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.
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
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