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2023 AMC 8 Problems/Problem 21

Revision as of 18:53, 24 January 2023 by Sohumuttamchandani (talk | contribs) (→‎Animated Video Solution)
  1. of ways to split (1,2 , 3, 4, 5, 6, 7, 8, 9) into 3 equal groups



# of ways to split (1,2 , 3, 4, 5, 6, 7, 8, 9) into 3 equal groups


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 $(9, 2, 4) (8, 1, 6) (7, 3 ,5)$


Animated Video Solution

https://youtu.be/_gpWj2lYers

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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 $(9, 2, 4) (8, 1, 6) (7, 3 ,5)$

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