2002 AMC 10P Problems/Problem 12
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.