Nine stones are arranged in a straight line. They are counted from
left to right as $ 1,2,3, \ldots, 9$, and then from right to left, so that
the stone previously counted as 8 is counted as 10. The pattern
is continued to the left until the stone previously counted as 1 is
counted as 17. The pattern then reverses so that the stone originally
counted as 2 is counted as 18, 3 as 19, and so on. The counting
continues in this manner. Which of the original stones is counted
as 99? Express your answer as a single digit which corresponds to
the first digit assigned to that stone.

The first time around, $ 9$ stones are counted. Each of the next times, however, the first stone has already been counted in the previous sweep, so that there are $ 8$ stones in one count across. Thus, let's find the [i]remainder[/i] of $ x$ in $ 8x\plus{}9\equal{}99$ to find which one on the moving across it is. Solving the above-stated, equation we find that $ 8x\equal{}90$, so that the number of eighths in the fractional part of $ x$ is $ 2$. Also, since there are $ 11$ eighths, we are going around for the $ 11\plus{}1$ st time (remember that we didn't count the first time across).

Thus, we are going from right to left. Counting $ 2$ from $ 9$, we find that the stone counted as $ 99$ is stone number $ \fbox{7}$.

I also put $7$, but the answer isn't $7$.

[hide="What the answer is"]
$3$[/hide]

Still don't get the problem, someone post a solution?

The reasoning above is correct, but however:

[hide="Solution"]We note that after the first pass to the 9th stone, we count 9 numbers. Then, as we go back, we skip the 9th AND the 1st stone. As we reach the first stone again, we start a new cycle. Thus, we have a cycle of $9+(9-2)= 16$ (not 17). Since:

$99 \equiv 3 \bmod (16)$

Our answer is \boxed{3}$.[/hide]


ratio
linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra unsolved

Hello,

I found this one very nice.

1) If $F\in\mathcal M_{n\times m}$ with $m\neq n$, show that $\rho(FF^T)=\rho(F^TF)$, where $\rho(X)$ prepresent the spectral ratio of the square matrix $X$ (take it general, because in the first case the matrix is $n\times n$ and in the second $m\times m$).

2) If $A,B\in\mathcal M_{n\times m}$, is it true that $\rho(AB^T)=\rho(A^TB)$ or $\rho(AB^T)=\rho(B^TA)$?

Best,

Anyone? It is a nice linear algebra problem!

[quote="Peter VDD"]what is spectral ratio? I thought the set of eigenvalues but then the thing is false, so what is it? :)[/quote]No, it is not the set of eighenvalues, it is the largest eighenvalue in absolute values. For example, if you have 1, 2, -3 as eighenvalues of a matrix A, then the spectral ratio is 3.

[quote="Peter VDD"]Aren't both a trivial consequence from {eigenvals(AB)} union {0} = {eigenvals(BA)} union {0}? :blush:

Or is that the theorem we have to prove? :)[/quote]Are you sure that is true for non square matrices? If you prove that, the theorem would be trivial!

Look: let k be an eigenvalue, x be an eigenvector:
ABx = kx, k<>0
BABx = Bkx = kBx
=> Bx is an eigenvector of BA, corresponding to eigenvalue k

:P

That's the simple version. If you don't have eigenvectors, use this.

Of course it is- just pad both matrices to square with zeros: $\begin{bmatrix}A&0\end{bmatrix}\cdot\begin{bmatrix}B\\0\end{bmatrix}= \begin{bmatrix}AB\end{bmatrix}$
and $\begin{bmatrix}B\\0\end{bmatrix}\cdot\begin{bmatrix}A&0\end{bmatrix}= \begin{bmatrix}BA&0\\0&0\end{bmatrix}$
(assuming WLOG that AB is larger than BA, and choosing the zero blocks to match dimensions).

This then reduces to the case for square matrices, in which $\det A\det B\det(\lambda I-AB)=\det(B(\lambda I-AB)A)=\det(\lambda BA-BABA)=\det BA\det (\lambda I-BA)=\det B\det A\det(\lambda I-BA)$. That says that for invertible $A,B$ the characteristic polynomial of $AB$ is the same as the characteristic polynomial of $BA$. Since the coefficients of the characteristic polynomial are polynomials in the coefficients of $A$ and $B$ which agree almost everywhere, they must be identical polynomials. (The coefficients of this polynomial are rational; work over $\mathbb{C}$ to make sure this works. Since the coefficients depend only on dimension and not on the field, this then applies in a general field).

[quote="jmerry"]That's the simple version. If you don't have eigenvectors, use this.[/quote]Isn't there always at least one eigenvector per eigenvalue? :?:

What if there aren't any eigenvalues at all? For all you know, I'm working over the rationals or some finite field. The spectral radius question assumes $\mathbb{C}$, but the fact that the characteristic polynomial of $AB$ is a power of $x$ times the characteristic polynomial of $BA$ doesn't.

[quote="djimenez"][quote="Peter VDD"]Look: let k be an eigenvalue, x be an eigenvector:
ABx = kx, k<>0
BABx = Bkx = kBx
=> Bx is an eigenvector of BA, corresponding to eigenvalue k[/quote]That works if your matrices are square matrices, but what if they are not..., then, there would not be any eighenvalue at all! In my case, I have $AB^T$ and $A^TB$, where one of them is $n\times n$ and the other is $m\times m$, where probably $m\neq n$. It is easy to prove if $A=B$, otherwise, it is not that simple.[/quote]Where in the proof am I using that A or B is square? :? [of course it depends on your definition of eigenvalues...]

At jmerry, could be... I just don't know a thing about finite fields :)

To make those unambiguous eigenvalues, pad the matrices out to square as in my argument.


Michigan Math League / Feb. 14

I thought it was pretty hard this time.  Usually I get 4 or 5 out of 6 but this time I for sure know I only got 2.

What'd you get for your answers?

[quote="chess64"]I don't think we're allowed to discuss answers yet, are we?

Anyway, I actually thought that this round was much easier than usual, I'm almost 100% sure that I got a 6.[/quote]

I edited my post just to be safe, but I'm pretty sure the contest has to be taken on the same time, and all participants from a school has to take it at the same time.  Either way, can you send me a PM explaning how you got your answers? Would love to know.

NYML is actually one of many regional contests that use the same questions. At the time you posted, several schools might still not have administered the test, like on the west coast. Note that there are also alternative test dates, so some schools might not have taken it yet.

Yes, I did get a six, so I can PM you with my solutions. :)

Technically the test should be done already because it can ONLY be administered officially on Feb. 14.

I got a 6. I started the year with 3 careless mistakes so I currently have 21/24.


vector
group theory
superior algebra
superior algebra unsolved

This is probably well known, but I could not find it by searching.

Let $G$ be a cyclic group of order 3. Show that the group ring $\mathbb{C}[G]$ is isomorphic to $\mathbb{C}\times \mathbb{C}\times \mathbb{C}$

A group ring FG is a vector space over the field F with basis G. So as vector spaces $\mathbb{C}G$ is isomorphic to $\mathbb{C}\times \mathbb{C}\times \mathbb{C}$.

If you want the ring structure you may as well artificially impose the multiplication (it will not be the usuall product multiplication)

It will turn out to be something like

$(a,b,c) \cdot (x,y,z)= (ax+bz+cy, ay+bx+cz, az+by+ca)$

Ok, I haven't seen group rings presented as vector spaces, but as rings. I want to show that they are isomorphic as rings. I don't understand what you mean by "artifically impose the multiplication", it is understood by the question that $\mathbb{C}\times\mathbb{C}\times \mathbb{C}$ is the direct product ring with component-wise operations.

first, the group ring is isomorphic to $\mathbb{C}\times \mathbb{C}\times \mathbb{C}$, if we define the multiplication according to qleak. but surprisingly, this ring is isomorphic to the direct product $P : = \mathbb{C}^{3}$ ! consider

$P \to \mathbb{C}\times \mathbb{C}\times \mathbb{C}, (1,0,0) \to (1,1,1), (0,1,0) \to (1,w,\overline{w}), (0,0,1) \to (1,\overline{w},w)$

where $w$ is a 3th primitive root.


Does anybody know results of Mediterranean M.O. 2007 and cutoffs for medals.
Also can you post results from your country.

delegat,you will have silver for sure.23 points (max.)arent good enough for gold ,sorry   :P

[quote]delegat,you will have silver for sure.23 points (max.)arent good enough for gold ,sorry[/quote]
 Ok but I can be sure that I will get silver :winner_second:

With your 20 points :clap2:   you can't be sure of getting silver medal...  :harhar:

28. (4x7)

In Croatia, winner had 27, I think after that came something like this 25, 24, 22, 21, 21...


Well I know the Holder's Inequality only as:
$ (a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} ......... \plus{} a_n)^(pq) \le {{(a_1)^p \plus{} (a_2)^p \plus{} ...... \plus{} (a_n)^p}^q }{{(a_1)^q \plus{} ........(a_n)^q}^p}$.
But iI do not know the inequality applied for the 3 rows of positive reals though I know it is very useful.
Could some one please help me out in this?
Futher, after explaining, could you also post some examples along with solutions so that I can understand better.

PS Here $ a_i , p , q$ have their assumed characters.

Thanks in advance.
Aravind.

Holder's Inequality can be applied to any number of rows.

Suppose $ \alpha \plus{} \beta \plus{} \gamma...\omega \equal{} 1$

$ (a_1 \plus{} a_2... \plus{} a_n)^{\alpha}(b_1 \plus{} b_2... \plus{} b_n)^{\beta}...(z_1 \plus{} z_2... \plus{} z_n)^{\omega} \ge a_1^{\alpha}b_1^{\beta}...z_1^{\omega} \plus{} a_2^{\alpha}b_2^{\beta}...z_2^{\omega}\plus{}... \plus{} a_n^{\alpha}b_n^{\beta}...z_n^{\omega}$


AMC
AIME
LaTeX
geometry
algebra
difference of squares
special factorizations

1983 #8

What is the largest 2-digit prime factor of the integer n = $\binom {200}{100}$?

1986 #2

Evaluate the product:

$(\sqrt 5 + \sqrt 6 + \sqrt 7)(\sqrt 5 + \sqrt 6 - \sqrt 7)(\sqrt 5 - \sqrt 6 + \sqrt 7)(-\sqrt 5 + \sqrt 6 + \sqrt 7)$.

[hide]Wouldn't the first one just be the largest 2 digit prime number?  Since 200C100 is 200*199*198...*101, multiplying the largest 2 digit prime number would be between 101-200.  So the answer would be [b]97[/b].  [/hide]
[hide]
((sqrt5+sqrt6)+sqrt7)((sqrt5+sqrt6)-sqrt7)((sqrt5-sqrt6)+sqrt7)((sqrt5-sqrt6)-sqrt7)*-1
using (a+b)(a-b)=a^2-b^2 on the first two terms and the last two terms we get:
(4+2 :sqrt: 30)(4-2 :sqrt: 30)*-1, for which using (a+b)(a-b)=a^2-b^2 again we get -(16-120) which equals [b]104[/b][/hide]

[quote="eryaman"]Since 200C100 is 200*199*198...*101,[/quote]

No.

Anyways, after some fiddling, the answer is $61$.

We observe that if $p$ is a 'large' 2 digit prime, then there exists one 'copy' of $p$ in the numerator and the denominator.  For example, if $p=97$, then $p\vert 194$ and $p\vert 97$ and no other terms are divisible by $p$, so $p\not\vert C^{200}_{100}$.  This clearly holds true for all $p$ such that $100>p>200/3$.  $p=61$ is the next smallest prime.  In the interval $[101,200]$, $p\vert 122,183$, and in $[1,100]$, $p\vert 61$, so $p=61$ works, and we're done.

He's right.

I can't think of better explanation than what he did. :)

Since I corrected him, I should probably add that, by definition,

$\binom{200}{100}=\frac{200\cdot 199\cdot\ldots\cdot 101}{100\cdot 99\cdot\ldots\cdot 1}$

i'm confused about why 61 works.  I got 47(i know that's wrong) b/c I thought every prime number greater than 101/2 (50) could cancel out with it times 2, for example, the one that blahblahblah gave was 97, with 194.
and in blahblahblah's answer, he clearly says that 61 has 122 and 183, so wouldn't it cancel out?  That's why I got 47, because it's double is too small to be in 101-200

122 and 183 are on the top of the fraction and 61 is on the bottom, so we get 61^2/61=61.

[quote="eryaman"]I feel stupid now, I don't know why I thought it was 200*199...101.  Thanks for your solution, I liked it . But what was the symbol you used that was like a straight like with a cross through it?  (Like an x)  I don't think I've seen it before.[/quote]

In $\LaTeX$, adding (\not) to certain things produces a slash through them.  So since $\vert$ is 'divides', I used the command \not\vert

e.g, $3\not\vert 6$

[hide]
Just kidding  :D
[/hide]

For the second problem, notice the strong resemblance to Heron's formula.  There is a nifty little way to make use of it if you like.  Otherwise, just do the algebra and it's simple enough.  Algebra might even be faster, but I just thought the equation of areas was cute. :)

probability1.01, it took me much less to expand using difference of squares then to find the square of the area of the triangle with sides $2\sqrt{5},2\sqrt{6},2\sqrt{7}$  :)

And in case Mr. Ocax still doesn't get why it is 61...

We have 200*199*....*101/100!
To find the largest 2 digit prime, the numerator must have the prime twince so that it does not cancel out from the denominator....
61*2=122
61*3=183
The next largest prime would be 67*3=201 and thats over 200....

Nevermind... I missed tetrahedr0n's post

By the way, ragging, that was a great post from you on the MI forum. Check it out.


function
LaTeX
integration
Pascal\u0027s Triangle
algebra
binomial theorem
combinatorics open

[hide="answer"]
Telescopes to $ \frac{1}{n\plus{}1}$
[/hide]

Can you show me how you arrived at that answer?

Sorry if these seem "easy" but I haven't had much work with them and I'm going to be gone from class when we discuss it.

$ \sum \limits_{x = 1}^n {( - 1)^{x+1} \binom{n}{x} = 1}$
This follows by picturing Pascal's Triangle and every term is added to its negative (except the very left element, ie. 1)

I am not exactly sure how to show this since I don't remember (:D) but

$ \sum \limits_{x = 1}^n \frac {{( - 1)^{x + 1} \binom{n}{x}}}{x + 1} = 1 - \frac {1}{x + 1}$

and so we get $ 1 - (1 - \frac {1}{x + 1}) = \frac {1}{x + 1}$

Mhmm well while I was at work I figured out how to solve the problem with a very basic example.
Take n = 6 and try it. You will get
 $ 1\minus{}\frac{1}{2}\binom{6}{1}\plus{}\frac{1}{3}\binom{6}{2}\minus{}\frac{1}{4}\binom{6}{3}\plus{}\frac{1}{5}\binom{6}{4}\minus{}\frac{1}{6}\binom{6}{5}\plus{}\frac{1}{7}\binom{6}{6}$

Now use the fact that $ \binom{n}{k} \equal{} \binom{n}{n\minus{}k}$ If you havn't learned, easily seen with the "function" for choose. $ \binom{n}{k} \equal{} \frac{n!}{k!(n\minus{}k)!}$. You can see its true because since $ AB \equal{} BA$ for the$ \mathbb{R}, (n\minus{}k)!(k!) \equal{} (k!)(n\minus{}k)!$

Anyways, given the "function" and the set up you can basically figure it out with case n=6 easily enough. I chose 6 because it is short, but still there is enough to see how it would all form.

Anyways you'll get $ 1\minus{}\frac{6}{2}\plus{}\frac{15}{3}\minus{}\frac{20}{4}\plus{}\frac{15}{5}\minus{}\frac{6}{6}\plus{}\frac{1}{6\plus{}1}$

where I wrote $ \frac{1}{6\plus{}1}$ in that form to see it would be $ \frac{1}{n\plus{}1} \forall n \in \mathbb{Z}$.

P.S. I am really beginning to enjoy Latex :)

the sum is:
$ \sum_{k\equal{}0}^{n} \frac{(\minus{}1)^k}{k\plus{}1}\binom{n}{k} \equal{} \frac{\sum_{k\equal{}0}^{n} (\minus{}1)^{k}\binom{n\plus{}1}{k\plus{}1}}{n\plus{}1} \equal{} \frac{\sum_{k\equal{}1}^{n\plus{}1} (\minus{}1)^{k\plus{}1}\binom{n\plus{}1}{k}}{n\plus{}1} \equal{} \frac{\sum_{k\equal{}0}^{n\plus{}1} (\minus{}1)^{k\plus{}1}\binom{n\plus{}1}{k} \plus{} 1}{n\plus{}1}\equal{} \frac{\minus{}(\minus{}1\plus{}1)^{n} \plus{} 1}{n\plus{}1}\equal{}\frac{1}{n\plus{}1}$

Thank you very much for all the help.  I think I'm getting the hang of this stuff.

another solution:

By the binomial theorem,
$ (x\plus{}1)^n \equal{} 1 \plus{} \binom{n}{1} x \plus{} \binom{n}{2} x^2 \plus{} \ldots \plus{} \binom{n}{n} x^n$

$ \int (x\plus{}1)^n \; dx \equal{} \int \left(1 \plus{} \binom{n}{1} x \plus{} \binom{n}{2} x^2 \plus{} \ldots \plus{} \binom{n}{n} x^n \right) \; dx$

$ \frac{(x\plus{}1)^{n\plus{}1}}{n\plus{}1} \equal{} C \plus{} x \plus{} \binom{n}{1} \frac{x^2}{2} \plus{} \binom{n}{2} \frac{x^3}{3} \plus{} \ldots \plus{} \binom{n}{n} \frac{x^{n\plus{}1}}{n\plus{}1}$

Letting x = 0 to determine the constant C we get $ C \equal{} \frac{1}{n\plus{}1}$, then letting x = -1 we get:

$ 0 \equal{} \frac{1}{n\plus{}1} \minus{}1 \plus{} \binom{n}{1} \frac{1}{2} \minus{} \binom{n}{2} \frac{1}{3} \plus{} \ldots \plus{} \binom{n}{n} \frac{(\minus{}1)^{n\plus{}1}}{n\plus{}1}$

From which we find: $ 1 \minus{} \binom{n}{1} \frac{1}{2} \plus{} \binom{n}{2} \frac{1}{3} \plus{} \ldots \plus{} \binom{n}{n} \frac{(\minus{}1)^n}{n\plus{}1} \equal{} \frac{1}{n\plus{}1}$


function
inequalities unsolved
inequalities

Let $a_1,a_2,\ldots$ be a sequence of real numbers.  Prove that
\[ 0\leqslant\sum_{i,j\geqslant 1} \frac{a_ia_j}{i+j}\leqslant \pi\sum_{k=1}^{\infty} a^2_k  \]

Consider the function given by \[ f(x) = \frac{1}{x} \left(\sum_{i = 1}^{\infty} a_ix^i\right)^2...  \] Can someone finish it from here? It's pretty easy from here. Don't have time for the rest now, I've got to go to class :)

I'm bumping this thread for the other half of the inequality.

Yeah, to be honest, I don't really know what to do with the $\pi$-part :)


The number $ (\frac{\sqrt{x}}{1\plus{}\sqrt{x}}\minus{}\frac{\sqrt{x}}{1\minus{}\sqrt{x}})$ is natural . Find the greatest value of $ x$.

$ (\frac{\sqrt{x}}{1\plus{}\sqrt{x}}\minus{}\frac{\sqrt{x}}{1\minus{}\sqrt{x}}) \equal{} \frac{2x}{x\minus{}1}$.

But $ 2x\equal{}2(x\minus{}1)\plus{}2$ which means that $ x\minus{}1|2$. We have $ x\minus{}1\equal{}1$ or $ x\minus{}1\equal{}2$ which means $ x\equal{}2$ or $ x\equal{}3$. Since $ 3>2$ and $ x\equal{}3$ satisfies the problem's conditions, we are done.


integration
function
calculus
derivative
algebra
rational function
probability and stats

let $ U$ be a positive valued random variable (its a sojourn time) with density $ u(t)$

and $ w(s) \equal{} \int_0^\infty e^{ \minus{} st}u(t)dt$ its laplace transform

im working to discover how a function $ \xi(s)$ approaching its zero at $ s_0$ and i have the equation

$ \xi(s) \plus{} 1 \minus{} s \equal{} w(\xi(s))$

so its there a nice form for each of $ \xi^{(n)}(s_0)$?

for instance.. $ \xi'(s) \minus{} 1 \equal{} w'(\xi(s))\xi'(s)$ so $ \xi'(s) \equal{} \frac {1}{1 \minus{} w'(\xi(s))}$
and since $ \xi(s_0) \equal{} 0$, $ \xi'(s_0) \equal{} \frac {1}{1 \minus{} w'(0)} \equal{} \frac {1}{1 \plus{} E[U]}$ which is nice
then $ \xi''(s) \equal{} \frac {w''(\xi(s))\xi'(s)}{(1 \minus{} w'(\xi(s)))^2} \equal{} [\xi'(s)]^3w''(\xi(s))$ gives $ \xi''(s_0) \equal{} \frac {E[U^2]}{(1 \plus{} E[U])^3}$

so by repeating this, we will get each derivative as some rational function of different moments of $ U$, i am just hoping for a nice formula for them? if there is one..

ps i would also love it if someone could point me in the direction of finding out these values of the moments for a simple random walk in $ Z^d$

Somehow I'm getting that $ s_0 \equal{} 1$. Look, $ w(0) \equal{} 1$, because $ u$ is a pdf, so putting $ s_0$ into your equation for $ \xi$ leads to $ 1 \minus{} s_0 \equal{} 0$.

$ w(0) \equal{} 1 \minus{} s_0$ actually.

so $ s_0\equal{}P(U\equal{}\infty)$ is the escape possibility..

So $ U$ can take infinite values, I see.

Anyway, the problem is just to find the derivatives of the function inverse to
\[ f(x)\equal{}x\minus{}w(x)\plus{}1.\]
There are no nice formulae, however.


Does anyone know where I can find past contests and solutions to ARMLs? I'v tried their site but I only got the 2007 contest there.

Thanks[/i]

Well, you can change the URL accordingly and get the 2006 and 2005 contests as well (maybe 2004 and 2003?). But for the other ones, you have to buy the books (or ask your coach to let you borrow their copy, if they have one), there's one from 1989-1994, one from 1995-2003 (this is offered as USAMTS prize, but will arrive like the day after you leave for ARML probably), and one from some time  before 1989.

Anyone know where to order the green book (I know the blue book is on amazon, but the others are out of print)? The link from the arml.com website redirects to a site which doesn't seem to have the arml books...

05,06,07,92 are available online. I also found the 04 individuals...somewhere...

Google Books gives 89,90 without solutions.

You can find other problems in the various marathons here on aops, like joml and towerfreak's "Arml Practice Series". Just search for it.

Otherwise, you can look around at other arml team websites for their practice materials. *cough*

[quote="Walk Around The River"][quote="CatalystOfNostalgia"]Well, you can change the URL accordingly and get the 2006 and 2005 contests [/quote]

Is it just me, or does that not work anymore?
I need the 2006 answers.[/quote]
All of 2004, 2005, 2006: [hide="Enjoy."]
[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Individuals.pdf]2004 Individual[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Individuals.pdf]2005 Individual[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Individuals.pdf]2006 Individual[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Individual%20Solutions.pdf]2004 Individual Solutions[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Individual%20Solutions.pdf]2005 Individual Solutions[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Individual%20Solutions.pdf]2006 Individual Solutions[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Team.pdf]2004 Team[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Team.pdf]2005 Team[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Team.pdf]2006 Team[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Team%20Solutions.pdf]2004 Team Solutions[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Team%20Solutions.pdf]2005 Team Solutions[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Team%20Solutions.pdf]2006 Team Solutions[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Relays.pdf]2004 Relay[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Relays.pdf]2005 Relay[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Relays.pdf]2006 Relay[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Relay%20Solutions.pdf]2004 Relay Solutions[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Relay%20Solutions.pdf]2005 Relay Solutions[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Relay%20Solutions.pdf]2006 Relay Solutions[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Super%20Relay.pdf]2004 Super Relay[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Super%20Relay.pdf]2005 Super Relay[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Super%20Relay.pdf]2006 Super Relay[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Super%20Relay%20Solutions.pdf]2004 Super Relay Solutions[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Super%20Relay%20Solutions.pdf]2005 Super Relay Solutions[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Super%20Relay%20Solutions.pdf]2006 Super Relay Solutions[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Tiebreaker%20Problems.pdf]2004 Tiebreaker[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Tiebreaker%20Problems.pdf]2005 Tiebreaker[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Tiebreaker%20Problems.pdf]2006 Tiebreaker[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Tiebreaker%20Solutions.pdf]2004 Tiebreaker Solutions[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Tiebreaker%20Solutions.pdf]2005 Tiebreaker Solutions[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Tiebreaker%20Solutions.pdf]2006 Tiebreaker Solutions[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Power%20Question.pdf]2004 Power Question[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Power%20Question.pdf]2005 Power Question[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Power%20Question.pdf]2006 Power Question[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20PQ%20Solution.pdf]2004 Power Question Solutions[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20PQ%20Solution.pdf]2005 Power Question Solutions[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20PQ%20Solutions.pdf]2006 Power Question Solutions[/url][/hide]

[quote="pianoforte"]thanks

04 pq sols is dead[/quote]
ok, fixed.  They aren't very consistent in their URLs  :P

The list has been updated for 2004 - 2008.  All of these problems are on the official ARML website.

[hide="ARML Contests 2004 - 2008"]
[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Individuals.pdf]2004 Individual[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Individuals.pdf]2005 Individual[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Individuals.pdf]2006 Individual[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Individual.pdf]2007 Individual[/url]
[url=http://arml.com/2008%20Contest/ARML%202008%20Individual.pdf]2008 Individual[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Individual%20Solutions.pdf]2004 Individual Solutions[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Individual%20Solutions.pdf]2005 Individual Solutions[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Individual%20Solutions.pdf]2006 Individual Solutions[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Individual%20Solutions.pdf]2007 Individual Solutions[/url]
[url=http://arml.com/2008%20Contest/ARML%202007%20Individual%20Solutions.pdf]2008 Individual Solutions[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Team.pdf]2004 Team[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Team.pdf]2005 Team[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Team.pdf]2006 Team[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Team.pdf]2007 Team[/url]
[url=http://arml.com/2008%20Contest/ARML%202008%20Team.pdf]2008 Team[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Team%20Solutions.pdf]2004 Team Solutions[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Team%20Solutions.pdf]2005 Team Solutions[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Team%20Solutions.pdf]2006 Team Solutions[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Team%20Solutions.pdf]2007 Team Solutions[/url]
[url=http://arml.com/2008%20Contest/ARML%202008%20Team%20Solutions.pdf]2008 Team Solutions[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Relays.pdf]2004 Relay[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Relays.pdf]2005 Relay[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Relays.pdf]2006 Relay[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Relays.pdf]2007 Relay[/url]
[url=http://arml.com/2008%20Contest/ARML%202008%20Relays.pdf]2008 Relay[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Relay%20Solutions.pdf]2004 Relay Solutions[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Relay%20Solutions.pdf]2005 Relay Solutions[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Relay%20Solutions.pdf]2006 Relay Solutions[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Relay%20Solutions.pdf]2007 Relay Solutions[/url]
[url=http://arml.com/2008%20Contest/ARML%202008%20Relay%20Solutions.pdf]2007 Relay Solutions[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Super%20Relay.pdf]2004 Super Relay[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Super%20Relay.pdf]2005 Super Relay[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Super%20Relay.pdf]2006 Super Relay[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Super%20Relay.pdf]2007 Super Relay[/url]
[url=http://arml.com/2008%20Contest/ARML%202008%20Super%20Relay.pdf]2008 Super Relay[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Super%20Relay%20Solutions.pdf]2004 Super Relay Solutions[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Super%20Relay%20Solutions.pdf]2005 Super Relay Solutions[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Super%20Relay%20Solutions.pdf]2006 Super Relay Solutions[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Super%20Relay%20Solutions.pdf]2007 Super Relay Solutions[/url]
[url=http://arml.com/2008%20Contest/ARML%202008%20Super%20Relay%20Solutions.pdf]2008 Super Relay Solutions[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Tiebreaker%20Problems.pdf]2004 Tiebreaker[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Tiebreaker%20Problems.pdf]2005 Tiebreaker[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Tiebreaker%20Problems.pdf]2006 Tiebreaker[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Tiebreakers.pdf]2007 Tiebreaker[/url]
[url=http://arml.com/2008%20Contest/ARML%202008%20Tiebreakers.pdf]2008 Tiebreaker[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Tiebreaker%20Solutions.pdf]2004 Tiebreaker Solutions[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Tiebreaker%20Solutions.pdf]2005 Tiebreaker Solutions[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Tiebreaker%20Solutions.pdf]2006 Tiebreaker Solutions[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Tiebreaker%20Solutions.pdf]2007 Tiebreaker Solutions[/url]
[url=http://arml.com/2008%20Contest/ARML%202008%20Tiebreaker%20Solutions.pdf]2008 Tiebreaker Solutions[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20Power%20Question.pdf]2004 Power Question[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20Power%20Question.pdf]2005 Power Question[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20Power%20Question.pdf]2006 Power Question[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Power%20Question.pdf]2007 Power Question[/url]
[url=http://arml.com/2008%20Contest/ARML%202008%20Power%20Question.pdf]2008 Power Question[/url]

[url=http://www.arml.com/2004%20Contest/2004%20ARML%20PQ%20Solution.pdf]2004 Power Question Solutions[/url]
[url=http://www.arml.com/2005%20Contest/2005%20ARML%20PQ%20Solution.pdf]2005 Power Question Solutions[/url]
[url=http://www.arml.com/2006%20Contest/2006%20ARML%20PQ%20Solutions.pdf]2006 Power Question Solutions[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Power%20Question%20Solutions.pdf]2007 Power Question Solutions[/url]
[url=http://arml.com/2007%20Contest/ARML%202007%20Power%20Question%20Solutions.pdf]2008 Power Question Solutions[/url]
[/hide]

The URLs for 2008 doesn't work. Here is 2008. (This is easy to find, since they have a link on ARML website)

[url=http://www.arml.com/2008_Contest/ARML_2008_Individual_Questions.pdf]2008 Individual[/url]
[url=http://www.arml.com/2008_Contest/ARML_2008_Individual_Questions_Solutions.pdf]Solutions[/url]

[url=http://www.arml.com/2008_Contest/ARML_2008_Tiebreaker.pdf]2008 Tiebreaker[/url]
[url=http://www.arml.com/2008_Contest/ARML_2008_Tiebreaker_Solutions.pdf]Solutions[/url]

[url=http://www.arml.com/2008_Contest/ARML_2008_Team.pdf]2008 Team[/url]
[url=http://www.arml.com/2008_Contest/ARML_2008_Team_solutions.pdf.pdf]Solutions[/url]

[url=http://www.arml.com/2008_Contest/ARML_2008_PQ.pdf]2008 Power[/url]
[url=http://www.arml.com/2008_Contest/ARML_2008_PQ_Solutions.pdf]Solutions[/url]

[url=http://www.arml.com/2008_Contest/ARML_2008_Relay.pdf]2008 Relay[/url]
[url=http://www.arml.com/2008_Contest/ARML_2008_Relay_Solutions.pdf]Solutions[/url]

[url=http://www.arml.com/2008_Contest/ARML_2008_Relay_Solutions.pdf]2008 Super Relay[/url]
[url=http://www.arml.com/2008_Contest/ARML_2008_Super_Relay%20_%20Solutions.pdf]Solutions[/url]

Hey, it looks like these links are dead.

The ARML people in general are very careful with their problems - for example, they won't let us post them on AoPS.

Yeah, those links dont work :(

Do we have to buy them or something:?:

They probably took those pages down now that there's a book out with 2004-2008 contests.;


I NEED HELP ON CHOOSING WHAT TO MAJOR IN, IF YOU WOULD PLEASE TAKE THE TIME TO TALK WITH ME AND HELP ME DECIDE I WOULD REALLY ENJOY THAT.  EMAIL ME @ ballerkidd4@yahoo.com

PLEASE AND THANK YOU

My first friendly suggestion is to make sure your online posts aren't posted in all capital letters, because ALL CAPS LOOKS LIKE SHOUTING to most people in online discussion.

My second friendly suggestion would be to leave this discussion of what major to choose on the public board, where we can all weigh in with advice based on what we know about you--which is next to nothing yet, but maybe you can help us out with that too.

Welcome to the AoPS site.

Yes - it would be helpful if you could state your general fields of interest and expertise so that more people could contribute their thoughts.

Also, how old are you?  Most colleges don't expect you to choose a major until sophmore year.  You shouldn't really need to think very hard about it until you actually start college, at worst.

You know, his caps suggested that it was urgent.

Something tells me he's gone for good.

[quote="Magnara"]You know, his caps suggested that it was urgent.[/quote]

Then maybe JBL's reply helped him the most. Declaring a major isn't an irreversible decision at most colleges, and doesn't need to be done until one is in college.


Let $ a, b, c$ be positive real numbers. Prove or disprove that

$ \sum\frac {a^2 \plus{} ab \plus{} b^2}{c^2 \plus{} ab}\ge \frac {21}{4} \minus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{4abc}$.

_______
emoticons

It's obviously trues because:
$ a^3\plus{}b^3\plus{}c^3 \geq\ 3abc$
and: $ \sum\ \frac{(a\plus{}b)^2}{ab\plus{}c^2} \geq\ 6$
 :)

[quote="nguoivn"]It's obviously trues because:
$ a^3 \plus{} b^3 \plus{} c^3 \geq\ 3abc$
and: $ \sum\ \frac {(a \plus{} b)^2}{ab \plus{} c^2} \geq\ 6$
 :)[/quote]

You want to write some passage. The method is good.
Thanks

But ,with this accounting items,solutions than I


There are $n$ red and $n$ blue points on a line. Prove that there is a point $M$ on that line such that the sum of the distances from $M$ to the blue points is equal to the sum of the distances from $M$ to the red points. Is this statement still true if the points are arbitrarily given on the plane?

[hide="answer"]Let $f(x)$ be the sum distance to the blue points minus the sum distance to the red points. It is a continuous function because it is the sum of continuous $|x-p|$ functions. Let $b$ be the average of all blue points, and let $r$ be the average of all red points. If $b<r$ then $f$ must be less then $0$ at the leftmost point and greater then $0$ at the rightmost point. Reverse this if $b>r$. If $b=r$ then $f$ must be $0$ at the leftmost point. If there is both a positive and a negative there must be a zero in between.[/hide]


2n white and 2n black balls are arranged in a line, prove some line of 2n balls contains exactly n white balls.

Consider the number of white balls within any given line of 2n balls. Now, consider the effect of moving from one line to the next (i.e. from 1, 2, ..., 2n to 2, 3, ..., 2n + 1). The number of white balls either stays constant or changes by 1. Thus, assume this value is never equal to n; if it is initially greater than n, it must forever remain greater than n, so just add this result for the first line of 2n balls and the last line of 2n balls to obtain a contradiction. A similar argument works for the "less than" case.

that's very easy !! let's solve it with invariance principle !!


1. Find all positive integers x,y,z satisfying:
1! + 2! + ... + y! = x^z

or 1! + 2! + ... + y! = y^z
sorry I can't remember it.

2. Find all positive integers m,n satisfying:
(m+1)! - 1 = m^n

Sorry for my english.

Edit:
2. Sorry, I was thinking about
(m-1)! - 1=m^n

2, if m is prime number , we use fermat's theorem
    if m isn't prime number , let p be the less prime divisor of m


For number 2, substitute $b_{n}=a_{n}-n$ and it reduces to number 1.


hi,
calculate the product

\[ P \equal{} (1 \plus{} tg1^{\circ})(1 \plus{} tg2^{\circ}) ... (1 \plus{} tg44^{\circ})\]

$ 1\plus{}tg x\equal{}\frac{\sqrt 2\sin{(x\plus{}\frac{\pi}{4})}}{\cos x}\equal{}\frac{cos(\frac{\pi}{4}\minus{}x)}{cos x}.$
Therefore $ \prod_{i\equal{}1}^{44}(1\plus{}tg i^0)\equal{}2^{22}.$


inequalities
function
inequalities proposed

[quote="dmitin"]
2) $ \frac {a}{2a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {b}{2b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac {c}{2c^2 \plus{} a^2 \plus{} b^2}\le\frac {9}{4(a \plus{} b \plus{} c)}$, $ a,b,c > 0$[/quote]

Let $ S \equal{} a^2 \plus{} b^2 \plus{} c^2$. Notice we can fix $ S \equal{} 1$ since the inequality is homogeneous $ \left(F(ka,kb,kc) \equal{} \frac {1}{k}F(a,b,c)\textrm{ where } F(a,b,c) \equal{} \sum_{\textrm{cyc}}\frac {a}{(S \plus{} a^2)} \minus{} \frac {9}{4(a \plus{} b \plus{} c)}\right)$. Therefor we can assume that $ 0 < a,b,c\le 1$.
Let $ f(x) \equal{} \frac {x}{S \plus{} x^2}$. $ f'(x) \equal{} \frac {S \minus{} x^2}{(S \plus{} x^2)^2}$. $ f''(\sqrt {x}) \equal{} \frac { \minus{} (S \plus{} x)^2 \minus{} 2(S \minus{} x)(S \plus{} x)}{(S \plus{} x)^4} \equal{} \minus{} \frac {3S^2 \plus{} 2Sx \minus{} x^2}{(S \plus{} x)^4} < 0$ when $ x\in(0;9S^2)$ which means the function is concave.

$ \Rightarrow \sum_{\textrm{cyc}}\frac {a}{(S \plus{} a^2)} \le 3\frac {\frac {1}{3}(a \plus{} b \plus{} c)}{S \plus{} \frac {1}{9}(a \plus{} b \plus{} c)^2}$ from Karamata (or more exactly its consequence Jensen).
But $ \frac {(a \plus{} b \plus{} c)}{S \plus{} \frac {1}{9}(a \plus{} b \plus{} c)^2}\le$ $ \frac {9}{4(a \plus{} b \plus{} c)}\Leftrightarrow (a \plus{} b \plus{} c)^2\le 3(a^2 \plus{} b^2 \plus{} c^2)$.

[quote="dmitin"]1) $ \frac {a}{3a^2 \plus{} b^2 \plus{} 2ac} \plus{} \frac {b}{3b^2 \plus{} c^2 \plus{} 2ab} \plus{} \frac {c}{3c^2 \plus{} a^2 \plus{} 2bc}\le\frac {3}{2(a \plus{} b \plus{} c)}$, $ a,b,c > 0$

2) $ \frac {a}{2a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {b}{2b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac {c}{2c^2 \plus{} a^2 \plus{} b^2}\le\frac {9}{4(a \plus{} b \plus{} c)}$, $ a,b,c > 0$[/quote]

For the first, by AM-GM
\[ \sum {\frac{a}{{3a^2  \plus{} b^2  \plus{} 2ac}}}  \equal{} \sum {\frac{a}{{\left( {a^2  \plus{} b^2 } \right) \plus{} 2a^2  \plus{} 2ac}}}  \le \sum {\frac{a}{{2a\left( {a \plus{} b \plus{} c} \right)}}}  \equal{} \frac{3}{{2\left( {a \plus{} b \plus{} c} \right)}}
\]
For the second by AM-GM

\[ \sum {\frac{a}{{2a^2  \plus{} b^2  \plus{} c^2 }}}  \equal{} \sum {\frac{a}{{\left( {a^2  \plus{} b^2 } \right) \plus{} \left( {a^2  \plus{} c^2 } \right)}}}  \le \sum {\frac{a}{{\frac{{\left( {a \plus{} c} \right)^2 }}{2} \plus{} \frac{{\left( {a \plus{} b} \right)^2 }}{2}}}}  \le \sum {\frac{a}{{\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}}}  \equal{} \frac{{2\left( {ab \plus{} bc \plus{} ca} \right)}}{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}} \le \frac{9}{{4\left( {a \plus{} b \plus{} c} \right)}}
\]


The Vietnamese team came 4th, with the score of 196 (four golds and two silvers).

there are :
+ Gold medals :
- Pham Kim Hung
- Nguyen Minh Truong
- Le Hung Viet Bao (perfect scorer IMO 2003)
- Nguyen kim Son
+ Silver medals :
- Nguyen Duc Thinh
- Hua Khac Nam


Putnam
number theory unsolved
number theory

we difine $f(n)$ is an non-zero end digit of $n!$. If $a_1,a_2,...,a_k$ are non-negative integer number . Show that $f(5^{a_1}+5^{a_2}+...+5^{a_k})$ is depended on $a_1+a_2+...+a_k$.

[quote]we difine f(n) is an non-zero end digit of n!.[/quote]

Is it an arbitrary digit or the first non-zero digit counting from the end?  :(

For example, what is f(6)?

[quote]al.M.V wrote:Is it an arbitrary digit or the first non-zero digit counting from the end?

[/quote]
It is the first non-zero digit and when $6!=720\rightarrow f(6)=2$

It's from the 1994 Putnam, problem A6: [url=http://www.kalva.demon.co.uk/putnam/putn84.html]www.kalva.demon.co.uk/putnam/putn84.html[/url].

And it is a nice problem.


[b]Let $k$  be a positive integer , $n: =k^2$ , and  $-\infty< x_1 \le x_2 \le ...\le x_n< \infty$   such that $x_n-x_1 = k$ and $x_1 + x_2 + ...+ x_n=0 \; ,\; x_1^2 +x_2^2+ ... +x_n^2 = 10 .$ [/b]   [color=green][b]Prove or disprove that $k=3$  or $k=4$. [/b]  [/color]

Is there something missing in the statement of the problem?

=========================
The case k>=5 is clearly impossible: Then x_n-x_1=k>=5, and hence
(x_n)^2+(x_1)^2 is at least 25/2>10.

=========================
The case k=1 is clearly impossible.

=========================
In the case k=2 we are looking for a>=b>=c>=d with d=a-2,
a+b+c+d=0 and sum of squares =10.

Without loss of generality, we may assume that |a|>=|a-2|
(otherwise replace a,b,c,d by -d,-c,-b,-a).

Then 10=a^2+b^2+c^2+d^2<=a^2+a^2+a^2+(a-2)^2
yields a>= 1.618 and d>=-0.38.  Then a+b+c+d=0 can
never hold true.
=========================

[quote="test20"]Is there something missing in the statement of the problem?
The case k>=5 is clearly impossible: Then $x_n-x_1=k\geq 5,$ and hence       $(x_n)^2+(x_1)^2$ is at least   $25/2>10.$
The case k=1 is clearly impossible. [/quote]
[color=red][b]  I do'nt understand !  Please excuse me !/Proposer. [/b] [/color]

what is it that you don't understand? $x_1^2+x_n^2 \ge \frac{(x_1-x_n)^2}2$ ... that's not a hard inequality?

[quote="Peter VDD"]what is it that you don't understand? [/quote]
In order to complete the solution , please [b]prove  that cases $k=3$ and/or $k=4$ are possible !
Find  at least a system of numbers $\{ x_1,x_2,...x_n\} , n\in \{9,16\}$ which satisfies  the conditions. [/b]

I remember that you posted a problem in algebra section. This is the same kind of it.


2 watches are set at 12:00. One of them drops back 2 minutes each day and the other goes forward 1 minute each day. After how many hours will the watches show a difference of 1 hour?

umm...if im not being completely stupid xD...they get 3 minutes farther apart every day so after 20 days they will be 1 hour apart so 20*24=480...so the answer is $ \boxed{480}$

well in 24 hours the two watches will be 1+2=3 minutes apart
so 60 / 3 = 20
so 20 * 24 = 480

so in 480 hours  :maybe: :)

[hide]
After 24 hours, you have a difference of 3 minutes.
You want 20 times this so the number of hours is 24*20 = 480. [/hide]

480 hours

The message is too small. Please make the message longer before submitting.

So each day, they get separated by 3 minutes more, so after 20 days, they'll have 1 hour difference.
20 x 24 = [u][b]480 hours[/b][/u]

guys, if you actually read the posts, this answer has been given 5 times already
also, most of the solution are exactly the same as well
plz see if the answer and solution u were going to give have already been posted, because if it has, plz don't post again
thanks

Could this problem be solved using algebra?

technically, you could, if you could define a variable and put in correct coefficients and constants

BTW, don't revive old threads
if you are really inspired to talk about something that came up over a month ago, make a new thread


[hide]the radius is 4, where the circumfrence is 8pi and the area is 16pi.[/hide]

[quote="236factorial"]Find the radius of a circle whose circumfrence is half the area.[/quote]

[hide]Set the radius as $x$

$2*\pi*{2x}=\pi{x^2}$

$4x=x^2$

$x=4$[/hide]

[quote="236factorial"]Find the radius of a circle whose circumfrence is half the area.[/quote]
[hide]
r=radius

pi r^2 = 4 pi r
r^2=4 r
r = 4

[/hide]


Dawn has $ \$$1.20 in nickels, dimes and quarters. For each nickel she has one dime and for each dime she has one quarter. How many total coins does she have?

For every nickel she has, she has 1 dime and quarter, so the ratio of n:d:q is 1:1:1. Thus, $ 1.20\equal{}x(0.05\plus{}0.10\plus{}0.25)$, where x is the number of nickels. so $ 120\equal{}40x$, 3 nickels, or a total of 9 coins.


For the following questions you may assume Einstein's two axioms of special relativity, and that Newton's first law of motion implies that any two inertial frames are linked by a linear map in 4-dimensional spacetime, and conversely moving between two such frames leaves Newton's law of motion unchanged. You may further assume (since the derivation is trivial) that an observor in an inertial frame, S, will observe any other inertial frame, S', as being Euclidian with all of it's defining particles (i.e points at rest in S' as observed in S') moving through S with the same velocity.

Problem 1.

Take two distinct (gravity-free) inertial frames: S and S'. The respective origins of the two frames are O and O'. A stationary observor in S observes that O and O' coincide at a point in time. Is it possible that an observor in S' will never see the O and and O' coinciding? Explain your answer.

Problem 2.

For the set up in "Problem 1" suppose that a stationary observor at O in S observes that O' is moving with speed v. What does a stationary observor at O' in S' measure the speed of O as? Explain your answer.

I managed to explain problem 1, but had limited success in tackling problem 2.

It's just v, right? If I'm reading the problem correctly, the observer at O and the one at O' are in symmetric situations.

(Note in Problem 2, the image of O' in S is moving along the x-axis of S)

This is where I got confused. Where does this symmetry come from? Say we are an observor at O in frame S. Now we see that the image of O' in S moves with speed v. So in S, the distance between the two origins increases at speed v, so their mutual velocity in this frame is v. So it would appear in this frame that if the observor is stationary, then O moves relative to the image of O' with speed v. But this is not what we are trying to aim for. What we want to know, is what a stationary observor at O' in S' observes the speed of the image of O in S' as.

So mathematically what I wan't to do, is to take is a linear map S to S' (between two inertial frames). Then I want to show under this map that if the preimage of a fixed point in space (not in time) in S' moves with a certain speed in S, then the image of a fixed point in space (not in time) in S moves with the same certain speed in S'.


Prove that the equation $ (x\minus{}y)^3\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^3\equal{}2006$ has no integer solutions . But the equation $ (x\minus{}y)^3\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^3\equal{}2007$ ?

Let $ x\minus{}y\equal{}a$ and $ z\minus{}x\equal{}b$. Then, $ (y\minus{}z)^2\equal{}(a\plus{}b)^2$.

Our equation becomes $ a^3 \plus{} (a\plus{}b)^2 \plus{} b^3 \equal{} 2006$ in the first case, and $ a^3 \plus{} (a\plus{}b)^2 \plus{} b^3 \equal{} 2007$ in the second case.

Notice that $ a^3 \plus{} b^3$ and $ a\plus{}b$ have the same parity, so the second equation can't have integer solutions since $ a^3 \plus{} (a\plus{}b)^2 \plus{} b^3$ is always even.

As for the first one, I don't know yet...

Substitute as you $ a\equal{}x\minus{}y\ ,\ b\equal{}y\minus{}z\ ,\ c\equal{}z\minus{}x$ we have $ a\plus{}b\plus{}c\equal{}0 \Rightarrow a^3\plus{}b^3\plus{}c^3\equal{}3abc$ . Here we can't use AM-GM do we'll say that the first equation is equivalent with $ 3(x\minus{}y)(y\minus{}z)(z\minus{}x)\equal{}2006$ , which has no integer solution since $ 3 \nmid 2006$ .

The second equation becomes $ (x\minus{}y)(y\minus{}z)(z\minus{}x)\equal{}669\equal{}3\cdot 13\cdot 17$ , which is impossible in integers since their sum is zhero .

[quote="facis"]alex: you misread the question.  One of the quantities is squared, not cubed.[/quote]

Obviously one of my typo  :D

Just wondering would $ (x\minus{}y)^{3}\plus{}(y\minus{}z)^{2}\plus{}(z\minus{}x)^{3}\equal{}2006$ have any solutions in integers?


algebra
polynomial
number theory unsolved
number theory

Assume WLOG x <= y <= z. Assume (x1,y1,z1,t1) is a solution to the system txyz = x^2 + y^2 + z^2 (which is our equation rewritten). Put t = t1, x = x1, y = y1 and view z^2 - (t1)*(x1)*(y1)z + (y1)^2 + (x1)^2 = 0 as a polynomial in z. Let z1 and z2 be it's roots. Then we have z1 + z2 = t1*x1*y1 and z1*z2 = (x1)^2 + (y1)^2. So z2 = t1*x1*y1 - z1 is also a root to that equation (easy to check) and therefore (x1,y1,z2,t1) is also a solution to our original equation. So we can assume z <= 0.5*txy. From this we get that t <= x/(yz) + y/(xz) + 0.5t => 0.5t <= 2 => t <= 4. Now you can probably continue from here...

You could modify the statement so that t= (x^2 + y^2 + z^2)/xyz. You can probably solve it from here.

[quote="Fitim"]If $x,y,z,t$ are natural numbers and $t=\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}$. Prove that $t=1$ or $t=3$[/quote]
My solution:Firstly,it easy to prove $x^{2}+y^{2}+z^{2}-2kxyz=0$have no solution except for $(0,0,0,0)$ because if $(x,y,z,k)$ is a soluiton,then $(\frac x2,\frac y2,\frac z2 ,2k)$is also one.
secondly, if $(x,y,z,t)$is a solution
of$x^{2}+y^{2}+z^{2}-xyzt=0$ then $(yzt-x,y,z,t)$ also is . so we can
suppose $z\leq y\leq x\leq \frac12yzt$ then $t=\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}\leq \frac{t}{2}+\frac1z+\frac1y\leq \frac{t}{2}+2$
so $t\leq 4$
so $t=1,3$


algorithm
number theory unsolved
number theory

I was doing a computer algorithm and then I came across this problem. How to find the maximum values of a and |b| in order to cover all numbers with 3 digits?
$ k \equal{} a^{2}\plus{}b^{3}$. For example if it was $ k \equal{} a^{3}\plus{}b^{3}$ I set $ a \equal{} 19$ and $ b \equal{}\minus{}18$ then $ k \equal{} 1027$. Hence it covers all numbers with 3 digits. In fact, I don't know if this limit exists for this problem in particular.

Be careful not all $ 3$ digits numbers can be written as $ a^{2}\plus{}b^{3}$ or $ a^{3}\plus{}b^{3}$ with $ a,b\in\mathbb Z^{2}$

Yes. I know it can't be written. But I want to find only those that can be written in the way I proposed.


Να υπολογιστει το ολοκληρωμα

$ \int \frac {x^2 e^{Arctanx}}{\sqrt {1 \plus{} x^2}}dx$

  :)

Νομίζω το 'χε προτείνει παλαιότερα ο φίλος Γιώργος Δέσκος.

Λύση είχε δοθεί από τον Καραμιχάλη Ράλλη.

Το ολοκληρωμα αυτο εχει πεσει στο τελικο γυρο της μαθηματικης ολυμπιαδας της Ρουμανιας το 1975.

Αν θελετε μπορω να γραψω την λυση.

Με έκανες και σκάλισα το φόρουμ, αλλά χαλάλι ! Η λύση επίσης από τον κύριο Γαλάνη.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131274

Για τους παλιούς μάλλον η άσκηση είναι ήδη γνωστή!

Στέλιος

:oops: sorry δεν το ειχα δει...οποτε το post μου μπορει να διαγραφει.

Eγώ πάλι σκέφτηκα μία άλλη λύση (αν και δε διάβασα όλες τις άλλες λύσεις του παλιότερου topic, αν είναι κάποια από αυτές sorry...)

Μετά από λίγες δοκιμές, μου βγήκε ότι βολεύει να ψάχνω για λύση της μορφής: $ P(x)\sqrt {1 \plus{} x^2}e^{\arctan x}$ όπου $ P$ πολυώνυμο.

Παραγώγιση οδηγεί στο ότι το $ P$ πρέπει να ικανοποιεί την: $ P'(x)(1 \plus{} x^2) \plus{} (x \plus{} 1)P(x) \equal{} x^2$.

Λύση πανεύκολη... και προκύπτει το ζητούμενο $ P(x) \equal{} \frac{1}{2}(x\minus{}1)$ :)

Το καλό είναι ότι έτσι μπορούμε να βγάλουμε το ολοκλήρωμα για οποιοδήποτε (σχεδόν) πολυώνυμο στον αριθμητή αντί του $ x^2$, γενικεύοντας έτσι αρκετά την άσκηση.

Cheerio,

Durandal 1707

Νομίζω έτσι την είχε λύσει ο Ράλλης

Ο Παναγιώτης(Durandal) παλιά μου είχε δώσει την παρακάτω άσκηση που βρίσκω σχετική για εδώ: να υπολογιστεί το$ \int \frac{x\minus{}1\plus{}sinx}{2x\plus{}sinx\plus{}cosx\plus{}e^x}\,dx$

An de kanw la8os prepei na einai
\[ \frac{1}{2}( x\minus{}ln|2x\plus{}sinx\plus{}cosx\plus{}e^x|) \plus{}c\]

[quote="mailo"][quote="Durandal"]
Λύση πανεύκολη... και προκύπτει το ζητούμενο $ P(x) \equal{} \frac {1}{2}(x \minus{} 1)$ :)
[/quote]

αν μπορεις πες μου πως το βρηκες αυτο.  :oops:[/quote]

Δύο τρόποι: ο γενικός είναι να λύσεις τη διαφορική (πχ με ολοκληρωτικό παράγοντα... bleh!). Ο δεύτερος τρόπος (που είναι και αυτό που έκανα) είναι να θυμηθείς ότι το P είναι πολυώνυμο. Βαθμός max 1 λόγω $ x^2$ δεξιά και... τέλος  :)

Cheerio,

Durandal 1707

Wow, ούτε που το θυμόμουν ότι είχα φτιάξει τέτοια άσκηση! Καλά, ντροπή μου... thanks for the memory trip good sir! :)

[quote="Linger"]An de kanw la8os prepei na einai
\[ \frac {1}{2}( x \minus{} ln|2x \plus{} sinx \plus{} cosx \plus{} e^x|) \plus{} c
\]
[/quote]

Yup, σωστό είναι... πώς οδηγήθηκες στο αποτέλεσμα είναι όμως το θέμα ;)

Cheerio,

Durandal 1707

apla paratirisa oti o ari8mitis einai sthn ousia o paronomasths meiwn thn paragwgo tou dia 2. Opote e8esa $ f(x) \equal{} 2x \plus{} sinx \plus{} cosx \plus{} e^x$ kai elysa to $ \int \frac {f(x) \minus{} f'(x)}{2f(x)}dx$ to opoio kataligei sto apotelesma pou edwsa prin

Ωραιος!  :)

Durandal, την διαφορικη εξισωση [img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/1/5/8/158626addf02a3ecc0c9581a9e5e23af4825d474.gif[/img] πως την λυνουμε, γιατι προσπαθησα και δεν μου βγαινει  :oops:

Meeeh... έχεις δίκιο, αν πας με ολοκληρωτικό παράγοντα και πολ/σεις με $ \sqrt {1 \plus{} x^2}e^{\arctan x}$ για να βγάλεις ένα τέλειο διαφορικό στην LHS, θα καταλήξεις στο αρχικό ολοκλήρωμα, οπότε δεν βολεύει. Και λύνοντας την ομογενή πρώτα, πάλι στο ίδιο ολ/μα θα καταλήξεις, οπότε nevermind.

Hmmm... πιθανώς οι γενικές μέθοδοι να σε γυρνάνε πίσω στο αρχικό ολοκλήρωμα όλες (να σου πω την αλήθεια βαριέμαι να τις κοιτάξω όλες). Αλλά, αφού έχεις πολυώνυμα σαν συντελεστές, δεν είναι τόσο παράλογο να αναζητήσεις πολυωνυμική λύση και έτσι βγαίνει σε δευτερόλεπτα. Και, καθώς είναι και πρωτοβάθμια, όποια λύση και να βρεις θα είναι μοναδική (up to a constant), οπότε τελέιωσες :)

Τι να σου πω... αν σκεφτώ κάτι πιο standard θα το γράψω.

Cheerio,

Durandal 1707


This proof is actually kind of simple, but it uses a theorom that I'm not sure that many Getting Started people would know, so here it is -

Power of a Point - This basically says that if you have two chords in a circle, AB and CD, that intersect at P, then ${AP}\cdot {BP} = {CP}\cdot {DP}$.  It also includes other things, but that isn't really needed for this.

And the proof -

Let's say you have two chords in the same circle that intersect.  Proove that the triangles formed by connecting the endpoints of the two circles are similar.

[hide="Hint"]Think about similar triangles.[/hide]

[hide="another hint"]
think about how we could say that the corresponding angles are the same...
[/hide]

i saw the proof of this one in the Elements but i don't remember it...
so can you post it?

I have to go right now so I'll post the proof I got later, but think about the congruency of vertical angles and the congruency of angles who subtend equal arcs.

P.S. Is congurency a word?

I believe congruency is a word.

And there's another way to proove it, without using power of a point.

OOO I thought we were supposed to prove the power of a point theorem.  The second proof is essentially the same as proving power of a point though.

No.  We are trying to prove the triangles similar not congruent.  Heres the proof:

[hide]Call the intersection of the traiangles P, call the vertices of the top triangle from left to right A and B.  Call the vertices of the bottom triangles C and D going from left to right.  So angle APB is congruent to angle CPD also we see the angles ABP and CDP are inscribed angles who subtend the same arc AC.  So angle ABP is congruent to angle CDP.  Since you have proved two of the angles in both triangles congruent the triangles are similar.  There is another way to do it using the Power of a Point theorem, if someone wants that I'll post it later.[/hide]

BTW, how do you get the sections you hide to be labeled something other than "Click Here to reveal hidden content"

If you want it to say My solution here or something you would have:

[Hide=My Solution Here]

.
.
.

[/Hide]

[quote="mna851"]No.  We are trying to prove the triangles similar not congruent.[/quote]

Actually, there is a way called side-angle-side to proove triangles congruent.  Basically, if two sides are proportional, and the angle in between them are congruent, then they are similar.  There's also side-side-side, where all the sides are proportional, and angle-angle where the angles are congruent.

[hide="Hint"]
Use side-angle-side to proove this with power of a point[/hide]

[hide]Just notice that the angles of the two triangles intercept the same arcs.[/hide]


There is an algebraic solution for this.
If an echiangular polygon (equal angles) with $n$ sides has the property that $a_{0}\geq a_{1}\geq a_{2}\geq ... \geq a_{n-1}$ where $a_{i},i\in \{0,1,...,n-1\}$ are the lengths of the sides then the polygon is regular.
I appologise if it was posted before.


Who is doing ARML this year?  I'm not, I have to watch my brother's graduation.

BTW, the "z" in the title is just to appease the lowercase rule in the title.

I don't know if my dad would feel like driving down so many times...

dude I should like join
I failed teh OCTM though so I'm not sure if I'll get in

I also failed the AMC massively

What, you got a 138?  I got that too and made the top ARML team last year!  You sure didn't fail. :P

[quote="#H34N1"]Probably. I failed OCTM so I just got in with AMC 10.
the OCTM is ridiculous in that I made so many stupiders[/quote]

well yeah there was so little time, so people who make mistakes fail..
and the desks are pathetically small at OSU
i had to put my calculator on the desk next to me, the problems on my lap, my answer sheet on my desk.
lol

wait so are the invitations out now?

I don't know. We had a really small space too


i was just wondering cuz apparently the largest named number is googol...i wa wondering what the smallest is

on a related note, there was a game where you are given 10 minutes to write a description of a number on a piece of paper. No references to the other name allowed (ie what my opponent choose +1), and whoever has the higher number wins. How do you win the game? btw, a googolplex is bigger than a googol.

Oh I just remember, there is a number even bigger, but I cannot remember what it was. It had to introuduce new notation.

yer i have the googleplex thing

but we needed to get the largest adn smallest...and i couldnt find anytihng about the smallest on google.com ...although now i know y its called google.com

The number is not "google", it's "googol".

Actual answer: YOU ARE ALL WRONG!

There is no largest number. DUHHHHH.  ;)

how about aleph_aleph_..., as many alephs as aleph_aleph..., which that number as many alephs as... etc.

If your talking about the smallest number with a name that would be "yotta". I'm pretty sure I my be confusing it with its counterpart tho'.

So, it's important to make a distinctions here: what type of numbers are you talking about?  If you mean positive integers, for example, the smallest positive integer is easy to find.

In general though, if you are talking about positive real numbers, the answer is "there is no smallest or largest number."  It's not too hard to see why: say you had a number that you claimed was the largest number.  I could always find a larger one, for example by adding 1.  So, that means that your number couldn't have been the largest.  Thus, there cannot be a largest number within our system.  Googol and googolplex are just very large numbers -- but there are plenty of numbers bigger than either of them.  Similarly, if you claimed you had the smallest (positive) number, I could simply divide it in half to get an even smaller number.

Oh, and for the record, if anyone tells you that "infinity" is the largest number:
(1) Infinity is not a number in the same sense that we are talking about.
(2) Ask them "Which infinity?"

kueh -- [hide]before you start taking aleph_aleph_..., you have to define the expression $\aleph_x$ where $x$ is not $\in \mathbb{Z}^{\geq 0}$.[/hide]

Ah, multiple infinities... we were having a seminar on that in one of our math meetings. I didn't really get everything, but it was an interesting seminar.

wait a second..........

before I say something, what's so important about knowing the smallest number in the world? :?

[quote="roxy_foxy_218"]...I was wondering what the smallest is?[/quote]
Depending how you are defining "smallest", some people would say the smallest number is named "zero".  Others might think "negative googolplex" is a very small number.

Largest and smallest numbers are an interesting topic of discussion for students of all abilities.  But let's not get too carried away with high-level topics.  For those who haven't seen it before, the aleph symbol ($\aleph$) is used in a mathematically precise (and rather advanced) way to describe "infinities" of different sizes.  For further discussion on infinities or $\aleph_0$, I would suggest those who are interested inquire in the "Intermediate" forum.

my sister and i are always saying... how can there be an 'infinite number' of something if numbers are finate!!  but also, a confusing thought to ponder: subinfinity! if there is an infinate amount of numbers from 1 onward, how is it infinity if there is also an infinate amount of numbers between 1 and 2?
also, dividing by zero.  its undefined, right?  just thinking, though, dividing by 1 tells you how many times 1 can go into a number, so dividing by zero is saying how many times you can put zero into a number.  technically you can put zeroes together infinately and never get anywhere, so is it undefined because dividing by zero it is infinate, or because zero is nothing and doesn't exist?

a quote my dad likes: "everything is nothing, and it is all the same"

To be exact, $\frac{x}{0}$ is undefined. And so is $0^0$.

Actually you can compare numbers. If you consider, how do you show that 3 apples is the same as 3 oranges? You pair off the apples and oranges. So since everything is paired and nothing is left out, they are the same. It is the same with infinities. $(0,1)$ and $(1, \inf)$ are the same because you can pair $x$ with $\frac{1}{x}$.

Hope that clears some things up.

yes that kind of helps :) but kind of odd to think that one infinity is more specific, like just whole numbers, and then another infinity is whole numbers and decimals between those whole numbers... kind of makes it seem like one infinity has more in it than another

grr must get out of linear thinking!!!

There Is no SMALLEST number! The closest you can get is negative infinity, but that is not a number, just an idea. The biggest number is not googol. There is no biggest number either.  :D

ooo negative infinity... my sisters going to love that one  :lol:

[quote="Magnara"]$.\bar{0}1$

 :P[/quote]

which limit is 0, but neverreaches it. so $.\bar{0}1=0<.\bar{0}1$  :D

[quote="Magnara"]$.\bar{0}1$

 :P[/quote]
I think she means the negative infinity, not the smallest positive rational number...

Actually, according to dilbert(!) the name of the biggest number is the frogopoopillion...

Yeah, but are we not talking about [b]named[/b] numbers?

Anyways, for that game, you could always say: Largest mersenne prime or something.


find all prime numbers like $p$ such that : $p|2^p+1$

[hide]If $p=2$, $2|5$ which is false.
If $p\neq 2$, $p|2^{p-1}-1$ (from Fermat). Then $p|2^p-2$, so $p|3$. Then $p=3$.[/hide]

I didn't follow you Scorpius119  :(

Alternative solution : we have $2^{2p} = 1$ in $\mathbb{Z}/p {\mathbb{Z}}^*$ so the order of $2$ is $> 1$ and divides both $2p$ and $p-1$ : $3 \mid p \Rightarrow p =3$


Of course not. But that can depend on the molecule.

The question that tested this concept was reduction of a simple ketone and I apparently was thinking of sharpless epoxidation instead of LiAlH4 lol, so alright that's good.

LiAlH4 can reduce a COOH all the way to an alcohol, NaBH4 and H2 cant.

[quote="Albert Einstien"]hey is LAH so highly reactive that isnt selective to anything special can it be used for any reduction???? coz other reducing agents like NaBH4 are selective no??? and even H2/Ni is not selective no it is highly reducing in nature??? :maybe:[/quote]

can't really understand what you're saying, but if you're asking if LAH reduces everything, that would clearly be no ... as far as I know, in most cases it does not reduce double or triple bonds, for example

[quote="VARUNARASU"]EINSTEIN PLZ CUT OUT THE NOS IN YOUR POST. IT MAKES COMPREHENDING YOUR POST A LITTLE DIFFICULT[/quote]
ok!!! :D

[quote="Albert Einstien"]yeah ofcourse but redP/HI is a better reducing agent no???[/quote]

Isn't that a set of reagants used in alpha-halogenation of the Hell-Volhard-Zalinsky reaction...where's the redox reaction?

You sure about that? What about PCl3 and PBr3, they can be used too...

In the HVZ reaction we use Cl2 or Br2 together with phosphorus: these reagents can then react to form PCl3 or PBr3.

and P/HI can be used to reduce acids to alkanes. dunno how feasible that is.
and carcul is rite abt the formation of pbr3 or pcl3.
on a side note, nice conceptual question:
predict the product with explanation ofc:
RCH2COOH + Cl2 (in presence of PBr3)->
not for carcul. :)


Suppose that $a,b\in \mathbb{R^+}$ and we define the sequence $(x_n)_{n\geq 0}$ s.t. $x_0=a , x_1=b $ and $\forall n\geq 1, x_{n+1}=\frac {x^2_n+x^2_{n-1}}{x_n+x_{n-1}}$.Prove that this sequence is convergent.

Clearly, the sequence is bounded, since $\frac{x^2+y^2}{x+y}\leq max(x,y)$. Next, observe that $ x_{n+1}\geq\frac{x_n+x_{n-1}}{2}$ and so $ 2 x_{n+1}+x_n$ is increasing and bounded and thus convergent. But we shoed several times that this implies the convergence of our initial sequence. By the way, where are all these problems taken from? I haven't seen them in our olympiads.  :?


The difference between two perfect squares is 133.  What is the
smallest possible sum of the two perfect squares?

the difference between two perfect square x and y is $ x^2\minus{}y^2\equal{}(x\plus{}y)(x\minus{}y)\equal{}133\equal{}19*7$ or $ 133*1$.

But if (x+y)=133 and (x-y)=1, x becomes 67 and y becomes 66.
Not that small, is it?

However, if (x+y)=19 and (x-y)=7, x becomes 13 and y becomes 6.
That's pretty small.

so $ x^2\plus{}y^2\equal{}169\plus{}36\equal{}205$

answer : 205

How come I found that 169 and 36 works too. The question didn't say consecutive squares. So this answer in my opinion is 13 + 3 = 16

[quote="needmorecoffee"]How come I found that 169 and 36 works too. [...] So this answer in my opinion is 13 + 3 = 16[/quote]

@needmorecoffee, we have $\sqrt{169}+\sqrt{36}=13+6=19$.

...the difference between 144 and 9 is 135, which is not 133.


Subsets
Set systems
combinatorics
IMO Shortlist

Let $U=\{1,2,\ldots ,n\}$, where $n\geq 3$. A subset $S$ of $U$ is said to be [i]split[/i] by an arrangement of the elements of $U$ if an element not in $S$ occurs in the arrangement somewhere between two elements of $S$. For example, 13542 splits $\{1,2,3\}$ but not $\{3,4,5\}$. Prove that for any $n-2$ subsets of $U$, each containing at least 2 and at most $n-1$ elements, there is an arrangement of the elements of $U$ which splits all of them.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Since official solution is easy to find,I will post an alternative approach:

[b]Proof[/b]:Of course,we use induction.Base case is trivial

Denote the subsets of $ U$ are $ A_1,A_2,...A_{n\minus{}2}$

Firstly,We can simply assume [u]$ |A_i|\equal{}2$[/u] for all $ i,1\le i \le n\minus{}2$

Because if $ |A_i|>2$ we can choose a $ 2\minus{}element$ subset $ A_i'$ of $ A_i$,and replace $ A_i$ by $ A_i'$

Secondly if every element of $ U$ appears in $ A_1,A_2,...A_{n\minus{}2}$ at least twice

Then $ 2n\minus{}4\equal{}|A_1|\plus{}...\plus{}|A_{n\minus{}2}|\ge 2n$ A contradiction!

This implies one element(WLOG,let it be $ 1$) appears in $ A_1,A_2,...A_{n\minus{}2}$ exactly once or never

So we divided the problem into two cases

[b]Case 1[/b]:$ 1$ doesn't appear in $ A_1,A_2,...A_{n\minus{}2}$

delete $ 1$ from $ U$ and applies induction hypothesis to $ U'$ and $ A_2,...,A_{n\minus{}2}$

We got an arrangement of $ 2,3,...,n$ splits $ A_2,...A_{n\minus{}2}$ and add $ 1$ between the elements of $ A_1$

get our result!

[b]Case 2[/b]:$ 1$ appears in $ A_1,A_2,...A_{n\minus{}2}$ exaxtly once (WLOG,let $ 1$ appear in $ A_1\equal{}\{1,a\}$

delete $ 1$ from $ U$ and applies induction hypothesis to $ U'$ and $ A_2,...,A_{n\minus{}2}$

We got an arrangement of $ 2,3,...,n$ splits $ A_2,...A_{n\minus{}2}$ and add $ 1$ far away from $ a$

get our result!

QED

I don't think you can assume $|A_i|=2$ because if $A'_i$ is split by the arrangement that doesn't necessarily mean $A_i$ is.

We use induction and prove base cases later in the proof.

First, suppose that there is some element in $U$ which isn't contained in the subsets. WLOG let this element be $n$. Suppose that the complement of $n$ (i.e. the set $\{1,2,\dots,n-1\}$) isn't one of the $n-2$ subsets. Then we can delete the element $n$ and note that all of the new subsets satisfy the size condition. Then delete one of these subsets so that there is now $n-3$ subsets. By the inductive hypothesis there is some arrangement of $\{1,2,\dots,n-1\}$ which splits all of these new subsets. Now we can insert the element $n$ into the arrangement so that it splits the subset that we deleted, giving a valid arrangement. The same logic holds when the complement of $n$ is one of the $n-2$ subsets (we can just delete it).

Now, suppose that any element of $U$ is contained in one of the subsets, and suppose that there exists an element such that all subsets containing this element have size at least $3$. WLOG, suppose that this element is $n$. Then if we delete $n$ from all subsets containing $n$ we get $n-2$ subsets, all of which have size at least $2$. Now if the complement of $n$ isn't one of the subsets then we can delete some arbitrary subset and note by the inductive hypothesis that there's some arrangement that splits all of these new subsets. Again, add $n$ back so that the subset we deleted is split by the arrangement (we take cases based on whether the original subset contained $n$), and note that we now have a valid arrangement. We can do the exact same thing if the complement of $n$ was one of the subsets, by deleting that subset.

Finally, suppose that for every element in $U$, there is a subset with size $2$ containing that element. Then if $n\ge 5$, suppose that there's some element (again, WLOG suppose it is $n$) which is only contained by a single subset with size $2$. Then note that we can delete $n$ from all subsets containing $n$, and then delete the complement of $n$ (if it's one of the subsets) as well as the new subset with size $1$. Clearly the size condition is satisfied and there's at most $n-3$ subsets so there's an arrangement that splits all of the new subsets. However, we can also always find a spot to insert $n$ such that it splits both the complement of $n$ and is not adjacent to the element that $n$ was paired up with. This implies the existence of a valid arrangement.

Then for all elements in $U$ there are at least two pairs which contain that element, a contradiction since it implies the existence of at least $n$ pairs. Now when $n=3$ we are immediately done, and finite case check for $n=4$ finishes. $\blacksquare$

(I assumed that subsets are distinct, but this is valid.)

We proceed with induction. If $n=3$ then simply arrange $U$ so that the second entry is the one not in the subset.

Assume the result is true for $n-1$. Call a subset $A$ small if it has two elements and large if it has $n-1$. Let $S$ and $L$ be the number of small and large subsets we need to split. We claim that we can find an element in every large one and less than two small ones. Let $X$ be the set of values in at least two small subsets and $Y$ be the the set of values not in some large subset. Note that \[|X\cup Y| \le |X| + |Y| \le S + L \le n-2\] so there exists at least one value in neither $X$ nor $Y$. Call this element $z$. Then, remove this value from every subset and remove one of the subsets which had $z$. By the induction hypothesis, we can split every remaining subset. Adding $z$ in will not affect any of these subsets. For the entire subset we just removed, we can split it no matter what by putting $z$ in the right place. We are done.


Let $x,\ y$ be real numbers such that $\sin x \cos y =\sin y+\cos x =k.$
Find the maximum and minimum value of $k.$

Regrettably your answer is incorrect. :(

No. :(

I think this problem is difficult as look.

[quote="jimhu"]Yea. i think i found an error in my solution.. is the answer $\sqrt{25/32}$and $-\sqrt{25/32}$[/quote]

make sure you show that there are some x and y such that your min/max occur

[quote="boxedexe"]You want to find the minimum/maximum of $f(x,y)=\sin x+\cos y$ such that $(\sin x-1)(\cos y-1)=1$.  This may be helpful.

Masoud Zargar[/quote]

Can you prove that?  $(\sin x-1)(\cos y-1)=1$

Anyways, can we use calculus. Ah, it makes life so much easier.
plzzzzz :D

$\sin x\cos y=k, \sin x+\cos y=k$.  Subtracting the first from the second, and adding one gives you the result.  I think that Calculus should do the job.

Masoud Zargar

Your answer is correct, pontios. :)
Now I have 3 solutions including yours.

[quote="neelnal"]Can you prove that?  $(\sin x-1)(\cos y-1)=1$[/quote]

That is an equation, not an identity.  That is, $x, y$ are specific values that satisfy that equation.


I've been having a hard time with this problem and was wondering if anyone had any suggestions to get me started.

There are three series of numbers $ a_n , b_n , c_n$ such that:
$ a_{n\plus{}1} \equal{} \frac{1}{2} (b_n \plus{} c_n) \\ b_{n\plus{}1} \equal{} \frac{1}{2} (a_n \plus{} c_n) \\ c_{n\plus{}1} \equal{} \frac{1}{2} (a_n \plus{} b_n)$

Prove that $ \lim_{n \to \infty} a_n \equal{} \lim_{n \to \infty} b_n \equal{} \lim_{n \to \infty} c_n \equal{} \frac{a_0 \plus{} b_0 \plus{} c_0}{3}$

[hide="A couple of ideas"]
The sum $ a_n + b_n + c_n$ is invariant. What happens to the quantities $ \text{min}\,(a_n,b_n,c_n)$  and $ \text{max}\,(a_n,b_n,c_n)$?

Or, \begin{align*}a_{n+1} &= \frac{1}{2}(b_n + c_n) = \frac{1}{2}\left(\frac 12 (a_{n-1} + b_{n-1}) + \frac 12 (a_{n-1} + b_{n-1})\right)\\ &= \frac{1}{2}\left(\frac{1}{2}(b_{n-1}+c_{n-1}) + a_{n-1}\right) = \frac{a_n+a_{n-1}}{2}\end{align*}, with initial conditions $ a_0 = a_0, a_1 = \frac{b_0 + c_0}{2}$.
[/hide]

Let $ S_n \equal{} (a_n \minus{} b_n)^2 \plus{} (a_n \minus{} c_n)^2 \plus{} (b_n \minus{} c_n)^2$.  What is the relationship between $ S_n$ and $ S_0$?

By the way, this is too hard for "Intermediate Topics."

Thanks for the suggestions!

Sorry about it being too hard,  didn't realize that it was actually that hard. I kinda just thought that I was missing something fairly obvious.

:maybe:  Why is this too hard for intermediate...

[b]Hint:[/b] What does this problem have to do with [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125811]this problem[/url]?

Altheman's solution to that particular problem is very nice.

[b]Hint 2:[/b] If you don't see how that problem (which has a given initial condition) applies [i]directly[/i] to this situation...

[hide]I'm going to use vector notation here for the sake of readability.

This is a linear recurrence. Write it as $ \mathbf{x}_n = (a_n, b_n, c_n)$, and $ \mathbf{x}_{n + 1} = A \mathbf{x}_n$, where $ A = \left( \begin{array}{ccc} 0 & 1/2 & 1/2 & 1/2 & 0 & 1/2 & 1/2 & 1/2 & 0 \end{array} \right)$

If $ \mathbf{x}_0 = (p, q, r)$, then we can write $ \mathbf{x}_n = p \mathbf{p}_n + q \mathbf{q}_n + r \mathbf{r}_n$, which each satisfy $ \mathbf{p}_{n + 1} = A \mathbf{p}_n$, etc, and such that $ \mathbf{p}_0 = (1,0,0)$, $ \mathbf{q}_0 = (0,1,0)$, and $ \mathbf{r}_0 = (0,0,1)$[/hide]


When a water tank is $ 30\%$ full, it contains 27 gallons less than when it is $ 20\%$ empty. How many gallons of water does the tank hold when it is full?

If it is 20% empty => 80% full
So diff = 50% of tank

50% x = 27
x = [b]54[/b]


calculus
integration
algebra
function
domain
abstract algebra
Ring Theory

My goal is to show that the characteristic of an integral domain is either zero or a prime number. I have the basic intuition of it I think, and think I should be able to show it pretty simply and easily. However, I'm new to rings and their notation, and I'm a bit confused in particular when dealing with integer multiples (multiple additions) of ring elements. The case I'm looking at right now is based on the following proof of my goal that I found online:

Let R be an integral domain. If the characteristic of R is 0, then we are done. So suppose the characteristic of R is n != 0. Then n*a = 0 for all a in R and so n*1 = 0 but m*1 != 0 for any m < n (m in R), for otherwise ma = 0 for all a in R, which would contradict the minimality of n. Now suppose n is composite (i.e., not prime). Then n = pq for some integers p,q where 1 < p,q < n. Now (p*1)*(q*1) = (p*q)*1 = n*1 = 0. Since R is an integral domain, either p*1 = 0 or q*1 = 0, none of which are true. Therefore n can never be composite and so n is prime. Since every field is an integral domain, the result is also true for a field.

Initially I had a similar thing planned out in my head, but when I started writing it out I ran into some confusion with the 1 business. (I run into the same confusion whilst trying to understand the above proof as well, but since "my" proof only works in my head, I figured I'd give the above one which supposedly is valid, and hopefully someone can help me understand it.)

What I get stuck on at first in the above is "n*1 = 0" in the fourth sentence. What this implies to me is that 1 is in R. But unless that's a property of integral domains with finite characteristic that I don't know about, I can't figure out where they're getting that from.

Now, I believe I just made that a whole bunch more complicated than it is in reality or in my head, but hopefully you get the gist of it. Any help is appreciated!

[quote="puuhikki"][quote="algebra-lover"]What this implies to me is that 1 is in R.[/quote]
The definition of integral domain.[/quote]

Please forgive my ignorance, it seems this question is probably as simple as I initially suspected it to be. But my understanding of the definition of an integral domain is a commutative ring with no zero-divisors. I assume the existence of 1 in R can be derived from this in an obvious fashion, since I can't see it..? :blush:

Thanks for the quick response!

EDIT: Nevermind, I think I got it. Thanks.

Ignore the edit to my previous post, it turns out I don't have it. I thought I could adapt a proof for finite integral domains, but it turns out the finiteness was actually necessary, imagine that.

[quote="puuhikki"]How do you define a ring?  :maybe:  I have read the definition where multiplication has a neutral element denoted by $ 1$. But if $ 1 \equal{} 0$ in a commutative ring then $ x \equal{} x\cdot 1 \equal{} x\cdot 0 \equal{} x\cdot (0 \plus{} 0) \equal{} x\cdot 0 \plus{} x\cdot 0$ so $ x \equal{} 0$. Thus every element is a zero-divisor.[/quote]

I'm not sure I understand what you're saying... Here are all the definitions I am using which should be required:

An "associative ring" is a nonempty set R with two operations + and * such that:
1) (R, +) forms an abelian group (with 0 as its identity element).
2) R is closed under *.
3) x*(y + z) = x*y + x*z AND (x + y)*z = x*z + y*z for all x,y,z in R.
4) * is associative.

All "rings" I am considering are synonymous with "associative rings".

A "ring with unit element" is a ring with an element 1 in R such that 1*x = x*1 = x for all x in R.

A "commutative ring" is a ring such that x*y = y*x for all x,y in R.

If R is a commutative ring, then x in R, x != 0 is a "zero divisor" if there exists y in R, y!= 0 such that x*y = 0.

A commutative ring is an "integral domain" if it has no zero divisors.

EDIT: The proof I was attempting for the existence of the unit element is as follows:

R is an integral domain. Choose x in R, x != 0. Then S := {y*x | y in R} is a subset of R since R is closed under *. Assume there exist a,b in R, a != b, such that a*x = b*x. But then a*x + (-b*x) = (a + (-b))*x = 0, and since x != 0, then we must have that a + (-b) = 0, or a = b. Contradiction. So a*x = b*x iff a = b, which means S = R (this conclusion is what I'm not sure is valid). Since x is in R, there exists c in R such that c*x = x*c = x (since R is commutative). Pick any z in R. There exists d in R such that z = d*x. Then since R is commutative, z*c = c*z = c*(d*x) = d*(c*x) = d*x = z. Since z was arbitrary, c = 1.

I'm confused by this discussion.  The standard definition of an integral domain is a commutative ring [i]with unity[/i] and without zero divisors.  In this case, you seem to understand the problem mostly, with a little hitch.

When you write $ p*1$, you really mean $ 1$ added to itself $ p$ times.  You are not given, a priori, some element $ p\in R$.  To avoid confusion, I would use a different symbol for repeated addition and multiplication.  So the notation in your proof would look like this: $ (p\cdot 1)*(q\cdot 1)\equal{}0$, implying $ p\cdot 1\equal{}0$ or $ q\cdot 1\equal{}0$.

If we are talking about whether the same theorem holds when $ R$ is a commutative ring without zero divisors that [i]does not necessarily have a unity[/i], then I suspect the answer is no, but at the very least it is a harder problem.  (maybe someone who knows more examples of unusual rings than I do can provide a counterexample)

[quote="LydianRain"]I'm confused by this discussion.[/quote]Yeah, I'm sorry about that. I was a bit unclear with my initial question, but it seems you've managed to accurately sort through my obscurities. :)

[quote="LydianRain"]The standard definition of an integral domain is a commutative ring [i]with unity[/i] and without zero divisors.[/quote][quote="LydianRain"]If we are talking about whether the same theorem holds when $ R$ is a commutative ring without zero divisors that [i]does not necessarily have a unity[/i], then I suspect the answer is no, but at the very least it is a harder problem.  (maybe someone who knows more examples of unusual rings than I do can provide a counterexample)[/quote]Yes, this is exactly the problem. I guess the definition I'm using is not standard then. The definitions I'm using are out of Herstein, Topics in Algebra, and the problem actually comes from there too, so I doubt the answer is no...

[quote="LydianRain"]In this case, you seem to understand the problem mostly, with a little hitch.

When you write $ p*1$, you really mean $ 1$ added to itself $ p$ times.  You are not given, a priori, some element $ p\in R$.  To avoid confusion, I would use a different symbol for repeated addition and multiplication.  So the notation in your proof would look like this: $ (p\cdot 1)*(q\cdot 1) \equal{} 0$, implying $ p\cdot 1 \equal{} 0$ or $ q\cdot 1 \equal{} 0$.[/quote]Thank you, that's an excellent idea. That [u]was[/u] the major thing confusing me initially with the notation. I figured out what was being talked about, but that's a good idea for keeping the two straight for myself, thanks!

[quote="t0rajir0u"][quote="algebra-lover"]R is an integral domain. Choose x in R, x != 0. Then S := {y*x | y in R} is a subset of R since R is closed under *. Assume there exist a,b in R, a != b, such that a*x = b*x. But then a*x + (-b*x) = (a + (-b))*x = 0, and since x != 0, then we must have that a + (-b) = 0, or a = b. Contradiction. So a*x = b*x iff a = b, which means S = R (this conclusion is what I'm not sure is valid). Since x is in R, there exists c in R such that c*x = x*c = x (since R is commutative). Pick any z in R. There exists d in R such that z = d*x. Then since R is commutative, z*c = c*z = c*(d*x) = d*(c*x) = d*x = z. Since z was arbitrary, c = 1.[/quote]
This works if and only if $ R$ is [i]finite[/i].  For example, the even integers form a commutative ring with no identity and no zero divisors.[/quote]Yeah, that's what I thought. I saw a similar thing for a finite ring, but wasn't sure if that part of the proof required the finiteness. Thanks for clarifying. :)

[quote="LydianRain"]Cute side question: why are the even integers modulo 30 [i]not[/i] a counterexample? :)[/quote]

Why should it be? It is not even an "integral domain" ($ 6\cdot 10\equiv 0\mod 30$). Am I misunderstanding something about the topic of this discussion? (Has it changed since post #1 ?) If not, the original question is rather easy. Note that I hesitate to call "integral domains" without unity integral domains (if things would go my way, they wouldn't be even called rings), and we don't even need commutativity, so I'll rather quote the necessary part of the definition of integral domains than talk about "integral domains".

[color=blue][b]Theorem.[/b] Let $ R$ be a ring (not necessarily commutative, not necessarily with unity). Consider $ R$ as a $ \mathbb{Z}$-algebra (not necessarily commutative, not necessarily with unity) in the canonical way. Assume that the set $ \left\{n\in\mathbb{N}^\plus{}\mid nR\equal{}0\right\}$ is nonempty (where $ \mathbb{N}^\plus{}$ means the set $ \left\{1,2,3,...\right\}$). Let $ m\equal{}\min\left\{n\in\mathbb{N}^\plus{}\mid nR\equal{}0\right\}$. Assume that $ R\neq 0$, and that

[/color][color=green][b](1)[/b][/color][color=blue] for any two elements $ a$ and $ b$ of $ R$ such that $ ab\equal{}0$, we have $ a\equal{}0$ or $ b\equal{}0$.

Then, $ m$ is prime.[/color]

[i]Proof.[/i] Clearly, $ m\equal{}\min\left\{n\in\mathbb{N}^\plus{}\mid nR\equal{}0\right\}$ yields $ mR\equal{}0$, so that $ m\neq 1$ (because $ m\equal{}1$ would lead to $ R\equal{}1R\equal{}mR\equal{}0$, but this would contradict $ R\neq 0$).

Assume that $ m$ is not prime. Then, there exist $ u\in\mathbb{N}^\plus{}$ and $ v\in\mathbb{N}^\plus{}$ such that $ u<m$ and $ v<m$ and $ m\equal{}uv$ (since  $ m\neq 1$ and $ m$ is not prime). Now, $ m\equal{}\min\left\{n\in\mathbb{N}^\plus{}\mid nR\equal{}0\right\}$ yields $ uR\neq 0$ (since $ u\in\mathbb{N}^\plus{}$ and $ u<m$) and $ vR\neq 0$ (since $ v\in\mathbb{N}^\plus{}$ and $ v<m$).

Since $ uR\neq 0$, there exists some $ x\in R$ such that $ ux\neq 0$. Similarly, there exists some $ y\in R$ such that $ vy\neq 0$. Now, $ uvR\equal{}0$ yields $ uvxy\equal{}0$, so that $ ux\cdot vy\equal{}uvxy\equal{}0$. Thus, [color=green][b](1)[/b][/color] (applied to $ a\equal{}ux$ and $ b\equal{}vy$) yields $ ux\equal{}0$ or $ vy\equal{}0$. This contradicts $ ux\neq 0$ and $ vy\neq 0$.

This contradiction shows that our assumption (the assumption that $ m$ is not a prime) was wrong. Hence, $ m$ is prime, qed.

[i]Side-note.[/i] Algebra-lover showed that any finite integral domain has a unity. This generalizes: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=228032]Any finite commutative ring (not a priori required to have a unity) that is not completely annihilated by one nonzero element has a unity.[/url] This is, however, a way harder problem.

  darij

Sorry, I was hallucinating before.  I meant to point out that my ring had a "hidden" unity--16--but it was a bad example.

Thanks very much guys, I really appreciate the discussion!

[quote="darij grinberg"]Am I misunderstanding something about the topic of this discussion? (Has it changed since post #1 ?)[/quote]Nope, you hit it right on the nose. :)

[quote="darij grinberg"]If not, the original question is rather easy. Note that I hesitate to call "integral domains" without unity integral domains (if things would go my way, they wouldn't be even called rings), and we don't even need commutativity, so I'll rather quote the necessary part of the definition of integral domains than talk about "integral domains".

[color=blue][b]Theorem.[/b] Let $ R$ be a ring (not necessarily commutative, not necessarily with unity). Consider $ R$ as a $ \mathbb{Z}$-algebra (not necessarily commutative, not necessarily with unity) in the canonical way. Assume that the set $ \left\{n\in\mathbb{N}^ \plus{} \mid nR \equal{} 0\right\}$ is nonempty (where $ \mathbb{N}^ \plus{}$ means the set $ \left\{1,2,3,...\right\}$). Let $ m \equal{} \min\left\{n\in\mathbb{N}^ \plus{} \mid nR \equal{} 0\right\}$. Assume that $ R\neq 0$, and that

[/color][color=green][b](1)[/b][/color][color=blue] for any two elements $ a$ and $ b$ of $ R$ such that $ ab \equal{} 0$, we have $ a \equal{} 0$ or $ b \equal{} 0$.

Then, $ m$ is prime.[/color]

[i]Proof.[/i] Clearly, $ m \equal{} \min\left\{n\in\mathbb{N}^ \plus{} \mid nR \equal{} 0\right\}$ yields $ mR \equal{} 0$, so that $ m\neq 1$ (because $ m \equal{} 1$ would lead to $ R \equal{} 1R \equal{} mR \equal{} 0$, but this would contradict $ R\neq 0$).

Assume that $ m$ is not prime. Then, there exist $ u\in\mathbb{N}^ \plus{}$ and $ v\in\mathbb{N}^ \plus{}$ such that $ u < m$ and $ v < m$ and $ m \equal{} uv$ (since  $ m\neq 1$ and $ m$ is not prime). Now, $ m \equal{} \min\left\{n\in\mathbb{N}^ \plus{} \mid nR \equal{} 0\right\}$ yields $ uR\neq 0$ (since $ u\in\mathbb{N}^ \plus{}$ and $ u < m$) and $ vR\neq 0$ (since $ v\in\mathbb{N}^ \plus{}$ and $ v < m$).

Since $ uR\neq 0$, there exists some $ x\in R$ such that $ ux\neq 0$. Similarly, there exists some $ y\in R$ such that $ vy\neq 0$. Now, $ uvR \equal{} 0$ yields $ uvxy \equal{} 0$, so that $ ux\cdot vy \equal{} uvxy \equal{} 0$. Thus, [color=green][b](1)[/b][/color] (applied to $ a \equal{} ux$ and $ b \equal{} vy$) yields $ ux \equal{} 0$ or $ vy \equal{} 0$. This contradicts $ ux\neq 0$ and $ vy\neq 0$.

This contradiction shows that our assumption (the assumption that $ m$ is not a prime) was wrong. Hence, $ m$ is prime, qed.[/quote]Thank you so much Darij, I understood just about all of that! There are two things which confuse me though.

First, though it does not seem that this line is used directly in either the theorem or your proof, I'm afraid I do not understand this line since I have not yet been formally introduced to even the concept of an "algebra", much less a Z-algebra, or how you can consider anything like one:[quote="darij grinberg"]Consider $ R$ as a $ \mathbb{Z}$-algebra (not necessarily commutative, not necessarily with unity) in the canonical way.[/quote]Unless you just meant it as you explained in parentheses, merely that it is "not necessarily commutative, not necessarily with unity". (Though I still don't technically understand why that is, I at least understand the consequences of what you mean by it then.)

Second, the part which confuses me about the actual proof, is this one: ux*vy = uvxy. This is more just a basic algebra thing I'm sure, but I can't seem to get it to work out explicitly. (It makes sense intuitively of course.) This relates to my original question about 'notation' and really gets at the root of the question I was trying to ask from the beginning. If R specifically has a unit element, then I can certainly see the above working out as follows: (u*x) * (v*y) = (u*1) * x * (v*1) * y, and since R is commutative, (u*1) * x * (v*1) * y = (u*1) * (v*1) * x * y = u*v*x*y. But without the unit element, I don't see how to make that connection. Am I simply missing something excruciatingly obvious?

[quote="algebra-lover"]I'm afraid I do not understand this line since I have not yet been formally introduced to even the concept of an "algebra", much less a Z-algebra[/quote]
All this means is that there is an operation $ \mathbb{Z} \times A \to A$ that satisfies certain axioms, and this operation is $ na \equal{} \underbrace{a \plus{} a \plus{} ...}_{n \text{ times}}$.  Think of this as a restricted form of scalar multiplication.

[quote="algebra-lover"]Second, the part which confuses me about the actual proof, is this one: ux*vy = uvxy. [/quote]
See the above!  $ u, v$ belong to $ \mathbb{Z}$ while $ x, y$ belong to $ R$, which is probably why you're confused.  All this expression is is $ (\underbrace{x \plus{} x \plus{} ...}_{u \text{ times}})(\underbrace{y \plus{} y \plus{} ...}_{v \text{ times}})$.

[quote="t0rajir0u"][quote="algebra-lover"]I'm afraid I do not understand this line since I have not yet been formally introduced to even the concept of an "algebra", much less a Z-algebra[/quote]
All this means is that there is an operation $ \mathbb{Z} \times A \to A$ that satisfies certain axioms, and this operation is $ na \equal{} \underbrace{a \plus{} a \plus{} ...}_{n \text{ times}}$.  Think of this as a restricted form of scalar multiplication.[/quote]Ah, perfect, thank you! So in other words I've been using this operation without realizing where it came from.

[quote="t0rajir0u"][quote="algebra-lover"]Second, the part which confuses me about the actual proof, is this one: ux*vy = uvxy. [/quote]
See the above!  $ u, v$ belong to $ \mathbb{Z}$ while $ x, y$ belong to $ R$, which is probably why you're confused.  All this expression is is $ (\underbrace{x \plus{} x \plus{} ...}_{u \text{ times}})(\underbrace{y \plus{} y \plus{} ...}_{v \text{ times}})$.[/quote]Yes, thank you, I actually did realize that. However, at your urging, I just tried working it out again and now see it too. Hard as it may be for you to believe, this thing which I'm sure is quite obvious to you was actually the source of my confusion. I couldn't get from $ (\underbrace{x \plus{} x \plus{} ...}_{u \text{ times}})(\underbrace{y \plus{} y \plus{} ...}_{v \text{ times}})$ to $ (\underbrace{xy \plus{} xy \plus{} ...}_{uv \text{ times}})$.
Trying for this very simple idea once more, I managed to get it:

$ (\underbrace{x \plus{} x \plus{} ...}_{u \text{ times}})(\underbrace{y \plus{} y \plus{} ...}_{v \text{ times}})$

$ \equal{} (\underbrace{x(\underbrace{y \plus{} y \plus{} ...}_{v \text{ times}}) \plus{} x(\underbrace{y \plus{} y \plus{} ...}_{v \text{ times}}) \plus{} ...}_{u \text{ times}})$

$ \equal{} (\underbrace{(\underbrace{xy \plus{} xy \plus{} ...}_{v \text{ times}}) \plus{} (\underbrace{xy \plus{} xy \plus{} ...}_{v \text{ times}}) \plus{} ...}_{u \text{ times}})$

$ \equal{} (\underbrace{xy \plus{} xy \plus{} ...}_{uv \text{ times}})$

I tend to miss very obvious things on an unfortunately rather frequent occasion. :blush: But when I can't get them on my own, I've got to try to have someone else help point them out for me, embarrassing as it may be.

Again, thanks so much to everyone who participated in this little venture. I'm sorry for making such a big thing out of it, but like I said, there comes a point when I know it's something small and stupid, but for the life of me I just can't get it on my own. Thank you!


probability
logarithms
conditional probability
probability and stats

imagine that you are the manager of a company and that you have to hire a secretary. This morning, $ n$ secretaries will come in your office, one after one (but you do not have to see all of them, if you do not want to). Suppose moreover that you have to give your answer (hired or not) immidiately after that you have interviewed a secretary. What would be your strategy in order to hire the best secretary ?

For example your strategy could be to see the first $ 3$ of them and then stop the first time (if it happens) that you interview a better secretary? The score of this strategy would be $ S$=(probability that you hire the best secretary)

Suppose we employ the strategy alekk mentions: we interview and dismiss the first $ k$ candidates, then select the first one after that who is better then the first $ k.$

(One thing to note: if the best candidate was among the first $ k,$ we will eventually be obligated to hire the $ n$th candidate, who may be the most incompetent of all of them. I will evaluate the strategy on the terms suggested, the probability of hiring the best, and not worry about the quality of the hire if not the best.)

What is the conditional probability of hiring the best candidate given that that best candidate is in the $ m$th interview position? Clearly that is zero if $ m\le k.$ For $ m > k,$ it is the probability that the best of the first $ m \minus{} 1$ is in fact in the first $ k,$ namely $ \frac {k}{m \minus{} 1}.$

If we then add up these conditional probabilities, we get that the overall probability of hiring the best is
\[ S \equal{} \frac1n\sum_{m \equal{} k \plus{} 1}^n\frac {k}{m \minus{} 1}
\]
To investigate this for small values of $ n,$ we should simply explicitly add that sum to see what we have. If we suppose that $ n$ and $ k$ are both large, that sum can be reasonably approximated by $ \frac {k}{n}\ln\left(\frac {n}{k}\right).$ In that case, we can maximize it by  maximizing $ f(x) \equal{} \minus{} x\ln x$ on $ (0,1).$ That maximum occurs when $ x \equal{} e^{ \minus{} 1},$ and the maximum value is $ e^{ \minus{} 1}.$

Hence, assuming $ n$ is large, our best strategy [i]from this family of strategies[/i] is to interview and dismiss about the first $ 37\%$ of the candidates and then take the next one better than those dismissed. This will result in the hiring of the best candidate about $ 37\%$ of the time. The interesting thing is that the probability of hiring the best tends to a positive limit as $ n\to\infty.$

Of course, there may be different strategies; I can't claim that this exhausts the possibilities.

A possibly more practical question: how to get the best chance of hiring one of the best - say, one of the top $ 10\%$ or so of candidates. That would be a different problem.

nice !  now the hard part is to effectively show that it is the best strategy  :)  cool, isn't it.

Assuming no prior knowledge about the distribution of qualifications, this family of strategies gives the best chance of hiring the best secretary. At each stage, there are two possibilities: either the current ($ m$th) candidate is the best so far, or he isn't. If the current candidate isn't the best, you obviously shouldn't hire. If he is, you have some probability $ p$ of winning if you hire now, and some probability $ q$ of winning if you don't hire now and continue your strategy. History is irrelevant- you can't hire any of the secretaries you passed over already and none of them were the best, so they don't matter. If $ p>q$, you should hire now. If $ p<q$, you should wait- and these are deterministic. Also, $ p\equal{}\frac{m}{n}$ always increases, while $ q$ always decreases since your options are reduced. This leads inevitably to a strategy in the family, interviewing some with no chance of hiring and then jumping at the first which could be best.