Someone owes $t_{1}, t_{2}, t_{3}, \ldots t_{n}$ forints at the end of the 1st, 2nd, 3rd, ..., nth year; when can they clear their entire debt at once with $t_{1}+t_{2}+t_{3}+\ldots+t_{n}$ forints?
At the end of the 1st, 2nd, $3 ., \ldots n$. year, instead of paying $t_{1}+t_{2}+t_{3} \ldots t_{n}$ forints, one could pay $\frac{t_{1}}{e}, \frac{t_{2}}{e^{2}}, \frac{t_{3}}{e^{3}}, \ldots \frac{t_{n}}{e^{n}}$ forints at the beginning of the 1st year, if $e$ denotes the interest factor. The present value of the entire debt, which could be settled at once at the end of the $x$. year, is:

$$
\frac{t_{1}+t_{2}+t_{3}+\ldots+t_{n}}{e^{x}}
$$

The average maturity can therefore be calculated from the following equation:

$$
\frac{t_{1}}{e}+\frac{t_{2}}{e^{2}}+\frac{t_{3}}{e^{3}}+\ldots++\frac{t_{n}}{e^{n}}=\frac{t_{1}+t_{2}+t_{3}+\ldots+t_{n}}{e^{x}}
$$

from which:

$$
e^{x}=\left(t_{1}+t_{2}+t_{3}+\ldots+t_{n}\right):\left(\frac{t_{1}}{e}+\frac{t_{2}}{e^{2}}+\frac{t_{3}}{e^{3}}+\ldots+\frac{t_{n}}{e^{n}}\right)
$$

and finally

$$
\begin{gathered}
x=\left[\log \left(t_{1}+t_{2}+t_{3}+\ldots+t_{n}\right)-\right. \\
\left.-\log \left(\frac{t_{1}}{e}+\frac{t_{2}}{e^{2}}+\frac{t_{3}}{e^{3}}+\ldots+\frac{t_{n}}{e^{n}}\right)\right]: \log e
\end{gathered}
$$

The problem was solved by: Hofbauer Ervin, Budapest; Visnya Aladár, Pécs; Friedmann Bernát, S.-A.-Ujhely; Grünhut Béla, Pécs; Suschnik József, Kecskemét.