quadratics
number theory unsolved
number theory

Find all integers $x,y$

$\frac{x^2+y^2}{xy-1}=k$

for wich $k$ is an integer.
My conjecture is that $k=5$, i.e. $(x,y)=(14,67)$.

Below (Find all integers) there is a similar problem, but I have made a mistake there.

Look at that : [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=quadratic+equation&t=22810][u]here[/u][/url]


If $ f(a)\equal{}a\minus{}2$ and $ F(a,b)\equal{}b^2\plus{}a$, then $ F(3,f(4))$ is:
$ \textbf{(A)}\ a^2\minus{}4a\plus{}7 \qquad\textbf{(B)}\ 28 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 11$

[hide="Solution"]$ f(4)\equal{}4\minus{}2\equal{}2$.
$ F(3,2)\equal{}2^2\plus{}3\equal{}4\plus{}3\equal{}\boxed{\textbf{(C)}\ 7}$.[/hide]


1.prove that loga+logax+logax.x+... to n terms=nloga+.5n(n-1)logx
2.how mny numbers of five or six digits can be formed from  the digits 1,1,2,2,3,4(i got 240 but the book says 360)
3.in a class of thirty one prize was awarded for english another for physics and a third for math in how many ways can the recipients be choosen?[quote]WARNING!WARNING!IFF YOUR IQ IS130 OR MORE[/quote]

For your first problem, I can't quite understand the syntax, but
$ \log{a}\plus{}\log{ax}\plus{}\log{ax^{2}}\plus{}\cdots\equal{}\log{a^{n}x^{\frac{n(n\plus{}1)}{2}}}$

And why are you putting on a restriction, iff your IQ is above 130?  That seems pretty pointless.

2.
[hide]5 digit numbers are split into three cases: 2 1's, 2 2's, or both.  If it is both, then we multiply by 2 since there are two other digits to choose from.  We have $ \frac{5!}{2!}\plus{}\frac{5!}{2!}\plus{}2\cdot\frac{5!}{2!2!}$.  Then, there are $ \frac{6!}{2!2!}$ six digit numbers.  This sums to $ 360$.[/hide]

3.
[hide]If people can win more than one prize, it is $ 30^3$.  If not, it is $ 30\cdot29\cdot28$.[/hide]

I've never had an IQ test.  :blush:

[quote="JRav"]For your first problem, I can't quite understand the syntax, but
$ \log{a} \plus{} \log{ax} \plus{} \log{ax^{2}} \plus{} \cdots \equal{} \log{a^{n}x^{\frac {n(n \plus{} 1)}{2}}}$
[/quote]

I think you misread, since that would be true for the sum

$ \overbrace{\log{ax} \plus{} \log{ax^2} \plus{} \log{ax^3} \plus{} \cdots \plus{} \log{ax^n}}^{n\text{ terms}}$.

heh. :P

Two things:

1) Please split different problems into different posts, rather than posting them all in one post.

2) Please make your subjects relevant to the posts.  This will attract many more readers, and will help people find the thread later if they want to look at it again.

[quote="JRav"]That's what it looks like to me.  What do you think it reads?  Again, that is why I pointed out that I couldn't quite understand the syntax.[/quote]

If we have $ n$ terms, the last term is $ \log ax^{n\minus{}1}$.  Then we have $ \sum_{i\equal{}1}^n \log ax^{i\minus{}1} \equal{} \sum_{i\equal{}1}^n \log a \plus{} (i\minus{}1)\log x \equal{} n\log a \plus{} n(n\minus{}1)/2 \log x$

Yes, I finally see it.  I was assuming for $ n\plus{}1$ terms, but for $ n$ terms, it obviously goes to $ n(n\minus{}1)/2$.


The Canadian IMO 2007 team has been announced today:
http://www.newswire.ca/en/releases/archive/May2007/29/c5844.html

The Canadian IMO 2007 team this year is:

Kent Huynh, University of Toronto Schools, Toronto
Steven Karp, Lord Byng Secondary School, Vancouver
Jonathan Schneider, University of Toronto Schools, Toronto
Yan (Cynthia) Li, Dr. Norman Bethune Collegiate Institute, Scarborough
Alexander Remorov, William Lyon Mackenzie Collegiate Institute, North York
Jarno (Chengyue) Sun, Western Canada High School, Calgary

newswire always announces the IMO team members..i think..they did last year..

Congratulations everyone!!! Make Canada Proud!!! :D  :D

[quote="cynthia1991"]Alex, how did you find the website?[/quote]
I found this website from the topic in which last year's team was announced. Then I found the link with this year's team.

I look forward to meeting you, Yufei. :lol:

congratulations to everyone
and good luck at IMO
make canada proud!

Congratulations!!!!

Steven Karp made our school proud  :lol:

how did you guys do so good? what's your secret? do you read a lot of book or do a lot of practice? anyway i'm so proud of you guys for making it to the IMO.

Congrats!
It's rare to see 6 new faces!
I'm looking forward to work with you in Calgary

[quote="billzhao"]Congratulations. I look forward to working with you guys in Calgary/Banff  :D[/quote]
Yufei!

I hope you live up to the legend. :)

Congrats to the team! Cynthia, Alex and Carpo, I am so proud of you guys!!! I guess I will see you all at seminars...too bad I don't live in Calgary....

Btw, Carpo, trust me, the Canadian Putnam fellow will live up to his legend. ( :first: Yufei)

WOW, good job guys, esp steven karp (from bc)

GOOD LUCK!!! :D

Aha....You guys are really fast....I just came online in order to post the new on IMO team, and I found 14 replies already....XD

Congrats to the new team. Make Canada proud. :)

Whee, you guys rock! I'll try to come by for the Seminar and see all of you. I still have the big Canadian flag that Elyot passed down to me for the IMO, so it can continue its glorious voyage with you guys. If you are lucky, you might also get the moose from Richard  :lol: .

the mexican team will join the canadian team for some training in calgary. i will be there!!! see you!!

[quote="YawningLion"]If you are lucky, you might also get the moose from Richard  :lol: .[/quote]This year I shall finally steal the  :moose: moose ...  :diablo:

lol...congrats u guys...see u all in calgary...

I got owned :(

Congrats to all IMO attendees.

[quote="WindingFunction"]PS - Yufei is a double-crosser.  He's also working with the American kids at MOSP!  :o[/quote]

WTF. What a player! ;)

[quote="WindingFunction"]Congratulations to all! Especially Jonathan Schneider, who I had the pleasure of meeting at National Camp 2006! Train them well, Yufei! :D

PS - Yufei is a double-crosser.  He's also working with the American kids at MOSP!  :o[/quote]

What? What is MOSP? :ninja:... anyways guys this is OK: He is YU FEI ZHAO.

Math Olympiad Summer Program.  The American IMO training program.

Yes, I know he is YUFEI ZHAO.  Met him four years ago, before he got all his medals.  :wink:

[quote="WindingFunction"]Math Olympiad Summer Program.  The American IMO training program.[/quote]Probability is just trying to make fun of someone :) He is in the US IMO Team (and is an IMO Medalist already :D). I think he knows what MOSP is.

[quote="WindingFunction"]PS - Yufei is a double-crosser.  He's also working with the American kids at MOSP!  :o[/quote]
Hey it can be worse: my two teammates from our Romanian team in 2001, + our deputy leader (!!) were training US students at MOSP in 02, while our team was trained by the Hungarians :D

[quote="Valentin Vornicu"][quote="WindingFunction"]Math Olympiad Summer Program.  The American IMO training program.[/quote]Probability is just trying to make fun of someone :) He is in the US IMO Team (and is an IMO Medalist already :D). I think he knows what MOSP is. [/quote]
Hmmm I actually do not know of any MOSP... but I do know of the perhaps related camp MOP.

Is that what they call it? Shows how much I learned being in this country for two years...

So I'm going to guess that Probability is Alex Zhai.


Given a function $f(x)=\frac{t^x}{{t^x}+\sqrt{t}}$ in which $t$ is a paramater.
Compute for the sum: $f(\frac{1}{2005}) + f(\frac{2}{2005}) +f(\frac{3}{2005}) +...+f(\frac{2004}{2005})$ .


function
limit
induction
real analysis
real analysis unsolved

Hello! I need for help for the following problem?
Let $ \varphi: [1,\infty]\rightarrow [0, \infty]$ be a positive continuous function such that $ \lim_{x\to\infty}\varphi(x) \equal{} \infty$. Is it true that one can find $ f: [0, 1]\rightarrow \mathbb{R}$ such that the following hold
i) $ f\in L^p([0, 1])$ for all $ 1\leq p < \infty$;
ii) $ \|f\|_p\leq \varphi(p)$ for all sufficiently large $ p$;
iii) $ \lim_{p\to\infty}\|f\|_p \equal{} \infty$?

Thanks!

[color=green]Moderator says: never quote the previous post in full, you are responding to it by default[/color]

Basic idea: Choose $ p_1 < p_2 < ...$ such that $ \varphi(p) > n$ on $ [p_n,\infty)$. Using induction, choose small positive "lengths" $ l_1, l_2,...$ so that $ \sum_{m \equal{} 1}^{\infty} m^pl_m < \varphi(p)^p$ on $ [p_1,\infty)$. Then define $ f \equal{} \sum_{m \equal{} 1}^{\infty} m\chi_{I_m}$, where the $ I_m$ are pairwise disjoint intervals of length $ l_m$ inside $ [0,1]$ .


search
number theory unsolved
number theory

Hi

Let  a  be  the  greatest  positive  root  of  the  equation  $ x^3 \minus{} 3x^2 \plus{} 1 \equal{} 0$. Show that $ [a^{1788}]$ and
$ [a^{1988}]$ are both divisible by $ 17$. ($ [x]$ denotes the integer part of $ x$).

Greetings  :D


calculus
real analysis
real analysis solved

Well, i came with the following problem while thinking of calculus problem... But i don't know whether it is true or provable.
Prove or disprove that every infinite and uncountable subset of the reals has the power of the continum.
At first sight i thought that this one is very similar to the Continum Hypothesis, but it realized that it is much more simple.

Thanks

p.s ofcourse where is no possibility to disprove it  :D But, can we prove it?

It's perfectly equivalent to the Continuum Hypothesis.

Well, i can't see the reason for their "perfectly" equivalence!
By proving the above result, i can't understand why the Continuum Hypothesis follows...
Can you be more specific ?

Thanks

Assume the Continuum Hypothesis is false, and you can find a set with cardinality strictly between $\aleph_0$ and $c$. Since its cardinal is $<c$, there is an injection from it to the reals, so you can identify it with a subset of the reals. This means that you can also find a real set with this cardinality.

On the other hand, "nice" infinite subsets of $\mathbb R$ are, indeed. either countable or of continuum cardinality. I do not remember all possible admissible interpretations of "nice" here, but, in particular, it is true for all Borel sets.

:blush:  :blush:
You are completly right!!
I was a dump that i didn't see that.  :blush:

Thanks


how come the chinese aare the best can any one tutor me to be the best :)

Not sure what you are asking or why I even bothered to browse the Chinese forum. In general just practice and work hard.


function
inequalities
inequalities unsolved

if a, b, c are positive, prove that

 :sqrt:  (a 2 +b 2 -ab) -  :sqrt: (b 2 +c 2 -bc) >  :sqrt: (c 2 +a 2 -ac)

I am sorry I don't know how to input the square root  properly.

That's a famous one
Take a quadrilateral $ABCD$ such that : $\angle{ADB}= \angle{BDC}=60$
put $a=AD$ and $b=BD$ and $c=DC$
Then calculate $AB$ and $BC$ in function of $a,b,c$
and the inequality becomes obvious  because : $AB+BC>AC$  ;)

if $a=b=c$
$\sqrt{a^2 + b^2 -ab} - \sqrt{b^2 + c^2 -bc} > \sqrt{c^2 + a^2 -ac}$
$\Rightarrow 0 > \sqrt{a^2}$

strange

the inital inequality is wrong. One - hv to be replaced by +. But I forgot which one.

i have seen this problems like this in kvant,and other russian books.
they all have the same idea.


An artist is designing a sphere which will have a wire around the sphere's equator with the wire being ten feet above the sphere's surface at all points. The circumference of the sphere is 50000 yards. How many feet of wire will the artist need? Express your answer to the nearest whole number.

Let the diameter of the circle be $ d$, so $ d\pi \equal{}50000$.  The length of the wire is $ (d\plus{}20)\pi \equal{}d\pi \plus{}20\pi$.

We know that $ d\pi \equal{}50000$, so the length of the wire is $ 50000\plus{}20\pi$ yards which in feet is $ \approx \boxed{150188}$

I am sorry to bring this up again, but I just had this question and the answer is 150063

the problem with the original solution was that it was [b]10 feet[/b] above, not 10 yards

which makes $\frac{3\text{ feet}}{1\text{ yard}}\cdot(50000\text{ yards}) + (20\pi\text{ feet})$ cause the circumference is linearly proportional to the radius, so an increase of 10 feet to the radius would result in an increase of $10\cdot2\pi$ of the circumference (can also be derived from circumference formulas for circles)

so that makes
$3\cdot50000+20\pi = 150000 + 20\cdot3.14 \\= 150000+62.8 \approx 150063$


inequalities
geometry
trigonometry
inequalities solved

Given is a triangle $ABC$, the inscribed circle $G$ of which has radius $r$. Let $r_a$ be the radius of the circle touching $AB$, $AC$ and $G$. [This circle lies inside triangle $ABC$.] Define $r_b$ and $r_c$ similarly. Prove that $r_a + r_b + r_c \geq r$ and find all cases in which equality occurs.

[i]Bosnia - Herzegovina Mathematical Olympiad 2002[/i]

Note that sin(A/2) = (r-r_a)/(r+r_a) => r_a = r(1-sin(A/2)/(1+sin(A/s). So we have to prove that
   Sum (1-sin(A/2))/(1+sin(A/2)) >=1
<=> Sum 1/(1+sin(A/2) >= 2
But it 's trivial because Sum 1/(1+sin(A/2) >= 9/(Sum (1+sin(A/2)) >= 9/(3+3/2) = 2 (we use the wellknowm fact: Sum sin(A/2) <= 3/2).
The equality holds iff ABC is equilateral.

Namdung

hmm. I like my solution here. For the incircle $\omega$, draw the tangent parallel to $BC$ and let it hit $AB, AC$ at $B', C'$, and let $r_a'$ be the radius of the incircle $\omega_a$ of triangle $AB'C'$. Similarly define $r_b', r_c'$. Well, we claim $r_a' \le r_a$, as line $B'C'$ separates $\omega_a$ and $\omega$, implying that those two circles cannot intersect (more than once) and thus the claim. However, $\frac{r_a}{r} \ge \frac{r_a'}{r} = \frac{r}{r'} = \frac{\frac{K}{s}}{\frac{K}{s-a}} = \frac{s-a}{s}$, where $r'$ is the radius of the excircle opposite $A$, by a simple similarity argument, and summing up both sides of this inequality yields the desired.

Remark a right triangle, $r-r_a=(r_a+r)\sin \frac{A}{2}\Longleftrightarrow \frac{r_a}{r}=\frac{1-\sin \frac{A}{2}}{1+\sin \frac{A}{2}}$, anagorously we have $\frac{r_b}{r}=\frac{1-\sin \frac{B}{2}}{1+\sin \frac{B}{2}}, \frac{r_c}{r}=\frac{1-\sin \frac{C}{2}}{1+\sin \frac{C}{2}}$,

thus we can rewrite the given inequality as follows.

$r_a+r_b+r_c\geq r\Longleftrightarrow \frac{r_a}{r}+\frac{r_b}{r}+\frac{r_c}{r}\geq 1$

$\Longleftrightarrow \frac{1-\sin \frac{A}{2}}{1+\sin \frac{A}{2}}+\frac{1-\sin \frac{B}{2}}{1+\sin \frac{B}{2}}+\frac{1-\sin \frac{C}{2}}{1+\sin \frac{C}{2}}\geq 1$

${{\Longleftrightarrow -1+\frac{2}{1+\sin \frac{A}{2}}-1+\frac{2}{1+\sin \frac{B}{2}}}}-1+\frac{2}{1+\sin \frac{C}{2}}\geq 1$

${\Longleftrightarrow \frac{1}{1+\sin \frac{A}{2}}+\frac{1}{1+\sin \frac{B}{2}}}+\frac{1}{1+\sin \frac{C}{2}}\geq 2$, which is derived by Namdung.

we can prove this inequality as follows.

Now for $y=\frac{1}{1+\sin x}\ (0<x<\frac{\pi}{2})$, we have $y'=-\frac{\cos x}{(1+\sin x)^2}<0,\ y''=\frac{2-\sin x}{(1+\sin x)^2}>0$ that is

$y=\frac{1}{1+\sin x}\ (0<x<\frac{\pi}{2})$ is concave down and decreasing function.

Thus use Jensen's inequality we have

$\frac{1}{1+\sin (\frac{\frac{A}{2}+\frac{B}{2}+\frac{C}{2}}{3})}\leq \frac{1}{3}(\frac{1}{1+\sin \frac{A}{2}}+\frac{1}{1+\sin \frac{B}{2}}+\frac{1}{1+\sin \frac{C}{2}})$

${\Longleftrightarrow \frac{1}{1+\sin \frac{A}{2}}+\frac{1}{1+\sin \frac{B}{2}}+\frac{1}{1+\sin \frac{C}{2}}\geq \frac{3}{1+\sin (\frac{A+B+C}{6}})}$

$=\frac{3}{\sin 30^\circ}=2\ (\because A+B+C=180^\circ)$,

the equality holds when $\frac{A}{2}=\frac{B}{2} =\frac{C}{2}\ and\ A+B+C=180^\circ \Longleftrightarrow A=B=C$

that is $\triangle{ABC}$ is a regular triangle, we are done.

kunny

Kunny, from Cauchy and from $\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\leq\frac{3}{2}$ follows immediately that $\frac{1}{2+\sin\frac{A}{2}}+\frac{1}{2+\sin\frac{B}{2}}+\frac{1}{2+\sin\frac{C}{2}}\geq 2.$ ;)

First I thought use of Cauchy-Schwalz's inequality, but I couldn't.
Could you show your solution? :)

Or you could have used the inequality $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9,$ which follows also from Cauchy-Schwarz. ;)

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=55605 .

  darij


I really good solutions to see if i am right.
1)Find the largest possible value of $ {m}^2\plus{}{n}^2$ if $ {m}$ and $ {n}$ are members of a set of numbers greater than zero but less than 1982 and $ {n}^2\minus{}{mn}\minus{}{m}^2\equal{}1$.
2)Mr arnold is offering money base on a set of given criterion.
Firstly,you are to give him 2000dollars.
Secondly,you are to find ten people to also give him 2000 dollars if you do so you will be given 13000 dolars so anybody that convinces 10 people get 13000 dollars so the question is how do you maximise you money

Part 2. Get as many people as you know to give him money!


[quote="nttu"]Let $k_n = min(k_1,k_2,...,k_n)$ and $k_i \in N \forall i$
  Prove that :
 $(a_1+k_1a_2).(a_2+k_2a_3)...(a_n+k_na_1) \geq (k_1+1)(k_2+1)...(k_n+1).a_1a_2...a_n$ , where $a_n \geq a_{n-1} \geq ... \geq a_1 \geq 0$  :)[/quote]
                      I thing that you are right .It is very easy


what are some of your favorite nintendo ds games? I was just introduced to it so i've only played mario cart so far, and it's pretty fun. I'm thinking about maybe getting a ds now... so what are some games you'd recommend? no street-fighter stuff, but i do like things like mario, super smash brothers, and the aforementioned mario cart.

also, anyone played nintendogs here before? is it actually fun? it seems pretty boring, to be honest...

I like Advance Wars: Dual Strike.............it's a must have for strategy buffs... :D

[quote]PSP way better graphics and more stylish[/quote]

I personally feel that stuff like the ninetendo ds are fun in their own way.......I agree that a PSP is more stylish and all that, but then after sometime, you tend to want something that gives an old touch.......Don't mistake me here -- I'm not saying that the ninetendo ds is something old -- it's just that it helps in giving a classic feeling to the game you're playing...


I was told this could be factored but could not figure it out.
If any one could help me or give me a hint I would be most greatfull.
 ;)

X^4 + 4

Thanks
Steve

basically, anything in the form of $a^2+b^2$ could be factored if $2ab$ is also a perfect square...

in this case, $x^4+4 = (x^2)^2 + 2^2$
since $2 \cdot 2x^2 = 4x^2=(2x)^2$, it is factorable.

I would like to say Thanks for the help on this problem

The other approach is to take the complex roots and multiply them together in pairs.

You are right, zhaoli.

In personally I would like to write the result to $(a^2+ab+b^2)(a^2-ab+b^2)(a^4-a^2b^2+b^4)$.

You mean Germain's identity.

There is also Euler, Fibonacci's identities and their obvious generalistions to 2^n squares.

Anmything i missed.

Bomb

I'm not very good at English now,therefore i don't want say a lot.


number theory
least common multiple
modular arithmetic
Crt
Good for your health
Euclidean algorithm

I saw this problem somewhere else, but I didn't get the explanation:

Find the smallest number which divided by 7 gives a remainder of 2, and by 5 gives a remainder of 3.

I know this involves the Chinese Remainder Theory.  I've googled Chinese Remainder Theory, and saw several explanations, including one at Cut-The-Knot, but I don't understand any of them.  Can someone please explain the Chinese Remainder Theorem very simply?

EDIT: I can do the problem without the Chinese Remainder Theorem, but I need to know how to do this type of problem in case a harder version shows up.

Don't get too caught up in the wording of the Chinese Remainder Theorem.  It deals with systems of modular congruences (just like systems of equations but with mods instead) and it tells you when you can be ensured a solution to a system.

It's more important to understand how to use it.

In your problem, we have the following equations:

\begin{eqnarray*} x &\equiv& 2\mod 7\\ x &\equiv& 3\mod 5 \end{eqnarray*}

From the first, we have that $x=7t+2$ (this comes from the definition of a mod).  So plug this into the second equation: $7t+2\equiv 3\mod 5$.  This simplifies to $7t\equiv 1\mod 5.$  So we need to find a value of $t$ that satisfies this.  Since we're in mod 5, we just need to try $t=0,1,2,3,4$.  We find that 3 works.  So the smallest $x$ is $7(3)+2=23$.

We can find larger solutions if we make $t=5k+3$.  So then $x=35k+23$.

Follow up problems:

1. Find the smallest number which gives a remainder of 1 when divided by 2, 3, 5, 6, 11, and 12.

2.  Find the smallest solution to this system of congruences:

\begin{eqnarray*} x &\equiv& 1\mod 2\\ x &\equiv& 2 \mod 3\\ x &\equiv& 3\mod 5\\ x &\equiv& 4\mod 7 \end{eqnarray*}

Thanks for the follow-up problems.  You really understood what I was saying earlier.  Lol.

[hide="1"]$2 \cdot 3 \cdot 5 \cdot 11 \cdot 2+1=\boxed{661}$ cuz we're trying to find the lcm of all the numbers plus $1$[/hide]

[hide="2"]Since by using mod congruences, we know for the first two conditions the number must be in the form $5+6n$, trying out a bunch of values of $n$, we see that $n=8$ works out, so the answer is $\boxed{53}$[/hide]

Joe,
Good job in posting follow up problems.  I think that will help a lot.  I will try to post some at the end of the day.

Yeah, that's a good idea, i need to work on Number Theory anyway...

hey my way was systematic....sort of  :roll:

A woman with a basket of eggs finds that if she removes
either 2, 3, 4, 5, or 6 at a time from the basket, there is always one
egg left over. If she removes 7 eggs at a time from the basket, there
are no eggs left over. If the basket holds up to 500 eggs, how many
eggs does she have?

basically, what we want is a multiple of $7$ with a units digit of $1$ and satisfies all those conditions above. So we want to multiply $7$ by $13,23,33....$, and we see $43$ works out, so $\boxed{301}$ is the answer.

it should probably be noted that for number 2, one can make the observation that $2x-1$ is congruent to 0 mod 3, 5, and 7 (and still 1 mod 2). From this, we solve for the least $2x-1$, which is the LCM of 3,5 and 7 (which will still be congruent to 1 mod 2, so it works).

From this, we have $2x-1 = 105$...$\boxed{x=53}$.

Always look out for these kinds of things, you dont always have to do it in a systematic way if you see something better. Math is meant for you to think creatively.

for mr r's problem:

we know that we have a number of eggs $x$ such that $x=60k+1$ for some integer $k$. (this is b/c 60 is the LCM of 2,3,4,5, and 6). we also have $x \equiv 0 \pmod{7}$.

from this:

$60k+1 \equiv 0 \pmod{7}$
$4k+1 \equiv 0 \pmod{7}$
$4k \equiv 6 \pmod{7}$
$k \equiv 5 \pmod{7}$.

So, plugging in $k=5$ into $x=60k+1$, we have $\boxed{x=301}$

Well...the systematic way provides a straightforward way to do the problem in general.  In case you can't find an elegant approach, you can be assured that you know the 'long' way around it.  Often times it's faster to just do the tedious work than to look for an insightful observation.  Don't get me wrong...look to see if you can find anything that will simplify the problem but if you can't start plugging away. I really don't think the math involved with the systematic way for the CRT takes too long anyway...

Wow, your method owns mine..., although mine might have been faster :lol:  :P

there actually are some tricks to implement the CRT and find a number, simplifying the modular arithmetic, but i never can remember them. its easier for me to work it out the 'long' way.

Can someone post these "tricks to implement CRT?"

[quote="mihail911"][quote]1. Find the smallest number which gives a remainder of 1 when divided by 2, 3, 5, 6, 11, and 12[/quote]

wouldn't it be 1?
some people said 661, but in fact, i think that this is the second smallest number which gives a remainder of 1. what do you guys think?  :huh:[/quote]

well if 1 works, then 1-660=-659 also gives a remainder of 1..any 1+660k does..
i think its bigger than its divisors..

"The solution exists if and only if  for all  where  stands for the greatest common divisor. Moreover, in the case when the problem is solvable, any two solutions differ by some common multiple of . "- aopswiki

Will someone show how we can us CRT to ensure there are solutions?
Since this is the thread from the aopswiki page this will make it complete. Sorry to revive  :P

mathemonster - You just kindof have to test it.  For instance, $ x \equiv 2 \pmod{4}$, $ x \equiv 3 \pmod{8}$ has no solutions because $ 2$ is not equivalent to $ 3$ in mod $ 4$ (the gcd of $ 4$ and $ 8$).  Alternatively, we can note that one of the equivalences requires x to be odd while the other requires x to be even.

So where does this condition come from?  Well for the system $ x \equiv a \pmod{p}$, $ x \equiv b \pmod{q}$ to have a solution we must have

$ a \plus{} kp \equal{} b \plus{} jq$ for some integers $ k$, $ j$.  Now let $ d$ be the minimum positive value (or the maximum negative value) that $ jq \minus{} kp$ takes for any choice of $ j,k$.  But $ jq \minus{} kp$ can only be a multiple of $ gcd(p,q)$ so $ d \equal{} gcd(p,q)$, which divides all $ jq\minus{}kp$.  Rearranging, $ a \equal{} b \plus{} (jq \minus{} kp)$, so $ a \equiv b \pmod{d}$.  Hence we have shown that $ a \equiv b \pmod{d}$ is necessary for the system to have a solution.

Alternatively, assume $ a \equiv b \pmod{d}$, $ a \equal{} b \plus{} id$ (d is still the gcd of $ p$ and $ q$) for some integer $ i$.  By the extended euclidean algorithm, there exists $ j,k$ such that $ jq \minus{} kp \equal{} d$, so $ a \equal{} b \plus{} (ij)q \minus{} (ik)p$ and hence $ a \plus{} ikp \equal{} b \plus{} ijq$.  Letting this equal $ x$, we see that $ x$ satisfies the given system of congruences.  Hence if the condition that $ a \equiv b \pmod{d}$ is met, the system has a solution.

So the system has a solution if and only if $ a \equiv b \pmod{gcd(p,q)}$.

I just learned about this today! This is truly awesome!

Yay, go amazing AZIANS

[quote]i just wanted to know, as chineese remainder theoram is described on the page https://artofproblemsolving.com/wiki/index.php?title=Chinese_Remainder_Theorem ,
can a(i) that satisfies ,
[b]a(i) * b(i) -1 =0(mod d(i))[/b]
be fraction number or it has to be an integer ?[/quote]

It has to be an integer. To find this, it is very easy. Put your equation in form of
${a_{i}}{b_{i}}=1\pmod{d_{i}}$
Since we are in $\pmod{d_{i}}$, we can add ${d_{i}}$ to either side. This gets us to
${a_{i}}{b_{i}} + {d_{i}}y=1\pmod{d_{i}}$ for some $b$.
Since $b_{i}$ is a constant, let it be ${a}$. For sake of simplicity, let $a_{i}$ and ${d_{i}}$ be $x$ and $y$ respectively. So we now have
$ax+by=1$ which can be found through the [url=https://brilliant.org/wiki/extended-euclidean-algorithm/#extended-euclidean-algorithm]Extended Euclidean Algorithmn[/url]
Note. This will take you to the brilliant website. They also have a section for the Chinese Remainder Theorem which is quite helpful - [url=https://brilliant.org/wiki/chinese-remainder-theorem/]Chinese Remainder Theorem[/url]

Does anyone know the origin of this theorem?

From [url]https://en.wikipedia.org/wiki/Chinese_remainder_theorem[/url]:
[quote]The earliest known statement of the theorem is by the Chinese mathematician Sun-tzu in the Sun-tzu Suan-ching in the 3rd century AD.[/quote]

[quote=joml88][hide="Systematic Way for #2"]
Start with the first equation, i.e. $x\equiv 1\mod 2$.  We turn this into $x=2p+1$.  Now plug this into the second equation: \[ 2p+1 \equiv 2\mod 3\Leftrightarrow 2p\equiv 1\mod 3. \]

Note that this means that $p\equiv 2\mod 3$.  (You can arrive at this by either testing $p=0,1,2$ or adding 3 to both sides of $2p\equiv 1\mod 3$ and dividing...you can divide without worrying since 2 and 3 are relatively prime...).  Therefore $p=3q+2$. So $x=2(3q+2)+1=6q+5$.

Again, plug this in for $x$ in the next equation: \[ 6q+5\equiv 3\mod 5. \]

Thus $q\equiv 3\mod 5$.  So $q=5r+3$ and $x=30r+23$.

Finally, we plug this into the last equation: \[ 30r+23\equiv 4\mod 7. \] So $r\equiv 1\mod 7$.   To find the smallest (positive was assumed in the question statement) $x$, we take $r=1$ which gives $x=53$.  More generally, we can take $r=7k+1$ and thus have $x=210k+53$.[/hide][/quote]

You can Also go from highest mod to lowest.  or mod the moduli together. 7 is 1 mod 3, 4 is 1 mod 3 , and so you need to be in a 1 mod 3 multiple of 7,  You can do the same for 5 and 3 to get it needs a 1 mod 3 multiplier as well. so a 1 mod 3 multiplier on 5 makes it 8 mod 15, a 1 mod 3 multiplier on 7 means 11 mod 21

as there's a 6x+3 difference and x=2 gives you 15 you'll be at 2 times 21+11 equals 3 times 15+8 equals 53  which then fits the mod 2 requirement   and we can mod by  210


calculus
integration
trigonometry
calculus computations

Wow. I have a guess - its based on heuristics (I will seek (or await) a proof)
[hide="Answer"]$\frac{\sec{a}}{1+\sin{a}}$[/hide]

[quote="liyi"]Yes, Mathematica can do it.[/quote]

Thumbs down for Maple then!!

[quote="dheardie"][quote="Jimmy"]Maybe this will help spark a solution

$\int \frac{dx}{\left(\sin(a+x)+\cos(x)\right)^2} = \frac{\sin(x)}{\left(\cos\left(\frac{a}{2}\right) +\sin\left(\frac{a}{2}\right)\right)^3\left(\cos\left(\frac{a}{2}+x\right)+\sin\left(\frac{a}{2}+x\right)\right)}$

(According to Mathematica)

This can probably be simplified, but I'm sure there is a more "elegant" solution than directly computing the anti-derivative.[/quote]

Can Mathematica do the integral? Maple cant![/quote]

I don't actually have Mathematica, I just used their online program, The Integrator. http://integrals.wolfram.com/


[quote="yurik"]We can transform it to (n!-1)!>(n-1)! n!-1. By AM-GM we have  n!-1  :sqrt: (n!-1)!  >=(1+2+...+(n!-1))/(n!-1)=n!/2; But n!/2>(n-1)! since n>2[/quote]

This is wrong. The sign of AM-GM is reversed.


First evaluate the exponent then solve.

\[ \binom{2}{0}\plus{}\binom{2}{1}\plus{}\binom{2}{2}\equal{}4\implies2^4\equal{}16
\]


Forgot to square all of them.

The volume is $ 1^2 \plus{} 3^2 \plus{} 5^2 \plus{} 7^2 \plus{} 9^2 \equal{} \boxed{165}$


AMC
USA(J)MO
USAMO
combinatorics unsolved
combinatorics

group X has n members. Given n+1 different subsets of X, each with 3 members, show that we can always find two which have just one member in common. (we say that two subsets are different if they have at least one different member)

I'm assuming that just one means at most one (i.e. the contrdiction of the statement is that all groups have none, as opposed to all groups have either 2 or 0)
Again, assume contradiciton (so for all n+1 subsets, no two members are the same)
And its basically done, this is way impossible for n members...

btw this is usamo 1979 #5

Well, construct the set by placing subsets one at a time.  We first know $n >3$.  So you place one set.  Each new set either shares exactly two elements with any previous set, or is disjoint from that previous set.  Suppose we have two sets which overlaps in two elements.  Then suppose we have a third set, which overlaps one of the sets in two elements.  It cannot be completely disjoint from the other one then, and it follows that the set overlaps in the same two elements.  Continuing this way, each new set adds at least one more element.  And we end up eith a contradiction.  I think you could even do it with $n-1$ sets.

[quote="me@home"]I'm assuming that just one means at most one (i.e. the contrdiction of the statement is that all groups have none, as opposed to all groups have either 2 or 0)
[/quote]

Since the above statement is true, I imagine that they meant what they said.

[quote]
Again, assume contradiciton (so for all n+1 subsets, no two members are the same)
And its basically done, this is way impossible for n members...

btw this is usamo 1979 #5[/quote]

It sounds like you reversed things... two members can be the same, just not one member.

[quote="K81o7"]Then suppose we have a third set, which overlaps one of the sets in two elements.  It cannot be completely disjoint from the other one then, and it follows that the set overlaps in the same two elements. [/quote]

This doesn't follow; the third set could overlap the second in two different elements.  So you do need $n+1$ subsets

[quote="Deedlit"]
This doesn't follow; the third set could overlap the second in two different elements.  So you do need $n+1$ subsets[/quote]

But then it would overlap the first set in 1 element.

No, it just has to overlap in two different elements.  I.e., {a,b,c}, {a,b,d}, and {b,c,d}.  (This example may lead you to the correct solution.)

[hide]So, to sum up, you can divide the $3$-element subsets of $X$ into equivalence classes such that $A,B\in C\leftrightarrow A\cap B\neq \emptyset$. These classes can be of 2 different kinds, left apart those containing just one member; either there are 2 elements $\{a,b\}$ of $X$ such that $A\in C \leftrightarrow a,b\in A$, either  $|\bigcup C|=4$ and $|C|\leq 4$. In any case  $|\bigcup C|\geq |C|$, so  $n=|X|=|\bigcup \{ C_{i}\}|\geq \sum|C_{i}|=n+1$,  a contradiction.  [/hide]


Find all pairs $ (a, b)$ of entire components, such that $ a ^ 2 \plus{} b ^ 2 \equal{} 650$

Saludos  :D


Don't know if this is exactly the right way to do it:

$\log_6$ everything and take out the exponent, thus you have:

$\log_{6}{x}(\log_{6}{3})+\log_{6}{x}(\log_{6}{4})+\log_{6}{x}(\log_{6}{4})=\log_{6}{x}$

$\log_{6}{x+3}+\log_{6}{x+4}+\log_{6}{x+4}=\log_{6}{x}$

$(x+3)(x+4)^2=x$ And then solve for that.

[quote="Iversonfan2005"]Don't know if this is exactly the right way to do it:

$\log_6$ everything and take out the exponent, thus you have:

$\log_{6}{x}(\log_{6}{3})+\log_{6}{x}(\log_{6}{4})+\log_{6}{x}(\log_{6}{4})=\log_{6}{x}$

[/quote]

That's not true. You can't log $LHS$ like that(you log the whole expresion, not each term apart)

[quote="Slizzel"]Let $M=\{x\in (0,\infty)| 3^{\log_{6}{x}}+4^{\log_{6}{x}}+4^{\log_{6}{x}}=x\}$. Find $M$[/quote]
[hide]We could base change...

$3^{\frac{\log_3x}{\log_36}}+2(4^{\frac{\log_4x}{\log_46}})=x$

$x^{\frac1{\log_36}}+2(x^{\frac1{\log_ 46}})=x$

or call $\log_6x=y$

$3^y+4^y+4^y=6^y$

that looks better but I don't know how to solve it[/hide]


How is the direction vector t(4, -5, 2) is parallel to the plane $2x + 2y + z = 7$?

In general, how do I find a vector parallel to a plane given the equation of the plane?

The plane $2x+2y+z=7$ is parallel to the plane $2x+2y+z=0$, and is therefore parallel to any line, ray, or vector on that plane.

The consider a line $l$ that goes from the origin (0, 0, 0) to (4, -5, 2). Since both points are on the plane $2x+2y+z=0$, that is contained in the plane $2x+2y+z=0$, and is therefore parallel to the plane $2x+2y+z=7$.

Since the direction vector t(4, -5, 2) is clearly parallel to line $l$, it is parallel to the plane.

In general, plane $Ax+By+Cz=D$ is parallel to plane $Ax+By+Cz=E$.

hmm...not sure i follow what you're saying.

wouldnt this also imply that (0,1.0) is also parallel to the plane if we use $2x+2y+z=2$ ?

edit; oh both points need to be on the equation...nvm.

Doesn't that mean any vector is parallel to any plane?

t(a,b,c), plug in the triplet to arbitrary linear equation of 3 variables and it will yield a solution. Now you can say that this ordered triplet is a solution to the yielded equation. Then you can say since a,b,c is parallel to that equation, it must be parallel to any plane parallel to that plane.

Don't forget about the origin, we need [b]both[/b] points to be contained in the appropriate plane.

any plane equation can be written in the form

$ax+by+cz+d=0$ with $(a,b,c)$ as the normal vector .

For $(p,q,r)$ to be parallel to the plane , the dot product of $(p,q,r)$ and the normal vector should be $0$ ( draw out and see that they are perpendicular to each other .)

So if $(p,q,r)\cdot (a,b,c)=0$ then they are parallel .

 :)

ediited some typo .

[quote="shyong"]
the cross product of $(p,q,r)$ and the normal vector should be $0$
 :)[/quote]
I think you mean the dot product  ;)

EDIT: nvm

At first, I thought you meant that dot product between the plane and the parallel vector, but I now realize that you mean the dot product of the normal vector and the parallel vector.


Let $ P(n)$ denote the set of first $ n$ primes. Let $ P_1$ and $ P_2$ be a partition of the set $ P(n)$ (by partition I mean $ P_1\cup P_2 \equal{} P(n)$ and $ P_1\cap P_2 \equal{} \emptyset$). Can we always find a partition of $ P(n)$ and positive integers $ a_i,i \equal{} 1,\dots,|P_1|$ and $ b_j,j \equal{} 1,\dots,|P_2|$ so that
\[ \exists k\in \mathbb{N} \textrm{ so that } k \equal{} \prod_{i \equal{} 1}^{|P_1|} p_i^{a_i} \textrm{ and } k \plus{} 1 \equal{} \prod_{j \equal{} 1}^{|P_2|}p_j^{b_j} \textrm{  where  } p_i\in P_1,P_j\in P_2
\]

ps: edited as suggested by Aziz

[quote="srikanth"]Let $ P(n)$ denote the set of first $ n$ primes. Let $ P_1$ and $ P_2$ be a partition of the set $ P(n)$ (by partition I mean $ P_1\cup P_2 \equal{} P(n)$ and $ P_1\cap P_2 \equal{} \emptyset$). Can we always find a partition of $ P(n)$ and positive integers $ a_i,i \equal{} 1,\dots,|P_1|$ and $ b_j,j \equal{} 1,\dots,|P_2|$ so that
\[ \exists k\in \mathbb{N} \textrm{ so that } k \equal{} \prod_{i \equal{} 1}^{|P_1|} p_i^{a_i} \textrm{ and } k \plus{} 1 \equal{} \prod_{j \equal{} 1}^{|P_2|}p_j^{b_j}
\]
[/quote]
here we have $ p_1|gcd(k,k \plus{} 1)$ so $ gcd(k,k \plus{} 1)\neq 1$

you want say that:
[quote="aviateur"]Can we always find a partition $ P_1,P_2$ of $ P(n)$ and positive integers $ a_i,i \equal{} 1,\dots,|P_1|$ and $ b_j,j \equal{} 1,\dots,|P_2|$ so that:
$ P_1 \equal{} \{p_1,p_2,...,p_{|P_1|}\},P_2 \equal{} \{q_1,q_2,...,q_{|P_1|}\}$
such that $ 1 \plus{} \prod_{i \equal{} 1}^{|P_1|}p_i^{a_i} \equal{} \prod_{i \equal{} 1}^{|P_2|}q_i^{b_i}$[/quote]
we must prove that $ \exists k\in\mathbb{N}: \ \prod_{p\in P(n)}p|k(k\plus{}1)\ and\ \forall p\not\in P(n): p\not|k(k\plus{}1)$


[quote="socrates"][quote="silouan"]Αν $a,b,c,d\geq 0$ τότε δείξτε ότι
$\frac{a-b}{a+2b+c}+\frac{b-c}{b+2c+d}+\frac{c-d}{c+2d+a}+\frac{d-a}{d+2a+b}\geq 0$[/quote]

Kανεις?...... :maybe:

ειναι $\frac{a}{a+2b+c}+\frac{b}{b+2c+d}+\frac{c}{c+2d+a}+\frac{d}{d+2a+b}\geq 1$ κτλ.....

τωρα?
 :D[/quote]

Τώρα ολοκλήρωσε την απόδειξή σου   :P  :D

οκ.Ας ειμαι λιγο αναλυτικότερος.... :D

$\frac{a}{a+2b+c}+\frac{b}{b+2c+d}+\frac{c}{c+2d+a}+\frac{d}{d+2a+b}=$
$=\frac{a^{2}}{a^{2}+2ab+ac}+\frac{b^{2}}{b^{2}+2bc+bd}+\frac{c^{2}}{c^{2}+2cd+ac}+\frac{d^{2}}{d^{2}+2ad+bd}\geq 1$ απο κοσι....

τωρα αρκει νδο $-\frac{b}{a+2b+c}-\frac{c}{b+2c+d}-\frac{d}{c+2d+a}-\frac{a}{d+2a+b}\geq-1$ ή ισοδυναμα

$\frac{a+c}{a+2b+c}+\frac{b+d}{b+2c+d}+\frac{c+a}{c+2d+a}+\frac{d+a}{d+2a+b}\geq 2$ και αυτο με....κοσι παλι.....

 :)


calculus
derivative
calculus computations

Ok, Odd Question that has no revelance to my work, just a Question that has me a little stumped.  I have a Bachelors Degree in Mathematics from The University of West Florida.  I graduated 9 Years ago, but do not have a job that uses my math background, so I have forgotten quite a bit of what I had learned.

My Question...
On a Propeller, I would like to be able to Find the speed of the Inner hub say radius 3 feet to the tip of the propeller at say radius 30 feet.  I know the inner hub and the propeller are fixed together, but the inner diameter travels less distance than the outer edge of the propeller.  Taking into consideration of an arbitrary speed, how would I find the speed of the Inner Hub compared to the Outer Edge of the Propeller.  The outer edge would have to travel faster than the inner hub just because it travels more distance in the same amount of time.  I'm pretty much having a Brain Freeze.
Thanks...
Any Help will be appreciated.

The key here is the definition of radian measure of an angle. On a circle of radius $r,$ we have $s=r\theta$ where $s$ is the length of an arc along the circle and $\theta$ is the angle, in radians, that subtends that arc.

Now assume we are looking at a fixed location on the propeller - since it stays the same distance from the hub, we're treating $r$ as constant. Differentiating our basic equation gives us $\frac{ds}{dt}=r\frac{d\theta}{dt}.$ (Do you remember "related rates" problems from your long-ago calculus course? That's the mindset I want you to be in.) The derivative $\frac{ds}{dt},$ which is in velocity units (such as meters per second) is the speed that you are talking about.

Now consider two points on the same propeller, at different distances $r_{1}$ and $r_{2}$ from the hub. Since the propeller spins rigidly, $\frac{d\theta}{dt}$ is the same for both points.

We have $\frac{ds_{1}}{dt}=r_{1}\frac{d\theta}{dt}$ and $\frac{ds_{2}}{dt}=r_{2}\frac{d\theta}{dt}$

Divide these equations by $r_{1}$ and $r_{2},$ respectively, and equate. We get

$\frac{\frac{ds_{1}}{dt}}{r_{1}}=\frac{\frac{ds_{2}}{dt}}{r_{2}}$

Hence, a point on the propeller that is 30 feet from the hub is moving 10 times as fast as a point that is 3 feet from the hub.


$(4x+6y+7+5n+2t)^{2}$

and If you can't figure it out, don't look at other peoples answers and post it as if you figured it out

Also, because the answer is so much longer then the question I have no clue why people would call it simplified

Just to post it in latex, he means $(4x+6y+7+5n+2t)^{2}$.

[quote="Ihatepie"][quote="13375P34K43V312"]Just to post it in latex, he means $(4x+6x+7+5n+2t)^{2}$.[/quote]

How do you do the squared sign?[/quote]
Simply use
[code]$x^{2}$[/code],
which produces $x^{2}$.
If you want to use a power that contains more than one character/digit in the exponent, you need to surround the exponent with braces; for example, "2^16" produces $2^{1}6$, while "2^{16}" produces $2^{16}$.

In general, if you'd like to know how someone typed something in LaTeX, you can either hover your mouse above their LaTeX (which will cause a little box containing the code to appear) or quote the person.  :)

[quote="Ihatepie"][quote="bpms"][quote="Ihatepie"]someone just the answer the question[/quote]
No need to get so antsy :)
[hide]$(10x+5n+2t+7)^{2}=100x^{2}+25n^{2}+4t^{2}+49+100xn+40xt+140x+20nt+70n+28t$[/hide]
1800...[/quote]

wrong problem...[/quote]
Did you read what was hidden?
1800 has nothing to do with the answer...

[quote="bpms"][/quote][quote="Ihatepie"][quote="bpms"][quote="Ihatepie"]someone just the answer the question[/quote]
No need to get so antsy :)
[hide]$(10x+5n+2t+7)^{2}=100x^{2}+25n^{2}+4t^{2}+49+100xn+40xt+140x+20nt+70n+28t$[/hide]
1800...[/quote]

wrong problem...[/quote][quote="bpms"]
Did you read what was hidden?
1800 has nothing to do with the answer...[/quote]
Ihatepie means you messed up with the variables. It's $4x+6y,$ not $4x+6x.$ Look at the first post.

[quote="i_like_pie"][/quote][quote="bpms"][/quote][quote="Ihatepie"][quote="bpms"][quote="Ihatepie"]someone just the answer the question[/quote]
No need to get so antsy :)
[hide]$(10x+5n+2t+7)^{2}=100x^{2}+25n^{2}+4t^{2}+49+100xn+40xt+140x+20nt+70n+28t$[/hide]
1800...[/quote]

wrong problem...[/quote][quote="bpms"]
Did you read what was hidden?
1800 has nothing to do with the answer...[/quote][quote="i_like_pie"]
Ihatepie means you messed up with the variables. It's $4x+6y,$ not $4x+6x.$ Look at the first post.[/quote]
Oh. I read leetspeaks post, in which he made a typo.
[hide]
$(4x+6y+5n+2t+7)^{2}=16x^{2}+36y^{2}+25n^{2}+4t^{2}+49+48xy+40xn+16xt+56xy+60yn+24ty+84y+70n+20tn+28t$[/hide]

[quote="bpms"][/quote][quote="i_like_pie"][/quote][quote="bpms"][/quote][quote="Ihatepie"][quote="bpms"][quote="Ihatepie"]someone just the answer the question[/quote]
No need to get so antsy :)
[hide]$(10x+5n+2t+7)^{2}=100x^{2}+25n^{2}+4t^{2}+49+100xn+40xt+140x+20nt+70n+28t$[/hide]
1800...[/quote]

wrong problem...[/quote][quote="bpms"]
Did you read what was hidden?
1800 has nothing to do with the answer...[/quote][quote="i_like_pie"]
Ihatepie means you messed up with the variables. It's $4x+6y,$ not $4x+6x.$ Look at the first post.[/quote][quote="bpms"]
Oh. I read leetspeaks post, in which he made a typo.
[hide]
$(4x+6y+5n+2t+7)^{2}=16x^{2}+36y^{2}+25n^{2}+4t^{2}+49+48xy+40xn+16xt+56xy+60yn+24ty+84y+70n+20tn+28t$[/hide][/quote]

wrong

[quote="nebula42"][quote="Ihatepie"][quote="13375P34K43V312"]Just to post it in latex, he means $(4x+6x+7+5n+2t)^{2}$.[/quote]

How do you do the squared sign?[/quote]
Simply use
[code]$\$$x^{2}$\$$[/code],
which produces $x^{2}$.
If you want to use a power that contains more than one character/digit in the exponent, you need to surround the exponent with braces; for example, "2^16" produces $2^{1}6$, while "2^{16}" produces $2^{16}$.

In general, if you'd like to know how someone typed something in LaTeX, you can either hover your mouse above their LaTeX (which will cause a little box containing the code to appear) or quote the person.  :)[/quote]

thanks

Please do not abuse the quoting feature and quote many times - a quote is not needed most of the time if you are posting directly after someone.  Lots of quoting inside quoting makes the page long and more difficult to read.  Thanks  :)


integration
calculus
trigonometry
real analysis
real analysis unsolved

For calculating the first sum I would do this:

$ \frac{2x\arcsin x}{\sqrt{1-x^{2}}}=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n\binom{2n}{n}}$.

it follows that

$ \int_{0}^{x}t(\frac{2t\arcsin(t)}{\sqrt{1-t^{2}}})^{'}dt=2x\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{(2n+1)\binom{2n}{n}}$.

Now if $ x=\frac{\sqrt{3}}{2}$ we get that

$ \int_{0}^{\sqrt{3}/2}t(\frac{t\arcsin(t)}{\sqrt{1-t^{2}}})^{'}dt=\sqrt{3}(S-1)$,

where $ S$ is your sum.

If you calculate the preceding integral by parts, twice, (no need to differentiate) I think you should get the result.

I hope I made no mistakes. Probably, there are simpler ways to find your sum...


function integration calculus limit trigonometry algebra domain

The Gamma function is defined as
$\Gamma(z) = \int_0^\infty t^{z -1} e^{-t} dt$
for real part of $z$ that is greater than 0 [that is ${\cal R}(z) > 0$]. Does anyone know the formula for ${\cal R}(z) \leq 0$?

I always see this relation $\Gamma(z + 1) = z \Gamma(z)$. Is this formula only true for ${\cal R}(z) > 0$?

[quote]I always see this relation $\Gamma(z + 1) = z \Gamma(z).$ Is this formula only true for ${\cal R}(z) > 0$ ?[/quote]
That recursive formula is used to [b]define[/b] the gamma function on the left half plane. In other words, $\Gamma(z)$ is the unique function, defined everywhere in the complex plane except for the nonpositive integers, which satisfies the integral formula on the the right hand half plane and the recursive formula everywhere.

We can see that $\lim_{z\to0}z\Gamma(z)=1.$ In other words (in the language of complex analysis), $\Gamma(z)$ has a simple pole at the origin with residue $1.$ It then follows from the recursive formula that it has a simple pole of residue $\frac{(-1)^n}{n!}$ at the point $z=-n.$

[quote="kaos"]Does anyone know the formula for $\Re z\le 0$? [/quote]
Another formula for $\Gamma$-function is
\[ \Gamma(z)=\lim_{n\to\infty}\frac{n^z n!}{z(z+1)\dots(z+n)} \]
This one is valid for all $z$ except nonpositive integers. It is easy to see from this expression that, if $z$ is a positive integer, then $\Gamma(z)=(z-1)!$ and that $z\Gamma(z)=\Gamma(z+1)$ for all $z\in\mathbb C$. On the other hand, this formula is nothing but an attempt to define $\Gamma(z)$ using its asymptotics at $+\infty$ and the property $\Gamma(z+1)=z\Gamma(z)$, so it agrees with what Kent said.

[quote="Kent Merryfield"]
We can see that $\lim_{z\to0}z\Gamma(z)=1.$ In other words (in the language of complex analysis), $\Gamma(z)$ has a simple pole at the origin with residue $1.$ It then follows from the recursive formula that it has a simple pole of residue $\frac{(-1)^n}{n!}$ at the point $z=-n.$[/quote]

Could you please elaborate further on how you obtain $\frac{(-1)^n}{n!}$ at the point $z=-n$?

[quote="fedja"]Another formula for $\Gamma$-function is
\[ \Gamma(z)=\lim_{n\to\infty}\frac{n^z n!}{z(z+1)\dots(z+n)}  \]
This one is valid for all $z$ except nonpositive integers.[/quote]

I am quite confused with so many formulae for Gamma function. The one I always see is the one I quoted, but it does not generalized to non-integer negatives. I wonder why it has become the main definition.

Saying that $\Gamma(z)$ has a simple pole of residue $\frac{(-1)^n}{n^!}$ at $z=-n$ is saying this:

$\lim_{z\to-n}(z+n)\Gamma(z)=\frac{(-1)^n}{n!}.$

We can prove this by induction. Rearrange the recursive identity to this: $\Gamma(z)=\frac1z\Gamma(z+1).$

Start with the $n=0$ case:

$\lim_{z\to0}z\Gamma(z)=\lim_{z\to0}\frac zz\Gamma(z+1)= \Gamma(1)=1.$

Now assume the statement about pole and residue is true for $n-1.$

$\lim_{z\to-n}(z+n)\Gamma(z)=\lim_{z\to-n}\frac{(z+n)}z\Gamma (z+1)$

Make the substitution $w=z+1,\,z=w-1.$

$=\lim_{w\to-(n-1)}\frac{w+n-1}{w-1}\Gamma(w)= \frac1{-(n-1)-1}\cdot\lim_{w\to-(n-1)}(w+n-1)\Gamma(w)$

This equals $-\frac1n\cdot\frac{(-1)^{n-1}}{(n-1)!}=\frac{(-1)^n} {n!}$ by the inductive assumption.

[quote="kaos"]I am quite confused with so many formulae for Gamma function. The one I always see is the one I quoted, but it does not generalized to non-integer negatives. I wonder why it has become the main definition.[/quote]
It's a feature, not a bug.

Consider the trigonometric functions. How do you define them? In terms of triangles? But that limits your domain. In terms of a "wrapping function?" And then there are so many different identities - consider all the ways you know to write $\sin^2\theta\cos^2\theta.$ Why do we need or want six different trigonometric functions? That's so redundant. And then we can express trigonometric functions in terms of power series; in terms of the complex exponential; in terms of infintite products. Each of these additional representations may be "confusing" but they also add to the richness of our understanding. Being able to represent the same functions in so many different ways gives us tools we can use to solve problems. Now, out of all of that, which is the "right" way to define and represent trigonmetric functions? I'd say, "all of them" or "it doesn't matter, except that logically you have to start somewhere."

The other special functions - the Gamma function, the various Bessel functions, and so forth and so on - are the same way.

Hi,

I can't seem to integrate this function for x in the range (0, 1)

$\int_0^\infty t^{x -1} e^{-t} dt = - t^{x - 1} e^{-t} \Big|_0^\infty + \int_0^\infty (x-1) t^{x-2} e^{-t} dt$.

The first term diverges as $t \to 0$. What have I done wrong?

Both terms on the right diverge. (Note the $t^{x-2}$ in the integral.) What you've done is to step outside the domain of the integral definition of the gamma functions - the very concern that caused kaos to start this thread. That particular integration by parts is only valid for $\mathcal{R}(x)>1.$ Complete it, and you will have a version of what we have been calling the recursive identity. You'll have it in the form $\Gamma(x)=(x-1)\Gamma(x-1).$


Pruebe que en la secuencia $ \left\{\left\lfloor n\sqrt{2003}\right\rfloor\right\}_{n\geq 1}$ hay progreciones geométricas arbitrariamente largas y con una razón arbitrariamente grande.

A menos que esté entendiendo mal, el problema puede ser redactado de la siguiente manera:

[b]Sean N>0 un numero entero y $ \lambda > 0$ un número real. Dada la secuencia $ \left\{\left\lfloor n\sqrt {2003}\right\rfloor\right\}_{n\geq 1}$, pruebe que existen números enteros $ 0 < n_1 < n_2 < \ldots < n_N$ tales que $ \left\{\left\lfloor n_k\sqrt {2003}\right\rfloor\right\}_{1\le k\le N}$ es una PG con razón $ \mu > \lambda$[/b]

Es definida la función $ F: \mathbb R\to[0,1)$ por la fórmula: $ F(x) = x - \lfloor x\rfloor$.
En esta solución es importante el siguiente teorema (en este momento no recuerdo su nombre):
[hide]Si $ \alpha$ es irracional, entonces la secuencia $ \{F(n\alpha)\}_{n\ge1}$[list][*]apenas asume valores irracionales[*]es densa en el intervalo abierto (0,1)[/list][/hide]
[b]Solución:[/b] Dados N y $ \lambda$, sea M>0 un número entero tal que $ 2^M > \lambda$. Por el teorema mencionado, existe un número entero $ m_1 > 0$ tal que $ 0 < F(m_1\sqrt {2003}) < 2^{1 - MN}$. Definiendo $ m_k = 2^{k - 1}\cdot m_1$, para $ 2\le k\le MN$, se tiene lo siguiente:

\begin{align*} m_k\sqrt {2003} & = 2^{k - 1}\cdot m_1\sqrt {2003} \\
m_k\sqrt {2003} & = 2^{k - 1}\cdot(\lfloor m_1\sqrt {2003}\rfloor + F(m_1\sqrt {2003})) \\
m_k\sqrt {2003} & = 2^{k - 1}\cdot\lfloor m_1\sqrt {2003}\rfloor + 2^{k - 1}\cdot F(m_1\sqrt {2003}) \end{align*}

Como $ 0 < 2^{k - 1}\cdot F(m_1\sqrt {2003}) \le 2^{MN - 1}\cdot F(m_1\sqrt {2003}) < 1$, entonces $ \lfloor m_k\sqrt {2003}\rfloor = 2^{k - 1}\cdot\lfloor m_1\sqrt {2003}\rfloor$, por lo tanto $ \left\{\left\lfloor m_k\sqrt {2003}\right\rfloor\right\}_{1\le k\le MN}$ es una PG con razón 2. De esta PG es posible obtener la sub-PG $ \left\{\left\lfloor n_k\sqrt {2003}\right\rfloor\right\}_{1\le k\le N}$, definiendo $ n_k = m_{kM}$. Esta es una PG con N términos y razón $ 2^M > \lambda$

Espero que no haya cometido algún error al escribir mis ideas  :)

[b]Observación:[/b] Este problema puede ser generalizado, cambiando $ \sqrt {2003}$ por cualquier número irracional positivo.

Muy bien. La solución es la misma que la mía.
Tienes una prueba para el teorema?
Creo que esa es la parte interesante de este problema.

Alguien de ustedes sabe donde se puede encontrar este tipo de teoremas.  :blush:


function
quadratics
inequalities
inequalities open

how many $ [x]$s that meet $ [x]^{2}\minus{}9[x\plus{}1]\minus{}1<0$, an inequality about $ x$?
($ [x]$ is maximal integer that isnt bigger than $ x$.)

[quote="Danielsinx"]how many $ [x]$s that meet $ [x]^{2} \minus{} 9[x \plus{} 1] \minus{} 1 < 0$, an inequality about $ x$?
($ [x]$ is maximal integer that isnt bigger than $ x$.)[/quote]
Let $ n \equal{} [x]$ then obviously $ [x\plus{}1] \equal{} n\plus{}1$. So $ [x]^{2} \minus{} 9[x \plus{} 1] \minus{} 1 \equal{} n^2 \minus{} 9(n\plus{}1) \minus{} 1 \equal{} (n\minus{}10)(n\plus{}1)$. So when $ n \in \{0,1,2,..,9\}$ the inequality is fulfilled. Then the answer is $ 10$. (This should be on the preolympiad or intermediate forum)


function
LaTeX
quadratics
parameterization
algebra
algebra solved

Solve the system of equations :
 \[ \begin{cases}x+y+(z^2-8z+14)\sqrt{x+y-z} =1 \\2x+5y+\sqrt{xy+z} = 3\end{cases} \]

[quote="shobber"][quote="Valiowk"][b][u]The author of this post is: cowboy.[/u][/b]
_____________________________________________________________

We have:
$z^2 - 8z+14=(z-4)^2-2$.  After that this problem is very easy.[/quote]
I can't see it............[/quote]

Neither do I.. Also, do you mean $z$ on the original equation? Because then what does $z$ equal?

The question asks to solve the system of equation, which means for $x,y,z$. Your equation only gives $z$ away..

Perhaps little clarification? Thank you.

maybe use the monotonicity of functions??
Or the fact that $A^2+\sqrt{B}=0$ then $A=0$ and $B=0$??
But i still don't see it...

i dont see it either... could some one post another clue??

.. i dont see how using (z-4)^2-2 helps...

[quote="Eluer000"]I can not do it.[/quote]
So please show your work.
Thanks.

[quote="math_friends"]is it ..  A square - 2AB +B square = (A+B) square

 !! but its so simple right!![/quote]
$a^2-2ab+b^2=(a-b)^2$??
How to use it?

[quote="tommy1729"]:rotfl:  :D

a simultane equation with n variables and (n-1) equations ,

isnt that asking for trouble ??

equations like that have normally no solution or infinite many.

look at

1)x+y=1

2)4x+5y+6z=1 ; 3x+2y+z=2

its obvious.

same thing for inequations or even a 'mix'

1)x+y>1

2)x+y+z<1 ; 4x+y²+z=3

so i dont see what the fuss is al about  :!:
even with conditions and stuff pfff  :P
pfffff  :roll:

and if your looking for boundarys of the possible solutions , just replace variables by others and compute equations or their derivates
pfff still simple
srr if im arrogant  :D  :P  :rotfl:  :roll:[/quote]

   Really my dear friend? Why i can not see it? Can u give me more than 2 set of root for this system?
It's much much differrent with the thing u have seemed...Plz think more,it's really an interesting problem,if u can not solve this i'll post my solution( But i need time to learn Latex )
2 small hint for its:
1) What's specialization?
2) Can u find any relation between equation 1 and 2...? The answer is no,so means with only 1 equation u can solve it completely,the second is used to substitute the result from 1 to find the root (1,01)=(x,y,z)..OK try it and good luck

[quote="tommy1729"]:rotfl:  :D

a simultane equation with n variables and (n-1) equations ,

isnt that asking for trouble ??

equations like that have normally no solution or infinite many.

look at

1)x+y=1

2)4x+5y+6z=1 ; 3x+2y+z=2

its obvious.

same thing for inequations or even a 'mix'

1)x+y>1

2)x+y+z<1 ; 4x+y²+z=3

so i dont see what the fuss is al about  :!:
even with conditions and stuff pfff  :P
pfffff  :roll:

and if your looking for boundarys of the possible solutions , just replace variables by others and compute equations or their derivates
pfff still simple
srr if im arrogant  :D  :P  :rotfl:  :roll:[/quote]
Well, using "$\text{number of eqautions}\leq \text{number of variables}$" as a reason to say that this system has none or infinitely many solutions is not always correct.  :rotfl:
For example, $x^2+y^2+5z^2-2x+2z+4yz+2=0$ has unique real solution.
Can you solve it? (I'm asking tommy1729 only.)

[quote="shobber"]I decide to remind you of this equation. I hope that someone would notice it and solve it.

Besides, I think this should be in Unsolved Problems or Open Problems sections.[/quote]

   Yeah really that i have created a non acceptable mistake....U are right Shobber it should be placed in Algebra open problem...I will try to solve it for a few day more ...My previous way was: from the equation (1) only, i put $\sqrt{x+y-z}=t$ and (1) become a quadratic equation of variable t if we treat z as parametter..and again treat t as parameter we have another quadratic equation of z...By the codition of positive delta we can determined the identical set of z...1 sign mistake made all my solution totally wrong....Really sorry...

( I seldom mistake in solving equation but "seldom" not means "never" )...How sad..

Kummer

hello, we get here $x=f_1(z),y=f_2(z)$
Sonnhard.


the graphs of y=x^3-3x+2 and x+4y=4 intersect at the points (x1,y1), (x2, y2) and (x3, y3).  If x1+x2+x3=A and y1+y2+y3=B, compute the ordered pair (A,B)


[quote="futuremontrealian"]This was a recent problem we did at my school.

Find the sum of:

$\frac{1}{10^{0}}+\frac{1}{10^{1}}+\frac{2}{10^{2}}+\frac{3}{10^{3}}+\frac{5}{10^{4}}+\frac{8}{10^{5}}+\frac{13}{10^{6}}+...$

Hint:
[hide]Find a way to express the sum in terms of a constant and itself[/hide][/quote]
[hide]We can notice that the numbers in the numerators are the Fibonacci numbers, and we can use this to our advantage to find the following equation (in terms of $S$, the sum):
$S=\frac{S}{10}+\frac{S}{100}+1$.
(This can be easily checked if you right out the first few terms of each of the terms.)
We can solve this equation to get
$\frac{89S}{100}=1$, or
$S=\boxed{\frac{100}{89}}$.[/hide]

Let the sum be x; since the numerator is Fibonacci we have $x=\frac{x}{10}+\frac{x}{100}+\frac{1}{1}\rightarrow 89x=100\rightarrow x\frac{100}{89}$


calculus
function
quadratics
algebra unsolved
algebra

hello, i have got with the help of calculus
$ \minus{}\infty<y \le\minus{}\frac{21}{4}$ or $ 0\le y<\plus{}\infty$
Sonnhard.

[quote="ayush_2008"]y=(x^2+2x+1)/(x^2-8x+12)
what is the range of this function.
i am getting (- infinity to -5] and  [0, infinity)
but graph does not show any gap.[/quote]

Let $ f(x)\equal{}\frac{x^2\plus{}2x\plus{}1}{x^2\minus{}8x\plus{}12}$ and $ g(x)\equal{}\frac 1{f(x\minus{}1)}\minus{}1\equal{}\frac{21\minus{}10x}{x^2}$

$ g'(x)\equal{}\frac{10x\minus{}42}{x^3}$ and so  :

$ g(x)$ is increasing on $ (\minus{}\infty,0)$ and $ g(\mathbb R^\minus{})\equal{}\mathbb R^\plus{}$

$ g(x)$ is decreasing on $ (0,\frac{21}{5})$ and increasing on $ (\frac{21}{5},\plus{}\infty)$ and $ g(\mathbb R\plus{})\equal{}[g(\frac{21}{5}),\plus{}\infty)$ $ \equal{}[\minus{}\frac{25}{21},\plus{}\infty)$

So $ g(\mathbb R)\equal{}[\minus{}\frac{25}{21},\plus{}\infty)$ $ \implies$ $ (\frac 1f)(\mathbb R)\equal{}[\minus{}\frac{4}{21},\plus{}\infty)$ $ \implies$  $ \boxed{f(\mathbb R)\equal{}(\minus{}\infty,\minus{}\frac{21}{4}]\cup[0,\plus{}\infty)}$

As Dr Sonnhard already said in the previous post.

[quote="ayush_2008"]patric when i draw the graph it shows all the value between -5.25 to infinity where is the problem.[/quote]

You fail to draw  :wink:

Look at your graph, choose $ x$ such that $ f(x)\approx \minus{}2$ and check computing the exact value for this $ x$. You'll see that $ (\minus{}5.25,0)\not \subset f(\mathbb R)$

i draw graph on computer the line pass through -5.25 to infinity. it means function will give the value from -5.25 to infinity.does the graph is wrong.

Huh ?

This graph confirms perfectly that $ f(\mathbb R) \equal{} ( \minus{} \infty, \minus{} \frac {21}{4}]\cup[0, \plus{} \infty)$  and that there is no value of $ f(x)$ in $ ( \minus{} 5.25,0)$

What is your problem ?

A short solution(without caculus)
On clearing the denominator we have $ x^{2}(y\minus{}1)\minus{}(8y\plus{}2)x\plus{}(12y\minus{}1)\equal{}0$
This is quadratic in $ x$ and its discriminant should be greater or equal to zero,From here you will get the range of $ y$.

[quote="Maths Mechanic"]A short solution(without caculus)
On clearing the denominator we have $ x^{2}(y \minus{} 1) \minus{} (8y \plus{} 2)x \plus{} (12y \minus{} 1) \equal{} 0$
This is quadratic in $ x$ and its discriminant should be greater or equal to zero,From here you will get the range of $ y$.[/quote]

Quite nice and quick! Congrats!
 :)

PATRIC I AM VERY SORRY I . what a silly mind i have .i am taking range along x axis.sorry again for wasting your your precious time .
once again thanks and regards.


quadratics
superior algebra
superior algebra unsolved

1. How many elements does have:
{$x^2 | x \in \mathbb{Z}/n^2\mathbb{Z}$}

2. Can we find the smallest $m(n)$ from $\mathbb{Z}/n^2\mathbb{Z}$, $m(n) \neq 0$ such that there is:
$x \in \mathbb{Z}/n^2\mathbb{Z}$ and $x^2 \equiv -m(n) \mod n^2$?

wouldnt the first problem be equivalent to
How many squares are there in the set $\left\{1,2,...,n^2 -1\right\}$ ?

for the second one i think $x^2+m(n)=0mod n^2$
and we can take $n^2=(x+(n-x))^2=x^2+2(n-x)x+(n-x)^2$
so we can have $m(n)=2(n-x)x+(n-x)^2$ no ?

No, the first question aks for the number of quadratic residues (included those not coprime to $n$ here) $\mod n^2$.
Let $f(n)$ be the number of squares $\mod n$.
Then $f(ab)=f(a)f(b)$ for coprime $a,b$ by the Chinese Remainder Theorem.
Thus it suffices to find the values for prime powers where it gets down to ugly casework, using that the multiplicative group $\mod p^n$ is cyclic for odd primes $p$ (the cases will concern the parity of $n$ and the cases with $p=2$; I will not do this SPAM).

Well this problem has one advantage: since we are looking $\mod n^2$, the cases with odd exponents of primes can be neglected, thus only half that much is to do.

i dont understand why zetax you say that im wrong
i still think im right :roll:

E.g. take $n=3$:
$7 \equiv 4^2 \mod 9$, but $7$ is not a square of your set.

ok then...if i correct the set a little bit
$\left\{1,2,...,(n^2 -1)^2\right\}$ is this good now ?
and how about the second problem ?

[quote="ZetaX"]
Thus it suffices to find the values for prime powers where it gets down to ugly casework, using that the multiplicative group $\mod p^n$ is cyclic for odd primes $p$ (the cases will concern the parity of $n$ and the cases with $p=2$; I will not do this SPAM).
[/quote]

Why do you think this is SPAM?  ;)

It's straightforward computation... (using known properties)
Also, I think have already done it somewhere here ;) (but no idea where... was about one month ago).


calculus
integration
algebra
function
domain
Ring Theory
number theory

Let m in [b]Z[/b], where m is not a square in [b]Z[/b]
Define N: [b]Z[/b][sqrt(m)]---> [b]N[/b]
Suppose that : a belongs to [b]Z[/b] [sqrt(m)]. If N(a) is a prime in [b]N[/b] . Show that a is an atom in [b]Z[/b] [sqrt(m)]. Give an example of an m and a in [b]Z[/b] [sqrt(m)] such that a is an atom but N(a) is not a prime

I dont understand the difference between atom and prime?

Thank you

Atom is a synonym for irreducible. prime => atom, but not conversely.

[quote="billwgd"] prime => atom, but not conversely.[/quote]

Not necessarily.

If $ R$ is an integral domain, then every prime element $ p \in R$ is irreducible (but not conversely, like you said).  But if $ R$ is a principal ideal domain, then an element $ p \in R$ is prime iff it is irreducible.  For example:

Let $ p \in R$ be an irreducible element, with $ R$ a principal ideal domain.  We show that the ideal generated by $ p$ is a maximal ideal, hence a prime ideal so that $ p$ is a prime element.  Suppose $ I$, an ideal, is such that $ (p) \subseteq I \subseteq R$.  Since $ R$ is a principal ideal domain $ I \equal{} (r)$, some $ r \in R$, so we have that $ (p) \subseteq (r)$.  Thus, there exists an $ s \in R$ such that $ p \equal{} rs$.  But $ p$ is irreducible, so either $ r$ or $ s$ is a unit.  If $ r$ is a unit, then $ I \equal{} (r) \equal{} R$.  If $ s$ is a unit, then $ (p) \equal{} (r) \equal{} I$.  Either way, $ (p)$ is maximal, hence prime, and $ p$ is a prime element.

So actually, in a lot of cases, prime and irreducible are interchangeable.

Of course. "Not conversely" simply means that there exists a domain where the converse fails.

Many domains have prime atoms,  e.g. PID,Bezout,UFD,GCD, Schreier, Gaussian. See my post [1] for much further discussion.

--Bill Dubuque

[1] http://google.com/groups?selm=y8z8yto8k85.fsf%40nestle.ai.mit.edu

PS a slicker proof of atom => prime that works for ideals/gcds

is by Euclid's Lemma: (p,a) = 1 = (p,b) => (p,ab) > (p,a)(p,b) = 1


Pos pigame sth junior?
exo poly douleia kai den parakoloutho to forum...sorry.alla thelo an mou peite apotelesmata.

Έχουμε και λέμε: 5 μετάλλια εκ των οποίων 2 αργυρά (Λογοθέτης, Παγωνάκης) και 3 χάλκινα (Παντέλα, Μπραζιτίκος, Ντριάνκος). Συγχαρητήρια στα παιδιά!!

Τερματίσαμε στην 5η θέση με 88 βαθμούς. Not Bad

Η επίσημη σελίδα της Βαλκανιάδας : http://www.jbmo.eu

Αλέξανδρος

Συγχαρητήρια στους συμμετέχοντες για τα καλά αποτελέσματα :thumbup:

5 ήταν τα 40αρια (2 ρουμανοι, 2 βουλγαροι και ενας τουρκος) και μαλιστα ο βουλγαρος βγηκε στη 1:20 λεπτα!  :o

Συγχαρητήρια στα παιδια και ιδιαιτερα στην Στέλλα που είναι απο το σχολείο μου :lol:  :lol:  και αν και 2α γυμνασίου τα καταφερε περίφημα!!! 5η η Ελλάδα ειναι πολύ καλή θεση!!! Μπράβο σε όλη την ομάδα!!!

Συγχαρητήρια σε όλους !

Καλή επιτυχία στην IMO !   :)

[quote="silouan"][quote="dr.koukoulas"]μαλιστα ο βουλγαρος βγηκε στη 1:20 λεπτα!  :o[/quote]

Στη Βουλγαρία δεν γινόταν ?  :wink:
Δε μπορούσε να περιμένει και πολύ ,έπρεπε να πάει για καφέ στο Shumen  :D[/quote]

κι μετά να διαβάσει τον πρωταθλητή της Βουλγαρίας :rotfl:  :rotfl:  :rotfl:

Sygharhthria paidia!!!!!Para poly kala-apsoga!!!

[quote="silouan"][quote="dr.koukoulas"]μαλιστα ο βουλγαρος βγηκε στη 1:20 λεπτα!  :o[/quote]

Στη Βουλγαρία δεν γινόταν ?  :wink:
Δε μπορούσε να περιμένει και πολύ ,έπρεπε να πάει για καφέ στο Shumen  :D[/quote]

Ela re silouane....afou etsi ki allios oi voilgaroi einai para poly kaloi.den yparxei logos na afhnoume iponooumena..
(ektos an ksereis kati pou den ksero...)

ps. congrats kai gia ton aderfo  sou -na tou ta metafereis bye

[quote="stergiu"][color=red]Συγχαρητήρια[/color] σε όλη την ομάδα !

 Έλλειπα και μόλις τώρα γύρισα. ΑΥτά είναι πολύ καλά αποτελέσματα. Μακάρι τα παιδιά αυτά να μπορέσουν να συνεχίσουν έτσι και στις ΒΜΟ , ΙΜΟ.
  Καλή συνάχεια και ακόμα μεγαλύτερη διάκριση εύχομαι και στους  ''μεγάλους''.

Μπάμπης[/quote]
Νομίζω ότι το τελευτσίο θέμα έχει λάθος ? εκφώνηση . Αντί να λεει $ 7p+3p-4$ θα έπρεπε να λεει $ 7^{p}+3^{p}-4$

Πράγματι, απ'ότι έμαθα η εκφώνηση στο εν λόγω θέμα ήταν $ 7p+3^{p}-4$. Μία λύση που έδωσα μπορείς να δεις στο

http://www.mathlinks.ro/Forum/viewtopic.php?p=878053#878053

Αλέξανδρος

Είχε μπει και νωρίτερα
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=874754#874754

Nick δεν αφήνω υπονοούμενο . Ξέρω ότι οι Βούλγαροι είναι πολύ καλοί απλά το είπα για πλάκα . :)
Πώς τα πας στο ΜΙΤ ? Είναι φοβερό ?

Silouan geia!
mia xara pao!akrivos opos ithela!to mit sygekrimena einai kati parapano apo fovero.de leo perissotera oxi giati den exo ti na po h giati variemai apla giati exo douleia na kano-sthn amerikh panta exeis kati na kaneis!apla h diafora einai sto kato apo poies proypotheseis to kaneis, me ti mesa kai sto oti kaneis me efxaristhsh (panta)

Χαίρομαι που τα περνάς καλά . Ήμουν σίγουρος δηλαδή  :)
Θα τα πούμε .....


When someone says $ 5$ per $ 8$, it means $ 5$ out of $ 8$, or $ \frac58$.

Is that what you were looking for?

well, kind of...
Thanks for responding, though :)

[quote][b]What does "per" mathematically mean (it would also be very helpful if someone could also back up their definition with a math book defintion)?

I think I saw it defined in the aops volume 1: the basics book, but can't seem to find it. CAN ANYONE GIVE ME THE PAGE NUMBER WHERE I CAN FIND THE DEFINTION? PLEASE?[/b][/quote]

I don't have any AoPS books, however.  :noo:

"per" means "for each". It gives rise to a conversion factor.

For example, 55 mpg, or 55 miles per gallon, means every gallon of gas is equivalent to 55 miles of travel of the particular car.

This means that $ \frac{55 \; \text{mi}}{1 \; \text{gal}}$ is a conversion factor.


2006 students sit a test with six questions.  Each solution is marked right or wrong.  Given is that every three students solved at least 5 different questions between them.  Find the minimum value of the sum of the scores of all the students.


trigonometry
calculus
integration
number theory solved
number theory

Quite a few of these have been posted and solved around here. Take the second one, for instance. Let $z=e^{i\frac {2\pi}n}$. It's an algebraic integer, because it satisfies $z^n=1$. This means that $2i\sin (\frac {2\pi}n)=z-\bar z$ is also an algebraic integer, and so is $2\sin(\frac {2\pi}n)$. Since it's also rational, it means that it's integral, and it's easy from here: being in $[-2,2]$ it can only be $0,\pm 1,\pm 2$, and we solve it for these values.

[quote="N.T.TUAN"]
I haven't got any solution.[/quote]

So, please post it in the 'Unsolved problems' section or in the 'Open problems' one.
Not in the 'Solved problems' section.... ;)

Pierre.

[quote="N.T.TUAN"]Find all numbers positive integer [tex]n[/tex] such that:
a)[tex]sin\frac{\pi}{n}[/tex] is rational number
b)[tex]sin\frac{2\pi}{n}[/tex] is rational number
I haven't got any solution.[/quote]
Well, I solved this problem but i can0t remember the method, but grobber failed, because $\sin(\pi/6)=1/2$. In fact I think the unique solutions was $n=1,\, 2, \, 6$. Good luck looking for a nice solution for this nice problem.

[quote="RobertuX"]Well, I solved this problem but i can0t remember the method, but grobber failed, because $\sin(\pi/6)=1/2$. In fact I think the unique solutions was $n=1,\, 2, \, 6$. Good luck looking for a nice solution for this nice problem.[/quote]You are hurrying up to judge people, without paying attention to their posts. Let me recap for you what grobber [b]said[/b]:

[quote="grobber"][u][b]Take the second one, for instance[/b][/u]. Let $z=e^{i\frac {2\pi}n}$. It's an algebraic integer, because it satisfies $z^n=1$. This means that $2i\sin (\frac {2\pi}n)=z-\bar z$ is also an algebraic integer, and so is $2\sin(\frac {2\pi}n)$. Since it's also rational, it means that it's integral, and it's easy from here: being in $[-2,2]$ it can only be $0,\pm 1,\pm 2$, and [u][b]we solve it for these values.[/b][/u][/quote]
Now if you have carefully read, and it is clear that you did [b]not[/b], you would have seen that grobber did not say that $ \{ 0,\pm 1,\pm 2 \} $ is the set of solutions for $n$, but the values that $2 \sin{\left( \frac {2\pi}n \right)}$ takes, and now we can easily solve each of the 5 equations
\[ \begin{array}{ccc} 2 \sin{\left( \dfrac {2\pi}n \right)} &=&  0 \\ \ & \ & \\
2 \sin{\left( \dfrac {2\pi}n \right)} &=& \pm 1 \\ \ & \ & \\
2 \sin{\left( \dfrac {2\pi}n \right)} &=& \pm 2
\end{array} \] giving all the solutions, including the one you specified above.

This is the [b]Olympiad[/b] section of the site. If needed more details on easy problems such as this one, one can use the [b][url=http://www.mathlinks.ro/Forum/index.php?f=214]High-School[/url][/b] forums.

PS Next time I will not waste anymore time explaining posts of other users in such detail, as for kids, but I will rather directly delete the additional balast ;)


AMC
USA(J)MO
USAMO
number theory proposed
number theory

Show that there are infinitely many positive rational solutions $(a_1,a_2,a_3)$ to:

\[ a_1+a_2+a_3=a_1a_2a_3=n\in \mathbb{N}  \]

Yes, it is related to http://www.mathlinks.ro/Forum/viewtopic.php?p=490647#p490647 but only by its aesthetics, not by its content.

Infinitely many sets $(a_1, a_2, a_3, n)$?  Or infinite for each $n$ such that there exist solutions?

Can you use ideas like moving to a side and factoring nice things that look like $abc-a-b-c+C$ using stuff like (a-1)(b-1)(c-1)?

[quote="ikap"]Show that there are infinitely many positive rational solutions $(a_{1},a_{2},a_{3})$ to: \[a_{1}+a_{2}+a_{3}=a_{1}a_{2}a_{3}=n\in \mathbb{N}\] Yes, it is related to http://www.mathlinks.ro/Forum/viewtopic.php?p=490647#p490647 but only by its aesthetics, not by its content.[/quote]
we know that there are infinitely rational solutions to a_1+a_2=a_1a_2 by basic factoring. Thus, since mulitplication is restrictive, a_1a_2 is rational, and a_1+a_2 is the same rational. Thus a_3+r=r*a_3, or (a_3)(r-1)=r, which means a_3 is (a1+a2)/(a1a2-1) when a1a2=a1+a2.
Thus there is at least a bijection from the set of (b_1, b_2) where b_1+b_2=b1b2, to the set of (a_1,a_2,a_3) such that a1a2a3=a1+a2+a3

Since there are infinitely b_1 b_2 that satisfy the first requirement, there must be infinitely many that satisfy the second.


[b]Welcome to Mafia Game 23![/b]

It is now Day Five. 12 players are alive, so approximately 7 votes are required to lynch, and 6 to acquit.

Day 1 begins on page 2 of the thread.
Day 2 begins on page 9.
Day 3 begins on page 20.
Day 4 begins on page 30.
Day 5 begins on page 33.

This section will contain updated information about the game (i.e. what day/night it is, who’s alive, etc.).

If you ever have any questions at all about your role, events in the game, the meaning of life, the structure of finitely generated torsion modules over a principal ideal domain, or anything else, please PM me. Some of the roles are kind of complicated and/or weird in general.

Living players (14):
Jashper
daine
kheg_k
pieterminate
Junggi [size=59]LadyKn1ght[/size]
Peter VDD
me@home
baseballfreak2
Koda
Vihang
jackdillon
Lena [size=59]jhredsox [/size]

Dead players (9):
[color=green]junggi INSPECTOR 2, lynched day 1
Ars DOCTOR MOD 3, died night 1
jli INSPECTOR 3, lynched day 2
dts MASON 5, died night 2
laughinghead505 CITIZEN MOD 3, lynched day 3
Ars [size=59](boesfx)[/size] DOCTOR MOD 2, died night 3[/color]
[color=red]Ignite168 NURSE --> AVERY, died night 3[/color]
[color=green]mustafa MASON 4, died night 3
besttate SUPER CITIZEN, died night 3[/color]
[color=red]Klebian LUCIUS, lynched day 4[/color]
[color=green]perfect628 CITIZEN MOD 2, died night 4 [/color]

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

[b]RULES[/b]

Anyone not posting for 72 hours during a day phase is subject to replacement.

Voting will follow the method that has become more or less standard: Accuse, second, defend, then vote.  If you accuse or second, you must still vote to be counted as an “aye”.

You may not PM anyone at night unless your role PM says so.

When someone dies, you will see their complete role PM except the names of other people they are affiliated with.  If your role PM has a line in the middle, the information below the line will not be made public when you die.

When people are nightkilled, you will not find out who did it.  You should never read anything into my stories about how people died.

If you have a night role, you are not required to exercise it.  You may also exercise it only partially (if this applies to you, the meaning should be clear).  There is one exception, and I am notifying that person by PM.

If everyone dies the same night, everyone is resurrected the next day.

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

[b]NOTES[/b]

There are many, many, many new roles.  There are also some old roles implemented in new ways.

Lynches may fail.  If a lynch fails, the day will continue unless you vote for “no lynch”.  If a lynch fails, you can’t try to lynch the same person again the same day.  You can still try to lynch them again on a later day.

There are some roles that mess with the voting.  Thus, some players may at times have zero votes, or two or more votes.  You won’t know this—in fact, it’s possible that even the player in question may not know about it!  For the purposes of vote counting, I’ll assume that everyone has one vote.  But I’ll take into account the actual vote numbers for determining if someone is lynched or acquitted.  Thus it’s possible that the vote count might read, say 11 “aye” votes out of 23, but the person under accusation would still be lynched.

There are no “secrets”.  Everything I know is known to at least one player.  For example, there are no millers—if you show up guilty to inspection, either you know it or someone else does.

In response to some PM's I've received, I'm changing the rule about not PMing night zero.  If your role allows night PMing, you may start that night zero.  However, I'll only guarantee you 24 hours of night to do that.

Night zero will end when the following two things have happened:
1. It's after 11 PM CST tomorrow (Tuesday)
2. All night zero choices are in

I'd also like to remind everyone that some, but not all, roles are themed.

yo wassup

[quote="Sly Si"]This section will contain updated information about the game (i.e. what day/night it is, who’s alive, etc.). [/quote]
That might be hard, since we can't edit posts after 24 hours of the post.

Silas can PM me or IM me what he wants to change and I will do as he says.

(I'm a moderator of this forum)

[quote="junggi"][quote="Koda"][quote="jli"]lol...i have no questions about the meaning of life[/quote]

Nor do I.[/quote]
is that supposed to mean something[/quote]
of course it is. it is supposed to mean that i have no questions about the meaning of life, not even with brother bear thrown in.

[quote="Koda"][/quote][quote="junggi"][quote="Koda"][quote="jli"]lol...i have no questions about the meaning of life[/quote]

Nor do I.[/quote]
is that supposed to mean something[/quote][quote="Koda"]
of course it is. it is supposed to mean that i have no questions about the meaning of life, not even with brother bear thrown in.[/quote]

Bubka please cut off this brother bear stuff.

[quote="Ignite168"]stfu it's night[/quote]

Sorry, anyway you said hi first. Well good night!

A couple of additions to the rules and notes have been posted.

--You don't have to exercise your night role, and you don't have to exercise it in full if you do.  The one exception has been notified by PM.

--If everyone dies the same night, everyone is resurrected the following day.

--There are no "secrets" (i.e. millers who don't know their identity, etc.).  If I know something, then at least one player knows it, though that doesn't say anything about which player (i.e. there could be millers whose identity is known to someone else).

[quote="Sly Si"]The [b]one[/b] exception has been notified by PM.[/quote]

thx for the info

[quote="Sly Si"]
There are some roles that mess with the voting.  Thus, some players may at times have zero votes, or two or more votes.  You won’t know this—[b]in fact, it’s possible that even the player in question may not know about it![/b]  For the purposes of vote counting, I’ll assume that everyone has one vote.  But I’ll take into account the actual vote numbers for determining if someone is lynched or acquitted.  Thus it’s possible that the vote count might read, say 11 “aye” votes out of 23, but the person under accusation would still be lynched.

There are no “secrets”.  [b]Everything I know is known to at least one player.[/b]  For example, there are no millers—if you show up guilty to inspection, either you know it or someone else does.[/quote]

???

I [color=red]accuse [/color] pioanoforte because of jasphers roleclaim.

screw the defense thing
[color=red]aye[/color] on pianoforte

OK... I don't know why pianoforte is giving up, but that is in fact a lynch.
[size=150]
[b]Game over.

Citizens win.[/b][/size]

Here's pianoforte's role PM:
[quote]You are the SERIAL KILLER MOD 3.  You win all by yourself by being the last person alive.  Your power is to kill any three people on nights with a number divisible by 3, beginning with night 3.[/quote]

Here's Jashper's:
[quote]You are the BLOODY BARON.  You win with the citizens.  However, if you are lynched, you survive and join the Death Eater team.  If lynched a second time or nightkilled at any time, you die.[/quote]

And here's jackdillon's:
[quote]You are SNAPE.  You are innocent and win with the town.  However, because everyone—innocent or guilty—thinks you’re on their side, you survive one lynching and one night kill.  Unfortunately, your past is somewhat checkered, so you show up guilty to inspection.[/quote]

And here's the list of everyone's roles:

[hide][u]Death Eater team:[/u]
jhredsox/Lena: Lord Voldemort (nightkills, recruits, makes Horcrux)
Klebian: Lucius (redirects anyone's night choice, makes them vote nay on DE's)
Koda: Bellatrix (roleblocks)
Ignite168: Formerly Nurse (replaces a doctor), converted to Avery (no special power)

[u]Serial killers (each wins alone):[/u]
Peter VDD (mod 2)
baseballfreak2/pianoforte (mod 3)

[u]Other special roles:[/u]
junggi: Inspector 2 (knows jli is naive)
jli: Inspector 3 (knows junggi is insane)
BoesFX/Ars: Doctor Mod 2
Ars: Doctor Mod 3
Vihang: Legilimens (checks veracity of one statement each night)
daine: Dumbledore (protects one person at a time)
jackdillon: Snape (survives one lynch and one nightkill)
Jashper: Bloody Baron (converts to DE if lynched)

[u]Semi-ordinary citizens:[/u]
perfect628: Citizen Mod 2
laughinghead505: Citizen Mod 3
besttate: Super Citizen (has two extra votes for the whole game)
me@home: Proxy Citizen (gives vote to someone else)
kheg_k: Sorting Hat (can make anyone lynch-immune the first day, but no power after that)

[u]Masons:[/u]
LadyKn1ght/junggi: Mason 1 (sends messages to Mustafa)
pieterminate: Mason 2 (sends to dts)
Mustafa: Mason 4 (sends to pieterminate)
dts: Mason 5 (sends to LadyKn1ght/junggi)[/hide]

Some special notes:
[hide]
1. me@home gave his vote to Peter VDD every day until he was lynched on day six.

2. besttate never used any extra votes.

3. Vihang was the Horcrux, created night one.

4. daine protected Koda until night four, when he switched to Peter VDD.  Peter was then protected on night five and day six.  On night five, daine switched to protecting no one, which took effect the following night.

5. On the last day, the only person who could have been lynched was pianoforte.  jackdillon would have survived, and Jashper would have survived but switched sides.  So there were three possible outcomes.  One is that pianoforte was lynched and citizens win.  Had you voted for no lynch, pianoforte would have won.  Had you tried to lynch Jashper, then lynched pianoforte, Jashper and the Death Eaters would have won.

6. The purpose of the Sorting Hat role was to give you something to go on for Day One.

7. There was no Inspector 1, and there was no Mason 3.  I expected this to throw you off more than it did, but apparently you figured out that the numbering wasn't really significant.
[/hide]

Finally, here's a summary of what happened each night and day:
[hide]Night Zero: kheg_k makes himself and besttate lynch-immune.

Day One:
Perfect628, laughinghead505, and me@home had no votes; Peter VDD had two votes.  11 votes were required to lynch, out of 21 total.
junggi was lynched.

Night One:
jli inspected Ignite.
jhredsox killed Ars, recruited Ignite, and made Vihang a Horcrux.
Klebian retargeted Peter to jackdillon.
Koda blocked Vihang.
Vihang checked "Ignite and Ars are fellow mafia members", but didn't get a result.

Day Two:
Peter had two votes, but couldn't vote aye on any DE.
Perfect628 had two votes.
Laughinghead505 and me@home had no votes.
11 votes to lynch out of 21 total.
jli was lynched.

Night Two:
DE's killed dts, block Peter, and send laughinghead to nowhere.
Peter VDD targeted Ars and Koda, but fails.
Ars protected himself and Peter.
Vihang checks laughinghead's roleclaim, gets true.

Day Three:
laughinghead had 3 votes, but voted nay on any DE.
Peter had 2 votes.
perfect628 and me@home had no votes.
11 votes to lynch out of 20 total.
laughinghead lynched at deadline.

Night Three:
baseballfreak2 killed Ars, Ignite, and besttate
Vihang checked Peter's roleclaim of "Inspector mod 2", got false.
DE's killed Mustafa, blocked pieterminate, and sent Ars to nowhere.

Day Four:
me@home had no votes.
Peter VDD and perfect628 had two votes.
Klebian was lynched.

Night Four:
Vihang checked jackdillon saying "I am innocent but turn up guilty to all inpsections", got true.
DE's killed perfect628, blocked Peter
Peter targeted Vihang and Koda, but failed again.

Day Five:
me@home had no votes; Peter VDD had two votes
7 votes to lynch out of 12 total.
Koda was lynched.

Night Five:
Lena targeted Peter, but fails due to daine's protection
Vihang checked Lena's claim of innocence, got false
No one died.

Day Six:
me@home had no votes; Peter VDD had two votes
6 votes to lynch out of 11 total.
me@home was lynched.

Night Six:
Lena killed pieterminate.
Peter killed Vihang and junggi.
pianoforte killed Peter, junggi, and daine.

Day Seven:
Lena was lynched.

Night Seven:
Nothing happened.

Day Eight:
kheg_k was lynched.

Night Eight:
Nothing happened.

Day Nine:
pianoforte was lynched; citizens won. [/hide]

Good game, everyone!  I hope you enjoyed the setup.

I also want to point out one of the best maneuvers I've ever seen in this game.  Laughinghead roleclaimed on day two.  The following night, the death eater's targeted her so that she would automatically vote nay on one of their own on day three.  Then they "put her claim to the test" by making her the last one to vote on Klebian--and of course, her vote didn't make a difference.  She was then lynched, as the assumption was that her roleclaim had been proved false.

Nice going, DE's, and sorry, laughinghead.

jhredsox and baseballfreak just moved to the top of my vig list,

also I was 99% sure peter was guilty but wnated to keep him alive in case he was saying the truth or would give us up his mafia to survive. also I was slighty suspicious of my own paranoia. It was sad though that I had to get killed the night after I joined twice in a row

ah, after Jashper roleclaimed, I was pretty sure I was gonna be lynched. Sly Si, in my role, typo: "kill any [b]three[/b] people".

I couldn't do much in the game early on but hopefully I helped later with jackdillion.
(Thanks to Jungii who thought I was innocent).

And I don't know what Peter's plan was but me being a legilimens and my claim that Jackdillion was innocent helped the citizens the last 2-3 days.
Of course I could never help with the laughinghead issue because I had no proof and roleclaiming could've ruined the game.

Anyway yay for the Innocents!

I really thought Pianoforte would use the missing mason number for an arguement that Jaspher's claim had to be false and pianoforte as a result innocent...


The two roots of the equation $ a(b\minus{}c)x^2\plus{}b(c\minus{}a)x\plus{}c(a\minus{}b)\equal{}0$ are $ 1$ and:

$ \textbf{(A)}\ \frac{b(c\minus{}a)}{a(b\minus{}c)} \qquad
\textbf{(B)}\ \frac{a(b\minus{}c)}{c(a\minus{}b)} \qquad
\textbf{(C)}\ \frac{a(b\minus{}c)}{b(c\minus{}a)} \qquad
\textbf{(D)}\ \frac{c(a\minus{}b)}{a(b\minus{}c)} \qquad
\textbf{(E)}\ \frac{c(a\minus{}b)}{b(c\minus{}a)}$

[hide="Click for solution"]
By Vieta's Formulas, the product of the roots is $ \frac{c(a\minus{}b)}{a(b\minus{}c)}$.  Since one root is $ 1$, the other is the fraction shown, or $ \boxed{\textbf{(D)}}$.
[/hide]


group theory
abstract algebra
geometry
geometric transformation
rotation
superior algebra
superior algebra unsolved

Hi,
How will you show that a Sylow p-subgroup of $D_{2n}$ is cyclic and normal for every odd prime p?

The cyclic part is easy.  The rotations in D_2n are a cyclic subgroup order n and any Sylow p-subgroup for p odd will be contained in that.

For showing that it is normal, let P be the Sylow p-subgroup.  Then wts xgx^-1 is in P for g in P.  If we pick x a 'rotation' then it commutes with P.  If not, recall the representation of D_2n = < r, s | r^2 = s^n = 1, rsr = s^-1>.  Thus xgx^-1 = g^-1 which is in P.

[quote="Aknoln"]D_2n = < r, s | r^2 = s^n = 1, rsr = s^-1>.  Thus xgx^-1 = g^-1 which is in P.[/quote]

I believe that you meant s^2=r^n=1  .

Also, this problem, while both fundamental to most collegiate algebra courses and quite simple, is almost indentical to one of the numbers on a set I turned in just yesterday.

I wonder where koola goes to school...


If one square of a 2^n x 2^n chessboard is removed, prove that you can tile the remainder with L-trominoes.

.
..

is what an L-tromino looks like, if you didn't know.

Yeah, this is a nice problem.  I gave it once to my discrete math class.  A couple of students thought that all they had to do was show that the number of remaining squares, 4n - 1, was divisible by 3.  It was hard to convince them that there was more to the problem than that...

Yeah, this is a nice problem - we used to ask it at interviews at DE Shaw occasionally.

Ravi - couldn't you convince them by asking them to tile a 1x9 or 3x3?

I tried that.  I think their counterargument was that was a different problem, and so was irrelevant.  Not the brightest students.  The class was half computer science grad students (who were in general weak/rusty at math), and half math grad students (for whom the class was too easy).  A tough mix.

hmm, i sense induction: [hide][hide]we cna do it no matter what for a 2*2, now lets prove that if we know how to fill a 2^n*2^n we can fill 2^(n+1)*2^(n+1). This is done by filling the 2^n with the taken away squrare and then taking away a square from the reamining three (L figure attached to corner of filled in.). Thus we can do it for all 2^n*2^n.[/hide][/hide]

I feel like every discrete math class I've ever seen has exactly the profile you describe, Ravi.  Looks like ml2 would have done just fine in that class.

Yes, ml2 gets an A.  To me, this is a great problem for illustrating the power of induction.  It's not just another numerical problem like the sum of the first n integers.


Fie $x_n = \{\sum_{k=1}^{n}\frac{1}{\sqrt{k}}\}$ cu n natural. Sa se demonstreze ca oricare ar fi x din [0, 1] exista un subsir al lui $x_n$ convergent la x. ({} reprezinta parte fractionara)

Incearca sa demonstrezi asta pentru orice sir $x_n$ care satisface $x_n\to\infty$ si $x_{n+1}-x_n\to 0$. Sunt sigur ca au fost discutate chestii din astea de multe ori pe aici. Cauta in subforumul "Calculus - Analysis".

Hm, e cam mare subforumul ala. Daca cumva dai peste ceva de genul asta, te-as ruga sa ma anunti.


function
topology
real analysis
real analysis unsolved

Have  $\Omega$ is open,bounded set in  $R^{n}$ and f is uniformly continous ;bounded function in  $\Omega$.Let construct an open set
$\Omega_{1}$ contain  $\overline \Omega$ and a function g is extension of  f so that  g is continous in   $\Omega_{1}$.

Help me. :roll:

Define $\omega(\delta)=\sup\{|f(a)-f(b)|\colon |a-b|\le\delta, \ a,b\in\Omega\}$. This is the modulus of continuity of $f$. For any $x$ in $\mathbb R^{n}$ let $F(x)=\sup_{a\in\Omega}(f(a)-\omega(|x-a|))$. I leave it for you to prove that

(1) $F(x)=f(x)$ for all $x$ in $\Omega$;
(2) $F$ is uniformly continuous on $\mathbb R^{n}$.

can anyone give some help on how to prove mlok's claim :maybe:


hello, we have $ \frac{10}{100}\cdot x\equal{}100$ which gives $ x\equal{}1000$.
Sonnhard.


For how many integral values of n are the numbers 2n, 3n, and 10-n thte lengths of the sides of an obtuse triangle?

[hide="Answer"]2[/hide]

ummm guess: 2

the longest sides can be either 10-n or 3n. thats it?


An arithmetic progression has 5 terms that have a sum of 200. The sum of the last three terms added to 6 times the sum of the first two terms equals zero. What is the first term?

Read the stickies. Particularly, "Write Your Solution WELL".

Anyone want to explain how to solve it?

The middle term is 200/5=40.
If the last two terms add up to x
then the first two terms add up to 160-x.
So, $ 40\plus{}x\plus{}960\minus{}6x\equal{}0 \implies 5x\equal{}1000$
x=200
So, the last two terms are 100-y and 100+y.
100-y-40 = 2y, so y=20.
Therefore the last two terms are 80 and 120.
So, difference=40 and firstterm=-40.


logarithms
real analysis
real analysis unsolved

Let $ a_1,a_2,\ldots$ be a bounded monotone increasing sequence of positive numbers and let $ D_n \equal{} a_1a_2\cdots a_n$. Show that the sequence $ D_n$ converges to a nonzero limit if and only if $ \sum_{n \equal{} 1}^\infty (1 \minus{} a_n) < \infty$.

Consider $ \log D_n$ as a partial sum of an infinite series; apply the limit comparison test.

How could the limit comparison test apply if $ (1\minus{}a_n)$ is negative?

If $ a_n>1$ for some $ n$, then the limit does not exist. Remember that the sequence is increasing.

What does monotonicity have to do with it? I think all you need is that the elements of your sequence have the same sign. Then the fact that they must tend to 1 is encoded in the necessary and sufficient condition.

ln (1-a(n)) ~ a(n) as n->infinity, if a(n) is an infinitesimal over the base.

The monotonicity does matter, because this is really about absolute convergence.
The product $ \frac12\cdot\frac21\cdot\frac12 \cdot\frac23\cdot\frac32\cdot\frac23\cdot\frac32\cdot\frac23 \cdot\frac34\cdot\frac43\cdot\frac34\cdot\frac43\cdot\frac34\cdot\frac43\cdot\frac34 \cdots$ diverges to zero, while the corresponding sum $ \minus{} \frac12 \plus{} 1 \minus{} \frac12 \minus{} \frac13 \plus{} \frac12 \minus{} \frac13 \plus{} \frac12 \minus{} \frac13 \minus{} \frac14 \plus{} \frac13 \minus{} \frac14 \plus{} \frac13 \minus{} \frac14 \plus{} \frac13 \minus{} \frac14 \plus{} \cdots$ or its negative) converges to zero.

By the way, this is a very bad and misleading title.

[quote="jmerry"]
By the way, this is a very bad and misleading title.[/quote]
OK, I changed it :lol:.

[b]§outh§tar[/b] had a point: monotonicity can be weaked to the assumption that all (but finitely many) numbers $ 1\minus{}a_n$ have the same sign.


trigonometry
function
algebra unsolved
algebra

For real numbers a. b, c, d not all equal to 0, define a real function

[b]f(x) = a + b*cos(2x) + c*sin(5x) + d*cos(8x)[/b]

Suppose f(t) = 4a for some real 't'.

Prove that [i]there exists a real number 's' such that f(s) < 0[/i]

it would be good if you could continue your discussions here

http://www.mathlinks.ro/Forum/viewtopic.php?p=333297#333297 :)

but that topic i suppose is lost in oblivion and also the question wasn't answered.
 :blush:
so i expect people here to post their solutions.

No. [b]pardesi[/b] is right, it's better to revive an old topic to discuss it than to post a new one.


[quote="dynamic03"]Using Inclusion-Exclusion, find the number of integers between 1 and 2000 (inclusive) which are divisible by at least 1 of 2,3,5 or 7.

Can someone solve?

thanks.[/quote]

I don't think this should be in Advanced Fields, but may be in one of the Middle or High School Forum.

Anyway, the principle of inclusion-exclusion stablish that you have to count the number of numbers that are divisible for each, substract the number that are divisible for two, add the ones that are divisible for three, and substract the ones that are divisible by four of them. If $N$ is the number you want, then

\[N=
\left\lfloor\frac{2000}{2}\right\rfloor+
\left\lfloor\frac{2000}{3}\right\rfloor+
\left\lfloor\frac{2000}{5}\right\rfloor+
\left\lfloor\frac{2000}{7}\right\rfloor-
\left\lfloor\frac{2000}{2\cdot3}\right\rfloor-
\left\lfloor\frac{2000}{2\cdot5}\right\rfloor-
\left\lfloor\frac{2000}{2\cdot7}\right\rfloor-
\left\lfloor\frac{2000}{3\cdot5}\right\rfloor-
\left\lfloor\frac{2000}{3\cdot7}\right\rfloor-
\left\lfloor\frac{2000}{5\cdot7}\right\rfloor+
\left\lfloor\frac{2000}{2\cdot3\cdot5}\right\rfloor+
\left\lfloor\frac{2000}{2\cdot3\cdot7}\right\rfloor+
\left\lfloor\frac{2000}{3\cdot5\cdot7}\right\rfloor-
\left\lfloor\frac{2000}{2\cdot3\cdot5\cdot7}\right\rfloor=\]
\[\left\lfloor\frac{2000}{2}\right\rfloor+
\left\lfloor\frac{2000}{3}\right\rfloor+
\left\lfloor\frac{2000}{5}\right\rfloor+
\left\lfloor\frac{2000}{7}\right\rfloor-
\left\lfloor\frac{2000}{6}\right\rfloor-
\left\lfloor\frac{2000}{10}\right\rfloor-
\left\lfloor\frac{2000}{14}\right\rfloor-
\left\lfloor\frac{2000}{15}\right\rfloor-
\left\lfloor\frac{2000}{21}\right\rfloor-
\left\lfloor\frac{2000}{35}\right\rfloor+
\left\lfloor\frac{2000}{30}\right\rfloor+
\left\lfloor\frac{2000}{42}\right\rfloor+
\left\lfloor\frac{2000}{105}\right\rfloor-
\left\lfloor\frac{2000}{210}\right\rfloor=1514\]

Best,


Which college in Cambridge or Oxford that is best for Math and/or Engineering? Please help! There are just too many colleges to choose from!

You have to check to see what is in each school.  I didn't apply to either but I think there is a school for math and whatever else.  For some reason, Trinity at Cambridge rings a bell.


Well, this is my Flash art showcase. Some of them might be a little violent, sexually explicit, etc., so you have been warned.

[url=http://photobucket.com/albums/v259/drspyder/?action=view&current=SpecialOlympics.swf]Special Olympics[/url]

[url=http://photobucket.com/albums/v259/drspyder/?action=view&current=magerampage5.swf]Insane Killer Wizard. (don't ask)[/url]

A little Viagra gadget thing. [b][i]Moderator Note: Link has been removed.[/i][/b]

EDIT: Do not go anywhere else in my Photobucket album. I repeat, do not. You might see very inappropriate and gory images. Only my newer ones are disturbing, though.

My mind's a little twisted for a 13-year-old. That's all I have to say for where I got the ideas for these things.

[b][i]Moderator Note: I would follow that advice.[/i][/b]

Uh, thanks for the warning. The links don't even work for me.