If p and q are distinct primes, find the number of distinct divisors of pmqn.
Would the answer just be
(p^m-m^(m-1))(q^n-q^(n-1))?
Ok I think I know where I went wrong. I got the total number of relatively prime numbers to (p^m)(q^n)
(p^m)(q^n)-(p^m-m^(m-1))(q^n-q^(n-1))
Amount of numbers from 1 to (p^m)(q^n) subtract the amount of numbers that are relatively prime.
Ya I was never thinking about this question properly.
ok so the number of distinct divisors of p and q are 2 each if they are prime right? p and 1, and q and 1 respectively
So the number of distinct divisors of p^k is k+1?
2^3 has four distinct divisors 4 right? 1, 2, 2^2, 2^3
so (p^m)(q^n) has mn many divisors?