2021 AMC 12A Spring Problem 16

Question

In the following list of numbers, the integer

n

appears

n

times in the list for

1 \le n \le 200.

1,

2,

2,

3,

3,

3,

4,

4,

4,

4,

\ldots,

200,

200,

\ldots,

200

What is the median of the numbers in this list?

100.5

134

142

150.5

167

The list has

1 + 2 + \cdots + 200 = 20100

terms, so the median averages the

10050

th and

10051

st

The numbers through

n

fill positions up to

\dfrac{n(n+1)}{2};

find the

n

containing position

10050

The list has

1 + 2 + \cdots + 200

= \dfrac{200\cdot 201}{2}

= 20100

terms, so the median is the average of the

10050

th and

10051

st terms.

The value

n

occupies positions up to

\dfrac{n(n+1)}{2}.

Since

\dfrac{141\cdot 142}{2} = 10011

and

\dfrac{142\cdot 143}{2} = 10153,

positions

10012

through

10153

all equal

142.

Both middle positions fall in this block, so the median is

142.

Thus, the correct answer is C.

Problem 16 in other years