Quote:
Originally Posted by prince_chittu @Srikar
Can u plz elaborate on the answers u hv given of the this question
There are 500 balls.A person can pick a minimum of 5 balls and maximum of 10 balls.A & B are playing this game.A starts the game.How many balls A should pick to win if:
i)Last picker wins the game
ii)Last picker looses the game. |
Okay Here is my brain work ..
A & B will always make a combination of 15 score if B picks first and A follows ie BA = 15. eg B = 5 , A=10 or B=10, A=5
Now In case I, let A pick x balls. so we have x + 15N = 500
At N=33 and x = 5 we have a solution i.e if A picks 5 Balls initially, A will always pick last ball. Sequence would be A BA BA BA BA ...
Now In case II, let A pick x balls in first turn. Assuming B will have to pick last set of balls. In this case, B should be left with 5 to 9 balls in the last turn. P.S - In case 9 balls are left, B cannot pick 5 cause this will make 4 balls left, which cannot be picked by A!
The sequence would be A BA BA BA BA B
The equation would be x + 15N + (5...9) = 500
i.e x + 15N = (491...495)
Now for N = 32, x = 11...15
N = 33 x = 0....5
So that means, If A starts, A cannot Win.
Others .. please put your thoughts on this.