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, 2019

Algebra Difficulty 4.9 AIME Prove it United States

Problem:

Let $S(x)$ denote the sum of the digits of a positive integer $x$ . Find the maximum possible value of $S(x+2019)-S(x)$ .

Solution

Solution:

We note that $S(a+b) \leq S(a)+S(b)$ for all positive $a$ and $b$ , since carrying over will only decrease the sum of digits. (A bit more rigorously, one can show that $S\left(x+a \cdot 10^{b}\right)-S(x) \leq a$ for $0 \leq a \leq 9$ .) Hence we have $S(x+2019)-S(x) \leq S(2019)=12$ , and equality can be achieved with $x=100000$ for example.

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