Difference between revisions of "2008 AIME II Problems/Problem 8"
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== Solution == | == Solution == | ||
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+ | ===Solution 1=== | ||
By the [[trigonometric identity|product-to-sum identities]], we have that <math>2\cos a \sin b = \sin (a+b) - \sin (a-b)</math>. Therefore, this reduces to a [[telescope|telescoping series]]: | By the [[trigonometric identity|product-to-sum identities]], we have that <math>2\cos a \sin b = \sin (a+b) - \sin (a-b)</math>. Therefore, this reduces to a [[telescope|telescoping series]]: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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Thus, we need <math>\sin \left(\frac{n(n+1)\pi}{2008}\right)</math> to be an integer; this can be only <math>\{-1,0,1\}</math>, which occur when <math>2 \cdot \frac{n(n+1)}{2008}</math> is an integer. Thus <math>1004 = 2^2 \cdot 251 | n(n+1) \Longrightarrow 251 | n, n+1</math>. It easily follows that <math>n = \boxed{251}</math> is the smallest such integer. | Thus, we need <math>\sin \left(\frac{n(n+1)\pi}{2008}\right)</math> to be an integer; this can be only <math>\{-1,0,1\}</math>, which occur when <math>2 \cdot \frac{n(n+1)}{2008}</math> is an integer. Thus <math>1004 = 2^2 \cdot 251 | n(n+1) \Longrightarrow 251 | n, n+1</math>. It easily follows that <math>n = \boxed{251}</math> is the smallest such integer. | ||
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+ | ===Solution 2=== | ||
+ | We proceed with complex trigonometry. We know that for all <math>\theta</math>, we have <math>\cos \theta = \dfrac{1}{2} \left( z + \dfrac{1}{z} \right)</math> and <math>\sin \theta = \dfrac{1}{2i} \left( z - \dfrac{1}{z} \right)</math> for some complex number <math>z</math>. Similarly, we have <math>\cos n \theta = \dfrac{1}{2} \left( z^n + \dfrac{1}{z^n} \right)</math> and <math>\sin n \theta = \dfrac{1}{2i} \left(z^n - \dfrac{1}{z^n} \right)</math>. Thus, we have <math>\cos n^2 a \sin n a = \dfrac{1}{4i} \left( z^{n^2} + \dfrac{1}{z^{n^2}} \right) \left( z^{n} - \dfrac{1}{z^n} \right)</math> | ||
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+ | <math>= \dfrac{1}{4i} \left( z^{n^2 + n} - \dfrac{1}{z^{n^2 + n}} - z^{n^2 - n} + \dfrac{1}{z^{n^2 - n}} \right)</math> | ||
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+ | <math>= \dfrac{1}{2} \left( \dfrac{1}{2i} \left(z^{n^2 + n} - \dfrac{1}{z^{n^2 + n}} \right) - \dfrac{1}{2i} \left(z^{n^2 - n} - \dfrac{1}{z^{n^2 - n}} \right) \right)</math> | ||
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+ | <math>= \dfrac{1}{2} \left( \sin ((n^2 + n)a) - \sin ((n^2 - n)a) \right)</math> | ||
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+ | <math>= \dfrac{1}{2} \left( \sin(((n+1)^2 - (n+1))a) - \sin((n^2 - n)a) \right)</math> | ||
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+ | which clearly telescopes! Since the <math>2</math> outside the brackets cancels with the <math>\dfrac{1}{2}</math> inside, we see that the sum up to <math>n</math> terms is | ||
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+ | <math>\sin ((2^2 - 2)a) - \sin ((1^2 - 1)a) + \sin ((3^3 - 3)a) - \sin ((2^2 - 2)a) \cdots + \sin (((n+1)^2 - (n+1))a) - \sin ((n^2 - n)a)</math> | ||
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+ | <math>= \sin (((n+1)^2 - (n+1))a) - \sin(0)</math> | ||
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+ | <math>= \sin ((n^2 + n)a) - 0</math> | ||
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+ | <math>= \sin \left( \dfrac{n(n+1) \pi}{2008} \right)</math>. | ||
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+ | This expression takes on an integer value iff <math>\dfrac{2n(n+1)}{2008} = \dfrac{n(n+1)}{1004}</math> is an integer; that is, <math>1004 \mid n(n+1)</math>. Clearly, <math>1004 = 2^2 \cdot 251</math>, implying that <math>251 \mid n(n+1)</math>. Since we want the smallest possible value of <math>n</math>, we see that we must have <math>{n,n+1} = 251</math>. If <math>n+1 = 251 \rightarrow n=250</math>, then we have <math>n(n+1) = 250(251)</math>, which is clearly not divisible by <math>1004</math>. However, if <math>n = 251</math>, then <math>1004 \mid n(n+1)</math>, so our answer is <math>\boxed{251}</math>. | ||
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+ | It should be noted that the product-to-sum rules follow directly from complex trigonometry, so this solution is essentially equivalent to the solution above. | ||
== See also == | == See also == |
Latest revision as of 19:35, 10 January 2016
Problem
Let . Find the smallest positive integer such that is an integer.
Solution
Solution 1
By the product-to-sum identities, we have that . Therefore, this reduces to a telescoping series:
Thus, we need to be an integer; this can be only , which occur when is an integer. Thus . It easily follows that is the smallest such integer.
Solution 2
We proceed with complex trigonometry. We know that for all , we have and for some complex number . Similarly, we have and . Thus, we have
which clearly telescopes! Since the outside the brackets cancels with the inside, we see that the sum up to terms is
.
This expression takes on an integer value iff is an integer; that is, . Clearly, , implying that . Since we want the smallest possible value of , we see that we must have . If , then we have , which is clearly not divisible by . However, if , then , so our answer is .
It should be noted that the product-to-sum rules follow directly from complex trigonometry, so this solution is essentially equivalent to the solution above.
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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