If $x$ leaves remainder $2$ when divided by $8$, what will the remainder be when $x + 9$ is divided by $8$? 
If the positive integer $x$ leaves a remainder of $2$ when divided by $8$, what will the remainder be when $x + 9$ is divided by $8$?

I love to put stuff into algebraic equations to make life easier; however, this simple (or at least that what it appears to be) problem stumps me (specifically the first part, regarding the remainder). How would I express this algebraically? 
Thanks! Sorry if this is a stupid question, but I'm studying to improve my math.
 A: The easiest way: 
Note that $x-2$ is a multiple of $8$, hence $x-2=8k,\ k\in\mathbb{Z}.$
Therefore $$x+9=(x-2)+11=(x-2)+8+3=8k+8+3=8(k+1)+3.$$
Now we can conclude the remaind is $3$.
The cool way:
We have that
$$x\equiv2\ (mod\ 8)$$
Add $9$ in both sides we have
$$x+9\equiv11\equiv3\ (mod \ 8).$$
A: Using notation more popular in Computer Science than in Mathematics, we could say
$x\bmod 8=2$. What is $(x+9)\bmod 8$?
We have $9\bmod{8}=1$. 
But in general 
$$(a+b)\bmod{m}=(a\bmod{m}+b\bmod{m})\bmod{m}.$$
It follows that 
$$(x+9)\bmod 8=(2+1)\bmod{8}=3.$$
Remark: An alternate computer oriented notation for $y\bmod m$ is $y\operatorname{\%}m$.  
A: $\begin{eqnarray}{\bf Hint}\qquad\ x &\!=\!& \ \ 2 + 8q\\
\Rightarrow\,\ 9+x &\!=\!& 11+8q\, =\, 3 + 8(1\!+\!q)\ \ \text{has remainder}\,\ 3\end{eqnarray}\ $ 
Or, $\,\ {\rm mod}\ 8\!:\,\ \begin{array}{}\color{#c00}{9\equiv 1}\\ \color{#0a0}{x\equiv 2}\end{array}\,\Rightarrow\, \color{#c00}9+\color{#0a0}x\equiv \color{#c00}1+\color{#0a0}2\equiv 3\,$ by the Congruence Sum Rule. 
A: I assure you this is definitely the most easiest answer for this question.
Q. If x leaves remainder 2 when divided by 8, what will the remainder be when x+9 is divided by 8?
x/8 gives remainder of 2 

eg:10/8 gives remainder of 2

to get remainder of x+9/8 :

substituting x=10  in x+9/8:

19/8 gives remainder 3,
so the answer is 3. 

its as simple as that!:)

