Let $A=\{1,2,3,\cdots ,17\}$. A mapping $f:A\rightarrow A$ is defined as follows: $f^{[1]}(x)=f(x), f^{[k+1]}(x)=f(f^{[k]}(x))$ for $k\in\mathbb{N}$. Suppose that $f$ is bijective and that there exists a natural number $M$ such that: i) when $m<M$ and $1\le i\le 16$, we have $f^{[m]}(i+1)- f^{[m]}(i) \not=\pm 1\pmod{17}$ and $f^{[m]}(1)- f^{[m]}(17) \not=\pm 1\pmod{17}$; ii) when $1\le i\le 16$, we have $f^{[M]}(i+1)- f^{[M]}(i)=\pm 1 \pmod{17}$ and $f^{[M]}(1)- f^{[M]}(17)=\pm 1\pmod{17}$. Find the maximal value of $M$.

Let $a$ and $n$ be two integers greater than $1$. Prove that if $n$ divides $(a-1)^k$ for some integer $k\geq 2$, then $n$ also divides $a^{n-1}+a^{n-2}+\cdots+a+1$.

Let $ m$ and $ n$ be positive integers such that $ 1 \le m < n$. In their decimal representations, the last three digits of $ 1978^m$ are equal, respectively, to the last three digits of $ 1978^n$. Find $ m$ and $ n$ such that $ m \plus{} n$ has its least value.

The sequence of integers $\{ x_{n}\}_{n\ge1}$ is defined as follows: \[x_{1}=1, \;\; x_{n+1}=1+{x_{1}}^{2}+\cdots+{x_{n}}^{2}\;(n=1,2,3 \cdots).\] Prove that there are no squares of natural numbers in this sequence except $x_{1}$.

Define a function $f:\mathbb{N}\rightarrow\mathbb{N}$, \[f(1)=p+1,\] \[f(n+1)=f(1)\cdot f(2)\cdots f(n)+p,\] where $p$ is a prime number. Find all $p$ such that there exists a natural number $k$ such that $f(k)$ is a perfect square.

Find a pair of relatively prime four digit natural numbers $A$ and $B$ such that for all natural numbers $m$ and $n$, $\vert A^m -B^n \vert \ge 400$.

Sequences $ (x_n)$ and $ (y_n)$ are constructed as follows: $ x_0 \equal{} 365$, $ x_{n\plus{}1} \equal{} x_n\left(x^{1986} \plus{} 1\right) \plus{} 1622$, and $ y_0 \equal{} 16$, $ y_{n\plus{}1} \equal{} y_n\left(y^3 \plus{} 1\right) \minus{} 1952$, for all $ n \ge 0$. Prove that $ \left|x_n\minus{} y_k\right|\neq 0$ for any positive integers $ n$, $ k$.

For a polynomial $ P$ of degree 2000 with distinct real coefficients let $ M(P)$ be the set of all polynomials that can be produced from $ P$ by permutation of its coefficients. A polynomial $ P$ will be called [b]$ n$-independent[/b] if $ P(n) \equal{} 0$ and we can get from any $ Q \in M(P)$ a polynomial $ Q_1$ such that $ Q_1(n) \equal{} 0$ by interchanging at most one pair of coefficients of $ Q.$ Find all integers $ n$ for which $ n$-independent polynomials exist.

How many integers $x$, from $10$ to $99$ inclusive, have the property that the remainder of $x^2$ divided by $100$ is equal to the square of the units digit of $x$?

Find all prime numbers $p, q$ such that$$p(p+1)(p^2+1) = q^2(q^2+q+1) + 2025.$$ [i]Proposed by Md. Fuad Al Alam[/i]

Let $n \geq 2$ be a positive integer. Suppose there exist non-negative integers ${n_1},{n_2},\ldots,{n_k}$ such that $2^n - 1 \mid \sum_{i = 1}^k {{2^{{n_i}}}}$. Prove that $k \ge n$.

Determine all positive integers $n$ such that $3^{n}+1$ is divisible by $n^{2}$.

Let $ p>2 $ be a prime number. For any permutation $ \pi = ( \pi(1) , \pi(2) , \cdots , \pi(p) ) $ of the set $ S = \{ 1, 2, \cdots , p \} $, let $ f( \pi ) $ denote the number of multiples of $ p $ among the following $ p $ numbers: \[ \pi(1) , \pi(1) + \pi(2) , \cdots , \pi(1) + \pi(2) + \cdots + \pi(p) \] Determine the average value of $ f( \pi) $ taken over all permutations $ \pi $ of $ S $.

Solve the equation $p^2-pq-q^3=1$ in prime numbers. [i]A. Golovanov[/i]

Are there $4$ distinct positive integers such that adding the product of any two of them to $2006$ yields a perfect square?

Determine the least odd number $a > 5$ satisfying the following conditions: There are positive integers $m_1,m_2, n_1, n_2$ such that $a=m_1^2+n_1^2$, $a^2=m_2^2+n_2^2$, and $m_1-n_1=m_2-n_2.$

How many positive integer multiples of 1001 can be expressed in the form $10^{j}-10^{i}$, where $i$ and $j$ are integers and $0\leq i < j \leq 99$?

Let $ \left(x_{n}\right)$ be a real sequence satisfying $ x_{0}=0$, $ x_{2}=\sqrt[3]{2}x_{1}$, and $ x_{n+1}=\frac{1}{\sqrt[3]{4}}x_{n}+\sqrt[3]{4}x_{n-1}+\frac{1}{2}x_{n-2}$ for every integer $ n\geq 2$, and such that $ x_{3}$ is a positive integer. Find the minimal number of integers belonging to this sequence.

Consider the assertion that for each positive integer $n\geq2$, the remainder upon dividing $2^{2^n}$ by $2^n-1$ is a power of $4$. Either prove the assertion or find (with proof) a counterexample.

Do there exist two quadratic trinomials $ax^2 +bx+c$ and $(a+1)x^2 +(b + 1)x + (c + 1)$ with integer coeficients, both of which have two integer roots? [i]N. Agakhanov[/i]

The diagram below shows the first three figures of a sequence of figures. The first figure shows an equilateral triangle $ABC$ with side length $1$. The leading edge of the triangle going in a clockwise direction around $A$ is labeled $\overline{AB}$ and is darkened in on the figure. The second figure shows the same equilateral triangle with a square with side length $1$ attached to the leading clockwise edge of the triangle. The third figure shows the same triangle and square with a regular pentagon with side length $1$ attached to the leading clockwise edge of the square. The fourth figure in the sequence will be formed by attaching a regular hexagon with side length $1$ to the leading clockwise edge of the pentagon. The hexagon will overlap the triangle. Continue this sequence through the eighth figure. After attaching the last regular figure (a regular decagon), its leading clockwise edge will form an angle of less than $180^\circ$ with the side $\overline{AC}$ of the equilateral triangle. The degree measure of that angle can be written in the form $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [asy] size(250); defaultpen(linewidth(0.7)+fontsize(10)); pair x[],y[],z[]; x[0]=origin; x[1]=(5,0); x[2]=rotate(60,x[0])*x[1]; draw(x[0]--x[1]--x[2]--cycle); for(int i=0;i<=2;i=i+1) { y[i]=x[i]+(15,0); } y[3]=rotate(90,y[0])*y[2]; y[4]=rotate(-90,y[2])*y[0]; draw(y[0]--y[1]--y[2]--y[0]--y[3]--y[4]--y[2]); for(int i=0;i<=4;i=i+1) { z[i]=y[i]+(15,0); } z[5]=rotate(108,z[4])*z[2]; z[6]=rotate(108,z[5])*z[4]; z[7]=rotate(108,z[6])*z[5]; draw(z[0]--z[1]--z[2]--z[0]--z[3]--z[4]--z[2]--z[7]--z[6]--z[5]--z[4]); dot(x[2]^^y[2]^^z[2],linewidth(3)); draw(x[2]--x[0]^^y[2]--y[4]^^z[2]--z[7],linewidth(1)); label("A",(x[2].x,x[2].y-.3),S); label("B",origin,S); label("C",x[1],S);[/asy]

Suppose we have distinct positive integers $a, b, c, d$, and an odd prime $p$ not dividing any of them, and an integer $M$ such that if one considers the infinite sequence \begin{align*} ca &- db \\ ca^2 &- db^2 \\ ca^3 &- db^3 \\ ca^4 &- db^4 \\ &\vdots \end{align*} and looks at the highest power of $p$ that divides each of them, these powers are not all zero, and are all at most $M$. Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$) such that whenever $p$ divides an element of this sequence, the maximum power of $p$ that divides that element is exactly $p^T$.

Find all positive integers $ n$ having the following properties:in two-dimensional Cartesian coordinates, there exists a convex $ n$ lattice polygon whose lengths of all sides are odd numbers, and unequal to each other. (where lattice polygon is defined as polygon whose coordinates of all vertices are integers in Cartesian coordinates.)

A student at Harvard named Kevin Was counting his stones by $11$ He messed up $n$ times And instead counted $9$s And wound up at $2007$. How many values of $n$ could make this limerick true?

$x,y,z,a,b,c$ are positive integers that satisfy $xy \equiv a \pmod z$, $yz \equiv b \pmod x$, $zx \equiv c \pmod y$. Prove that $\min{\{x,y,z\}} \le ab+bc+ca$.