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Whoever empties loses

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There are $3$ boxes holding $1$ , $4$ and $6$ marbles. $2$ players alternate; a move takes any positive number of marbles from exactly one box. The player who takes the very last marble loses. You move first and both players play optimally. Exactly one first move guarantees you a win. How many marbles does that move take?

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Hint

While some box still holds $2$ or more marbles, play as in the ordinary game where taking the last marble wins.

Solution

In this version the rule is the same as in the ordinary game as long as some box has at least $2$ marbles: move so that the XOR of the sizes is $0$ .
Only at the end, when your move would leave every box with at most $1$ marble, do you leave an odd number of single marbles instead.
Here $1$ XOR $4$ is $5$ , and $5$ XOR $6$ is $3$ , so the position is won for you.
Only the box of $6$ can be lowered to $6$ XOR $3$ , which is $5$ ; this leaves $1$ , $4$ and $5$ , with XOR $0$ and a box still larger than $1$ .
So the winning move takes $6 - 5 = 1$ marble, and the answer is $1$ .

Answer: $1$

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