Difference between revisions of "2016 AMC 12B Problems/Problem 3"
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==Solution== | ==Solution== | ||
| + | First of all, lets plug in all of the <math>x</math>'s into the equation. | ||
| + | |||
| + | <math>\bigg|</math> <math>||-2016|-(-2016)|-|-2016|</math> <math>\bigg|</math> <math>-(-2016)</math> | ||
| + | |||
| + | Then we simplify to get | ||
| + | |||
| + | <math>\bigg|</math> <math>|2016+2016|-2016</math> <math>\bigg|</math> <math>+2016</math> | ||
| + | |||
| + | which simplifies into | ||
| + | |||
| + | <math>\bigg|</math> <math>2016</math> <math>\bigg|</math> <math>+2016</math> | ||
| + | |||
| + | and finally we get <math>\boxed{\textbf{(D)}\ 4032}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=2|num-a=4}} | {{AMC12 box|year=2016|ab=B|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:08, 21 February 2016
Problem
Let
. What is the value of
?
Solution
First of all, lets plug in all of the
's into the equation.
Then we simplify to get
which simplifies into
and finally we get
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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.