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Math Help - Factors Of 3-digit Numbers

  1. #1
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    Factors Of 3-digit Numbers

    Hi all!

    I have this question:

    Find all positive integers n (n will be called special)
    such that if n divides a three digit number say ABC where A = hundreds, B = tens, C = ones. Then n also divides BCA and CAB.

    So far I have worked out that 1 is special of course, so is 3 (factor test sum of digits will always be the same) and 9 as well as 111,222,333....999 and any number greater than 1000 as it will only divide into the 3 digit number 000.

    I am told that there are two more such numbers n under 100. How do you find them? And please explan your reasoning. Thanks!
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  2. #2
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    Quote Originally Posted by Infinate Possibilities
    Hi all!

    I have this question:

    Find all positive integers n (n will be called special)
    such that if n divides a three digit number say ABC where A = hundreds, B = tens, C = ones. Then n also divides BCA and CAB.

    So far I have worked out that 1 is special of course, so is 3 (factor test sum of digits will always be the same) and 9 as well as 111,222,333....999 and any number greater than 1000 as it will only divide into the 3 digit number 000.

    I am told that there are two more such numbers n under 100. How do you find them? And please explan your reasoning. Thanks!
    Hello,

    interesting problem.
    First idea: The digits must be manifolds of your divisor.
    So 2 divides all numbers build of 2, 4, 6 or 8. (248, 862, 628, ...)
    So 5 divides all numbers build of 5 or 0: 550, 505, (maybe even 055).
    So 4 divides all numbers build of 4 or 8. (484, 444, 848, ...)

    There should be more numbers of the described properties, but I don't know how to get them. Good luck!

    Greetings

    EB
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  3. #3
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    We say that n is not special as long as we can find one example of it not being special.

    Thus for 5: See (455)

    5 divides 455
    But it does not divide 554 therefore 5 is not special.

    Do you kind of see what im after?
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  4. #4
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    Now 4 is not special either:

    4(312)

    4 divides 312
    But not 123.
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  5. #5
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    Interesting problem.
    Maybe you can do this:
    If ABC is a number then it by definition of digits (base 10) is expressable as ABC=100x+10y+z
    Thus, BCA=100y+10z+x
    and CAB=100z+10x+y.
    Now n divides these, thus (if and only if)
    100x+10y+z=nk
    100y+10z+x=nk
    100z+10x+y=nk For some k.

    Thus upon addition of equations,
    111(x+y+z)=3nk.
    Since \gcd(111,3)=1 Euclid's Lemmas tells us that 3|(x+y+z). Thus, the sum of the digits must be divisible by 3.
    Further information is that x,y,z are integers between 1 and 9 inclusive. Thus, you can go through all possibilities.
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  6. #6
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    Ok i'll thy that, thanks a lot. I'll post when I get some answers.
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  7. #7
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    I'm trying to do it, but their are too many numbers. As a lot of numbers add to 9 in their digits and divide 3. Any more ideas?
    Last edited by Infinate Possibilities; February 22nd 2006 at 12:45 AM.
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  8. #8
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    So far I have found 1 more : 27

    But tell me why 27 can be n. I can't just say I did guess + check or something like that.

    Also there is still 1 more n under 100.
    Last edited by Infinate Possibilities; February 22nd 2006 at 12:58 AM.
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