proof that number of sum-product numbers in any base is finite


Let b be the base of numeration.

Suppose that an integer n has m digits when expressed in base b (not counting leading zeros, of course). Then nbm-1.

Since each digit is at most b-1, we have that the sum of the digits is at most m(b-1) and the product is at most (b-1)m, hence the sum of the digits of n times the product of the digits of n is at most m(b-1)m+1.

If n is a sum-product number, then n equals the sum of its digits times the product of its digits. In light of the inequalitiesMathworldPlanetmath of the last two paragraphs, this implies that m(b-1)m+1nbm-1, so m(b-1)m+1bm-1. Dividing both sides, we obtain (b-1)2m(b/(b-1))m-1. By the growth of exponential function, there can only be a finite number of values of m for which this is true. Hence, there is a finite limit to the number of digits of n, so there can only be a finite number of sum-product numbers to any given base b.

Title proof that number of sum-product numbers in any base is finite
Canonical name ProofThatNumberOfSumproductNumbersInAnyBaseIsFinite
Date of creation 2013-03-22 15:47:06
Last modified on 2013-03-22 15:47:06
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 7
Author rspuzio (6075)
Entry type Proof
Classification msc 11A63