2002 AMC 12B Problems/Problem 3
- The following problem is from both the 2002 AMC 12B #3 and 2002 AMC 10B #6, so both problems redirect to this page.
Contents
Problem
For how many positive integers is a prime number?
Solution 1
Factoring, we get . Either or is odd, and the other is even. Their product must yield an even number. The only prime that is even is , which is when is or . Since is not a positive number, the answer is .
Solution 2
Considering parity, we see that is always even. The only even prime is , and so whence .
See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 2 |
Followed by Problem 4 |
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.
Usage of the Prime Number 2 Note that two is the only even prime number. So, in order for this expression to result in a prime number, the result must be 2. You can factor the quadratic equation and eventually see that 3 is the only number satisfying this condition. Hence, the answer is B.