Divisibility criterion of a number by $7$, $11$ or $13$

Question

To paraphase my textbook:

Use that $1001 = 7 \cdot 11 \cdot 13$ to prove that if $a= a_n a_{n-1}\dots a_1 a_0$ in base $10$, then, respectively, $7$ or $11$ or $13$ divides $a$ if and only if it divides $a_2 a_1 a_0- a_5 a_4 a_3 +a_8 a_7 a_6- \cdots $.

I'm on the chapter of my textbook in which modular congruences are introduced.

because $1000\equiv-1\bmod1001$, $a=a_2a_1a_0+1000 a_5a_4a_3+1000^2 a_8a_7a_6-...\equiv a_2a_1a_0-a_5a_4a_3+a_8a_7a_6-...\bmod1001$ and thus $\bmod7,11,13$ — J. W. Tanner, Sep 08 '20 at 21:26
This should remind you of the divisibility tests for $11$ and $101$ — Ross Millikan, Sep 08 '20 at 21:27

score 0 · Accepted Answer · edited Sep 08 '20 at 21:39

0

Hint:

Your equality shows that $1000\equiv -1\mod 7\;(\text{or }11,\;\text{or }13).$

Now writing $a$ in base $1000$ consists in grouping its digits by groups of $3$, starting from the right, namely: $$a=(a_2a_1a_0)1000^0+(a_5a_4a_3)1000+(a_8a_7a_6)1000^2+\dotsm$$ so that, for instance, $$a\bmod 7\equiv (a_2a_1a_0)-(a_5a_4a_3)+(a_8a_7a_6)-\dotsm$$

edited Sep 08 '20 at 21:39

J. W. Tanner

60,406

answered Sep 08 '20 at 21:30

Bernard

175,478

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I think you should remove $1000$ from the last line – J. W. Tanner Sep 08 '20 at 21:34
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@J.W.Tanner: Still another copy-paste accident… Thanks for warning me! – Bernard Sep 08 '20 at 21:38
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I definitely should reread what I post! ;o) – Bernard Sep 08 '20 at 21:40
Thanks for the help! – Pedro Cardoso Sep 08 '20 at 21:54

Divisibility criterion of a number by $7$, $11$ or $13$

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