Found problems: 2008
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]
1996 Niels Henrik Abels Math Contest (Norwegian Math Olympiad) Round 2, 8
Let $ x$ and $ y$ be positive integers. The least possible value of $ |11x^5 \minus{} 7y^3|$ is
A. 1
B. 2
C. 3
D. 4
E. None of these
1992 AMC 12/AHSME, 17
The two digit integers from $19$ to $92$ are written consecutively to form the larger integer $N = 19202122\ldots909192$. If $3^{k}$ is the highest power of $3$ that is a factor of $N$, then $k =$
$ \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \text{more than 3} $
2016 Iran MO (3rd Round), 3
A $30\times30$ table is given. We want to color some of it's unit squares such that any colored square has at most $k$ neighbors. ( Two squares $(i,j)$ and $(x,y)$ are called neighbors if $i-x,j-y\equiv0,-1,1 \pmod {30}$ and $(i,j)\neq(x,y)$. Therefore, each square has exactly $8$ neighbors)
What is the maximum possible number of colored squares if$:$
$a) k=6$
$b)k=1$
1990 China Team Selection Test, 3
Prove that for every integer power of 2, there exists a multiple of it with all digits (in decimal expression) not zero.
2013 International Zhautykov Olympiad, 2
Find all odd positive integers $n>1$ such that there is a permutation $a_1, a_2, a_3, \ldots, a_n$ of the numbers $1, 2,3, \ldots, n$ where $n$ divides one of the numbers $a_k^2 - a_{k+1} - 1$ and $a_k^2 - a_{k+1} + 1$ for each $k$, $1 \leq k \leq n$ (we assume $a_{n+1}=a_1$).
2012 ELMO Shortlist, 1
Find all positive integers $n$ such that $4^n+6^n+9^n$ is a square.
[i]David Yang, Alex Zhu.[/i]
2013 Kazakhstan National Olympiad, 1
On the board written numbers from 1 to 25 . Bob can pick any three of them say $a,b,c$ and replace by $a^3+b^3+c^3$ . Prove that last number on the board can not be $2013^3$.
2010 China Western Mathematical Olympiad, 1
Suppose that $m$ and $k$ are non-negative integers, and $p = 2^{2^m}+1$ is a prime number. Prove that
[b](a)[/b] $2^{2^{m+1}p^k} \equiv 1$ $(\text{mod } p^{k+1})$;
[b](b)[/b] $2^{m+1}p^k$ is the smallest positive integer $n$ satisfying the congruence equation $2^n \equiv 1$ $(\text{mod } p^{k+1})$.
2013 AIME Problems, 14
For positive integers $n$ and $k$, let $f(n,k)$ be the remainder when $n$ is divided by $k$, and for $n>1$ let $F(n) = \displaystyle\max_{1 \le k \le \frac{n}{2}} f(n,k)$. Find the remainder when $\displaystyle\sum_{n=20}^{100} F(n)$ is divided by $1000$.
2015 India National Olympiad, 6
Show that from a set of $11$ square integers one can select six numbers $a^2,b^2,c^2,d^2,e^2,f^2$ such that $a^2+b^2+c^2 \equiv d^2+e^2+f^2\pmod{12}$.
2004 Vietnam National Olympiad, 3
Let $ S(n)$ be the sum of decimal digits of a natural number $ n$. Find the least value of $ S(m)$ if $ m$ is an integral multiple of $ 2003$.
1991 AIME Problems, 6
Suppose $r$ is a real number for which \[ \left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546. \] Find $\lfloor 100r \rfloor$. (For real $x$, $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.)
2012 All-Russian Olympiad, 2
Does there exist natural numbers $a,b,c$ all greater than $10^{10}$ such that their product is divisible by each of these numbers increased by $2012$?
2011 Brazil Team Selection Test, 4
Let $a, b$ be integers, and let $P(x) = ax^3+bx.$ For any positive integer $n$ we say that the pair $(a,b)$ is $n$-good if $n | P(m)-P(k)$ implies $n | m - k$ for all integers $m, k.$ We say that $(a,b)$ is $very \ good$ if $(a,b)$ is $n$-good for infinitely many positive integers $n.$
[list][*][b](a)[/b] Find a pair $(a,b)$ which is 51-good, but not very good.
[*][b](b)[/b] Show that all 2010-good pairs are very good.[/list]
[i]Proposed by Okan Tekman, Turkey[/i]
1999 Romania Team Selection Test, 10
Determine all positive integers $n$ for which there exists an integer $m$ such that ${2^{n}-1}$ is a divisor of ${m^{2}+9}$.
1987 IMO Shortlist, 23
Prove that for every natural number $k$ ($k \geq 2$) there exists an irrational number $r$ such that for every natural number $m$,
\[[r^m] \equiv -1 \pmod k .\]
[i]Remark.[/i] An easier variant: Find $r$ as a root of a polynomial of second degree with integer coefficients.
[i]Proposed by Yugoslavia.[/i]
2009 China Team Selection Test, 3
Let $ f(x)$ be a $ n \minus{}$degree polynomial all of whose coefficients are equal to $ \pm 1$, and having $ x \equal{} 1$ as its $ m$ multiple root. If $ m\ge 2^k (k\ge 2,k\in N)$, then $ n\ge 2^{k \plus{} 1} \minus{} 1.$
2021 China Team Selection Test, 4
Let $f(x),g(x)$ be two polynomials with integer coefficients. It is known that for infinitely many prime $p$, there exist integer $m_p$ such that
$$f(a) \equiv g(a+m_p) \pmod p$$
holds for all $a \in \mathbb{Z}.$ Prove that there exists a rational number $r$ such that
$$f(x)=g(x+r).$$
2011 IMO Shortlist, 7
Let $p$ be an odd prime number. For every integer $a,$ define the number $S_a = \sum^{p-1}_{j=1} \frac{a^j}{j}.$ Let $m,n \in \mathbb{Z},$ such that $S_3 + S_4 - 3S_2 = \frac{m}{n}.$ Prove that $p$ divides $m.$
[i]Proposed by Romeo Meštrović, Montenegro[/i]
2013 Baltic Way, 19
Let $a_0$ be a positive integer and $a_n=5a_{n-1}+4$ for all $n\ge 1$. Can $a_0$ be chosen so that $a_{54}$ is a multiple of $2013$?
Oliforum Contest II 2009, 4
Let $ m$ a positive integer and $ p$ a prime number, both fixed. Define $ S$ the set of all $ m$-uple of positive integers $ \vec{v} \equal{} (v_1,v_2,\ldots,v_m)$ such that $ 1 \le v_i \le p$ for all $ 1 \le i \le m$. Define also the function $ f(\cdot): \mathbb{N}^m \to \mathbb{N}$, that associates every $ m$-upla of non negative integers $ (a_1,a_2,\ldots,a_m)$ to the integer $ \displaystyle f(a_1,a_2,\ldots,a_m) \equal{} \sum_{\vec{v} \in S} \left(\prod_{1 \le i \le m}{v_i^{a_i}} \right)$.
Find all $ m$-uple of non negative integers $ (a_1,a_2,\ldots,a_m)$ such that $ p \mid f(a_1,a_2,\ldots,a_m)$.
[i](Pierfrancesco Carlucci)[/i]
2016 Iran MO (3rd Round), 1
Let $p,q$ be prime numbers ($q$ is odd). Prove that there exists an integer $x$ such that:
$$q |(x+1)^p-x^p$$
If and only if $$q \equiv 1 \pmod p$$
2010 Princeton University Math Competition, 7
Let $f$ be a function such that $f(x)+f(x+1)=2^x$ and $f(0)=2010$. Find the last two digits of $f(2010)$.
2011 Switzerland - Final Round, 7
For a given rational number $r$, find all integers $z$ such that \[2^z + 2 = r^2\mbox{.}\]
[i](Swiss Mathematical Olympiad 2011, Final round, problem 7)[/i]