A two-digit number is such that the product of the digits is 8 . If 63 is added to this number, the digits interchange their places. What is the sum of the digits in the number ?
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 Eq. (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.