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I wanted to know about the $16$ digit numbers those are divisible by $17$ and when this $16$ digit number is broken in groups of $4$ those groups of four are also divisible by $17$ and a check to verify their occurrence.

Emma.

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This is neither about linear algebra nor division algebras. The divisibility tag is appropriate. –  Robert Israel Jul 23 '12 at 7:21
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$9996999699969996$ is the maximum. –  Quixotic Jul 23 '12 at 7:38

3 Answers 3

up vote 2 down vote accepted

Use a divisibility rule:

Subtract 5 times the last digit from the rest, e.g. $221: 22 − 1\times 5 = 17.$

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Take any $k$ $n$-digit numbers divisible by $d$ and concatenate them, and you have a $kn$-digit number divisible by $d$.

The $4$-digit numbers divisible by $17$ (if you don't allow leading zeros) are $17 j$ for $j$ from $59$ to $588$.

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$10^2≡-2(mod\ 17) => 10^4≡4 =>10^8≡16≡-1 $

So,$\sum_{0≤r≤15} a_r10^r=10^8\sum_{8≤r≤15} a_r10^{r-8}+ \sum_{0≤r≤7} a_r10^r$ ≡$ \sum_{0≤r≤7} a_r10^r- \sum_{8≤r≤15} a_r10^{r-8}$(mod 17)

Subtract the higher 8 digits from the lower 8 digits. If the difference is divisible by 17, the given number will also be.


Or, let the absolute value of the resultant 8 digit number(B, say) be $\sum_{0≤r≤7} b_r10^r$

B= $ \sum_{0≤r≤3} b_r10^r + 10^4\sum_{4≤r≤7} b_r10^{r-4}$ ≡$\sum_{0≤r≤3} b_r10^r+4\sum_{4≤r≤7} b_r10^{r-4}$(mod 17) as $10^4≡4(mod\ 17)$

B will be divisible by 17 iff the RHS is divisible by 17.

The absolute value of the RHS(C,say) is of digit 4 or 5.

It can be further reduced using the fact $10^2≡-2(mod\ 17)$

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