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 (\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.  
  
<math>zw = cis (\frac{\pi k_1}{9} + \frac{\pi k_2}{24}) = cis (\frac{8\pi k_1 + 3 \pi k_2}{72})</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.
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<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.
  
 
== See also ==
 
== See also ==

Revision as of 20:22, 26 October 2007

Problem

The sets $\displaystyle A = \{z : z^{18} = 1\}$ and $\displaystyle B = \{w : w^{48} = 1\}$ are both sets of complex roots of unity. The set $C = \{zw : z \in A ~ \mbox{and} ~ w \in B\}$ is also a set of complex roots of unity. How many distinct elements are in $C_{}^{}$?

Solution

Solution 1

The least common multiple of $18$ and $48$ is $144$, so define $\displaystyle n = e^{2\pi i/144}$. We can write the numbers of set $A$ as $\displaystyle \{n^8, n^{16}, \ldots n^{144}\}$ and of set $B$ as $\displaystyle \{n^3, n^6, \ldots n^{144}\}$. $n^x$ can yield at most $144$ different values. All solutions for $zw$ will be in the form of $n^{8k_1 + 3k_2}$. Since $8$ and $3$ are different $\pmod{3}$, all $144$ distinct elements will be covered.

Solution 2

The 18th and 48th roots of $1$ can be found using De Moivre's Theorem. They are $cis (\frac{2\pi k_1}{18})$ and $cis (\frac{2\pi k_2}{48})$ respectively, where $cis \theta = \cos \theta + i \sin \theta$ and $k_1$ and $k_2$ are integers from 0 to 17 and 0 to 47, respectively.

$zw = cis \left(\frac{\pi k_1}{9} + \frac{\pi k_2}{24}\right) = cis \left(\frac{8\pi k_1 + 3 \pi k_2}{72}\right)$. Since the trigonometric functions are periodic every $2\pi$, there are at most $72 \cdot 2 = 144$ distinct elements in $C$. As above, all of these will work.

See also

1990 AIME (ProblemsAnswer KeyResources)
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