group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

I'm not sure I understand what those symbols mean.  Does $ \subseteq$ mean "subset," $ \le$ mean "subgroup," and $ \unlhd$ mean "normal subgroup"?

(As a general comment, it is almost never the case that symbols make something easier to understand.)

[quote="Peter"]For $ X\subseteq G$ we define $ X^* \equal{} \{g\in G\mid \langle g\rangle\cap X \equal{} 1\}$.

(a) Let $ X\unlhd G$. Is $ X^*$ a subgroup of $ G$?
(b) Let $ X\unlhd G$. Is $ (X^*)^* \equal{} X$? What about $ X\le G$ or even $ X\subseteq G$?[/quote]

Did you write this down correctly?  This does not define a duality even in vector spaces.

If G is the Klein four-group = { 1, x, y, xy }, and X={1,x}, then X is a normal subgroup and X*={1,y,xy} is not a subgroup.

In the dihedral group of order 8 = { 1, r, rr, rrr, f, fr, frr, frrr } with X = { 1, rr, f, frr }, then X* = { 1, fr, frrr } is not a subgroup, and X** = { 1, r, rr, rrr, f, frr } is not equal to X and is not a subgroup.

You are right. My text book is wrong. Thank you and sorry for the stupid question.


$y^{-v}\frac{dy}{dt}+f(t)y^{1-v}= g(t)$
$w = y^{1-v}$
$dw = (1-v)y^{-v}dy$
$dy = \frac{y^{v}}{1-v}dw$
$\frac{1}{1-v}\frac{dw}{dt}+wf(t) = g(t)$

applying this to your particular DE

$-\frac{dw}{dt}+tw = t$
apply the formula for linear DE's, can you do the problem now?

thnx for the reply dvagner

i got lost after u said:

[quote]applying this to your particular DE
-\frac{dw}{dt}+tw=t
apply the formula for linear DE's, can you do the problem now?[/quote]

well now you have a linear differential equation if u multiply by -1
$\frac{dw}{dt}-tw =-t$
multiply by your integrating factor then group, are you familiar with this concept? if not then i will go over it

hey,

sorry - i meant to say - i didn't get how you got the equation after u said:

"applying this to your particular DE"

oh sorry, in your DE:
v = 2 (look at the exponent on the y)
f(t) = g(t) = t (that's the coefficient on the y and y^v terms respectively)

plugging those into the formula that the substitution yielded gives you that answer


so, what do you guys think about this year's NYSML?

personally, i didnt like it that much  - especially too easy power q's and individuals.
(even regardless of ugly #6)

any comments including some suggestions for ARML?

i agree completely, at least with the comments on the power question. Far too easy for any one person, even, to complete. Grading it was very easy, however. I think the last proofs should be more involved, and there shouldn't have been as many perfects as there were.

Individuals did seem easier than usual, but considering only 4 8's, i guess it wasn't too bad. I didn't get a chance to look at the team or relays yet, so I can't comment on that.

on a side note, doesnt it feel a bit wrong going to nysml and just dominating it completely? it's strange, since nyc kinda looks at it like practice for arml. I mean it was fine and dandy while nyc A kept winning first, but now it's come to the point where 2-3 (or this year, all 4) of the nyc teams beat all other teams. I mean it just makes me a bit uncomfortable. anyone have any comments on this?

[quote="penny77114"] Far too easy for any one person, even, to complete. [/quote]

"it was so easy, i couldn't even COMPLETE it"  -- hahaha.

This year is kind of too easy.....it's rare that team is the hardest round..at least in my opinion..and the individual round is like usual---many know how to do 8 question,but not much can get it on the contest.......

But for those of you who have the questions, can you please post them?? Just the indvidiual round would be fine!

Do you think it would be possible to ask JackPo or someone else for the questions if no one else has them?

we dont have the questions because a slight problem occurred with the number of answer slips.  so, for the sake of uniformity, they had us write the answers on the back of the problems, and hand them in.  iirc, the questions and answers will be posted up at [url]http://www.nycmathteam.com[/url] in the near future.

I agree with KD.

We all thought that we wont need practice for NYSML, and we didnt.
we all said that we wont need strong relays for NYSML, and we didnt.

maybe combine some other reagions to form A, B, C team so that at least A team can compete against us? or make NYC math team into queens team, manhattan team or so?

it was cool but kind of weird when all people in the auditorium were saying "New York City A" for winning team announcement altogether.

that's what  Jack suggest on the way back...to disolve the A team,
the reason is,all us A team people know we are winning long before the game started,while all the other teams,knowing they have almost no chance beating us,so they kinda of give up ,or at least did not drive each other as hard as possible.It will be nice to have every team start from same place,and have more competition..just like how stuy and TJ works.......

[quote="Geehoon"]it was cool but kind of weird when all people in the auditorium were saying "New York City A" for winning team announcement altogether.[/quote]

well i dunno, i kinda liked NYSML this year, only cuz i finally got my sweatshirt  :lol:

so i dunno, i think it's a cool idea to split up the NYC team. But we should only split into like, Manhattan, Queens, and maybe Bklyn, cuz Staten Island and Bronx will get killed. But of course, we reform a NYC A team when we go to ARML, right?

Is the proposal to split according to where the student lives or where the student goes to school?

should be by where he lives.

IF by school, it wouldn't be fair cuz Manhattan might always win cuz of Stuy....just a thought...

I was **SUGGESTING** the idea of simply forming mixed teams.  Currently NYC does not have the funds to support multiple practices.  I was hoping that if we assign 5 ambitious kids as captains in the beginning of the year (say September) and host one competition per month (rotate through the HS), that the ambitious captains would naturally begin to prep their kids outside the classroom (and thus the kids would also get to know each other fairly well prior to these competitions).

This model is purely based on the business settings (you have groups of comparable strengths internally competing for performance bonuses).

I think this has two benefits:

1)  The kids would be far better accustomed to their team members and the bottom line would be much stronger.
2)  Each year @ NYSML, it would be a free for all, between our 4 teams and any of hte other NYSML teams, leading to a much better atmosphere.

This is my opinion ONLY and likely will NOT be implemented.

[quote="ShadowLord"]the thing is..we still want to win..and as it on relays,we want to concentrate our best people not to spread them. i agree the fact it will rise competition,and make the average level of the team better,but i don't think one year of mixing team can improve a lot of thing,unless Mr linker decide to keep it as tradition......
It's fine to train that way or even NYSML ,but i still think we need to pull off the best people in order to win,besides,we are not ultimate in ARML,we don;t really need too much internal competition since we had enough externals.......[/quote]

Ah-ha! you missed my point!  I think the policy of conncentrating the "best" people is wrong.  I think what should happen is for the team to develop as a whole, which means the captains would naturally "bring up" the level of those who are struggling.  We do need the internal competition, because at this point... NYC has resigned themselves to not winning ARML.

i think the idea of splitting up the teams to increase competition level at NYSML sounds like a good idea, but does anyone agree that NYC practices way too much for NYSML? also, wouldn't those individual teams by borough by sort of like the IMLs?

jack o' lantern said, "NYC has resigned themselves to not winning ARML".

why, HUH?  in 2000 we were 1 point behind the first place team.  and we would have been first had linker not made such sub-optimal decisions about who to put on the A-team...


calculus
derivative
limit
trigonometry
algebra
binomial theorem
calculus computations

How would you show that $\frac{d}{dx}(e^{rx}) = re^{rx}$ where $r = a+bi$?

That depends, to a certain extent, on how you introduced the complex exponential in the first place.

Here's one path:

For all $z\in\mathbb{C},$ define $e^{z}=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}.$

Show that this series converges absolutely for all $z.$

Prove that for all $z,w\in\mathbb{C},\ e^{z+w}=e^{z}e^{w}.$

(This is a beautiful argument that everyone should do at least once - proving that just from the power series and the Binomial Theorem.)

Then $\frac{d}{dx}e^{rx}=\lim_{h\to0}\frac{e^{r(x+h)}-e^{rx}}h$

$=e^{rx}\lim_{h\to0}\frac{e^{rh}-1}h$ (This step requires the theorem above.)

$=e^{rx}\lim_{h\to0}\left(r+\frac{r^{2}h}2+\frac{r^{3}h^{2}}{3!}+\cdots\right) =re^{rx}$

----

Alternatively (and this really comes from a different initial approach) you can just write out

$e^{(a+bi)x}=e^{ax}\cos bx+ie^{ax}\sin bx,$

grind out the derivative of that, and compare that to the product of that with $a+bi.$

[quote="Kent Merryfield"]
(This is a beautiful argument that everyone should do at least once - proving that just from the power series and the Binomial Theorem.)
[/quote]
Differentiating $e^{-z}e^{z+w}$ w.r.t. $z$ has always struck me as easier.

Oh, I guess doing that uses the original question  :lol: Why don't we just differentiate term by term to show that $\frac{d}{dz}e^{z}= e^{z}$ and then use the chain rule to get the $r$ in? (Surely if you know enough about sums to formally motivate grouping the terms of the series together to use your argument with the binomial theorem you also know how to motivate differentiating inside the sum)

Kent, did you mean that: $e^{(a+ib)x}= e^{ax}(\cos bx+i\sin bx)$ (i.e. Euler's formula)?

Here is what I did:

$\frac{d}{dx}(e^{rx}) = \frac{d}{dx}(e^{(a+bi)x}) = \frac{d}{dx}(e^{ax+bix})$

$\frac{d}{dx}(e^{ax}\cdot e^{bix}) = \frac{d}{dx}(e^{ax}\cdot (\cos bx+i\sin bx))$

If $f(x) = f(x)+ig(x)$ then $f'(x) = f'(x)+ig'(x)$

$f(x) = e^{ax}\cos bx+ie^{ax}\sin bx$

$f'(x) = e^{ax}(-b\sin bx)+ae^{ax}\cos bx+i(a^{ax}(b\cos bx)+ae^{ax}\sin bx)$

How would I get this in the form of: $(a+bi)e^{(a+bi)x}= re^{rx}$

That's just algebraic manipulation. Start from the other end and multiply out $(a+bi)e^{ax}(\cos bx+i\sin bx).$


a) Prove that there are infinitely many  points between 2 given points on a line.

b)Prove there are infinitely many points on a line not between 2 given points.

[hide="The relevant intuition"] Every open interval is basically a copy of the real line.  (Consider, for example, the arctangent.)  This is the same intuition that allows you to prove that the rationals are dense in the reals. [/hide]

could you say the following?

let the two points be x1 and x2. take a third point x3=(x1+x2)/2. take a fourth point x4=(x3+x1)/2. in general, take a point xn such that xn=(x(n-1)+x1)/2

x(n-1) is a subscript not a multiplication....
wouldnt this generate infinite number of points between x1 and x2?

Edit: NVM, you need many of Hilberts axioms to prove this.

If we put the lines on a number line, this is true because between any two real numbers $ a$ and $ b$, there are infinitely many real numbers between $ a$ and $ b$.

If we scaled this so $ a \equal{} 0$ and $ b \equal{} 1$, then for every $ x$ in $ (a, b)$, there is a corresponding value $ \frac 1x$ that is not in $ (a, b)$, and since $ \frac 1x$ is a bijective (one-to-one) function, there are infinitely many points not between $ a$ and $ b$ as well.

[quote="Yongyi781"]If we put the lines on a number line, this is true because between any two real numbers $ a$ and $ b$, there are infinitely many real numbers between $ a$ and $ b$.[/quote]
Why?  (This is the content of the problem.)

[quote="mathking123"]let the two points be x1 and x2. take a third point x3=(x1+x2)/2. take a fourth point x4=(x3+x1)/2. in general, take a point xn such that xn=(x(n-1)+x1)/2[/quote]
This is not, by itself, enough.  What guarantees that these points are distinct?  (This might seem like nitpicking until you ask the same question but replace $ \mathbb{R}$ with a finite field.)

Consider the values of the points on the number line. Move to the last digit where they are the same, then go the the next digit of the lower one and change it from 1-9, then go to the next, and so on. Is this a valid argument?
For example, if we have 1.35622 and 1.35611
We would have 1.356111,1.356112,......1.3561111,1.3561112......

Assume there were finitely many real numbers between $ a$ and $ b$. However, we know that there must be at least 1 $ r$ because we can perform the operation dgreenb801 described. But now, we can do the same thing, except in the interval $ (a, r)$ and $ (r, b)$. We can do this infinitely many times, therefore, there are infinitely many real numbers between $ a$ and $ b$.

I know there is an axiom  "There is at least 1 point betweem 2 given point on  a line " but I am not sure for the second proof.

An axiom is a statement one takes to be true.  That is not an axiom of Euclidean geometry.

The answer I was looking for is very simple:  if $ x_1 < x_2$, then $ x_1 < \frac{x_1 \plus{} x_2}{2} < x_2$.

There is a postulate in my geometry book sayin that "There is at least 1 point between 2 given points"

and with this there are some others postulates such that " There are at least 2 points on a given line and at least 1 point outside it"


integration
calculus
function
limit
real analysis
real analysis unsolved

Let  {$ \ A_i$}  be a sequence of measurable subsets of a measure space $ (X,\mu)$. Suppose that
    $ \sum_{i \equal{} 1}^{\infty} \mu( \ A_i) < {\infty}$
Let $ Z$ be members of $ x$ in $ X$ such that $ x\in\ A_i$,for infinitely many $ i$.
prove that $ \mu(Z) \equal{} 0$ .

what kind of measure you are talking about?
I guess it is counting measure, isn't it ?

No, he's not talking about counting measure. The details of the measure are unimportant to this; it's broadly true.

Let $ g(x)\equal{}\sum_i\chi_{A_i}(x)$ where $ \chi_{A_i}(x)\equal{}\begin{cases}1&x\in A_i\\0&x\not\in A_i\end{cases}.$

We can precisely identify $ Z$ as $ Z\equal{}\{x\in X: g(x)\equal{}\infty\}.$

Now we claim that $ g$ is integrable on $ X:$

$ \int_Xg\equal{}\int_X\sum_i\chi_{A_i}\equal{}\sum_i\int_X\chi_{A_i}\equal{}\sum_i\mu(A_i)<\infty$

The interchange of sum and integral is justified by the Monotone Convergence Theorem.

We know that $ g$ is a nonnegative function that is integrable. It is known that such a function must be finite almost everywhere - that is, $ \mu(Z)\equal{}0.$

=============

Here's a second proof that does not involve either integration or the Monotone Convergence Theorem.

For any given $ n,$ if $ x\in Z,$ then $ x$ belongs to at least one (in fact, infinitely many) $ A_i$ with $ i\ge n.$ Hence $ Z\in \bigcup_{i\equal{}n}^{\infty}A_i$

But by countable subadditivity, $ \mu\left(\bigcup_{i\equal{}n}^{\infty}A_i\right)\le\sum_{i\equal{}n}^{\infty}\mu(A_i).$

Just from the definition of what it means to be a convergent series of nonnegative real numbers, we have that for any $ \epsilon>0,$ there exists $ n$ such that

$ \sum_{i\equal{}n}^{\infty}\mu(A_i)<\epsilon.$

Hence, $ \mu(Z)<\epsilon$ for all $ \epsilon>0,$ from which we deduce that $ \mu(Z)\equal{}0.$

==================

Note in both cases the centrality of the hypothesis that $ \sum_{i\equal{}1}^{\infty}\mu(A_i)<\infty.$ Would it be enough just to have $ \lim_{i\to\infty}\mu(A_i)\equal{}0?$

(Yes, I already know the answer to that; I'm looking for a response from a student, possibly one taking this course.)


I vote Bubala as funniest of the camp  :rotfl:

Team Contest 1: Amazing 11-variable bogus proof

Team Contest 2: My wrist!!!!!!!

:D

Vote in the poll...

I was there, LOL :D

DELTA team will rise again!!!!!!!

Gabriel wasnt that bad... :maybe:
not_trig, are you damien?

Yes, asdf4. LOOK AT THE YEARBOOK :D

Team $\gamma$ [b]PWNED[/b] team $\delta$. Don't try it, Ben.

BTW, who voted 10!?!?!?!?  Gabriel is a genius, but he's not that great a teacher.

dude gabriel and i love each other so much it's not even funny

"What the s***? I don't get it at all..."
"Who said that? Oh, you. You should just go back to sleep."

I gave Gabriel a 10 because he's hilarious and he laughed at the Dorin-speedo joke.

and because he curses all the time and yells at damien.

PS~ bubala SO should've won that bs-proof award...

NO, Vitek should have won the BS proof award,  :rotfl: .  Remember the team contests?

i think a certain BS proof that actually got credit beats them all?

Bubala should have won the BS proof; I agree. He didn't get any credit (1 -1)

You love Gabriel because he yelled at me?
@#$%^{&}amp;*)!@#
$%^&*(!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[quote="mathclass"]Gabriel and bubula rock my socks.[/quote]

OMG bubs you have an admirer! I wonder who this is hmmmm hyperness suggests Patricia Li but eh

whoa patricia and bubs would be interesting

nope, not patricia

i'm patricia's husband. or wife. either one works.

I heard Bubala was genuinely upset he didn't get any points on the giant bogus proof that he pulled an all-nighter to concoct ("z looks like 2, so let's let them be equal...") :D The wrist thing was a classic :rotfl:

Actually, it was "h looks like k, so I used them interchangeably".

z=2 for all z was attributed to him, though I stole it.  :ninja:

And also in all cases y=z since they look really similar in cursive (at least that's the way I see it).

Stop making fun of Gabriel by the way.

Gabriel's the man. He's got 11 votes for 10.


[quote="José"][quote="frt"]10 planes cut through a sphere so that the sphere is divided into $n$ pieces. What is the maximum possible value of $n$?[/quote]

[hide] $28?$[/hide][/quote]
Got proof?

[quote="frt"][quote="José"][hide] $28?$[/hide][/quote]
Sorry, that's not correct. The answer is more than that.[/quote]

Thanks, anyway.

i believe the answer is [hide]176[/hide]

generalization: try and find the number of sections using $k$ cuts


Let $\{u_{n}\}_{n \ge 0}$ be a sequence of integers satisfying the recurrence relation $u_{n+2}=u_{n+1}^2 -u_{n}$ $(n \in \mathbb{N})$. Suppose that $u_{0}=39$ and $u_{1}=45$. Prove that $1986$ divides infinitely many terms of this sequence.

$ r_n\equiv u_n (\mod 1986)$ which $ 0\leq r_n\leq 1985$
Because $ \{r_n\}$ take finite value so it has $ k,l\in N$ such that :
$ k > l$
$ r_k \equal{} r_l,r_{k \plus{} 1} \equal{} r_{l \plus{} 1}$ (1)
 We prove that $ r_{n} \equal{} r_{n \plus{} k \minus{} l},\forall n$
Because (1) we has :$ r_{k \minus{} 1}\equiv r_{l \minus{} 1} (\mod 1986)$ so $ r_{l \minus{} 1} \equal{} r_{k \minus{} 1}$
                               $ r_{k \plus{} 2} \equal{} \equiv r_{l \plus{} 2}(\mod 1986)$ so $ r_{k \plus{} 2} \equal{} r_{l \plus{} 2}$
Contine this progress we has $ r_{n} \equal{} r_{n \plus{} k \minus{} l},\forall n\in N$
Because $ r_2\equiv 0(\mod 1986)$ so $ r_{h(k \minus{} l) \plus{} 2}\equiv 0(\mod 1986)$
So has infinite $ u_n$ such that $ 1986|u_n$

this is not a problem in China 1991.
In fact, it is in Canada 1986.

maybe that is the reason why the sourse of M27 was shown as "Canada 1986":http://www.mathlinks.ro/viewtopic.php?p=849911#849911

Prove TTsphn is not complete. May be period $ r_{k\plus{}T}\equiv r_k \ \forall k>C$. In these case we can continue sequense $ n\to n\minus{}1$: $ u_n\equal{}u_{n\plus{}1}^2\minus{}u_{n\plus{}2}$, therefore $ C\equal{}0$.


If $s$ is the semiperimeter prove or disrove the following
$8s^3<9^s$

Er, does the fact its the semiperimeter have anything to do with anything???

Strange indeed. The question seems to be rubbish.

I think he means ... of a polygon with area 1. Or something.

As it's true from 3/2 onwards... (and the optimal is a circle, where the semiperimeter would only be pi/2)

Seems quite nice thus :)

It is a part of a problem which I tried to solve. I reach at this and I couldn't find an answer. I make some computations and this ineq is true I think but how could someone solve it? Help me.

It is not true for all s. For >3/2, it's sufficient to state that
$9^s-8s^3$ is increasing.

[quote="Peter VDD"]everything between 1.3 and 1.5 should be a counterexample?[/quote]

Ok thanks


AIME
modular arithmetic
function
algebra
binomial theorem
number theory

Let $a_n = 6^n + 8^n$.  Determine the remainder on dividing $a_{83}$ by 49.

[hide="Solution"]First, we try to find a relationship between the numbers we're provided with and 49. We realize that $49=7^2$ and both $6$ and $8$ greater or less than 7 by $1$.

Expressing the numbers in terms of $7$, we get $(7-1)^{83}+(7+1)^{83}$.

Applying the Binomial Theorem, half of our terms cancel out and we are left with $2(7^{83}+3403*7^{81}+\cdots + 83*7)$. We realize that all of the terms in this big jumble of numbers are divisible by 49 except the final term.

After some quick division, our answer is $35$.[/hide]

[hide]

The divisor can be rewritten as $7^2$.

Rewrite the dividend as: $(7-1)^{83}+(7+1)^{83}$

When we expand this out using the Binomial Theorem, we'll get the sum of a bunch of different powers of $7$ multiplied by the corresponding coefficient. We are dividing by $7^2$, so any term with $7$ raised to a power greater than or equal to $2$ is going to leave no remainder, so we're only concerned with the terms where $7$ is raised to either the first or zero power.

Those terms are: $\displaystyle{83\choose 82}7^1(-1)^{82}+\displaystyle{83\choose 83}7^0(-1)^{83}+\displaystyle{83\choose 82}7^1(1)^{82}+\displaystyle{83\choose 83}7^0(1)^{83}$

Their sum equals: $2\times83\times7=1162$
$1162\equiv \boxed{35} \bmod{49}$

[/hide]

Oh, I get it.

The $-1$ and $1$ terms will cancel out. The whole thing is multiplied by $2$ because there are two seperate "binomials".

Thanks.

[quote="towersfreak2006"][hide]That's really nice. You can also finish it by
\[ 6^{ \minus{} 1} \plus{} 8^{ \minus{} 1} \equal{} \frac {8 \plus{} 6}{48}\equiv\frac {14}{ \minus{} 1}\pmod{49} \equal{} \minus{} 14\equiv35\pmod{49},
\]
instead of finding inverses.
[/hide][/quote]Very clever!

[url=http://www.artofproblemsolving.com/Wiki/index.php/Totient_function]Euler's Totient Function[/url]

[url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_totient_theorem]Euler's Totient Theorem[/url]

[quote=brainiac1]Saying $\frac{}{48}$ doesn't really mean dividing by $48$; it's just a shorthand way of writing the inverse of $48 \bmod 49$. Which turns out to be $-1$.[/quote]

Ok got it .. so this means 48 ^ (-1) mod 49  ..and which is equal to 49 ..

[quote=surge]I know this is a really old bump and all, but I want to do the AIME problems systematically. Can someone explain what one has to divide to get 35?  :huh:[/quote]

[quote=mysmartmouth]Old topic, but I thought this was a particularly clever solution, and needed to be added.  It's probably the fastest way I've seen to do this problem.

[hide]
We are looking for $ 6^{83} \plus{} 8^{83} \pmod{49}$
We know $ \phi(49)\equal{}42$ so we can subtract out 42 from the exponents:
$ 6^{41} \plus{} 8^{41} \pmod{49}$
And again:
$ 6^{\minus{}1} \plus{} 8^{\minus{}1} \pmod{49}$

So now this becomes a question of finding inverses mod 49.
To find the inverse of $ a \bmod b$, we can solve the Diophantine equation $ ax \plus{} by \equal{} 1$.  The value we get for $ x$ is the inverse we are looking for.

So looking at $ 6x \plus{} 49y \equal{} 1$, we quickly see $ x\equal{}\minus{}8$ and $ y\equal{}1$.
Looking at $ 8x\plus{}49y\equal{}1$, we see $ x\equal{}\minus{}6$ and $ y\equal{}1$.

So we can simplify

$ 6^{\minus{}1} \plus{} 8^{\minus{}1} \equiv \minus{}8 \plus{} \minus{}6 \equiv \boxed{35}  \pmod{49}$
[/hide][/quote]

[quote=cdymdcool][quote="mysmartmouth"]Old topic, but I thought this was a particularly clever solution, and needed to be added.  It's probably the fastest way I've seen to do this problem.

[hide]
We are looking for $ 6^{83} \plus{} 8^{83} \pmod{49}$
We know $ \phi(49) \equal{} 42$ so we can subtract out 42 from the exponents:
$ 6^{41} \plus{} 8^{41} \pmod{49}$
And again:
$ 6^{ \minus{} 1} \plus{} 8^{ \minus{} 1} \pmod{49}$

So now this becomes a question of finding inverses mod 49.
To find the inverse of $ a \bmod b$, we can solve the Diophantine equation $ ax \plus{} by \equal{} 1$.  The value we get for $ x$ is the inverse we are looking for.

So looking at $ 6x \plus{} 49y \equal{} 1$, we quickly see $ x \equal{} \minus{} 8$ and $ y \equal{} 1$.
Looking at $ 8x \plus{} 49y \equal{} 1$, we see $ x \equal{} \minus{} 6$ and $ y \equal{} 1$.

So we can simplify

$ 6^{ \minus{} 1} \plus{} 8^{ \minus{} 1} \equiv \minus{} 8 \plus{} \minus{} 6 \equiv \boxed{35} \pmod{49}$
[/hide][/quote]

What do you mean by $ \phi(49) \equal{} 42$[/quote]

[quote=RamanConjecture][quote=towersfreak2006][hide]That's really nice. You can also finish it by
\[ 6^{ \minus{} 1} \plus{} 8^{ \minus{} 1} \equal{} \frac {8 \plus{} 6}{48}\equiv\frac {14}{ \minus{} 1}\pmod{49} \equal{} \minus{} 14\equiv35\pmod{49},
\] instead of finding inverses.
[/hide][/quote]

Massive bump, but how is 14/48 congruent to -14 in mod 49?[/quote]

[quote=targo___]but is it a correct way? 48 is below 14 , so how can we say this? 14/48 is congruent to -14 mod 49 ?[/quote]

[quote=Paragdey12]$\phi (49)=42$
$6^{–1}+8^{–1}\equiv 35 (mod49)$[/quote]

[quote=OlympusHero]hello,,, i saw one person on blog clowning this problem with very hard method,,, so this must be fixed,,, we will fix it now

[hide=Solution]From Euler's Theorem we have $6^{42} \equiv 1 \pmod {49}$ and $8^{42} \equiv 1 \pmod {49}$. This means our expression is equal to $6^{-1}+8^{-1} \equiv 7 \cdot 24^{-1} \pmod {49}$. Note that $24 \cdot 2 \cdot -1 \equiv 1 \pmod {49}$, so $24 \cdot 47 \equiv 1 \pmod {47}$. We now just need $7 \cdot 47 \pmod {49} = \boxed{35} \pmod {49}$.[/hide][/quote]

Every single post is a bump lol

Nice, problem, this was in my AMSP class. I took session 1.

We have $\phi(49)=42$, so by Euler's, $6^{83} + 8^{83} \equiv 6^{-1} + 8^{-1} \equiv 41 + 43 \equiv 35 \pmod{49}.$

this is an okayish problem having the euler totient function
 and at last we  have 35

[hide]Substituting $6=7-1$ and $8=7+1$ and using the binomial theorem will do the trick but Euler's totient offers a quicker solution. We know that $6^{84}=(6^{42})^2= 1 \mod 49$ and $8^{84}=(8^{42})^2= 1 \mod 49$. Therefore, $6^{83} = 6^{-1} \mod 49$ and $8^{83} = 8^{-1} \mod 49$. Finally, we can find that $6^{-1}+ 8^{-1}=41 + 43 \mod 49 = 35 \mod 49$.[/hide]

Actually everyone above is wrong

It's $\varphi$ not $\phi=\frac{1+\sqrt{5}}{2}$
:nerd:

Nope... both work... you're the one who's wrong...


geometry
3D geometry
calculus
calculus computations

My teacher gave me a related rates worksheets where we apply derivatives. I have solved all of them except this one:

Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 8 cm and depth of 10cm. As you suck on the straw, root beer leaves the cup at the rate 7 cm^3/sec. At what rate is the level of the liquid in the cup changing when the liquid is 6cm deep?

[quote="skimnc"]Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of $ 8 ~\text{cm}$ and depth of $ 10 ~\text{cm}$. As you suck on the straw, root beer leaves the cup at the rate $ 7 ~\frac {\text{cm}^3}{\text{sec}}$. At what rate is the level of the liquid in the cup changing when the liquid is $ 6 ~\text{cm}$ deep?

If my final answer is wrong I apologize, its been a while since I've had to do these.  I'm fairly positive it is correct though.

[b]Hint:[/b]
[hide]If you were able to solve others you should note that the radius is directly proportional to the height in a cone.  That is, $ r \equal{} k\cdot h \implies 4 \equal{} 10k \implies k \equal{} \frac {2}{5}$.  So $ \boxed{\frac {r}{h} \equal{} \frac {2}{5}}$ and use this in solving.[/hide]

[b]Solution:[/b]
[hide]Note that $ \frac {dV}{dt} \equal{} \minus{} 7$

$ V \equal{} \frac {1}{3} \pi r^2 h$
$ V \equal{} \frac {1}{3} \pi \left( \frac {2h}{5} \right)^2 h$
$ V \equal{} \frac {4 \pi h^3}{75}$

Thus:
$ \frac {dV}{dt} \equal{} \frac {4 \pi}{75} \cdot 3h^2 \cdot \frac {dh}{dt}$
$ \minus{} 7 \equal{} \frac {4 \pi}{25} \cdot 6^2 \cdot \frac {dh}{dt}$
$ \boxed{ \minus{} \frac {175}{144 \pi} \equal{} \frac {dh}{dt}}$[/hide][/quote]

I just checked over your work. Nice job!

Okay, thank you very much. Makes sense now. I just wasn't sure where to begin, and I'm still a little shaky on implicit differentiation.


I know nothing about c++ programming.Yet I need to program unknown math into a computer program.Can anyome help me
Thanks
unknowm math

You might want to ask some more specific questions to make it easier for others to help you. :wink:
Thanks.

Hi
What I need is to program numbers in the computer .As of now I have done this in excel.I had one of the best programmer in the world from a small island .He was over Boland and Microsoft as I said one of the best in the world.

He knew what I had really worked But the trust was only one way .I trusted him to do the right thing but he had stops in the program and when I found out I told him he was out. I know I would lose one of the best programmer in the world but I needed to have somebody that I could trust and he didn't fit that bill.

What the program can do is to pick an event before it will happen.This program would be worth many millions of dollars when finish. It's a self proven system that must work because it only uses only the Math Laws.The concept I invented myself.

I have the lay out still, so it wouldn't be that hard to finish.Who ever can finish it may use it at the end to keep.
Thanks


IMO LongList 1967 to 1977 problems and solutions.


Let a,b are natural numbers for which a|(b+1) and b|(a^2 -2). Prove that (b+1)/2 is a square of natural number.

 :flex:

The first condition implies the existence of integer $ c$ such that $ b\equal{}ac\minus{}1,\frac{b\plus{}1}{2}\equal{}\frac{ac}{2}$. So the second condition implies that $ ac\minus{}1$ divides $ a^2\minus{}2$. Therefore, $ ac\minus{}1$ also divides
\[ a(ac\minus{}1)\minus{}c(a^2\minus{}2)\equal{}2c\minus{}a\]
so that $ 2c\minus{}a\equal{}d(ac\minus{}1)$ for some integer $ d$. As $ a$ and $ b$ are positive, $ c$ is positive, and $ a\equal{}c\equal{}1$ does not satisfy the equation for any $ d$. So $ ac\minus{}1$ positive. We then have
\[ d\plus{}1\equal{}\frac{2c\minus{}a}{ac\minus{}1}\plus{}1\equal{}\frac{ac\minus{}a\plus{}2c\minus{}1}{ac\minus{}1}\equal{}\frac{(a\plus{}2)(c\minus{}1)\plus{}1}{ac\minus{}1}>0\]
and (unless $ a\equal{}1$)
\[ 1\minus{}d\equal{}1\minus{}\frac{2c\minus{}a}{ac\minus{}1}\equal{}\frac{ac\plus{}a\minus{}2c\minus{}1}{ac\minus{}1}\equal{}\frac{(a\minus{}2)(c\plus{}1)\plus{}1}{ac\minus{}1}>0\]
So there are two cases: $ a\equal{}1$ or $ d\equal{}0$. In the first case, $ c\minus{}1$ divides both $ 2c\minus{}1$ and $ 2c\minus{}2$, therefore 1, so that $ c\equal{}2$ giving $ \frac{ac}{2}\equal{}1$, a square. In the second, $ a\equal{}2c$ so $ \frac{ac}{2}\equal{}c^2$, as desired.


A guidance counselor is planning schedules for 30 students. Sixteen students say they want to take French, 16 want to take Spanish, and 11 want to take Latin. Five say they want to take both French and Latin, and of these, 3 wanted to take Spanish as well. Five want only Latin, and 8 want only Spanish. How many students want French only?
1 day ago - 2 days left to answer.

Are you going to get in trouble for cheating? It appears as though you've copy-pasted this from some sort of online math course, and have forgotten to delete the thing saying how long you have.

yeah, I had the same thoughts from the "1 day ago - 2 days left to answer"  How often are there deadlines for problems posted on AoPS?  Some problems probably have been discussed for a very long time before the answer was final, but all of a sudden there's this

This isn't even a hard problem. Just use deductive reasoning.

You might find [url=http://www.artofproblemsolving.com/Wiki/index.php/PIE]PIE[/url] useful in solving this type of problem.


blogs ARML AMC USA(J)MO USAMO HMMT MATHCOUNTS

currently, I've gotten 2 people in my school district to join AoPS.  4 more are joining very soon.  And once my math club gets rolling, my blog will be used to detail what goes on at each meeting and what to expect, so everyone will have to join.

Of course, none of these people are active.... :sad:

Nope. Not a single one. Not even me.
lol only one answer to the poll even makes sense for those who can answer it.

Come on, you know what he means :P.  Or maybe some people would have to answer "What's AoPS?" :D

um...... me???? duh.... but no one else, and I am not that interested in bringing others either, since they dont want to do, or be interested in the sort of competitions that I like. I dont like my school's policy of prioritizing Mu Alpha Theta competition, rather than something like ARML or USAMO, or even some other ones like HMMT or BAMO (although we cant participate). I am  beginning to restrain from letting my classmates know about the math i do out of school, because I know that they will soon be disinterested in the concept of Olympiad and hard math, and just be focused on Class subjects and et cetera, which I think in my school has a slow curriculum anyways.

So, no other kids from our school are AoPSer.

well, no one around at my school has an ounce of math talent (includes me) as only 1 person (or 2 or 3 max) got over 100 on the AMC's :( I don't tell anyone else about AoPS as they don't really care, but oh well... our math club isn't very constructive; we're more concerned with rearranging the keys on the keyboard than learning math :D

A couple people from my school are on AoPS, though it's not because I told them. I really don't do much to promote AoPS because I get the feeling that not very many people would appreciate what this community has to offer. However, our math team is pretty strong, so at least I'm not too alone.

I think I'm the only one from my school here, *shrugs* our math team is pretty good, but there aren't too many people who really do it because they like to do it, mostly it's because they have to or because they think colleges will like it *shrugs*

I know some people that are on AoPS as users, but they hardly ever post on the forum. They only have an AoPS account for USAMTS.

sheepwarrior made me get an account -_-;;

I am a "wanna be". I just talk on the forum. I have never read the AoPs books, I don't kinow why.  Another person at my school bought both AoPS books, and read both.  We are #1 and #2 in our school in math.

None that I know of.  It is a little too bad because there are alot of strong math kids in my school.  Although I do know of a couple kids who bought the Aops books and seem to like them.

I'm a newbie here and our school has no math team or math club or anything math related.. except for math classes.  Sucks.

You live in Calif and your school has no math team?  Where in California do you live????  Certainly not the SFBA, right?  Maybe LA or something...

heh, i live in a town...so I'm the only one in the town somewhat interested in math.  sad eh

[quote="bubala"]yeah, well how big is your town?  You live in Alberta...[/quote]

Middle of no where alberta actually.  15 000 people in my town, heh.  I'm lucky though, I live incredibly close to the city that its not a problem living in Okotoks

aznness/uberl33tmage and jimli go to the same school as I do. There's also someone who lurks occasionally but doesn't have an account. There may be others, I wouldn't know.

[quote="ambierona"]I used to live in SFBA! When I was like 2! But now I live in southern California.. in Santa Ana.. in Orange County, yeah, around LA.[/quote] Arent there still good math schools in SoCal like Mark Keppel?

[quote="math92"]My sister "colts18" goes to my school.[/quote]
Yes that is true.

Even though Edmonton has population of 1 million, I am the only AoPSer(I'm quite sure about this) :( . So, I assume no one at my school would be AoPSer....

I know two people in Aops in my school uberl33tmage and lynelle ye.  We all were on a Mathcounts team.  I have both Aops books but no independent study.  ccy is also a person in my school but he's not really on our team or anything much.

[quote="furious"]joml88 introduced me and some other members to AoPS. actually he tortured me until i joined. now our entire team is registered although most ppl only post in the blog, if ever[/quote]

Hehe, that just about says it for me....except I dont' remember "torturing" furious (to join AoPS that is... ;) )

ast92391 goes to my school, but I think he only joined cause our mathcounts coach said we should go to AoPS. ineedmathhelp used to go to my school, and will go to my school next year.

[quote="jb05"]lol only one answer to the poll even makes sense for those who can answer it.[/quote]

Technically, a MathLinkser who never looks at the top right corner could answer option 3.

I have a strange feeling that a decent amount of kids know about AoPS in miami...

I'm the only one at my school who's any good at competitive math. :( There are many people who are good at "school math" as in Algebra II, or whatever they're in, but they do not know... [hide]The Art of Problem Solving![/hide] So I'm the only AoPSer at my school. There's another kid I'm trying to convince, but...

I'm the only AoPSer who goes to my school.  I've told a couple people, but I don't think any of them signed up.

[quote="mathbomberII"]i think math92 should of put i don't know in there,too because i don't know, but i'm preety sure that there isn't anybody in my school[/quote]
colts18???she goes to my school

I'm probably the only one here.  There is another kid that I think is really good at problem solving type of stuff (he got a 97 on the AMC 10 without even knowing what it was 15 minutes before he took it).\, but I don't think he would want to join.

[quote="Scrambled"][quote="yif man12"]There's one other at my school. Of course, there aren't that many because we're mainly a humanities school...[/quote]
that would be jsrosen right?

In our school, everyone in our math enrichment class has to sign up. But there are only a handful of people who actually want to be here[/quote]

 :)  Yup. Though he doesn't really post anymore. He comes and looks around once every month or so.

We only have two people in Shanghai American School. Me and sheepwarrior.


symmetry
ratio
geometry
3D geometry
icosahedron
group theory
superior algebra

The symmetry group of a collection of points $ C$ in the plane is the set of orthogonal transformations in $ \mathbb{R}^2$ that restrict to a bijection on $ C$.  For example, the symmetry group of the vertices of a regular $ n$-gon is the dihedral group $ D_n$.

What is the maximal size of the symmetry group of a collection of $ n$ points?  (Can one do better than the dihedral group?)

How does the answer change in three dimensions?  $ d$ dimensions?

In two dimensions, we can embed the symmetry group in the dihedral group on the convex hull.

In three dimensions, the largest ratio of symmetry group size to set size comes for the icosahedron, with 12 vertices and 120 elements in the group. We're limited by the low number of regular polyhedra, and for large $ n$ the best possible is about 4n: the dihedral group plus flips, achieved for a double pyramid.

We should require that the collection encloses a $ d$-dimensional volume; otherwise, we get large groups acting on the empty dimensions.

You might want to read [url=http://en.wikipedia.org/wiki/Coxeter_group]here[/url] for more information on finite symmetry groups.


A pyramid with a square base is cut by a plane that is parallel to its base and is $ 2$ units from the base. The surface area of the smaller pyramid that is cut from the top is half the surface area of the original pyramid. What is the altitude of the original pyramid?

$ \textbf{(A)}\ 2\qquad
\textbf{(B)}\ 2 \plus{} \sqrt{2}\qquad
\textbf{(C)}\ 1 \plus{} 2\sqrt{2}\qquad
\textbf{(D)}\ 4\qquad
\textbf{(E)}\ 4 \plus{} 2\sqrt{2}$

[hide]Let $ h$ be the altitude of the original, and $ h \minus{} 2$ altitude of the smaller pyramid. The smaller pyramid and the original one are similar with coefficient $ k \equal{} \frac {h \minus{} 2}{h}$. Let surface area of the original be $ S_1$ and of the small one $ S_2$. Since all the areas are proportional with coefficient $ k^2 \implies S_2 \equal{} k^2S_1$. Let's substitute $ S_1$ using $ S_1 \equal{} 2S_2$ identity. We get $ S_2 \equal{} 2k^2S_2 \implies k^2 \equal{} \frac {1}{2}$. Using $ k \equal{} \frac {h \minus{} 2}{h}$, we get $ h \equal{} 4 \plus{} 2\sqrt {2}$.

The answer is $ \textbf{(E)}$.[/hide]

[hide="alternate solution"]
The two pyramids are similar therefore if their suface area has a raito $ \frac{1}{2}$, there legths have a ratio of $ \frac{1}{\sqrt{2}}$

$ \Rightarrow k\minus{}\frac{1}{\sqrt{2}}k \equal{} 2 \Leftrightarrow k \equal{} \frac{2\sqrt{2}}{\sqrt{2}\minus{}1}\cdot \frac{\sqrt{2}\plus{}1}{\sqrt{2}\plus{}1} \equal{} 4\plus{}2\sqrt{2} \longrightarrow \boxed{E}$[/hide]


I have an arithmetic sequence. The first term of the first sequence is $ 0$. The second term of the first sequence is the first term of the first sequence plus the first term of the second sequence. Similarly, the third term of the first sequence is the second term of the first sequence plus the second term of the second sequence. If the fifth term of the second sequence is $ 3$, what is the fifth term of the first sequence?

Suppose the first sequence went up by $ n$ - that is, the first term is 0, the second is $ n$, the third $ 2n$, and so on. Similarly, let the first term of the second sequence be $ k$, and let it go up by intervals of $ d$, so that the second term is $ k \plus{} d$ and the third $ k \plus{} 2d$.

We can write some equations: $ n \equal{} 0 \plus{} k$, from the second sentence, which implies that $ n \equal{} k$.

Now, the third sentence gives us $ 2n \equal{} n \plus{} k \plus{} d$, or $ n \equal{} k \plus{} d$, meaning $ 0 \equal{} d$. From the last sentence, we see that $ k \equal{} 3$, so the second sequence looks like 3, 3, 3, ...

Since $ n \equal{} k$, that means $ n \equal{} 3$, so the fifth term of the first sequence is $ 4 \times 3 \equal{} \boxed{12}$.


[hide]I did it by drawing a triangle with side lengths $\sqrt 5$, $2\sqrt5$, and 5...
$ \sin (\cos^{-1}  \frac {\sqrt 5}{5}) = \frac {2 \cdot \sqrt 5}{5}\\
a+b+c = 2 + 5 + 5 = 12$[/hide]

[quote="Ravi B"]The problem should have imposed some restrictions on a, b, and c, such as they are all integers, b is square-free, and a and c are relatively prime.[/quote]

Not to mention the sign .. as there can be two answers.

it is implied that $\cos^{-1} x$ is the principal value...
on another note, verify $\tan^2 x+1=\sec^2 x$.

[quote="chenwb"]on another note, verify $\tan^2 x+1=\sec^2 x$.[/quote]

Multiplying through by $\cos^2{x}$ gives the well-known trig identity $\sin^2{x}+\cos^2{x}=1$.

[hide][quote="Silverfalcon"]Here is an interesting but easy challege.

Please do without using calculator.

$\sin (\cos^{-1}  \frac {\sqrt 5}{5}) = \frac {a \cdot \sqrt b}{c}$

Where a,b, and c don't necessarily represent the different numbers.

Find $a+b+c$.[/quote]

$\frac{\sqrt 5}{5}=\frac{1}{\sqrt{5}}$

the hypotenus is $\sqrt 5$

the other side is $\sqrt{5-1}=2$

so $\sin \theta=\frac{2}{\sqrt 5}=\frac{2\cdot \sqrt 5}{5}$

$2+5+5=12$[/hide]

$\tan^2 x+1=\sec^2 x$

$\frac{opposite^2}{adjacent^2}+\frac{adjacent^2}{adjacent^2}=\frac{hypotenus^2}{adjacent^2}$

$\frac{opposite^2+adjacent^2}{adjacent^2}=\frac{hypotenus^2}{adjacent^2}$

$\frac{hypotenus^2}{adjacent^2}=\frac{hypotenus^2}{adjacent^2}$

TADA!


Find all $ n \in \mathbb{N}$ such that :
$ n2^{n\minus{}1} \plus{} 1$ is a perfect square.

[hide]Let the square be $ x^2$.  Then $ x^2\minus{}1\equal{}(x\plus{}1)(x\minus{}1)\equal{}n(2^{n\minus{}1})$.  But $ x\plus{}1$ and $ x\minus{}1$ are different mod 4, so one of them must not be divisible by 4, so one of them is divisible by $ 2^{n\minus{}2}$.  Thus x is at least $ 2^{n\minus{}2}\minus{}1$, so $ x^2\minus{}1\geq(x\plus{}1)(x\minus{}1)\equal{}2^{2n\minus{}4}\minus{}2^{n\minus{}1}$, so $ 2^{n\minus{}3}\minus{}1\leq{n}$.  Thus n<6, so we just have to try n=1, 2, 3, 4, 5.  n=1 yields 2, not a square.  n=2 yields 5, also not a square.  n=3 yields 13, also a nonsquare.  n=4 yields 33, a nonsquare, and n=5 yields 81, which is a square.  Thus the only solution is n=5.[/hide]


function
integration
calculus
real analysis
real analysis unsolved

I got this problem on an exam today: let $f : (a,b) \to \mathbb{R}$ be a function, with $a,b \in \mathbb{R} \cup \{\pm \infty\}$, which is derivable and convex on $(a,b)$. Prove that it has limits in $a$ and $b$.

[hide="Hint"]Use the fact that $f'$ exists and is non-decreasing, thus now allowing random 0-s to appear.[/hide]

Really quite straightforward. Pick $a_n\rightarrow a$. Then we have $f(a_n)=f(x)-\int_{a_n}^xf'(t)dt$. Since the integrals converge ($f'$ is non-decreasing) the left-hand side converges too. The same argument works for $b$.

Misha


linear algebra
matrix
linear algebra unsolved

Let $ A \equal{}[a_{i,j}]_{nxn}$ satisfying the following condition:

                                                      $ a_{i,j}\equal{}0$ if $ i\equal{}j$
                                           and
                                                      $ a_{i,j}\equal{}1$ or $ p$ if $ i$#$ j$. where $ p$ is a prime number $ >2$.
   Prove that : $ rank(A) \geq {n\minus{}1}$

You don't need $ p$ to be prime.
$ A\plus{}I_n$ or $ A\plus{}pI_n$ has rank 1, hence $ \minus{}1$ or $ \minus{}p$ is an eignevalue with geometric multiplicity $ n\minus{}1$, as a result $ 0$ is at most an eigenvalue with geometric multiplicity $ 1$, hence $ \mathrm{rank} A\geqslant n\minus{}1$.

JC_math, why "$ A \plus{} I_n$ or $ A \plus{} pI_n$ has rank 1"? The matrix

$ A \equal{} \begin{bmatrix} 0 & 3 & 1 \\
1 & 0 & 3 \\
3 & 3 & 0 \\
\end{bmatrix}$

is in the conditions of the problem and $ rank(A \plus{} I) \equal{} rank(A \plus{} 3I) \equal{} 3 \equal{} rank(A)$.

Sorry but I read this
[quote="minhthutt"] $ a_{i,j} \equal{} 1$ or $ p$ if $ i$#$ j$. where $ p$ is a prime number $ > 2$.[/quote]
diiferently, i.e. all the $ a_{i,j}$ for $ i\neq j$ are either equal to 1 or $ p$.

Here is another approach.
Consider $ A$ as an element of $ \mathcal{M}_n(\mathbb{Z})$, then reduce in $ \mathbb{Z}/(p\plus{}1)\mathbb{Z}$, you have $ \mathrm{rank}_{\mathbb{Z}/(p\plus{}1)\mathbb{Z}} (A) \equal{} n\minus{}1$, hence $ \mathrm{rank}_{\mathbb{Z}} (A) \geqslant n\minus{}1$,...

I have different version of solution, please check whether my answer is true (or not)
[hide="answer"]
Look the matrix over $ Z_2$ (So that if the number is even we give value $ 0$ and if the number is odd we take the value $ 1$), then the matrix will be : $ A\equal{}{a}_{ij}$ where $ {a}_{ij}\equal{}0$ for all $ i\equal{}j$ and $ {a}_{ij}\equal{}1$ if $ i$ isn't equal to $ j$ (By noting that all prime that is greater than $ 2$ is an odd number.
By some elementary row operations we can obtain that for the first $ n\minus{}1$ columns in that matrix we have that $ {a}_{ij}\equal{}1$ if $ i\equal{}j$ or $ i\equal{}j\plus{}1$ and $ 0$ in the other place (in the first $ n\minus{}1$ columns in $ A$).Check that all of the $ n\minus{}1$ rows is linearry independent so $ rank({A}_{{Z}_{2}}) \geq n\minus{}1$ and $ rank(A) \geq rank({A}_{{Z}_{2}})$ so that $ rank(A) \geq n\minus{}1$. So you don't need $ p$ to be prime number greater than $ 2$, you just need $ p$ to be an odd number.
I am sorry if my answer is wrong.
[/hide]


[quote="ch1n353ch3s54a1l"][quote="nonie"]Television screen sizes are the diagonal length of the rectangular screen. Hector recently changed from watching television witha 13-inch screen to a television with a similar 19-inch screen. If a boxcar appeared 8 inches long on the 13-inch screen how long, to the nearest inch, will it appear on the 19-inch screen?

source: 2005-06 Preparing for the ACT booklet.[/quote]

[hide]Uh...
NOt sure if i misunderstood this problem....
$\frac{13}{8}=\frac{19}{x}$
$x\approx 12$[/hide][/quote]

No , you're right.  :)

it is 12 inches

you need a proportion . 8/13= x/19
cross multiply
19x8= 152
then divided 152/13= 11.69 round to 12 :D

Yep, they've covered it above. Also, some notes on similar figures:
-You can use any common linear measure to set up a proportion (like we did in this problem)
-You can use the square of any common linear measure to relate areas. So if you were asked the ratio of the areas of the two screens, it would be (19/13)^2.
-You can use the cube of any common linear measure to relate volume (if you have similar prisms, for instance).

Those can be useful and can save a lot of work.


Hi I was wondering if there are any good books out there with a collection of differential equations problems which include ODEs and PDEs.  Currently, the only problem book that I have which have some differential equations problems is the "Berkeley Problems in Mathematics".  I was wondering if there are any other books that people here can recommend. Thanks.

Do you want exclusively a problem book devoted to differential equations?  I'm not sure one of those exists.  If you pick up a differential equations textbook, chances are, you'll get far more problems that just a problem book.

my class is currently using Partial Differential Equations by Evans and its a pretty reader friendly book. although i cant compare it to other books of its kind yet, i like it.


algebra
polynomial
linear algebra
matrix

Have anyone know if there is in Linear Algebra a theorem like this:
"Two matrices are similar [u]if and only if[/u] have the same minimal & charecteristic polynomials"
I need this theorem (if exists) in order to solve an exercise.
If exists a such theorem could someone post a link to its proof??

Thanks in advance

This doesn't hold. What I posted in http://www.mathlinks.ro/viewtopic.php?t=175802 may be the best result of such kind which holds over arbitrary fields. I think some other topic gives a detailed proof for the algebraically closed field case, but I can't find it.

  darij

[quote="hsiljak"]It does hold for $ 3\times 3$ matrices, if it can help you, but for $ n\times n$, $ n\geq 4$, no. (Just like Darij said, and shown in the linked thread)[/quote]

My case is 2x2 matrices. does it hold?? and Why??

Thank you

Yes, it does hold. I was always bad at algebra (mental note: study algebra this summer), but I think you could prove it on Jordan blocks... Correct me if I'm wrong  :)

The $ 2 \times 2$ case is degenerate because there are only two possible Jordan block decompositions:  one block or two blocks.  In the one-block case, the characteristic and minimal polynomials are both $ (x \minus{} \lambda)^2$.  In the two-block case, the characteristic and minimal polynomials coincide if eigenvalues are distinct and the minimal polynomial is of degree $ 1$ if they aren't.  That is, the map from Jordan blocks to characteristic and minimal polynomials is injective, which isn't true in general.

For what it's worth, it's pretty easy to show that a statement of this form can't hold for sufficiently large $ n$:  the number of Jordan block decompositions for an $ n \times n$ matrix with a single eigenvalue of multiplicity $ n$ is $ B_n$, which grows faster than exponentially in $ n$, whereas knowing characteristic and minimal polynomials only gets you polynomial information in $ n$.


advanced fields
advanced fields unsolved

Why is a compact operator automatically bounded?

Because the image of the unit ball is contained in a compact set and compact sets are always bounded.

and you should also ask yourself why it is not invertible when working with infinite dimensional spaces ..


Prove that :

$ \cos^7\ x \plus{} \cos^7\ \left(x \plus{} \frac {2\pi}{3}\right) \plus{} \cos^7 \ \left(x \plus{} \frac {4\pi}{3}\right) \equal{} \frac {63}{64}\cos \ 3x\ , \ (\forall)x\in \mathbb{R}$

[quote="Dr Sonnhard Graubner"]hello, i think it must be
$ \cos^7(x) \plus{} \cos^7(x \plus{} \frac {2\pi}{3}) \plus{} \cos^7(x \plus{} \frac {4\pi}{3}) \equal{} \frac {63}{64}\cos(3x)$
Sonnhard.[/quote]

You're right . It was an idiotic typo of mine .

$ a\equal{}cos(x),b\equal{}cos(x\plus{}\frac{2\pi}{3}),c\equal{}cos(x\plus{}\frac{4\pi}{3})$ are roots of $ f(y)\equal{}4y^3\minus{}3y\minus{}cos(3x)\equal{}0$.
Therefore $ P_m\equal{}a^m\plus{}b^m\plus{}c^m\equal{}cos^mx\plus{}cos^m(x\plus{}\frac{2\pi}{3})\plus{}cos^m(x\plus{}\frac{4\pi}{3})$ satisfyed
\[ P_{m\plus{}3}\minus{}3P_{m\plus{}1}\minus{}cos(3x)P_m\equiv 0.\]
We had $ P_0\equal{}3,P_1\equal{}0,P_2\equal{}(a\plus{}b\plus{}c)^2\minus{}2(ab\plus{}bc\plus{}ca)\equal{}0\minus{}2(\minus{}3/4)\equal{}3/2.$
Therefore $ P_7\equal{}\frac 14(3 P_5\plus{}cos(3x)P_4)\equal{}\frac{1}{16}(9P_3\plus{}6cos(3x)P_2\plus{}cos^2(3x)P_1)$ or
$ P_7\equal{}\frac{9}{16}(P_3\plus{}cos(3x))\equal{}\frac{9}{16}(\frac 34 \plus{}1)cos(3x)\equal{}\frac{63}{64}cos(3x)$.


Let x, y, and z be positive real numbers with $ x \plus{} y \plus{} z \equal{} 1$. Prove that
\[ \frac{xy}{\sqrt{xy\plus{}yz}} \plus{} \frac{yz}{\sqrt{yz\plus{}zx}} \plus{} \frac{zx}{\sqrt{zx\plus{}xy}} \leq \frac{1}{\sqrt{2}}.\]

$ \sum \frac{xy}{\sqrt{xy+yz}} = \sum x \sqrt{ \frac{y}{1-y}} \le \sqrt{\frac{\sum xy}{1-\sum xy}}$ (Jensen's on $ f(x) = \sqrt{\frac{y}{1-y}}$ with x,y,z as weights)

$ (\sum x)^2 \ge 3 \sum xy \Rightarrow \sum \xy \le \frac{1}{3}$

Hence $ \sqrt{\frac{\sum xy}{1-\sum xy}} = \sqrt{\frac{1}{1-\sum xy}-1} \le \frac{1}{\sqrt 2}$

For $ f(x) \equal{} \sqrt{\frac{x}{1\minus{}x}}$
$ f''(x) \equal{} \frac{4x\minus{}1}{4x^{3/2}(1\minus{}x)^{5/2}}$ is not concave on $ (0,1)$. Therefore you cannot apply Jensen's inequality.

[quote="MayankM"]Let x, y, and z be positive real numbers with $ x \plus{} y \plus{} z \equal{} 1$. Prove that
\[ \frac {xy}{\sqrt {xy \plus{} yz}} \plus{} \frac {yz}{\sqrt {yz \plus{} zx}} \plus{} \frac {zx}{\sqrt {zx \plus{} xy}} \leq \frac {1}{\sqrt {2}}.\]
[/quote]

$ \frac {xy}{\sqrt {xy \plus{} yz}} \plus{} \frac {yz}{\sqrt {yz \plus{} zx}} \plus{} \frac {zx}{\sqrt {zx \plus{} xy}}$

$ \le \frac {\sqrt{2}xy}{\sqrt {xy} \plus{} \sqrt{yz}} \plus{} \frac {\sqrt{2}yz}{\sqrt {yz} \plus{} \sqrt{zx}} \plus{} \frac {\sqrt{2}zx}{\sqrt {zx} \plus{} \sqrt{xy}}$

$ \equal{} \frac {xy\plus{}yz}{\sqrt{2}(\sqrt {xy} \plus{} \sqrt{yz})} \plus{} \frac {yz\plus{}zx}{\sqrt{2}(\sqrt {yz} \plus{} \sqrt{zx})} \plus{} \frac {zx\plus{}xy}{\sqrt{2}(\sqrt {zx} \plus{} \sqrt{xy})},$

\[ \sum {\frac{{xy \plus{} yz}}{{\sqrt {xy}  \plus{} \sqrt {yz} }}}  \le x \plus{} y \plus{} z\]
\[ \Leftrightarrow \sum {\left( {\sqrt x  \minus{} 2\sqrt y  \plus{} \sqrt z } \right)\left( {\frac{{z \plus{} x}}{{\sqrt z  \plus{} \sqrt x }}} \right)}  \ge 0\]
\[ \Leftrightarrow \sum {\left( {\sqrt x  \minus{} \sqrt y } \right)\left( {\frac{{z \plus{} x}}{{\sqrt z  \plus{} \sqrt x }} \minus{} \frac{{y \plus{} z}}{{\sqrt y  \plus{} \sqrt z }}} \right)}  \ge 0\]
\[ \Leftrightarrow \sum {\frac{{\left( {\sqrt x  \minus{} \sqrt y } \right)^2 \sqrt {xy} }}{{\left( {\sqrt z  \plus{} \sqrt x } \right)\left( {\sqrt y  \plus{} \sqrt z } \right)}}}  \ge 0.\]

i was kinda betting on f being concave  :lol:

Kuing I'm unable to follow this step.
\[ \frac{\sqrt{2}xy}{\sqrt{xy}\plus{}\sqrt{yz}}\plus{}\frac{\sqrt{2}yz}{\sqrt{yz}\plus{}\sqrt{zx}}\plus{}\frac{\sqrt{2}zx}{\sqrt{zx}\plus{}\sqrt{xy}}
\\
 \equal{}\frac{xy\plus{}yz}{\sqrt{2}(\sqrt{xy}\plus{}\sqrt{yz})}\plus{}\frac{yz\plus{}zx}{\sqrt{2}(\sqrt{yz}\plus{}\sqrt{zx})}\plus{}\frac{zx\plus{}xy}{\sqrt{2}(\sqrt{zx}\plus{}\sqrt{xy})},\]

[quote="MayankM"]Kuing I'm unable to follow this step.
\[ \frac {\sqrt {2}xy}{\sqrt {xy} \plus{} \sqrt {yz}} \plus{} \frac {\sqrt {2}yz}{\sqrt {yz} \plus{} \sqrt {zx}} \plus{} \frac {\sqrt {2}zx}{\sqrt {zx} \plus{} \sqrt {xy}} \\
\equal{} \frac {xy \plus{} yz}{\sqrt {2}(\sqrt {xy} \plus{} \sqrt {yz})} \plus{} \frac {yz \plus{} zx}{\sqrt {2}(\sqrt {yz} \plus{} \sqrt {zx})} \plus{} \frac {zx \plus{} xy}{\sqrt {2}(\sqrt {zx} \plus{} \sqrt {xy})},\]
[/quote]

\[ A \equal{} \sum {\frac{{xy}}{{\sqrt {xy}  \plus{} \sqrt {yz} }}} ,B \equal{} \sum {\frac{{yz}}{{\sqrt {xy}  \plus{} \sqrt {yz} }}}\]
\[ A \minus{} B \equal{} \sum {\frac{{xy \minus{} yz}}{{\sqrt {xy}  \plus{} \sqrt {yz} }}}  \equal{} \sum {\frac{{\left( {\sqrt {xy}  \plus{} \sqrt {yz} } \right)\left( {\sqrt {xy}  \minus{} \sqrt {yz} } \right)}}{{\sqrt {xy}  \plus{} \sqrt {yz} }}}  \equal{} \sum {\left( {\sqrt {xy}  \minus{} \sqrt {yz} } \right)}  \equal{} 0\]
\[ A \equal{} B\]


inequalities
analytic geometry
graphing lines
slope
absolute value

Alright, there's another problem that just stumped me this time.
I'm solving for $X$.

$|x + 1/3| = |x - 3|$
$-x...............-x$

$1/3 = -3$

:| That's not right....

The answer to the problem is $X=4/3$ but I don't see how that can be accomplished.

I got it solved, nevermind. I'd delete the thread if I could.

Are you sure? Why not post your solution?

Hehe, your subtracting x from both sides is pretty much just ignoring the absolute values all together. Why's the title have "inequalities" tho?

If one part of the absolution process does not work, that means there might be only one possible answer left for X.
There isn't always two answers you can get.
Either they can share the same answer, or there is only one answer for both and [b]not[/b] two answers by doing the problem [b]two different ways. [/b]

$|x + \frac{1}{3}| = |x - 3|$

SO:

$-x - \frac{1}{3} = x - 3$

$+x$

$-\frac{1}{3} = 2x - 3$

$+3$

$\frac{8}{3} = 2x$

$8/3/2 = \frac{2x}{2} = x = \frac{4}{3}$

$x = \frac{4}{3}$

Another way is by squaring both side ,

$x^2+\frac{2}{3}x+\frac 19 =x^2-6x+9$

Simplifying and get $x=\frac 43$  :D

Also, you can think of it graphically. Absolute value graphs are just where the line is "reflected" at the x-axis, cuz the y-value can't go below 0. They still keep their slope, so theres only one intersection of your two.


So much have been mentioned about the application of fractal geometry in physics. Anyone know if there are any applications of fractal geometry in economics?   :oops:

[url=http://classes.yale.edu/fractals/RandFrac/welcome.html]Random Fractals and the Stock Market[/url]
You can find more stuff like that with google. I'm not a big fan of it.

Application of fractal geometry in finance is well known.
for example, paths of Brownian Motion are fractals.
How about the other parts of economics apart from finance?
Do we have applications of fractal geometry in microeconomics or industrial economics? :roll:


Find all the real solutions $(x,y,z)$ of the system:

$xyz = 8$

$x^2y+y^2z+z^2x = 73$

$x(y-z)^2+y(z-x)^2+z(x-y)^2 = 98$

from the third equation , we have $xy^2+yz^2+zx^2 =73$
that implies $x^2y+y^2z+z^2x=xy^2+yz^2+zx^2$
thus $(x-y)(y-z)(z-x)=0$
and then it is easy to find roots


trigonometry
calculus
calculus computations

how would you sketch a phase portrait of

$ x''\equal{}\cos x$?

Thank you.

In Maple

[img]http://img508.imageshack.us/img508/9675/clipboard01hg0.jpg[/img]

 :roll:


[hide]
Define $x_k$ by $x_1=2$ and $x_{k+1}=x_k^2-x_k+1$.

Lemma 1: $\sum_{j=1}^{n-1} \frac{1}{x_j}=1-\frac{1}{x_n-1}$ for $n\geq 2$.

Proof: The base case checks because $\frac 12 =1-\frac 12 =1-\frac{1}{3-1}$ and $x_2=3$.

For the inductive step, $\sum_{j=1}^n \frac{1}{x_j}=1-\frac{1}{x_n-1}+\frac{1}{x_n}=1-\frac{1}{x_n(x_n-1)}=\frac{1}{x_{n+1}-1}$, so the identity holds.

Lemma 2: $x_{k+1}=1+\prod_{j=1}^k x_j$.

Proof: Again, clearly visible for $k=2$. The inductive step works because $x_{k+2}=1+x_{k+1}(x_{k+1}-1)=1+x_{k+1}\prod_{j=1}^k x_j=1+\prod_{j=1}^{k+1} x_j$.

Let $a_j=x_j$ for $j\leq n-1$. Applying Lemma 1, $\sum_{j=1}^n \frac{1}{a_j}=1-\frac{1}{x_n-1}+\frac{1}{a_n}$. The right side of the original equation equals $1+\frac{1}{(x_n-1)a_n}$ from Lemma 2.

Subtracting off 1 and multiplying to clear fractions, $x_n-1-a_n=1\Rightarrow a_n=x_n-2$.

Therefore the n-tuple $(x_1,x_2,\cdots ,x_{n-1},x_n-2)$ is a solution for any $n\geq 2$.
[/hide]

Thank you, nice solution. I think that Rust's construction almost works after all since you can take $(a_1,a_2,...a_{n-1},a_n-2)$, which have the desired property.

There are only one solution (obviosly):
$a_1=2,a_k=\prod_{i=1}^{k-1}a_i +1,1<k<n,a_n=\prod_{i=1}^{n-1}a_i -1.$, were $a_1<a_2<\dots a_{n-1}$,


fie A,B 2 matrici din Mn (R)  {n..matrice patratica de gradul n}
    se stie ca A^m*B=A^m +b  ,m numar natural nenul
se cere: A*B=B*A

Fie A,B 2 matrici din $M_n(\Re)$
Se stie ca $A^m*B=A^m +B$ ,m natural nenul
Se cere: $A*B=B*A$

Let A and B be 2 square matrices from M n (R)

and A^m * B =A^m + B ,m from N

Proove that A*B=B*A

...hope i got it well  :) [/quote][/code]

$A^m\cdot B=A^m +B$ atunci $A^m(B-I_n)=B$ ceea ce se poate scrie ca $(A^m-I_n)(B-I_n)=I_n$ adica $A^m$ si $B$ comuta deci $B$ se scrie ca polinom in  $A^m$ care e de fapt polinom in $A$ asa ca $AB=BA$

Mie partea cea mai grea mi se pare demonstrarea ca daca $A^m$ si $B$ comuta atunci $A$si $B$ comuta... desi nu e prea greu(cred). Anyway ... din $(A^m-I_n)(B-I_n)=I_n$ ar trebui sa iti dai seama ca, comuta(in acelasi timp $(B-I_n)(A^m-I_n)=I_n$ , poti sa faci calcule  ;) )

[quote="siriusra"]e evident ca daca A^m comuta cu B atunci A comuta cu B....cum ai ajuns insa la concluzia ca A^m comuta cu B ?[/quote]Chiar sunt curios cum anume arati tu ca daca $A^m$ comuta cu $B$ atunci $A$ comuta cu $B$ ...  :?

Imi dau seama ca am gresit mai sus ... probabil ca din graba  :blush: abordare corecta : din $(A^m-I_n)(B-I_n)=I_n\longrightarrow (A-I_n)X(B-I_n)=I_n$ unde $X=A^{m-1}+A^{m-2}+...+A+I_n$ este vizibil ca $X$ si $A-I_n$ comuta deci $X(A-I_n)(B-I_n)=I_n$ si pe de alta parte $(B-I_n)(A-I_n)X=I_n$. Din relatiile anterioare inmultind cu $X^{-1}$ in ambele parti obtinem ca $(A-I_n)(B-I_n)=(B-I_n)(A-I_n)$ de unde se obtine simplu ca $AB=BA$.

m-am prins....sa cresti mare..
valentin....se arata ca daca matricile sunt de ordin 2 si daca A^m*B^m=B^n*A^m rezulta ca A*B=B*A...pt ordin mai mare imi dau seama inductiv...chiar daca nu e riguros...acuma gandesc pt n

[quote="siriusra"]valentin....se arata ca daca matricile sunt de ordin 2 si daca A^m*B^m=B^n*A^m rezulta ca A*B=B*A...pt ordin mai mare imi dau seama inductiv...chiar daca nu e riguros...acuma gandesc pt n[/quote]Poate nu ai inteles. Intrebarea mea era una mai mult retorica. Evident ca nu este adevarat ca in general daca $A^mB=BA^m$ atunci $AB=BA$ pentru matrici $n\times n$, cu $n\geq 3$. Mai bine te gandesti cum sa construiesti un contra-exemplu :)

La $n=2$ merge pentru ca din relatia lui Cayley avem ca $A^m = a_mA+b_mI_2$ ...

iar pt n=3:

A^m=a[m]*A^2+b[m]*A+c[m]*I  .........(tot din cayley generalizat cred nu?)
inmultesti la stanga si la dreapta cu B....se reduce un termen si mai scrii odata pt a[1]=0 pentru ca aceasta conditie trebuie indeplinita pentru a scoate a[m] ,b[m] si c[m] ca termeni generali....
pana la urma iese ca AB=BA.....incearca si u si zi-mi daca am greshit...
cred ca pt n= 4 se continua....pana la p(n+1)....

Gresesti. $a_1$ nu inteleg de ce tebuie sa fie zero. Apoi nu inteleg dupa ce faci inductie dupa $m$ sau dupa $n$ . Daca inmultesti in ambele parti cum zici tu obtii $a_mA^2B+b_nAB=a_mBA^2+b_mBA$. :roll:

da mai ...am gresit...se pare ca beat nu fac bine probleme la matematica

Din $A^mB=A^m+B$ rezultă $A^{m+1}B=A^{m+1}+AB$ şi $A^mBA=A^{m+1}+BA$. Prin scadere, obţinem $A^{m+1}B-A^mBA=AB-BA$, sau $A^m\left(AB-BA\right)=AB-BA$, adică $(A^m-I_n)(AB-BA)=0_n$. Cum matricea $A^m-I_n$ este inversabilă (postul #5 de mai sus) rezultă $AB-BA=0_n$, deci $AB=BA$.

Nu! Se poate si gasi usor contraexemplu pt matrici de 2 pe 2

[quote="siriusra"]enescu...daca produsul a doua matrici este 0 nu inseamna ca una este 0 sau cealalta este 0[/quote]
Oare de ce am menţionat că $A^m-I_n$ este inversabilă?  :)

pentru ca e inversabila ca are determinantul nenul...din relatia de produsi daca treci la determinant si in stanga si in dreapta

E clar ca nu ai inteles solutia domnului Enescu care de altfel e frumoasa. Daca $(A^m-I_n)(AB-BA)=0$ si $A^m-I_n$ inversabila atunci poti sa inmultesti in ambele parti cu $(A^m-I_n)^{-1}$ si obtii ca $AB-BA=O_n$.


Set forth here to use for your self or distribute as you like.

Problems from the Harvard-MIT Math Tourney

http://web.mit.edu/hmmt/www/datafiles/problems/

Problems from around the world (probably the most comprehensive)

http://www.kalva.demon.co.uk/

Practice Geometry Problems:

http://155.225.48.46/statemat/stuf2004/Geoprobs.htm

Nice Problem Corner:

http://www.mandelbrot.org

Problems from the Duke Math Contest (somewhat similar, although it is more like ARML):

http://www.math.duke.edu/dumu/contests/problems/index.html

Feel free to post your own links.

Most of the links you posted don't work

Edit: I just realize you haven't been active for $9$ years


Find the remainder when  $ 2^{50}$ is divided by $ 7$.

\[ 2^3 \equiv 8 \equiv 1 (mod \ 7)\]
\[ \Rightarrow (2^3)^{16} \equiv 1^{16} \equiv 1 (mod \ 7)\]
\[ \Rightarrow 2^{50} \equiv 4 \cdot 2^{48} \equiv 4 (mod 7)\]
\[ \Rightarrow \ Ans \ : \ \boxed{4}\]

[quote="Dr Sonnhard Graubner"]hello, since $ 2^3 \equiv 1 \mod 7$ we get $ 2^{48}\equiv 1 \mod 7$ therefore we have $ 2^{50}\equiv 4 \mod 7$.
Sonnhard.[/quote]
Sonnhard, isn't this [b]EXACTLY[/b] the same as I've written?

[quote="Dr Sonnhard Graubner"]hello, hm, what do you mean with your comment?[/quote]
[b]i meant that what's the point of repeating the same answer that I've already given in detail?[/b]


[quote="mathfanatic"]It's a higher-quality format than PDF, and I think you can incorporate .eps files into it.  It also takes up a lot more space.[/quote]

And what programs can read it?

I use Ghostscript with Ghostview to read postscript files. Both of those programs can be downloaded for free on the web. Try a Google search for them.


calculus
integration
algebra
function
domain
analytic geometry
trigonometry

Hi!

I need to calculate the tripple integral I=80* (tripple integral of x dxdydz) and need help finding the limit in which between i should integrate.

The domain D is {(x,y,z): 16x^2+y^2+25z^2=<25 , 0=<y=<4x , z>=0}

The limits of the integral of dy is already clear (0=<y=<4x) as well as the lower limit of z (z=0?).
But how do i find the upper and lower limits of 16x^2+y^2+25z^2=<25 and the upper limit of z?

Do i need to change to polar coordinates when i integrate?

Please help

Peter

[quote="PeterS1974"]Hi!

I need to calculate the tripple integral I=80* (tripple integral of x dxdydz) and need help finding the limit in which between i should integrate.

The domain D is $\{(x,y,z): 16x^2+y^2+25z^2 \le 25 , 0 \le y \le 4x ,\, z\ge 0\}$
Do i need to change to polar coordinates when i integrate?
[/quote]
Begin by substituting $x' = 4x, \, z' = 5z, \, dx'dydz' = 20 dxdydz$. The domain becomes $D' = \{(x',y,z'): x'^2+y^2+ z'^2 \le 25 , 0 \le y \le x' ,\, z'\ge 0\}$. Then you can go to [b]spherical[/b] coordinates (and the integration domain becomes rectangular). Whether this works or not depends of course on your integrand.

Added in edition - I just noticed that your integrand is $x = r \cos \theta \sin \varphi$, so this should work.


vector
linear algebra
linear algebra unsolved

Prove that any subspace of a finite dimensional vector space is finite dimensional as well

Suppose that $ V$ is a finite dimensional vector space, and $ U$ is a subspace of $ V$.Since every vector space has a basis , let $ X$ be a basis for $ U$.Because the elements of $ X$ are linear independent, it follows that $ X$ is contained in some basis of $ V$, and since $ dim V$ is finite it follows that $ X$ is finite.Thus $ U$ is finite dimensional.

[quote="Svejk"]Since every vector space has a basis , [/quote]
Can we do this without that particular appeal to the axiom of choice?

Every finite dimensional space has a basis. We can do that without the axiom of choice. Take a finite spanning set (which exists by the definition of finite dimensionality.) Delete elements until you have an independent set with the same span. That's an argument by induction.

So $ V$ is finite dimensional and $ U$ is a subset of $ V.$ Suppose $ V$ has dimension $ n.$ Now take a finite independent set in $ U.$ Does it span $ U?$ If not, then append another element from $ U$ that lies outside the span. This becomes impossible as soon as (or before) the set has $ n$ elements, since any set of $ n\plus{}1$ elements in $ V$ must be dependent. Hence there is a finite set that spans $ U.$

In any vector space, if A is a finite spanning set and B is a linearly independent set, then the number of elements in B is <= the number of elements in A.  This would do it for you right?  Was this proved in your text?


1.  "Find the probability of getting a sum of four or doubles when two dice are tossed."
[hide="Answer"]2/9[/hide]

2.  "Find the probability of getting a sum greater than 5 when two dice are tossed."
[hide="Answer"]13/18[/hide]

[hide="1"]The ways to get a $ 4$ are $ (3,1)$, $ (1,3)$, $ (2,2)$. The ways to get doubles are $ (1,1)$, $ (2,2)$, $ (3,3)$, $ (4,4)$, $ (5,5)$, $ (6,6)$. But we counted $ (2,2)$ two times! So we subtract $ 1$. The total is $ 6\cdot6$. Hence, our answer is $ \frac {3 \plus{} 6 \minus{}1 }{36} \equal{} \frac {8}{36} \equal{} \frac {2}{9}$[/hide]

[hide="2"]We need to get a sum of $ 6 \minus{} 12$. Instead, let us find the probability of getting a sum $ 2 \minus{} 5$ and subtract it from $ 1$. There is only $ 1$ way to get $ 2$. There are $ 2$ ways to get $ 3$. There are $ 3$ ways to get $ 4$. And finally we have $ 4$ ways to get $ 5$. We have $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \equal{} 10$. Once again, there's a total of $ 6\cdot6$. Now $ \frac {10}{36} \equal{} \frac {5}{18}$. Next,
\[ 1 \minus{} \frac {5}{18} \equal{} \frac {13}{18}
\]
[/hide]

I'm sorry for acting like a mod.  :(


modular arithmetic
number theory
imo 2000
IMO
prime divisors
IMO Shortlist

For any $k\ge 1$ there is such an $n$ with exactly $k$ prime factors.

For $k=1,\ n=3^t$ works for every $t\ge 1$. Take $t$ s.t. for $n_1=3^t,\ 2^{n_1}+1$ has a prime factor $p_2$ larger than $3$. Now take $n_2=n_1p_2$. Then $n_2|2^{n_1}+1|2^{n_2}+1$, and $2^{p_2}+1$ has a prime factor $p_3\not|n_2$. This is because $(2^{n_1}+1,2^{p_2}+1)=3,\ p_2\not|2^{p_2}+1$, and $2^{p_2}+1$ cannot be a power of $3$, since $p_2>3$ (I'm using the well known and easy to prove fact that the only positive integer solution $(a,b),a>1$ to $3^a=2^b+1$ is $(a,b)=(2,3)$). Then take $n_3=n_2p_3$. Just like above, we deduce that $2^{p_3}+1$ has a prime factor $p_4$ which is coprime to $n_3$, and take $n_4=n_3p_4$, and so on. $n_k$ will have exactly $k$ prime factors and will satisfy $n_k|2^{n_k}+1$.

[quote="Valentin Vornicu"]Does there exist a positive integer $n$ such that $n$ has exactly 2000 prime divisors and $n$ divides $2^n + 1$?[/quote]

You can find my paper for a general problem of this problem in [url=http://www.math.ust.hk/excalibur/v9_n3.pdf]This file[/url]

Let $N=2^n+1$. We will assume for the sake of contradiction that $n|N$.

$2^n+1 \equiv 0$ (mod $n$) $\Rightarrow 2^n \equiv -1$ (mod $n$). So 2 does not divide $n$, and so $n$ is odd.

Select an arbitrary prime factor of $n$ and call it $p$. Let's represent $n$ in the form $p^am$, where $m$ is not divisible by $p$.

Note that $p$ and $m$ are both odd since $n$ is odd. By repeated applications of Fermat's Little Theorem:

$N = 2^n+1 = 2^{p^am} + 1 = (2^{p^{a-1}m})^p + 1 \equiv 2^{p^{a-1}m} + 1$ (mod $p$)

Continuing in this manner, and inducting on k from 1 to $a$,

$2^{p^{a-k}m}+1 \equiv (2^{p^{a-k-1}m})^p + 1$ (mod $p$) $\equiv 2^{p^{a-k-1}m} + 1$ (mod $p$)

So we have $N \equiv 2^m+1$ (mod $p$)

Since $p$ is relatively prime to $m$, $N \equiv 1+1$ (mod $p$) $\equiv 2$ (mod $p$)

Since $p$ is odd, $N$ is not divisible by $p$. Hence $N$ is not divisible by $n$. So we have a contradiction, and our original assumption was false, and therefore $N$ is still not divisible by $n$.

Hmmm... I made a mistake here somewhere but I do not see it.

Yes, since there are lots of integers $n$ such that $n$ divides $2^n + 1$, and the statement is true!

There exists a pretty beautiful generalization:

"
Let $s, a, b$ positive integers, such that $GCD(a,b) = 1$ and $a+b$ is not a 2-power.
Show that there exists infinitely many $n \in N$ such that

--- $n=p_1^{e_1} \cdot p_2^{e_2} \cdot p_3^{e_3} \cdots p_s^{e_s} \cdot$ is the canonical factoring of $n$.

--- $n|(a^n+b^n)$
"

[quote="Valentin Vornicu"]Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n \plus{} 1$?[/quote]
Here is a solution that I don't think has been mentioned yet:
[hide="Solution"]
We proceed with induction.  Clearly, $ n\equal{}9$ works for the base case.  Now, we will show that we can repeat an algorithm so that we can find a prime factor, $ p$, that is a factor of $ 2^n\plus{}1$, but not of $ n$.  Once this is done, we simply have that $ np|(2^n\plus{}1)|(2^{np}\plus{}1)$ and $ np$ has one more prime divisor than before.  We use induction to find this algorithm.  For the base case ($ n\equal{}9$), we can find $ 19$ is a factor of $ 2^9\plus{}1$, so the base case is proven.  Now, let us consider $ 2^{np}\plus{}1$, where $ p$ is a prime divisor if $ 2^n\plus{}1$ and not of $ n$ (which is the inductive hypothesis.)  $ 2^{np}\plus{}1\equal{}(2^n\plus{}1)(2^{(p\minus{}1)n}\minus{}2^{(p\minus{}2)n}\plus{}\cdots \plus{}1)$.  Set $ 2^n\equal{}x$.  Then, this is $ 2^{np}\plus{}1\equal{}(x\plus{}1)(x^{p\minus{}1}\minus{}x^{p\minus{}2}\plus{}\cdots \plus{}1)$.  We compute the $ \gcd$ of both factored terms by computing the remainder of the second when divided by the first.  Plug $ x\equal{}\minus{}1$ in, so we get that the remainder is $ p$, meaning that $ p$ divides the $ \gcd$ of $ x\plus{}1$ and $ x^{p\minus{}1}\minus{}x^{p\minus{}2}\plus{}\cdots \plus{}1$.  In fact, this $ \gcd$ is $ p$ since $ p|(x\plus{}1)$. Then, $ np^2|(x\plus{}1)(x^{p\minus{}1}\minus{}x^{p\minus{}2}\plus{}\cdots \plus{}1)\equal{}x^p\plus{}1$.  This means that $ np^2$ divides $ 2^{np}\plus{}1$, so $ np^2$ divides $ 2^{np^2}\plus{}1$.  We do this again, but now we set $ y\equal{}2^{np}$, so $ 2^{np^2}\plus{}1\equal{}(y\plus{}1)(y^{p\minus{}1}\minus{}y^{p\minus{}2}\plus{}\cdots \plus{}1)$.  Again, the $ \gcd$ between the factored terms is $ p$.  Since $ p^2|(y\plus{}1)$, we have that $ p$ can only divide $ y^{p\minus{}1}\minus{}y^{p\minus{}2}\plus{}\cdots \plus{}1$ once.  Yet, $ y^{p\minus{}1}\minus{}y^{p\minus{}2}\plus{}\cdots \plus{}1>p$ for $ p>3$, so another prime factor that is not a factor of $ np^2$ divides $ y^{p\minus{}1}\minus{}y^{p\minus{}2}\plus{}\cdots \plus{}1$, so we are done. [/hide]

For any i, [code]2^3^i+1[/code] is divisible by [code]3^i[/code] (the proof is easy with euler`s theorem+induction and maybe with primitive roots (2 is primitive root modulo 3^i for any i)). Hence, for i=1999, [code]3^1999[/code] has 2000 divisors and it satisfies the asked in the problem.

it is correct??

please , answer me.

bye

sorry for my english.

Well, the point is to find a number which has exactly $ 2000$ [b]prime[/b] divisiors, whereas $ 3^{1999}$ has only one ($ 3$).

But it is a very nice thought to look at powers of $ 3$, when the problem considers powers of $ 2$ (vide grobber's solution, which I do not completely understand yet, but it seems very nice and simple)

[hide="Solution overkilling with Zsigmondy's"]

By Zsigmondy's we know that there exists a prime $p$ such that $p\mid 2^i+1$ but $p\nmid 2^j+1$ for $j<i$ unless $i=3$.

Begin with noting that $9\mid 2^9+1$ so set $n=9n_1$ for $\gcd(n_1, 3)=1$.  We know that $2^9+1\mid 2^{n_1}+1$.  Now consider $p\mid 2^{9}+1$ for which we arrive at $p=19$.  Therefore set $n=3^2\cdot 19\cdot n_2$ for $\gcd(n_1, 3^2\cdot 19)=1$.

I claim that given $n$ in the form $n=9\cdot p_1\cdot p_2\cdots p_k$ with $p_i\neq p_j$ for $i\neq j$ and $n\mid 2^{9\cdot p_1\cdot p_2\cdot \cdots p_{k-1}}+1$ we can find $n=9\cdot p_1\cdot p_2\cdots p_{k+1}$ with $n\mid 2^{9\cdot p_1\cdot p_2\cdot \cdots p_k}+1$ By Zsigmondy's there exists a prime $q$ such that \[q\mid 2^{9\cdot p_1\cdot p_2\cdots p_k}+1, q\nmid 2^{9\cdot p_1\cdot p_2\cdots p_{k-1}}+1\]  Set $q=p_{k+1}$ which by Zsigmondy's doesn't equal $p_i$ for $1\le i\le k$.  Then we have $9\cdot p_1\cdot p_2\cdots p_{k+1}\mid 2^{9\cdot p_1\cdot p_2\cdots p_k}+1$ using the inductive hypothesis and Chinese Remainder Theorem.  We are done.  [/hide]

I feel like it isn't interesting to make any new remark on this problem, but anyway I'm posting my approach too.

If we show that for any positive integer k, there exists a positive integer n with exactly k distinct prime divisors such that n | 2^n + 1, then we are done, since the problem asks us to examine a special case, more exactly k = 2000. Furthermore, we can even show that we can find these n's divisible by a power of 3, which will help us on our proof.

We use induction on k. k = 1, we can choose n(1) = 3, which clearly satisfies the conditions. Assume that k >= 1, and there exists n(k) = 3^t * m, where gcd(3, m) = 1, and m has exactly (k - 1) distinct prime divisors. So, we have n(k) | 2^(n(k)) + 1.
Before generating n(k+1) from n(k), let us look at the number 3n(k), which clearly has k distinct prime divisors. 2^(3n(k)) + 1 = (2^(n(k)) + 1)(2^(2n(k)) - 2^(n(k)) + 1). Since we must have n(k) always odd because of the fact that n(k) = 1 (mod 2), we deduce that 3 | (2^(2n(k)) - 2^(n(k)) + 1), so we have that 3n(k) | 2^(3n(k)) + 1. It is enough to find a prime p, such that p | 2^(3n(k)) + 1 and p doesn't divide (2^(n(k)) + 1), which could guarantee us that p doesn't divide n(k) and consequently, we could generate n(k + 1) = 3n(k)*p, which could clearly work by observing that 2^(3n(k)*p) + 1 = (2^(n(k)) + 1)(2^(2n(k)) - 2^(n(k)) + 1)*A. But, since we can pick up this prime p by Zsigmondy, we are done.

Note that in case of not using Zsigmondy, we can observe that gcd(a^2 - a + 1, a + 1) = gcd(3, a + 1) = 1 if a is not 2 mod 3 and 3 if a = 3k + 2. But if a = 3k + 2, then a^2 - a + 1 is divisible by 3 but not by 9, so we could pick up any prime p that divides (a^2 - a + 1) / 3.

Let $n=\prod_{i=1}^{2000}{p_i}$ and wlog let $p_1<p_2<\dots<p_{2000}$.I dscribe a process on how to construct such an $n$.By the problem we have $2^{2n} \equiv 1\pmod{p_1} \Rightarrow ord_{p_1}{2} |2n$ and so by the minimality of $p_1$ we get that $ord_{p_1}=1$ or $ord_{p_1}=2$.In the first case we get $p_1|1$ which is absurd while in the second case we get $p_1|2^2-1=3 \Rightarrow \boxed{p_1=3}$.Similarly it is easy to note that $ord_{p_2}{2}|2n$ and so by the choice of $p_2$ we get $ord_{p_2}{2}|2*3=6$.Clearly there exists such a prime such that $ord_{p_2}{2}=6$(By Zsigmondy!!)In general we have $2^{2n} \equiv 1\pmod{p_k} \Rightarrow ord_{p_k}2|p_1p_2\dots p_{k-1}$.Clearly there exists a prime such that $ord_{p_k}2=p_1p_2\dots p_{k-1}$(Again Zsigmondy!!).So we are done!!!(In fact by this procedure we may fix any number of primes).

Since $N=3^k$ all satisfay the condition.Now,it is enough to prove that numbers of the form $N=2{}^3{}^k+1$ have infinitely many primes dividing them,but this is easy to prove,since we have for $n<m$ $2{}^3{}^n+1$ divides $2{}^3{}^m+1$ so suppose opposite.Now,we just need to prove that $a+1$ and $a^3+1$ can't have identical sets of primes for $a>2$,and this is true because $GCD(a+1,a^2-a+1)$ is at most $3$ and $a^2-a+1>3$ for $a>2$ so we are done.

[quote=Quidditch][quote=Msn05][quote=v_Enhance]Answer: Yes.

We say that $n$ is [i]Korean[/i] if $n \mid 2^n+1$. First, observe that $n=9$ is Korean. Now, the problem is solved upon the following claim:

[b]Claim[/b]:  	If $n > 3$ is Korean, 	there exists a prime $p$ not dividing $n$ 	such that $np$ is Korean too.

[i]Proof.[/i]  	I claim that one can take any primitive prime divisor $p$ of $2^{2n}-1$, 	which exists by Zsigmondy theorem.  	Obviously $p \neq 2$. 	Then: 	[list] 		[*]Since $p \nmid 2^{\varphi(n)}-1$ it follows then that $p \nmid n$. 		[*]Moreover, $p \mid 2^n+1$ since $p \nmid 2^n-1$; 	[/list] 	Hence $np \mid 2^{np} + 1$ by Chinese Theorem, 	since $\gcd(n,p) = 1$. $\blacksquare$

EDIT: The version of the proof I posted four years ago was incorrect. This one should work.[/quote]

Why $p \nmid 2^{\varphi(n)}-1$?[/quote]

Because $\varphi(n)<n<2n$ for an obvious reason, and $p$ is a primitive root hence it doesn't divide any $a^k-b^k$ for any $0<k<2n$.[/quote]

Thanks:)

The answer is yes. We will describe an algorithm to generate the wanted $n$. Use Zsigmondy to generate a "new"(not $3$) prime divisor of $2^{3^x}+1$ where $x\ge 2$, and call it $p_x$. Then
\[n=3^{2000}p_2p_3\dots p_{2000}\]
suffices.

To prove that this works, first note that $3^{2000}$ obviously divides $2^n+1$ by LTE. Otherwise,
\[p_i\mid 2^{3^i}+1\mid 2^n+1\] because $3^i\mid n$.

Let $$a_m=2^{3^m}+1.$$

Note that for all positive integers $m$, we have $$3^m\mid a_m,$$ (which can be shown by sum of cubes factorization and induction). Furthermore, due to sum of cubes, $a_m|a_{m+1}$, so $a_m$ is a divisor of $a_s$ for all $s\geq m.$ However, by Zsigmondy's, each $a_m$ other than $a_1$ has a new prime divisor, and since all old divisors are retained when increasing $m$, we can make $a_m$ have arbitrarily many prime divisors by making $m$ large enough.\\

Therefore, for sufficiently large $m$, $$3^m p_1p_2p_3\cdots p_{1999}\mid 2^{3^m}+1$$ for some odd primes $p_1\cdots p_{1999}.$ However, since $2^x+1$ divides $2^y+1$ if $y$ is an odd multiple of $x$, we clearly also have $$3^m p_1p_2p_3\cdots p_{1999}\mid 2^{3^m p_1p_2p_3\cdots p_{1999}}+1,$$ so we are done.

[hide=Solution]The answer is YES, We call a number nice if $n|2^n+1$,We prove a more general result that- there exists a nice number with $k$ divisors where $k$ is a natural number
[color=#f00]Claim 1[/color] If $n$ is nice, so is $n*p$ for some odd prime $p$
[hide=Proof]To show this, we choose our "primitive" "n" as one added to two's power, Now, we want to show $n*p|2^{n*p}+1$ for a prime $p$ we greedily choose $p$ as divisor of $2^n+1$ but not of $n$, Now, such an $p$ is guaranteed by Zsigmondy's theorem, so $n*p|2^n+1|2^{np}+1$ [/hide]

[quote=Valentin Vornicu]Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n \plus{} 1$?[/quote]
$n=3^2$ satisfies the condition
By Zsigmondy's Theorem:
$\exists$ prime $p_1\neq 3/ p|2^9-1$
We will prove that $n=3^2p_1:$
$9|2^9+1|2^{9p_1}$
$p_1|2^9+1|2^{9p_1}$
$\Rightarrow 3^2p_1|2^{9p_1}+1$
By induction, suppose that it satisfies for $n=3^2p_1p_2...p_k:$
By Zsigmondy's Theorem:
$\exists$ prime $p_{k+1}\neq 3,p_1,p_2,\ldots ,p_{k+1}/ p|2^{9p_1p_2\ldots p_k}+1$
$9|2^9+1|2^{9p_1p_2...p_k}+1$
$p_1|2^9+1|2^{9p_1p_2...p_k}+1$
$p_2|2^{9p_1}+1|2^{9p_1p_2...p_k}+1$
$\ldots$
$p_{k+1}|2^{9p_1p_2\ldots p_k}+1|2^{9p_1p_2...p_k}+1$
$\Rightarrow 3^2p_1p_2 \ldots p_k|2^{9p_1p_2...p_k}+1 $
the induction is complete.
$\Rightarrow$ we simply choose $n=3^2p_1p_2 \ldots {p_{1999}} _\blacksquare$

Note that n=9 satisfies the condition. By Zsigmondy, we can find a prime that is not included in the factorization of n, that divides 2^n+1. We can then add it in, and now our number is np_k|2^(np)+1, which is true since np_k|2^n+1|(2^n)^p+(1^n)^p, where the last step follows from as p is odd, we always can factor a+b|a^p+b^p where a=2^n and b=1. Then this adds one prime divisor in n, and we repeat this until we have 2000 of them.

ok ngl idt i understood what i was doing back then skull
heres the formally written; i reported my old post to not double post

We consider an $n$ such that $n |2^n+1$. We claim that for all such $n \neq 3$ there exists a prime $p$ such that $pn | 2^{pn} + 1$. We can see that this obviously means that the answer is $\boxed{\text{yes}}$.

We start this inductive building with $n = 9$. We can see that from Zsigmondy's there must be a prime $p$ such that $p \not | 2^n + 1$ and $p |2^{pn} + 1$. $\blacksquare$

Yes there does. We may construct one as follows: Define a sequence $a_i$ such that $a_1=9$ and $a_{i+1} \mid 2^{a_1a_2\dots a_i}+1$ and $a_{i+1} \nmid 2^j+1$ for $1 \le j < a_1a_2\dots a_i$ and $a_i$ prime for $i \ge 2$ (existence guaranteed by Zsigmondy). Take $n=a_1a_2\dots a_{2000}$ to finish, which works since $a_i$ are distinct, none of them are $3,$ and $9 \mid 2^9+1.$

Pain and suffering.

We will induct. For the base case take $n = 9$ which indeed satisfies $n \mid 2^n + 1$ and has exactly one prime factor.

Now assume that we have found $n$ such that $n \mid 2^n + 1$. Then we wish to find an odd prime $p$ not dividing $n$ such that  $$np \mid 2^{np} + 1$$
To do this we take $p$ as a primitive prime divisor of $2^{2n} - 1$, which is guaranteed to exist by Zsigmondy. Clearly this $p$ is relatively prime to $n$ as $n \mid 2^{\phi(n)} - 1$ yet $p \nmid 2^{\phi(n)} - 1$ from Zsigmondy. Also $p \mid 2^n + 1$ as we must have $p \nmid 2^n - 1$, once again from Zsigmondys.

 :coolspeak:
We claim that $n=3^\theta\cdot p_1p_2\cdots p_{2000}$ works, where $p_i$ are primitive prime divisors of some term in the sequence $2^{3^1}+1, 2^{3^2}+1, 2^{3^3}+1,\ldots$ which by Zsigmondy they exists; note that
$$2^{3^x}+1\mid 2^{3^\theta\cdot p_1p_2\cdots p_{2000}}+1 \hspace{0.5cm} \text{for enough big} \hspace{0.11cm}\theta$$
and also $3^\theta\mid 2^{3^\theta\cdot p_1p_2\cdots p_{2000}}+1$ by LTE; so $3^\theta\cdot p_1p_2\cdots p_{2000}\mid 2^{3^\theta\cdot p_1p_2\cdots p_{2000}}+1$ , as desired.

The answer is $\boxed{\text{yes}}$.

Let $p_x$ be a prime that divides $2^{3^x}+1$ but not $2^{3^{i}}+1$ for $i<x$. This prime will always exist by Zsigmondy, meaning our desired number is

\[n=3^{2000}p_2p_3\cdots p_{2000}.\]

Seeing that this works is trivial, so we are done.

We note $n_1=3^2$ satisfies the condition. We aim to find a prime $p$ such that $pn_i$ satisfies the condition and $\gcd(p,n_i)=1$, so it suffices to have
\[2^{pn} \equiv -1 \pmod{p} \implies 2^{2n} \equiv 1, 2^n \not\equiv 1 \pmod{p}.\]

This primitive root of 2 modulo $p$ exists by Zsigmondy, as desired. Hence we can induct to find $n_{2000}$ with the desired 2000 prime divisors. $\blacksquare$

Overkill?

In general, I claim that there exists a positive integer $n$ such that $n\mid 2^n+1$ with exactly $k$ distinct prime divisors for all positive integers $k$. \\\\
I prove this by claiming that if there exists a positive integer $n$ with $k$ distinct integer divisors in the form of
\[n=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k},\]
such that $p_1<p_2\dots<p_k$, then there exists a prime $p_{k+1}>p_k$ such that for all positive integers $e_{k+1}$, it is true that $np_k^mp_{k+1}^{e_{k+1}}$ satisfies the problem condition, where we can make $m$ any positive integer we want. \\\\
[color=#9a00ff][b]C1:[/b][/color] First, before we begin, I claim that if any prime $q\mid n$, then $nq$ also satisfies problem conditions. This will then prove the part where we can make $m$ whatever we want. This is because
\[n\mid 2^n+1 \iff \nu_q(n)\leq \nu_q(2^n+1),\]
and by LTE (which we can use, since $n\mid 2^n+1$ implies that $q\mid 2^n+1$), this gives us that
\[\nu_q(nq)=1+\nu_q(n)\leq 1+\nu_q(2^n+1)=\nu_q(2^{nq}+1),\]
which means that $nq$ also satisfies the problem conditions, as desired. Therefore, by induction, we also get that $nq^m$ also satisfies problem conditions. \\\\
[color=#9a00ff][b]C2:[/b][/color] Now, I claim that there exists a prime $p_{k+1}$ such that for all positive integers $e_{k+1}$, $np_k^mp_{k+1}^{e_{k+1}}$ satisfies the conditions. For simplicity, from here on out, I will refer to $p_{k+1}$ as $r$ and $p_k$ as $q$. Note that this implies that
\[r\mid 2^n+1 \iff ord_r(2)\mid 2n,\]
and since $ord_r(2)\mid \phi(r)=r-1$, we get that $ord_r(2)\mid \gcd(2n,r-1)$. First, I claim that there always exists an $r$ that divides $2^{q^c}+1$ for some $c$ we can choose such that $r>q$. This is clearly true by Zsigmondy's theorem. I now claim that if we take this $r$, it is true that $rn$ also satisfies problem conditions. This, combined with [color=#9a00ff][b](C1)[/b][/color], will prove our master claim, which states that $nr^e_{k+1}$ satisfies problem conditions for any $e_{k+1}$. \\\\
To prove this, first make sure that our previous $n$ is divisible by the $q^c$ we chose. We can do this by making another $n$ that satisfies problem conditions by multiplying it by a power of $q$, which we can do by [color=#9a00ff][b](C1)[/b][/color]. Next, by LTE, we have,
\[\nu_r(rn)=1\leq \nu_r(2^{rn}+1)=\nu_r(2^n+1)+1,\]
which we can do since we know that $q^c \mid n$ and $r\mid 2^{q^c}+1$, which gives that $r\mid 2^n+1$. This means that $rn$ satisfies problem conditions, since $rn\mid 2^{rn}+1$, which combined with [color=#9a00ff][b](C1)[/b][/color], proves our inductive step claim. \\\\
Finally, to complete our induction, note that $3\mid 2^3+1$. Therefore, there exists a positive integer $n$ such that $n\mid 2^n+1$ with exactly $k$ distinct prime divisors for all positive integers $k$, finishing the problem.


What is the inverse of the function $f: (2, +\infty)\to R$, $f(x)= \frac{2x^2-7x-9}{x-2}$. How can i use the fact that $(f^{-1}(y))' = \frac{1}{f'(x)}$ ?


Suppose I have an idea in mind. How will I know if its worth writing a paper on and not plain trash?

Suppose I ask for the opinion of a prof and he says its plain trash but uses it for his own interests.
So whose opinion should I consider relevant in evaluating my idea?

What is the formal procedure for sending papers/articles to a journal?

Does "research" mean doing "genius" stuff and out-of-the-blue stunning results only??

Thanks.

You could send it to a professor with a sufficiently well-worded non-disclosure agreement. Though that's probably not necessary. First of all, unless your idea is something really monumental (like a proof of the Riemann Hypothesis or something), most professors will not covet your result enough to even contemplate stealing it. Furthermore, most professors would not risk their integrity by stealing the work of someone who could easily prove that the idea was theirs (which you probably could if required).

So basically, if you genuinely believe that you have a noteworthy result, find a professor who will take you seriously. Then politely show them your work and ask for feedback. If it will make you feel more comfortable, put a simple statement of proprietary rights somewhere at the beginning of your paper.

After you've had someone look at it and give you feedback, then if they think it is journal worthy you can find an appropriate journal to submit it to. You'd contact the journal directly for the correct method for submitting it. Each journal will have different requirements for formatting and methods of submission.

Research doesn't have to be mind-blowingly out of the blue to be worthy of publication. It just has to be something with enough originality or new work to be worth other people to read.

[quote="romanovodka"]Suppose I have an idea in mind. How will I know if its worth writing a paper on and not plain trash?
...
So whose opinion should I consider relevant in evaluating my idea?[/quote]  The only way to do this is to talk to other people.  Professors are a good bet, because they're most likely to be familiar with the state of knowledge.  But any person with such familiarity would suffice.

[quote="romanovodka"]Suppose I ask for the opinion of a prof and he says its plain trash but uses it for his own interests.[/quote]  This is unlikely to happen.  But, there are several ways to protect yourself against it.  The most obvious is to stick a draft of a paper in some public place that demonstrates your priority.  An appropriate part of this forum, for example, but more usually the [url=http://arxiv.org]arXiv[/url].

[quote="romanovodka"]What is the formal procedure for sending papers/articles to a journal?[/quote]  This depends on the journal -- any journal website should have a link to "Instructions for Authors" or "Guide to submission" or something on the front page.

[quote="romanovodka"]Does "research" mean doing "genius" stuff and out-of-the-blue stunning results only??[/quote]  No.  But it does generally mean producing results that have not been proved before, or new proofs of known results that show something new and interesting.

[quote="Vinoth"]Another question:
Suggestions for journals for someone interested in Algebra generally (more specifically Representation Theory?) What low level journals would be good for someone to publish a paper in?[/quote]
One method to find a good journal to submit to:  Look at the bibliography of your paper, and see what journals recently published the related results you used.


How does one make amazing PDFs like [url=http://www.ctan.org/tex-archive/macros/latex/contrib/pdftricks/manual.pdf]this[/url] one?

Use [url=http://latex-beamer.sourceforge.net/]Beamer[/url]. Here's [url=http://www.sixthform.info/steve/Slideb.pdf]one[/url] that I did a couple of years ago using beamer - save it then run it (so it doesn't run inside your Browser). Press CTRL-L to start and then the arrow buttons (or mouse) to go through it. ESC goes back to the normal screen.
Beamer has lots of ready made templates but you can take a little while to get used to. There are lots of tutorials to be found on the net, for example, [url=http://www.math.umbc.edu/~rouben/beamer/]A Beamer Quickstart[/url].
Once you get used to it you'll love it and it can produce stunning presentations.

Alternatively, you can use prosper, HA-prosper, or the latest replacement of these, namely [url=http://www.ctan.org/tex-archive/macros/latex/contrib/powerdot/]powerdot[/url].  A nice introduction to powerdot can be found [url=http://csgsa.cs.clemson.edu/seminar/tools06/docbook/html/ch02s04.html]here[/url].

I personally use [url=http://amath.colorado.edu/documentation/LaTeX/prosper/]prosper[/url], because I found that there is currently more help available online for this package, as it is older and more widely used.  I am not sure what exactly has been fixed or upgraded in HA-prosper and powerdot; prosper is easy to use and works just fine for all practical purposes.


Let $x;y;z >0$ and satisfying $x+y+z=xyz$
Find max of $S=(1-x)(1-y)(1-z)$

See $x\rightarrow 0$ and $y,z \rightarrow \infty$.
But
$S \ge (1-\sqrt 3)^{3}$.

Other problem:
Let $x;y;z>0$ and satisfying $x+y+z=xyz$.Find max:
$A=(x-1)(y-1)(z-1)$


geometry
trapezoid
trigonometry
geometry unsolved

PQR is a triangle in which there is an inside trapezoid ABCD. AB parralel to QR, BC parralel PQ and AD parallel PR. Also, area of PQR is 1 unit. Find out maximum possible area of trapezoid ABCD

(from Indian Embassy practice papers)

[hide]Let $E$ be the area of $ABCD$ and $F=1-E$ the rest area of $\triangle PQR$

Let $m=\frac{PA}{PQ}=\frac{PB}{PR}$.
Then

$QA = (1-m)PQ$
$RB = (1-m)PR$
$QD = RC = (1-m)QR$

Also, it should be $m>1-m \Leftrightarrow \boxed{m>\frac{1}{2}}$ or else $ABCD$ will be not a trapezoid.

So $(PAB) = m^{2}, \ \ (QAD) = (RBC) = (1-m)^{2}\Rightarrow $

$F=m^{2}+2(1-m)^{2}$
$= 3m^{2}-4m+2$

Then $E=1-F =-3m^{2}+4m-1$

$E$ has a maximum value for $m=-\frac{b}{2a}= \frac{2}{3}$

So we find that the maximum value of the area is $E_{max}=\frac{1}{3}$[/hide]

It is obvious that the base of trapezoid has to pass centroid for max area.

  M.T.

[quote="shyamsundar23"]how does (PAB) = m^2?[/quote]

$(PQR) = \frac{1}{2}PQ\cdot PR\cdot \sin P$

$(PAB) = \frac{1}{2}PA\cdot PB\cdot \sin P = \frac{1}{2}(m\cdot PQ)\cdot (m\cdot PR)\cdot \sin P= m^{2}\cdot \frac{1}{2}PQ\cdot PR\cdot \sin P$

$= m^{2}\cdot (PQR)$


induction
number theory
prime factorization

Prove that a list can be made of all the subsets of a finite set in such a way that
(i) the empty set is frst in the list;
(ii) each subset occurs exactly once;
(iii) each subset in the list is obtained either by adding one element to the preceding subset or by deleting one element of the previous subset.

[hide="Solution"]
We'll give the construction for a countably infinite set such that truncating the list at $ 2^{n}$ gives the desired list for an $ n$ element set.

First of all, it's equivalent to solve the problem for the set $ \{0,1,2,3,\ldots,n\minus{}1\}$, so we'll only do it for that.

The first term in the list, $ S_{0}$ is the empty set, of course.

For $ i\geq 1$, let $ S_{i}\equal{} S_{i\minus{}1}\Delta\{a_{i}\}$, where $ \Delta$ is the [url=http://en.wikipedia.org/wiki/Symmetric_difference]symmetric difference[/url], and $ a_{i}$ is the exponent of the 2 in the prime factorization of $ i$ (so $ a_{2}\equal{} 1$, $ a_{24}\equal{} 3$, etc.). Then the desired list for an $ n$ element set is $ S_{0}, S_{1}, S_{2}\ldots S_{2^{n}\minus{}1}$.

Obviously (i) is satisfied, and (iii) is satisfied by the properties of the symmetric difference. We have $ 2^{n}$ elements in the list, the number of subsets of the set, so if we prove that no two $ S_{i}$ are the same, we have proven (ii).

Again by the properties of the symmetric difference, it suffices to show that no nonempty subsequence of $ \{a_{i}\}$ containing consecutive terms contains every term an even number of times. I will state without proof that if a consecutive subsequence of $ \{a_{i}\}$ contains at least two equal elements $ x$, it also contains a term that is at least $ x\plus{}1$. I might come back and prove this is true later. So suppose such a consecutive subsequence exists, and take the highest term. It must occur at least two times. But then there is another term that is higher. Contradiction. And we're done.

For those who want a more concrete look at the answer, here are the first few $ S_{i}$:

[hide="First 16 terms, showing how to do it for up to a 4 element set"]
$ S_{0}\equal{}\{\}$
$ S_{1}\equal{}\{0\}$
$ S_{2}\equal{}\{0,1\}$
$ S_{3}\equal{}\{1\}$
$ S_{4}\equal{}\{1,2\}$
$ S_{5}\equal{}\{0,1,2\}$
$ S_{6}\equal{}\{0,2\}$
$ S_{7}\equal{}\{2\}$
$ S_{8}\equal{}\{2,3\}$
$ S_{9}\equal{}\{0,2,3\}$
$ S_{10}\equal{}\{0,1,2,3\}$
$ S_{11}\equal{}\{1,2,3\}$
$ S_{12}\equal{}\{1,3\}$
$ S_{13}\equal{}\{0,1,3\}$
$ S_{14}\equal{}\{0,3\}$
$ S_{15}\equal{}\{3\}$
[/hide]
[/hide]

I prefer a nice
[hide="induction"]
Base case: $ |S|\equal{}0$
$ \{\}$ Done.

Inductive step. Take away one element, $ a$ from $ S$ to get $ S'$. Then, create the sequence of subsets of $ S'$, $ S_{0}, S_{1}, S_{2},\ldots, S_{m}$. Then you append $ S_{m}\cup\{a\}, S_{m\minus{}1}\cup\{a\},\ldots, S_{0}\cup\{a\}$. Done.

[hide="The first few of mine"]
Notice that they all begin the same way, so I'll just note where to stop for each size of $ S$.
{}
---- 0
{0}
---- 1
{0,1}
{1}
---- 2
{1,2}
{0,1,2}
{0,2}
{2}
---- 3
{2,3}
{0,2,3}
{0,1,2,3}
{1,2,3}
{1,3}
{0,1,3}
{0,3}
{3}
---- 4
[/hide]
These are actually the same sets as given above, but generated inductively instead of explicitly.
[/hide]

I would hardly say mine were generated explicitly; it was recursive, at best. :P

That's a much cleaner way than mine to do it; don't have to do any work to prove all of the terms in the sequence are different.