Difference between revisions of "2021 AMC 12A Problems/Problem 16"
Pi is 3.14 (talk | contribs) (→Video Solution by Hawk Math) |
Pi is 3.14 (talk | contribs) (→Video Solution by pi_is_3.14 (Using Vandermonde's Identity)) |
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https://www.youtube.com/watch?v=AjQARBvdZ20 | https://www.youtube.com/watch?v=AjQARBvdZ20 | ||
− | == Video Solution by pi_is_3.14 (Using | + | == Video Solution by pi_is_3.14 (Using Algebra) == |
− | https://youtu.be/ | + | https://youtu.be/HkwgH9Lc1hE |
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=15|num-a=17}} | {{AMC12 box|year=2021|ab=A|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:26, 11 February 2021
Contents
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
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. .
Solution 2
The th number of this sequence is obviously 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 pi_is_3.14 (Using Algebra)
See also
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.