2002 AMC 12P Problems/Problem 12
- The following problem is from both the 2002 AMC 12P #12 and 2002 AMC 10P #18, so both problems redirect to this page.
Problem
For how many positive integers 
is 
a prime number?
Solution 1
Since this is a number theory question, it is clear that the main challenge here is factoring the given cubic. In general, the rational root theorem will be very useful for these situations.
The rational root theorem states that all rational roots of will be among 
, and 
. Evaluating the cubic at these values will give 
as a root. Doing some synthetic division gives ![\[n^3 - 8n^2 + 20n - 13 = (n-1)(n^2 - 7n + 13)\]](http://latex.artofproblemsolving.com/4/7/b/47b27e3de3c19dab158a8f6b76725030f0ef6639.png)
Since 
, 
must be positive. Since 
evaluates to a prime, it is clear that exactly one of and 
is 
. We proceed by splitting the problem into 2 cases.
Case 1: 
It is clear that 
. We have 
, so this case yields as a solution.
Case 2: 
Solving for gives 
or 
. Therefore, 
or 
.
Since both 
and 
are prime, both and work, yielding 2 solutions.
Putting everything together, the answer is 
.
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 12P (Problems • Answer Key • Resources) | |
| Preceded by Problem 11 |
Followed by Problem 13 |
| 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. 
