The sum of digits in a 2-digit number 
The sum of digits in a two digit number formed by the two digits from $1$ to $9$ is $8$. If $9$ is added to the number then both the digits become equal. Find the number. 

My attempt:
Let the two digit number be $10x+y$ where, $x$ is a digit at tens place and $y$ is the digit at unit's place. According to question:
$$x+y=8$$
I could not figure out the other condition. Please help. Thanks in advance.

Let the $2$-digit number be $\overline{ab} = 10a + b$.

If $9$ is added to the number then both the digits become equal.

This means that $a = b$, so the number plus $9$ is $10a + a = 11a$, which is always $0 \text{ mod } {11}$ (divisible by $11$).
Therefore $\overline{ab} + 9 \equiv 0 \text{ mod } {11} \Rightarrow \overline{ab} \equiv 2 \text{ mod } {11}$.
But $a + b \equiv 8 \text{ mod } {11}$. Thus $(10a+b)-(a+b) \equiv 9a \equiv -2a \equiv 2-8 \text{ mod } {11}$, so $a \equiv 3 \text{ mod } {11}$.
As $a+b = 8$, hence $\overline{ab} = \boxed{35}$.
