Skip to main content

2002 AMC 12B Problem 11

Problem 11 of 25IntermediateNumber Theory

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

Answer choices

Show solution

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.

More practice

Concepts: parity · prime

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.