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 to point downwards?

The original puzzle [Mathematical Circles, p. 132] stated for N = 7 and M = 4 has negative solution. It is easily seen that, in this case, the number of inverted triangles is always even and, therefore, can't be 7.

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

References

1. D. Fomin, S. Genkin, I. Itenberg, Mathematical Circles (Russian Experience), AMS, 1996

Copyright © 1996-2009 Alexander Bogomolny

Quite obviously, the problem for a given pair M, N, is solvable iff the problem is solvable for the (reduced) pair M/gcd(M, N), N/gcd(M, N). Assume therefore that M and N are mutually prime. Then, the puzzle is solvable wherever M is odd, and unsolvable otherwise. Why?

Copyright © 1996-2009 Alexander Bogomolny