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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.
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