Difference between revisions of "1990 AIME Problems/Problem 10"
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=== Solution 3 === | === Solution 3 === | ||
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!!! == | ||
+ | https://www.youtube.com/watch?v=hdamWTu_F94 | ||
== See also == | == See also == |
Revision as of 08:28, 28 June 2022
Contents
[hide]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 |
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