An integer \(n \geq 2\) is said to be [i]tuanis[/i] if, when you add the smallest prime divisor of \(n\) and the largest prime divisor of \(n\) (these divisors can be the same), you obtain an odd result. Calculate the sum of all [i]tuanis[/i] numbers that are less or equal to \(2023\).

A prime number has the property that however its decimal digits are permuted, the obtained number is also prime. Prove that this number has at most three different digits. Also prove a stronger statement.

Find all primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$ Proposed by Stanislav Dimitrov,Bulgaria

Each cell of the infinite checkered plane contains one from the numbers $1, 2, 3, 4$ so that each number appears at least once. Let's call a cell [i]correct [/i] if the number of distinct numbers written in four adjacent (side) cells to it, equal to the number written in this cell. Can all the cells of the plane be [i]correct[/i]?

Let $P(x) = x^2 - 20x - 11$. If $a$ and $b$ are natural numbers such that $a$ is composite, $\gcd(a, b) = 1$, and $P(a) = P(b)$, compute $ab$. Note: $\gcd(m, n)$ denotes the greatest common divisor of $m$ and $n$. [i]Proposed by Aaron Lin [/i]

Determine all integers $n \geq 2$ such that there exists a permutation $x_0, x_1, \ldots, x_{n - 1}$ of the numbers $0, 1, \ldots, n - 1$ with the property that the $n$ numbers $$x_0, \hspace{0.3cm} x_0 + x_1, \hspace{0.3cm} \ldots, \hspace{0.3cm} x_0 + x_1 + \ldots + x_{n - 1}$$ are pairwise distinct modulo $n$.

Prove that $(2m)!(2n)!$ is a multiple of $m!n!(m+n)!$ for any non-negative integers $m$ and $n$.

Solve $ x^3-y^3=2xy+7 $ in integers.

Let $n$ be a positive integer. Let $S$ be a set of $n$ positive integers such that the greatest common divisors of all nonempty sets of $S$ are distinct. Determine the smallest possible number of distinct prime divisors of the product of the elements of $S$.

Find all triples of natural numbers $ a $, $ b $, $ c $ such that $$ \gcd (a ^ 2, b ^ 2) + \gcd (a, bc) +\gcd (b, ac) +\gcd (c, ab) = 239 ^ 2 = ab + c . $$

Let $m$ and $n$ be positive integers such that $5m+ n$ is a divisor of $5n +m$. Prove that $m$ is a divisor of $n$.

Given positive integers $a,b,c$ which are pairwise coprime. Let $f(n)$ denotes the number of the non-negative integer solution $(x,y,z)$ to the equation $$ax+by+cz=n.$$ Prove that there exists constants $\alpha, \beta, \gamma \in \mathbb{R}$ such that for any non-negative integer $n$, $$|f(n)- \left( \alpha n^2+ \beta n + \gamma \right) | < \frac{1}{12} \left( a+b+c \right).$$

Find all solutions for positive integers $(x,y,k,m)$ such that \[ 20x^k+24y^m = 2024\] with $k, m > 1$

Let $n$ be a positive integer. $S$ is the set of nonnegative integers $a$ such that $1<a<n$ and $a^{a-1}-1$ is divisible by $n$. Prove that if $S=\{ n-1 \}$ then $n=2p$ where $p$ is a prime number. [i]Mihai Cipu and Nicolae Ciprian Bonciocat[/i]

Find all the integer solutions of the equation $ m^4 + 2n^2 = 9mn$.

Does there exist a function $s\colon \mathbb{Q} \rightarrow \{-1,1\}$ such that if $x$ and $y$ are distinct rational numbers satisfying ${xy=1}$ or ${x+y\in \{0,1\}}$, then ${s(x)s(y)=-1}$? Justify your answer. [i]Proposed by Dan Brown, Canada[/i]

Find all natural numbers $n>3$, such that $2^{2000}$ is divisible by $1+C^1_n+C^2_n+C^3_n$.

Sequence $a_1, a_2, a_3, \cdots$ satisfies the following condition. [b](Condition)[/b] For all positive integer $n$, $\sum_{k=1}^{n}\frac{1}{2}\left(1 - (-1)^{\left[\frac{n}{k}\right]}\right)a_k=1$ holds. For a positive integer $m = 1001 \cdot 2^{2025}$, compute $a_m$.

Let $ABC$ be an isosceles triangle with $AB=4$, $BC=CA=6$. On the segment $AB$ consecutively lie points $X_{1},X_{2},X_{3},\ldots$ such that the lengths of the segments $AX_{1},X_{1}X_{2},X_{2}X_{3},\ldots$ form an infinite geometric progression with starting value $3$ and common ratio $\frac{1}{4}$. On the segment $CB$ consecutively lie points $Y_{1},Y_{2},Y_{3},\ldots$ such that the lengths of the segments $CY_{1},Y_{1}Y_{2},Y_{2}Y_{3},\ldots$ form an infinite geometric progression with starting value $3$ and common ratio $\frac{1}{2}$. On the segment $AC$ consecutively lie points $Z_{1},Z_{2},Z_{3},\ldots$ such that the lengths of the segments $AZ_{1},Z_{1}Z_{2},Z_{2}Z_{3},\ldots$ form an infinite geometric progression with starting value $3$ and common ratio $\frac{1}{2}$. Find all triplets of positive integers $(a,b,c)$ such that the segments $AY_{a}$, $BZ_{b}$ and $CX_{c}$ are concurrent.

Given $n \ge 3$ positive integers not exceeding $100$, let $d$ be their greatest common divisor. Show that there exist three of these numbers whose greatest common divisor is also equal to $d$.

Find the set of all ordered pairs $(s,t)$ of positive integers such that \[t^{2}+1=s(s+1).\]

Let $m>1$ be an integer. Find the smallest positive integer $n$, such that for any integers $a_1,a_2,\ldots ,a_n; b_1,b_2,\ldots ,b_n$ there exists integers $x_1,x_2,\ldots ,x_n$ satisfying the following two conditions: i) There exists $i\in \{1,2,\ldots ,n\}$ such that $x_i$ and $m$ are coprime ii) $\sum^n_{i=1} a_ix_i \equiv \sum^n_{i=1} b_ix_i \equiv 0 \pmod m$

An integer $c$ is [i]square-friendly[/i] if it has the following property: For every integer $m$, the number $m^2+18m+c$ is a perfect square. (A perfect square is a number of the form $n^2$, where $n$ is an integer. For example, $49 = 7^2$ is a perfect square while $46$ is not a perfect square. Further, as an example, $6$ is not [i]square-friendly[/i] because for $m = 2$, we have $(2)2 +(18)(2)+6 = 46$, and $46$ is not a perfect square.) In fact, exactly one square-friendly integer exists. Show that this is the case by doing the following: (a) Find a square-friendly integer, and prove that it is square-friendly. (b) Prove that there cannot be two different square-friendly integers.

Sasha multiplied all the divisors of the natural number $n$. Fedya increased each divider by $1$, and then multiplied the results. If the product found Fedya is divided by the product found by Sasha , what can $n$ be equal to ?

Fix integers $a$ and $b$ greater than $1$. For any positive integer $n$, let $r_n$ be the (non-negative) remainder that $b^n$ leaves upon division by $a^n$. Assume there exists a positive integer $N$ such that $r_n < \frac{2^n}{n}$ for all integers $n\geq N$. Prove that $a$ divides $b$. [i]Pouria Mahmoudkhan Shirazi, Iran[/i]