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?

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 
$$ = (a_{10}a_{11}a_{12} + a_{4} a_{5}a_{6} - a_{7}a_{8}a_{9} - a{1} a_{2} a_{3}) $$
