Question:
The square root of a perfect square of $n$ digits will have $\left(\frac{n+1}{2}\right)$ digits, if $n$ is odd.
Solution:
True
If the square has 3 digits, then its square root has 2 i.e. $\left(\frac{3+1}{2}\right)$ digits.
Similarly, if the square has 5 digits, then its square root has 3 i.e. $\left(\frac{5+1}{2}\right)$ digits.
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