Word problem regarding system of linear congruences... Full problem:

A hoard of gold pieces ‘comes into the possession of’ a band of $15$ pirates.
  When they come to divide up the coins, they find that three are left over.
  Their discussion of what to do with these extra coins becomes animated,
  and by the time some semblance of order returns there remain only $7$
  pirates capable of making an effective claim on the hoard. When, however,
  the hoard is divided between these seven it is found that two pieces are left
  over. There ensues an unfortunate repetition of the earlier disagreement,
  but this does at least have the consequence that the four pirates who remain
  are able to divide up the hoard evenly between them. What is the minimum
  number of gold pieces which could have been in the hoard?

So the information that I picked out from this was, using equivalence classes
$$[x]_{15}=[3]_{15}$$
$$[x]_7=[2]_7$$
$$[x]_4=[0]_4$$
I solved the system and obtained $[-390]_{420}=[30]_{420}$. The answer to our problem is supposedly $408$, but from the point that I've gotten to, I am not quite sure how to get this number. Did I overlook something big? Or am I on the right track?
EDIT:
I re-calculated and obtained $[12]_{420}$. Or $x \equiv 12 \mod 420$. If only it were $-12$...
EDIT 2:
Nevermind. I've got the answer. I had a sign error and indeed my answer is $[-12]_{420}$.
 A: Have 
$$x \equiv 3 \mod 15$$
$$x \equiv 2 \mod 7$$
$$x \equiv 0 \mod 4$$
Using Chinese Remainder Theorem, we first solve
$$x \equiv 3 \mod 15$$
$$x \equiv 2 \mod 7$$
Since $15,7$ are relatively prime, we have
$$15(1)+7(-2)=1$$
$$\implies 15(1)(2)+7(-2)(3)=30-42=-12$$
$$\implies x \equiv 93 \mod 105$$
Now we need to solve the system 
$$x \equiv 93 \mod 105$$
$$x \equiv 0 \mod 4$$
Since $105,4$ are relatively prime, we have
$$105(1)+4(-26)=1$$
$$\implies 105(1)(0)+4(-26)(93)=(-104)(93)=-9672$$
$$\implies x \equiv -9672 \mod 420$$
Note that $420\cdot 23=9660$, so $-9672+9660=-12$. Further, observe that $420-12=408$. Thus,
$$x \equiv 408 \mod 420$$
So the minimum number of gold pieces is 408.
A: $\begin{eqnarray}{\bf Hint}\   
&& {\rm mod}\ 15\!:\,\ {-x}\equiv \color{#c00}{12}\\
&& {\rm mod}\ \ \ 7\!:\,\ {-x}\equiv  5\equiv \color{#c00}{12}\\
&& {\rm mod}\ \ \ 4\!:\,\ {-x}\equiv  0\equiv\color{#c00}{12}\end{eqnarray}\ \iff\ {-}x\equiv{\color{#c00}{12}}
\,\ ({\rm mod\ 15\cdot 7\cdot 4)}$
Remark $\ $ The above is a special case of the following constant case CRT optimization
$\begin{eqnarray}\phantom{\bf Hint}\   
&& {\rm mod}\ m\!:\,\ {y}\equiv \color{#c00}{a}\\
&& {\rm mod}\ \  n\!:\,\ {y}\equiv  b\equiv \color{#c00}{a}\\
&& {\rm mod}\ \ \, k\!:\,\ {y}\equiv  c\equiv \color{#c00}{a}\end{eqnarray}\ \iff\ y\equiv{\color{#c00}{a}}
\,\ ({\rm mod\ lcm}(m,n,k))$
