AMC USA(J)MO USAMO geometry email ARML AIME

We also have to find a place to administer it, because we are a homeschool group and the AMC/AIME/USAMO can't be given in the home.

This year, the University of Maryland (UMBC) has agreed to let me proctor the test there.  Is there a college our university near you with a faculty member or graduate student who could proctor the test?  If it is okay with your school (and you should ask first), perhaps you could contact the chair of the math department.

If your school does go that route, they need to check with the AMC administration to make sure everything is done according to proper procedures.  This is important; you don't want anything to invalidate your results.

It can be difficult for a school to put aside time and space for a nine hour, two day test.  You're right, it would be good not to have to do it during a lecture, but they may be strapped for alternatives.  If you can offer them one, and get permission to be out of school during the USAMO times, they might be willing to consider it.

Good luck!

Best,
Dr. Asaro

[quote="pkerichang"]Can we possibly change the testing site to other places that administer USAMO???

Because our administrator told me that I'll have to take the USAMO test in his class WHILE he's lecturing, and when students walks in and out of the class...I don't think that's the best environment for working out nice proofs...

His justification is that the school can find no teacher and no room to be available for straight 4.5 hours.  When I try to protest he's like "real mathematicians won't be affected by the environment"... :o OK, I might not be good, but at least I want to make this experience worth it since I'm a senior already...

If we can, where/how do we change it??

Thank you for all your replys.[/quote]

Do *everything you can* to find a better site.  I took the USAMO my first year in the teacher's typing room.  That was awful.  My last two years, they gave me the conference room in the guidance department (which was the only good thing the guidance department did for me my senior year, but that's another story).

I believe there are other people in your area taking it, including a homeschooler (if I'm remembering your locale correctly).  Talk to some of them, see if you can take the test with them, and OK everything with the AMC.

Don't take it in a room where a lecture is being given.  This may even be against AMC rules. (I don't know for sure if it is, but I wouldn't chance it if I were you.)

Also, I forgot to mention in the above, it is possible to do it with another high school, too.  Or with a homeschool group, as Richard mentioned.  Are you in California?  (I'm guessing from the name Hacienda Heights :-)).

Last year, we did it at Montgomery Blair, and I helped proctor in return for them letting my student sit with theirs.  But the coach was willing to let us do it even if I hadn't been able to proctor.  Unfortunately, in Maryland, the Board of Education doesn't actually allow homeschool students to do such activities with the schools, so this year we are going to UMBC.  If you're already in a traditional school, though, you probably wouldn't need to worry about that.

Regardless of whether you go to a university or another high school, your contest manager should make sure he fills out the forms with his address and email, so he gets your results and your scores are properly credited to your school.

Best,
Catherine Asaro

I also posted in the So Cal ARML forum, looking for an alternate site.  Thanks for all the suggestions.

Yeah, during the AIME I got annoyed enough because it's right next to the quad, and they play BOOMING music during break...I have to force a 15 minute rest because I stopped thinking...

Here's a question: how many readers have ever actually taken the USAMO in a classroom with regular classes in progress? I'm curious as to how widespread this might be.

I think for one year, I took the AIME in a classroom that was occupied for part of the time, but that's it.  I also seem to recall that the class was doing something pretty quiet for most of the time.  And I may actually be making this all up; my memory's pretty fuzzy.

And certainly never the USAMO.  I got a conference room to myself for three years for that, and MIT classrooms for 2002.

I have heard of a student who had to take the USAMO in the detention/in-school suspension room!  (I.e., the room had been designated for use by students who were serving terms of in-school suspension.  The other students were supposed to be sitting there silently doing their assigned work from their regular classes.)

This was back in the day when USAMO was a six-hour one day exam.  The other students in the room expressed amazement that anyone would voluntarily take a six-hour math test!

It doesn't sound like the most pleasant atmosphere for a USAMO and although the other students were supposed to be working silently throughout the day, one imagines that the students assigned to the room may not always be entirely compliant.

Finding a good room available for 9 hours over two days is a real challenge.

The local colleges have been hospitable and helpful, but it's hard to find a suitable room that is free for the entire period needed.

Small group studies in the college library have been reliably available (especially now that USAMO is a bit earlier in April; there is more student demand for them as May finals approach, but since they are pretty vacant in the third week of April, librarians are generally amenable to allowing them to be reserved for USAMO.)   This can be a good solution if the number of students taking the USAMO is only one or two.

Some group studies have a somewhat "cave-like" atmosphere, i.e., no windows.  Others have windows, but they don't open (which can lead to uncomfortable fish-bowl greenhouse effects in late April, because temperature shifts in the northeast can be rapid and HVAC systems don't always immediately adjust.)

Last year, we were given a meeting room in the student union, which seemed nice enough when I checked it out a few weeks ahead of time.  But during the actual USAMO, it was uncomfortably warm unless we kept the door and windows opened, and it was a beautiful warm sunny day outside and there was lots of noise of students frolicking about right underneath the window.  (The room was located immediately over the entrance to the building, which had benches and a patio for student socializing.  The day I had checked out the room two weeks in advance had been cold and rainy--all was quiet near the room.  Very different from the actual USAMO days.)

Finding a good room for 9 hours in April really is a challenge, whether it's a high school or a college.

I took the AIME in my school's conference room (me and two other people)

Though, my principal is very considerate of the math team.
I actually thinks he's sometimes bias towards us (in a good way)

I don't know how your principal would feel about having one of the top 505 mathematicians in his school, but I know my principal would be ecstatic.

Approach your principal about this
I am sure he will find an alternative for you

I took the AMC in my teachers classroom.
For the AIME I took it in one of the almost never used rooms in the library.
For the USAMO, I'm not sure what I'm going to do.

I took the AMC in the theater, the AIME I in "an almost-never used room in the library" as well, and I didn't make USAMO, but I'm pretty sure they would've given me that room again. It's literally almost never used.

AMC = teachers room
AIME = teachers room, then had to move elsewhere, while pushing through crowd of students going to next period...
USAMO = I didn't get my score yet. don't know.

My USAMO room was "cave-like" as sophia described.  I actually liked it.  It helped me focus, I think.

We did the AMC-A in a classroom at UMBC, which was a bit cramped, but quiet (Edit: students sit alternating AMC-10/AMC-12, so no students taking the same test sit next to each other).  For the AMC-B, we took it in the county library, which had more room.  It was quiet, too.

For the AIME, we got a conference room in one of the nicer buildings on campus.  We also get to use those conference rooms for the Maryland Math League.  They are quiet and usually have plenty of space, enough space for 8-9 students.

For the USAMO, I don't know yet, but I suspect it will be one of the conference rooms.

When I proctored at Blair, they had a room like a lab.  It was a good room with plenty of space.  The school was noisy, though probably not more than any other school.  Bells going off, students talking outside, that sort of thing.  You could hear everyone talking in the hall outside the room.

Every now and then someone would stride into the room and insist he had a class there.  At one point, a class =did= come in, so all the USAMO students had to move (I think there were 11, most from MB).  We had warning ahead of time, and we all trooped to a computer lab in another part of the building, a cramped room with barely anywhere to put papers.  (Edit: Everyone just walked to the other room down the hall together, without talking or looking at papers.  They were cramped because of the equipment in the room; no students were sitting next to each other).  Everyone kept working diligently.

Then another bell rang and we all trooped back to the other room.  I had to politely convince the people lingering to talk from the class that they actually wanted to leave, because a test was going on.  It was ... interesting.

And in all that to-do, Brian Lawrence wrote a perfect paper, and Richard McCutchen was another of the USAMO winners. Pretty impressive.  :-)

Best,
Catherine

[quote="CatherineAsaro"]We did the AMC-A in a classroom at UMBC, which was a bit cramped, but quiet.  For the AMC-B, we took it in the county library, which had more room.  It was quiet, too.

For the AIME, we got a conference room in one of the nicer buildings on campus.  We also get to use those conference rooms for the Maryland Math League.  They are quiet and usually have enough space for 8-9 students.

For the USAMO, I don't know yet, but I suspect it will be one of the conference rooms.

When I proctored at Blair, they had a room like a lab.  It was a good room with plenty of space.  The school was noisy, though probably not more than any other school.  Bells going off, students talking outside, that sort of thing.  You could hear everyone talking in the hall outside the room.

Every now and then someone would stride into the room and insist he had a class there.  At one point, a class =did= come in, so all the USAMO students had to move (I think there were 11, most from MB).  We had warning ahead of time, and we all trooped to a computer lab in another part of the building, a cramped room with barely anywhere to put papers.  Everyone kept working diligently.

Then another bell rang and we all trooped back to the other room.  I had to politely convince the people lingering to talk from the class that they actually wanted to leave, because a test was going on.  It was ... interesting.

And in all that to-do, Brian Lawrence wrote a perfect paper, and Richard McCutchen was another of the USAMO winners. Pretty impressive.  :-)

Best,
Catherine[/quote]

Couldn't they have somewhat "cheated"
Though I highly doubt it (like a 95% doubt that they didn't cheat)
Just wondering...

When a teacher administers the USAMO, they need to sign a statement. Here are two of the points they are agreeing to:
[quote]
3. Arrange for the exclusive use of a room for four and a half hours (from 12:30 p.m. to 5:00 p.m., eastern
daylight time or your time zone equivalent, see below) on the designated days which will be used to administer
the USAMO.
4. Be present in the room to supervise the student(s) during the entire nine hours duration of the examination.[/quote]

I believe at some point in the earlier process, they agree to offer the USAMO if they have a qualifier.

Can you, or Jay, get your parents involved? I think someone needs to talk to the administration. You both qualified for a very prestigious exam. This is not something to be dealt with in such a shoddy fashion. I suspect if a parent, or trusted adult, was part of this process, your teacher might find a way to make this happen, even if it is a bit of an inconvenience for them. too bad.

If you families can afford it, offer to help pay for a proctor.

Good Luck!

[quote="CatherineAsaro"][quote="Aznwithabrain"]Couldn't they have somewhat "cheated"
Though I highly doubt it (like a 95% doubt that they didn't cheat)
Just wondering...[/quote]

No.  I was proctoring it the entire time.

That is truly a cheap shot, Aznwithabrain, for students who were at a distinct disadvantage due to their situation.[/quote]

oops i didn't word my phrase right
sorry
I mean that I was 95% sure they took the test without cheating

[quote="Aznwithabrain"][quote="CatherineAsaro"]No.  I was proctoring it the entire time.

That is truly a cheap shot, Aznwithabrain, for students who were at a distinct disadvantage due to their situation.[/quote]

oops i didn't word my phrase right
sorry
I mean that I was 95% sure they took the test without cheating[/quote]

Edit: It occurred to me you might have been referring to the students being cramped rather than their having to move.  They were cramped because of equipment in the room, which left little space to write.  No one was next to each other; they were all separated by equipment.

The post sounded as if you referred to their having to move.  That was why it sounded like a cheap shot; because they chose to do the right thing even though it put them at a disadvantage.  Moving was the appropriate response compared to staying in a classroom where a class was going on. It wasn't a huge disadvantage, but every moment matters.  It isn't only lost time, but also the disruption of concentration.  If you only meant the cramped space, though, that is a misunderstanding which is easy to clear up.

Also, for AMC exams, a lot of schools alternate AMC-10 and AMC-12 students, in addition to separating them by extra seats if such is available.

----
Original post ...

And it is a cheap shot.  I was proctoring it, and these are some of the top students in the world.  I am 100% sure.

Are you aware that one of these students you have made this suggestion about wrote a perfect paper at the IMO the year before?

I would suggest you think a bit more carefully before making such posts.

ok...
I meant the part where I put the "doubt" part
And I just put 95% cause I don't know any of your students or anything
I was just wondering
You don't have to get all defensive
It wasn't a shot at all at them

Aznwithabrain, I suspect you don't realize the seriousness of what you implied.  These aren't my students; they are internationally known in the math community.

I gather from your last post that you meant no offense.  Fair enough.  Sometimes these things happen, and especially when you are newer to posting, you may not realize how your words come across.

[off-topic] Yeah, posts can come across way differently than what you thought it would come off like, definitely. In particular, sarcasm doesn't really work at all. [/off-topic]

Slightly back on topic...
It was rather funny when I took the AIME, because I had four different proctors (they came one after the other, like in shifts), because previous three all had classes to teach, and no one was available for the entire duration.  :P

[quote="contradictory"]For the AIME, I was put into a NOISY AP Statistics classroom with constant chatter and unruly classroom discussion. :([/quote]

I think my friend had to take the AMC-10 during a morning Algebra 2 class because he had to play in a football game :roll:

Though his situation was a little better than yours

Edit:
[quote="davidyko"][off-topic] Yeah, posts can come across way differently than what you thought it would come off like, definitely. In particular, sarcasm doesn't really work at all. [/off-topic]
[/quote]

Yeah
Agreed Definitely
Thats why they should have Chatrooms instead :rotfl:
(Although these emoticons do help :rotfl: )

[quote="TimReyn"][quote="Aznwithabrain"][quote="contradictory"]For the AIME, I was put into a NOISY AP Statistics classroom with constant chatter and unruly classroom discussion. :([/quote]

I think my friend had to take the AMC-10 during a morning Algebra 2 class because he had to play in a football game :roll:
[/quote]

Let me preface this comment by saying that I do not suspect that either contradictory or Aznwithabrain's friend cheated or had any dishonest intentions whatsoever.

Hmm, something irks me about taking AMC's during a math class.  Besides the fact that it's distracting...well, they're discussing math.  Hearing (even inadvertently) what they're discussing could inspire a new idea for solving a problem.  For instance, next year, when calculators aren't allowed, if there's a trig problem and you can't remember if the sine of 120 degrees is $\frac{1}{2}$ or $\frac{\sqrt{3}}{2}$, there's the possibility that the class you're in is doing trig.  And if you overhear the correct answer, what are you supposed to do?  Derive the value yourself?  Intentionally get it wrong?  Leave it blank?  There's no good, honest answer.[/quote]

True your statement does have a point
but
my friend did horrible on it :rotfl:
I think a 100 something :rotfl:

I'll explain what happened with me. A few years ago, I was the only qualifier with the EPGY Stanford program. I was already the only person that they administered AIME for, and I felt quite grateful for the 3 hour proctoring they gave me. Needless to say, 9 more hours was pretty harsh. I spoke to the administrator about taking the exam at the local high school, and he got everything done.

So here's what I would do:
1. Find a high school/other group willing to proctor for you
2. Speak to the group who administered your AIME exam
3. If the above 2 fail, contact the AMC offices directly and explain your situation

(use e-mail, as time is pretty tight)

[quote="pkerichang"]So, um, to bring back kinda on topic, what will be the procedures necessary to change the testing site??? Do I just email them and ask??[/quote]

Oh that is pretty basic. All qualifiers have to fill out a form, an S-III. On it you list where you are going to take the USAMO. The person administering the test also needs to fill out a form. There are pointers to the forms in your qualifier email.

For most people they are going to just list the same place the qualified. You can change it. I think it is due on April 20th, so you have to  resolve this by then.


Daniel chooses a diagonal from a regular polygon having  8  sides . Celine chooses another diagonal from the same polygon. What is the probability that they chose two diagonals of the same length? Give your answer as a common fraction in lowest terms.

[hide]Since this is an octagon, there are 20 diagonals.  When you draw the diagonals, you see that there are eight diagonals that are from one vertex to another that is two vertices away.  There are eight diagonals whose endpoints are three vertices apart.  And finally, there are four diagonals that span the whole octagon.  This means that there are three types of diagonals Daniel can choose.  You can treat this problem like a probablity problem with eight green marbles, eight blue marbles, and four red marbles, with replacement.  I think that the answer is (8/20)^2 + (8/20)^2 + (4/20)^2 or 4/25+4/25+1/25=9/25[/hide]

They can't?  Does it say that in the problem or is it implied?

[quote="#H34N1"]Celine chooses [b]another [/b]diagonal from the same polygon. What is the probability that they chose [b]two diagonals[/b] of the same length? [/quote]

It's not explicit, but my reading, based on the bolded words above, is that they can't pick the same diagonal.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $f(n)$ denote the number of (isomorphism classes of) groups of order $n$. Given $m,n \in \mathbb{N}$, find a necessary and sufficient condition that $f(mn)=f(n)$.

--Vesselin

With the definition of $g(n)$ in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=automorphisms&t=45849]this thread[/url], I think a necessary and sufficient set of conditions (I can't call this "one condition" :)) would be $(*)\ (m,n)=(m,\varphi(m))=(m,g(n))=(n,g(m))=1$.

First, the necessity of $(*)$:

There are precisely $f(n)$ groups of the form $\mathbb Z_m\times G,\ |G|=n$, so these must be all of them. The expression of a finite group as a direct product of indecomposable factors is unique up to isomorphism, so, in particular, all our groups must have a direct factor isomorphic to $\mathbb Z_m$. If, say, $p^a\|m,p^b\|n,\ a,b\ge 1$, then the group $\mathbb Z_{p^{a+b}}\times\mathbb Z_{\frac{mn}{p^{a+b}}}$ has no such direct factors, so we get a contradiction. This shows that $(m,n)=1$.

Now, if $(m,\varphi(m))>1$, then there is a group $H$ of order $m$ which is not cyclic, and $H\times G,\ |G|=n$ would have no order $m$-cyclic indecomposable factors, thus yielding a contradiction.

Assume now that $p|m,q^b\|n,\ b\ge 1$ where $p,q$ are primes, and also $p|g(q^k)$ or $q|p-1$ (so that either $(m,g(n))>1$ or $(n,g(m))>1$). Then we can take our group to be the direct product between a semidirect product of $(\mathbb Z_q)^b$ by $\mathbb Z_p$ via a non-trivial homomorphism from $\mathbb Z_p$ to $\mbox{Aut}((\mathbb Z_q)^b)$ and some group of order $\frac{mn}{pq^b}$, or the other way around: the direct product between a semidirect product of $\mathbb Z_p$ by $\mathbb Z_q$ via a non-trivial homomorphism from $\mathbb Z_q$ to $\mbox{Aut}(\mathbb Z_p)$ and some group of order $\frac{mn}{pq}$. Either way, the group we get does not satisfy the requirement that every element of order dividing $m$ commute with every element of order dividing $n$, so it can't be of the form $\mathbb Z_m\times G,\ |G|=n$.

Now the sufficiency of $(*)$:

Suppose $n$ is even. Then, since $(n,g(m))=1$, we must have $m=2$, contradiction. In the exact same way we prove that $m$ is odd (now using $(m,g(n))=1$), so $mn$is odd. This means that all the groups we're working with are soluble (yeah, the Feit-Thompson Theorem :)). It suffices to prove that all subgroups of order $mn$ are of the form $\mathbb Z_m\times G,\ |G|=n$ in the case when $m$ is prime, since then we can "peel off" direct factors of prime orders dividing $m$ from the group of order $mn$ one by one. From now on, we consider $m$ to be prime.

An $m$-Sylow subgroup of a group $K,\ |K|=mn$ with $m,n$ satisfying $(*)$ is contained in the center of its normalizer $N$. This is because the normalizer is a semidirect product of $\mathbb Z_m$ by some group of order coprime to $m$, and this order is coprime to $m-1=|\mbox{Aut}(\mathbb Z_m)|$, so the normalizer must actually be a direct product between $\mathbb Z_m$ and some other group. Burnside's Normal Complement Theorem states that such an $m$-Sylow subgroup of $K$ (which is contained in the center of its normalizer) has a normal complement, or, in other words, that $K$ is a semidirect product of some group $G$ of order $n$ by $\mathbb Z_m$. According to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=automorphisms&t=45849]this[/url], all the prime divisors of $|\mbox{Aut}(G)|$ also divide $ng(n)$, since $G$ is soluble. This means that $(m,|\mbox{Aut}(G)|)=1$, so $K$ is actually a direct product $\mathbb Z_m\times G$, and we're done.

I hope it's correct.

P.S.

Somewhere in there, I used the fact, already discussed on the forum, that if $(m,\varphi(m))=1$, then $f(m)=1$ (i.e. all groups of order $m$ are cyclic).


My name is Aldo Pacchiano, I will be part of Mexico's IMO team 2008. We have thought and decided to challenge any other team going to IMO 2008 to an Age of Empires II the Conquerors Expansion battle (3 vs 3 or 4vs4) (if there are computers in the IMO hotel). We have been training hard both Math and Age, so be prepared.

 :lol:

Anyone interested please post a reply.

[u]Mexican team:[/u]
Aldo Pacchiano       (paqui_2590)
Guillermo Buenfil
Manuel Novelo Puc
Andres Campero     (Chingonkan999 and Mexico's Next Gold)
Rodrigo Mendoza
Malors Espinoza      (Camus)

I would much rather participate in a Dota competition if there was one :)

I am Manuel Candales from Cuba. I am the only member of my team, and I would like to participate in a Counter-Strike play.

Youre a pendejazo!!!!
Ea!!!
You should remember write "infinite descent" in your Paper exam for the Silver!!!
jajajajajaja
Ea!!!!

We all love you UADY. (Don't pay attention to this last thing, is just a local joke between conejita and us)
Conejita went to IMO 2007 Vietnam.

Manuel, what has happened with Otto?

Why didn't he manage to go?

we could prefer warcraft I think...
or call of duty

if i come to the IMO i would also play age of empires II - the conquerors expansion. we played it several times during the preparation week(end)s^^

Naphthalin

So Germany vs México would be great.

       See ya in Madrid

i hope so. but atm, there are only two AOE-players in the (provisional) team, so there couldn't be a 3vs3 ;)

Naphthalin

Ok, what can we do then, 2vs2 is fine. México has 3 players but we could choose 2.

   Bye

hey, m Cambodian IMO contestant... we'd rather play FIFA 2008 or real football... that's wil b competative  :lol: . I haven't played age since long... now just play some others like warcraft, CS ...

well, thnow the german team is selected, there are two players (but the other one plays just for 2 months^^). let's what's possible. do you have a "mobile" version of AOK?

Naphthalin

[quote="Naphthalin"]	well, thnow the german team is selected, there are two players (but the other one plays just for 2 months^^). let's what's possible. do you have a "mobile" version of AOK? [/quote]
I have the $ 1.0c$ patch, the no cd crack and installation cds in my iPod (and in cds). We can install the game there.

jajaja. My name is Juan Pablo. I remember that you were in Portugal. I was there. I think that Uruguay can play :lol: !!! We will see.. :wink:

Mexicovs Urugay vs Germany.

  By the way, i will play fifa and real life soccer too but i don´t think many of mexico want to play.

   See you in Madrid

cowards ;)
in fact there won't be much time - i leave at 19th from madrid...

Naphthalin

Why????

I meant I won't play fifa. (I don't know how to play)

ah that sounded different to me ;)
why i'm leaving - look to the IPhO 2008 Ha Noi thread^^

Naphthalin

I alsOo want to play!! :P I bet I'm better than all of mexicOo's team :lol:

which country are you from?
so the 2 of us against mexico - okay? ^^

Naphthalin

Play [url=http://www.artofproblemsolving.com/Edutainment/g1/index.php]this game[/url]. 6th graders will own you.


modular arithmetic
number theory unsolved
number theory

We say that a natural number $n$ is [i]charrua[/i] if it satisfy simultaneously the following conditions:
- Every digit of $n$ is greater than 1.
- Every time that four digits of $n$ are multiplied, it is obtained a divisor of $n$

Show that every natural number $k$ there exists a [i]charrua[/i] number with more than $k$ digits.

$10^{\phi (9^4)n}-1$ is charrua for every $n$.

[quote="carlosbr"]We say that a natural number $n$ is [i]charrua[/i] if it satisfy
simultaneously the next conditions:
- Every digit of $n$ is greater than 1.
- Every time that four digits of $n$ are multiplied, it is obtained a
divisor of $n$

Show that every natural number $k$ there exists a [i]charrua[/i] number with
more than $k$ digits.[/quote]

[b]Lemma:[/b] For all $n\in \mathbb {N}$ there is a $N\in \mathbb {N}$ such that $2^n|N$, $10^{n-1}\leq N<10^n$ and if $x$ is a digit on $N$, then $x\in \{1,2\}$

[b]Proof of the lemma:[/b] [hide]By induction. Let $a_{n}$ such number. For $n=1,2,3$ choose $a_1=1, a_2=12, a_3=112$.  If $a_{n}\equiv 0\pmod {2^{n+1}}$, choose $a_{n+1}=2\cdot 10^n+a_n$. In other case, if $a_n\equiv 2^n\pmod {2^{n+1}}$, choose $a_{n+1}=10^n+a_n$, and we are done with the lemma. $\square$[/hide]

We shall prove that $2a_{k+3}$ is a [i]charrúa[/i] number for $k$. In fact:

[list][*] Every digit of $2a_{k+3}$ belongs to $\{2,4\}$.
[*] The product of four digits of $n$ can be equal to $2^4,2^5,2^6,2^7,2^8$. Because $k\ge 4$ it holds that $2^8|2^{k+4}|2a_{k+3}$.

[*] $2a_{k+3}$ has $k+3>k$ digits.[/list]

And we finish the proof. $\blacksquare$

Take the number $M=2222....223232$ where the amount of two's is of the form $9n+6$ and there's only 2 three's. Denote by $S(n)$ the sum of the digits of $n$. It is well knows that $9$ divides $n$ iff $9$ divides $S(n)$.

Note that $S(M)= 2(9n+6)+3+3 = 18n+18 =>$ $9$ divides $M$.

It is also well knows that $2^k$ divides $n$ iff $2^k$ divides the last $k$ digits of $n$.

So $2^4$ divides $M$ iff $2^4$ divides $3232$ which is true since $3232= 202*16$

Therefore $3^2*2^4$ divides $M$.  Now the product of any $4$ digits of $M$ is of the form $2^a*3^b$ with $4\ge a \ge 2$ and

$2 \ge b\ge0$. So the product of any four digits of $M$ divides $3^2*2^4$ and therefore, the product of any four digits of $M$ divides $M$. Therefore $M$ is charrua. For any positive integer $k$ take $n=k$ and $M$ will have $9k+8$ numbers.

By the way, see here [url]http://en.wikipedia.org/wiki/Charr%C3%BAa[/url] what [i]charrúa[/i] means - the contest chairperson was maybe a descendant of them ...  :)

By the way, any number with prime divisors different from $2$ and $5$ has infinitely many[i] repunit [/i]multiples (numbers made of just digit $1$). The proof is folklore. Thus $3^4$, $7^4$ and $9^4$ do that, and then those repunits multiplied by $3$, $7$, respectively $9$, are charrúa. So we can do this for any other number than four digits.

The other solutions here are welcome if we impose as extra condition that not all digits be equal in a charrúa number.

it suffices to consider $22...23232$(9k+6 2s in total).

Motivated by the divisibility test of 9, by Euler's theorem, $n=\phi (243)=162$ satisfies $10^n \equiv 1(mod 243)$  so if we consider the number $333..333$ with 162 digits we can rewrite it as $3\cdot \frac{10^{162}-1}{9}$ meaning this number is divisible by $3^4=27$. So it is sufficient to pick a number of this form such that the number of its digits is a multiple of $162$.


algebra
polynomial
Vieta
algebra proposed

Let $M$ be a finite set of non-zero numbers. A polynomial $f(x)=\sum_{k=0}^n a_kx^k$ satisfies two conditions:

(i) it has $n$ real roots (possibly, multiple) and

(ii) all coefficients $a_k (k=0..n)$ belong to $M$.

Prove that there exists a constant $C(M)$ such that $n\le C(M)$.

Nice problem, remembers me of the problem of Moubinool and in fact the same idea works just fine. Using Vieta's relations we can see there exists a constant $ A(M)$ such that for any such polynomial $f$ with roots $ x_1,...,x_n$ we have $ A(M)\geq x_1^2+...+x_n^2$. Also there exists a constant $B(M)>0$ such that for any such polynomial $ x_1^2...x_n^2\geq B(M)$ (again Vieta's formulae). Thus $ A(M)^n\geq n^n B(M)^n$ and from here it easily follows that $n$ is bounded by a constant depending only on M.


vector
linear algebra
linear algebra unsolved

If dimV=n and S is a linearly independent set in V, then S is a basis for V.

True or False? Justification?

A linearly independent set can have any size less than or equal to $ n$.  For example, a single vector is a linearly independent set.

A basis is a [b]maximal[/b] independent set. As t0rajir0u has pointed out, leaving out the word "maximal" changes the meaning.

A basis for a given vector space $ V$ is a set of vectors $ v_1, v_2, ..., v_n$ such that:

I. The vectors $ v_1, v_2, ..., v_n$ are linearly independent.
II. $ v_1, v_2, ..., v_n$ span $ V$.

If dim V = n, that means that any basis of V will consist of n vectors. Therefore, if S is a linearly independent set in V then S will be a basis of V iff:

1. S has n vectors.
2. Those n vectors span V.


In the year 2004, there were 5 Sundays in February. What is the next year in which this will happen?

Source: Australian Math Competition, J26, 2005

[quote="shyong"]if you know something about gregorian calender , then you probably would know that the calender cycle every 28 years ....[/quote]

Is that so, even when we take out a leap year every 100 years and add one back every 400 years?

[hide="method"]The fact is that on every seventh year, the months start with the same day they started with seven years ago. Counting leap years, we add 28 to 2004 to get $\boxed{2032}$[/hide]

There's this thing about leap years.. every year which is a multiple of 4 is a leap year unless it is on the turn of the century, whence it must be divisible by 400. So we have to use these principles carefully. 236factorial certainly noticed something important.  :)

And 2032 is correct. Wasn't that hard eh.


Given are triangle two medians $m_{a},m_{b}$ and internal bisectors $l_{a},l_{b}$ find the limit
$\lim_{a\rightarrow b}\frac{m_{a}-m_{b}}{l_{a}-l_{b}}$

I found this topic unsolved and I tried to solve it :P
I think we can take a triangle $ _{\Delta}ABC$, where $ |AB|=a$, $ |BC|=b$, $ |CA|=c$, then we'll easy compute:
$ m_{a}=\frac{1}{2}\sqrt{2b^{2}+2c^{2}-a^{2}}$

$ m_{b}=\frac{1}{2}\sqrt{2a^{2}+2c^{2}-b^{2}}$

$ l_{a}=\sqrt{bc-\frac{a^{2}bc}{(b+c)^{2}}}=\sqrt{bc\left(1-\frac{a^{2}}{(b+c)^{2}}\right)}=\frac{\sqrt{bc(a+b+c)(-a+b+c)}}{b+c}$

$ l_{b}=\sqrt{ac-\frac{ab^{2}c}{(a+c)^{2}}}$

And if we put this into limit, we'll have something like this:
$ \lim_{a\to b}\frac{m_{a}-m_{b}}{l_{a}-l_{b}}=...=3b\sqrt{\frac{b^{2}+bc}{b^{2}+2c^{2}}}\frac{(b+c)^{2}}{(b+c)^{3}+b^{2}(3b+c)}$


[quote="pohoatza"]Let $a,b,c \geq 0$ with $abc=1$. Prove or disprove that:

$\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\leq 1$[/quote]

just consider $a=\frac{1}{10}, b=\frac{1}{10}, c=100$

[quote="lotrgreengrapes7926"]Reverse the inequality??[/quote]

$10,10,\frac{1}{100}$


How many integers with four different digits are there between 1,000 and 9,999 such that the absolute value of the difference between the first digit and the last digit is 2?

$\text{(A)} \ 672 \qquad \text{(B)} \ 784 \qquad \text{(C)} \ 840 \qquad \text{(D)} \ 896 \qquad \text{(E)} \ 1008$

[hide]There are 14 different possible arrangements for the first and last digit so that the absolute value is 2. For each of these, there are $7\cdot8$ arrangements of the middle 2 digits so that there are 4 different digits. $14\cdot7\cdot8=784\Rightarrow\boxed{B}$[/hide]

I got something different  :?:

[hide]
if first digit is 1, 8, 9 theres 1 possible last digit. if the first digit is anything else theres 2 possible last digit. 6 * 2 = 12 + 3 = 15. 8 * 7 possible middle digit. 56 * 15 = 840. So answer is C[/hide]

You are correct Javaman :yup:

There are the pairs $(9,7)(8,6)(7,5)(6,4).....(2,0)$ that can be used and their reverse order, with the exception of $(0,2)$, so a total of $15$ pairs of numbers for the first and last digits, and ofcourse, we want different numbers for the middle digits, so that $8 \cdot 7$, and so the total numbers of numbers is $\boxed{840}$

[quote="Iversonfan2005"][quote="catcurio"]There are 14 different possible arrangements for the first and last digit so that the absolute value is 2.[/quote]

You probably forgot $(2,0)$ :P[/quote]No, I forgot $(1,3)$... For some reason I completely skipped considering 1 as the first digit.  :blush:


limit
Euler
real analysis
real analysis unsolved

I think that the limit should be $ \minus{}\gamma$.

My solution is based on the following, probably not very well known, limit: $ \lim_{n\rightarrow\infty}n\minus{}\Gamma(1/n)\equal{}\gamma$.

Let $ x_{n}$ be your sequence. Then

$ x_{n}\equal{}\frac{(\Gamma(1/n)\minus{}n)\Gamma(n\plus{}1)n^{1\plus{}1/n}}{\Gamma(n\plus{}2\plus{}1/n)}\plus{}\frac{n^{2\plus{}1/n}\Gamma(n\plus{}1)}{\Gamma(n\plus{}2\plus{}1/n)}\minus{}\frac{n^{2}}{n\plus{}1}$

Using Stirling's formula we get that

$ x_{n}\approx\left(\Gamma(1/n)\minus{}n\right)(\frac{n}{n\plus{}1})^{1\plus{}1/n}\plus{}\frac{n}{n\plus{}1}\frac{(\frac{n}{n\plus{}1})^{1/n}\minus{}1}{1/n}\rightarrow\minus{}\gamma$.

I hope I made no computational mistakes.

I have other result. I can post my solution if somebody want it.
the left side of equation is equal to
$ \frac{\Gamma (\frac{1}{n})(1\plus{}\frac{1}{12n}\plus{}O(\frac{1}{n^{2}}))}{n\plus{}\frac{37}{12}\plus{}O(\frac{1}{n})}\minus{}\frac{n^{2}}{n\plus{}1}$

[quote="didilica"] $ \lim_{n\rightarrow\infty}n\minus{}\Gamma(1/n) \equal{}\gamma$.

[/quote]
Where can I see this fact?
Also aproxymation of $ \Gamma{(x)}$ when $ x\to 0$?

Some formulas related to the Euler-Mascheronni's constant can be found here:

http://mathworld.wolfram.com/Euler-MascheroniConstant.html

especially entries 35 and 50.

[quote="didilica"]I think that the limit should be $ \minus{}\gamma$.
[/quote]
you are correct, it [b]is [/b]$ \minus{}\gamma$, sorry about that...  :(


Guess the next number in the following series:

172, 468, 317, 246, ?

Belongs [url=http://www.artofproblemsolving.com/Forum/index.php?f=337]here[/url] (this is not a very mathematical question). I would guess $ 831$.


What is the greatest number of feet in the perimeter of a rectangular floor that can be covered by 30 whole 1' by 1' square tiles? :idea:

I do not get what he is saying. I put 30?????????????????????????????????????????????????????????????????!!!!!!!!!!!!!!!!!!!!!!!!!!!

The greatest possible perimeter is when the tiles are all lined up straight.
In that case, the perimeter is (30+1)*2=62ft.

Also the last possible perimeter is when they form a square (or a rectangle close to a square).
So here that would be a 6' by 5' rectangle, which has perimeter 22ft.


calculus
derivative
function
trigonometry
limit
real analysis
real analysis unsolved

if: $ f$ is differentiable, then: $ f'$ is continuous

You don't. Functions like
\[ f(x)\equal{}\left\{
\begin{array}{rl}
0,&x\equal{}0\\
x^2\sin \frac{1}{x},&x\neq 0\\
\end{array}\right.\]
may be differentiable, but not have a continuous derivative. In this particular example, we get
\[ f^\prime (x)\equal{}\left\{
\begin{array}{rl}
0,&x\equal{}0\\
2x\sin \frac{1}{x}\minus{}\cos \frac{1}{x},&x\neq 0\\
\end{array}\right.\]
(the chain rule is valid when $ x\neq 0$, and for $ x\equal{}0$ we can use the definition directly). Here $ \lim_{x\rightarrow 0}f^\prime (x)$ doesn't exist.


if [b]m^2 +1 -n^2[/b] divides [b]m^2[/b] then [b]m^2 +1 -n^2[/b] is a perfect square  :)

$ m,n$ must be odd  :wink:
$ a \equal{} \frac {m \minus{} n}{2}$ and $ b \equal{} \frac {m \plus{} n}{2}$
So $ 4ab \plus{} 1|(a \plus{} b)^2$
Exist a natural number k such that :
$ (a \plus{} b)^2 \equal{} 4k(ab \plus{} 1)$
$ \Leftrightarrow a^2 \minus{} 2a(2k \minus{} b) \plus{} b^2 \minus{} 4k$
This equation has solution in positive integer so exist a solution $ (a_0,b_0)$ such that $ a_0 \plus{} b_0$ min .
Not lost the general we suppose that $ b_0\leq a_0$
From Viete law we have the equation has an solution $ a_1$ .
$ a_1 \plus{} a_0 \equal{} 2k \minus{} b$ and $ a_1a_0 \equal{} b_0^2 \minus{} 4k$
From the equation we have $ a_0\geq 0$
If $ a_0 > 0$ then
$ a_1 \equal{} \frac {b_0^2 \minus{} k}{a_0} < a_0$ so $ a_1 \plus{} b_0 > a_0 \plus{} b_0$ it give contradition .
We must have $ 4k \equal{} b^2$
It mean that k is a perfect square .

[quote="SpongeBob"]$ m^2 \minus{} (m \minus{} 1)^2 \plus{} 1 \equal{} 2m|m^2$ for even $ m$. :wink:
edit: do you mean that $ m,n$ should be odd in the problem statement?[/quote]

 yes your right . they are odd in the problem statement

 i'm sorry


ceiling function
combinatorics proposed
combinatorics

Call the Graph the set which composed of several vertices $P_1,\ \cdots P_2$ and several edges $($segments$)$ connecting two points among these vertices. Now let $G$ be a graph with 9 vertices and satisfies the following condition.

Condition: Even if we select any five points from the vertices in $G,$ there exist at least two edges whose endpoints are included in the set of 5 points.

What is the minimum possible numbers of edges satisfying the condition?

Hi [b]Kunny![/b]
[b]My ans is $\boxed{9}$[/b].
[hide]+ Let $x_n$ is the $ \min$ number of edges, where $n$: vertices;
$\implies \frac{x_{n+1}}{x_n} \ge \frac{n+1}{n-1}$; (1)
+ Let $y_k$ is the number of edges of the graph obtained by removing vertex $k$
$\implies \sum_{k=1}^{n+1} y_k = (n-1)(x_{n+1})$; where $y_k \ge x_n$ (2)
+ Because: "[i]Even if we select any five points from the vertices in $G$ [u]there exist more than two edges [/u] whose endpoints are included in the set of $5$ points[/i]". (3)
Since $(1); (2)$ and $(3)$:  We have: $x_5=2 \implies  x_6 \ge 3  \implies  x_7 \ge 5  \implies  x_8 \ge 7  \implies  x_9 \ge 9$.[/hide]

The answer is correct, but I can't judge whether your solution is correct or not.
Because I am poor at something like this typed problems.

Let us put some precision here. First, if the answer $9$ is correct, the condition must read "among any five vertices of the graph there exist at least two edges", not "more than two edges".

tuan119's result (1) may be better explained like this. Let $x_n$ be the minimal number of edges of a graph on $n$ vertices having the mentioned property, thus $x_5 = 2$. For a minimal graph on $n+1$ vertices having the mentioned property, by removing any vertex  the mentioned property remains (it is clearly hereditary). But $\sum_{k=1}^{n+1} \deg v_k = 2x_{n+1}$, so for at least one vertex we must have $\deg v \geq \dfrac {2x_{n+1}} {n+1}$. By removing this vertex there remain $n$ vertices, and at most $x_n \leq x_{n+1} - \dfrac {2x_{n+1}} {n+1} = \dfrac {(n-1)x_{n+1}} {n+1}$ edges, thus $x_{n+1} \geq \left \lceil \dfrac {n+1} {n-1} x_n \right \rceil$. tuan119's result (2) only justifies this in a slightly different way.

So $x_6 \geq \left \lceil \dfrac {6} {4} \cdot 2 \right \rceil = 3$, $x_7 \geq \left \lceil \dfrac {7} {5} \cdot 3 \right \rceil = 5$, $x_8 \geq \left \lceil \dfrac {8} {6} \cdot 5 \right \rceil = 7$, finally $x_9 \geq \left \lceil \dfrac {9} {7} \cdot 7 \right \rceil = 9$.

A model with exactly $9$ edges is easily found - take as graph $G$ the union of three disjoint triangles $K_3$. Then any set of five vertices must either have two pairs, each in a different triangle (thus at least two edges), or a triplet making up one of the triangles (thus at least those three edges).


1) An object is doing uniform acceleration motion with initial velocity $v$, and it stops after some time. If the ratio of the displacement of the object in the first 5 seconds and the displacement in the last 5 seconds is 11:5, then find the time the object take from the beginning to the rest.

2) A body is doing uniform acceleration motion with initial velocity $v_0$ and, when it stops, the displacement is $s$. When the displacement is $\frac{3}{2}s$, the time used is $t_1$; when the velocity is $\frac{1}{3}v_0$, the time used is $t_2$.
Find $t_1:t_2$.

3) An object start an uniform accelerating motion from rest. The time used to pass the last $\frac{1}{3}$ displacement of the $\text{2nd}$ second is $t_1$; the time used to pass the last $\frac{1}{5}$ displacement of the $\text{3rd}$ second is $t_2$.
Find $t_1:t_2$.

4) A car start to do an uniform accelerating motion with acceleration $a_1$ from rest. After some time it begins to do another uniform accelerating motion with acceleration $a_2$ until it stops again. The total displacement is $L$.
Find the total time $t$ the car take to complete the motion.

Note: Problems are easy but interesting. Enjoy them.  :)

[b]First Question[/b]

Let $v_0 $ be the initial velocity and let $ -a $ be the uniform acceleration(retardation).

For the first 5 seconds :-
$ X_1 =\ V_0 t -\ \frac{1}{2} a t^2 = 5v_0 - 12.5a $

For the last 5 seconds :-
Since the body comes to rest after these 5 seconds , let it's velocity at the start of this period be $v_1$
$ \therefore v_1 - 5a =\ 0 \Rightarrow v_1 =\ 5a $ therefore , displacement during last 5 seconds $=\  5a  \times 5 -\ \frac{1}{2} a \times 25 =\  25a -\ 12.5a =\  12.5a $

Since the ratio of gicen displacements is $ \frac {11}{5} $ we've
$ \frac{5v_0 -\ 12.5a}{ 12.5a} =\ \frac{11}{5}  \\
\\ \Rightarrow 25v_0 =\ 200a \Rightarrow v_0 -\ 8a =0 $
and comparing with eqns of motion , we get that time of motion is $ 8 \  s\ $

The second one is a bit lenghty .I am just posting the answer.

Is it $\frac{3}{2} \times (\ \sqrt { \frac {3}{2} } -\ 1 )\ $ ??

[b]Answer of fourth question[/b]

\[ \sqrt {2L} \times \frac{ \sqrt{a_1} +\ \sqrt{a_2} }{ \sqrt{(a_1)^2 +\ (a_2)^2} } \]

[hide]I am sorry I have become so lazy as to not give the solns. They are calculative mainly. [/hide]

[quote="shobber"]Number one is correct... 2 and 3 seem to have some mistakes...[/quote]

Can you tell them to me as a PM


find the smallest positive integer $n$ such that if we take the last digit of $n$ to the front, the new number is $2n$.


calculus
integration
real analysis
real analysis unsolved

let f defined on[a,b] be real valued monotonically increasing riemann integrable function.let F(x)=integral of f from a to x.show F is differnciable at p implies f is continuous at p. :D

This is a fairly routine problem (and I'm pretty sure it generalizes to all f). I'll sketch a proof: suppose that f is discontinuous at p; there exists epsilon so that there exists t with f(t) > f(p) + epsilon in any neighborhood of p (or f(t) < f(p) - epsilon, but WLOG we'll use f(t) > f(p) + epsilon). Since F is differentiable at p, you can easily deduce that f is not always more than f(p) + epsilon/2 in (p, p+delta) for any given delta. Hence within any neighborhood of p, f attains values both greater than f(p) + epsilon and less than f(p) + epsilon/2. This contradicts f being Reimann integrable.


How many integers $ n$ satisfy
$ 15 \ge \lceil (n\minus{}1)/2 \rceil \ge \lceil \sqrt{n} \rceil \ge 0$, where $ \lceil n \rceil$ denotes the least integer greater than or equal to $ n$?

just find the largest, and the smallest values, then that is the total.

so it can easily be seen that 31 is the largest.

after some guess and check, you find that 5 is the smallest. so 5-31 inclusive is 27.

5 doesn't work. 7 does. So the answer is 25.

Whoops!

I remember typing in 6 earlier...then for some reason, I forgot about the ceilings, and moved straight on to 7, forgetting that 6 works.

Apologies.  :blush:

sorry if i am misunderstanding something but couldn't 4 work also?
if 4 worked though then their would be two possible outcomes for the ceiling function of its sqrt.

Actually, (thanks by the way for noticing that), I think there's going to be some guess and check work here. For numbers that don't have an integer square root, the ceiling makes the root increase, even though the second term in the inequality doesn't.

Really busy right now, so someone else can figure this out....

*sigh*  :blush:

whoops, I thought that it was floor function. Anyways, when it says sqrtn, it implies only the positive root.

[quote="archimedes1"]How many integers $ n$ satisfy
$ 15 \ge \lceil (n \minus{} 1)/2 \rceil \ge \lceil \sqrt {n} \rceil \ge 0$, where $ \lceil n \rceil$ denotes the least integer greater than or equal to $ n$?[/quote]

If $ 15\equal{}\lceil \frac{n\minus{}1}{2} \rceil, n$ is at most 31.

For the least value: $ \lceil \frac{n\minus{}1}{2} \rceil \ge \lceil \sqrt{n} \rceil$ ($ n \ge 0$)

We consider those numbers that are perfect squares.  Since we know that $ n\le 31$ from above, we only need to check $ 0, 1, 4, 9, 16, 25$. $ 0, 4, 9, 16,$ and $ 25$ work.

The largest $ \lceil \sqrt{n} \rceil$ can be is 6.  So any value of $ n$ greater than 11 will work.

Now the largest will be at most 4 for $ n<11$.  So $ n>7$ works.
Then, the largest possible value is 3.  So $ n>5$.

Largest is then $ 2$.  This gives $ n>3$.
Then we get $ n>1$.

The only values we need to check are 1, 3, 5, 7, and 11.

$ 11$ works.
$ 7$ works.
$ 5$ does not work.
$ 3$ does not work.
$ 1$ does not work.
Don't forget 0. (works)
A total of $ 29$ values. I think

wait, what else is there?

all I can see is 6-31 and 0 which gets 27.

4. The only ones which don't work are 1, 3, and 5
Lol 28 xD. I just realized two doesn't work. because the largest of the square root thing is still 3 for $ n<3$.


vector
LaTeX
function
real analysis
real analysis unsolved

$ \vec{a} \times (\nabla \times \vec{a}) \equal{} \nabla (a^2/2) \minus{} (\vec{a} \cdot \nabla)\vec{a}$

latex issues:
the triangle need to be flipped upside down (=gradient)
alle a need to have an arrow symbol (except a^2 thing) on top of them = vector

question.
- i don't see why this is identity holds :s neither do I now what a^2 should be ?  scalar product? But then i have a scalar and taking gradient does not make sense ? :s

(identity taken from course fluid dynamics)

[color=olive][LaTeX fixed][/color]

Gradient of a scalar function makes perfect sense.

Write $ \vec a\equal{}a_1\vec i\plus{}a_2\vec j\plus{}a_3\vec k$ and expand both sides carefully.


[quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
$ \sqrt [9]{\frac {a^3}{b \plus{} c}} \plus{} \sqrt [9]{\frac {b^3}{c \plus{} a}} \plus{} \sqrt [9]{\frac {c^3}{a \plus{} b}} \geq\ \frac {1}{\sqrt [9]{2}}(\sqrt [9]{ab} \plus{} \sqrt [9]{bc} \plus{} \sqrt [9]{ca})$

It exists for a long time (althought we can solve it by only Am-Gm). And I can't find the link of it. So, I reposted it here as a gift for my younger brother.  :)
@quykhtn-qa1: [hide]Hi vong. em thjk bai` nay`. Hiz, dot. nay` cong viec kinh doanh cua anh ban. ron. qua' nen ko con` tg de? sang' tac', danh`lay' tam. bai` nay` tang. em vay.  :(  [/hide][/quote]
Thank you for your gift,my brother :) ,but it is not nice
@nguoivn: [hide] Anh sap viet sach phai o a.?bao gio xong a.? [/hide]

[quote="quykhtn-qa1"]
Thank you for your gift,my brother :) ,but it is not nice
[/quote]
Are you sure? It looks short and not have any condition. Can you find an intersting ineq which have high root (and not easy) like it  :maybe:
@quykhtn-qa1: [hide]Truoc mat' thi` tui. anh dang lo cho cuon' sach' ve` MIC sap' toi'. Noi' chung ben nxb gd cung~ ok roi`.Tiec la` Loc va Luat dang thi dai hoc, n~ ng` con` lai nhu anh Can?, anh va` anh Hung` deu` kha' ban. ron. nen co le~ phai? den' thang' 7 tui anh moi' hoan` tat dc cuon sach'. Sau do', anh Can? va` anh se~ viet' cung` nhau 1 cuon' khac' la` "N~ loi` giai hay cho n~ bai toan dep". Bi gio` tren thuc te' anh cung~ ko lam` bdt nua~ roi`, thinh? thoang? hom nao` roi~ vao` chat. chem' ti' cho vui thoi. ^^!  :)[/hide]


[quote="cancer"]A piece stands on the leftmost cell of a $1$x$100$ board. Two players alternately move the piece by $1,10$ or $11$ cells to the right. The player who cannot move loses. Who can win?[/quote]

[size=150][hide="Preface"][/size]
Let $x$ be the position of the piece.

From the definition of the game, if some player moves the piece to the position $100$ then the other player loses.
So $x=100$ is a losing position. We'll find all the winning and losing positions.

We define the subsets $W,L$ of $\{1,2,...,100\}$ such that $W\cup \L = \{1,2,...,100\}$ and $W\cap L=\emptyset$

$W:$ The player who has turn will [u]definetely[/u] win
$L:$ The player who has turn will [u]definetely[/u] lose

The rule to construct these equivalence classes is simple:
If $x$ can lead to at least one $L$-element then $x\in W$
If $x$ can lead only to $W$-elements then $x\in L$

[u]Namely[/u]:
Take the numbers $x+1, x+10, x+11$
If at least one of those numbers is an $L$-element then $x\in W$
If not then $x\in L$[/hide]$\hline$

[size=134][hide="Some examples"][/size]
$100\in L$

Then, $99,90,89 \in W$ (because you can move the piece by $1,10,11$ cells respectively)

Then, $98$ could only lead to $99$, hence $98\in L$
But $97$ can lead to $98 \in L$, hence $97\in W$[/hide]$\hline$

[size=150]Theorem[/size] $\boxed{x\in L \iff x = 0,12,14,16,18\mod{20}}$

[size=150][hide="Proof"][/size]
$\blacktriangleright$ 1) $L$ cannot lead to $L$

Suppose that $x \in L$. Then $x \equiv 0,12,14,16,18\mod{20}$

Of course $x+1,x+11\not\in L$, because they are odd
On the other hand, if $x = 0,12,14,16,18\mod{20}$
then we have respectively $x+10\equiv 10,2,4,6,8 \mod{20}\Rightarrow x+10\not\in L$

So we have $x+1,\ x+10,\ x+11\in W$

$\blacktriangleright$ 2) $W$ leads to $L$

Suppose that $x \in W$. Then $x \equiv 1,2,3,4,5,6,7,8,9,10,11,13,15,17,19\mod{20}$

If $x \equiv 1,3,5,7,9\mod{20}$ then move the piece by $11$ cells

If $x \equiv 2,4,6,9,10\mod{20}$ then move the piece by $10$ cells

If $x \equiv 11,13,15,17,19\mod{20}$ then move the piece by $1$ cell

At any case, the result is an element of $L$[/hide]$\hline$

[color=green][u]Answer[/u]
At the beginning of the game we have $x=1\in W$, so the first player wins[/color]


Here is the problem:

In the centre of a square swimming pool is a boy, while his teacher (who cannot swim) is at one corner of the pool the teacher can run three times faster than the boy can swim, but the boy can run faster than the teacher can. Can the boy escape from the teacher? (Assume both persons are infinitely manoeuvrable).  :lol:

(*)
Suppose the teacher can run six times as fast as the boy can swim. Now show that the boy cannot escape. :)

(**)

Suppose the pool was circular instead of square. Now what is the minimum speed of the teacher (relative to the boy's swimming speed) needed to catch the boy? :(

(***)
What if the pool was square? :mad:

First part is easy,  second too.  I'm stuck on the third and fourth part.

Third part: Assume the pool has radius 1 unit.  If the teacher is less than $ k$ times as fast as the boy, the boy can swim out a distance of $ \frac{1}{k}$, and can then swim around in a circle with higher angular speed than the teacher can achieve.  If the boy is to escape, then the boy must be able to swim to the edge of the pool when the teacher is on the opposite side of the pool.  Then $ 1\minus{}\frac{1}{k}<\frac{\pi}{k}$.  Then $ k<\pi\plus{}1$.  So if the teacher has speed less than $ \pi\plus{}1$, the boy can escape.  This still has to be formalized...not to mention I have to prove the teacher can catch the boy is $ k \ge \pi\plus{}1$...but no time now...


function
real analysis
real analysis unsolved

Hey,
  Today my prof gave me a problem: if y(x) is a continously differentiable function on R, and (y')^2 + y^3 ---> 0 as x --> infinity, then show that y' --> 0 as x --> inf.

   Hints, (and solutions too are welcome)

Thank you

Cheers
Shek

Assume that y' doesn't tend towards 0.  What then does that imply about y?


Are you doing the Wisconsin Mathematics Talent Search? If so, how are you doing?

What is the website for this search?

I amn't going to do it, but I just want to know.

math.wisc.edu/~talent/

Yes, but I never dedicate much time to it- I'm usually TeXing until midnight. I got 23333 on the first set (didn't realize my system of equations had only one solution...), not sure what I'll get on the second.

[quote="now a ranger"]I would do it, except for one thing!

I'm from OMEGA BOOGA! not wisconsin[/quote]

Well, that's good for you.

Is a 33333, the one you want to get?

Yeah, there are 5 problems worth 3 points each.


could someone post the question and solution? :ninja:

Please use the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1514487479&t=75588]search function[/url].


inequalities
real analysis
real analysis unsolved

How do you know that such integers exist?

The integer a_i ( i = 1..10, I forgot the case i=10 in my initial post) corresponds to the number of elements of the set {f^(-1)(k/10); k = 1..10} contained in the interval [f((i-1)/10);f(i/10)]. The majoration of the sum I gave follows immediately.


Εστω η συναρτηση φ:(0,+απειρο)-->R  με τυπο
φ(χ)=χ^2+e^2-λ*lnχ με λ>2
δινεται οτι ισχυει:φ(χ)>=0 για καθε χ>0
1)βρες τη μεγαλυτερη τιμη που μπορει να παρει ο αριθμος λ
2)ενα κινητο κινειται πανω στη γραφικη παρασταση της φ.Τη χρονικη στιγμη που το κινητο διερχεται απο το Α(1,φ(1)) η τεταγμενη του ελαττωνεται με ρυθμο 2*λ-4 μοναδες ανα δευτερολεπτο.Να βρειτε τη χρονικη αυτη στιγμη τον ρυθμο μεταβολης της τετμημενης του κινητου.

Μάνο γεια σου.

 Καλή σχολική χρονιά! Λοιπόν , η άσκηση που έβαλες είναι καλή. Όμως θα σου πω, τελείως φιλικά, ότι με ασκήσεις ανάλυσης δε θα βρεις μεγάλη ανταπόκριση στο φόρουμ αυτό.Αν χρειαστείς βοήθεια ή κάτι άλλο στην ύλη της Γ΄Λυκείου , γράψε μου e mail

                              stergiu@otenet.gr

  Θα προσπαθήσω να βοηθήσω , όσο μου το επιτρέπει ο χρόνος μου.
  Καλή Μελέτη !

  Μπάμπης

Δεν νομίζω ότι οι ασκήσεις ανάλυσης μένουν άλυτες γενικά σε αυτό το forum. Φυσικά τις ασκήσεις που είναι τελείως στο βαρετό στυλ της Γ' Λυκείου (όπως και αυτή) δεν τις βλέπουν πολλοί.

Και για να μη μένει άλυτη: (ελέγξτε αν είναι σωστή γιατί δε με πολυεμπιστεύομαι στην ανάλυση)

$f(x) = x^{2}+e^{2}-\lambda \ln x \geq 0 \Leftrightarrow x^{2}+e^{2}\geq \lambda \ln x$

Αν $0 < x \leq 1$ τότε το $\lambda$ παίρνει αυθαίρετα μεγάλες τιμές και άρα περιοριζόμαστε στο $x > 1$ και παίρνουμε:

$\frac{x^{2}+e^{2}}{\ln x}\geq \lambda$

Θέλουμε να βρούμε το ελάχιστο της (παραγωγίσιμης) $g(x) = \frac{x^{2}+e^{2}}{\ln x}$ με $x > 1$.

$g'(x) = \frac{2x \ln x-x-\frac{e^{2}}{x}}{\ln^{2}x}$

Ο παρονομαστής είναι πάντα θετικός και άρα τον αγνούμε και πολλαπλασιάζοντας με $x$ θεωρούμε την $h(x) = 2 x^{2}\ln x-x^{2}-e^{2}$ που είναι ομόσημη της $g'(x)$ για $x > 1$.

Έχουμε ότι $h'(x) = 4 x \ln x > 0$ και άρα η $g'(x)$ γν. αύξ. Όμως $g'(e) = 0$ και αφού η $g'(x)$ είναι γν. αύξ. το $x_{0}= e$ είναι ολικό ελάχιστο για την $g(x), x > 1$ που σημαίνει ότι $\lambda_{\mbox{max}}= g(e) = \boxed{2 e^{2}}$

Και για το 2ο ερώτημα, αν δεν κάνω λάθος:

Δίνεται ότι: $\frac{dy}{dt}= 4-2 \lambda$ και θέλουμε να βρούμε το $\frac{dx}{dt}$. Ξέρουμε ότι $\left. \frac{dy}{dx}\right|_{x=1}= f'(1) = 2-\lambda$

Από τον κανόνα της αλυσίδας στο $x = 1$: $\frac{dy}{dt}= \frac{dy}{dx}\cdot \frac{dx}{dt}\Leftrightarrow 4-2 \lambda = (2-\lambda) \cdot \frac{dx}{dt}\Leftrightarrow \boxed{\frac{dx}{dt}= 2}$


Hi,

What are the best books on mathematical proof methods (For elementary, intermediate and professional/olympiad level)?

The books AoPS sells are good.  If  you want a rigorous textbook on proofs, you should probably look into an Analysis textbook, although you'd have to know the pre-requisite stuff.


I think it will be a 99 because the test was harder, but no calculators were allowed.

I'm the only one so far who thinks that the test was hard enough to have a 94.5 cut-off...

I did ALOT worse (28.5 pt) than last year, but mostly due to careless mistakes and a not having a calculator took up some extra time. I think it was harder than last years A, and they lowered the score on that one, and without a calculator I would think the lower scorers did worse. I'm guessing at least <96.

I hope it can creep down to 91.5 so that my school can have three qualifiers, but I kind of doubt it will.

I don't think the easy questions were that bad, so probably won't be happening.

96 seems fair to me. This test was somewhat harder than last year's 12A, which the cutoff was 97.5

My first comment on an AMC contest: I believe that this AMC contest is very easy if you only consider the level of the problems.  For example, 25 was very easy compared to the other years (however I thought is said (2,2) instead of (2,4) :P).  The only thing that made the problems time consuming was the lack of calculators.  Having said that, I believe that the floor would be 100.

Except that the relative ease of the last few problems doesn't really affect the majority of people, and it is how the bulk mass of lower scorers that determines where the 5% cutoff lies. And I'll say that the lack of a calculator, plus last year's 1.5 per blank rule, will affect the majority and bring the floor down.

As AoPSers, IMO we have a bias for considering only the harder questions, when it is the easier and moderate level questions that determine the cutoff.

i took a look at the test and i think it was clearly harder than last year's. i remember last year's cutoff was 99.5 (?).

so im guessing 97.5

it was 97.5, i dont think 99.5 is a possible achievable score...

im going to guess that the cutoff is <=96...

but i really dont care, im more concerned with the USAMO cutoff.

Remember, $ 3 | 2(SCORE)$ (necessary, but not sufficient condition)

I think the cutoff will be 97.5 or 99.

since the cutoff last year was 97.5 and there are no calcs this year i predicted 96.

i know it's too early in the game, but what would you GUESS is the USAMO cutoff =) if you ahd to guess.

[quote="xscapezaer"]100 is not a possible score; All scores have to be divisible by 3 :P[/quote]
[quote="E^(pi*i)=-1"]No, but it is the default if at least 5 percent of people got at least 100.[/quote]
Also, possible scores include scores which are half of a divisor of three, i.e. $ 97.5$.

if u have a # that has a .5, subtract 1.5 and if that score is divisible by 3 then it is possible.

altho doesn't work the other way around

when will the official cutoff be announced?

According to official AIME/AMC12 results for 2002-2007 the only time qualifying score was lowered in AMC12 was 12A 2007 when ave score was 66.8 and 97.5 cut-off gave 6.66% AIME qualifiers.
All other times with ave scores >72.7 qualifying score of 100 was giving >7%, up to 22% in 2006 12A AIME qualifiers.
So if this year's 12A was the same difficulty as 2007 12A and considering having no calculators at least slowed everybody down if not made it more difficult, 96 seems pretty reasonable to me for 2008 12A. (My assumption is that everybody slowing down means losing at least one problem).

Um, my school got our AMC results back a couple of days ago, and the cutoff was 97.5.


A 23x23 square is completely tiled by 1x1, 2x2 and 3x3 tiles. What is the smallest number of 1x1 tiles needed?

[hide]
If we color the rows of the 23x23 square alternating white and black (starting with white), there will be exactly 23 more white squares than black squares. On the other hand, 2x2 squares cover 2 of each color, while 3x3 squares cover 6 of one color and 3 of the other. So the square cannot be filled without using at least one 1x1 square (otherwise, the difference between number of white squares and number of black squares would be a multiple of 3).

On the other hand, we [b]can[/b] cover the square using a single 1x1 square. Put the 1x1 square in the center, then the remaining tiles can be split into four 12x11 rectangles. But a 12x11 rectangle can be further split into a 12x2 rectangle and a 12x9 rectangle. The first can be filled by 2x2 squares and the second can be filled by 3x3 squares. Therefore, a minimum of $\boxed{1}$ 1x1 square is needed.
[/hide]


Can anyone help me to prove this:

$ \frac11 \plus{} \frac19 \plus{} \frac{1}{25} \plus{}... \equal{}  \frac{\pi^2}{8}$

[quote="Gibbenergy"]Can anyone help me to prove this:

$ \frac11 \plus{} \frac19 \plus{} \frac {1}{25} \plus{} ... \equal{} \frac {\pi^2}{8}$[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159424

 :oops:


Carla rotated point $ A$ 420 degrees clockwise about point $ B$ and it landed at point $ C$. Devon rotated the original point $ A$ $ x$ degrees counterclockwise about point $ B$ and it also landed at point $ C$. If $ x<360$, what is the value of $ x$?

Carla rotated point $ A$ $ (420\minus{}360)\equal{}60^{\circ}$ degrees clockwise about $ B$, which is equivalent to $ (360\minus{}60)\equal{}\boxed{300^{\circ}}$ counterclockwise.


Galois Theory
superior algebra
superior algebra unsolved

(Problem taken from Eisenbud's Appendix 1, Problem 3)
Recall that if $ k\subset K$ is an algebraic extension, then there is a unique largest field $ L$ with $ k\subset L \subset K$ such that $ L$ is separable. Show that this result may fail for $ k\subset K$ even for finitely generated extensions of transcendence degree $ 1$

I'm quite a beginner to field theory, so I'm not exactly sure where to start. I think I should be considering fields of characteristic $ \neq 0$, but I'm not sure where to go from there. Also, I'm also not entirely familiar with the result the problem is referring to so if anyone knows where I can find a proof of that fact that would also be helpful. Any input is appreciated, thanks  :)

Then how does this contrast in the case of algebraic extensions? I'm guessing that the proof for the existence of a unique maximal separable field is an application of Zorn's lemma to the set of separable extensions partially ordered by inclusion. If that were indeed the case, then we would need to prove something like $ \bigcup C$ is a separable algebraic extension where $ C$ is a chain of separable algebraic extensions. If this were indeed the case, is the difference that when the extension is allowed to be transcendental, that union can in general be equal to $ k(x)$?

It's really much simpler than that.  Let $ L$ be the set of all elements of $ K$ that are separable over $ k$.  How difficult it will be to prove that this is what you want depends on what definition you have taken for a separable extension.

[quote="t0rajir0u"]If I'm not misunderstanding, take $ K \equal{} k[x]$; every intermediate field has transcendence degree $ 1$, so no such $ L$ exists.[/quote]

This doesn't seem to work as a counterexample.  The field we want is just $ L\equal{}k$. (I'm assuming you mean $ K\equal{}k(x)$ here)

[quote="LydianRain"][quote="t0rajir0u"]If I'm not misunderstanding, take $ K \equal{} k[x]$; every intermediate field has transcendence degree $ 1$, so no such $ L$ exists.[/quote]

This doesn't seem to work as a counterexample.  The field we want is just $ L \equal{} k$. (I'm assuming you mean $ K \equal{} k(x)$ here)[/quote]

I don't exactly get what you mean by this. Why does picking $ L\equal{}k$ show that there is no unique maximal separable subextension?

[quote="LkNsngth"][quote="LydianRain"][quote="t0rajir0u"]If I'm not misunderstanding, take $ K \equal{} k[x]$; every intermediate field has transcendence degree $ 1$, so no such $ L$ exists.[/quote]

This doesn't seem to work as a counterexample.  The field we want is just $ L \equal{} k$. (I'm assuming you mean $ K \equal{} k(x)$ here)[/quote]

I don't exactly get what you mean by this. Why does picking $ L \equal{} k$ show that there is no unique maximal separable subextension?[/quote]

You seem to have missed the double negative "doesn't work as a counterexample".  If t0rajir0u's claim is right, that any intermediate extension of $ k\subset k(x)$ is not algebraic, then $ k$ is the [i]only[/i] separable extension of $ k$, hence trivially the largest one.

Oh, I think we're talking about different things, when I say separable extension, I mean it like this:

$ k\subset K$ is a separable extension if for every finitely generated subextension $ k\subset K'$ has a transcendence base $ B$ such that $ K'$ is separable algebraic over $ k(B)$. Sorry about the miscommunication.

OK, I have only seen algebraic separable extensions, sorry for the confusion.

Anyway, with that definition, t0rajir0u's claims imply $ k\subset k(x)$ is already separable, so we can take $ L\equal{}k(x)$.  For a counterexample, we're going to need something more complicated.


There are $7$ girls and $7$ boys. Kids of the same sex do not know each other. Moreover, there are no two girls who know the same two boys. At most how many acquaintances can be among the kids?

i think it is $13$ . It is like drawing a Bipartitle graph with 7 vertices in each set and you dont want cycle to appear in the graph . So the most edge you can draw is connect one vertex from the first set too all vertices in another set which is 7 edges . Then every other vertices in the first set connect to only one distinct vertices in the second set which give us 6 edges . So total 13 edges .


search
inequalities proposed
inequalities

Prove that forall $a,b,c>0$ and $\alpha\in\mathbb{R}$,
1. if $\alpha\leq 0$ then
$\left(\frac{a}{b+c}\right)^\alpha+\left(\frac{b}{c+a}\right)^\alpha+\left(\frac{c}{a+b}\right)^\alpha\geq\frac{3}{2^\alpha}.$
2. if $\alpha>0$ then
$\left(\frac{a}{b+c}\right)^\alpha+\left(\frac{b}{c+a}\right)^\alpha+\left(\frac{c}{a+b}\right)^\alpha\geq\min\left\{2,\frac{3}{2^\alpha}\right\}.$

Problem 4.3.2, page 304, Hungkhtn's book. This one was a long time ago posted in Mathlinks also by Hung and solved by Mixing variable method. Search for it, with tile "Nesbitt".

Hope the following is useful

http://www.mathlinks.ro/Forum/viewtopic.php?t=30733
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=nesbitt&t=76277


function
complex analysis
complex analysis solved

If f is analytic at the origin and f ' (0) does not equal to 0, prove the existence of an analytic function g such that f(z^n) =f(0)+(g(z))^n in a neighborhood of 0.

Thx in advance

This is a complex version of Rank theorem.

I guess that you mean that $g(z)$ depends on $n$ and that everything is in complex dimension $1$.

Let us suppose that $\mathsf{Re}(f'(0))>0$. Write, for $|z|$ small enough,
\[f(z^n)=\sum_{k}a_kz^{nk}=a_0+z^n\sum_{k\geq1}a_kz^{n(k-1)}=a_0+z^nh(z^n)\]
then $h(0)=f'(0)\neq0$, therefore there exists a neighbourhood $U$ of $0$ such that
\[\mathsf{Re}(h(z^n))>0\qquad \forall\ z\in U\]
So the image of $U$ by $h$ lays in the complex plane minus the negative real axis, on which the $n-$th root can be unambiguously defined.
Hence we can produce a function $g(z)=\sqrt[n]{h(z^n)}$ for $z\in U$ (meaning that we chose a branch of the $n-$th root function), so that we have
\[f(z^n)=a_0+(zg(z))^n\]
But, obviously, $g$ depends on $n$.


The fraction $ \frac{a^{\minus{}4}\minus{}b^{\minus{}4}}{a^{\minus{}2}\minus{}b^{\minus{}2}}$ is equal to:

$ \textbf{(A)}\ a^{\minus{}6}\minus{}b^{\minus{}6} \qquad
\textbf{(B)}\ a^{\minus{}2}\minus{}b^{\minus{}2} \qquad
\textbf{(C)}\ a^{\minus{}2}\plus{}b^{\minus{}2} \\
\textbf{(D)}\ a^2\plus{}b^2 \qquad
\textbf{(E)}\ a^2\minus{}b^2$

[quote="AIME15"]The fraction $ \frac {a^{ \minus{} 4} \minus{} b^{ \minus{} 4}}{a^{ \minus{} 2} \minus{} b^{ \minus{} 2}}$ is equal to:

$ \textbf{(A)}\ a^{ \minus{} 6} \minus{} b^{ \minus{} 6} \qquad \textbf{(B)}\ a^{ \minus{} 2} \minus{} b^{ \minus{} 2} \qquad \textbf{(C)}\ a^{ \minus{} 2} \plus{} b^{ \minus{} 2} \\
\textbf{(D)}\ a^2 \plus{} b^2 \qquad \textbf{(E)}\ a^2 \minus{} b^2$[/quote]
[hide]$ \frac{a^{\minus{}4}\minus{}b^{\minus{}4}}{a^{\minus{}2}\minus{}b^{\minus{}2}}\equal{}\frac{(a^{\minus{}2}\plus{}b^{\minus{}2})(a^{\minus{}2}\minus{}b^{\minus{}2})}{a^{\minus{}2}\minus{}b^{\minus{}2}}\equal{}a^{\minus{}2}\plus{}b^{\minus{}2}$ or C[/hide]


[quote="manjil"]Prove that the square of two odd numbers can never be a square...[/quote]

Guess there's something wrong with the statement of your problem, [i]manjil[/i]... Could you please revise it?  :wink:

[quote="manjil"]Prove that the square of two odd numbers can never be a square[/quote]

Maybe you want to say "Prove that the [b][u]sum of the [/u][/b]square[b][u]s[/u][/b] of two odd numbers can never be a square.

If so, just look at the equation modulus $ 4$ and you get the result.

And ... you are not in the good forum.

[quote="pco"][quote="manjil"]Prove that the square of two odd numbers can never be a square[/quote]

Maybe you want to say "Prove that the [b][u]sum of the [/u][/b]square[b][u]s[/u][/b] of two odd numbers can never be a square.[/quote]

Yes thats what I meant.

[quote]And ... you are not in the good forum.[/quote]

Sorry I did not get you

[quote="manjil"][quote]And ... you are not in the good forum.[/quote]

Sorry I did not get you[/quote]

I just wanted to say that, IMHO, this is not an olympiad problem.


Larry traveled from home to the mall at 40 mph and returned along the same route at 60 mph. What was his average speed in miles per hour?

That makes absolutely no sense at all.

Don't you have to use harmonic mean?

Anyways, the harmonic mean of $ (x,y)$ is $ \frac{2}{\frac{1}{x} \plus{} \frac{1}{y}}$.

Thus, the harmonic mean of $ (40,60)$ is $ \frac{2}{\frac{1}{40} \plus{} \frac{1}{60}}$, which equals $ \boxed{48}$.

What's a harmonic mean?
And ernie, how does my explanation not make sense/

It does not make sense because $ 24$ is not between $ 40$ and $ 60$.

And for harmonic mean, google it. :wink:

lol... i didnt even know this kind of problem could be solved by harmonic mean

jongao is almost right, but the distance is 120*2=240, not 120.

its 240/5=48.

The reason that it can be solved by harmonic mean is that the distance is irrelevant, and you can set it to $ 1$. Thus, if we have $ n$ rates $ r_1,r_2,\ldots,r_n$, then our times are $ \frac1{r_1},\frac1{r_2},\ldots,\frac1{r_n}$, and our total time to go the whole distance of $ n$ is $ \frac1{r_1} \plus{} \frac1{r_2} \plus{} \cdots \plus{} \frac1{r_n}$. Since we have our distance and time, we know that
$ \boxed{R \equal{} \frac {n}{\frac1{r_1} \plus{} \frac1{r_2} \plus{} \cdots \plus{} \frac1{r_n}}}$.


vector
calculus
conics
ellipse
trigonometry
integration
calculus computations

i need help with this question:

F= (x+y)i+(y^(2)+z)j+(z-x)k
find the work done by F along the path C. C consists of the straight line segment from (1,6,3) to (3,0,0) followed by the ellipse (x^2)/9  + (y^2)/4) = 1
in the xy plane starting at (3,0,0) and taken once in an anti clockwise diection when viewed from above.

-i think firstly i should calculate the work done along the straight line 1st and then the work done in the ellipse and then add them together...

for the straight line segment from the equation
r= -i+6j+3k+t(2i-6j-3k) I get x=1+2t y=6-6t z=3-3t
and dz=2 dy=-6 and dz=-3

and when i sub these into F and integrate from 1 to 0 with respect to dt i get -75.5. is that right?

then, i'm not sure what to do with the ellipse,
should i let x=t y=sqrt(4 - ((4t^2)/9)) and go from there? It looks a bit messy to do, is there a simplier way?

The ellipse can be also parametrized by $x=3\cos t$, $y=2\sin t$, where $0\le t\le 2\pi$. But it's easier to use Green's theorem: the $\vec k$-component of $\vec F$ does zero work along the ellipse and can be ignored. Hence $\int_{\textrm{ellipse}}\vec F\cdot d\vec r=-\iint_{\textrm{int. of ellipse}}(-1)\,dA=6\pi$.


If $ x$ is the largest root of $ (\log x)^{2}\minus{}4\log x \equal{}7$, what is true about $ x/100$?

a) It is between 1000 and 2000
b) It is between 2000 and 3000
c) It is between 3000 and 4000
d) bigger than 4000
e) none of the above

[quote="Tenoreoz"]If $ x$ is the largest root of $ (\log x)^{2} \minus{} 4\log x \equal{} 7$, what is true about $ x/100$?

a) It is between 1000 and 2000
b) It is between 2000 and 3000
c) It is between 3000 and 4000
d) bigger than 4000
e) none of the above[/quote]

[hide="Wendy"]$ \log x\equal{}y$

$ y^2\minus{}4y\minus{}7\equal{}0$

$ y\equal{}2\plus{}\sqrt{11}$

$ x\equal{}10^{2\plus{}\sqrt{11}}$

$ \frac{x}{100}\equal{}10^{\sqrt{11}}$

We just need to approximate $ 10^{\sqrt{11}}$.[/hide]

[hide="Finishing up the above solution"]$ \sqrt {11} \approx \sqrt{\frac{100}{9}} \equal{} \frac {10}{3} \equal{} 3 \plus{} \frac {1}{3}$

$ 8 < 10 < 27 \implies 2 < 10^{\frac {1}{3}} < 3 \implies 2000 < 10^{3 \plus{} \frac {1}{3}} < 3000$

Hence, the answer is [b]B[/b][/hide]


Glucose sugar has the formula C6 H12 06. If you dissolved 15.0g of glucose in 100g of water, what would be the concentration of glucose, expressed as a percentage composition by mass?

Percentage composition by mass (assuming m/m) is defined as the mass (in grams) of solute that exists by every 100 grams of solution. So in your case the solution would be 13% by mass in glucose.

Hmm..15.0g of glucose in 100grams of water means 15 grams in every 100grams of water. Mass of solute (15)/solution(100). percentage is 15%. What equation did you employ to arrive at 10%?

Sorry, I made a little mistake above: the correct percentage is 13%. Look the definition I gave above: mass (in grams) of solute that exists by every 100 grams of [i][b]solution[/b][/i]. So when we mix 15 g of glucose with 100 g of water, the total mass of solution is 100 + 15 = 115 g. So we have 15 g of solute (glucose) by every 115 g of solution, so x grams of solute will exist in 100 g of solution, and $ x \equal{}\frac{15\times 100}{115}$ = 13%.

O, but of course! Your help has been very much appreciated.


floor function
number theory theorems
number theory

[quote="sinajackson"]There exist's a real number like $\mu$ exists such that every number in the form $\left\lfloor 2^{2^{.^{.^{.^{\mu}}}}}\right\rfloor$ is a prime.

Can anyone find  $\mu$ or Mr.Right just proved that it exist's and no one can find it....?[/quote]
Maybe this will help: [url]http://mathworld.wolfram.com/PrimeFormulas.html[/url].


Given an acute triangle $ ABC$ with $ AB \neq AC$, let the excircle for $ A$ meets lines $ AB,AC,BC$ at $ P,Q,R$. Let $ BC$ and $ PQ$ meet at $ X$, and the circumcircles of $ XCQ$ and $ BCP$ meet at $ Y \neq C$. Prove that $ \frac {RP}{PX} \equal{} \frac {RY}{YX}$.

$ \angle XQY\equal{}\angle XCY \equal{}\pi \minus{}\angle RPY\equal{}\pi\minus{}(\angle RPY\plus{}\angle RPB)$
$ \equal{}180\minus{}(\angle RPY \plus{}\angle RQP)\Leftrightarrow \angle RQY\plus{}\angle RPY \equal{}\pi \Leftrightarrow P,Q,R,Y$ is concyclic.
$ \angle RXY \equal{}\pi \minus{}\angle CQY \equal{}\angle QRY$
and $ \angle XRY\equal{}\angle RQY \Rightarrow \triangle XYR \sim \triangle RYQ$
$ \angle XRP\equal{}\angle XQR\Rightarrow \triangle XPR \sim \triangle XRQ$

We have $ \frac {RY}{YX}\frac{PX}{RP}\equal{}\frac{QY}{YR}\frac{RX}{QR}\equal{} \frac{QY}{QR} \frac{RX}{RY}\equal{}1$
$ \Leftrightarrow \frac{RY}{YX}\equal{}\frac{RP}{PX}$.


topology
function
analytic geometry
geometry
3D geometry
sphere

Let $ A\subset \mathbb{R}^n$ and $ B\subset \mathbb{R}^m$ be sets.

Prove that

i) If $ U\subset A$ and $ V\subset B$ are open subsets, then $ U\times V$ is an open subset of $ A\times B$

ii) If $ W\subset A\times B$ is an open set, then for every point $ (p_1,p_2)\in W$ tehre are numbers $ \varepsilon_1,\varepsilon_2>0$ such that $ O_{\varepsilon_1}(p_1,A)\times O_{\varepsilon_2}(p_2,B)\subset W$

Thank you

i) Suppose $ U$ is open in $ A$ and $ V$ is open in $ B$. Let $ (p_1,p_2)$ be a point of $ U \times V$. By openess of $ U$ we can find an open neighborhood $ D_U$ of $ p_1$ in $ U$ and by openess of $ V$, we can find an open neighborhood $ D_V$ of $ p_2$ in $ V$. By definition of the product topology, $ D_U \times D_V$ is open and is contained in $ U \times V$ so $ U \times V$ is also open.

ii) I'm not sure what the notation means, can you explain what $ O_\epsilon$ means?

Thank you very much sir. I really appreciate your response, the only problem is that the book states the problem although wer are NOT to assume that the cartesian products of open sets are open in the product topology..which I thought was defintion.

And I am sorry $ \mathcal{O}_{\varepsilon}$ is an open ball of radius $ \varepsilon$

[quote="Nukular"]What are you allowed to assume about the product? If you assume the projections $ X \times Y \to X$, $ X \times Y \to Y$ are continuous, it follows that the cartesian product of open sets must be open (Why?).[/quote]

I don't think we are supposed to do it in that manner. The book presents itself in a way such that we should be able to prove this only using the basic definitions of openedness (i.e. every point contains an open ball completely contained within the set) The answer to your "Why?" question would be because $ f: X\times Y\longmapsto Y$ is continuous and $ Y$ is open we must have that $ f^{\minus{} 1}(Y) \equal{} X\times Y$ is open...due to the properties of continuous functions.

Ok, let me try again to solve part one:

i) Let $ U$ be open in $ A$ and $ V$ be open in $ B$. By definition of openess, for any point $ p_1 \in U$we can find a length $ \epsilon_1$ such that $ O_\epsilon_1(p_1)$ is contained in $ U$. We can do the same thing for $ V$, call this other length $ \epsilon_2$. Take the smaller of the two and consider the open ball $ O_{min(\epsilon_1,\epsilon_2)}(p_1,p_2)$. This set should be contained in $ U\times V$ proving openess.

Also sorry again but what does your book mean by $ O_\epsilon(p_1,A)$?

I would guess that $ O_\epsilon (p_1,A)$ is just the open ball of radius $ \epsilon$ centered at $ p_1$ in the space $ A$.

@LkNsngth: Thank you very much! That is exactly what I had ironically...I though it was too simple to be true though  :blush: ...and yes as Roosta said $ \mathcal{O}_{\varepsilon}(p,A)$ is an open ball of radius $ \varepsilon$ around P, I must have erased this information when I edited my last post...sorry for the confusion.

Now doesn ii) follow directly from i) though?

It seems like it does, but not quite. This is how I would do it:

Since $ W$ is open, there exist open sets $ U\subset A,V\subset B$ s.t. $ (p_1,p_2)\in U\times V\subset W$. Since $ U$ and $ V$ are open, there exist $ \epsilon _1,\epsilon _2$ s.t. $ O_{\epsilon _1}(p_1,A)\subset U$, $ O_{\epsilon _2}(p_2,B)\subset V$. Then $ O_{\epsilon _1}(p_1,A)\times O_{\epsilon _2}(p_2,B)\subset W$, as was to be shown.

[quote="Mathstud28"]I don't think we are supposed to do it in that manner. ...[/quote]
There is a notion of the product topology, which is based on the [url=http://en.wikipedia.org/wiki/Universal_property]universal property[/url] that projection is continuous, and that the product of continuous functions into the components is continuous; if $ f_i$ maps $ X$ to $ Y_i$ continuously, $ f(x)\equal{}(f_1(x),f_2(x),\cdots)$ maps $ X$ to $ \prod Y_i$ continuously. In this model, the basic open sets are products of open sets*, and general open sets are unions of those.

This exercise is about showing that the usual metric topology on $ \mathbb{R}^{n\plus{}m}$ is the same as the product topology for $ \mathbb{R}^n\times \mathbb{R}^m$. That means we do in fact have to fit balls into boxes or boxes into balls.

*OK, that's for finite products only. For infinite products, it's an open set in finitely many coordinates and the whole space in the rest. There are often other reasonable topologies in this case, as well.

Is the euclidean metric equivalent to the metric from the sup norm:

$ d(\vec{x},\vec{y})\equal{}\mbox{min}\: \{ |x_1\minus{}y_1|,|x_2\minus{}y_2|...,|x_n\minus{}y_n|\}$?

Boxes fit inside of spheres, and spheres inside of boxes, so I'm pretty sure they are equivalent. If they are, then maybe the proof that $ \mathbb{R}^{n\plus{}m}$ is homeomorphic to $ \mathbb{R}^n\times \mathbb{R}^m$ (under the appropriate metrics) can be simplified, because instead of boxes inside of spheres, we will have boxes inside of boxes (the open balls of the sup norm-metric are box shaped).


Express cos$ 3\theta$ in terms of only cos$ \theta$.

Let $ a\equal{}2\theta$ and $ b\equal{}\theta$.

$ \cos(a\plus{}b) \equal{} \cos{a}\cos{b}\minus{}\sin{a}\sin{b}$
$ \equal{}  \cos{2b}\cos{b}\minus{}\sin{2b}\sin{b}$
$ \equal{} (\cos^2{b}\minus{}\sin^2{b})\cos{b}\minus{}(2\sin{b}\cos{b})\sin{b}$
$ \equal{} \cos^3{b}\minus{}\sin^2{b}\cos{b}\minus{}2\sin^2{b}\cos{b}$
$ \equal{} \cos^3{b}\minus{}3\sin^2{b}\cos{b}$
$ \equal{} \cos^3{\theta}\minus{}3\sin^2{\theta}\cos{\theta}$

To put $ sin^2{\theta}$ in terms of cosine, we have $ \sin^2{\theta} \equal{} 1\minus{}\cos^2{\theta}$
continuing..

$ \equal{} \cos^3{\theta}\minus{}(1\minus{}\cos^2{\theta})3\cos{\theta}$
$ \equal{} \cos^3{\theta}\minus{}3\cos{\theta}\plus{}3\cos^3{\theta}$
$ \equal{} \boxed{4\cos^3{\theta}\minus{}3\cos{\theta}}$

You should check in case I made any mistakes with signs, but there you have it :)

I get the same answer, albeit I use $ \cos 2\theta\equal{}1\minus{}2\sin^2 \theta$

[quote="happyme"]I get the same answer, albeit I use $ \cos 2\theta \equal{} 1 \minus{} 2\sin^2 \theta$[/quote]Or use $ \cos 2\theta \equal{} 2\cos^2\theta \minus{} 1$ ;)

Well, they're all based off of each other, so it's just using the same equation in different forms...

hello, and $ \cos(nx)$ can be written as $ \cos(nx)\equal{}2\cos((n\minus{}1)x)\cos(x)\minus{}\cos((n\minus{}2)x)$.
Sonnhard.


complex analysis
function
Stanford
college
integration
calculus
algebra

Construct a power series which diverges everywhere on the unit circle except at point $1$.

Given an integer $n$, let $f$ be a holomorphic function satisfying $\text{Re}(f(z))\leq |z|^n$ for all $z$.  Prove that $f(z)$ is a polynomial.

blah: an answer to this problem (which is a Stanford Prelim Exam question) has already been posted on the corresponding thread of sci.math.

alekk: if you are interested by a solution, check mathprepa.com. Oumpapah answered me.

@+

Hey Julian, I'm really sorry about that - I thought I started a new thread.  Yes, I cross-posted from sci.math because I thought it was a neat problem.  One of the moderators can come and split it into a new thread, or delete it entirely I guess.

Just to have a solution posted here as well (for blahblahblah's problem):

One notices that the function $\tilde f_r(z)=\frac 1{2\pi}\int^\pi_{-\pi}\frac{re^{it}+z}{re^{it}-z}\Re f(re^{it})\mbox dt\ (*)$ defined on the open disk of radius $r>0$ is a holomorphic function having $\Re f$ as its real part. This means that either it's constant (which we assume is not the case), or coincides with $f$ up to addition of a scalar. It's not that easy to "notice" this :); it's related to the Poisson integral, and, since I'm a total newbie in the field, check Rudin's "Real and Complex Analysis" (the chapter entitled "Harmonic Functions"). If we also subtract a constant from $f$, which will not matter in the following, and assume that $f(0)=0$, then for $z=0$ we have $\tilde f_r(z)=f(z)$, which means that in this case, $\tilde f$ and $f$ coincide. We'll make this assumption from now on.

Anyway, assume we know that for all complex $z$ we have $\left|\Re f(z)\right|\le M|z|^n$ for some $n$ and some constant $M>0$ (I suppose you intended for that real part to be inside an absolue value?). We fix some $r>0$, use $(*)$ to define $\tilde f$ inside the disk of radius $2r$ centered at the origin, and find an upper bound on the absolute values of $f(z)$ for $|z|\le r$.

If $|z|\le r$, we have $\left|\frac{2re^{it}+z}{2re^{it}-z}\right|\le 3$, and we get $|f(z)|=|\tilde f_{2r}(z)|\le 3Mr^n$ for $|z|\le r$. Since this holds for arbitrarily large $r$, it easily implies that $f$ is a polynomial of degree at most $n$. We can use Cauchy's inequality and Liouville's Theorem, for example: $\left|f^{(n)}(z)\right|\le\frac{n!\max_{|z|\le r}|f|}{r^n}\le\frac{3n!Mr^n}{r^n}=3n!M$, which means that the $n$'th derivative of $f$ is an entire bounded function, so it must be constant.

I hope it's at least remotely correct :).


[quote="Padmavathi"]2.[hide]Because it was covered in spots...[/hide]
3.[hide]All months... but Feb has ONLY 28 days[/hide][/quote]

padmavathi, ur #3 is correct,but #2 is not right!
Keep trying!

[quote="mathnerd314"][quote="klutzycherry14"][color=blue][b]1. Why was the baby ant confused?

2. Why did the banana go to the doctor?

3. What month has 28 days?[/b][/color][/quote]

1. Obviously, the baby ant was confused because the researcher painted it with that one chemical that smells exactly like dead ant (I forget the name). Therefore, the other ants in the anthill tried to carry her to the ant cemetery, despite her fervent protest. This was an actual study conducted by some researcher, which proved that ants rely on their sense of smell to determine whether other ants are alive or not.

2. The banana went to the doctor because a human was holding the banana, waving it around and pretending it was a gun. However, the receptionists were not fooled, although one unfortunate patient was scared badly and had to be hospitalized for two weeks.

3. No month has 28 days by the lunar calendar. Doesn't everyone use the lunar calendar?

</SARCASM>[/quote]

the lunar calendar is helpful when brewing potions

[/more sarcasm]

[quote="Padmavathi"]Oh Yeah...I remember...my bro told me the Banana one long back...but i 4got!
On the lines of silly riddles...here are some more-
1.What do injured pigs use?
2.Where do injured insects go?
3.What do frogs drink?
4.What do cows listen to?
5.Which stone has people under it?[/quote]

[hide="4"]moosic[/hide]

[quote="moogra"][quote="klutzycherry14"][color=blue][b]1. Why was the baby ant confused?

2. Why did the banana go to the doctor?

3. What month has 28 days?[/b][/color][/quote]

1. [hide]All his uncles were ants![/hide]
2. [hide]Because it wasn't peeling well[/hide]
3. [hide]All of them, question isn't phrased correctly though[/hide][/quote]

yup! :lol:
pretty silly, but they are fun!

1. Stretchers
2. The Ground
3. Clueless...
4. Babies :P

[quote="moogra"][quote="Padmavathi"]Oh Yeah...I remember...my bro told me the Banana one long back...but i 4got!
On the lines of silly riddles...here are some more-
1.What do injured pigs use?
2.Where do injured insects go?
3.What do frogs drink?
4.What do cows listen to?
5.Which stone has people under it?[/quote]

[hide="3 wild guess"]
croak[/hide][/quote]
Actually I think it's "croak-a cola"  :rotfl:

[hide="1"]OINKment (ointment)[/hide]
[hide="2"]HOPSpital (hospital)[/hide]
[hide="3"]CROAKa cola (coca cola)[/hide]
[hide="4"]MOOsic (music)[/hide]
:lol: :laugh:

[quote="rjt"][hide="5"]
A headstone
[/hide]

:P[/quote]
Ooh! I know what it is!
[hide]GraveSTONE[/hide] :roll:

Yeah ur all right!!!  :first:
Only....Injured Insects go to a WASPital!!! :P


Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were $71$, $76$, $80$, $82$, and $91$. What was the last score Mrs. Walter entered?

$\text{(A)}\ 71 \qquad\text{(B)}\ 76 \qquad\text{(C)}\ 80 \qquad\text{(D)}\ 82 \qquad\text{(E)}\ 91$

[hide]Since the average of the first three scores has to be a whole number, the sum of the first three scores must be a multiple of three (or congruent to 0 (mod 3).
Writing the scores in ascending order (mod 3): 2, 1, 2, 1, 1.
Only if we choose the three numbers that are congruent to 1 (mod 3) can the sum of these first three numbers be 0 (mod 3).

Thus the first three numbers are 76, 82, and 91. Since the first two scores have an average of a whole number, their sum must be even. Thus the first two scores are 76 and 82.

The fourth number has to make the sum of the current list be divisible by 4, which, out of the remaining two numbers, only 71 works. Thus the last number entered is 80. [/hide]

[hide=Solution]Taking $\pmod 3$ gives $2, 1, 2, 1, 1$, so our first three are $76, 82, 91$. We must have an even sum for the first two, so it is in that order. Only $71$ works for the next one, so the answer is $80 \implies \boxed{\text{(C)}}$.[/hide]


integration
trigonometry
calculus
calculus computations

For $ |k|<1,$ prove that:

$ \int_0^{\frac{\pi}{2}} \sqrt{1\minus{}k^2\sin ^ 2 \theta}\ d\theta \equal{}\frac{\pi}{2}\left(1\minus{}\frac{k^2}{4}\minus{}\frac{3}{64}k^4\minus{}\frac{5}{256}k^6\minus{}\cdots\right).$

This Integral is complete elliptic integral $ E(k^2)$?

:D
Kunny, these are entrance exam questions again?

Needless to say, Japanese high school students usually don't study Taylor expansion. :D


#1: Just $ \frac{1}{6}$

#2: We have

$ 4\plus{}1 \equal{}5$
$ 4\plus{}2 \equal{}6$
$ 4\plus{}3 \equal{}7$
$ 4\plus{}4\equal{}8$
$ 4\plus{}5\equal{}9$
$ 4\plus{}6\equal{}10$

Which is $ \frac{3}{6}$ outcomes, which is a 50% probability.

[quote="guile"]A pair of dice is thrown. If it is known that 1 die shows a four, what is the probability that
(a) the other die shows a 5.
(b) the total of both dice is greater than 7.[/quote]

[hide]
a. $ 1/6$ because there is only $ 1$ way to get a $ 5$ on $ 1$ die.
b. to be greater than $ 7$, $ 4\plus{}x>7$, so $ x>3$ so it can be $ 4,5,6$ so $ 3$ ways. Since it is 3 ways, and there are 6 possible ways, the probability is $ 1/2$ [/hide]

[quote="guile"]nope. it's conditional probability.[/quote]

wait how does that change anything? It's only 1 die getting rolled basically
unless if you want the probability of getting  4 too

[quote="guile"]nope. it's conditional probability.[/quote]

In that case, multiply both answers by $ \frac{1}{6}$ (the probabillity of getting a 4 on first roll). So

a) $ \frac{1}{36}$

b) $ \frac{1}{6}\times\frac{1}{2}\equal{}\frac{1}{12}$


Starting on June 1, on each odd-numbered date Kirsten earns 1 dollar and on each even-numbered date she earns 2 dollars. On what month and date will her total earnings first reach or exceed 50 dollars?

[hide] We may view each combination of even and odd numbered days as a unit. Therefore, in one of these "units", she earned $ 3$ dollars. $ \frac{50}{3}\equal{}16\frac23$, so in the $ 17$th unit she met her goal. Remembering that June has $ 30$ days, and the difference between even and odd, we see that the day is July 4.[/hide]

[hide]Every two days, she earns 3 dollars. After 32 days, she has 48 dollars. The answer is thus 34 days.[/hide]


Can anyone help with this?
[url=http://img105.imageshack.us/my.php?image=untitledfl5.png][img]http://img105.imageshack.us/img105/3299/untitledfl5.th.png[/img][/url]

Multiply and divide by $\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}$

The limit is 1. Multiply and divide by the sum of the two large square roots and then divide both terms of the fraction by square root of x.

Lance DuLac


(a) Deﬁne an equivalence relation on the plane $ X\equal{}\mathbb{R}^2$ as follows:
$ x_0 \times y_0$ ~ $ x_1 \times y_1$ if $ x_0 \plus{} y_0^2\equal{}x_1 \plus{} y_1^2.$
Let $ X^*$ be the corresponding quotient space. It is homeomorphic to a familiar space; what is it? [Hint: Set $ g(x \times y)\equal{}x\plus{}y^2.$]

(b) Repeat (a) for the equivalence relation
$ x_0 \times y_0$ ~ $ x_1 \times y_1$ if $ x_0^2\plus{}y_0^2\equal{}x_1^2\plus{}y_1^2.$
[this is Munkres §22.4]

the first is $ \mathbb{R}$ and the second is $ [0,\infty)$ with the subspace topology.

I'm not exactly sure what the Munkre's book is asking for, but here is a rather quick way of doing this problem:

Clearly the map $ g(x,y) \equal{} x\plus{}y^2$ induces the same partition on $ \mathbb{R}^2$ as the equivalence relation, so consider the surjective map $ g: \mathbb{R}^2\rightarrow \mathbb{R}$ given by that function. Now because we know $ g$ is a quotient map (verify this) we know that $ \mathbb{R}$ is homeomorphic to $ X^*$


function
set theory
superior algebra
superior algebra unsolved

Choosing a bijection $ f\colon S\to S$ is like drawing elements of $ S$ without replacement: you choose $ f(s_1)\in S$, then $ f(s_2)\in S\setminus\{f(s_1)\}$, etc. You have at most $ |S|$ choices on each step, so the cardinality of $ B$ is at most $ |S|^{|S|}$. On the other hand, you have at least two choices on each step but the last one, hence $ |B|\ge 2^{|S|\minus{}1}\equal{}2^{|S|}$. Under the axiom of choice $ |S|^{|S|}\equal{}2^{|S|}$ for any infinite $ S$. Thus, $ |B|\equal{}2^{|S|}$ (which is strictly greater than $ |S|$, hence no bijection between $ B$ and $ S$).

[quote="seamusoboyle"]Given an infinite set S (and its cardinality), define B to be the set of all bijections from S to S. Prove there is no bijection between B to S. What is the cardinality of B?[/quote]

Hello.
I find this in Bourbaki, Set theory, ch. III, § 6, exerc. 5.

First fact : if a set $ X$ has at least two elements, there is a permutation of this set without fixed point.
Proof.
If $ X$ is finite with cardinality $ n \geq 2$, take a $ n$-cycle.
If $ X$ is infinite denumerable, carry the permutation $ (\ (0,1), (2,3)\ ...\ )$ of $ \mathbb{N}$ onto $ X$.
If $ X$ is any infinite set, there exists a partition $ (X_{i})_{i \in I}$ of $ X$ with infinite denumerable sets; treat each $ X_{i}$ as above and paste the $ i$ permutations together.

Second fact : if $ Y$ is a set, if $ A$ is a subset of $ Y$ whose complement in $ Y$ has more than one element, then there exists a permutation of $ Y$ whose fixed points are exactly the elements of $ A$.
Proof. In view of the first fact, there exists a permutation of the complement with no fixed point. Complete it to a permutation $ f$ of $ Y$ by putting $ f(x) \equal{} x$ for each $ x \in A$. Clearly, $ f$ is as desired.

Third fact. If $ X$ is an infinite set, the set $ S_{X}$ of all permutations of $ X$ has the same cardinality as the set $ \mathcal{P}(X)$ of all subsets of $ X$.
Proof. To each element of $ S_{X}$ other than the identity, we let correspond the set of all fixed points of this permutation. In view of the second fact, we define thus a surjection of  $ S_{X} \ \minus{} \{id_{X}\}$ on the set $ \mathcal{P'}(X)$ of all subsets of $ X$ whose complement in $ X$ has at least two elements. Thus Card($ S_{X}$) $ \geq$ Card($ \mathcal{P'}(X)$). (Result 1.)
Now, $ \mathcal{P'}(X)$ is the complement in $ \mathcal{P}(X)$ of the union of the set  $ \{id_{X}\}$ with the set of all subsets of $ X$ with cardinality $ 1$.
Since $ X$ is infinite and $ \mathcal{P}(X)$ has cardinality strictly greater than $ X$, it follows easily that $ \mathcal{P'}(X)$ is equipotent to $ \mathcal{P}(X)$.
In view of (result 1), we have proved that Card($ S_{X}$) $ \geq \mathcal{P}(X)$.
Now, $ S_{X}$ is a subset of the set $ X^{X}$ of all functions of $ X$ into itself, and it is known that Card($ X^{X}$) = Card($ \mathcal{P}(X)$), thus Card($ S_{X}$) $ \leq$ Card($ \mathcal{P}(X)$).
Finally, Card($ S_{X}$) $ \equal{}$ Card($ \mathcal{P}(X)$).

Friendly,
P.

Edit : I had not seen the shorter proof of mlok.

[quote="mlok"] On the other hand, you have at least two choices on each step but the last one, hence $ |B|\ge 2^{|S| \minus{} 1} \equal{} 2^{|S|}$. Under the axiom of choice $ |S|^{|S|} \equal{} 2^{|S|}$ for any infinite $ S$. Thus, $ |B| \equal{} 2^{|S|}$ (which is strictly greater than $ |S|$, hence no bijection between $ B$ and $ S$).[/quote]

Hello Mlok.
I think this proof needs to be refined. We must be sure that the set will not be exhausted when we come at the last step. We have already chosen pairs of elements in quantity equal to the cardinality of $ S$ before we reach the last (or penultimate...) element to which we must attribute a value, and how are we sure that the union of these pairs is not already the whole $ S$ ?
So, I think it is better to begin with choosing a partition of $ S$ in subsets of two elements. There is a bijection between $ S$ and the partition (viewed as a set of pairs), and so the proof can be made rigorous, at least in order to prove that the set of injections of $ S$ into itself has cardinality $ \geq 2^{|S|}$. But does it prove that the set of permutations of $ S$ has cardinality $ \geq 2^{|S|}$ ?

Friendly,
P.


can anyone find malaysia national mathematics olympiad questions and solutions website to me?

Try for [hide="EATS"]eats.com.my/home/[/hide] that is all what I can help you  :wink: