Question

In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200.$ $1, 2, 2, 3, 3, 3, 4, 4, 4, 4,$ $\ldots, 200, 200, \ldots , 200$ What is the median of the numbers in this list?

$100.5$

$134$

$142$

$150.5$

$167$

Solution:

The total number of numbers in the list is $1 + 2 + \cdots + 200$ $= \dfrac{200 \cdot 201}{2} = 20100.$

We want to find the median $k$ such that $\dfrac{k(k + 1)}{2}$ is near $\dfrac{20100}{2}.$

Multiplying by $2,$ we want $k(k + 1)$ near $20100.$ Note that $\sqrt{20100} \approx 142.$

Plugging in $k = 142$ yields $\dfrac{1}{2} \cdot 142 \cdot 143 = 10153.$

$10153 - 142 < 10050,$ which shows that $142$ is our desired median ( $142$ is the $10049$ th and $10050$ th number).

Thus, C is the correct answer.

2021 AMC 10A Spring Problem 16

Solution:

Problem 16 in Other Years