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.

AND
OR
NO

Found problems: 700

Let $ p \geq 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1,2,\ldots, p-1 \}$ that was not chosen before by either of the two players and then chooses an element $a_i$ from the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices have been chosen .Then the following number is computed: $$M=a_0+a_110+a_210^2+\cdots+a_{p-1}10^{p-1}= \sum_{i=0}^{p-1}a_i.10^i$$. The goal of Eduardo is to make $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy. [i]Proposed by Amine Natik, Morocco[/i]
$ a \minus{} )$ Find all prime $ p$ such that $ \dfrac{7^{p \minus{} 1} \minus{} 1}{p}$ is a perfect square $ b \minus{} )$ Find all prime $ p$ such that $ \dfrac{11^{p \minus{} 1} \minus{} 1}{p}$ is a perfect square
Determine, with proof, if there exists an odd prime $p$ such that the following equation holds: $$\sum_{n = 1}^{\frac{p-1}{2}} \cot\left(\frac{\pi n^2}{p}\right) = 69\sqrt{p}$$ [i]Proposed by Chris Bao[/i]
Let $m, n, a, k$ be positive integers and $k>1$ such that the equality $$5^m+63n+49=a^k$$ holds. Find the minimum value of $k$.
We call a positive integer $n$ is $good$ , if there exist integers $a,b,c,x,y$ such that $n=ax^2+bxy+cy^2$ and $b^2-4ac=-20$. Prove that the product of any two good numbers is also a good number.
Find the smallest $x \in\mathbb{N}$ for which $\frac{7x^{25}-10}{83}$ is an integer.
S is the set of all ($a$, $b$, $c$, $d$, $e$, $f$) where $a$, $b$, $c$, $d$, $e$, $f$ are integers such that $a^2 + b^2 + c^2 + d^2 + e^2 = f^2$. Find the largest $k$ which divides abcdef for all members of $S$.
Find the total number of primes $p<100$ such that $\lfloor (2+\sqrt{5})^p \rfloor-2^{p+1}$ is divisible by $p$. Here $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.
Let $p$ and $q$ be prime numbers of the form $4k+3$. Suppose that there exist integers $x$ and $y$ such that $x^2-pqy^2=1$. Prove that there exist positive integers $a$ and $b$ such that $|pa^2-qb^2|=1$.
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]
Find all natural integers $n$ such that $(n^3 + 39n - 2)n! + 17\cdot 21^n + 5$ is a square.
Let $b \geqslant 2$ be a positive integer. Anu has an infinite collection of notes with exactly $b-1$ copies of a note worth $b^k-1$ rupees, for every integer $k\geqslant 1$. A positive integer $n$ is called payable if Anu can pay exactly $n^2+1$ rupees by using some collection of her notes. Prove that if there is a payable number, there are infinitely many payable numbers. [i]Proposed by Shantanu Nene[/i]
We say that an integer $a$ is a quadratic, cubic, or quintic residue modulo $n$ if there exists an integer $x$ such that $x^2\equiv a \pmod n$, $x^3 \equiv a \pmod n$, or $x^5 \equiv a \pmod n$, respectively. Further, an integer $a$ is a primitive residue modulo $n$ if it is exactly one of these three types of residues modulo $n$. How many integers $1 \le a \le 2015$ are primitive residues modulo $2015$? [i] Proposed by Michael Ren [/i]
We call a positive integer $n\ge 4$[i] beautiful[/i] if there exists some permutation $$\{x_1,x_2,\dots ,x_{n-1}\}$$ of $\{1,2,\dots ,n-1\}$ such that $\{x^1_1,\ x^2_2,\ \dots,x^{n-1}_{n-1}\}$ gives all the residues $\{1,2,\dots, n-1\}$ modulo $n$. Prove that if $n$ is beautiful then $n=2p,$ for some prime number $p.$
The numbers from 1 to $ 2013^2 $ are written row by row into a table consisting of $ 2013 \times 2013 $ cells. Afterwards, all columns and all rows containing at least one of the perfect squares $ 1, 4, 9, \cdots, 2013^2 $ are simultaneously deleted. How many cells remain?
Determine the number of all numbers which are represented as $x^2+y^2$ with $x, y \in \{1, 2, 3, \ldots, 1000\}$ and which are divisible by 121.
For any prime $p$, prove that there exists integer $x_0$ such that $p | (x^2_0 - x_0 + 3)$ $\Leftrightarrow$ there exists integer $y_0$ such that $p | (y^2_0 - y_0 + 25).$
Prove that there are infinite many positive integers $ n$ such that $ n^2\plus{}1\mid n!$, and infinite many of those for which $ n^2\plus{}1 \nmid n!$.
Given odd integers $a,b,c$ prove that the equation $ax^2+bx+c=0$ cannot have a solution $x$ which is a rational number.
A subset $S$ of the set $M=\{1,2,.....,p-1\}$,where $p$ is a prime number of the kind $12n+11$,is [i]essential[/i],if the product ${\Pi}_s$ of all elements of the subset is not less than the product $\bar{{\Pi}_s}$ of all other elements of the set.The [b]difference[/b] $\bigtriangleup_s=\Pi_s-\bar{{\Pi}_s}$ is called [i]the deviation[/i] of the subset $S$.Define the least possible remainder of division by $p$ of the deviation of an essential subset,containing $\frac{p-1}{2}$ elements.
Find all positive integers $n$ such that the set $\{n,n+1,n+2,n+3,n+4,n+5\}$ can be partitioned into two subsets so that the product of the numbers in each subset is equal.
Let $p$ be an odd prime number. Show that the smallest positive quadratic nonresidue of $p$ is smaller than $\sqrt{p}+1$.
Let $ p \geq 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1,2,\ldots, p-1 \}$ that was not chosen before by either of the two players and then chooses an element $a_i$ from the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices have been chosen .Then the following number is computed: $$M=a_0+a_110+a_210^2+\cdots+a_{p-1}10^{p-1}= \sum_{i=0}^{p-1}a_i.10^i$$. The goal of Eduardo is to make $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy. [i]Proposed by Amine Natik, Morocco[/i]