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 12:08, 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 |
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