Number of divisors of huge numbers

Question

How many positive integers n are there such that n is a divisor of at least one of the numbers $10^{40}$,$20^{30}$?

I'm having problems with this question. I know how to find the number of integers in a set which are divisible by a particular number, but not sure in this case.

EDIT:

The answer given is 2301.

Answer 1

Well if you factorise the numbers into prime factors you get $$10^{40}=2^{40}\times5^{40}$$ and $$20^{30}=5^{30}\times2^{60}$$ Therefore the divisors are the combinations of $2$ and $5$. There are a lot of different combinations because you have 40 2's and 40 5's to consider and then 30 5's and 60 2's to consider.

Answer 2

$10^{40}=2^{40}\cdot5^{40}$. Just compute the number of different combinations of $2$ and $5$ that are possible.

Answer 3

Well, you can use the Principle of Inclusion/Esclusion: Take the number of divisors of $10^{40}$ (41*41), sum to the number of divisors of $20^{30}$ (31*61), and then subtract the numbers that divide both $10^{40}$ and $20^{30}$, that are the ones that divide their maximum common divisor $5^{30}*2^{40}$ that are 31*41.

$$ 41^2+31*61-31*41=2301 $$