Question

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

Disclaimer: A lot of points to be given, Full explanation required. Not only answer. Remember the sum of the number must be divisible by 4. I think modular arithmetic is the way to solve it, but who knows???

Answer 1

• 55 ways

Step-by-step explanation:

Out of 12 consecutive integers:

• 3 - divide by 4, so the remainder is 0
• 3- give remainder of 1
• 3- give remainder of 2
• 3 - give remainder of 3

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

• 0 - 0 - 0 ⇒ 1 combination
• 0 - 1 - 3 ⇒ 3*3 = 9 combinations
• 0 - 3 - 1 ⇒ 3*3 = 9 combinations
• 1 - 1 - 2 ⇒ 2*3 = 6 combinations
• 1 - 2 - 1 ⇒ 2*3 = 6 combinations
• 2 - 1 - 1 ⇒ 2*3 = 6 combinations
• 3 - 0 - 1 ⇒ 3*3 = 9 combinations
• 3 - 1 - 0 ⇒ 3*3 = 9 combinations

So total number of combinations is:

• 1 + 9*4 + 6*3 = 55