2020 AMC 10A Problem 11

With $4040$ entries, the median is the average of the $2020$th and $2021$st smallest.

For an integer $m \le 2020$, the entries that are at most $m$ are the integers $1, \dots, m$ together with the perfect squares up to $m$, namely $\lfloor \sqrt{m} \rfloor$ of them. Since $44^2 = 1936 \le 2020 < 2025 = 45^2$, for $1936 \le m \le 2020$ this count is $m + 44$.

We want $m + 44 = 2020$, so $m = 1976$. Thus exactly $2020$ entries are $\le 1976$, and the $2020$th smallest entry is $1976$ itself. The next entry is $1977$ (not a square, so it appears once), which is the $2021$st.

Median $= \frac{1976 + 1977}{2} = 1976.5$.

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

To locate the $k$-th smallest element of a merged list, count how many entries lie at or below a threshold and adjust the threshold until the count hits $k$. Here the squares below $2020$ are few ($44$), so they merely push the median down from the halfway point $2020.5$ by $44$.

Estimate first: without the squares, the $2020$th and $2021$st entries would be $2020$ and $2021$; the $44$ small squares each displace the median one position down, landing near $1976$. Then confirm $1976$ and $1977$ are not squares.

• Using 45 squares instead of 44 (since 45^2 = 2025 > 2020), which shifts the answer to 1975.5.
• Taking the median as the single $2020$th entry instead of averaging two middle entries, or forgetting that the squares $1^2, \dots, 44^2$ duplicate values already in the list.