Question

The square root of a perfect square of $n$ digits will have $\frac{n+1}{2}$ digits, if $n$ is odd. State whether the statement is true (T) or false (F).

• A. True
• B. False

Answer 1

The correct option is A True

We are proving this statement by giving some examples.

Let us take a number $169$.

The number of digits in $169$ is $3$.

As $3$ is an odd number so according to the formula:

$\begin{array}{rcl}\frac{n+1}{2}& =& \frac{3+1}{2}\\ & =& 2\end{array}$

So, the square root of $169$ contains $2$digits which is True as square root of $169$ is $13$.

Taking another example let us take another number $15625$.

The number of digits in $15625$ is $5$.

As $5$ is an odd number so according to the formula:

$\begin{array}{rcl}\frac{n+1}{2}& =& \frac{5+1}{2}\\ & =& 3\end{array}$

So, the square root of $15625$ contains $3$ digits which is True as square root of $15625$ is $125$.

Hence, from the above examples, we prove that the statement is True.