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]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
− | 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, \ | + | 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? |
− | + | <math>\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167</math> | |
− | + | ||
+ | ==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> | ||
− | <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>. |
− | + | ||
− | The <math>x</math>th number of this sequence is | + | 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 | ||
+ | |||
+ | 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 <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 | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | -[[User:PureSwag|PureSwag]] | ||
+ | |||
+ | ==Solution 4 (Geometry)== | ||
+ | |||
+ | We can arrange the numbers in the following pattern: | ||
+ | <cmath>\[ | ||
+ | \begin{array}{cccccc} | ||
+ | \ &\ &\ &\ &\ 200 & \\ | ||
+ | \ &\ &\ &\ 199 & \ 200 & \\ | ||
+ | \ &\ &\ \iddots& \ \vdots& \ \vdots& \\ | ||
+ | \ &\ 2& \ \cdots& \ 199& \ 200& \\ | ||
+ | 1 & \ 2 & \ \cdots& \ 199& \ 200& | ||
+ | \end{array} | ||
+ | \]</cmath> | ||
+ | |||
+ | We can see this as a isosceles right triangle, with legs of length <math>200.</math> | ||
+ | <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 <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. | ||
+ | |||
+ | Solving for <math>x</math> yields: | ||
+ | <cmath>\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*}</cmath> | ||
+ | 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== | ==Video Solution by Hawk Math== | ||
https://www.youtube.com/watch?v=AjQARBvdZ20 | 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 22:56, 16 October 2023
- The following problem is from both the 2021 AMC 10A #16 and 2021 AMC 12A #16, so both problems redirect to this page.
Contents
Problem
In the following list of numbers, the integer appears
times in the list for
.
What is the median of the numbers in this list?
Solution 1
There are numbers in total. Let the median be
. We want to find the median
such that
or
Note that
. Plugging this value in as
gives
, so
is the
nd and
rd numbers, and hence, our desired answer.
.
Note that we can derive through the formula
where
is a perfect square less than or equal to
. We set
to
, so
, and
. We then have
. ~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 th number of this sequence is
via the quadratic formula. We can see that if we halve
we end up getting
. This is approximately the number divided by
.
and since
looks like the only number close to it, it is answer
~Lopkiloinm
Solution 3 (Answer Choices)
We can look at answer choice , which is
first. That means that the number of numbers from
to
is roughly the number of numbers from
to
.
The number of numbers from to
is
which is approximately
The number of numbers from
to
is
which is approximately
as well. Therefore, we can be relatively sure the answer choice is
Solution 4 (Geometry)
We can arrange the numbers in the following pattern:
We can see this as a isosceles right triangle, with legs of length
Let be the side length such that both sides of the triangle have the same area. The desired answer is then around
because about half of the numbers in the list fall on each side.
Solving for yields:
We see that
is the closest to
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)
~pi_is_3.14
Video Solution by TheBeautyofMath
~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.