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    (Original post by atsruser)
    All of the questions that I've posted can be done with GCSE knowledge + insight. It's the insight that makes them hard, of course.
    When I said the thread had gone off topic, I was referring to the posts above me e.g. your post #72.
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    A computer password for a top secret system comprises 50 randomly chosen digits, each in the range 0-9. There are 10^{50} distinct such passwords.

    A computer cracker builds a powerful computer to break into the system. It tries each password in turn until the correct one is found. It can try a million million million passwords per second, and it is left running for a million million million years.

    To the nearest one millionth of one percent, what percentage of passwords are still left to be tried, after this period of time?
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    Circle A is inscribed inside a square of side 2 so that it is tangent to all four sides. A smaller circle, B, is inscribed in the gap between circle A and the two sides of the square at one of the corners. Circle B is tangent to circle A and to the two sides of the square.

    Find the radius of circle B.
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    Solve n^3-4n-315=0
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    15 - 6 = 11

    I'm sorry. It just is. No, it just is.
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    (Original post by atsruser)
    Circle A is inscribed inside a square of side 2 so that it is tangent to all four sides. A smaller circle, B, is inscribed in the gap between circle A and the two sides of the square at one of the corners. Circle B is tangent to circle A and to the two sides of the square.

    Find the radius of circle B.
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    0.21?
    (Original post by atsruser)
    Solve n^3-4n-315=0
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    Is n just 7? Since sqrt(7^2 - 4*45) gives you an irrational number?

    (Original post by atsruser)
    A computer password for a top secret system comprises 50 randomly chosen digits, each in the range 0-9. There are 10^{50} distinct such passwords.


    A computer cracker builds a powerful computer to break into the system. It tries each password in turn until the correct one is found. It can try a million million million passwords per second, and it is left running for a million million million years.

    To the nearest one millionth of one percent, what percentage of passwords are still left to be tried, after this period of time?
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    I got 0.000032%
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    (Original post by TheOtherSide.)
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    Is n just 7? Since sqrt(7^2 - 4*45) gives you an irrational number?
    I would presume so, a small note however that the other roots are actually complex, not irrational.
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    (Original post by joostan)
    I would presume so, a small note however that the other roots are actually complex, not irrational.
    My bad! I'll keep that in mind when doing something like this again...
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    (Original post by TheOtherSide.)
    My bad! I'll keep that in mind when doing something like this again...
    It's not really an issue for GCSE level, I just thought I'd mention it for interest's sake, by definition an irrational number is by definition real.
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    (Original post by joostan)
    It's not really an issue for GCSE level, I just thought I'd mention it for interest's sake, by definition an irrational number is by definition real.
    If I'm being honest, I still don't get the difference between a complex number and an irrational number - do complex numbers only show up when working out quadratics?
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    (Original post by joostan)
    It's not really an issue for GCSE level, I just thought I'd mention it for interest's sake, by definition an irrational number is by definition real.
    To be fair, it's usually by convention that n \in \mathbb{N} or n \in \mathbb{Z} so being irrational would be a sufficient condition to discard those solutions, I think.
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    (Original post by TheOtherSide.)
    If I'm being honest, I still don't get the difference between a complex number and an irrational number - do complex numbers only show up when working out quadratics?
    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.

    (Original post by Zacken)
    To be fair, it's usually by convention that n \in \mathbb{N} or n \in \mathbb{Z} so being irrational would be a sufficient condition to discard those solutions, I think.
    Agreed, though I tend to be a bit nit-picky, guess the pure courses have rubbed off on me :/
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    (Original post by TheOtherSide.)
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    0.21?
    That's not what I got. I'll check my result - it came to a nice exact answer, with surds.

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    Is n just 7? Since sqrt(7^2 - 4*45) gives you an irrational number?
    Yes but...
    Hint:
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    rewrite as n^3-4n=315 then factorise both sides fully


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    I got 0.000032%
    I think that you misread the question. And that doesn't look quite right, even if you are trying to compute the % finished.

    It would be easier to comment if you put up some working.
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    (Original post by joostan)
    Essentially one defines an imaginary number i=\sqrt{-1} and proceed from there, though that is a discussion best suited to another thread.
    You can definitely ignore complex solutions here - it's a GCSE thread.
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    (Original post by atsruser)
    That's not what I got. I'll check my result - it came to a nice exact answer, with surds.
    I did get an exact answer with surds, except I wasn’t sure howto put it down: (-1 + sqrt2)/2

    Yes but...
    Hint:
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    rewrite as n^3-4n=315 then factorise both sides fully
    That’s what I did at the beginning, except I didn’t know whereto go from n(n+2)(n-2) = 315, so I ended up using the factor theorem to get(n-7) and then used long division to get a quadratic.
    I think that you misread the question. And that doesn't look quite right, even if you are trying to compute the % finished.

    It would be easier to comment if you put up some working.
    I still haven’t familiarised myself with latex, so apologiesabout not being able to display everything in a way that’s easy to understand. Amillion million million passwords is the same as 10^17 passwords per second. Ithen multiplied this by 31536000, which is the number of seconds in a year, andthen multiplied by another 10^17 for the number of years. I divided all this by10^50 and multiplied by 100 to get a percentage. I’m not sure what else I could have done, unless I didmisread the question, or maybe made a stupid mistake.
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    (Original post by TheOtherSide.)
    I did get an exact answer with surds, except I wasn’t sure howto put it down: (-1 + sqrt2)/2
    I get r = \frac{\sqrt{2}-1}{\sqrt{2}+1}

    That’s what I did at the beginning, except I didn’t know whereto go from n(n+2)(n-2) = 315, so I ended up using the factor theorem to get(n-7) and then used long division to get a quadratic.
    You didn't factorise 315 - do that, then use the fact that there is only one way to factorise both sides.

    You can't use the factor theorem here - this is a GCSE thread, so you must restrict yourself to GCSE methods.

    I still haven’t familiarised myself with latex, so apologiesabout not being able to display everything in a way that’s easy to understand. Amillion million million passwords is the same as 10^17 passwords per second. Ithen multiplied this by 31536000, which is the number of seconds in a year, andthen multiplied by another 10^17 for the number of years. I divided all this by10^50 and multiplied by 100 to get a percentage. I’m not sure what else I could have done, unless I didmisread the question, or maybe made a stupid mistake.
    Actually I have noticed a typo in my question so it didn't quite work out as nice as I wanted it. But yes, you have badly misread the question. What does it ask you to find?

    [edit: and you might want to check your powers - what is a million million million in powers of 10?]
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    (Original post by OGGUS)
    This
    Is this correct? I skipped some steps as it isn't an exam.
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