# Number of divisors of $20^{20}$ with exactly $20$ divisors [closed]

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

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

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• Hint : The number is equal to the number of the divisors of $20$ (Try to find out why!) – Peter Apr 13 at 11:48
• You might say that's a "perfect" hint. – Oscar Lanzi Apr 13 at 11:52

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. $$3\times2=6$$.