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Difference between revisions of "1990 AIME Problems/Problem 10"

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=== Solution 3 ===
 
=== Solution 3 ===
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The values in polar form will be <math>(1, 20x)</math> and <math>(1, 7.5x)</math>. Multiplying these gives <math>(1, 27.5x)</math>. Then, we get <math>27.5</math>, <math>55</math>, <math>82.5</math>, <math>110</math>, <math>\cdots</math>
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The values in polar form will be <math>(1, 20x)</math> and <math>(1, 7.5x)</math>. Multiplying these gives <math>(1, 27.5x)</math>. Then, we get <math>27.5</math>, <math>55</math>, <math>82.5</math>, <math>110</math>, <math>...</math> up to <math>3960</math> <math>(\text{lcm}(55,360)) \implies \frac{3960 \cdot 2}{55}=\boxed{144}</math>.
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up to <math>3960</math> <math>(lcm(55,360)) \implies \frac{3960 \cdot 2}{55}=\boxed{144}</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 00:21, 26 June 2022

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 $C_{}^{}$?

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 by the Chicken McNugget Theorem, 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.

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

Solution 3

The values in polar form will be $(1, 20x)$ and $(1, 7.5x)$. Multiplying these gives $(1, 27.5x)$. Then, we get $27.5$, $55$, $82.5$, $110$, $...$ up to $3960$ $(\text{lcm}(55,360)) \implies \frac{3960 \cdot 2}{55}=\boxed{144}$.

See also

1990 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AIME Problems and Solutions

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