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

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In the following list of numbers, the integer <math>n</math> appears <math>n</math> times in the list for <math>1 \leq n \leq 200</math>.<cmath>1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200</cmath>What is the median of the numbers in this list?
 
In the following list of numbers, the integer <math>n</math> appears <math>n</math> times in the list for <math>1 \leq n \leq 200</math>.<cmath>1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200</cmath>What is the median of the numbers in this list?
  
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==Solution==
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==Solution 1==
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===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
 
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>
 
<cmath> \frac{k(k+1)}{2}=20100/2,</cmath>
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<cmath>\frac{1}{2}(142)(143)=10153.</cmath>
 
<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>.
 
<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>.
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===Solution 2===
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==Solution 2==
 
The <math>x</math>th number of this sequence is <math>\left\lceil\frac{-1\pm\sqrt{1+8x}}{2}\right\rceil</math> via the quadratic formula. We can see that if we halve <math>x</math> we end up getting <math>\left\lceil\frac{-1\pm\sqrt{1+4x}}{2}\right\rceil</math>. This is approximately the number divided by <math>\sqrt{2}</math>. <math>\frac{200}{\sqrt{2}} = 141.4</math> and since <math>142</math> looks like the only number close to it, it is answer <math>\boxed{(C) 142}</math> ~Lopkiloinm
 
The <math>x</math>th number of this sequence is <math>\left\lceil\frac{-1\pm\sqrt{1+8x}}{2}\right\rceil</math> via the quadratic formula. We can see that if we halve <math>x</math> we end up getting <math>\left\lceil\frac{-1\pm\sqrt{1+4x}}{2}\right\rceil</math>. This is approximately the number divided by <math>\sqrt{2}</math>. <math>\frac{200}{\sqrt{2}} = 141.4</math> and since <math>142</math> looks like the only number close to it, it is answer <math>\boxed{(C) 142}</math> ~Lopkiloinm
  

Revision as of 23:25, 11 February 2021

Problem

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

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 $\left\lceil\frac{-1\pm\sqrt{1+8x}}{2}\right\rceil$ via the quadratic formula. We can see that if we halve $x$ we end up getting $\left\lceil\frac{-1\pm\sqrt{1+4x}}{2}\right\rceil$. 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

See also

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

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