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 and are all prime numbers. The sum of these four primes is
Solution
Solution1
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 .
However in either case, one of and can be expressed as which is a multiple of 3. Therefore the only possibility that works is when and
Which is a prime number.
~ 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.
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.