2023 AMC 8 Problems/Problem 21

Problem

Alina writes the numbers $1, 2, \dots, 9$ on separate cards, one number per card. She wishes to divide the cards into 3 groups of 3 cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?

$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$

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

Written Solution 2

The group with 5 must have the two other numbers adding up to 10, since the sum of each group is $\frac{1 + 2 \cdots + 9}{3}$ = (\frac{9(10)}{2}) = 15 $.. We can have$ (1, 5, 9) $,$ (2, 5, 8) $,$ (3, 5, 7) $, or$ (4, 5, 6) $. With the first group, we have$ (2, 3, 4, 6, 7, 8) $left over. The only way to form a group of 3 numbers that add up to 15 is with$ (3, 4, 8) $or$ (2, 6, 7) $. One of the possible arrangements is therefore$ (1, 5, 9) (3, 4, 8) (2, 6, 7) $. Then, with the second group, we have$ (1, 3, 4, 6, 7, 9) $left over. With these numbers, there is no way to form a group of 3 numbers adding to 15. Similarly, with the third group there is$ (1, 2, 4, 6, 8, 9) $left over and we can make a group of 3 numbers adding to 15 with$ (1, 6, 8) $or$ (2, 4, 9) $. Another arrangement is$ (3, 5, 7) (1, 6, 8) (2, 4, 9) $. Finally, the last group has$ (1, 2, 3, 7, 8, 9) $left over. There is no way to make a group of 3 numbers adding to 15 with this, so the arrangements are$ (1, 5, 9) (3, 4, 8) (2, 6, 7) $and$ (3, 5, 7) (1, 6, 8) (2, 4, 9) $. There are$ \boxed{\text{(C)}2}$ sets that can be formed. -Turtwig113

Video Solution 1 by OmegaLearn (Using Casework)

https://youtu.be/l1MfKj5MkWg

Animated Video Solution

https://youtu.be/_gpWj2lYers

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

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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

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