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A problem from The Strand

  1. Oct 22, 2006 #1
    I was reading Ramanujan’s biography (by Kanigel) and there was a mathematical problem in that book which was published in the Strand magazine during the First World War. This is the problem:

    "I was talking the other day," said William Rogers to the other villagers gathered around the inn fire, "to a gentleman about the place called Louvain, what the Germans have burnt down. He said he knowed it well – used to visit a Belgian friend there. He said the house of his friend was in a long street, numbered on this side one, two, three, and so on, and that all the numbers on one side of him added up exactly the same as all the numbers on the other side of him. Funny thing that! He said he knew there was more than fifty houses on that side of the street, but not so many as five hundred. I made mention of the matter to our parson, and he took a pencil and worked out the number of the house where the Belgian lived I don’t know how he done it."
    Perhaps the reader may like to discover the number of that house.

    Here is what I tried:
    Let n = the number of the house that the Belgian lived in
    m = the total number of houses in that street
    and 50<m<500

    [tex]1+2+ \cdots +(n-1) = (n+1)+(n+2)+ \cdots +m[/tex]

    [tex]\frac{n(n-1)}{2} = \frac{(n+1+m)(m-n)}{2}[/tex]

    [tex]n(n-1) = (n+1+m)(m-n)[/tex]

    [tex]2n^2 = m^2 + m[/tex]

    now how do i sove for n and m?
    Last edited: Oct 22, 2006
  2. jcsd
  3. Oct 22, 2006 #2

    matt grime

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    By using the restrictions placed on how many houses there are. It is just trial and error from there, plus some other observations, like m^2+m is the product of coprime integers m and m+1.
    Last edited: Oct 22, 2006
  4. Oct 22, 2006 #3
    but Ramanujan supposedly solved it using continued fractions. but how?

    i dont see how that helps.
  5. Oct 22, 2006 #4


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    Ramanujan solved every problem he met with continued fractions. Perhaps that's why we don't understand any longer how he got his results.

    Sorry, just kidding, I'm out of here..
  6. Oct 22, 2006 #5

    matt grime

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    So, if I tell you that ab is a perfect square and that a and b are coprime, you can't deduce anything at all about a and b?
  7. Oct 23, 2006 #6
    sorry, maybe it is incredibly stupid of me, but i can't see what you are suggesting.
  8. Oct 23, 2006 #7
    can anyone please help me a little more?
  9. Oct 23, 2006 #8

    matt grime

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    Didn't we have long thread on getting you to prove that if p is a prime and p divides a^2 then p divides a? Doesn't this spark something. Just think of a composite squared, like 6^2=36. Now, how can you write 36 as a product of comprime numbers? Notice anything about them?
  10. Oct 23, 2006 #9
    so the coprimes must be perfect squares too? is that it?
  11. Oct 23, 2006 #10
    What is the difference between a perfect square and a square? (if any)

    Its ambiguous to me. :frown:
  12. Oct 23, 2006 #11


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    "Perfect square" = "square of an integer"

    [itex]s^2[/itex], where [itex]s=\sqrt2[/itex], is a square but not a perfect square.
  13. Oct 23, 2006 #12

    matt grime

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    Now, you have 2n^2 = m(m+1).

    Either m is even and m/2 and m+1 are prefect squares, or m is odd and m and (m+1)/2 are prefect squares. There aren't many perfect squares in the region of 50 to 500 to check, are there? (Yes, 'perfect square' means 'square of an integer', it is perhaps a silly distinction).
  14. Oct 23, 2006 #13
    Are all real numbers then square?

    To me, square number = perfect square number.
    It seems pointless to give all real numbers the distinction square. I think square should mean all real numbers with integer roots and perfect squares should be conserved for the set of perfect numbers, squared.

    That is:
    Perfect Numbers = {6,28,496,....}
    Perfect square: {36,784,.......}

  15. Oct 23, 2006 #14
    The square of "perfect numbers" (integers) ARE perfect squares.
  16. Oct 23, 2006 #15

    matt grime

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    why are you introducing real numbers?
  17. Oct 23, 2006 #16
    but the square roots of all perfect squares are NOT perfect numbers. :)
  18. Oct 23, 2006 #17
    Example, please.

    (I thinlk this silly argument is irrelevant to your original post)
  19. Oct 23, 2006 #18

    matt grime

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    Ok, Murshid, you're free to tihnk that. No one else does, and it is common to conventional to mean something else. So you'd be pig-headed to carry on in that vein.
  20. Oct 23, 2006 #19
    Just limiting the possibility of mistakes :P
  21. Oct 23, 2006 #20
    the square root of 4 is 2 which is not a perfect number
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