Difference between revisions of "2023 AMC 8 Problems/Problem 21"

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==Written Solution==
  
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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. <math>1 + 2 \cdots + 9 = \frac{9(10)}{2} = 45</math>. Then dividing by <math>3</math> we have <math>\frac{45}{3} = 15</math> so each group of <math>3</math> 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 <math>(9, 2, 4)</math> and <math>(9, 1, 5)</math>. Going down each of these avenues we will repeat the same process for <math>8</math> using the remaining elements in the list. Where there is only 1 set of elements getting the sum of <math>7</math>, <math>8</math> needs in both cases. After <math>8</math> is decided the remaining 3 elements are forced in a group. Yielding us an answer of <math>\boxed{\text{(C)}2}</math> as our sets are <math>(9, 1, 5) (8, 3, 4) (7, 2, 6)</math> and <math>(9, 2, 4) (8, 1, 6) (7, 3 ,5)</math>
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
  
 
==Animated Video Solution==
 
==Animated Video Solution==
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~Star League (https://starleague.us)
 
~Star League (https://starleague.us)
==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. <math>1 + 2 \cdots + 9 = \frac{9(10)}{2} = 45</math>. Then dividing by <math>3</math> we have <math>\frac{45}{3} = 15</math> so each group of <math>3</math> 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 <math>(9, 2, 4)</math> and <math>(9, 1, 5)</math>. Going down each of these avenues we will repeat the same process for <math>8</math> using the remaining elements in the list. Where there is only 1 set of elements getting the sum of <math>7</math>, <math>8</math> needs in both cases. After <math>8</math> is decided the remaining 3 elements are forced in a group. Yielding us an answer of <math>\boxed{\text{(C)}2}</math> as our sets are <math>(9, 1, 5) (8, 3, 4) (7, 2, 6)</math> and <math>(9, 2, 4) (8, 1, 6) (7, 3 ,5)</math>
 
 
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
 

Revision as of 20:01, 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. $1 + 2 \cdots + 9 = \frac{9(10)}{2} = 45$. Then dividing by $3$ we have $\frac{45}{3} = 15$ so each group of $3$ 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 $(9, 2, 4)$ and $(9, 1, 5)$. Going down each of these avenues we will repeat the same process for $8$ using the remaining elements in the list. Where there is only 1 set of elements getting the sum of $7$, $8$ needs in both cases. After $8$ is decided the remaining 3 elements are forced in a group. Yielding us an answer of $\boxed{\text{(C)}2}$ as our sets are $(9, 1, 5) (8, 3, 4) (7, 2, 6)$ and $(9, 2, 4) (8, 1, 6) (7, 3 ,5)$

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat

Animated Video Solution

https://youtu.be/_gpWj2lYers

~Star League (https://starleague.us)