Prove:$\frac{n(n+1)}{2}\neq k^{2}$, for $n$, $k$ are integers and $n>1$.

Prove:$\frac{n(n+1)}{q}\neq k^{q}$, for the same conditions as the previous proof plus $q>1$ and $q$ is an integer.

Prove:$\frac{n(n+q-1)}{q}\neq k^{q}$, for the same conditions as the previous proofs.

NOTE: For clarification, all integers are positive!!!

umm for one if $n=8$ then we have

$\frac{n(n+1)}{2}=\frac{8\cdot9}{2}=36=6^{2}$

Waaaait a second.

When $n=8$, $\frac{8(8+1)}{2}=36=6^{2}$.
Sorry :wink:

The other two aren't true either.

Where did you find these problems?  Did you make them up?

EDIT:  Well, someone beat me to it.

AH! I thought I was just being the biggest newb ever in my induction... way to make me lose my sanity for about 30 seconds.

The first one actually has infinitely many solutions.  It's well-known, and is not hard to solve by Pell's.  ;)


what is a cows favorite class in school? :P
yes, let us assume that cows get an education...

[quote="mustafa"]Math. Because math is the best class in school. :P[/quote]

 :rotfl:

and it's Moo-sic...(music)

[quote="klutzycherry14"][quote="mustafa"]Math. Because math is the best class in school. :P[/quote]
agreed...and science, but im afraid that that is not the right answer....sorry mustafa....the correct answer is moosic...lol :rotfl:[/quote]

sweet  :lol:

Obviously it's computer science. Why else would USACO have problems like

Some of Farmer John's cows are having a bad hair day!

or

The cows are playing silly number games again.

or

Farmer John's cows and Farmer Bob's cows are having a contest to see who can make the largest square.

or

Farmer John knows that if N (1<=N<=500) cows lick sugar, they will produce sweet butter. Where should he place the sugar to minimize the total walking time of all the cows?

You get the idea.

The cows' favorite class is math because they get to use graphing COWculators!!! :rotfl: :roll:

This topic is getting spammy...

[quote="mathnerd314"]The cows' favorite class is math because they get to use graphing COWculators!!!

This topic is getting spammy...[/quote]
haha :P  :rotfl: ...nope! though it makes sense.... :maybe:
hmm?


number theory
relatively prime
number theory unsolved

Find all positive integer $ m$ if there exists prime number $ p$ such that $ n^m\minus{}m$ can not be divided by $ p$ for any integer $ n$.

This seems to generalize IMO 2003, number 6.

[hide="Solution"]
We claim that the answer is all $ m\ne 1$.  Clearly, $ m \equal{} 1$ yields $ n \minus{} 1$, which can clearly be a mutliple of any prime.  Now, we consider $ m > 1$.  Let $ p$ be an arbitrary prime factor of $ m$.  Let $ p^k \parallel{} m$, so let $ p^k l \equal{} m$, where $ \gcd (l, p) \equal{} 1$.  Now, assume that no prime exists so that $ n^m \minus{} m \equiv 0\bmod q$ has no solution for $ n$.  Set $ (p^kl)^{p \minus{} 1} \plus{} (p^kl)^{p \minus{} 2} \plus{} \cdots \plus{} p^kl \plus{} 1 \equiv 0 \bmod q$.  Since the Left Hand Side of this is not congruent to $ 1\bmod p^{k \plus{} 1}$, we may set $ q$ not congruent to $ 1\bmod p^{k \plus{} 1}$.  We will show that this $ q$ yields a contradiction.  First, note that the remainder of $ m^{p \minus{} 1} \plus{} m^{p \minus{} 2} \plus{} \cdots \plus{} 1$ when divided by $ m \minus{} 1$ is $ 1 \cdot p \equal{} p$, which is relatively prime with $ m \minus{} 1 \equal{} p^kl \minus{} 1$.  Thus, $ \gcd (q, m \minus{} 1) \equal{} 0$, so $ m$ is not congruent to $ 1\bmod q$.  Since $ n^m \equiv m \bmod q$, we have that $ n^{p^kl} \equiv p^kl\bmod q$, so $ n^{p^{k \plus{} 1}l}\equiv (p^kl)^p\equiv 1\bmod q$.  Let the order of $ n\bmod q$ be $ x$.  This means that $ x|(p^{k \plus{} 1}l)$.  However, since $ n^{p^kl}\equiv p^kl\bmod q$, which is not congruent to $ 1\bmod q$, we have that $ x$ is not a factor of $ p^kl$, so $ p^{k \plus{} 1}|x|(q \minus{} 1)$, so $ q\equiv 1\bmod p^{k \plus{} 1}$, which is a contradiction.  Thus, there exists a prime, $ q$, so that $ q$ is not a factor of $ n^m \minus{} m$ for all integers $ n$. [/hide]

[quote="The QuattoMaster 6000"]This seems to generalize IMO 2003, number 6.

[hide=Solution]
$ \gcd (q, m \minus{} 1) \equal{} 0$, so $ m$ is not congruent to $ 1\bmod q$.
...
...
However, since $ n^{p^kl}\equiv p^kl\bmod q$, which is not congruent to $ 1\bmod q$, we have that $ x$ is not a factor of $ p^kl$, [b]so $ p^{k \plus{} 1}|x|(q \minus{} 1)$, so $ q\equiv 1\bmod p^{k \plus{} 1}$[/b], which is a contradiction.[/quote]

Should it be $ \gcd (q, m \minus{} 1) \equal{} 1$ rather than 0?  And also, is the bold part really necessary?  Since you showed that $ x$ divides $ m$ but doesn't divide $ mp$, isn't that contradiction by itself?

[quote="ZzZzZzZzZzZz"]
Should it be $ \gcd (q, m \minus{} 1) \equal{} 1$ rather than 0?
[/quote]
Yes, sorry.

[quote="ZzZzZzZzZzZz"]
And also, is the bold part really necessary?  Since you showed that $ x$ divides $ m$ but doesn't divide $ mp$, isn't that contradiction by itself?[/quote]
I showed that $ x | pm$, but $ x$ is not a factor of $ m$, not the other way around; sorry for the confusion.

I think this is much much harder than IMO Shortlist..


read the description.  Is anyone else having the same problem? Ex.  Post something and the number is still the same.

A post in the Fun Factory does NOT count as a "post," so you do not get more posts when you post in here.

That's why you can have that type of funny thing where someone posts in Fun Factory and the panel shows "0 posts"  :!: .

um...so where do your posts actually count?

That is because the level of thread discourse in Fun Factory is of a much lower standard as compared to everywhere else; the rules here are much more lenient, and much of the stuff here doesn't really contribute to anything.

If you put something here that contributes, could the post count?

There doesn't exist a formula that can objectively determine whether if a post "counts" or not; thus it will not count.

Additionally, if your will is to contribute posts solely for increasing your number of posts, you shouldn't post at all.

What if you want to contribute to the math cause, like something math-related? Will it count then?

If you want to post seriously, don't post it here.  And if you care that much about post count, you shouldn't be posting at all.


which one is larger? A or B?
$ A\equal{} \sqrt[3]7 \plus{} \sqrt {15}$,
$ B\equal{} \sqrt[3]{28} \plus{} \sqrt{10}$
 :D
I still can't find it without a computer  :(

Actually do u know Calculus ? its can be easily found using approximations in calculus .

I get the answer as B . please do let me know if ur familiar with atleast some basic calculus and then i will post my solution

No calculus needed. $ A<2\plus{}4\equal{}6$ and $ B>3\plus{}3\equal{}6$.

oh thank you jmerry! i can't believe it is only a very simple innequality
 :)


calculus
vector
MIT
college
linear algebra
real analysis
real analysis theorems

What is a logical order for taking these:

Calculus AB
Calculus BC
Multivariable Calculus
Differential Equations
Linear Algebra
Vector Calculus

Usually the first two are high school courses whereas the last four are college courses.  The college equivalent of AB/BC is usually called "Calculus I" or "Single-variable calculus."  The MIT math department, to give an example of an order, orders them as follows:

Single variable Calculus (18.01)
Multivariable and Vector Calculus (18.02)
Differential Equations (18.03)
Linear Algebra (18.06).

In my opinion, though, it's quite useful to have some linear algebra background before learning multivariable.

Depending on what topics you do in Differential Equations it could be very useful to have Linear before it as well (not necessary, though).

You will find, in general, that the notion of a [url=http://en.wikipedia.org/wiki/Vector_space]vector space[/url] is very powerful.  For example, the solutions of homogeneous linear differential equations such as $ y'' \plus{} ky \equal{} 0$ form vector spaces, and you will find this notion very useful when studying DifEq.  It is not necessary, but it helps.


calculus
integration
number theory unsolved
number theory

(a)
Decide, e.g. through inspection, least positive solution to x^2 - 13*y^2 = -1, start by decide the class x modulo 13. deduce also a least positive solution to x^2 - 13*y^2 = +1.

(b)
Decide the least positive solution to x^2 - 30*y^2 = 1. Decide by using the result to see x^2 - 30*y^2 = -1 is solvable.

[b]a)[/b]
Looking $\mod 13$ we get that $x^2 \equiv -1 \mod 13 \implies x \equiv \pm 5 \mod 13$.
Checking $x=5,8,18$ shows that $x=18, y=5$ is the smallest solution.
Now by $(18+5 \sqrt{13})^2= 649 + 180 \sqrt{13}$ we see that $x=649, y= 180$ is the smallest (I think you posted another thread where this was proven) solution to $x^2-13y^2=1$.

[b]b)[/b]
As above (one gets $x \equiv \pm 1 \mod 2,3,5$, so $x \equiv \pm 1 , \pm 11 \mod 30$, so $x=1,11,19,29,...$), one can find $x=11,y=2$ to be the smallest solution to $x^2-30y^2=1$.
Now assume that there is a solution $a,b$ to $a^2-30b^2= -1$, then as before one would have $(a+b \sqrt{30})^2 = x + y \sqrt{30}$, yielding $11=x=a^2+30b^2 , 2=y=2ab$, thus $ab=1$, so $a = b = \pm 1$ resulting in $11=x=(\pm 1)^2+30 (\pm 1)^2 =31$, clearly a contradiction.

If a pell equation has no solutions, can you always prove it using congurences?

If you are looking for rational solutions, then yes by the theorem of Hasse-Minkowski.
But when you are looking for integral solutions, then not, e.g.: $x^2-34y^2=-1$.


I know this is of little or no significance, but I wanted to point out one broken link:

At [url]http://www.artofproblemsolving.com/Legal/AoPS_L_Privacy.php[/url], the word "here" in

[quote]A parent may modify or delete all information pertaining their child by logging in to the community [url=http://www.artofproblemsolving.com/wforum/login.php]here[/url] using the user name and password which the child selected or by contacting us directly via one of the methods listed below.[/quote]

links to [url]http://www.artofproblemsolving.com/wforum/login.php[/url], not [url]http://www.artofproblemsolving.com/Forum/index.php[/url].

[quote]I'll fix it when I get back from MATHCOUNTS[/quote]

You're at mathcounts nationals? Where is it this year?

DC?! My 8th grade year it was in chicago. Chicago is nice and all, but I would have much rathered DC. Oh well.

The marriot hotel for mathcounts in chichago is a piece of crap. I want a refund. I am wondering something. I went to Washington D.C. How come all the buildings are short and like less than 10 stories high.

[quote="churchilljrhigh"]How come all the buildings are short and like less than 10 stories high.[/quote]

I have heard that is because buildings in DC are not supposed to be higher than the Capitol, but I can't find a source to confirm whether or not there is really such a law. The uniformity of height among the buildings along K Street is quite remarkable.

[quote="rrusczyk"]Thanks much - I'll fix it when I get back from MATHCOUNTS.[/quote]

You're at MATHCOUNTS Nationals?  Cool.  I had forgotten that MATHCOUNTS was moving its dates up - in 2002, it was in June, I believe.

[quote="churchilljrhigh"]How come all the buildings are short and like less than 10 stories high.[/quote]
[quote="tokenadult"]... The uniformity of height among the buildings along K Street is quite remarkable.[/quote]
At the risk on continuing this (off-topic) thread..  More than most cities, The District of Columbia, has scads of laws regulating architecture due to the unique character of our Nation's Capital.  Check out [url=http://dccode.westgroup.com/home/dccodes/]DC Code[/url], Division I, Title 6, Chapter 6, Subchapter I, Section 6-601.05(b).  Once you find the section, scroll down to the historical notes, which may help put this in context.


abstract algebra
Ring Theory
superior algebra
superior algebra solved

hi,  :D

suppose R is noetherian ring

i need a simple argument for the fact that R^n is a noetherian module then

is it as simple as this : if A and B are noetherian R modules,
A*B is also a noetherian module

the argument my professor gave me is : every submodule of A*B can be written
as A'*B' with A' submodule of A and B' submodule of B
is this true?

thx

fred

Off topic question concerning commutative diagrams: Is xypic installed here?

In any case, show that if \[0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0\] is an exact sequence with L,N noetherian then M is noetherian. Your question follows.

thanks koo

i know this theorem :

if N is a submodule of M, and N is noetherian and M/N too, then M is noetherian

so with this i proved your question

so what are you implying, that R^n is noetherian if R is noetherian

because : 0-->R^(n-1)-->R^n-->R-->0

by taking morphisms (a,b,c)--->(a,b,c,0)
and (a,b,c,d)--->(0,0,0,d)

is that what you meant??

however, what about the argument of direct product of two noetherian modules
and about the submodules of the direct product
do you have any comments about that?

:?  :blush:

i can be such an egg sometimes :

of course you meant

0-->A--->A*B--->B-->0

with (a)--->(a,0)
and (a,b)-->(0,b)

so that A*B is also noetherian

but however, another question : i'd like to ask, i need the second isomorphism theorem of modules for proving that M/N and N noetherian implies M noetherian, is that too farfetched??
thx

fred

A typical approach works for proving the property in koo's post.

We have an exact sequence $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$. Let $u:M'\to M$ and $v:M\to M''$ be the injection and surjection represented by the second and third arrow respectively.

Assume now that $M',M''$ are Noetherian. We can use the other interpretation of Noetherian: any increasing sequence of submodules is stationary.

Let $M_1\subseteq M_2\subseteq\ldots$ be a sequence of submodules of $M$. We define $M'_i=u^{-1}(M_i),M''_i=v(M_i)$. Since $M',M''$ are Noetherian, the sequences $M'_i,M''_i$ are stationary (it's easy to see that these sequences are increasing). Take an $n$ s.t. $M'_n=M'_{n+1}$ and $M''_n=M''_{n+1}$. It is an easy exercise to show now that $M_n=M_{n+1}$ (we use the given properties, of course: the injectivity of $u$, the surjectivity of $v$, and the fact that $Im(u)=Ker(v)$).


A spider must eat 3 flies a day to survive , After he has taken his three , he quits for the day . For a fly that passes in his path ,  he has an even chance of nabbing the fly . given that 5 flies have sailed past the spider today ( some survived , some not ) , the chance of survival of the next fly is ?

[hide]Actually i get answer as $ \frac{9}{13}$ but i am not sure whether it is right .
[/hide]
so, please post ur answers .

but how ? can u please explain  :( .

[hide]Also the book gives  4 options , they are :
a) $ \frac{9}{16}$ b) $ \frac{9}{13}$    c)$ \frac{9}{11}$     d)$ \frac{9}{17}$ [/hide]

Well, I may misunderstand something, but the first thing is to find the probability that the spider is still hungry by the moment the 6th fly arrives. It is the probability that at most $ 2$ flies out of $ 5$ have been caught, which is $ \frac {{5\choose 0} \plus{} {5\choose 1} \plus{} {5\choose 2}}{2^5} \equal{} \frac {1 \plus{} 5 \plus{} 10}{32} \equal{} \frac 12$. Now, with probability $ \frac 12$ the spider attempts to eat the fly at all, in which case it has $ \frac 12$ chance to succeed. So the probability for the fly to die is $ \frac 14$.

The possible caveat here is that I considered the model when the spider tries to catch every fly but eats only the first 3 and releases the rest but I do not see how that is different from the conditions of the problem.

How did you get your answer?

here is how i got it :

let A be the event that the 6th fly will survive
$ E_1 :$  Out of the first 5 , 3 are already eaten by the spider
$ E_2 :$Out of the first 5 ,  2 are already eaten by the spider
$ E_3 :$ Out of the first 5 , 1 is already eaten by the spider .
$ E_4 :$ out of the first 5 , none eaten by the spider
therefore
$ P(A/E_1 ) \equal{} 1$
$ P(A/E_2 ) \equal{} \frac {1}{2}$
$ P(A/E_3 ) \equal{} \frac {1}{2}$
$ P(A/E_4 ) \equal{} \frac {1}{2}$

total no: of possibilities are:$ {5\choose 0} \plus{} {5\choose 1} \plus{} {5\choose 2} \plus{} {5\choose 3} \equal{} 26$

$ P(E_1) \equal{} \frac {{5\choose 3}}{26}$

$ P(E_2) \equal{} \frac {{5\choose 2}}{26}$

$ P(E_3) \equal{} \frac {{5\choose 1}}{26}$

$ P(E_4) \equal{} \frac {{5\choose 0}}{26}$

therefore $ P(A) \equal{} \frac {10}{26} \plus{} \frac {10}{52} \plus{} \frac {5}{52} \plus{} \frac {1}{52} \equal{} \frac {9}{13}$

is this correct ?

This seems to be a completely wrong probability space to me. Let's bring this idea to the extreme. Assume that the spider needs just one fly to get full, the chance to catch the fly is $ 1$, and 5 flies already passed by. Following your method, we will get

$ E_1$: Out of the first 5, 1 is eaten
$ E_2$: Out of the first 5, 0 are eaten.

$ P(A|E_1)\equal{}1$
$ P(A|E_2)\equal{}0$

Total number of possibilities: $ {5\choose 1}\plus{}{5\choose 0}\equal{}5\plus{}1\equal{}6$

$ P(E_1)\equal{}\frac{5}{6}$
$ P(E_2)\equal{}\frac 16$

$ P(A)\equal{}\frac 56\cdot 1\plus{}\frac 16\cdot 0\equal{}\frac 56$.

Now, I hope, you see all the absurdity of this answer. If the spider catches every fly passing its way for sure and 5 of them have already passed his way, how on Earth can he stay hungry by the moment the 6th one arrives?

The logic behind your solution is the same as behind the famous statement about the probability to meet a dinosaur on the street tomorrow: there are two possible outcomes (to meet and not to meet), one of them is favorable, the other one is not, so the probability is $ \frac 12$.

I am puzzled about what the book authors had in mind though... :?

Actually i am not thoroughly convinced about that solution u provided in post 4 . because you say that there are  $ 2^5$ possibilities . but does that not include all 5 flies being eaten , which is not possible .

Why can't the probability be 5/6 ?

let me write down a table of all the possibilities .
let 1 denote , that $ F_i$ escapes and 0 that it is caught .
$ F_1$              $ F_2$           $ F_3$           $ F_4$           $ F_5$         $ F_6$

   1                  1                1             1         1            0

   1                  1                 1            1        0             1

   1                  1                 1            0         1            1

   1                  1                 0           1          1            1

   1                   0                1            1          1           1

   0                    1              1              1          1          1

so no: 1's of $ F_6$  is 5 out of 6 possibilities .

Well, I described the model. The spider takes a go at each fly it sees but eats only $ 3$ first and releases the other ones. Then the probability space includes the event that 5 flies have been caught (but only 3 of them have been eaten and 2 have been released).

If it is unclear why this modification is the same as the original problem, consider the arriving flies one by one. Let $ S_k$ be the state in which $ k$ flies have been eaten. At the moment $ 0$, we are at $ S_0$. After the first fly, we are at $ S_0$ or $ S_1$ with probability $ \frac 12$ each. After the second fly, we are at $ S_0$ with probability $ \frac 14$, at $ S_1$ with probability $ \frac 12$ and at $ S_2$ with probability $ \frac 14$. After the third fly, we get $ \frac 18, \frac 38,\frac 38, \frac 18$ for $ S_0,S_1,S_2,S_3$ respectively. Up to this moment, we have binomial distributions but now the situation changes: we do not go anywhere from $ S_3$ but still continue to move to the right with probability $ \frac 12$ from $ S_0,S_1,S_2$, so after the 4th fly we have $ \frac 1{16},\frac 4{16},\frac 6{16}, \frac 5{16}$ and after the 5th fly we get the probabilities $ \frac 1{32},\frac 5{32},\frac {10}{32},\frac {16}{32}$ for these $ 4$ possible states, as promised.

As to your last question, why cannot the probability to meet the dinosaur on the street tomorrow be $ \frac 12$? If you think it is, indeed $ \frac 12$, then you'd better stay home (because, if you meet it, with probability $ \frac 12$ it'll be a T-rex (again, either it'll be a T-rex, or not: 2 outcomes, one favorable), so with probability $ \frac 14$ you'll be eaten tomorrow) :P.

I hope you see the absurdity of the dinosaur example but can you pinpoint where the flaw in the argument is?

Please don't think i am arguing .
I understand ur model. But there is no condition given that  the spider can catch and then leave  the fly if need be .
Isn't that an assumption  :(  ?  Also can u point out specifically the mistake in my solution . about the table in post 7 is that also wrong ?

whats the answer u get when the spider has to eat jus 1 fly an there is only 1 fly coming at first , and then the second fly . what will the probability tahat the 2nd will survive . i get it as 1/2 .

[hide]

about the dinosaur problem , why should we care what type of dinosaur it is ? whether it be a t-rex or any other kind it shouldn't be a problem for us . right?

[/hide]

[quote="vijaymenon"]Also can u point out specifically the mistake in my solution . about the table in post 7 is that also wrong ?[/quote]
It is [b]exactly[/b] the same as in the dinosaur example. It is true that the full event space is partitioned into the described events, but what on Earth makes you think that they are of equal probability? And if they aren't, the (number of favorable outcomes)/(number of all outcomes) formula for the probability may be completely wrong.

If the probability to catch the fly is $ 1$, one fly is enough to make the spider full and one fly has already passed by the spider, the probability of the survival of the second fly is $ 1$. The reason is that the event that $ 0$ flies were caught after the first fly passed by has then probability $ 0$, not $ \frac 12$ (i.e., $ E_0$ and $ E_1$ do [b]not[/b] have the same probability).

There is no difference between "try to catch and release if caught" and "not to try catch at all" models as far as survival of flies is concerned but the first one makes the computations simpler. Since the equivalence argument seems shaky to you, I gave a second solution. Do you have any trouble with it?

In the first post that is the question says there is an even chance of nabbing the fly .
[quote="vijaymenon"]For a fly that passes in his path ,  he has an even chance of nabbing the fly . [/quote]

No problem with the second solution , but still i am confused  and maybe i think i should be thinking about this again .

So do u say the book has given  the wrong options ? And the answer is 3/4  ?

[quote="fedja"]

If the probability to catch the fly is $ 1$, one fly is enough to make the spider full and one fly has already passed by the spider, the probability of the survival of the second fly is $ 1$. The reason is that the event that $ 0$ flies were caught after the first fly passed by has then probability $ 0$, not $ \frac 12$ (i.e., $ E_0$ and $ E_1$ do [b]not[/b] have the same probability).
[/quote]

actually regarding the above lines i don't get u . i asked in [b]post 9[/b] the probability for the second fly to survive given that the spider has to eat jus one fly.
is it always 1 ? does that mean the probability that second will survive is 100% .

[quote="vijaymenon"]
So do u say the book has given  the wrong options ? And the answer is 3/4  ?
[/quote]
Yes, that's what I'm saying.

[quote]
 i asked in [b]post 9[/b] the probability for the second fly to survive given that the spider has to eat just one fly.
is it always 1 ? does that mean the probability that second will survive is 100% .[/quote]
If the probability to catch the fly is $ 1$, then yes. Otherwise you have to count. So, if the probability to catch each fly is $ p$, and one fly is enough, what would be the chance of the survival for the second fly in your opinion?

I believe when fedja calculated 1/4 as his probability, that is the probability that the fly
survives if at most 2 were eaten; but the fly survives also if 3 were eaten with probability
\[ \frac {{5 \choose 3}}{ 2^5} *1 \equal{} .3125\]
.

Hence

$ P(survive) \equal{} P(survive|at most 2 eaten) \plus{} P(survive|3 eaten) \equal{} .3125 \plus{} .25 \equal{} 9/16$.

Nope. The probability to survive if at most 2 have been eaten is $ \frac 12$, and the probability to survive if 3 have been eaten is $ 1$. So, if $ q$ is the probability that at most 2 have been eaten, then the final answer is $ q\cdot \frac 12 \plus{} (1 \minus{} q)\cdot 1 \equal{} 1 \minus{} \frac q2$. We all agree upon that. The disagreement is about the value of $ q$. I claim it is $ \frac 12$, dallasboy seems to claim that it is $ \frac {2^5 \minus{} {5\choose 3}}{2^5}$, and vijaymenon's original claim was that it is $ \frac {26 \minus{} {5\choose 3}}{26}$.

To make things even clearer, let me notice that any formula with $ {5\choose 3}$ in the numerator implicitly assumes that all triples of flies are eaten with equal probability. My main point is that it isn't the case. Indeed, the probability to eat the first 3 flies out of 5 is $ \frac 18$ (three first successful catches) and the probability to eat the last 3 is $ \frac 1{32}$ (two unsuccessful catches followed by 3 successful ones). One can make the full table of probabilities here (it is somewhat tiresome, but is directly related to the point I'm trying to pass across, so I'll do it). 0 means failed catch, 1 means successful catch, X means no attempt at catching.

00000 ... $ \tfrac 1{32}$

00001 ... $ \tfrac 1{32}$
00010 ... $ \tfrac 1{32}$
00100 ... $ \tfrac 1{32}$
01000 ... $ \tfrac 1{32}$
10000 ... $ \tfrac 1{32}$

00011 ... $ \tfrac 1{32}$
00101 ... $ \tfrac 1{32}$
01001 ... $ \tfrac 1{32}$
10001 ... $ \tfrac 1{32}$
00110 ... $ \tfrac 1{32}$
01010 ... $ \tfrac 1{32}$
10010 ... $ \tfrac 1{32}$
01100 ... $ \tfrac 1{32}$
10100 ... $ \tfrac 1{32}$
11000 ... $ \tfrac 1{32}$

00111 ... $ \tfrac 1{32}$
01011 ... $ \tfrac 1{32}$
01101 ... $ \tfrac 1{32}$
10011 ... $ \tfrac 1{32}$
10101 ... $ \tfrac 1{32}$
11001 ... $ \tfrac 1{32}$
0111X ... $ \tfrac 1{16}$
1011X ... $ \tfrac 1{16}$
1101X ... $ \tfrac 1{16}$
111XX ... $ \tfrac 1{8}$

Now you can easily condition upon these 26 events and see that you get $ \frac34$ again.

Note also that the idea to assign equal probability to these 26 events will result in a wrong probability for the first fly to be caught merely because it is caught in 11 of these 26 cases, so you'll get $ \frac {11}{26}$ instead of $ \frac 12$ that was given in the statement of the problem.

[quote] To make things even clearer, let me notice that any formula with {5\choose 3} in the numerator implicitly assumes that all triples of flies are eaten with equal probability. My main point is that it isn't the case. Indeed, the probability to eat the first 3 flies out of 5 is \frac 18 (three first successful catches) and the probability to eat the last 3 is \frac 1{32} (two unsuccessful catches followed by 3 successful ones) [/quote]

 :!: Ahhh, that is a very good point.

Thanks Fedja, for all ur replies . But i asked my maths professor who trains me and problem was actually reffered my him . he also says it is $ \frac{9}{13}$ . Dunno who is correct . :(

You know my argument. Ask him for his. Let him see this thread and post his objections if he disagrees.

Not that this really needs asserting, but everything* fedja has said is right and the answer is $ \frac{3}{4}$.

* Well, I haven't checked every last detail, but all the important stuff is all correct.  (And very clearly explained -- anyone still doubting should probably *carefully* reread and think about the various explanations he made.)

As u said i will also put forward ur argument to him , and let me see what he says . i will post his reply to that soon in a weeks time ,sorry for the delay.

Since in real life a spider has a web and passively hunts (unless you assume a more rare spider) and a fly that is nabbed in the web is dead since it is not released by the spider or does not stick to the web if the spider is full, its always 50%

There are some spiders that hunt without webs (the [url=http://en.wikipedia.org/wiki/Wolf_spider]wolf spider[/url] is, probably, the best known one) but it is beside the point. The point is that both "spider" and "fly" in this problem are just words that the book authors found more colorful than more abstract words "predator" and "prey" and that even those would be completely irrelevant to the question. The textbook authors often try to use some "real life" characters and events to make the problem statements more appealing and I have to agree that some of such attempts are plainly pitiful. Still, one thing mathematics is about is the extraction of a well-posed abstract problem from (alas, more often than not) incomplete and confusing verbal descriptions of something. If everyone were as picky as you and tried to point out all minor inconsistencies, the communication would come to a screeching halt. ;) So, we will all appreciate if you try to comment upon the essense of the problem instead of nitpicking on some clearly irrelevant details in the future :).

Understood.  So perhaps when using "words" it would best to stay with Star Wars references so as not to imply any reality.  Especially when the reality of the "words" obviously enters into the probability.  I believe the Wolf spider and other ambush types would be considered by most as "a more rare spider".  It just seems reasonable to interject some reality for consideration of the obvious before launching into the complex.  The arguments for an airplane on a treadmill not flying is an example.  It seems as though this would widen the problem solving and reasoning abilities.  Cutting is almost always the solution if you ask a surgeon.


Post and tell me what image u like; I'm not changing my avatar anyway, but...

I'm keeping this one as my avatar... don't try to persuade me otherwise...

EDIT: I've made a better version, which is the next? picture.

This is an awesome book...

This is obviously too big, and I can't keep the title part, but I could have edited it to just show the good ship [i]Lydia[/i]

I like [url=http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=9089]this one[/url] better. Besides, more people could actually guess what it means.

I mean, who's going to guess that your avatar is actually a three-footed crow, let alone understand what the heck it means.

Hah, better crow (brighter, intact left wing)

Photo editing masters, I know there are plenty of problems left, but I'm okay with this anyway.


LaTeX
calculus
integration
function
real analysis
real analysis unsolved

given x(t) has a fourier transform
X(f) = 1 / b +j*2*pi*f

determine the fourier transform

1. x(t-4)
2. x(5t)
3. x(5t-4)

my answer is

1. e^-j*4*w
2. 5

using the table

x(t-a) = e^-j*a*w
x(t) = 1

how do i solve 3? and i'm not sure if the rest that i did is correct?
can any1 help?

Seems like you are taking some engineering course (nobody else uses $ j$ for $ \sqrt{-1}$ instead of the conventional $ i$) :)

1) Write you formulae in an unambiguous way. I cannot tell anything about your answers (except that they seem completely wrong) because I do not really understand what the formula in the problem is. Use parentheses if you don't know $ \text{\LaTeX}$.

2) The correct shift and scaling properties of the Fourier transform are
$ g(x)=f(x-a)\implies \widehat g(\xi)=e^{-ia\xi}\widehat f(\xi)$
$ g(x=f(ax)\implies \widehat g(\xi)=\frac 1a\widehat f(\xi/a)$
Your "table" as you wrote it is nonsense.

3) Assume now that $ \widehat f(\xi)=\frac 1{1+\xi^2}$. If $ g(x)=f(3x)$, then $ \widehat g(\xi)=\frac 13\cdot\frac 1{1+(\xi/3)^2}=\frac 3{9+\xi^2}$ and if $ h(x)=f(x-5)$, then $ \widehat h(\xi)=e^{-5i\xi}\cdot\frac 1{1+\xi^2}$. I hope you got the idea.

i've never done fourier transform so don't really understand what i am doing.

but now i think i get what you are trying to explain..

thanks.. :)

[quote="wyl86"]i've never done fourier transform so don't really understand what i am doing.
[/quote]
Essentially, you are doing the linear change of variable in the integral
\[ \widehat f(\xi)\equal{}\int_{\minus{}\infty}^{\infty}f(x)e^{\minus{}i\xi x}\,dx\]
(that, up to $ 2\pi$ for which everybody has his own favorite spot, is exactly what the Fourier transform is: you put in a function of $ x$ and you get another function of $ \xi$; why such a strange operation makes any sense is a separate question with a very long answer)


[quote="DiscreetFourierTransform"]Prove:

$\frac{e^{ix}-e^{-ix}}{2i}= \sum_{n=0}^{\infty}(-1)^{n}\cdot \frac{x^{2n+1}}{(2n+1)!}$[/quote]

Uh...
[hide]It is well known that the LHS is $\sin x$. It is also known that the RHS is $\sin x$.
BUT $\sin x=\sin x$ :!: so we're done...?[/hide]

Congrats! You're done.

That was a funny problem, eh?
Do you have an alternate solution?
Other then just going with definitions?

[quote]Do you have an alternate solution?
Other then just going with definitions?

[/quote]
Just expand it! ( $e^{x}= 1+x+x^{2}/2!+\dots$) ... remember that $i^{2}=-1$ and you are done.
(That is how one can expand sin(x))


You are at a point $(a,b)$ and you need to reach another point $(c,d)$. Both points are below the line $x = y$ and have integer coordinates. You can move in steps of length 1, either upwards of to the right, but you may not move to a point on the line $x = y$. How many different paths are there?

[quote="Diarmuid"]If we take a path from $(a,b)$ to $(x,x)$, then the path with the opposite steps (i.e. replace all ups with rights and rights with ups) will go from $(b,a)$ to $(x,x)$. So for each path from $(a,b)$ to $(c,d)$ passing through some point on the line $x=y$, there exists a path from $(b,a)$ to $(c,d)$.  It is easy to see that this is a bijection.

The total number of paths not passing though the line is therefore ${(c-a)+(d-b)\choose c-a}-{(c-b)+(d-a)\choose c-b}$ (by the normal path counting formula).[/quote]

But you're not allowed to even touch the line $x=y$. I think that number that you came up with, it counts the instances where the path from $(a,b)$ touches the line but not cross it.

The paths from $(a,b)$ to $(c,d)$ can be regarded as the path from $(0,0)$ to $(c-a,d-b)$.
The numbers of the paths from $(0,0)$ to $(c-a,d-b)$ is $\binom{(c-a)+(d-b)}{c-a}$.
We substract this with the number of the paths that crossing $y=x$,then we will get the number of the
ways from $(0,0)$ to $(c-a,d-b)$ not crossing $y=x$.

Consider the paths that crossing $y=x$.
We move the paths downward one unit,then the paths will be from $(0,-1)$ to $(c-a,d-b-1)$.
The path will intersect $y=x$.
Choose the point they first intersect $(d,d)$,and reflecting the subpath from $(0,-1)$ to $(d,d)$ with $y=x$
Then the paths become from $(-1,0)$ to $(c-a,d-b-1)$,and every paths must cross $y=x$.
Obviously,the paths from $(0,0)$ to $(c-a,d-b-1)$ crossing $y=x$ are one to one corresponding to the paths from $(-1,0)$ to $(c-a,d-b-1)$.
Thus,the number of the paths from $(0,0)$ to $(c-a,d-b-1)$ crossing $y=x$ is equal to the number of the paths from $(-1,0)$ to $(c-a,d-b-1)$.
the number of the paths from $(-1,0)$ to $(c-a,d-b-1)$ is $\binom{c-a+1+d-b-1}{c-a+1}=\binom{(c-a)+(d-b)}{c-a+1}$

$\binom{(c-a)+(d-b)}{c-a}-\binom{(c-a)+(d-b)}{c-a+1}=\frac{(c-a)+(d-b)+2}{c-a+1}\binom{(c-a)+(d-b)}{c-a}$.

Sorry for my poor English, it may be hard to understand.


Does anyone have any cool [b]ideas[/b] of programs or games that could possibly be programmed in TI-BASIC?

Why do you want that? Use a computer instead.

[quote="SplashD"]I need some thing to do during math class while the rest of class is reduced to drawing tree diagrams[/quote]

Computer? are you allowed?

phoenix? jupiter hard should keep you occupied for a while...

Here is a decent game with highscores that takes about 15 minutes to create (I created it during my excess time in the EOGs).

[hide="code"]
:While 1
:ClrHome
:Output(1,1,"CAVE GAME v1.3
:Output(2,1,"Penguin Studios
:Output(3,1,"(C) 5-23-2006
:Pause
:ClrHome
:Output(2,2,"Easy
:Output(3,2,"Moderate
:Output(4,2,"Hard
:Output(5,2,"Insane
:Output(6,2,"Diety
:Delvar $\theta$
:2$\rightarrow$x
:Repeat $\theta$=21 or $\theta$=105
:Getkey$\rightarrow\theta$
: x+($\theta$=34 and x<6)$\rightarrow$x
: x-($\theta$=25 and x>2)$\rightarrow$x
:Output(x,1,">
:Output(x-1,1," "
:Output(x+1,1," "
:If $\theta$=45:Stop
:End
:ClrHome
:Delvar S
:1$\rightarrow$A
:Output(8,2,"Score:-----
:3$\rightarrow$L
:While L>0
:Output(8,8,S
:RanInt(1,7$\rightarrow$H
:RanInt(1,7$\rightarrow$J
:RanInt(1,7$\rightarrow$C
:RanInt(1,7$\rightarrow$D
:RanInt(1,7$\rightarrow$E
:RanInt(1,7$\rightarrow$F
:RanInt(9-(x>3)-(x>4)-(x=6),11-(x>4)-(x=6)$\rightarrow$G
:For(I,G,1,-1
:GetKey$\rightarrow\theta$
:Output(I,H,"*
:Output(I,J,"*
:Output(I,C,"*
:Output(I,D,"*
:Output(I,E,"*
:Output(I,F,"*
:Output(A,1,">
:A$\rightarrow$B
:If A>1:Output(A-1,1," "
:Output(A+1,1," "
:A+($\theta$=34 and A<7)$\rightarrow$A
:A-($\theta$=25 and A>1)$\rightarrow$A
:S+RanInt(40,50)$\rightarrow$S
:End
:If B=H or B=J or B=C or B=D or B=E or B=F:L-1$\rightarrow$L
:Output(8,14,sub("  LLL",L,3
:End
:5$\rightarrow$dim(LCAVE
:If S>LCAVE(x-1
:S$\rightarrow$LCAVE(x-1
:Output(1,1,"SCORE:-----
:Output(1,7,S
:Output(2,1,"Easy:-----
:Output(2,6,LCAVE(1
:Output(3,1,"Moderate:-----
:Output(3,10,LCAVE(2
:Output(4,1,"Hard:-----
:Output(4,6,LCAVE(3
:Output(5,1,"Insane:-----
:Output(5,8,LCAVE(4
:Output(6,1,"DIETY:-----
:Output(6,7,LCAVE(5
:Output(8,1,sub("PPPPPPPPPPPPPPPP",1,1+ipart(S/2000
:Pause
:End[/hide]

I would really only like ideas so I can make them myself.

Well, this other kid and I used to program a lot on the 89. We would make sequals of games...sorta similar to pokemon on the gameboy.

[quote="SplashD"]I need some thing to do during math class while the rest of class is reduced to drawing tree diagrams[/quote]
I think you should also draw diagrams. Or if those are a piece of cake for you, get some more advanced math book and study that.

try rectangular to polar coordinates, and polar to rectangular coordinates.

[quote="qqwref"]Hey b-flat, your game doesn't work on my calculator. What did you use? I have an 83. I can only get it to actually show something resembling a game if I tell it to Output(H,I,"* [for example] instead of Output(I,H,"*. My calculator usually gets stuck on the Output(8,14,sub(" LLL",L,3 line, anyway. What do you think?[/quote]

Sorry, but I had to type the entire thing quickly from memory (my calculator is skrewed), so I made mistakes.

For the 1st problem, just switch them all.

On the second problem, change it to:
:If L
:Output(8,14,sub("  LLL",L,3

That should do it.

How does the game run on an 83?  I have an 84+ SE, so it runs really well on mine (well, it used to anyway).


analytic geometry
graphing lines
slope
geometry

lets say you find the slope to be t. then you sub it in the general to get
y=tx+b
now you sub in the x and y values using one of the points. thus you will find b.

[quote]Find the equation of each line.
how do i do this if i know 2 of the points
such as (-5,7) and (1,-3)
??[/quote]
I shall explain this in your scenario. We have the two points specified. we do (-3-7)/(1-(-5)) because of the slope formula specified earlier in this thread. Our slope is -10/6, or -5/3. Write out the equation y=(-5/3)x+b. Now substitute (1,-3) for x and y and solve for b. We have our equation.  :D

[quote="linzk3"]Find the equation of each line.
how do i do this if i know 2 of the points
such as (-5,7) and (1,-3)
??[/quote]

[hide]find slope
-3-7/1--5=-10/6=-5/3
then use pointslope y-y1=slope(x-x1)[/hide]

just to clear any confusion

(x1, y1)
(x2, y2)

those are the corresponding variables to your questions. either points can be used...the slope will be the same.

[quote="linzk3"]Find the equation of each line.
how do i do this if i know 2 of the points
such as (-5,7) and (1,-3)
??[/quote]

[hide]the way to find the slope is, (y2-y1) divides by (x2-x1). in these points, (-5,7) and (1,-3), you take -3-7 and divide that by 1-(-5). that is -10 divided by 6. that is -1.6 repeating. that is your slope. then u put that into the standard form equation, y=mx+b.
first, u take a coordinate, let's use (1,-3). so,
-3=-1.6(1)+b
-3=-1.6+b
b= -1.4
so, u put it all together, and ur equation is, y=-1.6x-1.4..i think that should be rihgt. [/hide]

[quote="linzk3"]Find the equation of each line.
how do i do this if i know 2 of the points
such as (-5,7) and (1,-3)
??[/quote]
In geometry there is a statement, "Any two points determine a line."

You can see this is true if you plot the points, and then connect them using a straight edge like a ruler.

Any line can be described by the equation $y = mx+b.$

The slope $m =\frac{\text{rise}}{\text{run}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}.$

It doesn't matter which point you pick to be $\left(x_{1},y_{1}\right)$ and which you pick to be $\left(x_{2},y_{2}\right)$, but you do have to be consistent. Once you've made the choice, stick with that.

For example, $m = \frac{7-(-3)}{-5-1}= \frac{10}{-6}=-\frac{5}{3}.$

You'll notice that since $y = mx+b$ that $b = y-mx.$

You've already computed the slope, so that $b = y-\left(-\frac{5}{3}\right)x.$

Now just plug in one of the points such as: $b = 7-\left(-\frac{5}{3}\right)(-5) = 7-\frac{25}{3}=-\frac{4}{3}.$

And you're done:

$\boxed{y =-\frac{5}{3}x-\frac{4}{3}.}$


calculus
integration
real analysis
real analysis solved

For $\varepsilon>0$ and $k>0$ denote by $A(k,\varepsilon)$, the set of $x\in\mathbb R$ such that $\left|x-\frac{p}{q}\right|\geq\frac{1}{k|q|^{2+\varepsilon}}$ for any integers $p$ and $q$, with $q\neq0$. Prove that $\mathbb R\setminus \bigcup_{k=1}^\infty A(k,\varepsilon)$ has Lebesgue measure zero.

Enjoy it. I really did!

It is a famous problem. The following problem was an exercise in my real analysis book.

Let $E=\left\{x\in\mathbb{R}: \mbox{there exists infinitely many } \frac{p}{q}\mbox{ with }(p,q)=1\mbox{ such that }\left|x-\frac{p}{q}\right| < \frac{1}{|q|^{2+\epsilon}}\right\}$ . Show that $m(E)=0$.

Notice that $E=\mathbb{R}\setminus \bigcup_k A(k,\epsilon)$.
We need only to show that $m(E_1)=0$, where $E_1=E\cap (0,1]$.
Denote $E_{p,q}$ by the reals in $E$ for fixed $p$ and $q$.
Let $E_q=\bigcup_{p=1}^q E_{p,q}$. Now it is sufficient to show that $\sum_q m(E_q)<\infty$ then we have $m(\varlimsup E_q)=0$ and hence $mE=0$ because $E\subset \varlimsup_{q\to\infty} E_q$.

we have $m(E_{p,q})=\frac{2}{q^{2+\epsilon}}$ because $\frac{p}{q}-\frac{1}{q^{2+\epsilon}}<x<\frac{p}{q}+\frac{1}{q^{2+\epsilon}}$ for $x\in E_{p,q}$. Hence $m(E_q)\leq \frac{2}{q^{1+\epsilon}}$, hence $\sum_q m(E_q)\leq \sum_q \frac{2}{q^{1+\epsilon}}<\infty$.


What is the least possible whole number value of $ n$ such that $ (\frac12)^n < 0.01$?

$ (\frac{1}{2})^n<\frac{1}{100}$

We need to find the minimum whole $ n$ such that

$ 2^n>100$

Note that $ 2^6\equal{}64$ and $ 2^7\equal{}128$.

Minimum $ n$ is thus $ \boxed{7}$.


To take quizzes, each of 30 students in a class is paired with
another student. If the pairing is done randomly, what is the
probability that Margo is paired with her best friend, Irma?
Express your answer as a common fraction.

There are two ways for the two friends to be together - Margo-Irma or Irma-Margo. In total, there are $ 30 \times 29 \equal{} 870$. Thus, we have $ \frac{2}{870} \equal{} \frac{1}{435}$.

actually, there are 15 total groups.
$ \frac{1}{435}\cdot 15\equal{}\frac{1}{29}$
another approach is to first know that margo will be paired up with a mysterious 'somebody'
there are 29 kids that have potential to be the 'somebody' so it's simply $ \frac{1}{29}$


Let $a, b$ and $c$ be real numbers s.t: $\large{b^{2}-4ac\geq 0}.$
Prove that: $\large{a+b+c\leq \frac{9}{4}\max(a,b,c)}$.

[quote="N.T.TUAN"]IF $a+b+c>\frac{9}{4}\max(a,b,c)$ then $4b+4c>5a\\4c+4a>5b\\4a+4b>5c$ therefore $a,b,c>0$, Can you help me?  :D[/quote]
I have a solution, but a little long. I'm looking for a solution using the discriminant of the polynom $ax^{2}+bx+c$..

PS: I don't have much time to type the solution..

http://www.mathlinks.ro/Forum/viewtopic.php?t=64712 has some proofs.

  darij


[url]http://www.spoj.pl/problems/COVER/[/url]

Well the idea came in my mind after reading the problem description is:

Split each node. Make Weighted Bipartite Matching size of which is "[b]K[/b]".

So if we think about finding K flows (mincost) we need K bellmen ford. K*V*E. thats too much. Even a single bellman ford is very costly. $10^{4}* 10^{6}= 10^{10}$.

So what is the way?

I think that if this problem is less than linear in $K$, then we can use it to solve general bipartite matching faster than the best known solution, because at most it is linear in $E$ I think, then you can binary search on $K$ to solve bipartite matching in $O(E\log(K)f(K))$, where $f(K)$ is our solution's runtime in $K$, which is less than linear. But I am pretty sure that this is too fast for general bipartite matching, though I could be mistaken.

As a note, I think you can at any rate beat $O(EVF)$ with min-cost max-flow, since the graph is so simple.

hmm... u have point!

bu still KVE seems too much to me! i hope after submitting the code it will surprize me giving ac  :lol:

Well my point was that you have to work on getting rid of the $V$ and $E$ parts, not the $K$ part. Probably $O(KV)$ or something similar will work.


Write each expression in the standard form a + bi.

(1) 13/(5 - 12i)

(2) (2 - i)/-2i

[hide="hint"]

Multiply the numerator and the denomenator by the conjugate of the denomenator. The conjugate of $a+bi$ is $a-bi$

[/hide]

[hide="solution for 1"]

The conjugate of $5-12i$ is $5+12i$

$\left(\frac{13}{5-12i}\right)\left(\frac{5+12i}{5+12i}\right)$
$\frac{65-156i}{25+60i-60i-144i^{2}}$
$\frac{65-156i}{25+144}$
$\frac{65-156i}{169}$
$\frac{65}{169}-\frac{156}{169}i$
$\boxed{\frac{5}{13}-\frac{12}{13}i}$

[/hide]

[quote="life=tennis+math"]tjhance already covered #1 so i'll get

[hide="2"]$(\frac{2-i}{-2i})(\frac{-2i}{-2i}) = \frac{-2-4i}{-4}={\frac{1+2i}{2}}$[/hide][/quote]

actually the answer should be $\frac{1}{2}+i$ because the form must be $a+b(i)$

Why is it necessary to multiply by (-2i/-2i) in question 2?

by doing that, the denominator is eliminated from the presence of i. then you can simplify to a+bi form.


How many possible Sudoku grids are there?

See [url=http://en.wikipedia.org/wiki/Mathematics_of_Sudoku]Mathematics of Sudoku[/url].


If you know [b]what is remainder [/b] you must know how to proceed.

The question is too simple for this section.

[quote="bgbgbgbg"]how can we know:
if the remainder of  f(x)/(x-4) is 3 and the remainder of  f(x)/(x-3) is 4 what is the remainder of f(x)/[(x-3)(x-4)][/quote]

We are given that

$ f(x)\equal{}Q_1(x)(x\minus{}4)\plus{}3$

$ f(x)\equal{}Q_2(x)(x\minus{}3)\plus{}4$,

where $ Q_1(x),Q_2(x)$ are the quotients when f(x) is divided by x-4 and x-3, respectively.

Since these are identities, we can plug in $ x\equal{}4$ into the first equation to get $ f(4)\equal{}3$ and $ x\equal{}3$ into the second equation to get $ f(3)\equal{}4$.

We know that $ Q_1,Q_2$ are linear polynomials, since the remainders are constant, making $ f(x)$ a quadratic.

Thus, letting $ f(x)\equal{}ax^2\plus{}bx\plus{}c$, we have:

$ 16a\plus{}4b\plus{}c\equal{}3$

$ 9a\plus{}3b\plus{}c\equal{}4$

or

$ 7a\plus{}b\equal{}\minus{}1$.

It doesn't matter much exactly which quadratic f(x) is, so we choose $ a\equal{}1, b\equal{}\minus{}8$, making $ c\equal{}19$.

Our quadratic is

$ x^2\minus{}8x\plus{}19$ (Checking, f(3)=4 and f(4)=3), and we want the remainder when it is divided by $ x^2\minus{}7x\plus{}12$.

Our answer is $ (x^2\minus{}8x\plus{}19)\minus{}(x^2\minus{}7x\plus{}12)\equal{}\minus{}x\plus{}7$.

Hope this helped.

Its true that the remainder is $ \minus{}x\plus{}7$ but  the proof is false here
[quote="BOGTRO"]
We know that $ Q_1,Q_2$ are linear polynomials [/quote]
The common degree of these polynomials may be any non-negative integer as large as you like.


geometry
3D geometry
group theory
abstract algebra
superior algebra
superior algebra solved

K is a finite field s.t. there is a $u\in K^*$ s.t. $u^3=2$ (2=1+1). Show that every element from K can be written as the sum of 2 cubes of elements in K.

Here are some of my ideas about this problem. I hope my efforts will convince you to tackle this problem.

Let p be the characteristic of K. Thus p is a prime. Let $C=(K^*)^3$ be the set of nonzero cubes in K.

Since K is finite, $\exists \ n\in\mathbb{N^*}$ such that $|K|=p^n$.

If $3\nmid (|K|-1)$ then in the group K* there is no element of order 3, and so the morphism $\varphi :K*\rightarrow K*$ defined by $\varphi (x)=x^3$ is injective because it has trivial ker and bijective because K* is finite. So all elements of K are cubes.

If 3|(|K|-1) then $\varphi (K^*)=C$ and |K*/C|=3 because in K* there is an element of order 3 by Cauchy.

Let $C'=C\cup \{0 \}$. Assume C'+C'=C'. Then it is easy to see that C' is a subfield of K and thus has pm elements for some $m\in\mathbb{N}$. So $p^m=\frac{p^n-1}3+1$. This yields the solutions m=2,n=1 and p=2. It is easy to see that the 4elements field verifies the conditions of the problem.

I'll be back.

Hi, I'm back.
If |K|>4 then C' is not a field, so there is a nonzero element in C+C (i.e. a sum of 2 cubes, but not a cube). Let this element be $x=a^3+b^3$. Then, because C is a subgroup of K*, $xC\cap C=\phi$. It is easy to see that the elements of xC are sums of 2 cubes. So the elements $\{ 0 \} \cup C\cup xC$ are cubes or sums of two cubes.

I don't know what to do about the rest.

Victory!!!

I will give here the complete proof, so you can disregard my previous posts.

$K$ is finite, so by Wederburn, commutative. Let $p=charK$, the prime number representing the characteristic of K.
$u^3=2$ and $u\in K^*$ imply $p\neq 2$.
Let $C=\{x^3|x\in K^*\}$ be the set of nonzero cubes of K and let $C'=C\cup {0}$.

If $3 \nmid |K^*|$ then $K^*$ has no element of order 3, so the multiplicative morphism $\phi:K^*\to K^*\ \phi(x)=x^3\ \forall x\in K^*$ is injective. Since $K$ is finite, $\phi$ is bijective, so all elements of $K^*$ are cubes, so all elements of $K$ are cubes and we are done.

If $3||K^*|$ then let $C=Im\phi$ and let $C'=C\cup\{0\}$.$(K^*,\cdot)$ has an element of order 3, call it $\xi$ and $\ker\phi=<\xi>$, so $K^*/<\xi>\simeq Im\phi=C$. So $\displaystyle |C'|=\frac{|K|-1}{3}+1$. Assume $C'+C'=C'$. Then $C'$ is a subfield of $K$. Let $|K|=p^n$ and $|C'|=p^m$. Then $\frac{p^n-1}{3}+1=p^m \Leftrightarrow p^n+2=3p^m \Rightarrow p|2$. So $p=2$ and we get a contradiction.

So $C'+C'\varsupsetneq C'$ so there exists $p,q\in K$ such that $p^3+q^3$ is a sum of two cubes, but not a cube. Let $a=p\cdot q^{-1}$. Then $x=1+a^3$ is a sum of two cubes but not a cube. In particular, this implies $x\in K^*$. $K^*/C$ is a group with 3 elements. $\hat x\neq \hat 1$ because $x$ is not a cube. This means that $K^*/C=\{\hat 1,\hat x,\hat x^2\}$. It follows that $K^*=C\sqcup xC\sqcup x^2C$. ($\sqcup$ means disjoint union)

If $1-a^3=0$ then $a^3=1$ and $x=2$. But by the hypothesis, 2 is a cube and $x$ is not so we get a contradiction.

If $1-a^3\in C$ then $x^2=1+2a^3+a^6=(1-a^3)^2+4a^3$. Because $1-a^3\in C$, $1-a^3$ is a cube. Also 4 is the square of a cube, so it is a cube. This means that $x^2$ is the sum of two cubes. 0 is a cube, the elements of $C$ are cubes and the elements of $xC$ and $x^2C$ are sums of two cubes and we are done.

If $1-a^3\in xC$ then $1-a^6=(1-a^3)(1+a^3)=x(1-a^3)\in xxC=x^2C$. Note that this implies that the elements of $x^2C$ are sums of two squares and, as before, we are done.

If $1-a^3\in x^2C$ then the elements of $x^2C$ are sums of two squares and we are done.

I think this is too much for district olympiad. It does not have lots of ideas in it but it takes a lot to link all the steps.


If Bill gives Nancy $ \$10$, they have the same amount of money. When Nancy gives Bill $ \$10$ instead, Bill has twice as much as Nancy. How many dollars do they have together?

Set up a system of equations.

$ b\minus{}10\equal{}n\plus{}10$

$ 2(n\minus{}10)\equal{}b\plus{}10$

Distribute in the second equation.

$ 2n\minus{}20\equal{}b\plus{}10$

$ 2n\minus{}30\equal{}b$

Substitute into the first.

$ (2n\minus{}30)\minus{}10\equal{}n\plus{}10$

$ 2n\minus{}40\equal{}n\plus{}10$

$ 2n\equal{}n\plus{}50$

$ n\equal{}50$

Thus, $ b \equal{} 70$.

The answer is $ 70\plus{}50$, or $ \boxed{120}$.


how would you find all values of $(x,y)$ that satisfyf the following system?

$x+y+\frac{x}{y}=10$
and
$\frac{x(x+y)}{y}=20$

any help would be grealy apreciated

[hide="hint"] try first to find $x+y$ and $\frac xy$, then its easy to finish .. [/hide]

doesnt anyone want to finish this: $x=?$ and $y=?$

its easy from here ...

[quote="dondigo"]Let
$x+y=a$
$\frac{x}{y}=b$
$a+b=10$
$ab=20$
$a$ and $b$ are the roots of the equation:
$x^2-10x+20=0$
$a=5-\sqrt{5}$
$b=5+\sqrt{5}$
so we have to systems of equations:
$x+y=5-\sqrt{5}$
$\frac{x}{y}=5+\sqrt{5}$
or
$x+y=5+\sqrt{5}$
$\frac{x}{y}=5-\sqrt{5}$[/quote]
lol I'll finish it.
$5+\sqrt{5} = (6-\sqrt{5})y \rightarrow y = \frac{5+\sqrt{5}}{6-\sqrt{5}} = \frac{35+11\sqrt{5}}{31} \rightarrow x = \frac{120+20\sqrt{5}}{31}$
or
$5-\sqrt{5} = (6+\sqrt{5})y \rightarrow y = \frac{5-\sqrt{5}}{6+\sqrt{5}} = \frac{35-11\sqrt{5}}{31} \rightarrow x = \frac{120-20\sqrt{5}}{31}$


algebra
function
domain
superior algebra
superior algebra solved

Does there exist a commutative ring $R$ with unity, and $x,y \in R$ such that $x|y$, $y|x$ and $x$ and $y$ are not associates?

Yes that's possible and I've posted it before, but it's hard to find.
Take any field $K$ and look at $x=x(1+y)(1-y)$ and $x(1+y)$ in $K[x,y]/(xy^2)$.

never thought of that, but I think it is because intuitively I think of rings with cancellation law  :oops:

I am correct that in the commutative case of a domain such example cannot be found?

[quote="fredbel6"]I am correct that in the commutative case of a domain such example cannot be found?[/quote]

Yes


How many posts are required to support a straight fence that is 100 yards long if a post is placed every 10  yards?

(A) 9 (B) 10 (C) 11 (D) 15 (E) 20

There will have to be $ \frac{100}{10}\equal{}10$ spaces between posts, so $ \boxed{11}$ posts will be needed to do this. The correct answer is $ \boxed{\text{C}}$.


Let $ a,b,c>0$. Prove that:

$ \dfrac{a^2}{b^2}\plus{}\dfrac{b^2}{c^2}\plus{}\dfrac{c^2}{a^2}\plus{}4(\dfrac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2})^2\ge 7$

[quote="Inequalities Master"]Let $ a,b,c > 0$. Prove that:

$ \dfrac{a^2}{b^2} \plus{} \dfrac{b^2}{c^2} \plus{} \dfrac{c^2}{a^2} \plus{} 4(\dfrac{ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2})^2\ge 7$[/quote]
Hint:
\[ \sum \frac{a^2}{b^2} \ge \frac{3}{2} \sum \frac{b\plus{}c}{a} \minus{}6\]


Let be $ a,b,c>0$ such that $ ab \plus{} bc\plus{}ca \equal{} 3$ . Prove that :

$ \frac {abc}{a \plus{} b \plus{} c} \plus{} \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b}\ge \frac {11}{6}$

Using CS  and Schur ineq, we can prove the stronger:
$ \sum\ \frac{a}{b\plus{}c}\plus{} \frac{9abc}{2(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)} \geq\ 2$
 :)

[quote="nguoivn"]Using CS  and Schur ineq, we can prove the stronger:
$ \sum\ \frac {a}{b \plus{} c} \plus{} \frac {9abc}{2(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \geq\ 2$
 :)[/quote]

$ \sum\ \frac {a}{b \plus{} c} \plus{} \frac {9abc}{2(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \geq \sum\ \frac {a}{b \plus{} c} \plus{}\frac {4abc}{(a \plus{} b)(b\plus{}c)(c\plus{}a)} \geq 2$


let $a_i,b_i \in Z$ with $i$ ranging from $1$ to $n$
can $\prod (a_i ^2+ b_i ^2)$ be written as $A^2+B^2$ with $A,B \in Z$ ?


algebra polynomial quadratics trigonometry AMC

Solve x^4 + x^3 + x^2 + x + 1 = 0.

(from Zeitz, if Im not mistaken)

O man, this is when i'm really glad i paid attention to Joel..

Anwway,

solution:

[hide]divide through by x^2, then regroup the terms as (x^2+1/x^2) + (x+1/x) + 1=0.

Now, let x+1/x=y. Thus, (x^2+1/x^2)=y^2-2.

Now, you get a quadratic in y, solve for y, then get two quadratics in x, and u get all 4 solutions[/hide]

Or if you want another way of doing it:

[hide]Multiply both sides by (x-1).[/hide]

That's not originally from Zeitz, but it was in his book. If you use darkquantum's method combined with TripleM's method, you can compute sine and cosine of multiples of 72. Isn't that neat?

It's not my method, its' Joel's.

And computing multiples of sine and cosine of 72...

is that by any chance related to Demoivre's on the 5th roots 1?

Cuz the arguments in that case are multiples of 72....

That's right. The fifth roots of 1 are cis(72), cis(144), cis(216), cis(288), and 1.

But that would work for say,
x^5+x^4+x^3+x^2+x+1. In that case, you can compute the sine and cosine of multiples of 60...which we already know....hmmm


Let $ a, b, c$  positives numbers such  $ a \geq b$ $ \geq c$
Prove that  $ a^2\frac{b}{c} \plus{} b^2\frac{c}{a} \plus{} c^2\frac{a}{b} \geq a^2\plus{} b^2\plus{} c^2$

Let $ S_1\equal{}a^{2}\frac{b}{c}\plus{}b^{2}\frac{c}{a}\plus{}c^{2}\frac{a}{b}$ and $ S_2\equal{}a^{2}\frac{c}{b}\plus{}b^{2}\frac{a}{c}\plus{}c^{2}\frac{b}{a}$
then $ S_1 \plus{} S_2 \geq 2(a^2\plus{}b^2\plus{}c^2)$
since $ S_1 \geq S_2$
we deduce the inequality.


real analysis
limit
function
integration
calculus
real analysis unsolved

This result is just a consequence of generalized dominated convergence theorem.
Fact: $ |f_n\minus{}f|^p\le 2^p( |f_n|^p\plus{}|f|^p)$

[quote="limsup"]This result is just a consequence of generalized dominated convergence theorem.
Fact: $ |f_n \minus{} f|^p\le 2^p( |f_n|^p \plus{} |f|^p)$[/quote]

Thanks. But the right hand side depends on $ n$, although it belongs to $ L^1$. How can the theorem work?

[quote="zhoraster"]No, it is not true. Take $ f_n \equal{} 1_{[n,n \plus{} 1)}$.[/quote]
$ f_n\to 0$ a.e. But see whether it satisfies the other hypothesis :)

Damn, messed up, you are right.

You can apply Fatou lemma to $ 2^p(|f_n|^p\plus{}|f|^p) \minus{} |f_n\minus{}f|^p$, which is positive, as [b]limsup[/b] wrote (accidentally, the Fatou lemma is about $ \liminf$ ;) ).

[quote="man.maiko"] But the right hand side depends on $ n$, although it belongs to $ L^1$. How can the theorem work?[/quote]
Look for [b]Generalised DCT[/b] not just DCT.

Thank  [b]zhoraster[/b] and [b]limsup[/b] much!

$ |f_n \minus{} f|^p\le 2^p( |f_n|^p \plus{} |f|^p)\le \lim\inf 2^p( |f_n|^p \plus{} |f|^p)\equal{}2^{p\plus{}1}|f|^p$ in regard with Fatou's lemma. Now the theorem works.

[quote="zhoraster"][quote="man.maiko"]
$ |f_n \minus{} f|^p\le 2^p( |f_n|^p \plus{} |f|^p)\le \lim\inf 2^p( |f_n|^p \plus{} |f|^p) \equal{} 2^{p \plus{} 1}|f|^p$ in regard with Fatou's lemma. Now the theorem works.[/quote]
Honestly, I don't understand what you mean. Hope you understand it better.[/quote]

I meant with $ |f_n|^p\to |f|^p$ a.e. and $ |f_n \minus{} f|^p\le 2^{p \plus{} 1}|f|^p\in L^1$ the dominated convergence theorem works, which means $ \int |f_n \minus{} f|^p\to 0.$ Then $ f_n\to f$ in $ L^p.$ Ok?

[quote="zhoraster"][quote="man.maiko"]
$ |f_n \minus{} f|^p\le 2^p( |f_n|^p \plus{} |f|^p)\le \lim\inf 2^p( |f_n|^p \plus{} |f|^p) \equal{} 2^{p \plus{} 1}|f|^p$ in regard with Fatou's lemma. Now the theorem works.[/quote]
Honestly, I don't understand what you mean. Hope you understand it better.[/quote]

Oh, I was wrong.

$ |f_n \minus{} f|^p\le 2^p( |f_n|^p \plus{} |f|^p)\le \lim\sup  2^p( |f_n|^p \plus{} |f|^p) \equal{} 2^{p \plus{} 1}|f|^p$. Now the dominated convergence theorem works.

[quote]$ 2^{p}( |f_{n}|^{p} \plus{} |f|^{p})\le\lim\sup 2^{p}( |f_{n}|^{p} \plus{} |f|^{p})$[/quote]
Unfortunately, it's not true (if it were, it would work for my "counterexample" as well).

By Fatou
\[ 2^{p \plus{} 1}\|f\|^p_p \equal{} \int_{\mathbb R}\liminf\big[(2^p(|f(x)|^p \plus{} |f_n(x)|^p) \minus{} |f(x) \minus{} f_n(x)|^p\big]dx\le \liminf\big[ 2^p(\|f\|_p^p \plus{} \|f_n\|_p^p) \minus{} \|f \minus{} f_n\|_p^p\big] \equal{} 2^{p \plus{} 1}\|f\|^p_p \minus{} \limsup\|f \minus{} f_n\|_p^p,\]
which yields $ \limsup\|f \minus{} f_n\|_p^p\le 0$.

[quote="man.maiko"][quote="CommandoGuard"]I believe that the theorem [b]limsup[/b] is referring to is the following:

Let $ f_1,f_2,...$ be a sequence of measurable functions which converges pointwise to a function $ f$. Assume that $ |f_k (x)| \le g_k (x)$ for all $ x$ and some sequence of functions $ g_k \in L^1$. If $ g_k \to g$ (pointwise) for $ g \in L^1$ and $ \int g_k \to \int g$ then $ f \in L^1$ and $ \int f_k \to \int f$. Obviously, all hypotheses related to the sequences $ f_k$ and $ g_k$ can be relaxed to hold almost everywhere.[/quote]

Please tell me where I can see this. I think [b]limsup[/b] reminded of [b]Vitali convergence theorem[/b]?[/quote]

It appears as an exercise (with a hint) in Jones' book on Lebesgue integration in $ \mathbb{R}^n$. The proof is basically [b]zhoraster[/b]'s argument, it's just a generalization. I don't see the relevance of Vitali's theorem here.

[quote="man.maiko"][I think [b]limsup[/b] reminded of [b]Vitali convergence theorem[/b]?[/quote]
No. I do not refer to Vitali convergence theorem.
What CommandoGuard says is right. [b]The Generalised DCT[/b] is exactly the following.
[quote="CommandoGuard"]I believe that the theorem [b]limsup[/b] is referring to is the following:

Let $ f_1,f_2,...$ be a sequence of measurable functions which converges pointwise to a function $ f$. Assume that $ |f_k (x)| \le g_k (x)$ for all $ x$ and some sequence of functions $ g_k \in L^1$. If $ g_k \to g$ (pointwise) for $ g \in L^1$ and $ \int g_k \to \int g$ then $ f \in L^1$ and $ \int f_k \to \int f$. Obviously, all hypotheses related to the sequences $ f_k$ and $ g_k$ can be relaxed to hold almost everywhere.[/quote]

Let me give the complete solution. It's just a simple application of the above theorem!
I think you agree that $ |f_{n}\minus{}f|^{p}\le 2^{p}( |f_{n}|^{p}\plus{}|f|^{p})$(if not I can give you a proof)
My $ g_n$ is $ 2^{p}( |f_{n}|^{p}\plus{}|f|^{p})$. Then $ g_n\to 2^{p\plus{}1}|f|^{p}$ a.e. and $ \int g_n\to 2^{p\plus{}1}\int |f|^{p}$      (by the given hypothesis)
Hence by the above theorem, $ \int|f_n\minus{}f|^p\to 0$

Thank [b]limsup[/b]! Your solution is nice.


How many zeros are there at the end of 100!?

General formula is:
[hide]$\sum_{i=1}^\infty \left\lfloor\frac{n}{5^{i}}\right\rfloor$[/hide]


Given $ n$ real numbers $ x_1,x_2,\cdots,x_n$, find the smallest $ m$ and real numbers $ y_1,\cdots,y_m$ such that each $ x_i$ can be represented as the sum of some $ y_j$.

[quote="Albanian Eagle"]by sum of some $ y_j$ do you mean distinct $ y_j$'s? or can one use the same $ y$ more than once?[/quote]

$ y_1,\cdots,y_m$ are fixed, but some of them can be the same. $ x_1,\cdots,x_n$ are distinct real numbers.

for example, $ x_i\equal{}i$, and let $ m\equal{}n$, then both $ y_i\equal{}i$ or $ y_i\equal{}1$ are solutions.

Well, if the $ x_i$'s are linearly independent over $ \mathbb{Q}$ (i.e. belong to a Hamel basis), then one needs $ m \geq n$, else the $ n$-dimensional subspace spanned by the $ x_i$'s would be included in the at most $ m$-dimensional subspace spanned by the $ y_j$'s.
But for $ m \equal{} n$ one can always take the $ y$'s exactly the $ x$'s. Therefore, in general, one can say nothing more than $ m \equal{} n$.


Does anybody know the winning strategy of the Super Nim game?

http://www.gameintellect.com/java-games/super-nim/#menu

For those who don't like clicking through: we have an $ n \times n$ rectangular board and some random subset of points are marked.  A "move" consists of erasing all the marks in an entire column or row, and the goal is to be the person to erase the last mark.

What do you mean by random subset of points ? Like are the vertices of unit squares the points that are marked or are the points the centres of the squares? Moreover, there is randomness related to selection of the vertixes in case it is the vertices of the squares{unit} because it depends on the number of unit squares chosen.
Someone please clarify.

For the purposes of a game, "random" means "given."  Whether the vertices or centers are marked is irrelevant; this just changes the effective value of $ n$.

As for the original question, the [url=http://en.wikipedia.org/wiki/Sprague%E2%80%93Grundy_theorem]Sprague-Grundy theorem[/url] says that any such game is equivalent to a nimber.


Rid-a-Spot comes in a 1-quart container. There are 4 cups in
1 quart. To wash windows, 2 cups of Rid-a-Spot are to be mixed
with 5 cups of water. How many quarts of this mixture can be made
from the 1-quart container of Rid-a-Spot? Express your answer as a
decimal to the nearest tenth.

For every two cups of rid-a-spot (RAS), 5 cups are needed. So in a quart there are 2 2 cups of RAS. From this, $ \frac{2(2\plus{}5)}{4}\equal{}\boxed{3.5}$


Could someone give me some relations for :

  z0 and  zH  ( in function of zA,zB,zC)


Ring Theory
linear algebra
matrix
number theory
least common multiple
search
group theory

Let $ A$ be a unitary finite ring with $ n$ elements, such that the equation $ x^n\equal{}1$ has a unique solution in $ A$, $ x\equal{}1$. Prove that

a) $ 0$ is the only nilpotent element of $ A$;

b) there exists an integer $ k\geq 2$, such that the equation $ x^k\equal{}x$ has $ n$ solutions in $ A$.

a) Let $ x,k$ be such that $ x^k\neq 0$ and $ x^{k\plus{}1}\equal{}0$. Then $ (1\plus{}x^k)^n \equal{} 1 \plus{}nx^k \equal{} 1$ because the ring has $ n$ elements, so $ 1\plus{}x^k\equal{}1$, so $ x^k\equal{}0$, contradiction, unless $ x\equal{}0$.

b) Consider the $ n^n \plus{} 1$ $ n$-tuplets: $ (a_1^k,\ldots, a_n^k)$, $ 1\leq k\leq n^n \plus{} 1$, where $ A \equal{} \{a_1,a_2,\ldots,a_n\}$. Then obviously, since there are only $ n^n$ distinct such $ n$-tuplets, two of them must be equal. Say the $ i$th and the $ j$th tuplets are equal, $ i > j$. Choose $ i,j$ such that $ j$ is minimal. Then $ (x^{i \minus{} 1} \minus{} x^{j \minus{} 1})^2 \equal{} x^{2i \minus{} 2} \plus{} x^{2j \minus{} 2} \minus{} 2x^{i \plus{} j \minus{} 2} \equal{} 2x^{i \plus{} j \minus{} 2} \minus{} 2x^{i \plus{} j \minus{} 2} \equal{} 0$, therefore $ x^{i \minus{} 1} \minus{} x^{j \minus{} 1} \equal{} 0$, which is a contradiction with the minimality of $ j$, unless $ j \equal{} 1$, in which case $ x^i \equal{} x$ for all $ x$.

In particular we only need that $ x^2 \equal{} 0 \Rightarrow x \equal{} 0$ from part a) for part b).

a) is similarly easy.  Let $ x$, $ k$ be such that $ x^k \neq 0$, $ x^{k\plus{}1}\equal{}0$, then $ (1\plus{}x^k)^n\equal{}1\plus{}nx^k\equal{}1$.

[quote="LydianRain"]a) is similarly easy.  Let $ x$, $ k$ be such that $ x^k \neq 0$, $ x^{k \plus{} 1} \equal{} 0$, then $ (1 \plus{} x^k)^n \equal{} 1 \plus{} nx^k \equal{} 1$.[/quote]Yeah, I figured that out after that, but I just got to post it now :D

[quote="Valentin Vornicu"]$ 1 \plus{} nx^k \equal{} 1$[/quote]
I think this should be explained.
If $ m\in\mathbb{N}$ is the characteristic of $ A$, i.e. minimal with $ m\cdot 1_A\equal{}0$,
then $ A$ has a unitary subring $ \cong \mathbb{Z}/m\mathbb{Z}$,
and using Lagrange on the additive groups we get $ m\equal{}|\mathbb{Z}/m\mathbb{Z}|$ divides $ n\equal{}|A|$.
And then $ nx\equal{}{n\over m}\cdot (m\cdot 1_A)\cdot x\equal{}0$ for all $ x\in A$.

[quote="olorin"][quote="Valentin Vornicu"]$ 1 \plus{} nx^k \equal{} 1$[/quote]
I think this should be explained.
If $ m\in\mathbb{N}$ is the characteristic of $ A$, i.e. minimal with $ m\cdot 1_A \equal{} 0$,
then $ A$ has a unitary subring $ \cong \mathbb{Z}/m\mathbb{Z}$,
and using Lagrange on the additive groups we get $ m \equal{} |\mathbb{Z}/m\mathbb{Z}|$ divides $ n \equal{} |A|$.
And then $ nx \equal{} {n\over m}\cdot (m\cdot 1_A)\cdot x \equal{} 0$ for all $ x\in A$.[/quote]The ring has $ n$ elements. It is obvious that its characteristic divides $ n$. In other words $ nx\equal{}0$ for all $ x \in A$.

If $ A$ has no nilpotent elements, then it is [url=http://en.wikipedia.org/wiki/Artinian_ring]artinian[/url] with [url=http://en.wikipedia.org/wiki/Jacobson_radical]Jacobson Radical[/url] $ J(A)=0$, and therefore [url=http://en.wikipedia.org/wiki/Semisimple]semisimple[/url].
Then [url=http://en.wikipedia.org/wiki/Artin-Wedderburn_theorem]Artin–Wedderburn[/url] and [url=http://en.wikipedia.org/wiki/Wedderburn_theorem]Wedderburn[/url] gives $ A$ a direct product of matrix algebras $ \mathcal{M}_l(\mathbb{F}_q),l\ge 1$ over some [url=http://en.wikipedia.org/wiki/Finite_field]finite fields[/url] $ \mathbb{F}_q$.
But this contains nilpotent elements for $ l>1$,
so $ A$ must be the direct product of some fields $ \mathbb{F}_{q_1},\ldots ,\mathbb{F}_{q_h}$ with some prime powers $ q_1,\ldots,q_h$.
This gives $ A$ commutative and $ n=q_1\cdots q_h$.
And then $ \mathbb{F}_{q_i}^\times$ cyclic for all $ i$ gives
$ x^k=x$ for all $ x\in A\Leftrightarrow k=1+r\cdot \text{lcm} \{q_1-1,\ldots ,q_h-1\}$ for some $ r\in\mathbb{N}$.

[quote="Valentin Vornicu"]The ring has $ n$ elements. It is obvious that its characteristic divides $ n$. In other words $ nx = 0$ for all $ x \in A$.[/quote]
That hurt.
It's obvious to you maybe, who already knew it, but it was not obvious to me (though easy),
and it's not obvious to those who never heard of the characteristic of a general ring.

[quote="olorin"][quote="Valentin Vornicu"]The ring has $ n$ elements. It is obvious that its characteristic divides $ n$. In other words $ nx \equal{} 0$ for all $ x \in A$.[/quote]
That hurt.
It's obvious to you maybe, who already knew it, but it was not obvious to me (though easy),
and it's not obvious to those who never heard of the characteristic of a general ring.[/quote]It is obvious for any student in Romania in 12th grade (that has anything to do with maths) :D (I'm saying it's in standard classroom textbook).

[quote="Valentin Vornicu"]It is obvious for any student in Romania in 12th grade (that has anything to do with maths) :D (I'm saying it's in standard classroom textbook).[/quote]
As I said, it's obvious to those, who read a book, where it is written in. :lol:
But the rest of us never heard of the characteristic of a ring (only of fields) unless in some little exercise like this.

[quote="olorin"]But the rest of us never heard of the characteristic of a ring (only of fields) unless in some little exercise like this.[/quote]Actually it's a simple thing related to commutative group theory which you find in any text book that talks about groups: $ x^{|G|} \equal{}e$, where $ G$ is the commutative group, $ x$ is random element, and $ e$ is the neutral element.

[quote="harazi"]Here's another solution. http://www.mathlinks.ro/viewtopic.php?search_id=45577503&t=46392[/quote]
This works only for $ A$ commutative.
Otherwise if $ M$ is a maximal ideal of $ A$, then $ A/M$ is a simple ring,
and therefore $ \cong \mathcal{M}_l(D)$ for some $ l\ge 1$ and some division ring $ D$ (i.e. not a field for $ l>1$)
and $ 1\minus{}xr$ not invertible doesn't mean it's in some maximal ideal.

Indeed, I hurried when  I made this assertion. However, it's not that bad since that problem shows what are all rings of the problem. And a solution to the mentioned problem is based on three facts: any finite ring is a product of rings without nontrivial central idempotents,  in any finite ring without nontrivial nilpotents any idempotent is central and finally any finite ring with no nontrivial nilpotents or idempotents is a field. Only the third assertion is not a formal computation, but Wedderburn's theorem.


trigonometry
number theory unsolved
number theory

Hello,

While working on my websites on 666 myth ( http://www.666myth.co.nr/ ) and on Isomorphous Triplets ( http://www.chez.com/cosmos2000/Numbers/IsomorphousTriplets.html ) I have found recently [b]direct links between the Beast Number 666 and the famous Golden Number Phi[/b], well-known by Pythagoras, Leonardo da Vinci ... and in Sacred Geometry. This unexpected and incredible relation 666 versus Φ may constitute a way for the [b]Rehabilitation of 666 !!! ...[/b]

First we have this curious relations:

[b]666 = 7^3 pi (Phi – 1) = 7^3 pi phi [/b]    or         [b]666 = (6/5) 7^3 Phi[/b]
with   [b] pi = 3.141593       Phi = 1.618034       phi = Phi -1 = 0.618034    7^3 = cubic of 7[/b]

Besides, after computing some sinus and cosines [a good online Trigonometry calculator available at http://www.1728.com/trigcalc.htm ], expressed in Degrees and absolute values, we obtain:

[b]Phi /2 = sin 666º[/b] = cos 324º = cos 216º = cos 144º = cos 36º = 0.80901699...
[b]Phi = 2sin 666º[/b] = 2cos 324º = 2cos 216º = 2cos 144º = 2cos 36º = 1,61803399...
[b]Phi = sin 666º + cos 216º [/b]= sin 666º + cos (6x6x6)º
[b]Phi = cos 144º + cos 36º[/b] = cos [(6+6) x (6+6)]º + cos (6 x 6)º
[b]Phi = sin 666º + cos 144º[/b] = sin 666º + cos [(6+6) x (6+6)]º

But, I am wondering if there could exist formal mathematical proofs for this 666 and Golden Number links.

Thanks for advice and help.

I don't think the first one containung $7^3$ and $\pi$ are exact, so there is no proof.

But the ones with $sin,cos$ are exact and can be proven by the addition theorems of these functions.

Well, they are NOT OK !!!
you just use some aproximations and surely it is $665.98... \neq 666$
You have to 'round' the expression to get an integer value here!
But the proof when rounding is obvious: get $\pi , \phi$ exact  enough (as you did) and calculate it


calculus
integration
limit
function
real analysis
ratio
geometric series

Find the value of

$\lim_{n\to\infty}\int^{1}_{0}\frac{ny^{n-1}}{1+y}\,dy$

[quote="Ottem"][hide="Answer"]1[/hide][/quote]
I don't think so...
$\int^{1}_{0}\frac{ny^{n-1}}{1+y}\,dy = n \sum_{k=0}^\infty \frac{(-1)^{k}}{n+k}$
Using properties of the digamma function, the limit is $\frac{1}{2}$.

Can you plz explain kalle. How did you get it?

Sorry, if it is very elementary.

Yes, Kalle you are right, I just mixed up the alternating summation for a minute there.. \[\lim\int_{0}^{1}\frac{ny^{(n-1)}}{1+y}= \lim \sum_{i=0}^{n}\frac{(-1)^{k}}{1+k/n}= \lim-1/2\,n \left(-\Psi \left( 1/2+1/2\,n \right)+\Psi \left( 1/2\,n \right) \right) = \frac12\] evidently, by a computation in maple..

Doesn't it have any method without using the digamma func?
Coz I don't know any properties of digamma func.

yes, sure..
i know at least two ways..

one of them it's been discussed somewhere in a long topic, which i answered to, too, and it's a kind of measure theoretic solution...
you are just saying that the sequence (of measures) $ny^{n-1}\mathcal{L}$ weakly converges to $\delta_{1}$ ($\mathcal{L}$ is the lebesgue measure).

the other one is pretty easier: just plug $y^{n}= t$ in your integral, and you should get $\int_{0}^{1}\frac{ny^{n-1}}{1+y}dy = \int_{0}^{1}\frac1{1+x^{1/n}}dx \to \int_{0}^{1}\frac12dx$, because you can show the convergence of the sequence $\frac1{1+x^{1/n}}$ to $1/2$ to be uniform...

anyway, i think there are plenties of way to prove that this limit is $1/2$ :)

My idea with the digamma function was that from the integral you show that the limit must exist. To find the value of the of the limit we use that
$n\sum_{k=0}^{\infty}\frac{(-1)^{k}}{n+k}= \frac{n}{2}\left( \Psi(\frac{n}{2}+\frac{1}{2})-\Psi(\frac{n}{2}) \right)$

For simplicity in notation let $n = 2m$. The formula $\Psi(x+1) = \Psi(x)+\frac{1}{x}$ now gives that
$m\left( \Psi(m+\frac{1}{2})-\Psi(m) \right) = m\left( \frac{2}{2m-1}+\Psi(m-\frac12)-\Psi(m) \right)$

If we denote the sought limit with $A$ we get $A = 1-A$, that is $A = \frac{1}{2}$.

Of course, I see now that this is a bad solution...

Just for completeness, here's a way to do it with only Calc 2 tools:

Integrate by parts twice:
$\int_{0}^{1}\frac{nx^{n-1}}{1+x}\,dx=\left[\frac{x^{n}}{1+x}\right]_{0}^{1}+\int_{0}^{1}\frac{x^{n}}{(1+x)^{2}}\,dx$
$\left[\frac{x^{n}}{1+x}\right]_{0}^{1}+\left[\frac{x^{n+1}}{(n+1)(1+x)^{2}}\right]_{0}^{1}+\int_{0}^{1}\frac{2x^{n+1}}{(n+1)(1+x)^{3}}\,dx$
$=\frac12+\frac1{4n+4}+\int_{0}^{1}\frac{2x^{n+1}}{(n+1)(1+x)^{3}}\,dx$

Using the simple estimate that $0\le x,\frac1{1+x}\le 1$ in the interval, we have
$\frac12+\frac1{4(n+1)}\le \int_{0}^{1}\frac{nx^{n-1}}{1+x}\,dx\le \frac12+\frac1{4(n+1)}+\frac2{n+1}$
It is clear that the limit as $n\to\infty$ is $\frac12$.

If we repeat the process indefinitely, we get a series expansion for the integral:
$\int_{0}^{1}\frac{nx^{n-1}}{1+x}\,dx=\sum_{k=0}^\infty \frac1{2^{k+1}\binom{n+k}{k}}$
By comparison to a geometric series with ratio $\frac12$, the error in any partial sum is no more than the last term added.

Calculus 2 is the second class in the introductory college calculus sequence. It's not universally the same (the meaning changes depending on whether it's a semester or quarter sequence), but it's first year college material for many students. I probably should have said something like "first-year calculus".

This argument is understandable to a student who just learned the techniques of integration (if I expand some of the details integrating by parts), even if that student would never think of the argument.
As a general point- integration by parts is a really powerful tool. It's introduced as a method of finding antiderivatives, but it also excels at estimating other integrals.

[quote="ma_go"]you can show the convergence of the sequence $\frac1{1+x^{1/n}}$ to $1/2$ to be uniform[/quote]
Not on the interval $[0,1]$. Here is a simpler proof: for every $\epsilon\in (0,1)$ we have \[\int_{\epsilon}^{1}\frac{1}{1+x^{1/n}}dx\le \frac{1-\epsilon}{1+\epsilon^{1/n}},\] hence \[\int_{0}^{1}\frac{1}{1+x^{1/n}}dx\le \varepsilon+\frac{1-\epsilon}{1+\epsilon^{1/n}}.\] Let $n\to \infty$ and then $\epsilon\to 0$ to obtain \[\limsup_{n\to\infty}\int_{0}^{1}\frac{1}{1+x^{1/n}}dx\le \frac12.\] On the other hand, $\int_{0}^{1}\frac{1}{1+x^{1/n}}dx\ge \frac12$ for every $n$.

Note that the order of taking limits matters. I would not get far by letting $\epsilon\to 0$ first.

We don't have uniform convergence, but we do have monotone, pointwise convergence...

sure, blahblahblah  :oops:
actually, i didn't check for the details... i thought of a proof like the one you wrote down, but after all i thought convergence was uniform, how a silly mistake!
anyway, i know dominated/uniform convergence does work, but i was looking for an elementary solution :wink:


Do there exist infinitely many positive integers such that a^2 +a + 1 is prime

This is an open problem.  A result of this type is not known for [i]any[/i] polynomials of higher than linear degree.  The result for linear polynomials is, of course, Dirichlet's theorem, and the conjecture for irreducible primitive polynomials of higher degree is the Cauchy-Schwarz-Buniakowski conjecture.


Euler
function
limit
number theory
relatively prime
number theory unsolved

hey .. i was wondering if the following was true .. well it belongs in analysis but since its got a number theory flavour ill post it here ..
prove that p(n)/n can be made arbitrarily small, where p(n) is the Euler phi function, the number of numbers less than n relatively prime to it.
alternatively (same problem obviously) prove that the sequence [(p_1-1)/p_1][(p_2-1)/p_2][(p_3-1)/p_3] ... converges to 0 as n approaches infinity, where here p_k is the k-th prime.
also determine whether or not p(n) is surjective(i.e. it assumes all integer values).

[quote="epitomy01"]
also determine whether or not p(n) is surjective(i.e. it assumes all integer values).[/quote]
No, it isn't, since we know that, for $n > 2$, $\varphi(n)$ is even. ;)

The standard argument is the following.
Let $p_k$ denote the $k_{\mbox{th}}$ prime. Then for all positive integer $n$,
\begin{eqnarray*} \prod_{k = 1}^n \left( 1 - \frac{1}{p_k} \right)^{- 1} & = & \prod_{k = 1}^n \left( 1 + \frac{1}{p_k} + \frac{1}{p_k^2} + \ldots \right)\\ & \geq & \sum_{m \leq p_n}^n \frac{1}{m} \end{eqnarray*}
(Observe that every positive integer $m \leq p_n$ can be written as a product $\prod_{j = 0}^n p_j^{\alpha_j}$ where $\alpha_j \geq 0$ are positive integers). By the infinitude of primes and the divergence of the harmonic series,
\[ \lim_{n \rightarrow \infty} \sum_{m \leq p_n}^n \frac{1}{m} = \infty  \]
Therefore
\begin{eqnarray*} \lim_{n \rightarrow \infty} \prod_{k = 1}^n \left( 1 - \frac{1}{p_k} \right)^{- 1} & = & + \infty\\ \lim_{n \rightarrow \infty} \prod_{k = 1}^n \left( 1 - \frac{1}{p_k} \right) & = & 0 \end{eqnarray*}
From now on observe that
\begin{eqnarray*} \frac{\varphi ( p_1 \cdot \ldots \cdot p_n )}{p_1 \cdot \ldots \cdot p_n} & = & \prod_{k = 1}^n \left( 1 - \frac{1}{p_k} \right) \end{eqnarray*}
And we showed that $\underset{n \longrightarrow \infty}{\lim \inf} \left( \frac{\varphi ( n )}{n} \right) = 0$, [i]c'est-a-dire[/i] $\frac{\varphi ( n )}{n}$ can be made as small as possible.

Homever $\underset{n \longrightarrow \infty}{\lim \sup} \left( \frac{\varphi ( n )}{n} \right) = 1$, (since for $n$ prime we have $\frac{\varphi ( n )}{n} = \frac{n - 1}{n}$), so the limit
\[ \lim_{n \rightarrow \infty} \frac{\varphi ( n )}{n}  \]
does not exist. (This is not related to the question, but a frequent mistake is to consider $\lim \inf = \lim$ as always true)


For $ x,y,z>0$ prove that the following inequality holds

$ \left.\frac{x}{ (x\plus{}z)(x\plus{}2y\plus{}z)\plus{}10y^2}\plus{}\frac{y}{ (y\plus{}x)(y\plus{}2z\plus{}x)\plus{}10z^2}\plus{}\frac{z}{(z\plus{}y)(z\plus{}2x\plus{}y)\plus{}10x^2}\geq\frac{1}{2\left(x\plus{}y\plus{}z\right)}\right.$

We can homogenize, so assume that $ x \plus{} y \plus{} z \equal{} 3$.  Then, the inequality becomes
$ \sum_{cyc} \frac{x}{1\plus{}y^2} \geq \frac{3}{2}$.
Multiply both sides by $ (1\plus{}x^2)(1\plus{}y^2)(1\plus{}z^2)$.  We need to show that
$ \sum_{cyc} x(1\plus{}x^2)(1\plus{}z^2) \geq \frac{3}{2} \plus{} \frac{3}{2} x^2 y^2 z^2 \plus{} \frac{3}{2} \sum_{cyc} (x^2 \plus{} x^2 y^2)$
Rearranging,
$ \sum_{cyc} (x \plus{} x^3 \plus{} x^2 y \plus{} x^2 y^3 \minus{} \frac{3}{2} x^2 \minus{} \frac{3}{2} x^2 y^2) \geq \frac{3}{2} \plus{} \frac{3}{2} x^2 y^2 z^2$.
Writing the left-hand side in terms of squares,
$ \sum_{cyc} (x(x\minus{}1)^2 \plus{} x^2 y (y \minus{} 1)^2) \plus{} \frac{1}{2} \sum_{cyc} x^2 \plus{} \frac{1}{2} \sum_{cyc} x^2 y^2 \geq \frac{3}{2} \plus{} \frac{3}{2} x^2 y^2 z^2$.
The terms that contain squares are clearly non-negative.  And, observe that $ \sum_{cyc} x^2 \geq 3$ by AM-GM, and that $ \sum_{cyc} x^2 y^2 \equal{} x^2 y^2 z^2 \sum_{cyc} 1/x^2 \geq 3 x^2 y^2 z^2$ also by AM-GM, so adding yields the desired result.

[quote="Vn2009"]We can homogenize, so assume that $ x \plus{} y \plus{} z \equal{} 3$.  Then, the inequality becomes
$ \sum_{cyc} \frac {x}{1 \plus{} y^2} \geq \frac {3}{2}$.
 [/quote]

Thanks, Vn2009!
I think that this inequality is already known. It looks familiar. Is there anyone who knows it?
My solution is different than yours. I'm still waiting for new ideas.


geometry
parallelogram
trigonometry
geometry proposed

Let be $ ABCD$ a parallelogram, with $ \angle D\equal{} 80^ \circ$. Let be $ M \in (AB)$ so that $ \angle MDA\equal{} \angle MDB$ and $ DM\equal{}AB\equal{}a$.
$ a)$ Fiind $ \angle MDB$.
$ b)$ Fiind the area of $ ABCD$.

$ \angle MDB \equal{} x$ is given by $ \sin(80\minus{}x) \sin 2x \equal{} \sin (80\minus{}2x) \sin 80$ that seems to not have integer solution...are you sure that there is an integer solution?


Έστω $ (X, \Vert \cdot \Vert)$ ένα infinite dimensional Banach space, και $ \{ e_n \}_{n\equal{}1}^\infty \subset X$ μια ακολουθία τέτοια ώστε $ \Vert e_n \Vert \equal{} 1$ για κάθε $ n$, και $ \Vert e_n \minus{} e_m \Vert \geq 1$ για κάθε $ n \neq m$ (*). Τότε φυσικά το $ S : \equal{} \left\{ \sum_{n\equal{}1}^N \lambda_n e_n \; : \; N \geq 1, \lambda_1, \ldots, \lambda_N \in \mathbb{C} \right\} \subset X$ δεν είναι κλειστό (**). Αλλά μπορούμε να αναρωτηθούμε: ανήκει η συγκλίνουσα σειρά $ \sum_{n\equal{}1}^\infty \frac{1}{n^2} e_n$ στο $ S$? (... σίγουρα ανήκει στο closure)

(*) Όσοι δεν γνωρίζετε, δοκιμάστε να κατασκευάσετε μια τέτοια ακολουθία, ακόμα και αν το $ X$ δεν είναι complete (ίσως είχε βάλει παρόμοιο πρόβλημα ο Παναγιώτης κάποτε, δε θυμάμαι)
(**) Γιατί;

Μου φαίνεται είχα βάλει κάποια στιγμή ένα πρόβλημα [hide] για τη συμπάγεια της μοναδιαίας σφαίρας σε απειροδιάστατους χώρους Banach,[/hide] αλλά δεν είμαι και σίγουρος. Επίσης, μου φαίνεται ότι ο Anto είχε βάλει αρκετά πιο παλιά ένα πρόβλημα που έμοιαζε... αλλά τώρα είναι αργά και δεν αντέχω να ψάξω.

Από αύριο :)

Cheerio,

Durandal 1707

Μπορεί και να ανήκει στο S. Παράδειγμα: $ X \equal{} \ell^{\infty}$ με
$ e_1 \equal{} (1,1,\minus{}1/9,\minus{}1/16,\minus{}1/25,\ldots)$
$ e_2 \equal{} (\minus{}1,\minus{}1,0,0,\ldots)$
$ e_3 \equal{} (0,0,1,0,\ldots)$
$ e_4 \equal{} (0,0,0,1,\ldots)$
κτλ

Τότε $ \sum\limits_{n\equal{}1}^{\infty}\frac{1}{n^2}e_n \equal{} (3/4,3/4,0,0,\ldots) \equal{} \frac{3}{4}e_2 \in S$.

Μια απορία όμως για το (**):

Πρωτού διαβάσω ολόκληρη την άσκηση προσπάθησα να αποδείξω το (**) δείχνοντας πως η $ e_1 \plus{} e_2/2 \plus{} e_3/4 \plus{} \ldots$ δεν ανήκει στο $ S$, πράγμα που τελικά δεν ισχύει πάντα. Τελικά για οποιοαδήποτε συγκλίνουσα σειρά, αν δουλέψουμε όπως στην λύση μου θα δούμε ότι μπορεί να ανήκει στο $ S$. Πως λοιπόν δείχνουμε την (**); Η μόνη λύση που βρήκα χρησιμοποιεί Baire Category. Μήπως έχετε βρει καμιά άλλη απόδειξη του (**);

Ναι η πιο γρήγορη λύση που έχω υπόψη μου χρησιμοποιεί το BCT. Αν το $ \Sigma$ ήταν κλειστό τότε θα ήταν Banach space, και θα γραφόταν ως αριθμήσιμη ένωση από finite-dimensional (άρα closed) proper (άρα nowhere dense) subspaces, που αντιφάσκει το BCT.

Αν σκεφτείτε γρηγορότερη λύση θα χαρώ να τη μάθω.

Το πιο απλο παραδειγμα ειναι στον l2(N) με τη συνηθη βαση.Αντιπαραδειγμα ειναι ο c00(N) με τη συνηθη βαση(που ειναι απλα Norm space).


algebra
function
domain
induction
combinatorics solved
combinatorics

In the plane we have n lines. Prove that we can color all the regions determined in the plane by these lines with two colors such that no two neighbour domains are coloured with the same colour.

We use induction on the number n > 0 of lines :
For n = 1, there are two regions, so the result is obvious.
Now, suppose that the result holds for some n> 0, and let's give n+1 lines of the plane. Let D be one of these lines.
First consider that D is omitted, and color the regions determined by the other ones as allowed by the induction hypothesis.
Now, let add D, which divides the plane into two half-planes, say P and P'. And some of the regions are also divided by D. Note that all these regions are colored. Then, do not change the color of any regions in P and change the color of any of the regions in P' to get a desired coloring.

Pierre.


Find the answer to this problem and prove your answer.

When 600 students are seated in m rows, each of which contains n chairs, every chair is occupied. If five more chairs were in each row, everyone could be seated in four fewer rows and again every chair would be occupied. What is the value of m + n ?

This problem was on this years mmpc exam.

Hikari

[hide]So mn=600 and (m-4)(n+5)=600. (m-4)(n+5)=mn-4n+5m-20 so substitution and get -4n+5m=20. However 5m=4n so 0=20. Therefore this is impossible...?[/hide]

[color=cyan]howd you get the 5m=4n part?[/color]

Draw two diagrams, before and after moving the chairs. Erase the overlap (the m by n rectangle). You're left with 4 by n rectangle and 5 by m rectangle. Since the number of chairs didn't change those two should be of equal size. Thus 4n=5m.

Tare, you need to try that diagram again -- it should give you
5*(m - 4) = 4n.  Note that the diagrams you drew are [i]exactly[/i] the same argument as the calculations you made, so they [i]have[/i] to give you the same result.

This problem looks familiar...wasn't it on the MMPC?

Oh poo, should've known it was solvable...
[quote="hikari"]
This problem was on this years mmpc exam.[/quote]

[quote="notfunny"]i keep on getting n as 25... then m = 24 and m+n = 49... is this correct?

mn=600
(m-4)(n+5) = 600 = [color=red]mn +5n - 4n - 20[/color]
m = 600/n

plug this into the red equation and simplify.

you get n = 25 and then solving for m gives m=24 so m+n = 49[/quote]

don't you mean mn+5m-4n-20 as your equation in red?

[quote="melg"][quote]mn=600
(m-4)(n+5) = 600 = mn +5n - 4n - 20
m = 600/n

plug this into the red equation and simplify.[/quote]

I'm having difuculty simplifying that. Could someone show me the steps?[/quote]

melg, we have two equations

mn=600
mn+5m-4n-20=600

we are taking the first one:
mn=600 and dividing by n to get
m=600/n  which is being plugged into the second equation to get the answer

no not really. I knew everything you just said. I am having trouble simplifying it after it is put into the other equation.

[quote="TripleM"]Once you have plugged it in, multiply everything by n to clear the fractions. You then have a quadratic equation.. try to factorise it.[/quote]
M+N=49
 :D  :D

[quote="TripleM"]Once you have plugged it in, multiply everything by n to clear the fractions. You then have a quadratic equation.. try to factorise it.[/quote]
[hide]$m=\frac{600}{n}$
$mn+5m-4n-20=600$
$(\frac{600}{n})(n)+5(\frac{600}{n})-4n=620$
$600+\frac{3000}{n}-4n=620$
$\frac{3000}{n}-4n=20$
$3000-4n^2=20n$
$n^2+5n-750=0$
$(n+30)(n-25)=0$
$n=25,-30$[/hide]

im getting that 5m=-4n, then im proving myself wrong.  Help!!!!
here was my reasoning:
m(n+5)=n(m+4)
mn+5m=mn-4n

Now that im thinking about it, I just found another way to prove myself wrong.  mn=600, but 600+5m=600-4n=600 can't be true unless m and n are both 0.  But then 0 x 0 isn't 600!!!! Help!!!

sry for overposting again this computer is CRAZY

oh oops [i]now[/i] i see what i did wrong!
i said n(m-4)=m(n+5)=600 instead of (n+5)(m-4)=600.  Now I agree with 49.


Hi dears

I want to know What is the name of below Chemical Equation:

  $\mbox{CH_4_{(g)}+Cl_2_{(g)}}\longrightarrow^{\Delta} \mbox{CH_3Cl_{(g)}+HCl_{(g)}}$

is it Cumbustion,Compound,Replacement,...?

Its a Free Radical Substitution /Replacement Reaction

yes, Rushil is Correct it is a free Radical replacement.

(Radical is when the $Cl-Cl$ bond is broken, and each $Cl$ atom is called a Radical)


[quote="gemath"]Given triangle $ABC$ find the locus of $M$ on plane such that
$\frac{\angle MBC}{\angle MCB}=k>0$[/quote]

What do you want to hear about the locus? For rational k, these will be algebraic curves, as you can easily see; for irrational k, the curves should be transcendental. Ratios of angles have never had any interesting meaning in geometry...

  darij

When I post this problem I don't wait for any reply!
It is pity that here is last post of me on mathlink, mathlink always is memories of me! Bye everybody !


Let $ a , b , c$ be positive numbers such that :$ a^2\plus{}b^2\plus{}c^2\plus{}abc\equal{}4$.
Prove that:
 $ \sqrt{(4\minus{}a^2)(4\minus{}b^2)}\plus{}\sqrt{(4\minus{}b^2)(4\minus{}c^2)}\plus{}\sqrt{(4\minus{}c^2)(4\minus{}a^2)}\geq\ (2\plus{}\sqrt{abc})^2$

[quote="Potla"]Do not double post please..... :blush:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1449380#1449380[/quote]
Sorry :wink:


[quote="mismatchtea"]sqrt (1 + 2sqrt(1 + 3sqrt(1 + 4sqrt(1 + ...))))

Find the sum of the infinite series.[/quote]

$ \sqrt {1 \plus{} 2\sqrt{1 \plus{} 3\sqrt{1 \plus{} 4\sqrt{1 \plus{} ...}}}}$

Is it?

This is a relatively famous problem, posed by Ramanujan (I think). Consider

$ (x + n)^2 = x^2 + 2xn + n^2$

this implies

$ x + n = \sqrt{n^2 + x[(x+n) + n]}$

Now let's calculate $ 2 + 1$ based on this identity:

$ 2 + 1 = \sqrt{1 + 2(3 + 1)}$

$ 3 + 1$ is a sum which we can "simplify" in the same manner:

$ 2 + 1 = \sqrt{1 + 2\sqrt{1 + 3[4 + 1]}}$

Keep going to get:

$ 2 + 1 = \sqrt {1 + 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{1 + \dots}}}}$

That solution is interesting, and is fairly easy to come up with if you know the answer ahead of time. But I couldn't imagine myself figuring that one out for a long time if nobody told me it was 3.

I've always wondered if there is a less contrived solution (perhaps one that uses bounding or something).

Well, [url=http://www.isibang.ac.in/~sury/ramanujanday.pdf]this talk[/url] is a solution that arrives by bounding $ f(x) \equal{} \sqrt{1 \plus{} xf(x\plus{}1)}$ (our infinite series is $ f(2)$), arriving on $ x \plus{} 1 \le f(x) \le x \plus{} 1$.

Yes! That's exactly what I've been looking for. Thanks!

[b]Edit:[/b] The original link doesn't seem to work. [url=http://74.125.47.132/search?q=cache:http://www.isibang.ac.in/~sury/ramanujanday.pdf]Here is a cache version[/url] of the same thing, although it has been translated into HTML. [b]ffao[/b], if you have the original, can you please upload it?

The original is still on my cache, so here it is.


geometry
geometric transformation
geometry unsolved

I have a problem with this task:
Two circles with radius $R$ are tangent at point $K$. Point $A$ is on the first circle and point $B$ is on the second one
such that angle$AKB=90$. Prove that $AB=2R$.

I would like a proof with translation

Let the center line $O_1O_2$ of the 2 congruent tangent circles $(O_1), (O_2)$ meet these circles at points M, L different from K. Translate the circle $(O_1)$ and the line AK by the distance 2R in the direction of the ray $(O_1O_2$ into a circle $(O_1')$ and the line A'K'. Obviously, $O_1O_1' = 2R = O_1O_2$, hence, the centers $O_1' \equiv O_2$ become identical, the circles $(O_1') \equiv (O_2)$ become identical, $KK' = 2R = KL$, hence, the points $K' \equiv L$ become identical, and the point A' lies on the circle $(O_1') \equiv (O_2)$. The lines $A'L \equiv A'K' \perp BK$ are perpendicular, because $AK \perp BK$ and a translation does not change angles of lines. Since KL is a diameter of the circle $(O_2)$, $A', B \in (O_2)$, and $A'L \perp BK$, the points $A' \equiv B$ are identical and $AB \equiv AA' = 2R$.


Freeway exits $A$ and $B$ are exactly 245 miles apart. Two cars, having their cruise controls set at different integral speeds, pass A simultaneously, both headed for $B$. Three hours later, the distance of the slower car from A was nine times the distance of the faster car from $B$. If the difference in their speeds is a prime number, how fast(mph) is the faster car traveling?

[hide="solution"]
let speeds be $x$ and $y$, so equation becomes
$3x=9(245-3y)$
$x=3(245-3y)$
$x=735-9y$

we also now that $y-x=P$

$y-P=735-9y$
$10y=735+P$
$P=5$ satisfys this only,
In this case,
$y=74$
and $x=69$

[/hide]


In the obtuse triangle $ABC$, $AM = MB, MD \perp BC, EC \perp BC$. If the area of $\triangle ABC$ is 24, then the area of $\triangle BED$ is

[asy]
size(200);
defaultpen(linewidth(0.8)+fontsize(11pt));
pair A = (6.5,3.2), B = origin, C = (5.0), D = (3.3,0);
pair Xc = (C.x,4), Xd = (D.x,4), E = intersectionpoint(A--B,C--Xc), M = intersectionpoint(D--Xd, A--B);
draw(C--A--B--C--E--D--M);
label("$A$",A,NE);
label("$B$",B,W);
label("$C$",C,SE);
label("$D$",D,S);
label("$E$",E,N);
label("$M$",M,N);
draw(rightanglemark(D,C,E,7)^^rightanglemark(B,D,M,7));
[/asy]

$\textbf{(A) }9\qquad
\textbf{(B) }12\qquad
\textbf{(C) }15\qquad
\textbf{(D) }18\qquad
\textbf{(E) }\text{not uniquely determined}$

By similarity, $ BD: BM=BC: BE\implies BD={BM\cdot BC\over BE}$, hence

\begin{eqnarray*}[BED]={BE\cdot BD\over 2}\sin B={BE\over 2}\sin B\cdot {BM\cdot BC\over BE}={AB\cdot BC\over 4}\sin B={1\over 2}[ABC]\end{eqnarray*}

So the answer is B.

[quote="Farenhajt"]By similarity, $ BD: BM = BC: BE\implies BD = {BM\cdot BC\over BE}$, hence

\begin{eqnarray*}[BED] = {BE\cdot BD\over 2}\sin B = {BE\over 2}\sin B\cdot {BM\cdot BC\over BE} = {AB\cdot BC\over 4}\sin B = {1\over 2}[ABC]\end{eqnarray*}
So the answer is B.[/quote]
Nice solution!

I have an alternate one.

[hide="My Solution"]
Notice that $ MD\parallel CE$, so $ [CED] = [CEM]$.  Thus, we have that $ 12 = \frac {[ABC]}{2} = [BCM] = [ABC] - [ACM] = [ABC] - ([ACE] + [CEM]) = [ABC] - ([ACE] + [CED]) = [ABC] - [ACDE] = [BED]$, so $ [BED] = \boxed{12}$. [/hide]


If $\tan x+\tan y=25$ and $\cot x + \cot y=30$, what is $\tan(x+y)$?

[hide]let $\tan{x}=a,\tan{y}=b$
we have: $(a+b)=25,\frac{1}{a}+\frac{1}{b}=30=\frac{a+b}{ab}\Rightarrow ab=\frac{5}{6}$
we want to find $\frac{a+b}{1-ab}=\boxed{150}$ [/hide]

Wow, this is by far the easiest AIME problem I've ever seen :o

[hide]$\tan{x}+\tan{y}=25$
$\frac{1}{\tan{x}}+\frac{1}{\tan{y}}=30 \Longrightarrow \frac{\tan{x}+\tan{y}}{\tan{x} \cdot \tan{y}}=30 \Longrightarrow \tan{x} \cdot \tan{y}=\frac{25}{30}$
Since $\tan{(x+y})=\frac{\tan{x}+\tan{y}}{1-\tan{x} \cdot \tan{y}}$, after plugging in what we know, we have $\boxed{150}$[/hide]

[hide]Letting $\tan{x}=a$ and $\tan{y}=b$, we see that the answer is $\frac{a+b}{1-ab}$.

We're given $(a+b)=25$ and $\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=30$. To find $ab$, $\frac{a+b}{ab}=\frac{25}{ab}=30 \implies ab=\frac{5}{6}$ Now, $\tan (x+y)=\frac{25}{1-\frac56}=\boxed{150}$.[/hide]

never turn down a chance to practice trig.

[hide=Solution]Note that $\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y}$. We have $\frac{1}{\tan x}+\frac{1}{\tan y}=30=\frac{\tan x + \tan y}{\tan x \tan y}$, so $\tan x \tan y = \frac{5}{6}. Upon plugging in we get $\boxed{150}$.[/hide]

Let $\tan(x)=a$, and $\tan(y)=b$ so $(a+b)=25$, and $ab=\frac{5}{6}$, now we want to find $\frac{a+b}{1-ab}=\boxed{150}.$

[hide = ez trig  :cool: ]
We can first substitute $p$ for $\tan x$ and $q$ for $\tan y$. Then we get the systems of equations $p+q = 25$ and $\frac{1}{p} + \frac{1}{q} = 30$. The second equation simplifies to $\frac{p+q}{pq} = 30$, so $pq = \frac{5}{6}$. Then, by using the difference identity for tangent on $\tan(x+y)$, we get $\frac{\tan{x} + \tan{y}}{1 - \tan{x}\tan{y}}$. Plug and chug on $\frac{p+q}{1-pq}$  :cool: to get $\boxed{150}$.

[/hide]


geometry
incenter
circumcircle
cyclic quadrilateral
geometry unsolved

Let $ ABCD$ be a cyclic quadrilateral, with $ AB>CD$. Denote by $ T$ the intersection point of $ AC$ and $ BD$. Consider a circle $ \Gamma$ that is tangent to $ [TA],[TB]$ and internally tangent to the circumcircle of the $ \triangle ABC$ at the points $ E,F$ and $ S$, respectively. Prove that the incenters of the triangles $ ABC$ and $ ABD$ are found on $ [EF]$

You can see [b]Lemma 1[/b] of the artice in Forum Geometricorum, Volume 3 (2003) 225–229,"Sawayama and Thebault’s theorem"- author: [b]Jean-Louis Ayme[/b] (maybe he is the member [b]jayme[/b] of our forum?)