Someone deposits $t$ forints in a savings bank; how much money will they have at the end of the $n$-th year, if after the deduction of tax they receive a $p_{1} \%$ interest rate, but after the annually due interest, they only receive a $p_{2} \%$ interest rate? For example, $t=5600 \mathrm{frt}$.

$$
p_{1}=4 \frac{1}{2}, p_{2}=4, n=12 \text { years. }
$$
The annual interest can be considered as an annually recurring deposit; the interest of $t$ forints at an annual rate of $p_{1}$ percent is:

$\frac{t p_{1}}{100};$ the value of the interest due at the end of the 1st, 2nd, 3rd, ..., $n$th year, at a rate of $p_{2}$ percent at the end of the $n$th year is:

$$
\frac{t p_{1}}{100}\left(1+\frac{p_{2}}{100}\right)^{n-1}, \frac{t p_{1}}{100}\left(1+\frac{p_{2}}{100}\right)^{n-2}, \frac{t p_{1}}{100}\left(1+\frac{p_{2}}{100}\right)^{n-3}, \ldots, \frac{t p_{1}}{100}\left(1+\frac{p_{2}}{100}\right), \frac{t p_{1}}{100}
$$

Thus, the value of the capital at the end of the $n$th year will be:

$$
T=t+\frac{t p_{1}}{100}\left[1+\left(1+\frac{p_{2}}{100}\right)+\left(1+\frac{p_{2}}{100}\right)^{2}+\ldots+\left(1+\frac{p_{2}}{100}\right)^{n-2}+\left(1+\frac{p_{2}}{100}\right)^{n-1}\right]
$$

Summing the geometric progression in the parentheses, we get

$$
T=t+\frac{t p_{1}}{100} \frac{\left(1+\frac{p_{2}}{100}\right)^{n}-1}{\frac{p_{2}}{100}}
$$

or

$$
T=t\left\{1+\frac{p_{1}}{p_{3}}\left[\left(1+\frac{p_{2}}{100}\right)^{n}-1\right]\right\}
$$

Substituting the given values, we obtain:

$$
T=9385.67 \, forints
$$

The problem was solved by: Goldstein Zsigmond, senior gym. VII. class, Nyíregyháza; Hofbauer Ervin, Budapest; Messinger Ábrahám, senior gym. VII. class, S.-A.-Ujhely; Szabó Isván, senior gym. VI. class, Debrecen; Visnya Aladár, Pécs; Friedmann Bernát, S.-A.-Ujhely.