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    (Original post by thefatone)
    i'm gonna slap me in the face, i filled the whole page with stuff and didn't get an answer -.-
    Arriving at a correct value for N takes seconds, and I'm sure you managed that, it is the process of showing how to get the answer beyond trial and error that is difficult. I'd need to see the mark scheme but I'd wager many people failed to answer it.

    I don't know though. I'm about as fart away from an expert in maths as you're likely to find on this forum :lol:
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    (Original post by ByronicHero)
    Arriving at a correct value for N takes seconds, and I'm sure you managed that, it is the process of showing how to get the answer beyond trial and error that is difficult. I'd need to see the mark scheme but I'd wager many people failed to answer it.

    I don't know though. I'm about as fart away from an expert in maths as you're likely to find on this forum :lol:

    Indeed. Many people found it difficult. There was a big cry out and hitler videos and hastags and everything.

    (Original post by thefatone)
    i'm afraid my page of work about a year ago during my GCSE has probably been binned
    Oh I see. Incorrectly though you did it recently as a mock.
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    (Original post by ByronicHero)
    Arriving at a correct value for N takes seconds, and I'm sure you managed that, it is the process of showing how to get the answer beyond trial and error that is difficult. I'd need to see the mark scheme but I'd wager many people failed to answer it.

    I don't know though. I'm about as fart away from an expert in maths as you're likely to find on this forum :lol:
    It's not hard, in fact, extremely trivial. If you can't do it, it's because you have just been drilled to answer questions like a monkey and don't actually understand any mathematics.
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    (Original post by ByronicHero)
    Arriving at a correct value for N takes seconds, and I'm sure you managed that, it is the process of showing how to get the answer beyond trial and error that is difficult. I'd need to see the mark scheme but I'd wager many people failed to answer it.

    I don't know though. I'm about as fart away from an expert in maths as you're likely to find on this forum :lol:
    that's pretty far you've diffused everywhere
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    (Original post by HapaxOromenon)
    It's not hard, in fact, extremely trivial. If you can't do it, it's because you have just been drilled to answer questions like a monkey and don't actually understand any mathematics.
    He did it mentally quite quickly.
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    (Original post by HapaxOromenon)
    It's not hard, in fact, extremely trivial. If you can't do it, it's because you have just been drilled to answer questions like a monkey and don't actually understand any mathematics.
    It is very easy if you know how to do it. I think the fact it deviates considerably from how most similar questions work means it can reasonably be called a hard question, in the context of GCSE exams. :dontknow:

    (Original post by Kvothe the arcane)
    Indeed. Many people found it difficult. There was a big cry out and hitler videos and hastags and everything.



    Oh I see. Incorrectly though you did it recently as a mock.
    Oh, it was one of those. It is always amusing when there is outcry about a difficult question - even if I couldn't always do them :lol:

    (Original post by thefatone)
    that's pretty far you've diffused everywhere
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    (Original post by HapaxOromenon)
    It's trivial, and if you can't do the derivation you're stupid. I guess you really are the opposite of an expert in maths...

    hey it's only 1 question

    (Original post by Kvothe the arcane)



    He did it mentally quite quickly.
    D isn;t it?
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    (Original post by HapaxOromenon)
    A function, defined on the set of positive integers, is such that f(xy)=f(x)+f(y) for all x and y. It is known that f(10)=14 and f(40)=20. What is the value of f(500)?
    sounds like maybe you need to make some simultaneous equations?
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    (Original post by thefatone)
    D isn;t it?
    Yes
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    (Original post by Kvothe the arcane)



    He did it mentally quite quickly.
    I would guess D on the basis of each exterior angle being 360/(number of sides) and that being two angles added together?
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    (Original post by HapaxOromenon)
    A function, defined on the set of positive integers, is such that f(xy)=f(x)+f(y) for all x and y. It is known that f(10)=14 and f(40)=20. What is the value of f(500)?
    so far i've tried xy=14 then x+y=10 and tried subsituting but i don't think that's right

    lets call Zacken Student403
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    Shanaynay and Dakwonda play a game in which they take turns filling entries of an initially empty 2008 × 2008 array. Shanaynay plays first. At each turn, a player chooses a real number and places it in a vacant entry. The game ends when all the entries are filled. Shanaynay wins if the determinant of the resulting matrix is non-zero; Dakwonda wins if it is zero. Which player has a winning strategy?
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    (Original post by HapaxOromenon)
    A function, defined on the set of positive integers, is such that f(xy)=f(x)+f(y) for all x and y. It is known that f(10)=14 and f(40)=20. What is the value of f(500)?
    Never done a spoiler on the mobile app so hopefully this works;

    Spoiler:
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    39?


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    (Original post by Kvothe the arcane)
    f(x)=c \log(x)
    It's usually written as f(x) = \log_a x, but that's not really the wanted solution here.
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    (Original post by Kvothe the arcane)
    f(x)=c \log(x)
    (Original post by Zacken)
    It's usually written as f(x) = \log_a x, but that's not really the wanted solution here.
    not quite sure if logs was the way to go :/
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    (Original post by thefatone)
    not quite sure if logs was the way to go :/
    It's not. f(500) = f(50 \times 10) = f(50) + f(10) = \cdots or something along those lines.
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    (Original post by Zacken)
    It's not. f(500) = f(50 \times 10) = f(50) + f(10) = \cdots or something along those lines.
    but don't the x and y values both have to satisfy both given equations?
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    (Original post by Zacken)
    It's not. f(500) = f(50 \times 10) = f(50) + f(10) = \cdots or something along those lines.
    I tried various combinations like that but couldn't reconcile the f(10) = 14 with any. This said, I have never covered this so was completely guessing as to how to solve it :lol:

    Try my Shanaynay question :ninja:
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    How do these compare to the BMO?

    http://www.edb.gov.hk/en/curriculum-...mo-papers.html
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    (Original post by thefatone)
    but don't the x and y values both have to satisfy both given equations?
    f(500) = f(2^2 \times 5^3) =2f(2) + 3f(5) but we know that f(2) + f(5) = 14 and that f(40) = f(2^3 \times 5) = 3f(2) + f(5)

    Solving simultaneously -2f(2) = -6 \Rightarrow f(2) = 3 \Rightarrow f(5) = 11 so \boxed{f(500) = 2\times 3 + 3\times 11 = 39}
 
 
 
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