2002 AMC 10B Problem 15

Try small examples first. $B = 2$, $A = 5$ gives primes $5, 2, 3, 7$, which works. The steps below show it is the only choice, so the sum is fixed.

1. $A - B$ and $A + B$ are both odd. They differ by $2B$, an even number, so they are both even or both odd. The only even prime is $2$, and they cannot both be $2$ (one is bigger). So both are odd.
2. $B = 2$. $A + B$ is odd, so one of $A$, $B$ is even and the other odd. The even one is a prime, so it is $2$. It cannot be $A$: then $A - B$ would be less than $2$ and could not be prime. So $B = 2$.
3. $A = 5$. Now $A - 2$, $A$, $A + 2$ are all prime. These are three odd numbers in a row, like $3, 5, 7$ or $9, 11, 13$. In any such run, one of the three is a multiple of $3$ (the run jumps by $2$, so it hits all three possible remainders when dividing by $3$). The only prime that is a multiple of $3$ is $3$ itself.
- If $A - 2 = 3$, then $A = 5$, and $3, 5, 7$ are all prime.
- If $A = 3$, then $A - 2 = 1$, not prime.
- If $A + 2 = 3$, then $A = 1$, not prime.
4. So $A = 5$, $B = 2$, and the four primes are $5, 2, 3, 7$. Their sum is $17$, which is prime.

The answer is $\boxed{\textbf{(E)}\ \text{prime}}$.

When primes are added or subtracted, two quick tests do most of the work. Even and odd: if a sum or difference of primes must be odd, one of them is $2$. Multiples of $3$: three numbers spaced $2$ apart always include a multiple of $3$, which must be $3$ itself if they are all prime. Check $2$ and $3$ first whenever primes form a pattern.

In the contest you can just try examples. $B = 2$ is the natural guess, and $A = 5$ works immediately. The sum $17$ is odd and not divisible by $3$, $5$ or $7$, so only (E) fits. The problem promises the answer does not depend on which primes you pick, so one working example is enough to choose.