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  • Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.

Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.

Answers (1)

Any positive integer can be written in form of 3m, 3m + 1 or 3m + 2
Case-1:
n = 3m
  3m is divisible by 3
Adding two on both sides
n + 2 = 3m + 2
Here when n + 2 is divided by 3 it leaves remainder 2
Adding four on both sides
  n + 4 = 3m + 4
= 3m + 3 + 1
 = 3(m + 1) + 1
Here on dividing by 3, it leaves remainder 1
In this case n is divisible by 3 but (n + 2) and (n + 4) are not divisible by 3

   Case-2 : 
n = 3m + 1
  on dividing by 3, it leaves remainder 1
adding two on both sides
n + 2 = 3m + 1 + 2
n + 2 = 3m + 3
= 3(m + 1)
  on dividing by 3 it leaves remainder 0
Hence it is divisible by 3
adding four on both sides
x + 4 = 3m + 1 + 4
x + 4 = 3m + 3 + 2
= 3(m + 1) + 2
on dividing by 3, it leaves remainder 2
In this (n + 2) is divisible by 3 but n and (n + 4) are not divisible by 3

Hence through both the cases we conclude that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.
Similarly, we can show for n = 3m + 2 form.

 

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infoexpert27

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