Question

If a number of n-digits is a perfect square and ‘n' is an even number, then which of the following

is the number of digits of its square root?

i) n-1/2
ii) n/2
iii) n+1/2
iv) 2n​

Answer 1

Answer:

n+1/2

Step-by-step explanation:

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Answer 2

The number of digits of its square root is $\frac{n}{2}$

Step-by-step explanation:

Given: Number of $n$-digits is a perfect square and $'n'$ is an even number

To find: The number of digits of its square root

Concept/Formula used: Hit and Trial Method

Let us assume the perfect square number whose $n$ is an even number be $81$

Number of digits $(n)$ are $2$ , here $n$ is an even number

The square root of $81$ is  $9$, here $n=1$

So, we can say that number of digits of its square root is $\frac{n}{2}$

Let us take another example and assume the value to be $9801$

Here, $n=4$  is an even number

The square root of $9801$ is $99$  ,

Thus, we can say that number of digits of its square root is $\frac{n}{2}$