We know that we currently have six denominations of banknotes in circulation: 500, 1000, 2000, 5000, 10000, and 20000 forints. How many different amounts can be paid with three banknotes?
Solution. Let's calculate how many 3-term sums there are with different total values. First, consider the case when we have 3 twenty-euro notes. This only occurs once. Leave out one of the twenty-euro notes and see how many different sums we can get by changing the third addend. For the third place, we can only choose values smaller than twenty euros, which gives 5 possibilities. Finally, if we have 1 twenty-euro note and a ten-euro note in the second place, we can have a ten-euro, five-euro, two-euro, one-euro, or five-hundred cent note in the third place, which is 5 more possibilities.

If a five-euro note is in the second place, then a five-euro, two-euro, one-euro, or five-hundred cent note can be in the third place, giving 4 values to choose from; if a two-euro note is there, then 3 values; if a one-euro note, then 2; and finally, we have the case of twenty-euro + five-hundred cent + five-hundred cent, which is 1 case. So far, we can pay 21 different sums.

If we have 2 ten-euro notes, the third place can have a ten-euro, five-euro, two-euro, one-euro, or five-hundred cent note, which is another $\mathbf{5}$ possibilities.

Finally, in the case of 1 ten-euro note, the following sums can occur:

$$
\begin{array}{llll}
10 \mathrm{e}+5 \mathrm{e}+5 \mathrm{e} & 10 \mathrm{e}+2 \mathrm{e}+2 \mathrm{e} & 10 \mathrm{e}+1 \mathrm{e}+1 \mathrm{e} & 10 \mathrm{e}+500+500 \\
10 \mathrm{e}+5 \mathrm{e}+2 \mathrm{e} & 10 \mathrm{e}+2 \mathrm{e}+1 \mathrm{e} & 10 \mathrm{e}+1 \mathrm{e}+500 & \\
10 \mathrm{e}+5 \mathrm{e}+1 \mathrm{e} & 10 \mathrm{e}+2 \mathrm{e}+500 & & \\
10 \mathrm{e}+5 \mathrm{e}+500 & & &
\end{array}
$$

This, together with the previous, totals $\mathbf{15}$ sums.

In addition to these, the following sums can occur:

$$
\begin{array}{llll}
5 \mathrm{e}+5 \mathrm{e}+5 \mathrm{e} & 5 \mathrm{e}+2 \mathrm{e}+2 \mathrm{e} & 5 \mathrm{e}+1 \mathrm{e}+1 \mathrm{e} & \\
5 \mathrm{e}+5 \mathrm{e}+2 \mathrm{e} & 5 \mathrm{e}+2 \mathrm{e}+1 \mathrm{e} & 5 \mathrm{e}+1 \mathrm{e}+500 & \\
5 \mathrm{e}+5 \mathrm{e}+1 \mathrm{e} & 5 \mathrm{e}+2 \mathrm{e}+500 & 5 \mathrm{e}+500+500 & \\
5 \mathrm{e}+5 \mathrm{e}+500 & & & 500+500+500 \\
2 \mathrm{e}+2 \mathrm{e}+2 \mathrm{e} & 2 \mathrm{e}+1 \mathrm{e}+1 \mathrm{e} & 1 \mathrm{e}+1 \mathrm{e}+1 \mathrm{e} & \\
2 \mathrm{e}+2 \mathrm{e}+1 \mathrm{e} & 2 \mathrm{e}+1 \mathrm{e}+500 & 1 \mathrm{e}+1 \mathrm{e}+500 & \\
2 \mathrm{e}+2 \mathrm{e}+500 & 2 \mathrm{e}+500+500 & 1 \mathrm{e}+500+500 &
\end{array}
$$

This is a total of 20 more possibilities.

Summing up the results: $21+15+20=56$ cases.

Notice that some of the sums appear twice. Let's find and list those:

$$
\left\{\begin{array} { l } 
{ 2 0 \mathrm { e } + 5 \mathrm { e } + 5 \mathrm { e } } \\
{ 1 0 \mathrm { e } + 1 0 \mathrm { e } + 1 0 \mathrm { e } } \\
{ 1 0 \mathrm { e } + 1 \mathrm { e } + 1 \mathrm { e } } \\
{ 5 \mathrm { e } + 5 \mathrm { e } + 2 \mathrm { e } }
\end{array} \left\{\begin{array} { l } 
{ 2 0 \mathrm { e } + 1 \mathrm { e } + 1 \mathrm { e } } \\
{ 1 0 \mathrm { e } + 1 0 \mathrm { e } + 2 \mathrm { e } } \\
{ 1 0 \mathrm { e } + 5 0 0 + 5 0 0 } \\
{ 5 \mathrm { e } + 5 \mathrm { e } + 1 \mathrm { e } }
\end{array} \left\{\begin{array} { l } 
{ 2 0 \mathrm { e } + 5 0 0 + 5 0 0 } \\
{ 1 0 \mathrm { e } + 1 0 \mathrm { e } + 1 \mathrm { e } } \\
{ 2 \mathrm { e } + 2 \mathrm { e } + 2 \mathrm { e } } \\
{ 5 \mathrm { e } + 5 0 0 + 5 0 0 }
\end{array} \quad \left\{\begin{array}{l}
2 \mathrm{e}+500+500 \\
1 \mathrm{e}+1 \mathrm{e}+1 \mathrm{e}
\end{array}\right.\right.\right.\right.
$$

In total, 7 sums appear twice. Subtracting this from 56, we get that the number of different sums is 49.