16 digit numbers divisible by 17 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.
 A: Use a divisibility rule:

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

A: 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 $588$.
A: $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  $10^2≡-2(mod\ 17)$
