the same polyomino.
|
By enumeration, the answer is 6 .
|
6
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Find the number of pentominoes (5-square polyominoes) that span a 3 -by- 3 rectangle, where polyominoes that are flips or rotations of each other are considered the same polyomino.
|
By enumeration, the answer is 6 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n17. ",
"solution_match": "\nSolution: "
}
|
c8eaca65-a94e-5079-a02f-c2921cbd66f7
| 611,025
|
For how many integers $a(1 \leq a \leq 200)$ is the number $a^{a}$ a square?
|
107 If $a$ is even, we have $a^{a}=\left(a^{a / 2}\right)^{2}$. If $a$ is odd, $a^{a}=\left(a^{(a-1) / 2}\right)^{2} \cdot a$, which is a square precisely when $a$ is. Thus we have 100 even values of $a$ and 7 odd square values $\left(1^{2}, 3^{2}, \ldots, 13^{2}\right)$ for a total of 107 .
|
107
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
For how many integers $a(1 \leq a \leq 200)$ is the number $a^{a}$ a square?
|
107 If $a$ is even, we have $a^{a}=\left(a^{a / 2}\right)^{2}$. If $a$ is odd, $a^{a}=\left(a^{(a-1) / 2}\right)^{2} \cdot a$, which is a square precisely when $a$ is. Thus we have 100 even values of $a$ and 7 odd square values $\left(1^{2}, 3^{2}, \ldots, 13^{2}\right)$ for a total of 107 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n19. ",
"solution_match": "\nSolution: "
}
|
83648f60-58c6-50bd-9b58-cb144999acba
| 611,027
|
The Antarctican language has an alphabet of just 16 letters. Interestingly, every word in the language has exactly 3 letters, and it is known that no word's first letter equals any word's last letter (for instance, if the alphabet were $\{a, b\}$ then $a a b$ and aaa could not both be words in the language because $a$ is the first letter of a word and the last letter of a word; in fact, just aaa alone couldn't be in the language). Given this, determine the maximum possible number of words in the language.
|
1024 Every letter can be the first letter of a word, or the last letter of a word, or possibly neither, but not both. If there are $a$ different first letters and $b$ different last letters, then we can form $a \cdot 16 \cdot b$ different words (and the desired conditions will be met). Given the constraints $0 \leq a, b ; a+b \leq 16$, this product is maximized when $a=b=8$, giving the answer.
|
1024
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
The Antarctican language has an alphabet of just 16 letters. Interestingly, every word in the language has exactly 3 letters, and it is known that no word's first letter equals any word's last letter (for instance, if the alphabet were $\{a, b\}$ then $a a b$ and aaa could not both be words in the language because $a$ is the first letter of a word and the last letter of a word; in fact, just aaa alone couldn't be in the language). Given this, determine the maximum possible number of words in the language.
|
1024 Every letter can be the first letter of a word, or the last letter of a word, or possibly neither, but not both. If there are $a$ different first letters and $b$ different last letters, then we can form $a \cdot 16 \cdot b$ different words (and the desired conditions will be met). Given the constraints $0 \leq a, b ; a+b \leq 16$, this product is maximized when $a=b=8$, giving the answer.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n20. ",
"solution_match": "\nSolution: "
}
|
d46272a9-7e26-56ce-b598-6a3e5249afee
| 611,028
|
The Dyslexian alphabet consists of consonants and vowels. It so happens that a finite sequence of letters is a word in Dyslexian precisely if it alternates between consonants and vowels (it may begin with either). There are 4800 five-letter words in Dyslexian. How many letters are in the alphabet?
|
12 Suppose there are consonants, $v$ vowels. Then there are $c \cdot v \cdot c \cdot v \cdot c+$ $v \cdot c \cdot v \cdot c \cdot v=(c v)^{2}(c+v)$ five-letter words. Thus, $c+v=4800 /(c v)^{2}=3 \cdot(40 / c v)^{2}$, so $c v$ is a divisor of 40 . If $c v \leq 10$, we have $c+v \geq 48$, impossible for $c, v$ integers; if $c v=40$, then $c+v=3$ which is again impossible. So $c v=20$, giving $c+v=12$, the answer. As a check, this does have integer solutions: $(c, v)=(2,10)$ or $(10,2)$.
|
12
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
The Dyslexian alphabet consists of consonants and vowels. It so happens that a finite sequence of letters is a word in Dyslexian precisely if it alternates between consonants and vowels (it may begin with either). There are 4800 five-letter words in Dyslexian. How many letters are in the alphabet?
|
12 Suppose there are consonants, $v$ vowels. Then there are $c \cdot v \cdot c \cdot v \cdot c+$ $v \cdot c \cdot v \cdot c \cdot v=(c v)^{2}(c+v)$ five-letter words. Thus, $c+v=4800 /(c v)^{2}=3 \cdot(40 / c v)^{2}$, so $c v$ is a divisor of 40 . If $c v \leq 10$, we have $c+v \geq 48$, impossible for $c, v$ integers; if $c v=40$, then $c+v=3$ which is again impossible. So $c v=20$, giving $c+v=12$, the answer. As a check, this does have integer solutions: $(c, v)=(2,10)$ or $(10,2)$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n21. ",
"solution_match": "\nSolution: "
}
|
0dc73851-0a6b-5930-8274-b39278fd8b1c
| 611,029
|
Find $P(7,3)$.
|
The number of paths that start at $(0,0)$ and end at $(n, m)$ is $\binom{n+m}{n}$, since we must choose $n$ of our $n+m$ steps to be rightward steps. In this case, the answer is 120 .
|
120
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Find $P(7,3)$.
|
The number of paths that start at $(0,0)$ and end at $(n, m)$ is $\binom{n+m}{n}$, since we must choose $n$ of our $n+m$ steps to be rightward steps. In this case, the answer is 120 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n23. ",
"solution_match": "\nSolution: "
}
|
6e587f89-b94f-514d-9b60-c25381648c59
| 611,031
|
A restricted path of length $n$ is a path of length $n$ such that for all $i$ between 1 and $n-2$ inclusive, if the $i$ th step is upward, the $i+1$ st step must be rightward.
Find the number of restricted paths that start at $(0,0)$ and end at $(7,3)$.
|
This is equal to the number of lattice paths from $(0,0)$ to $(7,3)$ that use only rightward and diagonal (upward+rightward) steps plus the number of lattice paths from $(0,0)$ to $(7,2)$ that use only rightward and diagonal steps, which is equal to the number of paths (as defined above) from $(0,0)$ to $(4,3)$ plus the number of paths from $(0,0)$ to $(5,2)$, or $\binom{4+3}{3}+\binom{5+2}{2}=56$.
|
56
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A restricted path of length $n$ is a path of length $n$ such that for all $i$ between 1 and $n-2$ inclusive, if the $i$ th step is upward, the $i+1$ st step must be rightward.
Find the number of restricted paths that start at $(0,0)$ and end at $(7,3)$.
|
This is equal to the number of lattice paths from $(0,0)$ to $(7,3)$ that use only rightward and diagonal (upward+rightward) steps plus the number of lattice paths from $(0,0)$ to $(7,2)$ that use only rightward and diagonal steps, which is equal to the number of paths (as defined above) from $(0,0)$ to $(4,3)$ plus the number of paths from $(0,0)$ to $(5,2)$, or $\binom{4+3}{3}+\binom{5+2}{2}=56$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n24. ",
"solution_match": "\nSolution: "
}
|
1a878f8b-b76a-50c5-a645-5b0ed0cb483e
| 611,032
|
A math professor stands up in front of a room containing 100 very smart math students and says, "Each of you has to write down an integer between 0 and 100, inclusive, to guess 'two-thirds of the average of all the responses.' Each student who guesses the highest integer that is not higher than two-thirds of the average of all responses will receive a prize." If among all the students it is common knowledge that everyone will write down the best response, and there is no communication between students, what single integer should each of the 100 students write down?
|
Since the average cannot be greater than 100, no student will write down a number greater than $\frac{2}{3} \cdot 100$. But then the average cannot be greater than $\frac{2}{3} \cdot 100$, and, realizing this, each student will write down a number no greater than $\left(\frac{2}{3}\right)^{2} \cdot 100$. Continuing in this manner, we eventually see that no student will write down an integer greater than 0 , so this is the answer.
|
0
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
A math professor stands up in front of a room containing 100 very smart math students and says, "Each of you has to write down an integer between 0 and 100, inclusive, to guess 'two-thirds of the average of all the responses.' Each student who guesses the highest integer that is not higher than two-thirds of the average of all responses will receive a prize." If among all the students it is common knowledge that everyone will write down the best response, and there is no communication between students, what single integer should each of the 100 students write down?
|
Since the average cannot be greater than 100, no student will write down a number greater than $\frac{2}{3} \cdot 100$. But then the average cannot be greater than $\frac{2}{3} \cdot 100$, and, realizing this, each student will write down a number no greater than $\left(\frac{2}{3}\right)^{2} \cdot 100$. Continuing in this manner, we eventually see that no student will write down an integer greater than 0 , so this is the answer.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n25. ",
"solution_match": "\nSolution: "
}
|
b6033d0b-6b7d-571f-8e5e-df0f9be47fc6
| 611,033
|
Consider the two hands of an analog clock, each of which moves with constant angular velocity. Certain positions of these hands are possible (e.g. the hour hand halfway between the 5 and 6 and the minute hand exactly at the 6 ), while others are impossible (e.g. the hour hand exactly at the 5 and the minute hand exactly at the 6 ). How many different positions are there that would remain possible if the hour and minute hands were switched?
|
143 We can look at the twelve-hour cycle beginning at midnight and ending just before noon, since during this time, the clock goes through each possible position exactly once. The minute hand has twelve times the angular velocity of the hour hand, so if the hour hand has made $t$ revolutions from its initial position $(0 \leq t<1)$, the minute hand has made $12 t$ revolutions. If the hour hand were to have made $12 t$ revolutions, the minute hand would have made $144 t$. So we get a valid configuration by reversing the hands precisely when $144 t$ revolutions land the hour hand in the same place as $t$ revolutions - i.e. when $143 t=144 t-t$ is an integer, which clearly occurs for exactly 143 values of $t$ corresponding to distinct positions on the clock ( $144-1=143$ ).
|
143
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
Consider the two hands of an analog clock, each of which moves with constant angular velocity. Certain positions of these hands are possible (e.g. the hour hand halfway between the 5 and 6 and the minute hand exactly at the 6 ), while others are impossible (e.g. the hour hand exactly at the 5 and the minute hand exactly at the 6 ). How many different positions are there that would remain possible if the hour and minute hands were switched?
|
143 We can look at the twelve-hour cycle beginning at midnight and ending just before noon, since during this time, the clock goes through each possible position exactly once. The minute hand has twelve times the angular velocity of the hour hand, so if the hour hand has made $t$ revolutions from its initial position $(0 \leq t<1)$, the minute hand has made $12 t$ revolutions. If the hour hand were to have made $12 t$ revolutions, the minute hand would have made $144 t$. So we get a valid configuration by reversing the hands precisely when $144 t$ revolutions land the hour hand in the same place as $t$ revolutions - i.e. when $143 t=144 t-t$ is an integer, which clearly occurs for exactly 143 values of $t$ corresponding to distinct positions on the clock ( $144-1=143$ ).
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n27. ",
"solution_match": "\nSolution: "
}
|
a536c1d5-4bd2-5f6f-9743-0bcd0202231b
| 611,035
|
Count how many 8-digit numbers there are that contain exactly four nines as digits.
|
There are $\binom{8}{4} \cdot 9^{4}$ sequences of 8 numbers with exactly four nines. A sequence of digits of length 8 is not an 8 -digit number, however, if and only if the first digit is zero. There are $\binom{7}{4} 9^{3}$-digit sequences that are not 8 -digit numbers. The answer is thus $\binom{8}{4} \cdot 9^{4}-\binom{7}{4} 9^{3}=433755$.
|
433755
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Count how many 8-digit numbers there are that contain exactly four nines as digits.
|
There are $\binom{8}{4} \cdot 9^{4}$ sequences of 8 numbers with exactly four nines. A sequence of digits of length 8 is not an 8 -digit number, however, if and only if the first digit is zero. There are $\binom{7}{4} 9^{3}$-digit sequences that are not 8 -digit numbers. The answer is thus $\binom{8}{4} \cdot 9^{4}-\binom{7}{4} 9^{3}=433755$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n28. ",
"solution_match": "\nSolution: "
}
|
e919de52-70cf-567f-a2aa-17257d9c6b5c
| 611,036
|
A sequence $s_{0}, s_{1}, s_{2}, s_{3}, \ldots$ is defined by $s_{0}=s_{1}=1$ and, for every positive integer $n, s_{2 n}=s_{n}, s_{4 n+1}=s_{2 n+1}, s_{4 n-1}=s_{2 n-1}+s_{2 n-1}^{2} / s_{n-1}$. What is the value of $s_{1000}$ ?
|
720 Some experimentation with small values may suggest that $s_{n}=k$ !, where $k$ is the number of ones in the binary representation of $n$, and this formula is in fact provable by a straightforward induction. Since $1000_{10}=1111101000_{2}$, with six ones, $s_{1000}=6!=720$.
|
720
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
A sequence $s_{0}, s_{1}, s_{2}, s_{3}, \ldots$ is defined by $s_{0}=s_{1}=1$ and, for every positive integer $n, s_{2 n}=s_{n}, s_{4 n+1}=s_{2 n+1}, s_{4 n-1}=s_{2 n-1}+s_{2 n-1}^{2} / s_{n-1}$. What is the value of $s_{1000}$ ?
|
720 Some experimentation with small values may suggest that $s_{n}=k$ !, where $k$ is the number of ones in the binary representation of $n$, and this formula is in fact provable by a straightforward induction. Since $1000_{10}=1111101000_{2}$, with six ones, $s_{1000}=6!=720$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n29. ",
"solution_match": "\nSolution: "
}
|
5b7c495d-eeee-506a-887a-155ebbdc263d
| 611,037
|
A conical flask contains some water. When the flask is oriented so that its base is horizontal and lies at the bottom (so that the vertex is at the top), the water is 1 inch deep. When the flask is turned upside-down, so that the vertex is at the bottom, the water is 2 inches deep. What is the height of the cone?
|
$\frac{1}{2}+\frac{\sqrt{93}}{6}$. Let $h$ be the height, and let $V$ be such that $V h^{3}$ equals the volume of the flask. When the base is at the bottom, the portion of the flask not occupied by water forms a cone similar to the entire flask, with a height of $h-1$; thus its volume is $V(h-1)^{3}$. When the base is at the top, the water occupies a cone with a height of 2 , so its volume is $V \cdot 2^{3}$. Since the water's volume does not change,
$$
\begin{gathered}
V h^{3}-V(h-1)^{3}=8 V \\
\Rightarrow 3 h^{2}-3 h+1=h^{3}-(h-1)^{3}=8 \\
\Rightarrow 3 h^{2}-3 h-7=0 .
\end{gathered}
$$
Solving via the quadratic formula and taking the positive root gives $h=\frac{1}{2}+\frac{\sqrt{93}}{6}$.
|
\frac{1}{2}+\frac{\sqrt{93}}{6}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A conical flask contains some water. When the flask is oriented so that its base is horizontal and lies at the bottom (so that the vertex is at the top), the water is 1 inch deep. When the flask is turned upside-down, so that the vertex is at the bottom, the water is 2 inches deep. What is the height of the cone?
|
$\frac{1}{2}+\frac{\sqrt{93}}{6}$. Let $h$ be the height, and let $V$ be such that $V h^{3}$ equals the volume of the flask. When the base is at the bottom, the portion of the flask not occupied by water forms a cone similar to the entire flask, with a height of $h-1$; thus its volume is $V(h-1)^{3}$. When the base is at the top, the water occupies a cone with a height of 2 , so its volume is $V \cdot 2^{3}$. Since the water's volume does not change,
$$
\begin{gathered}
V h^{3}-V(h-1)^{3}=8 V \\
\Rightarrow 3 h^{2}-3 h+1=h^{3}-(h-1)^{3}=8 \\
\Rightarrow 3 h^{2}-3 h-7=0 .
\end{gathered}
$$
Solving via the quadratic formula and taking the positive root gives $h=\frac{1}{2}+\frac{\sqrt{93}}{6}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n30. ",
"solution_match": "\nSolution: "
}
|
25cfcb7f-1f0e-5753-a0df-2a6e3bfa15b9
| 611,038
|
Express, as concisely as possible, the value of the product
$$
\left(0^{3}-350\right)\left(1^{3}-349\right)\left(2^{3}-348\right)\left(3^{3}-347\right) \cdots\left(349^{3}-1\right)\left(350^{3}-0\right)
$$
|
0 . One of the factors is $7^{3}-343=0$, so the whole product is zero.
|
0
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Express, as concisely as possible, the value of the product
$$
\left(0^{3}-350\right)\left(1^{3}-349\right)\left(2^{3}-348\right)\left(3^{3}-347\right) \cdots\left(349^{3}-1\right)\left(350^{3}-0\right)
$$
|
0 . One of the factors is $7^{3}-343=0$, so the whole product is zero.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n31. ",
"solution_match": "\nSolution: "
}
|
36288ba9-acbd-56a9-8237-d82a7b1c1f94
| 611,039
|
Two circles have radii 13 and 30 , and their centers are 41 units apart. The line through the centers of the two circles intersects the smaller circle at two points; let $A$ be the one outside the larger circle. Suppose $B$ is a point on the smaller circle and $C$ a point on the larger circle such that $B$ is the midpoint of $A C$. Compute the distance $A C$.
|
$12 \sqrt{13}$ Call the large circle's center $O_{1}$. Scale the small circle by a factor of 2 about $A$; we obtain a new circle whose center $O_{2}$ is at a distance of $41-13=28$ from $O_{1}$, and whose radius is 26 . Also, the dilation sends $B$ to $C$, which thus lies on circles $O_{1}$ and $O_{2}$. So points $O_{1}, O_{2}, C$ form a 26-28-30 triangle. Let $H$ be the foot of the altitude from $C$ to $O_{1} O_{2}$; we have $C H=24$ and $H O_{2}=10$. Thus, $H A=36$, and $A C=\sqrt{24^{2}+36^{2}}=12 \sqrt{13}$.
|
12 \sqrt{13}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Two circles have radii 13 and 30 , and their centers are 41 units apart. The line through the centers of the two circles intersects the smaller circle at two points; let $A$ be the one outside the larger circle. Suppose $B$ is a point on the smaller circle and $C$ a point on the larger circle such that $B$ is the midpoint of $A C$. Compute the distance $A C$.
|
$12 \sqrt{13}$ Call the large circle's center $O_{1}$. Scale the small circle by a factor of 2 about $A$; we obtain a new circle whose center $O_{2}$ is at a distance of $41-13=28$ from $O_{1}$, and whose radius is 26 . Also, the dilation sends $B$ to $C$, which thus lies on circles $O_{1}$ and $O_{2}$. So points $O_{1}, O_{2}, C$ form a 26-28-30 triangle. Let $H$ be the foot of the altitude from $C$ to $O_{1} O_{2}$; we have $C H=24$ and $H O_{2}=10$. Thus, $H A=36$, and $A C=\sqrt{24^{2}+36^{2}}=12 \sqrt{13}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n32. ",
"solution_match": "\nSolution: "
}
|
a561dbda-a3b1-54ba-babc-d8b69911cedf
| 611,040
|
The expression $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$. Find the value of
$$
\left\lfloor\frac{2002!}{2001!+2000!+1999!+\cdots+1!}\right\rfloor .
$$
|
2000 We break up 2002! $=2002(2001)$ ! as
$$
\begin{gathered}
2000(2001!)+2 \cdot 2001(2000!)=2000(2001!)+2000(2000!)+2002 \cdot 2000(1999!) \\
>2000(2001!+2000!+1999!+\cdots+1!)
\end{gathered}
$$
On the other hand,
$$
2001(2001!+2000!+\cdots+1!)>2001(2001!+2000!)=2001(2001!)+2001!=2002!
$$
Thus we have $2000<2002!/(2001!+\cdots+1!)<2001$, so the answer is 2000 .
|
2000
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
The expression $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$. Find the value of
$$
\left\lfloor\frac{2002!}{2001!+2000!+1999!+\cdots+1!}\right\rfloor .
$$
|
2000 We break up 2002! $=2002(2001)$ ! as
$$
\begin{gathered}
2000(2001!)+2 \cdot 2001(2000!)=2000(2001!)+2000(2000!)+2002 \cdot 2000(1999!) \\
>2000(2001!+2000!+1999!+\cdots+1!)
\end{gathered}
$$
On the other hand,
$$
2001(2001!+2000!+\cdots+1!)>2001(2001!+2000!)=2001(2001!)+2001!=2002!
$$
Thus we have $2000<2002!/(2001!+\cdots+1!)<2001$, so the answer is 2000 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n33. ",
"solution_match": "\nSolution: "
}
|
2c43d815-d1c3-5861-9edf-72d3663c766c
| 611,041
|
Suppose $a, b, c, d$ are real numbers such that
$$
|a-b|+|c-d|=99 ; \quad|a-c|+|b-d|=1
$$
Determine all possible values of $|a-d|+|b-c|$.
|
99 If $w \geq x \geq y \geq z$ are four arbitrary real numbers, then $|w-z|+|x-y|=$ $|w-y|+|x-z|=w+x-y-z \geq w-x+y-z=|w-x|+|y-z|$. Thus, in our case, two of the three numbers $|a-b|+|c-d|,|a-c|+|b-d|,|a-d|+|b-c|$ are equal, and the third one is less than or equal to these two. Since we have a 99 and a 1 , the third number must be 99 .
|
99
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose $a, b, c, d$ are real numbers such that
$$
|a-b|+|c-d|=99 ; \quad|a-c|+|b-d|=1
$$
Determine all possible values of $|a-d|+|b-c|$.
|
99 If $w \geq x \geq y \geq z$ are four arbitrary real numbers, then $|w-z|+|x-y|=$ $|w-y|+|x-z|=w+x-y-z \geq w-x+y-z=|w-x|+|y-z|$. Thus, in our case, two of the three numbers $|a-b|+|c-d|,|a-c|+|b-d|,|a-d|+|b-c|$ are equal, and the third one is less than or equal to these two. Since we have a 99 and a 1 , the third number must be 99 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n35. ",
"solution_match": "\nSolution: "
}
|
1cfc4fae-d511-5a67-ad2f-d018d71f193c
| 611,043
|
Call a positive integer "mild" if its base-3 representation never contains the digit 2 . How many values of $n(1 \leq n \leq 1000)$ have the property that $n$ and $n^{2}$ are both mild?
|
7 Such a number, which must consist entirely of 0 's and 1 's in base 3, can never have more than one 1. Indeed, if $n=3^{a}+3^{b}+$ higher powers where $b>a$, then $n^{2}=3^{2 a}+2 \cdot 3^{a+b}+$ higher powers which will not be mild. On the other hand, if $n$ does just have one 1 in base 3, then clearly $n$ and $n^{2}$ are mild. So the values of $n \leq 1000$ that work are $3^{0}, 3^{1}, \ldots, 3^{6}$; there are 7 of them.
|
7
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Call a positive integer "mild" if its base-3 representation never contains the digit 2 . How many values of $n(1 \leq n \leq 1000)$ have the property that $n$ and $n^{2}$ are both mild?
|
7 Such a number, which must consist entirely of 0 's and 1 's in base 3, can never have more than one 1. Indeed, if $n=3^{a}+3^{b}+$ higher powers where $b>a$, then $n^{2}=3^{2 a}+2 \cdot 3^{a+b}+$ higher powers which will not be mild. On the other hand, if $n$ does just have one 1 in base 3, then clearly $n$ and $n^{2}$ are mild. So the values of $n \leq 1000$ that work are $3^{0}, 3^{1}, \ldots, 3^{6}$; there are 7 of them.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n37. ",
"solution_match": "\nSolution: "
}
|
d82e470e-e509-5f67-8da4-19a05ffc0c91
| 611,045
|
Massachusetts Avenue is ten blocks long. One boy and one girl live on each block. They want to form friendships such that each boy is friends with exactly one girl and viceversa. Nobody wants a friend living more than one block away (but they may be on the same block). How many pairings are possible?
|
89 Let $a_{n}$ be the number of pairings if there are $n$ blocks; we have $a_{1}=$ $1, a_{2}=2$, and we claim the Fibonacci recurrence is satisfied. Indeed, if there are $n$ blocks, either the boy on block 1 is friends with the girl on block 1, leaving $a_{n-1}$ possible pairings for the people on the remaining $n-1$ blocks, or he is friends with the girl on block 2 , in which case the girl on block 1 must be friends with the boy on block 2 , and then there are $a_{n-2}$ possibilities for the friendships among the remaining $n-2$ blocks. So $a_{n}=a_{n-1}+a_{n-2}$, and we compute: $a_{3}=3, a_{4}=5, \ldots, a_{10}=89$.
|
89
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Massachusetts Avenue is ten blocks long. One boy and one girl live on each block. They want to form friendships such that each boy is friends with exactly one girl and viceversa. Nobody wants a friend living more than one block away (but they may be on the same block). How many pairings are possible?
|
89 Let $a_{n}$ be the number of pairings if there are $n$ blocks; we have $a_{1}=$ $1, a_{2}=2$, and we claim the Fibonacci recurrence is satisfied. Indeed, if there are $n$ blocks, either the boy on block 1 is friends with the girl on block 1, leaving $a_{n-1}$ possible pairings for the people on the remaining $n-1$ blocks, or he is friends with the girl on block 2 , in which case the girl on block 1 must be friends with the boy on block 2 , and then there are $a_{n-2}$ possibilities for the friendships among the remaining $n-2$ blocks. So $a_{n}=a_{n-1}+a_{n-2}$, and we compute: $a_{3}=3, a_{4}=5, \ldots, a_{10}=89$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n38. ",
"solution_match": "\nSolution: "
}
|
b11806f8-b3d8-5de4-b7dc-546224e09806
| 611,046
|
In the $x-y$ plane, draw a circle of radius 2 centered at $(0,0)$. Color the circle red above the line $y=1$, color the circle blue below the line $y=-1$, and color the rest of the circle white. Now consider an arbitrary straight line at distance 1 from the circle. We color each point $P$ of the line with the color of the closest point to $P$ on the circle. If we pick such an arbitrary line, randomly oriented, what is the probability that it contains red, white, and blue points?
|
Let $O=(0,0), P=(1,0)$, and $H$ the foot of the perpendicular from $O$ to the line. If $\angle P O H$ (as measured counterclockwise) lies between $\pi / 3$ and $2 \pi / 3$, the line will fail to contain blue points; if it lies between $4 \pi / 3$ and $5 \pi / 3$, the line will fail to contain red points. Otherwise, it has points of every color. Thus, the answer is $1-\frac{2 \pi}{3} / 2 \pi=\frac{2}{3}$.
|
\frac{2}{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In the $x-y$ plane, draw a circle of radius 2 centered at $(0,0)$. Color the circle red above the line $y=1$, color the circle blue below the line $y=-1$, and color the rest of the circle white. Now consider an arbitrary straight line at distance 1 from the circle. We color each point $P$ of the line with the color of the closest point to $P$ on the circle. If we pick such an arbitrary line, randomly oriented, what is the probability that it contains red, white, and blue points?
|
Let $O=(0,0), P=(1,0)$, and $H$ the foot of the perpendicular from $O$ to the line. If $\angle P O H$ (as measured counterclockwise) lies between $\pi / 3$ and $2 \pi / 3$, the line will fail to contain blue points; if it lies between $4 \pi / 3$ and $5 \pi / 3$, the line will fail to contain red points. Otherwise, it has points of every color. Thus, the answer is $1-\frac{2 \pi}{3} / 2 \pi=\frac{2}{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n39. ",
"solution_match": "\nSolution: "
}
|
ce3686cd-de38-581e-9f84-bb1c2ad121d2
| 611,047
|
Find the volume of the three-dimensional solid given by the inequality $\sqrt{x^{2}+y^{2}}+$ $|z| \leq 1$.
|
$2 \pi / 3$. The solid consists of two cones, one whose base is the circle $x^{2}+y^{2}=1$ in the $x y$-plane and whose vertex is $(0,0,1)$, and the other with the same base but vertex $(0,0,-1)$. Each cone has a base area of $\pi$ and a height of 1 , for a volume of $\pi / 3$, so the answer is $2 \pi / 3$.
|
\frac{2\pi}{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Find the volume of the three-dimensional solid given by the inequality $\sqrt{x^{2}+y^{2}}+$ $|z| \leq 1$.
|
$2 \pi / 3$. The solid consists of two cones, one whose base is the circle $x^{2}+y^{2}=1$ in the $x y$-plane and whose vertex is $(0,0,1)$, and the other with the same base but vertex $(0,0,-1)$. Each cone has a base area of $\pi$ and a height of 1 , for a volume of $\pi / 3$, so the answer is $2 \pi / 3$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n40. ",
"solution_match": "\nSolution: "
}
|
cbcc243e-3869-5c21-b4a0-7452c2cd1790
| 611,048
|
For any integer $n$, define $\lfloor n\rfloor$ as the greatest integer less than or equal to $n$. For any positive integer $n$, let
$$
f(n)=\lfloor n\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{n}{3}\right\rfloor+\cdots+\left\lfloor\frac{n}{n}\right\rfloor .
$$
For how many values of $n, 1 \leq n \leq 100$, is $f(n)$ odd?
|
55 Notice that, for fixed $a,\lfloor n / a\rfloor$ counts the number of integers $b \in$ $\{1,2, \ldots, n\}$ which are divisible by $a$; hence, $f(n)$ counts the number of pairs $(a, b), a, b \in$ $\{1,2, \ldots, n\}$ with $b$ divisible by $a$. For any fixed $b$, the number of such pairs is $d(b)$ (the number of divisors of $b$ ), so the total number of pairs $f(n)$ equals $d(1)+d(2)+\cdots+d(n)$. But $d(b)$ is odd precisely when $b$ is a square, so $f(n)$ is odd precisely when there are an odd number of squares in $\{1,2, \ldots, n\}$. This happens for $1 \leq n<4 ; 9 \leq n<16 ; \ldots ; 81 \leq n<100$. Adding these up gives 55 values of $n$.
|
55
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
For any integer $n$, define $\lfloor n\rfloor$ as the greatest integer less than or equal to $n$. For any positive integer $n$, let
$$
f(n)=\lfloor n\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{n}{3}\right\rfloor+\cdots+\left\lfloor\frac{n}{n}\right\rfloor .
$$
For how many values of $n, 1 \leq n \leq 100$, is $f(n)$ odd?
|
55 Notice that, for fixed $a,\lfloor n / a\rfloor$ counts the number of integers $b \in$ $\{1,2, \ldots, n\}$ which are divisible by $a$; hence, $f(n)$ counts the number of pairs $(a, b), a, b \in$ $\{1,2, \ldots, n\}$ with $b$ divisible by $a$. For any fixed $b$, the number of such pairs is $d(b)$ (the number of divisors of $b$ ), so the total number of pairs $f(n)$ equals $d(1)+d(2)+\cdots+d(n)$. But $d(b)$ is odd precisely when $b$ is a square, so $f(n)$ is odd precisely when there are an odd number of squares in $\{1,2, \ldots, n\}$. This happens for $1 \leq n<4 ; 9 \leq n<16 ; \ldots ; 81 \leq n<100$. Adding these up gives 55 values of $n$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n41. ",
"solution_match": "\nSolution: "
}
|
15ef13db-ab71-5d2f-8816-f2634449487b
| 611,049
|
Given that $a, b, c$ are positive integers satisfying
$$
a+b+c=\operatorname{gcd}(a, b)+\operatorname{gcd}(b, c)+\operatorname{gcd}(c, a)+120
$$
determine the maximum possible value of $a$.
|
240. Notice that $(a, b, c)=(240,120,120)$ achieves a value of 240 . To see that this is maximal, first suppose that $a>b$. Notice that $a+b+c=\operatorname{gcd}(a, b)+\operatorname{gcd}(b, c)+$ $\operatorname{gcd}(c, a)+120 \leq \operatorname{gcd}(a, b)+b+c+120$, or $a \leq \operatorname{gcd}(a, b)+120$. However, $\operatorname{gcd}(a, b)$ is a proper divisor of $a$, so $a \geq 2 \cdot \operatorname{gcd}(a, b)$. Thus, $a-120 \leq \operatorname{gcd}(a, b) \leq a / 2$, yielding $a \leq 240$. Now, if instead $a \leq b$, then either $b>c$ and the same logic shows that $b \leq 240 \Rightarrow a \leq 240$, or $b \leq c, c>a$ (since $a, b, c$ cannot all be equal) and then $c \leq 240 \Rightarrow a \leq b \leq c \leq 240$.
|
240
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Given that $a, b, c$ are positive integers satisfying
$$
a+b+c=\operatorname{gcd}(a, b)+\operatorname{gcd}(b, c)+\operatorname{gcd}(c, a)+120
$$
determine the maximum possible value of $a$.
|
240. Notice that $(a, b, c)=(240,120,120)$ achieves a value of 240 . To see that this is maximal, first suppose that $a>b$. Notice that $a+b+c=\operatorname{gcd}(a, b)+\operatorname{gcd}(b, c)+$ $\operatorname{gcd}(c, a)+120 \leq \operatorname{gcd}(a, b)+b+c+120$, or $a \leq \operatorname{gcd}(a, b)+120$. However, $\operatorname{gcd}(a, b)$ is a proper divisor of $a$, so $a \geq 2 \cdot \operatorname{gcd}(a, b)$. Thus, $a-120 \leq \operatorname{gcd}(a, b) \leq a / 2$, yielding $a \leq 240$. Now, if instead $a \leq b$, then either $b>c$ and the same logic shows that $b \leq 240 \Rightarrow a \leq 240$, or $b \leq c, c>a$ (since $a, b, c$ cannot all be equal) and then $c \leq 240 \Rightarrow a \leq b \leq c \leq 240$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n43. ",
"solution_match": "\nSolution: "
}
|
f9affd7f-95ba-5c90-a5b4-f36f8f877839
| 611,051
|
The unknown real numbers $x, y, z$ satisfy the equations
$$
\frac{x+y}{1+z}=\frac{1-z+z^{2}}{x^{2}-x y+y^{2}} ; \quad \frac{x-y}{3-z}=\frac{9+3 z+z^{2}}{x^{2}+x y+y^{2}}
$$
Find $x$.
|
$\sqrt[3]{14}$ Cross-multiplying in both equations, we get, respectively, $x^{3}+y^{3}=$ $1+z^{3}, x^{3}-y^{3}=27-z^{3}$. Now adding gives $2 x^{3}=28$, or $x=\sqrt[3]{14}$.
|
\sqrt[3]{14}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
The unknown real numbers $x, y, z$ satisfy the equations
$$
\frac{x+y}{1+z}=\frac{1-z+z^{2}}{x^{2}-x y+y^{2}} ; \quad \frac{x-y}{3-z}=\frac{9+3 z+z^{2}}{x^{2}+x y+y^{2}}
$$
Find $x$.
|
$\sqrt[3]{14}$ Cross-multiplying in both equations, we get, respectively, $x^{3}+y^{3}=$ $1+z^{3}, x^{3}-y^{3}=27-z^{3}$. Now adding gives $2 x^{3}=28$, or $x=\sqrt[3]{14}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n44. ",
"solution_match": "\nSolution: "
}
|
b35100f5-12c4-597c-aaf9-28f29d77af44
| 611,052
|
Find the number of sequences $a_{1}, a_{2}, \ldots, a_{10}$ of positive integers with the property that $a_{n+2}=a_{n+1}+a_{n}$ for $n=1,2, \ldots, 8$, and $a_{10}=2002$.
|
3 Let $a_{1}=a, a_{2}=b$; we successively compute $a_{3}=a+b ; \quad a_{4}=a+$ $2 b ; \quad \ldots ; \quad a_{10}=21 a+34 b$. The equation $2002=21 a+34 b$ has three positive integer solutions, namely $(84,7),(50,28),(16,49)$, and each of these gives a unique sequence.
|
3
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the number of sequences $a_{1}, a_{2}, \ldots, a_{10}$ of positive integers with the property that $a_{n+2}=a_{n+1}+a_{n}$ for $n=1,2, \ldots, 8$, and $a_{10}=2002$.
|
3 Let $a_{1}=a, a_{2}=b$; we successively compute $a_{3}=a+b ; \quad a_{4}=a+$ $2 b ; \quad \ldots ; \quad a_{10}=21 a+34 b$. The equation $2002=21 a+34 b$ has three positive integer solutions, namely $(84,7),(50,28),(16,49)$, and each of these gives a unique sequence.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n45. ",
"solution_match": "\nSolution: "
}
|
f48666ca-df41-5440-8cf4-a51882cb7d81
| 611,053
|
Points $A, B, C$ in the plane satisfy $\overline{A B}=2002, \overline{A C}=9999$. The circles with diameters $A B$ and $A C$ intersect at $A$ and $D$. If $\overline{A D}=37$, what is the shortest distance from point $A$ to line $B C$ ?
|
$\angle A D B=\angle A D C=\pi / 2$ since $D$ lies on the circles with $A B$ and $A C$ as diameters, so $D$ is the foot of the perpendicular from $A$ to line $B C$, and the answer is the given 37 .
|
37
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Points $A, B, C$ in the plane satisfy $\overline{A B}=2002, \overline{A C}=9999$. The circles with diameters $A B$ and $A C$ intersect at $A$ and $D$. If $\overline{A D}=37$, what is the shortest distance from point $A$ to line $B C$ ?
|
$\angle A D B=\angle A D C=\pi / 2$ since $D$ lies on the circles with $A B$ and $A C$ as diameters, so $D$ is the foot of the perpendicular from $A$ to line $B C$, and the answer is the given 37 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n46. ",
"solution_match": "\nSolution: "
}
|
8282c0ad-c243-5d66-8952-890b6be44f1c
| 611,054
|
The real function $f$ has the property that, whenever $a, b, n$ are positive integers such that $a+b=2^{n}$, the equation $f(a)+f(b)=n^{2}$ holds. What is $f(2002)$ ?
|
We know $f(a)=n^{2}-f\left(2^{n}-a\right)$ for any $a, n$ with $2^{n}>a$; repeated application gives
$$
\begin{gathered}
f(2002)=11^{2}-f(46)=11^{2}-\left(6^{2}-f(18)\right)=11^{2}-\left(6^{2}-\left(5^{2}-f(14)\right)\right) \\
=11^{2}-\left(6^{2}-\left(5^{2}-\left(4^{2}-f(2)\right)\right)\right) .
\end{gathered}
$$
But $f(2)=2^{2}-f(2)$, giving $f(2)=2$, so the above simplifies to $11^{2}-\left(6^{2}-\left(5^{2}-\left(4^{2}-\right.\right.\right.$ 2))) $=96$.
|
96
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
The real function $f$ has the property that, whenever $a, b, n$ are positive integers such that $a+b=2^{n}$, the equation $f(a)+f(b)=n^{2}$ holds. What is $f(2002)$ ?
|
We know $f(a)=n^{2}-f\left(2^{n}-a\right)$ for any $a, n$ with $2^{n}>a$; repeated application gives
$$
\begin{gathered}
f(2002)=11^{2}-f(46)=11^{2}-\left(6^{2}-f(18)\right)=11^{2}-\left(6^{2}-\left(5^{2}-f(14)\right)\right) \\
=11^{2}-\left(6^{2}-\left(5^{2}-\left(4^{2}-f(2)\right)\right)\right) .
\end{gathered}
$$
But $f(2)=2^{2}-f(2)$, giving $f(2)=2$, so the above simplifies to $11^{2}-\left(6^{2}-\left(5^{2}-\left(4^{2}-\right.\right.\right.$ 2))) $=96$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n47. ",
"solution_match": "\nSolution: "
}
|
d066e32f-fdf8-5ec6-9678-fd0f79436d4e
| 611,055
|
A permutation of a finite set is a one-to-one function from the set to itself; for instance, one permutation of $\{1,2,3,4\}$ is the function $\pi$ defined such that $\pi(1)=1, \pi(2)=3$, $\pi(3)=4$, and $\pi(4)=2$. How many permutations $\pi$ of the set $\{1,2, \ldots, 10\}$ have the property that $\pi(i) \neq i$ for each $i=1,2, \ldots, 10$, but $\pi(\pi(i))=i$ for each $i$ ?
|
For each such $\pi$, the elements of $\{1,2, \ldots, 10\}$ can be arranged into pairs $\{i, j\}$ such that $\pi(i)=j ; \pi(j)=i$. Choosing a permutation $\pi$ is thus tantamount to choosing a partition of $\{1,2, \ldots, 10\}$ into five disjoint pairs. There are 9 ways to pair off the number 1, then 7 ways to pair off the smallest number not yet paired, and so forth, so we have $9 \cdot 7 \cdot 5 \cdot 3 \cdot 1=945$ partitions into pairs.
|
945
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A permutation of a finite set is a one-to-one function from the set to itself; for instance, one permutation of $\{1,2,3,4\}$ is the function $\pi$ defined such that $\pi(1)=1, \pi(2)=3$, $\pi(3)=4$, and $\pi(4)=2$. How many permutations $\pi$ of the set $\{1,2, \ldots, 10\}$ have the property that $\pi(i) \neq i$ for each $i=1,2, \ldots, 10$, but $\pi(\pi(i))=i$ for each $i$ ?
|
For each such $\pi$, the elements of $\{1,2, \ldots, 10\}$ can be arranged into pairs $\{i, j\}$ such that $\pi(i)=j ; \pi(j)=i$. Choosing a permutation $\pi$ is thus tantamount to choosing a partition of $\{1,2, \ldots, 10\}$ into five disjoint pairs. There are 9 ways to pair off the number 1, then 7 ways to pair off the smallest number not yet paired, and so forth, so we have $9 \cdot 7 \cdot 5 \cdot 3 \cdot 1=945$ partitions into pairs.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n48. ",
"solution_match": "\nSolution: "
}
|
d3d743d7-6ba0-5cca-afcf-ae4b28a020f5
| 611,056
|
Two integers are relatively prime if they don't share any common factors, i.e. if their greatest common divisor is 1 . Define $\varphi(n)$ as the number of positive integers that are less than $n$ and relatively prime to $n$. Define $\varphi_{d}(n)$ as the number of positive integers that are less than $d n$ and relatively prime to $n$.
What is the least $n$ such that $\varphi_{x}(n)=64000$, where $x=\varphi_{y}(n)$, where $y=\varphi(n)$ ?
|
For fixed $n$, the pattern of integers relatively prime to $n$ repeats every $n$ integers, so $\varphi_{d}(n)=d \varphi(n)$. Therefore the expression in the problem equals $\varphi(n)^{3}$. The cube root of 64000 is $40 . \varphi(p)=p-1$ for any prime $p$. Since 40 is one less than a prime, the least $n$ such that $\varphi(n)=40$ is 41 .
|
41
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Two integers are relatively prime if they don't share any common factors, i.e. if their greatest common divisor is 1 . Define $\varphi(n)$ as the number of positive integers that are less than $n$ and relatively prime to $n$. Define $\varphi_{d}(n)$ as the number of positive integers that are less than $d n$ and relatively prime to $n$.
What is the least $n$ such that $\varphi_{x}(n)=64000$, where $x=\varphi_{y}(n)$, where $y=\varphi(n)$ ?
|
For fixed $n$, the pattern of integers relatively prime to $n$ repeats every $n$ integers, so $\varphi_{d}(n)=d \varphi(n)$. Therefore the expression in the problem equals $\varphi(n)^{3}$. The cube root of 64000 is $40 . \varphi(p)=p-1$ for any prime $p$. Since 40 is one less than a prime, the least $n$ such that $\varphi(n)=40$ is 41 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n49. ",
"solution_match": "\nSolution: "
}
|
c91079b7-1c83-5c72-9dda-e72b461aedd5
| 611,057
|
Define $\varphi^{k}(n)$ as the number of positive integers that are less than or equal to $n / k$ and relatively prime to $n$. Find $\phi^{2001}\left(2002^{2}-1\right)$. (Hint: $\phi(2003)=2002$.)
|
$\varphi^{2001}\left(2002^{2}-1\right)=\varphi^{2001}(2001 \cdot 2003)=$ the number of $m$ that are relatively prime to both 2001 and 2003, where $m \leq 2003$. Since $\phi(n)=n-1$ implies that $n$ is prime, we must only check for those $m$ relatively prime to 2001, except for 2002, which is relatively prime to $2002^{2}-1$. So $\varphi^{2001}\left(2002^{2}-1\right)=\varphi(2001)+1=\varphi(3 \cdot 23 \cdot 29)+1=$ $(3-1)(23-1)(29-1)+1=1233$.
|
1233
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Define $\varphi^{k}(n)$ as the number of positive integers that are less than or equal to $n / k$ and relatively prime to $n$. Find $\phi^{2001}\left(2002^{2}-1\right)$. (Hint: $\phi(2003)=2002$.)
|
$\varphi^{2001}\left(2002^{2}-1\right)=\varphi^{2001}(2001 \cdot 2003)=$ the number of $m$ that are relatively prime to both 2001 and 2003, where $m \leq 2003$. Since $\phi(n)=n-1$ implies that $n$ is prime, we must only check for those $m$ relatively prime to 2001, except for 2002, which is relatively prime to $2002^{2}-1$. So $\varphi^{2001}\left(2002^{2}-1\right)=\varphi(2001)+1=\varphi(3 \cdot 23 \cdot 29)+1=$ $(3-1)(23-1)(29-1)+1=1233$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n51. ",
"solution_match": "\nSolution: "
}
|
9f1210d3-ea11-5284-8daf-8e509846903b
| 611,059
|
Let $A B C D$ be a quadrilateral, and let $E, F, G, H$ be the respective midpoints of $A B, B C, C D, D A$. If $E G=12$ and $F H=15$, what is the maximum possible area of $A B C D$ ?
|
The area of $E F G H$ is $E G \cdot F H \sin \theta / 2$, where $\theta$ is the angle between $E G$ and $F H$. This is at most 90 . However, we claim the area of $A B C D$ is twice that of $E F G H$. To see this, notice that $E F=A C / 2=G H, F G=B D / 2=H E$, so $E F G H$ is a parallelogram. The half of this parallelogram lying inside triangle $D A B$ has area $(B D / 2)(h / 2)$, where $h$ is the height from $A$ to $B D$, and triangle $D A B$ itself has area $B D \cdot h / 2=2 \cdot(B D / 2)(h / 2)$. A similar computation holds in triangle $B C D$, proving the claim. Thus, the area of $A B C D$ is at most 180. And this maximum is attainable - just take a rectangle with $A B=C D=$ $15, B C=D A=12$.
|
180
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C D$ be a quadrilateral, and let $E, F, G, H$ be the respective midpoints of $A B, B C, C D, D A$. If $E G=12$ and $F H=15$, what is the maximum possible area of $A B C D$ ?
|
The area of $E F G H$ is $E G \cdot F H \sin \theta / 2$, where $\theta$ is the angle between $E G$ and $F H$. This is at most 90 . However, we claim the area of $A B C D$ is twice that of $E F G H$. To see this, notice that $E F=A C / 2=G H, F G=B D / 2=H E$, so $E F G H$ is a parallelogram. The half of this parallelogram lying inside triangle $D A B$ has area $(B D / 2)(h / 2)$, where $h$ is the height from $A$ to $B D$, and triangle $D A B$ itself has area $B D \cdot h / 2=2 \cdot(B D / 2)(h / 2)$. A similar computation holds in triangle $B C D$, proving the claim. Thus, the area of $A B C D$ is at most 180. And this maximum is attainable - just take a rectangle with $A B=C D=$ $15, B C=D A=12$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n52. ",
"solution_match": "\nSolution: "
}
|
2944aa69-18f1-5410-9b80-d345488e8e69
| 611,060
|
$A B C$ is a triangle with points $E, F$ on sides $A C, A B$, respectively. Suppose that $B E, C F$ intersect at $X$. It is given that $A F / F B=(A E / E C)^{2}$ and that $X$ is the midpoint of $B E$. Find the ratio $C X / X F$.
|
Let $x=A E / E C$. By Menelaus's theorem applied to triangle $A B E$ and line $C X F$,
$$
1=\frac{A F}{F B} \cdot \frac{B X}{X E} \cdot \frac{E C}{C A}=\frac{x^{2}}{x+1} .
$$
Thus, $x^{2}=x+1$, and $x$ must be positive, so $x=(1+\sqrt{5}) / 2$. Now apply Menelaus to triangle $A C F$ and line $B X E$, obtaining
$$
1=\frac{A E}{E C} \cdot \frac{C X}{X F} \cdot \frac{F B}{B A}=\frac{C X}{X F} \cdot \frac{x}{x^{2}+1},
$$
so $C X / X F=\left(x^{2}+1\right) / x=\left(2 x^{2}-x\right) / x=2 x-1=\sqrt{5}$.
|
\sqrt{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
$A B C$ is a triangle with points $E, F$ on sides $A C, A B$, respectively. Suppose that $B E, C F$ intersect at $X$. It is given that $A F / F B=(A E / E C)^{2}$ and that $X$ is the midpoint of $B E$. Find the ratio $C X / X F$.
|
Let $x=A E / E C$. By Menelaus's theorem applied to triangle $A B E$ and line $C X F$,
$$
1=\frac{A F}{F B} \cdot \frac{B X}{X E} \cdot \frac{E C}{C A}=\frac{x^{2}}{x+1} .
$$
Thus, $x^{2}=x+1$, and $x$ must be positive, so $x=(1+\sqrt{5}) / 2$. Now apply Menelaus to triangle $A C F$ and line $B X E$, obtaining
$$
1=\frac{A E}{E C} \cdot \frac{C X}{X F} \cdot \frac{F B}{B A}=\frac{C X}{X F} \cdot \frac{x}{x^{2}+1},
$$
so $C X / X F=\left(x^{2}+1\right) / x=\left(2 x^{2}-x\right) / x=2 x-1=\sqrt{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n53. ",
"solution_match": "\nSolution: "
}
|
f38c1584-b1f6-556e-b020-467df73115a6
| 611,061
|
How many pairs of integers $(a, b)$, with $1 \leq a \leq b \leq 60$, have the property that $b$ is divisible by $a$ and $b+1$ is divisible by $a+1$ ?
|
The divisibility condition is equivalent to $b-a$ being divisible by both $a$ and $a+1$, or, equivalently (since these are relatively prime), by $a(a+1)$. Any $b$ satisfying the condition is automatically $\geq a$, so it suffices to count the number of values $b-a \in$ $\{1-a, 2-a, \ldots, 60-a\}$ that are divisible by $a(a+1)$ and sum over all $a$. The number of such values will be precisely $60 /[a(a+1)]$ whenever this quantity is an integer, which fortunately happens for every $a \leq 5$; we count:
$a=1$ gives 30 values of $b$;
$a=2$ gives 10 values of $b$;
$a=3$ gives 5 values of $b$;
$a=4$ gives 3 values of $b$;
$a=5$ gives 2 values of $b$;
$a=6$ gives 2 values ( $b=6$ or 48);
any $a \geq 7$ gives only one value, namely $b=a$, since $b>a$ implies $b \geq a+a(a+1)>60$.
Adding these up, we get a total of 106 pairs.
|
106
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many pairs of integers $(a, b)$, with $1 \leq a \leq b \leq 60$, have the property that $b$ is divisible by $a$ and $b+1$ is divisible by $a+1$ ?
|
The divisibility condition is equivalent to $b-a$ being divisible by both $a$ and $a+1$, or, equivalently (since these are relatively prime), by $a(a+1)$. Any $b$ satisfying the condition is automatically $\geq a$, so it suffices to count the number of values $b-a \in$ $\{1-a, 2-a, \ldots, 60-a\}$ that are divisible by $a(a+1)$ and sum over all $a$. The number of such values will be precisely $60 /[a(a+1)]$ whenever this quantity is an integer, which fortunately happens for every $a \leq 5$; we count:
$a=1$ gives 30 values of $b$;
$a=2$ gives 10 values of $b$;
$a=3$ gives 5 values of $b$;
$a=4$ gives 3 values of $b$;
$a=5$ gives 2 values of $b$;
$a=6$ gives 2 values ( $b=6$ or 48);
any $a \geq 7$ gives only one value, namely $b=a$, since $b>a$ implies $b \geq a+a(a+1)>60$.
Adding these up, we get a total of 106 pairs.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n54. ",
"solution_match": "\nSolution: "
}
|
f894f36b-163c-5782-96d3-3adf75e52e9e
| 611,062
|
A sequence of positive integers is given by $a_{1}=1$ and $a_{n}=\operatorname{gcd}\left(a_{n-1}, n\right)+1$ for $n>1$. Calculate $a_{2002}$.
|
3. It is readily seen by induction that $a_{n} \leq n$ for all $n$. On the other hand, $a_{1999}$ is one greater than a divisor of 1999. Since 1999 is prime, we have $a_{1999}=2$ or 2000; the latter is not possible since $2000>1999$, so we have $a_{1999}=2$. Now we straightforwardly compute $a_{2000}=3, a_{2001}=4$, and $a_{2002}=3$.
|
3
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
A sequence of positive integers is given by $a_{1}=1$ and $a_{n}=\operatorname{gcd}\left(a_{n-1}, n\right)+1$ for $n>1$. Calculate $a_{2002}$.
|
3. It is readily seen by induction that $a_{n} \leq n$ for all $n$. On the other hand, $a_{1999}$ is one greater than a divisor of 1999. Since 1999 is prime, we have $a_{1999}=2$ or 2000; the latter is not possible since $2000>1999$, so we have $a_{1999}=2$. Now we straightforwardly compute $a_{2000}=3, a_{2001}=4$, and $a_{2002}=3$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n55. ",
"solution_match": "\nSolution: "
}
|
5a07ee50-435a-5789-a0e4-b73e53ba7704
| 611,063
|
$x, y$ are positive real numbers such that $x+y^{2}=x y$. What is the smallest possible value of $x$ ?
|
4 Notice that $x=y^{2} /(y-1)=2+(y-1)+1 /(y-1) \geq 2+2=4$. Conversely, $x=4$ is achievable, by taking $y=2$.
|
4
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
$x, y$ are positive real numbers such that $x+y^{2}=x y$. What is the smallest possible value of $x$ ?
|
4 Notice that $x=y^{2} /(y-1)=2+(y-1)+1 /(y-1) \geq 2+2=4$. Conversely, $x=4$ is achievable, by taking $y=2$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n56. ",
"solution_match": "\nSolution: "
}
|
9e1077c7-756c-5b4f-b157-7b85fbd1a35c
| 611,064
|
How many ways, without taking order into consideration, can 2002 be expressed as the sum of 3 positive integers (for instance, $1000+1000+2$ and $1000+2+1000$ are considered to be the same way)?
|
Call the three numbers that sum to $2002 A, B$, and $C$. In order to prevent redundancy, we will consider only cases where $A \leq B \leq C$. Then $A$ can range from 1 to 667 , inclusive. For odd $A$, there are $1000-\frac{3(A-1)}{2}$ possible values for $B$. For each choice of $A$ and $B$, there can only be one possible $C$, since the three numbers must add up to a fixed value. We can add up this arithmetic progression to find that there are 167167 possible combinations of $A, B, C$, for odd $A$. For each even $A$, there are $1002-\frac{3 A}{2}$ possible values for $B$. Therefore, there are 166833 possible combinations for even $A$. In total, this makes 334000 possibilities.
|
334000
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many ways, without taking order into consideration, can 2002 be expressed as the sum of 3 positive integers (for instance, $1000+1000+2$ and $1000+2+1000$ are considered to be the same way)?
|
Call the three numbers that sum to $2002 A, B$, and $C$. In order to prevent redundancy, we will consider only cases where $A \leq B \leq C$. Then $A$ can range from 1 to 667 , inclusive. For odd $A$, there are $1000-\frac{3(A-1)}{2}$ possible values for $B$. For each choice of $A$ and $B$, there can only be one possible $C$, since the three numbers must add up to a fixed value. We can add up this arithmetic progression to find that there are 167167 possible combinations of $A, B, C$, for odd $A$. For each even $A$, there are $1002-\frac{3 A}{2}$ possible values for $B$. Therefore, there are 166833 possible combinations for even $A$. In total, this makes 334000 possibilities.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n57. ",
"solution_match": "\nSolution: "
}
|
dab6414f-9502-5749-b4ad-d00ffc165814
| 611,065
|
A sequence is defined by $a_{0}=1$ and $a_{n}=2^{a_{n-1}}$ for $n \geq 1$. What is the last digit (in base 10) of $a_{15}$ ?
|
6. Certainly $a_{13} \geq 2$, so $a_{14}$ is divisible by $2^{2}=4$. Writing $a_{14}=4 k$, we have $a_{15}=2^{4 k}=16^{k}$. But every power of 16 ends in 6 , so this is the answer.
|
6
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
A sequence is defined by $a_{0}=1$ and $a_{n}=2^{a_{n-1}}$ for $n \geq 1$. What is the last digit (in base 10) of $a_{15}$ ?
|
6. Certainly $a_{13} \geq 2$, so $a_{14}$ is divisible by $2^{2}=4$. Writing $a_{14}=4 k$, we have $a_{15}=2^{4 k}=16^{k}$. But every power of 16 ends in 6 , so this is the answer.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n58. ",
"solution_match": "\nSolution: "
}
|
83372dc1-d990-5845-bef8-f3f0f9b3499a
| 611,066
|
Determine the value of
$$
1 \cdot 2-2 \cdot 3+3 \cdot 4-4 \cdot 5+\cdots+2001 \cdot 2002 .
$$
|
2004002. Rewrite the expression as
$$
2+3 \cdot(4-2)+5 \cdot(6-4)+\cdots+2001 \cdot(2002-2000)
$$
$$
=2+6+10+\cdots+4002 .
$$
This is an arithmetic progression with $(4002-2) / 4+1=1001$ terms and average 2002, so its sum is $1001 \cdot 2002=2004002$.
|
2004002
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Determine the value of
$$
1 \cdot 2-2 \cdot 3+3 \cdot 4-4 \cdot 5+\cdots+2001 \cdot 2002 .
$$
|
2004002. Rewrite the expression as
$$
2+3 \cdot(4-2)+5 \cdot(6-4)+\cdots+2001 \cdot(2002-2000)
$$
$$
=2+6+10+\cdots+4002 .
$$
This is an arithmetic progression with $(4002-2) / 4+1=1001$ terms and average 2002, so its sum is $1001 \cdot 2002=2004002$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n59. ",
"solution_match": "\nSolution: "
}
|
eadfbd18-fddc-54fa-81e0-adc38c53c441
| 611,067
|
A $5 \times 5$ square grid has the number -3 written in the upper-left square and the number 3 written in the lower-right square. In how many ways can the remaining squares be filled in with integers so that any two adjacent numbers differ by 1 , where two squares are adjacent if they share a common edge (but not if they share only a corner)?
|
250 If the square in row $i$, column $j$ contains the number $k$, let its "index" be $i+j-k$. The constraint on adjacent squares now says that if a square has index $r$, the squares to its right and below it each have index $r$ or $r+2$. The upper-left square has index 5 , and the lower-right square has index 7 , so every square must have index 5 or 7 . The boundary separating the two types of squares is a path consisting of upward and rightward steps; it can be extended along the grid's border so as to obtain a path between the lower-left and upper-right corners. Conversely, any such path uniquely determines each square's index and hence the entire array of numbers - except that the two paths lying entirely along the border of the grid fail to separate the upper-left from the lower-right square and thus do not create valid arrays (since these two squares should have different indices). Each path consists of 5 upward and 5 rightward steps, so there are $\binom{10}{5}=252$ paths, but two are impossible, so the answer is 250 .
|
250
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A $5 \times 5$ square grid has the number -3 written in the upper-left square and the number 3 written in the lower-right square. In how many ways can the remaining squares be filled in with integers so that any two adjacent numbers differ by 1 , where two squares are adjacent if they share a common edge (but not if they share only a corner)?
|
250 If the square in row $i$, column $j$ contains the number $k$, let its "index" be $i+j-k$. The constraint on adjacent squares now says that if a square has index $r$, the squares to its right and below it each have index $r$ or $r+2$. The upper-left square has index 5 , and the lower-right square has index 7 , so every square must have index 5 or 7 . The boundary separating the two types of squares is a path consisting of upward and rightward steps; it can be extended along the grid's border so as to obtain a path between the lower-left and upper-right corners. Conversely, any such path uniquely determines each square's index and hence the entire array of numbers - except that the two paths lying entirely along the border of the grid fail to separate the upper-left from the lower-right square and thus do not create valid arrays (since these two squares should have different indices). Each path consists of 5 upward and 5 rightward steps, so there are $\binom{10}{5}=252$ paths, but two are impossible, so the answer is 250 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n60. ",
"solution_match": "\nSolution: "
}
|
55e0f4b1-ffce-5b67-a4ac-101fb21936fc
| 611,068
|
Bob Barker went back to school for a PhD in math, and decided to raise the intellectual level of The Price is Right by having contestants guess how many objects exist of a certain type, without going over. The number of points you will get is the percentage of the correct answer, divided by 10 , with no points for going over (i.e. a maximum of 10 points).
Let's see the first object for our contestants...a table of shape (5, 4, 3, 2, 1) is an arrangement of the integers 1 through 15 with five numbers in the top row, four in the next, three in the next, two in the next, and one in the last, such that each row and each column is increasing (from left to right, and top to bottom, respectively). For instance:
```
1
6
10 11 12
13 14
15
```
is one table. How many tables are there?
|
$15!/\left(3^{4} \cdot 5^{3} \cdot 7^{2} \cdot 9\right)=292864$. These are Standard Young Tableaux.
|
292864
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Bob Barker went back to school for a PhD in math, and decided to raise the intellectual level of The Price is Right by having contestants guess how many objects exist of a certain type, without going over. The number of points you will get is the percentage of the correct answer, divided by 10 , with no points for going over (i.e. a maximum of 10 points).
Let's see the first object for our contestants...a table of shape (5, 4, 3, 2, 1) is an arrangement of the integers 1 through 15 with five numbers in the top row, four in the next, three in the next, two in the next, and one in the last, such that each row and each column is increasing (from left to right, and top to bottom, respectively). For instance:
```
1
6
10 11 12
13 14
15
```
is one table. How many tables are there?
|
$15!/\left(3^{4} \cdot 5^{3} \cdot 7^{2} \cdot 9\right)=292864$. These are Standard Young Tableaux.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n61. ",
"solution_match": "\nSolution: "
}
|
93c09ce1-8bed-56a1-ba0a-c02a94316e64
| 611,069
|
Our next object up for bid is an arithmetic progression of primes. For example, the primes 3,5 , and 7 form an arithmetic progression of length 3 . What is the largest possible length of an arithmetic progression formed of positive primes less than $1,000,000$ ? Be prepared to justify your answer.
|
12 . We can get 12 with 110437 and difference 13860 .
|
12
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Our next object up for bid is an arithmetic progression of primes. For example, the primes 3,5 , and 7 form an arithmetic progression of length 3 . What is the largest possible length of an arithmetic progression formed of positive primes less than $1,000,000$ ? Be prepared to justify your answer.
|
12 . We can get 12 with 110437 and difference 13860 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n62. ",
"solution_match": "\nSolution: "
}
|
fe810f6d-f12a-5612-ab2c-dca11fe8a5a6
| 611,070
|
Our third and final item comes to us from Germany, I mean Geometry. It is known that a regular $n$-gon can be constructed with straightedge and compass if $n$ is a prime that is 1 plus a power of 2 . It is also possible to construct a $2 n$-gon whenever an $n$-gon is constructible, or a $p_{1} p_{2} \cdots p_{m}$-gon where the $p_{i}$ 's are distinct primes of the above form. What is really interesting is that these conditions, together with the fact that we can construct a square, is that they give us all constructible regular $n$-gons. What is the largest $n$ less than $4,300,000,000$ such that a regular $n$-gon is constructible?
|
The known primes of this form (Fermat primes) are 3, 5, 17, 257, and 65537, and the result is due to Gauss (German). If there are other such primes (unknown), then they are much bigger than $10^{10}$. So for each product of these primes, we can divide $4.3 \cdot 10^{9}$ by that number and take $\log _{2}$ to find the largest power of 2 to multiply by, then compare the resulting numbers. There are 32 cases to check, or just observe that $2^{32}=4,294,967,296$ is so close that there's likely a shortcut. Note that $2^{32}+1$ is divisible by 641 , and hence not prime. $3 \cdot 5 \cdot 17 \cdot 257 \cdot 65537=2^{32}-1$ is smaller; replacing any of the factors by the closest power of 2 only decreases the product, and there's not enough room to squeeze in an extra factor of 2 without replacing all of them, and that gives us $2^{32}$, so indeed that it is the answer.
Help control the pet population. Have your pets spayed or neutered. Bye-bye.
|
4294967296
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Our third and final item comes to us from Germany, I mean Geometry. It is known that a regular $n$-gon can be constructed with straightedge and compass if $n$ is a prime that is 1 plus a power of 2 . It is also possible to construct a $2 n$-gon whenever an $n$-gon is constructible, or a $p_{1} p_{2} \cdots p_{m}$-gon where the $p_{i}$ 's are distinct primes of the above form. What is really interesting is that these conditions, together with the fact that we can construct a square, is that they give us all constructible regular $n$-gons. What is the largest $n$ less than $4,300,000,000$ such that a regular $n$-gon is constructible?
|
The known primes of this form (Fermat primes) are 3, 5, 17, 257, and 65537, and the result is due to Gauss (German). If there are other such primes (unknown), then they are much bigger than $10^{10}$. So for each product of these primes, we can divide $4.3 \cdot 10^{9}$ by that number and take $\log _{2}$ to find the largest power of 2 to multiply by, then compare the resulting numbers. There are 32 cases to check, or just observe that $2^{32}=4,294,967,296$ is so close that there's likely a shortcut. Note that $2^{32}+1$ is divisible by 641 , and hence not prime. $3 \cdot 5 \cdot 17 \cdot 257 \cdot 65537=2^{32}-1$ is smaller; replacing any of the factors by the closest power of 2 only decreases the product, and there's not enough room to squeeze in an extra factor of 2 without replacing all of them, and that gives us $2^{32}$, so indeed that it is the answer.
Help control the pet population. Have your pets spayed or neutered. Bye-bye.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n63. ",
"solution_match": "\nSolution: "
}
|
775e9ae2-f7d1-57fc-a33d-2e0ec203050d
| 611,071
|
Determine the number of palindromes that are less than 1000 .
|
Every one-digit number (there are nine) is a palindrome. The two-digit palindromes have the form $\underline{a} \underline{a}$ for a nonzero digit $a$, so there are nine of them. A three-digit palindrome is $\underline{a} \underline{b} \underline{a}$ with $a$ a nonzero digit and $b$ any digit, so there are $9 \times 10=90$ of these. Thus the number of palindromes less than 1000 is $9+9+90=\mathbf{1 0 8}$.
|
108
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Determine the number of palindromes that are less than 1000 .
|
Every one-digit number (there are nine) is a palindrome. The two-digit palindromes have the form $\underline{a} \underline{a}$ for a nonzero digit $a$, so there are nine of them. A three-digit palindrome is $\underline{a} \underline{b} \underline{a}$ with $a$ a nonzero digit and $b$ any digit, so there are $9 \times 10=90$ of these. Thus the number of palindromes less than 1000 is $9+9+90=\mathbf{1 0 8}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-team-solutions.jsonl",
"problem_match": "\n1. [15]",
"solution_match": "\nSolution. "
}
|
23c55222-0dda-5725-8cad-30b90b90b480
| 611,072
|
Determine the number of four-digit integers $n$ such that $n$ and $2 n$ are both palindromes.
|
Let $n=\underline{a} \underline{b} \underline{b} \underline{a}$. If $a, b \leq 4$ then there are no carries in the multiplication $n \times 2$, and $2 n=(2 a)(2 b)(2 b)(2 a)$ is a palindrome. We shall show conversely that if $n$ and $2 n$ are palindromes, then necessarily $a, b \leq 4$. Hence the answer to the problem is $4 \times 5=\mathbf{2 0}$ (because $a$ cannot be zero).
If $a \geq 5$ then $2 n$ is a five-digit number whose most significant digit is 1 , but because $2 n$ is even, its least significant digit is even, contradicting the assumption that $2 n$ is a palindrome. Therefore $a \leq 4$. Consequently $2 n$ is a four-digit number, and its tens and hundreds digits must be equal. Because $a \leq 4$, there is no carry out of the ones place in the multiplication $n \times 2$, and therefore the tens digit of $2 n$ is the ones digit of $2 b$. In particular, the tens digit of $2 n$ is even. But if $b \geq 5$, the carry out of the tens place makes the hundreds digit of $2 n$ odd, which is impossible. Hence $b \leq 4$ as well.
|
20
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Determine the number of four-digit integers $n$ such that $n$ and $2 n$ are both palindromes.
|
Let $n=\underline{a} \underline{b} \underline{b} \underline{a}$. If $a, b \leq 4$ then there are no carries in the multiplication $n \times 2$, and $2 n=(2 a)(2 b)(2 b)(2 a)$ is a palindrome. We shall show conversely that if $n$ and $2 n$ are palindromes, then necessarily $a, b \leq 4$. Hence the answer to the problem is $4 \times 5=\mathbf{2 0}$ (because $a$ cannot be zero).
If $a \geq 5$ then $2 n$ is a five-digit number whose most significant digit is 1 , but because $2 n$ is even, its least significant digit is even, contradicting the assumption that $2 n$ is a palindrome. Therefore $a \leq 4$. Consequently $2 n$ is a four-digit number, and its tens and hundreds digits must be equal. Because $a \leq 4$, there is no carry out of the ones place in the multiplication $n \times 2$, and therefore the tens digit of $2 n$ is the ones digit of $2 b$. In particular, the tens digit of $2 n$ is even. But if $b \geq 5$, the carry out of the tens place makes the hundreds digit of $2 n$ odd, which is impossible. Hence $b \leq 4$ as well.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-team-solutions.jsonl",
"problem_match": "\n2. [30]",
"solution_match": "\nSolution. "
}
|
837f465f-c493-50f5-8e06-931bd25eaafd
| 611,073
|
In this problem only, assume that $s_{1}=4$ and that exactly one board square, say square number $n$, is marked with an arrow. Determine all choices of $n$ that maximize the average distance in squares the first player will travel in his first two turns.
|
Because expectation is linear, the average distance the first player travels in his first two turns is the average sum of two rolls of his die (which does not depend on the board configuration) plus four times the probability that he lands on the arrow on one of his first two turns. Thus we just need to maximize the probability that player 1 lands on the arrow in his first two turns. If $n \geq 5$, player 1 cannot land on the arrow in his first turn, so he encounters the arrow with probability at most $1 / 4$. If instead $n \leq 4$, player 1 has a $1 / 4$ chance of landing on the arrow on his first turn. If he misses, then he has a $1 / 4$ chance of hitting the arrow on his second turn provided that he is not beyond square $n$ already. The chance that player 1's first roll left him on square $n-1$ or farther left is $(n-1) / 4$. Hence his probability of benefiting from the arrow in his first two turns is $1 / 4+(1 / 4)(n-1) / 4$, which is maximized for $n=4$, where it is greater than the value of $1 / 4$ that we get from $n \geq 5$. Hence the answer is $n=4$.
|
4
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In this problem only, assume that $s_{1}=4$ and that exactly one board square, say square number $n$, is marked with an arrow. Determine all choices of $n$ that maximize the average distance in squares the first player will travel in his first two turns.
|
Because expectation is linear, the average distance the first player travels in his first two turns is the average sum of two rolls of his die (which does not depend on the board configuration) plus four times the probability that he lands on the arrow on one of his first two turns. Thus we just need to maximize the probability that player 1 lands on the arrow in his first two turns. If $n \geq 5$, player 1 cannot land on the arrow in his first turn, so he encounters the arrow with probability at most $1 / 4$. If instead $n \leq 4$, player 1 has a $1 / 4$ chance of landing on the arrow on his first turn. If he misses, then he has a $1 / 4$ chance of hitting the arrow on his second turn provided that he is not beyond square $n$ already. The chance that player 1's first roll left him on square $n-1$ or farther left is $(n-1) / 4$. Hence his probability of benefiting from the arrow in his first two turns is $1 / 4+(1 / 4)(n-1) / 4$, which is maximized for $n=4$, where it is greater than the value of $1 / 4$ that we get from $n \geq 5$. Hence the answer is $n=4$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-team-solutions.jsonl",
"problem_match": "\n8. [35]",
"solution_match": "\nSolution. "
}
|
12f56826-323a-54a2-9b4a-2acdfc652d19
| 611,079
|
Exhibit a configuration of the board and a choice of $s_{1}$ and $s_{2}$ so that $s_{1}>s_{2}$, yet the second player wins with probability strictly greater than $\frac{1}{2}$.
|
Let $s_{1}=3$ and $s_{2}=2$ and place an arrow on all the even-numbered squares. In this configuration, player 1 can move at most six squares in a turn: up to three from his roll and an additional three if his roll landed him on an arrow. Hence player 1 cannot win on his first or second turn. Player 2, however, wins immediately if she ever lands on an arrow. Thus player 2 has probability $1 / 2$ of winning on her first turn, and failing that, she has probability $1 / 2$ of winning on her second turn. Hence player 2 wins with probability at least $1 / 2+(1 / 2)(1 / 2)=3 / 4$.
|
\frac{3}{4}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Exhibit a configuration of the board and a choice of $s_{1}$ and $s_{2}$ so that $s_{1}>s_{2}$, yet the second player wins with probability strictly greater than $\frac{1}{2}$.
|
Let $s_{1}=3$ and $s_{2}=2$ and place an arrow on all the even-numbered squares. In this configuration, player 1 can move at most six squares in a turn: up to three from his roll and an additional three if his roll landed him on an arrow. Hence player 1 cannot win on his first or second turn. Player 2, however, wins immediately if she ever lands on an arrow. Thus player 2 has probability $1 / 2$ of winning on her first turn, and failing that, she has probability $1 / 2$ of winning on her second turn. Hence player 2 wins with probability at least $1 / 2+(1 / 2)(1 / 2)=3 / 4$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-team-solutions.jsonl",
"problem_match": "\n10. [30]",
"solution_match": "\nSolution. "
}
|
3fb767e4-d8f3-59f0-a3c5-26f15010cfcc
| 611,081
|
Find the smallest value of $x$ such that $a \geq 14 \sqrt{a}-x$ for all nonnegative $a$.
|
49
We want to find the smallest value of $x$ such that $x \geq 14$ sqrta $-a$ for all $a$. This is just the maximum possible value of $14 \sqrt{a}-a=49-(\sqrt{a}-7)^{2}$, which is clearly 49 , achieved when $a=49$.
|
49
|
Yes
|
Yes
|
math-word-problem
|
Inequalities
|
Find the smallest value of $x$ such that $a \geq 14 \sqrt{a}-x$ for all nonnegative $a$.
|
49
We want to find the smallest value of $x$ such that $x \geq 14$ sqrta $-a$ for all $a$. This is just the maximum possible value of $14 \sqrt{a}-a=49-(\sqrt{a}-7)^{2}$, which is clearly 49 , achieved when $a=49$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-alg-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution: "
}
|
baf84fc2-3f49-5542-8d7a-f7cb3296ec4b
| 611,084
|
Compute $\frac{\tan ^{2}\left(20^{\circ}\right)-\sin ^{2}\left(20^{\circ}\right)}{\tan ^{2}\left(20^{\circ}\right) \sin ^{2}\left(20^{\circ}\right)}$.
|
1
If we multiply top and bottom by $\cos ^{2}\left(20^{\circ}\right)$, the numerator becomes $\sin ^{2}\left(20^{\circ}\right) \cdot(1-$ $\left.\cos ^{2} 20^{\circ}\right)=\sin ^{4}\left(20^{\circ}\right)$, while the denominator becomes $\sin ^{4}\left(20^{\circ}\right)$ also. So they are equal, and the ratio is 1 .
|
1
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Compute $\frac{\tan ^{2}\left(20^{\circ}\right)-\sin ^{2}\left(20^{\circ}\right)}{\tan ^{2}\left(20^{\circ}\right) \sin ^{2}\left(20^{\circ}\right)}$.
|
1
If we multiply top and bottom by $\cos ^{2}\left(20^{\circ}\right)$, the numerator becomes $\sin ^{2}\left(20^{\circ}\right) \cdot(1-$ $\left.\cos ^{2} 20^{\circ}\right)=\sin ^{4}\left(20^{\circ}\right)$, while the denominator becomes $\sin ^{4}\left(20^{\circ}\right)$ also. So they are equal, and the ratio is 1 .
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-alg-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\n## Solution: "
}
|
2579febe-ac02-543b-ab5b-93ef94fd01f9
| 611,085
|
Find the smallest $n$ such that $n$ ! ends in 290 zeroes.
|
1170
Each 0 represents a factor of $10=2 \cdot 5$. Thus, we wish to find the smallest factorial that contains at least 290 2's and 290 5's in its prime factorization. Let this number be $n$ !, so the factorization of $n$ ! contains 2 to the power $p$ and 5 to the power $q$, where
$$
p=\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{n}{2^{2}}\right\rfloor+\left\lfloor\frac{n}{2^{3}}\right\rfloor+\cdots \text { and } q=\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{5^{2}}\right\rfloor+\left\lfloor\frac{n}{5^{3}}\right\rfloor+\cdots
$$
(this takes into account one factor for each single multiple of 2 or 5 that is $\leq n$, an additional factor for each multiple of $2^{2}$ or $5^{2}$, and so on). Naturally, $p \geq q$ because 2 is smaller than 5 . Thus, we want to bring $q$ as low to 290 as possible. If $q=\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{5^{2}}\right\rfloor+\left\lfloor\frac{n}{5^{3}}\right\rfloor+\cdots$, we form a rough geometric sequence (by taking away the floor function) whose sum is represented by $290 \approx \frac{n / 5}{1-1 / 5}$. Hence we estimate $n=1160$, and this gives us $q=288$. Adding 10 to the value of $n$ gives the necessary two additional factors of 5 , and so the answer is 1170 .
|
1170
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the smallest $n$ such that $n$ ! ends in 290 zeroes.
|
1170
Each 0 represents a factor of $10=2 \cdot 5$. Thus, we wish to find the smallest factorial that contains at least 290 2's and 290 5's in its prime factorization. Let this number be $n$ !, so the factorization of $n$ ! contains 2 to the power $p$ and 5 to the power $q$, where
$$
p=\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{n}{2^{2}}\right\rfloor+\left\lfloor\frac{n}{2^{3}}\right\rfloor+\cdots \text { and } q=\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{5^{2}}\right\rfloor+\left\lfloor\frac{n}{5^{3}}\right\rfloor+\cdots
$$
(this takes into account one factor for each single multiple of 2 or 5 that is $\leq n$, an additional factor for each multiple of $2^{2}$ or $5^{2}$, and so on). Naturally, $p \geq q$ because 2 is smaller than 5 . Thus, we want to bring $q$ as low to 290 as possible. If $q=\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{5^{2}}\right\rfloor+\left\lfloor\frac{n}{5^{3}}\right\rfloor+\cdots$, we form a rough geometric sequence (by taking away the floor function) whose sum is represented by $290 \approx \frac{n / 5}{1-1 / 5}$. Hence we estimate $n=1160$, and this gives us $q=288$. Adding 10 to the value of $n$ gives the necessary two additional factors of 5 , and so the answer is 1170 .
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-alg-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution: "
}
|
03e4c8a8-8b38-50c2-982c-8ce6bcdbed1f
| 611,086
|
Simplify: $2 \sqrt{1.5+\sqrt{2}}-(1.5+\sqrt{2})$.
|
$1 / 2$
The given expression equals $\sqrt{6+4 \sqrt{2}}-(1.5+\sqrt{2})=\sqrt{6+2 \sqrt{8}}-(1.5+\sqrt{2})$. But on inspection, we see that $(\sqrt{2}+\sqrt{4})^{2}=6+2 \sqrt{8}$, so the answer is $(\sqrt{2}+\sqrt{4})-(1.5+\sqrt{2})=$ $2-3 / 2=1 / 2$.
|
\frac{1}{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Simplify: $2 \sqrt{1.5+\sqrt{2}}-(1.5+\sqrt{2})$.
|
$1 / 2$
The given expression equals $\sqrt{6+4 \sqrt{2}}-(1.5+\sqrt{2})=\sqrt{6+2 \sqrt{8}}-(1.5+\sqrt{2})$. But on inspection, we see that $(\sqrt{2}+\sqrt{4})^{2}=6+2 \sqrt{8}$, so the answer is $(\sqrt{2}+\sqrt{4})-(1.5+\sqrt{2})=$ $2-3 / 2=1 / 2$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-alg-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\n## Solution: "
}
|
13476d2f-5ba4-5e2c-bc92-281c22403344
| 611,087
|
Several positive integers are given, not necessarily all different. Their sum is 2003. Suppose that $n_{1}$ of the given numbers are equal to $1, n_{2}$ of them are equal to $2, \ldots$, $n_{2003}$ of them are equal to 2003 . Find the largest possible value of
$$
n_{2}+2 n_{3}+3 n_{4}+\cdots+2002 n_{2003} .
$$
|
2002
The sum of all the numbers is $n_{1}+2 n_{2}+\cdots+2003 n_{2003}$, while the number of numbers is $n_{1}+n_{2}+\cdots+n_{2003}$. Hence, the desired quantity equals
$$
\begin{gathered}
\left(n_{1}+2 n_{2}+\cdots+2003 n_{2003}\right)-\left(n_{1}+n_{2}+\cdots+n_{2003}\right) \\
=(\text { sum of the numbers })-(\text { number of numbers }) \\
=2003-(\text { number of numbers }),
\end{gathered}
$$
which is maximized when the number of numbers is minimized. Hence, we should have just one number, equal to 2003, and then the specified sum is $2003-1=2002$.
Comment: On the day of the contest, a protest was lodged (successfully) on the grounds that the use of the words "several" and "their" in the problem statement implies there must be at least 2 numbers. Then the answer is 2001, and this maximum is achieved by any two numbers whose sum is 2003 .me way.)
|
2002
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Several positive integers are given, not necessarily all different. Their sum is 2003. Suppose that $n_{1}$ of the given numbers are equal to $1, n_{2}$ of them are equal to $2, \ldots$, $n_{2003}$ of them are equal to 2003 . Find the largest possible value of
$$
n_{2}+2 n_{3}+3 n_{4}+\cdots+2002 n_{2003} .
$$
|
2002
The sum of all the numbers is $n_{1}+2 n_{2}+\cdots+2003 n_{2003}$, while the number of numbers is $n_{1}+n_{2}+\cdots+n_{2003}$. Hence, the desired quantity equals
$$
\begin{gathered}
\left(n_{1}+2 n_{2}+\cdots+2003 n_{2003}\right)-\left(n_{1}+n_{2}+\cdots+n_{2003}\right) \\
=(\text { sum of the numbers })-(\text { number of numbers }) \\
=2003-(\text { number of numbers }),
\end{gathered}
$$
which is maximized when the number of numbers is minimized. Hence, we should have just one number, equal to 2003, and then the specified sum is $2003-1=2002$.
Comment: On the day of the contest, a protest was lodged (successfully) on the grounds that the use of the words "several" and "their" in the problem statement implies there must be at least 2 numbers. Then the answer is 2001, and this maximum is achieved by any two numbers whose sum is 2003 .me way.)
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-alg-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\n## Solution: "
}
|
99e5d76f-c7c3-5460-b56b-34c10546805e
| 611,088
|
Let $a_{1}=1$, and let $a_{n}=\left\lfloor n^{3} / a_{n-1}\right\rfloor$ for $n>1$. Determine the value of $a_{999}$.
|
999
We claim that for any odd $n, a_{n}=n$. The proof is by induction. To get the base cases $n=1$, 3, we compute $a_{1}=1, a_{2}=\left\lfloor 2^{3} / 1\right\rfloor=8, a_{3}=\left\lfloor 3^{3} / 8\right\rfloor=3$. And if the claim holds for odd $n \geq 3$, then $a_{n+1}=\left\lfloor(n+1)^{3} / n\right\rfloor=n^{2}+3 n+3$, so $a_{n+2}=$ $\left\lfloor(n+2)^{3} /\left(n^{2}+3 n+3\right)\right\rfloor=\left\lfloor\left(n^{3}+6 n^{2}+12 n+8\right) /\left(n^{2}+3 n+2\right)\right\rfloor=\left\lfloor n+2+\frac{n^{2}+3 n+2}{n^{2}+3 n+3}\right\rfloor=n+2$.
So the claim holds, and in particular, $a_{999}=999$.
|
999
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $a_{1}=1$, and let $a_{n}=\left\lfloor n^{3} / a_{n-1}\right\rfloor$ for $n>1$. Determine the value of $a_{999}$.
|
999
We claim that for any odd $n, a_{n}=n$. The proof is by induction. To get the base cases $n=1$, 3, we compute $a_{1}=1, a_{2}=\left\lfloor 2^{3} / 1\right\rfloor=8, a_{3}=\left\lfloor 3^{3} / 8\right\rfloor=3$. And if the claim holds for odd $n \geq 3$, then $a_{n+1}=\left\lfloor(n+1)^{3} / n\right\rfloor=n^{2}+3 n+3$, so $a_{n+2}=$ $\left\lfloor(n+2)^{3} /\left(n^{2}+3 n+3\right)\right\rfloor=\left\lfloor\left(n^{3}+6 n^{2}+12 n+8\right) /\left(n^{2}+3 n+2\right)\right\rfloor=\left\lfloor n+2+\frac{n^{2}+3 n+2}{n^{2}+3 n+3}\right\rfloor=n+2$.
So the claim holds, and in particular, $a_{999}=999$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-alg-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nSolution: "
}
|
d66a1b34-036d-55fc-bb23-7055396986fa
| 611,089
|
Let $a, b, c$ be the three roots of $p(x)=x^{3}+x^{2}-333 x-1001$. Find $a^{3}+b^{3}+c^{3}$.
|
2003
We know that $x^{3}+x^{2}-333 x-1001=(x-a)(x-b)(x-c)=x^{3}-(a+b+c) x^{2}+$ $(a b+b c+c a) x-a b c$. Also, $(a+b+c)^{3}-3(a+b+c)(a b+b c+c a)+3 a b c=a^{3}+b^{3}+c^{3}$. Thus, $a^{3}+b^{3}+c^{3}=(-1)^{3}-3(-1)(-333)+3 \cdot 1001=2003$.
|
2003
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a, b, c$ be the three roots of $p(x)=x^{3}+x^{2}-333 x-1001$. Find $a^{3}+b^{3}+c^{3}$.
|
2003
We know that $x^{3}+x^{2}-333 x-1001=(x-a)(x-b)(x-c)=x^{3}-(a+b+c) x^{2}+$ $(a b+b c+c a) x-a b c$. Also, $(a+b+c)^{3}-3(a+b+c)(a b+b c+c a)+3 a b c=a^{3}+b^{3}+c^{3}$. Thus, $a^{3}+b^{3}+c^{3}=(-1)^{3}-3(-1)(-333)+3 \cdot 1001=2003$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-alg-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nSolution: "
}
|
4ff530c2-6867-5b97-aafa-c1a2a5b3d922
| 611,090
|
Find the value of $\frac{1}{3^{2}+1}+\frac{1}{4^{2}+2}+\frac{1}{5^{2}+3}+\cdots$.
|
$13 / 36$
Each term takes the form
$$
\frac{1}{n^{2}+(n-2)}=\frac{1}{(n+2) \cdot(n-1)}
$$
Using the method of partial fractions, we can write (for some constants $A, B$ )
$$
\begin{gathered}
\frac{1}{(n+2) \cdot(n-1)}=\frac{A}{(n+2)}+\frac{B}{(n-1)} \\
\quad \Rightarrow 1=A \cdot(n-1)+B \cdot(n+2)
\end{gathered}
$$
Setting $n=1$ we get $B=\frac{1}{3}$, and similarly with $n=-2$ we get $A=-\frac{1}{3}$. Hence the sum becomes
$$
\frac{1}{3} \cdot\left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\cdots\right] .
$$
Thus, it telescopes, and the only terms that do not cancel produce a sum of $\frac{1}{3} \cdot\left(\frac{1}{2}+\right.$ $\left.\frac{1}{3}+\frac{1}{4}\right)=\frac{13}{36}$.
|
\frac{13}{36}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the value of $\frac{1}{3^{2}+1}+\frac{1}{4^{2}+2}+\frac{1}{5^{2}+3}+\cdots$.
|
$13 / 36$
Each term takes the form
$$
\frac{1}{n^{2}+(n-2)}=\frac{1}{(n+2) \cdot(n-1)}
$$
Using the method of partial fractions, we can write (for some constants $A, B$ )
$$
\begin{gathered}
\frac{1}{(n+2) \cdot(n-1)}=\frac{A}{(n+2)}+\frac{B}{(n-1)} \\
\quad \Rightarrow 1=A \cdot(n-1)+B \cdot(n+2)
\end{gathered}
$$
Setting $n=1$ we get $B=\frac{1}{3}$, and similarly with $n=-2$ we get $A=-\frac{1}{3}$. Hence the sum becomes
$$
\frac{1}{3} \cdot\left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\cdots\right] .
$$
Thus, it telescopes, and the only terms that do not cancel produce a sum of $\frac{1}{3} \cdot\left(\frac{1}{2}+\right.$ $\left.\frac{1}{3}+\frac{1}{4}\right)=\frac{13}{36}$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-alg-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\n## Solution: "
}
|
adeaa20e-f41b-55d4-91a6-0fe0e1a2ad5f
| 611,091
|
For how many integers $n$, for $1 \leq n \leq 1000$, is the number $\frac{1}{2}\binom{2 n}{n}$ even?
|
990
In fact, the expression $\binom{2 n}{n}$ is always even, and it is not a multiple of four if and only if $n$ is a power of 2 , and there are 10 powers of 2 between 1 and 1000 .
Let $f(N)$ denote the number of factors of 2 in $N$. Thus,
$$
f(n!)=\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{n}{4}\right\rfloor+\left\lfloor\frac{n}{8}\right\rfloor+\cdots=\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^{k}}\right\rfloor .
$$
Also, it is clear that $f(a b)=f(a)+f(b)$ and $f\left(\frac{a}{b}\right)=f(a)-f(b)$ for integers $a, b$. Now for any positive integer $n$, let $m$ be the integer such that $2^{m} \leq n<2^{m+1}$. Then
$$
\begin{aligned}
f\left(\binom{2 n}{n}\right)=f\left(\frac{(2 n)!}{n!n!}\right) & =\sum_{k=1}^{\infty}\left\lfloor\frac{2 n}{2^{k}}\right\rfloor-2\left(\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^{k}}\right\rfloor\right) \\
& =\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^{k-1}}\right\rfloor-2\left(\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^{k}}\right\rfloor\right) \\
& =\lfloor n\rfloor-\left(\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^{k}}\right\rfloor\right) \\
& =n-\left(\sum_{k=1}^{m}\left\lfloor\frac{n}{2^{k}}\right\rfloor\right) \\
& \geq n-\left(\sum_{k=1}^{m} \frac{n}{2^{k}}\right) \\
& =n-n\left(\frac{2^{m}-1}{2^{m}}\right)=\frac{n}{2^{m}} \geq 1
\end{aligned}
$$
Both equalities hold when $n=2^{m}$, and otherwise, $f\left(\binom{2 n}{n}\right)>1$.
|
990
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
For how many integers $n$, for $1 \leq n \leq 1000$, is the number $\frac{1}{2}\binom{2 n}{n}$ even?
|
990
In fact, the expression $\binom{2 n}{n}$ is always even, and it is not a multiple of four if and only if $n$ is a power of 2 , and there are 10 powers of 2 between 1 and 1000 .
Let $f(N)$ denote the number of factors of 2 in $N$. Thus,
$$
f(n!)=\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{n}{4}\right\rfloor+\left\lfloor\frac{n}{8}\right\rfloor+\cdots=\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^{k}}\right\rfloor .
$$
Also, it is clear that $f(a b)=f(a)+f(b)$ and $f\left(\frac{a}{b}\right)=f(a)-f(b)$ for integers $a, b$. Now for any positive integer $n$, let $m$ be the integer such that $2^{m} \leq n<2^{m+1}$. Then
$$
\begin{aligned}
f\left(\binom{2 n}{n}\right)=f\left(\frac{(2 n)!}{n!n!}\right) & =\sum_{k=1}^{\infty}\left\lfloor\frac{2 n}{2^{k}}\right\rfloor-2\left(\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^{k}}\right\rfloor\right) \\
& =\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^{k-1}}\right\rfloor-2\left(\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^{k}}\right\rfloor\right) \\
& =\lfloor n\rfloor-\left(\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^{k}}\right\rfloor\right) \\
& =n-\left(\sum_{k=1}^{m}\left\lfloor\frac{n}{2^{k}}\right\rfloor\right) \\
& \geq n-\left(\sum_{k=1}^{m} \frac{n}{2^{k}}\right) \\
& =n-n\left(\frac{2^{m}-1}{2^{m}}\right)=\frac{n}{2^{m}} \geq 1
\end{aligned}
$$
Both equalities hold when $n=2^{m}$, and otherwise, $f\left(\binom{2 n}{n}\right)>1$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-alg-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nSolution: "
}
|
17fc4867-f607-533d-a270-33cba91ba1f3
| 611,092
|
Suppose $P(x)$ is a polynomial such that $P(1)=1$ and
$$
\frac{P(2 x)}{P(x+1)}=8-\frac{56}{x+7}
$$
for all real $x$ for which both sides are defined. Find $P(-1)$.
|
$\quad-5 / 21$
Cross-multiplying gives $(x+7) P(2 x)=8 x P(x+1)$. If $P$ has degree $n$ and leading coefficient $c$, then the leading coefficients of the two sides are $2^{n} c$ and $8 c$, so $n=3$. Now $x=0$ is a root of the right-hand side, so it's a root of the left-hand side, so that $P(x)=x Q(x)$ for some polynomial $Q \Rightarrow 2 x(x+7) Q(2 x)=8 x(x+1) Q(x+1)$ or $(x+7) Q(2 x)=4(x+1) Q(x+1)$. Similarly, we see that $x=-1$ is a root of the left-hand side, giving $Q(x)=(x+2) R(x)$ for some polynomial $R \Rightarrow 2(x+1)(x+7) R(2 x)=$ $4(x+1)(x+3) R(x+1)$, or $(x+7) R(2 x)=2(x+3) R(x+1)$. Now $x=-3$ is a root of the left-hand side, so $R(x)=(x+6) S(x)$ for some polynomial $S$.
At this point, $P(x)=x(x+2)(x+6) S(x)$, but $P$ has degree 3 , so $S$ must be a constant. Since $P(1)=1$, we get $S=1 / 21$, and then $P(-1)=(-1)(1)(5) / 21=-5 / 21$.
|
-\frac{5}{21}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose $P(x)$ is a polynomial such that $P(1)=1$ and
$$
\frac{P(2 x)}{P(x+1)}=8-\frac{56}{x+7}
$$
for all real $x$ for which both sides are defined. Find $P(-1)$.
|
$\quad-5 / 21$
Cross-multiplying gives $(x+7) P(2 x)=8 x P(x+1)$. If $P$ has degree $n$ and leading coefficient $c$, then the leading coefficients of the two sides are $2^{n} c$ and $8 c$, so $n=3$. Now $x=0$ is a root of the right-hand side, so it's a root of the left-hand side, so that $P(x)=x Q(x)$ for some polynomial $Q \Rightarrow 2 x(x+7) Q(2 x)=8 x(x+1) Q(x+1)$ or $(x+7) Q(2 x)=4(x+1) Q(x+1)$. Similarly, we see that $x=-1$ is a root of the left-hand side, giving $Q(x)=(x+2) R(x)$ for some polynomial $R \Rightarrow 2(x+1)(x+7) R(2 x)=$ $4(x+1)(x+3) R(x+1)$, or $(x+7) R(2 x)=2(x+3) R(x+1)$. Now $x=-3$ is a root of the left-hand side, so $R(x)=(x+6) S(x)$ for some polynomial $S$.
At this point, $P(x)=x(x+2)(x+6) S(x)$, but $P$ has degree 3 , so $S$ must be a constant. Since $P(1)=1$, we get $S=1 / 21$, and then $P(-1)=(-1)(1)(5) / 21=-5 / 21$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-alg-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nSolution: "
}
|
909c2f60-c580-5e4c-9412-a5e915be89f0
| 611,093
|
A particle moves along the $x$-axis in such a way that its velocity at position $x$ is given by the formula $v(x)=2+\sin x$. What is its acceleration at $x=\frac{\pi}{6}$ ?
|
$5 \sqrt{3} / 4$
Acceleration is given by $a=\frac{d v}{d t}=\frac{d v}{d x} \cdot \frac{d x}{d t}=\frac{d v}{d x} \cdot v=\cos x \cdot(2+\sin x)=5 \sqrt{3} / 4$.
|
\frac{5 \sqrt{3}}{4}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
A particle moves along the $x$-axis in such a way that its velocity at position $x$ is given by the formula $v(x)=2+\sin x$. What is its acceleration at $x=\frac{\pi}{6}$ ?
|
$5 \sqrt{3} / 4$
Acceleration is given by $a=\frac{d v}{d t}=\frac{d v}{d x} \cdot \frac{d x}{d t}=\frac{d v}{d x} \cdot v=\cos x \cdot(2+\sin x)=5 \sqrt{3} / 4$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-calc-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution: "
}
|
36a5a2e9-0354-540f-8cbb-e03607e05bc6
| 611,095
|
What is the area of the region bounded by the curves $y=x^{2003}$ and $y=x^{1 / 2003}$ and lying above the $x$-axis?
|
1001/1002
The two curves intersect at $(0,0)$ and $(1,1)$, so the desired area is
$$
\int_{0}^{1}\left(x^{1 / 2003}-x^{2003}\right) d x=\left[\frac{x^{2004 / 2003}}{2004 / 2003}-\frac{x^{2004}}{2004}\right]_{0}^{1}=\frac{1001}{1002}
$$
|
\frac{1001}{1002}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
What is the area of the region bounded by the curves $y=x^{2003}$ and $y=x^{1 / 2003}$ and lying above the $x$-axis?
|
1001/1002
The two curves intersect at $(0,0)$ and $(1,1)$, so the desired area is
$$
\int_{0}^{1}\left(x^{1 / 2003}-x^{2003}\right) d x=\left[\frac{x^{2004 / 2003}}{2004 / 2003}-\frac{x^{2004}}{2004}\right]_{0}^{1}=\frac{1001}{1002}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-calc-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution: "
}
|
ca790189-d441-56f3-aa96-b8b523a72009
| 611,096
|
The sequence of real numbers $x_{1}, x_{2}, x_{3}, \ldots$ satisfies $\lim _{n \rightarrow \infty}\left(x_{2 n}+x_{2 n+1}\right)=315$ and $\lim _{n \rightarrow \infty}\left(x_{2 n}+x_{2 n-1}\right)=2003$. Evaluate $\lim _{n \rightarrow \infty}\left(x_{2 n} / x_{2 n+1}\right)$.
|
-1
We have $\lim _{n \rightarrow \infty}\left(x_{2 n+1}-x_{2 n-1}\right)=\lim _{n \rightarrow \infty}\left[\left(x_{2 n}+x_{2 n+1}\right)-\left(x_{2 n}+x_{2 n-1}\right)\right]=315-2003=$ -1688 ; it follows that $x_{2 n+1} \rightarrow-\infty$ as $n \rightarrow \infty$. Then
$$
\lim _{n \rightarrow \infty} \frac{x_{2 n}}{x_{2 n+1}}=\lim _{n \rightarrow \infty} \frac{x_{2 n}+x_{2 n+1}}{x_{2 n+1}}-1=-1
$$
since $x_{2 n}+x_{2 n+1} \rightarrow 315$ while $x_{2 n+1} \rightarrow-\infty$.
|
-1
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
The sequence of real numbers $x_{1}, x_{2}, x_{3}, \ldots$ satisfies $\lim _{n \rightarrow \infty}\left(x_{2 n}+x_{2 n+1}\right)=315$ and $\lim _{n \rightarrow \infty}\left(x_{2 n}+x_{2 n-1}\right)=2003$. Evaluate $\lim _{n \rightarrow \infty}\left(x_{2 n} / x_{2 n+1}\right)$.
|
-1
We have $\lim _{n \rightarrow \infty}\left(x_{2 n+1}-x_{2 n-1}\right)=\lim _{n \rightarrow \infty}\left[\left(x_{2 n}+x_{2 n+1}\right)-\left(x_{2 n}+x_{2 n-1}\right)\right]=315-2003=$ -1688 ; it follows that $x_{2 n+1} \rightarrow-\infty$ as $n \rightarrow \infty$. Then
$$
\lim _{n \rightarrow \infty} \frac{x_{2 n}}{x_{2 n+1}}=\lim _{n \rightarrow \infty} \frac{x_{2 n}+x_{2 n+1}}{x_{2 n+1}}-1=-1
$$
since $x_{2 n}+x_{2 n+1} \rightarrow 315$ while $x_{2 n+1} \rightarrow-\infty$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-calc-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution: "
}
|
abc30e65-27f7-50ed-9d9e-b2f51928993e
| 611,097
|
Find the minimum distance from the point $(0,5 / 2)$ to the graph of $y=x^{4} / 8$.
|
$\sqrt{17} / 2$
We want to minimize $x^{2}+\left(x^{4} / 8-5 / 2\right)^{2}=x^{8} / 64-5 x^{4} / 8+x^{2}+25 / 4$, which is equivalent to minimizing $z^{4} / 4-10 z^{2}+16 z$, where we have set $z=x^{2}$. The derivative of this expression is $z^{3}-20 z+16$, which is seen on inspection to have 4 as a root, leading to the factorization $(z-4)(z+2-2 \sqrt{2})(z+2-2 \sqrt{2})$. Since $z=x^{2}$ ranges over $[0, \infty)$, the possible minima are at $z=0, z=-2+2 \sqrt{2}$, and $z=4$. However, the derivative is positive on $(0,-2+2 \sqrt{2})$, so this leaves only 0 and 4 to be tried. We find that the minimum is in fact achieved at $z=4$, so the closest point on the graph is given by $x= \pm 2$, with distance $\sqrt{2^{2}+\left(2^{4} / 8-5 / 2\right)^{2}}=\sqrt{17} / 2$.
|
\sqrt{17} / 2
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Find the minimum distance from the point $(0,5 / 2)$ to the graph of $y=x^{4} / 8$.
|
$\sqrt{17} / 2$
We want to minimize $x^{2}+\left(x^{4} / 8-5 / 2\right)^{2}=x^{8} / 64-5 x^{4} / 8+x^{2}+25 / 4$, which is equivalent to minimizing $z^{4} / 4-10 z^{2}+16 z$, where we have set $z=x^{2}$. The derivative of this expression is $z^{3}-20 z+16$, which is seen on inspection to have 4 as a root, leading to the factorization $(z-4)(z+2-2 \sqrt{2})(z+2-2 \sqrt{2})$. Since $z=x^{2}$ ranges over $[0, \infty)$, the possible minima are at $z=0, z=-2+2 \sqrt{2}$, and $z=4$. However, the derivative is positive on $(0,-2+2 \sqrt{2})$, so this leaves only 0 and 4 to be tried. We find that the minimum is in fact achieved at $z=4$, so the closest point on the graph is given by $x= \pm 2$, with distance $\sqrt{2^{2}+\left(2^{4} / 8-5 / 2\right)^{2}}=\sqrt{17} / 2$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-calc-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nSolution: "
}
|
894f7365-39ca-575c-9a4c-4ed12b8fbd91
| 611,098
|
For $n$ an integer, evaluate
$$
\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^{2}-0^{2}}}+\frac{1}{\sqrt{n^{2}-1^{2}}}+\cdots+\frac{1}{\sqrt{n^{2}-(n-1)^{2}}}\right)
$$
|
$\pi / 2$
Note that $\frac{1}{\sqrt{n^{2}-i^{2}}}=\frac{1}{n} \cdot \frac{1}{\sqrt{1-\left(\frac{i}{n}\right)^{2}}}$, so that the sum we wish to evaluate is just a Riemann sum. Then,
$$
\lim _{n \rightarrow \infty}\left(\frac{1}{n} \sum_{i=0}^{n-1} \frac{1}{\sqrt{1-\left(\frac{i}{n}\right)^{2}}}\right)=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} d x=\left[\sin ^{-1} x\right]_{0}^{1}=\frac{\pi}{2}
$$
|
\frac{\pi}{2}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
For $n$ an integer, evaluate
$$
\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^{2}-0^{2}}}+\frac{1}{\sqrt{n^{2}-1^{2}}}+\cdots+\frac{1}{\sqrt{n^{2}-(n-1)^{2}}}\right)
$$
|
$\pi / 2$
Note that $\frac{1}{\sqrt{n^{2}-i^{2}}}=\frac{1}{n} \cdot \frac{1}{\sqrt{1-\left(\frac{i}{n}\right)^{2}}}$, so that the sum we wish to evaluate is just a Riemann sum. Then,
$$
\lim _{n \rightarrow \infty}\left(\frac{1}{n} \sum_{i=0}^{n-1} \frac{1}{\sqrt{1-\left(\frac{i}{n}\right)^{2}}}\right)=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} d x=\left[\sin ^{-1} x\right]_{0}^{1}=\frac{\pi}{2}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-calc-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\n## Solution: "
}
|
d03bad15-8e09-5ef4-a1a9-af554ff38353
| 611,099
|
For what value of $a>1$ is
$$
\int_{a}^{a^{2}} \frac{1}{x} \log \frac{x-1}{32} d x
$$
minimum?
|
3
Let $f(a)=\int_{a}^{a^{2}} \frac{1}{x} \log \frac{x-1}{32} d x$. Then we want $\frac{d f}{d a}=0$; by the Fundamental Theorem of Calculus and the chain rule, this implies that
$2 a\left(\frac{1}{a^{2}} \log \frac{a^{2}-1}{32}\right)-\frac{1}{a} \log \frac{a-1}{32}=\frac{d}{d a}\left(\int_{c}^{a^{2}} \frac{1}{x} \log \frac{x-1}{32} d x-\int_{c}^{a} \frac{1}{x} \log \frac{x-1}{32} d x\right)=0$,
where $c$ is any constant with $1<c<a$. Then $2 \log \frac{a^{2}-1}{32}=\log \frac{a-1}{32}$, so that $\left(\frac{a^{2}-1}{32}\right)^{2}=$ $\frac{a-1}{32}$. After canceling factors of $(a-1) / 32($ since $a>1)$, this simplifies to $\left(a^{2}-1\right)(a+1)=$ $32 \Rightarrow a^{3}+a^{2}-a-33=0$, which in turn factors as $(a-3)\left(a^{2}+4 a+11\right)=0$. The quadratic factor has no real solutions, so this leaves only $a=3$. However, we have that $a>1$, and we can check that $f(1)=0, \lim _{a \rightarrow \infty} f(a)>0$, and $f(3)<0$, so the global minimum does occur at $a=3$.
|
3
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
For what value of $a>1$ is
$$
\int_{a}^{a^{2}} \frac{1}{x} \log \frac{x-1}{32} d x
$$
minimum?
|
3
Let $f(a)=\int_{a}^{a^{2}} \frac{1}{x} \log \frac{x-1}{32} d x$. Then we want $\frac{d f}{d a}=0$; by the Fundamental Theorem of Calculus and the chain rule, this implies that
$2 a\left(\frac{1}{a^{2}} \log \frac{a^{2}-1}{32}\right)-\frac{1}{a} \log \frac{a-1}{32}=\frac{d}{d a}\left(\int_{c}^{a^{2}} \frac{1}{x} \log \frac{x-1}{32} d x-\int_{c}^{a} \frac{1}{x} \log \frac{x-1}{32} d x\right)=0$,
where $c$ is any constant with $1<c<a$. Then $2 \log \frac{a^{2}-1}{32}=\log \frac{a-1}{32}$, so that $\left(\frac{a^{2}-1}{32}\right)^{2}=$ $\frac{a-1}{32}$. After canceling factors of $(a-1) / 32($ since $a>1)$, this simplifies to $\left(a^{2}-1\right)(a+1)=$ $32 \Rightarrow a^{3}+a^{2}-a-33=0$, which in turn factors as $(a-3)\left(a^{2}+4 a+11\right)=0$. The quadratic factor has no real solutions, so this leaves only $a=3$. However, we have that $a>1$, and we can check that $f(1)=0, \lim _{a \rightarrow \infty} f(a)>0$, and $f(3)<0$, so the global minimum does occur at $a=3$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-calc-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nSolution: "
}
|
bc8bc341-e0bf-5eab-84e7-76e594d1ee95
| 611,100
|
A right circular cone with a height of 12 inches and a base radius of 3 inches is filled with water and held with its vertex pointing downward. Water flows out through a hole at the vertex at a rate in cubic inches per second numerically equal to the height of the water in the cone. (For example, when the height of the water in the cone is 4 inches, water flows out at a rate of 4 cubic inches per second.) Determine how many seconds it will take for all of the water to flow out of the cone.
|
$9 \pi / 2$
When the water in the cone is $h$ inches high, it forms a cone similar to the original, so that its base has radius $h / 4$ and its volume is hence $\pi h^{3} / 48$. The given condition then states that
$$
\frac{d}{d t}\left(\frac{\pi h^{3}}{48}\right)=-h \Rightarrow \frac{\pi h^{2}}{16} \cdot \frac{d h}{d t}=-h \Rightarrow 2 h \cdot \frac{d h}{d t}=-\frac{32}{\pi} .
$$
Integrating with respect to $t$, we get that $h^{2}=-32 t / \pi+C$; setting $t=0, h=12$, we get $C=144$. The cone empties when $h=0$, so $0=-32 t / \pi+144 \Rightarrow t=9 \pi / 2$.
|
\frac{9\pi}{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
A right circular cone with a height of 12 inches and a base radius of 3 inches is filled with water and held with its vertex pointing downward. Water flows out through a hole at the vertex at a rate in cubic inches per second numerically equal to the height of the water in the cone. (For example, when the height of the water in the cone is 4 inches, water flows out at a rate of 4 cubic inches per second.) Determine how many seconds it will take for all of the water to flow out of the cone.
|
$9 \pi / 2$
When the water in the cone is $h$ inches high, it forms a cone similar to the original, so that its base has radius $h / 4$ and its volume is hence $\pi h^{3} / 48$. The given condition then states that
$$
\frac{d}{d t}\left(\frac{\pi h^{3}}{48}\right)=-h \Rightarrow \frac{\pi h^{2}}{16} \cdot \frac{d h}{d t}=-h \Rightarrow 2 h \cdot \frac{d h}{d t}=-\frac{32}{\pi} .
$$
Integrating with respect to $t$, we get that $h^{2}=-32 t / \pi+C$; setting $t=0, h=12$, we get $C=144$. The cone empties when $h=0$, so $0=-32 t / \pi+144 \Rightarrow t=9 \pi / 2$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-calc-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution: "
}
|
20570d49-de33-5ccf-8ee7-d3d3feb78ac9
| 611,101
|
Two differentiable real functions $f(x)$ and $g(x)$ satisfy
$$
\frac{f^{\prime}(x)}{g^{\prime}(x)}=e^{f(x)-g(x)}
$$
for all $x$, and $f(0)=g(2003)=1$. Find the largest constant $c$ such that $f(2003)>c$ for all such functions $f, g$.
|
$1-\ln 2$
Rearranging the given equation gives $f^{\prime}(x) e^{-f(x)}=g^{\prime}(x) e^{-g(x)}$ for all $x$, so $\frac{d}{d x}\left(e^{-f(x)}-\right.$ $\left.e^{-g(x)}\right)=-f^{\prime}(x) e^{-f(x)}+g^{\prime}(x) e^{-g(x)}=0$. Thus, $e^{-f(x)}-e^{-g(x)}$ is a constant, and it must be less than $e^{-f(0)}=e^{-1}$. Thus, $e^{-f(2003)}<e^{-g(2003)}+e^{-1}=2 e^{-1}=e^{\ln 2-1} \Rightarrow f(2003)>$ $1-\ln 2$. On the other hand, we can find positive-valued functions $e^{-f(x)}, e^{-g(x)}$ that take on the required values at 0 and 2003 and have constant difference arbitrarily close to $e^{-1}$. For example, for arbitrarily large $t$, we can set $e^{-f(x)}=e^{-(t(2003-x)+1)}+e^{-1}-$ $e^{-(2003 t+1)}$ and $e^{-g(x)}=e^{-(t(2003-x)+1)}$, and we can check that the resulting functions $f, g$ satisfy the required conditions. Thus, we can make $f(2003)$ arbitrarily close to $1-\ln 2$, so this is the answer.
|
1-\ln 2
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Two differentiable real functions $f(x)$ and $g(x)$ satisfy
$$
\frac{f^{\prime}(x)}{g^{\prime}(x)}=e^{f(x)-g(x)}
$$
for all $x$, and $f(0)=g(2003)=1$. Find the largest constant $c$ such that $f(2003)>c$ for all such functions $f, g$.
|
$1-\ln 2$
Rearranging the given equation gives $f^{\prime}(x) e^{-f(x)}=g^{\prime}(x) e^{-g(x)}$ for all $x$, so $\frac{d}{d x}\left(e^{-f(x)}-\right.$ $\left.e^{-g(x)}\right)=-f^{\prime}(x) e^{-f(x)}+g^{\prime}(x) e^{-g(x)}=0$. Thus, $e^{-f(x)}-e^{-g(x)}$ is a constant, and it must be less than $e^{-f(0)}=e^{-1}$. Thus, $e^{-f(2003)}<e^{-g(2003)}+e^{-1}=2 e^{-1}=e^{\ln 2-1} \Rightarrow f(2003)>$ $1-\ln 2$. On the other hand, we can find positive-valued functions $e^{-f(x)}, e^{-g(x)}$ that take on the required values at 0 and 2003 and have constant difference arbitrarily close to $e^{-1}$. For example, for arbitrarily large $t$, we can set $e^{-f(x)}=e^{-(t(2003-x)+1)}+e^{-1}-$ $e^{-(2003 t+1)}$ and $e^{-g(x)}=e^{-(t(2003-x)+1)}$, and we can check that the resulting functions $f, g$ satisfy the required conditions. Thus, we can make $f(2003)$ arbitrarily close to $1-\ln 2$, so this is the answer.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-calc-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nSolution: "
}
|
d143b4fa-3adc-54de-94e7-c00219d12a01
| 611,102
|
Evaluate
$$
\int_{-\infty}^{\infty} \frac{1-x^{2}}{1+x^{4}} d x
$$
|
0
Let $S=\int_{0}^{\infty} 1 /\left(x^{4}+1\right) d x$; note that the integral converges absolutely. Substituting $x=1 / u$, so that $d x=-1 / u^{2} d u$, we have
$$
\begin{gathered}
S=\int_{0}^{\infty} \frac{1}{1+x^{4}} d x=\int_{\infty}^{0} \frac{1}{1+u^{-4}} \frac{d u}{-u^{2}}=\int_{\infty}^{0} \frac{-u^{2}}{u^{4}+1} d u \\
=\int_{0}^{\infty} \frac{u^{2}}{1+u^{4}} d u=\int_{0}^{\infty} \frac{x^{2}}{1+x^{4}} d x
\end{gathered}
$$
(the manipulations are justified by absolute convergence), from which we see that $\int_{0}^{\infty}\left(1-x^{2}\right) /\left(1+x^{4}\right) d x=0$. Since the integrand is an even function, it follows that the integral from $-\infty$ to $\infty$ is zero as well.
|
0
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Evaluate
$$
\int_{-\infty}^{\infty} \frac{1-x^{2}}{1+x^{4}} d x
$$
|
0
Let $S=\int_{0}^{\infty} 1 /\left(x^{4}+1\right) d x$; note that the integral converges absolutely. Substituting $x=1 / u$, so that $d x=-1 / u^{2} d u$, we have
$$
\begin{gathered}
S=\int_{0}^{\infty} \frac{1}{1+x^{4}} d x=\int_{\infty}^{0} \frac{1}{1+u^{-4}} \frac{d u}{-u^{2}}=\int_{\infty}^{0} \frac{-u^{2}}{u^{4}+1} d u \\
=\int_{0}^{\infty} \frac{u^{2}}{1+u^{4}} d u=\int_{0}^{\infty} \frac{x^{2}}{1+x^{4}} d x
\end{gathered}
$$
(the manipulations are justified by absolute convergence), from which we see that $\int_{0}^{\infty}\left(1-x^{2}\right) /\left(1+x^{4}\right) d x=0$. Since the integrand is an even function, it follows that the integral from $-\infty$ to $\infty$ is zero as well.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-calc-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nSolution: "
}
|
9f9cdfd6-b02a-5c9c-b223-6bbcacd902e3
| 611,103
|
You have 2003 switches, numbered from 1 to 2003, arranged in a circle. Initially, each switch is either ON or OFF, and all configurations of switches are equally likely. You perform the following operation: for each switch $S$, if the two switches next to $S$ were initially in the same position, then you set $S$ to ON; otherwise, you set $S$ to OFF. What is the probability that all switches will now be ON?
|
$1 / 2^{2002}$
There are $2^{2003}$ equally likely starting configurations. All switches end up ON if and only if switches $1,3,5,7, \ldots, 2003,2,4, \ldots, 2002$ - i.e. all 2003 of them - were initially in the same position. This initial position can be ON or OFF, so this situation occurs with probability $2 / 2^{2003}=1 / 2^{2002}$.
|
\frac{1}{2^{2002}}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
You have 2003 switches, numbered from 1 to 2003, arranged in a circle. Initially, each switch is either ON or OFF, and all configurations of switches are equally likely. You perform the following operation: for each switch $S$, if the two switches next to $S$ were initially in the same position, then you set $S$ to ON; otherwise, you set $S$ to OFF. What is the probability that all switches will now be ON?
|
$1 / 2^{2002}$
There are $2^{2003}$ equally likely starting configurations. All switches end up ON if and only if switches $1,3,5,7, \ldots, 2003,2,4, \ldots, 2002$ - i.e. all 2003 of them - were initially in the same position. This initial position can be ON or OFF, so this situation occurs with probability $2 / 2^{2003}=1 / 2^{2002}$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-comb-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution: "
}
|
d1aa5a21-fcf7-5ef6-9012-209db1f7426f
| 611,104
|
You are given a $10 \times 2$ grid of unit squares. Two different squares are adjacent if they share a side. How many ways can one mark exactly nine of the squares so that no two marked squares are adjacent?
|
36
Since each row has only two squares, it is impossible for two marked squares to be in the same row. Therefore, exactly nine of the ten rows contain marked squares. Consider two cases:
Case 1: The first or last row is empty. These two cases are symmetrical, so assume without loss of generality that the first row is empty. There are two possibilities for the second row: either the first square is marked, or the second square is marked. Since the third row must contain a marked square, and it cannot be in the same column as the marked square in the second row, the third row is determined by the second. Similarly, all the remaining rows are determined. This leaves two possibilities if the first row is empty. Thus, there are four possibilities if the first or last row is empty.
Case 2: The empty row is not the first or last. Then, there are two blocks of (one of more) consecutive rows of marked squares. As above, the configuration of the rows in each of the two blocks is determined by the position of the marked square in the first of its rows. That makes $2 \times 2=4$ possible configurations. There are eight possibilities for the empty row, making a total of 32 possibilities in this case.
Together, there are 36 possible configurations of marked squares.
|
36
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
You are given a $10 \times 2$ grid of unit squares. Two different squares are adjacent if they share a side. How many ways can one mark exactly nine of the squares so that no two marked squares are adjacent?
|
36
Since each row has only two squares, it is impossible for two marked squares to be in the same row. Therefore, exactly nine of the ten rows contain marked squares. Consider two cases:
Case 1: The first or last row is empty. These two cases are symmetrical, so assume without loss of generality that the first row is empty. There are two possibilities for the second row: either the first square is marked, or the second square is marked. Since the third row must contain a marked square, and it cannot be in the same column as the marked square in the second row, the third row is determined by the second. Similarly, all the remaining rows are determined. This leaves two possibilities if the first row is empty. Thus, there are four possibilities if the first or last row is empty.
Case 2: The empty row is not the first or last. Then, there are two blocks of (one of more) consecutive rows of marked squares. As above, the configuration of the rows in each of the two blocks is determined by the position of the marked square in the first of its rows. That makes $2 \times 2=4$ possible configurations. There are eight possibilities for the empty row, making a total of 32 possibilities in this case.
Together, there are 36 possible configurations of marked squares.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-comb-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution: "
}
|
6fefc6dd-1706-5fa6-9179-a4d51f559987
| 611,105
|
Daniel and Scott are playing a game where a player wins as soon as he has two points more than his opponent. Both players start at par, and points are earned one at a time. If Daniel has a $60 \%$ chance of winning each point, what is the probability that he will win the game?
|
$9 / 13$
Consider the situation after two points. Daniel has a $9 / 25$ chance of winning, Scott, $4 / 25$, and there is a $12 / 25$ chance that the players will be tied. In the latter case, we revert to the original situation. In particular, after every two points, either the game
returns to the original situation, or one player wins. If it is given that the game lasts $2 k$ rounds, then the players must be at par after $2(k-1)$ rounds, and then Daniel wins with probability $(9 / 25) /(9 / 25+4 / 25)=9 / 13$. Since this holds for any $k$, we conclude that Daniel wins the game with probability $9 / 13$.
|
\frac{9}{13}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Daniel and Scott are playing a game where a player wins as soon as he has two points more than his opponent. Both players start at par, and points are earned one at a time. If Daniel has a $60 \%$ chance of winning each point, what is the probability that he will win the game?
|
$9 / 13$
Consider the situation after two points. Daniel has a $9 / 25$ chance of winning, Scott, $4 / 25$, and there is a $12 / 25$ chance that the players will be tied. In the latter case, we revert to the original situation. In particular, after every two points, either the game
returns to the original situation, or one player wins. If it is given that the game lasts $2 k$ rounds, then the players must be at par after $2(k-1)$ rounds, and then Daniel wins with probability $(9 / 25) /(9 / 25+4 / 25)=9 / 13$. Since this holds for any $k$, we conclude that Daniel wins the game with probability $9 / 13$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-comb-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution: "
}
|
8b7ca36f-1c74-5397-950a-cdd58057eab3
| 611,106
|
In a certain country, there are 100 senators, each of whom has 4 aides. These senators and aides serve on various committees. A committee may consist either of 5 senators, of 4 senators and 4 aides, or of 2 senators and 12 aides. Every senator serves on 5 committees, and every aide serves on 3 committees. How many committees are there altogether?
|
160
If each senator gets a point for every committee on which she serves, and every aide gets $1 / 4$ point for every committee on which he serves, then the 100 senators get 500 points altogether, and the 400 aides get 300 points altogether, for a total of 800 points. On the other hand, each committee contributes 5 points, so there must be $800 / 5=160$ committees.
|
160
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In a certain country, there are 100 senators, each of whom has 4 aides. These senators and aides serve on various committees. A committee may consist either of 5 senators, of 4 senators and 4 aides, or of 2 senators and 12 aides. Every senator serves on 5 committees, and every aide serves on 3 committees. How many committees are there altogether?
|
160
If each senator gets a point for every committee on which she serves, and every aide gets $1 / 4$ point for every committee on which he serves, then the 100 senators get 500 points altogether, and the 400 aides get 300 points altogether, for a total of 800 points. On the other hand, each committee contributes 5 points, so there must be $800 / 5=160$ committees.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-comb-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\n## Solution: "
}
|
f24351ce-958e-5b7b-bf9d-882e01e92eec
| 611,107
|
We wish to color the integers $1,2,3, \ldots, 10$ in red, green, and blue, so that no two numbers $a$ and $b$, with $a-b$ odd, have the same color. (We do not require that all three colors be used.) In how many ways can this be done?
|
186
The condition is equivalent to never having an odd number and an even number in the same color. We can choose one of the three colors for the odd numbers and distribute the other two colors freely among the 5 even numbers; this can be done in $3 \cdot 2^{5}=96$ ways. We can also choose one color for the even numbers and distribute the other two colors among the 5 odd numbers, again in 96 ways. This gives a total of 192 possibilities. However, we have double-counted the $3 \cdot 2=6$ cases where all odd numbers are the same color and all even numbers are the same color, so there are actually $192-6=186$ possible colorings.
|
186
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
We wish to color the integers $1,2,3, \ldots, 10$ in red, green, and blue, so that no two numbers $a$ and $b$, with $a-b$ odd, have the same color. (We do not require that all three colors be used.) In how many ways can this be done?
|
186
The condition is equivalent to never having an odd number and an even number in the same color. We can choose one of the three colors for the odd numbers and distribute the other two colors freely among the 5 even numbers; this can be done in $3 \cdot 2^{5}=96$ ways. We can also choose one color for the even numbers and distribute the other two colors among the 5 odd numbers, again in 96 ways. This gives a total of 192 possibilities. However, we have double-counted the $3 \cdot 2=6$ cases where all odd numbers are the same color and all even numbers are the same color, so there are actually $192-6=186$ possible colorings.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-comb-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\n## Solution: "
}
|
a89eed6a-e3be-506a-bb1e-0bf3a2df0881
| 611,108
|
In a classroom, 34 students are seated in 5 rows of 7 chairs. The place at the center of the room is unoccupied. A teacher decides to reassign the seats such that each student will occupy a chair adjacent to his/her present one (i.e. move one desk forward, back, left or right). In how many ways can this reassignment be made?
|
0
Color the chairs red and black in checkerboard fashion, with the center chair black. Then all 18 red chairs are initially occupied. Also notice that adjacent chairs have different colors. It follows that we need 18 black chairs to accommodate the reassignment, but there are only 17 of them. Thus, the answer is 0 .
|
0
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In a classroom, 34 students are seated in 5 rows of 7 chairs. The place at the center of the room is unoccupied. A teacher decides to reassign the seats such that each student will occupy a chair adjacent to his/her present one (i.e. move one desk forward, back, left or right). In how many ways can this reassignment be made?
|
0
Color the chairs red and black in checkerboard fashion, with the center chair black. Then all 18 red chairs are initially occupied. Also notice that adjacent chairs have different colors. It follows that we need 18 black chairs to accommodate the reassignment, but there are only 17 of them. Thus, the answer is 0 .
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-comb-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nSolution: "
}
|
8beaa1f0-778a-5dfb-b637-d945e6d5bc0f
| 611,109
|
You have infinitely many boxes, and you randomly put 3 balls into them. The boxes are labeled $1,2, \ldots$. Each ball has probability $1 / 2^{n}$ of being put into box $n$. The balls are placed independently of each other. What is the probability that some box will contain at least 2 balls?
|
$$
5 / 7
$$
Notice that the answer is the sum of the probabilities that boxes $1,2, \ldots$, respectively, contain at least 2 balls, since those events are mutually exclusive. For box $n$, the probability of having at least 2 balls is
$$
3\left[\left(1 / 2^{n}\right)^{2}\left(1-1 / 2^{n}\right)\right]+\left(1 / 2^{n}\right)^{3}=3 / 2^{2 n}-2 / 2^{3 n}=3 / 4^{n}-2 / 8^{n}
$$
Summing to infinity using the geometric series formula, we get the answer (3/4)/(1$1 / 4)-(2 / 8) /(1-1 / 8)$, which is equal to $5 / 7$.
|
\frac{5}{7}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
You have infinitely many boxes, and you randomly put 3 balls into them. The boxes are labeled $1,2, \ldots$. Each ball has probability $1 / 2^{n}$ of being put into box $n$. The balls are placed independently of each other. What is the probability that some box will contain at least 2 balls?
|
$$
5 / 7
$$
Notice that the answer is the sum of the probabilities that boxes $1,2, \ldots$, respectively, contain at least 2 balls, since those events are mutually exclusive. For box $n$, the probability of having at least 2 balls is
$$
3\left[\left(1 / 2^{n}\right)^{2}\left(1-1 / 2^{n}\right)\right]+\left(1 / 2^{n}\right)^{3}=3 / 2^{2 n}-2 / 2^{3 n}=3 / 4^{n}-2 / 8^{n}
$$
Summing to infinity using the geometric series formula, we get the answer (3/4)/(1$1 / 4)-(2 / 8) /(1-1 / 8)$, which is equal to $5 / 7$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-comb-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nSolution:\n\n"
}
|
b46f3866-5b82-5157-85b5-822a7c582b53
| 611,110
|
For any subset $S \subseteq\{1,2, \ldots, 15\}$, a number $n$ is called an "anchor" for $S$ if $n$ and $n+|S|$ are both members of $S$, where $|S|$ denotes the number of members of $S$. Find the average number of anchors over all possible subsets $S \subseteq\{1,2, \ldots, 15\}$.
|
$13 / 8$
We first find the sum of the numbers of anchors of all subsets $S$; this is equivalent to finding, for each $n$, the number of sets for which $n$ is an anchor, and then summing over all $n$. Suppose that $n$ is an anchor for $S$, and $S$ has $k$ elements. Then $n, n+k \in$ $S \Rightarrow k \geq 2$, and also $n+k \leq 15$, or $k \leq 15-n$. The remaining $k-2$ elements of $S$ (other than $n$ and $n+k$ ) may be freely chosen from the remaining 13 members of $\{1,2, \ldots, 15\}$, so we get $\binom{13}{k-2}$ possible sets $S$. Summing over all allowed values of $k$, we then have $\binom{13}{0}+\binom{13}{1}+\binom{13}{2}+\cdots+\binom{13}{13-n}$ sets with $n$ as an anchor. If we sum over all $n=1,2, \ldots, 13$ (since there are no possible values of $k$ when $n>13$ ), we get a total of
$$
13\binom{13}{0}+12\binom{13}{1}+11\binom{13}{2}+\cdots+\binom{13}{12}
$$
If we call this quantity $A$, then, by symmetry, $2 A$ equals
$$
\begin{aligned}
& 13\binom{13}{0}+12\binom{13}{1}+11\binom{13}{2}+\cdots+\binom{13}{12} \\
& \begin{array}{l}
+ \\
\left.+\left(\begin{array}{c}
13 \\
1 \\
13 \\
1
\end{array}\right)+2\binom{13}{2}+\cdots+12\left(\begin{array}{c}
13 \\
13 \\
2
\end{array}\right)+\cdots\left(\begin{array}{c}
13 \\
13 \\
12
\end{array}\right)+\left(\begin{array}{c}
13 \\
13 \\
13 \\
13
\end{array}\right)\right]=13 \cdot 2^{13} .
\end{array}
\end{aligned}
$$
So $A=13 \cdot 2^{12}$ is the total number of anchors over all possible sets $S$. Finally, to find the average number of anchors, we divide by the number of sets, which is $2^{15}$; thus, the answer is $13 \cdot 2^{12} / 2^{15}=13 / 8$.
|
\frac{13}{8}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
For any subset $S \subseteq\{1,2, \ldots, 15\}$, a number $n$ is called an "anchor" for $S$ if $n$ and $n+|S|$ are both members of $S$, where $|S|$ denotes the number of members of $S$. Find the average number of anchors over all possible subsets $S \subseteq\{1,2, \ldots, 15\}$.
|
$13 / 8$
We first find the sum of the numbers of anchors of all subsets $S$; this is equivalent to finding, for each $n$, the number of sets for which $n$ is an anchor, and then summing over all $n$. Suppose that $n$ is an anchor for $S$, and $S$ has $k$ elements. Then $n, n+k \in$ $S \Rightarrow k \geq 2$, and also $n+k \leq 15$, or $k \leq 15-n$. The remaining $k-2$ elements of $S$ (other than $n$ and $n+k$ ) may be freely chosen from the remaining 13 members of $\{1,2, \ldots, 15\}$, so we get $\binom{13}{k-2}$ possible sets $S$. Summing over all allowed values of $k$, we then have $\binom{13}{0}+\binom{13}{1}+\binom{13}{2}+\cdots+\binom{13}{13-n}$ sets with $n$ as an anchor. If we sum over all $n=1,2, \ldots, 13$ (since there are no possible values of $k$ when $n>13$ ), we get a total of
$$
13\binom{13}{0}+12\binom{13}{1}+11\binom{13}{2}+\cdots+\binom{13}{12}
$$
If we call this quantity $A$, then, by symmetry, $2 A$ equals
$$
\begin{aligned}
& 13\binom{13}{0}+12\binom{13}{1}+11\binom{13}{2}+\cdots+\binom{13}{12} \\
& \begin{array}{l}
+ \\
\left.+\left(\begin{array}{c}
13 \\
1 \\
13 \\
1
\end{array}\right)+2\binom{13}{2}+\cdots+12\left(\begin{array}{c}
13 \\
13 \\
2
\end{array}\right)+\cdots\left(\begin{array}{c}
13 \\
13 \\
12
\end{array}\right)+\left(\begin{array}{c}
13 \\
13 \\
13 \\
13
\end{array}\right)\right]=13 \cdot 2^{13} .
\end{array}
\end{aligned}
$$
So $A=13 \cdot 2^{12}$ is the total number of anchors over all possible sets $S$. Finally, to find the average number of anchors, we divide by the number of sets, which is $2^{15}$; thus, the answer is $13 \cdot 2^{12} / 2^{15}=13 / 8$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-comb-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\n## Solution: "
}
|
d048457f-54c2-5973-b215-cf9498e23229
| 611,111
|
At a certain college, there are 10 clubs and some number of students. For any two different students, there is some club such that exactly one of the two belongs to that club. For any three different students, there is some club such that either exactly one or all three belong to that club. What is the largest possible number of students?
|
513
Let $C$ be the set of clubs; each student then corresponds to a subset of $C$ (the clubs to which that student belongs). The two-student condition implies that these subsets must be all distinct. Now (assuming there is more than one student) some student belongs to a nonempty set $S$ of clubs. For every subset $T \subseteq C$, let $f(T)$ be the subset of $C$ consisting of those clubs that are in exactly one of $S$ and $T$ (so that $f(T)=$ $(S \cup T)-(S \cap T)$ ). It is straightforward to check that $f(f(T))=T$ and $f(T) \neq T$, so that the collection of all $2^{10}$ subsets of $C$ is partitioned into pairs $\{T, f(T)\}$. Moreover,
as long as $S$ is distinct from $T$ and $f(T)$, every club is in either none or exactly two of the sets $S, T$, and $f(T)$, so we cannot have a student corresponding to $T$ and another corresponding to $f(T)$. This puts an upper bound of 513 possible students (one for $S$, one for $\emptyset=f(S)$, and one for each of the 511 other pairs). On the other hand, if we take some club $c$, we can have one student belonging to no clubs and 512 other students all belonging to $c$ and to the 512 possible subsets of the other 9 clubs, respectively. It is readily checked that this arrangement meets the conditions - for the three-student condition, either all three students are in $c$, or one is the student who belongs to no clubs and we reduce to the two-student condition - so 513 is achievable.
|
513
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
At a certain college, there are 10 clubs and some number of students. For any two different students, there is some club such that exactly one of the two belongs to that club. For any three different students, there is some club such that either exactly one or all three belong to that club. What is the largest possible number of students?
|
513
Let $C$ be the set of clubs; each student then corresponds to a subset of $C$ (the clubs to which that student belongs). The two-student condition implies that these subsets must be all distinct. Now (assuming there is more than one student) some student belongs to a nonempty set $S$ of clubs. For every subset $T \subseteq C$, let $f(T)$ be the subset of $C$ consisting of those clubs that are in exactly one of $S$ and $T$ (so that $f(T)=$ $(S \cup T)-(S \cap T)$ ). It is straightforward to check that $f(f(T))=T$ and $f(T) \neq T$, so that the collection of all $2^{10}$ subsets of $C$ is partitioned into pairs $\{T, f(T)\}$. Moreover,
as long as $S$ is distinct from $T$ and $f(T)$, every club is in either none or exactly two of the sets $S, T$, and $f(T)$, so we cannot have a student corresponding to $T$ and another corresponding to $f(T)$. This puts an upper bound of 513 possible students (one for $S$, one for $\emptyset=f(S)$, and one for each of the 511 other pairs). On the other hand, if we take some club $c$, we can have one student belonging to no clubs and 512 other students all belonging to $c$ and to the 512 possible subsets of the other 9 clubs, respectively. It is readily checked that this arrangement meets the conditions - for the three-student condition, either all three students are in $c$, or one is the student who belongs to no clubs and we reduce to the two-student condition - so 513 is achievable.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-comb-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nSolution: "
}
|
82ad73a4-3a0e-5ef5-9e9b-44f63351152a
| 611,112
|
A calculator has a display, which shows a nonnegative integer $N$, and a button, which replaces $N$ by a random integer chosen uniformly from the set $\{0,1, \ldots, N-1\}$, provided that $N>0$. Initially, the display holds the number $N=2003$. If the button is pressed repeatedly until $N=0$, what is the probability that the numbers $1,10,100$, and 1000 will each show up on the display at some point?
|
$1 / 2224222$
First, we claim that if the display starts at some $N$, the probability that any given number $M<N$ will appear at some point is $1 /(M+1)$. We can show this by induction on $N$. If $N=M+1$ (the base case), $M$ can only be reached if it appears after the first step, and this occurs with probability $1 / N=1 /(M+1)$. If $N>M+1$ and the claim holds for $N-1$, then there are two possibilities starting from $N$. If the first step leads to $N-1$ (this occurs with probability $1 / N$ ), the probability of seeing $M$ subsequently is $1 /(M+1)$ by the induction hypothesis. If the first step leads to something less than $N-1$ (probability $(N-1) / N)$, then it leads to any of the integers $\{0,1, \ldots, N-2\}$ with equal probability. But this is exactly what the first step would have been if we had started from $N-1$; hence, the probability of seeing $M$ is again $1 /(M+1)$ by induction. Thus, the overall probability of seeing $M$ is $\frac{1}{N} \cdot \frac{1}{M+1}+\frac{N-1}{N} \cdot \frac{1}{M+1}=1 /(M+1)$, proving the induction step and the claim.
Now let $P(N, M)(M<N)$ be the probability of eventually seeing the number $M$ if we start at $N$; note that this is the same as the conditional probability of seeing $M$ given that we see $N$. Hence, the desired probability is
$$
P(2003,1000) \cdot P(1000,100) \cdot P(100,10) \cdot P(10,1)=\frac{1}{1001} \cdot \frac{1}{101} \cdot \frac{1}{11} \cdot \frac{1}{2}=\frac{1}{2224222} .
$$
|
\frac{1}{2224222}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A calculator has a display, which shows a nonnegative integer $N$, and a button, which replaces $N$ by a random integer chosen uniformly from the set $\{0,1, \ldots, N-1\}$, provided that $N>0$. Initially, the display holds the number $N=2003$. If the button is pressed repeatedly until $N=0$, what is the probability that the numbers $1,10,100$, and 1000 will each show up on the display at some point?
|
$1 / 2224222$
First, we claim that if the display starts at some $N$, the probability that any given number $M<N$ will appear at some point is $1 /(M+1)$. We can show this by induction on $N$. If $N=M+1$ (the base case), $M$ can only be reached if it appears after the first step, and this occurs with probability $1 / N=1 /(M+1)$. If $N>M+1$ and the claim holds for $N-1$, then there are two possibilities starting from $N$. If the first step leads to $N-1$ (this occurs with probability $1 / N$ ), the probability of seeing $M$ subsequently is $1 /(M+1)$ by the induction hypothesis. If the first step leads to something less than $N-1$ (probability $(N-1) / N)$, then it leads to any of the integers $\{0,1, \ldots, N-2\}$ with equal probability. But this is exactly what the first step would have been if we had started from $N-1$; hence, the probability of seeing $M$ is again $1 /(M+1)$ by induction. Thus, the overall probability of seeing $M$ is $\frac{1}{N} \cdot \frac{1}{M+1}+\frac{N-1}{N} \cdot \frac{1}{M+1}=1 /(M+1)$, proving the induction step and the claim.
Now let $P(N, M)(M<N)$ be the probability of eventually seeing the number $M$ if we start at $N$; note that this is the same as the conditional probability of seeing $M$ given that we see $N$. Hence, the desired probability is
$$
P(2003,1000) \cdot P(1000,100) \cdot P(100,10) \cdot P(10,1)=\frac{1}{1001} \cdot \frac{1}{101} \cdot \frac{1}{11} \cdot \frac{1}{2}=\frac{1}{2224222} .
$$
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-comb-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\n## Solution: "
}
|
727ea503-431a-58fd-8c6a-a257244e219d
| 611,113
|
10 people are playing musical chairs with $n$ chairs in a circle. They can be seated in 7 ! ways (assuming only one person fits on each chair, of course), where different arrangements of the same people on chairs, even rotations, are considered different. Find $n$.
|
4
The number of ways 10 people can be seated on $n$ chairs is $n!$ multiplied by the number of ways one can choose $n$ people out of 10 . Hence we must solve $7!=n!\cdot 10!/(n!\cdot(10-$ $n)!$ ). This is equivalent to $(10-n)!=10!/ 7!=8 \cdot 9 \cdot 10=720=6!$. We therefore have $n=4$.
|
4
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
10 people are playing musical chairs with $n$ chairs in a circle. They can be seated in 7 ! ways (assuming only one person fits on each chair, of course), where different arrangements of the same people on chairs, even rotations, are considered different. Find $n$.
|
4
The number of ways 10 people can be seated on $n$ chairs is $n!$ multiplied by the number of ways one can choose $n$ people out of 10 . Hence we must solve $7!=n!\cdot 10!/(n!\cdot(10-$ $n)!$ ). This is equivalent to $(10-n)!=10!/ 7!=8 \cdot 9 \cdot 10=720=6!$. We therefore have $n=4$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen1-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution: "
}
|
57ede10b-223b-568b-8b10-513afc245f4d
| 611,114
|
$O P E N$ is a square, and $T$ is a point on side $N O$, such that triangle $T O P$ has area 62 and triangle $T E N$ has area 10 . What is the length of a side of the square?
|
12
$62=P O \cdot O T / 2$ and $10=E N \cdot T N / 2=P O \cdot T N / 2$, so adding gives $72=P O \cdot(O T+$ $T N) / 2=P O \cdot O N / 2=P O^{2} / 2 \Rightarrow P O=12$.
|
12
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
$O P E N$ is a square, and $T$ is a point on side $N O$, such that triangle $T O P$ has area 62 and triangle $T E N$ has area 10 . What is the length of a side of the square?
|
12
$62=P O \cdot O T / 2$ and $10=E N \cdot T N / 2=P O \cdot T N / 2$, so adding gives $72=P O \cdot(O T+$ $T N) / 2=P O \cdot O N / 2=P O^{2} / 2 \Rightarrow P O=12$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen1-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution: "
}
|
198004d0-b69b-58ad-bee0-f5dcdc49cdf5
| 611,115
|
There are 16 members on the Height-Measurement Matching Team. Each member was asked, "How many other people on the team - not counting yourself - are exactly the same height as you?" The answers included six 1's, six 2's, and three 3's. What was the sixteenth answer? (Assume that everyone answered truthfully.)
|
3
For anyone to have answered 3 , there must have been exactly 4 people with the same height, and then each of them would have given the answer 3. Thus, we need at least four 3's, so 3 is the remaining answer. (More generally, a similar argument shows that the number of members answering $n$ must be divisible by $n+1$.)
|
3
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
There are 16 members on the Height-Measurement Matching Team. Each member was asked, "How many other people on the team - not counting yourself - are exactly the same height as you?" The answers included six 1's, six 2's, and three 3's. What was the sixteenth answer? (Assume that everyone answered truthfully.)
|
3
For anyone to have answered 3 , there must have been exactly 4 people with the same height, and then each of them would have given the answer 3. Thus, we need at least four 3's, so 3 is the remaining answer. (More generally, a similar argument shows that the number of members answering $n$ must be divisible by $n+1$.)
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen1-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\n## Solution: "
}
|
8a0f26b3-610e-5e2e-a338-876ba37361d2
| 611,116
|
How many 2-digit positive integers have an even number of positive divisors?
|
84
An integer has an odd number of divisors precisely if it is a square. So we take the 90 2 -digit numbers $(10,11, \ldots, 99)$ and remove the 6 squares $\left(4^{2}, 5^{2}, \ldots, 9^{2}\right)$, for a total of 84.
|
84
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many 2-digit positive integers have an even number of positive divisors?
|
84
An integer has an odd number of divisors precisely if it is a square. So we take the 90 2 -digit numbers $(10,11, \ldots, 99)$ and remove the 6 squares $\left(4^{2}, 5^{2}, \ldots, 9^{2}\right)$, for a total of 84.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen1-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\n## Solution: "
}
|
8e2924c1-f387-5264-99b3-be534ad3ae7c
| 611,117
|
A room is built in the shape of the region between two semicircles with the same center and parallel diameters. The farthest distance between two points with a clear line of sight is 12 m . What is the area (in $\mathrm{m}^{2}$ ) of the room?
|
$18 \pi$
The maximal distance is as shown in the figure (next page). Call the radii $R$ and $r$, $R>r$. Then $R^{2}-r^{2}=6^{2}$ by the Pythagorean theorem, so the area is $(\pi / 2) \cdot\left(R^{2}-r^{2}\right)=$ $18 \pi$.
|
18 \pi
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A room is built in the shape of the region between two semicircles with the same center and parallel diameters. The farthest distance between two points with a clear line of sight is 12 m . What is the area (in $\mathrm{m}^{2}$ ) of the room?
|
$18 \pi$
The maximal distance is as shown in the figure (next page). Call the radii $R$ and $r$, $R>r$. Then $R^{2}-r^{2}=6^{2}$ by the Pythagorean theorem, so the area is $(\pi / 2) \cdot\left(R^{2}-r^{2}\right)=$ $18 \pi$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen1-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nSolution: "
}
|
4aa60ba6-7197-5302-b66e-1d42aaed4a21
| 611,118
|
In how many ways can 3 bottles of ketchup and 7 bottles of mustard be arranged in a row so that no bottle of ketchup is immediately between two bottles of mustard? (The bottles of ketchup are mutually indistinguishable, as are the bottles of mustard.)
|
22
Consider the blocks of consecutive bottles of ketchup in such an arrangement. A block of just one bottle must occur at the beginning or the end of the row, or else it would be between two bottles of mustard. However, a block of two or three bottles can occur anywhere. We cannot have three blocks of one bottle each, since there are only two possible locations for such blocks. Thus, we either have a block of one bottle and a block of two, or one block of all three bottles. In the first case, if the single bottle occurs at the beginning of the row, then anywhere from 1 to 7 bottles of mustard may intervene before the block of 2 ketchup bottles, giving 7 possible arrangements. We likewise have 7 arrangements if the single bottle occurs at the end of the row. Finally, if there is just one block of three bottles, anywhere from 0 to 7 mustard bottles may precede it, giving 8 possible arrangements. So, altogether, we have $7+7+8=22$ configurations.
|
22
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In how many ways can 3 bottles of ketchup and 7 bottles of mustard be arranged in a row so that no bottle of ketchup is immediately between two bottles of mustard? (The bottles of ketchup are mutually indistinguishable, as are the bottles of mustard.)
|
22
Consider the blocks of consecutive bottles of ketchup in such an arrangement. A block of just one bottle must occur at the beginning or the end of the row, or else it would be between two bottles of mustard. However, a block of two or three bottles can occur anywhere. We cannot have three blocks of one bottle each, since there are only two possible locations for such blocks. Thus, we either have a block of one bottle and a block of two, or one block of all three bottles. In the first case, if the single bottle occurs at the beginning of the row, then anywhere from 1 to 7 bottles of mustard may intervene before the block of 2 ketchup bottles, giving 7 possible arrangements. We likewise have 7 arrangements if the single bottle occurs at the end of the row. Finally, if there is just one block of three bottles, anywhere from 0 to 7 mustard bottles may precede it, giving 8 possible arrangements. So, altogether, we have $7+7+8=22$ configurations.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen1-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nSolution: "
}
|
611fbdbd-703d-5cbf-a39e-5ca47a497a0a
| 611,119
|
Find the real value of $x$ such that $x^{3}+3 x^{2}+3 x+7=0$.
|
$\quad-1-\sqrt[3]{6}$
Rewrite the equation as $(x+1)^{3}+6=0$ to get $(x+1)^{3}=-6 \Rightarrow x+1=\sqrt[3]{-6} \Rightarrow x=$ $-1-\sqrt[3]{6}$.
|
-1-\sqrt[3]{6}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the real value of $x$ such that $x^{3}+3 x^{2}+3 x+7=0$.
|
$\quad-1-\sqrt[3]{6}$
Rewrite the equation as $(x+1)^{3}+6=0$ to get $(x+1)^{3}=-6 \Rightarrow x+1=\sqrt[3]{-6} \Rightarrow x=$ $-1-\sqrt[3]{6}$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen1-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nSolution: "
}
|
56ee9d98-107a-5cfd-9727-b0147a6632ac
| 611,120
|
A broken calculator has the + and $\times$ keys switched. For how many ordered pairs $(a, b)$ of integers will it correctly calculate $a+b$ using the labelled + key?
|
2
We need $a+b=a b$, or $a=\frac{b}{b-1}=1-\frac{1}{b-1}$, so $1 /(b-1)$ is an integer. Thus $b$ must be 0 or 2 , and $a$ is 0 or 2 , respectively. So there are 2 .
|
2
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
A broken calculator has the + and $\times$ keys switched. For how many ordered pairs $(a, b)$ of integers will it correctly calculate $a+b$ using the labelled + key?
|
2
We need $a+b=a b$, or $a=\frac{b}{b-1}=1-\frac{1}{b-1}$, so $1 /(b-1)$ is an integer. Thus $b$ must be 0 or 2 , and $a$ is 0 or 2 , respectively. So there are 2 .
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen1-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution: "
}
|
6d67ad51-fc69-5b85-8457-df719352cc45
| 611,121
|
Consider a 2003-gon inscribed in a circle and a triangulation of it with diagonals intersecting only at vertices. What is the smallest possible number of obtuse triangles in the triangulation?
|
1999
By induction, it follows easily that any triangulation of an $n$-gon inscribed in a circle has $n-2$ triangles. A triangle is obtuse unless it contains the center of the circle in its interior (in which case it is acute) or on one of its edges (in which case it is right). It is then clear that there are at most 2 non-obtuse triangles, and 2 is achieved when the center of the circle is on one of the diagonals of the triangulation. So the minimum number of obtuse triangles is $2001-2=1999$.
|
1999
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Consider a 2003-gon inscribed in a circle and a triangulation of it with diagonals intersecting only at vertices. What is the smallest possible number of obtuse triangles in the triangulation?
|
1999
By induction, it follows easily that any triangulation of an $n$-gon inscribed in a circle has $n-2$ triangles. A triangle is obtuse unless it contains the center of the circle in its interior (in which case it is acute) or on one of its edges (in which case it is right). It is then clear that there are at most 2 non-obtuse triangles, and 2 is achieved when the center of the circle is on one of the diagonals of the triangulation. So the minimum number of obtuse triangles is $2001-2=1999$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen1-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nSolution: "
}
|
717b1208-baec-5b2d-96b5-72159779c7fb
| 611,122
|
Bessie the cow is trying to navigate her way through a field. She can travel only from lattice point to adjacent lattice point, can turn only at lattice points, and can travel only to the east or north. (A lattice point is a point whose coordinates are both integers.) $(0,0)$ is the southwest corner of the field. $(5,5)$ is the northeast corner of the field. Due to large rocks, Bessie is unable to walk on the points $(1,1),(2,3)$, or $(3,2)$. How many ways are there for Bessie to travel from $(0,0)$ to $(5,5)$ under these constraints?
|
32
In the figure, each point is labeled with the number of ways to reach that point. The numbers are successively computed as follows: The point $(0,0)$ can trivially be reached in 1 way. When Bessie reaches any subsequent point $(x, y)$ (other than a rock), she can arrive either via a northward or an eastward step, so the number of ways she can reach that point equals the number of ways of reaching $(x-1, y)$ plus the number of ways of reaching $(x, y-1)$. By iterating this calculation, we eventually find that $(5,5)$ can be reached in 32 ways.
|
32
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Bessie the cow is trying to navigate her way through a field. She can travel only from lattice point to adjacent lattice point, can turn only at lattice points, and can travel only to the east or north. (A lattice point is a point whose coordinates are both integers.) $(0,0)$ is the southwest corner of the field. $(5,5)$ is the northeast corner of the field. Due to large rocks, Bessie is unable to walk on the points $(1,1),(2,3)$, or $(3,2)$. How many ways are there for Bessie to travel from $(0,0)$ to $(5,5)$ under these constraints?
|
32
In the figure, each point is labeled with the number of ways to reach that point. The numbers are successively computed as follows: The point $(0,0)$ can trivially be reached in 1 way. When Bessie reaches any subsequent point $(x, y)$ (other than a rock), she can arrive either via a northward or an eastward step, so the number of ways she can reach that point equals the number of ways of reaching $(x-1, y)$ plus the number of ways of reaching $(x, y-1)$. By iterating this calculation, we eventually find that $(5,5)$ can be reached in 32 ways.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen1-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nSolution: "
}
|
9df08165-f0da-5c2e-ba9f-be2e3d5723a9
| 611,123
|
A compact disc has the shape of a circle of diameter 5 inches with a 1 -inch-diameter circular hole in the center. Assuming the capacity of the CD is proportional to its area, how many inches would need to be added to the outer diameter to double the capacity?
|
2
Doubling the capacity is equivalent to doubling the area, which is initially $\pi\left[(5 / 2)^{2}\right.$ $\left.(1 / 2)^{2}\right]=6 \pi$. Thus we want to achieve an area of $12 \pi$, so if the new diameter is $d$, we want $\pi\left[(d / 2)^{2}-(1 / 2)^{2}\right]=12 \pi \Rightarrow d=7$. Thus we need to add 2 inches to the diameter.
|
2
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A compact disc has the shape of a circle of diameter 5 inches with a 1 -inch-diameter circular hole in the center. Assuming the capacity of the CD is proportional to its area, how many inches would need to be added to the outer diameter to double the capacity?
|
2
Doubling the capacity is equivalent to doubling the area, which is initially $\pi\left[(5 / 2)^{2}\right.$ $\left.(1 / 2)^{2}\right]=6 \pi$. Thus we want to achieve an area of $12 \pi$, so if the new diameter is $d$, we want $\pi\left[(d / 2)^{2}-(1 / 2)^{2}\right]=12 \pi \Rightarrow d=7$. Thus we need to add 2 inches to the diameter.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen2-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\n## Solution: "
}
|
05dff2ea-8d52-5dae-8694-46352eeabd96
| 611,124
|
You have a list of real numbers, whose sum is 40 . If you replace every number $x$ on the list by $1-x$, the sum of the new numbers will be 20 . If instead you had replaced every number $x$ by $1+x$, what would the sum then be?
|
100
Let $n$ be the number of numbers on the list. If each initial number is replaced by its negative, the sum will then be -40 , and adding 1 to every number on this list increases the sum by $n$, so $n-40=20 \Rightarrow n=60$. Then, if we had simply added 1 to each of the initial numbers (without negating first), the sum would increase to $40+n=40+60=100$.
|
100
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
You have a list of real numbers, whose sum is 40 . If you replace every number $x$ on the list by $1-x$, the sum of the new numbers will be 20 . If instead you had replaced every number $x$ by $1+x$, what would the sum then be?
|
100
Let $n$ be the number of numbers on the list. If each initial number is replaced by its negative, the sum will then be -40 , and adding 1 to every number on this list increases the sum by $n$, so $n-40=20 \Rightarrow n=60$. Then, if we had simply added 1 to each of the initial numbers (without negating first), the sum would increase to $40+n=40+60=100$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen2-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\n## Solution: "
}
|
403c9094-21bc-54f1-ac68-8991e02164b7
| 611,125
|
How many positive rational numbers less than $\pi$ have denominator at most 7 when written in lowest terms? (Integers have denominator 1.)
|
54
We can simply list them. The table shows that there are $3+3+6+6+12+6+18=54$.
| Denominator | Values |
| ---: | :--- |
| 1 | $\frac{1}{1}, \frac{2}{1}, \frac{3}{1}$ |
| 2 | $\frac{1}{2}, \frac{3}{2}, \frac{5}{2}$ |
| 3 | $\frac{1}{3}, \frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{7}{3}, \frac{8}{3}$ |
| 4 | $\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, \frac{11}{4}$ |
| 5 | $\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \frac{9}{5}, \frac{11}{5}, \frac{12}{5}, \frac{13}{5}, \frac{14}{5}$ |
| 6 | $\frac{1}{6}, \frac{5}{6}, \frac{7}{6}, \frac{11}{6}, \frac{13}{6}, \frac{17}{6}$ |
| 7 | $\frac{1}{7}, \frac{2}{7}, \ldots, \frac{6}{7}, \frac{8}{7}, \frac{9}{7}, \ldots, \frac{13}{7}, \frac{15}{7}, \frac{16}{7}, \ldots, \frac{20}{7}$ |
|
54
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many positive rational numbers less than $\pi$ have denominator at most 7 when written in lowest terms? (Integers have denominator 1.)
|
54
We can simply list them. The table shows that there are $3+3+6+6+12+6+18=54$.
| Denominator | Values |
| ---: | :--- |
| 1 | $\frac{1}{1}, \frac{2}{1}, \frac{3}{1}$ |
| 2 | $\frac{1}{2}, \frac{3}{2}, \frac{5}{2}$ |
| 3 | $\frac{1}{3}, \frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{7}{3}, \frac{8}{3}$ |
| 4 | $\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, \frac{11}{4}$ |
| 5 | $\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \frac{9}{5}, \frac{11}{5}, \frac{12}{5}, \frac{13}{5}, \frac{14}{5}$ |
| 6 | $\frac{1}{6}, \frac{5}{6}, \frac{7}{6}, \frac{11}{6}, \frac{13}{6}, \frac{17}{6}$ |
| 7 | $\frac{1}{7}, \frac{2}{7}, \ldots, \frac{6}{7}, \frac{8}{7}, \frac{9}{7}, \ldots, \frac{13}{7}, \frac{15}{7}, \frac{16}{7}, \ldots, \frac{20}{7}$ |
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen2-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution: "
}
|
de7fdbb5-3f42-54a7-8eff-afca1796e1d2
| 611,126
|
In triangle $A B C$ with area 51, points $D$ and $E$ trisect $A B$ and points $F$ and $G$ trisect $B C$. Find the largest possible area of quadrilateral $D E F G$.
|
17
Assume $E$ is between $D$ and $B$, and $F$ is between $G$ and $B$ (the alternative is to switch two points, say $D$ and $E$, which clearly gives a non-convex quadrilateral with smaller
area). If two triangles have their bases on the same line and the same opposite vertex, then it follows from the $\frac{1}{2} b h$ formula that their areas are in the same ratio as their bases. In particular (brackets denote areas),
$$
\frac{[D B G]}{[A B C]}=\frac{[D B G]}{[A B G]} \cdot \frac{[A B G]}{[A B C]}=\frac{D B}{A B} \cdot \frac{B G}{B C}=\frac{2}{3} \cdot \frac{2}{3}=\frac{4}{9},
$$
and similarly $[E B F] /[A B C]=1 / 9$. Subtracting gives $[D E F G] /[A B C]=1 / 3$, so the answer is $[A B C] / 3=17$.
|
17
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In triangle $A B C$ with area 51, points $D$ and $E$ trisect $A B$ and points $F$ and $G$ trisect $B C$. Find the largest possible area of quadrilateral $D E F G$.
|
17
Assume $E$ is between $D$ and $B$, and $F$ is between $G$ and $B$ (the alternative is to switch two points, say $D$ and $E$, which clearly gives a non-convex quadrilateral with smaller
area). If two triangles have their bases on the same line and the same opposite vertex, then it follows from the $\frac{1}{2} b h$ formula that their areas are in the same ratio as their bases. In particular (brackets denote areas),
$$
\frac{[D B G]}{[A B C]}=\frac{[D B G]}{[A B G]} \cdot \frac{[A B G]}{[A B C]}=\frac{D B}{A B} \cdot \frac{B G}{B C}=\frac{2}{3} \cdot \frac{2}{3}=\frac{4}{9},
$$
and similarly $[E B F] /[A B C]=1 / 9$. Subtracting gives $[D E F G] /[A B C]=1 / 3$, so the answer is $[A B C] / 3=17$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen2-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution: "
}
|
3c60ab73-b78a-5262-842d-2c4d1d6773eb
| 611,127
|
The numbers $112,121,123,153,243,313$, and 322 are among the rows, columns, and diagonals of a $3 \times 3$ square grid of digits (rows and diagonals read left-to-right, and columns read top-to-bottom). What 3-digit number completes the list?
|
524
| 1 | 1 | 2 |
| :--- | :--- | :--- |
| 5 | 2 | 4 |
| 3 | 1 | 3 |
The center digit is the middle digit of 4 numbers (hence at least 3 members of the above list), so it must be 2 . The top-left digit begins at least 2 members of the above list, so it must be 1 or 3 . If it is 3 , then after placing 313 we see that we need three more numbers starting with 3 , impossible; hence, it is 1 . So 243 and 313 must (in some order) be the last row and the last column, and now it is easy to complete the grid as shown; the answer is 524 .
|
524
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
The numbers $112,121,123,153,243,313$, and 322 are among the rows, columns, and diagonals of a $3 \times 3$ square grid of digits (rows and diagonals read left-to-right, and columns read top-to-bottom). What 3-digit number completes the list?
|
524
| 1 | 1 | 2 |
| :--- | :--- | :--- |
| 5 | 2 | 4 |
| 3 | 1 | 3 |
The center digit is the middle digit of 4 numbers (hence at least 3 members of the above list), so it must be 2 . The top-left digit begins at least 2 members of the above list, so it must be 1 or 3 . If it is 3 , then after placing 313 we see that we need three more numbers starting with 3 , impossible; hence, it is 1 . So 243 and 313 must (in some order) be the last row and the last column, and now it is easy to complete the grid as shown; the answer is 524 .
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen2-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nSolution: "
}
|
36500d75-9ef5-5757-a83e-343acaaf835b
| 611,128
|
If $x \geq 0, y \geq 0$ are integers, randomly chosen with the constraint $x+y \leq 10$, what is the probability that $x+y$ is even?
|
6/11
For each $p \leq 10$, if $x+y=p, x$ can range from 0 to $p$, yielding $p+1$ ordered pairs $(x, y)$. Thus there are a total of $1+2+3+\cdots+11$ allowable ordered pairs $(x, y)$, but $1+3+5+\cdots+11$ of these pairs have an even sum. So the desired probability is
$$
\frac{1+3+5+\cdots+11}{1+2+3+\cdots+11}=\frac{6^{2}}{11 \cdot 6}=\frac{6}{11} .
$$
|
\frac{6}{11}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
If $x \geq 0, y \geq 0$ are integers, randomly chosen with the constraint $x+y \leq 10$, what is the probability that $x+y$ is even?
|
6/11
For each $p \leq 10$, if $x+y=p, x$ can range from 0 to $p$, yielding $p+1$ ordered pairs $(x, y)$. Thus there are a total of $1+2+3+\cdots+11$ allowable ordered pairs $(x, y)$, but $1+3+5+\cdots+11$ of these pairs have an even sum. So the desired probability is
$$
\frac{1+3+5+\cdots+11}{1+2+3+\cdots+11}=\frac{6^{2}}{11 \cdot 6}=\frac{6}{11} .
$$
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen2-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\n## Solution: "
}
|
7b9187f4-37aa-5da7-b12e-e360f8fc852f
| 611,129
|
Several positive integers are given, not necessarily all different. Their sum is 2003. Suppose that $n_{1}$ of the given numbers are equal to $1, n_{2}$ of them are equal to $2, \ldots$, $n_{2003}$ of them are equal to 2003 . Find the largest possible value of
$$
n_{2}+2 n_{3}+3 n_{4}+\cdots+2002 n_{2003}
$$
|
2002
The sum of all the numbers is $n_{1}+2 n_{2}+\cdots+2003 n_{2003}$, while the number of numbers is $n_{1}+n_{2}+\cdots+n_{2003}$. Hence, the desired quantity equals
$$
\begin{gathered}
\left(n_{1}+2 n_{2}+\cdots+2003 n_{2003}\right)-\left(n_{1}+n_{2}+\cdots+n_{2003}\right) \\
=(\text { sum of the numbers })-(\text { number of numbers })
\end{gathered}
$$
$$
=2003-\text { (number of numbers), }
$$
which is maximized when the number of numbers is minimized. Hence, we should have just one number, equal to 2003 , and then the specified sum is $2003-1=2002$.
Comment: On the day of the contest, a protest was lodged (successfully) on the grounds that the use of the words "several" and "their" in the problem statement implies there must be at least 2 numbers. Then the answer is 2001 , and this maximum is achieved by any two numbers whose sum is 2003 .
|
2002
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Several positive integers are given, not necessarily all different. Their sum is 2003. Suppose that $n_{1}$ of the given numbers are equal to $1, n_{2}$ of them are equal to $2, \ldots$, $n_{2003}$ of them are equal to 2003 . Find the largest possible value of
$$
n_{2}+2 n_{3}+3 n_{4}+\cdots+2002 n_{2003}
$$
|
2002
The sum of all the numbers is $n_{1}+2 n_{2}+\cdots+2003 n_{2003}$, while the number of numbers is $n_{1}+n_{2}+\cdots+n_{2003}$. Hence, the desired quantity equals
$$
\begin{gathered}
\left(n_{1}+2 n_{2}+\cdots+2003 n_{2003}\right)-\left(n_{1}+n_{2}+\cdots+n_{2003}\right) \\
=(\text { sum of the numbers })-(\text { number of numbers })
\end{gathered}
$$
$$
=2003-\text { (number of numbers), }
$$
which is maximized when the number of numbers is minimized. Hence, we should have just one number, equal to 2003 , and then the specified sum is $2003-1=2002$.
Comment: On the day of the contest, a protest was lodged (successfully) on the grounds that the use of the words "several" and "their" in the problem statement implies there must be at least 2 numbers. Then the answer is 2001 , and this maximum is achieved by any two numbers whose sum is 2003 .
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-gen2-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nSolution: "
}
|
254a92af-9153-56a0-947b-79c07b3a2d6a
| 611,130
|
$A D$ and $B C$ are both perpendicular to $A B$, and $C D$ is perpendicular to $A C$. If $A B=4$ and $B C=3$, find $C D$.
|
$20 / 3$
By Pythagoras in $\triangle A B C, A C=5$. But $\angle C A D=90^{\circ}-\angle B A C=\angle A C B$, so right triangles $C A D, B C A$ are similar, and $C D / A C=B A / C B=4 / 3 \Rightarrow C D=20 / 3$.
|
\frac{20}{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
$A D$ and $B C$ are both perpendicular to $A B$, and $C D$ is perpendicular to $A C$. If $A B=4$ and $B C=3$, find $C D$.
|
$20 / 3$
By Pythagoras in $\triangle A B C, A C=5$. But $\angle C A D=90^{\circ}-\angle B A C=\angle A C B$, so right triangles $C A D, B C A$ are similar, and $C D / A C=B A / C B=4 / 3 \Rightarrow C D=20 / 3$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-geo-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution: "
}
|
842c7039-be12-5b18-a53c-2ff6c4b1bd30
| 611,131
|
As shown, $U$ and $C$ are points on the sides of triangle $M N H$ such that $M U=s$, $U N=6, N C=20, C H=s, H M=25$. If triangle $U N C$ and quadrilateral $M U C H$ have equal areas, what is $s$ ?
|
4
Using brackets to denote areas, we have $[M C H]=[U N C]+[M U C H]=2[U N C]$. On the other hand, triangles with equal altitudes have their areas in the same ratio as their bases, so
$$
2=\frac{[M N H]}{[U N C]}=\frac{[M N H]}{[M N C]} \cdot \frac{[M N C]}{[U N C]}=\frac{N H}{N C} \cdot \frac{M N}{U N}=\frac{s+20}{20} \cdot \frac{s+6}{6} .
$$
Clearing the denominator gives $(s+20)(s+6)=240$, and solving the quadratic gives $s=4$ or -30 . Since $s>0$, we must have $s=4$.
|
4
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
As shown, $U$ and $C$ are points on the sides of triangle $M N H$ such that $M U=s$, $U N=6, N C=20, C H=s, H M=25$. If triangle $U N C$ and quadrilateral $M U C H$ have equal areas, what is $s$ ?
|
4
Using brackets to denote areas, we have $[M C H]=[U N C]+[M U C H]=2[U N C]$. On the other hand, triangles with equal altitudes have their areas in the same ratio as their bases, so
$$
2=\frac{[M N H]}{[U N C]}=\frac{[M N H]}{[M N C]} \cdot \frac{[M N C]}{[U N C]}=\frac{N H}{N C} \cdot \frac{M N}{U N}=\frac{s+20}{20} \cdot \frac{s+6}{6} .
$$
Clearing the denominator gives $(s+20)(s+6)=240$, and solving the quadratic gives $s=4$ or -30 . Since $s>0$, we must have $s=4$.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-geo-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution: "
}
|
2606f5dd-9f20-5c3f-970d-88055b971e91
| 611,132
|
Take a clay sphere of radius 13 , and drill a circular hole of radius 5 through its center. Take the remaining "bead" and mold it into a new sphere. What is this sphere's radius?
|
12
Let $r$ be the radius of the sphere. We take cross sections of the bead perpendicular to the line of the drill and compare them to cross sections of the sphere at the same
distance from its center. At a height $h$, the cross section of the sphere is a circle with radius $\sqrt{r^{2}-h^{2}}$ and thus area $\pi\left(r^{2}-h^{2}\right)$. At the same height, the cross section of the bead is an annulus with outer radius $\sqrt{13^{2}-h^{2}}$ and inner radius 5 , for an area of $\pi\left(13^{2}-h^{2}\right)-\pi\left(5^{2}\right)=\pi\left(12^{2}-h^{2}\right)\left(\right.$ since $\left.13^{2}-5^{2}=12^{2}\right)$. Thus, if $r=12$, the sphere and the bead will have the same cross-sectional area $\pi\left(12^{2}-h^{2}\right)$ for $|h| \leq 12$ and 0 for $|h|>12$. Since all the cross sections have the same area, the two clay figures then have the same volume. And certainly there is only one value of $r$ for which the two volumes are equal, so $r=12$ is the answer.
|
12
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Take a clay sphere of radius 13 , and drill a circular hole of radius 5 through its center. Take the remaining "bead" and mold it into a new sphere. What is this sphere's radius?
|
12
Let $r$ be the radius of the sphere. We take cross sections of the bead perpendicular to the line of the drill and compare them to cross sections of the sphere at the same
distance from its center. At a height $h$, the cross section of the sphere is a circle with radius $\sqrt{r^{2}-h^{2}}$ and thus area $\pi\left(r^{2}-h^{2}\right)$. At the same height, the cross section of the bead is an annulus with outer radius $\sqrt{13^{2}-h^{2}}$ and inner radius 5 , for an area of $\pi\left(13^{2}-h^{2}\right)-\pi\left(5^{2}\right)=\pi\left(12^{2}-h^{2}\right)\left(\right.$ since $\left.13^{2}-5^{2}=12^{2}\right)$. Thus, if $r=12$, the sphere and the bead will have the same cross-sectional area $\pi\left(12^{2}-h^{2}\right)$ for $|h| \leq 12$ and 0 for $|h|>12$. Since all the cross sections have the same area, the two clay figures then have the same volume. And certainly there is only one value of $r$ for which the two volumes are equal, so $r=12$ is the answer.
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-geo-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nSolution: "
}
|
7b15b62f-6103-543a-9fee-0365de61ab51
| 611,134
|
Let $R S T U V$ be a regular pentagon. Construct an equilateral triangle $P R S$ with point $P$ inside the pentagon. Find the measure (in degrees) of angle $P T V$.
|
6
We have $\angle P R V=\angle S R V-\angle S R P=108^{\circ}-60^{\circ}=48^{\circ}$. Since $P R=R S=R V$, triangle $P R V$ is isosceles, so that $\angle V P R=\angle R V P=\left(180^{\circ}-\angle P R V\right) / 2=66^{\circ}$. Likewise, we have $\angle T P S=66^{\circ}$, so that
$$
\angle T P V=360^{\circ}-(\angle V P R+\angle R P S+\angle S P T)=360^{\circ}-\left(66^{\circ}+60^{\circ}+66^{\circ}\right)=168^{\circ}
$$
Finally, by symmetry, triangle $P T V$ is isosceles $(P T=T V)$, so $\angle P T V=\angle T V P=$ $\left(180^{\circ}-\angle T P V\right) / 2=6^{\circ}$. (See the figure.)
|
6
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $R S T U V$ be a regular pentagon. Construct an equilateral triangle $P R S$ with point $P$ inside the pentagon. Find the measure (in degrees) of angle $P T V$.
|
6
We have $\angle P R V=\angle S R V-\angle S R P=108^{\circ}-60^{\circ}=48^{\circ}$. Since $P R=R S=R V$, triangle $P R V$ is isosceles, so that $\angle V P R=\angle R V P=\left(180^{\circ}-\angle P R V\right) / 2=66^{\circ}$. Likewise, we have $\angle T P S=66^{\circ}$, so that
$$
\angle T P V=360^{\circ}-(\angle V P R+\angle R P S+\angle S P T)=360^{\circ}-\left(66^{\circ}+60^{\circ}+66^{\circ}\right)=168^{\circ}
$$
Finally, by symmetry, triangle $P T V$ is isosceles $(P T=T V)$, so $\angle P T V=\angle T V P=$ $\left(180^{\circ}-\angle T P V\right) / 2=6^{\circ}$. (See the figure.)
|
{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-geo-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nSolution: "
}
|
cc215d81-a312-535b-8953-4d49956be629
| 611,135
|
Let $A B C$ be an equilateral triangle of side length 2 . Let $\omega$ be its circumcircle, and let $\omega_{A}, \omega_{B}, \omega_{C}$ be circles congruent to $\omega$ centered at each of its vertices. Let $R$ be the set of all points in the plane contained in exactly two of these four circles. What is the area of $R$ ?
|
$2 \sqrt{3}$
$\omega_{A}, \omega_{B}, \omega_{C}$ intersect at the circumcenter; thus, every point within the circumcircle, and no point outside of it, is in two or more circles. The area inside exactly two circles is shaded in the figure. The two intersection points of $\omega_{A}$ and $\omega_{B}$, together with $A$, form the vertices of an equilateral triangle. As shown, this equilateral triangle cuts off a "lip" of $\omega$ (bounded by a $60^{\circ}$ arc of $\omega$ and the corresponding chord) and another, congruent lip of $\omega_{B}$ that is not part of the region of interest. By rotating the first lip to the position of the second, we can reassemble the equilateral triangle. Doing this for each of the 6 such triangles, we see that the desired area equals the area of a regular hexagon inscribed in $\omega$. The side length of this hexagon is $(2 / 3) \cdot(\sqrt{3} / 2) \cdot 2=2 \sqrt{3} / 3$, so its area is $6 \cdot(\sqrt{3} / 4) \cdot(2 \sqrt{3} / 3)^{2}=2 \sqrt{3}$, and this is the answer.
|
2 \sqrt{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be an equilateral triangle of side length 2 . Let $\omega$ be its circumcircle, and let $\omega_{A}, \omega_{B}, \omega_{C}$ be circles congruent to $\omega$ centered at each of its vertices. Let $R$ be the set of all points in the plane contained in exactly two of these four circles. What is the area of $R$ ?
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$2 \sqrt{3}$
$\omega_{A}, \omega_{B}, \omega_{C}$ intersect at the circumcenter; thus, every point within the circumcircle, and no point outside of it, is in two or more circles. The area inside exactly two circles is shaded in the figure. The two intersection points of $\omega_{A}$ and $\omega_{B}$, together with $A$, form the vertices of an equilateral triangle. As shown, this equilateral triangle cuts off a "lip" of $\omega$ (bounded by a $60^{\circ}$ arc of $\omega$ and the corresponding chord) and another, congruent lip of $\omega_{B}$ that is not part of the region of interest. By rotating the first lip to the position of the second, we can reassemble the equilateral triangle. Doing this for each of the 6 such triangles, we see that the desired area equals the area of a regular hexagon inscribed in $\omega$. The side length of this hexagon is $(2 / 3) \cdot(\sqrt{3} / 2) \cdot 2=2 \sqrt{3} / 3$, so its area is $6 \cdot(\sqrt{3} / 4) \cdot(2 \sqrt{3} / 3)^{2}=2 \sqrt{3}$, and this is the answer.
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{
"resource_path": "HarvardMIT/segmented/en-62-2003-feb-geo-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution: "
}
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21173bef-3e71-5498-97c1-bd6aadc95302
| 611,136
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