Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 3"
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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 </math> | + | 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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Revision as of 16:08, 21 August 2006
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