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Topic:
Tiling with M by N rectangles
Replies:
1
Last Post:
Dec 4, 2000 10:47 PM




Re: Tiling with M by N rectangles
Posted:
Dec 4, 2000 10:47 PM


George Gross wrote: > > Given N and M which A by B rectangles can be tiled by a finite number > of N by M rectangles ? > (with all A, B, M, N integeres ) .
When N=M, it is clear that NxM tiles AxB only if NA (N divides A) and NB, so let me assume N < M and A <= B for the rest.
It is clear that NxM tiles AxB if NA and MB, or vice versa. Let me call this condition P.
P is clearly also a necessary condition if A < M, or if N = 1 and M is prime. I suspect that the same is true for N = 1, M composite. However it is clear that P may not be necessary for larger N: for example 2x3 tiles all AxB for 1 < A <= B, 6AB.
I think that for gcd(N,M) = 1, NxM tiles AxB precisely when NMAB and either P or NMA and xN+yM=B (for some order of A, B) in nonnegative integers x, y. This is at least a sufficient condition, and I believe it is necessary.
When gcd(N,M) = d, NxM tiles AxB precisely if dA, dB, and (N/d)x(M/d) tiles (A/d)x(B/d).
Hugo



