Question 31433: Suppose there are N marbles in a bowl. Al, Bo, and Coy will take turns (Al will go 1st) removing marbles from the bowl, and the person to remove the last marble in the bowl loses. Each person is allowed to remove 1 or 2 marbles on his turn. For what values of N can Bo and Coy work together and force Al to lose.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! LET US ARGUE FROM END POSITION AND GO BACKWARDS TO START OF GAME.
E.P=END POSITION....-1P=1 POSITION BEFORE THAT ETC...
E.P......IF C LEAVES 1 COIN THEN A LOSES AS HE HAS TO TAKE THAT LAST ONE COIN.
-1P....SO C SHOULD LEAVE 1
-2P....SO B CAN LEAVE 2 OR 3 ON BOARD
WE HAVE TO TAKE 2 OPTIONS FOR A..HE CAN PICK ONE OR TWO..
-3P....SO C CAN LEAVE 4 ON THE BOARD SO THAT IF A PICKS ONE OR 2,B
WILL HAVE IN THE WINNING POSITION TO LEAVE 2 OR 3 AS PROVED ABOVE.
-4P....SO B CAN LEAVE 5 OR 6 ON THE BOARD.......ETC...ETC....
...........ETC.......ETC........
SO WE CONCLUDE THAT C SHOULD LEAVE 1,OR 4 OR 7 OR 10 ....IN GENERAL
3N-2 COINS ON THE BOARD.IN FACT AT THE BEGINING OF THE GAME THIS
SHOULD BE THE START POSITION AS A HAVE TO PLAY FROM THIS AND WE SHOWED
THAT IN SUCH A CASE BY PROPERLY KEEPING THE ABOVE COMBINATIONS
C......KEEPING NUMBER OF THE FORM 3N-2......AND B KEEPING THE NUMBER
IN THE FORM OF 3N-1 OR 3N ,THEY CAN MAKE A LOSE.
SO THE NUMBER OF COINS AT THE BEGINING SHALL BE
3N-2...THAT IS 1 OR 4 OR 7 OR 10....OR 28....OR 58..... OR 88......ETC.
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