Library / All-Soviet-Union Mathematical Olympiad /81 of 185

All-Soviet-Union Mathematical Olympiad

Number theory Difficulty 5.0 AIME, harder Prove it Soviet Union

Problem:
Given a number with 1998 digits which is divisible by 9. Let $x$ be the sum of its digits, let $y$ be the sum of the digits of $x$ , and $z$ the sum of the digits of $y$ . Find $z$ .

Solution

Solution:
$x \leq 9 \times 1998 = 17982$ . Hence $y \leq$ the greater of $1 + 7 + 9 + 9 + 9 = 35$ and $9 + 9 + 9 + 9 = 36$ . But $9$ divides the original number and hence also $x$ , $y$ and $z$ . Hence $z = 9$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.