1990 AIME Problems/Problem 10

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.

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1990 AIME Problems/Problem 10

Problem

Contents

Solution

Solution 1

Solution 2

See also