Difference between revisions of "2021 AMC 12A Problems/Problem 16"
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+ | {{duplicate|[[2021 AMC 10A Problems#Problem 16|2021 AMC 10A #16]] and [[2021 AMC 12A Problems#Problem 16|2021 AMC 12A #16]]}} | ||
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==Problem== | ==Problem== | ||
In the following list of numbers, the integer <math>n</math> appears <math>n</math> times in the list for <math>1 \leq n \leq 200</math>.<cmath>1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200</cmath>What is the median of the numbers in this list? | In the following list of numbers, the integer <math>n</math> appears <math>n</math> times in the list for <math>1 \leq n \leq 200</math>.<cmath>1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200</cmath>What is the median of the numbers in this list? |
Revision as of 17:43, 12 February 2021
- The following problem is from both the 2021 AMC 10A #16 and 2021 AMC 12A #16, so both problems redirect to this page.
Contents
[hide]Problem
In the following list of numbers, the integer appears times in the list for .What is the median of the numbers in this list?
Solution 1
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
Solution 2
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
Video Solution by Hawk Math
https://www.youtube.com/watch?v=AjQARBvdZ20
Video Solution by OmegaLearn (Using Algebra)
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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.