Found problems: 28
2023 LMT Fall, 5B
Bamal, Halvan, and Zuca are playing [i]The Game[/i]. To start, they‘re placed at random distinct vertices on regular
hexagon $ABCDEF$. Two or more players collide when they‘re on the same vertex. When this happens, all the colliding players lose and the game ends. Every second, Bamal and Halvan teleport to a random vertex adjacent to their current position (each with probability $\dfrac{1}{2}$), and Zuca teleports to a random vertex adjacent to his current position, or to the vertex directly opposite him (each with probability $\dfrac{1}{3}$). What is the probability that when [i]The Game[/i] ends Zuca hasn‘t lost?
[i]Proposed by Edwin Zhao[/i]
[hide=Solution][i]Solution.[/i] $\boxed{\dfrac{29}{90}}$
Color the vertices alternating black and white. By a parity argument if someone is on a different color than the other
two they will always win. Zuca will be on opposite parity from the others with probability $\dfrac{3}{10}$. They will all be on the same parity with probability $\dfrac{1}{10}$.
At this point there are $2 \cdot 2 \cdot 3$ possible moves. $3$ of these will lead to the same arrangement, so we disregard those. The other $9$ moves are all equally likely to end the game. Examining these, we see that Zuca will win in exactly $2$ cases (when Bamal and Halvan collide and Zuca goes to a neighboring vertex). Combining all of this, the answer is
$$\dfrac{3}{10}+\dfrac{2}{9} \cdot \dfrac{1}{10}=\boxed{\dfrac{29}{90}}$$
[/hide]
2012 Irish Math Olympiad, 3
Find, with proof, all polynomials $f$ such that $f$ has nonnegative integer coefficients, $f$($1$) = $8$ and $f$($2$) = $2012$.
2020 Harvest Math Invitational Team Round Problems, HMI Team #6
6. A triple of integers $(a,b,c)$ is said to be $\gamma$[i]-special[/i] if $a\le \gamma(b+c)$, $b\le \gamma(c+a)$ and $c\le\gamma(a+b)$. For each integer triple $(a,b,c)$ such that $1\le a,b,c \le 20$, Kodvick writes down the smallest value of $\gamma$ such that $(a,b,c)$ is $\gamma$-special. How many distinct values does he write down?
[i]Proposed by winnertakeover[/i]
2021 Harvard-MIT Mathematics Tournament., 10
Jude repeatedly flips a coin. If he has already flipped $n$ heads, the coin lands heads with probability $\tfrac{1}{n+2}$ and tails with probability $\tfrac{n+1}{n+2}.$ If Jude continues flipping forever, let $p$ be the probability that he flips $3$ heads in a row at some point. Compute $\lfloor 180p \rfloor.$
LMT Theme Rounds, 2023F 5B
Bamal, Halvan, and Zuca are playing [i]The Game[/i]. To start, they‘re placed at random distinct vertices on regular
hexagon $ABCDEF$. Two or more players collide when they‘re on the same vertex. When this happens, all the colliding players lose and the game ends. Every second, Bamal and Halvan teleport to a random vertex adjacent to their current position (each with probability $\dfrac{1}{2}$), and Zuca teleports to a random vertex adjacent to his current position, or to the vertex directly opposite him (each with probability $\dfrac{1}{3}$). What is the probability that when [i]The Game[/i] ends Zuca hasn‘t lost?
[i]Proposed by Edwin Zhao[/i]
[hide=Solution][i]Solution.[/i] $\boxed{\dfrac{29}{90}}$
Color the vertices alternating black and white. By a parity argument if someone is on a different color than the other
two they will always win. Zuca will be on opposite parity from the others with probability $\dfrac{3}{10}$. They will all be on the same parity with probability $\dfrac{1}{10}$.
At this point there are $2 \cdot 2 \cdot 3$ possible moves. $3$ of these will lead to the same arrangement, so we disregard those. The other $9$ moves are all equally likely to end the game. Examining these, we see that Zuca will win in exactly $2$ cases (when Bamal and Halvan collide and Zuca goes to a neighboring vertex). Combining all of this, the answer is
$$\dfrac{3}{10}+\dfrac{2}{9} \cdot \dfrac{1}{10}=\boxed{\dfrac{29}{90}}$$
[/hide]
2023 LMT Fall, 1B
Evaluate $\dbinom{6}{0}+\dbinom{6}{1}+\dbinom{6}{4}+\dbinom{6}{3}+\dbinom{6}{4}+\dbinom{6}{5}+\dbinom{6}{6}$
[i]Proposed by Jonathan Liu[/i]
[hide=Solution]
[i]Solution.[/i] $\boxed{64}$
We have that $\dbinom{6}{4}=\dbinom{6}{2}$, so $\displaystyle\sum_{n=0}^{6} \dbinom{6}{n}=2^6=\boxed{64}.$
[/hide]
2021 Harvard-MIT Mathematics Tournament., 6
A light pulse starts at a corner of a reflective square. It bounces around inside the square, reflecting off of the square’s perimeter $n$ times before ending in a different corner. The path of the light pulse, when traced, divides the square into exactly $2021$ regions. Compute the smallest possible value of $n$.
2023 LMT Fall, 4
The numbers $1$, $2$, $3$, and $4$ are randomly arranged in a $2$ by $2$ grid with one number in each cell. Find the probability the sum of two numbers in the top row of the grid is even.
[i]Proposed by Muztaba Syed and Derek Zhao[/i]
[hide=Solution]
[i]Solution. [/i]$\boxed{\dfrac{1}{3}}$
Pick a number for the top-left. There is one number that makes the sum even no matter what we pick. Therefore, the
answer is $\boxed{\dfrac{1}{3}}$.[/hide]
2022 Olimphíada, 3
Let $m$ and $n$ be positive integers. In Philand, the Kingdom of Olymphics, with $m$ cities, and the Kingdom of Mathematicians for Fun, with $n$ cities, fight a battle in rounds. Some cities in the country are connected by roads, so that it is possible to travel through all the cities via the roads. In each round of the battle, if all cities neighboring, that is, connected directly by a road, a city in one of the kingdoms are from the other kingdom, that city is conquered in the next round and switches to the other kingdom. Knowing that between the first and second round, at least one city is not conquered, show that at some point the battle must end, i.e., no city can be captured by another kingdom.
2021 Harvard-MIT Mathematics Tournament., 7
Let $S = \{1, 2, \dots , 2021\}$, and let $\mathcal{F}$ denote the set of functions $f : S \rightarrow S$. For a function $f \in \mathcal{F},$ let
\[T_f =\{f^{2021}(s) : s \in S\},\]
where $f^{2021}(s)$ denotes $f(f(\cdots(f(s))\cdots))$ with $2021$ copies of $f$. Compute the remainder when
\[\sum_{f \in \mathcal{F}} |T_f|\]
is divided by the prime $2017$, where the sum is over all functions $f$ in $\mathcal{F}$.
2021 Harvard-MIT Mathematics Tournament., 1
Leo the fox has a $5$ by $5$ checkerboard grid with alternating red and black squares. He fills in the
grid with the numbers $1, 2, 3, \dots, 25$ such that any two consecutive numbers are in adjacent squares
(sharing a side) and each number is used exactly once. He then computes the sum of the numbers in
the $13$ squares that are the same color as the center square. Compute the maximum possible sum Leo
can obtain.
2021 Harvard-MIT Mathematics Tournament., 5
Teresa the bunny has a fair $8$-sided die. Seven of its sides have fixed labels $1, 2, \cdots , 7,$ and the label on the eighth side can be changed and begins as $1$. She rolls it several times, until each of $1, 2, \dots, 7$ appears at least once. After each roll, if $k$ is the smallest positive integer that she has not rolled so far, she relabels the eighth side with $k$. The probability that $7$ is the last number she rolls is $\tfrac ab,$ where $a$ and $b$ are relatively prime positive integers. Compute $100a + b$.
2021 Harvard-MIT Mathematics Tournament., 3
Let $N$ be a positive integer. Brothers Michael and Kylo each select a positive integer less than or equal to $N$, independently and uniformly at random. Let $p_N$ denote the probability that the product of these two integers has a units digit of $0$. The maximum possible value of $p_N$ over all possible choices of $N$ can be written as $\tfrac ab,$ where $a$ and $b$ are relatively prime positive integers. Compute $100a + b$.
2020 Harvest Math Invitational Team Round Problems, HMI Team #4
4. There are 5 tables in a classroom. Each table has 4 chairs with a child sitting on it. All the children get up and randomly sit in a seat. Two people that sat at the same table before are not allowed to sit at the same table again. Assuming tables and chairs are distinguishable, if the number of different classroom arrangements can be written as $2^a3^b5^c$, what is $a+b+c$?
[i]Proposed by Tragic[/i]
2021 Harvard-MIT Mathematics Tournament., 9
An up-right path between two lattice points $P$ and $Q$ is a path from $P$ to $Q$ that takes steps of length $1$ unit either up or to the right.
How many up-right paths from $(0, 0)$ to $(7, 7),$ when drawn in the plane with the line $y = x - 2.021$,
enclose exactly one bounded region below that line?
LMT Speed Rounds, 4
The numbers $1$, $2$, $3$, and $4$ are randomly arranged in a $2$ by $2$ grid with one number in each cell. Find the probability the sum of two numbers in the top row of the grid is even.
[i]Proposed by Muztaba Syed and Derek Zhao[/i]
[hide=Solution]
[i]Solution. [/i]$\boxed{\dfrac{1}{3}}$
Pick a number for the top-left. There is one number that makes the sum even no matter what we pick. Therefore, the
answer is $\boxed{\dfrac{1}{3}}$.[/hide]
2021 Harvard-MIT Mathematics Tournament., 2
Ava and Tiffany participate in a knockout tournament consisting of a total of $32$ players. In each of $5$ rounds, the remaining players are paired uniformly at random. In each pair, both players are equally likely to win, and the loser is knocked out of the tournament. The probability that Ava and Tiffany play each other during the tournament is $\tfrac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. Compute $100a + b.$
2022 EGMO, 5
For all positive integers $n$, $k$, let $f(n, 2k)$ be the number of ways an $n \times 2k$ board can be fully covered by $nk$ dominoes of size $2 \times 1$. (For example, $f(2, 2)=2$ and $f(3, 2)=3$.) Find all positive integers $n$ such that for every positive integer $k$, the number $f(n, 2k)$ is odd.
LMT Theme Rounds, 2023F 1B
Evaluate $\dbinom{6}{0}+\dbinom{6}{1}+\dbinom{6}{4}+\dbinom{6}{3}+\dbinom{6}{4}+\dbinom{6}{5}+\dbinom{6}{6}$
[i]Proposed by Jonathan Liu[/i]
[hide=Solution]
[i]Solution.[/i] $\boxed{64}$
We have that $\dbinom{6}{4}=\dbinom{6}{2}$, so $\displaystyle\sum_{n=0}^{6} \dbinom{6}{n}=2^6=\boxed{64}.$
[/hide]
2023 LMT Fall, 3B
Evin and Jerry are playing a game with a pile of marbles. On each players' turn, they can remove $2$, $3$, $7$, or $8$
marbles. If they can’t make a move, because there's $0$ or $1$ marble left, they lose the game. Given that Evin goes first and both players play optimally, for how many values of $n$ from $1$ to $1434$ does Evin lose the game?
[i]Proposed by Evin Liang[/i]
[hide=Solution][i]Solution.[/i] $\boxed{573}$
Observe that no matter how many marbles a one of them removes, the next player can always remove marbles such
that the total number of marbles removed is $10$. Thus, when the number of marbles is a multiple of $10$, the first player loses the game. We analyse this game based on the number of marbles modulo $10$:
If the number of marbles is $0$ modulo $10$, the first player loses the game
If the number of marbles is $2$, $3$, $7$, or $8$ modulo $10$, the first player wins the game by moving to $0$ modulo 10
If the number of marbles is $5$ modulo $10$, the first player loses the game because every move leads to $2$, $3$, $7$, or $8$ modulo $10$
In summary, the first player loses if it is $0$ mod 5, and wins if it is $2$ or $3$ mod $5$. Now we solve the remaining cases by induction. The first player loses when it is $1$ modulo $5$ and wins when it is $4$ modulo $5$. The base case is when there is $1$ marble, where the first player loses because there is no move. When it is $4$ modulo $5$, then the first player can always remove $3$ marbles and win by the inductive hypothesis. When it is $1$ modulo $5$, every move results in $3$ or $4$ modulo $5$, which allows the other player to win by the inductive hypothesis.
Thus, Evin loses the game if n is $0$ or $1$ modulo $5$. There are $\boxed{573}$ such values of $n$ from $1$ to $1434$.[/hide]
LMT Speed Rounds, 6
Blue rolls a fair $n$-sided die that has sides its numbered with the integers from $1$ to $n$, and then he flips a coin. Blue knows that the coin is weighted to land heads either $\dfrac{1}{3}$ or $\dfrac{2}{3}$ of the time. Given that the probability of both rolling a $7$ and flipping heads is $\dfrac{1}{15}$, find $n$.
[i]Proposed by Jacob Xu[/i]
[hide=Solution][i]Solution[/i]. $\boxed{10}$
The chance of getting any given number is $\dfrac{1}{n}$
, so the probability of getting $7$ and heads is either $\dfrac{1}{n} \cdot \dfrac{1}{3}=\dfrac{1}{3n}$ or $\dfrac{1}{n} \cdot \dfrac{2}{3}=\dfrac{2}{3n}$. We get that either $n = 5$ or $n = 10$, but since rolling a $7$ is possible, only $n = \boxed{10}$ is a solution.[/hide]
2023 India Regional Mathematical Olympiad, 6
Consider a set of $16$ points arranged in $4 \times 4$ square grid formation. Prove that if any $7$ of these points are coloured blue, then there exists an isosceles right-angled triangle whose vertices are all blue.
2023 LMT Fall, 6
Blue rolls a fair $n$-sided die that has sides its numbered with the integers from $1$ to $n$, and then he flips a coin. Blue knows that the coin is weighted to land heads either $\dfrac{1}{3}$ or $\dfrac{2}{3}$ of the time. Given that the probability of both rolling a $7$ and flipping heads is $\dfrac{1}{15}$, find $n$.
[i]Proposed by Jacob Xu[/i]
[hide=Solution][i]Solution[/i]. $\boxed{10}$
The chance of getting any given number is $\dfrac{1}{n}$
, so the probability of getting $7$ and heads is either $\dfrac{1}{n} \cdot \dfrac{1}{3}=\dfrac{1}{3n}$ or $\dfrac{1}{n} \cdot \dfrac{2}{3}=\dfrac{2}{3n}$. We get that either $n = 5$ or $n = 10$, but since rolling a $7$ is possible, only $n = \boxed{10}$ is a solution.[/hide]
2021 Harvard-MIT Mathematics Tournament., 8
Compute the number of ways to fill each cell in a $8 \times 8$ square grid with one of the letters $H, M,$ or $T$ such that every $2 \times 2$ square in the grid contains the letters $H, M, M, T$ in some order.
2021 Harvard-MIT Mathematics Tournament., 4
Let $S = \{1, 2, \dots, 9\}.$ Compute the number of functions $f : S \rightarrow S$ such that, for all $s \in S, f(f(f(s))) =s$ and $f(s) - s$ is not divisible by $3$.