How many positive integers are divisors of at least one of $24^5, 20^6$, and $45^7$?

Question

I have a vague idea of prime factorizing 24, 20, and 45, then distributing the exponent, find how many numbers are divisible by them (factors), and then subtract the overcounts. How do I do this?

Answer 1

$24=2^3\times 3\\ \implies 24^5=2^{15}\times 3^5$

Positive divisors of $24^5=16 \times 6=96$

$20=2^2\times 5\\ \implies 20^6=2^{12}\times 5^6$

Positive divisors of $20^6=13 \times 7=91$

$45=3^2\times 5\\ \implies 45^7=3^{14}\times 5^7$

Positive divisors of $45^7=15 \times 8=120$

Positive divisors of both $24^5$ and $20^6=13$ (as $2^{12}$ is the only common factor)

Positive divisors of both $24^5$ and $45^7=6$

Positive divisors of both $20^6$ and $45^7=7$

Positive divisors of $24^5,20^6,45^7=1$

Required answer

$96+91+120-(13 + 6 + 7) + (1)$

Answer 2

You want to apply the Principle of Inclusion-Exclusion. In this context, it gives rise to the equation:

Number that divide any of the three = [number that divide at least one] - [number that divide at least two] + [number that divide at least three]

In order to count the number that divide two of the numbers, recall that $$a|b\text{ and }a|c\iff a|\gcd(b,c)$$

I'm getting $$96+91+120-(13+6+7)+1=282$$ as the answer.