Difference between revisions of "2002 AMC 10P Problems/Problem 12"
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<math>\text{I. } (f_{11}(a)f_{13}(a))^{14}</math> | <math>\text{I. } (f_{11}(a)f_{13}(a))^{14}</math> | ||
| − | (f_{11}(a)f_{13}(a))^{14} | + | <math>(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}</math> | |
| + | |||
| + | <math>\text{II. } f_{11}(a)f_{13}(a)f_{14}(a)</math> | ||
| + | |||
| + | <math>f_{11}(a)f_{13}(a)f_{14}(a) | ||
| + | =a^{11}a^{13}a^{14} | ||
| + | =a^{38} | ||
| + | \neq a^{2002}</math> | ||
| + | |||
| + | <math>\text{III. } (f_{11}(f_{13}(a)))^{14}</math> | ||
| + | |||
| + | <math>(f_{11}(f_{13}(a)))^{14} | ||
| + | =((a^{13})^{11})^{14} | ||
| + | =a^{13 \cdot 11 \cdot 14} | ||
| + | =a^{2002}</math> | ||
| + | |||
| + | <math>\text{IV. } f_{11}(f_{13}(f_{14}(a)))</math> | ||
| + | |||
| + | <math>f_{11}(f_{13}(f_{14}(a))) | ||
| + | =((a^{14})^{13})^{11} | ||
| + | =a^{14 \cdot 13 \cdot 11} | ||
| + | =a^{2002}</math> | ||
| + | |||
| + | Thus, our answer is <math>\boxed{\textbf{(C) }\text{ III and IV only}}.</math> | ||
| + | |||
| + | == Solution 2 == | ||
| + | This is the much more realistic and less-time-consuming approach. | ||
| + | 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: | ||
| + | |||
| + | <math>\text{II. } f_{11}(a)f_{13}(a)f_{14}(a)</math> | ||
| + | |||
| + | <math>f_{11}(a)f_{13}(a)f_{14}(a) | ||
| + | =a^{11}a^{13}a^{14} | ||
| + | =a^{38} | ||
| + | \neq a^{2002}</math> | ||
| + | |||
| + | 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 04:57, 15 July 2024
Contents
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 
Thus, our answer is 
Solution 2
This is the much more realistic and less-time-consuming approach.
Notice that all answer choices except 
include 
in them. Therefore, it is sufficient to prove that is false. Similar to solution 1, a quick glance tells us:
Therefore, by process of elimination, our answer is
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
