delete this

go here instead http://www.artofproblemsolving.com/Forum/viewtopic.php?p=276798#276798

cool game.  i have been trying for ten minutes and can't get it :lol:

[quote="jmadsen"]It's not very hard if you work backwards.[/quote]

post you answer then, we can see if it works

We want to cover all but one of the red dots. Suppose the one we will leave uncovered is the top one. Since we must place the balls by starting on an uncovered dot, the uncovered one at the top must serve as the "launching point" for the last ball we place. We can actually place 2 balls in the center using this as the launching point. Before we place those, we have 2 empty spots in the center, which we can use as the launching points for the 2 bottom points of the star. Before placing those, we can use those empty spaces to place 2 more in the center. Filling in the first 3 is easy.

For example, the balls could be placed in this order:

yes that is correct, it is also possible to make variations of the current position


modular arithmetic
number theory unsolved
number theory

Denote by $S(n)$ the sum of decimal digits of $n$. Prove that there doesn't exist any number such that it appears in the sequences $\left\{S(2^n),n>0\right\}$ and $\left\{S(n!),n>0\right\}$ infinitely many times.

Consider the equation

$(1) \; S(2^m) = S(n!)$.

Since these integers are represented in the decimal system,

$2^m \equiv n! \pmod{9},$

which implies that

$2^m \equiv n! \pmod{3}.$

Hence, if $n \geq 3$, then

$2^m \equiv n! \equiv 0 \pmod{3},$

thus $3 \,|\, 2$. This contradiction gives $n \leq 2$, so

$(2) \; S(2^m) \leq 2.$

according to (1). If  $2^m>10$, then the first and the last digit of $2^m$ are nonzero integers. The fact that the last digit is even implies that $S(2^m) \geq 3$ when $m \geq 4$. Moreover $S(2^m) = 2^m$ for $m < 4$, hence $m = 1$ by (2). In other words, the only solution of equation (1) is $(m,n) = (1,2)$. Consequently, the sequences $\{S(2^n)\}_{n \in {\bf N}}$ and $\{S(n!)\}_{n \in {\bf N}}$ have only the term 2 in common.


$ \lim_{n\rightarrow infinity } n_{C_{x}}\left(\frac{m}{n} \right)^{x}\left(1\minus{}\frac{m}{n} \right)^{n\minus{}x}$

I think you need to say what $ m$, $ x$, and $ C_x$ are...

n ,  m ,  x  are all natural numbers.    that $ n_{C_{x}}$  is the symbol we use in combinatorics..

yes modularmarc101.   i want to  say the same.

I do not know if this will help:
$ 1 \equal{} \left(1 \minus{} \frac{m}{n} \plus{} \frac{m}{n}\right)^{n} \equal{} \sum_{x \equal{} 0}^{n} \binom{n}{x}\left(1 \minus{} \frac{m}{n}\right)^{n \minus{} x} \left(\frac{m}{n}\right)^{x}$

Apparently that holds for all $ n$. So let $ n$ go to infinity. The series converges (to 1), so its terms must go to zero.

You can't conclude that; when the number of terms changes, so do the terms themselves.  Anyway, for $ n$ large we have $ {n \choose x} \sim \frac {n^x}{x!}$ so the limit simplifies to

$ \lim_{n \to \infty} \frac {m^x}{x!} \left( 1 \minus{} \frac {m}{n} \right)^n \left( 1 \minus{} \frac {m}{n} \right)^{ \minus{} x}$

where the left factor is a constant, the middle factor converges to $ e^{ \minus{} m}$, and the right factor converges to $ 1$.  This gives the limit

$ \frac {m^x e^{ \minus{} m}}{x!}$

that describes the [url=http://en.wikipedia.org/wiki/Poisson_distribution]Poisson distribution[/url].  Akech's observation is the observation that this is the limit of the binomial distribution as the number of trials tends to infinity.


Evaluate and express as a common fraction: $ \frac{\frac 12 \minus{} \frac 13}{\frac 18}$


Let $ ABC$ be an acute triangle inscribed in $ (O)$. Let $ AH$ be the altitude from $ A$ to $ BC$ ($ H$ belongs to $ [BC]$). Denote $ (C_1)$, $ (C_2)$ the circles tagent to $ AH$, $ BC$, and $ (O)$. Prove that the innertagent of $ (C_1)$ and $ (C_2)$ passes through the midpoint of $ BC$ if and only if $ AB\plus{}AC\equal{}2BC$. :D

It's Romanian TST 2006,day 5,problem 4.
See here:
http://www.mathlinks.ro/viewtopic.php?p=518139#518139


Source: Old AMC Test

1.  One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk.  The amounts of coffee and milk varied from cup to cup, but were never zero.  Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee.  How many people are in the family?

2.  When the mean, median, and mode of the list 10,2,5,2,4,2,x are arranged in increasing order, they form a non-constant arithmetic progression.  What is the sum of all possible real values of x?

3.  In year N, the 300th day of the year is a Tuesday.  In year N + 1, the 200th day is also a Tuesday.  On what day of the week did the 100th day of year N - 1 occur?

What's the source of these problems? All three look very familiar.

[quote="SnowStorm"]2.  When the mean, median, and mode of the list 10,2,5,2,4,2,x are arranged in increasing order, they form a non-constant arithmetic progression.  What is the sum of all possible real values of x?[/quote]Fun problem, I hope I got it right. [hide]Write the list without the $x$ in order: $2,2,2,4,5,10$. Clearly the mode is $2$. If $x\le2$ then the median is also $2$, but this contradicts the conditions of the problem, so $x>2$. If $x\le4$, then the median is equal to $x$. The mean will be equal to $\frac{25+x}{7}$, and $\frac{25+x}{7}>x\Leftrightarrow{x}<\frac{25}{6}$ shows that for $x\le$ 4, the mean is greater than the median. Thus by the conditions of the problem the following equality must be satisfied: $x-2=\frac{25+x}{7}-x\Rightarrow{x}=3$. For the other case say $4<x$; then the median is equal to $4$. $\frac{25+x}{7}>4\Leftrightarrow{x}>3$ shows that the mean is greater than the median for this case. Thus we must have $4-2=\frac{25+x}{7}-4\Rightarrow{x}=17$. The sum of all possible value for $x$ is $3+17=20$.[/hide]

[hide=""problem_1""]
This is not a very rigorous answer but it seems to work.  Looking at the problem we see that if there are 4 or fewer people, then angela will not have enough dring, and if their are six or more than 6 people she will have too much drink.  This leads us to think that there are probably five people.  With five people we find that in total there are 24 ounces of coffee and 16 ounces of milk, with Angela having 4 ounces of milk and 4 ounces of coffee, which gives the desired result. [/hide]


Problem:

If 4/2001 < a/(a+b) < 5/2001 , compute the number of integral values that b/a can take on.

Holler if you get [hide]100[/hide] as your answer, because that's what I got and I'm not sure.

Yes I get that.
[hide="(range)"]400 to 499[/hide]


can anybody please tell me what are the general cutoff scores for the objective paper in the national standard examination in physics(NSEP).some say that even if you are not in the  top 10% of yuour centre but your score in the objective paper is more than a certain cutoff fixed by IAPT,then your subjective paper will be corrected. i desperately want to know that cutoff score. does a score of 75 in the objective paper (out of a possible 180) stand anywhere near selection for the subjective papers to be corrected.
 pleaaaaaaaase tell me

hi ashwin

usually,centre cutoffs in obj r abt 40 mks(atleast in mine!!!).(i know b'cos i missed it by i mark last yr. :oops:  )
anyways, all da best.

siva

last years nsep paper was terribly difficult as compared to that 2yrs. ago. my friend got 34 correct in the objective paper 6 in the subjective section  yet he did not qualify(we had cross checked) It appears as if the writing skills are also awarded.


linear algebra
matrix
number theory
greatest common divisor
superior algebra
superior algebra solved

Let A be a euclidian ring. MM_n(A). I want to prove the existance of P and Q (matrices of det +1 or -1)  so that  PMQ=D : D is a diagonal matrix so that d_k divides d_(k+1) for all k. d_k are unique.
  Thanks

Sorry, det of P and Q is in the set of invertible elements of our ring.

I think that this is well known. It is used for the proof of the Diagonal form of a matrix, when we take the ring to be K[X], there K is the field from where A takes its entries.

I don't post the solution, but I'll give ideas. Uses elementary transformation. These have det +1 or -1. P and Q will be obtained from the composition of some of these, so ignore them.

We want to prove that using elementary transformations, we can transform M into D.
Apply euclid's alghorithm for lines and colums - This is the idea.
It will be proven that d_k is gcd of all minors of order k of M.


What is the sum of all the positive two-digit integers where one of the digits is four times the other?

Let's make a list of all such integers.
41
82
14
28
add them to get
$ \boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{165}}}}}}}}}}}}}}}$

Ahh! That's a lot of boxes...burned my eyes...not really necessary.

AHHHHHHHH
boxes
that hurt....
good job solving the question though


The horizontal line in my header does not go all the way across the text field.  And it's not centered?  Everything else is.
[code]
\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{fancyhdr}

\pagestyle{fancy}

\pdfpagewidth 8.5in
\pdfpageheight 11in
\setlength\topmargin{0in}
\setlength\headheight{0in}
\setlength\headsep{0in}
\setlength\textheight{9in}
\setlength\textwidth{6.5in}
\setlength\oddsidemargin{0in}
\setlength\evensidemargin{0in}
\setlength\headheight{81pt}
\setlength\headsep{0.25in}

\begin{document}

\chead{\parbox{1in}{}
\parbox{6.5in}{\begin{center}\large {\em Header}
\\  {\bf Line 2} \\Line 3
\\ {\normalsize Line 4}\end{center}}}

This is my practice document. This is my practice document.  This is my practice document.  This is my practice document.  This is my practice document.  This is my practice document.\end{document}[/code]

Hmm! The line [i]is[/i] centered because it goes the complete width of the text page, but you have made the text width shorter.
Unfortunately, the preamble is a mess as:
1. You are using \parbox twice which does nothing useful;
2. centering a header that is already centered by \chead;
3. using obsolete commands like \bf, \em;
4. using too many page sizing commands rather than using the simpler geometry package that does all the calculations.
So try this:
[code]\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{fancyhdr}
\usepackage[letterpaper,headheight=81pt,headsep=0.25in]{geometry}
\pagestyle{fancy}
\chead{%
{%
\large \emph{Header} \\
\textbf{Line 2} \\
Line 3
} \\
Line 4
}

\begin{document}
This is my practice document. This is my practice document.  This is my practice document.  This is my practice document.  This is my practice document.  This is my practice document.
\end{document}[/code]


[quote="Mathsniper"]just cauchy-schwarz's inequality can kill it \[\left( \sum a_{i}(2-a_{i}) \right)\left( \sum\frac{a_{i}}{2-a_{i}}\right) \geq \left( \sum a_{i}\right)^{2}\] simplify the inequality above. We only prove \[\sum a_{i}^{2}\geq \frac{1}{n}\] which is trivial to be proved by power mean.[/quote]
IT'S A BALKAN QESTİON


Let be a real positive x. Show that

$1+x^{2006}\geq\frac{(2x)^{2005}}{(1+{x)}^{2004}}$

[hide]
For all real $a$ we have

$1+a^{2}\geq2a$ (because $1+a^{2}-2a = (1-a)^{2}\geq0$)

Setting $a=\sqrt{x}$ we have,

$1+x\geq2\sqrt{x}\Leftrightarrow\frac{2\sqrt{x}}{x+1}\leq1\Rightarrow\frac{(2\sqrt{x})^{2004}}{(1+x)^{2004}}\leq1$ $(1)$

Setting $a=x^{1003}$ we have,

$1+x^{2006}\geq2x^{1003}$ $(2)$

Multiplying (1) and (2), we find the desired result
[/hide]

It can be seen also as a consequence of Holder's inequality :
\begin{align*}
(1+x^{2006})\underbrace{(1+x)(1+x)\cdots(1+x)}_{2004 times} &\geq (1+\sqrt[2005]{x^{2006+2004}})^{2005} \\
 &= (1+x^2)^{2005} \\
 &\geq (2x)^{2005}
\end{align*}
The inequality follows.


calculus integration algebra polynomial function domain geometry

Hi
Here is a problem i did not manage to solve

let P(z) be a Polynomial of degree >= 2.
Given that all its roots are inside the open unit circle prove that integral on |z|=1 of 1/P(z) is 0.

Thanks

Consider the integral of $ 1/P(z)$ over $ |z|\equal{}R$, where $ R\ge 1$. On one hand, it goes to zero as $ R\to \infty$ (consider the growth of $ P(z)$ at infinity). On the other hand, the integral does not depend on $ R$ by Cauchy's theorem.

Another way to put it: one can use the residue theorem on the [u]exterior[/u] of a closed curve. [url=http://eom.springer.de/r/r081560.htm]Source[/url]. The residue of $ 1/P$ at $ \infty$ is $ 0$ because its degree is more than $ 1$.

how u infer from Cauchy Theorem that the integral does not depend on R?
as far as i recall (this is what we have learned so far) is that Cauchy theorem says that if f is an analytic function on a "Star Like Domain" (i do not know the exact term in English so i used so i used word-word translation)  then integral over closed contour of f is 0.

Thanks

It's based on the Deformation of Path Theorem:

Let $ f$ be analytic on a region $ A$ and let $ \gamma$ be a simple closed curve in $ A$.  Suppose that $ \gamma$ can be continuously deformed to another simple closed curve $ \lambda$ without passing outside this analytic region.  Then:

$ \int_{\gamma}f\equal{}\int_{\lambda}f$

So, if we start with a closed contour $ \gamma$ surrounding all the zeros of the polynomial and $ \lambda$ is a larger one enclosing all the zeros and we know that between them $ 1/P(z)$ is still analytic since no more zeros, then:

$ \int_{\gamma}\frac{1}{P(z)}dz\equal{}\int_{\lambda}\frac{1}{P(z)}dz$

all the way up to $ R\to\infty$

since all we learned yet is Cauchy theorem for starry regions and Cauchy's integral formula on a circle when the function is analytic on the circle's inside.

We can generalize this problem to two polynomial $ p_1(z)$ and $ p_2(z)$ both of which are non-zero and $ \deg p_1(z) \leq \deg p_2(z) \plus{} 2$. That means the sum of residues of $ p_2(z)/p_1(z)$ will be zero.


Putnam MIT college geometry calculus integration inequalities

This year will be my first year in college, and I plan to take the Putnam for the first time in December.  Anyway, Matthew Crawford, mentioned here ([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=597751#597751[/url]), that if some more experienced problem solvers would be willing to host a math jam for the Putnam, he will set it up.  So, if anyone is interested let me know.

I don't think I'm good enough to host it, but I'd attend religiously! I think it's a good idea. :)

I'm totally in.  I think Prof. Merryfield should host it.  He's got t3h sk1llz.

I would certainly attend. I miss the days of having math jams with content (as opposed to just reviewing future class offerings) - of course, I understand this is probably due to the steady expansion of AoPS and the administrators' duties...

justdudxit: I've already been asked, and I've already said no. I appreciate you thinking of me, but I'd rather not.

im definitely in!

but, whats a math jam? ha  :maybe:
i've heard it around aops before but dunnoo what it is

[quote="Xevarion"]That's why we need to find somebody else to take the responsibility and teach us. Maybe I can coerce Melanie Wood to do it.... Someone at MIT should try some of the awesome people there, too.[/quote]

I don't think you need to go that far; there are several very capable regular posters here, some of which are in university and have been HM+ in the past.

Id like to list the names that came to mind and hopefully them or anyone else who's willing and capable would contribute. Maybe if one person cant do a bunch, then a few people could do one or two each.
JBL
ComplexZeta
Arne
Valentin Vornicu

You guys know you wanna :roll: :D  :lol:  :P
Anyway lets get this idea alive!

jmerry? blahblahblah?

mlok probably could (I know him, he's awesome) but I don't think he has time. Maybe he could give a "guest lecture" once in a while.

what does math jam mean? is it sort of an online discussion or is it an in-class stuff?

I guess it would be like an informal lecture or a directed problem-solving session. It's up to the instructor really. But to answer your question specifically, both! It would be online with AoPS' virtual classroom. :D

Jaimemorgan asked me if I would host, and I have (rudely) not responded yet (sorry!), but I think I'm just going to be too busy this semester (senior thesis, grad school aps, classes, etc.).  Sorry!  Best of luck finding someone, though.

[quote="Xevarion"]blahblahblah? [/quote]

I'd probably participate, but there are much more capable people than me in terms of running such a session (I was only top 200 last year).

courses i have take only include
calc 1 and 2. I am currently taking calc 3.
as you have probably guess i am a freshman.  :blush:
i have a general understanding of most topics on putnam that are not linear algebra. i just have no idea how to attack them.

I've taken Calc 1-3 and linear and am currently taking abstract algebra.  I would personally like to see lots of linear algebra, number theory, and calculus techs; the other stuff seems to be diluted USAMO stuff.

well number theory is also USAMO stuffs that you could find in probably many other resources; I agree with all the calculus stuffs (tricky integrals and whatnot?), though.

The reason I didn't suggest tricky integrals etc is that those happen on the forum a lot more often than lin alg type problems, and Olympiad combinatorics and NT are usually beyond me.... I think [b]FieryHydra[/b] and [b]benorin[/b] should teach a tricky integral/summation lesson. ;)

Ah, [b]Xevarion,[/b] you think too highly of me when it comes to integrals! :) Thanks!

However, I would strongly suggest that we get [b]benorin[/b], [b]Virgil Nicula [/b], [b]kunny[/b] and [b]prithwijit[/b] when it comes to tricky integrals/summations. I am sure there are other experts too, but these four seem to have a huge collection of such problems (and they are very good too.) As for me, I have none!

Is the class still going to be on the 19th at 2:00 i think i can make that. Anynews is mucho appreciated.

As far as I know everything is exactly as I said in my earlier post.

well i just realized that I will not be able to attend the math jam  :(  because i have class from 5:10-7:00pm. (in my time zone the math jam is at 6:00pm)
but i definitely plan to read the transcript and ask questions if i have any.

Im curious whether there will be a transcript posted with the math jam, i had trouble setting up my computer, and could only make the tail end.

i was wondering if there was going to be another math jam for putnam. if so could we please try and have it on a different day of the week (not tuesday)

Simon said he wants to wait until he knows how busy he'll be this semester before discussing scheduling another one.

If someone else wants to run one, of course the date and time is up to him or her.

On paper it appears that I have a busy schedule this coming quarter -- I'm taking 6 or 7 classes, two seminars, two research projects, and grad schools apps. But sometimes it turns out that there's more time available than I expect. So we'll see. If someone else wants to runs one, I'd be interested in that too.

[quote="jamiemorgan88"]Im curious whether there will be a transcript posted with the math jam[/quote]
The transcript has been posted.

I wish I had read this thread sooner.  Anyways, count me in from now on.

And I know a lot of the pros are busy, but couldn't they do one time specials?  And then we will have the regular hosts as well.

I'm in for any putnam problem solving groups.


function
number theory
least common multiple
algorithm
Euclidean algorithm

The function $f$, defined on the set of ordered pairs of positive integers, satisfies the following properties:
\begin{eqnarray*} f(x,x) &=& x, \\ f(x,y) &=& f(y,x), \quad \text{and} \\ (x + y) f(x,y) &=& yf(x,x + y). \end{eqnarray*}
Calculate $f(14,52)$.

[hide="Solution"]Analyzing the function's properties, we know that we must eventually use $f(x,x)=x$ in order to obtain an integer answer. However, we must first transform $f(14,52)$ to make it look like this. Since $f(14,52)=f(52,14)$ does us no good, we must find a way to exploit the third property.

Note that \[f(14,52)=\frac{38}{38}f(14,14+38)=\frac{52}{38}f(14,38).\] This is good progress. Repeating the process several times, \begin{eqnarray*}f(14,52) &=& \frac{38}{38}f(14,14+38)= \frac{52}{38}f(14,38) \\ &=& \frac{52}{38}\times \frac{24}{24}f(14,14+24)=\frac{52}{24}f(14,24) \\ &=& \frac{52}{10}f(10,14) \\ &=& \frac{52}{10}\times\frac{14}{4}f(10,4)=\frac{92}{5}f(4,10) \\ &=& \frac{91}{3}f(4,6)\\ &=& 91f(2,4)\\ &=& 91\times 2 f(2,2)=364. \end{eqnarray*} [/hide]

Hmm, I'm surprised no one has asked the obvious follow-up question: what is the function $f(x, y)$?

[hide="Hint"]I found it very helpful to compute the first few values and see what it looks like.
$\begin{array} {c|ccccc}
& 1 & 2 & 3 & 4 & 5\\
\hline
1 & 1 & 2 & 3 & 4 & 5\\
2 & 2 & 2 & 6 & 4 & 10\\
3 & 3 & 6 & 3 & 12 & 15\\
4 & 4 & 4 & 12 & 4 & 20
\end{array}$
[/hide]

Sorry for reviving but...
[quote="illazar"]4everwise i am quite confused with your solution.  How do you go from:

(38/38)f(14, 14+38)

to

(52/38)f(14, 38)

The property states that yf(x, x + y) = (x+y)f(x, y).  How do you get that 14+(38/38) = 52/38?....[/quote]
We see that $f(x, x+y)=\left(\frac{x+y}{y}\right)f(x,y)$ where $x=14$ and $y=38$.  We already have the value of $f(x,y)$ we just replace it with the RHS.

Also,
[quote="hrithikguy"][hide="hint"]$f(x,y) = lcm (x,y)$[/hide][/quote]
How would you go about proving this?

[hide="Proof"]
The first two rules are clearly true. Now we must prove
$(x+y)lcm (x,y)= ylcm (x,x+y)$
$(x+y)\cdot\frac{xy}{\gcd(x,y)}=y\cdot\frac{x(x+y)}{\gcd(x, x+y)}$
By the Euclidean Algorithm, $\gcd(x,y)=\gcd(x, x+y)$, so
$\frac{xy(x+y)}{\gcd(x,y)}=\frac{xy(x+y)}{\gcd(x,y)}$[/hide]

Thanks pi37.

[hide="One more question."]How would we prove that's the only function that satisfies those conditions?[/hide]

basically first note that f(2x,x)=2x then bash out from f(14,52) until you reach 91f(4,2) which is 91*4=364


$a_1=1$
$a_{n+1}=a_n\cdot\frac{n+2}{2n+2}+1$

Is there any way of solving this type of rucurrence, I mean : $a_{n+1}=a_ng(n)+1$


function
linear algebra
matrix
quadratics
complex analysis
number theory unsolved
number theory

Let p be a prime and let x_1, x_2 to be elements in (Z_p)*. Define M_n(x, p) to be the set of n x n matrices with coefficients in Z_p and determinant x. prove that |M_n(x_1, p)|=|M_n(x_2,p)|.

If I'm not making a stupid mistake, while this appears hard, it really is not.  If anyone sees a blunder, please tell me.  Since for any $x_{1}\in\mathbb{Z}_{p}^{*}$ the function $f: \mathbb{Z}_{p}^{*}\to\mathbb{Z}_{p}^{*}$ defined by $f(a)=ax_{1}$ is bijective, we have the result almost immediately.

First notice that because $\mathbb{Z}_{p}^{*}$ is a group, there exists a unique $X\in\mathbb{Z}_{p}^{*}$ such that $x_{1}X=x_{2}$.

Now, we have for our $X$ and $n\times n$ matrix $A$, let $B$ be the matrix defined having the same entries as $A$, except that in the first column, all the entries are mutliplied by $X$.  Then we have the $det(B)=X\cdot det(A)$.  Therefore, for every element of $M_{n}(x_{1},p)$, the function $F: M_{n}(x_{1},p)\to M_{n}(x_{2},p)$ defined above is bijective (each entry is mapped by a bijective function, so the entire function on the matrix is bijective), and hence the size of $M_{n}(x_{1},p)$ and $M_{n}(x_{2},p)$ are the same.

EDIT: THANKS FOR SEEING THE MISTAKE GUYS  :)

[quote="zgorkster"]Now, we have for any $r\in\mathbb{R}$ and $n\times n$ matrix $A$, the $det(rA)=r\cdot det(A)$.[/quote]
I think it rather should be $r^{n}\cdot \det (A)$. Correct me if I'm wrong, but then your reasoning doesn't work, since there doesn't necessairly exist such $X$ that $X^{n}=\frac{x_{2}}{x_{1}}$ (you can for example take $n=2$ and $X$ is not a quadratic residue modulo $p$).

[quote="Peter Scholze"]Well, the argument works if instead of multiplying the whole matrix by $X$, you only multiply e.g. the first column by $X$.[/quote]
Yes, you're right.  :blush:


abstract algebra
invariant
superior algebra
superior algebra unsolved

Show that $ \mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ and $ \mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ are both left $ \mathbb{R}$-modules but are not isomorphic as $ \mathbb{R}$-modules.

The tensor product $ \mathbb{C} \otimes_\mathbb{R} \mathbb{C}$ is a free module over $ \mathbb{R}$ of rank $ 4$. What can you say about $ \mathbb{C} \otimes_\mathbb{C} \mathbb{C}$? Is the rank invariant under an isomorphism?


[b]#7266 A square with side length 2 and a circle share the same cent [...][/b]
[416] USA - 2005 - AMC8

A square with side length 2 and a circle share the same center.
The total area of the regions that are inside the circle and
outside the square is equal to the total area of the regions that
are outside the circle and inside the square. What is the radius
of the circle?

so....
we have a circle and a square

area of circle-area of square = 0 because the total area of the regions that are inside the circle and
                                             outside the square is equal to the total area of the regions that
                                             are outside the circle and inside the square.

Therefore, area of circle=area of square

The area of the square is 2*2 or 4.

so...    pi*r^2=4
          r^2=4/pi       (divided both sides by pi)
          r=sqrt(4/pi)   (square root of both sides)
          r=2/sqrtpi

2/sqrtpi is the answer... i think ... to the non-answered problem

plzzzzz rate this post excellent with 6 cubes

You made this post twice...

Be careful when doing that.


[quote="Interval"]Chad had a garden that was in the shape of a rectangle. Its length was twice its width. He decided to make a new garden that was 2 feet longer and 2 feet wider than his first garden. If x represents the original width of the garden, which expression represents the difference between the area of his new garden and the  area of the original garden?

(1) 6x + 4         (3) x^2 + 3x + 2
(2) 2x^2              (4) 8[/quote]

$(x+2)(2x+2)-(x)(2x) = 2x^{2}+6x+4-2x^{2}= 6x+4 = \boxed{\text{(1)}}$

In place of simply giving me the answer, can you show how to get the answer, too?

What exactly did you do to find the answer?


I'm not sure if anyone has tried this before, but I want to try to start a game of Rumble.

Read the Rules: [url]http://kevan.org/rumble.cgi?genre=hero&mode=rules[/url]

Just post if you want to join, I will start a once there are 5 people.
For this game, I will make the power list, and as soon as 5 people have joined, I will start the auction for powers. Once the auction has started, PM me your bids. Also, every round, you need to PM your moves.

Please do not communicate outside this thread with other players until the auction phase is over.
You will need to be fairly active. The auction phase/every round will last two days.
I hope this works out. If you have questions, ask me.

I don't think this is working too well, but I'm going to try it... I played it on another forum and it's fun.

Current Players:
Klebian
Ignite168
AnonAmoz

Almost started a game of this earlier.

Why no outside communication?

/in anyway

So people know what you're up to. :lol:

Hmm... maybe I'll rethink that rule.

EDIT: Ok, you are only allowed to communicate AFTER the auction phase.

OK SINCE KLEBIAN HAS BEEN BEGGING ME ALL NIGHT

/IN

I'll play. What lvl is ur wiz on ur avatar and hwat world r u in? im in windia. 35 hunter is my main right now. hes a noob

Ok, I can't edit the first post anymore, so here's the list of players:

Klebian
Ignite168
AnonAmoz
azhang
Mustafa

We can start!

And @azhang, I'm level 64 in Scania.

If you'd like, I can edit the first post periodically, if you just PM me what you want the post to look like (so that I could just copy and paste what you wanted into the topic)

/in

!

Could we nominate powers? It makes strategizing better

Each person needs to PM me two powers within 24 hours (extensions possible). Remember, if you make a power that gives a huge advantage, it could easily be won by someone else. If you miss the deadline, I will make the two powers myself.

I'm going to make a new thread for the auction, it will be easier. My list of powers will be posted, but do not bid yet (more powers may be sent in). I will tell you when bidding can start.

we only choose 1 power?
isn't it 2 as per the rules?

EDIT: Everyone send in two powers. If you miss the deadline, I will make the two powers myself. (extensions possible on the deadline)


Let be $ a,b,c$ the sidelenghts of a triangle . Show that :

$ \frac{a}{b\plus{}c\minus{}a}\plus{}\frac{b}{c\plus{}a\minus{}b}\plus{}\frac{c}{a\plus{}b\minus{}c}\ge \frac{b\plus{}c}{2a}\plus{}\frac{c\plus{}a}{2b}\plus{}\frac{a\plus{}b}{2c}$

[quote="alex2008"]Let be $ a,b,c$ the sidelenghts of a triangle . Show that :

$ \frac {a}{b + c - a} + \frac {b}{c + a - b} + \frac {c}{a + b - c}\ge \frac {b + c}{2a} + \frac {c + a}{2b} + \frac {a + b}{2c}$[/quote]
let x=b+c-a,....
so we have to prove that:
$ \sum \frac{y+z}{2x}\geq \sum \frac{2x+y+z}{2(y+z)}$
$ \sum \frac{y+z}{x} \geq 2\sum\frac{x}{y+z}+3$
but
$ \sum \frac{y+z}{x}\geq 4\sum\frac{x}{y+z}\geq \2sum\frac{x}{y+z}+3$
q.e.d

[quote="Kenshin24"]Alex2008..is it necessary that a,b,c are sides of a triangle???[/quote]

Obviously , yes :huh: , try $ a\equal{}b\equal{}2$ and $ c\equal{}5$


geometry
3D geometry
geometric transformation
reflection
analytic geometry
calculus
integration

Consider the first quadrant cube defined by endpoints $( \{ 0,42 \}, \; \{ 0,42 \}, \; \{ 0,42 \})$.

A high-speed particle begins at point $(21,21,0)$ and begins with an x-direction speed of $6$, an y-direction speed

of $2$, and a z-direction speed of $3$. When a particle reaches either $0$ or $42$ in a given dimension, it changes

direction in that dimension.

What is the shortest possible distance between the particle and one of the corners, and what is the first point in time when the particle reaches this position?

[quote="The Zuton Force"]Consider the first quadrant cube defined by endpoints ${{{( \{ 0,42}, \; \{ 0,42}, \; \{ 0,42})$.

A high-speed particle begins at point $(21,21,0)$ and begins with an x-direction speed of $6$, an y-direction speed

of $2$, and a z-direction speed of $3$. When a particle reaches either $0$ or $42$ in a given dimension, it changes

direction in that dimension.

What is the shortest possible distance between the particle and one of the corners, and what is the first point in time when the particle reaches this position?[/quote]

What are the endpoints? $(0,0,0)$ and $(42,42,42)$?

[hide="Anyways, the trick to this problem, I think, is"]
that you can reflect all of the vertices over all of the edges/faces to make a lattice-like group of all edge-coordinates $(42x,42y,42z)$ for integral $x,y,z$. Then we just want the smallest positive $t$ such that \[42|21+6t , \ 42| 21+2t , \ 42|3t \]  From that we can also get the distance.

But there is no such $t$...Hmm. Can anyone help? [/hide]


I have just posted Tokyo University Entrance Exam/Science 2006.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=77163]www.mathlinks.ro/Forum/viewtopic.php?t=77163[/url]

Have fun! :)

kunny

[quote="spoudyal"]I got an error that says the topic does not exist.[/quote]

Sorry for that, Please take a look this.

[url=http://www.mathlinks.ro/Forum/resources.php?c=87]www.mathlinks.ro/Forum/resources.php?c=87[/url]

kunny

Can i know is this the entrance exam for Tokyo university? And this is for Maths courses? If so, undergraduate or postgraduate, thanks

[url=http://www.mathlinks.ro/Forum/resources.php?c=87&cid=74&year=2006]This is Tokyo university entrance exam/Science 2006 for undergraduate course.[/url]


It takes Mary 30 minutes to walk uphill 1 km from her home to school, but it takes her only 10 minutes to walk from school to home on the same route. What is her average speed, in km/hr, for the round trip?

$\left(A\right) 3$
$\left(B\right) 3.125$
$\left(C\right) 3.5$
$\left(D\right) 4$
$\left(E\right) 4.5$

[hide]
$\text{average rate}= \frac{\text{total distance}}{\text{total time}}= \frac{2\ \text{km}}{\tfrac{1}{2}\ \text{hr}+\tfrac{1}{6}\ \text{hr}}= 3\ \text{km/hr}\implies \boxed{\text{A}}$
[/hide]


I couldn't get this problem either... Can someone help me?  Thanks!

[hide="Solution"]
$ \overline{t324t}_7\equal{}t\cdot7^4\plus{}3\cdot7^3\plus{}2\cdot7^2\plus{}4\cdot7\plus{}t\equiv t(7^4\plus{}1)\plus{}1\cdot1^3\plus{}0\plus{}0\equiv(1^4\plus{}1)t\plus{}1$
$ \equiv 2t\plus{}1\equiv1\pmod2$, so this is always odd. The values $ \boxed{1,2,3,4,5,6}$ all work.
[/hide]

Wait, what does the line on the top mean and what does three lines mean? I get how to get to the solution but what do those symbols mean? Sorry, I'm not very smart... :|

The line on the top there is just to denote that $ \overline{t324t}_7$ is equal to $ t\cdot7^4\plus{}3\cdot7^3\plus{}2\cdot7^2\plus{}4\cdot7\plus{}t$ and not equivalent to $ 324t^2$. The three lines represent "equivalence" (in this case, $ \pmod2$), if you are familiar with modular arithmetic.

By the looks of it, you can't use calcs, so:

Alternate (sorry if it's confusing):

t324t (base7)=7^4*t(odd or even#)+7^3*3(odd)+7^2*2(even)+7^1*4(even)+7^0*t(odd or even)

If 7^4*t is odd, then 7^0*t is odd too, so the sum of these will be even.
If 7^4*t is even, then 7^0*t is also even, so the sum of these terms will too be even.
Therefore, the sum of these numbers will be odd, so all numbers that are 0<n<7 work, or [color=darkred]1,2,3,4,5,6[/color]

What did I even do that involved calculators? :|


vector
linear algebra
linear algebra unsolved

Can $ \mathbb{C}$ be visualised as a vector space over $ \mathbb{R}$?

That was a stupid one. I am very sorry.

I meant to ask,
Can $ \mathbb{R}$ be visualised as vector space over $ \mathbb{C}$?

[quote="Fermat's Little Turtle"]No since a scalar such as $ i$ will take any vector out of the reals.[/quote]

I think you are incorrect.
Scalar multiplication not necessarily means multiplying with the product rule of the field.
(Actually here you are able to do as your vector space is itself a field and the field over which you are considering is its extension)

"Scalar multiplication" is basically a "homomorphism from the vector space to itself" associated with the scalar satisfying some properties.

So, here's the question you're trying to ask:

Is there a scalar multiplication rule that makes $ \mathbb{R}$ into a vector space over $ \mathbb{C}$?

There is. It can't be continuous, it can't include the standard real scalar multiplication, and it can't be explicitly written down; it relies on the fact that both sets are vector spaces over $ \mathbb{Q}$ with the same uncountable dimension. It is so completely useless that "no" is a better answer.


Stanford
college
mathleague.org
ARML
AMC
AMC 12
AMC 8

I don't know if I posted this in a right place or not ... :P
I want to study really hard in this summer to get perfect scores for the California Math League (CAML) next year.
Can anyone recommend me any good books to study for this competition, and where to find the tests in previous years ? Thanks in advance.
PS: It doesn't have to be only CAML. Any resources with similar kind to this competition would be helpful as well. Is the GPML like CAML but in different places ? :?:

Tim Sanders started GPML while studying and later teaching in Missouri.  Missouri, Kansas, Iowa, and formerly Nebraska particpate.  Since going to Stanford, Tim is expanding it into California, so I assume that that is CAML.

On the mathleague.org http://mathleague.org/   website, you can order sets of past years' tests.  I have heard that they will be available for sale at 2006 ARML sites.  The level of the questions are about the same as AMC12, so studying those would be valuable as well.

I would actually like to say that the MathLeague problems are much easier than AMC12, more like intermediate between AMC8 and AMC10. For good all-around practice, try the AoPS Volumes 1+2

California Math League at present refers to something completely separate from GPML.  They're part of a national organization whose website is here: [url]http://www.mathleague.com[/url].  (The California results can be found here: [url]http://www.mathleague.com/reports/2007_08/CA6.HTM[/url])  I'm not sure how the level of difficulty compares.  Mathleague might have problems on their website.

The situation is actually pretty confusing: the two websites are mathleague.com (nationwide organization) and mathleague.org (Tim's leagues, including GPML).  We're trying to get a mathleague.org league (thus an analog of GPML) running in California, which could theoretically also be called California Math League if it takes on a full season structure like GPML has.

the website is now themathleague.com.
And that's...quite confusing  :help:


geometry
perpendicular bisector
geometry unsolved

Let $ ABC$ be a triangle with $ AC > AB$. The point $ X$ lies on the side $ BA$ extended through $ A$, and the point $ Y$ lies on the side $ CA$ in such a way that $ BX\equal{}CA$ and $ CY\equal{}BA$. The line $ XY$ meets the perpendicular bisector of side $ BC$ at $ P$. Show that

$ \angle BPC \plus{} \angle BAC \equal{} 180^{\circ}$.

[quote="livetolovemath030894"]Let $ H$ be midpoint arc $ BAC$.It's easy to see that $ HP$ is the perpendicular bisector of $ BC$ and $ XBPH,CYPH$ are cyclic.We have $ \triangle XBH \equal{} \triangle AHC$ => $ \angle BXH \equal{} \angle CHA \equal{} \dfrac {\angle B \plus{} \angle C}{2}$.Similarly, $ \angle HYC \equal{} \angle HAB$. Since we have the result  :)  :)[/quote]

Can you explain why $ XBPH$ is cyclic please !  and maybe you mean $ \angle BHX\equal{}\angle CHA$  :oops:

Sorry for my fault.$ \angle BXH \equal{} \angle CAH$
It's easy to see that $ XAY$ is isoceles triangle => $ \angle AXY \equal{} \dfrac {\angle A}{2}$ so $ XBPH$ is cyclic  :)


quadratics
calculus
integration
number theory
prime numbers
algebra
number theory proposed

If it had integer solution $x_{0}$, then $f(x)=(x-x_{0})(ax-y)$. Therefore $p=f(10)=(10-x_{0})(10a-y)\to x_{0}=10\pm 1$.
Because $c=x_{0}y\le 9$ may be only $x_{0}=9,y=1$. Therefore $p=10a-1$ had only 2 digits - contradition.

[quote="kunny"]Let $p$ be prime numbers with three digit such that $p=100a+10b+c$ for integers $a,\ b,\ c$.
Can the quadratic equation $ax^{2}+bx+c=0$ has integral solution?[/quote]
If $p=(\overline{a_{n}a_{n-1}...a_{0}})_{10}$ then $a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0}$ is irreducible in $\mathbb{Q}[x]$.


If I took the AMC 12B at a university, will I be able to take the AIME at my school or do I have to take it at the same place I took the AMC.

We will send the B-date contest results directly to the administering location and contest manager.  That includes your AIME contest as well as scores and awards.  So in your case, we will send your 12 B results to the university.

Then you can arrange (in a way mutually agreeable between you, the B-date administering university and your local high school teacher) to take the AIME at your high school.  Make arrangements for the university contest manager to get the AIME contest to the high school contest manager.  Have the high school contest manager send in your contest results with all the rest of the high school's AIMEs.

Steve Dunbar
Director, American Mathematics Competitions

If several students from the same high school take the AMC B at a university, will their scores be eligible for school team awards?

No.

In order to qualify for a school team award, the school has to register and administer the contest as a school, at the school.

Steve Dunbar
Director, American Mathematics Competitions


Ioana Dumitriu (1996)
Melanie Wood (2002)
Ana Caraiani (2003)

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-html/putnam2003results.html

wow great news. it seems that there are quite a few romanians on the putnam winners list :D

princeton as again all the team Romanian members, and with the leaving of Hornet (1st in the world in 1999, gold in 1997 and 1998) next year, Ana will most probably take his place, and the Priceton team will remain a Romanian bastion in the Ivy League :D:D:D;)

Gabriel Carroll is 4 times Putnam fellow. (2000,2001,2002,2003) In 2000, he was still in high school. Very amazing!

Wow Ana is amazing, the fact that there are only a few female mathematicians around the world that can achieve such high honors.
Someone got 110. I got 49/120 and got rank 76.5 hehe.

well I predict now that she is going to be the first girl to make it fellow (this her being a girl and coming out a fellow stuff is like really wierd :D) putnam for 4 consecutive times. :)

[quote="3X.lich"]I bet she'll set the first World Putnam record  :D[/quote]
I'M SURE SHE WILL!!!!!!!!!!!!!!

[quote="Valentin Vornicu"]well he (Gabriel Carroll) still has one year to go :D[/quote]

You can only take the Putnam 4 times officially. So his eligibility is over now.

oh and well another thing ... two of the girls that were putnam fellows are Romanian girls, isn't that a nice coincidence?

btw we do have not-so-smart-but-very-beautifull-girls here in romania too :P

[quote="Valentin Vornicu"]btw we do have not-so-smart-but-very-beautifull-girls here in romania too :P[/quote]
Can I visit you Valentin? :D  :blush:  :D

I am not stopping you :D but I suppose that you mean by "visit you" like can I "visit Romania?" :D

Please have a look here: http://www.unl.edu/amc/a-activities/a7-problems/putnam/-html/putnam2003stats.html . Does anyone want to guess who got what score among the Putnam Fellows ?

well Ana has 110p and it's the highest score. I know that because right after the contenst she told me that she had solved 11 problems. :)

Congratulations to Ana. Concerning the superb performances of Romanian Putnam competitors I have to say that they have quite an advantage. This is not meant to decrease the reputation of Ana who managed to become a Putnam fellow right at her first participation. But if you have a look at the Romanian olympiad curriculum for the upper grades you will notice that they handle a lot of stuff which is mostly handled in the undergraduate curriculum at the university. So I feel it is kind of advantage that Romanian Putnam participants are often familiar with some of the topics for a long time while others still have to develop a feeling for it. What do the other Romanian members think about it ?

Not necessarily, because the judges may deduct points for incomplete explainations. Each question is scored from 0 - 10. But it's certainly possible that she scored the 110. That would be very impressive. I don't think even Ciprian ever scored that high on the Putnam exam.

Well, I know Ana very well (we went to 3 imo's together) and each time she said she did x problems, she DID x problems (at the tst's also).

She said she solved 11, I am convinced that she would have figured out if she hadn't, and also there was a "leak" from one of the putnam official board members, which confirms that assumption, so I am 99,5% sure that she's the one with 110p :)

Long live the Romanian team of Princeton university ;) :D

Yay Princeton!!

And that is indeed impressive! No matter that she is a female, just the fact that she is a first year University student to score that high is great! Not to mention that it also means she scored higher than both Reid Barton and Gabriel Carroll.

yeah, but well it's understandable, since this year's putnam is really a romanian math olympiad job, many of the problems being considered too easy for our national olympiad for example.

and Gabriel and Reid are probably both involved with other projects, rather than problem solving.

I am curious if she will be the first female to get fellow 4 times ... but I guess only time will show us :D

I have compiled the TRUE list of Elizabeth Lowell Putnam Prize but I'll have to disappoint some people that Ana is [b]not[/b] the third female Putnam fellow :( , although I'm impressed by her performance :D .

1992--Dana Pascovici, Dartmouth College
1993--No award given
1994--Ruth A. Britto-Pacumio, Massachusetts Institute of Technology
1995--Ioana Dumitriu, New York University
1996--Ioana Dumitriu, New York University
1997--Ioana Dumitriu, New York University
1998--No award given
1999--Wai Ling Yee Univesity of Waterloo
2000--No award given
2001--Melanie E. Wood, Duke University
2002--Melanie E. Wood, Duke University
2003--Ana Caraiani, Princeton University

Sorry for breaking the calamities but a fact is a fact.

Arthur Rubin, Bjorn Poonen, Ravi Vakil and Gabriel Carroll are the only ones with 4 Putnam fellow titles.

Ana is the third female fellow, the Elizabeth Lowell Putnam Prize is just given out to the top scoring female, not necessarily to a female fellow.

Hmm, yeah you are right, I didn't disappoint anyone after all!!!
So...will she be the fifth person to get 4 Putnam fellow titles???? O_o
Looking at the names again, I see many of the Putnam fellows are actually famous people. :D


probability
inequalities
triangle inequality

Amanda has a 2-cm rod, a 5-cm rod, and a 9-cm rod. She will randomly choose a fourth rod from a box containing rods of lengths 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20 cm. What is the probability that Amanda's four rods will be able to form a convex quatrilateral, if the rods are to be connected endpoint to endpoint? Express your answer as a common fraction.

i think we can use the triangle inequality theorem and relate it to this. a+b+c<d b+c+d<a a+c+d<b a+b+d<c
the shortest possible side is as we find greater than 2 and the longest possible side is less than 16. so we have it is between 14 and 4. we get 6 possible sides. there are 10 total. $ \frac{6} {10}\equal{}\boxed{\frac{3} {5}}$

[hide]

Use the quadrilateral inequality theorem, same as triangle one except 3 sides have to be greater than the 4th side.

2, 16, 18, and 20 don't work so 4/10 or 2/5 don't work

so 1-2/5 makes it a 3/5 probability.[/hide]


Find all  n positive integers  that satisfy  sum of digits of  2^n is equal to 5   in base 10 .

Some reasoning:

[hide]The number $2^n$ must contain the possiblities.

(1) Just 5 ( and 0's).
(2) Just 4 and 1 ( and 0's).
(3) Just 3 and 2 ( and 0's).

For (1), assume there exists a $2^n$ in that form; we know a number satisfying (1) must be in the form $5 \cdot 10^x$, which is divisible by $5$, a contradiction.

For (2), assume the same. The number doesn't end in 1. It must end in 4 or 0. If it ends in 0, it's in the form $41 \cdot 10^x$, divisible by $41$, contradiction - or $14 \cdot 10^x$, divisible by $7$, contradiction. If it ends in 4, there can be any number of 0's between the 1 and 4. Factoring out such a number reveals $10^x + 4 = 2^n$, which is impossible ( if you factor out the LHS ).

For (3), $2^n = 32$. It can't end in 3. So either 2 or 0. If 0 and we have the power of 10 greater than 0, it is divisible by 10 - not possible. If 2, we can apply the same factoring in case (2). So, $3 \cdot 10^x + 2 = 2^n$, and factoring out the two reveals another divisibility contradiction.

P.S. I'm not sure this is pre-olympiad - maybe intermediate...[/hide]

 :blush:  Forgot other cases...

[quote="white_horse_king88"]For (3), $n = 32$[/hide][/quote]

You mean $2^n = 32$, be careful with what you are writing :)

You forgot a few cases - like, maybe there exists some power of 2, which consists of 2, 2, 1 and some zeros?

Hm, I'll just keep it in Pre-Olympiad for now.

You can just look at two-digit endings of $2^i$ there are 20 difrent endings so with a little discusion you obtain solution...
[/quote]

[quote="white_horse_king88"]Some reasoning:

The number $2^n$ must contain the possiblities.

(1) Just 5 ( and 0's).
(2) Just 4 and 1 ( and 0's).
(3) Just 3 and 2 ( and 0's).

ctoring out the two reveals another divisibility contradiction.

P[/quote]

How do you get possibilities why it can't have one 2, and three 1 that is also posibility

I found this problem very very difficult, and infact still havent solved.

Problem: Find all $n$ s.t. sum of digits of number $2^n$ is $5$.

Answer: $n=5$

First we will prove one well known theorem:

Lemma: Let $x=\overline{a_ka_{k-1}...a_1}$ then $2^m|x$ iff
$2^m|\overline{a_ma_{m-1}...a_1}$.

Proof: $2^m|x \Leftrightarrow 2^m|\overline{a_ka_{k-1}...a_{m+1}}\cdot 10^m
+\overline{a_ma_{m-1}...a_1}$. We se now that $2^m|10^m$ so
$2^m|\overline{a_ka_{k-1}...a_{m+1}}\cdot 10^m$ from that we obtain $2^m|x
\Leftrightarrow 2^m|\overline{a_ma_{m-1}...a_1}$

Now let's go back to problem.

See what residues can be obtain by dividing power of two with 100. $2^i\equiv_{100}
2,[4,8,16, 32, 64, 28, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52], 4, 8,
16...$ so as we can se every power of two ends with two digits from a period except
$2^1$ (where is one digit in period it ends with $0a$ where $a$ is that digit of
course) and also for $i\in \{2,3,...,20\}$ it stands $2^i\equiv_{100} 2^{20k+i+1}$
for every $k\in N$.

Now only two-digits endings from period that is less or equal with $5$ is $4, 32, 12$.

$1^o$ $2^1=4$ so that is not possible. Every other power of two witch ends in $04$
has form $20k+3$, in another words $2^{20k+4}\equiv_{100} 4$, but then it has more
then $3$ digits so we look in last three digits of number $2^{20k+4}$, last to of
them are $04$ and let's say that the third is $a$. By $lemma$ number
$\overline{a04}$ must be divisible with $8$, so from that $a>1$ (because $8$ doesn't
divide $4$  and $8$ doesn't divide $104$) from that sum of digit of $2^{20k+4}$ is
bigger then $5$.

$2^o$ $2^5=32$ and we have one solution $n=5$ every other power of two witch ends
with $32$ has more digits witch of course are not all zeros so sum of digits of
every other power of two ending like this one is bigger then $5.$

$3^o$ $2^9=512$ so this is not a solution. Others powers of two ending with $12$ are
all in form $2^{20k+10}$ so they are divisible with $8$ and $16$ and $32$ and....
From divisibility with $8$ and $lemma$ we see that the third digit from right can be
$1$ or bigger then $2$ (because $8$ divides $112$ and not divide $212$). If it is
bigger then $2$ then sum of digits is bigger then $5$ so let third digit from right
to be $1$. Then divisibility with $16$ come on scene. Let fourth digit from right to
be $b$. Then $16|\overline{b112}$, how $16$ doesn't divide $1112$ so $b$ can be $0$ or
digit bigger then $1$ so $b$ is $o$. Now divisibility with $32$. By $Lemma$ number
$\overline{c0112}$ must be divisible with $32$ where $c$ is fifth digit from right.
We see that it can't be zero (because of divisibility) and it can't be bigger then
$1$ (because of sum of digits) so it is $1$ but number $2^{20k+10}$ certainly has
more then $5$ digits so sum of digits is bigger then $5$ (because only the sum of
last five digits is $5$).

Proof complete!


[hide="hint"]try tu put the expression $ 4.(1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4)$ between two squares.[/hide]


[color=blue][b]Intr-un campionat de fotbal se inscriu 18 echipe. Fiecare echipa poate sa joace cel mult un meci cu o alta echipa. Spunem ca campionatul este bun daca la sfarsitul sau oricum am alege 12 echipe din cele 18, exista cel putin 6 meciuri disputate intre ele.
Determinati numarul minim posibil de meciuri care se desfasoara intr-un campionat bun.[/b][/color]

[color=green]Cred ca numarul este 18. Trebuie insa si o demonstratie. Nu stie nimeni cum se face sau nu vreti sa trimiteti solutia? [/color]

[color=red][size=150]Chiar nu stie nimeni cum se face?[/size][/color]

nu prea ma prind eu de ceva aici...
am abordat`o clasic, insa se pare ca inegalitatea care iese este prea slaba...
[hide="nu prea pot sa ii zic solutie"]
pentru inceput, voi reformula problema cu notatiile "clasice".
fie $G=(V,E)$ un graf cu $n$ varfuri, cu proprietatea ca orice subgraf al sau cu $k$ varfuri are cel putin $m$ muchii. se cere $\min |E|$.

voi nota cu $\deg x$ gradul varfului $x$ din $G$, si cu $\deg_{H}x$ gradul din $H$ al varfului $x$. voi conveni ca $\deg_{H}x=0$ daca $x\notin H$. fie $F$ familia de subgrafuri ale lui $G$ cu $k$ varfuri. mai notez si cu $E(H)$ multimea muchiilor din subgraful $H$.

ideea pe care am avut`o a fost sa exprim suma urmatoare in doua moduri :
$S=\sum_{H\in F}\sum_{x\in H}\deg_{H}x=\sum_{H\in F}2|E(H)|\geq \binom{n}{k}\cdot 2m$.
pe de alta parte (deoarece daca $x\notin H, \deg_{H}x=0$),
$S=\sum_{H\in F}\sum_{x\in G}\deg_{H}x=\sum_{x\in G}\sum_{H\in F}\deg_{H}x$.
acum voi exprima ultima parte a egalitatii de mai sus :
fie $x\in G$ fixat. voi calcula valoarea
$T(x)=\sum_{H\in F}\deg_{H}x$.
voi presupune daca $\deg x=p$ si fie $y_{1},y_{2},\ldots,y_{p}$ vecinii lui $x$. din nou voi aplica acelasi mecanism de a exprima o suma in doua moduri. pentru un subgraf $H$, voi nota cu $h(x)$ functia sa caracteristica ($h(x)=1$ daca si numai daca $x\in H$ si $h(x)=0$ in rest).
deci
$T(x)=\sum_{H\in F}\deg_{H}x=\sum_{H\in F}\sum_{j=1}^{p}h(v_{j})h(x)=\sum_{j=1}^{p}\sum_{H\in F}h(v_{j})h(x)$.
insa pentru $v_{j}$ si $x$ fixate, numarul de subgrafuri din $F$ ale lui $G$ ce le contin este chiar $\binom{n-2}{k-2}$. deci ultima suma este
$T(x)=\sum_{j=1}^{p}\binom{n-2}{k-2}=p\binom{n-2}{k-2}=\binom{n-2}{k-2}\deg x$.
deci $S=\sum_{x\in G}T(x)=\binom{n-2}{k-2}\sum_{x\in G}\deg x=2\binom{n-2}{k-2}|E|$.
acum mai ramane doar sa trag concluzia, si anume
$2\binom{n-2}{k-2}|E|\geq 2\binom{n}{k}m$, deci
$|E|\geq\frac{n(n-1)}{k(k-1)}\cdot m$.

pt problema de mai sus, se obtine $|E|\geq 13,9...$ , deci $|E|\geq 14$.
si nu ma prind de ce iese doar asa...
oricum cred ca asta este ideea, sper sa fi gresit eu la calcul si sa ma corecteze cineva

[/hide]

La o aproximare destul de directa nr e mai mare sau egal cu
6*C(18,12)/C(16,10) (cel putin 6 muchii pt oricare 12 echipe si fiecare muchie e numarata de cel mult C(16,10) ori). Totusi calculul da 14, deci acelasi raspuns cu maky. Inclin sa cred ca nu e corect raspunsul asta, mai ales pt ca exista cu siguranta 12 echipe intre care se joaca mai mult de 6 meciuri :(  Probabil ca 18 e raspunsul corect :)  Solutii?

Sper sa nu fie greseli prea mari in cele ce urmeaza.....
18 este evident bun (de ex se ia un ciclu). Presupunem ca putem avea mai putin de 18 meciuri si configuratia ceruta.
1) In orice subgraf cu 11 varfuri exista cel putin 4 muchii. Daca nu, aleg unul in care sunt mai putin de 4. Atunci din fiecare din cele 7 puncte ramase pornesc cel putin cate 3 muchii catre cele 11 puncte. Deci avem mai mult de 33 de muchii, contradictie.
2) In orice subgrafd cu 7 varfuri exista cel putin o muchie. Daca nu, are unu fara nici o muchie. Iau alte 5 puncte (din afara lui) si din ele trebuie sa am cel putin 6 muchii catre subgraful meu. Deci din unul din cele 5 pornesc cel putin 2 muchii. Il pun separat si reiterez procedeul. In final o sa obtin 7 punte (din afara subgrafului) cu gradul mai mare sau egal cu 2 si inca 4 puncte (tot din afara) ramase. Aceste 4 puncte impreuna cu subgraful formeaza un subgraf nou cu 11 varfuri, care conform cu 1) are cel putin 4 muchii. Deci in total avem cel putin 7*2+4=18 muchii, contradictie (am presupus ca sunt mai putine de 18)
3) Aleg complementarul grafului meu (adik graful obtinut prin stergerea muchiilor existente si adaugarea de muchii acolo unde nu au existat). Cum in graful initial aveam o muchie in fiecare subgraf de 7, in complementar nu vor exista K7 (complete cu 7 varfuri). Din Turan avem cel mult 135 de muchii. Deci in graful initial sunt cel putin 17*18/2-135=18 muchii. Din nou contradictie (sau cel putin asa pare :rotfl: )
Prin urmare presupunerea facuta este falsa si rezulta ca avem cel putin 18 muchii in graful initial...
Acum se prea poate sa existe multe greseli mai sus :( , mai ales in rationamentul logic (trebuia sa ma opresc mai devreme???)

io cred ca e bine....
nu prea mai stiu io cum vine turan ala  :blush: dar cred ca trebuie totusi sa dea, iar rationamentul imi pare corect

$\frac{k-2}{k-1}\frac{n^{2}-r^{2}}{2}+\binom{r}{2}$, unde r e restul impartirii lui n la k-1 e nr max de muchii (nici eu nu il tin minte niciodata :) )


Duke college topology calculus derivative probability algorithm

What are some good math/econ/cs courses to take if I am considering a career in hedge fund/trading/security analysis/etc etc. I am entering my third year of undergraduate at Duke, and I am wondering what courses I could potentially take to be more well-rounded for jobs in those fields. I am not planning on going to grad school so I would like to take a few grad-level courses before working. The reason I am asking is that I am not really interested in pure mathematics, but there are too many applied mathematics to choose from (i.e. is topology really useful? How about PDE?). Help me out. :)

Thanks!

It depends very much on what specifically you're looking to be doing in that field.  On one end of the spectrum there are the quantitative analysts that work on things like portfolio optimization, risk analysis, derivatives pricing modeling, etc.  For any of these you absolutely [i] must [/i] have a solid background in probability/statistics and linear algebra.  For the more theoretical stuff, especially derivatives modeling and hedging strategies and such, you will also need some background in PDE and in continuous-state stochastic processes (typically a graduate level course).  Although if you don't plan on getting a graduate degree, your chances of getting a derivatives position are pretty slim anyway -- I don't know of anyone working in those research groups without a Master's; most have PhD's.  This is not to discourage you, though.  Many of the people working in the quantitative finance industry didn't get their degrees via the traditional route, but rather worked on their PhD's while already working in the industry.

I wouldn't say any CS courses in particular are helpful, but a good amount of experience with the common programming languages is definitely needed.

As for econ, I think that's more on the financial analysis end of things; that's quite a different beast, and the overlap with the quantitative analysts is not that large.  There are, however, some 'in-between' positions for people well-versed in both economics/finance and mathematics/physical sciences that facilitate communication between the two groups; again though, these typically require graduate degrees and prior experience.

I very much agree with JoeBlow, a good course in probability theory, statistics, PDE will do you much better, then taking some fancy class like risk theory, stochastic calculus and so on, especially at undergraduate level, in my opinion a complete waste of time. Why? Well students don't understand the basics, ( the things I have mentioned in the beginning of the post ), so most of the course students spend "memorizing" formulas, theorems, that are given without any proof ( of sketch of a proof ), and you need to blindly believe what's the professor saying. Stick to the basics, everything else can be learned along the way.

I need to point out that I don't work in the finance industry, I'm a financial math major, so I have a good understanding what's going on at undergraduate level.

Absolutely learn how to code.  This doesn't mean "learn CS".  It means "learn how to program."

That last bit of advice applies to pretty much any career path coming out of a math degree: know algorithms and how to implement them, because you will be using computers to calculate and to manipulate data.


Αν a,b,c,d ακεραιοι και $(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^{2}=d$ τοτε ο d einai τετραγωνο ακεραιου

Στο ανωτέρω εννοώ ^ όπου *

[quote="DuLac"]Τετριμμένο: d =a*2 επειδή 1, 2*(1/3), 2*(2/3) είναι γραμμικώς ανεξάρτητα επί των ρητών.[/quote]

pirazi pu dn katalaveno to tetrimeno..?ine toso profanes...??

[quote="kostas"]Αν a,b,c,d ακεραιοι και (a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^{2}=d τοτε ο d einai τετραγωνο ακεραιου[/quote]

an katalava kala to $2^{\frac{1}{3}}$..dn ine i triti riza tu 2...?
tote omos gia a=b=c=1 8a eprepe to $(1+2^{\frac{1}{3}}+2^{\frac{2}{3}})^{2}$ na ine tetragono akereu....
omos to proto melos ine a8risma geometrikis proodu +pernume oti
$\frac{(2^{\frac{1}{3}})^{3}-1}{2^{\frac{1}{3}}-1}$ $=\frac{1}{2^{1/3}-1}$ ine akereos...
malon dn exo katalavi kati..... :maybe:

μιλαω για εκεινα τα $a,b,c$ για τα οποια το $d$ ειναι ακεραιος

Αντί για "ακέραιος" μπορείς να αντικαταστήσεις "ρητός". Ανάπτυξε το τετράγωνο, βάλε τους συντελεστές των 1, 2^(1/3) και 2^(2/3) ίσον με 0 και λύσε το σύτημα εξισώσεων.

ok..ok..diko mu la8os...dn ixa katalavi oti i askisi anaferotan se ekina ta a,b,c gia ta opia to d itan akereos...meta..ontos ine profanes...+ginete +ayti i <epektasi> ton akereon stus rirus...

 :D

επειδή για μενα δεν είναι τόσο προφανές μμπορείτε να βάλετε αναλυτική λύση??

Λοιπον κανουμε τις πραξεις  και βγαζουμε κοινους παραγωντες τα $2^{\frac{2}{3}}$ kai $2^{\frac{4}{3}}$ και εχουμε :

$2^{\frac{2}{3}}(b^{2}+2ac)+2^{\frac{4}{3}}(c^{2}+ab)+4ab+a^{2}=d$
ομως τα $2^{\frac{2}{3}},2^{\frac{4}{3}}$ ειναι αρρητοι αρα $2^{\frac{2}{3}}k+2^{\frac{4}{3}}m+n=w$ δινει m=k=0,n=d m,m,k,w ρητοι.
αρα :$b^{2}+2ac=c^{2}+ab=0$ kai $4bc+a^{2}=d$

[quote="kostas"]για την ιστορια ,ηταν θεμα της ΔΜΟ του 1980 η οποια δεν εγινε  :P[/quote]  :P ..kalo..ontos itan sta shortlist,i to ipes gia plaka?

a..gia na olkliroso +tin apodiksi ..apo ki pu to ftse o kostas exume
$a=\frac{-b^{2}}{2c}$ opote antika8istontas stin ali to a..pernume
$b^{3}=2c^{3}$ diladi gia $c$ diaforo tu midenos $\frac{b}{c}=2^{\frac{1}{3}}$ pu dini ritos=aritos,atopo
opote $b=c=0$ kai ara $d=a^{2}$

 :D

εκεινη τη χρονια εγιναν διαφορες ΔΜΟ σαν αντικατασταση της  μιας και παγκοσμιας ΔΜΟ.πχ καλεσε η Φιλανδια 5 χωρες και εκανα της δικη τους γιορτη  :D
ευχαριστω  για την λυση  ειχα κανει λαθος και νομιζα οτι $d=(3a)^{2}$


[quote="abdurashidjon"]Problem
Assume a triangle ABC if there is another triangle PQR which is similar to ABC and inside it’s the circumcircle such that two two points of PQR on sides ABC (at most two points) then the circumcenter of ABC lies on one of the altitudes of PQR.[/quote]

Posting the problem a second time doesn't make it better. I still have not the slightest idea of what you mean.

  darij


It is possible to earn 0, 1, 3, 7, or 10 points with each shot. How many positive scores less than 30 can't be made in 3 shots?

There's probably some other (mods, or something else) way to do it, but I Brute Forced it.

[hide]
19, 22, 25, 26, 28, 29
[/hide]

[quote="JesusFreak197"]Just make all possible combinations of those five scores, and find the ones that aren't there. That shouldn't take long.[/quote]

Want me to list 125 ways in 80 seconds?  :roll:

Well, take the order repeat stuff out and you dont' have as much, but good luck  :P

Well, I'm assuming that you're intelligent enough to not have to count $(0,0,1)$, $(0,1,0)$, and $(1,0,0)$. :roll:

Or you could just test every value and find the ones that aren't possible to make... :P Is there actually a better way to do it than that? If so, it would be nice to know.

Klebian: I don't think mods would help. There may be a way that would be useful, but mods will only tell you the first digit unless you're using a high value, which isn't really any benefit at that point.


A bin contains 20 red balls, 30 green balls, and 60 blue balls.  The balls are drawn randomly from the bin, one at a time, without replacement.  What is the probability that the color of the first ball that is exhausted is red?

$ P(red \ ball \ on \ first \ try) \equal{} \frac {\# \ red \ balls}{total \ \# \ of \ balls} \equal{} \frac {20}{110} \equal{} \boxed{\frac {2}{11}}$

No the question is asking what is the probability that the first color to run out of balls in the bag is red not what the probability is for the first ball picked to be red.

Oh really,  can someone explain it to me?

That doesn't seem right. Remember that you can still pick 29 green balls and 59 blue balls and 20 red balls in any order as long as the last ball is red in order to satisfy the requirement. The probability given doesn't encapsulate that case and the other numerous cases.

I attempted case work by encoding the balls you pick as a sequence. $ R$ for red, $ G$ for freen, and $ B$ for blue. For each case there will always be 1 red ball at the end, so you only have to arrange the $ R$'s, $ G$'s, and $ B$'s. There were so many cases that it seemed like there must be a better approach.

Ah yeah if the question was like what is the probability that red is the first or second color exhausted.  Then we could view this problem as a sequence of Rs,Gs, and Bs.  The probability that the last ball chosen is red would be 2/11.  This is the probability regardless of the order of colors before it since no matter what to be the last ball exhausted your color has to come up last.  Then the probability of the red ball being exhausted first or second would be 9/11.  Maybe you can use some sort of variation of this method to solve this problem?

@sunehra: try using smaller numbers with 2 red, 3 green and 6 blue balls... that should be a more reasonable size to work with.

[quote="Truffles"]lol that's what I tried doing.[/quote]So... what's the answer for the small case?  :D

There are still too many cases, I stopped a while ago.  :P

What do you mean by "the small case"?

I'm pretty sure the probability is high since it's obviously easier to exhaust the color which has the last amount of balls.

[hide]If the red balls are exhausted first, then there must be a run of at least one green ball and at least one blue ball after the last red ball is removed.  So, what is the probability that the last ball removed is blue?  What is the probability that, given the last ball removed is blue, there is a green ball removed after the last red ball?[/hide]

How about we think about it this way: Considering every ball independent regardless of what color it is, each ball is equally likely to picked, and once it is picked that particular ball is exhausted. Now, we can consider a certain group of balls that associate the "red" group. Since we are given that there are 20 such balls, and we compute the probability, that those 20 are picked out of the total 110 balls, to be 2/11. I am not so sure this is a good way think about it... :maybe:

Well I don't know if you already took this into account but anyway note that if the first ball picked is red there is 19/109 chance of a red ball to appear again.

$ \frac{2}{11}$ is the probability that the red balls are LAST to be exhausted.

Here is my attempt. I am not sure it's right though.

$ \frac{6}{11}$ is the probability that blue balls are last to be exhausted.

Suppose that is that case.

Now ignore blue balls. We have two kinds of balls left, and the probability that green balls are later to be exhausted is $ \frac{3}{5}$

So the probability that red balls are first to be exhausted is $ \frac{3\times6}{5\times11}\plus{}\frac{3\times3}{11\times4}\equal{}\frac{19}{44}$

That basically sounds right?  But I don't know if 3/5 would really be the probability green balls are exhausted last?

$ \frac {3}{5}$ in my solution is the probability the green balls are exhausted second give that blue balls are exhausted last.


Fie S o multime finita de numere intregi mai mari ca 1 astfel incat, oricare ar fi n natural, exista s $\in$ S cu c.m.m.d.c. (s,n)=1 sau c.m.m.d.c.(s,n)=s. Sa se arate ca exista t,s $\in$ S astfel incat c.m.m.d.c.(s,t) este numar prim.

Hint: fie $n$ cel mai mic numar natural astfel incat $gcd(n,s)>1$ pentru orice $s\in S$. Atunci $n$ e liber de patrate si exista $s\in S$ ai $s$ divide $n$.


group theory
abstract algebra
combinatorics unsolved
combinatorics

Suppose that there are are $n$ students in a class . Every boy knows at least $1$ girl . Prove that there is aset with at least $\frac{n}{2}$ of students such that in this set every boy knows odd girls .

I don't understand... ;)  You want each boy in the subgroup knows an odd number of girls? Is any girl one of the students?

Pierre.


analytic geometry
graphing lines
slope
parameterization

Find the slope of the "ray" defined by y=(t^2)-2, x=4-t^2

[hide]As stated, we eliminate the parameter.

$y+2=t^{2}$ also $4-x=t^{2}$

Equating the two,

$y+2=4-x$

Thus

$y=2-x$

So the slope is $-1$ when you graph it on the Cartesian plane.[/hide]


Find the greatest value of the expression
$ (x^a_1 \plus{} x^a_2 \plus{} ..x^a_n)(\frac {1}{x^{ \minus{} a_1}} \plus{}\frac {1}{x^{ \minus{} a_2}} ...\plus{}\frac {1}{x^{ \minus{} a_n}})$
where $ x$ is an arbitrary positive number $ x$#$ 1$,n is a given even natural number,$ a_1,..a_n$ are given numbers belonging to the segment $ [m,m \plus{} 1]$,there $ m$ is also a given natural number

[quote="Allnames"]Find the greatest value of the expression
$ (x^a_1 \plus{} x^a_2 \plus{} ..x^a_n)(\frac {1}{x^{  a_1}} \plus{} \frac {1}{x^{  a_2}} ... \plus{} \frac {1}{x^{  a_n}})$
where $ x$ is an arbitrary positive number $ x$#$ 1$,n is a given even natural number,$ a_1,..a_n$ are given numbers belonging to the segment $ [m,m \plus{} 1]$,there $ m$ is also a given natural number[/quote]
Sorry,the problem must be above,and I think it is result why no one help me
[quote='christ']What about x1,x2,...,xn[/quote]
I think you want to know $ a_i$,you can see it in my condition

][quote="Allnames"]Find the greatest value of the expression
$ (x^a_1 \plus{} x^a_2 \plus{} ..x^a_n)(\frac {1}{x^{ a_1}} \plus{} \frac {1}{x^{ a_2}} ... \plus{} \frac {1}{x^{ a_n}})$
where $ x$ is an arbitrary positive number $ x$#$ 1$,n is a given even natural number,$ a_1,..a_n$ are given numbers belonging to the segment $ [m,m \plus{} 1]$,there $ m$ is also a given natural number[/quote]
One month since I posted it,No one can solve ,
arqady,can you help me?


geometry
3D geometry
probability
expected value

Six number cubes with faces numbered 1-6 are rolled. The sum of the numbers on the top faces is calculated. What is the most likely sum?

The [url=http://www.artofproblemsolving.com/Wiki/index.php/Expected_value]expected value[/url] of one roll of the die is \[ 1\cdot\frac{1}{6}\plus{}2\cdot\frac{1}{6}\plus{}3\cdot\frac{1}{6}\plus{}4\cdot\frac{1}{6}\plus{}5\cdot\frac{1}{6}\plus{}6\cdot\frac{1}{6}\equal{}\frac{21}{6}.\] Since we are rolling 6 dice, the expected value is just 6 times that, or \[ 6\times\frac{21}{6}\equal{}\boxed{21}.\]


geometry
geometric transformation
reflection
Euler
geometry unsolved

Denote F by the Fermat point of the triangle ABC.Define Fa as the reflection of the side BC of the triangle.Similar we will have Fb,Fc.Prove that AFa,BFb,CFc are concurent.

A nice hint: When you draw the figure, join $ AF_c, CF_a, AF_b$ to get three isiosceles triangles. Also join $ AF_{c}B , BF_{a}C$, $ AF_{b}C$.

[b]Lemma.[/b] If $ P,Q$ are isogonal conjugates with respect to $ \triangle ABC,$ then the reflections of $Q$ across $BC,CA,AB$ form a triangle perspective with $\triangle ABC$ $\Longleftrightarrow$ $PQ$ is parallel to the Euler line of $ \triangle ABC.$

The line joining the first Fermat point and the first isodynamic point has equation

\[ \left[\begin{array}{ccc}a^2\sqrt {3}S_A \plus{} S & b^2\sqrt {3}S_B \plus{} S & c^2\sqrt {3}S_C \plus{} S \\
 \\
\frac {1}{\sqrt {3}S_A \plus{} S} & \frac {1}{\sqrt {3}S_B \plus{} S} & \frac {1}{\sqrt {3}S_C \plus{} S } \\
 \\
x & y & z\end{array}\right] \equal{} 0
\]
This line has infinite point $ X_{30}: (S_AS_B \plus{} S_AS_C \minus{} 2S_BS_C),$ which coincides with the infinite point of the Euler line of $\triangle ABC.$ Thus, reflections of the 1st Fermat point on $BC,CA,AB$ form a triangle perspective with $\triangle ABC.$


[quote="Sharparrow"]An isosceles triangle $ ABC$ where $ AB \equal{} AC \equal{} \sqrt 3$ & $ D$ is a point on $ BC$ such that $ AD \equal{} 1$. Find the value of $ BD \cdot DC$[/quote]
Draw the circle $ w$, which is centered at $ A$ and passes through the points $ B$ and $ C$.
Use power of the point $ D$ with respect to the circle $ w$

If you still have any problem, please write back again

[quote="Sharparrow"]An isosceles triangle ABC where AB=AC=sqrt3 & D is a point on BC such that AD=1. Find the value of BD x DC.?[/quote]
This is the solution using pontios's hint:
[hide="Solution"]
Consider the circle with center $ A$ and radius $ \sqrt {3}$.  Thus, we can locate $ B$ and $ C$ on this circle.  Now, we can also find $ D$.  Now, extend $ AD$ to meet the circumcircle at $ X$ and $ DA$ to meet it at $ Y$.  Hence, by Power of a Point, $ BD\cdot CD \equal{} XD\cdot YD \equal{} (AX \minus{} AD)\cdot (AY \plus{} AD) \equal{} (\sqrt {3} \minus{} 1)\cdot (\sqrt {3} \plus{} 1) \equal{} 3 \minus{} 1 \equal{} \boxed{2}$. [/hide]


group theory
abstract algebra
algebra proposed
algebra

Let G be a finite group. Let H1, H2, H3 be proper (different from G) subgroups which 'cover' G completely. prove that x^2 is in H1 \cap H2 \cap H3 for all x in G.

The problem is Ok, but it's too weak. The result is still true if you don't assume G to be finite.

I will show that all x 2 are in H1. It is the same for the others.
Take an arbitrary x.
If x is in H1, than x 2 is also in H1
If x is not in H1, suppose it is in H2 (H3 is similar:))
Choose y in H1-(H2 U H3) which is not void because the group cannot be written as the union of two subgroups.
xy not in H1 or H2 so xy in H3
x in H3 would give y in H3 also, so x not in H3
x*xy=x 2 y not in H3 or H2 so x 2 y in H1
y also in H1, so x 2 in H1

I guess this is all.:)

Is this thing with writing as the union of three subgroups possible for many groups? Looks kind of restrictive to me. :)

blondu I read your proof carefully, you  did a  good job :)

Try the following:

If G is the union of four proper subgroups H1, H2, H3, H4 such as the union of any three of them does not yield G, show that x 12  is in H1 \cap H2 \cap H3 \cap H4. Also give an example to show that 12 is the lowest number with this property.  :)


calculus
inequalities
function
inequalities proposed

[quote="Vasc"]Let $a,b,c,d$ be positive numbers such that $abcd=1$. Prove that

(1) $\frac 1{(3a-1)^{2}}+\frac 1{(3b-1)^{2}}+\frac 1{(3c-1)^{2}}+\frac 1{(3d-1)^{2}}\geq 1$;

(2) $\frac 1{2a^{2}+a+1}+\frac 1{2b^{2}+b+1}+\frac 1{2c^{2}+c+1}+\frac 1{2d^{2}+d+1}\geq 1$.[/quote]

 Vasc, I have a very ugly solution for your problem. Do you have an elementary solution without calculus?

I have for (1) three nice solutions, but for (2) only one nice solution. :P

[quote="Vasc"]Let $a,b,c,d$ be positive numbers such that $abcd=1$. Prove that

$\frac 1{(3a-1)^{2}}+\frac 1{(3b-1)^{2}}+\frac 1{(3c-1)^{2}}+\frac 1{(3d-1)^{2}}\geq 1$. [/quote]

Firstly, we have:

$(3a-1)^{2}\leq a^{3}+7a^{2}-5a+1$.

We are to prove:

$\sum_{cyc}{1 \over a^{3}+7a^{2}-5a+1}\geq 1.$

Replace all values with $a={yzt \over x^{3}},b={ztx \over y^{3}},c={txy \over z^{3}},d={xyz \over t^{3}}$, then we shall prove:

$\sum_{cyc}{x^{9}\over x^{9}+7x^{3}y^{2}z^{2}t^{2}-5x^{6}yzt+y^{3}z^{3}t^{3}}\geq 1.$

$\sum_{cyc}{x^{9}\over x^{9}+7x^{3}y^{2}z^{2}t^{2}-5x^{6}yzt+y^{3}z^{3}t^{3}}\geq{(x^{4.5}+y^{4.5}+z^{4.5}+t^{4.5})^{2}\over \sum x^{6}+\sum 7x^{3}y^{2}z^{2}t^{2}-\sum 5x^{6}yzt+\sum y^{3}z^{3}t^{3}}\geq 1.$

The last inequality is equvalent to :

$2\sum_{sym}x^{4.5}y^{4.5}+5\sum_{cyc}x^{6}yzt \geq 7\sum_{cyc}x^{3}y^{2}z^{2}t^{2}+\sum_{cyc}x^{3}y^{3}z^{3}.$

This is clearly true by Muirhead’s. The problem is proved.

[quote="Vasc"]Let $a,b,c,d$ be positive numbers such that $abcd=1$. Prove that

$\frac 1{2a^{2}+a+1}+\frac 1{2b^{2}+b+1}+\frac 1{2c^{2}+c+1}+\frac 1{2d^{2}+d+1}\geq 1$.[/quote]

A bit similarity, using the substitutions like above, we have:

$\sum_{cyc}{x^{6}\over x^{6}+x^{3}yzt+y^{2}z^{2}t^{2}}\geq 1.$

By the CS inequality:

$\sum_{cyc}{x^{6}\over x^{6}+x^{3}yzt+y^{2}z^{2}t^{2}}\geq{(x^{3}+y^{3}+z^{3}+t^{3}) \over \sum_{cyc}x^{6}+\sum_{cyc}x^{3}yzt+\sum_{cyc}2y^{2}z^{2}t^{2}}.$

Let the last term be not less than 1, it gives:

$2\sum_{sym}x^{3}y^{3}\geq xyzt(x^{2}+y^{2}+z^{2}+t^{2})+2(x^{2}y^{2}z^{2}+y^{2}z^{2}t^{2}+z^{2}t^{2}x^{2}+t^{2}x^{2}y^{2}).$

It reduces to Muirhead’s theorem.
The problem is proved.

These solutions proclaim that there’s some inequalities can be done simply by the CS inequality, for instance, with product of some positive numbers,

${1 \over (1+a)^{2}}+{1 \over (1+b)^{2}}+{1 \over (1+c)^{2}}\geq{3 \over 4},$

${1 \over (1+a)^{2}}+{1 \over (1+b)^{2}}+{1 \over (1+c)^{2}}+{1 \over (1+d)^{2}}\geq 1.$

Your solution is very interesting, Lovasz. :!:
I didn't use Muirhead in my solutions.

[quote="Lovasz"]

These solutions proclaim that there’s some inequalities can be done simply by the CS inequality, for instance, with product of some positive numbers,

${1 \over (1+a)^{2}}+{1 \over (1+b)^{2}}+{1 \over (1+c)^{2}}\geq{3 \over 4},$

${1 \over (1+a)^{2}}+{1 \over (1+b)^{2}}+{1 \over (1+c)^{2}}+{1 \over (1+d)^{2}}\geq 1.$[/quote]

I am too lazy to solve all of your problems Lovasz. However, for the first one:
$abc=1$ Prove:
${1 \over (1+a)^{2}}+{1 \over (1+b)^{2}}+{1 \over (1+c)^{2}}\geq{3 \over 4},$
Here is my solution:
Let $a=\frac{x}{y}$,$b=\frac{y}{z}$, $c=\frac{z}{x}$
Hence, your inequality becomes:
$\frac{y^{2}}{(x+y)^{2}}+\frac{z^{2}}{(y+z)^{2}}+\frac{x^{2}}{(x+z)^{2}}\geq \frac{3}{4}$

Now L-R=$\frac{(x-y)^{2}(y-z)^{2}(z-x)^{2}+4(x+y)(y+z)(z+x)(x^{2}z+y^{2}x+z^{2}y-3xyz)}{(x+y)^{2}(y+z)^{2}(z+x)^{2}}$

Edited:clean up the argument.

Note that
\[\frac{1}{(3a-1)^{2}}\geq 1\Leftrightarrow a(9a-6)\leq 0\Leftrightarrow 0\leq a\leq \frac{2}{3}\]
So we only need to consider the case when $\frac{2}{3}\leq a\leq b\leq c\leq d$. From this we have $abc\geq (2/3)^{3}$ and $d\leq (3/2)^{3}$.

Hence $S'=\{(a,b,c,d): a\leq b\leq c\leq d\leq (3/2)^{3}, abcd=1\}$ is compact.
Let $f(a,b,c,d)=\sum 1/(3a-1)^{2}$. This function is continuous so has a minimum on $S'$. Suppose the minimum attained at $(a,b,c,d)$ where  $a<d$.
Note that
\[f(a,b,c,d)-f(\sqrt{ad},b,c,\sqrt{ad})\geq \frac{2}{(3a-1)(3d-1)}-\frac{2}{(3\sqrt{ad}-1)^{2}}>0 \]
because
\[(3\sqrt{ad}-1)^{2}-(3a-1)(3d-1)=3a+3d-6\sqrt{ad}>0 \]
So $(\sqrt{ad},b,c,\sqrt{ad})\in S'$ and $f(a,b,c,d)>f(\sqrt{ad},b,c\sqrt{ad})$ (contradiction!). So we must have $a=d$, i.e., the minimum is attained at $(1,1,1,1)$ with value $f(1,1,1,1)=1$

Since Vasc is an expert, I would like to have a comment from him.

Comment: In SMV Theorem Khungth claims $\Delta^{k}(a,b,c,d)$ convergent to $(g,g,g,g)$ where $g$ is the geometric mean of the sequence and where the  transformation defined by $\Delta(a,b,c,d)=(\sqrt{ad},b,c,\sqrt{ad})$. (But I don't see a proof). If one accept this then we don't need to create the compact set $S'$ as above since
\[f(a,b,c,d)\geq f(\Delta^{k}(a,b,c,d)) \]
and upon taking limit we have
\[f(a,b,c,d)\geq f(1,1,1,1)=1 \]

[quote="Gibbenergy"][quote="Lovasz"]

These solutions proclaim that there’s some inequalities can be done simply by the CS inequality, for instance, with product of some positive numbers,

${1 \over (1+a)^{2}}+{1 \over (1+b)^{2}}+{1 \over (1+c)^{2}}\geq{3 \over 4},$

${1 \over (1+a)^{2}}+{1 \over (1+b)^{2}}+{1 \over (1+c)^{2}}+{1 \over (1+d)^{2}}\geq 1.$[/quote]

I am too lazy to solve all of your problems Lovasz.[/quote]

No, Gibbenergy, they were Vietnam and China TST, about which have many topics and solutions.

For the inequality \[\frac{1}{(a+1)^{2}}+\frac{1}{(b+1)^{2}}+\frac{1}{(c+1)^{2}}\geq\frac{3}{4}\], there are two nice snd short solutions, using my inequality from here: http://www.mathlinks.ro/Forum/viewtopic.php?&t=108710 :\[ 2\cdot\sum\left(\frac{b}{a+b}\right)^{2}\geq\sum\frac{a}{a+b}.\]

Actually, you used the inequality
$\frac 1{(3a-1)^{2}}\geq \frac 1{4a^{3}}$,
which is not  true for any positive number $a$ (see $a=\frac 1{6}$).
However, your solution works. :lol:

[quote="toanhocmuonmau"][quote="Vasc"]Let $a,b,c,d$ be positive numbers such that $abcd=1$. Prove that

(1) $\frac 1{(3a-1)^{2}}+\frac 1{(3b-1)^{2}}+\frac 1{(3c-1)^{2}}+\frac 1{(3d-1)^{2}}\geq 1$;

(2) $\frac 1{2a^{2}+a+1}+\frac 1{2b^{2}+b+1}+\frac 1{2c^{2}+c+1}+\frac 1{2d^{2}+d+1}\geq 1$.[/quote]
My solution:
(1) WLOG, $a=\min{a,b,c,d}$
+, If $a \ge 1/4$, then
\[\frac{1}{(3a-1)^{2}}\ge \frac{1}{4a^{3}}\]
+, If $a \le 1/4$, then
\[\frac{1}{(3a-1)^{2}}>1 \]
(2) can prove easily by Cauchy. :)

We are done. :)[/quote]
how can you use Cauchy to solve it  easily?


Let AH1,BH2,CH3 be the altitudes of  scalene triangle ABC. The incircle of triangle ABC is tangent to BC, CA, and AB at T1,T2, and T3, respectively. Line HiHi+1 (where H4=H1) intersects TiTi+1 at Pi. Prove that
(1):Three lines P1T3,P2T1,and P3T2 are concurrent.
(2):Three lines P1H3,P2H1,and P3H2 are concurrent.

Restate the problem as follows:
Let AH1,BH2,CH3 be the altitudes of scalene triangle ABC. The incircle of triangle ABC is tangent to BC, CA, and AB at T1,T2, and T3, respectively. Line HiHi+1 (where H4=H1) intersects TiTi+1 at Pi+2. Prove that
(I)Three lines P1T1,P2T2 and P3T3 are concurrent, and their intersection is on the incircle of triangle ABC.
(II) Three lines P1H1,P2H2, and P3H3 are concurrent.


inequalities
Cauchy Inequality
inequalities proposed

Given x,y are real number such that $ x\plus{}y\ge 1$ and $ |xy|\le 2$. Prove

$ x^3\plus{}y^3\ge \minus{}7$

I think it holds $ x^3 \plus{}y^3 \geq 7$

Let $ x>y$ , then the smallest possible value of $ y$ is $ y\equal{}\minus{}1$ , then $ x \geq 2$

[quote="enndb0x"]I think it holds $ x^3 \plus{} y^3 \geq 7$

Let $ x > y$ , then the smallest possible value of $ y$ is $ y \equal{} \minus{} 1$ , then $ x \geq 2$[/quote]

I don't think so :wink:

[quote="gigaman"]Given x,y are real number such that $ x \plus{} y\ge 1$ and $ |xy|\le 2$. Prove

$ x^3 \plus{} y^3\ge \minus{} 7$[/quote]
Case 1: If $ x,y$ are of the same sign they must be positive. So, $ x^3\plus{}y^3 \geq 0$
Case 2: If $ x,y$ are of the opposite sign, $ x^3\plus{}y^3\equal{}(x\plus{}y)(x^2\minus{}xy\plus{}y^2) \geq (x\plus{}y)( 2|xy|\minus{}xy)\equal{}3(x\plus{}y)|xy| \geq 0$

There must be something wrong.... Though I do not see any flaw in my proof.  BTW what is the equality case??

In fact, based on the given condition, we have
$ x^3\plus{}y^3\equal{}(x\plus{}y)(x^2\minus{}xy\plus{}y^2) \ge x^2\minus{}xy\plus{}y^2>0>\minus{}7$

There must be something missing here or incorrect.

[quote="gigaman"]Given x,y are real number such that $ x \plus{} y\ge 1$ and $ |xy|\le 2$. Prove

$ x^3 \plus{} y^3\ge \minus{} 7$[/quote]

Sorry for my mistake, this is the true ineq

Given x,y are real number such that $ x \plus{} y\ge \minus{} 1$ and $ |xy|\le 2$. Prove

$ x^3 \plus{} y^3\ge \minus{} 7$

[hide]Hình như với x+y\ge 1 thì hơi dễ nên có khá nhiều mem tham gia giải :D [/hide][/hide]

yeah, that is what I wanna say, a right problem comes out with a nice expression as follows

$ x^3\plus{}y^3\plus{}1\minus{}3xy\plus{}3(xy\plus{}2) \ge 0$

$ \Leftrightarrow \frac{(x\plus{}y\plus{}1)}{2}[(x\minus{}y)^2\plus{}(x\minus{}1)^2\plus{}(y\minus{}1)^2] \plus{} 3(xy\plus{}2) \ge 0$

We are done since $ |xy| \le 2 \Rightarrow \minus{}2\le xy \le 2$

[hide="Gigaman"]Bai toan lop 8 day ma  :) [/hide]

[quote="great math"][hide="Gigaman"]Bai toan lop 8 day ma  :) [/hide][/quote]
[hide="great math"]Hi, tuy la lop 8 nhung ko it nguoi quen cai dang thuc do  :D [/hide]

Next

Given a,b,c positive such that a+b+c=3. Prove that:

$ \frac{1}{ab\sqrt{(a\plus{}b)(c\plus{}1)}}\plus{}\frac{1}{bc\sqrt{(b\plus{}c)(c\plus{}1)}}\plus{}\frac{1}{ca\sqrt{(c\plus{}a)(b\plus{}1)}}\ge \frac{3}{2}$

We have

$ \sum_{cyc}\frac{1}{ab\sqrt{(a\plus{}b)(c\plus{}1)}} \ge \sum_{cyc}\frac{2}{ab(a\plus{}b\plus{}c\plus{}1)} \equal{}\sum_{cyc}\frac{1}{2ab}$

The last step is pretty much trivial by using Cauchy inequality for 3 real numbers.
 We are done!

[hide="gigaman"]Biet anh la ai o VUT khong?[/hide]

[quote="great math"][hide="gigaman"]Biet anh la ai o VUT khong?[/hide][/quote]

[hide="great math"]Tui kô biết  :D, chắc là 1 mem nào đó  :rotfl: [/hide]

[quote="great math"]Do not spam gigaman, please post your solution for the above problem  :D[/quote]

Sorry, but my solution is the same  :D

Next

Given a,b,c positive such that ab+bc+ca=3. Prove that:

$ \frac {1}{ab\sqrt {(a \plus{} b)(c \plus{} 1)}} \plus{} \frac {1}{bc\sqrt {(b \plus{} c)(c \plus{} 1)}} \plus{} \frac {1}{ca\sqrt {(c \plus{} a)(b \plus{} 1)}}\ge \frac {3}{2}$

This ineq is harder than that one  :D

[hide="great math"]có phải ông là vietmath kô nhỉ[/hide]

Hello, giga, you are partly right though.
However, I think it is harder in the case of the condition but the following demonstration has it all

$ \frac{1}{ab\sqrt{(a\plus{}b)(c\plus{}1)}} \equal{} \frac{\sqrt{c}}{\sqrt{ab}\sqrt{c(a\plus{}b)(abc\plus{}ab)}} \ge \frac{2\sqrt{c}}{\sqrt{ab}(3\plus{}abc)}$

Finally, we only need to apply cauchy inequality for 3 positive real numbers and we are done!


algebra
polynomial
ratio
geometry
angle bisector
geometry unsolved

Given two vertices of triangle B, C, and a line l which contain the 3rd vertex A.  Let v be an vector.  Construct A such that the angle bisector of BAC is in the same direction as v.

is it solvable?  I got a 3rd degree polynomial using the fact that the angle bisector divides the base into ratios same as sides of the triangle.  Doesn't that say it is unsolvable in most cases?


Let $ a_0,a_1,a_2,...$ be a sequence where $ a_0\equal{}a_1\equal{}1$ and $ a_n\equal{}4a_{n\minus{}1}\minus{}4a_{n\minus{}2}$,for $ n\geq2$.Express $ a_n$ in terms of $ n$.

Let $ g(x)$ be the generating function of $ \{a_n\}$, we have:
$ g(x) \equal{} a_0 \plus{} a_1x \plus{} a_2x^2 \plus{} ... \plus{} a_nx^n \plus{} ...$
$ 4xg(x) \equal{} 4a_0x \plus{} 4a_1x^2 \plus{} ... \plus{} 4a_{n \minus{} 1}x^n \plus{} ...$
$ 4x^2g(x)\equal{} 4a_0x^2 \plus{} ... \plus{} 4a_{n \minus{} 2}x^n \plus{} ...$
$ \Longrightarrow (1 \minus{} 4x \minus{} 4x^2)g(x) \equal{} a_0 \plus{} (a_1 \minus{} 4a_0x) \equal{} 1 \minus{} 3x\Longrightarrow g(x) \equal{} \frac {1 \minus{} 3x}{1 \minus{} 4x \minus{} 4x^2}$
...


hello
how we can find the sequence $ (u_n)$ such that $ a.u_{n\plus{}2}\plus{}b.u_{n\plus{}1}\plus{}c.u_n \equal{}0$ ?

$ u_{n\plus{}2}\equal{}\minus{}\frac{b\cdot u_{n\plus{}1}\plus{}c\cdot u_n}{a}$ Let $ u_0\equal{}r, u_1\equal{}t$, then use generating functions...


For all non-negative number $ \large x,y,z$, prove that:

$ \large\sum(\frac{x}{y\plus{}z}\plus{}\frac{1}{2})^{2}\plus{}\frac{3\sqrt{2}}{2}\geq (2\plus{}\sqrt{2})\sum\frac{x}{y\plus{}z}$

Could you replace $ \large\sqrt{2}$ with a better constant?

You have recognized that $ f(a,b,c) \equal{}\sum(\frac{a}{b\plus{}c}\plus{}\frac{1}{2})^{2}\plus{}\frac{3k}{2}\minus{}(2\plus{}k)\sum\frac{a}{b\plus{}c}\geq 0$ then at $ \large (a,b,c) \equal{} (1,1,\sqrt{2}\minus{}1)$ you have $ \large f(1, 1,\sqrt{2}\minus{}1)\geq 0\Rightarrow k\leq\sqrt{2}$, but how can you prove that the inequality holds at $ \large k\equal{}\sqrt{2}$ :maybe:


Let x,y,z >0. Prove that
         \sum (y+z)/x >= 2* \sqrt [( \sum x^2)( \sum x)/(xyz)]

Finally! Take a triangle whose angles A,B,C verify sinA= \sqrt x(x+y+z)/(x+y)(x+z) and apply to it the inequality:
  ( \sum (x+y)/z) 2  \geq 4 \sum (x+y)(x+z)/yz*(sinA)^2.

Your inequality is equal to:

 \sum {sym}x^4*y^2+2*( \sum x^3*y^3)+6*x 2
y 2 z 2      \geq         2*xyz( \sum x^3)+2*xyz*( \sum {sym}x 2 y)

Then by Cauchy:
x^4y^2 +(xyz) 2     \geq   2*x^3*y^2*z

and the same.

And    \sum x^3*y^3   \geq      \sum xyz*( \sum x^3)

I sure that the inequality will can be proved when it become:
 \sum (y 2 z 2 +yz^3+x 2 z 2 +xz^3-x 2 yz-xy 2 z){=A1}8
(x-y) 2  \geq 0.(1)
So let a=x-z;b=z-y and x \geq z \geq y.Then (1) will become:
a 2 (A2+A1)+b 2 (A3+A1)+2abA1 \geq 0.
But we can easily proved that(try it by yourself):
A2+A1 \geq 0,A3+A1 \geq 0,A1 \geq 0,ab \geq 0(with condition x \geq z \geq y).
So we have the result we want :) .


quadratics
modular arithmetic
number theory theorems
number theory

My question is to the solution of Hawk Tiger in :http://www.artofproblemsolving.com/Forum/viewtopic.php?p=860863#860863

[quote="Hawk Tiger"]
Case 2:$ n$ is even .
We have that $ (x \plus{} 1)(x^{2} \minus{} x \plus{} 3) \equal{} x^{3} \plus{} 2x \plus{} 3 \equal{} 2(2^{n \minus{} 1} \plus{} 1) \equal{} 2(2*(2^{\frac {n \minus{} 2}{2}})^{2} \plus{} 1)$
Because $ x\equiv 5(mod 8)$,$ x^{2} \minus{} x \plus{} 3\equiv \minus{} 1(\mod 8)$.So there exist a prime fractor$ q$ of $ x^{2} \minus{} x \plus{} 3$ such that $ q\not\equiv 1,3(\mod 8)$(Notice that $ 3^{l}\equiv 1 or 3(\mod 8)$)
Then we have that $ 2*(2^{\frac {n \minus{} 2}{2}})^{2}\equiv \minus{} 1(\mod 8).$ Then $ \left ( \frac { \minus{} 2}{q}\right ) \equal{} 1$
but this is a contradition with $ q\not\equiv 1,3(\mod 8)$
Now we deduce that when $ n\ge 3$,there is no solution. :)
[/quote]

The line where  "Then we have that $ 2*(2^{\frac {n \minus{} 2}{2}})^{2}\equiv \minus{} 1(\mod 8).$ Then $ \left ( \frac { \minus{} 2}{q}\right ) \equal{} 1$ "

I thought the reciprocity law covers cases of $ a^{\frac {p \minus{} 1}{2}}\pmod p$, not $ a^{\frac {p \minus{} 2}{2}}$?

That $ 8$ should be a $ q$ here: $ 2*(2^{\frac {n \minus{} 2}{2}})^{2}\equiv \minus{} 1(\mod q).$
But this doesn't answer your question.

Well, he has shown that $ \minus{} 2^{ \minus{} 1} \mod q$ is a square $ \mod q$, in fact that of $ 2^{\frac {n \minus{} 2}{2}}$. Euler's criterion isn't used here (note that it's $ n$ in the exponent, not $ p \equal{} q$), it's just that this number's square happens to be $ \minus{} 2^{ \minus{} 1} \mod q$.
(See the further discussion in that thread).

Oh! Oops  :oops:  thank you for your help.


[i]   Find all fuctions $ f \ : \ \mathbb{R} \ \mapsto \ \mathbb{R}$ such that :

 $ f(x \plus{} y) \equal{} f(x)f(y) \minus{} f(xy) \plus{} 1 \ \forall \ x;y \ \in \mathbb{R}$[/i]

[b]  Ps Pco :[/b]

  [hide]Have you tried this one ?

       http://www.mathlinks.ro/viewtopic.php?t=324522[/hide]

I found $ g(x)+g(y)=g(x+y),g(xy)=g(x)g(y),g(1)=1$ in one of my specialcases, but I don't know if this is enough to determine $ g(x) = x$.
[hide]
Let $ f(1) = a$ and $ P(x,y)$ be the assertion $ f(x+y) = f(x)f(y) - f(xy) + 1, x,y \in \mathbb{R}$.
Then $ P(1,1) \iff f(2) = a^2-a+1$
$ P(2,1) \iff f(3) = f(2)(a-1)+1 = a^3-2a^2+2a$
$ P(3,1) \iff f(4) = f(3)(a-1)+1 = (a^3-2a^2+2a)(a-1)+1 = a^4-3a^3+4a^2-2a+1$
$ P(2,2) \iff 2f(4) = (f(2))^2+1 = (a^2-a+1)^2+1 = a^4-2a^3+3a^2-2a+2$

So $ 2f(4) = 2f(4) \iff$

$ a^4-2a^3+3a^2-2a+2 = 2(a^4-3a^3+4a^2-2a+1) \iff$
$ a^4 - 4a^3+5a^2-2a = 0 \iff$
$ a(a-1)^2(a-2) = 0$
So $ a=0$ or $ a=1$ or $ a=2$.

[b]Case $ a = 0 \iff f(1) = 0$:[/b]
$ P(x,1) \iff f(x+1)  = f(x)f(1) - f(x) + 1 = -f(x)+1 \iff f(x) + f(x+1) = 1$
Hence $ f(x) = f(x+2)$
So: $ P(x+1,1) \iff f(x+2) = f(x)f(2) - f(2x) + 1 = f(x) - f(2x) +1 \iff$
$ f(2x) = 1$
Inserting $ x = \frac{1}{2}$ gives a contradiction so $ a \neq 0$.

[b]Case $ a = 1 \iff f(1) = 1$:[/b]
Then $ P(x,1) \iff f(x+1) = f(x)f(1) - f(x) + 1 = 1$ gives the solution $ f(x) = 1 \forall x \in \mathbb{R}$, which is easily seen to be a solution.

[b]Case $ a = 2 \iff f(1) = 2$:[/b]
Let $ g(x) = f(x) - 1$, then $ g(1) = 1$.
Let $ Q(x,y) \iff P(x,y) \iff g(x+y) = g(x)g(y) + g(x) + g(y) - g(xy)$
$ Q(x,1) \iff g(1+x) = g(x) + 1$
So by induction $ g(x+n) = g(x) + n$.
$ Q(0,0) \iff g(0) = 0$
So: $ g(n) = n \forall n \in \mathbb{N}$

Let $ T(n)$ be the assumption: $ g(nx) = ng(x) \forall x \in \mathbb{R}$.
$ T(1)$ is clearly true.
Assume that $ T(n)$ is true.
$ Q(x,n+1) \iff g(x+n+1) = g(x)g(n+1) - g(x(n+1)) + 1 \iff$
$ g(x(n+1)) = (n+1)g(x)$. (Since $ g(x+n+1) = g(x)+n+1$ and $ g(n+1) = n+1$)
So we get $ T(n) \Rightarrow T(n+1)$, and in the same way $ T(n) \Rightarrow T(n-1)$. Hence $ f(nx) = nf(x) \forall n \in \mathbb{Z}, x \in \mathbb{R}$, and we easily get:
$ g(kx) = k g(x) \forall k \in \mathbb{Q}, x \in \mathbb{R}$

Note that: $ Q(x,y) \iff g(x+y) - (g(x) + g(y)) = g(x)g(y) - g(xy)$
$ Q(x,y+1) \iff g(x+y+1) - g(x) - g(y+1) = g(x)g(y+1) -g(x(y+1))$
Since $ g(x+y+1) = g(x+y)+1, g(y+1) = g(y)+1$ we get:
$ g(x)g(y) - g(xy) = g(x)g(y+1) - g(x(y+1)) = g(x)g(y)+g(x) - g(xy+x) \iff$
$ g(x)+g(xy) = g(xy+x)$
Letting $ y = \frac{z}{x}$ we get:
$ g(x) + g(z) = g(z+x)$.
But then $ 0 = g(x+y) - g(x) - g(y) = g(x)g(y)-g(xy)$

So we have $ g(x+y) = g(x)+g(y), g(xy) = g(x)g(y), g(1)=1$. Is this enough to prove $ g(x) = x$?[/hide]

[quote="prester"]I was starting to solve it, but I saw that this is a gift to pco and so I prefer to let him to solve..
 :lol:[/quote]

Yes it is, from $ g(xy)\equal{}g(x)g(y)$ you have $ g(x^2)\equal{}g(x)^2\ge 0$ therefore if $ x\ge0$ then $ g(x)\ge 0$.
Then set $ y\ge0$, then you have $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)\ge g(x)$ hence $ g$ is monotonic increasing.

Also from $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ you get $ g(x)\equal{}x$ for all rational $ x$, and since its monotonic increasing you also have it for all reals.

[quote="Rafikafi"]

Yes it is, from $ g(xy) \equal{} g(x)g(y)$ you have $ g(x^2) \equal{} g(x)^2\ge 0$ therefore if $ x\ge0$ then $ g(x)\ge 0$.
Then set $ y\ge0$, then you have $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)\ge g(x)$ hence $ g$ is monotonic increasing.

Also from $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$ you get $ g(x) \equal{} x$ for all rational $ x$, and since its monotonic increasing you also have it for all reals.[/quote]
Thanks! :)

Then finishing my proof we find $ g(x) \equal{} x$ and therefore $ f(x) \equal{} x\plus{}1$, which is easily seen to be a solution. So two solutions:
$ f(x) \equal{} 1 \forall x \in \mathbb{R}$
$ f(x) \equal{} x\plus{}1 \forall x \in \mathbb{R}$


Consider a paper punch that can be centered at any point of the plane and that, when operated, removes from the plane precisely those points whose distance from the center is irrational.  How many punches are needed to remove every point?

The answer is 3.

To prove this, we have to show
1) It can not be done with fewer than 3 punches
2) It can be done with 3 punches.

First, we will show that if you use 2 punches, there will be some points left in the plane.  Let the points be $(A, B)$ and $(C,D)$.  Let the distance between these points be X.  Choose a ratoinal number, Q such that .5X < Q < X.

There exists a point in the plane that are exactly Q distance from both $(A, B)$ and $(C, D)$.  Thus, with two punches, the entire plane can not be removed.

Now, we will show that three punches can remove every point in the plane.  To do this, we will use punches centered at $(0,0), (\sqrt{2}, 0), (\Pi, 0)$.  If a point, $(X, Y)$ is not removed by the first punch (centered at 0,0), then we know that $\sqrt{x^2 + y^2}$ is a rational number.  This means that $x^2 + y^2$ is also a rational number.  Similarly, we know that for the point to not be removed by the second punch, $x^2 + 2*\sqrt{2}*X + 2 +y^2$ is rational.  Finally, if a point is not removed by the third punch $x^2 + 2*\Pi*X + \Pi^2+y^2$ is rational.

Thus, if there exists a point that is not removed by any of the three punches, the following three numbers must all be rational:
$x^2 + y^2$
$x^2 + 2*\sqrt{2}*X + 2 + y^2$
$x^2 + 2*\Pi*X + \Pi^2 + y^2$

The difference of two rational numbers is rational.  Thus if you sutract the first number from the second and subtract 2 you get that $2*\sqrt{2}*X$ is rational.  This implies, there exists integers, m and n such that X = $\sqrt{2} * \frac{m}{n}$

Similarily, taking the first from the third yields that $2*\Pi*X + \Pi^2 = 2*\Pi*\sqrt{2} * \frac{m}{n} + \Pi^2$ is rational.  Thus, there exists integers a and b such that
$\frac{a}{b} = 2*\Pi*\sqrt{2} * \frac{m}{n} + \Pi^2$

If we move the $\Pi^2$ to the other side, and sqaure both sides of the equation, we find that
$\frac{a^2}{b^2} - \frac{2*a*\Pi^2}{b} + \Pi^4 = 8 * \frac{m^2}{n^2} * \Pi^2$.

However, $\Pi$ is a transcendental number.  Thus, it can not be the root to any polynomial.  Thus, there can not exist any points $(X, Y)$ that are not removed by those three punches.

QED.

Dan Johnston

But if you center it at one point, say A, you only have points with a rational distance from A. Then you can punch all you like, and A will not be removed.

Also, even if you could remove A, you minimize everything down to a few points, and operate on one, and that point is not removed, even though it may remove all the other points.

:maybe:

The punch at $ (0, 0)$ removes $ (\sqrt {2}, 0), (\pi, 0)$, and the punch at $ (\sqrt {2}, 0)$ removes $ (0, 0)$.  (Note that you can center a punch at a point that's been removed.)

Edit:  In one dimension, the answer is two punches.  Can we generalize to $ n$ dimensions?

Three punches at $ (0, 0, \ldots, 0)$, $ (\sqrt {2}, 0, \ldots, 0)$ and $ (\pi, 0, \ldots, 0)$ suffice, no?

Ah, that did occur to me.  Suppose we strengthen the condition:  the punch removes all points such that the [i]square[/i] of the distance is irrational.  What then?

(I am trying to get at some deeper result here by considering more arbitrary metrics $ d(x, y)$:  in particular, what can we say about the number of punches required if we replace the Euclidean metric with a metric satisfying some polynomial condition $ P( d(x, y), x, y) \equal{} 0$ of degree at most $ n$?)


What is a 'component' of a vector? What is the 'magnitude' of a vector?

Say you have a vector <[b]a[/b],[b]b[/b]>.
So the a is the x-component, and the b is the y-component.
The magnitude is simply how long it is, or sqrt(a^2 + B^2) by the pythagoream theorem.

Example: Find the x and y components and magnitude of the vector <3,-4>.
x-component: 3
y-component: -4
Magnitude: sqrt(3^2 +(-4)^2) = 5

Another Example:
Say you have a vector with magnitude 6 and theta = 30 degrees.
Find the x and y components of the vector.
<x,y> = < 6 cos 30, 6 sin 30>
= <3sqrt(3), 3>

Hope that helps!


Find the sum upto n terms for

$ (1)(1!)\plus{}2(2!)\plus{}3(3!)\plus{}...\plus{}n(n!)$

and then find the minimum number of terms for the sum to exceed $ 10^9$

Add 1 to the beginning. We get $ 1\plus{}1!\equal{}2!$, $ 2!\plus{}2 \cdot 2!\equal{}3!$, $ 3!\plus{}3 \cdot 3!\equal{}4!$ etc, and it collapses all the way to $ (n\plus{}1)!$. So the answer is $ (n\plus{}1)!\minus{}1$.

And from this, it is easy to see that $ 12!$ is the first factorial to exceed $ 10^9$, so you need 11 terms.

Can the second question be solved without using a calculator or software?

You can remember $ 5! \equal{} 120$.

$ 12! \equal{} 12 \cdot 8 \cdot 11 \cdot 9 \cdot 10 \cdot 7! \equal{} 96 \cdot 99 \cdot 100 \cdot 42 \cdot 120$

This is going to be a bit over $ 100 \cdot 100 \cdot 100 \cdot 42 \cdot 100$, so the logarithm is more than $ 2 \plus{} 2 \plus{} 2 \plus{} 1 \plus{} 2 \equal{} 9$. It is also obviously less than $ 10$.

$ 11! \equal{} 12!/12 < 12!/10$, and its logarithm must be less than $ 10 \minus{} 1 \equal{} 9$, and it is not sufficient.

[quote="TZF"]You can remember $ 5! \equal{} 120$.

$ 12! \equal{} 12 \cdot 8 \cdot 11 \cdot 9 \cdot 10 \cdot 7! \equal{} 96 \cdot 99 \cdot 100 \cdot 42 \cdot 120$

[/quote] I think it should be 10 instead of 100 on the rhs.

Wow, $ 10 \neq 100$. Brilliant job on my part. So what I should have said was that:

$ 10^8 < 12! < 10^9$ because of what I wrote there. And then when you multiply by $ 13$, you get over $ 10^9$, and so we want the value $ 13!\minus{}1$ and $ n\equal{}12$