AMC
USA(J)MO
USAMO
email
AIME
AMC 10
MATHCOUNTS

I write to correct a number of false statements and incorrect
understandings in this forum about the time it takes the AMC office
to complete the USAMO selection process.

In the 2009 USAMO Teachers' Manual, on page 4, Section I, paragraph 6
it says:  "The selection of the USAMO participants will be made on or
before April 10th.  All students ... will receive information...no
later than April 10."  We published the USAMO invitation list and sent
the email invitations on April 10.  We met our announced date.  We
were not late or delayed in making the invitations.

The 2009 USAMO Teachers' Manual was written and printed months ago.
After more than 30 years of administering the USAMO, we know our
business and we know how long it takes to complete our business.  We
met our announced date.  We were not late or delayed in making our
invitations.

There are two primary reasons that it takes us approximately 9 days
from the alternate AIME date to the announcement of USAMO
invitations.
 (1)  We do not receive the AIME II score sheets from some schools
      until as late as 1 week after the AIME II.  On Friday, April 3
      we had received from about 60% of participating
      schools.  Even on Wednesday, April 8 the AMC office still
      received AIME score sheets from some schools.  This is in spite
      of requiring schools to send by some kind of express mail or
      package delivery.  In some cases, it still takes 5 calendar days
      or more for the answer forms to reach us.

  (2) Second, and far more time consuming for us is that fact that
      many students incorrectly fill out the answer sheets.  The most
      common error is to not use the same name when taking the AMC
      contest and the AIME contest, or when taking the AMC contests on
      both the A-date and the B-date.  If the student uses the same
      name each time, our scoring processes automatically match the
      names.  If a student did not use the same name, or incorrectly
      filled out the answer from, we have to laboriously match the
      student names and scores.  This is tedious and time-consuming,
      particularly if the students took the contest at different
      locations.  If students correctly and consistently filled out
      the answer forms, we might be able to save 2 to 3 days on the
      selection process.

  (3) A third time consuming effort is when students take the AMC 10
  or AMC 12 at more than one location, but request that scores and
  affiliations be reported from, say, their high school.  It takes a
  lot of searching, matching, cross-checking and validation to make
  sure that we have reported the scores correctly.

One potential solution to the second and third problems would be to allow
students to take only one AMC contest on the A-date or the B-date but
not both.  Of course, a simpler solution would be for all students to
consistently and correctly enter their information on the answer
forms.  The AMC office continues to explore technological solutions
to the first problem that would easily adapt to all schools nationally
and internationally.

Steve Dunbar
MAA Director, American Mathematics Competitions

Thanks for the time and effort you spend responding to concerns on this forum.

Yes, I agree with worthawholebean, thank you.

And thanks for all the effort on sorting out tests and making sure that the results were on time.  Just a thought, but maybe the AMC office was too nice last year with the preliminary list?  :D

Thank you for your efforts. Sometimes people are just a bit eager to find out whether they made USAMO =p but we appreciate your work.

[quote](3) A third time consuming effort is when students take the AMC 10
or AMC 12 at more than one location, but request that scores and
affiliations be reported from, say, their high school. It takes a
lot of searching, matching, cross-checking and validation to make
sure that we have reported the scores correctly. [/quote]
I'm sure this is a massive problem and a thankless one to try to overcome. And very, very, very hard to get everyone on the first pass. Do not be surprised if the list of USAMO qualifiers grows slightly as more multiple-site names are matched up. (In fact, I personally know of one such case, and I will inform the AMC office of what I know.)

[quote="AMCDirector"]
  (2) Second, and far more time consuming for us is that fact that
      many students incorrectly fill out the answer sheets.  The most
      common error is to not use the same name when taking the AMC
      contest and the AIME contest, or when taking the AMC contests on
      both the A-date and the B-date.  If the student uses the same
      name each time, our scoring processes automatically match the
      names.  If a student did not use the same name, or incorrectly
      filled out the answer from, we have to laboriously match the
      student names and scores.  This is tedious and time-consuming,
      particularly if the students took the contest at different
      locations.  If students correctly and consistently filled out
      the answer forms, we might be able to save 2 to 3 days on the
      selection process.

  (3) A third time consuming effort is when students take the AMC 10
  or AMC 12 at more than one location, but request that scores and
  affiliations be reported from, say, their high school.  It takes a
  lot of searching, matching, cross-checking and validation to make
  sure that we have reported the scores correctly.
[/quote]

Dear AMCDirector,
Have you thought about assigning an unique number for each student (like using SSN)? I know many students took AMC 10 / AMC 12 at more than one location, unique number for each student may help to identify the correct score for each student.

[quote="AMCDirector"]One potential solution to the second and third problems would be to allow
students to take only one AMC contest on the A-date or the B-date but
not both.  [/quote]

A better solution is to modifying the answer sheets for students to write in their CEEB# and scores earned at the specific CEEB location:

AMC 10/12A: CEEB#, AMC10 or AMC 12, your score
AMC 10/12B: CEEB#, AMC 10 or AMC 12, your score

It will make the automatic matching easier.

[quote="shtsxc12"][quote="AMCDirector"]
Dear AMCDirector,
Have you thought about assigning an unique number for each student (like using SSN)? I know many students took AMC 10 / AMC 12 at more than one location, unique number for each student may help to identify the correct score for each student.[/quote][/quote]

That will not work as AMC/AIME exams are international.  Students in China, Taiwan, Singapore, etc. do not have SSN!

[quote="zmli"][quote="shtsxc12"][quote="AMCDirector"]
Dear AMCDirector,
Have you thought about assigning an unique number for each student (like using SSN)? I know many students took AMC 10 / AMC 12 at more than one location, unique number for each student may help to identify the correct score for each student.[/quote]

That will not work as AMC/AIME exams are international.  Students in China, Taiwan, Singapore, etc. do not have SSN![/quote][/quote]

I think he means a special identification number just for AMC, not the actual SSN itself.

[quote="not_trig"]I think he means a special identification number just for AMC, not the actual SSN itself.[/quote]

So something along the lines of what they do with USAMO student numbers? The problem is that the numbers would have to be centrally managed, and that takes a long time to administer. An [i]extremely[/i] long time.

The problems with an AMC-assigned special identification number:

It would have to have enough digits to distinguish several hundred thousand individuals - so it would have to be a 6 or 7 digit number, not the easiest thing to remember.

The onus would be on the individual student to remember/bring that number to the B date. There would have to be a mechanism for giving each student a piece of paper or sticker with that number on it (extra printing cost), and even if you did that, many hundreds of students would forget to bring that sticker to the B exam, in which case you're no better off than before, perhaps worst off.

There are a dozen different reasons why the AMC shouldn't and won't use Social Security numbers. I won't elaborate, other than to say this is a complete non-starter.

The only scheme I can see with a slight amount of hope: phone numbers. Put a place for a 10 - digit number, and ask student to fill in a number of their own choosing that they will remember, with the suggestion that it be a phone number (but the student is allowed to invent some other number). There would be some duplicates, both among siblings and among people that invented numbers, and there would still be students forgetting which number it was they had used the first time.

I still think that there really isn't a better scheme than name-matching with address-matching as a backup. Unless the AMC wants to prohibit students from taking more than one AMC, or from taking AMC and AIME at different locations (which actions would ignite their own protests), then this problem is going to be there.

I would like to continue having the opportunity to take two AMCs. Apart from the mathematical side, AMC is very error-prone; by that, I include scantron error and teacher error. Being able to take two time greatly reduces such problems.

A possible modification: if one choose to take the second AMC, then he or she automatically forfeits the first AMC score. It would be best if the first AMC score comes out before this decision is made.

I can only imagine the logistical challenge facing the AMC every year!  Kudos to them for managing as well as they do.

According to the published statistics, thousands of students don't even bother to bubble in their grade level on the AMC exams, and hundreds of students don't bubble in their grade levels on the AIME.  Also, students with common names present challenges as well.  (At our state MATHCOUNTS competition, there were ~200 students competing, and when the emcee read off the list of the top twelve students to call them down for Countdown, 14 students materialized at the front of the room, due to multiplicity of names.)

So any suggestions we can make can only mitigate the logistical challenge, not make it go away.  I completely agree with Kent that the idea of SSN's is totally a non-starter.

[quote]The only scheme I can see with a slight amount of hope: phone numbers. Put a place for a 10 - digit number, and ask student to fill in a number of their own choosing that they will remember, with the suggestion that it be a phone number (but the student is allowed to invent some other number).[/quote]

How about including a field for date of birth month/day/year on the AMC and AIME answer sheets, as a supplement to the phone number?

The date of birth should at least be easy for students to remember and replicate consistently each time.   The phone number is a good idea because there may be some occasional instances when the easiest way for the AMC to get some piece of information is just to call and ask.

Also, how about designing the AIME answer form to ask the student about both their A date and B date scores, along with a place for the student to bubble in a field to designate if one of the AMC scores comes from a different location than the place where s/he is taking the AIME?

Again, these suggestions are only designed to make life a little bit easier for the AMC.   I am completely in awe of their ability to manage the process as quickly as they do.

One more thought:  perhaps there could be a special incentive for complete, consistent, and correctly filled out answer sheets.

Each year, students whose AMC and AIME forms provide clear, consistent, and correct information that cause no problems for the AMC staff could be eligible for a drawing for a small prize, for example, a subscription to the MAA publication, [url=http://www.maa.org/mathhorizons/]Math Horizons[/url].  Students whose AIME forms wound up in the "problem pile" would not be eligible for the drawing.

Alternatively, the incentive could operate at a school level.  Schools that had no students winding up in the "problem pile" could be eligible for a drawing for a bulk subscription of Math Horizon magazines.

The drawing could be done in late summer, when the AMC is not quite so busy as at other times of year.  The winning school could get recognized on the AMC website and in the MAA Messenger as a way to remind contest managers and students to be careful with their data entry.

I have previously donated to the MAA to recognize the hard work done by the AMC Lincoln staff--there's a brick in the [url=http://www.maa.org/development/riverofbricks.html]MAA headquarters "River of Bricks" walkway[/url]celebrating the AMC staff, with a special shout-out to Donita Bowers, who retired a couple years ago.

I would be happy to make an additional donation to fund subscriptions to a great magazine as a small incentive to students who do their part to make the AMC staff's work just a little bit easier.

[quote="timwu"]A possible modification: if one choose to take the second AMC, then he or she automatically forfeits the first AMC score. It would be best if the first AMC score comes out before this decision is made.[/quote]
Correction: this should only be applicable if the first score comes out before the second test. You need the information first. What if you got better on the first one and didn't know it until after you did worse on the second one? It's too risky to do it without the information (and isn't in the spirit of the AMC. I don't think it will promote dishonesty either).

[quote="sophia"]The date of birth should at least be easy for students to remember and replicate consistently each time. The phone number is a good idea because there may be some occasional instances when the easiest way for the AMC to get some piece of information is just to call and ask. [/quote]
This is an extremely good idea. They do this on the Euclid contest (not sure about other CEMC contests, but alright) as well.

[quote="sdkudrgn88"][quote="zmli"]A better solution is to modifying the answer sheets for students to write in their CEEB# and scores earned at the specific CEEB location:

AMC 10/12A: CEEB#, AMC10 or AMC 12, your score
AMC 10/12B: CEEB#, AMC 10 or AMC 12, your score

It will make the automatic matching easier.[/quote]

Problem is, people from International locations don't have a CEEB. (Or, could you explain what a CEEB is, if I'm wrong?)[/quote]

This is not a problem.  AMC office automatically assign a test site a CEEB number if they don't have one.  So, every test site has a CEEB number as far as AMC is concerned.

The problem is that the CEEB number of the location where the student took a previous exam is not something that most students carry around in their heads. The diligent ones will write it down and carry a note with them, but if you could trust everyone to do that, you could go ahead an assign a unique identifier to each individual.

Students don't always remember their previous scores correctly, either. I'm sure the AMC policy about that is "Thanks for the hint, but if you don't mind, we'll look that up ourselves."

For starters, why not just print a warning message?

"If you are also taking the AMC 10/12B, or if you qualify for the AIME, note that [b]you must give exactly the same name on each contest that you participate in[/b]."

And similarly for each contest.

[quote="worthawholebean"]Personally, I really like the idea of restricting the AMC to one taking.[/quote]

I certainly agree with you relative to taking 2 AMC-12s or 2 AMC-10s.  Getting two bites at the apple here doesn't seem to make things more fair for anyone.

However, given all the fuss on these forums about whether underclassmen should take the AMC-12 or the AMC-10 (maximizing AIME chances (and taking a more challenging test) vs. potentially maximizing USAMO chances), I think a great solution there is to have the gifted underclassmen take 1 AMC-12 test and and 1 AMC-10 test.  I realize this won't work for everyone (e.g. school doesn't/won't offer both test dates, money concerns, etc.) and some will have to choose (or have their school choose for them), but it does present a remedy for a large number of the borderline cases.  I wouldn't want to give this up unless there is a good solution to the other problem that doesn't compare AMC-10 scores as equal to AMC-12 scores for the underclassmen USAMO index calculation.

P.S. On a 'rabbit-trail' side note, I went through the system in the early 90s where we had 1 ASHME and 1 AIME (no alternate date tests, and no AMC-10), and consideration to underclassmen wasn't given until MOP selection.  I find it a bit ironic that the process has certainly bent over backwards to afford underclassmen a lot of opportunity (a whole AMC-10 test, a generous portion of 'underclassmen-only' slots in the USAMO, and the large number of MOP spots reserved for freshmen and underclassmen), yet we spend time on the forums debating how the process isn't fair to those underclassmen.  I'm not suggesting the debate isn't warranted, but there is a level of irony here that is quite inescapable.

I'm pretty sure we're debating on how it's not fair to the upperclassmen, not the other way around.

No, I thought everyone was saying that some underclassmen are kicked out of the second cut because of large amount of underclassmen who take the 10 instead of the 12.  The upperclassmen actually get the advantage of not having to compete with many underclassmen.

[quote]
could you explain what a CEEB is
[\quote]

CEEB stands for College Entrance Exam Board, now just known as the College Board.  They are the folks behind the SAT, the AP Exams and several other things.  They have a 6-digit code for each high school in the US, the CEEB number, that helps identify their information.  In the 1980s and 1990s when the AMC went to centralized scoring, we adopted the numbers as a way to uniquely identify high schools.  With many home schools, math circles, colleges and universities and schools outside the US now administering the AMC contests, we have created many new numbers, so the "CEEB"s that many of you use do not actually exist at the CB, just at the AMC.

[/quote][quote]
perhaps there could be a special incentive for complete, consistent, and correctly filled out answer sheets.
[\quote]

Interesting idea.  We'll explore it.  Philosophically, I like positive incentives rather than negative enforcements.  Might be worth considering.

[quote]I have previously donated to the MAA to recognize the hard work done by the AMC Lincoln staff--there's a brick in the MAA headquarters "River of Bricks" walkway celebrating the AMC staff, with a special shout-out to Donita Bowers, who retired a couple years ago.
[\quote]

Thanks!  Next time I'm in Washington at MAA HQ, I'll take a picture of it, and we'll post it at the website.

[quote]
field for date of birth month/day/year on the AMC and AIME answer sheets, as a supplement to the phone number
[/quote]

The date of birth and the phone number are both possibly feasible ideas.  We will consider using them.  I wonder about students outside the US, where phone numbers can be variable length.....

One thing to balance here is that we try to collect as little personal information as reasonably possible about each student.  That's better for you, and and better for us.  This maybe a place where we reasonably have to increase what we collect.

[quote]For starters, why not just print a warning message?

"If you are also taking the AMC 10/12B, or if you qualify for the AIME, note that you must give exactly the same name on each contest that you participate in."
[/quote][/quote][/quote][quote][/quote]

We will consider adding some kind of warning like that, perhaps in the instructions in the Teachers' Manual.

Our experience is that such warning messages are ignored.  Most students seem to be focused on solving the problems and getting the responses bubbled in, not on studying detailed instructions on the answer sheet.  And that's okay, I understand that.

In any case, the students who did not follow the instructions to use the same name each time would still write or call us, and ask that we match their scores.  In that case, we might not have a significant increase in efficiency.

We work about 12-24 months in advance planning for contests.  We might be able to make these changes for 2010, more likely for 2011.

Steve Dunbar
MAA Director, American Mathematics Competitions

I think the date of birth/phone number method is the best idea here so far.  I don't think people will mind having that extra personal information collected as long as they know it's only being used to match up tests.  And international phone numbers shouldn't be a problem--just allow the phone number to have variable length.

It still leaves room for ambiguity, though--you'd have to remember if you put your home or cell number.  And international students, I suppose, would have to remember if they included country code and all that.

[quote="m1sterzer0"]P.S. On a 'rabbit-trail' side note, I went through the system in the early 90s where we had 1 ASHME and 1 AIME (no alternate date tests, and no AMC-10), and consideration to underclassmen wasn't given until MOP selection.  [/quote]
In the late 90's, there was only 1 ASHME and 1 AIME, but there was definitely a multi-tiered selection system. A main selection ("first cut") with everyone eligible, followed by a few more 11th grade and below, followed by a few more 10th grade and below, followed by a few more 9th grade and below.

I still find that to be an attractive scheme, provided there is a single basis of selection for all of these tiers, such as "M.M. index."

Of course, when [i]I[/i] went through the system (1969-71), there was no AIME and no USAMO. Just the AHSME. And the U.S. did not participate in the IMO - that was something that communist countries in Eastern Europe did.

I think the change might be at 1996- that's when the one qualifier per state rule was introduced, at least. Since then, the AMC has had various systems, all of which made it easier to qualify as an underclassman.


So an interesting idea hit me the other day.  What if we as members of AOPS got together and formed a global secret society where we aspire to take over the world....muhuhahaha.  Think about it for a second; AOPS consists of perhaps some of the most intelligent and gifted young students in the world, and we could do some great things.  The society could consist of different chapters in various regions, with some sort of a hierarchy in which each region would have a head person and some sub-leaders.  We could have virtual meetings online via webcam/webchat (perhaps Skype?) where all of the leaders would meet each week online to discuss the "operations" of the society and to see how each region is doing.  It could be quite the organization.  We could even design some sort of a stock market operation for investments and whatnot.  What do you guys think of the idea?

hm.we probably could talk about random stuff for 5 seconds and then get bored and then go do the same on FTW...

how much time did you spend planning this thread and thinking about all the e-props that you were going to get for it?

Haha, no not at all.  I just found it to be quite an amusing idea.  Believe me, I definitely have a life ;)

[quote="Lazarus"]imo you need a life

the fact that you posted it in round table kinda implies that you're taking this seriously

which... is a scary thought[/quote]

This isn't round table. This is Games & Fun Factory.

It was originally posted in Round Table.

Yeah, that was my bad.  I was unsure of which one I should post it in.  Obviously I chose the wrong one though.

[quote="MysticTerminator"]hey neato I call dibs on ultimate world leader okay[/quote]

How did I know that MysticTerminator was going to be the first person to say this?

lol I'll join the ring... I call Planner!

I'll join ring... I call [i][b]Whatever[/b][/i].

I'll join, I'll ask people in EPGY OHS (gifted online high school run by stanford) to join :)

EHEM! this is [size=200][u][i][b][color=darkred]AOPS[/color][/b][/i][/u][/size] secret society!!! :mad:

I join! I can be the entering platform to Latin America

I call snacks provider. Anyone care for a cookie?

[quote="mathblitz"]i join as prez advisor me is so not smart! o and lol googlism is so funny!id join but my parents wont let me[/quote]

Just join it secretly.

um... wait so what was the original purpose of this discussion again???

anyways i is ok with revealing who i really am to the rest of the secret society. like who i am other than an anime-obsessed computer freak.

I was showing how great a religion Googlism was :)

hm...well, I guess I'll be the person who pretends not to be a part of this, therefore allowing me to retrieve outside information from certain connections... =D

jk, I'll do whatever.

An AOPS Secret Society FAILED!!!!!!!!

It is supposed to be SECRET. This is NO LONGER SECRET. So THIS FAILED!!! QED

Sorry

How do YOU know it's not secret...

You aren't in the society...

anyway, i also proclaim myself church co-minister (with chenhsi) and that the official religion is Googlism. Usually im an atheist, but googlism sounds good. :D

i think if we got any more members it REALLY wouldnt be secret any more...

I join!  But considering the fact that I don't exist, this secret organization is still a secret, right?  :roll:

[quote="MysticTerminator"]hey neato I call dibs on ultimate world leader okay[/quote]

OMG THAT IS NOT ARNAV HE IS NOT POSTING IN CAPS OMG HELP ARNAV!

Also, this is a dumb idea, since some people *cough*me*cough* live in the middle of nowhere, and the nearest user isn't even in their county.

[quote="splatyango"]um actually same here, don't get depressed
unless you count all the ppl in seattle
but there still in a different county[/quote]
Hey! What about me?!! (I'm her brother)
yeah........

oh sry lol  :blush:

but u dont come on often, excluding today

so yeah.

EDIT: putting subjects is ok, but most ppl dont, so unless you have a really important point, i would advise you not to...


What is the sum of the digits of the sum of the digits of the sum of the digits of $ 4444^{4444}$

Let $ A=4444^{4444}< 4444^{10000}$
Then $ A$ has less than $ 4444*\lg10000 = 17776$ digits.
$ \implies \text{the sum of digits of A}< 9*17776 = 159984$
So the sum of the digits of the sum of digits A is less than $ 9*6 = 54$
Then X is the sum of the digits of  the sum of the digits of the sum of digits A $ \leq 4+9 = 13$
We know that $ X \equiv A \equiv (-2)^{4444}\equiv 2^{3*1481+1}\equiv-2 (mod 9)$
That means $ X=7$

Hope I don't make any mistakes.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1299623712&t=133303

Please, if you're going to insist on posting so many questions, at least check to make sure they haven't been posted before.

[quote="bibobeo"]Let $ A = 4444^{4444}< 4444^{10000}$
Then $ A$ has less than $ 4444*\lg10000 = 17776$ digits.
$ \implies\text{the sum of digits of A}< 9*17776 = 159984$
So the sum of the digits of the sum of digits A is less than $ 9*6 = 54$
Then X is the sum of the digits of  the sum of the digits of the sum of digits A $ \leq 4+9 = 13$
We know that $ X\equiv A\equiv (-2)^{4444}\equiv 2^{3*1481+1}\equiv-2 (mod 9)$
That means $ X = 7$

Hope I don't make any mistakes.[/quote]

Where did you get $ X\equiv A\equiv (-2)^{4444}$ from?

Since $ 4446$ is a multiple of $ 9$, $ 4444$ must be equivalent to $ 7$, or $ \minus{}2\text{ mod }9$. Raising both $ 4444$ and $ \minus{}2$ to the power of $ 4444$, we must obtain that the results would still be congruent $ \text{ mod }9$. So $ 4444^{4444}\equiv (\minus{}2)^{4444}\text{ mod }9$.

Considering the numbers mod 9 in these types of problems is very helpful, because if you keep taking the sum of digits till there is only one digit left, you get the [i]digital root[/i], which is constant mod 9.


[url]http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=66008[/url]

It seems like this person is an impersonator of the user r15s11z55....(something) or some existing user's multi. Suspicions/thoughts?

Please pm an admin directly about these.

This user is just bored and has nothing better to do. Please ignore him.


function
trigonometry
algebra
functional equation
real analysis
real analysis unsolved

Prove or disprove the following:
let f,g be real functions s.t g(x-y)=g(x)g(y)+f(x)f(y) and g(0)=1 then g'(x)=-f(x)f'(0) & f'(x)=g(x)f'(0)

Let $\alpha\left(x\right)$ be any function satisfies $\alpha\left(x+y\right)=\alpha\left(x\right)+\alpha\left(y\right)$, then $f\left(x\right)=\sin\left(\alpha\left(x\right)\right),\ g\left(x\right)=\cos\left(\alpha\left(x\right)\right)$ is a solution, which is not necessarily continous.

Yes, but given $f$ and $g$ can one construct a solution of $\alpha(x+y) = \alpha(x) + \alpha(y)$?
This is the point mentioned above.

This problem is relatively easy if also given the functional equation for $f(x+y)$.  Given only the equation for $g(x+y)$, is it possible?

You're so greedy. The Captain wrote f'(0) in his post, so f'(0)=k exists, we're able show $f\left(x\right)=\sin\left(k x\right),\ g\left(x\right)=\cos\left(k x\right)$.

I interpreted the problem as to prove or disprove that the functional equation implies the differentiability and the formulas for $f', g'$.  Nowhere is it stated that $f', g'$ are assumed to exist, it makes things easier to assume that, but it makes things more interesting to not assume that.

fleeting_guest is right.
Given g(x-y)=g(x)g(y)+f(x)+f(y) where f,g are real functions can we say that
(( g(0)=1 )) & ((there exists an a s.t f(a)=1 & g(a)=0)) are equivalent ?
I think this will help a lot.

The condition on $g(0)$ or $f(0)$ is kinda redundant, we can derive this from $g(x-y)=g(x)g(y)+f(x)f(y)$ except for some trivial case, in which $f,\ g$ are constant functions.
Since we can find solution $g(x)\equiv 1,\ f(x)\equiv 0$, thus $g(0)=1$ and $\exists a$ s.t $f(a)=1,\ g(a)=0$ are not equivalent.
In order to get the differentiability, we need some extra conditions, for example, the existence of $f'(0)$.

[quote="fleeting_guest"]Yes, but can that be concluded from the functional equation (assuming continuity or measurability), for instance by constructing an additive function from $f$ and $g$? Some trigonometric equations are available, such as $f^2 + g^2 = 1$ and the angle-doubling formula for $g(x)$, but it is not immediately clear whether e.g. $f + ig$ is a homomorphism.[/quote]
I have a basic question.
what is the f+ig ? :?:

[quote="fleeting_guest"]$f+ig = f + g\sqrt{-1}$ is a homomorphism (from $\mathbb{R}^\ast \to \mathbb{C}$) if $ (f(x) + ig(x))(f(y) + ig(y)) = f(x+y) + ig(x+y)$.  This would be true if $f = \cos (kx), g = \sin (kx)$.[/quote]

Right ;but the captain said that f & g are real functions :!:  :huh:

If $f$ and $g$ are real functions, $f+ig$ is a complex function.  This complex function is often a useful thing to consider when solving problems about real $f,g$ satisfying functional equations similar to those of $\sin$ and $\cos$.

[quote="Buffalo"]Since we can find solution $g(x)\equiv 1,\ f(x)\equiv 0$, thus $g(0)=1$ and $\exists a$ s.t $f(a)=1,\ g(a)=0$ are not equivalent.
Buffalo!Is'nt it fairly clear that in the case of constant functions the question is trivial?


cho da thuc$P(x)$ he so nguyen thoa man :
           $P(2006)=2006!$
     $xP(x-1)=(x-2006)P(x)$ voi moi x thuoc $R$
   CMR: $G(x)=P(x)^2 +1$ bat kha qui  trong$Z[x]$

De dang chung minh duoc rang:$ P(0)\equal{}P(1)\equal{}P(2)\equal{}...\equal{}P(2005)\equal{}0$.
Dat $ P(x)\equal{}x(x\minus{}1)(x\minus{}2)...(x\minus{}2005)Q(x)$, thay vao gia thiet ta co $ Q(x)\equal{}Q(x\minus{}1)$
suy ra $ Q(x)\equal{}k$ voi k la hang so.
Mat khac $ P(2006)\equal{}2006!$ nen suy ra $ P(x)\equal{}x(x\minus{}1)(x\minus{}2)...(x\minus{}2005)$
Ta da dua ve bai toan quen thuoc:
"Chung minh rang: $ R(x)\equal{}x^2(x\minus{}1)^2(x\minus{}2)^2...(x\minus{}n)^2\plus{}1$ bat kha quy tren $ Z[x]$ voi moi n nguyen duong"


I'm  just wondering...
coz there's just a "small" little chance for me to get enroll...
lol...
I can't say no chance...but I know the truth :P

There have been lots of discussion on this topic.  Please check the American Mathematics Competitions board on this website.

If you're in 11th or 12th grade, then you will need a good index to qualify for the USAMO.  Your index score is AMC + 10*AIME, and you qualify for the AIME by scoring at least 100 on the AMC 12 (AMC 10 = 120).  It is in your best interest to do the best you can on the AMC 12.

However, if you're in 10th grade or below, then just getting a 100 on the AMC 12 or something will do.  Just make sure you do your best on the AIME.

I believe I've answered your question?

Bottom line, do well on the AMCs.

Does she mean if the AMC is good for CMO?

If that is what she asks I want to know the answer too.

I mean I may never can get enroll to the Olympiad
so...what's the point of doing the AMC??

Sigh, alright.  I'll give you some reasons.

Idealistically:  You enjoy doing Math and you want the opportunity to show all you've got, but also have fun and get a free pretty coloured pamphlet after the test.

Realistically:  A good AMC score can get the attention of college adcoms.  Making the AIME would be ideal.

Super Idealistically:  You never know, you just might make the USAMO.  I know a kid who went from a 61 on the COMC last year to USAMO qualifier that same spring.  Huge improvements can be made in a short amount of time; it just depends on how well you prepare I guess.  And evidently, Canadians are known very much so for improving vastly in very short amounts of time.

As for the COMC, if you didn't score about a 69 or a 68 this time...strive for a better future! :)

I wish it can be anything to canadian universities....

[quote="british8985"]since I've registered :P

goal for this yr is 152 hope it's not too hard for me :huh:[/quote]Oh 152....

[quote="british8985"]too low??
that's the best I can do I think..[/quote]Uh, no.
Isn't 150 the perfect score? :?

I think she means 152 index, AMC + 10*AIME score.  That's not a bad goal.

for me, i am wondering what is the right goal for a scarcely educated 12th grader, gone through little math study, first time writing AMC?  :P


Find the area of trapezoid ABCD if AB || CD, AB = 24, CD = 12, the measure of angle A is 30 and the measure of angle B is 60.

[hide="ans"]
we need to find the height of the trapezoid and in order to do that we need to find the side length of one of our 30-60-90 triangles

let x and y equal the top lengths of the two triangles, x is for the angle 30, and y the angle 60

$x+y = 12$ and $x\tan30 = y\tan60$
thus,

$x\tan30 = (12-x)\tan60$

$x\tan30 = 12\tan60-x\tan60$

$x(\tan30+\tan60) = 12\tan60$

$x = \frac{12\tan60}{\tan30+\tan60}= 9$

Now we can find the height of our trapezoid, using the 30-60-90 theory

since x = 9 and this side is $\sqrt{3}$ in the theorem, then it has been mutliplied by $3\sqrt{3}$, thus the height of the trapezoid is $3\sqrt{3}$

OR, we use the fact that $x\tan30$ = the height of the trapezoid which comes out to 5.196 which is exactly the same as $3\sqrt{3}$

computing the area is now easy

$area = 3\sqrt{3}* (\frac{24+12}{2})= 93.53$
[/hide]


Jim has a certain amount of people in front of him. First he calculates the number of ways he can take six people from the group in front of him. Then he calculates the number of ways he can take ten people from the group in front of him. He then realizes that the number of ways he can take six or ten people are exactly the same. How many people are in front of him?

[hide="hint"]
Remember that $\binom nr=\binom{n}{n-r}$[/hide]

eheehee is right
[hide="solution"]Take the mean of 10 and 6
8 is the center number. Doubling it brings [b]16[/b][/hide]

The correct answer is...

[hide="Solution"]
(n r) = (n n-r)

(n 6) = (n 10)
(n 6) = (n n - 10)
6 = n - 10
n = [b]16[/b]
[/hide]

(sorry about the lack of latex)


Solve for $ x: 3^{2x} \plus{} 19 \equal{} 10^x$.

[quote="nikeballa96"]look at it. x=2. simplest way to do it.[/quote]

I seriously do not see the simplicity of brute force/guessing here. How would you know to try x=2?

try multiples of 10. 10 gives a negative value for 3^2x. 10^2 gives an answer. also try math's method.

/bump can anyone explain how to get this? Logically.

[hide=One Solution]
We have $3^{2x}+19=10^x\implies 9^x+19=10^x$. Suppose $x=2y$. The equation becomes $9^{2y}+19=10^{2y}$. Subtracting $9^{2y}$ over and factoring, we get $19=10^{2y}-9^{2y} \implies 19=(10^y-9^y)(10^y+9^y)$. Since $19$ is prime, the only two positive integers that multiply to equal $19$ are $1$ and $19$. Since $1<19$, we have $10^y-9^y=1$ and $10^y+9^y=19$. It is clear that $y=1$ satisfies both of these, and since $x=2y=2(1)=\boxed{2}$.
We can check this by plugging in $x=2$ to the original equation: $3^4+19=10^2 \implies 81+19=100$, which is true. [/hide]


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just about 4 days left for the exam

i m a bit confused abt cut offs

LAst yr total marks 300 (part A-180,PART B-120)
this yr total marks 240 (part A-180,PART B-60)

some 1 told me cut offs were 120-130 and 150 safe zone
well was it out of 180 :P  , 240 or 300 :o

do u think HBCSE wud publish cutoffs?? :(

ofc i forgot NSEJS and NSEA :P

i had a doubt regarding NSEP
is it that we have $ 2$ and a $ \frac {1}{2}$ hours for both the sections A and B :( or only section A

u r giving the exam?? :oops:

sorry to disappoint u but we hav just 2 hrs ....for both section A and B

best of luck to everybody giving the exam :)

@rt:these exams are for students of class 12th,not for 11thies  :rotfl:

ofc eleventhies lik u and skand will get thru the 1st round :)

all the best to all for the test going on...................... :P  :rotfl:

srry for the late wishes :maybe:

well how many of you wrote NSEA????
and what are you expecting???
Mine expected score is anything between 100 - 120

thats really great.. :thumbup:

it became bad 2 worse frm me :(

,i answered 35+

in end it all went in hurry

@ritu,NSEP and NSEC were meant 4 timepass only 4 me

how did you all performed in them

congrats to those who are selected :)

as expected I am not selected

never mind,i have next chance :D
and one in INMO too :rotfl:

@anand...
nice to see u dude
thanks and grats to u
i lost ur phone no :(
and can u mail it to me?
and i have ur arihant organice chem...didnt do a page :((
btw..why din ya write the last base exam??

[quote="sumanth_d"]@anand...

and i have ur arihant organice chem...[b]didnt do a page[/b] :((
[/quote]

even i havent done a page.......and i dont plan to do it also  :P

Here is the site for chem papers. They are Icho but icho and incho are not very different.
[url=http://nhb.topcities.com/Chemclub/Icho.html]IChO[/url]

[hide]Made NSEC, didnt give astro, and failed in nsep[/hide]

[hide="@Da Vinci"]Don't give up on chem da, especially organic chem. I am sure you also are aware of your talent in chem. [/hide]

@ritu
Do not get confused betn ICO, INChO and INCO. The first one stands for International Chemistry Olympiad and rest two stands for Indian Chemistry Olympiad though INChO is more frequently used. Got it????

i  not getting confused neways :P ...there r millions of exceptions in chem :rotfl:

but just cant believe how i got thru Chem while other JOI'ians didn't :maybe:

Can any1 plz tell the following details of the INCho paper ????
1. Total marks
2. Subjective or objective
3. Any negative marking process followed
Plz. give me at least some idea. I'm totally blank!!!!

Is any1 interested in buying the book/s provided by HBCSE with contains the theory problems of past years INPho and INCho ? Btw what are your views regarding open tests held by various institutes like Fiitjee, Resonance, etc. for IITJEE???

I just got my NSEP certificate! :rotfl:  Statewise 1% thing.
Anyone getting it so late?
Edit: do they send 11th and 12th certificates separately?

[quote="VP_91"]i got to know that i have cleared neep/c/a only after the results of ino's were out[the equivalent ino concept]  :rotfl: i don't think there can b anything funnier than this[/quote]
Oh yeah! In comparison this is nothing....

I havent got NSEP certificate yet :rotfl:

and there was 1 more funnier thing
Tuhin got 71.5 in INChO(initially)
He didnt send the paper 4 reevaluation
The cutoff was written 72.5
Still his name was there on the list as selected :rotfl:

@rituraj
all papers were rechecked due to mistake in q7.8 soln

even i got a new performance card without sending for rechecking

i got also nsec certificate

[hide]
i want to tear it to pieces burn it and drink the ashes
olympiads are a big lie :furious:
being top 1% in chem olympiad doesn't guarantee being in even top 10% in jee chem
[/hide]

dont say u r not in top 10% in JEE chem :mad:

3.98 lakhs sat for JEE
that means top 10% implies being in top 40,000.. :rotfl:


Without dividing, determine the remainder when f(x) = x^3 + 5x^2 - x -21 is divided by x^2 + 4x - 10.

[quote="inom"][quote="king_23"]Without dividing, determine the remainder when
 $f(x) = x^3 + 5x^2 - x -21$
is divided by
$x^2 + 4x - 10$  [/quote]
sorry but i made it clearlier[/quote]

solution,
$x^2+4x^2-10=10$
$x^2=10-4x$
$F(x^2)=69-26x$ so
$R(x)=69-26x$[/quote]

Sorry, I don't quite understand how you did solution.  Your remainder is possible, but I don't understand how you derived it.

By division (as a check), I got 5x-1.

But I am intriqued by your method!  More info?


Solve the following simultaneous equation for all positive integer $a,b,c$

$a+b+c=38$

$a^2+b^2+c^2=722$

I got that much, but where do you go from there?

[quote="silouan"]We  see that $a^2+b^2+c^2=2ab+2ac+2bc$[/quote]

ya , silouan , can you show your solution ? ( I believe you wouldnt brute force it out ...)

[quote="krustyteklown"]
Well from brute force I got 4,9,25 as a solution. Iono if its coincidence or not that those are all perfect squares themselves.[/quote]

Well , I think it can be shown that they are all perfect square . :)

[quote="silouan"]We  see that $a^2+b^2+c^2=2ab+2ac+2bc$[/quote]

$ab+bc+ca=361$

$\Rightarrow ab + (a + b)( - a - b + 38) = 361$

$\Rightarrow (a + b)^2  - 38(a + b) + 19^2  = ab \Rightarrow (a + b - 19)^2  = ab$

$\Rightarrow (19 - c)^2  = ab$

This gives $(a - b)^2  = c(76 - 3c)$

Then I tried for values of $c$ which make $c(76 - 3c)$ a perfect square.

I found $c= 4, 9, 25$ (25 is maximum for c)are the only numbers which satisfy.

Hence the solution set is any permutation of the numbers $4, 9, 25$ and are $6$ in number.

well done [b]vidyamanohar[/b]  :)  I have a method to show that $a,b,c$ must be prefect square

WLOG let $a\geq b\geq c$ . Since we know

$a^2+b^2+c^2=2(ab+ac+bc)$
 :arrow: $c^2-2c(a+b)+(a-b)^2$   which mean

$c=(a+b)-2\sqrt{ab}$    *we take minus sign here since $c<a+b$

which is also $c=(\sqrt a-\sqrt b)^2$ .

Since $c$ is positive integer , the RHS must also be integer . So $a,b$ must either be perfect square or in the form $a=a'^2k$ , $b=b'^2k$ for positive integer $k$

which gives us $c=(a'-b')^2k$ . Substitute all this into first equation yields

$a'^2k+b'^2k+(a'-b')^2k=38$  :arrow: $k(a'^2+b'^2-a'b')=19$

Obviously $a'^2+b'^2-a'b'$ cannot be $1$ , or else $a'=b'=1$ will result $c=0$ Hence $k=1$ so $a,b,c$ are all perfect square .

Then we have $38\equiv 2 \mod 4$ show us the first equation must have two odd perfect and one even perfect .The even perfect can only be $4,16,36$ . By try and error , only $4$ satisfy and it can be obtained that the other two is $9$ and $25$ .

 :)

[quote="shyong"]well done [b]vidyamanohar[/b]  :)  I have a method to show that $a,b,c$ must be prefect square

WLOG let $a\geq b\geq c$ . Since we know

$a^2+b^2+c^2=2(ab+ac+bc)$
 :arrow: $c^2-2c(a+b)+(a-b)^2$   which mean

$c=(a+b)-2\sqrt{ab}$    *we take minus sign here since $c<a+b$

which is also $c=(\sqrt a-\sqrt b)^2$ .

 :)[/quote]

good one


i think i will get it wrong not my fav topic but anyway

$ 1\plus{}cos2x\equal{}|sinx\plus{}cosx|$
$ (1\plus{}cos2x)^2\equal{}(sinx\plus{}cosx)^2$
$ 4\minus{}8sin^{2}x\minus{}4sin^{4}x\equal{}sin^{2}\plus{}2sinxcosx\plus{}cos^{2}x$
$ 4\minus{}8sin^{2}x\minus{}4sin^{4}x\equal{}1\plus{}sin2x$
$ sin2x\plus{}8sin^{2}x\minus{}4sin^{4}x\equal{}3$
$ 2sinx(cosx\plus{}4sinx\minus{}2sin^{3}x)\equal{}3$

after this i dont know what to do

[quote="toancapba"]$ 1 \plus{} \cos 2x \equal{} |\sin x \plus{} \cos x|$[/quote]

$ 1 \plus{} \cos 2x \equal{} |\sin x \plus{} \cos x|$
$ (2\cos^2x)^2\equal{}1\plus{}\sin 2x$
$ \tan 2x\equal{}1\plus{}2cos^2x$
$ \tan 2x\equal{}1\plus{}\frac {2}{\sec^2x}\equal{}1\plus{}\frac {2}{1\plus{}\tan^2x}$
$ \frac {2\tan x}{1\minus{}\tan^2x}\equal{}1\plus{}\frac {2}{1\plus{} \tan^2x}$

You may continue from here onwards..... :P

[quote="toancapba"]$ 1 \plus{} \cos 2x \equal{} |\sin x \plus{} \cos x|$[/quote]
$ (1\plus{}\cos2x)^2\equal{}(1\plus{}sin2x)^2\\
2\cos2x\plus{}\cos^22x\equal{}2\sin2x\plus{}\sin^22x$
Put $ t\equal{}tanx$


Given are 100 real numbers whose sum is equal to zero. At least how many pairs can be chosen from them such that the sum of the numbers in each pair is not negative?

You never said whether they had to be distinct -
...so they could all just be 0.


absolute value
real analysis
real analysis unsolved

For any integer $a$ with $|a|\geq 2$ the number $\prod_{n\geq 1}{(1-\frac{1}{a^n})}$ is irrational. However, does anyone know if the transcendency is known or not, at least in some particular cases? I have no ideas for this last problem.

Suppose $(x_n)_n$ is an increasing sequence of positive integers with the property that $x_{n+1}-x_n\to\infty$, and $a\ge 2$ is a positive integer. Also assume that $(\varepsilon)_n$ is a sequence whose terms have the values $\pm 1$. Then we can show that $S=\sum_{n\ge 1}\frac {\varepsilon_n}{a^{x_n}}\ (*)$ is irrational:

Suppose $S=\frac pq$. Then $qa^{x_n}S$ is an integer for all $n$. At the same time, the difference between $qa^{x_n}S$ and the integer $q\cdot\sum_{i=1}^n\varepsilon_ia^{x_n-x_i}$ is $q\cdot\sum_{m\ge n+1}\frac{\varepsilon_m}{a^{x_m-x_n}}$. The absolute value of this last expression goes to $0$ as $n\to\infty$, and it's easy to see that it can't be $0$. The conclusion is that we can find positive integers which are arbitrarily close to each other, a contradiction.

Now we can apply the above to the problem here: by [url=http://www.google.ro/url?sa=t&ct=res&cd=5&url=http%3A//planetmath.org/encyclopedia/PentagonalNumbersTheorem.html&ei=OEgEQ_38JbSSiALY8LyODg]Euler's Pentagonal Number Theorem[/url], the given product equals $\sum_{n\in\mathbb Z}\frac {(-1)^n}{a^{\frac{3n^2+n}2}}$, which can be arranged as a series of the form $(*)$.

Yop, this is the idea. It's a very nice problem.


[quote="anirudh"]For a user, it says online under them, but if I check on the bottom of forum index, they are not here. :([/quote]
That's probably because he isn't in the same forum as you, but he's on AoPS. :roll:

Check the list at the bottom of the Forum Index.

maybe because they haven't been on for the past 5 minutes? it says something like that, even on the admin panel. its somewhat inaccurate, because it updates every 5 minutes... :| but im not too sure...

It doesn't update every second you know :roll:

apparently, they are not online for like the past like 1-2 hrs.


linear algebra
matrix
linear algebra unsolved

Here is a problem about invertibility:

Suppose $A$ is a square matrix such that $A^5 = 0$. Is $A + I$ invertible? Find its inverse if it is invertible, or provide a counterexample if it is not.

I suspect that the answer is yes, but I could not give a proof. (More honestly I think both choices are strong candidates.)

Can anyone help?

Try $I-A+A^{2}-A^{3}+A^{4}$ as an inverse. This is obtained by using geometrical series.


LaTeX
inequalities
inequalities proposed

He must mean + on the left hand side which was left out in the latex, I'm guessing :wink:

Yes thus you recognized it, I am finding a beautiful solution, I known solution of Vasc but it is same of you (about beauty) :D

[quote="skywalker"]I am finding a beautiful solution...[/quote]
It's very interesting.

I found a solution but perhaps it's ugly as well (but it's fast for me 'cause I'm familar with such kind of ways using Schur Inequality); let
$x=a+b+c,\ y=ab+bc+ca,\ z=abc$, and let
$a^{2}+b^{2}+c^{2}=x^{2}-2y=1$
then we need to prove
$(2-3ab)(2-3bc)(2-3ca)\geq0\Longleftrightarrow 7-12y+3\cdot 3z(2x-3z)\geq0$
And we see that if
$f(t)=t(u-t)\ \left(t\leq\frac{u}{2}\right)$
then
$f'(t)=u-2t\geq0$
So with
$x-3z=(a+b+c)(a^{2}+b^{2}+c^{2})-3abc\geq6abc\geq0$
and with Schur Inequality
$a^{3}+b^{3}+c^{3}+3abc\geq ab(a+b)+bc(b+c)+ca(c+a)\Longleftrightarrow z\geq\frac{4xy-x^{3}}{9}$
(notice that $x^{2}=1+2y$) we have
$7-12y+3\cdot 3z(2x-3z) \geq 7-12y+3\cdot\frac{4xy-x^{3}}{3}\left(2x-\frac{4xy-x^{3}}{3}\right)=7-12y+\frac{1}{3}(1+2y)(2y-1)(7-2y)=\frac{1}{3}(1-y)(4y^{2}-10y+7)\geq0$
(because
$y=ab+bc+ca\leq a^{2}+b^{2}+c^{2}=1$)
Then it's done.

I think, it's better:
$(2-3ab)(2-3ac)(2-3bc)\geq1\Leftrightarrow$
$\Leftrightarrow7-12(ab+ac+bc)+18(a+b+c)abc-27a^{2}b^{2}c^{2}\geq0.$
$3(a+b+c)abc\geq27a^{2}b^{2}c^{2}.$ Hence, remain to prove that
$7-12(ab+ac+bc)+15(a+b+c)abc\geq0,$ which is easily killed by SOS. :wink:

[quote="arqady"][quote="skywalker"]
Given $a,b,c>0$ prove that
$(2(b-c)^{2}+2a^{2}+bc)(2(c-a)^{2}+2b^{2}+ca)(2(a-b)^{2}+2c^{2}+ab)\ge (a^{2}+b^{2}+c^{2})^{3}$
 [/quote]
Let $a^{2}+b^{2}+c^{2}=1.$ Then your inequality equivalent to the following inequality:
$(2-3ab)(2-3ac)(2-3bc)\geq1,$ which was posted many time ago by Vasc or hungkhtn.
My ugly proof:$(2-3ab)(2-3ac)(2-3bc)\geq1\Leftrightarrow$
$\Leftrightarrow2\cdot\sum_{cyc}(a-b)^{6}+\frac{3}{2}\cdot\sum_{cyc}(a-b)^{2}(a^{4}+2a^{3}b-3a^{2}b^{2}+2ab^{3}+b^{4})+$
$+6\cdot\sum_{cyc}abc(a^{3}-a^{2}b-a^{2}c+abc)+\sum_{cyc}(a^{3}b^{3}-a^{2}b^{2}c^{2})\geq0.$[/quote]

You are right, Arqady. More exactly, this problem is of VASC and in Hungkhtn's book (vietnamese), following nice solution are show:

Let $a,b,c \ge 0$ verify that $a^{2}+b^{2}+c^{2}=3$. Prove that
\[(2-ab)(2-bc)(2-ca) \ge 1. \]
Proof.
Let $x=2-ab,y=2-bc,z=2-ca$.
We need to prove $xyz \ge 1$ if $x,y,z$ are three positive numbers satisfying that
\[\frac{(2-x)(2-y)}{2-z}+\frac{(2-y)(2-z)}{2-x}+\frac{(2-z)(2-x)}{2-y}=3. \]
The above condition can be rewritten as follow
\[(2-x)(2-y)+(2-y)(2-z)+(2-z)(2-x)\&=3(2-x)(2-y)(2-z)\]
\[\Leftrightarrow 8(x+y+z)+3xyz\&=12+5(xy+yz+zx).\]
Assume that it's false with some numbers $x,y,z$, or $xyz <1$. We will prove
\[8(x+y+z)+3xyz <12+5(xy+yz+zx). \]
This last one is easy by Mixing variable or SOS or Schur. Notice that we can assume $xyz=1$.

[quote="GacKiem"]
\]
Assume that it's false with some numbers $ x,y,z$, or $ xyz < 1$. We will prove
\[ 8(x \plus{} y \plus{} z) \plus{} 3xyz < 12 \plus{} 5(xy \plus{} yz \plus{} zx).
\]
This last one is easy by Mixing variable or SOS or Schur. Notice that we can assume $ xyz \equal{} 1$.[/quote]

I think it is wrong. :P

[quote="arqady"][quote="skywalker"]
Given $ a,b,c > 0$ prove that
$ (2(b \minus{} c)^{2} \plus{} 2a^{2} \plus{} bc)(2(c \minus{} a)^{2} \plus{} 2b^{2} \plus{} ca)(2(a \minus{} b)^{2} \plus{} 2c^{2} \plus{} ab)\ge (a^{2} \plus{} b^{2} \plus{} c^{2})^{3}$
 [/quote]
Let $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1.$ Then your inequality equivalent to the following inequality:
$ (2 \minus{} 3ab)(2 \minus{} 3ac)(2 \minus{} 3bc)\geq1,$ which was posted many time ago by Vasc or hungkhtn.
[/quote]
pqr kill it easily  :maybe:


Let $a$ and $b$ be positive integers. When $a^{2}+b^{2}$ is divided by $a+b,$ the quotient is $q$ and the remainder is $r.$ Find all pairs $(a,b)$ such that $q^{2}+r=1977$.

$ \frac{a\plus{}b}{2}\le [\frac{a^2\plus{}b^2}{a\plus{}b}]\equal{}q<a\plus{}b, 0\le r<a\plus{}b$, therefore $ 44\le a\plus{}b\le 88$.
If $ q<44$, then $ r\equal{}1977\minus{}q^2\ge 128$ - contradition, therefore $ q\equal{}44,r\equal{}41$. It give $ a^2\plus{}b^2\equal{}44(a\plus{}b)\plus{}41$ or $ (a\minus{}22)^2\plus{}(b\minus{}22)^2\equal{}1009\equal{}28^2\plus{}15^2$. Because $ 1009$ is prime one of them (a,b) $ 22\plus{}28\equal{}50$ second $ 22\pm 15$.


integration
trigonometry
calculus
function
real analysis
real analysis unsolved

Let $a_{n}=\int_{0}^{\pi}\frac{\sin nx}{\sqrt x}dx$,then the series:
$\sum_{n=1}^{\infty}a_{n}^{2}$ convergent or divergent?
Thanks for help

Change variable $nx=u$

$a_{n}=\frac{1}{\sqrt{n}}\int_{0}^{n\pi}\frac{\sin(u)}{\sqrt{u}}du$

another change variable $u=t^{2}$

$a_{n}=\frac{1}{\sqrt{n}}\int_{0}^{(n\pi)^{2}}{\sin(t^{2})}dt$

the integral is convergent it is Fresnels integral

$(a_{n})^{2}\sim \frac{C}{n}$

The serie is divergent

If someone has another solution please post it
thank's

[quote="Moubinool"]

$a_{n}=\frac{1}{\sqrt{n}}\int_{0}^{(n\pi)^{2}}{\sin(t^{2})}dt$

the integral is convergent it is Fresnels integral

$(a_{n})^{2}\sim \frac{C}{n}$

[/quote]
Can you tell me more about is,because I haven;t Fresnels integral before,Thank you.

Anyway I think we can prove:
$\int_{0}^{n\pi}\frac{\sin u}{\sqrt u}du$ convergent directly.(and not 0)
This we can apply Dirichlet's Rule quickly.
And the limit $\ge \int_{0}^{2\pi}\frac{\sin u}{\sqrt u}du>0$ :)

The sequence $a_{n}$ is (up to a constant multiple) the sine coefficients of the Fourier series for the function $f(x)$ such that $f(x)=\frac1{\sqrt{x}}$ for $0<x<\pi,$ extended to be odd and periodic of period $2\pi.$

By Parseval's identity, $\sum_{n=1}^{\infty}a_{n}^{2}=C\int_{0}^{\pi}f(x)^{2}\,dx.$

(We could work out exactly what the constant $C$ is but in this case we don't need to.)

Since the integral is infinite - the function is not square integrable - then the series in question also diverges.

[quote="Kent Merryfield"]
By Parseval's identity, $\sum_{n=1}^{\infty}a_{n}^{2}=C\int_{0}^{\pi}f(x)^{2}\,dx.$

.[/quote]
Hi,Kent,
but does Parseval's identity true for all function? :?:

Well, technically, Parseval only applies to $L^{2}$ functions, which this isn't. But we could argue like this (the constant $C$ is probably different each time I invoke it - it can be figured; paying attention to it only distracts you from the argument):

If $\sum_{n=1}^{\infty}a_{n}^{2}<\infty,$ then there would exist a function $g\in L^{2}$ such that the series $C\sum_{n=1}^{\infty}a_{n}\sin nx$ converges in the $L^{2}$ sense to $g.$

But then $g$ also belongs to $L^{1}$ and $a_{n}=C\int_{0}^{\pi}g(x)\sin nx\,dx.$

Since we also have $a_{n}=C\int_{0}^{\pi}f(x)\sin nx\,dx$ and Fourier coefficents are unique in $L^{1},$ we must have $f=g$ as an element of $L^{1},$ hence almost everywhere. But since $f$ does not belong to $L^{2},$ then our initial supposition that the coefficients are square summable must be incorrect.


linear algebra
matrix
vector
probability
expected value
linear algebra unsolved

How can i prove that a covariance matrix is always positive semidefinite ? Could someone give a proof ?
( The definition of covariance matrix is here [url]http://en.wikipedia.org/wiki/Covariance_matrix[/url] )

$X_i$ random variables, $Y=(y_1,y_2,..,y_n)$ a vector. If $K$ is the covariance matrix of the $X_i$ then $(Y,KY)=\sum y_iy_jK_{i,j}=Var(\sum y_iX_i) \geq 0$.

Please could you explain what have u written ? What means $(Y,KY)$ ? :blush:

Let $X$ be a random vector (a random variable taking values in $\mathbb{R}^n.)$ Write $X$ as a column vector. Then the covariance matrix is

$C=E(XX^T)-E(X)E(X)^T.$

Let $y$ be an arbitrary constant column vector.

$C$ is positive semidefinite iff $y^TCy\ge 0$ for all $y.$

Now define $U=y^TX.$ This gives us that $U$ is a one-dimensional random variable.

$\text{Var}(U)=E(U^2)-E(U)^2$

$=E(UU^T)-E(U)E(U^T)$ (using that the transpose of a $1\times 1$ matrix is itself)

$=E(y^TXX^Ty)-E(y^TX)E(X^Ty)$

Now use the linearity of expected value:

$=y^TE(XX^T)y-y^TE(X)E(X)^Ty$

$=y\left[E(XX^T)-E(X)E(X)^T\right]y =y^TCy.$

So $y^TCy$ is the variance of a random variable, hence nonnegative.

The covariance matrix might fail to be positive definite, but that would require the components of $X$ to be linearly dependent (a form of perfect correlation.)

This is exactly the same proof alekk gave; I was just trying to add more explanation.


So, who else has received their APMO invite?

Me and this guy from my school named Saroosh.

Ever increasing are Saroosh's odds for making the IMO team.

Cheers,

Hunter

Ah; there's going to be a fairly large group from Calgary writing.
Strange, apparently I write APMO at Ottawa University instead of my school...

I also got an invite as well...I will be writing across the border at Windsor.

Hello all

For those who have been invited to write the APMO, if you are looking for training material or practice competitions, you can visit my website here:

http://math.ucalgary.ca/~atang/mathcircles/

Check out the Mock Olympiads to practice some competition-like problems and check the notes/ problem sets to get better at certain topics.

If you want even more notes and problem sets, they can be found at the Winter Camp 2010 website:

http://math.ucalgary.ca/~atang/wc10/

See the links section to see past Winter camp websites with even more mathematical gems.

Best of luck training and writing all competitions this coming year!

Cheers,
Adrian

Does anyone know the criteria for one to be invited to write this contest? Like what ranks do you have to be on COMC to be invited?Or it uses different criteria from CMO?


algebra
polynomial
quadratics
Galois Theory
arithmetic sequence
number theory theorems
number theory

Can anyone give me some link or even post here something about cyclotomic polynomials.

Some statements about $\Phi_n(x)$ and roots of unity:
- $x^n-1 = \prod_{d|n} \Phi_d(x)$.
- irreducible over $\mathbb{Q}$ (see http://www.mathlinks.ro/Forum/viewtopic.php?t=15750 for a proof) with Galois group $\mathbb{Z}/n\mathbb{Z}^*$ with degree $\phi(n)$; similar properties can be said about other fields, e.g. for finite ones (but they aren't irreducible everytime then).
- the above implies indirectly which $n$-gons can be constructed (for ruler and compas they are those with $\phi(n)=2^k$ for some $k$).
- For any integer $k$, it holds that $\Phi_n(k)$ has only prime divisors $p$ with $p \equiv 1 \mod n$ or $p|n$; this gives a nice elementary proof for Dirichlets Theorem on primes in arithmetic progression for the residue $1$.
- they give a nice proof for Wedderburns Theorem (every finite division ring is already a field).
- there are also a lot of theorems about gaussian sums, e.g.: \[ \sum_{k=0}^{p-1} \zeta^{k^2} = \sum_{k=0}^{p-1} \left( \frac{k}{p} \right) \zeta^{k} = \begin{cases} \sqrt{p} & \text{ for } p \equiv 1 \mod 4 \\ i \cdot \sqrt{p} & \text{ for } p \equiv 3 \mod 4\end{cases}  \] where $p$ is prime and $\zeta$ is the "first" $p$-th root of unity. These sums can be used to show reciprocity theorems (especially the quadratic one) and to find explicit formulas for $\zeta$ only using roots.

I posted six problems on them some time ago:
http://www.mathlinks.ro/Forum/viewtopic.php?t=42609
http://www.mathlinks.ro/Forum/viewtopic.php?t=43028

Some others:
http://www.mathlinks.ro/Forum/viewtopic.php?t=27089 (very nice problem indeed!)
http://www.mathlinks.ro/Forum/viewtopic.php?t=62706
http://www.mathlinks.ro/Forum/viewtopic.php?t=62955

Some more could probably be found with Google.

There is also : http://www.mathlinks.ro/Forum/viewtopic.php?t=25794 . :)


if you want to look over the problems, pm me

i plan on posting them here in a few days

Oh, just to clarify I meant that if you want to edit the problems and find all the mistakes I made, you should PM me. So far nobody has volunteered :(

[quote="chess64"][quote="Phelpedo"]I'd be willing...can't make any guarantees about how much time I'll be able to spend though.[/quote]

You want to participate or edit?[/quote]

Edit.

[quote="chess64"]I find it quite humorous that I'm receiving PMs from people who want to participate while people who want to edit are posting in this thread :ninja:[/quote]
I'm not volunteering to edit, but you could ask some of the adults in this forum (the whole thing not just AMC) about editing...
Just a suggestion...

i would like to help edit a bit...all i ask is that you give me maybe a sentence giving the idea behind each problem,


Find all pairs of functions $f, g: \mathbb R \to \mathbb R$ such that
(a) If $ x < y$, then $ f(x) < f(y)$, and
(b) For all $ x,y \in \mathbb R$, \[ f(xy) = g(y)f(x) + f(y).\]

[quote="amparvardi"]Find all pairs of functions $ f,g : R\rightarrow R$ such that

(a) if $ x < y$, then $ f(x) < f(y)$;

(b) for all $ x,y \in R$, $ f(xy) \equal{} g(y)f(x) \plus{} f(y)$.[/quote]
Let $ P(x,y)$ be the assertion $ f(xy) \equal{} g(y)f(x) \plus{} f(y)$

$ P(x,1)$ $ \implies$ $ f(x)(1 \minus{} g(1)) \equal{} f(1)$ and so $ g(1) \equal{} 1$ (else $ f(x)$ is constant and condition (a) is no longer true) and so $ f(1) \equal{} 0$ and so $ f(0) < 0$ (using (a))

Since $ f(x)$ solution implies $ cf(x)$ is also solution (with $ c > 0$), wlog say $ f(0) \equal{} \minus{} 1$

Then $ P(0,x)$ $ \implies$ $ g(x) \equal{} f(x) \plus{} 1$ and $ P(x,y)$ becomes $ f(xy) \equal{} f(x)f(y) \plus{} f(x) \plus{} f(y)$ and so $ f(xy) \plus{} 1 \equal{} (f(x) \plus{} 1)(f(y) \plus{} 1)$

and so $ g(xy) \equal{} g(x)g(y)$ with $ g(x)$ strictly increasing and so $ g(x) \equal{} x^a$ with $ a > 0$ and such that $ x^a$ is an increasing function defined even for $ x < 0$

And it is easy to check back that this indeed is a solution.

Hence the result :

$ f(x) \equal{} \alpha(x^a \minus{} 1)$
$ g(x) \equal{} x^a$

For any $ \alpha > 0$ and any $ a > 0$ such that $ x^a$ is an increasing function defined even for $ x < 0$ (example : $ 2p \plus{} 1, \frac 1{2q \plus{} 1}, ...$)

(be careful that acceptance of $ x^a$ for $ x < 0$ when $ a\notin\mathbb N$ may vary in different countries. But $ f(x) \equal{} \sqrt [3] x \minus{} 1$ is indeed a solution, for example).

[quote="NikolayKaz"]It was supposed to be strictly increasing, i just mistranslated the task. I have a question regarding your solution.
[quote="pco"]
$ g(xy) \equal{} g(x)g(y)$ with $ g(x)$ strictly increasing and so $ g(x) \equal{} x^a$
[/quote]

How do we know that we can assume any strictly increasing function here? I mean what shows us that all strictly increasing functions g are the solution?[/quote]

I dont understand your question.

$ f(x)$ is increasing, so $ g(x)\equal{}f(x)\plus{}1$ is increasing too.

[quote="NikolayKaz"]I understand this but my question is how do we know that it holds for ANY increasing function.[/quote]

[u]Problem [/u]: $ g(xy) = g(x)g(y)$ and $ g(x)$ increasing.
Let $ P(x,y)$ be the assertion $ g(xy) = g(x)g(y)$

If $ g(u) = 0$ for some $ u\ne 0^$, $ P(u,\frac xu)$ $ \implies$ $ g(x) = 0$ $ \forall x$. Impossible since $ g(x)$ is increasing. So $ g(x)\ne 0$ $ \forall x\ne 0$
$ P(x,1)$ $ \implies$ $ g(x)(g(1) - 1) = 0$ and so $ g(1) = 1$ else $ g(x) = 0$ $ \forall x$, in contradiction with $ g(x)$ increasing.
$ P(x,0)$ $ \implies$ $ g(0)(g(x) - 1) = 0$ and so $ g(0) = 0$ else $ g(x) = 1$ $ \forall x$, in contradiction with $ g(x)$ increasing.
$ P( - 1, - 1)$ $ \implies$ $ 1 = g( - 1)^2$ and so $ g( - 1) = - 1$ since $ g(x)$ is increasing and $ g(0) = 0$
$ P(x, - 1)$ $ \implies$ $ g( - x) = - g(x)$

Let then $ h(x)$ : $ \mathbb R\to\mathbb R$ such that $ h(x) = \ln(g(e^x))$ : $ h(x)$ is increasing and $ h(x + y) = h(x) + h(y)$
This is Cauchy's equation and the only monotonic solutions are $ h(x) = cx$ for any $ c > 0$

So $ g(x) = x^c$ $ \forall x > 0$ and $ g(x) = - ( - x)^c$ $ \forall x < 0$

And so [u][b]I was wrong : there are some other solutions [/b][/u]!

The general solution for $ g(x)$ is $ g(x) = sign(x)|x|^c$ for any $ c > 0$

So for example $ g(x) = sign(x)x^2$ is a solution (that I did not give in my previous posts).

Hence the result for the original problem :

$ f(x) = \alpha(sign(x)|x|^a - 1)$
$ g(x) = sign(x)|x|^a$

For any $ \alpha > 0$ and any $ a > 0$

And thanks, NikolayKaz, for your question.  :)

@pco sorry to revive... but how do you get $g(x)=x^a$ for some $a>0$ with $g(xy)=g(x)g(y)$ and $g$ monotonic?

I dont see where.
We are speaking, I suppose about post #5 (and not post #2)
Dont hesitate to quote the first two or three lines you dont understand.


Say I want to create a document with a lot of problems, each followed by a solution. I want to make it so that I can print
(1) The document with both problems and solutions (interleaved) and
(2) The document with only the problems, and the solutions taken out.

Now, is there an easy way to do this WITHOUT having to create two LaTeX documents?

I thought that something like

\newenvironment{solution}{ \begin{comment} } {\end{comment} }

might work, but it gives an error (try it yourself). So any other suggestions?

You could always type out the entire document, and then copy/cut/paste the solutions.  I think it might be a better idea to just type out all the problems, start a new page, and then post the solutions.

I believe the [i]comment[/i] package (different from the command), the [i]version [/i]package, and the [i]versions[/i] package can do what you want.  See
[url]http://www.tex.ac.uk/cgi-bin/texfaq2html?label=conditional[/url].

Here's what's up I think with the comment package:

1.  First load the package:

[code]\usepackage{comment}
[/code]
2.  Define inclusion and exclusion environment names.  For example, suppose want to selectively include/exclude all the problems.  Then define a "problems" environment name and add the command:

[code]\includecomment{problems}
[/code]
Now can create an environment like:

[code]\begin{problems}
This is problem 1:
$$\int_0^1 xdx$$
\end{problems}[/code]

Since the environment "problems" is an included environment, this block of latex is included in the document.

3.  Now define exclusion environments.  For example, define an "answers" environment:

[code]\excludecomment{answers}
[/code]
Now can place "answers" environments with the problems like:

[code]\begin{answers}
This is answer to problem 1
\end{answers}[/code]

and since "answers" is an excluded environment, it won't be included in the document.

4.  If want to disable excluding all the answers and print them out, just comment out the \excludecommand{answers}

[code]%\excludecommand{answers}[/code]

Perhaps another solution is to use test style file
http://www.math.rochester.edu/people/faculty/cmlr/Latex-exam-ur/
or
answers.sty
mechanics.civil.tohoku.ac.jp/~bear/bear-collections/style-files/answers.sty
or

http://www-hep2.fzu.cz/tex/texmf-dist/doc/latex/probsoln/probsoln.html

I hope that it help

Genick Bar-Meir
http://www.potto.org
where free textbooks roam!


calculus
integration
real analysis
real analysis theorems

From what I've seen in learning numerical analysis, it looks like a lot of numerical integration methods are based on the principle that you can use your k-step methods for solving ODEs to approximate Riemann integrals. This got me thinking about the possibility of a numerical Lebesgue integral but that sounds ridiculous. However, I don't see any obstruction to numerical Henstock-Kurzweil integration. I don't know anything about adaptive numerical methods but it sounds like this would be a natural framework for them. Has this ever been done or is it stupid?

In a way, this is already done by [url=http://en.wikipedia.org/wiki/Adaptive_quadrature]adaptive quadrature[/url] methods, which evaluate the integrand on a mesh that becomes denser near a singularity.
[quote="Maple 10"]evalf(Int((1/x)*sin(1/x^3),x=0..1));
0.2082377521[/quote]


calculus
integration
function
analytic geometry
geometry
calculus computations

Can anyone help me evaluate the following improper integral?:

$\int_{0}^{\infty} x e^{-x^2} dx$

I tried using integration by parts along with the error function but I am getting $\infty$ for an answer. May be because I don't have much experience in using integration by parts for improper integrals.

Try substitution. What might be a good choice of a function to substitute? A common strategy is to split your integrand into $f(u)$ and $du$ so that $f(u)$ is something you know how to integrate and $du$ equals $dx$ times the leftover junk not included in $f(u)$.

Yeah, I did that. I had $u = x \Rightarrow du = dx$ and $dv = e^{-x^2} dx \Rightarrow v = \int_{0}^{\infty} e^{-x^2} dx$. Thus $v = \frac{{\pi}^{1/2}}{2}$ by the error function. But then you get $uv = x\frac{{\pi}^{1/2}}{2}$ evaluated from $0$ to $\infty$ which is equal to $\infty$!!

I think you want to make the substitution $u = x^2$.

I don't mean integrating by parts, either. I mean plain old substitution. Try it, see what happens! :)

Oh Wow! I never realized that. I was unnecessarily trying to make the problem hard!  :roll:

Well, now that this problem is solved, what about this one:

$\int_{0}^{\infty} x^2 e^{-x^2} dx$

I don't think any u substitution would do the trick this time.  :(

You're right. See the "marathon" thread -- we've been discussing a very similar problem.

I read the thread, but that just labels this function as non-elementary and moves on. I wanna know how you would use the error function to evalue this integral.

Ah, that's a commonly discussed integral. Have you had calculus of several variables yet?

The simplest way to evaluate the integral is to say

$I = \int_0^\infty e^{-x^2}dx$

so

$I^2 = \int_0^\infty e^{-x^2}dx \cdot \int_0^\infty e^{-y^2}dy = \iint_{QI}e^{-x^2-y^2}dA$

That is, we write the two integrals as the integral over the first quadrant of the product of the two functions. This is an application of Fubini's theorem, which you can't always use with impunity, but I don't think this is the right place for such subtleties. Anyway, then we change variables to polar coordinates. The limits of integration in polar coordinates are $0$ to $\infty$ for $r$ and $0$ to $\frac{\pi}{2}$ for $\theta$. The differential area in polar coordinates is $rdrd\theta$ (which you can obtain from the Jacobian of the change of variables, but you can also see it geometrically).

Then we have

$I^2 = \int_0^\frac{\pi}{2}\int_0^\infty e^{-r^2}rdrd\theta = \frac{\pi}{4}$.

So $I = \frac{\sqrt{\pi}}{2}$, and you can get the integral of your function using integration by parts and that formula.

That is my problem. When I use integration by parts along with the error function on this new integral (like I used it on the previous integral), I get infinity for an answer. I mean, I have always been used to do integration by parts on indefinte integrals but for some reason I am getting confused using integration by parts for definite integrals.
:huh:

We have $I = \int_0^\infty x^2e^{-x^2}dx$.

Let $u = x$ and $dv = xe^{-x^2}dx$. Then we get

$I = \left[\frac{-x}{2}e^{-x^2}\right]_0^\infty + \frac{1}{2}\int e^{-x^2}dx$

I think your mistake was in the first part, where you evaluate at the bounds after you integrate as if it were an indefinite integral. You should be able to finish it from here. Nothing diverges.

Thanks a lot Xevarion!  :lol: I never realized that setting $u = x$ would simplify the problem so elegantly. I was actually setting $u = x^2$ and so I just wansn't able to evalute the integral.

(Also, you were right about me being evaluating at the bounds after me integrate :oops: )

-Swapnil


maybe you can use the fact that

$qa \equiv qb  (mod p) \Rightarrow a \equiv b (mod \displaystyle\frac{p}{\gcd(p,q)})$

or... at least i think that's how the division thing goes...

Since $1776$ and $2000$ are both divisible by $16$, the remainder left over must also be divisible by $16$. By looking at the possibilities ($126, 251, 376,...$) we see that it must be $1376$.

I heard this question has something to do with Chinese remainder theorem, can somebody fill me in?

The Chinese Remainder Theorem states that given simultaneous congruences

$ x \equiv a_i (mod m_i)$ for i=1,...r, and $m_i$ relatively prime,

the set of solutions:

$x= a_1 b_1 \frac{M}{m_1}+...a_r b_r \frac{M}{m_r}$

where $ M \equiv m_1  m_2 ...m_r$

and $b_i \frac{M}{m_i} \equiv 1 (mod m_i)$

which is equivalent to solving

$x\equiv 1\mod 125$
$x\equiv 0\mod 16$

where $x=1776^{1492!}$

By using Chinese Remainder Theorem

we have $x=125p+1$ and $x=16q$ for some integer $p,q$

Solving $125p+1=16q$ by diophantine gives us one of the solution is $(p,q)=(11,86)$

So $p=11+16k$ , $q=86+125k$ for some integer $k$ .Hence

$x=16(86+125k)=1376+2000k$ which is $x\equiv 1376 \mod 2000$


I started out with 1 and 2 and took and took the arithmetic mean and the geometric mean of both. Then I did the same with the resulting two numbers. I kept doing the same over and over. Both numbers approached
[hide="Click"] 1.456791031...[/hide]
Can anyone figure out what this number is equal to, or is there no way to express it?

Well, what you discovered is called the arithmetic-geometric mean of the two numbers (1 and 2 in this case). As far as I know, those sequences converge, but the closed form is in terms of elliptic integrals.
I found about this at http://mathworld.wolfram.com/Arithmetic-GeometricMean.html, and you might want to take a look ( or maybe not  :P )

Interesting.... Thank you for the link. And from the same website, the limit of the arithmetic mean and the harmonic mean is the geometric mean. There is also the geometric-harmonic mean on wikipedia [url]http://en.wikipedia.org/wiki/Geometric-harmonic_mean[/url].


modular arithmetic
number theory open
number theory

Is it true that if $ q$ is prime, $ n$ is prime, and $ q | 3^n \minus{} 2^n$, then $ n | q \minus{} 1$?

work fermat's little theorem,i feel u ll get it

I can't seem to get far with Fermat's little theorem. $ q | 3^n \minus{} 2^n \implies \left(3/2\right)^n \equiv 1 \pmod q$. From this, we get that $ \mbox{ord}_q(3/2) | n$, but if I'm not mistaken, $ q \minus{} 1 | n$ only follows if $ \frac {3}{2}$ is a primitive root modulo $ q$. Am I missing something?

When you have $ \textrm{ord} (3/2)\mid n$, since $ n$ is prime, it follows that $ \textrm{ord}(3/2)\equal{}n$. Since the order of anything divides $ q\minus{}1$, we have $ n\mid q\minus{}1$.

$ n\equal{}p$,$ q|3^{p}\minus{}2^{p}$

$ q>3$ which is obvious

$ 3^{p}\equiv 2^{p}\pmod{q}$

$ 3^{q\minus{}1}\equiv 2^{q\minus{}1}\pmod{q}$

Claim:$ x^{a}\equiv y^{a}\pmod{q},x^{b}\equiv y^{b}\pmod{q}$,$ (x,q)\equal{}(y,q)\equal{}1$

$ \implies$  $ x^{(a,b)}\equiv y^{(a,b)}\pmod{q}$

$ \frac{x}{y}\equal{}w$  $ \implies$  $ w^{a}\equiv 1\pmod{q},w^{b}\equiv 1\pmod{q}$

$ w^{(a,b)}\equiv 1\pmod{q}$

$ \implies$  $ 3^{(p,q\minus{}1)}\equiv 2^{(p,q\minus{}1)}\pmod{q}$

$ (p,q\minus{}1)\equal{}1$  or $ (p,q\minus{}1)\equal{}p$

if $ (p,q\minus{}1)\equal{}1$  $ \implies$  $ q|1$  Contradiction!

$ \implies$  $ p|q\minus{}1$ :wink:

Oh wow, I completely forgot the condition that $ n$ was prime. :blush: How silly of me. Thanks for all the responses.


algebra
polynomial
superior algebra
superior algebra unsolved

Let $p$ be a prime,$k$ a field, $\beta$ a non-zero elemnt of $k$, and $K$ a splitting filed over $k$ for the polynomial $f(x)=X^p-\beta$.
(i) Prove that $K$ contains a subfield $K_1$ which is a splitting field over $k$ for the polynomial $f(X)=X^p-1$, and that if $\alpha$ is a root of $f(X)$ in $K$, then $K=K_1(\alpha)$.
(ii) Show that eother $f(X)$ is irredicuble in $K[x]$, or $f(X)$ has a root in $k$.

Let $a_1,a_2...a_p$ be the roots of $x^p-B$ in $L$. Then $a_ia_1^{-1}$ is a root for $x^p-1$ and they are all in $L$, so $x^p-1$ has its $p$ roots on $L$, So $L$ contains its splitting field. Also if $L: M: K$ and $M$ is the splitting field for $x^p-1$ then it is easy to see that $M=K(a)$ where $a$ acts on $K$ as an n primitive root of unity (because all other roots are powers of $a$ then)

Note we asume Char $K$ is not $p$

Second part is Known as Abels theorem.


This problem is from back in the old days when Martin was around...(no offense :| )

How many cards must be selected from a standard deck to be certain that three cards from the same suit are drawn?

hint: Pigeon Hole Principle

[hide]
9
[/hide]

[quote="xpmath"]This is for their mathcount members :wink:
But since my school only makes us do workouts and warm ups :sleeping: , and since you already posted the answer, I'll provide a solution.
[hide]Since there are 4 suits, the worst case scenario is when you have 8 cards, 2 from each suit. Thus, the smallest is 9. Which  is also true by Pidgeonhole.[/hide][/quote]

Thanks xpmath! :lol:

[quote="dan80"]By the way, you will be back with Martin again starting next week and will probably be there for most of the rest of the time.
[/quote]
Alrighty..  :lol:

[quote="Chinaboy"]lol, that makes me feel old!

But frankly, I don't remember the question because there are so many like that.[/quote]

Yeah.. I noticed that there were whole bunch of those in back in the old days....(no offense, again..) And you answered almost all of them..!! :lol:


algebra
system of equations
algebra unsolved


logarithms
Gauss
real analysis
real analysis unsolved

$ \sum_{n\equal{}1}^{\infty}(2\minus{}x)(2\minus{}x^\frac{1}{2})...(2\minus{}x^{\frac{1}{n}})$

where $ x>0$

Discuss the above series converges or not

Pag.49 of Bromwich's book (not exactly the same exercise but very close).  If  $ a_n \equal{} (2 \minus{} x)(2 \minus{} x^\frac {1}{2})...(2 \minus{} x^{\frac {1}{n}}),$
$ {a_n\over a_{n \minus{} 1}} \equal{} (2 \minus{} x^{\frac {1}{n}}) \equal{} 1 \minus{} {\ln x\over n} \plus{} O(1/n^2)$ and by Gauss' criterion the series converges
if and only if $ \ln x > 1$ hence $ x > e.$

[color=green]Redundant quote deleted.[/color]

From $ x^{\frac1k}\approx 1 \plus{} \frac1k\ln x$, we have $ (2 \minus{} x)(2 \minus{} x^{\frac12})\cdots (2 \minus{} x^{\frac1n})\approx c\cdot \exp\left( \minus{} \ln x \minus{} \frac {\ln x}{2} \minus{} \cdots \minus{} \frac {\ln x}{n}\right)$
$ \approx c'\exp( \minus{} \ln x\ln n) \equal{} c'\cdot n^{ \minus{} \ln x}$; that $ c$ comes from the deviations in the early terms from $ \exp( \minus{} \frac1k\ln x)$.
Those terms certainly go to zero for all $ x > 1$.

Now, there is one value missed. At $ x \equal{} 2$, all terms are equal to zero, and the sum converges trivially.

If $ x\equal{}2^N$ for some integer $ N,$ the general term is definitively zero. Otherwise
$ \prod_{k\equal{}1}^n (2\minus{}x^{1/k}) \to0\ \iff\  \prod_{k\equal{}1}^n \vert (2\minus{}x^{1/k})\vert \to 0 \ \iff\
\sum_{k\equal{}1}^n \ln \vert (2\minus{}x^{1/k}) \vert \to\minus{}\infty.$ Now break
$ \sum_{k\equal{}1}^n \ln \vert (2\minus{}x^{1/k}) \vert \equal{} \sum_{k\equal{}1}^{n_0} \ln \vert (2\minus{}x^{1/k})\vert  \plus{}
\sum_{k\equal{}n_0\plus{}1}^n \ln (2\minus{}x^{1/k})$ where evidently $ 2>x^{1/k}$ for any $ n_0\plus{}1 \le k \le n.$ Now
$ \sum_{k\equal{}n_0\plus{}1}^n \ln (2\minus{}x^{1/k}) \equal{} \sum_{k\equal{}n_0\plus{}1}^n (\minus{}{\ln x\over k} \plus{} O(\ln^2 x/k^2))$ and this quantity goes to $ \minus{}\infty$ if and only if $ x>e.$ Moreover the general term has definite sign (a fact that I skipped when applying the Gauss's criterion) because $ 2>x^{1/k}$ definitively.

Do you agree?

[quote="milin"]Not so simple. When does the general term tend to $ 0$? It certainly does not for all $ x > e$.[/quote]


Let $a_i=x_i+1\ge 0,\;1=1,2,\ldots,n$ so
$ \sum^n_{i=1}(a_i-1)^3=0 \Longleftrightarrow \sum^n_{i=1}a_i^3-3\sum^n_{i=1}a_i^2+3\sum^n_{i=1}a_i-n=0$
We must prove that
$\sum^n_{i=1}a_i\le \frac{4n}{3}\Longleftrightarrow 3\sum^n_{i=1}a_i\le 4\sum^n_{i=1}a_i^3-12\sum^n_{i=1}a_i^2+12\sum^n_{i=1}a_i\Longleftrightarrow$ $\sum^n_{i=1}a_i(2a_i-3)^2\ge 0$
which is true.


calculus
integration
geometry
algebra
function
domain
real analysis

attached the question, and what progress I had towards a solution.
now I'm not sure how to calculate the last integral.
is it the surface of the torus?
should it be taken as a torus of 2 unit circles?

it [i]is[/i] the area of the torus, which is 1 (with respect to this coordinates).
you can see it by considering a fundamental domain (e.g. $ Q\equal{}[0,1]\times[0,1]\subset\mathbb{R}^2$) for the projection $ \mathbb{R}^2\to T$, and observing that integrating the pull-back of $ dx_1\wedge dx_2$ over $ Q$ gives you the lebesgue measure of $ Q$ (or, if you prefer, the "standard" area of $ Q$).


calculus
derivative
calculus computations

for the life of me I can't solve it.

It's

y=(x+1)^(x+1)

I did
lny= (x+1)ln(x+1)
then
y'/y= (1)ln(x+1)+ 1/X (x+1)
y'= [ln(x+10) +(x+1)/x) ] (x+1)^(x+1)

the solution in my textbook says something else...what did i do wrong?

You are wrong! Here is it :)
[quote="rayray123"]
y'/y= (1)ln(x+1)+ 1/X (x+1)

[/quote]

If you know the derivative of $x^{x}$, you can find the derivative of $(x+1)^{x+1}$ by replacing $x$ with $x+1$. (A special case of the Chain Rule).


function
superior algebra
superior algebra unsolved

I'm sorry if this is an absurd question, but I haven't had any class in algebra (especially not noncommutative algebra) so I have no idea what's what.

Anyway, de Bruijn is explaining big-O notation, and he explains that sometimes you write $A=B$ when you really mean $A$ is of type $B$. Then he gives the example of $n=L(m)$ meaning $n < m$ (which you might write if you didn't have "<" in your typewriter, he says). Then we have some interesting properties of this, such as $L(3) = L(5)$ but $L(5) \ne L(3)$. Is there any interest or value at all in this, or am I just entertained by a parlor novelty that doesn't have any substance?

edit: I guess it isn't really noncommutative either. I was just thinking of the fact that $a=b$ is not the same as $b=a$.... Not quite the same I guess.

Hi.

You have a binary relation "a = L(b)" or "a = O(b)" and are thinking about the property $a = L(b) \Rightarrow b = L(a)$, which is called "symmetry". The O notation is not symmetric, and it is also not antisymmetric: The property $a = O(b) \wedge b = O(a) \Rightarrow a = b$ is called "antisymmetry".

Therefore some authors prefer to use the notation $a \in O(b)$, which makes it clear that we do not have an equation, but some arbitrary relation (indeed any binary relation $\sim$ can be written in the form $a \sim b \Leftrightarrow a \in L(b)$ with an appropriate function $L$).
But most of the time you will see $a = O(b)$, and have to remember that "$=O$" is one single relation symbol.

LiMa


Let $AA'$ be the line tangent in $A$ to the circumcircle of $ABC$ ($A'\in BC$). Show that the orthic axis of $ABC$ passes through the midpoint of $AA'$.

[The orthic axis is the following line: let $H_a,H_b,H_c$ be the feet of the altitudes from $A,B,C$ respectively. Now let $A^*=H_bH_c\cap BC$, and define $B^*,C^*$ similarly. $A^*,B^*,C^*$ are collinear, and the line they lie on is called the orthic axis]

That is on page 6 (at the very end of paragraph 11) of

William Gallatly, [i]The Modern Geometry of the Triangle[/i], Second Edition London 1913.

There is an affine generalization of the problem:

[i]Let P be a point in the plane of a triangle, and let the lines AP, BP, CP intersect the lines BC, CA, AB at the points X, Y, Z, respectively. The parallel to the line YZ through the point A intersects the line BC at A'. Let the lines YZ, ZX, XY meet the lines BC, CA, AB at the points X', Y', Z', respectively. It is well-known that the points X', Y', Z' lie on one line, the so-called [b]tripolar[/b] of the point P with respect to triangle ABC. Prove that this tripolar passes through the midpoint of the segment AA'.[/i]

Here is my solution:

Since the lines AX, BY, CZ concur at one point (namely, the point P), we have by Ceva

$\frac{BZ}{ZA}\cdot \frac{AY}{YC}\cdot \frac{CX}{XB}=1$.

Since the points X, Y, Z' are collinear, Menelaos yields

$\frac{BZ^{\prime }}{Z^{\prime }A}\cdot \frac{AY}{YC}\cdot \frac{CX}{XB}=-1$.

Comparing these two equations, we see that

$\frac{BZ^{\prime }}{Z^{\prime }A}=-\frac{BZ}{ZA}$.

Now, since AA' || YZ, we have by Thales

$\frac{BZ}{ZA}=\frac{BX^{\prime }}{X^{\prime }A^{\prime }}$.

Thus,

$\frac{BZ^{\prime }}{Z^{\prime }A}=-\frac{BZ}{ZA}=-\frac{BX^{\prime }}{X^{\prime }A^{\prime }}=-\frac{X^{\prime }B}{A^{\prime }X^{\prime }}$.

Now, if M is the midpoint of the segment AA', then we clearly have $\frac{AM}{MA^{\prime }}=1$. Hence,

$\frac{BZ^{\prime }}{Z^{\prime }A}\cdot \frac{AM}{MA^{\prime }}\cdot \frac{A^{\prime }X^{\prime }}{X^{\prime }B}=\left( -\frac{X^{\prime }B}{A^{\prime }X^{\prime }}\right) \cdot 1\cdot \frac{A^{\prime }X^{\prime }}{X^{\prime }B}=-1$.

Thus, by Menelaos, the points Z', X' and M are collinear, i. e. the line Z'X' passes through the point M. But the line Z'X' is the tripolar of the point P with respect to triangle ABC, and the point M is the midpoint of the segment AA'. Hence, we see that the tripolar of the point P with respect to triangle ABC passes through the midpoint of the segment AA'. Proof complete.

Now, your problem is a special case of this since, if we take P to be the orthocenter of triangle ABC, then the tripolar of P is the orthic axis of triangle ABC, while the line AA' is the tangent to the circumcircle of triangle ABC at the vertex A (in fact, the line AA' is parallel to the line YZ, and since the line YZ, being a side of the orthic triangle XYZ of triangle ABC, is antiparallel to the side BC, the line AA' must be antiparallel to the side BC, too; but the antiparallel to BC through A is the tangent to the circumcircle at A, and therefore, the line AA' is the tangent to the circumcircle at A). And hence we get your problem.

  Darij

Well, my solution to the initial problem fits the generalization as well: $(A,C;Y,Y')=-1$, so the lines $X'A,X'C,X'Y,X'Y'$ form a harmonic quadruple, and $AA'$ is parallel to one of the lines of this quadruple, so the other three cut equal segments on it.


integration
vector
calculus
real analysis
real analysis theorems

I've noticed that $\int_{x_{a}}^{x_{b}}y dx+\int_{y_{a}}^{y_{b}}x dy = x_{b}y_{b}-x_{a}y_{a}$, and I think I can prove that it holds for vector line integrals when $x$ and $y$ are vector fields on the plane. Is this a well-known theorem? Is there a multivariable generalization? (like n variables instead of x,y)

Yes, this is well known. It's a special case of the fundamental theorem for line integrals; note that $(y,x)$ is the gradient of $xy$.

There is a large family of "fundamental theorem" results in multivariable calculus- most of them go by the name of Stokes' theorem.


For how many positive integers, $n < 1000$, is $n^4 \equiv 9 \pmod {16}$?

None. Since $2^4=16$, any even integer will automatically result in $n^4\equiv 0 \bmod{16}$. Next, we check odd integers, using the form $(n^2)^2\equiv x \bmod{16}$ for easier calculations, and we see that all odd integers squared will result in either a $1$ or a $9$ when taken $\bmod{16}$; from there, we have either $1\bmod{16}$ or $81\bmod{16}$, both of which will always be $1$. Therefore, if $n^4\equiv x\bmod{16}$, then $x=0$ for all even integers or $x=1$ for all odd integers.

I agree with JesusFreak197.

If $n$ is even, then $n=2k$ for some integer $k$.
$(2k)^4=16k^4 \equiv 0\pmod{16}$

If $n$ is odd, then $n=2k+1$ for some integer $k$.
$(2k+1)^4=16k^4+32k^3+24k^2+8k+1=16(k^4+2k^3)+8k(3k+1)+1$
Note that $k(3k+1)$ is always even.
[hide="reason"]when $k$ is even, $k(3k+1)$ is $(even)(odd)=(even)$
when $k$ is odd, $k(3k+1)$ is $(odd)(even)=(even)$[/hide]
Thus $8k(3k+1)$ is a multiple of 16 and we have
$16(k^4+2k^3)+8k(3k+1)+1 \equiv 1\pmod{16}$

So there does not exist any integer $n$ such that $n^4 \equiv 9\pmod{16}$.

Now, if it was $n^2\equiv 9 \bmod{16}$, you would actually have some that work. :) I challenge people to find all integers $n<1000$ that satisfy this equation. ;)

My pathetic attempt so far:

$n^2\equiv9(mod 16)$
$n^2-9\equiv0(mod 16)$
$(n+3)(n-3)\equiv0(mod 16)$

What do do now, is just a mystery :( (without trial and error that is) ooh...$5$ works!

[hide]$(n+3)(n-3)$ must be a multiple of $16$, so it must contain at least four $2$'s.[/hide]

[hide="Hint"]There might be a formula for finding the answer, but I just found a pattern.[/hide]

I used the same method as I solved the previous problem :)

Clearly, even number doesn't work, so we just have to check for odd numbers.
All odd numbers are in one of the following forms:
$8k+1$, $8k+3$, $8k+5$, or $8k+7$.

If $n=8k+1$, then $(8k+1)^2=64k^2+16k+1 \equiv 1\pmod{16}$
If $n=8k+3$, then $(8k+3)^2=64k^2+48k+9 \equiv 9\pmod{16}$
If $n=8k+5$, then $(8k+5)^2=64k^2+80k+25 \equiv 9\pmod{16}$
If $n=8k+7$, then $(8k+7)^2=64k^2+112k+49 \equiv 1\pmod{16}$

Therefore, all numbers in the form $8k+3$ or $8k+5$ ($3$, $5$, $11$, $13$,...) are equivalent to 9 in mod 16, and there are $1000\times \frac{2}{8}=250$ of them for $n<1000$.

Alright, now that the answer has been posted... the pattern that I found is essentially the one that you found, although I didn't think about it in the same way; I took consecutive odd numbers, squared them, divided by 16, and found the pattern $1$, $9$, $9$, $1$, $1$, $9$, etc.


[quote="futurus"]Let x, y be positive integers such that 3x + 7y is divisible by 11. Prove that 4x - 9y is also divisible by 11.

Thanks,[/quote]
$3(4x-9y)=4(3x+7y)-55y.$


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Let $ p\text{ be a prime},n\in\mathbb{N}, (p,n)\equal{}1$,
and let $ o_n(p)$ be the order of $ p\in(\mathbb{Z}/n\mathbb{Z})^*$
Prove that the $ n$-th cyclotomic polynomial $ \Phi_n(x)$ over $ \mathbb{F}_p$
factors into irreducible factors of degree $ o_n(p)$.

pick a primitive $ n$-root of unity $ \zeta$ in $ \overline{\mathbb{F}_p}$ and let $ G$ be the galois group of $ \Phi_n$ over $ \mathbb{F}_p$, which is generated by the frobenius $ x \mapsto x^p$. the equation $ g(\zeta) \equal{} \zeta^{\alpha(g)}$ yields an injective homomorphism $ \alpha : G \to (\mathbb{Z}/n)^*$, which maps the frobenius to $ p$. thus $ ord(G) \equal{} o_n(p)$.

now if $ f$ is an irreducible factor of $ \Phi_n$, $ f$ is the minimal polynomial of $ \zeta$ (or another choice of $ \zeta$), so that its degree equals $ ord(G) \equal{} o_n(p)$.


For solving the Riccati equations, can u guess the particular solution or it should be given in the question.

dy/dx= y + y^2/x^2 - x(2x-1)

If it is not given, how can I find a particular solution for this equation?

If you can give me a particular solution, that would be helpful and can u tell me why too...

Try the variable substitution defined by $y = -x^2\frac{w'}{w}$ where $w(x)$ is some function.

hello, Maple 13 gives the following solution
$ y \left( x \right) \equal{}1/2\,{\frac { \left( \minus{}7\,x{\it \_C1}\plus{}ix\sqrt {7}{
{\rm e}^{\minus{}i\sqrt {7}x}}\plus{}14\,{\it \_C1}\minus{}2\,i\sqrt {7}{{\rm e}^{\minus{}i\sqrt
{7}x}}\minus{}7\,i\sqrt {7}x{\it \_C1}\plus{}7\,x{{\rm e}^{\minus{}i\sqrt {7}x}} \right) x
}{\minus{}7\,{\it \_C1}\plus{}i\sqrt {7}{{\rm e}^{\minus{}i\sqrt {7}x}}}}$
Sonnhard.


Let A be an nxn matrix of rank (n-1). So some row/column a_i will be a linear combination of the rest. Why would the value of every determinant of size (n-1)x(n-1) formed from the cofactor expansion on A would be zero, except when we expand along a_i? Why is the linear dependence of a_i preserved even when we take out one of the columns/rows that participates in forming a_i?

Ok, can I reason myself in the following way:

If A is an nxn of rank (n-1), then there are n-1 vectors that can serve as a basis of the space of dimension n-1. So, if one of these vectors is taken out, there will be n-2 linearly independent vectors in the subspace of dimension n-2 which will span it. So, from this it follows that the linearly dependent a_i in A will still be linearly dependent in the subspace of dimension n-2 (which will be any cofactor matrix not created with expansion on a_i) and could be expressed as a combination of the vectors left.

I think you're trying to prove something that isn't true. We can easily construct $n$ vectors so that any $n-1$ are linearly independent.


Try solving this please beacause I want to know your opinion.

Let ABC be a triangle with the standard notations.
sin2A+sin2B+sin2C=1/2
Prove that:

p/3 \leq  \sqrt 3S

Solution:

we know that:
sin 2A+sin 2B+sin 2C=4sinA*sinB*sinC =>
sinA*sinB*sinC=1/8

now we use this identity: S=2R^2*sinA*sinB*sinC => S=R^2/4
and the inequality becomes:
p/3<=sqrt(3*R^2/4) => p<=3Rsqrt(3)/2 which is Mitrinovic's inequality
so p/3<=sqrt(3S) is proved.

bye!
:D :D


A number consists of 3 different digits. The sum of the 5 other numbers formed with those digits is 2003. Find the number.

217.

Trivial casework...did I misunderstand the question?

Trivial casework is possible indeed. Although there is a nicer solution.
Don't expect it to be too hard anyways.

But the other problems is not open for posting to non-mods, and I guess they'll complain about this in intermediate :?

[quote="pbornsztein"]Well, I don't there is a problem to post easy problems into the advanced section especially if they are coming from a national olympiad (in that case, the mods do not want them into the intermediate forum).

Dont worry about it, there are many others superhard problems here to maintain the average level  :D

Pierre.[/quote]

Pierre, I don't think Flanders MO is the level people expect to get when they hear "National Olympiad" :D

But you're right. Anyway someone got ideas for a beatiful solution here?

not beautiful, perhaps the same thought process:

the six numbers formed by the digits a,b,c are
[tex]100a+10b+c[/tex]
[tex]100a+10c+b[/tex]
[tex]100b+10a+c[/tex]
[tex]100b+10c+a[/tex]
[tex]100c+10a+b[/tex]
[tex]100c+10b+a[/tex]

Adding these 6 gives [tex]222(a+b+c)[/tex]

Since the 5 arrangements other than our solution adds to 2003, we have that [tex]10 \leq a+b+c \leq 13[/tex]

From here we test the number forced by a+b+c=10,11,12,13
a+b+c=10 --> 2220-2003 = 217 (our solution)
a+b+c=11--> 2442-2003 = 439 (sum is 16 - bad)
a+b+c=12 --> 2664-2003 = 661 (sum is 13 - close, but bad)
a+b+c=13 --> 2886-2003 = 883 (sum is 19 - worst)

That's the best solution indeed. But instead of case checking you can show the sum doesn't divide 3, doesn't divide 11, etc. But that one is far the most efficient.

Let the number be abc.
Then 2003=acb+bac+bca+cab+cba=122a+212b+221c.
Since a is not equal to 0,easy calculation shows that a=2,b=1 and c=7.
Thus,the number is 217.  :lol:

Then another problem similar to that.Hope it would not be so easy.Find
$\overline {abcd}$ such that $4|\overline {abcd},7|\overline {bacd},$$9|\overline {abdc},5|\overline {acbd}$


Reaper
combinatorics unsolved
combinatorics

We have a stack of $ n$ books piles on each other, and labeled by $ 1,2,...,n$. In each round we make $ n$ moves in the following manner: In the $ i$-th move of each turn, we turn over the $ i$ books at the top, as a single book. After each round we start a new round similar to the previous one. Show that after some moves, we will reach the initial arrangement.

this one is from [b]iran 2006 second round[/b] to,like the other problmes you posted...
just note that we have $ n!$ different arrangements,and also using the moves you told,there are at most $ 2^n.n!$ possibilities...
now try to use pigeon hole principle to complete the proof...

Well, Mr.Babak Ghalebi, can you please provide the full solution?  :)

ok here it goes,I'd try to make it easier to understand...
assume the sequence $ \{a_j^i\}$ we will give any POSITION of books a member of this sequence in the following way:

the index $ j$ in $ a_j^i$ introduces the arrangement of the books,or in other words it shows that [b]which book is on top,which one is second,which one is third,...[/b] and also each book is faced up or faced down...
hence $ 1\leq j\leq 2^n.n!$ (I)...

the index $ i$ in $ a_j^i$ introduces the MOVE which was taken to reach this POSITION,in other words we have [b]turned the $ i$ books at top[/b] to reach this POSITION...
hence $ 1\leq i\leq n$ (II)...

so now every POSITION of books has it's member of the sequence,now lets start to solve the problem...

as the problem says,we start the moves,in each step write down the [b]member of the sequence[/b] which is given to this position...
so we would have an infinite sequence of $ a_j^i$s,now this special sequence has the following properties:

$ i)$the index $ i$ in this sequence starts from $ 1$ goes on to $ n$ and then turns $ 1$ again,like the following one:

$ a_{j_1}^1,a_{j_2}^2,a_{j_3}^3,\ldots,a_{j_n}^n,a_{j_{n \plus{} 1}}^1,...$

$ ii)$let the $ k^{th}$ member of this sequence be $ a_{j}^i$ then the $ k \plus{} 1^{th}$ member is unique...

$ iii)$if $ a_j^i \equal{} a_{j}^{i'}$ then THE two positions which these members introduce,are the same...

now note that we have $ n.2^n.n!$ different $ a_j^i$s (according to (I),(II)) but our sequence is infinite so there must be a repeat in our sequence,let the first repeat be $ a_s^r$ i.e.:

$ a_{j_1}^1,\ldots,a_k^{r \minus{} 1},a_s^r,\ldots,a_t^{r \minus{} 1},a_s^r,\ldots$

and assume that $ r > 1$...

now according to the property $ ii)$ we must have:

$ a_k^{r \minus{} 1} \equal{} a_t^{r \minus{} 1}$

but this contradicts the fact that $ a_s^r$ was the first reapet,so the assumption $ r > 1$ is incorrect so we must have $ r \equal{} 1$ and also from $ iii)$ we get that $ s\equal{}j_1$,which means that the first POSITION repeats after finitiely many moves...


If the smaller angle between the hour and minute hands of a 12 hour analog clock is 84 degrees and the sum of the hours and minutes is 17, what time between noon and midnight is it?

Use $ |5.5m\minus{}30h|$ is angle where $ m$ is minutes and $ h$ hours.

$ |5.5m\minus{}30h|\equal{}84$ or $ 360\minus{}84$
and $ m\plus{}h\equal{}17$
$ |5.5m\minus{}510\plus{}30m|\equal{}|35.5m\minus{}510|$
Solutions are $ m\equal{}12$ and $ m\equal{}6.59...$ so :
$ m\equal{}12$ and $ h\equal{}5$
Time: $ 5: 12am$