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As with pretty much all the nim variants, this one can be solved by starting from the end and working backwards. With the original total number of stones being an odd number (15, as given in the title) the players will have the same parity whenever there's an odd number of pebbles left, so it's easy to work out the best strategies: they are the ones that put ...


The strategy is to do a move that In particular, for $15$ pebbles, your first move would be The reason this works is more interesting than with other single-pile nim variants.

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