# Flipping Items Simultaneously

The applet displays several (N) triangles, all pointing upwards initially. On any move, you can turn over any M of them. (You do that by clicking on M triangles in turn.) The question is, Is it possible to have all N triangles point downwards?

What if applet does not run? |

The original puzzle [*Mathematical Circles*, p. 132] stated for

The solution begs for a generalization: what can be said about other pairs of M and N?

## References

- D. Fomin, S. Genkin, I. Itenberg,
*Mathematical Circles (Russian Experience)*, AMS, 1996 - C. W. Trigg,
*Mathematical Quickies*, Dover, 1985, #22.

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Quite obviously, if the problem is solvable for a given pair M, N then it is solvable for the pair qN, qM, with q a positive integer. For a long time I thought that the converse is also true, i.e. that the problems for

For M and N are mutually prime, the puzzle is solvable wherever M is odd, and unsolvable otherwise. Why?

It is easy to see that when N is odd and M is even, the puzzle is unsolvable. Indeed, assign number ±1 to each of the triangles depending whether it points up or down. Note the product Π of all the assigned numbers. When all triangles point upwards,

If M = N - 1, there are C(N, N-1) = N combinations of N-1 elements out of N. Each of N elements enters N-1 of the combinations. Carrying out the flips corresponding to all N combinations, will turn each of the N triangles N-1 times, which is an odd number, and therefore leave it in a position different from the one it was originally in, i.e., upside down.

It remains to be shown that when gcd(M, N) = 1 and M odd the puzzle is always solvable. One solution can be found at my blog.

It so happens that the case of an odd N is much easier than the case where N is even. For odd N, the problem can always be solved in 3 steps. When N is even, one needs at least 4 moves.

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