As mentioned in another question, the number of factors of 2^(a1) * 3^(a2) * 5^(a3) * ... is (a1+1)(a2+1)(...
Now we just try out possibilities - note that to make the number of factors as big as possible, the more primes we have in the number the better.
We can't get 5 different primes as 2*3*5*7*11 is too big, but if we start with 2*3*5*7, that gives us 210 which has 16 factors. We can multiply that by a couple more 2s, which gives 840 = 2^3 * 3 * 5 * 7 which has 4*2*2*2 = 32 factors.
I don't think you can beat that.
(edit - confirmed, you *can't* beat that :) 512, mentioned below, only has 10 factors.)
2006-12-04 09:37:14
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answer #1
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answered by stephen m 4
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The logic above is sound, I would go with 840 with factors of
1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840
2006-12-04 17:37:39
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answer #2
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answered by Puzzling 7
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