AMC AIME USA(J)MO USAMO Harvard college MIT \/closed

So I really want to take WOOT... but I doubt my parents are going to be willing to pay the $ \$800$ (after the $ \$200$ I can pay, $ \$600$) it requires, and, unless I get REALLY REALLY lucky there's *no* way I'm making MOP this year. I typically solve very few olympiad-level problems, but I consistently get 10+ on the AIME in practice. I might be able to get credit for it at school, which would certainly help. Anything anyone thinks I should say and/or said to their parents to convince them?

Well, I don't have too much experience on this, since my parents are pretty willing to pay for AoPS classes, but I'm betting my brother might be more experienced, because of his love for music that my parents do not completely agree with.

Say that spending $ \$$800 will not have a permanent or significant affect on either one of your lives, but taking the WOOT could help you do well on Olympiad problems, thus getting you in to top tier colleges (if you have a college obsessed parent, this works really well), eventually leading to a higher education and better job. Not spending the 800 might permanently affect both of you lives (negatively).

That's what I would do. But I'm not that experienced. Don't take my word for it.

[quote="Nerd_of_the_Ages"]
Say that spending $ \$$800 will not have a permanent or significant affect on either one of your lives, but taking the WOOT could help you do well on Olympiad problems, thus getting you in to top tier colleges (if you have a college obsessed parent, this works really well), eventually leading to a higher education and better job. Not spending the 800 might permanently affect both of you lives (negatively).[/quote]
I've spent the last 2 or 3 years telling my parents how much I actively don't care about college, so that would be a bit hypocritical. But that's a topic for another thread.

I would basically say the thing the second poster put. My parents have their own way of doing things, which is pretty hard to influence. I could tell my mom was a little uneasy with spending 30 dollars on aops vol2 so I think I might have some trouble convincing her to pay 800 for woot, if I don't make MOP, so good luck with your parents. If you can't do woot, just do some older USAMO problems, or something.

I think its 700 if you apply before a certain date...

Just be realllllllly good for a few days, absolutely no misbehaving and when your parents are commenting and encouraging ask them if you can take the course. Thats how i got my wii.

[quote="serialk11r"]I would basically say the thing the second poster put. My parents have their own way of doing things, which is pretty hard to influence. I could tell my mom was a little uneasy with spending 30 dollars on aops vol2 s[/quote]

Ditto. My parents were uneasy about spending 70 bucks for aops 1 and 2 w/ solutions lol. I think it really depends on what your parents really think, and how far they are willing to go, as I know for a fact my parents would NEVER spend 800 bucks for math. =()

I don't think you should use a "WOOT = Harvard" type argument. Parents are not stupid. They know that something like WOOT has at best a very indirect effect on college admissions. I think the best way is to convince them that WOOT will really, um, nurture your passion for mathematics  (very corny, I know :P) And if they still don't budge, well, you don't HAVE to take it. There are other ways of preparing for olympiads.

If they don't really care about what you are interested in (and therefore wouldn't spend money on WOOT), you can act really sad for a long period of time after they say "no" and pretend to not care about math. Don't make it obvious, like you suddenly hate math with a burning passion now, but make it subtle, and just don't really care about it as much before.

[quote="7h3.D3m0n.117"][quote="worthawholebean"]I've spent the last 2 or 3 years telling my parents how much I actively don't care about college,[/quote]

That won't get their attention.

You can say that if you did really well on AIME to get to USAMO, colleges would take that into consideration (resume). However because you currently do not have any education in Olympiad problems, it is unlikely that you will do well (just say this - I'm not saying it's true).

My parents were forced to spend some odd $ \$ 600$ for a Chemistry CTY course so I could take AP chem next year as a sophmore. They were actually quite pleased at the price since the alternative on-site price was around $ \$ 2000$.[/quote]

Hm, I didn't know that they spent $ \$ 600$ on your Chemistry CTY. And I got a free "scholarship" to a $ \$ 375$ writing camp last summer from my school. They were really happy. But if it's something like an art class that costs..say $ \$ 162$, and you probably 99% chance won't get a career involving art, they won't provide it for you. I know math is involved with everything, but if you've already taken the WOOT class, or any other advanced class involving math, your parents would probably think it's the best to take a break from it and broaden your interests.

[quote="junggi"][quote="Phelpedo"]I don't think you should use a "WOOT = Harvard" type argument. Parents are not stupid. They know that something like WOOT has at best a very indirect effect on college admissions. [/quote]
it worked for me
actually, my mom was the first one to come to me and suggest that i take it, with her "good results on contests=harvard" mentality[/quote]
Taylor that's your mom remember...and according to you your mom is some freakish version of asian parent that is obsessed with the getting into Harvard thing.

Yeah it always bugs me about how asian parents always have to associate a math competition or soemthing like that with resume's and colleges... Can't we just do them because, you know, they're fun?  :huh:

[quote="7h3.D3m0n.117"]Actually, math competitions are actually benefit your resume when applying to colleges. Applying to MIT with a first place in national Mathcounts, or a participant in the IMO, would surely tip the registration in your favor. Of course, most mathletes do math for fun, and competition second  :)[/quote]

Yes of course, but that's not the POINT. I don't do them BECAUSE of colleges, I do them because I like math... There shouldn't always be a college reason for EVERY competition. It gets old  :(

[quote="crazychinesemaniac"]Everyone keeps saying Harvard...Sure there the best school in the world but that doesn't mean jump at the opportuninty to go to Harvard like a mad bean...There are many things to look at when going to college such as the city, crime rate, graduation rate, teacher to student ratio, housing, college lifestyle and hardness to get in. Harvard ranks in the top of most of those tho....well what I'm trying to say is that Harvard is way over rated.[/quote]
Eh speaking of that, even though I'm only in 9th grade, I think that its not about getting into the most prestigious college that matters most, but rather getting into the college you want to. Some colleges are better for some things than others, or as I heard...eh I don't know anything. But still, being pro at math helps you get into the college you want, because of obvious reasons (its just not Harvard though).

[quote="Anaxerzia"]Yes of course, but that's not the POINT. I don't do them BECAUSE of colleges, I do them because I like math... There shouldn't always be a college reason for EVERY competition. It gets old  :([/quote]
Yes. My philosophy: I'll get into a college I want to go to. That's it. End of story. And I'll like it there; if I don't, I'll transfer. I don't need to make every decision in my life be in terms of some specter of college hanging over my head because that's simply a waste of time. If I appropriately challenge myself in school and elsewhere, which I do, I will make the same decisions as if I were thinking about college. I'm not going to drown myself in work, but I won't slack off, either. I'm not going to do 5,000,000 extracurricular activities; I do what I do because I want to do it. Especially schools like mine, people are *encouraged* to go crazy, and when they freak out completely come senior fall, no one is surprised. Why is this necessary? I don't think it is. So I'm not going to do it.

I am somewhat lucky in that my parents are mostly pretty cool about the subject.

[quote="davidyko"][quote="worthawholebean"]
I am somewhat lucky in that my parents are mostly pretty cool about the subject.[/quote]

I, too, am fortunate enough to have Asian parents that understand this point.[/quote]
i wish to kill you guys

So do I.
My parents care too much about college.

After MathCounts Nationals last year, I got this "Scholarship" to WOOT, you know, those "100 dollars off before such and such date", and it was pretty funny, Dad wanted me to take it and I wasn't sure...

Now that I am taking it, I'm learning!

[quote="7h3.D3m0n.117"]Actually, math competitions are actually benefit your resume when applying to colleges. Applying to MIT with a first place in national Mathcounts, or a participant in the IMO, would surely tip the registration in your favor. Of course, most mathletes do math for fun, and competition second  :)[/quote]A lot of people in eastern Europe see the Olympiad (the IMO especially) as a ticket to getting admitted into an Ivy League school. It does benefit a lot your resume, though it doesn't ensure your admission (you have to have other stuff on the side).


logarithms
calculus
calculus computations

Find the limit:	\[lim_{n\rightarrow\infty}\sum_{n=1}^{n}(-1)^{v+1}\left(\frac{1+\frac{1}{2}+..+\frac{1}{n}}{n}\right) \]

I haven't found a solution..

have a error

what is n!?

n = 1  in the sum to  n

but n tend to infinity

it its impossible

please use other letters and correct this

I assume that you are looking for the following sum,

$\sum_{n=1}^{\infty}(-1)^{n}\frac{H_{n}}{n}$,

where $H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$ is the $n^{th}$ harmonic number.

The sum equals: $\frac{\ln^{2}2}{2}-\frac{\pi^{2}}{12}$.

For solving the problem use the following formula:

$\frac{1}{2}\ln^{2}(1-z)=\sum_{n=1}^{\infty}\frac{H_{n}}{n+1}z^{n+1}$, for $|z|<1$.

Now, set $z=-1$ and after some rearrangements you should get it.

We want to find the limit of this series..
We have that $n\in{N^{\ast}}$..
Also the power of (-1) is n+1 and not v+1,I made a mistake before..
I believe Thales418 understand now what I want to saw...

i understand, but with that mistake the solution was another

hehe


geometry
3D geometry
sphere
calculus
integration
function

I have a question about the surface area of a sphere. Our teacher told us that this is how it works. She cut out one of the "flat spheres" (the ones in the atlas where it's like a map and if you cut it out and put all the ends together you'd get a sphere) and she cut in half length wise, then she cut it in half height wise. She said that one fourth was the area of a circle, and there are 4 of these fourths (no duh) so the surface area is 4 pi r^2 (I have forgotten how to use latex...).
The part I'm confused about is WHY the fourth is equal to the area of a circle. Is it because of the first cut (since you'd be cutting the "circumference" of the sphere")?

From what I remember, my friend told me to find that formula, it requires calculus.  But i don't know really.

Unless I'm being dumb, it's not at all obvious how the areas of those $ 4$ regions are each $ \pi r^2$. It requires calculus.

It requires calculus

$ \int_a^b2\pi f(x)\sqrt {1 \plus{} f'(x)^2}dx$ is the general case for surface area for a function $ f(x)$ when rotated.

And I'm pretty sure your teacher is wrong. It doesn't seem true that that is 4 circles.

Is there a simpler way like drawing a diagram? Also, the thing I drew isn't the sphere she got. It's just an example if you didn't know what I was talking about. :)

Actually I'm sure 1/4 of a sphere is the area of a circle but it certainly isn't a circle and it isn't very obvious.

You can't draw a sphere well. Just look at the world maps; there is always distortion. It's because sphere "folds" mess things up.

:rotfl:
  :ninja:  :D
Well it's like you said. It's in the shape of a cats eye. But if you flatten it then it becomes a circle.

isn't the surface area pi*radius*lateral hight?

That's the lateral surface area of a cone, isn't it?


Let a triangle $ ABC$ with Lemoine point $ L$, circumcenter $ O$. $ AL$ cut $ BC$ at $ A_1$, $ B_a\in AC, C_a\in AB$ so that $ AB_aA_1C_a$ is a parallelogram. Then $ B_aC_aBC$ is concylic, call it's center is $ O_a$. Prove that: $ AO_a$ pass over the midpoint of $ OL$.


Let $ n\ge 2$ be an integer. Prove that
                  $ \sum \frac {p}{q}  \equal{} \frac {1}{2}$
where the sum is over all integers p,q satisfying $ 0 < p < q \le n, p\plus{}q>n$ and  (p,q)=1.

Could some mod, edit the post? I know chapplikka meant $ \frac{1}{pq}$.

And 'n' can be any integer >=2.


We have an arithmetic sequence a(1) + a(2) + a(3) … + a(98) = 137, with a common difference of 1. Find a(2) + a(4) + a(6) + … a(98).

By the way, Z = mod 2Z, if you want to edit a post because you forgot something, you can use the EDIT button. If you post too many times in a row, people can get mad and think you're spamming. No offense.

That's one way you can do it. I did it a different way.
[hide="Another Solution"] You can see that a(2) + a(4) + ... + a(98) = a(1) + a(3) + ... + a(97) + 49 because each series has 49 terms, and the common difference is 1. Call a(2) + a(4)... = x. Then a(1) + a(3)... = x - 49. We have 2x - 49 = 137, so x = [b]93[/b][/hide] By the way, Z = mod 2Z, you should hide your solutions. (You do this by clicking the hide button when you post.)


articles algebra function domain logarithms LaTeX calculus

Given the identity: (T/T)a^x=T(a/T)^((xlna/lnT-1)/(lna/lnT-1)), can we "cross out" the T's and make them "disappear"? Don.

The things you've written on your website about cohesive terms are nonsense.  The correct answer to your question in the open letter is just $ S \equal{} 2 \sqrt {N}$ with ordinary reasoning; the second answer involving some unknown $ X$ is garbage.  At best, what you're doing is using some (very poorly defined) axiomatic system that [b]isn't[/b] the standard system we use to prove results about the standard system, and I don't understand how you could possibly consider that valid.  Most of what you write is rhetoric, not mathematics.

Well t0rajir0u, you are both rude and wrong. For one thing, I do not write about the the equation: S=2(N^(1/2)) in the "Open Letter", but in the article "Cohesive Terms vs. Arabic Terms". For another thing, the question in that article implies taking the square root of a square, so again, you are wrong in your assertion that S=2(N^(1/2)) is the correct answer. For yet another thing, you are clealy wrong about "cohesive terms" having a "poorly defined axiomatic system" because unlike "non-cohesive terms", "cohesive terms" have perfectly defined variables which makes them the most powerfull tool in all of mathematics! Most importantly however, you have not displayed the necessary reasoning power to stay "on topic" and answer the simple yes or no question that I asked in my first post, so I will ask it again. Given the identity: (T/T)a^x=T(a/T)^((xlna/lnT-1)/(lna/lnT-1)), can we "cross out" the T's and make them "disappear"? This question, like the rest of my work, is not "rhetoric", but as anyone can see, purely mathematical. Go ahead genius, answer it! Don.

This is hardly worth replying to, but...

[quote="Open letter"]... if two different variables have the same definition, then they are, in essence, the same variable...[/quote]
The word definition is misused and should be domain. Why are two variables with the same domain the same value?

I agree with t0rajir0u in that you are wrong here. First of all, I see no sensible definition of a "cohesive term," but instead you write about how everyone should use them. It would be easier for everyone if you could clearly define your terms.

As for your initial "formula", it is unclear what you mean. I am going to say that you mean the following, since you seem to want to exclude $ T\equal{}0$ and $ T\equal{}1$.

$ (\frac{T}{T})a^x \equal{} T(\frac{a}{T})^{\frac{\frac{x\ln a}{\ln(T \minus{} 1)}}{\frac{\ln a}{\ln(T \minus{} 1)}}}$

Whether or not you allow "crossing out" the $ T$s, your "formula" is [i]wrong[/i]. In fact, this formula only holds true if $ x\equal{}1$.

After looking at your website, I think that your theory, if it truly deserves that name, is ridiculus. The article about "Defining the Undefined" is purely based on your own set of axioms (which you have yet to define). The notion of calling a number multiplied by one or divided by one undefined is senseless. I can't see a good reason for doing so. What angers me more than this mess of "rhetoric", to borrow a phrase, is the fact that you don't acknowledge that algebra is a field of math that has been studied throughout history and is mathematically perfect. You still haven't explained the problems you see in mathematics and seem to babble on about the benifit illogical mathematics. You ask for answers but refuse to see your own flaws. tOrajirOu was not being rude, he was speaking truth, which you seem to hate. Also, you said: "Given the identity: (T/T)a^x=T(a/T)^((xlna/lnT-1)/(lna/lnT-1)), can we "cross out" the T's and make them "disappear"? This question, like the rest of my work, is not "rhetoric", but as anyone can see, purely mathematical. Go ahead genius, answer it! Don".

I can't how someone can answer a question about your fake math and just because there are numbers and mathematical operations doesn't make it "purely mathematical". The "most important equation ever" lol.

By the way, for those who haven't seen it, see his website ([url]http://donblazys.com/[/url]).

To: Mellow Mellon. Variables are defined by their domains. Given the expression: x , If I asked you to define the vaiable x , what would you say? Well, if you were smart, then you would say something like, "x is a non-negative integer", and you would write something like, x={0, 1, 2, 3...}, so you see, the definition of a variable is its domain, and since in the case of variables domain=definition, it logically follows that two different variables should not have the same definition, just as the two different words "car" and "tree" should not have the same definition. You see, up until a decade ago, the only basic (one multiplication, one exponentiation) term that mathematicians were aware of was Ta^x, whose variables are intrinsically defined as T=a=x={0, 1, 2, 3...}. Thus, mathematicians had no choice but to assume that elementary terms whose variables are defined by their own unique domains did not exist in algebra. As it turns out, that assumption was wrong, because about a decade ago, I discovered the incredible identity: Ta^x=(Ta)^((xlna/lnT+1)/(lna/lnT+1)), which shows us that there is indeed an algebraic term whose variables are perfectly defined by their own unique domains (T={2, 3, 4...}, a={1, 2, 3...} and x={0, 1, 2...}), so now, since the expression on the left is a part of that identity, it too can be legitimately viewed as having those same, perfectly defined variables, and that's a very, very good thing, because in and of itself, the symbolic representation: T=a=x={0, 1, 2, 3...} does indeed imply that T, a, and x have the same value. Don.

To: JZambrano201. You know, one really cool thing about my incredible equation: (T/T)a^x=T(a/T)^((xlna/lnT-1)/(lna/lnT-1)), is that no matter how much venom you spit, or how much hot air you belch, it remains perfectly true! Another really cool thing about it is that it can do all my talking for me! You see, I don't really have to tell anyone that multiplication and/or division by one automatically results in division by zero, because anyone can simply let T=1 and see for themselves! No science is perfect, and algebra has a long way to go before it is "complete". You also have a long way to go before you grow up. Only a child would use a phrase like "fake math" without any logic or reason to back it up. Don.

To: All four of you. You know, anybody can look at your posts and see that instead of answering the simple yes or no question that I asked in my very first post, you would all rather make a mad dash to my website and look for the slightest excuse to "bad mouth" my work. Well, mere "bad mouthing" can never negate the truth, and the truth is that you have yet to present me with a cogent argument that points to some flaw in my reasoning. I, on the other hand, have pointed out all kinds of flaws in you collective reasoning and will continue to note that you haven't answered the question! Don.

[quote="Don Blazys"]To: Hamster1800. I have defined "cohesive terms" plenty of times throughout my articles, but for those of you who have not yet visited my website, I will present the definition here. Cohesive term: Any algebraic term comprised of three or more variables that has the following properties. (1) Each and every symbol is defined by its own unique domain. (2) Unit coefficients and exclusive multiplications by unity result in division by zero, and are thus prohibited. (3) The order of operations is such that cancelled factors never form unity, and are therefore never evaluated as unity, which precludes them from getting "crossed out", and "disappearing".________________________________ My "initial formula", as you call it, is: (T/T)a^x=T(a/T)^((xlna/lnT-1)/(lna/lnT-1)). Anyone can see that it is absolutely correct, and that it is your "version" which is wrong. Don.[/quote]

How about you let $ a\equal{}2, x\equal{}2, T\equal{}2$, and show me that your "formula" gives two equal values on both sides? As far as I can tell, what you have written gives $ 4 \equal{} something\ undefined$, which is why I say it is wrong.

For your initial question: You can cross out some of the Ts, but not others. You most certainly can cross out the Ts on the left. If lnT-1 means ln(T-1), then you can cross out the ln a and the lnT-1 (but if it means (lnT) - 1, then you can't) In either case, your formula is still wrong.

Hmm, it seems to me that you mean to have the -1 outside the fraction. In that case, I don't see how you can possibly think that this is correct.

Also, I suggest you read this: [url]http://www.catb.org/~esr/faqs/smart-questions.html[/url]
When asking a "simple yes or no question", explain why we should take our time to step into what is most likely a trap to say "AHAH You're wrong!"

Lastly, in response to your "same domain means same value" thing, what you should actually be writing is $ a, x, T \in \{0,1,2,\ldots\}$. This does not at all mean that they [i]equal[/i] the set. In particular, two things can be elements of the same set without being equal.

Just to be clear, everyone: you're talking to someone who thinks multiplication by 1 is the same as division by 0, who thinks we're only allowed to consider one variable in a given domain at a time because if I define two variables in the same domain they must be equal, who has serious issues with proper parenthization (leading, for example, to his interchanging $ T(a^x)$ with $ (Ta)^x$ at will and thereby drawing wrong conclusions, such as the "identity" in his original post that he calls "the most important equation ever"), and so on.  I think the best thing to do is refer them to posts 28 and 30-33 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28607&search_id=9485350&start=28]here[/url].

To Don, if you don't mind a little of my ribbing: why don't you show the derivation of your identity?  The forum is equipped with LaTeX, so it's easy to write legible mathematics.  (See [[LaTeX]] or http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 for an introduction.)  Please, while you're writing it up, be careful with your use of parentheses: your original formula could mean a variety of different things (none of which appear to be correct, but that's not the key issue at the moment) depending on how we choose to disambiguate it.

Here, I'll set this straight:

In the "Open Letter", you talked about and equation similar to this:

$ Ta^x\plus{}Tb^y\equal{}Tc^z$ where $ T>1$

Then dividing by $ T$ yields, $ \frac{T}{T}a^x\plus{}\frac{T}{T}b^x\equal{}\frac{T}{T}c^z$

You say that $ T$ was supposed to be greater than one, but your not realizing that $ \frac{T}{T}\ne T$ therefore, if $ \frac{T}{T}\equal{}p$, then $ p$ is not necessarily going to be greater than one. And by the rules of division, $ \frac{T}{T}\equal{}1$. Also, the question that asks, "If a number has only two factors that are equal to one another, how shall we write a formula for their sum $ S$?". The answer is cut and dry, $ S\equal{}2N^{1/2}$. I don't see why that could be wrong (I'd like to see an explanation).

If I were you, I would invest some time in relearning/learning algebra, calculus, and possibly higher mathematics. Then maybe you'll see why this "math" is nonsense.

To: Hamster. The identity: (T/T)a^x=T(a/T)^((xlna/lnT-1)/(lna/lnT-1)), represents a "higher order of logic". Therefore, it requires that the cancelled variable T represents the greatest possible factor of the term as a whole. Letting T=2, a=2 and x=2 in: Ta^x=(Ta)^((xlna/lnT+1)/(lna/lnT+1)) gives us 8=8, and the largest factor of 8 that leaves a prime when cancelled is 4, (not 2). Thus, if we were now to divide by T, then we must let T=4, a=2 and x=1 in: (T/T)a^x=T(a/T)^((xlna/lnT-1)/(lna/lnT-1)), which gives us 2=2. The result 4=UNDEFINED occured only because you were trying to cancel a factor that was not the greatest, and didn't leave a prime when cancelled. You see, "cohesive terms" are "smart". They "know" that cancelling the largest factor from a number must leave a prime, and that primes don't have exponents. Hence, "cohesive terms" do not allow x>1 at T=a, whereas whereas "non-cohesive terms" do, so you tell me, which term constitutes the more logical construct? Now, on the "same domain implies same value" issue, if T,a,x, E {0, 1, 2, 3....}, then T,a,x, E {0, 1, 2, 3, T, a, x, 4, 5, 6...}, where it is also true that 0, 1, 2, E {0, 1, 2, T, a, x 1, 2, 3...}, and you are now implying that integers have the power to function as variables! Don.

To: JBL. The identity: (T/T)a^x=T(a/T)^((xlna/lnT-1)/(lna/lnT-1)), demonstrates that multiplication and/or division by one automatically results in division by zero. Simply let T=1, and see for yourself! Its parenthesis are perfectly in place and absolutely unambiguous. If you are unable to evaluate it, then please get someone to help you. Now, if you don't mind a little ribbing in return, are you hallucinating?! Where do you find (Ta)^x on any of my posts? By the way, if you want to see how "cohesive terms" are derived, then go to my website (donblazys.com). I also have some very popular topics on the Marilyn Vos Savant website. Don.

Could you show [i]how[/i] you got 8=8 from T=2, a=2, x=2? Also, $ 0,1,2 \in \{0,1,2,T,a,x,,3,4,5,\ldots\}$ is a [i]true statement[/i]. It does [i]not[/i] imply that $ 0, 1, 2$ are variables.

Hmm, I should have taken my own advice and stopped with "don't feed the trolls."

[quote="Don Blazys"]Its parenthesis are perfectly in place and absolutely unambiguous. If you are unable to evaluate it, then please get someone to help you.[/quote]

Really?  How would you unambiguously evaluate the term $ \ln a / \ln T \minus{} 1$?  Because of your failure to parenthesize, this could mean any of $ \frac {\ln a}{\ln (T \minus{} 1)}$, $ \frac {\ln a}{\ln T} \minus{} 1$ or $ \frac {\ln a}{(\ln T) \minus{} 1}$.  The fact that you think you're revolutionizing mathematics isn't a good excuse for sloppy notation.

In fact it seems your entire work involves some misunderstanding or misuse of common notations: in particular, the form (Ta)^x [i]should[/i] appear in several of your works, assuming what you were doing made sense.  It (or something like it) [i]should[/i] appear in the left-hand side of your "most important equation ever," if we want the left- and right-hand sides to actually agree.  I'm with Hamster: do a step-by-step exposition of how to derive your equation up at the top, and let's see what it looks like.

But, actually, I personally would prefer (and of course, you are under no obligations to spend your time catering to my preferences) if you would answer the following nonmathematical question: the modern European tradition (including the use of letters for variables) goes back several hundred years -- why hasn't anyone before you noticed that multiplying by 1 is the same as dividing by 0?

By the way, to others, if you're interested in the mathematics that drives this particular idea, it's the following observation:  if you have an identity, you can derive new identities from this identity that won't be valid at all values of one of the variables involved.

To: Hamster1800. My first explanation was quite clear. If Ta^x=(Ta)^((xlna/lnT+1)/(lna/lnT+1)), then letting T=2, a=2 and x=2 gives us 8=8. Now, like I said before, "cohesive terms" represent a "higher order of logic" and therefore require that any cancellation of factors be complete. This means that we must now let T=4, a=2 and x=1 when we write (T/T)a^x=T(a/T)^((xlna/lnT-1)/(lna/lnT-1)). In other words, we can no longer let T=2, a=2 and x=2 because doing so would constitute an incomplete cancellation of the factors of 8. You see, cancelling each and every factor from a number must leave a prime, and when you got the result 4=UNDEFINED, it's because you cancelled the smallest factor 2 instead of the largest factor 4. (4=UNDEFINED is actually a true statement because 4 can't be defined as a prime!) Cohesive terms are awesome because they "tell us" when the cancellation of factors is incomplete! As for 0,1,2 E{0,1,2,T,a,x,3,4,5...} being a true statement, well, maybe not. You see, sets are formed in order to classify members or "elements"  into categories according to their properties. The properties of the variables T, a and x are not the same as the properties of the non-negative integers 0, 1, 2... because variables can assume different values whereas non-negative integers can't. Therefore, the statement: 0,1,2 E{0,1,2,T,a,x,3,4,5...} does indeed imply that either the non-negative integers 0, 1, 2...are variables, or that the variables T, a and x can not assume different values. Don.

Peter (or some other one who is able to): please lock this thread.
Enough food was given...

[quote]To:JZambrano201. In the open letter, we begin with: Ta^x+Tb^y=Tc^z, where T>1. However, dividing by T yields: (T/T)a^x+(T/T)b^y=(T/T)c^z, and clearly, this implies that T=1 because symbolically, all we really did was substitute the "unit coefficient" (T/T)=1, for the original coefficient T>1. Why is this so hard for you to understand? As for the other question, if a number has only two factors that are equal to one another, then, by definition, that number is automatically a square, and we can write it as N^2. The formula S, for the sum of those two factors would then be twice the square root of N^2, which is S=2(N^2)^(1/2)=2N, which is the exact same formula that we got using "Cohesive terms". Now can you see why it is you who is wrong? Don.
[/quote]

I can't see why I am wrong because I'm not. Now get real you pompous idiot. You haven't discovered anything, your mathematics are wrong, and you have shown me that you don't even understand the foundations of math. Now get
over yourself and read a book on fundemental arithemetic.

Edit: Also the answer to that question is $ S\equal{}2\sqrt{N}$ not $ S\equal{}2n$.

To: JBL. How would I unambiguously evaluate the expression lna/lnT-1 ? Why, I would follow the "standard order of operations", of course! They are: (1) Powers, roots and logarithms, (2) Multiplication and division, and (3), addition and subtraction. Now, if you perform the operations in that order, then I can guarantee you that the expression lna/lnT-1 is absolutely unambiguous, and can only mean (((lna)/(lnT))-1). You see, the parenthesis that I just put in are really quite unnecessary, because most kids have the "order of operations" memorized by the time they are out of high school. (Now that you read them here, you got a "head start" on the rest of your class!) Anyway, first you said that the expression (Ta)^x "does" appear in my work, and now you are saying that it "should" appear in my work. Sorry, but you are wrong on both counts. Both sides of the identity: (T/T)a^x=T(a/T)^((xlna/lnT-1)/(lna/lnT-1)) are in perfect agreement, and as I told you before, you can find the "step by step" derivation of the unbelievable term on the right on my website (donblazys.com). As for your question, " Why hasn't anyone before me noticed that multiplying and/or dividing by one results in division by zero?" Well, I really don"t know. (Nobody does.) If I had to guess, then I would say that the thousand year old practice of "crossing out" factors and making them "vanish" became so common and ubiquitous, so routine and automatic, that no one other than myself ever thought that it just might be possible to develop a superior algebraic term that actually prevents cancelled variables from getting "crossed out" and  "disappearing". As it turned out, those same terms also showed that multiplying and/or dividing by one results in division by zero.  Don.

To: ZetaX. Lock my thread!? Why? Because you can't handle the truth? You know, if my fantastic equation: (T/T)a^x=T(a/T)^((xlna/lnT-1)/(lna/lnT-1)), was not true, then nobody would care because things that are not true, especially in mathematics, are not very interesting. Someone would just come along, point out the mistake, and my topic would be over. It would not get another visit. However, just the opposite happened here. Everyone who tried to refute my work was shown that it is they who are in error, and went slinking away with their tails between their legs! The readers know this, and so they keep coming back to my topic, every day, just to see who else will make fools of themselves. It is, after all, very entertaining and lots of fun. I mean, it sure beats discussing some boring topic where the answers can be found in some textbook or on the internet. You see, any fool can pretend to be smart by going to Mathworld or Wikipedia, finding the answer, then posting it as their own. Thinking for oneself, however, requires talent. That's probably one of the main reasons why ZetaX and others like him are so angry and frustrated. They can no longer run to Mathworld or Wikipedia or some other site as if they were running to mommy and daddy! Now they are literally being forced to think for themselves, and are proving themselves to be like fish out of water, just flopping about aimlessly and in great desperation. Don.

To: JZambrano201. You don't sound very sure of yourself, do you?

Where on your website is there the derivation? I have no desire to search every single "paper" that you have written.

Also, the order of operations gets considerably more fuzzy when you use functions like logarithms that can exclude the use of parentheses. For example, $ \sin 2x$ means $ \sin(2x)$, not $ (\sin 2)\cdot x$ as your order of operations would suggest.

Finally, please learn some mathematical logic. In no way does $ 1 \in \{1,T\}$ and $ T \in \{1,T\}$ mean that they have the same properties! Do you know the definition of a set?

I love this guy. He has his head shoved so far up his ***, it hilarious. Heres why division by one is justified captain crunch:

Let division be defined in terms of multiplication such that,

$ \frac{p}{q}\equal{}x\rightarrow p\equal{}qx$

If $ q\equal{}1$, then $ \frac{p}{1}\equal{}p\rightarrow p\equal{}p(1)$ which is true. (and not undefined)

I would invest some time in to learning what you preach clown.

By the way Captain Crunch, why don't you put this in a college board if you don't think we understand (also, the people posting here know their stuff i.e. tOrajirOu, JBL, Hamster)

[quote]You see, the parenthesis that I just put in are really quite unnecessary, because most kids have the "order of operations" memorized by the time they are out of high school. (Now that you read them here, you got a "head start" on the rest of your class!)[/quote]

Why do you assume he hasn't learned them, the equation is not well written. Use latex. In my book JBL is many levels beyond you so you should havesome repect. Post it in a higher level board, unless your afraid of the big bad college people making fun of you. Well, there's nothing else to say but,

CRUNCHITIZE ME CAPTAIN~!                                   lol

[quote="Don Blazys"]Everyone who tried to refute my work was shown that it is they who are in error, ...[/quote]
To be more accurate: You believe they are in error.
And I will now get this locked myself  :wink:

Don blazys, why dont you use latex and explain that "important equation" step by step for us [size=150]please [/size]

Hamster is right why dont you show us that the equation works [color=red]step by step[/color] for $ T \equal{} 2,a \equal{} 2,x \equal{} 2$. please

Please write your Equation in Latex, and explain each term clearly, and show us some examples of what you think is right.
i am not saying that your equation is wrong because i am wondering how come you went so far, and you still insist you are right  :maybe:  :?: there must be some misundrestanding with us or with you.

atleast show us one example of what you think is right with LATEX.

Zetax please dont lock ,let him explain his equation clearly

Binomial: he's a troll and crank. Not more. The only thing missing is a claim that we belong to some conspiration...


Find the smallest positive integer having divisors ending in every decimal digit.

[hide="Solution"] Let the smallest such number be $ n$.  We need divisors ending in $ 0, 2, 5$, so we know $ 10 |n$.  The odd digits $ 3, 7, 9$ must come from factors relatively prime to $ 10$, so $ n$ must have at least three factors ending in three different odd digits.  Let $ d$ be the factors of $ n$ relatively prime to $ 10$.  Then the smallest possible value of $ d$ is $ 27$.  In fact, $ \boxed{270}$ has the divisors $ 1, 2, 3, 54, 5, 6, 27, 18, 9, 10$. [/hide]


consider the equation $ x^2 \minus{} ax \minus{} 1 \equal{} 0 (a > 0)$ and $ \alpha$ is the positive root of this equation. Consider the sequence $ (x_n)$: $ x_0 \equal{} a$; $ x_{n \plus{} 1} \equal{} [\alpha x_n], n \equal{} 0,1,2...$. Prove that there are infinitely many integers $ n$ such that $ x_n$ is a multiple of $ a$.

I'm sorry, but what's $ \alpha$ got to do with it?

$ \alpha$ is the positive root of this equation :)

Sorry for the question. I missed to notice it in the $ [\alpha x_n]$ part so was confused since it wasn't mentioned anywhere else!

Already posted:

[url]http://www.mathlinks.ro/viewtopic.php?t=208198[/url]


MATHCOUNTS
Ring Theory
geometry
AMC
AIME
AIME I

Does anyone know where I can find the 2000-2009 state and nats sprint,target, cd, masters rounds with solutions?

Also does anyone know a website that has a most of AMC tests?

MATHCOUNTS doesn't put out nationals online. You have to get them from other people. Older states are also gotten by trading from other people. I believe [url=http://www.mathacle.com/assessments/]Mathacle[/url] has some previous state tests. Also, check out [url=http://www.mnspe.org/mathcounts/coach.htm]the right hand side of this page.[/url] Obviously, to get the 2009 tests, you would go to the [url=https://mathcounts.org/Page.aspx?pid=295]MATHCOUNTS website.[/url] I can donate some tests so you can start trading, if you want :D

As for the AMC's check [url=http://www.math.ksu.edu/main/handbook/ProblemSets]here,[/url] [url=http://www.agmath.com/23806.html]here.[/url]

I'm pretty sure your last link is illegal.

Why not just use the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182]Resources Section[/url]?

Most of them aren't printable because of the geometry pictures.

Sorry if the last link wasn't supposed to be there. :oops:

Oh! Wow that's awesome RoFlLoLcOpT! Thanks alot! I think that's enough material to keep me occupied till the end of the summer, so you don't have donate anything to me(I appreciate the offer though). :)  :)

@AIME, I don't think the resources section has MC stuff...

Sorry - was referring to the AMC's.

By the way, Master's Rounds are not available for sale, even through the MATHCOUNTS site (well, fine, there's one in the All-Time Greatest book), so you have to trade for them.

Keep discussion of Mathcounts in the Mathcounts forum, or whatever it's called now.


How do you solve this system of equations?

$2x+y-3z+w=6$
$x+y+2z+w=-1$
$y-z+w=0$
$x+z-w=5$

I kept combining and eliminating but it was a mess...
Thanks!

You can also use Gaussian Elimination to solve a system of equations

http://mathworld.wolfram.com/GaussianElimination.html


Does anyone over here have AOPS books but does not need them anymore?I mean the college going people?I need them please :(

I know i can order it but i don't trust giving away my credit card number nor sending a check and wasting away 54$(54$ is a lot in rupees)

Any benevolent soul out there willing to help me?

But how do you think OTHER people are going to send their copies, that is, if they want to? It is going to cost money to send the copies. ;)

Do you have the AOPS books,riddler?No,right?

I will send the money to them and i know it will be tough for them to believe me but i swear i will other wise you can send someone to kill me at the address i'll give you people to whoever is interested on sending me the book(s).Pleaase do if you can.

I don't know whyno one's replying to me...please i need help fellow brothers.

No one's going to send you AoPS books for free.

At first you said you couldn't afford the $\$$54 but then you said you said you would pay whoever sent you books. :?: You need to be more specific in what you are proposing to do.

It is safe sending your credit card number. Actually, its probably safer than paper, because you can keep a record of the sale. You can't do that by sending your only copy of the check.

AoPS isn't going to rip you off. I personally know this.


calculus
integration
function
calculus computations

I was just wondering what would be good examples for these two cases:

1) $ f: [a,b]\rightarrow \mathbb{R}$ is discontinuous at some point of the interval, then there is exists a monotone increasing function $ g$ such that $ f$ is not $ g$- integrable.

2)$ f$ is not Riemann integrable over some interval $ J$ but $ |f|$ and $ f^{2}$ are Riemann integrable over $ J$

for 1) I was thinking of a Dirichlet function and just taking $ g(x)\equal{} x$ since the Dirichlet function is not Riemann integrable anyway.

for 2) I would like a hint, since I think I have an idea, but I 'm not sure if my idea is right.

lastly, out of curiosity, since a differentiable function is continuous, does that mean a Riemann integrable function is continuous( since I was just wondering that a differentiable function has to be Riemann integrable ...right?)

Thanks

2) Consider $ f(x)=\{ \begin{array}{c l}
1 & x\in \mathbb{R}-\mathbb{Q} \\
-1 & x\in \mathbb{Q} \\
\end{array} $.
Then $ f$ is not Riemann-integrable on $ J=[0,1]$, but $ |f|=f^2=1$ are Riemann-integrable.

Curiosity: A Riemann-integrable functions needs not to be continuous.
Consider $ f(x)=\{ \begin{array}{c l}
1 & x>0 \\
0 & x\leq 0 \\
\end{array} $

#1: That doesn't work, since you don't get to choose $ f$. I prefer the following example: $ f\equal{}g\equal{}H$, where $ H$ is the standard step function $ H(x)\equal{}\begin{cases}0&x<0\\1&x\ge 0\end{cases}$. For continuous $ f$, $ \int f\,dH\equal{}f(0)$. As a Riemann-Stieltjes integral, $ \int f\,dH$ exists only if $ f$ is continuous at zero.

To adapt this to the problem, suppose $ f$ is discontinuous at $ c$. Let $ g(x)\equal{}H(x\minus{}c)$.

I do have one question though, for the suggestion for 1): I thought the step function can't be monotone increasing?

I really wish someone would answer my question... :(

the step function is not strictly monotone increasing, but it is monotone increasing--understood.


If $ |4x\plus{}2|\equal{}10$ and $ x<0$, what is the value of $ x$?

[hide="Solution"]
Since we know that $ x$ is negative we need to solve the equation $ 4x\plus{}2\equal{}10\implies x\equal{}\boxed{\minus{}3}$
[/hide]

[quote="hunter34"]Since we know that $ x$ is negative we need to solve the equation $ 4x \plus{} 2 \equal{} {\color{red}\minus{}} 10\implies x \equal{} \boxed{ \minus{} 3}$[/quote]

Minor correction.

hello, we have to distinguish two cases. $ 4x\plus{}2\geq0 \Leftrightarrow x\geq\minus{}\frac{1}{2}$, so we have $ 4x\plus{}2\equal{}10 \Leftrightarrow x\equal{}3$. In the other case we have $ x<\minus{}\frac{1}{2}$ and it follows $ \minus{}(4x\plus{}2)\equal{}10
\Leftrightarrow x\equal{}\minus{}3$. Hence our solution is $ \boxed{\minus{}3}$.
Sonnhard.


function
integration
real analysis
real analysis unsolved

Just as I thought! Apply Cauchy in the equality $\int_{0}^{1}{ x(1-x)f'(x)dx}=1$.

that's exactly my solution.apply Cauchy and then integrate by part. ;)


"Twin primes” are a pair of prime numbers that differ by 2.  How many pairs of twin primes are between 10 and 50?

A.  1	B.  2	C.  3	D.  4	E.   5


geometry
integration
real analysis
calculus
calculus computations

Find the area bounded by two curves in terms of x and y:
$ x \equal{} y^2 \minus{} 4$ and $ x \equal{} y \plus{} 2$
I solved it in terms of y and got $ \frac {125}{6}$, but got $ \frac {19}{2}$ in terms of x? Does anyone clarify it for me which one is correct?? I think the result calculated in terms of y is correct though.

EDIT: Thank jmerry. I made a typo and now I fixed it. Never mind, everybody. I already saw what I got wrong after a few minutes posting this problem. The answer is actually $ \frac{125}{6}$ :)

Why don't you show what you did, and then we can take a look.

Well, here's a start: there is no such region, because the curves don't intersect.

[quote="ghjk"]Find the area bounded by two curves in terms of x and y:
$ x \equal{} y^2 \minus{} 4$ and $ x \equal{} y \plus{} 2$[/quote]

Japanese high school students are trained to solve the problem in 30 seconds.

$ S\equal{}\minus{}\int_{\minus{}2}^3 (y\plus{}2)(y\minus{}3)dy\equal{}\frac{1}{6}(3\plus{}2)^3\equal{}\frac{125}{6}$

[quote="kunny"][quote="ghjk"]Find the area bounded by two curves in terms of x and y:
$ x \equal{} y^2 \minus{} 4$ and $ x \equal{} y \plus{} 2$[/quote]

Japanese high school students are trained to solve the problem in 30 seconds.

$ S \equal{} \minus{} \int_{ \minus{} 2}^3 (y \plus{} 2)(y \minus{} 3)dy \equal{} \frac {1}{6}(3 \plus{} 2)^3 \equal{} \frac {125}{6}$[/quote]
I know it's quite easy to solve by Definite Integral. Solving by Riemann Sum probably requires a more calculation, but it's still easy after I realize my mistake in Riemann Sum
Btw, why you always talk as if just Japanese students can solve these problems in 30 seconds, Mr.Kunny? No comparison should be made between the level of Mathematics between my country and yours though. And 30seconds is still a long time :P.  Anyway, thank you for your solution!

You are welcome. To tell the truth, I wanted to say, in 10 seconds. :lol:

I have almost found my post, Here is [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=170850[/url]


topology
advanced fields
advanced fields unsolved

In Armstrong's book, a locally compact space is defined as each point has a compact neighbourhood.
How to prove that any closed set of a locally compact space is locally compact?

Another question:
Is this definition equivalent to say that each point has a open neighbourhood whose closure is compact?

[quote]How to prove that any closed set of a locally compact space is compact? [/quote]
This is false. $\mathbb{R}^n$ is locally compact; a closed set in it is compact only if it is also bounded. There are plenty of unbounded closed sets.

[quote]Is this definition equivalent to say that each point has a open neighbourhood whose closure is closed?[/quote]
I assume you meant to say "each point has an open neighborhood whose closure is compact."

[quote="Kent Merryfield"][quote]How to prove that any closed set of a locally compact space is compact? [/quote]
This is false. $\mathbb{R}^n$ is locally compact; a closed set in it is compact only if it is also bounded. There are plenty of unbounded closed sets.[/quote]
Oh, Sorry for the typo. It should be "any closed set of a locally compact space is locally compact". Now I have corrected the problem. And I have proved it.

[quote="liyi"]Another question:
Is this definition equivalent to say that each point has a open neighbourhood whose closure is compact?[/quote]
Not quite. If the topology isn't Hausdorff, compact sets may fail to be closed. A silly example:
Let $X=\mathbb{Z}$, with the topology that $U\subset X$ is open when $U$ is empty or $0\in U$. Then the pairs $\{0,a\}$ are a basis for the topology, and each such pair is obviously compact. On the other hand, the closure of any such pair is $X$, which is not compact.

On the first question: If $K$ is a compact neighborhood of $x$ in $X$ and $C$ is a closed set containing $x$, then any open cover of $K\cap C$ may be completed to an open cover of $K$ by addind the set $X\setminus C$. This has a finite subcover; removing $X\setminus C$,  the original open cover of $K\cap C$ has a finite subcover. Of course, the interior of $K\cap C$ in $C$ still contains $x$, since it is the intersection of the interior of $K$ with $C$.


algebra
polynomial
roots
equation
IMO Shortlist
Diophantine equation

Find the roots $ r_i \in \mathbb{R}$  of the polynomial  \[ p(x) \equal{} x^n \plus{} n \cdot x^{n\minus{}1} \plus{} a_2 \cdot x^{n\minus{}2} \plus{} \ldots \plus{} a_n\] satisfying \[ \sum^{16}_{k\equal{}1} r^{16}_k \equal{} n.\]

First, it's Colombia (COL), not ILL (which does not exist).

From Viete's relation we have $ \sum_{i\equal{}1}^n r_i \equal{} \minus{}n$. Then, from Holder (or repeated applications of Cauchy-Schwartz) we have $ n^{16} \equal{} \left(\sum_{i\equal{}1}^n r_i^{16}\right)\cdot n^{15} \geq \left(\sum_{i\equal{}1}^n r_i\right)^{16} \equal{} n^{16}$, hence equality, hence all roots $ r_i$ are equal (to $ \minus{}1$).

Therefore the polynomial is $ p(x) \equal{} (x\plus{}1)^n$.

The latter (the one I used). It was a typo by orl, the same as ILL instead of COL. I know because I own the 1989 booklet containing the IMO Longlist.

Solution: We repeatedly apply Cauchy-Schwartz to obtain $(\sum r_i^{16})(1+1+...+1)\ge (\sum r_i^{8})^2\ge n^2$. Since $n$ is positive ($\text{deg}(p)>0$), we have that $(\sum r_i^{16})\ge n$. Hence we have the equality case of Cauchy, so $\boxed{r_1=r_2=...=r_n=-1}$. Our $p(x)=(x+1)^n$. $\blacksquare$


Typing up my latest reals assignment, I have the line:

$ \prod_{i\equal{}1}^N \xi_i^{\alpha_i} \equal{} e^{\sum_{i\equal{}1}^N \alpha_i \ln(\xi_i)} \equal{} e^{\textstyle\int_0^1 \ln(f_N(t))dt}$

The product extends the line down (this is in display mode, not inline), but the e^{sum} does not compensate for this.  As a result, it looks rather off.  What I would rather have is that the e^{sum} and e^{int} be shifted down to line up with the bottom of the product, and enlarged.

A second thing with this equation is that I want the indices of the sum to appear above and below the summation symbol, as you would typically get.  However, throwing a \displaystyle in there makes the sum far too large.

$ \prod_{i\equal{}1}^N \xi_i^{\alpha_i} \equal{} e^{\displaystyle\sum_{i\equal{}1}^N \alpha_i \ln(\xi_i)} \equal{} e^{\textstyle\int_0^1 \ln(f_N(t))dt}$

Hmm.. oddly enough on this forum, it does not change the size of the summation symbol.

Any recommendations on how to vertically enlarge or space this better are welcome.

On this foum \displaystyle is automatic but you can use inline mode with \textstyle:
$ \textstyle \prod_{i=1}^N \xi_i^{\alpha_i} = e^{\sum_{i=1}^N \alpha_i \ln(\xi_i)} = e^{\textstyle\int_0^1 \ln(f_N(t))}$.

1. To lower the formula use \raisebox with a negative value
2. As you know \textstyle in the exponents makes it larger
3. To put the limits at top and bottom use \limits
This is in display mode:
\[ \prod_{i=1}^N \xi_i^{\alpha_i} = \raisebox{-3mm}{e\: ^{{\textstyle\sum\limits_{i=1}^N \alpha_i \ln(\xi_i)}}} = \raisebox{-3mm}{e\: ^{\textstyle\int\limits_0^1 \ln(f_N(t))}}\]
So play around with this to get the effect that you want.

Of course many people solve the problem by using exp(x) instead of e^x
\[ \prod_{i=1}^N \xi_i^{\alpha_i} = \exp\left(\sum_{i=1}^N \alpha_i \ln(\xi_i)\right) = \exp\left(\int_0^1 \ln(f_N(t))\right)\]


Assuming x means multiply, and not a variable:  and also following order of operations:
$ 557 \frac{3}{7}$

[quote="the1wholovestexting"]3+9-15+3/7+16x 35=[/quote]

First I'll LaTexify this:

$ 3 \plus{} 9 \minus{} 15 \plus{} 3 \div 7 \plus{} 16 \times 35$
$ 3 \plus{} 9 \minus{} 15 \plus{} \frac37 \plus{} 560$
$ 12 \minus{} 15 \plus{} \frac37 \plus{} 560$
$ \minus{} 3 \plus{} \frac37 \plus{} 560$

$ \boxed{557 \frac 37}$

EDIT: typo  :blush:   i think i added the frac to -3 ,BTW i was sleepy

Well classroom math is the easiest section on this website, and I did this last year in my normal math class. it is not really that easy, just to us it is.


$ z, z'$ are complex numbers. If $ f(z)\equal{} \displaystyle\sum_{n\equal{}0}^\infty \frac{z^n}{n!}$, prove that $ f(z)f(z')\equal{}f(z\plus{}z')$ for arbitrary $ z$ and $ z'$

well, what you have is $ e^z$ which then makes it all too obvious, since $ e^A\cdot e^B\; \equal{} \;e^{A \plus{} B}$

a rigorous proof   involves uncle cauchy  :D

$ \textbf f(z)\cdot\textbf f(z')\; \equal{} \;\left(\sum_{n \equal{} 0}^\infty\dfrac{z^n}{n!}\right)\cdot\left(\sum_{n \equal{} 0}^\infty\dfrac{(z')^n}{n!}\right)\; \equal{} \; \sum_{n \equal{} 0}^\infty\sum_{k \equal{} 0}^n\left(\dfrac{z^k}{k!}\cdot \dfrac{(z')^{n \minus{} k}}{(n \minus{} k)!}\right)$

$ \equal{} \; \sum_{n \equal{} 0}^\infty\dfrac{1}{n!}\sum_{k \equal{} 0}^n\left(z^k\cdot (z')^{n \minus{} k}\cdot \dfrac{n!}{k!\,(n \minus{} k)!}\right)\; \equal{} \; \sum_{n \equal{} 0}^\infty\dfrac{1}{n!}\sum_{k \equal{} 0}^n\left(z^k\cdot (z')^{n \minus{} k}\cdot \dbinom{n}{k}\right)$

$ \equal{} \; \sum_{n \equal{} 0}^\infty\dfrac{1}{n!}(z' \plus{} z)^n\; \equal{} \;\textbf f(z' \plus{} z)$

That's the computation. Now: how can you justify scrambling the order? You need to prove absolute convergence of the whole double sum.


conics
parabola
combinatorics unsolved
combinatorics

The expression $ x^3 \plus{} . . . x^2 \plus{} . . . x \plus{} ... \equal{} 0$ is written on the blackboard. Two pupils alternately replace the dots by real numbers. The first pupil
attempts to obtain an equation having exactly one real root. Can his opponent spoil his efforts?

Here is my solution.  I am not sure if it is right or not, so please inform me of this.

The answer is that the opponent cannot spoil his efforts.  Here is the first pupil's strategy:  First he assigns some positive real number to the constant term.  Now his opponent chooses either the $ x^2$ or $ x$ term to attach a coefficient to.  This leaves the first pupil with the task of choosing a real number $ a$ such that $ p(x)\equal{}ax^2$ or $ p(x)\equal{}ax$, depending on what the opponent did previously, has only one real solution, where $ p(x)$ is a cubic with all the chosen coefficients.

Now we know for $ p(x)$ the relative minimum occurs to the right of the y-axis, since $ p(0)>0$ and $ p(x)\to \infty$ as $ x\to \infty$ and vice versa.  If the first pupil must choose a real $ a$ for the equation $ p(x)\equal{}ax$ to have one unique real solution, he does this by taking $ a$ to be negative so that we know the line intersects the cubic at some point $ (x_1,y_1)$ with $ x_1<0$.  The pupil also must make $ |a|$ large enough so that the line $ y\equal{}ax$ does not intersect the cubic at on the interval between $ x_1$ and the relative minimum.  This way, the line only intersects $ p(x)$ once.

The pupil works similarly in the case of $ p(x)\equal{}ax^2$ having a unique real solution.  First, the pupil chooses $ a$ to be negative so that we know that the parabola and the cubic intersect at some point $ (x_1,y_1)$ for $ x_1<0$ (since the cubic decreases faster than the parabola for $ x<0$).  The pupil also chooses $ |a|$ to be large enough so that the parabola does not intersect the cubic on the interval between $ x_1$ and the relative minimum.

also works if the first pupil assigns at the constant term 0????? the second pupl should to assign 0 to the second dots.

Here is the first pupil's strategy: the first pupil assign 0 at the first dots, if the second pupil assigns at the constant term 'a' then the first pupil chooses 1, therefore P'(x)=3x*x+1>0 ( P is increasant) since degree(P)=3:odd then P(-  \infinite )=- \infinite and P( \infinite)= \infinite
hence P has exactly one real root;
if the secondpupil assign 0 at the seconddots, the first pupil chooses 1, so -1 is the one real root of P
if the second pupil chooses 'a' >0 therefore P'(x)=3x*x+a>0(P is increasant) since degree(P)=3:odd then $ P(\minus{}infinite)\equal{}\minus{}infinite$ and $ P(infinite)\equal{}infinite$ hence P has exactly one real root;
if the second pupil chooses '-a'; a>0 then the first pupil choose $ b>(2a/9) *\sqrt{3a}$ so the local minimum is positive, and p has one real root.


If a+b+c=0 and a,b,c are not equal then find the value a^2/ (2a^2+bc) +b^2/ (2b^2+ac) +c^2/ (2c^2+ab)

hello, why you don't use $ \text{\LaTeX}$? so it looks much better
If $ a+b+c=0$ and $ a,b,c$ are not equal then find the value $ \frac{a^2}{ 2a^2+bc} +\frac{b^2}{2b^2+ac} +\frac{c^2} {2c^2+ab}$.
Sonnhard.

hello, plugging $ c\equal{}\minus{}a\minus{}b$ in the term above and simplifying we get $ 1$ as result.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, plugging $ c \equal{} \minus{} a \minus{} b$ in the term above and simplifying we get $ 1$ as result.
Sonnhard.[/quote]
sir pl clear the method


ratio
geometry
geometric transformation
reflection
circumcircle
parallelogram

Prove the following: if, in $\triangle ABC$, median $\overline{AM}$ is such that $m\angle BAC$ is divided in the ratio $1: 2$, and $\overline{AM}$ is extended through $M$ to $D$ so that $\angle DBA$ is a right angle, then $AC=\frac{1}{2}AD$.

[hide]
Reflect $\triangle ABD$ over $AD$, and let $B'$ be the new location of $B$.
Extend $DB'$ and $AC$ to meet at $E$.
Clearly, $\triangle ADB'$ and $\triangle AEB'$ are congruent, which means $AD=AE$.

Since $\triangle AB'D$ is the reflection of $\triangle ABD$, $MB=MB'=MC$.
Therefore, $\triangle BB'C$ is a right triangle with $\angle BB'C=90^\circ$. ($\because$ $M$ is the circumcenter of $\triangle BB'C$ and is on $BC$)

Let $F$ be the intersection of $\overrightarrow{AM}$ and $BB'$.
Since $AB=AB'$ and $\angle BAF=\angle B'AF$, $AF\perp BB'$.

Now we see that both $AF$ and $CB'$ are perpendicular to $BB'$, which implies $AF\parallel CB'$.

$AD\parallel CB'$ and $DB'=B'E$, so $CB'=\frac{1}{2}AD$ (Midline Theorem).
Also, $\angle DAB'=\angle AB'C=\angle B'AC$, so $CA=CB'$.

Therefore, $CA=CB'=\frac{1}{2}AD$.
[/hide]

i posted it in geometry mrathon. :)

[hide="hint"]form parallelogram by extending median. Find isosceles triangles. mdian in right triangle divides hypotenuse in half. This should suffice. :D[/hide]


I solved that problem, but with a horrible solution, that is different from the very smaller official solution.I just want to konw ifmy solution is right.If you note something wrong, please tell me!
Thanks :lol:

Problem: Let $m, n$ be positive intengers.Show that $4mn-m-n$ can never be a square.

Solution:Lets suposse that it s a square for some pair (m,n), so we can write: $4mn-m-n = a^{2}$.
And since ${(m+n)}^{2}-(m^{2}+n^{2})+2nm= 4nm$ we can do ${(m+n)}^{2}-(m^{2}+n^{2})+2nm-(m+n) =a^{2}$, so $(m+n)(m+n-1) = a^{2}+{(m-n)}^{2}$.Now, by a well know theorem we know that a  positive intenger can be expressed by a sum of two squares if each prime factor of the form $4k+3$ appears an even number of times.And now, see that $mdc(m+n, m+n-1)=1$.So if the prime $p=4k+3$ appears in the product (n+m)(n+m - 1) it appears an even number of times in one of the factor n+m, n+m -1   and zero times at the another.But that means that each factor n+m, n+m -1 can be expressed by a sum of two squares(because of the same theorem used before).So we can write $n+m= x^{2}+y^{2}$ and $n+m-1 = r^{2}+s^{2}$.
But now, see that $a^{2}= 4mn-m-n \equiv-n-m (mod 4)$ So $-n-m \equiv 0(mod 4)$ or $-n-m \equiv 1(mod 4)$

First case: $x^{2}+y^{2}= n+m \equiv 0 (mod 4)$.In that case we have that $n+m-1 = r^{2}+s^{2}\equiv-1 (mod 4)$, that is contradition!!

Second case: $n+m \equiv-1 (mod 4)$. So  $n+m =x^{2}+y^{2}\equiv-1 (mod 4)$. Contradition!!
 Q.E.D

Also have a look here :)
http://www.mathlinks.ro/Forum/viewtopic.php?t=85789


I want to post a solution for #15 but when I got to the part of AoPS that has the questions and solutions of this year's AIME, #15 isn't there. What should I do if I want to post a solution?

Oh shoot, sorry about that.  It was in there, then I was trying to fix something and it somehow went away....  It's back in there now but here's the link to #15 anyway http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78383


Show that for any integer n > 10000, there are integers a, b such that n < a2 + b2 < n + 3n1/4.

Let us write k=[ \sqrt n] and n=k 2+m, where m is 0,1,...or 2k. Now, let m=A 2 +B, where A=[ \sqrt m] and B is between 0 and 2A. So, A \leq  \sqrt 2*n^1/4. Let us take now a=k and b=A+1. We obviously have a 2 +b 2 >n (since it reduces to A+1> \sqrt m, clear). We prove now that a 2 +b 2 <n+3n^1/4. Since A \leq  \sqrt m, it is enough to prove that 1+2 \sqrt m<3n^1/4. Since  \sqrt m \leq  \sqrt (2k) \leq  \sqrt 2 \sqrt n, it suffices to have (3-2 \sqrt 2)n^1/4>1, which comes down to n>1500, which is true.


There are lots of different interesting sequences, so Ithought it would be nice if we had a marathon where people would try to solve various sequences.
There are several rules. First do not spam, and if you solve the sequence, then please write the next term(s) and specify how you got the answer. Please do not write things like "this sequence can be represented as the roots of an n-degree polynomial and so the next term can be any term", this totally ruins the sequence. Also, to make sure this thread does not die like many other marathons, the author of a sequence that was not solved for a week, please post the answer, and for a sequence not solved in three days, please post a hint and hide it. If the author doesn't reply, then a new sequence should be posted in two weeks since the last sequence was posted.
The sequences can contain anything: letters, numbers, symbols, etc.

Ok, now let's start. Here's an easy one:
S, S, F, T, W, ...

[quote="rem"]Ok, now let's start. Here's an easy one:
S, S, F, T, W, ...[/quote]

Sunday, Saturday, Friday, Thursday, Wednesday, Tuesday
$\boxed{T}$

Someone else can post the next sequence, I don't know any.

[quote="t0rajir0u"]$1, \infty, 5, 6, 3, 3, 3, ...$

:P

I found this on AoPS once.  It's my favorite sequence.[/quote]

I believe I posted this here, after being introduced to it by Jan Siwanowicz.  It is a nice sequence :-)

Ok three days have passed since lotgreenapes and t0rajir0u's sequences were posted, so if those two ppl could please write small hints to the sequences, that would be very helpful. :)

[hide="Hint 1"] The answer only depends on the index; in other words,

$1 \to 1$
$2 \to \infty$
$3 \to 5$
$4 \to 6$
$5 \to 3$
$6 \to 3$
$7 \to 3$

And so forth.  [/hide][hide="Hint 2"] Geometry. [/hide][hide="Hint 3"] Dimensions. [/hide][hide="Hint 4"] There are only five Platonic solids. [/hide]

[quote="lotrgreengrapes7926"]
14, 91, 62, 53, 64, _, _

6, 2, 9, 5, 1, 4, _

1, 5, 9, 15, _[/quote]
[hide="1"] $\boxed{}$ [/hide]
[hide="2"] Mmm, ___. [/hide]
[hide="3.1"] It has nothing to do with math. [/hide]
[hide="3.2"] It has to do with letters. [/hide]
[hide="3.3 (Big Hint)"] There are only 5 numbers in the sequence. [/hide]
And 4 was just stupid.

[hide]3. 21-vowel's positions in alphabet[/hide]

[hide]1. all the squares written out, take digits 2 by 2[/hide]

btw that was a really good hint lol/

[quote="IntrepidMath"]Number of objects that have all sides and edges and faces equal (you know what I mean...). I don't know how to find the next term though...[/quote]

This is correct.  I should tell you that every other term of the sequence is $3$.   :rotfl:

[quote="junggi"][quote="rem"]Just because nutz has it in his signature does not mean it is e. http://www.google.ca/search?hl=en&q=e&btnG=Google+Search&meta=[/quote]
nutz is sometimes right  :wink:
[url]http://antwrp.gsfc.nasa.gov/htmltest/gifcity/e.5mil[/url]  :wink:
p.s. google rounded up.
wait nutz has e in his sig?
his username you mean?[/quote]
Sorry  :blush:  :wallbash_red:
I will never trust google anymore.
Yes nutz has it in his sig.

[quote="mathnerd314"]Try typing in "the answer to life, the universe and everything" into Google and see what it gives... :P

Here's a few:

31, 62, 93, 25, 56, 87, 19, 401, ?

C, A, D, ?, E, I, B, F, ?, C, E, ...[/quote]

$31\mod99,31*2\mod99,31*3\mod99,31*4\mod99,$
$31*5\mod99,31*6\mod99,31*7\mod99.....$,uhhh what?

[quote="laughinghead505"]How did you figure that out :?:   It seems like it would take forever to figure out an answer like that...[/quote]
uhhh me?
I don't think I got the answer because
$31*8\mod99\not=401$

You have the right idea with the 31's, though. Keep thinking. :D

Anyone solved the one with the letters?

[quote="esti"][quote="lingomaniac88"]Look for those terms in mine.[/quote]
Yep,I was inspired by your sequence lingo,when creating mine  :wink:

Lighthree figured it out first (only except the  answer was 25,but that doesn't matter much to me).
BTW rem :it's not a question wether you can trust uncle Google or not   but if you can trust your pocket calculator  :lol:  :roll: .[/quote]

That's a big hint to mine.

Here it was again:

[b]1 2 3 6 7 1 1 1 4 8 8 1 0 2 5 _ _ _ _ _ _ 9 8 6 8 2 5 9 8 4

Six numbers are missing from this sequence. What are these numbers? Why?[/b]

[quote="lotrgreengrapes7926"][quote="junggi"][quote="thatin"]Thats right junggi!
try this: 64, 289, 625, ???[/quote]
[hide]$8^{2},(2\cdot8+1)^{2},(3\cdot8+1)^{2},(4\cdot8+1)^{2}$
so 1089??[/hide][/quote]
[hide] But wouldn't the first one be $(1\cdot8+1)^{2}=81$? [/hide][/quote]
Τhats wrong junggi...the first one would be $(1\cdot8+1)^{2}=81$


1. Sea $n$ un entero positivo, pruebe (sin asumir el T de Dirichlet) que hay infinitos primos de la forma $nk+1$ con $k$ entero positivo.

2.  Los enteros $a$ y $b$ tienen la propiedad de que $2^na+b$ es un cuadrado perfecto para todo entero positivo $n$. Pruebe que $a=0$.

3. Sean $m$ y $n$ enteros positivos tales que: $A=\frac{(m+3)^n+1}{3m}$ es un entero. Pruebe que $A$ es impar.

4. Sea $f(x)=x^3+17$ pruebe que para cada $n$ existe un entero $t_n$ tal que $3^n||f(t_n)$.

5. Muestre que existen infinitos enteros positivos n tales que $n^2+1$ divide a $n!$.

6. Sean $p_1,p_2...,p_n$ primos distintos mayores que $3$. muestre que el numero $2^{p_1p_2...p_n}+1$ tiene por lo menos $4^n$ divisores.

Algunos Hints, a ver si alguien postea alguna solucion:

1. Considere el n-esimo polinomio ciclotomico.

2. Considere la ecuacion de Pell $x^2-2y^2=-b$ y mire que los numeros $2^na+b$ son soluciones consecutivas de un punto en adelante.

3. Mire los casos en los que A puede ser par, ataquelos usando reciprocidad cuadratica y congruencias.

4. Haga induccion, teniendo $t_n$ construya $t_{n+1}$.

5. tome un $n$ tal que $n^2+1$ tenga divisores conocidos o por lo menos acotables y menores que $n$.

6. HAga una induccion cuidadosa, ademas $(2^p+1,2^q+1)|3$ para $p$ y $q$ primos distintos y mayores que $3$.


For $ n\geq2$, prove that:
$ \sqrt[n]{n\plus{}\sqrt[n]{n}}\plus{}\sqrt[n]{n\minus{}\sqrt[n]{n}}<2\sqrt[n]{n}$

Let $ a\equal{}\sqrt[n]{n\plus{}\sqrt[n]{n}}$ and $ b\equal{}\sqrt[n]{n\minus{}\sqrt[n]{n}}$
$ \frac {a \plus{} b}{2} \le \sqrt [n]{\frac {a^n \plus{} b^n}{2}}$ With equality when a=b, so we are done.


Can u qualify for USAMO if u have an index of 220 or higher
but didn't make the cut off line for 160 highest AMC 12 score?
(I took AMC 12)

Most certainly. There is no "cutoff" for AMC12 scores iirc. You do need to make the cutoff for AIME scores, and an index of 220 or above should get you through fine.


now since olympic has started,i felt the need for a thread to discuss about some things:

1.Will india get even a single medal?????
(i have hopes from rathore only)

2.Will record of Mark Spitz will be broken??

i wud say gagan narang is only other hope

and Phelps,well doesn't the man get tired of winning golds????
i can't say abt him breaking spitz's record so early,but surely he wud be at top of all time gold medallists list :D

hey no non on  i am astrict follower of the olympics and would like to all of u that hopes are from
RAJYAVARDHAN RATHORE
GAGAN NARANG
SAINA NEHWAL
DUO OF PAES AND BHUPATHI
DOLA BANERJEEE
SKIJHU MAKHEJA
AND ONLY SOME OTHERS...............

didn't india get a medal in weighlifting, of running???? one time??? my dad says so. i hope they win again, is there anyone in swimming. michael phelps broke world record!!!!!

there is one but he lost in 200m and will battle in another round?

YAY whoever he is, at least he got in!!!!!

Abhinav Bindra won  :first:
hats off to him  :winner_first:

by the way,i saw Nehwal's match(free time for me,festive season going on  :P )
awesome performance,she defeated world rank 5  :clap2:

another hope shaping up for india(at least she is only 18 and thus we can see her play in next two olympics as well)
go ahead
she made me feel bad for her,compare her position with sania mirza
so much attention and so bad performance from sania :mad:
and so little media attention but so stable performance from Nehwal :D

i don't have hopes from Paes -Bhupati duo this time :(

the main thing in Sania is her image management and certainly,she is more glamorous(you know what it means :wink: ) than saina

Hats off to Bindra

I bet Phelps will win 8 golds

watch out for my favourite CHAMPIA and RATHORE they r bound to do well!

[quote="rt_08"]hey no non on  i am astrict follower of the olympics and would like to all of u that hopes are from
RAJYAVARDHAN RATHORE
GAGAN NARANG
SAINA NEHWAL
DUO OF PAES AND BHUPATHI
DOLA BANERJEEE
SKIJHU MAKHEJA
[/quote]

that doesn't includes Bindra "THE CHAMP" :rotfl:
who is makhija :huh:

Champia is archerer
he has qualified 2nd for finals :roll:
expected to win a medal :rotfl:

my bloody :rotfl:  :rotfl:  :rotfl:

Champia is "HE"

INDIA ROX

12th in standings currently
best ever....

MICHAEL PHELPS MADE LEGEND. YAY. :first:  :winner_first:  :10:

akhil lost :(  :(

anyways good job!!!!!!

Phelps rocked!!!!!
hats off to him

YAY!!!!!!!

We won!!!!!!!!

another medal :winner_third:
Htas off to I don't know the name :oops: (i saw just on TV)

sushil kumar and also a medal assurance for vijender(my favourite)

vijeneder lost to his cuban world no 1 8-5 it was close but he still he did his level best  :(
but still bronze to him :)
also waiting for 4*400m relay to start on at 5:10 pm today.

hey
paddlers are TT guys. :P

u don't even understand jokes :rotfl:

thank god--the bloody server was down 4 4 days :mad:

good

mine was also down for some days

bsnl s**ks :mad:

well played Vijender
 :clap: Better luck next time :P

hey womens relay team came seventh in the heats and failed to qualify :(
yay jamaicans(mens) got another gold (4*100m) relay!

r u a jamaican supporter now????? :rotfl:

Bolted ,isn't it :P

hey i respect the jamaicans especially their will to always win and also their hard work
and by the way i support virgin islands :rotfl:  :rotfl:  :rotfl:


inequalities
function
Cauchy Inequality
inequalities proposed

[quote="zhaobin"]$a+b+c=3,a,b,c \ge 0$
find the minimum of $\frac{a^2+b^2+c^2}{a^2b^2+b^2c^2+c^2a^2}$
maybe it is easy hope you like it[/quote]

  Ok i'll produce a solution for your problem.
From $a+b+c=3$ we have : $b+c=3-a$(1)
Also from Cauchy inequality: $b^2c^2 \leq ({ \frac{b+c}{2}})^4=({ \frac{3-a}{2}})^4$
Assume that a=min{a,b,c} so 0 < a < 1
So we can rewrite : $M= \frac{a^2+x}{a^2x+b^2c^2} \geq F(x)= \frac{a^2+x}{a^2x+({ \frac{3-a}{2}})^4}$
with $x=b^2+c^2 \geq \frac{(b+c)^2}{2}= \frac{(3-a)^2}{2}$
But $F'(x)>0$( for all 0<a < 1 )( self check ) so $F(x) \geq F( \frac{(3-a)^2}{2})$
From here i think it's easy,i found that the minimum value occured at b=c and a is the root of  a 4th degree equation( sorry for my lazyness really that i do not wish to solve it) $3a^4-10a^3+24a^2+18a-27=0$ hope no mistake will be found from my solution

[quote="Kummer"][quote="zhaobin"][quote="Kummer"]
But $F'(x)>0$( for all 0<a < 1 )( self check ) [/quote]
I think it is wrong here  ;)
check yourself ;)
thanks for reply :)[/quote]

  Sorry i don't think i was wrong in this step...[/quote]
sorry,then I was wrong here. :(
but I think you must be something(I will check it later)
because my answer is very simple. ;)

[quote="Kummer"] $3a^4-10a^3+24a^2+18a-27=0$ hope no mistake will be found from my solution[/quote]
ok,now I think $(a)$ get its maximum when satisfy $3a^4-10a^3+24a^2+18a-27=0$ :)

We can also find maximum , the equality occurs when $a=b$ too. Let have a try.  :)

[quote="Kummer"]Sorry but i think u again mistake Zhaobin,minimum only not maximum[/quote]
Here is a nice picture for you.hope you like it  ;)

to AC ;)
for the orginal problem the maximum is $+\infty$

[quote="zhaobin"]

to AC ;)
for the orginal problem the maximum is $+\infty$[/quote]
Ompp.., yes, $a=b=0$ and $c=3$  :blush:
About the original problem, I find that $\frac{a^2+b^2+c^2}{a^2b^2+b^2c^2+c^2a^2} \geq \frac{8}{9}$, equality occurs when $a=0,b=c=\frac{3}{2}$  :)

Ok,my solution must work( because it's mine ) ,a compliment for AnhCuongs solution,really it's hard for me to find a such nice and elementary proof like this.I have  forgotten manything  :blush: but i do not wish to study again :D .With a lie on a closed interval [0,1] and a function is continuos there must be exist a minimum value.Oops( go to sleep,i'm too sleepy )


algebra
polynomial
induction
quadratics
algebra proposed

Let {P_n} be a sequence polynominal defined by
      P_0(x)=P_1(x)=1/2
      A(x)=2x 2 +12x+17
      P_n+2(x)=2A(x)P_n+1(x)-P_n(x)
  Prove that 2(x+3)P_n(x)-1=(x+2)H 2 (x)  with H(x) is a polynominal with integer cofficients

Not difficult!
The polynomial sequence is similar the sequence on real number .
We use the result :
  a_(n+2)=a*a_(n+1)+b*a_n then a_(n+1)^2 -a*a_(n+1)*a_n  -b*a_n^2=c*(-b)^n, where c is const (*)
denote A=A(x),P_n=P_n(x)
It is easy to have :2(x+3)P_n(x)-1=(x+2)Q(x) , Q in Z[x]
Applying (*) we have : (P_(n+1),P_n) is root of equation :
        X^2 - 2AXY + Y^2 - (1-A)/2 = 0
Hence delta =(AP_n)^2 - P_n^2 + (1-A)/2 = R^2 ,    R in Z[x]
or        :2(x+2)(x+4)[4(x+3)^2P_n^2 -1)=2R^2
Note that x+2 , x+4 are irreducible polynomial so :
        [2(x+3)P_n-1 ][2(x+3)P_n+1] =(x+2)K^2      K in Z[x]
Hence
    2(x+3)P_n-1=(x+2)H^2

That's right.
I have this problem after solving a problem about sequence of integer in Rom 2000
Polynominal and integer have some similar property.
Sequence is too.

1/I think (*) is very well-known result:
 Proof : a_(n+2)^2 - a*a_(n+2)*a_(n+1)-b*a_(n+1)^2
      = (a*a_(n+1)+b*a_n)^2 -a*(a*a_(n+1)+b*a_n)-b*a_(n+1)^2
      =...
      =-b(a_(n+1)^2 - a*a_(n+1)*a_n-b*a_n^2)
      =...
      =(-b)^(n+1)(u1^2 - a*u1*u0-b*u0^2)
      =(-b)^(n+1)*c (we known : u1 and u0)
2/prove :Q(x) is in Z[x]
  clearly we have 2(x+3)P_n(x)-1 is in Z[x] hence Q(x) in Z[x]
3/prove R(x) is in Z[x]
  I said that :"The polynomial sequence is similar the sequence on real number"
Note that :  P(x) is in Z[x] equivallent a in Z
We always have X^2 - 2AXP_n + P_n^2 - (1-A)/2 has root P_(n+1)is polynomial with integers coefficients
(equivallent the quadratic equation has root integer number)
Thus delta must is a square of plynomial with integers coefficients
so R is in Z[x]

Now,everything is clear!! :)  :)

Thank's A1lqdSchool for your anwsers
you are a master  :D  of sequence polynomial


Let $a,b,c\leq 4$ be integers  . Solve the equation
$ab+bc+ca=abc$

[hide]

$ab+bc+ac=abc$
$\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

for $a=b=c=3$
we also have $a=2,b=c=4$

also if $a=1$ then $b=-c$, so we have
$a=1,b=\pm1,c=\mp1$
$a=1,b=\pm2,c=\mp2$
$a=1,b=\pm3,c=\mp3$
$a=1,b=\pm4,c=\mp4$

Hmm thats what i have! Any suggestions..
[/hide]

Yay snow day!!! (lol I dont even know why our county gave us the day off its not even that bad..but our universities are closed too)


for distinct natural number $a_1,a_2,..,a_n (n >1)$ and $i=1,2,3,...,n$ we assume :
                        $p_i= \prod(a_i-a_j)$ ($1\leq j\leq n  ,j <> i$)
prove that for any $k\in N$ :
                     $\sum \frac{a_i^k}{p_i}$  is natural.

Yes, very nice. Let us take $ f(X)=(X-a_1)\cdot(X-a_2)\...\cdot (X-a_n) $ and let r(X) be the remainder of $ X^k $ when divided by $ f(X)$. Let us decompose into simple fractions the expression $\frac{r(X)}{f(X)}=\frac{A_1}{X-a_1}+...+\frac{A_n}{X-a_n} $. Then we find by identifying that $ A_i=\frac{a^k_i}{p_i}$. Thus we have $\frac{ Xr(X)}{f(X)}=\frac{a^k_1}{p_1}\cdot\frac{X}{X-a_1}+...$ and all we need now is to make X very large and to see that r(X) has integer coeficients.


A turn consists of rolling a standard die and tossing a fair coin. The game is won when the die shows a 1 or a 6 and the coin shows a head. What is the probability the game will be won on or before the fourth turn? Express your answer as a common fraction.

On any given turn, the probability that you win is $ \frac13\cdot\frac12\equal{}\frac16$, so the probability that you don't win is $ \frac56$.

The probability that you don't win within the first four turns is $ \left(\frac56\right)^4\equal{}\frac{625}{1296}$.

Since you must eventually win, the answer is $ 1\minus{}\frac{625}{1296}\equal{}\boxed{\frac{671}{1296}}$.


Red, blue, green arcs are used to join pairs of $2n$ points such that there is exactly one arc of each colour at each point. Let $a,b,c$ be the numbers of red-blue, red-green and blue-green cycles respectively. Prove that $n+a \geq b + c.$

Pierre.

Do you mean we have 3-regular graph, edges of which are properly colored?

I have absolutely idiotic idea.

Let's remove all red edges. Then remaining graph consists of vertices of degree 2 and all eges are properly colored with blue and green colors. It follows that this graph is a union of cycles of length at least [b]4[/b], so number of blue-green cycles is at most $2n/4$, i.e. $b,c\leq n/2$.

That's all.

What is wrong this time?  :?


Show that the square of any integer is congruent to either 0,1, or 4 mod 8

[hide="lemma"]
$(k^{2}\mod 8)\equiv (k \mod 8)^{2}) \mod 8$
We just need to show that $(8n+k)^{2}\equiv k^{2}\mod 8$
$64n^{2}+16nk+k^{2}\equiv k^{2}\mod 8$
$k^{2}\equiv k^{2}\mod 8$
Q.E.D.[/hide]
[hide="the remainder"]Using our lemma, we merely need to check the squares of 0-7.
0-0
1-1
2-4
3-1
4-0
note that 5=-3, 6=-2, 7=-1
So we're done![/hide]

QED=quod erat demonstrandum ("which was to be demonstrated")

You generally use it at the end of a proof to show that something has been proven.


logarithms
number theory open
number theory

Is there infinite prime numbers $ p$ which have the form: $ p=n^2+1$ with $ n\in N$?

You are right : It is an open problem.  :D

Hardy and Littlewood have conjectured that the number $P(n)$ of such primes less than $n$ is :
\[P(n) \sim c \frac {\sqrt n} {\log n} \]
with $c = \prod (1 - \frac {(-1)^{\frac {p-1} 2}} {p-1}) = 1,3727...$ where the product is over al odd primes.

Pierre.

PNT has great ideas! This remembers me of another problem of him, that with $ 2^n+3^n$. By the way, did he solve it?  :D


Now, halfway through 2009, let's take a look back to the leaderboard at the very end of 2008 (a few minutes before the new year)...

...things have dramatically changed.

and now Jongy has 400 more countdowns won.

[quote="Mewto55555"]Of those top 10, 7 are now done with MC which explains their not appearing again.[/quote]

When you say "of those top 10", which catagory do you have in mind?

heh, maybe [hide="this?"] games won?[/hide]

I meant top 10 in the rating category which is the oft-agreed indicator of who is the best (versus who merely played the most).

In games won, 8/10 (9?) are done.

In cds won 8 (9?) are done.

In games played, 7(8/9?) are done.


inequalities
algebra
function
domain
email
real analysis
real analysis unsolved

$ U \in R^n$ is bounded, connected and $ \partial U$ is smooth.
$ m \equal{} [\frac {n}{2}] \plus{} 2$, $ X \equal{} H^m (U) \cap H^1_0 (U)$,
How to show: $ X \subset C^2(\overline U)$
(Evans said use the fact that any $ f \in W^k_p (U)$ can be approxomated by $ f_m \in C^{\infty}( \overline U)$ in  $ W^k_p (U)$ and I don't know how to use this hint)

I think I need to show  $ Df_m$ and $ D^2 f_m$ converge to $ Df$ and $ D^2 f$ uniformaly? But $ f_m \minus{} > f$ in  $ W^k_p (U)$ only means converge in $ L^2$, not uniform convergence.

or using sobolev's inequality, we can see if $ f \in X$, then $ f \in C^{{m \minus{} [\frac {n}{2}] \minus{} 1}, \gamma} (U) \equal{} C^{1, \gamma} (U)$, with $ 0 < \gamma < 1$,i.e. $ Df$ is holder continuous, but this doesn't imply $ f \in C^2(U)$...

Since $ X \in H_0^1$ then $ X \equal{} 0$ on $ \partial \Omega$. Can you prove that: for a $ C^{1,1}$ domain $ \Omega$ the subspace
\[ \left\{ {u \in C^2 \left( {\overline \Omega  } \right),u\left| {_{\partial \Omega }  \equal{} 0} \right.} \right\}
\]is dense in $ W^{2,p} \left( \Omega  \right) \cap W_0^{1,p} \left( \Omega  \right)$ for each $ 1<p<\infty$.  :D

[quote="bookworm_vn"]Since $ X \in H_0^1$ then $ X \equal{} 0$ on $ \partial \Omega$. Can you prove that: for a $ C^{1,1}$ domain $ \Omega$ the subspace
\[ \left\{ {u \in C^2 \left( {\overline \Omega } \right),u\left| {_{\partial \Omega } \equal{} 0} \right.} \right\}
\]
is dense in $ W^{2,p} \left( \Omega \right) \cap W_0^{1,p} \left( \Omega \right)$ for each $ 1 < p < \infty$.  :D[/quote]

This is easy, Evans book"PDE" Thm3 in 5.3.3 (p252) is stronger: $ \partial \Omega$ only needs to be $ C^1$, no need to be $ C^{1,1}$.

How do we proceed from here?

[quote="bookworm_vn"]Oops, I cannot find your question in the book by Evans :-?[/quote]
it is in the proof of Thm3 in 6.5.2 (Principal eigenvalue for nonsymmetric elliptic operators).

The statement appears in p341 top.

[quote="bookworm_vn"]Oh  :o , there is no Thm 3 in subsection 5.3.2. Something wrongs  :?:[/quote]

It is in p252, section 5.3.3, Thm3: (Global approximation by functions smooth up to the
boundary).

I have an idea. This comes from Schauder's estimate. I refer you to a book of H. Brezis, [i]Analyse fonctionnelle theorie et applications[/i],  Massion, Paris, 1992.

In this theorem, we assume $ \Omega$ is bounded and in $ C^{m\plus{}2, \alpha}$ for some $ 0 < \alpha <1$. For each $ f \in C^{m, \alpha}(\overline \Omega)$. Consider the following bvp
\[ \left\{ \begin{gathered}
   \minus{} \Delta u \plus{} u \equal{} f\left( x \right) \hfill \\
  u\left| {_{\partial \Omega }  \equal{} 0}. \right. \hfill \\
\end{gathered}  \right.
\]
Then
\[ u \in C^{m \plus{} 2,\alpha } \left( {\overline \Omega  } \right).
\]
Now, in order to apply this theorem, consider the above problem where $ f (x) : \equal{} \minus{} \Delta u(x) \plus{} u(x)$. When $ u \in H^m \equiv W^{m,2}$ we get $ f \in W^{m\minus{}2,2}$.

Since
\[ m \equal{} \left[ {\frac{N}
{2}} \right] \plus{} n
\]
then
\[ 0 < 1 < \left( {\left[ {\frac{N}
{2}} \right] \plus{} 2} \right) \minus{} \frac{N}
{2} < 2.
\]
This allows us to apply Sobolev embedding theorem which yields
\[ f \in C^{0,\alpha } ( \overline \Omega  ).\]
This means that $ u \in C^{2,\alpha } ( \overline \Omega )$. OK?

Oops, there is a mistake in my previous post. After reading the whole proof in Evans's book, I think that we can increase $ m$, I mean we can choose $ m : \equal{} {\left[ {\frac{N}
{2}} \right] \plus{} 3}$.

[quote="bookworm_vn"]Oops, there is a mistake in my previous post. After reading the whole proof in Evans's book, I think that we can increase $ m$, I mean we can choose $ m : \equal{} {\left[ {\frac {N} {2}} \right] \plus{} 3}$.[/quote]

I was thinking the same thing: use $ m : \equal{} {\left[ {\frac {N} {2}} \right] \plus{} 3}$ , then "general sobolev inequality" in Evans Thm6 in 5.6.3. immediatly imply $ X \subset C^2(\overline U)$. So I guess the book has a typo here.

Another two question,
1. what do you think about the statement after this: (same page p341 Evans), he claimed that the PDE $ Lu \equal{} f$ in $ U$ and $ u \equal{} 0$ on $ \partial U$ has unique weak solution in $ X$. This is true by strong maximum principle (because we just proved that $ X \subset C^2(\overline U)$, so we can use strong maximum principle ) and Fredholm alternative ( maximum principle imply there is no non-zero solution to homogenous equation, so...)

so we can define operator $ A: X \minus{} > X$ by $ A(f) \equal{} u$. (i.e. $ A \equal{} L^{ \minus{} 1}$). He then claimed that $ A$ is compact operator. He didn't give details why.
According to Evans book's earlier section, 6.2.3 Thm4 , in the proof of Fredholm alternative, he proved $ L_{\gamma}^{ \minus{} 1} \equal{} (L \plus{} \gamma)^{ \minus{} 1}$ is compact for some positive constant $ \gamma$.
This of course doesn't imply $ L^{ \minus{} 1}$ is also compact. I wanted to mimic his proof to proof but failed. Any idea?

2. still same page in Evans book, (p341), how to get equation (25) from (24)?

Since (22), then "Energy estimates" holds true in case $ \gamma \equal{} 0$, that is to say
\[ \beta \left\| u \right\|_{H_0^1 }^2  \leqslant B\left[ {u,u} \right]
\]and
\[ B\left[ {u,v} \right] \leqslant \alpha \left\| u \right\|_{H_0^1 }^{} \left\| v \right\|_{H_0^1 }^{}
\]for all $ u, v \in H_0^1$.

Now, $ u \in H_0^1$ is a weak solution of
\[ \left\{ \begin{gathered}
  Lu \equal{} f \hfill \\
  u\left| {_{\partial \Omega }  \equal{} 0} \right. \hfill \\
\end{gathered}  \right.
\]iff
\[ B\left[ {u,v} \right] \equal{} \left( {f,v} \right) \quad , \quad \forall v \in H_0^1 \left( \Omega  \right).
\]Put $ A \equal{} L^{\minus{}1}$. We will prove $ A$ is compact. Indead,
\[ \beta \left\| u \right\|_{H_0^1 }^2  \leqslant B\left[ {u,u} \right] \equal{} \left( {f,u} \right) \leqslant \left\| f \right\|_{L^2 } \left\| u \right\|_{L^2 } \mathop  \leqslant \limits^{Poincare} c_1\left\| f \right\|_{L^2 } \left\| u \right\|_{H_0^1 } .
\]This means that
\[ \left\| u \right\|_{H_0^1 }^{}  \leqslant c_2 \left\| f \right\|_{L^2 } .
\]Now
\[ \left\| {Af} \right\|_{H_0^1 }^{}  \equal{} \left\| u \right\|_{H_0^1 }^{}  \leqslant c_2 \left\| f \right\|_{L^2 }
\]means that the following
\[ L^2 \left( \Omega  \right)\xrightarrow{A}H_0^1 \left( \Omega  \right)
\]is continuous. Moreover,
\[ H_0^1 \left( \Omega  \right)\mathop \hookrightarrow L^2 \left( \Omega  \right)
\]is compact. Thus $ A$ is compact.

[quote="bookworm_vn"]Since (22), then "Energy estimates" holds true in case $ \gamma = 0$, that is to say
\[ \beta \| u \|_{H_0^1 }^2 \leqslant B[ {u,u} ]
\]
[/quote]

This is where I don't understand: Evans 6.2.2 Thm2 (ii) in p300. said

$ \beta \| u \|_{H_0^1 }^2 \leqslant B[ {u,u}] + \gamma \|u\|_{L^2}$ for some positive constant $ \gamma$. In his proof in p300,
I can't see why the condistion $ c > = 0$ will imply $ \gamma = 0$ works.

For example: in equation (6) in p300, if we know $ b^i=0$ for all $ i$, then $ c>=0$ will imply $ \gamma = 0$ works.  But in genetal $ b^i \not =0$, Can you give details? thanks.

[quote="kalazealot"][quote="bookworm_vn"]Since (22), then "Energy estimates" holds true in case $ \gamma = 0$, that is to say
\[ \beta \| u \|_{H_0^1 }^2 \leqslant B[ {u,u} ]
\]
[/quote]

This is where I don't understand: Evans 6.2.2 Thm2 (ii) in p300. said

$ \beta \| u \|_{H_0^1 }^2 \leqslant B[ {u,u}] + \gamma \|u\|_{L^2}$ for some positive constant $ \gamma$. In his proof in p300,
I can't see why the condistion $ c > = 0$ will imply $ \gamma = 0$ works.

For example: in equation (6) in p300, if we know $ b^i = 0$ for all $ i$, then $ c > = 0$ will imply $ \gamma = 0$ works.  But in genetal $ b^i \not = 0$, Can you give details? thanks.[/quote]

finnaly, I got a hint:

since $ c>= 0$, (by maximum principle), $ 0$ is not an eigen value of $ L$, so we can apply Thm6 in p306 to get :

$ \|u\|_{L^2}<=C\|Lu\|_{L^2}$.

So energy estimate (ii) in p300 gives:

$ \beta \|u\|_{H^1_0}^2 <= B[u, u] +\gamma \|u\|^2_{L^2}=<Lu,u> +\gamma \|u\|^2_{L^2}
<= \|Lu\|_{L^2} \|u\|_{L^2}+C^2 \gamma \|Lu\|^2_{L^2}<=C\|Lu\|_{L^2} ^2+C^2 \gamma \|Lu\|^2_{L^2}=C\|Lu\|_{L^2}^2$

so $ \|u\|_{H^1_0}<=C\|Lu\|_{L^2}$

i.e $ \|Au\|_{H^1_0}<=C\|u\|_{L^2}$ and since $ H^1_0$ compactly imbed in $ L^2$, $ A$ is compact.

[quote="bookworm_vn"]By concerning maximum principle, I have a question for you: why the interior ball condition automatically holds if $ \partial \Omega$ is of class $ C^2$?[/quote]

I had the same question when I read Hopf lemma proof on Evans p330. He tersely mentioned this on p330 without proof. So I did some sreach on internet: I found this paper has proof.

"Boundary Harnack Principle for p-harmonic Functions in Smooth Euclidean Domains", Potential Analysis, Volume 26, Number 3 / May, 2007, p281-301

Lemma 2.2: Let $ D \subset R^n$ be a bounded domain. Then $ D$ is a $ C^{1,1}$ domain if and only
if $ D$ satisfies the ball condition.

Here "ball condition" means both interior and exterior ball condition (with same radius $ r$).

$ C^2$ boundary is stronger than $ C^{1,1}$ domain, of course.

If you can't download this paper, give me your email. I can email you.


not at all

hint* (i done this one before)

[hide]just use congujates[/hide]

Yes, as zratonx said, you can use conjugates, reduce the equation to x+:sqrt:(x2-1) = 10, solve for x with algebra and plug it in to the other equation. However, if you have little time, you can also notice that the first equation is very similar to 4x = 20, and the last one is very similar to 2x2. Since the choices are pretty far apart, this will get you a number quite close to the correct answer: [hide]50 is the number you get, and 51.005 is the choice[/hide]

[hide]
Hmm...

$x + \sqrt{x^2-1} + \frac{1}{x - \sqrt{x^2-1}} = 20$
$(x + \sqrt{x^2-1} )+( \frac{1}{x - \sqrt{x^2-1}}) = 20$
$(\frac{2}{x - \sqrt{x^2-1}}) = 20$
$(\frac{1}{x - \sqrt{x^2-1}}) = 10$
$x + \sqrt{x^2-1} = 10$

Solving for x, we get x = 5.05

Plug in x into equation 2 to arrive at the answer ... someone else can do that... [/hide][/hide]


Has anyone been watching Jeopardy lately?  If you have you know that Ken Jennings is already a legend.  He has won for over three weeks and has accumulated over $500,000!  I wasn't able to watch Jeopardy last night but I heard he won again.  So much longer do you think he will last?
Personally I think he is smart but I think he will go down if the right person goes on there. I give him 2 more weeks.

I don't expect anyone to beat Ken Jennings.  I've seen a lot of movie categories on Jeopardy, and if you watched a couple of days ago, you would have heard him say that he had a website with his 2,000 favorite movies.  I couldn't watch last night, though, becuase some county far away from us had a tornado warning and there were covering it on the same channel the Jeopardy is on and at the same time.

I just found out that Jeporady is taped like 3 months in advance: http://entertainment.msn.com/tv/article.aspx?news=162272

So, his streak is over by now, but i think he might keep going for 2 more weeks.

lol, i liked that idea that he and his friends do at the movies. when alex was talking to the players, he told him that at the movies, he and his friend always clap when the characters say the name of the movie. that'd be great to try out in Garfield

[quote="hello"]For those that watch jeoprady how much do you usually get correct? What 's the record for the most appearences?[/quote]
by how many do you usually get correct, i usually don't pay much attention until final jeopardy. so the few times i do watch it, i can usually get at least 7 or 8 each round. more if there's a category on sports or video games. as for the record, i've gotta believe that he's setting the record right now, because they only changed the system a few months ago.

[quote="hello"]For those that watch jeoprady how much do you usually get correct? What 's the record for the most appearences?[/quote]

I usually don't watch Jeopardy either, but I can usually get 2 or 3 correct.

I must be real smart.

I can get maybe , eh , 10 on a good day? At least 10 faster than the other contestants - that is.

Ho hum.  Ken Jennings has won 27 days and has over $880,000.  Unsurprising.

[quote="h_s_potter2002"]Since Jeopordy is like taped 3 months or something before it's aired,  some newspaper has found out tha Ken Jennings has lost.

I'd like to se the guy who beat him.  He must either be really smart, or really lucky.[/quote]

Are you sure it was taped 3 months before?  A couple days ago, the host (I forget his name) commented about it being summer.

He still hasn't lost. He's only back home becuase the current season of Jeopardy is over and we have to wait like 5 months till the new season and KJ is back in action.

Yeah I know.  That is so crazy.  I was almost sure that he would lose at some time(soon that is).

[quote="h_s_potter2002"]It says [url=http://www.mercurynews.com/mld/mercurynews/entertainment/television/9117613.htm?1c]here[/url]   that the show's taped in Febuary, so he's already at his home, and has lost.[/quote]

lol, looks like he's still going.  ;)

Yeah, I saw the thing where he would be back for antoher season.

I wonder when he's going to lose though.

KJ's biggest enemy: his greed of money

Cuz you can't get your prize money until you lose, so he can't touch that money just yet.
Go here: http://boards.sonypictures.com/boards/showthread.php?s=a51c5ebde2ab5a246e9468c4a33513ea&threadid=83360 for a really funny KJ parody

I actually dreamt a couple nights ago that he'd lose after being about $1000 behind and missing his chance in the final round.

Of course, I also had predicted someone like Ken Jennings long before he showed up. So I don't know.... :D

Actually he will be back in 6 weeks as Alex said in the last episode. How did somebody here know that so far he would win 38 times? And where did you get all that February tapings stuff from? Don't you think that there will be many people having so many appearences because of the new rule that makes more then 5 appearences allowed. Ken started winning really soon after it happened.

This was announced yesterday but I'll put it in spoiler for those who don't want to know when Ken Jennings loses[hide]he will lose on his 74th night which will air sometime in October[/hide]

I heard that he lost on his [hide]75th[/hide]...

Anybody here read [i]Time[/i]?  They had a funny essay in the back about what might happen if he goes on winning for months and months.

[quote=mathleticguyyy]LONG BUMP[/quote]

JEEZ WHAT A BUMP

/bump................... also I still watch Jeopardy


What is the smallest positive integer than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?

[hide="Maybe you could call this a hint, but just the very first step"]The lowest number which can be expressed as $\frac{a(a+1)}{2}-\frac{(a-9)(a-8)}{2}$, $\frac{b(b+1)}{2}-\frac{(b-10)(b-9)}{2}$, and $\frac{c(c+1)}{2}-\frac{(c-11)(c-10)}{2}$ for integers a, b, c.[/hide]

[hide]
We have three sequences:

$a+(a+1)+(a+2)+...+(a+10)=11a+55$
$b+(b+1)+(b+2)+...+(b+9)=10b+45$
$c+(c+1)+(c+2)+...+(c+8)=9c+36$

All three sequences equal each other, so:
$11a+55=10b+45=9c+36$
$11a+19=10b+9=9c$

Since the sum is a multiple of $9$, we know that $b$ must also be a multiple of $9$.

$11a+19=10b+9$
$11a=10(b-1)$

$b-1$ must be a multiple of $11$, smallest possible vlaue of $b$ is $45$. The integer we are looking for is then $10(45)+45=\boxed{495}$.

[/hide]

[quote="chess64"][hide]
\[ n = a_k + (a_k+1) + \cdots + (a_k+k-1) = \frac{k}{2}(2a_k+k-1).  \]
\[ n = 9(a_9+4)=5(2a_{10}+9)=11(a_{11}+5).  \]

$\therefore \text{lcm}(9,5,11) = \boxed {495}$.
[/hide][/quote]What's with the Kalva Solution? At least make it more detailed...

[hide="Detailed Kalva Solution"]As shown in the previous solution, $n=9a+36=10b+45=11c+55$. This can also be written as $n=9(a+4)=5(2b+9)=11(c+5)$. Therefore, $n$ must be divisible by $9$, $5$, and $11$.

However, notice that $a+4=5(11)$ when $a=51$, $2b+9=99$ when $b=45$, and $c+5=45$ when $c=40$. This means that the smallest LCM of $(9)(5)(11)$, $9(a+4)$, $5(2b+9)$, and $11(c+5)$ is $(9)(5)(11)$, implying that $n=(9)(5)(11)=495$.[/hide]

Smallest LCM? Umm redudancy (LCM=least common multiple)

But yeah Mr. Scholes doesn't use LaTeX.

[quote="chess64"]Smallest LCM? Umm redudancy (LCM=least common multiple)[/quote]The LCM is dependant on the values of a, b, and c.  ;)

Well I think your whole second paragraph is unnecessary because if a number is the product is a multiple of a,b,c it must be divisible by LCM(a,b,c). You don't actually have to find a,b,c, since they're obviously positive integers. (anyway, the answer is between 0 and 999 :lol:)

Err..perhaps the least tedious and easy to see solution, idk

If the number is the sum of nine consecutive integers, it must be in the form:

$(x-4)+(x-3)+(x-2)+(x-1)+x+(x+1)+(x+2)+(x+3)+(x+4)=9x$

If the number is the sum of ten consecutive integers, it must be in the form:

$(y-4)+(y-3)+(y-2)+(y-1)+y+(y+1)+(y+2)+(y+3)+(y+4)+(y+5)=10y+5$

If the number is the sum of eleven consecutive integers, it must be in the form:

$(z-5)+(z-4)+(z-3)+(z-2)+(z-1)+z+(z+1)+(z+2)+(z+3)+(z+4)+(z+5)=11z$

This means that the number must end in $5$ and be divisible by both $9$ and $11$. Of course, the smallest number possible must be $11 \cdot 9 \cdot 5=\boxed{495}$

Least tedious, eh?

I think chess's solution is the best.  It could some words to explain, but otherwise it takes a nice general approach.

[hide]
try sum with fewer terms and do it like this: x-1, x, x+1, so the sum of consecutive three terms will be 3x.  Following this fashion, we have $9a = 10b+5 = 11c$.  So this number must be divisible by 9, 5, and 11.  Multiply these together and check: it works, so the answer is $495$.
[/hide]

This doesn't seem very hard for a #6, and why the fuss about this problem?

[b]SOLUTION[/b]

The number is the sum of nine consecutive terms if and only if it can be expressed as $x+(x+1)+(x+2)...+(x+8)=9x+36$. Similarly, a number is the sum of ten consecutive terms if and only if it can be expressed as $y+(y+1)+(y+2)+...+(y+9)=10y+45$. Lastly, a number is a sum of eleven consecutive integer if and only if it can be expressed as $z+(z+1)+(z+2)...+(z+10)=11z+55$. In other words, $9$, $5$, and $11$ must divide the number, so the it must be $\text{lcm} (9,11,5)=\boxed{495}$. We can see that $x=51,y=45,$ and $z=40$ satisfies the equation as well.