Let the number be 10x + y.∴ 10x + y - (10y + x) = 27or 9x - 9y = 27∴ x - y = 3 .......(I)x + y = 13 .......(II)From eqn (I) and (II)x = 8 and y = 5.So the number = 85Therefore, product = 8 x 5 = 40
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