Difference between revisions of "2023 AMC 8 Problems/Problem 21"
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| + | # of ways to split (1,2 , 3, 4, 5, 6, 7, 8, 9) into 3 equal groups | ||
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| + | 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{\text{(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 19:24, 24 January 2023
- of ways to split (1,2 , 3, 4, 5, 6, 7, 8, 9) into 3 equal groups
# of ways to split (1,2 , 3, 4, 5, 6, 7, 8, 9) into 3 equal groups
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 
