Someone for 10 years deposits $500 \mathrm{~K}-\mathrm{t}$ in the savings bank at the beginning of each year; then for another 10 years, withdraws $500 \mathrm{~K}-\mathrm{t}$ from the savings bank at the beginning of each year. How much money will they have at the end of the 20th year? $p=4$.
$$
500\left(\frac{e^{10}-1}{e-1} e^{11}-\frac{e^{10}-1}{e-1} e\right)=500 e \cdot \frac{\left(e^{10}-1\right)^{2}}{e-1}=2996.4 \mathrm{~K}
$$

(Kürth Richárd, Nyitra.)

The problem was also solved by: Csada I., Dömény I., Ehrenfeld N., Fekete M., Frank A., Freund E., Haar A., Heimlich P., Jánosy Gy., Kiss J., Krampera Gy., Merse P., Pető L., Pözel T., Ruvald S., Schuster Gy., Schwarz O., Szécsi I., Wottitz R.