1. Every perfect square ends with \(0, 1, 4, 5, 6\) and \(9\).
2. No perfect square ends with \(2, 3, 7\) and \(8\) or odd number of zeroes
3. No perfect square can be of the form of 
   \(4k + 2, 4k + 3,3k + 2, 5k + 2\) and  \(5k + 3\).
4. Last two digits of \({n^2},{\rm{ }}{(50\; - \;n)^2},{\rm{ }}{\left( {50\; + \;n} \right)^2},{(100 - n)^2},{\rm{ }}{\left( {100{\rm{ }} + n} \right)^2}\) etc are same. For example last two digits of \(12^2\), \(38^2\) , \(62^2\), \(88^2\) (and so on) are same i.e. 44.
5. The last two digits of any square are the last two digits of one of the squares from 1 to 24
6. Square of a perfect square ends only with 1, 5, 6 or \(4n\) zeroes.

Example 5: A four-digit perfect square is of the form \(aabb\), where \(a\) and \(b\) are the single digits. How many such numbers are possible?

Solution: Since the last two digits are the same, \(b\) can be 0 or 4.  Now \(b\) can not be  0, because in that first two digits are not the same. So \(b = 4\) or number is ending with 44, which means number is of the form of \({(50 \pm 12)^2}\) or \({\left( {100\;-\;12} \right)^2}\). Since aabb is also a multiple of 11. so the number can be only \({\left( {100\;-\;12} \right)^2} = {\rm{ }}{88^2} = 7744\). There is only one such number.

Example 6: There is a five-digit perfect square in which the last two digits are the same and non-zero, and the number formed by the first three digits is a perfect cube. How many such numbers exist?

Solution: Since the last two digits are the same, the last digit can be only 4. So the number can be a square of 112, 138, 162, 188, 212, 238, 262, 288, 312. Out of these numbers, the first three digits of only \({112^2}\)\(\) is a perfect cube. \({112^2} = {\rm{ }}12544\).\(\)