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.