Question

Prove that one and only one out of any three consecutive positive integers is divisible by $3$.

Answer 1

Let the three consecutive positive integers be $n,n+1,n+2$

We know that $n$ is of the form $3q,3q+1or,3q+2$

Hence we can consider the following three cases:

Case:$\mathrm{I}$ When$n=3q$

$n$ is divisible by $3$ but it is not possible for $n+1andn+2$

Case:$\mathrm{II}$ When $n=3q+1,n+2=3q+1+2=3q+3$ is divisible by $3$ but it is not possible for $n$ and $n+1$

Case:$\mathrm{III}$ When $n=3q+2,n+1=3q+2+1=3(q+1)$ is divisible by $3$ but it is not possible for $n$ and$n+2$

Hence one amongst $n,n+1,andn+2\mathrm{is}\mathrm{divisible}by3.$