A bag of 2000 chips is used in a game for two players, who take turns drawing 2, 3, 4, 5 or 6 chips each, but no other amount. The player to draw last loses. I can always win if i go first. How many chips do i need to draw first to guarente that i win? what is my susequent strategy?
Im not that fast with math when a math teacher gives me word problems i get all confused so if you could explain to me how i figure this out that would be great! ^.^
2007-10-01 11:22:11 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You want to leave your opponent with 2 chips so he/she loses. Note that no matter how many chips (x) your opponent takes on a particular turn, you can take 8-x chips on the next turn so that you have taken 8 chips between you. Therefore, if you leave your opponent with 10 chips, it doesn't matter how many he/she takes, you can leave him/her with 2. If you leave your opponent with 18 chips, you can ensure that he/she will have 10 left on his/her next turn. And so on. In general, if you leave your opponent with 8k+2 chips on his/her turn, you can win (k is an integer). Therefore, you want to take 6 chips on the first turn, to leave your opponent with 1994 chips (k=249). You will make a total of 250 moves and defeat your opponent.
2007-10-01 11:36:35 · answer #1 · answered by 7 · 1⤊ 0⤋
When you work backwards the trend becomes clear.
You want to leave your opponent with 2 on his last play, that way he has no choice but to lose. To accomplish that, you must have 4-8 on your last play. That ways you choose enough to leave him with 2.
This means that on his second to last play, he must have 10. This means on your second to last play, you must have 12-16.
This means that on his third to last play, he must have 18. And so on. Keep adding 8, and this pattern repeats itself.
Strategy: On every turn leave your opponent 1994 - 8n.
So on your first turn, choose 6. Then whatever he does, make sure he gets 1986 on his next try. And so on. That way you're guaranteed to win.
If however you make a mistake, and he knows how to play, you'll never be able to recover.
2007-10-01 19:31:19 · answer #2 · answered by Dr D 7 · 0⤊ 0⤋
the place * potential multiply: [- 5*y^4*y^(5*2) ] / [ 15*y^7*y^(2*3) ] <-- multiply powers [- 5*y^4*y^(10) ] / [ 15*y^7*y^(6) ] [- 5*y^(4+10) ] / [ 15*y^(7+6) ] <-- upload powers [- 5*y^(14) ] / [ 15*y^(13) ] cancelling: (5/15 = a million/3 & y^(14)/y^(13) = y <-- subtract powers) -a million*y /3 or -y/3 or -a million/3 *y
2016-10-10 03:13:31 · answer #3 · answered by ? 4 · 0⤊ 0⤋