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Explain how any unit fraction $\frac{1}{n}$ can be decomposed into other unit fractions.
|
$\frac{1}{2 n}+\frac{1}{3 n}+\frac{1}{6 n}$
|
\frac{1}{2 n}+\frac{1}{3 n}+\frac{1}{6 n}
|
Yes
|
Problem not solved
|
math-word-problem
|
Number Theory
|
Explain how any unit fraction $\frac{1}{n}$ can be decomposed into other unit fractions.
|
$\frac{1}{2 n}+\frac{1}{3 n}+\frac{1}{6 n}$
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. ",
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2008"
}
|
(a) Write 1 as a sum of 4 distinct unit fractions.
|
$\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}$
(b) Write 1 as a sum of 5 distinct unit fractions.
|
\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
(a) Write 1 as a sum of 4 distinct unit fractions.
|
$\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}$
(b) Write 1 as a sum of 5 distinct unit fractions.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. ",
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2008"
}
|
(a) Write 1 as a sum of 4 distinct unit fractions.
|
$\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\frac{1}{43 \cdot 42}$
(c) Show that, for any integer $k>3,1$ can be decomposed into $k$ unit fractions.
|
\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\frac{1}{43 \cdot 42}
|
Yes
|
Incomplete
|
math-word-problem
|
Number Theory
|
(a) Write 1 as a sum of 4 distinct unit fractions.
|
$\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\frac{1}{43 \cdot 42}$
(c) Show that, for any integer $k>3,1$ can be decomposed into $k$ unit fractions.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. ",
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2008"
}
|
(a) Write 1 as a sum of 4 distinct unit fractions.
|
If we can do it for $k$ fractions, simply replace the last one (say $\frac{1}{n}$ ) with $\frac{1}{n+1}+\frac{1}{n(n+1)}$. Then we can do it for $k+1$ fractions. So, since we can do it for $k=3$, we can do it for any $k>3$.
|
not found
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
(a) Write 1 as a sum of 4 distinct unit fractions.
|
If we can do it for $k$ fractions, simply replace the last one (say $\frac{1}{n}$ ) with $\frac{1}{n+1}+\frac{1}{n(n+1)}$. Then we can do it for $k+1$ fractions. So, since we can do it for $k=3$, we can do it for any $k>3$.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. ",
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2008"
}
|
Say that $\frac{a}{b}$ is a positive rational number in simplest form, with $a \neq 1$. Further, say that $n$ is an integer such that:
$$
\frac{1}{n}>\frac{a}{b}>\frac{1}{n+1}
$$
Show that when $\frac{a}{b}-\frac{1}{n+1}$ is written in simplest form, its numerator is smaller than $a$.
|
$\quad \frac{a}{b}-\frac{1}{n+1}=\frac{a(n+1)-b}{b(n+1)}$. Therefore, when we write it in simplest form, its numerator will be at most $a(n+1)-b$. We claim that $a(n+1)-b<a$. Indeed, this is the same as $a n-b<0 \Longleftrightarrow a n<b \Longleftrightarrow \frac{b}{a}>n$, which is given.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Say that $\frac{a}{b}$ is a positive rational number in simplest form, with $a \neq 1$. Further, say that $n$ is an integer such that:
$$
\frac{1}{n}>\frac{a}{b}>\frac{1}{n+1}
$$
Show that when $\frac{a}{b}-\frac{1}{n+1}$ is written in simplest form, its numerator is smaller than $a$.
|
$\quad \frac{a}{b}-\frac{1}{n+1}=\frac{a(n+1)-b}{b(n+1)}$. Therefore, when we write it in simplest form, its numerator will be at most $a(n+1)-b$. We claim that $a(n+1)-b<a$. Indeed, this is the same as $a n-b<0 \Longleftrightarrow a n<b \Longleftrightarrow \frac{b}{a}>n$, which is given.
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. ",
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2008"
}
|
An aside: the sum of all the unit fractions
It is possible to show that, given any real M , there exists a positive integer $k$ large enough that:
$$
\sum_{n=1}^{k} \frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3} \ldots>M
$$
Note that this statement means that the infinite harmonic series, $\sum_{n=1}^{\infty} \frac{1}{n}$, grows without bound, or diverges. For the specific example $\mathrm{M}=5$, find a value of $k$, not necessarily the smallest, such that the inequality holds. Justify your answer.
|
Note that $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{1}{2 n}+\ldots+\frac{1}{2 n}=\frac{1}{2}$. Therefore, if we apply this to $n=1,2,4,8,16,32,64,128$, we get
$$
\left(\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\ldots+\left(\frac{1}{129}+\ldots+\frac{1}{256}\right)>\frac{1}{2}+\ldots+\frac{1}{2}=4
$$
so, adding in $\frac{1}{1}$, we get
$$
\sum_{n=1}^{256} \frac{1}{n}>5
$$
so $k=256$ will suffice.
|
256
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
An aside: the sum of all the unit fractions
It is possible to show that, given any real M , there exists a positive integer $k$ large enough that:
$$
\sum_{n=1}^{k} \frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3} \ldots>M
$$
Note that this statement means that the infinite harmonic series, $\sum_{n=1}^{\infty} \frac{1}{n}$, grows without bound, or diverges. For the specific example $\mathrm{M}=5$, find a value of $k$, not necessarily the smallest, such that the inequality holds. Justify your answer.
|
Note that $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{1}{2 n}+\ldots+\frac{1}{2 n}=\frac{1}{2}$. Therefore, if we apply this to $n=1,2,4,8,16,32,64,128$, we get
$$
\left(\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\ldots+\left(\frac{1}{129}+\ldots+\frac{1}{256}\right)>\frac{1}{2}+\ldots+\frac{1}{2}=4
$$
so, adding in $\frac{1}{1}$, we get
$$
\sum_{n=1}^{256} \frac{1}{n}>5
$$
so $k=256$ will suffice.
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n## 5. ",
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2008"
}
|
Now, using information from problems 4 and 5 , prove that the following method to decompose any positive rational number will always terminate:
Step 1. Start with the fraction $\frac{a}{b}$. Let $t_{1}$ be the largest unit fraction $\frac{1}{n}$ which is less than or equal to $\frac{a}{b}$.
Step 2. If we have already chosen $t_{1}$ through $t_{k}$, and if $t_{1}+t_{2}+\ldots+t_{k}$ is still less than $\frac{a}{b}$, then let $t_{k+1}$ be the largest unit fraction less than both $t_{k}$ and $\frac{a}{b}$.
Step 3. If $t_{1}+\ldots+t_{k+1}$ equals $\frac{a}{b}$, the decomposition is found. Otherwise, repeat step 2 .
Why does this method never result in an infinite sequence of $t_{i}$ ?
|
Let $\frac{a_{k}}{b_{k}}=\frac{a}{b}-t_{1}-\ldots-t_{k}$, where $\frac{a_{k}}{b_{k}}$ is a fraction in simplest terms. Initially, this algorithm will have $t_{1}=1, t_{2}=\frac{1}{2}, t_{3}=\frac{1}{3}$, etc. until $\frac{a_{k}}{b_{k}}<\frac{1}{k+1}$. This will eventually happen by problem 5 , since there exists a $k$ such that $\frac{1}{1}+\ldots+\frac{1}{k+1}>\frac{a_{k}}{b_{k}}$. At that point, there is some $n$ with $\frac{1}{n}<t_{k}$ such that $\frac{1}{n}>\frac{a_{k}}{b_{k}}>\frac{1}{n+1}$. In this case, $t_{k+1}=\frac{1}{n+1}$.
Suppose that there exists $n_{k}$ such that $\frac{1}{n_{k}}>\frac{a_{k}}{b_{k}}>\frac{1}{n_{k}+1}$ for some $k$. Then we have $t_{k+1}=\frac{1}{n_{k}+1}$ and $\frac{a_{k+1}}{b_{k+1}}<\frac{1}{n_{k}\left(n_{k}+1\right)}$. This shows that once we have found $n_{k}$ such that $\frac{1}{n_{k}}>\frac{a_{k}}{b_{k}}>\frac{1}{n_{k}+1}$ and $\frac{1}{n_{k}} \leq t_{k}$, we no longer have to worry about $t_{k+1}$ being less than $t_{k}$, since $t_{k+1}=\frac{1}{n_{k}+1}<\frac{1}{n_{k}}<$ $t_{k}$, and also $n_{k+1} \geq n_{k}\left(n_{k}+1\right)$ while $\frac{1}{n_{k}\left(n_{k}+1\right)} \leq \frac{1}{n_{k}+1}=t_{k+1}$.
On the other hand, once we have found such an $n_{k}$, the sequence $\left\{a_{k}\right\}$ must be decreasing by problem 4. Since the $a_{k}$ are all integers, we eventually have to get to 0 (as there is no infinite decreasing sequence of positive integers). Therefore, after some finite number of steps the algorithm terminates with $a_{k+1}=0$, so $0=\frac{a_{k}}{b_{k}}=\frac{a}{b}-t_{1}-\ldots-t_{k}$, so $\frac{a}{b}=t_{1}+\ldots+t_{k}$, which is what we wanted.
## Juicy Numbers [100]
A juicy number is an integer $j>1$ for which there is a sequence $a_{1}<a_{2}<\ldots<a_{k}$ of positive integers such that $a_{k}=j$ and such that the sum of the reciprocals of all the $a_{i}$ is 1 . For example, 6 is a juicy number because $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$, but 2 is not juicy.
In this part, you will investigate some of the properties of juicy numbers. Remember that if you do not solve a question, you can still use its result on later questions.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Now, using information from problems 4 and 5 , prove that the following method to decompose any positive rational number will always terminate:
Step 1. Start with the fraction $\frac{a}{b}$. Let $t_{1}$ be the largest unit fraction $\frac{1}{n}$ which is less than or equal to $\frac{a}{b}$.
Step 2. If we have already chosen $t_{1}$ through $t_{k}$, and if $t_{1}+t_{2}+\ldots+t_{k}$ is still less than $\frac{a}{b}$, then let $t_{k+1}$ be the largest unit fraction less than both $t_{k}$ and $\frac{a}{b}$.
Step 3. If $t_{1}+\ldots+t_{k+1}$ equals $\frac{a}{b}$, the decomposition is found. Otherwise, repeat step 2 .
Why does this method never result in an infinite sequence of $t_{i}$ ?
|
Let $\frac{a_{k}}{b_{k}}=\frac{a}{b}-t_{1}-\ldots-t_{k}$, where $\frac{a_{k}}{b_{k}}$ is a fraction in simplest terms. Initially, this algorithm will have $t_{1}=1, t_{2}=\frac{1}{2}, t_{3}=\frac{1}{3}$, etc. until $\frac{a_{k}}{b_{k}}<\frac{1}{k+1}$. This will eventually happen by problem 5 , since there exists a $k$ such that $\frac{1}{1}+\ldots+\frac{1}{k+1}>\frac{a_{k}}{b_{k}}$. At that point, there is some $n$ with $\frac{1}{n}<t_{k}$ such that $\frac{1}{n}>\frac{a_{k}}{b_{k}}>\frac{1}{n+1}$. In this case, $t_{k+1}=\frac{1}{n+1}$.
Suppose that there exists $n_{k}$ such that $\frac{1}{n_{k}}>\frac{a_{k}}{b_{k}}>\frac{1}{n_{k}+1}$ for some $k$. Then we have $t_{k+1}=\frac{1}{n_{k}+1}$ and $\frac{a_{k+1}}{b_{k+1}}<\frac{1}{n_{k}\left(n_{k}+1\right)}$. This shows that once we have found $n_{k}$ such that $\frac{1}{n_{k}}>\frac{a_{k}}{b_{k}}>\frac{1}{n_{k}+1}$ and $\frac{1}{n_{k}} \leq t_{k}$, we no longer have to worry about $t_{k+1}$ being less than $t_{k}$, since $t_{k+1}=\frac{1}{n_{k}+1}<\frac{1}{n_{k}}<$ $t_{k}$, and also $n_{k+1} \geq n_{k}\left(n_{k}+1\right)$ while $\frac{1}{n_{k}\left(n_{k}+1\right)} \leq \frac{1}{n_{k}+1}=t_{k+1}$.
On the other hand, once we have found such an $n_{k}$, the sequence $\left\{a_{k}\right\}$ must be decreasing by problem 4. Since the $a_{k}$ are all integers, we eventually have to get to 0 (as there is no infinite decreasing sequence of positive integers). Therefore, after some finite number of steps the algorithm terminates with $a_{k+1}=0$, so $0=\frac{a_{k}}{b_{k}}=\frac{a}{b}-t_{1}-\ldots-t_{k}$, so $\frac{a}{b}=t_{1}+\ldots+t_{k}$, which is what we wanted.
## Juicy Numbers [100]
A juicy number is an integer $j>1$ for which there is a sequence $a_{1}<a_{2}<\ldots<a_{k}$ of positive integers such that $a_{k}=j$ and such that the sum of the reciprocals of all the $a_{i}$ is 1 . For example, 6 is a juicy number because $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$, but 2 is not juicy.
In this part, you will investigate some of the properties of juicy numbers. Remember that if you do not solve a question, you can still use its result on later questions.
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n6. ",
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2008"
}
|
Explain why 4 is not a juicy number.
|
If 4 were juicy, then we would have $1=\ldots+\frac{1}{4}$. The $\ldots$ can only possible contain $\frac{1}{2}$ and $\frac{1}{3}$, but it is clear that $\frac{1}{2}+\frac{1}{4}, \frac{1}{3}+\frac{1}{4}$, and $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$ are all not equal to 1 .
|
proof
|
Incomplete
|
Yes
|
math-word-problem
|
Number Theory
|
Explain why 4 is not a juicy number.
|
If 4 were juicy, then we would have $1=\ldots+\frac{1}{4}$. The $\ldots$ can only possible contain $\frac{1}{2}$ and $\frac{1}{3}$, but it is clear that $\frac{1}{2}+\frac{1}{4}, \frac{1}{3}+\frac{1}{4}$, and $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$ are all not equal to 1 .
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. ",
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2008"
}
|
It turns out that 6 is the smallest juicy integer. Find the next two smallest juicy numbers, and show a decomposition of 1 into unit fractions for each of these numbers. You do not need to prove that no smaller numbers are juicy.
|
12 and $151=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}, 1=\frac{1}{2}+\frac{1}{3}+\frac{1}{10}+\frac{1}{15}$.
|
12 \text{ and } 15
|
Yes
|
Incomplete
|
math-word-problem
|
Number Theory
|
It turns out that 6 is the smallest juicy integer. Find the next two smallest juicy numbers, and show a decomposition of 1 into unit fractions for each of these numbers. You do not need to prove that no smaller numbers are juicy.
|
12 and $151=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}, 1=\frac{1}{2}+\frac{1}{3}+\frac{1}{10}+\frac{1}{15}$.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. ",
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2008"
}
|
Let $p$ be a prime. Given a sequence of positive integers $b_{1}$ through $b_{n}$, exactly one of which is divisible by $p$, show that when
$$
\frac{1}{b_{1}}+\frac{1}{b_{2}}+\ldots+\frac{1}{b_{n}}
$$
is written as a fraction in lowest terms, then its denominator is divisible by $p$. Use this fact to explain why no prime $p$ is ever juicy.
|
We can assume that $b_{n}$ is the term divisible by $p$ (i.e. $b_{n}=k p$ ) since the order of addition doesn't matter. We can then write
$$
\frac{1}{b_{1}}+\frac{1}{b_{2}}+\ldots+\frac{1}{b_{n-1}}=\frac{a}{b}
$$
where $b$ is not divisible by $p$ (since none of the $b_{i}$ are). But then $\frac{a}{b}+\frac{1}{k p}=\frac{k p a+b}{k p b}$. Since $b$ is not divisible by $p, k p a+b$ is not divisible by $p$, so we cannot remove the factor of $p$ from the denominator. In particular, $p$ cannot be juicy as 1 can be written as $\frac{1}{1}$, which has a denominator not divisible by $p$, whereas being juicy means we have a sum $\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n}}=1$, where $b_{1}<b_{2}<\ldots<b_{n}=p$, and so in particular none of the $b_{i}$ with $i<n$ are divisible by p.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $p$ be a prime. Given a sequence of positive integers $b_{1}$ through $b_{n}$, exactly one of which is divisible by $p$, show that when
$$
\frac{1}{b_{1}}+\frac{1}{b_{2}}+\ldots+\frac{1}{b_{n}}
$$
is written as a fraction in lowest terms, then its denominator is divisible by $p$. Use this fact to explain why no prime $p$ is ever juicy.
|
We can assume that $b_{n}$ is the term divisible by $p$ (i.e. $b_{n}=k p$ ) since the order of addition doesn't matter. We can then write
$$
\frac{1}{b_{1}}+\frac{1}{b_{2}}+\ldots+\frac{1}{b_{n-1}}=\frac{a}{b}
$$
where $b$ is not divisible by $p$ (since none of the $b_{i}$ are). But then $\frac{a}{b}+\frac{1}{k p}=\frac{k p a+b}{k p b}$. Since $b$ is not divisible by $p, k p a+b$ is not divisible by $p$, so we cannot remove the factor of $p$ from the denominator. In particular, $p$ cannot be juicy as 1 can be written as $\frac{1}{1}$, which has a denominator not divisible by $p$, whereas being juicy means we have a sum $\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n}}=1$, where $b_{1}<b_{2}<\ldots<b_{n}=p$, and so in particular none of the $b_{i}$ with $i<n$ are divisible by p.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. ",
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2008"
}
|
Show that if $j$ is a juicy integer, then $2 j$ is juicy as well.
|
Just replace $\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n}}$ with $\frac{1}{2}+\frac{1}{2 b_{1}}+\frac{1}{2 b_{2}}+\ldots+\frac{1}{2 b_{n}}$. Since $n>1,2 b_{1}>2$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Show that if $j$ is a juicy integer, then $2 j$ is juicy as well.
|
Just replace $\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n}}$ with $\frac{1}{2}+\frac{1}{2 b_{1}}+\frac{1}{2 b_{2}}+\ldots+\frac{1}{2 b_{n}}$. Since $n>1,2 b_{1}>2$.
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. ",
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2008"
}
|
Prove that the product of two juicy numbers (not necessarily distinct) is always a juicy number. Hint: if $j_{1}$ and $j_{2}$ are the two numbers, how can you change the decompositions of 1 ending in $\frac{1}{j_{1}}$ or $\frac{1}{j_{2}}$ to make them end in $\frac{1}{j_{1} j_{2}}$ ?
|
Let $1=\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n}}=\frac{1}{c_{1}}+\ldots+\frac{1}{c_{m}}$, where $b_{n}=j_{1}$ and $c_{m}=j_{2}$. Then
$$
1=\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n-1}}+\left(\frac{1}{b_{n} c_{1}}+\frac{1}{b_{n} c_{2}}+\ldots+\frac{1}{b_{n} c_{m}}\right)
$$
and so $j_{1} j_{2}$ is juicy.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Prove that the product of two juicy numbers (not necessarily distinct) is always a juicy number. Hint: if $j_{1}$ and $j_{2}$ are the two numbers, how can you change the decompositions of 1 ending in $\frac{1}{j_{1}}$ or $\frac{1}{j_{2}}$ to make them end in $\frac{1}{j_{1} j_{2}}$ ?
|
Let $1=\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n}}=\frac{1}{c_{1}}+\ldots+\frac{1}{c_{m}}$, where $b_{n}=j_{1}$ and $c_{m}=j_{2}$. Then
$$
1=\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n-1}}+\left(\frac{1}{b_{n} c_{1}}+\frac{1}{b_{n} c_{2}}+\ldots+\frac{1}{b_{n} c_{m}}\right)
$$
and so $j_{1} j_{2}$ is juicy.
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n5. ",
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2008"
}
|
If $a$ and $b$ are positive integers such that $a^{2}-b^{4}=2009$, find $a+b$.
|
47
|
47
|
Yes
|
Problem not solved
|
math-word-problem
|
Number Theory
|
If $a$ and $b$ are positive integers such that $a^{2}-b^{4}=2009$, find $a+b$.
|
47
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
If $a$ and $b$ are positive integers such that $a^{2}-b^{4}=2009$, find $a+b$.
|
We can factor the equation as $\left(a-b^{2}\right)\left(a+b^{2}\right)=41 \cdot 49$, from which it is evident that $a=45$ and $b=2$ is a possible solution. By examining the factors of 2009 , one can see that there are no other solutions.
|
47
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
If $a$ and $b$ are positive integers such that $a^{2}-b^{4}=2009$, find $a+b$.
|
We can factor the equation as $\left(a-b^{2}\right)\left(a+b^{2}\right)=41 \cdot 49$, from which it is evident that $a=45$ and $b=2$ is a possible solution. By examining the factors of 2009 , one can see that there are no other solutions.
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $S$ be the sum of all the real coefficients of the expansion of $(1+i x)^{2009}$. What is $\log _{2}(S)$ ?
|
1004
|
1004
|
Yes
|
Incomplete
|
math-word-problem
|
Algebra
|
Let $S$ be the sum of all the real coefficients of the expansion of $(1+i x)^{2009}$. What is $\log _{2}(S)$ ?
|
1004
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n2. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Let $S$ be the sum of all the real coefficients of the expansion of $(1+i x)^{2009}$. What is $\log _{2}(S)$ ?
|
The sum of all the coefficients is $(1+i)^{2009}$, and the sum of the real coefficients is the real part of this, which is $\frac{1}{2}\left((1+i)^{2009}+(1-i)^{2009}\right)=2^{1004}$. Thus $\log _{2}(S)=1004$.
|
1004
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $S$ be the sum of all the real coefficients of the expansion of $(1+i x)^{2009}$. What is $\log _{2}(S)$ ?
|
The sum of all the coefficients is $(1+i)^{2009}$, and the sum of the real coefficients is the real part of this, which is $\frac{1}{2}\left((1+i)^{2009}+(1-i)^{2009}\right)=2^{1004}$. Thus $\log _{2}(S)=1004$.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n2. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
If $\tan x+\tan y=4$ and $\cot x+\cot y=5$, compute $\tan (x+y)$.
|
20
|
20
|
Yes
|
Problem not solved
|
math-word-problem
|
Algebra
|
If $\tan x+\tan y=4$ and $\cot x+\cot y=5$, compute $\tan (x+y)$.
|
20
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n3. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
If $\tan x+\tan y=4$ and $\cot x+\cot y=5$, compute $\tan (x+y)$.
|
We have $\cot x+\cot y=\frac{\tan x+\tan y}{\tan x \tan y}$, so $\tan x \tan y=\frac{4}{5}$. Thus, by the tan sum formula, $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=20$.
|
20
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
If $\tan x+\tan y=4$ and $\cot x+\cot y=5$, compute $\tan (x+y)$.
|
We have $\cot x+\cot y=\frac{\tan x+\tan y}{\tan x \tan y}$, so $\tan x \tan y=\frac{4}{5}$. Thus, by the tan sum formula, $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=20$.
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n3. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Suppose $a, b$ and $c$ are integers such that the greatest common divisor of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x+1$ (in the ring of polynomials in $x$ with integer coefficients), and the least common multiple of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x^{3}-4 x^{2}+x+6$. Find $a+b+c$.
|
$\quad-6$
|
-6
|
Yes
|
Incomplete
|
math-word-problem
|
Algebra
|
Suppose $a, b$ and $c$ are integers such that the greatest common divisor of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x+1$ (in the ring of polynomials in $x$ with integer coefficients), and the least common multiple of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x^{3}-4 x^{2}+x+6$. Find $a+b+c$.
|
$\quad-6$
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Suppose $a, b$ and $c$ are integers such that the greatest common divisor of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x+1$ (in the ring of polynomials in $x$ with integer coefficients), and the least common multiple of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x^{3}-4 x^{2}+x+6$. Find $a+b+c$.
|
Since $x+1$ divides $x^{2}+a x+b$ and the constant term is $b$, we have $x^{2}+a x+b=(x+1)(x+b)$, and similarly $x^{2}+b x+c=(x+1)(x+c)$. Therefore, $a=b+1=c+2$. Furthermore, the least common multiple of the two polynomials is $(x+1)(x+b)(x+b-1)=x^{3}-4 x^{2}+x+6$, so $b=-2$. Thus $a=-1$ and $c=-3$, and $a+b+c=-6$.
|
-6
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose $a, b$ and $c$ are integers such that the greatest common divisor of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x+1$ (in the ring of polynomials in $x$ with integer coefficients), and the least common multiple of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x^{3}-4 x^{2}+x+6$. Find $a+b+c$.
|
Since $x+1$ divides $x^{2}+a x+b$ and the constant term is $b$, we have $x^{2}+a x+b=(x+1)(x+b)$, and similarly $x^{2}+b x+c=(x+1)(x+c)$. Therefore, $a=b+1=c+2$. Furthermore, the least common multiple of the two polynomials is $(x+1)(x+b)(x+b-1)=x^{3}-4 x^{2}+x+6$, so $b=-2$. Thus $a=-1$ and $c=-3$, and $a+b+c=-6$.
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $a, b$, and $c$ be the 3 roots of $x^{3}-x+1=0$. Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$.
|
$\quad-2$
|
-2
|
Yes
|
Incomplete
|
math-word-problem
|
Algebra
|
Let $a, b$, and $c$ be the 3 roots of $x^{3}-x+1=0$. Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$.
|
$\quad-2$
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n5. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Let $a, b$, and $c$ be the 3 roots of $x^{3}-x+1=0$. Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$.
|
We can substitute $x=y-1$ to obtain a polynomial having roots $a+1, b+1, c+1$, namely, $(y-1)^{3}-(y-1)+1=y^{3}-3 y^{2}+2 y+1$. The sum of the reciprocals of the roots of this polynomial is, by Viete's formulas, $\frac{2}{-1}=-2$.
|
-2
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a, b$, and $c$ be the 3 roots of $x^{3}-x+1=0$. Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$.
|
We can substitute $x=y-1$ to obtain a polynomial having roots $a+1, b+1, c+1$, namely, $(y-1)^{3}-(y-1)+1=y^{3}-3 y^{2}+2 y+1$. The sum of the reciprocals of the roots of this polynomial is, by Viete's formulas, $\frac{2}{-1}=-2$.
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n5. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $x$ and $y$ be positive real numbers and $\theta$ an angle such that $\theta \neq \frac{\pi}{2} n$ for any integer $n$. Suppose
$$
\frac{\sin \theta}{x}=\frac{\cos \theta}{y}
$$
and
$$
\frac{\cos ^{4} \theta}{x^{4}}+\frac{\sin ^{4} \theta}{y^{4}}=\frac{97 \sin 2 \theta}{x^{3} y+y^{3} x}
$$
Compute $\frac{x}{y}+\frac{y}{x}$.
|
4
|
4
|
Yes
|
Incomplete
|
math-word-problem
|
Algebra
|
Let $x$ and $y$ be positive real numbers and $\theta$ an angle such that $\theta \neq \frac{\pi}{2} n$ for any integer $n$. Suppose
$$
\frac{\sin \theta}{x}=\frac{\cos \theta}{y}
$$
and
$$
\frac{\cos ^{4} \theta}{x^{4}}+\frac{\sin ^{4} \theta}{y^{4}}=\frac{97 \sin 2 \theta}{x^{3} y+y^{3} x}
$$
Compute $\frac{x}{y}+\frac{y}{x}$.
|
4
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n6. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Let $x$ and $y$ be positive real numbers and $\theta$ an angle such that $\theta \neq \frac{\pi}{2} n$ for any integer $n$. Suppose
$$
\frac{\sin \theta}{x}=\frac{\cos \theta}{y}
$$
and
$$
\frac{\cos ^{4} \theta}{x^{4}}+\frac{\sin ^{4} \theta}{y^{4}}=\frac{97 \sin 2 \theta}{x^{3} y+y^{3} x}
$$
Compute $\frac{x}{y}+\frac{y}{x}$.
|
From the first relation, there exists a real number $k$ such that $x=k \sin \theta$ and $y=k \cos \theta$. Then we have
$$
\frac{\cos ^{4} \theta}{\sin ^{4} \theta}+\frac{\sin ^{4} \theta}{\cos ^{4} \theta}=\frac{194 \sin \theta \cos \theta}{\sin \theta \cos \theta\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}=194
$$
Notice that if $t=\frac{x}{y}+\frac{y}{x}$ then $\left(t^{2}-2\right)^{2}-2=\frac{\cos ^{4} \theta}{\sin ^{4} \theta}+\frac{\sin ^{4} \theta}{\cos ^{4} \theta}=194$ and so $t=4$.
|
4
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $x$ and $y$ be positive real numbers and $\theta$ an angle such that $\theta \neq \frac{\pi}{2} n$ for any integer $n$. Suppose
$$
\frac{\sin \theta}{x}=\frac{\cos \theta}{y}
$$
and
$$
\frac{\cos ^{4} \theta}{x^{4}}+\frac{\sin ^{4} \theta}{y^{4}}=\frac{97 \sin 2 \theta}{x^{3} y+y^{3} x}
$$
Compute $\frac{x}{y}+\frac{y}{x}$.
|
From the first relation, there exists a real number $k$ such that $x=k \sin \theta$ and $y=k \cos \theta$. Then we have
$$
\frac{\cos ^{4} \theta}{\sin ^{4} \theta}+\frac{\sin ^{4} \theta}{\cos ^{4} \theta}=\frac{194 \sin \theta \cos \theta}{\sin \theta \cos \theta\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}=194
$$
Notice that if $t=\frac{x}{y}+\frac{y}{x}$ then $\left(t^{2}-2\right)^{2}-2=\frac{\cos ^{4} \theta}{\sin ^{4} \theta}+\frac{\sin ^{4} \theta}{\cos ^{4} \theta}=194$ and so $t=4$.
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n6. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Simplify the product
$$
\prod_{m=1}^{100} \prod_{n=1}^{100} \frac{x^{n+m}+x^{n+m+2}+x^{2 n+1}+x^{2 m+1}}{x^{2 n}+2 x^{n+m}+x^{2 m}}
$$
Express your answer in terms of $x$.
|
$x^{9900}\left(\frac{1+x^{100}}{2}\right)^{2}\left(\right.$ OR $\left.\frac{1}{4} x^{9900}+\frac{1}{2} x^{10000}+\frac{1}{4} x^{10100}\right)$
|
x^{9900}\left(\frac{1+x^{100}}{2}\right)^{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Simplify the product
$$
\prod_{m=1}^{100} \prod_{n=1}^{100} \frac{x^{n+m}+x^{n+m+2}+x^{2 n+1}+x^{2 m+1}}{x^{2 n}+2 x^{n+m}+x^{2 m}}
$$
Express your answer in terms of $x$.
|
$x^{9900}\left(\frac{1+x^{100}}{2}\right)^{2}\left(\right.$ OR $\left.\frac{1}{4} x^{9900}+\frac{1}{2} x^{10000}+\frac{1}{4} x^{10100}\right)$
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Simplify the product
$$
\prod_{m=1}^{100} \prod_{n=1}^{100} \frac{x^{n+m}+x^{n+m+2}+x^{2 n+1}+x^{2 m+1}}{x^{2 n}+2 x^{n+m}+x^{2 m}}
$$
Express your answer in terms of $x$.
|
We notice that the numerator and denominator of each term factors, so the product is equal to
$$
\prod_{m=1}^{100} \prod_{n=1}^{100} \frac{\left(x^{m}+x^{n+1}\right)\left(x^{m+1}+x^{n}\right)}{\left(x^{m}+x^{n}\right)^{2}}
$$
Each term of the numerator cancels with a term of the denominator except for those of the form $\left(x^{m}+x^{101}\right)$ and $\left(x^{101}+x^{n}\right)$ for $m, n=1, \ldots, 100$, and the terms in the denominator which remain are of the form $\left(x^{1}+x^{n}\right)$ and $\left(x^{1}+x^{m}\right)$ for $m, n=1, \ldots, 100$. Thus the product simplifies to
$$
\left(\prod_{m=1}^{100} \frac{x^{m}+x^{101}}{x^{1}+x^{m}}\right)^{2}
$$
Reversing the order of the factors of the numerator, we find this is equal to
$$
\begin{aligned}
\left(\prod_{m=1}^{100} \frac{x^{101-m}+x^{101}}{x^{1}+x^{m}}\right)^{2} & =\left(\prod_{m=1}^{100} x^{100-m} \frac{x^{1}+x^{m+1}}{x^{1}+x^{m}}\right)^{2} \\
& =\left(\frac{x^{1}+x^{1} 01}{x^{1}+x^{1}} \prod_{m=1}^{100} x^{100-m}\right)^{2} \\
& =\left(x^{\frac{99 \cdot 100}{2}}\right)^{2}\left(\frac{1+x^{100}}{2}\right)^{2}
\end{aligned}
$$
as desired.
|
x^{9900} \left(\frac{1+x^{100}}{2}\right)^{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Simplify the product
$$
\prod_{m=1}^{100} \prod_{n=1}^{100} \frac{x^{n+m}+x^{n+m+2}+x^{2 n+1}+x^{2 m+1}}{x^{2 n}+2 x^{n+m}+x^{2 m}}
$$
Express your answer in terms of $x$.
|
We notice that the numerator and denominator of each term factors, so the product is equal to
$$
\prod_{m=1}^{100} \prod_{n=1}^{100} \frac{\left(x^{m}+x^{n+1}\right)\left(x^{m+1}+x^{n}\right)}{\left(x^{m}+x^{n}\right)^{2}}
$$
Each term of the numerator cancels with a term of the denominator except for those of the form $\left(x^{m}+x^{101}\right)$ and $\left(x^{101}+x^{n}\right)$ for $m, n=1, \ldots, 100$, and the terms in the denominator which remain are of the form $\left(x^{1}+x^{n}\right)$ and $\left(x^{1}+x^{m}\right)$ for $m, n=1, \ldots, 100$. Thus the product simplifies to
$$
\left(\prod_{m=1}^{100} \frac{x^{m}+x^{101}}{x^{1}+x^{m}}\right)^{2}
$$
Reversing the order of the factors of the numerator, we find this is equal to
$$
\begin{aligned}
\left(\prod_{m=1}^{100} \frac{x^{101-m}+x^{101}}{x^{1}+x^{m}}\right)^{2} & =\left(\prod_{m=1}^{100} x^{100-m} \frac{x^{1}+x^{m+1}}{x^{1}+x^{m}}\right)^{2} \\
& =\left(\frac{x^{1}+x^{1} 01}{x^{1}+x^{1}} \prod_{m=1}^{100} x^{100-m}\right)^{2} \\
& =\left(x^{\frac{99 \cdot 100}{2}}\right)^{2}\left(\frac{1+x^{100}}{2}\right)^{2}
\end{aligned}
$$
as desired.
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
If $a, b, x$, and $y$ are real numbers such that $a x+b y=3, a x^{2}+b y^{2}=7, a x^{3}+b y^{3}=16$, and $a x^{4}+b y^{4}=42$, find $a x^{5}+b y^{5}$.
|
20.
|
20
|
Yes
|
Problem not solved
|
math-word-problem
|
Algebra
|
If $a, b, x$, and $y$ are real numbers such that $a x+b y=3, a x^{2}+b y^{2}=7, a x^{3}+b y^{3}=16$, and $a x^{4}+b y^{4}=42$, find $a x^{5}+b y^{5}$.
|
20.
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [7]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
If $a, b, x$, and $y$ are real numbers such that $a x+b y=3, a x^{2}+b y^{2}=7, a x^{3}+b y^{3}=16$, and $a x^{4}+b y^{4}=42$, find $a x^{5}+b y^{5}$.
|
We have $a x^{3}+b y^{3}=16$, so $\left(a x^{3}+b y^{3}\right)(x+y)=16(x+y)$ and thus
$$
a x^{4}+b y^{4}+x y\left(a x^{2}+b y^{2}\right)=16(x+y)
$$
It follows that
$$
42+7 x y=16(x+y)
$$
From $a x^{2}+b y^{2}=7$, we have $\left(a x^{2}+b y^{2}\right)(x+y)=7(x+y)$ so $a x^{3}+b y^{3}+x y\left(a x^{2}+b y^{2}\right)=7(x+y)$. This simplifies to
$$
16+3 x y=7(x+y)
$$
We can now solve for $x+y$ and $x y$ from (1) and (2) to find $x+y=-14$ and $x y=-38$. Thus we have $\left(a x^{4}+b y^{4}\right)(x+y)=42(x+y)$, and so $a x^{5}+b y^{5}+x y\left(a x^{3}+b y^{3}\right)=42(x+y)$. Finally, it follows that $a x^{5}+b y^{5}=42(x+y)-16 x y=20$ as desired.
|
20
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
If $a, b, x$, and $y$ are real numbers such that $a x+b y=3, a x^{2}+b y^{2}=7, a x^{3}+b y^{3}=16$, and $a x^{4}+b y^{4}=42$, find $a x^{5}+b y^{5}$.
|
We have $a x^{3}+b y^{3}=16$, so $\left(a x^{3}+b y^{3}\right)(x+y)=16(x+y)$ and thus
$$
a x^{4}+b y^{4}+x y\left(a x^{2}+b y^{2}\right)=16(x+y)
$$
It follows that
$$
42+7 x y=16(x+y)
$$
From $a x^{2}+b y^{2}=7$, we have $\left(a x^{2}+b y^{2}\right)(x+y)=7(x+y)$ so $a x^{3}+b y^{3}+x y\left(a x^{2}+b y^{2}\right)=7(x+y)$. This simplifies to
$$
16+3 x y=7(x+y)
$$
We can now solve for $x+y$ and $x y$ from (1) and (2) to find $x+y=-14$ and $x y=-38$. Thus we have $\left(a x^{4}+b y^{4}\right)(x+y)=42(x+y)$, and so $a x^{5}+b y^{5}+x y\left(a x^{3}+b y^{3}\right)=42(x+y)$. Finally, it follows that $a x^{5}+b y^{5}=42(x+y)-16 x y=20$ as desired.
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [7]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $f(x)=x^{4}+14 x^{3}+52 x^{2}+56 x+16$. Let $z_{1}, z_{2}, z_{3}, z_{4}$ be the four roots of $f$. Find the smallest possible value of $\left|z_{a} z_{b}+z_{c} z_{d}\right|$ where $\{a, b, c, d\}=\{1,2,3,4\}$.
|
8
|
8
|
Yes
|
Incomplete
|
math-word-problem
|
Algebra
|
Let $f(x)=x^{4}+14 x^{3}+52 x^{2}+56 x+16$. Let $z_{1}, z_{2}, z_{3}, z_{4}$ be the four roots of $f$. Find the smallest possible value of $\left|z_{a} z_{b}+z_{c} z_{d}\right|$ where $\{a, b, c, d\}=\{1,2,3,4\}$.
|
8
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [7]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Let $f(x)=x^{4}+14 x^{3}+52 x^{2}+56 x+16$. Let $z_{1}, z_{2}, z_{3}, z_{4}$ be the four roots of $f$. Find the smallest possible value of $\left|z_{a} z_{b}+z_{c} z_{d}\right|$ where $\{a, b, c, d\}=\{1,2,3,4\}$.
|
Note that $\frac{1}{16} f(2 x)=x^{4}+7 x^{3}+13 x^{2}+7 x+1$. Because the coefficients of this polynomial are symmetric, if $r$ is a root of $f(x)$ then $\frac{4}{r}$ is as well. Further, $f(-1)=-1$ and $f(-2)=16$ so $f(x)$ has two distinct roots on $(-2,0)$ and two more roots on $(-\infty,-2)$. Now, if $\sigma$ is a permutation of $\{1,2,3,4\}$ :
$\left|z_{\sigma(1)} z_{\sigma(2)}+z_{\sigma(3)} z_{\sigma(4)}\right| \leq \frac{1}{2}\left(z_{\sigma(1)} z_{\sigma(2)}+z_{\sigma(3)} z_{\sigma(4)}+z_{\sigma(4)} z_{\sigma(3)}+z_{\sigma(2)} z_{\sigma(1)}\right)$
Let the roots be ordered $z_{1} \leq z_{2} \leq z_{3} \leq z_{4}$, then by rearrangement the last expression is at least:
$\frac{1}{2}\left(z_{1} z_{4}+z_{2} z_{3}+z_{3} z_{2}+z_{4} z_{1}\right)$
Since the roots come in pairs $z_{1} z_{4}=z_{2} z_{3}=4$, our expression is minimized when $\sigma(1)=1, \sigma(2)=$ $4, \sigma(3)=3, \sigma(4)=2$ and its minimum value is 8 .
|
8
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $f(x)=x^{4}+14 x^{3}+52 x^{2}+56 x+16$. Let $z_{1}, z_{2}, z_{3}, z_{4}$ be the four roots of $f$. Find the smallest possible value of $\left|z_{a} z_{b}+z_{c} z_{d}\right|$ where $\{a, b, c, d\}=\{1,2,3,4\}$.
|
Note that $\frac{1}{16} f(2 x)=x^{4}+7 x^{3}+13 x^{2}+7 x+1$. Because the coefficients of this polynomial are symmetric, if $r$ is a root of $f(x)$ then $\frac{4}{r}$ is as well. Further, $f(-1)=-1$ and $f(-2)=16$ so $f(x)$ has two distinct roots on $(-2,0)$ and two more roots on $(-\infty,-2)$. Now, if $\sigma$ is a permutation of $\{1,2,3,4\}$ :
$\left|z_{\sigma(1)} z_{\sigma(2)}+z_{\sigma(3)} z_{\sigma(4)}\right| \leq \frac{1}{2}\left(z_{\sigma(1)} z_{\sigma(2)}+z_{\sigma(3)} z_{\sigma(4)}+z_{\sigma(4)} z_{\sigma(3)}+z_{\sigma(2)} z_{\sigma(1)}\right)$
Let the roots be ordered $z_{1} \leq z_{2} \leq z_{3} \leq z_{4}$, then by rearrangement the last expression is at least:
$\frac{1}{2}\left(z_{1} z_{4}+z_{2} z_{3}+z_{3} z_{2}+z_{4} z_{1}\right)$
Since the roots come in pairs $z_{1} z_{4}=z_{2} z_{3}=4$, our expression is minimized when $\sigma(1)=1, \sigma(2)=$ $4, \sigma(3)=3, \sigma(4)=2$ and its minimum value is 8 .
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [7]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $f(x)=2 x^{3}-2 x$. For what positive values of $a$ do there exist distinct $b, c, d$ such that $(a, f(a))$, $(b, f(b)),(c, f(c)),(d, f(d))$ is a rectangle?
|
$\left[\frac{\sqrt{3}}{3}, 1\right]$
|
\left[\frac{\sqrt{3}}{3}, 1\right]
|
Yes
|
Incomplete
|
math-word-problem
|
Algebra
|
Let $f(x)=2 x^{3}-2 x$. For what positive values of $a$ do there exist distinct $b, c, d$ such that $(a, f(a))$, $(b, f(b)),(c, f(c)),(d, f(d))$ is a rectangle?
|
$\left[\frac{\sqrt{3}}{3}, 1\right]$
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Let $f(x)=2 x^{3}-2 x$. For what positive values of $a$ do there exist distinct $b, c, d$ such that $(a, f(a))$, $(b, f(b)),(c, f(c)),(d, f(d))$ is a rectangle?
|
Say we have four points $(a, f(a)),(b, f(b)),(c, f(c)),(d, f(d))$ on the curve which form a rectangle. If we interpolate a cubic through these points, that cubic will be symmetric around the center of the rectangle. But the unique cubic through the four points is $f(x)$, and $f(x)$ has only one point of symmetry, the point $(0,0)$.
So every rectangle with all four points on $f(x)$ is of the form $(a, f(a)),(b, f(b)),(-a, f(-a)),(-b, f(-b))$, and without loss of generality we let $a, b>0$. Then for any choice of $a$ and $b$ these points form a parallelogram, which is a rectangle if and only if the distance from $(a, f(a))$ to $(0,0)$ is equal to the distance from $(b, f(b))$ to $(0,0)$. Let $g(x)=x^{2}+(f(x))^{2}=4 x^{6}-8 x^{4}+5 x^{2}$, and consider $g(x)$ restricted to $x \geq 0$. We are looking for all the values of $a$ such that $g(x)=g(a)$ has solutions other than $a$.
Note that $g(x)=h\left(x^{2}\right)$ where $h(x)=4 x^{3}-8 x^{2}+5 x$. This polynomial $h(x)$ has a relative maximum of 1 at $x=\frac{1}{2}$ and a relative minimum of $25 / 27$ at $x=\frac{5}{6}$. Thus the polynomial $h(x)-h(1 / 2)$ has the double root $1 / 2$ and factors as $\left(4 x^{2}-4 x+1\right)(x-1)$, the largest possible value of $a^{2}$ for which $h\left(x^{2}\right)=h\left(a^{2}\right)$ is $a^{2}=1$, or $a=1$. The smallest such value is that which evaluates to $25 / 27$ other than $5 / 6$, which is similarly found to be $a^{2}=1 / 3$, or $a=\frac{\sqrt{3}}{3}$. Thus, for $a$ in the range $\frac{\sqrt{3}}{3} \leq a \leq 1$ the equation $g(x)=g(a)$ has nontrivial solutions and hence an inscribed rectangle exists.
|
\frac{\sqrt{3}}{3} \leq a \leq 1
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $f(x)=2 x^{3}-2 x$. For what positive values of $a$ do there exist distinct $b, c, d$ such that $(a, f(a))$, $(b, f(b)),(c, f(c)),(d, f(d))$ is a rectangle?
|
Say we have four points $(a, f(a)),(b, f(b)),(c, f(c)),(d, f(d))$ on the curve which form a rectangle. If we interpolate a cubic through these points, that cubic will be symmetric around the center of the rectangle. But the unique cubic through the four points is $f(x)$, and $f(x)$ has only one point of symmetry, the point $(0,0)$.
So every rectangle with all four points on $f(x)$ is of the form $(a, f(a)),(b, f(b)),(-a, f(-a)),(-b, f(-b))$, and without loss of generality we let $a, b>0$. Then for any choice of $a$ and $b$ these points form a parallelogram, which is a rectangle if and only if the distance from $(a, f(a))$ to $(0,0)$ is equal to the distance from $(b, f(b))$ to $(0,0)$. Let $g(x)=x^{2}+(f(x))^{2}=4 x^{6}-8 x^{4}+5 x^{2}$, and consider $g(x)$ restricted to $x \geq 0$. We are looking for all the values of $a$ such that $g(x)=g(a)$ has solutions other than $a$.
Note that $g(x)=h\left(x^{2}\right)$ where $h(x)=4 x^{3}-8 x^{2}+5 x$. This polynomial $h(x)$ has a relative maximum of 1 at $x=\frac{1}{2}$ and a relative minimum of $25 / 27$ at $x=\frac{5}{6}$. Thus the polynomial $h(x)-h(1 / 2)$ has the double root $1 / 2$ and factors as $\left(4 x^{2}-4 x+1\right)(x-1)$, the largest possible value of $a^{2}$ for which $h\left(x^{2}\right)=h\left(a^{2}\right)$ is $a^{2}=1$, or $a=1$. The smallest such value is that which evaluates to $25 / 27$ other than $5 / 6$, which is similarly found to be $a^{2}=1 / 3$, or $a=\frac{\sqrt{3}}{3}$. Thus, for $a$ in the range $\frac{\sqrt{3}}{3} \leq a \leq 1$ the equation $g(x)=g(a)$ has nontrivial solutions and hence an inscribed rectangle exists.
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $f$ be a differentiable real-valued function defined on the positive real numbers. The tangent lines to the graph of $f$ always meet the $y$-axis 1 unit lower than where they meet the function. If $f(1)=0$, what is $f(2)$ ?
|
$\ln 2$
|
\ln 2
|
Yes
|
Incomplete
|
math-word-problem
|
Calculus
|
Let $f$ be a differentiable real-valued function defined on the positive real numbers. The tangent lines to the graph of $f$ always meet the $y$-axis 1 unit lower than where they meet the function. If $f(1)=0$, what is $f(2)$ ?
|
$\ln 2$
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Let $f$ be a differentiable real-valued function defined on the positive real numbers. The tangent lines to the graph of $f$ always meet the $y$-axis 1 unit lower than where they meet the function. If $f(1)=0$, what is $f(2)$ ?
|
The tangent line to $f$ at $x$ meets the $y$-axis at $f(x)-1$ for any $x$, so the slope of the tangent line is $f^{\prime}(x)=\frac{1}{x}$, and so $f(x)=\ln (x)+C$ for some $a$. Since $f(1)=0$, we have $C=0$, and so $f(x)=\ln (x)$. Thus $f(2)=\ln (2)$.
|
\ln (2)
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Let $f$ be a differentiable real-valued function defined on the positive real numbers. The tangent lines to the graph of $f$ always meet the $y$-axis 1 unit lower than where they meet the function. If $f(1)=0$, what is $f(2)$ ?
|
The tangent line to $f$ at $x$ meets the $y$-axis at $f(x)-1$ for any $x$, so the slope of the tangent line is $f^{\prime}(x)=\frac{1}{x}$, and so $f(x)=\ln (x)+C$ for some $a$. Since $f(1)=0$, we have $C=0$, and so $f(x)=\ln (x)$. Thus $f(2)=\ln (2)$.
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
The differentiable function $F: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $F(0)=-1$ and
$$
\frac{d}{d x} F(x)=\sin (\sin (\sin (\sin (x)))) \cdot \cos (\sin (\sin (x))) \cdot \cos (\sin (x)) \cdot \cos (x)
$$
Find $F(x)$ as a function of $x$.
|
$-\cos (\sin (\sin (\sin (x))))$
|
-\cos (\sin (\sin (\sin (x))))
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
The differentiable function $F: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $F(0)=-1$ and
$$
\frac{d}{d x} F(x)=\sin (\sin (\sin (\sin (x)))) \cdot \cos (\sin (\sin (x))) \cdot \cos (\sin (x)) \cdot \cos (x)
$$
Find $F(x)$ as a function of $x$.
|
$-\cos (\sin (\sin (\sin (x))))$
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
The differentiable function $F: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $F(0)=-1$ and
$$
\frac{d}{d x} F(x)=\sin (\sin (\sin (\sin (x)))) \cdot \cos (\sin (\sin (x))) \cdot \cos (\sin (x)) \cdot \cos (x)
$$
Find $F(x)$ as a function of $x$.
|
$\quad$ Substituting $u=\sin (\sin (\sin (x)))$, we find
$$
F(x)=\int \sin (u) d u=-\cos (u)+C
$$
for some $C$. Since $F(0)=1$ we find $C=0$.
|
F(x) = -\cos(\sin(\sin(\sin(x))))
|
Yes
|
Incomplete
|
math-word-problem
|
Calculus
|
The differentiable function $F: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $F(0)=-1$ and
$$
\frac{d}{d x} F(x)=\sin (\sin (\sin (\sin (x)))) \cdot \cos (\sin (\sin (x))) \cdot \cos (\sin (x)) \cdot \cos (x)
$$
Find $F(x)$ as a function of $x$.
|
$\quad$ Substituting $u=\sin (\sin (\sin (x)))$, we find
$$
F(x)=\int \sin (u) d u=-\cos (u)+C
$$
for some $C$. Since $F(0)=1$ we find $C=0$.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Compute $e^{A}$ where $A$ is defined as
$$
\int_{3 / 4}^{4 / 3} \frac{2 x^{2}+x+1}{x^{3}+x^{2}+x+1} d x
$$
|
$\frac{16}{9}$
|
\frac{16}{9}
|
Yes
|
Problem not solved
|
math-word-problem
|
Calculus
|
Compute $e^{A}$ where $A$ is defined as
$$
\int_{3 / 4}^{4 / 3} \frac{2 x^{2}+x+1}{x^{3}+x^{2}+x+1} d x
$$
|
$\frac{16}{9}$
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Compute $e^{A}$ where $A$ is defined as
$$
\int_{3 / 4}^{4 / 3} \frac{2 x^{2}+x+1}{x^{3}+x^{2}+x+1} d x
$$
|
We can use partial fractions to decompose the integrand to $\frac{1}{x+1}+\frac{x}{x^{2}+1}$, and then integrate the addends separately by substituting $u=x+1$ for the former and $u=x^{2}+1$ for latter, to obtain $\ln (x+1)+\left.\frac{1}{2} \ln \left(x^{2}+1\right)\right|_{3 / 4} ^{4 / 3}=\left.\ln \left((x+1) \sqrt{x^{2}+1}\right)\right|_{3 / 4} ^{4 / 3}=\ln \frac{16}{9}$. Thus $e^{A}=16 / 9$.
Alternate solution: Substituting $u=1 / x$, we find
$$
A=\int_{4 / 3}^{3 / 4} \frac{2 u+u^{2}+u^{3}}{1+u+u^{2}+u^{3}}\left(-\frac{1}{u^{2}}\right) d u=\int_{3 / 4}^{4 / 3} \frac{2 / u+1+u}{1+u+u^{2}+u^{3}} d u
$$
Adding this to the original integral, we find
$$
2 A=\int_{3 / 4}^{4 / 3} \frac{2 / u+2+2 u+2 u^{2}}{1+u+u^{2}+u^{3}} d u=\int_{3 / 4}^{4 / 3} \frac{2}{u} d u
$$
Thus $A=\ln \frac{16}{9}$ and $e^{A}=\frac{16}{9}$.
|
\frac{16}{9}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Compute $e^{A}$ where $A$ is defined as
$$
\int_{3 / 4}^{4 / 3} \frac{2 x^{2}+x+1}{x^{3}+x^{2}+x+1} d x
$$
|
We can use partial fractions to decompose the integrand to $\frac{1}{x+1}+\frac{x}{x^{2}+1}$, and then integrate the addends separately by substituting $u=x+1$ for the former and $u=x^{2}+1$ for latter, to obtain $\ln (x+1)+\left.\frac{1}{2} \ln \left(x^{2}+1\right)\right|_{3 / 4} ^{4 / 3}=\left.\ln \left((x+1) \sqrt{x^{2}+1}\right)\right|_{3 / 4} ^{4 / 3}=\ln \frac{16}{9}$. Thus $e^{A}=16 / 9$.
Alternate solution: Substituting $u=1 / x$, we find
$$
A=\int_{4 / 3}^{3 / 4} \frac{2 u+u^{2}+u^{3}}{1+u+u^{2}+u^{3}}\left(-\frac{1}{u^{2}}\right) d u=\int_{3 / 4}^{4 / 3} \frac{2 / u+1+u}{1+u+u^{2}+u^{3}} d u
$$
Adding this to the original integral, we find
$$
2 A=\int_{3 / 4}^{4 / 3} \frac{2 / u+2+2 u+2 u^{2}}{1+u+u^{2}+u^{3}} d u=\int_{3 / 4}^{4 / 3} \frac{2}{u} d u
$$
Thus $A=\ln \frac{16}{9}$ and $e^{A}=\frac{16}{9}$.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $P$ be a fourth degree polynomial, with derivative $P^{\prime}$, such that $P(1)=P(3)=P(5)=P^{\prime}(7)=0$. Find the real number $x \neq 1,3,5$ such that $P(x)=0$.
|
$\frac{89}{11}$
|
\frac{89}{11}
|
Yes
|
Incomplete
|
math-word-problem
|
Algebra
|
Let $P$ be a fourth degree polynomial, with derivative $P^{\prime}$, such that $P(1)=P(3)=P(5)=P^{\prime}(7)=0$. Find the real number $x \neq 1,3,5$ such that $P(x)=0$.
|
$\frac{89}{11}$
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Let $P$ be a fourth degree polynomial, with derivative $P^{\prime}$, such that $P(1)=P(3)=P(5)=P^{\prime}(7)=0$. Find the real number $x \neq 1,3,5$ such that $P(x)=0$.
|
Observe that 7 is not a root of $P$. If $r_{1}, r_{2}, r_{3}, r_{4}$ are the roots of $P$, then $\frac{P^{\prime}(7)}{P(7)}=$ $\sum_{i} \frac{1}{7-r_{i}}=0$. Thus $r_{4}=7-\left(\sum_{i \neq 4} \frac{1}{7-r_{i}}\right)^{-1}=7+\left(\frac{1}{6}+\frac{1}{4}+\frac{1}{2}\right)^{-1}=7+12 / 11=89 / 11$.
|
\frac{89}{11}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $P$ be a fourth degree polynomial, with derivative $P^{\prime}$, such that $P(1)=P(3)=P(5)=P^{\prime}(7)=0$. Find the real number $x \neq 1,3,5$ such that $P(x)=0$.
|
Observe that 7 is not a root of $P$. If $r_{1}, r_{2}, r_{3}, r_{4}$ are the roots of $P$, then $\frac{P^{\prime}(7)}{P(7)}=$ $\sum_{i} \frac{1}{7-r_{i}}=0$. Thus $r_{4}=7-\left(\sum_{i \neq 4} \frac{1}{7-r_{i}}\right)^{-1}=7+\left(\frac{1}{6}+\frac{1}{4}+\frac{1}{2}\right)^{-1}=7+12 / 11=89 / 11$.
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Compute
$$
\lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{3}+4 h\right)-4 \sin \left(\frac{\pi}{3}+3 h\right)+6 \sin \left(\frac{\pi}{3}+2 h\right)-4 \sin \left(\frac{\pi}{3}+h\right)+\sin \left(\frac{\pi}{3}\right)}{h^{4}}
$$
|
$\frac{\sqrt{3}}{2}$
|
\frac{\sqrt{3}}{2}
|
Yes
|
Incomplete
|
math-word-problem
|
Calculus
|
Compute
$$
\lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{3}+4 h\right)-4 \sin \left(\frac{\pi}{3}+3 h\right)+6 \sin \left(\frac{\pi}{3}+2 h\right)-4 \sin \left(\frac{\pi}{3}+h\right)+\sin \left(\frac{\pi}{3}\right)}{h^{4}}
$$
|
$\frac{\sqrt{3}}{2}$
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n5. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Compute
$$
\lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{3}+4 h\right)-4 \sin \left(\frac{\pi}{3}+3 h\right)+6 \sin \left(\frac{\pi}{3}+2 h\right)-4 \sin \left(\frac{\pi}{3}+h\right)+\sin \left(\frac{\pi}{3}\right)}{h^{4}}
$$
|
The derivative of a function is defined as $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$. Iterating this formula four times yields
$$
f^{(4)}(x)=\lim _{h \rightarrow 0} \frac{f(x+4 h)-4 f(x+3 h)+6 f(x+2 h)-4 f(x+h)+f(x)}{h^{4}} .
$$
Substituting $f=\sin$ and $x=\pi / 3$, the expression is equal to $\sin ^{(4)}(\pi / 3)=\sin (\pi / 3)=\frac{\sqrt{3}}{2}$.
|
\frac{\sqrt{3}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Compute
$$
\lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{3}+4 h\right)-4 \sin \left(\frac{\pi}{3}+3 h\right)+6 \sin \left(\frac{\pi}{3}+2 h\right)-4 \sin \left(\frac{\pi}{3}+h\right)+\sin \left(\frac{\pi}{3}\right)}{h^{4}}
$$
|
The derivative of a function is defined as $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$. Iterating this formula four times yields
$$
f^{(4)}(x)=\lim _{h \rightarrow 0} \frac{f(x+4 h)-4 f(x+3 h)+6 f(x+2 h)-4 f(x+h)+f(x)}{h^{4}} .
$$
Substituting $f=\sin$ and $x=\pi / 3$, the expression is equal to $\sin ^{(4)}(\pi / 3)=\sin (\pi / 3)=\frac{\sqrt{3}}{2}$.
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n5. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $p_{0}(x), p_{1}(x), p_{2}(x), \ldots$ be polynomials such that $p_{0}(x)=x$ and for all positive integers $n$, $\frac{d}{d x} p_{n}(x)=p_{n-1}(x)$. Define the function $p(x):[0, \infty) \rightarrow \mathbb{R} x$ by $p(x)=p_{n}(x)$ for all $x \in[n, n+1]$. Given that $p(x)$ is continuous on $[0, \infty)$, compute
$$
\sum_{n=0}^{\infty} p_{n}(2009)
$$
|
$e^{2010}-e^{2009}-1$
|
e^{2010}-e^{2009}-1
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Let $p_{0}(x), p_{1}(x), p_{2}(x), \ldots$ be polynomials such that $p_{0}(x)=x$ and for all positive integers $n$, $\frac{d}{d x} p_{n}(x)=p_{n-1}(x)$. Define the function $p(x):[0, \infty) \rightarrow \mathbb{R} x$ by $p(x)=p_{n}(x)$ for all $x \in[n, n+1]$. Given that $p(x)$ is continuous on $[0, \infty)$, compute
$$
\sum_{n=0}^{\infty} p_{n}(2009)
$$
|
$e^{2010}-e^{2009}-1$
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n6. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Let $p_{0}(x), p_{1}(x), p_{2}(x), \ldots$ be polynomials such that $p_{0}(x)=x$ and for all positive integers $n$, $\frac{d}{d x} p_{n}(x)=p_{n-1}(x)$. Define the function $p(x):[0, \infty) \rightarrow \mathbb{R} x$ by $p(x)=p_{n}(x)$ for all $x \in[n, n+1]$. Given that $p(x)$ is continuous on $[0, \infty)$, compute
$$
\sum_{n=0}^{\infty} p_{n}(2009)
$$
|
By writing out the first few polynomials, one can guess and then show by induction that $p_{n}(x)=\frac{1}{(n+1)!}(x+1)^{n+1}-\frac{1}{n!} x^{n}$. Thus the sum evaluates to $e^{2010}-e^{2009}-1$ by the series expansion of $e^{x}$.
|
e^{2010}-e^{2009}-1
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Let $p_{0}(x), p_{1}(x), p_{2}(x), \ldots$ be polynomials such that $p_{0}(x)=x$ and for all positive integers $n$, $\frac{d}{d x} p_{n}(x)=p_{n-1}(x)$. Define the function $p(x):[0, \infty) \rightarrow \mathbb{R} x$ by $p(x)=p_{n}(x)$ for all $x \in[n, n+1]$. Given that $p(x)$ is continuous on $[0, \infty)$, compute
$$
\sum_{n=0}^{\infty} p_{n}(2009)
$$
|
By writing out the first few polynomials, one can guess and then show by induction that $p_{n}(x)=\frac{1}{(n+1)!}(x+1)^{n+1}-\frac{1}{n!} x^{n}$. Thus the sum evaluates to $e^{2010}-e^{2009}-1$ by the series expansion of $e^{x}$.
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n6. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
A line in the plane is called strange if it passes through $(a, 0)$ and $(0,10-a)$ for some $a$ in the interval $[0,10]$. A point in the plane is called charming if it lies in the first quadrant and also lies below some strange line. What is the area of the set of all charming points?
|
$50 / 3$
|
\frac{50}{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A line in the plane is called strange if it passes through $(a, 0)$ and $(0,10-a)$ for some $a$ in the interval $[0,10]$. A point in the plane is called charming if it lies in the first quadrant and also lies below some strange line. What is the area of the set of all charming points?
|
$50 / 3$
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
A line in the plane is called strange if it passes through $(a, 0)$ and $(0,10-a)$ for some $a$ in the interval $[0,10]$. A point in the plane is called charming if it lies in the first quadrant and also lies below some strange line. What is the area of the set of all charming points?
|
The strange lines form an envelope (set of tangent lines) of a curve $f(x)$, and we first find the equation for $f$ on $[0,10]$. Assuming the derivative $f^{\prime}$ is continuous, the point of tangency of the line $\ell$ through $(a, 0)$ and $(0, b)$ to $f$ is the limit of the intersection points of this line with the lines $\ell_{\epsilon}$ passing through $(a+\epsilon, 0)$ and $(0, b-\epsilon)$ as $\epsilon \rightarrow 0$. If these limits exist, then the derivative is indeed continuous and we can calculate the function from the points of tangency.
The intersection point of $\ell$ and $\ell_{\epsilon}$ can be calculated to have $x$-coordinate $\frac{a(a-\epsilon)}{a+b}$, so the tangent point of $\ell$ has $x$-coordinate $\lim _{\epsilon \rightarrow 0} \frac{a(a-\epsilon)}{a+b}=\frac{a^{2}}{a+b}=\frac{a^{2}}{10}$. Similarly, the $y$-coordinate is $\frac{b^{2}}{10}=\frac{(10-a)^{2}}{10}$. Thus,
solving for the $y$ coordinate in terms of the $x$ coordinate for $a \in[0,10]$, we find $f(x)=10-2 \sqrt{10} \sqrt{x}+x$, and so the area of the set of charming points is
$$
\int_{0}^{10}(10-2 \sqrt{10} \sqrt{x}+x) d x=50 / 3
$$
|
\frac{50}{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A line in the plane is called strange if it passes through $(a, 0)$ and $(0,10-a)$ for some $a$ in the interval $[0,10]$. A point in the plane is called charming if it lies in the first quadrant and also lies below some strange line. What is the area of the set of all charming points?
|
The strange lines form an envelope (set of tangent lines) of a curve $f(x)$, and we first find the equation for $f$ on $[0,10]$. Assuming the derivative $f^{\prime}$ is continuous, the point of tangency of the line $\ell$ through $(a, 0)$ and $(0, b)$ to $f$ is the limit of the intersection points of this line with the lines $\ell_{\epsilon}$ passing through $(a+\epsilon, 0)$ and $(0, b-\epsilon)$ as $\epsilon \rightarrow 0$. If these limits exist, then the derivative is indeed continuous and we can calculate the function from the points of tangency.
The intersection point of $\ell$ and $\ell_{\epsilon}$ can be calculated to have $x$-coordinate $\frac{a(a-\epsilon)}{a+b}$, so the tangent point of $\ell$ has $x$-coordinate $\lim _{\epsilon \rightarrow 0} \frac{a(a-\epsilon)}{a+b}=\frac{a^{2}}{a+b}=\frac{a^{2}}{10}$. Similarly, the $y$-coordinate is $\frac{b^{2}}{10}=\frac{(10-a)^{2}}{10}$. Thus,
solving for the $y$ coordinate in terms of the $x$ coordinate for $a \in[0,10]$, we find $f(x)=10-2 \sqrt{10} \sqrt{x}+x$, and so the area of the set of charming points is
$$
\int_{0}^{10}(10-2 \sqrt{10} \sqrt{x}+x) d x=50 / 3
$$
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Compute
$$
\int_{1}^{\sqrt{3}} x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x
$$
|
13
|
13
|
Yes
|
Problem not solved
|
math-word-problem
|
Calculus
|
Compute
$$
\int_{1}^{\sqrt{3}} x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x
$$
|
13
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [7]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Compute
$$
\int_{1}^{\sqrt{3}} x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x
$$
|
Using the fact that $x=e^{\ln (x)}$, we evaluate the integral as follows:
$$
\begin{aligned}
\int x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x & =\int x^{2 x^{2}+1}+x^{2 x^{2}+1} \ln \left(x^{2}\right) d x \\
& =\int e^{\ln (x)\left(2 x^{2}+1\right)}\left(1+\ln \left(x^{2}\right)\right) d x \\
& =\int x e^{x^{2} \ln \left(x^{2}\right)}\left(1+\ln \left(x^{2}\right)\right) d x
\end{aligned}
$$
Noticing that the derivative of $x^{2} \ln \left(x^{2}\right)$ is $2 x\left(1+\ln \left(x^{2}\right)\right)$, it follows that the integral evaluates to
$$
\frac{1}{2} e^{x^{2} \ln \left(x^{2}\right)}=\frac{1}{2} x^{2 x^{2}}
$$
Evaluating this from 1 to $\sqrt{3}$ we obtain the answer.
|
\frac{1}{2} \left( (\sqrt{3})^{2 (\sqrt{3})^{2}} - 1^{2 \cdot 1^{2}} \right)
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Compute
$$
\int_{1}^{\sqrt{3}} x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x
$$
|
Using the fact that $x=e^{\ln (x)}$, we evaluate the integral as follows:
$$
\begin{aligned}
\int x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x & =\int x^{2 x^{2}+1}+x^{2 x^{2}+1} \ln \left(x^{2}\right) d x \\
& =\int e^{\ln (x)\left(2 x^{2}+1\right)}\left(1+\ln \left(x^{2}\right)\right) d x \\
& =\int x e^{x^{2} \ln \left(x^{2}\right)}\left(1+\ln \left(x^{2}\right)\right) d x
\end{aligned}
$$
Noticing that the derivative of $x^{2} \ln \left(x^{2}\right)$ is $2 x\left(1+\ln \left(x^{2}\right)\right)$, it follows that the integral evaluates to
$$
\frac{1}{2} e^{x^{2} \ln \left(x^{2}\right)}=\frac{1}{2} x^{2 x^{2}}
$$
Evaluating this from 1 to $\sqrt{3}$ we obtain the answer.
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [7]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
let $\mathcal{R}$ be the region in the plane bounded by the graphs of $y=x$ and $y=x^{2}$. Compute the volume of the region formed by revolving $\mathcal{R}$ around the line $y=x$.
|
$\square$
|
not found
|
Yes
|
Incomplete
|
math-word-problem
|
Calculus
|
let $\mathcal{R}$ be the region in the plane bounded by the graphs of $y=x$ and $y=x^{2}$. Compute the volume of the region formed by revolving $\mathcal{R}$ around the line $y=x$.
|
$\square$
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [7]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
let $\mathcal{R}$ be the region in the plane bounded by the graphs of $y=x$ and $y=x^{2}$. Compute the volume of the region formed by revolving $\mathcal{R}$ around the line $y=x$.
|
We integrate from 0 to 1 using the method of washers. Fix $d$ between 0 and 1. Let the line $x=d$ intersect the graph of $y=x^{2}$ at $Q$, and let the line $x=d$ intersect the graph of $y=x$ at $P$. Then $P=(d, d)$, and $Q=\left(d, d^{2}\right)$. Now drop a perpendicular from $Q$ to the line $y=x$, and let $R$ be the foot of this perpendicular. Because $P Q R$ is a $45-45-90$ triangle, $Q R=\left(d-d^{2}\right) / \sqrt{2}$. So the differential washer has a radius of $\left(d-d^{2}\right) / \sqrt{2}$ and a height of $\sqrt{2} d x$. So we integrate (from 0 to 1 ) the expression $\left[\left(x-x^{2}\right) / \sqrt{2}\right]^{2} \sqrt{2} d x$, and the answer follows.
|
\frac{\pi}{6}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
let $\mathcal{R}$ be the region in the plane bounded by the graphs of $y=x$ and $y=x^{2}$. Compute the volume of the region formed by revolving $\mathcal{R}$ around the line $y=x$.
|
We integrate from 0 to 1 using the method of washers. Fix $d$ between 0 and 1. Let the line $x=d$ intersect the graph of $y=x^{2}$ at $Q$, and let the line $x=d$ intersect the graph of $y=x$ at $P$. Then $P=(d, d)$, and $Q=\left(d, d^{2}\right)$. Now drop a perpendicular from $Q$ to the line $y=x$, and let $R$ be the foot of this perpendicular. Because $P Q R$ is a $45-45-90$ triangle, $Q R=\left(d-d^{2}\right) / \sqrt{2}$. So the differential washer has a radius of $\left(d-d^{2}\right) / \sqrt{2}$ and a height of $\sqrt{2} d x$. So we integrate (from 0 to 1 ) the expression $\left[\left(x-x^{2}\right) / \sqrt{2}\right]^{2} \sqrt{2} d x$, and the answer follows.
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [7]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $a$ and $b$ be real numbers satisfying $a>b>0$. Evaluate
$$
\int_{0}^{2 \pi} \frac{1}{a+b \cos (\theta)} d \theta
$$
Express your answer in terms of $a$ and $b$.
|
$\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}$
|
\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}
|
Yes
|
Incomplete
|
math-word-problem
|
Calculus
|
Let $a$ and $b$ be real numbers satisfying $a>b>0$. Evaluate
$$
\int_{0}^{2 \pi} \frac{1}{a+b \cos (\theta)} d \theta
$$
Express your answer in terms of $a$ and $b$.
|
$\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}$
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Let $a$ and $b$ be real numbers satisfying $a>b>0$. Evaluate
$$
\int_{0}^{2 \pi} \frac{1}{a+b \cos (\theta)} d \theta
$$
Express your answer in terms of $a$ and $b$.
|
Using the geometric series formula, we can expand the integral as follows:
$$
\begin{aligned}
\int_{0}^{2 \pi} \frac{1}{a+b \cos (\theta)} d \theta & =\frac{1}{a} \int_{0}^{2 \pi} 1+\frac{b}{a} \cos (\theta)+\left(\frac{b}{a}\right)^{2} \cos ^{2}(\theta) d \theta \\
& =\frac{1}{a} \sum_{n=0}^{\infty} \int_{0}^{2 \pi}\left(\frac{b}{a}\right)^{n}\left(\frac{e^{i \theta}+e^{-i \theta}}{2}\right)^{n} d \theta \\
& =\frac{2 \pi}{a} \sum_{n=0}^{\infty}\left(\frac{b^{2}}{a^{2}}\right)^{n} \frac{\binom{2 n}{n}}{2^{2 n}} d \theta
\end{aligned}
$$
To evaluate this sum, recall that $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$th Catalan number. The generating function for the Catalan numbers is
$$
\sum_{n=0}^{\infty} C_{n} x^{n}=\frac{1-\sqrt{1-4 x}}{2 x}
$$
and taking the derivative of $x$ times this generating function yields $\sum\binom{2 n}{n} x^{n}=\frac{1}{\sqrt{1-4 x}}$. Thus the integral evaluates to $\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}$, as desired.
|
\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Let $a$ and $b$ be real numbers satisfying $a>b>0$. Evaluate
$$
\int_{0}^{2 \pi} \frac{1}{a+b \cos (\theta)} d \theta
$$
Express your answer in terms of $a$ and $b$.
|
Using the geometric series formula, we can expand the integral as follows:
$$
\begin{aligned}
\int_{0}^{2 \pi} \frac{1}{a+b \cos (\theta)} d \theta & =\frac{1}{a} \int_{0}^{2 \pi} 1+\frac{b}{a} \cos (\theta)+\left(\frac{b}{a}\right)^{2} \cos ^{2}(\theta) d \theta \\
& =\frac{1}{a} \sum_{n=0}^{\infty} \int_{0}^{2 \pi}\left(\frac{b}{a}\right)^{n}\left(\frac{e^{i \theta}+e^{-i \theta}}{2}\right)^{n} d \theta \\
& =\frac{2 \pi}{a} \sum_{n=0}^{\infty}\left(\frac{b^{2}}{a^{2}}\right)^{n} \frac{\binom{2 n}{n}}{2^{2 n}} d \theta
\end{aligned}
$$
To evaluate this sum, recall that $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$th Catalan number. The generating function for the Catalan numbers is
$$
\sum_{n=0}^{\infty} C_{n} x^{n}=\frac{1-\sqrt{1-4 x}}{2 x}
$$
and taking the derivative of $x$ times this generating function yields $\sum\binom{2 n}{n} x^{n}=\frac{1}{\sqrt{1-4 x}}$. Thus the integral evaluates to $\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}$, as desired.
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute value of the numbers in the sequence is nondecreasing?
|
128
|
128
|
Yes
|
Problem not solved
|
math-word-problem
|
Combinatorics
|
How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute value of the numbers in the sequence is nondecreasing?
|
128
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute value of the numbers in the sequence is nondecreasing?
|
Each of the pairs $a,-a$ must occur in increasing order of $a$ for $a=1, \ldots, 7$, but $a$ can either occur before or after $-a$, for a total of $2^{7}=128$ possible sequences.
|
128
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute value of the numbers in the sequence is nondecreasing?
|
Each of the pairs $a,-a$ must occur in increasing order of $a$ for $a=1, \ldots, 7$, but $a$ can either occur before or after $-a$, for a total of $2^{7}=128$ possible sequences.
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. What is the expected number of cards that will be between the two jokers?
|
$52 / 3$
|
52 / 3
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. What is the expected number of cards that will be between the two jokers?
|
$52 / 3$
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. What is the expected number of cards that will be between the two jokers?
|
Each card has an equal likelihood of being either on top of the jokers, in between them, or below the jokers. Thus, on average, $1 / 3$ of them will land between the two jokers.
|
\frac{52}{3}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. What is the expected number of cards that will be between the two jokers?
|
Each card has an equal likelihood of being either on top of the jokers, in between them, or below the jokers. Thus, on average, $1 / 3$ of them will land between the two jokers.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
In how many ways can you rearrange the letters of "HMMTHMMT" such that the consecutive substring "HMMT" does not appear?
|
361
|
361
|
Yes
|
Problem not solved
|
math-word-problem
|
Combinatorics
|
In how many ways can you rearrange the letters of "HMMTHMMT" such that the consecutive substring "HMMT" does not appear?
|
361
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n3. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
In how many ways can you rearrange the letters of "HMMTHMMT" such that the consecutive substring "HMMT" does not appear?
|
There are $8!/(4!2!2!)=420$ ways to order the letters. If the permuted letters contain "HMMT", there are $5 \cdot 4!/ 2!=60$ ways to order the other letters, so we subtract these. However, we have subtracted "HMMTHMMT" twice, so we add it back once to obtain 361 possibilities.
|
361
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In how many ways can you rearrange the letters of "HMMTHMMT" such that the consecutive substring "HMMT" does not appear?
|
There are $8!/(4!2!2!)=420$ ways to order the letters. If the permuted letters contain "HMMT", there are $5 \cdot 4!/ 2!=60$ ways to order the other letters, so we subtract these. However, we have subtracted "HMMTHMMT" twice, so we add it back once to obtain 361 possibilities.
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n3. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ satisfy $f(f(x))=f(x)$ for all $x \in\{1,2,3,4,5\}$ ?
|
196
|
196
|
Yes
|
Incomplete
|
math-word-problem
|
Combinatorics
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ satisfy $f(f(x))=f(x)$ for all $x \in\{1,2,3,4,5\}$ ?
|
196
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n4. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ satisfy $f(f(x))=f(x)$ for all $x \in\{1,2,3,4,5\}$ ?
|
A fixed point of a function $f$ is an element $a$ such that $f(a)=a$. The condition is equivalent to the property that $f$ maps every number to a fixed point. Counting by the number of fixed points of $f$, the total number of such functions is
$$
\begin{aligned}
\sum_{k=1}^{5}\binom{5}{k} k^{5-k} & =1 \cdot\left(5^{0}\right)+5 \cdot\left(1^{4}+4^{1}\right)+10 \cdot\left(2^{3}+3^{2}\right) \\
& =1+25+10 \cdot 17 \\
& =196
\end{aligned}
$$
|
196
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ satisfy $f(f(x))=f(x)$ for all $x \in\{1,2,3,4,5\}$ ?
|
A fixed point of a function $f$ is an element $a$ such that $f(a)=a$. The condition is equivalent to the property that $f$ maps every number to a fixed point. Counting by the number of fixed points of $f$, the total number of such functions is
$$
\begin{aligned}
\sum_{k=1}^{5}\binom{5}{k} k^{5-k} & =1 \cdot\left(5^{0}\right)+5 \cdot\left(1^{4}+4^{1}\right)+10 \cdot\left(2^{3}+3^{2}\right) \\
& =1+25+10 \cdot 17 \\
& =196
\end{aligned}
$$
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n4. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $s(n)$ denote the number of 1's in the binary representation of $n$. Compute
$$
\frac{1}{255} \sum_{0 \leq n<16} 2^{n}(-1)^{s(n)}
$$
|
45
|
45
|
Yes
|
Problem not solved
|
math-word-problem
|
Number Theory
|
Let $s(n)$ denote the number of 1's in the binary representation of $n$. Compute
$$
\frac{1}{255} \sum_{0 \leq n<16} 2^{n}(-1)^{s(n)}
$$
|
45
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n5. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Let $s(n)$ denote the number of 1's in the binary representation of $n$. Compute
$$
\frac{1}{255} \sum_{0 \leq n<16} 2^{n}(-1)^{s(n)}
$$
|
Notice that if $n<8,(-1)^{s(n)}=(-1) \cdot(-1)^{s(n+8)}$ so the sum becomes $\frac{1}{255}\left(1-2^{8}\right) \sum_{0 \leq n<8} 2^{n}(-1)^{s(n)}=$ 45.
|
45
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $s(n)$ denote the number of 1's in the binary representation of $n$. Compute
$$
\frac{1}{255} \sum_{0 \leq n<16} 2^{n}(-1)^{s(n)}
$$
|
Notice that if $n<8,(-1)^{s(n)}=(-1) \cdot(-1)^{s(n+8)}$ so the sum becomes $\frac{1}{255}\left(1-2^{8}\right) \sum_{0 \leq n<8} 2^{n}(-1)^{s(n)}=$ 45.
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n5. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
How many sequences of 5 positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$ ?
|
116
|
116
|
Yes
|
Problem not solved
|
math-word-problem
|
Combinatorics
|
How many sequences of 5 positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$ ?
|
116
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n6. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
How many sequences of 5 positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$ ?
|
We count based on how many 1's the sequence contains. If $a=b=c=d=e=1$ then this gives us 1 possibility. If $a=b=c=d=1$ and $e \neq 1$, $e$ can be $2,3,4,5,6$. Each such sequence $(1,1,1,1, e)$ can be arranged in 5 different ways, for a total of $5 \cdot 5=25$ ways in this case.
If three of the numbers are 1 , the last two can be $(2,2),(3,3),(2,3),(2,4)$, or $(2,5)$. Counting ordering, this gives a total of $2 \cdot 10+3 \cdot 20=80$ possibilities.
If two of the numbers are 1 , the other three must be equal to 2 for the product to be under 10 , and this yields 10 more possibilities.
Thus there are $1+25+80+10=116$ such sequences.
|
116
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many sequences of 5 positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$ ?
|
We count based on how many 1's the sequence contains. If $a=b=c=d=e=1$ then this gives us 1 possibility. If $a=b=c=d=1$ and $e \neq 1$, $e$ can be $2,3,4,5,6$. Each such sequence $(1,1,1,1, e)$ can be arranged in 5 different ways, for a total of $5 \cdot 5=25$ ways in this case.
If three of the numbers are 1 , the last two can be $(2,2),(3,3),(2,3),(2,4)$, or $(2,5)$. Counting ordering, this gives a total of $2 \cdot 10+3 \cdot 20=80$ possibilities.
If two of the numbers are 1 , the other three must be equal to 2 for the product to be under 10 , and this yields 10 more possibilities.
Thus there are $1+25+80+10=116$ such sequences.
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n6. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Paul fills in a $7 \times 7$ grid with the numbers 1 through 49 in a random arrangement. He then erases his work and does the same thing again (to obtain two different random arrangements of the numbers in the grid). What is the expected number of pairs of numbers that occur in either the same row as each other or the same column as each other in both of the two arrangements?
|
$147 / 2$
|
\frac{147}{2}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Paul fills in a $7 \times 7$ grid with the numbers 1 through 49 in a random arrangement. He then erases his work and does the same thing again (to obtain two different random arrangements of the numbers in the grid). What is the expected number of pairs of numbers that occur in either the same row as each other or the same column as each other in both of the two arrangements?
|
$147 / 2$
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [7]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Paul fills in a $7 \times 7$ grid with the numbers 1 through 49 in a random arrangement. He then erases his work and does the same thing again (to obtain two different random arrangements of the numbers in the grid). What is the expected number of pairs of numbers that occur in either the same row as each other or the same column as each other in both of the two arrangements?
|
Each of the $\binom{49}{2}$ pairs of numbers has a probability of $\frac{14 \cdot\binom{7}{2}}{\binom{49}{2}}=1 / 4$ of being in the same row or column in one of the arrangements, so the expected number that are in the same row or column in both arrangements is
$$
\binom{49}{2} \cdot(1 / 4)^{2}=\frac{147}{2}
$$
|
\frac{147}{2}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Paul fills in a $7 \times 7$ grid with the numbers 1 through 49 in a random arrangement. He then erases his work and does the same thing again (to obtain two different random arrangements of the numbers in the grid). What is the expected number of pairs of numbers that occur in either the same row as each other or the same column as each other in both of the two arrangements?
|
Each of the $\binom{49}{2}$ pairs of numbers has a probability of $\frac{14 \cdot\binom{7}{2}}{\binom{49}{2}}=1 / 4$ of being in the same row or column in one of the arrangements, so the expected number that are in the same row or column in both arrangements is
$$
\binom{49}{2} \cdot(1 / 4)^{2}=\frac{147}{2}
$$
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [7]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
There are 5 students on a team for a math competition. The math competition has 5 subject tests. Each student on the team must choose 2 distinct tests, and each test must be taken by exactly two people. In how many ways can this be done?
|
2040
|
2040
|
Yes
|
Incomplete
|
math-word-problem
|
Combinatorics
|
There are 5 students on a team for a math competition. The math competition has 5 subject tests. Each student on the team must choose 2 distinct tests, and each test must be taken by exactly two people. In how many ways can this be done?
|
2040
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [7]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
There are 5 students on a team for a math competition. The math competition has 5 subject tests. Each student on the team must choose 2 distinct tests, and each test must be taken by exactly two people. In how many ways can this be done?
|
We can model the situation as a bipartite graph on 10 vertices, with 5 nodes representing the students and the other 5 representing the tests. We now simply want to count the number of bipartite graphs on these two sets such that there are two edges incident on each vertex.
Notice that in such a graph, we can start at any vertex and follow one of the edges eminating from it, then follow the other edge eminating from the second vertex, etc, and in this manner we must eventually end up back at the starting vertex, so the graph is partitioned into even cycles. Since each vertex has degree two, we cannot have a 2 -cycle, so we must have either a 10 -cycle or a 4 -cycle and a 6 -cycle.
In the former case, starting with Person $A$, there are 5 ways to choose one of his tests. This test can be taken by one of 4 other people, who can take one of 4 other tests, which can be taken by one of 3 other people, etc, so the number of 10 -cycles we obtain in this way is $5!\cdot 4!$. However, it does not matter which of the first person's tests we choose first in a given 10 -cycle, so we overcounted by a factor of 2 . Thus there are $5!\cdot 4!/ 2=1440$ possibilities in this case.
In the latter case, there are $\binom{5}{3}^{2}=100$ ways to choose which three people and which three tests are in the 6 -cycle. After choosing this, a similar argument to that above shows there are $2!\cdot 1!/ 2$ possible 4 -cycles and $3!\cdot 2!/ 2$ possible 6 -cycles, for a total of $100 \cdot 1 \cdot 6=600$ possibilities in this case.
Thus there are a total of 2040 ways they can take the tests.
|
2040
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
There are 5 students on a team for a math competition. The math competition has 5 subject tests. Each student on the team must choose 2 distinct tests, and each test must be taken by exactly two people. In how many ways can this be done?
|
We can model the situation as a bipartite graph on 10 vertices, with 5 nodes representing the students and the other 5 representing the tests. We now simply want to count the number of bipartite graphs on these two sets such that there are two edges incident on each vertex.
Notice that in such a graph, we can start at any vertex and follow one of the edges eminating from it, then follow the other edge eminating from the second vertex, etc, and in this manner we must eventually end up back at the starting vertex, so the graph is partitioned into even cycles. Since each vertex has degree two, we cannot have a 2 -cycle, so we must have either a 10 -cycle or a 4 -cycle and a 6 -cycle.
In the former case, starting with Person $A$, there are 5 ways to choose one of his tests. This test can be taken by one of 4 other people, who can take one of 4 other tests, which can be taken by one of 3 other people, etc, so the number of 10 -cycles we obtain in this way is $5!\cdot 4!$. However, it does not matter which of the first person's tests we choose first in a given 10 -cycle, so we overcounted by a factor of 2 . Thus there are $5!\cdot 4!/ 2=1440$ possibilities in this case.
In the latter case, there are $\binom{5}{3}^{2}=100$ ways to choose which three people and which three tests are in the 6 -cycle. After choosing this, a similar argument to that above shows there are $2!\cdot 1!/ 2$ possible 4 -cycles and $3!\cdot 2!/ 2$ possible 6 -cycles, for a total of $100 \cdot 1 \cdot 6=600$ possibilities in this case.
Thus there are a total of 2040 ways they can take the tests.
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [7]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
The squares of a $3 \times 3$ grid are filled with positive integers such that 1 is the label of the upperleftmost square, 2009 is the label of the lower-rightmost square, and the label of each square divides the one directly to the right of it and the one directly below it. How many such labelings are possible?
|
2448
|
2448
|
Yes
|
Incomplete
|
math-word-problem
|
Combinatorics
|
The squares of a $3 \times 3$ grid are filled with positive integers such that 1 is the label of the upperleftmost square, 2009 is the label of the lower-rightmost square, and the label of each square divides the one directly to the right of it and the one directly below it. How many such labelings are possible?
|
2448
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
The squares of a $3 \times 3$ grid are filled with positive integers such that 1 is the label of the upperleftmost square, 2009 is the label of the lower-rightmost square, and the label of each square divides the one directly to the right of it and the one directly below it. How many such labelings are possible?
|
We factor 2009 as $7^{2} \cdot 41$ and place the 41 's and the 7 's in the squares separately. The number of ways to fill the grid with 1 's and 41 's so that the divisibility property is satisfied is equal to the number of nondecreasing sequences $a_{1}, a_{2}, a_{3}$ where each $a_{i} \in\{0,1,2,3\}$ and the sequence is not $0,0,0$ and not $1,1,1$ (here $a_{i}$ corresponds to the number of 41 's in the $i$ th column.) Thus there are $\binom{3+4-1}{3}-2=18$ ways to choose which squares are divisible by 41 .
To count the arrangements of divisibility by 7 and 49 , we consider three cases.
If 49 divides the middle square, then each of the squares to the right and below it are divisible 49. The two squares in the top row (besides the upper left) can be $(1,1),(1,7),(1,49),(7,7),(7,49)$, or $(49,49)$ (in terms of the highest power of 7 dividing the square). The same is true, independently, for the two blank squares on the left column, for a total of $6^{2}=36$ possibilities in this case.
If 1 is the highest power of 7 dividing the middle square, there are also 36 possibilities by a similar argument.
If 7 is the highest power of 7 dividing the middle square, there are 8 possibilities for the upper right three squares. Thus there are 64 possibilities in this case.
Thus there are a total of 136 options for the divisibility of each number by 7 and $7^{2}$, and 18 options for the divisibility of the numbers by 41 . Since each number divides 2009 , this uniquely determines the numbers, and so there are a total of $18 \cdot 136=2448$ possibilities.
|
2448
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
The squares of a $3 \times 3$ grid are filled with positive integers such that 1 is the label of the upperleftmost square, 2009 is the label of the lower-rightmost square, and the label of each square divides the one directly to the right of it and the one directly below it. How many such labelings are possible?
|
We factor 2009 as $7^{2} \cdot 41$ and place the 41 's and the 7 's in the squares separately. The number of ways to fill the grid with 1 's and 41 's so that the divisibility property is satisfied is equal to the number of nondecreasing sequences $a_{1}, a_{2}, a_{3}$ where each $a_{i} \in\{0,1,2,3\}$ and the sequence is not $0,0,0$ and not $1,1,1$ (here $a_{i}$ corresponds to the number of 41 's in the $i$ th column.) Thus there are $\binom{3+4-1}{3}-2=18$ ways to choose which squares are divisible by 41 .
To count the arrangements of divisibility by 7 and 49 , we consider three cases.
If 49 divides the middle square, then each of the squares to the right and below it are divisible 49. The two squares in the top row (besides the upper left) can be $(1,1),(1,7),(1,49),(7,7),(7,49)$, or $(49,49)$ (in terms of the highest power of 7 dividing the square). The same is true, independently, for the two blank squares on the left column, for a total of $6^{2}=36$ possibilities in this case.
If 1 is the highest power of 7 dividing the middle square, there are also 36 possibilities by a similar argument.
If 7 is the highest power of 7 dividing the middle square, there are 8 possibilities for the upper right three squares. Thus there are 64 possibilities in this case.
Thus there are a total of 136 options for the divisibility of each number by 7 and $7^{2}$, and 18 options for the divisibility of the numbers by 41 . Since each number divides 2009 , this uniquely determines the numbers, and so there are a total of $18 \cdot 136=2448$ possibilities.
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Given a rearrangement of the numbers from 1 to $n$, each pair of consecutive elements $a$ and $b$ of the sequence can be either increasing (if $a<b$ ) or decreasing (if $b<a$ ). How many rearrangements of the numbers from 1 to $n$ have exactly two increasing pairs of consecutive elements?
|
$3^{n}-(n+1) \cdot 2^{n}+n(n+1) / 2$ or equivalent
|
3^{n}-(n+1) \cdot 2^{n}+n(n+1) / 2
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Given a rearrangement of the numbers from 1 to $n$, each pair of consecutive elements $a$ and $b$ of the sequence can be either increasing (if $a<b$ ) or decreasing (if $b<a$ ). How many rearrangements of the numbers from 1 to $n$ have exactly two increasing pairs of consecutive elements?
|
$3^{n}-(n+1) \cdot 2^{n}+n(n+1) / 2$ or equivalent
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Given a rearrangement of the numbers from 1 to $n$, each pair of consecutive elements $a$ and $b$ of the sequence can be either increasing (if $a<b$ ) or decreasing (if $b<a$ ). How many rearrangements of the numbers from 1 to $n$ have exactly two increasing pairs of consecutive elements?
|
Notice that each such permutation consists of 3 disjoint subsets of $\{1, \ldots, n\}$ whose union is $\{1, \ldots, n\}$, each arranged in decreasing order. For instance, if $n=6$, in the permutation 415326 (which has the two increasing pairs 15 and 26), the three sets are $\{4,1\},\{5,3,2\}$, and 6 . There are $3^{n}$ ways to choose which of the first, second, or third set each element is in. However, we have overcounted: some choices of these subsets result in permutations with 1 or 0 increasing pairs, such as $\{6,5,4\},\{3,2\},\{1\}$.
Thus, we must subtract the number of ordered partitions of $\{1,2, \ldots, n\}$ into 3 subsets for which the minimum value of the first is not less than the maximum of the second, or the minimum value of the second is not less than the maximum of the third.
We first prove that the number of permutations having exactly one increasing consecutive pair of elements is $2^{n}-(n+1)$. To do so, note that there are $2^{n}$ ways to choose which elements occur before the increasing pair, and upon choosing this set we must arrange them in decreasing order, followed by the remaining elements arranged in decreasing order. The resulting permutation will have either one increasing pair or none. There are exactly $n+1$ subsets for which the resulting permutation has none, namely, $\},\{n\},\{n, n-1\},\{n, n-1, n-2\}$, etc. Thus the total number of permutations having one increasing pair is $2^{n}-(n+1)$ as desired.
We now count the partitions of $\{1,2, \ldots, n\}$ whose associated permutation has exactly one increasing pair. For each of the $2^{n}-(n+1)$ permutations $p$ having exacly one increasing pair, there are $n+1$ partitions of $\{1,2, \ldots, n\}$ into 3 subsets whose associated permutation is $p$. This is because there are $n+1$ ways to choose the "breaking point" to split one of the subsets into two. Thus there are a total of $(n+1)\left(2^{n}-(n+1)\right)$ partitions whose associaated permutation has exactly one increasing pair.
Finally, we must count the number of partitions whose associated permutation is $n, n-1, \ldots, 3,2,1$, i.e. has no increasing pair. There are $\frac{n+2}{2}$ ways of placing two barriers between these elements to split the numbers into three subsets, and so there are $\frac{n+2}{2}$ such partitions of $\{1,2, \ldots, n\}$ into three subsets.
Thus, subtracting off the partitions we did not want to count, the answer is $3^{n}-(n+1)\left(2^{n}-(n+\right.$ 1) $)-\binom{n+2}{2}=3^{n}-(n+1) \cdot 2^{n}+n(n+1) / 2$.
|
3^{n}-(n+1) \cdot 2^{n}+n(n+1) / 2
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Given a rearrangement of the numbers from 1 to $n$, each pair of consecutive elements $a$ and $b$ of the sequence can be either increasing (if $a<b$ ) or decreasing (if $b<a$ ). How many rearrangements of the numbers from 1 to $n$ have exactly two increasing pairs of consecutive elements?
|
Notice that each such permutation consists of 3 disjoint subsets of $\{1, \ldots, n\}$ whose union is $\{1, \ldots, n\}$, each arranged in decreasing order. For instance, if $n=6$, in the permutation 415326 (which has the two increasing pairs 15 and 26), the three sets are $\{4,1\},\{5,3,2\}$, and 6 . There are $3^{n}$ ways to choose which of the first, second, or third set each element is in. However, we have overcounted: some choices of these subsets result in permutations with 1 or 0 increasing pairs, such as $\{6,5,4\},\{3,2\},\{1\}$.
Thus, we must subtract the number of ordered partitions of $\{1,2, \ldots, n\}$ into 3 subsets for which the minimum value of the first is not less than the maximum of the second, or the minimum value of the second is not less than the maximum of the third.
We first prove that the number of permutations having exactly one increasing consecutive pair of elements is $2^{n}-(n+1)$. To do so, note that there are $2^{n}$ ways to choose which elements occur before the increasing pair, and upon choosing this set we must arrange them in decreasing order, followed by the remaining elements arranged in decreasing order. The resulting permutation will have either one increasing pair or none. There are exactly $n+1$ subsets for which the resulting permutation has none, namely, $\},\{n\},\{n, n-1\},\{n, n-1, n-2\}$, etc. Thus the total number of permutations having one increasing pair is $2^{n}-(n+1)$ as desired.
We now count the partitions of $\{1,2, \ldots, n\}$ whose associated permutation has exactly one increasing pair. For each of the $2^{n}-(n+1)$ permutations $p$ having exacly one increasing pair, there are $n+1$ partitions of $\{1,2, \ldots, n\}$ into 3 subsets whose associated permutation is $p$. This is because there are $n+1$ ways to choose the "breaking point" to split one of the subsets into two. Thus there are a total of $(n+1)\left(2^{n}-(n+1)\right)$ partitions whose associaated permutation has exactly one increasing pair.
Finally, we must count the number of partitions whose associated permutation is $n, n-1, \ldots, 3,2,1$, i.e. has no increasing pair. There are $\frac{n+2}{2}$ ways of placing two barriers between these elements to split the numbers into three subsets, and so there are $\frac{n+2}{2}$ such partitions of $\{1,2, \ldots, n\}$ into three subsets.
Thus, subtracting off the partitions we did not want to count, the answer is $3^{n}-(n+1)\left(2^{n}-(n+\right.$ 1) $)-\binom{n+2}{2}=3^{n}-(n+1) \cdot 2^{n}+n(n+1) / 2$.
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
If $a$ and $b$ are positive integers such that $a^{2}-b^{4}=2009$, find $a+b$.
|
47
|
47
|
Yes
|
Problem not solved
|
math-word-problem
|
Number Theory
|
If $a$ and $b$ are positive integers such that $a^{2}-b^{4}=2009$, find $a+b$.
|
47
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [2]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
If $a$ and $b$ are positive integers such that $a^{2}-b^{4}=2009$, find $a+b$.
|
We can factor the equation as $\left(a-b^{2}\right)\left(a+b^{2}\right)=41 \cdot 49$, from which it is evident that $a=45$ and $b=2$ is a possible solution. By examining the factors of 2009 , one can see that there are no other solutions.
|
47
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
If $a$ and $b$ are positive integers such that $a^{2}-b^{4}=2009$, find $a+b$.
|
We can factor the equation as $\left(a-b^{2}\right)\left(a+b^{2}\right)=41 \cdot 49$, from which it is evident that $a=45$ and $b=2$ is a possible solution. By examining the factors of 2009 , one can see that there are no other solutions.
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [2]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Suppose $N$ is a 6-digit number having base-10 representation $\underline{a} \underline{b} \underline{c} \underline{d} \underline{e} \underline{f}$. If $N$ is $6 / 7$ of the number having base-10 representation $\underline{d} \underline{e} \underline{f} \underline{a} \underline{b} \underline{c}$, find $N$.
|
461538
|
461538
|
Yes
|
Incomplete
|
math-word-problem
|
Number Theory
|
Suppose $N$ is a 6-digit number having base-10 representation $\underline{a} \underline{b} \underline{c} \underline{d} \underline{e} \underline{f}$. If $N$ is $6 / 7$ of the number having base-10 representation $\underline{d} \underline{e} \underline{f} \underline{a} \underline{b} \underline{c}$, find $N$.
|
461538
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [2]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Suppose $N$ is a 6-digit number having base-10 representation $\underline{a} \underline{b} \underline{c} \underline{d} \underline{e} \underline{f}$. If $N$ is $6 / 7$ of the number having base-10 representation $\underline{d} \underline{e} \underline{f} \underline{a} \underline{b} \underline{c}$, find $N$.
|
We have $7(a b c d e f)_{10}=6(\text { defabc })_{10}$, so $699400 a+69940 b+6994 c=599300 d+59930 e+$ $5993 f$. We can factor this equation as $6994(100 a+10 b+c)=5993(100 d+10 e+f)$, which yields $538(a b c)_{10}=461(d e f)_{10}$. Since $\operatorname{gcd}(538,461)=1$, we must have $(a b c)_{10}=461$ and $(d e f)_{10}=538$.
|
461538
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Suppose $N$ is a 6-digit number having base-10 representation $\underline{a} \underline{b} \underline{c} \underline{d} \underline{e} \underline{f}$. If $N$ is $6 / 7$ of the number having base-10 representation $\underline{d} \underline{e} \underline{f} \underline{a} \underline{b} \underline{c}$, find $N$.
|
We have $7(a b c d e f)_{10}=6(\text { defabc })_{10}$, so $699400 a+69940 b+6994 c=599300 d+59930 e+$ $5993 f$. We can factor this equation as $6994(100 a+10 b+c)=5993(100 d+10 e+f)$, which yields $538(a b c)_{10}=461(d e f)_{10}$. Since $\operatorname{gcd}(538,461)=1$, we must have $(a b c)_{10}=461$ and $(d e f)_{10}=538$.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [2]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below, so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of the resulting trapezoid.
|
$55 / 2$
|
\frac{55}{2}
|
Incomplete
|
Incomplete
|
math-word-problem
|
Geometry
|
A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below, so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of the resulting trapezoid.
|
$55 / 2$
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below, so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of the resulting trapezoid.
|
Drawing the perpendiculars from the point of intersection of the corners to the bases of the trapezoid, we see that we have similar $3-4-5$ right triangles, and we can calculate that the length of the smaller base is 3 . Thus the area of the trapezoid is $\frac{8+3}{2} \cdot 5=55 / 2$.
|
\frac{55}{2}
|
Incomplete
|
Yes
|
math-word-problem
|
Geometry
|
A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below, so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of the resulting trapezoid.
|
Drawing the perpendiculars from the point of intersection of the corners to the bases of the trapezoid, we see that we have similar $3-4-5$ right triangles, and we can calculate that the length of the smaller base is 3 . Thus the area of the trapezoid is $\frac{8+3}{2} \cdot 5=55 / 2$.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
If $\tan x+\tan y=4$ and $\cot x+\cot y=5$, compute $\tan (x+y)$.
|
20
|
20
|
Yes
|
Problem not solved
|
math-word-problem
|
Algebra
|
If $\tan x+\tan y=4$ and $\cot x+\cot y=5$, compute $\tan (x+y)$.
|
20
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
If $\tan x+\tan y=4$ and $\cot x+\cot y=5$, compute $\tan (x+y)$.
|
We have $\cot x+\cot y=\frac{\tan x+\tan y}{\tan x \tan y}$, so $\tan x \tan y=\frac{4}{5}$. Thus, by the tan addition formula, $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=20$.
|
20
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
If $\tan x+\tan y=4$ and $\cot x+\cot y=5$, compute $\tan (x+y)$.
|
We have $\cot x+\cot y=\frac{\tan x+\tan y}{\tan x \tan y}$, so $\tan x \tan y=\frac{4}{5}$. Thus, by the tan addition formula, $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=20$.
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. What is the expected number of cards that will be strictly between the two jokers?
|
$52 / 3$
|
\frac{52}{3}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. What is the expected number of cards that will be strictly between the two jokers?
|
$52 / 3$
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n5. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. What is the expected number of cards that will be strictly between the two jokers?
|
Each card has an equal likelihood of being either on top of the jokers, in between them, or below the jokers. Thus, on average, $1 / 3$ of them will land between the two jokers.
|
\frac{52}{3}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. What is the expected number of cards that will be strictly between the two jokers?
|
Each card has an equal likelihood of being either on top of the jokers, in between them, or below the jokers. Thus, on average, $1 / 3$ of them will land between the two jokers.
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n5. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacent to the vertex of the chosen corner. What is the height of the cube when the freshly-cut face is placed on a table?
|
$2 \sqrt{3} / 3$
|
\frac{2 \sqrt{3}}{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacent to the vertex of the chosen corner. What is the height of the cube when the freshly-cut face is placed on a table?
|
$2 \sqrt{3} / 3$
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n6. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacent to the vertex of the chosen corner. What is the height of the cube when the freshly-cut face is placed on a table?
|
The major diagonal has a length of $\sqrt{3}$. The volume of the pyramid is $1 / 6$, and so its height $h$ satisfies $\frac{1}{3} \cdot h \cdot \frac{\sqrt{3}}{4}(\sqrt{2})^{2}=1 / 6$ since the freshly cut face is an equilateral triangle of side length $\sqrt{2}$. Thus $h=\sqrt{3} / 3$, and the answer follows.
|
\sqrt{3} / 3
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacent to the vertex of the chosen corner. What is the height of the cube when the freshly-cut face is placed on a table?
|
The major diagonal has a length of $\sqrt{3}$. The volume of the pyramid is $1 / 6$, and so its height $h$ satisfies $\frac{1}{3} \cdot h \cdot \frac{\sqrt{3}}{4}(\sqrt{2})^{2}=1 / 6$ since the freshly cut face is an equilateral triangle of side length $\sqrt{2}$. Thus $h=\sqrt{3} / 3$, and the answer follows.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n6. [4]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $s(n)$ denote the number of 1's in the binary representation of $n$. Compute
$$
\frac{1}{255} \sum_{0 \leq n<16} 2^{n}(-1)^{s(n)}
$$
|
45
|
45
|
Yes
|
Problem not solved
|
math-word-problem
|
Number Theory
|
Let $s(n)$ denote the number of 1's in the binary representation of $n$. Compute
$$
\frac{1}{255} \sum_{0 \leq n<16} 2^{n}(-1)^{s(n)}
$$
|
45
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\n## Answer: ",
"tier": "T4",
"year": "2009"
}
|
Let $s(n)$ denote the number of 1's in the binary representation of $n$. Compute
$$
\frac{1}{255} \sum_{0 \leq n<16} 2^{n}(-1)^{s(n)}
$$
|
Notice that if $n<8,(-1)^{s(n)}=(-1) \cdot(-1)^{s(n+8)}$ so the sum becomes $\frac{1}{255}\left(1-2^{8}\right) \sum_{0 \leq n<8} 2^{n}(-1)^{s(n)}=$ 45.
|
45
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $s(n)$ denote the number of 1's in the binary representation of $n$. Compute
$$
\frac{1}{255} \sum_{0 \leq n<16} 2^{n}(-1)^{s(n)}
$$
|
Notice that if $n<8,(-1)^{s(n)}=(-1) \cdot(-1)^{s(n+8)}$ so the sum becomes $\frac{1}{255}\left(1-2^{8}\right) \sum_{0 \leq n<8} 2^{n}(-1)^{s(n)}=$ 45.
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $a, b$, and $c$ be the 3 roots of $x^{3}-x+1=0$. Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$.
|
$\quad-2$
|
-2
|
Yes
|
Incomplete
|
math-word-problem
|
Algebra
|
Let $a, b$, and $c$ be the 3 roots of $x^{3}-x+1=0$. Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$.
|
$\quad-2$
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Let $a, b$, and $c$ be the 3 roots of $x^{3}-x+1=0$. Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$.
|
We can substitute $x=y-1$ to obtain a polynomial having roots $a+1, b+1, c+1$, namely, $(y-1)^{3}-(y-1)+1=y^{3}-3 y^{2}+2 y+1$. The sum of the reciprocals of the roots of this polynomial is, by Viete's formulas, $\frac{2}{-1}=-2$.
|
-2
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a, b$, and $c$ be the 3 roots of $x^{3}-x+1=0$. Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$.
|
We can substitute $x=y-1$ to obtain a polynomial having roots $a+1, b+1, c+1$, namely, $(y-1)^{3}-(y-1)+1=y^{3}-3 y^{2}+2 y+1$. The sum of the reciprocals of the roots of this polynomial is, by Viete's formulas, $\frac{2}{-1}=-2$.
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [5]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ satisfy $f(f(x))=f(x)$ for all $x \in\{1,2,3,4,5\}$ ?
|
196
|
196
|
Yes
|
Incomplete
|
math-word-problem
|
Combinatorics
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ satisfy $f(f(x))=f(x)$ for all $x \in\{1,2,3,4,5\}$ ?
|
196
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [6]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ satisfy $f(f(x))=f(x)$ for all $x \in\{1,2,3,4,5\}$ ?
|
A fixed point of a function $f$ is an element $a$ such that $f(a)=a$. The condition is equivalent to the property that $f$ maps every number to a fixed point. Counting by the number of fixed points of $f$, the total number of such functions is
$$
\begin{aligned}
\sum_{k=1}^{5}\binom{5}{k} k^{5-k} & =1 \cdot\left(0^{5}+5^{0}\right)+5 \cdot\left(1^{4}+4^{1}\right)+10 \cdot\left(2^{3}+3^{2}\right) \\
& =1+25+10 \cdot 17 \\
& =196
\end{aligned}
$$
|
196
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ satisfy $f(f(x))=f(x)$ for all $x \in\{1,2,3,4,5\}$ ?
|
A fixed point of a function $f$ is an element $a$ such that $f(a)=a$. The condition is equivalent to the property that $f$ maps every number to a fixed point. Counting by the number of fixed points of $f$, the total number of such functions is
$$
\begin{aligned}
\sum_{k=1}^{5}\binom{5}{k} k^{5-k} & =1 \cdot\left(0^{5}+5^{0}\right)+5 \cdot\left(1^{4}+4^{1}\right)+10 \cdot\left(2^{3}+3^{2}\right) \\
& =1+25+10 \cdot 17 \\
& =196
\end{aligned}
$$
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [6]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
A kite is a quadrilateral whose diagonals are perpendicular. Let kite $A B C D$ be such that $\angle B=$ $\angle D=90^{\circ}$. Let $M$ and $N$ be the points of tangency of the incircle of $A B C D$ to $A B$ and $B C$ respectively. Let $\omega$ be the circle centered at $C$ and tangent to $A B$ and $A D$. Construct another kite $A B^{\prime} C^{\prime} D^{\prime}$ that is similar to $A B C D$ and whose incircle is $\omega$. Let $N^{\prime}$ be the point of tangency of $B^{\prime} C^{\prime}$ to $\omega$. If $M N^{\prime} \| A C$, then what is the ratio of $A B: B C$ ?
|
$\frac{1+\sqrt{5}}{2}$
|
\frac{1+\sqrt{5}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A kite is a quadrilateral whose diagonals are perpendicular. Let kite $A B C D$ be such that $\angle B=$ $\angle D=90^{\circ}$. Let $M$ and $N$ be the points of tangency of the incircle of $A B C D$ to $A B$ and $B C$ respectively. Let $\omega$ be the circle centered at $C$ and tangent to $A B$ and $A D$. Construct another kite $A B^{\prime} C^{\prime} D^{\prime}$ that is similar to $A B C D$ and whose incircle is $\omega$. Let $N^{\prime}$ be the point of tangency of $B^{\prime} C^{\prime}$ to $\omega$. If $M N^{\prime} \| A C$, then what is the ratio of $A B: B C$ ?
|
$\frac{1+\sqrt{5}}{2}$
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n10. [6]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
A kite is a quadrilateral whose diagonals are perpendicular. Let kite $A B C D$ be such that $\angle B=$ $\angle D=90^{\circ}$. Let $M$ and $N$ be the points of tangency of the incircle of $A B C D$ to $A B$ and $B C$ respectively. Let $\omega$ be the circle centered at $C$ and tangent to $A B$ and $A D$. Construct another kite $A B^{\prime} C^{\prime} D^{\prime}$ that is similar to $A B C D$ and whose incircle is $\omega$. Let $N^{\prime}$ be the point of tangency of $B^{\prime} C^{\prime}$ to $\omega$. If $M N^{\prime} \| A C$, then what is the ratio of $A B: B C$ ?
|
Let's focus on the right triangle $A B C$ and the semicircle inscribed in it since the situation is symmetric about $A C$. First we find the radius $a$ of circle $O$. Let $A B=x$ and $B C=y$. Drawing the radii $O M$ and $O N$, we see that $A M=x-a$ and $\triangle A M O \sim \triangle A B C$. In other words,
$$
\begin{aligned}
\frac{A M}{M O} & =\frac{A B}{B C} \\
\frac{x-a}{a} & =\frac{x}{y} \\
a & =\frac{x y}{x+y}
\end{aligned}
$$
Now we notice that the situation is homothetic about $A$. In particular,
$$
\triangle A M O \sim \triangle O N C \sim \triangle C N^{\prime} C^{\prime}
$$
Also, $C B$ and $C N^{\prime}$ are both radii of circle $C$. Thus, when $M N^{\prime} \| A C^{\prime}$, we have
$$
\begin{aligned}
A M & =C N^{\prime}=C B \\
x-a & =y \\
a=\frac{x y}{x+y} & =x-y \\
x^{2}-x y-y^{2} & =0 \\
x & =\frac{y}{2} \pm \sqrt{\frac{y^{2}}{4}+y^{2}} \\
\frac{A B}{B C}=\frac{x}{y} & =\frac{1+\sqrt{5}}{2} .
\end{aligned}
$$
|
\frac{1+\sqrt{5}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A kite is a quadrilateral whose diagonals are perpendicular. Let kite $A B C D$ be such that $\angle B=$ $\angle D=90^{\circ}$. Let $M$ and $N$ be the points of tangency of the incircle of $A B C D$ to $A B$ and $B C$ respectively. Let $\omega$ be the circle centered at $C$ and tangent to $A B$ and $A D$. Construct another kite $A B^{\prime} C^{\prime} D^{\prime}$ that is similar to $A B C D$ and whose incircle is $\omega$. Let $N^{\prime}$ be the point of tangency of $B^{\prime} C^{\prime}$ to $\omega$. If $M N^{\prime} \| A C$, then what is the ratio of $A B: B C$ ?
|
Let's focus on the right triangle $A B C$ and the semicircle inscribed in it since the situation is symmetric about $A C$. First we find the radius $a$ of circle $O$. Let $A B=x$ and $B C=y$. Drawing the radii $O M$ and $O N$, we see that $A M=x-a$ and $\triangle A M O \sim \triangle A B C$. In other words,
$$
\begin{aligned}
\frac{A M}{M O} & =\frac{A B}{B C} \\
\frac{x-a}{a} & =\frac{x}{y} \\
a & =\frac{x y}{x+y}
\end{aligned}
$$
Now we notice that the situation is homothetic about $A$. In particular,
$$
\triangle A M O \sim \triangle O N C \sim \triangle C N^{\prime} C^{\prime}
$$
Also, $C B$ and $C N^{\prime}$ are both radii of circle $C$. Thus, when $M N^{\prime} \| A C^{\prime}$, we have
$$
\begin{aligned}
A M & =C N^{\prime}=C B \\
x-a & =y \\
a=\frac{x y}{x+y} & =x-y \\
x^{2}-x y-y^{2} & =0 \\
x & =\frac{y}{2} \pm \sqrt{\frac{y^{2}}{4}+y^{2}} \\
\frac{A B}{B C}=\frac{x}{y} & =\frac{1+\sqrt{5}}{2} .
\end{aligned}
$$
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n10. [6]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute values of the numbers in the sequence are nonincreasing?
|
128
|
128
|
Yes
|
Problem not solved
|
math-word-problem
|
Combinatorics
|
How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute values of the numbers in the sequence are nonincreasing?
|
128
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [2]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute values of the numbers in the sequence are nonincreasing?
|
Each of the pairs $a,-a$ must occur in increasing order of $a$ for $a=1, \ldots, 7$, but $a$ can either occur before or after $-a$, for a total of $2^{7}=128$ possible sequences.
|
128
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute values of the numbers in the sequence are nonincreasing?
|
Each of the pairs $a,-a$ must occur in increasing order of $a$ for $a=1, \ldots, 7$, but $a$ can either occur before or after $-a$, for a total of $2^{7}=128$ possible sequences.
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [2]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
How many ways can you tile the white squares of following $2 \times 24$ grid with dominoes? (A domino covers two adjacent squares, and a tiling is a non-overlapping arrangement of dominoes that covers every white square and does not intersect any black square.)
|
27
|
27
|
Incomplete
|
Problem not solved
|
math-word-problem
|
Combinatorics
|
How many ways can you tile the white squares of following $2 \times 24$ grid with dominoes? (A domino covers two adjacent squares, and a tiling is a non-overlapping arrangement of dominoes that covers every white square and does not intersect any black square.)
|
27
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [2]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
How many ways can you tile the white squares of following $2 \times 24$ grid with dominoes? (A domino covers two adjacent squares, and a tiling is a non-overlapping arrangement of dominoes that covers every white square and does not intersect any black square.)
|
Divide the rectangle into three $3 \times 8$ sub-rectangles. It is easy to count that there are 3 ways of tiling each of these sub-rectangles independently, for a total of $3^{3}=27$ possibilities.
|
27
|
Incomplete
|
Yes
|
math-word-problem
|
Combinatorics
|
How many ways can you tile the white squares of following $2 \times 24$ grid with dominoes? (A domino covers two adjacent squares, and a tiling is a non-overlapping arrangement of dominoes that covers every white square and does not intersect any black square.)
|
Divide the rectangle into three $3 \times 8$ sub-rectangles. It is easy to count that there are 3 ways of tiling each of these sub-rectangles independently, for a total of $3^{3}=27$ possibilities.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [2]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
Let $S$ be the sum of all the real coefficients of the expansion of $(1+i x)^{2009}$. What is $\log _{2}(S)$ ?
|
1004
|
1004
|
Yes
|
Incomplete
|
math-word-problem
|
Algebra
|
Let $S$ be the sum of all the real coefficients of the expansion of $(1+i x)^{2009}$. What is $\log _{2}(S)$ ?
|
1004
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
Let $S$ be the sum of all the real coefficients of the expansion of $(1+i x)^{2009}$. What is $\log _{2}(S)$ ?
|
The sum of all the coefficients is $(1+i)^{2009}$, and the sum of the real coefficients is the real part of this, which is $\frac{1}{2}\left((1+i)^{2009}+(1-i)^{2009}\right)=2^{1004}$. Thus $\log _{2}(S)=1004$.
|
1004
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $S$ be the sum of all the real coefficients of the expansion of $(1+i x)^{2009}$. What is $\log _{2}(S)$ ?
|
The sum of all the coefficients is $(1+i)^{2009}$, and the sum of the real coefficients is the real part of this, which is $\frac{1}{2}\left((1+i)^{2009}+(1-i)^{2009}\right)=2^{1004}$. Thus $\log _{2}(S)=1004$.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
A torus (donut) having inner radius 2 and outer radius 4 sits on a flat table. What is the radius of the largest spherical ball that can be placed on top of the center torus so that the ball still touches the horizontal plane? (If the $x-y$ plane is the table, the torus is formed by revolving the circle in the $x-z$ plane centered at $(3,0,1)$ with radius 1 about the $z$ axis. The spherical ball has its center on the $z$-axis and rests on either the table or the donut.)
|
$9 / 4$
|
\frac{9}{4}
|
Yes
|
Incomplete
|
math-word-problem
|
Geometry
|
A torus (donut) having inner radius 2 and outer radius 4 sits on a flat table. What is the radius of the largest spherical ball that can be placed on top of the center torus so that the ball still touches the horizontal plane? (If the $x-y$ plane is the table, the torus is formed by revolving the circle in the $x-z$ plane centered at $(3,0,1)$ with radius 1 about the $z$ axis. The spherical ball has its center on the $z$-axis and rests on either the table or the donut.)
|
$9 / 4$
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2009"
}
|
A torus (donut) having inner radius 2 and outer radius 4 sits on a flat table. What is the radius of the largest spherical ball that can be placed on top of the center torus so that the ball still touches the horizontal plane? (If the $x-y$ plane is the table, the torus is formed by revolving the circle in the $x-z$ plane centered at $(3,0,1)$ with radius 1 about the $z$ axis. The spherical ball has its center on the $z$-axis and rests on either the table or the donut.)
|
Let $r$ be the radius of the sphere. One can see that it satisfies $(r+1)^{2}=(r-1)^{2}+3^{2}$ by the Pythagorean Theorem, so $r=9 / 4$.
|
\frac{9}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A torus (donut) having inner radius 2 and outer radius 4 sits on a flat table. What is the radius of the largest spherical ball that can be placed on top of the center torus so that the ball still touches the horizontal plane? (If the $x-y$ plane is the table, the torus is formed by revolving the circle in the $x-z$ plane centered at $(3,0,1)$ with radius 1 about the $z$ axis. The spherical ball has its center on the $z$-axis and rests on either the table or the donut.)
|
Let $r$ be the radius of the sphere. One can see that it satisfies $(r+1)^{2}=(r-1)^{2}+3^{2}$ by the Pythagorean Theorem, so $r=9 / 4$.
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [3]",
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2009"
}
|
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