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Difference between revisions of "2002 AMC 12B Problems/Problem 11"

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\qquad\mathrm{(E)}\ \mathrm{prime}</math>
 
\qquad\mathrm{(E)}\ \mathrm{prime}</math>
  
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== Solution ==
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== Solution 1 ==
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=== Solution1 1 ===
 
  
 
Since <math>A-B</math> and <math>A+B</math> must have the same [[parity]], and since there is only one even prime number, it follows that <math>A-B</math> and <math>A+B</math> are both odd. Thus one of <math>A, B</math> is odd and the other even. Since <math>A+B > A > A-B > 2</math>, it follows that <math>A</math> (as a prime greater than <math>2</math>) is odd. Thus <math>B = 2</math>, and <math>A-2, A, A+2</math> are consecutive odd primes. At least one of <math>A-2, A, A+2</math> is divisible by <math>3</math>, from which it follows that <math>A-2 = 3</math> and <math>A = 5</math>. The sum of these numbers is thus <math>17</math>, a prime, so the answer is <math>\boxed{\mathrm{(E)}\ \text{prime}}</math>.
 
Since <math>A-B</math> and <math>A+B</math> must have the same [[parity]], and since there is only one even prime number, it follows that <math>A-B</math> and <math>A+B</math> are both odd. Thus one of <math>A, B</math> is odd and the other even. Since <math>A+B > A > A-B > 2</math>, it follows that <math>A</math> (as a prime greater than <math>2</math>) is odd. Thus <math>B = 2</math>, and <math>A-2, A, A+2</math> are consecutive odd primes. At least one of <math>A-2, A, A+2</math> is divisible by <math>3</math>, from which it follows that <math>A-2 = 3</math> and <math>A = 5</math>. The sum of these numbers is thus <math>17</math>, a prime, so the answer is <math>\boxed{\mathrm{(E)}\ \text{prime}}</math>.
  
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=== Solution 2 ===
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== Solution 2 ==
  
 
In order for both <math>A - B</math> and <math>A + B</math> to be prime, one of <math>A, B</math> must be 2, or else both <math>A - B</math>, <math>A + B</math> would be even numbers.  
 
In order for both <math>A - B</math> and <math>A + B</math> to be prime, one of <math>A, B</math> must be 2, or else both <math>A - B</math>, <math>A + B</math> would be even numbers.  
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If <math>A = 2</math>, then <math>A < B</math> and <math>A - B < 0</math>, which is not possible. Thus <math>B = 2</math>.
 
If <math>A = 2</math>, then <math>A < B</math> and <math>A - B < 0</math>, which is not possible. Thus <math>B = 2</math>.
  
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Since <math>A</math> is prime, then for all <math>A > 3</math>, <math>A</math> can be expressed as <math>6n /pm 1</math> for some natural number <math>n</math>.
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Since <math>A</math> is prime and <math>A > A - B > 2</math>, we can infer that <math>A > 3</math> and thus <math>A</math> can be expressed as <math>6n \pm 1</math> for some natural number <math>n</math>.
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However in either case, one of <math>A - B</math> and <math>A + B</math> can be expressed as <math>6n \pm 3 = 3(2n \pm 1)</math> which is a multiple of 3. Therefore the only possibility that works is when <math>A - B = 3</math> and <cmath>A + B + (A - B) + (A + B) = 5 + 2 + 3 + 7 = 17</cmath>
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Which is a prime number. <math>\boxed{(E)}</math>
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~ Nafer
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== Solution 3 (intuitive)==
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Trying out some primes for <math>A</math> and <math>B</math> such that <math>A-B</math> and <math>A+B</math> are prime, <math>A=5</math> and <math>B=2</math> can be found almost immediately. Summing the four primes, the result is <math>5+2+3+7=17</math>, which is <math>\boxed{\mathrm{(E)}\ \text{prime}}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}
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Note:
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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.

Latest revision as of 10:57, 10 August 2022

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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