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^n 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>. |
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 22:32, 20 February 2017
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 .
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
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 | ||
