\section*{Exercise 3 - 290923}

Determine the smallest natural number \(n\), for which (when represented in the decimal positional system) 5 is both a divisor of the cross sum of \(n\) and a divisor of the cross sum of \(n+1\).
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If the number \(n\) ends with exactly \(k\) nines and has the digit \(z<9\) before them (possibly with a leading zero), then the number \(n+1\) ends with exactly \(k\) zeros and has the digit \(z+1 \leq 9\) before them. (The digits before \(z\) and \(z+1\) are identical in both numbers.) Let \(q_{n}\) and \(q_{n+1}\) be the digit sums of \(n\) and \(n+1\), respectively. Then it follows that \(q_{n+1}=q_{n}-k \cdot 9+1\).

For both digit sums to be divisible by 5, \(k \cdot 9-1\) must be divisible by 5. This first occurs for \(k=4\), meaning that any number \(n\) that satisfies the problem must end with at least four nines.

In fact, the first such numbers, namely 9999, 19999, 29999, and 39999, do not satisfy the condition that their digit sum is divisible by 5. The next such number \(n=49999\) does satisfy the problem, as both its digit sum 40 and the digit sum 5 of \(n+1=50000\) are divisible by 5.