Find the largest 2-digit number N which is divisible by 4 , such that all integral powers of N end with N .

Step-by-step solution

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. b=6N is divisible by 4 . Possibilities of N are 16,36,56,76,96 . Among these only 76 satisfies the required condition N=76