Recently I was thinking about the IMO and realizing that IMO problems seem much easier now than back when I was in high school.  Well, some of you may be thinking "well, of course they are, you know and understand math better".  But that doesn't paint the whole picture.  In high school I could solve some but not all of the IMO problems.  However, I rarely came across a solution to an IMO problem that I couldn't understand.  In other words, it's not just that I know more about math now than I did then.  I was equipped with nearly all the theorems and problem solving tools that I use to solve the same problems now.  Sure, I have more experience using them, but that doesn't explain the whole difference.

The difference that I am driving at is [i]investigation[/i].  What do I mean by that?  Think about the national olympiad problems you have worked through.  How often do these require more than 1 or 2 serious insights?  They rarely do because that would put solving them out of reach of too large a portion of the students.  Sure, there is usually a killer problem or two, but everything at the IMO seems like a killer problem to those who struggle to get 2 or 3 problems on the USAMO.

At the IMO, the problems often require a series of insights as serious as those that would individually crack open a national olympiad problem.  This means that you're not often going to sit down and recognize the solution to an IMO problem after 2 minutes of scratch work.  You must be patient with an IMO problem.  Try several avenues.  Write down everything you can figure out.  CONDUCT AN INVESTIGATION of the problem!

This kind of investigation is the first step toward doing serious research in mathematics or any mathematical field.  Today's unsolved problems have been attacked by many smart people before you.  In order to solve them you will have to be experienced at sitting in front of a problem and exhausting your toolbox of knowledge and tactics in an effort to crack them open.

Think about the way you felt when you first moved from simple contests to the AMC, then to the AIME, then to the USAMO (or other similar paths of problem solving).  Each step of the way you felt like you were undergoing a metamorphosis in your outlook of mathematics.  Part of what was happening is that you were maturing in your ability to investigate a problem and find a solution.  Learning to solve an IMO problem is no different.  Just be patient and investigate.

I definitely agree. In 9th grade, I knew all the math I needed to know for the AIME, but I still could scarecely get 4 or 5 on a given test. I could read all the solutions and understand them, however. Now I'm at the point where AIME questions seem easy, even trivial, USAMO problems seem difficult, and all but numbers 1 and 4 on the IMO seem impossible. But I can still understand all the solutions.

The problem is that there are very few places that one can read investigations of problems. Certainly there is no shortage of proofs to problems, but proofs do not help one who wishes to know how they could have solved the problem. This is the value of books like AoPS and ACoPS, as well as this website; they allow people to see examples of investigations, not just solutions. And of course solutions are more use when read between the lines for the investigation behind it, as I believe Rusczyk once told someone on this forum.

It is certainly a matter of patience to learn to investigate, and there is a lot of mental training as well. You have to go in believing it is possible, as corny as that sounds. For example, when I first read AoPS, a long time ago, I would never be able to to solve any of the IMO problems included. This was not necessarily because they were hard, it was because I would see IMO and think "IMO! IMPOSSIBLE! ABORT NOW!" It is a first step in itself to see the solution as something you really can find if you look for it properly. In that respect it is important to realize from the beginning that you know all the math necessary to solve the hardest problems.

Philosophical tangent: That to me is what makes math so beautiful as an academic subject. The truths of math can be found with the mind alone, without any special aids beyond a pencil and paper. So, for example, one who forgets a particular formula on a math test might rederive it themself on scratch paper, but one who does not remember the succession of British Stuart monarchs can not rediscover such information without a book.

There is a book called "Seeking Solutions" that goes over investigations of IMO problems over 3 consecutive IMOs (88 - 90).
It's well written and understandable.

Isn't that a TV show? (Maybe it only airs in New England since it's on the Comcast Channel)

ive never heard of it but it probably is a common alliteration.

I found out something about the IMO when I was brousing through the USAMTS website. I shall say no more 8-) (seriously)


function
algebra
polynomial
real analysis
real analysis unsolved

Can we find a continuous function g(x,y) on [0,1]^2,
such that given any continuous f(x) on [0,1] and any e>0, there are points y0,y1,...yk in [0,1] and real coefficients a0,a1,...ak, such that
|f(x) - Sum{ak*g(x,yk)}| < e for all x in [0,1]?

[hide="almost, but..."]$g(x, y) = e^{-2\pi i x/y}$

the $y_{k}$ will be $\frac{1}{j}$ and the $a_{k}$ will be Fourier coefficients of $f$.[/hide]
...my function has a problem at (0, 0)... :P :(

Xevarion: multiply your example by $y$. Also, Fourier coefficients will not work, since the Fourier series of a continuous function need not converge to it. There are better choices of coefficients.

With the modification to make it continuous, Xevarion's function works; the trigonometric polynomials are dense in $C[0,1]$. If you prefer polynomials, you can use something like $g(x,y)=yx^{\frac1y}$.


A play has two male roles, two female roles and two roles that can
be either gender. Only a man can be assigned to a male role, and
only a woman can be assigned to a female role. If five men and six
women audition, in how many ways can the six roles be assigned?

First we will take care of the male roles.

$ 5 \times 4 \equal{} 20$, since the roles are distinct.

Now for the female roles.

$ 6 \times 5 \equal{} 30$, because once more the roles are distinct.

Finally, the roles that can be either gender. We have $ 11 \minus{} 4 \equal{} 7$ people left.

$ 7 \times 6 \equal{} 42$

$ 20 \cdot 30 \cdot 42 \equal{} \boxed{25200}$


I have a question about magnetic field lines in a motor-like situation. Suppose I have two horizontal bar magnets, with one north end parallel to x=-X and one south end parallel to x=X, as here:

-- --
| |
| |
N | | S
| |
| |
-- --

(forget the 'as here', the board is stripping out my nice spacing...)

If I have one or more current-carrying wires in the intervening space (not near the poles), can I still be sure that the magnetic field line from (-X,y) will end at (X,y)? In my model, this isn't happening, but I'm not sure if this is due to numerical error, some other error in the model, or whether it's actually expected real-life behaviour. I've made every correction I can think of to the line integration, and the lines are still warping significantly.

Thanks

The only thing you can be generally sure of is that the magnetic field lines are closed - if you start at (-X,y) and make one circuit along the magfield lines you'll again end up in (-X,y). Without the wire, you know that you'll be going through point (X,y) due to certain symmetry of the system. If you insert the wire so that the system remains symmetrical in the proper way (it's up to you to figure out what that means :)), this should still hold, otherwise it probably won't.

Just imagine a situation where the wires cause the magfield to be weaker near one end of the magnet and stronger near the other. Then the magfield lines are denser near one end and thinner near the other end so it is quite obvious that the same line won't start at (-X,y) and end at (X,y).

The magnets are collinear , with S facing N. The wires are looped, but  ignore that, so consider them as two infinitely long wires with oppositely facing currents. (incidentally, I think I know the answer to my question: the fields don't return to the same place, but are drawn up or down)


topology
advanced fields
advanced fields unsolved

Do there exist a topological space which is Hausdorff, path connected and countable infinite?

Nope :). See for example post #14 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=hahn+mazurkievicz&t=34104]this thread[/url] for a proof that a Hausdorff continuous image of $[0,1]$ must be metrizable. If we restrict our attention to a path in our space which has at least two points, we thus find it to be a metrizable countable space (with at least two points), and such a space cannot be connected:

Take a point $x$ in such a space $X$, and let $(r_n)_n$ be the sequence of distances from the other points to it. Then pick $r>0$ different from all the $r_n$ but for which there are both larger and smaller $r_n$'s, and $X=\{y\in X\ |\ d(x,y)<r\}\cup\{y\in X\ |\ d(x,y)>r\}$ will be partition of $X$ into disjoint open sets.

Doesnt Hahn-Mazurkievicz theorem characterizes Peano Spaces.

P.S.: Can any body post any reading material on peano spaces.

What about if our space is non-metrizable?

I looked only at a subspace consisting of a path, and that space [b]is[/b] metrizable, as shown in that other thread. Like I said in my previous message: a Hausdorff continuous image of $[0,1]$ is metrizable.

So you proved that if space is metrizable it can't satisfy all the properties I gave. OK, now I know such a space is non-metrizable. I also heard from a friend that he had heard that there is a space satisfying the properties I gave but he don't know an explicit construction for such a space.

I think you’re still misunderstanding. What I proved is that there are no spaces with all the properties you mentioned. I’ll try to make that clear (once more :)).

Let your space be called $X$. It’s path connected and countably infinite (all we need is that it has more that one point), so there is a path $f: [0,1]\to X$ in it consisting of at least two points. Now look at the subspace $Y=f([0,1])$ of $X$. It’s a Hausdorff continuous image of $[0,1]$, so, by that theorem that I mentioned, it is metrizable. The rest of the proof takes place in $Y$: we simply forget about the whole of $X$ and restrict our attention to $Y$, which we know for sure is metrizable.

Is it clear now?

Another argument: If the closed unit interval $[0,1]$ is the disjoint union of countably many closed sets, one of those sets is the whole interval $[0,1]$.
Using this powerful theorem, we have as a corollary that if $X$ is a path-connected space with countably many elements, and single-points subsets of $X$ are closed, $X$ has only one element.

[quote="jmerry"]Another argument: If the closed unit interval $[0,1]$ is the disjoint union of countably many closed sets, one of those sets is the whole interval $[0,1]$.
.[/quote]
You are using Baire category theorem rt.It says that one of these sets will contain an interval but how can you say that it is [0,1]

Let $X_n$ be the closed sets. For each subinterval $I$ of $[0,1]$, some $X_n$ contains a subinterval of $I$ by restriction of the category theorem. Let $Y=\cup\text{ Int}(X_n)$. $Y$ is a dense open set, so it is a countable union of open intervals $(a_n,b_n)$. No two of those intervals can have the same endpoint because $[b_n,c_n]$ is in some $X_n$, so $A=[0,1]\setminus Y$ has no isolated points. In addition, if $\alpha\in A$, every neighborhood of $\alpha$ contains members of at least two of the sets $A_n=X_n\cap A$ (since $\alpha$ is not in the interior of any $X_n$, the neighborhood intersects at least two sets, and each set either intersects $A$ or contains an interval of $Y$, and then the endpoints are in $A$).

Now the kicker: we apply the Baire theorem to $A_n$ in $A$. The $A_n$ are nowhere dense, so their union cannot be all of $A$. This contradiction shows that $A$ is empty, and some $X_n$ is $[0,1]$.

I didn't know a proof until I put it together just now; I'm not really sure why I was so sure it was true before.

It's been discussed before, in a more general form. I don't have the time to recheck the proofs in the following thread, but I hope they are correct:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=compact+AND+connected&t=25908[/url]


algorithm
number theory
relatively prime
greatest common divisor

Sure, it's IMO.

Prove that $\frac{21n+4}{14n+3}$ is irreducible for all natural numbers n.

There are two ways to do it.

[hide="IMO problems were once this easy?!"]
Say that a prime $p$ divides both $21n+4$ and $14n+3$.  Then it also divides their difference, $7n+1$.  It thus divides $2(7n+1)=14n+2$.  But it also divides $14n+3$.  Consecutive integers are relatively prime, so this is a contradiction.  Thus the fraction is irreducible.
[/hide]

Wow. A third way! :o

Here's my way:

[hide="second solution"]Let's just say that 21n+4 and 14n+3 have a greatest common factor greater than 1. We can use Euclid's algorithm:

(21n+4,14n+3)
(14n+3,7n+1)
(7n+2,7n+1)
(1,7n+1)

uh-oh. The fraction can't be reduced.[/hide]

Lol my original solution had this tons of ugly stuff and it was super long and then the officers of math team were like

[hide]$p|21n+4$ and $p|14n+3$ implies $p|3(14n+3)-2(21n+4)$ implies $p|1$ which is a contradiction.[/hide]


Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
 \[ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac 1{c} \plus{} \frac {8}{a \plus{} b \plus{} c} \ge \frac {17}{3\sqrt[3] {abc}}\]

Well, put a+b+c=p,ab+bc+ca=q, and abc=r
sorry for no latex. I will learn it a bit later.

We have to prove q/r+8/p>=17/3*r^(1/3)
or q/r-17/3*r^(1/3)+8/p>0

Let us fix p,q and move r.
differienting by r, we can easily see that it is a increasing function on r.
with fixed p and q, r takes extremal values on a=b or cyclic(can be shown by many arguments)

so it is sufficient to put a=b
because of homogeneousity, putting a=b=1 and c=t and analyzing function f(t) will work.

(sorry for the incomplete solution :( , It's just idea)


I suggest that discussion pages must be started for the articles before anyone starts contributing to the articles.

This way the articles can be written in a more organised way.

Also the articles will be more uniform if a single person has written it although there might be minor edits and additions in the future.

A little organization is good, and I recommend more of it for the future.

However, I think we're currently in a stage in which users are learning to use the wiki by creating article stubs.  They can be rewritten and reorganized later, which is part of the beauty of the wiki.  Otherwise, we could just develop permanent resources pages, organized with content through the forums.

Uniformity doesn't seem to be such a big issue in wikis so far as I can tell.


Bueno el que sea colombiano hable aqui!!! es solo para ver quienes estan y quienes faltan por llegar y asi hacer la comunidad colombiana aca!

Suerte! :D

Yo estoy aca vivo, y atento a lo que pase, toca seguir insistiendo a todos los de olimpiadas en colombia de unirse aca, para tener el orden y dedicacion que el grupito yahoo no tiene
 :|
Y pues los que estemos, vamos que nos quedan todavia muchas olimpiadas y tiempo para prepararnos.

QUe mas! Espero que se metan mas a esta vaina. Bueno pues suerte,

JAS

Soy Juan Sebastián Marulanda Puldio del Colegio San Pedro Claver de Bucaramanga, voy para centro este año

Bueno ya que la IMO terminó tenemos un año completo para seguir entrenando así que deberíamos organizarnos para entrenar. Ya todos deberían estar participando en el reto de la semana y los invito también a organizar algo para estudiar teoría.

Saludes a todos.

excelente tenerlos a todos aqui, aun no veo a varios, Acosta, federico, fabian, Camacho, Cifuentes, ochoa, faltan muchos

Yo se que acosta, federico y ochoa estan en el foro. A los demas les e dicho muchas veces que se metan y ninguno a querido ensayar. Toca seguir molestandolos hasta que se decidan a meterse. Pero no seria malo si Pascual les dice porque a Cuellar y a mi no nos hacen caso.

Bueno cree un blog (De mathlinks) para los colombianos para que montemos material de entrenamiento aver si nos ponemos serios con esto. Porfavor todos los articulos que valgan la pena mandenmelos para subirlos al Blog.

http://www.mathlinks.ro/Forum/weblog.php?w=493


calculus integration factorial function search algebra domain

I'm new to this forum, so first of all .... HI EVERYBODY !!!

Yesterday I was following some calculation on a paper and I couldn't get rid of some steps, there should be some rule or theorem I' ve totally forgotten.... Here the incriminated equivalence:

[tex] \int_0^T dx \int_0^T dy f(y,x)  =  2! \int_0^T dx \int_0^x dy f(y,x)  [/tex]

Where comes from the factorial?????

Many many thanks

Vale :bye:

Thanks for your fast replay!
Actually my function takes exponential form.. , but could you give me any reference or just a name of what I have to search for... I gues that the factorial has to play the role of a correction factor... but I don't see clearly how it comes out  :?

Maybe I can easly guess that if I wuold have three integrals, I would end up with a [b]3! [/b]......

many tanks vale!!

sorry,does you function such that $f(x,y)=f(y,x)$?

Actually it is a trick to rewrite the square of an integral... maybe my notation hasn't been so correct... :oops:

Let's suppose we have
[tex] \left(\int_0^T f(x) dx \right)^2 [/tex]
This is equal to
[tex] \left(\int_0^T f(x) dx \right) \left(\int_0^T f(y) dy \right)  = \int_0^T \int_0^T f(x)f(y) dy dx [/tex]

So why should be right to write

[tex] \int_0^T \int_0^T f(x)f(y) dy dx = 2! \int_0^T \int_0^x f(x)f(y) dy dx  [/tex]

vale

This rule is becoming really an obsession... I mean it seams everybody applaied it, and so you can tell me, do as well , I want to know why... :(

by the way I've found another application in the general case of the power n of an integral... now I better get the rule, but no mention of the theorem that enables you do to that way  :huh:  ... Maybe it's trivial for evrybody except me   :stink:

[tex]  \left(\int_0^T f(x) dx \right)^n  =  \int_0^T \int_0^T ... \int_0^T f(x) f(z)... f(y) dy...dz dx  = n! \int_0^T \int_0^x ... \int_0^z f(x) f(z) ...  f(y) dy ...  dz dx   [/tex]

where it is specified that the integrating variables are in descending order!!!! That is : [tex]y< ... < z < x  [/tex]

Can you tell me what's going on, or simply give me references?  :?
Any idea?
Many many many ... thanks!
vale

What's going on is that you split the domain of integration $[0,T]^n$ into $n!$ subdomains each of which corresponds to some particular ordering of $x,y,\dots,z$. For instance, if $n=3$, you get $6$ possible orderings: $x>y>z$, $x>z>y$, $y>x>z$, $y>z>x$, $z>x>y$, and $z>y>x$. (the set where two of your variables are equal has measure $0$ and can be safely forgotten). Now, according to this partition, the initial integral is represented as a sum of $n!$ integrals. The last thing is to notice that all those integrals are actually the same up to the notation of the variables (for $n=2$,say, you have $\iint_{0<x<y<T}f(x)f(y)\,dx\,dy=\int_0^T f(y) \int_0^y f(x) \,dx\,dy$ and $\iint_{0<y<x<T}f(x)f(y)\,dx\,dy=\int_0^T f(x) \int_0^x f(y) \,dy\,dx$; this integrals are, clearly, equal (just denote $y$ by $x$ and $x$ by $y$ in the first one and you'll get the second one). Thus, the sum of $n!$ integrals is the same as $n!$ times one of them, for example, the one corresponding to the ordering $x>z>\dots>y$. Hope this helps :).

[size=200][color=red][i] Thank You ! [/i][/color][/size]

Lovely explained :thumbup: .... Now seams so obvious .. :oops:

Bye bye and thanks to you all,
Vale

Hi guys!

I was just wandering if it exists some sort of similar rearrengement integral formula, also in the case of [i]fractional[/i] power of a given integral...

In short, what can we say about $ \left(\int_0^T f(x) dx \right)^\alpha$ where $\alpha$ is not an integer ? I was traying a sort of similar reasoning (integer case), adjusting with the gamma funtion in place of the factorial... but i don't think it could work ... and I'm not so sure what happens to the domain of integration, when for instance you take its square root .....

Well, I don't know.... I put it to you as I was thinkin it... If you have any clues  :idea:  about it ... I will appreciate so much...

:bye: vale


complex numbers
algebra unsolved
algebra

Suppose $z_1, z_2 , \cdots z_n$ are $n$ complex numbers such that $min_{j \not= k} | z_{j} - z_{k} | \geq max_{1 \leq j \leq n} |z_j|$. Find the maximum possible value of $n$. Further characterise all such maximal configurations.

consider all lines passing through $(0,0)$ and at least one complex point $z_i$ (which is not $(0,0))$. These lines can't pass through two distinct points $z_i,z_j$ because obviously we would have $|z_i-z_j| < max\{|z_i|,|z_j|\}$.
Let $p_i$ be the line passing through $(0,0)$ and $z_i$. We can see that the angle between two consecutive lines
$p_i,p_j$ is not smaller than $\frac{\pi}{3}$. Otherwise we would obtain a triangle $(0,0),z_i,z_j$ where the longest
(other sides are shorter) side is equal to $\max\{|z_i|,|z_j|\}$. Thus we can have at least $[\frac{2\pi}{\frac{\pi}{3}}] \leq 6$, but if there are such $6$ points then they form a regular hexagon ( go to (i) ). Adding point $(0,0)$ which suffices our conditions, we obtain $7$ complex points - $6$ are vertices of a regular hexagon.

(i) And finally, if each angle between two consecutive lines is $\frac{\pi}{3}$ and $|z_k|=max\{|z_i| : 1 \geq i \geq 6\}$ (let's say $k \geq 2$)
and we name consecutive points $z_1,z_2,...,z_6$ then  if $|z_{k-1}-z_k| \geq |z_k|$ then $|z_{k-1}| \geq |z_k|$ so $|z_{k-1}| = |z_k|$ and this way we get that $|z_1|=|z_2|=...=|z_6|$ - so it must be a regular hexagon.


Find all pairs of positive integers $ (x,y)$ such that $ (xy^2\minus{}1)|(x^2\plus{}y)$

$ xy^{2}\minus{}1|x^{2}\plus{}y$

$ x^{2}\plus{}y\equal{}kxy^{2}\minus{}k$,$ x^{2}\minus{}kxy^{2}\plus{}k\plus{}y\equal{}0$

$ k^{2}y^{4}\minus{}4(k\plus{}y)$ must be perfect square;

if ;$ (ky^{2}\minus{}1)\equal{}k^{2}y^{4}\minus{}2ky^{2}\plus{}1<k^{2}y^{4}\minus{}4(k\plus{}y)<(ky^{2})^{2}$  $ \implies$  We get contradiction

$ \implies$  $ (ky^{2}\minus{}1)^{2}\geq k^{2}y^{4}\minus{}4(k\plus{}y)$

$ \implies$  $ 4(k\plus{}y)\geq 2ky^{2}\minus{}1$

$ k,y$ $ \in\mathbb{Z}$ $ \implies$  $ 2(k\plus{}y)\geq ky^{2}$

$ k(2\minus{}y^{2})\geq \minus{}2y$  if $ y^{2}>2$  $ \implies$  $ 2y\geq (y^{2}\minus{}2)k\geq y^{2}\minus{}2$

$ \implies$  $ (y^{2}\minus{}2y\minus{}2)\leq 0$  $ \implies$  $ (y\minus{}1)^{2}\leq 3$,$ (y>1)$  $ \implies$  $ y\equal{}2$

$ y^{2}\leq 2$  $ \implies$  $ y\equal{}1$   $ \implies$   $ y\equal{}1$  or $ y\equal{}2$

if $ y\equal{}1$  $ \implies$  $ x\minus{}1|x^{2}\plus{}1$  $ \implies$  $ x\minus{}1|2$

if $ y\equal{}2$  $ \implies$   $ 4x\minus{}1|x^{2}\plus{}2$

$ 4x\equiv 1$  $ \implies$  $ x^{2}\plus{}2\equiv 33$  $ \implies$  $ 4x\minus{}1|33$  :wink:


AMC
AMC 8
geometry
calculus
AMC 10
MATHCOUNTS
AIME

My Algebra class is taking the AMC8 this Friday, and everyone, including my teacher, expects me to get a perfect. Since I don't wan to let them down, how can I study? What do I need to know? What areas of mathematics are covered?

The AMC 8 is a middle school test -- it does not go up to Algebra 2.  I would just say relax and get a good nights sleep -- you can't really cram for this kind of test.  Have breakfast, too.

[quote="ml2"]friday? i thought it was tuesday. anyways AMC8 is pretty much basic math, probably up to Algebra II, and then it is just working fast enough, and not making mistakes (no calculator).[/quote]
a.) Today's Tuesday (and you probably knew this :wink: </no offence>)
b.) Tuesday (Today) is Veteran's Day

AMC 8? Phht, it'd be a breeze to you. Just be in good condition, don't panic, eat candy :) and you'll do great!

it is Tuesday...we're taking one from several years ago for practice on Friday...I just got confused...eat candy??? I don't get it??? Plus, how do I eat candy before a class and not get into trouble???

Um, do it discretely? :lol:
The extra sugar boost will definitely help you, at least statistically it will :)

It really came through for me, though I'm not sure if that is the only thing that contributed...

Oooh, [b]next[/b] Tuesday...:oops:

i dont think eating sugar will make a big difference... you really dont need to.  I think this is more of Tare's pre-contest preparation than anything else.  do what works for you.

[quote="MathFiend"]My Algebra class is taking the AMC8 this Friday,[/quote]

I see it was cleared up later in the thread that you are taking an OLD AMC 8 for practice on Friday. (Taking the AMC 8 early invalidates the score. The first test date this year is Tuesday 18 November 2003, when the people in my test-taking group take it.)

[quote="MathFiend"]and everyone, including my teacher, expects me to get a perfect. Since I don't wan to let them down, how can I study? What do I need to know? What areas of mathematics are covered?[/quote]

Perfect scores on the AMC 8 are rather rare outside Taiwan. (Taiwan last year had half as many perfect scores as the whole United States, with less than one-tenth the population.) I know someone locally who is in her last year of eligibility, and would dearly like to ace the AMC 8 this year. (I think she got a 23 last year, which is about as well as anyone in this whole state, especially anyone with a year of eligibility left, did.)

The thing to study to do well on the AMC 8 is old AMC 8 tests. The tests have their own style of questions. I know students of calculus who get flummoxed by some of those questions, and on the other hand know kids who are only just starting algebra who do unexpectedly well on the AMC 8. ALL of the math topics on the AMC 8 belong strictly to prealgebra mathematics, but many of the AMC 8 questions have twists that test one's mathematical sophistication. It's very important simply to know how to count well to do well on the AMC 8.

Good luck on Tuesday 18 November 2003. Hope this helps!

[quote="tokenadult"](Taiwan last year had half as many perfect scores as the whole United States, with less than one-tenth the population.)[/quote]

That's not a fair comparison without knowing the relative numbers of students taking the exam in both places, as well as the particular distribution of students who take the exam within their own school systems.  That is, it might be a small group of very good (for Taiwan) Taiwanese students taking the test compared with a much larger, but only average (for the U.S.) group of American students taking the exam.  Not that your implication is necessarily false -- just that it wasn't merited by the information available.

On a more relevant note, I do agree that AMC 8 perfect scores are quite difficult to achieve.  Don't feel pressured about it -- you almost certainly will do well, and there is certainly no point in worrying yourself overly much about it.

[quote="mathfanatic"][quote="ml2"]where can i get old AMC 8 tests?[/quote]

You can order them from [url]http://www.unl.edu/amc[/url].

On a different note, getting a perfect score on the AMC8 is difficult.  For that matter, perfection on anything is a pain.  When I took the AMC8 in 8th grade I missed 1 problem and was consequently rather annoyed.  So in response to your question, MathFiend, I recommend that you check your answers as much as you can in the allotted time - which isn't much (45 minutes for 25 questions!)[/quote]

which problem was that? i missed the one where it asked how much of the floor was tiled because i didn't see the NOTCHES IN THE FRICKIN TILE ERG! besides, what difference does it make if you get a perfect or not? you'll advance to AMC 10 (or whatever you advance to)

[quote="tetrahedr0n"]cant you just take the amc 10 anyway through the local high school? is there a minumum for qualification for amc 10 from amc 8?[/quote]

No, AMC8 and AMC10 have nothing to do with each other.  Just like you don't have to advance from the 10 to the 12.

But as for getting a perfect, you get a couple of math books for getting a perfect score.  I noticed in the results booklet.

Sigh.

I don't know which one I missed, either.  The exam was pretty informal - I had to take it during lunch, and we ended up having to switch rooms in the middle of the test because people kept passing through the room.  If I'd had that extra time or less distraction...maybe...

This guy from my school got a perfect last year as an 8th grader...on the AMC8...he got like a $2000 college scholarship or something like that...I got the notches one right...=) I got a 20 on the AMC8 last year when I took it as a 7th grader...

amc8 is pretty much like mathcounts. nuff said

i dont agree. i think amc 8 depends more on logic and common sense than mc. mc is more math based, or at least more formula based. an mc sprint round, which is most comparable to amc 8 has 5 more problems, but its also more work to do in that time (since there are no choices and no calc.) speed is of the essence is mc, but in amc 8 its hard to run out of time. personally, i did well in amc 8 all three years i took it, getting school winner from 6 th grade, but i barely made mc 7th grade. i account this to my poor speed.

well amc 8 is on the same level as a chapter sprint mc i think.

Mystic Terminator: so youre saying that your school just let kids take only the amc8?

haha, my former school was far worse, you can only take amc 8 if you are in geometry or higher, so it was very hard to take it as a 6th grader. also, you cannot take amc 10 unless you go to take math at the high school.

no, i'm saying that only kids who do well on the AMC 8 take the AMC 10.

I vaguely remember having to have done well on the AJHSME in order to take the AHSME.

Caffeine an hour before, sugar 10-15 minutes before, but the sugar will wear off rather quickly. Don't drink too much caffeine or you'll need the bathroom. Also, of course, don't drink so much it makes you jittery. You just want to be wide awake. (I'm used to math competitions in the early morning, and I don't generally get enough sleep. So I need some caffeine to keep me awake/at my highest potential)

[quote="MathFiend"]how soon before do you want to take the sugar???[/quote]

Just figure out what works for you. I recommend caffiene (about half a can of Coke) for my son. This works for him because he hardly ever drinks caffiene except on a big test day. I just have him take that at the last meal before the test, with a slightly better amount of food than usual.

Good luck on Tuesday.

[quote]grr. now i feel bad FRICKIN NOTCHES were they good book prizes?

sorry, my school advanced people to AMC 10 through AMC 8 i assumed everyone did that sorry.[/quote]

I don't know.  It just says "The students listed below are designated as National Student Award Recipients for achieving a perfect score of 25 on the 2002 AMC 8.  They will recieve an award consisting of a Certificate of distinction and two donor mathematics books".

The thing that makes me mad is that the little sister of my mathcounts teammate got a perfect score, thereby beating me and getting the prize.  Grr.....stupid room-switching...

just wondering. what do you need on the amc10 in order to go on to the next level. is 107.5 enough or do you need more?

The lowest the cutoff has ever been is 100. However, it seems to be getting higher almost every test, so if you want to be sure you get to the AIME, you have to get mid-120s or higher.

Go here, Lee:
http://www.santarosa.edu/mathematics/AMCPages/CAHONTMA10.HTML
It has the scores and names of people who qualified last year for the AIME in our region or possibly in the state. Five people from Miller!!!! :shock:  :shock:

$2000 scholarship or prizes?  I got perfect last year and didnt get anything at all.  (or i think i got perfect)


calculus
limit
trigonometry
logarithms
calculus computations

Calculus

1) $ I\equal{}\displaystyle{\lim_{x\rightarrow 0}\frac{\sqrt{1\minus{}2x}\sqrt[3]{1\minus{}3x}\ldots \sqrt[n]{1\minus{}nx}\minus{}1}{\sin x}}$

2) $ J\equal{}\displaystyle{\lim_{x\rightarrow
0}\frac{\sqrt{\cos2x}\sqrt[3]{\cos3x}\ldots\sqrt[n]{\cos
nx}\minus{}1}{x^2}}$

[hide="solution"]
1.Observe that $ \frac {d \sqrt [n]{1 \minus{} nx}}{dx} \equal{} \minus{} (1 \minus{} nx)^{\frac {1 \minus{} n}{n}},x\neq 1$
Apply L'Hospital's rule on the numerator and denominator
if $ f_{n}(x) \equal{} \frac {f(x) \plus{} 1}{\sqrt [n]{1 \minus{} nx}}$
where $ f(x)$ is the numerator of the expression.Then,
$ g(x) \equal{} \frac {df(x)}{dx} \equal{} \minus{} \sum_{m \equal{} 2}^{n}f_{m}(x)(1 \minus{} mx)^{\frac {1 \minus{} m}{m}}$.Now $ \sin x$ in the denominator goes as $ x$ so divide and multiply by $ x$.So in the denominator we have $ x$ now which vanishes after differentiating.
So the limit is $ g(0) \equal{} \minus{} \sum_{m \equal{} 2}^{n}f_{m}(0) \equal{} \minus{} (n \minus{} 1)$.

2.Proceeding similarly,$ \frac {d \sqrt [n]{\cos nx}}{dx} \equal{} \minus{} \sin nx \cos^{\frac {1 \minus{} n}{n}} nx,x \neq \frac {\pi}{2}$
Applying L'Hospital's rule
if $ f_{n}(x) \equal{} \frac {f(x) \plus{} 1}{\sqrt [n]{\cos nx}}$
where $ f(x)$ is the numerator of the expression.Then,
$ g(x) \equal{} \frac {df(x)}{dx} \equal{} \minus{} \sum_{m \equal{} 2}^{n}f_{m}(x) \sin nx \cos^{\frac {1 \minus{} n}{n}}nx$.The denominator has $ 2x$ the $ x$ goes as $ \sin nx$ in each term in the mumerator above and gives $ n$ while the $ \cos$ part gives $ 1$.
Thus the limit is $ \minus{} \frac {n(n \plus{} 1)}{4} \plus{} \frac {1}{2}$
[/hide]

Let $ f(x) \equal{} \prod_{k\equal{}2}^n (1\minus{}kx)^{1/k}$.  Then $ f(0) \equal{} 1$, and therefore

$ I \equal{} \lim_{x \rightarrow 0} \frac{f(x) \minus{} f(0)}{x \minus{} 0} \cdot \frac{x}{\sin x} \equal{} f'(0) \lim_{x \rightarrow 0} \frac{x}{\sin x} \equal{} f'(0)$.

But

$ \log f(x) \equal{} \sum_{k\equal{}2}^n \frac{1}{k} \log (1\minus{}kx)$,

from which it follows that

$ \frac{d}{dx} \left[ \log f(x) \right] \equal{} \frac{f'(x)}{f(x)} \equal{} \sum_{k\equal{}2}^n \frac{\minus{}1}{1\minus{}kx}$.

Consequently,

$ I \equal{} f'(0) \equal{} f(0) \sum_{k\equal{}2}^n (\minus{}1) \equal{} 1\minus{}n$.

Let $ f(x) \equal{} \prod_{k \equal{} 2}^n \cos^{1/k} kx$.  Then

$ \log f(x) \equal{} \sum_{k \equal{} 2}^n \frac {1}{k} \log \cos kx$,

and

$ \frac {d}{dx}\left[\log f(x)\right] \equal{} \frac {f'(x)}{f(x)} \equal{} \minus{} \sum_{k \equal{} 2}^n \tan kx$.

Differentiating the above gives

$ \frac {f''(x)f(x) \minus{} f'(x)^2}{f(x)^2} \equal{} \minus{} \sum_{k \equal{} 2}^n k \sec^2 kx$,

and evaluating at $ x \equal{} 0$ gives

$ f''(0) \equal{} \minus{} \sum_{k \equal{} 2}^n k \equal{} \minus{} \frac {n(n \plus{} 1)}{2} \plus{} 1 \equal{} \minus{}\frac {(n \minus{} 1)(n \plus{} 2)}{2}$.

Therefore, applying L'Hopital's rule twice gives

$ J \equal{} \lim_{x \rightarrow 0} \frac {f''(x)}{2} \equal{} \minus{}\frac {(n \minus{} 1)(n \plus{} 2)}{4}$.

[quote="atomicwedgie"]
$ \frac {d}{dx}\left[\log f(x)\right] \equal{} \frac {f'(x)}{f(x)} \equal{} \sum_{k \equal{} 2}^n \tan kx$.

[/quote]
You missed a minus sign here.

Oops, indeed I did.  I'll go back and fix it.   :blush:


I'm currently co-teaching a class in which students submit new to LaTeX submit files which we then comment on, a lot.  I was wondering if there are any good packages for creating fancy .pdf files which link the comments to the text being commented on, in the same way that one can do with Acrobat.  I'm aware of the \marginpar{} command, but
(1) it isn't so great for indicating what you're commenting on (though this is manageable), and
(2) I'm worried that we may have more comments than there is margin space, in which case having comments that can "open up" would be helpful.

Does such a thing exist?

There are three ways to approach this problem:

1) Annotate the submitted pdf as Acrobat Pro can do, but not at the price they charge. [url=http://www.docu-track.com/]PDF XChange Viewer[/url] does the annotations very well, is easy to use, and best of all is free (make sure you download Viewer not Viewer Pro or any of the software they offer, unless you wish to pay for them).

Or add the note to the tex file and compile:

2) Use \pdfannot command supported by pdflatex. It's not a particularly well-known command but you will find it in section 4.5 of [url=http://had.fmf.uni-lj.si/~horvat/publishing/latex/general/latex2pdf.pdf]How to Create PDF from LaTeX[/url].

3) Use a LaTeX package. Such packages include [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=pdfcomment]pdfcomment[/url], [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=cooltooltips]cooltooltips[/url] and [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=fancytooltips]fancytooltips[/url]. They give popups, so space is not a problem but some may not be printable.

I imagine you will need to choose whatever method suits your situation. If you do decide to use a particular route it would be interesting for others to hear what you chose and why.


parameterization
algebra proposed
algebra

Find all real number $ a$ for which the equation:

$ 2^{x\minus{}1}\plus{}2^{x^2\minus{}1}\equal{}\dfrac{a^2\plus{}a\plus{}1}{a^2\plus{}1}$

has integer solution.

$ 2^{x\minus{}1}\plus{}2^{x^2\minus{}1}\equal{}\dfrac{a^2\plus{}a\plus{}1}{a^2\plus{}1}\equal{}1\plus{}\dfrac{a}{a^2\plus{}1} \le \dfrac{3}{2}$
But $ 2^{x^2\minus{}1}>\dfrac{3}{2}$ for $ x$ integer and $ |x|>1$
When $ x\equal{}0$, $ a\equal{}0$ is a solution.
When $ x\equal{}1$, there is no solution for $ a$
When $ x\equal{}\minus{}1$, we get $ a^2\minus{}4a\plus{}1$, so $ a\equal{}\pm \sqrt{3}$

[quote="stephencheng"]$ 2^{x \minus{} 1} \plus{} 2^{x^2 \minus{} 1} \equal{} \dfrac{a^2 \plus{} a \plus{} 1}{a^2 \plus{} 1} \equal{} 1 \plus{} \dfrac{a}{a^2 \plus{} 1} \le \dfrac{3}{2}$
But $ 2^{x^2 \minus{} 1} > \dfrac{3}{2}$ for $ x$ integer and $ |x| > 1$
When $ x \equal{} 0$, $ a \equal{} 0$ is a solution.
When $ x \equal{} 1$, there is no solution for $ a$
When $ x \equal{} \minus{} 1$, we get $ a^2 \minus{} 4a \plus{} 1$, so $ a \equal{} \pm \sqrt {3}$[/quote]

It's true except the last last solution of $ a$. It must be $ a\equal{}2\pm\sqrt3$


i need a good way of taking sqrt w/o calculator for the AIME or something.. and feel free to insert other skill i might be able to use.. ty

Just memorize the basic square roots.

root 2 is about 1.414
root 3 is about 1.732
root 5 is about 2.235

Note that I'm reciting these from memory, so they might be wrong...

I don't know if they still teach the  method to claculate the square-root by hand  which is similar to long divison.

That method is not very hard, another practical way is seriies expansion or iteration:

Iteration: Say you want to find sqrt(24) it is very near 5;   but 24/5= 4.8 so you take mean of 5.0 and 4.8 and get 4.9... etc. (ok for 2 sig figure)

Or say you are asked sqrt(120) and  you start with 10; 120/10 = 12; Average of 10,12 =11
Now 120/11 =  10.91 and average of 10.91 and 11 is 10.95 etc.. (Ok 4 sig figure)

Series: Say you want to find sqrt of  172.4 ... You  happen to notice 13^2=169,  172.4 is about 2%  more than 169 so squrt will be about 1% more, or (13.13)  (which is correct up to 4 sig figure)

Basically divide the number into (A+ (or -) B) where you know the square-root of A ; and B is small compared to A. .. expand in terms of ( B/A):

Then sqrt(A+B) =  sqrt(A) + (1/2)  sqrt(A)*(B/A)  -(1/8)*sqrt(A)(B/A)^2

generally one or two terms are more than enough -

You can apply similar logic for cube-root or sin or cos , log etc.
(Of course, in our times we remembered log tables and it was just easy to divide the log by 2 and take antilog) .

OK i learned this way a long time ago but it works if i dont make a stupid mistake
so you have $ \sqrt{256} $
so you split the number into pairs of two from the right side $ \sqrt{2   56} $
than you find the closest interger to the root of the first number in this case 1 and you take this away from 2 to get 156, no you double the number you previously squared (1) and you now need to find a digit A so that 2A*A=156, the number is 6 since 26*6=156
this means that $ \sqrt{256} = 16 $

hope you understand that, i was never really able to explain that process well

Check out this site: it explains it pretty well.

By the way, if you were told the number was a perfect square beforehand, it makes it easier as you can use tricks like working out the last digit, using division by 9, 11, 13 rules etc. In fact if I remember quickly, its extremely simple to work out the cube root of (a three digit number)^3, in your head.

TripleM, which site are you citing?  Hm, a bad pun.

I found this way that works as long as the number you want to take the square root of is rational and positive.  Let the number you want to take the square root of be a/b.  Start with 1/1 for the first estimate, and multiply the denominator by a and add the numerator of the first estimate to obtain the numberator for the second estimate.  To obtain the denominator of the second estimate, multiply the numerator to the first estimate by b and add the denominator of the first estimate.  Continuing this way except replacing second with third and first with second, etc. you obtain a method that gives a close approximate that converges on the number.  Don't try this for finding the square root of perfect squares.

Wow, did I really do that :S
Ok, here it is this time :D
http://www.nist.gov/dads/HTML/squareRoot.html

Hello, kentliu!

[quote]i need a good way of taking sqrt w/o calculator for the AIME or something[/quote]
If you don't mind doing a bit of arithmetic, there is an elementary approximation  method.

Suppose we are seeking $\sqrt{N}$.
Let $a_1$ be our first approximation . . . Then: $\frac{N + a_1^2}{2a_1} \;= \;a_2$ is a better approximation.
For an even better approximation, repeat the procedure with $a_2$.

Example: $\sqrt{2}\qquad a_1 = 1.5 = \frac{3}{2}$

Then: $\displaystyle{a_2 = \frac{2 + (\frac{3}{2})^2}{2(\frac{3}{2})} \;= \;\frac{17}{12} \;= \;1.41\overline{6} }$

 . . .  Check: $\left(\frac{17}{12}\right)^2 = 2.0069\overline{4}$ . . . [i]Not bad![/i]

Repeat: $\displaystyle{a_3 \;= \;\frac{2 + (\frac{17}{12})^2}{2(\frac{17}{12})} \;= \; \frac{577}{24} = 1.414215686 }$

. . . Check: $\left(\frac{577}{24}\right)^2 = 2.000006007$ . . . [i]an excellent result from simple pencil-pushing![/i]

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here is the source of that approximation formula.

We are hoping to find: $\sqrt{N} = a$ . . . or: $a^2 = N\;\;\Rightarrow\;\;a = \frac{N}{a}$

If our first approximation, $a$, is correct, then $a$ and $\frac{N}{a}$ are equal.

If the two factors are not equal, a better approximation is their [b]average[/b].

And their average, of course, is: $\displaystyle{\frac{\frac{N}{a} + a}{2} \;= \;\frac{N + a^2}{2a}   }$


See posts $8$, $13$ and $17$ in http://www.mathlinks.ro/Forum/viewtopic.php?t=59905 .


The sequence 2, 3, 5, 6, 7, 10, 11, $ \ldots$ contains all the positive integers from least to greatest that are neither squares nor cubes. What is the $ 400^{\mathrm{th}}$ term of the sequence.

ans=425
 [b]CASE 1[/b]
IF SERIES STARTS FROM 1
i.e.set is
[1,2,3,4,5........}
then in first 400 no.s 20 are squares and 7 are cubes
that makes the 400th term=427
but, 1^2=1^3 and similarly for 8^2 and4^3
so 2 no.s are repeated. so ans is 425

[b]CASE 2 [/b]
HERE SET OF NO.S CAN BE
{2,3,4,5,6......} i.e. starting from 2
then from 427 no.s 1^2 and 1^3 are deleted
so, we get 427-2 =425
moreover 4^3 and 8^2 are again repeated
so, we get 424.
       :maybe:  but , as the set of numbers has shifted from {1,2,3,4,5.} to {2,3,4,5.] which means it has shifted by 1
  so, we get our ans.=424+1 =425. :idea:  :)

From one to 400, there are 20 perfect squares (20x20=400) and 7 perfect cubes (7x7x7=343), so you must add 27 to 400 because you skipped those numbers. But you counted both the perfect 6th powers from one to 400 (1 and 64) twice, so you must subtract 2 from 427 so that you count the perfect sixth powers once. There are no other perfect squares or cubes between 400 and 425, so the answer is 425.

Let's try counting the number of perfect squares and cubes less than $441=21^2$. There are twenty perfect squares less than 441: $1^2, 2^2, \ldots, 20^2$. There are also seven perfect cubes less than 441: $1^3, 2^3, \ldots, 7^3$. So there would seem to be 20+7=27 numbers less than 441 which are either perfect squares and perfect cubes.

But wait! $1=1^2=1^3$ is both a perfect square and a perfect cube, so we've accidentally counted it twice. Similarly, we've counted any sixth power less than 441 twice because any sixth power is both a square and a cube at the same time. Fortunately, the only other such one is $2^6=64$. Thus, there are 27-2=25 numbers less than 441 that are perfect squares or perfect cubes. Also, since $20^2=400$ and $7^3=343$, then all 25 of these numbers are no more than 400. To compensate for these twenty-five numbers missing from the list, we need to add the next twenty-five numbers: 401, 402, $\ldots$, 424, 425, none of which are perfect square or perfect cubes. Thus, the $400^{\text{th}}$ term is $\boxed{425}$.

All of these comments are helpful, but use [ hide ]


limit
induction
geometry
calculus
calculus computations

Form easy induction we can see that all the terms of $a_n$ are positive. So the $a_n$ is descrising, that's why the limit exist and equal to 0.
Now $\lim_{n\to\infty} \frac{1}{n} (a_1+a_2+\cdots +a_n)$ follows from Cezaro-Stolz.
the first part comes out from induction.
the third part follows from  Cezaro-Stolz:
$\lim_{n\to\infty}\frac{n}{\frac{1}{a_n}}=\lim_{n\to\infty}\frac{1}{\frac{1}{a_{n+1}}-\frac{1}{a_n}}=\lim_{n\to\infty}\frac{a_n}{(1+a_n)^2-1}$

Thank you for your reply, rustam. Japanese highschool students don't study Cezaro-Stolz.

kunny

they should d oit with $\epsilon -\delta$?

As to $(2),$ we are to compare the area of certain figure, or to remark the Riemman's sum.

we have that
$a_n \to 0 \Leftrightarrow \forall \varepsilon > 0{\rm{ }}\exists {\rm{ }}N_1 : \forall n > N_1 \Rightarrow a_n < \frac{\epsilon }{2} \\ \frac{{a_1 + \cdots + a_n }}{n} = \frac{{a_1 + \cdots + a_{N_1 } }}{n} + \frac{{a_{N_1 } + \cdots + a_n }}{n} < \frac{c}{n} + \frac{{n - N_1 }}{n}\cdot\frac{\epsilon }{2}$
where
$c: = a_1 + \cdots + a_{N_1} \forall \epsilon > 0 { } \exists N_2 : \forall n > N_2 \Rightarrow \frac{c}{n} < \frac{\varepsilon }{2} \Rightarrow \forall \epsilon > 0{\rm{ }}\exists N = \max \left\{ {N_1, N_2 } \right\} \\ \frac{{a_1 + \cdots + a_n }}{n} < \frac{\epsilon }{2} + \left( {1 - \frac{{N_1 }}{n}} \right)\frac{\epsilon }{2} < \epsilon \Rightarrow \ \lim_{n \to \infty } \frac{a_1 + \cdots + a_n }{n} =0$

kunny i did not understand ur hint and cant do the (3) without the Cezaro-Stolz?


floor function
advanced fields
advanced fields unsolved

Hello,

Does anyone knows how to compute $\sup_{\alpha\in\mathbb{R}}[d(L_\alpha,\mathbb{Z}\times\mathbb{Z}-\{(0,0)\})]$ where $L_\alpha=\{(x,y):y=\alpha x\}$ and for $A,B\subseteq\mathbb{R}\times\mathbb{R}$, $d(A,B)=\inf\{\|a-b\|:a\in A, b\in B\}$, and the norm is the euclidean norm. My hypothesis is that such number is zero, but I can't prove it.

Best,

Or, in other words, you want to prove that the distance between $L_{\alpha}$ and $\mathbb Z^2$ is $0$ for all $\alpha$? :?

If $\alpha\in\mathbb Q$, then there exist $x,y\in\mathbb Z$ s.t. $y=\alpha x$, so the distance is obviously $0$. On the other hand, if $\alpha\in\mathbb R\setminus\mathbb Q$, then look at the sequence of points $(n,\alpha n),\ n\in\mathbb Z$. What you want to show follows rather easily from the fact that the fractional part of $n\alpha$ is dense in $(0,1)$ (as a sequence indexed by $n$), so there are points of the type $(n,\alpha n)$ which are arbitrarily close to $(n,\lfloor\alpha n\rfloor)\in\mathbb Z^2$.

Did I misunderstand what you were asking? :?

[quote="grobber"]:D Didn't see that :). Of course, we can look at $d(L_{\alpha},\mathbb Z^2\setminus \{(0,0)\})$ :).[/quote]

Yes please, I made that stupid mistake (already edited), but I am interested obviously in the case of $\mathbb{Z}^2-\{(0,0)\}$!!

Best,

It is a very old problem. Imagine that you stay at (0,0) and there are circles of radii $\epsilon$ with centers at each lattice point. Then you can see only finite part of the plane.

[quote="grobber"][quote="djimenez"]
Yes please, I made that stupid mistake (already edited), but I am interested obviously in the case of $\mathbb{Z}^2-\{(0,0)\}$!!
Best,[/quote]

But did I not answer that question in post number $2$ on this topic? :?[/quote]

Yes, you did.

Best,


Hello,
given two complex numbers r,s we know that arg(rs*)=arg(r)-arg(s).

Now I don't know the complex numbers r,s but all I know are the two values arg(r) and arg(s) (that are both in the range [-pi,pi]).

How can I get only from arg(r) and arg(s) the exact value in [-pi,pi] that would return arg(rs*) ?

Thanks!

well, in case someone is interested I answer to myself,

if A is an angle given by A=arg(r)+arg(s), then

arg(rs) = A (for |A|<= pi) ;  A-2pi*sgn(A)  (for |A|>pi)

Hopefully I haven't made any mistake...sorry for the bad type-format!


AMC
AMC 10
AMC 10 A
AMC 10 B
calculus
email
AIME

So i really want to take the amc10a a exam because i won't make the amc10b exam, but i went to the website it says the registration ends on Feb 3rd.  Is there any way that i can pay extra late fee for taking the test?  Please reply!
btw, I'm pretty close to the Hartford, CT area.

Check with the local high schools and see if one of them is giving the test.  Maybe they will let you sit in and take the test with them. If you have a middle school that is strong they MAY be giving it.  check with them also.

i contacted them last week, but they haven't sent the reply yet....

A phone call to the Math Chair or AP Calculus teacher will be more effective than an email at this late time.

You may try contacting the person at Fairfield University in CT listed on the College and University administration site.  I see that it is grayed out indicating FU has not registered for 2009, but maybe they would be willing to schedule something special for you, if you ask them politely and are willing to work with their schedule.

It is still possible to register with the AMC office, but at this late time, it takes extra effort by our staff and it takes special arrangements to get the contest to you.  You will have to call the AMC office and start making the special arrangements it is going to take.

Remember:  The AMC 10 (and the AMC 12, and the AIME) are school based activities, an individual cannot register himself or herself.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="cdymdcool"]they said that their spaces are full, is there any way that they can order on more test for me?[/quote]The tests come in batches of 10.


Mary will pick a positive integer less than 80 that is a multiple of 7.  Susan will pick a positive integer less than 80 that is a multiple of 9.  What is the probability that they both will pick the same number?  Express your answer as a common fraction.

11 multiples of 7 under 80
8 multiples of 9 under 80

so 88 possible combos of choices. only one, 63 would constitute a match while staying in accordance to the guidelines. Therefore, our answer is $ \boxed{\frac{1}{88}}$

[quote="AruKasera"]Isn't 63 counted twice? So wouldn't the answer be 1/87?[/quote]
No. 7*9 is the way.


geometry
parallelogram
algebra
polynomial
graph theory

[quote="M4RI0"]Show that for any convex polygon with area $1$, there is a parallelogram with area $2$ that contains it.[/quote]
nice problem, did you get it from Projecto Delfos?

:thumbup: Nice, I think this is a solution
Call AB is the longest distance from a summit to the another.
C is the farthest point from AB on the left and  D is the farthest point from AB on the right.
Draw two lines run through C,D such that parallel with AB and two parallel lines run through A,B.
The four intersection points of these lines we call that is M,N,P,Q such that MNPQ contains the convex polygon with area 1 and $S_{MNPQ}=2S_{ADBC}\leq\ 2\times\ 1$

It's from the Iberoamerican Olympiad(1990). Also what's project Delfos :?:

http://www.mat.uc.pt/~delfos/PROB-OIAM.pdf
It has some very nice questions, including yours. :)

Very nice page. Have you solved 1990/6 :?:  there's no solution in the page and I can't solve it.

1990/6. $f(x)$ is a polynomial of degree 3 with rational coefficients. If its graph touches the x-axis, show that it has three rational roots.
good question, i'm not sure how to approach it.

1990/5 is a very nice and hard problem. If I solve it I will post the solution here. :)

[quote]$1990/5$. A and B are opposite corners of an n x n board, divided into $n^{2}$ squares by lines parallel to the sides. In each square the diagonal parallel to AB is drawn, so that the board is divided into $2n^{2}$ small triangles. The board has $(n+1)^{2}$ nodes and large number of line segments, each of length $1$ or $\sqrt{2}$. A piece moves from A to B along the line segments. It never moves along the same segment twice and its path includes exactly two sides of every small triangle on the board. For which n is this possible?[/quote]
That's a hard question. What are nodes?
[quote="M4RI0"]Show that for any convex polygon with area $1$, there is a parallelogram with area $2$ that contains it.[/quote]
I will post a solution to this soon (if i get it done). :D

Are nodes the same thing as in bridge building, where members meet?, in this case, line segments?

Yes, that's what the problem means by nodes.

Node (or "vertex") has a particular meaning in graph theory and it is used to differentiate graph theory from, say, geometry.  They look like the same thing, but graph theory is a very different theory.  :)


TI-89 can not factor it and the solution it gets after it solves it is $x \approx 19.5177975$, which it later says is wrong.

Only solution I can find (checking interval $[-100,100]$) is ~$19.517797516533$. That is obviously irrational.

If you'd like to learn about methods of solution for large-degree polynomials, head over to Wikipedia. They've got some good articles there.


The houses on the East side of Elm Street are consecutive odd numbers. Find the sum of the house numbers which are exactly two doors away from number 15.

Note that the average of the house numbers is 15, so we double it, getting 30.

Or u could do it the slow way...
Find the odd integer before and after 15 which are 17 and 13. 17+13=30


:idea: a beaker of water has a piece of ice floating on its surface.When the ice melts,show that (by logic or by equations)
 :arrow: 1.that the water level remains unchanged.
 :arrow: 2.If a coin was initially kept above the ice block what change will be there in answer to 1?
 :arrow: 3.if the ice piece initially had an air cavity inside it how will the answer change?
 :arrow: 4.if there is a miscible liquid TRAPPED inside the cavity then what will happen to the level of the water?
 :P

I like to pose problems like these to my Early History of Mathematics classes when we're talking about Archimedes.

We'll assume that the liquid water remains at $0^{\circ}\text{C}$ throughout, with constant density (and constant dimensions of the glass.)

1. As long is the ice cube is floating, it displaces a volume of water equivalent to the mass of the ice. Once it melts, it becomes a volume of water equivalent to its mass. These two are the same and the water level remains constant.

2. (In the version I tell, we have a boulder in a rowboat is a swimming pool to start with; we lift the boulder out of the boat and let it fall to the bottom of the pool.) As long as it is resting on the ice cube, the coin causes the cube to displace a volume of water equivalent to the mass of the coin. Once the ice melts and the coin sinks to the bottom of the glass, the coin is displacing a volume of water equivalent to the volume of the coin. Since the coin is heavier than water, a volume equivalent to its mass is greater than its volume. Hence, the level of water in the glass falls.

2.5 Let me add another question: make that a chip of wood rather than a coin. While resting on the ice cube, it displaces a volume of water equivalent to its mass; once the cube melts, the wood floats, and still displaces a volume of water equivalent to its mass. The water level remains unchanged.

3. The trapped air is not changing the mass of the ice cube. (To a first approximation, neglecting the mass of the air. But a second approximation would also have to include air pressure and (in effect) bouyancy in air - it still all washes out.) Hence the answer is the same as in #1 - the water level remains unchanged.

4. Let's first address what happens with an immiscible trapped fluid. If that fluid is denser than water, then we repeat our answer to #2: the water level falls as the dense fluid forms a layer accross the bottom.

But a lighter-than-water immiscible fluid is more complicated. We start with our answer to #2.5. Suppose the water level in $h$ and the surface area at height $h$ in the glass is $A$ and we have a chip of wood of mass $m$ and density $\rho.$ The volume of wood below the water line is $m$ (taking the density of water as $1.$) If we redistributed the water so that it was level, the top of the water would be at $h-\frac{m}{A}.$ Now turn the wood chip into an immiscible liquid, and spread it on top of the water. It occupies volume $\frac{m}{\rho},$ and hence spreads to a depth of $\frac{m}{\rho A}.$ The top on the combined fluids is now at $h-\frac{m}{A}+\frac{m}{\rho A}=h+\left(\frac1{\rho}-1\right)\frac{m}{A}.$ Hence the combined level of all the fluids rises.

So, assuming no unusual or nonlinear things happen when these fluids combine, if the trapped fluid is miscible and denser than water, then the fluid level in the glass falls. If it is lighter than water, then the fluid level in the glass rises.

There is much talk of the potential effect of melting ice in Greenland and Antarctica on global sea levels. In both places, but especially in Antarctica, ice pushes from the land out into the ocean as glacier tounges and ice shelves. What is the effect of those on sea level?

Any ice that is "grounded," resting on bedrock, is not yet part of the ocean, but any ice that is floating is already part of the ocean as far as sea-level calculations are concerned - it doesn't matter whether it's melted yet.

(Actually, it does make a little bit of difference whether it's melted, because glacial ice melts to become fresh water, which is lighter than salt water.)


Find all possitive values of the integer $ n$ , such that the nubmer

 $ A\equal{} \sqrt \frac {9n\minus{}1}{n\plus{}7}$ is rational .

Babis


prove or disprove:

let $ \sum a_n$ be a convergent series, then so does $ \sum a_n^2.$

i know that $ a_n\to0$ as $ n\to\infty,$ but i'm not getting anywhere with that assumption.

This is false.  Let $ a_n \equal{} \frac{(\minus{}1)^n}{ \sqrt[3]{n} }$.

Suppose you are also given absolute convergence.  Then does the result hold?

Let's construct some counterexamples.

Let $ \{ c_n \}$ be a sequence of positive real numbers so that $ c_n$ decreases to 0 as $ n \to \infty$ but $ \textstyle \sum c_n \equal{} \plus{}\infty$. There are plenty of such examples such as $ c_n \equal{} n^{\minus{}1}$. Then put $ a_n \equal{} (\minus{}1)^{n} \sqrt{c_n}$. You can easily show by alternating series test that $ \textstyle \sum a_n$ converges, while $ \textstyle \sum a_n^2 \equal{} \sum c_n \equal{} \plus{}\infty$.


Let $ T$ be real number satisfying the property: For any nonnegative  real numbers $ a,b,c,d,e$with their sum equal 1, it is possible to arrange them around a circle such that the products of any neighboring numbers are no greater than $ T$. Determine the minimum value of $ T$

Consider the arrangement of numbers $ \epsilon, \frac{1}{2} \minus{} \frac{3\epsilon}{2}, \epsilon, \epsilon, \frac{1}{2} \minus{} \frac{3\epsilon}{2}$ on the circumference of the circle in this order. As $ \epsilon \rightarrow 0$ the largest possible neighboring product $ (\frac{1}{2} \minus{} \frac{3\epsilon}{2}).\epsilon$ also $ \rightarrow 0$ and hence does $ T$.

You've completely misunderstood the problem: $ T$ should be the smallest number that simultaneously serves as a bound for [i]all[/i] collections of 5 numbers, not the smallest number that serves as a bound for some collection.  I think the obvious guess is $ \frac {1}{9}$, but I haven't tried to prove it  :oops:

Good guess! :D

[hide]WLOG $ a \geq b \geq c \geq d \geq e$. Optimal arrangement will be $ d,a,e,b,c$ or $ e,a,d,b,c$ (it's irrelevant) when the largest product is either $ bc$ or $ ad$. Any other arrangement up to reflection/cyclic permutation will put $ a$ adjacent to $ b$ or $ c$ and give a larger product.

If the largest product is $ ad$, then to maximize $ a$ and $ d$ we set $ b \equal{} c \equal{} d$ and $ e \equal{} 0$. Now we get $ a \plus{} 3d \equal{} 1$ and AM-GM gives $ 1/2 \geq \sqrt{3}\sqrt{ad}$ or $ ad \leq \frac{1}{12}$. This is obtainable when $ a \equal{} 1/2, d \equal{} 1/6$.

If the largest product is $ bc$, then to maximize $ b$ and $ c$ we set $ a \equal{} b$ and $ d \equal{} e \equal{} 0$. Now we get $ 2b \plus{} c \equal{} 1$. Let $ b \equal{} 1/3 \plus{} t$ and $ c \equal{} 1/3 \minus{} 2t$ for some nonnegative $ t$. Then $ bc \equal{} 1/9 \minus{} t/3 \minus{} t^2$, which on $ t \geq 0$ is obviously maximized at $ t \equal{} 0$, giving $ bc \equal{} 1/9$ and $ b \equal{} c \equal{} 1/3$. This is a greater product than $ 1/12$, so $ T \equal{} 1/9$ is the minimum.[/hide]


function
calculus
derivative
logarithms
complex analysis
complex analysis unsolved

I am sorry if i posted a lot, but complex analysis is killing me!!! We arent being given enough worked examples to know how to approach these problems. And this is still the beginning of the year course  :o

a) Use Taylors Theorem to find a power series expansion for f(z)= $z^{-1}$ where [i]z[/i] lies in the disk $D ={z\in \mathbb{C}: |z-1|<1}$. Given that

 $Ln z= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$$(z-1)^{2}$
for all $z \in D$ , show that $\frac{d}{dx}\left(z\right) = Lnz^{-}1$

b) Using power series for the exponential function , show that for any $a,b \in \mathbb{C}$

$e^{a}e^{b}$=$e^{a+b}$

[Hint: Evaluate the product of the series expressions for $e^{az}$ and $e^{bz}$ , then make substitution z=1]

a) The $n$th derivative of $f$ is $(-1)^{n}n! z^{-n-1}$. Plugging this into the Taylor formula, find the expansion $\sum_{n=0}^{\infty}\frac{(-1)^{n}n!}{n!}(z-1)^{n}=\sum_{n=0}^{\infty}(-1)^{n}(z-1)^{n}$. But there is a better way, which also takes care of the convergence issue: $\frac{1}{z}=\frac{1}{1-(1-z)}=\sum_{n=0}^{\infty}(1-z)^{n}$.
Now differentiate the series for $z^{-1}$ (with respect to $z$, not $x$!) and see that it matches the expansion for $\ln$.

 You may want to check if your library has books with more examples like (a) and (b).


analytic geometry
geometry unsolved
geometry

hi im a newbie, sry if this is in the wrong section, but here its the question:

for this assignment, i have to write a program that lets the user to enter the length of the 3 sides for a triangle and i must draw it out, now the tricky part is to draw obtuse/acute triangles

so my question is, is there a way i can figure out the coordinates of the triangle (right-angle, accute, or obtuse) when given the lengths of 3 sides...i thnk cos/sin laws are needed here, but im having trouble implementing them

i'll be really appreciated if anyone can help me out....thankyou in advance

p.s.: the reason why i need coordinates is because i must draw the triangle so that its centered in the plane

$a^2+b^2-c^2>0$ implies acute.
$a^2+b^2-c^2<0$ implies obtuse.
Is that what you want?


calculus
function
trigonometry
real analysis
real analysis unsolved

Let f(x) = x cos (1/x) , x>=1

Then,
which of the following is true?

1) for atleast one x in [1,∞)  , f(x+2) - f(x) < 2

2) for  all x in [1, ∞)  , f(x+2) - f(x) > 2

Just  give a minor hint  or mention the theorem.

Thanks in advance

Because of the cyclical nature of the function, I suspect that there is an infinite number of values of $ x \in [1, \infty)$ for which $ f(x\plus{}2) \minus{} f(x) > 2$ and an infinite number for which it is $ < 2$.

Have you tried differentiating or graphing the curve?

$ x\cos(1/x)\approx x \minus{} 1/(2x)$. So I suspect that $ f(x \plus{} 2) \minus{} f(x) > 2$ for all $ x > 1$. Though a more rigorous argument is required. You might try to prove that $ \cos x \plus{} x\sin x > 1$ for $ x\in(0,1)$ (which is true; consider the derivative).

[quote="zhoraster"]$ x\cos(1/x)\approx x \minus{} 1/(2x)$. So I suspect that $ f(x \plus{} 2) \minus{} f(x) > 2$ for all $ x > 1$. Though a more rigorous argument is required. [/quote]

In fact for |u| <= 1,  1 - u^2/2 <= cos(u) <= 1 - u^2/2 + u^4/24. (Just apply Taylor's theorem.) This will solve the problem.

using squeeze theorem i  get

       2  <=    f(x+2) - f(x) <= 2

Write $ f(x\plus{}2)\minus{}f(x) \equal{} 2f'(\theta)$ and read my post again (the last sentence).

Can't we solve it using squeeze theorem?

[quote="zhoraster"]Write $ f(x \plus{} 2) \minus{} f(x) \equal{} 2f'(\theta)$ and read my post again (the last sentence).[/quote]

Right! I didn't follow you the first time.


anybody know when we're going to get them? haha
i'm a bit jumpy for the calc.

what prizes does arml give?

there are arml teams for schools, regions, AND states?

uh, this year we got a few books I think and a TI-NSpire, but of course, the calculator is all that really matters :D

you can form your own arml team, so yes, there are teams from schools, regions (although I don't know what you mean by "regions"), and states. but in IL there are really only two teams, Chicago A/B/C/D and [i don't know what the other one is], but Chicago is pretty much the only good one. we go to the University of Iowa in Iowa City for the competition, but the rankings are nationwide. Chicago A tied for seventh this year, so we got a prize (i think the prizes are top 8 in the nation)

Pretty sure the other teams in the top 8 haven't gotten their prizes yet eitherp


Hi everyone. This problem was written by Hyunsoo Kim, a member of this year's USA IMO team, for the ELMO, which is an olympiad written by the veterans each year, at the USA's Math Olympiad Program (MOP), for the rookies to take. This was the fourth and most difficult problem of the contest.

Given positive reals $a, b, c$ such that $abc \ge 1$, prove \[\frac{1}{a + b^{2005} + c^{2005}} + \frac{1}{a^{2005} + b + c^{2005}} + \frac{1}{a^{2005} + b^{2005} + c} \le 1.\]

Nice problem. Choose $p$ such that $ 1+\frac{p}{2}=2005-p$ and observe that ${b^{2005}+c^{2005}\geq \frac{b^p+c^p}{2}\cdot (b^{2005-p}+c^{2005-p})\geq\frac{b^{2005-p}+c^{2005-p}}{a^{\frac{p}{2}}}}$. Thus $\frac{1}{a+b^{2005}+c^{2005}}\leq\frac{a^{r-1}}{ a^r+b^r+c^r}$. It is straightforward using Chebyshev to prove the last sum is at most 1.

[quote="darktreb"]Hi everyone. This problem was written by Hyunsoo Kim, a member of this year's USA IMO team, for the ELMO, which is an olympiad written by the veterans each year, at the USA's Math Olympiad Program (MOP), for the rookies to take. This was the fourth and most difficult problem of the contest.
[/quote]

Why don't you guys don't post the USA TSTs of the previous two years and some MOP problems and tests?  :)

Is Harazi 's solution the same as Hyunsoo Kim's ? If not , could you please post Kim's solution here ?

I would be interested to know how this one is similar to IMO 2005 #3

Because in both cases you manipulate the denominators nontrivially (with Cauchy) to get the proof to fall out.


limit
logarithms
trigonometry
calculus
derivative
calculus computations

hi,

i will be glad if somebody shows me how to compute this limit without using hospital's theorem .

${f(x)= \frac{1}{x^{4}}Ln(\frac{\sqrt{1-x^{2}}}{cos(x)}})$
${\lim_{x\rightarrow 0}}f(x)$

thanks in advance.

That is wrong.  Post a solution, and perhaps I can show you why it's wrong.

please ,do not give the result .i can do that using the hospital theorem .what is i want is the proof ,please.

sorry,i didn't want to double the post.i would like to delete the other one but i don't know how . :blush:

netcat, your solution is wrong because the derivative of $\ln\frac{\sqrt{1-x^{2}}}{\cos x}$ is $\tan x-\frac{x}{1-x^{2}}$, not $\sin x-\frac{x}{\sqrt{1-x^{2}}}$.

Also, toetoe, a proof with L'hopital is a proof; why do you insist on another method?

sorry,but i don't understand how did you get the second step.please ,can anyone explain to me the answer ?

They hasn't taught us this theorem in classroom so it is forbiden in my level to use it .

toetoe, you did double post. You posted the first topic in "proposed and unsolved problems." I decided it belonged in "computations and tutorials" so I moved it. You then re-posted the same problem back in "proposed and unsolved problems." That's what we don't want you to do - follow  your topic to where a moderator moved it. Don't try to put it back where it doesn't belong. This topic now has the responses on both original topics jumbled together, so the chronology might not make sense.

Elaborating: we have $\frac1{x^{4}}\left[\frac12\ln(1-x^{2})-\ln\cos x\right]$

From $\ln(1+u)=u-\frac12u^{2}+O(u^{3})$ we have the following:

(1) $\frac12\ln(1-x^{2})=\frac12\left(-x^{2}-\frac12x^{4}+O(x^{6})\right)=-\frac12x^{2}-\frac14x^{4}+O(x^{6}).$

Then from $\cos x=1-\frac12x^{2}+\frac1{24}x^{4}+O(x^{6})$ we get

(2) $\ln\cos x=\left(-\frac12x^{2}+\frac1{24}x^{4}+O(x^{6})\right)-\frac12\left(\frac14x^{4}+O(x^{6})\right)$

$=-\frac12x^{2}-\frac1{12}x^{4}+O(x^{6})$

Subtracting (2) from (1) we get:

$\frac12\ln(1-x^{2})-\ln\cos x=-\frac16x^{4}+O(x^{6})$

$\frac1{x^{4}}\ln\left(\frac{\sqrt{1-x^{2}}}{\cos x}\right) =-\frac16+O(x^{2})$

The limit is $-\frac16.$

This might be what Extremal did; I wouldn't know since I didn't download his file.

[quote="toetoe"]thanks so much but i still don't understand what O(x) means ?[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31517

I will say by this way: In math very often use g(x)= o(f(x)) and g(x)=O(f(x)) in point 0. I use O(f(x)) because it give us more good rezult than o(f(x)).
1.g(x)=O(f(x)) mean, exist some C=const(C is not equal 0) :  g(x)<=Cf(x) when x->0.
2.g(x)=o(f(x)) mean, exist lim g(x)/f(x)->0 in point 0.


The sum to infinity of the terms of an infinite geometric progression is $ 6$. The sum of the first two terms is $ 4\frac {1}{2}$. The first term of the progression is:

$ \textbf{(A)}\ 3 \text{ or } 1\frac {1}{2} \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2\frac {1}{2} \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 9 \text{ or } 3$

[hide="Solution"]
Since the infinite sum converges, we know that common ratio is smaller than 1.

Thus, we have $ \frac{a}{1\minus{}r}\equal{}6\implies a\equal{}6\minus{}6r$ where $ a$ is the first term in the sequence.

And, we have $ a\plus{}ar\equal{}\frac{9}{2}$

Solving the system, we have $ r\equal{}\frac{1}{2}, \minus{}\frac{1}{2}$

Thus, $ a\equal{}3,9$

Answer: E[/hide]


what is the maximum value of $x$ in which $x^2+y^2=x+y$ for $x,y\in\mathbb{R}$?

$y^2-y+x^2-x=0$
Let $f(y)=y^2-y+x^2-x$, $f(y)$ is a quadratic function, and this function has root(s), so
$\Delta\geq 0$
$1-4(x^2-x)\geq 0$
$4x^2-4x-1\leq 0$
$x\in [\frac{1}{2}(1-\sqrt{2}) , \frac{1}{2}(1+\sqrt{2})]$
So the maximum value of $x$ is $\frac{1}{2}(1+\sqrt{2})$

[hide="More generally"]For $n$ variables the maximum of one of them is given by $\frac{1+\sqrt{n}}{2}$.

If $\sum_{k=1}^n a_k^2=\sum_{k=1}^n a_k$, move everything to the left and complete the square to get $\sum_{k=1}^n \left(a_k-\frac{1}{2}\right)^2=\frac{n}{4}$. Suppose $a_1$ is the variable to maximize. Set the other squares equal to 0 (this can be done by setting $a_k=\frac{1}{2}$ for $k>1$) to get $\left(a_1-\frac{1}{2}\right)^2=\frac{n}{4}$. Taking square root and adding $\frac{1}{2}$ gives $a_1=\frac{1+\sqrt{n}}{2}$.[/hide]

well done scorpius119!  I really like general solutions.

A very eloquent solution indeed, scorpius.


geometry
projective geometry
superior algebra
superior algebra unsolved

I see that nobody is interrested in these equations as applications of $\mathbb{Z}[i]$, etc are euclidians, etc. :)

you are wrong
  :)
i think you can solve the first one by pure projective geometry in the plane over the rationals, knowing that if you have one point you can determine the full quadric

there are many many other and better ways

but do you know how to deal with the rest?

I know what I'm saying. Anyway, can you please post your proof using projective geometry? :)

He probably meant http://www.mathlinks.ro/Forum/viewtopic.php?t=65751 (the one missing solution comes from infinity).

yeah that's it
I like it when theory of algebra/geometry kicks in  :)

I showed interest and proposed a solution, please reveal your cards :ninja:  for the other problems


function
topology
advanced fields
advanced fields unsolved

I proved half of the problem, but I still get stuck in the last one (probably it is easy for you, but not to me as I am still a bit nervous about these stuffs). Anyway, here it is

Let $F(R)$ be the set of all function on R. We define the set $B_{M}(f)$ for every set $M$ and for every function $f$ as the set of all functions $g$ such that $f|_{M}= g|_{M}$.

Let $\omega$ be the collection of all subsets $U$ of $F(R)$ with the property: for every $f$ in $U$, there is a countable set $M$ such that $B_{M}(f)$ is a subset of $U$.

a) Show that every subset $S$ of $F(R)$ is sequentially closed relatively to $\omega$.

 [hide="My proof, please help me check if it is right"] Consider the collection $\alpha = \{B_{M}(f)| M$ is countable and $f$ in $F(R) \}$. Then $\alpha$ is a base for $\omega$. However, $\omega$ is a topology on $F(R)$. That means $\omega$ is the collection of all $\alpha$-open sets. Thus it doesn't matter whether we talk about $\omega$-closed or $\alpha$-closed. They are the same. If $f_{1}, f_{2},....$ are $\alpha$ converging to $f$ and $f$ is different from $f_{i}$ for all $i$, then for every $i$, there is an $x_{i}$ such that $f(x_{i}) <> f_{i}(x_{i})$. By taking $M =\{x_{1}, x_{2},....\}$, then we will have $B_{M}(f)$ doesn't contain any $f_{i}$ (contradiction). [/hide]

b) There are $S$ that are not closed relative to $\omega$.

Thank you very much for your help.

I think your proof works, but I don't quite follow the first part and don't see why it is needed. You can begin the proof with "If $f_{i}\to f$ and $f$ is different from $f_{i}$ for all $i$, then...".

(b) is equivalent to the statement that not all subsets of $F(R)$ are open. Is a singleton $\{f\}$ open?

To mlok,

Thank you very much. I think you are right. My first part is just redundant (This is mainly because of my uneasiness about this topic. I just want to change to $\alpha$ convergence so that I can easily work).

I think singleton subset isn't open. But must this information be given in the problem?

[quote="treegoner"]I think singleton subset isn't open. But must this information be given in the problem?[/quote]
I don't really understand this question. To solve (b), you must demonstrate that there is a subset that is not closed. I suggested that you prove that a singleton is not open; this immediately implies that the complement of a singleton is not closed, solving your problem.

To mlok.

Thank you for your suggestion. Actually, I misunderstood your question.


algebra
polynomial
calculus
calculus computations

The particular solution is easy to find, just guess.

[hide]It's $-\frac{1}{2}x^2+\frac{1}{2}x-\frac{1}{4}e^x$, right?[/hide]

Some nifty trick ought to be in order for the homogenous part though. I can't solve that polynomial. :\

[quote="kunny"][quote="Beginner"]This ODE is from my exam, several hours ago. The only problem that i didn't  finish. I thought it is Typo problem.
Solve
\[ \frac{d^3 y}{dx^3} - 2 \frac{d^2 y}{dx^2} - \frac{d y}{dx} -2 y = e^x + x^2  \]

 :?[/quote]

Sorry, but are you sure that this problem?

I think the real problems are the following two possibilities.

1.Solve
\[ \frac{d^3 y}{dx^3} - 2 \frac{d^2 y}{dx^2} - \frac{d y}{dx} +2 y = e^x + x^2  \]

2.Solve
\[ \frac{d^3 y}{dx^3}+ 2 \frac{d^2 y}{dx^2} - \frac{d y}{dx} -2 y = e^x + x^2  \]

kunny[/quote]

That's why I said this is typo problem. The caracteristic polynomial has one real root in $(2,3)$. It is still posible to solve it, but require some technique to find the numerical value of roots (we can also know that two of them are imaginer).
About your "remake" problem from Kunny
1) The roots of characteristic polynomial $r^3 - 2r^2 -r +2 = 0$ are $r_1=1$, $r_2=2$ and $r_3=-1$ So the homogenous solution is $y_h = C_1 e^{-x} + C_2 e^{x} + C_3 e^{2x}$. The  particular solution is $y_p = Axe^x + Bx^2+Cx+D$.
2)  The roots of characteristic polynomial $r^3 + 2r^2 -r -2 = 0$ are $r_1=1$, $r_2=-2$, and $r_3 = -1$. And the particular solution is the same as part 1, but posiblely different $A$, $B$,  $C$ , and $D$ .


As I traveled up and down our great and glorious country, I found myself in a place where the temperature goes up sharply in the day and down at night. This had an affect on my watch. I noticed it was $ \frac{1}{2}$ minute fast at nightfall, but at dawn it had lost $ \frac{1}{3}$ minute, making it only $ \frac{1}{6}$ minute fast.
One morning - May 1 - my watch showed the right time. By what date was it 5 minutes fast ?

[size=75]
from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide
[/size]

No!!!!!!!!!!!!!!!!!
[hide="Oh Well"]
In one day, the watch gained $ \frac{1}{6}$ of a minute. It would look like the watch would be 5 minutes fast in 30 days. However, on May 28, the watch was already 4.5 minutes fast. At the end of the day, the watch gained a bit more, so it was 5 minutes fast on May 28.[/hide]

I would just like to note that you, Small Unicorn, pay WAY to much attention to what time your watch displays.  And anyway, if you know the correct time so precisely, why wear a watch?

I enjoyed doing your puzzles, small_unicorn. Do you have any that don't involve telling the time? Thanks.

[hide]5/ 1/6 = 30.  Therefore, May 1 + 30 days = May 30.[/hide]

EDIT: I'm probably wrong, but can you explain your reasoning for this: [quote="$ LaTeX$"]No!!!!!!!!!!!!!!!!!
[hide="Oh Well"]
In one day, the watch gained $ \frac {1}{6}$ of a minute. It would look like the watch would be 5 minutes fast in 30 days. However, on May 28, the watch was already 4.5 minutes fast. At the end of the day, the watch gained a bit more, so it was 5 minutes fast on May 28.[/hide][/quote]?

Err, 30+1=31 and since May has 30 days, it is June 1, right?

It is [b]May 28[/b].

We know that on May 28, it was 4.5 minutes fast. But before nightfall, it gained .5, making it 5 minutes fast.

Where does it say that? I don't see anything about May 28 in the problem.

May 28 is the answer.....  :yup:

Conversation over...  :lol:

Seriously, where did you get May 28 from? Is it your birthday or something?

It is the answer, what do you not get....  :mad:  See my reasoning (the second post) for how I got it...

People these days, and no it is [u]not[/u] my birthday. :wacko:

Mods lock.

wow...[hide] The answer is May 28th and May 29th because it would be 5 minutes ahead during nightfall on May 28th and before dawn on May 29th[/hide]

I agree with LaTeX.

The watch gains $ 1$ minute every $ 6$ days (at dawn). So after $ 27$ days, at dawn on May $ 28$th, it gained $ 27/6 \equal{} 4.5$ minutes. Then by nightfall (still the $ 28$th) it gained another $ 0.5$ minutes, so that it was then $ 5$ minutes fast - on the $ 28$ of May.

Sorry, but I can't see how to improve on LaTeX's explanation.

Hi small_unicorn. Would still like to see other puzzles from your book (that aren't about time).


Three chords are drawn in a circle to create seven regions. The numbers from 1 to 7 are to be placed, one in each region, so that, for each chord, the total numbers in the circle on one side of the chord is equal to the total of the numbers on the other side.
How many possible values are there for the number $x$ in the central region?

[hide]Let the areas of the 3 regions bounded by two chords and part of the circumference be $a,b,c$.

We have $(28-x)+a+b+c=42\Rightarrow (a+b+c)-x=14$

the maximum of $(a+b+c)$ is $5+6+7=18$, so the maxmimum of $x$ is $18-14=4$

Therefore, $x$ has 4 possible values and they are 1,2,3,4[/hide]

quick sol
numbers from 1 to 7 are placed on the regions. It is nothing to do with area although you could consider it.(no point)
Your solution is close to the right answer.

Like I said it is close to the right answer. so no.
[hide="why"]
x=3 is not a solution because it does not satisfy the question
[/hide]


Does anyone know of a good Mathcounts coach for a homeschool mathcounts group in NYC?

I might have some leads.  Send me a PM about your group.

wait. you don't have to go to a school to get on a mathcounts team. how do u enter like this?
-jorian

well we registered our math group as a school, even though we are just a group, so, I guess you can do it in with anyone.

i do!!!!!!!

 :rotfl:  :rotfl: [hide]mickey elias[/hide]


What are some good Wii Games to buy?
I need to know

The only games I have are:
Madden 2007
Wii Sports

and a bunch of GameCube games when I get bored

Please don't include Virtual Console games.

Thanks in Advance

[quote="7h3.D3m0n.117"]I'm an Xbox fan. I heard the launch titles for Wii are dissapointing. But try Zelda. I heard it was pretty good. Also, I think I saw Call of Duty for the Wii at Target or something.[/quote]

XBox is pretty raw with Halo 2
but
I'm a fanboy

The launch titles weren't bad. They just included WAY TOO MANY VIRTUAL CONSOLE GAMES (aka old nintendo games that are played on the Wii emulator). I have almost every single system made by Nintendo, so no need for those virtual console games (there are like 500 of them too).

I hate playing Zelda (mainly because I don't have the time to learn the controls :rotfl: )(and because it's just too childish for me :ninja: )

Call of Duty 3, I will consider
Haha, I love shooting/killing video games :D

[quote="ra5249"]Not liking Zelda? You are missing out on so much.
Anyways, after twilight princess, I would recommend Warioware and Sonic/Secret Rings.[/quote]

Yeah...I just wished I liked RPG's (I'm not into complex video games. I like pick up and play games. Sports games, Shooting Games, and etc)

Warioware I might consider (it gets boring sometimes)
Sonic/Secret Rings I am so not going to buy (mainly because it's not worth the time for me. Probably for others though)


There is a three way duel between persons A, B, and C. When person A tries to shoot someone, he has a 1/3 chance of hitting; when person B tries to shoot someone, he has a 1/2 chance of hitting; and when person C tries, he has a 100% chance of hitting. The duel proceeds by each person getting one shot in the order A, B, then C. On each person's turn, they decide who they want to take a shot at and then they get one shot. If someone is eliminated, the duel continues, in order, between the remaining persons.

The question is, what is person A's best strategy on his first shot (assuming all players are rational and intent on winning), and what is the probability of A winning the duel following the best strategy possible?

I've seen one problem similarly recently. :D

Btw: I'm new here, how am I supposed to post my attempt to the solution?

Just type away! If you want to display a mathematical expression its best to use LaTeX code. You don't need to download anything and it doesn't take long to learn. You can even click on other people's expressions and copy and paste the commands that you want from it. Try it here:

$ \dfrac{1 \plus{} \frac {1}{x}}{\pi^2 \plus{} \sqrt {x \plus{} 1}}$

A quick primer can be found in the LaTeX link to the left, but the most important commands are here: http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php

Who are you, by the way?

This is Philip.
[hide="Answer"]Shoot C, 13/36 probability of living[/hide]
[hide="Solution"]Let's consider the possibility of A choosing to shoot B:
-If Successful (1/3), C shoots next and wins.
-If Unsuccessful (2/3), choice goes to B. As C's winning possibility by shooting B is greater, B must first deal with C or certainly die with C's shot:
---If Successful (1/2), A and B will continue to shoot at each other, with (1/3) probability A is next successful, (2/3*1/2=1/3) probability B is next successful, and the rest if everybody misses and we try again, with the same probabilities as before.  Both are equally likely to win, so A wins with (1/2) probability.
---If Unsuccessful (1/2), B is shot next and A wins with (1/3) probability.

Let's now consider the possibility of A choosing to shoot C:
-If Successful (1/3), A and B shoot at each other again only now, B shoots first, with (1/2) probability B wins first round, (1/2*1/3) probability that A wins next round, and the rest again being if everyone tries again with the same probabilities. A's probability is 1/3 B's, so the probability A wins is (1/4).
-If Unsuccessful (2/3), choice goes to B as above, with (1/4+1/6) Probability.

Quickly, we find that A's initial failure leads to the same events regardless of choice, while A's initial success lends more chances for survival only when A is shooting at C. Thus, A should shoot at C first.

A's survival probability is (1/3*1/4)+(2/3*(1/4+1/6))=[b]13/36[/b]
I hope.[/hide]

I agree with Philip. Just saved myself 20 minutes of typing. :blush:
Also, A would have a higher survival probability if he aims at the sky first.  :)

Btw, I'm Qianli.

[hide]Ah. Very insightful analysis Phillip - with the correct probability computed for that strategy. But ultimately missing a better option for A!

[hide]Qian Li is correct. :) Although a proof is needed to verify that it is in deed a better strategy.[/hide][/hide]

I thought of that just a little too late. Failure. :(

[hide="Further Work"]
Refer to the previous solution.  Whenever A is unsuccessful, regardless of who he tries to shoot, the choice goes to B.  In this scenario, the probability that A will survive is (1/4 + 1/6)=5/12.  This probability has less impact, however, because A is unsuccessful 'only' 2/3 of the time. By shooting at the sky, moon, trees, ext, A guarantees himself of this 5/12 probability. Because (5/12)>(13/36), the wild-shot strategy is indeed the best.
[/hide]

Yay! More new people!

Hpoefully I'll be there on the 9th to meet everyone: I've only met Mr. R and seen Sean. I think I met Qianli for a minute too...


Given that $P=\{p_1, p_2, ... , p_k\}$ is a set of distinct, not necessarily consecutive primes, prove that $\frac{1}{p_1}+\frac{1}{p_2}+...\frac{1}{p_k}$ is never integer.

Take any prime from the list.  Let that prime be $x$.
Thn assume that the sum is an integer, and let that integer be equal to $y$.
If you remove $x$ from the list, the sum is now $\frac{yx-1}{x}$.
Now take another prime from the list, and let that be $z$.
If we remove this from the list, the sum is now $\frac{z(yx-1)-1}{xz}$.
Continuing in this manner, eventually we will have one term left.  Its sum must then have a denominator of $p_1p_2p_3\ldots p_k$, where one number is missing from the product. Let that term be called $p_i$.
Then since this term is equal to the sum of the list, we have that $p_i=p_1p_2p_3\ldots p_k$.  But since all $p_k$ are prime, this is impossible because it would mean that the product of multiple primes is anothe prime, which cannot happen.  Therefore our assumption was false and the prduct cannot be an integer.


Yes I know this is satirical.
Rules: +1 heal, -2 hurt. You may place your -2 hurt in different areas.

Top COUNTRY on AoPS [Wickedestjr]: 25
top sport on AOPS [jjx1]: 25
Top Staff of AoPS (Elimination Game) [Nexmus]: 26
Top Video Game Companies [Wickedestjr]: 25
top soda on aops [jjx1]: 24
Top TV show elimation game on AoPS [crazychinesemaniac]: 25
Top Math Subject Hurt&Heal [james4l]: 25
Civil War 1/2/3 [tinytim/james4l/chenhsi]: 25
3 team tug of war [(^_^)]: 25
Top Chip on AOPS [jjx1]: 24
Top AOPSer on AOPS [(^_^)]: 25
Prez Game [jjx1]: 25
Top Hurt and Heal-ish Game on AOPS [james4l]: 25

+1 Staff of AOPS, -1 on top soda and top chip

OMG PLEASE!!! STOP THIS!!! NOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Top COUNTRY on AoPS [Wickedestjr]: 25
top sport on AOPS [jjx1]: 25
Top Staff of AoPS (Elimination Game) [Nexmus]: 26
Top Video Game Companies [Wickedestjr]: 25
top soda on aops [jjx1]: 24-1=23
Top TV show elimation game on AoPS [crazychinesemaniac]: 25
Top Math Subject Hurt&Heal [james4l]: 25
Civil War 1/2/3 [tinytim/james4l/chenhsi]: 25+1=26
3 team tug of war [(^_^)]: 25
Top Chip on AOPS [jjx1]: 24
Top AOPSer on AOPS [(^_^)]: 25
Prez Game [jjx1]: 25-1=24
Top Hurt and Heal-ish Game on AOPS [james4l]: 25

Top COUNTRY on AoPS [Wickedestjr]: 25
top sport on AOPS [jjx1]: 23
Top Staff of AoPS (Elimination Game) [Nexmus]: 26
Top Video Game Companies [Wickedestjr]: 25
top soda on aops [jjx1]: 23
Top TV show elimation game on AoPS [crazychinesemaniac]: 25
Top Math Subject Hurt&Heal [james4l]: 25
Civil War 1/2/3 [tinytim/james4l/chenhsi]: 27
3 team tug of war [(^_^)]: 25
Top Chip on AOPS [jjx1]: 24
Top AOPSer on AOPS [(^_^)]: 25
Prez Game [jjx1]: 24
Top Hurt and Heal-ish Game on AOPS [james4l]: 25

At least make it obvious which one you subtract/add to...
Proposal for alliance: We make this game win, first eliminating any game being made by jjx1, Wickedestjr, and (^_^) (in that order).

[color=red]Top COUNTRY on AoPS [Wickedestjr]: 23[/color]
top sport on AOPS [jjx1]: 23
Top Staff of AoPS (Elimination Game) [Nexmus]: 26
Top Video Game Companies [Wickedestjr]: 25
top soda on aops [jjx1]: 23
Top TV show elimation game on AoPS [crazychinesemaniac]: 25
Top Math Subject Hurt&Heal [james4l]: 25
Civil War 1/2/3 [tinytim/james4l/chenhsi]: 27
3 team tug of war [(^_^)]: 25
Top Chip on AOPS [jjx1]: 24
Top AOPSer on AOPS [(^_^)]: 25
Prez Game [jjx1]: 24
[color=green]Top Hurt and Heal-ish Game on AOPS [james4l]: 26[/color]

[i]Something[/i] seems off, but I can't put my finger on what...  :huh:

[quote="perfect628"]Top COUNTRY on AoPS [Wickedestjr]: 0
top sport on AOPS [jjx1]: 0
Top Staff of AoPS (Elimination Game) [Nexmus]: 0
Top Video Game Companies [Wickedestjr]: 0
top soda on aops [jjx1]: 0
Top TV show elimation game on AoPS [crazychinesemaniac]: 0
Top Math Subject Hurt&Heal [james4l]: 0
Civil War 1/2/3 [tinytim/james4l/chenhsi]: 0
3 team tug of war [(^_^)]: 0
Top Chip on AOPS [jjx1]: 0
Top AOPSer on AOPS [(^_^)]: 0
Prez Game [jjx1]: 0
Top Hurt and Heal-ish Game on AOPS [james4l]: 0[/quote]

Game over! :P

This should be like a voodoo game where when the game dies here, the corresponding topic is locked. And deleted. Forever. And overwritten so recovery systems won't work.

Or, even better, we skip the "game" altogether, but still do that.

I think someone wishes he had mod powers...  :ninja:

It [i]would[/i] make resolving these issues easier... :roll:

Mwahahaha?

I would love at least mod powers on my own topics... (thinking of stupid Russian Roulette where the first post is frozen on signups)

Top COUNTRY on AoPS [Wickedestjr]: 24
top sport on AOPS [jjx1]: 23
Top Staff of AoPS (Elimination Game) [Nexmus]: 24
Top Video Game Companies [Wickedestjr]: 25
top soda on aops [jjx1]: 23
Top TV show elimation game on AoPS [crazychinesemaniac]: 25
Top Math Subject Hurt&Heal [james4l]: 25
Civil War 1/2/3 [tinytim/james4l/chenhsi]: 27
3 team tug of war [(^_^)]: 25
Top Chip on AOPS [jjx1]: 24
Top AOPSer on AOPS [(^_^)]: 25
Prez Game [jjx1]: 24
Top Hurt and Heal-ish Game on AOPS [james4l]: 26

Top COUNTRY on AoPS [Wickedestjr]: 24
top sport on AOPS [jjx1]: 23
Top Staff of AoPS (Elimination Game) [Nexmus]: 24
Top Video Game Companies [Wickedestjr]: 25
top soda on aops [jjx1]: 23
Top TV show elimation game on AoPS [crazychinesemaniac]: 25
Top Math Subject Hurt&Heal [james4l]: 25
Civil War 1/2/3 [tinytim/james4l/chenhsi]: 25
3 team tug of war [(^_^)]: 25
Top Chip on AOPS [jjx1]: 24
Top AOPSer on AOPS [(^_^)]: 25
Prez Game [jjx1]: 25
Top Hurt and Heal-ish Game on AOPS [james4l]: 26

Top COUNTRY on AoPS [Wickedestjr]: 24
top sport on AOPS [jjx1]: 23
Top Staff of AoPS (Elimination Game) [Nexmus]: 24
Top Video Game Companies [Wickedestjr]: 24
top soda on aops [jjx1]: 23
Top TV show elimation game on AoPS [crazychinesemaniac]: 25
Top Math Subject Hurt&Heal [james4l]: 25
Civil War 1/2/3 [tinytim/james4l/chenhsi]: 25
3 team tug of war [(^_^)]: 25
Top Chip on AOPS [jjx1]: 24
Top AOPSer on AOPS [(^_^)]: 25
Prez Game [jjx1]: 25
Top Hurt and Heal-ish Game on AOPS [james4l]: 27

-video game companies
+ top hurt and heal-ish

SERIOUSLY
THIS
IS
ANNOYING
PLEASE
LOCK
THIS
NOOOOOOOO!!!

Top COUNTRY on AoPS [Wickedestjr]: 24
top sport on AOPS [jjx1]: 23
Top Staff of AoPS (Elimination Game) [Nexmus]: 24
Top Video Game Companies [Wickedestjr]: 24
top soda on aops [jjx1]: 23
Top TV show elimation game on AoPS [crazychinesemaniac]: 25
Top Math Subject Hurt&Heal [james4l]: 25
[color=green][b]Civil War 1/2/3 [tinytim/james4l/chenhsi]: 26[/b][/color]
3 team tug of war [(^_^)]: 25
Top Chip on AOPS [jjx1]: 24
[color=red][b]Top AOPSer on AOPS [(^_^)]: 23[/b][/color]
Prez Game [jjx1]: 25
Top Hurt and Heal-ish Game on AOPS [james4l]: 27
[color=yellow][b]Top Brawl Character [erabel]: 24[/b][/color]

Brawl Character Has Officially Joined. Notice that each time a new Top game arrives it starts with 1 point less, so the next Top game that arrives has 23 initial points.

Top COUNTRY on AoPS [Wickedestjr]: 24
top sport on AOPS [jjx1]: 23
Top Staff of AoPS (Elimination Game) [Nexmus]: 24
Top Video Game Companies [Wickedestjr]: 24
top soda on aops [jjx1]: 23
Top TV show elimation game on AoPS [crazychinesemaniac]: 25
Top Math Subject Hurt&Heal [james4l]: 25
[color=green][b]Civil War 1/2/3 [tinytim/james4l/chenhsi]: 27[/b][/color]
3 team tug of war [(^_^)]: 25
Top Chip on AOPS [jjx1]: 24
[color=red][b]Top AOPSer on AOPS [(^_^)]: 21[/b][/color]
Prez Game [jjx1]: 25
Top Hurt and Heal-ish Game on AOPS [james4l]: 27
[b]Top Brawl Character [erabel]: 24[/b]

[quote="james4l"]Brawl Character Has Officially Joined. Notice that each time a new Top game arrives it starts with 1 point less, so the next Top game that arrives has 23 initial points.[/quote]
okay.

seriously.
we have established the fact that we don't want any more hurt and heals.
the first couple about food and stuff were fine. but seriously.
can't we just have a POLL?????

that's what they're for.
and this is getting insanely annoying.
i come and check the games forum for the games that I'm in, and I see
"TOP SODA ON AOPS"
"TOP CHIP ON AOPS"
"TOP TV SHOW ON AOPS"
"TOP WHATEVER ON AOPS"
we really need to stop.

well yes Thats why I wanted to start FFC to control all these hurt and heals and stuff Its somewhere at the bottom of the page or the 2nd page. But since a mod locked that people can have as many hurt and heals until a mod comes on and locks it.

Top COUNTRY on AoPS [Wickedestjr]: 24
top sport on AOPS [jjx1]: 23
Top Staff of AoPS (Elimination Game) [Nexmus]: 24
Top Video Game Companies [Wickedestjr]: 24
top soda on aops [jjx1]: 23
Top TV show elimation game on AoPS [crazychinesemaniac]: 25
Top Math Subject Hurt&Heal [james4l]: 25
Civil War 1/2/3 [tinytim/james4l/chenhsi]: 25
3 team tug of war [(^_^)]: 25
Top Chip on AOPS [jjx1]: 24
Top AOPSer on AOPS [(^_^)]: 21
Prez Game [jjx1]: 26
Top Hurt and Heal-ish Game on AOPS [james4l]: 27
Top Brawl Character [erabel]: 24

we have  mods, they're enough.
and FFC, from my point of view, is just trying to overthrow the mods of Fun Factory, because the mods are here to control what's going on.

-_-

mods, please do something about this.

BUt see, tahts the problem. Mods are people too. they can't be on 24/7. they're probably really busy and can't come on a lot. I mean klebian is never on fun factory.

Top COUNTRY on AoPS [Wickedestjr]: 24
top sport on AOPS [jjx1]: 23
Top Staff of AoPS (Elimination Game) [Nexmus]: 24
[color=red][b]Top Video Game Companies [Wickedestjr]: 23[/b][/color]
top soda on aops [jjx1]: 23
Top TV show elimation game on AoPS [crazychinesemaniac]: 25
Top Math Subject Hurt&Heal [james4l]: 25
[color=green][b]Civil War 1/2/3 [tinytim/james4l/chenhsi]: 26[/b][/color]
3 team tug of war [(^_^)]: 25
Top Chip on AOPS [jjx1]: 24
Top AOPSer on AOPS [(^_^)]: 21
Prez Game [jjx1]: 26
Top Hurt and Heal-ish Game on AOPS [james4l]: 27
Top Brawl Character [erabel]: 24


[quote="chien than"]Continue....
Find $x;y;z \in Z$ such that:
$x^{y}+1=z$[/quote]

i think you've written something wrong. because for each pair of $x\in \mathbb Z,y\in \mathbb N\cup \{0\}$, there exists a $z\in \mathbb Z$. if $y$ is negative it has no solutions.

[quote="chien than"]Find $x;y \in P$ such that:$2^{x}+1=y$[/quote]

Here is another approach.

A "Fermat prime" is a prime of the form $2^{n}+1$, it can be shown (like Rust) that it is necessary that $n=2^{m}$.

Note, $2^{x}+1=y$ hence $2^{x}+1$ is a prime because $y$ is a prime. Hence it is a Fermat prime. Which implies that $x=2^{m}$, but it is also a prime. Which only happens when $m=1$ thus $x=2 , y=5$

I think he meant in the second question that z had to be prime, like the first question.


If $ f(1)\equal{} 5$ and $ f(n\plus{}1) \equal{} 2f(n)$, what is $ f(7)$?

hello, easy we get $ y_n\equal{}5\cdot2^{n\minus{}1}$, therefore is $ y_7\equal{}320$.
Sonnhard.

[quote="AndrewTom"]If $ f(1) \equal{} 5$ and $ f(n \plus{} 1) \equal{} 2f(n)$, what is $ f(7)$?[/quote]

$ 5\cdot2^6\equal{}64\cdot5\equal{}320.$

Sorry, kunny, $ n$ is a positive integer.

There are 6 n's less than 7. So 2^6*5=64*5=320