Difference between revisions of "1985 AHSME Problems/Problem 23"
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<math> \mathrm{(D) \ } x^{11}+y^{11}=-1 \qquad \mathrm{(E) \ }x^{13}+y^{13}=-1 </math> | <math> \mathrm{(D) \ } x^{11}+y^{11}=-1 \qquad \mathrm{(E) \ }x^{13}+y^{13}=-1 </math> | ||
− | ==Solution== | + | ==Solution 1== |
Notice that <math> x+y=-1 </math> and <math> xy=1 </math>. We have <math> 1=(x+y)^2=x^2+2xy+y^2=x^2+y^2+2\implies x^2+y^2=-1 </math>. | Notice that <math> x+y=-1 </math> and <math> xy=1 </math>. We have <math> 1=(x+y)^2=x^2+2xy+y^2=x^2+y^2+2\implies x^2+y^2=-1 </math>. | ||
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<math> 1=x^{13}+y^{13}+x^9+y^9=x^{13}+y^{13}+2\implies x^{13}+y^{13}=-1 </math>, and this is true for all other answer choices. | <math> 1=x^{13}+y^{13}+x^9+y^9=x^{13}+y^{13}+2\implies x^{13}+y^{13}=-1 </math>, and this is true for all other answer choices. | ||
+ | |||
+ | ==Solution 2 (using polar complex numbers)== | ||
+ | Note that <math>x=\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})=e^{\frac{2\pi}{3}i}</math> and that <math>y=\cos(-\frac{2\pi}{3})+i\sin(-\frac{2\pi}{3})=e^{-\frac{2\pi}{3}i}</math>. | ||
+ | |||
+ | Then <math>x^k=e^{\frac{2\pi k}{3}i}=\cos(\frac{2\pi k}{3})+i\sin(\frac{2\pi k}{3})</math> and <math>y^k=e^{-\frac{2\pi k}{3}i}=\cos(-\frac{2\pi k}{3})+i\sin(-\frac{2\pi k}{3})=\cos(\frac{2\pi k}{3})-i\sin(\frac{2\pi k}{3})</math>. | ||
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+ | Thus, <math>x^k+y^k=2\cos(\frac{2\pi k}{3})</math>. Testing <math>k = 5,7,9,11,13</math> for choices A, B, C, D, and E, respectively, we find that for <math>k=9</math>, <math>x^9+y^9=2\cos(6\pi)=2\neq -1</math>. The answer is <math>\boxed{C}</math>. | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=22|num-a=24}} | {{AHSME box|year=1985|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:48, 5 November 2013
Problem
If and , where , then which of the following is not correct?
Solution 1
Notice that and . We have .
We also have .
Finally,
.
Let . We have
.
For , so
.
Therefore, for , and
, so .
As a check, we have for , and
.
Finally, for we have , and
, and this is true for all other answer choices.
Solution 2 (using polar complex numbers)
Note that and that .
Then and .
Thus, . Testing for choices A, B, C, D, and E, respectively, we find that for , . The answer is .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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