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Difference between revisions of "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 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 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 $2021$th term will be $1977$.
Solving for the median of the two numbers, we get $\boxed{\textbf{(C) } 1976.5}$ ~toastybaker

Video Solution

~IceMatrix

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