\section*{Problem 6 - 011016}

A six-digit number begins with the digit 1 at the highest place. If this digit is removed and appended to the end of the number, the result is three times the original number.
}

a) If one divides this six-digit number \(x\) into its first digit 1 and the remaining five-digit number \(a\), this can be expressed as follows: \(x=100000+a\).

At the same time, for the second six-digit number \(y\), it is stated that it is formed by removing the first digit of \(x\) and appending this digit at the end of the number, thus: \(y=10 \cdot a+1\).

Furthermore, it is said that \(y=3x\) and therefore \(10 \cdot a+1=3 \cdot (100000+a)\). After rearranging, one gets \(7a=3 \cdot 100000-1\), which results in: \(a=42857\). For \(x\) and \(y\), this yields: \(x=142857, y=428571\).

b) The following statements apply:

If one removes the first two digits 14 from a six-digit number and appends them to the remaining four-digit number, the resulting number is twice as large as the original number.

If one removes the last two digits 57 from a six-digit number and places them in front of the remaining four-digit number, the resulting number is four times as large as the original number.

If one removes the last digit 7 from a six-digit number and places it in front of the remaining five-digit number, the resulting number is five times as large as the original number.

If one removes the first three digits 142 from a six-digit number and appends them to the remaining three-digit number, the resulting number is six times as large as the original number.

c) The problem can be derived from the answer to part a):

\[
x=100000+a=100000+\frac{3 \cdot 100000-1}{7} \quad \text{and} \quad y=10 \cdot a+1=10 \cdot \frac{3 \cdot 100000-1}{7}+1
\]