Question number 15. How many three-digit prime numbers p have the property that p minus 1 has exactly one prime divisor? So let's gather the information we have about p. We know that p is prime, so in this case p has to be odd. That is because simply p is not equal to 2. It's a three-digit number and the only even prime is 2, so p has to be odd. Then we know that p minus 1 is even, and from the sentence here that p minus 1 has exactly one prime divisor, we conclude that p minus 1 is even and only divisible by 2. So that means p minus 1 is a power of 2. Otherwise it will have other prime divisors than 2. And so we are looking now for three-digit numbers that are powers of 2. So p minus 1 could be equal to 2 to the power 9, which is 512, so that p is equal to 513. But 513 is not prime because it is in fact divisible by 3, and we can factor it if we really like. This is 3 times 171, which is equal to 3 times 3 times 57, and 57 is also divisible by 3, so this is 3 cubed times 19, so definitely not a prime number. And then the next candidate is that p minus 1 is equal to 2 to the power 8, so 256, and in that case p would be 257, and 257 is prime. I guess we just have to know that. And this is our number that satisfies all the given conditions, so we have at least one. And then there is one more number to check, so let me write that here in the remaining space. So then we have p minus 1 can also be 2 to the power 7, 128, and in that case p would be equal to 129, and 129 can be factored as 3, because the sum of the digits is divisible by 3, times 43. And 43 is prime, so we have here two prime factors, two prime divisors. So 513 is not a possibility, 129 is not a possibility, so there is just one choice, 257, with just one number having this property, so the answer is B.

three-digit prime

prime divisor

power of 2

prime number

257