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-cfrac-n-2-cfrac-n-1-2-cfrac-n-1-2-2n

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Toppr Admin · Accepted Answer

No of digits in a perfect square is n-xA0-If n is odd then no of digits in its square roots is n-12

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

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

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?n2n+12n−122n

no of digits in a perfect square is n If n is odd then no of digits in its square roots is n+12

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

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