SOLUTION: Candace and Darryl are playing a game with piles of beans. On each player’s turn, they must remove at least one, but at most 7 beans from one of the piles. It is Candace’s turn and

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Candace and Darryl are playing a game with piles of beans. On each player’s turn, they must remove at least one, but at most 7 beans from one of the piles. It is Candace’s turn and      Log On


   

Question 1074179: Candace and Darryl are playing a game with piles of beans. On each player’s turn, they must remove at least one, but at most 7 beans from one of the piles. It is Candace’s turn and there are
two piles of beans with 25 and 27 beans in each. The winner is the player to take the last bean. What is Candace’s best move?
Answer by ikleyn(52813) About Me  (Show Source):
You can put this solution on YOUR website!
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I think that Candace's best strategy (winning strategy) is to maintain the difference of 0 between the number of beans in two piles.


Following to this strategy, Candace's next (closest) step is to remove 2 beans from the 27-beans pile.


So, Candace will achieve his strategic goal at the first step, and after any step of his partner Candace can maintain the difference of 0.


By doing in this way, whichever number of beans will Darryl remove from one container, the same number of beans will Candace 
remove from the other container, maintaining the number of beans the same in both containers.


By doing in this way, he will FORCE Darryl to remove, finally, the last bean (or the last portion of beans) from one of 
the two containers at Darryl's current step. 


Then Candace wins the game at his next (=last) turn removing the same number of  remaining beans from the other container.

Solved.