# 2004 AMC 12A Problems/Problem 23

## Problem

$$P(x) = c_{2004}x^{2004} + c_{2003}x^{2003} + ... + c_1x + c_0$$

has real coefficients with $c_{2004}\not = 0$ and $2004$ distinct complex zeroes $z_k = a_k + b_ki$, $1\leq k\leq 2004$ with $a_k$ and $b_k$ real, $a_1 = b_1 = 0$, and

$$\sum_{k = 1}^{2004}{a_k} = \sum_{k = 1}^{2004}{b_k}.$$

Which of the following quantities can be a non zero number?

$\text {(A)} c_0 \qquad \text {(B)} c_{2003} \qquad \text {(C)} b_2b_3...b_{2004} \qquad \text {(D)} \sum_{k = 1}^{2004}{a_k} \qquad \text {(E)}\sum_{k = 1}^{2004}{c_k}$

## Solution 1

We have to evaluate the answer choices and use process of elimination:

• $\mathrm{(A)}$: We are given that $a_1 = b_1 = 0$, so $z_1 = 0$. If one of the roots is zero, then $P(0) = c_0 = 0$.
• $\mathrm{(B)}$: By Vieta's formulas, we know that $-\frac{c_{2003}}{c_{2004}}$ is the sum of all of the roots of $P(x)$. Since that is real, $\sum_{k = 1}^{2004}{b_k}=0=\sum_{k = 1}^{2004}{a_k}$, and $\frac{c_{2003}}{c_{2004}}=0$, so $c_{2003}=0$.
• $\mathrm{(C)}$: All of the coefficients are real. For sake of contradiction suppose none of $b_{2\ldots 2004}$ are zero. Then for each complex root $z_k$, its complex conjugate $\overline{z_k} = a_k - b_k i$ is also a root. So the roots should pair up, but we have an odd number of imaginary roots! (Remember that $b_1 = 0$.) This gives us the contradiction, and therefore the product is equal to zero.
• $\mathrm{(D)}$: We are given that $\sum_{k = 1}^{2004}{a_k} = \sum_{k = 1}^{2004}{b_k}$. Since the coefficients are real, it follows that if a root is complex, its conjugate is also a root; and the sum of the imaginary parts of complex conjugates is zero. Hence the RHS is zero.

There is, however, no reason to believe that $\boxed{\mathrm{E}}$ should be zero (in fact, that quantity is $P(1)$, and there is no evidence that $1$ is a root of $P(x)$).

## Solution 2(cheap method using answer choices)

Rule out answer choices $A, B, D$ as done in solution 1. Assume for the sake of contradiction that $\mathrm{(E)}$ is $0$. Then $P(1)=0$, so there is some $m\neq 1$ such that $z_m=1+0i$, which implies at least one $b_i$, $i\neq 1$ is $0$ SO if $\sum_{k=1}^{2004}c_k=0$, then $\prod_{i=2}^{2004}b_i=0$. So we have that $\sum_{k=1}^{2004}c_k\neq 0$, because then that would rule out all the answer choices. Hence $\boxed{\mathrm{E}}$ must be nonzero, so the answer is $\mathrm{(E)}$. -vsamc