Relationship between the number of digits of a natural number and that of its factors

Question

I come from primarily a programming background, relatively weak theoretical math background, hoping some of you mathematicians could help me out here.

I was wondering (from a purely recreational perspective) what the relationship between the number of digits of a natural number N, to the sum of the number of digits of its factors.

For example, if N has K factors (K_1, K_2, ....K_n), it appears that the number of digits required to represent the factors individually is greater than the number of digits required to represent N, but i am interested in the general relationship (i.e. is it logarithmic, linear, etc)

Does the relationship differ if we are only talking about prime factors?

Would appreciate any help analyzing this :)

Thanks in advance!

(P.S. sorta guessing this is related to number theory in the tags here, correct me if i am wrong P.P.S I assume this is the same as the relationship between the number of bits - please correct me if this is a special case)

Answer 1

Consideration of examples like $1\times1=1$ and $9\times9=81$ show that the product of a $a$-digit and a $b$-digit number can have as few as $a+b-1$ digits and no more than $a+b$. So if $N=K_1\times K_2\times\cdots\times K_n$, where $K_i$ has $d_i$ digits, the number of digits in $N$ can be no bigger than $D=d_1+d_2+\cdots+d_n$ and no smaller than $D-(n-1)$.

Answer 2

Consider representation in base $b$ the number of digits required is the ceiling of $\log_b(N)$ and due to the properties of logarithms, if $N = v\cdot u$ then the bits required is $\lceil\log_b(v)\rceil + \lceil\log_b(u)\rceil$. Due to the ceiling operation this is at best an equality and in some cases will increase the number of digits required to represent the factors individually. For instance, representing 999 in binary requires 10 bits, but to represent its factors individually requires 12 bits. In the first million numbers, 31 of them will use 5 more bits.

Not a complete answer, unfortunately, because I don't know how to determine the growth of the number of extra bits.