The same number of prime factors and decimal digits. 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.
 A: 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?
A: How about a different approach?
To make it easier, I'm going to define a number as wonderful iff the number of its decimal digits (base 10) equals the number of its distinct prime factors.
Then,


*

*Let x be some wonderful number with n digits.

*We know that x is wonderful, so it has n distinct prime factors.

*To maximize the number of distinct prime factors in x, we should have x be the multiplication of the smallest prime factors, in order of increasing numbers (i.e. x = 2*3*5*7*11 ... n primes)

*However, primes after 7 have at least 2 digits, so multiplying them into x will cause x to increase by at least 1 digit in length.

*This can keep going until after 100, where primes after 100 will cause x to increase by at least 2 digits in length, but only add 1 to the amount of distinct prime factorization of x.

*Clearly, this is unsustainable, and eventually the wonderful number x cannot add another unique prime factorization as it would cause x to gain more digits in length than the single distinct prime factorization would add to the number of distinct prime factors of x.

*If x is multiplied by a non-distinct prime factor, then eventually it will also increase in length, due to the nature of multiplication, but will not gain an extra distinct prime factorization, so x will also no longer be wonderful.

*At some point, x cannot be increased any further and still stay wonderful.

*Thus, wonderful numbers are upper bounded.

