Problem 4 - Entrance Test

A positive integer has exactly three divisors if and only if it is the square of a prime number. Let the integer be n = p^2, where p is a prime number. We are looking for integers n such that 100 < n < 200. So, we need to find prime numbers p such that 100 < p^2 < 200. Taking the square root of the inequality: sqrt(100) < p < sqrt(200) 10 < p < sqrt(200). We know that 14^2 = 196, and 15^2 = 225. So, sqrt(200) is approximately 14.14. Therefore, we are looking for prime numbers p such that 10 < p < 14.14. The prime numbers in this range are: - p = 11 (since 11 is prime and 10 < 11 < 14.14) - p = 13 (since 13 is prime and 10 < 13 < 14.14) Now we find the squares of these primes: - For p = 11, n = 11^2 = 121. The divisors of 121 are 1, 11, 121 (exactly three divisors). 100 < 121 < 200. - For p = 13, n = 13^2 = 169. The divisors of 169 are 1, 13, 169 (exactly three divisors). 100 < 169 < 200. For the next prime, p = 17, n = 17^2 = 289, which is greater than 200. Thus, there are two such positive integers: 121 and 169.