Zero Two Seven Nine

Two players alternate. A pile holds $n$ stones; a move removes exactly 1, 3, or 4 stones. Whoever cannot move loses. You cannot see a pattern, so you brute force the first ten values, marking W if the player to move wins:

$n$	0	1	2	3	4	5	6	7	8	9
result	L	W	L	W	W	W	W	L	W	L

The problem asks about $n = 100$ . What do you take from the table?

Ten values is not enough to justify any rule; the honest move is to extend the brute force to $n = 100$ and read the answer off directly.

The losing positions are exactly $n \equiv 0$ or $2 \pmod 7$ . Since $100 = 14 \cdot 7 + 2$ , the player to move at $n = 100$ loses.

The losing positions are exactly $n \equiv 0$ or $2 \pmod 7$ , and $100 \bmod 7 = 3$ , so the player to move wins.

The losing positions are the even numbers below 3 and then every odd number from 7 on, which is too irregular to extend; compute $n = 100$ with a $O(n)$ dynamic program instead.