AlgebraBasic AMC10/12 Intermediate
2020


Problem - 4733

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

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


Answer     $1976.5$

The answer will be the average of the $2020^{th}$ and $2021^{th}$ numbers when these numbers are sorted. Therefore, the key is to find these two "magic" numbers.

Because $44^2 < 2020 < 45^2$, therefore the two "magic" numbers must be among the following $2020+44=2064$ numbers: $$1, \cdots, 2020 ,1^2, \cdots, 44^2$$

We need to eliminate the biggest $44$ numbers from the above sequence in order to find the $2020^{th}$ number. Because $2020 - 44^2 = 84 > 64$, it is now clear that the $2020^{th}$ number is just $2020-44=1976$ and the $2021^{th}$ number is $1977$. Therefore the final answer is $\boxed{1976.5}$.

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