Difference between revisions of "2002 AMC 10A Problems/Problem 2"
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Alternate solution for the lazy: Without computing the answer exactly, we see that <math>2/12=\text{a little}</math>, <math>12/9=\text{more than }1</math>, and <math>9/2=4.5</math>. | Alternate solution for the lazy: Without computing the answer exactly, we see that <math>2/12=\text{a little}</math>, <math>12/9=\text{more than }1</math>, and <math>9/2=4.5</math>. | ||
− | The sum is <math>4.5 + (\text{more than }1) + (\text{a little}) = (\text{more than }5.5) + (\text{a little})</math>, and as all the options are integers, the correct one is | + | The sum is <math>4.5 + (\text{more than }1) + (\text{a little}) = (\text{more than }5.5) + (\text{a little})</math>, and as all the options are integers, the correct one is <math>6</math>. |
==See Also== | ==See Also== |
Revision as of 18:07, 21 November 2018
Problem
Given that a, b, and c are non-zero real numbers, define , find
.
Solution
. Our answer is then
.
Alternate solution for the lazy: Without computing the answer exactly, we see that ,
, and
.
The sum is
, and as all the options are integers, the correct one is
.
See Also
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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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