2023 AMC 8 Problems/Problem 21
Contents
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 we will just see the possible groups 9 can be with. The possible groups 9 can be with 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
~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 . There are sets that can be formed.
~Turtwig113
Video Solution 1 by OmegaLearn (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/1bA7fD7Lg54?t=2062
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.