Prove the divisibility test by $7,11,13$ for numbers more than six digits Prove the divisibility test by $7,11,13$ for numbers more than six digits

Attempt:
We know that $7\cdot 11 \cdot 13 = 1001$. The for a six-digit number, for example, $120544$, we write it as
$$ 120544 = 120120 + 424 = 120\cdot1001 + 424 $$
thus we just check the divisibility of $424$ by $7,11,13$.
Know for a number with more than six digits,  for example: $270060340$,
$$270060340 = 270270270 - 209930$$
$$ = 270 \cdot (1001001) - 209930 $$
$$ = 270 \cdot (1001000) + (270 - 209930) =270 \cdot (1001000) - 209660$$
so we check the divisibility of $209660 = 209209 + 451$, or just $451$.
But the test states that: for $270060340$, we group three digits from the right:
$$ 270, 60, 340$$
then check divisibility of $340+270 - (60)$.
How to prove this?
 A: It's the radix$+1$ divisibility test for radix $10^3,\,$ i.e. the analog of casting out $11's$ in radix $10,\,$ viz.
$\!\!\!\begin{align}\bmod 10^{\large 3}\!+\!1\!:\,\ \color{#0a0}{10^{\large 3}}\equiv \color{#c00}{\bf -1}\, \ \Rightarrow\!\!\!\!  &\ \ \ \ \ \ \overbrace{d_0 + d_1 \ \color{#0a0}{10^{\large 3}} +\, d_2(\color{#0a0}{10^{\large 3}})^{\large 2}\! + d_3(\color{#0a0}{10^{\large 3}})^{\large 3}+\,\cdots }^{\!\!\!\!\!\!\!\textstyle\text{integer in radix $\color{#0a0}{10^{\large 3}}$ with digits $\,d_i$}} \\[.3em] 
&\equiv\, d_0\!+d_1(\color{#c00}{\bf -1})\!+d_2(\color{#c00}{\bf -1})^{\large 2}\! + d_3(\color{#c00}{\bf -1})^{\large 3} +\,\cdots \\[.3em]
&\equiv\, d_0\  \ \color{#c00}{\bf -}\ \  d_1\ \  +\ \  d_2\ \  \color{#c00}{\bf -}\ \  d_3\ +\, \cdots\\[.2em] 
&\equiv\, \color{#c00}{\text {alternating}}\text{ digit sum}\end{align}$
where we employed the Congruence Sum & Product Rules (or Polynomial Rule)
$\!\begin{align}\text{E.g. in your 2nd example: }\ \ \ \ \ \ \ &\overbrace{270\,,\,060\,,\,340}^{\textstyle d_2,\ \ d_1,\ \ d_0}\\[.2em] 
\equiv\ &270\! -\! 060\! +\! 340\, \equiv\, 550\!\pmod{\!1001}\end{align}$
A: $1001$ divides $999,999$ and $999,999 = 1,000,000 - 1$
$270,060,340 = (270,000,000 - 27) + (60,000 + 60) + 340 + 27 -60$ 
The first two terms are divisible by $1001$ (and hence 7,11, and 13)
A: For 9-digit number: $abcdefghi$, 
$$ abcdefghi = abc000000 + defghi = abc000000 + def \cdot 1001 - def +  ghi $$
so we check divisibility of 
$$ abc000000 - def + ghi = abc \cdot 1001000 - abc000 - def + ghi$$
so we cehck divisibility of 
$$ -abc000 - def + ghi = -(abc \cdot 1001 - abc) - def + ghi$$
in the end we only check $abc + ghi - def$.
If 12-digit number:
$$ a_{1} a_{2} ... a_{11}a_{12} = a_{1} a_{2} a_{3} 000000000 + \underbrace{a_{4} ... a_{12}}_{9-digit}$$
so we check the divisibility of 
$$ a_{1} a_{2} a_{3} 000000000 + ( a_{10}a_{11}a_{12} + a_{4} a_{5}a_{6} - a_{7}a_{8}a_{9})  $$
or just 
$$ = (a_{10}a_{11}a_{12} + a_{4} a_{5}a_{6} - a_{7}a_{8}a_{9} - a{1} a_{2} a_{3}) $$
