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

See problem 1.