Let the no. be 10x + y.No. formed by the interchange of digits = 10y + xWe have y - x = 2 .........(i)y + x = 14 .........(ii)Solving (i) and (ii), we get x = 6, and y = 8∴ the no. is 68.
Since the difference between the divisors and the respective remainders is not constant, back substitution is the convenient method. None of the given numbers is satisfying the condition.
EXAMIANS