Quote:
Originally Posted by cdcox
You're doing it wrong.
Go through the sums, starting with zero.
0, none
1, 1+0, none
2, 1+1 and 2+0, bam 20
3, 1+2, 2+1, 3+0 bam 12, 21, and 30
4, 1+3, 3+1, 2+2, 4+0, bam 40
that is going to be quicker than looking at all 90. For instance 5 you can see in about an instant that 50 is the only one. Its reasonable to do the calculations in a minute or so if you have the concept down.
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In the end I think this is the only route to go.
Again, have to finish dinner then I'm thinking their may be a formulaic way to do this but not anything that would be any faster.
This is the next step btw... (10x+y) = p^n * q^m and (n+1)(m+1) = # of factors