Hundreds’ place digit of $1993^3 – 913^3 – 1083^3$ How can I find the hundreds’ place digit of the following number:

$$1993^3 – 913^3 – 1083^3$$

I have not tried this one since I don't know how to begin.
I can tell the units digit of this but how to approach for hundred's place.
I am looking for approaches on how to proceed with these types of questions.
 A: They key here is to use modular arithmetic. What you want to find is the last three digits of $1993^3-913^3-1083^3$, and to do so you can take advantage of the fact that if $x-y$ is a multiple of $1000$ (in which case we say $x\equiv y\mod 1000$), then $x$ and $y$ have the same last three digits. In your case, we have
$$\begin{eqnarray}1993^3-913^3-1083^3&\equiv&(-7)^3-(-87)^3-83^3\mod 1000\\
&\equiv&-7^3+87^3-83^3\mod 1000\\
&\equiv&-343+658503-571787\mod 1000\\
&\equiv&-343+503-787\mod 1000\\
&\equiv&-627\mod 1000\\
&\equiv& 373\mod 1000\end{eqnarray}$$
so the last 3 digits are $373$. This approach works for finding other places (replace $10^3$ with $10^n$) and in other bases as well (replace $10^n$ with $b^n$).
A: One can do this with mental arithmetic if one first employs the binomial theorem, viz.
$\begin{eqnarray} (100 a + 10 b + c)^3\  &\equiv& &\ \ \ c^3& + 30 c (bc &+& 10 (ac +\ b^2))\pmod{1000},\quad \rm therefore \\

1993^3\equiv 993^3\ &\equiv& &\ \ \ 27& +\ 90 (27 &+& 10 (27 + 81)) \\

 913^3\ &\equiv& &\ \ \ 27& +\ 90 (\ 3 &+& 10 (27 +\ 1\ )) \\

 1083^3\equiv 083^3 \ &\equiv& &\ \ \ 27& +\ 90 (24 &+& 10 (\ 0\  + 64)) \\

1993^3-913^3-1083^3\ &\equiv& &-27& +\ 90 (\ 0 &+& 10(81-65))\equiv  -27 - 100\cdot 16 \equiv -627 \equiv 373
\end{eqnarray}$
