Answer
There are four base cases. If n = 1 = 4·0+1, then clearly the
second player wins.
-If there are two, three, or four matches
(n = 4·0+2, n = 4·0+3, or n = 4·1),
-then the first player can win by removing all but one match.
Work Step by Step
--Basic step:
There are four base cases. If n = 1 = 4·0+1, then clearly the
second player wins.
-If there are two, three, or four matches
(n = 4·0+2, n = 4·0+3, or n = 4·1),
-then the first player can win by removing all but one match.
-- Inductive step:
-Assume the strong inductive hypothesis, that in games with k or fewer matches, the first player can win if k ≡ 0, 2, or 3 (mod 4) and
the second player can win if k ≡ 1 (mod 4).
Suppose we have a game with k+1 matches, with k≥4. If k+1 ≡ 0 (mod 4),
then the first player can remove three matches, leaving k − 2
matches for the other player.
Because k −2 ≡ 1 (mod 4),
- by the inductive hypothesis,
this is a game that the second player at that point (who is the first player in our game) can win.
-Similarly,
if k + 1 ≡ 2 (mod 4),
then
-the first player can remove
one match; and if k + 1 ≡ 3 (mod 4),
then
- the first player
can remove two matches. Finally, if k +1 ≡ 1 (mod 4), then
the first player must leave k, k − 1, or k − 2 matches for the
other player.
-Because k ≡ 0 (mod 4), k − 1 ≡ 3 (mod 4),
and k − 2 ≡ 2 (mod 4),
- by the inductive hypothesis,
this is a game that the first player at that point (who is the second
player in our game) can win.
