<s>[INST] 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!
 [/INST] Case 1: factors that are perfect squares
\begin{align}
A & =\{2^0,2^2,2^4,2^6,2^8,2^{10},2^{12},2^{14}\} \\
B & =\{3^0,3^2,3^4,3^6,3^8,3^{10}\} \\
C & =\{5^0,5^2,6^4,5^6\}
\end{align}
Choose 1 from each set. We have $8\times6\times4=192$ choices.
Case 2: factors that are perfect cubes
\begin{align}
A& =\{2^0,2^3,2^6,2^9,2^{12},2^{15}\} \\
B & =\{3^0,3^3,3^6,3^9\} \\
C & =\{5^0,5^3,5^6\}
\end{align}
Choose 1 from each set. We have $6\times4\times3=72$ choices.
Therefore, we have $72+192-3\times2\times2=252$ factors in total.
Note: subtract $3\times2\times2$ from the total because some sixth powers have been counted twice.
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