What is the maximum number of rectangles into which a 17x24-rectangle can be partitioned, when all rectangles must be incongruent and have integer side lengths?
(In reply to
Ceiling by Eric)
The ceiling is the solution. It is easy to tescalate 35 incongruent inegral-side-length rectangles with areas less than or equal to 20 which sum to 408 to fit inside a 17x24 rectangle.
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Posted by Eric
on 2006-12-20 15:54:43 |
re: Ceiling
