Question

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

Answer 1

Let tens digit be x
And, Ones digit $=9-x$ [$\because$ Sum of digits $=9$]

$\therefore$ Original number $=10x+(9-x)=9x+9$

When the digits are reversed, the reversed number $=10(9-x)+x=90-9x$

According to question, we have

$9\times$ Original Number $=2\times$ Reversed Number

$\Rightarrow 9(9x+9)=2(90-9x)$

$\Rightarrow 81x+81=180-18x$

$\Rightarrow 81x+18x=180-81$

$\Rightarrow 99x=99$

$\therefore x=1$

Tens digit $=x=1$

Ones digit $=(9-x)=9-1=8$

Hence, the original number is $18.$