Results 1 to 5 of 5

Math Help - Today's problem that I don't understand.

  1. #1
    Member billym's Avatar
    Joined
    Feb 2008
    Posts
    183

    Today's problem that I don't understand.

    "Show that there is a multiple of 2004 which has the form 22...200...0, i.e., a sequence of digits 2 followed by a sequence of digits 0."

    These seem so easy after I see the answer, but I never know exactly what is being asked.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    645
    Hello, billym!

    A fascinating problem . . . I haven't solved it yet.
    But I have a start . . . maybe someone can follow through?


    Show that there is a multiple of 2004 which has the form 22...200...0,
    that is, a sequence of 2's, followed by a sequence of 0's.

    \text{We have: }\;N \;=\;\underbrace{222\hdots2}_{m\text{ digits}} \underbrace{000\hdots0}_{n\text{ digits}}\;=\;2004k

    \text{This can be written: }\;N \;=\;2\underbrace{(111\hdots1)}_{m\text{ digits}} \cdot 10^n \;=\;2^2\cdot501\cdot k

    \text{Divide by 2: }\;10^n\cdot(111\hdots1) \;=\;2\cdot501\cdot k

    \text{Divide by 2: }\;5\cdot 10^{n-1}(111\hdots1) \;=\;501\cdot k


    It seems that a number of the form (111\hdots1) is a multiple of 501.

    And all we have to do is find it . . . or prove that it exists.


    Anyone? .Anyone?

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member billym's Avatar
    Joined
    Feb 2008
    Posts
    183
    Apparently helpful hint:

    Consider the 2005 “pigeons” 2, 22, 222, . . . , 22...2 and put them into holes according to their remainder upon division by 2004. Then argue as in here.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,658
    Thanks
    1615
    Awards
    1
    Quote Originally Posted by billym View Post
    Apparently helpful hint:
    Consider the 2005 “pigeons” 2, 22, 222, . . . , 22...2 and put them into holes according to their remainder upon division by 2004. Then argue as in here.
    What a completely surprising application of the pigeon hole principle.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Apr 2009
    Posts
    677
    This was really nice !! Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Today's Putnam Problem of the Day
    Posted in the Math Challenge Problems Forum
    Replies: 1
    Last Post: May 19th 2010, 04:39 AM
  2. Replies: 0
    Last Post: February 27th 2010, 11:57 AM
  3. Replies: 8
    Last Post: January 20th 2009, 09:49 PM
  4. Don't understand this problem.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 4th 2008, 05:59 PM
  5. Replies: 2
    Last Post: April 30th 2008, 10:12 AM

Search Tags


/mathhelpforum @mathhelpforum