Difference between revisions of "2021 AMC 12A Problems/Problem 16"
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<math>10153-142<10050</math>, so <math>142</math> is the <math>152</math>nd and <math>153</math>rd numbers, and hence, our desired answer. <math>\fbox{(C) 142}.</math>. | <math>10153-142<10050</math>, so <math>142</math> is the <math>152</math>nd and <math>153</math>rd numbers, and hence, our desired answer. <math>\fbox{(C) 142}.</math>. | ||
===Solution 2=== | ===Solution 2=== | ||
− | The <math>x</math>th number of this sequence is | + | The <math>x</math>th number of this sequence is <math>\frac{-1\pm\sqrt{1+8x}}{2}</math> via the quadratic formula. We can see that if we halve <math>x</math> we end up getting <math>\frac{-1\pm\sqrt{1+4x}}{2}</math>. This is approximately the number divided by <math>\sqrt{2}</math>. <math>\frac{200}{\sqrt{2}} = 141.4</math> and since <math>142</math> looks like the only number close to it, it is answer <math>\boxed{(C) 142}</math> ~Lopkiloinm |
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== |
Revision as of 00:02, 12 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 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 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.