[hide="Cliche to reveal hidden content"]1. glass
2. Humans
3. ???
4. Pencil sharpening
5. Un-manned space expeditions
6. ??? My dog can do it with four paws on the ground.(one-up-manship)
7. uuuhhh... potato?
8. violin
9. wth?!
10. ticks?[/hide]

[hide]1. glass [b]yes[/b]
2. Humans [b]no[/b]
3. ??? [b]no[/b]
4. Pencil sharpening [b]yes[/b]
5. Un-manned space expeditions [b]not what I was looking for, but still a possible explanation[/b]
6. ??? My dog can do it with four paws on the ground.(one-up-manship) [b]no[/b]
7. uuuhhh... potato? [b]no[/b]
8. violin [b]no, it is not [i]bowed or plucked[/i][/b]
9. wth?! [b]no. quite simple, actually...[/b]
10. ticks? [b]a possible explanation, but like I said there are a lot of possibilities and ticks do not live at the [i]top[/i][/b][/hide]

[hide="3"]One of them was an unusually active grandmother, a mother, and a daughter.[/hide]

[hide] One of them was an unusually active grandmother, a mother, and a daughter. [b]yes[/b][/hide]

[hide]

5. your age
7. a towel
9. a blind donkey[/hide]

[hide]2. a dog

8. your voice singing

10. pollution?[/hide]

[quote="D3m0n Shad0w"]1. A red house is made out of red bricks. A blue house is made out of blue bricks. A yellow house is made out of yellow bricks. What is a green house made out of?

2. What wears a coat in the winter, and pants in the summer?

3. Two mothers and two daughters went fishing. They each caught two fish. By the end of the day, they had 6 fish. They did not throw any back, they did not eat them, they did not lose them. How is this possible?

4. What wood-working is done in most offices?

5. What goes up but never comes down?

6. What does a man do standing, a woman do sitting down, and a dog do with three paws on the ground? (It is not what you first think, trust me. My teacher got us with that one) Hint: Something from the old days.

7. What gets wetter as it dries?

8. What orchestral device is not plucked, bowed, shook, or struck?

9. What eats oats and hay and sees equally well from both ends?

10. What can live on top of you and infest? (Many answers, one specific one)[/quote]

1.glass or greenbricks, depending on wheter the space was intended
2. a dog, humans should also be acceptable
3. there were 2 moms and 2 daughters but only 3 people; a grandma, a mom, and a daughter.
4.pencil sharpening
5.age, knowledge...etc.
6.pee
7. a towel
8. the conductor's baton
9. a blind horse
10. lice?

please tell me if any of these are wrong, and if so which ones

6. can be a handshake. According to common etiquette.

[quote="1=2"]my mom doesn't shake hands only when she sits...[/quote]

 :lol:  Men are obliged to stand up, women are not. Ofcourse, that doesn't mean that they have to sit when they shake hands, just that they don't have to stand up  :wink:

[quote="Arrange your tan"]D3m0n Shad0w and everyone else,

   for the first item, the correct answer

is not "glass."  It would be "green bricks"

in keeping with the pattern of the other

(example) answers.  If you had instead

typed "greenhouse," THEN the answer

would be (primarily) "glass."  However,

you typed "green house," and that is

a house that is green.  Not having the

space between the two words matters.[/quote]

Question number 1 was intended verbally, if you had said it out loud. "A red house is built with red bricks. A blue house is built with blue bricks. A yellow house is built out of yellow bricks. A green-house is made out of what?"

And, it is indeed shaking hands. Nothing inappropriate, as I already told you.  :no:

Mathking123 got every single one right, except for number 6. Hsiljak got that one.

IT'S A FREAKING RIDDLE! IT'S SUPPOSED TO BE CHALLENGING! AND IF IT WAS JUST "GREEN BRICKS" THEN IT WOULDN'T BE CHALLENGING! DEAL WITH IT!

All women are daughters, so the answer to 3 might not work (depending on whether it was implied that it was exactly two).

[quote="D3m0n Shad0w"][quote="Arrange your tan"]D3m0n Shad0w and everyone else,

   for the first item, the correct answer

is not "glass."  It would be "green bricks"

in keeping with the pattern of the other

(example) answers.  If you had instead

typed "greenhouse," THEN the answer

would be (primarily) "glass."  However,

you typed "green house," and that is

a house that is green.  Not having the

space between the two words matters.[/quote]

Question number 1 was intended verbally, if you had said it out loud. "A red house is built with red bricks. A blue house is built with blue bricks. A yellow house is built out of yellow bricks. A green-house is made out of what?"

And, it is indeed shaking hands. Nothing inappropriate, as I already told you.  :no:

[b]Mathking123 got every single one right, except for number 6. Hsiljak got that one.[/b] peeing is somewhat inappropriate but i think it should still be correct and accepted. a man pees standing up, a woman pees sitting down, and a dog pees lifting up one of his legs...[/quote]

[quote="Arrange your tan"]No, if you had intended it verbally, THEN you had to indicate it as such.

You're caught with an error that I pointed out, so now you're trying to change

an aspect of the puzzle to try to cover yourself.  Just come clean and admit

it.  All of those items were intended for reading directly off of the screen to

be worked out by that same participant.

And in this most recent post you typed "green-house."  However, there is no

hyphen in the original post between those two words.  You have to make up

your mind as to which form you're presenting to people![/quote]

Shut the heck up! Its a frickin riddle!!!

You are caught with an error that I point out:

Why should you care, unless you can't get it?

1. Glass, I've heard this before
2. Humans?
3. Grandmother is there making a mother a daughter?
4. Pencil sharpening?
5. Age, very well-known.
6. Heard this, handshake
7. Sponge? Towel?
8. Trumpet? Tuba? ... Oh wait the baton thingy?... I suck at music.
9. Blind Horse, Blindfolded horse, etc.
10. ??????? Something in your hair? Head lice??

[quote="BOGTRO"][quote="Arrange your tan"]No, if you had intended it verbally, THEN you had to indicate it as such.

You're caught with an error that I pointed out, so now you're trying to change

an aspect of the puzzle to try to cover yourself.  Just come clean and admit

it.  All of those items were intended for reading directly off of the screen to

be worked out by that same participant.

And in this most recent post you typed "green-house."  However, there is no

hyphen in the original post between those two words.  You have to make up

your mind as to which form you're presenting to people![/quote]

Shut the heck up! Its a frickin riddle!!!

You are caught with an error that I point out:

Why should you care, unless you can't get it?

[/quote]
Guys, let's not get overexcited.  I think Arrange your tan was not being entirely serious.
[hide="8"]Music stand?  Metronome? Baton? Microphone?[/hide]

My dog doesn't lift up one of it's legs unless his paws are cold.

Just telling you, this is not the first time Arrange your tan has argued a riddle *cough* 1+2=e *cough*

Oh-my-god, even first-graders [b][u]I[/u][/b] know don't even argue as much as you, Arrange Your Tan! It's just a riddle. My error. I didn't catch that because it was in the evening, I was tired and bored, and I was doing it quickly. You don't have to yell and cause a near-flame war. I really thought that by coming to this forum I'd communicate with [i]mature[/i] people.

And 1=2, your dog could be an exception  :P What kind of dog is it, anyway, not trying to get off track?[/u]

A cross between german shepherd and labrador.

well if pee isnt correct, then i can argue that my mom stands up when she shakes hands...and either way, women dont shake hands in proper ettiquette, they kiss on the cheek....

Some women shake hands standing up, some shake hands standing down, and some kiss cheeks standing up, and some kiss cheeks standing down.

You have to remember.....

[quote="D3m0n Shad0w"]6. What does a man do standing, a woman do sitting down, and a dog do with three paws on the ground? (It is not what you first think, trust me. My teacher got us with that one) [b][u]Hint: Something from the old days[/u][/b]. [/quote]

This is not the old days.

[quote="D3m0n Shad0w"][quote="dynamo729"]You have to remember....
This is not the old days.[/quote]

Yes, however, it is a [u]riddle[/u] that can make you think one thing, but it is really another.[/quote]

I was trying to explain to him why his answer wasn't correct, or why he wouldn't have thought of that because we aren't in the old days.

[quote="dynamo729"]
I was trying to explain to him why his answer wasn't correct, or why he wouldn't have thought of that because we aren't in the old days.[/quote]

Oh. Didn't see that.

[quote="D3m0n Shad0w"]1. A red house is made out of red bricks. A blue house is made out of blue bricks. A yellow house is made out of yellow bricks. What is a green house made out of?

2. What wears a coat in the winter, and pants in the summer?

3. Two mothers and two daughters went fishing. They each caught two fish. By the end of the day, they had 6 fish. They did not throw any back, they did not eat them, they did not lose them. How is this possible?

4. What wood-working is done in most offices?

5. What goes up but never comes down?

6. What does a man do standing, a woman do sitting down, and a dog do with three paws on the ground? (It is not what you first think, trust me. My teacher got us with that one) Hint: Something from the old days.

7. What gets wetter as it dries?

8. What orchestral device is not plucked, bowed, shook, or struck?

9. What eats oats and hay and sees equally well from both ends?

10. What can live on top of you and infest? (Many answers, one specific one)[/quote]

1. Glass, I think. Because its a greenhouse...

2. Doggy!

3. Grandmother, mother, daughter

4. Um...I'm hazarding a guess, pencil-sharpening?

5. I've heard this one, it's growth, right? Or age, something like that, they both work.

6. We already got the answer to this one.

7. A towel!

8. Well, it could be a flute or clarinet.

9. Blind horsie?

10. Um, I think lice....

Yep  :lol: Answers scrambled have already been posted though, but you have made it well organized. Well done.

Oh, and it's not flute or clarinet. I skimmed through your answers. I should have included on number 8, "blown". It was actually conductor's baton.

:huh: How does a towel get wetter as it dries? I must be missing something obvious  :blush: .

It's okay. I was thinking about this one sleepless night, and I suddenly got it. A towel [b]dries[/b] other things, like spills(I'm very experienced with spills -- I have caused many permanent stains *cough cough*) and it gets wetter as it soaks up whatever is spilled.

I'm still not convinced that #6 isn't pee...

:rotfl:

[quote="Arrange your tan"]This is try to correct D3m0n Shad0w's pattern of deception:
[/quote]

D3m0n Shad0w is in 5th grade (going into 6th?). She is very mature for her age. I don't really see how the puzzle was written really matters. It's a classic problem. I don't think she's trying to trick you into thinking that she is great.

[quote="Arrange your tan"]This is try to correct D3m0n Shad0w's pattern of deception:
===================================================

quote: "even first graders I know don't even argue as much you"

So what, that's changing the subject.

---------------------------------------------------------------------------

quote:  "It's just a riddle."

No, that's changing the subject.  You were told about the correction,
and you denied it, and made up excuses for it and held to your story.
Also, you expect to "fool" people with ambiguities in your riddles and reap the satisfaction, but when there are attacks on the questions you put out there,
you change your tune, disassociate yourself from it/them, and try to
wash yourself by hiding behind, "It's just a riddle."

--------------------------------------------------------------------------

quote:  "I didn't catch that because (blah, blah, blah)."

No, as I've already ponted out,  I corrected you early on, but you
stayed with your wrong scenario, rather than be big about it and
come out with any correction and/or "I'm wrong." statement sooner.

--------------------------------------------------------------------------

quote:  "You don't have to..."

That's right, because it was all up to you in this case.
To contrast with what you did here, keep
"all your ducks in the row," and just be upfront with
other users as soon as possible when you
goof up with a riddle or problem.

-------------------------------------------------------------------------

quote:  "I really thought that by coming to this forum
I'd communicate with *mature* people."

Well, then, start with yourself.  A mature person would
not have made/typed the above quote you did because
it is childish, especially of anpassive/aggressive type.
I'm persistent because I don't like lies being passed off
to cover flawed riddles and other problems.[/quote]

If you have some personal conflict with D3m0n Shad0w, here is not the place to argue.

1. What goes back when coming forward?
2. What plays games and breaks when thrown?

A short quiz.


linear algebra
matrix
linear algebra theorems

i need find the determinent of  not square matriz by blocks, what book do you know or maybe an article? because i was searching in the web and some books but i dont find anything. please send me information about that because i will have an exam. send me the information to jwsanchezo@unal.edu.co

Two points:
- Non-square matrices don't have determinants.
- Blocks are not a generally useful technique for finding determinants; they're only worth anything in a few special cases.

That's why you didn't find anything.

The closest to what you ask for is, probably, [url=http://en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula]Binet-Cauchy[/url] formula (at least, it is related to the key words you mentioned). But it is just a wild guess. As jmerry pointed out, the question, as posed, hardly makes any sense.

Yeah, I'm pretty sure determinants are applicable only to square matrices.


New Format:
.
.
.
Grade: 11
Taking: AMC 12 A and AMC 12 B

I planned on taking the 10 twice at first. But somehow, I can always get a better score on the 12 than the 10 ( :lol: ) when I practice, so I'll be doing both this year. Also, I think 10 questions are on both the 10 and 12, so doing those 10 plus 2 more on the 12 will qualify someone for AIME.

Grade: 10
Taking: AMC-10 (don't know if it's A or B)

My school gives both the AMC-10 and AMC-12 at the same time, so I can't take both. They would probably only let me take the AMC-10 anyway.

Could everyone post in the new format (see first post)?

Thanks. :)

Grade: 11
Taking AMC 12 A and B.

Won't improve much from last year... if I don't do any preparation I would only have a 10 point improvement from last year at max.

Grade: 10
Taking: 12A probs, dunno if its A or B.

If I could also take the 10 again I would, and I'd aim for a 0.

[quote="yif man 12"]If I could also take the 10 again I would, and I'd aim for a 0.[/quote]

Haha.. It would be *real* hard for you to score that after what you did last year.  :rotfl:  :rotfl:

Grade: 9
Taking AMC 10A (and possibly AMC 12B at another school)

grade: 8
AMC12 A and B

my district only gives out 12's and 8's  :(  ( but I took 8 already so, 12)

coz I always get simile marks when I'm doing 10 and 12
and...........you can get enroll to the AIME by lower mark on AMC12
so...............
but I hav no choice............... I'm in gr 11


algebra
polynomial
function
number theory proposed
number theory

I have searched quite extensively on the internet for some site that would have information on whether or not there exists a closed form expression or polynomial that generates every non prime , but alas I could not find what I was looking for , as a matter of fact I could not find a lot on non primes in general.

I made a formula to generate non primes but it wont suffice (1.)

What I need is a polynomial or closed form expression for generating all non primes . I hope you folks could help me .

(1.)I am not very fond of sharing my research but okay , here is the formula for generating every non prime .

(n + 2)^Floor[((n + 2)*Floor[Pochhammer[1, n + 1]/
              (n + 2)])/Pochhammer[1, n + 1]]

As you can see it can also very easily be adapted to generate only primes . You will also notice why I need a better formula once you test it .

Evolve

As $ \pi(x) \in o(x)$, such a polynomial must be of type $ x\plus{}a$, which doesn't work.
In general, it should be not much easier to find such a function instead of a prime-generating function.


in the set $A$, the sum of every three distictive number, is a prime number. what is the maximum number of members that $A$ can have?

use the fact that in any set of 5 numbers, there's a set of 3 numbers whose sum is divisible by 3.

here is my attempt:

[hide]
Lemma: If there were to be 5 numbers in set $A$, then there would be a sum of three numbers which is a multiple of 3:

Proof of Lemma: Since numbers are either $0,1,2 \mod{3}$, and it would take:

3 $1\mod{3}$'s
one each of $0,1,2 \mod{3}$
3 $2\mod{3}$'s
3 $0\mod{3}$'s

to sum up to a multiple of 3. it is impossible to achieve the given condition, because to avoid having 3 or more of the same modulos, we have to have all three, which is one of the ways. So we cannot have 5 numbers where every sum of three numbers are prime.

We have proven that A cannot have 5. the largest number of element in $A$ then is 4
[/hide]

Mods are only good for showing that something necessarily doesn't work -- they aren't helpful for showing that things do in fact work outside of the mod.  For example, I can find a solution to Fermat's Last Theorem mod 72 with very little trouble (for example, try n = 3, x = 6, y = 18, z = 12), but that doesn't mean that an actual solution to the equation exists.

[quote="xxreddevilzxx"][quote="paladin8"][hide]
We should probably give an example to finish up the proof: $\{3,5,11,15\}$ works.

The sums are $19, 23, 29, 31$.
[/hide][/quote]

well... we sort of know that there are infinite sets that could work, because we can use modulus to find the set that works.

ex: (1,2,2,1) (where the number 1 means 1mod3)... from there we can give numerous examples...[/quote]

They don't ALWAYS work, though, so maybe by some stroke of chance NONE of them work... thus you have to give an example. The set (1,2,2,1) itself does not work, so we aren't sure that there indeed exists one.

xxreddevilzxx's solution is correct.
assume $|A|\ge 5$ then
either $A$ has at least one member of the remaiders $mod 3$, which implies that the sum of these three numbers would be divisible by $3$.
or that there exist an $r$, that $0\le r\le 2$ which $A$ has three members such as $x_1, x_2, x_3$ where $x_i\equiv r (mod 3)$ for $i=1,2,3$, this implies that $3|x_1+x_2+x_3$ which is not correct so the first assumption was incorrect, so the max number would be four.
there is also another way to solve this problem, which is a longer methode, i can post the proof if you guys want to...

well actually the actual problem that i posted was in a form of multiple choice, and the choices that were there, were these:
$A.4$                                      $B.5$                          $C.6$                           $D.7$
but any ways.
Second way of solution:
for this problem we  prove that the set $A$ can not have more than $5$ members. assume $A=(a_1,a_2,a_3,a_4,a_5)$ exists, in which the sume any three chosen members, is a prime number. therefore it can easily be conluded all of the $a{}_i{}_{i=1}^{5}$ must be odd numbers, because if three or more of these numbers were even that if you choose those three even numbers, then the sum would be an even number, therefore it will not be a prime (but we must also prove that it is not possible for the sum to equal  $2$. to prove this we say if there exist a sum of three chosen members, is equal to $2$. then there must be another another set of $3$, members in which the sum is an even number, therefore the sum can not equal to $2$.)
if we have $2$, even numbers, then if you choose two odd numbers and an even the number, would give you an even number, therefore this cannot exist, this is the same if you have one even number in the set. so we have proven that all of $a{}_i{}_{i=1}^{5}$must be odds. so we can say the members have  to be in a form of $6k+1$ or $6k+3$ or $6k+5$, they also have to be less than $3$, because if they're not, then the sum of the numbers will be divisible by $3$. there fore we can conclude that in this set there is maximum one member that is in form of $6k+1$, maximum two members that are in a form of $6k+3$, and maximum two members that are in a form of $6k+5$. therefore we [u]must[/u] have at least one number in a forms of $6k+1, 6k+3,6k+5$ and since: $6k+1+6k+3+6k+5=3t$ therefore the sum of these three numbers can not be prime, so our original assumption: $A=(a_1,a_2,a_3,a_4,a_5)$ was wrong, there fore the maximum number of members would be $4$.


inequalities
blogs
geometry
algebra
polynomial
number theory
functional equation

How did everybody do in the first contest day of IMO?

I'm asking this question because as long as I know today, is 1-st day of the IMO contest and I'm very interested in everybody's participation.

Moderators, if you created a similar topic, then please delete this one.

Of course, it is always a good idea to occupy #3 with an ugly combi geo which can be owned without much troubles if one is patient enough to fight one's way through the 1000 cases... I mean, the Chinese team will always be on top, and what more do we want?
Problem #2 is a really nice number theory, however somewhat inappropriate because of some $ p$-adics theory being a bit too useful.
Yeah for #1, a functional equation consisting of more than substituting some values, then substituting the results and then substituting the results again ;) - I never liked functional equations, but this one is quite an exception. I am wondering whether we will are getting to see a hard (and not Muirheadable) inequality tomorrow... And maybe, just maybe, we will have a less idiot geometry.

By the way, kudos to the official IMO site with its very precise and timely information about the location of the contest and its problems!

  darij

The second and third problem were chosen badly:
Number 2 is an immediate consequence of the Cassels-Lindemann theorem (on multivariate forms of genus $ 5$ over the $ p$-adics), and number 3 was already mentioned in Terence Tao's blog half a year ago.
Number 1 in contrary, was an intriguing and extremely hard functional equation. Maybe the hardest number 1 ever.

I've heard rumors that G9 was chosen, so we are going to see a very hard geometry problem tomorrow.

Well, they are being posted on the official site in the very next time, so I doubt we will be faster. Besides, the bus is going in 5 minutes.

  dg

can you please post a link to the problems, because i can't find them, tnx

We're all still waiting for them...

P. S. Nice sense of humor shown here by some posters  :wink:

P.P.S:
No. 3: [url]http://www.mathlinks.ro/viewtopic.php?t=215221[/url]
No. 2: [url]http://www.mathlinks.ro/viewtopic.php?t=215222[/url]
No. 1: [url]http://www.mathlinks.ro/viewtopic.php?t=215228[/url]

[quote="ZetaX"]... Number 1 in contrary, was an intriguing and extremely hard functional equation. Maybe the hardest number 1 ever ....[/quote]

I'm condused, the geometry problem was Fuctional Equation?how?

Sorry, I was talking about day 2  :wink:

isn't it obvious? telekinesis...
And problem 6 is trivial by aplying the well known Priston-Mannxin principle to the polynomials involved ( obviously you have to apply it 2008 times)

[quote="greentreeroad"][quote="ZetaX"]Sorry, I was talking about day 2  :wink:[/quote]

But how can you talk about Day 2 problem before it start?(I guess it will start like several hours later)[/quote]

Insider knowlegde...  :D

I wonder how many credulous contestants they could gull into searching the internet for those theorems. :P

See. Do you understand now the potential of insider's knowledge?

hmmm. I see this thread is not exactly serving its purpose...

[quote="greentreeroad"][quote="Albanian Eagle"][quote="greentreeroad"]What do insider knowledge mean? Also some contestant may see the message via internet. Right?[/quote]
he has a point here... :!:  :D[/quote]

Why not speaking directly? I don't know anything you are talking about :!:

P.S. Is this just kidding?
P.P.S It seems to be a joke indeed![/quote]
This wasn't a joke!
2ZetaX: I think what you have done is a certain mistake.  :(

Pavel, I think that ZetaX's joke just accidentally appeared to be truth. And it was natural to expect hard geometry after first day.


To the nearest hundredth of a gallon, how many gallons of gasoline can be saved each year by driving a car which gets $ 32$ miles per gallon as opposed to a car which gets $ 18$ miles per gallon? Assume the number of miles traveled each year $ 9,000$.

[quote="bobbym"]Hi;

[hide]218.75 gallons are saved.[/hide][/quote]
To be more specific, we arrive to that by subtracting the number of gallons used by a car which gets 18  miles per gallon by the number of gallons used by a car which gets  32 miles per gallon which is:

$\frac{9000}{18}-\frac{9000}{32}=\boxed{218.75}\text{ gallons}$


algebra
function
domain
Support
Ring Theory
advanced fields
advanced fields unsolved

Let $ \mathcal{D}$ be a Dedekind domain. Show that every ideal of $ \mathcal{D}$ can be generated by (at most) two elements of $ \mathcal{D}$.

Is there something wrong in this solution because I found it in two second by Google? http://planetmath.org/encyclopedia/IdealOfDedekindDomain.html.

A result due to Steinitz: Let $ R$ be a Dedekind domain, and let $ \mathfrak{a}$ and $ \mathfrak{b}$ be ideals of $ R$. Then there is an element $ \omega$ and an ideal $ \mathfrak{c}$ of $ R$ such that $ \mathfrak{ac} \equal{} (\omega)$ and $ \mathfrak{b\plus{}c} \equal{} R$.

Let $ \mathfrak{a}$ be an arbitrary ideal of a Dedekind domain $ R$. Let $ \mathfrak{b}$ be such an ideal of $ R$ that $ \mathfrak{ab}$ is a principal ideal $ (\beta)$. The lemma to which this entry is attached gives also an element $ \gamma$ and an ideal $ \mathfrak{c}$ of $ R$ such that $ \mathfrak{ac} \equal{} (\gamma)$ and $ \mathfrak{b \plus{} c} \equal{} R$. Then we have
\[ \mathfrak{a} \equal{} \gcd(\mathfrak{ab},\,\mathfrak{ac}) \equal{} \gcd((\beta),\,(\gamma)) \equal{} (\beta,\,\gamma)
\]
because $ \gcd(\mathfrak{b},\,\mathfrak{c}) \equal{} \mathfrak{b \plus{} c} \equal{} R \equal{} (1)$.

The following solution uses the technique of localisations of $ D$, the structure theory of ideals in local Dedekind rings and the Chinese remainder theorem.

Let $ \mathfrak{a}$ be a non-zero ideal of $ D$ and pick an arbitrary non-zero element $ a\in\mathfrak{a}$. Then the inclusion $ f: Da \rightarrow \mathfrak{a}$ might fail to be surjective (as $ D$-modules).

To amend this we need a second element $ b\in \mathfrak{a}$ which is constructed as follows: Let $ \mathfrak{m}_1,\ldots,\mathfrak{m}_s$ be the finitely many maximal ideals where $ f$ fails to be surjective (i.e. they are precisely the maximal ideals of the support of $ coker(f)$). If not mentioned in the list add the (finitely many) maximal ideals containing $ \mathfrak{a}$. Let $ \pi_i\in\mathfrak{m}_i\setminus\mathfrak{m}_i^2$ and choose the natural number $ e_i > 0$ such that $ \pi_i^{e_i}$ generates the localisation of $ \mathfrak{a}$ at $ \mathfrak{m}_i$ (by the structure theory of ideals in local Dedekind domains). Pick $ r > 0$ such that the ideal product $ (\mathfrak{m}_1\cdots\mathfrak{m}_s)^r$ is contained in $ \mathfrak{a}$.

Use the Chinese remainder theorem to choose an element $ b\in D$ such that
\[ b\equiv\pi_i^{e_i} \mod \mathfrak{m}_i^r
\]
for all $ i$. Show that $ b$ is contained in $ \mathfrak{a}$ and that $ b$ generates $ \mathfrak{a}$ at the maximal ideals $ \mathfrak{m}_i$, i.e. that the inclusion $ g: Db \rightarrow \mathfrak{a}$ is surjective at all $ \mathfrak{m}_i$.

This implies that the inclusion $ Da \plus{} Db \rightarrow \mathfrak{a}$ is surjective which means that $ \mathfrak{a}$ is generated by $ a,b$. (Check this at every maximal ideal using localisations.)

Basically $ b$ is chosen in that way that $ b$ generates the ideal $ \mathfrak{a}$ where $ a$ fails to do so.


Find all functions $ f: R\minus{}\minus{}> R$ which satisfy the equation
$ f(x^2\plus{}y\plus{}f(y))\equal{}2y\plus{}f(x)^2$
for all real numbers $ x,y$

[quote="dragon1"]Find all functions $ f: R \minus{} \minus{} > R$ which satisfy the equation
$ f(x^2 \plus{} y \plus{} f(y)) \equal{} 2y \plus{} f(x)^2$
for all real numbers $ x,y$[/quote]

A trivial solution by observation,
$ f(x)\equal{}x$

I think everyone realizes that; if you're going to post some observations make them nontrivial.

$ f$ is clearly surjective. Observe that (with $ y \equal{} 0$) $ f(x)^2 \equal{} f(x^2\plus{}f(0)) \equal{} f((\minus{}x)^2\plus{}f(0)) \equal{} f(\minus{}x)^2$. Therefore $ f(x) \equal{} \pm f(\minus{}x)$ (where the sign can depend on $ x$). Let $ c$ be such that $ f(c) \equal{} 0$. Then $ x \equal{} 0, y \equal{} c$ gives $ f(0)^2 \equal{} \minus{}2c$. $ x \equal{} y \equal{} 0$ gives $ f(f(0)) \equal{} f(0)^2 \equal{} \minus{}2c$. Now put in $ x \equal{} f(0), y \equal{} c$. We get $ f(\minus{}c) \equal{} 2c\plus{}4c^2$. But $ f(\minus{}c) \equal{} \pm f(c)$, so $ 2c \plus{} 4c^2 \equal{} 0$ (1a). Replacing all of this with $ \minus{}c$, since $ f(\minus{}c) \equal{} 0$, gives $ \minus{}2c \plus{} 4c^2 \equal{} 0$ (1b). Adding (1a) and (1b) gives $ c \equal{} 0$ and $ f(0) \equal{} 0$.

Putting $ y \equal{} 0$ in gives $ f(x^2) \equal{} f(x)^2$, which implies $ f$ of any positive number is another positive number and $ f(x\plus{}y\plus{}f(y)) \equal{} 2y \plus{} f(x)$ (2) for any $ x \geq 0$. Suppose $ f(a) \equal{} f(b)$ for $ a,b \geq 0$. Then $ 2a\plus{}f(b) \equal{} f(b\plus{}a\plus{}f(a)) \equal{} f(a\plus{}b\plus{}f(b)) \equal{} 2b\plus{}f(a)$, so $ a \equal{} b$. Thus when we restrict $ f$ to nonnegative reals, $ f$ is injective. Recall we had $ f(x) \equal{} \pm f(\minus{}x)$. Suppose $ f(a) \equal{} b$ for $ a,b > 0$. Since $ f$ is injective, there is some $ c$ such that $ f(c) \equal{} \minus{}b$, and since $ \minus{}b < 0$ we must have $ c < 0$. But then $ f(\minus{}c) > 0$, so $ f(\minus{}c) \equal{} b$, $ a \equal{} \minus{}c$ by injective on nonnegative reals, and $ f(\minus{}a) \equal{} \minus{}b$. Thus $ f(x) \equal{} \minus{}f(\minus{}x)$ for all $ x$.

Let $ y \equal{} \minus{}x$ in (2). We get $ f(f(x)) \equal{} 2x \minus{} f(x)$. Now turn $ y$ into $ f(y)$ in (2) to get $ f(x\plus{}f(y)\plus{}f(f(y))) \equal{} 2f(y)\plus{}f(x)$, or $ f(x\plus{}2y) \equal{} 2f(y)\plus{}f(x)$. $ x \equal{} 0$ in this gives $ f(2y) \equal{} 2f(y)$, so we can change it to $ f(x\plus{}y) \equal{} f(x)\plus{}f(y)$ (where $ x \geq 0$). Since $ f(x) > 0$ for all $ x > 0$, we get $ f$ monotonic, which combined with this being Cauchy's equation gives $ f(x) \equal{} cx$ on nonnegative reals (and then all reals by $ f$ odd). Verify that only $ c \equal{} 1$ works. $ f(x) \equal{} x$ is the only solution.

[i]Yes , it is an old problem and has been posted many times in Mathlinks

    I haven't checked Palmer 's solution but here is two way to attack that one [/i]

 [i]     Enjoy it  :D [/i]


trigonometry
inequalities proposed
inequalities

If $ x,y,z$ be real numbers such that $ 2xyz \plus{} x^2 \plus{} y^2 \plus{} z^2\geq 1,$ then

$ x^2\plus{}y^2\plus{}z^2\plus{}yz\plus{}zx\plus{}xy\geq x\plus{}y\plus{}z.$

[quote="Ji Chen"]If $ x,y,z$ be real numbers such that $ 2xyz \plus{} x^2 \plus{} y^2 \plus{} z^2\geq 1,$ then

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} yz \plus{} zx \plus{} xy\geq x \plus{} y \plus{} z.$[/quote]
frist we proof $ 2xyz \plus{} x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$ ,then

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} yz \plus{} zx \plus{} xy\geq x \plus{} y \plus{} z.$
let $ x \equal{} \cos A ,y \equal{} \cos B,z \equal{} \cos C$
$ \iff \sum_{cyc}\cos^2 A \plus{} \sum_{cyc}\cos A\cos B \geq \sum_{cyc}\cos A$
$ (\sum_{cyc} \cos A\minus{} 1)^2 \geq 2\prod\cos A$
$ \sum_{cyc}\cos A \geq 1 \plus{} \sqrt {2\prod\cos A}$
see here
http://www.mathlinks.ro/viewtopic.php?t=242015
when
$ 2xyz \plus{} x^2 \plus{} y^2 \plus{} z^2 \geq 1$
let $ x \minus{} > xt,y \minus{} > yt,z \minus{} > zt,t \leq 1$
Satisfing $ 2xtytzt \plus{} (xt)^2 \plus{} (yt)^2 \plus{} (zt)^2 \equal{} 1$ ,
So $ (xt)^2 \plus{} (yt)^2 \plus{} (zt)^2 \plus{} ytzt \plus{} ztxt \plus{} xtyt\geq xt \plus{} yt \plus{} zt$
$ \iff t(x^2 \plus{} y^2 \plus{} z^2 \plus{} yz \plus{} zx \plus{} xy)\geq x \plus{} y \plus{} z,t \leq 1$
So $ x^2 \plus{} y^2 \plus{} z^2 \plus{} yz \plus{} zx \plus{} xy\geq x \plus{} y \plus{} z$
done~


$(a+b+c)^{2}\geq 3(ab+bc+ca)$ so the maximum value is $\frac{1}{3}$

[quote="HTA"]how about the generlization with $x_{i}(i=\bar{1,n})$ and $x_{1}\geq x_{2}\geq .. \geq x_{n}$[/quote]

what should we consider? cyclic or symmetric sum?

[b]Solution.[/b]  Let $\sum x_{i}=k$.  We are to find the maximum of $\sum_{1\leq i<j\leq n}x_{i}x_{j}$.  Notice that
\[\sum x_{i}^{2}\geq \sum_{1\leq i<j\leq n}x_{i}x_{j}.\]
Adding $2\sum_{1\leq i<j\leq n}x_{i}x_{j}$ to both sides gives us
\[\left(\sum x_{i}\right)^{2}\geq 3\sum_{1\leq i<j\leq n}x_{i}x_{j}.\]
As a result,
\[\sum_{1\leq i<j\leq n}x_{i}x_{j}\leq \frac{k^{2}}{3}.\]
And the conclusion follows.$\Box$

(I hope I understood the problem correctly.)


What is your typing speed?
I'm in 5th grade, just before people type around 60+ wpm.
My homerow keys: 70-100

Really? When I was in elementary school my school's goal was 30 wpm, LOL. Now I'm about 60, yeah, I'm really slow. But I usually type slower than that, since I think as I type (unlike typing tests, where you just copy things).

14 wpm  :( i know really im slow. i think i actually type faster than that, since on the typing test we werent allowed to look at the keyboard, and we had to use the right position. i'm very good at just using my index fingers, u know!

It depends where Im typing. Like Nerd_of_the_ages, at school I typed 31 wpm max. And then for the typing test I scored like 54 because I memorized it since 4th grade. On AoPS, I usually type maybe 40-50 wpm? On FTW, I type with 2 fingers. :D I usually have to look at my fangers when typing though. Sometimes I can look at screen and type, but can't type when looking at something else.

I'm a ok typer... I don't know my typing speed...

I can type pretty fast, but I spell a lot of words wrong, and I hate spelling words wrong, so I have to go back and fix :) so that balances out...

I can't type fast without looking at my fingers.

I'm about 50wpm. I hate the stupid typing tests becuase they are so random! (Quincey likes to wear red cowboy boots; he likes the color red.)
I can look anywhere while I'm typing, I usually catch the mistakes in my head (I make a lot of them)

probably its ok

for me, i have like a vocabulary of 1000 words stored into my hands and they know how to type those words faster than others

like i type frac and sqrt (and my computer pass, which is, btw, 38 characters long) more quickly that other words

ptherwise, i have to look at the keyboard and i type with whichever fingers i want

take the word "word"
i use my left ring finger for w (standard), right middle finger for o (not standard), and index finger for r and d (not standard)

I mostly use my second finger and third fingers for typing although I occasionally use my fourth finger. I use my thumb for pressing space.

[quote="crazychinesemaniac"]I use whatever finger is most accessible at that time to reach the key.[/quote]

Me too!

[quote="hunter34"]I don't know what I type with I just type.[/quote]

Exactly!

i'm about 80-90 wpm. with 97% accuracy..

I usually average 60-65 and 95-100% accuracy, but if the words are really short I can hit 75-80 wpm. :) There are still like 5 keys that I use like once a year that I have to look for... but other than that I can type without even looking at the computer.

I type aruond 70 wpm i think, I usually use the normal fingers for the normal keys.  After learning touch typing in 6th grade, I was typing like 30 wpm, then somehow, I more than doubled it...  I think it's all my computer/aim use  :D

I am around 60 wpm and I have horrible accuracy(maybe only like 60%-90% depending on how my day is going.) The only thing that I can't type very fast is the numbers. I can type a series of numbers really fast but I can't alternate words numbers words numbers. Guess my fingers aren't that flexible...

[url]http://www.typeracer.com[/url] is a fun thing to do to test typing speed. There I average 95 wpm (it uses 5 char/word rather than actual words though).

[quote]www.typeracer.com is a fun thing to do to test typing speed. There I average 95 wpm (it uses 5 char/word rather than actual words though). [/quote]

I just went there and I averaged around 70-75 wpm. Lol but I kept misstyping and going like two words farther before I realized I had typoed.

I type with my middle fingers instead of my index fingers--it just seems more natural that way--except I use my thumb to press space.

I can get 100 wpm max, but my accuracy is good.

I can get 80 max. Dang I feel slow.

@Lazarus: Then submit to the Guiness book of world records...

[quote="mgao"]Average: 82 wpm, 98% accuracy +- 1%
[quote="stevenmeow"]for me, i have like a vocabulary of 1000 words stored into my hands and they know how to type those words faster than others[/quote]
So true.
On a test where there's lots of words stored in my hands i average 93 wpm, on a test without that many words stored in my hands i only get about 65-70.  Average typing speed ( i chose an average test) is 82wpm like i said above.[/quote]

 :o I thought at first that was me because she had the same avatar as me! So yeah... around 100 to 150 WPM, at 98% accuracy (more like 80% accuracy when I type at 150 WPM)
But that's because I type like a million words every day :D

I don't have words stored in my hands. I have letters. So I can spell out something.

c h r y s a n t h e m u m

And I average 66. :( I can get more than 70.

What is the fastest speed of the worlds fastest typer. That would interesting to know.

Definitely not. If that were so, I would be terribly close. But I know I'm not close. Contradiction.

I could probably go in the sixties and low seventies by now. i haven't timed myself in a very long time.

[quote="dragon96"]If you go to typing test.com, I always use the same test so I have like 54 wpm for that. I even memorized part of it, even though I used to know more. By the way, in the process of posting this, I did not go to the site:

Dorothy lived in the midst of the great Kansas prairies, with Uncle Henry, who was a farmer, and Aunt Em, who was the farmer's wife. Their house was small, for the lumber used to build it had to be carried by wagon many miles.

I forgot most of it already.  :oops:[/quote]

It is a strange fact, but I am not taken to romancing...

Yes, I used this to get 82 wpm, 98% accuracy!

Usually if I go as fast as I can, then I get percentages in the low 90's.


linear algebra
matrix
linear algebra unsolved

When it is given that any n x n matrices A and B are nilpotent, how can you show that AB is nilpotent for matrices A and B which satisfy AB=BA?

Is is possible that AB is invertible?

thnx

(1) If $ AB \equal{} BA$ then $ (AB)^n \equal{} A^n B^n$.
(2) If $ B$ is nilpotent, it is not injective and neither is $ AB$. Same goes for $ A$ and surjectivity.

What happens if you take away the hypothesis that $ AB\equal{}BA?$

Try this: $ A\equal{}\begin{bmatrix}0&1\\0&0\end{bmatrix}$ and $ B\equal{}\begin{bmatrix}2&\minus{}4\\1&\minus{}2\end{bmatrix}.$

$ A$ is obviously nilpotent, and if you multiply it out, you'll see that $ B$ is also nilpotent.

But $ AB\equal{}\begin{bmatrix}1&\minus{}2\\0&0\end{bmatrix}.$

That's not nilpotent. In fact, it's idempotent.


[quote="Dimitris X"][quote="4865550150"]
[b](2)[/b]  If $ a,b,c$ are three positive numbers such that $ abc \equal{} 1$. Let $ S \equal{} a^2 \plus{} b^2 \plus{} c^2, S' \equal{} ab \plus{} bc \plus{} ca$. Prove or disprove

$ (S' \plus{} 1)(S' \plus{} 12) \leq 2(S \plus{} 3)(S \plus{} 2)$.[/quote]
I don't think we need the condition for it :wink:[/quote]
You're wrong,just try to make $ S'$ less than $ 3$ and you'll see that the inequality is no more true.

[quote="Bugi"]Condition is here probably then to make it easier...[/quote]
Nope...
I think it is for misleading....
The solution is 2 lines...:
$ S'^2 + 13 S' + 12 \leq 2 S^2 + 10S + 12$
But $ S \geq S'$, and $ S^2\geq 3S'$ $ \Box$

To rachid:are you sure???Can you give an example??

[quote="Dimitris X"][quote="Bugi"]Condition is here probably then to make it easier...[/quote]

To rachid:are you sure???Can you give an example??[/quote]
Yes,I am sure Dimitris X,next time check your inequalities since $ S^2 \geq 3S'$ is not always true  :lol:  (try $ S \equal{} S' \equal{} 2$)

[quote="rachid"][quote="Dimitris X"][quote="Bugi"]Condition is here probably then to make it easier...[/quote]

To rachid:are you sure???Can you give an example??[/quote]
Yes,I am sure Dimitris X,next time check your inequalities since $ S^2 \geq 3S'$ in not always true  :lol:  (try $ S \equal{} S' \equal{} 2$)[/quote]
Ohhh yes.... :oops:
You are right...
We have first to prove that $ a^2\plus{}b^2\plus{}c^2 \geq a\plus{}b\plus{}c$,where we need the condition $ abc\equal{}1$

[quote="4865550150"][b](1)[/b]  Let $ p,q,r$ be positive numbers. Prove or disprove

$ 9 \plus{} p \plus{} q \plus{} r \leq \sqrt {6(p^2 \plus{} q^2 \plus{} r^2) \plus{} 15(p \plus{} q \plus{} r) \plus{} 18} \plus{} \frac {18}{p \plus{} q \plus{} r \plus{} 3}$.

[b](2)[/b]  If $ a,b,c$ are three positive numbers such that $ abc \equal{} 1$. Let $ S \equal{} a^2 \plus{} b^2 \plus{} c^2, S' \equal{} ab \plus{} bc \plus{} ca$. Prove or disprove

$ (S' \plus{} 1)(S' \plus{} 12) \leq 2(S \plus{} 3)(S \plus{} 2)$.[/quote]

$ 1)$ Let $ p \plus{} q \plus{} r \equal{} a$ and $ p^2 \plus{} q^2 \plus{} r^2 \equal{} b$ then $ b \geq \frac {a^2}{3}$,

The inequality is equivalent after squaring to:

$ 6b(a^2 \plus{} 6a \plus{} 9) \plus{} 27a \plus{} 81 \geq a^4 \plus{} 9a^3 \plus{} 54a^2$

replacing $ b$ by $ \frac {a^2}{3}$ in the $ LHS$ it suffices to prove that:

$ a^4 \plus{} 3a^3 \plus{} 27a \plus{} 81 \geq 36a^2$

which is clearly true by AM-GM.

$ 2)$ We know that $ S \geq S'$,then replacing $ S$ by $ S'$ in the $ RHS$ it suffices to prove that $ S' \geq 3$ which is clearly true by AM-GM whith $ abc \equal{} 1$.

 :)

[quote="4865550150"]If $ a, b, c \geq 1$. Prove or disprove

$ 3(\sqrt {a \minus{} 1} \plus{} \sqrt {b \minus{} 1} \plus{} \sqrt {b \minus{} 1}) \leq a \plus{} b \plus{} c \plus{} \sqrt {3(a \plus{} b \plus{} c \minus{} 3)} \leq \frac {3\sqrt {3}}{2} \sqrt {a^2 \plus{} b^2 \plus{} c^2}$[/quote]
$ (*)$ We'll prove first that $ 3(\sqrt {a \minus{} 1} \plus{} \sqrt {b \minus{} 1} \plus{} \sqrt {b \minus{} 1}) \leq a \plus{} b \plus{} c \plus{} \sqrt {3(a \plus{} b \plus{} c \minus{} 3)}$

By AM-GM we have:

$ (\sqrt {a \minus{} 1} \plus{} \sqrt {b \minus{} 1} \plus{} \sqrt {b \minus{} 1}) \leq \sqrt {3( (a \minus{} 1) \plus{} (b \minus{} 1) \plus{} (c \minus{} 1) ) } \equal{} \sqrt {3(a \plus{} b \plus{} c \minus{} 3) }$     $ (1)$

also we have that:

$ a \equal{} (a \minus{} 1) \plus{} 1 \geq 2\sqrt {a \minus{} 1}$

similarly for $ b$ and $ c$,then adding cyclically the three inequalities we have:

$ a \plus{} b \plus{} c \geq 2(\sqrt {a \minus{} 1} \plus{} \sqrt {b \minus{} 1} \plus{} \sqrt {b \minus{} 1})$     $ (2)$

adding the inequalities $ (1)$ and $ (2)$ we get the desired result.

$ (*)$ Let us prove now that $ a \plus{} b \plus{} c \plus{} \sqrt {3(a \plus{} b \plus{} c \minus{} 3)} \leq \frac {3\sqrt {3}}{2} \sqrt {a^2 \plus{} b^2 \plus{} c^2}$

It's easy to see from AM-GM that:

$ \frac {3\sqrt {3}}{2} \sqrt {a^2 \plus{} b^2 \plus{} c^2} \geq \frac {3}{2}.(a \plus{} b \plus{} c)$

It suffices to prove that:

$ \frac {3}{2}.(a \plus{} b \plus{} c) \geq a \plus{} b \plus{} c \plus{} \sqrt {3(a \plus{} b \plus{} c \minus{} 3)}$

which is equivalent to:

$ (a \plus{} b \plus{} c \minus{} 6)^2 \geq 0$

which is true.

 :)

Let $ a,b,c \in R$, prove or disprove

$ 3(a \plus{} b \plus{} c) \leq \sqrt {25(a^2 \plus{} b^2 \plus{} c^2) \plus{} 2(ab \plus{} bc \plus{} ca)} \leq 3 \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}$

It seem stonger than CS inequality.  :blush:

[hide="Prove of LHS"]
The inequality is equivalent to $ 9(a\plus{}b\plus{}c)^2 \leq 24(a^2\plus{}b^2\plus{}c^2)\plus{}(a\plus{}b\plus{}c)^2$.
$ \Leftrightarrow  (a\plus{}b\plus{}c)^2 \leq 3(a^2 \plus{}b^2\plus{}c^2)$
[/hide]

[hide="Prove of RHS"]
The inequality is equivalent to $ 24(a^2\plus{}b^2\plus{}c^2)\plus{}(a\plus{}b\plus{}c)^2 \leq 27(a^2\plus{}b^2\plus{}c^2)$.
$ \Leftrightarrow (a\plus{}b\plus{}c)^2 \leq 3(a^2\plus{}b^2\plus{}c^2)$
[/hide]

Do anyone have its generalized from?


That's what Mr. Fish said anyways =D.

1.  There are 5 girls and one boy who have 4 reserved rooms.  If the boy must sleep alone, and two of the rooms contains two girls each, how many possible arrangements are there.  One possible way is shown.

Room 1( two girls)    Room 2 (two girls)         Room 3 (1 girl)     Room 4 (1 boy)
A                              C                               E                           F
B                              D

I got 720 -_-, which is wrong.  I'm kinda lazy though, so I dun wanna go look for another answer =D.

2. Two circles with radii of different lengths are in the first quadrant.  They are both tangent to the x- and y- axes.  One of the intersection points is $ (1,5)$.  Find the product of the lengths of their radii.

This problem is fun. =D.

3.  You flip a fair coin 4 times, but at least one of the flips must be heads.  What is the probability that you flip exactly 2 heads?

I have guesses for problem 1, but I think I have the answer for #3.

[hide]If 1 must be heads, then u can assume that u only need to calculate probability for only 3 flips. The chance of getting 1 more head is 3/8.
Well, at least I think its 3/8. [/hide]

Actually, for 3...
[hide="3"]
There are 16 distinct ways to flip a coin 4 times, where order matters. Only one case (TTTT) does not have at least one head. So there are 15 ways. 6 of these have exactly 2 heads. So the probability is 6/15 = 2/5.
[/hide]

Meh

[hide="3"]
You can not get 4 tails. So there is a probability of $ \frac{15\minus{}\binom{4}{2}}{16}\equal{}\boxed{\frac{9}{16}}$
[/hide]

I am not entirely sure though...

For #1
[hide="might not be right"]
Assume rooms 1 and 2 have two girls each, and the boy is in room 4.
To choose two girls for room 1, take 2C5, which is 10 ways.
To choose two girls for room 2, take 2C3 (since there are 3 girls left), which is 3 ways.
The last girl has to be in room 3.
So, we have 10*3=30 ways to arrange 5 girls in 2 rooms.
Since two girls can be in rooms 1 and 2, or 1 and 3, or 2 and 3, multiply the 30 ways by 3.
30*3=90
The boy can be in any of the four rooms, so multiply by four
$ 90*4\equal{}\boxed{360}$
[/hide]

#2

[hide="btw, i am in 8th grade"]
Label the centers of the smaller circle and bigger circle $ (r_1,r_1)$ and $ (r_2,r_2)$, respectively, with radii of $ r_1$ and $ r_2$ (obviously).
The distances from the centers of the circles to the point $ (1,5)$ is equal to the radii of the circles, so we can write (with the distance formula)
$ \sqrt{(r_1\minus{}1)^2\plus{}(r_1\minus{}5)^2}\equal{}r_1$
Expanding and simplifying...
$ \sqrt{r_1^2\minus{}2r_1\plus{}1\plus{}r_1^2\minus{}10r_1\plus{}25}\equal{}r_1$
$ 2r_1^2\minus{}12r_1\plus{}26\equal{}r_1^2$
$ r_1^2\minus{}12r_1\plus{}26\equal{}0$
Note that this equation gives us the length of both radii since using the distance formula with $ (r_2,r_2)$ would produce the same thing.

You are asking for the product of the lengths. Since that is the same as asking for the product of the roots of the quadratic, you can use $ c/a\equal{}26/1\equal{}\boxed{26}$
[/hide]

[hide]im almost(100%) sure the answer is 360[/hide

for number 1

[quote="hunter34"]Meh

You can not get 4 tails. So there is a probability of $ \frac {15 \minus{} \binom{4}{2}}{16} \equal{} \boxed{\frac {9}{16}}$

I am not entirely sure though...[/quote]

Actually, there are 15 possibilities, and there are $ {\binom{4}{2}} \equal{} 6$ ways there can be two heads, so the probability is $ \frac {6}{15} \equal{} \boxed{\frac {2}{5}}$ that there is exactly two heads


Given the system of equations
\[ ax \plus{} (a \minus{} 1)y \equal{} 1 \\
(a \plus{} 1)x \minus{} ay \equal{} 1.
\]
For which one of the following values of $ a$ is there no solution $ x$ and $ y$?

$ \textbf{(A)}\ 1\qquad \textbf{(B)}\ 0\qquad \textbf{(C)}\ \minus{} 1\qquad \textbf{(D)}\ \frac {\pm \sqrt {2}}{2}\qquad \textbf{(E)}\ \pm\sqrt {2}$

Adding the two equations together gives

$ (2a \plus{} 1)x \minus{} y \equal{} 2$
$ y \equal{} (2a \plus{} 1)x \minus{} 2$

Substituting that in the second equation gives

$ (a \plus{} 1)x \minus{} (2a^2 \plus{} a)x \minus{} 2a \equal{} 1$
$ ( \minus{} 2a^2 \plus{} 1)x \equal{} 1 \plus{} 2a$
$ x \equal{} \frac {1 \plus{} 2a}{ \minus{} 2a^2 \plus{} 1}$

We must find the $ a$ values where $ x$ is undefined, that is, when $ 1 \minus{} 2a^2 \equal{} 0$, or $ a^2 \equal{} \frac {1}{2}$. So $ a$ can be $ \frac {\sqrt {2}}{2}$ or - $ \frac {\sqrt {2}}{2}$.

Answer: D

[hide="Quicker solution"]Let $ A\equal{}\left[\begin{array}{cc}
a&a\minus{}1\\
a\plus{}1&\minus{}a
\end{array}\right]$.

Write it as $ A\times \left[
\begin{array}{c}
x\\
y
\end{array}
\right]
\equal{}
\left[
\begin{array}{c}
1\\
1
\end{array}
\right]$.

This has no solution if $ \det A\equal{}0\implies \minus{}a^2\minus{}(a\plus{}1)(a\minus{}1)\equal{}\minus{}a^2\minus{}a^2\plus{}1\equal{}\minus{}2a^2\plus{}1\equal{}0$

$ a\equal{}\pm\frac{1}{\sqrt{2}}\implies\boxed{\textbf{(D)}}$.[/hide]

hello, solving the equation system we get
$ x\equal{}\minus{}\frac{1\minus{}2a}{\minus{}1\plus{}2a^2}$
$ y\equal{}\minus{}\frac{1}{1\minus{}2a^2}$
so our solution ist not defined if $ a\equal{}\pm\frac{1}{\sqrt{2}}\equal{}\pm\frac{\sqrt{2}}{2}$, so the answer is $ \boxed{D}$.
Sonnhard.


geometry
3D geometry
search
complex numbers
algebra
binomial theorem

What is the value of $ \binom{2000}{2}\plus{}\binom{2000}{5}\plus{}\binom{2000}{8}\plus{}\cdots\plus{}\binom{2000}{2000}$?

I know the solution uses complex numbers, could someone explain? Thanks

Is the sequence in the bottom parts $ 3n\plus{}2$?

I think this might help:   [hide]Let $ \omega$ be a cube root of unity, and use the binomial theorem to expand an expression involving $ \omega$, such as $ (1\plus{}\omega)^{2}000$[/hide]

A search for the words "binomial complex" (the very first words I tried; no quotes while searching) gives http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1954541149&t=154452

Thanks for the reference JBL, great help. And also for the solution, p4fn2w.  :)


Find all positive integers with two or more digits such that if we insert a 0 between the units and tens digits we get a multiple of the original number.

[hide]Write the number as 10a+b where a is a positive integer and b is a digit from 0 to 9. Then inserting a 0 as described changes the number to 100a+b. Since (100a+b)/(10a+b) = 10 - 9b/(10a+b) is an integer if and only if 9b/(10a+b) is, we need to find solutions to 10a+b | 9b. We need a divisor of 9b greater than or equal to 10 that ends in the digit b. There are 10 cases, and we check each separately (if there's a better way to do this i dont know it):

b=0: any a works

b=1: nothing

b=2: nothing

b=3: nothing

b=4: nothing

b=5: a=4

b=6: nothing

b=7: nothing

b=8: a=1

b=9: nothing

So, our final answers are: any multiple of 10, 45, and 18.[/hide]


I am looking for more information on Kingdom Fungi...
especially the way the reproduce...suggestions?


Let $ ABC$ be an acute-angled triangle, and let $ D, E$ be the feet of the perpendiculars from $ A, B$ to $ BC, CA$ respectively. Let $ P$ be the point where $ AD$ meets the semi-circle constructed outwardly on $ BC$ and $ Q$ be the point where $ BE$ meets the semi-circle constructed outwardly on $ AC$. Prove that $ CP\equal{}CQ$.

[hide]Note that $ \angle CPB$ is a right angle as it is inscribed in a semicircle, it follows that $ \triangle CDP \sim  \triangle BPC$, or $ \frac{CD}{CP}\equal{}\frac{CP}{BC}$, or $ CP^2\equal{} CD \cdot CB$, similarly $ CQ^2\equal{}CE \cdot CA$. So we have to show $ CD \cdot CB\equal{}CE \cdot CA$. Note that since $ \angle AEB\equal{}\angle ADB\equal{}90$, AEDB is cyclic, thus by power of a point $ CD \cdot CB\equal{}CE \cdot CA$.[/hide]

Yes.

We could also use Pythagoras' theorem, or the dot product.


inequalities
integration
inequalities unsolved

[b]Problem[/b]

[i] If $m$ and $n$ are natural numbers then the following inequality holds:

$\sum _{k=1}^n \frac{1}{k^m} < \sum _{k=1}^{m-1} \frac{1}{k^m} + \frac{1}{m-1} \frac{1}{(m-1)!}$ [/i]

Myth: I guess you don't need pass to the limit.  I didn't see at first glance that this problem is rather easy -$\int _{m-1}^n$ finishishes it. And it is really strict inequality. I'd try to optimizate it a little bit.


What is the value of the expression $ \left(2\frac{1}{2}\right) \left(4\frac{2}{3}\right) \plus{} 2\frac{1}{3} \plus{}4\frac{2}{3} \plus{}1$ when expressed as a mixed number?

change the 2 mixed numbers in the beginning to improper fractions
$ \left(\frac{5}{2}\right)\left(\frac{14}{3}\right)\plus{}7\plus{}1\equal{}\left(\frac{35}{3}\right)\plus{}8\equal{}11\frac{2}{3}\plus{}8\equal{}19\frac{2}{3}$


ARML
summer program
MathZoom
counting
distinguishability
AMC
AIME

This year's ARML Local questions and solutions are posted at our website: http://www.arml.com/arml_local/

The results will be finalized and posted by the end of the week.

Hope folks enjoyed it.

I dunno, but I thought that those were just about the rockingest problems that ever rocked.  :lol:

::possibly biased, for some reason::

Seriously though, that balls-in-boxes problem from the team question sounds neat. Someone should make a point out of systematically modifying and experimenting with it in large-group settings!  :)

Results are now posted [url=http://www.arml.com/arml_local/arml_local_main_page.php]here.[/url]

The ballsin boxes team round problem I believe ws ambiguous. It was not stated whether or not the boxes were distinguisable or not (unless I'm mssing something). Am I right?

2nd Place Bernoulli Division, did 1 place worse than last year, of course, this year there were more than 3 teams  :D

Meh.  12th place Archimedes...  The sad part is we would've had a 101 if, on the theme round problem 4, we would've correctly written (39, 7) instead of (3, 97), but w/e.  I also was pleasantly surprised to get an 8 on the individual round, as I expected to make stupid mistakes as badly as on the AIME I a few days prior and wind up with something <5.

I had revised the balls-in-boxes problem to reflect that the boxes were distinct, so that Pr([2,3,0,0,0]) = Pr([3,2,0,0,0]) = Pr([0,5,0,0,0]), and updated it everywhere (solutions, compressed problems, etc.) EXCEPT on the problem sheets that got printed for the students.  Apologies for that.

We moved from 2nd to 3rd in Bernoulli  :(

[quote="PaulDreyer"]The final results are on the ARML Local webpage (including some teams that I had accidentally left out), as are the results from the tiebreaker.  In-Sung Na won the tiebreaker among the individuals who scored 10 on the round.  The explanation for the tiebreaker is on the site as well.

http://arml.com/arml_local/[/quote]

That's an extremely interesting way to resolve a tiebreaker. 10 points for creativity! :P

Am I allowed to give a partial "you're welcome"?  :lol:

[quote="Differ"]
That's an extremely interesting way to resolve a tiebreaker. 10 points for creativity! :P[/quote]
The tiebreaker problem seems to be interesting....  kinda reminds me of programming contest problems.


The sum of the roots of the equation $4x^2 + 5 - 8x = 0$ is equal to:
\[ \text{(A)}\ 8 \qquad \text{(B)}\ -5 \qquad \text{(C)}\ -\frac{5}{4} \qquad \text{(D)}\ -2 \qquad \text{(E) none of these} \]

are you sure that there are real roots to this equation? Using the quadratic equation, I get (8+(or -) :sqrt:-16)/8
I should try guess and check... E

[hide]Regardless of wether there are real roots or not,
the sum of the roots of a quadratic equation in the form 0=ax2+bx+c
is equal to b/a
So the answer is -8/4= [b]-2 [/b]D
The roots of a quadratic are [-b+sqrt(b2-4ac)]/2a and [-b-sqrt(b2-4ac)]/2a
This adds up to -2b/2a=-b/a[/hide]

[quote="4everwise"]I didn't get any real roots either, so E.[/quote]

The roots occur in conjugate pairs

They occur in what?

Although the answers are not real, they can still be added together. Oh yeah that's right. Maybe it's -2...

Not sure what I was thinking earlier. There's a formula for this and I learned it in my first semmester of Algebra 2.

The formula is -b/a , so...

the answer is b (-5/4)


can someone figure this out and show [b]ALL [/b]the work..thanx

15/18 + 1/2 - 1/3 - 2/6 =

1) Learn common denominators.

$ \frac{15}{18}\plus{}\frac{9}{18}\minus{}\frac{6}{18}\minus{}\frac{6}{18}\equal{}\frac{15\plus{}9\minus{}6\minus{}6}{18}\equal{}\frac{12}{18}\equal{}\frac{6}{9}\equal{}\frac{2}{3}$

I did this quickly in my head so there might be a computational error but you get the idea.

Yeah. It's really easy to do fractions when you're in middle school honors. Just use common denominators, add/subtract, simplify, and there you have it. Easy as peas. :D  :)  :P

This looks like REALLY easy math
[hide="Solution"]
Convert everything to $ \frac{x}{18}$
$ \frac{15}{18}\plus{}\frac{1}{2}\minus{}\frac{1}{3}\minus{}\frac{2}{6}$
$ \equal{}\frac{15}{18}\plus{}\frac{9}{18}\minus{}\frac{6}{18}\minus{}\frac{6}{18}$
$ \equal{}\frac{15\plus{}9\minus{}6\minus{}6}{18}$
$ \equal{}\frac{24\minus{}12}{18}$
$ \equal{}\frac{12}{18}$
$ \equal{}\frac{2}{3}$
[/hide]

[quote="minicon"]This looks like REALLY easy math[/quote]
[quote="levans"]Please refrain from stating problems are easy.  While they may be easy for you, they may not be easy for others.[/quote]


Guys what your favorite Millenium Problem?
Even though Poincare was proven, I still be use it.

Mine is conjecture of Birch and Swinnerton-Dyer. I see that is also the highest rating a memeber can get. So I guess you agree with me.

Reimann's hypothesis. it's pretty kool, and that's my rating on AoPS. :D  :P  :thumbup:


[color=blue]Two circles C1 and C2 with centers O1 and O2 respectively intersect at A and B.O1B extended intersects C2 at E.O2B extended intersects C1 at F.MN is drawn parallel to EF through B which intersects C1 at N and C2 at M.AE and AF are joined. Proove that MN = AE + AF.[/color]

It is easy to acquire O1,F,E,O2 are concyclic
∠O1EF=∠O1O2F=∠AMB=∠AEB   because we have MN∥EF  =〉 ∠BEF=∠O1EF=∠EBM
AB=EM=BE1=BF1=NF     BEMA is an isosceles trapezoid  =〉 AE=BM
Analogously  BFNA is an isosceles trapezoid  =〉 AF=BN
=〉MN=BM+BN=AE+AF

Can you please explain How they are concyclic ?
And the rest of the part seems quite hazy to me
Could you please explain it to me in more detail ?
 :oops:  :blush:

∠O1BF=∠O2BE=∠O1FB=∠O2EB    So O1,F,E,O2 are concyclic
In my figure , E1,F1 are point that on line EF, so E1F1∥MN  =〉EE1MB,FF1NB are  isosceles trapezoids

[quote="A K Bhattacharyya"]Another thing.How can you say that The circle will always pass through A ?[/quote]
I don't use the condition A lies on the circle in my solution

[quote="A K Bhattacharyya"]then How can you say that O1O2B = AMB ?[/quote]
O1O2B=1/2 AO2B =AMB

But your diagram doesn't show AO2 joined.But Thanks anyway! :)  :wink:

Yes, I think too many lines will make the problem look confusion

If you wanna know why A lies on the circlr , look at this:
 ∠FAE= ∠FAB+ ∠EAB= ∠BF1F+ ∠BE1E= ∠FO1B= ∠EO2B
this means A,O1,F,E,O2 are concyclic
 :wink:


The average of the values 23 and $ x$ is 27. What is the positive difference of these two values?

$ 23\plus{}x\equal{}27(2)\equal{}54$, so $ x\equal{}54\minus{}23\equal{}31$.

Positive difference between 23 and 31 is $ 31\minus{}23\equal{}\boxed{8}$.


conics
parabola
geometry unsolved
geometry

Here is a parabola C:y^2=2x,two lines L1(y1=mx+p) and L2 (y2=nx+q)pass point P(1,1);L1 meet the parabola at point A,B;L2 meet the parabola at C and D ;Prove that A,B,C,D are concyclic if and only if m+n=0.

HERE IS AN IMAGE . MAYBE IT CAN HELP.. :D


calculus integration function logarithms limit algebra polynomial

Just as a reminder for those of you participating in kunny's "Today's Calculation of Integral" topics: being able to find elementary antiderivatives of selected elementary functions is a useful skill, but it is by no means all there is to calculus, or even all there is to integration.

As a change of pace, try this:

a) Prove that the improper integral $\int_0^1\frac{-\ln(1-x)}x\,dx$ converges.

b) Express $\int_0^1\frac{-\ln(1-x)}x\,dx$ as a series - a fairly familiar series, whose exact sum is known.

[quote="zhaobin"]the answer is $\frac{\pi ^2}{6}$? :)[/quote] Correct!

i knew there is a direct way to calculate $\int_0^1\frac{-\ln(1-x)}x\,dx $,
but I forget the detail,can somebody post it?

Thanks, Jimmy, I was about to ask for that myself. There is a part a), guys. And a lot of details on b) that haven't been shown here yet.

[quote="zhaobin"][quote="Jimmy"]Can someone post a solution, and proof that the integral converges?

Thanks.[/quote]
we have $\lim_{x \to 0}  \frac{-\ln(1-x)}x =1$
and $\lim_{x \to 1}  \frac{-\ln(1-x)(x-1)^k}{x} = 0$
k is any positive number.
is it enough? :)
indeed,I think we only prove   $\sum_{j}\frac{x^{j-1}}{j}$,converges,when $x \in (0,1)$[/quote]
 :) that's right.

There's one piece of justification I would add to what zhaobin wrote.

We have that $\int_0^1\frac{-\ln(1-x)}x\,dx$ converges and that

$\int_0^y\frac{-\ln(1-x)}x\,dx=\sum_{j=1}^{\infty}\frac{y^j}{j^2}.$

But how to we know that $\lim_{y\to1^-}\sum_{j=1}^{\infty}\frac{y^j}{j^2} = \sum_{j=1}^{\infty}\frac1{j^2}$ ?

The answer to that is [b]Abel's Theorem.[/b]

For the first part, since $-\ln(1-x)/x\leq -2\ln(1-x)$ for $x$ sufficiently close to $1$, the behavior of of the integral near $1$ is determined by the integral of, say

$\int^a_0 \ln xdx=(a\ln a -a)-\lim_{x\to 0} x\ln x$

Since $\ln x$ goes to infinity slower than any polynomial, it is clear that $x \ln x\to 0$ which establishes that that integral exists.  Since the integral exists over small closed intervals with right endpoint at $1$ and also over small closed intervals with left endpoint at $0$ (because the function goes to $1$ there) and the function is continuous over everywhere else on $[0,1]$, the integral exists.

As for the second part, we need some justification for the term by term integration.  The function we are integrating converges uniformly on $[\epsilon, 1-\epsilon]$ and by the previous remarks we can make the integrals over $[0,\epsilon]$, $[1-\epsilon,1]$ negligible.  By uniform convergence, we may interchange summation and integration.  This gives the desired result.


function
limit
calculus
calculus computations

Give an example of a bounded, continuous function $ f$ on $ (0,\infty)$ such that $ \lim_{x\rightarrow\infty}f(x)\equal{}0$ but $ f\notin L^{p}(0,\infty)$ for any $ p>0$.

All you need is sufficiently slow decay at infinity. $ f(x)\equal{}\frac1{\ln(x\plus{}2)}$ ought to do it.


Help me please!

[b]Find all natural numbers N the product of whose decimal digits is n^2 - 10n - 22. [/b]

Denote the product of the decimal digits of $n$ by $p(n)$. Let $n = \overline{a_{k}a_{k-1}\ldots a_{1}}$. Then \[p(n) = a_{k}a_{k-1}\ldots a_{1}\leq a_{k}\cdot 9^{k-1}\leq a_{k}\cdot 10^{k-1}\leq n.\]We are given that $p(n) = n^{2}-10n-22$, so $0 \leq n^{2}-10n-22 \leq n$. With the restriction that $n\geq 0$ and $n$ integer, we deduce that $11,855... \leq n \leq 12,728...$, so $n = 12$. And it's easy to check that $12^{2}-10\cdot 12-22 = 2 = p(12)$.

[quote="kay_genius253"][quote="Kurt Gödel"]Denote the product of the decimal digits of $n$ by $p(n)$. Let $n = \overline{a_{k}a_{k-1}\ldots a_{1}}$. Then
\[p(n) = a_{k}a_{k-1}\ldots a_{1}\leq a_{k}\cdot 9^{k-1}\leq a_{k}\cdot 10^{k-1}\leq n. \]
We are given that $p(n) = n^{2}-10n-22$, so $0 \leq n^{2}-10n-22 \leq n$. With the restriction that $n\geq 0$ and $n$ integer, we deduce that $11,855... \leq n \leq 12,728...$, so $n = 12$. And it's easy to check that $12^{2}-10\cdot 12-22 = 2 = p(12)$.[/quote]

I have a question. How did you get

p(n) = a_{k}a_{k-1}\ldots a_{1}\leq a_{k}\cdot 9^{k-1}\leq a_{k}\cdot 10^{k-1}\leq n. or in other words your first inequality?

Thanks![/quote]

The first inequality follows from the fact that $a_{i}\leq 9$, for $i = 1,2,\ldots, k-1$. (Because they're digits.) Multiply those $k-1$ inequalities.


inequalities
logarithms
limit
real analysis
real analysis solved

Let A_n be given a positive sequence.K_n is the smallest natural number such that the inequality :1+(1/2)+...+(1/K_n) >= A_1+...+A_n holds.
Prove that:there exits lim(K_(n+1)/K_n) iff exits limA_n.
Hint:lim(A_1+...+A_n-lnK_n)=0 and lim(A_n-ln(K_(n+1)/K_n))=0. :P  :cool:

I put $u_n=A_n$ (I wrote it in Word and I'm just pasting it).

Let $\displaystyle H_n=\sum_{i=1}^n \frac 1i$ be the $n$’th harmonic number and let $\displaystyle S_n=\sum_{i=1}^n u_i$. Let $c$ be Euler’s constant. Then $H_n=\ln n+c+\varepsilon_n\ (*)$, with $\displaystyle \lim_{n\to\infty}\varepsilon_n=0$.

Let’s observe that $\displaystyle S=\sum_{i=1}^{\infty}u_n<\infty\iff$ the sequence $k_n$ becomes constant for large enough $n$, so let’s assume that this is not the case (so we’re assuming $S=\infty$).

It’s easy to see that $\displaystyle \lim_{n\to\infty}H_{k_n}-S_n=0$, so $\displaystyle \lim_{n\to\infty}H_{k_{n+1}}-H_{k_n}-u_{n+1}=0$, meaning that $\displaystyle \lim_{n\to\infty}\ln\frac{k_{n+1}}{k_n}-u_{n+1}=0$ (by using $(*)$), and this proves what we want.


integration
trigonometry
logarithms
function
calculus
calculus computations

[quote="tdl"]Calculate $ \int{x\tan{x}dx}$[/quote]

http://integrals.wolfram.com/index.jsp

hello, we have
$ \int x\tan(x)\,dx \equal{}$
$ 1/2\,i{\it polylog} \left( 2, \minus{} \cos \left( 2\,x \right) \minus{} i\sin \left( 2 \,x \right) \right) \minus{} 1/2\,x\ln \left( \left( \cos \left( 2\,x \right) \plus{} 1 \right) ^{2} \plus{} \left( \sin \left( 2\,x \right) \right) ^{2} \right) \plus{} i \left( 1/2\,{x}^{2} \minus{} x\arctan \left( \sin \left( 2\,x \right) ,\cos \left( 2\,x \right) \plus{} 1 \right) \right)$.
Sonnhard.

Could somebody have a synthentic solution?

Use series for $ \tan(x).$
$ \int{x\tan (ax)dx} \equal{} \sum_{n \equal{} 1}^{\infty}{\frac {2^{2n}(2^{2n} \minus{} 1)B_{n}a^{2n \minus{} 1}x^{2n \plus{} 1}}{(2n \plus{} 1)!}}.$

$ B_{n}$ - Bernoulli numbers.

In summary, although the function $ f(x) \equal{} x \tan x$ is continous in any interval where $ cos x \neq 0$, its primitive cannot be expressed as a finite sum of elementary functions. Observe that this is more "the rule than the exception".


vector
combinatorics unsolved
combinatorics

If S is a finite set of non-zero vectors in the plane, then a maximal subset is a subset whose vector sum has the largest possible magnitude. Show that if S has n vectors, then there are at most 2n maximal subsets of S. Give a set of 4 vectors with 8 maximal subsets and a set of 5 vectors with 10 maximal subsets.

[hide]Suppose $ M$ be a maximal subset of $ S$(we assume that $ S\equal{}\{ \vec{OP_i} \}$. let $ \vec{l}\equal{}sum_{\vec{v}\in M} \vec{v}$. Then consider the distance form the end-points of the vectors in $ M$ to the line which is perpendicular to $ \vec{l}$. It is true that there exist $ \alpha$ such that all the vectors which has the distance larger than $ \alpha$ belong to $ M$ and the other not belong to $ M$.
In other words, for any maximal subset $ M$ we sort all the vectors in order of the distance from the end-points to the line which is perpendicular to $ \vec{l}$. Then there exists a $ k$ such that the first $ k$ vectors belong to $ M$ and others belong to $ S\minus{}M$.
Since there are at most $ n$ lines being able to gvie different sorted results of the vectors, there are at most $ 2n$ different maximal subsets.  [/hide]
Sorry for my implicit English..


How many sets of integers (a,b,c) are there such that abc=4000.

(10,10,40) is the same as (40,10,10)

The old AHSME problem distinguished (10,10,40) and (40,10,10) but now with this modification the problem is a little harder.

[quote="Tenoreoz"][hide="Pretty quick"]
Let $ a \equal{} 2^{a_{1}}5^{a_{2}}$, and similarly for the others. Thus, we wish to find
\[ \#(a_1 \plus{} b_1 \plus{} c_1 \equal{} 5)\times \#(a_2 \plus{} b_2 \plus{} c_2 \equal{} 3)
\]
Then ball in urns finishes it off. [/hide][/quote]

wrong because its not distinguishable.

i purposely altered the question.

in your case we would have (10,10,40) and (40,10,10) as different but in the new question, they are the same---so the answer is going to be a lot less.

for exmaple we could have (1,1,3) and (1,1,1) in your case
and also (1,3,1) and (1,1,1) for (a1,b1,c1) and (a2,b2,c2).  They are considered the same but in balls and urns, you made them different.

The solution does involve balls and urns but you have to count for some overcounts and then divide them.

[hide="Balls and urns"]First, let's name the exponents.  $ a_i$ is the exponent of the 2 in the $ i$th position of the solution. $ b_i$ is named likewise, but for 5.
Balls and urns gives us that there are 7 choose 2 or 21 ways to pick the $ a_i$s, not necessarily distinct, and 5 choose 2 or 10 ways to pick the $ b_i$s. This gives 210 nondistinct ways to pick the ways. This is way too many, so let's try to narrow it down.
There are 3 cases:
1. All the numbers are the same.
2. 2 numbers are the same.
3. All the numbers are different.
#1 is impossible because the cube root of 4000 is not an integer.
#2: We must have $ 5^a2^b, 5^a2^b, 5^{3\minus{}2a}2^{5\minus{}2b}$ where a is less than 2 and b is less than 3.  This gives 0 and 1 as answers for $ a$ and 0, 1, and 2 as answers for b, or $ 2\times 3\equal{}6$ solutions, namely
(10, 10, 40)
(5, 5, 160)
(20, 20, 10)
(4, 4, 250)
(2, 2, 1000)
(1, 1, 4000)
#3. If all 3 numbers are different, then there are a few subcases.
3a. 5 distribution: (1, 1, 1)
3b. 5 distribution: (0, 1, 2)
3c. 5 distribution: (0, 0, 3)

3a.  In this subcase, all the $ a_i$s must be different.  There is
(0, 1, 4) and
(0, 2, 3), corresponding to
(5, 10, 80)
(5, 20, 40)
3b. In this subcase, all are distinct.  So there are 7 choose 2 ways to divide the 2s up, or 21.
3c. In this subcase, 2 of the $ a_i$s must be different, and the third one does not matter.  So there are 9 answers:
(0, 1, 4)
(0, 2, 3)
(0, 3, 2)
(0, 4, 1)
(0, 5, 0)
(1, 2, 2)
(1, 3, 1)
(1, 4, 0)
(2, 3, 0)
Adding all these up, we get 9+2+21+6=38.[/hide]
Whew! Not easy.

[quote="Brut3Forc3"]This, of course, could be a lot harder, considering that you haven't specified that all the integers must be positive.  :P[/quote]
Umm I wouldnt say its a lot harder since you would just quadruple the answer if negatives were accounted for

since for every (a,b,c) that could work an (-a,-b,c) (a,-b,-c) or (-a,b,-c) would work so its 38*4=152.


n2 coins, of which exactly n are silver, are arranged at random in n rows, each containing n coins. What is the chance that at least one row occurs in which there is no silver coin?

please solve this problem as soon as possible

What is the chance that there is exactly one silver coin in each row? ;)

pretty sure you got it from here, but just for completeness

P(empty row)=1-P(no empty row)=1-P(at least one in each row)=1-P(1 per row)

so we want to solve probability for row j=1,2,..n that we have just one silver coin

$ P_j \equal{} \frac {n(n \minus{} (j \minus{} 1))[(n \minus{} j)n]![(n \plus{} 1 \minus{} j)(n \minus{} 1)]!}{[(n \minus{} j)(n \minus{} 1)]![(n \plus{} 1 \minus{} j)n]!} \equal{}$

$ \equal{} \frac {(n^2 \minus{} nj)\cdot (n^2 \minus{} nj \minus{} 1) ... (n^2 \minus{} nj \minus{} n \plus{} j \plus{} 1)} {(n^2 \minus{} jn \plus{} n \minus{} 1) \cdot (n^2 \minus{} jn \plus{} n \minus{} 2) ... (n^2 \minus{} jn \plus{} j)}$

and $ P \equal{} 1 \minus{} \prod_{j \equal{} 1}^n P_j$

this.. doesnt look to have an easy closed form?

You make the things overly complicated: just look at how the silver coins are placed. The total number of placements in the table is $ {n^2\choose n}$ and the number of placements when you have one coin in each row is $ n^n$. ;)


Prove that given any real numbers $\alpha <\beta$, there exists infinitely many rational numbers in the interval $[\alpha,\beta]$.

[hide]Take the average of $\alpha$ and $\beta$, then take the average of that number and $\alpha$ (and/or $\beta$), etc. Continuing this pattern infinitely gives an infinite amount of rational numbers.[/hide]
Btw, proofs don't belong in Classroom Math, even if they're this easy.

but how can you be sure that the average of $\alpha$ and $\beta$ will be rational?

and for the latter part, sorry :blush:  i didn't know that.

[hide]
If they are not rational here's something(it's not a proof, though)
There will be a difference of at least 0.0(put as many 0's as necessary)1
Add to $\alpha$ rounded to the nearest something Then you have the number 0.0(necessary 0's +1 0)1
Add to $\alpha$rounded to the nearest something You can infinitely continue this.
[/hide]

[hide="my solution"]
[b]Lemma:[/b] For any two real numbers $\alpha$ and $\beta$, there exists a rational number in $[\alpha,\beta]$
[b]Proof:[/b] consider the difference $d=\beta-\alpha$.
we can choose an integer $n$ large enough such that $nd>1$
that is $n\beta-n\alpha>1$.
so there exists an integer $k$ between $n\alpha$ and $n\beta$(think about number line)
then $n\alpha< k< n\beta$
thus $\alpha< \frac{k}{n}< \beta$ and the lemma is proved.$\blacksquare$
now for our problem, by the lemma, there exists a rational number $r$in $[\alpha,\beta]$
but there exists anothere one in $[\alpha,r]$ and another one in $[r,\beta]$
continuing this process, we find infinitely many rational numbers in $[\alpha,\beta]$[/hide]

[hide="A silly approach"] If either $\alpha, \beta$ are irrational consider rational convergents.  Since they get arbitrarily close from both sides, one of them must be within the range.

If neither $\alpha$ nor $\beta$ are irrational then their average is rational. [/hide]

[hide="Yet ANOTHER approach"]

We get a fraction inbetween a and b, juse multiply the numerator and denominator by n, and just add or subtract 1 from the numerator, and then multiply the numerator and denominator by ANOTHER number, then do it all over again.[/hide]

[quote="1=2"]We get a fraction inbetween a and b[/quote]

how can you be sure that there exists one?

there's a delta for every epsilon
It's a fact that you can always count upon
there's a $\delta$ for every $\epsilon$

and once and again
there's also an $N$

[quote="Altheman"][quote="Teki-Teki"][hide="a sillier approach"]the rational numbers $\mathbb{Q}$ are dense in $\mathbb{R}$  :rotfl: [/hide][/quote]

i have heard that term thrown around...what does it mean precisely? what are the applications of proving a set is dense in another set?[/quote]

I believe that dense means that there's one of one set between any two in the other set (there's a rational between any two reals in this case)

[quote="me@home"][hide="this problem has gone on for too long with incomplete proofs"]
We show that there are an arbitrarily large and hence infinite amount of rationals in the interval.
Consider the interval from $0$ to arbitrarily large $N$. There are like $N+1$ integers in this interval. Now scale down the interval to be the same size as the given interval, but starting at zero. The $N$ ish integers become $N$ ish rationals. Now we can "drag" the interval over by a constant until it coincides with previous interval and the rationals are clearly mapped to new rationals. But $N$ is indeterminately large; thus there are at least $N$ rationals in any interval implying there are arbitrarily many implying there are infinite.[/hide][/quote]

why? what's wrong with my proof? :huh: and anothere fact, its high school basics. so we are only allowed to use elementary facts here.

well what if we replace [i]rational[/i] by [i]irrational[/i] in the original problem?
i found nothing yet. :( and please no more silly or sillier approaches. :rotfl:

[b]Lemma:[/b]  A real number is irrational iff its decimal representation is aperiodic.

Clearly there exist infinite such numbers between any two reals.

[quote="nayel"]but how can you be sure that the average of $\alpha$ and $\beta$ will be rational?[/quote]The sum of two rationals is rational, the quotient of two rationals is rational, therefore the average of two rationals is rational.

$\alpha, \beta$ are two given [i]reals[/i], not two given rationals.


[quote="vishalarul"]Here is Problem #4 from the 2001 State Target Round.[/quote]
1 and 3 couldn't happen at the same time and 1 and 4 couldn't either
from there it's pretty easy to see that the answer is 3 with the set 2,8,35

what does this question have to do with external circles?

Oops. I forgot to change the title.  :blush: Also, ragnarok23 is correct. Well anyway, here is the original "external circles problem":

[hide]The belt touches each circle on $\frac{1}{3}$ of it's circumference and the part of the belt that doesn't touch any circle is 2r or 4 so then $4\pi$ for the parts touching the circle and then $4*3=12$ and then the answer is $12+4\pi$ :P [/hide]

i remember doing this problem w/ dan tobias. he got the right answer, i got the wrong one, and we were doing it together. i thought he was randomly putting down a number :( . anyways, how do you know the belt part between the circles are 4 each?

-jorian

Take the top part, between the two circles at the top. Translate that part (which is a line segment) straight down, and you'll see that its endpoints become the centers of the two top circles.


The three roots of the cubic equation $9x^{3}-36x^{2}+cx-16\;=\;0$ are in arithmetic progression. Determine the value of $c$ and the three roots.

[hide="Hint"]Vieta+system of 2 equations[/hide]

i saw this on nyc chapter tournament questions. IS this part of geometry? or is it on a harder level for which I don't understand.

could you explain further how you got the answers? I don't get these polynomial things like this...

He just used Vieta's formulas for a cubic.


In a certain lottery, 7 balls are drawn at random from n balls numbered from 1 to n. If the probability that no pair of consecutive numbers is drawn is the same as the probability of drawing exactly one pair of consecutive numbers, then determine n.

The given solution has a shortcut for doing it, but I don't understand how it works...

[hide="given solution"]
watered down to the main parts:
number of ways to choose 7 balls that are no consecutive:
$ \binom{n \minus{} 6}{7}$
number of ways to chose 7 balls with exactly one pair of consecutive numbers:
$ \binom{n \minus{} 6}{6}*6$
Equate and solve...
how did they get those formulas?
[/hide]

[quote="alkjash"]Well, I guess you could view it as a distribution problem...
So distribute (n-7) balls into 8 slots, such that the middle six slots all have a positive number of balls.
So you put 6 balls into these slots, and you have a simple distribution problem of (n-13) balls into 8 slots.
The number of ways to do this is $ \binom{n \minus{} 6}{7}$.

For the second part, the number of ways to choose 7 balls with exactly one pair is basically distributing (n-7) balls into 7 slots, and then choosing one of the dividers to have 2 in 6 ways, which gives $ \binom{n \minus{} 6}{6}*6$ different ways.[/quote]

I don't understand how distributing the balls into the different slots models the given situation with the consecutive integers on the balls. Could you elaborate more on that?

Thanks!

Well, now, you have them order for example like this:
oooo|oo|ooo|o|o|oo|o|ooo
So the first ball chosen is the 5th one, then the 8th, then 12th, 14th, 16th, 19th, and 21st.

Thanks so much for your helpful replies!
... so n=
[hide] 54 [/hide]
, right?

I have no clue; didn't actually do the problem.  :P


Combinatorics are by far my worst subject within math.  I could use some help on this question...

What is the number of ways to order the 26 letters of the alphabet so that no two vowels (a, e, i, o, and u) occur consecutively?

Then look at all possible ways to choose two vowels - they are $ \binom{5}{2}$.
And then we may look at two sequential vowels as an one element.
But we need to take in mind the order of the vowels ... so we must multiply by 2.
Total number of sequences of 26 letters is 26!.
So Answer in my opinion should be: $ 26!\minus{}2.\binom{5}{2}.25! \equal{} 6.25!$
Am I right?

P.S. If I'm not right - you may take smaller sets ot letters and to verify your hypothesis.

[quote="patham"]You are overcounting those arrangements in which there are more than 1 pairs of vowels.
The right answer is $ 5!*21!*($No of solutions for $ x_{1}\plus{}x_{2}\plus{}x_{3}\plus{}x_{4}\plus{}x_{5}\plus{}x_{6}\equal{} 21)$
with each $ x_{i}\geq 1$
So answer is $ 5!*21!*C^{20}_{5}$[/quote]

Can you elaborate on your answer?  Like explain the reasoning behind each number.  I just don't quite understand.

No of ways of arranging the vowels only is $ 5!$
No of ways of arranging the consonants only is $ 21!$
Now we have to mix the two arrangements
Let $ x_{1}$ be the no. of consonants before 1st vowel, $ x_{2}$ be the no. of consonants between the first and second vowels and so on
Hence $ \sum_{i\equal{}1}^{6}{x_{i}}\equal{}21$
Since no two vowels are consecutive hence each $ x_{i}\geq 1$
Let $ y_{i}\equal{}x_{i}\minus{}1$
Now $ \sum_{i\equal{}1}^{6}{y_{i}}\equal{}15$
Hence each $ y_{i}>0$
No of solutions of this equation is $ {n\plus{}r\minus{}1\choose r\minus{}1}$ where $ n\equal{}15$ and $ r\equal{}6$
Hence multiplying these we get $ 5!*21!*{20\choose 5}$


ceiling function
algebra unsolved
algebra

Let $P\in\mathbb{R}[x]$ such that for infty of integer number $c$ : Equation $P(x)=c$ has more than one integer root. Prove that $P(x)=Q((x-a)^{2})$, where $a\in\mathbb R$ and $Q$ is a polynomial.

First this is clear for $P$ constant, so assume from now on $\deg P\geq 1$.

By assumption $P(x)\in\mathbb{Z}$ for an infinite number of $x\in\mathbb{Z}$.
So using [url=http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html]Lagrange Interpolation[/url] we get $P(x)\in\mathbb{Q}[x]$ and $mP(x)\in\mathbb{Z}[x]$ for some $m\in\mathbb{N}$.

If $P(x)=c\in\mathbb{Z}$ has more than one integer root, then so does $mP(x)=mc\in\mathbb{Z}$,
and if $mP(x)=Q((x-a)^{2})$ for some $Q\in\mathbb{R}[x]$ and $a\in\mathbb{R}$, then $P(x)={1\over m}Q((x-a)^{2})$ with ${1\over m}Q(x)\in\mathbb{R}[x]$.
So we can replace $P(x)$ with $mP(x)$ and assume from now on $P(x)\in\mathbb{Z}[x]$.

Say $P(x)=\alpha x^{n}+\beta x^{n-1}+\cdots$ with $n=\deg P\geq 1$ and $\alpha,\beta\in\mathbb{Z}, \alpha \not =0$.
Then $\tilde{P}(x): =n^{n}\alpha^{n-1}P({x\over n\alpha})=x^{n}+\beta nx^{n-1}+\cdots\in\mathbb{Z}[x]$
and if $P(x)=c\in\mathbb{Z}$ has integer roots $k\not =l$, then $\tilde{P}(x)=n^{n}\alpha^{n-1}c\in\mathbb{Z}$ has integer roots $n\alpha k\not =n\alpha l$.
Also if $\tilde{P}(x)=Q((x-a)^{2})$ for some $Q\in\mathbb{R}[x]$ and $a\in\mathbb{R}$, then $P(x)={Q(n^{2}\alpha^{2}(x-{a\over n\alpha})^{2})\over n^{n}\alpha^{n-1}}$ with ${Q(n^{2}\alpha^{2}x)\over n^{n}\alpha^{n-1}}\in\mathbb{R}[x]$.
So we can replace $P(x)$ with $\tilde{P}(x)$ and assume $P(x)=x^{n}+\beta nx^{n-1}+\cdots\in\mathbb{Z}[x]$ for some $\beta\in\mathbb{Z}$.

Then $P(x-\beta)=x^{n}+R(x)$ for some $R(x)\in\mathbb{Z}[x]$ with $\deg R\leq n-2$.
If $P(x)=c\in\mathbb{Z}$ has integer roots $k\not =l$, then $P(x-\beta)=c$ has integer roots $k+\beta\not =l+\beta$,
and if $P(x-\beta)=Q((x-a)^{2})$ for some $Q\in\mathbb{R}[x]$ and $a\in\mathbb{R}$, then also $P(x)=Q((x-(a-\beta))^{2})$.
So we can replace $P(x)$ with $P(x-\beta)$ and assume $P(x)=x^{n}+R(x)$ for some $R(x)\in\mathbb{Z}[x]$ with $\deg R\leq n-2$.

And now write $R(x)=a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}$ and set $S(x)=|a_{n-2}|x^{n-2}+\cdots+|a_{1}|x+|a_{0}|$.
Then $|R(x)|\leq S(|x|)$ for all $x\in\mathbb{R}$ and $S(x)\leq S(y)$ for all $0\leq x\leq y$.
Also $\deg S\leq n-2$ and therefore ${2S(x)\over n(x-1)^{n-1}}\mathop{\longrightarrow}_{x\to\infty}0$. (also true for $n=1$ and $S\equiv 0$)
So we find $C>1$ with $2S(x)<n(x-1)^{n-1}$ for all $x>C$.
$[-C,C]$ contains only finitely many integers with only finitely many images under $P$.
So by assumption there are infinitely many $c\in\mathbb{Z}$ with integer roots $k\not =l$ of $P(x)=c$ s.t. $|k|\geq |l|>C$.
But then for each such $c$ we have $-k=l$ and $c=P(k)=P(-k)$,
or otherwise $P(k)=P(l)$ for some $k,l\in\mathbb{Z}$ with $|k|>|l|>C$.
But then $k^{n}-l^{n}=R(l)-R(k)$ with
$|R(l)-R(k)|\leq |R(l)|+|R(k)|\leq S(|l|)+S(|k|)\leq 2S(|k|)<n(|k|-1)^{n-1}$
and
$|k^{n}-l^{n}|\geq |\ |k^{n}|-|l^{n}|\ |=|k|^{n}-|l|^{n}\geq |k|^{n}-(|k|-1)^{n}\geq n(|k|-1)^{n-1}$.
Contradiction!
So $P(x)-P(-x)$ has an infinite number of roots,
and therefore $P(x)$ is even and $P(x)=Q(x^{2})$ for some $Q(x)\in\mathbb{R}[x]$.

You are true and your solution is nice! Thank you very much!
Part 1
[quote="olorin"]First this is clear for $P$ constant, so assume from now on $\deg P\geq 1$.

By assumption $P(x)\in\mathbb{Z}$ for an infinite number of $x\in\mathbb{Z}$.
So using [url=http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html]Lagrange Interpolation[/url] we get $P(x)\in\mathbb{Q}[x]$ and $mP(x)\in\mathbb{Z}[x]$ for some $m\in\mathbb{N}$.

If $P(x)=c\in\mathbb{Z}$ has more than one integer root, then so does $mP(x)=mc\in\mathbb{Z}$,
and if $mP(x)=Q((x-a)^{2})$ for some $Q\in\mathbb{R}[x]$ and $a\in\mathbb{R}$, then $P(x)={1\over m}Q((x-a)^{2})$ with ${1\over m}Q(x)\in\mathbb{R}[x]$.
So we can replace $P(x)$ with $mP(x)$ and assume from now on $P(x)\in\mathbb{Z}[x]$.

Say $P(x)=\alpha x^{n}+\beta x^{n-1}+\cdots$ with $n=\deg P\geq 1$ and $\alpha,\beta\in\mathbb{Z}, \alpha \not =0$.
Then $\tilde{P}(x): =n^{n}\alpha^{n-1}P({x\over n\alpha})=x^{n}+\beta nx^{n-1}+\cdots\in\mathbb{Z}[x]$
and if $P(x)=c\in\mathbb{Z}$ has integer roots $k\not =l$, then $\tilde{P}(x)=n^{n}\alpha^{n-1}c\in\mathbb{Z}$ has integer roots $n\alpha k\not =n\alpha l$.
Also if $\tilde{P}(x)=Q((x-a)^{2})$ for some $Q\in\mathbb{R}[x]$ and $a\in\mathbb{R}$, then $P(x)={Q(n^{2}\alpha^{2}(x-{a\over n\alpha})^{2})\over n^{n}\alpha^{n-1}}$ with ${Q(n^{2}\alpha^{2}x)\over n^{n}\alpha^{n-1}}\in\mathbb{R}[x]$.
So we can replace $P(x)$ with $\tilde{P}(x)$ and assume $P(x)=x^{n}+\beta nx^{n-1}+\cdots\in\mathbb{Z}[x]$ for some $\beta\in\mathbb{Z}$.

Then $P(x-\beta)=x^{n}+R(x)$ for some $R(x)\in\mathbb{Z}[x]$ with $\deg R\leq n-2$.
If $P(x)=c\in\mathbb{Z}$ has integer roots $k\not =l$, then $P(x-\beta)=c$ has integer roots $k+\beta\not =l+\beta$,
and if $P(x-\beta)=Q((x-a)^{2})$ for some $Q\in\mathbb{R}[x]$ and $a\in\mathbb{R}$, then also $P(x)=Q((x-(a-\beta))^{2})$.
So we can replace $P(x)$ with $P(x-\beta)$ and assume $P(x)=x^{n}+R(x)$ for some $R(x)\in\mathbb{Z}[x]$ with $\deg R\leq n-2$.[/quote]

Part 2

[quote]And now write $R(x)=a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}$ and set $S(x)=|a_{n-2}|x^{n-2}+\cdots+|a_{1}|x+|a_{0}|$.
Then $|R(x)|\leq S(|x|)$ for all $x\in\mathbb{R}$ and $S(x)\leq S(y)$ for all $0\leq x\leq y$.
Also $\deg S\leq n-2$ and therefore ${2S(x)\over n(x-1)^{n-1}}\mathop{\longrightarrow}_{x\to\infty}0$. (also true for $n=1$ and $S\equiv 0$)
So we find $C>1$ with $2S(x)<n(x-1)^{n-1}$ for all $x>C$.
$[-C,C]$ contains only finitely many integers with only finitely many images under $P$.
So by assumption there are infinitely many $c\in\mathbb{Z}$ with integer roots $k\not =l$ of $P(x)=c$ s.t. $|k|\geq |l|>C$.
But then for each such $c$ we have $-k=l$ and $c=P(k)=P(-k)$,
or otherwise $P(k)=P(l)$ for some $k,l\in\mathbb{Z}$ with $|k|>|l|>C$.
But then $k^{n}-l^{n}=R(l)-R(k)$ with
$|R(l)-R(k)|\leq |R(l)|+|R(k)|\leq S(|l|)+S(|k|)\leq 2S(|k|)<n(|k|-1)^{n-1}$
and
$|k^{n}-l^{n}|\geq |\ |k^{n}|-|l^{n}|\ |=|k|^{n}-|l|^{n}\geq |k|^{n}-(|k|-1)^{n}\geq n(|k|-1)^{n-1}$.
Contradiction!
So $P(x)-P(-x)$ has an infinite number of roots,
and therefore $P(x)$ is even and $P(x)=Q(x^{2})$ for some $Q(x)\in\mathbb{R}[x]$.[/quote]
But why you found it? You start  for polynomias  where $a$ must is $0$?  :maybe:

[quote="N.T.TUAN"]But why you found it? You start  for polynomias  where $a$ must is $0$?  :maybe:[/quote]
In Part 1 I simply made $P(x)$ more "likeable" step by step, not knowing where it's going.
And I also did a few more things, which I didn't need in the end, like $\deg P$ even and $2a\in\mathbb{Z}$.
I wanted the symmetrie point $a$ to move to $0$, knowing that it's pretty much the second coefficient of $Q((x-a)^{2})$.
And in Part 2 I had an equation $k^{n}-l^{n}=R(l)-R(k)$ with one side growing stronger than the other,
knowing $(x+1)^{n}-x^{n}$ has degree $n-1$.
Easy to get a contradiction from there.
Does that help?  :maybe:

This problem can in fact be generalised with shorter proof.  :D

If $P(x)\in\mathbb{R}[x]$ with $P(k)=P(l)$ for infinitely many $(k,l)\in\mathbb{Z}^{2},k\not =l$,
then $P(x)=Q((x-{a\over 2})^{2})$ for some $Q(x)\in\mathbb{R}[x]$ and $a\in\mathbb{Z}$.

[u]Proof:[/u]
Again this is clear for $P$ constant, so assume from now on $\deg P\geq 1$.

We can replace $P(x)$ with $\lambda P(x)$ for any $\lambda\in\mathbb{R}^{*}$ and assume $P(x)$ has leading coefficient $1$.
Say $P(x)=x^{n}+\alpha x^{n-1}+\cdots$ with $n=\deg P\geq 1$ and set $a: =-{2\alpha\over n}$.
Then $P(x+{a\over 2})=x^{n}+R(x)$ and $P(x)=(x-{a\over 2})^{n}+R(x-{a\over 2})$ for some $R(x)\in\mathbb{R}[x]$ of degree $\leq n-2$.
So $P(k)=P(l)$ gives $(k-{a\over 2})^{n}-(l-{a\over 2})^{n}=R(l-{a\over 2})-R(k-{a\over 2})$ for all $(k,l)\in\mathbb{Z}^{2}$.
Write $R(x)=a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}$ and set $S(x)=|a_{n-2}|x^{n-2}+\cdots+|a_{1}|x+|a_{0}|$.
Then $|R(x)|\leq S(|x|)$ for all $x\in\mathbb{R}$ and $S(x)\leq S(y)$ for all $0\leq x\leq y$.
Set $\mu: =\min\{1,[a]+1-a,a+1-\lceil a \rceil\}>0$.  ($\lceil a \rceil$ is the smallest integer $\geq a$)

Then for all $(k,l)\in\mathbb{Z}^{2}$ with $k\not =l$ and $k+l\not =a$ we have $|\ |k-{a\over 2}|-|l-{a\over 2}|\ |\geq \mu$:
If $k-{a\over 2}$ and $l-{a\over 2}$ have the same signs, then $|\ |k-{a\over 2}|-|l-{a\over 2}|\ |=|k-l|\geq 1$.
Otherwise $|\ |k-{a\over 2}|-|l-{a\over 2}|\ |=|k+l-a|$.
Then if $k+l>a\geq [a]$ we have $|k+l-a|=k+l-a\geq [a]+1-a$,
and if $k+l<a\leq \lceil a \rceil$ we have $|k+l-a|=a-k-l\geq a+1-\lceil a \rceil$.

Now $x^{n}-(x-\mu)^{n}\in\mathbb{R}[x]$ has degree $n-1$ and leading coefficient $n\mu>0$. Also $\deg S\leq n-2$.
So ${2S(x)\over x^{n}-(x-\mu)^{n}}\mathop{\longrightarrow}_{x\to\infty}0$ and there exists $C>0$ with $2S(x)<x^{n}-(x-\mu)^{n}$ for all $x>C$.

$[{a\over 2}-C,{a\over 2}+C]$ contains only finitely many integers and each $P(x)=c$ has only finitely many roots.
So by assumption there are infinitely many $(k,l)\in\mathbb{Z}^{2},k\not =l$ with $P(k)=P(l)$ and $|k-{a\over 2}|\geq |l-{a\over 2}|>C$.
Then for all such $(k,l)$ we have $k+l=a$,
or otherwise $|k-{a\over 2}|-|l-{a\over 2}|\geq \mu$, and then for $y=k-{a\over 2}$ and $z=l-{a\over 2}$
we have $y^{n}-z^{n}=R(z)-R(y)$ and $|z|\leq |y|-\mu$.
But $|y^{n}-z^{n}|\geq |\ |y^{n}|-|z^{n}|\ |=|y|^{n}-|z|^{n}\geq |y|^{n}-(|y|-\mu)^{n}$
and $|R(z)-R(y)|\leq |R(z)|+|R(y)|\leq S(|z|)+S(|y|)\leq 2S(|y|)<|y|^{n}-(|y|-\mu)^{n}$,
as $|y|>C$. Contradiction!

So $k+l=a$ and $P(k)=P(l)$ for infinitely many $(k,l)\in\mathbb{Z}^{2}$,
and then $a\in\mathbb{Z}$ and $P(x)-P(a-x)$ has an infinite number of roots.
So $P(x)=P(a-x)$ and $P({a\over 2}+x)$ is even.
Therefore $P({a\over 2}+x)=Q(x^{2})$ and $P(x)=Q((x-{a\over 2})^{2})$ for some $Q(x)\in\mathbb{R}[x]$.


If $ a,b,c\ge 0$ and $ a\plus{}b\plus{}c\equal{}\frac{3}{\sqrt{2}}$, prove that:
$ \frac{a^{3}\plus{}b^{3}\plus{}c^{3}}{3}\plus{}\frac{\sqrt{5\left(a^{2}\plus{}b^{2}\plus{}c^{2}\right)}}{\sqrt{6}}\plus{}abc\plus{}\frac{2\sqrt{2}\minus{}\sqrt{5}}{2}\ge a\sqrt[3]{ab\plus{}ac}\plus{}b\sqrt[3]{ba\plus{}bc}\plus{}\plus{}c\sqrt[3]{ca\plus{}cb}$.

http://www.math10.com/forum/viewtopic.php?p=557#557
Here's a solution :)


Hmm, I thought about holding a different kind of math jam than the usual ones...Here's what we'd probably do:
1 hr. countdown
1 hr. "problem discussion"
The rest of the time just miscellaneous stuff. It seems kinda long though, what do you guys think?
It would probably start around 6:30 EST, or 3:30 PST.

starts a little early for me, but it sounds cool.
1 hour countdown
1 hour problem discussion
is really enough. The miscellaneous=spam.

also, what happened to typing up the national tests?

i couldn't go to the one tonight. i had to go to youth group.  they should schedule the math jams on days other than wednesdays, maybe tuesdays. i never seem to have anything on tuesdays.

Hey you guys should post AIM math Jams here[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79720]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79720[/url]

I put something on my blog about that coach needing it til saturday...
I added this to the sticky topic.