2020 AMC 12A Problem 8

Question

What is the median of the following list of

4040

numbers?

1,

2,

3,

\ldots,

2020,

1^2,

2^2,

3^2,

\ldots,

2020^2

1974.5

1975.5

1976.5

1977.5

1978.5

With

4040

numbers the median averages the

2020

th and

2021

st smallest

Squares up to

2020

come from

1^2

through

44^2;

count how many list entries are at most a given integer

The median is the average of the

2020

th and

2021

st smallest values.

The perfect squares that are at most

2020

are

1^2, \ldots, 44^2

(since

44^2 = 1936

and

45^2 = 2025

), so there are

44

of them.

Among the list, the numbers

\le 1976

are the

1976

integers

1, \ldots, 1976

together with those

44

squares, totaling

1976 + 44 = 2020.

Thus the

2020

th value is

1976

and the

2021

st value is

1977,

making the median

\dfrac{1976 + 1977}{2} = 1976.5.

Thus, C is the correct answer.

Problem 8 in other years