give 12 consecutive integers, in how many ways can three of these integers be selected to give a sum which divides by 4?

55 ways

Step-by-step explanation:

Out of 12 consecutive integers:

3 - divide by 4, so the remainder is 03- give remainder of 13- give remainder of 23 - give remainder of 3

Sum of 3 integers will be divisible by 4 if the remainders are:

0 - 0 - 0 ⇒ 1 combination0 - 1 - 3 ⇒ 3*3 = 9 combinations0 - 3 - 1 ⇒ 3*3 = 9  combinations1 - 1 - 2  ⇒ 2*3 = 6  combinations1 - 2 - 1  ⇒ 2*3 = 6  combinations2 - 1 - 1 ⇒ 2*3 = 6 combinations3 - 0 - 1 ⇒ 3*3 = 9 combinations3 - 1 - 0 ⇒ 3*3 = 9 combinations

So total number of combinations is:

1 + 4*9 + 3*6 = 55

Were is the problem