Amar and Bimal play a game with a single pile of N identical coins placed on a table. They take turns, with Amar always moving first. On each turn, a player must remove exactly 1, 3 or 4 coins from the pile (a player cannot remove more coins than remain). In the standard version of the game, the player who removes the last coin wins.
Both players always play optimally: if a player can force a win from a position, he will do so, whatever the opponent does. A position (number of coins left, with a given player to move) is called a *losing position* for the player to move if the opponent can force a win from it.
Some questions describe variations of these rules; each variation applies only to the question in which it is described.
Both players always play optimally: if a player can force a win from a position, he will do so, whatever the opponent does. A position (number of coins left, with a given player to move) is called a *losing position* for the player to move if the opponent can force a win from it.
Some questions describe variations of these rules; each variation applies only to the question in which it is described.
Question 1
Single correct option
If N = 20, how many coins should Amar remove on his first move to be sure of winning?
- A. 1
- B. 3
- C. 4
- D. Amar cannot force a win
Show answer and solution
Answer: (C) 4
Work backwards. A position is losing (L) if every move leads to a winning (W) position for the opponent.
| Coins left (n) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|---|
| Player to move | L | W | L | W | W | W | W | L | W | L |
The pattern repeats with period 7: losing positions are n ≡ 0 or 2 (mod 7).
20 ≡ 6 (mod 7) is winning. Amar's options: remove 1 → 19 (≡ 5, W for Bimal), remove 3 → 17 (≡ 3, W for Bimal), remove 4 → 16 (≡ 2, L for Bimal).
So Amar must remove 4 coins.
| Coins left (n) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|---|
| Player to move | L | W | L | W | W | W | W | L | W | L |
The pattern repeats with period 7: losing positions are n ≡ 0 or 2 (mod 7).
20 ≡ 6 (mod 7) is winning. Amar's options: remove 1 → 19 (≡ 5, W for Bimal), remove 3 → 17 (≡ 3, W for Bimal), remove 4 → 16 (≡ 2, L for Bimal).
So Amar must remove 4 coins.
Question 2
Numerical answer
For how many values of N from 1 to 50 (both inclusive) can Bimal force a win?
Show answer and solution
Answer: 14
Bimal (second player) wins exactly when N is a losing position for Amar, i.e. N ≡ 0 or 2 (mod 7).
Why period 7? From n ≡ 0 or 2, every move (−1, −3, −4) lands on residues 6, 4, 3 or 1, 6, 5 — all winning; from every other residue some move reaches 0 or 2.
Values: 2, 7, 9, 14, 16, 21, 23, 28, 30, 35, 37, 42, 44, 49 → 14 values.
Why period 7? From n ≡ 0 or 2, every move (−1, −3, −4) lands on residues 6, 4, 3 or 1, 6, 5 — all winning; from every other residue some move reaches 0 or 2.
Values: 2, 7, 9, 14, 16, 21, 23, 28, 30, 35, 37, 42, 44, 49 → 14 values.
Question 3
Single correct option
Consider a variation in which the player who removes the last coin loses. For which of the following values of N will Bimal win?
- A. 21
- B. 22
- C. 23
- D. 25
Show answer and solution
Answer: (B) 22
Now a player facing 0 coins has won (the opponent took the last coin). Recompute: n = 1 is losing (you must take the last coin); n = 2: take 1 → opponent faces 1 → W; n = 3: moves to 2 or 0, both good for the opponent → L; n = 4 → take 1 to leave 3 → W; 5, 6, 7 → W; n = 8: moves to 7, 5, 4, all W → L; n = 9 → W; n = 10 → L.
Losing positions: n ≡ 1 or 3 (mod 7).
- 21 ≡ 0 → W for Amar
- 22 ≡ 1 → L for Amar, so Bimal wins
- 23 ≡ 2 → W
- 25 ≡ 4 → W
Losing positions: n ≡ 1 or 3 (mod 7).
- 21 ≡ 0 → W for Amar
- 22 ≡ 1 → L for Amar, so Bimal wins
- 23 ≡ 2 → W
- 25 ≡ 4 → W
Question 4
Single correct option
Consider a variation in which each player may remove exactly 1, 2 or 4 coins per turn (the player removing the last coin wins). For which of the following values of N will Bimal win?
- A. 20
- B. 25
- C. 27
- D. 29
Show answer and solution
Answer: (C) 27
With moves {1, 2, 4}: n = 0 is L; 1, 2 are W; 3 is L (moves to 2 or 1, both W); 4 → take 4 → W; 5 → take 2 to leave 3 → W; 6: moves to 5, 4, 2 → all W → L.
Since 1, 2 and 4 are all non-multiples of 3, from a multiple of 3 every move leaves a non-multiple, and from a non-multiple one can always move (by 1 or 2) to a multiple. So losing positions are exactly the multiples of 3.
Among the options, only 27 is a multiple of 3, so Bimal wins for N = 27.
Since 1, 2 and 4 are all non-multiples of 3, from a multiple of 3 every move leaves a non-multiple, and from a non-multiple one can always move (by 1 or 2) to a multiple. So losing positions are exactly the multiples of 3.
Among the options, only 27 is a multiple of 3, so Bimal wins for N = 27.
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