Using scrap metal collection, a group of pioneer boys challenged their sister troop to a competition. At the end of the competition, it turned out that they collected a total of $28,31 \mathrm{q}-\mathrm{-t}$ of metal and that on average, each boy collected $95 \mathrm{~kg}$, while each girl collected $74 \mathrm{~kg}$. Which group won and how many boys and how many girls participated in the competition?
I. Solution: The problem in equation form:

$$
95 x+74 y=2831
$$

where $x$ is the number of boys, and $y$ is the number of girls.

$$
\begin{aligned}
& y=38-x+\frac{19-21 x}{74}=38-x+t \\
& x=\frac{19-74 t}{21}=-3 t+\frac{19-11 t}{21}=-3 t+u \\
& t=\frac{19-21 u}{11}=1-2 u+\frac{8+u}{11}=1-2 u+v \\
& u=11 v-8
\end{aligned}
$$

Substituting the value of $u$ back in,

$$
\begin{aligned}
& t=1-22 v+16+v=17-21 v \\
& x=-51+63 v+11 v-8=-59+74 v \\
& y=38+59-74 v+17-21 v=114-95 v
\end{aligned}
$$

Since only positive integer solutions make sense,

$$
x=-59+74 v>0, \quad \text { which implies } \quad v>\frac{59}{74}>0
$$

and

$$
y=114-95 v>0, \quad \text { which implies } \quad v<\frac{114}{95}<2,
$$

thus

$$
0<v<2, \quad \text { or } \quad v=1
$$

and hence

$$
x=-59+74 v=15, \quad y=114-95 v=19
$$

Therefore, 15 boys collected a total of $15 \cdot 95=1425 \mathrm{~kg}$, while 19 girls collected a total of $19 \cdot 74=1406 \mathrm{~kg}$. Thus, the boys won the competition.

Zsolt Csapó (Kecskemét, Katona József g. II. o. t.)

II. Solution:

Using the previous notation,

$$
74 y=2831-95 x=19(149-5 x)
$$

$x$ and $y$ can only be integers if the left side is also divisible by 19, and since 74 and 19 are relatively prime, $y$ must be divisible by 19. Additionally, $y<\frac{2831}{74}<39$, so $y$ can only be 19 or 38. In the latter case, $x$ is not an integer, but when $y=19$, $x=15$. This is therefore the only solution to the problem.

Tibor Harza (Székesfehérvár, József A. g. II. o. t.)