2020 AMC 10A Problem 11

There are $4040$ numbers, an even count, so the median is the average of the $2020$th and $2021$st smallest. The job is to find those two numbers.

1. Without the squares. The list $1, 2, \ldots, 2020$ alone would have $2020$ as its $2020$th number.
2. The squares that get in the way. Some squares are small and slip in among the numbers below $2020$. Which ones? $44^2 = 1936$ and $45^2 = 2025$. So the squares $1^2, 2^2, \ldots, 44^2$ are all at most $1936$: that is $44$ small squares. Every other square is at least $2025$.
3. Count up to a number $m$ between $1936$ and $2020$. The numbers in the list that are at most $m$ are the ordinary numbers $1, 2, \ldots, m$ (that is $m$ of them) plus the $44$ small squares. That is $m + 44$ numbers.
4. We want this count to be $2020$: $m + 44 = 2020$ gives $m = 1976$. So exactly $2020$ numbers are at most $1976$, and the biggest of them is $1976$. That makes $1976$ the $2020$th number.
5. The next number is $1977$. It is not a square, so it appears only once, and it is the $2021$st.
6. The median is $\frac{1976 + 1977}{2} = 1976.5$.

The answer is $\boxed{\textbf{(C)}\ 1976.5}$.

To find the number in a given position of a sorted list, count how many entries are at or below a guess, and adjust the guess until the count is right. Here the $44$ small squares crowd in at the bottom, so everything moves $44$ places down compared with the plain list $1$ to $2020$.