2002 AMC 12B Problem 11

Attempt Problem 11 of the 2002 AMC 12B below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

The positive integers A,A, B,B, A−B,A-B, and A+BA+B are all prime numbers. The sum of these four primes is
even
divisible by 33
divisible by 55
divisible by 77
prime
Answer: E
Concepts:parityprime
Small hint:
A−BA-B and A+BA+B have the same parity, and both are prime, so both are odd
Big hint:
Then B=2,B=2, making A−2,A-2, A,A, A+2A+2 three primes in arithmetic progression
Solution:
A−BA-B and A+BA+B have the same parity; being prime, both are odd, so AA and BB have opposite parity. If AA were even, then the prime AA would equal 2,2, but the positive prime BB would satisfy B≥2B\ge2 and make A−B≤0.A-B\le0. Hence AA is odd and the even prime BB is 2.2.
Then A−2,A-2, A,A, A+2A+2 are three primes. One is divisible by 3,3, so that one must equal 3;3; the triple is 3,3, 5,5, 7.7. Their sum together with 22 is 2+3+5+7=17,2+3+5+7=17, a prime.
Thus, the correct answer is E.
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