OK, since no one wants to answer my post in the other thread... I'll make a new one.

So how EXACTLY does the US Open work (for those of you who have taken it or are more knowledgeable about it than me)? Is it like the other contests throughout the year where we send in the programs online except just proctored? Or is it something completely different?

Thanks in advance for your replies.

The U.S. Open is conducted pretty much like the other contests (but proctored and on a specific day). Additionally, the contest is 5 hours long and, as I recall, has stronger limits on what resources can be used.

On another note, can Canadian students also write the U.S. Open proctored on the earlier date or is it just for the american students?

All students trying for spots on the US team are supposed to take it on the April 28th date, with a proctor. The proctor has to attest to who you are, who they are, and help you comply with some extra security measures in order for anyone to log on on that date. Everyone else (including Canadians) takes it that week-end, like a regular internet contest (no proctor).

One more question: we theoretically take this at school right? Does that mean we install our compiler there? Because I don't want to waste time trying to figure out how a different compiler works...

[quote="paladin8"]One more question: we theoretically take this at school right? Does that mean we install our compiler there? [/quote]

You can take it wherever you want to, as long as you have a proctor who is approved in advance by the USACO coach.

also, any friendly adult, including one of your parents can proctor if you would rather do this at home rather than at school. They have to send an email to the USACO coach sort of introducing themselves. He may call them to confirm that there is a responsible adult vouching for the exam conditions and not a student pretending to be a proctor.

Thanks for your help, texas137! I'm gonna be taking it at home with my mom as proctor. Good luck to everyone who is taking it!


Find length of normals drawn from the point on the axis of parabolay2=8ax whose distance from focus is 8a.

Is the answer $ 8|a|$? If it is, I'll post my solution. If not, I just want some clarification. The normal line to a point on the parabola is the line perpendicular to the tangent line right? And is the length the question is asking for the distance between the point on the axis mentioned and the point on the parabola?

the answer given is 10a,8a and 8a.
yes,the normal is perpendicular to the tangent.
yes,that's the distance asked.


Following is a recap of yesterday's MMC meeting posted at http://www.metroplexmathcircle.org

The first Math Circle of the 2008-2009 season succeeded on multiple levels. In its third year, Metroplex Math Circle finally exceeded the capacity of its current (large) lecture hall with over 100 students, teachers and parents in attendance. We were pleased to see many friends return, but even more new faces who were discovering Math Circles for the first time.

Comparative Advantage

Richard Rusczyk’s talks also exceeded all expectations. His first lecture was on the concept of comparative advantage and the benefits of free trade which many adults fail to grasp. Like many lessons in Math Circle, Mr. Rusczyk allowed the students to discover the meaning of the concept by playing out a game between two imagined countries. “Games” and the importance of “play” would be stressed throughout the day.

Option Pricing

The second segment introduced the audience to the principles of market prices and call options. With the current turmoil on Wall Street, parents were just as interested in this topic as the students. Mr. Rusczyk presented the students with a problem that he would use when he was recruiting for D.E. Shaw. Only the best students from elite schools ever made it to this stage in the interview process and apparently all but a handful (accomplished problem solvers) ever answered the problem correctly.

Life After MATHCOUNTS

In his final lecture, Richard Rusczyk sent students, parents and teachers home with a great deal to consider. He spoke very frankly and persuasively about the short comings of the standard school curriculum. But rather than just criticize, he also laid out concrete ways that students, parents and teachers can all improve the situation.

We won’t attempt to summarize this excellent talk particularly since an early version of the slides can be accessed below. However, a couple of themes should be highlight. First was his emphasis on play and the importance of giving students the time and freedom to work on challenging problems. A second theme was the importance of building a math community for young problem solvers, and the central role that a Math Circle can play. Finally, Mr. Rusczyk endorsed what many of us have already discovered, that Dr. Titu Andreescu’s books and leadership have been critical to the Renaissance in global problem solving.

how often do they hold a MMC meeting?  :)

MMC meetings are basically every Saturday at 2:00 while UT Dallas is in session (Fall and Spring semesters).  Details on the schedule and each speaker are available at:  http://www.metroplexmathcircle.org

We do not do an official sign up, so anyone traveling through the area or interested in trying it out is welcome to attend.


Hi,
# 40 on the 2006 Algebra II State Math Contest is:

[b]Let the function A(x,y) be defined by the following rule:[/b]
\[ A(0,n) \equal{} n \plus{} 1
\]

\[ A(m,0) \equal{} A(m \minus{} 1,1)
\]

\[ A(m,n) \equal{} A(m \minus{} 1, A(m,n \minus{} 1))
\]
[b]where m and n are positive integers.
Find the numerical value of A(2,3)[/b]

I later found that this is the Ackermann Function, but it seems the only way to solve it is using the pre-made table on wikipedia... Anybody have a better way to find the answer (without brute force)?

Thanks damien.
I gave up on it after the fourth substitution. lol.
i guess if even you don't have a better way, nobody i'd be competing with would either....


(1)Prove that any rational number can be written as a sum of reciprocals of finitely many DISTINCT positive integers.

(2). Find the sum to infinity:-

	$1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + .....$

1) $\frac pq=\frac 1q+\frac 1q+\dots \frac 1q$
2) $1-\frac 13 +\frac 15 -\frac 17+\dots =Im(ln(1+i))=\frac{\pi}{4}.$

Well for 1): I think it shall be distinct integers (and positive rationals).

@rohitsingh0812
Please only one problem per thread.
And 2) is really not Number Theory.


I'm not sure we have many/any middle schoolers from wisconsin, but please say how you placed, not scored at chapters in '07

6th written, 1st cd

 :mad:  :mad:  :mad:

jorian

How bad did you crush the spirits of the other poor middle-schoolers kstan?


modular arithmetic
number theory
number theory proposed

[color=blue]Find all the integer $ n$ such at $ n \mid (2^n\plus{}1)$[/color]

[hide="incorrect solution"]First of all $ 2^n \plus{} 1$ is odd, so $ n$ must be odd, but then $ gcd(2,n) \equal{} 1$ and using Fermat's little theorem we get that $ 2^n \plus{} 1 \equiv 3$ (mod $ n$) so the only "good" $ n$'s are $ n \equal{} 1$ or $ n \equal{} 3$ and we may check, that $ 1|2^1 \plus{} 1$ and also $ 3|2^3 \plus{} 1$[/hide]

[edit] @Rust: thanks for pointing out my mistake :)

[quote="Ereinion"]First of all $ 2^n \plus{} 1$ is odd, so $ n$ must be odd, but then $ gcd(2,n) \equal{} 1$ and using Fermat's little theorem we get that $ 2^n \plus{} 1 \equiv 3$ (mod $ n$) so the only "good" $ n$'s are $ n \equal{} 1$ or $ n \equal{} 3$ and we may check, that $ 1|2^1 \plus{} 1$ and also $ 3|2^3 \plus{} 1$[/quote]
It is not true/. If $ n|2^n\plus{}1$, then $ n|2^{nm}\plus{}1$ for any odd $ m$. For example $ N\equal{}2^n\plus{}1\to N|2^N\plus{}1$ ($ 9|2^9\plus{}1$), for any odd $ n$, suth that exist $ n|2^m\plus{}1$ exist $ n|N|2^N\plus{}1$.

$ 3^{k}|2^{3^{k}}\plus{}1$

proceed by induction.
Alternatively,
This is a direct consequence of the lifing the exponent lemma.
http://docs.google.com/viewer?a=v&q=cache:93o6nb9DW4gJ:reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf+exponent+lifting&hl=en&gl=in&sig=AHIEtbT1lViijOH2j8bsRZ1BkfM5zKKL_w

This doesnt solve your problem in its entirety though.

[quote="nolove"][color=blue]Find all the integer $ n$ such at $ n \mid (2^n \plus{} 1)$[/color][/quote]

I think that the solution of this problem is infinite.

Let us consider the sequence $ \{x_n\}$ defined by $ x_1 \equal{} 1$, $ x_{n \plus{} 1} \equal{} 2^{x_n} \plus{} 1$. We will prove that $ x_n|x_{n \plus{} 1}$ for all $ n\in \mathbb {N}^{*}$. Indeed, for $ n \equal{} 1$, this is obviously true. Now suppose that for some $ k\ge 1$, $ x_k|x_{k \plus{} 1}$. We will prove that $ x_{k \plus{} 1}|x_{k \plus{} 2}$, then we will complete the deduction. Since $ 2^{x_k} \plus{} 1$ and $ x_k$ are odd then $ \frac {2^{x_k} \plus{} 1}{x_k}$ is odd also. Then:
\[ x_{k \plus{} 2} \equal{} 2^{x_{k \plus{} 1}} \plus{} 1 \equal{} 2^{2^{x_k} \plus{} 1} \plus{} 1 \equal{}\left (\left (2^{x_k}\right )^{\frac {2^{x_k} \plus{} 1}{x_k}} \plus{} 1\right ) \vdots \left (2^{x_k} \plus{} 1\right ) \equal{} x_{k \plus{} 1}\]
Hence, $ x_n|x_{n \plus{} 1}$ for all $ n$ is positive integer numbers. And the result is lead as followed.

@mathVNpro : I don't think your prove is enough :)

Note that [b]Find all [/b]

[quote="nolove"][color=blue]plz help me  :( ... anyone had files about [b]ord , Legendre’s symbols [/b]...

How about this one:

Find $ n \in N^*$ such at $ n^2 \mid (3^{n} \plus{} 1)$[/color][/quote]

We have $ n \equal{} 1$ is trivial.
If $ n$ is even $ \Rightarrow 3^n \plus{} 1 \equiv 2( mod 4 )$. It is impossible because $ n^2 | 3^n \plus{} 1$
So , $ n$ is odd.
Defining $ p$ as the least prime divisor of $ n$ $ \Rightarrow p$ is odd. We have $ p|3^{2n} \minus{} 1$
Let $ ord_p3 \equal{} d$
$ \Rightarrow d|2n$ and $ d|p \minus{} 1$
If $ gcd(d,n) > 1$ then exist $ p'|n$ satisfy $ p' < p$, contradiction.
Hence, $ d \equal{} 2$ or $ d \equal{} 1$ , which both are impossible because $ p$ is odd.
In conclusion, $ n \equal{} 1$ is the unique solution.

[quote]Find $ n \in N^*$ such at $ n^2 \mid (3^{n} + 1)$[/quote]
Just another way but maybe the same idea. I hope there is no serious mistake.

Since $ n^2 \mid (3^{n} + 1)$, thus $ n \mid (3^{n} + 1)$. For $ n > 1$, then let us denote $ p$ by the least prime divisor of $ n$, we have $ 3^n\equiv - 1 {\pmod p}$. Squaring both side, we obtain, $ 3^{2n}\equiv 1 {\pmod p}$. Let $ d = ord_{p}(3)$ and suppose $ d > 1$. It is implied that $ d \mid 2n$. If $ d < 2$ then $ d\in \{1,2\}$, but in both cases and also note that $ n$ must be odd, it is not hard to check that $ p = 2$, which leads to the contradiction. Therefore, it is enough to consider the case $ d\ge 2$. Suppose $ d = 2^{k}.d'$, where $ \gcd (2,d') = 1$. We have
\[ p \mid  (3^{2^{k}.d'} - 1 ) =  (3^{2^k} - 1 ) (\sum_{i = 0}^{d' - 1} (3^{2^k})^{i} )  )\]
Since $ 2^{k} < d$ $ \Longrightarrow$ $ 3^{2^k}\not \equiv 1 {\pmod p}$. Thus $ A =\sum_{i = 0}^{d' - 1} (3^{2^k})^{i} )$ $ \equiv 0 {\pmod p}$. Then $ s = v_p(A)$ $ \le$ $ \max \{v_p  ( (3^{2^k} )^{i} ) \}$, with $ i = \overline {0,d' - 1}$. But $ p$ cannot be 3 $ \Longrightarrow$ $ v_p ( (3^{2^k} )^{i} ) = 0$, $ \forall i$. As the consequence $ s\le 0$ $ \Longrightarrow$ $ s = 0$. Which leads to the contradiction. As the result, $ \gcd (2,d) = 1$, hence $ d\mid n$. But $ d$ also divides $ p - 1$ (By Fermat's little theorem), which implies that $ n$ has another prime divisor that smaller than $ p$, which also leads to the contradiction. Hence, for the case $ n > 1$, there is no such $ n$ satisfies the problem. Hence $ n = 1$. It is easy to check again that $ n = 1$ is satisfied. To sum up, $ n = 1$ is the only solution to the problem.


Παιδιά φίλοι μου απεχθείς φήμες κυκλοφορουν οτι ο Θαλής θα πραγατοποιηθει Δεκέμβριο ούτε που θέλω να σκέφτομαι να αλλάξει ημερομηνια τι λέτε για αυτό???

Σύμφωνα πάντοτε με τη μαθηματική εταιρία, ο Θαλής θα διεξαχθεί κανονικά, το Σάββατο 4 Νοεμβρίου στις 9:00 π.μ.

Check it out here:
http://www.hms.gr/gr/activities/competitions/0607/diag.html

υπαρχει ενας φοβος λογω καταληψεων και   απεργιων .εχει αναβληθει ποτε μαθηματικος διαγωνισμος στην Ελλαδα?

φίλε μου κι εγώ την ίδια πληροφόρηση έχω
αλλά από κύκλους της εταιρείας έμαθα τις φήμες

Δεν πιστεύω ότι θα ναβληθεί γιατί εντάξει κάποιο χώρο θα βρούνε να γίνει ο διαγωνισμός .........

μα το πρόβλημα φίλε μου Σιλουανέ ΔΕΝ είναι ο ΄χωρος αλλά ο χρόνος

http://www.hms.gr/gr/activities/index.html
Απο εδώ φίλοι μου θα δούμε την λύση στο πρόβλημα αυτό
Ο θαλής διεξάγεται κανονικότατα συγχαρητήρια σε όσους συνέβαλλαν σε αυτό
αλλά μομφή σε όλους όσους συνέβαλλαν στο χάλι του ότι ακόμα δεν έχουν διεξαχθει μαθήματα

Μαθήματα; Δεν τη γνωρίζω αυτή τη λέξη... Εμείς στην πόλη όπου κατοικώ, το Κιλκίς, έχουμε παράρτημα της Ελληνικής Μαθηματικής Εταιρίας, το οποίο [b]ΔΕΝ[/b] έχει κάνει τίποτα για να βοηθήσει τους μαθητές σε ό,τι αφορά την προετοιμασία για τους μαθηματικούς διαγωνισμούς. Ο πρόεδρος του παραρτήματός μας εκλέγεται μόνο και μόνο για τη συγκέντρωση εξουσιών. Οι περισσότεροι που θέλουν να έχουν ελπίδες για να περάσουν στην επόμενη φάση κάνουν "ιδιαίτερα". Κάτι τέτοιο το βρίσκω απαράδεκτο και καθόλου αξιοκρατικό και ούτε πρόκειται να κάνω φροντιστήριο για να περάσω σε ένα διαγωνισμό (που στην πόλη μας τουλάχιστον) μας έχει κάνει να πιστεύουμε πως φέρει το σλόγκαν "Κάντε ιδιαίτερα για να περάσετε...". Επομένως, οι διακρίσεις των μαθητών δε γίνονται μόνο βάσει των ικανοτήτων τους, αλλά δεν πρέπει να παραλείψουμε και τους δευτερογενείς (?) παράγοντες όπως:

α) Έλλειψη μαθηματικής υποστήριξης σε σχέση με άλλους νομούς (μας υποσχέθηκαν και βιβλιοθήκη, αλλά...)
β) Υποβολή θεμάτων που ως σχολική ύλη τα είχαμε εκτός και δεν μας είπε κανείς διαβάστε τα (π.χ. πέρσι βάλαν στη 2α Λυκείου τριγωνομετρική ανισότητα που δεν την είχαμε διδαχθεί εμείς στην ύλη της 1ης λυκείου)
γ) Πραγματική "άγνοια" των καθηγητών του σχολείου για τη Μαθηματική Εταιρία και καθαρή αδιαφορία.

Φανταστείτε πόσα παιδιά χάνονται έτσι και δεν αναδεικνύονται.

[b][i]Ως πότε αυτή η κατάσταση?[/b][/i]

(Η ερώτηση μάλλον είναι ρητορική)

Φιλε Μοστελ, ειναι ολοφανερο οτι το σχολειο δεν σε θελει
να αγαπησεις τα μαθηματικα (δεν θελω να το σχολιασω περαιτερω).
Δεν φτανει μονο αυτο αλλα εμμεσα σου λεει οτι τα μαθηματικα
δεν ειναι για τον οποιονδηποτε ειναι για ευφειες που μπορουν
και λυνουν ασκησεις επι φυτιση αγιου πνευματος!
Εγω θεωρω τη διεξαγωγη των μαθηματικων ολυμπιαδων στην
ελλαδα μια πληρη αποτυχια, ειναι δυνατον να γραφει στο
site της η ε.μ.ε οτι τα θεματα των διαγωνισμων αναφερονται στη
σχολικη διδακτεα υλη, επιτρεπεται να μην προτεινουν βιβλιογραφια?
Παντως οποιος εχει παθος για τα μαθηματικα δεν πρεπει να
απογοητευεται απο αυτη την κατασταση, γιατι τα μαθηματικα ειναι
ειναι ενας κοσμος μονξιας οπου κυριαρχει η ελλογη σκεψη κανεις
δεν προκειται να σε μαθει αν δεν κοπιασεις μονος σου, μπορει να
σε μαθει μηχανισμους για να λυνεις ακησεις αλλα οχι το να ερευνας
αυτο ειναι καθαρα προσωπικη διαδικασια.

ομως χωρις διαγωνισνους πως θα ηταν  τα μαθηματικα στα σχολεια?οσο για αυτο που λες μοστελ δεν ξερω ισως τα οικονομικα να μην επιτρεπουν μια καλητερη οργανωση  στην ΕΜΕ.ευτηχως υπαρχει ακομα η αγαπη την επιστημη και συνεχιζουν καποιοι ανθρωποι  :roll:

Ο ΘΑΛΗΣ ΘΑ ΓΙΝΕΙ ΜΑΛΟΝ ΣΤΙΣ 9 ΔΕΚΕΜΒΡΙΟΥ

Το ΔΣ της ΕΜΕ ζήτησε σήμερα από το ΥΠΕΠΘ την αναβολή του ΘΑΛΗ για τις 9 Δεκεμβρίου 2006.

Lance

Οριστικά στις [color=red]9 Δεκεμβρίου [/color].

 Ήρθε σήμερα πρωϊ  η ανακοίνωση από το Υπουργείο.

 [u]Μπάμπης [/u]

Λοιπόν παιδιά αυτό είναι το αποκορύφωμα της σαπίλας που κυβερνά την παιδεία μας (υπερβάλλω αλλά...) Δεν μπορεί να υπάρχει αυτή ή κατάσταση δη΄λαδή για κάποια κωλοτσογλάνια τεμπερχανάδες που απλώς κωλοβαράνε κι κάνουν καταλήψες να την πληρώνουν οι λίγοι που θέλουν να κάνουν παιδεία (πώς τολμάνε έ???) ε βέβαια για αυτό μας περνάνε σε όλα οι Τούρκοι εκεί τα τσογλάνια τρωνε ξυλίκι από την αστυνομία κι τελειώνει η ιστορία
αυτά ειναι λίγα από αυτά που θέλω να πώ είναι όμως τα πιο αντιπροσωπευτικά τα θέτω προς συζήτηση

ευχαριστώ πάρα πολύ που συμφωνείς μαζί μου φίλε μου
κι λογικά θα πρέπει οι πιο ένθερμοι οπαδοί των απόψεων μου να επαυξάνετε με αυτές
τέλος πάντων μια έκφραση του hilbert θα πώ
Sie mussen Vissen Wien Werden Vissen
(πρέπει να μάθουμε κ ιμε κάθε τρόπο θα μάθουμε)


[url=http://www.icho39.chem.msu.ru/html/english/ICHO-2007/Events.htm]here[/url] are the results of ICho 2007.india has got 2 golds,1 silver and a Bronze  :)

Do you have also the results of ICHO2008? Thank you.

I think India won 3 silver and 1 bronze.

absolutely correct chemrock
and the detailed results on
http://www.icho.hu/files/40thIChO_list.pdf


probability
modular arithmetic
number theory
greatest common divisor
relatively prime

Let $S$ be the set of integers between $1$ and $2^{40}$ whose binary expansions have exactly two $1$'s. If a number is chosen at random from $S$, the probability that it is divisible by $9$ is $p/q$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

[hide="hint"]The decimal expansion of integers between $1$ and $2^{40}$ with two $1 's$ in it binary expansion is defined like:

$\{2^n.1+2^{n-k}.1=2^{n-k}(2^k+1)|\forall n,k\ 1\le k \le n\ \wedge 1\le n <40\}$

the set of integers that's divisible by $9$ contain all the numbers such that $(2^k+1)$ is a consecutive of a even number and a multiple of nine and since the four possible unit digit of this number is:$3,5,9,7$ we need to take the numbers that finish in $3$ or $9$
 [/hide]

[hide]of the two $1$'s, one of them must be on the left, and the other one can be anywhere among the remaining digits.
The smallest number consists of one digit on the left and one slot.
The largest number consists of one digit on the left and 39 slots.
Thus we have $1+2+\dots+39 = \frac{39*40}{2}=780$ possible values.
These numbers are the sum of a smaller power of 2 and a larger power of 2.
We can always factor out the smaller one, which is never divisible by 9.
Now we have to consider the other factor, $1+2^x \equiv 0 \mod 9$
The obvious scenario is when $x=3$.  There are $40-4+1 = 37$ multiples of 9.
However, this does not give all the cases.
$2^{\phi(9)} \equiv 1 \mod 9$, so we have to consider all $x=3+6n$
Each subsequent $x$ yields 6 fewer multiples.
Thus the sum is $37+31+\dots + 1 = \frac{38*7}{2} = 133$
The probability is $\frac{133}{780}$ and $133+780=\boxed{913}$[/hide]

[quote="alan"][hide]of the two $ 1$'s, one of them must be on the left, and the other one can be anywhere among the remaining digits.
The smallest number consists of one digit on the left and one slot.
The largest number consists of one digit on the left and 39 slots.
Thus we have $ 1 \plus{} 2 \plus{} \dots \plus{} 39 \equal{} \frac {39*40}{2} \equal{} 780$ possible values.
These numbers are the sum of a smaller power of 2 and a larger power of 2.
We can always factor out the smaller one, which is never divisible by 9.
Now we have to consider the other factor, $ 1 \plus{} 2^x \equiv 0 \mod 9$
The obvious scenario is when $ x \equal{} 3$.  There are $ 40 \minus{} 4 \plus{} 1 \equal{} 37$ multiples of 9.
However, this does not give all the cases.
$ 2^{\phi(9)} \equiv 1 \mod 9$, so we have to consider all $ x \equal{} 3 \plus{} 6n$
Each subsequent $ x$ yields 6 fewer multiples.
Thus the sum is $ 37 \plus{} 31 \plus{} \dots \plus{} 1 \equal{} \frac {38*7}{2} \equal{} 133$
The probability is $ \frac {133}{780}$ and $ 133 \plus{} 780 \equal{} \boxed{913}$[/hide][/quote]

Why do we have to consider all $ x \equal{} 3 \plus{} 6n$?  This step is really confusing...

The idea is that if you have some $ 2^a \plus{} 2^b, a<b$ then $ 2^{b\minus{}a}\plus{}1$ is a multiple of $ 9$. To find $ 2^k$ that are $ \minus{}1 mod 9$, first find the trivial case $ 2^3$. $ 2^6$ is $ 1$ mod $ 9$, and we check that there is no lower identity (none of $ 2^1...2^5$ are 1 mod 9). So that's how u get $ 3\plus{}6k$.

[hide="Perhaps an easier method"]
We start calculating the powers of $ 2$ mod $ 9$:

$ 2^0\equiv 1\pmod 9$

$ 2^1\equiv 2\pmod 9$

$ 2^2\equiv 4\pmod 9$

$ 2^3\equiv 8\pmod 9$

$ 2^4\equiv 7\pmod 9$

$ 2^5\equiv 5\pmod 9$

$ 2^6\equiv 1\pmod 9$

So the powers cycle with period $ 6$.  Between $ 0$ and $ 39$ inclusive there are $ 7$ numbers congruent to $ 0,1,2,3$ mod $ 9$ and $ 6$ numbers congruent to $ 4,5$.

For the two powers to sum to $ 0$, they must be $ 8$ and $ 1$, $ 7$ and $ 2$, or $ 5$ and $ 4$ mod $ 9$.  The total number of possibilities is thus

$ 7(7)\plus{}7(6)\plus{}7(6)\equal{}133$.

The number of ways to form a number with $ 2$ $ 1$'s in its binary representation without exceeding $ 2^{40}$ is simply $ \binom{40}{2}\equal{}780$.

So the probability is $ \frac{133}{780}$ and the answer is $ 913$.
[/hide]

[quote="cdymdcool"]Why do we have to consider all $ x \equal{} 3 \plus{} 6n$?  This step is really confusing...[/quote]

Perhaps a more clear explanation:
By Euler's totient theorem,
\[ a^{\phi(n)}\equiv 1\pmod n\]
if $ \text{gcd}(a,n)\equal{}1$, where \[ \phi(n)\equal{}n\prod_{p|n}\left(1\minus{}\frac{1}{p}\right),\] where $ p$ is prime.

\[ \phi(9)\equal{}9\cdot\frac{2}{3}\equal{}6,\] and $ 2$ is coprime to $ 9$, so \[ 2^{6}\equiv 1\pmod 9.\] However, we want values of, say, $ x$, such that \[ 2^{x}\equiv \minus{}1\equiv 8\pmod 9\]. $ 8$ is a power of $ 2$, $ 2^3$.

Therefore, multiplying both sides by $ 2^3$, we get:
\[ 2^{x'\plus{}3}\equiv 2^3\equal{}8\pmod 9,\] where $ x'\equiv 0\pmod 6.$

So $ x'\plus{}3$ is equivalent to $ 6n\plus{}3$ for some integer $ n$.

My Solution:
Since the binary has two ones only, we know that the number is in form 2^a+2^b.
Let $2^x$ be the smaller of the two and $2^ (x+y)$ be the larger. So you can add those together then factor to
$2^x (1+2^y)$. We know 2^x cannot be multiple of 9. So $(1+2^y)$ has to. See if we can find pattern:
If $y=0$ then  $(1+2^y)=2$.
If $y=1$ then  $(1+2^y)=3$.
If $y=2$ then  $(1+2^y)=5$.
If $y=3$ then  $(1+2^y)=0$.
If $y=4$ then  $(1+2^y)=8$.
If $y=5$ then  $(1+2^y)=6$.
If $y=6$ then  $(1+2^y)=2$.
As you can see repeats from $y=0 to y=5$. Its repetitive cycle. As you can see $y=3$ is 0 mod 9 which means its divisible. Each 3+6x is. Lets count possibilities.
If $y=3$ then  $x$ can be from (0 to 36). Because it can't get over 40.
If $y=9$ then  $x$ can be from (0 to 30).
If $y=15$ then  $x$ can be from (0 to 24).
If $y=21$ then  $x$ can be from (0 to 18).
If $y=27$ then  $x$ can be from (0 to 12).
If $y=33$ then  $x$ can be from (0 to 6).
If $y=39$ then  $x$ can be only (0).
Add 37,31,25,19, 13,7 and 1 together. Thats 133. Numerator.
Denominator is 40 choose 2 because there are 40 possible slots and two 1's to insert in. Thats 780.
So we have 133/780 which is 133+780=913. 913 is final answer!


function
search
algebra unsolved
algebra

Given $a\in R$ and $f_{1}, f_{2},..., f_{n}: R \to R$ additive functions such that $f_{1}(x).f_{2}(x)...f_{n}(x)=ax^{n}$ for all $x\in R$. Prove that there exists $b\in R$ and $i\in{1,2,...,n}$ such that $f_{i}(x)=bx$ for all $x\in R$

[quote="fermat3"]romania 1995 :wink:[/quote]
Solution?

[quote="fermat3"]romania 1995 :wink:[/quote]
Can you post the solution of this problem?

http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc96-97-01feb.pdf.  search romanian olympiad :lol:


factorial
combinatorics unsolved
combinatorics

How many ways can 8 mutually non-attacking rooks be placed on a 9  9 chessboard so that all 8 rooks are on squares of the same colour?

---

This is a very easy problem. The reason I put under the Advanced Section is because I heard that there's a very elegent bijection solution. Of course, during the contest, no one would bother to look for a bijection when an obvious divide-and-conquer solution is present.
Considering the answer is 8!, there must be some kind of bijection with the 8  8 board. Can anyone find it?

are you sure the answer is 8!? :?

I have not proven it yet, but it seems a bit weird
I'd have expected $81 . 8!$ options...

(9 options for the first tower, 8 for the second, ... 2 for the last and dividing by 8! cuz the towers are not numbered...)

All 8 rooks have to be on the same coloured squares. (where the board is coloured as usual - no two black squares are adjacent and no two white squares are adjacent)

Peter: I made exactly the same mistake as you on the contest.  I didn't read that the rooks had to be on the same colour.  This cost me an HM!

Yufei: I actually talked with some of the CMO markers.  When I mentioned 8!, they told me they found no bijection.  A number of them tried unsuccessfully, and I'm quite sure 8! is just a pleasant coincidence.  At the same time, that the CMO markers couldn't find a bijection doesn't mean that it doesn't exist.  But I'm particularly skeptical because when you try it out generally for (2n+1) x (2n+1) chessboards, you don't get a factorial number.


Let $ a,b,c >0$. Prove that :
\[ \frac{1}{a^{2}\plus{}ab\plus{}b^2}\plus{}\frac{1}{b^{2}\plus{}bc\plus{}c^2}\plus{}\frac{1}{c^2\plus{}ca\plus{}a^2}\geq \frac{9}{(a\plus{}b\plus{}c)^2}\]

[quote="Sergic Primazon"]Let $ a,b,c > 0$. Prove that :
\[ \frac {1}{a^{2} \plus{} ab \plus{} b^2} \plus{} \frac {1}{b^{2} \plus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \plus{} ca \plus{} a^2}\geq \frac {9}{(a \plus{} b \plus{} c)^2}\]
[/quote]

Solution
$ 4(a^2\plus{}ab\plus{}b^2)(ab\plus{}bc\plus{}ca) \le (a\plus{}b)^2(a\plus{}b\plus{}c)^2$
Then Use Iran 96
Done!  :lol:


Find the product when
$ \text{6,6 di \minus{} methyl cyclohex \minus{} 3 \minus{} en \minus{} 1 \minus{} ol}$ is treated with $ \ce{conc.H2SO4 /\Delta}$
$ \text{6,6 di \minus{} methyl 1 \minus{} chloro cyclohex \minus{} 3 \minus{} en}$ is treated with $ \ce{Alc.KOH}$

[hide="Opinion"]This is just an opinion:

1) In the first reaction, I think there is the formation of 4,5-dimethylcyclohexene. However, the formation of o-xylene is perhaps also possible.

2) Elimination to form 5,5-dimethylcyclohexa-1,4-diene.[/hide]


λεω να βαλουμε τα θεματα του θαλη εδω.ηρθε η ωρα ,εχω απορια ναδω τα θεματα που θα πεσουν.καλη επιτυχια σε ολους... :D

[quote="kostas"]εχω απορια ναδω τα θεματα που θα πεσουν....[/quote]
den nomizo na diaferun poli apo ta 8emata alon xrono tu 8ali.... :P
[quote="kostas"]καλη επιτυχια σε ολους... :D[/quote]
kali epitixia...

 :D

θελω να πω οτι εχω πατατηρησει οτι μια πεφτει δυσκολος θαλης (με ωραια θεματα ) και την επομενη χρονια ευκολος με οχι ομορφα  :)

Λοιπόν . Ο διαγωνισμός τελείωσε (σε 40 λεπτά για εμένα και 50 για τον αδερφό μου  :) )
Τα θέματα από όλες τις τάξεις ήταν κλασσικά αλλά πολύ ωραία (εμένα μου άρεσαν τουλάχιστον) .Θα βάλλω θέματα και τις λύσεις μου στην Β Λυκείου (Να σημειώσω ότι ίσως η πιο ωραία άσκηση ήταν στην Α Λυκείου το τελευταίο θέμα )

Β ΛΥΚΕΙΟΥ
1) Να εξεταστεί αν η εξίσωση $x^{2}-(2006k+1)x+2007=0$  όπου k ακέραιος , έχει δύο ακέραιες ρίζες .
[color=brown]Η λύση μου[/color]
Έστω ότι έχει την $p$ . Τότε είτε ο $p$ είναι άρτιος είτε αν είναι περιττός η παράσταση $p^{2}-(2006k+1)p$ είναι άρτιος +2007 είναι περιττός 0=περιττός , άτοπο.

2)Δίνεται ορθογώνιο $ABCD$ με $AB=4$ , $BC=2$ και σημείο $M$ στο εσωτερικό του με $MC=1$ και $MB=\sqrt 3$ .
Nα βρείτε το εμβαδό του τριγώνου $MAB$ .
[color=brown]H λύση μου [/color]

Από το νόμο των συνημιτόνων sto $MBC$  βρίσκουμε ότι $cos\widehat{MBC}=\frac{3}{2\sqrt 3}=sin\widehat{MBA}$
Άρα από $(MAB)=\frac{AB\cdot MB\cdot sin\widehat{MBA}}{2}=3$

3)Έστω $k=2+2^{2}+2^{3}+....+2^{2008}$ . Να δείξετε ότι ο 30 διαιρεί τον $k$ .
[color=brown]H λύση μου [/color]

Παρατηρούμε ότι $2+2^{2}+2^{3}+2^{4}=30$ άρα
$k=2+2^{2}+2^{3}+2^{4}+2^{4}(2+2^{2}+2^{3}+2^{4})+...+2^{2004}(2+2^{2}+2^{3}+2^{4})=30m$

4) a) Να δείξετε ότι $\sqrt[3]{2}+\sqrt[3]{3}>\sqrt[3]{19}$
4)b)Να λύσετε την εξίσωση $2^{-1}x+x^{-1}=\frac{\sqrt[3]{38}}{\sqrt[3]{2}+\sqrt[3]{3}}$

[color=brown]H λύση μου [/color]
4)a) Από AM-GM είναι $\sqrt[3]{2}+\sqrt[3]{3}>2\sqrt[6]{6}$
Άρα αρκεί να δείξουμε ότι $2\sqrt[6]{6}>\sqrt[3]{19}$
ή $384>361$  που ισχύει (Είμαι σχεδόν βέβαιος ότι η ΕΜΕ θα το λύνει με την ταυτότητα Εuler :P )

4)b) Προφανώς πρέπει το $x>0$ αφού αν $x<0$ δεν έχουμε λύσεις .
Από το α) ερώτημα παίρνουμε ότι
$\frac{x}{2}+\frac{1}{x}<\sqrt[3]{2}$  (1)
Όμως από AM-GM $\frac{x}{2}+\frac{1}{x}\geq \sqrt{2}$ (2)
Άρα από (1),(2) => $\sqrt{2}<\sqrt[3]{2}$ άτοπο άρα δεν έχει λύσεις

Kalispera se olous apo Hrakleio,

Thewrw oti ta themata fetos itan arketa vata kai arketa eukolotera se sxesi me proigoumenes xronies! Eidika oson afora ti gewmetria pou paradosiaka stin B Likeiou empaine sxetika diskolo thema. Gia tin 3h Likeiou den exw na pw tipota kathws pisteuw oti itan arketa antimetwpisima.

Gia tin prwti askisi tis B, mporeis na efarmoseis tipous vieta. To athroisma einai $2006k+1$=perittos kai to ginomeno $2007$, episis perittos kai oles oi periptwseis kataligoun se atopo.

Exw ksekinisei vathmologisi graptwn tis A' Likeiou kai pros to paron (mesa sta prwta 10 pou exw koitaksei) iparxei ena strogilo $20$. Gia na doume...

Alexandros

[quote="silouan"]
2)Δίνεται ορθογώνιο $ABCD$ με $AB=4$ , $BC=2$ και σημείο $M$ στο εσωτερικό του με $MC=1$ και $MB=\sqrt 3$ .
Nα βρείτε το εμβαδό του τριγώνου $MAB$ .
[/quote]

 Το τρίγωνο $MBC$ είναι ορθογώνιο (από το αντίστροφο του Πυθαγόρα).Αν $ME \perp BC$ τότε $( \sqrt3 )^{2}= EB \cdot BC$ ,  οπότε βρίσκουμε   $BE = \frac{3}{2}$ .Άρα το ύψος του τριωγώνου $AMB$ είναι  $\frac{3}{2}$ κλπ.

    ΜΠΑΜΠΗΣ

 [i] α) Από την Γ΄ τάξη(και τη Β΄) έλειπε ίσως ένα θέμα με το κάτι παραπάνω.Γενικά όμως τα θέματα ήταν μέσης δυσκολίας και καλά.Αν η επιτροπή είχε χρόνο , θα μπορούσε με δύο - τρεις αλλαγές να κάνει τα θέματα να αγγίξουν το άριστα !
 β)  Δε στέλνω λύσεις θεμάτων άλλων τάξεων.Αυτό ας γίνει καλύτερα από εσάς για να έχει και ενδιαφέρον ![/i]

gia tin geometria....an ME ine  ka8eti sti BC kai MT ka8eti stin AB,  tote 1/ME^2=1/MC^2+1/MB^2<=>ME=(riza3)/2 opote (pi8agorio)MT=3/2 kai ara (MAB)=1/2*4*3/2=3....

Mipos mpori na postari kanis ta 8emata tis c likiu....?

 :D

1) Έστω η συνάρτηση $f$ από τους πραγματικούς στους πραγματικούς και τέτοια ώστε $f(f(x+y))=x-f(y)$ για κάθε $x,y\in R$ .Να αποδειχθεί ότι η $h$ από τους πραγματικούς στους πραγματικούς με $h(x)=f(x)+f(-x)$ είναι σταθερή .

2) Να βρεθούν οι ακέραιες λύσεις της εξίσωσης
$3^{x+1}-x\cdot 3^{x}-4x-1=0$

3)Έστω οι μιγαδικοί αριθμοί $z_{1},\ z_{2}$ και $w\in\left( 0,\frac{\pi}{2}\right)$ να δειχθεί ότι
$\frac{\left|z_{1}\right|^{2}}{(cos\ w)^{2}}+\frac{\left|z_{2}\right|^{2}}{(sin\ w)^{2}}\geq \left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+2Re\left(\bar{z_{1}}z_{2}\right)$

4)Έστω ευθύγραμμο τμήμα BC και σημεία $K,\ L,\ M$ προς το ίδιο μέρος τη BC ώστε $\angle{BKC}=\angle{BLC}=\angle{BMC}$
τότε να δειχθεί ότι τουλάχιστον δύο από τα γινόμενα   $KB\cdot KC$ , $LB\cdot LC$ , $MB\cdot MC$

Η τελευταία είναι πολύ πιο εύκολη από αυτή της Α Λυκείου. :)
Πάντως τα θέματα ήταν ωραία .

Σιλουανέ , όντως !

 Αν θέλουμε ο Θαλής να έχει συμμετοχή , έτσι πρέπει να είναι τα θέματα ! Με αυτή περίπου τη δυσκολία και καμιά φορά και με κανένα εξυπνούτσικο, έτσι για να ξυπνάνε τα αίματα !
  Αφού να σκεφτείς ότι φέτος όλα τα θέματα λύθηκαν και από τους καθηγητές που εμπλέχτηκαν στο διαγωνισμό(από δυο τρία ο καθένας , αλλά λύθηκαν όλα !!!).Όλοι δείχνουν ευχαριστημένοι. Ο Ευκλείδης όμως ίσως να ανέβει αισθητά.Και έτσι πρέπει !

όμορφα θέματα του Θαλή αλλά πολύ εύκολα (καλό αυτό όπως το βλέπει ο κ Στεργίου για να προσελκύσει κόσμο)

μαλλον ευκολα τα θεματα...ετοιμαστειτε για δυσκολο Ευκλειδη. :)

[quote][b]Α ΛΥΚΕΙΟΥ ΘΕΜΑ 4[/b]
Έστω τρίγωνο $ABC$ με $\hat{A}>\hat{B}$ και έγκεντρο το $I$. Στην πλευρά $BA$ παίρνουμε τμήμα $BD=BC-AC$. Να αποδείξετε ότι $ID=IA$[/quote]
Δε μου άρεσε η λύση που δίνει η ΕΜΕ.

Έστω σημείο $E$ στην πλευρά $CB$ τέτοιο ώστε $CA=CE$. Τότε θα είναι $BE=BC-AC=BD$. Επομένως τα τρίγωνα $CAE,BED$ είναι ισοσκελή. Οι διχοτόμοι $CI,BI$ των τριγώνων αυτών είναι ταυτόχρονα μεσοκάθετοι των βάσεών τους $AE,ED$. Άρα το $I$ περίκεντρο του $\triangle ADE$

Me ton idio akrivws tropo to antimetwpisa ki egw. Kai o tropos autos simfwnei me ton tropo pou skeftikan kai ta perissotera paidia pou elisan 'h en pasei periptwsei proseggisan poli to thema auto!

Alexandros

έχω απλά να πω κάτι .Τα θέματα είναι εύκολα για αυτούς που ασχολούνται έτσι ?(Η θεωρία αριθμών δε διδάσκεται και τόσο πολύ στο σχολείο) . Επίσης τα χαρακτήρισα ωραία επειδή ήταν καλό το επίπεδό τους αλλά και για το ότι δεν απαιτούσαν στριφνές , περίεργες λύσεις (όπως παλαιότερα ειδικά σε φάσεις πιο πέρα )
Τέλος πάντων .Ας περιμένουμε έναν Ευκλείδη επίσης καλό

[quote="silouan"]έχω απλά να πω κάτι .Τα θέματα είναι εύκολα για αυτούς που ασχολούνται έτσι?[/quote]

Πολύ σωστά Σιλουανέ. Αν δεν είχα στα χέρια μου το βιβλίο "Θεωρία αριθμών" από την Ε.Μ.Ε. δύσκολα θα προσέγγιζα τα θέματα που απαιτούσαν (έστω και στοιχειώδεις) γνώσεις θεωρίας αριθμών. Επίσης, περιττό να σας πω πως από τα 15 (μόλις) παιδιά που δώσαμε από όλο το νομό στη Β' Λυκείου, κανένα δεν κατάλαβε τη λόγικη του "παίρνω σχέσεις μεταξύ περιττών-άρτιων" (αναφέρομαι σε άριστους μαθητές). Συνοψίζοντας, το εύκολο στην εκάστοτε περίπτωση είναι [u]υποκειμενικό[/u].

Επίσης, στο 4ο θέμα της 3ης γυμνάσιου η λύση που δίνεται δε μου άρεσε... ορίστε μία άλλη δική μου (νομίζω απλούστερη και περισσότερο κατανοητή):

Να εξετάσετε αν υπάρχουν πραγματικοί α,β διάφοροι του μηδενός, τέτοιοι ώστε:

$\frac{3}{2}ab^{-1}+\frac{10}{3}a^{-1}b=3$

Έχουμε λοιπόν:

$\frac{3a}{2b}+\frac{10b}{3a}=3\implies\frac{9a^{2}+20b^{2}}{6ab}=3\implies9a^{2}+20b^{2}-18ab=0\implies(3a-3b)^{2}+11b^{2}=0$

Δηλαδή θα πρέπει και $(3a-3b)^{2}=0$ και $11b^{2}=0$ .Δηλαδή $b=0$, το οποίο είναι άτοπο.

Έτσι την έλυσε και ο αδερφός μου  :lol:
Πάνω στα παραπάνω έχω να προσθέσω κάτι .Ο θείος μου μου λέει πάντα το εξής
<<Εύκολο είναι κάθετι το οποίο ξέρουμε και έστω και με οποιονδήποτε τρόπο μπορούμε να το λύσουμε . Κάτι το οποίο δεν ξέρουμε όσο στοιχειώδες και να είναι για κάποιους άλλους είναι δύσκολο >>
Και εξάλλου πάντα υπάρχουν θέματα που μας <<πάνε>> .Άλλοι είναι αλγεβριστες άλλοι αριθμοθεωρητικοί άλλοι ανισοτικοί άλλοι γεωμέτρες κτλ ....Είναι να <<σε θέλει>>  :D

Πολύ σωστά Σιλουανέ, και ακόμη πιο σωστά και σοφά τα λόγια του θείου σου :wink:

Δυο παρατηρήσεις
Για το γεωμετρικό θέμα της Α λυκείου καλό θα ήταν να παράτηρήσετε ότι Αν ΙΚ κάθετη στην ΑΒ είναι ΙΚ μεσοκάθετος υπολογίζοντας το με την βοήθεια της μερικής σχέσης εφαπτόμενων του εγγεγραμμένου κύκλου στο ΑΒΓ πάντως κι η΄λύση σας είναι εξαιρετική
Κι επίσης ο Ευκλείδης δεν θα διαφέρει τρομακτικά σε δυσκολία από τον Θαλή κι από τις προηγούμενες χρονιές έτσι ώστε να έχουμε ΟΛΟΥΣ όσους ασχολούνται πραγματικά με μαθηματικά στον Αρχιμήδη
Χαιρετώ

[quote]Α ΛΥΚΕΙΟΥ ΘΕΜΑ 4

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

Quote:
Α ΛΥΚΕΙΟΥ ΘΕΜΑ 4
Έστω τρίγωνο  με  και έγκεντρο το . Στην πλευρά  παίρνουμε τμήμα . Να αποδείξετε ότι

Δε μου άρεσε η λύση που δίνει η ΕΜΕ.
[/quote]

Έχω διαφορετική λύση αλλά δεν ξέρω αν θα σου αρέσει κι αυτή...
Λοιπόν εγώ θεωρώ το σημείο Ε πάνω στην ΒΓ τ.ω. ΒΕ=ΒΔ.Έτσι προκύπτει ΓΕ=ΓΑ, άρα ΓΙ κάθετη στην ΑΕ συνεπώς ΙΑ=ΙΕ.Επίσης με γωνιακές σχέσεις και χρησιμοποιώντας το γνωστο θεωρηματάκι που λέει ότι ΕΑΒ=(Α-Β)/2 δείχνουμε ότι ΙΕ=ΙΔ.

Γενικώς αν το ψάξουμε λίγο περαιτέρω, είμαι σίγουρος ότι θα βρούμε πάνω από 10 διαφορετικές λύσεις για κάθε γεωμετρικό θέμα. Στη Β' Λυκείου το 2ο θέμα π.χ. μπορούσε να το προσεγγίσει κάποιος και με τον εξής τρόπο: Να βρει την τρίτη πλευρά του τριγώνου με νόμο συνημιτόνων, και μετά να κάνει εφαρμογή του τύπου:

$E=\sqrt{\tau(\tau-a)(\tau-b)(\tau-c)}$

[i]Συμπέρασμα[/i]: Η ευκλείδια γεωμετρία ζει και βασιλεύει!

[quote="kostas"]παντως δυσκολα θα περναγε καποιος το θαλη της Β και της Γ λυκειου αν δεν ασχολειται με μαθηματικα εκτος σχολειου... :wink:[/quote]

Tha pw tin proswpiki mou apopsi oso ki an fainetai kapws adiki gia kapoious mathites!

Katarxin thewrw dedomeno oti vazontas 4 kai mono themata, einai diskolo na figoun oi perissoteroi mathites euxaristimenoi! Sigoura oi perissoteroi den tha mporesoun na kanoun sxedon tipota!! Opote tha figoun disarestimenoi!!

Episis dedomeno thewrw oti stis mikres takseis (2a kai 3h Gimnasiou) prepei na einai poli eukola toulaxiston 2 themata, na exoun prakseis kai genika pragmata me ta opoia enas mathitis na mporesei na asxolithei kai na figei euxaristimenos ki etsi tin epomeni xronia na exei parei mia geusi kai an tou aresei auto pou ekane na asxolithei perissotero monos tou (oloi mas panw katw to idio den kaname?). To kalo tha einai omws oti tha ton exei elkisei auto me to opoio asxolithike dioti den apogoiteutike!! Pisteuw oti auto ginetai kai stin EME kai poli kala ginetai...

Ara loipon stin 2a kai 3h Likeiou o mathitis pleon den exei alla perithwria na diavasei monos tou gia tous mathimatikous diagwnismous an idi den to exei kanei apo mikroteres takseis kai auto simvainei gia pollous logous o prwtos ek twn opoiwn einai oi panellinies eksetaseis pou ton exoume mathei (kakws) na ton apasxoloun monomerws ksexnontas o,tidipote allo. Etsi to mono pou ginetai stous diagwnismous tis EME twn teleutaiwn taksewn einai oi mathites na apogoiteuontai sxedon plirws. An lavoume ipopsin mas omws tis parapanw dikes mou (den kserw an einai kai dikes sas) paradoxes tote mallon kala kanei i EME kai vazei ligo pio tsimpimena themata stis dio teleutaies takseis. Prepei opwsdipote PANTA stin prwti fasi na iparxei ena eukolo thema (pou sxedon panta iparxei) kai apo ekei kai pera ta themata na einai gia ekeinous tous mathites pou exoun ksekinisei tin proetoimasia tous se mikroteres takseis. Eksallou apo ekeina ta paidia tha epilegoun ki ekeina pou tha mas ekproswpisoun sto ekswteriko. Den to thewrw teleiws adiko auto pou ginetai me to iparxon sistima tis EME (Diladi 4 askiseis plirous anaptiksis). Ti na prwtovaleis tote?

Enw antitheta nomizw oti to sistima pou ipirxe mexri kai to 1994 ston Thali kai ston Eukleidi me peripou 10 erwtiseis pollaplis-epilogis kai 2-3 themata plirous anaptiksis nomizw oti itan POLI POLI kalitera! Afenos dioti KANENAS mathitis den efeuge apogoiteumenos kai afeterou dioti eksetazotan megalitero euros ilis me eksipnoulika mikra thematakia... (Eksalou o kalos fainotan kai stis erwtiseis plirous anaptiksis). Nomizw oti autos tha itan kai o kaliteros tropos eksetasis!!

Epeidi ki egw eimoun arnitikos me tis erwtiseis pollaplwn epilogwn mexri prosfata stous mathimatikous diagwnismous, thelw na anaferw oti diavasa prosfata ena arthro pou edeixne statistika oti opoioi mathites ta pigainan kala se tetoious diagwnismous (me pollaplwn epilogwn erwtiseis) itan kai ekeinoi pou epairnan diakriseis se diethneis diagwnismous tipou IMO ktl... Dioti opws kai na to kanoume einai diskolo na apantiseis swsta se 20-30 sxetika aples erwtiseis se 1,5 wra kai apo tin alli akoma kai o adinamos mathitis tha mporesei na apantisei se 5-6 erwtiseis den tha apogoiteutei na pei oti einai "xazos" opws polles fores akougetai apo tous mathites otan feugoun apo mathimatikous diagwnismous, toso wste na asxolithei kai monos tou meta gia na vgalei kai tis ipoloipes (apla tha xreiastei 4-5 wres anti gia 1,5). Thelei texni na apantiseis se 1,5 wra tetoia thematakia... Den nomizw na diafwneitai etsi?

Akoma kai apo tixi na perasoun merikoi mathites (epeidi evalan stin tixi mia apantisi 'h epeidi aperripsan kapoies apantiseis (akoma ki auto thelei eksipnada)) etsi ki alliws tha ginei i diakrisi kai o diaxwrismos ston Eukleidi kai poli perissotero ston Arximidi pou epivalletai na einai plirous anaptiksis pleon gia na eksetasoun ki alla pragmata (tin krisi tou mathiti, ton tropo diatipwsis mias askisis ktl). Akoma ki auto den to katigorw. Isa isa tha anapterwsei to ithiko kapoiwn mathitwn pou (estw kai tixaia) diakrithikan se ena diagwnismo tis Mathimatikis etaireias... Akougetai poli omorfi sta autia enos mathiti i leksi "diakrisi" den simfwneite? Etsi nomizw tha anevei kata poli to epipedo twn mathimatikwn kai sigoura tha exoume akomi megaliteri proseleusi molis matheutei oti merika pragmata mporoun OLOI oi mathites na ta kanoun... Twra oi perissoteroi den pigainoun giati fovountai tin apotixia les kai kanoume anartisi twn apotelesmatwn stin kentriki plateia tis polis kai me megala entona grammata!!

Eksalou se polles xwres opws Kanada, Ameriki kai diaforous diagwnismous sti Roumania eksetazoun me tetoiou tipou erwtiseis. Pistepste me oti einai arketa petiximeno to sistima auto... Kante mia anazitisi na vreite sxetika themata gia na pisteite (min viasteite na peite oti einai poli eukola an den deite to xroniko perithwrio pou dinoun gia tin apantisi tous).

Perimenw tin gnwmi sas...

Alexandros

Αλέξανδρε συμφωνω με τα περισσότερα . Δεν συμφωνώ στο ότι θα πρέπει να υπάρχουν πολλαπλής επιλογής θέματα .Η προσωπική μου άποψη πάνω σε αυτό είναι ότι αν γίνει κάτι τέτοιο θα γίνει λιγότερο <<μαθηματικός>> ο διαγωνισμός .Πιστεύω ότι κάτι τέτοιο πρέπει να το ξεκινήσουν τα σχολεία ανεξάρτητα από την ΕΜΕ .
[color=red]Και κάποιες άλλες ιδέες[/color]
Πρώτον πιστεύω ότι θα πρέπει να γίνονται περισσοτερα τέστ επιλογής (πουθενά δεν κάνουν μόνο ένα) .
Δεύτερον πιστεύω ότι ο χρόνος  εξέτασης στον Αρχιμήδη πρέπει να γίνει 4 ώρες .
Να δημιουργηθεί διαγωνισμός από το Δημοτικό .

Περιμένω την γνώμη σας για τα παραπάνω.

Αγαπητοί φίλοι ,

  μόλις γύρισα από την Κάρυστο όπου είχαμε πάει ( το ΔΣ του παραρτήματος Εύβοιας)  για να απονείμουμε τα βραβεία στους μαθητές του διαγωνισμού για το Δημοτικό που κάνουμε κάθε χρόνο.
  Διάβασα με προσοχή τις σκέψεις σας για τη μορφή του διαγωνισμού ΘΑΛΗΣ.
Λοιπόν , στον πρώτο γύρο ενός διαγωνισμού , συνηθίζεται να τίθενται [b]θέματα πολλπλής[/b] .ΑΥτή είναι η καλύτερη μέθοδος που ανακαλύφτηκε μέχρι τώρα !
[color=red]  α)[/color] Με 20 τέτοιες ερωτήσεις πετυχαίνεις σχεδόν όλους τους στόχους: προσελκύεις κόσμο , διαφημίζεις και προβάλεις τα μαθηματικά , δεν απογοητεύεις τους μαθητές , διότι βάζεις θέματα όλων των επιπέδων , εντοπίζεις τους καλύτερους μαθητές ( σίγουρο αυτό !) , διορθώνεις γρήγορα και δίκαια .
  [color=red] β) [/color]Η μορφή αυτή έχει την εξής  ''αδυναμία'' : Δε μπορείς να φτιάξεις τα θέματα σε μία ώρα ή να το κάνεις ως αγγαρία !!!
    Τώρα , το μαθητή με τη σύνθετη σκέψη και το φαινόμενο , θα τους επιλέξεις τελικά στην τελευταία φάση με θέματα ανάπτυξης.Εκεί οι μαθητές είναι λίγοι και δεν υπάρχει πρόβλημα.
  Για τα test επιλογής ο Σιλουανός έχει απόλυτο δίκαιο! Χρειάζονται τουλάχιστον δύο (μην πω τρία!).
 Νομίζω όμως ότι η δομή της δικής μας ΕΜΕ δε βοηθάει τη μορφή με τα 20 θέματα πολλαπλής επιλογής , διότι δεν υπάρχει ειδική ομοιόμορφη επιτροπή που να ζει σχεδόν με αποκλειστικό μέλημα το διαγωνισμό ΘΑΛΗΣ.Το να φτιάξεις 20 καλά θέματα είναι πολύ πιο δύσκολο από το να φτιάξεις  τέσσερα και ...ότι βγει !
  Εμείς εδώ στο παράρτημα που ετοιμάζουμε τα 10 +10 = 20 ερωτήματα πολλαπλής επιλογής για το Δημοτικό αφιερώνουμε πολύ χρόνο και πάλι δεν είμαστε απόλυτα ευχαριστημένοι από το αποτέλεσμα.
 Να τονίσω ότι ακόμα και στη θεωρία αριθμών ή τη γεωμετρία μπορείς να διατυπώσεις ερώτηση πολλαπλής επιλογής επιπέδου ΙΜΟ !  Φυσικά οι ερωτήσεις πολλαπλής επιλογής δε γεννήθηκαν για να εξετάζουν δύσκολα πράγματα , αλλά πολλά και γρήγορα.Πιστεύω ότι σε ένα διαγωνισμό με τρεις φάσεις , ο πρώτος γύρος μπορεί να γίνει με αυτό το σύστημα με απόλυτη επιτυχία , ταχύτητα και αξιοπιστία.Φυσικά , οι απαντήσεις των μαθητών που θα περάσουν στο β΄γύρω , μπορούν να ελεχθουν ως προς την ορθότητα, με βάση αυτά που γράφει ο μαθητης στο τετράδιό του.Το κάνουμε και εμείς στο Δημοτικό και κανένας μαθητής δεν παίρνει βραβείο , αν δεν έχει τη λύση στο τετράδιο.
  Όλα αυτά είναι σκέψεις και τίποτα παραπάνω. Εμείς τις δικιμάσαμε στην πράξη και αποδίδουν απόλυτα!.Όσοι ενδιαφέρονται , ας τις μελετήσουν.

                 Μπάμπης

συμφωνω οτι το συστημα με 20 ερωτησεις πολαπλης επιλογης και 2 ερωτησεις αναπτηξης ειναι πολυ καλο για Θαλη. αλλα γιατι λοιπον το αλλαξαν το 1994? και κατι αλλο ο Αρχημηδης πρεπει σιγουρα να γινει 4 ωρες με θεματα παντα πρωτοτυπα....,
δεν υπαρχει λογος να μιλαμε για οικονομια ξανα :D

Η ιστορία με τις ερωτήσεις πολλάπλής επιλογής θα είναι το μόνο σημείο που θα διαφοροποιηθώ από όλους σας Θεωρώ άδικο αυτοό το σύστημα  πολλαπλής επιλογής αν βάλεις μικρά χρονικά περιθώρια Γενικά παραδεκτό είναι οτι σε μαθηματικό διαγωνισμό ΠΡΕΠΕΙ Ννα δίνεις άπλετο χρονο να ξεδιπλώσει κάποιος το ταλέντο του
αναφέρομαι σε πρόταση Σιλουανού
αυτά κι άλλα στην συνέχεια

simfono apolita sta 8emata polaplis epilogis (mon ston 8ali omos+isos ena 8ema tu eyklidi..)...+panta anarotiomun gt den ixe ayti tin morfi o diagonismos...
+proselkii kosmo ala kai den isteri se tipota apo ta 8emata plirus anaptiksis (s ayto to epipedo tu diagonismu..)
kai ontos exo vri tetia 8emata polaplis epilogis epipedu imo pu lei o k.Stergiu
kai ine pragmatika entiposiaka...aplos nomizo oti i perisoteri imaste arnitiki sta test polaplis epilogisite giati ta exume sindesi me kati ite to opio eksartate merikes fores apo ton paragonta tixi ite me kati poli aplo....proxirodulia....
den ine etsi omos.....

parola ayta i me8odos ayti pisteuo oti armozi apolita mon sta prota stadia ton diagonismon 8ali-euklidi,+gt den mporun toso eykola na kataskeyastun 8emata polaplis iyiloteru epipedu ala k gt prepi o ma8itis na eksetasti kai stin pliri anaptiksi....eki 8a gini i diakrisi metaksi tu kalu kai tu poli kalu ma8iti...

den simfono ka8olu omos s ayto pu ipe o ambros,oti prepi na iparxi apletos xronos,xoris kanena periorismo...oxi...gt poli apla..ena 8ema geometrias xoris xroniko peri8orio mpori na to lisi opiosdipote me analitiki geometria..kanontas praksis,praksis.praksis...den ine omos ayti i usia tu diagonismu...kai malista se megales taksis...na dis an o ma8itis kseri na kani praksis...

metrai i kali skeyi i omorfi idea...ala pu na exi vgi mesa sto 3oro tu diagonismu....den 8imame pos legotan ala enas (nomizo filosofos itane) ixe orisi tin efiia os tin dinatotita prosarmogis se nees sin8ikes....kai ayto epiveveonete...to na <piesis>ton eayto sy na dosi to 100% se ekino to 3oro gia na lisis 4 provlimata ine ikanotita prosarmogis stis sin8ikes <diagonismu>.... akoma kai ekseliktika na to dis,me to perasma ton xronon epivionun mono osa zoa mporun na prosarmostun stis nees perivalontikes sin8ikes....ta ipolipa pe8enun...

kai sto kato-kato an kapios nomizi oti o xronos gia kapion diagonismo tu ine ligos ,den eparki.. as skefti oti diagonizete me...........an8ropus kai gia tus opius den ftani o xronos...den 8a pi kanis tipota an to kalitero grapto itan 3 sta 4 kai ola ta ipolipa itan apo 2 kai kato.........(akoma kai an xriazotan mono kana misaoro akomi gia na li8i to tetarto 8ema...),,ala 8a parapone8i o poli kalos ma8itis an ipirxan 15 isova8mies sto arista...gt o enas 8a ixe teliosi stis diomisi ores eno o alos 5 min prin to telos...
tote i eystrofia tu protu den fenete pu8ena...
pantos gia na gini aytos o diaxorismos metaksi poli kalu kai aristu den mpori na gini sigura me mon ena test epilogis...se kamia periptosi....

 :D

[quote="RAKAR"]simfono apolita sta 8emata polaplis epilogis (mon ston 8ali omos+isos ena 8ema tu eyklidi..)...+panta anarotiomun gt den ixe ayti tin morfi o diagonismos...
+proselkii kosmo ala kai den isteri se tipota apo ta 8emata plirus anaptiksis (s ayto to epipedo tu diagonismu..)
kai ontos exo vri tetia 8emata polaplis epilogis epipedu imo pu lei o k.Stergiu
kai ine pragmatika entiposiaka...aplos nomizo oti i perisoteri [color=red]imaste arnitiki sta test polaplis epilogisite giati ta exume sindesi me kati ite to opio eksartate merikes fores apo ton paragonta tixi ite me kati poli aplo....proxirodulia....
den ine etsi omos.....[/color]
 :D[/quote]

 [color=blue]Σε θαυμάζω γιατί  , αν και νέος ,  έχεις καταλάβει αυτό που εγώ κατάλαβα σχεδόν στα 48 μου ! Έχεις συλλάβει ακριβώς την καρδιά του προβλήματος.Ας το σκεφτούν οι αρμόδιοι ξανά. [/color]

Xairomai idiaitera pou ginetai toso epoikodomitiki sizitisi sto forum mas kai mas vriskei olous panw katw me tis idies apopseis! Sigoura opoios exei themata prin apo to 1994 mporei na dei oti to epipedo itan para poli kalo stous diagwnismous me tis erwtiseis tetoiou eidous. Gia paradeigma o Thalis kai o Eukleidis tou 94 eixe 10 erwtiseis pollaplis epilogis kai 3 themata plirous anaptiksis, enw to 1993 itan 25 erwtiseis pollaplis epilogis. Ontws einai omws poli diskolo, opws leei kai o kos Stergiou na vgaleis tosa themata gia kathe taksi. Gia paradeigma to 1993 evazan mazi 3h Gimnasiou-1h Likeiou kai 2a kai 3h Likeiou (mallon logw elleipsis thematwn).

 Episis tha simfwnisw me ton Silouano oti eimaste apo tis liges xwres pou enw exoume poli kosmo pou dinei se mathimatikous diagwnismous, exoume mono ena diagwnismo kai mono 3 faseis ti stigmi kata tin opoia h Roumania exei panw apo 10 diaforetikous diagwnismous gia ta mathimatika. Den tha mporouse gia paradeigma na ginetai o diagwnismos polewn (Tournament of the Towns) stin xwra mas 'h kapoios allos diagwnismos diethnous epipedou, na mas stelnoun ta themata kai na grafoun oi mathites? Mallon omws kapoio simferon tha eksipiretoun merika pragmata!!

Xreiazetai ananewsi giati edw kai tosa xronia ginontai ta idia kai ta idia! Kai to thewrw apolita logiko ean skeftw oti pleon gia tous diagwnismous tis EME asxolountai mono 1-2 atoma (p.x. o kos Louridas kai 1-2 alloi). Ti na prwtoprolavoun omws ki ekeinoi? Prepei na iparxei arketos kosmos pou na asxoleitai me tous diagwnismous!! Diladi an den ipirxan ekeinoi oi 2-3 "romantikoi" ti tha ginotan me tous diagwnismous stin Ellada? Nomizw omws oti eimaste se kalo dromo giati erxontai kapoia kala nea ta opoia tha sas ta anakoinwsw se merikes meres gia na vevaiwthw oti isxioun... Tote pragmatika tha ksekinisei anavathmisi tis Mathimatikis Paideias sti xwra mas oson afora tous diagwnismous... Gia na doume!!!

Alexandros

[color=red]EDIT:[/color]

Kai gia na mi milame me fantasmata ekana scan  kai sas episinaptw ta themata "THALI" 1992 kai 1993 (gia na deite kai ta dio eidi erwtisewn : a)10 erwtiseis pollaplis epilogis+ 3 plirous anaptiksis, b) 25 erwtiseis pollaplis epilogis) kathws episis kai themata tou "EUKLEIDH" tou 1993.

Ti lete? Den itan mia xara ta themata tote parolo pou itan pollaplis?

Αλεξανδρε με φτιαξες  :D ελπιζω να μας πεις καλα νεα λοιπον... :)

Ωραίες οι ερωτήσεις πολλαπλής επιλογής. Μερικές από αυτές μάλιστα εξετάζουν και το IQ και δεν απαιτούν γνώσεις μαθηματικών, παρά μόνο ευστροφία.

φίλε RAKAR με παρεξήγησες όταν λέω άπλετος χρόνος στις πολλαπλής επιλογής εννοώ τόσος ώστε να είναι ικανός κάθε μαθητής χωρίς να πιέζεται με τον χρόνο κι αρα να καταφύγει σε  τύχη να απαντά σε όλες τις ερωτήσεις (αν τις ξέρει) κι βεβαίως στον Αρχιμήδη αν κάποιος μπορεί θα λύσει τα θέματα στις 3 ώρες επίσης ο πολύ καλός στις ανάπτυξης ερωτήσεις φαίνεται από τον τρόπο λύσης ΚΙ ΟΧΙ ΚΑΤ ΑΝΑΓΚΗ ΑΠΟ ΤΟΝ ΧΡΟΝΟ ΛΥΣΗΣ
αυτά χαίρομαι που κάνουμε εποικοδομητικό διάλογο

η αντίστροφη μέτρηση άρχισε...
σύμφωνα με έγκυρες πηγές τα αποτελέσματα θα βγουμ σε καμιά εβδομάδα...

και ένα αστείο για να ξεχάσετε το άγχος... :P
-Κύριε το 0/0 να το αφήσω έτσι ή να το βάλλω 0?
(απάντηση που θα έπρεπε να δοθεί)-όχι αφού τα 0 απλοποιούνται, πρεπει να το βάλλεις 1

Καιρός ήταν  :P

Τα αποτελέσματα του Θαλή ανακοινώθηκαν στη γνωστή ιστοσελίδα.

Συγχαρητήρια σε όλα τα παιδιά από το φόρουμ μας!

@Silouan
[hide]Άντε και με το τιμημένο φέτος  :) [/hide]

ΜΠΡΑΒΟ ΣΕ ΟΛΑ ΤΑ ΠΑΙΔΙΑ .ΚΑΛΗ ΣΥΝΕΧΕΙΑ  :)

Ξερει κανεις ποιες ηταν οι βασεις σε καθε ταξη?

 :)

Στην Β Λυκείου μπορεί να  ήταν και κάτω από 10 αφού κάποια παιδιά που  πέρασαν από το σχολείο μου είχαν δεν είχαν γράψει 10 .
@mostel
[hide]Και εγώ το ελπίζω Στέλιο αλλά τα πράγματα είναι δύσκολα και ο συναγωνισμός μεγάλος[/hide]

Confindential information: b Gym: base grade 11; c Gym: base grade 8; A, B, C  Lyc: base grade 6. Pretty low, aren't they?

Ginger Rogers

Σχετικά πολύ χαμηλά, όντως.

Στον Ευκλείδη σε τι βαθμό κυμαίνονται οι βάσεις;

δεν ηταν παραξενο να ειναι τοσο χαμηλες οι βασεις σε αυτον τον ευκολο Θαλη ?

Στις μεγάλες τάξεις , δηλαδή μετά την Α΄Λυκείου , το ολυμπιακό επίπεδο των πολλών μαθητών πέφτει,διότι οι περισσότεροι καλοί μαθητές αρχίζουν φροντιστήρια για τη Γ΄. Έτσι  , πολλοί πάνε στο δδιαγωνισμό ΘΑΛΗΣ μόνο με ότι απέκτησαν μέχρι τη Γ΄Γυμνασίου. Βλέπω στη Χαλκίδα , ότι ενώ μέχρι πριν από 5 χρόνια υπήρχαν μαθητές που φτάνανε στν Αρχιμήδη , τώρα δεν περνάνε ούτε το ΘΑΛΗ ! Και μη νομίζετε ότι δεν υπάρχουν παιδιά με ταλλέντο ! Απλά δεν ασχολούνται με τα μαθηματικά , διότι οι γονείς τους τους ρίχνουν στα μαθηματα για τα ΑΕΙ.
 Έτσι , σε ορισμένους νομούς μπορεί να υπάρξει και βάση το 7. Δεν ξέρω ακριβώς ποια πολιτική ακολουθεί η ΕΜΕ στις βάσεις, αλλά μη σας φαίνεται περίεργο στη Β΄και Γ΄Λυκείου να ήταν πολύ λίγοι οι μαθητές που έγραψαν 10.Εγώ που διόρθωσα τη Β΄Λυκείου στη Χαλκίδα βρήκα γραπτά μέχρι το 6. Και η μία μονάδα (6-5) ήταν σε προσπάθεια και όχι σε λύση.Εύχομαι σε όλους σας :

                  [color=red]ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΣΤΟΝ ΕΥΚΛΕΙΔΗ !!![/color]

Η βάσεις είναι σχετικά χαμηλές (για το Θαλή καλό αυτό)
γιατί όταν περνάς έναν τέτοιο διαγωνισμό το ηθικό σου γίνεται ακμαιότατο  :wink:  .
Ας πούμε ένα παιδί που πέρασε με κάτω από 10 εδώ στα Τρίκαλα μου ζήτησε σημειώσεις ασκήσεις κτλ άρα μαθηματική πρόοδος...
 :)

Πάντως συνολικά η πόλη μας είχε 49 επιτυχόντες  :D

Τέλος πιστεύω ότι ο Ευκλείδης θα πρέπει να είναι μετρίου επιπέδου ώστε να δοθεί η ευκαιρία σε παιδιά να πάνε στον Αρχιμήδη (και εκεί ότι ήθελε προκύψει )


limit
integration
trigonometry
logarithms
calculus
calculus computations

calculat  the exact value of the limit :

\[ L \equal{} \lim_{x\to 0}S(x).C(x)\]

\[ where :\]

\[ C(x)  \equal{}  \int_{x}^{\infty }\frac{cosy}{y} dy\] ,    \[ S(x)  \equal{}  \int_{0}^{x}\frac{siny}{y} dy\]

[quote="Dr Sonnhard Graubner"]hello, i have not a proof yet, but i think the searched limit is zero.
Sonnhard.[/quote]

may be you put your think , please .

High time to use L'Hopital's rule,...

$ \lim_{x\rightarrow 0^{ \plus{} }}\left( \int_{0}^{x}\frac {\sin y}{y}dy\right) \left( \int_{x}^{\infty }\frac {\cos y}{y}dy\right)$

$ \equal{} \lim_{x\rightarrow 0^{ \plus{} }}\frac {\int_{x}^{\infty }\frac {\cos y}{y}dy}{\left( \int_{0}^{x}\frac {\sin y}{y}dy\right) ^{ \minus{} 1}}$

$ \equal{} \lim_{x\rightarrow 0^{ \plus{} }}\frac { \minus{} \frac {\cos x}{x}}{ \minus{} \left( \int_{0}^{x}\frac {\sin y}{y}dy\right) ^{ \minus{} 2}\frac {\sin x}{x}}$

$ \equal{} \lim_{x\rightarrow 0^{ \plus{} }}\frac {\left( \int_{0}^{x}\frac {\sin y}{y}dy\right) ^{2}}{\sin x}$

$ \equal{} \lim_{x\rightarrow 0^{ \plus{} }}\frac {\frac {2\sin x}{x}\int_{0}^{x}\frac {\sin y}{y}dy}{\cos x}$

$ \equal{} \frac {2(0)}{1} \equal{} 0$

I'm not entirely convinced by this, but for $ x$ positive and near 0, shouldn't we have $ \int_{x}^\infty \frac{\cos y}{y} \, dy \equal{} \minus{} \ln x \minus{} \frac{x^2}{4} \plus{} O(x^4)$ and $ \int_{0}^x \frac{\sin y}{y} \, dy \equal{} \Theta(x)$ or something like that?  (Mathematica is telling me to put in a $ \gamma$ somewhere in the first expression, but I don't believe it.)

I'd say that $ \int_x^{\infty}\frac{\cos y}y\,dy\equal{}\minus{}\ln x\plus{}O(1)$ as $ x\to 0^\plus{}.$

After all, for any fixed $ b,$ we have that $ \int_b^{\infty}\frac{\cos y}y\,dy$ converges, so just pick some specific $ b.$

Anyway, that settles it: the limit looks like $ (\minus{}\ln x)\cdot x$ and tends to zero as $ x\to0.$


[quote="zhaobin"][quote="siuhochung"][quote="zhaobin"]I'm not sure about my idea.
let $f(a,b,c)=a+b+c-ab^2-bc^2-ca^2=-ca^2-(b^2-1)+b+c-bc^2$
then when $f(a,b,c)$ be maximum $a=0$ or $a=1$
we do like this then come the result.
is it right? ;)[/quote]
I don't know how zhaobin arrives at "when $f(a,b,c)$ be maximum $a=0$ or $a=1$".
I considered the discriminant and i got $b^4-2b^2 \le 0 \le (1-c)(1+c)(4-bc)$ so it's true.[/quote]
I think I'm wrong,can you complete your solution?[/quote]
I'm wrong too.
Let me try again.

Hmm,the problem is really hard,i have a very long long solution of it,but i do not satisfy it.Go on! :)

I agree this problem is not easy now ;)
and I have an approach (haven't complete now).
let $t=a+b+c$
can you find the minimum of $ab^2+bc^2+ca^2$ when $a+b+c=t$
for explame when $0 \leq t \leq 1$ the minimum of $ab^2+bc^2+ca^2$ is 0.
and what about $1 \leq t \leq 3$
hope somebody can understand what i mean and help me.

because nobody answer my question,so i made another topic

good news ;)
you can read the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=251657#p251657]here[/url]
then your problem is sloved,too :)

Only now I am seeing what a nice result came from ineq I was proving in other thread :)
It looks nice. Also Equality case is strange: $a=1b=\frac{1}{2},c=0$ :)


For a given real number $ a > 0$, how many solutions for real $ x > 0$ has the equation $ x^{x\plus{}1} \equal{} a^{2x \minus{} 2}$ ?

Partial solution, need help completing.  Excluding the trivial $ x\equal{}1,$ let $ f(x) \equal{} \ln{x}\cdot \frac{x\plus{}1}{2x\minus{}2}.$
Then for fixed $ a, x$ is a solution $ \Leftrightarrow \ln{a} \equal{} f(x).$
Consequently, the graph of $ f(x)$ must be visualized to determine which values of $ a$ will have 0, 1, or more than 1 solution value $ x.$  Since $ f(x)$ -> infinity as $ x$ -> 0 and as $ x$ -> infinity, there is at least 1 value of $ x$ for which $ f^{\prime}(x) \equal{} 0.$

$ f^{\prime}(x) \equal{} \frac{1}{x} \cdot \frac{x\plus{}1}{2x\minus{}2} \plus{} \ln{x} \frac{(2x\minus{}2) \minus{} 2(x\plus{}1)} {(2x\minus{}2)^2} \equal{} \frac{1}{x} \cdot \frac{x\plus{}1}{2x\minus{}2} \plus{} \ln{x} \frac{\minus{}4}{(2x\minus{}2)^2} \implies
f^{\prime}(x) \equal{} 0 \Leftrightarrow \frac{x\plus{}1}{x} \equal{} \ln{x} \frac{4}{(2x\minus{}2)} \Leftrightarrow
\ln{x} \equal{} \frac{(x\plus{}1)(x\minus{}1)}{2x}.$

Let $ g(x) \equal{} \ln{x} \minus{} \frac{(x\plus{}1)(x\minus{}1)}{2x} : 0<x \text{ not } \equal{} 1.$
$ g(x)$ must be graphed to determine how many times it intercepts the x-axis.  My blind guess is that $ g(x)$ only hits the x-axis once, which means that when ln($ a$) is above the minimum value of f(x), $ a$ will have two solutions (besides the trivial $ x\equal{}1$).


Suppose that $ a,b,c$ be 3 sides of an acute triangle.Prove that:
$ (a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)(a^3\plus{}b^3\plus{}c^3)$≥$ 4(a^6\plus{}b^6\plus{}c^6)$

[quote="SUPERMAN2"]Suppose that $ a,b,c$ be 3 sides of an acute triangle.Prove that:
$ (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)(a^3 \plus{} b^3 \plus{} c^3)$≥$ 4(a^6 \plus{} b^6 \plus{} c^6)$[/quote]

It's equivolent to
$ a^6\plus{}b^6\plus{}c^6\plus{}\sum_{sym}{a^3b^3}\plus{}\sum_{sym}{a^5b}\plus{}\sum_{sym}{a^4b^2}\plus{}\sum_{sym}{a^3b^2c} \ge 4(a^6\plus{}b^6\plus{}c^6)$
or
$ \sum_{sym}{a^3b^3}\plus{}\sum_{sym}{a^5b}\plus{}\sum_{sym}{a^4b^2}\plus{}\sum_{sym}{a^3b^2c} \ge 3(a^6\plus{}b^6\plus{}c^6)$

We have that :
$ cos\alpha \cdot cos \beta \cdot cos \gamma \equal{} \frac{(a^2\plus{}b^2\minus{}c^2)(b^2\plus{}c^2\minus{}a^2)(a^2\plus{}c^2\minus{}b^2)}{8a^2b^2c^2}>0$
(becouse $ \alpha$, $ \beta$, $ \gamma$ are acute angles.)
or
$ (a^2\plus{}b^2\minus{}c^2)(b^2\plus{}c^2\minus{}a^2)(a^2\plus{}c^2\minus{}b^2)>0$
or
$ \sum_{sym}{a^4b^2} > a^6\plus{}b^6\plus{}c^6\plus{}2abc$
Using this inequality we will have:
$ \sum_{sym}{a^3b^3}\plus{}\sum_{sym}{a^5b}\plus{}\sum_{sym}{a^4b^2}\plus{}\sum_{sym}{a^3b^2c} \ge 3\sum_{sym}{a^4b^2}\plus{}\sum_{sym}{a^3b^2c} >3(a^6\plus{}b^6\plus{}c^6) \plus{}6abc\plus{}\sum_{sym}{a^3b^2c}>3(a^6\plus{}b^6\plus{}c^6)$
which was desired.

[quote="SUPERMAN2"]Suppose that $ a,b,c$ be 3 sides of an acute triangle.Prove that:
$ (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)(a^3 \plus{} b^3 \plus{} c^3)$≥$ 4(a^6 \plus{} b^6 \plus{} c^6)$[/quote]
 $ (a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)>a^3\plus{}b^3\plus{}c^3\plus{}c(a^2\plus{}b^2)\plus{}b(c^2\plus{}a^2)\plus{}a(b^2\plus{}c^2) >2(a^3\plus{}b^3\plus{}c^3)$
So,I wiil must prove that:
$ (a^3\plus{}b^3\plus{}c^3)^2>2(a^6\plus{}b^6\plus{}c^6)$
$ <\equal{}> (\sqrt{a^3}\plus{}\sqrt{b^3}\plus{}\sqrt{c^3})(\sqrt{a^3}\plus{}\sqrt{b^3}\minus{}\sqrt{c^3})(\sqrt{b^3}\plus{}\sqrt{c^3}\minus{}\sqrt{a^3})(\sqrt{a^3}\plus{}\sqrt{c^3}\minus{}\sqrt{b^3}) >0$
Assume $ a\ge b \ge c$.So we will must prove that:$ \sqrt{b^3}\plus{}\sqrt{c^3} > \sqrt{a^3}$
Since $ a\ge b \ge c$then $ \sqrt{\frac{b^3}{a^3}}\plus{}\sqrt{\frac{c^3}{a^3}} >\frac{b^2}{a^2}\plus{}\frac{c^2}{a^2} >1$
So, we have Q,E,D :)


[quote="solafidefarms"]If we represent a date as mm/dd/yy, how many dates in the 20th century were palindromes? (A palindrome reads the same forwards as backwards, i.e. February 11th, 1920, is a palindromatic date in this system.)[/quote]
[hide]The left (mm) can be 01-12. The middle (dd) can only be 11 and 22. The right (yy) is the reverse of the left (mm).

An entire year has 2.

There are 12 months, so there are also 12 years. In the 20th century, there were $12\cdot2=\boxed{24}$ palindrome dates.

The same goes for every century.[/hide]

[hide]mm/dd/yy

so basically mmyy has to be a palindrome and then dd has to equal 11 or 22. Since all months have more than 22 days, we can just multiply the number of mmyy palindromes by two.

Listing each months' palindrome

10
20
30
40
50
60
70
80
90
01
11
21

Since each of these months correspond to one palindrome, there are only 12 palindrome years.
$12\times2=24$

[b]24 palindromes[/b][/hide]

[hide]Just list the palindromes:
[list]
10
20
30
40
50
60
70
80
90
01
11
21
[/list]

There are 12. $12\cdot 2=\boxed{24\text{ palindromes}}$.[/hide]


My three favorite Beethoven Sonatas (all movements together):
1) No. 23
2) No. 8
3) No. 14

Individually, I really like the second movement of No. 18 and the first movement of No. 31.

What are your three favorite sonatas by Beethoven? Also please vote in the poll.

Unless this is part of ATTOS or something, shouldn't this be in the Fun Factory?

Supposedly, discussion of games belongs here. But this is neither a game nor a frivolous marathon thing. I'll move it to FF.


function
floor function
LaTeX
inequalities
algebra

Find the value of
\[ \Big\lfloor \frac{ 2004!}{ 2003! + 2002! + 2001! + \ldots + 1!} \Big\rfloor. \]

I'd be glad to see a solution for this beautiful problem.

What is the answer?

[hide]My solution:
We will prove that the fraction above is between 2003 and 2002.
First, we see that 2004!/(2003!+2002!+2001!+2000!+....+1!)<2003, which is equivalent to: 2002!<2002!+2001!+...+1(this one is obviously true).
Now, we just need to prove that: 2004!/(2003!+2002!+...+1!)>2002, which is equivalent to: 2001!*2003>2000!+1999!+...+1!(this one is also true.If we want to factorize to get the difference, we will have a positive integer).
Therefore, 2002<this fraction<2003.Therefore, the greatest integer function is 2002(Q.E.D)
@Mr. MCrawford: Indeed, a very nice problem! :wink: [/hide]

Nobody in this forum minds giving a comment on my solution?It seems like a month ago :( .Or is it very hard to read?If it is, pleased help me fix it by adding LATEX to it.

I am interested in having feedback on the concision (i.e. if its unnecessarily wordy/long) of this solution. Then again, I suppose this depends on the level of my audience. Also, I think I should probably change my solution so that it reads forwards, instead of backwards, but I am feeling lazy right now.

[hide="solution"]
To solve the general, that is, finding the value of ${ \lfloor x \rfloor =  \left\lfloor \frac{n!}{(n - 1)! + (n - 2)! + \cdots + 2! + 1!} \right\rfloor}$, we can first easily see that for small values of $ n$, it is always $ n - 2$. Now we want to try to prove this for all $ n$. This amounts to proving two facts: (1) $ x < n - 1$ and  (2) $ x > n- 2$

(1) this statement is equivalent to \[ x  =  \frac{n!}{(n - 1)! + (n - 2)! + \cdots + 2! + 1!} < n - 1 \\ \Longleftrightarrow  n! < (n - 1)((n - 1)! + (n - 2)! + \cdots + 1!) \\ \Longleftrightarrow n! < (n - 1)((n - 2)!(n) + k) \\ \Longleftrightarrow n! < n! + k\] where $ k = (n - 3)! + (n - 4)! + \cdots + 1!$. Since $ k$ is clearly positive, we are done.

(2) Ok now writing my proof forwards:
Since $ n$ is positive, we have \[ n^2 - n > n^2 - 4n + 5 \\ \Longleftrightarrow n! > (n - 2)!(n^2 - 4n + 5) \\ \Longleftrightarrow n! > (n - 2)! + (n - 2)!(n - 2)^2 \\ \Longleftrightarrow n! > (n - 2)((n - 1)! + (n - 2)(n - 2)!)\] Now,  we have \[ n! > (n - 2)((n - 1)! + (n - 2)(n - 2)!) > (n - 2)((n - 1)! + (n - 2)! + \cdots + 1) \Longleftrightarrow x > n - 2.\]
[/hide]

[quote="ghjk"]
Now, we just need to prove that: 2004!/(2003!+2002!+...+1!)>2002, which is equivalent to: 2001!*2003>2000!+1999!+...+1!(this one is also true.If we want to factorize to get the difference, we will have a positive integer).
[/quote] Can you please explain? I am not seeing where the positive integer is.

[quote="hurdler"]I am interested in having feedback on the concision (i.e. if its unnecessarily wordy/long) of this solution. Then again, I suppose this depends on the level of my audience. Also, I think I should probably change my solution so that it reads forwards, instead of backwards, but I am feeling lazy right now.

[hide="solution"]
To solve the general, that is, finding the value of ${ \lfloor x \rfloor = \left\lfloor \frac {n!}{(n - 1)! + (n - 2)! + \cdots + 2! + 1!} \right\rfloor}$, we can first easily see that for small values of $ n$, it is always $ n - 2$. Now we want to try to prove this for all $ n$. This amounts to proving two facts: (1) $ x < n - 1$ and  (2) $ x > n - 2$

(1) this statement is equivalent to
\[ x = \frac {n!}{(n - 1)! + (n - 2)! + \cdots + 2! + 1!} < n - 1 \\
\Longleftrightarrow n! < (n - 1)((n - 1)! + (n - 2)! + \cdots + 1!) \\
\Longleftrightarrow n! < (n - 1)((n - 2)!(n) + k) \\
\Longleftrightarrow n! < n! + k
\]
where $ k = (n - 3)! + (n - 4)! + \cdots + 1!$. Since $ k$ is clearly positive, we are done.

(2) Ok now writing my proof forwards:
Since $ n$ is positive, we have
\[ n^2 - n > n^2 - 4n + 5 \\
\Longleftrightarrow n! > (n - 2)!(n^2 - 4n + 5) \\
\Longleftrightarrow n! > (n - 2)! + (n - 2)!(n - 2)^2 \\
\Longleftrightarrow n! > (n - 2)((n - 1)! + (n - 2)(n - 2)!)
\]
Now,  we have
\[ n! > (n - 2)((n - 1)! + (n - 2)(n - 2)!) > (n - 2)((n - 1)! + (n - 2)! + \cdots + 1) \Longleftrightarrow x > n - 2.
\]
[/hide]

[quote="ghjk"]
Now, we just need to prove that: 2004!/(2003!+2002!+...+1!)>2002, which is equivalent to: 2001!*2003>2000!+1999!+...+1!(this one is also true.If we want to factorize to get the difference, we will have a positive integer).
[/quote] Can you please explain? I am not seeing where the positive integer is.[/quote]
OK!I will try to explain it more clearly:
Since 2001!>=2000!
        2001!>=1999!
        2001!>=1998!
        ......
        2001!>=1!
Adding all these inequalities above, we have: 2003.2001!>2000.2001!>2000!+...+1!(Q.E.D) :)

could you please explain the step before that too? like how $ \frac{2004!}{2003! \plus{} 2002! \plus{} \cdots \plus{} 1!} > 2002$ is equivalent to $ 2001! \times 2003 > 2000! \plus{} 1999! \plus{} \cdots \plus{} 1!$? Thanks.

Oh ok thanks. You know if you wish to learn some LaTeX, you can always put your mouse over some LaTeX text and click on it, and a browser should pop up with how the person typed that bit of text in LaTeX. This way, you can learn little by little, and it doesn't seem hard at all once you get used to it. The only thing you need is just to put dollar signs around your text. Of course, I, like most other people, can read math without LaTeX just fine, but its always helpful for it to be in latex for the audience.


number theory
greatest common divisor
modular arithmetic
abstract algebra
quadratics
number theory proposed

Let $ p \equiv 1 \mod 4$ be prime. Show that $ \frac{p^p\minus{}1}{p\minus{}1}$ is not prime.

i mad an error, the right demonstration is
since $ p\equiv 1 \bmod 4$, so $ p \geq 5$.
if we note $ q\equal{}\frac{p^p\minus{}1}{p\minus{}1}$, and since $ p\equiv 1 \bmod 4$, we can write $ q\minus{}1\equal{}p(p\plus{}1)(p^2\plus{}1)(p^{p\minus{}5}\plus{}p^{p\plus{}9}\plus{}p^{p\minus{}13}\plus{}\cdot\cdot\cdot\plus{}1)$ then $ p\mid q\minus{}1$.
we have also  $ p^p\minus{}1\equal{}q(p\minus{}1)$, then $ p^p\equiv 1\bmod q$. If q is a prime, by fermat theorem, we obtain $ p^{q\minus{}1}\equiv 1\bmod q$, then $ q\minus{}1\mid p$,and hence $ q\minus{}1\equal{}p$. so $ 1\equal{}(p\plus{}1)(p^2\plus{}1)(p^{p\minus{}5}\plus{}p^{p\plus{}9}\plus{}p^{p\minus{}13}\plus{}\cdot\cdot\cdot\plus{}1)$, it's impossible.
then q is not prime.

[quote="NA_219"]we obtain $ p^{q \minus{} 1}\equiv 1\bmod q$, then $ q \minus{} 1\mid p$[/quote]
Why $ q\minus{}1|p$¿

$ q$ is prime so $ Z/qZ\equal{}{1,p,p^{2},\cdot\cdot\cdot,p^{q\minus{}1}}$ and hence any number $ k$ with the relation $ p^k\equiv 1\mod q$ must be a multiple of q-1

in fact $ Z/qZ*$ is a multiplicatif group, so if $ \gcd(k,q)=1$ so $ Z/qZ*=(1,k,k^{2},\cdot\cdot\cdot,k^{p-2})$. in this case $ k=p$ we have $ p^p\equiv 1\bmod q$ let $ p=s(q-1)+r$ where $ 0\geq r <q-1$ so $ p^r\equiv 1\bmod q$ if $ r$ is not null then $ Z/qZ*=(1,p,p^{2},\cdot\cdot\cdot,p^{r-1})$ it's impossible so $ r=0$ (because $ %Error. "midZ" is a bad command.
/qZ*\mid=q-1>r-1$

$ q\equal{}7,p\equal{}2$.
Your argument that $ (\mathbb Z / q \mathbb Z)^* \equal{} \{ 1, p ,p^2, p^3, ...\}$ is wrong.

I'm looking for the generalization. Find the condition of prime $ q$ such that $ \frac {p^q \minus{} 1}{p \minus{} 1}$ is not prime for $ p \equiv 1 (mod 4)$.
Could anyone help me find them please. :blush:

you are right (i am pitful today!!=
since $ p\equiv 1 \bmod 4$, so $ p \geq 5$.
if we note $ q\equal{}\frac{p^p\minus{}1}{p\minus{}1}$, and since $ p\equiv 1 \bmod 4$, we can write $ q\minus{}1\equal{}p(p\plus{}1)(p^2\plus{}1)(p^{p\minus{}5}\plus{}p^{p\plus{}9}\plus{}p^{p\minus{}13}\plus{}\cdot\cdot\cdot\plus{}1)$ then $ p\mid q\minus{}1$.
we have also  $ p^p\minus{}1\equal{}q(p\minus{}1)$, then $ p^p\equiv 1\bmod q$. If q is a prime, by fermat theorem, we obtain $ p^{q\minus{}1}\equiv 1\bmod q$. [color=red]Let $ \delta$ the order of $ p$ in $ (\mathbb Z/q\mathbb Z)^*$, so $ \delta\mid q\minus{}1$.and
$ p\geq\delta$. let $ p\equal{}s\delta\plus{}r$ where $ 0\leq r <\delta$, we have $ p^{r}\equiv 1\bmod q$ and  $ r <\delta$ this is a contradiction with $ \delta$ the order of $ p$ ( minimal element) so $ r\equal{}0$, $ s\equal{}1$ and $ p\equal{}\delta$[/color]
hence $ p\mid q\minus{}1$.

You've already shown this before, the task was to prove $ q\minus{}1|p$ and not $ p|q\minus{}1$ !

ZetaX, Could you show your solution please. :P

Let me try for a while

It's known that: $ p\mid q \minus{} 1$, $ \text{gcd}(q,p \minus{} 1) \equal{} 1$. Now take $ g$ is primitive root of $ q$ then $ g\equiv p^s\pmod q$ so $ g^p\equiv 1\pmod q$. This implies $ q \minus{} 1\mid p$ because $ q \minus{} 1$ is order of $ g$ modulo $ q$. Okey ? :D

[quote="pluricomplex"]Now take $ g$ is primitive root of $ q$ then $ g\equiv p^s\pmod q$...[/quote]
Why $ g\equiv p^s\pmod q$ for some $ s$¿

[quote="madshiet"]If p = 1 (mod 4) and 1 is not a prime number, so p >= 5 then we can think of proceeding... If p >= 5 then p is always odd obviously.
Yeah, now (at the moment) think p as an odd number. p^p - 1 = (p - 1) * (p^(p - 1) + p^(p - 2) + ... + p + 1). So the number we are searching for is (p^(p - 1) + p^(p - 2) + ... + p + 1). Let me show it by using of q. It is the time now for us to prove that q is not a prime.
Now let us use that p = 1 (mod 4). That is
                                                                 q = (p - 2)*1 + 5 + 1 (mod 16)
                                                                 q = p + 4 (mod 16)
                                                                 q = 9 (mod 16).
As all you know that... a prime is never 9 in module 16.
Proven in a clear and clever way.[/quote]

$ 41 \equiv 9 \mod 16$ and $ 41$ is prime. But I already do not understand how you get that $ q$ must be $ \equiv 9 \mod 16$.

Yes ZetaX I made a very trivial mistake. I'm trying to use cyclotomic but I still don't know how use $ p\equiv1\pmod4$ in here :blush:

@ZetaX could You post your solution or some hint?

I can prove it when $ p$ is Germaine prime, that is when $ q\equal{}2p\plus{}1$ is a prime.

If $ q\equal{}2p\plus{}1$ is prime then $ q|p^p\minus{}1$. Indeed by quadratic recoprity law $ (\frac pq)\equal{}(\frac qp)\equal{}(\frac 1p)\equal{}1$, thus $ p^p\equal{}p^{(q\minus{}1)/2}\equiv 1(mod q)$.

As I'm repeatedly asked to give a hint, here is a big one:
[hide="Click for a hint"] http://www.mathlinks.ro/viewtopic.php?t=146020 [/hide]

By the way, a bit more general than Baysa (but I don't know how to use this for a solution):
A prime $ q$ divides $ \frac{p^p\minus{}1}{p\minus{}1}$ if and only if $ q\equal{}pn\plus{}1$ is prime and $ p$ is a $ n$-th power $ \mod q$.

[quote="ZetaX"]As I'm repeatedly asked to give a hint, here is a big one:
[hide="Click for a hint"] http://www.mathlinks.ro/viewtopic.php?t=146020 [/hide]

[/quote]
Let $ p\equal{}1\mod 4$, then $ \frac{p^p\minus{}1}{p\minus{}1}\equal{}\Phi_p(p)\equal{}(R(p)\minus{}pS(p))(R(p)\plus{}pS(p))$.
But may be $ R(p)\minus{}pS(p)\equal{}\pm 1$ or $ R(p)\plus{}pS(p)\equal{}\pm 1$.

Yes.
Thus the main goal is to show the formula and that the factors are not $ \pm 1$.

Any progress, now that Lucas' theorem was proved¿

If $ R(p) \minus{} pS(p) \equal{} \pm 1$

then  $ R(p) \plus{} pS(p) \equal{} \pm \frac {p^p \minus{} 1}{p \minus{} 1}$

and thus $ 2R(p) \equal{} \pm \left(\frac {p^p \minus{} 1}{p \minus{} 1} \plus{} 1\right)$

Which is a contradiction since $ R$ has integer co-efficients. Same for the other one

I don't get why this is a contradiction: both sides are even.

[quote="Rust"][quote="ZetaX"]As I'm repeatedly asked to give a hint, here is a big one:
[hide="Click for a hint"] http://www.mathlinks.ro/viewtopic.php?t=146020 [/hide]

[/quote]
Let $ p \equal{} 1\mod 4$, then $ \frac {p^p \minus{} 1}{p \minus{} 1} \equal{} \Phi_p(p) \equal{} (R(p) \minus{} pS(p))(R(p) \plus{} pS(p))$.
But may be $ R(p) \minus{} pS(p) \equal{} \pm 1$ or $ R(p) \plus{} pS(p) \equal{} \pm 1$.[/quote]

This right? You end up with $ p^2S(p)$

Rust's comment is correct, but I don't understand your question.

Excuse me, im not having a good day. Think ill stop posting here.

Occasionally, I've came to these two reviews:
http://www.ams.org/mathscinet-getitem?mr=MR1367409
http://www.ams.org/mathscinet-getitem?mr=MR1371000
Do they answer the original question?  :wink:

As I can't take a look at them no, I can't decide  :wink:
But do the really show that no factor is $ 1$¿

I've attached these reviews in PDF to my message above.


calculus
integration
function
calculus computations

Let $ f,g$ be two singly differentiable functions
Find
$ \int\frac{f(x)g(x)\plus{}f'(x)g'(x)}{(f(x)g(x)\plus{}1)\sqrt{f(x)g(x)\minus{}1}}\text{dx}$

Just a question: doesn't the answer depends on the nature of f(x) and g(x)?

like what i mean assume the square root can be taken and the quantity in the denominator doesn't vanish
the question should [b]if[/b] it exists then what is it


geometry
circumcircle
incenter
inequalities
angle bisector
triangle inequality
inequalities unsolved

Let the internal angle bisector of $ A,B,C$ meet circumcircle of $ ABC$ at $ D,E,F$
Prove that $ AD\plus{}BE\plus{}CF>AB\plus{}BC\plus{}CA$

let $ I$ be the incenter of $ \triangle ABC$ now from a well-known theorem we know that $ DI\equal{}DB\equal{}DC$ hence by triangle inequality we get that:

$ 2ID\equal{}DB\plus{}DC\geq BC$

$ 2IE\equal{}EC\plus{}EA\geq CA$

$ 2IF\equal{}FA\plus{}FB\geq AB$

also by triangle inequality we have:

$ BI\plus{}IC\geq BC$

$ IC\plus{}IA\geq AC$

$ IA\plus{}IB\geq AB$

now add the inequlities above to get that:

$ 2(AI\plus{}ID\plus{}BI\plus{}IE\plus{}CI\plus{}IF)\geq 2(AB\plus{}BC\plus{}CA)$

$ \Rightarrow AD\plus{}BE\plus{}CF\geq AB\plus{}BC\plus{}CA$

as wanted...


Suppose that $E$ is a separable Banach space. Does there always exist a compact $K$ such that the closure of the vectorial space generated by $K$ is $E$ ?

Yes. Just take a dense countable set $\{x_j\}_{j\in\mathbb N}$ and put $K=\{0\}\cup \left\{\frac{x_j}{j\|x_j\|}\right\}_{j\in\mathbb N}$


geometry
ratio
similar triangles
area of a triangle
geometry solved

Let T = (a; b; c) be a triangle with sides a, b, and c and area M. Denote by T' = (a' ;  b'; c' ) the triangle whose sides are the altitudes of T (i.e., $a^{\prime} = h_a$, $b^{\prime} = h_b$, $c^{\prime} = h_c$) and denote its area by M. Similarly, let T'' = (a''; b''; c'') be the triangle formed from the altitudes of T', and denote its area by M''. Given that M' = 30 and M'' = 20, find M.

Frank Morgan

The area of a triangle equals $\frac12 \cdot \text{ sidelength }\cdot\text{ corresponding altitude}$. Applying this to the triangle T, we get $M = \frac12 a h_a$. Knowing that $a^{\prime} = h_a$, we rewrite this as $M = \frac12 a a^{\prime}$, so that $a^{\prime} = \frac{2M}{a}$. Similarly, $a^{\prime\prime}=\frac{2M^{\prime}}{a^{\prime}}$, so that $a^{\prime\prime}=\frac{2M^{\prime}}{\left(\frac{2M}{a}\right)}=\frac{M^{\prime}}{M}\cdot a$. Similarly, $b^{\prime\prime}=\frac{M^{\prime}}{M}\cdot b$ and $c^{\prime\prime}=\frac{M^{\prime}}{M}\cdot c$, so that the triangles T = (a; b; c) and T" = (a"; b"; c") turn out to be similar, the ratio of similitude being $\frac{M^{\prime}}{M}$. But the areas of these triangles T and T" are M and M", respectively, and we know that the ratio of the areas of two similar triangles equals the squared ratio of similitude. Hence, $\frac{M^{\prime\prime}}{M}=\left(\frac{M^{\prime}}{M}\right)^2$, what quickly simplifies to $M^{\prime}\ ^2 = MM^{\prime\prime}$ and thus to $M = \frac{M^{\prime}\ ^2}{M^{\prime\prime}}$. So we can compute M from M' and M".

  Darij


superior algebra
superior algebra unsolved

(1) Suppose $ G = < x >$ is infinite.
[list]
(a) If $ m \ne k$ in $ \mathbb{Z}$, show that $ x^m \ne x^k.$

(b) Show that $ G = < x > = < x^ {-1} >$, but that $ G \ne < x^k >$ if $ k \ne 1, - 1.$
[/list]

(2) Suppose $ G = < x >$ is finite of order $ n.$
[list]
(a) If $ m,k \in \mathbb{Z},$ show that $ x^m = x^k$ if and only if $ m \equiv k \(mod n),$ i.e., $ n|m - k.$

(b) Show that $ G = < x^m >$ if and only if $ (m,n) = 1.$
[/list]

These are elementary results on cyclic groups that one can find in almost any abstract algebra book.


calculus
integration
trigonometry
calculus computations

[hide="My Solution"]

$ \sqrt{x} \equal{} t \\ \Rightarrow \ I \equal{} 2\int_1^e(e^t \minus{} e^{\minus{}t})cos(e^t \plus{} e^{\minus{}t} \plus{} \frac{\pi}{4})dt  \plus{} 2\int_1^e(e^t \plus{} e^{\minus{}t})cos(e^t \minus{} e^{\minus{}t} \plus{} \frac{\pi}{4})dt$

these 2 integrals can be calculated by substituting the argument of cosine

$ I \equal{} 2 [ sin( e^t \plus{} e^{\minus{}t} \plus{} \frac{\pi}{4} ) \plus{} sin ( e^t \minus{} e^{\minus{}t} \plus{} \frac{\pi}{4} ) ]_1^e$

$ \equal{} 2 ( sin( e^e \plus{} e^{\minus{}e} \plus{} \frac{\pi}{4} ) \plus{} sin ( e^e \minus{} e^{\minus{}e} \plus{} \frac{\pi}{4} ) \minus{} sin( e \plus{} e^{\minus{}1} \plus{} \frac{\pi}{4} ) \minus{} sin ( e \minus{} e^{\minus{}1} \plus{} \frac{\pi}{4} )  )$

hmm.... Is there further simplification?

[/hide]

Using $ \sin(u)\plus{}\sin(v) \equal{} 2\sin(\frac{u\plus{}v}{2})\cos(\frac{u\plus{}v}{2})$ I got down to
\[ 4[\sin(e^e\plus{}\frac{\pi}{4})\cos(e^e\plus{}\frac{\pi}{4})\minus{}\sin(e\plus{}\frac{\pi}{4})\cos(e\plus{}\frac{\pi}{4})]\]

Edit:This also comes to ($ 2\sin(u)\cos(u)\equal{}\sin(2u)$ \[ 2[\sin(2e^e\plus{}\frac{\pi}{2})\minus{}\sin(2e\plus{}\frac{\pi}{2})] \equal{} 2[\cos(2e^2)\minus{}\cos(2e)]\]
Hopefully I did not err ;)

Em... Sorry Kunny, I saw your problem too late.. :P
Thanks Nora and Fermat for their solutions :lol:

This is my version...

$ \int_{1}^{e^{2}}\frac{\left( e^{\sqrt{x}}\minus{}e^{\minus{}\sqrt{x}}\right) \cos \left( e^{\sqrt{x}}\plus{}e^{\minus{}\sqrt{x}}\plus{}\frac{\pi }{4}\right) \plus{}\left( e^{\sqrt{x}}\plus{}e^{\minus{}\sqrt{x}}\right) \cos \left( e^{\sqrt{x}}\minus{}e^{\minus{}\sqrt{x}}\plus{}\frac{\pi }{4}\right) }{\sqrt{x}}dx$

$ \equal{}2\int_{1}^{e^{2}}\frac{\left( e^{\sqrt{x}}\minus{}e^{\minus{}\sqrt{x}}\right) \cos \left( e^{\sqrt{x}}\plus{}e^{\minus{}\sqrt{x}}\plus{}\frac{\pi }{4}\right) }{2\sqrt{x}}dx\plus{}2\int_{1}^{e^{2}}\frac{\left( e^{\sqrt{x}}\plus{}e^{\minus{}\sqrt{x}}\right) \cos \left( e^{\sqrt{x}}\minus{}e^{\minus{}\sqrt{x}}\plus{}\frac{\pi }{4}\right) }{2\sqrt{x}}dx$

$ \equal{}2\left[ \sin \left( e^{\sqrt{x}}\plus{}e^{\minus{}\sqrt{x}}\plus{}\frac{\pi }{4}\right) \plus{}\sin \left( e^{\sqrt{x}}\minus{}e^{\minus{}\sqrt{x}}\plus{}\frac{\pi }{4}\right) \right] _{1}^{e^{2}}$

$ \equal{}2\sqrt{2}\left[ \left( \cos e^{\minus{}\sqrt{x}}\right) \left( \sin \left( e^{\sqrt{x}}\right) \plus{}\cos \left( e^{\sqrt{x}}\right) \right) \right] _{1}^{e^{2}}$

$ \equal{}2\sqrt{2}\left[ \left( \cos e^{\minus{}e}\right) \left( \sin e^{e}\plus{}\cos e^{e}\right) \minus{}\left( \cos e^{\minus{}1}\right) \left( \sin e\plus{}\cos e\right) \right]$


group theory
abstract algebra
superior algebra
superior algebra solved

The 6th dihedral group is as follows:

D6={e, a, a^2, a^3, a^4, a^4, a^5, b, ab, a^2b, a^3b, a^4b, a^5b}
where a^6=b^2=e and ba^i for all i in Z. Now I need to show whether D6 is isomorphic to G:
Here is G:

a) G= Z2 X Z2
b)G=Z4

if they are isomorphic I need to list the elements if they are not I need to prove that no such subgroups exists.

so here's what I think:
neither of these are isomorphic to D6 since their order differ but how do I prove that

Consider the definition of group isomorphism: a structure preserving, [b]bijective[/b] map.

mikejones ... think about the number of elements


Denote by $a \bmod b$ the remainder of the euclidean division of $a$ by $b$. Determine all pairs of positive integers $(a,p)$ such that $p$ is prime and \[ a \bmod p + a\bmod 2p + a\bmod 3p + a\bmod 4p = a + p. \]

Oh yes, $p$ is a prime. Sorry.

I edited the statement.

modulo $p$ we get $3a=0$ so $p=3$ or $a=0$ modp

With $p=3$ we need to check cases, since our expression is at most $2+5+8+11=26$ so a is at most $23$ (I checked them manually actually)

With $a=0$ mod p, we can use that $ap$ mod $bp$=$p(a$mod$b)$.

And it also reduces to check some cases.

The solutions $(p,3p)$ and $(3,1)$, $(3,17)$

[hide]
The idea is to consider cases of $a$ that depend on $p$. Notice that the remainders can be at most $p-1, 2p-1, 3p-1, 4p-1$ respectively. and thus $a$ is at most $9p-4$ for a given prime. Now consider cases $a<p, p\leq a<2p, 2p \leq a<3p,..., 8p \leq a<9p$. For each case because of division algorithm, $a mod kp$ is equal to the difference between $a$ and the greatest multiple of $k$ that is less or equal than the lower bound, so for example if we're considering the case $5p \leq a<6p$ we would have,
\[a mod p = a-5p, a mod 2p=a-4p, a mod 3p= a-3p, a mod 4p=a-4p\]
when examining all the cases we see that there are valid solutions only when $a<p$ and $5p\leq a<6p$ namely, $(1,3)$ and $(17,3)$.


inequalities
search
inequalities proposed

If $ x$ $ y$ and $ z$ are positive reals then prove the inequality:

$ \frac{(2x\plus{}y\plus{}z)^{2}}{(y\plus{}z)^{2}\plus{}(x\plus{}y)(x\plus{}z)}\plus{}\frac{(x\plus{}2y\plus{}z)^{2}}{(x\plus{}z)^{2}\plus{}(y\plus{}z)(y\plus{}x)}\plus{}\frac{(x\plus{}y\plus{}2z)^{2}}{(x\plus{}y)^{2}\plus{}(z\plus{}x)(z\plus{}y)}$ $ \geq6$

Let $ x \plus{} y \equal{} a,\ y \plus{} z \equal{} b,\ z \plus{} x \equal{} c\ (a,\ b,\ c > 0)$, using C.S. and $ a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca$ gives proof.

Oh[b] kunny[/b] I think that you are wrong
If you use C.S. you will get that $ ab \plus{} bc \plus{} ca\geq a^{2} \plus{} b^{2} \plus{} c^{2}$ which is obviously incorrect.

This is an old inequality posted by Peter Scholze and Darij Grinberg in 2005 on this forum, for any positive numbers $ a,b,c$.

Here is the link
http://www.mathlinks.ro/viewtopic.php?search_id=7020805&t=60128


Does anyone have a file of any national competition rankings before 2004?

There's a list of winners somewhere on the MathCounts website, it shouldn't be hard to find. As for the top placings and stuff, I doubt those are anywhere on the internet (after all, the internet wasn't readily available in the 80's and early 90's), so for all those competitions, you would have to ask someone who was actually there. come to think of it, lists like they distribute now may not even be distributed anymore.

I believe the 2002 and 2003 top 10 teams and individuals are in the newsletters that had the press release about the national champions.

2007 -- Kevin Chen
2006 -- Daesun Yim
2005 -- Neal Wu
2004 -- Greg Gauthier
2003 -- Adam Hesterberg
2002 -- Albert Ni
2001 -- Ryan Ko

I actually have that memorized...

Let's see...

2000-Joe Jia
1999-Po-Ru Loh
1998-Ricky Liu

...

1991-Jonathan Weinstein

hmm i think that's right

They have the list of winners in the handbook (from last year at least)

And on wikipedia

[quote="soxcubs104"]How about PDF files of the top 57?[/quote]

Read my first post

Here are all the winners up through 2006.

How about files or lists of past countdown round particpants or the top 57 ranked students prior to 2004?

[quote="13375P34K43V312"] As for the top placings and stuff, I doubt those are anywhere on the internet (after all, the internet wasn't readily available in the 80's and early 90's), so for all those competitions, you would have to ask someone who was actually there. come to think of it, lists like they distribute now may not even be distributed anymore.
[/quote]


Three congruent circles intersect at a point H and again
in pairs at A, B, and C . Prove that H is the
orthocenter of ∆ABC .

Let $ O_1, O_2, O_3$ be three centers of the circles, and let $ P,Q,R$ be midpoints of $ \overline{O_1O_2},\overline{O_2},\overline{O_3O_1}$. It is easy to proove that $ H$ is the orthocenter of $ \triangle{PQR}$, because $ \overline{HP}$ is perpendicular to $ \overline{O_1O_2}$ and so on. And because $ \triangle{PQR}$ and $ \triangle{ABC}$ are on the position of similarity, and the center of similarity is $ H$(ratio is $ 1: 2$), $ H$ is also an orthocenter of $ \triangle{ABC}$.

Sorry for my clumsy English.


I'm a student from Greece and 1,5 years i'm finishing the high school......
I love maths so i'm looking for a good university about mathematics in America...
But i dont know the requirements to pass in a university in America.
So can you tell me about this?
And of course can you suggest me a good university...........

Dimitris

[quote="tokenadult"]Most United States universities describe their admission requirements fairly clearly on the university websites. A typical research university with a strong program in mathematics will want to see your record of high school classes (called a "transcript" in the United States), college entrance test scores from either the SAT or ACT examinations, and possibly letters of recommendation from one or two teachers, and possibly essays you write describing yourself and your interests. State universities usually have simpler application requirements and are sometimes easier to get into than privately operated universities. To study in the United States you will also need to get a United States student visa, and that has its own application procedure.

Please ask follow-up questions as needed. Good luck.[/quote]
Thanks....
Can you tell me now some good universities in math?????
And what is 'privately operated universities'


I was wondering how to use Fermats Little Theorem in general. Like I know that the theorem states $ a^{p\minus{}1} \equiv 1 (mod p)$, but how can you use that to solve problems like, say, "Find $ 2^{10} \plus{} 4^{20} \plus{} 6^{30} \plus{} 12^{40}$ mod $ 13$" (as an example)? If someone could post a step by step solution to solving those types of problems that would be great.

[quote="Thunder365"]I was wondering how to use Fermats Little Theorem in general. Like I know that the theorem states $ a^{p \minus{} 1} \equiv 1 (mod p)$, but how can you use that to solve problems like, say, "Find $ 2^{10} \plus{} 4^{20} \plus{} 6^{30} \plus{} 12^{40}$ mod $ 13$" (as an example)? If someone could post a step by step solution to solving those types of problems that would be great.[/quote]
Taking your example, any number $ a$ that is relatively prime to $ p$ would be in the form $ a^{p\minus{}1}\equiv 1\bmod{p}$, so for $ 2^{10}\plus{}4^{20}\plus{}6^{30}\plus{}12^{40}$ we can take each term separately in mod 13.
$ 2^{12}\equiv 1\bmod{13}$, $ 4^{12}\equiv 1\bmod{13}$, $ 6^{12}\equiv 1\bmod{13}$, and $ 12^{12}\equiv 1\bmod{13}$.
From this information we can determine that
$ 2^{12}\equiv 2^{10}\cdot2^{2}\equiv 1\cdot4\bmod{13}$
$ 4^{20}\equiv 4^{12}\cdot4^{8}\equiv 1\cdot3^{4}\equiv 3\bmod{13}$ ( I used the information that $ 4^{2}\equiv 3\bmod{13}$)
$ 6^{30}\equiv 6^{12}\cdot6^{12}\cdot6^{4}\equiv1\cdot1\cdot(\minus{}3)^{2}\equiv 9\bmod{13}$ ($ 6^{2}\equiv\minus{}3\bmod{13}$)
$ 12^{40}\equiv 12^{12}\cdot12^{12}\cdot12^{12}\cdot12^{4}\equiv 1\cdot1\cdot1\cdot(\minus{}1)^{4}\equiv 1\bmod{13}$
Adding those up we have $ 4\plus{}3\plus{}9\plus{}1\equal{}4\bmod{13}$

Thanks for the clear explanation!  :lol:


$ a,b,c \ge 0$ and $ a\plus{}b\plus{}c\equal{}1$,prove that:
$ \frac {1}{\sqrt {a^2\plus{}ab\plus{}b^2}}\plus{}\frac {1}{\sqrt {b^2\plus{}bc\plus{}c^2}}\plus{}\frac {1}{\sqrt {a^2\plus{}ac\plus{}c^2}} \ge \frac {2(6\plus{}\sqrt {3})}{3}$

[quote="rachid"]$ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 1$,prove that:
$ \frac {1}{\sqrt {a^2 \plus{} ab \plus{} b^2}} \plus{} \frac {1}{\sqrt {b^2 \plus{} bc \plus{} c^2}} \plus{} \frac {1}{\sqrt {a^2 \plus{} ac \plus{} c^2}} \ge \frac {2(6 \plus{} \sqrt {3})}{3}$[/quote]

http://www.toanthpt.net/forums/showthread.php?t=11166


Greetings. Emory will offer the AMC 10B/12B this year on February 25 at 5:00.  Due to the late notice however, I need a head count for ordering tests by Friday afternoon,  Jan 30.  I am sorry that I cannot give you more time than this to ask your students and friends.  The cost is 2.10 per student for the AMC 10B and 2.20 per student for the AMC 12B.

You are only eligible to take it here if your school is not already administering it and you may only take one test as they will be offered at the same time.  The AMC 10 is open to freshmen and sophomores (or exceptional middle schoolers), the AMC 12 is open to all high schoolers.  Freshmen and sophomores may thus take either.

For more information see the AMC website:
http://www.unl.edu/amc/e-exams/e6-amc12/amc12.shtml

or their brochure
http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/2009-H-Brochure.pdf

-Paul Wrayno
Emory Graduate Student
pwrayno (at) emory (dot) edu

Paul graduated from Duke and ran their math meet a couple of years ago.~Mrs D/Billy's mom

My son wanted to participate in today's AMC. Is it too late?

I'm running the AMC 10/12B at Emory this year, and we still have 8 AMC 12B and 5 AMC 10B tests available for the 24th.  Anyone interested in taking it whose school is not offering it please email me at pwrayno@emory.edu if interested as I don't come on AOPS very often.


Hello,

I am using Internet Explorer (6.0).
Yesterday I logged in and everything was fine.
Today I keep getting this weird (and very difficult) layout, but ONLY on one specific page in one specific topic in Games and Fun( Movies you just saw)

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=34602&start=500[/url]

[img]http://i116.photobucket.com/albums/o39/evilbu/BADAOPS.jpg[/img]

[quote="alanchou"]It also happens in IE7, just a little different.  I dunno, maybe someone used \pir?[/quote]
Maybe I should add that everything is fine when I use Opera 8.2?

But Opera doesn't let me log in onto this site? What's going on? :o

Check Opera's cookie settings.

As usual, everything is fine in firefox, and safari too.

Hmmm....... That happened to me yesterday but is' all fine today.

Yeah I think that fixes it. It was my bad trying to fix another bug, I created another one :D