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. $\fbox{(C) 142}.$ .

Solution 2

The $x$ th number of this sequence is obviously $\frac{-1\pm\sqrt{1+8x}}{2}$ via the quadratic formula. We can see that if we halve $x$ we end up getting $\frac{-1\pm\sqrt{1+4x}}{2}$ . This is approximately the number divided by $\sqrt{2}$ . $\frac{200}{\sqrt{2}} = 141.4$ and since $142$ looks like the only number close to it, it is answer $\boxed{(C) 142}$ ~Lopkiloinm

Video Solution by Hawk Math

https://www.youtube.com/watch?v=AjQARBvdZ20

Video Solution by OmegaLearn (Using Algebra)

https://youtu.be/HkwgH9Lc1hE

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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