10.4. Every day, from Monday to Friday, the old man went to the blue sea and cast his net into the water. Each day, the net caught no more fish than the previous day. In total, over the five days, the old man caught exactly 100 fish. What is the smallest total number of fish he could have caught on the three days - Monday, Wednesday, and Friday?
Answer: 50.

Solution. If the old man caught 25 fish each of the first four days and caught nothing on Friday, the conditions of the problem are met, and exactly 50 fish were caught over the specified three days.

Let's prove that in the specified days, fewer than 50 fish could not have been caught. Indeed, suppose fewer than 50 fish were caught in these days. Since on Tuesday and Thursday the old man caught no more fish than on Monday and Wednesday, on Tuesday and Thursday fewer than 50 fish were also caught. Thus, in total over the five days, fewer than 100 fish were caught, which contradicts the condition.

The estimate (the second part of the solution) can be written algebraically in various ways. Let the old man catch $a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant a_{4} \geqslant a_{5}$ fish from Monday to Friday. Then:

1) If $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=100$, then $a_{1}+a_{3}+a_{5}=100-\left(a_{2}+a_{4}\right) \geqslant 100-\left(a_{1}+a_{3}\right)$. Therefore, $2 a_{1}+2 a_{3}+a_{5} \geqslant 100$. Since $2 a_{5} \geqslant a_{5}$, then $a_{1}+a_{3}+a_{5} \geqslant 50$.
2) Since $a_{1} \geqslant \frac{a_{1}+a_{2}}{2}, a_{3} \geqslant \frac{a_{3}+a_{4}}{2}, a_{5} \geqslant \frac{a_{5}}{2}$, then $a_{1}+a_{3}+a_{5} \geqslant \frac{a_{1}+a_{2}}{2}+\frac{a_{3}+a_{4}}{2}+\frac{a_{5}}{2}=$ $=\frac{a_{1}+a_{2}+a_{3}+a_{4}+a_{5}}{2}=50$.

## Grading Criteria

+ Correct answer and a fully justified solution are provided

干 Only the correct answer and an example are provided

$\mp$ Only the estimate is proven

- Only the answer is provided
- The problem is not solved or is solved incorrectly

The jury believes that the plot of the problem should not lead the solver to think that the old man necessarily catches at least one fish each day (and in Pushkin's fairy tale, the old man only pulled out mud). However, if the solver clearly demonstrates this understanding and justifiably finds the answer 51, they should be given a grade of +. Intermediate criteria do not apply to this situation.