Question

In how many ways can three different numbers be selected from the numbers 1 -12, so that their sum can be exactly divided by 3?

Please give a good way of doing it.

Thanks

Answer 1

Every third number would be divisible by 3. i.e. 1+1+1=3, 1+1+2=4, 1+1+3=5 etc.

Answer 2

Do you know how to find the total number of numbers?

Answer 3

That's what I want to know.

Answer 4

Each number has 12 options and there are three numbers. So 12^3.

Answer 5

The number of combinations in which 3 (k) things can be selected from 12 (n) things equals
n! / k! • (n - k)! =
12! / 3! • (9!) =
12 • 11 • 10 • 9! / 6 • 9! =
2 • 11 • 10 =
220 ways in which 3 numbers can be selected from 12
These would go from (1, 2, 3) all the way to (10, 11, 12)
If the sum of these 3 numbers is divisible by 3, 6, 9, 12, 15, etc then the sum is divisible by 3.
So, this narrows down the choices even further to 220/3 or 73 groups of numbers.

And heck, if you want to know what those 73 groups of numbers are, just look here:

1 2 3
1 2 6
1 2 9
1 2 12
1 3 5
1 3 8
1 3 11
1 4 7
1 4 10
1 5 6
1 5 9
1 5 12
1 6 8
1 6 11
1 7 10
1 8 9
1 8 12
1 9 11
1 11 12

2 3 4
2 3 7
2 3 10
2 4 9
2 4 12
2 5 8
2 6 7
2 6 10
2 7 9
2 7 12
2 8 11
2 9 10
2 10 12

3 4 5
3 4 8
3 4 11
3 5 7
3 5 10
3 6 9
3 6 12
3 7 8
3 7 11
3 8 10
3 9 12
3 10 11

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

5 6 10
5 7 9
5 7 12
5 8 11
5 9 10
5 10 12

6 7 8
6 7 11
6 8 10
6 9 12
6 10 11

7 8 9
7 8 12
7 9 11
7 11 12

8 9 10
8 10 12

9 10 11

10 11 12

Answer 6

Thanks very much

Answer 7

Do you know a little bit of modular arithmetic?