With a 2-3-5-7 configuration, it would seem you have to go first to win. You start the game unbalanced--three groups of 2, two 4's and four 1's. When you go first, you are able to get the Nim sum to zero. After that, assuming the other person knows what she's doing, I don't see how you could win. Every move after your move, the opponent could restore the sum to zero.

A separate Nim question: At the end you are down to 1-1-2. At this point, to get the Nim sum to zero, one must eliminate the row of two it would seem. But that is of course a losing strategy if in order to win, you must not be the one to remove the last coin. Yet, if you employ the winning strategy of leaving the opponent 1-1-1, the Nim sum is no longer zero, which is supposedly what is needed to ensure victory. Any explanation? This is not a logic problem I'm putting out there--I don't know the answer.

## I don't see it

With a 2-3-5-7 configuration, it would seem you have to go first to win. You start the game unbalanced--three groups of 2, two 4's and four 1's. When you go first, you are able to get the Nim sum to zero. After that, assuming the other person knows what she's doing, I don't see how you could win. Every move after your move, the opponent could restore the sum to zero.

A separate Nim question: At the end you are down to 1-1-2. At this point, to get the Nim sum to zero, one must eliminate the row of two it would seem. But that is of course a losing strategy if in order to win, you must not be the one to remove the last coin. Yet, if you employ the winning strategy of leaving the opponent 1-1-1, the Nim sum is no longer zero, which is supposedly what is needed to ensure victory. Any explanation? This is not a logic problem I'm putting out there--I don't know the answer.