Let $X$ be a set containing $n$ elements. Find the number of ordered triples $(A,B, C)$ of subsets of $X$ such that $A$ is a subset of $B$ and $B$ is a proper subset of $C$.

this one is straightforward. count the number of triples $(A\subseteq B\subseteq C)$ and then subtract the triples corresponding to pairs, $(A\subseteq B=C)$. The first there are $4^n$ such triples(for each $x\in X$ we can decide for one of $4$ options-belongs to all of them,belongs to $B,C$ only, belongs to $C$ only, belongs to none). Likewise there are $3^n$ options to subtract.


geometry
rectangle
3D geometry
sphere
pyramid
conics
ellipse

A water tank is 150 cm long, 40 cm high and 40 cm deep. How many mL of water will it hold?

Can this problem be solved? I thought that high and deep meant the same thing. Volume = l x w x h.... But in this case, is there a width?!

Unless the deep part is the width, this problem can't be determined.

[quote="serenityx33"][quote="Iversonfan2005"]Area of cylinder=$\pi r^2 h$.

Is it a cylinder?[/quote]

OH they never said it wasn't a cylinder. THey just said it was a water tank... If it is a cylinder, can the answer be determined with the information in the problem?[/quote]

Yes, because then the radius would be $75$, and you can solve it from there.

[quote="h78reg"]well, if the tank is a cylinder, then it holds 706.5L of water. if it's a rectangle, we need to know the width.[/quote]

How? The length is $150$, which is equal to the diameter of the circle.

[quote="h78reg"][quote]How? The length is , which is equal to the diameter of the circle.[/quote]

so then the radius is 75 and the height is 40.[/quote]

Yes, and so $75^2 \pi (40)=\boxed{225000\pi}$

[quote="h78reg"]and 225000pi is equal to 706500 if we use 3.14 as pi. 706500cm^3 is equal to 706500mL, or 706.5L[/quote]

Oops, sorry, I didn't see it was expressed in L :blush:

no problem. we still don't know if this is the correct volume though because we don't know the shape of the container. what if the container's a sphere? :D

I don't think it can be a sphere because the length and width are different.

I don't think a container that can hold water can be a pyramid. Can it?

[quote="h78reg"]oh, sorry, didn't notice that :blush: wait, it could be a pyramid couldn't it?[/quote]

That's if the bottom of the pyramid is a square.

Uhh I'm going with cylinder. Since you can find the volume using the (Pi r^2 h ) formula. It can't be a rectangle, because they don't have the width.

okay, so, your answer is 706.5L of water.

I searched online.. And it says that the diameter is equal to the width? But I thought someone said the diameter is equal to the length?

"Because the width is the diameter of the cylinder, the radius will equal the width divided by two."

Very vague, do they mean the same thing by height and depth?  doesn't make much sense.

*Sigh.

Then perhaps it's a no solution problem...

well, a cylinder is constructed by taking a circle and giving it height. a circle has the same width and diameter because that's its diameter. for other ellipses though, you can have different lengths and widths.

The problem was poorly worded. Do you have a solution manual to the problem?

Unfortunately not.

It's okay though. Thanks to everyone's contributions =)


49 students took a test with 3 questions and in each question they scored an integer from 0 to 7. Prove there are two students A and B such that A scored more in each question than B.

Does $A$ have to score strictly more than $B$, or at least as much as $B$?

Let us assume the contrary. We can show that the existence of a counter-example implies the existence of one in which the total scores of all the students are the same. But there are at most 48 distinct possibilities for any given total (attained when the total is 10 or 11). Let P be the contradiction of the required property (that is, no student has all three scores greater than or equal to another).

We proceed in a series of steps; in each step, we take the students with the lowest total scores and increase the highest score of each by one. It is clear that this will eventually result in a situation with equal totals. It remains to show that if the situation before each step satisfies P, then so does the situation afterwards.

Let $a_1,a_2,a_3$ be the scores of a student A with a (joint) lowest total. Let $b_1,b_2,b_3$ be the scores of any other student B. Wlog suppose $a_1 \geq a_2,a_3$, so in the step we increase $a_1$. This violates P only if $a_1 +1=b_1$.

Since P was true before, either $a_2 > b_2$ or $a_3 > b_3$. Suppose wlog that $a_2 > b_2$. Since B has the lower or equal total, $a_3 \leq b_3$, with equality only if A and B have equal totals. But in this case $b_1 > a_1 \geq a_2,a_3 \geq b_2,b_3$, so $b_1$ is also increased, preserving P, as required. If $a_3 < b_3$, then it is clear that P is also preserved.


Given a convex quadrilateral $ ABCD$ and let $ P$ be the intersection of $ AC$ and $ BD$
Denote by $ I_1,I_2,I_3,I_4$ the incenter of $ APB,BPC,CPD,DPA$ respectively
Prove:$ I_1,I_2,I_3,I_4$is concyclic iff $ ABCD$ has incircle.

Dear Mathlinkers,
a nice synthetic proof has been given by Darij Grinberg:

Grinberg D., Message Hyacinthos # 9022 du 10/01/2004 ;
http://tech.groups.yahoo.com/group/Hyacinthos/

I think that this problem has been presented on Mathlinks.
Sincerely
Jean-Louis

Thank you very much for the reference :)
This is the solution from Hyacinthos
[quote="darij grinberg"]We will make use of a well-known result about circumscribed
quadrilaterals:

Theorem 6. The incircles (A') and (C') of triangles DAB and BCD touch
BD at one point T.

The incircles (B') and (D') of triangles ABC and CDA touch AC at one
point U.

Doubters can perform an easy computation of lengths (making use of
the fact that AB + CD = BC + DA in the circumscribed quadrilateral
ABCD).

Of course, the point T lies on A'C'; moreover, the points T and R
divide the segment A'C' harmonically, since T is the internal center
of similtude of (A') and (C') and R is their external center of
similtude.

Now, we can apply Theorem 1 to triangle ABC and the point P on its
side AC. Consequently, the points P and U lie on the circle with
diameter XY. Hence, the points X, Y, P, U are concylic, and RX * RY =
RP * RU. Similarly, RZ * RW = RP * RU, and thus RX * RY = RZ * RW,
and the points X, Y, Z, W are concyclic. This proves Theorem 3. The
last part of this elegant proof is due to Juan Carlos Salazar.
[/quote]

This nice part proved if $ ABCD$ has incircle,then $ I_1,I_2,I_3,I_4$is concyclic
But I didn't see the proof for converse.Could you help me to find it? :lol:

p.s.Another proof for $ I_1,I_2,I_3,I_4$is concyclic from Kvant is pasted below.


advanced fields
advanced fields unsolved

Hi,

I'm trying to show that linear order is not FO^2 definable over finite directed graphs.  I know that I need to find structures A,B where only A exhibits linear order, and then show for each phi in FO^2, that phi cannot distinguish between A and B, but I'm pretty much stuck.  Any ideas?  Thanks.

Remind me again what FO^2 means.  I recall FO means "first order", but I don't recall the 2.  Also, when you say "linear order", does that mean the graph is just a long directed path, or that the graph has a topological ordering (acyclic)?

FO^k = first order with a maximum of k variables.

A graph that exhibits linear order is of the form <A,E> where E is an irreflexive, transitive, total relation on A.


I say we have a Sanskrit forum....

It is the first langauge ever....

And very useful in computers....

And yea....  :!:

Of course, I am just kidding. I don't think many people speak this fluently anymore as it was spoken a veeerrrryy long time ago.

there is an india forum, but i still support, because sanskrit was the [u]first [/u]language

no one else knows how to speak sanskrit?

haha, maybe they still speak it in Gondor, Middle Earth!

 :rotfl:

[quote="anirudh"]No. Sanskrit was the first language ever. Even older than latin and greek...[/quote]
Please try not to be so definite on matters with which you are not well acquainted. :wink:

Really, "first language" is a concept like "first human"—there is no scientific way to determine what it was, and it certainly isn't around any more.

[quote="Boy Soprano II"][quote="neelnal"][quote="jb05"]First language ever? Right...

Speaking of Gondor, what about a Quenya forum? Or more realistically, a Latin forum (after all, there are all those old papers in Latin). :)[/quote]

You are clearly ignorant of all history outside of European. Your kind of view is called Eurocentric. Very narrowed.

Sanskrit was the first language, with ancient tales dating back to about I'd say 30,000 BC. Texts seem to have withered until the Vedas, which using dating tech is about 15,000 to 10,000 years old.[/quote]
So there was no human speech before 30,000 BCE? :wink:[/quote]

so the cavemen didn't talk :rotfl:  :rotfl:  :rotfl:

[quote="neelnal"]the cavemen wrote on walls with pictures.

no really, there was no real written language before Sanskrit with a proper grammatical, etc. system[/quote]
Here you are arguing something very difficult.  In the first place, whether a language is written or not has very little to do with how complete of a language it is.  For instance, the Celtic languages existed for thousands of years before being written down (apart from a few rudimentary Ogham inscriptions on graves, etc.).  In the second place, your claim is still invalid, as, for example, the writings of the Harrapan civilization are indisputably older (since the Harrapan civilization was around before the Aryan migration and therefore before Indo-european tongues such as Sanskrit or its predecessors would have reached India), although linguists have not yet deciphered them, and the same is probably true of the strikingly similar writings that have been found on Easter Island.

Wow, you malinformed Asiacentric minds... Not all Americans/Europeans are as ignorant as you claim.

I believe Devanagari was the first phonetic alphabet (?), but plenty (as in thousands) of languages existed before Sanskrit. Sanskrit comes from the same place as most European languages, which was decidedly itself a language. Sanskrit is actually very recent linguistically.

[quote="anirudh"]I thought the language the assyrians used were wrong though[/quote]
What does this even mean?  How can a language be "right" or "wrong"?

That problem even arises with character map at times. By the way, is there anyone who's studying sanskrit here ?

I [i]was[/i] studying Sanskrit as my 2nd language when I lived in India for 6th and 7th grade/standard.

i studied it in my 9th and 10th standards .it was boring then but now i feel gr8 to have studed it :D

yes i am studying sanskrit from the past 4 years  :D  :P

So....can I safely say now that Sanskrit is the earliest language still to be spoken widely?

BTW I too studied sanskrit in 6,7,8th grades... :)

me too know Sanskrit a bit..n have studied it in 6-9th  :lol:  :maybe:

Sanskrit Exam on wednesday.

http://www.spokensanskrit.de/

This is a good website for learning sanskrit.

[quote="Rijul saini"]So....can I safely say now that Sanskrit is the earliest language still to be spoken widely?[/quote]
I don't think so.  A quick Google search leads to dubious claims for Tamil, Lithuanian, and various Australian aboriginal languages, among others.  (I'm surprised that I didn't see any for Hebrew.)  Also, it probably depends on what you mean by "still spoken".  Any language will change over thousands of years---look at how much English has changed since the time of Chaucer, or even Shakespeare.  One could say that "Greek" has been spoken for at least three thousand years or so, but my Greek friends have told me that the Greek of Homer and Sophocles and Euclid is rather different from the Greek spoken today.

By the way, it's not really cool to revive a thread from three or four years ago unless you have a good reason.  (Good reasons include posting a new solution to a problem on a forum.)  Start your own thread instead.

At this point the discussion should probably be in a different forum anyway---perhaps the Round Table or linguistics.  It looks like nobody was [i]ever[/i] talking seriously about anything relevant to this forum (National and Local Communities Round Table), even when the first post was made years and years ago.

I think it is a big nonsence to discuss against such anceint language.no can oppose that sanskrit was & still it is the most scientific language.for many many other causes it couldn't develop further.i stuied about sankrit & i beleive if it  would taken as universal language, our world would have more developed science ,maths.
It seems difficult but we have time to follow sanskrit.
>otherwise language is only a medium of conveing our veiws not else.

I personally think it is bad to revive posts that are 2 years old. Please stop it (reviving posts).
It just is spammy and we really don't need more spam.

I have been learning sanskrit for one year  I speak sanskrit quiet fluently
I think so it is a nice idea to have a sanskrit forum
I think we ought to vote for this sanskrit forum

Um... how's it the first language? Might be the first written language but not necessarily the first language.

^cuenifrom or hieroglyphics

[quote="math26"]Please don't bump a thread.[/quote]
Please don't backseat moderate

[color=#FF0000][mod: "Please don't backseat moderate" to you, too.][/color]


If the temperature of an aqueous solution of NaCl is increased from 20C to 90C, which of the following are unchanged?
density
molality
molarity

the property tyhat will remain unchange dopn changing the temperature is molality

as u change the temperature of the substance, the volume of the substance changes.

Since the density is dependent on volume,the density changes with temperature.
Also,writeing the formula for molarity we get that it is dependent on volume and so molarity also changes with temperature.

But write the molality formula, we get that it is dependent only upon the mass and not on the volume.

Since mass does not change with temperature,the molality is constant and does not change with temperature.

But theoretically, shouldn't the solvent evaporate more to reach equilibrium as the temperature increases? Thus none would remain unchanged?


algebra
function
domain
real analysis
real analysis unsolved

[quote="swag"]I know that the maximum (strong or weak) principle needs to be used in this problem, just can't get started:

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

On a bounded domain $\Omega$ with smooth boundary, let u be a solution to the problem:

$\Delta u$ + $a_{i}(x)\frac{\partial u}{\partial x_{i}} = f(x)$, $\frac{\partial u}{\partial n}=0$ on $\partial \Omega$ (the boundary of $\Omega$)

Assume that $f \geq 0$ in $\Omega$ and here n is the normal vector.

Show that u is constant and that $f=0$.[/quote]
I'm not sure but I don't think this problem deals with principle maximum modulus. If the boundary condition were $u=g$ on $\partial \Omega$ then it will be Dirichlet problem, the unicity of which is easily can be proved using principle maximum modulus.
In our problem with that stating, I think you should you Green's formules

[quote="swag"]On a bounded domain $\Omega$ with smooth boundary, let u be a solution to the problem:

$Lu = \Delta u$ + $a_{i}(x)\frac{\partial u}{\partial x_{i}} = f(x) \ge 0$, $\frac{\partial u}{\partial n}=0$ on $\partial \Omega$ (the boundary of $\Omega$)
Show that u is constant and that $f=0$.[/quote]
Use the strong maximum principle (Gilbarg-Trudinger 2nd ed. p. 34-35). If $u$ attains ist maximum in the interior of $\Omega$, and $Lu \ge 0$, then $u$ is constant and therefore $f = 0$, see Thm. 3.5. Suppose therefore that $u$ is not constant, then $u$ attains its maximum on $\partial \Omega$ at, say, $x_0$, and then $u(x_0) > u(x)$ for all $x \in \Omega$. But then $\partial_n u(x_0) > 0$ by Lemma 3.4, so this is impossible.

Yes, I was thinking about that solution. Seems to work, I just thought that it would be more complicated.

Thanks.


Η ασκηση βασιζεται σε μια ενδιαφερουσα προταση η οποια κιολας την ξεκλειδωνει.

Εστω $ ABCD$ περιγεγραμενο τετραπλευρο σε κυκλο που εφαπτεται στις $ AB,BC,CD,DA$ στα $ K,L,M,N$ αντιστοιχως. Εστω $ X$ το σημειο τομης των $ NK,ML$.
Φερω την $ AC$, η οποια τεμνει τον κυκλο στα $ P,Q$. ΝΔΟ $ XP\equal{}XQ$.

Ευχομαι καλη επιτυχια στα παιδια της εθνικης ομαδας αφιερωνοντας τους αυτην την ασκηση.

[quote="Athinaios"]thnx ρε Σακη.
Η ασκηση πιστευω οτι λυνεται γρηγορα αν παρατηρησουμε οτι,
το $ X$ ανηκει στην πολικη του $ A$  και στην πολικη του $ C$ ως προς τον κυκλο αρα η πολικη του $ X$ ειναι η $ AC$ δηλαδη η $ PQ$ αρα $ XP \equal{} XQ$ ως εφαπτομενα τμηματα που αγονται απο το Χ στον κυκλο.[/quote]

m areseis hlia!

[hide]La Hire's Theorem![/hide]

Αυτη ηταν η προταση που ειχα και εγω στο μυαλο μου.
Η ολοκληρωμενη διατυπωση του θεωρηματος ειναι η εξης:

Αν η ευθεια $ e$ ειναι η πολικη του σημειου $ A$ ως προς κυκλο και η ευθεια $ d$ η πολικη του σημειου $ B$
τοτε η $ AB$ ειναι η πολικη του σημειου τομης των $ e,d$.

Το θεωρημα αυτο εχει πολυ ωραιες εφαρμογες ειδικα οταν η μια ευθεια απο τις δυο
ειναι εφαπτομενη του κυκλου και εχει πολο το σημειο επαφης.


Asymptote LaTeX articles integration trigonometry

[quote="NeverOddOrEven"]I typed the following command into TeXnic Center:

draw((0,0)--(100,100));
draw((0,100)--(100,0));
dot((50,50));
label("$P$",(50,50),S);

I saved my file as text.asy, but when i got my output in GSview, the label P didn't show up. Does anyone know where I could've gone wrong?[/quote]

It seems to work:
[asy]draw((0,0)--(100,100)); draw((0,100)--(100,0)); dot((50,50)); label("$P$",(50,50),S);[/asy]

Do you see the label if you replace
label("$\$$P$\$$",(50,50),S);
by
label("P",(50,50),S);
in other words remove the dollar signs?
If you do see the label then Asymptote may not be finding LaTeX and that is dealt with in the last Problem in [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Help]Asymptote: Help[/url]

sry for not being clear the first time. i typed the following into TeXnicCenter

draw((0,0)--(100,100));
draw((0,100)--(100,0));
dot((50,50));
label("$P$",(50,50),S);

Then when i pressed my asymptote command thing (ctrl + F8), nothing opened so i couldn't see my image. i just see the asymptote program pop up for about 1 second and then disappear. I saved my file as text.asy and i configured my asymptote tool. what could be wrong?

another thing i am having problems with (probably related to my problem above), when i type the following into TeXnicCenter

\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\begin{document}
Hello.
I like to make pics with Asymptote like this one:
\begin{figure}[h]
 \begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{figure}
\par It makes me happy,
since I can still type my normal LaTeX stuff around it:
\(\displaystyle\int_0^{\pi}{\sin{x}}\,dx=2\)
\end{document}

I can see everything but the yellow smiley face.

The problem is that Asymptote is probably giving you an error message but the window closes too quickly to see it. It is possible to set things up so that the error message appears in TeXnicCenter, and once you have read it, you may know what to do, or can ask here. As you say the two problems you have are probably related.
Have a look at [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Showing_Asymptote_error_messages_in_TeXnicCenter]Showing Asymptote error messages in TeXnicCenter[/url] and set it up according to the instructions there (it is independent of your current Asymptote setup in TeXnicCenter so won't interfere with it).
Let us know how you get on.

i'm getting nothing to show in TeXnic Center

That is strange, as if it . Did Asymptote work before?
At the end of asymptote.bat add the line
[code]pause[/code]That will mean the window won't close instantly and gives you the opportunity to report what it says.

If the file was saved as file.tex did you get an empty file_asy.log? What other file.* files did you get? Was there a file.asy and if so, what was in it? Finally, what does file.log contain?


calculus
integration
real analysis
real analysis unsolved

Find:
$ \int \frac {6x^{10} \plus{} 25x^{9} \minus{} 4x^{8} \minus{} 32x^{7} \plus{} 44x^{6} \plus{} 63x^{5} \minus{} 8x^{4} \plus{} 76x^{3} \plus{} 72x^{2} \minus{} 16x \minus{} 6}{2x^{11} \plus{} 10x^{10} \minus{} 3x^{9} \minus{} 15x^{8} \plus{} 36x^{7} \plus{} 44x^{6} \minus{} 47x^{5} \minus{} 31x^{4} \plus{} 64x^{3} \plus{} 48x^2 \plus{} 16x \plus{} 12} \,dx$

It is not intended for solving it is a response to bookworm_vn and Extremal (http://www.mathlinks.ro/Forum/viewtopic.php?t=169577)... try to solve this ... to be more interesting ... try to solve without software :wink: ... I think it is an ugly problem and I may invent far more difficult but I think goal here is to spend good time in solving interesting problems. You are right ... my problems weren't difficult enough but my goal were to show something interesting available to more people. If you are interested in more difficult problems - check other my integrals as well as misan's integrals. I think there are many not solved integrals in mathlinks...

Will you solve this or to say the solution?

It was just a joke. It comes to show that sometimes it is too stupid to solve problems and there will be many problems that we never will solve :-). Of course I like good short and interesting problems that make you smarter. If you really want solution to this integral ...
See:
http://www.mathlinks.ro/Forum/viewtopic.php?t=167800 and
http://www.mathlinks.ro/Forum/viewtopic.php?t=169021 it is a sum of them.
So please if someone may solve: http://www.mathlinks.ro/Forum/viewtopic.php?t=169021 to solve it. Sometimes I cannot solve problems that are elementary to other people.
--
Success

OMG...HAHA :rotfl:  :rotfl:  :rotfl:

thats the funniest thing ive ever read on this forum.

[color=darkblue][b]borislav_mirchev[/b], you run over stock and block or wander in one's speech !

Do as you like and ... a rest ([b]pause[/b] or [b]stand by[/b] or [b]stop[/b]) .[/color]

[quote="Virgil Nicula"][size=200][color=darkred][b]GOOD NIGHT ![/b][/color][/size][/quote]
[color=darkblue][b]P.S.[/b] Please, give me a prognostication for the football-match [b]BULGARIA - ROMANIA [/b]![/color]

Bulgaria - Romania
12!        - 0
;) ... or 3 - one for Valentin Vornicu, Virgil Nicula and Carcul

Good Night

[color=darkred]Here is [b]1:33 AM[/b] - good night ![/color]

I think the time in bulgaria and romania is the same ;-)


[quote="bpms"][quote="4everwise"]Find the least positive integer $n$ such that no matter how $10^{n}$ is expressed as the product of any two positive integers, at least one of these two integers contains the digit $0.$[/quote]
[hide]If the two numbers that multiply to $10^{n}$ both have a 2 and a 5 then it'll automatically have a 0, so the factors are $2^{n}$ and $5^{n}$. So we need to find the smallest n so that one of them has a 0. For 2 it is 10. For 5 it is 7, so our answer is 7.[/hide][/quote]
[hide]
Do you mean $8$? $5^{7}= 78125$.
[/hide]

[quote="The Zuton Force"]I don't understand... wouldn't it always be divisible by 1, 2, and 5?[/quote]

What if we split it into factors that are only divisible by $2$ or only divisible by $5$?

My bad, I thought both the factors had to have a 0 at the end  :wallbash:

Yeah, but luckily it is 5^8, which is 390625, so there.

[quote="bpms"]Oh I thought $5^{7}$ was 80125 :blush:[/quote]

Ah, computation error.

5^7=78125, right.

Just wondering-if only one of the 2 integers have to have zero as a digit and not both, wouldn't the answer be 3

1000:
2x500
4x250
10x100
1x1000
20x50
40x25

All of these mathematical equations have at least 1 integer as 0.
So...........either I understood it wrong or your answers  aren't right. Um, could someone tell me what it is?

[quote="perfect628"]You've missed some factors. Consider 8 and 125...[/quote]
oops :blush:

Im still confused on what the answer is.
Basically, my thought process was pigeon hole rule (kinda).
To avoid getting a 0 in a product you group the 5's together and 2's together. (5^n) * (2^n). It was sorta intuition, but I realized that a 0 shows up in 2^10=1024. So I figured the answer was n=10.

Is this right? I'm not very good at explaining things, but I feel like my logic makes some sense.  :oops:

You have the right idea, but note that $ 5^8\equal{}390625$ and clearly $ 8<10$. Also, remember,

[quote="Altheman"]this problem was a pain...2^10 is an obvious upper bound, but you have to multiply out 5^k for k=4,5,6,7,8...[/quote]

both of the integers have no $0$ means one looks like $2^e$ while the other looks like $5^e$ so we have to find the least multiple of 2or 5 that has a 0 in it which is $5^8$  therefore,  $n=8$

is there any way to do this other than just calculating powers???

@above: I don't think so. We're left calculating the smallest $n$ such that $2^n$ or $5^n$ has a $0.$ And the $0$ is not at the end, so there isn't much we can do.

Well, good thing it's an AIME problem, so you could (theoretically) be computing up until 999.

[quote=4everwise]Find the least positive integer $n$ such that no matter how [quote=4everwise]Find the least positive integer $n$ such that no matter how $10^{n}$ is expressed as the product of any two positive integers, at least one of these two integers contains the digit $0.$[/quote]

We can always Express $10^n$ as $2^{n}  \times  5^{n}$ so no such n exists .

0 IQ solution
$5^8=390,625$
So the smallest integer is $n=8$

lol i thought 10 cause 2^10 = 1028 and its an aime problem so i thought its supposed to be clean somewhat, rip

[quote=think4l][hide=Solution]
Note that $10^n=(2^n)(5^n)$. Now, if we were to split up $10^n$ into two factors, in which one factor includes both a $2$ and a $5$, then we would have a $0$ in that factor. Therefore, the only case that remains is when we split it into $5^n$ and $2^n$. We see that the first time we get a digit $0$ in the two is $5^8$ and $2^{10}$. At $n=8$, even if we split it into only a power of 2 and only a power of 5, we would still end up having a digit of $0$ in the power of 5. Hence, the answer is $n=\boxed{008}$.
[/hide][/quote]

Nice one!

[hide=sol]$10^n = 5^n \cdot 2^n.$ We know that if one of the positive integer contains both factors $2$ and $5,$ then its always contains the digit $0.$ Hence, we have to find the minimum $n$ such that either $2^n$ or $5^n$ contains the digit $0$ to account for the case where one factor is $2^n$ and the other being $5^n.$ After some calculating, we see that since $5^8$ contains the digit $0,$ the answer is $\boxed{008}.$[/hide]

Note, that $10^n=5^n\cdot 2^n.$ For, all values of $n=1,2,3,4,5,6,7$, there is no digit $0,$ but for $8,$ we have that $5^8$, has digit $0,$ hence, the answer is $\boxed{8}.$


Of course we want! Please, write them!  :lol:

[quote="N.T.TUAN"]Yes, I very want  :wink:[/quote]
ok, these are Mongolian national olympiad2005's problems. i'll post 2006's problems soon.
[b]Problem1[/b]
On plane located 2005 white circle and 2005 black circle. Any black circle intersects at least 1003 white circle, any white circle intersects at least 1003 black circle. Prove that there exists a line which intersects 1 black circle and 2005 white circle.
[b]Problem2[/b]
The circle w has diametr BC side of acute angle triangle ABC. The circle which externally tangents w tangents at A1 tangents AB and AC. and defined B1 and C1 similar to A1. Prove that AA1, BB1, CC1 are concurrent

[b]Problem3[/b]
Let a.b.c are real such that a>=b>=c>0, a+c>=2b.
Prove that 4(a^4+b^4+c^4)+6abc>=5(a^3*b+a^3*c+b^3*c+b^3*a+c^3*a+c^3*b)

[b]Problem4[/b]
Acute angle triangle ABC has perpendiculars AH1, BH2 and angle bisectors AL1, AL2. Let O is the circumcenter and I is the incenter. Prove that if and if only L1, O, L2 are collinear then H1, I, H2 are collinear.

[b]Problem5[/b]
On a 12*12 board located queens such that any queen to threaten exactly 3 queen. How many queen in a board at most?

[b]Problem6[/b]
m1, m2 , ...., mk are distinct positive integers. If their sum is n(n+1)/2 Can divide set (1, 2,3,....,n) into subsets A1, A2,...,Ak such that the sum of all elements of subset Ai(1<=i<=k) is Mi?
[b]1[/b][url]http://www.mathlinks.ro/Forum/viewtopic.php?p=788140#788140[/url]
2[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=139265[/url]
3 [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=788142#788142[/url]
4 [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=788148#788148[/url]
5 [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=788152#788152[/url]
6 [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=788154#788154[/url]

I am sorry. I wrote 3rd problem wrong. It was
Problem3
Let a.b.c are real such that a>=b>=c>0, a+c>=2b.
Prove that 4(a^4+b^4+c^4)+6abc(a+b+c)>=5(a^3*b+a^3*c+b^3*c+b^3*a+c^3*a+c^3*b)


ESPN2 is scheduled to rebroadcast the 2004 MATHCOUNTS National Competition Countdown Round on Friday, August 20 at 11 am ET and again at 4 pm ET.

According to MATHCOUNTS, "ESPN has granted permission for the program to be taped for viewing by your class at a later time".

Thanks for the notice.  I missed the initial broadcast because I didn't have cable, but I'll tune in this time.

How irritating.  ESPN2 decided to keep showing golf - it's 10:10 and it's still showing.

In the end, the show started 20 minutes late.  The good part was that they cut out a lot of the commercials to make up for it, so there were very few commercials interrupting it.

[quote=sleepypuppy]I did not watch this live.[/quote]

this is from 2004 leave it alone
someone lock this pls

[quote=sleepypuppy]I did not watch this live.[/quote]

Actually. Stop bumping forums from over 10 years back.

[quote=sleepypuppy]I did not watch this live.[/quote]
bro you weren't even born back then

also why was mathcounts over the summer back then


geometry
perimeter
analytic geometry
combinatorics unsolved
combinatorics

All vertices of a polygon in a coordinate plane are integer points , and all its sides have integer lengths. Prove that its perimeter is even.

Going clockwise around the polygon, let its $ n $ vertices have coordinates $ ( x_i, y_i ) $. Now $ \sum ( x_i - x_{i+1} ) = 0 $, (where $ x_{n+1} = x_1 $) and since taking the square of each term preserves it's parity, the sum $ \sum (x_i - x_{i+1})^2 $ must be even, and similarly with $ \sum (y_i - y_{i+1})^2 $. So the sum of the squares of the side lengths, $ \sum ( (x_i - x_{i+1})^2 + (y_i - y_{i+1})^2) $ must also be even, and taking the square root of each term, we find that the perimeter must be even as well.

[quote]and taking the square root of each term, we find that the perimeter must be even as well.[/quote]

I am sorry, but why is that?

Isn't it obvious? :)

The square of a natural number has the same parity as the natural itself. This means that the sum of several squares has the same parity as the sum of their square roots.

OH! I forgot that all the side lengths are integers, sorry...yeah, it is obvious. :lol:


Hello!
I'm alex and live in germany. But I am half British and half American. I would like to apply for Cambridge to study maths or maybe physics.
Is there anyone here you has already been through the application process and can explain how it went. I've taken part in several math competitions (USAMTS, Bundeswettbewerb Mathematik, National Competitions). How much will this help me for cambridge? I've also heard that you might be asked to take a STEP exam. Does anyone have experience with this?

Online I read that you first need to send in an application to UCAS where you also need to state which college you want to be in. For which college are the chances of being accepted the greatest? There is also the option for an "open" application, where they assign you to a college. Do you reckon that this might be a good option in order to be accepted?

I would really appreciate your help!
thanks
alex

I asked a friend of mine, Stergios, to reply you in the [url=http://www.mathlinks.ro/viewtopic.php?p=1251896#1251896]British forum.[/url] I think this forum is strongly US-biased, mostly undergraduate (Bachelor). It should rather be called "US Colleges". MIT and standardized tests are a popular topic. But then sometimes people discuss the Budapest Seminar or IIT from India.  :lol:

From all British universities, Cambridge and its rival Oxford are government-funded, and under pressure to become more inclusive. The government wants half of all young people to attend college by 2010, which means universities need to target all economic backgrounds. In some ways Oxford and Cambridge -- elegant, affluent universities known collectively as "Oxbridge" resemble U.S. Ivy League schools, which have long tried to attract minority and less well-off students through scholarships and outreach programs.
---------------------------
marita

[url=http://www.drivenwide.com]Guaranteed ROI[/url]

I would have a question regarding Cambridge Univ., and since this thread is about Cambridge, I thought I would put it here. I wasn't invited to interview by Trinity College and I don't know if my chances of getting there got lower or they just thought that I live too far (I'm from Romania) to come for an interview. I'm a bit worried and if any of you knows how things work, I would be really grateful if you told me. Thanks and I'm waiting for your replies.


Consider a n*m grid. a particle starts from [0,0] and reaches [n,m] by moving either up, down, right or left every move(if possible, I mean for example only up or right are possible moves from[0,0]).

 Enumerate the possible paths given
[i] No portion of the path is re traced.
[ii] Path passes through the point only once.

[img]http://bp3.blogger.com/_j64KsKfZBQk/RuUTW5CJzKI/AAAAAAAAAIg/IpCJZQFCV5U/s1600-h/latice.JPG[/img]

this is one example of the path[/img]

It seems hard -- even the case $ n \equal{} 2$ is nontrivial.


Prove for all real numbers $ \not= 0$:

$ x^{8}-x^{5}-\frac{1}{x}+\frac{1}{x^{4}}\geq 0$

[hide="Not a nice solution"]
$ x^{8}-x^{5}-\frac{1}{x}+\frac{1}{x^{4}}\geq 0 ~|\cdot x^{4}$

By polynomial long division with $ (x-1)$ we get:

$ (x-1)(x^{11}+x^{10}+x^{9}-x^{2}-x-1)\geq 0$ (*)

a) $ x \geq 1$ : both factors are positiv; (*) is nonnegative.

b) $ x \leq 1$ : both factors are negativ; (*) is nonnegative.

[/hide]

[quote="MellowMelon"][hide="Better way"]Assume $ x$ is positive.
$ x^{8}+\frac1{x^{4}}-(x^{5}+\frac1x) \geq 2x^{2}-2x^{2}= 0$ by AM-GM.

Now assume $ x$ is negative
$ x^{8}+x^{5}+\frac1x+\frac1{x^{4}}\geq 4x^{2}\geq 0$ by AM-GM.[/hide][/quote]

You just showed the minimum of $ x^{8}+\frac1{x^{4}}$ is greater than or equal to the minimum of $ x^{5}+\frac1x$.  :huh:

[quote="April"]Thanks. My stupidity!  :blush:[/quote]

No problem, your solution is elegant. :)

This leads me to think: if a polynomial is non-negative for all values can it always be arranged into a sum of squares?


a,b,c>0
prove that:
$ \sum(b\plus{}c)\sqrt{\frac{(b\plus{}a)(c\plus{}a)}{bc}}\geq 4(a\plus{}b\plus{}c)$

It's very easy but nice  :lol:
By cauchy-swarchz anh Am-Gm we have:
$ LHS \ge \sum (b \plus{} c)(1 \plus{} \frac {a}{\sqrt {bc}}) \equal{} 2(a \plus{} b \plus{} c) \plus{} \sum\frac {a(b \plus{} c)}{\sqrt {bc}} \ge 4(a \plus{} b \plus{} c)$  :P


MATHCOUNTS
geometry
trigonometry
probability
AMC 8
AMC 10

Ok, who from the Bay Area does Mathcounts here? I just want to see how badly I'll fail this year. I even have doubts about making it to chapter.  :o

Post how smart you are and discourage me please. :P

if you suck at mathcounts, i suggest you just dont do it at all, and just do more aime/usamo/olympiad stuff
i mean just dont practice for it
its a waste of time

I think Mathcounts is highly biased towards speed (sprint and countdown), and it doesn't have enough questions that actually make you think (target and team). Perfecting mental computation can sometimes be useful later on in life, but actual problem solving skills (AMC 8 and higher) are much more important.

You can do Mathcounts to make sure you can do easy problems quickly, but don't waste too much time on it. Instead, focus on more difficult contests (AMC 10 and higher), where mental computation isn't the most important thing.

One final note (on harder competitions): Don't start practicing hard unless you're sure you'll be able to cover all the topics that could be on the competition. It is really frustrating to solve all the problems that you practiced for, and then be stuck with a couple which you have no idea how to handle.

Once you get to the point where you can easily solve any MATHCOUNTS problem, I suggest moving on to other things, and not spending time getting faster and more robot-like.

I can't easily solve all Mathcounts problems, maybe about 3/4 of them. The rest either take some time, or I can't do them at all. Speed is not what I'm worried about right now.

Seriously, after seeing some of the solutions in the Mathcounts forums for this long, I feel like I'm going to fail pretty badly this year.

Does anyone know where I can get past AMC problems?

[quote="CheeseIsGood"][quote="junggi"][quote="CheeseIsGood"]I can't easily solve all Mathcounts problems, maybe about 3/4 of them. The rest either take some time, or I can't do them at all. Speed is not what I'm worried about right now.
[/quote]
its called "The Art of Problem Solving"
get both books
should be great intro to stuff[/quote]

I have the first one. What's in the second one?[/quote]
1) do the first one. read it, do the problems, etc etc
2) then buy volume 2. its more advanced stuff

[quote="CircleSquared"]Rule of thumb: A post on AoPS MC forums is usually around nationals level.

Anyways I do mathcounts, probably anyone could have told you that though.

You have to buy AMC problems.  And they're MUCH easier than MC.  Trust me.

I recommend buying the book "First Steps for Math Olympians" if you really want to prepare with AMC's.[/quote]

Are you talking AMC 8, 10, or 12?

8 is way easier.
10 is comparable.
12 is way harder.

[quote="CircleSquared"]AMC $ 2^3$[/quote]
oh snap, that was clever

Based on the problems I've done, I find the AMC 10 (and the last few problems of AMC 8) more closely mirrors the tougher problems in MATHCOUNTS.

[quote="CircleSquared"]Okay, who gave my post a 1 rating?
If I get one more useless rating like that I'll petition to have ratings removed.[/quote]
dude shut up ratings dont matter at all

[quote="CircleSquared"]Okay, who gave my post a 1 rating?
If I get one more useless rating like that I'll petition to have ratings removed.[/quote]

Get used to it...

Btw, have you asked Michelle out for me?

[quote="Ubemaya"][quote="CircleSquared"][quote="dwatrgrl"]i don't understand why people are saying that AMCs are for wimps and Mathcounts are for noobs and all. True, mathcounts really does emphasize on speed and it doesn't require much thinking than the AMC. But really, if u really look into the State Competition or the National rounds of Sprint, there are many problems that include "thinking" and "problem solving." It's rare enough for any one to get some National Sprint Round all right, why? Because its either (a) silly mistake(s) or that it's hard. People that win Nationals are people who spent countless hours practicing Mathcounts, learning in depth questions, and taking time to study the main topics of MC...So basically, if ur an expert at Mathcounts, guaranteed, in AMC 8 and 10, 1-15 will be a breeze, AND because of so much timing practice, you would be able to finish them in 20-35 minutes!  And then u guys are talking about how IMO uses a variety of questions and Mathcounts doesn't, did u guys ever notice that at least 1 person in the U.S. IMO team always was a National Mathcounts participant? What u guys are saying doesn't make much sense.[/quote]

AHH! A BUMP!  *skids to a halt*[/quote]

... Huh? I don't get it.[/quote]
A car hits a bump and it skids to a halt?

I never openly told them they were wrong, I just said that they didn't make much sense, meaning (if u didn't realize) that i didn't understand them, not that they were plain wrong. wth?

[quote="junggi"][quote="serialk11r"]Look our friend Ubey here just likes to kid around and say random stuff like that, don't worry about it, and don't take it seriously.[/quote]
actually, do take it seriously, what he said made perfect sense, and, if you are still doing mathcounts, i suggest you move on if you find math interesting

you never explicitly stated they were wrong, but your post obviously imply that they were wrong

anyway, that wasnt the important part of ubey's post. the important part was the other 90% which shot down your entire post[/quote]
I meant the "AMCs are for whimps. Girls are where its at" posts are not to be taken seriously. His argument on why Mathcounts shouldn't be taken too seriously is correct though.

look, junggi, i don't want to sound so defensive or anything, but wat i posted up wasn't really to offend math veterans, i was just really really confused. And if u think that i was implying that u guys are wrong, i guess u never heard of something called a good argument or discussion. i posted up that up to get wat others thought and feedback, which someone did...

It's okay, we're not mad at you or anything (at least, most of us aren't...).

By the way, it would be nice if you typed in real English, because personally I can't read chatspeak very well, especially in huge blocks of prose. I dunno about everyone else here, but that's just a pet peeve of mine...
It probably helps that I type at 110 wpm so I can type huge blocks of text faster than most people can type it in chatspeak.

[quote="CircleSquared"][quote="junggi"][quote="serialk11r"]Look our friend Ubey here just likes to kid around and say random stuff like that, don't worry about it, and don't take it seriously.[/quote]
actually, do take it seriously, what he said made perfect sense, and, if you are still doing mathcounts, i suggest you move on if you find math interesting

you never explicitly stated they were wrong, but your post obviously imply that they were wrong

anyway, that wasnt the important part of ubey's post. the important part was the other 90% which shot down your entire post[/quote]

don't be mean
ubey's trying to help for once
seriously[/quote]
It would be really helpful if you would read the post after that and not post stuff like this.

[quote="serialk11r"][/quote][quote="CircleSquared"][quote="junggi"][quote="serialk11r"]Look our friend Ubey here just likes to kid around and say random stuff like that, don't worry about it, and don't take it seriously.[/quote]
actually, do take it seriously, what he said made perfect sense, and, if you are still doing mathcounts, i suggest you move on if you find math interesting

you never explicitly stated they were wrong, but your post obviously imply that they were wrong

anyway, that wasnt the important part of ubey's post. the important part was the other 90% which shot down your entire post[/quote]

don't be mean
ubey's trying to help for once
seriously[/quote][quote="serialk11r"]
It would be really helpful if you would read the post after that and not post stuff like this.[/quote]

uhh i was talking to junggi
and i know you were referring to ubey's girls posts, i saw your post.

[quote="CircleSquared"][/quote][quote="serialk11r"][/quote][quote="CircleSquared"][quote="junggi"][quote="serialk11r"]Look our friend Ubey here just likes to kid around and say random stuff like that, don't worry about it, and don't take it seriously.[/quote]
actually, do take it seriously, what he said made perfect sense, and, if you are still doing mathcounts, i suggest you move on if you find math interesting

you never explicitly stated they were wrong, but your post obviously imply that they were wrong

anyway, that wasnt the important part of ubey's post. the important part was the other 90% which shot down your entire post[/quote]

don't be mean
ubey's trying to help for once
seriously[/quote][quote="serialk11r"]
It would be really helpful if you would read the post after that and not post stuff like this.[/quote][quote="CircleSquared"]

uhh i was talking to junggi
and i know you were referring to ubey's girls posts, i saw your post.[/quote]
Well then read his post more carefully, since junggi is supporting ubey, and not saying a single bad thing.

[quote="serialk11r"][/quote][quote="CircleSquared"][/quote][quote="serialk11r"][/quote][quote="CircleSquared"][quote="junggi"][quote="serialk11r"]Look our friend Ubey here just likes to kid around and say random stuff like that, don't worry about it, and don't take it seriously.[/quote]
actually, do take it seriously, what he said made perfect sense, and, if you are still doing mathcounts, i suggest you move on if you find math interesting

you never explicitly stated they were wrong, but your post obviously imply that they were wrong

anyway, that wasnt the important part of ubey's post. the important part was the other 90% which shot down your entire post[/quote]

don't be mean
ubey's trying to help for once
seriously[/quote][quote="serialk11r"]
It would be really helpful if you would read the post after that and not post stuff like this.[/quote][quote="CircleSquared"]

uhh i was talking to junggi
and i know you were referring to ubey's girls posts, i saw your post.[/quote][quote="serialk11r"]
Well then read his post more carefully, since junggi is supporting ubey, and not saying a single bad thing.[/quote]

personally the part about "the important part was the 90% that shot down your entire post" was pretty mean to me, for lack of a better word  :(

[quote="CheeseIsGood"]This thread was revived, and has now turned into an argument. Here are the teams that I can distinguish.

Team 1:
dwatrgrl

Team 2:
Ubemaya
junggi
CircleSquared

Team 3:
serialk11r

I'm bad at this. :ninja:[/quote]

NOT MY FAULT PLS
/KILL THREAD


I will post some nice and relatively simple results from 'Fundamentals of Number Theory' by William J. LeVeque. Here goes the first one:

Let a be a positive integer such that gcd(a, 561) = 1. Show that 561 divides a560 - 1.

[quote="ComplexZeta"]No, 561 isn't prime. It's a Carmichael number.[/quote]

Oh, poo, you're right.

it's 3.11.17

but I think that won't help a lot here :?

As was mentioned, 561 is the product of 3, 11, and 17.  Also, 560 happens to be a multiple of 3-1, 11-1, and 17-1.  Thus by Fermat's Little Theorem, a560 is 1 mod 3, 11, and 17.  Combining, we get a560 is 1 mod 561.

:? thought the art was just to proof that that followed from FLT ?

or was I sleeping once again when I tried it?


i know that somewhere on AoPS i saw a theorem called the "Chicken McNugget Theorem" that said that, if chicken mcnuggets are sold in sets of m and n, then mn-m-n is the largest number that can't be sold.  how would one prove this?  is there a version for more sets?

search it, there are dozens of proofs already on aops.

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=mcnugget+chicken&t=129763

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=mcnugget+chicken&t=114774

j/w why is it called the chicken mcnugget theorem?

1. search it, it's on aops about 6 hundred times.
2. refer to the topic's first post.


In trapezoid ABCD, AB is parallel to CD and angles B and D are a cute. if sin of angle b= cos of angle D, compute angle a- angle b in degrees.

[hide]Since $\sin B=\cos D$, $B+D=90$.

Also, $A+D=180$ because of parallel lines.

Subtract the first equation from the the second to get $A-B=90$.[/hide]

[hide]If $\sin B=\cos D$, then they are complementary.
If B=x, D=90-x, A=90+x (from same-side interior angles(?)), A-B=90 degrees.[/hide]


An $ a,b,c>0$ kai $ ab\plus{}bc\plus{}ca\plus{}2abc\equal{}1$ na apodeixete oti

$ \sum\frac{(a\plus{}ab)^2}{(b\plus{}ab)^2\plus{}(1\minus{}ab)^2}\geq a\plus{}b\plus{}c$

Ωραία και ψαρωτική Κώστα.

Θέτουμε $ a\equal{}\frac{1\minus{}x}{x}$ κτλ . Τότε $ x\plus{}y\plus{}z\equal{}2$. Και η ανισότητα είναι ισοδύναμη με

$ \sum\frac{(1\minus{}x)^2}{(1\minus{}y)^2\plus{}(1\minus{}z)^2}\geq\sum\frac{1\minus{}x}{1\minus{}y\plus{}1\minus{}z}$

Δηλαδή αν θέσουμε $ 1\minus{}x\equal{}m$ , $ 1\minus{}y\equal{}n$ και $ 1\minus{}z\equal{}s$ η ανισότητα είναι ισοδύναμη με

$ \sum\frac{m^2}{n^2\plus{}s^2}\geq\sum\frac{m}{n\plus{}s}$

Η τελευταία όμως είναι ισοδύναμη με

$ (m^2\plus{}n^2\plus{}s^2\plus{}mn\plus{}ns\plus{}ms)\sum\frac{mn(m\minus{}n)^2}{(n\plus{}s)(s\plus{}m)(n^2\plus{}s^2)(s^2\plus{}m^2)}\geq 0$

που προφανώς ισχύει


ratio
geometry
geometric transformation
reflection
MATHCOUNTS
geometric sequence
AMC 8

We will call a sequence of 5 numbers, {a, b, c, d, e} "special" if it has all of the following properties:

1) It is composed entirely of positive integers
2) it has a unique mode
3) it contains its mean
4) The mean, median and mode are three distinct numbers (that is, no two of them are the same)
5) b, c, d are in geometric progression with ratio 2 (that is, c is twice b and d is twice c)
6) a :le: b :le: c :le: d :le: e = 72

Find all such sequences.

What if we allow a to be negative (but require b, c, d and e to be positive)?  Are there any other solutions?

:Note: edited to reflect tokenadult's comments.

[quote="JBL"]We will call a sequence of 5 numbers, {a, b, c, d, e} "special" if it has all of the following properties:

5) b, c, d are in geometric progression with ratio 2

[/quote]

Hi, JBL, thanks for sharing a challenge problem. Now, here I will explain why I think this is probably an intermediate level problem (although anyone else posting here is welcome to prove me wrong). I think in most mathematics syllabuses, an explicit explanation to the student of what a geometric progression is would come at a HIGHER curriculum level than the level assumed by Mathcounts or by AMC 8, which I am taking as being indisputably at getting started level. In most syllabuses I have seen, the algebra of geometric progressions follows a first course in geometry (in other words, it is an "algebra 2" topic) and thus this would be AMC 10 level stuff, or what I am taking to be intermediate level.

I'm just trying to keep a dialog going about where our level boundaries are, until I'm straight about where to post problems on the board.

I was under the impression that geometric sequences were taught fairly early, but I honestly couldn't say for sure.  Are you sure that that terminology doesn't make it on to mathcounts or AMC 8?

Either way, if that's the only complaint, I'll just edit the question.

[quote="JBL"]I was under the impression that geometric sequences were taught fairly early, but I honestly couldn't say for sure.  Are you sure that that terminology doesn't make it on to mathcounts or AMC 8?[/quote]

I'm less clear about what's on and off the assumed syllabus for Mathcounts than I am for AMC tests, but [referring to reference material] the Mathcounts 2003-2004 School Handbook includes a glossary, and that glossary includes the terms "sequence," "arithmetic sequence," and "geometric sequence." So I guess Mathcounts, which I will accept as an example of a getting started level competition program, uses that terminology in problems.

By contrast, AMC 8 doesn't use terminology like that. It might have such a sequence, defined by listing terms, but it wouldn't use that terminology and expect the student to know what that terminology means, I'm awfully sure. (I've been looking at quite a few old AMC 8s in the last few months, and have more recently been looking at all of the old AMC 10s.)

In my son's current Algebra II class, geometric sequences come in the last month of the course. They are near the end of Gelfand and Shen's textbook Algebra (which has both "algebra 1" and "algebra 2" topics) and weren't thoroughly presented in my high school courses until the algebra 2 course. So I guess on this basis (and on the basis of other facts I have observed), Mathcounts can be taken to be a little higher in curriculum level than AMC 8, which is, after all, strictly prealgebra mathematics really.

Having seen the Mathcounts handbook, I'll withdraw my objection to the challenge question, and from now on I'll know where to turn for another test of question levels.

[hide]

Rewrite the sequence as

a, b, 2b, 4b, 72

The mean is (a+b+2b+4b+72)/5.  Since it contains its mean, this is either equal to b, 2b, or 4b.

Unique mode guarantees that a=b or d=e.

Median is obviously 2b. This means the mean is either b or 4b.

1 :le: a :le: 16 (since d is at most, 68).

Consider the mean b:

(a+b+2b+4b+72) = 5b

-2b = a + 72, contradiction

Consider the mean 4b:

a+b+2b+4b+72 = 20b

13b = 72 + a

if b=5, 13b = 65

if b=6, 13b = 78

if b=7, 13b =  91

We see that a=6, b=6 will work.

Thus, the only positive series is 6, 6, 12, 24, 72.

If a can be negative, then we return to the mean=4b case:

Fixed (I think):

13b = 72 + a

The sequence must satisfy a=b or 4b=72 (b=16). Obviously a is not b (negative and positive) and b=16 is far too big, so there are no sequences.

From the mean=b case:

-2b = a + 72

If a=b (to satisfy the unique mode) but a is not equal to b (since one is negative while the other is positive). Thus, no solution.

So, there are no additional sets.

[/hide]

Edited because I read the question incorrectly in the beginning.

Good job on the first one, although you can do it without testing specific numbers.  In fact, it's plugging in specific numbers that causes your problem in the second part.  So half good job, half keep trying :)


As happened in the new LaTeX forum, guests are able to post in the International Contests forum.  Apparently someone has taken advantage of this and spammed the place up by linking to some site or another.

You've probably seen this already, Mr. Rusczyk, but if you haven't, please delete the spam messages and disable guest posting.

[And to everyone else]: Don't go to whatever site the guest poster linked to.

You actually missed when there were three spammerbots, covering all the forums with that same trash.  I think Mathew is working on it.

This afternoon. They posted like the same stuff 10 times in each forum.

I banned the users, their IP address, and all 150ish of their posts have been deleted.

At first I thought the site got a lot more popular since yesterday. :D .  Then I saw that it was just a hacker.  I think hacking is stupid especially when it's something like that.  Thanks for cleaning it up.

I don't want to start a new topic for this (especially in the "spam report" forum LOL), but...

Can we put a line limit on signature length? Because there's a new user (I don't want to mention his name, but he's posted several times in the MathCounts forum) whose signature is one digit per line, and his sig takes up more space than his posts, usually.

[quote="Fermatprime"]his sig takes up more space than his posts, usually.[/quote]

Always.  It's very annoying.  I would also appreciate it if you either shrunk his sig manually or limited its length.

Thank you, Mr. Crawford.

And I have no qualms about saying [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=85679#85679]who[/url] has the offensive signature, since it will speed up the process of ``encouraging'' that person to shorten it.

Um, I was sort of hoping for a more "generalized" solution, because otherwise, the same thing could happen again, and we don't want to have Mr. Rusczyk and Mr. Crawford having to shorten every single offender's sig manually...

Hmm, FICS was just hacked as well. Perhaps it's some sort of Hacking challenge (yes, those happen from time to time. Usually it's a bunch of kiddie hackers who want to brag about their exploits in some hacking forum. lame)


Select 35 numbers from  the first 70 positive whole numbers ( which are 1, 2, ............, 70)  such that none of the selected numbers  is a multiple or a divisor of any of the remaining selected numbers.

Is it possible  to select more than 35 such numbers satisfying all the conditions of the problem?

1 is a divisor of all numbers, and any number is a multiple of 1, so it's impossible.

1) select the numbers 36-70.

2) hmm....

Any number can be written in the form $2^ab$ with b odd. The only possible odd parts for the question are $1, 3, \cdots , 69$ which is 35 numbers. Thus if 36 numbers are selected, two have the same odd part and thus one is a multiple of another.

This can easily be generalized to selecting n+1 numbers for any group of 2n numbers from 1 to 2n.

[quote="filletwho"]9 isn't a multiple of 6.[/quote]

$9$ and $6$ don't have the same odd part. If $2^ab$ and $2^cb$ are two numbers with $a < c$, we indeed do have $2^ab | 2^cb$.


Find all functions f:N (to) N
  that what:
                    f(x + f(y)) = f(x) + y; for all x,y positive integer numbers.

[quote="al.M.V."]In fact, it is slightly more interesting to consider the following:

Find all f: R->R such that f(x+f(y))=f(x)+y.

First show that f is one to one. (Suppose there exists i\=j such that f(i)=f(j). Then f(x+f(i))=f(x+f(j)) implies i=j, contradiction.)
Now setting y=0 gives f(x+f(0))=f(x). It follows that f(0)=0 (recall f is one to one.)
Observe f(f(x+f(y)))=f(f(x)+y)=x+f(y), now setting y=0 gives f(f(x))=x.
Moreover, f(f(x)+f(y))=x+y, and f(x+y)=f(x)+f(y). (This is known as Cauchy's Equation.)
Hence all solutions take the form f(x)=cx, where f(1)=c is a constant.
We trivially check to see that c=1 or c=-1. Hence f(x)=x and f(x)=-x are the only solutions.

Now if you insist on f: N->N, then f(x)=x is the only solution. :D[/quote]

This is wrong for restriction to [tex]\mathbb{N}[/tex]. You cannot extend the range of function because you may lose some solutions. Another flaw is that you cannot put [tex]y=0[/tex] under restriction.

Put [tex]y=t+f(z)[/tex], so [tex]f(x+f(t+f(z)))=f(x)+f(z)+t[/tex] or [tex]f(x+z+f(t))=f(x)+f(z)+t[/tex]. Put [tex]y=t[/tex] and change [tex]x[/tex] to [tex]x+z[/tex]. We get [tex]f(x+z+f(t))=f(x+z)+t=f(x)+f(z)+t[/tex]. Hence [tex]f(x+y)=f(x)+f(y)[/tex]. Thus [tex]f(n)=n\cdot f(1)[/tex] and, putting into main equality, we get [tex]f(1)=1[/tex]. So [tex]f(n)=n[/tex]


How do you prove these two Fibonacci identities?

$ F_{n\minus{}1}\cdot F_{n\plus{}1} \equal{} F_{n}^2 \plus{} (\minus{}1)^n$
$ F_{m\plus{}n\plus{}1} \equal{} F_{m\plus{}1} \cdot F_{n\plus{}1} \plus{} F_{m} \cdot F_n$

Use:
$ F_n \equal{} \frac {1}{\sqrt 5}\left(\left(\frac {1 \plus{} \sqrt 5}{2}\right)^n \minus{} \left(\frac {1 \minus{} \sqrt 5}{2}\right)^n\right) n\in\mathbb{N}_0$

rest you need to do is calculation.

Alternatively, use induction, because about every fibonacci problem can be done by it ;)

The squence $ F_{n \minus{} 1}\cdot F_{n \plus{} 1} \minus{} F_{n}^2$ is a geomteric sequence with common ratio $ \minus{}1.$

You can use Matrix. I have posted the problem or the solution, I don't remember well, I'm sure that darij has posted the solution using Matrix.

[quote="kunny"]The squence $ F_{n \minus{} 1}\cdot F_{n \plus{} 1} \minus{} F_{n}^2$ is a geomteric sequence with common ratio $ \minus{} 1.$
[/quote]
I don't understand how to get this result?  :huh:

[quote="t0rajir0u"]A discussion of the matrix technique can be found [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=175257]here[/url].[/quote]
I haven't really learned anything about matrices beyond simple addition, subtraction, multiplication..., so are there any other ways besides matrices?

Induction.  Alternately, consult the link.  It follows directly by taking the determinant of the "Fibonacci matrix."

"Your" Fibonacci numbers are defined by $ F_0 \equal{} 0, F_1 \equal{} 1$ so $ F_2 \equal{} 1$, etc.  If we change to a differently-indexed set of Fibonacci numbers, defined by $ f_0 \equal{} f_1 \equal{} 1$, so $ f_2 \equal{} 2$, then the second equation to be proven can be rewritten as $ f_{m \plus{} n} \equal{} f_{m}f_{n} \plus{} f_{m \minus{} 1}f_{n \minus{} 1}$.  From there, we can use a [hide="bijective proof."]
Note that $ f_n$ is the number of ways of tiling a $ 1 \times n$ strip with $ 1\times 1$ and $ 1 \times 2$ tiles.  ($ f_2 \equal{} 2$ and the two ways to tile a $ 1 \times 2$ strip are with two $ 1 \times 1$ or with one $ 1 \times 2$.)  Thus, we can tile a $ 1 \times(m \plus{} n)$ strip in $ f_{m \plus{} n}$ ways.  We can also tile it as follows: start at the $ m$th square.  If we cover the $ m$th and $ m \plus{} 1$st squares with a single $ 1 \times 2$ tile, we're left with a strip of length $ m \minus{} 1$ and a strip of length $ n \minus{} 1$ to tile, so $ f_{m \minus{} 1}f_{n \minus{} 1}$ ways.  If we [i]don't[/i] cover the $ m$th and $ m \plus{} 1$st squares with a single tile, we must tile the first $ m$ and last $ n$ tiles independently, so $ f_{m}f_n$ ways.  These possibilities are exhaustive, so we have our result.[/hide]

Ten brownie points to anyone who can adjust this or similar method to the second problem.


Prove that $ \varphi (n!) \equal{} \frac {1}{\prod\limits_{p \equal{} 1,\text{p is composite}}^{n}\left(1 \minus{} \frac {1}{p}\right)}$ for all $ n\in \mathbb{N}$.

[quote="Temperal"]Prove that $ \varphi (n!) = \frac {1}{\prod\limits_{p = 1,\text{p is composite}}^{n}\left(1 - \frac {1}{p}\right)}$ for all $ n\in \mathbb{N}$.[/quote]

[hide="Are you sure that this works?"]
Unless I'm missing something, ${ \varphi (4! = 24) = 8 \neq \frac {1}{\prod\limits_{p = 1,\text{p is composite}}^4 \left(1 - \frac {1}{p}\right)}} = \frac {1}{\left(1 - \frac {1}{4}\right)} = \frac 43$, for example.[/hide]

Hmmm... you're right, it doesn't.

[hide="Hm, problem with the proof...?"]
In general, $ \varphi (n)\equal{}\varphi (p_1^{e_1}...p_n^{e_n})\equal{}n\left(1\minus{}\frac{1}{p_1}\right)\left(1\minus{}\frac{1}{p_2}\right)....\left(1\minus{}\frac{1}{p_n}\right)$.

For $ n!$ the factors (excluding 1) are $ 1,2,3,4,5...n$. So if we do $ n\left(1\minus{}\frac{1}{2}\right)\left(1\minus{}\frac{1}{3}\right)....\left(1\minus{}\frac{1}{n}\right)$ and then divide by the expressions for the numbers that are composite. It's easy to see that $ \left(1\minus{}\frac{1}{2}\right)\left(1\minus{}\frac{1}{3}\right)....\left(1\minus{}\frac{1}{n}\right)\equal{}\frac{1}{n}$, and thus the expression is correct.

I might have remembered the expression for $ \varphi (n)$ wrong, though.
[/hide]

Edit: Haven't checked carefully, but I think the problem is that you are mixing up $ n!$ and $ n$.

[quote="Temperal"]
For $ n!$ the factors (excluding 1) are $ 1,2,3,4,5...n$. So if we do $ n\boxed{!}\left(1 \minus{} \frac {1}{2}\right)\left(1 \minus{} \frac {1}{3}\right)....\left(1 \minus{} \frac {1}{n}\right)$ and then divide by the expressions for the numbers that are composite. It's easy to see that $ \boxed{n! \cdot }\left(1 \minus{} \frac {1}{2}\right)\left(1 \minus{} \frac {1}{3}\right)....\left(1 \minus{} \frac {1}{n}\right) \equal{} \frac {1}{n} \boxed{\cdot n! \equal{} (n\minus{}1)!}$, and thus the expression is correct.[/hide][/quote]

Then it should make sense.

The problem is that $ 1, 2, 3, 4, ... n$ aren't the [b]prime[/b] factors of $ n!$.

Edit:  Sorry, I misread the argument.   :oops:

Yes, I know, but I divided by the composite terms $ \left(1\minus{}\frac{1}{c_n}\right)$ in the product to get rid of them. I think azjps found the problem.


If a triangle having base $ a$ and the ratio of other two sides is $ r<1$ , show that altitude of the triangle is $ \frac{ar}{1\minus{}r^2}$

Review the statement as it doesn't hold. Take $ 3\minus{}4\minus{}5$ triangle. If we take $ a\equal{}5$, then by the given formula $ h\equal{}{5\cdot{3\over 4}\over{1\minus{}{9\over 16}}}\equal{}{60\over 7}$, but $ h_a\equal{}{3\cdot 4\over 5}\equal{}{12\over 5}$


Let $x_1,x_2, \ldots, x_n$ and $y_1,y_2, \ldots, y_n$ be reals s.t. \[ \sum\limits_{k=1}^n x_k=1, ~~~ \sum\limits_{k=1}^n |x_k|=2 .  \]

Prove that \[ \left| \sum\limits_{k=1}^n x_k y_k \right| \leq \dfrac32 \max\limits_k y_k - \dfrac12 \min \limits_k y_k .  \]

[i]Marcel Chiriţă[/i]

Suppose $x+1, \dots, x_n$ = $a_1, \dots, a_i, b_1, \dots, b_j$ where the a terms are nonnegative, the b terms are negative,
$i + j = n$, and $a_1 \geq a_2 \geq \dots \geq a_i \geq b_1 \geq b_2 \geq \dots \geq b_j$.
Further suppose $y_1 \geq \dots \geq y_n$.

Then

$\sum a_i + \sum b_i = 1$
$\sum a_i - \sum b_i = 2$

so that $\sum a_i = 3/2$
and $\sum b_i = 1/2$.

Now, we maximize $\sum a_iy + \sum b_iy$ by putting the largest $y$ terms on the a_i and the smallest $y$ terms on the $b_i$.  So:

$|\sum a_ky + \sum b_ky|$(for all permutations y)

$\leq |\sum a_ky_k + \sum b_ky_{k+i}|$

$\leq |\sum a_ky_1 + \sum b_ky_n| $

$\leq |(3/2)y_1 - (1/2)y_n|$

$= \dfrac32 \max\limits_k y_k - \dfrac12 \min \limits_k y_k $

as desired.


How many solutions are there in [b]N x N [/b]to the equation

1/x  + 1/y  =  1/1995 ?

[hide]Answer : 16[/hide]

That is a pretty hard question but 1995 has 12 factors if that helps.
1/1995*2+1/1995*2=1995. Well that is only 1. I don't no. I think this has a quick solution but i do not no the formula for these kind of unit fractions.

The answer turns out to be $ d(1995)$, where $ d(n)$ is the number of divisors function.
Can you see why this is? (Use SFFT)

Thanks for the quick replies !

[quote="AIME15"]I have to leave soon, but this should get you started.  Multiply by $ xy$ on both sides and use SFFT.  You can generalize also :D[/quote]

umm.. whats the full form of SFFT ?

SFFT comes from the expansion $ (a \plus{} b)(c \plus{} d) \equal{} ac \plus{} bc \plus{} ad \plus{} bd$, or more commonly $ (a \plus{} 1)(b \plus{} 1) \equal{} ab \plus{} a \plus{} b \plus{} 1$. AoPSWiki calls adding in the $ bd$ term "completing the rectangle."

Here's how I would find the solutions:
[hide="Generalization"]
$ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{a}$

$ ax\plus{}ay\equal{}xy$

$ ay\equal{}x(y\minus{}a)$

$ x\equal{}\frac{ay}{y\minus{}a}$.

$ x$ needs to be an integer, so $ y\minus{}a|ay$; that is, $ (ay,y\minus{}a)\ne 1$.

$ (ay,y\minus{}a)\equal{}(ay\minus{}ay\plus{}a^2,y\minus{}a)\equal{}(a^2,y\minus{}a)$.

$ x\equal{}a\plus{}\frac{a^2}{y\minus{}a}$

At this point, you can find values of $ y\minus{}a$ that divide $ a^2$ pretty easily.

In your case, $ y\minus{}1995|1995^2$, so either $ y\equal{}1996,...$
[/hide]

But getting the number of solutions need not be that hard...


A pentagon train is made by attaching regular pentagons with one-inch sides so that each pentagon, except the two on the ends, is attached to exactly two other pentagons along sides as shown. How many inches are in the perimeter of a pentagon train made from 85 pentagons?
[asy]import olympiad; size(200); import geometry_dev;
import graph; defaultpen(linewidth(0.8));
void drawPentagon(real side, pair blCorner, real degrees = 0){
  pair c1 = blCorner;
  pair c2 = c1 + side*dir(degrees);
  pair c3 = c2 + side*dir(72 + degrees);
  pair c4 = c3 + side*dir(144 + degrees);
  pair c5 = c4 + side*dir(216 + degrees);
  draw(c1--c2--c3--c4--c5--cycle);
}
pair p1 = origin;
pair p2 = origin + (1,0) + dir(72) + (1,0);
pair p3 = p2 + dir(-72);
pair p4 = p3 + (1,0) + dir(72) + (1,0);
drawPentagon(1,p1);
drawPentagon(1,p2,180);
drawPentagon(1,p3);
drawPentagon(1,p4,180);[/asy]

Observation shows that the perimeter is $ 3x\plus{}2$, where $ x$ is the number of pentagons, thus making the perimeter of $ 85$ pentagons arranged in this way $ 3(85)\plus{}2\equal{}255\plus{}2\equal{}\boxed{257}$.


geometry
geometric transformation
reflection
geometry solved

let C1 and C2 be two circles with centers O1 and O2. F is a
fixed point on O1O2 and A and B points on C1 and C2. E is
on AB such that EA/EB =FO1/FO2. A and B vary on C1 and
C2 such that E moves on a circle with center F. find the locus
 of the intersection of AO1 and BO2.

I'm pretty sure the locus consists of two circles passing through $O_1,O_2$, but I didn't show it.

yeah that's it. I used spiral similarity to show that :)

Suppose spiral similarity of center T takes A to O_1 and B to O_2(Also this spiral similarity takes E to F)
Easily because TO_1/TF and TF/TO_2 are constant,locus of T is two points that are reflection of each other in line O_1O_2
Now intersection of AO_1 and BO_2,is on the circle TO_1O_2.
So the locus is two circles passing through O_1 and O_2 and these circles are reflection of each other in line O_1O_2


This is a long shot but has there ever been a reduction of polythene to ringed compounds, maybe 5/6 membered observed? If there is, can somebody post the link for further information? Any help would be appreciated.

By "polythene" do you mean polyethylene? Also, wouldn't that be an oxidation?

Yes, it is polyethylene. I made a typo error. Sorry. Any help? Please.

Well, polyethylene is made up of long and saturated carbon chains (with more or less ramification, depending on the production process), but what you are looking for should be equivalent to the same experiments with simple alcanes. Another tip: make a search in Polymer Chemistry books.


[quote="randomdragoon"][quote="jli"][quote="ckck"]Find the area of a regular dodecagon with side of 7. Rounded to the nearest tenth.
How do you find the area of a regular dodecadon?[/quote]

[hide]draw a hexagon wth side 21.  Now cut off 30-30-150 triangls with the short sides=7 from the corners of the hexagon.  this gets you a dodecagon.

You can easily figure out the area of the hexagon and the cut off triangles should be easy to find too.[/hide][/quote]
30-30-150 triangles? :huh:

[hide]You can also divide the dodecagon into 12 isosceles triangles  with base length 7 and base angle 75 degrees (sort of like how you would slice a pie). The height of this triangle would be $3.5 \tan 75^\circ$, so the area of one triangle would be

$\frac{1}{2}\cdot{7}\cdot 3.5 \tan 75^\circ$

Multiplying this result by 12 will give you the final answer. (Since they asked you to round, I'm assuming a calculator is allowed -- otherwise, you could use the trig identities to get the exact value of $\tan 75^\circ$.)[/hide][/quote]

oh crap... 30-30-120 triangles.


calculus
integration
trigonometry
calculus computations

$ \boxed1$ $ \int {\frac {{\sin (x \plus{} \alpha )}} {{\cos ^3 x}}} \sqrt {\frac {{\cos ecx \plus{} \sec x}} {{\cos ecx \minus{} \sec x}}} dx$

$ \boxed2$ $ \int {\frac {{\sin 2x \minus{} \sin 2\alpha }} {{(\sin x \minus{} \sin \alpha ) \plus{} (\cos x \minus{} \cos \alpha )}}} dx$

$ \boxed3$ $ \int {\frac {{dx}} {{\sqrt {\cos ^3 (x \plus{} \alpha )\sin (x \plus{} \beta )} }}}$

$ \boxed4$ $ \int {\frac {{x^2 \minus{} 1}} {{1 \plus{} x^2 }}} \frac {{dx}} {{\sqrt {1 \plus{} x^2 \plus{} x^4 } }}$

$ \boxed5$ $ \int {\frac {{x^3 }} {{(ax^3 \plus{} b)^{7/3} }}} dx$

[hide="3"]
$ \int{\frac{{dx}}{{\sqrt{\cos^{3}(x+\alpha )\sin (x+\beta )}}}}$
=${ \int {\frac {dx}{\sqrt{\cos^{3}(x+\alpha )\sin (x+\alpha+\beta-\alpha )}}}}$
=${ \int {\frac {dx}{\sqrt{\cos^{3}(x+\alpha )[sin(x+\alpha)cos(\beta-\alpha)+cos(x+\alpha)sin(\beta-\alpha)])}}}}$
=$ \int \frac{sec^{2}(x) dx }{ \sqrt{tan(x+\alpha)cos(\beta-\alpha)+sin(\beta-\alpha)}}dx$
put $ tan(x+\alpha)cos(\beta-\alpha)+sin(\beta-\alpha) =t$
$ cos(\beta -\alpha) sec^2(x+\alpha)dx =dt$
$ =\int \frac{dt}{cos(\beta-\alpha )\sqrt{t}}$

[/hide]

[hide="5"]
just take $ x^{3}$ common from $ (ax^{3}\plus{}b)$
$ \equal{}\int \frac{dx}{x^{4} (a\plus{}\frac{b}{x^{3}})}$
put $ a \plus{}\frac{b}{x^{3}} \equal{}t$
 [/hide]
[/hide]

[hide="2"]
 $ I = \int {\frac {{\sin 2x - \sin 2\alpha }} {{(\sin x - \sin \alpha ) + (\cos x - \cos \alpha )}}} dx$

$ I = \int {\frac {2sin(x - \alpha)cos(x + \alpha)}{2sin\frac {x - \alpha}{2}cos\frac {x + \alpha}{2} + 2sin\frac {x + \alpha}{2}sin\frac {\alpha - x}{2}}dx}$

$ I = \int {\frac {2sin\frac {x - \alpha}{2}cos\frac {x - \alpha}{2}cos(x + \alpha)}{sin\frac {x - \alpha}{2}cos\frac {x + \alpha}{2} - sin\frac {x + \alpha}{2}sin\frac {x - \alpha}{2}}dx}$

$ I = \int [{\frac {2cos\frac {x - \alpha}{2}cos(x + \alpha)}{cos\frac {x + \alpha}{2} - sin\frac {x + \alpha}{2}}]dx}$

$ I = \int [{\frac {2cos\frac {x - \alpha}{2}cos(x + \alpha)}{cos\frac {x + \alpha}{2} - sin\frac {x + \alpha}{2}}](\frac {cos\frac {x + \alpha}{2} + sin\frac {x + \alpha}{2}}{cos\frac {x + \alpha}{2} + sin\frac {x + \alpha}{2}})dx}$

${ I = \int 2cos\frac {x - \alpha}{2}[cos\frac {x + \alpha}{2} + sin\frac {x + \alpha}{2}}]dx$

$ I = \int 2cos\frac {x - \alpha}{2}cos\frac {x + \alpha}{2} + 2sin\frac {x + \alpha}{2}os\frac {x - \alpha}{2}dx$

$ I = \int [cosx + cos\alpha + sinx + sin \alpha]dx$

$ \boxed{I = sinx - cosx + x(cos\alpha + sin\alpha)+ C}$

[/hide]

hello, for 1) we have
$ \sqrt{\frac{%Error. "cosec" is a bad command.
(x)+\sec(x)}{%Error. "cosec" is a bad command.
(x)-\sec(x)}}=\sqrt{\frac{\cos(x)+\sin(x)}{\cos(x)-\sin(x)}}=t$
from here we get
$ x=\pm\arccos\left(\pm\frac{1+t^2}{\sqrt{2(1+t^4)}}\right)$
now we assume that
$ x=\arccos\left(\frac{1+t^2}{\sqrt{2(1+t^4)}}\right)$
and $ t^2>1$ and from here we get
$ \frac{dx}{dt}=\frac{2t}{1+t^4}$
and our integral will be
$ \int\frac{4t^2(t^2-1)}{(1+t^2)^3}\cos(\alpha)+\frac{\sin(\alpha)}{(1+t^2)^2}\,dt$
Sonnhard.

hello, for 5) we rewrite the integrand in the form
$ \frac{x^3}{(x^3)^{\frac{7}{3}}\left(a\plus{}\frac{b}{x^3}\right)^{\frac{7}{3}}}\equal{}\frac{1}{x^4\left(a\plus{}\frac{b}{x^3}\right)^{\frac{7}{3}}}\equal{}\minus{}\frac{1}{3b}\frac{\minus{}3b}{x^4\left(a\plus{}\frac{b}{x^3}\right)^{\frac{7}{3}}}$
Now we set after the idea from kabi $ t\equal{}a\plus{}\frac{b}{x^3}$ and we get $ dt\equal{}\minus{}\frac{3b}{x^4}\,dx$.
Sonnhard.


In the not convex hexagon every angle are equal either $90^{0}$ or $270^{0}$.Is that true that for some length of sides the hexagon can be cat into the two not equal hexagons similar to him?

If it can help you,I can say that the answer of this problem is YES!! :wink:

The attached figure shows an arrangement not leading to immediate contradiction, where the hexagon ABCZYX is cut into the hexagons BCZPQR and RAXYPQ. Denote the 4 independent sides of ABCDZYX as $ a \equal{} AB,\ b \equal{} BC,\ x \equal{} XY,\ y \equal{} YZ.$ Assume $ BCZPQR \sim ABCZYX$ with similarity coefficient $ 0 < p < 1$ and $ RAXYPQ \sim BCZPQR$ with similarity coefficient $ 0 < q \neq 1.$ Denote $ a' \equal{} pa,\ b' \equal{} pb,\ x' \equal{} px,\ y' \equal{} py$ and $ a'' \equal{} qa',\ b'' \equal{} qb',\ x'' \equal{} qx',\ y'' \equal{} qy'$ the sides of BCZPQR and RAXYPQ corresponding to the sides a, b, x, y of ABCZYX. From the lines $ BC,\ CD \equiv CZ,\ VZ \equiv YZ,\ AB,\ RS \equiv RQ,\ XU \equiv XY,$ we can read the following equations:

$ BC: \ \ b \equal{} a' \equal{} pa$

$ CD: \ \ a \equal{} b' \plus{} x \equal{} pb \plus{} x$

$ VZ: \ \ b \equal{} b'' \plus{} y \equal{} pqb \plus{} y$

$ AB: \ \ a \equal{} a'' \plus{} (b' \minus{} y') \equal{} pqa \plus{} p(b \minus{} y)$

$ RS: \ \ b \equal{} b'' \minus{} y'' \plus{} a' \minus{} x' \equal{} pq(b \minus{} y) \plus{} p(a \minus{} x)$

$ XU: \ \ a \equal{} a'' \minus{} x'' \plus{} b' \equal{} pq(a \minus{} x) \plus{} pb$

Using the 1st equation in the above set of 6 to eliminate $ b \equal{} pa,$ we get

$ a \equal{} p^{2}a \plus{} x$

$ pa \equal{} p^{2}qa \plus{} y$

$ a \equal{} pqa \plus{} p^{2}a \minus{} py$

$ pa \equal{} p^{2}qa \minus{} pqy \plus{} pa \minus{} px\ \Longrightarrow\ pqa \equal{} x \plus{} qy$

$ a \equal{} pqa \minus{} pqx \plus{} p^{2}a$

Using the 1st and 2nd equations in the above set of 5 to eliminate $ x \equal{} a(1 \minus{} p^{2}),$ $ y \equal{} pa(1 \minus{} pq),$ we get

$ a \equal{} pqa \plus{} p^{2}a \minus{} p^{2}a \plus{} p^{3}qa\ \Longrightarrow\ pq(1 \plus{} p^{2}) \equal{} 1$

$ pqa \equal{} a \minus{} p^{2}a \plus{} pqa \minus{} p^{2}q^{2}a\ \Longrightarrow\ p^{2}(1 \plus{} q^{2}) \equal{} 1$

$ a \equal{} pqa \minus{} pqa \plus{} p^{3}qa \plus{} p^{2}a\ \Longrightarrow\ p^{2}(1 \plus{} pq) \equal{} 1$

Comparing the 2nd and 3rd equations in the above set of 3 immediately yields p = q. This turns each of the 3 equations into the same quartic equation $ p^{4} \plus{} p^{2} \minus{} 1 \equal{} 0,$ quadratic in $ p^2,$ with the only positive root $ p \equal{} \frac {1}{\sqrt {G}} < 1,$ where G is the golden ratio. Then also $ q \equal{} p \equal{} \frac {1}{\sqrt {G}}.$ Going back,

$ \frac {a}{b} \equal{} \frac {1}{p} \equal{} \sqrt {G}$

$ \frac {a}{x} \equal{} \frac {1}{1 \minus{} p^{2}} \equal{} \frac {G}{G \minus{} 1} \equal{} G^{2} \equal{} G \plus{} 1$

$ \frac {b}{y} \equal{} \frac {1}{1 \minus{} pq} \equal{} \frac {G}{G \minus{} 1} \equal{} G^{2} \equal{} G \plus{} 1$

Nice job dear Yetti!

In fact this problem was solved  only by one person(a girl :) ) during the competition.But,in fact, she solved only this problem and got a honorable mention.

Thank you for your solution!!!


$20)$ Find the radius of a sphere inscribed in a regular tetrahedron which has a height of $8$.

 :lol:

it is 2...consider disecting the tetrahedron into 4 little tetrahedrons, where each small tetrahedron has teh center as a vertex, and a side of the larger tetrahedron as a the oppsite base...the volume computed both ways is height large*base =4height small*base, giving height large=4height small

yeah I found out that

sphere inscribed in tetrahedron: radius of sphere is $\frac{1}{4}$ of the height of tetrahedron

sphere circumscribing a tetrahedron: radius of sphere is $\frac{3}{4 }$the altitude of tetrahedron.

 :lol:


AMC
AIME
MATHCOUNTS
AMC 12
AMC 12 A
AMC 12 B

If it is okay with the AMC folks could someone please post the answers to the AMC 10B?
Thanks

they're availabe online, officially:

http://www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2006-10a/06AMC10AB-answers.html

AMC12A: 99.5
AMC12B: 92.5
This is my first time taking it as a 6th grader.
$NO$ $WONDER!$
 :fool:

aw man .5 points from AIME qualification then 7.5 from it. That really sucks

Still REALLY good for a 6th grader... congrats.

Yeah but he could still make it through the 5% cutoff, right?

[quote="pkothari13"]Oh that's true... and apparently the 12B was harder than others... cross your fingers![/quote]

REALLY? I thought that it was comparable, if not easier. And, the number of AIME qualifiers (score of 100+) in our school has quadrupled this year compared to the average of past years.

Eh yeah qualifying for AIME is easier, but the last 5 problems were MUCH harder than usual.

So do you think that the AMC is trying to shift the score distribution?

Hmmm...I wonder what this means about the AIME and what this will do to the index...

Four people at my school aced one of the AMC 10s, myself included  :lol:

[quote="qweretyq"]my school had 74[/quote]

74 what? Test-takers? AIME-qualifiers? Or...*gasp*...74 people who aced the 10? (I'm not even sure there are that many perfect scores in all, let alone in one school.)

if your a 5th/6th grader and you got a 89 (I guessed trying to get to AIME, would have had a 94) on the AMC 12B, do you think I can get in. My math teacher (Mr. Eric Walstein from Mountgomery Blair Highschool) said that sometimes they invite young people with potential to the AIME, how do you find out if you took it as a visitor?

It is actually eaiser to get to AIME with the 12 as long as you can do 11-16 of the problems.  I got 112.5 on the 12B this year as a seventh grader.  Me and my MathCounts team all got to the AIME.  Two took 10 two took 12.  Last year I got 96 on the 10B so your doing better than me mbabbitt2003.

[quote="qweretyq"]sorry for not being clear
74 AIME qualifiers

and lyra, sorry you are not going to get to AIME
why did you take the 12 by the way? shouldnt you have taken the 10?
was it because you thought it would be easier to qualify for AIME?[/quote]

HOLY CRAP.  My school only had 13.. I thought 40 was pretty high too when I heard.... wow....
What school do you go to?!

And to mr. babbit, don't worry about it man.  Now you know exactly what to expect next year, and you're like 2-3 years ahead of the game now.  I didn't really know much about the AMC until my freshman year, but I made a pretty drastic improvement in one year (I'm a sophomore).  What I suggest you do is start taking a few practice tests like 3-4 weeks before the test next year, and making sure you can do like 14-18 problems on each test.  Plus, by then, you'll be in a higher math and thus, you'll be able to solve more problems.  If you're feeling extra motivated, you could probably take 1-2 AIMEs just to know what to expect.

hey anyone if I scored a 108 on the AMC 10B do I qualify for the AME?

no you normaly need a 120 about to get to the AIME but you might of because I think this year was a harder test

I can guarantee that 120 is the floor.
Even if it was really hard, there would have to be less than 1,000 people with 120 or better, on either. Very unlikely

holy @*%# my school had 4 middle schoolers and like 6 high schoolers.

my school had 98 qualifiers from the 10/12 B...

How do you even get that many people to take it? Last year, my school had noticeably fewer than that even take any of the 10A/B or 12A/B.

[quote="JesusFreak197"]How do you even get that many people to take it? .[/quote]
this is exactly what i thought ....when i read it...
my school had two records this year...
highest AIME qualifications: 8 :lol:  (only 15 people took the test)
and highest AMC 12 score: 150 (me :)  )

[quote="Kent Merryfield"][quote]74 AIME qualifiers [/quote]
I've heard from the teacher (Tom Armbruster) who supervises the AMC exams at San Marino HS in Southern California. According to him, the school had 89 AIME qualifiers based on the 12A/10A. Since they also took the 12B/10B, there might even be a few more.[/quote]

Test A had 74. Test B had 94. I am not sure how many overlap but I'm pretty sure theres a few people that qualified on the B that didn't on A because they usually take the 10 A and 12 B. So I'm estimating a total of 100-110 qualifiers in total.
AAST

[quote="mrosett"]I really get the sense that the AMCs this year were much easier to qualify off of. I think my school had 17 people qualify for sure (plus any 5% qualifiers- and we have two 99.5s). Last year there were four, and I think the most ever was seven. We do have a new coach, but most of the qualifiers aren't actually Math Clubbers- half of them have never worked with the coach. That said, the last few 12B problems were absurdly hard. I, of course, choked. I answered 22, but missed a fairly easy one and then bubbled two of them in wrong. Oops  :blush: . So I got a 121.5, which is decent, but still lower than the 126 I got in 7th grade. (That was 3 years ago. I keep choking when I take the 12). On the flip side, I aced the 10A for the second year in a row. :lol:[/quote]

Our school had ~350.
New record.

[quote="JesusFreak197"]How do you even get that many people to take it? [/quote]

In our school, people who take the AMC get 1% in extra credit that goes towards their quarter grade, AIME-qualifiers get 2%, and the school winner gets 3%. (This makes a big difference in a competitive high school.) So almost all the students in advanced math classes take the AMC's, giving us around 150 test-takers out of a high school of approximately 500.


vector
geometry
3D geometry
sphere
calculus
integration
trigonometry

[quote="venkata"]hello everybdy,
well, when a spherical metallic conductor is placed in a uniform electric field, its charges redistribute such that the net E inside the conductor is 0. I was told that the charge distribution is such that the surface charge density at any point on the surface takes the form $ a.cos x$ where $ a$ is a constant and $ x$ is the angle between E and the radius vector at that point. could smbdy pls giv a mathematical explanation 4 this?
thanks in advance.[/quote]

The phenomenon that you describe can be explained like this:

Maxwell’s curl equation for magnetic field (Ampere’s law) is:

	 \[ \nabla\times\vec{H}\equal{} \vec{J} \plus{} \frac{\partial\vec{D}}{\partial t}
\]

If  we take the divergence of both sides of this equation, we obtain:

	\[ \nabla\cdot\left(\nabla\times\vec{H}\right)\equal{} \nabla\cdot\vec{J} \plus{} \frac{\partial\nabla\cdot\vec{D}}{\partial t}\equal{}0
\]

This is because the divergence of any curl is always zero.

Maxwell’s divergence equation for D-field (Gauss’ law) gives:

	\[ \nabla\cdot\vec{D} \equal{} \rho
\]

Likewise, $ \vec{J} \equal{} \sigma\vec{E}$  , where $ \sigma$  is the conductivity. Also $ \vec{D}\equal{}\epsilon_0\vec{E}$ .

Thus,  $ \nabla\cdot\vec{J}\equal{}\sigma\nabla\cdot\vec{E}\equal{}\frac{\sigma}{\epsilon_0}\nabla\cdot\vec{D}$

That is $ \nabla\cdot\vec{J} \equal{} \frac{\sigma}{\epsilon_0}\rho$ .

If we substitute these results into the equation above:

	 \[ \nabla\cdot\vec{J} \plus{} \frac{\partial\nabla\cdot\vec{D}}{\partial t}\equal{} \frac{\sigma}{\epsilon_0}\rho \plus{}\frac{\partial\rho}{\partial t} \equal{} 0
\]

If we now stipulate that $ \rho\left(t\right) \equal{}\rho_0e^{st}\Rightarrow \left(\frac{\sigma}{\epsilon_0}\plus{}s\right)\rho\left(t\right)\equal{}0$  .

Hence we can see that $ s \equal{} \minus{}\frac{\sigma}{\epsilon_0}$  .  It follows that $ \rho\left(t\right) \equal{}\rho_0\exp\left(\minus{}\frac{\sigma}{\epsilon_0}t\right)$ .

Since the sphere is a metallic conductor, $ \frac{\sigma}{\epsilon_0}\to \infty$ , and thus $ \rho\left(t\right)$  vanishes immediately.

That means that the whole charge resides on the surface of the sphere.

Gauss’ divergence theorem says that:

	 \[ \iint_{\mathbb{R}} \vec{D}\cdot\hat{n}\, da \equal{}\iiint \nabla\cdot\vec{D}\, dV \equal{} Q
\]

The triple integral over volume on the right hand side clearly gives the total charge because $ \nabla\cdot\vec{D} \equal{}\rho$ , the volume charge density, but the continuity equation – the differential equation for $ \rho$  -- goes to zero immediately, meaning that the total charge rests on the surface.

Because $ \vec{D}\equal{}\epsilon_0\vec{E}$ ,  both $ \vec{D}\text{ and }\vec{E}$  point in the same direction. The unit vector $ \hat{n}$  is the vector normal to the surface of the sphere, i.e. pointing in the same direction as the radius vector.

Differentiating under the integral on the left hand side we find:

	\[ \vec{D}\cdot\hat{n} \equal{}\left|\vec{D}\right|\cos x \equal{}\rho_s
\]  , where $ \rho_s$  is the surface charge density and $ x$  is the angle between the radius vector and the E-field as you described it.


Whenever I build a PDF, the bar at the bottom gets stuck at

[code]C:\Program Files\MiKTeX 2.7\tex\latex\amsfonts\umsa.fd[/code]

I don't know if this helps, but that is indented and above it, it says "Loading MPS to PDF converter".

I really need help to have it stop doing that and make it compile the PDF.

Try going to Start, MikTex 2.7, Settings and pressing Refresh FNDB then Update Formats buttons. Then try and compile this simple document as LaTeX=>PDF
[code]\documentclass{article}
\begin{document}
Hello world
\end{document}[/code]

I didn't do what yuo said, but when I saved that same document as a different filename and then built it, it worked.  It's that specific file that didn't work, and both the files were the exact same and 1 worked, so I don't know what was the problem.  Thanks anyway!


geometry
3D geometry
tetrahedron
induction
combinatorics proposed
combinatorics

All the faces of a convex polyhedron are triangles. Show that we can color the edges in black and white s.t. between any two points there is a path formed with black edges, and another path formed with white edges.

I Just have a small clue: If the figure is not a tetraedron, then we must be able to color the poliedron in 3 colors, such that there are no 2 adjacent faces with same color... Now, I was trying to find some  way to direct edges, and use the direction and the colors of faces to define the color of an edge...However it is not working...! :(

Note: We will be using PLANAR graphs in the entire solution.

We first need to prove a lemma. Consider a graph with the following description:
(1) For $a\ge3$ there is a loop $X_1X_2\cdots X_aX_1$ such that all other edges and all other vertices (which we'll label $Y_1,\ldots,Y_b$, $b\ge0$) are inside this loop (i.e. $X_1X_2\cdots X_aX_1$ is the outer face of the graph),
(2) all other faces of this graph are triangles,
(3) all of the Ys have degree ${}\ge3$, and
(4) $a+b\ge4$.
[b]LEMMA 1: Such a graph contains a spanning tree so that every vertex has an edge not in the tree. (We call such a tree nice)[/b]

Proof by induction on the number of vertices: If there are four vertices, then there are two possible configurations. The first is a square $X_1X_2X_3X_4$ with a diagonal $X_1X_3$, and the tree with edges $X_1X_2$, $X_1X_3$, and $X_3X_4$ works. The second is a triangle $X_1X_2X_3$  and one interior point $Y_1$, and the tree with edges $X_1Y_1$, $Y_1X_2$, and $X_2X_3$ works.

Now we consider the inductive step, $a+b\ge5$. Define a chord to be an edge between two non-adjacent X's. We'll prove there exists an X with no chords. Assume otherwise, and begin with the chord $X_{1}X_{p_1}$. $X_{2}$ must also have a chord, so call it $X_2X_{p_2}$ and notice that $p_2\le p_1$. Likewise, $X_3$ must have a chord $X_3X_{p_3}$ with $p_3\le p_2$, and so on. The value $p_i-i$ strictly decreases as i increases, and so there must exist an $i$ so that $p_i-i=2$. But then the only neighbors of $X_{i+1}$ are $X_i$ and $X_{p_i}$, meaning $X_i$ has no chord, contradiction.

Thus, we may assume that $X_1$ has no chords. Say that the neighbors of $X_1$ are, in order, $X_a,Y_1,\ldots,Y_k,X_2$ (where $k\ge0$). ($X_1$ has no other Xs as neighbors, by design.) Notice that since all inner faces are triangles, all of the edges $X_aY_1,Y_1Y_2,\ldots,Y_kX_2$ are in the graph. Consider what happens when $X_1$ and all of its edges are deleted from the graph. We then have a graph satisfying properties (1), (2), (3), and (4) with the loop $X_2,\ldots,X_n,Y_1,\ldots,Y_k,X_2$. Since this new graph has $a+b-1$ vertices, by the induction hypothesis, it has nice spanning tree. If we then add $X_1$ and its edges back to the graph, and then include $X_1X_2$ in the tree, we have a nice tree. Thus, the lemma is proved. $\blacksquare$

If we take a=3 in this lemma, we find that our polyhedron has a nice spanning tree. Color all of the edges in the tree white, and the rest black. We claim that this coloring works. Since all of the faces of the polyhedron are triangles, we easily find that 3f=2e, and so by Euler's theorem, 2e/3+v=e+2, i.e. e=3v-6. There are thus v-1 white edges and 2v-5 black edges. Furthermore, every vertex has at least one black edge. Since trees are connected, there is a white path between any two points. It remains to show that the black edges also form a connected graph.

Assume not, and split the vertices into sets R and S, with r and s vertices respectively ($r+s=v,r\ge2,s\ge2$), so that there is no black edge with a vertex in each set.

[b]LEMMA 2: There are at most $2r-3$ edges with vertices in set R, and likewise for set S.[/b]

Proof: Since the white edges span the entire graph, there must be at least one white edge with one vertex in R and the other vertex in S. With the vertices in R, form a new graph as follows: add a new point T, and replace every white edge $R_iS_j$ with $R_iT$. All other edges with both vertices in R remain unchanged. (We can now disregard set S.) It is clear that the white edges form a connected subgraph that includes all the vertices $R_1,\ldots,R_r,T$, and thus there are at least (r+1)-1=r white edges in this graph. Furthermore, this graph has only triangular faces, and so by the same reasoning as above, there are a total of 3(r+1)-6=3r-3 edges. Thus, there are at most (3r-3)-(r)=2r-3 black edges, qed. The proof for set S is similar. $\blacksquare$

With lemma 2, we see that there are at most (2r-3)+(2s-3)=2(r+s)-6=2v-6 black edges in our polyhedron. But this is a contradiction becase we have 2v-5 black edges. So the black edges must form a connected subgraph, and since every vertex has at least one black edge, there exists a black path between any two vertices, as required.

First just look at a vertex and the vertices connected to it. It's not hard to color only the edges between these so that both the resulting graphs are connected (both the white one and the black one). After this, at each step, we say that a vertex of the polyhedron is reachable if it's part of the graph that we are building, which is connected in both situations (when we look at its white edges, and when we look at its black edges).

At each step, take an edge $xy$ which is already colored (so that $x,y$ are reachable), s.t. there is a vertex $z$ with the property that $xyz$ is a face and $z$ is not yet reachable (I think it's pretty clear that this is always possible, unless we have colored all the edges). We color the edges $xz,yz$ in different colors, so we have added one more vertex to our doubly connected graph (at most one of the edges $xz,yz$ is colored, so if this is the case, we color the other one with the other color). We continue like this until we finish all the vertices.

Is it ok?

At least, it is written in the official solution.

I was sure I had stumbled over the official solution :). It just "felt right" :D.


complex analysis
real analysis
real analysis unsolved

What is the dual space of [b]Hardy[/b]-space in the case $ 1<p<\infty$

Depends on whether you're using a real or complex definition. The real definition: $ f\in H^p$ if $ f\in L^p$ and $ \widetilde{f}\in L^p$ (that being the Hilbert transform or conjugate function.) If this is what we're doing, then for $ 1<p<\infty,$ $ H^p\equal{}L^p,$ the norms are equivalent, and the duality is the same as the duality for $ L^p.$

(If that's not what you meant, then please take the question to the Complex Analysis forum.)

Much more entertaining: $ H^1$ is not the same space as $ L^1,$ and the dual of $ H^1$ is $ BMO.$


inequalities
geometry
3D geometry
tetrahedron
perimeter
geometry proposed

Show that any intersection of a tetrahedron with a plane has a perimeter which is less then the perimeter of one of the faces.

I'm not sure what you are asking... :)
is the statement not clear?

Yeah maybe I misunderstood, can you translate that. :D


How do you define $ 1$, $ 2$ and $ \plus{}$?

You have one apple then someone gives you an apple. How many apples do you have? That's the best I can think of. Some people just spend their whole lives trying to prove this. I think that is silly.

Also, proofs don't belong here.

This has something to do with logic or sets? I forget but heres the proof

http://mathforum.org/library/drmath/view/51551.html

yeah.... i know that proving dosent belong here but i too want to know how do i prove 1+1=2 and 1+1 is not = to 2

Maybe you should go look at my link...

Heres another one
http://www.idt.mdh.se/~icc/1+1=2.htm


Let k* mean k+(k-1)+(k-2)...1. Prove that (2k)*=2(k*)+k^2.

Edit: thanks xpmath and grn turtle

Do you mean k*=k+(k-1)+(k-2)+...+1? Otherwise I don't get what you mean by k+(k+1)+...+1.

Am I misreading something?

[hide]Clearly $ k*\equal{}\frac{k(k\plus{}1)}{2}$.

We have to prove $ {k(2k\plus{}1)}\cdot\frac{2}{k(k\plus{}1)}\equal{}k(k\plus{}1)\plus{}4k^2$

$ 2(2k\plus{}1)\equal{}k(k\plus{}1)^2\plus{}4k^2(k\plus{}1)$

$ 4k\plus{}2\equal{}k^3\plus{}2k^2\plus{}k\plus{}4k^3\plus{}4k^2$

But this is not necessarily true.[/hide]

Wow, I have two more mistakes.  :maybe:

I'll edit again.  :blush:

lol, ok

Let's try that again.
[hide="Solution"]
Clearly, $ k*\equal{}\frac{k(k\plus{}1)}{2}$.
[hide="Proof of that"]
Use induction, or let $ S\equal{}k\plus{}(k\minus{}1)\plus{}\cdots\plus{}1\equal{}1\plus{}\cdots\plus{}(k\minus{}1)\plus{}k$. Add the two representations of $ S$ together to get $ 2S\equal{}\underbrace{(k\plus{}1)\plus{}(k\plus{}1)\plus{}\cdots\plus{}(k\plus{}1)}_{k\text{ times}}\equal{}k(k\plus{}1)$

Therefore $ S\equal{}\frac{k(k\plus{}1)}{2}$.
[/hide]

$ k(2k\plus{}1)\equal{}k(k\plus{}1)\plus{}k^2\equal{}2k^2\plus{}k\equal{}k(2k\plus{}1)$.
[/hide]

[hide]
Notice that $ k* \equal{} k \plus{} k \minus{} 1 \plus{} k \minus{} 2... \plus{} k \minus{} (k \minus{} 1) \equal{} k^2 \minus{} \dfrac{k(k \minus{} 1)}{2} \equal{} \dfrac{k(k \plus{} 1)}{2}$

Substitute that into $ (2k)* \equal{} 2(k*) \plus{} k^2$

$ (2k)* \equal{} ?2(k*) \plus{} k^2\implies \dfrac{2k(2k \plus{} 1)}{2} \equal{} ?2(\dfrac{k(k \plus{} 1)}{2}) \plus{} k^2\implies$
$ 2k^2 \plus{} k \equal{} 2k^2 \plus{} k$

 :thumbup: [/hide]

How is this number theory? I'm not trying to condescend or anything, just trying to understand since i recently finished my first dose of number theory in AOPS vol.1

It's not, it's algebra. Ihatepie was mistaken I guess.

My original problem was gonna be number theory. But I edited when I realized it wouldn't work and forgot to change the title (before I posted.) Hence all of the little mistakes in my new problem.


Hey guys,
im gonna be in the bay area (San Francisco) this Friday-Sunday for some school musical ensemble field trip(some competition we're performing in). maybe ill see some of you guys. i won't be able to recognize you, but maybe you'll recognize me (ill try and wear one of my two mathcounts t-shirts).
see you there, hopefully

Most of us here live in fremont, palo alto, etc. I think jungi is the only one in San Francisco, so you probably won't see any of us. D'oh.

junggi doesn't live in SF either. I'm guessing you can always travel around though, come to the other cities.

that stinks.... *sigh
i guess ill just have to see you guys at competitions, like ARML. (not this year)

[quote="Nerd_of_the_Ages"]I think jungi is the only one in San Francisco[/quote]
huh what
where do you get that

daermon/Eric goes to SF Day, if that means anything.


For $a\in\mathbb{N}$ ,$a>1$ , prove that

$\frac{a^{2000}-1}{a-1}$ is not a perfect square

Edit : I change the word " square free " to " not perfect square "

Without any restriction on $a$ we can say that $a \geq -1$.
If $a = -1$ or $a = 0$ there are two perfect squares (0 and 1).
Now consider $a > 0$ with $a \neq 1$.

$\frac{(a^{500} + 1)(a^{500} - 1)(a^{1000} + 1)}{a - 1}$.
Clearly $a - 1 | a^{500} - 1$.
Note that $(a^{1000} + 1, a^{500} - 1) \leq 2$ and in general the $\gcd$ between any two of the three factors $a^{500} + 1$, $a^{1000} + 1$ and $\frac{a^{500} - 1}{a - 1}$ is $\leq 2$.
There are two cases:
I. $\gcd = 1$ So the three factors are all perfect squares. But $a^{1000} + 1$ and $a^{500} + 1$ are not perfect squares.
II. $(a^{1000} + 1, a^{500} + 1) = 2$. So the product of this two factors must be a double square.
Hence $4a^{1500} + 4a^{1000} + 4a^{500} + 4 = k^2$.
But $k^2 > (a^{250} + 2a^{750})^2$ and $k^2 < (a^{250} + 2a^{750} + 1)^2$. And this is impossible for $a \geq 2$

[quote="{x}"]The problem is [b]wrong[/b] : $5^2 |2^{2000} - 1$
We have $\varphi ( 25 ) = 20$. By Euler's theorem $2^{2000} \equiv \left(2^{20^{}} \right)^{100} \equiv 1 ( \mbox{mod } 5^2 )$. Hence $5^2 |2^{2000} - 1$ and $\frac{a^{2000}-1}{a-1}$ for $a=2$ is [b]not[/b] squarefree.[/quote]

sorry , I thought that square-free = not perfect square  :blush: What I mean is perfect square

Here you'll see what a squarefree number is ;) :
[quote]A number is said to be squarefree if its prime decomposition contains no repeated factors. [/quote]
http://mathworld.wolfram.com/Squarefree.html