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1990 AIME Problems/Problem 10

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Problem

The sets A = \{z : z^{18} = 1\} and 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_{}^{}?

Contents

Solution

Solution 1

The least common multiple of 18 and 48 is 144, so define n = e^{2\pi i/144}. We can write the numbers of set A as \{n^8, n^{16}, \ldots n^{144}\} and of set B as \{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}.

8 and 3 are relatively prime, and it is well known that for two relatively prime integers a,b, the largest number that cannot be expressed as the sum of multiples of a,b is a \cdot b - a - b. For 3,8, this is 13; however, we can easily see that the numbers 145 to 157 can be written in terms of 3,8. Since the exponents are of roots of unities, they reduce \mod{144}, so all numbers in the range are covered. Thus the answer is \boxed{144}.

Solution 2

The 18 and 48th roots of 1 can be found by De Moivre's Theorem. They are \text{cis}\,\left(\frac{2\pi k_1}{18}\right) and \text{cis}\,\left(\frac{2\pi k_2}{48}\right) respectively, where \text{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 = \text{cis}\,\left(\frac{\pi k_1}{9} + \frac{\pi k_2}{24}\right) = \text{cis}\,\left(\frac{8\pi k_1 + 3 \pi k_2}{72}\righ.... 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 (ProblemsResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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