This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 536

2010 Postal Coaching, 1

Let $n \in \mathbb N$ and $A_n$ set of all permutations $(a_1, \ldots, a_n)$ of the set $\{1, 2, \ldots , n\}$ for which \[k|2(a_1 + \cdots+ a_k), \text{ for all } 1 \leq k \leq n.\] Find the number of elements of the set $A_n$. [i]Proposed by Vidan Govedarica, Serbia[/i]

2023 Bulgaria JBMO TST, 2

Determine the smallest positive integer $n\geq 2$ for which there exists a positive integer $m$ such that $mn$ divides $m^{2023} + n^{2023} + n$.

2006 France Team Selection Test, 3

Let $a$, $b$ be positive integers such that $b^n+n$ is a multiple of $a^n+n$ for all positive integers $n$. Prove that $a=b$. [i]Proposed by Mohsen Jamali, Iran[/i]

1988 IMO Longlists, 1

An integer sequence is defined by \[{ a_n = 2 a_{n-1} + a_{n-2}}, \quad (n > 1), \quad a_0 = 0, a_1 = 1.\] Prove that $2^k$ divides $a_n$ if and only if $2^k$ divides $n$.

1989 IMO Longlists, 86

Let $ m$ be a positive odd integer, $ m > 2.$ Find the smallest positive integer $ n$ such that $ 2^{1989}$ divides $ m^n \minus{} 1.$

2016 Brazil Team Selection Test, 3

Let $m$ and $n$ be positive integers such that $m>n$. Define $x_k=\frac{m+k}{n+k}$ for $k=1,2,\ldots,n+1$. Prove that if all the numbers $x_1,x_2,\ldots,x_{n+1}$ are integers, then $x_1x_2\ldots x_{n+1}-1$ is divisible by an odd prime.

2015 IMO Shortlist, N3

Let $m$ and $n$ be positive integers such that $m>n$. Define $x_k=\frac{m+k}{n+k}$ for $k=1,2,\ldots,n+1$. Prove that if all the numbers $x_1,x_2,\ldots,x_{n+1}$ are integers, then $x_1x_2\ldots x_{n+1}-1$ is divisible by an odd prime.

2024 Balkan MO, 3

Let $a$ and $b$ be distinct positive integers such that $3^a + 2$ is divisible by $3^b + 2$. Prove that $a > b^2$. [i]Proposed by Tynyshbek Anuarbekov, Kazakhstan[/i]

2002 VJIMC, Problem 2

Let $p>3$ be a prime number and $n=\frac{2^{2p}-1}3$. Show that $n$ divides $2^n-2$.

2015 Indonesia MO, 5

Given positive integers $a,b,c,d$ such that $a\mid c^d$ and $b\mid d^c$. Prove that \[ ab\mid (cd)^{max(a,b)} \]

2006 Taiwan TST Round 1, 3

Let $a$, $b$ be positive integers such that $b^n+n$ is a multiple of $a^n+n$ for all positive integers $n$. Prove that $a=b$. [i]Proposed by Mohsen Jamali, Iran[/i]

2005 IberoAmerican, 3

Let $p > 3$ be a prime. Prove that if \[ \sum_{i=1 }^{p-1}{1\over i^p} = {n\over m}, \] with $\gdc(n,m) = 1$, then $p^3$ divides $n$.

2022 China Team Selection Test, 6

Given a positive integer $n$, let $D$ be the set of all positive divisors of $n$. The subsets $A,B$ of $D$ satisfies that for any $a \in A$ and $b \in B$, it holds that $a \nmid b$ and $b \nmid a$. Show that \[ \sqrt{|A|}+\sqrt{|B|} \le \sqrt{|D|}. \]

2021 Taiwan TST Round 1, N

Given a positive integer $k$ show that there exists a prime $p$ such that one can choose distinct integers $a_1,a_2\cdots, a_{k+3} \in \{1, 2, \cdots ,p-1\}$ such that p divides $a_ia_{i+1}a_{i+2}a_{i+3}-i$ for all $i= 1, 2, \cdots, k$. [i]South Africa [/i]

1978 IMO Longlists, 52

Let $p$ be a prime and $A = \{a_1, \ldots , a_{p-1} \}$ an arbitrary subset of the set of natural numbers such that none of its elements is divisible by $p$. Let us define a mapping $f$ from $\mathcal P(A)$ (the set of all subsets of $A$) to the set $P = \{0, 1, \ldots, p - 1\}$ in the following way: $(i)$ if $B = \{a_{i_{1}}, \ldots , a_{i_{k}} \} \subset A$ and $\sum_{j=1}^k a_{i_{j}} \equiv n \pmod p$, then $f(B) = n,$ $(ii)$ $f(\emptyset) = 0$, $\emptyset$ being the empty set. Prove that for each $n \in P$ there exists $B \subset A$ such that $f(B) = n.$

2024 Kyiv City MO Round 2, Problem 2

Mykhailo wants to arrange all positive integers from $1$ to $2024$ in a circle so that each number is used exactly once and for any three consecutive numbers $a, b, c$ the number $a + c$ is divisible by $b + 1$. Can he do it? [i]Proposed by Fedir Yudin[/i]

2022 IMO, 5

Find all triples $(a,b,p)$ of positive integers with $p$ prime and \[ a^p=b!+p. \]

2016 Switzerland Team Selection Test, Problem 11

Let $m$ and $n$ be positive integers such that $m>n$. Define $x_k=\frac{m+k}{n+k}$ for $k=1,2,\ldots,n+1$. Prove that if all the numbers $x_1,x_2,\ldots,x_{n+1}$ are integers, then $x_1x_2\ldots x_{n+1}-1$ is divisible by an odd prime.

2024 Polish Junior MO Finals, 5

Let $S=\underbrace{111\dots 1}_{19}\underbrace{999\dots 9}_{19}$. Show that the $2S$-digit number \[\underbrace{111\dots 1}_{S}\underbrace{999\dots 9}_{S}\] is a multiple of $19$.

2016 Saudi Arabia IMO TST, 2

Tags: SAU , Divisibility
Let $a$ be a positive integer. Find all prime numbers $ p $ with the following property: there exist exactly $ p $ ordered pairs of integers $ (x, y)$, with $ 0 \leq  x, y \leq p - 1 $, such that $ p $ divides $ y^2 - x^3 - a^2x $.

1969 IMO Shortlist, 34

$(HUN 1)$ Let $a$ and $b$ be arbitrary integers. Prove that if $k$ is an integer not divisible by $3$, then $(a + b)^{2k}+ a^{2k} +b^{2k}$ is divisible by $a^2 +ab+ b^2$

1969 IMO Shortlist, 24

$(GBR 1)$ The polynomial $P(x) = a_0x^k + a_1x^{k-1} + \cdots + a_k$, where $a_0,\cdots, a_k$ are integers, is said to be divisible by an integer $m$ if $P(x)$ is a multiple of $m$ for every integral value of $x$. Show that if $P(x)$ is divisible by $m$, then $a_0 \cdot k!$ is a multiple of $m$. Also prove that if $a, k,m$ are positive integers such that $ak!$ is a multiple of $m$, then a polynomial $P(x)$ with leading term $ax^k$can be found that is divisible by $m.$

2023-24 IOQM India, 4

Let $x, y$ be positive integers such that $$ x^4=(x-1)\left(y^3-23\right)-1 . $$ Find the maximum possible value of $x+y$.

1977 Germany Team Selection Test, 4

When $4444^{4444}$ is written in decimal notation, the sum of its digits is $ A.$ Let $B$ be the sum of the digits of $A.$ Find the sum of the digits of $ B.$ ($A$ and $B$ are written in decimal notation.)

2008 Germany Team Selection Test, 1

Find all pairs of natural numbers $ (a, b)$ such that $ 7^a \minus{} 3^b$ divides $ a^4 \plus{} b^2$. [i]Author: Stephan Wagner, Austria[/i]