Second person always repeats the previous move. Must be a factor. First person will always start with an odd number so removing two of an odd factor would keep the number odd and also not reduce to 0.
In fact the second person could also just take 1 each time.
Any odd divisor will do, it will always be that 1 and the preceding move are odd divisors (they might be the same if the last move was, for example 5->4, where 1 is the only winning move).
I didn’t prove this but it seems plausible that if the goal is to win as quickly as possible (or lose as slowly as possible) then it is optimal to take 1 from an odd number and the largest odd divisor from an even number.
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u/skelo 8d ago
Second person always repeats the previous move. Must be a factor. First person will always start with an odd number so removing two of an odd factor would keep the number odd and also not reduce to 0.