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

Revision as of 01:56, 2 February 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 $1936$ 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 $2021$ th 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

Video Solution

https://youtu.be/ZGwAasE32Y4

~IceMatrix

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 23: / Line 23: @@
 <br>
 We know that <math>44^2=1936</math> <br>
-So numbers <math>1^2, 2^2, ...,44^2</math> are between <math>1</math> to <math>1936</math>. <br>
+So numbers <math>1^2, 2^2, ...,44^2</math> are in between <math>1</math> to <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>
 Also, notice that <math>45^2=2025</math>, which is larger than <math>2021</math>. <br>

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