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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 1

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

Solution 3 (intuitive)

Trying out some primes for $A$ and $B$ such that $A-B$ and $A+B$ are prime, $A=5$ and $B=2$ can be found almost immediately. Summing the four primes, the result is $5+2+3+7=17$, which is $\boxed{\mathrm{(E)}\ \text{prime}}$.

See also

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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

Note: Simple trail and error gives us the primes 5 and 2 which fits the description the question asks for; 5, 2, 3, 7 are all primes.

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