games

Help you solve math 24/08/2017 at 19:23
10000+20000+2543+125444
=30000+2543+125444
=32543+125444
=157987

Dao Trong Luan 24/08/2017 at 16:47
10000 + 20000 + 2543 + 125444
= 30000 + 2543 + 125444
= 32543 + 125444
= 157987

tgrtrdt 20/01 at 15:30
157987

Help you solve math 24/08/2017 at 19:25
4+2+3+200+200+300+12345
=9+400+300+12345
=9+700+12345
=709+12345
=13054

Dao Trong Luan 24/08/2017 at 16:46
4 + 2 + 3 + 200 + 200 + 300 + 12345
= 9 + 400 + 300 + 12345
= 9 + 700 + 12345
= 709 + 12345
= 13054

tgrtrdt 20/01 at 15:31
13054



Help you solve math 24/08/2017 at 19:21
Đao Trong Luan was wrong

Help you solve math 22/08/2017 at 20:17
Human. You should write in english


tgrtrdt 20/01 at 15:32
first is human
Jaki Natsumi
15/07/2017 at 13:03

tranthimai 19/08/2017 at 21:19
hi . me too

Koroshimasu 19/07/2017 at 15:53
me too

tgrtrdt 20/01 at 15:33
me too

Phan Minh Anh 11/06/2017 at 14:58
History is the longest. ^_^
tk mk nha!
Tiểu Thư Họ Phan selected this answer. 
hatsunemiku 03/10/2017 at 22:03
history is the longest

Nguyễn Thị Thanh Hiền 12/12/2017 at 09:20
history
Phan Minh Anh
10/06/2017 at 16:20

Phan Minh Anh 14/06/2017 at 12:58
It Thai Son mountain.

Đỗ Thanh Hải 29/06/2017 at 09:51
Núi Thái Sơn nha Bạn

Phan Minh Anh 10/06/2017 at 16:21
Sorry: What the mountains cut into pieces?


tgrtrdt 20/01 at 15:34
four ducks

Carter
11/04/2017 at 08:14

Nhat Lee 16/04/2017 at 17:16
Pigeonhole principle ?

Lương Trí Dũng 17/05/2017 at 19:17
Pigeonhole principle

Help you solve math 12/08/2017 at 16:59
Principle Pigeonhole
Carter
18/04/2017 at 06:50

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.

Nguyễn Thị Thu Thủy 18/05/2017 at 21:47
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.

Duy Trần Đức 03/05/2017 at 18:46
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.
1
falcon handsome moderators
09/03/2017 at 12:27

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.

Ace Legona 09/03/2017 at 20:09
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 at 12:19
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.

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.
falcon handsome moderators
09/03/2017 at 10:26

Ace Legona 09/03/2017 at 19:37
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?
falcon handsome moderators
09/03/2017 at 10:20

Faded 22/01 at 12:19
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.

Vũ Việt Vương 01/04/2017 at 16:56
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.

Ace Legona 09/03/2017 at 20:12
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.
American
09/03/2017 at 09:51

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 8k 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.

Faded 22/01 at 12:17
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 8k 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.

Vũ Việt Vương 01/04/2017 at 16:57
My answer is 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 8k 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.
American
09/03/2017 at 09:39

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).

Faded 19/01 at 14:52
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). [haha]

An Duong 10/03/2017 at 14:14
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).

Ace Legona 09/03/2017 at 19:14
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.

Faded 19/01 at 14:52
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.

Vũ Việt Vương 01/04/2017 at 16:59
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.