Difference between revisions of "2002 AMC 12B Problems/Problem 11"
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| + | {{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #11]] and [[2002 AMC 10B Problems|2002 AMC 10B #15]]}} | ||
== Problem == | == Problem == | ||
The positive integers <math>A, B, A-B, </math> and <math>A+B</math> are all prime numbers. The sum of these four primes is | The positive integers <math>A, B, A-B, </math> and <math>A+B</math> are all prime numbers. The sum of these four primes is | ||
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== Solution == | == Solution == | ||
| − | 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>, | + | 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>. |
== See also == | == See also == | ||
| + | {{AMC10 box|year=2002|ab=B|num-b=14|num-a=16}} | ||
{{AMC12 box|year=2002|ab=B|num-b=10|num-a=12}} | {{AMC12 box|year=2002|ab=B|num-b=10|num-a=12}} | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
Revision as of 17:14, 28 July 2011
- 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 
and 
are all prime numbers. The sum of these four primes is
Solution
Since 
and must have the same parity, and since there is only one even prime number, it follows that and are both odd. Thus one of 
is odd and the other even. Since 
, it follows that 
(as a prime greater than 
) is odd. Thus 
, and 
are consecutive odd primes. At least one of is divisible by 
, from which it follows that 
and 
. The sum of these numbers is thus 
, a prime, so the answer is 
.
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 | |
