Prove that one and only one out of any three consecutive positive integers is divisible by 3.

The problem is based on the concept of divisibility by 3.

Answer: Out of any three consecutive positive integers only one number is divisible by 3. The proof is given below.

Let's explore the nature of multiples of 3.

Explanation:

Let n, n+1, n+ 2 be the three consecutive positive integers.

• case1: if n is a multiple of 3
  • n = 3q
  • n + 1 = 3q +1 [ not divisible by 3 ]
  • n + 2 = 3q + 2  [ not divisible by 3 ]
• case2: if n+1 is a multiple of 3
  • n+1 = 3q
  • n = 3q - 1 [ not divisible by 3 ]
  • n + 2 = 3q + 1  [ not divisible by 3 ]
• case3: if n+2 is a multiple of 3 
  • n+2 = 3q
  • n = 3q - 2 [ not divisible by 3 ]
  • n + 1 = 3q - 1  [ not divisible by 3 ]

Thus, only one among the three consecutive positive integers only one number is divisible by 3.