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?

The ceiling on the number of incongruent integral length rectangles is found by choosing rectangles with minimal area whose areas sum to 408 (17*24). At most 35 incongruent rectangles achieve this goal. Rectangles of height 1 and width 1 to 20. Rectangles of height 2 and width 2 to 10. Rectangles of height 3 and width 3 to 6. And rectangles of height 4 and width 4 and 5. It remains to be shown that these such rectangles can be tescalated to form a 17x24 rectangle...

Posted by Eric on 2006-12-20 15:34:14