Mekkora $B$ betétet kell öt éven át minden év elején a bankban elhelyeznünk, hogy évi $20 \%$-os kamat mellett az ötödik év végén ugyanakkora legyen a követelésünk, mintha az első év elején egyszerre $100000 \mathrm{Ft}$-ot tettünk volna a bankba?

What amount $B$ must we deposit at the beginning of each year for five years in the bank, so that with an annual interest rate of $20 \%$, our claim at the end of the fifth year is the same as if we had deposited $100000 \mathrm{Ft}$ at the beginning of the first year?
If we deposit 100,000 Ft at an annual interest rate of 20% for 5 years in the bank, at the end of the fifth year,

$$
10^{5}\left(1+\frac{1}{5}\right)^{5}=6^{5} \cdot 2^{5}=248832 \text{ Ft}
$$

we will receive. The question is, at the same 20% interest rate, how much money should be deposited at the beginning of each year in the bank to receive the above amount?

If we placed \( B \) Ft, at the end of the first year we would have \( 1.2B \) forints, and then we would deposit another \( B \) Ft. At the end of the second year, we would have \( (B + 1.2B) \cdot 1.2 = 1.2B + 1.2^2B \) forints, and so on. By the end of the fifth year,

$$
1.2B + 1.2^2B + 1.2^3B + 1.2^4B + 1.2^5B = B(1.2 + 1.2^2 + 1.2^3 + 1.2^4 + 1.2^5)
$$

Notice that the sum in the parentheses is the sum of a geometric series whose first term and common ratio are both 1.2. Using the formula for the sum of a geometric series and the given condition, we can write:

$$
B \cdot 1.2 \cdot \frac{1.2^5 - 1}{1.2 - 1} = 248832
$$

From which, after rearranging,

$$
B = \frac{248832}{6(1.2^5 - 1)} = 27864.98
$$

Therefore, we need to deposit 27,864.98 forints at the beginning of each year.