Question

A two digit number is such that the product of its digit is 8 when 63 is subtracted from the number the digit interchange their places .find the number

Answer 1

hello there!!!

let the two digit number be 10x+ y

ATQ,. x × y= 18..........(1)

10x + y - 63 = 10 y + x

9x - 9y - 63= 0

9x - 9y = 63

x-y = 7..........(.2)

x= 7+y

put value of x in eq 1b, we get

( 7+y ). y = 18

y² + 7y = 18

y² +7y-18=0

y² +9y -2y -18= 0

y(y+9) -2 ( y +9) =0

( y+9) (y-2). = 0

y=-9 , y = 2

y= 2........

so, x= 7+2=9

therefore the two digit number is 10(9)+2

= 92.....

HOPE IT HELPS YOU

Answer 2

$\huge\pink{ \underline{ \overline{ \tt \: solution}} \mid}$

Let the digit at the ten's place be x.

Given, Digit at the ten's place × Digit at the one's place = 8

∴ Digit at the one's place = 8 / x

Original number = 10x + 8/x

Reverse number = 10 (8/x) + x

Given, Original number – 63 = Reverse number

10x + (8/x) - 63 = 10 (8/x) + x

10x'2 + 8 -63x divide X = 80 + X'2/x

⇒ 10x2 + 8 - 63x - 80 - x2 = 0

⇒ 9x2 - 63x - 72 = 0

⇒ x2 - 7x - 8 = 0

⇒ x2 - 8x + x - 8 = 0

⇒ x(x - 8) + 1 (x - 8) = 0

⇒ (x + 1) (x - 8) = 0

⇒ x = 8 or x = -1(rejected)

so the original number = 10 × 8 + 8/8 = 80 + 1 = 81

$\red {\bold{answer \: is \: 81}}$