Difference between revisions of "2021 AMC 12A Problems/Problem 16"
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==Solution== | ==Solution== | ||
− | + | ===Solution 1=== | |
+ | There are <math>1+2+..+199+200=\frac{(200)(201)}{2}=20100</math> numbers in total. Let the median be <math>k</math>. We want to find the median <math>k</math> such that | ||
+ | <cmath> \frac{k(k+1)}{2}=20100/2,</cmath> | ||
+ | or | ||
+ | <cmath> k(k+1)=20100.</cmath> | ||
+ | Note that <math>\sqrt{20100} \approx 142</math>. Plugging this value in as <math>k</math> gives | ||
+ | <cmath>\frac{1}{2}(142)(143)=10153.</cmath> | ||
+ | <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>. | ||
==Note== | ==Note== | ||
See [[2021 AMC 12A Problems/Problem 1|problem 1]]. | See [[2021 AMC 12A Problems/Problem 1|problem 1]]. |
Revision as of 15:37, 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. .
Note
See problem 1.
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 |
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