2023 AMC 8 Problems/Problem 20: Difference between revisions
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==Solution 1 == | ==Solution 1 == | ||
To double the range, we must find the current range, which is <math>28 - 3 = 25</math>, to then double to: <math>2(25) = 50</math>. Since we do not want to change the median, we need to get a value less than <math>8</math> (as <math>8</math> would change the mode) for the smaller, making <math>53</math> fixed for the larger. | To double the range, we must find the current range, which is <math>28 - 3 = 25</math>, to then double to: <math>2(25) = 50</math>. Since we do not want to change the median, we need to get a value less than <math>8</math> (as <math>8</math> would change the mode) for the smaller, making <math>53</math> fixed for the larger. Anything less than <math>3</math> is not beneficial to the optimization because you want to get the largest range without changing the mode. So, taking our optimal values of <math>7</math> and <math>53</math>, we have an answer of <math>7 + 53 = \boxed{\textbf{(D)}\ 60}</math>. | ||
~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower, CHECKMATE2021 | |||
==Video Solution by STEM Station(Quick and Easy to Understand)== | |||
https://www.youtube.com/watch?v=qZO9JLezr-U | |||
==Video Solution (A Clever Explanation You’ll Get Instantly)== | |||
https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=2294 | |||
~hsnacademy | |||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/BdkBQppueWY | https://youtu.be/BdkBQppueWY | ||
| Line 13: | Line 21: | ||
Please like and subscribe | Please like and subscribe | ||
== | ==Solution for AMC 8 Problem 20. Easy and understandable. Enjoy! Type in the comments if you have any questions! (Sridipto Banerjee)== | ||
https://youtu.be/ | |||
https://youtu.be/YeCldZn5qD8 | |||
Subscribe for more videos and please like! | |||
==Video Solution (CREATIVE THINKING!!!)== | ==Video Solution (CREATIVE THINKING) & (Very fast paced!!!!!!)== | ||
https://youtu.be/NpVLhU3AgNg | https://youtu.be/NpVLhU3AgNg | ||
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~savannahsolver | ~savannahsolver | ||
==Video Solution by harungurcan== | ==Video Solution by harungurcan== | ||
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~harungurcan | ~harungurcan | ||
==Video Solution by Dr. David== | |||
https://youtu.be/mMU-uvgwVzg | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2023|num-b=19|num-a=21}} | {{AMC8 box|year=2023|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Introductory Number Theory Problems]] | |||
Latest revision as of 05:51, 29 August 2025
Problem
Two integers are inserted into the list 
to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?
Solution 1
To double the range, we must find the current range, which is 
, to then double to: 
. Since we do not want to change the median, we need to get a value less than 
(as would change the mode) for the smaller, making 
fixed for the larger. Anything less than 
is not beneficial to the optimization because you want to get the largest range without changing the mode. So, taking our optimal values of 
and , we have an answer of 
.
~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower, CHECKMATE2021
Video Solution by STEM Station(Quick and Easy to Understand)
https://www.youtube.com/watch?v=qZO9JLezr-U
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=2294 ~hsnacademy
Video Solution
Please like and subscribe
Solution for AMC 8 Problem 20. Easy and understandable. Enjoy! Type in the comments if you have any questions! (Sridipto Banerjee)
Subscribe for more videos and please like!
Video Solution (CREATIVE THINKING) & (Very fast paced!!!!!!)
~Education, the Study of Everything
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by OmegaLearn (Using Smart Sequence Analysis)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=3136
Video Solution by Interstigation
https://youtu.be/DBqko2xATxs&t=2625
Video Solution by WhyMath
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=534s
~harungurcan
Video Solution by Dr. David
See Also
| 2023 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
