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    (Original post by JAKstriked)
    I like this one, it's number theoryish. (obviously)
    Are you asking for help or as a question to try?
    Spoiler:
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    Easiest way to start is to let 10^N-N=(10^N-1)-(N-1)
    This is actually one of the very few I have managed to do, so I was suggesting it as a question to try.

    What are everyone's favourite question types?
    I guess I like the geometry and algebra....and I hate the combinatorics.

    Okay another question to try:
    "Circle A lies inside circle B and touches it at M. A straight line touches A at P and B at Q & R. Prove angle QMP = angle RMP."

    Some of you may have seen it before...
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    (Original post by refref)
    *edit*

    Actually it came tody....I really dont know how thats possible.

    But awesome.
    awesome!
    i think the boundaries would be up on UKMT website in a couple of days hopefully tomorrow~
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    (Original post by lovmaths)
    awesome!
    i think the boundaries would be up on UKMT website in a couple of days hopefully tomorrow~
    I know I haven't made it I still might get best in the school though, which would be nice.

    I think I can still enter anyway though.
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    (Original post by refref)
    I know I haven't made it I still might get best in the school though, which would be nice.

    I think I can still enter anyway though.
    I got either 92 or 87 if i got 92, i think i’m in, but if i got 87… i’m not sure
    If the boundary for BMO is lower than 87, i’m definitely in, but i don’t think that’s going to happen…. The boundary always seems to be somewhere around 90…
    Hope i make it

    Do you happen to have 'Plane Euclidean Geometry' and 'Introductions to Number Theory and Inequalities'?
    Do they go in more depth than Olympiad Primer?

    Never mind if you don’t have them thanks
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    (Original post by lovmaths)
    I got either 92 or 87 if i got 92, i think i’m in, but if i got 87… i’m not sure
    If the boundary for BMO is lower than 87, i’m definitely in, but i don’t think that’s going to happen…. The boundary always seems to be somewhere around 90…
    Hope i make it

    Do you happen to have 'Plane Euclidean Geometry' and 'Introductions to Number Theory and Inequalities'?
    Do they go in more depth than Olympiad Primer?

    Never mind if you don’t have them thanks
    No I dont. The primer only has a short chapter on theory (a few pages on geometry/algebra/combinatorics/number theory, about 40 pages altogether) but then the rest is solutions. The solutions arent just solutions though, there is quite a bit of commentary with them which makes them alot more useful.
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    (Original post by refref)
    No I dont. The primer only has a short chapter on theory (a few pages on geometry/algebra/combinatorics/number theory, about 40 pages altogether) but then the rest is solutions. The solutions arent just solutions though, there is quite a bit of commentary with them which makes them alot more useful.
    hey thanks for always answering my posts
    *rep*
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    Into to NT&I has a poor number theory but good inequaltiy section.
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    (Original post by DeanK22)
    Into to NT&I has a poor number theory but good inequaltiy section.
    I don't think I really enjoyed the number theory section in it that much either - I've hardly looked at the inequalities section actually, but it looks much better. I'm glad I forced myself through it though, because I feel I gained a lot of new ideas from it, and became much more proficient.
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    (Original post by Hughh)
    This is actually one of the very few I have managed to do, so I was suggesting it as a question to try.

    What are everyone's favourite question types?
    I guess I like the geometry and algebra....and I hate the combinatorics.
    I personally prefer the number theory and algebra questions, not that I always can solve them (in fact, I rarely can!), and the odd bit of combinatorics, if it's one of the first couple of questions :p: I quite dislike geometry questions though :o:
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    (Original post by refref)
    I have all the answers now (I was only expecting answers for a few years...but it seems it has them all. We were right (well, obviously) on Q's 3 and 4. Do you have this book? If not, if you want any answers nudge me.
    I do have the book, but I'd lent it to a friend to borrow; I only just got it back yesterday in fact.
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    (Original post by GHOSH-5)
    I personally prefer the number theory and algebra questions, not that I always can solve them (in fact, I rarely can!), and the odd bit of combinatorics, if it's one of the first couple of questions :p: I quite dislike geometry questions though :o:
    Yeah well I'm hoping for 2 geometry questions this year and one easy algebra question. No doubt there will be a difficult combinatorics one which I will avoid. I would be very happy if I got into BMO2... but I am not expecting to.

    Update from UKMT: Bronze 54-63 Silver 64 - 78 Gold 79+ BMO1 91+

    Wahey
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    1997 question 2

    http://www.bmoc.maths.org/home/bmo1-1997.pdf

    I've read the solution, still dont get it.

    !!!
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    (Original post by refref)
    1997 question 2

    http://www.bmoc.maths.org/home/bmo1-1997.pdf

    I've read the solution, still dont get it.

    !!!
    Spoiler:
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    I started this question by finding some a_i and seeing if I could see a nice pattern; so let's do that:

     a_1 = 1

     a_2 = 3

     a_3 = \frac{4}{2} (4) = 8

     a_4 = \dfrac{5}{3} (12) = 20

     a_5 = \frac{6}{4} (32) = 48

    And I noticed that, to get the next term, you essentially double the previous term and add an appropriate power of 2 . So after conjecturing that  a_n = 2a_{n-1} + 2^{n-2} it can be easily proved by induction.

    Now consider that  a_n = 2(2a_{n-2} + 2^{n-3}) + 2^{n-2} = 4a_{n-2} + 2 \times 2^{n-2}

     \equiv 4(2a_{n-3} + 2^{n-4}) + 2 \times 2^{n-2} = 8a_{n-3} + 3 \times 2^{n-2}

     \equiv 8(2a_{n-4} + 2^{n-5}) + 3 \times 2^{n-2} = 16a_{n-4} + 4 \times 2^{n-2}

    And so I now noticed and conjectured that  a_n = 2^k a_{n-k} + k2^{n-2} , which can also be proven by induction. From there, setting n = 1997, k = 1996, the question drops out quite quickly.
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    (Original post by GHOSH-5)
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    I started this question by finding some a_i and seeing if I could see a nice pattern; so let's do that:

     a_1 = 1

     a_2 = 3

     a_3 = \frac{4}{2} (4) = 8

     a_4 = \dfrac{5}{3} (12) = 20

     a_5 = \frac{6}{4} (32) = 48

    And I noticed that, to get the next term, you essentially double the previous term and add an appropriate power of 2 . So after conjecturing that  a_n = 2a_{n-1} + 2^{n-2} it can be easily proved by induction.

    Now consider that  a_n = 2(2a_{n-2} + 2^{n-3}) + 2^{n-2} = 4a_{n-2} + 2 \times 2^{n-2}

     \equiv 4(2a_{n-3} + 2^{n-4}) + 2 \times 2^{n-2} = 8a_{n-3} + 3 \times 2^{n-2}

     \equiv 8(2a_{n-4} + 2^{n-5}) + 3 \times 2^{n-2} = 16a_{n-4} + 4 \times 2^{n-2}

    And so I now noticed and conjectured that  a_n = 2^k a_{n-k} + k2^{n-2} , which can also be proven by induction. From there, setting n = 1997, k = 1996, the question drops out quite quickly.
    Hello, we meet again! Really don't know how one is supposed to spot that, but okay then! Thanks.

    Also I cant seem to get going on question 2 1998 (http://www.bmoc.maths.org/home/bmo1-1998.pdf). I've read the solution but it still makes little sense! And I love sequences/series so much
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    (Original post by refref)
    Hello, we meet again! Really don't know how one is supposed to spot that, but okay then! Thanks.

    Also I cant seem to get going on question 2 1998 (http://www.bmoc.maths.org/home/bmo1-1998.pdf). I've read the solution but it still makes little sense! And I love sequences/series so much
    It's essentially just looking for patterns in the sequence...

    Note that you can 'ignore' the fact about it being the remainder when divided by 100, as suppose  a_n + a_{n+1} = 100k + r where r is between 0 and 99 inclusive.

    Hence  a_{n+2} = r by definition. However  a_{n+2}^2 = r^2 \equiv (r+100k)^2 \pmod{8} , which is what we are interested in.

    Try and write each term of the series in terms of  a_2 and  a_1 . You should find that it is related to Fibonacci numbers. After that, try and look for patterns in the Fibonacci sequence in mod 4 arithmetic.

    By 'I've read the solution' do you mean the one in the primer? Does that vary much to my solution on page 2 of this thread? If not, perhaps read mine for an alternative view.
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    (Original post by GHOSH-5)
    It's essentially just looking for patterns in the sequence...

    Note that you can 'ignore' the fact about it being the remainder when divided by 100, as suppose  a_n + a_{n+1} = 100k + r where r is between 0 and 99 inclusive.

    Hence  a_{n+2} = r by definition. However  a_{n+2}^2 = r^2 \equiv (r+100k)^2 \pmod{8} , which is what we are interested in.

    Try and write each term of the series in terms of  a_2 and  a_1 . You should find that it is related to Fibonacci numbers. After that, try and look for patterns in the Fibonacci sequence in mod 4 arithmetic.

    By 'I've read the solution' do you mean the one in the primer? Does that vary much to my solution on page 2 of this thread? If not, perhaps read mine for an alternative view.
    Thanks for that by the way, I think I get it now

    My confidence is slowly deteriorating now. I'm on Q2 on 1999 http://www.bmoc.maths.org/home/bmo1-1999.pdf.

    It took me about 40 minutes to get to AM/MP must be constant, by some ridiculous long method, so I thought I had a breakthrough but I couldnt prove that this must be constant. So I looked at the solution and I dont really get his conclusion. Have you done this paper?
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    (Original post by refref)
    Thanks for that by the way, I think I get it now

    My confidence is slowly deteriorating now. I'm on Q2 on 1999 http://www.bmoc.maths.org/home/bmo1-1999.pdf.

    It took me about 40 minutes to get to AM/MP must be constant, by some ridiculous long method, so I thought I had a breakthrough but I couldnt prove that this must be constant. So I looked at the solution and I dont really get his conclusion. Have you done this paper?
    I haven't actually looked at that question, or indeed many geometry questions. Sorry.
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    (Original post by GHOSH-5)
    I haven't actually looked at that question, or indeed many geometry questions. Sorry.
    Oh ok. Thanks anyway.

    I'll have to neg you though!

    Actually I get it now.
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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.
    First of all, you notice that $k^2\equiv (k-100)^2\pmod 8$. This tells you that you can ignore the remainder 100 part of the problem all together, and focus on modulo 8. By the pigeonhole principle, the sequence a_n in mod 8 is periodic. This means we can find that period with sufficient patience. We find that it is very small (the period), and then we find that every 6 terms sum to 0 mod 8 due to the period. Therefore, since 1998 is divisible by 6, the whole sum is divisible by 8. QED.
 
 
 
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