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

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 Hugo van der Sanden Posts: 234 Registered: 12/8/04
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 N|A (N divides A) and
N|B, so let me assume N < M and A <= B for the rest.

It is clear that NxM tiles AxB if N|A and M|B, 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, 6|AB.

I think that for gcd(N,M) = 1, NxM tiles AxB precisely when NM|AB and
either P or NM|A 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 d|A, d|B, and (N/d)x(M/d)
tiles (A/d)x(B/d).

Hugo

Date Subject Author
12/3/00 George Gross
12/4/00 Hugo van der Sanden