A fisherman put the three largest fish, which accounted for $35 \%$ of the total weight of his daily catch, into the freezer. The three smallest fish, which together made up $\frac{5}{13}$ of the remaining part, were taken by the cat, and the rest were frozen for lunch. How many fish did the fisherman catch?
Let the total weight of the catch be $M$. The total weight of the three largest fish placed in the deep freezer is: $0.35 M$.

The total weight of the three smallest fish - which the cat took away - is $\frac{5}{13}$ of the remaining part, that is,

$$
\frac{5}{13}(M-0.35 M)=0.25 M
$$

The total weight of the fish cooked for lunch is:

$$
M-(0.35+0.25) M=0.4 M
$$

This represents at least 4 fish, since the total weight of the three largest fish is only $0.35 M$. Therefore, there were at least 10 fish in total.

Suppose the fisherman caught $n$ fish ($n \geq 10$). Since the total weight of the three largest fish is greater than that of the fish cooked, which in turn is greater than that of the fish taken by the cat, the following inequalities hold for their averages:

$$
\frac{0.35 M}{3} \geq \frac{0.4 M}{n-6} \geq \frac{0.25 M}{3}
$$

From the first inequality, $n \geq 9.428$, and from the second inequality, $n<10.8$. Since $n$ is an integer, $n=10$.

György Csató (Hajdúszoboszló, Hőgyes E. Gymnasium, 11th grade)