At least one of 3 consecutive odd numbers is a multiple of 3

The way to approach this problem is to show that there will always be a number among three consecutive odd numbers that is divisible by $3$. Let three consecutive odd numbers be $2n+1,2n+3,2n+5$. If $2n+1$ is a multiple of $3$ then we are done. Else the remainder when divided by $3$ is either $1$ or $2$. Suppose its $1$ then it means $2x$ is divisible by $3$ which means $2x+3$ is divisible by $3$. Or suppose the remainder is $2$, which means $2x-1$ is divisible by $3$ which means $2x+2$ and $2x+5$ are divisible by $3$. Hence proved.