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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