Difference between revisions of "2023 AMC 8 Problems/Problem 21"
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat | ||
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| + | ==Video Solution 1 by OmegaLearn (Using Casework)== | ||
| + | https://youtu.be/l1MfKj5MkWg | ||
==Animated Video Solution== | ==Animated Video Solution== | ||
Revision as of 20:49, 24 January 2023
Written Solution
First we need to find the sum of each group when split. This is the total sum of all the elements divided by the # of groups. 
. Then dividing by 
we have 
so each group of must have a sum of 15. To make the counting easier we we will just see the possible groups 9 can be with. The posible groups 9 can be with with 2 distinct numbers are 
and 
. Going down each of these avenues we will repeat the same process for 
using the remaining elements in the list. Where there is only 1 set of elements getting the sum of 
, needs in both cases. After is decided the remaining 3 elements are forced in a group. Yielding us an answer of 
as our sets are 
and 
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
Video Solution 1 by OmegaLearn (Using Casework)
Animated Video Solution
~Star League (https://starleague.us)
