Difference between revisions of "2014 AMC 12B Problems/Problem 25"
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<math> 2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1</math> | <math> 2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1</math> | ||
− | <math> \textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D) | + | <math> \textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi </math> |
==Solution== | ==Solution== |
Revision as of 09:18, 3 March 2015
Problem
Find the sum of all the positive solutions of
Solution
Rewrite as
. Now let
, and let
. We have
Notice that either
and
or
and
. For the first case,
only when
and
is an integer.
when
is an even multiple of
, and since
,
only when
is an odd divisor of
. This gives us these possible values for
:
For the case where
,
, so
, where m is odd.
must also be an odd multiple of
in order for
to equal
, so
must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for
, and therefore no cases where
and
. Therefore, the sum of all our possible values for
is
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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