Question

What is the median of the following list of $4040$ numbers $?$ $\begin{align*} &1, 2, 3, \ldots, 2020, \\&1^2, 2^2, 3^2, \ldots, 2020^2 \end{align*}$

$1974.5$

$1975.5$

$1976.5$

$1977.5$

$1978.5$

Video solution:

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Written solution:

For a number near the median, the sorted list includes all ordinary integers up to that number and all squares up to that number. Since $44^2=1936$ and $45^2=2025$ , there are $44$ squares not exceeding any number from $1936$ through $2020$ .

At $1975$ , there are $1975+44=2019$ list entries at most $1975$ . At $1976$ , there are $2020$ entries at most $1976$ , so the $2020$ th entry is $1976$ , and the next is $1977$ . The median is $1976.5$ . Thus, C is the correct answer.

2020 AMC 10A Problem 11

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Written solution:

Problem 11 in Other Years