2020 AMC 10A Problem 17

Question

Define

\begin{aligned} P(x)={}&(x-1^2)(x-2^2)\\ &\cdots(x-100^2). \end{aligned}

How many integers

n

are there such that

P(n)\leq 0?

4900

4950

5000

5050

5100

Make a sign chart across the square roots

1^2,2^2,\ldots,100^2

Count integer points in every other interval, including endpoints

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The polynomial changes sign at each square

1^2,2^2,\ldots,100^2

, and its leading coefficient is positive. Thus

P(n)\le0

for integers in the intervals

[1^2,2^2]

,

[3^2,4^2]

,

\ldots

,

[99^2,100^2]

.

For odd

k

, the interval

[k^2,(k+1)^2]

contains

(k+1)^2-k^2+1=2k+2

integers. Summing over odd

k=1,3,\ldots,99

gives

2(1+3+\cdots+99)

+2\cdot50=5000+100

=5100

. Thus, E is the correct answer.

Problem 17 in other years