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Abba divisible by 11

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A four digit number can be represented by 'abba'. (where a and b are digits)

    1) show that abba is always divisible by 11.
    2) show that abbbba is always divisible by 11.
    3) is abbba divisble by 11?

    2. Relevant equations


    3. The attempt at a solution

    i honestly have no idea how to begin solving this, i understand what its asking but have no idea how to prove it.
  2. jcsd
  3. Jan 10, 2010 #2
    What are the rules of divisibility for 11?
  4. Jan 10, 2010 #3


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    Homework Helper

    You can write the number abba with its digits as N=1000*a+100*b+10b+a.

    Collect the terms containing a and b:

    N=1001*a +110 *b.

    See if the factors 1001 and 110 are divisible by 11.

  5. Jan 10, 2010 #4
    I have a easier way.

    you can express N as :
    ==(-1)³a+(-1)²b+(-1)b+a mod(11)
    ==-a+a+b-b mod(11)
    ==0 mod(11)

    I believe the others can be proved analogously.
  6. Jan 10, 2010 #5
    What icystrike said, can be generalized for any integer N.



    Then, since 10=(-1)(mod11) it follows that 10^n=(-1)^n(mod11), and similarly for others, so:

    [tex]a_n10^n+...+a_110+a_0=a_n(-1)^n+a_{n-1}(-1)^{n-1}+...+a_1(-1)+a_0(mod 11)[/tex]

    Which basically tells you that a number N is divisible by 11 iff when adding and subtracting its digits alternatively gives you zero.

    now 2) -a+b-b+b-b+a=0. so yes!
    Last edited: Jan 10, 2010
  7. Jan 10, 2010 #6
    we havent learned modulo yet so im not sure how to use them. I understand what you guys mean when you say it can be represented by 1000a + 100b + 10b + a but i dont understand how to use 11 to solve it.
  8. Jan 10, 2010 #7
    ehild explained it without modulo.

    Have a look at his/her post! (post #3)
  9. Jan 10, 2010 #8
    thanks to everyone, i figured it out and gave an explanation stating that if both nomials are divisible by a common factor then the sum of the two must also be.
  10. Jan 10, 2010 #9
    It can easily be proved that if d|a and d|b then d|(ah+bk), where h,k are integers.
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