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Show that there exists finitely many numbers $n$ satisfying:

"the number of digits" $=$ total number of its prime divisor"

For instance, $18 = 3^2*2$ satisfies, while $27 = 3^3$ does not.

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There are few primes less than 10... –  user641 Nov 12 '12 at 5:24
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up vote 6 down vote accepted

The main idea is that the primorial $n\#$ grows much faster than the powers of $10$. If you have a number $x$ which is a product of $n$ distinct primes, then it is at least as large as the product of the first $n$ primes.

If a number is a product of $11$ or more distinct prime factors, then it is at least $12$ digits long because $31\# = 200560490130$ ($31$ is the $11$th prime) is twelve digits long. Therefore any number which satisfies your criteria must be less than $10$ digits long and the set of such numbers is finite.

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