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Consider the following game played on triples of integers modulo $3n$. Player $B$ writes down the triple $(a,a+n,a+2n)$. Starting with player $A$, they then take it in turns to add $1$ to one of the components of the triple, so that all three components are distinct. Player $A$ wins if at any stage, two of the components are in the set $\{0,n,2n\}$. For which $n$ does player $A$ have a winning strategy?

For $n=1,2,3$, player $A$ wins. I think you can show that player 1 can always force a series of moves to go from a triple of the form $(a,b,b+n)$ to $(c,b+1,b+n+1)$, which should imply the result inductively, but the cases are annoying me.

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    Why does it work if $n=2$, say? Suppose $B$ starts with three odd numbers. Then $A$ needs to produce two evens. But whichever one $A$ makes even, $B$ can promptly make it odd again...so I'm not seeing how $A$ wins. – 2017-01-21
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    Oh, I suppose the "three components are distinct" rule messes up my simple strategy. – 2017-01-21

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