Mock AIME I 2012 Problems/Problem 15
Problem
Paula the Painter initially paints every complex number black. When Paula toggles a complex number, she paints it white if it was previously black, and black if it was previously white. For each , Paula progressively toggles the roots of
. Let
be the number of complex numbers are white at the end of this process. Find
.
Solution
First, note that . Let
be the distinct complex numbers that were toggled, where
is a positive integer, and for each complex number
, let
be the number of times
was toggled. Then, we have this relation:
\begin{equation}\label{eq:product} \prod_{k=1}^{20}(x^{2k}+x^k+1)=\prod_{k=1}^{20}\frac{x^{3k}-1}{x^k-1}=\prod_{i=1}^n (x-z_i)^{t(z_i)}. \end{equation}
The problem is equivalent to finding the number of complex numbers such that
is odd.
We now focus on the second formula in \eqref{eq:product}. From this formula, we know that all of the must be roots of unity. Furthermore, for each
,
is the number of factors in the numerator that have
as a root minus the number of factors in the denominator that have
as a root, since no polynomial of the form
has a repeated root.
Now let denote a primitive
th root of unity. First, when
, there are an equal number of factors in the numerator of the fraction as in the denominator of the fraction which have
as a root, so
in this case.
We now consider the case when . The numerator has
factors with
as a root, while the denominator has
factors with
as a root. Therefore, in this case,
. As there are
primitive
th roots of unity for each
, every
with
odd will contribute
to the sum. The table below shows the calculations for
.
Adding up the entries in the last column of the table gives the final answer
.