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

Revision as of 02:07, 28 January 2023

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?

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

Solution 1

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 $\boxed{\textbf{(C)}\ 2}$ 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

Solution 2

The group with 5 must have the two other numbers adding up to $10$ , since the sum of all the numbers is $(1 + 2 \cdots + 9)$ = $\frac{9(10)}{2}$ = $45$ . The sum of the numbers in each group must therefore be $\frac{45}{3}=15$ . 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 $\boxed{\textbf{(C)}\ 2}$ 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)

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=2853

@@ Line 11: / Line 11: @@
 ==Solution 2==
-The group with 5 must have the two other numbers adding up to 10, since the sum of all the numbers is <math>(1 + 2 \cdots + 9)</math> = <math>\frac{9(10)}{2}</math> = <math>45</math>. The sum of the numbers in each group must therefore be <math>\frac{45}{3}</math>=<math>15</math>. We can have <math>(1, 5, 9)</math>, <math>(2, 5, 8)</math>, <math>(3, 5, 7)</math>, or <math>(4, 5, 6)</math>. With the first group, we have <math>(2, 3, 4, 6, 7, 8)</math> left over. The only way to form a group of 3 numbers that add up to 15 is with <math>(3, 4, 8)</math> or <math>(2, 6, 7)</math>. One of the possible arrangements is therefore <math>(1, 5, 9) (3, 4, 8) (2, 6, 7)</math>. Then, with the second group, we have <math>(1, 3, 4, 6, 7, 9)</math> 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 <math>(1, 2, 4, 6, 8, 9)</math> left over and we can make a group of 3 numbers adding to 15 with <math>(1, 6, 8)</math> or <math>(2, 4, 9)</math>. Another arrangement is <math>(3, 5, 7) (1, 6, 8) (2, 4, 9)</math>. Finally, the last group has <math>(1, 2, 3, 7, 8, 9)</math> left over. There is no way to make a group of 3 numbers adding to 15 with this, so the arrangements are <math>(1, 5, 9) (3, 4, 8) (2, 6, 7)</math> and <math>(3, 5, 7) (1, 6, 8) (2, 4, 9)</math>. There are <math>\boxed{\textbf{(C)}\ 2}</math> sets that can be formed.
+The group with 5 must have the two other numbers adding up to <math>10</math>, since the sum of all the numbers is <math>(1 + 2 \cdots + 9)</math> = <math>\frac{9(10)}{2}</math> = <math>45</math>. The sum of the numbers in each group must therefore be <math>\frac{45}{3}=15</math>. We can have <math>(1, 5, 9)</math>, <math>(2, 5, 8)</math>, <math>(3, 5, 7)</math>, or <math>(4, 5, 6)</math>. With the first group, we have <math>(2, 3, 4, 6, 7, 8)</math> left over. The only way to form a group of 3 numbers that add up to <math>15</math> is with <math>(3, 4, 8)</math> or <math>(2, 6, 7)</math>. One of the possible arrangements is therefore <math>(1, 5, 9) (3, 4, 8) (2, 6, 7)</math>. Then, with the second group, we have <math>(1, 3, 4, 6, 7, 9)</math> left over. With these numbers, there is no way to form a group of 3 numbers adding to <math>15</math>. Similarly, with the third group there is <math>(1, 2, 4, 6, 8, 9)</math> left over and we can make a group of 3 numbers adding to <math>15</math> with <math>(1, 6, 8)</math> or <math>(2, 4, 9)</math>. Another arrangement is <math>(3, 5, 7) (1, 6, 8) (2, 4, 9)</math>. Finally, the last group has <math>(1, 2, 3, 7, 8, 9)</math> left over. There is no way to make a group of 3 numbers adding to <math>15</math> with this, so the arrangements are <math>(1, 5, 9) (3, 4, 8) (2, 6, 7)</math> and <math>(3, 5, 7) (1, 6, 8) (2, 4, 9)</math>. There are <math>\boxed{\textbf{(C)}\ 2}</math> sets that can be formed.
 ~Turtwig113

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