If a number of $n$ -digits is perfect square and $n$ is an odd number, then which of the following is the number of digits of its square root?

\frac{n - 1}{2}

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\frac{n}{2}

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\frac{n + 1}{2}

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2 n

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Solution

The correct option is B $\frac{n + 1}{2}$
no of digits in a perfect square is $n$
If $n$ is odd then no of digits in its square roots is $\frac{n + 1}{2}$

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Question 21

Write the correct answer from the given four options.

A perfect square number having n digits, where n is even, will have square root with

a) $(n + 1)$ digit
b) $\frac{n}{2}$ digit
c) $\frac{n}{3}$ digit
d) $(\frac{n + 1}{2})$ digit

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