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This same question to check divisibility by 10 is quite straightforward but when we have to check whether it is divisible by 11 there is one case that has to be excluded, I would like it if someone explains it intuitively.

edit: Nvm its not very different from 10 or any other number n

Saahil Dhar
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  • Please, explain what is your problem in more details. There are usually several different proofs for a single result, and not everyboby know every possible proof – Exodd Jun 18 '20 at 10:15
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    There is no notion of selecting natural numbers in a (uniformly) random fashion, as is explained here – Ben Grossmann Jun 18 '20 at 10:17
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    In what way is the problem easier for division by 10? – Ben Grossmann Jun 18 '20 at 10:18
  • If you replace $10$ by any other $n$, the result is $\frac{1}{n}$. So, $10$ is nothing special. Another thing is if you want a divisibility-rule. – Peter Jun 18 '20 at 10:21
  • wait yeah, i understand now, my bad its nothing special i was thinking a little too much, thanks! – Saahil Dhar Jun 18 '20 at 18:25

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If you have fixed one number, the other can have $11$ possible residues mod $11$, one of which is suitable. Hence the answer is simply $\frac{1}{11}$

Peter
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  • This can be even more illustrative if you draw a table with 11 rows and cells where rows stand for a mod 11 and columns - for b mod 11 (where a and b are your numbers), and each cell is a sum of its row and column number. You will see one cell with 0, 9 cells with 11 and 1 cell with 22 (out of 121), giving you 11/121 = 1/11, as @Peter said – Andrei Rykhalski Jun 18 '20 at 10:28
  • Thank you @Andrei Rykhalski , that was what i was searching for, Peter that also makes sense, Thank you! – Saahil Dhar Jun 18 '20 at 18:28