2021 AMC 12A Problems/Problem 16

Problem

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

Solution

Solution 1

There are $1+2+..+199+200=\frac{(200)(201)}{2}=20100$ numbers in total. Let the median be $k$ . We want to find the median $k$ such that $\frac{k(k+1)}{2}=20100/2,$ or $k(k+1)=20100.$ Note that $\sqrt{20100} \approx 142$ . Plugging this value in as $k$ gives $\frac{1}{2}(142)(143)=10153.$ $10153-142<10050$ , so $142$ is the $152$ nd and $153$ rd numbers, and hence, our desired answer. $\fbox{(C) 142}.$ .

Note

See problem 1.

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 12 Problems and Solutions

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2021 AMC 12A Problems/Problem 16

Contents

Problem

Solution

Solution 1

Note

See also