Difference between revisions of "2020 AMC 10A Problems/Problem 11"

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We can see that <math>44^2=1936</math> which is less than 2020. Therefore, there are <math>2020-44=1976</math> of the <math>4040</math> numbers greater than <math>2020</math>. Also, there are <math>2020+44=2064</math> numbers that are less than or equal to <math>2020</math>.  
 
We can see that <math>44^2=1936</math> which is less than 2020. Therefore, there are <math>2020-44=1976</math> of the <math>4040</math> numbers greater than <math>2020</math>. Also, there are <math>2020+44=2064</math> numbers that are less than or equal to <math>2020</math>.  
  
Since there are 44 duplicates/extras, it will shift up our median's placement up <math>44</math>. Had the list of numbers been <math>1,2,3, \dots, 4040</math>, the median of the whole set would be <math>\dfrac{1+4040}{2}=2020.5</math>.  
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Since there are 44 duplicates/extras, it will shift up our median's placement down <math>44</math>. Had the list of numbers been <math>1,2,3, \dots, 4040</math>, the median of the whole set would be <math>\dfrac{1+4040}{2}=2020.5</math>.  
  
 
<math>2020.5-44</math> gives us <math>1976.5</math>. Thus, our answer is <math>\boxed{\textbf{(C) } 1976.5}</math>.
 
<math>2020.5-44</math> gives us <math>1976.5</math>. Thus, our answer is <math>\boxed{\textbf{(C) } 1976.5}</math>.
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== Solution 3 ==
 
== Solution 3 ==
 
We want to know the <math>2020</math>th term and the <math>2021</math>st term to get the median. <br>
 
We want to know the <math>2020</math>th term and the <math>2021</math>st term to get the median. <br>
<br>
 
 
We know that <math>44^2=1936</math> <br>
 
We know that <math>44^2=1936</math> <br>
So numbers <math>1^2, 2^2, ...,44^2</math> are in between <math>1</math> to <math>1936</math>. <br>
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So numbers <math>1^2, 2^2, ...,44^2</math> are in between <math>1</math> and <math>1936</math>. <br>
 
So the sum of <math>44</math> and <math>1936</math> will result in <math>1980</math>, which means that <math>1936</math> is the <math>1980</math>th number. <br>
 
So the sum of <math>44</math> and <math>1936</math> will result in <math>1980</math>, which means that <math>1936</math> is the <math>1980</math>th number. <br>
 
Also, notice that <math>45^2=2025</math>, which is larger than <math>2021</math>. <br>
 
Also, notice that <math>45^2=2025</math>, which is larger than <math>2021</math>. <br>
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We note that <math>44^2 = 1936</math>, which is the first square less than <math>2020</math>, which means that there are 44 addition terms before <math>2020</math>. This makes <math>2020</math> the 2064th term. To find the median, we need the 2020th and 2021th term. We note that every term before <math>2020</math> is one less than the previous term (that is, we subtract 1 to get the previous term). If <math>2020</math> is the 2064th term, than <math>2020 - 44</math> is the (2064 - 44)th term. So, the 2020th term is <math>1976</math>. The next term (term 2021) is <math>1977</math>, and the average of these two terms is the median, or <math>\boxed{\textbf{(C) } 1976.5}</math>.
 
We note that <math>44^2 = 1936</math>, which is the first square less than <math>2020</math>, which means that there are 44 addition terms before <math>2020</math>. This makes <math>2020</math> the 2064th term. To find the median, we need the 2020th and 2021th term. We note that every term before <math>2020</math> is one less than the previous term (that is, we subtract 1 to get the previous term). If <math>2020</math> is the 2064th term, than <math>2020 - 44</math> is the (2064 - 44)th term. So, the 2020th term is <math>1976</math>. The next term (term 2021) is <math>1977</math>, and the average of these two terms is the median, or <math>\boxed{\textbf{(C) } 1976.5}</math>.
 
~ primegn
 
~ primegn
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==Solution 5 (Decreasing Order)==
 +
To find the median, we sort the <math>4040</math> numbers in decreasing order, then average the <math>2020</math>th and the <math>2021</math>st numbers of the sorted list.
 +
 +
Since <math>45^2=2025</math> and <math>44^2=1936,</math> the first <math>2021</math> numbers of the sorted list are <cmath>\underbrace{2020^2,2019^2,2018^2,\cdots,46^2,45^2}_{1976\mathrm{ \ numbers}}\phantom{ },\phantom{ }\underbrace{2020,2019,2018,\cdots,1977,1976}_{45\mathrm{ \ numbers}}\phantom{ },</cmath> from which the answer is <math>\frac{1977+1976}{2}=\boxed{\textbf{(C)}\ 1976.5}.</math>
 +
 +
~MRENTHUSIASM
  
 
==Video Solution 1==
 
==Video Solution 1==
 +
https://youtu.be/luMQHhp_Rfk
  
 
Education, The Study of Everything
 
Education, The Study of Everything
 
https://youtu.be/luMQHhp_Rfk
 
  
 
==Video Solution 2==
 
==Video Solution 2==

Revision as of 19:48, 8 May 2021

The following problem is from both the 2020 AMC 12A #8 and 2020 AMC 10A #11, so both problems redirect to this page.

Problem

What is the median of the following list of $4040$ numbers$?$

\[1, 2, 3, ..., 2020, 1^2, 2^2, 3^2, ..., 2020^2\]

$\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$

Solution 1

We can see that $44^2=1936$ which is less than 2020. Therefore, there are $2020-44=1976$ of the $4040$ numbers greater than $2020$. Also, there are $2020+44=2064$ numbers that are less than or equal to $2020$.

Since there are 44 duplicates/extras, it will shift up our median's placement down $44$. Had the list of numbers been $1,2,3, \dots, 4040$, the median of the whole set would be $\dfrac{1+4040}{2}=2020.5$.

$2020.5-44$ gives us $1976.5$. Thus, our answer is $\boxed{\textbf{(C) } 1976.5}$.

~aryam

~Additions by BakedPotato66

Solution 2

As we are trying to find the median of a $4040$-term set, we must find the average of the $2020$th and $2021$st terms.

Since $45^2 = 2025$ is slightly greater than $2020$, we know that the $44$ perfect squares $1^2$ through $44^2$ are less than $2020$, and the rest are greater. Thus, from the number $1$ to the number $2020$, there are $2020 + 44 = 2064$ terms. Since $44^2$ is $44 + 45 = 89$ less than $45^2 = 2025$ and $84$ less than $2020$, we will only need to consider the perfect square terms going down from the $2064$th term, $2020$, after going down $84$ terms. Since the $2020$th and $2021$st terms are only $44$ and $43$ terms away from the $2064$th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$. Averaging the two, we get $\boxed{\textbf{(C) } 1976.5}.$ ~emerald_block

Solution 3

We want to know the $2020$th term and the $2021$st term to get the median.
We know that $44^2=1936$
So numbers $1^2, 2^2, ...,44^2$ are in between $1$ and $1936$.
So the sum of $44$ and $1936$ will result in $1980$, which means that $1936$ is the $1980$th number.
Also, notice that $45^2=2025$, which is larger than $2021$.
Then the $2020$th term will be $1936+40 = 1976$, and similarly the $2021$th term will be $1977$.
Solving for the median of the two numbers, we get $\boxed{\textbf{(C) } 1976.5}$
~toastybaker

Solution 4

We note that $44^2 = 1936$, which is the first square less than $2020$, which means that there are 44 addition terms before $2020$. This makes $2020$ the 2064th term. To find the median, we need the 2020th and 2021th term. We note that every term before $2020$ is one less than the previous term (that is, we subtract 1 to get the previous term). If $2020$ is the 2064th term, than $2020 - 44$ is the (2064 - 44)th term. So, the 2020th term is $1976$. The next term (term 2021) is $1977$, and the average of these two terms is the median, or $\boxed{\textbf{(C) } 1976.5}$. ~ primegn

Solution 5 (Decreasing Order)

To find the median, we sort the $4040$ numbers in decreasing order, then average the $2020$th and the $2021$st numbers of the sorted list.

Since $45^2=2025$ and $44^2=1936,$ the first $2021$ numbers of the sorted list are \[\underbrace{2020^2,2019^2,2018^2,\cdots,46^2,45^2}_{1976\mathrm{ \ numbers}}\phantom{ },\phantom{ }\underbrace{2020,2019,2018,\cdots,1977,1976}_{45\mathrm{ \ numbers}}\phantom{ },\] from which the answer is $\frac{1977+1976}{2}=\boxed{\textbf{(C)}\ 1976.5}.$

~MRENTHUSIASM

Video Solution 1

https://youtu.be/luMQHhp_Rfk

Education, The Study of Everything

Video Solution 2

https://youtu.be/ZGwAasE32Y4

~IceMatrix

Video Solution 3

https://youtu.be/B0RPkcjdkPU

~savannahsolver

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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
2020 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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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