Abstract Algebra: Congruence and The Division Algorithm 1

Hello Mathematicians,

I've come to bother you again. This is a homework problem, so please, offer only hints if you can, especially since I think this problem should be easy for me (it has the stench of something that is straight forward).

**Problem:**

Prove that if $\displaystyle n$ is a positive integer, and $\displaystyle a,b \in \mathbb{Z}$, then there is an integer $\displaystyle x$ such that

$\displaystyle a + x \equiv b~(\mbox{mod }n)$

__Things that may come in handy:__

As the title suggests, I have a strong gut feeling that I'm supposed to use the Division Algorithm here. But I can't seem to make it fit together nicely.

**The Division Algorithm:** If $\displaystyle a,b \in \mathbb{Z}$ with $\displaystyle b>0$, then there exists unique integers $\displaystyle q$ and $\displaystyle r$such that

$\displaystyle a = bq + r,~~~~~0 \le r < b$

__What I've Tried:__

Okay, so I decided to try and make this work by the division algorithm.

Now, $\displaystyle a + x \equiv b~(\mbox{mod }n)$ means that $\displaystyle n | (a - b + x) \Longleftrightarrow a - b + x = kn$ for $\displaystyle k \in \mathbb{Z}$

By the Division Algorithm, if $\displaystyle a_1, b_1 \in \mathbb{Z}$ with $\displaystyle b_1 > 0$, then there are unique integers $\displaystyle q$ and $\displaystyle r$ such that

$\displaystyle a_1 = b_1q + r$ for $\displaystyle 0 \le r < b_1$

...and I'm stuck there...

I was thinking of choosing $\displaystyle n = b_1$, and $\displaystyle a - b = a_1$. so the integer $\displaystyle x$ I am looking for would be $\displaystyle -r$

but I don't think that proves anything, nor am I sure that I can actually choose them like that...

Help :D

Thanks guys (and gals -- dedicated to JaneBennet)