Am using dialup right now, but by tonight should have DSL.

w00t im using DSL! Leaving all you snails in my dust!

lol jp

No option for T1, T2, T3, or Internet 2???

LOL jk, I highly doubt anyone here has that in their home, unless you are some one with too much money for you own good.

Cable. *nods*

Bah I need to get rid of Time Warner. Any suggestions?

DSL!  Used to be cable...but DSL is cheaper.  :D

dial up!
the sad thing is... before i got my new laptop the computer i was using was too slow for the dial-up data transfer speed anyway...

cable! more expensive, but faster than dsl =D

[quote="shobber"]Dialup......30 kb/ sec......So sometimes I am afraid to open the topics which have been replied by darij or Yetti.[/quote]

Why?

I have cable too, getting expensive...

[quote="236factorial"][quote="shobber"]Dialup......30 kb/ sec......So sometimes I am afraid to open the topics which have been replied by darij or Yetti.[/quote]

Why? [/quote]
I think that's because they often post very long (and complete!) solutions, as well as some large geometric figures.

I use dialup (AOL). I've got a new version of AOL recently, but it's still kinda slow :(

Go SBC Yahoo DSL.  That one's best cause of all the acronyms.

[quote="math92"]what does DSL stand for?
[hide]i don't know, i'm asking you!![/hide][/quote]
Digital Subscriber Lines

dial up!!! *grumble grumble* ...no matter, i'll be off to college in a year...

w00t cable...altho i live in the middle of nowhere and it's kinda spotty everynow and then...by which i mean all of the bloody time

but hey it's cable  :P

[quote="math92"]what does DSL stand for?
[hide]i don't know, i'm asking you!![/hide][/quote]

digital subscriber line. Also, what about SDSL and ADSL? You know, SDSL is faster than DSL, and ADSL is slower.  :?:

i have Internet Explorer, rarely use AOL

But SDSL is over 10MBPS true connection speed.  :)


geometry
combinatorics unsolved
combinatorics

A tramp has a coat of area 1 with 5 patches; each patch has area 1/2. Prove that two patches exist with common area>=1/5

It has almost certainly appeared repeatedly ... do a search.


Let $b^n=b^{n-1}+1$ ($n\in N^*, b\geq 1, n\geq 3$)
Prove that $b^n-nb+1<0$

[b]EDIT:[/b] I did not see the first condition :oops:


Okay, I just got very confused by a very ambiguous site from mathworld (http://mathworld.wolfram.com/WholeNumber.html)[url=http://mathworld.wolfram.com/WholeNumber.html]mathworld.wolfram.com/WholeNumber.html[/url]
For the purposes of mathcompetitions, can someone provide me with a definition for natural number, whole number, and counting number? Thanks.

The natural and counting numbers are the same.
They are 1,2,3,4,5...
Whole numbers are the same except they include 0
They are
0,1,2,3,4,5,6,7...
You rarely hear the term natural or counting numbers on MATHCOUNTS, though. It is almost always "positive integers"

Think natrual/counting numbers as the numbers they taught you in preschool.
1,2,3,4,5... (The idea of zero and negetive numbers is coufusing to preschool kids)

Think whole numbers as the numbers they taught you in kindergarden/1st grade.
0,1,2,3,4... (The idea of zero should be understandable by kindergarden/1st grade kids, but not negetive numbers)

I don't know if that helps. That's what they made us do in preschool and kindergarden/1st grade. But I knew about the integers when i was 3.

Life was so easy back then.

I should add, always look for Non-negative (0 included) and Positive (no 0).

So, Counting/Natural=positive

Whole number=non-negative.

natural=counting numbers: so 1,2,3,4,5,6,7,8 well you get the idea
whole numbers just have 0 and all the natural numbers.


function
calculus
integration
induction
real analysis
real analysis solved

Note R+ = the non negative real numbers set and R = the real numbers set.
Let A be a positive real number.
Find all the continuous functions f such that :
- f : R+ ---> R
- f(x) = Integral(0 =< t =< A*x){f(t).dt} , for all x of R+

See you

Sorry Moubi, I dont agree your proof ..  It's Ok if A =< 1 ..
But could you explain  the case : A > 1 ?!  I suppose it's very easy for the famous "preparatory classes" ... !

I didn't it is easy,  i considered  the case 0<=A<=1
------------------------------------------------------------

Another way:

The function g:R+--->R,

g(A) = int_[0;Ax]f(t)dt - f(x) is derivable as

a function of A variable,   g(A) = 0 by the hypothesis

g'(A) = x.f(Ax) - 0 = 0

for x>0 =>for any A >= 0 f(Ax) = 0

Sorry Guest , I dont agree your proof .. Dont forget : f might depend on A ..
See you ..

For 0 \leq A \leq 1 see the post Fri sept 05, 2002 11:46 pm

For A >1 it seems more interesting, have a look on Fabius function

http://www.math.ohio-state.edu/~edgar/selfdiff

Thank you Diogene  for this problem i never  studied the case A>1

oubi

Let F distribution function  of
X= \sum_{k \geq 1}2 -nR_k , R_k are independant random
variables, uniformly distributed on [0;1]
Consider f the unique function on [0;+\infty[ such that

f(x) = F(x) on [0;1]

f(x) = 1- f(1-x) on ]1,2]

f(x) = - f(x-2 n)  on ]2 n;2 n+1]

Exercice : Check f(x) = \int_[0;2x] f(t) dt for any x  \in R+

reference:
J.Fabius, "A probalistics example of nowhere analytic C \infty function", Z.Wahrsch.Verw.Geb.5 (1966) 173-174

(in math research library)

"This exercice is well known in preparatory classes (math-sup; math-sp)" -- What classes are that?

Great Moubi .. I didnt know these  Fabius  functions , and I was looking for a such function !!!  Thank you very much ..
Just one remark : if F is the Fabius function , from X = Sum[k>=1]{R(k)/(a^k)}  with a > 1,
I think its more convenient  to define f such that
f(x) = F(x)       , for 0 =< x =< min((a-1)/a ,1/a)
f(x) = f(x/a)/a , for  min((a-1)/a ,1/a) =< x
Thank you again for the reference.   Im still looking for  the book


geometry
circumcircle
radical axis
geometry proposed

Let $ABC$ be a triangle. Let $D,E,F$ be points on $\overline{BC},\overline{CA}$ and $\overline{AB}$ respectively, such that $AD,BE$ and $CF$ meet in a single point. Suppose that $ACDF$ and $BCEF$ are cyclic quadrilaterals. Prove that $AD\perp BC, BE\perp AC$ and $CF\perp AB$.

someone post a solution please? i know it's an easy problem... :maybe:

Some hints by Menelaus theorem we can show $AEBD$ is cyclic too, we well know Miquel theorem circumcircle of triangl $AEF,CDE,BFD$ are concurrent at a point $M$ call concurrent point of $AD,BE,CF$ be $M'$ we will show $M\equiv M'$ indeed we have $A$ is radical center of the circles $(BFD),(CDE),(BCEF)\Rightarrow\ A,M',D$ are colinear similarly we get $M\equiv M'$ thus the quadrilaterals $AEMF,CDME,BFMD$ are cyclic, too.By calculate the angles we can show $M$ is othorcenter.

Suppose cevians$AD,BE,CF$ meet in $H$

because $ACBF$ and $CEBF$ cyclic so $\angle ADC=\angle AFC$,$\angle FBE=\angle FCE$ and $\angle CEB=\angle CFB=\pi-\angle AFC$

so $EHDC$ is cyclic too..

because $\angle FBE=\angle FCE=\angle HDE$ so $ABED$ cyclic or $\angle AEB=\angle ADE \rightarrow \angle ADB =\frac{\pi}{2}$

the conclusion follows

here's my solution: first verify that if one of $AD\perp BC, BE\perp CA, CF\perp AB$ holds then all three holds(easy). so we need only show that $AD\perp BC$. let $AD,BE,CF$ intersect at $O$. then $\angle OEC=\angle BFC=180-\angle AFC=180-\angle ODC\quad (1)$. So $ODCE$ is cyclic $\iff AO\cdot AD=AE\cdot AC$. but as $BCEF$ is cyclic we have $AE\cdot AC=AF\cdot AB$. so $AO\cdot AD=AF\cdot AB\iff ODBF$ is cyclic hence $\angle ODB+\angle OFB=180\quad (2)$. but $\angle OFB=\angle OEC$ so from $(1)$ and $(2)$, we get $\angle ODB=\angle ODC$ and the conclusion follows immediately.


Find all nondecreasing functions $ f : R\rightarrow R$ such that $ f(0) \equal{} 0, f(1) \equal{} 1$ and $ f(a) \plus{} f(b) \equal{} f(a)f(b) \plus{} f(a \plus{} b \minus{} ab)$ for all real numbers $ a,b$ such that $ a < 1 < b$

Answer:
[hide]
I.
\[ f(t) \equal{} \begin{cases} 1 \minus{} (1 \minus{} t)^p, & \text{ if } t \leq 1; \\
1 \plus{} C |t \minus{} 1|^p , & \text{ if } t \geq 1. \end{cases}
\]
where $ p > 0$, $ C > 0$.

II. $ f(t) \equal{} 1$ when $ t\geq 1$; $ f(t)$ is an arbitrary non-decreasing function when $ t\leq 1$ s.t. $ f(0) \equal{} 0$, $ f(1) \equal{} 1$.

[/hide]
Hint:
[hide]
Consider the function $ g(s) \equal{} 1 \minus{} f(1 \minus{} s)$. Observe that $ g(s) g(t) \equal{} g (st)$ when $ s < 0$ and $ t > 0$.
[/hide]

Consider the function $ g(s) \equal{} 1 \minus{} f(1 \minus{} s)$. For every $ s < 0$ and $ t > 0$,  we have
\[ g(s) g(t) \equal{} (1 \minus{} f(1 \minus{} s)) (1 \minus{} f(1 \minus{} t)) \equal{} 1 \minus{} f(1 \minus{} s) \minus{} f(1 \minus{} t) \plus{} f(1 \minus{} s) f(1 \minus{} t) \equal{} 1 \minus{} f((1 \minus{} s) \plus{} (1 \minus{} t) \minus{} (1 \minus{} s) (1 \minus{} t)) \equal{} 1 \minus{} f(1 \minus{} st) \equal{} g(st).
\]

\[ g(0) \equal{} 1 \minus{} f(1) \equal{} 0.
\]

\[ g(1) \equal{} 1 \minus{} f(0) \equal{} 1.
\]
Denote $ C \equal{} \minus{} f( \minus{} 1)$. For every $ t > 0$,  we have $ f( \minus{} t) \equal{} f( \minus{} 1) f(t) \equal{} \minus{} C f(t)$. Now we consider two cases: $ C \neq 0$ and $ C \equal{} 0$.
I. First, let us assume that $ C \neq 0$. Then for every $ s, t > 0$,
\[ f(s) f(t) \equal{} \minus{} f( \minus{} s) f(t) /C \equal{} \minus{} f( \minus{} st) / C \equal{} f(st).
\]
It is well known (and it was discussed at this forum many times) that every monotone function $ f(s)\not\equiv 0$ that satisfies this identity equals $ s^p$ for some $ p$ when $ s > 0$. Now, $ f( \minus{} s) \equal{} \minus{} C f(s) \equal{} \minus{} C s^p$ (when $ s > 0$). We conclude that
\[ f(s) \equal{} \begin{cases} s^p, & \text{ if } s \geq 0; \\
\minus{} C |s|^p, & \text{ if } s < 0. \end{cases}
\]
Plugging this expression into the formula $ f(a) \equal{} 1 \minus{} f(1 \minus{} a)$, we get the answer (see my previous post). It is easy to check that it satisfies the condition of the problem.

II. Assume $ C \equal{} 0$. Then $ f(s) \equal{} 0$ when $ s < 0$. It is easy to see that when $ s > 0$ the function $ f(s)$ can be an arbitrary non-decreasing function s.t. $ f(1) \equal{} 1$.


I can't say this in spanish so I'll just say in english.

Anyone compete in spanish olympiad?

Yes o No

¡ Eso es un buen ejercicio de traducción !

[quote]Eso es un buen ejercicio de traducción ![/quote]

=

[quote="Translator"]That's a good exercise of translation![/quote]

Gracias.

P.S. Does that mean you compete for the high school math competition? If so, can you tell what kind and some rules?

I'll try to take the idea by translator if I don't get it.

Thnx.

No !
I'm French (as you can see with the flag)...
So...

I don't even participate in French Olympiad...

Yo tengo amigos que participan, que quiere saber? igual creo que existe una buena pagina web, busque: "olimpiada española de matematicas"


The circles $C_{1}$ and $C_{2}$ touch externally at $M$ and the radius of $C_{2}$ is larger than that of $C_{1}$. $A$ is any point on $C_{2}$ which does not lie on the line joining the centers of the circles. $B$ and $C$ are points on $C_{1}$ such that $AB$ and $AC$ are tangent to $C_{1}$. The lines $BM, CM$ intersect $C_{2}$ again at $E, F$ respectively. $D$ is the intersection of the tangent at $A$ and the line $EF$. Show that the locus of $D$ as $A$ varies is a straight line.

It has already been posted before at http://www.mathlinks.ro/Forum/viewtopic.php?t=16007

Kostas Vittas.


I have been at the classroom as of about 7:55. The gray little button that says connect is grayed out, however. I'm pretty sure I'm logged in, so why can't I connect? BTW, it is 8:07 EST right now.

Sorry I missed this earlier, Mystic!

When the little button is grayed out, that means the class is in the process of loading for you.  (The button is grayed out for everyone when they go there logged in and class is open.)  It sometimes takes several minutes for the classroom to load, but it shouldn't take 12 minutes.

Are you sure your computer is Java enabled?  You should see an interactive Mandelbrot set on the Virtual Classroom page (even now, when class is closed).

Again, sorry I didn't see this and answer it sooner.  If you like, you can send me an email (rusczyk@artofproblemsolving.com) and we can set up a time to test your connection (I'll create a page for you to go see and try to connect - if it fails again, we'll get together & figure it out.).

ok, I went over there and checked. I was able to see the Mandelbrot, and zoom up on it with the mouse, if that's what you mean.

I'll try again at the upcoming mathcounts mathjam. If that doesn't work, I'll send an e-mail.

I made a little test page for you; I sent you an email with details.  Please email me back and let me know what happens...

Thanks.

I am having a problem with joining the Math Jams.  It seems that everytime I click on the Math Jams link (leave the Forum) I get logged off.  When I go back to the Forum, I have to log in again.

Peculiar error.  What browser/platform are you using?  I saw you (I think) in class today towards the end - did you solve your problem?

Are you sure you saw me?  I didn't go in the classroom.  I'm using Microsoft Internet Explorer.

I sent you an email yesterday about what to try - please send me an email back after you try it out.


induction
superior algebra
superior algebra solved

[quote="Myth"]Let $S$ be finite commutative semigroup wrt multiplication consisting of integer-valued non-invertable matrices of order 2. Prove that if $|S|>5$ then $AB=0$ for all $A,B\in S$.[/quote]

Myth: Nice problem, it seems trivial when you look at it, but it is not, actually, I haven't been able to find a counterexample, but I am not 100% is sure. the condition $AB=0$ is for $A\neq B$ or in general? If it is in general, I think it should be enough to prove that the matrices should be nilpotent. Any way, I have to run, need to give a class in five minutes, I'll try it later!

We can assume that $S$ is such that $0\in S$ and $X\in S\Rightarrow -X\in S$. This is because adding $0$ and $-S$ to $S$ does not change the property of $S$ of being a finite semigroup.

Let $A\in S$. Because $A,A^2,A^3,\ldots$ is a finite set, so is $\det A, \det A^2,\ldots$. But $\{n^k|k\in\mathbb{N}\}$ is finite iff $n\in\{-1,0,1\}$.
If $\det A=\pm 1$ then $A$ is invertible which is false by hypothesis. So $\det A=0\ \forall\ A\in S$.
$A^2=aA$ (Hamilton-Cayley), where $\mathbb{Z}\ni a=Tr(A)=Trace(A)$.
By induction, $A^n=a^{n-1}A\ \forall n\in\mathbb{N}^*$. For the set $\{a^nA|n\in\mathbb{N}\}$ to be finite we must have $a\in\{-1,0,1\}$ or $A=0$(which in turn leads to $a=0$). So $Tr(A)\in\{-1,0,1\}\ \forall\ A\in S$.

Assume there exists $A\in S$ such that $A^2\neq 0$. Then $a=Tr(A)\neq 0$, so $a=\pm 1$.
Let $X\in S$. Because $AX=XA$ and $A\not\in\mathbb{Z}\cdot I_2$, there exist $f(X),g(X)\in\mathbb{Q}$ such that $X=f(X)A+g(X)I_2$.
Because the eigenvalues of $X$ are $0$ and $Tr(X)$, we get $\{0,Tr(X)\}=\{f(X)\cdot a+g(X), g(X)\}$.
If $g(X)=0$ then $f(X)=aTr(X)=\pm Tr(X)$, so $X=\pm A$ or $f(X)=0\Rightarrow X=0_2$.
If $a\cdot f(X)+g(X)=0$ and $g(X)=Tr(X)=\pm 1$. Then $f(X)=-a\cdot g(X)$ and $g(X)=\pm 1$, leading to the solutions $X=\pm (I_2-a\cdot A)$.

So if $a=1$ then $S\subset\{0_2,A,-A,I_2-A,A-I_2\}$. If $a=-1$ then $S\subset\{0_2,A,-A,I_2+A,-I_2-A\}$. Both cases contradict $|S|>5$.
So $A^2=0_2\ \forall\A\in S$.
Choose $0_2\neq A\in S$. For any $X\in S$ consider $f(X),g(X)$ as above. Then $X=f(X)A+g(X)I_2$.
Eigenvalues reasoning leads to $g(X)=0$, so $X=f(X)A$. Multiply by $X$ to obtain $f(X)\cdot (AX)=0_2\Rightarrow$ $(f(X)=0\Rightarrow X=0_2)$ or $AX=0$.
This immediately implies $AX=0_2\ \forall\ A,X\in S$.

As an example of such $S$, take $S=\{f\cdot A|f\in F\}$, where $F$ is a finite set containing $0$ and $A$ is a nilpotent matrix.


[quote="Soarer"]Is there any difference between this one and vasc's one?! :?[/quote]

Yes .Vasc's one is  $(a^2+b^2+c^2)^2\geq 3(a^3b+b^3c+c^3a)$

...so how is that different?  You're just switching c and b...

[quote="K81o7"] You're just switching c and b...[/quote]

Oh yes....It is easy.....


geometry AMC USA(J)MO USAMO search articles induction

Can someone tell me how to get good at proofs?

I can't write a good proof if my life depended on it... I need some dire help...

 :noo:  :help:  :ewpu:

Take Geometry. When you are proving something, you really can't use the theorem that you are proving inside your proof. Geometry is the main thing with proofs. You prove SAS, ASA, SSS, and a lot more congruences. You'll learn. Don't worry. If you're worrying about it because you are proving stuff for USAMO and stuff, then you need to seriously search up how to prove things.

Proofs is mostly about geometry in middle school and stuff, but it can get much harder. I started learning proofs this year, and I'm understanding a bit since they teach you geometric proofs in geometry.

There are some easy proofs like proving that any odd number is a difference between 2 squares. To prove this i would say that any odd number is in form 2a+1. And $ 2a \plus{} 1$=$ (a \plus{} 1)^2$-$ a^2$ because $ (a \plus{} 1)^2$ is $ a^2$+$ 2a \plus{} 1$ and subtract $ a^2$ from that and you get the equation 2a+1=2a+1.

However, there are very complicated proofs. My suggestion would be start from the very basic and work your way up. Here is a great site to start learning proofs: [url]http://zimmer.csufresno.edu/~larryc/proofs/proofs.html[/url]

Good luck!!

eh... I guess I'll take intro to geometry!

read this [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php]article[/url] on writing proofs.

The more advanced AoPS books cover proof by induction, contradiction, pigeonhole, and direct proof

[quote="jonathanchou711"]Wait! They might not have proofs in online classes. It's actually better to take it in real person with a teacher that you can see and a class that is actually the full length. Never skip Geometry. Only skip Algebra 1 or Algebra 2. As I am skipping Algebra 2.[/quote]

Er, yes they do.  Also, proofs you learn at school are just two-column trivial proofs where you have to show about 30 steps in a 1-line proof.  AoPS Intro to Geometry is a much better course than school geometry (took both).

eh... AIME15 is the [i]expert[/i]... I'll agree with him

[quote="Isabella2296"]-- Before posting, think hard whether your post is necessary or not. Remember, you do not have to push the "Submit" button. [/quote]

Think about your post. Was it really necessary?

---EDIT---
I just realized that my post was spam, too. Sorry!  :blush:


Explain why Isoborneol ([url]http://en.wikipedia.org/wiki/Isoborneol[/url]) on treatment with 50% sulphuric acid gives Camphene ([url]http://en.wikipedia.org/wiki/Camphene[/url]) and not Bornylene ([url]http://www.pherobase.com/database/compound/compounds-detail-bornylene.php[/url])

Give a mechanism if possible.

[hide="My Opinion"]Of course the reason should be that the double bond would rather be outside the ring than within it due to high angle strain. In bornylene, 2 carbon atoms are under angle strain, whereas in camphene, only one atom is under angle strain.
But who can give the mechanism? It seems very different, to me atleast.  :maybe: [/hide]


AMC
USA(J)MO
USAMO
AMC 12
AIME
USAJMO
AMC 10

OK, so I am kind of confused on how they do the USAMO Index.  I understand that they will take your [AMC12]*10[AIME] scores, but I don't know where to see the floor scores from past years.  What score should I be shooting for?  I've heard 200, but will it increase or decrease because of the USAJMO?  Does anyone know where I can find the floor scores or the Index scores by state?

Well, they shouldn't have any floor scores this year for either USAMO or USAJMO because you can only make USAMO by taking the AMC12 and make USAJMO by taking the AMC10 (the only exception is if you solely take the AMC10 and get an 11+ on the AIME)

Previous approximates were at 210 because that is how it's been in the past when there were fewer qualifiers. Before there was Red MOP qualification through USAJMO for 10th graders as well, a better estimate would have been 220, since many of the ambitious underclassmen who took the AMC10 last year will likely take the AMC12 this year to get to USAMO, and they are not factored in.

Of course, with Red MOP qualification, there will likely be a decrease in AMC12 takers. I'd say you are probably safe with a 215 index.

lol i hope this helps

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=321488

Is there any site that has records of the indices of the USAMO participants from past years?

http://en.wikipedia.org/wiki/USAMO#Recent_qualification_indices

Not a very useful list anymore, considering all the changes that have been put in place between this year and last year.

If you take the average of the USAMO invitation indices listed, the result is 213.  If you take the average of the USAMO invitation indices for 2000, 2001, 2003, 2004, 2005 when we invited about 250-260 participants the result is 217.  Using the average like this is not perfect, but year-by-year values and the overall average are consistent with my estimate based on 2008 and 2009 results that an index of 210 will be the approximate invitation level for 2010.  Of course, the precise value will depend on the variability in difficulty of  the 2010 AMC 10 and AMC 12 and the AIME compared to previous years.  The committees do an excellent job of calibrating the difficulty level given that there are 6 contests each year, each with new problems, but the contests do vary in overall difficulty with a standard deviation which is about 1 problem from an overall average.

-- Steve Dunbar
MAA Director, American Mathematics Competitions

To Steve,

Is it true that the average AIME score in 2008 and 2009 are higher than that of 2000 - 2005?  If true, what is the impact of higher AIME score on the USAMO index estimate?

The information is at:
General Statistics
on
http://www.unl.edu/amc/e-exams/e7-aime/archiveaime.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions

everyone in usajmo will have an AIME score 10 or less right?

Oh yeah, I didn't notice that.  It's still going to be 120 to qualify for AIME if you take the amc 10 right?

its going to be 120 or the minimum score of the top 1%, whichever is lower
i remember 2007 AMC10B the cutoff for AIME was 117, so occasionsly it is less than 120

What is the USAJMO index probably going to be then?

it will be at most 250   lol

but im guessing usajmo will be around 210.


search
number theory
prime numbers
number theory proposed

Prove that there exists an infinity of pairs of primes $ (p, q)$ such that $ p|2^{q\minus{}1}\minus{}1$ and $ q|2^{p\minus{}1}\minus{}1$

[quote="andyciup"]Prove that there exists an infinity of pairs of primes $ (p, q)$ such that $ p|2^{q \minus{} 1} \minus{} 1$ and $ q|2^{p \minus{} 1} \minus{} 1$[/quote]

I would suggest $ (p,p)$ for any odd prime $ p$

:)

It should be ''different prime numbers '' and have a look here :
 http://www.mathlinks.ro/viewtopic.php?search_id=1054974708&t=204055


Let $p_n$ be the $n$-th prime. ($p_1=2$)
Define the sequence $(f_j)$ as follows:
- $f_1=1, f_2=2$
- $\forall j\ge 2$: if $f_j = kp_n$ for $k<p_n$ then $f_{j+1}=(k+1)p_n$
- $\forall j\ge 2$: if $f_j = p_n^2$ then $f_{j+1}=p_{n+1}$

(a) Show that all $f_i$ are different
(b) from which index onwards are all $f_i$ at least 3 digits?
(c) which integers do not appear in the sequence?
(d) how many numbers with less than 3 digits appear in the sequence?

The sequence is $1, 1*p_1, 2*p_1, p_2, 2*p_2, 3p_2, p_3, \dots, p_k, 2*p_k, \dots, (p_k -1)(p_k), p_k^2, p_{k+1}, \dots$.

(a)  Suppose two f_i are the same, say $ap_x$ and $bp_y$.  (a < p_x ; b < p_y).  Suppose $p_x > p_y$.  Then $p_x$ divides into $b$ implying b is a multiple of $p_x$, contradiction $b < p_y$.  Similarily for $p_y > p_x$.  For $p_x = p_y$ we get $a = b$, so that they are the same term.

(b) When we get to the prime number 101 (26th prime), we will have used 1 +  the sum of the prime numbers up to 101.  Afterwards, we always have a factor > 100, so we are guaranteed.  We are also guaranteed previous up to 2 * 97, where before that we have 97, which is 2 digit.  So we go back 96 numbers.

(c) any number where dividing by the largest prime yields a number larger than that prime.

(d)  We go up to 9*11; then 7*13; 5* 17,19 ; 4 * 23; 3 * 29, 31; 2 * 37, 41, 43, 47; 1 * 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.  This can be counted as 1 + the sum of the primes up to 11 exclusive, plus 9 + 7 + 5*2 + 4 + 3*2 + 2*4 + 10.

[quote="Singular"](a)  Suppose two f_i are the same, say $ap_x$ and $bp_y$.  (a < p_x ; b < p_y).  Suppose $p_x > p_y$.  Then $p_x$ divides into $b$ implying b is a multiple of $p_x$, contradiction $b < p_y$.  Similarily for $p_y > p_x$.  For $p_x = p_y$ we get $a = b$, so that they are the same term.[/quote]Correct. :)

[quote="Singular"](b) When we get to the prime number 101 (26th prime), we will have used 1 +  the sum of the prime numbers up to 101.  Afterwards, we always have a factor > 100, so we are guaranteed.  We are also guaranteed previous up to 2 * 97, where before that we have 97, which is 2 digit.  So we go back 96 numbers.[/quote]So the explicit index number? ...

[quote="Singular"](c) any number where dividing by the largest prime yields a number larger than that prime.[/quote]Correct. :)

[quote="Singular"](d)  We go up to 9*11; then 7*13; 5* 17,19 ; 4 * 23; 3 * 29, 31; 2 * 37, 41, 43, 47; 1 * 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.  This can be counted as 1 + the sum of the primes up to 11 exclusive, plus 9 + 7 + 5*2 + 4 + 3*2 + 2*4 + 10.[/quote]So... you say 2+3+5+7+9+7+5*2+4+3*2+2*4+10 = 69? Not correct in that case.


function
real analysis
real analysis solved

grobber wrote:
4)
f:[0,1]->[0,1] is continuous and bijective. Determine the set A={f(x)-f(y) | x, y are in [0,1]\Q}.

(Students may use the result: there are no injective functions from [a,]\Q to Q)
----------------------------------------------------------------------

Very good job, Harazi

If I'm not wrong, my solution was the following:
  Clearly, f must bestrtly monotonuous. Since changing f to -f doesn't moify anythind, let us assume that f is strictly increasing. Then f(0)=0 and f(1)=1 and also for each x and y we have 1 \geq f(x)-f(y) \geq -1. Thus, A is in (-1,1). Now let us take t in (-1,0). Suppose t is not in A.Let us take the interval I=[0, f^(-1)(t+1)] . For each x in I we have 0 \leq f(x)-t \leq 1. THus, we can define the function g from I\Q to Q, g(x)=f^(-1)(f(x)-t). This is well defined because if there is x such that g(x) belongs to R\Q, then t=f(x)-f(g(x)). And g is clearly injective, contradiction. So, we have that (-1,0) is in A. But if x is in A, so is -x. So, (-1,1) is in A and A=(-1,1).


inequalities
function
inequalities solved

Write the inequality in the form  \sum xi*ln(xi) \geq  (\sum xi)*ln(\sum xi/n) and then apply Jensen's inequality for the convex function f(x)=x*ln(x).


Wich point $ P$ inside a triangle $ ABC$ minimizes the sum $ PA\plus{}PB\plus{}PC$

 :)

[quote="AoPSWiki"]A method to find the point is to construct three equilateral triangles out of the three sides from $ \triangle ABC$, then connect each new vertex to each opposite vertex, as these three lines will concur at first Fermat point.[/quote]

Take directly from [[Fermat point]].


function
number theory
algebra unsolved
algebra

Let a_1, a_2, ..., a_n be a string of symbols from the set {a,b,...,j}. Prove that there is a bijective function f : {a,b,...,j} -> {0,1,...,9} such that the string f(a_1)f(a_2)...f(a_n) represents an n-digit integer in decimal expansion which is divisible by 3.

Let symbols be $f_{0}$, $f_{1}$, ..., $f_{9}$, and they counts in the initial string are $b_{0}$, ..., $b_{9}$ respectively. We may assume $b_{i}\in\{0,1,2\}$.

Case 1. All $b_{i}$ are equal modulo 3. Then any bijection satisfy the condition.

Case 2. There are $b_{i}=0$, $b_{j}=1$ and $b_{k}=1$. Then consider arbitrary bijection where $f_{i}=0$, $f_{j}=1$ and $f_{k}=2$. If it doesn't satisfy the condition then consider the same bijection but $f_{i}=1$ and $f_{j}=0$ OR $f_{i}=2$ and $f_{k}=0$. One of these two bijections will satisfy the condition.

Case 3. All other cases are analoguosly to Case 2.

P.S. Please, prove the general statement: $3\to p$, $9\to p^{2}$.

P.P.S. Clearly, it is a Number Theory problem :)


algebra polynomial function calculus integration

Let $f \in \mathbb{C} [X]$ and degree $f=n$. We know that $f(0),f(1),\ldots,f(n) \in \mathbb{Z}$.

Prove that $f(a) \in \mathbb{Z}$, $\forall a \in \mathbb{Z}$

n+1 points uniquely determines a polynomial of degree n?

(I dont understand the notation $grad f = n$ for a function of one variable.)

It follows from difference tables and what Singular said.

For any polynomial $f$ of degree $n$ on the integers, the function $\delta f(n) = f(n + 1) - f(n)$ is a polynomial of degree $n - 1$.  You can make this into a pretty-looking table, which is why I said "difference table," but I probably should have said "finite differences."  In any event, it follows that we can keep doing this, $n$ times, until we get a bottom row that will be constant.  This row, and every other entry in the table at that point, will be integral.  Thus, building back up from the bottom, the function always takes integral values.  (This was not a good description -- can someone do a better job?)

Here's a little picture, though, for a known cubic with 4 integral values:

[code]f:   2   9  28   65
df:    7  19   37
ddf:    12  18
dddf:      6[/code]
[code]f:   2   9  28   65
df:    7  19   37
ddf:    12  18
dddf:      6    6    6    6  ...[/code]
[code]f:   2   9  28   65   126   217  ...
df:    7  19   37   61    91  ...
ddf:    12  18   24    30  ...
dddf:      6    6    6     6  ...[/code]


could nebdy please give a brief review on the use of NBS in the formation of allyl bromides.
i mean, for example, if we have 2,4 dimethyl hex-3-ene, to which carbon does the bromine radikl get added? y?
if possibl, pls suggest some references.
thanks in advance

$ \text{NBS}$ is a very useful comound for replacing an allylic $ \text{H}$ by $ \text{Br}$.
it is used with high temperatures.In high temepratures the alyylic $ \text{H}$ is captured by $ \text{Br}^.$ which may come from $ \text{NBS}$ or otherwise.The important function of $ \text{NBS}$ is to react with so formed $ \text{HBr}$ and convert to succinimide nad step up the forward reaction
References-Morrison&Boyd.
In $ \text{2,4 - Dimethyl Hex - 3 - ene}$ there are two allylic $ H$ but the $ 4 - \text{H}$ is  resonance , hyperconjugation and inductively stabilized so the $ \text{H}$ is captured from there

[b]NOTE[/b]:the name of the compound is $ \text{N - Bromo Succinimide}$ and not otherwise

thanks pardesi for the soln, reference as well as correction ( truly spkin, i was actually confused whether its succinate or succinamide)

The allylic bromination of compounds (specially alkenes, but also alkynes, carbonyls, and benzylic positions) with NBS is commonly called the Wohl-Ziegler bromination. The reaction is often performed in a non polar solvent such as CCl4, and some heating is always involved (refluxing for some time) but not very high temperatures. Regarding the regioselectivity, it is known that a secondary position is more readily substituted than a primary one; however, the relative reactivity of tertiary carbons is not clear, with sometimes they being substituted faster than secondary, but quite often secondary is substituted faster than tertiary. So, regarding your compound, 2,4-dimethylhex-3-ene, there are three allylic positions - carbon 2 (tertiary), the 4-methyl group (primary), and carbon 5 (secondary) -, so I would perhaps vote on carbon 5 to be the more readily substituted.
An important issue in this reaction is its mechanism. The first step is formation of some bromine atoms from NBS via the cleavage of the weak nitrogen-bromine bond - mediated by heat or by some radical initiator present in the mixture:

$ (NBS)N\minus{}Br \longrightarrow (NBS)N \cdot \plus{} \cdot Br$

Now the bromine atom will abstract a hydrogen atom from an alkene molecule:

$ R\minus{}H \plus{} \cdot Br \longrightarrow HBr \plus{} R \cdot$

Now there is a fast ionic reaction between NBS and the HBr liberated in the last step:

$ NBS \plus{} HBr \longrightarrow (NBS)\minus{}H \plus{} Br_2$,

and this step ilustrates that the main function of NBS is to consume the HBr liberated and provide a low but constant concentration of Br2. Now, in the next step the allyl radical reacts with bromine molecule to form the allyl bromide,

$ R \cdot \plus{} Br_2 \longrightarrow R\minus{}Br \plus{} Br \cdot$

and the chain repeats. Now, an important question: if bromine (Br2) is one of the reacting species why it does not add to the double bond?

1. NBS gives a low concentration of Br.
2. at high temperatures, radical reaction competes`successfully with the addition reaction
and moreover,
3. in the presence of a non polar solvent, since there aren't any polar molecules to solvate and thus stabilise the bromide ion formed in the first step of the addition reaction, the bromide ion uses a bromine atom as a substitute. So, in a non polar solvent the rate equation is second order w.r.t. bromine, so the low concentration (point 1.) has an even more pronounced effect in slowing the reaction.

[quote="Cyprusad"]at high temperatures, radical reaction competes`successfully with the addition reaction[/quote]

And that's because the first step of the addition reaction is reversible.


How many different combinations of $ \$5$ bills and $ \$2$ bills can
be used to make a total of $ \$17$?  Order does not matter in this
problem.

two ways: 3 5s and one 2
or 1 5 and 6 2s


algebra
polynomial
LaTeX
probability
linear algebra
matrix

Prove that determinant

col 1           col2       col3

2bc-a^2 ,   c^2 ,       b^2

c^2 ,        2ac-b^2 , a^2          ==a^3+b^3+c^3-3abc

                                                                                b^2 ,        a^2 ,       2ab-c^2
please refine your language .this does not look nice for a person, intelligent like you. apology if my words hurts you

,

This is complete nonsense. I have no idea what you even intend that statement to mean.

Presumably that

\[ \det\left( \begin{array} {ccc}
2bc \minus{} a^2 & c^2 & b^2 \\
c^2 & 2ac \minus{} b^2 & a^2 \\
b^2 & a^2 & 2ab \minus{} c^2
\end{array} \right) \equal{} a^3\plus{}b^3\plus{}c^3\minus{}3abc
\]

Of course, this isn't true since the LHS is a homogeneous polynomial of degree $ 6$ and the RHs is a polynomial of degree $ 3$.

[b] ayush_2008 [/b]

I think it would be much better if you learned LATEX (I really can't offer you such an advice because I myself am learning LATEX). If you managed to learn LATEX, it would be much better because it would make your technical posts more legible. It also increases the probability of you getting substantive replies/solutions.

The original version of this didn't include the word "determinant". I saw it anyway, and dismissed it as obviously wrong for the same reasons as blahblahblah.
On the $ \text{\LaTeX}$, both this thread and your previous thread contain examples of how to build a matrix. Use them; I learned a great deal of what I know of the language by looking at other people's constructions on this forum.

On my language- "nonsense" is a dangerous word, and I'm aware that I shouldn't use it around my students. Still, some statements truly are nonsense, and this isn't a classroom. Stating that a matrix is equal to a number rises to that level, especially when the obvious choices of taking a trace or a determinant make the statement false.

Anyways, it appears that the correct problem is the 'obvious' correction;

\[ \det\left(\begin{array}{ccc}2bc\minus{}a^{2}& c^{2}& b^{2}\\ c^{2}& 2ac\minus{}b^{2}& a^{2}\\ b^{2}& a^{2}& 2ab\minus{}c^{2}\end{array}\right) \equal{} (a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc)^2 \]

Perhaps the identity $ a^3\plus{}b^3\plus{}c^3\minus{}3abc \equal{} (a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)$ is useful here?

[quote="blahblahblah"]Anyways, it appears that the correct problem is the 'obvious' correction;
\[ \det\left(\begin{array}{ccc}2bc \minus{} a^{2} & c^{2} & b^{2} \\
c^{2} & 2ac \minus{} b^{2} & a^{2} \\
b^{2} & a^{2} & 2ab \minus{} c^{2}\end{array}\right) \equal{} (a^{3} \plus{} b^{3} \plus{} c^{3} \minus{} 3abc)^2
\]
Perhaps the identity $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)$ is useful here?[/quote]

blahblahblah is right, I just have not found an elegant way yet to prove the result, just develop the determinant and identify the result as the square...

It is well-known that

$ \det \left( \begin{array}{ccc} a & b & c \\
c & a & b \\
b & c & a \end{array} \right) \equal{} a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc$

due to the properties of the eigenvalues of a circulant matrix.  I'm reasonably certain there's a nice way to express the given matrix as a product of two matrices similar to the above.  Alternately, it's a symmetric matrix, so it should have three real eigenvalues (for real $ a, b, c$), which might be something like

$ (a \plus{} b \plus{} c)^2, (a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca), (a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)$.

(but this choice isn't consistent with the trace.)  I'll try to see if either of these calculations work out.

[b]Edit:[/b]  Got it.  The given matrix is equal to

$ \left( \begin{array}{ccc} a & c & b \\
b & a & c \\
c & b & a \end{array} \right) \left( \begin{array}{ccc} \minus{} a & \minus{} b & \minus{} c \\
b & c & a \\
c & a & b \end{array} \right).$

[b]Edit #2:[/b]  If it's not clear why the above works (in particular, why the sign is correct), multiply on the right by $ \left( \begin{array}{ccc} 0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \end{array} \right)$.  Anyway, I'm wondering what the source of this problem is.  The matrix wants to be a [url=http://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant]Gram matrix[/url], but isn't.

After some experimenting, this seems to be the "right" solution:

$ \left(
\begin{array}
[c]{ccc}
2bc\minus{}a^{2} & c^{2} & b^{2}\\
c^{2} & 2ca\minus{}b^{2} & a^{2}\\
b^{2} & a^{2} & 2ab\minus{}c^{2}
\end{array}
\right)  \equal{}\left(
\begin{array}
[c]{ccc}
1 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}
a & b & c\\
c & a & b\\
b & c & a
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}
\minus{}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}
a & b & c\\
c & a & b\\
b & c & a
\end{array}
\right)$,

and

$ \det\left(
\begin{array}
[c]{ccc}
1 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0
\end{array}
\right)  \equal{}\minus{}1$,

$ \det\left(
\begin{array}
[c]{ccc}
\minus{}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}
\right)  \equal{}\minus{}1$ and

$ \det\left(
\begin{array}
[c]{ccc}
a & b & c\\
c & a & b\\
b & c & a
\end{array}
\right)  \equal{}\left(  a\plus{}b\plus{}c\right)  \left(  a\plus{}\zeta b\plus{}\zeta^{2}c\right)  \left(
a\plus{}\zeta^{2}b\plus{}\zeta c\right)$ (by the circulant formula, where $ \zeta$ is a primitive $ 3$-rd root of unity)
$ \equal{}a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc$.

  darij


From 6:00pm until midnight the temperature dropped $ 2^\circ$ per hour. The temperature was $ 5^\circ$ at 6:00pm. In degrees, what was the temperature at midnight?

There are $ 6$ hours from $ 6: 00$pm to midnight, so the temperature is $ 5\minus{}2(6)\equal{}\boxed{\minus{}7^\circ}$.


Precisely three of the following ten collections of subsets of $ \mathbb R$ are topologies. Identify these and justify your answer.

(1) $ \tau_1$ consists of $ \mathbb R$, $ \emptyset$, and every $ (a,b)$ for $ a,b\in\mathbb R$ with $ a < b$.

(2) $ \tau_2$ consists of $ \mathbb R$, $ \emptyset$, and every $ ( \minus{} r,r)$ for $ r\in\mathbb R^ \plus{}$.

(3) $ \tau_3$ consists of $ \mathbb R$, $ \emptyset$, and every $ ( \minus{} r,r)$ for $ r\in\mathbb Q^ \plus{}$.

(4) $ \tau_4$ consists of $ \mathbb R$, $ \emptyset$, and every $ [ \minus{} r,r]$ for $ r\in\mathbb Q^ \plus{}$.

(5) $ \tau_5$ consists of $ \mathbb R$, $ \emptyset$, and every $ ( \minus{} r,r)$ for $ r\in\mathbb R^ \plus{} \backslash \mathbb Q$.

(6) $ \tau_6$ consists of $ \mathbb R$, $ \emptyset$, and every $ [ \minus{} r,r]$ for $ r\in\mathbb R^ \plus{} \backslash\mathbb Q$.

(7) $ \tau_7$ consists of $ \mathbb R$, $ \emptyset$, and every $ [ \minus{} r,r)$ for $ r\in\mathbb R^ \plus{}$.

(8) $ \tau_8$ consists of $ \mathbb R$, $ \emptyset$, and every $ ( \minus{} r,r]$ for $ r\in\mathbb R^ \plus{}$.

(9) $ \tau_9$ consists of $ \mathbb R$, $ \emptyset$, every $ [ \minus{} r,r]$, and every $ ( \minus{} r,r)$ for $ r\in\mathbb R^ \plus{}$.

(10) $ \tau_{10}$ consists of $ \mathbb R$, $ \emptyset$, every $ [ \minus{} n,n]$, and every $ ( \minus{} r,r)$ for $ n\in\mathbb Z^ \plus{}$ and $ r\in\mathbb R^ \plus{}$.

[hide]For (1) - (6), I think only $ \tau_2$ is a topology. But for (7) - (10) I am not sure.[/hide]

[quote="JRav"]Isn't (1) the standard topology?  Why don't you think that counts?[/quote]

$ \tau_1$ is a [b]base[/b] for the standard topology. But It's not a topology. The union of $ (1,2)$ and $ (3,4)$ is not contained...

In (1)-(6) only $ \tau_2$ is a topology. $ \tau_9$ and $ \tau_{10}$ are topology too...

If $ \tau_7$ is a topology $ \tau_8$ must be a topology too. So I have simply to show that $ \tau_7$ is not a topology... Consider an infinite series with limits $ \lim_{n\to\infty} r_n \equal{} l$ then the union of the sets in the form $ [r_n,r_n)$ is $ (l,l)$ that it's not in $ \tau_7$.

[quote="vict85"]Consider an infinite series with limits $ \lim_{n\to\infty} r_n \equal{} l$ then the union of the sets in the form $ [\minus{}r_n,r_n)$ is $ (\minus{}l,l)$ that it's not in $ \tau_7$.[/quote]
I also think that $ \tau_7$ and $ \tau_8$ are not topologies. But for your reason, could you please explain $ \bigcup_{n\equal{}1}^{\infty} [\minus{}r_n,r_n)\equal{}(\minus{}l,l)$ in detail?

[quote="ifai"][quote="vict85"]Consider an infinite series with limits $ \lim_{n\to\infty} r_n \equal{} l$ then the union of the sets in the form $ [ \minus{} r_n,r_n)$ is $ ( \minus{} l,l)$ that it's not in $ \tau_7$.[/quote]
I also think that $ \tau_7$ and $ \tau_8$ are not topologies. But for your reason, could you please explain $ \bigcup_{n \equal{} 1}^{\infty} [ \minus{} r_n,r_n) \equal{} ( \minus{} l,l)$ in detail?[/quote]

Ok, I forgot to write that $ r_n < l$ for any $ n$ (it's a necessary condiction). With this condiction $ \bigcup_{n \equal{} 1}^{\infty} [ \minus{} r_n,r_n) \subseteq [ \minus{} l,l)$.

The points $ \minus{} l$ and $ l$ are not in any of the $ [ \minus{} r_n,r_n)$ and so $ \bigcup_{n \equal{} 1}^{\infty} [ \minus{} r_n,r_n) \subseteq ( \minus{} l,l) \subset [ \minus{} l,l)$. Now consider a real number $ \epsilon > 0$, then you know that $ \bigcup_{n \equal{} 1}^{\infty} [ \minus{} r_n,r_n) \supseteq [ \minus{} l \plus{} \epsilon,l \minus{} \epsilon)$ for any $ \epsilon$ and so $ \bigcup_{n \equal{} 1}^{\infty} [ \minus{} r_n,r_n) \equal{} ( \minus{} l,l)$

P.S: the sequences $ r_n \equal{} r \minus{} \frac {1}{n}$ and $ r_n \equal{} r \minus{} \frac {1}{2^n}$ are common choices...

Oh, I see. Thank you very much, vict85. Originally I was confused since $ \lim_{n\to\infty} r_n\equal{}l$ even if $ r_n\notin l$ for all $ n$. Hence I thought $ \minus{}l$ should be in $ \bigcup_{n\equal{}1}^{\infty}[\minus{}r_n,r_n)$.

Now I know I was wrong. Note that $ \bigcup_{S\in\mathcal F} S\equal{}\{x: x\in S \text{ for some } S\in\mathcal F\}$. But $ \minus{}l\notin [\minus{}r_n,r_n)$ for all $ n$, thus we have $ \minus{}l\notin\bigcup_{n\equal{}1}^{\infty} [\minus{}r_n,r_n)$.

[quote="JRav"]I agree that testing topologies can be tricky (as you saw, I forgot the definition above), and I got an A in my topology course!  Perhaps I'm rusty... :)[/quote]

Well, I simply gave more attention... At a first look it's normal to see in the first one the standard topology...

Also there are some books which define topological space as what in other books is the fundamental neighborhoods system, and the more common definition using open sets is regarded as an alternative definition which does not alters the subject of study (together with the other alternatives definitions using closed sets, closure operation, etc...).

So I guess it is kind of a justified "mistake"...

:)


Prove that if p is a prime then
(p-1)!≡p-1(mod p(p-1)/2)

Hi,

[hide]

Using the fact that p is relatively prime to (p-1)/2 and Wilson's theorem:

p | (p-1)!-(p-1)

(1/2)(p-1) | (p-1)!-(p-1)

implies

(1/2)p(p-1) | (p-1)!-(p-1)

[/hide]

I don't understand something in your solution.
Why (1/2)(p-1) | (p-1)!-(p-1).
Maybe it's trivial, but...


How many $ (x,y)$ solution in positive integers of the given equation :
$ \sqrt{xy}\minus{}71\sqrt{x}\plus{}30\equal{}0$

[hide="is there something wrong with this?"]
$ \sqrt{xy}\minus{}71\sqrt{x}\plus{}30\equal{}0 \rightarrow \sqrt{x}*(\sqrt{y}\minus{}71)\equal{}\minus{}30$.
This suggests that we should pair up factors of $ 30$.
$ \sqrt{x}>0 \rightarrow \sqrt{x}\equal{}1,2,3,5,6,10,15,30$.
This suggests there are 8 solutions.

[/hide]

Wouldn't it have to be?

Assume that $ \sqrt{x}$ is not an integer, which suggests it is irrational.  Then we still have $ \sqrt{x}(\sqrt {y} \minus{}71)\equal{}\minus{}30$.  $ \sqrt y\equal{}\frac{\minus{}30}{\sqrt x}\plus{}71$, we will get a rational and irrational portion. But y is an integer,  contradiction.

[quote="azn6021023"]Wouldn't it have to be?

Assume that $ \sqrt {x}$ is not an integer, which suggests it is irrational.  Then we still have $ \sqrt {x}(\sqrt {y} \minus{} 71) \equal{} \minus{} 30$.  $ \sqrt y \equal{} \frac { \minus{} 30}{\sqrt x} \plus{} 71$, we will get a rational and irrational portion. But y is an integer,  contradiction.[/quote]
Is $ \frac { \minus{} 30}{\sqrt x} \plus{} 71$ an integer if $ \sqrt x$ is irrational?


Prove that $n$ is prime if and only if:
$\phi(n) + \sigma(n) = n \cdot d(n)$,
where $\phi(n)$ is the number of numbers $m\leq n$ : $gcd(m,n) = 1$, $\sigma(n)$ is the sum of the divisors of $n$,
$d(n)$ is the number of divisors of $n$.

If $n$ is prime then $\phi(n) + \sigma(n) = n \cdot d(n)$ takes a trivial form $(n-1)+(n+1)=n\cdot 2$.

Suppose that $\phi(n) + \sigma(n) = n \cdot d(n)$. Then $\phi(n)\leq n-1$ and
\[ \phi(n) = n d(n) - \sigma(n) = n \sum_{q\mid n} 1 - \sum_{q\mid n} q = \sum_{q\mid n} (n - q) \geq n-1. \]
Therefore, $\phi(n)=n-1$ meanning that $n$ is prime.


For a positive integer $ n$, find

$ \binom{2n}{0}^2\plus{}\binom{2n}{2}^2\plus{}\binom{2n}{4}^2\plus{}...\plus{}\binom{2n}{2n}^2$

Upon introspection, the same sum without the squared terms seems to be $ 2^{2n \minus{} 1}$. With the squared terms, it goes like 2, 38, 452, ...

[quote="Yongyi781"]Upon introspection, [/quote] You looked deep into your soul, and discovered a classic problem on binomial series?

The answer is [url=http://www.research.att.com/~njas/sequences/A098772]A098772[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url], with a nice simple closed form, but I'm still looking for the combinatorial proof.

A related sum:
$ \sum_{k\equal{}0}^{n} \ {n\choose k}^{2}\equal{}{2n\choose n}$
This one has a very easy combinatorial proof :wink:

Well, I think there is a combinatorial proof to be had, but I don't really want do the work necessary to see if the outline of an idea I have can be turned into an actual proof.  So, instead, I give you a generating functional proof:

[hide]$ 2\sum_{k \equal{} 0}^n \binom{2n}{2k} x^{2k} \equal{} (1 \plus{} x)^{2n} \plus{} (1 \minus{} x)^{2n}$.  The expression we want is $ \sum_{k \equal{} 0}^n \binom{2n}{2k} \binom{2n}{2n \minus{} 2k}$, and this is the coefficient of $ x^{2n}$ in the product $ \left(\sum_{k \equal{} 0}^n \binom{2n}{2k} x^{2k}\right)^2$.  So we want one quarter of the coefficient of $ x^{2n}$ in $ \left((1 \plus{} x)^{2n} \plus{} (1 \minus{} x)^{2n}\right)^2 \equal{} (1 \plus{} x)^{4n} \plus{} (1 \minus{} x)^{4n} \plus{} 2(1 \minus{} x^2)^{2n}$, and this is $ \frac{1}{4}\left(\binom{4n}{2n} \plus{} \binom{4n}{2n} \plus{} 2(\minus{}1)^n \binom{2n}{n}\right) \equal{} \frac{\binom{4n}{2n} \plus{} (\minus{}1)^n \binom{2n}{n}}{2}$.[/hide]

Consider all sequences of $ n$ $ A$'s and $ n$ $ B$'s. Obviously there are $ \binom{2n}{n}$ ways. Now lets count this differently. Suppose there are $ k$ $ A$'s in the first half of the sequence. Then there are $ \binom{n}{k}$ ways for the first half. The the other half has $ n \minus{} k$ $ A$'s, and $ k$ $ B$'s so there are $ \binom{n}{k}$ ways for that two. Summing from $ k \equal{} 0$ to $ k \equal{} n$ establishes the identity posted by [polskimisiek].


The newly published book:
First Steps for Math Olympians:  Using the American Mathematics Competitions, by J. Dougles Faires
is now available.  You can purchase it through the MAA on-line bookstore (http://www.maa.org, then choose Bookstore) or by calling 1-800-331-1622.  The catalog code is PSC/SF06, the ISBN is 0-88385-824-X.

You can also find it listed in the Art of Problem Solving Bookstore, and at the major web bookstores.

It is 320 pages and covers mostly problem solving techniques for the AMC 10, some for the AMC 12.  Doug Faires has been the AMC 10 SubCommittee Chair since the inception of the AMC 10.  Faires is a very talented writer and teacher.  This would be an excellent book to use to prepare for the AMC 10, and the AMC 12.

Steve Dunbar
AMC Director

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=52]Click here for information about the book on the AoPS site[/url].


Let $ a,b,c,d$ are non-negative real numbers. Prove that
\[ a^{4}+b^{4}+c^{4}+d^{4}+8abcd \ge \sum_{cyclic}abc(a+b+c)+\frac{1}{6}\sum_{cyclic}ab(a-b)^{2}\]


Se aleg trei numere la intamplare din intervaul (o,1). Care este probabilitatea ca diferenta dintre cel mai mare dintre ele si cel mai mic sa fie mai mica decat 1/3?

Problema este propusa [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=614198#614198]aici[/url]. Eu am postat acolo o rezolvare geometrica, m-ar interesa daca stie cineva si alta rezolvare.

am incercat eu sa fac ceva aici, nu stiu cat de bine e, macar rezultatu e la fel :)
[hide="solutie"]fie numerele $a,b,c$.
il aleg pe $c$ sa fie cel din mijloc.
apoi, pentru $a$, ma uit in ce interval trebuie sa fie $b$ :
pt $a\in[0,1/3]$, rezulta ca $b\in[0,a+1/3]$. (i)
pt $a\in[1/3,2/3]$, rezulta ca $b\in[a-1/3,a+1/3]$. (ii)
pt $a\in[2/3,1]$, rezulta ca $b\in[a-1/3,1]$. (iii)
evident, $c\in[a,b]$ sau $c\in[b,a]$.
pentru cazul (i) :
pentru un $a$ fix, probabilitatea ca $c$ sa fie intre $a$ si $b$ este :
$\int_{0}^{a+1/3}|x-a|\,dx=\int_{0}^{a}(a-x)\,dx+\int_{a}^{a+1/3}(x-a)dx=\frac{a^{2}}{2}+\frac{1}{18}$.
acum integrez si dupa $a$ :
$\int_{0}^{1/3}\left(\frac{a^{2}}{2}+\frac{1}{18}\right)\,da=\frac{1}{27\cdot 6}+\frac{1}{27\cdot 2}=\frac{2}{27\cdot 3}$.
deci probabilitatea cazului (i), cu conditia din ipoteza si presupunerile facute, este $\frac{2}{27\cdot 3}$.
cazul (iii) este analog, deoarece este "simetric". (se obtine aceeasi probabilitate).
pentru cazul (ii) :
il fixez pe $a$. probabilitatea ca $c$ sa fie intre $a$ si $b$ este :
$\int_{a-1/3}^{a+1/3}|x-a|\,dx=2\int_{a}^{a+1/3}(x-a)\, dx=\frac{1}{9}$.
apoi integrez dupa $a$ :
$\int_{1/3}^{2/3}\frac{1}{18}\, da=\frac{1}{27}$.
deoarece evenimentele (i), (ii), (iii) sunt independente (nu se poate simultan $a\in[0,1/3]$ si $a\in[1/3,2/3]$), voi aduna probabilitatile, si voi obtine rezultatul :
$p_{c\ mijloc}=\frac{2}{27\cdot 3}+\frac{1}{27}+\frac{2}{27\cdot 3}=\frac{7}{27\cdot 3}$.
am scris indicele "c mijloc" deoarece asa am presupus de la inceput. cum combinatiile cu $c$ in "mijloc" sunt $1/3$ din total, rezulta ca probabilitatea dorita este chiar :
$p=3p_{c\ mijloc}=\frac{7}{27}$.[/hide]

Great!  :)

Mi se pare ok rezolvarea ta si cred ca, spre deosebire de mea, poate fi generalizata pentru n numere (vad ca a ridicat cineva intre timp problema asta pe topicul la care am dat link mai sus).

Imi mirosea si mie a integrale, da' nu stiam cum sa le aranjez.

solutia ta are avantajul elegantei. felicitari :)
acum ma gandesc la generalizare, sper sa nu fie prea mult de caculat :)

edit :

[hide="generalizare"]procedez exact la fel :
$a$ cel mai mare, $b$ cel mai mic, restul sunt $c_{1},c_{2},\ldots,c_{n-2}$.
am trei cazuri :
$a\in\left[0,\frac{1}{n}\right]$.
$a\in\left[\frac{1}{n},\frac{n-1}{n}\right]$.
$a\in\left[\frac{n-1}{n},1\right]$.
de aici analog ca pentru cazul $n=3$.
bla bla am gresit la calcule :)[/hide]

E o greseala pe undeva, fiindca nu se potriveste raspunsul pentru n=3  :huh: (nici pentru n=2, de altfel).

Nu stiu ce functie ai integrat tu.

Pentru $c$ si $d$ fixate, probabilitatea ca alte $n-2$ numere sa fie intre $c$ si $d$ este $|c-d|^{n-2}$ (sunt $n-2$ evenimente independente cu aceeasi probabilitate), deci cred ca trebuie integrata initial functia $|x-a|^{n-2}$, in loc de $|x-a|$.

Procedand exact cum ai procedat tu in cazul $n=3$, dar cu functia asta in loc de cea veche, mi-a dat in final:

$p_{n}=\frac{n^{2}-n+1}{n^{n}}$

Pentru $n=2$ obtinem $3/4$, iar pentru $n=3$ obtinem $7/27$, adica exact cat trebuia sa dea.

da, si eu cred ca am gresit la calcule :)
eu sunt mai neatent de obicei, eh..macar am gasit ideea, dar n`am avut rabdare sa le fac cum trebe.
te cred pe cuvant ca ai dreptate :)


Prove that the fraction $ \dfrac{21n \plus{} 4}{14n \plus{} 3}$ is irreducible for every natural number $ n$.

Let $k$ be a common factor, greater than zero, of both the numerator and the denominator. Let
\[ 21n+4=Ak  \]
and
\[ 13n+3=Bk  \]
Then
\[ k(3B-2A)=3Bk-2Ak=(42n+9)-(42n+8)=1  \]
The first expression is divisible by $k$, so $k$ must be a divisor of 1. Thus the greatest common factor of the numerator and the denominator is one, and the fraction is not reducible.

Using Euclid's algorithm for determining the greatest common divisor of the two numbers we obtain

\[ 21n+4=1 \cdot (14n+3)+(7n+1)  \]
\[ 14n+3=2 \cdot (7n+1)+1  \]
\[ 7n+1=1 \cdot (2n+1)+0  \]

The greatest common divisor of the numerator and denominator is one. Therefore the given fraction is irreducible.

Easiest IMO prob. ever.  :D
We have 2(21n+4)-3(14n+3)=-1, o we are done...

Bomb

[quote="bomb"]Easiest IMO prob. ever.  :D
[/quote]Well it was the [b]first[/b] imo problem .... [i]ever[/i] :)

Of course it seems easy to you since everyone knows how to use Euclid's algorithm, but 49 years ago it was a new method, probably.

Probably, it was new 2300 years ago :wink:

[quote="darij grinberg"]49 years ago people DID learn the Euclidean algorithm at school; it's rather nowadays that they don't. You will be amazed at what people learnt at school 49 years ago; the textbooks contained things that are meanwhile forgotten enough that you can pose them as IMO problems...[/quote]

Could you provide some examples for such IMO problems?

Hi, this is my first ever post here so please forgive me for being a little stupid (perhaps), but all I did was:

$ \dfrac{21n \plus{} 4}{14n \plus{} 3} \equal{} \dfrac{14n \plus{} 3 \plus{} 7n \plus{} 1}{14n \plus{} 3} \equal{} 1 \plus{} \dfrac{7n \plus{} 1}{14n \plus{} 3} \equal{} 1 \plus{} \dfrac{7n \plus{} 1}{2(7n \plus{} 1) \plus{} 1}$

and so we see that the part by which the fraction exceeds 1 can never be reduced because the numerator does not divide the denominator.

EDIT: Sorry, what I mean is that nothing divides both the numerator and the denominator because of the extra "1" hanging around.

I'm not really sure if this is a solution.  I'm not too smart so I might be wrong.  What  I did was assume that the fraction is reducible.  Thus

(21n+4)/(14n+3)=k for some positive integer k.

Thus 21n+4=14nk+3k
       7n(3-2k)=3k-4
       7n        =(3k-4)/(3-2k)
Since n is a natural number (3k-4)/(3-2k) must be a positive integer for some k.  Unfortunately, (3k-4)/(3-2k)>0 only in the interval (4/3,3/2).  We see that for (3k-4)/(3-2k) to be an integer k must be between 3/2 and 4/3.  But then, k isn't an integer and we reach a contradiction.  Thus (21n+4)/(14n+3) is irreducible.
I apologize for the lack of LaTeX in advance.  I still have to learn.

[quote="darij grinberg"][quote="apiarist1"]I mean that an ordinary kid didn't learn this method at school. I'm not ignorant - I know that Euclid lived in Ancient Greece. :lol:[/quote]

1) Ordinary kids don't go to IMO's.
2) 49 years ago people DID learn the Euclidean algorithm at school; it's rather nowadays that they don't. You will be amazed at what people learnt at school 49 years ago; the textbooks contained things that are meanwhile forgotten enough that you can pose them as IMO problems...

  dg[/quote]

People do realize this was in the soviet Union, america didn't enter till 1970's, with Paul Zeitz being first IMO member for the US (author of "Arts and Crafts of Problem Solving").

[moderator says: do not quote the entire post before yours]

... and your point is?__________________________

[b]Solution 1.[/b] Note that

\[3(14n+3)-2(21n+4)=1,\]

so by Bézout's lemma, we have $\gcd(21n+4,14n+3)=1,$ meaning that $21n+4$ and $14n+3$ are coprime, so the fraction $\frac{21n + 4}{14n + 3}$ is irreducible.

[b]Solution 2.[/b] Repeated applications of the Euclidean algorithm yield,

$\gcd(21n+4,14n+3)=\gcd(14n+3,7n+1)=\gcd(7n+2,7n+1)=\gcd(7n+1,1)=1,$

and thus the fraction $\frac{21n + 4}{14n + 3}$ is irreducible.

It's not like the solutions here are actually different from one another, so I might as well post mine.

Observe that$$3(14n+3)-2(21n+4)=1,$$so by Bézout's Identity we have $\gcd (21n+4,14n+3)=1$ (alternatively, any common divisor of $21n+4$ and $14n+3$ divides any integer linear combination of them, so it divides $1$), hence $\frac{21n+4}{14n+3}$ is irreducible.

First Post and first IMO problem solved by me
[hide=Solution] Observe that when we have a fraction $\frac{m}{f}$, if the greatest common divisor of $m$ and $f$ is 1 then the fraction can't be reduced.
$$gcd(14n+3, 21n+4)=gcd(7n+1, 14n+3)=gcd(1, 7n+1)= 1$$
By the Euclidean Algorithm $\frac{21n+4}{14n+3}$ is irreducible for every natural number $n$.

[b]Remark[/b]: Notice how most of the solutions to this problem is very similar.
[/hide]

Because $-2(21n+4)+3(14n+3)=1$, $21n+1$ and $14n+3$ are prime (bezout)
So the fraction is not reducible

This is so easy. We need to use Euclid's algorithm. So $\gcd(21n+4,14n+3)=\gcd(7n+1,14n+3)=\gcd(7n+1,7n+2)=1.$ So since the gcd is 1 the fraction is irreducible.

[hide=solution] Note that $$-2(21n+4) + 3(14n+3)=1,$$ so by Bezout's we have $$\gcd(21n+4, 14n+3)=1,$$ as desired. $\square$[/hide]

[hide=solution]Note that the fraction is irreducible if and only if gcd(21n+4,14n+3) = 1.  Using the Euclidean algorithm, we get gcd(21n+4,14n+3) = gcd(7n+1,14n+3) = gcd(7n+1,1) which is obviously 1.  Thus, the fraction is irreducible over all natural numbers n.[/hide]

oh look its the imo problem of all time

Euclidean Algorithm suffices. $\blacksquare$

[quote=apiarist1]I mean that an ordinary kid didn't learn this method at school. I'm not ignorant - I know that Euclid lived in Ancient Greece. :lol:[/quote]

No one learns this at school nowdays though

By Euclidean Algorithm, $$\text{gcd}(21n + 4, 14n + 3) = \text{gcd}(14n + 3, 7n + 1) = \text{gcd}(7n + 1, 1) = 1.$$ Hence, $ \dfrac{21n + 4}{14n + 3}$ is irreducible for every natural number $ n$.

Simple one:

[hide]
https://youtu.be/hORB1K2F1sY[/hide]

By Euclidean division Algorithm,

$$gcd(21n+4,14n+3) = gcd(7n+1,14n+3) = gcd(7n+1,7n+2) = gcd(7n+1,1) = 1.$$

  Hence, $\frac{21n+4}{14n+3}$ is irreducible for natural ${n}$.

Assuming that the fraction $\frac{21n+4}{14n+3}$ is reducible for some $n$, we get that $(21n+4,14n+3)\neq1$. $\Rightarrow \exists$ $d\neq1 \in \mathbb{N}$ such that $d \vert 21n+4$, $d\vert 14n+3$.

Now, $d\vert21n+4$ $\Rightarrow$ $d\vert2(21n+4)$ $\Rightarrow d\vert 42n+8$.

Also, $d\vert14n+3 \Rightarrow d\vert3(14n+3) \Rightarrow d\vert42n+9$

$\Rightarrow d\vert(42n+9)-(42n+8) \Rightarrow d\vert1$

As $d \in \mathbb{N}$, $d=1$, which implies that $21n+4$ and $14n+3$ have no common factor other than 1.
$(21n+4,14n+3)=1$, which contradicts our assumption. $(\Rightarrow\Leftarrow)$. Proved.

you can just Euclidean Algorithm this

alternatively

Let $k = 7n + \frac{3}{2}$.

This transforms our fraction into $\frac{3k - \frac{1}{2}}{2k}$ and since $2k \nmid 3k - \frac{1}{2}$, this fraction is irreducible. $\blacksquare$


I just wanted to point out something.In the next few years, a majority 49% of students passing out of institutes of higher education in India will be underserving candidates.The fact is that 49% of the seats are reserved for students of some "backward classes". This decision is based on a data of 1930.Wow! That's called rubbing salt on an already-wounded India. This is politically motivated(that's what I think).

You think? Obviously there isn't a single advantage to the system, and the sole reason is to gain votes.

1. Non-SC/ST/OBC (from hereon, SSO) students will have a harder time gaining access to these prestigious universities. When they do make it there, there won't be as much synergy. Unlike American affirmative action, where the goal is to bring variform viewpoints and skills to the table, a non-ST/ST/OBC student will probably have little to gain from someone who is allowed in because of class-status.

2. SSO students might not be prepared for the education ahead of them. There is a good reason why non-SSO students have to grind it out to get in. They might also be the targets of discrimination among their peers, or otherwise, just might not fit in. Finally, employers will know that they got in because of class-status.

Also consider that the reservation allows many students to get in because of their ancestry, even though they themselves might not necessarily be disadvantaged.

3. The institution loses out because of a diminished student body caliber.

And wait... that's not all... what's this I hear about a law to institute reservations among the [b]faculty [/b]in IIT's? Talk about shooting their institutions in the foot! I was planning to study abroad at IIT Kanpur for a semester... obviously I'm not going to do that anymore.

Same thinking here.

Politicians never understand what IIT is for general indian students.
The dream for which students are more than willing to sacrifice 2 years.

they are diluting standards of IITs just for votes thats wrong.

Reservatin is not going to help the deserving people.Most of students who'll get admission in good streams(eg,CSE,IIT mumbai,etc)will be students who would have got ranks below the requirement but they'll be selected just because of the fact that they were lucky to be born in OBC family.

Also the definition of creamy layer has gone up to Rs.450,000.(do you think people with family incomes like Rs.400,000 deserve to get reservations?????)

about standard there is not any second thought,STANDARD WILL FALL DOWN.
Earlier good IITs had students of high caliber,it was something that made IIT famous(even today infrastructure,etc of IITs is not that good as many institutions),one can jus assume that IIT studen will be certainly good(it proved true in majority of cases)
but now before deciding about caliber of student beforehand,you'll have to ask his caste(it divides the society)

@TZF, you seem to have good idea about India as well as these things

That is why,the craze for IIT is now a little less. :(

@TZF: You are a very motivating person,with very good thoughts.Kudos to you man :)

another discussion!! :)

well i think u can go to the other thread--dont remember the name...

[quote="TZF"] I was planning to study abroad at IIT Kanpur for a semester... obviously I'm not going to do that anymore.[/quote]
I'd advise you not to be hasty  :maybe:  Even though there is some reservation among students, there are still bright students in IIT's and after all, everything depends upon you and the type of people you hang with wherever you study. And i can ascertain you that there are still bright and focussed people in the IIT's (not to mention that [url=http://indiaedunews.net/IIT/IIT-JEE_topper_bags_gold_at_International_Physics_Olympiad_5413/]people like this[/url] are at IITK too :P)

but these are only exceptions to the pool of applicant's out of which some who just get in IIT but don't deserve to get in(This is basically about the reservation system//////)
According to me,a lot of politics is engulfing IIT(not just IITK..............)which should be stopped immediately

[quote="TZF"]
I was planning to study abroad at IIT Kanpur for a semester... obviously I'm not going to do that anymore.[/quote]

Don't even think about it. It is a breeding ground for hindu extremist policies. Some professors there are core hindu extremists, who hate islam and muslims., and are recruiting hindu students into their fray.

This is a true story and has appeared in many major newspapers.


induction
inequalities
combinatorics unsolved
combinatorics

In a country there are several cities; some of these cities are connected by airlines, so that an airline connects exactly two cities in each case and both flight directions are possible. Each airline belongs to one of $k$ flight companies; two airlines of the same flight company have always a common final point. Show that one can partition all cities in $k+2$ groups in such a way that two cities from exactly the same group are never connected by an airline with each other.

Observe that airlines of each company compose a triangle or a star. (Triangle is a graph $(\{u,v,w\},\{uv,vw,wu\})$ and star is a graph $(\{c,u_1,\ldots,u_l\},\{cu_1,\ldots,cu_l\})$).

Induction by k.

The statement of the problem is obvious for $k=1$.

[b]Inductive step.[/b]If for some company airlines compose a star then delete the center vertex $u$ of the star (with all incident edges). Then for the rest cities we may apply inductive conjecture and put $u$ in new group. We are done in this case.

Now consider a case when all companies are triangles. Consider any company composing a triangle $uvw$. For the rest cities we may apply inductive conjecture and put them into $k+1$ groups in a proper way. And now we have to put two vertices from $\{u,v,w\}$ into some old groups. Suppose we can't do it. It means that there exists $x,y \in \{u,v,w\}$ such that there are edges from $x$ to each group and from $y$ to each group. Totally $2(k+1)$ edges incident to $u$, $v$ or $w$. But we have $k-1$ companies exept $uvw$. Each company is a triangle, and it is easy to see that no company can have all 3 edges incident to $u$, $v$ or $w$. There are at most two such edges. Totally we may have $2(k-1)$ edges incident to $u$, $v$ or $w$. Since $2(k+1)>2(k-1)$, we have a contradiction. So, we can put some two vertices from $\{u,v,w\}$ to old groups. The rest vertex we will put into new group. In this case we are also done.

The problem is solved.

[quote="barasawala"][quote="Remike"] it is easy to see that no company can have all 3 edges incident to $u$, $v$ or $w$. [/quote]

Can anyone tell me why this is?[/quote]

Because all the companies have to be triangles?

OK, I misunderstood something. Sorry.

Thanks for your reply anyway, weiquan.

For the case where everything is a triangle, we can do it with the following inequality: $\binom{\chi(G)}{2} \le |E|$. $\chi(G)$ = minimum number of colors you need to color the vertices such that no adjacent vertices have the same color (aka chromatic number).

Proof: Color it with the minimum amount of color. Assume that $\binom{\chi(G)}{2} > |E|$. Then some pair of colors is not connected by an edge. (Say 1, 2). Then just change all the 1's to 2's, and since they aren't adjacent, this new coloring works, contradicts minimality. $\blacksquare$

So assume $\chi(G) \ge k+3$. Then $\binom{k+3}{2} \le |E| = 3k$, a contradiction for any $k \ge 1$.

Note that the set of edges of the same airline forms either a star or a triangle.

Let $x$ be the number of stars. For each star, let the central vertex be the vertex of degree $\ge 2$. Then I color this vertex with a color that is different from all other vertices. This takes at most $x$ colors.

Let $y$ be the number of remaining triangles, which is at most the number of original triangles. There are $3y$ edges.

Claim: $E\ge \binom{\chi(G)}{2}$.

Proof: There exists an edge between two vertices of any pair of colors $(i,j)$; if $(i,j)$ is a counterexample, $i,j$ can be merged to be the same color.

If follows that if the graph needs $y+3$ colors, then we need at least $\binom{y+3}{2}$ edges, but $\binom{y+3}{2} > 3y$ for all $y\ge 1$, contradiction.

We do a little trolling

I assume that no two flight companies can operate a flight between the same two cities (this seems to be what the wording indicates). We are essentially being asked to show that it is possible to $k+2$-color such a graph.

It is clear that the set of edges of some flight company is either a triangle or a star. We now use induction, with the case of $k=1$ being obvious. If a flight company has edges forming a star, then temporarily delete the company and color the resulting graph with $k+1$ colors. Then add it back, and and recolor the "center" of the star with the $k+2$-nd color, which clearly works. Therefore, suppose that all of the flight companies are triangles.

If some non-isolated vertex $u$ has degree $d \leq k+1$, then temporarily delete the $\tfrac{d}{2}$ flight companies with an edge attached to the city and color the remaining graph. When we add the $\tfrac{d}{2}$ flight companies back, assign a unique one of our new $\tfrac{d}{2}$ colors to each flight company containing $u$. If a flight company on vertices $\{u,v,w\}$ is assigned a color, then change the color of vertex $u$ to the assigned color. Finally, since $d \leq k+1$, one of the $k+2$ available colors is not present among the neighbors of $u$, so we can change the color of $u$ to that. It is clear that the resulting graph coloring is valid.

Therefore, suppose that every vertex has degree at least $k+2$. On the other hand, we have $3k$ edges, hence there are at most $\frac{6k}{k+2}$ vertices, i.e. at most $5$ vertices. Since we supposed $k \geq 2$, we actually need $5$ vertices and for the graph to be a $K_5$, but this is clearly impossible. Hence we are done. $\blacksquare$

[b]Remark:[/b] [url]https://en.wikipedia.org/wiki/Spoke%E2%80%93hub_distribution_paradigm[/url]

funny
[hide = solution]
We know that the graph can be edge decomposed into a bunch of triangles and stars, so that there are exactly $k$ of them together. We've to show that $\chi(G) \leq k+2$. To prove this, we induct on $n$. Bases cases are easy to deal with. For the inductive step, if there's a star, we delete the central vertex and all the edges it is part of, induct down to the left over graph and then add in the central vertex as a new group all by itself. If there's only triangles, there are at most $3k$ edges. Since this implies ${\chi(G) \choose 2} \leq 3k$ the result follows quickly now.
[/hide]


number theory
least common multiple
greatest common divisor

three numbers a, b, and c have a LCM of 60. numbers a and c have a GCF of 6. numbers a and b have a GCF of 2. numbers b and c have a GCF of 4.

1. what is the smallest possible sum of a, b, and c?

2. what is the largest possible sum of a, b, and c?

3. how many possible sums are there?

to do the first in 5 minutes:
three numbers a, b, and c have a LCM of 60. numbers a and c have a GCF of 6. numbers a and b have a GCF of 2. numbers b and c have a GCF of 4.

1. what is the smallest possible sum of a, b, and c?
[hide]6*X

4*x

12*X

60=2^2*3*5

so we already have 2^2, and we already have 3, but we don't have 5 so we add this to b(for smallest value)

6x
20x
12x
and for the smallest value x=1

38?!?[/hide]

[hide]The smallest sum is $6+20+12=38$

The largest sum is $6+4+60$.

There are three possible sums. [/hide]


calculus
integration
trigonometry
function
calculus computations

Hi,
I need help to solve these examples:

1) $ \int \frac{cosx \, dx}{2sin^2x\plus{}5}$

2)$ \int(x^2\plus{}3)sin 2x\,dx$

3)$ \int(4x^3\plus{}2x)arctgx\, dx$

Thanks in advance.

First one

substitute sinx as t

this gives

$ \int\frac{dt}{2t^2\plus{}5} \equal{} \frac{1}{2} \int\frac{dt}{t^2\plus{}(\sqrt{\frac{5}{2}})^2}$

$ \equal{} \frac{1}{2.\sqrt{\frac{5}{2}}} tan^{\minus{}1}(\frac{t\sqrt{2}}{\sqrt{5}}) \plus{} C$

$ \equal{} \frac{1}{\sqrt{10}} \ tan^{\minus{}1}(\sqrt{\frac{2}{5}}sinx) \plus{} C$

Second one, try Integration by Parts ...

hello, for 1), use the substitution $ z\equal{}\tan(\frac{x}{2})$ with $ dx\equal{}\frac{2dz}{1\plus{}z^2}$, $ \cos(x)\equal{}\frac{1\minus{}z^2}{1\plus{}z^2}$, $ \sin(x)\equal{}\frac{2z}{1\plus{}z^2}$.
Sonnhard.

[quote="Ligeia"]Hi,
I need help to solve these examples:

1) $ \int \frac {cosx \, dx}{2sin^2x \plus{} 5}$

2)$ \int(x^2 \plus{} 3)sin 2x\,dx$

3)$ \int(4x^3 \plus{} 2x)arctgx\, dx$

Thanks in advance.[/quote]
1)
 $ \int \frac {cosx \, dx}{2sin^2x \plus{} 5}\equal{}\int \frac {d\sin x}{2\sin^2x \plus{} 5}\equal{}\sqrt {\frac 25}\int \frac {d\sqrt {\frac 25}\sin x}{2\sin^2x \plus{} 5}\equal{}\dots$
2)
$ \int(x^2 \plus{} 3)sin 2x\,dx\equal{}\int x^2 \sin 2x\,dx\plus{}\int 3\sin 2x\,dx\equal{}\frac 12\int x^2 d\cos 2x\plus{}3\int \sin 2x\,dx\equal{}\dots$
3)
$ \int(4x^3 \plus{} 2x)\arctan x\, dx\equal{}\int 4x^3 \arctan x\, dx\plus{}\int 2x \arctan x dx\equal{}\int \arctan x dx^4\plus{}\int \arctan x dx^2\equal{}\dots$

hello, in the second case write the integral in the form
$ \int x^2\sin(2x)\,dx\plus{}3\int\sin(3x)\,dx$ and use integration by parts for the first one.
Sonnhard.

the Third one

u can write  x = tan z

Integral reduces to

$ 2 \int tanz(2tan^2z \plus{} 1)sec^2z.z.dz$

which can be integarted by parts taking z as first function ...


Last year Mr. John Q. Public received an inheritance. He paid $ 20\%$ in federal taxes on the inheritance, and paid $ 10\%$ of what he had left in state taxes. He paid a total of $ \$10,500$ for both taxes. How many dollars was 	the inheritance?

$ \textbf{(A)}\ 30,000 \qquad \textbf{(B)}\ 32,500 \qquad \textbf{(C)}\ 35,000 \qquad \textbf{(D)}\ 37,500 \qquad \textbf{(E)}\ 40,000$

[hide] Say that John got 100x dollars of inheritance.  We know that 20% of this, or 20x of it went to federal taxes.  Thus, he has 80x dollars left.  10% of this, or 8x went to state taxes.  Thus, a total of 28x was what he paid.  Hence, 28x=10500, so x=375.  Therefore, John got 375*100=37500 dollars from inheritance, so the answer is $D$. [/hide]

[quote="4everwise"]Last year Mr. John Q. Public received an inheritance. He paid $20\%$ in federal taxes on the inheritance, and paid $10\%$ of what he had left in state taxes. He paid a total of $\$10,500$ for both taxes. How many dollars was 	the inheritance?

$\text{(A)}\ 30,000 \qquad \text{(B)}\ 32,500 \qquad \text{(C)}\ 35,000 \qquad \text{(D)}\ 37,500 \qquad \text{(E)}\ 40,000$[/quote]
Me remember this problem, I did it the guess and check way, but I know the real way.
[hide="The real way"]So he has x amount of money. So he pays .2x for federal taxes. he has .8x left. Then he pays .08x to pay for state taxes. So in total he pays 10500 so the equation is .28x=10500. Solving this you get that x=37500.[/hide]

[quote="4everwise"]Last year Mr. John Q. Public received an inheritance. He paid $20\%$ in federal taxes on the inheritance, and paid $10\%$ of what he had left in state taxes. He paid a total of $\$10,500$ for both taxes. How many dollars was 	the inheritance?

$\text{(A)}\ 30,000 \qquad \text{(B)}\ 32,500 \qquad \text{(C)}\ 35,000 \qquad \text{(D)}\ 37,500 \qquad \text{(E)}\ 40,000$[/quote]

[hide="solution"]Let the inheritance be $x$.

$.2x+.8\cdot .1x=10500$.

$x=\boxed{37500}$.[/hide]

[quote="King_James23"]Also on the 10A
 The answer is [b]3500[/b][/quote]Answer is wrong, see above solutions, and we don't care what you got.

[quote="4everwise"]Last year Mr. John Q. Public received an inheritance. He paid $20\%$ in federal taxes on the inheritance, and paid $10\%$ of what he had left in state taxes. He paid a total of $\$10,500$ for both taxes. How many dollars was 	the inheritance?

$\text{(A)}\ 30,000 \qquad \text{(B)}\ 32,500 \qquad \text{(C)}\ 35,000 \qquad \text{(D)}\ 37,500 \qquad \text{(E)}\ 40,000$[/quote]
[hide="Answer"]20% off 100% is 80%. 10% off 80% is 72%. 100% -72% = 28%. Let the inheritance be x. Then, 10500 = 28% of x. Thus, x equals 37500, or D.[/hide]


Let a,b,c are lengths sides of triangel. Prove that
$ \sum \frac {a}{\sqrt{5a^2\plus{}13bc}}\ge \frac {1}{\sqrt {2}}$
Find the best constant for this inequality and prove it. :)

No anyone prove this inequality??? :( argady, can-hang, where are you?

We can use Holder inquality. Like
\[ \begin{array}{l}
 (a^3  \plus{} b^3  \plus{} c^3 )(m^3  \plus{} n^3  \plus{} p^3 )(x^3  \plus{} y^3  \plus{} z^3 ) \ge (amx \plus{} bny \plus{} cpz)^3  \\
 With \\
 a^3  \plus{} b^3  \plus{} c^3  \equal{} m^3  \plus{} n^3  \plus{} p^3  \equal{} \sum\limits_{cyc} {(\frac{{a^{\frac{1}{3}} }}{{(5a^2  \plus{} 13bc)^{\frac{1}{6}} }}} )^3  \\
 x^3  \plus{} y^3  \plus{} z^3  \equal{} \sum\limits_{cyc} {(a^{\frac{1}{3}} } (5a^2  \plus{} 13bc)^{\frac{1}{3}} )^3  \\
 \end{array}
\]
 :wink:

Yes, but it is not finish. :D
Who can finish it?? :lol:

[quote="thegod277"]Let a,b,c are lengths sides of triangel. Prove that
$ \sum \frac {a}{\sqrt {5a^2 \plus{} 13bc}}\ge \frac {1}{\sqrt {2}}$
Find the best constant for this inequality and prove it. :)[/quote]

By Holder inequality we have
\[ \left( {\sum {\frac {a}{{\sqrt {5a^2 \plus{} 13bc} }}} } \right)^2 \left( {\sum {a\left( {5a^2 \plus{} 13bc} \right)} } \right) \ge \left( {\sum a } \right)^3
\]
Then it suffices to show that
\[ \Leftrightarrow a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc \le 2\left( {ab\left( {a \plus{} b} \right) \plus{} bc\left( {b \plus{} c} \right) \plus{} ca\left( {c \plus{} a} \right)} \right)
\]
which is an old problem of nhocnhoc. :)

I think the followed holds for $ a,b,c$ be nonnegative numbers
\[ \frac{a}{{\sqrt {4a^2  \plus{} 5bc} }} \plus{} \frac{b}{{\sqrt {4b^2  \plus{} 5ca} }} \plus{} \frac{c}{{\sqrt {4c^2  \plus{} 5ab} }} \le 1
\]

[quote="thegod277"]:D oh, yes. A finish inequality is
$ \sum (a \plus{} b \minus{} c)(a \minus{} c)(b \minus{} c)\ge 0$
which is true by vornicur schur. :)
And the best constan for this inequality is 2,53...[/quote]

Show your solution if you can, the god.


trigonometry
complex analysis
complex analysis unsolved

Another problem from Conway:

I'm trying to show that the image of $ \frac{\cos z}{z}$ outside of any ball around the origin is $ \mathbb{C}$.  This is of course very easy to do using Picard's Theorem, but surely there is a more elementary way to see it?

Or did he just put it in his book to make us realize how hard it is without Picard?  :P

I wouldn't put it past him...


algebra
polynomial
search
number theory
prime numbers
arithmetic sequence
number theory proposed

Prove that there are an infinite number of primes $ p$ such that neither $ p \minus{} 2$ nor $ p \plus{} 2$ is a prime, i.e., there is an infinite number of prime numbers not belonging to any pair of twin primes.

$ p \equiv 2 \bmod 3, p \equiv \minus{} 2 \bmod 5$ is an arithmetic progression with infinitely many primes by Dirichlet's theorem.  (CRT lets you generalize as follows:  there are infinitely many primes $ p$ such that $ p \minus{} 2$ and $ p \plus{} 2$ both have at least $ n$ prime factors for any $ n$.)

(The strongest result I know about elementary proofs of special cases of Dirichlet's theorem couldn't cover this case as far as I know, but I hesitate to claim that there isn't a simpler proof.)

It would be sufficient to prove there are infinitely many primes of the form $ 15n\plus{}8$, by elementary methods.

I don't see how you are doing $ 15n\plus{}8$ on elementary ways (note that $ 8^2 \not \equiv 1 \mod 15$).

I was not stating that it can be done for sure (only a suggestion).  Does there exist a solution without the use of Dirichlet's Theorem?

Not that I know of.  As ZetaX has hinted, the strongest result (at least to my knowledge) is that there exists a "polynomial" proof (like the proof for $ 4n \plus{} 1$) about primes congruent to $ a \bmod m$ if and only if $ a^2 \equiv 1 \bmod m$.  See [url=http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf]this paper[/url].

This is the only elementary method to prove [i]any[/i] special cases of Dirichlet's method I know of.  Therefore if there's an elementary solution to this problem one might have to prove compositeness without exhibiting an explicit divisor.

That doesn't rule out other approaches entirely, such as showing that there aren't very many twin primes compared to primes.

Asuming the Schinzel hypothesis, one can even prove that there is no "euclid-style" proof if $ a^2 \not\equiv 1 \mod n$. The paper was posted here about a year ago, I will take a search if someone wants it.


Solve x and y

Problem 1
$\displaystyle x^2y+xy^2=70$
$\displaystyle (x+y)(x^2+y^2)=203$

Problem 2
$\displaystyle (x^2+y^2)x/y=6$
$\displaystyle (x^2-y^2)y/x=1$

Problem 3
$\displaystyle x^4+y^4-x^2y^2=13$
$\displaystyle x^2-y^2+2xy=1$

Problem 4
$x^3-y^3=19(x-y)$
$x^3+y^3=7(x+y)$

Problem 5
$x^2/y+y^2/x=12$
$1/x+1/y=1/3$

[hide] Let $xy=a$
$x+y=b$
and $x^2+y^2=b^2-2a$
then we have
$ab=70$
$b(b^2-2a)=203$
or
$b^3-2ab=203$
substituting $70$ for $ab$ gives $b^3=343$ or $b=7$
so $xy=10$
plugging in some numbers, we see that
$x=5$, $y=2$ [/hide]

[hide]
#4
$x^3-y^3\Rightarrow(x-y)(x^2+xy+y^2)=19(x-y)
also
x^3+y^3\Rightarrow(x+y)(x^2-xy+y^2)=7(x+y)$

we can cross out $(x-y)$ and $(x+y)$ from the equations respectively to get...

$(x^2+xy+y^2)=19
and
(x^2-xy+y^2)=7$

subtract the equations to get...

$2xy=12\Rightarrow xy=6$

substituting xy=6 to either equation, we get $x^2+y^2=13$

we can add and subtract 2xy into $x^2+y^2=13$ to eventually get

$(x+y)^2=25 and (x-y)^2=1$

therefore $x+y=\pm5 and x-y=\pm1$

then we get $x=\pm3 and y=\pm2$

#5

we can multiply the second equation by 36 and set it equal to equation 1 to get...

\[\frac{36}{x}+\frac{36}{y}=\frac{x}{y}(x+y)\]

this is also equal to \[\frac{36}{xy}=\frac{x}{y}\]

cancelling, we get $x=\pm6$

plugging this into equation 2, we get $x,y=6$ or $x=-6, y=2$

we plug these into equation 1 and see that only $x,y=6$ works so that is our answer.
[/hide]

The easiest way to do all of these is to use this
[hide="Problem5"]
Using $s=x+y$ and $p=xy$
So substituting in the equation, $\frac{s(s^2-3p)}{p}=12$-$\boxed2$
and then the $2nd$ we get $3s=p$ -$\boxed1$
so substituting $\boxed1$ in $\boxed2$ we get the quadratic equation,
$s^2-9s-36=0$ which gives the roots $s=12$ and $s=-3$
[/hide]

[quote="felipesa"]Problem 2
$\displaystyle (x^2+y^2)x/y=6$
$\displaystyle (x^2-y^2)y/x=1$
[/quote]
Divide them, we get
$\frac{x^2+y^2}{x^2-y^2}\frac{x^2}{y^2}=6$
or
$\frac{\frac{x^2}{y^2}+1}{\frac{x^2}{y^2}-1}\frac{x^2}{y^2}=6$

Let $t=x^2/y^2$,
$\frac{t+1}{t-1}t=6$, we get $t=2$ or $t=3$.

I think you can finish it from this.

[hide="partial solution for problem 3"]
$x^2-y^2=1-2xy$
square both sides
$x^4+y^4+2x^2y^2=1-4xy+4x^2y^2$
subtract $3x^2y^2$ from both sides, and now you can use make it fit into the other equation
$x^2y^2-4xy+1=13$
$(xy-6)(xy+2)=0$
substitution and manipulation will probably get to the answer from here :)
[/hide]

[quote="felipesa"]
Problem 3
$\displaystyle x^4+y^4-x^2y^2=13$
$\displaystyle x^2-y^2+2xy=1$
[/quote]
the first equation is $(x^2-y^2)^2+x^2y^2=13$
and from the second equation we have $x^2-y^2 = 1-2xy$
hence
$(1-2xy)^2+x^2y^2=13$
Let $t=xy$ and it becomes $(1-2t)^2+t^2=13$ and we get $t=-\frac{6}{5}$ or $t=2$.

For $t=2$, we have
$x^4+y^4=17$ and $x^2-y^2=-3$
Notice that $x^4+y^4 = (x^2+y^2)^2-2x^2y^2=(x^2+y^2)^2-8=17$
hence $x^2+y^2=5$.
Now we can solve out $x=1,y=2$, or $x=-1,y=-2$. (Since $xy>0$, $x$ and $y$ have the same sign)

You can do it in a similar way for $t=-\frac{6}{5}$ but the answers are complicated.


function
vector
trigonometry
calculus
calculus computations

Find all continuous functions f:R->R, if $ f\left( x \right) \plus{} f\left( {x \plus{} 1} \right) \plus{} f\left( {x \plus{} 2} \right) \equal{} 6x \plus{} 6$
for every real number x

There's far too many to give "all". It's a translate of an infinite-dimensional vector space.

First, let $ g(t)\equal{}f(t)\minus{}2t$. We have $ g(x)\plus{}g(x\plus{}1)\plus{}g(x\plus{}2)\equal{}0$, hence $ g(x\plus{}3)\equal{}g(x)$. Write its Fourier series $ g(t)\equal{}\sum_n a_n e^{2\pi i n t/3}$. From $ g(x)\plus{}g(x\plus{}1)\plus{}g(x\plus{}2)\equal{}0$, $ a_n\equal{}0$ for each $ n$ divisible by 3. From $ f$ being real-valued, $ a_n\equal{}\overline{a_{\minus{}n}}$ (or we could do the whole thing with sines and cosines).
That's basically all we can say.


A cube of cheese $ C\equal{}${$ (x,y,z)|0\leq x,y,z\leq 1$} is cut along the planes $ x\equal{}y$, $ y\equal{}z$, and $ z\equal{}x$.  How many pieces are there?  (No cheese is moved until all three cuts are made.)

(a) 5
(b) 6
(c) 7
(d) 8
(e) 9

I don't get what $ C\equal{}${$ (x,y,z)|0\leq x,y,z\leq 1$} and "planes $ x\equal{}y$, $ y\equal{}z$, and $ z\equal{}x$" means.  What does the graph look like?  Could someone please help me?  Thanks!

The problem is talking about a 3-dimensional coordinate system with perpendicular axes $ x, y, z$ intersecting at $ (0,0,0)$, the origin. The graph of an equation like $ x\equal{}y$ would be the set of all points with coordinates satisfying the equation. It will take some imagination to get the picture in your head if you aren't familiar with the 3D system yet.

$ x\equal{}y$ is a plane that includes the entire $ z$-axis.

I'll give you another opportunity to try to solve the problem.  :)

I still don't get it.  Could someone please provide me with a solution?  Thanks!

[hide="Approach A"]$ x \equal{} y$ partitions space into two sections $ x > y$ and $ x < y$.
Similarly for $ x \equal{} z$ and $ y \equal{} z$.

So, the three cuts together partition the space into sections $ x < y < z$ and permutations. There are $ 6$ such possibilities, which correspond to the permutations of $ (x,y,z)$, so the answer is $ \textbf{(B)}$[/hide]

[hide="Approach B"]All three planes contain the line $ x \equal{} y \equal{} z$. When three different planes all intersect in a common line, they will divide space into 6 parts.

[url=http://www.josechu.com/planes_in_3d/pl6_gif_p.gif]Here is an image[/url] showing three planes with a common line dividing space into 6 sections.[/hide]