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

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Found problems: 1048

Prove that the product of numbers $1, 2, ..., n$ ($n \ge 2$) is divisible by their sum if and only if the number $n + 1$ is not prime.
Let $(a_i)_{i\in \mathbb{N}}$ and $(p_i)_{i\in \mathbb{N}}$ be two sequences of positive integers such that the following conditions hold: $\bullet ~~a_1\ge 2$. $\bullet~~ p_n$ is the smallest prime divisor of $a_n$ for every integer $n\ge 1$ $\bullet~~ a_{n+1}=a_n+\frac{a_n}{p_n}$ for every integer $n\ge 1$ Prove that there is a positive integer $N$ such that $a_{n+3}=3a_n$ for every integer $n>N$
Find all prime numbers $p$ and $q$ such that $p^q + 5q - 2$ is also a prime number.
We say that a prime $p$ is $n$-$\textit{rephinado}$ if $n | p - 1$ and all $1, 2, \ldots , \lfloor \sqrt[\delta]{p}\rfloor$ are $n$-th residuals modulo $p$, where $\delta = \varphi+1$. Are there infinitely many $n$ for which there are infinitely many $n$-$\textit{rephinado}$ primes? Notes: $\varphi =\frac{1+\sqrt{5}}{2}$. We say that an integer $a$ is a $n$-th residue modulo $p$ if there is an integer $x$ such that $$x^n \equiv a \text{ (mod } p\text{)}.$$
Let $n\ge 2$ be a positive integer. For any integer $a$, let $P_a(x)$ denote the polynomial $x^n+ax$. Let $p$ be a prime number and define the set $S_a$ as the set of residues mod $p$ that $P_a(x)$ attains. That is, $$S_a=\{b\mid 0\le b\le p-1,\text{ and there is }c\text{ such that }P_a(c)\equiv b \pmod{p}\}.$$Show that the expression $\frac{1}{p-1}\sum\limits_{a=1}^{p-1}|S_a|$ is an integer. [i]Proposed by fattypiggy123[/i]
Find all triples $(x,n,p)$ of positive integers $x$ and $n$ and primes $p$ for which the following holds $x^3 + 3x + 14 = 2 p^n$
We call a natural number $p{}$ [i]simple[/i] if for any natural number $k{}$ such that $2\leqslant k\leqslant \sqrt{p}$ the inequality $\{p/k\}\geqslant 0,01$ holds. Is the set of simple prime numbers finite? [i]Proposed by M. Didin[/i]
Let $p\geq 5$ be a prime number, and set $M=\{1,2,\cdots,p-1\}.$ Define $$T=\{(n,x_n):p|nx_n-1\ \textup{and}\ n,x_n\in M\}.$$ If $\sum_{(n,x_n)\in T}n\left[\dfrac{nx_n}{p}\right]\equiv k \pmod {p},$ with $0\leq k\leq p-1,$ where $\left[\alpha\right]$ denotes the largest integer that does not exceed $\alpha,$ determine the value of $k.$
Let $n \geq 4$ be an integer. Consider a regular $2n-$gon for which to every vertex, an integer is assigned, which we call the value of said vertex. If four distinct vertices of this $2n-$gon form a rectangle, we say that the sum of the values of these vertices is a rectangular sum. Determine for which (not necessarily positive) integers $m$ the integers $m + 1, m + 2, . . . , m + 2n$ can be assigned to the vertices (in some order) in such a way that every rectangular sum is a prime number. (Prime numbers are positive by definition.)
Let $f$ be the function of the set of positive integers into itself, defi ned by $f(1) = 1$, $f(2n) = f(n)$ and $f(2n + 1) = f(n) + f(n + 1)$. Show that, for any positive integer $n$, the number of positive odd integers m such that $f(m) = n$ is equal to the number of positive integers[color=#0000FF][b] less or equal to [/b][/color]$n$ and coprime to $n$. [color=#FF0000][mod: the initial statement said less than $n$, which is wrong.][/color]
Determine all the pairs $ (p , n )$ of a prime number $ p$ and a positive integer $ n$ for which $ \frac{ n^p + 1 }{p^n + 1} $ is an integer.
Let $S$ be the set of positive integers. For any $a$ and $b$ in the set we have $GCD(a, b)>1$. For any $a$, $b$ and $c$ in the set we have $GCD(a, b, c)=1$. Is it possible that $S$ has $2012$ elements? [i]Proposed by Ognjen Stipetić.[/i]
Fix a sequence $a_1,a_2,a_3\ldots$ of integers satisfying the following condition:for all prime numbers $p$ and all positive integers $k$,we have $a_{pk+1}=pa_k-3a_p+13$.Determine all possible values of $a_{2013}$.
Find all primes $p,q,r$ and natural numbers $n$ such that $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{1}{n}$.
Let $n$ be a positive integer. Show that there is a prime $p$ and a sequence $\left(a_k\right)_{k\ge1}$ of positive integers such that the sequence $\left(p+na_k\right)_{k\ge1}$ consists of distinct primes.
Find all prime numbers $p$, $q$ and $r$ such that the fourth power of any of them minus one is divisible by the product of the other two. (Author: V. Senderov)
Let $p$ be a given prime number. For positive integers $n,k\geq2$ let $S_1, S_2,\dots, S_n$ be unit square sets constructed by choosing exactly one unit square from each of the columns from $p\times k$ chess board. If $|S_i \cap S_j|=1$ for all $1\leq i < j \leq n$ and for any duo of unit squares which are located at different columns there exists $S_i$ such that both of these unit squares are in $S_i$ find all duos of $(n,k)$ in terms of $p$. Note: Here we denote the number of rows by $p$ and the number of columns by $k$.
Prove that if $p$ is an odd prime, then $p^2(p^2 -1999)$ is divisible by $6$ but not by $12$.
What is the sum of the digits of the smallest prime that can be written as a sum of $5$ distinct primes? $\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$
$p$ is an odd prime number. Find all $\frac{p-1}2$-tuples $\left(x_1,x_2,\dots,x_{\frac{p-1}2}\right)\in \mathbb{Z}_p^{\frac{p-1}2}$ such that $$\sum_{i = 1}^{\frac{p-1}{2}} x_{i} \equiv \sum_{i = 1}^{\frac{p-1}{2}} x_{i}^{2} \equiv \cdots \equiv \sum_{i = 1}^{\frac{p-1}{2}} x_{i}^{\frac{p - 1}{2}} \pmod p.$$ [i]Proposed by Ali Partofard[/i]
Prove that for each prime number distinct from $2$ and $5$ there exist infinitely many multiples of $p$ of the form $1111...1$.
Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around the circle such that the product of any two neighbors is of the form $x^2+x+k$ for some positive integer $x$.
Find all (finite) increasing arithmetic progressions, consisting only of prime numbers, such that the number of terms is larger than the common difference.