Difference between revisions of "2006 AMC 12A Problems/Problem 15"
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<!-- explanation needed-->The smallest possible value of <math>z</math> will be that of <math>\frac{5\pi}{3} - \frac{3\pi}{2} = \frac{\pi}{6} \Rightarrow \mathrm{(A)}</math>. | <!-- explanation needed-->The smallest possible value of <math>z</math> will be that of <math>\frac{5\pi}{3} - \frac{3\pi}{2} = \frac{\pi}{6} \Rightarrow \mathrm{(A)}</math>. | ||
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== See also == | == See also == |
Revision as of 16:54, 12 October 2016
Contents
Problem
Suppose and
. What is the smallest possible positive value of
?
Solutions
Solution 1
- For
, x must be in the form of
, where
denotes any integer.
- For
,
.
The smallest possible value of will be that of
.
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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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