A positive integer, n, where the number of its decimal digits (base 10) equals the number of its distinct prime factors has an upper-bound (there is a maximum n).
Anyone knows the proof for this?
Thinking of establishing a relationship between number of digits and number of decimal places (and $10^x < n$) and proving n has an upper bound. I have tried linking the fact that the number of primes $\leq n$ is $\frac n{\ln n}$, but can't seem to find a good link.
Best Answer
For any $n$-digit number $x$ we have $x<10^n$. Let $p_i$ be the $i$-th prime. If $x$ has $n$ digits and $n$ distinct prime factors, we have that $$p_1p_2\ldots p_n\leq x< 10^n.$$
Thus, all it takes to finish the proof is to show that $$p_1p_2\ldots p_n \geq 10^n$$ for all but finitely many positive integers $n$. Can you do that?