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Math Help - Decimal Palindromic Integer

  1. #1
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    Decimal Palindromic Integer

    Show every dec. palind. int. that has an even # of digits is div. by 11.

    For example,

    18833881 is divisible by 11; why?
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  2. #2
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    Quote Originally Posted by Ideasman View Post
    Show every dec. palind. int. that has an even # of digits is div. by 11.

    For example,

    18833881 is divisible by 11; why?
    Proof is by induction.

    Every 2 digit palindrome is obviously divisible by 11.

    Suppose every 2n digit palindrome is divisible by 11.

    Consider any 2(n+1) digit palindrome P. The central 2n digits define a 2n
    digit palindrome PP divisible by 11, so write:

    P=PP*10 + (10^(2n+1)+1)Q

    where Q is the most (and least) significant digit of P. Then P is divisible by
    11 because PP is and so is 10^(2n+1)+1.

    Then the proof follows by mathematical induction.

    RonL

    (You may feel the need to prove that 10^(2n+1)+1 is divisible by 11.

    This would proceed along the lines of observing that 10=-1 mod 11
    so any odd power of 10 is also congruent to -1 modulo 11, which is
    equivalent to any odd power of 10 plus 1 is divisible by 11)
    Last edited by CaptainBlack; January 28th 2007 at 10:42 PM.
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  3. #3
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    Quote Originally Posted by Ideasman View Post
    Show every dec. palind. int. that has an even # of digits is div. by 11.

    For example,

    18833881 is divisible by 11; why?
    This problem appeared as an excerise in my number theory book about 3 years ago, since it was such an interesting fact, I memorized it ever since.

    The simplest way to see this is by using the rule that to determine if a number is divisible by 11 if and only if the "alternating sum" is divisible by 11.

    Thus, take for example 12344321
    The alternating sum is,
    1-2+3-4+4-3+2-1=0
    And 0 is divisible by 11.

    In fact, any even palindrom is divisible by 11 because the alternating sum (from right to left) is going to be zero.

    Thus, if a number is an ODD palindrome with zero as its middle then it too will be divisible by 11. For example, 12021 then alternating sum is,
    1-2+0+2-1=0.

    Furthermore, the middle digit in an odd palindrome determines the the remainder upon division (excerise).
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Thus, if a number is an ODD palindrome with zero as its middle then it too will be divisible by 11. For example, 12021 then alternating sum is,
    1-2+0+2-1=0.
    Uh, how? 12021 is not divisible by 11.

    It should be:

    1 - 2 + 0 - 2 + 1 = -2, which does not divide 11.
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  5. #5
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    Quote Originally Posted by Ideasman View Post
    Uh, how? 12021 is not divisible by 11.

    It should be:

    1 - 2 + 0 - 2 + 1 = -2, which does not divide 11.
    Okay, relax, thus I made a mistake.
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