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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?

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  • $\begingroup$ You can ignore those eight nines, at least. :) $\endgroup$ Commented Oct 16, 2016 at 23:40
  • $\begingroup$ Use the criterion of divisibility by $11$. $\endgroup$ Commented Oct 16, 2016 at 23:41

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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 $11$).

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    $\begingroup$ (+1) ... since $10^k\equiv (-1)^k\pmod{11}$. $\endgroup$ Commented Oct 16, 2016 at 23:06
  • $\begingroup$ I would think this is a bit too much to give away to someone who wants just a hint, and not the full solution minus a few small details. $\endgroup$ Commented Oct 16, 2016 at 23:08
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    $\begingroup$ How less than this is it possible to give a hint after the OP said he tried what he did? This is just fine, imo. +1 $\endgroup$ Commented Oct 16, 2016 at 23:10
  • $\begingroup$ @Arthur: Hmm, it's possible. It's also possible that this is more or less what the OP wants. Unfortunately, I've already let the cat out of the bag, so to speak. $\endgroup$ Commented Oct 16, 2016 at 23:11
  • $\begingroup$ True. But I was contemplating how to hint at the "every other digit" without actually saying it outright, and you took that away from me. *Sulking in a corner* Maybe something like Jack's comment above would've been vaguer. I dunno. $\endgroup$ Commented Oct 16, 2016 at 23:16
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Idea: $1001 = 7 \times 11 \times 13$.

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  • $\begingroup$ This is the source of a reasonable divisibility hack for $7$. $\endgroup$ Commented Oct 17, 2016 at 2:08
  • $\begingroup$ Indeed! And for $13$ too. And it involves about the same amount of work as the well-known test for divisibility by $11$ to get the problem down to $3$ digits. $\endgroup$ Commented Oct 17, 2016 at 2:20
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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?

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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....)

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  • $\begingroup$ Hint 4: 10 isn't divisible by 11. a*10 is divisible by 11 iff and only if .... what? $\endgroup$ Commented Oct 17, 2016 at 0:37
  • $\begingroup$ Hint 5: $100c + 10b + a \mod 11 \equiv 99c + c + 11b - b +a \mod 11 \equiv c - b + a \mod 11$. $\endgroup$ Commented Oct 17, 2016 at 0:41

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