Difference between revisions of "1990 AIME Problems/Problem 10"
m (→Solution) |
Kcbhatraju (talk | contribs) m (→Solution 2) |
||
Line 12: | Line 12: | ||
The 18 and 48th roots of <math>1</math> can be found by [[De Moivre's Theorem]]. They are <math>\text{cis}\,\left(\frac{2\pi k_1}{18}\right)</math> and <math>\text{cis}\,\left(\frac{2\pi k_2}{48}\right)</math> respectively, where <math>\text{cis}\,\theta = \cos \theta + i \sin \theta</math> and <math>k_1</math> and <math>k_2</math> are integers from <math>0</math> to <math>17</math> and <math>0</math> to <math>47</math>, respectively. | The 18 and 48th roots of <math>1</math> can be found by [[De Moivre's Theorem]]. They are <math>\text{cis}\,\left(\frac{2\pi k_1}{18}\right)</math> and <math>\text{cis}\,\left(\frac{2\pi k_2}{48}\right)</math> respectively, where <math>\text{cis}\,\theta = \cos \theta + i \sin \theta</math> and <math>k_1</math> and <math>k_2</math> are integers from <math>0</math> to <math>17</math> and <math>0</math> to <math>47</math>, respectively. | ||
− | <math>zw = \text{cis}\,\left(\frac{\pi k_1}{9} + \frac{\pi k_2}{24}\right) = \text{cis}\,\left(\frac{8\pi k_1 + 3 \pi k_2}{72}\right)</math>. Since the [[trigonometry|trigonometric]] functions are [[periodic function|periodic]] every <math>2\pi</math>, there are at most <math>72 \cdot 2 = 144</math> distinct elements in <math>C</math>. As above, all of these will work. | + | <math>zw = \text{cis}\,\left(\frac{\pi k_1}{9} + \frac{\pi k_2}{24}\right) = \text{cis}\,\left(\frac{8\pi k_1 + 3 \pi k_2}{72}\right)</math>. Since the [[trigonometry|trigonometric]] functions are [[periodic function|periodic]] every <math>2\pi</math>, there are at most <math>72 \cdot 2 = \boxed{144}</math> distinct elements in <math>C</math>. As above, all of these will work. |
=== Solution 3 === | === Solution 3 === |
Revision as of 17:39, 10 July 2020
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 (1, 20x) and (1, 7.5x). Multiplying these gives (1, 27.5x). Then, we get 27.5, 55, 82.5, 110, ... up to 3960 (lcm(55,360)) => 3960*2/55=144.
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.