Question

The $7928$ number is not a perfect square. Give reason.

Answer 1

Perfect square numbers:

$\begin{array}{rcl}{1}^2& =& 00\left[1\right]6^2=03\left[6\right]\\ 2^2& =& 00\left[4\right]7^2=04\left[9\right]\\ 3^2& =& 00\left[9\right]8^2=06\left[4\right]\\ 4^2& =& 01\left[6\right]9^2=08\left[1\right]\\ 5^2& =& 02\left[5\right]{10}^2=1\left[00\right]\end{array}$

The square of a number having$0,1,4,5,6$ or $9$ at its unit place is a perfect square. Also, the square of a number can only have an even number of zeros at the end.

Hence, $7928$ is not a perfect square as it does not end with the digits $0,1,4,5,6$ or $9$nor end with an even number of zeroes.