Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
There are few primes less than 10... – user641 Nov 12 '12 at 5:24
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.