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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.