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    (Original post by Zacken)
    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}
    oh....... how does f(2)+f(5)=14? assuming x=2 and y=5 5x2 doesn't= 14
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    (Original post by thefatone)
    oh....... how does f(2)+f(5)=14? assuming x=2 and y=5 5x2 doesn't= 14
    f(10) = f(2 \times 5) = f(2) + f(5) = 14
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    (Original post by Zacken)
    f(10) = f(2 \times 5) = f(2) + f(5) = 14
    there's something i'm seriously not understanding here where does the 14 come from?

    does the function have a different value or something?
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    (Original post by thefatone)
    there's something i'm seriously not understanding here where does the 14 come from?

    does the function have a different value or something?
    The question says "given f(10) = 14 and f(40) = 20".
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    (Original post by Zacken)
    The question says "given f(10) = 14 and f(40) = 20".
    oh ok but surely you could have f(10)+f(1) then aswell?
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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)?
    Nice question. I'm tempted to think up others along these lines. Also very suitable for a hard GCSE thread.
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    (Original post by DylanJ42)
    Never done a spoiler on the mobile app so hopefully this works;
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    39?
    Posted from TSR Mobile
    (Original post by Zacken)
    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}
    Correct. It was from the 2015 Senior Mathematical Challenge.
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    (Original post by HapaxOromenon)
    Correct. It was from the 2015 Senior Mathematical Challenge.
    ah cool do you do competition maths annually?
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    (Original post by DylanJ42)
    ah cool do you do competition maths annually?
    Yes.
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    (Original post by joostan)
    An irrational number is a number which cannot be expressed as the ratio, \frac{a}{b} of two integers.
    A complex number is a generalisation of the real numbers which gives every polynomial of order n exactly n solutions in the complex plane. Essentially one defines an imaginary number i=\sqrt{-1} and proceed from there, though that is a discussion best suited to another thread.



    Agreed, though I tend to be a bit nit-picky, guess the pure courses have rubbed off on me :/
    Oh right! I think I meant to say it was a complex number at the beginning, but for some reason, I thought you called it an irrational number...

    Thanks for clearing things up!
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    Forget Hannah's sweets, here is the worst answered question in the last 10 years for all the KS4 maths exams:



    This is a treat for all the GCSE students revising hard
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    (Original post by notnek)
    Forget Hannah's sweets, here is the worst answered question in the last 10 years for all the KS4 maths exams:



    This is a treat for all the GCSE students revising hard
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    no sorry I worked it out wrong, one minute
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    (Original post by surina16)
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    I got 53.55 minutes, so rounded it to 54?
    That's not correct.
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    (Original post by notnek)
    That's not correct.
    Spoiler:
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    54.31 minutes?
    Spoiler:
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    :lol:
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    (Original post by surina16)
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    54.31 minutes?
    Spoiler:
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    :lol:
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    That's not correct unfortunately but keep trying


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    The mark scheme allows an answer between 55 and 56 with correct working. You can post your working if you like.
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    (Original post by notnek)
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    That's not correct unfortunately but keep trying
    Spoiler:
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    The mark scheme allows an answer between 55 and 56 with correct working. You can post your working if you like.
    oh dear

    working:
    Spoiler:
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    worked out how many small squares there were all together:
    (5*18) + (10*35) + (7*20) + (15*12) + (7*5) = 795 squares

    25% of 795 = 198.75
    So I thought that the 25% mark would be at the 198.75th square?

    198.75 - 90 = 108.75 because it wouldn't be in the first group. I then saw that it had to be in the second group and so I did:

    108.75/350 = 0.310... to work out how far the value would be into the group (the way we were taught how to find the median)
    and then added this to the lower class boundary of 54 to get 54.310...

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    (Original post by surina16)
    oh dear

    working:
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    worked out how many small squares there were all together:
    (5*18) + (10*35) + (7*20) + (15*12) + (7*5) = 795 squares

    25% of 795 = 198.75
    So I thought that the 25% mark would be at the 198.75th square?

    198.75 - 90 = 108.75 because it wouldn't be in the first group. I then saw that it had to be in the second group and so I did:

    108.75/350 = 0.310... to work out how far the value would be into the group (the way we were taught how to find the median)
    and then added this to the lower class boundary of 54 to get 54.310...

    This is kind of what I did (except I assigned numbers to the y axis 'cos I thought that would make it easier to visualise :dontknow:) and I ended up with 56.09.. Dunno how, though.
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    (Original post by surina16)
    oh dear

    working:
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    worked out how many small squares there were all together:
    (5*18) + (10*35) + (7*20) + (15*12) + (7*5) = 795 squares

    25% of 795 = 198.75
    So I thought that the 25% mark would be at the 198.75th square?

    198.75 - 90 = 108.75 because it wouldn't be in the first group. I then saw that it had to be in the second group and so I did:

    108.75/350 = 0.310... to work out how far the value would be into the group (the way we were taught how to find the median)
    and then added this to the lower class boundary of 54 to get 54.310...

    Spoiler:
    Show

    You're very close.

    108.75/350 = 0.310

    This is how far along the second bar that T lies.

    But this is just a fraction of the second bar and not the time.

    So adding this to 54 won't give the answer.
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    (Original post by notnek)
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    You're very close.

    108.75/350 = 0.310

    This is how far along the second bar that T lies.

    But this is just a fraction of the second bar and not the time.

    So adding this to 54 won't give the answer.
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    would you multiply 0.3107 by the class width and then add it to 54 or...?
    (because it gives you 55.243 )
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    (Original post by notnek)
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    You're very close.

    108.75/350 = 0.310

    This is how far along the second bar that T lies.

    But this is just a fraction of the second bar and not the time.

    So adding this to 54 won't give the answer.
    The answer to that question seems strangely satisfying now.
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    Unless I made another stupid mistake..
 
 
 
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