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The number of factors of $2^{15} ×3^{10}×5^6$ which are either perfect squares or perfect cubes (or both) is?

There would more than $100$ such factors. What should be my approach? I don't want the answer but a hint or approach.

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Hint: Write every factor as $d =2^a3^b5^c$. Now, what does it mean for $d$ to be a perfect square, in terms of $a,b,c$? What restrictions do we have on $a,b,c$ for $d$ to be a factor?

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    Is the answer 256 or 258? – 2017-01-17
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    Answer is 252, inclusion exclusion works. – 2017-01-17
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HINT:

Suppose that $n = a^x\times b^y\times c^z$

therefore $n^2 = a^{2x}\times b^{2y} \times c^{2z}$ and $n^3 = a^{3x}\times b^{3y} \times c^{3z}$ and $n^2$ is a perfect square, $n^3$ is a perfect cube.

It should be easy to see that if a number is a perfect square, then all the exponents of its prime factors are even. What can you say about the exponents of a perfect cube? What if the number is both a perfect square and a perfect cube?

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    Is the answer 256 or 258? – 2017-01-17
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    @Resorcinol I counted 252 but I am quite sleepy. The number of perfect squares is $8*6*4$ (the exponents can be 0 as well); perfect cubes is $3*4*6$ (again exponents can be 0) and then subtract the ones that are both things simultaneously, because we counted them twice: 2*2*3 – 2017-01-17