## Task 6 - 150736

If $z$ is a natural number, let $a$ be the cross sum of $z$, $b$ be the cross sum of $a$, and $c$ be the cross sum of $b$.

Determine $c$ for every 1000000000-digit number $z$ that is divisible by 9!
Due to $z>0$, it also holds that $a>0, b>0, c>0$.

If a number is divisible by 9, then its cross sum is also divisible by 9. Therefore, all numbers $a, b$, and $c$ in this problem are divisible by 9.

Since each of the 1000000000 digits of $z$ is at most 9, we have

$$
a \leq 9 \cdot 1000000000=9000000000
$$

so $a$ is at most ten digits long.

Therefore, $b<9 \cdot 10=90$, so $b$ is one of the numbers $81, 72, 63, 54, 45, 36, 27, 18$, 9. The cross sum of each of these 9 numbers is 9.

Thus, $c=9$ for all numbers $z$ under consideration.

Solutions of the III. Round 1975 taken from [5]

### 4.18 XVI. Olympiad 1976

### 4.18.1 I. Round 1976, Class 7