Difference between revisions of "2002 AMC 12B Problems/Problem 19"
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== Solution 2 (Answer Choices) == | == Solution 2 (Answer Choices) == | ||
| − | Expand the system and subtract equation 3 from equation 2 to get <math>ab-ac=-8</math>, and add it to the first equation to get <math>2ab=144</math> \Longrightarrow <math>ab=72.</math> Since we want <math>abc,</math> it's trivial that 72 is a factor of this. We can quickly see that none of the answer choices have the factor <math>72</math> except choice <math>\boxed | + | Expand the system and subtract equation 3 from equation 2 to get <math>ab-ac=-8</math>, and add it to the first equation to get <math>2ab=144</math> <math>\Longrightarrow</math> <math>ab=72.</math> Since we want <math>abc,</math> it's trivial that 72 is a factor of this. We can quickly see that none of the answer choices have the factor <math>72</math> except choice <math>\boxed{\mathrm{(D)}}.</math> |
~dbnl | ~dbnl | ||
Latest revision as of 00:08, 5 February 2025
Problem
If 
and 
are positive real numbers such that 
and 
, then 
is
Solution
Adding up the three equations gives 
. Subtracting each of the above equations from this yields, respectively, 
. Taking their product, 
.
Solution 2 (Answer Choices)
Expand the system and subtract equation 3 from equation 2 to get 
, and add it to the first equation to get 
Since we want 
it's trivial that 72 is a factor of this. We can quickly see that none of the answer choices have the factor 
except choice 
~dbnl
See also
| 2002 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 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 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
