I believe it is $2^4*5^3=2000$.

Somewhat related to this question (but not really) is this year's USAMTS R3 P3

Ha_ha_ha wrote: I believe it is $2^4*5^3=2000$.

There are smaller values. The smallest is Click to reveal hidden text

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ah whoops i miscalculated it should be $2^4*5*3$ thank you for problem

DylanN wrote: Find the smallest positive multiple of $20$ with exactly $20$ positive divisors. Headsolve of 15 seconds. So basicaly note that $20=2^2 \cdot 5$ so to make some $a_j+1$ multiple of 5 where the number we are looking for is $N=\prod_{i=1}^{k} p_i^{a_i}$ we see that the way to make it without making $N$ grow to much is making $2^4 \mid N$ becuase we are had $2^2 \mid N$ so the $2$ was basicaly made for that, now just $2^4 \cdot 5=80$ has 10 divisors, multiplying by the second smallest prime (i,e. 3) we have that $N=240=2^4 \cdot 3 \cdot 5$ has $(4+1)(1+1)(1+1)=5 \cdot 2 \cdot 2=20$ divisors, so the smallest number with such conditions is $240$ and we are done