The Page used for that. A number like 1000 has prime factors 2^3 x 5^3. 
Take the factor 2 = 0 1 2 3 times = 4 choices
Take the factor 4 = 0 1 2 3 times = 4 choices
So all together there could be 4*4 = 16 distinct factors.
Thanks to Zaxo for making me google for that :)

In reply to Re: Re: Subset Sum Problem by OverlordQ
in thread Subset Sum Problem by beretboy

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