Difference between revisions of "2005 AIME II Problems/Problem 9"
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=== Solution 2 === | === Solution 2 === | ||
This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</math>. Notice, <math>\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}</math> since <math>\sin(-t) = -\sin(t)</math>. Using this, <math>(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)</math> is recast as <math>(i e^{-it})^n = i e^{-itn}</math>. Hence we must have <math>i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}</math>. Thus since <math>1000</math> is a multiple of <math>4</math> exactly one quarter of the residues are congruent to <math>1</math> hence we have <math>\boxed{250}</math>. | This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</math>. Notice, <math>\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}</math> since <math>\sin(-t) = -\sin(t)</math>. Using this, <math>(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)</math> is recast as <math>(i e^{-it})^n = i e^{-itn}</math>. Hence we must have <math>i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}</math>. Thus since <math>1000</math> is a multiple of <math>4</math> exactly one quarter of the residues are congruent to <math>1</math> hence we have <math>\boxed{250}</math>. | ||
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+ | === Solution 3 === | ||
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+ | De Moivre's Theorem states that <math>(\cos(t) + i \sin(t))^n = \cos(nt) + i \sin(nt).</math> However, our given expression is <math>(\sin (t) + i \cos (t))^n</math>. We use the identity <math>\sin(x) = \cos(90-x)</math> to substitute things in, getting <math>(\cos(90-t) + i \sin(90-t))^n</math>. We now use De Moivre's, getting <math>\cos(90n-nt) + i \sin(90n-nt)</math>. We want this to be equal to <math>\cos(90-nt) + i \sin(90-nt)</math>. This is only true when the two angles are the same, or when their difference is a multiple of <math>90</math>. <math>(90n - nt) - (90 - nt) = 90(n-1)</math>. <math>90(n-1)</math> is only a multiple of <math>360</math> when <math>n-1</math> is a multiple of <math>4</math>, or when <math>n \equiv 1 \bmod {4}</math>. There are therefore <math>\boxed{250}</math> possible values of <math>n</math>. | ||
== See also == | == See also == |
Revision as of 19:07, 10 March 2016
Problem
For how many positive integers less than or equal to is true for all real ?
Solution
Solution 1
We know by De Moivre's Theorem that for all real numbers and all integers . So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem.
Recall the trigonometric identities and hold for all real . If our original equation holds for all , it must certainly hold for . Thus, the question is equivalent to asking for how many positive integers we have that holds for all real .
. We know that two complex numbers are equal if and only if both their real part and imaginary part are equal. Thus, we need to find all such that and hold for all real .
if and only if either or for some integer . So from the equality of the real parts we need either , in which case , or we need , in which case will depend on and so the equation will not hold for all real values of . Checking in the equation for the imaginary parts, we see that it works there as well, so exactly those values of congruent to work. There are of them in the given range.
Solution 2
This problem begs us to use the familiar identity . Notice, since . Using this, is recast as . Hence we must have . Thus since is a multiple of exactly one quarter of the residues are congruent to hence we have .
Solution 3
De Moivre's Theorem states that However, our given expression is . We use the identity to substitute things in, getting . We now use De Moivre's, getting . We want this to be equal to . This is only true when the two angles are the same, or when their difference is a multiple of . . is only a multiple of when is a multiple of , or when . There are therefore possible values of .
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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