Difference between revisions of "2020 AMC 10A Problems/Problem 11"
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| + | ~IceMatrix | ||
==See Also== | ==See Also== | ||
Revision as of 23:15, 31 January 2020
- The following problem is from both the 2020 AMC 12A #8 and 2020 AMC 10A #11, so both problems redirect to this page.
Contents
Problem 11
What is the median of the following list of 
numbers
![\[1, 2, 3, ..., 2020, 1^2, 2^2, 3^2, ..., 2020^2\]](http://latex.artofproblemsolving.com/9/2/a/92a67eb9979c30da61da9773b88914c0df6de226.png)
Solution
We can see that 
is less than 2020. Therefore, there are 
of the numbers after 
. Also, there are 
numbers that are under and equal to . Since is 
it, with the other squares will shift our median's placement up 
. We can find that the median of the whole set is 
, and 
gives us 
. Our answer is 
.
~aryam
Video Solution
~IceMatrix
See Also
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
