Given distinct primes $a, b, c$ such that $a+bc$ is odd, which of the following expressions is always even?

Question

Suppose that $a, b, c$ are distinct primes such that $a+bc$ is an odd integer. Which of the following expressions is always even?

1. $(a-c)b$
2. $ab-c$
3. $abc$
4. $(a+b)c$
5. $ac+b$

By definition, there's only one prime that is even, which is $2$. So if $a+bc$ is odd, then either $a$ or $bc$ is odd and not both. If $a = 2$, then $2+bc$ is odd if and only if $bc$ is odd which is true since $b, c$ are primes. If $a\neq 2$, then $bc$ is even and so either $b = 2$ or $c = 2$. From which we conclude that one of the primes $a, b, c$ is $2$, and that $abc$ is always even.

Answer 1

Your answer is correct. It might be worth checking that none of the other options would be correct. As you say, exactly one of $a,b,c$ is $2$ and the others are odd. Therefore:

1. $(a-c)b$ is odd if $a=2$ or $c=2$;
2. $ab-c$ is always odd, because $2$ is a factor of either the first term or the second term but not both;
3. $(a+b)c$ is odd if $a=2$ or $b=2$;
4. $ac+b$ is always odd for the same reason as 2.