\section*{Exercise 2 - 330932}

For each positive integer \(n\), imagine another number \(n^{\prime}\) formed according to the following rule: From the digit representation of \(n\) in the decimal system, the first digit is removed and appended to the end of the last digit.

Then let \(n^{\prime}\) be the number with the resulting digit representation. Investigate whether there are numbers \(n\) divisible by 7 for which \(n^{\prime}=n / 7\) holds!
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We write \(n\) in the form \(n=a \cdot 10^{x}+b\), where \(a \in\{1, \ldots, 9\}, x, b \in \mathbb{N}\) and \(b<10^{x}\). Then \(n^{\prime}=10 b+a\).

For \(n^{\prime}=n: 7\), i.e., \(7 \cdot(10 b+a)=a \cdot 10^{x}+b\) to hold, it must be that \(a\left(10^{x}-7\right)=69 b\).

We choose \(a=1\) and look for an \(x \in \mathbb{N}\) such that \(10^{x}-7\) is divisible by \(69=3 \cdot 23\). Since \(10^{x}-7 \equiv 1^{x}-1 \equiv 0\) \((\bmod 3)\) is always satisfied, it suffices to find an \(x\) such that \(10^{x} \equiv 7(\bmod 23)\). By Fermat's Little Theorem, \(1 \equiv 10^{22}(\bmod 23)\). Multiplying by 7 gives \(7 \equiv(7 \cdot 10) \cdot 10^{21} \equiv 10^{21}(\bmod 23)\). Therefore, we can choose \(x=21\).

For \(b=\frac{10^{x}-7}{69}\), it is clear that \(b<10^{x}\), so we can conclude that \(n=10^{21}+\frac{10^{21}-7}{69}\) is a number with the required properties.