Found problems: 259
There is a pile with $15$ coins on a table. At each step, Pedro choses one of the piles in the table with $a>1$ coins and divides it in two piles with $b\geq1$ and $c\geq1$ coins and writes in the board the product $abc$. He continues until there are $15$ piles with $1$ coin each. Determine all possible values that the final sum of the numbers in the board can have.
Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$ initially contains one coin. The following operations are allowed
Type 1) Choose a non-empty box $B_j$, $1\leq j \leq 5$, remove one coin from $B_j$ and add two coins to $B_{j+1}$;
Type 2) Choose a non-empty box $B_k$, $1\leq k \leq 4$, remove one coin from $B_k$ and swap the contents (maybe empty) of the boxes $B_{k+1}$ and $B_{k+2}$.
Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ become empty, while box $B_6$ contains exactly $2010^{2010^{2010}}$ coins.
[i]Proposed by Hans Zantema, Netherlands[/i]
The vertices of a regular $n$-gon are initially marked with one of the signs $+$ or $-$ . A [i]move[/i] consists in choosing three consecutive vertices and changing the signs from the vertices , from $+$ to $-$ and from $-$ to $+$.
[b]a)[/b] Prove that if $n=2015$ then for any initial configuration of signs , there exists a sequence of [i]moves[/i] such that we'll arrive at a configuration with only $+$ signs.
[b]b)[/b] Prove that if $n=2016$ , then there exists an initial configuration of signs such that no matter how we make the [i]moves[/i] we'll never arrive at a configuration with only $+$ signs.
Given a positive integer $k$ and other two integers $b > w > 1.$ There are two strings of pearls, a string of $b$ black pearls and a string of $w$ white pearls. The length of a string is the number of pearls on it. One cuts these strings in some steps by the following rules. In each step:
[b](i)[/b] The strings are ordered by their lengths in a non-increasing order. If there are some strings of equal lengths, then the white ones precede the black ones. Then $k$ first ones (if they consist of more than one pearl) are chosen; if there are less than $k$ strings longer than 1, then one chooses all of them.
[b](ii)[/b] Next, one cuts each chosen string into two parts differing in length by at most one. (For instance, if there are strings of $5, 4, 4, 2$ black pearls, strings of $8, 4, 3$ white pearls and $k = 4,$ then the strings of 8 white, 5 black, 4 white and 4 black pearls are cut into the parts $(4,4), (3,2), (2,2)$ and $(2,2)$ respectively.) The process stops immediately after the step when a first isolated white pearl appears.
Prove that at this stage, there will still exist a string of at least two black pearls.
[i]Proposed by Bill Sands, Thao Do, Canada[/i]
The numbers $1$ through $4^{n}$ are written on a board. In each step, Pedro erases two numbers $a$ and $b$ from the board, and writes instead the number $\frac{ab}{\sqrt{2a^2+2b^2}}$. Pedro repeats this procedure until only one number remains. Prove that this number is less than $\frac{1}{n}$, no matter what numbers Pedro chose in each step.
A hexagonal table is given, as the one on the drawing, which has $~$ $2012$ $~$ columns. There are $~$ $2012$ $~$ hexagons in each of the odd columns, and there are $~$ $2013$ $~$ hexagons in each of the even columns. The number $~$ $i$ $~$ is written in each hexagon from the $~$ $i$-th column. Changing the numbers in the table is allowed in the following way: We arbitrarily select three adjacent hexagons, we rotate the numbers, and if the rotation is clockwise then the three numbers decrease by one, and if we rotate them counterclockwise the three numbers increase by one (see the drawing below). What's the maximum number of zeros that can be obtained in the table by using the above-defined steps.
An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer $m$ from each of the integers at two neighboring vertices and adding $2m$ to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount $m$ and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integers, there is exactly one vertex at which the game can be won.
The numbers $1,2,\dots,10$ are written on a board. Every minute, one can select three numbers $a$, $b$, $c$ on the board, erase them, and write $\sqrt{a^2+b^2+c^2}$ in their place. This process continues until no more numbers can be erased. What is the largest possible number that can remain on the board at this point?
[i]Proposed by Evan Chen[/i]
There are three boxes of stones. Sisyphus moves stones one by one between the boxes. Whenever he moves a stone, Zeus gives him the number of coins that is equal to the difference between the number of stones in the box the stone was put in, and that in the box the stone was taken from (the moved stone does not count). If this difference is negative, then Sisyphus returns the corresponding amount to Zeus (if Sisyphus cannot pay, generous Zeus allows him to make the move and pay later).
After some time all the stones lie in their initial boxes. What is the greatest possible earning of Sisyphus at that moment?
[i]I. Izmest’ev[/i]
Given any set of $9$ points in the plane such that there is no $3$ of them collinear, show that for each point $P$ of the set, the number of triangles with its vertices on the other $8$ points and that contain $P$ on its interior is even.
Let $m$ boxes be given, with some balls in each box. Let $n < m$ be a given integer. The following operation is performed: choose $n$ of the boxes and put $1$ ball in each of them. Prove:
[i](a) [/i]If $m$ and $n$ are relatively prime, then it is possible, by performing the operation a finite number of times, to arrive at the situation that all the boxes contain an equal number of balls.
[i](b)[/i] If $m$ and $n$ are not relatively prime, there exist initial distributions of balls in the boxes such that an equal distribution is not possible to achieve.
Mad scientist Kyouma writes $N$ positive integers on a board. Each second, he chooses two numbers $x, y$ written on the board with $x > y$, and writes the number $x^2-y^2$ on the board. After some time, he sends the list of all the numbers on the board to Christina. She notices that all the numbers from 1 to 1000 are present on the list. Aid Christina in finding the minimum possible value of N.
Let $ S $ be the first quadrant and $ T:S\longrightarrow S $ be a transformation that takes the reciprocal of the coordinates of the points that belong to its domain. Define an [i]S-line[/i] to be the intersection of a line with $ S. $
[b]a)[/b] Show that the fixed points of $ T $ lie on any fixed S-line of $ T. $
[b]b)[/b] Find all fixed S-lines of $ T. $
[i]Gabriel Popa[/i]
Let $n$ be an integer greater than $1.$ $n$ pupils are seated around a round table, each having a certain number of candies (it is possible that some pupils don't have a candy) such that the sum of all the candies they possess is a multiple of $n.$ They exchange their candies as follows: For each student's candies at first, there is at least a student who has more candies than the student sitting to his/her right side, in which case, the student on the right side is given a candy by that student. After a round of exchanging, if there is at least a student who has candies greater than the right side student, then he/she will give a candy to the next student sitting to his/her right side. Prove that after the exchange of candies is completed (ie, when it reaches equilibrium), all students have the same number of candies.
We call the three variable polynomial $P$ cyclic if $P(x,y,z)=P(y,z,x)$. Prove that cyclic three variable polynomials $P_1,P_2,P_3$ and $P_4$ exist such that for each cyclic three variable polynomial $P$, there exists a four variable polynomial $Q$ such that $P(x,y,z)=Q(P_1(x,y,z),P_2(x,y,z),P_3(x,y,z),P_4(x,y,z))$.
[i]Solution by Mostafa Eynollahzade and Erfan Salavati[/i]
Written on a blackboard are $n$ nonnegative integers whose greatest common divisor is $1$. A [i]move[/i] consists of erasing two numbers $x$ and $y$, where $x\ge y$, on the blackboard and replacing them with the numbers $x-y$ and $2y$. Determine for which original $n$-tuples of numbers on the blackboard is it possible to reach a point, after some number of moves, where $n-1$ of the numbers of the blackboard are zeroes.
For any integer $n\geq 2$, we compute the integer $h(n)$ by applying the following procedure to its decimal representation. Let $r$ be the rightmost digit of $n$.
[list][*]If $r=0$, then the decimal representation of $h(n)$ results from the decimal representation of $n$ by removing this rightmost digit $0$.
[*]If $1\leq r \leq 9$ we split the decimal representation of $n$ into a maximal right part $R$ that solely consists of digits not less than $r$ and into a left part $L$ that either is empty or ends with a digit strictly smaller than $r$. Then the decimal representation of $h(n)$ consists of the decimal representation of $L$, followed by two copies of the decimal representation of $R-1$. For instance, for the number $17,151,345,543$, we will have $L=17,151$, $R=345,543$ and $h(n)=17,151,345,542,345,542$.[/list]
Prove that, starting with an arbitrary integer $n\geq 2$, iterated application of $h$ produces the integer $1$ after finitely many steps.
[i]Proposed by Gerhard Woeginger, Austria[/i]
In a simple graph $G$, an operation is defined as taking two neighbor vertices $u,v$ which have a common neighbor, deleting the edge between $u,v$ and adding a new vertex $w$ whose neighbors are exactly the common neighbors of $u$ and $v$. Starting with the complete graph $G=K_n$ where $n\ge 3$ is a positive integer, find the maximum number of operations that can be applied.
Proposed by[i] Deniz Can Karaçelebi[/i]
Consider a table with one real number in each cell. In one step, one may switch the sign of the numbers in one row or one column simultaneously. Prove that one can obtain a table with non-negative sums in each row and each column.
In a room there are several children and a pile of 1000 sweets. The children come to the pile one after another in some order. Upon reaching the pile each of them divides the current number of sweets in the pile by the number of children in the room, rounds the result if it is not integer, takes the resulting number of sweets from the pile and leaves the room. All the boys round upwards and all the girls round downwards. The process continues until everyone leaves the room. Prove that the total number of sweets received by the boys does not depend on the order in which the children reach the pile.
[i]Maxim Didin[/i]
Let $ M$ and $ N$ be points on the side $ BC$ of triangle $ ABC$, with the point $ M$ lying on the segment $ BN$, such that $ BM \equal{} CN$. Let $ P$ and $ Q$ be points on the segments $ AN$ and $ AM$, respectively, such that $ \measuredangle PMC \equal{}\measuredangle MAB$ and $ \measuredangle QNB \equal{}\measuredangle NAC$. Prove that $ \measuredangle QBC \equal{}\measuredangle PCB$.
Some positive integers are initially written on a board, where each $2$ of them are different.
Each time we can do the following moves:
(1) If there are 2 numbers (written in the board) in the form $n, n+1$ we can erase them and write down $n-2$
(2) If there are 2 numbers (written in the board) in the form $n, n+4$ we can erase them and write down $n-1$
After some moves, there might appear negative numbers. Find the maximum value of the integer $c$ such that:
Independetly of the starting numbers, each number which appears in any move is greater or equal to $c$
Let $n > 3$ be a positive integer. Suppose that $n$ children are arranged in a circle, and $n$ coins are distributed between them (some children may have no coins). At every step, a child with at least 2 coins may give 1 coin to each of their immediate neighbors on the right and left. Determine all initial distributions of the coins from which it is possible that, after a finite number of steps, each child has exactly one coin.
There are infinitely many boxes - initially one of them contains $n$ balls and all others are empty. On a single move we take some balls from a non-empty box and put them into an empty box and on a sheet of paper we write down the product of the resulting amount of balls in the two boxes. After some moves, the sum of all numbers on the sheet of paper became $2023$. What is the smallest possible value of $n$?
Let $M$ be a set of points in a plane with at least two elements. Prove that if $M$ has two axes of symmetry $g_1$ and $g_2$ intersecting at an angle $\alpha = q\pi$, where $q$ is irrational, then $M$ must be infinite.