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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 ==
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How many positive [[integer]]s <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer?
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For how many positive integers <math>n</math> is <math>n^3 - 8n^2 + 20n - 13</math> a prime number?
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<math>
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\text{(A) 1}
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\qquad
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\text{(B) 2}
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\qquad
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\text{(C) 3}
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\qquad
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\text{(D) 4}
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\qquad
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\text{(E) more than 4}
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</math>
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== Solution 1 ==
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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.
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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>
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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.
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Case 1: <math>n-1 = 1</math>
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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.
  
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<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 }  </math>
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Case 2: <math>n^2 - 7n + 13 = 1</math>
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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>.
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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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== Solution ==
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Putting everything together, the answer is <math>1 + 2 = \boxed{\textbf{(C) }3}</math>.
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If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
 
  
 
== See also ==
 
== See also ==
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{{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 $n$ 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

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