\section*{Problem 4 - 261014}

Jürgen claims that there is a positional system with base \(m\) in which the following calculation is correct:

\begin{tabular}{lllllll} 
& 7 & 0 & 1 &. & 3 & 4 \\
\hline 2 & 5 & 0 & 3 & & & \\
& 3 & 4 & 0 & 4 & & \\
\hline 3 & 0 & 4 & 3 & 4 & &
\end{tabular}

Determine all natural numbers \(m\) for which this is true!

Hint: In a positional system with base \(m\), there are exactly the digits \(0,1, \ldots, m-2, m-1\). Each natural number is represented as a sum of products of a power of \(m\) with one of the digits; the powers are ordered by decreasing exponents. The sequence of digits is then written as it is known for \(m=10\) in the decimal notation of natural numbers.
}

Since the largest digit used in the calculation is 7, a natural number \(m\) can only be the base of a positional system in which the calculation is correct if \(m \geq 8\). Since in the partial calculation

\[
\begin{array}{ccccc}
2 & 5 & 0 & 3 & \\
& 3 & 4 & 0 & 4 \\
\hline 3 & 0 & 4 & 3 & 4
\end{array}
\]

a carry occurs, which has the meaning \(5+3=0+1 \cdot m\), it follows that \(m=8\). Therefore, the calculation can only be correct in the octal system.

II. Indeed, when converting the factors and the product contained in the calculation from the octal system to the decimal system, we get

\[
\begin{aligned}
{[701]_{8} } & =7 \cdot 8^{2}+0 \cdot 8^{1}+1 \cdot 8^{0}=449 \\
{[34]_{8} } & =3 \cdot 8^{1}+4 \cdot 8^{0}=28 \\
{[30434]_{8} } & =3 \cdot 8^{4}+0 \cdot 8^{3}+4 \cdot 8^{2}+3 \cdot 8^{1}+4 \cdot 8^{0}=12572
\end{aligned}
\]

and it holds that \(449 \cdot 28=12572\).

From I. and II. it follows: The given calculation is correct exactly in the positional system with the base \(m=8\).

Solutions of the 1st Round 1986 adopted from [5]

\subsection*{7.28.2 2nd Round 1986, Class 10}