Review question

# When will $N$ definitely be a square number? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9800

## Solution

Let $N=2^k\times4^m\times8^n$where $k,m,n$ are positive whole numbers. Then $N$ will definitely be a square number whenever

1. $k$ is even;

2. $k+n$ is odd;

3. $k$ is odd but $m+n$ is even;

4. $k+n$ is even.

We’ll start by rewriting all the terms as powers of $2$.

By doing this, we will get $N=2^k\times2^{2m}\times 2^{3n}\Longrightarrow N=2^{k+2m+3n}.$

Since square numbers are all of the form $x^{y}$, where $y$ is an even number, $N$ will be a square number when $k+2m+3n$ is even.

That is, as $2m+2n$ is always even, when $k+n$ is even so the answer is (d).

Alternatively, we could have noticed that $4^m$ is already a square, as $4^m=\left(2^m\right)^2$, and rewrite $8^n$ as $4^n\times2^n$.

Since $4^n$ is also a square, for $N$ to be a square we will need only $2^k\times 2^n=2^{k+n}$ to be a square, and, therefore, $k+n$ to be even.