1990 AIME Problems/Problem 10

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.

Comment on S2

By the property of Diophantine equation, given a problem to find integers x and y so that ax + by = c for some integer constants a, b, c: if gcd(a, b) = 1, then any arbitrary integer c could by formed with some combination of integers (x, y). Double checking the constraints of k_1 and k_2, we realize that all integers of [0, 143] can be formed by 8 * k1 + 3 * k2, yielding the answer of 144. ~Will_Dai

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 $3960$ $(\text{lcm}(55,360)) \implies \frac{3960 \cdot 2}{55}=\boxed{144}$ .

Video Solution!

https://www.youtube.com/watch?v=hdamWTu_F94

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

Page

Toolbox

Search