Difference between revisions of "1990 AIME Problems/Problem 10"
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| − | The 18th and 48th roots of <math>1</math> can be found using [[De Moivre's Theorem]]. They are <math>cis (\frac{2\pi k_1}{18})</math> and <math>cis (\frac{2\pi k_2}{48})</math> respectively, where <math>cis \theta = \cos \theta + i \sin \theta</math> and <math>k_1</math> and <math>k_2</math> are integers from 0 to 17 and 0 to 47, respectively. | + | The 18th and 48th roots of <math>1</math> can be found using [[De Moivre's Theorem]]. They are <math>cis \left(\frac{2\pi k_1}{18}\right)</math> and <math>cis \left(\frac{2\pi k_2}{48}\right)</math> respectively, where <math>cis \theta = \cos \theta + i \sin \theta</math> and <math>k_1</math> and <math>k_2</math> are integers from 0 to 17 and 0 to 47, respectively. |
<math>zw = cis \left(\frac{\pi k_1}{9} + \frac{\pi k_2}{24}\right) = cis \left(\frac{8\pi k_1 + 3 \pi k_2}{72}\right)</math>. Since the [[trigonometry|trigonometric]] functions are [[periodic function|periodic]] every <math>2\pi</math>, there are at most <math>72 \cdot 2 = 144</math> distinct elements in <math>C</math>. As above, all of these will work. | <math>zw = cis \left(\frac{\pi k_1}{9} + \frac{\pi k_2}{24}\right) = cis \left(\frac{8\pi k_1 + 3 \pi k_2}{72}\right)</math>. Since the [[trigonometry|trigonometric]] functions are [[periodic function|periodic]] every <math>2\pi</math>, there are at most <math>72 \cdot 2 = 144</math> distinct elements in <math>C</math>. As above, all of these will work. | ||
Revision as of 20:23, 26 October 2007
Problem
The sets 
and 
are both sets of complex roots of unity. The set 
is also a set of complex roots of unity. How many distinct elements are in 
?
Solution
Solution 1
The least common multiple of 
and 
is 
, so define 
. We can write the numbers of set 
as 
and of set 
as 
. 
can yield at most different values. All solutions for 
will be in the form of 
. Since 
and 
are different 
, all distinct elements will be covered.
Solution 2
The 18th and 48th roots of 
can be found using De Moivre's Theorem. They are 
and 
respectively, where 
and 
and 
are integers from 0 to 17 and 0 to 47, respectively.
. Since the trigonometric functions are periodic every 
, there are at most 
distinct elements in 
. As above, all of these will work.
See also
| 1990 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
