Aren't there several values for $abc$? I mean, it can't be a definite answer!

Question

This is a problem from the National Mathematics Olympiad:

There are three prime numbers $a$, $b$ and $c$; such that $a>b>c$ and $a-b-c=22$. What is the value of $abc$?

I've got more than one value for $abc$.

• Assuming that $a=31$, $b=7$, $c=2$, then $abc=434$
• Assuming that $a=37$, $b=13$, $c=2$, then $abc=962$
• Assuming that $a=47$, $b=23$, $c=2$, then $abc=2162$

and so on. Thus, I see that $abc$ can have more than one value. Please tell me if I misunderstood something about the problem.

Answer 1

I am curious to hear which kind of math olympiads you are attending.

If $a>b>c>2$ then all of them are odd, so $a-b-c$ should be odd. This implies that $c=2$. Hence, we have to find all primes $a>b>2$ such that $a-b=24$, i.e., all odd primes $b$ such that $b+24$ is another prime. This is an open problem as far as I know.