2005 AIME II Problems/Problem 9
Contents
Problem
For how many positive integers less than or equal to 1000 is
true for all real
?
Solution
We know by De Moivre's Theorem that for all real numbers
and all integers
. So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem. Recall the trigonometric identities
and
hold for all real
. If our original equation holds for all
, it must certainly hold for
. Thus, the question is equivalent to asking for how many positive integers
we have that
holds for all real
.
. We know that two complex numbers are equal if and only if both their real part and imaginary part are equal. Thus, we need to find all
such that
and
hold for all real
.
if and only if either
or
for some integer
. So from the equality of the real parts we need either
, in which case
, or we need
, in which case
will depend on
and so the equation will not hold for all real values of
. Checking
in the equation for the imaginary parts, we see that it works there as well, so exactly those values of
congruent to
work. There are 250 of them in the given range.
Solution 2
We can rewrite as
which, by De Moivre's Theorem is equal to
, but we know that is is equal to
, now if we replace
with
. This gives us the equation:
Equating the real parts or the imaginary parts will give the same solution set, so we will equate the real parts. So we get
but in the right hand side of the equation is just the principal value, but we can have any equivalent value. So our new equation is:
for
being an integer.
we know that
so we want all numbers that are less than or equal to and also
and there are
such numbers
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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