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

Revision as of 03:13, 5 January 2022

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 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 $C$ , 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 $\boxed{(C) \text{ } 142}.$

-PureSwag

Solution 4

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 pi_is_3.14 (Using Algebra)

https://youtu.be/HkwgH9Lc1hE

Video Solution by TheBeautyofMath

https://youtu.be/CTXQunZpBA4

~IceMatrix

Revision as of 21:26, 6 November 2021 (view source) Thinker123 (talk \| contribs) (→‎Solution 4) ← Older edit		Revision as of 03:13, 5 January 2022 (view source) MRENTHUSIASM (talk \| contribs) m Newer edit →
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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]]}}	+	{{duplicate\|[[2021 AMC 10A Problems/Problem 16\|2021 AMC 10A #16]] and [[2021 AMC 12A Problems/Problem 16\|2021 AMC 12A #16]]}}

	==Problem==		==Problem==

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