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Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 3"

 
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3. 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>.
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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>.
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==Solution==
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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 14:53, 3 April 2012

Let $\triangle ABC$ have $BC=\sqrt{7}$, $CA=1$, and $AB=3$. If $\angle A=\frac{\pi}{n}$ where $n$ is an integer, find the remainder when $n^{2007}$ is divided by $1000$.

Solution

By the Law of Cosines, $\cos A = \frac{3^2 + 1^2 - \sqrt{7}^2}{2\cdot3\cdot1} = \frac12$. Since $A$ is an angle in a triangle the only possibility is $A = \frac{\pi}{3}$. Since $\gcd(3, 1000) = 1$ we may apply Euler's totient theorem: $\phi(1000) = 400$ so $3^{400} \equiv 1 \pmod{1000}$ and so $3^{2000}\equiv 1 \pmod{1000}$ and so $3^{2007} \equiv 3^7 \equiv 2187 \equiv 187 \pmod{1000}$

So the answer is $187$

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