Difference between revisions of "2023 AMC 8 Problems/Problem 21"
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| + | ==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? | ||
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| + | <math>\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4</math> | ||
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==Written Solution== | ==Written Solution== | ||
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~Star League (https://starleague.us) | ~Star League (https://starleague.us) | ||
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| + | ==See Also== | ||
| + | {{AMC8 box|year=2023|num-b=20|num-a=22}} | ||
| + | {{MAA Notice}} | ||
Revision as of 22:42, 24 January 2023
Contents
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
