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How many positive integers $x$ with $x\mid 20^{20}$ have exactly $20$ divisors ?

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closed as off-topic by Gregory J. Puleo, Eevee Trainer, Alexander Gruber Apr 29 at 23:47

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    $\begingroup$ Hint : The number is equal to the number of the divisors of $20$ (Try to find out why!) $\endgroup$ – Peter Apr 13 at 11:48
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    $\begingroup$ You might say that's a "perfect" hint. $\endgroup$ – Oscar Lanzi Apr 13 at 11:52
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Since $20^{20}=2^{40}5^{20}$, the general form of such a factor is $2^a5^b$ for non-negative integers $a,\,b$ with $(a+1)(b+1)=20$. Note in particular the ordered pair $(a,\,b)$ is what matters, not the unordered pair $\{a,\,b\}$. There are exactly as many of these as there are factors of $20=2^25$, i.e. $3\times2=6$.

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