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A number is divisible by $11$, when the difference between the sum of the digits in the odd positions counting from the left (the first, third, ....) and the sum of the remaining digits is either 0 or divisible by 11. Why is that?

Martin Sleziak
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N Chikama
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    Are you familiar with modulo arithmetic?

    You can note a number with $\sum_{k=0}^n a_k\cdot 10^k$ with $a_k\in{0,...,9}$ Now observe this term $\mod 11$. Hence $\sum_{k=0}^n a_k\cdot 10^k\mod 11=\sum_{k=0}^n a_k\cdot (-1)^k$.

     – Mr.Topology Jun 19 '16 at 02:34
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    Are you familiar with basic arithmetic modulo $11$? It turns out that even powers of ten are congruent to $1$ mod $11$, while odd powers are congruent to $-1$ mod $11$. If you are not familiar with such topics, perhaps a proof by induction would be understandable? – hardmath Jun 19 '16 at 02:36
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    See http://math.stackexchange.com/questions/328562/divisibility-criteria-for-7-11-13-17-19 – lab bhattacharjee Jun 19 '16 at 03:34
  • @hardmath I had a doubt from a long time, Does every prime number have divisibility criteria ? i think it is too stupid to post it as a independent question. –  Jun 19 '16 at 09:53
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    @ritwiksinha, yes, but citeria for individual primes need not be terribly fancy. Those for $2$ and $5$ are clear. If $p \ne 2, 5$ is a prime, you then compute the powers of $10$ modulo $p$. They will keep repeating, and this will yield a criterion. For instance when $p = 7$ you get that $a = a_{0} + a_{1} 10 + a_{2} 10^2 + \dots$ (where $0 \le a_{i} < 10$) is divisible by $7$ is only if $a_{0} + 3 a_{1} + 2 a_{2} - a_{3} - 3 a_{4} - 2 a_{5} + a_{6} + \dots$ is. – Andreas Caranti Jun 19 '16 at 10:36
  • @AndreasCaranti So given a number, i can check if it is prime or not by testing it with criteria of every number less than half of it. Is this how they find large numbers ? –  Jun 19 '16 at 14:03
  • @ritwiksinha: Don't confuse criteria for numbers divisible by $p$ with criteria for $p$ is divisible by a smaler number. For the former, see the previous Question (and its answers) Divisibility criterian for 7, 11, 13, 17, 19. – hardmath Jun 19 '16 at 14:13
  • @hardmath does the latter even exists ? –  Jun 19 '16 at 14:18
  • @ritwiksinha: It is testing for compositeness, equivalently testing for primality. – hardmath Jun 19 '16 at 14:29

2 Answers2

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\begin{align} abcde_{10} &= 10000a + 1000b + 100c + 10d + e \\ &= (9999a + a) + (1001b - b) + (99c + c) + (11d - d) + e \\ &= (9999a + 1001b + 99c + 11d) + (a-b+c-d+e) \\ &= 11(101a + 91b + 11c + d) + (a-b+c-d+e) \\ &\equiv a-b+c-d+e \pmod{11} \end{align}

Another way to look at this problem is to note that $10^{2n} \equiv (-1)^{2n} \equiv 1 \pmod{11}$

So $10^{2n}(10m + n) \equiv 1(-m + n) \equiv -m + n \pmod{11}$

So \begin{align} abcdef_{10} &\equiv 10^4(10a + b) + 10^2(10c + d) + 10^0(10e + f) \\ &\equiv (-a+b) + (-c + d) + (-e + f) \\ &\equiv -a+b-c+d-e+f \end{align}

Steven Alexis Gregory
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Let $a$ be a whole number. We'll write $a$ as $a=a_0+a_1\cdot10+_{\cdots}+a_k\cdot 10^k$.

Let's check when $a (mod 11)=\overline{a}=0$ ($\overline a$ is just to make the writing more comfortable.

$\overline a=\overline{a_0+a_1\cdot10+_{\cdots}+a_k\cdot 10^k}=\overline{a_0+a_1\cdot(-1)+_{\cdots}+a_k\cdot (-1)^k}=\overline{a_0-a_1+_\cdots+a_k\cdot(-1)^k}$

and that equals zero only when $a_0-a_1+_\cdots+a_k\cdot(-1)^k$ equals to zero or something that is divisible by 11.

nono
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