2020 AMC 10A Problems/Problem 17

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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 $\boxed{\textbf{(E) } 5100}$ total possible values of $n$.


See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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