TITLE: How/why does this "divisible by 17" gimmick work?
QUESTION [7 upvotes]: Take any 16 digit number, where the first group of 4 digits is repeated 3 more times, e.g.
1,234,123,412,341,234 is such a number.  Such a number is always divisible by 17.  It even works when starting with zero:  0432043204320432  !  I did not expect that!
How is it that this is true?  I've got as far as noting that such a number, if the digits are i,j,k and l, is composed of:
i,jkl,000,000,000,000 + ijk,l00,000,000 + ij,kl0,000 + i,jkl
So, what is special about adding up these terms:
i,jkl +  (i,jkl * 10,000) + (i,jkl * 100,000,000) + (i,jkl * 1,000,000,000,000) ?
Can someone with more math skills explain (slowly) why this works?  Thank you!

REPLY [4 votes]: No gimmicks
$$10^2\equiv-2\pmod{17}$$
$$\implies10^4\equiv(-2)^2\equiv4$$
$$10^8\equiv\cdots\equiv-1$$
$$1+10^4+10^8+10^{12}\equiv1+4+(-1)+(-4)\equiv0\pmod{17}$$
More generally for $a^4\ne1,$
$$1+a^4+a^8+a^{12}=\dfrac{a^{16}-1}{a^4-1}$$
Now for $a^4\not\equiv1\pmod{17},$ use https://mathworld.wolfram.com/FermatsLittleTheorem.html