Given a 50 card deck with cards numbered from 1 through 10 in each of 5 suits, how many 5 card hands are there that include exactly one pair of two cards that have the same numeric value?

The answer is $ \dbinom52\cdot10\cdot\dbinom93\cdot5^3\equal{}\boxed{1050000}$. This is because we first choose two of the five suits and one of the ten numbers to be the pair, and then, since the rest must all be of different numbers, we choose three from the remaining nine choices, and then for each choice, there are five suits to choose from.

[hide="Alternative Solution"]We can choose the 2 cards that have the same number in $\frac{50\cdot 4}{2}$, since the first card has 50 choices and the second card has 5-1=4 choices, and order doesn't matter.

We can choose the remaining 3 cards in $\frac{45\cdot 40\cdot 35}{3!}$ ways, since there are 50-5=45 choices for the fourth card, 50-10=40 choices for the fifth card, 50-15=35 choices for the sixth card, and order doesn't matter.

Hence the answer is $\frac{50\cdot 4}{2}\cdot \frac{45\cdot 40\cdot 35}{3!}=\boxed{1050000}$.[/hide]

[quote=TheMaskedMagician]Can someone tell me what I did wrong?

$\dfrac{50\cdot45\cdot 40\cdot 35\cdot 4}{5!}=105000$. Did I need to multiply by 10?[/quote]

This took me half an hour to figure this out.  :)  Hope this helps.

There are two mistakes in this equation.

First, the $4$ in the numerator should be $16$ since there are $4$ numbers available (the one chosen in the first number $50$, second one chosen by $45$, the third chosen by $40$, and the forth chosen by $35$ )and $4$ suits for each number.

Second, this equation originally has an permutation hidden in it. Let's do it case by case.
The first $4$ numbers has a permutation of $4!$.
The $2$ cards that has the same number has a permutation of $2$.
The denominator is actually $4!*2$

LOL, I calculated the probability instead, and got 3750/7567 and was bashing my head for so long trying to figure out the answer  :blush:  :wacko:


I'll assume that Circe A and Circle B are concentric...

[hide]Since the radius of Circe B is $ 8$, the area is $ 64pi$.
If we call the radius of Circle A $ x$, we can set up a proportion.

$ \dfrac{64 \minus{} x^2}{x^2}$ = $ \dfrac{3}{1}$

Solve for $ x$, and $ 2x$ = $ 8$[/hide]

[quote="sri340"]Circle A is in the interior of circle B. The diameter of circle B is 16 cm. What is the diameter of circle A for which the ratio of the shaded area to the area of circle A is 3:1?[/quote]

 [Small Circle]/[Big Circle]=1/4 --> [Radius of SC]/[Radius of BC] = sqrt(1/4)= 1/2

Diameter of Circle A= 1/2 * 16 = [b]8[/b]

I think someone posted a solution the same time I did :|


I was wondering if your ability to solve math problems correlated with artstic ability? For example, if we were to train people like Picasso to do math contests, is it likely he succeed?

I undersand art and math are similar aesthetic pursuits, and one can often help bring out the abstraction of the other; but whether it trains the mind in the art of problem solving I do not know one could say for sure without doing some extensive testing.

Artistic ability would probably help with some kinds of math problems, but not all.

IMO, artistic ability can help solve some math problems - but it is generally not a major factor in whether you are good at math or not. It probably does help in geometry, as an artist can visualize the problem as an art project where he must draw it to dimensions n such.. but id unno/

Artistic ability would promote creativity, which would help to think of math problems in different ways, which could help to solve the problem. I think this is related to how the left and right hemispheres of the brain work together...

Well there's deductive reasoning, whcih is good for contest problems and problems which simply require math that exists to solve.
And there's inductive reasoning, which is useful in using new approaches, creating new math, etc.

The inductive comes from creativity, and is also related to fuzzy logic.

Could this be why my geometry skills are nearly as bad as my artistic ones?  :huh:

Oh, I can't draw or do anything with art either.  :D But I still think that math and art are somewhat related...

I can draw pretty well with a ruler and a compass :)

I don't think art helps algebra at all... I'm pretty good at drawing and geometry is my favorite kind of math.

Look at the name of this website: The [b]Art[/b] of Problem Solving. :)
I think it depends what kind of art you're talking about. Many musicians are also adept at mathematics, but I doubt that visual art has very much to do with mathematical problem solving ability.

[quote="trisana"]I think that math (at least algebra) and art couldn't be further apart. I don't mean to brag, but I got really good grades in algebra I, and I have issues drawing a stick figure, while one of my best friends, who is like, TEH ARTIST, not so much.[/quote]

Well, I draw and won several competition, and I still have good grade in Alg II/Trig, and managed to get on math team... :P XD

I think math and artistic ability would kinda link... if u know how to use it, i mean... well.. i haven't found it out yet though.. XD

I'm pretty good at art and I'm allright in mathematics so I guess it's connected a little bit. I think artistic ability is required for mathematics, not necessarily as in drawing or anything, but just creativity in general, especially when working with proofs.

There are some types of art where it makes no difference, and some that it does, like depth, perceptive, spacial vs. drawing a self-portrait.

Well...

If you are actually able to solve a hard math problem,
where a lot of steps in your mind are involved, then in my opinion this is indeed very creative.
Because you have to be able "think" about a problem in a lot of different ways, to figure it out.

When you draw a picture you use creativity too, but not in as well defined limits compared to math.
In art one uses more intuition and feelings, and there are truly infinite solutions, and usually one doesn't exactly know what will come out at the end (ok... sometimes that's the same with math:))

So i think:

Art and Math are both related but what you use your
mind for in Math, you use your heart for in Art.

JAN :)

[quote="Melissa"]If you include musical abilities in artistic abilities (music is a form of art after all) then yes, definitely there is a correlation.[/quote]Most of the people I know that are really good at math also play a musical instrument. :)

I don't know about you guys, but art used to ALWAYS help me solve math problems (geometry).

On almost any problem I got, I'd draw a nice big diagram before tackling the problem, and that would help me tremendously. :yup:

Now, I only kinda use it on harder problems, such as Aime, Usamts, and etc, but thte ability to draw is very important.

I draw nice big diagrams too sometimes lol  :lol:

Do you people think that Artistic Ability really helps you think creatively in problem solving?

I'm not sure because i'm pretty good at math but i suck at art. But I would think that being an artist would help in math but not the other way around.

I think art and problem solving skills are somewhat connected, although art tends to be much more open-ended than math, and so you think about them in different ways. An important part of both fields is the capacity for imagination, especially visualization. I find that many I can put many math problems in a form where I can visualize them. This is especially useful when I am experimenting with a problem, finding different representations and ways of thinking about it. I'm not very good at art (although I have been practicing recently), but I am quite good at visualizing things, and this helps me a lot in art and math.

if it includes poisoning people(*cough cough* i meant scheming to poison them) yes I always come out with better plots after mathing alot

I wouldn't say that there is necessary connection between mathematics and art. Mathematicians and artists need fair amount of creativity and imagination, but there is huge difference in the way they are using it.
One isn't mathematician because he has good marks, neither one can call himself an artist because he draws nice sketches for geometry class...
Being mathematician lies more in manner of thinking and discovering the truth (in science and in everyday life), being an artist means capturing a moment, feeling and understanding life with heart rather than mind.

Good job reviving a year-old topic! Anyhow, I have very poor artistic skills. My finest work of art is probably a stick man with a wooden leg. :blush:

I am the best problem solver in my school because i am creative when i solve the problems.
however, my most interesting pictures include a stick man with a clock for a face and also a picture of a flock of McDonald's logos flying through a building.  i guess some people might call that artistic. :maybe:

Well...it's said that the left side of the brain is associated with math and logic while the right has to do with art,creativity and emotion. That way, there seems to be no connection between the two.
Yet, if u can't draw a circle or a triangle in art class, then how will u construct it in geometry?
Somehow most of the math-oriented people in my school are awful at art but there are exceptions!
However an art teacher of ours says that art and music slows ur brainwaves and helps u think better and gives u stability, which will come in useful in other walks of life...opening up new ideas and logic...weird...
If there is no connection then how come Da Vinci was good at both??

It just also happens to be said that there is a bridge between the two hemispheres of the brain called the corpus callosum. It is vital for the exchange of information within the brain; therefore, we conclude that artistic and mathematical abilities are indeed connected. Da Vinci was good at both because there IS a connection. There are also math-oriented people here who are extremely talented in art, I bet. In fact, I'm 100% sure of it. There is no principle stating that being good at math makes you suck at art.

Ok I'm convinced now...there IS a connection... u do require geometry and math in art...for symmetry in sculptures and stuff...and 2 make sure that ur painting doesn't look unreal..u need math...I guess my vote is a YES then!

[quote] There is no principle stating that being good at math makes you suck at art.
[/quote]

Naturally, but there is no principle stating that being good at math makes you good at art.

Do you really think that artists and mathematicians observe the world in the same way? What does "space" or "geometry" mean for mahtematician and what does it mean for an artist?  :?:

[quote="Nienna"]Do you really think that artists and mathematicians observe the world in the same way? What does "space" or "geometry" mean for mahtematician and what does it mean for an artist?[/quote]For the artist, it's an aesthetic concept.  For the mathematician, it's something that must be rigorously defined and you have several dozen versions running around of.

i think that contest math is different than regular classroom math in its respect to art.  contest math is out-of-the-box, so it involves thinking creativley sometimes, while textbook math is often churning out problems.  yeah..


limit
trigonometry
calculus
calculus computations

Calculate the limit
$ lim \lim_{x \to0^{\plus{}} } ( \frac{1}{x}) ^{tgx}$

Do you mean

$ \displaystyle\lim_{x\rightarrow0} (\frac{1}{x})^x$ ?

hello,applying the L'Hospitals rule we get
${ {\lim_{x \to 0+}\left(\frac{1}{x}\right)^{\tan(x)}=e^{\lim_{x \to 0+}-\frac{\ln(x)}{\cot(x)}}=e^{\lim_{x\to 0+}\frac{-{\frac{1}{x}}}{\frac{1}{\sin^2(x)}}}=e^{\lim_{x \to 0+}-\frac{\sin^2(x)}{x}}}=e^0=1}$
so the searched limit is $ 1$.
Sonnhard.

what are the modst upto date and comprehensive and philosophical books on limits
and supreme algebra


to set up a isomorphous of group $\mathbb{Z}_{12 \simeq }\mathbb{Z}_{3}*\mathbb{Z}_{4}$

 :(  :)

If you mean direct product of $\mathbb{Z}_{3}$ and $\mathbb{Z}_{4}$ ($\mathbb{Z}_{3}\times \mathbb{Z}_{4}$) it is simple:
For $x\in\mathbb{Z}_{12}$ we define $\varphi (x)=(2x\pmod 3,3x\pmod 4)$. It's easy to see that it is an isomorphus.


Hi, I have purchased AOPS Volume 1 a few days ago and started workin on the problems. Working trough the exercises was fairly simple but as I got to the Chapter questions of the first chapter things got different. For the last 6 questions I did not have a clue of where to start or the apporach to the questions. Is this a problem? Are those questions ment to be solved by people who can easily do the exercises?

I'm talkin about questions like $\log_{a}{B}=\log_{b}{A}$ find $AB$

And what steps should I take in oreder to understand the approach to the questions?

Thanks!!!

Jeremic

possibly work through the book and try to understand the text and see tricks? and maybe ask people online questions?  :lol:

[quote="Jeremic"]Hi, I have purchased AOPS Volume 1 a few days ago and started workin on the problems. Working trough the exercises was fairly simple but as I got to the Chapter questions of the first chapter things got different. For the last 6 questions I did not have a clue of where to start or the apporach to the questions. Is this a problem? Are those questions ment to be solved by people who can easily do the exercises?

I'm talkin about questions like $\log_{a}{B}=\log_{b}{A}$ find $AB$

And what steps should I take in oreder to understand the approach to the questions?
[/quote]

This is not a problem - the books are designed to be like this, instead of your normal textbooks, in which all the problems are easy for you.

You're not alone - the books are meant to have very challenging problems for top students.

One suggestion is to try the problems on your own, and if you can't solve them after trying for a little while, read the solutions.  Mark the problem, then come back in 2-3 weeks and try the problem again.  When you read the solution, think about what clues in the problem might suggest to try solving the problem in the way presented.  This will help you locate these clues in the future with other problems.

Thank you both for the quick replies. I will take your guidance in concideration when I start Chapter 2    :D


As of April 14 there are close to 300 teams signed up to compete in this year's [color=indigo][b]Purple Comet Math Meet[/b][/color].  There is still plenty of time for you to sign up a team of from 1 to 6 students to compete beginning at any time you find convenient between NOON GMT MONDAY APRIL 16 to NOON GMT FRIDAY APRIL 20 (that is, 7 AM Eastern Daylight Time Monday to 7 AM Eastern Daylight Time Friday).

Contest Rules, Registration, and Copies of old contests are available on-line at [url]http://purplecomet.org[/url].  IT'S FREE!

TWO LEVEL OF COMPETITION: Middle School (15 problems in 60 minutes) or High School (25 problems in 90 minutes)

THREE CATEGORIES: Your team can be made up of students from the same Large School, students from the same Small School, or students from differnt schools.

Each team requires an adult supervisor and an internet connection.

The Purple Comet Math Meet for 2007 has ended.  Over 600 teams participated.  You are welcome to discuss the problems openly.  Answers and winners will be posted on the web site [url]http://purplecomet.org[/url] by sometime Monday afternoon (Central Time).  Thanks to all who participated.

let's see if we can get an unofficial answer key going

[hide="My answers (worked out in the two days after the contest)"]
Q1=231
Q2=7
Q3=45
Q4=120
Q5=383
Q6=770
Q7=65
Q8=67
Q9=1094
Q10=125
Q11=617
Q12=216
Q13=65
Q14=341
Q15=252
Q16=161? (I do not claim to be good with casework)
Q17=320
Q18=??? (I didn't want to devote huge amounts of time to analytic geo)
Q19=1051? (see Q16)
Q20=???
Q21=4
Q22=???
Q23=192
Q24=554
Q25=??? (I tried to brute-force it, but didn't get any answers)
[/hide]

This should go a long way towards that unofficial answer key.

The middle school and high school levels almost certainly used different questions.  I guess it looks like we will just have to have two answer keys instead of one :jump:

[quote="xpmath"][hide]I took the middle school, and we got:
Q1=9
Q2=1500
Q3=144
Q4=780
Q5=1461
Q6=360
Q7=12
Q8=97
Q9=1185
Q10=72
Q11=252
Q12=240
Q13=3003
Q14=280
Q15=124 (We guessed :P )[/hide]

Edit: If there are overlapping questions, the only one where we got the same answer as you was Q15 on yours. Are you sure about those?
Hopefully there aren't very many common questions :maybe:[/quote]

We are almost certain the one with a series summation is 1001.  (i think #13)

For 14, we also got 280 (or rather, I).  15, I brute forced it and got 70.

For 11, (permutations) I also got 252, and 12 was 240.   I remember 9 and 1461 as well.

Nice job! I didn't do #13, so I don't know about that one. We'll find out tomorrow :jump:

Answers are up on the website!

woot, I got #15 right.

Nice :clap2:   If I remember the answers we got, I think we got 13 of them right. We got #13 and #15 wrong.
When will  the winners be posted?

#22 should be thrown out.

Our test papers had

$f(n)={n\choose0}a^{n-1}-{n\choose1}a^{n-2}+{n\choose3}a^{n-3}$

when it should have been

$f(n)={n\choose0}a^{n-1}-{n\choose1}a^{n-2}+{n\choose2}a^{n-3}$

and wow. I totally guessed 7 on #21.

Well, my team scored a $14$ (High School)  :|

We were tied for $13^{th}$ in small high schools, so I guess we didn't do all that poorly.

Wait, how are we supposed to know our School ID?  The score breakdown is up in terms of that.  Nice job on 13/15.  Who was on your team?  WOOOO I Got 15 right.  brute force skillz...its how you do the AIME.

Our teacher told us our ID number. Thanks, even though it isn't that good. Good Job on the perfect.
Our team members: Me,Disha,Allen Chen,Rolland He,Parry Zong,Mason Liang. Respectively, 6th grade,8th grade,7th grade,7th grade,8th grade, and 8th grade.
Edit: @roadnottaken, how do you know you came 13th?

whoa wtf our team got 20...that's not 1st even though we were a mixed team?
we thought we got 23....but our guess on #18 was correct, so eh no idea what happened....
man we were supposed to be the palo alto all-star team...

whoa only 16 teams got #24 right...I solved that one! woot

Note that the 13th is only among small high schools $(\le1200)$.  I know my team's score because I saved a copy of our answers.  As to our placement, look [url=http://purplecomet.org/2071results.htm#numcorrect]here[/url].

That means we came first among the large teams in middle school? Only 3 teams got #15 right? We weren't one of them  :(

Woah, I think we got full score... I know for a fact that 11-15 were the answers we put down, but only 1 school had all those right - that school got all of the rest right too.  wohoo.

And 13 is really good, we got lucky!

Thanks. You guys weren't lucky, you're just smart. Who was on your team?
I can't wait 'til the official winners get put up.

Here are the winners for the Spring 2007 Meet:[url]http://purplecomet.org/2071winners.htm[/url]
Good job to everyone who competed!

Yay. I didn't know our team number, so I was unsure of what place we got, but it turns out I was more or less right and we got 3rd (TJ Frosh). The team composed of two freshmen.

#24 was actually one of the ones that I did personally. Having a computer for this test is very helpful. To do it, you consider the number that you can make with some number of 1s. Suppose you can make $n$ with $k$ 1s. Then you can make $n+1$ with $k+1$ ones, $n+11$ with $k+2$ ones, $n+111$ with $k+3$ ones, or $n+1111$ with $k+4$ ones. Putting this into a program is rather easy, and you can get the 121 answers very fast.

Huh. We won, for high schools of less than 1200 students.

Heck, we could win for high schools with less than 20
students.  :-) Our whole "school," from first to twelfth, only has
about 100.

Catherine Asaro, Coach
Howard Area Homeschoolers

Yay! Tied for first place, middle school division! :) We got #11 and #15 wrong - both of which I helped work on... :(

You tied with us then. We got #13 and #15 wrong.

[quote="roadnottaken"]Yay, we made honorable mention (hey, I'll take what I can get :D )![/quote]

Amen.  Go team  :ninja:  penguin.  :P


inequalities
function
logarithms
inequalities unsolved

[quote="Megus"]It immediately follows from Jensen as function $f(x)=\ln (\frac{1}{\sqrt{x}}-1)$ is convex[/quote]
No,I don't think that  $f(x)=\ln (\frac{1}{\sqrt{x}}-1)$ is convex. ;)

Yeah,it's convex in $(0,\frac{4}{9})$ but not convex in $(0,1)$. :(

[quote="zhaobin"][quote="Kummer"][quote="zhaobin"]$x+y+z=1,x,y,z>0$
can we prove:
$(\frac 1{\sqrt x}-1)(\frac 1{\sqrt y}-1)(\frac 1{\sqrt z}-1) \ge \left( \sqrt 3 -1\right)^3$[/quote]

  Ofcourse by taking ln both two side of the inequality we have to prove:
${\sum ln( \frac{1}{ \sqrt{x}} -1) \geq 3( \sqrt{3}}-1)$ with x+y+z=1
The rest i think can be solved by Lagrange multipliers[/quote]
are you sure?with x+y+z=1,$(x,y,z)$ is only a open interval...
if you finished,please post :)
thanks[/quote]

  Zhaobin,ofcourse for x+y+z=1,(x,y,z) is only the open interval,but the existing condition for root $\sqrt{x}$, $\sqrt{y}$, $\sqrt{z}$ can made it become closed interval..I hope i'm not wrong( if u agree with me,i'll produce a solution for this inequality )

Kummer

OK,Kummer,I think if you like the  Lagrange multipliers.It will be not easy than use the theorem[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64933]www.mathlinks.ro/Forum/viewtopic.php?t=64933[/url]
Here by the remind of harazi(I will post a proof in old and new inequalities,but I also see you Kummer's solution,because I am poor in Lagrange multipliers,thanks)
proof:
$(\frac 1{\sqrt x}-1)(\frac 1{\sqrt y}-1)(\frac 1{\sqrt z}-1) \ge \left( \sqrt 3 -1\right)^3 \iff (1-\sqrt x)(1-\sqrt y)(1-\sqrt z) \ge \left( \sqrt 3 -1\right)^3 \sqrt x \sqrt y \sqrt z$
$\iff (1- x)(1- y)(1- z) \ge \left( \sqrt 3 -1\right)^3 \sqrt x \sqrt y \sqrt z(1+\sqrt x)(1+\sqrt y)(1+\sqrt z)$ $(1)$
but we have:
$(1- x)^2(1- y)^2(1- z)^2 \ge (y+z)^2(z+x)^2(x+y)^2 \ge \frac{64} {81}(x+y+z)^2(xy+yz+zx)^2 \ge\frac{64}{ 81}(x+y+z)^2 3(x+y+z)xyz=\frac{64} {27}(x+y+z)^2 3(x+y+z)xyz$
and
$(1+\sqrt x)(1+\sqrt y)(1+\sqrt z) \le \left(\frac{3+\sqrt x+\sqrt y+\sqrt z}{3} \right)^3 \le \left(\frac{3+ \sqrt 3}{3} \right)^3$
so the inequality $(1)$ is ture,
we have proved

[quote="Vasc"]Actually, the more general and stronger inequality holds:
$(\frac{1}{\sqrt x_1}-\sqrt x_1)(\frac{1}{\sqrt x_2}-\sqrt x_2)...(\frac{1}{\sqrt x_n}-\sqrt x_n) \geq (\sqrt n-\frac{1}{\sqrt n})^n$
for $x_1+x_2+...+x_n=1$.
From this inequality, the inequalities follow
$(\frac{1}{\sqrt x_1}-1)(\frac{1}{\sqrt x_2}-1)...(\frac{1}{\sqrt x_n}-1) \geq (\sqrt n-1)^n$
and
$(\frac{1}{x_1}-x_1)(\frac{1}{x_2}-x_2)...(\frac{1}{x_n}-x_n) \geq (n-\frac{1}{n})^n$. ;)[/quote]
yes,And it can be sloved by Generalized Popoviciu   easily.
And yesterday I post another proof for Generalized Popoviciu.Does that right,vasc?and Is it the proof you sent to the crux?

[quote="Vasc"]Actually, the more general and stronger inequality holds:
$(\frac{1}{\sqrt x_1}-\sqrt x_1)(\frac{1}{\sqrt x_2}-\sqrt x_2)...(\frac{1}{\sqrt x_n}-\sqrt x_n) \geq (\sqrt n-\frac{1}{\sqrt n})^n$
for $x_1+x_2+...+x_n=1$.
From this inequality, the inequalities follow
$(\frac{1}{\sqrt x_1}-1)(\frac{1}{\sqrt x_2}-1)...(\frac{1}{\sqrt x_n}-1) \geq (\sqrt n-1)^n$
and
$(\frac{1}{x_1}-x_1)(\frac{1}{x_2}-x_2)...(\frac{1}{x_n}-x_n) \geq (n-\frac{1}{n})^n$. ;)[/quote]

I think we can prove:
$(\sqrt{x_1^k } - \frac{1}{{\sqrt{x_1^k } }})(\sqrt{x_2^k } - \frac{1}{{\sqrt{x_2^k } }}) \cdots (\sqrt{x_n^k } - \frac{1}{{\sqrt{x_n^k } }}) \ge (\sqrt{(\frac{n}{s})^k } - \sqrt{(\frac{s}{n})^k } )^n$

for $x_1+x_2+...+x_n=s,0 < s \le 1,n,k \in N,n \ge 3,k > 0$.

Yeah! I see all of it in simple idea. It's the same method with the following of Vasc :

$(2-ab)(2-bc)(2-ca) \ge 1$

If $a^2+b^2+c^2=3$.

Bur, for the permission, I won't post now. You'll se the full idea and method in my E-Book I'll complete next 2-month.

[quote="Vasc"]Yes, my proof is very similar.
My article was published in Crux Mathematicorum, in Sepember 2005 (no. 6).
But the starting idea of the proof appeared in 2003 or 2002, in an article from Gazeta Matematica-A.[/quote]
thanks,ok,It is not bad for me to prove it inedepent(although the proof is the same as you) ;)
congratulation to myself :)


Let $ n$ be an integer with $ n \geq 2$. Suppose that for $ k \equal{} 0,1,...,n \minus{} 2$ we have:

$ \binom {n \minus{} 2}{k} \equiv ( \minus{} 1)^{k}(k \plus{} 1) (\mod n)$.

Show that $ n$ is prime.

It give $ \binom{n \minus{} 1}{k} \equal{} \binom{n \minus{} 2}{k} \plus{} \binom{n \minus{} 2}{k \minus{} 1} \equal{} ( \minus{} 1)^k\mod n, k \equal{} 0,1,..,n \minus{} 2$ and
$ \binom{n}{k} \equal{} \binom{n \minus{} 1}{k} \plus{} \binom{n \minus{} 1}{k \minus{} 1} \equal{} 0\mod n, k \equal{} 1,...,n \minus{} 1$. Therefore $ n$ is prime.

I think that pretty simple and well-known idea works here,consider the least prime divisor of $ n$...

[b]Solution:[/b]
Suppose to the contrary that $ n$ is composite  and let $ p$ is the least prime divisor of $ n$.Since $ p\leq\frac {n}{2}$ we conclude that
$ p\leq n \minus{} 2$,otherwise $ n \equal{} 2$.
As we know
$ \binom{n \minus{} 2}{p \minus{} 1}\equiv ( \minus{} 1)^{p \minus{} 1}(p)$ mod $ n$
On the other hand:
$ \frac {n \minus{} 2}{2}\cdot\frac {n \minus{} 3}{3}\dots\frac {n \minus{} p \plus{} 1}{p \minus{} 1}\cdot\frac {n \minus{} p \plus{} 2}{1}\equiv ( \minus{} 1)^{p \minus{} 2}(n \minus{} p \plus{} 2)$ mod $ n$.
Here we use that all numbers $ 1,2,\dots,p\minus{}1$ are coprime with $ n$.
From previous two relations we conclude that
$ n \plus{} 2$ is divisible by $ n$, therefore $ n \equal{} 2$...

[quote="Ibn Rochde"]hi sorry guys but what does this symbole mean  $ \binom {n \minus{} 2}{k}$  :oops:[/quote]

It's a combination, or binomial coefficient.  You might know it as $ C^{n\minus{}2}_k$ or $ C^k_{n\minus{}2}$.  In any case, here's the wiki article for it:
http://www.artofproblemsolving.com/Wiki/index.php/Combination

[quote="Sepp"][quote="Erken"]From previous two relations we conclude that
$ n \plus{} 2$ is divisible by $ n$, therefore $ n \equal{} 2$...[/quote]
Sorry but I don't see why, could you please elaborate?[/quote]
The problem's statement asserts that
$ \binom{n \minus{} 2}{p \minus{} 1}\equiv ( \minus{} 1)^{p \minus{} 1}\cdot p$ mod $ n$.
On the other hand,as it was established:
$ \binom{n \minus{} 2}{p \minus{} 1}\equiv ( \minus{} 1)^{p \minus{} 2}(n \minus{} p \plus{} 2)$ mod $ n$.
It means that numbers $ ( \minus{} 1)^{p \minus{} 1}\cdot p$ and $ ( \minus{} 1)^{p \minus{} 2}(n \minus{} p \plus{} 2)$ are equal modulo $ n$,so
$ \minus{} p\equiv n \minus{} p \plus{} 2$ mod $ n$,it means that $ n \plus{} 2\equiv 0$ mod $ n$.

[quote="Sepp"]True, but here
[quote="Erken"]
On the other hand:
$ \frac {n \minus{} 2}{2}\cdot\frac {n \minus{} 3}{3}\dots\frac {n \minus{} p \plus{} 1}{p \minus{} 1}\cdot\frac {n \minus{} p \plus{} 2}{1}\equiv ( \minus{} 1)^{p \minus{} 2}(n \minus{} p \plus{} 2)$ mod $ n$.
[/quote]
I think it's missed the factor $ n \minus{} p$ to get equality with $ \binom {n \minus{} 2}{p \minus{} 1}$. Sorry if I'm making confusion  :maybe:[/quote]
Hmm...solution seems to be wrong,thanks,for pointing it out...Do we have another solution for that problem?I mean is Rust's solution correct?


A pie is cut into 6 slices. Five slices are equal in size while the remaining slice is half as large as the others. What fraction of the whole pie is the smallest slice?

[hide="Yay, :winner_first:"]

Let, the smallest piece be equal to $ x$. Therefore, each of the other pieces is equal to $ 2x$.

Adding it all together, we see that $ 5(2x)\plus{}x\equal{}11x$.

Therefore, $ \frac{x}{11x}\equal{}\boxed{\frac{1}{11}}$.

[/hide]


linear algebra
matrix
Galois Theory
superior algebra
superior algebra theorems

Considering a finite separable extension $E$ of a field $F$, a base $a_{1},\dots,a_{n}$ of $E$ over $F$, and disjoint $F$-homomorphisms $\sigma_{1},\dots,\sigma_{n}$ from $E$ into another field $\Omega$ containing $F$, the matrix $(\sigma_{i}(a_{j}))$ is invertible.

What I do not understand is why we need the extension to be separable ... seems to be working fine without this assumption !?

If $E$ is not seperable over $F$ then there wo'nt be $[E: F]$ homomorphisms(over F),they will be less than
$[E: F]$,perhaps this clears your doubt. :)

Thank you, it clears my doubts completely ...


Let a,b,c are non-negative such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that
$ 3\plus{}3abc\ge 2(ab\plus{}bc\plus{}ca)$
Equality occurs when $ a\equal{}b\equal{}c\equal{}1$ or $ a\equal{}0,b\equal{}c\equal{}\sqrt \frac 32$ and pemurtation.

[quote="thegod277"]Let a,b,c are non-negative such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that
$ 3 \plus{} 3abc\ge 2(ab \plus{} bc \plus{} ca)$
Equality occurs when $ a \equal{} b \equal{} c \equal{} 1$ or $ a \equal{} 0,b \equal{} c \equal{} \sqrt \frac 32$ and pemurtation.[/quote]
It's wrong, see $ a\equal{}b\equal{}1.1,c\equal{}\sqrt{3\minus{}2a^2}$.


limit
calculus
integration
function
algebra
polynomial
binomial coefficients

Hi can anyone help me with these problems please:

1. Find the sum of the first five terms of ∑ n=1 to infinity

(-1)^(n-1) (n+1)/(2)^n

How large is the error in using this approximation of the sum?

i think i found the sum to be .53125 but i don't know if i'm doing it correctly..and i have no idea how to find the error in using the sum.

2. ∑ n=0 to infinity n^(1/2) / ((n^2)+3) tell whether this converges absolutely, conditionally, or diverges. show what test you used.

3. find the interval of convergence for this series. be sure to check the endpoints and tell where it converges conditionally

∑ n=1 to infinity (3^n)(x^n) / (n+1)^2

Nice calculations Thales, however here is a problem which generalizes your result.

Generalization. Let $a>1$ and let $k\geq 1$ be a natural number. Find

$\sum_{n=1}^{\infty}\frac{n^{k}}{a^{n}}=?$.

http://mathworld.wolfram.com/Polylogarithm.html

this is the polylog

and i do not understand what you said

im wrong in someting?

No, you are not wrong. What I wanted to say is that you have calculated,nicely, a sum of the following form

$\sum \frac{n-1}{a^{n}}$, and that this series can be generalized to the series I have writen in a previous post. That is it.
I do not think that the series is a polylogarithm.

hm.. understand

and for the polylog

sorry

hahau

i switch  N for K  but the variable is Z

hauhau

sorry

This series is absolutely difficult to calculate, so, let'se generalize using mathematica.

${s\to \infty }$
${\sum_{n=1}^{s}\frac{n^{k}}{a^{n}}=\left(\text{Li}_{-k}\left(\frac{1}{a}\right)\right)-\left(\left(\frac{1}{a}\right)^{s+1}\Phi \left(\frac{1}{a},-k,s+1\right)\right)}$

LerchPhi is a function
http://mathworld.wolfram.com/LerchTranscendent.html

PolyLog is another function
http://mathworld.wolfram.com/Polylogarithm.html

After see the result we can solve this, because it's a case of manipulation of variables.

But, find the limit of ${s\to \infty }$ it's a impossible thing

Mathematica does not solve this

so, i will never try solve this hand-writing
hhe

It actually is a polylogarithm, but with a negative exponent.  The basis $\{n^{k}\}_{k}$ for polynomials in $n$ is actually not the nicest for this question; it's prettier to consider the basis $\left\{{n \choose k}\right\}_{k}$, for which the answer has a very simple form.

JBL

what is ${f}_{n}$

i dont know a notation using {}

The gist was, instead of calculating $\sum_{n=1}^{\infty}\frac{n^{k}}{a^{n}}$, it is simpler (and equally useful) to calculate $\sum_{n=1}^{\infty}\frac{{n\choose k}}{a^{n}}$, where $\binom{n}{k}= \frac{n!}{k!(n-k)!}$ is a binomial coefficient.  (The reason equally useful: because the binomial coefficients also form a basis for polynomials.)

That is the idea.

Consider the following decomposition in simple fractions.

$\frac{n^{k}}{a^{n}}=\frac{P_{k}(n+1)}{a^{n}}-\frac{P_{k}(n)}{a^{n-1}}$,

where $P_{k}$ is a polynomial of degree $k$ which can be found.

Thus,

$\sum_{n=1}^{\infty}\frac{n^{k}}{a^{n}}=-P_{k}(1)$.

Play with $k=2$, $k=3$ and $a=2$ and see what happens.


Set A has 5 consecutive numbers that have the sum of 15.  Set B has 4 consecutive numbers that have the sum of 18.  How many integers are in both Set A and Set B?

[hide]Set A with 5 consec numbers w/ sum 15

Set B with 4 cosec numbers w/ sum 18

In Set A are 1, 2, 3, 4, and 5

In Set B are 3, 4, 5, and 6

So...

6 integers in all.

I know there's a formula for finding the sum of consecutive numbers, but I forgot it...[/hide]

Hehe this is fun.

Good job!  Now, let me ask you this way, do you think my questions are pretty good questions to solve?  You can be critic...

Hmm... tough question.

Your questions are definitely fun  :)  but maybe not too hard... From what I've seen they seem pretty easy to me. But to answer your question, yes, I think they are good questions to solve. Although I may not learn too much, I don't get to try too many of these kinds of questions. It's always good to go over previously learned stuff.  :D

I think your questions are mostly good questions at an appropriate level for where you are posting them.

Duck-billed platypus: you may wish to review your answer and find out exactly what the question asked for.

P.S. for n consecutive numbers a, a+1, a+2, ..., a+n-1, the formula for finding their sum is

(first term + last term) * (# terms) / 2
= (a + (a + n - 1)) * n / 2
= (2a + n - 1) * n / 2.

I liked all the replies about my question's level because I myself think my questions are only appropriate for getting started level...
I don't usually post anything for the advanced level because I'm not there yet....

Osiris wrote:Duck-billed platypus: you may wish to review your answer and find out exactly what the question asked for.

Oops...

[hide]Is 3 better?[/hide]


the rules of hide and go seek are simple. the first person goes on a random thread in the fun factory (must be in top 5 threads) and posts 3 Mr. Greens, each on top of each other, like so:
  [hide=" :D "]this is for the hide and seek game.[/hide]
 :D
 :D

but, to determine that the 3 Mr. Greens are for this game, you must put one Mr. Green as a hide tag. click the top Mr. Green to find out what you should put.

the next person must then find the smiley and return to this thread and post the exact whereabouts of these smileys
eg. corrupt a wish thread, page 23
then that second person must post another 3 smileys somewhere else.
YOU CANNOT POST THE 3 SMILEYS IN THIS THREAD!!!!!!!

now, let's begin.


algebra
polynomial
linear algebra
matrix
function
linear algebra unsolved

Find matrices $ A, B,$ and $ C$ that satisfy the following [b]three[/b] conditions:

[b][color=red]1.[/color][/b] $ A, B, C \in \mathcal{M}_{5 \times 5} (\mathbb{Q}).$
[b][color=red]2.[/color][/b] The matrices have only nonzero eigenvalues.
[b][color=red]3.[/color][/b] $ [p_{A}(t)]^{2009} \plus{} [p_{B}(t)]^{2009} \equal{} [p_{C}(t)]^{2009}$

Hope you all like it.  :D

Best of luck...

P.S. Why, yes... The expression $ p_{A}(t)$ denotes the characteristic polynomial of matrix $ A$.

I don't follow.  The third condition has only trivial solutions by Fermat's Last Theorem for rational functions.

Good... Can you use now that information to derive a conclusion on the possibilities for the triple (A,B,C)?

By the way, there is no need to appeal to FLT for rational functions, [b]t0rajir0u[/b]...

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

The result that underlies this proposal is commonly referred to as the [color=red]coquitao-Wiles[/color] theorem.  :rotfl:

I still don't understand.  Whichever way you reach the conclusion, one of $ A, B, C$ needs to have a characteristic polynomial that's identically zero, so how can it have nonzero eigenvalues?

[quote="JBL"]Congratulations on post 10,000[/quote]
On that subject, it seems that 10000 posts comes with the title of "Moderator". This has nothing to do with actual moderator status anywhere.

Oh, and this is #6000 for me.

Whoa. Grats to you guys for large integer multiples of 1000.


trigonometry
function
algebra unsolved
algebra

Calculator
1) $ A\equal{}sin^2x\plus{}sin^22x\plus{}.....\plus{}sin^2nx$
2) $ B\equal{}cos^2x\plus{}cos^22x\plus{}cos^23x\plus{}...\plus{}cos^2nx$

I'm very interested in an explaination of the title "quickly!!!!!!!!!!!!!"...

hello, he wants to find the sums quickly.
Sonnhard.

Maybe I know the derivation of these identities?  What about higher powers of these two trig functions (in this case it is 2)?

$ sin^2kx \plus{} cos^2kx \equal{} 1\Rightarrow A \plus{} B \equal{} n$
and:
$ B \minus{} A \equal{} \sum_{k \equal{} 1}^n(cos^2kx \minus{} sin^2kx) \equal{} \sum_{k \equal{} 1}^ncos2kx \equal{} \sum_{k \equal{} 1}^n\frac {2.sinx.cos2kx}{2.sinx} \equal{} \\
\equal{} \frac {1}{2.sinx}.\sum_{k \equal{} 1}^n2.sinx.cos2kx \equal{} \frac {1}{2.sinx}.\sum_{k \equal{} 1}^n[sin(2k \plus{} 1)x \minus{} sin(2k \minus{} 1)x] \equal{} \\
\equal{} \frac {sin(2k \plus{} 1)x \minus{} sinx}{2.sinx} \equal{} \frac {sinnx.cos(n \plus{} 1)x}{sinx}\Rightarrow$

$ \Rightarrow A\equal{}\frac{(A\plus{}B)\minus{}(B\minus{}A)}{2}\equal{}\dots; B\equal{}\frac{(A\plus{}B)\plus{}(B\minus{}A)}{2}\equal{}\dots$


Prove the product of 5 consecutive natural numbers can not be a squared number

let's take the $ 5$ consecutive numbers $ (x \minus{} 2),(x \minus{} 1),(x),(x \plus{} 1),(x \plus{} 2)$

we want to prove that the product of these numbers cannot be a squared number

let's multiply it

$ (x \minus{} 2)(x \minus{} 1)(x)(x \plus{} 1)(x \plus{} 2)$

now, let's rearrange it

$ ((x \minus{} 2)(x \plus{} 2))((x \minus{} 1)(x \plus{} 1))(x)$

now we can see we have two product of difference and sum
that makes it easier to multiply

$ (x^2 \minus{} 4)(x^2 \minus{} 1)(x)$

$ (x^4 \minus{} 5x^2 \plus{} 4)(x)$

$ x^5 \minus{} 5x^3 \plus{} 4x$

as you can see, all the exponents are odd, which means that for any $ 5$ consecutive numbers, it is impossible for it to be a perfect square

[hide="the response to what i predict will be the next few posts if i dont post this"]^^okay, i admit, with clever manipulation, we might be able to turn an expression with odd exponents into a perfect square, but this is not one of those manipulations
if anyone has a better conclusion, i'm glad to hear it[/hide]

edit: don't bother looking at my last line of text and the hidden text
Math Champion's conclusion is so much better

You could also say that if $ x$ is not a perfect square, $ x^4\minus{}5x^2\plus{}4$ must contain a factor of $ x$, but it doesn't because you cannot factor $ x$ out. If $ x$ is a perfect square, then you prove that $ x^4\minus{}5x^2\plus{}4$ cannot be a perfect square.


AMC
USA(J)MO
USAMO
MIT
college
Harvard
summer program

[quote="mustafa"]How would you suggest convincing my parents to let me go to Mathcamp/Awesomemath?

They say that it is expensive, far from home, and there are perfectly good things to do closer to home.[/quote]
say something like this:
"but if i go to those camps, i'll get smarter and better at math, then i'll get into USAMO and MOP, and eventually, Ill do super good on USAMO, and ill be one of the hon. mentions or winners and then ill go to MIT,Harvard, etc"
thats why my mom is forcing me to go to one of those camps next summer(she wants me to go to promys i think.), which is very good for me.

With regards to Phelpedo's point, I give it a major second.  In fact, the attitude of campers couldn't be farther from the make MOP/get into MIT/Harvard mindset.  This was obvious enough that a reporter for the Globe and Mail, doing a story on mathcamp, noted that:

"The application process deliberately discourages meddling mothers and fathers who want either to unload their hyperdimensional teenagers for more than a month or to mould their straight-A prodigies into the next Bill Gates."

MathCamp days are filled with Math (and many, many other crazy activities) for its own sake, not for the sake of some parents or college admissions officers.  It makes it oh so much more fun!

As for convincing your parents, I'd suggest a couple of things.

First off, ask yourself why you want to go to MathCamp (or AwesomeMath).  Convince yourself as thuroughly as if you were doing a proof for (since I know you're participating) USAMTS if need be.  It'll make it that much easier to convince them when you know exactly why you want to go and what you expect to get out of the camp.  This will also make it easier to choose between MathCamp and AwesomeMath (which, from what I've gathered, are very different math programs.  Do some research, choose the one that would suit you best.)

Second, have them read the Globe and Mail article.  Reading about MathCamp culture from a non-mathematicians viewpoint might help convince them, and, if nothing else, it'll probably amuse you.  This article definitely sealed the deal for me (in particular, the above "meddling parents" quote stuck in my mind.  I know way too many people with such parents.)  The article is at http://www.mathcamp.org/GlobeandMail.htm

Finally, while finances will obviously be a question, they should not deter you from applying.  I too was worried whether my family would be able to afford the camp fee last year, but, fortunately enough, through a need-based scholarship, my tuition dropped from the listed $\$$3,195 to around $\$$700.  The USAMTS scholarship then reduced it by a further $\$$200, so don't let your parents get scared off by the ticket price.  I also know of at least one student who was given a full scholarship last year.  (Note that, apart from the 'MTS scholarship, as far as I know, all of MathCamp's financial aid is need-based).  As DanZ said in another thread, not a single person was unable to attend due to financial reasons last year.

If nothing else, convince them that applying can't hurt.  Assuming you're accepted, the costs, location, and details for the camp will be sent to you fairly quickly, and you can make the decision then, instead of at such an early point in the year.

[quote="Phelpedo"]Actually, I think that's the wrong idea. If you're going because it's a nice shiny thing to put on your resume, you're going for the wrong reasons.[/quote]
No I mean that it'll convince your parents. It convinced mine. :D
And duh why would any aopser want to go to those camps to get into mop or good colleges? of course it's because we like to do math.

Unfortunately the same attitude then leads them to say "No, you can't go back to Mathcamp ever again because it's too much fun and not enough hardcore training you for the AIME/USAMO which is what you really need for college. If you don't make MOP next year you're going to AwesomeMath since it's all about training." Sigh.

Okay, so first off, people at Mathcamp are not perverts. If The Globe and Mail wants to yank out the most "interesting" suggestions from what I suspect was a large pool of more politically correct ones, that's their business. Secondly, it's just a normal teenage sense of humor. Overall, it is simply amazing how mature Mathcampers are in comparison to an average high schooler. There are exceptions, of course, but in general, if you want friendly, polite, helpful, trustworthy peers, I don't know if there's a better place to go.

That [i]Globe and Mail[/i] article gave us lots of problems.  We were very unsure if we should post it online, because while it was positive about Mathcamp, it also gave an impression of the camp that is, to say the least, highly misleading, and has occasionally led to some confused e-mails from concerned parents/others.  I'd forgotten about that particular unfortunate quote.  When I first read the article, I totally didn't notice all of that --- I was in my "ah, Mathcamp!" mode and only got the good bits out of the article!

So we all breathed a sigh of relief when we were able to put a new article up on our website just yesterday.  It's an article written by a Mathcamp alumna for [i]Imagine[/i], a magazine published by Johns Hopkins University's Center for Talented Youth for gifted and talented precollege students.  It does a much better job of reflecting what Mathcamp is like.  I can assure you that Sara (who wrote this article) would not have written about camp as she did if that bit you excerpted from the [i]Globe and Mail[/i] article was anything like what camp is like!  (For that matter, I wouldn't be [b]working[/b] for the camp if that's what it was like!)

What I would say about finances has already been said; we should be able to meet your need.  If your family can afford it but just doesn't want to pay anything, then we won't be able to help as much, but if you have genuine financial difficulties we'll make it possible for you to come.

One thing that can help with convincing parents is getting put in touch with current instructors (such as myself) or alumni of the program.  That's something we'd be glad to do.

[quote="solafidefarms"]@Phelpedo: And is this supposed to help the case for Mathcamp?

[quote="The Globe and Mail"]That IS a Pencil in My Pocket (for "Abstinence Camp," the two nuns recently spotted in the rose garden)

How to Cheer in a Non-Sexual Manner (for the cheerleaders)

A sign-up sheet solicited suggestions for this year's Mathcamp T-shirt slogan. The time-honoured T-shirt slogan, "A mathematician is cautious in the presence of the obvious," was met with a scribbled "YAWN!" Some of the alternatives:

Math: Because when you can't get laid, you might as well be good at something[/quote][/quote][/quote]

What exactly is your point? That we should set false pretenses to be more appealing to the rest of society?

Ah, so now my mother knows who to look out for?

I'll pull out that Imagine issue to show my parents;;; we'll see if it helps any. Perhaps I shall apply...

Ah, now I understand what you are getting at.

yeah well at least you aren't a <insert funcia's age and gender>
that shouldn't be <insert miles away from home> all alone
my mom thought of that camp that violinists can go to for violin though since i can go with my sister and they have a math programme

Fortunately I don't have to convince my parents to let me go to MathCamp :) . My brother went to MathCamp twice, and he never had any problem with some of the stuff mentioned in the article. He had a great time both years, and most of his best friends at college now are MathCampers :) .

Hah! I've finally convinced my parents to send me to Mathcamp this summer. Thanks for the help :w00tb:!


Given two circle (O) and (O') intersect at A and B. In circle (O) contruct chord AE which ia tangented with (O') at A. In circle (O') contruct chord AF which is tangented with (O') at B. Prove that $ \frac{BE}{BF}\equal{}\frac{AE^2}{AF^2}$

[color=darkblue]Because $ AE$ and $ AF$ are tangents we have $ \left\|\begin{array}{cc} \angle{BFA} \equal{} \angle{EAB} \\
\angle{AEB} \equal{} \angle{BAF} \end{array} \right\| \Rightarrow \triangle{ABE}\sim \triangle{FBA}$.

Then $ \frac {BE}{AB} \equal{} \frac {AB}{BF} \Leftrightarrow \boxed{\frac {BE}{BF} \equal{} \frac {AB^2}{BF^2} \equal{} \frac {AE^2}{AF^2}}$[/color]


MAFIA GAME 43

It was a quiet night in Mewtoville. The children were sleeping, as were the adults. But then a bunch of bad guy who looked all clichéd with their ski masks and whatnot, went and killed the mayor.

“Oh noez” cried Mewto. “Whatever shall we do!?!”

“I know” suggested a townie. “You can give us all intriguing powers!”

“Indeed. How about I mess with your mind!”

Mewto waved his wand and pwned them all. Everyone (or almost everyone) changed. The good guys all started talking funny. Screams of "Pikachu!!!" were heard in the crowd.

“Now go ahead. Vote on who you want to kill…”

It is Day 1. 9 votes to kill.

[size=200][b]The Rules(read them)[/b][/size]
[list][*]Votes must be in [b]bold[/b].  If you do not bold your vote, it will not be counted.
[*]Once a player has reached lynching state, his pleas are useless and any attempts to unvote will be unheeded. You may [b]vote: no lynch[/b] – majority votes of this kind are necessary to end the day without a death.
[*]The game is not to be discussed outside of the thread unless your role specifically states that you may do so.
[*]Once your death scene has been posted, you’re dead.  Stop typing. The only thing you can type is a "bah" post.
[*]Don’t edit/delete previously submitted posts
[*]Don’t quote any PMs from me. Neither true, nor false. Seriously.
[*]If you have a night choice to make, it is due by the posted deadline.  I will not wait for you – if you do not submit a choice to me, tough!  No choice will be made.
[*]Treat this game as a commitment.  Be considerate – don’t leave us hanging.
[*]Remember, it’s a game.  It’s supposed to played.  It’s supposed to be [i]fun[/i].  Do your part.[/list]

The Living:

1. powerofpi
2. algebra2000
3. crazypianist1116
4. eggylv999
5. Bugi
6. isabella2296
7. alexhhmun
8. Abnormal Distribution
9. iLord
10. aleph0
11. tennis123
12. meewhee009
13. brut3forc3
14. james4l
15. pythag011
16. cyberspace

Many of you have special post restrictions or similar. Don't try to be fancy and look for loopholes. It's not cute, and I will modkill you.

Also mafia, a link to your discussion QuickTopic would be nice. PM me it please. You can't discuss until tonight though.

Confirm in thread.

Oh, this game started. I confirm.

Now, where is that chocolate? I want chocolate!!!!.... I must have chocolate to pwn it.... I don't even know what I am saying

Confirmed. In the spirit of pythag011, allow me to claim that I am the unlynchable unlimited-daykill serial killer.

Oh well, time to check my PM now. :P

[quote="meewhee009"]I like this theme... confirm.[/quote]
I'm claiming my role too... sorry forgot to read the PM first D:
I should really start reading my PM's first

/confirm....

I'm assuming that the game hasn't started yet, so this isn't dangerous.

1. Please answer the clarification i sent to you.
2. If this is dangerous, please just give me a warning.

All restrictions on posting of any type still apply and will apply. I will forgive any infractions thus far, but thats as much as I do. If, for example, I gave you the restriction of typing everything in iambic pentameter (would I be that mean?) and you wish to PM me your confirm, as opposed to composing a whole paragraph of funness, that is OK. All PM's to me can be freely written. Any further questions should be PMed to me. If you have a qualm, ask first, then post.

[quote="O' Glorious Mod"]
If you have any specific questions you can ask.[/quote]

Can you post [i]my[/i] role, I still don't know much about it.  :(
How did I turn in to an SK?

The Mod shared a lot with me, you turned into one when me/james4l died, we hypnotized you as well as each other.

james/pythag were in control of each other's posts, one being mafia, one town.
They also controlled meewhee, which was ***forced*** to be town. But with them gone, he turned into a SK.

But imagine this ending:
james/meewhee/townie (for example AD). Day time. They lynch james, meewhee wins because he turns into a SK? A little bit weird ending...

And, what if pythag and meewhee stayed alive? Town win?

You [i]hypnotized[/i] me?
Then what does iLord have to do with this situation?
[quote]Investigates meewhee (town) <----- THIS ACTION CHANGED THE COURSE OF THE GAME. MORE ON IT LATER.[/quote]
[quote="Bugi"]But with them gone, [b]he[/b] turned into a SK. [/quote]
[i]she[/i]

You thought you were town, so you decided that your house full or serial killery objects was a bit weird and traded it for a town house.

Thus, iLord investigating you thought you were town, Also, that action SHOULDN"T have changed the course of the game, but town played very badly endgame...

Town played a horrible endgame. iLord was the only real town brain. So, if we didn't kill iLord, I bet it would be a town win.
That action DID change the course. Now iLord sold the story (true) that meewhee is confirmed town, and he slipped to the end.

Well, you were lucky, but still your win is fantastic!!!  Well, as a mafia, I have to say: ''Haha town''

[quote="Bugi"]Well, as a mafia, I have to say: ''Haha town''[/quote]

LOL.  Me too.

Clarification:

Meewhee was essentially vanilla townie until [b]BOTH[/b] pythag and james were dead. Once they were both dead, the hypnotism broke and she IMMEDIATELY became neutral.

Hehe... neutral.
So I became violent when pythag and james died. Okay, okay... I'm getting it.

Basically like a reversed psychiatrist/psychopath role?

lastpost (this is used to preserve order of locked topic)


inequalities
induction
inequalities solved

Let a1,a2,..,an,b1,b2,...,bn be real numbers.
Let sqr(x)=x^2.
Prove that
 \sqrt (sqr(a1)+sqr(b1))+ \sqrt(sqr(a2)+sqr(b2))+..+ \sqrt (sqr(an)+sqr(bn)) \geq  \sqrt (sqr(a1+a2+...+an)+sqr(b1+b2+...+bn)). :)

Am I wrong ot this is exactly Minkowski inequality?

I know i different form of Minkowski's inequality but maybe it is.anyway try to solve it and try to find a beutiful solution.I have one.

Upps, I apologise. It isn't Minkowski inequality. I hurried. Anyway, I do have a beautiful solution for this one. Put z_k=a_k+i*b_k. It becomes
 sum |z_k|>=|sum z_k|, which is true.

And, in fact, it can be proved by induction using Minkowski inequality for n=2.


If you have an array with random odd and even numbers, what is the most efficient algorithm you can think of to put all even numbers on one side- and all odd numbers on the other side in this array?

I presume you mean with just O(1) storage space, since otherwise it's trivial.

Though that isn't much harder - just have two pointers, one starting at the start going forwards, one at the end going backwards. Swap the first 1 you find with the last 0 you find, and repeat.

Does it mean that all you have to do is find the modulus of the values in the first array to distinguish between the odd and even, store odd in an array, even in another array, then merge them?

Well, if you're allowed an extra array, just loop through the initial array and put all the even ones at the start of the new one, and all the odd ones at the end of the new one. Eg (java):
[code]int[] oddeven(int[] array) {
	int[] ret = new int[array.length];
	int start = 0, end = ret.length-1;
	for (int x : array) {
		if (x%2==0) ret[start++]=x;
		else ret[end--]=x;
	}
	return ret;
}[/code]

If you weren't allowed to have an extra array (O(1) storage as I mentioned before), you would do something like this:
[code]void oddeven(int[] array) {
	int start=0, end=array.length-1;
	while (start<end) {
		if (array[start]%2==0) start++;
		else if (array[end]%2==1) end--;
		else {
			int temp = array[start];
			array[start] = array[end];
			array[end] = temp;
			start++;
			end--;
		}
	}
}[/code]

You could also use two iterators from the beginning. Increment one of them until the two are in the wrong places (odd before even or vice versa). Then swap those two. increment the other until it is the other one. Keep going.

For example, say we want odds before evens in the array
1 3 4 5 8 1 2 3 5 2

Array: 1 3 4 5 8 1 2 3 5 2
Iterator 1: 1 Iterator 2: 1
Iterator 1: 3 Iterator 2: 1
Iterator 1: 4 Iterator 2: 1
Iterator 1: 4 Iterator 2: 3
Iterator 1: 4 Iterator 2: 4
Iterator 1: 4 Iterator 2: 5
SWAP
Array: 1 3 5 4 8 1 2 3 5 2
Iterator 1: 5 Iterator 2: 4
Iterator 1: 4 Iterator 2: 4
Iterator 1: 4 Iterator 2: 8
Iterator 1: 4 Iterator 2: 1
SWAP
Array: 1 3 5 1 8 4 2 3 5 2

You should be able to see how this is going. Increment iterator 1 until an even is found, then start iterator 2 and go until an odd is found after iterator 1, then swap them.

Could we just run an outline using a list of N randomly placed even and odd numbers.

(1) Set TOP to lowest index.
(2) Set BOT tp highest index.
(3) Assume ODD numbers start at top.
(4) Assume EVEN numbers start at bottom.
(5) Here is your O(1) loop: for 1..N
(6) If TOP index points to ODD number (then it is in place) so move on one past TOP then (5).
(7) If TOP index points to EVEN number (out of place) so process further
(8) IF BOT index points to EVEN number (in place) so move on one above BOT then (5).
otherwise Exchange TOP and BOT and move 1 past TOP OR one above BOT then (5)
(9) ODDS are TOP..N (if N minus TOP .gt. zero)and EVENS are where ODDS are not..
(10) end?

[quote="LeoYard"]Thanks. Could you give examples of scripts in JavaScript, PHP?[/quote]

The given C examples, unless I am mistaken, are also proper PHP syntax.


inequalities
logarithms
calculus
calculus computations

Prove that, for any $x \geq 0$, we have:
\[\log_{2}(1+2^{x}) \geq \log_{3}\left(3^{x}+2^{x/2}\right). \]


inequalities
function
inequalities proposed

For reals $ a,b,c\ge0$, prove

\[ \sum_{cyc} c(a\plus{}b)(c\minus{}a)(c\minus{}b)(c^2\plus{}ab\minus{}a^2\minus{}b^2) \ge 0\]

Can you write a solution then? explaining what function you are using?


A 4-digit number $ MEEP$ in base 5, 7 and 8 exists such that:

                          $ MEEP_5 \plus{} MEEP_7 \equal{} MEEP_8$.

Find $ MEEP$.

$ P\plus{}5E\plus{}5^2E\plus{}5^3M\plus{}P\plus{}7E\plus{}7^2E\plus{}7^3M\equal{}P\plus{}8E\plus{}8^2E\plus{}8^3M$
$ P\plus{}14E\equal{}44M$
Also $ P,E,M<5,P\not \equal{}E\not \equal{}M\not \equal{}P$
So $ \boxed{(M,E,P)\equal{}(1,3,2)}$


[size=150]Determine all pair (m,n) of positive integers such that m(n+1) and n(m+1) are perfect square[/size]

See also : http://www.mathlinks.ro/Forum/viewtopic.php?t=6505


inequalities
induction
inequalities proposed

(a) Prove that if $n$ is a integer such that $n \geq 4011^2$ then there exists an integer $l$ such that \[ n < l^2 < (1 + \frac{1}{{2005}})n .  \]

(b) Find the smallest positive integer $M$ for which whenever an integer $n$ is such that $n \geq M$
then there exists an integer $l$ such that \[ n < l^2 < (1 + \frac{1}{{2005}})n .  \]

i think this one is rather a number theory problem then an inequality

You do not require any number theory for the first part atleast.

Does somebody have a solution for the (B) part of this question... I think that's the only 'unsolved' problem by anyone from INMO 2006!!

For part (b)
We try \[ M = 4010^2  \]
But then when n=M we have
\[ \begin{array}{l} 4010^2 < l^2 < \frac{{2006}}{{2005}}*4010^2 \\ \Rightarrow 4010^2 < l^2 < 4012*4010 \\ \Rightarrow 4010^2 < l^2 < 4011^2 - 1 \\ \end{array}  \]

Giving us no solution
When \[ M = 4010^2 +1  \] we have
\[ l = 4011  \]
satisfying the conditions for \[ 4010^2 + 1 < n < 4011^2  \]
and we have already proved the existence of l for \[ n \ge 4011^2  \]
in part (a)
So \[ M = 4010^2 + 1  \]

[hide=My solution]
We require
\[\sqrt{\left(1 + \frac{1}{2005}\right)n} - \sqrt{n} > 1\]
This reduces to
\[\sqrt{n}(\sqrt{n} - 4010) > 0\]
which solves the first part. For the second part, just notice that this quadratic inequality in \(\sqrt{n}\) is satisfied only for \(\sqrt{n} < \alpha_1\) and \(\sqrt{n} > \alpha_2\), where \(\alpha_1,\alpha_2\) are the roots. This readily gives \(M = 4010^2 + 1\)
[/hide]

The part 1. of the question is asking us to prove :
((1+1/2005)n)^(1/2)-(n)^(1/2)>1.

Consider a function f(x),x>0
f(x)=x+ x/2005
f'(x)=1+1/2005 which is clearly greater than zero. Therefore f is an increasing function.
so plugging n=4011^2 we get,
((1+1/2005)4011^2)^(1/2)-(4011^2)^(1/2)~1.00012>1.

Now for the second part write the equation as:
((1+1/2005)n)^(1/2)>1+(n)^(1/2) and then squaring both sides we get a quadratic in n:
n^2+2005^2-4010^2 n-4010n >0 solving this we get,

M=[b][u]4010^2+1[/u][/b]


a)  Find the largest value  of $ x^{9}\plus{}y^{9}$.
b)  Find the pairs $ (x;y)$ that $ x^{4}\plus{}y^{4}\equal{}1$

For a, based on the way you wrote it, there is no maximum.  I can choose any $ x$ I want, and any $ y$ I want and it could be infinitely large.  For b, write $ y^{4}\equal{}1\minus{}x^{4}\Rightarrow y\equal{}\pm\sqrt[4]{1\minus{}x^{4}}$.  Then for any $ x\leq 1$, we have a pair of solutions.

Part b is a Diophantine equation in the form of $ x^4 \plus{} y^4 \equal{} 1^2$, which has [b]no integer solutions[/b].

[url]http://www.artofproblemsolving.com/Wiki/index.php/Diophantine#Sum_of_Fourth_Powers[/url]

Moved to Intermediate.

If you write it as you did $ x^{4} \plus{} y^{4} \equal{} 1^{2}$  there are two pairs of integer  that sodisfy  it, $ (0;1)$ ^ $ (1;0)$

      $ 1$ is not like other numbers becouse $ 1 \equal{} 1^{2} \equal{} 1^{3} \equal{} ........$ so if you  look $ 1^{4}$ you can expres it as you want

 a power with base 1 . And there is not true that there is no solution $ (x; y)$ for the equation,if you refere to Fermat Last Theorem.

 This is what I think.     (friendly)


floor function
function
algebra
ceiling function

If f(x) = int(x/2), find a - c below.

NOTE: What on earth is int(x)?  What does it stand for?

FIND:

(a) f(1.2)

(b) f(1.6)

(c) f(-1.8)

$\text{int}(x)= \left\lfloor x \right\rfloor$ is the function for the greatest integer less than or equal to $x$, also known as the "floor function" (imagine $x$ was plotted in a building with integer-numbered floors; $\left\lfloor x \right\rfloor$ would be dropping down to the first possible solid floor.)

An example:
$\left\lfloor \pi \right\rfloor = 3$

EDIT: Oops, wrong function. $\text{int(x)}$ is the [i]integer[/i] part of $x$. So for positive numbers it is the floor function; for negative integers it is the "ceiling function," which is related to the floor function. (e.g. $\text{int(-7.2)}=-7$)

[quote="Interval"]Sorry but I don't follow.

Can you answer my question and tell mw what you are doing each step of the way?

Thanks![/quote]

Sure.

(a)
Plug in the value for $x$: $f(1.2) = \text{int}(\frac{1.2}{2})$
Divide: $\text{int}(0.6)$
The function $\text{int}(x)$ gives the integer part of $x$, i.e. $x$ without any of the decimal places.
So, $\text{int}(0.6) = \boxed{0}$

Follow the same steps for (b) and (c).
If, though, you get something like $\text{int}(7.123 \ldots)$, just chop off the fractional part (to the right of the decimal), to get $7$.


Harvard
college
probability
geometry
geometric transformation
reflection
search

I know the subject of Harvard has been discussed before, but I don't think there has been much discussion on the direct relationship of the two.. If anyone got accepted to Harvard, and doesn't mind talking about their SAT scores etc, that would be very helpful.

Thanks!

im only going into 10th grade, but the senior in my school that got accepted to harvard had an sat score of 1570, 800 math and 770 verbal.

The book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0674016203/learninfreed]The Early Admissions Game[/url] includes a detailed study of this issue. The short answer is that the PROBABILITY of admission goes upward as SAT scores increase. That probability gets to be near 50:50 at the highest level of SAT scores, which means someone with the highest grades in the toughest courses, great extracurricular activities, and those peak SAT scores has a reasonable, but STILL not certain, chance of admission.

Harvard acceptance depends on many other things. there are thousands of people who apply to Harvard with perfect or near perfect SAT's yet only 2000 get in.

Here's a link to year-old information from the College Board about the number of people who received various single-sitting SAT scores in a recent testing year. Note that by Harvard's methodology, which gives you the benefit of the highest score you have ever obtained on each section of the SAT considered separately, there are more 1600 scorers than by the College Board's records of single-sitting scores.

[url=http://www.collegeboard.com/prod_downloads/about/news_info/cbsenior/yr2003/pdf/2003CBSVM.pdf]SAT Percentile Ranks for Males, Females, and Total Group[/url]

I have not seen a document like this yet that reflects the "new" SAT with its total possible score of 2400.

The basic picture continues to be that it is advantageous to score as high as you can on the SAT, but also to have good grades in high school  :mad:  and extracurricular activities in which you achieve well. And for all that, admission still isn't a sure thing.

Good luck on the applications.

For college application do they only take your highest grade on SAT's and all SAT II's, or do they notice if you take the SAT's like, 5 times before getting a decent score?

Harvard's answer specifically is that Harvard always gives you the benefit of the highest score you achieve on each test section, no matter how many times you have taken the test. But of course that means that all the people against whom you are competing for admission have the same advantage. (Didn't I just answer this question in this thread, implicitly, in my last reply?) Most highly selective colleges have exactly the same policy. Most applicants to most of the really selective colleges have taken the SAT more than once, for sure. (For one thing, a lot of highly able students have taken the SAT in talent search testing at middle-school age, although those scores don't remain on their permanent College Board records.)

However, if you took the SAT five times in your junior and senior years, most college admission officers would start to wonder what else you do with your Saturday mornings. Everyone is allowed to make up for a bad day, or for inexperience in taking that kind of test, but you really shouldn't plan on making taking the SAT your main extracurricular activity. For many people here on AoPS, doing more AMC preparation might be a better use of time than doing a lot more SAT I preparation. The Harvard supplement to the Common Application has an optional question asking what books you have read in the last year, so it might be a good idea to choose some worthwhile books to read rather than signing up for two more sessions of the SAT.

The SATs have been watered down, not to mention "race-normed", over the years and are no longer an effective discriminator at the top levels.   The number of people scoring 800+800 is several orders of magnitude higher than 30-40 years ago, when it was quite rare and of greater prestige than, say, USAMO qualification today. The publication of past SATs and the larger number of college applicants has something to do with this, but the tests are obviously much easier.

Most of the elite US universities have made public statements over the years about reducing the weight given to SAT scores.   Some schools, I think including Harvard, do not require the exam at all.

It is a step backward, in my opinion.

my friend got accepted to harvard with a 1550 SAT but i'm sure he got in because of all the extra-curricular activities he did and his great acedemic grades.

Exactly. You cannot make a causal statement from correlational evidence.

[quote="fleeting_guest"]Most of the elite US universities have made public statements over the years about reducing the weight given to SAT scores.   Some schools, I think including Harvard, do not require the exam at all.[/quote]

I think they still require the tests.. However, I think MIT doesn't look at your writing score. But Harvard, they require SAT I and 3 SAT IIs at least.

[quote][quote]Most of the elite US universities have made public statements over the years about reducing the weight given to SAT scores.   Some schools, I think including Harvard, do not require the exam at all.[/quote]
I think they still require the tests.. However, I think MIT doesn't look at your writing score. But Harvard, they require SAT I and 3 SAT IIs at least.[/quote]

That can't be true as written, because almost any US school would accept the ACT (a different exam used in the Midwestern states) in place of the SAT.  Also, it is probably illegal for a nonprofit organization such as a university to solicit business for a specific corporation such as the Educational Testing Service.    So the question is really "is some SAT-like standard test required".

I think that what you call SAT II is what were formerly called College Board Achievement Tests and cover standard high school subjects.    These tests were never as controversial as the SAT itself, and when universities talk(ed) about reducing the weight of the SAT they meant the general-purpose exam.   Actually, I think that is still what people usually refer to as "SAT", with its spectre of evil multiple choice questions and sharply pointing discriminatory fingers.

About MIT, I doubt very much that they discount any writing scores, and in fact they increased the weight of verbal scores compared to the olden days.  One reason (among several) was that it promotes the admission of more women.

Edit:   I see from the web site that Harvard does now require some version of SAT/ACT presumably including the general-purpose exam.   I think that is a change from earlier years, but don't know that for certain.

Here's a link to Harvard's policy on admissions tests:

[url=http://www.admissions.college.harvard.edu/faqs/admissions/tests/]Harvard College Undergraduate Admissions: Frequently Asked Questions: Admissions[/url]

"As always, students must also submit three SAT II exams of their choosing. The Admissions Committee will continue to consider all test results in light of students' educational opportunities."

MIT's site generates links that break AoPS's coding for link text, but the relevant place on its site says:

 "Standardized tests are required for any freshman application to MIT. . . .

. . . .

"For entry year 2006: SAT I (either the old version or the new version, available beginning March 2005) or the ACT (with or without the optional writing test). In addition, we require three SAT II Subject Tests: one in math (level 1 (Ic) or 2 (IIc)), one in science (physics, chemistry, or biology e/m), and the third in any area of your choosing.

"For entry year 2007 and beyond: SAT I or the ACT with the writing test. In addition, we require two SAT II Subject Tests: one in math (level 1 (Ic) or 2 (IIc)), one in science (physics, chemistry, or biology e/m)."

(MIT's Web site designers need some SERIOUS usability training about how to build links that can be emailed and pasted into online message boards.   :mad:    )

BTW, I should note that the SAT is NOT "race-normed" as that term is usually understood. In other words, anyone with the same item content score will get the same scaled score for a given test administration, regardless of the test-taker's "race." (A different, NOT "color-blind" scoring rule applies [or has applied] in United States Department of Labor job-proficiency testing, which is the context in which the term "race-norming" originated.) The College Board does take great care to experiment with new item content for the SAT test in broad sample populations that include all United States ethic groups, which is something it formerly did not do.

the only person from my high school who got into harvard last year had a 1460 total SAT score. she also had the 4th highest average in the graduating class though! admission might also depend on who is competing against you from your school or your geografic area...

Well my brother is going to Harvard, he got a 1600 on his SAT. But although SAT scores are important, extracurriculars are extremely important, but you have to be very good at something. Also, because we have to be racially equal or something like that, some people can get lower scores and get in, or if a person is in a worse financial position, etc.etc. There are many factors that an admissions council looks at, not just at SAT score. But if you have perfect score it does help a lot.

If you live in SoCal area you might want to look into this one guy, Mr.Kwon, SAT tutor.

Hi,

I think that Harvard, Yale, Princeton, and MIT work in very mysterious ways in terms of their admissions... There isn't any 1 thing that you can do or write that will get you in. I'm not sure I bye the whole "well-rounded" idea, or that you have to be great at something, etc. Actually, I have a little more information then most would, for Yale at least. I talked to my admission's officer and he basically told me why I got in. On my application, I wrote an essay about how I am an active participant at my local synagogue. So my admissions officer told me that what helped me was my passion for math combined with my passion for my religion. I know that that wasn't all that factored in... I had good grades, good extra curriculars including ambulance corps, etc. Now that's funny because I'm not a religious person...  I"m obviously amongst a lot of people, and I can give examples of why they got in: One's a trapeze artist, another traveled to Cambodia in order to work in an emergency room there for 2 months, another's a Canadian chess champion, and the list goes on.

I don't think SAT scores really matter that much at all... Just as long as you're in a certain acceptable range. The admissions process is just random. Funny thing is that you can do just as well at a school without as big a name. And from what I hear, in the real world, no one cares what school you go to; it's how you actually do at work. If anything, your grad school makes all the difference... so while it might help you land that first job, for all your subsequent jobs, no one cares about your schooling, it's all about how you did in that first job. So what I'm saying is that while the college admissions process is important, it is not the deciding factor in anything.


Can we prove $ (a^3\plus{}b^3\plus{}c^3)^2\geq abc(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)$ for positive reals $ a,\ b,\ c$?

By Holder,

$ \sum\frac {a^6}{b^2 \plus{} c^2}\ge \frac {(a^2 \plus{} b^2 \plus{} c^2)^3}{3\cdot 2(a^2 \plus{} b^2 \plus{} c^2)}$

$ \equal{} \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{6}\ge \frac {a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}{2}$

$ \ge \frac {(ab)(bc) \plus{} (bc)(ca) \plus{} (ca)(ab)}{2}$

$ \equal{} \frac {abc(a \plus{} b \plus{} c)}{2}$.

EDIT: [quote="kunny"]Can we prove $ (a^3 \plus{} b^3 \plus{} c^3)^2\geq abc(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)$ for positive reals $ a,\ b,\ c$?[/quote]

Yes. Use $ 3(a^3 \plus{} b^3 \plus{} c^3)\ge (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)$, square both sides and apply $ (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)\ge 9abc$.

[quote="nayel"]EDIT: [quote="kunny"]Can we prove $ (a^3 \plus{} b^3 \plus{} c^3)^2\geq abc(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)$ for positive reals $ a,\ b,\ c$?[/quote]

Yes. Use $ 3(a^3 \plus{} b^3 \plus{} c^3)\ge (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)$, square both sides and apply $ (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)\ge 9abc$.[/quote]

Thanks! We can use Cauchy-Schwarz,

$ \frac{a^6}{b^2\plus{}c^2}\plus{}\frac{b^6}{c^2\plus{}a^2}\plus{}\frac{c^6}{a^2\plus{}b^2}$

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

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

$ \geq \frac{(a^3\plus{}b^3\plus{}c^3)^2}{2(a^2\plus{}b^2\plus{}c^2)}\geq \frac{abc(a\plus{}b\plus{}c)}{2}$ which is true by nayel.

[quote="Yuriy Solovyov"]Let $ a, b,$ and $ c$ be positive real numbers. Prove the inequality
$ \frac {a^6}{b^2 \plus{} c^2} \plus{} \frac {b^6}{a^2 \plus{} c^2} \plus{} \frac {c^6}{a^2 \plus{} b^2}\geq \frac {abc(a \plus{} b \plus{} c)}{2}.$[/quote]
using tchebchev is more easier
s>=(a^4+b^4+c^4)/2>(abc(a+b+c)/2)
n.b a^n/(b+c)+b^n/(a+c)+c^n/(a+b)>(a^(n-1)+b^(n-1)+c^(n-1))/2
االه اكبر


inequalities
geometry
inequalities solved

If $ m_{a}$, $ m_{b}$, $ m_{c}$ are the medians of a triangle $ ABC$, then prove that

$ \frac{m_{b}\cdot m_{c}}{b\cdot c}\plus{}\frac{m_{c}\cdot m_{a}}{c\cdot a}\plus{}\frac{m_{a}\cdot m_{b}}{a\cdot b}\geq\frac{9}4$.

I presume ma is the median.

I think this works :

A classical result is Sum ma = 3/4 Sum a

C.S : Sum ma*mb/a*b Sum ab >= (Sum sqrt(ma*mb))

So it's enough to prove : 4 (Sum sqrt(ma*mb)) >= 9 Sum a*b

Rearrangement + C.S : 4 (Sum sqrt(ma*mb))) <= 4 (Sum ma) <= 4*3 Sum ma = 9 Sum a

Now Sum a >= Sum ab (by rearrangement for instance)

QED and equality iff ma=mb=mc (= iff a=b=c)  :)

Don't you see that the solution is wrong? The sign is reversed at least two times.
   This problem is very hard and surely it cannot be proved with Cauchy.

I think this ineq has elementary proof, but I will give nice proof using complex number:

From (m-a)(m-b)/(c-a)(c-b) + (m-b)(m-c)/(a-b)(a-c) + (m-c)(m-a)/(b-c)(b-a) = 1 we have  |(m-a)(m-b)/(c-a)(c-b)| + |(m-b)(m-c)/(a-b)(a-c)| +   |(m-c)(m-a)/(b-c)(b-a)| = 1, i.e. MA.MB/CA.CB + MB.MC/AB.AC + MC.MA/BC.BA >=1 for all M, A, B, C. Using this fact for triangle ABC and M = center of mass and we have done.

Namdung

Splendid solution. I also tried a lot to find an algebraic proof, but I didn't succeed. Again, this is memorable solution and also a memorable problem.

Thanks!!  I  spend  too  much  time  but  ...nothing  at all.
Thanks  again.

This one is easy. Take a+b+c=2 and x=1-a, y=1-b, z=1-c. After some computations you can reduce it to xy+yz+zx<=(9xyz+1)/4, which holds by Schur. I think that none of these two implyes the other. But I'm not sure.

Here  is  what  i  think :
Let  ma , mb , mc  meet  at  G.Then [ABG]=2/9ma*mb*sin(AGB) , but [ABG]=1/3[ABC]=1/6absinC , so  we  have (ma*mb)/(a*b)=(3sinC)/(4sin(AGB)) . So  we  only need  to  prove  sinC>=sin(AGB)  or
Σ[sinA/sin(AGB)]>=3.  But....??

how about this inequality?
mamb+mbmc+mcma>=3sqrt3 S
S is the area of triangle.
can it be used for this one?

S is a very small quantity in a triangle, so I really doubt that this could help. Anyway, Namdung solution is far too beautiful.

From Ptolemy we have mb*mc \geq bc/4+a 2 /2
The same for the others and adding we get:

ma*mb/ab + ..... \geq 3/4+  \sum a^2/2bc \geq 9/4
 \sum a 2 /2bc \geq 3/2 which is obvious

Again, this can be derived as an application of:

http://www.mathlinks.ro/Forum/viewtopic.php?t=15558

Let $\displaystyle x=\frac{a}{m_a}, y=\frac{b}{m_b}, z=\frac{c}{m_c}$ and M be the centroid, then

$\displaystyle
\left(\frac{a}{m_a} + \frac{b}{m_b} + \frac{c}{m_c}\right)\cdot \frac{4}{9}\cdot\left({a}{m_a} + {b}{m_b} + {c}{m_c}\right)\geq \frac{abc}{m_am_bm_c}\left({a}{m_a} + {b}{m_b} + {c}{m_c}\right)
$

which is equivalent to the desired result.

[quote="manlio"]In my notesbook I found a proof for

 (m_a)^2/bc  +(m_b)^2/ca + (m_c)^2/ab \geq 9/4

Do you think it is related with the forum inequality?[/quote]

This looks like  the same problem as the one posted at:

http://www.mathlinks.ro/Forum/viewtopic.php?t=2897

[quote="manlio"]In my notesbook I found a proof for

 (m_a)^2/bc  +(m_b)^2/ca + (m_c)^2/ab \geq 9/4

Do you think it is related with the forum inequality?[/quote]

Yes, it is related. Actually,

$\displaystyle \frac{m_a m_b}{ab} + \frac{m_b m_c}{bc} + \frac{m_c m_a}{ca} \geq \frac{m_a^2}{bc}  + \frac{m_b^2}{ca}  + \frac{m_c^2}{ab}$.

So your inequality is stronger than the posted inequality. Can you prove this?

It seems like a lot of problems have been posted here with only topic number 15558 needed for solution :)

besides fuzzy, I think it's ma[b]n[/b]lio ;)

[quote="manlio"]Harazi, can you please explain to me how to obtain the equivalent form

$\displaystyle \sum{\frac{m_a\cdot m_b}{ab}\geq\frac{9}{4} }$

Thank you very much.[/quote]
I think The general ineq is:
$\sum{\frac{MA.MB}{ab}} \ge 1$ for all M inside triangle ABC.
Let M be G.

[quote="zhaobin"]for P is any point in the triangle we can prove that PA*PB/a*b+PB*PC/b*c+PC*PA/ca >= 1
let P=G(G is barycenter of the triangle),use GA=2/3MB
is that right? :)[/quote]

PA*PB/a*b+PB*PC/b*c+PC*PA/ca  >= 1
can you kindly show me the proof?
[Moderator edit: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=6037 for proofs of this inequality.]

let Z_1=A,Z_2=B,Z_3=C,we have the equation sum(Z-Z_1)(Z-Z_2)/(Z_3-Z_1)(Z_3-Z_2)=1
is it enough? :)

After multiplying by $\frac{abc}{PA.PB.PC}(a.PA+b.PB+c.PC )$, we get

$a^2+b^2+c^2+\frac{bc}{PB.PC}(PB^2+PC^2-a^2)+...\geq 0$,

$a^2+b^2+c^2+2bc.cosBPC+2ca.cosCPA+2ab.cosAPB\geq 0$,

$(a+b.cosAPB+c.cosCPA)^2+(b.sinAPB-c.sinCPA)^2\geq 0$.


analytic geometry
trigonometry
complex numbers

You may call it a complex plane. It's a plane with a real axis and a imaginary axis.

If $w=x+yi$, then

\begin{eqnarray*}|x^{2}+2xyi-y^{2}-i|=1 &\iff& (x^{2}-y^{2})^{2}+(2xy-1)^{2}=1\\ &\iff& x^{4}-2x^{2}y^{2}+y^{4}+4x^{2}y^{2}-4xy+1=1\\ &\iff& x^{4}+2x^{2}y^{2}+y^{4}-4xy=0\\ &\iff& (x^{2}+y^{2})^{2}-4xy=0\end{eqnarray*}

Switching to polar coordinates $x=r\cos\theta, y=r\sin\theta$, we get

$r^{4}-4r^{2}\sin\theta\cos\theta=0\iff r^{2}=2\sin 2\theta$

which is an equation of a [i]lemniscate[/i] (more on that curve [url=http://mathworld.wolfram.com/Lemniscate.html]here[/url]) which has a basic shape resembling a number 8. In this particular case, the "bulbs" of the "eight" are located in first and third quadrants (the curve's symmetrial axes are $y=\pm x$)

[quote="drunner2007"][quote="OHO"]If $|z-i|=1$ and $w^{2}=z$, describe the locus of $w$ in the Argand diagram.[/quote]

can you clarify what the argand diagram is?[/quote]

The Argand plane is an x-y plane in which one may graph complex functions. The real part is plotted as the x coordinate, and the imaginary part is plotted as the y coordinate.

To solve the problem at hand find w in polar coordinates and then convert them to Cartesian, x-y, coordinates. Pick several angles around the unit circle, find x and y for these angles, and then connect the points like a dot-to-dot picture.


Let $S(n)$ be the sum of the digits of $n$. If for some integer $n$ we have that:

$S(n)=50$ and $S(15n)=300$

Find $S(4n)$

[quote="me@home"]$n \equiv 5 \pmod 9.$
$15n \equiv 3 \pmod 9.$
$4n \equiv 2 \pmod 9.$
construct $\frac{10^{51}-1}{9}$
now $15n=16666....66665$, digit sum 300 (we multiply by 6 from previous so a row of 6's makes sense, but 1 and 5 balance it.
now $4n=\frac{4(10^{51}-1)}{9}, s(4n)=200$[/quote]

well ,,, how do u arrive at such a nu.. is der ne specific waqy.. will it always work?????

The way I did it is certain to not always work, I just noticed some thing about the numbers they gave... I think that is how the problem was intended to be solved. I"m not sure if there is a very general method... if there is it probably uses modular arithmetic stuff like he said mod 9.


Determine $m$ so that $4x^2-6x+m$ is divisible by $x-3$. The obtained value, $m$, is an exact divisor of:

A. 12
B. 20
C. 36
D. 48
E. 64

[hide="Answer"]By polynomial division, we have $(6x+m)|(x-3)\Rightarrow m=-18$, which is a divisor of $36\Rightarrow \boxed{C}$.[/hide]

I hope I got the notation correct for divisibility. :?

[hide]By synthetic division, we find that $m+18=0$, and $m=-18$, and is a divisor of 36 (negative divisor? first time I've heard of that  :P )[/hide]

[quote="236factorial"] we find that $m+10=0$, and $m=-18$[/quote]
You know all the times chess64 and I told you to get out of classroom math?
$Go$ $Back$

 :P

Alright, calm down guys, he probably wrote $m+10=0$ instead of $m+18=0$. :roll:

[quote="JesusFreak197"][hide="Huh?"]$-18+10=0$? :what?:[/hide][/quote]

Sorry, I was into my own math world for a second  :rotfl:


can u guess the next number?

$10, 20 , 3 , 15 , 1060 ,...$

Is this meant to be a clever puzzle or is it just too hard for a middle school forum?  Let me know so that I know which forum to move it to.

dear mathew i choose this simple problem for the  middle school...


Let a right triangle  $ AB\Gamma$, με $ \widehat{A} \equal{} 90^{\circ}.$ Find the minimum of the expression  $ \displaystyle P \equal{} \frac {\beta^2(\gamma \plus{} \alpha) \plus{} \gamma^2(\beta \plus{} \alpha)}{\alpha\beta\gamma}$

Please define alpha beta and gamma. Are they the sides or angles ?

$ \mbox{If } \alpha\equal{}\angle{A}\equal{}90^0, \beta\equal{}\angle{B} \mbox{ and } \gamma\equal{}\angle{B} \mbox{, then: } P\ge 3.$

hello, i think $ \alpha,\beta,\gamma$ are the side length of the given triangle.
Sonnhard.

Mmm !

  Yes ! $ \alpha , \beta  , \gamma$ are the sides of the triangle. I had the text in Latex (in  greek) and I could not image

that this would cause confusion. In Greece the angles are noted with kapital letters.

Really sorry - Babis

[b]Answer : Min = $ 2\plus{} \sqrt 2$ [/b]

OK ! It is form th journal Math and Yought and in the new issue there is also a new proof.

 Babis


There are $ 6n \plus{} 4$ mathematicians participating in a conference which includes $ 2n \plus{} 1$ meetings. Each meeting has one round table that suits for $ 4$ people and $ n$ round tables that each table suits for $ 6$ people. We have known that two arbitrary people sit next to or have opposite places doesn't exceed one time.
1. Determine whether or not there is the case $ n \equal{} 1$.
2. Determine whether or not there is the case $ n > 1$.

I think 6n+4 mathematicians is more adequate?

Oh, yep, I am sorry for my blindness  :blush:  It must be $ 6n\plus{}4$ instead of $ 6n\plus{}1$. I will tell Orlando to edit the problem statement. Thank you!

I think that problem is quite easy.
Both answer are "yes"

Divide $ 6n\plus{}4$ people into $ 2n\plus{}2$ "teams": one team of one member(for convenience we call him $ A$), and $ 2n\plus{}1$ teams of three member.

It is obvious that we can arrange a tournament of $ 2n\plus{}1$ rounds such that any two teams exactly meet once in the tournament.

Now we use the arrangement of $ 2n\plus{}1$ rounds to arrange the $ 2n\plus{}1$ meetings:

In each round there is a team meeting $ A$. Let these four people sitting at the 4-person table; other $ 2n$ teams take part in $ n$ matches. For any two these teams meeting each other, let the six people sitting at the 6-person table such that everyone's both neighbors are of the other team.

Finally check that this arrangement satisfies the condition.
We claim that for any two people, they sit next to or have opposite places exactly once.(for convenience we call this relationship "meet")

For $ A$ and another one, their unique "meet" occurs when they both sit at the 4-person table.

For any two people other than $ A$,
if they are of the same team, their unique "meet" occurs when they both sit at the 4-person table.
if they are of different teams, their unique "meet" occurs when their teams meet. In other words, it occurs when their two teams sit at one 6-person table together.
So our claim is proved and our arrangement is enable.
Q.E.D.


Prove that for every [img]http://www.mathlinks.ro/Forum/latexrender/pictures/dde189e713fc6dd76fedcc9c6f155100.gif[/img]

we have

[img]http://www.mathlinks.ro/Forum/latexrender/pictures/0d0622f1b7cf0e72f2c6a6f252043808.gif[/img]

Alexandros

We have \[a^3 + b^3 + 4 = a^3 + b^3 + 1 + 1 + 1 + 1 \geq 6\sqrt[6]{a^3b^3} = 6\sqrt{ab}\] and hence \[\frac{1}{a^3 + b^3 + 4} \leq \frac{1}{6\sqrt{ab}}.\] Add this inequality and the 2 analogous inequalities. We need to prove that \[\frac{1}{\sqrt{ab}} + \frac{1}{\sqrt{bc}} + \frac{1}{\sqrt{ca}} \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\] or \[\left(\frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}}\right)^2 + \left(\frac{1}{\sqrt{b}} - \frac{1}{\sqrt{c}}\right)^2 + \left(\frac{1}{\sqrt{c}} - \frac{1}{\sqrt{a}}\right)^2 \geq 0\] and we're done.

This is the best solution .I have done the same solution as you !

[u]To the greek students:[/u]

      See the solution for problem 4.61 , page 206 ,  in the book << N. Skombris and B. Stergiou :  Algebraic Inequalities , Edition Savvalas , Athens 2004 >>


In each post you may only post one new record in each category. It must have fewer

[b]3-letter "word":[/b]
267000 [u]xqz[/u] leoxnlin

[b]4-letter "word":[/b]
28400 [u]xqxq[/u] leoxnlin

[b]5-letter "word":[/b]
17000 [u]zzxxx[/u] leoxnlin

meh didn't spend much time testing.

[b]no-letter word[/b]

-_- has 2 hits on google

mwahahaha

i win

i think if you have no results it shouldnt count

We're only doing 3, 4, and 5 letter words. And they must be all letters. And I'm pretty convinced that there are no 3, 4, or 5 letter sequences that have no results, after 10 seconds of searching. now THAT'S convincing. And if you find one, YOU WIN!!!!!!! Anyways, try to find the lowest.

[b]3-letter "word":[/b]
267000 [u]xqz[/u] leoxnlin
 :arrow: 122000 [u]zqx[/u] leoxnlin

[b]4-letter "word":[/b]
28400 [u]xqxq[/u] leoxnlin
 :arrow: 14100 [u]qxqx[/u] leoxnlin

[b]5-letter "word":[/b]
17000 [u]zzxxx[/u] leoxnlin
 :arrow: 492 [u]qzqzx[/u] leoxnlin

[b]3 letter word [u]vzq[/u]: 90,000[/b]
112,000 for qmy
102,000 for qvz
90,000 for vzq
98,700 for zvj

[b]4 letter word [u]zhvq[/u]: 1,110[/b]
1,310 for vjzq
1,690 for hvzq
1,490 for hzvq
1,600 for hvqz
1,280 for dzvq
1,280 for vjzp
1,260 for hzqv
1,210 for qzhv
1,110 for zhvq

[b]5 letter word [u]vqzjj[/u]: 10[/b]
24 for vqzph
42 for vqhzp
38 for hvqzp
30 for vqzjh
28 for zvjqh
17 for zqhvj
20 for qzhvv
21 for vvqhz
10 for vqzjj

the first four-letter word i tried had 184 million hits

:\

[quote="herefishyfishy1"]qzbyx returns only 4 results.

This reminds me of one time when I tried to find a search on Google consisting of two real, unquotationmarked words that returned exactly one result.[/quote]

i found 58 for qzbyx.

EDIT: also, [url]http://www.google.com/search?hl=en&q=lxhlq[/url] is 40

Accomplishments:
55 for qozxy
65 for prnqx
15,200,000 for pwn :(
56 for yxzoq
44 for xqnrp
38 for oxezq!
32 for qsrdz
28 for zkqmd
8 for ozcqu!!!

sorry thats the best I can do.

[quote="ZetaX"]You know, you are actively enlarging the number of hits  :wink:[/quote]Apparently Google likes searching through AoPS :D

[quote="isabella2296"]5 for xyqwer[/quote]

xyqwer has six letters. :wink:

31 for qarzx
53 for vzqkz
22 for qzxkx
25 for fkzxq
26 for qovnz

these were my best. obviusly i fail

EPIC PWN!
dgujz has 4 hits... maybe now 5 cuz of this post

pkxuz is glitching out... it says 1-5 of 0

this was a good use of 10 minutes...


The problem gives the following:
Bayes Theorem:
$ P (A|B) \equal{} \frac {P(B|A) P(A)} {P(B|A) P(A) \plus{} P(B|A') P(A')}$

P(S) = 30%
P(M|S) = 40%
P(M|S') = 1%

a) P(M' | S')?
[i]b) P(S|M)?[/i]
c) P(S|M')?

"b" seems obvious...
I couldn't do "a" and "c".
I suppose they are related and to solve them I would need to solve a systems with 4 equations or something like that, is that the way?

OBS: P (A) + P(A') = 1 (I think)

EDIT: Sorry, latex ate the % signs....

first of all, $ P(S) \equal{} 30$, $ P(M|S) \equal{} 40$ don't make any sense because a probability is always $ \leq 1$.

It says 30 PERCENT and 40 PERCENT.

 :huh:

Yes, but as the initial post notes, it was edited because "latex ate the % signs."   :roll:

@Original poster: What have you tried?  If you show some initial effort, we can see what you need help on.
[quote]OBS: P (A) + P(A') = 1 (I think) [/quote]  Yes, this is right.  Note that also $ P(X \mid Y) \plus{} P(X' \mid Y) \equal{} 1$.  You can get a percent sign in LaTeX using \%, but really, knowing those values is completely pointless and we could just as easily proceed with the values $ x$, $ y$ and $ z$ for those probabilities.

Sorry, I thought it would be silly to post my fruitless efforts  :(

Well, I think I got b right, because it is quite obvious, since Bayes theorems lets us calculate P (A|B) knowing P (B|A), right? Is was just a matter of doing it.

The problem I found with "a" and "c" was such that, when you apply bayes theorem it seemed we were lacking information (might be a wrong impression though)...

I mean applying Bayes to a and c pairs:

I. $ P (M'|S') \equal{} \frac {P (S'|M') P (M')} {P (S'|M') P (M') \plus{} P (S'|M) P(M)}$
II. $ P (S'|M') \equal{} \frac {P (M'|S') P(S')}{P (M'|S') P(S') \plus{} P (M'|S) P(S) }$
____

III. $ P (S|M') \equal{} \frac {P (M' | S) P (S)}{P (M' | S) P (S) \plus{} P (M' | S') P (S')}$
IV. $ P (M' | S) \equal{} \frac {P (S|M') P(M')}{P (S|M') P(M') \plus{} P (S|M) P(M)}$

I suppose this was why I thought the solution would be a system with 4 equations, but I not sure about it. The quantity of P's kinda of confused me...

EDIT: Oh yeah, I think I calculated P(M) and P(S'|M) by using the initial values and looking from an equation to another. What I did mostly was write the things (I,II... x) and look for relations...

Okay, if the number of P's is confusing you, don't start this way.  Instead, start by making a table of quantities you [i]do[/i] know.  Then, every time you apply Bayes' theorem, you can substitute, building up from the pieces you already understand.  For example, you wrote down lots of instances of $ P(S)$ and $ P(S')$, and these are numbers you already know (either are given, or can be trivially computed from them).  So, why don't you try that and see where it gets you?  Here are the first few things I would write down:

$ P(S) \equal{} \frac {3}{10}$
$ P(S') \equal{} ?$
$ P(M \mid S) \equal{} \frac {2}{5}$
$ P(M \mid S') \equal{} \frac {1}{100}$ (or was this a typo for $ \frac {1}{10}$?)
$ P(M' \mid S) \equal{} ?$
etc.
(though again note that the arithmetic with the numbers you're given may get messy, and it might actually be [i]nicer[/i] to use variables $ x, y, z$ in place of the three given numbers.)

One takeaway here is that you have a very powerful tool (Bayes' theorem) for computing certain kinds of probabilities, but it (like all other formulas in mathematics) isn't magic: there's a reason that it works, and there are circumstances under which it works, and it's much more important to understand why it's true and how to apply it than it is to memorize the magic formula of $ P$s, $ A$s and $ B$s.  (After all, in "real life," whatever that may mean for you, if you ever needed Bayes' theorem, you could always go look it up -- it's understanding the general principles of probability, and how to use it correctly, that are the important parts.)  You have a hammer -- now you need to get used to recognizing what things are nails ;)


A is  a binary compound of a univalent metal .1.422 g of A reacts completetely with 0.321g of Sulphur in an evacuated sealed tube to give 1.743 g of white crystalline Solid B that forms a hydrated double salt C with a compound D.
D shows the following reactions
D reacts with common salt to give E which
1)forms a Gelatinous white  preciptate with $ \ce{NaOH}$
2)it gives a white preciptate with $ \ce{BaCl2}$ which is insoluble in $ \ce{HNO3}$
Identify all the variables with necessary equations

By "univalent metal" you mean a metal that forms only monovalent cations (and so an alcaline metal), or do you mean a metal that in compound A is in oxidation state +1?

is it some alum?
Saw the same question in my Profs question list ( answers not discussed yet  :D  )

The white precipitate with barium chloride may indicate that a sulphate ion is involved.


I don't have an answer to this.  I came across the following series in calculating the density of a certain set of positive integers.

Can you find the value (or maybe just simplify) the following series (where $b>1$ is a positive integer) ?

\[\sum_{n=1}^\infty \frac{ (-1)^{n}}{ b^{2^{n+1}-1}-1 }\prod_{k=1}^{n}\frac{ b^{2^{k}-1}}{ b^{2^{k}-1}-1 }\]

Thanks for any input.

Tom

I see nothing.  What is the sequence of integers whose density you're trying to calculate?

It calculates (if I computed correctly) the density of Generalized Smarandache Palindromes (see, e.g., http://www.gallup.unm.edu/~smarandache/GSP.htm) in base $b$.

This problem is of the recreational number thoery variety.


For each positive integer $\,n,\;S(n)\,$ is defined to be the greatest integer such that, for every positive integer $\,k\leq S(n),\;n^{2}\,$ can be written as the sum of $\,k\,$ positive squares. [list=a] [*] Prove that $S(n)\leq n^{2}-14$ for each $n\geq 4$. [*] Find an integer $n$ such that $S(n)=n^{2}-14$. [*] Prove that there are infinitely many integers $n$ such that $S(n)=n^{2}-14$. [/list]

Solution at http://www.kalva.demon.co.uk/imo/isoln/isoln926.html
(a)  Let $ N \equal{} n^2$. Suppose we could express N as a sum of N - 13 squares. Let the number of 4s be a, the number of 9s be b and so on. Then we have 13 = 3a + 8b + 15c + ... . Hence c, d, ... must all be zero. But neither 13 nor 8 is a multiple of 3, so there are no solutions. Hence S(n) ≤ N - 14.

A little experimentation shows that the problem is getting started. Most squares cannot be expressed as a sum of two squares. For $ N \equal{} 13^2 \equal{} 169$, we find: 169 = 9 + 4 + 4 + 152 1s, a sum of 155 = N - 14 squares. By grouping four 1s into a 4 repeatedly, we obtain all multiples of 3 plus 2 down to 41 (169 = 9 + 40 4s). Then grouping four 4s into a 16 gives us 38, 35, ... , 11 (169 = 10 16s + 9). Grouping four 16s into a 64 gives us 8 and 5. We obtain the last number congruent to 2 mod 3 by the decomposition: $ 169 \equal{} 12^2 \plus{} 5^2$.

For the numbers congruent to 1 mod 3, we start with N - 15 = 154 squares: 169 = 5 4s + 149 1s. Grouping as before gives us all 3m + 1 down to 7: 169 = 64 + 64 + 16 + 16 + 4 + 4 + 1. We may use $ 169 \equal{} 10^2 \plus{} 8^2 \plus{} 2^2 \plus{} 1^2$ for 4.

For multiples of 3, we start with N - 16 = 153 squares: 169 = 9 + 9 + 151 1s. Grouping as before gives us all multiples of 3 down to 9: 169 = 64 + 64 + 16 + 9 + 9 + 4 + 1 + 1 + 1. Finally, we may take $ 169 \equal{} 12^2 \plus{} 4^2 \plus{} 3^2$ for 3 and split the $ 4^2$ to get $ 169 \equal{} 12^2 \plus{} 3^2 \plus{} 2^2 \plus{} 2^2 \plus{} 2^2 \plus{} 2^2$ for 6. That completes the demonstration that we can write $ 13^2$ as a sum of k positive squares for all k <= $ S(13) \equal{} 13^2 \minus{} 14$.

We now show how to use the expressions for $ 13^2$ to derive further N. For any N, the grouping technique gives us the high k. Simply grouping 1s into 4s takes us down: from 9 + 4 + 4 + (N-17) 1s to (N-14)/4 + 6 < N/2 or below; from 4 + 4 + 4 + 4 + 4 + (N-20) 1s to (N-23)/4 + 8 < N/2 or below; from 9 + 9 + (N-18) 1s to (N-21)/4 + 5 < N/2 or below. So we can certainly get all k in the range (N/2 to N-14) by this approach. Now suppose that we already have a complete set of expressions for $ N_1$ and for $ N_2$ (where we may have $ N_1 \equal{} N_2$). Consider $ N_3 \equal{} N_1N_2$. Writing $ N_3 \equal{} N_1$( an expression for $ N_2$ as a sum of k squares) gives $ N_3$ as a sum of 1 thru k_2 squares, where $ k_2 \equal{} N_2 \minus{} 14$ squares (since $ N_1$ is a square). Now express $ N_1$ as a sum of two squares: $ n_1^2 \plus{} n_2^2$. We have $ N_3 \equal{} n_1^2$(a sum of k_2 squares) + $ n_2^2$(a sum of k squares). This gives N_3 as a sum of $ k_2 \plus{} 1$ thru $ 2k_2$ squares. Continuing in this way gives N_3 as a sum of 1 thru $ k_1k_2$ squares. But $ k_i \equal{} N_i \minus{} 14 > 2/3 N_i$, so $ k_1k_2 > N_3/2$. So when combined with the top down grouping we get a complete set of expressions for N3.

This shows that there are infinitely many squares N with a complete set of expressions, for example we may take N = the squares of $ 13, 13^2, 13^3, ... .$

Solved c [url=https://artofproblemsolving.com/community/c6h60721p25022761]here[/url], I believe the others are not that hard.