Suppose that $n$ people $A_1$, $A_2$, $\ldots$, $A_n$, ($n \geq 3$) are seated in a circle and that $A_i$ has $a_i$ objects such that \[ a_1 + a_2 + \cdots + a_n = nN \] where $N$ is a positive integer. In order that each person has the same number of objects, each person $A_i$ is to give or to receive a certain number of objects to or from its two neighbours $A_{i-1}$ and $A_{i+1}$. (Here $A_{n+1}$ means $A_1$ and $A_n$ means $A_0$.) How should this redistribution be performed so that the total number of objects transferred is minimum?

Let $k$ be a positive real. $A$ and $B$ play the following game: at the start, there are $80$ zeroes arrange around a circle. Each turn, $A$ increases some of these $80$ numbers, such that the total sum added is $1$. Next, $B$ selects ten consecutive numbers with the largest sum, and reduces them all to $0$. $A$ then wins the game if he/she can ensure that at least one of the number is $\geq k$ at some finite point of time. Determine all $k$ such that $A$ can always win the game.

Let $k\geq4$ be an integer. Sunny and Ming play a game with strings. A string is a sequence that every element of it is an integer between $1$ and $k$, inclusive. At first, Sunny chooses two positive integers $N,L\geq2$ and write down $N$ strings, each having length $L$. Then Ming mark at most $\frac{N}{2}$ strings. Then Sunny chooses an unmarked string $s$ and calculate the biggest integer $n$ such that there exists another string satisfying its first $n$ element is the same as the first $n$ element of $s$. Then Sunny burn down all strings which first $n$ element if different from the first $n$ element of $s$, leaving only the ones which have the same first $n$ element of $s$. Finally, Ming chooses an integer $d$ between $1$ and $k$, inclusive, and remove all strings which $(n+1)$th element is $d$. Sunny's score would be the number of strings left. Find the maximum score that Sunny can guarantee to get. [i]Proposed by USJL[/i]

A cloakroom in a cinema works with some breaks. The total time cloakroom worked today is 8 hours. The schedule of the cloakroom is such that it is possible to show any film of duration at most 12 hours such that the cloakroom will be open at least one hour before and after the film (the films are shown without breaks). Find the minimal possible amount of breaks in the schedule of cloakroom. [i]A. Voidelevich[/i]

Let $ n$ and $ k$ be positive integers. Please, find an explicit formula for \[ \sum y_1y_2 \dots y_k,\] where the summation runs through all $ k\minus{}$tuples positive integers $ (y_1,y_2,\dots,y_k)$ satisfying $ y_1\plus{}y_2\plus{}\dots\plus{}y_k\equal{}n$.

[b]p1.[/b] Four positive numbers are placed at the vertices of a rectangle. Each number is at least as large as the average of the two numbers at the adjacent vertices. Prove that all four numbers are equal. [b]p2.[/b] The sum $498+499+500+501=1998$ is one way of expressing $1998$ as a sum of consecutive positive integers. Find all ways of expressing $1998$ as a sum of two or more consecutive positive integers. Prove your list is complete. [b]p3.[/b] An infinite strip (two parallel lines and the region between them) has a width of $1$ inch. What is the largest value of $A$ such that every triangle with area $A$ square inches can be placed on this strip? Justify your answer. [b]p4.[/b] A plane divides space into two regions. Two planes that intersect in a line divide space into four regions. Now suppose that twelve planes are given in space so that a) every two of them intersect in a line, b) every three of them intersect in a point, and c) no four of them have a common point. Into how many regions is space divided? Justify your answer. [b]p5.[/b] Five robbers have stolen $1998$ identical gold coins. They agree to the following: The youngest robber proposes a division of the loot. All robbers, including the proposer, vote on the proposal. If at least half the robbers vote yes, then that proposal is accepted. If not, the proposer is sent away with no loot and the next youngest robber makes a new proposal to be voted on by the four remaining robbers, with the same rules as above. This continues until a proposed division is accepted by at least half the remaining robbers. Each robber guards his best interests: He will vote for a proposal if and only if it will give him more coins than he will acquire by rejecting it, and the proposer will keep as many coins for himself as he can. How will the coins be distributed? Explain your reasoning. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

Consider a game played on the integers in the closed interval $[1, n]$. The game begins with some tokens placed in $[1, n]$. At each turn, tokens are added or removed from$ [1, n]$ using the following rule: For each integer $k \in [1, n]$, if exactly one of $k - 1$ and $k + 1$ has a token, place a token at $k$ for the next turn, otherwise leave k blank for the next turn. We call a position [i]static [/i] if no changes to the interval occur after one turn. For instance, the trivial position with no tokens is static because no tokens are added or removed after a turn (because there are no tokens). Find all non-trivial static positions.

Let $n$ be a positive integer and $A=\{ 1,2,\ldots ,n\}$. A subset of $A$ is said to be connected if it consists of one element or several consecutive elements. Determine the maximum $k$ for which there exist $k$ distinct subsets of $A$ such that the intersection of any two of them is connected.

For a given integer $n \ge 4$ we examine whether there exists a table with three rows and $n$ columns which can be filled by the numbers $1, 2,...,, 3n$ such that $\bullet$ each row totals to the same sum $z$ and $\bullet$ each column totals to the same sum $s$. Prove: (a) If $n$ is even, such a table does not exist. (b) If $n = 5$, such a table does exist. (Gerhard J. Woeginger)

Let $n$ be a positive integer. A [i]Japanese triangle[/i] consists of $1 + 2 + \dots + n$ circles arranged in an equilateral triangular shape such that for each $i = 1$, $2$, $\dots$, $n$, the $i^{th}$ row contains exactly $i$ circles, exactly one of which is coloured red. A [i]ninja path[/i] in a Japanese triangle is a sequence of $n$ circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with $n = 6$, along with a ninja path in that triangle containing two red circles. [asy] // credit to vEnhance for the diagram (which was better than my original asy): size(4cm); pair X = dir(240); pair Y = dir(0); path c = scale(0.5)*unitcircle; int[] t = {0,0,2,2,3,0}; for (int i=0; i<=5; ++i) { for (int j=0; j<=i; ++j) { filldraw(shift(i*X+j*Y)*c, (t[i]==j) ? lightred : white); draw(shift(i*X+j*Y)*c); } } draw((0,0)--(X+Y)--(2*X+Y)--(3*X+2*Y)--(4*X+2*Y)--(5*X+2*Y),linewidth(1.5)); path q = (3,-3sqrt(3))--(-3,-3sqrt(3)); draw(q,Arrows(TeXHead, 1)); label("$n = 6$", q, S); label("$n = 6$", q, S); [/asy] In terms of $n$, find the greatest $k$ such that in each Japanese triangle there is a ninja path containing at least $k$ red circles.

There are $998$ cities in a country. Some pairs of cities are connected by two-way flights. According to the law, between any pair cities should be no more than one flight. Another law requires that for any group of cities there will be no more than $5k+10$ flights connecting two cities from this group, where $k$ is the number number of cities in the group. Prove that several new flights can be introduced so that laws still hold and the total number of flights in the country is equal to $5000$.

A container contains five red balls. On each turn, one of the balls is selected at random, painted blue, and returned to the container. The expected number of turns it will take before all five balls are colored blue is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

There were finitely many persons at a party among whom some were friends. Among any $4$ of them there were either $3$ who were all friends among each other or $3$ who weren't friend with each other. Prove that you can separate all the people at the party in two groups in such a way that in the first group everyone is friends with each other and that all the people in the second group are not friends to anyone else in second group. (Friendship is a mutual relation).

The game of [b]Hive [/b]is played on a regular hexagonal grid (as shown in the figure) by 3 players. The grid consists of $k$ layers (where $k$ is a natural number) surrounding a regular hexagon, with each layer constructed around the previous layer. The figure below shows a grid with 2 layers. The players, [i]Ali[/i], [i]Shayan[/i], and [i]Sajad[/i], take turns playing the game. In each turn, a player places a tile, similar to the one shown in the figure, on the empty cells of the grid (rotation of the tile is also allowed). The first player who is unable to place a tile on the grid loses the game. Prove that two players can collaborate in such a way that the third player always loses. Proposed by [size=110]Pouria Mahmoudkhan Shirazi[/size].

Suppose that each of the 5 persons knows a piece of information, each piece is different, about a certain event. Each time person $A$ calls person $B$, $A$ gives $B$ all the information that $A$ knows at that moment about the event, while $B$ does not say to $A$ anything that he knew. (a) What is the minimum number of calls are necessary so that everyone knows about the event? (b) How many calls are necessary if there were $n$ persons?

How many six-digit perfect squares can be formed using all the numbers $1,2,3,4,5,6$ as digits? $\textbf{(A)}~5$ $\textbf{(B)}~19$ $\textbf{(C)}~7$ $\textbf{(D)}~\text{None of the above}$

$k$ is an integer, there exists a triangulation for a regular polygon with $2024$ sides and $2024$ colored dots with $k$ different colors meeting $(1)$ each color will be used at least once $(2)$ every small triangle will have at least $2$ dots that will be in the same color. Try to find the maximum value of$k$

You have a collection of at least two tokens where each one has a number less than or equal to 10 written on it. The sum of the numbers on the tokens is \( S \). Find all possible values of \( S \) that guarantee that the tokens can be separated into two groups such that the sum of each group does not exceed 80.

For a sequence of $10$ coin flips, each pair of consecutive flips and count the number of “Heads-Heads”, “Heads-Tails”, “Tails-Heads”, and “Tails-Tails” sequences is recorded. These four numbers are then multiplied to get the [i]Tiger number[/i] of the sequence of flips. How many such sequences have a [i]Tiger number [/i] of $24$?

At a meeting of students of the 9th "G" class, it was decided to declare the 9th "G" a presidential republic. Four blocs nominated their candidates for the presidency: “Our Street”, “Our Yard”, “Our House” and “Our Entrance”. When discussing how to select a president four proposals were made. A. “What is there to think about! Have each student place a piece of paper with the name of the candidate they support in the box. Whoever gets the most votes is the president.” B. “No, you can’t do that. If no one gets more than half the votes, a repeat vote must be held, in which the top two from the first vote must participate.” C. “We must choose the one who is better than anyone else. How to do it? Let each person make a list: in the first place on his list he should put the best in his opinion, in the second place - the second, etc. If in most lists B is higher than A, then he is better than B. So, B is better than everyone if he is better than A, better than B and better than D.” D. “Let everyone really make their own list, as V said. For the first place on the list, the candidate receives three points, for the second - 2, for the third - 1 point, and for the fourth - 0. Whoever scores the most points will he’s the president.” As you can see, all four proposed methods are quite democratic. And yet, can it turn out that with method A one candidate wins, with method B another candidate wins, with C a third one, and in option D a fourth one? It is known that there are 29 people in the class, but the applicants (there are four of them) do not participate in the voting. Each student votes strictly according to his list (see speech B). After a heated discussion, option B was adopted. Interestingly, if elections had been held immediately, then after two rounds a representative of the Our Yard bloc would have become president. However, elections were scheduled for a week later. Students belonging to the “Our Yard” bloc, not knowing the true state of affairs, based on the principle “you can’t spoil porridge with butter,” launched a vigorous campaign in support of their candidate. As a result of this agitation, many students did not change their opinion. True, in some lists the position of the representative of “Our Yard” has improved. (All the changes boiled down to the fact that only this candidate’s position improved.) But as a result, another was elected president. How could this happen? [hide=might have typos, here is the original wording] На собрании учеников 9-го «Г» класса было принято решение — объявить 9-й «Г» президентской республикой. Своих кандидатов на пост президента выдвинули четыре блока: «Наша улица», «Наш двор», «Наш дом» и «Наш подъезд». При обсуждении способов выбора президента прозвучало четыре предложения. А. «Чего здесь думать! Пусть каждый ученик опустит в ящик бумажку с фамилией поддерживаемого им претендента. Кто наберет больше голосов, тот и президент.» Б. «Нет, так нельзя. Если никто не наберет больше половины голосов, надо устроить повторное голосование, в котором должны участвовать двое лучших по результатам первого голосования.» В. «Надо выбрать того, кто лучше любого другого. Как это сделать? Пусть каждый человек составит список: на первое место в своем списке он должен поставить самого лучшего по его мнению, на второе — второго и т.д. Если в большинстве списков В стоит выше А, то он лучше В. Значит, В лучше всех, если он лучше А, лучше Б и лучше Г.» Г. «Пусть и в самом деле каждый составит свой список, как сказал В. За первое место в списке кандидат получает три очка, за второе — 2, за третье — 1 очко, а за четвертое — 0. Кто наберет больше всех очков, тот и президент.» Как видим, все четыре предложенных способа вполне демократичны. И все же, может ли получиться так, что при способе А побеждает один кандидат, при способе Б — другой кандидат, при В — третий, ну, а в варианте Г — четвертый? Известно, что в классе 29 человек, но претенденты (их четверо) в голосовании не участвуют. Каждый ученик голосует строго в соответствии со своим списком (см. выступление В). После бурного обсуждения был принят вариант Б. Интересно, что если бы сразу же были проведены выборы, то после двух туров президентом стал бы представитель блока «Наш двор». Однако выборы были назначены на неделю позже. Ученики, входящие в блок «Наш двор», не зная ис тинного положения дел, исходя из принципа «кашу маслом не испортишь», развернули бурную агитацию в поддержку своего кандидата. В результате этой агитации многие ученики никак не изменили своего мнения. Правда, в некоторых списках улучшилось положение представителя «Наш двор». (Все изменения свелись к тому, что улучшилось положение только этого претендента.) Но в результате президентом был избран другой. Как такое могло случиться? [\hide]

Prove that in the set $ \{1,2, \ldots, 1989\}$ can be expressed as the disjoint union of subsets $ A_i, \{i \equal{} 1,2, \ldots, 117\}$ such that [b]i.)[/b] each $ A_i$ contains 17 elements [b]ii.)[/b] the sum of all the elements in each $ A_i$ is the same.

There are $12$ mathematicians living in a village, each of whom belongs to the $\sqrt2$-clan or belong to the $\pi$-clan. Moreover every mathematician's birthday is in a different month and every mathematician has an odd number of friends among them the mathematicians. We agree that if mathematician $A$ is a friend of mathematician $B$, then so is $B$ is a friend of $A$. On his birthday, every mathematician looks at which clan the majority of his friends belong to, and decides to join that clan until his next birthday. Prove that the mathematicians no longer change clans after a certain point.

Towns $A_1, A_2, . . . , A_{30}$ lie on line $MN$. The distances between the consecutive towns are equal. Each of the towns is the point of origin of a straight highway. The highways are on the same side of $MN$ and form the following angles with it: [img]https://cdn.artofproblemsolving.com/attachments/a/f/6cfcac497bdd729b966705f1060bd4b1caba25.png[/img] Thirty cars start simultaneously from these towns along the highway at the same constant speed. Each intersection has a gate. As soon as the first (in time, not in number) car passes the intersection the gate closes and blocks the way for all other cars approaching this intersection. Which cars will pass all intersections and which will be stopped? Note: This refers to angles measured counterclockwise from straight MN to the corresponding road.

In a room, there are $100$ light switches, labeled with the positive integers ${1,2, . . . ,100}$. They’re all initially turned off. On the $i$ th day for $1 \le i \le 100$, Bob flips every light switch with label number $k$ divisible by $i$ a total of $\frac{k}{i}$ times. Find the sum of the labels of the light switches that are turned on at the end of the $100$th day.

A journalist is looking for a person $Z$ at a meeting of $n$ persons. He has been told that $Z$ knows all the other people at the meeting but none of them knows $Z$. The journalist may ask any person about any other person: “Do you know that person?” One person can be questioned many times. All answers are truthful. (a) Can the journalist always find $Z$ by asking less than $n$ questions? (b) What is the minimal number of questions which are needed to find $Z$? (G Galperin)