A grant award had 4200 Ft available. The judging committee recommended awarding 7 prizes, with the I., II., and III. prizes being 800, 700, and 300 Ft, respectively. How many people won the I., II., and III. prizes, if the available amount was completely distributed?
I. solution. The average of the awarded prizes is $4200: 7=600 \mathrm{Ft}$, which is lower than the second prize, so they must have awarded a third prize. In fact, at least two, because after awarding one third prize, the average of the rest would be: $(4200-300): 6=650 \mathrm{Ft}$, still lower than the second prize. However, if three third prizes were awarded, the average of the remaining prizes would be: $(4200-3 \cdot 300) / 4$, which would be higher than even the first prize. This is impossible, so two third prizes were awarded.

The average of the remaining five prizes is $3600: 5=720 \mathrm{Ft}$, which is between the first and second prizes, so at least one first prize and one second prize must have been awarded. Taking one of each, the remaining three prizes amount to $2100=3 \cdot 700 \mathrm{Ft}$. This can only be distributed between the first and second prizes by awarding three more second prizes. Thus, one first, four second, and two third prizes were awarded. Together, they indeed amount to $1 \cdot 800+4 \cdot 700+2 \cdot 300=4200$ forints.

II. solution. Denoting the number of applicants awarded the first, second, and third prizes by $x, y, z$ respectively, and simplifying the second equation by 100:

$$
\begin{aligned}
& x+y+z=7 \quad \text { and } \\
& 8 x+7 y+3 z=42
\end{aligned}
$$

Subtracting three times (1) from (2):

$$
5 x+4 y=21, \quad \text { or } 4(x+y)+x=4 \cdot 5+1
$$

According to this, dividing $x$ by 4 leaves a remainder of 1. Thus, $x=1$, because the only other possible value, $x=5$, would leave 2 prizes with only $200 \mathrm{Ft}$, which is impossible. From (3), $y=4$ and from (1), $z=2$.

It is easy to verify that these values satisfy equation (2) as well.