# Difference between revisions of "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 OmegaLearn (Using Algebra)

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