Question

The smallest value of N (greater than 5) that ensures a win for B is:

• A. 7
• B. 6
• C. 10
• D. 8
• A. 46
• B. 47
• C. 48
• D. 49

Answer 1

The Correct Option is B

We are given a two-player game with $N$ matchsticks where each player can remove either 1 or 2 matchsticks per turn. The player who takes the last stick wins. Player A moves first, Player B second. We need the smallest $N>5$ such that B has a guaranteed winning strategy.
The concept here is based on identifying "winning" and "losing" positions. A winning position is one where the player whose turn it is can force a win regardless of the opponent’s play. A losing position is one where the player’s move will inevitably allow the opponent to force a win.
We start from small values and build upward:
- $N=1$: A can take 1 stick and win instantly — winning position for current player.
- $N=2$: A can take 2 sticks and win instantly — winning position.
- $N=3$: Whatever A does (remove 1 or 2), B will be left with 2 or 1 sticks and win — losing position for current player.
- $N=4$: A can remove 1 stick (leaving 3, which is losing for B) — winning position.
- $N=5$: A can remove 2 sticks (leaving 3) — winning position.
- $N=6$: If A starts, any move leaves B with $N=4$ or $N=5$ — both winning positions for B. Hence, $N=6$ is losing for the starter (A) and winning for B.
This pattern shows that positions where $N$ is a multiple of 3 are losing positions for the current player. Since we want smallest $N>5$ that is multiple of 3, $N=6$ fits.
Thus, answer is (B).

Answer 2

From Q59, we already established that the losing positions for the current player occur when $N \equiv 0 \ (\text{mod } 3)$. In such cases, the player about to move is forced into a losing strategy because whatever move they make, they leave a winning position for the opponent.
For B to win, A must start in such a losing position. Therefore, $N$ must be a multiple of 3 when the game begins.
We list the multiples of 3 below 50: $3, 6, 9, 12, , 48$. The largest is 48.
When $N=48$, A starts at a losing position and any move A makes will hand B a winning position. Hence, B can win by always returning the pile to a multiple of 3 after each of A’s moves.
Thus, answer is (C) 48.