Difference between revisions of "2023 AMC 8 Problems/Problem 21"
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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> | <math>\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4</math> | ||
− | ==Solution== | + | ==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 we we will just see the possible groups 9 can be with. The posible groups 9 can be with 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> | 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 we will just see the possible groups 9 can be with. The posible groups 9 can be with 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> |
Revision as of 02:07, 28 January 2023
Contents
[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?
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 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
Solution 2
The group with 5 must have the two other numbers adding up to 10, 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 3 numbers that add up to 15 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 3 numbers adding to 15. Similarly, with the third group there is left over and we can make a group of 3 numbers adding to 15 with or . Another arrangement is . Finally, the last group has left over. There is no way to make a group of 3 numbers adding to 15 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
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.