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.
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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