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I'm talking about the winning and losing positions. I know that, if the xor of the heap sizes or the grundy numbers is 0, it's a losing position for the normal game play. I can also understand how we can only go to non-zero (winning) nim-sum positions from zero nim-sum; and, to zero (losing) nim-sum position from winning positions.

But how can someone think about it or arrive at this conclusion? It's kind of easy to prove when you know the losing and winning positions. But how do I determine it? How we would have thought about it in case we knew nothing about binary representation of numbers? (this game doesn't have any dependency on it)

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    Possible duplicate of [Why can a Nim sum be written as powers of 2?](http://math.stackexchange.com/questions/1178163/why-can-a-nim-sum-be-written-as-powers-of-2) – 2017-03-09
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    You may also be interested in http://mathoverflow.net/questions/209925/why-does-the-bitxor-function-appear-in-nim – 2017-03-09

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But how can someone think about it or arrive at this conclusion?

It has been concluded long time ago. A method to go from lose to win is found. Just search wikipedia or other way. Thinking about nim without binary brain and rather 10 fingers goes by splitting numbers in sums of powers of two. Yes, that's binary calculus.

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    There are lots of similar games where breaking things up into powers of $2$ is not the answer, and I think [ptntialunrlsd](http://math.stackexchange.com/users/239499/ptntialunrlsd) already understands why the winning strategy for nim works. – 2017-03-09