# 2020 AMC 10A Problems/Problem 17

Define$$P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).$$How many integers $n$ are there such that $P(n)\leq 0$?

$\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100$

## Solution

Notice that $P(x)$ is a product of many integers. We either need one factor to be 0 or an odd number of negative factors. Case 1: There are 100 integers $n$ for which $P(x)=0 Case 2: For there to be an odd number of negative factors,$n$must be between an odd number squared and an even number squared. This means that there are$2+6+\dots+10$total possible values of$n$. Simplifying, there are$5000$possible numbers. Summing, there are$ (Error compiling LaTeX. ! Missing $inserted.)\boxed{\textbf{(E) }} 5100$total possible values of$n$.