Difference between revisions of "1990 AIME Problems/Problem 10"
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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>. | 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>. | ||
| − | == Video Solution | + | == Video Solution! == |
https://www.youtube.com/watch?v=hdamWTu_F94 | https://www.youtube.com/watch?v=hdamWTu_F94 | ||
Revision as of 20:44, 2 June 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.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
