8.5. The numbers from 1 to 10 were written in some order and resulted in the numbers \(a_{1}, a_{2}, a_{3}, \ldots, a_{10}\), and then the sums \(S_{1}=a_{1}\), \(S_{2}=a_{1}+a_{2}\), \(S_{3}=a_{1}+a_{2}+a_{3}\), \ldots, \(S_{10}=a_{1}+a_{2}+a_{3}+\ldots+a_{10}\) were calculated. What is the maximum number of prime numbers that could be among the numbers \(S_{1}, S_{2}\), \(S_{3}, \ldots, S_{10}\)?
Answer: 7.

Among the numbers from 1 to 10, there are five odd numbers. Adding an odd number changes the parity of the sum. Let $y_{1}^{\prime}, y_{2}, y_{3}, y_{4}, y_{5}$ be the odd numbers, in the order they appear on the board. After adding $y_{2}$, the sum will become even and greater than 2, as it will after adding $y_{4}$. These sums are composite numbers, as is $S_{10}=55$. Therefore, among the sums, there are no more than 7 prime numbers.

Example of writing the numbers: $2,1,4,3,7,6,8,10,5,9$. The resulting sums: $2,3,7,10,17,23,31,41,46,55$.

Comment. An answer without justification - 0 points. A correct answer with an example - 3 points (the example provided in the solution is not unique). An estimate of the number of prime sums (without an example) - 3 points.