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.
 A: The number is $35$, since $$x+y=8$$ and if $9$ is added to any number the ones digit must decrease by $1$ and the tens digit must increase by $1$ if and only if the unit digit is not $0$ hence by adding $9$ $$x+1=y-1$$ $$x-y=-2.$$ By solving both the equations we have $$x=3$$ and $$y=5$$ $$**OR**$$ The numbers whose sum of digits is equal to $8$ the numbers are $$17,26,35,44,53,62,71,80$$ and in these only $35$ is the number whose digits become equal on adding $9$
A: First note that $y \ne 0$ since otherwise we would have $x + y = x + 0 = 8$, and so $10x + y = 80$, but $80$ doesn't satisfy the second condition.
Therefore we must have $1 \le y \le 9$.  This means that when we add $9$ to $10x + y$, the tens digit must increase by $1$ and the ones digit decreases by $1$.  So then $10x + y + 9 = 10(x+1) + (y-1)$.  Since the digits are equal, we have $x+1 = y-1$.  Now you just have a system of two equations in two variables:
\begin{align*}
  x+y &= 8\\
  x+1 &= y-1
\end{align*}
A: Let, number is of the form $10a+b$, then according to question:
$a+b=8$ and digits of $10a+b+9$ are equal.
Notice that adding $9$ to the give number will increase its tens digit by $1$ and decrease its unit digit by $1$. So, 
$a+1=b-1\implies a+2=b$.
Hence, we have $a+2+a=8\implies a=3\implies b=5 \implies 10a+b=35$
A: Let the number be $10x + y$. 
Adding 9 to the number will make both the digits equal. Let that digit be $m$. Then,
$$10x+y+9=11m$$
Since $x+y=8$,
$$9x=11m-17$$
Since x is an integer, 
$$(11m-17)\%9=0$$
$$\Rightarrow 2m\%9=8$$
$$\Rightarrow m=\{4,8.5,13 \ldots \}$$
But since $m$ is a non negative integer less than 9,
$$m=4$$
And therefore, original number $=44-9=35$.
A: Solving by observation and little mathematics;
let the number be
$$\overline{xy}:x,y\in [1,9]  \text{ and } x,y \in \text{Integers}$$
Given, $x+y=8$ and $\overline{xy} +9=\overline{zz}$
We know that,  $\overline{zz} \in \{11,22,33,44,55,66,77,88,99 \} $
So, $(\overline{zz}-9=\overline{xy})\in \{11-9,22-9,33-9,44-9,55-9,66-9,77-9,88-9,99-9 \} $
or, $(\overline{zz}-9=\overline{xy})\in \{13,24,35,55-9,66-9,77-9,88-9,99-9 \} $
We can see our answer.
A: 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 $\overline{ab} = \boxed{35}$.
