Is this number divisible by $11$? 
Is  $73864589999999923243431$ divisible by $11$? Show and justify?

Usually I would take modulo 11, but it doesnt seem like a reasonable option here. I dont want the answer, just a hint?
 A: Hint. Numbers are divisible by $11$ whenever the sum of the digits in odd positions differs from the sum of the digits in even positions by a multiple of $11$ (including zero and negative multiples of $11$).
A: Idea: $1001 = 7 \times 11 \times 13$.
A: Hint 1 (mouseover to show):

 $10^k \equiv (-1)^k \pmod{11}$

Hint 2 (mouseover to show): 

 $$ 73864589999999923243431 =  (1\times10^0) + (3\times 10^1) + (4\times 10^2) + (3\times 10^3) + (4\times 10^4) \\  \qquad \qquad \qquad \qquad \qquad \qquad + (2\times 10^5 \ldots) + (3\times 10^{21}) + (7\times 10^{22})$$

Hint 3 (mouseover to show):

 Apply Hint 1 to the sequence in Hint 2. What must the summation satisfy under mod 11 to be divisible by 11?

A: Well. Hint 1: $73864589999999923243431$ is divisible by $11$ if and only if 
$73864589999999923243431 - 11= 73864589999999923243420$ is divisible by $11$.
And $73864589999999923243420$ is divisible by $11$ if and only if $73864589999999923243420 - 220 = 73864589999999923243200$ is divisible by $11$ and so on....
Hint 2:  $11*abcdefg = a(a+b)(b+c)(c+d)(d+e)(e+f)(f+g)g$ almost sort of...
So $hijklmnop/11 = h(h-i)(i-j)(j-k)(k-l)(l-m)(m-n)(n-o)(o-p)p$ almost sort of, kind of, not really but in a way it can be almost, not really....
Hint 3:  If $11*abcdefg = a(a+b)(b+c)(c+d)(d+e)(e+f)(f+g)g$ then $a + (b+c) + (d+e) + (f+ g) = (a+b) + (c+d)+(e+f) + g$.  (Actually that is always true....)
