This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

AND
OR
NO

Found problems: 259

Let $ a, b, c$ be positive integers for which $ abc \equal{} 1$. Prove that $ \sum \frac{1}{b(a\plus{}b)} \ge \frac{3}{2}$.
Consider a $ 7\times 7$ numbers table $ a_{ij} \equal{} (i^2 \plus{} j)(i \plus{} j^2), 1\le i,j\le 7.$ When we add arbitrarily each term of an arithmetical progression consisting of $ 7$ integers to corresponding to term of certain row (or column) in turn, call it an operation. Determine whether such that each row of numbers table is an arithmetical progression, after a finite number of operations.
Five identical empty buckets of $2$-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Cinderella chooses a pair of neighbouring buckets, empties them to the river and puts them back. Then the next round begins. The Stepmother goal's is to make one of these buckets overflow. Cinderella's goal is to prevent this. Can the wicked Stepmother enforce a bucket overflow? [i]Proposed by Gerhard Woeginger, Netherlands[/i]
It is given $5$ numbers $1$, $3$, $5$, $7$, $9$. We get the new $5$ numbers such that we take arbitrary $4$ numbers(out of current $5$ numbers) $a$, $b$, $c$ and $d$ and replace them with $\frac{a+b+c-d}{2}$, $\frac{a+b-c+d}{2}$, $\frac{a-b+c+d}{2}$ and $\frac{-a+b+c+d}{2}$. Can we, with repeated iterations, get numbers: $a)$ $0$, $2$, $4$, $6$ and $8$ $b)$ $3$, $4$, $5$, $6$ and $7$
We are given an $n \times n$ board, where $n$ is an odd number. In each cell of the board either $+1$ or $-1$ is written. Let $a_k$ and $b_k$ denote them products of numbers in the $k^{th}$ row and in the $k^{th}$ column respectively. Prove that the sum $a_1 +a_2 +\cdots+a_n +b_1 +b_2 +\cdots+b_n$ cannot be equal to zero.
Consider $n$ students with numbers $1, 2, \ldots, n$ standing in the order $1, 2, \ldots, n.$ Upon a command, any of the students either remains on his place or switches his place with another student. (Actually, if student $A$ switches his place with student $B,$ then $B$ cannot switch his place with any other student $C$ any more until the next command comes.) Is it possible to arrange the students in the order $n,1, 2, \ldots, n-1$ after two commands ?
Let $a,b$ be two positive integers, such that $ab\neq 1$. Find all the integer values that $f(a,b)$ can take, where \[ f(a,b) = \frac { a^2+ab+b^2} { ab- 1} . \]
Triangles $ABC$ and $ABD$ are isosceles with $AB =AC = BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC$, then $\angle C + \angle D$ is [asy] size(130); defaultpen(linewidth(0.8) + fontsize(11pt)); pair A, B, C, D, E; real angle = 70; B = origin; A = dir(angle); D = dir(90-angle); C = rotate(2*(90-angle), A) * B; draw(A--B--C--cycle); draw(B--D--A); E = extension(B, D, C, A); draw(rightanglemark(B, E, A, 1.5)); label("$A$", A, dir(90)); label("$B$", B, dir(210)); label("$C$", C, dir(330)); label("$D$", D, dir(0)); label("$E$", E, 1.5*dir(340)); [/asy] $\textbf{(A)}\ 115^\circ \qquad \textbf{(B)}\ 120^\circ \qquad \textbf{(C)}\ 130^\circ \qquad \textbf{(D)}\ 135^\circ \qquad \textbf{(E)}\ \text{not uniquely determined}$
There are $ n$ markers, each with one side white and the other side black. In the beginning, these $ n$ markers are aligned in a row so that their white sides are all up. In each step, if possible, we choose a marker whose white side is up (but not one of the outermost markers), remove it, and reverse the closest marker to the left of it and also reverse the closest marker to the right of it. Prove that, by a finite sequence of such steps, one can achieve a state with only two markers remaining if and only if $ n \minus{} 1$ is not divisible by $ 3$. [i]Proposed by Dusan Dukic, Serbia[/i]
Numbers $\frac{49}{1}, \frac{49}{2}, ... , \frac{49}{97}$ are writen on a blackboard. Each time, we can replace two numbers (like $a, b$) with $2ab-a-b+1$. After $96$ times doing that prenominate action, one number will be left on the board. Find all the possible values fot that number.
Players $A$ and $B$ play a "paintful" game on the real line. Player $A$ has a pot of paint with four units of black ink. A quantity $p$ of this ink suffices to blacken a (closed) real interval of length $p$. In every round, player $A$ picks some positive integer $m$ and provides $1/2^m $ units of ink from the pot. Player $B$ then picks an integer $k$ and blackens the interval from $k/2^m$ to $(k+1)/2^m$ (some parts of this interval may have been blackened before). The goal of player $A$ is to reach a situation where the pot is empty and the interval $[0,1]$ is not completely blackened. Decide whether there exists a strategy for player $A$ to win in a finite number of moves.
There are $2019$ students sitting around circular table. Initially each of them have one candy. Teacher is allowed to pick one student, who has at least one can candy, and this student can decide, whether he gives his candy to his neighbour on the right or on the left. Prove that no matter what students teacher picks during the process, students can always ensure that any point of time no student has more than $2$ candies.
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.
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]
A finite set $S$ of unit squares is chosen out of a large grid of unit squares. The squares of $S$ are tiled with isosceles right triangles of hypotenuse $2$ so that the triangles do not overlap each other, do not extend past $S$, and all of $S$ is fully covered by the triangles. Additionally, the hypotenuse of each triangle lies along a grid line, and the vertices of the triangles lie at the corners of the squares. Show that the number of triangles must be a multiple of $4$.
Given $ a_0 \equal{} 1$, $ a_1 \equal{} 3$, and the general relation $ a_n^2 \minus{} a_{n \minus{} 1}a_{n \plus{} 1} \equal{} (\minus{}1)^n$ for $ n \ge 1$. Then $ a_3$ equals: $ \textbf{(A)}\ \frac{13}{27}\qquad \textbf{(B)}\ 33\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ \minus{}17$
Prove that each finite set of integers can be arranged without intersection.
A single player game has the following rules: initially, there are $10$ piles of stones with $1,2,...,10$ stones, respectively. A movement consists on making one of the following operations: i) to choose $2$ piles, both of them with at least $2$ stones, combine them and then add $2$ stones to the new pile; ii) to choose a pile with at least $4$ stones, remove $2$ stones from it, and then split it into two piles with amount of piles to be chosen by the player. The game continues until is not possible to make an operation. a) Give an example of a sequence of moves leading to the end of the game. b) Make a table with the total number of stones and the number of piles before and after the first 5 operations in your example above. c) Show that the number of piles with one stone in the end of the game is always the same, no matter how the movements are made.
A board of $2n$ x $2n$ is colored chess style, a movement is the changing of colors of a $2$ x $2$ square. For what integers $n$ is possible to complete the board with one color using a finite number of movements?
There are $ n$ markers, each with one side white and the other side black. In the beginning, these $ n$ markers are aligned in a row so that their white sides are all up. In each step, if possible, we choose a marker whose white side is up (but not one of the outermost markers), remove it, and reverse the closest marker to the left of it and also reverse the closest marker to the right of it. Prove that, by a finite sequence of such steps, one can achieve a state with only two markers remaining if and only if $ n \minus{} 1$ is not divisible by $ 3$. [i]Proposed by Dusan Dukic, Serbia[/i]
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.
There are $n$ children around a round table. Erika is the oldest among them and she has $n$ candies, while no other child has any candy. Erika decided to distribute the candies according to the following rules. In every round, she chooses a child with at least two candies and the chosen child sends a candy to each of his/her two neighbors. (So in the first round Erika must choose herself). For which $n \ge 3$ is it possible to end the distribution after a finite number of rounds with every child having exactly one candy?
To each vertex of a regular pentagon an integer is assigned, so that the sum of all five numbers is positive. If three consecutive vertices are assigned the numbers $x,y,z$ respectively, and $y<0$, then the following operation is allowed: $x,y,z$ are replaced by $x+y,-y,z+y$ respectively. Such an operation is performed repeatedly as long as at least one of the five numbers is negative. Determine whether this procedure necessarily comes to an end after a finite number of steps.
Consider 2n+1 coins lying in a circle. At the beginning, all the coins are heads up. Moving clockwise, 2n+1 flips are performed: one coin is flipped, the next coin is skipped, the next coin is flipped, the next two coins are skipped, the next coin is flipped,the next three coins are skipped and so on, until finally 2n coins are skipped and the next coin is flipped.Prove that at the end of this procedure,exactly one coin is heads down.
Objects $A$ and $B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $A$ starts at $(0,0)$ and each of its steps is either right or up, both equally likely. Object $B$ starts at $(5,7)$ and each of its steps is either left or down, both equally likely. Which of the following is closest to the probability that the objects meet? $ \textbf{(A)}\ 0.10 \qquad \textbf{(B)}\ 0.15 \qquad \textbf{(C)}\ 0.20 \qquad \textbf{(D)}\ 0.25 \qquad \textbf{(E)}\ 0.30$