1.) You and your coworker together make $16 per hour. You know your coworker earns 10 percent more than you do. Your hourly wage is $. 7.62
After taking Math 1010 your hourly wage is raised to $12. This is a raise of %. _________
After returning to work you can't help mentioning casually to your coworker that now you make_______ % more than he does.
2.)You can paint a certain room in 7 hours. Your brother can do it in 6 hours. How long does it take the two of you working together? (hours)
3.)How much 90 percent vinegar do you have to add to a gallon of 5 percent vinegar to get 20 percent vinegar?  (gallons)
4.)Enter an algebraic expression that gives the area of a circle with a circumference.
5.)You buy a house including the land it sits on for $155000. The real estate agent tells you that the land costs $24000 more than the house.
The price of the house is $____  and the price of the land is $_____ .

6.)You figure that you will buy property many times and you endeavor to work out a formula that will tell you the price of the land and the price of the buildings any time.

So suppose the price for the land and buildings combined is  (for "sum") and the land costs  (for "difference") more than the buildings.

Then the price of the buildings is _____   dollars
and the price of the land is ______  dollars.

7.)Suppose the next time you buy a house including the land it sits on for 120000 Dollars. The real estate agent tells you that the land costs 25000 more than the house.

You apply your formula derived in the previous problem.
and you find out that the price of the house is ____  dollars and the price of the land is _____ dollars.

8.)The next time you buy a house including the land for 156000 Dollars. This time the real estate agent tells you that the land costs 13000 Dollars less than the house.

You realize that you can still apply your formula. You set S=____  dollars and D=____  dollars.
and you find out that the price of the house is _____dollars and the price of the land is____dollars.

Sorry for all the problems i cant seem to figure them out

I'm pretty sure someone will be [size=75]helping[/size] you with the homework shortly. Just sit tight and relax.  :huh:  :rotfl:
EDIT: Ahh,  :spam:  detected in the other post here ->[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=748168#748168[/url]

Ohh pick me!...  :D

... to lock the topic.  :wink:


what textboooks should i buy????
any other sources?

Barron's is a pretty good buy: my teacher skipped around in our book and didn't teach the concepts so i bought barron's and got a 5 with that.


algebra
polynomial
calculus
integration
function
difference of squares
special factorizations

Here's how this works. One person posts a problem, a second posts a solution [b](not just an answer, a solution)[/b] and posts a second problem, a third posts a solution and posts a third problem, etc. These problems should be all ALGEBRA. I shall start.

[b][color=olive]Problem 1[/color][/b]
My eraser is in the shape of a rectangular prism. The three distinct lengths of the face diagonals are $ 13$ millimeters, $ \sqrt{221}$ millimeters, and $ 2\sqrt{85}$ millimeters. Every erasure uses up $ 2$ cubic millimeters of eraser. At every contest I go to, I erase 21 times. How many contests do I need to go to to use up my eraser entirely?

[hide]Let $ a$, $ b$, and $ c$ be the dimensions, in millimeters, of the eraser. Then $ a^2 \plus{} b^2 \equal{} 13^2 \equal{} 169$, $ b^2 \plus{} c^2 \equal{} 221$, and $ a^2 \plus{} c^2 \equal{} 4 * 85 \equal{} 340$. Adding all of the equations, we get $ 2a^2 \plus{} 2b^2 \plus{} 2c^2 \equal{} 730$, so $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 365$. Thus $ a^2 \equal{} 365 \minus{} 221 \equal{} 144$, $ b^2 \equal{} 365 \minus{} 340 \equal{} 25$, $ c^2 \equal{} 365 \minus{} 169 \equal{} 196$, and the volume of the eraser, in cubic millimeters, is $ abc \equal{} 12 * 5 * 14 \equal{} 840$. 42 cubic millimeters are used up at every contest, so it takes $ 840 / 42 \equal{} 20$ contests.[/hide]

Those are weird dimensions for a rectangular eraser...

[color=olive][b]Problem 2[/b][/color]

Find a polynomial with integral coefficients whose zeros include $ \sqrt{221} \plus{} 2\sqrt{85}$.

[hide]If $ x\minus{}(\sqrt{221}\plus{}2\sqrt{85})$ is a factor, then another factor can be $ x^2\minus{}(561\plus{}68\sqrt{65})$, by squaring the given root. We don't want the square root, so we multiply by $ x^2\minus{}(561\minus{}68\sqrt{65})$ and use the difference of squares formula to get

$ x^4\minus{}1122x^2\plus{}14161$[/hide]

[b]Problem 3[/b]

Find $ 2\sqrt{2\sqrt[3]{2\sqrt[4]{2...}}}$

Solution to last problem

[hide]
The Taylor series for the exponential function $e^x$ where $x=1$ gives a well-known identity:

$e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots$

This implies the following convenient expressions:

$e=1+1+\frac{1}{2}\left(1+\frac{1}{3}\left(1+\frac{1}{4}\left(1+\frac{1}{5}\left(1+\cdots \right)\right)\right)\right)$
and
$e-1=1+\frac{1}{2}\left(1+\frac{1}{3}\left(1+\frac{1}{4}\left(1+\frac{1}{5}\left(1+\cdots \right)\right)\right)\right)$

This, in turn, yields
$x^{e-1}=x^{\left(1+\frac{1}{2}\left(1+\frac{1}{3}\left(1+\frac{1}{4}\left(1+\frac{1}{5}\left(1+\cdots \right)\right)\right)\right)\right)}$

And with $x=2$, we get, quite amazingly:

$ 2\sqrt{2\sqrt[3]{2\sqrt[4]{2 \cdots}}} = 2^{e-1} \approx \framebox{3.29044299550896048542577...}$
[/hide]

[b]Problem 4[/b]

Find $ 2\sqrt{2\sqrt[5]{2\sqrt[8]{2\sqrt[11]{2 \cdots}}}} $

:)


Here's a quiz -- it's sort of like trivia.

1. How many edges does a stop sign have?

2. What century was the year 2000 in?

3. On what side do women have buttons on? What side do men have buttons on?

4. Name all of the 7 dwarfs from Snow White.

5. What is a rhino's horn made out of?

[hide="answers"]1. 8
2. 20th
3. women have it on the side sometimes, men have it on the front
4. sleepy, dopey, grumpy, sneezy, happy, doc, bashful
5. calcium[/hide]

[hide]
1. 2: The front and back
2. 21st century
3. women-left, men-right
4. doc, dopey, happy, sleepy, sneezy, grumpy, bashful
5. ummm, atoms.... :P
[/hide]

Yes, #1 is 8. @cyberspace: I said edges, not sides -- good try, though!

1=2 is correct -- the year 2000 is actually in the 20th century, the reason being that there was no year 0(and that's because the people who created our year system thingy didn't like the idea of 0 -- "Zero -- A dangerous idea" is a good book on the subject.)

27r and cyberspace got #3 correct - woman on the left, men on the right. No one really knows why though there are abundant theories.

You all know your dwarfs!  :P

None of you got #5 correct(but I like your answer, cyberspace!  :) )

[hide]
I'm kinda wondering what the intended answer for #1 is.  I would argue that a stop sign is a 3D object and therefore it has 24 edges (8 for the octagon on the front, 8 for the octagon on the back, and 8 for the connecting edges).[/hide]

Well, stop-signs are pretty much flat...

Is #5 Ivory? I know that's an elephant but...

[quote="SMCount"]Unless it's a trick question I believe 27r got #5 correct.[/quote]

It's not a trick question, and that's what I used to think too, but the answer's really surprising.

And no, not ivory -- nice try!

[quote="1=2"]it's a bone... what are bones made of again??? *thinks* RUBBER!!![/quote]

Not a bone.

[hide="Only Reasoning For 3 I Have Seen"]Men have their buttons on the right because it's easy to button this way for righties, which most people are. When buttons were first being used commonly, only the rich could afford clothing with buttons. At this time, rich women also had maid servants who would dress them, and therefore putting buttons on the left made it easier.[/hide]

[quote="27r"][hide]1. 8
2. 21st
3. women: left men: right
4. doc, dopey, sleepy, sneezy, bashful, grumpy, happy
5. the stuff your fingernails are made of[/hide][/quote]

Ya she did get it right... It's keratin with melanin and calcium somewhere in there. All common to your skin and bones.

A rhino's horn is made of something different. Should I just say it?

[hide="The Answer"]

Surprisingly, a rhino's horn is made of [b]hair.[/b]

[/hide]

On wikipedia it says keratin, which makes up hair and fingernails. But the horn itself is not made of hair..

[quote="isabella2296"]Rhino horns are made of hair...[/quote]

"The horns of a rhinoceros are made of keratin, the same type of protein that makes up hair and fingernails, but the horn is not itself made of hair as previously believed"


[quote="gtf"]OK, so this will be my 100th post and it's mostly gratuitous so that I get to my 100th post, but who's going and what have you done to prepare for the tournament this weekend?

Fulton[/quote]
I'm going.  To prepare for varsity state I prepared for USAMO.  And I also worked on not getting a cold.  The latter alone should get me some improvement from last year. :roll:

Eeh, I agree with Miles... last year, I spent half my time removing/putting on layers of clothing and asking for tissues. Not a pleasant time at all. But making ARML was nice  :)

Yay. My teacher got an e-mail, and I'm a "person of interest." Unfortunately, I stayed up until 4 A.M. Monday morning doing a term paper, and I've been really tired this week. I hope I've recovered by Saturday.

[quote="Boy Soprano II"][quote="Arvind_sn"]So how exactly does this work in conjunction to ARML? I mean, there are 11 USAMO qualifyers right? so is it top 19 non-USAMO-qualifiers? i know its based on other tournaments (and on past ARML experience, which doesn't really fit me), but which ones stand out? (not that i have any chance of hoping of considering of thinking about dreaming about making ARML this year, i just want to know for next year). :D[/quote]
Hey, don't sell yourself short.  Hm, about tournaments, we had a pretty good thread about them [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84622]here[/url].  (Whoa, do you realize this forum now has three pages?)  Except you might as well add North Fulton to those lists, if it continues to draw the crowd it did this year.  Anyhow, I refer you to the [url=http://paideiaschool.org/TeacherPages/Steve_Sigur/resources/GAARML/GEORGIAARML/Welcome.html]GA ARML website[/url].  Varsity State does not determine the ARML team completely.  You could not put everybody's records into a computer and use a simple program to determine the ARML team.  Humans decide it.  The coaches select the 33 people they want.[/quote]

Thanks, but here's the catch: I'm not going this year (other ppl on math team want to go, and only 4 ppl can. Why was that rule even created? :D

@Twiz: I read in an article in Time magazine that if you sleep late on one day, say Monday, and then wake up early on Tuesday, then you won't get affected until Wednesday, and if you sleep well (i.e. 9-10 hours) for like 3 days or something, then it goes away. So I guess you'll do fine on Saturday. :)

Eh, don't necessarily trust time magazine. No, I'm not going. It is enough to do math a lot. Maybe next year I will do contests in Georgia more.

dude why arent you going to varsity state its the biggest tournament all year plus its like two hours away from where i live dont spoil the fun man i mean seriously

and yay northview has five people coming

we were supposed to have six but then vibin couldnt go cuz of track which is a bit annoying since i suck at running and he is amazing at math

so blah we lose vibin

but we still have five people going

dude we're gonna pwn all who dare to oppose us at ultimate

maybe apart from lassiter

or maybe if phillip decides not to play

or heng may pwn him nevertheless since hes like 5 feet 7 inches but can catch the frisbee even when guarded by like e^i * pi * -12345 people

i learned that at mathfest that was really interesting

So how does ciphering work. I assume that individual is just the standard five points for one minute, ten points for two, but what about pair ciphering. How much time, how many questions at a time, how many points, etc.? Is there anything besides written, individual ciphering, and pair ciphering?

Or it could be 10 points for 1 minute and 5 for 2, but I know a bunch of people that would vote for it your way! :-)

Pair ciphering is 4 questions and periods of 2 minutes and 4 minutes. 20 points for every question the pair gets in the first 2 minutes and 10 points for every question in the 2nd 2 minutes. Two rounds of 4 questions each.

The written and individual ciphering are the only part that count for the individual scores.

Those 3 events are the only ones.

um i forgot from last year but was pair ciphering harder?

I don't think that it's significantly harder. At least I don't think that it's supposed to be.

Oh no! I have the hiccups. When I get them, they usually come back frequently. I hope I don't have them during the tournament.
Anyways, good luck everyone tomorrow.

I'll be there with a blue shirt and black pants. Tell me if you see me. I wear no glasses and will most probably be playing ultimate if there's a game.

Claus didn't go to varsity state. (But probably next year he will.) His primary qualification came from the USAMO index.

How many middle schoolers made ARML this year?

Just Claus. I can think of 2 or 3 others that are "watchable", but Claus is the one that has the AMC + AIME scores.

Fulton

I'm not sure that information is in the public domain.

Karth's list is still missing Michael Lai. It's also missing Kevin Sun from Centennial and Maja Wichrowska from Alpharetta. Which means we're only short one person. Anyone know who?

Jordan Ou I seem to remember, but don't shoot me if I'm wrong...

[quote="gtf"]Scott Brothers from Walton.

Karthik, please change Klaus to Claus Zheng

And we are still 3 short. There are 33 on the team.[/quote]

I count 29 currently on the list, which means that with Michael, Kevin, Maja and Scott, we have 33. (Unless, of course, there are people on the list who aren't doing it this year, which seems entirely possible.)

So the full list seems to be:

Miles Edwards (The Miles Edwards)
Phillip Mote (The P-Mote)
Wes Brown (The Wes Brown)
Harrison Brown ([b]The [/b]Harrison Brown... bolded because he made RSI [Ed: haha])
William Drobny,
Linda Liu
Peter Huang
Karthik Narayan
Benjamin Hu
Tony Huang
Brayden Ware
Amy Lanchester
Nayoon Kim
Sam Brotherton
Sagar Patel
Billy Dorminy
Nathan Hipsman ([b]The[/b] Nathan Hipsman... he made RSI too)
Eric Morphis
Eddy Ferreira
Sean Lee
Santosh Karnik
Mark Doss
Ethan Sawyer
Ruby Lai
Claus Zheng
Jordan Ou
Carol Wang
Tomas Leon
Robert Grosse
Scott Brothers
Kevin Sun
Maja Wichrowska
Michael Lai

[quote="gtf"]Scott Brothers from Walton.

Karthik, please change Klaus to Claus Zheng

And we are still 3 short. There are 33 on the team.[/quote]

I can't edit my previous post anymore  :( But here are the new ARML members:

New ARML members:

Benjamin Hu - 9th
Tony Huang - 10th
Sagar Patel - 9th
Eric Morphis - 10th (I think?)
Claus Zheng - 7th (zomg)
Tomas Leon (former - not in 2006, but in 2005) - 11th

Wow, so we only have six new members (5 excluding Tomas)? That's an amazing retainment rate  :D

It also makes it so that I may very well be an alternate (can former members be alternates?).

Scott Brothers, Robert Grosse, Kevin Sun, and
Maja Wichrowska are also new, correct?

Yes, I am in 10th grade.

As far as alternates go, they are full members of the team. We will choose 15 members of the 33 to be Ga. A, 15 members to be Ga. B, and 3 alternates. The alternates will compete with other alternates from other teams and very frequently lead those teams! In addition, we've had several alternates move up to the A team before they are done. So alternates do get PLENTY of work!

Fulton

:!: According to Carol Wang, who I spoke to earlier today, Ethan Sawyer will not be on the ARML team because he declined the offer to go. Will there be someone else, or has there already been someone chosen, or was he even on the original list?

Boy, do I miss out on things. I didn't notice this until I got to China. :(

By the way, Karthik, it's "to whom," not "who." And yes, I am a grammar freak most of the time(though I do make mistakes myself fairly often).


Vieta algebra polynomial function calculus articles

How would one prove the Newton sums? (I see the first one $ a_n S_1 \plus{} a_{n\minus{}1} \equal{} 0$ is a trivial consequence of Vieta's formula, though.)

http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums

Here's the basic idea:

We have $ P(x)\equal{}a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots \plus{}a_{0}$.  Multiplying the whole polynomial by $ x^{k}$ gives us $ x^{k}P(x)\equal{}a_{n}x^{n\plus{}k}\plus{}a_{n\minus{}1}x^{n\plus{}k\minus{}1}\plus{}\cdots \plus{}a_{0}x^{k}$.  Now sum the right side of the equation over all the roots of $ P$ to get $ a_{n}S_{n\plus{}k}\plus{}a_{n\minus{}1}S_{n\plus{}k\minus{}1}\plus{}\cdots \plus{}a_{0}S_{k}$.  This equals zero since if $ x$ is a root of $ P$ then $ x^{k}P(x)$ is 0.

For the first few Newton's sums this doesn't work.  But the $ k$th Newton's sum can be expressed in terms of the first $ k$ symmetric sums, which in turn are uniquely determined by the first $ k\plus{}1$ coefficients of the polynomial.  So the polynomial $ Q(x)\equal{}a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots \plus{}a_{n\minus{}k}x^{n\minus{}k}$ has the same $ S_{1},\ldots S_{k}$ as $ P(x)$.  Then we divide by $ x^{n\minus{}k}$ to get $ Q(x)/x^{n\minus{}k}\equal{}a_{n}x^{k}\plus{}a_{n\minus{}1}x^{k\minus{}1}\plus{}\cdots \plus{}a_{n\minus{}k}$.  Now we can just use the same trick as above, summing the right side of the equation over all non-zero roots of $ Q$ (the zero roots don't change the $ S_{i}$ anyway) to get $ a_{n}S_{k}\plus{}a_{n\minus{}1}S_{k\minus{}1}\plus{}\cdots \plus{}a_{n\minus{}k}k$.  The last term has a $ k$ since there are $ k$ equations we have to sum, and this whole expression again equals zero since if $ x$ is a non-zero root of $ Q$ then $ Q(x)/x^{n\minus{}k}$ is zero.

Actually when I'm doing a problem that could be solved using Newton's sums, I just use this trick rather than trying to remember the Newton sum formulas, since they're pretty much equivalent.  :)

I gave another proof based on generating functions here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159852[/url].

matt276eagles this only proves a few cases of newton's sums...

nice calculus proof: [url]http://planetmath.org/encyclopedia/ProofOfNewtonGirardFormulaForSymmetricPolynomials.html[/url]

[quote="facis"]matt276eagles this only proves a few cases of newton's sums...[/url][/quote]
What do you mean?  Read the proof again.

Yeah I guess I didn't read carefully enough.  But I don't quite understand your claim:

[quote]But the $ k$th Newton's sum can be expressed in terms of the first $ k$ symmetric sums, which in turn are uniquely determined by the first $ k\plus{}1$ coefficients of the polynomial. So the polynomial $ Q(x)\equal{}a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots \plus{}a_{n\minus{}k}x^{n\minus{}k}$ has the same $ S_1, \ldots S_k$ as $ P(x)$.[/quote]

[b]Theorem:[/b]  Any symmetric polynomial in $ n$ variables can be written as a polynomial in the [i]symmetric sums[/i] of those variables (which are essentially the coefficients of a polynomial with those variables as its roots).  In particular, a symmetric polynomial of degree $ k$ can be written as a polynomial in the first $ k$ symmetric sums (because higher symmetric sums are of higher degree).

For example, $ x^2 \plus{} y^2 \plus{} z^2 \equal{} (x \plus{} y \plus{} z)^2 \minus{} 2(xy \plus{} yz \plus{} zx)$.  This is a polynomial of degree $ 2$ that therefore does not depend on the third-degree symmetric sum $ xyz$.  What matt276eagles is saying, rather elegantly, is that (in this case) we can set $ xyz \equal{} 0$ without changing the value of $ x^2 \plus{} y^2 \plus{} z^2$; in other words, we can discard coefficients higher than the degree of the sum we want.

thanks...  thinking about it now, in terms of veita's formulas, it makes some sense... I think I'm just not convinced about your claim that "the $ k$th Newton's sum can be expressed in terms of the first $ k$ symmetric sums."

Looking at it now, I suppose I would like to see a proof of that theorem, t0rajir0u.

[quote="t0rajir0u"]A proof is available [url=http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#The_fundamental_theorem_of_symmetric_polynomials]here[/url].  The [url=http://mathworld.wolfram.com/SymmetricPolynomial.html]mathworld article[/url] tosses out a dimensionality argument which captures the essence of what's really going on.[/quote]

Can this be proved without using those advanced theories (as I presume Olympiad Algebra doesn't really include advanced topics like abstract algebra).

The inductive proof is more notation-heavy than actually technical.  This theorem appears without proof in Engel (where he uses it often), so I presume it can be cited on an Olympiad if that's your concern.

Just wondering, what is the correct formulation for $ S_{n\plus{}k}$ with $ k\ge1$ for a $ n$th degree polynomial?
Let's say for a $ 4$th degree polynomial, it seems to me from the wiki example that $ a_4S_6\plus{}a_3S_5\plus{}a_2S_4\plus{}a_1S_3\plus{}a_0S_2\equal{}0$, but how would we prove this is true?

I encountered this thought while trying to find $ p(a)\plus{}p(b)\plus{}p(c)\plus{}p(d)$ where $ p(x)\equal{}x^6\minus{}x^5\minus{}x^3\minus{}x^2\minus{}x$ and $ a,b,c,d$ are the roots of $ x^4\minus{}x^3\minus{}x^2\minus{}1\equal{}0$. I got $ 6$ using the newton sums btw.

That's correct.  All of the proofs given in this thread extend to that case.

Oh yeah... my bad - I only remembered the proof of the 2nd part in matt276eagle's proof and forgot about the first paragraph in his proof when the first part is much more straightforward than the second part... how ironic (seriously I need to get to bed before 11pm now... or otherwise my brain's going to break down...) :blush:

By the way, can that problem be solved without using newton sums (I guess newton sums weren't the intended solution)?


Greeting,

This is a thread telling that I am holding a math chat in level of AMC-10 for the people who are taking AMC-10 next February. What we basically do is just like in the AMC-10 Problem Series, look on many problems, solve them, and learn how to do them, and get transcripts so you can keep them as your independent study guide.

This chat is free and I maintains it. There is a restriction though. I am not going to accept anymore than 10 people on the chat. So, if you want to do it, you better sign up fast.

The people who participated in the previous one are automatically in for any further ones.

*theone853
*..um.. one newbie.. I can't remember your ID. Can you tell me what was your ID in AoPS? Thanks.
*Treething
*B Horse Math
*whitehorseking

And I talked with MathFiend and Yif Man 12 and both of them wanted to attend but couldn't attend so basically, we have 7 people on the list.

Only thing that I meant by restricting to 10 is that those 10 people will get announcement telling when the math chat will be held. Others, you just have to look up on the Art of Problem Solving regularly.

If you're not so confident on whether you should do this, look on the previous math chat.

http://dhost.info/jwec05/Silverfalcon/t1.PDF

We mostly focus on the major types of problems that are likely to appear on the AMC-10. Anyone in level of AMC-12 or even close to it aren't accepted.

Here are the benefits:
1. Free lecture like AMC-10 Chat
2. Transcript Available Online Although I don't have the copyright to the source, since it's available on the internet without buying it.
3. Not all problems will be easy. Some problems we will deal require thinking and therefore, develop your math skill.
4. I haven't decided yet but maybe, if this gets regular like, we might have something like end-of-math-chat session exam where questions are available on the my website and you're given a chance to solve them and submit the solutions. This is not confirmed. I don't know whether I'm going to have a time for this or not.

Questions? Post here. It's usually Friday 6:00 EST. About an hour like a math jam or class.

Regards and Sincerely,

Silverfalcon

can i participate in the math chat.  I'm taking the amc 10 B in feb

where would you go for the math chat?  Where would it be held

I want to participate. To Demji: I think he uses AIM
SN: mathlinksguy

Yes, this is held at IM.

You can get IM by free on going to:

http://www.aim.com

You don't have to download the program and uses something call AIM express but I would suggest you downloading it because it's more convenient.

Ok. So we got three more people.

demji
RC-7th
h_s_potter2002

I need you guys' screennames.

Post here. Thanks.

Next one will be upcoming Friday, 17th.

These math chats are always a great way to learn new problem solving strategies, so after math chats, if individuals have problems which weren't discussed, that would be a great time to get help and insight.

 :D  :D  :D

Yes. So, if you have any questions after math chat, you can always post here or ask someone else that was on the math chat.

 :D  :D Learning "how" is most important benefit. You might see the problem that was on the math chat somewhere in the test in future. Who knows how this life goes? :P

how do you know who to get into the chat since we don't know your IM name,  i'm demjis

I think there are like whole posts on this but my sn is jwec05.

Yours is demji or demjis?

AMC Math Chat has been scheduled. I told some people already but it's going to be Friday EST 6:00.

Be there or view the transcript. The topic will be generally in level of AMC competition. (AMC-10 of course)

 :lol:

its demjis at least thats what it was before i uninstalled aim.  i reinstalled it just for this so i'll keep you posted

/\/\1|\|45 /\/\012G|_|1 LOL. I can't speak leet, but oh well... nice SN.

There was unannounced math chat yesterday.

Transcript is:

http://dhost.info/jwec05/Silverfalcon/t2.PDF

These unannounced math chats happen sometimes, it depends on how bored I am so if I'm bored and have nothing to do, I might host one for like 30 min.

Usually, it's the Friday.

How would we know if the unanounced math chats happen?

Well, missing an unannounced math chat wouldn't hurt you a lot anyway. And if you want to keep in touch, it would be a good idea to have signed on for AIM for now and then (but not continously cuz that might influence you for signing and out) so that if happens, you can participate.

I usually do when there are enough people. There were quite people yesterday and that's why it happened.

I'll be there at around 6:15 PM on Friday, just like last week.

do you have a website that has all your math chat transcipts on it?

I don't have a webpage for the transcript yet.

Only ones are the ones available here.

does anyone know a good hosting service that is free so we could use it for the transcripts and other information?

That is useless because my website has all the things (transcripts) in it. Even the rounds of COP can be found there.

If you have rolled down the Getting Started page, you should've found the thread "COP MAIN WEBSITE).


Call $ f(x)\equal{}\frac{x^2\minus{}4}{x\plus{}1}\minus{}\sqrt{x}$. $ f(0)\equal{}\minus{}4, f(4)\equal{}12/5\minus{}2\equal{}4/20$. By the intermediate value theorem, somewhere in between, we must have $ f(x)\equal{}0$. More intuitively, the graph of $ f$ (which I haven't looked at :)) goes from a negative value to a positive value. Since the function is continuous, it must pass zero somewhere in between.

It is important to note that the function is [b]continuous[/b] for all $ x \in [a,b]$


limit
real analysis
real analysis solved

Original problem : 11th grade problem 2

Consider $a_n$ sequence of positive real numbers such that
$(a_{n+1}-a_n)^2=a_n$ for all $n\geq 1$ Find $\lim a_n =L$
-----------------------------------------------------------------------------------------------

Here we revisited

With the same statement give an equivalent of $a_n$ when n tends to infty

Is the sequence increasing? (because otherwise we can go either back or forth during each step and the resulting behavior can be very wild).  :?


[quote="zanttrang"][quote="rrusczyk"][quote="seesaw"]Can we register now for year 18?

Also, will there be a specific topic for year 18?[/quote]

If you register now, you will get a poster with Round 1 questions on it this summer (and an email when the questions are posted on the USAMTS site).  We haven't decided if we're going to repeat the 'topic' experiment.[/quote]
I thought the topic idea was a good one - it allowed us to hone our skills on subjects we may not have been familiar with over the course of the contest.  Why are you considering not trying it again?[/quote]

The two main reasons are difficulty finding a suitable topic and time to prepare the article.

So its best to go ahead and sign up now?

[quote="rrusczyk"]
If you register now, you will get a poster with Round 1 questions on it this summer (and an email when the questions are posted on the USAMTS site). [/quote]

Can we choose not to get the poster and only to get the e-mail?

I registered on the usamts website, but when i tried to print the entry form, it was for year 17, which is over.

should i send that in, or wait for a new entry form for year 18?

Wait until Year 18.  We should have all those forms up by the beginning of July.

The topic questions also mean less variety of questions overall, and give a disproportionate advantage to those one of whose favorite subjects the topic happens to be :).

Presumably the entire contest hones your skills at various areas of math--is the marginal benefit of a fourth problem of the same type really greater than that of the first?


For prime number $p$, let $f_p(x)=x^{p-1} +x^{p-2} + \cdots + x + 1$.

(1) When $p$ divides $m$, prove that there exists a prime number that is coprime with $m(m-1)$ and divides $f_p(m)$.

(2) Prove that there are infinitely many positive integers $n$ such that $pn+1$ is prime number.

$f_p(m)=\frac{m^p-1}{m-1}$. It's easy to show that $(\frac{m^p-1}{m-1},m-1)=(p,m-1)$, and since $p$ doesn't divide $m-1$, it means that $(\frac{m^p-1}{m-1},m-1)=1$, so the prime factors of $f_p(m)$ don't divide $m-1$. At the same time, the prime factors of $f_p(m)$ also divide $m^p-1$, so they can't divide $m$, and this solves $(1)$.

I'll be back for $(2)$. :)

For $(2)$ we look at the prime divisors $q$ of $f_p(kp)$ which do not divide $kp-1$ (such divisors exist, because of $(1)$). $q|(kp)^p-1,\ q|(kp)^{q-1}-1\Rightarrow q|((kp)^p-1,\ (kp)^{q-1}-1)=(kp)^{(q-1,p)}-1$. If $p$ doesn't divide $q-1$, then $q|kp-1$, which is false, so $p|q-1\Rightarrow q=M_p+1$.

It's not hard to show that a sequence of numbers $(k_n)_{n\geq 1}$ exists s.t. $((k_ip)^p-1,\ (k_jp)^p-1)=1\forall \ i\neq j$. We thus get a different prime $q_n$ as above for each $k_n$. In order to show the existence of such a sequence of $k$'s we can use, for example, the Chinese Remainder Theorem to make sure that we can suitably choose $k_{n+1}$ s.t. none of the prime divisors of $(k_ip)^p-1\ i\in \overline{1,n}$ divides $(k_{n+1}p)^p-1$. This means that we construct $(k_n)_{n\geq 1}$ inductively.

for the last part of (2) _make a sequence $k_{n}$ with the property of $((K_{i}p)^{p}-1,(k_{j}p)^{p}-1)=1 \forall i\neq j$_ also we could find $k_{i+1}$ from $k_{i}$  by choosing $k_{i+1}=(k_{i}p)^{p}-1$ which is also what we wanted to show.

[hide="Chinese Remainder Theorem"]how can we use this Theorem to make a sequence such grobber said?[/hide]

With Regards.

[quote="mathbuzz"](2) gcd(p,1)=1 , so , by DRICHLET's theorem , we are done.[/quote]
the proof of  DRICHLET's theorem is very long

(2)[size=150]lemma[/size]->ther is$ x $such that for any $fixed$ numbers $a,b,c $ we have$ c $ divised$ ax-b$ if and only if$ b$ is divisel by$ gcd(a,c)$.

Also see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=722&t=393347 .


If Nonie's mom spends 20 mins on the treadmill and burns 300 calories, how many calories does she burn per second?

What if she had worked a little harder and burned 350 calories in the same amount of time?

[hide="Click to see Answer 1 "]
$300\div20=15$ 15 calories a minute.
$15\div60=2.5$
[b]2.5 Calories per second[/b][/hide]
[hide="Click to see follow up answer"]
$350\div20=17.5$ 17.5 calories a minute.
$17.5\div60=.291666\infty$
[b]$.291666\infty$ Calories per second[/b][/hide]

P.S. Is there a way to put the infinity sign above the number?


A triangle has angles that measure n degrees, n + 20 degrees, and 3n - 15 degrees. What are their measures.

I keep ending up with a total of 160 when I add up all the measurements that I get. I end up getting 31, 51 and 78 as the angles, but that can't be right, since it adds up to 160 instead of 180. Maybe I'm doing it wrong. How should I set up the problem to figure out what n is equal to?

Thanks.

ummm... This isn't supposed to be in high school basics, but anyways:

5n+5=180, 5n=175, n=35, so the answers are: 35, 55, and 90.

Thanks. I don't know why I kept getting the wrong numbers. It's pretty simple.

Where was this thread supposed to go? I don't know where the line is drawn.

Actually, this does belong here, unless batteredbutnotdefeated thinks that this belongs in middle school math.


Let $ AB$ be a diameter of a circle and let $ C$ be a point of $ AB$ with $ 2\cdot AC \equal{} BC$. Let $ D$ and $ E$ be points on the circle such that $ DC \perp AB$ and $ DE$ is a second diameter. What is the ratio of the area of $ \triangle DCE$ to the area of $ \triangle ABD$?

$ \textbf{(A)}\ \frac{1}{6}\qquad
\textbf{(B)}\ \frac{1}{4}\qquad
\textbf{(C)}\ \frac{1}{3}\qquad
\textbf{(D)}\ \frac{1}{2}\qquad
\textbf{(E)}\ \frac{2}{3}$

Not too bad.  In fact, we don't even have to draw a line:
[hide]Let the center of the circle be $ O$.  Let the radius of the circle be $ r$.  Therefore, we have that $ 3AC\equal{}2r$, or
$ r\equal{}\frac{3AC}{2}$. Let $ AC\equal{}a$ and $ CD\equal{}h$. We see that the area of $ \triangle{ABD}$ is $ \frac{3ha}{2}$.  This is where the insight comes in.  We see that, since $ DE$ is a diameter of the circle, it goes through $ O$.  Therefore, the area of $ \triangle {ECD}$=$ 2 \bullet \triangle{DCO}$ (why?).  But the area of $ \triangle {DCO}$ is easy to find!  It's just $ r\minus{}AC\equal{}
\frac{3a}{2}\minus{}a\equal{}\frac{a}{2}$.  Hence, the area of $ \triangle {DOC}\equal{}[\frac{(\frac{a}{2})(h)}{2}] \bullet 2\equal{}\frac{ah}{2}$.  Thus, the ratio of areas is $ \frac{\frac{ah}{2}}{\frac{3ha}{2}}\equal{}\boxed {\frac{1}{3}} \longrightarrow (C)$.[/hide]


geometry
inradius
trigonometry
geometry proposed

Let $ABC$ be a triangle with $AB=AC$ and let $M$ be a mobile point of the line $BC$, such that $B$ lies between $M$ and $C$. Prove that the sum of the inradius of the triangle $AMB$ and the sum of the exradius of the excircle tangent to $AC$ of the triangle $AMC$ is constant.

Note that [tex]r_{AMB}=\frac12(AM+MB-AB)\tan\frac{\angle AMB}2[/tex], and [tex]r_{AMC}^M=\frac12(AM+MC+CA)\tan\frac{\angle AMB}2[/tex].
That gives the sum of the radii is [tex]\tan\frac{\angle BMA}2(AM+MD)[/tex].
Let [tex]\frac12\angle BMA=\theta[/tex], we have [tex]\tan\theta(AM+MD)=AD\tan\theta(\csc2\theta+\cot2\theta)=AD[/tex].
Q. E. D.

(Note: D is the midpoint of BC)


vector
function
algebra unsolved
algebra

Find all functions $f: \mathbb R \to \mathbb R$ such that
\[ f(x+f(y))=y+f(x+1),\]
for all $x,y \in \mathbb R$.

[quote="Anni"]find all f:R-R such that

    f(x+f(y))=y+f(x+1)[/quote]

Let $ g(x) \equal{} f(x) \minus{} 1$ the equation becomes $ g(x \plus{} g(y)) \equal{} y \plus{} g(x)$

$ x \equal{} 0$ $ \implies$ $ g(g(y)) \equal{} y \plus{} g(0)$ and so $ g(x)$ is bijective.
Then $ y \equal{} 0$ $ \implies$ $ g(x \plus{} g(0)) \equal{} g(x)$ and so $ g(0) \equal{} 0$ since $ g(x)$ is bijective and we have $ g(g(x)) \equal{} x$

Then $ y \equal{} g(z)$ $ \implies$ $ g(x \plus{} z) \equal{} g(x) \plus{} g(z)$

So we have $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$ and $ g(g(x)) \equal{} x$

The first part is a classical Cauchy equation and we conclude :

The only continuous solutions are $ f(x) \equal{} x \plus{} 1$ and $ f(x) \equal{} 1 \minus{} x$
Infinitely many non continuous solutions (with AC) exist.

In fact all solutions may be presented in a unique method :
Let $ \mathbb R$ considered as $ \mathbb Q$-vector space and $ A$ any $ \mathbb Q$-vector subspace.
Let $ B$ any supplementary (hope this is the good english word) vector subspace of $ A$ ($ B$ always exists with Axiom of Choice)

Let $ a(x)$ and $ b(x)$ the projection functions of $ x$ in $ A$ and $ B$.

Then $ f(x) \equal{} 1 \plus{} a(x) \minus{} b(x)$

$ A \equal{} \mathbb R$ gives $ f(x) \equal{} x \plus{} 1$
$ A \equal{} \{0\}$ gives $ f(x) \equal{} 1 \minus{} x$
any other $ A$ give non continuous solutions.

[u]edit [/u]: sorry, but there are a lot of other non continuous solutions

Fix $x$ and vary $y$. This shows that $f$ is surjective. Next since
\[y = f(x+f(y))-f(x+1) \implies f \text{ is injective}\]
Take $x = x+1$ to get
\[f(x-1+f(y)) = y+f(x) \implies f(x+y) = f(x)+f(y)-1\]
where we took $y = f(y)-1$ since $f$ is surjective. Now let $g(x) = f(x)-1$, so $g$ satisfies the additive Cauchy FE. Furthermore, by plugging in $g(x) = f(x)+1$ into the given assertion tells us that $g(g(y)) = y$ for all real $y$. Now one can easily verify all
\[\boxed{f(x) = g(x)+1} \text{ where $g$ is an involution that satisfies $g(x+y) = g(x)+g(y)$}\]
works.

[quote=tadpoleloop][quote=naonaoaz]Fix $x$ and vary $y$. This shows that $f$ is surjective. Next since
\[y = f(x+f(y))-f(x+1) \implies f \text{ is injective}\]
Take $x = x+1$ to get
\[f(x-1+f(y)) = y+f(x) \implies f(x+y) = f(x)+f(y)-1\]
where we took $y = f(y)-1$ since $f$ is surjective. Now let $g(x) = f(x)-1$, so $g$ satisfies the additive Cauchy FE. Now one can easily verify all
\[\boxed{f(x) = g(x)+1} \text{ where $g$ satisfies $g(x+y) = g(x)+g(y)$}\]
works.[/quote]

$f(x) = 1$ Clearly doesn't work.

Edit: If you changed it so that $g$ is any additive involution, then it is correct.[/quote]

Thanks for pointing this out! I totally missed that condition on $g$.


Let's start up a thread for asking how to do specific problems.

I'll start off by asking for number 12, the only one I have left:

The roots of $x^2+mx+n=0$ are double the roots of $x^2+px+m=0$ and $m$, $n$, and $p$ are not zero. What is $n/p$?

Let's say the roots of x^2+px+m=0 are a and b, so that p=-(a+b) and m=ab. Then the roots of x^2+mx+n=0 are 2a and 2b, so that m=-2(a+b) and n=4ab. Then, $\frac{n}{p}=\frac{4ab}{-(a+b)}=\frac{4m}{\frac{m}{2}}=\mathbf{8}$.

I really liked that problem.  Unfortunately I made a careless error on it and realized how to do the problem about 5 minutes after the test ended.  So I got a 96.  I knew I was going to be the kid that missed it by one question.  Anyway I'm  only a freshman.  So I was confused about #11, can someone help me with it.

You need to use combinations.

You have 8 bills. You need to add up two of them to over 20. Therefore, there are five possibilities for successful combinations, the combinations are listed in brackets next to the fraction for that probability. Add up the probabilities for all of these events to get one half.

(2c2)/(8c2) + (2c1)(2c1)/(8c2) + (2c1)(2c1)/(8c2) + (2c1)(2c1)/(8c2) + (2c2)/(8c2)

When you add up all of these fractions, it adds to 1/2.

This is also a pretty easy one to do on paper.

If you just list out all of the possible orders, you can find that there are really only 16 possible outcomes, of these possible outcomes, 8 will add up to over $20.

20-20
20-10
20-5
20-1
10-20
10-10
10-5
10-1
5-20
5-10

...and so on. This method would take much longer if there were additional bills, but this was a pretty easy one that could be done without combinations.

[quote="Zratonx"]i dunno if i consider that one nice, yet it did kill off some scores at my school...

personally i like 21 or 22 (can't remember) where they asked about the logs and replcing them into x^3+y^3[/quote]

It was 23. I really liked that problem (partially because I solved it =P).

yeah #11 sent me home. i got a 98.5. not one question...one OMITTANCE off from the AIME. lol

A quicker way to solve #11 is with the equation P(two 10's)+1-P(no 20's).  This turns out to be 1/4*1/7+1-3/4*5/7=1/28+28/28-15/28=1/2.

P.S.  Frankly, I didn't answer this problem correctly on the test.  I got the solution from one of my peers.

[quote="Wumbate"]How about the question with the circle with center (k,0) for k>4 I think.  Can someone rewrite the question and then solve it please?[/quote]

Q: Circle with center $(0,k)$ where $k>6$. It's tangent to $y=6$, $y=x$, and $y=-x$. Find the radius.

[hide]
Since the point is $k$ above the origin and is tangent to $y=6$, the radius is $k-6$. Draw the radius perpendicular to $y=x$. Then you have a right triangle with sides $k-6$, $k-6$, and $k$. So $(k-6)\sqrt{2} = k \Rightarrow k=12+6\sqrt{2}$. So $r = k-6 = 6+6\sqrt{2}$.[/hide]

Does anyone have nice solution to 24, I cant get it. aaaaaargh :(

can some one type up 24...cause i dunoo remember...and would like to solve it again

Alternate sol for 23 (the log one):

Notice that if z=0, then x^3+y^3=a+b. So set z=0, to find x+y=1 and x^2+y^2=10. This gives 10=(x^2+y^2)(x+y)=(x^3+y^3)+(x^2y+xy^2)=A+B. Also, 1=(x+y)^3=(x^3+y^3)+3(x^2y+xy^2)=A+3B. Solving this system for A gives A=[b]29/2[/b].

A solution to #17 would be appreciated.

Thanks.

The solution to 24 is in the math jam transcript. Go to the tab at the top and click Transcripts.

For 17, write the equation out in its exponential form. After some simplification:
$2^a3^b5^c7^d=10^{2005}$

Since $10^{2005}$ has only factors of 2 and 5 and a, b, c, and d are restricted to the rational numbers, the only solution is {2005, 0, 2005, 0}. If you have b and d not equal to 0, there will be extra factors and roots to deal with.


1) there is a parallelogram with area 2. prove that any triangle made by any 3 points in the parallelogram has an area smaller or equal to 1

2) there is a triangle with area 4. prove that if we choose 7 points in the triangle and make triangles with any 3 of them, at least 1 triangle has an area smaller than 1.

[hide="1"]
Suppose we have a triangle ABC in the parallelogram. If we can move C so that it is as distant as possible from line AB but still stays within the parallelogram, we will increase the area. So draw a line parallel to AB, and keep moving it away from AB until it goes through a vertex and can't go any further or it would be outside the parallelogram.Then if we move C to this vertex, the area of the triangle will increase. We can make the same argument with point B, so the area will increase if we move B to another vertex. Finally, we can increase the area by moving A to the opposite side of the parallelogram that is BC. Thus, we have shown that for any triangle for which any vertices aren't at a vertex of the parallelogram, we can move them to some vertices of the parallelogram and move the final point to the opposite side to increase the area to half that of the parallelogram, since all triangles of this form have this area. So a triangle of greatest area must be of this form, or otherwise we could could increase it as I mentioned to half the area[/hide]

[hide="1. Not a real proof, but..."]
1. If you choose three points in a parallelogram, this means that the area is maximized when you choose three different points in the actual shape itself (i.e. on the perimeter). This means that you will use exactly half of the parallelogram, meaning that you use at most an area of 1. QED
[/hide]

[hide="2. Again, not a real solution..."]
Suppose you are allowed to choose points on the triangle. If we choose all the corners of the triangle and the midpoints of the triangle, we have four triangles with area 1. However, we are not allowed to choose points on the triangle, only points within the triangle, so at least one of the triangles has to have area under one. This is especially true when we choose seven points instead of six, because it will be much harder to place the seventh point such that it cannot be used to make a triangle with area less than one. QED
[/hide]

As I mentioned, these aren't "real" proofs; I'm not to good at those, I guess. Is there a proof-worthy way to solve this?

2: [hide]If you split the triangle into 4 smaller triangles, notice that by joining the middle triangle with an edge one you create, most conveniently, a parrallelogram with area 2  :lol: . A bit of poking will show you that at least one of these parrallelograms must have at least 3 points in it. And thus it reduces to problem 1  :)[/hide]


Let $ n>2, n\in \mathbb N$ and $ a_1,a_2,...,a_n>0$ with $ a_1a_2...a_n\equal{}1$. Find the smallest constant $ k$ so that the following  inequality holds:

$ \dfrac{a_1a_2}{(a_1^2\plus{}a_2)(a_1\plus{}a_2^2)}\plus{}\dfrac{a_2a_3}{(a_2^2\plus{}a_3)(a_2\plus{}a_3^2)}\plus{}...\plus{}\dfrac{a_na_1}{(a_n^2\plus{}a_1)(a_n\plus{}a_1^2)}\leq k$

[quote="Inequalities Master"]Let $ n > 2, n\in \mathbb N$ and $ a_1,a_2,...,a_n > 0$ with $ a_1a_2...a_n \equal{} 1$. Find the smallest constant $ k$ so that the following  inequality holds:

$ \dfrac{a_1a_2}{(a_1^2 \plus{} a_2)(a_1 \plus{} a_2^2)} \plus{} \dfrac{a_2a_3}{(a_2^2 \plus{} a_3)(a_2 \plus{} a_3^2)} \plus{} ... \plus{} \dfrac{a_na_1}{(a_n^2 \plus{} a_1)(a_n \plus{} a_1^2)}\leq k$[/quote]

$ k\equal{}n\minus{}2$ is the best constant for this, see also in [b]Old and New Inequality I[/b]. :)

[quote="Honey_S"][quote="Inequalities Master"]Let $ n > 2, n\in \mathbb N$ and $ a_1,a_2,...,a_n > 0$ with $ a_1a_2...a_n \equal{} 1$. Find the smallest constant $ k$ so that the following  inequality holds:

$ \dfrac{a_1a_2}{(a_1^2 \plus{} a_2)(a_1 \plus{} a_2^2)} \plus{} \dfrac{a_2a_3}{(a_2^2 \plus{} a_3)(a_2 \plus{} a_3^2)} \plus{} ... \plus{} \dfrac{a_na_1}{(a_n^2 \plus{} a_1)(a_n \plus{} a_1^2)}\leq k$[/quote]

$ k \equal{} n \minus{} 2$ is the best constant for this, see also in [b]Old and New Inequality I[/b]. :)[/quote]

But I would like to see a proof, not just an answer.


modular arithmetic
quadratics
number theory unsolved
number theory

determine all positive integers n such that $ \frac{3^n \minus{}1}{2^n \minus{}1}$ is a integer :roll:

[quote="kihe_freety5"]determine all positive integers n such that $ \frac {3^n \minus{} 1}{2^n \minus{} 1}$ is a integer :roll:[/quote]
for $ n\equal{}1$ we have $ \frac{3\minus{}1}{2\minus{}1}\in \mathbb{N}$
$ 2^n\minus{}1$ is odd for all $ n\in \mathbb{N}$
Moreover
$ 3^{n}\minus{}1$ is even
therefore $ n\equal{}1$ is only solution satisfies $ \frac{3^n\minus{}1}{2^n\minus{}1}\in \mathbb{N}$

your solution is wroong!!!
example
6=3.2!!  :mad:

Why are you so ignorantly stressing his wrong solution. Just say you made a mistake at *point :P* and don't be so rude.
Until you offer the solution, you shouldn't say WRONG!!.|

I barely understood what he was talking about. Anyways I think he means an odd number can divide into an even number. Just the converse isn't true.
For instance 5 divides 10. 2 doesn't divide... well any odd.

Maybe this will help you a bit.

$ \frac{3^n\minus{}1}{2^n\minus{}1} \equal{} \frac{\left(3\minus{}1\right)\left[3^{n\minus{}1}\plus{}...\plus{}1\right]}{\left(2\minus{}1\right)\left[2^{n\minus{}1}\plus{}...\plus{}1\right]}$

$ \equal{} 2\frac{\left[3^{n\minus{}1}\plus{}...\plus{}1\right]}{\left[2^{n\minus{}1}\plus{}...\plus{}1\right]}$

[quote="kihe_freety5"]determine all positive integers n such that $ \frac {3^n \minus{} 1}{2^n \minus{} 1}$ is a integer :roll:[/quote]

hi  $ 3^{n}\equiv 1[2]$ ==> $ 3^{n} \minus{} 1 \equiv 0[2]$ and $ 2^{n} \minus{} 1 \equiv 1[2]$ so $ 2^{n} \minus{} 1 \equal{} 1$ that mean there is one value of $ n$ it^$ 1$

Ibn Rochde, you have made a mistake. If we go by your idea, we must observe some number that divides the denominator, and it doesn't divide the numerator. You switched sides. If the question was:

Is $ \frac {4^n\minus{}1}{3^n\minus{}1}$ a integer, then you can go by yours.

Can anyone finish this question? >_<
So far we have two ways, but both are wrong with a coincidently right answer for one case ><

Here's a solution, although it's not entirely elementary. Hopefully someone can simplify it. The answer is, indeed, that the only such $ n$ is 1.

[hide="Solution."]
First suppose that $ n$ is even. Then $ 2^n\minus{}1$ is divisible by 3, but $ 3^n\minus{}1$ obviously isn't, so $ \frac{3^n\minus{}1}{2^n\minus{}1}$ cannot be an integer. Therefore we may assume $ n$ is odd. We will also assume that $ n > 1$.

Let $ M\equal{}2^n\minus{}1$, and assume by way of contradiction that $ 3^n\minus{}1$ is divisible by $ M$. This means that

$ 3^n \equiv 1 \pmod{M}$

so in particular, for every prime divisor $ p$ of $ M$ we must also have

(*) $ 3^n \equiv 1 \pmod{p}$.

Consider the residue of $ M$ modulo 12. $ M$ must be congruent to 3 mod 4 since $ n>1$, and since $ n$ is odd it must also be congruent to 1 mod 3. Therefore $ M \equiv 7 \pmod{12}$. From this it follows that $ M$ must have a prime factor congruent to 5 or 7 modulo 12: if all the prime factors were 1 or 11 modulo 12, $ M$ would also be 1 or 11 modulo 12, which it isn't.

So let $ p$ be a prime factor of $ M$ such that $ p \equiv \pm 5 \pmod{12}$ . To finish the proof we will show that the order of 3 modulo $ p$ must be even. This means that whenever $ 3^m \equiv 1 \pmod{p}$ we must have $ m$ even, but this contradicts equation (*) above since $ n$ is odd.

I may be missing an elementary way of showing this, but here's a quick proof: using the quadratic reciprocity theorem it's easy to see that $ \left(\frac{3}{p}\right)$ is 1 precisely when $ p$ is 1 or 11 modulo 12, while if $ p$ is 5 or 7 modulo 12 than 3 is a quadratic non-residue. In other words,

$ 3^\frac{p\minus{}1}{2} \equiv \minus{}1 \pmod{p}$

so the order of 3 modulo p, which is a divisor of $ p\minus{}1$, is [b]not [/b]a divisor of $ (p\minus{}1)/2$, so it must be an even number. QED.

[/hide]

[quote="Ibn Rochde"][quote="kihe_freety5"]determine all positive integers n such that $ \frac {3^n \minus{} 1}{2^n \minus{} 1}$ is a integer :roll:[/quote]

hi  $ 3^{n} \equiv 1[2]$ ==> $ 3^{n} \minus{} 1 \equiv 0[2]$ and $ 2^{n} \minus{} 1 \equiv 1[2]$ so $ 2^{n} \minus{} 1 \equal{} 1$ that mean there is one value of $ n$ it^$ 1$[/quote]
I don't think it is true. If you write $ 3^{n} \minus{} 1 \equiv 1[k]$ and $ 2^{n} \minus{} 1 \equiv 0[k]$, it'll true. But as you say, i think it's false


AMC USA(J)MO USAMO LaTeX invariant induction symmetry

At the vertices of a regular hexagon are written six nonnegative integers whose sum is $2003^{2003}$. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Bert can make a sequence of moves, after which the number 0 appears at all six vertices.

When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Please use LaTeX for posting solutions. Thanks.

Just to make a small correction, the original text of the problem (via Kalva) says $2003^{2003}$

We do the following moves alternately: (a) from a position with odd sum to a position with exactly one odd number; (b) from a position with exactly one odd number to a position with odd sum, and with smaller maximum too, or to the all-zero position. The maximum value never increases and move (b) decreases the maximum, so this process must terminate at the all-zero position. Now we shall prove that the moves are always possible. Assume we have a position with odd sum $ (a,b,c,d,e,f)$, so either $ a+c+e$ is odd or $ b+d+f$ is odd. Without loss of generality assume $ a+c+e$ is odd. Now consider everything modulo 2. If one of them is odd, say $ a$, we can do the following moves: $ (1,b,0,d,0,f)\rightarrow(1,1,0,0,0,1)\rightarrow(0,1,0,0,0,1)\rightarrow(0,1,0,0,0,0)$. If all three of them are odd, we do the following: $ (1,b,1,d,1,f)\rightarrow(1,0,1,0,1,0)\rightarrow(1,0,0,0,0,0)$. Therefore move (a) is always possible. Now assume we have a position with exactly one odd number, say $ a$, and the maximum is $ m$.

(i) First case, $ m$ is even. The following moves give us a position with odd sum:
$ (1,0,0,0,0,0)\rightarrow(1,1,0,0,0,0)\rightarrow(1,1,1,0,0,0)\rightarrow(1,1,1,1,0,0)\rightarrow(1,1,1,1,1,0)$
Call this last position $ (a',b',c',d',e',f')$. Note that $ a',b',c',d',e'$ are odd. But $ m$ is even. Also, $ f'=|a'-e'|\le\max\{{a',e'\}<m}$. So the maximum value must be decreased.

(ii) Second case, $ m$ is odd. That is, $ m=a$. Assume $ c>0$. We do the moves:
$ (1,0,0,0,0,0)\rightarrow(1,1,0,0,0,0)\rightarrow(1,1,0,0,0,1)\rightarrow(0,1,0,0,0,1)\rightarrow(0,1,0,0,0,0)$
Call the last position $ (a',b',c',d',e',f')$. If the sum does not decrease, then the maximum is the odd number, that is $ b'$. But $ b'=|a-c|<a$. If $ e>0$, we do the same thing, interchanging $ b$ and $ c$ with $ f$ and $ e$ respectively. If $ c=e=0$, we can reach the desired position $ (a,b,0,d,0,f)\rightarrow(a,a,0,0,0,a)\rightarrow(0,a,0,0,0,a)\rightarrow(0,0,0,0,0,0)$.

So our proof is complete.

I'm not sure if this solution is correct because it seems too easy for a #6, but here goes:

We will show that Bert can make a sequence of moves such that the sum of all six integers, which we will denote in order as $ x_1, x_2, x_3, x_4, x_5, x_6$ will only decrease until it equals zero.  Since each move involves an absolute value, no integer will ever be negative, so when the sum of all six equals zero, it follows that each integer itself also equals zero.

Assume that at some point Bert cannot make a move that will decrease the sum of the integers.  In other words, he can only make moves that either do not  change or increase the sum.  Choose an arbitrary integer to perform the move on, say $ x_2$, and without loss of generality, let $ x_3 \ge x_1$.  It follows that $ x_3\minus{}x_1 \ge x_2$ and $ x_3 \ge x_2$.

Since every possible move will not decrease the sum of the integers, we also perform the move on $ x_3$.  Because $ x_3 \ge x_2$, we must have $ x_4 \ge x_3$ otherwise $ x_3$ will necessarily decrease when the move is performed.  So far we have $ x_2 \le x_3 \le x_4$, and by repeating the same argument over and over, we will have $ x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_1 \le x_2$ implying that $ x_1\equal{}x_2\equal{}x_3\equal{}x_4\equal{}x_5\equal{}x_6\equal{}c$.  If $ c>0$, then any move that Bert makes will obviously decrease the sum of the integers by creating a zero.  The sum will only stop decreasing once $ c\equal{}0$.  Contradiction, so each time it will be possible for Bert to make a move that decreases the sum of the integers until they all eventually equal 0.

Unfortunately, $ \{k,k,0,k,k,0\}$ is problematic (the $ x_4 \ge x_3$ assumption doesn't hold here).

any other holes?
I guess that one that azjps pointed out is pretty easy to correct since 2003 is odd.

It needs a [i]bit[/i] more fixing than that (perhaps a parity argument like Johan Gunardi's). One could start with $ (1,1,0,1,1,1999)$ and on the first move, remove $ 1999$ (in fact, your argument would remove $ 1999$, the only move that immediately decreases the sum). The only (afaik) case that messes up a simple decreasing invariant solution would be $ (k_1,k_1,0,k_2,k_2,0)$, but it's enough.

Hello,

Here is my solution which also uses a monovarient.

 [geogebra]8844b4a5554a71a813f9548c10f71ccbe4a80741[/geogebra]

Ok, let us use a,b,c,d,e,f to represent the values of the vertices. Now I claim that there always exists a move the strictly decreases the total sum a+b+c+d+e+f UNLESS one or more of the values is zero. To prove this, consider the following.

(>= means greater than or equal to)

b >= d,f.

Now if a + f > b then we can simply make the move a -> b-f which will decrease the total sum.

Likewise, if c+d > b then we can make a similar move that decreases the total sum. So assume that both of those equations are false, which means that

b >= a + f
b >= c + d

Now, wlog c >= a. So by the two equations above, a + b > c which means that the move b -> c-a will decrease the total sum. So every time, you can always decrease the total sum unless someone is 0 in which case you are already done.

QED

That solution is not complete - the question asks about ALL points being 0, not just one.

oh thanks.. Darn

well, I guess casework?? I haven't completely worked the cases out yet though

Hello,

We first prove that Bert is able to make a sequence of moves such that the numbers on the hexagon are alternately odd and even. This can be done with a lot of casework:

[b]Case 1:[/b] one vertex $P$ contains an odd number.
Then, apply a move to the vertices adjacent to $P$. Then, apply a move to the vertices which are two edges away from $P$. Then, again apply a move to the vertices adjacent to $P$, resulting in a hexagon with alternately odd and even numbers.

[b]Case 2:[/b] three vertices contain an odd number.
There are then two adjacent odd numbers (say, at $P$ and $Q$), else the numbers on the hexagon are already alternately odd and even. Let the third odd number be located at $R$. Then, $R$ is either adjacent to one of $P$, $Q$ or it isn't.

[b]Case 2a:[/b] $R$ is adjacent to one of $P$, $Q$.
WLOG, $R$ is adjacent to $Q$. Then, apply a move to the vertices which are two edges away from $Q$. Then, apply a move to $P$. Then, apply a move to $R$, resulting in a hexagon with alternately odd and even numbers.

[b]Case 2b:[/b] $R$ is not adjacent to one of $P$, $Q$.
WLOG, $R$ is closer to $Q$ than to $P$. Let $S \neq Q$ be adjacent to $P$. Then, apply a move to $S$. Then, apply a move to $P$, resulting in a hexagon with alternately even numbers.

[b]Case 3:[/b] five vertices contain an odd number.
Then there is a regular triangle $ABC$ such that $A, B, C$ contain odd numbers. A move can be applied to every vertex not in this regular triangle, resulting in a hexagon with alternately odd and even numbers.

Now, after applying three more moves, we arrive at a hexagon with numbers $(a, |a-b|, b, |b-c|, c, |c-a|)$ (in that order) written at its vertices, where $a, b, c$ are odd.

We will prove that it is possible to make a sequence of moves after which the number $0$ appears at all six vertices by strong induction on $3 \max\{a, b, c\} + f(a, b, c)$ where $f(a, b, c)$ is the number of elements of $\{a, b, c\}$ which are equal to $\max\{a, b, c\}$.

In the base case $3 \max\{a, b, c\} + f(a, b, c) \leq 6$, we must have $\max\{a, b, c\} = 1$, the hexagon has the numbers $(1, 0, 1, 0, 1, 0)$ written at its vertices, and moves can be applied to each vertex containing the number $1$.

Otherwise, $\max\{a, b, c\} > 1$. Because of the rotational and reflectional symmetry of the hexagon and the moves, we can WLOG assume $a \geq b \geq c$. If $a = b = c$, we simply apply a move to the vertices containing $a, b, c$, resulting in a hexagon containing all $0$'s.

Otherwise, $a > c$, and we can perform the moves: \[\begin{align*}
(a, a-b, b, b-c, c, a-c) \mapsto & (b-c, a-b, b, b-c, c, a-c) \\
\mapsto & (b-c, c, b, b-c, c, a-c) \\
\mapsto & (|a-2c|, c, b, b-c, c, a-c) \\
\mapsto & (|a-2c|, |b - |a-2c| |, b, b-c, c, a-c) \\
\end{align*}.
\]
This is again a hexagon containing numbers $(a', |a'-b'|, b', |b'-c'|, c', |c'-a'|)$ (where $a' = |a - 2c|, b' = b, c' = c$). It is also easy to see that $\max\{a', b', c'\} \leq \max\{a, b, c\}$. Also, since $a' < a$, if $\max\{a', b', c'\} = \max\{a, b, c\}$ then $b = a$ and we get $1 = f(a', b', c') < f(a, b, c) = 2$. Therefore, $3 \max\{a', b', c'\} + f(a', b', c') < 3 \max\{a, b, c\} + f(a, b, c)$. We can apply the inductive hypothesis to complete the induction.

Call a configuration of six numbers nice if its elements are $(0,1,0,1,0,1) \pmod{2}$.Our effort is to keep the sum of the numbers odd at every stage after the transformations.

I will first show that if the sum of the six numbers are odd,then we can reach a nice configuration from any such configuration.As the sum is odd,we have the following cases with the initial configurations at the start (ignoring symmetry) :

>>
 $(1,0,0,0,0,0) \rightarrow (1,1,0,0,0,0) \rightarrow (1,1,1,0,0,0) \rightarrow (1,1,1,1,0,0) \rightarrow (1,1,1,1,1,0) \rightarrow (1,0,1,0,1,0)$

>> $(1,0,1,0,1,0)$

>> $(1,1,0,0,1,0) \rightarrow (1,1,1,0,1,0) \rightarrow (1,0,1,0,1,0)$

>>
 $(1,1,1,0,0,0) \rightarrow (1,0,1,0,0,0) \rightarrow (1,0,1,1,0,0) \rightarrow (1,0,1,1,1,0) \rightarrow (1,0,1,0,1,0)$

We come back to our main problem.Taking a configuration with odd sum,translate it by a sequence of steps to a nice configuration.Now just consider the maximum number of the final configuration.If it is even,then it can be transformed so that the maximum is made smaller while keeping the sum odd.Consider the case when the maximum number $M$ is odd.We have the following cases:

<> If all the odd numbers are equal then the configuration is like $(M,a,M,b,M,c)$.Then $(M,a,M,b,M,c) \rightarrow (M,0,M,0,M,0) \rightarrow (0,0,0,0,0,0)$

<>If two odd numbers equal $M$,then we have a confsiguration like $(M,a,M,b,P,c)$.Then$(M,a,M,b,P,c) \rightarrow (|a-c|,a,M,b,P,c) \rightarrow (|a-c|,a,|a-b|,b,P,c)$

<>If there is only one maximum,then the configuration is like $(M,a,P,b,Q,c)$,Then $(M,a,P,b,Q,c) \rightarrow (|a-c|,a,P,b,Q,c) \rightarrow (|a-c|,a,|a-b|,b,Q,c)$

In each case we see that either the final configuration $(0,0,0,0,0,0)$ can be reached,or the maximum can be made smaller while keeping the sum odd at the same time.In the former case we are done,while in the later case we can again translate the configuration into a nice one and then making the maximum still smaller.We shall continue this algorithm as long as the maximum reaches $0$,when we must have $(0,0,0,0,0,0)$

[quote=funstar007]sorry for reviving the thread.
but can any body tell me if there's a flaw in my proof.
let the number on each vertices be $a,b,c,d,e,f$
maximum of $a,b,c,d,e,f$ would monotonically decrease at each step as if $a$ is the max then at the next move at A, it would change to $ |b-c|<a $ decreasing the a maximum . hence maximum would become 0 after finitely many move , implies all the numbers change to 0[/quote]

Same as mine ;)

[quote=funstar007]it would change to $ |b-c|<a $ decreasing the a maximum[/quote]

$|b-c|<a$ is not always true, say for $b=a,c=0$. Well it can be handled with some cases (and an extra move to reduce the maximum).

[quote=funstar007]sorry for reviving the thread.
but can any body tell me if there's a flaw in my proof.
let the nuber on each vertices be $a,b,c,d,e,f$
maximum of $a,b,c,d,e,f$ would monotonically decrease at each step as if $a$ is the max then at the next move at A, it would change to $ |b-c|<a $ decreasing the a maximum . hence maximum would become 0 after finitely many move , implies all the numbers change to 0[/quote]
Not at all.Consider this configuration (a,a,0,a,a,0).This configuration is a point of no return in the sense that Bert can no longer perform any move that will reduce the sum of the numbers on the hexagon or turn all of them into zero

I'm astonished that no one has mentioned in this thread that more generally
\[ (1 \bmod 2, 1 \bmod 2, 0 \bmod 2, 1 \bmod 2, 1 \bmod 2, 0 \bmod 2) \]
is a fatal configuration: one cannot change the parities. This is as I see it the core difficulty of the problem, it means that if you are not careful with parity (yes, we all want to overwrite the maximum) then you can walk Bert into a trap well before reaching the $(t,t,0,t,t,0)$ configuration. Thus the mistake pointed out several times in this thread (in #7, #16, #17) is [i]actually fatal[/i], you cannot just fix it at the end; it seems any correct solution has to worry about parity the entire time.

If $a \le b \le c$ are [i]odd[/i] integers, the configuration which has $(a,b-a,b,c-b,c,c-a)$ around the hexagon in some order (up to cyclic permutation and reflection) is said to be [i]great[/i] (picture below).

[b][color=red]Claim:[/color][/b]  	We can reach a great configuration 	from any configuration with odd sum.

[i]Proof.[/i]  	We should be able to find an equilateral triangle 	whose vertices have odd sum. 	If all three vertices are odd, then we are already done. 	Otherwise, operate as in the following picture (modulo $2$).

[asy] 	size(10cm); 	picture[] h = new picture[4]; 	h[0] = new picture; 	h[1] = new picture; 	h[2] = new picture; 	h[3] = new picture;

pen cu = blue + fontsize(9pt); 	pen cd = deepgreen + fontsize(7pt); 	pen mu = lightred + fontsize(9pt); 	pen md = orange + fontsize(7pt);

for (int i=0; i<=3; ++i) { 	draw(h[i], dir(30)--dir(150)--dir(270)--cycle, cd); 	draw(h[i], dir(90)--dir(210)--dir(330)--cycle, cu); 	draw(h[i], dir(30)--dir(90)--dir(150)--dir(210)--dir(270)--dir(330)--cycle, lightgrey); 	}

label(h[0], "$1$", dir(90), dir(90), cu); 	label(h[0], "$\ast$", dir(150), dir(150), cd); 	label(h[0], "$0$", dir(210), dir(210), cu); 	label(h[0], "$\ast$", dir(270), dir(270), cd); 	label(h[0], "$0$", dir(330), dir(330), cu); 	label(h[0], "$\ast$", dir(30), dir(30), cd);

label(h[1], "$1$", dir(90), dir(90), cu); 	label(h[1], "$1$", dir(150), dir(150), md); 	label(h[1], "$0$", dir(210), dir(210), cu); 	label(h[1], "$0$", dir(270), dir(270), md); 	label(h[1], "$0$", dir(330), dir(330), cu); 	label(h[1], "$1$", dir(30), dir(30), md);

label(h[2], "$1$", dir(90), dir(90), cu); 	label(h[2], "$1$", dir(150), dir(150), cd); 	label(h[2], "$1$", dir(210), dir(210), mu); 	label(h[2], "$0$", dir(270), dir(270), cd); 	label(h[2], "$1$", dir(330), dir(330), mu); 	label(h[2], "$1$", dir(30), dir(30), cd);

label(h[3], "$1$", dir(90), dir(90), cu); 	label(h[3], "$0$", dir(150), dir(150), md); 	label(h[3], "$1$", dir(210), dir(210), cu); 	label(h[3], "$0$", dir(270), dir(270), md); 	label(h[3], "$1$", dir(330), dir(330), cu); 	label(h[3], "$0$", dir(30), dir(30), md);

add(shift(0,0)*h[0]); 	add(shift(3,0)*h[1]); 	add(shift(6,0)*h[2]); 	add(shift(9,0)*h[3]); 	[/asy]

Thus we arrived at a great configuration. $\blacksquare$

[b][color=red]Claim:[/color][/b]  	Bert's goal is possible for all great configurations.

[i]Proof.[/i]  	If $a=b=c$ then we have $(t,0,t,0,t,0)$ which is obviously winnable.

Otherwise, 	perform six moves as shown in the diagram to reach a new 	great configuration whose odd entries 	are $b$, $|c-2a|$, $\left\lvert |c-2b|-(c-a) \right\rvert$ 	(and perform three more moves to get the even numbers). 	The idea is to show the largest odd entry has decreased.

[asy] 	size(12.5cm); 	picture[] h = new picture[5]; 	h[0] = new picture; 	h[1] = new picture; 	h[2] = new picture; 	h[3] = new picture; 	h[4] = new picture;

pen cu = blue + fontsize(9pt); 	pen cd = deepgreen + fontsize(7pt); 	pen mu = lightred + fontsize(9pt); 	pen md = orange + fontsize(7pt);

for (int i=0; i<=4; ++i) { 	draw(h[i], dir(30)--dir(150)--dir(270)--cycle, cd); 	draw(h[i], dir(90)--dir(210)--dir(330)--cycle, cu); 	draw(h[i], dir(30)--dir(90)--dir(150)--dir(210)--dir(270)--dir(330)--cycle, lightgrey); 	}

label(h[0], "$a$", dir(90), dir(90), cu); 	label(h[0], "$b-a$", dir(150), dir(150), cd); 	label(h[0], "$b$", dir(210), dir(210), cu); 	label(h[0], "$c-b$", dir(270), dir(270), cd); 	label(h[0], "$c$", dir(330), dir(330), cu); 	label(h[0], "$c-a$", dir(30), dir(30), cd);

label(h[1], "$c-b$", dir(90), dir(90), mu); 	label(h[1], "$b-a$", dir(150), dir(150), cd); 	label(h[1], "$b$", dir(210), dir(210), cu); 	label(h[1], "$c-b$", dir(270), dir(270), cd); 	label(h[1], "$b-a$", dir(330), dir(330), mu); 	label(h[1], "$c-a$", dir(30), dir(30), cd);

label(h[2], "$c-b$", dir(90), dir(90), cu); 	label(h[2], "$|c-2b|$", dir(150), dir(150), md); 	label(h[2], "$b$", dir(210), dir(210), cu); 	label(h[2], "$a$", dir(270), dir(270), md); 	label(h[2], "$b-a$", dir(330), dir(330), cu); 	label(h[2], "$c-a$", dir(30), dir(30), cd);

label(h[3], "$||c-2b|-(c-a)|$", dir(90), dir(90), mu); 	label(h[3], "$|c-2b|$", dir(150), dir(150), cd); 	label(h[3], "$b$", dir(210), dir(210), cu); 	label(h[3], "$a$", dir(270), dir(270), cd); 	label(h[3], "$|c-2a|$", dir(330), dir(330), mu); 	label(h[3], "$c-a$", dir(30), dir(30), cd);

label(h[4], "$||c-2b|-(c-a)|$", dir(90), dir(90), cu); 	label(h[4], "$b$", dir(210), dir(210), cu); 	label(h[4], "$|c-2a|$", dir(330), dir(330), cu);

add(shift(0,3)*h[0]); 	add(shift(3,3)*h[1]); 	add(shift(6,3)*h[2]); 	add(shift(1,0)*h[3]); 	add(shift(5,0)*h[4]); 	[/asy]

This is annoying, but straightforward. 	Our standing assumption is $a \ne c$ (but possibly $b=c$). 	It's already obvious that $|c-2a| < c$, 	so focus on the last term. 	If $c > 2b$, then $\left\lvert (c-2b)-(c-a) \right\rvert = 	\left\lvert 2b-a \right\rvert < c$ as well for $a \ne c$. 	When $c \le 2b$ we instead have 	$\left\lvert (2b-c)-(c-a) \right\rvert 	\le \max \left( 2b-c, c-a \right)$ 	with equality if and only if $c-a = 0$; but $2b-c \le c$ as needed. 	Thus, in all situations we have 	\[ c \ne a \implies 		\max\left( \left\lvert \left\lvert c-2b \right\rvert - (c-a) 		\right\rvert, \left\lvert c-2a \right\rvert \right) < c. \]

Now denote the new odd entries by $a' \le b' \le c'$ (in some order). 	If $b < c$ then $c' < c$, 	while if $b = c$ then $c' = b$ but $b' < c = b$. 	Thus $(c', b', a')$ precedes $(c, b, a)$ lexicographically, and we 	can induct down. $\blacksquare$

In any configuration, we label the vertices $1$ to $6$ and let $a_i$ be the value of the number at the vertex labelled $i$ and $a_{i+6} = a_i$ Call a configuration "bad" if there exists an $i$ such that $v_2(a_i) = v_2(a_{i+1}) = v_2(a_{i+3}) = v_2(a_{i+4}) = a$ for some $a$, and $v_2(a_{i+2}), v_2(a_{i+5}) > a.$ We claim that a configuration can reach all 0's if and only if it is not bad. The proof is kind of long but the main idea is to show that if our configuration is not of the above type, we can make a sequence of moves such that $\sum_{i=1}^6 a_i$ decreases and the configuration is still not of that type. This also solves the problem.

Outline (Incomplete):

1. Show that if a configuration is not bad, we can always decrease $\sum_{i=1}^6 a_i$ via a sequence of moves. In fact, we can even do this in one move unless, for some $a$ and $b,$ there exists an $i$ such that $a_i = a_{i+1} = a,$ $a_{i+2} = a_{i+5} = 0,$ and $a_{i+3} = a_{i+4} = b,$ in which case we can manually find a sequence of moves to decrease the total sum such that the resulting configuration is not bad, given that the original configuration is not bad, so we assume for the rest of the proof that we can decrease the total sum in one move from now on.
2. Now, assume that by making a move decreasing the sum, we result in a bad configuration such that $v_2(a_1) = v_2(a_2) = v_2(a_4) = v_2(a_5).$ It remains to consider 2 cases - if the last move made was at vertex $1$ and if the last move made was at vertex $6$ (other cases follow by symmetry).
2.1. If the last move made was at vertex $1,$ we split into cases based on whether $a_1$ is maximal or not.
2.2. If the last move made was at vertex $6,$ the proof is similar to above and is omitted for now.

Please help me and tell me why my solution is wrong.

[b] Algorithm: [/b] At every step we choose the maximum of those numbers and apply the process (i.e. write the absolute value of the different of the adjacent numbers).
It clearly decrease the sum of the numbers and it is monovariant. Hence, At some moment we will reach $0$.

Where is the mistake? :(
@below Thanks! Actually I tried it and thought that this can be solution. I haven't read the previous posts because I want to do it on my own. :)

[quote=ayan_mathematics_king]Please help me and tell me why my solution is wrong.

[b] Algorithm: [/b] At every step we choose the maximum of those numbers and apply the process (i.e. write the absolute value of the different of the adjacent numbers).
It clearly decrease the sum of the numbers and it is monovariant. Hence, At some moment we will reach $0$.

Where is the mistake? :([/quote]

Read posts 6, 16, 17, 18 above.

I appear to have a different solution, that is essentially the natural "remove maximum but avoid the fatal configuration mentioned in #18". It's possible it's just wrong, so could someone please check? Thanks!

The idea is to decrease the maximum value of the configuration while avoiding the patterns $(0,0,0,0,0,0)$ and $(0,1,1,0,1,1)$ mod $2$. Call a configuration $(x_1,x_2,x_3,x_4,x_5,x_6)$ [i]bad[/i] if it is part of one of the two previously mentioned patterns. Call a configuration $(x_1,x_2,x_3,x_4,x_5,x_6)$ [i]reducible[/i] if it is good and we can apply a sequence of moves to it such that the maximum value of the configuration strictly decreases (or the number of common maximums decreases) and the final configuration is good.

We have the following claim which shows that basically all good configurations are reducible.

[b]Claim:[/b]  Any good configuration $(x_1,\ldots,x_6)$ that is not of the form $(M,x,0,x,0,x)$ (or cyclic shifts) is good.

[i]Proof:[/i]  We have the following cases.

[b]Case 1[/b]: Suppose that all the $x_i$ are odd. In this case, simply apply the operation to the maximum value, since there is no way the resulting configuration can be bad. Thus, the configuration is reducible.

[b]Case 2[/b]: Suppose that five out of six of the $x_i$ are odd. WLOG assume $x_1$ is the only even one, so the parity configuration is $(1,0,0,0,0,0)$. The only move that would make the configuration bad is applying it on $x_4$, so if a maximum is not at $x_4$, the configuration is clearly reducible.

Now suppose the only maximum is at $x_4$. Note that we can make the sequence of moves \[(x_1,x_2,x_3,x_4,x_5,x_6)\to(x_1,x_2,x_4-x_2,x_4,x_5,x_6)\to(x_1,x_2,x_4-x_2,|(x_4-x_2)-x_5|,x_5,x_6),\] which clearly reduces the maximum since $x_2$ is odd, so at least $1$. Furthermore, the parity of this final configuration is $(0,1,0,1,1,1)$, so the configuration is reducible, as desired.

[b]Case 3[/b]: Suppose that four out of six of the $x_i$ are odd. Since the configuration is good, its parity is either $(1,1,1,0,1,0)$ or $(1,1,1,1,0,0)$ up to cyclic shifts. Applying a single move to either of these leads to a good configuration, so the configuration is reducible, as in case 1.

[b]Case 4[/b]: Suppose that three out of six of the $x_i$ are odd. Up to cyclic shifts and reflections, the only possible parity configurations are $(1,1,1,0,0,0)$, $(1,0,1,0,1,0)$, or $(1,1,0,1,0,0)$. In the first two cases, any move leads to a good configuration, so we may simply apply a move on the maximum to reduce. In the final case, the only bad move is on $x_5$, so WLOG assume that the only maximum is at $x_5$.

We may now apply the sequence of moves \[(x_1,x_2,x_3,x_4,x_5,x_6)\to(x_1,x_2,x_3,x_4,x_5,x_5-x_1)\to(x_1,x_2,x_3,x_4,|(x_5-x_1)-x_4|,x_5-x_1),\] which clearly reduces the maximum since $x_1$ is odd, so at least $1$. Furthermore, the parity of this final configuration is $(1,1,0,1,0,1)$, so the configuration is reducible, as desired.

[b]Case 5[/b]: Suppose that two out of six of the $x_i$ are odd. It is easy to see that any move leads to a good configuration, so the configuration is reducible simply by applying a move on the maximum.

[b]Case 6[/b]: Suppose that one out of size of the $x_i$ are odd. WLOG suppose $x_1$ is odd and the rest are even. Applying a move to anything other than $x_1$ leads to a good configuration, so suppose WLOG that the only maximum is at $x_1$.

If $x_3$ is nonzero, then we may apply the sequence of moves \[(x_1,x_2,x_3,x_4,x_5,x_6)\to(x_1,x_1-x_3,x_3,x_4,x_5,x_6)\to(|(x_1-x_3)-x_6)|,x_1-x_3,x_3,x_4,x_5,x_6).\] Since $x_3>0$, the maximum has decreased, and the new parity configuration is $(1,1,0,0,0,0)$, so the configuration is reducible.

Similarly if $x_5$ is nonzero, we can essentially flip the sequence and do the same thing as above, so WLOG suppose that $x_3=x_5=0$. If $x_2\ne x_4$, then we may do the move \[(x_1,x_2,x_3,x_4,x_5,x_6)\to(x_1,x_2,|x_4-x_2|,x_4,x_5,x_6).\] This has parity $(1,0,0,0,0,0)$, and the third element is nonzero, so it is reducible by the above logic. Thus, WLOG we have $x_2=x_4$, and similarly $x_6=x_4$.

Thus, the configuration is reducible as long as it is not of the form $(M,x,0,x,0,x)$, as desired. $\blacksquare$

Note that out original configuration has an odd sum, so it is good. By the above claim, we may keep reducing it until it is not longer reducible, so eventually we reach $(M,x,0,x,0,x)$. It is easy to see that we win in this configuration by applying \[(M,x,0,x,0,x)\to(0,x,0,x,0,x)\to(0,0,0,0,0,0),\] where the last step is three moves applied individually to each of the $x$s. This shows that the configuration can be reduced down to $(0,0,0,0,0,0)$, as desired.

Roughly the same as other solutions in the thread.

The only important fact about $2003^{2003}$ is that it is odd, which means that an odd number of the elements on the board are odd.
We will begin by reducing the given position to having exactly $1$ odd entry. If there is exactly $1$ odd entry to begin with, there is nothing to do. If there are $3$ odd entries, we have the following cases, up to reflection and rotation (where the entries are taken $\pmod 2$):
\begin{align*}
	(1, \mathbf 1, 1, 0, 0, 0) \to (1, 0, \mathbf 1, 0, 0, 0) \to (1, 0, 0, 0, 0, 0) \\
	(1, 1, 0, 1, 0, \mathbf 0) \to (\mathbf 1, 1, 0, 1, 0, 1) \to (0, \mathbf 1, 0, \mathbf 1, 0, 1) \to (0, 0, 0, 0, 0, 1) \\
	(1, 0, 1, 0, 1, 0) \to (1, 0, \mathbf 1, 0, \mathbf 1, 0) \to (1, 0, 0, 0, 0, 0).
\end{align*}
Finally, if there are $5$ odd entries, we have
\begin{align*}
	(1, \mathbf 1, 1, \mathbf 1, 1, 0) \to (1, 0, 1, 0, 1, 0),
\end{align*}
whence we can finish as before.

Henceforth, we assume that there is exactly one odd entry on the board. Let the hexagon be $ABCDEF$, and suppose the number at $A$ is odd while the rest are even. We show an algorithm which strictly decreases the maximum element while maintaining the fact that the only odd number on the board is at $A$.

First, if $3$ of the entries on the board are zero, we are able to finish directly. Indeed, suppose there are at least $3$ zeroes, and note that $A$ is nonzero since it is odd. If there are some $3$ zeros which are pairwise all non-consecutive, we have
\begin{align*}
	(\mathbf A, 0, \mathbf C, 0, \mathbf E, 0) \to (0, 0, 0, 0, 0, 0).
\end{align*}
Otherwise, if, out of any $3$ zeros, some two are consecutive, we have, up to reflection, the following cases:
\begin{align*}
	(A, 0, C, 0, 0, \mathbf F) \to (A, 0, C, 0, \mathbf 0, A) \to (A, 0, C, 0, A, \mathbf A) \to (A, 0, C, 0, A, 0) \\
	(A, \mathbf B, 0, 0, 0, \mathbf F) \to (\mathbf A, A, 0, 0, 0, A) \to (0, A, 0, 0, 0, A).
\end{align*}
In either case, we arrive at a state where we have $3$ zeros no two of which are consecutive, whence we may finish as before.

Henceforth, we suppose at most $2$ of the entries on the board are $0$. WLOG, suppose $B \geq F$ (if not, reflect the board). If there are two diametrically opposed zeros, these are hence at $C$ and $F$ (since $B \geq F$). \\

If the maximum number is at $A$, we go by the following moves:
\begin{align*}
	(A, B, C, D, E, \mathbf F) \to (A, B, C, D, \mathbf E, F_1) \to (A, B, C, \mathbf D, E_2, F_1) \to \\
	(A, B, \mathbf C, D_3, E_2, F_1) \to (A, B, C_4, D_3, E_2, F_1).
\end{align*}
Note that $C_4, D_3, E_2, F_1$ are all odd, and that each of them is at most $A$ (since $B, C, D, E, F \leq A$, the absolute difference between any two of them is also at most $A$; then we can just iterate this). Now, from the position $(A, B, C_4, D_3, E_2, F_1)$, we operate on $A$ so as to arrive at the position $(A', B, C_4, D_3, E_2, F_1)$. Note that $A' = |B - F_1| \leq \max(B, F_1) \leq A$. Suppose $A' = A$. In this case, one of $B$ and $F_1$ is zero, while the other is equal to $A$. Since $B$ is even, in fact $B = 0$ and $F_1 = A$. Then, we have $F_1 = |A - E| = A$. As $E < A$ (since $E$ is even, $A$ is odd, and $E \leq A$), we require $E = 0$. Hence, we have diametrically opposed zeros at $B$ and $E$, impossible as we previously reflected the board so that any diametrically opposed zeros are at $C$ and $F$. Therefore, $A' < A$. Now, we do the following sequence of moves:
\begin{align*}
	(A', B, C_4, \mathbf D_3, E_2, \mathbf F_1) \to (A', B, \mathbf C_4, D', \mathbf E_2, F') \to (A', B, C', D', E', F').
\end{align*}
Now, $C', D', E', F'$ are all even numbers, and $0 < C_4, D_3, E_2, F_1 \leq A \implies C', D', E', F' < A$ as well. Therefore, the maximal element on the board is now less than $A$, and, as before, the only odd number on the board is at $A$, which completes this step of the algorithm. \\

Otherwise, suppose there is a maximum number at $B$. As before, we begin with the moves
\begin{align*}
	(A, B, C, D, E, \mathbf F) \to (A, B, C, D, \mathbf E, F_1) \to (A, B, C, \mathbf D, E_2, F_1) \to \\
	(A, B, \mathbf C, D_3, E_2, F_1) \to (A, B, C_4, D_3, E_2, F_1).
\end{align*}
Note that $B > A$ since $B$ is even and $A$ is odd. Hence, $F_1 = |A - E| < B$ is odd, and similarly $C_4, D_3, E_2$ are also odd numbers less than $B$. Now, we operate on $B$ so that $(A, B, C_4, D_3, E_2, F_1) \to (A, B', C_4, D_3, E_2, F_1)$, where $B' = |A - C_4| < B$. We then do the sequence of moves
\begin{align*}
	(A, B', C_4, \mathbf D_3, E_2, \mathbf F_1) \to (A, B', \mathbf C_4, D', \mathbf E_2, F') \to (A, B', C', D', E', F').
\end{align*}
Note that $C', D', E', F'$ are all even and less than $B$ (since $C_4, D_3, E_2, F_1$ are all odd and less than $B$), so, once again, we have maintained the invariant that $A$ is the only odd number, while strictly reducing the maximum.

Otherwise, suppose there is a maximum at $C$ or $E$; if necessary, reflect the board so that there is a maximum at $C$. Note that $C > \max(B, F)$: if $C = \max(B, F)$, we would have done the procedure in the previous paragraph. In the case that $(B, D) \neq (0, C)$, replacing $C$ by $|B - D|$ does the job. Suppose $(B, D) = (0, C)$. We have two cases: when $E = 0$, and when $E \neq 0$. If $E \neq 0$, we have
\begin{align*}
	(A, 0, C, \mathbf C, E, F) \to (A, 0, \mathbf C, C - E, E, F) \to (A, 0, C - E, C - E, E, F).
\end{align*}
If $C \neq E$, we have reduced the maximum while maintaining our invariant, while if $C = E$, we are done as we have $3$ zeros. If $E = 0$, we have
\begin{align*}
	&(A, \mathbf 0, C, C, 0, F) \to (A, C - A, \mathbf C, C, 0, F) \to (A, C - A, A, \mathbf C, 0, F) \to (A, C - A, \mathbf A, A, 0, F) \to \\
	&(A, \mathbf{C - A}, |C - 2A|, \mathbf A, 0, F) \to (A, |A - |C - 2A||, |C - 2A|, |C - 2A|, 0, F).
\end{align*}
As $0 < A < C$, we have $\max(A, |A - |C - 2A||, |C - 2A|, F) < C$. \\

Otherwise, suppose there is a maximum at $D$; this maximum must be unique, as otherwise we would have done the procedures detailed in one of the two prior paragraphs. As $\max(C, E) < D$, we go by $(A, B, C, \mathbf D, E, F) \to (A, B, C, |C - E|, E, F)$, which is enough as $|C - E| < D$. \\

Thus, at each step of the algorithm, we either finish, or we decrease the maximum number while keeping invariant the fact that the only odd number is at $A$. As the maximum number must always be nonnegative, the algorithm eventually terminates, so we are done. $\Box$

For conciseness, for a position $(a,b,c,d,e,f)$ on board, let $\max = \max(a,b,c,d,e,f)$ and $\Sigma = a+b+c+d+e+f$. We give a strategy that always decreases $\max + \Sigma$, which means that the board will eventually become all zero.

We use the following terminologies. A position is [i]bad[/i] if after dividing all numbers by their gcd, it is $(0,1,1,0,1,1)$ or its cyclic/symmetric variations modulo 2. A position is [i]dangerous[/i] if it is not bad and operating on any maximal number results in a bad position. A position is [i]stagnant[/i] if it is not bad and operating on any maximal number does not change the position. Notice that the starting position is not bad since $\Sigma$ is odd. The strategy is: if the position is not dangerous or stagnant, then operate on the largest number so that the resulting position is not bad and $\Sigma$ decreases; if the position is dangerous, we perform the moves specified in part 1 to get a position that is not dangerous. If the position is stagnant, we perform the moves specified in part 2 to get a dangerous position.

[b]Part 1.[/b] Because we use it repeatedly, the sequence of moves
\[(a,b,c,d,e,f) \to (a,b,d-b,d,d-f,f) \to (a,b,d-b,|b-f|,d-f,f)\]
is called a [i]moov centered at[/i] $d$.

It's not hard to verify that the only dangerous positions are the following four.
[list]
[*] $(a,b,c,d,e,f) \equiv (0,1,1,1,1,1)$ modulo 2 and $d$ is the unique maximal number. In this case, we do a moov centered at $d$, which results in $(0,1,0,0,0,1)$ modulo 2. Notice that $\max$ decreases since $b,f$ are both odd.
[*] $(a,b,c,d,e,f) \equiv (1,0,1,0,0,1)$ modulo 2 and $d$ is the unique maximal number. In this case, we do a moov centered at $d$, which results in $(1,0,0,1,1,1)$ modulo 2. Notice that $\max$ decreases unless $b=0$. If $b = 0$, we continue to do a moov centered at the third entry (which is equal to $d-b=d$), which results in $(1,1,0,1,1,1)$ modulo 2. Notice that now $\max$ strictly decreases since the first and fifth entries are both odd. Now we just use the strategy in the first item if necessary.
[*] $(a,b,c,d,e,f) \equiv (0,0,0,1,0,0)$ modulo 2 and $d$ is the unique maximal number. We do a moov centered at $d$, which results in $(0,0,1,0,1,0)$ modulo 2. Clearly $b$ and $f$ cannot both be zero, since otherwise $(a,b,c,|c-e|,e,f) = (a,0,c,|c-e|,e,0)$ does not result in a bad position after we divide all components by their common gcd. If $b,f$ are both nonzero, then $\max$ decreases. Otherwise, WLOG $b=0$. Then we do a moov centered at the third entry (which is $d-b=d$), which results in $(0,1,1,0,1,0)$ modulo 2. We know that $a\neq 0$, since otherwise $(a,b,c,|c-e|,e,f) = (0,0,c,|c-e|,e,f)$ does not result in a bad position after we divide all components by their common gcd. Therefore $\max$ decreases. Now we just use the strategy in the second item if necessary.
[*] $a,b,c,d,e,f$ are all even. In this case, just divide all numbers by their $\gcd$ if it exists, and if it does not, we're automatically done since there are at least five 0's.
[/list]
These all strictly decrease $\max$ until we win.

[b]Part 2.[/b] For a stagnant configuration $(a,b,c,d,e,f)$, WLOG $a = b = \max$ and $c=f=0$. In this case, we claim that we can move to a dangerous position. WLOG $d\ge e$. If $d,e$ do not have the same parity, we do
\[(a,a,0,d,e,0) \to (a,a,0,d,e,a-e) \to (e,a,0,d,e,a-e) \to (e,e,0,d,e,a-e) \to (e,e,0,d,e,0).\]
Otherwise, $a,d$ are of different parity, and we do
\[(a,a,0,d,e,0) \to (a,a,a-d,d,e,0) \to (a,d,a-d,d,e,0) \to (a,d,0,d,e,0).\]
The two cases we obtained are both dangerous, and furthermore this strictly decreases $\Sigma$.

We can also note that there's nothing special about the given number that was used in the solution unless it's odd.
    Note that the maximum never increases, and we want to have a strategy that monotonically decreases it. Our strategy will be an algorithm that 1. makes only one number remain odd and then 2. if the sum is odd, decreases the maximum, while preserving odd sum until we get all 0s. It's clear that if this works but maximum never increases, we would be done. Suppose the vertices are ABCDEF cyclically.
        Since A+B+C+D+E+F is odd, WLOG A+C+E is odd (why would we think of this? because you want to break it down into smaller parts and use an algorithm to guarantee parity of B D F).
        If only, say, A, is odd, in A,C,E, then I let everyone verify that BFDAF makes only B odd. (to find this just use intuition with easy testing, since parity is not guaranteed of B D F we need to change them, and A C E "hug" B D F) If A,C,E are all odd, then do BDF (in any order,) CE (in any order lol). Hence we have established our first goal again of making only one odd, this time is namely A. (this one's obvious to find on your own ;)) $(Algorithm 1)$ Now for the second step, rename them s.t. only A is odd.
        Case 1. Maximum is even.
In this case the maximum is one of BCDEF, then make moves BCDEF, whence it's obvious the maximum strictly decreases (unless there are two maximums, in which case we remove one of them, and then apply it again).
        Case 2. Maximum is odd (namely A).
In this case if C=E=0 the sequence BFDABF leaves all numbers 0 (the most annoying edge case, hardest to find but pretty easy to prove, since two zeros are a lot to work with and do obvious subtracting into 0 or making terms equal)
Otherwise WLOG C>0 and make the moves BFAF, from where it's easily seen that the maximum decreases with odd sum. (it's relatively easy to find this one, too) $(Algorithm 2)$
        Hence we do this until we get all 0s, applying algorithm 1 if needed. $\blacksquare$

An exercise in WISHFUL THINKING. Not an optimized solution

Let the state of the hexagon be $(a,b,c,d,e,f)$. We freely refer to "moves on $a$" and similar terms, which mean what you think they do. Refer to $a+b+c+d+e+f$ as the [i]sum[/i] and the parity of the sum as the [i]parity[/i]. Finally, use $\equiv$ to mean equivalent modulo $2$.

[b][color=#00f]Claim 1:[/color][/b] If all our numbers are nonzero at some point and have odd sum, then we can get some of them to equal zero such that no diametrically opposed numbers are both zero, while preserving the odd sum.
[color=#00f]Proof:[/color] If all our numbers are nonzero at some point and the maximum of the numbers is $M$, then $M$ cannot appear as the result of any operation. WLOG suppose that $a=M$. We will decrease the number of times $M$ occurs, while ensuring that we will not generate two diametrically opposed zeroes, and also preserving the fact that the six integers have odd sum, with the following caveat: if, in the middle of one run of the procedure, we generate a zero, then we no longer case about decreasing the number of times $M$ occurs (we still wish to avoid diametrically opposed zeros and preserve the parity). If a single run of the procedure does not involve moving on the same point twice, then any zeroes we generate will still be there. Otherwise, I will specifically explain why our procedure either leaves us with a zero or decreases the occurrences of the maximum.
[list]
[*] If moving on $a$ preserves the parity of $a$ (removing an occurrence of $M$), then do this and repeat/finish.
[*] Otherwise, if we may move on $a$ (removing an occurrence of $M$), and then move on either $b$ or $f$, and find that the parity is preserved, then do this and repeat/finish. Only $f,a,b$ can become zero, so no diametrically opposed zeroes appear.
[*] Otherwise, if $a$ is even, then by the first assumption, up to reflection we can uniquely determine $(f,b) \equiv (1,0)$, and then by the second we can uniquely determine $(e,c)\equiv (0,1)$, so then for parity reasons $d \equiv 1$. Then move on $f$, then $a$ (removing an occurrence of $M$), then on $e$, then repeat/finish. This will not generate any diametrically opposed zeroes.
[*] Otherwise, if $a$ is odd and $b$ is even, then we must have $f$ even by the first assumption, so then $(e,c) \equiv (0,0)$ as well by the second, and thus $d \equiv 0$ for parity reasons. Then move on $b$ and $f$, then $a$, then $b$ again, then repeat/finish. The first two moves cannot possibly generate zeroes for parity reasons, so we will remove an occurrence of $M$ when moving on $a$, and then after this we no longer care. Since we never move on diametrically opposed points we never generate any diametrically opposed zeroes.
[*] On the other hand, if $a$ and $b$ are both odd, then we can similarly get $f \equiv 1$, hence $(e,c) \equiv (1,1)$ and then $d \equiv 0$. It turns out we can still repeat the same order as before: $b$ and $f$, then $a$, then $b$ again, and repeat/finish. This works for the same reason as before.
[/list]
It is clear that this operation cannot be repeated indefinitely, so at some point we must end up with at least one zero, but never any diametrically opposed zeroes.

[b][color=#00f]Claim 2:[/color][/b] With one zero, WLOG at $a$, and odd sum, we can make $b=c$, $e=f$, $b \not \equiv f$, and preserve odd sum, $a=0$, and $d \neq 0$.
[color=#00f]Proof:[/color] If $c \not \equiv e$, then we can move on $f$ (making it odd, thus nonzero), then $b$, then $f$ and finish. Otherwise, if $d \equiv 0$, then we must have $b \not \equiv f$, so either $b \not \equiv c$ or $e \not \equiv f$; WLOG the former. Then move on $c$, then $d$. If $d \equiv 1$, then first move on $b$ and $f$ to make $b \equiv c \equiv e \equiv f$. Then move on $c$, then $b$, then $f$, which can be seen to work.

[b][color=#00f]Claim 3:[/color][/b] With one zero, WLOG at $a$, odd sum, and $b=c$, $e=f$ (so $d \equiv 1$ is nonzero), but $b \not \equiv f$, we can either make $b=c=0$ or $e=f=0$.
[color=#00f]Proof:[/color] First move on $d$, so it now equals $|b-f| \equiv 1 \neq 0$. Now WLOG let $b>f$, and move on $c$, $b$, $c$, $b$, after which both $b$ and $c$ now equal $|b-2d|$ (where we use the "starting" value of $b$ but the "post-operation" value of $d$), which is strictly less than $b$ since $d \leq b$. Furthermore, all the conditions are preserved, so eventually we will either reach $b=c=0$ or $e=f=0$, at which point we can skip ahead to claim 6.

[b][color=#00f]Claim 4:[/color][/b] If we have two diametrically opposed zeroes, odd sum, then we can return to all nonzero values with odd sum (and then apply claim 1).
[color=#00f]Proof:[/color] WLOG let $a=d=0$. Then we cannot have $b \equiv f$ and $c \equiv e$, so WLOG the former is false and additionally that $b \equiv 0$. If $c \equiv d \equiv 0$, then move on $a,f,e,d$, making them all odd, which is fine. Otherwise, $c \equiv d \equiv 1$, so move on $a,e,d,e$, making them all odd, which is also fine.

[b][color=#00f]Claim 5:[/color][/b] If we have exactly two zeroes which are adjacent, then we can return to one zero and odd sum (and then apply claim 2).
[color=#00f]Proof:[/color] WLOG let the two zeroes be $a$ and $b$. Then if either $f$ is even, we move at $a$, and likewise if $c$ is even we move at $b$. Thus suppose $c \equiv f \equiv 1$. Then one of $d$ or $e$, WLOG $d$, is even. Then move at $c$, which makes it even but nonzero, and then move at $b$, making it nonzero (and even), then at $d$, making it odd.

[b][color=#00f]Claim 6:[/color][/b] Otherwise (i.e. exactly two zeroes separated by exactly one point, or at least three zeroes), then we can make everything zero.
[color=#00f]Proof:[/color] First, if we have exactly two zeroes separated by exactly one point, then we move on that point and create three zeroes, so assume we have at least three zeroes. For any nonzero value adjacent to two zeroes, move on that value, making it zero. After we do this until we cannot, we are either left with six zeroes, in which case we're just done, or either three or four adjacent zeroes (and no others). In the former case, WLOG let $a=b=c=0$; then move on $d$ and $f$ to make $d=e=f$. In the latter, WLOG let $a=b=c=d=0$. Then move on $f$, then $d$, to make $d=e=f$ as well. Then, given the state $(0,0,0,k,k,k)$, move on $a$ and $c$, making them both equal to $k$, then on $f$ and $d$, making them both $0$, so our state is now $(k,0,k,0,k,0)$, and we win by moving on $a,c,e$.

These six claims finish the proof. $\blacksquare$

[b]Remark:[/b] The "flow" of the claims is claim 4 -> claim 1, claim 1 -> claim 2/3/5/6, claim 5 -> claim 2, claim 2 -> claim 3, claim 3 -> claim 6. Everything ends up at claim 6 eventually.


In a triangle with sides $ a, b, c$, $ m_{1}, m_{2}, m_{3}$ are the length of the three medians. Prove that $ (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(m_{1}+m_{2}+m_{3})>\frac{15}{2}$.

[quote="sxcxc"]In a triangle with sides $a, b, c$, $m_{1}, m_{2}, m_{3}$ are the length of the three mediums. Prove that $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(m_{1}+m_{2}+m_{3})>\frac{15}{2}$.[/quote]

What is a medium?

By "medium", I mean "median". Sorry for my poor English.

See my post in http://www.mathlinks.ro/Forum/viewtopic.php?t=5209 .

  darij


function
induction
algebra unsolved
algebra

Let $ Q^\plus{}$ denote the set of positive rational numbers. Show that there exist one and only one

function $ f$ takes $ Q^\plus{}$ to $ Q^\plus{}$ satisfing the following conditions:

(i)If $ 0<q<1/2$ then $ f(q)\equal{}1\plus{}f(q/(1\minus{}2q))$,

(ii)If $ 1<q\leq 2$ then $ f(q)\equal{}1\plus{}f(q\minus{}1)$,

(iii)$ f(q).f(1/q)\equal{}1$ for all $ q \in Q^\plus{}$

I'm sorry ,the following solution may be incomplete or even wrong .

    Let's suppose that there is at least one such function .

f(1)f(1) = 1 then  f(1)^2 = 1 ,since range of f is Q+  ,  f(1) = 1  .
f(2) = 1+f(1) = 2

If m is an integer and m> 2   then    1/m <1/2   so:
f(1/m) = 1+1/f(m-2)

Since all of such functions take equal values in q= 1 and q=2 it is provable by induction that for every integer values all functions take equal values .On the other hand for integers m :

f(1/m) = 1/f(m) so we have proved all function take equal values in reciprocal of positive integers as well .

Let's q >2 . 0<1/q <1/2
f(1/q) = 1+f(1/(q-2) )
1/f(q) =  1+1/ f(q-2)
f(q) =  1 /(1+1/ f(q-2))

     The last result show that we can shift the intended independent variable step by step such that it get to the interval [0,2] . 0 is meaningless and 2 was considered previously .If    1<q-2<= 2  ,then f(q-2) = 1+ f(q-3)    0<q-3<= 1 .

   What be considered so far means that we can compute all values of functions according to their values in interval (0,1] by a unique manner .All functions take unique values in 1/m which m is a positive integer while 1/m is a member of (0,1] .
Thus what is left to do is the considering of positive rational  such as q= m/n which m<n .

q= m/n  , m<n  then   n/m >1
Dividing n by m yields :
n = mq1 + r1    0 <= r1 < m
n/m  = q1 +  r1 /m

     As mentioned earlier we can shift this independent variable by an arbitrary integer (odd or even ) ,q1 , to move it in interval (0,1] ,that is r1 /m . Let's continue this procedure .

m =  r1 q2 + r2      0 <= r2   < r1
m / r1=  q2 + r2 / r1
r1=  r2 q3 + r3     0 <= r3 < r2 < r1

It is evident that after a finite steps ri gets to  1 or 0 .In that case we must compute the value of function in a point such as 1/n or an integer (n is an integer) .This procedure is unique for all possible functions and the value all of them take in 1/n or an integer is also one .

   On the other hand we can allocate proper values to integer values or 1/n of independent variable such that intended conditions be true .This results in that there is only and only one function satisfying these conditions .