Question

Show that one and only one out of n, n+1 or n+2 is divisible by 3, where n is any positive integer.

Answer 1

Hint: We are going to consider three cases as if we divide any number by 3 we can get remainder as 0, 1, 2. So, we are going to consider n as these 3 cases and then we will write the other two numbers and see which one is divisible by 3.

Complete step-by-step answer:
Let’s start our solution by taking these cases one by one:
Case 1: When n gives remainder 0 when divided by 3.
Therefore,
n = 3k
n+1 = 3k+1
n+2 = 3k+2
Hence in this case we have seen that n is divisible by 3.
Case 2: When n gives remainder 1 when divided by 3.
Therefore,
n = 3k+1
n+1 = 3k+2
n+2 = 3k+3 = $3\left( k+1 \right)$
Hence in this case we have seen that (n+2) is divisible by 3.
Case 3: When n gives remainder 2 when divided by 3.
Therefore,
n = 3k+2
n+1 = 3k+3 = $3\left( k+1 \right)$
n+2 = 3k+4 = $3\left( k+1 \right)+1$
Hence in this case we have seen that (n+1) is divisible by 3.
So, we have seen that one and only one out of n, n+1 or n+2 is divisible by 3 in each of the three cases.
Hence Proved.

Note: One can also solve this question by thinking it in that way as in multiple of 3 we get every third number divisible by 3 and hence here is the same case we are taking any 3 consecutive numbers and hence for the same reason we one of those three numbers must be divisible by 3.