Difference between revisions of "2002 AMC 10P Problems/Problem 12"
(→Solution 2) |
(→Solution 2) |
||
Line 61: | Line 61: | ||
== Solution 2 == | == Solution 2 == | ||
This is the much more realistic and less-time-consuming approach. | 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: | + | 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>\text{II. } f_{11}(a)f_{13}(a)f_{14}(a)</math> |
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.