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Math Help - Sequence Problem

  1. #1
    Junior Member
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    Sequence Problem

    Hello everyone!!

    The following sequence continues indefinitely:

    27 = 3 x 3 x 3
    207 = 3 x 3 x 23
    2,007 = 3 x 3 x 223
    20,007 = 3 x 3 x 2223.......

    Which of the following integers is a multiple of 81?

    A: 200,007
    B:20,000,007
    C: 2,000,000,007
    D: 200,000,000,007
    E: 20,000,000,000,007

    I know what the answer is, can someone show me how it is done? And how similar problems are done. Thanks.
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Do you remember your divisibility rule for 9...

    if the sum of the digits is divisibile by 9 then so is the number...


    ex...2223 2+2+2+3=9 and 9 is divisibile by nine so

    so 2223 can be factored to equal 9*247

    so then

    20,007=3*3*9*247

    since 20,007 = 81*247 it is divisible by 81.

    so the next 22...3 will be divisible by nine is when the sum of the digits is divisivle by 9.
    2222222222223 would be the next one...

    since there are twelve 2's in the list it will be the number with twelve zero's

    20,000,000,000,007

    I hope this helps
    Last edited by TheEmptySet; February 23rd 2008 at 08:54 AM. Reason: I wanted to be clearer
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  3. #3
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    That was very clear!! Thank you so much.
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