Question

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?

• A. $\frac{n-1}{2}$
• B. $\frac{n}{2}$
• C. $\frac{n+1}{2}$
• D. $2n$

Answer 1

Answer: (C)

$\frac{n+1}{2}$

no of digits in a perfect square is $n$

If $n$ is odd then no of digits in its square roots is $\frac{n+1}{2}$