Prove that the number $ 9^{(a_1 + a_2)(a_2 + a_3)(a_3 + a_4)...(a_{98} + a_{99})(a_{99} + a_1)}$ − $1$ is divisible by $10$, for any choice of positive integers $a_1, a_2, a_3, . . . , a_{99}$.

There are $n$ holes in a circle. The holes are numbered $1,2,3$ and so on to $n$. In the beginning, there is a peg in every hole except for hole $1$. A peg can jump in either direction over one adjacent peg to an empty hole immediately on the other side. After a peg moves, the peg it jumped over is removed. The puzzle will be solved if all pegs disappear except for one. For example, if $n=4$ the puzzle can be solved in two jumps: peg $3$ jumps peg $4$ to hole $1$, then peg $2$ jumps the peg in $1$ to hole $4$. (See illustration below, in which black circles indicate pegs and white circles are holes.) [center][img]http://i.imgur.com/4ggOa8m.png[/img][/center] [list=a] [*]Can the puzzle be solved for $n=5$? [*]Can the puzzle be solved for $n=2014$? [/list] In each part (a) and (b) either describe a sequence of moves to solve the puzzle or explain why it is impossible to solve the puzzle.

There are $n$ holes in a circle. The holes are numbered $1,2,3$ and so on to $n$. In the beginning, there is a peg in every hole except for hole $1$. A peg can jump in either direction over one adjacent peg to an empty hole immediately on the other side. After a peg moves, the peg it jumped over is removed. The puzzle will be solved if all pegs disappear except for one. For example, if $n=4$ the puzzle can be solved in two jumps: peg $3$ jumps peg $4$ to hole $1$, then peg $2$ jumps the peg in $1$ to hole $4$. (See illustration below, in which black circles indicate pegs and white circles are holes.) [center][img]http://i.imgur.com/4ggOa8m.png[/img][/center] [list=a] [*]Can the puzzle be solved for $n=5$? [*]Can the puzzle be solved for $n=2014$? [/list] In each part (a) and (b) either describe a sequence of moves to solve the puzzle or explain why it is impossible to solve the puzzle.

A subset of $\{1, 2, 3, ... ... , 2015\}$ is called good if the following condition is fulfilled: for any element $x$ of the subset, the sum of all the other elements in the subset has the same last digit as $x$. For example, $\{10, 20, 30\}$ is a good subset since $10$ has the same last digit as $20 + 30 = 50$, $20$ has the same last digit as $10 + 30 = 40$, and $30$ has the same last digit as $10 + 20 = 30$. (a) Find an example of a good subset with 400 elements. (b) Prove that there is no good subset with 405 elements.

Let $ABC$ be an acute triangle. Let $D$ be the reflection of point $B$ with respect to the line $AC$. Let $E$ be the reflection of point $C$ with respect to the line $AB$. Let $\Gamma_1$ be the circle that passes through $A, B$, and $D$. Let $\Gamma_2$ be the circle that passes through $A, C$, and $E$. Let $P$ be the intersection of $\Gamma_1$ and $\Gamma_2$ , other than $A$. Let $\Gamma$ be the circle that passes through $A, B$, and $C$. Show that the center of $\Gamma$ lies on line $AP$.

Prove that there are exactly $2(2^{n-1}-1)$ ways of dealing $n$ cards to two persons. (The persons may receive unequal numbers of cards.)

Find a positive integer $N$ and $a_1, a_2, \cdots, a_N$ where $a_k = 1$ or $a_k = -1$, for each $k=1,2,\cdots,N,$ such that $$a_1 \cdot 1^3 + a_2 \cdot 2^3 + a_3 \cdot 3^3 \cdots + a_N \cdot N^3 = 20162016$$ or show that this is impossible.

Amy and Bob play a game. They alternate turns, with Amy going first. At the start of the game, there are $20$ cookies on a red plate and $14$ on a blue plate. A legal move consists of eating two cookies taken from one plate, or moving one cookie from the red plate to the blue plate (but never from the blue plate to the red plate). The last player to make a legal move wins; in other words, if it is your turn and you cannot make a legal move, you lose, and the other player has won. Which player can guarantee that they win no matter what strategy their opponent chooses? Prove that your answer is correct.

Given two triangles with the same perimeter. Both triangles have integer side lengths. The first triangle is an equilateral triangle. The second triangle has a side with length 1 and a side with length $d$. Prove that when $d$ is divided by 3, the remainder is 1.