4. We will call a number greater than 25 semi-prime if it is the sum of some two distinct prime numbers. What is the maximum number of consecutive natural numbers that can be semi-prime?
4. Answer. 5.

Note that an odd semiprime number can only be the sum of two and an odd prime number.

Let's show that three consecutive odd numbers \(2n+1, 2n+3\), and \(2n+5\), greater than 25, cannot all be semiprimes simultaneously. Assuming the contrary, we get that the numbers \(2n-1, 2n+1\), and \(2n+3\) are prime, and all of them are greater than 3. But one of these three numbers is divisible by 3, which is a contradiction. Note that among any six consecutive numbers, there are three consecutive odd numbers; therefore, there cannot be more than five consecutive semiprime numbers. Five consecutive numbers can be semiprimes; for example, \(30 = 17 + 13\),

![](https://cdn.mathpix.com/cropped/2024_05_06_08150518f205927190b8g-3.jpg?height=60&width=1645&top_left_y=1369&top_left_x=291)

The segment \(C_1M\) is the median of the right triangle \(CC_1A\), so \(CM = \underline{AC} = MA\) (see the figure). Then \(\angle CMA = \pi - 2\angle BAC\). Similarly, \(\angle AMC = \pi - 2\angle BCA\).

From this, \(\angle C_1MA + \angle A_1MC = 2(\pi - \angle BAC - \angle BCA) = 2\angle ABC\),

\(\angle A_1MC_1 = \pi - (\angle AMC_1 + \angle CMA_1) = \pi - 2\angle ABC\).