Show that player 1 can always win a Nim game in which the number of heaps with an odd number of coins is odd. This question is provided in Richard A. Brauldi's book on Introductory Combinatorics. I attempted the problem as follows:
A winning strategy provided in the book was to arrange the given heaps into subheaps of powers of two. Then, it is obvious enough to see that if there are an even number of like subheaps (for all like subheaps), then the game is balanced (first player loses); and if there is an odd number of like subheaps (for at least one like subheap), then the game is unbalanced (first player wins). Then the strategy is to balance the game every time the opponent unbalances it. If both players are intelligent, the game is just a matter of whether the game is initially balanced or unbalanced.
In the problem given, it is the case that each heap has an odd number of coins. It can be shown that every heap, when organized into subheaps of powers of two, will contain the subheap $2^0$. This is because, when forming subheaps, we will first form the greatest power of two first, then from what remains we form the next greatest power of two, and so on. Hence we will take the greatest power of two from the initial odd quantity. An odd minus an even is always an odd number. From this remaining odd number we will subtract the next greatest power of two, and again resulting in an odd number. So for $2^0$ to not be the last part of the subheap, it must be true that some greater power of two must be. Yet, all other powers of two greater than $2^0$ are even. And, as we've seen earlier, since we started with an odd number, and continuously subtract (even) powers of two, we will always end with an odd number. This shows that all heaps must have the subheap $2^0$. Since there are an odd number of heap, there are also an odd number of $2^0$ subheaps; hence, the game starts out unbalanced, and the first player is always the one who will win.
Is this proof valid? Does this mathematically convince the reader? I ask because I am still yet an amateur in combinatorial reasoning; hence I require constant affirmation from those more professional. Thank you in advance.
This is false. There could several heaps with an even number of coins.
Note: Re the condition of "number of heaps with an odd number of coins is odd",
In the comments, OP stated that their interpretation is "That literally means that we have no even number of heaps, and no even number of coins in any of the heaps". To them, $\{ 1, 2, 4,6 \}$ does not satisfy the condition.
I unsuccessfully tried explaining that it means "That literally means we have no information about heaps with an even number of coins". To me $\{ 1, 2, 4, 6 \}$ satisfies the condition.