Is it possible to tile $2003 \times 2003$ board by $1 \times 2$ dominoes placed horizontally and $1 \times 3$ rectangles placed vertically?

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 $a_1,a_2,a_3,\ldots$.

Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$. Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle.

What is the area of the region defined by the inequality $ |3x\minus{}18|\plus{}|2y\plus{}7|\le 3$? $ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac{7}{2} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{9}{2} \qquad \textbf{(E)}\ 5$

A 2-player game is played on $n\geq 3$ points, where no 3 points are collinear. Each move consists of selecting 2 of the points and drawing a new line segment connecting them. The first player to draw a line segment that creates an odd cycle loses. (An odd cycle must have all its vertices among the $n$ points from the start, so the vertices of the cycle cannot be the intersections of the lines drawn.) Find all $n$ such that the player to move first wins.

Let $N$ be a positive integer, and consider an $N \times N$ grid. A [i]right-down path[/i] is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A [i]right-up path[/i] is a sequence of grid cells such that each cell is either one cell to the right of or one cell above the previous cell in the sequence. Prove that the cells of the $N \times N$ grid cannot be partitioned into less than $N$ right-down or right-up paths. For example, the following partition of the $5 \times 5$ grid uses $5$ paths. [asy] size(4cm); draw((5,-1)--(0,-1)--(0,-2)--(5,-2)--(5,-3)--(0,-3)--(0,-4)--(5,-4),gray+linewidth(0.5)+miterjoin); draw((1,-5)--(1,0)--(2,0)--(2,-5)--(3,-5)--(3,0)--(4,0)--(4,-5),gray+linewidth(0.5)+miterjoin); draw((0,0)--(5,0)--(5,-5)--(0,-5)--cycle,black+linewidth(2.5)+miterjoin); draw((0,-1)--(3,-1)--(3,-2)--(1,-2)--(1,-4)--(4,-4)--(4,-3)--(2,-3)--(2,-2),black+linewidth(2.5)+miterjoin); draw((3,0)--(3,-1),black+linewidth(2.5)+miterjoin); draw((1,-4)--(1,-5),black+linewidth(2.5)+miterjoin); draw((4,-3)--(4,-1)--(5,-1),black+linewidth(2.5)+miterjoin); [/asy] [i]Proposed by Zixiang Zhou, Canada[/i]

Determine all finite groups $G$ that have an automorphism $f$ such that $H\not\subseteq f(H)$ for all proper subgroups $H$ of $G$. [B. Kovacs]

Suppose $n > 1$ is an odd integer satisfying $n \mid 2^{\frac{n-1}{2}} + 1$. Prove [color=red]or disprove[/color] that $n$ is prime. [i] Note: unfortunately, the original form of this problem did not include the red text, rendering it unsolvable. We sincerely apologize for this error and are taking concrete steps to prevent similar issues from reoccurring, including computer-verifying problems where possible. All teams will receive full credit for the question.[/i]

Prove that $ \forall x\in \mathbb{R} $ , $ \cos ^7x+\cos ^7(x+\frac {2\pi}{3})+\cos ^7(x+\frac {4\pi}{3})=\frac {63}{64}\cos 3x $

Let $ n \geq 2$ be a positive integer and $ \lambda$ a positive real number. Initially there are $ n$ fleas on a horizontal line, not all at the same point. We define a move as choosing two fleas at some points $ A$ and $ B$, with $ A$ to the left of $ B$, and letting the flea from $ A$ jump over the flea from $ B$ to the point $ C$ so that $ \frac {BC}{AB} \equal{} \lambda$. Determine all values of $ \lambda$ such that, for any point $ M$ on the line and for any initial position of the $ n$ fleas, there exists a sequence of moves that will take them all to the position right of $ M$.

Let $n\geqslant 2$ be a positive integer. Paul has a $1\times n^2$ rectangular strip consisting of $n^2$ unit squares, where the $i^{\text{th}}$ square is labelled with $i$ for all $1\leqslant i\leqslant n^2$. He wishes to cut the strip into several pieces, where each piece consists of a number of consecutive unit squares, and then [i]translate[/i] (without rotating or flipping) the pieces to obtain an $n\times n$ square satisfying the following property: if the unit square in the $i^{\text{th}}$ row and $j^{\text{th}}$ column is labelled with $a_{ij}$, then $a_{ij}-(i+j-1)$ is divisible by $n$. Determine the smallest number of pieces Paul needs to make in order to accomplish this.

Let $n\ge 2$ be an integer. A sequence $\alpha = (a_1, a_2,..., a_n)$ of $n$ integers is called [i]Lima [/i] if $\gcd \{a_i - a_j \text{ such that } a_i> a_j \text{ and } 1\le i, j\le n\} = 1$, that is, if the greatest common divisor of all the differences $a_i - a_j$ with $a_i> a_j$ is $1$. One operation consists of choosing two elements $a_k$ and $a_{\ell}$ from a sequence, with $k\ne \ell $ , and replacing $a_{\ell}$ by $a'_{\ell} = 2a_k - a_{\ell}$ . Show that, given a collection of $2^n - 1$ Lima sequences, each one formed by $n$ integers, there are two of them, say $\beta$ and $\gamma$, such that it is possible to transform $\beta$ into $\gamma$ through a finite number of operations. Notes. The sequences $(1,2,2,7)$ and $(2,7,2,1)$ have the same elements but are different. If all the elements of a sequence are equal, then that sequence is not Lima.

The numbers $1, 2, 3, \ldots , 10$ are written on the blackboard. In each step, Andrew chooses two numbers $a, b$ which are written on the blackboard such that $a\geqslant 2b$, he erases them, and in their place writes the number $a-2b$. Find all numbers $n$, such that after a sequence of steps as above, at the end only the number $n$ will remain on the blackboard.

Let $n>1$ be an integer. Suppose we are given $2n$ points in the plane such that no three of them are collinear. The points are to be labelled $A_1, A_2, \dots , A_{2n}$ in some order. We then consider the $2n$ angles $\angle A_1A_2A_3, \angle A_2A_3A_4, \dots , \angle A_{2n-2}A_{2n-1}A_{2n}, \angle A_{2n-1}A_{2n}A_1, \angle A_{2n}A_1A_2$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\circ}$ and $180^{\circ}$). Prove that there exists an ordering of the given points such that the resulting $2n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group.