4. The results of the fishing competition are summed up based on the points in three categories (float, feeder, spinner), and in each category, the first three places, the next three, the next three, and so on are awarded. The first place gives 1 point, the second place gives 2 points, and so on (the fewer points, the better the result). What is the lowest place that a participant who came first in the float category, second in the feeder category, and third in the spinner category could have taken in the overall competition?
Solution. Let $\mathrm{N}$ participants have overtaken him. This means that each of them scored no more than 5. Therefore, all together they ($\mathrm{N}+1$ people) scored no more than $5 \mathrm{~N}+6$. On the other hand, in this sum, there are only 9 ones and 9 twos, and the remaining $(3 \mathrm{~N}+3)-18$ results are at least 3, so this sum is no less than $9 \cdot 1+9 \cdot 2+(3 \mathrm{~N}-15) \cdot 3=9 \mathrm{~N}-18$. From the double inequality $9 \mathrm{~N}-18 \leq \mathrm{S} \leq 5 \mathrm{~N}+6$, we get $4 \mathrm{~N} \leq 24$, hence $\mathrm{N} \leq 6$.

Thus, the lowest possible place is seventh. Example (the only one!): 3+1+1, 1+3+1, 1+2+2, 2+1+2, 2+1+2, 2+2+1.

## 6th grade. Lecture hall.

## Solutions