Difference between revisions of "2023 AMC 8 Problems/Problem 21"
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==Problem== | ==Problem== | ||
− | Alina writes the numbers <math>1, 2, \dots, 9</math> 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? | + | 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? |
<math>\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4</math> | <math>\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4</math> | ||
− | == | + | ==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. <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 | + | 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 | + | ~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 <math>(1 + 2 \cdots + 9) | + | 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. |
− | |||
− | ==Video Solution 1 | + | ~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 | ||
+ | |||
+ | ==Solution 4== | ||
+ | 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>) | ||
+ | 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. | ||
+ | |||
+ | |||
+ | (*)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. | ||
+ | |||
+ | ~ Brian__Liu | ||
+ | |||
+ | ==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== | ||
+ | https://youtu.be/Ex54LNNPAwY | ||
+ | |||
+ | Please like and subscribe | ||
+ | |||
+ | ==Video Solution (THINKING CREATIVELY!!!)== | ||
+ | |||
+ | 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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==Video Solution by Magic Square== | ==Video Solution by Magic Square== | ||
https://youtu.be/-N46BeEKaCQ?t=2853 | 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== | ==See Also== | ||
{{AMC8 box|year=2023|num-b=20|num-a=22}} | {{AMC8 box|year=2023|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:14, 17 June 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution by Math-X (Let's first Understand the question)
- 7 Video Solution
- 8 Video Solution (THINKING CREATIVELY!!!)
- 9 Video Solution 1 (Using Casework)
- 10 Animated Video Solution
- 11 Video Solution by Magic Square
- 12 Video Solution by Interstigation
- 13 Video Solution by WhyMath
- 14 Video Solution by harungurcan
- 15 See Also
Problem
Alina writes the numbers on separate cards, one number per card. She wishes to divide the cards into
groups of
cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?
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. . Then, dividing by
, we have
, so each group of
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
and
. Going down each of these avenues, we will repeat the same process for
using the remaining elements in the list. Where there is only 1 set of elements getting the sum of
,
needs in both cases. After
is decided, the remaining 3 elements are forced in a group, yielding us an answer of
as our sets are
and
~CHECKMATE2021, apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
Solution 2
The group with must have the two other numbers adding up to
, since the sum of all the numbers is
. The sum of the numbers in each group must therefore be
. We can have
,
,
, or
. With the first group, we have
left over. The only way to form a group of
numbers that add up to
is with
or
. One of the possible arrangements is therefore
. Then, with the second group, we have
left over. With these numbers, there is no way to form a group of
numbers adding to
. Similarly, with the third group there is
left over and we can make a group of
numbers adding to
with
or
. Another arrangement is
. Finally, the last group has
left over. There is no way to make a group of
numbers adding to
with this, so the arrangements are
and
. So,there are
sets that can be formed.
~Turtwig113
Solution 3
The sum of the numbers across all equally valued sets is . The value of the numbers in each set would be
. We know that the numbers
,
, and
must belong in different sets, as putting any
numbers in
set will either pass or match the limit of
per set, and we would then still need to add
more number after that. Note that these numbers must be distinct, as Alina only has
of each number, and order does not matter in the sets. Starting with the set that includes the number
, the next two numbers must add up to
, and there are
ways of doing this
. Note we cannot use any number past
, as those numbers must be used in the other sets. The next set, which includes the number
, must have two numbers that add up to
, and there are
ways to do this
. The final set, which includes the number
, must have
numbers that sum up to
, and there are
ways to do this
. Now we have found the number of ways in which each set sums up to
. To find the number of ways in which all three sets sum up to
concurrently, we must take the minimum of
,
, and
, which gives us an answer of
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 Thus, this is equivalent to asking, “How many ways can you fill in a three by three magic square with the integers
through
?” 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
through
, the center of the square would be
(the average of the numbers), and the numbers in the corners should be even(*). The such pairs (disregarding order) are
and
Let’s fix the position of
to be the top left corner. This would make
in the bottom right corner. We can have either
or
to be in the top right corner, for a total of
such groups of three. (The groups are
and
)
Note that if we had instead fixed the position of
,
, or
, 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 to
from left to right, top to bottom. Firstly, we know that
and
sum to
since the center square is
. Thus,
and
must have the same parity, and so must
and
. Suppose that
and
have different parity. Since
,
must be even. By a similar argument,
must also be even, and so must
and
. Our initial assumption is that one of
and
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
to
. Now suppose that all of
and
are odd. This would make each of
and
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 by Math-X (Let's first Understand the question)
https://youtu.be/Ku_c1YHnLt0?si=S79t9BmOmSb-ACds&t=4641
~Math-X
Video Solution
Please like and subscribe
Video Solution (THINKING CREATIVELY!!!)
~Education, the Study of Everything
Video Solution 1 (Using Casework)
Animated Video Solution
~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
~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.