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Challenge VII: A bit of number theory solved by Boorglar

  1. Aug 8, 2013 #1
    This new challenge was suggested by jostpuur. It is rather number theoretic.

    Assume that [itex]q\in \mathbb{Q}[/itex] is an arbitrary positive rational number. Does there exist a natural number [itex]L\in \mathbb{N}[/itex] such that

    [tex]Lq=99…9900…00[/tex]

    with some amounts of nines and zeros? Prove or find a counterexample.
     
    Last edited by a moderator: Aug 8, 2013
  2. jcsd
  3. Aug 8, 2013 #2
    So we want to get a number of the form [itex] L \frac{a}{b} = 10^m(10^n-1) [/itex].
    Write [itex] a = 2^x5^yd [/itex], where d is relatively prime with 10.

    Then [itex] L = \frac{10^m}{2^x5^y} \frac{10^n-1}{d}[/itex].
    The left fraction is obviously an integer if we choose m larger than max( x, y ).
    The right fraction can be made an integer since d is relatively prime with 10, and therefore 10 is in the multiplicative group modulo d. Let n be the order of 10 in U(d), then [itex]10^n-1[/itex] is divisible by d so L is an integer.
     
  4. Aug 8, 2013 #3
    Well, that was fast. :tongue:
     
  5. Aug 8, 2013 #4
    Yeah I tend to be good at those types of problems haha.
    Ah by the way I forgot the b multiplying the fractions but it doesn't really matter.
     
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