2020 AMC 10A Problems/Problem 11

The following problem is from both the 2020 AMC 12A #8 and 2020 AMC 10A #11, so both problems redirect to this page.

Problem 11

What is the median of the following list of $4040$ numbers $?$

$1, 2, 3, ..., 2020, 1^2, 2^2, 3^2, ..., 2020^2$

$\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$

Solution

We can see that $44^2$ is less than 2020. Therefore, there are $1976$ of the $4040$ numbers after $2020$ . Also, there are $2064$ numbers that are under and equal to $2020$ . Since $44^2$ is $1936$ it, with the other squares will shift our median's placement up $44$ . We can find that the median of the whole set is $2020.5$ , and $2020.5-44$ gives us $1976.5$ . Our answer is $\boxed{\textbf{(D) } 1976.5}$ .

~aryam

2020 AMC 10A (Problems • Answer Key • Resources)
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2020 AMC 10A Problems/Problem 11

Problem 11

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