2023 AMC 8 Problems/Problem 21
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution
- 7 Video Solution by Math-X (Let's first Understand the question)
- 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
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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!!!)
~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.