We bought three types of gifts for 720 Ft, a total of $20 \mathrm{db}$, with unit prices of 60, 50, and $10 \mathrm{Ft}$. How many of each did we buy?
Let the number of items priced at $60, 50, 10 \mathrm{Ft}$ be $x, y, z$ respectively, thus we need to solve the following system of indeterminate equations:

$$
\begin{gathered}
x+y+z=20 \\
60 x+50 y+10 z=720
\end{gathered}
$$

However, we only accept positive integer solutions. Simplify (2) by dividing by 10 and subtract (1) from it:

$$
5 x+4 y=52 \text {. }
$$

Rewriting (3) as $x=4(13-x-y)$, it is clear that $x$ must be divisible by 4: $x=4 t$, and thus from (3) we get $y=13-5 t$.

Now, from the conditions $x>0, y>0$, we see that on one hand $t>0$, and on the other hand $t<13 / 5$, so only the values $t=1$ and 2 are possible. With these values,

$$
\begin{array}{rlrl}
x & =4 t \quad=4, & 8 \\
y & =13-5 t=8, & 3 \\
\text { and thus } \quad z & =20-x-y & =8, & 9
\end{array}
$$

Therefore, the purchase could have been made in two different ways.

László Gerencsér (Budapest, Rákóczi F. Gymnasium, 1st year student). Note. We can get simpler equations if we eliminate $x$, or $y$. Subtracting 5 times (1) from the simplified (2), or 6 times (1) from the simplified (2),

$$
\begin{gathered}
y+5 z=48, \quad-x+4 z=28, \quad \text { i.e., } \\
y=5(9-z)+3 \\
x=4(z-7)
\end{gathered}
$$

From (5), we get $z \geq 8$, and from (4), we get $z \leq 9$.