## Task 23/85

Given the $n$-digit natural number $z_{n}=1985$!. Form the natural number $z_{n-1}$ by removing the units digit of $z_{n}$ and subtracting it from the remaining $(n-1)$-digit number. Continue this process until a single-digit number $z$ is obtained. What is $z$?
Due to $z_{n}=1985!>11!$, $z_{n}$ is certainly divisible by 11 without a remainder. Now let $z_{k}=10 a+b$ with $a ; b \in N, 0 \leq b \leq 9$ be a number formed according to the problem statement and divisible by 11 without a remainder. Then,

$$
z_{k-1}=a-b=11 a-z_{k}
$$

is certainly also divisible by 11. This implies (by the principle of complete induction - applied in reverse), that all $z_{k}$, specifically $z$, are divisible by 11 without a remainder. Since $z$ is a single digit, it can only be $z=0$.