Question

In how many ways can three numbers be chosen from the set {1,2,3,...,12} such that their sum is divisible by 3? ​

Answer 1

Given: a set {1,2,3,4,5,6,7,8,9,10,11,12}.

To find: how many ways can three numbers be chosen from the set such that their sum is divisible by 3.

Solution:

• A Set is a collection of numbers or mathematical objects like symbols, lines etc.
• By the given Data, Sum of three numbers must be divisible by 3.
• From the set the numbers that are divisible by 3 is 3, 6, 9, 12.
• Calculating three numbers sum, Sum of three numbers cannot be the result for 3.
• For 6 ; Only with one way we can get 6. The sum of three numbers are 1+2+3= 6.
• For 9 ; There are Three ways to get 9 they are, The sum of three numbers are 1+2+6=9, 1+3+5=9, 2+3+4=9.
• For 12 ; There are Five ways to get 12 they are, The sum of three numbers are 1+2+9=12, 4+3+5=12, 2+3+7=12, 4+6+2=12, 5+6+1=12.
• Hence there are 9 ways (1 + 3 + 5) where three numbers can be chosen from the set {1,2,3,...,12} such that their sum is divisible by 3.

Answer 2

There are 76 ways to choose three numbers rom the set {1,2,3,...,12} such that their sum is divisible by 3

Choosing  3 numbers from the set  {1,2,3,...,12}  can have

Minimum sum 6

Maximum Sum 33

Divisible by 3  are : 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30 , 33

Case 1 :  Sum 6

1 + 2 + 3 = 6  

Only 1 combination

Case 2 :  Sum 9

1 + 2 + 6   , 1 + 3 + 5 , 2 + 3 + 4

3 combination

Case 3 :  Sum 12

1 + 2 + 9  , 1 + 3 + 8 , 1 + 4 + 7, 1 + 5 + 6 , 2 + 3 + 7 , 2 + 4 + 6 , 3 + 4 + 5

7 combination

Case 4 :  Sum 15

1 + 2 + 12  , 1 + 3 + 11 , 1 + 4 + 10, 1 + 5 + 9 , 1+6 + 8 ,

2 + 3 + 10 , 2 + 4 + 9 , 2 + 5 + 8 , 2 + 6 + 7 ,

3 + 4 + 8 , 3 + 5 + 7 ,

4 + 5 + 6

12 combination

Case 5 :  Sum 18

1 + 5 + 12  , 1 + 6 + 11 , 1 + 7 + 10, 1 + 8 + 10 ,

2 +4 + 12 , 2 + 5 + 11 , 2 + 6 + 10 , 2 + 7 +9 ,

3 + 4 + 11 , 3 + 5+ 10 , 3 + 6 + 9 , 3 + 7 + 8

4 + 5 + 9 , 4 + 6 + 8 ,

5 + 6 + 7

15 combination

Case 6 :  Sum 21

1 + 8 + 12  , 1 + 9 + 11  

2 +7 + 12 , 2 + 8 + 11 , 2 + 9 + 10

3 + 6 + 12, 3 + 7 + 11 , 3 + 8+ 10

4 + 5 + 12 , 4 + 6 + 11 , 4 + 7 + 10 , 4 + 8 + 9

5 + 6 + 10 , 5 + 7 + 9

6 + 7 + 8

15 combination

Case 7 :  Sum 24

1 + 11 + 12

2 + 10 + 12  

3 + 9 + 12 , 3 + 10 + 11 ,

4 + 8 +12 , 4 + 9 + 11

5 + 7 + 12 , 5 + 8 + 11 , 5 + 9 + 10

6 + 7 + 11 , 6 + 8 + 10

7 + 8 + 9

12 combination

Case 8 :  Sum 27

4 + 11 + 12  

5 + 10 + 12

6 + 9 +12 , 6 + 10 + 11

7 + 8 + 12 , 7 + 9 + 11

8 + 9 + 10

7 combination

Case 9 :  Sum 30

7 + 11 + 12  

8 + 10 + 12

9 + 10 + 11

3 combination

Case 10 :  Sum 33

10+ 11 + 12  

1 combination

Total Combinations

1 + 3 + 7 + 12 + 15 + 15 + 12 + 7 + 3 + 1 = 76

Hence There are 76 ways to choose three numbers rom the set {1,2,3,...,12} such that their sum is divisible by 3

Another Method:

set {1,2,3,...,12}

Set with mod 3

{1 , 2 , 0 , 1 , 2 , 0 , 1 , 2 , 0 , 1 , 2 , 0 }

4 Times 0 , 4 times 1 , 4 times 2

Sum Drivable by 3  , 0 + 0 + 0 , 0 + 1 + 2 , 1+ 1 + 1 , 2 + 2 + 2

0 + 0 + 0  = ⁴C₃  = 4    ( 3 zero to be selected out of 4)

0 + 1 + 2 =  ⁴C₁⁴C₁⁴C₁ = 64  (1 zero out of 4 , 1 one out of 4 and 1 two out of 4 to be selected)

1 + 1 + 1  = ⁴C₃  = 4   ( 3 one to be selected out of 4)

2 + 2 + 2  = ⁴C₃  = 4   ( 3 two to be selected out of 4)

Hence Total ways = 4 + 64 + 4 + 4 = 76

There are 76 ways to choose three numbers rom the set {1,2,3,...,12} such that their sum is divisible by 3