Difference between revisions of "2002 AMC 10P Problems/Problem 12"
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== Solution 1== | == Solution 1== | ||
| − | 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}, all cases must equal < | + | 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> |
| − | < | + | <math>\text{I. } (f_{11}(a)f_{13}(a))^{14}</math> |
| − | < | + | <math>\begin{align*} |
(f_{11}(a)f_{13}(a))^{14} \\ | (f_{11}(a)f_{13}(a))^{14} \\ | ||
&= (a^{11}a^{13})^{14} | &= (a^{11}a^{13})^{14} | ||
&= (a^{24})^14 | &= (a^{24})^14 | ||
&= a^{336} | &= a^{336} | ||
| − | &\neq a^{2002} | + | &\neq a^{2002}</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}} | ||
Revision as of 04:37, 15 July 2024
Problem 12
For 
and 
consider
Which of these equal 
Solution 1
We can solve this problem with a case by case check of 
and 
Since 
all cases must equal 
$\begin{align*} (f_{11}(a)f_{13}(a))^{14} \\ &= (a^{11}a^{13})^{14} &= (a^{24})^14 &= a^{336} &\neq a^{2002}$ (Error compiling LaTeX. Unknown error_msg)
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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