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.

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11.

The positive integers A,A, B,B, AB,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
Difficulty rating: 1430
Solution:

ABA-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 B2B\ge2 and make AB0.A-B\le0. Hence AA is odd and the even prime BB is 2.2.

Then A2,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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