Does there exist six 6-digit numbers with digits in the set $\{1,2,\dots,6\}$ (no digit come 2 times) that we can obtain each 3-digit number with digits in $\{1,2,\dots,6\}$ (no digit come two times) by removing 3 digits from one of the numbers?

Do you mean can we choose six permutations of 1..6 so that we can get every one of the 6c3 possible choices of three of those by removing digits in betwee? (i.e. 123456 can not become 561)

Yes, It's correct. You must give me six 6-digit numbers that make all $\binom 63$ 3-digit numbers.

Not exactly.
There are ${}_{6}P_{3}=6\times 5\times 4=120$ different numbers with $3$ different digits in $\{1,2,3,4,5,6\}$, because the ordering of the $3$ digits is important here.
Eliminating $3$ digits from a permutation of $123456$, we can obtain ${}_{6}C_{3}=\binom{6}{3}=\frac{6\times 5\times 4}{3\times 2\times 1}=20$ different 3-digit numbers.
So, if we could find $\frac{120}{20}=6$ permutations that solve the problem, it would show that $6$ is the minimum.
In that case, no $3$-digit combination should be repeated inside a permutation.

I think that if you pick any three permutations, you'll be able to find three digits that appear in the same order in two of the permutations.
In this case, according to the last post, the answer to the question of this problem would be negative.


calculus
integration
function
real analysis
ratio
derivative
real analysis solved

Let the function $ f:[0,\frac{1}{2}] \to R $ such that:
\[
f(x) = \left\{ \begin{array}{l}
 0,a_2 a_4 a_6 ....a_{2n} ....\quad if\quad x = 0,a_1 a_2 a_3 .... \in [0,\frac{1}{2}] \cap Q \\
 0,a_3 a_6 a_9 ....a_{3n} ....\quad if\quad x = 0,a_1 a_2 a_3 .... \in [0,\frac{1}{2}]\backslash Q \\
 \end{array} \right.
\]
Prove that the function isn't integrable

It's clear that the word "integrable" in this problem must refer to the Riemann integral. Since the function is constructed as a pointwise limit of measurable functions, it is Lebesgue measurable, and as a bounded function of a bounded interval, it is Lebesgue integrable.

Three questions Kent:

I) There are points in which the function is continuus?

II) If yes, is the function differentiable in these points?

III) O.K the function is Lesbegue integrable, but which is the value of integral?

[quote] I) There are points in which the function is continuus? [/quote]

Yes. For instance, 0.111111111....
By the way, the definition is ambiguous for some rational $x$:
if $x=0.09999999...=0.10000...$ what is $f(x)$: $0.99999...$ or $0.00000...$?

[quote]II) If yes, is the function differentiable in these points? [/quote]

No way: if you change the $6n$-th digit of $x$ by $1$ leaving all other digits intact,
it preserves the rationality of $x$, so, as the result, you change the $2n$-th or the $3n$-th digit of $f(x)$ by $1$. Thus the ratio of the increment of $f(x)$ to the increment of $x$ is either $10^{4n}$ or $10^{3n}$, i.e., as large as you want.
So there is no derivative anywhere.

[quote] III) O.K the function is Lesbegue integrable, but which is the value of integral?[/quote]

$\frac14$. For the Lebesgue integral the rational points do not matter.  Now
note that $f(0.5-x)+f(x)=1$ for all irrational $x$.


geometry
circumcircle
Euler
rhombus
radical axis
geometry proposed

[quote="campos"]Let $ ABC$ be a non-isosceles triangle with $ \angle A \equal{} 60^\circ$. Let $ \mathcal{O}$ be the circumcircle and $ H$ the orthocenter of the triangle $ ABC$. The bisector of angle $ A$ intersects $ \mathcal{O}$ again at $ M$, and $ MH$ intersects $ \mathcal{O}$ again at $ D$. Prove that $ AD, BC$ and the Euler line of triangle $ ABC$ are concurrent.[/quote]

[u]Hint\Solution[/u]

First of all, $ \angle BOC\equal{}\angle BMC\Longrightarrow BOC\equal{}BMC$ and $ HA\equal{}2ON\equal{}OM$ where $ N$ is the midpoint of $ BC$. Therefore we have shown that $ AHMO$ is a rhombus, which implies that $ ADHO$ is cyclic ($ C_1$). It is also well known that $ BHOC$ it is cyclic ($ C_2$). Now it is easy to show that the three lines are concurrent at a point $ P$ which is the radical center of $ O,C_1,C_2$).

Nick Rapanos


calculus
integration
logarithms
CMU
college
calculus computations

In this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=164417]topic[/url] about Vardi's integral it was shown

$ \int_{ \minus{} \infty}^\infty \frac {\ln(\alpha^2 \plus{} x^2)}{\cosh \pi x} dx \equal{} 4\ln\left(\sqrt {2} \, \frac {\Gamma\left(\frac34 \plus{} \frac {\alpha}{2}\right)}{\Gamma\left(\frac14 \plus{} \frac {\alpha}{2}\right)} \right)$

So $ \int_{ \minus{} \infty}^\infty \frac {\ln\big((\alpha\pi)^2 \plus{} t^2\big)}{\cosh t} dt \equal{} 4\pi \ln\left(\sqrt {2\pi} \,\frac {\Gamma\left(\frac34 \plus{} \frac {\alpha}{2}\right)}{\Gamma\left(\frac14 \plus{} \frac {\alpha}{2}\right)}\right)$

In particular for $ \alpha \equal{} 1$ :

$ \int_{ \minus{} \infty}^\infty \frac {\ln(\pi^2 \plus{} t^2)}{\cosh t}\, dt \equal{} 4\pi \, \ln\left(\sqrt {2\pi}\, \frac {\frac14\,\Gamma\left(\frac14\right)}{\Gamma\left(\frac34\right)}\right)$

It follows $ \int_0^\infty \frac {\ln(\pi^2 \plus{} t^2)}{e^t \plus{} e^{ \minus{} t}} \, dt \equal{} \pi\, \ln\left(\frac {\sqrt {\frac {\pi}{2}}}{2}\, \frac {\Gamma\left(\frac14\right)}{\Gamma\left(\frac34\right)}\right)$

Finally substitute $ t$ with $ \ln x$ .


Ross Mathematics Program
HCSSiM
number theory

I don't really think anyone has compared Ross and HCSSIM before & I've heard such good comments about both that I'm really in a muddle... If you have been to either one or possibly even both (which would be great  ;) ) please post comments on positives and negatives so I can choose, he he. Have a great summer and good luck on finals, everyone!!!  :lol:  :P

I have been to HCSSiM, and I have to say, it is a wonderful program!!It's so relaxing...You get to refer to the staff by their first name, and one by his last :D ...Anyone who goes is not disappointed...You do wonderful mathematics, and also have time to relax, hang out, go on trips,...The only bad thing about the program is that it ends...You will have met great people, made friends with everyone, and learned a great deal of mathematics...

There's info on HCSSiM in [url=http://artofproblemsolving.com/Forum/post-201306.html][u]this thread[/u][/url].

thanxx... how does it compare to ross though? i know hcssim is 6 wks & ross is 8 but did anyone have a great experience w// ross or is it more boring? thanxx in advance!!!

There are two major differences between HCSSiM and Ross, to my knowledge.

(1) Material covered.  HCSSiM does a survey for the first half, and students choose topics to concentrate in during the second half.  Ross centers the first-year curriculum around Number Theory, and does it very deeply.  (Ross' theme was to "think deeply of simple things.")
(2) Daily structure.  Ross has lecture in the morning, and then much of the day is unstructured; students work on problem sets and meet with junior counselors to talk about problem sets on their own schedule.  HCSSiM has interactive class in the morning, free afternoon, interactive lecture before dinner, and everyone does problem sets together in the evening.
So one might say that the time is less structured at Ross, and the material is less structured at HCSSiM :) though that might not actually be accurate.

Perhaps you prefer one of these structures or one of the contents to the other.  Certainly, there are some students for whom one program is a much better fit (and which one depends on the student, of course).  But as far as I know, participants love each of the programs and I know of excellent staff who've worked at each program (and some who've worked at both).

thank u smbelcas, u rock!!  :first:   Anyone else who has heard something or had an interesting experience, PLEASE POST!!! Thank you!  :P
ps. i LOVE these smileys!!!  :rotfl:  :what?:  :idea:  :!:

I have Ross right now and I'm waiting for a reply from HCSSiM.  If I do get both, which one should I go to?

Maybe it's better not to worry yourself right now, and actually wait to get a reply from Hampshire first...But, if you do get both,you're decision should be based alot on what type of surroundings and atmosphere you would prefer...I know that Hampshire has a more relaxing atmosphere...For Ross, not sure...You can read the opinions of smbelcas above though...

[quote="Stan17"]But, if you do get both,you're decision should be based alot on what type of surroundings and atmosphere you would prefer...[/quote]

Yes, I think there comes a point where the benefit of one program over another is limited to what you'll get out of it, however you determine that criteria to be.


geometry
vector
trigonometry
circumcircle
cyclic quadrilateral
geometry proposed

The points K,L,M are elements of sides AB,BC,CA of one triangle ABC such that:
AK/KB = BL/LC = CM/MA.
Prove that  the orthocenters of the triangles ABC and KLM are the same if ,and only if, ABC is one regular triangle.

Vector works fine for this problem :)
This is from the Czech-Slovak Polish Match.

Can you post Czech-Slovak Polish Match?
If you find one complete solution, please, post.

Firstly, the if part is quite easy, so I will just leave it here.
Let H be the orthocenter of $\triangle ABC$(so H is also the orthocenter of $\triangle KLM$).
$KM\cdot LH=(KA+AM)\cdot LH=KA\cdot(LC+CH)+AM(LB+BH)=KA\cdot LC+AM\cdot LB=KA\cdot LC\cos B-AM\cdot LB\cos C$. Thus, $KA\cdot LC\cosh=AM\cdot LB\cos C$. Similarly, $AM\cdot LB\cos C=BK\cdot CM\cos A$. Simplify it, we get $AM\cos C=BK\cos B$ and $BK\cos A=LC\cos C$(why ;) )
The conclusion follows then.
Sorry, but I'm not familiar with typing vector in LaTex... :blush:

Here is another geometry problem from Czech-Slovak-Polish Match 2004, which is much easier than the one above.

In the interior of a cyclic quadrilateral ABCD, a point P is given such that $\angle BPC=\angle BAP+\angle PDC$. Denote by E,F,G the feet of the perpendiculars from P to the lines AB,AD,DC respectively. Show that triangles FEG and PBC are similar.

This one is quite straightforward, so I'm not going to post the solution here :)

mecrazywong: thanks  :)

If you find, in english,Czech-Slovak-Polish Match 2004, please, post  :D

[quote="Tales"]mecrazywong: thanks  :) [/quote]
Thanks for your "thanks", Tales ;) (I really grow bored of saying welcome)
[quote="Tales"]If you find, in english,Czech-Slovak-Polish Match 2004, please, post  :D[/quote]
You can find all the problems from Czech-Slovak-Polish Match 2004 [url=http://www.math.ust.hk/excalibur/v9_n4.pdf]here[/url].

I don´t understand mecrazywong´s solution.
please, give me one other solution

thanks   :D

chavez

I'm sure you'll find it easy to show that they have the same centroid $G$. If they also have the same orthocenter $H$, then, since $\overrightarrow{OH}=3\overrightarrow{OG}$ (when $O,G,H$ are the circumcenter, centroid, and orthocenter of a triangle, respectively), they also have the same circumcenter $O$.

If $O$ is the circumcenter of $ABC$, we now know that $OK=OL=OM$. I think it's rather easy from here.


function
calculus
integration
complex analysis
calculus computations

The Gamma function, is the integral from 0 to infinity of

y^(a-1) e^(-y)dy, where a is alpha.

Why the Gamma function define for only a > 0 ?

I think the integral diverges for $\alpha \le 0$.

I don't think it is only defined for $\alpha>0$. As long as $\alpha\notin\mathbb{Z}^{-}$ it is defined.

The graph is here:
http://en.wikipedia.org/wiki/Image:Gamma_plot.svg

According to [url=http://mathworld.wolfram.com/GammaFunction.html]MathWorld[/url], the definite integral form only works for when $R[z] > 0$.

That's true - but the integral form is not the only form.

For $\text{Re}(z)>0,$ we can prove (by integration by parts) the gamma recursion:
\[\Gamma(z)=\frac{\Gamma(z+1)}{z}\]
But then for the strip $-1<\text{Re}(z)\le0$ (except for $0$), we can use this recursion to define $\Gamma(z).$ And having done that, to use the recursion to define $\Gamma$ in the next strip to the left, as so on.

As I said, there are other forms for the definition - some can be found in complex analysis textbooks. All the various forms can be shown to agree on the right hand half plane, and the principle of analytic continuation shows that the extension to the left must be unique.


Hallo,

ich halte einen Vortrag über Zerlegungsäquivalenz und wollte auf Hilberts drittes Problem und seine Lösung durch Dehn-Invarianten eingehen. Kennt ihr vielleicht eine leichtverständliche Einführung zu dem Thema?
Danke schon mal im Voraus für alle Vorschläge.

             Markus

kannst du den name den Problem sagen?...entsuldige fur mein deutsch :P

Das Problem heißt einfach drittes Hilbertsches Problem.
Es lautet:
Man gebe zwei Tetraeder an, die nicht ergänzungsgleich sind, oder beiweise, das ein solches Paar nicht existiert.

Ergänzungsgleich heißt, das man so äquivalente Tetraeder zu zwei gegebenen Polyedern hinzufügen kann, dass man so in Tetraeder zerlegen kann, dass jeder Tetraeder der einen Zerlegung genau einen äquivalenten  in der anderen hat (äquivalente Tetraeder in einer Zerlegung zählen mehrfach).

Ich hoffe das ist jetzt nicht allzu konfus.

Markus


geometry
inequalities
perimeter
inequalities unsolved

:huh:  I think an approach with Lagrange multipliers is possible for the worst case...

I have the feeling that a geometric interpretation might work but I am not so sure. Anyone?

Well, yeah.  It was originally a geometry problem :).  Here's the original problem.  I just thought it was an interesting inequality, and I wanted to see if any of you guys could use your crazy inequality skills to do it like that.

You're given a triangle ABC with a right angle at C and a 60 degree angle at B.  X is on BC, Y is on AC, Z is on BC.  Find the minimum perimeter of triangle ABC.

Edit: It would seem I read the problem wrong.  The goal was to minimize the longest of XY, XZ, and YZ.  Still, this inequality could be interesting anyway....


Euler
function
number theory
relatively prime
number theory unsolved

Find all $ q\in\mathbb{Q}$ such that $ q\equal{}\frac{2\phi \left( xy\right) }{\phi \left( x\right) \phi \left( y\right) }$ and $ x,y\in\mathbb{N}$

Notes: $ \phi \left( n\right)$ is the numbers of positive integers $ \geq n$ relatively prime with $ n$

it is straightforward to see that

\[ \frac{2\phi(xy)}{\phi(x)\phi(y)} \equal{} 2\cdot \prod_{p\text{ prime, }p|x \wedge p|y}\left(1\minus{}\frac{1}{p}\right)^{\minus{}1}
\]

so the set of $ q$ in question is the set of numbers of the form $ \frac{2p_1p_2\cdots p_k}{(p_1\minus{}1)\cdots(p_k\minus{}1)}$, where $ p_1, ..., p_k$ are distinct primes.


MATHCOUNTS
AMC
AIME
USA(J)MO
USAMO
email
AMC 10

Just out of curiosity, does anybody have the problems from the list below. If so, please post or pm them to me;

1984 Sprint
1984 Target
2008 Target

Please help!!!

Do you want State, Chapter, or Nationals?

Oh, oops, I forgot to include that information.

Nationals for all of those.

i really want some tests (i don't care what level but nationals would be preferred but state and chapter would be fine) can someone please pm them to me?

I have many problems. What problems have you already done?

Poincare: Google minnesota mathcounts. You can find states and chapters from 2008 to 2006 with full solutions. (You should come to a MathCounts websites - click "Coaches" and then "Resources".)

I have the following problems;

1988 School Target
1988 School Team
1988 Chapter Target
1989 State Sprint
1990 State Sprint
1991 Chapter Team
1991 State Sprint
1991 State Team
1991 National Sprint
1991 National Team
1992 National Sprint
1992 National Team
1993 State Team
1993 National Sprint
1993 National Target
1994 State Sprint
1995 School Sprint
1995 School Target
1995 Chapter Team
1995 State Sprint
1995 State Team
1996 Chapter Team
1996 State Sprint
1996 State Target
1996 State Team
1996 National Sprint
1996 National Target
1997 Chapter Sprint
1997 Chapter Target
1997 Chapter Team
1997 National Sprint
1997 National Target
1998 State Sprint
1998 National Team
1999 School Target
1999 Chapter Team
1999 State Team
1999 National Sprint
2000 Chapter Team
2000 State Team
2000 National Sprint
2000 National Target
2000 National Team
2001 Chapter Sprint
2001 Chapter Target
2001 State Sprint
2001 State Team
2001 National Sprint
2001 National Target
2001 National Team
2002 Chapter Team
2002 State Sprint
2002 State Target
2002 State Team
2003 Chapter Sprint
2003 Chapter Target
2003 Chapter Team
2003 State Sprint
2003 State Target
2003 State Team
2003 National Sprint
2003 National Target
2004 Chapter Sprint
2004 State Sprint
2004 State Target
2005 Chapter Sprint
2005 Chapter Target
2005 Chapter Team
2005 State Sprint
2005 State Target
2005 State Team
2006 School Team
2006 State Sprint
2006 State Target
2006 State Team
2006 National Sprint
2006 National Team
2007 School Team
2007 Chapter Sprint
2007 Chapter Target
2007 State Sprint
2008 National Sprint #'s 1-24 (lost the last page)
2008 National Team

Whew, that took a long time to type.

Oh, well I had a really good MC coach, and in the 1 year I was with him, he had me take all of those. That's how I got so good at math.

I think we can all pm our requests to him. :wink:

But since I already said that, I'd [i]love[/i] to see the 2008 National Sprint problems. Thanks!

[hide]You people should really think past Mathcounts...so let's say you get all those problems, waste/use 1000 hours, and get really good at Mathcounts...did you really improve so much in actual math?[/hide]

The point is that before 2000, the problems weren't that good...so your time could be used better doing other problems instead of finding/doing old tests...

most people who are really in need of MC problems can pwn in higher mathematics such as AIME/USAMO.

OMG dang it's my last year doing MC and i really wanna make it to nats.....
I would really appreciate if you PM me any national round/rounds before 2001
But any MC problems would really, REALLY help.

Thanks a lot

[quote="math154"]Hm? Are you saying those who spend all of their time on MC in middle school are almost guaranteed to do well in AIME/USAMo?[/quote]
No, but many people i know can do many AIME problems but can't do very many MC problems (me, for example)

Do you mean MC problems under time constraint? Then of course that's true...

Ernie, you can get 2006-2008 states by googling "minnesota mathcounts", going to the first website, click on Coaches' Resources or something like that, and you should find them

math154 has a good point, although mathcounts does help with amc 10, you should really spent more time on harder stuff

i regretted spending so much time on doing mathcount problems over and over again, and starting later than i should have on amc/aime/usamo

but if you really really really wanna make nats, go for it, you'll love it

You forgot to say, Wickedestjr, that I want the national tests from the 21st century. :wink:
And how long would it take you to email them to me?

That makes me think I still have a problem with my brain...

I don't have much time to prepare for USAMO...so even if I make it via AMC 10 and AIME, I would still fail.
But even so, I'm sacrificing that to make National MATHCOUNTS :roll:

[quote="Wickedestjr"]List;

isabella2296 wants '08 National Sprint
Unemployed wants National Rounds before 2001
LadyKn1ght wants '08 National Sprint
Poincare wants anything[/quote]
can you correct the anything to "possibly all"? :P

[quote="isabella2296"][quote="(^_^)"]math154 has a good point, although mathcounts does help with amc 10, you should really spent more time on harder stuff

i regretted spending so much time on doing mathcount problems over and over again, and starting later than i should have on amc/aime/usamo.[/quote]

I think MathCounts is a good level for me. I'm embarrassed to admit it, but I've done quite a few AMC 10's and I still can't get a score about 100. MathCounts is a good challenge for me, because I'm not really that good at math, even though I do put a lot of time and effort into it.[/quote]
sorry to double post and no offense but you should spend more time on the books than on the forums!

I want nats for all years 1990+, if you can get that.

[quote="AIME15"][quote="Wickedestjr"]List;

isabella2296 wants '08 National Sprint
Unemployed wants National Rounds before 2001
LadyKn1ght wants '08 National Sprint
Poincare wants anything[/quote]
I also said I wanted everything :D[/quote]

Well, you already pmed me, but I guess later I will update this post by putting the people that pmed me on the list;

isabella2296 wants '08 National Sprint
Unemployed wants National Rounds before 2001
LadyKn1ght wants '08 National Sprint
Poincare wants possibly everything
ernie wants Nationals from 2001 to 2008
gauss1181 wants Nationals 2000 to 2008
Textangle wants Nationals after 1990
rt_08 wants Nationals 2008

Okay, this is going to take a while, because I have so much going on lately.  :cleaning:

lol nevermind i just want 2006 and 2007 nats

sorry if i screwed u wasting ur time typing them on email...

isabella2296 wants '08 National Sprint
Unemployed wants National Rounds before 2001
LadyKn1ght wants '08 National Sprint
Poincare wants possibly everything
ernie wants Nationals from 2001 to 2008
gauss1181 wants Nationals 2006 and 2007
Textangle wants Nationals after 1990
rt_08 wants Nationals 2008

Lol, is anybody going to give me problems I don't have, which was the purpose of this topic in the first place?

Well, I can give you 2003 Nats Team and Countdown :D

That's all i have that you don't though...
And you STILL didn't put me on that list!

I am going to in a couple days. Just be patient. I am really busy.  :juggle:


$a_n=1+2^n+4^n$, $n$  is  element  of  natural  numbers.
If $a_n$ [b]is  prime [/b], then prove  that   $n=3^k$ for  $k>-1$(integer).

We already had that some times before:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47701
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27620


algorithm
combinatorics proposed
combinatorics

There is a WORD game with the following rules. There are finite number of relations $U_{i}\longrightarrow V_{i}$($U_{i},V_{i}$ are words). There is are two words $A,B$. We start from $A$, and we want to reach to $B$. At each step we can change one subword $U_{i}$ to $V_{i}$. Prove that there does not exist an algorithm that picks up $A,B$ and $U_{i}$'s,$V_{i}$'s and decides whether we can reach from $A$ to $B$ or not.

Please, could you re-state the problem? I think it is interesting, but:
- what is a word? Is it a finite sequence of elements prom a fixed set A? Is the alphabet (A) finite?
- what are relations? Just ordered couples between words?
- is the algorithm just given two words (A,B) and a finite set of couples of words?

- Each word is a finite sequence. of an alphabet. (for problem it does not differ that alphabet is finite or not, because each word is finite and we have infinite number of relations.)
- Just ordered couples between words.
- Yes, finite number of operations


Find the smallest  number of the form $ n^{3}\minus{}5n^{2}\minus{}8n\minus{}4$ that  is divisible by 173.  :)

[hide="(Wrong) Answer"]$ n^3 \minus{} 5n^2 \minus{} 8n \minus{} 4 \equal{} (n \minus{} 1)((n \minus{} 2)^2 \minus{} 8)$ and $ a^2\not\equiv 8\pmod{173}$ by law of quadratic reciprocity, so $ n \equal{} 174$.[/hide]

[quote="bertram"]$ n^3 \minus{} 5n^2 \minus{} 8n \minus{} 4 \equal{} (n \minus{} 1)((n \minus{} 2)^2 \minus{} 8)$[/quote]
But,
$ (n\minus{}1)((n\minus{}2)^2 \minus{} 8) \equal{} (n\minus{}1)(n^2 \minus{} 4n \minus{} 4) \equal{} n^3 \minus{} 4n^2 \minus{} 4n \minus{} n^2 \plus{} 4n \plus{} 4 \equal{} n^3 \minus{} 5n^2 \plus{} 4$

Ok it should be $ n^{3}\minus{}5n^{5}\plus{}8n\minus{}4$

Here is my solution:

We can express $ n^{3} \minus{} 5n^{2} \plus{} 8n \minus{} 4$ as $ (n \minus{} 1)(n \minus{} 2)^{2}$ so or $ (n \minus{} 1)$ or $ (n \minus{} 2)$ should be   173(as a prime number)

If $ (n \minus{} 1) \equal{} 173$ than $ n \equal{} 174$ and we have $ (174 \minus{} 1)(174 \minus{} 2)^{2}$

If $ (n \minus{} 2) \equal{} 173$ than $ n \equal{} 175$ and we have $ (175 \minus{} 1)(175 \minus{} 2)^{2}$

$ (175 \minus{} 1)(175 \minus{} 2)^{2} > (174 \minus{} 1)(174 \minus{} 2)^{2}$

So the smallest value is $ (174 \minus{} 1)(174 \minus{} 2)^{2}$ which is realized for $ n \equal{} 174$

 So bertram answer is fortunatelly true  :lol:

But isn't (n-1)^2(n-1)=n^3-5n^2+8n-4 :huh:

Check n=2, if what you said is true then it is a root.


[quote="tungtn0811"]Let a,b,c be positive numbers such that a+b+c =1 . Prove that
        81$ \sqrt\frac {a^3b^3}{c^3}$ + 32$ \sqrt\frac {c^3a^3}{b^3}$ + 5$ \sqrt\frac {b^3c^3}{a^3}$ $ \geq{12}$[/quote]
Put $ \sqrt{\frac{bc}{a}}\equal{}x,\sqrt{\frac{ca}{b}}\equal{}y,\sqrt{\frac{ab}{c}} \equal{}z$,then $ xy\plus{}yz\plus{}zx\equal{}1$
We have: $ 3x^3\plus{} 24y^3\plus{}3 \ge 6xy$,$ 2x^3\plus{}54z^3\plus{}2 \ge 6xz$,$ 8y^3\plus{}27z^3\plus{}1 \ge 6yz$
Then,we have Q.E.D

[quote="tungtn0811"][quote="quykhtn-qa1"][quote="tungtn0811"]Let a,b,c be positive numbers such that a+b+c =1 . Prove that
        81$ \sqrt\frac {a^3b^3}{c^3}$ + 32$ \sqrt\frac {c^3a^3}{b^3}$ + 5$ \sqrt\frac {b^3c^3}{a^3}$ $ \geq{12}$[/quote]
Put $ \sqrt {\frac {bc}{a}} \equal{} x,\sqrt {\frac {ca}{b}} \equal{} y,\sqrt {\frac {ab}{c}} \equal{} z$,then $ xy \plus{} yz \plus{} zx \equal{} 1$
We have: $ 3x^3 \plus{} 24y^3 \plus{} 3 \ge 6xy$,$ 2x^3 \plus{} 54z^3 \plus{} 2 \ge 6xz$,$ 8y^3 \plus{} 27z^3 \plus{} 1 \ge 6yz$
Then,we have Q.E.D[/quote]

$ 3x^3 \plus{} 24y^3 \plus{} 3 \ge 6xy$ ???? !        $ 3x^3 \plus{} 24y^3 \plus{} 3 \ge 18xy$ ! I had solution 1 !

Solution 2 ! ???????[/quote]
What do you mean?
Use AM-GM :wink:

[quote="quykhtn-qa1"][/quote][quote="tungtn0811"][quote="quykhtn-qa1"][quote="tungtn0811"]Let a,b,c be positive numbers such that a+b+c =1 . Prove that
        81$ \sqrt\frac {a^3b^3}{c^3}$ + 32$ \sqrt\frac {c^3a^3}{b^3}$ + 5$ \sqrt\frac {b^3c^3}{a^3}$ $ \geq{12}$[/quote]
Put $ \sqrt {\frac {bc}{a}} = x,\sqrt {\frac {ca}{b}} = y,\sqrt {\frac {ab}{c}} = z$,then $ xy + yz + zx = 1$
We have: $ 3x^3 + 24y^3 + 3 \ge 6xy$,$ 2x^3 + 54z^3 + 2 \ge 6xz$,$ 8y^3 + 27z^3 + 1 \ge 6yz$
Then,we have Q.E.D[/quote]

$ 3x^3 + 24y^3 + 3 \ge 6xy$ ???? !        $ 3x^3 + 24y^3 + 3 \ge 18xy$ ! I had solution 1 !

Solution 2 ! ???????[/quote][quote="quykhtn-qa1"]
What do you mean?
Use AM-GM :wink:[/quote]

$ 3x^3 + 24y^3 + 3 \ge 6xy$ ???? !      apply AMGM(cauchy) , we have : $ 3x^3 + 24y^3 + 3 \ge 18xy$! DONE

[quote="tungtn0811"][/quote][quote="quykhtn-qa1"][/quote][quote="tungtn0811"][quote="quykhtn-qa1"][quote="tungtn0811"]Let a,b,c be positive numbers such that a+b+c =1 . Prove that
        81$ \sqrt\frac {a^3b^3}{c^3}$ + 32$ \sqrt\frac {c^3a^3}{b^3}$ + 5$ \sqrt\frac {b^3c^3}{a^3}$ $ \geq{12}$[/quote]
Put $ \sqrt {\frac {bc}{a}} = x,\sqrt {\frac {ca}{b}} = y,\sqrt {\frac {ab}{c}} = z$,then $ xy + yz + zx = 1$
We have: $ 3x^3 + 24y^3 + 3 \ge 6xy$,$ 2x^3 + 54z^3 + 2 \ge 6xz$,$ 8y^3 + 27z^3 + 1 \ge 6yz$
Then,we have Q.E.D[/quote]

$ 3x^3 + 24y^3 + 3 \ge 6xy$ ???? !        $ 3x^3 + 24y^3 + 3 \ge 18xy$ ! I had solution 1 !

Solution 2 ! ???????[/quote][quote="quykhtn-qa1"]
What do you mean?
Use AM-GM :wink:[/quote][quote="tungtn0811"]

$ 3x^3 + 24y^3 + 3 \ge 6xy$ ???? !      apply AMGM(cauchy) , we have : $ 3x^3 + 24y^3 + 3 \ge 18xy$! DONE[/quote]
Ok,it must be: $ .... \ge 18 xy$ :oops:

[quote="quykhtn-qa1"][/quote][quote="tungtn0811"][/quote][quote="quykhtn-qa1"][/quote][quote="tungtn0811"][quote="quykhtn-qa1"][quote="tungtn0811"]Let a,b,c be positive numbers such that a+b+c =1 . Prove that
        81$ \sqrt\frac {a^3b^3}{c^3}$ + 32$ \sqrt\frac {c^3a^3}{b^3}$ + 5$ \sqrt\frac {b^3c^3}{a^3}$ $ \geq{12}$[/quote]
Put $ \sqrt {\frac {bc}{a}} = x,\sqrt {\frac {ca}{b}} = y,\sqrt {\frac {ab}{c}} = z$,then $ xy + yz + zx = 1$
We have: $ 3x^3 + 24y^3 + 3 \ge 6xy$,$ 2x^3 + 54z^3 + 2 \ge 6xz$,$ 8y^3 + 27z^3 + 1 \ge 6yz$
Then,we have Q.E.D[/quote]

$ 3x^3 + 24y^3 + 3 \ge 6xy$ ???? !        $ 3x^3 + 24y^3 + 3 \ge 18xy$ ! I had solution 1 !

Solution 2 ! ???????[/quote][quote="quykhtn-qa1"]
What do you mean?
Use AM-GM :wink:[/quote][quote="tungtn0811"]

$ 3x^3 + 24y^3 + 3 \ge 6xy$ ???? !      apply AMGM(cauchy) , we have : $ 3x^3 + 24y^3 + 3 \ge 18xy$! DONE[/quote][quote="quykhtn-qa1"]
Ok,it must be: $ .... \ge 18 xy$ :oops:[/quote]

$ 3x^3 + 24y^3 + 3 \ge 3\sqrt [3]{3x^3.3.24y^3} = 3.6xy = 18xy$

[quote="tungtn0811"][/quote][quote="quykhtn-qa1"][/quote][quote="tungtn0811"][/quote][quote="quykhtn-qa1"][/quote][quote="tungtn0811"][quote="quykhtn-qa1"][quote="tungtn0811"]Let a,b,c be positive numbers such that a+b+c =1 . Prove that
        81$ \sqrt\frac {a^3b^3}{c^3}$ + 32$ \sqrt\frac {c^3a^3}{b^3}$ + 5$ \sqrt\frac {b^3c^3}{a^3}$ $ \geq{12}$[/quote]
Put $ \sqrt {\frac {bc}{a}} = x,\sqrt {\frac {ca}{b}} = y,\sqrt {\frac {ab}{c}} = z$,then $ xy + yz + zx = 1$
We have: $ 3x^3 + 24y^3 + 3 \ge 6xy$,$ 2x^3 + 54z^3 + 2 \ge 6xz$,$ 8y^3 + 27z^3 + 1 \ge 6yz$
Then,we have Q.E.D[/quote]

$ 3x^3 + 24y^3 + 3 \ge 6xy$ ???? !        $ 3x^3 + 24y^3 + 3 \ge 18xy$ ! I had solution 1 !

Solution 2 ! ???????[/quote][quote="quykhtn-qa1"]
What do you mean?
Use AM-GM :wink:[/quote][quote="tungtn0811"]

$ 3x^3 + 24y^3 + 3 \ge 6xy$ ???? !      apply AMGM(cauchy) , we have : $ 3x^3 + 24y^3 + 3 \ge 18xy$! DONE[/quote][quote="quykhtn-qa1"]
Ok,it must be: $ .... \ge 18 xy$ :oops:[/quote][quote="tungtn0811"]

$ 3x^3 + 24y^3 + 3 \ge 3sqrt[3]{3.x^3.3.24y^3} = 3.6xy = 18$[/quote]
I know it :)

[quote="tungtn0811"][/quote][quote="quykhtn-qa1"][/quote][quote="tungtn0811"][/quote][quote="quykhtn-qa1"][/quote][quote="tungtn0811"][quote="quykhtn-qa1"][quote="tungtn0811"]Let a,b,c be positive numbers such that a+b+c =1 . Prove that
        81$ \sqrt\frac {a^3b^3}{c^3}$ + 32$ \sqrt\frac {c^3a^3}{b^3}$ + 5$ \sqrt\frac {b^3c^3}{a^3}$ $ \geq{12}$[/quote]
Put $ \sqrt {\frac {bc}{a}} = x,\sqrt {\frac {ca}{b}} = y,\sqrt {\frac {ab}{c}} = z$,then $ xy + yz + zx = 1$
We have: $ 3x^3 + 24y^3 + 3 \ge 6xy$,$ 2x^3 + 54z^3 + 2 \ge 6xz$,$ 8y^3 + 27z^3 + 1 \ge 6yz$
Then,we have Q.E.D[/quote]

$ 3x^3 + 24y^3 + 3 \ge 6xy$ ???? !        $ 3x^3 + 24y^3 + 3 \ge 18xy$ ! I had solution 1 !

Solution 2 ! ???????[/quote][quote="quykhtn-qa1"]
What do you mean?
Use AM-GM :wink:[/quote][quote="tungtn0811"]

You write $ 3x^3 + 24y^3 + 3 \ge 6xy$ is False
$ 3x^3 + 24y^3 + 3 \ge 6xy$ ???? !      apply AMGM(cauchy) , we have : $ 3x^3 + 24y^3 + 3 \ge 18xy$! DONE[/quote][quote="quykhtn-qa1"]
Ok,it must be: $ .... \ge 18 xy$ :oops:[/quote][quote="tungtn0811"]

$ 3x^3 + 24y^3 + 3 \ge 3\sqrt [3]{3x^3.3.24y^3} = 3.6xy = 18xy$[/quote]

you write $ 3x^3 + 24y^3 + 3 \ge 6xy$ is False

[quote="tungtn0811"]Let a,b,c be positive numbers such that a+b+c =1 . Prove that
        81$ \sqrt\frac {a^3b^3}{c^3}$ + 32$ \sqrt\frac {c^3a^3}{b^3}$ + 5$ \sqrt\frac {b^3c^3}{a^3}$ $ \geq{12}$[/quote]
it was a problem in Olympic Hanoi ams grade 10


inequalities
symmetry
inequalities unsolved

[quote="gemath"]Given $a,b,c,x,y,z>0$ such that
$a^{2}+b^{2}+c^{2}\ge x^{2}+y^{2}+z^{2}$ prove that
$ax+by+cz\ge x^{2}+y^{2}+z^{2}$  :?:[/quote]
Let $b=c \rightarrow 0, x \rightarrow 0$ :wink:

I don't understand can you show counterexample ?  :huh:

okie, let $b=c \rightarrow 0, y=z=1, a=\sqrt{2},x \rightarrow 0$ :wink:

I don't understand can you clearly $b=c\to 0$ :huh:

Probably this inequality is not true.
But may be true if a>b>c>x>y>z.(i havent checked yet)
ok...Let,(a,b,c,x,y,z)=(1,2,4,3,2,1)
then 11>=14 that is wrong.

Ok thank you Moonmathpi496 for your example, :)  I want to find counter example only,I think toanhocmuonmau is quite haughty when he doesn't want to write clearly an example here! :wink:

An inequality of this form is impossible, just consider the case $x^{2}+y^{2}+z^{2}=a^{2}+b^{2}+c^{2}$ then by symmetry if it holds then we have \[ax+by+cz\geq max\left(\sum x^{2},\sum a^{2}\right)\]
Which is impossible since by Cauchy, \[\left(\sum x^{2}\right)\left(\sum a^{2}\right)\geq \left(\sum ax\right)^{2}\] :(

ok thanks for your interest me@home, that result is useful for me ! :)


inequalities
geometry
circumcircle
inequalities unsolved

[quote="milkyway_00"]Given a triangle ABC with R,r are the incircle and circumcircle radii Bc=a,CA=b,AB=c .prove
   $cosAcosAcosC\leq\frac{r^{2}}{3R^{2}-2Rr}$[/quote]

probably you meant $cosAcosBcosC$ there  :wink:

Oh sorry you say that true that is cosAcosBcosC thank you
  And now post your solution


I was wondering how one could approach this infinite sum:
$ \sum_{k=2}^{\infty} \frac{a^k}{k!}$ when $ \abs{a}<1.$ If it is too hard to find a formula for a general $ a$, I am curious when $ a=.6$. However, if a general formula could be found that would be best. At first glance I figured that it looks similar to the Taylor series of $ e^k$, but I'm not sure how to use that.

It's the Taylor series of $ e^a$ minus two terms.

hello, we get $ \sum_{k\equal{}2}^{\infty}\frac{a^k}{k!}\equal{}e^a\minus{}1\minus{}a$
Sonnhard.


Let ABCD be a convex circumscriptible quadrilateral. Let E, F be the midpoints of the diagonals AC, BD. Show that the incenter of ABCD is on EF.

Oh, I think it received some solutions here. Try [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20264]this thread[/url].


Care a fost rezultatul vostru la ONM 2006?

lol, 60% din concurenti au avut sub 10 puncte. de ce n-ai bagat si varianta aia ? sau asta e doar pt VIP only  :ninja:

Tudor Micu, Cluj Napoca, actualmente clasa a VIII-a, premiul I 2005, premiul II 2006, deocamdata atat, cu ajutorul lui Dumnezeu poate va fi mai mult anii viitori  :) ...

Aha, bravo. Esti inca mic, dar oricum felicitari pentru rezultate. ;)

Multumesc frumos...de anul acesta ma apuc serios de treaba, nu mai pierd timpul ca in anii trecuti  :D

Felicitari si din partea mea!Voiam si eu sa ma duc la ONM, dar sincer sa fiu oricum nu ma lasa mama... :(  :(

De-ai sti cat imi pare de rau pentru tine...E trist ca nu te-a lasat mama ta dar probabil se gandea sa nu patesti ceva la asa departare. Stiu, pentru ca si Clujul se afla la mare distanta de Iasi.
Stai linistit ca anul viitor, cand vei fi a opta, nationala va fi la Pitesti si sigur te va lasa mama ta.

Sper din toata inima sa ma duc anul viitor la nationala, mai ales ca, dupa cum zici tu, se desfasoara la Pitesti si ca in a VIII-a inspectorul nostru de matematica va lua la ONM primii 4 copii.

Asta-i cel mai trist: Cand nu participi si vezi subiectele si vezi ca le puteai face...Asta e, eu iti urez succes de acuma si Doamne ajuta sa faci ceva frumos la ONM 2007   :lol:

Mult succes si tie in a IX-a.Am vazut ca ai luat locul II.Bravo!

Mersi frumos...si tie felicitari pentru problemele pe care le-ai propus pana acum, pentru varsta ta sunt foarte bune...

Iata ca am ajuns la un simpatic schimb de amabilitati intre matematicieni juniori... :thumbup:  :thumbup:

Fii fara griji. Am una in shortlist la clasa a-VIII-a. Mai propusesem vreo 2, dar erau prea grele. ;)

Da, dar pentru clasa a VII-a, nimik?Eu sunt in a VII-a. ;) O sa ma uit in lista scurta de a VIII-a sa vad ce ai propus...

[quote="Stormraven"]Hehe, frumos spus, frumos punctat. De aceea sugeram eu o injumatatire a loturilor, fiindca acestia de sub 10 puncte sunt cei turbulenti care cauzeaza probleme invariabil. Un om destept are si ceva bun simt.[/quote]

uite, iar ma faci prost ... oricum ai intoarce-o, tot prost ma faci . eu am fost la XI-a unde am luat 9 p . ceea ce e bine :D cu 8 p s-a luat mentiune cu argint, ceea ce e bine pt unii pamanteni . la brotaceii astia ca tine daca iei 15 p , nu esti sigur ca iei ceva . ca olimpiadele sunt din ce in ce mai grele pe an ce trece. cine stie, cunoaste .

deci ai o logica de [ cenzurat ] . fiti atenti logica aici la raven :

Cine ia mai putin de 10 p la nationala de matematica e nesimtit si prost !

ce parere aveti ? tu ce parere ai spix ? :P si pot da exemple de multi castigatori de interjudetene, premiati cu aur in anii trecuti and so on

Hooo, mai usor cu cearta!Stormraven n-a vrut sa te insulte!Nu stia cat ai luat tu la nationala.

Bine, bine...reformulez, reformulez...Cei cu mai putin de 10 p, ori nu se pricep la problemele de finala, si vor face mai bine la baraj, sau n-au fost la momentul respectiv in forma, intr-adevar nu e bun adjectivul "prosti" avand in vedere ca sunt multi in tara mult mai putin inzestrati decat aceia.
Stimabile "Negrupur", dupa cum a spus si Sorin, eu n-am dorit nicidecum sa te jignesc si n-aveam cum sa ghicesc ce punctaj ai luat...putea fi orice numar natural intre 0 si 28  :D  Ma rog, nu ma gandeam ca ai luat 0 p pentru ca mi-ai parut o persoana simpatica si, dupa raspunsurile pe care le-ai dat, nu mi-ai parut olimpic de 0 puncte... :D

am luat 9 p :) la clasa a 11-a , asta a insemnat locul 18 din 92 ... si am fost la prima participare . pt mine e ceva ff fff mare , ca n-am mai fost niciodata ... am fost la chimie 3 ani la nationala , mereu in primii 10 ... am luat 9 p ca am facut prea multe, nu ca n-am stiut sa le fac :) sincer sa fiu , problemele mi s-au parut f grele ... am facut una perfect, la una am avut ideea si nu am finalizat si nici nu am continuat, si anume 3 , iar la restu cate o ciupeala .  putea fi mult mai rau de tot fara absolut nici o problema ... un baiat care m-a facut in sange la o interjudeteana, ceva de genu el 29 din 30, eu 20 ... a luat doar 4 p . baiatu ala e prost ? ca a mai castigat inca 3-4 interjudetene . asa a fost sa fie ... din milioanele de probleme posibile, se dau numa 4 . daca stii sa faci doar 4 probleme si alea iti pica, iesi peste niste baieti care dorm cu matematica sub perna . eu sunt f multumit, as fi absurd sa zic ca se putea mai mult . se putea si mai putin , muuuuuuuult mai putin . cine stie, oi fi prost ... nu ma supar. poti sa ma faci cum vrei . 8-9p la nationala e destul ca sa iei ceva , ma refer la clasele 10-12 cel putin . la sectia brotacei , cum imi place sa le zic, se iau punctaje mai mari ... sunt destule motive pt asta

Au fost ani cand cu opt puncte se lua premiu....

Intr-adevar, oricat nu mi-ar placea sa recunosc, nationala e si o chestie de noroc si de forma momentana. Poti sa fii intr-o forma extraordinara (ca mine in a VII-a  :D ) si in doua zile sa prinzi o forma jalnica (ca mine la baraj in a VII-a  :( ). Plus, sunt si tipurile de probleme...daca ai facut multe de tot dintr-un tip si tipul ala iti vine, ai dat lovitura...

chestia asta e chiar ff adevarata, reyultatele tin si de dispozitia din acea zi. intr-o zi poti sa nu faci  o pb, si peste cateva zile sa gasesti chiar maimulte solutii la aceeasi pb :lol: (asa mi s-a intamplat mie anul asta...si la baraje si la nat, - m-am simtit f prost (prost chiar si in sensul propriu al cuvantului) cand am vazut baremul de la baraje in care era o solutie care imi venise si mie in minte numai ca de emotie am uitat+o fiindca m+am apucat de celelalte pb :oops:   )

btw, valentin sa stii ca majoritatea celor de la nationala stiu de site-ul tau dar nu toti scriu de  spre exemplu eu intru 0de anul trecut (am aflat tot de la nat de la niste fete
) , dar n-am scris foarte des. probabil k stiu mai multi.

[quote="omicroncz"]btw, valentin sa stii ca majoritatea celor de la nationala stiu de site-ul tau dar nu toti scriu de  spre exemplu eu intru 0de anul trecut (am aflat tot de la nat de la niste fete
) , dar n-am scris foarte des. probabil k stiu mai multi.[/quote]Ma bucura sa aud chestia asta :) Bafta la baraje in continuare daca ai intrat in lot (seniori, juniori) ;)

[quote]Intr-adevar, oricat nu mi-ar placea sa recunosc, nationala e si o chestie de noroc si de forma momentana. Poti sa fii intr-o forma extraordinara (ca mine in a VII-a  ) si in doua zile sa prinzi o forma jalnica (ca mine la baraj in a VII-a  ). Plus, sunt si tipurile de probleme...daca ai facut multe de tot dintr-un tip si tipul ala iti vine, ai dat lovitura... [/quote]

Eu nu cred ca exista forma 'de moment' (in sensul ca azi esti intr-o forma maine in alta). Esti in forma daca ai lucrat in anul respectiv si poate daca nu mai faci nimic vreo doua luni nu mai esti in forma.Norocul la nationala tine numai de problemele selectate.Oricum acel noroc nu poate fi prea mare.De exemplu eu(cls X ) am inceput cu pb3, n-am stiut de unde sa o iau, m-am enervat rau pt ca eu sunt (sau cel putin ma consider) bun la teoria numerelor si am abordat-o pe a 2-a ,am facut-o foarte elegant in 20 min(am luat 6,5 pt ca am uitat combinarile la binomul lui Newton lol) dupa care m-a surprins prima ce simpla e(am facut-o in 10 min).Apoi m-am dus la baie si am avut o revelatie la pb 3 am redactat-o si am luat 7 si pe asta.La ultima am avut niste idei dar nu le-am scris pt ca nu e stilu meu sa aburesc.Am luat un surprinzator loc 3(atat pt mine cat si pt ceilalti de a Xa).La baraj am facut varza n-am stiut sa fac nici o pb, prima nici n-am incercat-o, nu stiu polinoame asa ca nici a doua n-am abordat-o.M-am chinuit la 3 si 4 dar  la 3 s-au dat 4 punctaje de 7 si in rest 0p.In a doua zi n-am stiut deloc pb1, ce sa mai zic de pb2(eu sunt f slab la geometrie) am luat 6p la pb3(putea sa fie 7 dar am gresit la ce era mai simplu) si 2p la pb4.Deci am facut penibil. Intr-a 9a am fost pe 16 la natio si primii 15 s-au dus la baraj.As fi stiut 2pb la baraj si m-as fi calificat.Ghinionul meu...Sorry de post prea lung  :blush:

Intr-adevar, fara munca nu reusesti nimic si daca nu lucrezi o luna te-ai dus pe pluta. Eu totusi raman la parerea ca si forma de moment conteaza dar si, cum ai spus tu foarte bine, modul in care abordezi problemele. Cel mai important e sa NU intri in panica...Eu am aflat-o pe propria piele la baraje... :(

[quote="stancioiu sorin"]Sunt de acord cu Tudor.Totul depinde de forma pe care o ai in ziua respectiva.Am si eu un exemplu in judetul meu, ocupantul primului loc la judeteana a luat foarte bine, 21p din 28p a vut un parcurs excelent, obtinand aproape la toate concursurile locul I(eu dupa el si dupa inca cineva), dar acum la nationala nu cred ca a fost in forma, pentru ca aluat cam putin(17p din 28p), mai ales ca el ia la toate concursurile aproape maxim.  :cursing: Acum, sa vad ce facem la Ion Ciolac.Macar acum sa-l intrec...[/quote]
Ma scuzti ca intervin in aceasta discutie aprinsa da acest mesaj mi-a atras in mod deosebit atentia...Nu cred ca ultimele 2 propozitii isi au locul...Adik cel putin eu nu privesc olimpiada ca pe o competitie cu altii ci ca pe o competitie cu mine...iubesc momentul cand rezolvi o problema grea te simti foarte bine probabil stiti asta.Nu trebuie sa avem gandirea asta ma duc sa ies primu la concurs.Ca performante eu nu prea am facut nimik la viata mea chiar dak am participat la onm dintr-a saptea pana intr-a noua(acum trec a zecea).Cea mai mare realizare a vietii male de "matematician" este rezolvarea unei probleme care mi-a picat in clasa a saptea la barajul pt juniori , o problema de geometri...cand am fost atunci nu am rezolvato...totusi am tinut mortis so rezolv pe propiile forte fara sa ma uit la vreo rezolvare sau sa vad vreo idee.Desi toti care veti citi postul veti spune ca mananc(cenzurat) timp de doi ani mam gandit la ea si abia intr-a noua dupa onm am reusit so rezolv.desi o problema simpla eu nu am putut so rezolv!miau trebuit doi ani sa reusesc asta!!!Aici apare chalengul in matemtica,nu sa te clasezi mai bine decat unu sau altu...

[quote="Salajan Sabin"][quote]Intr-adevar, oricat nu mi-ar placea sa recunosc, nationala e si o chestie de noroc si de forma momentana. Poti sa fii intr-o forma extraordinara (ca mine in a VII-a  ) si in doua zile sa prinzi o forma jalnica (ca mine la baraj in a VII-a  ). Plus, sunt si tipurile de probleme...daca ai facut multe de tot dintr-un tip si tipul ala iti vine, ai dat lovitura... [/quote]

Eu nu cred ca exista forma 'de moment' (in sensul ca azi esti intr-o forma maine in alta). Esti in forma daca ai lucrat in anul respectiv si poate daca nu mai faci nimic vreo doua luni nu mai esti in forma.Norocul la nationala tine numai de problemele selectate.Oricum acel noroc nu poate fi prea mare.De exemplu eu(cls X ) am inceput cu pb3, n-am stiut de unde sa o iau, m-am enervat rau pt ca eu sunt (sau cel putin ma consider) bun la teoria numerelor si am abordat-o pe a 2-a ,am facut-o foarte elegant in 20 min(am luat 6,5 pt ca am uitat combinarile la binomul lui Newton lol) dupa care m-a surprins prima ce simpla e(am facut-o in 10 min).[b]Apoi m-am dus la baie si am avut o revelatie[/b] la pb 3 am redactat-o si am luat 7 si pe asta.La ultima am avut niste idei dar nu le-am scris pt ca nu e stilu meu sa aburesc.Am luat un surprinzator loc 3(atat pt mine cat si pt ceilalti de a Xa).La baraj am facut varza n-am stiut sa fac nici o pb, prima nici n-am incercat-o, nu stiu polinoame asa ca nici a doua n-am abordat-o.M-am chinuit la 3 si 4 dar  la 3 s-au dat 4 punctaje de 7 si in rest 0p.In a doua zi n-am stiut deloc pb1, ce sa mai zic de pb2(eu sunt f slab la geometrie) am luat 6p la pb3(putea sa fie 7 dar am gresit la ce era mai simplu) si 2p la pb4.Deci am facut penibil. Intr-a 9a am fost pe 16 la natio si primii 15 s-au dus la baraj.As fi stiut 2pb la baraj si m-as fi calificat.Ghinionul meu...Sorry de post prea lung  :blush:[/quote]

 :rotfl:  :rotfl: super tare asta cu `m-am dus la baie si am avut o revelatie` :)))


Hi.
I want to ask about an iqueality.
I work on it a lot but I was not suceed in it.
Now look up it.
[img]http://img19.imageshack.us/img19/295/problemnt3.png[/img]
xcuse me for my broken English... :)

The problem in the title of your message is different from the problem in your message.  Which is the correct inequality?

i am sorry. the correct one is in the title..
I will correct another too..
thanks..

[hide]You can subtract everything from the right side RHS leaving you with
$ \frac{a^{2}}{4}+b^{2}+c^{2}-ab+ac-2bc$. and we need to prove that is greater than or equal to $ 0$.
This can be factored as
$ (\frac{a}{2}-b+c)^{2}$, so for all reals, that expression is always nonnegative. [/hide]

ooh thanks..
i didn't think that the sollution could be as easy as this..
i as applying the Cauchy Schwarz, Jung and. others..
As a conclusion i understood that the big things could be hidden behind the details  :lol:

Xcuse for my broken english...


Ok, pretty much we just got to say mathematical tools that we use. Such as combinatorics or summation. If you quickly want to learn the LaTex symbols try this site:http://en.wikibooks.org/wiki/LaTeX/Mathematics

I'll start with this useful symbol/tool/technique

$ \prod$

My favorite and most useful tool is $ \pi$ (hence, my username).

$ \mathfrak {mathfrak (which I learned from algebra!)}$

$ \mathbf{mathbf}$ is another text style. (see my signature)

oh yea btw i forgot to credit dragon96's sig for teaching me $ \mathfrak{mathfrak}$

another tool: $ \pm$

it was useful when i typed out my mathpath application test answers in $ \text{\LaTeX}$ :P

another one: $ \binom{42}{42}$

Here's my representation and my favorite $ \mathfrak{greek}$ letter: $ \Omega$

To revive this thread: $ \phi$

phi is only one letter away from pi!


if $ d_n \equal{} \overbrace{33....3}^n$, then the value of $ S_n \equal{} \sum\limits_{i \equal{} 1}^n d_i$ IS??

This does not belong here.  Hint: Can you write $ d_n$ involving a power of 10?

[Moderator: moved]

Okay ;

Give me the result of sum at least please ..

Thank you

[hide]Using the formula for the sum of a geometric series: $ d_n\equal{}3\cdot10^{n\minus{}1}\plus{}3\cdot10^{n\minus{}2}\plus{}\cdots\plus{}3\cdot10^1\plus{}3\cdot10^0\equal{}\frac13(10^n\minus{}1)$.

So, $ S_n\equal{}\sum_{i\equal{}1}^{n}\frac13(10^i\minus{}1) \equal{} \frac13\left(\sum_{i\equal{}1}^{n}10^i\minus{}\sum_{i\equal{}1}^{n}1\right)\equal{}\frac13\left[\left(\frac{10^{n\plus{}1}\minus{}10}{9}\right)\minus{}n\right]\equal{}$ $ \frac{10^{n\plus{}1}\minus{}10\minus{}9n}{27}$[/hide]


[quote="pohoatza"]Positive integers $ a<b$ are given. Prove that among every $ b$ consecutive positive integers there are two numbers whose product is divisible by $ ab$.[/quote]

Let $ S = \{n+1, n+2, ..., n+b\}$.
Clearly there is a single $ (n+i)$ in $ S$ that is divisible by $ b$ because the remainders of all elements of $ S$ divided by $ b$ must belong to the set $ \{0, 1, ..., b-1\}$. Taking $ a$ consecutive integers in $ S$ we get some $ (n+j)$ divisible by $ a$.
If $ i \neq j$ then we are done.
If $ i = j$ then it means that $ b$ is a multiple of $ a$ because $ a<b$. So let $ b = ka$ for some $ k$. We can now partition $ S$ into $ k$ blocks so that each block has some element $ (n+j)$ divisible by $ a$. Just pick one of these $ (n+j)$ different from the $ (n+i)$ and we are done.

Why $ a|b$ ? I can't understand this.
$ a,b|n+i$ just imply $ lcm(a,b)|n+i$...

Yeah, it's wrong. However, that mistake is very easy to fix. You just have to find another number divisible by $ \gcd (a,b)$ :D

And it's obvious it exists since $ gcd(a, b) \leq \frac{b}{2}$.

O.K. I'm going to complete everyone's idea. :flex:

Obviously among $b$ consecutive positive integers there are two numbers $p$ which is divisible by $a$ and $q$ which is divisible by $b$.

[b]Case1:[/b]$p\neq q$
Then $ab|pq$,and the proof is completed.

[b]Case2:[/b] $p=q$
Then $lcm(a,b)|p$.Since $gcd(a,b)\le \frac{1}{2}b$,there are at least two numbers which are divisible by $gcd(a,b)$.Then ∃$x\neq p$ s.t. $gcd(a,b)|x$.Therefore $ab=lcm(a,b)gcd(a,b)|px$,and the proof is completed.

From case$1,2$,completing the proof.$\blacksquare$ :coolspeak:


I have defined a function to draw arrays of boxes and it calls an array and an integer. Something fouls up when I try and execute it, though.

[hide="code is here"]

void tabloid(int[] n, int ht) {
	int hor = 0;
	for (int i: n) {
		for(int j=1; j<i; ++j) {
			draw(boxy(ht, (hor, 0)), linetype("8 8"));
			hor = hor+ht;
		}

	}
	draw((0,0)--(0,ht)--(hor,ht)--(hor,0)--(0,0), linewidth(1));
}

tabloid({1,2,3,4}, 30);[/hide]

The error is always located at the last line, but I've done some googling and I thought I was using correct syntax  :maybe:

Any ideas would be appreciated  :)

[code]tabloid(new int[] {1,2,3,4}, 30);[/code]
What's boxy ?


A 25-foot ladder is placed against a vertical wall of a building. The foot of the ladder is 15 feet from the
base of the building. If the top of the ladder needs to be 4 feet higher, then the foot of the ladder needs to
be moved in how many feet?

A) 4 feet B) 5 feet C) 7 feet D) 8 feet E) 11 feet

By the Pyhtagorean Theorem (or a general knowledge of Pythagorean triples), the height of the wall is $ 20$. Thus the height must be rasied $ 20\plus{}4\equal{}24$ ft. Once again with a general knowledge of Pythagorean triples. We see that the distance from the wall to the base is $ 7$. Thus, we have $ 15\minus{}7\equal{}8$. So the answer is D.


integration
inequalities unsolved
inequalities

this sequence need more terms...

${1,2,\ldots ,15^{4}}$

humpf

In general, we can prove such problem in this way

By considering a skecth of $f(x) = \frac{1}{\sqrt[4] x}$, we have
$\int_{a-1}^{a}f(x) dx < f(x) < \int_{a}^{a+1}f(x) dx$.
Furthermore, by telescoping,
$\frac{4}{3}(a^{\frac{3}{4}}-(a-1)^{\frac{3}{4}}) < \sum_{i=1}^{a}f(i) < f(a+1)-f(1) = \frac{4}{3}((a+1)^{\frac{3}{4}}-a^{\frac{3}{4}})-1$.

RDeepMath91 what do you mean "telescoping"? Maybe in In my country this sum called in other word ..

[quote="RDeepMath91"]Hm.. For me, telescoping sum can be seen clearly in this simple example :

$\sum_{i=1}^{n}\frac{1}{i(i+1)}= \sum_{i=1}^{n}[\frac{1}{i}-\frac{1}{i+1}] = 1-\frac{1}{n+1}$

I can't explain so clearly, but the main idea is there exists some terms which "destroy each other"..  :lol:[/quote]
I know all of them. but I don't know :this called telescoping sum  :lol: .thanks anyway


Let x,y and z be integers such that $ S\equal{}x^{4}\plus{}y^{4}\plus{}z^{4}$ is divisible by 29.Show that S is divisible by $ 29^{4}$

See here http://www.mathlinks.ro/Forum/viewtopic.php?p=903344#903344 .


inequalities
calculus
derivative
logarithms
inequalities proposed

If $ a,b\in (0,1]$, then

\[ a^{b\minus{}a}\plus{}b^{a\minus{}b}\le 2.\]

[quote="Vasc"]If $ a,b\in (0,1]$, then
\[ a^{b \minus{} a} \plus{} b^{a \minus{} b}\le 2.\]
[/quote]
Why didn't you post if $ a,b\in(0,2]$,then$ a^{b \minus{} a} \plus{} b^{a \minus{} b}\le 2$:maybe:

also there is $ \sum _{i\equal{}1}^n x_i^{x_{(i \bmod n)\plus{}1}\minus{}x_i}$ for $ x_i \in(0,2]$

[quote="Vasc"]If $ a,b\in (0,1]$, then
\[ a^{b \minus{} a} \plus{} b^{a \minus{} b}\le 2.\]
[/quote]
Very nice your inequality, Mr Vasile. I have found a nice solution for this inequality. I will post it soon.

Your proof is very beautiful, Can.  :lol:
It is nicer than mine, since I used again derivative to show that $ b \plus{} x\ge b^\frac {1 \minus{} x}{1 \plus{} x}$.
We can also use the same method to prove the sharper inequality

$ a^{b \minus{} a} \plus{} b^{a \minus{} b} \plus{} (a \minus{} b)^2\le 2$,

where equality holds for $ a \equal{} b$ and for $ a \equal{} 1$, $ b \equal{} 0$.

On the other hand, we can write the original inequality as

$ 2a^ab^b\ge (ab)^a \plus{} (ab)^b$.

Notice that for the following nice generalization I have no solution.
If $ a,b,c\in (0,1]$, then

$ 3a^ab^bc^c\ge (abc)^a \plus{} (abc)^b \plus{} (abc)^c.$

For $ c\equal{}a$, we get the following nice inequality

$ a^{2(b\minus{}a)}\plus{}2b^{a\minus{}b}\le 3.$

[quote="Tourish"][quote="Vasc"]If $ a,b\in (0,1]$, then
\[ a^{b \minus{} a} \plus{} b^{a \minus{} b}\le 2.\]
[/quote]Why didn't you post if $ a,b\in(0,2]$,then$ a^{b \minus{} a} \plus{} b^{a \minus{} b}\le 2$:maybe:[/quote]
Indeed, this is true.
There is a nice proof for this case, except the proof of the necessary inequality
$ (2\minus{}x)^x\plus{}2^{\minus{}x}\le 2$,
where $ x\in [0,1]$.

[quote="Vasc"]If $ a,b\in (0,1]$, then
\[ a^{b \minus{} a} \plus{} b^{a \minus{} b}\le 2.\]
[/quote]
We can also prove this inequality as follows.
Assume that $ a\ge b$, and write the inequality as
$ (a^{b\minus{}a}\minus{}1)^2\plus{}(\frac b{a})^{a\minus{}b}\le 1$.
This follows from
$ (\frac b{a})^{a\minus{}b}\le 1\minus{}\frac {(a\minus{}b)^2}{a}$
and
$ a^{b\minus{}a}\le 1\plus{}\frac {a\minus{}b}{\sqrt a}$
(see http://www.mathlinks.ro/viewtopic.php?t=328058).

[quote="can_hang2007"][quote="Vasc"]If $ a,b\in (0,1]$, then
\[ a^{b \minus{} a} \plus{} b^{a \minus{} b}\le 2.\]
[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-63-v-cirtoaje.html[/quote]

very very nice!

BQ

A long time ago, I proved on this forum that

$ a^{2a}\plus{}b^{2b}\ge a^{2b}\plus{}b^{2a}$.

Moreover, I have conjectured that the inequality below holds for any positive $ k$, $ a$ and $ b$ if and only if $ 0<k\le e$:

$ a^{ka}\plus{}b^{kb}\ge a^{kb}\plus{}b^{ka}$.

Actually, I think that for $ a,b\in (0,1]$, the following open sharper inequalities hold

$ 2a^ab^b\ge a^{2b}\plus{}b^{2a}$,

$ 2\sqrt{a^{ka}b^{kb}}\ge a^{kb}\plus{}b^{ka}, \ \ 0<k\le e.$

[quote="Vasc"] The inequality below holds for any positive $ k$, $ a$ and $ b$ if and only if $ 0<k\le e$:

$ a^{ka}\plus{}b^{kb}\ge a^{kb}\plus{}b^{ka}$.

Actually, I think that for $ a,b\in (0,1]$, the following open sharper inequalities hold

$ 2a^ab^b\ge a^{2b}\plus{}b^{2a}$,

$ 2\sqrt{a^{ka}b^{kb}}\ge a^{kb}\plus{}b^{ka}, \ \ 0<k\le e.$[/quote]
A long and ugly  proof for the last inequality is given by L. Matejicka in the paper [i]Proof of One Open Inequality [/i]from [i]Journal of Nonlinear Science and Applications[/i], see
http://www.tjnsa.com/index.php?volume=7&issue=27
I think there is a nicer proof for the elegant inequality
$ 2a^ab^b\ge a^{2b}\plus{}b^{2a}$.

[quote="Vasc"][quote="Vasc"]This is a nice inequality
\[2a^ab^b\ge a^2+b^2, \ \ \ \ a,b>0.\][/quote] For $a,b\in (0,1]$ there is a nice proof. :wink:[/quote]
Also, for $a+b=2$ there is a nice proof.
On the other hand, a sharper inequality for all $a,b\in (0,1]$ is the following
\[(a^2+b^2)\left(\frac 1{a^{2a}}+\frac 1{b^{2b}}\right)\le 4.\]

Vasc, with all my respect, why do you set up such a bad example for other people by

a) reviving a [i]very[/i] old thread by adding a new problem to it instead of starting a new discussion
b) turning the thread into a monologue
c) changing problem statements faster than people can possibly think of them
d) quoting your own post that immediately precedes the current post and also quotes your own post that immediately precedes the post you quote in almost every post

?

I do not have mod privileges in the Inequalities, but such behavior in CP from anybody any less senior and revered than you would definitely warrant a warning. ;)

On the bright side, I like your three ($n?$) variable version of $a^{b-a}+b^{a-b}\le 2$ :). Also I wonder whether you are aware of the fact that
\[
a^{b-a}\le 1+a(1-\log a)-b(1-\log b)
\]
as long as $a,b\le q$ where $q\log^2 q=1$ ($q=2.0207\dots$) and the proof for $a,b\in[0,1]$ is not hard at all?

Dear fedja. I don't agree with you. Be well!

[i]Fedja[/i], I wonder why you didn't see the teaching side of my repeated posts in this topic.
Starting from the very beautiful and hard inequality $2a^ab^b\ge a^{2b}+b^{2a}$ for $a,b\in (0,1]$, I posted the less difficult and also beautiful inequality $2a^ab^b\ge a^{2}+b^{2}$ for $a,b>0$. However, since this is not easy to prove, I suggested that there are easy proofs for the cases $a,b\in (0,1]$ and $a+b=2$. Now, I think it is a good occasion  :wink:  to propose the following generalization of the last case $a+b=2$, namely
\[a^ab^b\ge \left(\frac{a^2+b^2}{2}\right)^{\frac{a+b}{2}}, \ \ \ a,b>0,\] for which there is nice proof.
Anyhow, your post is welcome, because it broke my monologue. :(
In addition, your proposed inequality $a^{b-a}\le 1+a(1-\log a)-b(1-\log b)$, helpful for $a^{b-a}+b^{a-b}\le 2, \ \ \ a,b\le 2.0207...$, is very nice and interesting. :lol:

[quote="Vasc"]Now, I think it is a good occasion  :wink:  to propose the following generalization of the last case $a+b=2$, namely
\[a^ab^b\ge \left(\frac{a^2+b^2}{2}\right)^{\frac{a+b}{2}}, \ \ \ a,b>0,\] for which there is nice proof.
[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=67464

Thank you, Arqady!
On the other hand, the following reverse inequality hold
\[a^ab^b\le (a^2-ab+b^2)^{\frac{a+b}{2}}, \ \ \ a,b>0.\]
Fedja, I found in my computer a proof of the inequality $a^{b-a}+b^{a-b}\le 2$ for $a+b\le 4$, but my proof (given in 2010) is not very nice (is based on three lemmas). I think that my method can be extended to the case $a+b\le 2q$, where $q\log^2 q=1$ $(q=2.0207\dots)$, which is more general than the case  $a,b\le q$. Have you a proof for your nice inequality  $a^{b-a}\le 1+a(1-\log a)-b(1-\log b)$ ?

[quote="Vasc"]
On the other hand, the following reverse inequality hold
\[a^ab^b\le (a^2-ab+b^2)^{\frac{a+b}{2}}, \ \ \ a,b>0.\]
[/quote]
For the proof we need to prove that $f(x)\geq0$ for $x>0$, where $f(x)=(x+1)\ln(x^2-x+1)-2x\ln{x}$, which is easy.


probability
induction
symmetry
combinatorics unsolved
combinatorics

In airplanes, the ticket indicates, among other things, the number of the chair that the passenger should sit.
John, the first passenger to enter the plane, with 100 chairs, lost his ticket, and so he sat in a random chair. After this, each one of the other 99 passengers sat in it's chair if it was empty or in a random empty chair if it's own chair was not empty.
Let $ P(k)$ be the probability of the $ k$-th passenger to enter the plane and sit in his own chair. Find an expression for $ P(k)$, $ 2 \equal{}< k \equal{}< 100$.

By induction on $ n \geq 2$, we will prove that for a group of $ n$ persons under the conditions of the statement, we have $ P(k) \equal{} \frac {n \plus{} 1 \minus{} k}{n \plus{} 2 \minus{} k}$ for $ k \equal{} 2, \cdots, n$.

We assume that the pre-assigned place of passenger $ i$ is seat number $ i$.
Let $ a$ be the number of the place chosen by the first passenger. Since he chooses his place at random, with equal probability, we will call him "a fool".

For $ n \equal{} 2$, it is clear that $ P(2) \equal{} \frac 1 2$, since the second passenger sits to his own place if and only if the fool chooses place number $ 1$.

Assume that the desired result holds for any group of $ p$ persons with $ 2 \leq p \leq n$.
Consider a group of $ n \plus{} 1$ persons, with a fool first entering.
Let $ k \in \{2,\cdots, n \plus{} 1 \}$.
We clearly have $ P(a \equal{} i) \equal{} \frac {1}{n \plus{} 1}.$

Now,
- If $ a \equal{} k$, we have $ P(k/a \equal{} k) \equal{} 0$.
- If $ a \equal{} i > k$ or $ a \equal{} 1$ then each of the persons $ 2,\cdots, k \minus{} 1$ seats on his own places, so $ P(k/a \equal{} i) \equal{} 1$.
- If $ a \equal{} i$ where $ i\in \{2,\cdots,k \minus{} 1\}$ :
Then, passengers $ 2,\cdots, i \minus{} 1$ are seating each on his own places and when passenger number $ i$ is entering on the plane we are in the initial conditions of the statement but with person number $ i$ acting as a fool, and in a group of $ n \plus{} 1 \minus{} (i \minus{} 1) \equal{} n \plus{} 2 \minus{} i \leq n$ persons. Note that this remaining group has at least two persons since $ i < k$. Moreover, in this group, person number $ k$ has now number $ k \minus{} (i \minus{} 1) \equal{} k \plus{} 1 \minus{} i > 1$.
Therefore, we may use the induction hypothesis to deduce that
$ P(k/a \equal{} i) \equal{} \frac {n \plus{} 2 \minus{} i \plus{} 1 \minus{} (k \minus{} i \plus{} 1)}{n \plus{} 2 \minus{} i \plus{} 2 \minus{} (k \minus{} i \plus{} 1)} \equal{} \frac {n \plus{} 2 \minus{} k}{n \plus{} 3 \minus{} k}.$

Thus
$ P(k) \equal{} \sum_{i \equal{} 1}^{n \plus{} 1} P(a \equal{} i)P(k/a \equal{} i)$
$ \equal{} \frac {1}{n \plus{} 1}[(n \plus{} 2 \minus{} k)\times 1 \plus{} 1\times 0 \plus{} \frac {n \plus{} 2 \minus{} k}{n \plus{} 3 \minus{} k} \times (k \minus{} 2)]$
$ \equal{} \frac {n \plus{} 2 \minus{} k}{n \plus{} 3 \minus{} k}$
as desired.
This ends the induction step and the proof.

Pierre.

[quote="pbornsztein"]
For $ n \equal{} 2$, it is clear that $ P(2) \equal{} \frac 1 2$, since the second passenger sits to his own place if and only if the fool chooses place number $ 1$.
[/quote]

But... since there are 100 places, the probability of choosing the right seat, for the first passenger, is $ \frac{1}{100}$... and then $ P(2)\equal{}\frac{1}{100}$ as well... isn't it?

No, for $ n \equal{} 2$ there are two persons. You are confusing $ n$ and $ k$.

Pierre.

Hey, thank you Pierre, I solved that and got same answer, but your solution is better. Really thanks.

An alternative ([b]not[/b] my own) way of looking at the problem:

[hide]
Let us suppose that instead of sitting in a random seat, each person after the first person kicks the other guy out of his seat, and the other guy goes and sits in a new random seat.  This leads to the same seat occupancy distribution as before, with the difference that persons $ 2$ through $ k \minus{} 1$ are automatically in their own seat.

By symmetry, person $ 1$ is equally likely to be in any seat at this point, so we are done.[/hide]


A certain sporting goods store imported 100 basketballs and 80 volleyballs for $ 2800. When selling the products, the price for each basketball increased by 5% and for each volleyball by 10%. After all the products are sold, the store collected
$3020. How much is the sum of the prices of one basketball and one volleyball?

The full problem is: "A certain sporting goods store imported 100 basketballs and 80 volleyballs for
2800 dollars. When selling the products, the price for each basketball increased by 5% and for each volleyball by 10%. After all the products are sold, the store collected 3020 dollars. How
much is the sum of the prices of one basketball and one volleyball?"

[hide]We set up two equations:

$ 100b \plus{} 80v \equal{} 2800$ -- 1

$ 105b \plus{} 88v \equal{} 3020$ -- 2

Solving for b and v, we get $ b \equal{} 12$ and $ v \equal{} 20$.

Therefore, the sum of the prices of one basketball and one volleyball is $ 12 \plus{} 20 \equal{} \fbox{32}$.[/hide]

[quote="sri340"]A certain sporting goods store imported 100 basketballs and 80 volleyballs for $ \$2800$. When selling the products, the price for each basketball increased by 5% and for each volleyball by 10%. After all the products are sold, the store collected $ \$3020$. How much is the sum of the prices of one basketball and one volleyball?[/quote]

Fixed LaTeX.

We have:

$ 100B\plus{}80V\equal{}2800$

$ 105B\plus{}88V\equal{}3020$

This gives

$ 5B\plus{}8V\equal{}220$

Multiplying this by 10 gives

$ 50B\plus{}80V\equal{}2200 \implies 50B\equal{}600 \implies B\equal{}12 \implies V\equal{}20$.

Our answer is $ 32$.


How many numbers are in the list \[ 6,7,10,11,14,15,\ldots,94,95,98?\]

Since every two terms increase it by 4, 98-6=92=4*23, and the answer is 23*2+1, or [b]47[/b].