Proof that a number cannot have more base b digits than its prime factors

let N represent a natural number with p_0,p_1...p_(k-1) representing the prime factors of N

let log(x) represent the logarithm base b of x

let D(x) represent the number of base b digits of x. D(x) = floor(log(x))+1

thus the proof can be restated as D(N)<=sum(i=0,k-1,D(p_i))

Suppose there existed an N such that D(N)>sum(i=0,k-1,D(p_i)).

Because log(N)=sum(i=0,k-1,log(p_i)), this inequality can be replaced with

floor(sum(i=0,k-1,log(p_i)))+1>sum(i=0,k-1,floor(log(p_i))+1)

because 1 is being added on the right hand side k times, it can be replaced with

floor(sum(i=0,k-1,log(p_i)))+1>sum(i=0,k-1,floor(log(p_i)))+k

rearranging gives

floor(sum(i=0,k-1,log(p_i)))-sum(i=0,k-1,floor(log(p_i)))>k-1

Because x-floor(x)<1, log(p_i)-floor(log(p_i))<1 for 0<=i<k. Thus

floor(sum(i=0,k-1,log(p_i)))-sum(i=0,k-1,floor(log(p_i)))<k

because the difference must be an integer, this can also be stated as

floor(sum(i=0,k-1,log(p_i)))-sum(i=0,k-1,floor(log(p_i)))<=k-1

However, in order for D(N)>sum(i=0,k-1,D(p_i)), it is mecessary that

floor(sum(i=0,k-1,log(p_i)))-sum(i=0,k-1,floor(log(p_i)))>k-1. Since

floor(sum(i=0,k-1,log(p_i)))-sum(i=0,k-1,floor(log(p_i)))<=k-1,

it cannot be that D(N)>sum(i=0,k-1,D(p_i)). Thus D(N)<=sum(i=0,k-1,D(p_i)). (QED)

It is important to note that this only holds for k>=1. For k=0, N must be equal to 1 (product of 0 primes), which indeed has more base b digits (1 digit) than its prime factors (0 digits).

For a number M with the same nonzero base b digits as its prime factors, this result means that the number of zero digits in M cannot exceed the number of zero digits in its prime factors. Since the nonzero digits are the same, M having more zero digits than its prime factors implies it has more digits, which is impossible.