1) Given $ x,y,z$ non-negative and $ xyz(x\plus{}y\plus{}z)\le 3$. Prove

$ 5(x^2\plus{}y^2\plus{}z^2)\plus{}\frac{72}{xy\plus{}yz\plus{}zx}\ge 39$

2) Given $ x,y,z\in R; x\plus{}y\plus{}z\equal{}xy\plus{}yz\plus{}zx$ and $ xyz\ge 1$. Prove

$ (x^5\minus{}1)(y^5\minus{}1)(z^5\minus{}1)\ge 0$

I solved the second one as follows

$ (x\minus{}1)(y\minus{}1)(z\minus{}1) \ge 0$

We need to prove that

$ x^4\plus{} x^3\plus{}x^2\plus{}x\plus{}1 \ge 0$

This inequality is obvious after we complete the square.

We are done!

[hide="Gigaman"] Ban tu VUT phai khong?[/hide]

[quote="great math"]I solved the second one as follows

$ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1) \ge 0$

We need to prove that

$ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \ge 0$

This inequality is obvious after we complete the square.

We are done!

[hide="Gigaman"] Ban tu VUT phai khong?[/hide][/quote]

This is my solution

We have

$ (x\minus{}1)(y\minus{}1)(z\minus{}1)\equal{}xyz\minus{}1\plus{}x\plus{}y\plus{}z\minus{}(xy\plus{}yz\plus{}zx)\equal{}xyz\minus{}1$

$ \equal{}> (x^5\minus{}1)(y^5\minus{}1)(z^5\minus{}1)\ge 0$
[hide="great math"] HI, tui tu diendan3t, nhung goi la VUT cung dc :D[/hide]

I think the first ineq isn't hard
we must prove
$ 5(\sum xy(x^2 \plus{} y^2) \plus{} xyz(x \plus{} y \plus{} z)) \plus{} 72 \geq 39xyz(x \plus{} y \plus{} z)$
let $ p \equal{} xy \plus{} yz \plus{} zx; q \equal{} xyz(x \plus{} y \plus{} z)$
$ 5(\sum xy(x^2 \plus{} y^2) \plus{} xyz(x \plus{} y \plus{} z)) \plus{} 72 \geq 10(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \plus{} 5q \plus{} 72$
$ \equal{} 10(p^2 \minus{} 2q) \plus{} 5q \plus{} 72 \geq 39q$ because $ q \leq 3$
where is wrong ^^

[quote="DangChienbn"]I think the first ineq isn'y hard
we must prove
$ 5(\sum xy(x^2 \plus{} y^2) \plus{} 5xyz(x \plus{} y \plus{} z)) \plus{} 72 \geq 39xyz(x \plus{} y \plus{} z)$
let $ p \equal{} xy \plus{} yz \plus{} zx; q \equal{} xyz(x \plus{} y \plus{} z)$
$ 5(\sum xy(x^2 \plus{} y^2) \plus{} 5xyz(x \plus{} y \plus{} z)) \plus{} 72 \geq 10(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \plus{} 5q \plus{} 72$
$ \equal{} 10(p^2 \minus{} 2q) \plus{} 5q \plus{} 72 \geq 39q$ because $ q \leq 3$[/quote]

So, how about this ineq :D

$ \frac{5(a\plus{}b\plus{}c)^2}{3}\plus{}\frac{72}{ab\plus{}bc\plus{}ca}\ge 39$

with the same condition :D

Dangchienbn, I think you are wrong since we don't have any element of xyz on the righ hand side. Please check your solution again. Thank you. :)

[quote="DangChienbn"]I think the first ineq isn'y hard
we must prove
[b][color=red]$ 5(\sum xy(x^2 \plus{} y^2) \plus{} 5xyz(x \plus{} y \plus{} z)) \plus{} 72 \geq 39xyz(x \plus{} y \plus{} z)$ [/color][/b]  :?:
let $ p \equal{} xy \plus{} yz \plus{} zx; q \equal{} xyz(x \plus{} y \plus{} z)$
$ 5(\sum xy(x^2 \plus{} y^2) \plus{} 5xyz(x \plus{} y \plus{} z)) \plus{} 72 \geq 10(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \plus{} 5q \plus{} 72$
$ \equal{} 10(p^2 \minus{} 2q) \plus{} 5q \plus{} 72 \geq 39q$ because $ q \leq 3$[/quote]

You are wrong

 :)

Oh, I am wrong in tex  :blush:  But my solution ís true :P

[quote="gigaman"][quote="DangChienbn"]I think the first ineq isn'y hard
we must prove
$ 5(\sum xy(x^2 \plus{} y^2) \plus{} 5xyz(x \plus{} y \plus{} z)) \plus{} 72 \geq 39xyz(x \plus{} y \plus{} z)$
let $ p \equal{} xy \plus{} yz \plus{} zx; q \equal{} xyz(x \plus{} y \plus{} z)$
$ 5(\sum xy(x^2 \plus{} y^2) \plus{} 5xyz(x \plus{} y \plus{} z)) \plus{} 72 \geq 10(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \plus{} 5q \plus{} 72$
$ \equal{} 10(p^2 \minus{} 2q) \plus{} 5q \plus{} 72 \geq 39q$ because $ q \leq 3$[/quote]

So, how about this ineq :D

$ \frac {5(a \plus{} b \plus{} c)^2}{3} \plus{} \frac {72}{ab \plus{} bc \plus{} ca}\ge 39$

with the same condition :D[/quote]
with $ abc(a\plus{}b\plus{}c) \leq 3 ?$

[quote="gigaman"]1) Given $ x,y,z$ non-negative and $ xyz(x \plus{} y \plus{} z)\le 3$. Prove

$ 5(x^2 \plus{} y^2 \plus{} z^2) \plus{} \frac {72}{xy \plus{} yz \plus{} zx}\ge 39$

2) Given $ x,y,z\in R; x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx$ and $ xyz\ge 1$. Prove

$ (x^5 \minus{} 1)(y^5 \minus{} 1)(z^5 \minus{} 1)\ge 0$[/quote]
Lemma: Let $ n \in [1;\plus{}\infty)$. Then $ a^n > 1 \iff a > 1$. The proof is obvious.
So $ (x^5 \minus{} 1)(y^5\minus{}1)(z^5\minus{}1) \ge 0 \iff (x\minus{}1)(y\minus{}1)(z\minus{}1) \ge 0 \iff$
$ xyz \ge 1$. Done.

Number 1 we only have to prove when $ y\equal{}z$, and when $ z\equal{}0$, by using the uvw-theorem.

[quote="DangChienbn"][quote="gigaman"][quote="DangChienbn"]I think the first ineq isn'y hard
we must prove
$ 5(\sum xy(x^2 \plus{} y^2) \plus{} 5xyz(x \plus{} y \plus{} z)) \plus{} 72 \geq 39xyz(x \plus{} y \plus{} z)$
let $ p \equal{} xy \plus{} yz \plus{} zx; q \equal{} xyz(x \plus{} y \plus{} z)$
$ 5(\sum xy(x^2 \plus{} y^2) \plus{} 5xyz(x \plus{} y \plus{} z)) \plus{} 72 \geq 10(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \plus{} 5q \plus{} 72$
$ \equal{} 10(p^2 \minus{} 2q) \plus{} 5q \plus{} 72 \geq 39q$ because $ q \leq 3$[/quote]

So, how about this ineq :D

$ \frac {5(a \plus{} b \plus{} c)^2}{3} \plus{} \frac {72}{ab \plus{} bc \plus{} ca}\ge 39$

with the same condition :D[/quote]
with $ abc(a \plus{} b \plus{} c) \leq 3 ?$[/quote]

Yes, with $ abc(a\plus{}b\plus{}c)\le 3$

[quote="devil_maths"]gigaman:Nguyễn Việt Hưng
đúng rồi còn gì  :huh:[/quote]
Bạn thử với n=2, khi đó $ a^2\ge 1\Leftrightarrow a\ge 1$ còn đúng nữa không :wink:

[quote="Mathias_DK"] $ (x^5 \minus{} 1)(y^5 \minus{} 1)(z^5 \minus{} 1) \ge 0 \iff (x \minus{} 1)(y \minus{} 1)(z \minus{} 1) \ge 0 \iff$
$ xyz \ge 1$. Done.

[/quote]
Sorry ,but I'm not sure that
 $ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1) \geq 0 \implies xyz \geq 1$ and $ xyz \geq 1 \implies (x \minus{} 1)(y \minus{} 1)(z \minus{} 1) \geq 0$

Take  $ x \equal{} y \equal{} 0$ and $ z \equal{} 10$ then $ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1)\geq 0$ but $ xyz \equal{} 0 < 1$

and yes,the lemma is obviously true  :)

you just look at the condition given for the original problem.

[quote="gigaman"][quote="Mathias_DK"]
Lemma: Let $ n \in [1; \plus{} \infty)$. Then $ a^n > 1 \iff a > 1$. The proof is obvious.
[/quote]

Are you sure about this lenma[/quote]
It should be $ n \in \{1,3,5,7,...\}$


Is true that if $ gcd(p,q)\equal{}1$ and $ pq\equal{}a^2, \, a \in N$ then $ p\equal{}m^2, \, q\equal{}n^2$???

[quote="jonysatie"]Is true that if $ gcd(p,q) \equal{} 1$ and $ pq \equal{} a^2, \, a \in N$ then $ p \equal{} m^2, \, q \equal{} n^2$???[/quote]

Yes, consider the prime factors

$ p\equal{}p_1^{m_1}p_2^{m_2}p_3^{m_3}\ldots p_n^{m_n}$ and
$ q\equal{}q_1^{m_1}q_2^{m_2}q_3^{m_3}\ldots q_n^{m_n}$

Since $ GCD(p,q)\equal{}1$. Therefore all $ q_n \neq p_m$

Since $ pq\equal{}m^2$ then the prime powers are even. Since $ p$ and $ q$ have distinct prime factors, they too must be even powers, hence they are squares


Recently, I've needed the following to solve a different problem, but I do not know a solution: (sums are cyclic)

$x^4 + y^4 + z^4 + xyz(x + y + z) \geq \sum x^3y + \sum xy^3.$

The following is a related statement, and I do not know if it is true:

$x^4 + y^4 + z^4 + 2xyz(x + y + z) \geq 3\sum x^2y^2.$

The first one is Schur's inequality with exponent 2.

Recall that Schurs ineq states that if r is a real, and a, b, c are positive, then
\[
\sum a^r (a-b)(a-c) \geq 0
\]

[quote="StRyKeR"]The following is a related statement, and I do not know if it is true:

$x^4 + y^4 + z^4 + 2xyz(x + y + z) \geq 3\sum x^2y^2.$[/quote]

This is not true. Counterexample: x = y = 100, z = 1.

  Darij


Finn alle $f : R \to  R$ slik at
$f(x-a) = f(x) + a$
Jeg vil gjerne se en fullstendig løsning...

Vel jeg har en løsning som jeg tror er eneste løsningen. Men jeg greier ikke å bevise at det ikke er flere løsninger...[hide]$f(x) = -x$[/hide]
$a$ er konstant (kanskje reel ?? ) Jeg er usikker.
Spørsmål:  jeg har en funksjonalligning og jeg fant en løsning og prøver å bevise at det er eneste løsningen som finnes.
F.eks. Jeg fant at  $f(x) = x^2$. Jeg tar og definerer $g(x) = f(x)-x^2$ som medfører at $f(x) = g(x) +x^2$... Hvis $f(x) = x^2$ er eneste løsningen da er $g(x) = 0$. Men når jeg erstatter alle $f(x)$ med $g(x)$ i selve funksjonalligningen kommer jeg vanligvis tilbake til samme likning bare med $g(x)$ istedenfor $f(x)$....
Hva betyr det? Har jeg bevist noe? Betyr det at det er flere løsninger?

Hvis $a$ er en gitt konstant og ikke 0 er det uendelig mange løsninger. Ta en vilkårlig funksjon $g : [0,a) \rightarrow \mathbb{R}$. Ethvert reelt tall $x$ kan skrives som $x = aq + r$, hvor $q$ er et heltall og $r\in [0, a)$ ($r = x \mod a$). La $f$ være gitt ved at $f(x) = f(aq + r) = g(r) - aq.$ Ligningen i oppgaven blir da
\[ f((aq+r)-a) = f(aq+r) + a \Leftrightarrow g(r)-a(q-1) = g(r)-aq+a,  \]
som stemmer...


algebra
polynomial
ratio
algebra unsolved

Let $ f(x)$ is a Polynomial with ration coefficients i,ts dergreeis greater than 2,and the infinite sequence $ a_1 , a_2 , ......$ there $ a_i$ are ration such that $ f(a_{n \plus{} 1}) \equal{} a_n$ for all $ n\in N$
Prove that there exists $ k\ge 1$ satisfying $ a_{n \plus{} k} \equal{} a_n$
Help me :blush:
Sorry,ihave edited it (because while i posed it,i was doing my homework),sorry :blush:

[quote="Allnames"]Let $ f(x)$ is a Polynomial with ration coefficients i,ts dergreeis greater than 2,and the infinite sequence $ a_1 , a_2 , ......$ there $ a_i$ are ration such that $ f(a_{n \plus{} 1} \equal{} a_n$ for all $ n\in N$
Prove that there exists $ k\ge 1$ satisfying $ a_{n \plus{} k} \equal{} a_n$
Help me :blush:
Sorry,ihave edited it (because while i posed it,i was doing my homework),sorry :blush:[/quote]
it's fals
we take $ a_n \equal{} n$ and $ f(x) \equal{} x \minus{} 1$
we have $ (a_{n \plus{} 1}) \equal{} a_n$but $ \forall k\ge 1: \ a_{n \plus{} k}\neq a_n$

aviateurpilot,
[quote="Allnames"]i,ts dergreeis greater than 2[/quote]

[quote="aviateurpilot"][quote="Allnames"]Let $ f(x)$ is a Polynomial with ration coefficients ,its dergreeis greater than 2,and the infinite sequence $ a_1 , a_2 , ......$ there $ a_i$ are ration such that $ f(a_{n \plus{} 1} \equal{} a_n$ for all $ n\in N$
Prove that there exists $ k\ge 1$ satisfying $ a_{n \plus{} k} \equal{} a_n$
Help me :blush:
Sorry,ihave edited it (because while i posed it,i was doing my homework),sorry :blush:[/quote]
it's fals
we take $ a_n \equal{} n$ and $ f(x) \equal{} x \minus{} 1$
we have $ (a_{n \plus{} 1}) \equal{} a_n$but $ \forall k\ge 1: \ a_{n \plus{} k}\neq a_n$[/quote]
Oh,my problem is
 [quote=Allnames ]its dergreeis greater than 2,[/quote],so your example is not true
But at all thank  for your care


function
linear algebra
matrix
algebra proposed
algebra

Let $\mathbb{N}=\{1,2,\ldots\}$. Find all functions $f: \mathbb{N}\to\mathbb{N}$ such that for all $m,n\in \mathbb{N}$ the number $f^2(m)+f(n)$ is  a divisor of $(m^2+n)^2$.

BTW, check out [url=http://www.animath.fr/tutorat/dossier_04054.html]this test[/url], problem 3 ;)
It seems that this is an ISL 2004 problem.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31951]All the questions of the test[/url]

No that Pierre has solved this not so long time ago. THis test seemed to be a great joke.  :?

Upps, sorry, I did not see you precised the link.  :D

Can someone post a link with Pierre's solution? :) :D It seems the Romanian students are not paying attention to the site (dzeta included!). I can count at least 3 problems from these TSTs that are posted with full solutions on the site  :(  :(

Valentin, on the other hand I do not think that Romanian TSTs should show who is paying attention to the site, although they seem to...  :(  :(  :(

I have proved that $ f(p-1)=p-1 $ for all prime number $p$ and I d'ont know if it can help. ;)

Look some three pages before in the proposed problems section at one of piglfly's posts with a very suggestive title $ f^2(m)+f(n)| (m^2+n)^2$ and you will see how close you are!

Valentine,let's have a look here: http://mathlinks.ro/Forum/topic-27188.html

[quote="nalpaction"]I'm really sorry  for I don't know how to creat a link  :?[/quote]Just write it down. It will be transformed automatically in a link. Or read the [url=http://www.mathlinks.ro/Forum/faq.php?mode=bbcode]FAQ of the BBCode[/url] :)

This problem is from Mohsen Jamali designer of problem 6 of IMO 2004.


here is a question, if there is a answer for it plz answer it:
 which one was in the begining, hen or egg? :?:

Hmm... this is not the philosophy forum ;)
Please define accuratly:
- beginning
- egg
- hen

By some genetic reasons I would say that the egg was earlier:
assuming:
- there's a sharp definition of 'hen'
- you call an animal 'hen' or 'egg for a hen' (speaking of the 'egg of a hen' is nonsense) when all his cells have the genetic information for a 'hen'
then:
The last mutation from the pre-hen to the hen must, when it shall give a 'new' animal, happen in the cells for reproduction. But then the first one having the necessary DNA is the egg.

But what if the first 'hen' was indeed a rooster¿  :lol:

Like ZetaX says, this is not a philosophy forum.

i though you can answer thisquestion,but may be you can't.

We can discuss it biologically (more or less) when you to define 'hen' and 'egg' so that we can speak about it.  :P


Hi all, can anyone help me solve this?

"A box with a square base and no lid is to be made from a square piece of metal by cutting squares from the corners and folding up the sides. The cut-off squares are 5 cm on a side. If the volume of the box is
100 cm to the power of 3, find the dimensions of the original piece of metal.

Thanks in advance

[hide="Solution"]
So, we have a square of metal. Call each side S.
Then, after cutting out a corner, one side loses two 5 cm segments, so the side length becomes $(S-10)$ The volume is given as 100. The base has a width and length of $(S-10)$, and the height (when folded up) is 5. Therefore the volume is $(S-10)^{2}\cdot 5 = 100$
$(S-10)^{2}= 20$
$S-10 = 2\sqrt{5}$
$S = 10+2\sqrt{5}$

[/hide]


a hint:
try to actually construct the sequence, inductively :)

I tried to do it inductively for a long time (some weeks ago) but I failed ...
Am I missing something here?

Inductively, we can also construst the sequence as follows. Assume the sequence constructed till $ n$. Let $ A=a_{1}^{2}+\dots+a_{n}^{2}$ and $ C=a_{1}+\dots+a_{n}$.
We want to find an $ B=a_{n+1}$ such that $ C+B|A+B^{2}$. Let $ A=uC$. Let $ s>1$ be a number such that $ s+1|u+C$. Then $ s+1|u+s^{2}C$ or $ (s+1)C|uC+s^{2}C^{2}$, or $ C+sC|A+(sC)^{2}$, so we take $ B=a_{n+1}=sC>a_{n}$.

[quote="TTsphn"]Easy sequence
m,2m,....,nm .
$ \sum_{1 = 1}^{n}\im = m.\frac{n(n+1)}{2}$
$ \sum_{i = 1}^{n}\ (mi)^{2}= m^{2}.\frac{n(n+1)(2n+1)}{6}$
Chose $ 3|m$[/quote]
How can you assume that $ 3|m$?I think that you must prove it for all $ m\in N$.

[quote="TTsphn"]No we  can chose m.Because I want  to find a sequence.[/quote]
You must prove that for each $ m\in N$,there exists sequence.So you can't chose $ m$,you must only show the sequence or show that there exist sequence.Maybe i misunderstand. :maybe:

I prove there exist a sequence .
$ a_{n}\equal{}mn$ which 3|m and $ m>a_{0}$
This sequence satisfy this conditional.

[quote="TTsphn"]I prove there exist a sequence .
$ a_{n}\equal{} mn$ which 3|m and $ m > a_{0}$
This sequence satisfy this conditional.[/quote]
Show that for each positive integers [u] $ a_{1}$ [/u] there exist a strictly increasing sequence of positive integers,such that
$ \sum_{i\equal{}1}^{n}a_{i}|\sum_{i\equal{}1}^{n}a_{i}^{2}$.
What is $ a_{o}$ mean?I think that you just misunderstood. :wink:

I saw that .
With the second we can induction.

[quote="Arne"]Prove that for each given positive integer a0, there exists a strictly increasing sequence {an} of positive integers such that
a1 + a2 + ... + an|a12 + a2 2 + ... + an2
for all positive integers n.[/quote]

It was posted before in PEN website,i think so :P  :)

I don't think that
For example $ a_{1}\equal{}1$

[quote="TTsphn"]I don't think that
For example $ a_{1}\equal{} 1$[/quote]

But $ a_{0}\ge 1$ and the integers sequence$ (a_{n})$ is a increasing sequence ,so $ a_{1}\ge 2$

More general
Let k is a natural number $ k>1$
Prove that there exist an strictly increasing $ a_{n}$
$ \sum_{i\equal{}1}^{n}{a_{i}}|\sum_{i\equal{}1}^{n}{a_{i}^{k}}$

[quote="TTsphn"]More general
Let k is a natural number $ k > 1$
Prove that there exist an strictly increasing $ a_{n}$
$ \sum_{i \equal{} 1}^{n}{a_{i}}|\sum_{i \equal{} 1}^{n}{a_{i}^{k}}$[/quote]

-If $ k$ is a even number.

Let :

 $ a_{n\plus{}1}\equal{}\sum_{i\equal{}1}^{n}. a_{i}^{k}\plus{}(\sum_{i\equal{}1}^{n}a_{i})^{k}\minus{}\sum_{i\equal{}1}^{n}a_{i}$.

-If $ k$ is a odd number .I don't have solution now. :blush: .I will try...  :arrow:  :P

[quote="Myth"]Only 2 minutes and I have solution.

an+1 = a12 + a22 + ... + an2 + (a1+...an)2 - (a1+...an)[/quote]

Myth, dear, can you explain why? You know, nowaays i have some problems with my ability of understanding.

[quote="TTsphn"]More general
Let k is a natural number $ k > 1$
Prove that there exist an strictly increasing $ a_{n}$
$ \sum_{i \equal{} 1}^{n}{a_{i}}|\sum_{i \equal{} 1}^{n}{a_{i}^{k}}$[/quote]

Hi. Can you expand? Or give it up... I didn't understand anyway.

TTsphn's solution is right , isnt it
given a0 , choose an = k*n*a0 , where 3 | k ( take k = 3)


limit
algebra
polynomial
linear algebra
matrix
vector
complex numbers

Since no-one seems interested in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26046&highlight=]this[/url] problem that I posted some time ago, I'll repeat the relevant problem which was the starting point there (and which I called 'absolutely trivial'  :D ). Quoting Myth, [i]let's organize a test and see who will solve it in 5 minutes![/i]  :P

Let $a_1,\ldots,a_m;b_1,\ldots,b_n \in \mathbb{C}^*$ be non-zero complex numbers such that
\[
a_1^k + \cdots + a_m^k = b_1^k + \cdots + b_n^k
\]
for all sufficiently large positive integers $k$. Prove that $m=n$, and the multisets $\{a_i\}$ and $\{b_i\}$ are identical (that is, after a suitable relabeling of the $b_i$, we have $a_i=b_i$ for all $i$).

 ;)

--Vesselin

It took 10 minutes from me to find nice solution :(

Suppose WLOG $|a_i|, |b_j|\leq 1$, $a_1=1$ and equality hods for all $k\geq k_0$. It is clear $\lim \frac{1}{n}\sum_{k=k_0}^{k_0+n-1} (a_1^k+...+a_m^k)=I$, where $I=|\{a_i=1\}|>0$. And the same for $J=|\{b_j=1\}|$. So $I=J$.
That's all.

My solution: multiply, for each $k \geq k_0$, the identity by $x^k$, and sum the resulting identities to obtain that
\[
\sum \frac{a_i^{k_0}}{1 - a_ix} = \sum \frac{b_i^{k_0}}{1 - b_ix}.
\]
Comparing the two partial fraction expansions, the result follows.  :)

[quote="vess"]My solution: multiply, for each $k \geq k_0$, the identity by $x^k$, and sum the resulting identities to obtain that
\[
\sum \frac{a_i^{k_0}}{1 - a_ix} = \sum \frac{b_i^{k_0}}{1 - b_ix}.
\] [/quote]
That is correct for some small $x$. What next? :?

Simply, clear denominators to obtain a polynomial identity. It is valid for infinitely many $x$ (all $x$ in some small neighborhood of the origin), and is therefore valid for [i]all[/i] $x \in \mathbb{C}$ (because it is a polynomial identity). Then put $x = a_i^{-1}$ in this polynomial to deduce that some $b_j$ equals $a_i$.

Or, we may directly proceed as follows. Assume WLG that $|a_1| = \max \{ |a_i|,|b_j|\}$; then, the identity
\[
\sum \frac{a_i^{k_0}}{1-xa_i} = \sum \frac{b_i^{k_0}}{1-xb_i}
\]
holds for all $x$ in the disk $|x|  < |a_1|^{-1}$. Letting $x \to a_1^{-1}$ via the segment joining the origin with $\frac{1}{a_1}$, we get that some $b_i$ equals $a_1$ (for otherwise the l.h.s. is unbounded and the r.h.s. is bounded).

:lol:  Haha, here's a one-line proof that $m=n$: as it is so well-known, the numbers $x_k := a_1^k + \cdots + a_m^k - b_1^k - \cdots - b_n^k$ satisfy a linear recurrence relation; because $x_k = 0$ for $k > k_0$, it follows that $x_k=0$ for all (positive or negative) integers $k$, so in particular $x_0=0$, i.e. $m=n$. :lol:

Mine solution (how about yours, Myth?) can easily be modified to imply the following generalization:

[b]Generalization. [/b] Let $\alpha_1,\ldots,\alpha_k \in \mathbb{C}^*$ be pairwise different non-zero complex numbers and $P_1,\ldots,P_k \in \mathbb{C}[z]$ be complex polynomials such that
\[
P_1(n) \alpha_1^n + P_2(n)c_2\alpha_2^n + \cdots + P_k(n)\alpha_k^n = 0
\]
for all $n \in \mathbb{N}$. Then, $P_1 \equiv \cdots \equiv P_k \equiv 0$.

And [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=166530#p166530]here[/url] is a further generalization (I don't claim originality, because this result is an easy consequence of the celebrated [url=http://mathworld.wolfram.com/Skolem-Mahler-LechTheorem.html]Skolem-Mahler-Lech theorem[/url]).

--Vesselin

[quote="vess"] Let $a_1,\ldots,a_m;b_1,\ldots,b_n \in \mathbb{C}^*$ be non-zero complex numbers such that
\[
a_1^k + \cdots + a_m^k = b_1^k + \cdots + b_n^k
\]
for all sufficiently large positive integers $k$. Prove that $m=n$, and the multisets $\{a_i\}$ and $\{b_i\}$ are identical [/quote]

Are you asking whether a linear recursive sequence (in this case, $\Sigma a_i^k$ starting at large enough $k$) has at most one expression as $q_k = \Sigma c_i {{\alpha}_i^k}$?  For simple roots the Vandermonde matrix is invertible so this is obvious.   For multiple roots, any such expression leads to a polynomial satisfied by the associated matrix (acting on the space of solutions of the recurrence). If there is a mismatch between the canonical expression and some other expression (using the $b_i$) you get a polynomial of smaller-than-minimal degree annihilating the given sequence (considered as vector in the space of solutions).

[quote="vess"]Mine solution (how about yours, Myth?) can easily be modified to imply the following generalization:

[b]Generalization. [/b] Let $\alpha_1,\ldots,\alpha_k \in \mathbb{C}^*$ be pairwise different non-zero complex numbers and $P_1,\ldots,P_k \in \mathbb{C}[z]$ be complex polynomials such that
\[
P_1(n) \alpha_1^n + P_2(n)c_2\alpha_2^n + \cdots + P_k(n)\alpha_k^n = 0
\]
for all $n \in \mathbb{N}$. Then, $P_1 \equiv \cdots \equiv P_k \equiv 0$.[/quote]
It is an absolutly trivial conclusion from my method.


geometry
geometric transformation
reflection
videos
blogs

[hide="First, an anecdote"] When my father asked my older brother if you can download music of youtube, my older brother said "yes, especially if it's popular." My father scolded him for his illogical answer. I think that his subconscious attitude that *if many other people are doing something illegal on the internet, then it must be okay * reflects what many other people think, given the ridiculously high download counts for certain leaked content such as movies and last in the series books about teenage wizards. [/hide]
[b]It seems in this age where teenagers like myself are growing up with computers and internet, many of us don't understand how laws apply to the internet.
What common teenager activities on the net do you think are illegal?
Which of those do you think should be legal?
Which activities do you think are almost illegal but are technically allowed?[/b]

[hide="Which of the hidden am I allowed to do? There is no need to answer any of these."]Download a music file off youtube.
Download an entire video off youtube or video google or any other
Download a music file using a file sharing thing without checking whether I would normally have to pay for the music?
Use clipart in my school presentations.
Use clipart in my school presentations that promote something that people pay for.
Ripping "sprites" or character symbols from video games such as the mario character from an old NES game and using the symbols in your own online comics.
Copying and pasting the lyrics to a song in your blog.
Copying a several paragraph expert from a book.[/hide]

Why doesn't my second "hide, work?"
[color=green]Fixed by moderator.[/color]

Answer to the technical question: ' is a special character which interrupts BBCode.

To Pul de Algodoncito,

Yes, but I can't understand any of that legal jargon I read. Why doesn't some government agency put it all out in plain English so I can better follow your advice?

To jmerry,

What is a technical character? Do you mean an apostrophe as you wrote? Did I use one in a bad place? What interrupted the BBCode?

The issue here is not so much one of legal jargon as it is one of [b]enforcement.[/b]  I'm sure most teenagers are aware, on some level, that downloading mp3s constitutes copyright infringement, but most of the ones I know (myself included) rationalize it in some way with no particular ill effects.  Do you know why?  It's because it is [b]very hard to punish.[/b]

The situation is even worse than in Prohibition, and if you've studied that particular era of our history you know how well that turned out.  When a law cannot be consistently and systematically applied, it is not followed.  It's as simple as that.

I don't think that people would have the time to look through the millions of websites in the world to find examples of copyright infringement, although there are some popular websites where this occurs frequently, such as YouTube.  And if an example of copyright infringement is found, then in order to file the appropriate legal action, they would have to track down the guilty party.  There are probably ways to do this, but then again, do these people have the time?

[quote="aufha"]Use clipart in my school presentations.
Use clipart in my school presentations that promote something that people pay for.[/quote]

This unquestionably should and do qualify as fair use.  They also qualify as "things not worth suing about."

@Mlok: See youtube-dl. And then you can get just the music part of the video by transcoding it. :)

This "juerge" is for specifying cases which some clausules apply. It refers to some special considerations. But every single case is told on the "Police Privacy" and all of it. It must say something like: blablabalbal----you can download things from here------ and blablabla (a million af restrictions but that doesnt forbid you to download it into your own computer) or something like that. That should allow you to download what you want or everything.

About things on internet like ppt presentations or that stuff online, is not ruled if it doesnt have copyright or something like that, you also can download it.

[quote="Pul de Algodoncito"]is not ruled if it doesnt have copyright or something like that, you also can download it.[/quote]
Almost everything is copyrighted, actually.  (There is an "implicit copyright," i.e. I don't need any official status or documentation to claim copyright on my work.  Of course, most people don't actually care if you use things they could have copyrighted.)  A good general policy (to avoid copyright violations, plagiarism and general aura of  impropriety) is to indicate your sources in a clear and useful way (which depends a great deal on the particular context that you happen to be in).

[quote="mlok"]How do you download an entire video off YouTube?  :maybe:

Quoting song lyrics in your blog would be fair use, I suppose. Just do it in moderation, without turning your blog into an unauthorized lyrics depository.[/quote]

Find appropriate downloader on web.
Keyword: flash video

[quote="Pul de Algodoncito"]"What is not forbidden is allowed" If an activity is explicity forbidden you must not do it. If not, you cant be breakin a law if it is not written  :lol:[/quote]
Not knowing that something is forbidden doesn't protect you!
That's one of the main problems of this: you may do something illegal online without even knowing. Additionally, there are idiots searching for stupid reasons (e.g. on webpages) to sue others because of really dumb reasons (e.g. wrongly written addresses in impressum). They do this for advantages and money only.


I just have to post this: http://www.workjoke.com/projoke.htm

hahahha they're so funny!!
and they're easy to get for someone like me!!

 :rotfl:  :rotfl:  :rotfl:


inequalities calculus function trigonometry

Show that cos(sin(x)) > sin(cos(x)) for all x

This took me really long to graph (I don't have a graphing calculator);
I hope it helps, I can't think of a way to prove the inequality analytically.

hello
[hide="hint by calculas"]
consider a function $ cos(sin(x)) \minus{} sin(cos(x))$
differentiate it u and u will find at 0 $ f'(x) \equal{} 0$ treat it differently for $ \minus{} ve$ and $ \plus{} ve$ nos.
U well get the result.
 [/hide]
Thank u.
Well I m posting the graph too to show the work of ManuelS.

The minimum of $ \cos(\sin(x))$ occurs where $ x \equal{}\frac{\pi}{2}$.

$ \cos(\sin(\frac{\pi}{2})) > \sin(\cos(\frac{\pi}{2})) \\ \\
\Rightarrow \cos(1) > \sin(0) \\ \\
\Rightarrow 0.5403 > 0$

Likewise, the maximum of $ \sin(\cos(x))$ occurs where $ x \equal{} 0$.

$ \cos(\sin(0)) > \sin(\cos(0)) \\ \\
\Rightarrow \cos(0) > \sin(1) \\ \\
\Rightarrow 1 > 0.8415$

Therefore $ \forall x,\,\cos(\sin x) > \sin(\cos x)$

[quote="Differ"][quote="Dr. No"]The minimum of $ \cos(\sin(x))$ occurs where $ x \equal{} \frac {\pi}{2}$.

$ \cos(\sin(\frac {\pi}{2})) > \sin(\cos(\frac {\pi}{2})) \\
 \\
\Rightarrow \cos(1) > \sin(0) \\
 \\
\Rightarrow 0.5403 > 0$

Likewise, the maximum of $ \sin(\cos(x))$ occurs where $ x \equal{} 0$.

$ \cos(\sin(0)) > \sin(\cos(0)) \\
 \\
\Rightarrow \cos(0) > \sin(1) \\
 \\
\Rightarrow 1 > 0.8415$

Therefore $ \forall x,\,\cos(\sin x) > \sin(\cos x)$[/quote]

I'm not sure how that works...[/quote]

$ \cos(\sin x) > \sin(\cos x) \leftrightarrow \cos(\sin x) \minus{}\sin(\cos x) > 0$

$ f(0) \equal{} \cos(\sin 0) \minus{}\sin(\cos 0)$ is a minimum, and $ f(0) >0$.

Therefore, $ \cos(\sin x) >\sin(\cos x)\quad \forall x$.

$ \cos(\sin(\theta))- \sin(\cos(\theta)) > 0$

$ \Rightarrow \cos(\sin(\theta))- \cos(\frac{\pi}{2} -\cos(\theta)) > 0$

$ \Rightarrow -2 \sin (\frac{(\sin(\theta)+ \frac{\pi}{2} -\cos(\theta)}{2} ). \sin (\frac{(\sin(\theta) +\cos(\theta)- \frac{\pi}{2} }{2} )  >0$

Note that:
$ \sin(\theta) + \cos(\theta) = \sqrt{2} \sin (\theta + \frac{\pi}{4}) \qquad(1)$
$ \sin(\theta) - \cos(\theta) = \sin(\theta - \frac{\pi}{4} ) \qquad(2)$

From $ (1)$;
$ \sin (\frac{(\sin(\theta) +\cos(\theta)- \frac{\pi}{2} }{2}  =\sin (\frac{\sqrt{2}}{2}\sin(\theta + \frac{pi}{4}) - \frac{\pi}{4})$

Since $ 2\sqrt{2} < \pi \longrightarrow     -\frac{\pi}{2}< \frac{\sqrt{2}}{2}\sin(\theta + \frac{pi}{4}) - \frac{\pi}{4} < 0$

$ \Rightarrow \sin (\frac{\sqrt{2}}{2}\sin(\theta + \frac{pi}{4}) - \frac{\pi}{4})<0$

From $ (2)$; (and for similar reasons)
${ \sin (\frac{(\sin(\theta)+ \frac{\pi}{2} -\cos(\theta)}{2} ) = \sin(\sin(\theta - \frac{\pi}{4}) + \frac{\pi}(4} ) >0$

Therefore, from $ (1)$ and $ (2)$
$ \Rightarrow -2 \sin (\frac{(\sin(\theta)+ \frac{\pi}{2} -\cos(\theta)}{2} ). \sin (\frac{(\sin(\theta) +\cos(\theta)- \frac{\pi}{2} }{2} )  >0$

$ \Rightarrow \cos(\sin(\theta))- \sin(\cos(\theta)) > 0$
QED

[quote="Dr. No"]The minimum of $ \cos(\sin(x))$ occurs where $ x \equal{} \frac {\pi}{2}$.

$ \cos(\sin(\frac {\pi}{2})) > \sin(\cos(\frac {\pi}{2})) \\
 \\
\Rightarrow \cos(1) > \sin(0) \\
 \\
\Rightarrow 0.5403 > 0$

Likewise, the maximum of $ \sin(\cos(x))$ occurs where $ x \equal{} 0$.

$ \cos(\sin(0)) > \sin(\cos(0)) \\
 \\
\Rightarrow \cos(0) > \sin(1) \\
 \\
\Rightarrow 1 > 0.8415$

Therefore $ \forall x,\,\cos(\sin x) > \sin(\cos x)$[/quote]

This seems incorrect to me. Basically, you used this fact. Let $ f(x_1)$ be the min of $ f$. Let $ g(x_2)$ be the max of $ g$. Then $ f(x_1)\ge g(x_1)$ and $ f(x_2)\ge g(x_2)$ implies that $ f(x)\ge g(x)$ for all $ x$.

I think you might be confusing this with $ f(x_1)\ge g(x_2)\implies f(x)\ge g(x)\forall x$.

I know its not good form, but i think i will give the link to a solution in another forum

http://goiit.com/posts/list/trignometry-which-is-great-sin-cos-x-or-cos-sin-x-give-me-proof-74807.htm#368352

[quote="Dr. No"]

The function $ h(x) \equal{} \cos(\sin x) \minus{} \sin(\cos x)$ has a minimum at $ x \equal{} 0$.

[/quote]

Altheman was referring to the first solution you posted. This statement on the second one, however, requires proof.

EDIT: Even more so because the graph above suggests this is not the minimum.

[quote="Dr. No"][quote="Altheman"]
I think you might be confusing this with $ f(x_1)\ge g(x_2)\implies f(x)\ge g(x)\forall x$.[/quote]

No, this particular problem is not of the form $ f(x_1)\ge g(x_2)$; it is of the form $ f(x)\ge g(x)$. The argument $ x$ is the same on both sides of the inequality.

The function $ h(x) \equal{} \cos(\sin x) \minus{} \sin(\cos x)$ has a minimum at $ x \equal{} 0$. Furthermore, $ h(0) > 0$.

If $ \exists y \ni \sin(\cos y) \ge \cos(\sin y)$ then $ \min (h(x)) \le 0$

It is not the case that $ \min (h(x)) \le 0$ thus it is also not the case that $ \exists y \ni \sin(\cos y) \ge \cos(\sin y)$.[/quote]

This is already wrong...
H[x_] := Cos[Sin[x]] - Sin[Cos[x]]
In[6]:= H[0.]

Out[6]= 0.158529

In[8]:= H[.2]

Out[8]= 0.149796

So H[0]>H[0.2] So H does not have a minimum at 0.

[quote="ffao"]
Altheman was referring to the first solution you posted. This statement on the second one, however, requires proof.[/quote]
I didn't intend them to be separate proofs.

[quote="ffao"] Even more so because the graph above suggests this is not the minimum.[/quote]

You're quite correct. $ h(x)$ does not have a minimum at $ x \equal{} 0$. I made an assumption without checking the second derivative. (I tried to avoid an appeal to the calculus at all -- it's not really appropriate for a high school audience.) Nonetheless,

$ \text{Let }h(x) \equal{}\cos(\sin x) \minus{}\sin(\cos x)\text{  then} \\ \\ h'(x) \equal{} \sin x\cdot \cos(\cos x)\minus{}\cos x\cdot\sin(\sin x)\text{ and } \\ \\ h''(x) \equal{} \cos x\cdot\cos(\cos x) \plus{}\sin^2 x\cdot\sin(\cos x)\minus{}\cos^2 x\cdot\cos(\sin x)\plus{}\sin x\cdot \sin(\sin x)$

The zeros of $ h'(x)$ are $ x \equal{} \pm \pi, x \equal{} \pm 0.6927, \& \, x \equal{} 0$.

Of these, $ x \equal{} \pm \pi \text{ and }x \equal{}0$ are relative maxima because $ h''(0) < 0\text{ and } h''(\pm \pi) < 0$ indicating a downward curvature at these points.

By contrast, $ h''(\pm 0.6927) > 0$, indicating upward curvature at these points; implying $ h(\pm 0.6927)$ are minima of $ h(x)$.

In any case, $ h(\pm 0.6927) > 0$ and the remainder of the proof holds: $ \forall x, \cos(\sin x) > \sin(\cos x)$ QED.


Find the median of all positive odd multiples of 3 which are less
than 55.

the odd multiples of 3 are 3,9,15,21,27,33,39,45,51 so the median is 27


I'll probably end up with a Silver or a Bronze and I was wondering what prize books AoPS be offering. Do you have a specific list?

[quote="NoSoupForYou"]Well, I got an autographed copy of [i]Countdown [/i]for my Round 2 score. Not sure what the other prizes are though.[/quote]
The random prizes are fun surprises  :lol: but I think the selection of prizes varies each year depending on what's available.  Usually the list includes both math topic books (ex. geometry, diophantine equations), generally problem-solving books (like books with Olympiad problems or old contests), general interest math books (things like Countdown), and software (last year Geometer's Sketchpad was available, iirc).  Hopefully that will give you an idea of what sort of prizes are available; unfortunately, I don't think a specific list exists at this point.

Hmm.... how are the random prizes distributed?
Does anyone remember the books from last year?

[quote="1234567890"]Hmm.... how are the random prizes distributed?[/quote]
Your name is entered in a random draw once for every five you receive on a problem in a round, and again for each commended score.  Some names (I don't know how many) are then drawn, and prizes are offered to each of those drawn; this is repeated on each round.  So, you have a better chance of winning a random prize if you score better and are commended more often.

Out of curiosity, then what's up with the "gold," "silver" or "bronze" level prizes?

JB


Find all polynomials $p(x)$, with real coeffcients, satisfying
$(x-1)^2p(x)=(x-3)^2p(x+2)$


algorithm
logarithms
number theory
prime factorization
number theory unsolved

Use the Fast Exponential algorithm to determinate

$38^{67}mod 77$

The most important problem is that I don't have any idea about this algorithm and I can't find anything about it that help me to solve the problem :(

The algorithm being referred to is probably based on this principle:
Decompose the exponent 67 as a sum of powers of two.
Repeatedly square the base 38 mod 77.
Multiply the appropriate terms of the $38^{2^{k}}$ sequence together mod 77.

This can be optimized, of course: the ideal version uses a constant amount of memory (it doesn't store the sequence of powers of the base or the details of the binary expansion permanently) and a number of multiplications proportional to the logarithm of the exponent.

A short demonstration.  $67 = 64+2+1$.  Now,

$38^{2}\equiv 58 \bmod 77$
$38^{4}\equiv 58^{2}\equiv 19^{2}\equiv 53 \bmod 77$
$38^{8}\equiv 53^{2}\equiv 24^{2}\equiv 37 \bmod 77$
$38^{16}\equiv 37^{2}\equiv 40^{2}\equiv 60 \bmod 77$
$38^{32}\equiv 60^{2}\equiv 58 \bmod 77$
$38^{64}\equiv 58^{2}\equiv 53 \bmod 77$

Hence

$38^{67}= 38^{64+2+1}\equiv 53 \cdot 58 \cdot 38 \equiv \boxed{ 3 }\bmod 77$

This is not really ideal, however, since $\lambda(77) = 30$ which shortens your work considerably.

[quote="t0rajir0u"]This is not really ideal, however, since $\phi(77) = 30$ which shortens your work considerably.[/quote]

But ... to take advantage of Euler's Theorem, you would have to know what $\phi(n)$ is, which is essentially the same as knowing the prime factorization of $n.$ Of course, you know the prime factorization of $77,$ but what happens if we make all of the numbers in this exercise 100-digit numbers? You're not going to do this calculation by hand, but with proper programming, a computer will do this calculation rapidly. (The size of the "constant memory" jmerry referred to does depend on the size of the numbers involved, but only linearly.) I'm pretty sure the algorithm is built into [i]Maple[/i] and [i]Mathematica,[/i] as well as the more specialized number theory packages.

Thank you for the answers.

Can anybody explain me how the algorithm would be using $\phi(n)$?

I have another exercise where I have to determine $3^{1386}\mod{1387}$ using the Fast Exponential Algorithm and then use the result obtained to deduce that 1387 is a composite integer and I think maybe it will be with something relative to $\phi(n)$

With paper and pen (I can't use computer, only non programmable calculador and pen)

[quote="charo"]Can anybody explain me how the algorithm would be using $\phi(n)$? [/quote]

As Kent points out, my comment is irrelevant as far as the uses to which this algorithm is normally put.

[quote="charo"]I have another exercise where I have to determine $3^{1386}\mod{1387}$ using the Fast Exponential Algorithm and then use the result obtained to deduce that 1387 is a composite integer[/quote]

A corollary of Fermat's Little Theorem is that if $a^{p-1}\not \equiv 1 \bmod p$ then $p$ cannot be prime.  (The inverse, however, is not true; composite numbers $p$ such that $a^{p-1}\equiv 1 \bmod p$ for all $(a, p) = 1$ are called Carmichael numbers and are a kind of pseudoprime.)

So once you calculate $3^{1386}\bmod 1387$, any result other than $1$ shows that $1387$ cannot be prime.

At SIMUW last summer William Stein, the creator of SAGE, was implementing this method of calculating powers quickly in SAGE in between teaching us elliptic curves/number theory/encryption.

He inspired at least one camper to create a similar program on his TI-89 :P


Prove or disprove

Exist a function $f: N\rightarrow N$ such that $f(f(n))=2n$

Yes, it is $f(x) = x\sqrt 2$.

EDIT: noo, mistake. it's $N \to N$ :)

[hide]
The following function works for nonnegative integers $a,b$:
\[f(2^{a}(4b+1))=2^{a}(4b+3)\]
\[f(2^{a}(4b+3))=2^{a+1}(4b+1)\]
[/hide]

[hide="what I did to arrive at my construction"]
I first noted that any function $f$ must satisfy
\[f(2n)=2f(n)\qquad (*)\]
for all positive integers $n$ (consider $f(f(f(n)))$).

By (*), it suffices to look at $f(n)$ for odd $n$. My next idea was to look at $f(1)$. I found that $f(1)$ could not be a multiple of 4, otherwise
\[2=f(f(1))=4f\left( \frac{f(1)}{4}\right)\]
(applying (*) twice) which is not possible. Therefore, the exists an odd integer $m$ such that $f(1)=m$ or $f(1)=2m$. In the second case, $f(2m)=2$ so $f(m)=1$. So the first case results in
\[f(1)=m, f(m)=2, f(2)=2m, f(2m)=4,...\]
and the second results in
\[f(m)=1,f(1)=2m,f(2m)=2,f(2)=4m,...\]
When we apply $f$ repeatedly, the greatest odd divisor of the result switches between two numbers (namely $1$ and $m$).

I then reasoned that I could split the odd numbers into pairs construct a function that switches the greatest odd divisor between two elements in a pair. My answer above is an example of this (here $4b+1$ is paired with $4b+3$).
[/hide]

[quote="scorpius119"][hide="what I did to arrive at my construction"]
I first noted that any function $f$ must satisfy
\[f(2n)=2f(n)\qquad (*) \]
for all positive integers $n$ (consider $f(f(f(n)))$).

By (*), it suffices to look at $f(n)$ for odd $n$. My next idea was to look at $f(1)$. I found that $f(1)$ could not be a multiple of 4, otherwise
\[2=f(f(1))=4f\left( \frac{f(1)}{4}\right) \]
(applying (*) twice) which is not possible. Therefore, the exists an odd integer $m$ such that $f(1)=m$ or $f(1)=2m$. In the second case, $f(2m)=2$ so $f(m)=1$. So the first case results in
\[f(1)=m, f(m)=2, f(2)=2m, f(2m)=4,... \]
and the second results in
\[f(m)=1,f(1)=2m,f(2m)=2,f(2)=4m,... \]
When we apply $f$ repeatedly, the greatest odd divisor of the result switches between two numbers (namely $1$ and $m$).

I then reasoned that I could split the odd numbers into pairs construct a function that switches the greatest odd divisor between two elements in a pair. My answer above is an example of this (here $4b+1$ is paired with $4b+3$).
[/hide][/quote]

How exactly did you get $f(2n)=2f(n)$? :maybe:

Both expressions equal $f(f(f(n)))$:
\[2n=f(f(n))\Rightarrow f(2n)=f(f(f(n)))\]
\[y=f(n),\ 2y=f(f(y))\Rightarrow 2f(n)=f(f(f(n)))\]


calculus
integration
algebra unsolved
algebra

[b](1) If the equation x^3+kx^2+k^2 x=7 has at least one real root and whose integral part is one. then find the range of k.[/b]

Let $ f(x)\equal{}x^3\plus{}kx^2\plus{}k^2 x\minus{}7\equal{}0$
To satisfy the conditions:
    [b]Case 1[/b]:f(1)<0 and f(2)>0
            $ k^2\plus{}k\minus{}6<0$ and $ 2k^2\plus{}4k\plus{}1>0$
            Solve them and we get $ 2>k> \frac {1}{\sqrt{2}}\minus{}1$
    [b]Case 2[/b]:f(1)>0 and f(2)<0
            which has no solution
    [b]Case 3[/b]:There is a min.pt. or max pt. in (a , b)
             So $ f'(x)\equal{}3x^2\plus{}2kx\plus{}k^2\equal{}0$ for some x in (a, b) , but discriminant of f'(x)=-8k^2<0
              unless k=0 which is included in the range in Case 1
    [b]Case 4[/b]:f(1)=0
             , i.e.$ k^2\plus{}k\minus{}6\equal{}0, k\equal{}2 or \minus{}3$
            So  $ 2 \ge k> \frac {1}{\sqrt{2}}\minus{}1 , k\equal{}\minus{}3$ is the range of k .


function
algebra
domain
algebra unsolved

1, find all real function satisfy

$ f(f(x) \plus{} y) \equal{} f(f(x) \minus{} y) \plus{} 4f(x)y$.

Easy to observe that $ f(x)\equal{}x^2,\: f\equiv 0,$ is solution, and if $ z\in Im f$ we have
\[ f(z\plus{}y)\equal{}f(z\minus{}y)\plus{}4zy\Rightarrow f(2z)\equal{}(2z)^2\plus{}f(0)\], so can be deduced that $ f(x)\equal{}x^2,\forall x\in 2Im f$.

[quote="cckek"]Easy to observe that $ f(x) \equal{} x^2,\: f\equiv 0,$ is solution, and if $ z\in Im f$ we have
\[ f(z \plus{} y) \equal{} f(z \minus{} y) \plus{} 4zy\Rightarrow f(2z) \equal{} (2z)^2 \plus{} f(0)
\]
, so can be deduced that $ f(x) \equal{} x^2,\forall x\in 2Im f$.[/quote]
i have found that $ f(x)\equal{}x^2,\ \forall x\in \{2f(a),2f(a)\minus{}2f(b)|\ a,b\in\mathbb{R}\}$

:maybe:  I don't really understand your solution ? because I want to find real function but you only find it with x $ \in$ im$ f$ .

[quote="cckek"]Easy to observe that $ f(x) \equal{} x^2,\: f\equiv 0,$ is solution, and if $ z\in Im f$ we have
\[ f(z \plus{} y) \equal{} f(z \minus{} y) \plus{} 4zy\Rightarrow f(2z) \equal{} (2z)^2 \plus{} f(0)
\]
, so can be deduced that $ f(x) \equal{} x^2,\forall x\in 2Im f$.[/quote]
What is the meaning of Im f?

Im $ f \equal{} \{f(x): x\in D\}$ where D is the domain of definition of f

nobody help me ? :(

I need  your help ,now  :(

I have just read solution of my brother .it 's interesting  :)

[b]solution[/b]

give$ g(x) \equal{} f(x) \minus{} x^{2} . \Rightarrow g(f(x) \plus{} y) \equal{} g(f(x) \minus{} y).$  with all $ x_{1},x_{2}$ $ \in R$ we have $ y_{1},y_{2}$ such that $ f(x_{1}) \plus{} y_{1} \equal{} f(x_{2}) \plus{} y_{2} \equal{} a$ (a : const)
$ \Rightarrow g(f(x_{1}) \plus{} y_{1}) \equal{} g(f(x_{2}) \plus{} y_{2})$
 $ \Rightarrow g(f(x_{1}) \minus{} y_{1}) \equal{} g(f(x_{2}) \minus{} y_{2})$
$ \Rightarrow g(2f(x_{1}) \minus{} a) \equal{} g(2f(x_{2}) \minus{} a)$
$ \Rightarrow g(2f(x) \minus{} a) \equal{} g(a)$ = const with all$ x \in R$
with $ x_{0}$  we give$ 2f(x_{0}) \minus{} a \equal{} b$
$ \Rightarrow g(a) \equal{} g( 2f(x_{0}) \minus{} a ) \equal{} g(b) \equal{} g(2f(y_{0}) \minus{} b) \equal{} g(b)$
$ \Rightarrow g(a) \equal{} g( 2f(x) \minus{} a ) \equal{} g(b) \equal{} g(2f(y) \minus{} b) \equal{} g(b) \equal{} g(2f(y) \minus{} 2f(x) \plus{} a )$  with all x,y

 $ \Rightarrow g(2x \plus{} a )$= const    $ \Rightarrow g(x)$ =const  $ \Rightarrow f(x) \equal{} x^{2} \plus{} c$


Half lives of two uranium isotopes $ U^{238}$ and $ U^{235}$ are known to be $ 4.5\times 109$ yr and $ 7.1\times 108$ yr, respectively. If the earth was formed with equal amounts of the two isotopes, estimate the current age of the earth given that uranium ores are $ 99.29%
$ of $ U^{238}$ and $ 0.71 %
$ of $ U^{235}$ by number.

Since we have the half lives for both ispotopes, we can find decay constants for each by using k= ln(2)/T where k is decay constant and T is half-life. Now we have k for both isotopes.

Assume, t years has passed since this decay started i.e. earth is t years old. Also, the question says start with equal amount of these isotopes, (by number) so say N.

Now, use the forumla: N(present)=N(start)*e^(-kt) where N(present) is the current number of isotopes and N(start) is the number we started with.

N(start) will be the same for both isotopes. Set two equations for 2 isotopes. Remember k will be different but t will be the same. Now, divide one by another....N(start) cancels off. As we have the data for N(present) for both isotopes, substitute them in the fraction....then its just a matter of arithmetic to solve for t.

I got t= 6*10^9 years which I think is pretty close.

Correct me if i am mistaken somewhere during the calculation.

Thanks.

The solution is correct. but the resultant is incorrect.

  With my calculations, i found 4.16 * 10^9 for this problem.

  As you know in reality the estimated age of the earth is around 4.5 * 10^9 years ,

 http://en.wikipedia.org/wiki/Age_of_the_Earth

 And it supports my answer.

Ok, my calculations might have been off.

actually the almost precise age of the earth is 4.6*10^9 years so u also got lost!
 :?:  :roll


Dear mathlinkers, consider the following equation

$ x^y\equal{}y^x$ ($ x,y\in\mathbb{R}$)

Obviously, for $ \forall k\in\mathbb{R}$, $ (k,k)$ is its solution

I wonder if their exists other solutions s.t. $ x\neq y$ just like$ (2,4)$ and $ (4,2)$? If so, is the number of this kind of solution $ (x,y)$ infinite?

Further more, I tried to graph this equation in Mathematca 7 but failed, I wonder how can I graph the equation? :blush:


Given set $ S$ satisfy $ |S| \equal{} n$ and $ n\geq 7$. Let $ A_1,A_2,\ldots,A_k$ be $ k$ subsets of $ S$ satisfy:
a. $ |A_i| \equal{} 3,\forall i \equal{} \overline{1,k}$
b. $ |A_i\cap A_j| \equal{} 1,\forall i,j \equal{} \overline{1,k},i\neq j$
Find $ \max{k}$.

My result is $ \max{k} \equal{} \max{\left(7;\left[\frac {n \minus{} 1}{2}\right]\right)}$, could you please check it :blush:

In general, it's hard to check without first having an explanation from you as to why you think your answer is correct.

For this problem, however, you may want to take a closer look at the case $ n\equal{}5$ (it says $ n\equal{}7$ in the problem, but surely the answer can't decrease as $ n$ increases...)

You are right,mashimaru
If we let $ k$ instead of $ 3$ in your first condition,the result will be $ \max{k} \equal{} \max{\left(7;\left[\frac {n \minus{} 1}{k\minus{}1}\right]\right)}$,The proof,I think ,will be same to your


$ ABC$ is a triangle with the vertices having as complex coordinates $ A(a)$, $ B(b)$, $ C(c)$, such that $ |a|\equal{}|b|\equal{}|c|\equal{}r$ such that $ a\plus{}b\plus{}c\equal{}r^{3}$. Find the ratio between the areas of $ \triangle{MNP}$ and $ \triangle{ABC}$, if $ M(m)$, $ N(n)$, $ P(p)$ have as complex coordinates $ m\equal{}\frac{ab\plus{}ac}{bc}$, $ n\equal{}\frac{bc\plus{}ba}{ca}$, $ n\equal{}\frac{ca\plus{}cb}{ab}$.

We use the fact that given a triangle $ XYZ$ with $ X(x), Y(y), Z(z)$ we have that $ S_{[XYZ]}\equal{}\frac{1}{2}Im(\overline{x}y\plus{}\overline{y}z\plus{}\overline{z}x)$,
where $ S_{[XYZ]}$ is the area of $ XYZ$
So, $ S_{[ABC]}\equal{}\frac{1}{2}Im (\overline{a}b\plus{}\overline{b}c\plus{}\overline{c}a)$.

$ a\plus{}b\plus{}c \equal{} r^{3}$ so $ \overline{a}\plus{}\overline{b}\plus{}\overline{c}\equal{} r^{3}$ $ \leftrightarrow$ $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{} r$ $ \leftrightarrow$ $ \frac{ab\plus{}bc\plus{}ca}{abc}\equal{} r$.
That means $ m \equal{} ar\minus{}1$ and the corresponding identities...
$ S_{[MNP]}\equal{}\frac{1}{2}Im((\overline{a}r\minus{}1)(br\minus{}1)\plus{}(\overline{b}r\minus{}1)(cr\minus{}1)\plus{}(\overline{c}r\minus{}1)(ar\minus{}1))$ $ \leftrightarrow$ $ S_{([MNP])}\equal{}\frac{1}{2}r^{2}Im (\overline{a}b\plus{}\overline{b}c\plus{}\overline{c}a).$

The ratio is $ r^{2}$.


superior algebra
superior algebra theorems

Perhaps I am missing something very basic, but why is the following a group?

({a+b(2)^(1/2) |  a,b is an element of [b]Q[/b]}, +)

Doesn't it fail closure?

Thanks.


geometry
AMC
AIME
number theory
AMC 10
AMC 8

After tomorrow, which is 1/24, the COP The Second will end. Then, perhaps a week after AMC, I'll write new COP, which is the COP The Third. Before I can even do that, I need to see who are interested and anything else that I need. So, if you're interested, please sign up here by "In" or any words that indicates you that you're interested.

Few confirmed test information:

COP, stands for Contest of Opportunity, is a contest that is open to EVERYONE in the Earth. This is a private messaging sending contest where contestants are given 1 week to do each round. Sometimes, it may little bit longer or lesser than 1 week but average is one week.

Rules for each round

Round 1 consists of very basic 5 questions from at least 5 different branches of math. Often times, the questions are coming from geometry, algebra, basic number theory, counting, sequence and series. The best way to study for this round is to just learn a lot of stufffs. The contestants take 1 week to solve the problems. The only way you can get to the next round is if and only if you get a perfect score. This is not hard since all you have to do is solve 5 basic problems in 1 week.

Round 2 consists of 7 questions that are in level of Getting Started and Intermediate. Some questions just require basic knowledge while some questions, you actually have to work out. This round is also about one week. The qualifying is only two top score. If there's a tie, we have a tiebreaker round, or round 3.

Round 3, if it's tiebreaker, consists of two proof for a week. The two proof questions are basic but I look for elegant solutions. There is no specific rules for how scoring works but simply, I grade them.

Round 4, is the Championship Round. In the First COP, there were 5 questions but from the Second COP and so on, there will be 15 questions, approximately in level of AMC-10 to AMC-12, perhaps easy AIME style questions. The question is not impossible but they require thinking and specific knoledge.

How to Prepare for This?

The best way to prepare is to study many math competitions. Especially for Round 4, it is highly recommend that one contestant look over AIME or old AHSMEs for practice because that helps a lot. For other rounds, look over easy AMC-8 or any other elementary math contests. I'm planning to write a math problem book, just for this competition, and only thing is I'm not sure how long it'll take. If I publish one, it would contain rougly 100 problems. But that's for later on. :)

Any questions are welcome here. Make sure, if you're interested, sign up here.

I would like to join :gerbil:

(inside joke for any of you who might've gone to the iDB at some point)

[quote="white_horse_king88"]Am I still allowed to do it?  :D[/quote]

It is open to everybody in this planet.

So far, there's no age limit. I might make one later but oh well.

Well, it should only be for people at a getting started level.  It would be unfair if somone like ComplexZeta, or TripleM decided to join.

I hardly assume either of them would compete in this competition. If they do, it would be my honor.

 :)

But if they really are, I'll maybe make special Olympiad Style Problems just for them. Of course, if I CAN make one.

Just announcement:

For middle schoolers, you may take an easier test called COPSD.

More detailed information can be found here:

http://www.artofproblemsolving.com/Forum/post-155670.html#p155670

But this is only if you want to. This doesn't mean you CAN"T attend regular COP Round 1. I suggest you do Round 1 if you're strong and confident on this round but if not, I think you're better off with COPSD.

is it too late to join? because if not, i sign up too

I guess I can do it, just hope I don't come last  :( .

Is it too late to join for the round 3? I have missed the first two rounds.

No, it's the Third COP, not third round.

You may join, yes, you can.

Each COP has five rounds (can be four if there's no tie) and there were two pass COPs (First COP and Second COP).

Third COP will begin in March 7th. So, you may join. :)

this might be really challenging for me, but I'll give it a try :blush:

So far the contestants:

There are 25 contestants.

More would make this contest more exciting! Sign-up whether you're confident or not!

Experience is what matters.

$\cal{MESINESUP}$

 :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P
 :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P
 :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P
 :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P
 :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P
 :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P v
 :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P
 :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P  :P

This topic will be locked to prevent anyone who's going to sign-up but not participate in the contest.

However, you may still send the answers without signing up. :)


Coulomb's law prescribes an infinitely large force to make two point-like charges of identical sign touch. What is the situation if the charges are uniformly distributed on two circular plates? What is the force required to make two circular plates of charge $Q$ and radius $r$ touch?

There is a duplicate topic in Advanced Physics by the same author.  Mods please delete this one. Thanks!  :wink:

I posted again because there were no replies

Thanks


geometry
3D geometry
summer program
PROMYS
algebra
system of equations
number theory

This is very easy but I'm posting it anyway :

Show that there exist infinitely many positive integers A such that
(1) 2A is a perfect square ;
(2) 3A is a perfect cube ;
(3) 5A is a perfect fifth power.

You can use "sub" and "sup" now belenos, so this looks better :

[hide]u(a, b, c) = 215(2a - 1)310(3b + 2)56(5c + 4)[/hide]

[color=cyan]Can't you also say that after you've got one, just multiply it by n^30 for all natural numbers n?[/color]

why did you rip that problem off the promys application?

if not where else did you get that

I'm not too familiar with the PROMYS program, but if those are THIS YEAR's problems then you should know not to post them.

However, if they are last year's then I apologize-please know that this forum is not to cheat on any contest under any cicrumstances.

Thank you,
The management.

"the management" part sounds oh so professional

What is "promys application" ? Never heard of it.

I took the problem from a regional Indian competition.

(Karnataka regional math olympiad (KRMO) 1995, problem 1)

You can check this at http://144.16.74.145/ if you want.

PROMYS is a summer math program specifically focusing on number theory. Here is its website: http://www.promys.org/

has anyone done it?  im thinking of, but not sure...

yeah thats what i meant.  How was it?  Learn a lot?  Would you recommend it?  Im concerned about the cost, is there a substantial amount of financial aid?

[quote="Arne"]It is natural to consider numbers of the form A = 2x3y5z.
Then we have
2A = 2x+13y5z.
3A = 2x3y+15z.
5A = 2x3y5z+1.
Now 2A must be a square so x + 1, y and z must be even.
3A must be a cube so x, y + 1 and z must be multiples of 3.
5A must be a fifth power so x, y and z + 1 must be multiples of 5.
Like this you can derive a system of equations ... which can be solved (a bit nasty), or you can just do some fiddling (like me :mrgreen:).[/quote]

It is interesting to note that the demonstration of the infinite number of solutions based on the Chinese Remainder Theorem.  The exponents of the primes in the prime factorization of a solution are just solutions to a series of linear congruences.


$ n\in\mathbb N.$ Solve the following equation:
$ 13^{n\plus{}1}\minus{}14^n\equal{}2001$

[quote="Daan"]2[/quote]
Maybe there is more roots? :wink:

of course there are one natural root
[hide="because"]
because : $ f$ is increasing ($ f(n) \equal{} 13^{n \plus{} 1} \minus{} 14^n$)
[/hide]
sorry for bad english

[quote="mathson"]of course there are one natural root
because : $ f$ is increasing ($ f(n) \equal{} 13^{n \plus{} 1} \minus{} 14^n$)[/quote]
Are you sure?

$ f$ is, in fact, decreasing if we choose $ n$ large enough. Thus, we only have to consider $ n$'s with $ n<\log_\frac{14}{13} 13$ ;)


Dear all,

I am sorry to announce that we are preparing to disqualify several (at least five) Mathcamp applications due to various violations of the rules of the qualifying quiz.   The violations include plagiarism -- copying solutions (or parts of solutions) from other sources without proper attribution -- and asking for outside help on the qualifying quiz.

If you have applied to Mathcamp and have not properly attributed everything in your application, now is the time to write us and rectify that -- we don't want to penalize an honest mistake so harshly.   Within a couple of days it may be too late.  Please note that we reserve the right to reject anyone who we believe to have plagiarized, incuding rescinding an acceptance if the offense is discovered at a later time.

I apologize for this situation.   This is part of the reason for the delays that I mentioned earlier today.  Though the problem only involves a small number of people relative to the total number of applications, unfortunately it is slowing things down for everyone.

best,
Dave

EDIT: For now, I will be unable to answer individual inquiries about whether or not one is involved.

how would we know if we are disqualified or not...

i am sure i didnt...

[quote="xxreddevilzxx"]how would we know if we are disqualified or not...[/quote]

No one has been disqualified yet.  The problem only involves a small number of people and the problem in each case is clear, so if you are sure you are not involved, you are almost certainly right.

I will be away from the computer for the next couple of hours.

Shouldn't it be immediately obvious though if a student can't explain their approach and/or result?

actually ive had those unproven conjectures... on some of my solutions... because i saw the pattern but i could not explain why. so would that be cheating??

and plaigiarism is more like copying someone elses work, so probably the solution graders have found out exact style of writing, approaches, etc., or even exact words, not necessarily connected to the unexplained result/approach.

[quote="xxreddevilzxx"]actually ive had those unproven conjectures... on some of my solutions... because i saw the pattern but i could not explain why. so would that be cheating??[/quote]

No, certainly not.

[quote]and plaigiarism is more like copying someone elses work, so probably the solution graders have found out exact style of writing, approaches, etc., or even exact words, not necessarily connected to the unexplained result/approach.[/quote]

Precisely.

P.S.  I put it into my own words.  Just forgot to cite the source! :(  :(  :(  :(  :(

[quote="fanaticsm"]Is it possible to find out if an individual is in violation of the aforementioned rules?[/quote]

[quote="ztbb"]EDIT: For now, I will be unable to answer individual inquiries about whether or not one is involved.[/quote]

[quote="mli"]AAAAAAAAARRRRGGGGGGG!!!! I think I might be one of those people![/quote]

Thanks for coming forward, Chenchuan.  We will add the source to your quiz, and continue to consider your application.

Note that it's not necessary to have posted anything publicly.  Private message would have been fine.

[quote="mli"]AAAAAAAAARRRRGGGGGGG!!!! I think I might be one of those people!  My name is Chenchuan Li (Age 16 Address: 3200 Drew Street, Downers Grove, IL 60515), and for quiz problem number 10, I saw a solution on the internet at http://www.cut-the-knot.org/blue/EuclidAlg.shtml
but I forgot to cite it!  SORRY!!!! This was an honest mistake!  Please send me an email if I am still to be disqualified!  My email: asiaknight001@yahoo.com[/quote]

But this game is different from the one on the quiz, right? otherwise.... :(

[quote="mcalderbank"]But this game is different from the one on the quiz, right? otherwise.... :([/quote]

Yes, it's different.


modular arithmetic
number theory
relatively prime

Find primes $p,q,r$ and positive integer $a$ that satisfies $\frac{1}{p}+\frac{1}{q}+\frac{1}{r} = \frac{1}{a}$

Do they have to be distinct? If not,

[hide]$p=q=r=3$
$a=1$[/hide]

Oops, I misread  :blush:  I found all integers...

[hide]
Rewriting:
a = (pqr)/(pq + qr + pr)
Let t = gcd(p,q,r).  So let p = tx, q=ty, r=tz, with gcd(x,y,z) = 1.  Simplifying
a = (txyz)/(xy + yz + xz).  Now we will show that xy+yz+xz is relatively prime to xyz.  Let n be any prime factor of xyz.  n must be a factor of only one of x,y,z, WLOG let it be y.  xy+yz+xz = xz mod n.  However, xz is not 0 mod n since neither x nor z has a factor of n.  Doing this for each prime factor, we find that no factor of xyz is a factor of xy+yz+xz, and thus they are relatively prime.  So xy+yz+xz must divide t.  However, t is independent of x,y, and z, so we can easily write a formula for p,q, and r.

Take arbitrary x,y,z,d.  t = (xy+yz+xz)d, p = tx, q = ty, r = tz, and a = txyz.
[/hide]

OK, using this, I'll solve the actual problem :D

[hide]
Recall: Take arbitrary integers x,y,z,d.  t = (xy+yz+xz)d, p = tx, q = ty, r = tz, and a = txyz make the equation true.
For p to be prime, either t = 1 and x = prime or t = prime or x=1.  t obviously can't be 1 (has to be at least 3), so we must have x=1.  Similarly x=y=z=1.  Thus p=q=r=3d.  So we must have d=1.  So
p=q=r=3
a=1.[/hide]

if im right the only solution is $(3,3,3,1)$:

we easily get $a(pq+pr+qr)=pqr$
because $p,q,r\in\mathbb P$ there are WLOG 4 possible values for $a$: $1$, $p$, $pq$, $pqr$

1. $a=1$:
$pq+pr+qr=pqr$
$r(p+q)=pq(r-1)$
since $\gcd(r,r-1)=1$ we get WLOG either $r=p$ or $r=pq$
1.1. $r=p$:
$r(r+q)=rq(r-1)$
$r+2q=qr$
hence $r$ be divisable by $q$, and since $q,r\in\mathbb P$, $r=q$
$q+2q=q^2$
$q(q-3)=0$ hence $p=q=r=3$ and $a=1$, works out in the equation as well.
1.2 $r=pq$:
this is not possible since $r$ is prime.

2. $a=p$:
$p(pq+pr+qr)=pqr$
$pq+pr=0$ is not possible since all terms of the sum are positive

3. $a=pq$:
$pq(pq+pr+qr)=pqr$
$pq+r(p+q-1)=0$ all terms positiv again

4. $a=pqr$:
$pqr(pq+pr+qr)=pqr$ how should that be possible ??

so we get: only solution is $(p,q,r,a)=(3,3,3,1)$

Here's a shorter solution based on my last post :)

[hide]
Suppose p,q,r are distinct.
(pq+pr+qr)/(pqr) = 1/a
Because the numerator is conguent to $qr \not\equiv 0 \pmod p$, the numerator shares no factor of p with the denominator. Doing the same for q and r, we find that the LHS is fully simplified.
If p and q are the same but r is different, the denominator has a factor of r but the numerator doesn't, so we cannot possibly simplify (pq+pr+qr)/(pqr) so that the numerator is 1.
Thus p=q=r.
3p^2/p^3 = 3/p = 1/a.  The only prime that is a multiple of 3 is p=3.  So (3,3,3,1) is our only solution.[/hide]


How can we prove that for any given matrices $A,B$ , then

$det(AB)=det(A)det(B)$ ?

You can find Kent Merryfield's proof [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=determinants&t=56815]here[/url].

Here is my proof :
Consider the the $n$-linear alternating form $f(X)=\det(AX)$. Because for each $n$-linear alternating function $f(X)=\det(X)f(I)$, so $\det(AX)=\det(X)f(I)=\det(X)\det(A)$

[quote="Omid Hatami"]Consider the the $n$-linear alternating form $f(X)=\det(AX)$.[/quote]

Can you tell me what exactly $n$-linear alternating form means? Thanks.

Look at [url=http://en.wikipedia.org/wiki/Multilinear]here[/url]. You see an $n$-linear function is a function that is linear in rows of matrix $A$ and is alternating iff for a matrix $A$ with 2 equal rows $f(A)=0$.
There is a theorem saying that for each n-linear alternating $f(X)=\det(X)f(I)$


Are you constantly correcting your friends for simple grammar mistakes?

For example, you correct

"Me and Drew are going to the mall."

"*cough cough*. It's actually 'Drew and I are goign to the mall.' "

Most of the time I just do it to annoy them:

Every time a friend says "I did good", I say "Well", which really annoys them.  :lol:

Just one person, and we do that to help him improve.

Why is it my place to correct people? Just last night, I said

"Someone pushed the lutton bights."

Then I said

"Someone pushed the button lights."

THEN I said

"Someone pushed the light buttons"

This question may be offensive to some as the question only applies to people who [i]have[/i] friends. :lol:

[quote="1=2"][quote="bubka"][quote="miyomiyo"]This question may be offensive to some as the question only applies to people who [i]ave[/i] friends. :lol:[/quote]

*cough cough* It's actually "... people who [i]have[/i] friends", it's important not to miss that 'h' in 'have'. :rotfl:[/quote]

*cough cough* It's actually ""This question may be offensive to some as the question only applies to people who [i]have[/i] friends. " [b]It's[/b] important not to miss that 'h' in 'have'."

 :rotfl:  :rotfl:  :rotfl:[/quote]

Actually, you are not supposed to use double quotation marks within other double quotation marks; instead, you should put single quotation marks inside the double quotation marks.

[quote="tjhance"][/quote][quote="1=2"][quote="bubka"][quote="miyomiyo"]This question may be offensive to some as the question only applies to people who [i]ave[/i] friends. :lol:[/quote]

*cough cough* It's actually "... people who [i]have[/i] friends", it's important not to miss that 'h' in 'have'. :rotfl:[/quote]

*cough cough* It's actually ""This question may be offensive to some as the question only applies to people who [i]have[/i] friends. " [b]It's[/b] important not to miss that 'h' in 'have'."

 :rotfl:  :rotfl:  :rotfl:[/quote][quote="tjhance"]

Actually, you are not supposed to use double quotation marks within other double quotation marks; instead, you should put single quotation marks inside the double quotation marks.

This part was absolutely not added by miyomiyo just so he could point out that tjhance made a grammar mistake, it was actually written by tjhance.[/quote]You're supposed to put a semi-colon as opposed to a comma in that last sentence.

Guys guys (and gals), we're not supposed to be correcting each others' grammar...

Was that not the purpose of this thread?  If not, huh?

[quote="7h3.D3m0n.117"]Are you constantly correcting your friends for simple grammar mistakes?[/quote]

Which means relate to past experiances...

I am always correcting my friends' grammar.  i may be a little hypocritical, though, since i always say things like "those was the most best of the two" (which contains at least three blatant grammar errors)

you know hibbett sports?  they're a national sports equipment chain, I guess.  we had to go there so my sister could get tennis shoes. (they didn't have any.)
anyway, the have a mens' section, a womens' section, and a kid's section.

did you notice that?  I guess the entire section is for one kid.

[quote="1=2"][quote="lingomaniac88"][quote="Captain Sugar"]did you notice that?  I guess the entire section is for one kid.[/quote]
Actually, you're supposed to capitalize the first letter in "did," because it's the first letter of a sentence.[/quote]

You're supposed to put the comma out of the quotation marks.[/quote]
nope wrong

Reminds me of a quote from a movie:

Rocco Dillon: I've been watching you McGirk. You handle yourself really good.
Frank Drebin: Really well.
Rocco Dillon: Whatever.

umm no i use terribly grammer all the time

Sometimes, when someone is having trouble to come up with a word, and said what it is not, I just use the opposite of that word, like this:

Mom: Nothing big going on, just......
Me: Small?


i considered this neat figure into 3D. Not 2D.
What.. um. just i did.


geometry
geometric transformation
reflection
Support
probability
conics
parabola

Rep, I think this is a great idea.

Here's another problem people might want to look at:

Find the sum of

[tex]1+3+5+...+41[/tex]

[tex]+6+9+12+...+66[/tex]

[tex]+7+11+15+...+87[/tex]

You can find the number of terms of an arithmetic sequence easily by taking [tex]\frac{z-a}{d}+1[/tex] because if we add d-a to every term (not changing the number of terms), the last term:

[tex]\displaystyle a+(n-1)d = z[/tex]

adding d-a to both sides

[tex]\displaystyle nd = z+d-a [/tex]

therefore,

[tex]\displaystyle n = \frac{z+d-a}{d} \mbox{ or, equivalently, }\frac{z-a}{d}+1[/tex]

Hint:  [hide]Can you make one big arithmetic sequence instead of three?[/hide]

I want to briefly revive this so that beginners who like to hang out in F&G mostly can come and have a look.  If you have been complaining that Getting Started is too hard, well, here's your chance, which means you don't have any more excuses :-) (jk)

I just thought I would make a quick note on terminology.  A [i]sequence[/i] is a list of numbers.  Thus, 2, 5, 8, 11, ... is the beginning of an infinite arithmetic sequence.

A [i]series[/i] is the sum of a sequence.  Thus, for the sequence 2, 5, 8, 11, ..., we have the associated [i]partial sums of the series[/i], 2, 7, 15, 26, ..., which we get by adding successive terms of the sequence.  Every sequence has an associated series and every sequence can be written as the partial sums of another sequence -- for arithmetic sequences, we always have that they are the partial sums of a sequence that looks like a, d, d, d, d, ...

So, sequence = list.
Series = sum of terms of a list.

You can read more about them at:
http://mathworld.wolfram.com/ArithmeticProgression.html
http://mathworld.wolfram.com/ArithmeticSeries.html
http://mathworld.wolfram.com/Series.html

[quote="JBL"]I just thought I would make a quick note on terminology.  A [i]sequence[/i] is a list of numbers.  Thus, 2, 5, 8, 11, ... is the beginning of an infinite arithmetic sequence.

A [i]series[/i] is the sum of a sequence. . . .

So, sequence = list.
Series = sum of terms of a list.

You can read more about them at:
http://mathworld.wolfram.com/ArithmeticProgression.html
http://mathworld.wolfram.com/ArithmeticSeries.html
http://mathworld.wolfram.com/Series.html[/quote]

Thank you, JBL. That terminological distinction is indeed very important to mathematicians. Students of middle school age have undoubtedly met smart adults who mistake this distinction, so it's no embarrassment for a young person to confuse those terms at first--I certainly did until into my thirties, or older. But the younger we can practice making these distinctions in terminology, the better. A lot of mathematics is built on carefully defined terms used in a consistent manner.

(After edit:) Perhaps various persons who posted above may like to edit their own posts (and the title of this thread?) to reflect the difference between the term "sequence" and the term "series." A great thing about AoPS is that we can edit posts.

Oh, it's a mistake I make all the time.  Having tried to teach lessons on sequences and series, I know how confusing it can be.  I actually had to look up the definition of arithmetic series before I was absolutely certain what it was :)

Question!

For arithmetic sequence, I used this formula:

[tex]L=F+M(N-1)[/tex] where L is the last term, F is the first term, M is the common increase, and N is number of terms.

Is this right formula?

Will you do a getting started lesson on modular arithmetic?  It would be very helpful.

This is a good idea.  I would also support such a lesson.  I second the motion.  (of course, this is all up to Rep, and I appreciate what he has done already)

I think other people did too but I'll just do too.

Exercise 1:

[hide]2,5,8 --> x,x+3,x+3(2) then tenth is x+3(9) --> 2+27 --> 29[/hide]

Exercise 2:

[hide]3,7,11 --> L=F+M(N-1) --> L=3+4(1099-1) --> L=3+4392 --> L=4395[/hide]

I tried to do by using formulas and logics so tell me if I'm wrong or correct.

Yes, please!!! Modular arithmetic!!!  I have not answered any of these questions because I am not a beginner in these type of problems, but mod ones I will be.

Probablility -- is it "I second the motion" or "I second the notion" ?  Just wondering b/c that sounds sortof funny.  BTW I am not a grammar expert as is probably evident.

[quote="JS1527"]Yes, please!!! Modular arithmetic!!!  I have not answered any of these questions because I am not a beginner in these type of problems, but mod ones I will be.

Probablility -- is it "I second the motion" or "I second the notion"?[/quote]

The stock phrase is "I second the motion," because a "motion" is a proposal to do something in a formal meeting. There is a Motown record that includes the love song line, "I second that emotion," which I have always thought was a funny joke.

I second the motion, notion, [b]and[/b] emotion :-)

Rep, this seems quite popular...

I don't know if it'll help, but maybe you could teach finite differences for those tougher arithmetic sequences like

1, 2, 8, 19, 35, 56, 82.....

when the constant difference does come immediately between the terms, but i don't even know if the sequence is classified as an arithmetic though. :oops:

That's not considered an arithmetic sequence.  I'm not sure what that's called, except that it fits a parabola if you give the abscissa (x-coordinate) of the nth term to be n.  Those can be calculated if you know that 1^2 + 2^2 + 3^3 + ... + n^2 = n(n+1)(2n+1)/6

Haha...I third the motion for modular arithmetic!  Also, could we have an extension on this for like, on the Getting Started Contest # 5, there was a problem that said something, "If these parts of an arithmetic series form a geometric series, what is the ratio of the geometric series..."  I've had experience with both...and I couldn't get it!  Also, maybe the other questions (I think it was questions 9 and 10) about what is the sum of this series.   I only know of the formula a1/1-r, but that didn't look like it applied to those questions on the contest.

There was a big thread on mods here [url]http://artofproblemsolving.com/Forum/viewtopic.php?t=6603[/url]


hello.  try this problem.

Let $ f(x)$ be a function such that $ f(a\plus{}a^{\minus{}1})\equal{}a^3\plus{}a^{\minus{}3}$ for all $ a$ in the domain of $ f$.  What is $ f(x)$ ?

[hide]As a trial function, use $ f(x) \equal{} x^3$. Therefore, $ f(a\plus{}a^{\minus{}1}) \equal{} a^3 \plus{} a^{\minus{}3} \plus{} 3(a \plus{} a^{\minus{}1})$. Therefore, $ f(x) \equal{} x^3 \minus{} 3x$.[/hide]

I got a different answer.

[hide="My Solution"]
$ f(a \plus{} a^{ \minus{} 1}) \equal{} a^3 \plus{} a^{ \minus{} 3} \equal{} (a \plus{} a^{ \minus{} 1})(a^2 \minus{} 1 \plus{} a^{ \minus{} 2}) \equal{} (a \plus{} a^{ \minus{} 1})((a \plus{} a^{ \minus{} 1})^2 \minus{} 2).$

Therefore, taking $ x \equal{} a \plus{} a^{ \minus{} 1}$, we obtain $ f(x) \equal{} x(x^2 \minus{} 2) \equal{} \boxed{x^3 \minus{} 2x}.$
[/hide]

[hide="The Mistake"]$ a^2 \minus{} 1 \plus{} a^{ \minus{} 2} \equal{} a^2 \plus{} 2 \plus{} a^{ \minus{} 2} \minus{} 3 \equal{} (a \plus{} a^{ \minus{} 1})^2 \minus{} 3$.[/hide]

it is easy

we have \[ f(a\plus{}a^{\minus{}1})\equal{}a^3\plus{}a^{\minus{}3}\equal{}(a\plus{}a^{\minus{}1})(a^2\minus{}1\plus{}a^{\minus{}2})\equal{}(a\plus{}a^{\minus{}1})((a\plus{}a^{\minus{}1})^2\minus{}3)\;\;\Rightarrow \;\;f(x)\equal{}x(x^2\minus{}3)\]
the end.


[quote="Interval"]A pond is enclosed by a wooden deck that is 3 feet wide.  The fence surrounding the deck is 100 feet long.

(a) If the pond is a square, what are its dimensions?

(b) If the pond is rectangular and the length of the pond is 3 times its width, what are the dimensions of the pond?

(c) If the pond is circular, what is the diameter of the pond?

(d) Which pond has the MOST AREA?[/quote]
[hide]My picture explains it all, the cirular one has the most area[/hide]

[quote="Interval"]An absolutely great reply.

I really like the pictures![/quote]
With problems like these, it really helps to draw all the different scenarios.

[quote="Interval"]To the Boston Red Sox fan:

My reply to your question is: NO IT IS NOT a homework question.

I am not in school.

I am over 40 trying to prepare for the upcoming state exam to finally get my teaching credentials.

So, are my questions homework related???

NO!

Thanks.[/quote]

OH!  I DONT THINK HE WAS TRYING TO ACCUSE YOU OR ANYTHING.  GOOD LUCK ON THE STATE EXAM THING!

I never said he was trying to accuse me of posting homework questions.  In fact, he [13375P34K43V312] has been very helpful to me on this forum.  I thank him for his great help.

I just want to make it plain as day for all tutors here.

My questions have ABSOLUTELY NOTHING to do with homework or school assignments.

Thanks for reading.


I need to take the SAT before I apply to college, but I can't do the April or May tests because I have various activities going on those days. That leaves me with June 3rd to take the SAT, but ARML is that day (if I make the team, which I think I will).
I could take the SAT during October, though, but that's kind of late for college admissions, in my opinion.
What do you think I should do?

Take it in June on the alternate day.  They can arrange that for ARML students.  arml.com has more information on that.

The October SAT your senior year is timely even for early round applications, so just don't worry about this at all. Do make sure you get the SAT in SOMETIME--it's penny wise and pound foolish to do activities on so many SAT dates that you forget to take the SAT in time for your college application.

Good luck on your activities, and on the SAT whenever you take it.

[quote="1234567890"]Yeah, I was leaning towards doing ARML.
Hmm...... Sly Si, can you go into more detail about the alternate test date in June? As in give me a step-by-step procedure for taking it the alternate date? The ARML site's info is kind of confusing.
So first I register for the June SAT, then......? How do I get the alternate test date?[/quote]I'd leave 2 dates open for Sat taking, in case you didn't perform your best on the first day.

That's just me though...   :D

1234567890, you should talk to your ARML coach about this, i know some teams (at least NYC) run an alternate test day for the people they send to ARML. i would personally vote for arml (my junior year, i took the SAT's in January and SAT IIs in May cause i had NYSML during the march SAT date and ARML during the june one- that was kind of interesting) but if you feel that the october test is too late i agree with that too! really do look into the alternate date- the website (last time i checked) just says that it is possible to arrange for an alternate test date, but i think it has to be set up through your team coach, so maybe talk to him/her during some practice or try to get in touch ASAP

The alternate date is usually two weeks after, right?
Because I have the ACT on June 10th.......

Too much stuff to do...... :stretcher:

Yup, 2 weeks after. I didn't take the alternate SAT, but I did take the ACT the week after.

I took the SAT for the first time (since 8th grade for HS admissions) in October of my senior year and it worked out with plenty of time to spare. I guess an early admission may be affected though? (I didn't apply early anywhere)

The October SAT is timely for early admission decisions everywhere. Whether or not one applies during the early round is a separate issue, but any SAT taken in October of the last year of high school or earlier can be used in either application round.


Let $x,y,z$ be real numbers greater than 1. Prove that
\[ x^{x^2+2yz} y^{y^2+2zx} z^{z^2+2xy} \geq (xyz)^{xy+yz+zx}. \]

The desired inequality is equivalent to
$(x-y)(x-z)ln(x) + (y-z)(y-x)ln(y) + (z-x)(z-y)ln(z) \geq 0$

which is a Schur's inequality type with $f(x) = ln(x)$ instead of $f(x) = x^r$ in the usual case. But, as discussed before on the forum, the proof can be managed in exactly the same way as the usual proof.

Note that a slightly more detailed solution of this problem and to many others can be found (in french) in the inequality course I wrote at :
http://www.animath.fr

Pierre.


1) A group of boys and girls are in line to buy tickets for a dance. The first person in the line is a boy and the last in line is a girl. Prove that somwhere in the line a boy must be standing directly in front of a girl.

Prove by simple induction that for all n belonging to the Natural #s, n >= 2, in a line of length n, where person 1 is a boy and person n is a girl theres exists a k where 1 <= k < n, such that person k is a boy and (k+1) is a girl.

2) Define a sequence of natural #s, a0,a1,a2,..., as follows:
a0 = 1
a1 = 0
a2 = 0
a(n) = a(n-1) + a(n-2) + a(n-3), for n >= 3.

Prove by complete induction that a(n) >n^2 for all integers n >= 12.

any help would be appreciated!

[quote]
2) Define a sequence of natural #s, a0,a1,a2,..., as follows:
a0 = 1
a1 = 0
a2 = 0
a(n) = a(n-1) + a(n-2) + a(n-3), for n >= 3.

Prove by complete induction that a(n) >n^2 for all integers n >= 12. [/quote]

for this problem, just verfify the inequality for a_12, a_13, a_14, then, use the inequality
(n-1)^2+(n-2)^2+(n-3)^2>n^2 for all n>=15
ps: I did not verify the inequality above, actually if it doesn't hold for all n>=15, just verify more by your hands, since this inequality must have some solution n>=some number.

[quote]

1) A group of boys and girls are in line to buy tickets for a dance. The first person in the line is a boy and the last in line is a girl. Prove that somwhere in the line a boy must be standing directly in front of a girl.

Prove by simple induction that for all n belonging to the Natural #s, n >= 2, in a line of length n, where person 1 is a boy and person n is a girl theres exists a k where 1 <= k < n, such that person k is a boy and (k+1) is a girl. [/quote]

I don't think in this problem--if you call it problem--you will need any help, do you? Spend more time and you'll get it.

[quote="Victor96"][quote]
2) Define a sequence of natural #s, a0,a1,a2,..., as follows:
a0 = 1
a1 = 0
a2 = 0
a(n) = a(n-1) + a(n-2) + a(n-3), for n >= 3.

Prove by complete induction that a(n) >n^2 for all integers n >= 12. [/quote]

for this problem, just verfify the inequality for a_12, a_13, a_14, then, use the inequality
(n-1)^2+(n-2)^2+(n-3)^2>n^2 for all n>=15
ps: I did not verify the inequality above, actually if it doesn't hold for all n>=15, just verify more by your hands, since this inequality must have some solution n>=some number.
[/quote]
hmmm...i get how you need to verify for a_12, a_13, a_14. But how did you come up with the inequality (n-1)^2+(n-2)^2+(n-3)^2>n^2 for all n>=15 ??

[quote="Victor96"][quote]
1) A group of boys and girls are in line to buy tickets for a dance. The first person in the line is a boy and the last in line is a girl. Prove that somwhere in the line a boy must be standing directly in front of a girl.

Prove by simple induction that for all n belonging to the Natural #s, n >= 2, in a line of length n, where person 1 is a boy and person n is a girl theres exists a k where 1 <= k < n, such that person k is a boy and (k+1) is a girl. [/quote]

I don't think in this problem--if you call it problem--you will need any help, do you? Spend more time and you'll get it.[/quote]

for this...do you do it by cases where case 1 is when n is even and case 2 is when n is odd?

thank you very much.

[quote]hmmm...i get how you need to verify for a_12, a_13, a_14. But how did you come up with the inequality (n-1)^2+(n-2)^2+(n-3)^2>n^2 for all n>=15 ?? [/quote]

the inequality is equivalent to n^2-6n+7>0, you can solve it yourself.

for the second problem, you don't need to assume n to be odd or even numbers.
suppose the statement holds for n
in the case of n+1, suppose the second place is a boy, then the children in the first and second place are what you want; if not, for subline counting from the girl in the second place to the last boy you can use induction


In the figure shown, friction is absent everywhere. The massless hinged rod is released with a negligible push at $ \theta $ = 0, with $m$ touching the vertical face of $4m$. As the rod rotates, it pushes the bigger block horizontally. Show that $m$ looses contact with $4m$ when $ \theta = 60 \ degrees $[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=257[/img]

I found a short solution using energy and calculating accleration but I think you know a better one. Could you tell me?

I yhink mine too is based on simple conservation of energy. Please post yours!!!

[hide="OK, here it is."]At the beginning, both rod and box are stationary. At 60°, box is moving at speed $v$ and the weight on the rod at speed $\frac{v}{\cos 60^\circ}=2v$.
We lost $\frac{1}{2}mL$ potential energy, so:

$\frac{1}{2}mgL=\frac{1}{2}(m(2v)^2+4mv^2)$

$gL=8v^2$

$\frac{1}{8}gL=v^2$

The weight keeps accelerating with acceleration $\frac{g}{\sin 60^\circ}=\frac{\sqrt{3}}{2}g$. Horizontal part of this is $\frac{\frac{\sqrt{3}}{2}g}{\cos 60^\circ}=\frac{\sqrt{3}}{4}g$.

There's also centripedal acceleration of the weight. It's $\frac{(2v)^2}{L}=\frac{1}{2}g$, the horizontal part is $\frac{\frac{1}{2}g}{\sin 60^\circ}=\frac{\sqrt{3}}{4}g$.

So at 60°, weight won't accelerate in horizontal direction any more, it will even start decelerating (this should be proven but it's kind of obvious) and it loses contact with the box.[/hide]

Nice problem. Here is my solution

Let the masses of small and large body be m & M.
Draw the FBD and you will see that

mv^2/L = Nsin(phi) + mgcos(phi)

When it loses contact N=0

so v=sqrt[glcos(phi)]
Conserving angular momentum about point where m is hinged

mvL + MVLcos(phi)=0

Now write eqn for conservation of energy. I wont write it down coz its too long to type it.
Maybe later when I figure out how to use the code thingy. Put values for m & M.
you should get 12cos^2(phi) - 8cos(phi) + 1 = 0
cos(phi)=1/2 & 1/6 so phi=60 and phi=arccos(1/6).

Any idea why there are 2 values of phi?  :?:


Just before the admins unlock the forum after the AMC B, can they create 4 threads like these? I think they did that last year.

AMC 12 B Score Poll
AMC 12 B Discussion
AMC 10 B Score Poll
AMC 10 B Discussion

It would be an improvement to the cluttered threads now. Thanks!

ah yes, no one listened to my proclamation that my thread is official, nor the request to not clutter the board in it, lol
anyways, maybe the admins can do it, and it will have a bigger voice

[quote="MCrawford"]I would prefer if the moderators took on that task from this point on.

Also, I am looking for a new moderator for this forum to replace one who has become inactive.  PM me with requests -- but you must be a student who spends time in this forum in particular and who plans to continue doing so for another year or so.[/quote]

Request accepted. I'll try to post those four threads after AoPS forum is unlocked.


ratio
geometry
angle bisector
geometry unsolved

Let $A,C,B,D$ be points on a line (in this order) such that $\frac{AC}{CB}=\frac{AD}{DB}$. Point $P$ is such that $\angle APB$ is a right angle. Show that $BP$ bisects angle $CPD$.

Let's consider an Apollonius circle wrt points $C$ and $D$ and ratio $\lambda=\frac{AC}{AD}=\frac{BC}{BD}$. This circle has segment $AB$ as a diameter so point $P$ lies on this circle. It follows that $\frac{PC}{PD}=\lambda$. Now it suffices to apply angle bisector theorem and we're done

[quote="dondigo"]Let's consider an Apollonius circle wrt points $C$ and $D$ [/quote]

What is the Apollonius circle?  :blush:


abstract algebra
superior algebra
superior algebra solved

Let $ G$ be a group s.t. $ x^2\equal{}1$ for all $ x\in G$. Prove that $ G$ is abelian.

$ xyxy\equal{}1$, hence $ xy\equal{}y^{\minus{}1}x^{\minus{}1}\equal{}yx$.


One thing that confuses me a lot is the way Chinese people eat their noodles, you know, even if it's a formal dinner, they'd suck in the noodles with tremendous sound! Isn't that kind of impolite? Later I found a Chinese learning site [url]http://www.activechinese.com[/url], it says:
 "Though in western culture, it’s considered impolite to make any sort of noise while eating, it’s very acceptable in China to “slurp” your noodles. This is the case not only in casual settings with friends, but even in formal settings or business meetings. So when you order noodles, don’t be shy! Slurp away!"
Is this true? I also heard the same habit in Japan. Anyway,I like it,cause now I can throw away all that "pretend-to-be-a-lady"thing and enjoy my favorite noodles!

haha I'm chinese too

I wouldn't call it a tremendous sound. It's just their active culture.

People in America here think I am WEIRD or ABNORMAL to do that. It's the same as eating everything with a fork.

every country has different cultures and meanings for literally everything  :P


geometry
perimeter
combinatorics unsolved
combinatorics

Consider $h+1$ chess boards. Number the squares of each board from 1 to 64 in such a way that when the perimeters of any two boards of the collection are brought into coincidence in any possible manner, no two squares in the same position have the same number. What is the maximum value of $h?$

fixing a table I obtained only $ h \leq 15 $... does somebody have a solution?

Well, 15 is definitely the best, as if $h \ge 16$, $h+1 \ge 17$, and thus there are a total of $68$ corners, while only numbers from 1 to 64 may appear. Corners on the same board cannot have the same number as specified by the numbering system, while corners on different board cannot have the same number as specified by the condition. So contradiction arises.

It's easy to do $h = 15$:
Let $A,B,C,D,E,F,G$ be the sets of squares in the following positions of each of the 16 boards:
[code]
ABCDEFGA
G......B
F......C
E......D
D......E
C......F
B......G
AGFEDCBA
[/code]
Put the numbers $4i+1,4i+2,4i+3,4i+4$ on board $i+1$ for set $A$, on board $i+2$ for set $B$, on board $i+3$ for set $C$, and so on until on board $i+7$ for set $G$, assuming board $16+i$ is board $i$. This makes the perimeter has no coincidental numbers in any position.

If the boards may be reflected, then $h=7$ is the maximum, with a similar argument as above.


[quote="kunny"]Let $ p$ be a prime, $ n$ be a postive integer, how many times can $ p$ devide $ (p^n)!$?[/quote]

$ p^{n \minus{} 1} \plus{} p^{n \minus{} 2} \plus{} .... \plus{} p^2 \plus{} p \plus{} 1 \equal{}\frac{ p^n \minus{} 1}{p\minus{}1}$

edit: thanks, yes i do=)

You meant, $ \frac{p^n\minus{}1}{p\minus{}1}$


I have just finished high school, but because of some reasons I didn't managed to do my SATs this year, so I will have an additional year before I go to college in US.

My question is: should I be home with my parents and study for my SATs there (option A), or go to some big city and live there (option B).

option A would be quite limited, mainly because my parents will move in September to a village, but they are still my parents..
option B would mean a many new friends, many things to do, many sports to do; big city offer many things to a young person, you know..

Which one do you think is a better choice?

If you and your family are agreed that you can live in the big city, that should be a good choice for showing a college admission office that you have practiced living on your own--which is another issue besides SAT scores that a college will evaluate when deciding whom to admit.

Good luck in making this decision, and best wishes for your application.

I think Option A is the best.- not many distractions.  :lol:

SAT is really about discipline than intelligence. I've seen smart people doing poorly and not so smart people scoring very highly. If you think you can put yourself to study SAT hardcore, go for it. In that perspective, city is a good thing except as you said, you seem to be more into meeting new things, and that shouldn't be your priority.

stay at home, man! you'll be so focused on the move that you'll totally ignore your sat's, trust me :)

--
Joanna Georgie
[url=http://www.nextstudent.com/]student loan consolidation[/url]

Silverfalcon, what do you mean by poorly(what score ranges)?


[quote="dnkywin"]... especially since you were on the Texas mathcounts team...[/quote]

What's your point? Actually, what's the point of this topic?

This is the MO forum coughc gouh

ouch

typoed badly on that last cough

ouch

this ends my spam post

MA!!!!!!!!! just change the a to an o and you get MO.

man people are so protective of their state's forum...if they don't want random randomness sneaking around their forum they should make a private forum.

btw, I am from MO just like ra

actually, we are not protective at all. 10 different states have posted in this forum.

I live in MO too!  Well kind of.. I did last year and I'm stayed in MO this summer..

Lol this is a melting pot of people form other states!

wait, MO is the postal abbreviation for Missouri? I always thought MI was the abbreviation for Missouri.... :huh:

It seems that it's not always the first two letters? That's strange. :o

Dude AIME stop spamming. THere are 50 states. Google is your friend.

[quote="CA Math"]lol you guys are so funny.

mA massachussets
mS mississippi
mO missouri
mT montana
mN minnesota
mI michigan
mD maryland

just look it up, it's on the internet.[/quote]

YAY! We are so good at acting like idiots!!! CA Math thought we were being serious when we pretended not to know the abbreviations of the states that started with M. :lol:

or CA Math is good at being an actual idiot...

Thats a stupid question... Obviously he lives THERE

I will interpret your statement as a sarcastic one... Am I right? (that's not a rhetorical question(my statement that my previous question was not rhetorical is not sarcastic))

No, you are not right.
The previous statement is false.
Actually, the previous statement is true.
I meant the other previous statement.
No, the above statment is actually false.
The above statement is true.

Wow, how good am I at writing poetry!!!!!

[hide="Simply put"]
Yongyi781 is invading the Missouri Forum. Thank you.[/hide]

Yongyi is a Missourian from anotha' motha'.