Question

Find the greatest four-digit number which is a perfect square.

Answer 1

Hint: Perfect square is an integer which produces an integer when we do square root (Evolution) of the number. The greatest four-digit number is 9999 so the greatest four-digit number which is a perfect square is less than 9999.Also the greatest four-digit number which is a perfect square is square of greatest two-digit number.

Complete step-by-step answer:
Let us assume that $n$is the square root of the greatest four-digit numbers.
As we know that greatest four-digit number is 9999, so we can say that square of $n$is less than or equal to 9999
Hence, we can write it in mathematical form as
${{n}^{2}}\le 9999$
Now as we know that if we square 100, we get 10000, which is the lowest five-digit number. So the square root of the greatest four-digit number is a two-digit number. So, if we multiply the greatest two-digit number by itself we get the greatest four-digit number which is a perfect square.
The greatest two-digit number is 99.
So, if we multiply 99 by itself, we get the greatest four-digit number which is a perfect square.
Let $N$be the greatest four-digit number which is a perfect square, so we can write
$N={{99}^{2}}$
Upon squaring 99 we get the greatest four-digit number which is a perfect square
So, we can write
$N=9801$
Hence, we can say that 9801 is the greatest four-digit number which is a perfect square.

Note: it is especially worthy of remark that every positive quantity has two square roots equal in magnitude, but opposite in sign. In this question the square root of 9801 is +99 and -99. When we square either +99 or -99, we get the same result that is 9801.
Any number when we do square root, we get a value in which we have integer parts and fractional parts. If fractional parts are zero then the given number is a perfect square.