Difference between revisions of "1996 AIME Problems/Problem 10"
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So <math>19x = 141 +180n</math>, for some integer <math>n</math>. | So <math>19x = 141 +180n</math>, for some integer <math>n</math>. | ||
− | + | Multiplying by <math>19</math> gives <math>x \equiv 141 \cdot 19 \equiv 2679 \equiv 159 \pmod{180}</math>. | |
+ | The smallest positive solution of this is <math>x = \boxed{159}</math> | ||
== See also == | == See also == |
Revision as of 04:39, 18 August 2015
Contents
Problem
Find the smallest positive integer solution to .
Solution
.
The period of the tangent function is , and the tangent function is one-to-one over each period of its domain.
Thus, .
Since , multiplying both sides by yields .
Therefore, the smallest positive solution is .
Solution 2
which is the same as i.e.
So , for some integer . Multiplying by gives . The smallest positive solution of this is
See also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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