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

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Found problems: 815

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.
Let $p$ be a prime number. Find all integers $k$ for which $\sqrt{k^2 -pk}$ is a positive integer.
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]
Let $n>1$ be a positive integer. Each cell of an $n\times n$ table contains an integer. Suppose that the following conditions are satisfied: [list=1] [*] Each number in the table is congruent to $1$ modulo $n$. [*] The sum of numbers in any row, as well as the sum of numbers in any column, is congruent to $n$ modulo $n^2$. [/list] Let $R_i$ be the product of the numbers in the $i^{\text{th}}$ row, and $C_j$ be the product of the number in the $j^{\text{th}}$ column. Prove that the sums $R_1+\hdots R_n$ and $C_1+\hdots C_n$ are congruent modulo $n^4$.
$n(>1)$ lotus leaves are arranged in a circle. A frog jumps from a particular leaf from another under the following rule: [list] [*]It always moves clockwise. [*]From starting it skips one leaf and then jumps to the next. After that it skips two leaves and jumps to the following. And the process continues. (Remember the frog might come back on a leaf twice or more.)[/list] Given that it reaches all leaves at least once. Show $n$ cannot be odd.
There are $n \geq 1$ notebooks, numbered from $1$ to $n$, stacked in a pile. Zahar repeats the following operation: he randomly chooses a notebook whose number $k$ does not correspond to its location in this stack, counting from top to bottom, and returns it to the $k$th position, counting from the top, without changing the location of the other notebooks. If there is no such notebook, he stops. Is it guaranteed that Zahar will arrange all the notebooks in ascending order of numbers in a finite number of operations? [i]Proposed by Zahar Naumets[/i]
Let $\mathcal{A}$ denote the set of all polynomials in three variables $x, y, z$ with integer coefficients. Let $\mathcal{B}$ denote the subset of $\mathcal{A}$ formed by all polynomials which can be expressed as \begin{align*} (x + y + z)P(x, y, z) + (xy + yz + zx)Q(x, y, z) + xyzR(x, y, z) \end{align*} with $P, Q, R \in \mathcal{A}$. Find the smallest non-negative integer $n$ such that $x^i y^j z^k \in \mathcal{B}$ for all non-negative integers $i, j, k$ satisfying $i + j + k \geq n$.
Let $ ABC$ be an acute-angled triangle. $ C_{1}$ and $ C_{2}$ are two circles of diameters $ AB$ and $ AC$, respectively. $ C_{2}$ and $ AB$ intersect again at $ F$, and $ C_{1}$ and $ AC$ intersect again at $ E$. Also, $ BE$ meets $ C_{2}$ at $ P$ and $ CF$ meets $ C_{1}$ at $ Q$. Prove that $ AP=AQ$.
Let $n$ be a positive integer. Tasty and Stacy are given a circular necklace with $3n$ sapphire beads and $3n$ turquoise beads, such that no three consecutive beads have the same color. They play a cooperative game where they alternate turns removing three consecutive beads, subject to the following conditions: [list] [*]Tasty must remove three consecutive beads which are turquoise, sapphire, and turquoise, in that order, on each of his turns. [*]Stacy must remove three consecutive beads which are sapphire, turquoise, and sapphire, in that order, on each of her turns. [/list] They win if all the beads are removed in $2n$ turns. Prove that if they can win with Tasty going first, they can also win with Stacy going first. [i]Yannick Yao[/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 $\nu$ be an irrational positive number, and let $m$ be a positive integer. A pair of $(a,b)$ of positive integers is called [i]good[/i] if \[a \left \lceil b\nu \right \rceil - b \left \lfloor a \nu \right \rfloor = m.\] A good pair $(a,b)$ is called [i]excellent[/i] if neither of the pair $(a-b,b)$ and $(a,b-a)$ is good. Prove that the number of excellent pairs is equal to the sum of the positive divisors of $m$.
Let $a,b$ be two positive integers, such that $ab\neq 1$. Find all the integer values that $f(a,b)$ can take, where \[ f(a,b) = \frac { a^2+ab+b^2} { ab- 1} . \]
Sheldon was really annoying Leonard. So to keep him quiet, Leonard decided to do something. He gave Sheldon the following grid $\begin{tabular}{|c|c|c|c|c|c|} \hline 1 & 1 & 1 & 1 & 1 & 0\\ \hline 1 & 1 & 1 & 1 & 0 & 0\\ \hline 1 & 1 & 1 & 0 & 0 & 0\\ \hline 1 & 1 & 0 & 0 & 0 & 1\\ \hline 1 & 0 & 0 & 0 & 1 & 0\\ \hline 0 & 0 & 0 & 1 & 0 & 0\\ \hline \end{tabular}$ and asked him to transform it to the new grid below $\begin{tabular}{|c|c|c|c|c|c|} \hline 1 & 2 & 18 &24 &28 &30\\ \hline 21 & 3 & 4 &16 &22 &26\\ \hline 23 &19 & 5 & 6 &14 &20\\ \hline 32 &25 &17 & 7 & 8 &12\\ \hline 33 &34 &27 &15 & 9 &10\\ \hline 35 &31 &36 &29 &13 &11\\ \hline \end{tabular}$ by only applying the following algorithm: $\bullet$ At each step, Sheldon must choose either two rows or two columns. $\bullet$ For two columns $c_1, c_2$, if $a,b$ are entries in $c_1, c_2$ respectively, then we say that $a$ and $b$ are corresponding if they belong to the same row. Similarly we define corresponding entries of two rows. So for Sheldon's choice, if two corresponding entries have the same parity, he should do nothing to them, but if they have different parities, he should add 1 to both of them. Leonard hoped this would keep Sheldon occupied for some time, but Sheldon immediately said, "But this is impossible!". Was Sheldon right? Justify.
For a sequence, one can perform the following operation: select three adjacent terms $a,b,c,$ and change it into $b,c,a.$ Determine all the possible positive integers $n\geq 3,$ such that after finite number of operation, the sequence $1,2,\cdots, n$ can be changed into $n,n-1,\cdots,1$ finally.
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.
Let $m$ and $n$ be fixed positive integers. Tsvety and Freyja play a game on an infinite grid of unit square cells. Tsvety has secretly written a real number inside of each cell so that the sum of the numbers within every rectangle of size either $m$ by $n$ or $n$ by $m$ is zero. Freyja wants to learn all of these numbers. One by one, Freyja asks Tsvety about some cell in the grid, and Tsvety truthfully reveals what number is written in it. Freyja wins if, at any point, Freyja can simultaneously deduce the number written in every cell of the entire infinite grid (If this never occurs, Freyja has lost the game and Tsvety wins). In terms of $m$ and $n$, find the smallest number of questions that Freyja must ask to win, or show that no finite number of questions suffice. [i]Nikolai Beluhov[/i]
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 a positive integer. Each cell of an $n\times n$ table contains an integer. Suppose that the following conditions are satisfied: [list=1] [*] Each number in the table is congruent to $1$ modulo $n$. [*] The sum of numbers in any row, as well as the sum of numbers in any column, is congruent to $n$ modulo $n^2$. [/list] Let $R_i$ be the product of the numbers in the $i^{\text{th}}$ row, and $C_j$ be the product of the number in the $j^{\text{th}}$ column. Prove that the sums $R_1+\hdots R_n$ and $C_1+\hdots C_n$ are congruent modulo $n^4$.
Let $FIG$ be a triangle and let $D$ be a point on $\overline{FG}$. The line perpendicular to $\overline{FI}$ passing through the midpoint of $\overline{FD}$ and the line perpendicular to $\overline{IG}$ passing through the midpoint of $\overline{DG}$ intersect at $T$. Prove that $FT = GT$ if and only if $\overline{ID}$ is perpendicular to $\overline{FG}$.