Question

For every natural number $n$, $n(n+3)$ is always :

• A. multiple of $4$
• B. multiple of $5$
• C. even
• D. odd

Answer 1

Answer: (C)

even

$n(n+3)=n^2+3n+2-2=(n+1)(n+2)-2\equiv \text{even}-\text{even}\equiv \text{even}$

Since for any natural number $n$, $(n+1)(n+2)$ will always be an even number.

OR

We know there are two types of natural numbers even and odd

Case 1. If $n$ is even then $n+3$ will be odd

Thus $n(n+3)=$ even $\times$ odd $=$ even

Case 2. If $n$ is odd then $n+3$ will be even

Thus $n(n+3)=$ odd $\times$ even $=$ even

Hence $n(n+3)$ will always be even number