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    (Original post by DFranklin)
    Just grinding through options is doable (and what I think most people who get it out under time pressure end up doing - it's what I did).

    There's some discussion of this question here: http://www.thestudentroom.co.uk/show...=475982&page=2
    Thanks for that.
    (Original post by DeanK22)
    that is a quadratic diophantine equation that you can solve. Have you come across quadratic diophantine equations before?

    infact, here is a site with step-by-step method with the above problem;

    http://www.alpertron.com.ar/QUAD.HTM
    I've seen some basic Pell equations, but I didn't realise we could extend the idea like that. Cheers.
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    doing BMO 1 1996.

    f(1)=1996 f(1)+f(2)+...+f(n)=n^2.f(n)

    find f(1996) i get
    Spoiler:
    Show
    2/1997
    can anyone verify?
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    Agreed.
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    (Original post by DFranklin)
    Agreed.
    yay

    when answering i didn't justify that it was the sum of reciprocal triangle numbers, would i have been penalised for this?
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    (Original post by Totally Tom)
    yay

    when answering i didn't justify that it was the sum of reciprocal triangle numbers, would i have been penalised for this?
    Probably.
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    (Original post by Totally Tom)
    yay

    when answering i didn't justify that it was the sum of reciprocal triangle numbers, would i have been penalised for this?
    This was quite a nice question. I'm wondering how you do it using the sum of reciprocal triangular numbers though.
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    (Original post by GHOSH-5)
    This was quite a nice question. I'm wondering how you do it using the sum of reciprocal triangular numbers though.
    make triangle number partial fraction and stuff cancels leaving the sum to n to equal 1-1/n
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    (Original post by Totally Tom)
    make triangle number partial fraction and stuff cancels leaving the sum to n to equal 1-1/n
    Ah right. I just went with:
    Spoiler:
    Show
     n^2 f(n) = f(1) + f(2) + \cdots + f(n) \iff (n^2-1)f(n) = (n-1)^2 f(n-1)

     \iff f(n-1) = f(n)\dfrac{n+1}{n-1}

     \iff f(n-2) = f(n) \dfrac{n+1}{n-1} \times \dfrac{n}{n-2}

    Generalising this:

     \iff f(n-k) = f(n) \dfrac{n+1}{n-1} \times \cdots \times \dfrac{n-k+2}{n-k}

    But we get lots of things cancelling...

     \therefore f(n-k) = f(n) \times \dfrac{n(n+1)}{(n-k)(n-k+1)}

    Setting k = 1995, n = 1996, we get the answer.
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    (Original post by GHOSH-5)
    Ah right. I just went with:
    Spoiler:
    Show
     n^2 f(n) = f(1) + f(2) + \cdots + f(n) \iff (n^2-1)f(n) = (n-1)^2 f(n-1)

     \iff f(n-1) = f(n)\dfrac{n+1}{n-1}

     \iff f(n-2) = f(n) \dfrac{n+1}{n-1} \times \dfrac{n}{n-2}

    Generalising this:

     \iff f(n-k) = f(n) \dfrac{n+1}{n-1} \times \cdots \times \dfrac{n-k+2}{n-k}

    But we get lots of things cancelling...

     \therefore f(n-k) = f(n) \times \dfrac{n(n+1)}{(n-k)(n-k+1)}

    Setting k = 1995, n = 1996, we get the answer.
    nice, i'm just going through papers looking for accessible questions atm strictly avoiding any geometry
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    (Original post by Totally Tom)
    nice, i'm just going through papers looking for accessible questions atm strictly avoiding any geometry
    :tongue: ditto :o:
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    Hi,

    Thought I might use my existing thread, rather than start another. Just attempting Q2, BMO-1, 1998, and I think I've managed to prove that the remainder is 0, any chance someone could check, as I'm a little unsure? :o:
    (Original post by Question 2, BMO-2, 1998)
    Let a_1 = 19, a_2 = 98. For n\geq 1 define  a_{n+2} to be the remainder of a_{n+1} + a_n when it is divided by 100. What is the remainder when

     a_1^2 + a_2^2 + \cdots + a_{1998}^2

    is divided by 8?
    Attempt
    We find that  a_n = F_{n-2}a_2 + F_{n-3}a_1 [Property 1] where F_n is the nth Fibonacci number ( F_0 = 1, F_1 = 1 and for n \geq 0, F_{n+2} = F_{n+1} + F_n ).

    Edit: we can ignore the fact that a_{n+2} is the remainder of a_{n+1} + a_n on division by 100 as, suppose l is that remainder, and so a_{n+2} = l+100k for some k; it is clear that  (l+100k)^2 \equiv l^2 \pmod{8} .

    Let us show that F_n has a period of 6 in modulo 4 arithmetic. The first 6 Fibonacci numbers are 1, 1, 2, 3, 5, 8. Their remainders on division by 4 are 1, 1, 2, 3, 1, 0 respectively. Due to the recurrence defining Fibonacci numbers, it is clearly true that, for all k

     F_{6k} \equiv 1 \pmod{4}, F_{6k+1} \equiv 1 \pmod{4}, F_{6k+2} \equiv 2 \pmod{4}

     F_{6k+3} \equiv 3 \pmod{4}, F_{6k+4} \equiv 1 \pmod{4}, F_{6k+5} \equiv 0 \pmod{4} .

    Notice that this implies that for all  n \equiv 0, 1 \pmod{3} \iff a_n \equiv 1 \pmod{2} , and therefore that for all such n,  a_n^2 \equiv 1 \pmod{8} .

    What we have shown about Fibonacci numbers also implies that (due to Property 1)  a_{6k+2}^2 \equiv 4 \pmod{8} and also that  a_{6k+5}^2 \equiv 0 \pmod{8} .

    Note that there are 1332 terms of the form  a_{3m} or  a_{3m+1} , and so their quadratic residues modulo 8 sum to 1332, and there are 333 terms of the form  a_{6k+2} so their quadratic residues modulo 8 sum to 1332, and all the other terms' quadratic residues (terms of the form  a_{6k+5} leave no remainder on division by 8, so adding everything up, we get

     a_1^2 + a_2^2 + \cdots + a_{1998}^2 \equiv 1332 + 1332 \pmod{8} \equiv 0 \pmod{8} .
    Thanks in advance for any help :smile:
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    A brute force computation says the remainder is definitely 0 (the sum before division is 6594784).

    The one thing I don't like about what you've done is "Due to the recurrence defining Fibonacci numbers, it is clearly true that, for all k ..." Reading it, I'm not sure why you think it's true - that is, I can think of several ways of proving it's true, but I don't know which method you actually have in mind. I don't know enough about BMO marking to say if you'd get away with it though.
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    (Original post by DFranklin)
    A brute force computation says the remainder is definitely 0 (the sum before division is 6594784).

    The one thing I don't like about what you've done is "Due to the recurrence defining Fibonacci numbers, it is clearly true that, for all k ..." Reading it, I'm not sure why you think it's true - that is, I can think of several ways of proving it's true, but I don't know which method you actually have in mind. I don't know enough about BMO marking to say if you'd get away with it though.
    Cheers.

    Hmm, maybe I ought to make it more rigorous. I suppose we could prove it* inductively. It is true for the first case as above, and suppose it is true for some  k = s . Therefore:

     F_{6(s+1)} = F_{6s+5} + F_{6s+4} \equiv 0+1 \equiv 1 \pmod{4}

     F_{6(s+1)+1} = F_{6(s+1)} + F_{6s+5} \equiv 1 + 0 \equiv 1 \pmod{4}

    F_{6(s+1)+2} = F_{6(s+1)+1} + F_{6(s+1)} \equiv 1+1 \equiv 2 \pmod{4}

    F_{6(s+1)+3} = F_{6(s+1)+2} + F_{6(s+1)+1} \equiv 1+2 \equiv 3 \pmod{4}

    F_{6(s+1)+4} = F_{6(s+1)+3} + F_{6(s+1)+2} \equiv 3+2 \equiv 1 \pmod{4}

    F_{6(s+1)+5} = F_{6(s+1)+4} + F_{6(s+1)+3} \equiv 1+3 \equiv 0 \pmod{4}

    And so by induction, the statement is true. Would that be enough?

    *It being the following for  k \geq 0 :

     F_{6k} \equiv 1 \pmod{4}, F_{6k+1} \equiv 1 \pmod{4}, F_{6k+2} \equiv 2 \pmod{4}

     F_{6k+3} \equiv 3 \pmod{4}, F_{6k+4} \equiv 1 \pmod{4}, F_{6k+5} \equiv 0 \pmod{4} .
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    Yes, that would be fine. It feels a bit overkill, but I didn't see anything markedly better/clearer myself.

    What you did before might have been OK - it's in the "it's obvious how to prove this if you are competent" category.
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    (Original post by DFranklin)
    Yes, that would be fine. It feels a bit overkill, but I didn't see anything markedly better/clearer myself.

    What you did before might have been OK - it's in the "it's obvious how to prove this if you are competent" category.
    Thanks again :smile:
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    Just had a quick question, so thought I might ask it here; for Q3 BMO-1 2006 [the combinatorics question], could anyone confirm that the answer is 2520?

    :smile:
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    I seem to recall that being the answer - haven't actually redone the question to make sure, but I have a reasonable memory for these things. (and from "2520" I could guess which question you meant, so that's a good sign).
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    (Original post by DFranklin)
    I seem to recall that being the answer - haven't actually redone the question to make sure, but I have a reasonable memory for these things. (and from "2520" I could guess which question you meant, so that's a good sign).
    Cheers :smile: I'm a little confused with Q4 of the same paper. I've attached a drawing of what I think the situation is like, but for the tangent at P to touch T at some point Q, surely it's obvious after this that S and T have the same radius; hence PQ = AP immediately? I think I'm missing something or assuming something I shouldn't :confused:
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    Don't ask me about BMO geometry (or indeed anything that involves drawing diagrams...)
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    The line PQ is tangent to the smaller circle at Q.

    Name:  circles.bmp
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