finding the number of factors $2^{15}\times3^{10}\times5^6$ The number of factors of $2^{15}\times3^{10}\times5^6$ which are either perfect square or perfect cubes(or both)
I don't know how to start this even!
Plz solve this!

A factor can be described by a tuple of exponents
$f \in 
\{ 0, \dotsc, 15 \} 
\times 
\{ 0, \dotsc, 10 \} 
\times 
\{ 0, \dotsc, 6 \}
$.
As
$$
(2^{f_1} 3^{f_2} 5^{f_3})^{1/n} =
2^{f_1/n} \, 3^{f_2/n} \, 5^{f_3/n}
$$
we need $n$ to divide the exponents $f_i$.
The perfect squares ($n=2$) are combined from 
$
S = \{ 0, 2, 4, 6, 8, 10, 12, 14 \} \times
\{ 0, 2, 4, 6, 8, 10 \} \times
\{ 0, 2, 4, 6 \}
$, the perfect cubes ($n=3$) from
$
C = \{ 0, 3, 6, 9, 12, 15 \} \times
\{ 0, 3, 6, 9 \} \times
\{ 0, 3, 6 \}
$.
Then 
$S \cap C = 
\{ 0, 6, 12 \} \times \{ 0, 6 \} \times \{ 0, 6 \}
$.
This gives
$$
\lvert S \rvert + \lvert C \rvert - \lvert S \cap C \rvert
= 8 \cdot 6 \cdot 4 + 6 \cdot 4 \cdot 3 - 3 \cdot 2 \cdot 2
= 252
$$
