I tackled the problem in the following way but i'm not sure if i'm correct.

I need the count of the numbers that have in their prime factorization only primes p such that $p \lt 20$ and those numbers can't be more than once in the prime factorization (right?)

So, the amount of numbers that can be expressed this way are all the subsets of the set $\{2,3,5,7,11,13,17,19\} = 2^8$.

Correct me if I'm wrong.

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    $\begingroup$ If the question was asking about natural numbers, you would be correct. $\endgroup$ – Don Thousand Oct 6 '19 at 20:20

Your method is perfectly correct, but if it is actually about integers then you need to include negatives as well to have twice as many.

I suspect this is not the case however since then if you said "primes $p<20$" you'd probably have to include negative primes as well, in which case there would be infinitely many.


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