Find the number of positive integers that divide (10)^999 but not (10)^998.

For a number to evenly divide the one but not by the other, it must be divisible by either 5^999 or by 2 ^ 999.  (Note, this includes the possibility it is divisible by both.)

So the numbers are of the form 5^999 * 2^n, where n is from zero to 999 inclusive, or 2^999 * 5 ^n, where n is from zero to 999 inclusive.  Each of these sets has 1000 members, but one of the members, 10^999 itself, is a member of both sets and has been counted twice, so the number sought is 1000+1000-1 = 1999.

Edited on May 18, 2006, 9:15 am

Posted by Charlie on 2006-05-18 09:11:23