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Bowl of sweets

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A bowl holds $40$ sweets. $2$ players take turns eating $1$ , $2$ or $3$ sweets from the bowl. The player who eats the last sweet loses. You move first and both players play optimally. How many sweets must you eat on your first move to guarantee a win?

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Hint

Decide how many sweets you want to leave your opponent at the very end, then step back.

Solution

If your opponent faces $1$ sweet, they must eat it and lose, so you want to leave $1$ .
Whenever the opponent eats $a$ sweets, you eat $4 - a$ , so each full round removes $4$ sweets.
So the positions to leave are $1$ , $5$ , $9$ and so on, all of the form $4 k + 1$ , and the largest below $40$ is $37$ .
You therefore eat $40 - 37 = 3$ sweets first, so the answer is $3$ .

Answer: $3$

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