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We know,

A number is divisible by $11$ if the difference of the sum of the digits in the odd places and the sum of the digits in the even places is divisible by $11$.

For example, Let's consider $814$:

Sum of the digits in the odd places $= 8 + 4 = 12$

Sum of the digits in the even place $= 1$

Difference between the two sums $= 12 - 1 = 11$

$11$ is divisible by $11$.

Hence, $814$ is divisible by $11$.

What is the proof of this?

Nur Mohammed Mehedy Hassan
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  • Did you try looking it up? Proofs of this can be found both on this site and elsewhere on the internet. – A. Goodier Sep 27 '21 at 19:42
  • $814=8\times10^2+1\times10^1+4\times 10^0 \equiv 8\times(-1)^2+1\times(-1)^1+4\times(-1)^0\equiv8-1+4\equiv11\equiv0 \pmod{11}$. Perhaps you can take it from here and make a generalization. $\sum a_n \times 10^n \equiv \sum (-1)^n a_n \equiv \sum a_{2n} - \sum a_{2n+1} \pmod{11}$ – Evariste Sep 27 '21 at 19:44
  • https://artofproblemsolving.com/wiki/index.php/Divisibility_rules/Rule_for_11_proof – Bertrand Einstein IV Sep 27 '21 at 19:44
  • After this, try that of $3$ and $7$. – homosapien Sep 27 '21 at 19:58
  • It is based on the idea that if $n,m \in \Bbb{Z^+}$, then $(10n + m) \equiv (m - n) \pmod{11}$. This is easily shown by noting that $11 ~| ~(11n) ~= ~[(10n + m) - (m - n)].$ – user2661923 Sep 27 '21 at 23:27

2 Answers2

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If you know modular arithmetic, here is a hint:

Write any $n \in \mathbb{N}$ in base $10$ so let $n=\sum_{i=0}^{k}(10)^i c_i$. Now use the fact that $10\equiv -1\pmod{11}$. Then clearly

$\sum_{i=0}^{k}(10)^i c_i \equiv \sum_{i=0}^{k}(-1)^i c_i \quad (\mod 11). $

Of course, the last sum is the sum you mentioned.

BR Pahari
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Note that $$ 1,100,10000,...$$ are all congruent to $1$ in $\pmod {11}$

Also $$ 10, 1000,100000,...$$ are all congruent to $-1 $ in $\pmod {11}$

$$814 = 8(100)+1(10)+4(1) \equiv 8(1)+1(-1)+4(1)\equiv 0 \pmod {11}$$ You can take it from there and finish the proof.

Mohammad Riazi-Kermani
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