The Proof is Trivial!

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.

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.

Well the question must have a solution by nature because either you can consistently predict the coin is counterfeit or you cannot

...

Spoken like a true Cambridge mathmo.

Spoken like a true Cambridge mathmo.

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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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.

x

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.

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.

How many 6th powers are there of form $2013n+9, n \in N$? is more what I was aiming for.

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?...

So, you say Cambridge drilling (as William Hopkins' pupils described it) is quite good.

Maybe $2014n + 9$ as 6-th powers.

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.

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.

C'mon $2014 \equiv 0 \pmod {19}$; thus our problem has no solutions as $\pm 3$ are not cubic residues $\pmod {19}$.

C'mon $2014 \equiv 0 \pmod {19}$; thus our problem has no solutions as $\pm 3$ are not cubic residues $\pmod {19}$.

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