year stringdate 1961-01-01 00:00:00 2025-01-01 00:00:00 ⌀ | tier stringclasses 5
values | problem_label stringclasses 119
values | problem_type stringclasses 13
values | exam stringclasses 28
values | problem stringlengths 87 2.77k | solution stringlengths 834 13k | metadata dict | problem_tokens int64 50 903 | solution_tokens int64 500 3.93k |
|---|---|---|---|---|---|---|---|---|---|
2013 | T4 | 33 | null | HMMT | On each side of a 6 by 8 rectangle, construct an equilateral triangle with that side as one edge such that the interior of the triangle intersects the interior of the rectangle. What is the total area of all regions that are contained in exactly 3 of the 4 equilateral triangles? | $\frac{96 \sqrt{3}-154}{\sqrt{3}}$ OR $\frac{288-154 \sqrt{3}}{3}$ OR $96-\frac{154}{\sqrt{3}}$ OR $96-\frac{154 \sqrt{3}}{3}$ Let the rectangle be $A B C D$ with $A B=8$ and $B C=6$. Let the four equilateral triangles be $A B P_{1}, B C P_{2}, C D P_{3}$, and $D A P_{4}$ (for convenience, call them the $P_{1^{-}}, P_{... | {
"problem_match": "\n33. [17]",
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 60 | 1,029 |
2013 | T4 | 34 | null | HMMT | Find the number of positive integers less than 1000000 that are divisible by some perfect cube greater than 1. Your score will be $\max \left\{0,\left\lfloor 20-200\left|1-\frac{k}{S}\right|\right\rfloor\right\}$, where $k$ is your answer and $S$ is the actual answer. | 168089 Using the following code, we get the answer (denoted by the variable ans):
ans $=0$
for $n$ in xrange $(1,1000000)$ :
```
divisible_by_cube = True
for i in xrange(2,101):
if n%(i*i*i)==0:
divisible_by_cube = False
break
if divisible_by_cube: ans = ans + 1
print ans
```
This gives the output
168089
Altern... | {
"problem_match": "\n34. [20]",
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 88 | 1,178 |
2013 | T4 | 4 | null | HMMT | There are 2 runners on the perimeter of a regular hexagon, initially located at adjacent vertices. Every second, each of the runners independently moves either one vertex to the left, with probability $\frac{1}{2}$, or one vertex to the right, also with probability $\frac{1}{2}$. Find the probability that after a 2013 ... | $\quad \frac{2}{3}+\frac{1}{3}\left(\frac{1}{4}\right)^{2013}$ OR $\frac{2^{4027}+1}{3 \cdot 2^{4026}}$ OR $\frac{2}{3}+\frac{1}{3}\left(\frac{1}{2}\right)^{4026}$ OR $\frac{2}{3}+\frac{1}{3}\left(\frac{1}{64}\right)^{671}$ Label the runners $A$ and $B$ and arbitrarily fix an orientation of the hexagon. Let $p_{t}(i)$ ... | {
"problem_match": "\n4. [7]",
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-thm-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 102 | 546 |
2013 | T4 | 10 | null | HMMT | Find the sum of all positive integers $n$ such that there exists an integer $b$ with $|b| \neq 4$ such that the base -4 representation of $n$ is the same as the base $b$ representation of $n$. | 1026 All 1 digit numbers, $0,1,2,3$, are solutions when, say, $b=5$. (Of course, $d \in$ $\{0,1,2,3\}$ works for any base $b$ of absolute value greater than $d$ but not equal to 4.)
Consider now positive integers $n=\left(a_{d} \ldots a_{1} a_{0}\right)_{4}$ with more than one digit, so $d \geq 1, a_{d} \neq 0$, and $0... | {
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-thm-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 55 | 2,010 |
2014 | T4 | 7 | Algebra | HMMT | Find the largest real number $c$ such that
$$
\sum_{i=1}^{101} x_{i}^{2} \geq c M^{2}
$$
whenever $x_{1}, \ldots, x_{101}$ are real numbers such that $x_{1}+\cdots+x_{101}=0$ and $M$ is the median of $x_{1}, \ldots, x_{101}$. | $\frac{5151}{50}$ OR 103.02 OR $103 \frac{1}{50}$ Suppose without loss of generality that $x_{1} \leq \cdots \leq x_{101}$ and $M=x_{51} \geq 0$.
Note that $f(t)=t^{2}$ is a convex function over the reals, so we may "smooth" to the case $x_{1}=\cdots=$ $x_{5} 0 \leq x_{51}=\cdots=x_{101}$ (the $x_{51}=\cdots$ is why we... | {
"problem_match": "\n7. ",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 104 | 604 |
2014 | T4 | 10 | Algebra | HMMT | For an integer $n$, let $f_{9}(n)$ denote the number of positive integers $d \leq 9$ dividing $n$. Suppose that $m$ is a positive integer and $b_{1}, b_{2}, \ldots, b_{m}$ are real numbers such that $f_{9}(n)=\sum_{j=1}^{m} b_{j} f_{9}(n-j)$ for all $n>m$. Find the smallest possible value of $m$. | 28 Let $M=9$. Consider the generating function
$$
F(x)=\sum_{n \geq 1} f_{M}(n) x^{n}=\sum_{d=1}^{M} \sum_{k \geq 1} x^{d k}=\sum_{d=1}^{M} \frac{x^{d}}{1-x^{d}}
$$
Observe that $f_{M}(n)=f_{M}(n+M!)$ for all $n \geq 1$ (in fact, all $n \leq 0$ as well). Thus $f_{M}(n)$ satisfies a degree $m$ linear recurrence if and... | {
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 107 | 975 |
2014 | T4 | 4 | Combinatorics | HMMT | Find the number of triples of sets $(A, B, C)$ such that:
(a) $A, B, C \subseteq\{1,2,3, \ldots, 8\}$.
(b) $|A \cap B|=|B \cap C|=|C \cap A|=2$.
(c) $|A|=|B|=|C|=4$.
Here, $|S|$ denotes the number of elements in the set $S$. | 45360 We consider the sets drawn in a Venn diagram.
Note that each element that is in at least one of the subsets lies in these seven possible spaces. We split by casework, with the cases based on $N=\left|R_{7}\right|=|A \cap B \cap C|$.
Case 1: $N=2$
Because we are given that $\left|R_{4}\right|+N=\left|R_{5}\right... | {
"problem_match": "\n4. ",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 100 | 601 |
2014 | T4 | 7 | Combinatorics | HMMT | Six distinguishable players are participating in a tennis tournament. Each player plays one match of tennis against every other player. There are no ties in this tournament- each tennis match results in a win for one player and a loss for the other. Suppose that whenever $A$ and $B$ are players in the tournament such t... | 2048 We first group the players by wins, so let $G_{1}$ be the set of all players with the most wins, $G_{2}$ be the set of all players with the second most wins, $\ldots, G_{n}$ be the set of all players with the least wins. By the condition in the problem, everyone in group $G_{i}$ must beat everyone in group $G_{j}$... | {
"problem_match": "\n7. ",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 112 | 811 |
2014 | T4 | 8 | Combinatorics | HMMT | The integers $1,2, \ldots, 64$ are written in the squares of a $8 \times 8$ chess board, such that for each $1 \leq i<64$, the numbers $i$ and $i+1$ are in squares that share an edge. What is the largest possible sum that can appear along one of the diagonals? | 432 Our answer is $26+52+54+56+58+60+62+64$.
One possible configuration:
WLOG, we seek to maximize the sum of the numbers on the main diagonal (top left to bottom right). If we color the squares in a checker-board pattern and use the fact that $a$ and $a+1$ lie on different colored squares, we notice that all numbers ... | {
"problem_match": "\n8. ",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 81 | 1,089 |
2014 | T4 | 9 | Combinatorics | HMMT | There is a heads up coin on every integer of the number line. Lucky is initially standing on the zero point of the number line facing in the positive direction. Lucky performs the following procedure: he looks at the coin (or lack thereof) underneath him, and then,
- If the coin is heads up, Lucky flips it to tails up... | 6098 We keep track of the following quantities: Let $N$ be the sum of $2^{k}$, where $k$ ranges over all nonnegative integers such that position $-1-k$ on the number line contains a tails-up coin. Let $M$ be the sum of $2^{k}$, where $k$ ranges over all nonnegative integers such that position $k$ contains a tails-up co... | {
"problem_match": "\n9. ",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 165 | 1,052 |
2014 | T4 | 10 | Combinatorics | HMMT | An up-right path from $(a, b) \in \mathbb{R}^{2}$ to $(c, d) \in \mathbb{R}^{2}$ is a finite sequence $\left(x_{1}, y_{1}\right), \ldots,\left(x_{k}, y_{k}\right)$ of points in $\mathbb{R}^{2}$ such that $(a, b)=\left(x_{1}, y_{1}\right),(c, d)=\left(x_{k}, y_{k}\right)$, and for each $1 \leq i<k$ we have that either $... | 1750 The number of up-right paths from $(0,0)$ to $(4,4)$ is $\binom{8}{4}$ because any such upright path is identical to a sequence of 4 U's and 4 R's, where $U$ corresponds to a step upwards and R corresponds to a step rightwards. Therefore, the total number of pairs of (possibly intersecting) up-right paths from $(0... | {
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 273 | 508 |
2014 | T4 | 21 | null | HMMT | Compute the number of ordered quintuples of nonnegative integers ( $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ ) such that $0 \leq a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \leq 7$ and 5 divides $2^{a_{1}}+2^{a_{2}}+2^{a_{3}}+2^{a_{4}}+2^{a_{5}}$. | 6528 Let $f(n)$ denote the number of $n$-tuples $\left(a_{1}, \ldots, a_{n}\right)$ such that $0 \leq a_{1}, \ldots, a_{n} \leq 7$ and $5 \mid 2^{a_{1}}+\ldots+2^{a_{n}}$. To compute $f(n+1)$ from $f(n)$, we note that given any $n$-tuple $\left(a_{1}, \ldots, a_{n}\right)$ such that $0 \leq a_{1}, \ldots, a_{n} \leq 7$... | {
"problem_match": "\n21. [14]",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 107 | 583 |
2014 | T4 | 23 | null | HMMT | Let $S=\{-100,-99,-98, \ldots, 99,100\}$. Choose a 50 -element subset $T$ of $S$ at random. Find the expected number of elements of the set $\{|x|: x \in T\}$. | $\square$ Let us solve a more generalized version of the problem: Let $S$ be a set with $2 n+1$ elements, and partition $S$ into sets $A_{0}, A_{1}, \ldots, A_{n}$ such that $\left|A_{0}\right|=1$ and $\left|A_{1}\right|=\left|A_{2}\right|=\cdots=\left|A_{n}\right|=2$. (In this problem, we have $A_{0}=\{0\}$ and $A_{k}... | {
"problem_match": "\n23. [14]",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 68 | 796 |
2014 | T4 | 27 | null | HMMT | Suppose that $\left(a_{1}, \ldots, a_{20}\right)$ and $\left(b_{1}, \ldots, b_{20}\right)$ are two sequences of integers such that the sequence $\left(a_{1}, \ldots, a_{20}, b_{1}, \ldots, b_{20}\right)$ contains each of the numbers $1, \ldots, 40$ exactly once. What is the maximum possible value of the sum
$$
\sum_{i... | 5530 Let $x_{k}$, for $1 \leq k \leq 40$, be the number of integers $i$ with $1 \leq i \leq 20$ such that $a_{i} \geq k$. Let $y_{k}$, for $1 \leq k \leq 40$, be the number of integers $j$ with $1 \leq j \leq 20$ such that $b_{j} \geq k$. It follows from the problem statement that $x_{k}+y_{k}$ is the number of element... | {
"problem_match": "\n27. [17]",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 146 | 651 |
2014 | T4 | 28 | null | HMMT | Let $f(n)$ and $g(n)$ be polynomials of degree 2014 such that $f(n)+(-1)^{n} g(n)=2^{n}$ for $n=$ $1,2, \ldots, 4030$. Find the coefficient of $x^{2014}$ in $g(x)$. | $\frac{3^{2014}}{2^{2014} \cdot 2014!}$ Define the polynomial functions $h_{1}$ and $h_{2}$ by $h_{1}(x)=f(2 x)+g(2 x)$ and $h_{2}(x)=f(2 x-1)-g(2 x-1)$. Then, the problem conditions tell us that $h_{1}(x)=2^{2 x}$ and $h_{2}(x)=2^{2 x-1}$ for $x=1,2, \ldots, 2015$.
By the Lagrange interpolation formula, the polynomia... | {
"problem_match": "\n28. [17]",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 76 | 582 |
2014 | T4 | 29 | null | HMMT | Natalie has a copy of the unit interval $[0,1]$ that is colored white. She also has a black marker, and she colors the interval in the following manner: at each step, she selects a value $x \in[0,1]$ uniformly at random, and
(a) If $x \leq \frac{1}{2}$ she colors the interval $\left[x, x+\frac{1}{2}\right]$ with her ma... | 5 The first choice always wipes out half the interval. So we calculate the expected value of the amount of time needed to wipe out the other half.
Solution 1 (non-calculus):
We assume the interval has $2 n$ points and we start with the last $n$ colored black. We let $f(k)$ be the expected value of the number of turns w... | {
"problem_match": "\n29. [20]",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 161 | 572 |
2014 | T4 | 32 | null | HMMT | Find all ordered pairs $(a, b)$ of complex numbers with $a^{2}+b^{2} \neq 0, a+\frac{10 b}{a^{2}+b^{2}}=5$, and $b+\frac{10 a}{a^{2}+b^{2}}=4$. | The first equation plus $i$ times the second yields $5+4 i=a+b i+\frac{10(b+a i)}{a^{2}+b^{2}}=a+b i-\frac{10 i}{a+b i}$, which is equivalent to $a+b i=\frac{(5 \pm 3)+4 i}{2}$ by the quadratic formula.
Similarly, the second equation plus $i$ times the first yields $4+5 i=b+a i-\frac{10 i}{b+a i}$, which is equivalent ... | {
"problem_match": "\n32. [20]",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"solution_match": "\nSolution 2. "
} | 71 | 508 |
2014 | T4 | 33 | null | HMMT | An up-right path from $(a, b) \in \mathbb{R}^{2}$ to $(c, d) \in \mathbb{R}^{2}$ is a finite sequence $\left(x_{1}, y_{1}\right), \ldots,\left(x_{k}, y_{k}\right)$ of points in $\mathbb{R}^{2}$ such that $(a, b)=\left(x_{1}, y_{1}\right),(c, d)=\left(x_{k}, y_{k}\right)$, and for each $1 \leq i<k$ we have that either $... | $0.2937156494680644 \ldots$ Note that any up-right path must pass through exactly one point of the form $(n,-n)$ (i.e. a point on the upper-left to lower-right diagonal), and the number of such paths is $\binom{800}{400-n}^{2}$ because there are $\binom{800}{400-n}$ up-right paths from $(-400,-400)$ to $(n,-n)$ and ano... | {
"problem_match": "\n33. [25]",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 333 | 630 |
2014 | T4 | 34 | null | HMMT | Consider a number line, with a lily pad placed at each integer point. A frog is standing at the lily pad at the point 0 on the number line, and wants to reach the lily pad at the point 2014 on the number line. If the frog stands at the point $n$ on the number line, it can jump directly to either point
$n+2$ or point $n... | 0.9102805441016536
First, we establish a rough upper bound for the probability $p$. Let $q$ be the probability that the frog can reach the lily pad at the point 2014 on the number line if it is allowed to jump from a point $n$ on the number line to the point $n+1$, in addition to the points $n+2$ and $n+3$. Clearly, $p... | {
"problem_match": "\n34. [25]",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 257 | 1,809 |
2014 | T4 | 1 | null | HMMT | Let $\omega$ be a circle, and let $A$ and $B$ be two points in its interior. Prove that there exists a circle passing through $A$ and $B$ that is contained in the interior of $\omega$. | N/A WLOG, suppose $O A \geq O B$. Let $\omega^{\prime}$ be the circle of radius $O A$ centered at $O$. We have that $B$ lies inside $\omega^{\prime}$. Thus, it is possible to scale $\omega^{\prime}$ down about the point $A$ to get a circle $\omega^{\prime \prime}$ passing through both $A$ and $B$. Since $\omega^{\prime... | {
"problem_match": "\n1. [10]",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-team-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 50 | 539 |
2014 | T4 | 2 | null | HMMT | Let $a_{1}, a_{2}, \ldots$ be an infinite sequence of integers such that $a_{i}$ divides $a_{i+1}$ for all $i \geq 1$, and let $b_{i}$ be the remainder when $a_{i}$ is divided by 210 . What is the maximal number of distinct terms in the sequence $b_{1}, b_{2}, \ldots$ ? | $\quad 127$ It is clear that the sequence $\left\{a_{i}\right\}$ will be a concatenation of sequences of the form $\left\{v_{i}\right\}_{i=1}^{N_{0}},\left\{w_{i} \cdot p_{1}\right\}_{i=1}^{N_{1}},\left\{x_{i} \cdot p_{1} p_{2}\right\}_{i=1}^{N_{2}},\left\{y_{i} \cdot p_{1} p_{2} p_{3}\right\}_{i=1}^{N_{3}}$, and $\lef... | {
"problem_match": "\n2. [15]",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-team-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 94 | 891 |
2014 | T4 | 9 | null | HMMT | For integers $m, n \geq 1$, let $A(n, m)$ be the number of sequences $\left(a_{1}, \cdots, a_{n m}\right)$ of integers satisfying the following two properties:
(a) Each integer $k$ with $1 \leq k \leq n$ occurs exactly $m$ times in the sequence $\left(a_{1}, \cdots, a_{n m}\right)$.
(b) If $i, j$, and $k$ are integers ... | We can also form a direct bijection to show $A(n, m)=A(m, n)$, as follows. Suppose that $a=\left(a_{1}, \ldots, a_{m n}\right)$ is a sequence satisfying properties 1 and 2 . We will define a sequence $f(a)=$ $\left(b_{1}, \ldots, b_{n m}\right)$ satisfying the same properties 1 and 2 , but with $m$ and $n$ switched.
Th... | {
"problem_match": "\n9. [35]",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-team-solutions.jsonl",
"solution_match": "\nSolution 2: "
} | 314 | 1,147 |
2014 | T4 | 10 | null | HMMT | Fix a positive real number $c>1$ and positive integer $n$. Initially, a blackboard contains the numbers $1, c, \ldots, c^{n-1}$. Every minute, Bob chooses two numbers $a, b$ on the board and replaces them with $c a+c^{2} b$. Prove that after $n-1$ minutes, the blackboard contains a single number no less than
$$
\left(... | N/A By a simple reverse induction, we can show that at any instant, any number on the board takes the form $\sum_{\alpha \in A} c^{r_{\alpha}} c^{\alpha}$ for a certain $A \subseteq\{0, \ldots, n-1\}$ for non-negative integer weights $r_{\alpha}$ satisfying $\sum_{\alpha \in A} \phi^{-r_{\alpha}}=1$, where these subset... | {
"problem_match": "\n10. [40]",
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-team-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 149 | 896 |
2014 | T4 | 1 | null | HMMT | Consider a regular $n$-gon with $n>3$, and call a line acceptable if it passes through the interior of this $n$-gon. Draw $m$ different acceptable lines, so that the $n$-gon is divided into several smaller polygons.
(a) Prove that there exists an $m$, depending only on $n$, such that any collection of $m$ acceptable li... | N/A We will prove that if $m \geq n-4$, then there is guaranteed to be a smaller polygon with 3 or 4 sides, while if $m \leq n-5$, there might not be a polygon with 3 or 4 sides. This will solve both parts of the problem.
Given a configuration of lines, let $P_{1}, \ldots, P_{k}$ be all of the resulting smaller polygon... | {
"problem_match": "\n1. [20]",
"resource_path": "HarvardMIT/segmented/en-174-tournaments-2014-hmic-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 126 | 642 |
2014 | T4 | 3 | null | HMMT | Fix positive integers $m$ and $n$. Suppose that $a_{1}, a_{2}, \ldots, a_{m}$ are reals, and that pairwise distinct vectors $v_{1}, \ldots, v_{m} \in \mathbb{R}^{n}$ satisfy
$$
\sum_{j \neq i} a_{j} \frac{v_{j}-v_{i}}{\left\|v_{j}-v_{i}\right\|^{3}}=0
$$
for $i=1,2, \ldots, m$.
Prove that
$$
\sum_{1 \leq i<j \leq m}... | N/A Since $v_{i} \cdot\left(v_{j}-v_{i}\right)+v_{j} \cdot\left(v_{i}-v_{j}\right)=-\left\|v_{j}-v_{i}\right\|^{2}$ for any $1 \leq i<j \leq m$, we have
$$
0=\sum_{i=1}^{m} a_{i} v_{i} \cdot 0=\sum_{i=1}^{m} a_{i} v_{i} \sum_{j \neq i} a_{j} \frac{v_{j}-v_{i}}{\left\|v_{j}-v_{i}\right\|^{3}}=-\sum_{1 \leq i<j \leq m} ... | {
"problem_match": "\n3. [30]",
"resource_path": "HarvardMIT/segmented/en-174-tournaments-2014-hmic-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 178 | 804 |
2014 | T4 | 5 | null | HMMT | Let $n$ be a positive integer, and let $A$ and $B$ be $n \times n$ matrices with complex entries such that $A^{2}=B^{2}$. Show that there exists an $n \times n$ invertible matrix $S$ with complex entries that satisfies $S(A B-B A)=(B A-A B) S$. | N/A Let $X=A+B$ and $Y=A-B$, so $X Y=B A-A B$ and $Y X=A B-B A$. Note that $X Y=-Y X$.
It suffices (actually is equivalent) to show that $A B-B A$ and $B A-A B$ have the same Jordan forms. In other words, we need to show that for any complex number $\lambda$, the Jordan $\lambda$-block decompositions of $A B-B A$ and ... | {
"problem_match": "\n5. [40]",
"resource_path": "HarvardMIT/segmented/en-174-tournaments-2014-hmic-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 77 | 972 |
2014 | T4 | 29 | null | HMMT | Let $\omega$ be a fixed circle with radius 1 , and let $B C$ be a fixed chord of $\omega$ such that $B C=1$. The locus of the incenter of $A B C$ as $A$ varies along the circumference of $\omega$ bounds a region $\mathcal{R}$ in the plane. Find the area of $\mathcal{R}$. | $\pi\left(\frac{3-\sqrt{3}}{3}\right)-1$ We will make use of the following lemmas.
Lemma 1: If $A B C$ is a triangle with incenter $I$, then $\angle B I C=90+\frac{A}{2}$.
Proof: Consider triangle $B I C$. Since $I$ is the intersection of the angle bisectors, $\angle I B C=\frac{B}{2}$ and $\angle I C B=\frac{C}{2}$. I... | {
"problem_match": "\n29. [15]",
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 83 | 591 |
2014 | T4 | 34 | null | HMMT | Let $M$ denote the number of positive integers which divide 2014!, and let $N$ be the integer closest to $\ln (M)$. Estimate the value of $N$. If your answer is a positive integer $A$, your score on this problem will be the larger of 0 and $\left\lfloor 20-\frac{1}{8}|A-N|\right\rfloor$. Otherwise, your score will be z... | 439 Combining Legendre's Formula and the standard prime approximations, the answer is
$$
\prod_{p}\left(1+\frac{2014-s_{p}(2014)}{p-1}\right)
$$
where $s_{p}(n)$ denotes the sum of the base $p$-digits of $n$.
Estimate $\ln 1000 \approx 8$, and $\ln 2014 \approx 9$. Using the Prime Number Theorem or otherwise, one mig... | {
"problem_match": "\n34. [20]",
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 95 | 564 |
2014 | T4 | 5 | null | HMMT | Let $A, B, C, D, E$ be five points on a circle; some segments are drawn between the points so that each of the $\binom{5}{2}=10$ pairs of points is connected by either zero or one segments. Determine the number of sets of segments that can be drawn such that:
- It is possible to travel from any of the five points to a... | 195 First we show that we can divide the five points into sets $S$ and $T$ according to the second condition in only one way. Assume that we can divide the five points into $S \cup T$ and $S^{\prime} \cup T^{\prime}$. Then, let $A=S^{\prime} \cap S, B=S^{\prime} \cap T, C=T^{\prime} \cap S$, and $D=T^{\prime} \cap T$. ... | {
"problem_match": "\n5. [5]",
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-team-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 131 | 1,039 |
2014 | T4 | 6 | null | HMMT | Find the number of strictly increasing sequences of nonnegative integers with the following properties:
- The first term is 0 and the last term is 12 . In particular, the sequence has at least two terms.
- Among any two consecutive terms, exactly one of them is even. | 144 For a natural number $n$, let $A_{n}$ be a set containing all sequences which satisfy the problem conditions but which 12 is replaced by $n$. Also, let $a_{n}$ be the size of $A_{n}$.
We first consider $a_{1}$ and $a_{2}$. We get $a_{1}=1$, as the only sequence satisfying the problem conditions is 0,1 . We also ge... | {
"problem_match": "\n6. [6]",
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-team-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 57 | 545 |
2014 | T4 | 7 | null | HMMT | Sammy has a wooden board, shaped as a rectangle with length $2^{2014}$ and height $3^{2014}$. The board is divided into a grid of unit squares. A termite starts at either the left or bottom edge of the rectangle, and walks along the gridlines by moving either to the right or upwards, until it reaches an edge opposite t... | 4 Let $R$ be the original rectangle and $R^{\prime}$ the new rectangle which is different from $R$. We see that the perimeter of $R^{\prime}$ depends on the possibilities for the side lengths of $R^{\prime}$.
We will prove that the dividing line must have the following characterization: starting from the lower left cor... | {
"problem_match": "\n7. [7]",
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-team-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 229 | 3,748 |
2014 | T4 | 9 | null | HMMT | How many lines pass through exactly two points in the following hexagonal grid?
 | 60 First solution. From a total of 19 points, there are $\binom{19}{2}=171$ ways to choose two points. We consider lines that pass through more than 2 points.
- There are $6+6+3=15$ lines that pass through exactly three points. These are: the six sides of the largest hexagon, three lines through the center (perpendicu... | {
"problem_match": "\n9. [5]",
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-team-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 88 | 689 |
2014 | T4 | 10 | null | HMMT | Let $A B C D E F$ be a convex hexagon with the following properties.
(a) $\overline{A C}$ and $\overline{A E}$ trisect $\angle B A F$.
(b) $\overline{B E} \| \overline{C D}$ and $\overline{C F} \| \overline{D E}$.
(c) $A B=2 A C=4 A E=8 A F$.
Suppose that quadrilaterals $A C D E$ and $A D E F$ have area 2014 and 1400,... | 7295 From conditions (a) and (c), we know that triangles $A F E, A E C$ and $A C B$ are similar to one another, each being twice as large as the preceding one in each dimension. Let $\overline{A E} \cap \overline{F C}=P$ and $\overline{A C} \cap \overline{E B}=Q$. Then, since the quadrilaterals $A F E C$ and $A E C B$ ... | {
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-team-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 146 | 513 |
2014 | T4 | 10 | null | HMMT | Let $z$ be a complex number and $k$ a positive integer such that $z^{k}$ is a positive real number other than 1. Let $f(n)$ denote the real part of the complex number $z^{n}$. Assume the parabola $p(n)=a n^{2}+b n+c$ intersects $f(n)$ four times, at $n=0,1,2,3$. Assuming the smallest possible value of $k$, find the lar... | $\boxed{\frac{1}{3}}$ Let $r=|z|$, $\theta = \arg z$, and $C = \frac{\Re z}{|z|} = \cos \theta = \cos \frac{2\pi j}{k}$ for some $j$ with $\gcd(j, k) = 1$. The condition of the four consecutive points lying on a parabola is equivalent to having the finite difference
$$
f(3)-3 f(2)+3 f(1)-f(0)=0
$$
This implies
$$
\be... | {
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-thm-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 108 | 556 |
2015 | T4 | 3 | Algebra | HMMT | Let $p$ be a real number and $c \neq 0$ an integer such that
$$
c-0.1<x^{p}\left(\frac{1-(1+x)^{10}}{1+(1+x)^{10}}\right)<c+0.1
$$
for all (positive) real numbers $x$ with $0<x<10^{-100}$. (The exact value $10^{-100}$ is not important. You could replace it with any "sufficiently small number".)
Find the ordered pair ... | $(-1,-5)$ This is essentially a problem about limits, but phrased concretely in terms of "small numbers" (like 0.1 and $10^{-100}$ ).
We are essentially studying the rational function $f(x):=\frac{1-(1+x)^{10}}{1+(1+x)^{10}}=\frac{-10 x+O\left(x^{2}\right)}{2+O(x)}$, where the "big-O" notation simply make precise the n... | {
"problem_match": "\n3. ",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 130 | 535 |
2015 | T4 | 6 | Algebra | HMMT | Let $a, b, c, d, e$ be nonnegative integers such that $625 a+250 b+100 c+40 d+16 e=15^{3}$. What is the maximum possible value of $a+b+c+d+e$ ? | 153 The intuition is that as much should be in $e$ as possible. But divisibility obstructions like $16 \nmid 15^{3}$ are in our way. However, the way the coefficients $5^{4}>5^{3} \cdot 2>\cdots$ are set up, we can at least easily avoid having $a, b, c, d$ too large (speifically, $\geq 2$ ). This is formalized below.
F... | {
"problem_match": "\n6. ",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 64 | 556 |
2015 | T4 | 7 | Algebra | HMMT | Suppose $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ is a 4 -term sequence of real numbers satisfying the following two conditions:
- $a_{3}=a_{2}+a_{1}$ and $a_{4}=a_{3}+a_{2}$;
- there exist real numbers $a, b, c$ such that
$$
a n^{2}+b n+c=\cos \left(a_{n}\right)
$$
for all $n \in\{1,2,3,4\}$.
Compute the maximum po... | $\quad-9+3 \sqrt{13}$ Let $f(n)=\cos a_{n}$ and $m=1$. The second ("quadratic interpolation") condition on $f(m), f(m+1), f(m+2), f(m+3)$ is equivalent to having a vanishing third finite difference
$$
f(m+3)-3 f(m+2)+3 f(m+1)-f(m)=0
$$
[^2]This is equivalent to
$$
\begin{aligned}
f(m+3)-f(m) & =3[f(m+2)-f(m+1)] \\
\... | {
"problem_match": "\n7. ",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-alg-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 181 | 539 |
2015 | T4 | 10 | Algebra | HMMT | Find all ordered 4 -tuples of integers $(a, b, c, d)$ (not necessarily distinct) satisfying the following system of equations:
\[
$$
\begin{aligned}
a^{2}-b^{2}-c^{2}-d^{2} & =c-b-2 \\
2 a b & =a-d-32 \\
2 a c & =28-a-d \\
2 a d & =b+c+31
\end{aligned}
$$
\] | The left hand sides clue us in to the fact that this problem is secretly about quaternions. Indeed, we see that letting $z=a+b i+c j+d k$ gives
$$
(z-i+j) z=-2-32 i+28 j+31 k
$$
Taking norms gives $N(z-i+j) N(z)=2^{2}+32^{2}+28^{2}+31^{2}=2773=47 \cdot 59$. By the triangle inequality, $N(z), N(z-i+j)$ aren't too far ... | {
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-alg-solutions.jsonl",
"solution_match": "\nSolution 3. "
} | 104 | 2,010 |
2015 | T4 | 4 | Combinatorics | HMMT | Alice Czarina is bored and is playing a game with a pile of rocks. The pile initially contains 2015 rocks. At each round, if the pile has $N$ rocks, she removes $k$ of them, where $1 \leq k \leq N$, with each possible $k$ having equal probability. Alice Czarina continues until there are no more rocks in the pile. Let $... | -501 We claim that $p=\frac{1}{5} \frac{6}{10} \frac{11}{15} \frac{16}{20} \cdots \frac{2006}{2010} \frac{2011}{2015}$. Let $p_{n}$ be the probability that, starting with $n$ rocks, the number of rocks left after each round is a multiple of 5 . Indeed, using recursions we have
$$
p_{5 k}=\frac{p_{5 k-5}+p_{5 k-10}+\cd... | {
"problem_match": "\n4. ",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 176 | 679 |
2015 | T4 | 7 | Combinatorics | HMMT | 2015 people sit down at a restaurant. Each person orders a soup with probability $\frac{1}{2}$. Independently, each person orders a salad with probability $\frac{1}{2}$. What is the probability that the number of people who ordered a soup is exactly one more than the number of people who ordered a salad? | To count the number of possibilities, we can directly evaluate the sum $\sum_{i=0}^{2014}\binom{2015}{i}\binom{2015}{i+1}$. One way is to note $\binom{2015}{i+1}=\binom{2015}{2014-i}$, and finish with Vandermonde's identity: $\sum_{i=0}^{2014}\binom{2015}{i}\binom{2015}{2014-i}=$ $\binom{2015+2015}{2014}=\binom{4030}{2... | {
"problem_match": "\n7. ",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-comb-solutions.jsonl",
"solution_match": "\nSolution 2. "
} | 71 | 504 |
2015 | T4 | 10 | Combinatorics | HMMT | A group of friends, numbered $1,2,3, \ldots, 16$, take turns picking random numbers. Person 1 picks a number uniformly (at random) in [0, 1], then person 2 picks a number uniformly (at random) in [0, 2], and so on, with person $k$ picking a number uniformly (at random) in $[0, k]$. What is the probability that the 16 n... | Here we present a discrete version of the previous solution.
To do this, we consider several related events.
Let $X$ be a 16 -tuple chosen uniformly and randomly from $[0,17]^{16}$ (used to define events $A, B, C$ ). Let $Z$ be a 16 -tuple chosen uniformly and randomly from $\{1,2, \ldots, 17\}^{16}$ (used to define ev... | {
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-comb-solutions.jsonl",
"solution_match": "\nSolution 2. "
} | 104 | 1,310 |
2015 | T4 | 10 | Combinatorics | HMMT | A group of friends, numbered $1,2,3, \ldots, 16$, take turns picking random numbers. Person 1 picks a number uniformly (at random) in [0, 1], then person 2 picks a number uniformly (at random) in [0, 2], and so on, with person $k$ picking a number uniformly (at random) in $[0, k]$. What is the probability that the 16 n... | Suppose that person $i$ picks a number in the interval $\left[b_{i}-1, b_{i}\right]$ where $b_{i} \leq i$. Then we have the condition: $b_{1} \leq b_{2} \leq \cdots \leq b_{16}$. Let $c_{i}$ be the number of $b_{j}$ 's such that $b_{j}=i$. Then, for each admissible sequence $b_{1}, b_{2}, \ldots, b_{16}$, there is the ... | {
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-comb-solutions.jsonl",
"solution_match": "\nSolution 3. "
} | 104 | 946 |
2015 | T4 | 5 | Geometry | HMMT | Let $I$ be the set of points $(x, y)$ in the Cartesian plane such that
$$
x>\left(\frac{y^{4}}{9}+2015\right)^{1 / 4}
$$
Let $f(r)$ denote the area of the intersection of $I$ and the disk $x^{2}+y^{2} \leq r^{2}$ of radius $r>0$ centered at the origin $(0,0)$. Determine the minimum possible real number $L$ such that ... | $\quad \frac{\pi}{3}$ Let $B(P, r)$ be the (closed) disc centered at $P$ with radius $r$. Note that for all $(x, y) \in I, x>0$, and $x>\left(\frac{y^{4}}{9}+2015\right)^{1 / 4}>\frac{|y|}{\sqrt{3}}$. Let $I^{\prime}=\{(x, y): x \sqrt{3}>|y|\}$. Then $I \subseteq I^{\prime}$ and the intersection of $I^{\prime}$ with $B... | {
"problem_match": "\n5. ",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-geo-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 131 | 589 |
2015 | T4 | 10 | Geometry | HMMT | Let $\mathcal{G}$ be the set of all points $(x, y)$ in the Cartesian plane such that $0 \leq y \leq 8$ and
$$
(x-3)^{2}+31=(y-4)^{2}+8 \sqrt{y(8-y)}
$$
There exists a unique line $\ell$ of negative slope tangent to $\mathcal{G}$ and passing through the point $(0,4)$. Suppose $\ell$ is tangent to $\mathcal{G}$ at a un... | $\left(\frac{12}{5}, \frac{8}{5}\right)$ Let $G$ be $\mathcal{G}$ restricted to the strip of plane $0 \leq y \leq 4$ (we only care about this region since $\ell$ has negative slope going down from $(0,4)$ ). By completing the square, the original equation rearranges to $(x-3)^{2}+(\sqrt{y(8-y)}-4)^{2}=1$. One could fin... | {
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-geo-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 134 | 608 |
2015 | T4 | 15 | null | HMMT | Find the maximum possible value of $H \cdot M \cdot M \cdot T$ over all ordered triples $(H, M, T)$ of integers such that $H \cdot M \cdot M \cdot T=H+M+M+T$. | 8 If any of $H, M, T$ are zero, the product is 0 . We can do better (examples below), so we may now restrict attention to the case when $H, M, T \neq 0$.
When $M \in\{-2,-1,1,2\}$, a little casework gives all the possible $(H, M, T)=(2,1,4),(4,1,2),(-1,-2,1),(1,-2,-1)$.
- If $M=-2$, i.e. $H-4+T=4 H T$, then $-15=(4 H... | {
"problem_match": "\n15. [8]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 53 | 767 |
2015 | T4 | 16 | null | HMMT | Determine the number of unordered triples of distinct points in the $4 \times 4 \times 4$ lattice grid $\{0,1,2,3\}^{3}$ that are collinear in $\mathbb{R}^{3}$ (i.e. there exists a line passing through the three points). | 376 Define a main plane to be one of the $x y, y z, z x$ planes. Define a space diagonal to be a set of collinear points not parallel to a main plane. We classify the lines as follows:
(a) Lines parallel to two axes (i.e. orthogonal to a main plane). Notice that given a plane of the form $v=k$, where $v \in\{x, y, z\},... | {
"problem_match": "\n16. [8]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 67 | 551 |
2015 | T4 | 18 | null | HMMT | Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be a function such that for any integers $x, y$, we have
$$
f\left(x^{2}-3 y^{2}\right)+f\left(x^{2}+y^{2}\right)=2(x+y) f(x-y)
$$
Suppose that $f(n)>0$ for all $n>0$ and that $f(2015) \cdot f(2016)$ is a perfect square. Find the minimum possible value of $f(1)+f(2)$. | 246 Plugging in $-y$ in place of $y$ in the equation and comparing the result with the original equation gives
$$
(x-y) f(x+y)=(x+y) f(x-y)
$$
This shows that whenever $a, b \in \mathbb{Z}-\{0\}$ with $a \equiv b(\bmod 2)$, we have
$$
\frac{f(a)}{a}=\frac{f(b)}{b}
$$
which implies that there are constants $\alpha=f... | {
"problem_match": "\n18. [11]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 130 | 695 |
2015 | T4 | 20 | null | HMMT | What is the largest real number $\theta$ less than $\pi$ (i.e. $\theta<\pi$ ) such that
$$
\prod_{k=0}^{10} \cos \left(2^{k} \theta\right) \neq 0
$$
and
$$
\prod_{k=0}^{10}\left(1+\frac{1}{\cos \left(2^{k} \theta\right)}\right)=1 ?
$$ | $\frac{\frac{2046 \pi}{2047}}{}$ For equality to hold, note that $\theta$ cannot be an integer multiple of $\pi$ (or else $\sin =0$ and $\cos = \pm 1$ ).
Let $z=e^{i \theta / 2} \neq \pm 1$. Then in terms of complex numbers, we want
$$
\prod_{k=0}^{10}\left(1+\frac{2}{z^{2^{k+1}}+z^{-2^{k+1}}}\right)=\prod_{k=0}^{10} ... | {
"problem_match": "\n20. [11]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 106 | 666 |
2015 | T4 | 21 | null | HMMT | Define a sequence $a_{i, j}$ of integers such that $a_{1, n}=n^{n}$ for $n \geq 1$ and $a_{i, j}=a_{i-1, j}+a_{i-1, j+1}$ for all $i, j \geq 1$. Find the last (decimal) digit of $a_{128,1}$. | 4 By applying the recursion multiple times, we find that $a_{1,1}=1, a_{2, n}=n^{n}+(n+1)^{n+1}$, and $a_{3, n}=n^{n}+2(n+1)^{n+1}+(n+2)^{n+2}$. At this point, we can conjecture and prove by induction that
$$
a_{m, n}=\sum_{k=0}^{m-1}\binom{m-1}{k}(n+k)^{n+k}=\sum_{k \geq 0}\binom{m-1}{k}(n+k)^{n+k}
$$
(The second ex... | {
"problem_match": "\n21. [14]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 91 | 500 |
2015 | T4 | 28 | null | HMMT | Let $w, x, y$, and $z$ be positive real numbers such that
$$
\begin{aligned}
0 & \neq \cos w \cos x \cos y \cos z \\
2 \pi & =w+x+y+z \\
3 \tan w & =k(1+\sec w) \\
4 \tan x & =k(1+\sec x) \\
5 \tan y & =k(1+\sec y) \\
6 \tan z & =k(1+\sec z)
\end{aligned}
$$
(Here $\sec t$ denotes $\frac{1}{\cos t}$ when $\cos t \neq... | $\sqrt{\sqrt{19}}$ From the identity $\tan \frac{u}{2}=\frac{\sin u}{1+\cos u}$, the conditions work out to $3 \tan \frac{w}{2}=4 \tan \frac{x}{2}=$ $5 \tan \frac{y}{2}=6 \tan \frac{z}{2}=k$. Let $a=\tan \frac{w}{2}, b=\tan \frac{x}{2}, c=\tan \frac{y}{2}$, and $d=\tan \frac{z}{2}$. Using the identity $\tan (M+N)=\frac... | {
"problem_match": "\n28. [17]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 147 | 520 |
2015 | T4 | 30 | null | HMMT | Find the sum of squares of all distinct complex numbers $x$ satisfying the equation
$$
0=4 x^{10}-7 x^{9}+5 x^{8}-8 x^{7}+12 x^{6}-12 x^{5}+12 x^{4}-8 x^{3}+5 x^{2}-7 x+4
$$ | $\quad-\frac{7}{16}$ For convenience denote the polynomial by $P(x)$. Notice $4+8=7+5=12$ and that the consecutive terms $12 x^{6}-12 x^{5}+12 x^{4}$ are the leading terms of $12 \Phi_{14}(x)$, which is suggestive. Indeed, consider $\omega$ a primitive 14 -th root of unity; since $\omega^{7}=-1$, we have $4 \omega^{10}... | {
"problem_match": "\n30. [20]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 80 | 583 |
2015 | T4 | 31 | null | HMMT | Define a power cycle to be a set $S$ consisting of the nonnegative integer powers of an integer $a$, i.e. $S=\left\{1, a, a^{2}, \ldots\right\}$ for some integer $a$. What is the minimum number of power cycles required such that given any odd integer $n$, there exists some integer $k$ in one of the power cycles such th... | 10 Solution 1. Partition the odd residues mod 1024 into 10 classes:
- Class 1: $1(\bmod 4)$.
- Class $n(2 \leq n \leq 9): 2^{n}-1\left(\bmod 2^{n+1}\right)$.
- Class 10: $-1(\bmod 1024)$.
Let $S_{a}$ be the power cycle generated by $a$. If $a$ is in class 1 , all of $S_{a}$ is in class 1 . If a is in class $n$ $(2 \l... | {
"problem_match": "\n31. [20]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 105 | 707 |
2015 | T4 | 31 | null | HMMT | Define a power cycle to be a set $S$ consisting of the nonnegative integer powers of an integer $a$, i.e. $S=\left\{1, a, a^{2}, \ldots\right\}$ for some integer $a$. What is the minimum number of power cycles required such that given any odd integer $n$, there exists some integer $k$ in one of the power cycles such th... | Lemma. Given a positive integer $n \geq 3$, there exists an odd integer $x$ such that the order of $x$ modulo $2^{n}$ is $2^{n-2}$.
Proof. We apply induction on $n$. The base cases of $n=3,4$ are clearly true with $x=3$, so suppose we have the statement holds for $n-1$ and we wish to show it for $n$ where $n \geq 5$. ... | {
"problem_match": "\n31. [20]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nSolution 2. "
} | 105 | 926 |
2015 | T4 | 32 | null | HMMT | A wealthy king has his blacksmith fashion him a large cup, whose inside is a cone of height 9 inches and base diameter 6 inches (that is, the opening at the top of the cup is 6 inches in diameter). At one of his many feasts, he orders the mug to be filled to the brim with cranberry juice.
For each positive integer $n$,... | $\frac{216 \pi^{3}-2187 \sqrt{3}}{8 \pi^{2}}$ First, we find the total amount of juice consumed. We can simply subtract the amount of juice remaining at infinity from the initial amount of juice in the cup, which of course is simply the volume of the cup; we'll denote this value by $V$.
Since volume in the cup varies a... | {
"problem_match": "\n32. [20]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 195 | 1,251 |
2015 | T4 | 34 | null | HMMT | For an integer $n$, let $f(n)$ denote the number of pairs $(x, y)$ of integers such that $x^{2}+x y+y^{2}=n$. Compute the sum
$$
\sum_{n=1}^{10^{6}} n f(n)
$$
Write your answer in the form $a \cdot 10^{b}$, where $b$ is an integer and $1 \leq a<10$ is a decimal number.
If your answer is written in this form, your sco... | $1.813759629294 \cdot 10^{12}$ Rewrite the sum as
$$
\sum_{x^{2}+x y+y^{2} \leq 10^{6}}\left(x^{2}+x y+y^{2}\right),
$$
where the sum is over all pairs $(x, y)$ of integers with $x^{2}+x y+y^{2} \leq 10^{6}$. We can find a crude upper bound for this sum by noting that
$$
x^{2}+x y+y^{2}=\frac{3}{4} x^{2}+\left(\frac... | {
"problem_match": "\n34. [25]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 199 | 917 |
2015 | T4 | 35 | null | HMMT | Let $P$ denote the set of all subsets of $\{1, \ldots, 23\}$. A subset $S \subseteq P$ is called good if whenever $A, B$ are sets in $S$, the set $(A \backslash B) \cup(B \backslash A)$ is also in $S$. (Here, $A \backslash B$ denotes the set of all elements in $A$ that are not in $B$, and $B \backslash A$ denotes the s... | $\quad \frac{18839183877670041942218307147122500601235}{47691684840486192422095701784512492731212} \approx 0.3950203047068107$ Let $n=23$, and $\ell=$ $\lfloor n / 2\rfloor=11$.
We use the well-known rephrasing of the symmetric difference $((A \backslash B) \cup(B \backslash A))$ in terms of addition modulo 2 of "indic... | {
"problem_match": "\n35. [25]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 242 | 1,210 |
2015 | T4 | 36 | null | HMMT | A prime number $p$ is twin if at least one of $p+2$ or $p-2$ is prime and sexy if at least one of $p+6$ and $p-6$ is prime.
How many sexy twin primes (i.e. primes that are both twin and sexy) are there less than $10^{9}$ ? Express your answer as a positive integer $N$ in decimal notation; for example, 521495223 . If yo... | 1462105 The Hardy-Littlewood conjecture states that given a set $A$ of integers, the number of integers $x$ such that $x+a$ is a prime for all $a \in A$ is
$$
\frac{x}{(\ln x)^{|A|}} \prod_{p} \frac{1-\frac{w(p ; A)}{p}}{\left(1-\frac{1}{p}\right)^{k}}(1+o(1))
$$
where $w(p ; A)$ is the number of distinct residues of... | {
"problem_match": "\n36. [25]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 176 | 1,156 |
2015 | T4 | 6 | null | HMMT | $\$$ indy has $\$ 100$ in pennies (worth $\$ 0.01$ each), nickels (worth $\$ 0.05$ each), dimes (worth $\$ 0.10$ each), and quarters (worth $\$ 0.25$ each). Prove that she can split her coins into two piles, each with total value exactly $\$ 50$. | N/A Solution 1. First, observe that if there are pennies in the mix, there must be a multiple of 5 pennies $($ since $5,10,25 \equiv 0(\bmod 5)$ ), so we can treat each group of 5 pennies as nickels and reduce the problem to the case of no pennies, treated below.
Indeed, suppose there are no pennies and that we are sol... | {
"problem_match": "\n6. [30]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-team-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 92 | 581 |
2015 | T4 | 8 | null | HMMT | Let $\pi$ be a permutation of $\{1,2, \ldots, 2015\}$. With proof, determine the maximum possible number of ordered pairs $(i, j) \in\{1,2, \ldots, 2015\}^{2}$ with $i<j$ such that $\pi(i) \cdot \pi(j)>i \cdot j$. | $\left.\begin{array}{c}2014 \\ 2\end{array}\right)$ Let $n=2015$. The only information we will need about $n$ is that $n>54$ For the construction, take $\pi$ to be the $n$-cycle defined by
$$
\pi(k)= \begin{cases}k+1 & \text { if } 1 \leq k \leq n-1 \\ 1 & \text { if } k=n\end{cases}
$$
Then $\pi(i)>i$ for $1 \leq i ... | {
"problem_match": "\n8. [40]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-team-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 86 | 997 |
2015 | T4 | 9 | null | HMMT | Let $z=e^{\frac{2 \pi i}{101}}$ and let $\omega=e^{\frac{2 \pi i}{10}}$. Prove that
$$
\prod_{a=0}^{9} \prod_{b=0}^{100} \prod_{c=0}^{100}\left(\omega^{a}+z^{b}+z^{c}\right)
$$
is an integer and find (with proof) its remainder upon division by 101. | 13 Solution 1. Let $p=101$ and $r=10$. Note that $p \nmid r$.
In the sequel, we will repeatedly use the polynomial identities $\prod_{k(\bmod p)}\left(x-z^{k}\right)=x^{p}-1$, and $\prod_{j(\bmod r)}\left(x-\omega^{j}\right)=x^{r}-1$.
The product is an integer by standard symmetric sum theory (concrete precursor to Gal... | {
"problem_match": "\n9. $[40]$",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-team-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 115 | 1,489 |
2015 | T4 | 10 | null | HMMT | The sequences of real numbers $\left\{a_{i}\right\}_{i=1}^{\infty}$ and $\left\{b_{i}\right\}_{i=1}^{\infty}$ satisfy $a_{n+1}=\left(a_{n-1}-1\right)\left(b_{n}+1\right)$ and $b_{n+1}=a_{n} b_{n-1}-1$ for $n \geq 2$, with $a_{1}=a_{2}=2015$ and $b_{1}=b_{2}=2013$. Evaluate, with proof, the infinite sum
$$
\sum_{n=1}^{... | $\quad 1+\frac{1}{2014 \cdot 2015}$ OR $\frac{4058211}{4058210}$ First note that $a_{n}$ and $b_{n}$ are weakly increasing and tend to infinity. In particular, $a_{n}, b_{n} \notin\{0,-1,1\}$ for all $n$.
For $n \geq 1$, we have $a_{n+3}=\left(a_{n+1}-1\right)\left(b_{n+2}+1\right)=\left(a_{n+1}-1\right)\left(a_{n+1} b... | {
"problem_match": "\n10. [40]",
"resource_path": "HarvardMIT/segmented/en-182-2015-feb-team-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 189 | 1,104 |
2015 | T4 | 1 | null | HMMT | Let $S$ be the set of positive integers $n$ such that the inequality
$$
\phi(n) \cdot \tau(n) \geq \sqrt{\frac{n^{3}}{3}}
$$
holds, where $\phi(n)$ is the number of positive integers $k \leq n$ that are relatively prime to $n$, and $\tau(n)$ is the number of positive divisors of $n$. Prove that $S$ is finite. | N/A
Proof. Let $S$ be the set of all positive integers $n$ such that
$$
\phi(n) \cdot \tau(n) \geq \sqrt{\frac{n^{3}}{3}}
$$
Define a function $\Phi$ on all positive integers $n$ by
$$
\Phi(n)=\frac{\phi(n)^{2} \cdot \tau(n)^{2}}{n^{3}}
$$
An important observation is that $\Phi$ has the property that for every rela... | {
"problem_match": "\n1. [20]",
"resource_path": "HarvardMIT/segmented/en-184-tournaments-2015-hmic-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 99 | 934 |
2015 | T4 | 2 | null | HMMT | Let $m, n$ be positive integers with $m \geq n$. Let $S$ be the set of pairs $(a, b)$ of relatively prime positive integers such that $a, b \leq m$ and $a+b>m$.
For each pair $(a, b) \in S$, consider the nonnegative integer solution $(u, v)$ to the equation $a u-b v=n$ chosen with $v \geq 0$ minimal, and let $I(a, b)$... | N/A
Proof. The fact that $I(a, b) \subseteq(0,1)$ follows from the small-ness of $n \leq m<a+b$ : the smallest solution $(u, v)$ has $0 \leq v \leq a-1$, so $u=\frac{n+b v}{a}<\frac{(a+b)+b(a-1)}{a}=b+1$ forces $1 \leq u \leq b$.
For the main part of the problem, it suffices (actually, is equivalent) to show that
(i) ... | {
"problem_match": "\n2. [25]",
"resource_path": "HarvardMIT/segmented/en-184-tournaments-2015-hmic-solutions.jsonl",
"solution_match": "\n## Answer: "
} | 189 | 1,074 |
2015 | T4 | 3 | null | HMMT | Let $M$ be a $2014 \times 2014$ invertible matrix, and let $\mathcal{F}(M)$ denote the set of matrices whose rows are a permutation of the rows of $M$. Find the number of matrices $F \in \mathcal{F}(M)$ such that $\operatorname{det}(M+F) \neq 0$. | $2013!!^{2}$
Proof. Number the rows of $M$ as $r_{1}, r_{2}, \ldots, r_{2014}$, and remark that by invertibility, these rows are linearly independent. We can write $\operatorname{det}(M+F)$ as the sum of the determinants of the $2^{2014}$ matrices which can be obtained by replacing some set of the rows of $M$ with the... | {
"problem_match": "\n3. [30]",
"resource_path": "HarvardMIT/segmented/en-184-tournaments-2015-hmic-solutions.jsonl",
"solution_match": "\n## Answer: "
} | 84 | 1,341 |
2015 | T4 | 4 | null | HMMT | Prove that there exists a positive integer $N$ such that for any positive integer $n \geq N$, there are at least 2015 non-empty subsets $S$ of $\left\{n^{2}+1, n^{2}+2, \ldots, n^{2}+3 n\right\}$ with the property that the product of the elements of $S$ is a perfect square. | N/A
Lemma 0.1. Let $n, k$ be integers with $1 \leq k \leq n$. Then
$$
\binom{n}{k} \leq\left(\frac{e n}{k}\right)^{k}
$$
where $e \approx 2.71828 \ldots$ is Euler's number.
Proof. We have
$$
\begin{aligned}
\binom{n}{k} & =\frac{n(n-1) \cdots(n-k+1)}{k!} \\
& \leq \frac{n^{k}}{k!} \\
& =\left(\frac{n}{k}\right)^{k} ... | {
"problem_match": "\n4. [35]",
"resource_path": "HarvardMIT/segmented/en-184-tournaments-2015-hmic-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 91 | 1,731 |
2015 | T4 | 5 | null | HMMT | Let $\omega=e^{2 \pi i / 5}$ be a primitive fifth root of unity. Prove that there do not exist integers $a, b, c, d, k$ with $k>1$ such that
$$
\left(a+b \omega+c \omega^{2}+d \omega^{3}\right)^{k}=1+\omega
$$ | N/A
Proof. Let $\zeta=\omega=e^{2 \pi i / 5}$ be a primitive fifth root of unity. If $x=a+b \zeta+c \zeta^{2}+d \zeta^{3}(a, b, c, d \in \mathbb{Z})$, then from $\zeta+\zeta^{-1}=\frac{\sqrt{5}-1}{2}$ and $\zeta^{2}+\zeta^{-2}=\zeta^{3}+\zeta^{-3}=-\frac{\sqrt{5}+1}{2}$, we deduce $|x|^{2}=\left(a^{2}+b^{2}+c^{2}+d^{2}... | {
"problem_match": "\n5. [40]",
"resource_path": "HarvardMIT/segmented/en-184-tournaments-2015-hmic-solutions.jsonl",
"solution_match": "\nAnswer: "
} | 80 | 1,167 |
2015 | T4 | 20 | null | HMMT | Let $n$ be a three-digit integer with nonzero digits, not all of which are the same. Define $f(n)$ to be the greatest common divisor of the six integers formed by any permutation of $n$ s digits. For example, $f(123)=3$, because $\operatorname{gcd}(123,132,213,231,312,321)=3$. Let the maximum possible value of $f(n)$ b... | Answer: 5994
Let $n=\overline{a b c}$, and assume without loss of generality that $a \geq b \geq c$. We have $k \mid 100 a+10 b+c$ and $k \mid 100 a+10 c+b$, so $k \mid 9(b-c)$. Analogously, $k \mid 9(a-c)$ and $k \mid 9(a-b)$. Note that if $9 \mid n$, then 9 also divides any permutation of $n$ s digits, so $9 \mid f(n... | {
"problem_match": "\n20. [11]",
"resource_path": "HarvardMIT/segmented/en-191-2015-nov-guts-solutions.jsonl",
"solution_match": "\nProposed by: Alexander Katz\n"
} | 122 | 768 |
2015 | T4 | 10 | null | HMMT | A number $n$ is $b a d$ if there exists some integer $c$ for which $x^{x} \equiv c(\bmod n)$ has no integer solutions for $x$. Find the number of bad integers between 2 and 42 inclusive. | Answer: 25
Call a number good if it is not bad. We claim all good numbers are products of distinct primes, none of which are equivalent to 1 modulo another.
We first show that all such numbers are good. Consider $n=p_{1} p_{2} \ldots p_{k}$, and let $x$ be a number satisfying $x \equiv c\left(\bmod p_{1} p_{2} \ldots ... | {
"problem_match": "\n10. [7]",
"resource_path": "HarvardMIT/segmented/en-191-2015-nov-team-solutions.jsonl",
"solution_match": "\nProposed by: Sam Korsky\n"
} | 57 | 702 |
2015 | T4 | 4 | null | HMMT | Consider a $4 \times 4$ grid of squares. Aziraphale and Crowley play a game on this grid, alternating turns, with Aziraphale going first. On Aziraphales turn, he may color any uncolored square red, and on Crowleys turn, he may color any uncolored square blue. The game ends when all the squares are colored, and Azirapha... | ## Answer: 6
We claim that the answer is 6 .
On Aziraphale's first two turns, it is always possible for him to take 2 adjacent squares from the central four; without loss of generality, suppose they are the squares at $(1,1)$ and $(1,2)$. If allowed, Aziraphale's next turn will be to take one of the remaining squares ... | {
"problem_match": "\n4. ",
"resource_path": "HarvardMIT/segmented/en-191-2015-nov-thm-solutions.jsonl",
"solution_match": "\nProposed by: Alexander Katz\n\n"
} | 132 | 505 |
2015 | T4 | 5 | null | HMMT | Consider a $5 \times 5$ grid of squares. Vladimir colors some of these squares red, such that the centers of any four red squares do not form an axis-parallel rectangle (i.e. a rectangle whose sides are parallel to those of the squares). What is the maximum number of squares he could have colored red? | Answer: 12
We claim that the answer is 12 . We first show that if 13 squares are colored red, then some four form an axis-parallel rectangle. Note that we can swap both columns and rows without affecting whether four squares form a rectangle, so we may assume without loss of generality that the top row has the most red... | {
"problem_match": "\n5. ",
"resource_path": "HarvardMIT/segmented/en-191-2015-nov-thm-solutions.jsonl",
"solution_match": "\nProposed by: Sam Korsky\n"
} | 67 | 549 |
2015 | T4 | 7 | null | HMMT | Consider a $7 \times 7$ grid of squares. Let $f:\{1,2,3,4,5,6,7\} \rightarrow\{1,2,3,4,5,6,7\}$ be a function; in other words, $f(1), f(2), \ldots, f(7)$ are each (not necessarily distinct) integers from 1 to 7 . In the top row of the grid, the numbers from 1 to 7 are written in order; in every other square, $f(x)$ is ... | Answer: 1470
Consider the directed graph with $1,2,3,4,5,6,7$ as vertices, and there is an edge from $i$ to $j$ if and only if $f(i)=j$. Since the bottom row is equivalent to the top one, we have $f^{6}(x)=x$. Therefore, the graph must decompose into cycles of length $6,3,2$, or 1 . Furthermore, since no other row is e... | {
"problem_match": "\n7. ",
"resource_path": "HarvardMIT/segmented/en-191-2015-nov-thm-solutions.jsonl",
"solution_match": "\n## Proposed by: Alexander Katz\n\n"
} | 163 | 538 |
2015 | T4 | 10 | null | HMMT | Consider a $10 \times 10$ grid of squares. One day, Daniel drops a burrito in the top left square, where a wingless pigeon happens to be looking for food. Every minute, if the pigeon and the burrito are in the same square, the pigeon will eat $10 \%$ of the burrito's original size and accidentally throw it into a rando... | Answer: 71.8
Label the squares using coordinates, letting the top left corner be ( 0,0 ). The burrito will end up in 10 (not necessarily different) squares. Call them $p_{1}=\left(x_{1}, y_{1}\right)=(0,0), p_{2}=\left(x_{2}, y_{2}\right), \ldots, p_{10}=\left(x_{10}, y_{10}\right)$. $p_{2}$ through $p_{10}$ are unifor... | {
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-191-2015-nov-thm-solutions.jsonl",
"solution_match": "\nProposed by: Sam Korsky\n"
} | 128 | 599 |
2016 | T4 | 4 | Algebra | HMMT | Determine the remainder when
$$
\sum_{i=0}^{2015}\left\lfloor\frac{2^{i}}{25}\right\rfloor
$$
is divided by 100 , where $\lfloor x\rfloor$ denotes the largest integer not greater than $x$. | Answer: 14
Let $r_{i}$ denote the remainder when $2^{i}$ is divided by 25 . Note that because $2^{\phi(25)} \equiv 2^{20} \equiv 1(\bmod 25)$, $r$ is periodic with length 20 . In addition, we find that 20 is the order of $2 \bmod 25$. Since $2^{i}$ is never a multiple of 5 , all possible integers from 1 to 24 are repre... | {
"problem_match": "\n4. ",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-alg-solutions.jsonl",
"solution_match": "\nProposed by: Alexander Katz\n"
} | 70 | 772 |
2016 | T4 | 6 | Algebra | HMMT | Call a positive integer $N \geq 2$ "special" if for every $k$ such that $2 \leq k \leq N, N$ can be expressed as a sum of $k$ positive integers that are relatively prime to $N$ (although not necessarily relatively prime to each other). How many special integers are there less than $100 ?$ | Answer: 50
We claim that all odd numbers are special, and the only special even number is 2 . For any even $N>2$, the numbers relatively prime to $N$ must be odd. When we consider $k=3$, we see that $N$ can't be expressed as a sum of 3 odd numbers.
Now suppose that $N$ is odd, and we look at the binary decomposition o... | {
"problem_match": "\n6. ",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-alg-solutions.jsonl",
"solution_match": "\nProposed by: Casey Fu\n"
} | 80 | 586 |
2016 | T4 | 7 | Algebra | HMMT | Determine the smallest positive integer $n \geq 3$ for which
$$
A \equiv 2^{10 n} \quad\left(\bmod 2^{170}\right)
$$
where $A$ denotes the result when the numbers $2^{10}, 2^{20}, \ldots, 2^{10 n}$ are written in decimal notation and concatenated (for example, if $n=2$ we have $A=10241048576$ ). | Answer: 14
Note that
$$
2^{10 n}=1024^{n}=1.024^{n} \times 10^{3 n}
$$
So $2^{10 n}$ has roughly $3 n+1$ digits for relatively small $n$ 's. (Actually we have that for $0<x<1$,
$$
(1+x)^{2}=1+2 x+x^{2}<1+3 x
$$
Therefore, $1.024^{2}<1.03^{2}<1.09,1.09^{2}<1.27,1.27^{2}<1.81<2$, and $2^{2}=4$, so $1.024^{16}<4$. Thu... | {
"problem_match": "\n7. ",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-alg-solutions.jsonl",
"solution_match": "\nProposed by: Evan Chen\n"
} | 115 | 792 |
2016 | T4 | 10 | Algebra | HMMT | Let $a, b$ and $c$ be positive real numbers such that
$$
\begin{aligned}
a^{2}+a b+b^{2} & =9 \\
b^{2}+b c+c^{2} & =52 \\
c^{2}+c a+a^{2} & =49
\end{aligned}
$$
Compute the value of $\frac{49 b^{2}-33 b c+9 c^{2}}{a^{2}}$. | Answer: 52
Consider a triangle $A B C$ with Fermat point $P$ such that $A P=a, B P=b, C P=c$. Then
$$
A B^{2}=A P^{2}+B P^{2}-2 A P \cdot B P \cos \left(120^{\circ}\right)
$$
by the Law of Cosines, which becomes
$$
A B^{2}=a^{2}+a b+b^{2}
$$
and hence $A B=3$. Similarly, $B C=\sqrt{52}$ and $A C=7$.
Furthermore, we... | {
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-alg-solutions.jsonl",
"solution_match": "\nProposed by: Alexander Katz\n"
} | 107 | 1,204 |
2016 | T4 | 6 | Combinatorics | HMMT | Define the sequence $a_{1}, a_{2} \ldots$ as follows: $a_{1}=1$ and for every $n \geq 2$,
$$
a_{n}= \begin{cases}n-2 & \text { if } a_{n-1}=0 \\ a_{n-1}-1 & \text { if } a_{n-1} \neq 0\end{cases}
$$
A non-negative integer $d$ is said to be jet-lagged if there are non-negative integers $r, s$ and a positive integer $n... | Let $N=n+r$, and $M=n$. Then $r=N-M$, and $s=a_{N}-a_{M}$, and $d=r+s=\left(a_{N}+N\right)-\left(a_{M}+M\right)$. So we are trying to find the number of possible values of $\left(a_{N}+N\right)-\left(a_{M}+M\right)$, subject to $N \geq M$ and $a_{N} \geq a_{M}$.
Divide the $a_{i}$ into the following "blocks":
- $a_{1}... | {
"problem_match": "\n6. ",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-comb-solutions.jsonl",
"solution_match": "\n## Proposed by: Pakawut Jiradilok\n\n"
} | 177 | 643 |
2016 | T4 | 7 | Combinatorics | HMMT | Kelvin the Frog has a pair of standard fair 8 -sided dice (each labelled from 1 to 8 ). Alex the sketchy Kat also has a pair of fair 8-sided dice, but whose faces are labelled differently (the integers on each Alex's dice need not be distinct). To Alex's dismay, when both Kelvin and Alex roll their dice, the probabilit... | Answer: 24, 28, 32
Ed. note: I'm probably horribly abusing notation
Define the generating function of an event $A$ as the polynomial
$$
g(A, x)=\sum p_{i} x^{i}
$$
where $p_{i}$ denotes the probability that $i$ occurs during event $A$. We note that the generating is multiplicative; i.e.
$$
g(A \text { AND } B, x)=g(... | {
"problem_match": "\n7. ",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-comb-solutions.jsonl",
"solution_match": "\nProposed by: Alexander Katz\n"
} | 132 | 1,632 |
2016 | T4 | 8 | Combinatorics | HMMT | Let $X$ be the collection of all functions $f:\{0,1, \ldots, 2016\} \rightarrow\{0,1, \ldots, 2016\}$. Compute the number of functions $f \in X$ such that
$$
\max _{g \in X}\left(\min _{0 \leq i \leq 2016}(\max (f(i), g(i)))-\max _{0 \leq i \leq 2016}(\min (f(i), g(i)))\right)=2015 .
$$ | Answer: $2 \cdot\left(3^{2017}-2^{2017}\right)$
For each $f, g \in X$, we define
$$
d(f, g):=\min _{0 \leq i \leq 2016}(\max (f(i), g(i)))-\max _{0 \leq i \leq 2016}(\min (f(i), g(i)))
$$
Thus we desire $\max _{g \in X} d(f, g)=2015$.
First, we count the number of functions $f \in X$ such that
$$
\exists g: \min _{i... | {
"problem_match": "\n8. ",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-comb-solutions.jsonl",
"solution_match": "\nProposed by:\n"
} | 139 | 540 |
2016 | T4 | 9 | Combinatorics | HMMT | Let $V=\{1, \ldots, 8\}$. How many permutations $\sigma: V \rightarrow V$ are automorphisms of some tree?
(A graph consists of a some set of vertices and some edges between pairs of distinct vertices. It is connected if every two vertices in it are connected by some path of one or more edges. A tree $G$ on $V$ is a con... | Answer: 30212
We decompose into cycle types of $\sigma$. Note that within each cycle, all vertices have the same degree; also note that the tree has total degree 14 across its vertices (by all its seven edges).
For any permutation that has a 1 in its cycle type (i.e it has a fixed point), let $1 \leq a \leq 8$ be a fix... | {
"problem_match": "\n9. ",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-comb-solutions.jsonl",
"solution_match": "\nProposed by: Mitchell Lee\n"
} | 152 | 877 |
2016 | T4 | 10 | Combinatorics | HMMT | Kristoff is planning to transport a number of indivisible ice blocks with positive integer weights from the north mountain to Arendelle. He knows that when he reaches Arendelle, Princess Anna and Queen Elsa will name an ordered pair $(p, q)$ of nonnegative integers satisfying $p+q \leq 2016$. Kristoff must then give Pr... | Answer: 18
The answer is 18 .
First, we will show that Kristoff must carry at least 18 ice blocks. Let
$$
0<x_{1} \leq x_{2} \leq \cdots \leq x_{n}
$$
be the weights of ice blocks he carries which satisfy the condition that for any $p, q \in \mathbb{Z}_{\geq 0}$ such that $p+q \leq 2016$, there are disjoint subsets ... | {
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-comb-solutions.jsonl",
"solution_match": "\nProposed by: Pakawut Jiradilok\n"
} | 139 | 1,523 |
2016 | T4 | 9 | Geometry | HMMT | In cyclic quadrilateral $A B C D$ with $A B=A D=49$ and $A C=73$, let $I$ and $J$ denote the incenters of triangles $A B D$ and $C B D$. If diagonal $\overline{B D}$ bisects $\overline{I J}$, find the length of $I J$. | Answer: $\frac{28}{5} \sqrt{69}$
Let $O$ be circumcenter, $R$ the circumradius and $r$ the common inradius. We have $I O^{2}=J O^{2}=$ $R(R-2 r)$ by a result of Euler; denote $x$ for the common value of $I O$ and $J O$. Additionally, we know $A J=A B=A D=49$ (angle chase to find that $\angle B J A=\angle J B A$ ). Sinc... | {
"problem_match": "\n9. ",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-geo-solutions.jsonl",
"solution_match": "\nProposed by: Evan Chen\n"
} | 81 | 542 |
2016 | T4 | 10 | Geometry | HMMT | The incircle of a triangle $A B C$ is tangent to $B C$ at $D$. Let $H$ and $\Gamma$ denote the orthocenter and circumcircle of $\triangle A B C$. The $B$-mixtilinear incircle, centered at $O_{B}$, is tangent to lines $B A$ and $B C$ and internally tangent to $\Gamma$. The $C$-mixtilinear incircle, centered at $O_{C}$, ... | Answer: $\sqrt{\frac{1}{3}(7+2 \sqrt{13})}$
Let the $B$-mixtilinear incircle $\omega_{B}$ touch $\Gamma$ at $T_{B}, B A$ at $B_{1}$ and $B C$ at $B_{2}$. Define $T_{C} \in \Gamma, C_{1} \in C B$, $C_{2} \in C A$, and $\omega_{C}$ similarly. Call $I$ the incenter of triangle $A B C$, and $\gamma$ the incircle.
We first ... | {
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-geo-solutions.jsonl",
"solution_match": "\nProposed by: Evan Chen\n"
} | 150 | 814 |
2016 | T4 | 22 | null | HMMT | On the Cartesian plane $\mathbb{R}^{2}$, a circle is said to be nice if its center is at the origin $(0,0)$ and it passes through at least one lattice point (i.e. a point with integer coordinates). Define the points $A=(20,15)$ and $B=(20,16)$. How many nice circles intersect the open segment $A B$ ?
For reference, the... | Answer: 10
The square of the radius of a nice circle is the sum of the square of two integers.
The nice circle of radius $r$ intersects (the open segment) $\overline{A B}$ if and only if a point on $\overline{A B}$ is a distance $r$ from the origin. $\overline{A B}$ consists of the points $(20, t)$ where $t$ ranges ove... | {
"problem_match": "\n22. [12]",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-guts-solutions.jsonl",
"solution_match": "\nProposed by:\n"
} | 174 | 628 |
2016 | T4 | 25 | null | HMMT | A particular coin can land on heads (H), on tails ( T ), or in the middle (M), each with probability $\frac{1}{3}$. Find the expected number of flips necessary to observe the contiguous sequence HMMTHMMT...HMMT, where the sequence HMMT is repeated 2016 times. | Answer: $\frac{3^{8068}-81}{80}$
Let $E_{0}$ be the expected number of flips needed. Let $E_{1}$ be the expected number more of flips needed if the first flip landed on H . Let $E_{2}$ be the expected number more if the first two landed on HM. In general, let $E_{k}$ be the expected number more of flips needed if the f... | {
"problem_match": "\n25. [14]",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-guts-solutions.jsonl",
"solution_match": "\nProposed by: Ritesh Ragavender\n"
} | 70 | 607 |
2016 | T4 | 26 | null | HMMT | For positive integers $a, b, a \uparrow \uparrow b$ is defined as follows: $a \uparrow \uparrow 1=a$, and $a \uparrow \uparrow b=a^{a \uparrow \uparrow(b-1)}$ if $b>1$.
Find the smallest positive integer $n$ for which there exists a positive integer $a$ such that $a \uparrow \uparrow 6 \not \equiv a \uparrow \uparrow ... | Answer: 283
We see that the smallest such $n$ must be a prime power, because if two numbers are distinct mod $n$, they must be distinct mod at least one of the prime powers that divide $n$. For $k \geq 2$, if $a \uparrow \uparrow k$ and $a \uparrow \uparrow(k+1)$ are distinct $\bmod p^{r}$, then $a \uparrow \uparrow(k-... | {
"problem_match": "\n26. [14]",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-guts-solutions.jsonl",
"solution_match": "\nProposed by: Sammy Luo\n"
} | 114 | 571 |
2016 | T4 | 31 | null | HMMT | For a positive integer $n$, denote by $\tau(n)$ the number of positive integer divisors of $n$, and denote by $\phi(n)$ the number of positive integers that are less than or equal to $n$ and relatively prime to $n$. Call a positive integer $n$ good if $\varphi(n)+4 \tau(n)=n$. For example, the number 44 is good because... | Answer: 172
We claim that $44,56,72$ are the only good numbers. It is easy to check that these numbers work.
Now we prove none others work. First, remark that as $n=1,2$ fail so we have $\varphi(n)$ is even, thus $n$ is even. This gives us $\varphi(n) \leq n / 2$. Now remark that $\tau(n)<2 \sqrt{n}$, so it follows we ... | {
"problem_match": "\n31. [16]",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-guts-solutions.jsonl",
"solution_match": "\nProposed by: Lawrence Sun\n"
} | 114 | 1,089 |
2016 | T4 | 2 | null | HMMT | For positive integers $n$, let $c_{n}$ be the smallest positive integer for which $n^{c_{n}}-1$ is divisible by 210, if such a positive integer exists, and $c_{n}=0$ otherwise. What is $c_{1}+c_{2}+\cdots+c_{210}$ ? | Answer: 329
In order for $c_{n} \neq 0$, we must have $\operatorname{gcd}(n, 210)=1$, so we need only consider such $n$. The number $n^{c_{n}}-1$ is divisible by 210 iff it is divisible by each of $2,3,5$, and 7 , and we can consider the order of $n$ modulo each modulus separately; $c_{n}$ will simply be the LCM of the... | {
"problem_match": "\n2. [25]",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-team-solutions.jsonl",
"solution_match": "\nProposed by: Joy Zheng\n"
} | 76 | 887 |
2016 | T4 | 3 | null | HMMT | Let $A B C$ be an acute triangle with incenter $I$ and circumcenter $O$. Assume that $\angle O I A=90^{\circ}$. Given that $A I=97$ and $B C=144$, compute the area of $\triangle A B C$. | Answer: 14040
We present five different solutions and outline a sixth and seventh one. In what follows, let $a=B C$, $b=C A, c=A B$ as usual, and denote by $r$ and $R$ the inradius and circumradius. Let $s=\frac{1}{2}(a+b+c)$. In the first five solutions we will only prove that
$$
\angle A I O=90^{\circ} \Longrightarr... | {
"problem_match": "\n3. [30]",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-team-solutions.jsonl",
"solution_match": "\nProposed by: Evan Chen\n"
} | 65 | 1,860 |
2016 | T4 | 6 | null | HMMT | A nonempty set $S$ is called well-filled if for every $m \in S$, there are fewer than $\frac{1}{2} m$ elements of $S$ which are less than $m$. Determine the number of well-filled subsets of $\{1,2, \ldots, 42\}$. | Answer: $\binom{43}{21}-1$
Let $a_{n}$ be the number of well-filled subsets whose maximum element is $n$ (setting $a_{0}=1$ ). Then it's easy to see that
$$
\begin{aligned}
& a_{2 k+1}=a_{2 k}+a_{2 k-1}+\cdots+a_{0} \\
& a_{2 k+2}=\left(a_{2 k+1}-C_{k}\right)+a_{2 k}+\cdots+a_{0}
\end{aligned}
$$
where $C_{k}$ is the... | {
"problem_match": "\n6. [35]",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-team-solutions.jsonl",
"solution_match": "\nProposed by: Casey Fu\n"
} | 69 | 507 |
2016 | T4 | 7 | null | HMMT | Let $q(x)=q^{1}(x)=2 x^{2}+2 x-1$, and let $q^{n}(x)=q\left(q^{n-1}(x)\right)$ for $n>1$. How many negative real roots does $q^{2016}(x)$ have? | Answer: $\frac{2^{2017}+1}{3}$
Define $g(x)=2 x^{2}-1$, so that $q(x)=-\frac{1}{2}+g\left(x+\frac{1}{2}\right)$. Thus
$$
q^{N}(x)=0 \Longleftrightarrow \frac{1}{2}=g^{N}\left(x+\frac{1}{2}\right)
$$
where $N=2016$.
But, viewed as function $g:[-1,1] \rightarrow[-1,1]$ we have that $g(x)=\cos (2 \arccos (x))$. Thus, th... | {
"problem_match": "\n7. [40]",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-team-solutions.jsonl",
"solution_match": "\nProposed by: Ernest Chiu\n"
} | 68 | 500 |
2016 | T4 | 9 | null | HMMT | Fix positive integers $r>s$, and let $F$ be an infinite family of sets, each of size $r$, no two of which share fewer than $s$ elements. Prove that there exists a set of size $r-1$ that shares at least $s$ elements with each set in $F$. | We can also use a more indirect approach (where the use of contradiction is actually essential).
Fix $S \in F$ and $a \in S$. By assumption, $S \backslash\{a\}$ does not $s$-meet $F$, so there exists $S^{\prime} \in F$ such that $S^{\prime}$ contains at most $s-1$ elements of $S \backslash\{a\}$, whence $S \cap S^{\pri... | {
"problem_match": "\n9. [40]",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-team-solutions.jsonl",
"solution_match": "\nSolution 2. "
} | 66 | 700 |
2016 | T4 | 10 | null | HMMT | Let $A B C$ be a triangle with incenter $I$ whose incircle is tangent to $\overline{B C}, \overline{C A}, \overline{A B}$ at $D, E, F$. Point $P$ lies on $\overline{E F}$ such that $\overline{D P} \perp \overline{E F}$. Ray $B P$ meets $\overline{A C}$ at $Y$ and ray $C P$ meets $\overline{A B}$ at $Z$. Point $Q$ is se... | The proof proceeds through a series of seven lemmas.
Lemma 1. Lines $D P$ and $E F$ are the internal and external angle bisectors of $\angle B P C$.
Proof. Since $D E F$ the cevian triangle of $A B C$ with respect to its Gregonne point, we have that
$$
-1=(\overline{E F} \cap \overline{B C}, D ; B, C)
$$
Then since $... | {
"problem_match": "\n10. [50]",
"resource_path": "HarvardMIT/segmented/en-192-2016-feb-team-solutions.jsonl",
"solution_match": "\nProposed by: Evan Chen\n"
} | 169 | 839 |
2016 | T4 | 1 | null | HMMT | Theseus starts at the point $(0,0)$ in the plane. If Theseus is standing at the point $(x, y)$ in the plane, he can step one unit to the north to point $(x, y+1)$, one unit to the west to point $(x-1, y)$, one unit to the south to point $(x, y-1)$, or one unit to the east to point $(x+1, y)$. After a sequence of more t... | The path Theseus traces out is a closed, non-self-intersecting path in the plane. Since each move is along a line segment, the path forms the boundary of a polygon in the plane. WLOG this polygon is traversed counterclockwise (i.e. with the interior of the polygon always on Theseus's left); the argument in the clockwis... | {
"problem_match": "\n1. [5]",
"resource_path": "HarvardMIT/segmented/en-194-tournaments-2016-hmic-solutions.jsonl",
"solution_match": "\nProposed by: Mitchell Lee\n"
} | 252 | 591 |
2016 | T4 | 4 | null | HMMT | Let $P$ be an odd-degree integer-coefficient polynomial. Suppose that $x P(x)=y P(y)$ for infinitely many pairs $x, y$ of integers with $x \neq y$. Prove that the equation $P(x)=0$ has an integer root. | Proof. Let $n$ be the (odd) degree of $P$. Suppose, for contradiction, that $P$ has no integer roots, and let $Q(x)=x P(x)=a_{n} x^{n+1}+a_{n-1} x^{n}+\cdots+a_{0} x^{1}$. WLOG $a_{n}>0$, so there exists $M>0$ such that $Q(M)<Q(M+1)<\cdots$ and $Q(-M)<Q(-M-1)<\cdots$ (as $n+1$ is even).
There exist finitely many soluti... | {
"problem_match": "\n4. [10]",
"resource_path": "HarvardMIT/segmented/en-194-tournaments-2016-hmic-solutions.jsonl",
"solution_match": "\nProposed by: Victor Wang\n"
} | 59 | 979 |
2016 | T4 | 5 | null | HMMT | Let $S=\left\{a_{1}, \ldots, a_{n}\right\}$ be a finite set of positive integers of size $n \geq 1$, and let $T$ be the set of all positive integers that can be expressed as sums of perfect powers (including 1) of distinct numbers in $S$, meaning
$$
T=\left\{\sum_{i=1}^{n} a_{i}^{e_{i}} \mid e_{1}, e_{2}, \ldots, e_{n... | Answer: N/A
In general we can assume that each $a_{i}>1$, since replacing $a_{i}=1$ by some large integer $a$ creates a set $T$ containing the original $T$ as a subset (by setting $e_{i}=0$ ).
We proceed by induction on $n$. For the base case $n=1$, an arithmetic progression of length at least 3 would give $a_{1}^{e_{1... | {
"problem_match": "\n5. [12]",
"resource_path": "HarvardMIT/segmented/en-194-tournaments-2016-hmic-solutions.jsonl",
"solution_match": "\nProposed by: Yang Liu\n"
} | 160 | 940 |
2016 | T4 | 7 | null | HMMT | Let $A B C$ be a triangle with $A B=13, B C=14, C A=15$. The altitude from $A$ intersects $B C$ at $D$. Let $\omega_{1}$ and $\omega_{2}$ be the incircles of $A B D$ and $A C D$, and let the common external tangent of $\omega_{1}$ and $\omega_{2}$ (other than $B C$ ) intersect $A D$ at $E$. Compute the length of $A E$. | Let $I_{1}, I_{2}$ be the centers of $\omega_{1}, \omega_{2}$, respectively, $X_{1}, X_{2}$ be the tangency points of $\omega_{1}, \omega_{2}$ with $B C$, respectively, and $Y_{1}, Y_{2}$ be the tangency points of $\omega_{1}, \omega_{2}$ with $A D$, respectively. Let the two common external tangents of $\omega_{1}, \o... | {
"problem_match": "\n7. ",
"resource_path": "HarvardMIT/segmented/en-201-2016-nov-gen-solutions.jsonl",
"solution_match": "\n## Solution 1:\n\n"
} | 115 | 867 |
2016 | T4 | 13 | null | HMMT | How many functions $f:\{0,1\}^{3} \rightarrow\{0,1\}$ satisfy the property that, for all ordered triples $\left(a_{1}, a_{2}, a_{3}\right)$ and $\left(b_{1}, b_{2}, b_{3}\right)$ such that $a_{i} \geq b_{i}$ for all $i, f\left(a_{1}, a_{2}, a_{3}\right) \geq f\left(b_{1}, b_{2}, b_{3}\right)$ ? | ## Answer: 20
Consider the unit cube with vertices $\{0,1\}^{3}$. Let $O=(0,0,0), A=(1,0,0), B=(0,1,0), C=(0,0,1)$, $D=(0,1,1), E=(1,0,1), F=(1,1,0)$, and $P=(1,1,1)$. We want to find a function $f$ on these vertices such that $f(1, y, z) \geq f(0, y, z)$ (and symmetric representations). For instance, if $f(A)=1$, the... | {
"problem_match": "\n13. [9]",
"resource_path": "HarvardMIT/segmented/en-201-2016-nov-guts-solutions.jsonl",
"solution_match": "\nProposed by: Eshaan Nichani\n\n"
} | 123 | 675 |
2016 | T4 | 20 | null | HMMT | Let $\mathcal{V}$ be the volume enclosed by the graph
$$
x^{2016}+y^{2016}+z^{2}=2016
$$
Find $\mathcal{V}$ rounded to the nearest multiple of ten. | Answer: 360
Let $R$ be the region in question. Then we have
$$
[-1,1]^{2} \times[-\sqrt{2014}, \sqrt{2014}] \subset R \subset[-\sqrt[2016]{2016}, \sqrt[2016]{2016}]^{2} \times[\sqrt{2016}, \sqrt{2016}]
$$
We find some bounds: we have
$$
\sqrt{2016}<\sqrt{2025}=45
$$
By concavity of $\sqrt{\cdot}$, we have the bound... | {
"problem_match": "\n20. [11]",
"resource_path": "HarvardMIT/segmented/en-201-2016-nov-guts-solutions.jsonl",
"solution_match": "\nProposed by: Henrik Boecken\n"
} | 59 | 527 |
2016 | T4 | 5 | null | HMMT | Allen and Brian are playing a game in which they roll a 6 -sided die until one of them wins. Allen wins if two consecutive rolls are equal and at most 3 . Brian wins if two consecutive rolls add up to 7 and the latter is at most 3 . What is the probability that Allen wins? | Answer: $5 / 12$
Note that at any point in the game after the first roll, the probability that Allen wins depends only on the most recent roll, and not on any rolls before that one. So we may define $p$ as the probability that Allen wins at any point in the game, given that the last roll was a 1,2 , or 3 , and $q$ as t... | {
"problem_match": "\n5. [5]",
"resource_path": "HarvardMIT/segmented/en-201-2016-nov-team-solutions.jsonl",
"solution_match": "\nProposed by: Eshaan Nichani\n"
} | 66 | 578 |
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