Found problems: 1385
(a) Two people perform a card trick. The first performer takes $5$ cards from a $52$-card deck (previously shuffled by a member of the audience) , looks at them, and arranges them in a row from left to right: one face down (not necessarily the first one) , the others face up . The second performer guesses correctly the card which is face down. Prove that the performers can agree on a system which always makes this possible.
(b) For their second trick, the first performer arranges four cards in a row, face up, the fifth card is kept hidden. Can they still agree on a system which enables the second performer to correctly guess the hidden card?
(G Galperin)
We will say that a natural number is [i]acceptable[/i] if it has at most $9$ distinct prime divisors. There is a stack of $100!=1\times2\times\cdots\times100$ stones. A [i]legal move[/i] consists in removing $k$ stones from the stack, where $k$ is an acceptable number. Two players, Lucas and Nicolas, take turns making legal moves; Lucas starts the game. The one who removes the last stone wins. Determine which of the players has a winning strategy and describe this strategy.
Two players are writting in turn natural numbers not exceeding $p$. The rules forbid to write the divisors of the numbers already having been written. Those who cannot make his move looses.
a) Who, and how, can win if $p=10$?
b) Who wins if $p=1000$?
On each of the 2006 cards a natural number is written. Cards are placed arbitrarily in a row. 2 players take in turns a card from any end of the row until all the cards are taken. After that each player calculates sum of the numbers written of his cards. If the sum of the first player is not less then the sum of the second one then the first player wins. Show that there's a winning strategy for the first player.
There are $11$ empty boxes and a pile of stones. Two players play the following game by alternating moves: In one move a player takes $10$ stones from the pile and places them into boxes, taking care to place no more than one stone in any box. The winner is the player after whose move there appear $21$ stones in one of the boxes for the first time. If a player wants to guarantee that they win the game, should they go first or second? Explain your reasoning.
Andrew and Barry play the following game: there are two heaps with $a$ and $b$ pebbles, respectively. In the first round Barry chooses a positive integer $k,$ and Andrew takes away $k$ pebbles from one of the two heaps (if $k$ is bigger than the number of pebbles in the heap, he takes away the complete heap). In the second round, the roles are reversed: Andrew chooses a positive integer and Barry takes away the pebbles from one of the two heaps. This goes on, in each round the two players are reversing the roles. The player that takes the last pebble loses the game.
Which player has a winning strategy?
[i]Submitted by András Imolay, Budapest[/i]
$ A$ and $ B$ play the following game with a polynomial of degree at least 4:
\[ x^{2n} \plus{} \_x^{2n \minus{} 1} \plus{} \_x^{2n \minus{} 2} \plus{} \ldots \plus{} \_x \plus{} 1 \equal{} 0
\]
$ A$ and $ B$ take turns to fill in one of the blanks with a real number until all the blanks are filled up. If the resulting polynomial has no real roots, $ A$ wins. Otherwise, $ B$ wins. If $ A$ begins, which player has a winning strategy?
Let $n$ be a positive integer. Humanity will begin to colonize Mars. The SpaceY and SpaceZ agencies will be responsible for traveling between the planets. To prevent the rockets from colliding, they will travel alternately, with SpaceY making the first trip. On each trip, the responsible agency will do one of two types of mission:
(i) choose a positive integer $k$ and take $k$ people to Mars, creating a new colony on the planet and settling them in that colony;
(ii) choose some existing colony on Mars and a positive integer $k$ strictly smaller than the population of that colony, and bring $k$ people from that colony back to Earth.
To maintain the organization on Mars, a mission cannot result in two colonies with the same population and the number of colonies must be at most $n$. The first agency that cannot carry out a mission will go bankrupt. Determine, in terms of $n$, which agency can guarantee that it will not go bankrupt first.
There are $n$ coins in the pile. Two players play a game by alternately performing a move. A move consists of taking $5,7$ or $11$ coins away from the pile. The player unable to perform a move loses the game. Which player - the one playing first or second - has the winning strategy if:
(a) $n=2001$;
(b) $n=5000$?
Given integer $n \geq 3$. There are $n$ dots marked $1$ to $n$ clockwise on a big circle. And between every two neighboring dots, there is a light. At first, every light were dark.
A and B are playing a game, A pick up $n$ pairs from $\{ (i,j)|1 \leq i < j \leq n \}$ and for every pairs $(i,j)$. B starts from the point marked $i$ and choose to walk clockwise or counterclockwise to the point marked $j$. And B invert the status of all passing lights (bright $\leftrightarrow$ dark)
A hopes the number of dark light can be as much as possible while B hopes the number of bright light can be as much as possible. Suppose A, B are both smart, how many lights are bright in the end?
[i]Proposed by BlessingOfHeaven[/i]
[img]https://pbs.twimg.com/profile_images/1014932415201120256/u9KAaMZ4_400x400.jpg[/img]
Three persons $A,B,C$, are playing the following game:
A $k$-element subset of the set $\{1, . . . , 1986\}$ is randomly chosen, with an equal probability of each choice, where $k$ is a fixed positive integer less than or equal to $1986$. The winner is $A,B$ or $C$, respectively, if the sum of the chosen numbers leaves a remainder of $0, 1$, or $2$ when divided by $3$.
For what values of $k$ is this game a fair one? (A game is fair if the three outcomes are equally probable.)
Andy, Bella, and Chris are playing in a trivia contest. Andy has $21,200$ points, Bella has $23,600$ points, and Chris has $11,200$ points. They have reached the final round, which works as follows:
[list]
[*] they are given a hint as to what the only question of the round will be about;
[*] then, each of them must bet some amount of their points---this bet must be a nonnegative integer (a player does not know any of the other players' bets, and this bet cannot be changed later on);
[*] then, they will be shown the question, where they will have $30$ seconds to individually submit a response (a player does not know any of the other players' answers);
[*] finally, once time runs out, their responses and bets are revealed---if a player's response is correct, then the number of points they bet will be added to their score, otherwise, it will be subtracted from their score, and whoever ends up having the most points wins the contest (if there is a tie for the win, then the winner is randomly decided).
[/list]
Suppose that the contestants are currently deciding their bets based on the hint that the question will be about history. Bella knows that she will likely get the question wrong, but she also knows that Andy, who dislikes history, will definitely get it wrong. Knowing this, Bella wagers an amount that will guarantee her a win. Find the maximum number of points Bella could have ended up with.
A frog jumps on the coordinate lattice, starting from the point $(1,1)$, according to the following rules:
(i) From point $(a,b)$ the frog can jump to either $(2a,b)$ or $(a,2b)$;
(ii) If $a>b$, the frog can also jump from $(a,b)$ to $(a-b,b)$, while for $a<b$ it can jump from $(a,b)$ to $(a,b-a)$.
Can the frog get to the point: (a) $(24,40)$; (b) $(40,60)$; (c) $(24,60)$; (d) $(200,4)$?
In a chess tournament there are $n>2$ players. Every two players play against each other exactly once. It is known that exactly $n$ games end as a tie. For any set $S$ of players, including $A$ and $B$, we say that $A$ [i]admires[/i] $B$ [i]in that set [/i]if
i) $A$ does not beat $B$; or
ii) there exists a sequence of players $C_1,C_2,\ldots,C_k$ in $S$, such that $A$ does not beat $C_1$, $C_k$ does not beat $B$, and $C_i$ does not beat $C_{i+1}$ for $1\le i\le k-1$.
A set of four players is said to be [i]harmonic[/i] if each of the four players admires everyone else in the set. Find, in terms of $n$, the largest possible number of harmonic sets.
If you take a subset of $4002$ numbers from the whole numbers $1$ to $6003$, then there is always a subset of $2001$ numbers within that subset with the following property:
If you order the $2001$ numbers from small to large, the numbers are alternately even and odd (or odd and even).
Prove this.
A pile of $n$ pebbles is placed in a vertical column. This configuration is modified according to the following rules. A pebble can be moved if it is at the top of a column which contains at least two more pebbles than the column immediately to its right. (If there are no pebbles to the right, think of this as a column with 0 pebbles.) At each stage, choose a pebble from among those that can be moved (if there are any) and place it at the top of the column to its right. If no pebbles can be moved, the configuration is called a [i]final configuration[/i]. For each $n$, show that, no matter what choices are made at each stage, the final configuration obtained is unique. Describe that configuration in terms of $n$.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=119189]IMO ShortList 2001, combinatorics problem 7, alternative[/url]
An angle of size $n \times m$, where $m, n \ge 2$, is called a figure, resulting from a rectangle of size $n \times m$ cells by removing the rectangle size $(n - 1) \times (m - 1)$ cells. Two players take turns making moves consisting in painting in a corner an arbitrary non-zero number of cells forming a rectangle or square.
Written on a blackboard is the polynomial $x^2+x+2014$. Calvin and Hobbes take turns alternately (starting with Calvin) in the following game. At his turn, Calvin should either increase or decrease the coefficient of $x$ by $1$. And at this turn, Hobbes should either increase or decrease the constant coefficient by $1$. Calvin wins if at any point of time the polynomial on the blackboard at that instant has integer roots. Prove that Calvin has a winning stratergy.
Jesse and Tjeerd are playing a game. Jesse has access to $n\ge 2$ stones. There are two boxes: in the black box there is room for half of the stones (rounded down) and in the white box there is room for half of the stones (rounded up). Jesse and Tjeerd take turns, with Jesse starting. Jesse grabs in his turn, always one new stone, writes a positive real number on the stone and places put him in one of the boxes that isn't full yet. Tjeerd sees all these numbers on the stones in the boxes and on his turn may move any stone from one box to the other box if it is not yet full, but he may also choose to do nothing. The game stops when both boxes are full. If then the total value of the stones in the black box is greater than the total value of the stones in the white box, Jesse wins; otherwise win Tjeerd. For every $n \ge 2$, determine who can definitely win (and give a corresponding winning strategy).
$100$ heaps of stones lie on a table. Two players make moves in turn. At each move, a player can remove any non-zero number of stones from the table, so that at least one heap is left untouched. The player that cannot move loses. Determine, for each initial position, which of the players, the first or the second, has a winning strategy.
[i]K. Kokhas[/i]
[b]EDIT.[/b] It is indeed confirmed by the sender that empty heaps are still heaps, so the third post contains the right guess of an interpretation.
The Bank of Pittsburgh issues coins that have a heads side and a tails side. Vera has a row of 2023 such coins alternately tails-up and heads-up, with the leftmost coin tails-up.
In a [i]move[/i], Vera may flip over one of the coins in the row, subject to the following rules:
[list=disc]
[*] On the first move, Vera may flip over any of the $2023$ coins.
[*] On all subsequent moves, Vera may only flip over a coin adjacent to the coin she flipped on the previous move. (We do not consider a coin to be adjacent to itself.)
[/list]
Determine the smallest possible number of moves Vera can make to reach a state in which every coin is heads-up.
[i]Luke Robitaille[/i]
Alice has 8 coins. She knows for sure only that 7 of these coins are genuine and weigh the same, while the remaining one is counterfeit and is either heavier or lighter than any of the other 7. Bob has a balance scale. The scale shows which plate is heavier but does not show by how much. For each measurement, Alice pays Bob beforehand a fee of one coin. If a genuine coin has been paid, Bob tells Alice the correct weighing outcome, but if a counterfeit coin has been paid, he gives a random answer. Alice wants to identify 5 genuine coins and not to give any of these coins to Bob. Can Alice achieve this result for sure?
Albert and Brita play a game with a bar of $19$ adjacent squares. Initially, there is a button on the middle square of the bar. At every turn Albert mentions one positive integer less than $5$, and Brita moves button a number of squares in the direction of her choice - while doing so however, Brita must not move the button more than twice in one direction order. Prove that Albert can choose the numbers so that by the $19$th turn, Brita to be forced to move the button out of the bar.
For an integer $n \geq 5,$ two players play the following game on a regular $n$-gon. Initially, three consecutive vertices are chosen, and one counter is placed on each. A move consists of one player sliding one counter along any number of edges to another vertex of the $n$-gon without jumping over another counter. A move is legal if the area of the triangle formed by the counters is strictly greater after the move than before. The players take turns to make legal moves, and if a player cannot make a legal move, that player loses. For which values of $n$ does the player making the first move have a winning strategy?
$n$ players are playing a game. Each player has $n$ tokens. Every turn, two players with at least one token are randomly selected. The player with less tokens gives one token to the player with more tokens. If both players have the same number of tokens, a coin flip decides which player receives a token and which player gives a token. The game ends when one player has all the tokens. If $n = 2019$, suppose the maximum number of turns the game could take to end can be written as $\frac{1}{d} (a \cdot 2019^3 + b \cdot 2019^2 + c \cdot 2019)$ for integers $a, b, c, d$. Find $\frac{abc}{d}$
.