## Task $7 / 76$

The smallest natural number $n$ is sought, with the property that both the cross sum $Q(n)$ of the number $n$ and the cross sum $Q(n+1)$ of the successor of $n$ are divisible by 5.
The number $n$ must end in the digit 9; for with any other digit, $Q(n+1)=Q(n)+1$, and it would not be possible for both $Q(n)$ and $Q(n+1)$ to be divisible by 5.

Let $n=10a+9$, where $a$ is a natural number with $a>0$. It is easy to see that the number $a$ must also end in the digit 9; otherwise, $Q(n+1)=Q(a)+1=Q(n)-8$ because $Q(n)=Q(a)+9$. Therefore, $a=10b+9$ with $b \neq 0$, of course, so $n=100b+99$.

Similarly, one can see that $b$ must end in the digit 9, otherwise $Q(n+1)=Q(b)+1=Q(n)-17$ would result. Thus, $b=10c+9$ and hence $n=1000c+999$.

Continuing this reasoning leads to $c=10d+9$ with $n=10000d+9999$ and $Q(n+1)=Q(d)+1=Q(n)-35$. Here the chain breaks off. One recognizes that $Q(n)$ and $Q(n+1)$ are always both either divisible by 5 or not divisible by 5 if and only if the last four digits of the number $n$ are all 9.

The smallest natural number with the required property is thus $n=49999$. It is $Q(n)=40$ and $Q(n+1)=5$.