Problem
NT-B2-M11-P018 The Digit Sum of a Factorial
#18
★★★★★ Level 5 of 5
Prove that for every positive integer \(A\) there exists a positive integer \(B\) such that for all \(n\ge B\), the sum of the decimal digits of \(n!\) is at least \(A\).
Inspired by regional olympiad method · 2022 · Grade 9 · Problem 10
Choose \(k\) such that \(9k\ge A\), and force \(n!\) to be divisible by \(10^k-1\).
Choose \(k\) such that \(9k\ge A\), and set \(B=10^k-1\). If \(n\ge B\), then \(10^k-1\) is one of the factors in \(1\cdot2\cdots n=n!\). Therefore \(10^k-1\mid n!\).
By the previous block estimate, every positive multiple of \(10^k-1\) has digit sum at least \(9k\). Hence the digit sum of \(n!\) is at least \(9k\), and therefore at least \(A\).
This is strong regional-style: the key is not computing the factorial, but choosing a large divisor of the form \(10^k-1\).