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The sum of ABCD and DCBA is always divisible by 11, where A, B, C and D are digits of a number. I understood that, ABCD = A(1000) + B (100) + C(10) + D(1) DCBA = D(1000) + C(100) + B(10) + A (1) Then ABCD + DBCA = A(1001) + B(110) + C (110) + D(1001) =11[91A + 10B +10C +91D] which is divisible by 11.

I want a generalized proof for the above problem. Also, I want why it is applicable only for 11 and why it is valid for even digit of numbers? Thanks in advance

  • 2
    If you ask a question that's related to another question you've asked (in this case http://math.stackexchange.com/questions/52361/base-system-and-divisibility), it makes sense to link to the question so people get a better idea of the context; e.g. in this case Gortaur wouldn't have had to bother to tell you about the divisibility rule for 11.2011-07-20
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    For an odd number of digits the middle digit will lead to a summand `2 * d * 10^k`, so it can only work for divisibility by multiples of 2 and/or 5.2011-07-20
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    In what direction do you want to generalize your result?2011-07-21

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