Problem Solving

If the integer x is greater than 1, does x = 2?
A) x is evenly divisible by exactly two positive integers.
B) The sum of any two distinct positive factors of x is odd.


The Answer I had was C
but shouldn't it be B

The only option possible for B is 2

I got the answer as B. Let me explain:

We are given an integer x, which is greater than one. So x can either be 2,3,4,5,6....etc.

1) x is evenly divisible by exactly 2 positive integers.
Since everything is divisible by 1, this statement simply tells us that x is a prime. (Only prime numbers have exactly two positive factors). But there are infinite number of primes (2,3,5,7.....etc). Hence we can't determine what x is from here. Insufficient.

2) We are told that the sum of "any" two distinct positive factors is odd.
Since 1 is a factor of every integer, let's check the case for the first 3 numbers greater than 1.

a. x=2, in this case we only have 2 factors, 1 and 2 and their sum is always odd. A contender.
b. x=3, in this case we have only 2 factors, 1 and 3, but their sum is not odd. Not a contender.
c. x=4, 1,2,4 as factors. Since 4+2 is not odd. This is not a contender.

Since all primes other than 2 are odd:
a. None of the odd primes can be x (because sum of 1 + odd is even).
b. None of the odd composite numbers from odd primes can be a contender since the prime itself and the number are always factors and odd + odd = even.
c. None of the composite numbers which are even can be a contender, since it will always have 2 and the even number itself as a factor, and 2+even = even.

Hence the only option left is x=2, and B is the correct answer.

Let me know if this helps