Solution:Let the digits at units and tens place of the given number be x and y respectively.Thus, the number is 10y + x.
The product of the digits is 8.
∴ xy = 8 … (i)
After interchanging the digits, the number becomes 10x + y.
If 63 is added to the number, the digits interchange their places.
Thus,
(10y + x) + 63 = 10x + y
⇒ 10x + y - 10y - x = 63
⇒ 9x - 9y = 63
⇒ x - y = 7 … (ii)
From Eq. (ii), put y = x - 7 into Eq. (i), we get
x(x - 7) = 8
⇒ x² - 7x - 8 = 0
⇒ x² - 8x + x - 8 = 0
⇒ x(x - 8) + 1(x - 8) = 0
⇒ (x - 8) (x + 1) = 0 ⇒ x = 8 or -1
From Equation (ii),
when x = 8
⇒ y = 8 - 7 = 1
When x = - 1
⇒ y = - 1 - 7 = - 8
We get (x, y) = (8, 1) and (x, y) = (- 1, - 8) Since, the digits of the number can't be negative.
So, we must remove second pair. Therefore, the number is 10 × 1 + 8 = 18 Hence, sum of the digits in the number is 1 + 8 = 9.