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Old 12-01-2007, 11:04 AM
bigpooch bigpooch is offline
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Join Date: Sep 2003
Location: Hong Kong
Posts: 1,330
Default Re: Hyperfactorial Problem

Okay, the problem is equivalent to showing:
<font color="white">
C(x+b,b)C(x+b+1,b)...C(x+b+m,b) is divisible by
C(b,b)C(b+1,b)...C(b+m,b) for any integer x&gt;=0.

This can be shown, but is almost certainly not the simplest solution.

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