1990 AIME Problems/Problem 10
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) | ||
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