# 2002 AMC 12B Problems/Problem 11

The following problem is from both the 2002 AMC 12B #11 and 2002 AMC 10B #15, so both problems redirect to this page.

## Problem

The positive integers $A, B, A-B,$ and $A+B$ are all prime numbers. The sum of these four primes is $\mathrm{(A)}\ \mathrm{even} \qquad\mathrm{(B)}\ \mathrm{divisible\ by\ }3 \qquad\mathrm{(C)}\ \mathrm{divisible\ by\ }5 \qquad\mathrm{(D)}\ \mathrm{divisible\ by\ }7 \qquad\mathrm{(E)}\ \mathrm{prime}$

## Solution

### Solution1

Since $A-B$ and $A+B$ must have the same parity, and since there is only one even prime number, it follows that $A-B$ and $A+B$ are both odd. Thus one of $A, B$ is odd and the other even. Since $A+B > A > A-B > 2$, it follows that $A$ (as a prime greater than $2$) is odd. Thus $B = 2$, and $A-2, A, A+2$ are consecutive odd primes. At least one of $A-2, A, A+2$ is divisible by $3$, from which it follows that $A-2 = 3$ and $A = 5$. The sum of these numbers is thus $17$, a prime, so the answer is $\boxed{\mathrm{(E)}\ \text{prime}}$.

### Solution 2

In order for both $A - B$ and $A + B$ to be prime, one of $A, B$ must be 2, or else both $A - B$, $A + B$ would be even numbers.

If $A = 2$, then $A < B$ and $A - B < 0$, which is not possible. Thus $B = 2$.

Since $A$ is prime and $A > A - B > 2$, we can infer that $A > 3$ and thus $A$ can be expressed as $6n \pm 1$ for some natural number $n$.

However in either case, one of $A - B$ and $A + B$ can be expressed as $6n \pm 3 = 3(2n \pm 1)$ which is a multiple of 3. Therefore the only possibility that works is when $A - B = 3$ and $$A + B + (A - B) + (A + B) = 5 + 2 + 3 + 7 = 17$$

Which is a prime number. $\boxed{(E)}$

~ Nafer

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 