Let $ R_1,R_2, \ldots$ be the family of finite sequences of positive integers defined by the following rules: $ R_1 \equal{} (1),$ and if $ R_{n - 1} \equal{} (x_1, \ldots, x_s),$ then \[ R_n \equal{} (1, 2, \ldots, x_1, 1, 2, \ldots, x_2, \ldots, 1, 2, \ldots, x_s, n).\] For example, $ R_2 \equal{} (1, 2),$ $ R_3 \equal{} (1, 1, 2, 3),$ $ R_4 \equal{} (1, 1, 1, 2, 1, 2, 3, 4).$ Prove that if $ n > 1,$ then the $ k$th term from the left in $ R_n$ is equal to 1 if and only if the $ k$th term from the right in $ R_n$ is different from 1.

$L$ is a fullrank lattice in $\mathbb R^{2}$ and $K$ is a sub-lattice of $L$, that $\frac{A(K)}{A(L)}=m$. If $m$ is the least number that for each $x\in L$, $mx$ is in $K$. Prove that there exists a basis $\{x_{1},x_{2}\}$ for $L$ that $\{x_{1},mx_{2}\}$ is a basis for $K$.

Let $S_i$ be the sum of the first $i$ terms of the arithmetic sequence $a_1,a_2,a_3\ldots $. Show that the value of the expression \[\frac{S_i}{i}(j-k) + \frac{S_j}{j}(k-i) +\ \frac{S_k}{k}(i-j)\] does not depend on the numbers $i,j,k$ nor on the choice of the arithmetic sequence $a_1,a_2,a_3,\ldots$.

Find all linear homogeneous differential equations with continuous coefficients (on the whole real line) such that for any solution $ f(t)$ and any real number $ c,f(t\plus{}c)$ is also a solution.

The cirumcentre of the cyclic quadrilateral $ABCD$ is $O$. The second intersection point of the circles $ABO$ and $CDO$, other than $O$, is $P$, which lies in the interior of the triangle $DAO$. Choose a point $Q$ on the extension of $OP$ beyond $P$, and a point $R$ on the extension of $OP$ beyond $O$. Prove that $\angle QAP=\angle OBR$ if and only if $\angle PDQ=\angle RCO$.

Let $n > 1$ be an integer. In a circular arrangement of $n$ lamps $L_0, \ldots, L_{n-1},$ each of of which can either ON or OFF, we start with the situation where all lamps are ON, and then carry out a sequence of steps, $Step_0, Step_1, \ldots .$ If $L_{j-1}$ ($j$ is taken mod $n$) is ON then $Step_j$ changes the state of $L_j$ (it goes from ON to OFF or from OFF to ON) but does not change the state of any of the other lamps. If $L_{j-1}$ is OFF then $Step_j$ does not change anything at all. Show that: (i) There is a positive integer $M(n)$ such that after $M(n)$ steps all lamps are ON again, (ii) If $n$ has the form $2^k$ then all the lamps are ON after $n^2-1$ steps, (iii) If $n$ has the form $2^k + 1$ then all lamps are ON after $n^2 - n + 1$ steps.

We have $2^m$ sheets of paper, with the number $1$ written on each of them. We perform the following operation. In every step we choose two distinct sheets; if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a + b$ on both sheets. Prove that after $m2^{m -1}$ steps, the sum of the numbers on all the sheets is at least $4^m$ . [i]Proposed by Abbas Mehrabian, Iran[/i]

At the vertices $A, B, C, D, E, F, G, H$ of a cube, $2001, 2002, 2003, 2004, 2005, 2008, 2007$ and $2006$ stones respectively are placed. It is allowed to move a stone from a vertex to each of its three neighbours, or to move a stone to a vertex from each of its three neighbours. Which of the following arrangements of stones at $A, B, \ldots , H$ can be obtained? $(\text{a})\quad 2001, 2002, 2003, 2004, 2006, 2007, 2008, 2005;$ $(\text{b})\quad 2002, 2003, 2004, 2001, 2006, 2005, 2008, 2007;$ $(\text{c})\quad 2004, 2002, 2003, 2001, 2005, 2008, 2007, 2006.$

The numbers $\frac 32$, $\frac 43$ and $\frac 65$ are intially written on the blackboard. A move consists of erasing one of the numbers from the blackboard, call it $a$, and replacing it with $bc-b-c+2$, where $b,c$ are the other two numbers currently written on the blackboard. Is it possible that $\frac{1000}{999}$ would eventually appear on the blackboard? What about $\frac{113}{108}$? [i] (Andrei Bâra)[/i]