10^40 has 41*41 integers that can divide it, because the integers can be 2^(0 to 40)*5^(0 to 40). Similarly, 20^30 has 61*31 integers that can divide it.
To get rid of the ones counted twice, I find the GFC of the two, which is 2^40*5^30. The number of integers that divide this is 41*31.
So the total number of integers that divide at least one is 41*41+61*31-41*31=1681+620=2301
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Posted by Tristan
on 2004-02-29 12:18:43 |
solution
