2002 AMC 10P Problems/Problem 12

Revision as of 05:52, 15 July 2024 by Wes (talk | contribs) (Solution 1)

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 $\box{\textbf{(C) }\text{ III and IV only}}}.$ (Error compiling LaTeX. Unknown error_msg)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png