## Task $1 / 88$

Let $x$ be the sum of the digits of $1988^{8891}$, $y$ the sum of the digits of $x$, and $z$ the sum of the digits of $y$. Determine $z$!
First, we limit the digit sums from above. Since $1988^{8891}<10000^{8891}$, the number of digits of $1988^{8891}$ is less than $1+\lg 10000^{8891}=1+8891 \cdot \lg 10000=1+8891 \cdot 4=35565$.

Since for all digits $a_{i}$ of $1988^{8891}$ it holds that $a_{i} \leq 9$, we have $x<35565 \cdot 9=320085$. The number with the largest digit sum in the interval [0; 320085] is 299999. Thus, it follows that $y \leq 47$. For $z$, this results in $z \leq 12$, since 39 is the number with the largest digit sum in the interval $[0 ; 47]$.

Now we use the theorem: A number $n$ and its digit sum leave the same remainder when divided by 9.

Proof: 1. The theorem holds for $n=1$.

2. Assuming the theorem holds for $n=k \geq 1$, then it also holds for $n=k+1$. If the last $i$ digits $(i \geq 0)$ of $k$ are 9, these become 0 when transitioning to $k+1$, and the first digit that is not 9 increases by 1.

The digit sum of $k+1$ is thus different from that of $k$ by $1-i \cdot 9 \equiv 9(\bmod 9)$, so its remainder is also increased by 1.

From this theorem, it follows that since $1988^{8891} \equiv(-1)^{8891} \equiv(-1) \equiv 8(\bmod 9)$ (since $1988=221 \cdot 9-1$), we also have $x \equiv y \equiv z \equiv 8(\bmod 9)$. Given that $z \leq 12$, the only possible value for $z$ is 8: $z=8$.