Question

Find the greatest $4$ digits number which is a perfect square.

• A. $9801$
• B. $9999$
• C. $1000$
• D. $9898$

Answer 1

The correct option is A $9801$

Greatest number of $4$ digits $=9999$.

We find $\sqrt{9999}$ by long division method.

The remainder is $198$.

This shows ${99}^2$ is less than $9999$ by $198$.
This means, if we subtract the remainder from the number we get a perfect square.

Therefore the required perfects square is $9999-198=9801$.
And $\sqrt{9801}=99$