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This is the statement for Divisibility test of 11.

Add and subtract digits in an alternating pattern (add digit, subtract next digit, add next digit, etc). Then check if that answer is divisible by 11.

This is the proof that I found :

 If x is divisible by 11, then x ≡ 0 (mod 11). Thus $x_{n} + 10x_{n−1} + 100x_{n−2} + . . . + 10^{n-1}x_{1} ≡ 0 (mod 11)$.

But $x_{n} + 10x_{n−1} + 100x_{n−2} +. . .+ 10^{n−1}x_{1} ≡ x_{n} −x_{n−1} +x_{n−2} −. . .+ (−1)^{n−1}x_{1} ≡ 0 (mod 11)$. Which means that the alternating 2 sum of the digits of the number x,$x_{n} − x_{n−1} + x_{n−2}$ − . . ., is divisible by 11. The argument can be easily reversed.

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Now, I didn't understand the proof starting from But.

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What couldn't you understand in the above reverse part?It is very much in line with the statement of the Theorem of Divisibility of a Number by 11.

In the reverse, every alternate term has been reversed. That is what the statement states. When you'll test divisibility of any given no, then just apply the reverse part.

Let's take an example, we've got to test the divisibility of 121. So this can be written as

 

1X1+10X2+100X1. The break up is done.[See reverse logic].  Now in d reverse logic ignoring d coefficients we have, going with the rule of every alternate term negative, 1-2+1, so 2-2=0.

Finally, goes your answer. :)

If still u don't get go thru the comments. :)

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