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You play the following game with a friend. You share a pile of chips, and you take turns removing between one and four chips from the pile. (In particular, at least one chip must be removed on each turn.) The game ends when the last chip is removed from the pile; the one who removes it is the loser.
It is your turn, and there are 2014 chips in the pile. How many chips should you remove to guarantee that you win, assuming you then make the best moves until the game is over?
london 26/04/2017 at 10:09
Each time, a person takes 1, 2, 3, or 4 chips. The strategy for a player to win the game is taking the number of chips so that the number of chips left is the form of 5k + 1 .
If It is my turn, and there are 2014 chips in the pile, I shoud take 3 chips and the left is 2011 (=5.k +1). And if the opponent takes x chips (x = 1, 2, 3, 4) , i win take 5 - x chips to guarantee the left is always in the form 5k + 1. And finally, the opponent will take the last chip.
Trần Đức Huy 15/05/2018 at 08:37
Nguyễn Thị Thu Thủy 18/05/2017 at 21:47
falcon handsome	moderators
Real numbers are written in an m x n table. Is is permissible to reverse the signs of all the numbers in any row or column, Prove that after a number of these operations we can make them sum of the numbers along each line (row or column) nonnegative
Ace Legona 09/03/2017 at 19:22
Let \(S\) be the sum of all the \(mn\) numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most \(2^{mn}\) tables. So \(S\) can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then \(S\) increases. Since \(S\) has finitely many possible values, \(S\) can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative.
FA FIFA Club World Cup 2018 16/01/2018 at 21:57
Let SS be the sum of all the mnmn numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most 2mn2mn tables. So SS can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then SS increases. Since SS has finitely many possible values, SS can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative
Nguyễn Anh Tuấn 24/03/2017 at 17:49
Let SS be the sum of all the mnmn numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most 2mn2mn tables. So SS can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then SS increases. Since SS has finitely many possible values, SS can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative.
Each member of a club has at most three enemies in the club. (Here enemies are mutual). Show that the members can be divided into two so that each member in each member in each group has at most one enemy in the group
In the beginning, randomly divide the members into two groups. Let S be the sum of the number of the pairs of enemies in each group. If a member has at least two enemies in the same group, then the member has at most one enemy in the other group. Transferring the member to the other group, we will decrease S by at least one. Since S is a nonnegative integer, it cannot be decreased forever. So after finitely many transfers, each member can have at most one enemy in the same group.
Faded 22/01/2018 at 12:19
Vũ Việt Vương 01/04/2017 at 16:55
n the beginning, randomly divide the members into two groups. Let S be the sum of the number of the pairs of enemies in each group. If a member has at least two enemies in the same group, then the member has at most one enemy in the other group. Transferring the member to the other group, we will decrease S by at least one. Since S is a nonnegative integer, it cannot be decreased forever. So after finitely many transfers, each member can have at most one enemy in the same group.
There are three piles of stone numbering 19,8 and 9, respectively . You are allowed to choose two piles and transfer one stone from each of these two piles to the third piles . After several of these operations, is it possible that each of three piles has 12 stones ?
No. Let the number of stones
in the three piles be \(a, b\) and \(c\),
respectively. Consider (mod \(3\)) of these
numbers. In the beginning, they are \(1, 2, 0\). After one operation, they become \(0, 1, 2\) no matter which two piles have stones
transfer to the third pile. So the
remainders are always \(0, 1, 2\) in some
order. Therefore, all piles having \(12 \) stones are impossible.
Neymar Jr 08/04/2017 at 17:06
i don't know.and you?
Four x's and five o's are written around the circle in an arbitraty order . If two consecutive symbols are the same then insert a new x between them. Otherwise insert a new o between them. Remove the old x's and o's . Keep on repeating this operation .It is possible to get nine o's ?
If we let x=1x=1 and o=–1o=–1, then note that consecutive symbols are replaced by their product. If we consider the product PP of the nine values before and after each operation, we will see that the new PP is the square of the old PP. Hence, PP will always equal 11 after an operation. So nine oo's yielding P=–1P=–1 can never happen.
If we let \(x = 1\) and \(o = – 1\), then note that consecutive symbols are replaced by their product. If we consider the product \(P \) of the nine values before and after each operation, we will see that the new \(P\) is the square of the old \(P\). Hence, \(P \) will always equal \(1 \) after an operation. So nine \(o\)'s yielding \(P = – 1\) can never happen.
In an 8x8 board, there are 32 white pieces and 32 black pieces, one piece in each square. If a player can change all the white pieces to the black and all the black to the white in any row or column in a single move, then is it possible that after finitely many moves, there will be exactly one black piece left on the board?
An Duong 10/03/2017 at 14:28
The answer id NO.
At the begining there are 32 black pieces. We will prove that the number of black pieces left on the board is always even.
In the single move, we change a row or column, assume that row or column has k black pieces and 8 - k while pieces, after change all black to white and while to black, that row or column will has 8-k black and k white. The difference between black pieces after a single move is (8 - k) - k = 8 - 2k. The difference is a even (8 - 2k). Therefore, after every move, the black pieces left is alaways even and cannot be one black piece at any time.
My answer is no.
(1974 Kiew Math Olympiad)
Numbers 1, 2, 3, ..., 1974 are written on the board. You are allowed to replace any two of these numbers by one number, which is either the sum or the difference of these two numbers. Show that after 1973 times performing this operations, the only number left on the board cannot be 0.
hghfghfgh 26/03/2017 at 20:16
At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a - b, we see that:
+ if a odd and b even, or a even and b odd then a + b or a - b is still odd
+ if a and b are both even then a + b or a -b is still even
+ If a and are both odd then a + b or a - b is even
So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even).
So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even). [haha]
Given 2n points in a plane with no three of them collinear. Show that thay can be divided into n pairs such that the n segments joining each pair do not intersect
In the beginning randomly pair the points and join the segments. Let \(S \) be the sum of the lengths of the segments. (Note that since there are finitely many ways of connecting \(2n \) points by \(n \) segments, there are finitely many possible values of \(S\).) If two segments \(AB\)and \(CD\)intersect at \(O\), then replace pairs \(AB\)and \(CD\)by \(AC\)and \(BD\). Since
\(AB + CD = AO + OB + CO + OD > AC + BD\)
by the triangle inequality, whenever there is an intersection, doing this replacement will always decrease \(S\). Since there are only finitely many possible values of \(S\), so eventually there will not be any intersection.
In the beginning randomly pair the points and join the segments. Let S be the sum of the lengths of the segments. (Note that since there are finitely many ways of connecting 2n points by n segments, there are finitely many possible values of S.) If two segments ABand CDintersect at O, then replace pairs ABand CDby ACand BD
. Since
AB+CD=AO+OB+CO+OD>AC+BD
by the triangle inequality, whenever there is an intersection, doing this replacement will always decrease S
. Since there are only finitely many possible values of S, so eventually there will not be any intersection.
In the beginning randomly pair the points and join the segments. Let SS be the sum of the lengths of the segments. (Note that since there are finitely many ways of connecting 2n2n points by nn segments, there are finitely many possible values of SS.) If two segments ABABand CDCDintersect at OO, then replace pairs ABABand CDCDby ACACand BDBD. Since
AB+CD=AO+OB+CO+OD>AC+BDAB+CD=AO+OB+CO+OD>AC+BD
by the triangle inequality, whenever there is an intersection, doing this replacement will always decrease SS. Since there are only finitely many possible values of SS, so eventually there will not be any intersection.
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