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Difference between revisions of "2021 AMC 12A Problems/Problem 16"
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{{duplicate|[[2021 AMC 10A Problems/Problem 16|2021 AMC 10A #16]] and [[2021 AMC 12A Problems/Problem 16|2021 AMC 12A #16]]}}
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==Problem==
 
==Problem==
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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, \cdot, 200, 200, \cdot , 200</cmath>What is the median of the numbers in this list?
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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?
  
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==Solution==
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<math>\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167</math>
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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>
Line 10: Line 13:
 
Note that <math>\sqrt{20100} \approx 142</math>. Plugging this value in as <math>k</math> gives
 
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>
 
<cmath>\frac{1}{2}(142)(143)=10153.</cmath>
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<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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<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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==Note==
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−
See [[2021 AMC 12A Problems/Problem 1|problem 1]].
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Note that we can derive <math>\sqrt{20100} \approx 142</math> through the formula <cmath>\sqrt{n} = \sqrt{a+b} \approx \sqrt{a} + \frac{b}{2\sqrt{a} + 1},</cmath>
 +
where <math>a</math> is a perfect square less than or equal to <math>n</math>. We set <math>a</math> to <math>19600</math>, so <math>\sqrt{a} = 140</math>, and <math>b = 500</math>. We then have <math>n \approx 140 + \frac{500}{2(140)+1} \approx 142</math>. ~approximation by ciceronii
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 +
Note by Fasolinka (use answer choices): Once you know that the answer is in the 140s range (200,000 is around 14^2 times 10^2) by the approximation, it is highly improbable for the answer to be anything but C.
 +
 
 +
==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
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 +
==Solution 3 (Answer Choices)==
 +
We can look at answer choice <math>\textbf{(C)}</math>, which is <math>142</math> first. That means that the number of numbers from <math>1</math> to <math>142</math> is roughly the number of numbers from <math>143</math> to <math>200</math>.
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 +
The number of numbers from <math>1</math> to <math>142</math> is <math>\frac{142(142+1)}{2}</math> which is approximately <math>10000.</math> The number of numbers from <math>143</math> to <math>200</math> is <math>\frac{200(200+1)}{2}-\frac{142(142+1)}{2}</math> which is approximately <math>10000</math> as well. Therefore, we can be relatively sure the answer choice is <math>\boxed{\textbf{(C)} ~142}.</math>
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 +
-[[User:PureSwag|PureSwag]]
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 +
==Solution 4 (Geometry)==
 +
 
 +
We can arrange the numbers in the following pattern:
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<cmath>\[
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\begin{array}{cccccc}
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\ &\ &\ &\ &\ 200 & \\
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\ &\ &\ &\ 199 & \ 200 & \\
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\ &\ &\ \iddots& \ \vdots& \ \vdots& \\
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\ &\ 2& \ \cdots& \ 199& \ 200& \\
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1 & \ 2 & \ \cdots& \ 199& \ 200&
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\end{array}
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\]</cmath>
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We can see this as a isosceles right triangle, with legs of length <math>200.</math>
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<asy>draw((0,0)--(200,200)--(200,0)--cycle);
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draw((142,0)--(142,142));
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label("$x$",(142,0)--(142,142),E);
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label("$x$",(0,0)--(142,0),S);
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label("$200$",(200,0)--(200,200),E);
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</asy>
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Let <math>x</math> be the side length such that both sides of the triangle have the same area. The desired answer is then around <math>x</math> because about half of the numbers in the list fall on each side.
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Solving for <math>x</math> yields:
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<cmath>\begin{align*}
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\frac{x^2}{2} =& \:\frac{1}{2} \cdot \frac{200^2}{2} \\
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x^2 =& \:\frac{1}{2}\cdot 200^2 \\
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x =& \:\frac{200}{\sqrt{2}} = \: 100\sqrt{2} \approx 141.
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\end{align*}</cmath>
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We see that <math>\boxed{(C) \: 142}</math> is the closest to <math>x</math> by far, and thus, can be relatively certain this is the answer.
 +
 
 +
~thinker123
 +
 
 +
==Video Solution by Punxsutawney Phil==
 +
https://youtube.com/watch?v=vsE_ezaV4Xs
 +
 
 +
==Video Solution by Hawk Math==
 +
https://www.youtube.com/watch?v=AjQARBvdZ20
 +
 
 +
==Video Solution by Answer Choice==
 +
https://www.youtube.com/watch?v=YxWjDcUcaeQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=13
 +
~North America Math Contest Go Go Go
 +
 
 +
== Video Solution by OmegaLearn (Using Algebra) ==
 +
https://youtu.be/HkwgH9Lc1hE
 +
 
 +
~pi_is_3.14
 +
 
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/CTXQunZpBA4
 +
 
 +
~IceMatrix
 +
 
 
==See also==
 
==See also==
 +
{{AMC10 box|year=2021|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2021|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2021|ab=A|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:01, 21 October 2024

The following problem is from both the 2021 AMC 10A #16 and 2021 AMC 12A #16, so both problems redirect to this page.

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?

$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$

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 that we can derive $\sqrt{20100} \approx 142$ through the formula \[\sqrt{n} = \sqrt{a+b} \approx \sqrt{a} + \frac{b}{2\sqrt{a} + 1},\] where $a$ is a perfect square less than or equal to $n$. We set $a$ to $19600$, so $\sqrt{a} = 140$, and $b = 500$. We then have $n \approx 140 + \frac{500}{2(140)+1} \approx 142$. ~approximation by ciceronii

Note by Fasolinka (use answer choices): Once you know that the answer is in the 140s range (200,000 is around 14^2 times 10^2) by the approximation, it is highly improbable for the answer to be anything but C.

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

Solution 3 (Answer Choices)

We can look at answer choice $\textbf{(C)}$, which is $142$ first. That means that the number of numbers from $1$ to $142$ is roughly the number of numbers from $143$ to $200$.

The number of numbers from $1$ to $142$ is $\frac{142(142+1)}{2}$ which is approximately $10000.$ The number of numbers from $143$ to $200$ is $\frac{200(200+1)}{2}-\frac{142(142+1)}{2}$ which is approximately $10000$ as well. Therefore, we can be relatively sure the answer choice is $\boxed{\textbf{(C)} ~142}.$

-PureSwag

Solution 4 (Geometry)

We can arrange the numbers in the following pattern: \[ \begin{array}{cccccc} \ &\ &\ &\ &\ 200 & \\ \ &\ &\ &\ 199 & \ 200 & \\ \ &\ &\ \iddots& \ \vdots& \ \vdots& \\ \ &\ 2& \ \cdots& \ 199& \ 200& \\ 1 & \ 2 & \ \cdots& \ 199& \ 200& \end{array} \]

We can see this as a isosceles right triangle, with legs of length $200.$ [asy]draw((0,0)--(200,200)--(200,0)--cycle); draw((142,0)--(142,142)); label("$x$",(142,0)--(142,142),E); label("$x$",(0,0)--(142,0),S); label("$200$",(200,0)--(200,200),E); [/asy]

Let $x$ be the side length such that both sides of the triangle have the same area. The desired answer is then around $x$ because about half of the numbers in the list fall on each side.

Solving for $x$ yields: \begin{align*} \frac{x^2}{2} =& \:\frac{1}{2} \cdot \frac{200^2}{2} \\ x^2 =& \:\frac{1}{2}\cdot 200^2 \\ x =& \:\frac{200}{\sqrt{2}} = \: 100\sqrt{2} \approx 141. \end{align*} We see that $\boxed{(C) \: 142}$ is the closest to $x$ by far, and thus, can be relatively certain this is the answer.

~thinker123

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=vsE_ezaV4Xs

Video Solution by Hawk Math

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

Video Solution by Answer Choice

https://www.youtube.com/watch?v=YxWjDcUcaeQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=13 ~North America Math Contest Go Go Go

Video Solution by OmegaLearn (Using Algebra)

https://youtu.be/HkwgH9Lc1hE

~pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/CTXQunZpBA4

~IceMatrix

See also

2021 AMC 10A (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 10 Problems and Solutions
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. AMC logo.png

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