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    (Original post by theone)
    That'll do, although i was thinking of a nicer method imho.
    That'll do for you then it'll do for me, yay I got one right! Well sort of... please enlighten us theone but don't be all "haha I know everything and you suck" about it
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    (Original post by ZJuwelH)
    That'll do for you then it'll do for me, yay I got one right! Well sort of... please enlighten us theone but don't be all "haha I know everything and you suck" about it
    Oh and thanks for the rep!
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    Why are we looking for a proof? It is bloody obvious that there are no solutions when there a 4 or more digits.
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    (Original post by mikesgt2)
    Why are we looking for a proof? It is bloody obvious that there are no solutions when there a 4 or more digits.
    i feel a big proof coming up.........
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    (Original post by mikesgt2)
    Why are we looking for a proof? It is bloody obvious that there are no solutions when there a 4 or more digits.
    Well not really, so the proof is supposed to make it obvious.
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    (Original post by elpaw)
    so what's your method?
    Let n be the integer and s(n) the sum of the digits of n.

    Let n have x digits.

    It is clear that s(n) =< 9x and n >= 10^(x-1)

    So if 10^x-1 > 99x (i.e. 11s(n)) then n > 11s(n) and the inequality holds with x >=4 (you can prove this by induction if u need convincing)
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    (Original post by theone)
    It is clear that s(n) =< 9x and n >= 10^(x-1)
    Umm how is that clear?
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    (Original post by 2776)
    Umm how is that clear?
    Because n has x digits. and s(n) is maximised with each digit being 9, so s(x) <= 9x. Also we know n has x digits and so must be greater than or equal to 10^(x-1).
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    (Original post by theone)
    Let n be the integer and s(n) the sum of the digits of n.

    Let n have x digits.

    It is clear that s(n) =< 9x and n >= 10^(x-1)

    So if 10^x-1 > 99x (i.e. 11s(n)) then n > 11s(n) and the inequality holds with x >=4 (you can prove this by induction if u need convincing)
    Too small and concise for me, please explain to a future UEL graduate <vomits> how all that works...
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    (Original post by ZJuwelH)
    Too small and concise for me, please explain to a future UEL graduate <vomits> how all that works...
    *see above* and ask if you need further explanation
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    (Original post by theone)
    Because n has x digits. and s(n) is maximised with each digit being 9, so s(x) <= 9x. Also we know n has x digits and so must be greater than or equal to 10^(x-1).
    umm how is it that n=> 10 ^(x-1) ?
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    (Original post by 2776)
    umm how is it that n=> 10 ^(x-1) ?
    n = a.10^x-1 + b.10^x-2 ... + (some constant).

    Now n is minimised with b=c=d=... = 0 and a=1 since a must be > 0.

    So n is minimised with n = 10^x-1.
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    (Original post by theone)
    So if 10^x-1 > 99x (i.e. 11s(n)) then n > 11s(n) and the inequality holds with x >=4 (you can prove this by induction if u need convincing)
    Can u explain this bit please oh angel of maths
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    (Original post by 2776)
    Can u explain this bit please oh angel of maths
    bump
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    (Original post by 2776)
    Can u explain this bit please oh angel of maths
    So if 10^x-1 > 99x (i.e. 11s(n)) then n > 11s(n) and the inequality holds with x >=4 (you can prove this by induction if u need convincing)

    10^x-1 is the minimum value of the number n with x digits. 99x is the maximum value of 11 times the sum of the digits. Therefore, the number must be greater than 11 times the sum of its digit as the minimum value of n is greater than the maximum value of 11 times the sum of the digits:

    10^x-1 > 99x

    This implies n>11S(n)

    When x=4, then 10^x-1=10^3 and 99x=396.

    Therefore, the basis case works.

    Assume P(k) is true and prove P(k+1)

    Therefore, 10^x-1>99x

    If 10^x-10^(x-1)>99(x+1)-99x then P(k+1) will be true if P(k) is true:

    10^x-10^(x-1)=[10^(x-1)](10-1)
    =9(10^(x-1))
    >9*99x

    As x>=4, then 9*99x>=3564

    99(x+1)-99x=99x+99-99x=99
    => 99(x+1)-99x<9*99x

    Put back above, this gives 10^x-10^(x-1)>99(x+1)-99x

    Therefore, if P(k) is true then P(k+1) is true. As the basis case P(4) it true, by induction P(x) is true where x is a positive integer greater than or equal to four.

    Therefore, n>11S(n) when x>=4, so there are no solutions when the number n has more than three digits.

    Nice proof theone and who have you got interviewing you at Trinity?
 
 
 
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