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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?

RobPratt
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Phaggot
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4 Answers4

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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 $11$).

Brian Tung
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    (+1) ... since $10^k\equiv (-1)^k\pmod{11}$. – Jack D'Aurizio Oct 16 '16 at 23:06
  • 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. – Arthur Oct 16 '16 at 23:08
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    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 – DonAntonio Oct 16 '16 at 23:10
  • @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. – Brian Tung Oct 16 '16 at 23:11
  • 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. – Arthur Oct 16 '16 at 23:16
  • @Arthur: Sorry about that! :-o – Brian Tung Oct 16 '16 at 23:52
  • Well, for a hint... Let $x = \sum b_i10^i$ so $11x = \sum b_i10^{i+1} + \sum b_i10^i = b_n10^{n+1} + \sum( b_{i+1}+b_i)10^i + b_0$. – fleablood Oct 17 '16 at 00:19
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Idea: $1001 = 7 \times 11 \times 13$.

Rob Arthan
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  • This is the source of a reasonable divisibility hack for $7$. – Brian Tung Oct 17 '16 at 02:08
  • 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. – Rob Arthan Oct 17 '16 at 02:20
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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?

gowrath
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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....)

fleablood
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