Which one can be a square number?
$1)\overline{ab\cdots18}$
$2)\overline{ab\cdots42}$
$3)\overline{ab\cdots46}$
$4)\overline{ab\cdots89}$
To solve this question I considered $17^2=289$ and last two digits of this perfect square is $89$ therefor I conclude the answer is $\overline{ab\cdots89}$. but I couldn't find a mathematically proof for that (I just tried a number and find out last digits can be $89$) . is there other approach to solve the qustion?
Yes, there is a way. If you find the last two digits of every square from $0$ to $50$, you will find this pattern: $$00, 01, 04, 09, 16, 25, 36, 49, 64, 81, 00, 21, 44, 69, 96, 25, 56, 89, 24, 61, 00, 41, 84, 29, 76, 25, 76, 29, 84, 41, 00, 61, 24, 89, 56, 25, 96, 69, 44, 21, 00, 81, 64, 49, 36, 25, 16, 09, 04, 01, 00$$
This pattern of last two digits repeats every $50$ numbers. $89$ is your only two digits that appears in this set, so $89$ is the answer.