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Prove Number in decimal representation $N=abc,def,ghi,\cdots ,xyz$ is divisible by $7$. Iff $abc-def+ghi-\cdots+xyz$, alternating sum of numbers formed by dividing the string $N$ into $3$ digit pairs of consecutive digits. Is divisible by $7$.

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closed as off-topic by Théophile, Namaste, Leucippus, Rolf Hoyer, Shailesh Nov 20 '17 at 2:24

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  • $\begingroup$ Works for 13 too. $\endgroup$ – Oscar Lanzi Nov 19 '17 at 22:23
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Let $N=\sum_{i=0}^{3n-1} a_i\cdot 10^{i}$. Since $10^3+1$ is divisible by $7$, modulo $7$ for each $k$ we have $$a_{3k+2}\cdot 10^{3k+2}+a_{3k+1}\cdot 10^{3k+1}+a_{3k}\cdot 10^{3k}\equiv$$ $$(a_{3k+2}\cdot 100+a_{3k+1}\cdot 10+a_{3k})\cdot 10^{3k}\equiv$$

$$(a_{3k+2}\cdot 100+a_{3k+1}\cdot 10+a_{3k})\cdot (-1)^{k},$$

which implies the required claim.

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  • $\begingroup$ $(-1)^k$ at the end, I guess $\endgroup$ – Peter Franek Nov 19 '17 at 22:05
  • $\begingroup$ @PeterFranek Thanks, corrected. $\endgroup$ – Alex Ravsky Nov 19 '17 at 22:07
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Note that $7$ divides $10^{3(2n+1)}+1$ and $10^{3(2n)}-1$, this is easy to see by Fermat \begin{eqnarray*} 10^{6n} \equiv 1 \pmod{7} \\ 10^{6n+3} \equiv -1 \pmod{7}. \end{eqnarray*} So $7 \mid 10^{3n} a_n b_n c_n + 10^{3(n-1)} a_{n-1} b_{n-1} c_{n-1} +\cdots +10^{3} a_1 b_1 c_1 + a_0 b_0 c_0 $ if and only if $7 \mid (-1)^{n} a_n b_n c_n + (-1)^{n-1} a_{n-1} b_{n-1} c_{n-1} +\cdots - a_1 b_1 c_1 + a_0 b_0 c_0 $.

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