2002 AMC 10B Problem 15

Below is the professionally curated solution for Problem 15 of the 2002 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10B solutions, or check the answer key.

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Concepts:primeparity

Difficulty rating: 1480

15.

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

Solution:

The numbers ABA - B and A+BA + B differ by 2B,2B, so they have the same parity. Being prime, they must both be odd, which forces AA and BB to have opposite parity.

Since 22 is the only even prime, B=2.B = 2. Then A2,A - 2, A,A, A+2A + 2 are three primes forming an arithmetic progression of odd numbers, which must be 3,5,7.3, 5, 7.

The four primes are 2,3,5,7,2, 3, 5, 7, and their sum is 17,17, which is prime.

Thus, the correct answer is E.

Problem 15 in Other Years