Maths Olympiad Prep

Library / /3 of 5

, 2003

Number theory Difficulty 5.0 AIME, harder Prove it Italy

Problem:

We take a positive integer nn, we sum its digits, and then we again sum the digits of that sum, obtaining an integer SS. What is the smallest nn that allows obtaining S10S \geq 10?

Solution

Solution:

The answer is 199. Indeed, if nn had 1 or 2 digits, n99n \leq 99, and then the sum of its digits would be 18\leq 18 and the sum of the digits of the number equal to the sum of its digits would be 9\leq 9, so nn has at least three digits. We observe that for n198n \leq 198, the sum of the digits of the number is still 18\leq 18. Instead if n=199n=199, the sum of its digits is 19 and the sum of the digits of 19 is 10, as required.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.