2021 AMC 12A Problems/Problem 16
- The following problem is from both the 2021 AMC 10A #16 and 2021 AMC 12A #16, so both problems redirect to this page.
In the following list of numbers, the integer appears times in the list for .What is the median of the numbers in this list?
There are numbers in total. Let the median be . We want to find the median such that or Note that . Plugging this value in as gives , so is the nd and rd numbers, and hence, our desired answer. .
Note that we can derive through the formula where is a perfect square less than or equal to . We set to , so , and . We then have . ~approximation by ciceronii
Note by Fasolinka (use answer choices): Once you know that the answer is in the 140s range by the approximation, it is highly improbable for the answer to be anything but C.
The th number of this sequence is via the quadratic formula. We can see that if we halve we end up getting . This is approximately the number divided by . and since looks like the only number close to it, it is answer ~Lopkiloinm
Solution 3 (answer choices)
We can look at answer choice , which is first. That means that the number of numbers from to is roughly the number of numbers from to .
The number of numbers from to is which is approximately The number of numbers from to is which is approximately as well. Therefore, we can be relatively sure the answer choice is
We can arrange the numbers in the following pattern:
We can see this as a isosceles right triangle, with legs of length
Let be the side length such that both sides of the triangle have the same area. The desired answer is then around because about half of the numbers in the list fall on each side.
Solving for yields: We see that is the closest to by far, and thus, can be relatively certain this is the answer.
Video Solution by Punxsutawney Phil
Video Solution by Hawk Math
Video Solution by Answer Choice
https://www.youtube.com/watch?v=YxWjDcUcaeQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=13 ~North America Math Contest Go Go Go
Video Solution by pi_is_3.14 (Using Algebra)
Video Solution by TheBeautyofMath
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