How many integers between 250 and 300, inclusive, can be evenly divided by NEITHER 3 NOR 5?

Explanation: The range is 250 to 300, inclusive. Total numbers = \(300 - 250 + 1 = 51\). We need to find the number of integers divisible by 3 OR 5, and subtract this from the total. The number of integers divisible by A or B is: \(N(A \cup B) = N(A) + N(B) - N(A \cap B)\). **Divisible by 3 (N(3))**: The first multiple of 3 is 252. The last is 300. \(N(3) = \frac{300}{3} - \lfloor \frac{250-1}{3} \rfloor = 100 - 83 = 17\). **Divisible by 5 (N(5))**: The first multiple of 5 is 250. The last is 300. \(N(5) = \frac{300}{5} - \lfloor \frac{250-1}{5} \rfloor = 60 - 49 = 11\). **Divisible by 3 and 5 (N(15))**: Divisible by 15. The first multiple is 255. The last is 300. \(N(15) = \frac{300}{15} - \lfloor \frac{250-1}{15} \rfloor = 20 - 16 = 4\). **Divisible by 3 OR 5 (\(N(3 \cup 5)\))**: \(N(3 \cup 5) = 17 + 11 - 4 = 24\). **Divisible by NEITHER 3 NOR 5**: Total numbers - \(N(3 \cup 5)\). \(51 - 24 = 27\). (The solution correctly uses $51 - (17 + 11 - 4) = 27$).