$S= \{1,4,8,9,16,...\} $is the set of perfect integer power. ( $S=\{ n^k| n, k \in Z, k \ge 2 \}$. )We arrange the elements in $S$ into an increasing sequence $\{a_i\}$ . Show that there are infinite many $n$, such that $9999|a_{n+1}-a_n$

We are given a non-infinite sequence $a_1,a_2…a_n$ of natural numbers. While it is possible, on each turn are chosen two arbitrary indexes $i<j$ such that $a_i \nmid a_j$, and then $a_i$ and $a_j$ are changed with their $gcd$ and $lcm$. Prove that this process is non-infinite and the created sequence doesn’t depend on the made choices.

Let $(a_n)_{n \ge 1}$ be a sequence of positive real numbers such that the sequence $(a_{n+1}-a_n)_{n \ge 1}$ is convergent to a non-zero real number. Evaluate the limit $$ \lim_{n \to \infty} \left( \frac{a_{n+1}}{a_n} \right)^n.$$

Show that if $(a_n)$ is an infinite sequence of distinct positive integers, neither of which contains digit $0$ in the decimal expansion, then $$\sum_{n=1}^{\infty} \frac{1}{a_n}< 29.$$

Let $a, \overline{bcd}, \overline{aef}, \overline{cfg}, \overline{hci}, \overline{dea}, \overline{ifd}, \overline{jgf}, \overline{bfeg},\ldots$ be an increasing arithmetic progression. Find the $16$th term of this sequence.

Alexa wrote the first $16$ numbers of a sequence: \[1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 8, 8, 9, 10, 10, 11, …\] Then she continued following the same pattern, until she had $2015$ numbers in total. What was the last number she wrote?

A sequence $a_1,a_2,a_3,\ldots $ of non-negative integers is such that $a_{n+1}$ is the last digit of $a_n^n+a_{n-1}$ for all $n>2$. Is it always true that for some $n_0$ the sequence $a_{n_0},a_{n_0+1},a_{n_0+2},\ldots$ is periodic?

For a fixed positive integer $k$, there are two sequences $A_n$ and $B_n$. They are defined inductively, by the following recurrences. $A_1 = k$, $A_2 = k$, $A_{n+2} = A_{n}A_{n+1}$ $B_1 = 1$, $B_2 = k$, $B_{n+2} = \frac{B^3_{n+1}+1}{B_{n}}$ Prove that for all positive integers $n$, $A_{2n}B_{n+3}$ is an integer.

The sequence $a_1,a_2,a_3\ldots $, consisting of natural numbers, is defined by the rule: \[a_{n+1}=a_{n}+2t(n)\] for every natural number $n$, where $t(n)$ is the number of the different divisors of $n$ (including $1$ and $n$). Is it possible that two consecutive members of the sequence are squares of natural numbers?

Given a triangle $ABC$ and a point $P_0$ on the side $AB$. Construct points $P_i, Q_i, R_i $ as follows. $Q_i$ is the foot of the perpendicular from $P_i$ to $BC, R_i$ is the foot of the perpendicular from $Q_i$ to $AC$ and $P_i$ is the foot of the perpendicular from $R_{i-1}$ to $AB$. Show that the points $P_i$ converge to a point $P$ on $AB$ and show how to construct $P$.

Prove that for any positive integer $ k$, there exists an arithmetic sequence $ \frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, ... ,\frac{a_k}{b_k}$ of rational numbers, where $ a_i, b_i$ are relatively prime positive integers for each $ i \equal{} 1,2,...,k$ such that the positive integers $ a_1, b_1, a_2, b_2, ..., a_k, b_k$ are all distinct.

For positive integral $k>1$, we let $p(k)$ be its smallest prime divisor. Given an integer $a_1>2$, we define an infinite sequence $a_n$ by $a_{n+1}=a_n^n-1$ for each $n\geq 1$. For which values of $a_1$ is the sequence $p(a_n)$ bounded?

Let $n$ be a positive integer. Given a sequence of nonnegative real numbers $x_1,\ldots ,x_n$ we define the [i]transformed sequence[/i] $y_1,\ldots ,y_n$ as follows: the number $y_i$ is the greatest possible value of the average of consecutive terms of the sequence that contain $x_i$. For example, the transformed sequence of $2,4,1,4,1$ is $3,4,3,4,5/2$. Prove that a) For every positive real number $t$, the number of $y_i$ such that $y_i>t$ is less than or equal to $\frac{2}{t}(x_1+\cdots +x_n)$. b) The inequality $\frac{y_1+\cdots +y_n}{32n}\leq \sqrt{\frac{x_1^2+\cdots +x_n^2}{32n}}$ holds.

Circle $C_0$ has radius $1$, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circles $C_1,C_2,C_3,...$ and a sequence of points on the circles $A_1,A_2,A_3,...$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^{\circ}$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r=\frac{11}{60}$, the distance from the center of $C_0$ to $B$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. [asy] size(6cm); real r = 0.8; pair nthCircCent(int n){ pair ans = (0, 0); for(int i = 1; i <= n; ++i) ans += rotate(90 * i - 90) * (r^(i - 1) - r^i, 0); return ans; } void dNthCirc(int n){ draw(circle(nthCircCent(n), r^n)); } dNthCirc(0); dNthCirc(1); dNthCirc(2); dNthCirc(3); dot("$A_0$", (1, 0), dir(0)); dot("$A_1$", nthCircCent(1) + (0, r), dir(135)); dot("$A_2$", nthCircCent(2) + (-r^2, 0), dir(0)); [/asy]

Let $N$ be a positive integer such that the sum of the squares of all positive divisors of $N$ is equal to the product $N(N+3)$. Prove that there exist two indices $i$ and $j$ such that $N=F_iF_j$ where $(F_i)_{n=1}^{\infty}$ is the Fibonacci sequence defined as $F_1=F_2=1$ and $F_n=F_{n-1}+F_{n-2}$ for $n\geq 3$. [i]Proposed by Alain Rossier, Switzerland[/i]