If you form the three heaps (and are therefore the second to draw), any
one of the following thirteen groupings will give you a win if you play
correctly: 15, 14, 1; 15, 13, 2; 15, 12, 3; 15, 11, 4; 15, 10, 5; 15, 9,
6; 15, 8, 7; 14, 13, 3; 14, 11, 5; 14, 9, 7; 13, 11, 6; 13, 10, 7; 12,
11, 7.
The beautiful general solution of this problem is as follows. Express
the number in every heap in powers of 2, avoiding repetitions and
remembering that 2^0 = 1. Then if you so leave the matches to your
opponent that there is an even number of every power, you can win. And
if at the start you leave the powers even, you can always continue to do
so throughout the game. Take, as example, the last grouping given
above--12, 11, 7. Expressed in powers of 2 we have--
12 = 8 4 - -
11 = 8 - 2 1
7 = - 4 2 1
-------
2 2 2 2
-------
As there are thus two of every power, you must win. Say your opponent
takes 7 from the 12 heap. He then leaves--
5 = - 4 - 1
11 = 8 - 2 1
7 = - 4 2 1
-------
1 2 2 3
-------
Here the powers are not all even in number, but by taking 9 from the 11
heap you immediately restore your winning position, thus--
5 = - 4 - 1
2 = - - 2 -
7 = - 4 2 1
-------
- 2 2 2
-------
And so on to the end. This solution is quite general, and applies to any
number of matches and any number of heaps. A correspondent informs me
that this puzzle game was first propounded by Mr. W.M.F. Mellor, but
when or where it was published I have not been able to ascertain.