## Task 36/78

Choose any 1979-digit natural number $a$ that is divisible by 9, and form the cross sum $z$ from the cross sum $y$ of the cross sum $x$ of $a$. How large is $z$?
If a number $a$ is divisible by 9, then the cross sum $Q(a)$ of this number is also divisible by 9. From this, it follows that $x$, $y$, and $z$ are divisible by 9. Furthermore, it holds that:

$x \leq 9 \cdot 1979 = 17811$; that is, $x$ is at most five digits.

This further implies

$y \leq 9 \cdot 5 = 45$; that is, for $y$ only the numbers $9 ; 18 ; 27 ; 36 ; 45$ are possible. In each case, this means $z = 9$.

### 2.19 Problems and Solutions 1979