Mock AIME 6 2006-2007 Problems/Problem 7

Problem

Let $P_n(x)=1+x+x^2+\cdots+x^n$ and $Q_n(x)=P_1\cdot P_2\cdots P_n$ for all integers $n\ge 1$ . How many more distinct complex roots does $Q_{1004}$ have than $Q_{1003}$ ?

Solution

The roots of $P_n(x)$ will be in the form $x=e^{\frac{k}{n+1}2\pi i}$ for $k=1,2,\cdots,n$ with the only real solution when $n$ is odd and $k=\frac{n+1}{2}$ and the rest are complex.

Therefore, each $P_n(x)$ will have $n$ distinct complex roots when $n$ is even and $n-1$ distinct complex roots when $n$ is odd.

The roots of $Q_n(x)$ will be all of the roots of $P_1,P_2,\cdots, P_n$ which will include several repeated roots.

To get how many more complex roots does $Q_{1004}$ have than $Q_{1003}$ that will be the number of complex roots of $P_{1004}$ .

But to get how many more distinct complex roots, we must subtract the complex roots of $P_{1004}$ that can be found in $Q_{1003}$

complex roots of $P_{1004}:\; x=e^{\frac{1}{1005}2\pi i},e^{\frac{2}{1005}2\pi i},e^{\frac{3}{1005}2\pi i},\cdots,e^{\frac{1004}{1005}2\pi i}$ for a total of $1004$ complex roots.

Now we subtract the common complex roots of $P_{1004}$ with $Q_{1003}$ by finding how many reducible fractions are there in $\frac{k}{1005}$ for $k=1,2,\cdots,1005$

Since $1005=(3)(5)(67)$ then $\frac{k}{1005}$ is reducible when $k \equiv 0\;(mod\;3)$ or $k \equiv 0\;(mod\;5)$ or $k \equiv 0\;(mod\;67)$

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Page

Toolbox

Search

Mock AIME 6 2006-2007 Problems/Problem 7

Problem

Solution