Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 AMC 12P Problems/Problem 12"

(→‎See also)
(→‎Solution 1)
 
(2 intermediate revisions by the same user not shown)
Line 21: Line 21:
 
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>
 
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 nonnegative. 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.
+
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>
 
Case 1: <math>n-1 = 1</math>
Line 30: Line 30:
 
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.
 
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{\text{(C) 3}}</math>.
+
Putting everything together, the answer is <math>1 + 2 = \boxed{\textbf{(C) }3}</math>.
  
 
== See also ==
 
== See also ==

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 $n$ is $n^3 - 8n^2 + 20n - 13$ a prime number?

$\text{(A) 1} \qquad \text{(B) 2} \qquad \text{(C) 3} \qquad \text{(D) 4} \qquad \text{(E) more than 4}$

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 $n^3 - 8n^2 + 20n - 13$ will be among $1, 13, -1$, and $-13$. Evaluating the cubic at these values will give $n = 1$ as a root. Doing some synthetic division gives \[n^3 - 8n^2 + 20n - 13 = (n-1)(n^2 - 7n + 13)\]

Since $n > 0$, $n-1$ must be positive. Since $(n-1)(n^2 - 7n + 13)$ evaluates to a prime, it is clear that exactly one of $n-1$ and $n^2 - 7n - 13$ is $1$. We proceed by splitting the problem into 2 cases.

Case 1: $n-1 = 1$ It is clear that $n = 2$. We have $2^2 - 7(2) + 13 = 3$, so this case yields $n = 2$ as a solution.

Case 2: $n^2 - 7n + 13 = 1$ Solving for $n$ gives $n^2 - 7n + 12 = 0$ or $(n-3)(n-4) = 0$. Therefore, $n = 3$ or $n = 4$. Since both $3-1 = 2$ and $4-1 = 3$ are prime, both $n = 3$ and $n = 4$ work, yielding 2 solutions.

Putting everything together, the answer is $1 + 2 = \boxed{\textbf{(C) }3}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12P_Problems/Problem_12&oldid=223676"