Task 7. On a circle, there are 25 non-overlapping arcs, and on each of them, two arbitrary prime numbers are written. The sum of the numbers on each arc is not less than the product of the numbers on the arc following it in a clockwise direction. What can the sum of all the numbers be?
Problem 7. On a circle, there are 25 non-intersecting arcs, and on each of them, two arbitrary prime numbers are written. The sum of the numbers on each arc is not less than the product of the numbers on the next arc clockwise. What can the sum of all the numbers be?

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| Let the numbers written be $\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right), \ldots,\left(a_{25}, b_{25}\right)$. Note that if $a_{i} b_{i} \geq a_{i}+b_{i}$, where $a_{i}$ and $b_{i}$ are prime numbers, since prime numbers are at least 2. Now, let's write down the condition of the problem: $a_{1}+b_{1} \geq a_{2} b_{2} \geq a_{2}+b_{2} \geq a_{3} b_{3} \geq$ $a_{3}+b_{3} \geq \cdots \geq a_{1}+b_{1}$. That is, the inequality holds only in the case of equality. Thus, $a_{i} b_{i}=a_{i}+b_{i}$ for all $i$. This equality is only satisfied if $a_{i}=b_{i}=2$. Therefore, all the numbers written on the circle are equal to 2. Consequently, the sum of the numbers is $2 \times 2 \times 25=100$. Answer: 100. | Only the correct answer - 1 point. Noted that the product of two prime numbers is not less than their sum - 3 points. |