Question

Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer. [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Proof : 2 Marks

Consider any two positive integers a and b such that a is greater than b.

Then according to Euclid's division algorithm

$a=bq+r$ where $q$ and $r$ positive integers and $0\le r<b$

Let $a=n,b=3$

$n=3q,3q+1,3q+2$ .....(i)

$n+2=3q+2,3q+3,3q+4$ .....(ii)

$n+4=3q+4,3q+5,3q+6$ .....(iii)

In (i), only $n=3q$ is divisible by 3

In (ii), only $n+2=3q+3$ is divisible by 3

In (iii), only $n+4=3q+6$ is divisible by 3

Thus only one of them is divisible by 3 in each case.