Difference between revisions of "2023 AMC 8 Problems/Problem 20"
m (→Solution) |
MRENTHUSIASM (talk | contribs) (→Video Solution by WhyMath) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 8: | Line 8: | ||
~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower | ||
| − | |||
==Animated Video Solution== | ==Animated Video Solution== | ||
https://youtu.be/ItntB7vEafM | https://youtu.be/ItntB7vEafM | ||
| Line 26: | Line 25: | ||
~savannahsolver | ~savannahsolver | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/FsT5WyQJbQ0 | ||
| + | |||
| + | Please like and subscribe!! | ||
==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}} | ||
Revision as of 04:14, 20 March 2023
Contents
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
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. Remember, anything less than 
is not beneficial to the optimization. So, taking our optimal values of 
and , we have an answer of 
.
~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower
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/1bA7fD7Lg54?t=1970
Video Solution by WhyMath
~savannahsolver
Video Solution
Please like and subscribe!!
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
