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Difference between revisions of "1990 AIME Problems/Problem 10"
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The [[least common multiple]] of <math>18</math> and <math>48</math> is <math>144</math>, so define <math>n = e^{2\pi i/144}</math>. We can write the numbers of set <math>A</math> as <math>\{n^8, n^{16}, \ldots n^{144}\}</math> and of set <math>B</math> as <math>\{n^3, n^6, \ldots n^{144}\}</math>. <math>n^x</math> can yield at most <math>144</math> different values. All solutions for <math>zw</math> will be in the form of <math>n^{8k_1 + 3k_2}</math>.  
 
The [[least common multiple]] of <math>18</math> and <math>48</math> is <math>144</math>, so define <math>n = e^{2\pi i/144}</math>. We can write the numbers of set <math>A</math> as <math>\{n^8, n^{16}, \ldots n^{144}\}</math> and of set <math>B</math> as <math>\{n^3, n^6, \ldots n^{144}\}</math>. <math>n^x</math> can yield at most <math>144</math> different values. All solutions for <math>zw</math> will be in the form of <math>n^{8k_1 + 3k_2}</math>.  
  
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<math>8</math> and <math>3</math> are relatively prime, and it is well known that for two relatively prime integers <math>a,b</math>, the largest number that cannot be expressed as the sum of multiples of <math>a,b</math> is <math>a \cdot b - a - b</math>. For <math>3,8</math>, this is <math>13</math>; however, we can easily see that the numbers <math>145</math> to <math>157</math> can be written in terms of <math>3,8</math>. Since the exponents are of roots of unities, they reduce <math>\mod{144}</math>, so all numbers in the range are covered. Thus the answer is <math>\boxed{144}</math>.
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<math>8</math> and <math>3</math> are relatively prime, and by the Chicken McNugget Theorem, for two relatively prime integers <math>a,b</math>, the largest number that cannot be expressed as the sum of multiples of <math>a,b</math> is <math>a \cdot b - a - b</math>. For <math>3,8</math>, this is <math>13</math>; however, we can easily see that the numbers <math>145</math> to <math>157</math> can be written in terms of <math>3,8</math>. Since the exponents are of roots of unities, they reduce <math>\mod{144}</math>, so all numbers in the range are covered. Thus the answer is <math>\boxed{144}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 02:39, 21 March 2018

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 $72 \cdot 2 = 144$ distinct elements in $C$. 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

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