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visits member for 3 years, 7 months
seen Jul 25 at 16:16

Jul
16
comment Interview riddle
Might be worth seeing if this would be a good fit for puzzling.stackexchange.com (I don't know if it is since although I've popped over there occasionally I've never wanted to post so haven't read their guidelines). Should have people who might be up for the challenge from a less mathematical point of view (which might be appropriate here).
Jul
15
asked Incorrect proof of the infinities between 0 and 1 and 0 and 2
Apr
28
comment Give the remainder of $x^{100}$ divided by $(x-2)(x-1)$.
Out of interest what is the proof that the remainder will be of the form ax+b? It seems like there should be a simple answer but I can't quite see it (happy to ask as a separate question if its not a comment length answer). Oh - or is it just that it must be a smaller order polynomial than what you are dividing by?
Apr
25
comment How to solve for a variable that is only in exponents?
@MichaelHardy: sorry, I meant abiessu whose answer it was. I'm not sure how I got confused about it. :)
Apr
25
comment How to solve for a variable that is only in exponents?
@MichaelHardy: Cool. It might be worth noting that in your question. For two reasons. Firstly I was initially suprised and confused by where the x=2 came from. Secondly my reading of the question is that the title is about solving for a variable in exponents and feels like a question about the general case. All the work you have done is obviously relevant for proving there is only one solution but it might be worth noting that there you'd have to use numerical techniques to get the actual answer. Just my thoughts. :)
Apr
24
comment How to solve for a variable that is only in exponents?
How do you get the solution x=2? You might be able to guess it in this case but you might not have so much luck for different solutions so how would you then solve it?
Apr
4
comment How can a piece of A4 paper be folded in exactly three equal parts?
@StevenTaschuk: Ah, that makes sense and is pretty obvious now you say it and makes a lot of sense.
Apr
4
comment How can a piece of A4 paper be folded in exactly three equal parts?
@StevenTaschuk: What does it mean then?
Apr
3
comment Fibonacci trick and proving it.
+1 for the geometric interpretation (and of course the correct proof).
Mar
28
comment Coin problem with 6 and 10
@MarcvanLeeuwen: Perhaps I should have said the correct response is that there is no valid answer.
Mar
28
comment Coin problem with 6 and 10
I think technically the answer is that there is no answer (ie no largest value). Giving you an infinite amount of 10s would give you infinity so you should likely avoid answers like that (and indeed any answer that pretends that infinity is an answer).
Mar
26
comment Probability that random moves in the game 2048 will win
I think your number of moves is wrong. ie its 1024 moves to get to 2048. Each move spawns a two or a four, if you assume its a two every time then the total on the board increases by two each move which means that 1024 moves gives a total of 2048 on the board.
Dec
20
comment Show that a continuous function with a certain integral property must be f(x)=x.
Ah yes, of course. I was thinking that it relied on the fact f(x)=x but it just relies on the fact that if f(x)=x then the integral is the same which is different.
Dec
20
comment Show that a continuous function with a certain integral property must be f(x)=x.
Where does the first line come from?
Nov
10
comment Is my proof correct for: $\sqrt[7]{7!} < \sqrt[8]{8!}$
Is his proof correct? Isn't there a mistake when he takes the 8! to the power of thing outside of the thing? Certainly the fact that you have 7!<8^7 and he gets 7!<8! makes me think there is a problem somewhere... Though of course this does demonstrate why getting rid of the roots is such a good idea because they confuse things. :)
Nov
1
comment How to weigh up to 100kg with 5 weights
@BrianM.Scott: Cool. I should have known it was a well studied thing. A name means I can read more about it. :)
Oct
31
comment How to weigh up to 100kg with 5 weights
@Cruncher: yeah, its more the mathematical analogy that is easy to imagine. I imagine I would get completely muddled if I tried to actually do maths like that... :)
Oct
31
comment How to weigh up to 100kg with 5 weights
@self.: No. This can be thought of as an expression of base 3 in a strange way. When counting in base 3 you have for each power of 3 a value of either zero, one or two. With this solution we are doing the same thing effectively but we give them values of one (its on the oposite side to the item being weighed), zero (not on the scales) or minus one (its on the same side as the weight being measured. If you wanted higher powers you'd need multiples of some weights. eg if you always had two of each size then powers of five would work. Powers of four would work as well but be inefficient.
Oct
7
comment Fastest way to check if $x^y > y^x$?
so can you do similar optimisations for if x and y are both less than e?
Oct
6
comment Proof that $n^3 + 3n^2 + 2n$ is a multiple of $3$.
Its not a bad way. I just got to the end and thought "oooh, that looks like a nice result. Oooh, look it factorises nicely to a really nice result! oh wait, that's what we started with..."