The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Let unit digit  = x

Tens digit  = y

Number will 10 times the tens digit + unit times the unit digit

Hence number will 10 y + x

Sum of digits are 9

So that

X + y  = 9        ………….(1)

 nine times this number is twice the number obtained by reversing the order of the digits

9 (10 y + x )  =  2 (10 x + y )

90 y + 9 x  = 20 x + 2y

88 y – 11 x  = 0

Divide by 11 we get

8 y  - x  = 0     …………..(2)

X + y  = 9        ………….(1)

Adding both equations we get

9 y = 9

Y = 9/9 = 1

Plug this value in equation first we get

X+ y = 9

X + 1 = 9

X = 8

So our original number is 10 y  + x    = 10*1 + 8  = 18