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

$24=2^3\times 3\\
\implies 24^5=2^{15}\times 3^5$
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
$96+91+120-(13 + 6 + 7) + (1)$
