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

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==Solution 1==
 
==Solution 1==
  
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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 possible 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{\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>
+
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>
  
 
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
 
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
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~Turtwig113
 
~Turtwig113
 +
 +
==Solution 3==
 +
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.
 +
 +
~Fernat123
  
 
==Video Solution (THINKING CREATIVELY!!!)==
 
==Video Solution (THINKING CREATIVELY!!!)==

Revision as of 11:48, 1 August 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 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

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/1bA7fD7Lg54?t=2062

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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