Given a sequence $\{a(n)\}$ s.t.
1) $a(0)=a(1)=a(2)=a(3)=1$;
2) $a(n)\cdot a(n-4)=a(n-1)\cdot a(n-3)+(a(n-2))^2$.
Prove that $a(n)$ is integer for every $n$.

Can you solve it but not used induction.

[i]Edited by Myth[/i]

everybody know :D : shortlist02 have problem: prove that there is many triples integer a,b,c satisfy: 1/a+1/b+1/c+1/abc=12/a+b+c
for solve this pro, need use sequence (a(n)): a(1)=a(2)=a(3)=1
 a(n+3).a(n)=a(n+1).a(n+2)+1, and can pro a(n) is integer by induction-(similar my pro) but sol very long, i have other sol very short and not use induction
and can you solved, not use induction? :(
  :P sorry for my poor english  :P

I saw non-unduction solution some time ago, but I can't remember it.

[Moderator edit: This post refers to a different version of the initial problem (actually, a particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=49408 for $ m\equal{}k\equal{}1$).]

[quote="nguyenquockhanh"]...for solve this pro, need use sequence (a(n)):
a(1)=a(2)=a(3)=1
a(n+3).a(n)=a(n+1).a(n+2)+1, and can pro a(n) is integer by induction-(similar my pro) but sol very long, i have other sol very short and not use induction
and can you solved, not use induction? :( [/quote]
Yes I can.
It immediatelly follows from identity:
$ \frac {a_{n \plus{} 3} \plus{} a_{n \plus{} 1}}{a_{n \plus{} 2}} \equal{} \frac {a_{n \plus{} 1} \plus{} a_{n \minus{} 1}}{a_n}$.

Please, do not change the initial problem next time. Create a new topic instead ;)

[quote="Myth"][quote="nguyenquockhanh"]...for solve this pro, need use sequence (a(n)):
a(1)=a(2)=a(3)=1
a(n+3).a(n)=a(n+1).a(n+2)+1, and can pro a(n) is integer by induction-(similar my pro) but sol very long, i have other sol very short and not use induction
and can you solved, not use induction? :( [/quote]
Yes I can.
It immediatelly follows from identity:
$\frac{a_{n+3}+a_{n+1}}{a_{n+2}}=\frac{a_{n+1}+a_{n-1}}{a_n}$.[/quote]
yes,but i have no solution not use induction :D
if you prove by induction- you must prove: for all n: a(n) is integer,(a(n),a(n-1))=1 and(a(n),a(n-2))=1. do you agree with me: like that very long, not nice :?

He just asks if you agree with him that proof by induction would be long and not nice  :)

Maybe you are right ;)
Did you solve his new problem, which, actually, is first in this thread now?


triangle inequality
IMO
algebra
polynomial
rearrangement inequality
Inequality
IMO 1983

Let $ a$, $ b$ and $ c$ be the lengths of the sides of a triangle. Prove that
\[ a^{2}b(a \minus{} b) \plus{} b^{2}c(b \minus{} c) \plus{} c^{2}a(c \minus{} a)\ge 0.
\]
Determine when equality occurs.

Come on, not this again. it has been posted many times before, I think. From the subtitutions $a=y+z, b=z+x, c=x+y$ and the inequality is equivalent with $\sum\frac{x^{2}}{y}\geq\sum x.$ From here it is clear how to proceed. :D

Sure enough, that's what i did.I was wondering if you could use Schur to solve it?
 Anyone?
   Ashwath

[quote="Gibbenergy"]Hi Cezar Lupu!
   As I do this IMO again, I found the very interesting solution for it and it is only 1 line . Can u believe it?
  He he . Ok ! Here is mine ( Any please don't ask me why I found it!);
IMO inequality is equivalent to this "obvious inequality":
$\frac12((a-b)^2(a+b-c)(b+c-a)+(b-c)^2(b+c-a)(a+c-b)+(c-a)^2(a+c-b)(a+b-c))\geq 0$
Right ![/quote]

  It's not new and obviously it's not your solution, Gibb . The ineq is homogeneous , so i can easily change the ineq into the fomula: $A(b-c)^2+B(c-a)^2+C(a-b)^2 \geq 0$    ;)

Hey Gibb this one-line solution had been given by an IMO contestant. :D  ;)

[quote="Gibbenergy"]Hey guys!
   That is my sol and I have found it by myself. I don't care who has posted it before.
    I found my sols because as I know the IMO US team at that time was also use the trick that put $x=u+v...$ It is obviously unnatural. So it is not important with the way they showed. The question is how can the author generate it? I believed that the author of IMO 83 used the same technique that I have discovered  now to make it. So it doesn't matter that my sols now is posted before.
    If you say you can make $A (b-c)^2$ i believed it is the theory. The important thing is how can you find $A$? $(b-c)^2....$ and also their permutation. If you can show me that technique, I would admire you as my best teacher.
    I feel that I don't want to share any my technique to you guys either. I will keep it as my secrect.
 Thanks.[/quote]

Ha! It's really funny. If you are willing to search on the forum, you can see that many people here prove inequalities by rewriting the inequality into sum of squares. Obviously they have a technique, or at least a general way to write inequalities in that form. Although it may not be able to be written rigorously, a general approach does exist I think.
Of course you have the right not to share your technique and keep it as your secret. But I just wonder, if you don't even want to share, why do you hope others are willing to share with you? It's just like you want others to share something with you while you are reluctant to hand out anything. What a joke.

What a ridiculous conversation. Well, there are plenty of people willing to share NEW and interesting techniques with us, so we shouldn't worry at all.

And, oh yes, topic locked.

Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=55406 and http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 (posts #8-#10) .

  darij

The inequality equivalent to:
\[a^{3}b+b^{3}c+c^{3}a \ge a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\]
And this is a result of Moorhead Inequality
\[\left({3,1,0}\right) > \left({2,2,0}\right) \]

This is an old topic but I think that it's worth pointing out that the LHS of your inequality is cyclic, not symmetric. Muirhead only works on symmetric sums, so the rearrangement method is needed.

We can use the Entirely Mixing variable method to solve this problem (but this method is so strong to solve such an easy problem). It's sufficient to consider one case $a=b+c$ and the rest is easy.

[quote="GacKiem"]We can use the Entirely Mixing variable method to solve this problem (but this method is so strong to solve such an easy problem). It's sufficient to consider one case $ a \equal{} b \plus{} c$ and the rest is easy.[/quote]

Sorry about reviving an old thread, but after reading http://reflections.awesomemath.org/2006_5/2006_5_entirelymixing.pdf, I still don't get the method. I can't show that replacing $ (a,b,c)$ with $ (a\minus{}c,b\minus{}c,0)$ will necessarily decrease LHS.

a^3b+b^3c+c^3a>=a^2b^2+c^2b^2+c^2a^2  это верно по rearrangement inequality .

[quote=lotharek]I use Ravi's substitution.

Let $a=x+y, b=y+z, c=z+x$.

Some computation give...:

\[ x^3z+xy^3+yz^3\geq xyz(x+y+z),  \]
\[ \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z,  \] what is true by rearrangement inequality.

Equality holds, when $x=y=z$, so when $a=b=c$.[/quote]

can you please explain the "some computation" part, I've been trying to solve this problem for hours and I don't understand how to get to \[ x^3z+xy^3+yz^3\geq xyz(x+y+z),  \]

Making a Ravi Substitution yields
\begin{align*}
\text{ LHS  } &= (y+z)^{2}(x+z)((y+z) - (x+z)) + (x+z)^{2}(x+y)((x+z) - (x+y)) + (x+y)^{2}(y+z)((x+y) - (y+z)) \\
&=2(xy^3 + x^3z - x^2yz - xy^2z - xyz^2 + yz^3) \\
&\implies \text{ LHS } \Leftrightarrow xy^3 + x^3z - x^2yz - xy^2z - xyz^2 + yz^3 \geq 0\\
&\implies \text{ LHS } \Leftrightarrow xy^3 + x^3z + yz^3 \geq x^2yz+xy^2z+xyz^2.\\
\end{align*}
This inequality is homogenous, therefore we let $xyz=1$ to get $$\text{ LHS } \Leftrightarrow \frac{y^2}{z} + \frac{x^2}{y} + \frac{z^2}{x} \geq x+y+z,$$ which is true by the Rearrangement Inequality on sequences $(x^2,y^2,z^2)$ and $(\tfrac{1}{x},\tfrac{1}{y},\tfrac{1}{z}).$ Equality holds when either of these sequences are constant, or whenever $x=y=z \implies a=b=c.$

$\mathbb{Q.E.D.}$

Insta-killed by Ravi!

Ravi sub, we get that we want to show that $\sum_{\text{sym}}x^3y\geq \sum_{\text{sym}}x^2yz,$ which is just Muirhead.

[color=#f00]Why does everyone overcomplicate these type of problems and use rearrangement and homogenizing when it's just a direct Muirhead application? I know that Mildorf said not to cite Muirhead, but I am pretty sure its well known enough by now(Mildorf wrote his article in 2005)[/color]

[quote=gold46]Muirhead and Rearrangement inequalities don't work here.[/quote]

Is this dude just clowning
A few others also say that Muirhead doesn't work because the inequality is cyclic, not symmetric, but idek :shrug:

[quote=JustKeepRunning]
[color=#f00]Why does everyone overcomplicate these type of problems and use rearrangement and homogenizing when it's just a direct Muirhead application? I know that Mildorf said not to cite Muirhead, but I am pretty sure its well known enough by now(Mildorf wrote his article in 2005)[/color][/quote]

Well homogenization was just a useless step because dividing by $xyz$ in my expression performs the same task. Tbh I have no idea why I used it, but either way it's automatic.

For reference [url=https://artofproblemsolving.com/articles/files/MildorfInequalities.pdf]here is the Mildorf article.[/url]

Also idk Muirhead that well and this is a direct rearrangement application too right?

[quote=JustKeepRunning]@above what do you mean? The symemtric is just $2$ times the cyclic...[/quote]

I don't know why - I just remarked that some of the other posts stated that rearrangement/muirhead don't work.

[quote=JustKeepRunning]@above what do you mean? The symemtric is just $2$ times the cyclic...[/quote]

For instance $\sum_{\text{sym}}a^2b=a^2b+ab^2+b^2c+bc^2+c^2a+ca^2$ but $2\sum_{\text{cyc}}a^2b=2a^2b+2b^2c+2c^2a$.

Making Muirhead work isn't trivial as far as I can tell.

With Ravi ($a=x+y,b=y+z,c=z+x$) we have to prove:
$$(x+y)^2(y+z)(x-z)+(y+z)^2(z+x)(y-x)+(z+x)^2(x+y)(z-y)\ge0$$
$$\Leftrightarrow2x^3z+2xy^3+2yz^3\ge2x^2yz+2xy^2z+2xyz^2,$$
which is true since $\frac{x^2}y+\frac{y^2}z+\frac{z^2}x\ge x+y+z$ by C-S.

Since sides of triangle it is true if we WLOG, $$a\geq b\geq c \implies a^2+bc \geq b^2+ca\geq c^2+ab.$$ By rearrangement inequality, \begin{align*} \sum_{\text{cyc}} \frac{a^2+bc}{c} \geq \sum_{\text{cyc}} \frac{a^2+bc}{a}\\ \implies \sum_{\text{cyc}} \frac{a^2}{c} \geq \sum_{\text{cyc}} \frac{bc}{a} \\ \implies \sum_{\text{cyc}} a^3b \geq \sum_{\text{cyc}} a^2b^2.\end{align*} Equality when $a=b=c.$ And the claim follows. $\square$

[quote=ZETA_in_olympiad]Ravi inequality seems cool, could I have a introductory note on it?[/quote]

[url=https://brilliant.org/wiki/ravi-substitution/]here

Rewrite the inequality as $$\sum_{cyc}a^2b(a-b)\geq 0.$$ Let $a=y+z,b=x+z,c=x+y$. Then, this becomes $$\sum_{cyc}(y+z)^2(x+z)(y-x)\geq 0.$$ Expanding, this is $$\sum_{cyc}(y^2+2yz+z^2)(xy-x^2+yz-xz)\geq 0$$ $$\sum_{cyc}xy^3+y^3z+yz^3-xz^3-2x^2yz+xy^2z-xyz^2-x^2y^2-x^2z^2+2y^2z^2\geq 0.$$ Note that $$\sum_{cyc}xy^2z-xyz^2=\sum_{cyc} -x^2y^2-x^2z^2+2y^2z^2=\sum_{cyc}y^3z-xz^3=0,$$ so it suffices to show that $$\sum_{cyc}xy^3+yz^3-2x^2yz\geq 0.$$ However, this is just $$\sum_{sym}x^3y\geq \sum_{sym} x^2yz,$$ which is just Muirhead's since $(3,1,0)\succ (2,1,1)$ so we are done. Equality occurs when $x=y=z$ or equivalently $a=b=c$ from the Muirhead step.

ugh.. i just had to expand it bruh

After my attachment, the inequality follows from C-S since $LHS(y+z+x)\ge (x+y+z)(y+z+x). \blacksquare$

oops other sol less expanding but much harder to find


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why?

Take a look at the last line of the first block of the Profile page - there's something there for you to type that we added to help prevent the automatic registrations that spam the site.  I'll look into getting rid of that for profile edits.


[quote="xirti"]Are given two sequences $a_n\to a$ and $b_n \to b$. Prove that $\frac{a_nb_1+a_{n-1}b_2+...+a_2b_{n-1}+a_1b_n} {n}$ is convergent and find it's limit.[/quote]
Let $x_n=a+\alpha_n$,$y_n=b+\beta_n$, thus $\alpha_n$,$\beta_n \to 0$.
Then
$z_n=\frac{a_nb_1+a_{n-1}b_2+...+a_2b_{n-1}+a_1b_n} {n}=ab+a\frac{\beta_1+...+\beta_n}{n}+b\frac{\alpha_1+...+\alpha_n}{n}+\frac{\alpha_1 \beta_n+\alpha_2 \beta_{n-1}+...+\alpha_n \beta_1}{n}=ab+a\gamma_{n1}+b\gamma_{n2}+\gamma{n3}$.
It is easy to verify that $\gamma_{n1},\gamma_{n2},\gamma{n3} \to 0$ as $n \to \infty$.

$\gamma_{n1}$ and $\gamma_{n2}$ tends to 0 by Cesaro-Stoltz theorem. But what about $\gamma_{n3}$?? this is my big problem.

For the third one apply Cauchy-Scharz and then Cesaro-Stolz.  ;)

Since $\alpha_n \to 0$ we can conclude that $\alpha_n$ is bounded and there exists $C>0$ s.t. $|\alpha_n| \leq C$ , thus $| \gamma_{n3}|\leq C\frac{|\beta_1|+...+|\beta_n|}{n}$ which comverges to $0$ by Cezaro-Stolz(as you concluded above)


Consider the number $N = 2^{2000} \cdot 5^{1990}$ written in decimal notation.  How many digits does $N$ have?
\[ \text{(A)}\ 3990 \qquad \text{(B)}\ 2000 \qquad \text{(C)}\ 1990 \qquad \text{(D)}\ 1994 \qquad \text{(E)}\ 1999 \]

[hide]We can rewrite this as $10^{1990}\times2^{10}$.  Obviously $10^{1990}$ has $1991$ digits, and multiplying it by $2^{10}=1024$ will give this three more digits for a total of $1994$.  So the answer is [b]D[/b].[/hide]


Sa se determine cifrele $ \{b,o,i\}\subset\overline {0,9}$ so that $ \overline {oi}^2\equal{}\overline {boi}$ , adica

$ (10\cdot o\plus{}i)^2\equal{}100\cdot b\plus{}10\cdot o\plus{}i$ .

[quote="Virgil Nicula"]Sa se determine cifrele $ \{a,b,c\}\subset\overline {0,9}$ so that :

$ 1\blacktriangleright\ \ \overline {ab}^2 \equal{} \overline {abc}$ , adica $ (10\cdot a \plus{} b)^2 \equal{} 100\cdot a \plus{} 10\cdot b \plus{} c$ .

$ 2\blacktriangleright\ \ \overline {ac}^2 \equal{} \overline {abc}$ , adica $ (10\cdot a \plus{} c)^2 \equal{} 100\cdot a \plus{} 10\cdot b \plus{} c$ .

$ 3\blacktriangleright\ \ \overline {bc}^2 \equal{} \overline {abc}$ , adica $ (10\cdot b \plus{} c)^2 \equal{} 100\cdot a \plus{} 10\cdot b \plus{} c$ .  [/quote]


$ ABCD$ is a cyclic Quadrilateral such that $ AB \equal{} 4$ , $ BC \equal{} 5$ , $ CD \equal{} 4.5$ , $ DA \equal{} 3.5$ . $ M$ and $ N$ are the midpoints of the diagonals $ AC$ and $ BD$ . Compute the lenght of $ MN$.

To solve this problem I use the STEWART theorema in triangles ABC ,ABD,BCD and ANC.The MN value is then found with the formulae for cyclic quadrilateral :AB*CD+AD*BC=p^2+q^2 where p and q are the diagonals.After this I found MN=38^.5/2

Actually, the formulae used above are not OK!
Use the [b][u]first Ptolemy theorem[/u][/b], and get $AC\cdot BD$: i.e. $AB\cdot CD+BC\cdot AD=AC\cdot BD$ and the [b][u]second Ptolemy theorem[/u][/b] to find $\frac{AC}{BD}$ : i.e. $\frac{AC}{BD}=\frac{AB\cdot BC+CD\cdot DA}{AB\cdot AD+BC\cdot CD}$.
From the above two relations we find the lengths of diagonals and, once they are known, we can apply [b][u]Newton formula[/u][/b]: $AB^2+BC^2+CD^2+DA^2=AC^2+BD^2+4\cdot MN^2$.

Best regards,
sunken rock


Πώς αποδεικνύεται αυτό; :maybe:
Για $ r\equal{}1$ υπάρχει κλειστή μορφή;

Για $ r\equal{}1$ ο τύπος που έγραψα (έκανα καποιες διορθώσεις) δίνει $ n! \binom{n\plus{}1}{k}$. Μήπως ο τύπος βοηθάει για να βρεις γιατί είναι αυτή η απάντηση; Αν όχι πες μου και θα εξηγήσω γιατί σε επόμενο post.

Ναι για n=1 είναι διωνυμικό  ανάπτυγμα οπότε προκύπτει η κλειστή μορφή αν βέβαια αποδείξουμε πώς συνδέεται το πρόβλημα με το ανάπτυγμα που λες.Αλλά αυτό πώς γίνεται;..δεν μου έρχεται κάποια ιδέα.

Για $ r \equal{} 1$: (Ξέχνα προς το παρών το ανάπτυγμα)

Έχουμε $ n$ γράμματα διαφορετικά από το $ A$ (και διαφορετικά μεταξύ τους). Βάζουμε αυτά τα γράμματα στην σειρά με $ n!$ διαφορετικούς τρόπους και τώρα πρέπει να βάλουμε τα $ A$ ενδιάμεσά τους.

Έχουμε π.χ. *Β*Γ*...*Ω* όπου σε κάθε αστεράκι μπορούμε να τοποθετήσουμε το πολύ 1 Α (αφού r=1). Έχουμε $ n \plus{} 1$ αστεράκια και $ k$ $ A$, άρα υπάρχουν ακριβώς $ \binom{n\plus{}1}{k}$ τρόποι για να το πετύχουμε αυτό.

Πολύ ωραία λύση Δημήτρη ευχαριστώ πολύ :)

[quote="Demetres"]Είναι ο συντελεστής του $ x^k$ στο ανάπτυγμα $ n!(1 \plus{} x \plus{} x^2 \plus{} \ldots \plus{} x^r)^{n \plus{} 1}$. (Απόδειξη; ) Τώρα αν υπάρχει κάποια κλειστή μορφή για κάθε $ r$ δεν είμαι σίγουρος αλλά θα το κοιτάξω.[/quote]
Δεν βάζεις και την απόδειξη σε αυτό;(εκτός αν συνδέεται κάπως με την προηγούμενη λύση και δεν το βλέπω)

[quote="giannis18"]Έστω μία λέξη με το γράμμα $ A$ να επαναλαμβάνεται ακριβώς $ k$ φορές και $ n$ διαφορετικά μεταξύ τους γράμματα.Αν $ r < k$ να υπολογιστεί το πλήθος των λέξεων που περιέχουν ακριβώς τα παραπάνω γράμματα και υπάρχουν το πολύ $ r$ συνεχόμενες θέσεις που περιέχουν το γράμμα $ A$.
Παιδιά μπορεί κάποιος να βοηθήσει γιατί έχω κολλήσει με αυτήν την άσκηση; :([/quote]

Παιδια δεν ξερω αν ειναι σωστο αλλα εγω βρηκα:

$ \sum_{i = 1}^{r}{n(n - 1)(k + n - i + 1)(n + k - i - 2)!}$

@giannis18: Ουσιαστικά η ίδια απόδειξη είναι. Τώρα στο κάθε αστεράκι μπορείς να βάλεις μέχρι $ r$ από τα $ A$. Η δύναμη του $ x$ που θα χρησιμοποιηθεί από κάθε ένα από τα $ (1\plus{}x\plus{}\ldots\plus{}x^r)$ θα σου πει πόσα $ A$ πάνε στο αντίστοιχο αστεράκι. (Αν θες περισσότερη επεξήγηση πες μου.)

@NickNafplio: Μάλλον κάτι διαφορετικό υπολογίζεις. Η τελική απάντηση πρέπει να είναι συνάρτηση των $ k,n$ και $ r$. Η δική σου απάντηση είναι μόνο συνάρτηση των $ k$ και $ n$.

[quote="NickNafplio"][quote="Demetres"]@giannis18: Ουσιαστικά η ίδια απόδειξη είναι. Τώρα στο κάθε αστεράκι μπορείς να βάλεις μέχρι $ r$ από τα $ A$. Η δύναμη του $ x$ που θα χρησιμοποιηθεί από κάθε ένα από τα $ (1 + x + \ldots + x^r)$ θα σου πει πόσα $ A$ πάνε στο αντίστοιχο αστεράκι. (Αν θες περισσότερη επεξήγηση πες μου.)

@NickNafplio: Μάλλον κάτι διαφορετικό υπολογίζεις. Η τελική απάντηση πρέπει να είναι συνάρτηση των $ k,n$ και $ r$. Η δική σου απάντηση είναι μόνο συνάρτηση των $ k$ και $ n$.[/quote]

Εχεις δικιο, ειχα κανει ενα μικρο τυπογραφικο στο αθροισμα, αυτο που ηθελα να γραψω ειναι:
$ \sum_{i = 1}^{r}{n(n - 1)(k + n - i + 1)(n + k - i - 2)!}$[/quote]

Βασικα η λυση ειναι λαθος
Υπαρχει ενα λαθος το οποιο δεν ειδα λογο κεκτημενης ταχυτητας.
Συγνωμη αν ταλαιπωρησα καποιον

[quote="Demetres"]@giannis18: Ουσιαστικά η ίδια απόδειξη είναι. Τώρα στο κάθε αστεράκι μπορείς να βάλεις μέχρι $ r$ από τα $ A$. Η δύναμη του $ x$ που θα χρησιμοποιηθεί από κάθε ένα από τα $ (1 \plus{} x \plus{} \ldots \plus{} x^r)$ θα σου πει πόσα $ A$ πάνε στο αντίστοιχο αστεράκι. (Αν θες περισσότερη επεξήγηση πες μου.)
[/quote]
Ok το κατάλαβα τώρα και ευχαριστώ πολύ για τη βοήθεια.
Με τη λύση σου μου ήρθε μία λύση στο μυαλό για το r γενικά αλλά δεν ξέρω κατά πόσο είναι σωστή.
Έστω $ x_1,x_2,...,x_{n \plus{} 1}$ το πλήθος των Α στις αντίστοιχες θέσεις-αστεράκια.Τότε $ \displaystyle\sum_{i \equal{} 1}^{n \plus{} 1}x_i \equal{} k$ όπου $ 0 \leq x_i \leq r$.Θέτουμε $ A_j \equal{} \{(x_1,....,x_{n \plus{} 1})|\displaystyle\sum_{i \equal{} 1}^{n \plus{} 1}x_i \equal{} k,x_j \geq r \plus{} 1\}$ και θα υπολογίσουμε το $ \left | \cap_{i \equal{} 1}^m A_{j_i} \right|$.
Αν θέσουμε $ y_i \equal{} x_i$ για $ i \neq j_1,...,j_m$ και $ x_i \equal{} y_i \plus{} r \plus{} 1$ για $ i \equal{} j_1,...,j_m$ τότε εύκολα προκύπτει ότι αρκεί να προσδιορίσουμε το πλήθος λύσεων της $ \displaystyle \sum_{i \equal{} 1}^{n \plus{} 1}y_i \equal{} k \minus{} m(r \plus{} 1)$ όπου $ 0 \leq y_i$ και αυτό ως γνωστό είναι $ \binom{n \plus{} 1 \plus{} k \minus{} m(r \plus{} 1) \minus{} 1}{n \plus{} 1 \minus{} 1}$ άρα $ \left | \cap_{i \equal{} 1}^m A_{j_i} \right| \equal{} \binom{n \plus{} k \minus{} m(r \plus{} 1)}{n}$.
Άρα inclusion-exclusion και τελικά το ζητούμενο είναι
$ n! \binom{n \plus{} k}{k} \minus{} n!\sum_{m \equal{} 1}^{n \plus{} 1}( \minus{} 1)^{m \minus{} 1}\binom{n \plus{} 1}{m}\binom{n \plus{} k \minus{} m(r \plus{} 1)}{n} \equal{} n! \binom{n \plus{} k}{k} \plus{} n!\sum_{m \equal{} 1}^{n \plus{} 1}( \minus{} 1)^m\binom{n \plus{} 1}{m}\binom{n \plus{} k \minus{} m(r \plus{} 1)}{n} \equal{} n!\sum_{m \equal{} 0}^{n \plus{} 1}( \minus{} 1)^m\binom{n \plus{} 1}{m}\binom{n \plus{} k \minus{} m(r \plus{} 1)}{n}$
Μπορούμε και να περιορίσουμε κάπως το τελευταίο άθροισμα αν δούμε ότι για $ m > \left \lfloor{\frac {k}{r \plus{} 1}} \right \rfloor$ μηδενίζονται οι όροι.
αλλά δεν είμαι σίγουρος αν είναι σωστό ούτε κατά πόσο το πιο πάνω(αν είναι σωστό) "βελτιώνεται" οπτικά.
Θεωρείται κλειστή μορφή το άθροισμα αυτό ή όχι; :maybe:

Πρέπει να είναι σωστή. Νομίζω πως θεωρείται κλειστή μορφή αν και παραπάνω εννοούσα αν μπορούμε να βρούμε κάτι απλό όπως το $ \binom{n \plus{} 1}{k}$ στην περίπτωση $ r \equal{} 1$.

Την ίδια απάντηση μπορείς να την βρεις και ως εξής: Παίρνεις το $ n!(1 \plus{} x \plus{} \ldots \plus{} x^r)^n$ και θες να βρεις τον συντελεστή του $ x^k$. Άρα αρκεί να το παραγωγίσεις $ k$ φορές, να πολλαπλασιάσεις επί $ k!$ και να θέσεις $ x \equal{} 0$. Όπως είναι η συνάρτηση είναι λίγο δύσκολο να παραγωγιστεί αλλά μπορείς να την γράψεις σαν $ (1 \minus{} x^{r \plus{} 1})^{n \plus{} 1} (1 \minus{} x)^{ \minus{} (n \plus{} 1)}$ και μετά παραγωγίζεται εύκολα   :lol:  με διωνυμικό και την μέθοδο του Leibnitz.

EDIT: Βλακεία μου. Δεν χρειάζεται να παραγωγίσεις. Η συνάρτηση ισούται με

\[ n!\sum_{i\equal{}0}^{n\plus{}1} \binom{n\plus{}1}{i}(\minus{}1)^i x^{i(r\plus{}1)} \sum_{j\equal{}0}^{\infty}\binom{n\plus{}j}{j}x^j  \equal{} n! \sum_{i,j}(\minus{}1)^i \binom{n\plus{}1}{i}\binom{n\plus{}j}{j}x^{i(r\plus{}1)\plus{}j} \equal{} n!\sum_k \sum_i (\minus{}1)^i \binom{n\plus{}1}{i} \binom{n\plus{}k\minus{}i(r\plus{}1)}{k\minus{}i(r\plus{}1)}x^k\]

και από αυτό διαβάζουμε τον συντελεστή του $ x^k$ που βρήκες και εσύ.


function
geometry
geometric transformation
reflection

Ok, i just have a quick question about inverse functions. I know how to find them, and i understand the concept of inverse functions ( reflection across y= x etc..), but i just need to know one thing. Is the following statement correct?:

given a function f(x), then f^-1(x) = f(y)

A stupid question, i know, but i just want to be sure.

That's not it... If $y=f(x)$, then $x=f^{-1}(y)$.

(If $f^{-1}(x)=f(y)$, then $x=f(f(y))\rightarrow f(x)=f(f(f(y)))=y$, which is false for most functions.)


How many real roots does the equation have?
a) sinx=x/100
b)sinx= logx

For the first one:
If you draw a graph you'll see that it has 1 solution in interval $(0,2\pi)$, 2 solutions in $(2\pi,4\pi),(4\pi,6\pi),\ldots,(15\pi,100)$. in $(100,\infty)$ it has no solutions. Both $\sin(x),\frac{x}{100}$ are even so you must double  the number of solutions and add solution x=0 to get all solutions. Completely it's $2\cdot (2\cdot 15+1)+1=63$

In second one I used similar method.

Solutions to these problems can be written completely in such way that you cut sin(x) to intervals where it's decreasing and interval where it's increasing.

For example if you know that f(x),g(x) are continuous on interval (a,b), f(x) is increasing and g(x) is decreasing. Then if you know that f(a) < g(a) and f(b) > g(b) then you know that there's exactly one solution in (a,b).


AMC
AIME
geometry
circumcircle
ratio
trigonometry
similar triangles

#13 on the AIME 1 2002:

The triangle ABC has AB = 24. The median CE is extended to meet the circumcircle at F. CE = 27, and the median AD = 18. Find area ABF.

Here's what I have so far, can somebody read what I've done and give me a hint or two (I haven't accomplished a whole lot)?  Thanks.

[hide]
So quadrilateral BFAC is cyclic.  Draw the median from B to side AC.  Let the medians intersect at G.  Since the medians are in a 2:3 ratio, EG = 9, GC = 18, AG = 6, and GD = 12.  Also, because triangles BFE and CAE are similar (since their angles are equal due to same arc degree), let FE be x, so we have 12/x = 27/12, so FE or x = 16/3.

That's all I have so far.  I drew the altitude from F to AB so I can find the area of triangle ABF.  I also labeled all the angles that are congruent (there are 4 pairs I think, not counting vertical angles and such, just arc types).

Maybe I can use Ptolemy somehow since I know the length of the two diagonals of BFAC.

Also, to find the area of ABF, I could use Heron on BAC and then subtract that area from BFAC's area.  I don't know much trig, but I don't think it helps here anyway, and neither does Ceva's Theorem or Menelaus' Theorem.

[/hide]

[hide]You could always use formulas involving trig.[/hide]

A hint:[hide]
You can determine $AC$ and $BC$ by Stewart's theorem applied twice. For example, it gives $4AD^2 = 2BC^2 + 2AC^2 - BC^2$.  $FE$ follows from Power of a point about $E$.  Then you can use $\frac{1}{2}bh$ and similar triangles to determine the ratio of the area of $ABF$ to the area of $ABC$.[/hide]

Try checking this post:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=7652]www.mathlinks.ro/Forum/viewtopic.php?t=7652[/url]

[hide]So you already found some important things about the picture: EF=16/3, AG=12, GD=6, CG=18, and GE=9. If we could find  :theta: =<FEB, then we'd know [FEB]=1/2*12*16/3*sin :theta: and then [AFB]=2[FEB]. So let's find something about  :theta: = <FEB = <AEG. Triangle EAG is isosceles, so cos :theta: = (9/2)/(12) = 3/8. Can you finish?[/hide]

EDIT: Ah, i see that there have been some posts while i was writing  :blush:


integration logarithms calculus derivative

A rectangular loop of dimensions l and w moves with a constant velocity v away from
a long wire that carries a current I in the plane of the loop. The total
resistance of the loop is R. Derive an expression that gives the current in the loop at the
instant the near side is a distance r from the wire.

Consider a small rectangular section of the loop of length $ l$ and width $ dx$ at a distance $ r\plus{}x$ from the wire.
Flux of the magnetic field through this section is $ \frac{\mu_0 I}{2\pi (r\plus{}x)}ldx$

Integrating over the loop, we have $ \Phi_B\equal{}\int_r^{r\plus{}w} \frac{\mu_0 I}{2\pi (r\plus{}x)}ldx\equal{}\frac{\mu_0 Il}{2\pi}\ln \displaystyle (\frac{r\plus{}w}{r})$

Taking the time derivative, $ \mathcal{E}\equal{}\minus{}\dot{\Phi}_B\equal{}\frac{\mu_0 Il}{2\pi}\frac{wv}{r(r\plus{}w)}$

Hence,current $ i\equal{}\frac{\mathcal{E}}{R}\equal{}\frac{\mu_0 Il}{2\pi R}\frac{wv}{r(r\plus{}w)}$


geometry
incenter
linear algebra
matrix
geometry unsolved

let P be an interior point of a triangle ABC such that for any line passing
through P and intersecting the sides AB and AC at distinct points E and F
respectively, we have

1/AE + 1/AF = (a + b + c)/bc

where BC = a, CA = b, and AB = c.

proof that P is actually the incenter of the triangle ABC.

this was an open contest problem until today, and its deadline has passed now.

Somebody has a proof that the deadline is over¿

@BaBaK Ghalebi: Please delete/edit your post until we know for sure.

If $ E\in$($ AB$) and $ F\in$($ AC$), then in barycentric coordonate $ E$($ BE$, $ AE$, 0); $ F$($ FC$, 0, $ FA$) and incentre $ I$($ {a}$, $ {b}$, $ {c}$).
$ I\in\(EF\)$ $ \Longleftrightarrow$ $ \begin{equation*} \begin{vmatrix} {BE} & {AE} & 0 \\
{FC} & 0 & {AF} \\
a & b & c \end{vmatrix} = 0 \end{equation*}$ $ \Longleftrightarrow$
$ \Longleftrightarrow$ $ \begin{equation*} \begin{vmatrix} c & {AE} & 0 \\
b & 0 & {AF} \\
{a + b + c} & b & c \end{vmatrix} = 0 \end{equation*}$
$ \Longleftrightarrow$$ \frac {1}{AE} + \frac {1}{AF} = \frac {a + b + c}{bc}$, $ {etc}$, $ {etc}$...

I'm willing to testify that all the deadline is indeed over. I guess discussions are acceptable now. I can send you the proof that the deadline is over by sending in a copy of the problem set where it states that it is due on the 20th of October.

sorry for the interruption, but in my opinion, itd b nice if we wait for a few more days b4 the posting of solns.
i bliv Sirs do accept solns 2-3 days late too( i guess that much tolerance is reqd, bcoz of other priorities . i myself havnt bn able to send the solns on time due to the Dussehra celebrations out here).
so, if possibl, just look into this. not necessary though, bcoz most wud hav sent their solns by now. just a prec. measure.
thanks.


[quote="MCrawford"]The arithmetic mean of the set $\{90, 85, 95, x, 88, 94\}$ is 90.  What is the value of $x$?[/quote]
[hide](90+85+95+x+88+94)/6=90
(452+x)/6=90
452+x=540
x=88
The value of x is 88. :)
[/hide]

Do any of you know any alternate methods for solving that problem?

Here is one method that you might want to take a look at, it saves you from doing some extra arithmatic!

[hide]The average of ${90, 85, 95, x, 88, 94}$ is 90.

Let's find out how many up or down each number is from nintey.

90 is up by [b]0[/b]
85 is down by 5, so [b]-5[/b]
95 is up by [b]5[/b]
88 is down by 2, so [b]-2[/b]
94 is up by [b]4[/b]

Add all of those values together, $0-5+5-2+4 \Rightarrow 2$ So what we have is two more than nintey, meaning that x must be 2 less than nintey to balance everything out, giving us an answer of [b]88[/b].

This method is particularly helpful in countdown...While the other person is busy multiplying and subracting numbers...You can get the answer in a couple of seconds.  ;)[/hide]

[hide]If the mean of 6 numbers is 90, then the sum of those numbers has to be 540.
The sum of the numbers that are given to you is 452; 540-452=88
So x=88 [/hide]

[hide]the sum of the six numbers is 540. Than u add all the numbers and u will get 452.
than to get X subract 452 from 540. :D
which will give u 88.[/hide]


Sorry, if this sounds unreasonable or has already been posted but I want to clear up some things because of the use of a tiebraker index system for 10th grade and below

1.  Lets say a person gets a 114 on amc 10a (which was not passing score for aime), but he gets a 102 on 12b (which allowed him to take the aime).  Now on the aime he gets a 7.  However, for 10th grade and below, tnhey used this tiebraker system of 6+ on AIME and an index of 183.5 and above.  Well, for this case, can you use the index of 114+10(7 on AIME)=184 (which would pass him), even though that 114 on the amc 10 wasnt even a passing score to take the AIME?

2.  Also, does the tiebraker say 6+ on AIME and 183.5 on index amc+10*AIME?  Well, does that mean if someone gets a 102 on amc 12 and 8 on AIME (for 10th grade and below), that person doesnt even pass because he didnt meet the index requirements (even though he got a 8 on AIME).  Isnt that kind of unfair because the only reason he wanst invited for USAMO because he took the amc 12 and the only reason that index tiebraker was used because there were too many 6s, not 7s and 8s, and that person got an 8.

I can't answer the first, but the 2nd - 102/8 would not make it. I have had two friends with 7's who didn't make it due to a low AMC12.

I'm not sure about your first case...I've never seen the idea of a non-AIME-qualifying AMC-10 score being used for the index discussed before. That's kind of interesting and now I'm surprised that no one else seems to have thought of it. But yes, your second person doesn't qualify for the USAMO, and there are a couple other threads in this forum discussing how that should be fixed.

for the first case i would assume the 114 would be counted but im not sure you would have to ask the AMCDirector

From the rules in the teacher's manual, it sounds like the 114 would be used.

Hmm... number 1 is a good question.  This is by far the lowest the cutoff has ever been, so this is the first time a situation like that could be feasible (it's not likely that someone would get 114 on the AMC 10 but then 10 on AIME).

Number 2:
As of AMCDirector's latest post regarding the 5% club, this person would make it.  However, let's suppose that this person got a 97.5 instead (12A).  Then 177.5 would miss the tiebreaker index.  This does seem unfair.  Many people, including myself, have suggested revisions to rectify this issue.  They are being currently discussed in other threads.

Uh..cause if they do count that 114 on the amc 10..thats exaclty the reason why the index for 10th and below is kinda flawed..

I do think that this is unfair (perhaps because I'm in that very situation). I got a 109.5 on the AMC 12 (as a sophomore) and a 7 on the AIME, for an index of 179.5. Of course, I don't know what the AMC could do to rectify this issue. They'd either have to
a) eliminate the index factor for younger students, relying on solely the AIME floor value
b) arbitrarily choose different floor values for younger students based on whether they took the AMC 10 or AMC 12
c) mandate that all students 10th grade and younger MUST take the AMC 10
d) completely change the whole contest system, eliminating the dual contest format and resorting to one contest in lieu of both the AMC 10 and AMC 12.
None of these is perfect, but I do think that option a) is the most fair of the three (and certainly more fair than what they're doing now). Yet, my view is going to be biased because I'm getting left out with the way things work now......

[quote="Ignite168"]It doesn't matter if it's unfair.

The procedure is outlined so if it's unfair, you take the test that helps you the most. I like how people are complaining about it now when it has been outlined in the Teacher's Manual for a while rather than earlier (maybe because they didn't qualify due to this? If you made the wrong choice of test, it's not the CAMC's fault, it's yours, you shouldn't be trying to make it so you have insurance regardless of which test you take)[/quote]

To clarify this, I don't think the AMC10 vs AMC12 issue is unfair but I do think that: 121.5 AMC 10 + 6 AIME > 111 AMC 12 + 7 AIME > 100.5 AMC 12 + 8 AIME is unfair.

Dude, if the amcs is going to go into all that, just make one test (no amc 10/12) and isnt the reason they use the index tiebrakers because of too many 6s????? not 7s or 8s, so it is unfair for those people (10th grade and under) who got 7s but didnt qualify because of the index.  If they are going to use the index, use it ONLY on the people with 6s.

[quote="ra5249"]For complete fairness, you should only decide by the aime score. that would eliminate 10/12 differences.[/quote]
for those eligible for AMC 10, then all they do care about is floor.
The problem arises when it comes to tiebrakers.

Wow...now I'm really confused as to whether I should do 10, 12, or both.


For each positive integer $n$, let $S(n)$ denote the sum of the digits of $n$. For how many values of $n$ is $n+S(n)+S(S(n)) = 2007$?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

What's the best way to do it other than brute-forcing? I omitted it on the test because I knew I would miss one of them and get it wrong.

I ended up brute forcing it.

I set $n=1000+x$, and then narrowed down what values $x$ should take. So I guess you could then conclude something like $x=900+y$, and so on. You slowly narrow it down, and then try out some cases, keeping in mind that $n$ is divisible by $3$.


If February is a month that contains Friday the $ 13^{\text{th}}$, what day of the week is February $ 1$?

$ \textbf{(A)}\ \text{Sunday}\qquad \textbf{(B)}\ \text{Monday}\qquad \textbf{(C)}\ \text{Wednesday}\qquad \textbf{(D)}\ \text{Thursday}\qquad \textbf{(E)}\ \text{Saturday}$

[hide]

Friday the $ 13$ is Friday.  Subtracting $ 7$, we get February $ 6$ is also Friday.  February $ 7$ will be Saturday,  and February $ 8$ will be Sunday.  If we subtract $ 7$ form $ 8$, we get February $ 1$, which will then be $ \boxed{\textbf{(A)}\ \text{Sunday}}$[/hide]


calculus
integration
trigonometry
quadratics
calculus computations

For $ f(x)\equal{}\sin ^ 3 x$, let $ p(x)$ be quadratic polynomial.
Evaluate $ \int_0^{2\pi} p(x)f''(x)\ dx$.

Integrate by parts twice, and note that both $ f(x)$ and $ f'(x)$ are zero at both $ 0$ and $ 2\pi,$ since they contain positive powers of $ \sin x.$

$ \int_0^{2\pi}p(x)f''(x)\,dx\equal{}\left(p(x)f'(x)\minus{}p'(x)f(x)\right)\bigg|_0^{2\pi}\plus{}\int_0^{2\pi}p''(x)f(x)\,dx$

The boundary terms evaluate to zero at both ends, and $ p''(x)\equal{}2a$ is constant.

$ \equal{}2a\int_0^{2\pi}\sin^3x\,dx\equal{}0.$

Equivalently, we could have integrated by parts a third time, noting that $ g(x)\equal{}\int \sin^3x\,dx$ is periodic of period $ 2\pi$ (that's the same fact as $ \sin^3x$ integrating to zero over one period) and $ p'''(x)\equal{}0.$ Either way, we get zero.


logarithms
limit
calculus
calculus computations

what the "Nth Term Test" indicates.

2/1 - 3/2 + 4/3 - 5/4 + ... =

4/2 - 3/2 + 16/12 - 15/12 + ... =

1/2 + 1/12 + ... =

1/(1*2) + 1/(3*4) + ... =

1 - 1/2 + 1/3 - 1/4 + ... = ln (2)

This is my answer above.

However the Nth Term Test states, in part, if the lim of a(sub n) is not equal to 0, then the summation
of the infinite series a(sub n) does not converge.

I would think the Nth Term Test doesn't apply here.

The $ n^{th}$ term test always applies.  The first series has a sum that oscillates between $ \ln 2$ and $ \ln 2 \plus{} 1$.  The partial sums do not diverge; however, they do not converge to a limit, either.

I never assumed I could manipulate those terms at without care which is what that meant.

[color=green][Deleted][/color]

[quote="box of rocks"]However the Nth Term Test states, in part, if the lim of a(sub n) is not equal to 0, then the summation
of the infinite series a(sub n) does not converge.
[/quote]

This is an obvious consequence of the definition of a limit!  The 'nth term test' isn't some sort of deep theorem.  However, your sum is cesaro convergent to $ \log 2$.

[color=green][Deleted.][/color]

4/2 - 3/2 + 16/12 - 15/12 + ... =

1/2 + 1/12 + ... =

which is more or less manipulating terms without care.  In particular you've shown that the sequence of partial sums s_{2n} converges to log 2, but that's not really the desired result.

No, if it were "without care," that would have allowed any ordering of the fractions, but I did not do that.  I kept them in their same order, which showed
the ln(2) sum, but not the 1 + ln(2) sum.

[color=green][Inappropriate name-calling edited out][/color]

I am sorry that you are unable to realize that mathematically, the sums
\[ \sum_{n\in\mathbb{N}} a_{n}
\]
and
\[ \sum_{n\in\mathbb{N}} b_{n}
\]
where $ b_n \equal{} a_{2n} \plus{} a_{2n \plus{} 1}$ are not the same despite the fact that they contain the same terms.  In fact you have more or less proven this ... [color=green][deleted][/color]

Let's keep the personal stuff out of this, ok?
I'm more interested, as a student, to learn why the sum cannot be re-arranged if some terms are negative. I assume this has something to do with the absolute convergence, but maybe someone could clarify it for me.

I flagged this thread for moderator attention the moment I saw it...

Theorem: Any rearrangement of an absolutely convergent series converges to the same limit.
Theorem: For any conditionally convergent series of real numbers and any real $ L$, there is some rearrangement of the series that converges to $ L$.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=90256]Two[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=114905]links[/url].

The manipulations in the original post don't even qualify as a rearrangement.

The first theorem makes perfect sense, but the second one is a bit confusing.
You're saying that I can arrange any conditional series (what's the plural of series? serieses? xD) to reach ANY finite limit?
That's a bit odd, I'm not quite sure I understand how that works.
I'll look at your links I guess.

[quote="DiscreetFourierTransform"]You're saying that I can arrange any conditional series (what's the plural of series? serieses? xD) to reach ANY finite limit?
That's a bit odd, I'm not quite sure I understand how that works.
I'll look at your links I guess.[/quote]

Roughly, the rearrangement looks like this:  keep adding positive terms until your partial sum is above $ L$, then add negative terms until your partial sum is below $ L$, and repeat.  If you allow arbitrary rearrangement of terms, this is perfectly valid, and moreover a conditionally convergent series contains infinitely many positive and negative terms.

[quote="JBL"][quote="t0rajir0u"]The partial sums do not diverge; however, they do not converge to a limit, either.[/quote]
Of course they diverge: anything that doesn't converge (like these partial sums) diverges.

The original poster's mistake is in assuming that it's okay to group terms arbitrarily, which is false in general.  (It's true if our series has all positive terms, but this one doesn't.)[/quote]

Sorry; by "diverge" I meant the partial sums do not diverge to infinity.

To the original poster:  let $ S_n \equal{} \sum_{k\equal{}1}^{n} (\minus{}1)^{k\plus{}1} \frac{k\plus{}1}{k}$, which is the sequence of partial sums of the first series.  A correct statement would be that

$ \lim_{k \to \infty} S_{2k} \equal{} \left( \frac{2}{1} \minus{} \frac{3}{2} \right) \plus{} \left( \frac{4}{3} \minus{} \frac{5}{4} \right) \plus{} ... \equal{} \ln 2$
$ \lim_{k \to \infty} S_{2k\plus{}1} \equal{} \frac{2}{1} \plus{} \left( \minus{}\frac{3}{2} \plus{} \frac{4}{3} \right) \plus{} \left( \minus{}\frac{5}{4} \plus{} \frac{6}{5} \right) \plus{} ... \equal{} \ln 2 \plus{} 1$

Hence by the definition of a limit $ \lim_{k \to \infty} S_k$ is not defined.

A more concrete example:  let $ S_n \equal{} \sum_{k\equal{}1}^{n} (\minus{}1)^{k\plus{}1}$.  Then

$ \lim_{k \to \infty} S_{2k} \equal{} (1 \minus{} 1) \plus{} (1 \minus{} 1) \plus{} ... \equal{} 0$
$ \lim_{k \to \infty} S_{2k\plus{}1} \equal{} 1 \plus{} (\minus{}1 \plus{} 1) \plus{} (\minus{}1 \plus{} 1) \plus{} ... \equal{} 1$

so again $ \lim_{k \to \infty} S_k$ is not defined.


Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two.  The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two.  Given that each monkey reveives a whole number of bananas whenever the bananas are divided, and the numbers of bananas in the first, second, and third monkeys have at the end of the process are in the ratio $ 3: 2: 1$, what is the least possible total for the number of bananas?

$ AIME$ $ \#6$

If you want to discuss the problem, use the topic linked to from the contests section: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1294061725&t=61901]this topic[/url] which is linked to from [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2004]this page[/url]. This way all the discussion is collected in one place, so the contests section is as comprehensive as possible.


I put a choice "15 and above: What do you expect, I'm Russian" it didn't show up however

I'm Russian, and I'm pretty sure $ 0\ne15$...

I am 19, and in America it is underage. And I drink a lot. Im not tryint to be cool or anything like that, nor am I addicted. I just like the burning sensation as it goes down your throat. I don't know how much I drink, but I can tell you a lot. (I'm careful though, not to have too much).

Um, are you sure a poll involving potentially illegal activity should be on AoPS?

In Christian Education the teacher polled the class to see how many drink alcohol at least yearly. Out of 26 students, only 2 (including me) were abstinent, which I found very surprising. (Mean age in my class is 15)

well, so far everyone single person never drinks. I put 0 as well, since I don't drink. I probably never will, since once I had 1 chocolate with beer in it and it was so nasty I spit it out.

I've drank twice in my life, got drunk both times. All I have to say is that there's not really any point in "getting wasted" as it takes very long to do and isn't that satisfying. Besides, I don't want to start having a drinking problem and then throwing my life away because of some short substance induced pleasure that I might crave. That is just a waste of a person's life potential.

[quote="Prime factorization"]All I have to say is that there's not really any point in "getting wasted" as it takes very long to do and isn't that satisfying. Besides, I don't want to start having a drinking problem and then throwing my life away because of some short substance induced pleasure that I might crave. That is just a waste of a person's life potential.[/quote]
Quoted for truth.

[quote="Kamil Witek"]Identity: were you suprised about the fact that there was another abstinent besides you, or that 92% of the class wasn't abstinent?[/quote]

more surprised about the 92% actually

[quote="Nerd_of_the_Ages"]Would drinking include things used in cooking? For example, many foods use cooking wine. A common chinese dish is chicken which has been sunk in this alcohol thingy, and another one is a kind of desert, which is rice with wine.[/quote]

No. Definitely not, or I'd have drank underage several thousand times before.

Lol, ok. Does it count if you ever had like a chocolate with wine or something in it? If so, I've drank once (three times, if eating three chocolates count as seperate)

I have never drunk. However, I eat foods cooked with wine.

Never real drinking.
But the chocolates, yes.

The only "drinking" I do is Hot Chocolate; one teaspoon of vanilla each huge mugful. There is some vodka in it, but not much(the vanilla, not HC).

I once has Saki... I forgot. Was it my 10th Bday, 11th NewYearEve, or NYDay? eh whatever.

It tasted terrible. I never had alcohol again(if it was more than 20%)

The only time I got drunk was when I had Saki. My parents kept refilling my glass. :(

Saki is basically a japanese rice wine; it looks like water because it's really "pure."
It's considered to not be very strong, but to me (after smelling it and asking my mom to compare it with another alcoholic food I had before) it's already pretty bad.

I don't drink. The closest I ever got to drunk was with a really cool water bottle. Not with any particular pattern, just really like glass. Became an aquaholic for a while. Then I forgot about water expanding after freezing and stuck it in the freezer for two days. *sigh*

And the first thing I think of when you mentioned vodka was an episode of Mythbusters.  :maybe:


floor function
number theory unsolved
number theory

Denote by $ a_n$ the integer closest to $ \sqrt{n}$. Determine $ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\cdots\plus{}\frac{1}{a_{1980}}$

An integer $ m$ is closest to $ \sqrt {n}$ iff $ m \minus{} \frac {1}{2} < \sqrt {n} < m \plus{} \frac {1}{2}$ or alternatively
\[ m^2 \minus{} m \plus{} \frac {1}{4} \equal{} \left(m \minus{} \frac {1}{2}\right)^2 < n < \left(m \plus{} \frac {1}{2}\right)^2 \equal{} m^2 \plus{} m \plus{} \frac {1}{4}.
\]
Therefore, each particular integer $ m$ is closest to exactly
\[ \left\lfloor m^2 \plus{} m \plus{} \frac {1}{4} \right\rfloor \minus{} \left\lfloor m^2 \minus{} m \plus{} \frac {1}{4} \right\rfloor \equal{} 2m
\]
numbers of the form $ \sqrt {n}$.

Taking into account that $ 1980 \equal{} 44(44 \plus{} 1)$, we get the answer
\[ \sum_{n \equal{} 1}^{1980} \frac {1}{a_n} \equal{} \sum_{m \equal{} 1}^{44} \frac {2m}{m} \equal{} 88.
\]

The true source of this problem is All-Soviet Union Olympiad.


What's the best subway sandwich in your opinion? What meat, bread, veggies, condiments and other stuff go on it?

White bread, Jack cheese, ham, honey mustard, lettuce, tomato, jalapeno and cheddar cheese. Yum  :thumbup: :thumbup: :thumbup:

[quote="CatalystOfNostalgia"]MAYBE YOU ALL SHOULD DO YOUR 1304910TH HURT AND HEAL ON THIS OMG[/quote]

? What does that mean?

Look at like most of the recent FF threads to find out...

I especially like the Subway melt with american cheese, toasted. The veggies don't matter, the sandwich still tastes good. Oh, please don't make a hurt and heal thing.

[quote="CatalystOfNostalgia"]MAYBE YOU ALL SHOULD DO YOUR 1304910TH HURT AND HEAL ON THIS OMG[/quote]
Agreed. >.> By that, I mean OMG please don't.
You know there was a FFC thing going on before... except the person making it was the creator of several of the spammy topics...  :wink:

I like the Sweet Onion Chicken Teriyaki one, mostly because of the sweet onion sauce.  :D

[b]Length:[/b] Footlong
[b]Bread:[/b] Italian Herbs and Cheese
[b]Type:[/b] Oven roasted chicken breast
[b]Cheese:[/b] Provolone
[b]Vegetables:[/b] Lettuce, tomatoes, onions, olives, green peppers, cucumbers, (jalapeños)
[b]Dressing:[/b] None

[quote="7h3.D3m0n.117"][quote="Brut3Forc3"]I like the Sweet Onion Chicken Teriyaki one, mostly because of the sweet onion sauce.  :D[/quote]

Agreed. The Chicken Teriyaki one is my favorite.

Although the "Five-Dollar Footlongs" have drawn a lot of "THAT'S WHAT SHE SAID!" remarks.[/quote]

lol same at my school. I used to acctually bring foot long subs to school and say "mmm...footlong" but then i kept setting my self up to get dissed. I hate the chicken teriyaki and all chicken subs for that matter. Chicken just doesnt go well with subs in my opinion.

ty very much for that compliment Brut.

My favorite sandwich is the tuna with american cheese toasted. The vegatbles on there include lettuce, tomatoes and thats it.

[quote]
Meh, I rarely eat subway, and when I for some reason eat a sandwich like that, I usually go to Quiznos instead.
[/quote]
me, too but when i eat subway, meatball marinara.  :)

I would just like to announce that I had my first ever meatball marinara sub today, and it was [u]awesome[/u]!

mmm meatball marinara....*drools* I had a hamburger  :rotfl:


Let $a,b,c,d$ be positive real numbers such that $abcd=1$ Prove that
$\displaystyle \left(\frac{1+ab}{1+a}\right)^4+\left(\frac{1+bc}{1+b}\right)^4+\left(\frac{1+cd}{1+c}\right)^4+\left(\frac{1+da}{1+d}\right)^4\geq 4$

Because $f(x)=x^4$ is convex and increasing for $x>0$, its enough to prove that for $abcd=1$ we have

$\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+cd}{1+c}+\frac{1+da}{1+d} \geq 4.$

Now using $cd=1/ab$ and $da=1/bc$, the left hand side equals

$\displaystyle (1+ab)\left(\frac{1}{1+a}+\frac{1}{ab+abc}\right)+(1+bc)\left(\frac{1}{1+b}+\frac{1}{bc+bcd}\right)$

Apply AM-HM to each of the large brackets, this is at least

$\displaystyle 4\left(\frac{1+ab}{1+a+ab+abc}+\frac{1+bc}{1+b+bc+bcd}\right)=4$

(extend the second fraction with $a$ and use $abcd=1$).


Day 1 began on page [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=178191]1[/url].
Day 2 began on page [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=987548#987548]5[/url]
Day 3 began on page [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=999462#999462]17[/url]
Day 4 began on page [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=178191&start=340]18[/url]
Welcome to Mushroom Kingdom Mafia!

Peaceful members of the Mushroom Kingdom, you've recently been, once again, overcome by evil forces. It's all of your jobs to stop them.

Still Alive (5):
- jnub
- woser456
- BoesFX
- Peter
- Ars

Dead:
- alanchou, Goomba, Lynched Day 1
- CheeseIsGood, Shy Guy, Killed Night 1
- 1=2 [size=59]BlueMouse[/size], Koopa, Lynched Day 2
- [color=red]miyomiyo, Pacman, Killed night 2[/color]
- xpmath, Toad, Killed night 2
- ZhangPeijin [size=59]mnmath[/size], Iggy, Lynched Day 3
[color=red]- DonkeyKong, Bowser, Lynched Day 4[/color]
- Hamster1800, Mario, Killed Night 4

Here is the townie pm:
[quote="perfect628"]You are *[b]Townie Name[/b]*. You are a peaceful member of the Mushroom Kingdom, which has recently(again) been overcome by evil. You don't have much power at your disposal, but you do have your vote. You are protown. You win when only innocents are living.
[/quote]

All of your pixels are scrambled, so it's impossible to tell who's evil and who's not. You're unsure as to how long you can survive here in the Mushroom Kingdom, and have all decided to, during each day, vote on someone to lynch, or kill. During the night, some of you may have an action. This could mean anything.

Good luck. Press start to begin.

Please post here to confirm that you're here. No voting yet, please, until further notice.

[b]The Rules (PLEASE READ THESE):
Votes must be in bold. If you do not bold your vote, it will not be counted.

Please be attentive and unvote, if necessary, before casting a new vote. This is not required, but I’d appreciate it.

Executions will require a simple majority of votes. Once a player has reached the necessary majority, his pleas are useless and any attempts to unvote will be unheeded. You may vote: no lynch – majority votes of this kind are necessary to end the day without a death.

The game is not to be discussed outside of the thread unless your role specifically states that you may do so.

Once your death scene has been posted, you’re dead. Stop typing.

Don’t edit/delete previously submitted posts

Don’t quote any PMs from me.

If you have a night choice to make, it is due by the posted deadline. I will not wait for you – if you do not submit a choice to me, tough! No choice will be made.

[color=red]New Rule: NO HIDE TAGS. Also, no changing font size, or posting in any other way that obscures something you're saying. Anything you say should be clearly visible to everyone without having to click a button or enlarge your text or anything like that.[/color]

If you anticipate being unavailable for more than a 48-hour period, please post a notice to that effect in the thread. Treat this game as a commitment. Be considerate – don’t leave us hanging.

Remember, it’s a game. It’s supposed to played. It’s supposed to be fun. Do your part.[/b]

I am here.

So we are allowed to post between getting a majority vote and having our death scene? Okay.

Je suis ici.

Sorry, too much French homework.

And as they say in France...
I am here you silly fool.

I am here, too..

This is an awesome topic to play mafia with.

[quote="perfect628"]The other two have until tomorrow to confirm; they won't be replaced yet, but the actual game will start.[/quote]

So, the game begins from today,right?

wait so Perfect what was that pm you sent me about it depends on your choice
and he was the day before
in fact he was confirmed scum
and I said you were innocent Peter based on your code of peach
not on my inspections
I never have faith in them

Actually, miyomiyo, we killed the same person that you did on night one. We never targeted Peter (I think) and every time except that really short night our kills were the ones that went through. Did we kill the same people almost every night!?  :rotfl:

I'm not sure how I got that conclusion from what I just wrote...  :huh:

OK, more info.

[hide="DEATH SUMMARY"]alanchou, Goomba, Lynched Day 1
CheeseIsGood, Shy Guy, Killed Night 1
1=2 BlueMouse, Koopa, Lynched Day 2
[color=darkred]miyomiyo, Pacman, Killed night 2 [/color]
xpmath, Toad, Killed night 2
ZhangPeijin mnmath, Iggy, Lynched Day 3
[color=red]DonkeyKong, Bowser, Lynched Day 4 [/color]
Hamster1800, Mario, Killed Night 4
Ars, Yoshi, Lynched Day 5
jnub, Professor E. Gadd, Killed in Endgame
Peter, Princess Peach, Killed in Endgame
[color=red]woser456, King Boo, survived
BoesFX, Petey Piranha, survived[/color][/hide]

[hide="ACTIONS"][hide="NIGHT 1"]MAFIA: CheeseIsGood
PETEY: cursed no one(no submission)
PACMAN: CheeseIsGood
YOSHI: Peter(received innocent)
TOAD: miyomiyo
E GADD: didn't use item[/hide]
[hide="NIGHT 2"]MAFIA: miyomiyo
PETEY: ZhangPeijin
PACMAN: xpmath
YOSHI: miyomiyo(received innocent)
TOAD: Hamster1800
E GADD: used item[/hide]
[hide="NIGHT 3"]MAFIA: Peter
PETEY: jnub
YOSHI: DonkeyKong(received innocent)[/hide]
[hide="NIGHT 4"]MAFIA: Hamster1800
PETEY: Ars
YOSHI: woser456(redeived guilty)[/hide][/hide]

[hide="NOTES"]-First 3 kills=townies lol
-In a votecount(somewhere), the title was "Slimy Grain," which anagrams to "Ars I'm Lying."
-The mafia had a [i]very[/i] slight advantage at the beginning. It got worse with mislynches.
-Luigi was intentionally left out. I assumed you'd all assume he was in the game. I was very close to making Mario-Peach-Luigi the mafia.
-I'm absolutely shocked no one mentioned that Peter posted an anagram of "Peach" in thread.
-I foresaw something like jnub's use of his item. I think it was executed well.
-Mafia should have won day 4. I think it was because of (woser's aforementioned) their lack of communication.
-The game really would have changed had Peter been lynched.
[/hide]

Analysis later/tomorrow

What I don't like about the mario/peach combo is that the town seems to benefit from a townie dying...

Of course, this was made even worse by all of us reading that Peter was innocent...

How much did peter know about my role? I knew nothing...my role PM was literally the townie PM from the first thread.

He was told that as long as mario was alive, he was night kill immune. I really wish I hadn't cleared my save box....Sorry.

Wait... why was I such a popular kill target night 1?

And perfect I did decode that as I knew the inspections were crap so I went off other things and I had seen that so I mentioned that I decoded it
then of course you screwed me over by lying
I thought you were just being a bastard with my role
it turns out you were being one through and through

and sorry to double post but the 2 things you posted perfect about the fact you were lying to me were 2 days before the situation
and I already knew that from the Miyomiyo investigation

and jnub you had no clue
you just are acting like a sore loser cause you didn't believe me
and Peter I told you like 8 times before you finally started to lynch me that I decoded the code
which is why I didn't care about my role

I at the time believed I was a roleblocker which is why you all suck
and Peter the reason for lynching me over Boes was the stupidist thing you have ever done
kill Boes the guarenteed mafia
and then worry
jnub's not stupid
and I doubt that the mafia had a clue your unnightkillability went away

anyways the mafia deserved the win
the setup was pretty harsh against the town
especially with the excess bastard modery
I mean its one thing to lie about a pm
and let the person discover it throughout the game
its another to lie again midgame and screw them over more

It made me come to the conclusion that I block the person with the egg
a bad conclusion but I doubt anyone else would have come up with better

[quote="Hamster1800"]How much did peter know about my role? I knew nothing...my role PM was literally the townie PM from the first thread.[/quote]
This was my doing. I still think it was a good decision to send you the same role PM

Oh, I didn't mind - I just wanted to know whether Peter knew that I would become a doctor.

[quote="perfect628"]-The game really would have changed had Peter been lynched.[/quote]You gave me night immortality, and I've never been daylynched in any game as far as I remember. Why do you assume I'll be killed?

[quote="Ars"]and Peter I told you like 8 times before you finally started to lynch me that I decoded the code
which is why I didn't care about my role[/quote]The point is: I didn't lynch you. You got killed only by mafia votes. I wanted to discuss with you and after that with woser - but we didn't get the chance.

I admit I had a slight preference for Luigi over Yoshi, but there were many other things you both did suspicious.


Is there any good program i can download that will let me play chess against a coumputer?

i did but i didn't find any thing good. :(

I've found that if you find a free chess program online, it will probably only have one difficulty level - extremely hard.  :lol:

First lookup Shredder clasic on google download it... then lookup Fruit 2.2.1 and download that. On the fruit website look how to install the engine with the shredder clasic and play ( btw fruit is capable of beating the grandmaster easily so i doubt you will be any good agains the computer)

[quote="ShoeFactory"]I wouldn't recommend playing vs computers.

play on yahoo vs humans.[/quote]

I Agree . I used to play online and i use [b]fritz[/b] .


[quote="duc 95"]can you solve it?[/quote]

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

I don't have other ways.Can you help me?


What's the difference between Socialism, Communism, Imperialism, Dictatorship, Absolute Monarchy, Tyranny, Marxism, and other words that kinda sound the same but is technically different? Which of these are political and which are philosophical?

Tare are you having us do your Social Studies homework? =)

Is this a joke, cause it could be a good one :lol:

{Socialism, Communism, Marxism} represent a different set of ideas than {Imperialism, Dictatorship, Absolute Monarchy, Tyranny}.  The first group is primarily economic philosophy -- that is, they are best contrasted with capitalism.  The second group is composed of political systems; they are best contrasted with democracy, republic, constitutional monarchy, and so on.  In modern history, governments with a Marxist economic philosophy (Communisms) have often (perhaps always) been totalitarian (another word that can go in the second group).  There have also been a large number of imperialist states, dictatorships, and tyranies that one would not describe as communist, socialist, or Marxist.  You might go read your encyclopedia or dictionary for good definitions.

[quote]Is this a joke, cause it could be a good one.[/quote]
Lol.

[quote="Tare"]I'm just wondering because i always get them confused and stuff... like what was the Soviet Union? Russia? Nazis Germany? China? Cuba?[/quote]

So:
The Soviet Union, Cuba, and what China used to be are all communisms -- basically, there is little or no free enterprise (privately owned businesses).  One might also describe Cuba and the Soviet Union (at least under Stalin and Lenin -- maybe not so much under later leaders) as dictatorships, authoritarian, or totalitarian.

Nazi Germany and China today were "totalitarian" or "authoritarian."  (I haven't ever seen anyone give a good differentiation between the two terms, except that we used totalitarian for people we didn't like and authoritarian for people we did like, or something.)  Nazi Germany also could have been described as a dictatorship.

One might also say that Nazi Germany was a "Fascist state," a designation which one might also use to describe at various times Mussolini's Italy, Franco's Spain, Greece, and (I think) Portugal.

Russia today is essentially totalitarian, although it maintains the trappings of democracy to some extent.  Italy is somewhat this way as well.  One might also describe Russia as an "oligarchy," in reference to the small number of hugely wealthy and powerful individuals who have significant holds on pieces of power in Russia.  Putin has done a good job of centralizing and consolidating his power, however.

[i]Nazism[/i] can be considered a more extreme version of Fascism - i.e. [i]fascism[/i] is fanatically belligerent nationalism (state above all, democracy and peace are for the weak) and Nazism is fascism with a very healthy dose of anti-Semitism. -.-" Doesn't being a totalitarian state necessitate having single-party dictatorship? I'm not too sure about that, though.

[i]Communism[/i] and [i]Marxism[/i] are pretty much 20th century concepts...I've seen them used interchangeably. In Marxist theory, communism refers to the 'final stage of human development' when the people govern and government withers away. Everyone works as much as he can and gets as much as he needs. There is no private property. Marx considers socialism to a step on the way to communism - it's meant to baby us from capitalism to communism. [i]Socialism[/i] consists of a "dictatorship of the proleteriat", cooperation and equality and peace. To reach this end, we have mass nationalization of industry and the virtual disappearance of a private sector. This is the stage at which most "communist" nations stop, and over time it's come to be our definition of communism. The present day definition of socialism is mostly economic and rather variable, depending on who you're speaking to. Socialism basically emphasizes the welfare of the community (society above self). Not all socialists want to become communist, however; Sweden is a prime example of a social democracy today.

[i]Dictatorship[/i] = absolute rule by one. [i]Tyranny[/i] = temporary dictatorship, according to the Romans. Though it's taken on a more negative connotation now, and a tyrant has become a cruel, despotic dictator. [i]Absolute Monarchy[/i] = absolute rule by a king/queen. Technically speaking an absolute monarchy is a dictatorship, though the latter is usually applied to non-royalty. Example of an absolute monarch would be Louis XIV, the "Sun King".

As far as I know, [i]imperialism[/i] is a national policy of taking over and exploiting another region's resources for one's own purposes. Google'll tell you "The policy, practice, or advocacy of extending the power and dominion of a nation especially by direct territorial acquisitions or by gaining indirect control over the political or economic life of other areas; extension or imposition of power, authority, or influence." The study of post-Industrial-Revolution Western imperialism is pretty interesting; you might consider looking that up.

Hrm. If I made any mistakes in that, someone tell me. The World Hist AP is in two weeks. Heh.

What is the difference between Marxism and Leninism?

Haha.  Leninism is a "modification" of Marxism.  They shared many beliefs, but they were not entirely overlapping.  In particular, Marx believed that Communism would come to industrialized nations, while Lenin attempted to communize Russia, a very agrarian nation.

[quote]Doesn't being a totalitarian state necessitate having single-party dictatorship?[/quote]
Dictatorship implies the existence of a single dictator -- Stalin, Hitler, Bush (joking, joking ... at least, joking until November).  Totalitarianism doesn't have that built-in requirement: China, for example, I would describe as totalitarian but not dictatorial.

Feel free to contradict me if you know what you're talking about very well.

[quote="Tare"][quote="rippledance"]As far as I know, [i]imperialism[/i] is a national policy of taking over and exploiting another region's resources for one's own purposes. Google'll tell you "The policy, practice, or advocacy of extending the power and dominion of a nation especially by direct territorial acquisitions or by gaining indirect control over the political or economic life of other areas; extension or imposition of power, authority, or influence." The study of post-Industrial-Revolution Western imperialism is pretty interesting; you might consider looking that up.[/quote]
So imperialism doesn't have anything to do with an emperor? interesting...but then again countries that fit(ted) that description like England, France, Dutch, and Spanish all had Kings and Queens and whatnots...of course the US is an exception :wink:

In Japan there is a group of people who believe in communism and so by excersizing their freedom of assembly and religious beliefs guaranteed by the Japanese constitution they formed a farm where everyone there works the same, eats the same, clothes the same. However people call it a "cult" and refers to them as "bad people" as people tend to do with "cult" people. Any thoughts on this?

Was Nazi Germany "democratic" to begin with? When exactly did it turn from democracy to monarchy?

Is Great Britain a real "democratic" nation? (now don't laugh but:) In Johnny English the main point was that the queen had some juice (power) left in herself. So even if the political part is democratic won't it be part-monarchy too since the royal family lives on? (Unlike in Japan where the royal family has no power...well have even less than no power, they have to do what they are told by the gov't mostly)

So the modern day definition of "socialism" can coexist with "democracy"?

Can there be an oligarchic tyranny? What about a absolute monarch who's a dictator too?[/quote]

Any country could use the policy of imperialism.  Dictatorship, communist, democratic, doesn't matter.  All that matters is that you expand, conquer, and control.

Actually, those Japanese communists probably aren't a cult (unless they are involved wiith something that you haven't informed me of)

Nazi Germany was democratic to being with.  Hitler was elected chancellor, and proceeded to remove it's democracy.

Of course it's a democratic nation.  The queen's influence is somewhat like that of a powerful businessman -- many friends in high places, and can influence some of them, but I don't think she has any real governmental power.

Yes it can, and that is what the USA Socialist party wishes to do (and Communist party with communism, though they wish to take minor steps).  However, it probably wouldn't work out too well because powerful people might not support it and lead to a coup.  However, there's no reason why it can't.

If you take the first definition, (with one all-powerful ruler) then you can't though you can have a tyrannic ogliarchy.
If you take the second definition, with absolute power, yes, becuase a few people could rule all.

And yes, you could.

[quote]Was Nazi Germany "democratic" to begin with? When exactly did it turn from democracy to monarchy?

Is Great Britain a real "democratic" nation? (now don't laugh but:) In Johnny ?[/quote]
Nazi germany turned autocratic because the wienmar democracy had rigged elections and was very unstable.  The people were naturally welcome to a firm stable government.

With Germany's defeat after WWI, we have the creation of a Weimar Republic that was a liberal democracy. The Treaty of Versailles, which called for Germany to pay $30 billion in Allied reparations and placed the blame for the entire WWI on Germany, created a great deal of resentment and the Weimar government bore a lot of blame for signing the treaty. Furthermore, the Weimar govt was an uneasy coalition of left-wing and right-wing people to begin with, and it got battered by extremists on both sides. The Great Depression worsened matters and support grew both for the Nazis and the Communists, who seemed to offer alternatives. In the Jan 1933 elections, Hindenberg (president at the time) agreed behind closed doors to have Hitler elected as Chancellor. Nazi appeal had been on the wane at that time and Hindenburg (with others in the government) believed that they could control Hitler. Oops. By March 1933 Hitler had passed the Enabling Acts which basically made him dictator of Germany. So you could say that Nazi Germany was democratic for all of two months, I guess.

Britain is certainly a democracy; the queen is more of a figurehead than anything else.

I'm not sure if you could have a tyranny that's an oligarchy as well... if you go by the Aristotelian definition of each, tyranny refers to one while oligarchy refers to a group. If you go by the modern definition of tyranny, then yes, you could have an oligarchy of tyrants.

And as for Marxism vs. Leninism, Joel's basically right; Marxism applies to industrialized countries while Leninism believes that socialism can be applied to underdeveloped countries as well. Lenin argued that Marx had not taken the effects of imperialism into consideration and that colonies would be where the revolution had the strongest chance at success. There's Maoism as well, which places the revolution entirely in the countryside.
By the by, Stalinism isn't related to any of those terms... it's used to refer to a tyrannical regime exemplified by Stalin's Soviet Union.

Thanks Joel for clearing that up. =)


MATHCOUNTS
probability
number theory
greatest common divisor
Ring Theory

[b]1.[/b] The mean of six positive integers is 5, and the median is 6.  What is the largest the mode could be, given that the mode is unique?

[b]2.[/b] Alexio has 100 cards numbered 1-100, inclusive, and places them in a box.  Alexio then chooses a card from the box at random.  What is the probability that the number on the card he chooses is a multiple of 2, 3, or 5?  Express your answer as common fraction.

[b]3.[/b] For how many integers [i]n[/i] between 1 and 100 is the greatest common divisor of 15 and [i]n[/i] equal to 3?

[b]4.[/b] The inside wheels of a car traveling on a circular path are rotating half as fast as the outside wheels.  The front two wheels are six feet apart.  What is the number of feet in the path traced by the inside front wheel in one trip around the circle?  Express your answer in terms of [i]:pi:[/i].

what do you mean?
how can a mode be unique if it has all the same numbers?

ok

[hide]1. the total of the number is 30, the middle numbers /2=6. the largest mode we could have could be 2 or 3 numbers, but if it (the mode) is two numbers than the smallest number we could have are therefore 1,2,3 and then the mode would be at most 10. on the other hand if it is three numbers than the smallest number we could have which when taken away from 30 are divisible by 3 are again 1,2, and 3. so the largest mode would be 8. this means the largest mode comes from the numbers 1,2,3,4,10,10.

2. multiples of 5:20 of 3:33 of 2:50. odd multiples of 5 and 3:10+17=27. odd multiples of 5 and 3 (the ones we overcounted within the 27):3 (15,45, and 75.) that means we have 50+27-3=74. chance is 74/100[/hide]

3. what do ou mean n is equal to 3?

[hide]4. pi*d?[/hide]

ml2, in number 1 you didn't take into account that the median is 5.

Leeyuhang -- in this set, the mode is unique: {1, 1, 1, 2, 4, 4, 5}.  There is exactly one number which appears the most number of times.
In this set the mode is not unique: {1, 1, 2, 4, 4, 5}.
In this one either: {1, 2, 3}.

ml2 -- he's asking how many values of n between 1 and 100 have the property that GCD(15, n) = 3

For number 4, you're supposed to use the information given to find out the value of d.

whoops, forgot about median of 6, oh well same arguement except we get the numbers [hide]1,2,3,6,9,9. and median 9[/hide]

[hide] 3.So we have numbers whose GCD is 3, this means the number is divisble by 3 and not by 5, this adds up to 33-5=28

4. well the outside wheel are going pi*(d+6) and the inside wheel goes pi(d) and we have pi(d+6)/2=pi*d so, d=6. so the inside wheel travels 6pi.

[/hide]

You didn't read number 1 carefully enough, again.

Why did you subtract 5 in number 3?  Other than the 5, you're correct.

Be more careful on number 4 -- is it (d + 6)?  Maybe if you did it in terms of the radius it would be easier.

well in number 3 i subtracted five because multiples of 3 such as 15,30 and 45 .... don't work.

what did i miss in number 1??

4. [hide]oh, yes, the radius is 6, the diameter is 12 inside wheel travels 12 pi.

.[/hide]

The same mistake again: The median is 6. Start from there instead...

ok, i need to be more careful.. anyways the median of 6 numbers is the two number in the middle divided by 2, so the 2 numbers in the middle add up to 12, leaving 18 for the other numbers, 8,8,1,1 doesn't work because mode isn't unique  :arrow: the mode is 7, the number are 7,7,7,5,3,1.

[quote="Leeyuhang"]i'm not sure if this makes sense but i got
1,3,5,7,7,7 where the mean is 5 median is 6 and mode is 7? it wouldn't work if you had 8,9,or greater as the mode. you can't have 2 6s because in the problem it said the mode had to be unique. and one 6 wouldn't be the median.

tare, what do you mean the solution bites?[/quote]
 :lol:, oh that was a great one, unless you didn't know German, then it wouldn't be funny at all, and even if you do get it it might even be funni[b]er[/b] than the level in which I'm laughing at because it might be wrong (I believe "bitte" in German means "please" but I'm not sure)

Mmm, there's a better answer than that... :wink: :wink:
I thought you'd get it now by elimination :lol: </jk>

ml2, bitte is German for please.

Number 1 is a Sprint Round problem from last year, I think. Here is my solution:

[hide] Since there are 6 elements with a mean of 5, the sum of the elements=5(6)=30. Since the mode must be as large as possible, the first 2 numbers should be 1. The second 2 number's sum should be 12. 30-2-12=16. SO the last 2 numbers should sum to 16. 16/2=8. So we have {1,1,x,y,8,8}. For the mode to be unique, there must be 3 8's. Since x and y add to 12, the set with the greatest mode is {1,1,4,8,8,8}. So the greatest possible mode is 8.[/hide]

Pointless green guy: :mrgreen:

actually, i finnaly realized the correct solution. the two middle number add up to 12, 9 and 3 doesn't work. 8 and 4 does. we get
1,1,4,8,8,8.


[quote="guyinPA"]I've never been to or seen a National (or even State) Mathcounts competition -- for anyone who's actually Been there (not just speculation, please) but ABOUT how long - how many seconds does it really take for the contestants to answer the question correctly once it pops up on the screen? I mean, I know it must vary from question to question, but on average, does it take like 5 secs? 10? 30? Do a lot of people take almost the whole 45 secs before buzzing in? Does that number change a lot between State Countdowns and Nat. ones?  Also, is there "strategy involved"? - like if you think the other person is going to answer in 10 secs, do you say Anything in 5, even if it's a guess?[/quote]
All I know that in Nationals when as soon as question pops up  the person buzzes in and usually gets it right.

[quote="Quevvy"][quote="Ravi B"]When the national countdown was broadcast on TV a couple of years ago, I timed how long the students took.  On average, the first student buzzed in after about 8 seconds.  If that student got the question wrong, the second student should take his or her time to answer.[/quote]
Yeah, if the other person gets it wrong, take as much time as you can to double check your work. Most of the time that won't happen though.[/quote]
Actually I'm going to have to disagree. When I was in Countdown in states last year I made a bunch of careless mistakes (including forgetting the volume of a pyramid). My opponent (who was a decent bit slower than me) still beat me.

@ the original poster: you'll find that as the ranks get higher the participants get a significant bit faster.

[quote="ehehheehee"]Time seems to run a lot slower when you are actually up there.[/quote]
Yeah, I always solve the problems much much quicker up there than when I'm watching. THat's partly because I get possessed.

[quote="kstan013"][quote="ehehheehee"]Time seems to run a lot slower when you are actually up there.

I was 3rd in state and top scorer for the Ohio team at nationals, and I never got past the 2nd round, even in chapter.

I don't think there is enough correlation between rank and countdown mightiness as there should be.  And I think they should do away with that bracket style nationals countdown round.[/quote]I know. State Countdown format rules! I think it gives the significant advnatage to the people who score higher. So for all of you nay-sayers who say the countdown corrupts your mind and is useless, it's not as if speed alone will get you to the top, at least not with State.[/quote]
Unless you live in a state like deleware, get a 24 on the written (sprint and target combined!), get 9th written, and countdown your way up to 2nd.

lol yeah that's kinda cheap, but not to diss anyone from Delaware or any small state, but there are people who can't even make it to states in CA that could probably p00wn the entire national team from a small state/territory. Any how, back on topic, I think that it is fair because if you have 1 standout, or even 3 standouts, they are guaranteed to make it to Nats. You could have someone ace the test, but with the bracket style they could lose right away and that 10th place person with a 24 could come and make nats :?:  :huh: Which is why the ladder is fairer since only 1 non-top-3er is in, which has been me for the last 2 years :D

sort of off topic:CD should DEFINITELY be unofficial everywhere. I'm lucky-I'm not very ood at it, but, in MA, it's unofficial. I've actually gotten much better, but I'm not awesome. Anyway, some Qs will take about 20 secs, but on some, b4 the guy starts reading the qustion, people will answer. I've never seen the full 45 secs go by.

I saw it a couple times in my chapter this year on particularly confusingly worded questions, but all 14 of the questions I was involved in were answered within 20 seconds [approx], either by me or him... hey i didn't face any girls and faced 6 guys :o.

I find that time runs slower when you're doing countdown...

I get most questions within 5 seconds, but it feels like a minute...

Thats why I do better on Countdown than on written!

About 1/3 the Qs were answered within 5 secs. They kept getting mad at me b/c I was the only one answering b4 he started reading.

except that sometimes your state of mind or determination is not enough.

For example, you would crush me in countdown no matter what I thought.

I freak out at countdown even though all the ones I participated have been unofficial.

[quote="Walk Around The River"][quote="Fanatic"]For me, time runs super fast. When I am up there, after I answer a question, it seems like only 0.5 to 2 seconds pass, when in reality it may have been anywhere from 3-8. No joke.[/quote]

agreeed.[/quote]
Really?
FOr me it seems like 1.5 min. have passed when the whole time is done. My brain goes into hypermode, and occasionally I get an answer in my head with no clue how it got there (that happened at states last year once).

[quote="Treething"]I imagine somebody at Nationals will have to take the place of me, Neal, and Ardito.

=X

Pressure gets to you if you're one of those people who freak out when something's at stake. You should be fine if you don't get overwhelmed by the fact that if you win you get goodies.[/quote]

Haha...goodies.  Be happy that you are in countdown!

I freak out like crazy...I got nervous for the School Competition, and almost had a heart attack last year at states (countdown was official, so they were doing the ladder system - I died when they were about to call 4th place).

[quote="bpms"][quote="kyyuanmathcount"]Haha...goodies.  Be happy that you are in countdown!

I freak out like crazy...I got nervous for the School Competition, and almost had a heart attack last year at states (countdown was official, so they were doing the ladder system - I died when they were about to call 4th place).[/quote]
Dude, how do [b]you[/b] get nervous at school competition?[/quote]

Don't think like me, but I was thinking:  Hm...what if...I totally make 10 stupid mistakes on sprint and then I pull 2 "treethings" and switch two pairs of target problems...

[quote="kyyuanmathcount"]Don't think like me, but I was thinking:  Hm...what if...I totally make 10 stupid mistakes on sprint and then I pull 2 "treethings" and switch two pairs of target problems...[/quote]
Hah, a 20 on the sprint would put you on my school's team :D

"countdown gurus"??!!! OK, I believe you - but what's their secret?? I mean, I looked over some vague state stats, and I noticed that people scoring low (or lower - say 5th, 6th, or 7th) in chapter were scoring number 1 in countdown! But I don't get it! It's not LOGICAL to me -  If you know the material - the math - well enough to score a top score on 30 probs in 40 mins, how could you know be able to solve somewhat easier problems quicker than the next person?  Do you think many have Equal knowledge, but some people's brains just automatically "click" a few seconds faster? Do you think someone can improve this? HOW?? How can I practice a countdown round -- no one I know is in the competition :o  :( ?

In my chapter countdown is unofficial but I still got nervous.  I agree with bpms, time seems like it's going faster, but that makes me seem like I'm going too slowly which makes me even more nervous.  Anyway it didn't matter because the desk for countdown was so close to the projected problem that you had to stare straight up to have any hope of seeing the problem and even then it was hard.
Apparently my opponent (who goes to my school so it's okay) had better luck since she got the first 2 questions right and then I was like "what it's over?".  Countdown is very unpredictable since there are so few problems.
But it's not like the written round isn't unpredictable, I was 7th on my school's team and then came in first in my chapter (Manhattan)...

[quote="guyinPA"]It's not LOGICAL to me -  If you know the material - the math - well enough to score a top score on 30 probs in 40 mins, how could you know be able to solve somewhat easier problems quicker than the next person?[/quote]

There's a big difference between being able to do a problem in a minute or so, and being able to do an easier problem in 5 seconds or so.  Some people who don't know all the math to be top on the sprint/target might still know the easy stuff better/faster and be able to just get right in there.  Also, some people who do know the math may make more mistakes under pressure, or buzz in just a bit too quickly, etc.  Finally keep in mind that the difference between 1st and 6th in written may be a very small number of problems (or even just tie-breakers), and someone may get lucky to get the few problem types they are really good at in countdown.  In our chapter, there are twins and both last year and this year, one twin came first in written, and the other was several places back in the written but came first in countdown.  They are different skills.

If you try a bunch of countdowns, I think you will see certain problem types and tricks popping up frequently.  Even if you don't have someone to practice against, try using a stopwatch and seeing how fast you can get each problem.


[quote="perfect628"]I don't think you're supposed to discuss this, given that it is not only given in NE (it's called something different in each region), and you can take it late I believe.[/quote]

You are correct.  The submission deadline for round one scores is Friday, so schools could still be taking the contest this week.

guys, on contests where there is a "window" for taking the test at your school, you can not discuss the test at all until after the deadline for submission of scores.  That means no giving answers, no saying a certain problem is hard, no saying you got perfect.  Nothing.  If you are unsure of the deadline, ask your coach or post here to check.

Thanks!

Terribly off topic. In the future, if you a comment that relates specifically to a username (such as an avatar or a signature), please PM them instead of posting a message that... er... will lead to less than useful discussion.

And no, it was the other way around :D

I'm not sure if this needs locking until Friday?

dude get out of the MA forum and go back to your Rhode Island forum!!!

Got a 6 anyways...

those are some ambitious goals...btw if there was one i would

lol

Even more ambitious is the merging of Massachusetts and Rhode Island. We're just too good :wink:

As for score, I'll be a good person and tell after the deadline. I will, however, tell you that it is an integer between 0 and 6. Is that a good enough range from which to guess?

lol
i don't care its not like i cant come here

maybe a mod should ban mathemonster from the MA forum lol

[quote="mathemonster"]if you took it (dont discuss just say a score) what did you get?
I got a 6/6[/quote]

Discussing scores is included in discussing the test. If you're not allowed to discuss the problems, you can't discuss your scores, because that's pretty much the same as discussing the difficulty.

well the window is waaaaay over so why did u just post that today?


Let function $ f : [0;1] \minus{}> R$ such that :
$ (i) f(0)\equal{}f(1)\equal{}0$
$ (ii)|f(x)\minus{}f(y)|<|x\minus{}y|$ for all $ x,y \in [0;1]$ and $ x \neq y$
Prove that : $ |f(x)\minus{}f(y)|<\frac{1}{2}$

If $ |x \minus{} y|\le \frac 12$, then obviosly $ |f(x) \minus{} f(y)| < \frac 12$.
Let $ x > y$ and $ |x \minus{} y| > \frac 12$, then $ 1 \minus{} x \plus{} y < \frac 12\to |1 \minus{} x|\pm |y\parallel{} < \frac 12\to |f(x) \minus{} f(y)| < \frac 12$.

[quote="Rust"]If $ |x \minus{} y|\le \frac 12$, then obviosly $ |f(x) \minus{} f(y)| < \frac 12$.
Let $ x > y$ and $ |x \minus{} y| > \frac 12$, then $ 1 \minus{} x \plus{} y < frac 12\to |1 \minus{} x|\pm |y\parallel{} < \frac 12\to |f(x) \minus{} f(y)| < \frac 12$.[/quote]
Why "$ x > y$ and $ |x \minus{} y| > \frac 12$, then $ 1 \minus{} x \plus{} y < frac 12\to |1 \minus{} x|\pm |y\parallel{} < \frac 12\to |f(x) \minus{} f(y)| < \frac 12$"
I think :"$ x > y$ and $ |x \minus{} y| > \frac {1}{2}$, then $ 1 \minus{} x \plus{} y < frac {1}{2}\to |1 \minus{} x|\pm |y\parallel{} < \frac 12\to |f(1 \minus{} x) \minus{} f(y)| < \frac {1}{2}$"
and $ (i) f(0)\equal{}f(1)\equal{}0$  :?:

Lemma:

f:[a,b] -> R and f(a)=f(b)=0 and f(x) does not equal to 0 for x on (a,b)

|f(x)-f(y)| <|x-y|

then |f(x)-f(y)| < (b-a)/2

Proof:

Without loss of generalities , let 's assume that f(x)>0 for all x in (a,b)

 From |f(x)-f(y)|< |x-y| we get f is continous. As [a,b] is compact, there exits M=max f(x) on [a,b]

There exist x0 such that f(x0)=M

if x0> (b+a)/2, we have M= | M- f(b)| < |b-x0| <(b-a)/2

if x0 =<(b+a)/2 , we have M=|M-f(a) | < |x0 -a |

In both cases , we get   M <(b-a)/2

|f(x)-f(y)| =< M < (b-a)/2

The main problem is a corrolaire of this lemma when we divide [0,1] in smaller intervals by zero points

[quote="mto"]Lemma:

f:[a,b] -> R and f(a)=f(b)=0 and f(x) does not equal to 0 for x on (a,b)

|f(x)-f(y)| <|x-y|

then |f(x)-f(y)| < b-a/2

Proof:

Without loss of generalities , let 's assume that f(x)>0 for all x in (a,b)

 From |f(x)-f(y)|< |x-y| we get f is continous. As [a,b] is compact, there exits M=max f(x) on [a,b]

There exist x0 such that f(x0)=M

if x0> b-a/2, we have M= | M- f(b)| < |b-x0| <b-a/2

if x0 =<b-a/2 , we have M=|M-f(a) | < |x0 -a |

In both cases , we get   M <b-a/2

The main problem is a corrolaire of this lemma[/quote]
I think you wrong because you wrote $ |f(x) \minus{} f(y)| < b \minus{} \frac {a}{2}$= >$ |f(x) \minus{} f(y)| < |1 \minus{} \frac {0}{2}| \equal{} 1$. It 's wrong !!! and $ b\minus{}\frac{a}{2}$ no in $ [a,b]$

[quote="isomorphism"][quote="conan_naruto236"]
I think you wrong because my solution is $ |f(x) \minus{} f(y)| < |\frac {b}{2} \minus{} a|$
=>$ |f(x) \minus{} f(y)| < |\frac {1}{2} \minus{} 0| \equal{} \frac {1}{2}$
and you $ |f(x) \minus{} f(y)| < b \minus{} \frac {a}{2}$= >$ |f(x) \minus{} f(y)| < |1 \minus{} \frac {0}{2}| \equal{} 1$. It 's wrong !!![/quote]

He means
[quote]$ f: [a,b] \to \mathbb{R}$ and $ f(a) \equal{} f(b) \equal{} 0$ and $ f(x)\neq 0, \forall x \in (a,b)$

$ |f(x) \minus{} f(y)| < |x \minus{} y|$

then $ |f(x) \minus{} f(y)| < \frac {b \minus{} a}2$[/quote][/quote]
It 's is right $ <\equal{}> \frac{b\minus{}a}{2} \in [a,b]$ <=> $ b > |a|$ and $ 3a < b$
So I think lemma add contention $ 0\leq a < b , 3a < b$ to right

I corrected my lemma

Just a mistake when I move from [0,1] to [a,b]  , when i thought that the midpoint of [a,b] is (b-a) /2

These problem is obviosly. If $ x\minus{}y\le \frac{b\minus{}a}{2}$, then $ |f(x\minus{}f(y)|<\frac{b\minus{}a}{2}$, if $ x\minus{}y>\frac{b\minus{}a}{2}$, then (by use $ f(a)\equal{}f(b)$)
$ |f(x)\minus{}f(y)|\equal{}|f(x)\minus{}f(b)\plus{}f(a)\minus{}f(y)|\le |f(x)\minus{}f(b)|\plus{}|f(y)\minus{}f(a)|<b\minus{}x\plus{}y\minus{}a<\frac{b\minus{}a}{2}.$


calculus
integration
analytic geometry
derivative
function

Here's the thing: I'd go on a date with a girl only if the girl is [b]better[/b] than I am in multivariable calculus. If the girl is no good in multivariable calculus I would reject her without a second thought (as a lover, though I may still try to befriend her). On the other hand, if the girl was a lot better, it would be harder to sustain a romantic relationship than if she was only a bit better.

Also, I would be tempted to tell her that she is prettier than:

Lagrange multipliers
Double/triple definite integrals in rectangular coordinates
Partial/implicit derivatives
Vectorial products
Variation rates (linked or not)
Trigonometrical functions
Limit at + infinity
Order-4 determinant
Definite integral
Revolution solid

[quote="CEM_College_Champ"]Here's the thing: I'd go on a date with a girl only if the girl is [b]better[/b] than I am in multivariable calculus.[/quote]

don't think you'll need to worry about this anytime soon buddy

Please read http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214779 with special attention to points 3), [b]1[/b] and [b]2[/b].


Find all pairs of positive integers $ (x,y)$ such that $ y^x\minus{}1\equal{}(y\minus{}1)!$

[quote="J.Mayers"][hide="hint"]For sufficiently large $ y$ and $ x$, can $ y$ be even?[/hide][/quote]
For even number $ y$ greater than 4, it's impossible to satisfy the equation. Because RHS is even, and LHS is odd.
For even number $ y \leq 4$, the only solution is $ (1,2)$.

Any other hint?

Yes:

[hide]Clearly, $ x > 0$. Taking this equation mod y, we see that $ (y \minus{} 1)! \equiv \minus{}1 \mod y$, so y must be prime by Wilson's theorem. [/hide]


Let $ p_i$ for $ i \equal{} 1,2,...,n$ be distinct prime numbers. Show that $ \sqrt {p_i}$s are linearly independent over $ \mathbb{Q}$.

See my post [url=http://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/]here[/url] (as well as the link to Iurie Boreico's article on the subject).

thank you very much for your informative blog.