Difference between revisions of "2002 AMC 12B Problems/Problem 11"
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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>. | ||
| − | Since <math>A</math> is prime and <math>A > A - B > 2</math>, <math>A > 3</math> and <math>A</math> can be expressed as <math>6n \pm 1</math> for some natural number <math>n</math>. Moreover for all <math>A > 5</math>, <math>A + B</math>, <math>A - B</math> | + | 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>. Moreover for all <math>A > 5</math>, <math>A + B</math>, <math>A - B</math> |
== See also == | == See also == | ||
Revision as of 13:55, 2 July 2019
- 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
Solution1 1
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 
.
Solution 2
In order for both 
and 
to be prime, one of must be 2, or else both , would be even numbers.
If 
, then 
and 
, which is not possible. Thus .
Since is prime and 
, we can infer that 
and thus can be expressed as 
for some natural number 
. Moreover for all 
, ,
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. 
