Difference between revisions of "2002 AMC 12P Problems/Problem 12"
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| + | {{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #12]] and [[2002 AMC 10P Problems|2002 AMC 10P #18]]}} | ||
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== Problem == | == Problem == | ||
For how many positive integers <math>n</math> is <math>n^3 - 8n^2 + 20n - 13</math> a prime number? | For how many positive integers <math>n</math> is <math>n^3 - 8n^2 + 20n - 13</math> a prime number? | ||
<math> | <math> | ||
| − | \text{(A) | + | \text{(A) 1} |
\qquad | \qquad | ||
| − | \text{(B) | + | \text{(B) 2} |
\qquad | \qquad | ||
| − | \text{(C) | + | \text{(C) 3} |
\qquad | \qquad | ||
| − | \text{(D) | + | \text{(D) 4} |
\qquad | \qquad | ||
| − | \text{(E) more than | + | \text{(E) more than 4} |
</math> | </math> | ||
| − | == Solution == | + | == 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. | 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 <math>n^3 - 8n^2 + 20n - 13</math> will be among <math>1, 13, -1</math>, and <math>-13</math>. | + | The rational root theorem states that all rational roots of <math>n^3 - 8n^2 + 20n - 13</math> will be among <math>1, 13, -1</math>, and <math>-13</math>. Evaluating the cubic at these values will give <math>n = 1</math> as a root. Doing some synthetic division gives <cmath>n^3 - 8n^2 + 20n - 13 = (n-1)(n^2 - 7n + 13)</cmath> |
| + | |||
| + | Since <math>n > 0</math>, <math>n-1</math> must be positive. Since <math>(n-1)(n^2 - 7n + 13)</math> evaluates to a prime, it is clear that exactly one of <math>n-1</math> and <math>n^2 - 7n - 13</math> is <math>1</math>. We proceed by splitting the problem into 2 cases. | ||
| + | |||
| + | Case 1: <math>n-1 = 1</math> | ||
| + | It is clear that <math>n = 2</math>. We have <math>2^2 - 7(2) + 13 = 3</math>, so this case yields <math>n = 2</math> as a solution. | ||
| + | |||
| + | Case 2: <math>n^2 - 7n + 13 = 1</math> | ||
| + | Solving for <math>n</math> gives <math>n^2 - 7n + 12 = 0</math> or <math>(n-3)(n-4) = 0</math>. Therefore, <math>n = 3</math> or <math>n = 4</math>. | ||
| + | Since both <math>3-1 = 2</math> and <math>4-1 = 3</math> are prime, both <math>n = 3</math> and <math>n = 4</math> work, yielding 2 solutions. | ||
| + | |||
| + | Putting everything together, the answer is <math>1 + 2 = \boxed{\textbf{(C) }3}</math>. | ||
== See also == | == See also == | ||
| + | {{AMC10 box|year=2002|ab=P|num-b=17|num-a=19}} | ||
{{AMC12 box|year=2002|ab=P|num-b=11|num-a=13}} | {{AMC12 box|year=2002|ab=P|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 23:11, 25 July 2024
- 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. 
