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.
If we have a number Y = p1^n1 * p2^n2 * p3^n3 * ... it will have (n1 + 1) * (n2 + 1) * (n3 + 1) * ... divisors
The largest number of distinct prime divisors is 4, with powers 1,1,1 and 2. Table of possible powers of prime factors, together with the number of possible ways to construct these from elements of X:
* 1,1,1,2 --> 25*24*23*22/6 = 50600
* 1,2,3 --> 25*24*23 = 13800
* 1,1,5 --> 25*24*23/2 = 6900
* 3,5 --> 25*24 = 600
* 2,7 --> 25*24 = 600
* 1,11 --> 25*24 = 600
* 23 --> 25
If all these numbers are equally probable, we just need the fraction (13800/6900)/S where S is the sum of the last column of this table. This gives:
28.3 %
Solution
