## Problem 9

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 $\mathrm{x}$, and $\mathrm{z}$ the sum of the digits of $\mathrm{y}$. Find $\mathrm{z}$.


## Solution

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

