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

Question

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

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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?

Answer 1

It's a 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 above mod arithmetic uses Congruence Sum & Product Rules (or Polynomial Rule), and where we use the method of simpler multiples to compute $\,n\bmod 7\,$ by first reducing $\,n\,$ mod the simpler multiple $1001$ of $7$ (which is "simpler" because it enables the above easy mod arithmetic).

$\!\begin{align}\text{E.g. in your 2nd example: }\ \ \ \ \ \ \ &\overbrace{270\,,\,060\,,\,340}^{\textstyle d_2,\ \ d_1,\ \ d_0}\\[.2em] \equiv\ &270\! -\! 060\! +\! 340\, \equiv\, 550\!\pmod{\!1001}\end{align}$

Answer 2

$1001$ 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)

Answer 3

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