## Task 14/88

We are looking for the smallest number $n \in N$, for which both the cross sum $q(n)$ and the cross sum $q(n+1)$ of the successor $n+1$ are divisible by 11.
The sought number $n$ must end in at least one digit 9, because otherwise the cross sum would change by 1 when transitioning from $n$ to $n+1$, and thus both cross sums could not be divisible by 11. Therefore, $n$ can be represented in the form

$$
n=a \cdot 10^{i}+\sum_{k=0}^{i-1} 9 \cdot 10^{k}
$$

with $a, i \in \mathbb{N}, i>0$. Then the following equation holds: $q(n+1)=q(n)-9i+1$.

For both $q(n)$ and $q(n+1)$ to be divisible by 11, $9i-1$ must also be divisible by 11. The smallest number $i$ that satisfies this is $i=5: 9 \cdot 5-1=44=4 \cdot 11$. Thus, $q(n)=q(a)+5 \cdot 9 \geq 55$, i.e., $q(a) \geq 10$. The smallest number $a$ whose cross sum is not less than 10 is $a=19$. Then $n=1999999, n+1=2000000$, and thus $q(n+1)=2$ is not divisible by 11. The next possible number $a$ is $a=28$. With this, we get $n=2899999, q(n)=55, n+1=2900000, q(n+1)=11$.

The solution path excludes smaller numbers for $n$.