1512 Ft consists of 2, 5, 10, 20, 50, 100, and 200 Ft banknotes. (Each denomination is present.) The money can be divided into the right and left pocket in 1512 different ways, including the two cases where one of the pockets is empty. How many banknotes of each denomination does the total amount contain? (Identical denomination banknotes are not distinguished.)
If we have $n$ two-forint coins, then $0, 1, \ldots, n$ coins can go into our left pocket, so the two-forint coins can be distributed in $(n+1)$ ways between the two pockets. The distribution of each denomination is independent of the others, so the number of ways to distribute the seven denominations is

$$
\left(c_{2}+1\right)\left(c_{5}+1\right)\left(c_{10}+1\right)\left(c_{20}+1\right)\left(c_{50}+1\right)\left(c_{100}+1\right)\left(c_{200}+1\right)
$$

where $c_{2}, c_{5}, \ldots, c_{200}$ are the numbers of 2, 5, ..., 200 forint coins, respectively. According to the condition, each denomination must appear, so in this factorization, each of the seven factors must be greater than 1.

On the other hand, $1512=2^{3} \cdot 3^{3} \cdot 7$ is exactly the product of seven prime numbers, so the product (1) is precisely the prime factorization of 1512. The numbers $c_{2}, c_{5}, c_{10}, c_{20}, c_{50}, c_{100}, c_{200}$ are exactly the numbers $1,1,1,2,2,2,6$ in some order, each one less than the prime factors. From this, the $c$ numbers can also be determined.

If $c_{200} \neq 6$, then the largest possible sum that can be made is

$$
6 \cdot 100 + 2 \cdot 200 + 2 \cdot 50 + 2 \cdot 20 + 1 \cdot 10 + 1 \cdot 5 + 1 \cdot 2 = 1168
$$

which is too small. Thus, we have 6 two-hundred-forint coins, and the remaining sum is 312. In this sum, every denomination from 2 to 100 appears - three of them twice. Therefore, by taking one of each denomination, three denominations remain, and their sum is

$$
312 - (100 + 50 + 20 + 10 + 5 + 2) = 125
$$

It is clear that 125 can only be expressed as $(100 + 20 + 5)$ as a three-term sum from the given six numbers, and thus the solution is:

$$
c_{1}=1, \quad c_{5}=2, \quad c_{10}=1, \quad c_{20}=2, \quad c_{50}=1, \quad c_{100}=2, \quad c_{200}=6
$$

Indeed, $6 \cdot 200 + 2 \cdot 100 + 1 \cdot 50 + 2 \cdot 20 + 1 \cdot 10 + 2 \cdot 5 + 1 \cdot 2 = 1512$.

Based on the work of Márton Hablicsek (Fazekas M. Főv. Gyak. Gimn., 7th grade)