modular arithmetic - Find the inverse modulo of a number - got a negative result

I'm trying to find the inverse modulo of $17\pmod{3120}$

I've tried:

$$\begin{eqnarray} 3120 =& 17\cdot 183 &+ 9\\ 17 =& 9\cdot 1 &+ 8\\ 9 =& 8\cdot 1 &+ 1\\ 8 =& 8\cdot1 \end{eqnarray}$$

and then do it from the the bottom:

$$\begin{align} 1 = & 9 - 8\cdot1 \\ = & 9 - (17 - 9\cdot1)\\ = & 9\cdot2 - 17\cdot1\\ = & 2 (3120 - 17\cdot183) -17\cdot1\\ = & 3120\cdot2 -17\cdot367 \end{align}$$
This means that $-367$ is the inverse modulo. Is it the correct result? Can a reverse modulo be negative? Using this website it gives the inverse modulo to be $2753$ instead, how can I get that?

Answer

You have obtained the correct answer. Note that
$$-367 \equiv (3120-367) \pmod{3120} \equiv 2753 \pmod{3120}$$

In general, if you do not like negative modulo, you can always make it into a positive one. If you have $-a \pmod{n}$, where $a \in \{1,2,\ldots,n\}$, then you can make it into a positive one as follows.
$$-a \pmod{n} \equiv (n-a) \pmod{n}$$