The numbers $ 1, 2,\ldots, 50 $ are written on a blackboard. Each minute any two numbers are erased and their positive difference is written instead. At the end one number remains. Which values can take this number?

A music streaming service proposes songs classified in $10$ musical genres, so that each song belong to one and only one gender. The songs are played one after the other: the first $17$ are chosen by the user, but starting from the eighteenth the service automatically determines which song to play. Elisabetta has noticed that, if one makes the classification of which genres they appear several times during the last $17$ songs played, the new song always belongs to the genre at the top of the ranking or, in case of same merit, at one of the first genres. Prove that, however, the first $17$ tracks are chosen, from a certain point onwards the songs proposed are all of the same kind.

Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers.

What is that largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?

Let $a, b, c, p, q, r$ be positive integers with $p, q, r \ge 2$. Denote \[Q=\{(x, y, z)\in \mathbb{Z}^3 : 0 \le x \le a, 0 \le y \le b , 0 \le z \le c \}. \] Initially, some pieces are put on the each point in $Q$, with a total of $M$ pieces. Then, one can perform the following three types of operations repeatedly: (1) Remove $p$ pieces on $(x, y, z)$ and place a piece on $(x-1, y, z)$ ; (2) Remove $q$ pieces on $(x, y, z)$ and place a piece on $(x, y-1, z)$ ; (3) Remove $r$ pieces on $(x, y, z)$ and place a piece on $(x, y, z-1)$. Find the smallest positive integer $M$ such that one can always perform a sequence of operations, making a piece placed on $(0,0,0)$, no matter how the pieces are distributed initially.

Let $ A_0 \equal{} (a_1,\dots,a_n)$ be a finite sequence of real numbers. For each $ k\geq 0$, from the sequence $ A_k \equal{} (x_1,\dots,x_k)$ we construct a new sequence $ A_{k \plus{} 1}$ in the following way. 1. We choose a partition $ \{1,\dots,n\} \equal{} I\cup J$, where $ I$ and $ J$ are two disjoint sets, such that the expression \[ \left|\sum_{i\in I}x_i \minus{} \sum_{j\in J}x_j\right| \] attains the smallest value. (We allow $ I$ or $ J$ to be empty; in this case the corresponding sum is 0.) If there are several such partitions, one is chosen arbitrarily. 2. We set $ A_{k \plus{} 1} \equal{} (y_1,\dots,y_n)$ where $ y_i \equal{} x_i \plus{} 1$ if $ i\in I$, and $ y_i \equal{} x_i \minus{} 1$ if $ i\in J$. Prove that for some $ k$, the sequence $ A_k$ contains an element $ x$ such that $ |x|\geq\frac n2$. [i]Author: Omid Hatami, Iran[/i]

Let $a$, $b$, $c$ be positive integers such that $29a + 30b + 31c = 366$. Find $19a + 20b + 21c$.

Given an initial integer $ n_0 > 1$, two players, $ {\mathcal A}$ and $ {\mathcal B}$, choose integers $ n_1$, $ n_2$, $ n_3$, $ \ldots$ alternately according to the following rules : [b]I.)[/b] Knowing $ n_{2k}$, $ {\mathcal A}$ chooses any integer $ n_{2k \plus{} 1}$ such that \[ n_{2k} \leq n_{2k \plus{} 1} \leq n_{2k}^2. \] [b]II.)[/b] Knowing $ n_{2k \plus{} 1}$, $ {\mathcal B}$ chooses any integer $ n_{2k \plus{} 2}$ such that \[ \frac {n_{2k \plus{} 1}}{n_{2k \plus{} 2}} \] is a prime raised to a positive integer power. Player $ {\mathcal A}$ wins the game by choosing the number 1990; player $ {\mathcal B}$ wins by choosing the number 1. For which $ n_0$ does : [b]a.)[/b] $ {\mathcal A}$ have a winning strategy? [b]b.)[/b] $ {\mathcal B}$ have a winning strategy? [b]c.)[/b] Neither player have a winning strategy?

We define a [i]chessboard polygon[/i] to be a polygon whose sides are situated along lines of the form $ x \equal{} a$ or $ y \equal{} b$, where $ a$ and $ b$ are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping $ 1 \times 2$ rectangles. Finally, a [i]tasteful tiling[/i] is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a $ 3 \times 4$ rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner. [asy]size(300); pathpen = linewidth(2.5); void chessboard(int a, int b, pair P){ for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j) if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6)); D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle); } chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12)); /* draw lines */ D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3));[/asy] a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully. b) Prove that such a tasteful tiling is unique.

Qing initially writes the ordered pair $(1,0)$ on a blackboard. Each minute, if the pair $(a,b)$ is on the board, she erases it and replaces it with one of the pairs $(2a-b,a)$, $(2a+b+2,a)$ or $(a+2b+2,b)$. Eventually, the board reads $(2014,k)$ for some nonnegative integer $k$. How many possible values of $k$ are there? [i]Proposed by Evan Chen[/i]

Find the number of ordered triples of positive integers $(a, b, c)$ such that $abc$ divides $(ab + 1)(bc + 1)(ca + 1)$.

The polynomial $P(x)$ is cubic. What is the largest value of $k$ for which the polynomials $Q_{1}(x) = x^{2}+(k-29)x-k$ and $Q_{2}(x) = 2x^{2}+(2k-43)x+k$ are both factors of $P(x)$?

Alice and the Cheshire Cat play a game. At each step, Alice either (1) gives the cat a penny, which causes the cat to change the number of (magic) beans that Alice has from $ n$ to $ 5n$ or (2) gives the cat a nickel, which causes the cat to give Alice another bean. Alice wins (and the cat disappears) as soon as the number of beans Alice has is greater than $ 2008$ and has last two digits $ 42$. What is the minimum number of cents Alice can spend to win the game, assuming she starts with 0 beans?

Each positive integer $a$ undergoes the following procedure in order to obtain the number $d = d\left(a\right)$: (i) move the last digit of $a$ to the first position to obtain the numb er $b$; (ii) square $b$ to obtain the number $c$; (iii) move the first digit of $c$ to the end to obtain the number $d$. (All the numbers in the problem are considered to be represented in base $10$.) For example, for $a=2003$, we get $b=3200$, $c=10240000$, and $d = 02400001 = 2400001 = d(2003)$.) Find all numbers $a$ for which $d\left( a\right) =a^2$. [i]Proposed by Zoran Sunic, USA[/i]

Let $G$ be a simple graph with $3n^2$ vertices ($n\geq 2$). It is known that the degree of each vertex of $G$ is not greater than $4n$, there exists at least a vertex of degree one, and between any two vertices, there is a path of length $\leq 3$. Prove that the minimum number of edges that $G$ might have is equal to $\frac{(7n^2- 3n)}{2}$.

Find a method by which one can compute the coefficients of $P(x) = x^6 + a_1x^5 + \cdots+ a_6$ from the roots of $P(x) = 0$ by performing not more than $15$ additions and $15$ multiplications.

Find in explicit form all ordered pairs of positive integers $(m, n)$ such that $mn-1$ divides $m^2 + n^2$.

On each day of their tour of the West Indies, Sourav and Srinath have either an apple or an orange for breakfast. Sourav has oranges for the first $m$ days, apples for the next $m$ days, followed by oranges for the next $m$ days, and so on. Srinath has oranges for the first $n$ days, apples for the next $n$ days, followed by oranges for the next $n$ days, and so on. If $\gcd(m,n)=1$, and if the tour lasted for $mn$ days, on how many days did they eat the same kind of fruit?

Several policemen try to catch a thief who has $2m$ accomplices. To that end they place the accomplices under surveillance. In the beginning, the policemen shadow nobody. Every morning each policeman places under his surveillance one of the accomplices. Every evening the thief stops trusting one of his accomplices The thief is caught if by the $m$-th evening some policeman shadows exactly those $m$ accomplices who are still trusted by the thief. Prove that to guarantee the capture of the thief at least $2^m$ policemen are needed.

For any integer $d > 0,$ let $f(d)$ be the smallest possible integer that has exactly $d$ positive divisors (so for example we have $f(1)=1, f(5)=16,$ and $f(6)=12$). Prove that for every integer $k \geq 0$ the number $f\left(2^k\right)$ divides $f\left(2^{k+1}\right).$ [i]Proposed by Suhaimi Ramly, Malaysia[/i]

Find all $ f: Q^ \plus{} \to\ Z$ functions that satisfy $ f \left(\frac {1}{x} \right) \equal{} f(x)$ and $ (x \plus{} 1)f(x \minus{} 1) \equal{} xf(x)$ for all rational numbers that are bigger than 1.

Let $a,b,c>0$ such that $a+b+c=1$. Prove: \[\frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a \ge 3(a^{2}+b^{2}+c^{2}) \]

The lock of a safe consists of 3 wheels, each of which may be set in 8 different ways positions. Due to a defect in the safe mechanism the door will open if any two of the three wheels are in the correct position. What is the smallest number of combinations which must be tried if one is to guarantee being able to open the safe (assuming the "right combination" is not known)?

Prove there exists a constant $c$ (independent of $n$) such that for any graph $G$ with $n>2$ vertices, we can split $G$ into a forest and at most $cf(n)$ disjoint cycles, where a) $f(n)=n\ln{n}$; b) $f(n)=n$. [i]David Yang.[/i]

There are 2002 towns in a kingdom. Some of the towns are connected by roads in such a manner that, if all roads from one city closed, one can still travel between any two cities. Every year, the kingdom chooses a non-self-intersecting cycle of roads, founds a new town, connects it by roads with each city from the chosen cycle, and closes all the roads from the original cycle. After several years, no non-self-intersecting cycles remained. Prove that at that moment there are at least 2002 towns, exactly one road going out from each of them.