An amount not exceeding $100000 P$ is made up of 5-penny coins. If we divide these coins into groups of 12, 18, or 45, there is always a remainder of 3. However, if we divide them into groups of 11, there is no remainder.

Which amounts meet these requirements, and how many 5-penny coins make up these amounts?
Let $N$ denote the number of 5-penny coins; according to our data, this is a multiple of 11, i.e., $N=11 x$. The other properties are expressed by the following equations:

$$
11 x=12 n+3, \quad 11 x=18 p+3, \quad 11 x=45 q+3
$$

which means that $11 x-3$ is a multiple of 12, 18, 45, and hence of their least common multiple, which is 180. Therefore,

$$
11 x-3=180 y \text { or } 11 x-180 y=3 \text {. }
$$

We need to find the positive integer solutions to this equation. First, we see that $x$ must be a multiple of 3, i.e., $x=3 z$, and thus after dividing by 3,

$$
11 z-60 y=1 \text { or } 11(z-5 y)-5 y=1
$$

One solution to the latter equation is: $y=2, z-5 y=1$, i.e., $z=11$. All solutions can be written in the form

$$
y=2+11 m, \quad z=11+60 m \quad \text { or } \quad x=33+180 m
$$

thus,

$$
N=363+1980 m
$$

However,

$$
\begin{gathered}
363+1980 m \leqq \frac{100000}{5}=20000 \\
1980 m \leqq 19637 \text { and } 0 \leqq m \leqq 9
\end{gathered}
$$

The number of solutions is 10, and they form an arithmetic progression,

$$
N=363, \quad 2343, \quad 4323, \ldots 18183
$$

The multiples of five of these numbers are the amounts of money in question.

László Szalay (Ref. Sándor Baksay g.VI. o., Kunszentmiklós)