2002 AMC 10B Problem 15

The primes $A - B$ and $A + B$ differ by $2B$, an even number, so they have the same parity. They cannot both be $2$ (they are different), so both are odd. An odd difference $A - B$ means $A$ and $B$ have opposite parity, so one of them is the even prime $2$. Since $A > A - B \ge 2$, we have $A > 2$, hence $B = 2$.

Now $A - 2$, $A$, $A + 2$ are three primes, and among any three consecutive odd numbers one is a multiple of $3$. A prime multiple of $3$ is $3$ itself, and $A = 3$ would make $A - 2 = 1$ (not prime), so $A - 2 = 3$ and $A = 5$. The four primes are $5, 2, 3, 7$, all valid.

Their sum is $5 + 2 + 3 + 7 = 17$, a prime.

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

Two standard filters do all the work: parity (any set of primes with an odd sum or difference must include $2$) and residues mod $3$ (three primes spaced by $2$ must include $3$). Together they force a unique configuration, and only then is the sum computed. When primes appear in arithmetic patterns, check $2$ and $3$ first.

Search directly: $B = 2$ is forced by parity, and trying $A = 3, 5, 7, \ldots$ finds $A = 5$ almost immediately ($3, 5, 7$ all prime). The sum $17$ is odd, not divisible by $3$, $5$ or $7$, and prime, so (E) is the only fitting choice.

• Concluding B = 2 and then assuming A can be any odd prime, without using the mod 3 argument that pins A = 5 (and the sum 17).
• Trying $A = 2$, which makes $A - B$ nonpositive for every prime $B$, and wasting time on that dead end.