10.5. We will call a natural number semi-prime if it is greater than 25 and is the sum of two distinct prime numbers. What is the maximum number of consecutive natural numbers that can be semi-prime? Justify your answer.
Solution: There are many sets of five consecutive semiprime numbers, for example $30(=13+17), 31(=2+29), 32(=3+29), 33(=2+31), 34$ $(=5+29)$ or $102(=5+97), 103(=2+101), 104(=31+73), 105(=2+103)$, $106(=47+59)$. We will show that there are no six consecutive semiprime numbers.

Indeed, among any six consecutive numbers, exactly three are odd, and these odd numbers increase by a step of 2. If an odd number is represented as the sum of two natural numbers, then one of the addends is even, and the other is odd. If both addends are prime, then the even one must be 2. Therefore, these odd semiprime numbers have the form $2+p_{1}, 2+p_{2}$, and $2+p_{3}$. In this case, the numbers $p_{1}, p_{2}$, $p_{3}$ are consecutive odd numbers. Among any three consecutive odd numbers, one is divisible by 3. Since it is prime, it must be 3. Then one of the numbers is 5, and it is not semiprime because it is less than 25. Contradiction.

Answer: 5 numbers.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| If the proof that there are no more than 5 numbers is correct, but the example of 5 consecutive semiprime numbers is missing | 4 points |
| Only an example (or examples) of 5 consecutive semiprime numbers is provided | 2 points |
| Correct answer without justification and/or examples of consecutive semiprime numbers in a quantity less than 5, and/or a more crude estimate of their quantity. | 0 points |