Question

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

Answer 1

Let the numbers be x, x+1, x+2.
when any of these numbers is divided by 3, the remainder obtained is either 0,1, or 2.
Hence, n = 3k, 3k+1, 3k+2 for some number k.
1st case:
If x = 3p, then it is clear that 3 divides x, that is, the first number

2nd case:
If x = 3p+1,
then x+2 = 3p+1+2
= 3p+3
=3(p+1), which is divisible by 3 and hence, 3 divides x+2, that is , the third number

3rd case:
If x = 3p+2,
then x+1 = 3p+2+1
= 3p+3
= 3(p+1), which is divisible by 3 and hence, 3 divides x+1, that is, the second number.

Hence in all the cases, any one number is divisible by 3. Hence proved.
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Answer 2

Step-by-step explanation:

Let 3 consecutive positive integers be n, n + 1 and n + 2 .

Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2.

:

Therefore:

n = 3p or 3p+1 or 3p+2, where p is some integer

If n = 3p = 3(p) , then n is divisible by 3

If n = 3p + 1, then n + 2 = 3p +1 + 2 = 3 p + 3 = 3 ( p + 1 ) is divisible by 3

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3

Thus, we can state that one of the numbers among n, n+1 and n+2 is always divisible by 3

Hence it is solved.