Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 3"
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− | + | Let <math>\triangle ABC</math> have <math>BC=\sqrt{7}</math>, <math>CA=1</math>, and <math>AB=3</math>. If <math>\angle A=\frac{\pi}{n}</math> where <math>n</math> is an integer, find the remainder when <math>n^{2007}</math> is divided by <math>1000</math>. | |
− | [[Mock AIME 1 2006-2007]] | + | ==Solution== |
+ | By the [[Law of Cosines]], <math>\cos A = \frac{3^2 + 1^2 - \sqrt{7}^2}{2\cdot3\cdot1} = \frac12</math>. Since <math>A</math> is an [[angle]] in a [[triangle]] the only possibility is <math>A = \frac{\pi}{3}</math>. Since <math>\gcd(3, 1000) = 1</math> we may apply [[Euler's totient theorem]]: <math>\phi(1000) = 400</math> so <math>3^{400} \equiv 1 \pmod{1000}</math> and so <math>3^{2000}\equiv 1 \pmod{1000}</math> and so <math>3^{2007} \equiv 3^7 \equiv 2187 \equiv 187 \pmod{1000}</math> | ||
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+ | So the answer is <math>187</math> | ||
+ | ---- | ||
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+ | *[[Mock AIME 1 2006-2007 Problems/Problem 2 | Previous Problem]] | ||
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+ | *[[Mock AIME 1 2006-2007 Problems/Problem 4 | Next Problem]] | ||
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+ | *[[Mock AIME 1 2006-2007]] |
Latest revision as of 15:53, 3 April 2012
Let have
,
, and
. If
where
is an integer, find the remainder when
is divided by
.
Solution
By the Law of Cosines, . Since
is an angle in a triangle the only possibility is
. Since
we may apply Euler's totient theorem:
so
and so
and so
So the answer is