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 = 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 = 144</math> distinct elements in <math>C</math>. 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 (lcm(55,360)) => 3960*2/55=144. | ||
== See also == | == See also == | ||
Revision as of 21:28, 19 December 2019
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.
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 (lcm(55,360)) => 3960*2/55=144.
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. 
