Difference between revisions of "1990 AIME Problems/Problem 10"
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<math>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)</math>. Since the [[trigonometry|trigonometric]] functions are [[periodic function|periodic]] every <math>2\pi</math>, there are at most <math>72 \cdot 2 = \boxed{144}</math> distinct elements in <math>C</math>. As above, all of these will work. | <math>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)</math>. Since the [[trigonometry|trigonometric]] functions are [[periodic function|periodic]] every <math>2\pi</math>, there are at most <math>72 \cdot 2 = \boxed{144}</math> distinct elements in <math>C</math>. 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 === | === Solution 3 === | ||
Revision as of 04:37, 4 August 2023
Contents
Problem
The sets 
and 
are both sets of complex roots of unity. The set 
is also a set of complex roots of unity. How many distinct elements are in 
?
Solution
Solution 1
The least common multiple of 
and 
is 
, so define 
. We can write the numbers of set 
as 
and of set 
as 
. 
can yield at most different values. All solutions for 
will be in the form of 
.
and 
are relatively prime, and by the Chicken McNugget Theorem, for two relatively prime integers 
, the largest number that cannot be expressed as the sum of multiples of is 
. For 
, this is 
; however, we can easily see that the numbers 
to 
can be written in terms of . Since the exponents are of roots of unities, they reduce 
, so all numbers in the range are covered. Thus the answer is 
.
Solution 2
The 18 and 48th roots of 
can be found by De Moivre's Theorem. They are 
and 
respectively, where 
and 
and 
are integers from 
to 
and to 
, respectively.
. Since the trigonometric functions are periodic every 
, there are at most 
distinct elements in 
. 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 
and 
. Multiplying these gives 
. Then, we get 
, 
, 
, 
, 
up to 
.
Video Solution!
https://www.youtube.com/watch?v=hdamWTu_F94
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
