Question

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

(n-1)/2

n/2

(n+1)/2

2n​

Answer 1

Answer:

$the \: answer \: is \: \frac{n}{2}$

Step-by-step explanation:

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