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The BMO2 Thread watch

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    Hi there, did anyone here take BMO2 today? Now that it's 8:00, we can discuss the paper. What did you think?

    EDIT: Sorry, we can't discuss it until 8:30. I obviously can't read.
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    Is that Edexcel? I took MD02 today, on AQA.
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    (Original post by orionmoo)
    Is that Edexcel? I took MD02 today, on AQA.
    British Mathematical Olympiad. Nothing to do with A levels
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    (Original post by orionmoo)
    Is that Edexcel? I took MD02 today, on AQA.
    No, the second round of the British Mathematical Olympiad. It's a bit harder than Edexcel :p:
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    Oops sorry, lol. Never heard of it
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    Ok, so the paper is up.
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    I just answered the first two questions.But I don't know it's right or not
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    I answered the first 2, have a shaky solution of 3, and 2 pages of meaningless ramble on 4. It's better than I thought I'd do!
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    I thought it was a nicer paper than usual. But that means the boundaries will be up...
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    Well, question 1 was a much easier than normal. Although I suppose question 4 kind of made up for it. I still haven't quite got it yet. I'm all out of ideas :huff:

    I answered 1 and 2 in the exam, and have since answered 3.
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    Yay, we got one. :woo: and it only took five minutes. :woo: :woo:

    Something along the lines of d(c+e)^2=2009 with d=41 and c+e=7?

    If we don't post again that means we can only do that one
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    (Original post by DeanK22)
    erm. Don't know what you did but heres what I would have done;

    Spoiler:
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     \sqrt{a} + \sqrt{b} = 2009 \Rightarrow a,b > 0 We know that a and b must also be perfect squares. WLOG let b > a.

    c has the property.

     a = c^2 and  b = k^2c^2

    This results in;

     c(k + 1) = 2009

    Fomo factorising 2009  2009 = 7 \times 7 \times 41

    Therefore we obtain;

    For integer solutions (a,b) =  {49^2, (49 \times 40)^2} and  {7^2, (7 \times 286)^2}

    but the question was  \sqrt{a} + \sqrt{b} = \sqrt{2009}

    not \sqrt{a} + \sqrt{b} = 2009
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    (Original post by DeanK22)
    ffs
    unlucky :p:

    and is this DeanK2? why the change in user?
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    (Original post by DeanK22)
    Spend too much time on here. Don't do homework

    Solution;

    Spoiler:
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     \sqrt{a} + \sqrt{b} = \sqrt{2009} *

    wlog let  b > a \Rightarrow b = c^2 a for an integer c.

    Shoving this into equation * and squaring;

     a(1 + 2c + c^2) = 49 \times 41 = 287 \times 7

    After a quick rinse.

     a = 41 \Rightarrow 1 + 2c + c^2 = 49 \Rightarrow c = 6

    So (a,b)  (41 , 1476

    you'll find c isn't necessarily an integer

    (369, 656) happens to be a solution and a contradiction to your statement...

    also, if you closed your other username, why did you open a new one? :lolwut:
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    (Original post by DeanK22)
    Spend too much time on here. Don't do homework

    Solution;

    Spoiler:
    Show


     \sqrt{a} + \sqrt{b} = \sqrt{2009} *

    wlog let  b > a \Rightarrow b = c^2 a for an integer c.

    Shoving this into equation * and squaring;

     a(1 + 2c + c^2) = 49 \times 41 = 287 \times 7

    After a quick rinse.

     a = 41 \Rightarrow 1 + 2c + c^2 = 49 \Rightarrow c = 6

    So (a,b)  (41 , 1476


    There are two more pairs.

    Edit: three more pairs. Missed the trivial pair. :blushing:
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    (Original post by Linda and Suzanne)
    There are two more pairs.
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    i found 8 in total...
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    (Original post by GHOSH-5)
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    i found 8 in total...
    :eek: So we didn't even get one right ? :eek:

    We got

    41,1476
    164,1025
    369,656

    What are the others?
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    ha no didnt get into this, i only got 1 mark on BMO1 !!!
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     \sqrt{2009} = 7\sqrt{41}

    7= 1 + 6 = 2 + 5 = 3 + 4

     7\sqrt{41} = d\sqrt{41} + c\sqrt{41}

    Sub away and you get 4. How did you manage 8?
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    (Original post by DeanK22)
     \sqrt{2009} = 7\sqrt{41}

    7= 1 + 6 = 2 + 5 = 3 + 4

     7\sqrt{41} = d\sqrt{41} + c\sqrt{41}

    Sub away and you get 3. How did you manage 8?
    you have 4... the other 4 are the same, merely a is flipped with b...

    you do need to show that these are the only solutions though...
 
 
 
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