what is sum divisible by 3

Question

In how many ways can three numbers be chosen from the set {1,2,3,...,12} such that their sum is divisible by3 ?

Answer 1

Let A = {1, 2, 3, ......, 12}.

Let's define a new set B, where B[n] = A[n] mod 3

Then B = {1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0}

For the sake of clarity, I'm going to lebel each number with a subscript letter.

(like 4 mod 3 = (1 x 3 + 1) mod 3 = l_b and 7 mod 3 = (2 x 3 + 1) mod 3 = 1)

Then B = {1_a, 2_a, 0_a, 1_b, 2_b, 0_b, 1_c, 2_c, 0_c, 1_d, 2_d, 0_d}.

There are 4 zeroes, 4 ones and 4 twos.

Case I : 0 + 0 + 0 = 0 (mod 3)

Total possibilities of that case are \(4_{c_3}\) = 4

Case II : 0 + 1 + 2 = 3 mod 3 = 0

Total possibilities of that case are \(4_{c_1}\times4_{c_1}\times4_{c_1}\) = 4 x 4 x 4 = 64

Case III : 1 + 1 + 1 = 3 mod 3 = 0

Total possibilities of that case are \(4_{c_3}\) = 4

Case IV : 2 + 2 + 2 = 6 mod 3 = 0

Total possibilities of that case are \(4_{c_3}\) = 4

Therefore, total ways to selecting 3 numbers from set {1, 2, 3 ..... 12} whose sum is divisible by 3 = 4 + 64 + 4 + 4 = 76