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Difference between revisions of "2023 AMC 8 Problems/Problem 21"
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==Written Solution==
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==Problem==
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Alina writes the numbers <math>1, 2, \dots , 9</math> on separate cards, one number per card. She wishes to divide the cards into <math>3</math> groups of <math>3</math> cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?
  
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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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<math>\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4</math>
  
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
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==Solution 1==
  
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==Video Solution 1 by OmegaLearn (Using Casework)==
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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 will just see the possible groups 9 can be with. The possible groups 9 can be 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{\textbf{(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>
 +
 
 +
~CHECKMATE2021, apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
 +
 
 +
==Solution 2==
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The group with <math>5</math> 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) = \frac{9(10)}{2} = 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 <math>3</math> 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 <math>3</math> 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 <math>3</math> 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 <math>3</math> 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>. So，there are <math>\boxed{\textbf{(C)}\ 2}</math> sets that can be formed.
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~Turtwig113
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==Solution 3==
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The sum of the numbers across all equally valued sets is <math>(1 + 2 \cdots + 9) = \frac{9(10)}{2} = 45</math>. The value of the numbers in each set would be <math>\frac{45}{3} = \textbf{15}</math>. We know that the numbers <math>9</math>, <math>8</math>, and <math>7</math> must belong in different sets, as putting any <math>2</math> numbers in <math>1</math> set will either pass or match the limit of <math>15</math> per set, and we would then still need to add <math>1</math> more number after that. Note that these numbers must be distinct, as Alina only has <math>1</math> of each number, and order does not matter in the sets. Starting with the set that includes the number <math>9</math>, the next two numbers must add up to <math>6</math>, and there are <math>\textbf{2}</math> ways of doing this <math>(2,4) (1,5)</math>. Note we cannot use any number past <math>6</math>, as those numbers must be used in the other sets. The next set, which includes the number <math>8</math>, must have two numbers that add up to <math>7</math>, and there are <math>\textbf{3}</math> ways to do this <math>(2,5) (1,6) (3,4)</math>. The final set, which includes the number <math>7</math>, must have <math>2</math> numbers that sum up to <math>8</math>, and there are <math>\textbf{2}</math> ways to do this <math>(2,6) (3,5)</math>. Now we have found the number of ways in which each set sums up to <math>15</math>. To find the number of ways in which all three sets sum up to <math>15</math> concurrently, we must take the minimum of <math>2</math>, <math>3</math>, and <math>2</math>, which gives us an answer of <math>\boxed{\textbf{(C)}\ 2}</math> triplets of sets with 3 values, in which each set sum to the same amount.
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~Fernat123
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==Solution 4==
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Note that each group of numbers should sum to <math>\frac{1+2+3+4+5+6+7+8+9}{3} = 15.</math> Thus, this is equivalent to asking, “How many ways can you fill in a three by three magic square with the integers <math>1</math> through <math>9</math>?” 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 <math>1</math> through <math>9</math>, the center of the square would be <math>5</math> (the average of the numbers), and the numbers in the corners should be even(*). The such pairs (disregarding order) are <math>(2,8)</math> and <math>(4,6).</math> Let’s fix the position of <math>2</math> to be the top left corner. This would make <math>8</math> in the bottom right corner. We can have either <math>4</math> or <math>6</math> to be in the top right corner, for a total of <math>\boxed{\textbf{(C)}\ 2}</math> such groups of three. (The groups are <math>(8,3,4) (1,5,7) (6,9,2)</math> and <math>(8,1,6) (3,5,7) (4,9,2).</math>)
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Note that if we had instead fixed the position of <math>4</math>, <math>6</math>, or <math>8</math>, they would correspond to one of the two cases, just in a different configuration.
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(*)We can prove this using proof by contradiction. Label the nine small squares within the magic square from <math>a</math> to <math>i</math> from left to right, top to bottom. Firstly, we know that <math>a+i</math> and <math>c+g</math> sum to <math>10</math> since the center square is <math>5</math>. Thus, <math>a</math> and <math>i</math> must have the same parity, and so must <math>c</math> and <math>g</math>. Suppose that <math>a</math> and <math>c</math> have different parity. Since <math>a+b+c=15</math>, <math>b</math> must be even. By a similar argument, <math>h</math> must also be even, and so must <math>d</math> and <math>f</math>. Our initial assumption is that one of <math>a</math> and <math>c</math> 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 <math>1</math> to <math>9</math>. Now suppose that all of <math>a, c, g,</math> and <math>i</math> are odd. This would make each of <math>b, d, f,</math> and <math>h</math> odd, but clearly we do not have enough odd numbers to make all nine numbers odd. Thus, each corner square must be even.
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~ Brian__Liu
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==Video Solution==
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https://youtu.be/Ex54LNNPAwY
 +
 
 +
Please like and subscribe
 +
 
 +
==Video Solution by Math-X (Let's first Understand the question)==
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https://youtu.be/Ku_c1YHnLt0?si=S79t9BmOmSb-ACds&t=4641 ~Math-X
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 +
 
 +
==Video Solution (THINKING CREATIVELY!!!)==
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 +
https://youtu.be/egXB9xayUF8
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
==Video Solution 1 (Using Casework)==
 
https://youtu.be/l1MfKj5MkWg
 
https://youtu.be/l1MfKj5MkWg
  
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~Star League (https://starleague.us)
 
~Star League (https://starleague.us)
 +
 +
==Video Solution by Magic Square==
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https://youtu.be/-N46BeEKaCQ?t=2853
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==Video Solution by Interstigation==
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https://youtu.be/DBqko2xATxs&t=2747
 +
 +
==Video Solution by WhyMath==
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https://youtu.be/l9zexK9hiBo
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 +
~savannahsolver
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==Video Solution by harungurcan==
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https://www.youtube.com/watch?v=Ki4tPSGAapU&t=872s
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 +
~harungurcan
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 +
==See Also==
 +
{{AMC8 box|year=2023|num-b=20|num-a=22}}
 +
{{MAA Notice}}

Latest revision as of 21:16, 19 January 2024

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

~CHECKMATE2021, 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

Video Solution

https://youtu.be/Ex54LNNPAwY

Please like and subscribe

Video Solution by Math-X (Let's first Understand the question)

https://youtu.be/Ku_c1YHnLt0?si=S79t9BmOmSb-ACds&t=4641 ~Math-X


Video Solution (THINKING CREATIVELY!!!)

https://youtu.be/egXB9xayUF8

~Education, the Study of Everything


Video Solution 1 (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

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=2747

Video Solution by WhyMath

https://youtu.be/l9zexK9hiBo

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=Ki4tPSGAapU&t=872s

~harungurcan

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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