A game of solitaire strats of with $25$ cards. Some are facing up and sum are facing down. In each move a card that's facing up should me choosen, taken away, and turning over the cards next to it (if there are cards next to it). The game is won when you have accomplished to take all the $25$ cards from the table. If you initially start with $n$ cards facing up, find all the values of $n$ such that the game can be won. Explain how to win the game, independently from the initial placement of the cards facing up, justify your answer for why it is impossible to win with other values of $n$. Two cards are neighboring when one is immediately next to the other, to the left or right. Example: The card marked $A$ has two neighboring cards and the one marked with only a $B$ has only one neighboring card. After taking a card there is a hole left, such that the card marked $C$ has only one neighboring card, and the one marked $D$ does'nt have any.

For any finite sets $X$ and $Y$ of positive integers, denote by $f_X(k)$ the $k^{\text{th}}$ smallest positive integer not in $X$, and let $$X*Y=X\cup \{ f_X(y):y\in Y\}.$$Let $A$ be a set of $a>0$ positive integers and let $B$ be a set of $b>0$ positive integers. Prove that if $A*B=B*A$, then $$\underbrace{A*(A*\cdots (A*(A*A))\cdots )}_{\text{ A appears $b$ times}}=\underbrace{B*(B*\cdots (B*(B*B))\cdots )}_{\text{ B appears $a$ times}}.$$ [i]Proposed by Alex Zhai, United States[/i]

Some paper squares of $k$ distinct colors are placed on a rectangular table, with sides parallel to the sides of the table. Suppose that for any $k$ squares of distinct colors, some two of them can be nailed on the table with only one nail. Prove that there is a color such that all squares of that color can be nailed with $2k-2$ nails.

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]

Addison and Emerson are playing a card game with three rounds. Addison has the cards $1, 3$, and $5$, and Emerson has the cards $2, 4$, and $6$. In advance of the game, both designate each one of their cards to be played for either round one, two, or three. Cards cannot be played for multiple rounds. In each round, both show each other their designated card for that round, and the person with the higher-numbered card wins the round. The person who wins the most rounds wins the game. Let $m/n$ be the probability that Emerson wins, where $m$ and $n$ are relatively prime positive integers. Find $m +n$. [i]Proposed by Ada Tsui[/i]

In the country some mathematicians know each other and any division of them into two sets contain 2 friends from different sets.It is known that if you put any set of four or more mathematicians at a round table so that any two neighbours know each other , then at the table there are two friends not sitting next to each other.We denote by $c_i $ the number of sets of $i$ pairwise familiar mathematicians(by saying "familiar" it means know each other).Prove that $c_1-c_2+c_3-c_4+...=1$

Let there be $n$ students and $m$ clubs. The students joined the clubs so that the following is true: - For all students $x$, you can choose some clubs such that $x$ is the only student who joined all of the chosen clubs. Let the number of clubs each student joined be $a_1,a_2,...,a_m$. Prove that $$a_1!(m - a_1)! + a_2!(m - a_2)! + ... + a_n!(m -a_n)! \le m!$$

Fernanda decided to decorate a square blanket with a ribbon and buttons, placing a button in the center of each square where the ribbon passes and forming the design indicated in the figure. If Fernanda sews the first button in the shaded square on line $0$, on which line does she sew the $2007$th button? [img]https://cdn.artofproblemsolving.com/attachments/2/9/0c9c85ec6448ee3f6f363c8f4bcdd5209f53f6.png[/img]

$3m$ balls numbered $1, 1, 1, 2, 2, 2, 3, 3, 3, \ldots, m, m, m$ are distributed into $8$ boxes so that any two boxes contain identical balls. Find the minimal possible value of $m$.

Lucy starts by writing $s$ integer-valued $2022$-tuples on a blackboard. After doing that, she can take any two (not necessarily distinct) tuples $\mathbf{v}=(v_1,\ldots,v_{2022})$ and $\mathbf{w}=(w_1,\ldots,w_{2022})$ that she has already written, and apply one of the following operations to obtain a new tuple: \begin{align*} \mathbf{v}+\mathbf{w}&=(v_1+w_1,\ldots,v_{2022}+w_{2022}) \\ \mathbf{v} \lor \mathbf{w}&=(\max(v_1,w_1),\ldots,\max(v_{2022},w_{2022})) \end{align*} and then write this tuple on the blackboard. It turns out that, in this way, Lucy can write any integer-valued $2022$-tuple on the blackboard after finitely many steps. What is the smallest possible number $s$ of tuples that she initially wrote?

There is a sequence of integers $ a_1, a_2, \ldots, a_{2n+1} $ with the following property: after eliminating any term, the remaining ones can be divided into two groups of $ n $ terms such that the sum of the terms in the first group is equal to the sum words in the second. Prove that all terms of the sequence are equal.

Prove that the number of ways of choosing $6$ among the first $49$ positive integers, at least two of which are consecutive, is equal to $\binom{49}6-\binom{44}6$.

[b]p1.[/b] Find $$2^{\left(0^{\left(2^3\right)}\right)}$$ [b]p2.[/b] Amy likes to spin pencils. She has an $n\%$ probability of dropping the $n$th pencil. If she makes $100$ attempts, the expected number of pencils Amy will drop is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$. [b]p3.[/b] Determine the units digit of $3 + 3^2 + 3^3 + 3^4 +....+ 3^{2022} + 3^{2023}$. [b]p4.[/b] Cyclic quadrilateral $ABCD$ is inscribed in circle $\omega$ with center $O$ and radius $20$. Let the intersection of $AC$ and $BD$ be $E$, and let the inradius of $\vartriangle AEB$ and $\vartriangle CED$ both be equal to $7$. Find $AE^2 - BE^2$. [b]p5.[/b] An isosceles right triangle is inscribed in a circle which is inscribed in an isosceles right triangle that is inscribed in another circle. This larger circle is inscribed in another isosceles right triangle. If the ratio of the area of the largest triangle to the area of the smallest triangle can be expressed as $a+b\sqrt{c}$, such that $a, b$ and $c$ are positive integers and no square divides $c$ except $1$, find $a + b + c$. [b]p6.[/b] Jonny has three days to solve as many ISL problems as he can. If the amount of problems he solves is equal to the maximum possible value of $gcd \left(f(x), f(x+1) \right)$ for $f(x) = x^3 +2$ over all positive integer values of $x$, then find the amount of problems Jonny solves. [b]p7.[/b] Three points $X$, $Y$, and $Z$ are randomly placed on the sides of a square such that $X$ and $Y$ are always on the same side of the square. The probability that non-degenerate triangle $\vartriangle XYZ$ contains the center of the square can be written as $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Find $a + b$. [b]p8.[/b] Compute the largest integer less than $(\sqrt7 +\sqrt3)^6$. [b]p9.[/b] Find the minimum value of the expression $\frac{(x+y)^2}{x-y}$ given $x > y > 0$ are real numbers and $xy = 2209$. [b]p10.[/b] Find the number of nonnegative integers $n \le 6561$ such that the sum of the digits of $n$ in base $9$ is exactly $4$ greater than the sum of the digits of $n$ in base $3$. [b]p11.[/b] Estimation (Tiebreaker) Estimate the product of the number of people who took the December contest, the sum of all scores in the November contest, and the number of incorrect responses for Problem $1$ and Problem $2$ on the October Contest. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

To connect to the OFM site, Alice must choose a password. The latter must be consisting of $n$ characters among the following $27$ characters: $$A, B, C, . . ., Y , Z, \#$$ We say that a password $m$ is [i]redundant [/i] if we can color in red and blue a block of consecutive letters of $m$ in such a way that the word formed from the red letters is identical to the word formed from blue letters. For example, the password $H\#ZBZJBJZ$ is redundant, because it contains the [color=#00f]ZB[/color][color=#f00]Z[/color][color=#00f]J[/color][color=#f00]BJ[/color] block, where the word $ZBJ$ appears in both blue and red. At otherwise, the $ABCACB$ password is not redundant. Show that, for any integer $n \ge 1$, there exist at least $18^n$ passwords of length $n$, that is to say formed of $n$ characters each, which are not redundant.

For integers $m,n \ge 3$ we consider a $m \times n$ rectangular frame, consisting of the $2m+2n-4$ boundary squares of a $m \times n$ rectangle. Renate and Erhard play the following game on this frame, with Renate to start the game. In a move, a player colours a rectangular area consisting of a single or several white squares. If there are any more white squares, they have to form a connected region. The player who moves last wins the game. Determine all pairs $(m,n)$ for which Renate has a winning strategy.

On a board with $3$ columns and $4$ rows, each of the $12$ squares will be painted green or white. In the first and last row, the number of squares painted green must be the same. Furthermore, in the first and last column, the number of squares painted green must also be unequal. How many different ways can you paint the board?

For positive integers $n$, let $S_n$ be the set of integers $x$ such that $n$ distinct lines, no three concurrent, can divide a plane into $x$ regions (for example, $S_2=\{3,4\}$, because the plane is divided into 3 regions if the two lines are parallel, and 4 regions otherwise). What is the minimum $i$ such that $S_i$ contains at least 4 elements?

Let $L$ be a figure made of $3$ squares, a right isosceles triangle and a quarter circle (all unit sized) as shown below: [img]https://wiki-images.artofproblemsolving.com//f/f9/Weirwiueripo.png[/img] Prove that any $18$ points in the plane can be covered with copies of $L$, which don't overlap (copies of $L$ may be rotated or flipped)

Given $n \in N$, let $A$ be a family of subsets of $\{1,2,...,n\}$. If for every two sets $B,C \in A$ the set $(B \cup C) -(B \cap C)$ has an even number of elements, find the largest possible number of elements of $A$ .

You are at a point $(a,b)$ and you need to reach another point $(c,d)$. Both points are below the line $x = y$ and have integer coordinates. You can move in steps of length 1, either upwards of to the right, but you may not move to a point on the line $x = y$. How many different paths are there?

Let $n$ a positive integer. In a $2n\times 2n$ board, $1\times n$ and $n\times 1$ pieces are arranged without overlap. Call an arrangement [b]maximal[/b] if it is impossible to put a new piece in the board without overlapping the previous ones. Find the least $k$ such that there is a [b]maximal[/b] arrangement that uses $k$ pieces.

Some squares of a $n \times n$ tabel ($n>2$) are black, the rest are withe. In every white square we write the number of all the black squares having at least one common vertex with it. Find the maximum possible sum of all these numbers.

Let $n\ge 4$ be an integer. You have two $n\times n$ boards. Each board contains the numbers $1$ to $n^2$ inclusive, one number per square, arbitrarily arranged on each board. A move consists of exchanging two rows or two columns on the first board (no moves can be made on the second board). Show that it is possible to make a sequence of moves such that for all $1 \le i \le n$ and $1 \le j \le n$, the number that is in the $i-th$ row and $j-th$ column of the first board is different from the number that is in the $i-th$ row and $j-th$ column of the second board.

We have $k$ switches arranged in a row, and each switch points up, down, left, or right. Whenever three successive switches all point in different directions, all three may be simultaneously turned so as to point in the fourth direction. Prove that this operation cannot be repeated infinitely many times.

Kolya and Dima play a game on an $8\times 8$ board, making moves in turn. During his turn, Kolya must put one cross in any empty cell (i.e., in a cell in which a cross has not yet been drawn and which has not yet been covered with a domino). Dima must cover two adjacent cells with a domino (which are not yet covered with other dominoes), in which there are an even number of crosses in total (0 or 2). The one who can't make a move loses. Which of does the player have a winning strategy, if [list=a] [*]Dima makes the first move? [*]Kolya makes the first move? [/list] [i]Proposed by M. Didin[/i]