\section*{Problem 7 - V01007}

A six-digit whole number ends at the lowest place (E) with 1. If you remove this last digit and place it at the front again, you get one third of the original number.

a) What are the two numbers?

b) Explain the reasoning that led you to the solution.
}

The sought number \(z\) has the form \(z=10 x+1\), where \(x\) is the remainder after removing the 1 at the end. Then,

\[
z=10 x+1=3 \cdot(100000+x) \quad \Rightarrow \quad x=42857
\]

The sought numbers are thus 428571 and 142857. The verification confirms the result.