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

(Solution 1)
(Solution 1)
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&= a^{336} \\
 
&= a^{336} \\
 
&\neq a^{2002} \\
 
&\neq a^{2002} \\
 +
\end{align*}
  
 
== 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}}

Revision as of 04:38, 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 $a^{2002}.$

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

\begin{align*} (f_{11}(a)f_{13}(a))^{14} \\ &= (a^{11}a^{13})^{14} \\ &= (a^{24})^14 \\ &= a^{336} \\ &\neq a^{2002} \\ \end{align*}

See also

2002 AMC 10P (ProblemsAnswer KeyResources)
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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