Suppose that you are on a TV show.  Suppose that Monty Hall(the host of this TV show) asks you to choose between three doors: #1, #2, and #3. Behind a random door is a new Rolls Royce. Behind each of the other two doors is a goat. Let's assume that you would prefer a Rolls Royce to a goat. You choose a door. Now, Monty, who knows which door hides the Rolls Royce, shows you a goat behind one of the two doors that you did not choose. He then gives you the opportunity to change your choice. Assume that Monty always does this, regardless of your guess. Should you change your choice? What is the probablity that there is a Rolls Royce in the door if you switch?

[hide]Switch, and you have a 2/3 chance.  :) [/hide]

obviously he has to show you a goat.  1/3 chance you got it right but then if you switch profit:0 cars. Then if you picked the wrong one, he shows you a goat, you get a car if you switch.  Since the probabilities aren't influenced by the previous, switch.  It can be done with chocolates and boxes 2 empty 1 with chocolate.

if someone doesnt know how to do it, heres an interactive and fun thing

[url=http://math.ucsd.edu/~crypto/Monty/Montytitle.html]math.ucsd.edu/~crypto/Monty/Montytitle.html[/url]

That game's rigged.  I click 1 car in 1.  Click 2 car in 3.  Click 3 car in 2.  Make it random.

I chose 2 the first time both times I could, then I got the car. Then the next 2 times, I chose the wrong one both times. ... Not rigged.

That explanation on math.ucsd.edu/~crypto/Monty/Montytitle.html is very interesting.  you should take a look

Now what happens if you have four doors and still only one car? :lol:

A different way of looking at the first problem is that you would only stay if you thought you won the car which is a 1/3 chance, so if you switch you have a 2/3 chance

Here's another way to think about this problem that might be more intuitive for some people.

Suppose there are not 3, but 1000 doors.  Not 2, but 999 goats!  Still just one car.

You pick one door.  Monty opens 998 doors revealing 998 goats.

Do you keep your door or do you switch to the other unopened door?

You switch.  The probability that you even got the car in the first place is 1/1000, that would be the probability that you win if ou don't switch.

So the probability if you switch is 999/1000

999 is a lot of goats!!!  :D

So the answer to JesusFreak197's question is:

The probability of winning if you stay is 1/4 and the probability of winning if you switch is 3/4

?

Objection to nat mc. You said that P(win if don't switch) is 1/1000, and that's fine. But you said that P(win if switch) is 999/1000 because it's 1 - 999/1000.

But you don't have 100% certainty that you will win. I think it's 998/1000.

you switch doors.  Probiltiy is 998/1000

There are two ways one can think about it:

One: The probability of having the door you just picked being correct is 1/1000. Because all probabilities in this case have to add up to 1, the other is 999/1000.

Two: Another way to think about it is having the door you would switch to as "representing" 999 of the doors.

thats what I thought too at first, but when I thought again, I came up with that if you switch, the probabilities of winning and losing add up to 1, and if you don't switch the probabilities of winning and losing add up to 1, they don't necessarily cross to each other, meaning the probability of winning does not have to add up to 1 if you switch or don't switch

Yes it does: think about it this way. Obviously, the car is behind one of the two doors. There is a 1/1000 probability that it is behind the one you picked. Now, because the car is either behind that one or the other one, there is a 999/1000 probability that it is behind the other.

Ok I see what your saying, and now it makes sense.  My answer confirmed  :D


Let A be a subset which is dense in the plane. Does A have to be dense in some line segment?

no, because one can construct a dense subset of the plane with no $3$ collinear points. It has already been discuss on the forum.


function
calculus
derivative
vector
absolute value
linear algebra

I have a question regarding the non-differentiability of some functions like absolute value $ |\cdot|$ and L2 norm $ \parallel{}\cdot\parallel{}$:

- For example I have a function
\[ f(b) \equal{} b^2 \minus{} 2yb \plus{} 2c|b|\]
of scalars $ b,y,c$ which I want to minimize with respect to  $ b.$ The first derivative I compute as
\[ \frac {df}{db} \equal{} 2b \minus{} 2y \plus{} 2c \cdot sign(b)\]
and setting zero we have as estimate
\[ \hat{b} \equal{} y \minus{} c \cdot sign(b),\]

\[ \hat{b} \equal{} sign(y)|y| \minus{} c \cdot sign(b)\]
and
\[ \hat{b} \equal{} sign(y) \left( |y| \minus{} c \frac {sign(b)}{sign(y)} \right).\]
But due to $ sign(b)$ this estimate is still in terms of $ b.$ I heard the correct explicit solution is
\[ \hat{b} \equal{} sign(y) \left[|y| \minus{} c \right]_{ \plus{} }\]
where $ [x]_{ \plus{} }$ is $ x$ for greater zero otherwise zero. Apparently the $ [x]_{ \plus{} }$ function has to do with the non-differentiability. I confirmed that the expression is true for the 8 combinations $ \{(|y| > c, |y| < c), (y > 0, y < 0), (b > 0, b < 0) \}$ but I do not know how to derive this explicit solution.

The second equation is
\[ g(b) \equal{} b^T b \minus{} 2 z^T b \plus{} 2c \parallel{}b\parallel{}\]
where $ b$ and $ z$ are $ m$-dimensional vectors and $ c$ is a scalar. Again I want to minimize in terms of $ b$ and I get
\[ \frac {dg}{db} \equal{} 2b \minus{} 2z \plus{} 2c \frac {b}{\parallel{}b\parallel{}}.\]
I set it zero again to yield
\[ b \plus{} c \frac {b}{\parallel{}b\parallel{}} \minus{} z \equal{} 0.\]
I guess one could find the solution by some numerical approach but can one write the solution in terms $ [x]_{ \plus{} }?$

There is no norm differentiable at 0. This fact is due to the non-diffenrentibility of the absvalue at 0 and the fact that any norm is "positivement homogène". All the directional derivatives are different and to be differentible it is necessary that they must be equal.


Can any1 tell me where I can find online the 2003 amc 10s ive done all the ones on aops.
mods feel free to move this to the AMC section if thats where it goes.

Thanks!

[quote="JRav"]You can order (buy) old exams from the site that actually sponsors the test.  Can't remember the name.[/quote]

Search amc on google and you will get there.

yeah worthawolebean pm-d me a good site, Kansas U has all the AMC 10's.

Could you post a direct link here? I couldn't find anything.

All of the AMC 10s from 2000-2007 are in the new [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=3013]Contest Problem Book VIII[/url].

More generally all of the past AMC/AHSME contests are in the [url=http://www.artofproblemsolving.com/Books/AoPS_B_CP_AMC.php]MAA Contest Problem Books[/url].

If you can't buy the books for whatever reason, Google is your friend. Just know that you might not be able to find everything/might not be able to find solutions/whatever. I've been busy posting lots of old AMCs but it's nowhere near complete.

The AoPS books are a great source of problems and almost all of the problems in those are from old AHSME's, Mandelbrots, Mu Alpha Theta's, ARML, IMO,...

uhh try this:
http://www.math.ksu.edu/main/handbook/ProblemSets
*Also its a good idea to save some of these tests on your hard drive as sites are unpredictable

Thanks! That's the one that I was looking for. I remember seeing it a long time ago...[downloaded everything yay]=P


There will be 6 rounds.

Then there will be one gun with 6 slots, 1 of the slots with a bullet, which will be my choice. Based on the order you sign up, you will have one slot. The person with the bullet slot will die. We then go on with 5,4,3,2, then the champion.


Euler
real analysis
real analysis theorems

for sure, at last Euler could ... Now it seems that Robin Chapman is among the choosen one who still can ;)
Homever if you are asking for a proof you can find it here : http://www.maths.ex.ac.uk/~rjc/etc/zeta2.pdf

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=50418]this thread[/url], with that same file linked.


Once upon a time, there were 4 people. (Yes, this has to be a fairy tale.) They weighed themselves in pairs. All but one pair weighed themselves. The resulting totals were 99, 113, 125, 130, and 144. What is the weight of the heavier person of the pair that did not weigh? (Is weigh a word?)

I guess this could be another problem, although harder than the previous.

If you have 25 distinct positive numbers. Prove you can divde one number into two halves (they don't have to be equal), so that you end up into 26 numbers, and you can put them into 2 groups of 13 numbers such that each group has the same sum.

[quote="Chinaboy"]Once upon a time, there were 4 people. (Yes, this has to be a fairy tale.) They weighed themselves in pairs. All but one pair weighed themselves. The resulting totals were 99, 113, 125, 130, and 144. What is the weight of the heavier person of the pair that did not weigh? (Is weigh a word?)[/quote]

4 people, numbers ending in 9,3,5,0,4.

then they have to be..

oh ** I have to go I'll finish this later

The second problem is easy to solve. Let the 25 numbers be A, B, C, D, ... Y in increasing order, so A < B < C < D etc. Then start out by putting A in the first group, B in the second, C in the first, D in the second, etc.
LEMMA: The difference between the two groups at this stage is less than Y.
PROOF: The second group is B + D + F + H ... etc. and the first group is A + C + E + G .. etc. Therefore, the difference between the two groups is equal to (B-A) + (D-C) + (F-E) + (H-G) ... + (X-W). Because the numbers are in numerical order, this is less than (B-0) + (D-B) + (F-D) + (H-F) ... + (X-V) and so is less than X, which is less than W. Therefore, the difference between the two groups is less than W.
END LEMMA
Let the difference between the two groups be Z. Then we can split up the number Y into two parts: (Y+Z)/2 and (Y-Z)/2. We know neither of these two are negative because Z is less than Y, as the lemma shows. If the (Y+Z)/2 is put into the maller group and the (Y-z)/2 is put into the larger group, the two groups now have the same sum.

is the answer [hide]111[/hide]?

Here's the weights of my four people: [hide]33, 66, 47, 78[/hide].  But, is there a more sophisticated way than educated guess and check?


topology
Functional Analysis
advanced fields
advanced fields unsolved

How do I prove that the ultraweak topology (sometimes called $ \sigma$-weak topology) (on f.e. the Banach space $ B(H)$ of bounded functionals on a Hilbert space $ H$) is stronger than the "usual" weak (operator) topology?

Apparently, at this point we are all lost ;). It will be great if you remind the definitions (I'm too lazy to google them up and I suspect that other people have a similar atittude).

I will make it a bit more concrete:

Let $ H$ be a Hilbert space and consider the Banach space $ B(H)$ of bounded operators on $ H$.

The weak topology on $ B(H)$ is defined by the family of semi-norms $ T \mapsto \left|\langle Tx, y \rangle \right|$ (where $ x,y \in H$). The ultraweak topology on $ B(H)$ is defined as the weak* topology on $ TC(H)^*$ (which is isometrically isomorphic to $ B(H)$). So it is given by the family of semi-norms $ T \mapsto \left|\text{Tr}(ST)\right|$ (where $ S$ is trace-class).

So I need to show that any weakly open set contains a ultraweakly open set. But how?

Well, just consider the operator $ Su\equal{}\langle u,y\rangle x$ which is a rank one operator (so, it is certainly in the trace class). Then $ \text{Tr\,}(ST)\equal{}\langle Tx,y\rangle$


[url]clubgauss.net.tc[/url]
Esta recién fundado y se encuentra en una etapa crecimiento.
Nos ayudaría mucho que te dieras una vuelta :)

[quote="Riquelme"]¿Qué países de América Latina cuentan con buenos foros de Matemáticas?

Yo conozco sólo dos... Uno chileno y uno de España (creo).

He aquí las ligas.

Chile: [url]http://www.fmat.cl[/url]

España: [url]http://rinconmatematico.com/foros[/url]

P.D. ¿Sabe alguien de algún otro?  :?:[/quote]

Hay otro foro chileno enfocado a olimpiadas [url]http://www.elnumerodeoro.cl/foro[/url].

[b]Hay uno de geoemtria, y es de Perú: Se llama:[/b]

[url]http://www.forogeometras.com[/url]

Un excelente foro:

[url]http://mundogeometria.wordpress.com/[/url]


Hi everybody,

I would be very grateful to get solutions of Rumanian TSTs from the following years:
1997, 1998, 2000, 2001 and 2002 (or at least for some of these years)

Any suggestions for the source  :)

Can somebody post Romanian NMO 2008 and Romanian TST 2008 problems?

The NMO (and thus TST) are coming at the end of April. Until then there is nothing to post ;)

[quote="joao1"]Is there a book with the solutions of Romanian TSTs?
Can they be found in "Gazeta Matematica"?[/quote]

Who know something about it?   :?:  :)

After the TST's the leaders of the Romanian olympiad comitee publish a booklet abreviated as the RMC (Romanian Mathematical Competitions). This anual booklet contains the solutions for the problems from the district, national round, together with the solutions for the problems for the team selection tests (seniors and juniors). It also contains the official solutions of the BMO problems.

Unfortunately, the RMC is not available on the internet (though I know some leaked versions). Valentin co-authored some of the booklets. Maybe he can give you more details.

[quote="joao1"]Can they be found in "Gazeta Matematica"?[/quote]
The answer is no.


The sides of triangle $BAC$ are in the ratio 2:3:4.  $BD$ is the angle-bisector drawn to the shortest side $AC$, dividing it into segments $AD$ and $CD$.  If the length of $AC$ is 10, then the length of the longer segment of $AC$ is:
\[ \text{(A)}\ 3\ \frac{1}{2}\ \ \ \ \text{(B)}\ 5\ \ \ \ \text{(C)}\ 5\ \frac{5}{7}\ \ \ \ \text{(D)}\ 6\ \ \ \ \text{(E)}\ 7\ \frac{1}{2} \]

[hide]
the length of the sides of the triangle are: 2k, 3k, and 4k (using the given ratio). since AC is the shortest side, and is also equal to 10, therefore: 2k=10 and k=5. Using this, we can find the other two sides of the triangle, they are: 15 and 20. then, by designating one segment of AC as x and the other as 10-x and by using the angle bisector theorem, we have the equation:

x/(10-x)=15/20

solving, we get x=30/7. Thus the longer segment is of length 10-30/7=40/7=C[/hide]


Do you know who's on the IOI (computer) team for Iran?

I think Iran's result in IMO was very good! Do you know what was Iran's place in team scores?

The Iranian IOI team's memebers are: Shayan Oveysgharan, Hossein Kaffash, Aidin Nasiri, Alborz Mazloomian

The result of Iran ioi team :

Rank	Country   Name	                                            Medal	 TOTAL

17	Iran	      Shayan Oveis Gharan	                 Gold	 460
40	Iran	      Aidin Nassiri Shargh	                          Silver	 415
44	Iran	      Seyed Hossein Kaffash Bokharaei        Silver	 410
67	Iran	      Seyed Alborz Mazloomian	                 Silver   370

Best wishes for the next Iran ioi team;  :D


a,b,c greater or equal to 1.

Prove:

sqrt(a-1) + sqrt(b-1) + sqrt(c-1) less that or equal to sqrt(abc+c)

It has been discussed before.
See [url]http://www.mathlinks.ro/Forum/topic-43276.html[/url]
(post #13)

hi
i looked on the inequality to be solved and the inequality u solved and i cant come to an end there. socrates could u or somebody else write down the complete solution, plz?

heres the inequality to be solved again in $\LaTeX$:

Prove: $a,b,c\ge 1 \implies \sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\le \sqrt{abc+c}$

peeta

Excuse me peeta!!
It was not the same.They just resemble each other.
Nothing more.
I 'll try to give a solution to this one!! :)  :lol:


Points M and N lie on a semicircle of diameter AB such that AM-BM=3 and AN-BN=7. Let P be the point of intersection of AN and BM. Evaluate area AMP-area BNP.

alrighty so i think i solved this, but im not entirely sure. and if a more "elegant" solution exists, please post! thanks!

[hide]

after drawing the picture i said that
[AMP]-[BNP] is equal to [triangle ABM]-[triangle ABN]
which equals (AM*BM/2)-(AN*BN/2)

ok the i squared the equation (AM-BM)=3
and got AM^2-2AMBM+BM^2=9
and by pythag and other stuff got
AM*BM/2= (AB^2-9)/4

and did the same thing with AN-BN=7
getting AN*BN/2=(AB^2-49)/4

so i subbed those back into the first equation
and got
(AB^2-9)/4    -    (AB^2-49)/4
which is 40/4
which is 10
[/hide]

Yep, thats how I did it, I don't think it gets more elegant than that  :D .

[hide]We note that $ [AMP]\minus{}[BNP]\equal{}\left([AMP]\plus{}[ABP]\right)\minus{}\left([BNP]\plus{}[ABP]\right)\equal{}[AMB]\minus{}[ANB]$. If we let $ x\equal{}MB$ and $ y\equal{}NB$, we can express the area as:
\[ \displaystyle \frac{x\left(x\plus{}3\right)\minus{}y\left(y\plus{}7\right)}{2}\equal{}\frac{\left(x^2\minus{}y^2\right)\plus{}3x\minus{}7y}{2}\]
Since the two triangles are right triangles, $ x^2\plus{}\left(x\plus{}3\right)^2\equal{}y^2\plus{}\left(y\plus{}7\right)^2 \implies x^2\minus{}y^2\equal{}\minus{}3x\plus{}7y\plus{}20$. Plugging this into the area equation, some terms cancel out, and we are left with $ \displaystyle \frac{20}{2}\equal{}\boxed{10}$.[/hide]


Hi folks!

It's time to announce year 2 of the [url=http://JRMathFestival.org]Julia Robinson Mathematics Festival[/url].

This year, we have two major components.

New for 2008 is the Problem Solving Challenge.  You'll find the round 1 problems on our website.  There's a set of short, warmup problems, that you can use at your school to generate some interest among your friends. Then there are some longer, more involved problems, closer to USAMTS level perhaps, that you can submit to us.  The due date was January 31, but we're going to extend that at least through the end of February, so get to work!  You can work in teams of from 1 to 4 students, and it's open library, open computer, and so on.  In fact, we'll give prizes in categories including best mathematical solutions of course, but also best writing, best use of library, best use of a computer, most interesting conjecture, and anything else we can think of.

We'll have a second round, coming soon and due in mid-April.

The Festival will be organized similarly to last year, with many tables of problems, lunch, a math talk, raffle prizes, free Tshirts, and a cool location.  Due to the demand last year, we were able to persuade our hosts to make it a two day event -- you can choose one of the two.  It'll be May 4 (Sunday) at Google in Mountain View, as last year, and also May 10 (Saturday) at Pixar in Emeryville.  Details will be coming soon on our web site.

It's been fun, and helpful, reading the comments on the 2007 festival.  As you can see (above) we're finding a way to hope to alleviate the overcrowding by hosting two events.  The other major concern was the handling of the raffle: we have a bunch of ideas on how to do it better, and we want to still do it as a raffle (rather than having the usual suspects winning all the prizes as they do at all the other contests), but of course we welcome your input.

If you have any questions or suggestions, I'll be reading this topic.  And please do download the problem solving challenge round 1 and get to work!

Thanks!

--Joshua Zucker
Director, Julia Robinson Mathematics Festival

Bad form to follow up my own post, I guess, but here goes anyway.

There's round 2 of the problem solving challenge now available at [url]http://jrmathfestival.org[/url]

More importantly, registration is now open for the festival itself -- follow the links at the same site.  We're hosting two sites this year, May 4 at Google in Mountain View and May 10 at Pixar in Emeryville.

Questions?  Send me a message (but please read the info on our web site first).

Hope to see you there!

Hmmmm...do we just go to Google tomorrow and meet up with our teams?

Is this mostly diophantine problems? All the warm-up problems are diophantine.

can we go if we haven't registered?
and when will it be?


Euler
number theory
prime numbers
relatively prime

How many positive integers less than or equal to 250 are relatively prime with 250?

I am not sure if this is correct but:

[hide="try this"]
The two prime numbers that divide into 250 are 2 & 5. 250/2 = 125

Next we divide 250 by 5 = 50, we know that half of those 50 are overlapping and have already been covered with the division of two, So there are really 25.

Adding 25+125= 150 , these are all of the numbers that are [b]not[/b] relatively prime, so 250-150 = $ \boxed{100}$ relatively prime numbers. [/hide]

$ 250\equal{}2*5^3$
So by Euler function(which shows how many numbers are relatively prime with n)
$ \varphi(n)\equal{}n(1\minus{}\frac{1}{p_1})(1\minus{}\frac{1}{p_2})\ldots(1\minus{}\frac{1}{p_k})$, where $ n\equal{}p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_k^{\alpha_k}$ and $ p_1,p_2,\ldots ,p_k$ are prime numbers.
So $ \varphi(250)\equal{}250*\frac{1}{2}*\frac{4}{5}\equal{}100.$


search
inequalities
inequalities unsolved

$x,y,z$ are real numbers verifiing : $x^{2}+y^{2}+z^{2}=2$
prove that : $x+y+z\leq xyz+2$

Inequalities Solved problems, I guess. Though I realize that it is quite hard.
It is better to wait for the new solution :P


find a function f(x)
such that
$f(x)+f(1-\frac{1}{x})=x$

[hide]
Let $g(x)=1-\frac{1}{x}$. Then applying $g$ multiple times,
\[g(g(x))=1-\frac{1}{1-\frac{1}{x}}=1-\frac{x}{x-1}=\frac{1}{1-x}\]
\[g(g(g(x)))=1-(1-x)=x\]
Since $f(y)+f(g(y))=y$ for all $y$, substituting in $y=x,\ y=g(x),\ y=g(g(x))$ gives the system of equations
\[f(x)+f(g(x))=x\]
\[f(g(x))+f(g(g(x)))=g(x)\]
\[f(x)+f(g(g(x)))=g(g(x))\]
Adding the three and dividing by 2,
\[f(x)+f(g(x))+f(g(g(x)))=\frac{x+g(x)+g(g(x))}{2}\]
Subtract the second equation to get
\[f(x)=\frac{x-g(x)+g(g(x))}{2}=\boxed{\frac{x+\frac{1}{x}+\frac{1}{1-x}-1}{2}}\]
[/hide]

well this is an easy case of http://www.mathlinks.ro/Forum/viewtopic.php?t=126510 . I didn´t find the topic that fermat has talked about and couldnt solve the problem of this link, so, i will just put my solution, since it is quite easy and short.

It´s trivial that:
$f(x+1)+f(\frac{x}{x+1})=x+1 (I)$
also is trivial that
$f(\frac{x}{x+1})+f(\frac{-x}{1}) = \frac{x}{x+1}(II)$
and also:
$f(\frac{-1}{x})+f(x+1)=\frac{-1}{x}(III)$
Subtracting I-II :
$f(x+1)-f(\frac{-x}{1}) =x+1-\frac{x}{x+1}$

Adding with III:

$2f(x+1)=\frac{-1}{x}-\frac{x}{x+1}+x+1$
substituting x+1=u .... the rest is just dumbass work.

... wops... almost the same thing that is posted above ( i just saw it after that i´ve posted! right in the same time man eheheh) maybe this is more clear considering that i didn´t use any additional definition


AMC AIME USA(J)MO USAMO logarithms trigonometry algebra

:arrow:  :huh:  :maybe: Hello, I have a question about if I should take AMC 10 or 12. Unlike you guys, I'm not that great at math (in my school I am). I was wondering what you guys reccomend I take to have the best chance at making AIME. On practice tests, I get about the eqivalent on the two tests. I dont think my school offers the option of AMC 10/12 B. Thanks in advance, and good luck to all on math competitions! :alien:  :)

If you just want to make AIME, take the 12.

This really depends on 1. what your goals are, 2. what topics you've been exposed to, and 3. what you are good at.

1. If your goal is to make the AIME, take the AMC 12 because the cutoff is only 100 which you can make by getting the first 14 questions right (rest blank) which are all amc 10 level questions.

If your goal is to make the USAMO, then take the AMC 10 because it is much easier to get a high score, thus a higher USAMO index.

2. If you have experience handling logarithims, trig, complex numbers, graphs of polynomial functions, and basically the curriculum up to and including precalculus (by knowing I mean you have pretty much mastered it and can solve difficult problems about these), then you are better off taking the AMC 12. If you have been exposed to these, then the amc 12 becomes a harder amc 10 with more topics covered.

If have had no experience with the aforementioned topics, then I would recommend taking the AMC 10 as it only covers first year algebra and geometry (although the last few geometry problems can be easily trig bashed  :P  ). Also, the AMC 12 relies heavily on algebra 2 and precalculus topics, and if you have never seen these, it will be a pretty uphill battle. Of course, with a SOLID mastery of algebra 1 and geometry, you possibly could get 115-120+, it depends on how the year is..

3. As has been said before, the AMC 12 relies heavily on second-year algebra, precalculus topics, and trig (not even close to as much as AIME though). Therefore, if your strength is algebra, the AMC 12 will feature many questions you will love.

However, if your strength is geometry, the AMC 10 is for you. It relies heavily on geometry, and no geometry topics (such as the geometry of complex numbers that occasionally show up on the AMC 12) that can't be learned in AoPS 1 and Intro to Geometry will be covered.

That's my two cents, I hope it helps!

Thanks. :thumbup: I'll consider that and also talk to my math teachers/coaches


limit
function
induction
inequalities
calculus
calculus computations

Let $ a_{0}\equal{}\sqrt{6}$ and $ a_{n}\equal{}\sqrt{6\plus{}a_{n\minus{}1}}$. Find $ \lim_{n\to\infty}a_{n}$.

Is there a way that does not involve letting $ \sqrt{6\plus{}\sqrt{6\plus{}\sqrt{6\plus{}\ldots}}}$ equal $ x$ and then solving the equation?

There's a general technique that works here.
Suppose $ f(x)$ is a continuous function from $ \mathbb{R}$ to itself, and the sequence $ a_{n}$ is defined by choosing a constant $ a_{0}$ and letting $ a_{n}\equal{}f(a_{n\minus{}1})$ for $ n\ge 1$. We want to find the limit of the $ a_{n}$, if it exists.

First, note that if $ \lim a_{n}\equal{}L$, $ f(L)\equal{}L$ by the sequential definition of continuity.

If $ f$ is an increasing function, with $ f(x)>x$ for $ x<A$ and $ f(x)<x$ for $ x>A$, then $ \lim a_{n}\equal{}A$.
Why is this true? We split into two cases.
Case 1: $ a_{0}\le A$. Then $ a_{n}\le A$ by induction; $ f(x)\le A$ for all $ x\le A$. Since $ a_{n\minus{}1}\le A$, $ a_{n}>a_{n\minus{}1}$ and the sequence is increasing. A sequence which is increasing and bounded above has a limit; this limit can't be greater than $ A$, and it can't be less than $ A$ by our observation on fixed points.
Case 2: $ a_{0}\ge A$. Reverse the inequality signs, and switch "increasing" to "decreasing". The argument is otherwise identical.

There are other criteria for convergence, like the contraction principle: if $ f$ maps some closed interval $ I$ to itself with $ |f(x)\minus{}f(y)|\le M|x\minus{}y|$ for some constant $ M<1$ and all $ x,y\in I$, there is a unique fixed point in $ I$ and the $ a_{n}$ converge to it if any $ a_{k}$ is in $ I$.

So here by induction $ \sqrt{6}\le a_{n}\le 3$ is bounded, it will be increasing, so it has a $ L$ limit. Solving $ L\equal{}\sqrt{6\plus{}L}$ you get $ L\equal{}3$

[quote="aspiringchemlete"]What is the value of $ \sqrt{0\plus{}\sqrt{0\plus{}\sqrt{0\plus{}\ldots}}}$?[/quote]
It is $ 0$, whats is the problem?
$ s_{1}\equal{}0,s_{2}\equal{}0,s_{3}\equal{}0,...$ so this is a konstant sequence of 0's.

[quote="sylow_theory"][quote="aspiringchemlete"]What is the value of $ \sqrt{0\plus{}\sqrt{0\plus{}\sqrt{0\plus{}\ldots}}}$?[/quote]
It is $ 0$, whats is the problem?
$ s_{1}\equal{} 0,s_{2}\equal{} 0,s_{3}\equal{} 0,...$ so this is a konstant sequence of 0's.[/quote]
$ 0$ is obvious, but when you square it and solve for $ x$, you find that the value can be  $ 1$... how come?

Going back to jmerry's language: we have a continuous function $ f$ and we choose a starting $ a_{0}$ and then define $ a_{n\plus{}1}\equal{}f(a_{n}).$

Here's what we can say, and be careful of the logic here:

[b]If[/b] the sequence converges, then the number it converges to must be [b]among[/b] the solutions to the equation $ x\equal{}f(x).$

For instance: define a sequence recursively by $ a_{n\plus{}1}\equal{}a_{n}^{2}.$ The equation $ x\equal{}x^{2}$ has two roots, $ 0$ and $ 1.$ Which one of these does the sequence converge to? Well, that depends on where you start. It turns out to be very difficult to get the sequence to converge to $ 1,$ because that fixed point is unstable. If $ a_{0}\equal{}1,$ or if $ a_{0}\equal{}\minus{}1,$ then the sequence becomes constantly $ 1.$ Otherwise, there's no way to get there in the limit. If $ |a_{0}|<1,$ then the sequence converges to $ 0.$ That point is stable. And if $ |a_{0}|>1,$ then the sequence tends to infinity.

Finding solutions to an equation of the form $ x\equal{}f(x)$ does not tell you what a limit is, it merely sets out some possibilities.


Let $ n$ be an odd positive integer. If $ \frac{2^{n}\plus{}1}{3}$ is prime, is n prime also ?

YES. Proof by contradiction. Assume $ n$ is not prime, so $ n \equal{} ab$, $ a,b > 1$ odd positive integers. Then $ \frac {2^n \plus{} 1} {3} \equal{} \frac {(2^a)^b \plus{} 1} {3} \equal{} \frac {2^a \plus{} 1} {3} ((2^a)^{b\minus{}1} \minus{} (2^a)^{b\minus{}2} \plus{} \cdots \plus{}1)$, with both factors greater than $ 1$, hence not prime.


inequalities
geometry
geometric transformation
reflection
geometry unsolved

let ABC, G is intersection of medians. M is in ABC. GM cut AB,BC,CA at I,J, K. prove that:
   IM/IG+JM/JG+KM/KG =3
I am bad at English!

I wonder why he had to move his problem to the Inequalities section.

The problem is:

Given a triangle $ ABC$, its centroid $ G$ and an arbitrary point $ M$ inside $ ABC$. $ GM$ cut $ AB,BC,CA$ at $ I,J, K$. Prove that:
$ \frac{IM}{IG} \plus{} \frac{JM}{JG}\plus{} \frac{KM}{KG} \equal{} 3$

To [b]loc1991[/b] : It is not very difficult. You should try to figure it out.

To [b]madshiet[/b]: Why do you want to show his nationality ? He has his own privacy that you should respect.

[quote="mathangel"]I wonder why he had to move his problem to the Inequalities section. [...] To [b]madshiet[/b]: Why do you want to show his nationality ? He has his own privacy that you should respect.[/quote]

Don't feed the troll, please. And as for the problem, $ \frac{IM}{IG} \plus{} \frac{JM}{JG}\plus{} \frac{KM}{KG} \equal{} 3$ rewrites as $ \frac{IG\plus{}GM}{IG} \plus{} \frac{JG\plus{}GM}{JG}\plus{} \frac{KG\plus{}GM}{KG} \equal{} 3$, what is equivalent to $ \left(1\plus{}\frac{GM}{IG}\right)\plus{}\left(1\plus{}\frac{GM}{JG}\right)\plus{}\left(1\plus{}\frac{GM}{KG}\right)\equal{}3$, what is equivalent to $ \frac{GM}{IG}\plus{}\frac{GM}{JG}\plus{}\frac{GM}{KG}\equal{}0$, what is equivalent to $ \frac{1}{IG}\plus{}\frac{1}{JG}\plus{}\frac{1}{KG}\equal{}0$, what was proven at ... some topic started by Ashegh(?) which I cannot find at the moment. Anyway, it should be noticed that the problem only makes sense if the segments are directed.

  darij

[quote="yetti"]It was solved by [b]gemath[/b] in [url=http://www.mathlinks.ro/viewtopic.php?t=108719]Line through centroid[/url] posted by [b]Jan[/b].[/quote]

Thanks for the link. Besides, it is Theorem 1 in my [url=http://www.cip.ifi.lmu.de/~grinberg/S64.pdf]solution of Mathematical Reflections problem S64[/url] which I just uploaded on my website because the deadline for this problem is over for some days.

  Darij

For cevians AA1, BB1,CC1  concurrent at point M we have two Gergonne relations

   MA1 / AM + MB1 / BM  +MC1 / CM = 1

   AM / AA1 + BM / BB1 + CM / CC1 = 2

  Adding them solves the problem.

   T.Y.

   M.T.

[quote="armpist"]For cevians AA1, BB1,CC1  concurrent at point M we have two Gergonne relations

   MA1 / AM + MB1 / BM  +MC1 / CM = 1[/quote]

You mean $ \frac{MA_1}{AA_1}\plus{}\frac{MB_1}{BB_1}\plus{}\frac{MC_1}{CC_1}\equal{}1$.

[quote="armpist"]   AM / AA1 + BM / BB1 + CM / CC1 = 2

  Adding them solves the problem.[/quote]

Non sequitur. At least I don't see how.

  darij


I would like to know the Vietas Formula and others that are helpful for math competitions.
If you know any , please post them below ; examples would be a great explanation

Another way to find these is by expanding $ (x \minus{} r_1)(x \minus{} r_2) \equal{} 0$. It can be helpful on problems where you are given roots and have to find coefficients.

[quote="bulldog23"]Sure...

The quadratic is written in the form of ...  ax^2+bx+c=0.

The sum of the roots is (-b/a)

The product of the roots is (c/a)

You can easily prove these two statements by using the quadratic formula.[/quote]

Expanding on what he said...

The sum of the roots for [b]any [/b]polynomial, no matter what degree, is always $ \frac { \minus{} b}{a}$.

The product of the roots is $ \frac {c}{a}$ if the polynomial degree is even, if odd it is $ \frac { \minus{} c}{a}$.

While Vieta's applies to quadratics, it's more useful in the context of $ n$-degree polynomials.

If $ P(x)\equal{}\sum a_ix^i\equal{}a_n\prod(x\minus{}r_i)$, we have the following relationships:

$ s_1\equal{}\sum r_i\equal{}\minus{}\frac{a_{n\minus{}1}}{a_n}$

$ s_2\equal{}\sum r_ir_{i\plus{}1}\equal{}\frac{a_{n\minus{}2}}{a_n}$

...

$ s_k\equal{}(\minus{}1)^k\frac{a_{n\minus{}k}}{a_n}$

You can see why this is true simply by expanding the polynomial. (The coefficient of $ x^{n\minus{}1}$ is clearly $ \minus{}(r_0\plus{}r_1\plus{}\cdots\plus{}r_n)$, etc).

And as for other formulas that are helpful, a related method is using Newton's sums to evaluate the sum of the $ m$th powers of the roots of a polynomial.

So, as before, let $ P(x)\equal{}\sum a_ix^i\equal{}a_n\prod(x\minus{}r_i)$.

Newton's sums state the following:

Define $ S_k\equal{}\sum r_i^k$. Then the following relationships hold:

$ a_nS_1\plus{}1a_{n\minus{}1}\equal{}0$

$ a_nS_2\plus{}a_{n\minus{}1}S_1\plus{}2a_{n\minus{}2}\equal{}0$

$ a_nS_3\plus{}a_{n\minus{}1}S_2\plus{}a_{n\minus{}2}S_3\plus{}3a_{n\minus{}3}\equal{}0$

...

For example, consider the following problem: Find the sum of the squares of the roots of $ 4x^3\plus{}3x^2\plus{}2x\plus{}1$. (Totally just made this up, so the answer might be ugly).

We apply Newton's sums like so:

$ 4S_1\plus{}3\equal{}0\implies S_1\equal{}\minus{}\frac 34$ (this is also true by Vieta's).

$ 4S_2\plus{}3S_1\plus{}2\cdot 2\equal{}0\implies S_2\equal{}\boxed{\minus{}\frac 7{16}}$

[quote="limac"][quote="007math"][quote="bulldog23"]Sure...

The quadratic is written in the form of ...  ax^2+bx+c=0.

The sum of the roots is (-b/a)

The product of the roots is (c/a)

You can easily prove these two statements by using the quadratic formula.[/quote]

Expanding on what he said...

The sum of the roots for [b]any [/b]polynomial, no matter what degree, is always $ \frac { \minus{} b}{a}$.

The product of the roots is $ \frac {c}{a}$ if the polynomial degree is [b]positive[/b], if [b]negative[/b] it is $ \frac { \minus{} c}{a}$.[/quote]

I think you mean even instead of positive, and odd instead of negative.[/quote]

Yeah...I was thinking about the positive and negatives and typed that up.


How do you get to Lexington via MBTA?

Buses? If so, what #'s?

Thanks, and do reply.
You would want me on the team.

There's a bus that takes you from Alewife to Lexington Center, and the school is about a 5 minute's walk from there. I want to say it's #62 but I'm not sure. Just look it up.


Let $ a_n$ be the last non-zero digit in the decimal representation of the number $ n!$.Is the sequence $ a_1,a_2,...$ eventually periodic?

Please continue discussing [url=http://www.mathlinks.ro/viewtopic.php?t=220913]here[/url].


inequalities
Cauchy Inequality
inequalities proposed

How about this
$ \frac{a^2}{(2c\minus{}b)(2b\minus{}a)} \ge \frac{a^3}{c^2b}$

[quote="great math"]How about this
$ \frac {a^2}{(2c \minus{} b)(2b \minus{} a)} \ge \frac {a^3}{c^2b}$[/quote]

I don't know what you mean , Great Math . :wink:

If you have solution, I appreciate if you post it, thanks . :roll:

[quote="forevercontinue"]Let $ a,b,c \geq 0$ and asume that the inequality is valid. Prove that :

$ \frac {a^2}{(2c \minus{} b)(2b \minus{} a)} \plus{} \frac {b^2}{(2a \minus{} c)(2a \minus{} b)} \plus{} \frac {c^2}{(2b \minus{} a)(2a \minus{} c)} \geq 3$

 :roll:  :P[/quote]
It's obvious because your inequality is valid.

[quote="forevercontinue"]Let $ a,b,c \geq 0$ and asume that the inequality is valid. Prove that :

$ \frac {a^2}{(2c \minus{} b)(2b \minus{} a)} \plus{} \frac {b^2}{(2a \minus{} c)(2a \minus{} b)} \plus{} \frac {c^2}{(2b \minus{} a)(2a \minus{} c)} \geq 3$

 :roll:  :P[/quote]
Assume what is valid? :huh:
Sorry but I did not understand the question at all.... :(
Can anyone elaborate a little? :roll:

My inequality is trivial after the expansion. You might check it out with the assumption that all the conditions are valid.
The last step is obvious, we only use Cauchy inequality to prove it.
Have fun!

[quote="great math"]How about this
$ \frac {a^2}{(2c \minus{} b)(2b \minus{} a)} \ge \frac {a^3}{c^2b}$[/quote]
This is wrong, take a=5 b=1 c=0.01.

[quote="great math"]No, the question says all the inequality is valid and I used Cauchy to prove the inequality that you spotted.[/quote]

Oh, no great math. I think there is a misunderstanding here. Because of my bad in English , what I want to mean here is : The condition of this inequality satisfy $ a \neq 2b , b \neq 2a , b \neq 2c , c \neq 2b,a \neq 2c , c \neq 2a$  :(

I am sorry for my silly, I also have a mistake in typing . The inequality must be:

$ \frac {a^2}{(2c \minus{} b)(2b \minus{} a)} \plus{} \frac {b^2}{(2a \minus{} c)(2c \minus{} b)} \plus{} \frac {c^2}{(2b \minus{} a)(2a \minus{} c)} \geq 3$

which is $ a,b,c \geq 0$ and satisfying  $ a \neq 2b , b \neq 2a , b \neq 2c , c \neq 2b,a \neq 2c , c \neq 2a$

[quote="quykhtn-qa1"][quote="arqady"]It's still wrong. :wink:[/quote]
Yes,this inequality is false :oops:[/quote]
[quote="arqady"]It's still wrong. :wink:[/quote]

Thanks, quykhtn-ga1 and arqady. I have a trouble with the way created this inequality . :(
Anyone can check this inequality for me :
Let $ a,b,c\geq 0$ satisfy $ a \neq 2b , b \neq 2a , b \neq 2c , c \neq 2b,a \neq 2c , c \neq 2a$. Prove that :
$ \frac {a^2}{|(2c \minus{} b)(2b \minus{} a)|} \plus{} \frac {b^2}{|(2a \minus{} c)(2c \minus{} b)|} \plus{} \frac {|c^2}{|(2b \minus{} a)(2a \minus{} c)|} \geq 3$


group theory
abstract algebra
superior algebra
superior algebra unsolved

I need a little help figuring this one out.

Let H be a normal subgroup of G group. If H and G/H are p-groups, then G is a p-group

[quote="ZetaX"]$ |G| \equal{} |H| \cdot |G/H|$.[/quote]
That only works in the finite case. A $ p$-group is usually defined as one where all elements have order which is a power of $ p$. There certainly are infinite $ p$-groups, and the question makes perfect sense in this case, too.

It's the same proof (if using supernatural numbers).


probability
geometry
3D geometry
geometric transformation
rotation
reflection
AMC

2006 Nats sprint...

25. A house with a square floor plan is to have each of its four outer walls finished with brick, stone, stucco, or wood. If no two adjacent outer walls are to have the same finish, in how many ways can the house's walls be constructed? (One such way is brcik on the front and back and stone on the left and right.)

I got 84, but I'm not sure if it's right, because I usually miss some cases.

28. The four dominoes below are used as fractions. For example, 2 dot, 3 dot (Two dots on top, 3 dots on bottom) can represent 2/3 or 3/2. One of these dominoes: 2 dot, 3 dot...4 dot, 5 dot...2 dot, 2 dot...6 dot, 3 dot is placed in each empty space of the expression below such that all four dominoes are used exactly once. What is the greatest possible value for the expression. Express your answer as a simplified mixed number.

[Blank space to put a domino] - [Blank] X [Blank] + [Blank]

I got 2 7/10, but once again, I don't trust myself.

29. A bag contains 3 tan, 2 pink, and 4 violet chips. If the 9 chips are randomly drawn from the bag, one at a time and without replacement, what is the probability that the chips are drawn in such a way that the 3 tan chips are drawn consecutively, the 2 pink chips are drawn consecutively, and the 4 violet chips are drawn consecutively, but not necessarily in the tan-pink-violet order? Express your answer as a common fraction.

I got 1/210, but counting +probability is my weakest subject...

2007 Nats Sprint...

16. Kyle will use four identical unit cubes to create a solid. each cube must be glued to at least one other cube. Two cubes may only be glued together in such a way that a face of one cube exactly covers a face of the other cube. How many distinct solids could Kyle create? Two solids are considered to be the same if one solid can be repositioned to match the other solid.

I don't really get the problem. My answer is 5 (the number of tetrominoes), but I'm not sure if rotations are counted or not. I would really appreciate clarification of this problem.

23. The variable x varies directly as the cube of y, and y varies directly as the square root of z. If x equals 1 when z equals 4, what is the value of z when x equals 27.

I got 36, but i'm not sure if I did the variation stuff right.

25. How many even, three-digit positive integers have the property that exactly two of the integer's digits are equal?

I got like ten different answers using complementary and constructive counting, brute force, etc... NO CLUE on this one. PLZ HELP

27. If each letter in the expression (a + (b/c))(d+e) is replaced by a different digit from 1 through 9, inclusive, what is the smallest possible integer value of the expression.

I got 8, but I hate these type of problems, so yea...

28. For any a and b, we will define a "move that adds (a,b)" as a move from any point (c,d) to the point (c+a,d+b). A particle moves according to the following rules. If a move added (p,q) to get to the current position, the next move will add either (p-1,q) or (p,q-1) to move to the next position. Neither p nor q can ever be negative. Particle A starts at (0,0) and moves to (3,4). Particle A will continue to move according to the preceding rules. When the particle can no longer move according to the preceding rules, it is located at point (r,r). What is the value of r for particle A.

I got 14, but I'm pretty sure I didn't understand the problem correctly.

30. An equilateral triangle is placed in its starting position on a table with its vertices labeled as shown (I'm really bad at this)
            TOP
             C

A                         B

On each move, it can either be reflected over a vertical axis through the top vertex (switching the location of the bottom two vertices), or it can be rotated clockwise 120 degrees about its center. How many different sequences of five moves will return the triangle to its original starting position?
               Reflection
                  TOP
                    C

B                                   A

               Rotation
                 TOP
                   A

B                               C

And that's about it (sorry if my drawings were really messed up) and I hope you can help. Thanks in advance,

EP

06-25

 u are right

[hide]
front has 4 choices

if L and R not the same , then L has 3 and R has 2 choices, then back will have 2 choices;

if L and R are same (3choices) , then Back will have 3 choices

so total:

4*3*2*2 + 4*3*3=84[/hide]

Thanks =P
I really appreciate the help. =)
One down, 10 to go, =(.

23. The variable x varies directly as the cube of y, and y varies directly as the square root of z. If x equals 1 when z equals 4, what is the value of z when x equals 27.

$ x\equal{}ky^3$ and $ y\equal{}l\sqrt{z}$.
Therefore $ x\equal{}mz^\frac{3}{2}$, where $ m\equal{}kl$.
$ m$ must be $ \frac{1}{8}$. The answer is $ \frac{9}{4}$.

Thanks aleph, I had a feeling it was wrong...

and thanks to icecap for verifying my answer. =) Thanks to all

wait aleph... if x=ky^3 and y\sqrt z then doesn't x=k(l^3)z^3/2? cuz you have to cube the L too, right?
so m=k(l^3) and x=mz^3/2 so z^3/2=27 X 8 so z^3=27^2 X 8^2 and if you take the cube root of both sides... z=3^2 X 2^2 =36...

what did i do wrong?

right...l is cubed, but that doesn't change anything

you're right it's 36  :blush:
my mistake

thanks for clarifying number 28, i was afraid it was wrong, cuz i blanked out while trying to do it...it took me 3 days to UNDERSTAND the problem =P

oh ok yea so it is 36 thx for your help i REALLY appreciate it =)

AIME, we all know ur a genius! now solve some problems =p

That problem is on mathcounts marathon we don't need it here

Yea...AIME go take your problems somewhere else =P

still...solve teh problems AIME u can do it!
=P

Does any1 get #16, even if u don't have an answer, i would like to hear your interpretation of the problem

#16:

[hide]all 4 cubes in one line: 1way

3 line up, 1 on side: 2 ways

2 line up: 1way

total 4 ways[/hide]

I could be wrong :(

25

[hide]1)	a,a,b type:  b=0,2,4,6,8 ---5ways, a can be 9 (if b=0);  8 (if b not zero)
  so total: 9+4*8=41
2)	b,a,a, type: if a=0, b has 9 ways, if a=2,4,6,8 b has 8 ways
         so total 9+4*8=41
3)	aba type:  a=2,4,6,8  4 ways;  b has 9 ways
   so 36 ways

[u][b]total: 41+41+36=138[/b][/u][/hide]

For numer 16, I think you're wrong...you could have stacked cubes.

#16:
Could also be wrong:
4 in a line,
3, on the central 1
3, on the corner 1
Snake,
Cube

5 ways?

[quote="Bugi"]
3, on the corner 1
[/quote]

there's 2 different ways to do this.

Yes 123
         4
and 123
      4

Then 6?

[quote="Bugi"]My last post should have been L shape (sorry)
2 L
1 snake
1 cube
1 odd t
1 line[/quote]

what about the other 3? (3 dimensional ones)? And only 1 L shape

#16 seems really weird...

ty to policecap for doing #25 thx a lot!

imma try brute force wait a sec...


AMC
USA(J)MO
USAMO
geometry
3D geometry
algorithm
number theory

We will show that any composite positive integer $n$ with mroe than 4 divisors, inluding 1 and itself, will work. We will also show that any $n$ with four divisors or less will not work unless "n" is a perfect square.

First consider $n=pq$where $p,q$ are primes. Here $n$ has exactly four divisors: $1, p, q, pq$. When we arrange them in a circle we only use $p, q, pq$. Since there are only 3 divisors , $p$ and $q$ will be adjacent to one another. However, since $p$ and $q$ are prime, then $gcd(p,q)=1$. Thus, if $p$ does not equal $q$ then $p$ and $q$ will be relatively prime and the arrangement will not work and if $p=q$ then $n$ will be a perfect square.

Now consider the case where $n=p_1 \cdot p_2 \cdot p_3 ... p_w$. We will arrange the divisors of $n$ in the following way:

First arrange all of the primes $p_1, p_2, p_3, ..., p_w$ in a circle. Then, in between each pair of primes, $p_i$ and $p-J$, where $1 \leq i,q \leq w$, insert $p_i \cdot p_j$. Now in between $p_i$ and $p_i \cdot p_j$ insert the remaining divisors which have $p_i$ as a factor but are less than $n$. You will end with some divisor which has $p_i$ as a factor along with some $p_m$ where $1 \leq m \leq w$. Continue this sam process with $p_m$ until you have exhausted all divisors containing it, except $n$, and do this for every prime until you are left with only $n$. Now insert $n$ in between your last written divisor and $p_i \cdot p_j$. This arrangement guarantees that each divisor of $n$ is adjacent to antoher divisor which contains at least one of the same prime factors so no two divisors are relatively prime.

Using the same reasoning as in the case where $n=p_1 \cdot p_2 ... \cdot p_w$ we can prove that any $n$ in the form $n=p_1^{k_1} \cdot p_2^{k_2} ... \cdot p_w^{k_w}$, where $w>2$ will satisfy the conditions of the problem.

Dimitar, *bop* quit worrying. I don't think anyone here except a USAMO grader could give you a score anyways since it depends on the rubric they use on this problem.

It's my job to worry. Bah, I can't believe how stupid I was  :(

The given solution to this problem is really confusing me.  The answer is all composites with the exception of the semi-primes, correct?

JB

Ahem. The cube of a prime has 4 divisors and it works. Minus 1 point!  :mad:

Haha...jkjk. That's clearly a typo, and even if that is what you wrote, it would not merit a deduction normally (at least to my experience).

[quote="DPopov"]My solution is confusing?  :([/quote]

DPopov
Sorry, the solution on the [i]AMC Website[/i] is confusing, not yours.  Yours looks fine.

JB

Not to be mean or anything, but I would guess there would be a deduction.  The solution doesn't state a correct characterization of valid n.  It never handles the case of a prime power.  It doesn't properly handle the case when n has exactly two prime factors.

Even the case when n is the product of 3 or more primes is confusing.  Are i and j supposed to be consecutive?  In other words, is j = i + 1 (mod w)?  The solution never says.  At first, the solution seems to be using all (consecutive) i and j, for example, when placing $p_i p_j$ between $p_i$ and $p_j$.  But later, when the algorithm places all remaining divisors between $p_i$ and $p_i p_j$, the solution seems to be assuming a fixed choice of i and j (say i = 1 and j = 2).  Otherwise, the same divisors would be placed multiple times throughout the circle.

I think you have the core of a correct idea, but the presentation was off.

Yeah, it could be explained a lot more clearly.  But stop worrying!  You'll get what you'll get, and next year (if you're not a senior) you'll do better.  If you do well this year, great.  If you don't, you'll have something to make you want to do better next year (hehe.. applicable to me :D).

It's saying any divisor that contains $p_i$ where $p_i$ is a prime factor of $n$ and $i$ is just the subscript of the prime indicating it's different from $p_j$. Note how I defined it previously. Yes, I admit it may be a little confusing :(


Below in the given figure the columns form a pattern (arithmetic series) and the rows form pythagorean triplets. After finding the nth term of each of these column. Prove that for every general term in the row there will be a pythagorean triplet.
PS: Pretty good eh for OUR school :D  but still $\text{very}^ {\text{very}}$ easy. :D

well here is a hint:
[hide]$a= 2n+1$ :)
no for real here is a hint:
[hide]
one can see that it must be $(2n+1)^2+x^2=(x+1)^2$. its quite easy from there to find the right sequences ;)
[/hide][/hide]

[hide="hints"]
[hide="left column"]odd[/hide]
[hide="middle column"]divide by 4...notice anything?[/hide]
[hide="right column"]what are the differences between consecutive terms?[/hide]
[/hide]


You are invited to Chow Ka Fat's "Three-way Artificial Intelligence Mastermind" Web Page, the most feature-rich on-line version of the board game classic. (Or being another classic itself?) By playing this game, you can have a good training of your logical skill. In fact, the process of breaking a Mastermind code is similar to cracking a Mathematical puzzle.
 :D
Web Links:
[url]http://home.pacific.net.hk/~kfzhou/mmf.html[/url]
or
[url]http://home.pacific.net.hk/~kfzhou/mmh.html[/url]

MAIN FEATURES
-There are billions of possible code combinations (or permutations, using the proper mathematical term). You have to take a life's (or several lives') time to break all the codes.

-There are altogether 1,424 different game settings to suit players of different levels.

-You may choose to be the Codebreaker or the Codemaker. You may also choose to watch the computer play with itself. (Hence the phrase "Three-Way" in the title of this homepage.)

-Equipped with Artificial Intelligence (A.I.) of this programme, the computer can break the most difficult code using an incredibly small number of steps within seconds! (Believe it or not?)

-There are different versions of my games, including VB, VBScript, JavaScript, ActionScript (i.e. Flash)

-You may play online and you may also download one to play on your PC.

[img]http://home.pacific.net.hk/~kfzhou/flaplay.gif[/img]


floor function
AMC
AIME
geometry
logarithms
trigonometry
function

Your solutions look fine from a brief glance, except you didn't show that the values 40 and 60 were attainable in problem 4. (I don't know how strictly they grade, though.)

I can't believe I got the wrong answer (84) to number 1! I counted a two cases where the 7 and 8 were connected, which then became multiplied by 12 to increase the answer by 24. However, I think my other solutions are okay.

Darn for #2 your solutions look so complicated.

What I did was I took the square roots of the numbers in the set as $ a$ and $ b$, wlog a>b, so we have $ a^2 \minus{} b^2 \equal{} 11111 \equal{} > (a \plus{} b) \equal{} \frac {11111}{a \minus{} b}$ and from there we use the factors of 11111 to determine a and b, rejecting the values of $ a^2$ and $ b^2$ that are not 5 digits.

And same for the 6 digit one, where neither $ a^2$ or $ b^2$ are 6 digits.

I would post my solution, but.....I epicly bombed #1 and only got 1/2 of #4...

FAIL!!!!!

EDIT: If only I knew how to program.....

Dang so my explanations on #4 as to why my constructions were convex were pretty hurried...hopefully they won't take off points?

Darn, so apparently they are looking for explicit construction? Hopefully they will not take more than 1 point off.

Will my "equilateral triangle" example for #4 be insufficient?

I hope points can't be taken off for ugliness.
Or addition mistakes (i.e. 1488+4=1490 :oops: )
5/5/5/5/5
(Solutions are to ugly to post.)

5/5/5/1/5
BS-ed #4 prob a point for effort  :D
I had a nice soln for 5 b):
$ r^{a_1} \plus{} r^{a_2}... \equal{} r^{b_1} \plus{} r^{b_2}... \implies {\lfloor {log_rLHS}\rfloor} \equal{} {\lfloor {log_rRHS}\rfloor}$. Then let $ a_k$ be the first $ a_i \neq b_i$. Canceling and factoring out the first term we show that $ a_k \equal{} b_k$ which contradicts. I'm too lazy to do this, its trivial tho. Shoot I have to be perfect now If I want to get gold  :wink:  prb need 21/23/24 if I want to make AIME through USAMTS.

[quote="andersonw"]Darn, I completely missed the 40 case on problem 4. Maybe 2-3/5 then? And maybe another point off for not doing the construction? Oh well. Also, my #3 solution was really convoluted, although I'm pretty sure it still works, just not well explained.[/quote]

Wait dude, will your #3 count? I think my solution is somewhat similar, and I wasn't very sure. Also, like you, I was up the night before writing out my solutions (until about 3:30 am)  :P

Hmm, so I'll post my pdf. I only answered 1-3 because I am not very good at math. Also, 4 is a geometry, and 5 is too difficult for me to understand completely.

Also, because of that time, I apparently completely missed #1. Perhaps 2 points for correct method?

Actually, how do you attach a file?  :|  :maybe:

Just because $ a_1\equal{}1$ is required doesn't mean $ 1,4,7,\ldots$ is the only working sequence. But you can easily extend the idea you used...

Oh darn, missed number 4's 40 case completely.  I must have made some very horrible assumptions...

This week is the week of Columbus Day!  So #3's answer must be a horrible coincidence.

@feuxfollets: I look forward to taking you on in your signature challenge.

[quote="Thunder365"]Darn for #2 your solutions look so complicated.

What I did was I took the square roots of the numbers in the set as $ a$ and $ b$, wlog a>b, so we have $ a^2 \minus{} b^2 \equal{} 11111 \equal{} > (a \plus{} b) \equal{} \frac {11111}{a \minus{} b}$ and from there we use the factors of 11111 to determine a and b, rejecting the values of $ a^2$ and $ b^2$ that are not 5 digits.
[/quote]

Exactly what I did.

I totally screwed up #1 and really failed #3 (I never liked Columbus much anyway).

[quote="mathemonster"]I had a nice soln for 5 b):
$ r^{a_1} \plus{} r^{a_2}... \equal{} r^{b_1} \plus{} r^{b_2}... \implies {\lfloor {log_rLHS}\rfloor} \equal{} {\lfloor {log_rRHS}\rfloor}$. Then let $ a_k$ be the first $ a_i \neq b_i$. Canceling and factoring out the first term we show that $ a_k \equal{} b_k$ which contradicts. I'm too lazy to do this, its trivial tho.[/quote]
Wait, I don't see why taking the floor of the $ \log_r$ of both sides makes things simpler—could someone explain?

1. Had a slightly different way of solving, basically the same. Probably 5.
2. I think that without all of their weird wording my solution is the same. Probably 5 again.
3. I put 1492 and an explanation, but it wasn't really a proof... maybe 3 or at best 4
4. No work. I suck at Geometry.
5. My solution was to assume that there were two sequences then prove that if a_i>b_i for some i then the RHS must be bigger then the LHS, so probably 5

Most likely:

5/5/3/0/5: 18.

I'm happy (only 7th grade)
Off to a quick start on rd. 2, have 2 probs done already

1. Woohoo, 3 pages of UGLY casework.  Correct answer, however - almost certainly 5.

2. Was going to set bounds for the second case, but decided not to, so I had 16 cases.  However, it worked and I got the right answer - probably 5.

3. Well, I got the correct spots of the boundary points :D and I got the right answer, but I totally BS'ed the proof.  Probably 1 or 2.

4. My trig totally died.  Hopefully 1, most likely 0.

5. Yay, my solution was exactly the same as official one.  5.

So, something like 5/5/2/0/5, 17?

EDIT: Ok, I'm uploading my solutions for #1 and #2.

[quote="007math"]AIME, your answer for #2 is wrong...you put it in the opposite order, since the problem stated that the second number has each digit one more than the first.[/quote]

I am pretty sure they won't take off for this...

Aha, I found the attachment button. My solutions to 1, 2, 3 with 1 incorrect. Darn.

1: Fine up to the point where I am putting in the 4-8, then pretty bad.
2: solved
3,4,5: probably one point each at most...

Hoping for 2/5/1/1/1=10.

when do they usually post your score on the website?

[quote="DPatrick"]Scores for Round 1 should be released in early November.[/quote]

Here is mine; judging from the looks of all your papers, mine is terrible, but I still want to share my solutions.

[quote="MathWise"][quote="mathemonster"]I had a nice soln for 5 b):
$ r^{a_1} + r^{a_2}... = r^{b_1} + r^{b_2}... \implies {\lfloor {log_rLHS}\rfloor} = {\lfloor {log_rRHS}\rfloor}$. Then let $ a_k$ be the first $ a_i \neq b_i$. Canceling and factoring out the first term we show that $ a_k = b_k$ which contradicts. I'm too lazy to do this, its trivial tho.[/quote]
Wait, I don't see why taking the floor of the $ \log_r$ of both sides makes things simpler—could someone explain?[/quote]
Ok factoring and log rules inside the floor functions gives : $ a_k + log_r({1 + r^{a_{k + 1} - a_k} + r^{a_{k + 2} - a_k})... = b_k + log_r({1 + r^{b_{k + 1} - b_k} + r^{b_{k + 2} - b_k})...}}$ Then $ log_r({1 + r^{b_{k + 1} - b_k} + r^{b_{k + 2} - b_k})...}$ is also strictly decreasing and its easy to show that its logarithm in base r is less than 0 but greater than -1 no matter what the $ a_k's$ and $ b_k's$ are, so we have that the floors of the LHS and RHS are $ a_k$ and $ b_k$ respectively thus $ a_k = b_k$ contradiction

[quote="mathemonster"]Ok factoring and log rules inside the floor functions gives : $ a_k + log_r({1 + r^{a_{k + 1} - a_k} + r^{a_{k + 2} - a_k})... = b_k + log_r({1 + r^{b_{k + 1} - b_k} + r^{b_{k + 2} - b_k})...}}$ Then $ log_r({1 + r^{b_{k + 1} - b_k} + r^{b_{k + 2} - b_k})...}$ is also strictly decreasing and its easy to show that its logarithm in base r is less than 0 but greater than -1 no matter what the $ a_k's$ and $ b_k's$ are, so we have that the floors of the LHS and RHS are $ a_k$ and $ b_k$ respectively thus $ a_k = b_k$ contradiction[/quote]

Thanks for explaining; now it makes sense. :)  It does look like a creative way to solve the problem!

[quote="feuxfollets"]My proof for 5b is like 4 lines... um did I screw something up?
Attached is just p5.[/quote]

The sequence isn't nessacarily geometric. For example, $ r^2\plus{}r^{99}\plus{}r^{100}\plus{}r^{346953476984357290347}\plus{}\hdots$

[quote="math154"]Just because $ a_1 \equal{} 1$ is required doesn't mean $ 1,4,7,\ldots$ is the only working sequence. But you can easily extend the idea you used...[/quote]

... in fact, nowhere does it even say it's a geometric series... :maybe:

I messed up number one and didn't finish number three (I guessed it was about 1489 and put down 1488 as the answer :) ) because it was 4 am. I should not have procrastinated. Darn, this is the first time I have gotten lazy on USAMTS.

I like your addendum anderson :)

[quote="veezbo"][quote="AIME15"]You have a small typo at the end of #3 - it's 1492, not 1452.[/quote]

 :(  :(  :(  They won't take off for this right? Is it obvious that my answer is supposed to be 1492?[/quote]

No, they almost definitely wont

I probably gave an incorrect solution to #4 because my solution proved only 1 solution.. unless I made a bad assumption.

#1, #2 should be right.. I showed 2 solutions to #2, so for being "extra" I'll probably get 2 points off for irrelevant info
#3 I totally made up random stuff for a proof. a 1 or 2
#5 I used logic to prove it.. so I'll probably get a 2.

Edit: I just read solutions. I think mine are all worth like 1 point each
except the 1st and second one. and in the second one, I didnt state there were no 6-digit pairs of numbers

so 5/4/1/1/1 = 12 which isn't what I wanted...

i mailed mine but here are the outlines:
1. casework stuff
2. basically same as soln
3. 2 coordinate planes, one rotated...then do nt stuff
4. consider where reflection is...basically same as soln
5. yeah basically same as soln

hmm. my #1-3 are right (except #2 i was like ok here's a chart (cause I didn't want to show my work for 7 lineraly independent systems of two equations) which finishes our solution and answers the problem), #4 i forgot the existence proof. #5 i hope is right; i didn't look at the soln yet.

i shall perhaps post my solutions on my blog sometime later.


MIT
college
videos
linear algebra
matrix
calculus

Most of you probably know about this, but for those of you who don't...MIT is doing a project called open course ware where they're trying to put all of their courses online. A good number are up, and some even have video lectures of every class. Linear Algebra is the only math one that has videos, I think, and I learned some matrix stuff from it.

http://ocw.mit.edu/index.html

The Physics E&M course has videos as well, which are good if you know calculus and some physics.

Thanks for pointing us towards that material.  There's a lot of good stuff there.


Hi,

could anyone help with this question. I have my exam tomorrow.

George had a collection of 40cds.
The mean playing time for these cds was 49 minutes.

George gave 8 of his cds to a friend.
The mean playing time for the 32 cds left was 47 minutes.
Find the mean time for the cds George gave to his friend.

My attempt.

40x49=1960 minutes all cds.

32x47=1504 minutes for 32 cds left.

the difference between these two

1906-1504=456 minutes

456 divide by 8=57 minutes.

Thanks for your help.

Charikaar

Did you get the problem wrong? Your steps are all very logical. I think it is right.

You confused me a little at first when you said that 1906-1504 = 456. I know you meant that [b]1960[/b]-1504 = 456.

Hi

thanks for your prompt reply.

yes i meant 1960 not 1906.

I don't have the answer to check whether i am wrong or right but i get confused over this bit of question

"The mean playing time for these cds was 49 minutes"

is the 49 minutes for all 40 cds or just a single cd.

Thanks once again.

Sainzay

[quote="236factorial"]When the problem says that the mean of a set of numbers is a specific number, like 49, it doesn't mean that all the parts of the data is the number 49. But, you can assume that every number in the data is 49. Isn't the average of forty 49's still 49?

This is a popular subject on mathcounts problems :) .[/quote]
Especially in the earlier problems, as knowing the strategy makes this problem easy but not knowing it makes it tough... the usual for sprints...


inequalities
function
calculus
integration
search
geometry unsolved
geometry

Let $ G$ be the centroid of a triangle $ ABC$. Denote by $ A_{1}$, $ B_{1}$, $ C_{1}$ the midpoints of the sides $ BC$, $ CA$, $ AB$. Prove that:
$ PA\plus{}PB\plus{}PC\plus{}3\cdot PG\geq 2\cdot\left(PA_{1}\plus{}PB_{1}\plus{}PC_{1}\right)$,
where $ P$ is an arbitrary point in the plane of triangle $ ABC$.

Denote $ \overrightarrow{a}\equal{}\overrightarrow{PA}$, $ \overrightarrow{b}\equal{}\overrightarrow{PB}$ and $ \overrightarrow{c}\equal{}\overrightarrow{PC}$. Then, since $ A_{1}$ is the midpoint of the segment $ BC$, we have $ \overrightarrow{PA_{1}}\equal{}\frac{\overrightarrow{PB}\plus{}\overrightarrow{PC}}{2}$, so that $ 2\cdot\overrightarrow{PA_{1}}\equal{}\overrightarrow{PB}\plus{}\overrightarrow{PC}\equal{}\overrightarrow{b}\plus{}\overrightarrow{c}$. Similarly, $ 2\cdot\overrightarrow{PB_{1}}\equal{}\overrightarrow{c}\plus{}\overrightarrow{a}$ and $ 2\cdot\overrightarrow{PC_{1}}\equal{}\overrightarrow{a}\plus{}\overrightarrow{b}$. Finally, since $ G$ is the centroid of triangle $ ABC$, we have $ \overrightarrow{PG}\equal{}\frac{\overrightarrow{PA}\plus{}\overrightarrow{PB}\plus{}\overrightarrow{PC}}{3}$, so that $ 3\cdot\overrightarrow{PG}\equal{}\overrightarrow{PA}\plus{}\overrightarrow{PB}\plus{}\overrightarrow{PC}\equal{}\overrightarrow{a}\plus{}\overrightarrow{b}\plus{}\overrightarrow{c}$.

Now, the inequality in question,

$ PA\plus{}PB\plus{}PC\plus{}3\cdot PG\geq 2\cdot\left(PA_{1}\plus{}PB_{1}\plus{}PC_{1}\right)$,

rewrites as

$ PA\plus{}PB\plus{}PC\plus{}3\cdot PG\geq 2\cdot PA_{1}\plus{}2\cdot PB_{1}\plus{}2\cdot PC_{1}$, what is equivalent to
$ \left|\overrightarrow{PA}\right|\plus{}\left|\overrightarrow{PB}\right|\plus{}\left|\overrightarrow{PC}\right|\plus{}3\cdot\left|\overrightarrow{PG}\right|\geq 2\cdot\left|\overrightarrow{PA_{1}}\right|\plus{}2\cdot\left|\overrightarrow{PB_{1}}\right|\plus{}2\cdot\left|\overrightarrow{PC_{1}}\right|$, what is equivalent to
$ \left|\overrightarrow{PA}\right|\plus{}\left|\overrightarrow{PB}\right|\plus{}\left|\overrightarrow{PC}\right|\plus{}\left|3\cdot\overrightarrow{PG}\right|\geq\left|2\cdot\overrightarrow{PA_{1}}\right|\plus{}\left|2\cdot\overrightarrow{PB_{1}}\right|\plus{}\left|2\cdot\overrightarrow{PC_{1}}\right|$, what is equivalent to
$ \left|\overrightarrow{a}\right|\plus{}\left|\overrightarrow{b}\right|\plus{}\left|\overrightarrow{c}\right|\plus{}\left|\overrightarrow{a}\plus{}\overrightarrow{b}\plus{}\overrightarrow{c}\right|\geq\left|\overrightarrow{b}\plus{}\overrightarrow{c}\right|\plus{}\left|\overrightarrow{c}\plus{}\overrightarrow{a}\right|\plus{}\left|\overrightarrow{a}\plus{}\overrightarrow{b}\right|$.

But this is the so-called Hlawka inequality, proved in http://www.mathlinks.ro/Forum/viewtopic.php?t=16600 post #4.

This proof shows that the point $ P$ needs not lie in the plane of triangle $ ABC$, but can lie anywhere in space. But some questions remain:

- How can one come up with Hlawka's identity at http://www.mathlinks.ro/Forum/viewtopic.php?t=16600 ?
- We can easily prove that
$ PA^{2}\plus{}PB^{2}\plus{}PC^{2}\plus{}3\cdot PG^{2}\geq 2\cdot\left(PA_{1}^{2}\plus{}PB_{1}^{2}\plus{}PC_{1}^{2}\right)$.
For what other $ t$ do we have
$ PA^{t}\plus{}PB^{t}\plus{}PC^{t}\plus{}3\cdot PG^{t}\geq 2\cdot\left(PA_{1}^{t}\plus{}PB_{1}^{t}\plus{}PC_{1}^{t}\right)$ ?
Generally, for what functions $ f:\mathbb{R}^{\plus{}}\to\mathbb{R}$ do we have
$ f\left(PA\right)\plus{}f\left(PB\right)\plus{}f\left(PC\right)\plus{}3\cdot f\left(PG\right)\geq 2\left(f\left(PA_{1}\right)\plus{}f\left(PB_{1}\right)\plus{}f\left(PC_{1}\right)\right)$ ?

  Darij

Besides the solution with the identity and the one with the integrals (which I consider to be very nice), there's also the one with complex numbers. See http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1542221167&t=39208.
(This is also based on a identity, but it looks more natural.)


1) Find a formula for $ max(a,b)$ in terms of $ a$ and $ b$.
2) Find a formula for $ min(a,b)$ in terms of $ a$ and $ b$.
3) Find formulas for $ max(a_1,a_2,\dots ,a_n)$ and $ min(a_1,a_2,\dots ,a_n)$ like the above.

This is a fun, quick problem.  :wink:

For positive $ a_j$ we can use Power Mean:

$ \blacktriangleright \; \; \text{max} \{ a_j \} \equal{} \lim _{k \to \plus{} \infty} \sqrt [k]{\frac {\sum_{j \equal{} 1}^n a_j}{n}}$

$ \blacktriangleright \; \; \text{min} \{ a_j \} \equal{} \lim _{k \to \minus{} \infty} \sqrt [k]{\frac {\sum_{j \equal{} 1}^n a_j}{n}}$


[hide="Hint"]When stuck, the quadratic formula would serve to be [i]very[/i] helpful and efficient.[/hide]


Let $ a,b,c,x,y,z > 0$ satisfy $ a \plus{} x \equal{} b \plus{} y \equal{} c \plus{} z \equal{} 1$.Prove that:
$ (abc \plus{} xyz)(\frac {1}{ax} \plus{} \frac {1}{by} \plus{} \frac {1}{cz}) \geq 3$

[quote="leedt26"]Let $ a,b,c,x,y,z > 0$ satisfy $ a \plus{} y \equal{} b \plus{} z \equal{} c \plus{} x \equal{} 1$.Prove that:
$ (abc \plus{} xyz)(\frac {1}{ax} \plus{} \frac {1}{by} \plus{} \frac {1}{cz}) \geq 3$[/quote]
I've edited it


Nan has art class every sixth day of school. Assume that school is held Monday through Friday for 36 consecutive weeks. School starts on a Monday and Nan has art the first day of school. How many times during the school year will Nan have art on Monday?

there are 5 days in a week, and school is held for 36 weeks
therefore, there are $ 5\times36\equal{}180$ days of school

since Nan has art class every sixth day, he would have $ 180\div6\equal{}30$ days of art class

correct, but they are asking for when the art class would be on monday.

We have the 30 days of art class divided by the 5 days of the week = 6 art classes on monday

Even though the answer says 6, I think the answer is 7. Here is my reasoning:
Here are the days of the week during which the art class is held
Week   Day
[b]   1       Mon[/b]
[b]   2       Mon[/b]
   3       Tues
   4       Wed
   5      Thurs
   6       Fri
[b]   8       Mon[/b]
   9       Tues
  10      Wed
  11      Thurs
  12      Fri
[b]  14      Mon[/b]
  15      Tues
  16      Wed
  17      Thurs
  18      Fri
[b]  20      Mon[/b]
  21      Tues
  22      Wed
  23      Thurs
  24      Fri
[b]  26      Mon[/b]
  27      Tues
  28      Wed
  29      Thurs
  30      Fri
[b]  32      Mon[/b]
  33      Tues
  34      Wed
  35      Thurs
  36      Fri

So it would be $ \boxed{7} $

Here is some proof for 7:

The first day of school is Monday and then the sixth day of school is also Monday,

Therefore the above posts are correct [i]except[/i] they forgot to add the art class on the [b]first day of school[/b], therefore it is 6+1=7

I'm not sure what planet your date assignment is from; I guess your "proof" is a bit more clarifying but does not capture the meaning of the problem. If you argue on persnickety bases, you have to consider that the problem asked for a certain amount, not a maximum.

What do you mean? The problem is asking for [b][i]all[/i][/b] the art classes on Monday, so it definitely is the maximum, and that is 7


complex numbers
inequalities proposed
inequalities

[quote="Chang Woo-JIn"]$ \sum_1^{n} |z_j| \ge 1$ where $ z_i$ are complex numbers

then there exist a subcollection $ \{ z{}_j{}_1 , \cdots , z{}_j{}_k \} \subseteq \{ z_1 , \cdots, z_n \}$ s.t

$ | \sum_1 ^k z{}_j{}_m | \ge C$

find best constant $ C$[/quote]
Can you be more precise? Is $ k$ also fixed? Is it correct to restate it like that:

(1) For given $ k\le n$, find the maximal constant $ C = C(n,k)$ such that the following holds:
for all $ z_1, \dots ,z_n\in\mathbb C$ with $ \sum_{j = 1}^{n} |z_j| \ge 1$, there is a subset  $ \{ j_1 , \dots , j_k \} \subseteq \{ 1 , \dots, n \}$ s.t.
\[ \left| \sum_{m = 1} ^k z{}_{j{}_m} \right| \ge C\]
We may then always look for the maximum over all $ k$, i.e.

(2) For given $ n$, find the maximal constant $ C = C(n)$ such that the following holds:
for all $ z_1, \dots,z_n\in\mathbb C$ with $ \sum_{j = 1}^{n} |z_j| \ge 1$, there is a subset  $ S \subseteq \{ 1 , \dots, n \}$ s.t.
\[ \left| \sum_{m\in S} z{}_{j{}_m} \right| \ge C\]

I'm sorry about that

$ k$ is not fixed. Your second restatement is what I wanted to say.


In triangle $ \triangle ABC$, a circle through $ A$ touches the side $ BC$ at $ D$ and intersects the sides $ AB$ and $ AC$ again at $ E, F$ respectively. $ EF$ bisects $ \angle AFD$ and $ \angle ADC \equal{} 80^{\circ}$. Find $ \angle ABC$.

[edit: LaTeXed]

Given <AFE=<EFD=<EAD=<EDA=y...(angles subtending the same chord)
<AED=80o from Alternate Segment Theorem ..
therefore y=50 from triangle AED......therefore <ABD=30o.


function
superior algebra
superior algebra unsolved

Let $R$ be an i.d and let $(a)$ and $(b)$ be principle ideals can I always say that $(a)\cap (b)$ is finitely generated? And if the answer is no can someone show a counterexample. And what are the minimal conditions I have to add to the hypothesis to make it work?

I think I have a counter example:

Consider the ring of continuous real valued  functions over $\mathbb{R}$:
$A$ the ideal generated by the function: $x \to x$
$B$ the ideal generated by the function: $x \to xsin(\frac{1}{x})$

If $F \in A \cap B$, then $F(x)=xf(x)=xsin(\frac{1}{x})g(x)$ where $f$ and $g$ are $2$ continuous functions.
Hence $f(x)=sin(\frac{1}{x})g(x)$ and is continuous at the point $x=0$. So $g(0)=0$. Conversely, take a function $h(x)=xsin(\frac{1}{x})g(x)$ where $g(0)=0$. Then $h(x) \in A \cap B$.

Now, if $A \cap B$ were generated by only finitely function $f_1(x)=xsin(\frac{1}{x})g_1(x),..,f_n(x)=xsin(\frac{1}{x})g_n(x)$, then the ideal $\{f|f(0)=0\}$ would be generated by $(g_1,..,g_n)$.

But this ideal is not finitely generated. Suppose it is generated by $g_1,..,g_n$. Take for example:
 $G(x)=max(|x|,|g_1(x)|,..,|g_n(x)|)$. Then $x \to \sqrt{G(x)}$ is a continuous function that is null in $0$ but that does not belong to the ideal generated by $(g_1,..,g_n)$.


How many $ m$th root of 1 does $ F_{p^n}$ have, provided that $ p$ is some prime and $ n$ is some positive integer?

This follows from the fact that any finite multiplicative subgroup of a field is cyclic.


Let 1<a,b,c<2.Prove that
           SUM {b\sqrt(a)/4b\sqrt(c)-c\sqrt(a)} >=1

It is  mean this$\sum{b\sqrt{a}/(4b\sqrt{c}-c\sqrt{a})} \geq 1$ :?:

Ok,now looking at the solution
I think you know this ineq.
${b_1}^2/a_1+{b_2}^2/a_2+{b_3}^2/a_3 \geq (a_1+a_2+a_3)^2/(b_1+b_2+b_3)$ for
$a_i,b_i>0$.

We have that $1<a,b,c<2$so $4b\sqrt{c}-c\sqrt{a}>0$=>

$\sum{b\sqrt{a}/(4b\sqrt{c}-c\sqrt{a})}=\sum{ab/(4b\sqrt{ac}-ac)}\geq (\sqrt{ab}+\sqrt{ac}+\sqrt{bc})^2/(4\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})-(ab+bc+ac))\geq 1$

becous $ab+bc+ac\geq \sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})$<=>$(\sqrt{ab}-\sqrt{bc})^2+(\sqrt{ab}-\sqrt{ac})^2+(\sqrt{bc}-\sqrt{ac})^2\geq 0$  ;)


function
floor function
logarithms
algebra

Hi, I couldn't get this problem, AMC-12 from 1986 #25, and i couldn't find it under olympic resources either.
[img]http://img167.imageshack.us/img167/9202/sdfla5.jpg[/img]

[hide="Hint"]
There are several "intervals" to consider. On what interval is the value inside the floor function 1? 2? 3? Etc? Find that and you get the answer. [/hide]

[hide="Almost entire solution"]We can rewrite the summation as:
\[ \displaystyle\sum_{n\equal{}1}^{2^{10}}\lfloor \log_2 n \rfloor\]
When $ 1\le n \le 1$, $ \lfloor \log_2 n \rfloor$ equals 0.
When $ 2\le n \le 3$, $ \lfloor \log_2 n \rfloor$ equals 1.
$ \cdots$
When $ 512\le n \le 1023$, $ \lfloor \log_2 n \rfloor$ equals 9.
When $ 1024\le n \le 1024$, $ \lfloor \log_2 n \rfloor$ equals 10.

So our sum is:
\[ 1 \cdot 0 \plus{} 2 \cdot 1 \plus{} 4 \cdot 2 \plus{} 8 \cdot 3 \plus{} \cdots \plus{} 512 \cdot 9 \plus{} 1 \cdot 10 \\
\Rightarrow \displaystyle\sum_{n\equal{}0}^{9}2^n\cdot n
\]
Evaluate the final summation and you're done.[/hide]

thanks, both of you- but how do you write those symbols?  i had to go use flash 8 and upload it on imageshack...
Also, do you know how long should be spent on a problem for the AMC-12 test?

Click "quote" on my post and you'll see the code. You can also click the symbols I typed to see the code.

Anyways, it's called Latex.

[url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]


A box contains 2 pennies, 4 nickels, and 6 dimes.  Six coins are drawn without replacement, with each coin having an equal probability of being drawn.  What is the probability that the value of the coins is at least 50 cents?

Note: infinity4ever got this right, but i keep getting 31/132

My answer is... kinda off? But I think that this approach should work.
I dunno, maybe the answers right :).

[hide="Answer"]$ \frac {127} {924}$[/hide]

[hide="Solution"]We proceed by finding the possible combinations.
This is extremely simple as there must be at least $ 4$ dimes for the conditions to be satisfied.

So: $ 4$ dimes and $ 2$ nickels.
$ 5$ dimes and $ 1$ nickel OR penny.
$ 6$ dimes.

The probability that we get the first case is: $ \frac {\binom {6} {2}*4*3*6*5*4*3} {12*11*10*9*8*7}$
with the $ 4*3$ coming from nickels, $ 6*5*4*3$ from dimes, and $ \binom {6} {2}$ coming from the possible ways to arrange these $ 6$ coins.
Then simply divide by $ 12$ permute $ 6$, because we know those will always be in the denominator.

Simplifying the expression above, we get that the probability is $ \frac {15} {154}$

The second case's probability is: $ \frac {\binom {6} {1}*6!*(4\plus{}2)} {12*11*\hdots}$
Simplifying this, we obtain the probability as $ \frac {6} {154}$

The third case is the simplest, being $ \frac {6!} {12*11*\hdots}\equal{}\frac {1} {924}$

Adding all of these up , we get $ \frac {21*6\plus{}1} {924}\equal{} \frac {127} {924}$.[/hide]

I get $ \frac {115}{924}$, i listed, including all 924 possiblilties, then checked list..only takes 30 minutes.

I just want to ask why everyone rates others' posts as spam. Even though I clearly explained my reasoning, you guys go on and rate it as spam? You rate Yuqing's problem as junk also? Seriously, wth? If you can't do it, don't get mad at people who actually can.

To firecricket: Listing them out will usually lead to some error.  Can you just check my logic and see if you agree? Maybe I added wrong or something, but I don't think 'listing' can really be classified as a solution that can be used, at least for this problem, during a contest.

[hide="hmmm....."]6 dimes-1
5 dimes and one penny-$ \binom{6} {5}*2\equal{}12$
5 dimes and one nickel-see above-12
4 dimes and 2 nickels-$ \binom{6} {4}*\binom{4} {2}\equal{}90$ i think
add them up...
115[/hide]

[b]firecricket[/b]-[quote]30 minutes.[/quote]
thats a lot of time, especially at the competition....

[b]mathcrazed[/b]-i dont unstand your solution. could you see what i did wrong.

[quote="abacadaea"][hide=hmmm.....]
5 dimes and one penny-$ \binom{6} {5}*2 \equal{} 12$
5 dimes and one nickel-see above-12[/quote]

this is incorrect because there are more nickels than pennies.

As stated by unimpossible, when solving that particular case, you must take into account that there are different numbers of nickels than dimes.

Unimpossible, are there any correct solutions so far? Or any correct answers?
Also, I believe that casework is the best way to approach this problem. However, if anyone has a better solution, feel free to post it up :). Just because we already have solutions doesn't mean that you can't provide an alternative method. The more the merrier :D.

[quote="abacadaea"]i didnt need much casework, and i think my way is faster...i dont know...[/quote]

getting a wrong answer fast is definitely not as respectable as getting the right answer slowly.

Also, in mathcounts, i've found that when i "find" a quick way, i often spend too much time checking it anyway.

alright then...

[hide="this was my old solution"]
6 dimes-1
5 dimes and one penny-$ \binom{6} {5}*2 \equal{} 12$
5 dimes and one nickel-see above-12
4 dimes and 2 nickels-$ \binom{6} {4}*\binom{4} {2} \equal{} 90$ i think
add them up...
115
[/hide]

[hide="this was my new solution"]
6 dimes-1
5 dimes and one penny-$ \binom{6} {5}*\binom{2} {1} \equal{} 12$
5 dimes and one nickel-$ \binom{6} {5}*\binom{4} {1}\equal{}24$
4 dimes and 2 nickels-$ \binom{6} {4}*\binom{4} {2} \equal{} 90$ i think
add them up...
127
[/hide]

lets just say i got a little HASTY with it...shall we....

Abc... Your way was simply my way, except I elaborated on my procedure for those who don't think as quickly or deep.  In reality, ours was the same exact thing. You say that you didn't need much casework, but you still needed the same as mine.  And yours was not faster, it was exactly the same, except without finding probabilities. What you found was the number of cases, and yes, that can be slightly faster, but my way is the same, except with fractions.  If you don't want to work with those, that's fine. So there wasn't much of a difference.
Unimpossible stated that getting the wrong answer fast is not as good as getting the right answer slowly.
However, in some cases, such as Mathcounts Sprint, or maybe AMC 10/12, you don't want to get the right answer slowly. So, instead, just find the problems that you like, and do those.  Also, don't work on extra problems you know you can't do. CHECK the rest of your work. I usually find 4-5 mistakes when doing this, at least during practice.

[quote]Also, don't work on extra problems you know you can't do. CHECK the rest of your work. I usually find 4-5 mistakes when doing this, at least during practice.
[/quote]
hmm...ill use that. usually on mathcounts sprint rounds, i do all of them and usually dont skip any. usually i finish on time. state, i usually barley finish. so i have no time to look for careless mistakes i always make.

and during your practice, do you time yourself?

Yes, if I ever practice.  Timing yourself is essential.  Also, try to finish what you want to do within 30-35 min. So if you work fast, about one problem a minute if you want to finish the whole test. Other ways, aim to get about 25 problems.  During Nationals, Kevin Chen only answered 25, and guessed 5 others.  He checked the 25 he did a ton, and he ended up getting a score of 25 or 26, forget which one. [/quote]

how do you know what kevin chen did?

usually at states, i know how to do each problem, but i cant finish all of them in 35 minutes. at chapter, i can finish in 30.

wow, i just scored 27 on a state sprint(2003)! i know its not much of an acheivement for you guys......i suck at math..

Um...maybe because he was there.

I can usually solve around 27-28 sprints, but make stupid mistakes, usually 7-8 targets, but again stupid mistakes. 2003 was my best practice, I went 28/14.

He might have asked, or someone could have told him.
Oh, I mistyped. I had 14 on target, missed number 2 :o

[quote="abacadaea"][quote="xpmath"]Um...maybe because he was there.
[/quote]

still, how would he know that he got that. event though he won, how would he know that kevin guessed 5 out of 30 and did the other 25. mathcrazed couldnt have been sitting right behind him watching.

and i just got a 16 on the target. 41![/quote]

ummm 27+16=43.

not 41 :)

well i know i got a 41. i think i fot a 14 or 25 on the spritn idk. ill have to look at it again, but i have to sleep now so i cant


Two cars are initially $ 36m$ apart travelling in the same direction along a stright horizontal road. the car in front is iinitially travelling at $ 10 ms^{\minus{}1}$, but decelerating at $ 2ms^{\minus{}2}$. the other car travels at a constant $ 15ms^{\minus{}1}$.
[color=red]
[size=150](1)[/size][/color]
Model the cars as particles. by finding the distance traveled by each car after $ t$ seconds, show that the distance between the two cars is $ 36\minus{}5t\minus{}t^{2}$ metres.find when they would collide if neither car takes avoiding action.

[color=red][size=150](2)[/size][/color]
would it be necessary to revise your answers to part (1) if the cars were not modelled as particles? give reasons to support your answer.

This is more appropriate for the [url=http://www.artofproblemsolving.com/Forum/index.php?f=405]introductory physics[/url] forum.

Solution

(1)Lets call the car with constant speed $ A$ and the other car $ B$. Let all displacements be with respect to the inital position of $ A$.

$ s_{A}\equal{} vt \equal{} 15t$

$ s_{B}\equal{} s_{0}\plus{}ut\plus{}\frac{1}{2}at^{2}\equal{} 36\plus{}10t\minus{}t^{2}$

The distance between, or relative displacement of the cars is:

$ d \equal{} s_{B}\minus{}s_{A}\equal{} 36\minus{}5t\minus{}t^{2}$

Cars collide when $ t \equal{} 0$

$ \Rightarrow 36\minus{}5t\minus{}t^{2}\equal{} 0$

$ \Rightarrow t^{2}\plus{}5t\minus{}36 \equal{} 0$

$ \Rightarrow (t\minus{}4)(t\plus{}9) \equal{} 0$

$ \Rightarrow t \equal{} 4\,s$

(2)We cannot choose where the particles are. Their position depends on how the initial $ 36\,m$ is measured. If measured from the front of $ A$, to the rear of $ B$ then the "particles" will be in these positions. Thus the cars will collide when these particles collide, so no revisions have to be made. However, if the $ 36\,m$ was measured from two other points (e.g. the car's centres of mass) then the front of car $ A$ and the rear of car $ B$ will collide before these two particles meet, and the answer will be different. At collision, the distance between the particles is equal to the sum of the distance of particle $ A$ from the front of car $ A$ and the distance of partcle $ B$ from the rear of car $ B$. It is this value which will have to be equated to $ 36\minus{}5t\minus{}t^{2}$ to find the time of collision.


geometry
geometric transformation
dilation

a) Give the formula of 4 acids that could have been in the solution if the pH changed one unit after 10 fold dilution

b) Could an acid be sulfuric acid? (pKa2 = 1.99)

c) Could an acid be acetic acid? (pKa = 4.76)

d) Could an acid be EDTA? (pKa1 = 1.7, pKa2=2.6, pKa3=6.3, pKa4=10.6)
Use reasonable approximations.

Can you please define a ten-fold dilution?

You can always check the solution on the IChO 2008 website. By the way diluting by ten-fold just means diluting to ten times its original concentration (so it becomes one tenth of the original concentration).

^.... Im not asking how to solve it.... Im  posing the question for people who are looking for interesting problems....

I HATE HATE HATE HATE HATE this problem.

aka I misread it......................................when it was most crucial


function
real analysis
real analysis unsolved

A sentence of the following type if often heard in Hungarian weather reports: "Last night's minimum temperatures took all values between $ \minus{}3$ degrees and $ \plus{}5$ degrees." Show that it would suffice to say, "Both $ \minus{}3$ degrees and $ \plus{}5$ degrees occurred among last night's minimum temperatures." (Assume that temperature as a two-variable function of place and time is continuous.)

[i]A.Csaszar[/i]

i dont its a normal question :wink:  :wink:

Almost all of the questions of this competition are not normal and this makes it unique.  :)


algebra
polynomial
function
algebra proposed

If f(x) is polynomial so that its coefficients are integers then deg f>1,then  in the solution set of the f:Z-Z function at most how many consecutive integers can be found?

the answer is $ n$, where $ n$ is the degree of $ f$.

by considering $ f(x \minus{} h)$ for an appropriate $ h$, we may assume that the consecutive integer values of $ f$ are $ f(a_1) \equal{} 1, f(a_2) \equal{} 2, ..., f(a_k) \equal{} k$. In general, if $ f$ has integer coefficients and $ a$ and $ b$ are integers, we have $ a \minus{} b|f(a) \minus{} f(b)$. so $ a_{i \plus{} 1} \minus{} a_i|1$ for $ i \equal{} 1, ..., k \minus{} 1$. thus $ a_i, a_{i \plus{} 1}$ are consecutive integers for all $ i \equal{} 1, ..., k \minus{} 1$. it follows that $ a_1, ..., a_k$ form a set of consecutive integers. thus there is an integer $ a$ such that $ a_i \equal{} a \plus{} i$ for all $ i$ or $ a_i \equal{} a \minus{} i$ for all $ i$. since both cases follow the same reasoning we'll assume $ a_i \equal{} a \plus{} i$. Then $ f(a \plus{} i) \equal{} i$ for $ i \equal{} 1, ..., k$, which implies that the polynomial $ f(a \plus{} x) \minus{} x$ has at least $ k$ different roots. since $ f(a \plus{} x) \minus{} x$ is not constant (because deg($ f$)>1), this means that $ k\leq n$. to see that $ n$ is attainable, take the polynomial $ f(x) \equal{} x \plus{} (x \minus{} 1)(x \minus{} 2)\cdots (x \minus{} n)$.


conics ellipse rotation geometry rectangle

On the blackboard the teacher drows an ellips and afterword wipes off its focuses.
A question for the classe;can you construct them? :P

[quote="Altheman"][hide="ideas"]a lame solution, take an arbitrary point as the origin, and draw arbitrary axes, then take 5 points, and solve the system for the ellipse, this involves all operations so we can construct the result

the other item that i am toying with is using the locus definition of an ellipse, if $F$ is a focus, and $L$ is the corresponding directrix, then $\frac{OP}{d(O,L)}=e$ where $e$ is the eccentricity of the ellipse[/hide][/quote]
In fact,for your constructions you only have standart instruments(a ruler and a compasses);can you pereform all the operations you said with only these instruments? :P

Somebody construct the focuses!!!!!!!!! :ninja:

[quote="cincodemayo5590"][quote="perfectnumber628"][hide]assuming you have the endpoints of the major and minor axes, just bisect them to find the origin.  and the sum of the distances to each focus from a point on the ellipse is constant, and also equal to the length of the major axis.  at an endpoint of the minor axis, the distance to one focus will be the same as the ditance to the other, because of symmetry.  so just use the compass to get a length equal to half the major axis, and draw a circle centered at an endpoint of a minor axis.  the foci are the points where this circle hits the major axis.[/hide][/quote]

But what if you don't have those endpoints?[/quote]
Yeah,what then? :P

[quote="Tiks"][quote="Altheman"]if you really want a solution, then draw a coorinate axis, and take 5 points on the conic, then use algebra to solve the linear system, and it is trivial to find the focus given the equation...obviously this is messy, but the lengths are trivial to construct[/quote]
More detals,please.And tell me what is "coorinate axis"?[/quote]

draw a line, call it the x-axis

draw another line, perpendicular to the x-axis, call it the y-axis...

take a point on the ellipse, and drop perpendiculars to the x and y axis, call these lengths x_1 and y_1 respectively

remember, an ellipse is in teh form ax^2+bxy+cy^2+dx+ey+f=0

in terms of 5 points, we can easily solve for the coefficients,

(once you have the equation, rotate it back, find the focus, rotate back...and these lengths are constructable because they are just product, sum and quotients, square roots, etc...)

so i agree this is a totally lame solution...but it is doable...perhaps you had a nicer solution in mind

1.Construct the centre of the ellipse:
THEOREM The midpoints of parallel chords of an ellipse lie on a line which passes through the centre of the ellipse

2.Construct the major and minor axes:
THEOREM If any circle with the same centre as the ellipse cuts the ellipse again in four points,then these points are vertices of a rectangle.
The lines joining midpoints of opposite sides of the rectangle are the major and minor axes.

We can now find the endpoints of the major and minor axes and construct the foci according to the post by perfectnumber628.


limit
calculus
inequalities
ratio
integration
geometric series
absolute value

What is an infinte series? Does it only apply to addition and subtraction?

An infinite series is the summation of an infinite number of terms (which can be negative, in response to your question about addition and subtraction, which is really addition), usually written \[\sum_{k=0}^\infty{a_k}=a_0+a_1+a_2+a_3+a_4+a_5+\ldots.\]

If ${\lim_{n\to\infty}\sum_{k=0}^n{a_k}}$ exists, then this limit is the value of the infinite series, and it $\text{converges}$. Otherwise it $\text{diverges}$ and has no value.

I just remembered that there isn't supposed to be any calculus posted in this forum, so here's another way of explaining convergence and divergence.

The series converges to (is equal to) $S$ if and only if for any $\epsilon>0$ there is a positive integer $N$ such that $n\ge{N}\Longrightarrow|\sum_{k=0}^n{a_k}-S|<\epsilon$. The series diverges (and has no value) if and only if for any positive integer $M$ there is a positive integer $N$ such that $n\ge{N}\Longrightarrow|\sum_{k=0}^n{a_k}|>M$.

Most of the exposure I've had to infinite series is from Calculus BC, whose treatment of them leaves much to be desired. Please feel free to revise or improve my explanation in any way..

[quote="gauss202"]As a teacher I am curious: what leaves you unsatisfied with the treatment of infinite series in BC Calculus?[/quote]Basically, nothing was proved. If you were to ask any of the other five students in the class in my high school to supply the definitions above (which I hope are correct) you would most likely get blank stares. I'm one of the few students in my school, maybe the only, to read or study any math outside of the high school cirriculum. Our teacher generally just gives us formulas that we use to do little computational problems, and that's about it. Next year as a senior I'll most likely be taking an honors introductory analysis 3-quarter series at OSU (I live in Columbus), and I'm fairly certain that the treatment of infinite series in this course will be [i]much[/i] more rigorous than in Calculus BC. For those not interested in pure math or planning to be a math major however, I would say Calculus BC is perfectly sufficient.

[quote="AntonioMainenti"]The series diverges (and has no value) if and only if for any positive integer M there is a positive integer N such that $n\ge{N}\Longrightarrow|\sum_{k=0}^n{a_k}|>M$[/quote]
No, you can't say that. The series $1-1+1-1+1-1+\dots$ diverges despite the fact that the partial sums are bounded.
A true negation of the definition: first let $s_n=\sum_{k=0}^na_k.$
The series diverges if and only if for any $L$ and any $\epsilon>0,$ $|s_n-L|>\epsilon$ for infinitely many $n.$
Since this definition is hard to use directly, we find some sufficient conditions for divergence: a series diverges if the partial sums are unbounded, and it diverges if the terms do not tend to zero.

Also, some additional definitions:
The series $\sum a_k$ [b]diverges to infinity[/b] if for any $M$ there exists a positive integer $N$ such that $n\ge{N}\Longrightarrow\sum_{k=0}^n{a_k}>M.$
Divergence to $-\infty$ is defined similarly, only with the inequality reversed.
All other forms of divergence for series are merely divergence.

[quote="AntonioMainenti"][quote="jmerry"][quote="AntonioMainenti"]The series diverges (and has no value) if and only if for any positive integer M there is a positive integer N such that $n\ge{N}\Longrightarrow|\sum_{k=0}^n{a_k}|>M$[/quote]
No, you can't say that. The series $1-1+1-1+1-1+\dots$ diverges despite the fact that the partial sums are bounded.[/quote]Oops :blush:. My definition applies to series that tend to $\pm\infty$...[/quote]

Another little thing with your definition is that it's a bit unnecessary to require $m\in\mathbb{N}$.  We have $n\in\mathbb{N}$ because $\{a_n\}$ is a sequence and we require sequences to be indexed by the naturals, but we may as well say that $m\in\mathbb{R}$ (by the Archimedian property, these are equivalent anyways, but still).

All of the talk on this topic is calculus-level, or maybe advanced calculus, and has probably gone well beyond what Tokyo meant when he asked the question.

Some things to remember.

A series is a sum. (So the answer to Tokyo's question is "yes.")

Don't confuse the series with its sequence of terms. For instance consider this:

$\sum_{k=0}^{\infty}\frac{k}{\sqrt{k^2+1}}.$ We can see that $\lim_{k\to\infty}\frac{k}{\sqrt{k^2+1}}=1.$

Does that mean that the series converges? No, it doesn't; this is the sequence of terms converging to 1, but the sequence of terms is not the series. After a while in this series we might as well be adding $\cdots+1+1+1+1+\cdots$ and the sum is infinite: the series diverges.

If the series converges, then the terms tend to zero. Contrapositive (also true): if the terms don't tend to zero, the series diverges. The converse of this statement is false. A famous example is the harmonic series

$\sum_{n=1}^{\infty}\frac1n$ which diverges even though the terms do tend to zero.

--

Finally: if you were to know only one thing about series what would it be? What would I say the single most important piece of information?

[b]The geometric series.[/b]

A series is geometric if there's a "common ratio": in order to get the next term from the one you have, you always multiply by the same quantity. A geometric series converges if and only if the absolute value of its ratio is less than 1. The sum of a convergent geometric series is given by

Sum $=\,\frac{\text{first term}}{1-\text{ratio}}.$

Example: $3+\frac32+\frac34+\frac38+\cdots.$

The common ratio is $\frac12$. That is, we have to multiply each term by $\frac12$ to get the next term. Since that's less than 1, the series converges, and the sum is:

$\frac3{1-\frac12}=6.$

hehe.. i just finished 29 homework problems on determining convergence/divergence
integral tests, comparison tests etc

The sum $1-1+1-1+1-1+\cdots$ diverges.  But what is the best mathematical reason for this?  Can you just say that it's a geometric series with $r=-1$ and geometric series diverge when $|r|\ge 1$?

From Kent's test, we see that the limit of the sequence is not 0, so the series diverges. In fact, the sequence has no limit because it goes back and forth between two values.


linear algebra
matrix
algebra
polynomial
function

Can u tell me how to prove not calculating the determinant in the following:
                          1          a        a^2
                          1          b         b^2               = (a-b)(b-c)(c-a) ?
                          1           c         c^2

This is a Vandermonde determinant. The simplest appoach is to use row elimination. The first steps: subtract the first row from each of the second and third rows:

$\begin{bmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{bmatrix}\sim \begin{bmatrix}1&a&a^2 \\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{bmatrix}$

Now subtract $\frac{c-a}{b-a}$ times the second row from the third row. Here's the key piece of arithmetic:

${c^2-a^2-\frac{c-a}{b-a}\cdot(b^2-a^2}=(c-a)\left[c+a-\frac{b^2-a^2}{b-a}\right]$

$=(c-a)[(c+a)-(b+a)]=(c-a)(c-b)]$

Thus the original matrix is row equivalent to $\begin{bmatrix}1&a&a^2\\0&b-a&b^2-a^2 \\ 0&0&(c-a)(c-b)\end{bmatrix}.$

From this, we can read off the determinant as $(b-a)(c-a)(c-b).$

We did implicitly assume in the computation that $b\ne a.$ But the determinant is a polynomial function of its entries, and thus our formula may be extended by continuity to include those cases as well.

Doing column elimination (or, equivalently, row elimination on the transpose) would also work well.

Test of technique (well-known, of course). What is

$\det\begin{bmatrix}1&1&1&1\\a&b&c&d\\a^2&b^2&c^2&d^2\\a^3&b^3&c^3&d^3 \end{bmatrix}$ ?

The other classic argument: Treat $a,b,c$ as variables. The determinant is a polynomial of total degree 3 (in the general case, $\frac{n(n-1)}2$). What can we say about the zeros/factors of this polynomial?