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    (Original post by TheMagicMan)
    I believe a necessary and sufficient condition would be something along the lines of having a measure 0 set of discontinuities on the domain of y in the solution.
    Spoken like a true Cambridge mathmo.
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    (Original post by TheMagicMan)
    Difficult. Didn't do as much work as I probably should have in the first two terms but I somewhat pulled it together for the exams.
    Glad to hear you did well though.

    (Original post by Jkn)
    Well the question must have a solution by nature because either you can consistently predict the coin is counterfeit or you cannot
    Telling that to a modern logician is like saying to a quantum physicist "Newtonian physics applies everywhere, including the atomic level."

    (Original post by Jkn)
    ...
    You haven't actually answered my problem. Recall that your question asked "Whether a method exists", not "How will such a method work". The assumption that air resistance is not negligible is just as non-negligible as the assuming the opposite, since it predisposes the answer. Hence you cannot say that an answer which utilises the air resistance assumption, in either direction by itself, is valid. Which is why I said unless you make it clearer then the only solution would be to give you the document listing all the possible assumptions and solutions.

    EDIT: I.e. what I'm saying is answering "yes" to whether the method exists is not inherently more preferable to answering "no", and so there is no reason why you should prefer to say air resistance exists over it not existing as this is an answer-influencing assumption.
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    (Original post by und)
    Spoken like a true Cambridge mathmo.
    It's terrifying how much I've changed over the last year...and I've got so much better: sometimes I wonder how I ever found STEP hard (and I think most others in my year would say the same).
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    (Original post by und)
    Spoken like a true Cambridge mathmo.
    can you do my problem?
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    (Original post by ukdragon37)
    Glad to hear you did well though.



    Telling that to a modern logician is like saying to a quantum physicist "Newtonian physics applies everywhere, including the atomic level."



    You haven't actually answered my problem. Recall that your question asked "Whether a method exists", not "How will such a method work". The assumption that air resistance is not negligible is just as non-negligible as the assuming the opposite, since it predisposes the answer. Hence you cannot say that an answer which utilises the air resistance assumption, in either direction by itself, is valid. Which is why I said unless you make it clearer then the only solution would be to give you the document listing all the possible assumptions and solutions.
    Honestly. It's good that you arent in year three. I couldn't put up with this every time you two met

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    (Original post by TheMagicMan)
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    I haven't posted on here in ages, but my first offering will be the following, which is a slight adaptation of a question from the cambridge entrance exam in 1981 (so roughly STEP level)...

    How many fifth powers are there of form 2013n+11, n \in N?

    It requires AS knowledge at the most, and I know of no solution that uses more complex methods.

    Hint:

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    161,051=11^5
    It clearly has infinitely many solutions.

    Take 2013n+11=k^{5}, it is obvious that k \equiv 0 \pmod {11}, so 183n+1=11^{4}k_{1}^{5} and note that 11^{4} \equiv 1 \pmod {183}. Hence, it is sufficient to take k_{1} \equiv 1 \pmod {183}, then \displaystyle n = \frac{11^{4}k_{1}^{5} -1}{183}.
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    (Original post by bananarama2)
    Honestly. It's good that you arent in year three. I couldn't put up with this every time you two met

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    Just like what MagicMan said, I swear I wasn't this pedantic before my mind was ravished by the beast that is logic.
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    (Original post by Mladenov)
    It clearly has infinitely many solutions.

    Take 2013n+11=k^{5}, it is obvious that k \equiv 0 \pmod {11}, so 183n+1=11^{4}k_{1}^{5} and note that 11^{4} \equiv 1 \pmod {183}. Hence, it is sufficient to take k_{1} \equiv 1 \pmod {183}, then \displaystyle n = \frac{11^{4}k_{1}^{5} -1}{183}.
    I didn't realise that 2013 was divisible by 11, which kind of destroys it. I was aiming to incorporate 2013 in my adaptation.

    How many 6th powers are there of form 2013n+9, n \in N? is more what I was aiming for.
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    (Original post by TheMagicMan)
    I didn't realise that 2013 was divisible by 11, which kind of destroys it. I was aiming to incorporate 2013 in my adaptation.

    How many 6th powers are there of form 2013n+9, n \in N? is more what I was aiming for.
    Then again, by chinese reminder theorem, we can choose k_{1} such that k_{1} \equiv 3 \pmod {11} and k_{1} \equiv 3 \pmod {61}. Thus the equation still has infinitely many solutions.
    To clarify, 2013\times 3\times n_{1} = (3^{3}k_{1}^{3}-3)(3^{3}k_{1}^{3}+3)
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    (Original post by Mladenov)
    x
    As usual, you are way too fast fml! :lol:
    (Original post by ukdragon37)
    Telling that to a modern logician is like saying to a quantum physicist "Newtonian physics applies everywhere, including the atomic level."

    You haven't actually answered my problem. Recall that your question asked "Whether a method exists", not "How will such a method work". The assumption that air resistance is not negligible is just as non-negligible as the assuming the opposite, since it predisposes the answer. Hence you cannot say that an answer which utilises the air resistance assumption, in either direction by itself, is valid. Which is why I said unless you make it clearer then the only solution would be to give you the document listing all the possible assumptions and solutions.

    EDIT: I.e. what I'm saying is answering "yes" to whether the method exists is not inherently more preferable to answering "no", and so there is no reason why you should prefer to say air resistance exists over it not existing as this is an answer-influencing assumption.
    But air resistance exists in the real world and the problem is taking place in that same world and not the mathematical world until we chose to create a model. Is it's current state, the problem exists only in the real world.

    Proving that air resistance is non-negligible is the objective of the first part of the problem (to allow us to consider the 'drops' as analogous to 'scales'). We prove it is non-negligible by first assuming assuming all of the law of Physics and then approximating each law as is appropriate for the problem. We decide what is appropriate by the situation; if we wanted to we could use all of the compacted laws, it would just take longer and mean we must seek further information by going out and doing experiments or modelling the coins and the bag and certain objects whilst remarking the validity of this assumption. Is Physics, this is the way every problem must be done and you are certainly correct in saying that this method would require a document listing all assumptions along with experimental evidence as to their validity. When we do Physics, we do not write such documents because we call on experiments performs in similar situations to tell us things about what might happen in our current situations. We establish mathematical relationships between the variables recorded in an experiment and this is how we get our Physical laws. Note that all laws must be accompanied with an analysis of how well this correlates with experiment as well as when it is valid to apply these laws. These documents make up the body of Physics and are the very documents to which you refer.

    Whilst, in mathematics, we happily build our knowledge on top of the axioms. We also don't require pages of proofs for every theorem we use in every other proof we use them for. In Physics, whilst the body of knowledge is not perfect and is based on estimation, it is fundamental to the scientific method to assume the current body of knowledge we have is correct because it is all based on experimental evidence.
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    (Original post by Mladenov)
    Then again, by chinese reminder theorem, we can choose k such that k \equiv 3 \pmod {11} and k \equiv 3 \pmod {61}. Thus the equation still has infinitely many solutions.
    It is not entirely clear how this solves the problem. The CRT guarantees existence and uniqueness of a solution mod (mn) to a system
    x=a(n)
    x=b(m) but it is not generally ab.

    In particular I could replace the 9 above with any other number in {1,2....2012} and you could still run the above CRT argument, but there are not always infinitely many solutions.

    So obviously you mean something different to what I'm inferring?...

    EDIT: Your edit makes it much clearer. Can you find an 'elementary' solution?
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    (Original post by TheMagicMan)
    How many 6th powers are there of form 2013n+9, n \in N? is more what I was aiming for.
    So, you say Cambridge drilling (as William Hopkins' pupils described it) is quite good.

    Maybe 2014n + 9 as 6-th powers.
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    (Original post by TheMagicMan)
    It is not entirely clear how this solves the problem. The CRT guarantees existence and uniqueness of a solution mod (mn) to a system
    x=a(n)
    x=b(m) but it is not generally ab.

    In particular I could replace the 9 above with any other number in {1,2....2012} and you could still run the above CRT argument, but there are not always infinitely many solutions.

    So obviously you mean something different to what I'm inferring?...
    Actually, what I am doing is:
    Set n=3n_{1}, k=3k_{1}. Then 671n_{1}=(9k_{1}^{3}-1)(9k_{1}^{3}+1). It is sufficient to show that 9k_{1}^{3}-1 \equiv 0 \pmod {671} is solvable.

    C'mon 2014 \equiv 0 \pmod {19}; thus our problem has no solutions as \pm  3 are not cubic residues \pmod {19}.
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    (Original post by jack.hadamard)
    So, you say Cambridge drilling (as William Hopkins' pupils described it) is quite good.

    Maybe 2014n + 9 as 6-th powers.
    Infinitely many by my method (which doesn't hold up well for general an+b tbh)? I did some off the cuff calculation in my head so could be wrong.
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    (Original post by Mladenov)
    Actually, what I am doing is:
    Set n=3n_{1}, k=3k_{1}. Then 671n_{1}=(9k_{1}^{3}-1)(9k_{1}^{3}+1). It is sufficient to show that 9k_{1}^{3}-1 \equiv 0 \pmod {671} is solvable.
    Yeah your edit of your first post cleared it up...can you do it using C1 maths and before (so very basic stuff only)?
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    (Original post by Jkn)
    But air resistance exists in the real world and the problem is taking place in that same world and not the mathematical world until we chose to create a model. Is it's current state, the problem exists only in the real world.

    Proving that air resistance is non-negligible is the objective of the first part of the problem (to allow us to consider the 'drops' as analogous to 'scales'). We prove it is non-negligible by first assuming assuming all of the law of Physics and then approximating each law as is appropriate for the problem.
    So you are essentially justifying existence of air resistance because the frame of the question is in the real world. But why stop there? Just as und says, why not model changing air density too? Or the many other variables which we could refine? For all we know if we take changing air density into account it may affect the ability for drops to be scales. It still sounds very much like you have a solution with an arbitrary (and unstated) set of assumptions you want people to take up so they could give you the exact solution you had in mind.

    (Original post by Jkn)
    We decide what is appropriate by the situation; if we wanted to we could use all of the compacted laws, it would just take longer and mean we must seek further information by going out and doing experiments or modelling the coins and the bag and certain objects whilst remarking the validity of this assumption. Is Physics, this is the way every problem must be done and you are certainly correct in saying that this method would require a document listing all assumptions along with experimental evidence as to their validity. When we do Physics, we do not write such documents because we call on experiments performs in similar situations to tell us things about what might happen in our current situations. We establish mathematical relationships between the variables recorded in an experiment and this is how we get our Physical laws. Note that all laws must be accompanied with an analysis of how well this correlates with experiment as well as when it is valid to apply these laws. These documents make up the body of Physics and are the very documents to which you refer.
    Again sure we don't have to take everything into account, but you seem to have set an arbitrary boundary on what you find essential to consider (air resistance) but what is not (changing air density). As I have been saying, how are we supposed to adequately answer the problem if we don't know to what detail you want to hear?

    Also I'm very much aware of at least how physical laws relate to experiments thank you very much, having taken the same Physics as NatScis in first year.
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    (Original post by Mladenov)

    C'mon 2014 \equiv 0 \pmod {19}; thus our problem has no solutions as \pm  3 are not cubic residues \pmod {19}.
    What is this world where cubic residues mod 19 are quotable?
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    (Original post by Mladenov)
    C'mon 2014 \equiv 0 \pmod {19}; thus our problem has no solutions as \pm  3 are not cubic residues \pmod {19}.
    Well, that was the intended outcome.
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    (Original post by ukdragon37)
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    I've decided to give you a taste of what I do

    Problem 224*/**:

    Whenever we say "function" we mean total function.

    Denote for every n \in \mathbb N _0 the set \underline n = \{ m \in \mathbb N _0 | 0 < m \leq n\}.

    The mystery concept of a strange product \underline m \otimes \underline n for two such sets \underline m and \underline n satisfies the following property:



    There exist functions \pi_1 : \underline m \otimes \underline n \to \underline m and \pi_2 : \underline m \otimes \underline n \to \underline n such that:

    For each k \in \mathbb N_0 and each pair of functions f: \underline k \to \underline m and g: \underline k \to \underline n there exists a unique function h : \underline k \to \underline m \otimes \underline n such that for all x\in \underline k we have f(x) = \pi_1(h(x)) and g(x) = \pi_2(h(x))



    Show that regardless of what the exact definition for \otimes is, if it satisfies the above property then for all sets X and (non-zero) natural n, |X \otimes \underline n| = n iff |X| = 1 (in other words X = \underline 1).

    EDIT1: made it clearer, but question is still the same.

    EDIT2: Hmmm I just realised that the elegant solution is far beyond *-level, and the inelegant one may be quite long.

    EDIT3: Actually for the above reason I'm going to withdraw it as a formal problem for this thread, but it's open to anyone who wants to try it.
    Firstly, note that the cartesian product satisfies the conditions; it makes the corresponding diagram commute.
    It turns out that, if A= \underline m \otimes \underline n and B = \underline m \times \underline n are two different products, then A \cong B. Let the projective maps be \pi_{1} : A \to \underline m; \pi_{2}: A \to \underline n; q_{1}: B \to \underline m; q_{2}: B \to \underline n. Further, there are two uniquely determined mappings f : A \to B and g: B \to A.
    Therefore,
    \pi_{1} \circ f = q_{1}
    \pi_{2} \circ f = q_{2}
    q_{1} \circ g = \pi_{1}
    q_{2} \circ g = \pi_{2}
    It follows, from the uniqueness of f, that f \circ g = id_{A}; similarly, g \circ f = id_{B}.
    Hence the result.

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    Mac Lane is sublime!
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    (Original post by ukdragon37)
    EDIT2: Hmmm I just realised that the elegant solution is far beyond *-level, and the inelegant one may be quite long.
    Category theory, essentially, labelled A-level knowledge? I wish Mac Lane and Eilenberg could see this.

    So, are you already at Cornell or are you about to go there?
 
 
 
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