Difference between revisions of "2021 AMC 12A Problems/Problem 16"

(Problem)
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
The solutions will be posted once the problems are posted.
+
===Solution 1===
 +
There are <math>1+2+..+199+200=\frac{(200)(201)}{2}=20100</math> numbers in total. Let the median be <math>k</math>. We want to find the median <math>k</math> such that
 +
<cmath> \frac{k(k+1)}{2}=20100/2,</cmath>
 +
or
 +
<cmath> k(k+1)=20100.</cmath>
 +
Note that <math>\sqrt{20100} \approx 142</math>. Plugging this value in as <math>k</math> gives
 +
<cmath>\frac{1}{2}(142)(143)=10153.</cmath>
 +
<math>10153-142<10050</math>, so <math>142</math> is the <math>152</math>nd and <math>153</math>rd numbers, and hence, our desired answer. <math>\fbox{(C) 142}.</math>.
 
==Note==
 
==Note==
 
See [[2021 AMC 12A Problems/Problem 1|problem 1]].
 
See [[2021 AMC 12A Problems/Problem 1|problem 1]].

Revision as of 15:37, 11 February 2021

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.

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS