Question 31433
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.