## Nim

### Problem 301

Published on Saturday, 11th September 2010, 04:00 pm; Solved by 3115*Nim* is a game played with heaps of stones, where two players take it in turn to remove any number of stones from any heap until no stones remain.

We'll consider the three-heap normal-play version of Nim, which works as follows:

- At the start of the game there are three heaps of stones.

- On his turn the player removes any positive number of stones from any single heap.

- The first player unable to move (because no stones remain) loses.

If (`n`_{1},`n`_{2},`n`_{3}) indicates a Nim position consisting of heaps of size `n`_{1}, `n`_{2} and `n`_{3} then there is a simple function `X`(`n`_{1},`n`_{2},`n`_{3}) — that you may look up or attempt to deduce for yourself — that returns:

- zero if, with perfect strategy, the player about to move will eventually lose; or
- non-zero if, with perfect strategy, the player about to move will eventually win.

For example `X`(1,2,3) = 0 because, no matter what the current player does, his opponent can respond with a move that leaves two heaps of equal size, at which point every move by the current player can be mirrored by his opponent until no stones remain; so the current player loses. To illustrate:

- current player moves to (1,2,1)

- opponent moves to (1,0,1)

- current player moves to (0,0,1)

- opponent moves to (0,0,0), and so wins.

For how many positive integers `n` ≤ 2^{30} does `X`(`n`,2`n`,3`n`) = 0 ?