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Find the number of positive integers $n=2^a3^b\, \, (a,b\geq 0)$ such that $n^6$ doesn't divide $6^n$.

I have not encountered such types of question before and so I donot know how to even approach this problem. Can this question be solved using basic combinatorics and number theory? Thanks.

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    Write 6 as 2x3. 6^n is 2^n * 3^n. Proceed with this. – 2017-02-21
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    Do $a$ and $b$ have to be integers? Or are they any positive real numbers? – 2017-02-21
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    @TimThayer $a,b$ are integers too, otherwise $n$ could assume any prime number, leading to an infinite number of $n$ with $n^6\nmid6^n$ - and that would be boring. – 2017-02-21

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Let $n=2^a\cdot 3^b$. Then $$n^6 \not|\ \ 6^n\quad\Longleftrightarrow\quad 2^{6a}\cdot 3^{6b}\ \not|\ \ 2^n\cdot 3^n\ ,$$ and this is equivalent with $$6a>n\quad \vee\quad 6b>n\ ,$$ or $$6a>2^a\cdot 3^b\quad \vee\quad 6b>2^a\cdot 3^b\ .$$ The first of these conditions is fulfilled iff $a\in\{1,2,3,4\}$ and $b=0$, and the second is fulfilled iff $b\in\{1,2\}$ and $a=0$. It follows that there are $6$ numbers $n$ with the desired property, namely$$2, 3,4,8,9,16\ .$$