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    (Original post by elpaw)
    *cough* 198 *cough*
    But if 89 = 10b + c how is b = 9?
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    (Original post by elpaw)
    you're not going to even get into QM
    Thanks for your support (!), I'll go and see if UEL have a place for me...
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    (Original post by ZJuwelH)
    But if 89 = 10b + c how is b = 9?
    its not, 89 = b + 10c
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    (Original post by ZJuwelH)
    But if 89 = 10b + c how is b = 9?
    Because it's 10c +b
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    (Original post by ZJuwelH)
    Thanks for your support (!), I'll go and see if UEL have a place for me...
    i wouldn't even bother going to the exam hall.....
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    (Original post by elpaw)
    i wouldn't even bother going to the exam hall.....
    lol
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    (Original post by mikesgt2)
    198 is the only one

    4 digit numbers gives:

    989a + 89b = 10d + c

    Which is impossible for a,b,c,d < 10 because max of 10d + c is 99 but the minimum of the lhs is 989 because a has to be at least 1
    And 5 digit numbers? 6,7,8?
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    (Original post by theone)
    And 5 digit numbers? 6,7,8?
    do you have to use some sort of proof by induction?
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    <sits in corner of infant school classroom and sobs> Leave me alone!

    Bloody football result messed my brain up. There's new champions on the block and they play in blue...
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    (Original post by elpaw)
    do you have to use some sort of proof by induction?
    Would be easier than going all the way to infinity-digit numbers...
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    (Original post by elpaw)
    do you have to use some sort of proof by induction?
    You can but don't really have to...
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    (Original post by theone)
    You can but don't really have to...
    9989a + 989b + 89c = d +10e

    Since least no. for rhs is 9989 when a is 1. it would be impossible, as d +10e can only go up to 99.
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    (Original post by 2776)
    9989a + 989b + 89c = d +10e

    Since least no. for rhs is 9989 when a is 1. it would be impossible, as d +10e can only go up to 99.
    what about k digit numbers, where k > 5
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    (Original post by theone)
    what about k digit numbers, where k > 5
    are you sadistic or something?
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    (Original post by theone)
    what about k digit numbers, where k > 5
    Surely u can't go any further. Because u will always meet 10a + b for the lhs. And where (k-1)9's89 is the minimum, so you will never get another answer, surely?
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    (Original post by 2776)
    Surely u can't go any further. Because u will always meet 10a + b for the lhs. And where (k-1)9's89 is the minimum, so you will never get another answer, surely?
    not rigorous enough for the rep
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    (Original post by theone)
    not rigorous enough for the rep
    tell me which part is not rigerous enough
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    (Original post by 2776)
    9989a + 989b + 89c = d +10e

    Since least no. for rhs is 9989 when a is 1. it would be impossible, as d +10e can only go up to 99.
    ((10^n)a-11)+((10^(n-1))b-11)...((10^2)z) = m+10n

    Since smallest possibility of LHS is ((10^n)a-11) it's impossible as m+10n only returns 99 at best...

    Really babyish I think but...
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    (Original post by ZJuwelH)
    ((10^n)a-11)+((10^(n-1))b-11)...((10^2)z) = m+10n

    Since smallest possibility of LHS is ((10^n)a-11) it's impossible as m+10n only returns 99 at best...

    Really babyish I think but...
    That'll do, although i was thinking of a nicer method imho.
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    (Original post by theone)
    That'll do, although i was thinking of a nicer method imho.
    so what's your method?
 
 
 
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