\section*{Exercise 6 - V01006}

What is the smallest number with the leftmost starting digit 7, which becomes one third of its original value when this 7 is removed from the front and appended to the remaining number as the rightmost digit?
}

Let \(x\) be the number without the leading digit 7. Then we have

\[
\begin{aligned}
& 7 \cdot 10^{n}+x=3(10 x+7)=30 x+21 \\
& 7 \cdot 10^{n}-21=7\left(10^{n}-3\right)=29 x
\end{aligned}
\]

Since \(x\) is a natural number, \(10^{n}-3\) must have 29 as a divisor.

\section*{1. Solution:}

Set \(r_{0}=-2\) and \(r_{n+1}=10 r_{n}-2 \bmod 29\), then \(r_{n} \equiv 10^{n}-3(\bmod 29)\).

By systematically calculating the sequence terms, we get for \(n=1,2, \ldots\)

\(-22,-19,-18,-8,-24,-10,-15,-7,-14,-26,-1,-12,-6,-4,-13,-16,-17,-27,-11,-25,-20,-28-21,-9,-5\), \(-23,0\)

Thus, \(10^{27}-3\) is a multiple of 29. The desired number is therefore

\[
7 \cdot 10^{27}+\frac{7 \cdot 10^{27}-3}{29}=\frac{3}{29}\left(7 \cdot 10^{28}-1\right)
\]