Anne and Brenda play a game which begins with a pile of $n$ toothpicks. They alternate turns with Anne going first. On each player's turn, she must remove 1,3 or 4 toothpicks from the pile. The player who removes the last toothpick wins the game. For which of the following values of $n$ does Brenda have a winning strategy? (In a game, a player has a winning strategy if, regardless of what the other player does, there are moves that she can make which guarantee that she will win.)
(A) 31
(B) 32
(C) 33
(D) 34
(E) 35
Let us consider some small values for $n$.

If $n=1$, Anne wins by taking the first toothpick.

If $n=2$, Anne must take 1 toothpick (since she cannot take 3 or 4 ), so Brenda takes 1 toothpick and wins.

If $n=3$ or $n=4$, Anne can win by taking all of the toothpicks.

When $n=5$, if Anne takes 3 toothpicks to start, then Brenda is left with 2 toothpicks, and so cannot win, since Anne could not win when starting with 2 toothpicks. So Anne wins.

When $n=6$, if Anne takes 4 toothpicks to start, then Anne will win in the same way as in the $n=5$ case. So Anne wins.

When $n=7$, if Anne takes 1, 3 or 4 toothpicks, then 6,4 or 3 toothpicks will be left for Brenda. But choosing from 6, 4 or 3 toothpicks is a winning position for the first person choosing (Brenda in this case). So $n=7$ is a losing position for Anne, since Brenda can use Anne's winning strategy in each case.

In general, Brenda will have a winning strategy for $n$ if $n-1, n-3$ and $n-4$ all give Anne a winning strategy (since if Anne starts with $n$ toothpicks an chooses 1, 3 or 4 , then Brenda will start choosing from $n-1, n-3$ or $n-4$ toothpicks).

Anne will have a winning strategy otherwise (since if Anne would lose starting with an initial position of $n-1, n-3$ or $n-4$, then Anne should win starting with an initial position of $n$ because she can leave Brenda with whichever of $n-1, n-3$ or $n-4$ is a losing position for the first person choosing).

We construct lists of the winning positions for each, adding a new $n$ to Brenda's list if $n-1$, $n-3$ and $n-4$ are all in Anne's list, and to Anne's list otherwise:

A's winning positions: $1,3,4,5,6,8,10,11,12,13,15,17,18,19,20,22,24,25$,

$26,27,29,31,32,33,34$

B's winning positions: 2, 7, 9, 14, 16, 21, 23, 28, 30, 35

So of the given possibilities, Brenda has a winning strategy when $n=35$.

ANSWER: (E)