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2023 AMC 8 Problems/Problem 21

Revision as of 10:03, 2 January 2024 by Rohinishu (talk | contribs) (→‎Solution 1)

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 will just see the possible groups 9 can be with. The possible groups 9 can be 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 $(3, 5, 7) (1, 6, 8) (2, 4, 9)$. So，there are $\boxed{\textbf{(C)}\ 2}$ sets that can be formed.

~Turtwig113

Solution 3

The sum of the numbers across all equally valued sets is $(1 + 2 \cdots + 9) = \frac{9(10)}{2} = 45$. The value of the numbers in each set would be $\frac{45}{3} = \textbf{15}$. We know that the numbers $9$, $8$, and $7$ must belong in different sets, as putting any $2$ numbers in $1$ set will either pass or match the limit of $15$ per set, and we would then still need to add $1$ more number after that. Note that these numbers must be distinct, as Alina only has $1$ of each number, and order does not matter in the sets. Starting with the set that includes the number $9$, the next two numbers must add up to $6$, and there are $\textbf{2}$ ways of doing this $(2,4) (1,5)$. Note we cannot use any number past $6$, as those numbers must be used in the other sets. The next set, which includes the number $8$, must have two numbers that add up to $7$, and there are $\textbf{3}$ ways to do this $(2,5) (1,6) (3,4)$. The final set, which includes the number $7$, must have $2$ numbers that sum up to $8$, and there are $\textbf{2}$ ways to do this $(2,6) (3,5)$. Now we have found the number of ways in which each set sums up to $15$. To find the number of ways in which all three sets sum up to $15$ concurrently, we must take the minimum of $2$, $3$, and $2$, which gives us an answer of $\boxed{\textbf{(C)}\ 2}$ triplets of sets with 3 values, in which each set sum to the same amount.

~Fernat123

Solution 4

Note that each group of numbers should sum to $\frac{1+2+3+4+5+6+7+8+9}{3} = 15.$ Thus, this is equivalent to asking, “How many ways can you fill in a three by three magic square with the integers $1$ through $9$?” since we can take the three rows of the magic square as our three groups. If you have closely studied magic squares, you might know that in a three by three magic square that is to be filled in with the integers $1$ through $9$, the center of the square would be $5$ (the average of the numbers), and the numbers in the corners should be even(*). The such pairs (disregarding order) are $(2,8)$ and $(4,6).$ Let’s fix the position of $2$ to be the top left corner. This would make $8$ in the bottom right corner. We can have either $4$ or $6$ to be in the top right corner, for a total of $\boxed{\textbf{(C)}\ 2}$ such groups of three. (The groups are $(8,3,4) (1,5,7) (6,9,2)$ and $(8,1,6) (3,5,7) (4,9,2).$) Note that if we had instead fixed the position of $4$, $6$, or $8$, they would correspond to one of the two cases, just in a different configuration.


(*)We can prove this using proof by contradiction. Label the nine small squares within the magic square from $a$ to $i$ from left to right, top to bottom. Firstly, we know that $a+i$ and $c+g$ sum to $10$ since the center square is $5$. Thus, $a$ and $i$ must have the same parity, and so must $c$ and $g$. Suppose that $a$ and $c$ have different parity. Since $a+b+c=15$, $b$ must be even. By a similar argument, $h$ must also be even, and so must $d$ and $f$. Our initial assumption is that one of $a$ and $c$ is odd and the other is even; however, we end up with six even numbers needed to fill in the square, but there are only four even integers from $1$ to $9$. Now suppose that all of $a, c, g,$ and $i$ are odd. This would make each of $b, d, f,$ and $h$ odd, but clearly we do not have enough odd numbers to make all nine numbers odd. Thus, each corner square must be even.

~ Brian__Liu

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See Also

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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