2023 AMC 8 Problems/Problem 21
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[hide]Problem
Alina writes the numbers 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?
Written Solution
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 posible 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
Written Solution 2
The group with 5 must have the two other numbers adding up to 10, since the sum of each group is = (\frac{9(10)}{2}) = 15(1, 5, 9)(2, 5, 8)(3, 5, 7)(4, 5, 6)(2, 3, 4, 6, 7, 8)(3, 4, 8)(2, 6, 7)(1, 5, 9) (3, 4, 8) (2, 6, 7)(1, 3, 4, 6, 7, 9)(1, 2, 4, 6, 8, 9)(1, 6, 8)(2, 4, 9)(3, 5, 7) (1, 6, 8) (2, 4, 9)(1, 2, 3, 7, 8, 9)(1, 5, 9) (3, 4, 8) (2, 6, 7)(3, 5, 7) (1, 6, 8) (2, 4, 9)\boxed{\text{(C)}2}$ sets that can be formed. -Turtwig113
Video Solution 1 by OmegaLearn (Using Casework)
Animated Video Solution
~Star League (https://starleague.us)
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 |
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