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    (Original post by physicsmaths)
    I will write something up about them later when I am free .


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    Thnx xD
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    (Original post by gasfxekl)
    give an example? you must always plug in easy values like x,0, x,-x etc. then another principle is to make things cancel- for example if u have xf(y)+y somewhere try subbing x=a/f(y) so that u get a+y, etc.
    if you have a symmetric expression switch around x and y.
    Note that f(f(x))=x or 6x or whatever implies injectivity! try to get injectivity as it is often very useful.
    could you help with bmo1 1992 q5 pls
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    (Original post by 11234)
    could you help with bmo1 1992 q5 pls
    So another thing you should keep in mind is finding important values (like f(1)).

    In this case observe that f(n+1)>f(n) implies that f(x+n)\geq f(x)+n, and since f(1) \geq 1 we have f(n) \geq n. In fact
    f(n) > n since f(n)=n \Leftrightarrow f(f(n))=f(n)=3n=n, contradiction.
    Therefore 3n=f(f(n)) > f(n).
    Furthermore f(3n)=f(f(f(n))=3f(n).
    Now 1<f(1)<3, so f(1)=2, f(2)=f(f(1))=3, f(3)=3f(1)=6, f(6)=3f(2)=9, so f(6)=9>f(5)>f(4)>f(3)=6 gives f(4)=7, f(5)=8

    We now spot something interesting, and prove it by induction:
    3^k\leq n \leq 2\cdot 3^k \Rightarrow f(n)=3^k+n.
    For multiples of 3 we have 3^k \leq 3l \leq 2\cdot 3^k \Leftrightarrow 3^{k-1}\leq l \leq 2\cdot 3^{k-1}, so f(3l)=3f(l)=3(l+3^{k-1}). Now we have shown it for the multiples of 3. Now f(3l)=3l+3^k <f(3l+1)<f(3l+2)<f(3l+3)=3l+3+3^  k so we have it for all values.

    Finally, if 2\cdot 3^k<n<3^{k+1}, we obtain f(n)=3(n-3^k) by using the previous part and f(f(n))=3n.

    So f(1992)=3(1992-3^6)=3789.
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    (Original post by gasfxekl)
    So another thing you should keep in mind is finding important values (like f(1)).

    In this case observe that f(n+1)>f(n) implies that f(x+n)\geq f(x)+n, and since f(1) \geq 1 we have f(n) \geq n. In fact
    f(n) > n since f(n)=n \Leftrightarrow f(f(n))=f(n)=3n=n, contradiction.
    Therefore 3n=f(f(n)) > f(n).
    Furthermore f(3n)=f(f(f(n))=3f(n).
    Now 1<f(1)<3, so f(1)=2, f(2)=f(f(1))=3, f(3)=3f(1)=6, f(6)=3f(2)=9, so f(6)=9>f(5)>f(4)>f(3)=6 gives f(4)=7, f(5)=8

    We now spot something interesting, and prove it by induction:
    3^k\leq n \leq 2\cdot 3^k \Rightarrow f(n)=3^k+n.
    For multiples of 3 we have 3^k \leq 3l \leq 2\cdot 3^k \Leftrightarrow 3^{k-1}\leq l \leq 2\cdot 3^{k-1}, so f(3l)=3f(l)=3(l+3^{k-1}). Now we have shown it for the multiples of 3. Now f(3l)=3l+3^k <f(3l+1)<f(3l+2)<f(3l+3)=3l+3+3^  k so we have it for all values.

    Finally, if 2\cdot 3^k<n<3^{k+1}, we obtain f(n)=3(n-3^k) by using the previous part and f(f(n))=3n.

    So f(1992)=3(1992-3^6)=3789.
    Thanks for the reply but Im still quite confused as to how it implies nf(n) dop you have any more questions i could try
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    (Original post by 11234)
    Thanks for the reply but Im still quite confused as to how it implies nf(n) dop you have any more questions i could try
    what do you mean how it implies nf(n)?
    yes you can try: find all functions
    f: \mathbb{Q} \to \mathbb{R}: f(x+y)=f(x)f(y) or
    find all functions
    f: \mathbb{R} \to \mathbb{R}: f(xy)=xf(y)-yf(x) or
    find all functions
    f: \mathbb{Q} \to \mathbb{Q}: f(x+y)=f(x)f(y)-f(x)-f(y)+1, f(1)=2
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    (Original post by gasfxekl)
    what do you mean how it implies nf(n)?
    yes you can try: find all functions
    f: \mathbb{Q} \to \mathbb{R}: f(x+y)=f(x)f(y) or
    find all functions
    f: \mathbb{R} \to \mathbb{R}: f(xy)=xf(y)-yf(x) or
    find all functions
    f: \mathbb{Q} \to \mathbb{Q}: f(x+y)=f(x)f(y)-f(x)-f(y)+1, f(1)=2
    How does f(3n) imply 3f(n)
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    (Original post by 11234)
    How does f(3n) imply 3f(n)
    well f(f(a))=3a, if a=f(n) this implies f(f(f(n)))=3f(n), on the other hand f(f(f(n)))=f(3n) since f(f(n))=3n
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    (Original post by gasfxekl)
    well f(f(a))=3a, if a=f(n) this implies f(f(f(n)))=3f(n), on the other hand f(f(f(n)))=f(3n) since f(f(n))=3n
    Could you help with f(xy)=f(x)+f(y) Ive got to f(x^n)=nf(x) but how do I solve for x. I cant see a substitution...
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    (Original post by 11234)
    Could you help with f(xy)=f(x)+f(y) Ive got to f(x^n)=nf(x) but how do I solve for x. I cant see a substitution...
    its the other way around! its f(x+y)=f(x)f(y)
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    (Original post by gasfxekl)
    its the other way around! its f(x+y)=f(x)f(y)
    How would I solve the other one apparently the answer is f(x)=f(1)log(x)
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    Anyone else planning on spending the summer doing BMO problems?
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    (Original post by Maths465Man)
    Anyone else planning on spending the summer doing BMO problems?
    Yep definitely need to improve do you wanna discuss any problems
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    (Original post by 11234)
    Yep definitely need to improve do you wanna discuss any problems
    Definitely. I going to make my summer plan tomorrow which should work out at about three hours a day on BMO problems. I'm thinking of starting at some of the older papers as there are less BMO1 questions on each older paper (only 5) and I'm guessing the latest papers will be harder. For BMO2, I'll start as soon as I'm getting 50+ on each BMO1 paper. I've managed it on a few BMO1 papers, but I've been really inconsistent lately with my scores making me think I need to focus more time on BMO1 again.
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    (Original post by Maths465Man)
    Definitely. I going to make my summer plan tomorrow which should work out at about three hours a day on BMO problems. I'm thinking of starting at some of the older papers as there are less BMO1 questions on each older paper (only 5) and I'm guessing the latest papers will be harder. For BMO2, I'll start as soon as I'm getting 50+ on each BMO1 paper. I've managed it on a few BMO1 papers, but I've been really inconsistent lately with my scores making me think I need to focus more time on BMO1 again.
    Silly question but how can you get 50+ if there are only questions each carrying 10 marks
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    (Original post by 11234)
    Silly question but how can you get 50+ if there are only questions each carrying 10 marks
    Well there are 6 questions each worth 10 marks and I'm hoping to get at least 50 or more from the possible 60 marks.
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    (Original post by Maths465Man)
    Well there are 6 questions each worth 10 marks and I'm hoping to get at least 50 or more from the possible 60 marks.
    Oh I'm sure some papers had 5...
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    (Original post by 11234)
    Oh I'm sure some papers had 5...
    No those are rather old.


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    @Maths465Man Nice little thread you've started here, you seem to be pretty active in olympiad maths, out of curiosity did you get into any of the summer schools?
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    (Original post by physicsmaths)
    No those are rather old.


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    I know this is unrelated but what jobs are you looking at after a maths degree because I really wanna do a maths degree but Im not sure on employment opportunities
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    (Original post by Aldrstream)
    @Maths465Man Nice little thread you've started here, you seem to be pretty active in olympiad maths, out of curiosity did you get into any of the summer schools?
    No. I've not made it to any camps yet but next year I'm hoping to go to the Hungary camp and maybe even try for the Trinity camp (not too sure if that will happen though). I'm going into year 11 next year so I'm hoping to make the national summer school then.

    Have you been to any? What are your Olympiad experiences and what year are you going into?
 
 
 
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