Find the largest 2-digit number N which is divisible by 4 , such...

If given condition is satisfied for power 2 (i.e. for square), then it will hold for all (+ve) integral powers. Let the number $N$ be " $a b$ ". Units place remains same on squaring $b=0$ or 6 If $b=0$, then on squaring even ten's place will become 0 . So $b=0$ is rejected. $\Rightarrow b=6$ $N$ is divisible by 4 . Possibilities of $N$ are $16,36,56,76,96$. Among these only 76 satisfies the required condition $\Rightarrow N=76$