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

Revision as of 17:22, 15 August 2020

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

Problem 11

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 $44^2$ is less than 2020. Therefore, there are $1976$ of the $4040$ numbers after $2020$ . Also, there are $2064$ numbers that are under and equal to $2020$ . Since $44^2$ is equal to $1936$ , it, with the other squares, will shift our median's placement up $44$ . We can find that the median of the whole set is $2020.5$ , and $2020.5-44$ gives us $1976.5$ . Our answer is $\boxed{\textbf{(C) } 1976.5}$ .

~aryam

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 $1$ to $1936$ .
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 $2020 - 34$ is the (2064 - 34)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

Video Solution

https://youtu.be/ZGwAasE32Y4

~IceMatrix

@@ Line 29: / Line 29: @@
 Solving for the median of the two numbers, we get <math>\boxed{\textbf{(C) } 1976.5}</math> <br>
 ~toastybaker
+== Solution 4 ==
+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 - 34</math> is the (2064 - 34)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
 ==Video Solution==

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

Page

Toolbox

Search