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Difference between revisions of "2002 AMC 10P Problems/Problem 12"

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== Problem 12 ==
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For <math>f_n(x)=x^n</math> and <math>a \neq 1</math> consider
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<math>\text{I. } (f_{11}(a)f_{13}(a))^{14}</math>
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<math>\text{II. } f_{11}(a)f_{13}(a)f_{14}(a)</math>
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<math>\text{III. } (f_{11}(f_{13}(a)))^{14}</math>
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<math>\text{IV. } f_{11}(f_{13}(f_{14}(a)))</math>
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Which of these equal <math>f_{2002}(a)?</math>
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<math>
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\text{(A) I and II only}
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\qquad
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\text{(B) II and III only}
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\qquad
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\text{(C) III and IV only}
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\qquad
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\text{(D) II, III, and IV only}
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\qquad
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\text{(E) all of them}
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</math>
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== Solution 1==
 
== Solution 1==
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We can solve this problem with a case by case check of <math>\text{I., II., III.,}</math> and <math>\text{IV.}</math> Since <math>f_n=x^n,</math> <math>f_{2002}(a)=a^{2002},</math> all cases must equal <math>a^{2002}.</math>
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<math>\text{I. } (f_{11}(a)f_{13}(a))^{14}</math>
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<math>(f_{11}(a)f_{13}(a))^{14}
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=(a^{11}a^{13})^{14}
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= (a^{24})^14
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= a^{336}
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\neq a^{2002}</math>
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<math>\text{II. } f_{11}(a)f_{13}(a)f_{14}(a)</math>
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<math>f_{11}(a)f_{13}(a)f_{14}(a)
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=a^{11}a^{13}a^{14}
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=a^{38}
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\neq a^{2002}</math>
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<math>\text{III. } (f_{11}(f_{13}(a)))^{14}</math>
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<math>(f_{11}(f_{13}(a)))^{14}
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=((a^{13})^{11})^{14}
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=a^{13 \cdot 11 \cdot 14}
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=a^{2002}</math>
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<math>\text{IV. } f_{11}(f_{13}(f_{14}(a)))</math>
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<math>f_{11}(f_{13}(f_{14}(a)))
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=((a^{14})^{13})^{11}
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=a^{14 \cdot 13 \cdot 11}
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=a^{2002}</math>
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Thus, our answer is <math>\boxed{\textbf{(C) }\text{ III and IV only}}.</math>
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== Solution 2 ==
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This is the much more realistic and less-time-consuming approach.
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Notice that all answer choices except <math>\text{(C)}</math> include <math>\text{II}.</math> in them. Therefore, it is sufficient to prove that <math>\text{II}.</math> is false. Similar to solution 1, a quick glance tells us:
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<math>\text{II. } f_{11}(a)f_{13}(a)f_{14}(a)</math>
  
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<math>f_{11}(a)f_{13}(a)f_{14}(a)
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=a^{11}a^{13}a^{14}
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=a^{38}
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\neq a^{2002}</math>
  
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Therefore, by process of elimination, our answer is <math>\boxed{\textbf{(C) }\text{ III and IV only}}.</math>
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2002|ab=P|num-b=11|num-a=13}}
 
{{AMC10 box|year=2002|ab=P|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 05:57, 15 July 2024

Problem 12

For $f_n(x)=x^n$ and $a \neq 1$ consider

$\text{I. } (f_{11}(a)f_{13}(a))^{14}$

$\text{II. } f_{11}(a)f_{13}(a)f_{14}(a)$

$\text{III. } (f_{11}(f_{13}(a)))^{14}$

$\text{IV. } f_{11}(f_{13}(f_{14}(a)))$

Which of these equal $f_{2002}(a)?$

$\text{(A) I and II only} \qquad \text{(B) II and III only} \qquad \text{(C) III and IV only} \qquad \text{(D) II, III, and IV only} \qquad \text{(E) all of them}$

Solution 1

We can solve this problem with a case by case check of $\text{I., II., III.,}$ and $\text{IV.}$ Since $f_n=x^n,$ $f_{2002}(a)=a^{2002},$ all cases must equal $a^{2002}.$

$\text{I. } (f_{11}(a)f_{13}(a))^{14}$

$(f_{11}(a)f_{13}(a))^{14} =(a^{11}a^{13})^{14} = (a^{24})^14 = a^{336} \neq a^{2002}$

$\text{II. } f_{11}(a)f_{13}(a)f_{14}(a)$

$f_{11}(a)f_{13}(a)f_{14}(a) =a^{11}a^{13}a^{14} =a^{38} \neq a^{2002}$

$\text{III. } (f_{11}(f_{13}(a)))^{14}$

$(f_{11}(f_{13}(a)))^{14} =((a^{13})^{11})^{14} =a^{13 \cdot 11 \cdot 14} =a^{2002}$

$\text{IV. } f_{11}(f_{13}(f_{14}(a)))$

$f_{11}(f_{13}(f_{14}(a))) =((a^{14})^{13})^{11} =a^{14 \cdot 13 \cdot 11} =a^{2002}$

Thus, our answer is $\boxed{\textbf{(C) }\text{ III and IV only}}.$

Solution 2

This is the much more realistic and less-time-consuming approach. Notice that all answer choices except $\text{(C)}$ include $\text{II}.$ in them. Therefore, it is sufficient to prove that $\text{II}.$ is false. Similar to solution 1, a quick glance tells us:

$\text{II. } f_{11}(a)f_{13}(a)f_{14}(a)$

$f_{11}(a)f_{13}(a)f_{14}(a) =a^{11}a^{13}a^{14} =a^{38} \neq a^{2002}$

Therefore, by process of elimination, our answer is $\boxed{\textbf{(C) }\text{ III and IV only}}.$

See also

2002 AMC 10P (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 10 Problems and Solutions

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