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2002 AMC 12B Problems/Problem 11

Revision as of 20:15, 18 February 2021 by Andrew maybe (talk | contribs) (→‎See also)
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

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

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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