G285 Summer Problem Set Problem 8

Problem 8

Let $p(x)=x^{12}-9x^{11}+16x^{10}+256x^5+1$ , Let $r_1, r_2, r_3, r_4, r_5, r_6, ..., r_{12}$ be the twelve roots that satisfies $p(x)=0$ , find the least possible value of $\left \lfloor \sum_{n=1}^{12}\sum_{k=1}^{12} r_nr_k-\sum_{s=1}^{11} r_s \right \rfloor$

$\textbf{(A)}\ 67 \qquad\textbf{(B)}\ 69 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 71 \qquad\textbf{(E)}\ 72$

Solution

The first summation is simply $9 \cdot 9 = 81$ by Vieta's. The second summation is $-9+r_{12}$ . The minimum possible value is $72+r_{12}$ , so we need to minimize $r_{12}$ . If we do bounding, when $x=0$ we have $p(x)=1$ , and when $x=-1$ we have $p(x)=-230$ . The shift implies there is a root $r_k$ where $0 \le k \le 12$ such that $-1 < r_k < 0$ . However, $x=-1.5$ seems very close to $0$ , and $x<-1.5$ approaches infinity, so there is another root $-2 < r_k < -1$ . Therefore, we have the smallest root $r_{12}$ must be $72+(-2+\{r_{12} \})$ , where $\{x \}$ is the fractional part. The answer is $\boxed{\textbf{(C)}\ 70}$

Page

Toolbox

Search

G285 Summer Problem Set Problem 8

Problem 8

Solution