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

m (→‎Solution)
m (→‎Solution 1)
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Case 1: <math>n-1 = 1</math>
 
Case 1: <math>n-1 = 1</math>
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It is clear that <math>n = 2</math>. However, <math>2^2 - 7(2) + 13 = -11</math>, so this case does not yield any solutions.
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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>
 
Case 2: <math>n^2 - 7n + 13 = 1</math>
Line 28: Line 28:
 
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.
  
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Putting everything together, the answer is <math>\boxed{\text{(B) 2}}</math>.
+
Putting everything together, the answer is <math>1 + 2 = \boxed{\text{(C) 3}}</math>.
  
 
== See also ==
 
== See also ==
 
{{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}}

Revision as of 19:10, 10 March 2024

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 nonnegative. 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 $1 + 2 = \boxed{\text{(C) 3}}$.

See also

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

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