2002 AMC 12B Problem 11

Below is the professionally curated solution for Problem 11 of the 2002 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12B solutions, or check the answer key.

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Concepts:parity prime

Difficulty rating: 1430

11.

The positive integers $A,$ $B,$ $A-B,$ and $A+B$ are all prime numbers. The sum of these four primes is

even

divisible by $3$

divisible by $5$

divisible by $7$

prime

Solution:

$A-B$ and $A+B$ have the same parity; being prime, both are odd, so one of $A,B$ is even. Since $A$ lies between the two odd primes $A-B$ and $A+B,$ it is the odd one, forcing $B=2.$

Then $A-2,$ $A,$ $A+2$ are three primes, which must be $3,$ $5,$ $7.$ Their sum together with $2$ is $2+3+5+7=17,$ a prime.

Thus, the correct answer is E.

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