2020 AMC 10A Problems/Problem 11

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 of the list, 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 $\underbrace{2020^2,2019^2,2018^2,\cdots,45^2}_{1976\text{ numbers}}\phantom{ },\phantom{ }\underbrace{2020,2019,2018,\cdots,1977}_{44\text{ numbers}}\phantom{ },\phantom{ }1976.$ Finally, the answer is $\frac{1977+1976}{2}=\boxed{\textbf{(C)}\ 1976.5}.$

~MRENTHUSIASM

Video Solution 1

Education, The Study of Everything

https://youtu.be/luMQHhp_Rfk

Video Solution 2

https://youtu.be/ZGwAasE32Y4

~IceMatrix

Video Solution 3

https://youtu.be/B0RPkcjdkPU

~savannahsolver

2020 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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