Let X be set of primes from 1 to 100, i.e., X: {2,3,5,7,....,97} and Y be set of numbers whose every prime divisor belongs to set X. A number from Y has exactly 24 positive divisors, find the probability it has exactly 3 distinct prime divisors.

for those who are not aware of the theorem to find the number of positive divisors of a number simply take the power of each of its prime factors, add 1 to them, then multiply them togeather.  For example if you used this on 12 you would get 12=2^2*3^1 so we have (2+1)*(1+1)=3*2=6 and indeed there are six divisors of 12 namely 1,2,3,4,6,12.  So using this we can find the number of ways for this number to have 1,2,3 and 4 distinct prime divisors.

1: is p^23 for some prime p, and there are 25 ways of picking this p from the primes less than 100.

2: is p1^1*p2^11 or p1^2*p2^7 or p1^3*p2^5 and this gives another 1800

3: is p1^1*p2^1*p3^5 or p1^1*p2^2*p3^3 and this gives another 20700

4: is p1^1*p2^1*p3^1*p4^2 and this gives another 50600 numbers

this gives a total of 73125  possibilities for a number that has 24 divisors and is in Y.  20700 of these have exactly 3 distinct prime divisors thus the probability is 20700/73125=92/325 or approxmiately 28.3%

Edited on May 18, 2009, 12:07 pm

Posted by Daniel on 2009-05-18 11:56:47