Difference between revisions of "2016 AMC 10B Problems/Problem 3"
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becomes <math>|4032-2016|+2016=2016+2016=4032</math> which is <math>\textbf{(D)}</math>. | becomes <math>|4032-2016|+2016=2016+2016=4032</math> which is <math>\textbf{(D)}</math>. | ||
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| + | ==Solution 2== | ||
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| + | Solution by e_power_pi_times_i | ||
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| + | Substitute <math>-y = x = -2016</math> into the equation. Now, it is <math>\Bigg\vert\Big\vert |y|+y\Big\vert-|y|\Bigg\vert+y</math>. Since <math>y = 2016</math>, it is a positive number, so <math>|y| = y</math>. Now the equation is <math>\Bigg\vert\Big\vert y+y\Big\vert-y\Bigg\vert+y</math>. This further simplifies to <math>2y-y+y = 2y</math>, so the answer is <math>\textbf{(D)} 4032</math> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=2|num-a=4}} | {{AMC10 box|year=2016|ab=B|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:12, 21 June 2016
Contents
Problem
Let 
. What is the value of 
?
Solution
Substituting carefully, 
becomes 
which is 
.
Solution 2
Solution by e_power_pi_times_i
Substitute 
into the equation. Now, it is 
. Since 
, it is a positive number, so 
. Now the equation is 
. This further simplifies to 
, so the answer is 
See Also
| 2016 AMC 10B (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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
