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.
(In reply to
solution by Daniel)
Hi Daniel,
I fear you made some mistakes is your calculation.
You say that "2: is p1^2*p2^12 or p1^3*p2^8 or p1^4*p2^6 and this gives another 900 numbers". This is, however, not true.
For the first part, let's say the one with power 2, there are 25 possibilities. For the second there are 24. You seem to have divided by 2, but switching the two numbers would give a different result. Something similar happens in (3) and (4).
re: solution
