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For all positive integers m,n show that [(3m)!(3n)!]/[m!n!(m+n)!(n+m)!] is always an integer.

[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=141316]A former blog post of mine[/url] hints at a generalization (which I have not proven; it might be false  :| ).

[hide="This is the generalization"] For two sequences of positive integers $ a_{1}, ... a_{n}, b_{1}, ... b_{n}$ such that $ \sum a_{i}= \sum b_{i}$, we have

$ \{ a_{i}\}\succ \{ b_{i}\}\Leftrightarrow \frac{ \prod (a_{i})! }{ \prod (b_{i})! }\in \mathbb{N}$ [/hide]

Aviateurpilot, the second line of your solution is exactly what we are trying to prove...

[hide="hint"]
You want to show $ \lfloor{\frac{3m}{p}\rfloor}+\lfloor{\frac{3n}{p}\rfloor}-\lfloor{\frac{m}{p}\rfloor}-\lfloor{\frac{n}{p}\rfloor}-2\lfloor{\frac{m+n}{p}\rfloor}$ for all primes p.  Show this for all real m, n in the interval $ [0, p)$ as follows...this will imply it for all real $ m, n$

Consider a square of dimensions p, placed with its axes parallel to the x and y axes.  Label the horizontal axis "m" and the vertical one "n".  We have a 3m term, so draw three vertical lines which trisect the area.  Similarly, we have a 3n term, so put three horizontal lines which trisect the area.  We have an m+n term, so place a line as the diagonal m+n=p.  Now we can number each region so that for an integer pair $ (m, n)$ in that region (modulo p), the number is the number of times p divides the expression is that number.  You'll find all regions have a nonnegative number, and you're done. [/hide]

T0r, tell me you remember this from the PROMYS research labs presentation last year... :wink:

[quote="K81o7"]tell me you remember this from the PROMYS research labs presentation last year... :wink:[/quote]

Honestly, I don't remember much from PROMYS besides the problem sets  :|


Polynomial $ f(x) \equal{} x^{1983} \plus{} 3x^8 \minus{} x \plus{} 1$ .Find the remainder
(1)$ f(x)$ is divided by $ x^2 \plus{} x \plus{} 1$
(2)$ f(x)$ is divided by $ x^3 \plus{} 2x^2 \plus{} 2x \plus{} 1$

[quote="math154"][hide="1"]
Since $ x^{1983} \minus{} 1\equiv x^3 \minus{} 1\equiv0\pmod{x^2 \plus{} x \plus{} 1}$, we have

$ x^{1983} \plus{} 3x^8 \minus{} x \plus{} 1 \equal{} x^{1983} \minus{} 1 \plus{} 3x^8 \minus{} x \plus{} 2$

$ \equiv3x^8 \minus{} x \plus{} 2\pmod{x^2 \plus{} x \plus{} 1}$.

Also, $ 3x^8 \minus{} x \minus{} 2 \equal{} (x \minus{} 1)(3x^7 \plus{} 3x^6 \plus{} 3x^5 \plus{} 3x^4 \plus{} 3x^3 \plus{} 3x^2 \plus{} 3x \plus{} 2)$, so

$ 3x^8 \minus{} x \plus{} 2\equiv(x \minus{} 1)(0 \plus{} 0 \plus{} 3x \plus{} 2)\equiv3x^2 \minus{} x \minus{} 2$

$ \equiv\boxed{ \minus{} 4x \minus{} 5}\pmod{x^2 \plus{} x \plus{} 1}$.
[/hide]
[hide="2"]
$ x^3 \plus{} 2x^2 \plus{} 2x \plus{} 1 \equal{} (x \plus{} 1)(x^2 \plus{} x \plus{} 1)$, and by the Remainder Theorem,

$ x^{1983} \plus{} 3x^8 \minus{} x \plus{} 1\equiv( \minus{} 1)^{1983} \plus{} 3( \minus{} 1)^8 \minus{} ( \minus{} 1) \plus{} 1\equiv4\pmod{x \plus{} 1}$.

We also have $ x^{1983} \plus{} 3x^8 \minus{} x \plus{} 1\equiv \minus{} 4x \minus{} 5\pmod{x^2 \plus{} x \plus{} 1}$, so all solutions to these satisfy

$ a \equal{} m(x \plus{} 1) \plus{} 4 \equal{} n(x^2 \plus{} x \plus{} 1) \minus{} 4x \minus{} 5$.

Taking this $ \pmod{x \plus{} 1}$, we get

$ x^2n\equiv5\equiv5 \plus{} 5(x \minus{} 1)(x \plus{} 1) \equal{} 5x^2\pmod{x \plus{} 1}$

$ \implies n\equiv5\pmod{x \plus{} 1}\implies n \equal{} b(x \plus{} 1) \plus{} 5$, so

$ a \equal{} b(x \plus{} 1)(x^2 \plus{} x \plus{} 1) \plus{} \boxed{5x^2 \plus{} x}$.
[/hide][/quote]
Hello, Math154
Although your work is very elegant, your ans.(1) &(2) are not correct , according  to our discussed solutions.I think the mistakes might come from the following boxed material.
In.1
$ \boxed{Also, 3x^8 \minus{} x \minus{} 2 \equal{} (x \minus{} 1)(3x^7 \plus{} 3x^6 \plus{} 3x^5 \plus{} 3x^4 \plus{} 3x^3 \plus{} 3x^2 \plus{} 3x \plus{} 2), so}$

$ \boxed{3x^8 \minus{} x \plus{} 2\equiv(x \minus{} 1)(0 \plus{} 0 \plus{} 3x \plus{} 2)\equiv3x^2 \minus{} x \minus{} 2}$

The different type in $ {\minus{}2,\plus{}2}$

[quote="math154"][hide="1"]
Since $ x^{1983} \minus{} 1\equiv x^3 \minus{} 1\equiv0\pmod{x^2 \plus{} x \plus{} 1}$, we have

$ x^{1983} \plus{} 3x^8 \minus{} x \plus{} 1 \equal{} x^{1983} \minus{} 1 \plus{} 3x^8 \minus{} x \plus{} 2$

$ \equiv3x^8 \minus{} x \plus{} 2\pmod{x^2 \plus{} x \plus{} 1}$.

Also, $ 3x^8 \minus{} x \minus{} 2 \equal{} (x \minus{} 1)(3x^7 \plus{} 3x^6 \plus{} 3x^5 \plus{} 3x^4 \plus{} 3x^3 \plus{} 3x^2 \plus{} 3x \plus{} 2)$, so

$ 3x^8 \minus{} x \plus{} 2\equiv(x \minus{} 1)(0 \plus{} 0 \plus{} 3x \plus{} 2) \plus{} 4\equiv3x^2 \minus{} x \plus{} 2$

$ \equiv\boxed{ \minus{} 4x \minus{} 1}\pmod{x^2 \plus{} x \plus{} 1}$.
[/hide]
[hide="2"]
$ x^3 \plus{} 2x^2 \plus{} 2x \plus{} 1 \equal{} (x \plus{} 1)(x^2 \plus{} x \plus{} 1)$, and by the Remainder Theorem,

$ x^{1983} \plus{} 3x^8 \minus{} x \plus{} 1\equiv( \minus{} 1)^{1983} \plus{} 3( \minus{} 1)^8 \minus{} ( \minus{} 1) \plus{} 1\equiv4\pmod{x \plus{} 1}$.

We also have $ x^{1983} \plus{} 3x^8 \minus{} x \plus{} 1\equiv \minus{} 4x \minus{} 1\pmod{x^2 \plus{} x \plus{} 1}$, so all solutions to these satisfy

$ a \equal{} m(x \plus{} 1) \plus{} 4 \equal{} n(x^2 \plus{} x \plus{} 1) \minus{} 4x \minus{} 1$.

Taking this $ \pmod{x \plus{} 1}$, we get

$ x^2n\equiv1\equiv1 \plus{} (x \minus{} 1)(x \plus{} 1) \equal{} x^2\pmod{x \plus{} 1}$

$ \implies n\equiv1\pmod{x \plus{} 1}\implies n \equal{} b(x \plus{} 1) \plus{} 1$, so

$ a \equal{} b(x \plus{} 1)(x^2 \plus{} x \plus{} 1) \plus{} \boxed{x^2 \plus{} 3x}$.
[/hide]

Edit: Thank you, vinskman, for the corrections. :) I should have caught my mistake earlier, though, if I had tested $ x \equal{} 1$ on problem one.[/quote]

$ \boxed{a \equal{} b(x \plus{} 1)(x^2 \plus{} x \plus{} 1) \plus{} \boxed{x^2 \plus{} 3x}}$.

Sorry, I think you have typo , since $ (x^2 \plus{} x \plus{} 1)\plus{}(\minus{}4x\minus{}1)\equal{}x^2 \minus{}3 x$ :D

[quote="math154"][hide="1"]
Since $ x^{1983} \minus{} 1\equiv x^3 \minus{} 1\equiv0\pmod{x^2 \plus{} x \plus{} 1}$, we have

$ x^{1983} \plus{} 3x^8 \minus{} x \plus{} 1 \equal{} x^{1983} \minus{} 1 \plus{} 3x^8 \minus{} x \plus{} 2$

$ \equiv3x^8 \minus{} x \plus{} 2\pmod{x^2 \plus{} x \plus{} 1}$.

Also, $ 3x^8 \minus{} x \minus{} 2 \equal{} (x \minus{} 1)(3x^7 \plus{} 3x^6 \plus{} 3x^5 \plus{} 3x^4 \plus{} 3x^3 \plus{} 3x^2 \plus{} 3x \plus{} 2)$, so

$ 3x^8 \minus{} x \plus{} 2\equiv(x \minus{} 1)(0 \plus{} 0 \plus{} 3x \plus{} 2) \plus{} 4\equiv3x^2 \minus{} x \plus{} 2$

$ \equiv\boxed{ \minus{} 4x \minus{} 1}\pmod{x^2 \plus{} x \plus{} 1}$.
[/hide]
[hide="2"]
$ x^3 \plus{} 2x^2 \plus{} 2x \plus{} 1 \equal{} (x \plus{} 1)(x^2 \plus{} x \plus{} 1)$, and by the Remainder Theorem,

$ x^{1983} \plus{} 3x^8 \minus{} x \plus{} 1\equiv( \minus{} 1)^{1983} \plus{} 3( \minus{} 1)^8 \minus{} ( \minus{} 1) \plus{} 1\equiv4\pmod{x \plus{} 1}$.

We also have $ x^{1983} \plus{} 3x^8 \minus{} x \plus{} 1\equiv \minus{} 4x \minus{} 1\pmod{x^2 \plus{} x \plus{} 1}$, so all solutions to these satisfy

$ a \equal{} m(x \plus{} 1) \plus{} 4 \equal{} n(x^2 \plus{} x \plus{} 1) \minus{} 4x \minus{} 1$.

Taking this $ \pmod{x \plus{} 1}$, we get

$ x^2n\equiv1\equiv1 \plus{} (x \minus{} 1)(x \plus{} 1) \equal{} x^2\pmod{x \plus{} 1}$

$ \implies n\equiv1\pmod{x \plus{} 1}\implies n \equal{} b(x \plus{} 1) \plus{} 1$, so

$ a \equal{} b(x \plus{} 1)(x^2 \plus{} x \plus{} 1) \plus{} \boxed{x^2 \minus{} 3x}$.
[/hide]

Edit: Thank you, vinskman, for the corrections. :) I should have caught my mistake earlier, though, if I had tested $ x \equal{} 1$ on problem one.

Edit2: *Sigh*...I can't do anything without making careless mistakes, can I? :D[/quote]

no worries, thanks to your work indeed.

minor mistakes don't be serious, my guy. :D  :lol:

[b]@ math154:[/b] You correctly observed that $ x^3 \minus{} 1 \equiv 0 \pmod{x^2\plus{}x\plus{}1}$. What happens when you rewrite this as $ x^3 \equiv 1 \pmod{x^2\plus{}x\plus{}1}$?  :)

[quote="TZF"][b]@ math154:[/b] You correctly observed that $ x^3 \minus{} 1 \equiv 0 \pmod{x^2 \plus{} x \plus{} 1}$. What happens when you rewrite this as $ x^3 \equiv 1 \pmod{x^2 \plus{} x \plus{} 1}$?  :)[/quote]

I am sure it becomes more convenient  to find ans. ' cause I have tried before your remind. :trampoline:


Solve the equation $2^x -5 =11^{y}$ in positive integers.

[quote="Peter"]Solve the equation $ 2^x \minus{} 5 \equal{} 11^{y}$ in positive integers.[/quote]
$ x \geq 2$
We have:
$ 11^y \plus{}5 \equiv (\minus{}1)^y\plus{}1 \equiv 2^x \equiv 0(mod 4)$
So $ y\equal{}2k\plus{}1$ with $ k \in N$
$ 2^x \equiv 5\plus{}11^y \equiv (\minus{}1)\plus{}(\minus{}1)^y \equiv \minus{}2 \equiv 1(mod 3)$
So $ x\equal{}2z$ with $ z \in N$
Then $ 4^z\minus{}5\equal{}11^y$
$ 4^z \equiv 11^y\plus{}5 \equiv 1(mod 5)$
=>$ z\equal{}2t$
=>$ 16^t\equal{}11^y\plus{}5$
So $ 11^y \equiv \minus{}5(mod 16)$
Otherhand $ 11^y \equiv (\minus{}5)^y (mod 16)$
So $ y\equal{}1$ then $ x\equal{}4$
I hope my solution is true :blush: !

I follow completely until the second last line: how does it follow $ y\equal{}1$?

i don't think the solution isn't true because you made a mistake : $ ( \minus{} 5)^y \equiv  \minus{} 5 (mod16)\Rightarrow y \equal{} 1$ it's not true !!!

[quote="Peter"]Solve the equation $ 2^x - 5 = 11^{y}$ in positive integers.[/quote]
It's easy t prove that $ x \vdots 4$ and y is odd.So $ x=4x';y=2y'+1$.
And $ 2^{2x'}=a;11^{y'}=b$.We have:
$ a^2-11b\&2=5. (1)$
It's is Pell Equation.All solutions of (1) are given by(note that we have$ a \vdots 2$ )
\[ \left\{ \begin{array}{l}
 a_1  = 4,b_1  = 1 \\
 a_{n + 1}  = 10a_n  + 33b_n  \\
 b_{n + 1}  = 3a_n  + 10b_n  \\
 \end{array} \right.\]
But $ a_j (j \ge 1)$is odd.So we have a solution $ a=4,b=1.$
Thus,we infer that $ x=4,y=1$

[quote="tronghieu"][quote="Peter"]Solve the equation $ 2^x - 5 = 11^{y}$ in positive integers.[/quote]
It's easy t prove that $ x \vdots 4$ and y is odd.So $ x = 4x';y = 2y' + 1$.
And $ 2^{2x'} = a;11^{y'} = b$.We have:
$ a^2 - 11b \& 2 = 5. (1)$
It's is Pell Equation.All solutions of (1) are given by(note that we have$ a \vdots 2$ )
\[ \left\{\begin{array}{l} a_1 = 4,b_1 = 1 \\
a_{n + 1} = 10a_n + 33b_n \\
b_{n + 1} = 3a_n + 10b_n \\
\end{array} \right.
\]
But $ a_j (j \ge 1)$is odd.So we have a solution $ a = 4,b = 1.$
Thus,we infer that $ x = 4,y = 1$[/quote]
Sorry I don't think you're right.I think we can't conclude that  $ a_j (j \ge 1)$ right?

[quote="hxy09"]
Sorry I don't think you're right.I think we can't conclude that  $ a_j (j \ge 1)$ right?[/quote]
Thanks hxy09. I'm not careful in some calculation  :( . This is my solution. :)
Consider modulo $ 5$, $ 2^x \equiv 1(\bmod 4) \Rightarrow x \vdots 4.$
and modulo $ 16$, we have $ y \equiv 1(\bmod 4).$
If $ y \equal{} 1$ then $ x \equal{} 4$. We'll prove $ (1;4)$ is unique solution. In fact, if $ y \ge 2$ , with some check calculation, we can prove $ x \ge 6$. Thus, $ 2^x \vdots 2^6$ . Morever, we have:
$ 2^x \equal{} 11^y \plus{} 5 \equal{} (2^3 \plus{} 3)^y \plus{} 5 \equal{} (2^3 )^y \plus{} \left( {\begin{array}{*{20}c} y \\
1 \\
\end{array}} \right)(2^3 )^{y \minus{} 1} .3 \plus{} ... \plus{} \left( {\begin{array}{*{20}c} y \\
{y \minus{} 1} \\
\end{array}} \right)2^3 .3^{y \minus{} 1} \plus{} \plus{} 3^y \plus{} 5.$
So $ 8y.3^{y \minus{} 1} \plus{} 3^y \plus{} 5 \vdots 2^6 .$
Consider $ y \equiv 1(\bmod 2^3 )$ then $ 8y \equiv 8(\bmod 2^6 )$. So $ 8y.3^{y \minus{} 1} \plus{} 3^y \plus{} 5\equiv 3^{y \plus{} 1} \plus{} 5(\bmod 2^6 )$. However, $ \left( {\frac {{ \minus{} 5}}{{2^6 }}} \right) \equal{} \minus{} 1$. That gives contracdition. Thus, $ y \equiv 5(\bmod 2^3 ). (1)$
Setting $ x \equal{} 4x' ( x' \in N, x' > 1)$. Rewrite our equation:
$ 16(16^{x' \minus{} 1} \minus{} 1) \equal{} 11(11^{y \minus{} 1} \minus{} 1).$
Because $ gcd(16,11) \equal{} 1,$ so $ 16^{x' \minus{} 1} \minus{} 1\vdots1$1. That gives $ x' \minus{} 1\vdots5.$So $ 16^{x' \minus{} 1} \minus{} 1\vdots 16^5 \minus{} 1\vdots 41$. From that, we have $ 11^{y \minus{} 1} \minus{} 1 \vdots 41.$ So$ y \minus{} 1 \vdots 40. (2)$
From $ (1)$ and $ (2)$ we have the contracdition. That finishes proff.

Very nice your solution :)
I am impressed by your setting $ 11\equal{}2^3\plus{}3$
This is a very efficient way!

thanks hxy09. I've edited some mistakes in typing. :blush:

[quote=tronghieu]Consider $ y \equiv 1(\bmod 2^3 )$ then $ 8y \equiv 8(\bmod 2^6 )$. So $ 8y.3^{y - 1} + 3^y + 5\equiv 3^{y + 1} + 5(\bmod 2^6 )$. However, $ \left( {\frac {{ - 5}}{{2^6 }}} \right) = - 1$. That gives contracdition. Thus, $ y \equiv 5(\bmod 2^3 ). (1)$[/quote]

another way to prove this is by lifting the exponent

in fact,we have $v_2(y-1)=2$

[quote=tronghieu]$ 11^{y - 1} - 1 \vdots 41.$ So$ y - 1 \vdots 40. (2)$[/quote]

to conclude this we must prove that $ord_{41}(11)=40$ . Am I missing something? :maybe:


calculus
integration
inequalities
function
real analysis
real analysis solved

Your function is always positive in the intervall so that the integral is larger than the integral form e^2004-1 to e^2004 and it is larger than f(e^2004-1) because the function is increasing between e^2004-1 and e^2004. Because of
(e^2004-2)^3=>(e^2004-1)e^2004 you have to show
(e^2004-2)^1999=>2003^2002 which follows for example from
e^2004-2=>2003^2.

Misha

To Misha123,thank you for your reply.You are right.
Kunny


geometry
3D geometry
tetrahedron
inequalities
email
linear algebra
matrix

Would someone post a solution for #2 please (the tetrahedron one)?

Thanks in advance.

I'll give you a hint: the solution I thought when composing this problem is based on the following idea: there exists a vertex in the tetrahedron such that the 3 sides emerging from it can form a triangle ...

Of course I checked and saw that the contestants gaved other solutions, much different from this one.

well actually there are lots of triangles in a tetrahedron. Are the solutions going to be discussed here?

After a round is finished, you can discuss solutions in these forums. I am very busy, so write-ups of the solutions will lag :(

Six sides of tetrahedron satisfy triangle inequality (i.e. each sides is less than sum of others). It is sufficient condition for constructing convex polygon.

[quote="pascual2004"]Yes, thats the same i did, but i didnt send solution (cant send emails) Myth: did you solve Question 1?[/quote]You should have sent a PM :)

As for question 1, which I am sorry now that I thought about it  :blush:, I am sorry for what I am about to tell you, but as pointed out to me by Myth before the starting of the round as a hunch, and by Harazi during the round confirming, I discovered that the solution I had for this question had a fault, thus it is still an open question.

Jacob Tsimermann send me a nice idea that potentially proves that are an infinity of such composite numbers. Let $p=2^k-3$ be a prime, and let $n=p(p+3)$. Obviously $\sigma(n) = 2(n+1)$ so $n$ is a good number. All that is left is to prove that there is an infinity of such numbers. Unfortunately I know of no reference of this matter, and neither does Jacob.

He did a construction similar to the perfect numbers one! Anyway dont worry Valentin, you are doing a great job! If someone finish this question please let all of us know.

But round's duration is two weeks! I can't believe that you are unable to invent appropriate solutions for pr2 and pr3.

[quote="darij grinberg"]I didn't participate in this MLC round, since I couldn't solve #1 (well, it's clear that all prime numbers are solutions, but there seem to be lots of other solutions, as you noted in the previous posts - will the problem be validated anyway?), #2 made me lots of troubles (do you have a solution without continuity?), and about #3, it was clear to me that any solution I would come up with would inevitably be a mess. Anyway, the new round seems to be nicer!

  Darij[/quote]My solution avoids all continuity arguments, and it is based on the fact I exposed above.

As for no. 1, since obviously no one solved it, it will not be considered. We still await a genius to solve it :)

Thanks Pascual :)

In problem 3, what does those symbols mean: are they matrixs 1x2 or Lagrangia numbers or smth. else?

These are binomial coefficients: $C_n^m=\frac{n!}{m!(n-m)!}$.

For #1, after seeing [url=http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A045768]this[/url] entry, I decided not to pour too much time into it.

For #3, I found two really short solutions, one using finite differences and one using a counting argument.

Nice link, so there is a solution for this problem, or a least a sequence of solutions has been classified :)

I wonder if anyone here knows Benoit Cloitre, maybe Pierre? He might give us a hint.


let ABCD be the convex quadrilateral insribed in the circle (O) .
    let S be a point in ABCD and satisfy  : angle SAD= angle SCD ; SDA=SBC
 the bisector of angle ASB cuts the circle (O) at P;Q respectively .
        prove that : SQ=SP

First, check your problems before sending them. This one contained a typo.
Then, specify the sources of your problem. "Given by a friend" is OK if you really have obtained the problem from a friend; "from journal" is OK if you say what journal it is and, even better, what issue; "from olympiad" if you tell us which olympiad it is from. But not "****".
Finally, doing some search on ML is not a discrace either. Try http://www.mathlinks.ro/Forum/viewtopic.php?t=54036 .

  darij

the problem is really nice...

and i solved it by inversion...

does it help u ,if i write the solution?


Ο αριθμός τηλεφώνου του Γιώργου είναι 324527 (όλα τα νούμερα 6ψήφια).Ποιά είναι η πιθανότητα χτυπώντας τυχαία νούμερα στο τηλέφωνο να χτυπήσουμε ακριβώς τα ίδια ψηφία αλλά να μην βρίσκεται κανένα στη σωστή θέση?(Κανονική απόδειξη παρακαλώ, όχι σκέτους τύπους)

Δεν καταλαβαινω τι εννοεις με το: κανονικη αποδειξη... Η μονη λυση που μπορω να σκευτω ειναι απλη εφαρμογη του βασικου τυπου που δινει τις αναδιαταξεις (μεταθεσεις με απαγορευμενες τις φυσικες θεσεις των στοιχειων) 6 στοιχειων. Η αποδειξη του τυπου αυτου γινεται με την Αρχη Εγγλισμου-Αποκλεισμου. Στη συγκεκριμενη περιπτωση ο τυπος δινει την πιθανοτητα: $ P \equal{} \frac{6!(1 \minus{} \frac{1}{1!} \plus{} \frac{1}{2!} \minus{} \frac{1}{3!} \plus{} \frac{1}{4!} \minus{} \frac{1}{5!} \plus{} \frac{1}{6!})}{9^6} \equal{} \frac{265}{9^6}$

Σωστός ο τύπος ωστόσο ξέχασες οτι ο αριθμός έχει κάποια ψηφία διπλά. Λέγοντας ''απόδειξη'' εννοούσα να βγει η λύση (ή έστω ο τύπος) με καθαρή συνδιαστική. Η αρχή που αναφέρεις έχει κάποια σχέση με τις λεγόμενες διαταραχές.

Ναι εχεις δικιο, απο βιασινη δεν προσεξα οτι εχει 2 "2".

Προσφατα αρχισα να διαβαζω συνδιαστικη και δεν εχω εξασκηθει καθολου με ασκησεις ακομα... βαζω μια λυση και ελπιζω να μην εχω κανει καμια βλακεια:

Στο πληθος 265 συμπεριλαμβανονται και οι αριθμοι που εχουν το προτελευταιο ψηφιο στη θεση του δευτερου, καθος και αυτοι που εχουν το δευτερο ψηφιο τους στη θεση του προ-τελευταιου, το πληθος τους (με βαση τον ιδιο τυπο) ειναι: $ 2(5!(1 \minus{} \frac {1}{1!} \plus{} \frac {1}{2!} \minus{} \frac{1}{3!} \plus{} \frac{1}{4!} \minus{} \frac{1}{5!})) \minus{} 4!(1 \minus{} \frac {1}{1!} \plus{} \frac {1}{2!} \minus{} \frac {1}{3!} \plus{} \frac {1}{4!})$ (λαμβανοντας υπ οψην οτι οι αριθμοι που εχουν τα 2 2αρια στις αντιθετες θεσεις απο τις κανονικες τους, εχουν μετριθει 2 φορες), ομως οι αριθμοι αυτοι πρεπει να αφαιρεθουν απο το συνολικο πληθος. Αρα η πιθανοτητα ειναι:
$ P \equal{} \frac {6!(1 \minus{} \frac {1}{1!} \plus{} \frac {1}{2!} \minus{} \frac {1}{3!} \plus{} \frac {1}{4!} \minus{} \frac {1}{5!} \plus{} \frac {1}{6!}) \minus{} 2(5!(1 \minus{} \frac {1}{1!} \plus{} \frac {1}{2!} \minus{} \frac {1}{3!} \plus{} \frac {1}{4!} \minus{} \frac {1}{5!})) \plus{} 4!(1 \minus{} \frac {1}{1!} \plus{} \frac {1}{2!} \minus{} \frac {1}{3!} \plus{} \frac {1}{4!})}{9^6} \equal{} \frac {186}{9^6}$

Ο τυπος τωρα απ οσο ξερω θεωρειται στοιχειωδης και η αποδειξη του ειναι αρκετα μεγαλη (αν ειναι να βαλουμε και αποδειξη της αρχης του Εγγλισμου-Αποκλεισμου...), οποτε δε νομιζω αν εμπαινε το θεμα αυτο σε καποιο διαγωνισμο να ζητουσαν αποδειξη του τυπου...

Μόνο παρατήρηση ότι είναι $ 10^6$, αφού περιλαμβάνεται και το 0 για ψηφίο.


Suppose $0 \leq x \leq a \leq y \leq b \leq z$

Determine the minimum value of $x+y+z$ and maximum value of $x+y+z$

if $x+a+y+b+z=2004$

[hide]Rewrite this as $x+y+z=2004-a-b$. The maximum is clearly 2004, because $a$ and $b$ are nonnegative and the solution $x=a=y=b=0,\ z=2004$ works.

Since $a\leq y$, $b\leq z$, and $0\leq x$, $x+y+z\geq 2004-x-y-z\Rightarrow x+y+z\geq 1002$. The solution $x=0,\ a=y=b=z=501$ attains the minimum.

Therefore the minimum is 1002 and the maximum is 2004.[/hide]


inequalities
trigonometry
inequalities proposed

1st inequality is equivalent to $ \prod \cos A\leq \prod (1\minus{}\cos A)$, this is well-known.

I know it, [b]N.T.Tuan[/b] . See http://mateforum.ro/viewtopic.php?t=443&highlight=
I posted and it only for helping. The second is the problem !

[quote="Virgil Nicula"]Prove that in any triangle $ ABC$ exists the following inequalites :

$ 1\blacktriangleright$ $ \prod\cos A\le\frac {r^2}{2R^2}\ .$

$ 2\blacktriangleright$ $ \boxed {\sum\cos B\cos C\le\frac {r(R \plus{} r)}{R^2}}\ .$[/quote]
2. $ \boxed {\sum\cos B\cos C \equal{} \frac {p^2 \plus{} r^2 \minus{} 4R^2}{4R^2} \le \frac {r(R \plus{} r)}{R^2}}$
$ \Leftrightarrow p^2 \plus{} r^2 \minus{} 4R^2 \le 4r(R \plus{} r)$
$ \Leftrightarrow p^2 \le 4r^2 \plus{} 4Rr \plus{} 3r^2$
Which is obivious true by Gerretsen inequality!


inequalities
arithmetic sequence
advanced fields
advanced fields unsolved

It might sound stupid, but what's $\mathbf{P}$ ?

[quote="Mindspa"]Let $\mathbf{P}$ be the set of primes, is the set

\[
\{\sin(p) \mid p \in \mathbf{P}\},
\]
dense in $[0,1]$?[/quote]

You mean $[-1,1]$ no ?

[quote="arthur"]excuse that question from a little sophomore, but what is the meaning of "dense" in mathematics ?[/quote]

[joke] A set X is "dense" in mathematics if $\forall\ m\in mathematics$ there exists a sequence in X that converges to m. [/joke]

From topology's point of view, A is dense in X if $\bar{A}=X$ (i.e. the closure of A is X).

You may regard this as: A is dense in X if every element from X is the limit of a sequence from A.

http://les-mathematiques.u-strasbg.fr/phorum/read.php?f=2&i=80992&t=80992#re

You will find some idea to write a proof of this problem, the author on the french
forum said it was a Romanian Olympiad problem.

Wich year ? local ? or national ?

That's not a proof, right? From what I can understand, that looks like an obvious argument, but without the proof. Maybe I don't know where to look? Maybe there are some other pages or something like that?

Yes it only says what you already know, that is :
We have to prove that for each $\alpha \in [-1;1]$ there is a sequence $\{ p_n \}$of primes such that $\alpha$ is the limit of the sequence $sin(p_n)$.
For that you have to look of the primes which are close to $\alpha + kp$ and to use the continuity of sin.

Well, I'm agree with that, but I think anyone has had the same idea. I think you probably have to use some number-theoretic results as Dirichlet's theorem about primes in arithmetic sequence or something like that.
To be continued....

Pierre.

This problem has meet some interest from many people here in France. At last, it has been solved in a french math forum, with an affirmative answer. A solution is available in french here :

http://www.gyptis.org/modules.php?name=Newsportal&op=article&id=26019&group=fr.sci.maths

The solution is not elementary, and uses tools as Fourier analysis, Vinogradov's theorem, Erds-Turan inequality and many others things (that I do not know myself...thus I will not translate it, sorry).

The question is now : Where did you find this problem? Is this some own question from you, from a book, from a competition?
Did someone seen this problem before?

Pierre.

Does any one know the source of this nice problem ?

I think it was problem without any prehistory.

I just thought of it after working with $\{ \sin(n) \mid n \in \mathbf{N}\}$. So there might not be any more elementary solution. Me and my friend thought of something relating $\sin(2^n)$ with Bertrands theorem, or something stronger, but we ran out of time so we didn't prove anything.


It's pretty obvious that the minimum is $n*(\frac{1}{n}+n) = n^{2}+1$, which is obtained by letting $a_{i}= \frac{1}{n}$, $\forall i \in \mathbb{N}_{n}$.

Proof:

The expression $= 1+\Sigma \frac{1}{a_{i}}$.

And, $\Sigma \frac{1}{a_{i}}\geq  \frac{n^{2}}{\Sigma a_{i}}= n^{2}$, by AM-HM.

Thus, the lower bound is $1+n^{2}$.

And since the value of $n^{2}+1$ can be obtained, as demonstrated before, it is in fact the minimum.

That is the minimum of:
$\left(a_{1}+\frac{1}{a_{1}}\right)+\left(a_{2}+\frac{1}{a_{2}}\right)+\dots+\left(a_{n}+\frac{1}{a_{n}}\right)$.

Sorry, I wrote the wrong problem before but now it's correct.Look the problem above!!

It's the same : just use cauchy and the previous inequality and find out that the minimum is $\frac{(n^{2}+1)^{2}}{n}$


How much time do you invest on solving math problems on average everyday?  When did you start doing the contest math problems?

I suggest at least 5-10 hours a week. If you have time.

You should definitely be practicing for AIME more than AMC. I made USAMO last year by taking a few AIMEs, generally with the whole 3 hours, to get the feel for the test. You will probably find yourself very tired 2 hours in, but you should attempt to keep working problems.

I'm on AoPS for something like 5-10 hours a week.  I don't do entire contest sets anymore; I just find hard problems to do on the forums.  It keeps me sharp :)

I do nothing outside of homework for my precalc class at school and the pretty simple problems our math team head gives us to practice for the local math league except until a couple days before the AMC 10/12 where I'll do two or three practice tests and look over solutions. (This year the AMC is on Tuesday and our school gave us Friday and Monday off as well as forbade teachers from assigning homework so I am able to do a couple of tests)

Then since the AIME II is almost immediately after 3 weeks of spring break where teachers are forbidden from assigning homework I can do 2-3 practice AIMEs and look over solutions.

I would do more but I am extremely busy with coursework


Seedings are based on highest rating.
[size=150][b]Bracket[/b][/size]

Quarterfinals:
1 AIME15 vs 8 cgyao15 (QF-1)
7 arb95 vs 2 Mewto55555 (QF-2) Mewto55555 wins 3-1
3 Jongao vs 6 jxl28 (QF-3) Jongao wins 3-1
5 yaofan vs 4 theprodigy (QF-4)

Semifinals:
Winner of QF-1 vs Jongao (SF-1)
Mewto55555 vs Winner of QF-4 (SF-2)

Finals:
Winner of SF-1 vs Winner of SF-2 (F)

3rd Place Match:
Loser of SF-1 vs Loser of SF-2 (3PM)

[quote="jjx1"]umm thats not how a typical bracket would work for the semis. it would be unfair for the 1st and 2nd seed.[/quote]

I don't tell you how to run your tournament; don't tell him how to run his. Whatever floats his boat.

Mewto vs. arb95:

Game 1: 4-1 (Mewto wins)
Game 2: 4-1 (Mewto wins)
Game 3- 2-4 (arb wins)
Game 4: 4-1 (Mewto wins)

oh, sorry  :blush:  yeah, it's best of 5.

theprodigy defeats yaofan, 4-3, 4-1, 4-1.

next match is going to be pretty hard
hope i can win :)

Quarterfinals:
1 AIME15 vs 8 cgyao15 (QF-1) AIME15 wins 3-0
7 arb95 vs 2 Mewto55555 (QF-2) Mewto55555 wins 3-1
3 Jongao vs 6 jxl28 (QF-3) Jongao wins 3-1
5 yaofan vs 4 theprodigy (QF-4) theprodigy wins 3-0

Semifinals:
AIME15 vs Jongao (SF-1)
Mewto55555 vs theprodigy (SF-2)

Finals:
Winner of SF-1 vs Winner of SF-2 (F)

3rd Place Match:
Loser of SF-1 vs Loser of SF-2 (3PM)

Places:
1st: ?
2nd: ?
3rd: ?
4th: ?
5th: yaofan
5th: jxl28
5th: arb95
5th: cgyao15

Quarterfinals:
1 AIME15 vs 8 cgyao15 (QF-1) AIME15 wins 3-0
7 arb95 vs 2 Mewto55555 (QF-2) Mewto55555 wins 3-1
3 Jongao vs 6 jxl28 (QF-3) Jongao wins 3-1
5 yaofan vs 4 theprodigy (QF-4) theprodigy wins 3-0

Semifinals:
AIME15 vs Jongao (SF-1) AIME15 wins 3-2
Mewto55555 vs theprodigy (SF-2)

Finals:
AIME15 vs Winner of SF-2 (F)

3rd Place Match:
Jongao vs Loser of SF-2 (3PM)

Places:
1st: ?
2nd: ?
3rd: ?
4th: ?
5th: yaofan
5th: jxl28
5th: arb95
5th: cgyao15

Theprodigy has informed me that Mewto55555 won't be able to play for a week, so theprodigy will win his match by default.

1 AIME15 vs 8 cgyao15 (QF-1) AIME15 wins 3-0
7 arb95 vs 2 Mewto55555 (QF-2) Mewto55555 wins 3-1
3 Jongao vs 6 jxl28 (QF-3) Jongao wins 3-1
5 yaofan vs 4 theprodigy (QF-4) theprodigy wins 3-0

Semifinals:
AIME15 vs Jongao (SF-1) AIME15 wins 3-2
Mewto55555 vs theprodigy (SF-2) theprodigy wins by default

Finals:
AIME15 vs theprodigy (F)

3rd Place Match:
Jongao vs Mewto55555 (3PM)

oh...what about that pm?

meh... i'll lose to u or AIME15 anyway

Deadline is Friday at 6:00 pm central time.

Deadline has passed, so this tournament is pretty much abandoned.


vector
abstract algebra
Ring Theory
linear algebra
linear algebra unsolved

Assume that $T\in\mathcal M_{n}(\mathbb C)$. Prove that if for each $v\in\mathbb C^{n}$ vectors $v,Tv,T^{2}v,\dots,T^{n}v$ are linearly independent then $I,T,T^{2},\dots,T^{n}$ are linearly dependent.

oh,It is obviouse with rational canonical form.(Every F[x] module defined on PID is direct sum of cyclic modules)


A number cube, with the integers 1-6 on its faces, is rolled. The result is used as the tens digit. The cube is rolled again, and the second result is used as the units digit. What is the probability that the two-digit number formed is prime? Express your answer as a common fraction.

In order to be prime, the units digit can only be $ 1$ or $ 3$, so we find the possible primes to be $ 11,13,23,31,41,43,53,61$, and the probability is $ \frac8{6\cdot6}\equal{}\boxed{\frac29}$.


Most ppl here have a lot of experience with math team.  I'm sure of that.  So as a sophomore in high school does anyone have any suggestions on how I could encourage more people to join math team?  Our program could use more ppl.  I know of stuff like telling ppl to join and stuff like that but I was wondering if anyone had any better advice.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11977

Take a look at that, it talks a little about how to get a math team started.  As far as motivating people to come... I think if they're interested, they'll come; you can't really force someone to like math.  Try enticing them with free food ;)

Offer food. Preferably donuts. Where there is food - people will come.
Second to that, try to make it on par with the school's sports. Ask your principal if they can offer a letter and a jacket to members of the math team who participate in a certain number of events. You can call it the Varsity Math Team.

Do most people's math teams have jackets?  I go to an all guy school, so there's no chance that math would be held on the same level as sports, but we're supposedly getting jackets if we win the NEAML next week.

AAST doesn't have jackets, but we do get t-shirts for ARML and HMMT.

Yeah if there's a reward for winning at contests, some people will be encouraged to work on them.  Unfortunately, most math contests just give a little plaque/certificate :-(.

We get sweaters with skulls on the back. I don't know what that has to do with Skyline Eagles.

I definitely agree.  There is little motivation for most people to do math contests or practice for them.

[quote="ComplexZeta"]I think a more annoying problem is that a lot of people won't work for math contests because they assume that everyone else is better, so they won't have any chance to win anything.[/quote]

Yes, I think the lack of "team spirit" is a problem in forming a math team. Here, I have a pretty good group of "ringers" in the form of a homeschooling support group for gifted children, so I can recruit a math team from a bunch of people who have a good degree of "natural" talent to make up for my pathetically bad coaching. But, sure enough, as it has become apparent which kids tend to have the highest scores in most math contests, the kids who do well, but not at the top level, tend to lose interest in being on the math team. People who play basketball, baseball, football, etc. understand better than many mathletes that some people have to show up for a team event just to fill out the team--not everyone can be a star all the time.

I'm glad that [url=http://members.aol.com/mathleague]Minnesota Junior High Mathematics League[/url] and [url=http://www.mathcounts.org]Mathcounts[/url] both have team scoring for some events, so that there is an incentive to recruit a whole team. But I wish that gathering homeschoolers were not so much like the "herding cats" that it is often analogized to. Forming a team to begin with is hard, finding meeting space is REALLY hard, and then keeping good second-stringers on the team is also very hard. I'm gonna have to buy a lot of donuts next year.

the math team at my school is really really small.  We only do one competition though.  We have like 10 people and several of them were forced to be on our team so they don't try at all.  Our team seems to have grown a little this year though we got 5 new people on it but half of them will quit anyway.

[quote="ComplexZeta"]I think a more annoying problem is that a lot of people won't work for math contests because they assume that everyone else is better, so they won't have any chance to win anything.[/quote]

I feel exactly that way, but I love math, so I think that the actual contests and preparing for them are fun.  And though I doubt I'll get first, it's a good feeling not to get last.  I always work as hard as I can at them, but without the expectation of winning.

Hey rep watch it.  U know I can't get a 50 on the test most of the time.

anyhow I'm gonna tell my coach about some of these ideas to get people interested in the team.

[quote="joml88"]Yeah my math team is pretty pathetic.  We only had 10 ppl who went to practice all year.  At a few of the competitions we only had a JV team and when we did have a Varsity team they didn't come close to placing.  The upside to this all was that the JV was dominating.  We got first in everything we competed in(that had a JV division) and I'm proud to say I was a part of the JV team all year.[/quote]

:) Nice job.

When you say "JV" or "varsity" in relation to a math team - do you just mean, for competitions like the Mandelbrot where there are Regional and National divisions?

umm... I mean for competitions like ones at nearby universities not ones that are national.  So we beat all the ppl in our area.  There may not be much competion in South Carolina(compared to other states) but it still is something to brag about.


So, I was looking around and a proof of Lucas' Theorem was nowhere to be found on this site.

Suppose that in base $p$, $n = n_{0}+n_{1}p+n_{2}p^{2}+\cdots+n_{k}p^{k}$ and $a = a_{0}+a_{1}p+a_{2}p^{2}+\cdots+a_{k}p^{k}$. Show that

$\binom{n}{a}= \prod_{i=0}^{k}\binom{n_{i}}{a_{i}}\pmod{p}$.

[hide="hint"] Consider the coefficient of $x^{m}$ in

$(1+x)^{p}= \prod_{i=0}^{k}\left((1+x)^{p^{i}}\right)^{n_{i}}$.[/hide]

EDIT: Birch & Swinnerton Dyer!

Bleh, I wrote a proof of this out two weeks ago on some sheet of paper...

I'm too lazy to get it

I think the only hard part is showing that $x^{k}$ has unique representation in the expansion of $(1+x)^{a_{0}}(1+x^{p})^{a_{1}}\ldots$.

[quote="paladin8"][hide="hint"] Consider the coefficient of $x^{m}$ in

$(1+x)^{p}= \prod_{i=0}^{k}\left((1+x)^{p^{i}}\right)^{n_{i}}$.[/hide][/quote]

Come on, this practically gives it away  :wink:

[hide]
If you say the hint gives it away, then I say the problem gives the solution away  :P any experienced solver (who would be able to apply hint) would realize that binomial is the only way to go on this problem (only nice, elementary way)[/hide]

[hide="combinatorial proof!"]
It's equivalent to proving that
\[\binom{qp+r}{xp+y}\equiv \binom{q}{x}\binom{r}{y}\pmod{p}\]
where $r,y<p$. We can create a $p$ row, $q$ column grid along with $r$ other squares for $qp+r$ total, and we need to consider the number of ways to fill in $xp+y$ of them. But $\binom{q}{x}\binom{r}{y}$ is the number of ways of choosing the squares such that the filled squares in the grid form complete columns. This is so because the number of squares filled in among the $r$ not in the grid is congruent to $y$ modulo $p$ but also less than or equal to $r$, which is less than $p$. So there are exactly $y$ filled that are not in the grid, and $xp$ squares ($x$ columns) in the grid.

Now we only need to show that the other combinations of filling squares can be placed into groups of $p$ combinations each. But there is a very simple way to do this; if at least one column in a grid is not completely empty or completely full, moving the bottom row to the top and shifting the rest down $p-1$ times (while not changing the other $r$ squares) will result in $p$ different arrangements.
[/hide]


modular arithmetic
Diophantine equation
number theory
relatively prime
number theory proposed

What is the smallest integer $n$, greater than one, for whice the root-mean-square of the first $n$ positive integers is an integer?

[hide="note"]
The root-mean-square of $n$  numbers $a_1,a_2,...,a_n$ is defined as following:
$\sqrt{\frac{a_1^2+a_2^2+...+a_n^2}{n}}$[/hide]

It equavalent to solve Pell equation:
$\frac{1+2^2+\dots +x^2}{x}=y^2$.
(1) $z^2-3t^2=-7,t=4y, z=4x+3$.
First solution is z=7,t=4(x=y=1). All solutions given by:
$z+t\sqrt 3=(7+4\sqrt 3)(2+\sqrt 3)^k$, or $z_0=7,t_0=4,z_{k+1}=2z_k+3t_k,t_{k+1}=2t_k+z_k$. But if k is odd, then it give not solution. $(2+\sqrt 3)^2=7+4\sqrt 3$, therefore if $4\not|k \ z\not=3(mod \ 4)$. Therefore general solution given by:
$x_0=1,y_0=1,x_{k+1}=97x_k+168y_k+72,y_{k+1}=56x_k+97y_k+42$.

[hide="solution"]
it is known that $1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$
thus we are looking for $n>1$ satisfying:
$(n+1)(2n+1)=6m^2$
but $6|(n+1)(2n+1) iff n\equiv 5$ or $1\pmod{6}$

Case 1:
$n=6k+5$
$\Rightarrow m^2=(k+1)(12k+11)$ since the two factorsa re relatively prime each must be a square: $(k+1)=a^2, (12k+11)=b^2\Rightarrow 12a^2=b^2+1$ the latter equation is impossible modulo $4$ since $b^2+1\equiv 1,2 \pmod {4}$

Case 2:
$n=6k+1$
$\Rightarrow m^2=(3k+1)(4k+1)\Rightarrow (3k+1)=a^2, (4k+1)=b^2\Rightarrow (2a-1)(2a+1)=3b^2$
any prime factor of the RHS except 3 must have an even pwer.  trying successive powers of $a$  starting with $1$ such that $2a\pm 1$ is not a prime other than $3$;
$a=1\Rightarrow b=1\Rightarrow n=1$ not good since $n>1$
$a=13\Rightarrow n=337$  as deseired [/hide]


A person cashed a cheque for $ 200$ pounds, asking for some one-pound coins, $ 10$ times as many two-pound coins and the balance in five-pound notes. How many of each did the person get?

I assume some one-pound coins means a number $ >1$.

Let $ a$ be the number of one-pound coins, $ b$ be the number of two-pound coins, and $ c$ be the number of five-pound notes.

Then,
$ b\equal{}10*a$ and $ c\equal{}\frac{200\minus{}2*10*a\minus{}a}{5}$

Thus, $ a$ must be a multiple of $ 5$.

If $ a\equal{}5$, then $ b\equal{}50$ and $ c\equal{}19$.
This is the only solution, as $ a<10$.


Putnam
modular arithmetic
number theory
Divisibility Theory

Let $p$ be the smallest prime divisor of $n$ and let $k$ be the order of $n$ mod $p$. Then of course: $k>1$, $k|n$, $k|p-1$ which implies that $n$ has a prime divisor smaller than $p$, contradiction.

Solution by Christian Hirsch (similar, but still different):

Assume that $n$ is the smallest number $>1$ with that property. Then let $k = \gcd(n,\varphi(n)) <n$ and see that $2^k \equiv 1 \mod n$ (by $2^n , 2^{\varphi(n)} \equiv 1 \mod n$), thus $k|2^k-1$.
By minimality we have $k=1$, giving $2 \equiv 1 \mod n$, contradiction.

We apply Fermat’s method of infinite descent to the prime factors of n . Let p1 be prime divisor of n and q be the smallest positive integer for which p1 divides 2q-1. From Fermat’s Little theorem it follows that p1 divides 2p1-1-1. Also we have q< p1-1<p
Let us prove that q divides n. If not let n=kq+r, where 0<r<q. Then 2n-1=2kq2r-1=(2q-1+1)2r-1 which is obviously equal to 2r-1 modulo p1.
It follows that p1 divides 2r-1, contradicting the minimality of q. Hence q divides n and 2q<p1. Let p2 be a prime divisor of q. Then it is also a divisor of n and hence p2<p1. Repeating the argument we construct an infinite sequence of prime divisors.
Hence the proof follows.
This problem is from the 1st Putnam in 1939 most probably!

Let $ p$ be the smallest prime divisor of $ n$. So we have $ 2^n\equiv_p1$, which means that $ ord_p2|n$, but by the Fermat's Little Theorem, $ \mbox{ord}_p2\le p \minus{} 1$, so $ \mbox{ord}_p2 \equal{} 1$, so $ 2\equiv_p1$, which is impossible.

[quote="Peter"]Let $n$ be an integer with $n \ge 2$. Show that $n$ does not divide $2^{n}-1$.[/quote]
Assume that $n|2^n-1$. Let $p$ be the smallest prime factor of $n$. Therefore $2^n-1 \equiv 0 \pmod{p}$. By Fermat's Little Theorem we have $2^{p-1} \equiv 1 \pmod{p}$. Thus, $\text{ord}_p(2)|n$ and $\text{ord}_p(2)|p-1$.
It follows that $p>\text{ord}_p(2)$.
If $\gcd ( \text{ord}_p(2),n)>1$ then $r|n, r| \text{ord}_p(2)$. It follows that $r|n$ and $p>\text{ord}_p(2)>r$, a contradiction since $p$ is the smallest prime factor of $n$. Thus $\gcd (n, \text{ord}_p(2))=1$. But $\text{ord}_p(2)|n$ then $\text{ord}_p(2)=1$. Hence $p|2-1$ or $p|1$, a contradiction.
Thus, $n \nmid 2^n-1$ with $n \ge 2$.  :idea:

I apologize to everyone for my solution being very similar to that of [b]ZetaX[/b]. :D

We suppose that there is $n\ge 2$ s.t. $n | 2^n-1$.We consider the minimum such $n$.Obviously $n$ is odd.Then by Euler's theorem,$2^{\varphi (n)}\equiv 1$ mod $n$.
Let the order of $2$ in mod $n$ be $k$.$k$ is common divisor of $n$ and $\varphi (n)$.Thus $2^k\equiv 1$ mod $k$.Since $k\le \varphi (n)<n$ and minimality of $n$,then $k=1$.Hence $2\equiv 1$ mod $n$ which is absurd.Therefore $n\nmid 2^n-1(n\ge 2)$.$\blacksquare$ :coolspeak:

[quote=Peter]Let $n$ be an integer with $n \ge 2$. Show that $n$ does not divide $2^{n}-1$.[/quote]

Assume that,p is the least prime factor of $2^n-1$.p here can't be equal to 2 since $2^n-1$ is a odd number.so,$p>2$

by FLT,$2^{p-1} \equiv 1(modp)$

By fundamental theorem of order,we can say,$Ord_{p}{2}\mid n$

On the other hand,$Ord_{p}{2}\leq {p-1}$ which contradicts our assumption that,p is the least prime factor of n. So,there does'nt exist such n.
Now we are done.


a, b, c,  and d are positive numbers.

Prove that (a+c)/(a+d) +(b+d)/(b+c) +(c+a)/(c+d) +(d+b)/(d+a) >= 4

I assume you mean that the denominator of the first term is a+b.

=========
Ineq is:

$(a+c)[ \frac{1}{a+b} + \frac{1}{c+d}] + (b+d)[\frac{1}{b+c} + \frac{1}{d+a}]$ (*)

We have $\frac{1}{x} + \frac{1}{y} \ge \frac{4}{x+y}$ (**)

So from (**), the LHS of (*) is $\ge \frac{4(a+c) + (b+d)}{a+b+c+d} = 4$ as desired.

$\square$


logarithms
induction
modular arithmetic
ceiling function
floor function
pigeonhole principle
inequalities

For $m$ a positive integer, let $s(m)$ be the sum of the digits of $m$. For $n\ge 2$, let $f(n)$ be the minimal $k$ for which there  exists a set $S$ of $n$ positive integers such that  $s\left(\sum_{x\in X} x\right)=k$ for any nonempty subset $X\subset S$.  Prove that there are constants $0<C_1<C_2$ with
\[C_1 \log_{10} n \le f(n) \le C_2 \log_{10} n.\]

Who is the author of this problem? I have one very strong supposition...

I am not sure what the AMC shortcuts as "AND" but it might be Titu Andreescu ...  :?  :?

I have another opinion. But I don't want to reveal it till the author will confirm I am right.

[quote="harazi"]You are right.  ;)[/quote]You??  :D  :D

[quote="harazi"]No, Titu Andreescu and me. :P[/quote]You assured yourself of some cursing :)

[quote="harazi"]You are right.  ;)[/quote]

Wow, you  and Titu proposed this problem? From the people I talked to, many people thought it was a hard, but good problem :D

I am glad to hear that, but I would prefer to see a nice solution to this problem.

Is there something unsatisfactory to the solution on the UNL website?

I think a lot of people got the RHS looking at sets consisting of all the same integer = 10^k - 1 for whatever k.

Did I say there is something unsatisfactory about the solution presented there? I just wanted to see many solutions.  :( . Indeed, the part with the construction was not difficult, the trick with $10^k-1$ being quite standard. But the other part was not easy (I think), even if it also used only standard arguments and lemmas. The main difficulty is to put everything together, I thinK.

[quote="darktreb"]Is there something unsatisfactory to the solution on the UNL website?[/quote]I belive it is the author's (ie Harazi's) solution :D

No, Titu has found this final version of the solution, a little bit easier than my initial solution, but on the same way. I think that when we created the problem, we found nice bounds: the smallest constant can be taken 8 and the largest one 19 or something like this.

[quote]But the other part was not easy (I think), even if it also used only standard arguments and lemmas. The main difficulty is to put everything together, I thinK.[/quote]

I agree completely. It was a beautiful problem in that way. I had Lemma 1 of the "standard solution" (the one posted on the website) with at least 2 1/2 hours to go. Putting everything together really didn't work out for me at all. It was a great problem, and definitely very difficult.

So...any solutions that woul denlighten me  :)?

Masoud Zargar

[quote="Zhero"]...
Letting $C_1 = 9 - \frac{18}{\log_{10} 2} > 0$ gives $(9-C_1) \log_{10} n \geq \frac{18}{\log_{10} 2} \log_{10} 2 \geq 18$, so $f(n) \geq 9 \log_{10} n - 18 \geq C_1 \log_{10} n$, which completes our proof.[/quote]
[hide="Unfortunately, however:"]
[url]http://www.wolframalpha.com/input/?i=9-18%2Flog_10%20%282%29&t=ff3tb01[/url]
$9 - \frac{18}{\log_{10} 2} \le -50.7947$...[/hide]
[hide="easy fix though"]
First, instead of $j=\lfloor \log_{10} n \rfloor - 1$ you can take $j = \lfloor \log_{10} (n+1) \rfloor$. Then you get $k\ge 9 \log_{10}\frac{(n+1)}{10}\ge(9-\frac{9}{\log_{10}11} )  \log_{10}n$ for $n>10$. Then let the maximum value of $k$ when $1<n<11$ be $t$. Taking $C_1=\min((9-\frac{9}{\log_{10}11} ), t)>0$ works since for $n\le 10$, $C_1\log_{10}n\le t$ and for $n>10$, we have what we had above.
So there exist such constants $C_1, C_2$.

EDIT2: Yep, this should work.

[/hide]
By the way, very cool solution. I got the first lemma part but couldn't think to use an inequality variant of it for the second part.

Let $\Sigma(X)$ denote the sum of the elements in a set $X$.

Obviously $f$ is nondecreasing. Therefore, to prove the upper bound it suffices to show that $f(10^k) \leq 2k$ for all $k$. We first have the following lemma.

[b][color=#00f]Lemma:[/color][/b] Let $e$ be a positive integer. Then for any $1 \leq n<10^e$, $s(n(10^e-1))=9e$.
[color=#00f]Proof:[/color] Motivated by how "subtraction by hand" is done, rewrite the number as $(n-1)10^e+(10^e-n)$ and note that
$$s(n(10^e-1))=s(n-1)+s(10^e-n)=s(n-1)+s((10^e-1)-(n-1)).$$
Since $n-1$ and $(10^e-1)-(n-1)$ sum to $10^e-1$ without any carries, the conclusion is clear. $\blacksquare$

I claim that the set $S=\{1(10^{2k}-1),\ldots,(10^k)(10^{2k}-1)\}$ works. Indeed, the sum of any of its nonempty subsets is of the form $n(10^{2k}-1)$ where $n \leq \frac{10^k(10^k+1)}{2}<10^{2k}$, hence we are done by the above lemma.

To prove the lower bound, I will prove the more general claim that for any nonnegative integer $N$ and any multiset $T$ containing at least $1000^k$ positive integers, there exists some $B \subseteq T$ such that $N+\Sigma(B)$ contains at least $k$ nonzero digits. This is by induction on $k$, with the base case of $k=1$ being obvious.

For the inductive step, we may WLOG assume that either $N$ or at least one of the elements of $T$ is not divisible by $10$. If $10 \mid N$, then remove some element $e \in T$ with $10 \nmid e$, and add it to $N$; this only makes our task harder. Therefore assume that $10 \nmid N$ and $|T| \geq 1000^k-1$. Then split the elements of $T$ into as many "blocks" of size $10$ as possible, where elements within each block are congruent modulo $10$. At most $90$ elements are not part of a block, so we have at least $\frac{1000^k-91}{10}$ such blocks. Then let $T'$ be the multiset of size at least $\frac{1000^k-91}{10}$ formed by taking the sum of each of the blocks, so every element in $T'$ is divisible by $10$, and let $N'$ be the result of replacing the units digit of $N$ with a zero. Since $\frac{1000^k-91}{10} \geq 1000^{k-1}$, it follows by inductive hypothesis that there exists some $B' \subseteq T$ such that $N'+\Sigma(B')$ has at least $k-1$ nonzero digits. Since this number is always divisible by $10$, it follows that $N+\Sigma(B')$ has at least $k$ nonzero digits, so by letting $B$ be the multiset of elements in the blocks chosen to form $B'$ we are done.

This proves the lower bound due to the non-decreasingness of $f$, since by taking $N=0$ we have $f(1000^k) \geq k$. $\blacksquare$

[b]Remark:[/b] first post in this thread since 11 years lmao. not as slick of a proof of the lower bound as the above but I think this works too. there are a lot of ideas for the lower bound that don't work but I think the ideas vaguely got refined into what I have here. in particular it would be nice to induct down somehow (and there are lots of possible setups but I think most straightforward approaches don't work), and one possible way to do this would be to find some number not divisible by $10$ and then force yourself to include it, and then try not to disturb the units digit as you choose the rest of the subsets, but it took me a while to realize that this generalization was possible since there are so many other possibilities. also doing math at midnight is not efficient

First, notice that $s(n) \equiv n \pmod{9}$, so all elements of $X$ are congruent to $k$ modulo $9$. Since the sum of any two elements of $X$ is also congruent to $k$ modulo $9$, we must have $9 \mid k$. Let a set be [i]$c$-regular[/i] if it satisfies the problem conditions for $k=9c$. It suffices to prove that
[list]
[*] there exists a $c$-regular set of size at least an exponential function of $c$
[*] all $c$-regular sets have size at most an exponential function of $c$.
[/list]

For the construction, consider $X=\{10^c-1,2(10^c-1),\ldots,a(10^c-1)\}$ for $a$ with $\tfrac{a(a+1)}{2} \le 10^c$. All subsets of $X$ have elements that sum to a multiple of $10^c-1$ no greater than $10^c(10^c-1)$. For a positive integer $b \le 10^c$, the digits of $b(10^c-1)$ consist of the concatenation of the digits of $b-1$ and the digits of $10^c-b$, possibly with leading zeros to make a $c$-digit integer. Since $(b-1)+(10^c-b)=10^c-1$ and  both $b-1$ and $10^c-b$ are nonnegative, the corresponding digits of $b-1$ and $10^c-b$ sum to $9$, so the digit sums of $b-1$ and $10^c-b$ add up to $9c$. Thus, $X$ is $c$-regular. We can set $a=2^c$ to give a $c$-regular set with exponential size, as desired.

For the upper bound, we claim that any set $X$ whose subsets of size $1$, $|X|-1$, and $|X|$ contain elements summing to numbers with a digit sum $9c$ has at most exponential size. Let $X=\{a_1,\ldots,a_n\}$, and let $s=a_1+\cdots+a_n$. The idea is to limit the number of digit places that can hold nonzero digits. Suppose that $s$ has nonzero digits on the $10^{b_1},10^{b_2},\ldots,10^{b_m}$ digit places, where $b_1<\cdots<b_m$. Notice that $m \le 9c$. We can check that $s(a-b)=s(a)-s(b)+9d$, where $d$ is the number of carries when doing standard decimal subtraction. Thus, the number of carries when subtracting $s-a_i$ is $c$ for $i=1,\ldots,n$. Fix $i$, and let $c_j$ be
[list]
[*] the largest nonnegative integer less than $b_j-b_{j-1}$ (where $b_0=-1$) such that the $10^{b_j-c_j}$ digit place of $a_i$ is nonzero, if there exists one
[*] $0$ otherwise
[/list]
for $j=1,\ldots,m$. The crux of the problem is the following.

[b][color=#07f]Claim:[/color][/b] $c_1+\cdots+c_m \le c$.

[i]Proof:[/i] There will be carries on the \[10^{b_1-c_1},10^{b_1-c_1+1},\ldots,10^{b_1-1},10^{b_2-c_2},10^{b_2-c_2+1},\ldots,10^{b_2-1},\ldots,10^{b_m-c_m},10^{b_m-c_m+1},\ldots,10^{b_m-1}\] digit places when subtracting $s-b_j$. This gives $c_1+\cdots+c_m$ carries, so $c_1+\cdots+c_m \le c$. $\square$

By stars and bars, there are at most $\tbinom{c+m}{c} \le \tbinom{10c}{c}$ ways to choose $c_1,\ldots,c_m$. After choosing $c_1,\ldots,c_m$, we have fixed $m+c_1+\cdots+c_m \le 10c$ digit places as the only possible nonzero digit places of $a_i$ with nonzero digits. Thus, there are at most $10^{10c}$ choices of $a_i$ given $c_1,\ldots,c_m$. We conclude that there are at most \[\tbinom{10c}{c}10^{10c} \le \frac{(10c)^c}{c!}10^{10c} \sim \frac{10^cc^c}{\frac{c^c}{e^c}\sqrt{2\pi c}}10^{10c} \le (10^{11}e)^c\] possible values of $a_i$, where the second last equivalence is by Stirling. Thus, $n$ is bounded above by an exponential function of $c$, as desired. $\square$


Prove that
$\frac{gcd(m,n)}{n}\binom{n}{m} \in Z^+$ for all $n\geq m\in Z^+$

Well, $\gcd{(m,n)}$ is a linear combination of $m$ and $n$, and $\frac{m}{n} {n \choose m} = {n - 1 \choose m - 1}$ I think, so the given number is some linear combination of ${n - 1 \choose m - 1}$ and ${n \choose m}$ and hence an integer.


Day 1
http://www.mathlinks.ro/Forum/viewtopic.php?t=161570
http://www.mathlinks.ro/Forum/viewtopic.php?t=161564
http://www.mathlinks.ro/Forum/viewtopic.php?t=161571
http://www.mathlinks.ro/Forum/viewtopic.php?t=161572

Day 2
http://www.mathlinks.ro/Forum/viewtopic.php?t=161574
http://www.mathlinks.ro/Forum/viewtopic.php?t=161565
http://www.mathlinks.ro/Forum/viewtopic.php?t=161573
http://www.mathlinks.ro/Forum/viewtopic.php?t=161575


If f(x) is a monic quartic polynomial such that f(-1) = -1, f(2) = -4, f(-3) = -9, and
f(4) = -6, nd f(1).

$ f(x)\equal{}x^4\plus{}ax^3\plus{}bx^2\plus{}cx\plus{}d$.  Plug in the given information and solve the resulting system of equations:

$ \minus{}1 \equal{} 1\minus{}a\plus{}b\minus{}c\plus{}d$
$ \minus{}4 \equal{} 16\plus{}8a\plus{}4b\plus{}2c\plus{}d$
$ \minus{}9 \equal{} 81\minus{}27a\plus{}9b\minus{}3c\plus{}d$
$ \minus{}6 \equal{} 256 \plus{} 64a\plus{}16b\plus{}4c\plus{}d$

The solution is $ a\equal{}\minus{}13/7$, $ b\equal{}\minus{}96/7$, $ c\equal{}93/7$, and $ d\equal{}162/7$, so $ f(1)\equal{}1\plus{}a\plus{}b\plus{}c\plus{}d\equal{}153/7$

perhaps less computation? :
let $ g(x)\equal{}\minus{}f(x)\minus{}x^2$
we then have $ g(\minus{}1)\equal{}g(2)\equal{}g(\minus{}3)\equal{}0$ (these are 3 roots), $ g(4)\equal{}\minus{}10$, and the leading coefficient is -1
therefore, g(x) is in the form $ g(x)\equal{}\minus{}1(x\plus{}1)(x\minus{}2)(x\plus{}3)(x\minus{}n)$ and $ g(4)\equal{}\minus{}10$
we plug in 4 for x: $ g(4)\equal{}\minus{}1(5)(2)(7)(4\minus{}n)\equal{}\minus{}10 \implies \minus{}70(4\minus{}n)\equal{}\minus{}10 \implies 4\minus{}n\equal{}\frac17 \implies n\equal{}\frac{27}{7}$
so $ g(x)$ is $ \minus{}1(x\plus{}1)(x\minus{}2)(x\plus{}3)(x\minus{}\frac{27}{7})$
we know $ g(1)\equal{}\minus{}f(1)\minus{}1^2 \implies f(1)\equal{}\minus{}g(1)\minus{}1$ so we just evaluate $ \minus{}g(1)\minus{}1$ and that will be our solution.
$ \minus{}g(1)\minus{}1\equal{}\minus{}[\minus{}1(1\plus{}1)(1\minus{}2)(1\plus{}3)(1\minus{}\frac{27}{7})]\minus{}1\equal{}(2)(\minus{}1)(4)(\minus{}\frac{20}{7})\minus{}1\equal{}\frac{160}{7}\minus{}1\equal{}\frac{153}{7}$


Se considera o sfera de raza R si centru O. Consideram trei puncte pe sfera care sa formeze un triunghi echilateral de latura R. Fie OO' proectia lui O peplanul triunghiului. Sa se afle $\frac{OO'}{R}$.

pai $OABC$ e tetraedru regulat si $OO'$ e inaltimea...iese repede.

ar trebui sa se mai dea ceva.altfel $OO'/R$ e variabil


I am starting a thread about interesting facts about AoPS. When posting in this thread, please number your fact so that we can refer to it easily. Some posts will need to be updated as more users join and your number may become off target.  If you think that someone is not planning to edit their post on a fact, you can redo that fact. I hope this will be fun!

1. AoPS has 12620 users, 4811 of whom have posted
2. Only 38.12% of AoPS users have posted
3. Only 524 users of AoPS have posted 100+ times
4. Only 4.15% of all users of AoPS have posted 100+ times
5. Only 137 (1.09%) of all users of AoPS have posted 500+ times

So few have actually posted?  ...why?
Hmm...here's one you didn't mention...
6. AoPS is a forum.

It's like that in most forums.  Usually 10% percent of all the people actually do anything with AoPS.  And there are some with no posts that still participate in all the AoPS stuff, like the Mock AMC's, and classes.

7. More than 300,000 articles have been posted (319475 articles as of Oct 11, 3:40) :D

8. Over 5 people on AoPS like the game mafia.

http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=2064

http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=2067

some info  :D

Thats not a fact.  I refute question 10.  In its place, I will post a more appropriate statement.

10. AOPS is the best forum ever  :D

11. AoPS has online classes.
12. AoPS is a forum designed for math.

How do we know AoPS was originally designed for math?
13. AoPS exists.
14. AoPS = AoPS = Art of Problem Solving.
15. The AoPS database contains more than 1 byte of data. (I think)
16. I visited AoPS today

17. there are 103 aopsers with a username that startes with something other than a letter (numbers, symbols

whoops :blush:

[quote="aznmonkey1992"]13. there are 103 aopsers with a username that startes with something other than a letter (numbers, symbols[/quote]
Wow, that's way more than I thought. Interesting.

BTW your fact should be #17.

umm, could we have a few real facts here?  Not obvious facts.  Real ones like aznmonkeys.

jhollen I don't think "Only" should be included in numbers 2-5, it downplays the facts that we AOPSers should be proud of (what other math site has anywhere close to this level of activity?).

18. 1027 AOPSers that usernames start with the letter A (including me!)  :D

OmG, I LyKe nEvEr nEw LyKe hAlF tHe StUfF iN hErE!!!!!!!!!!!!!!!!!!!!

22

the 34th highest poster is JLI?!?!

as of today

24. 650 Aopsers that start with letter B
25. 744 start with C
26. 736 start with D

27. This page had 48507 characters in the html before I posted it


Find all pairs of positive integers $ m$ and $ n$ that satisfy (both) following conditions:

(i) $ m^{2}\minus{}n$ divides $ m\plus{}n^{2}$

(ii) $ n^{2}\minus{}m$ divides $ n\plus{}m^{2}$

Here is my solution :
1)First assume that $ m\geq n$ and $ m\geq 4$ then we have :
$ m^{2} \geq n^{2}$ and $ m(m \minus{} 1) > 2n$ because $ m\geq n$ and $ m \minus{} 1\geq 3$
so we have $ m^{2} \plus{} m(m \minus{} 1) > n^{2} \plus{} 2n \longrightarrow 2(m^2 \minus{} n) > n^{2} \plus{} m$ but $ n^{2} \plus{} m|m^{2} \minus{} n$ :
$ n^{2} \plus{} m \equal{} m^{2} \minus{} n$ which is equilvalent to $ (m \plus{} n)(m \minus{} n \minus{} 1) \equal{} 0$ but $ m,n > 0$ and we get $ m \equal{} n \plus{} 1$ .
now put this in the second condition we get : $ n^{2} \minus{} n \minus{} 1|n^{2} \plus{} 3n \plus{} 1$ and $ n^{2} \minus{} n \minus{} 1|n^{2} \minus{} n \minus{} 1$ and then :
$ n^{2} \minus{} n \minus{} 1|4n \plus{} 2$ but $ n^{2} \minus{} n \minus{} 1$ is odd , so we have : $ n^{2} \minus{} n \minus{} 1|2n \plus{} 1$
$ n^{2} \minus{} n \minus{} 1\leq 2n \plus{} 1 \longrightarrow n\leq 3$ but $ m \equal{} n \plus{} 1$ and $ m\geq 4$ so we have $ m \equal{} 4$ , $ n \equal{} 3$ . but it isn't a solution.
2)$ m,n\leq3$ : we check all cases and we have this solution :
$ (m,n) \equal{} (1,1),(1,2),(2,1),(3,2),(2,3),(2,2),(3,3)$

Be careful with this $ (m,n)\equal{}(1,1)$  :) It can not be accepted as solution.

If $m=n$:
$m^2-m\mid m^2+m \Rightarrow m-1\mid m+1 \Rightarrow m-1\mid 2 \Rightarrow m=2$ or $m=3. $
Suppose $m=n+k$, where $k\in\mathbb{N}$.
From: $m^2-n\mid m+n^2$, we have: $m^2-n\leq m+n^2 \Leftrightarrow (k-1)(2n+k+1)\leq 0 \Rightarrow k=1$
Now we have $m^2-m-1\mid m^2+3m+1 \Leftrightarrow m^2-m-1\mid 4m+2 \Rightarrow m\leq 5$
By checking, we get for solutions:
$ (m,n) = (1,2),(2,1),(3,2),(2,3),(2,2),(3,3) $


[quote="likoucalvinhao"]Not to sound impatient, but when are we going to get our stuff back? Also, when are the next rounds going to be posted? I just wish to know so I can better plan out future stuff.[/quote]

Results will be out near the end of October.  The next round will be up next week.  We will send emails out for each.


I want to know is it true that

for any positive real $a,b,c$ where $abc = 1$
$a^n + b^n + c^n \geq a + b + c ,\, n \geq 1$

If it's true.Can someone prove it.?  :blush:

As you can see it can be even generalized to $a^n+b^n+c^n\geq a^k+b^k+c^k$ for $n\geq k$ and can similiarly be extended to $t$ variables (instead of $t=3$ in your problem).

Misha

Thanks ! Misha

I forgot to apply Chebyshev ineq.  :P


calculus
integration
trigonometry
limit
complex analysis

Here is something I don't understand at Complex Analisys. For example I saw in a book that the following integral \[\int_{|z-2|=3}\frac{e^{z^{2}}}{z^{2}-6z}dz\] can be done using one of Cauchy's theorems but the followong integral can be solved with Residue theorem \[\int_{|z|=3}\frac{1}{z(z+1)^{2}}\cos\frac{1}{z}dz.\] Can someone please solve them and please enlighten me with this? :)

P. S Actually I would like to see a proof for them, although I have solved the first one with Cauchy's theorem, but I'm really interrested in
when we apply these theorems. :)


inequalities function Princeton college quadratics logarithms integration

[quote="Cezar Lupu"]Let $a, b, c, d$ three strictly positive real numbers such that
$a^{2}+b^{2}+c^{2}=d^{2}+e^{2},$ $a^{4}+b^{4}+c^{4}=d^{4}+e^{4}.$
Compare $a^{3}+b^{3}+c^{3}$ with $d^{3}+e^{3}.$
[/quote]
I think,  $a, b, c, d, e$ are [b]five[/b] strictly positive real numbers :)

I am the author and proposer of this problem and shall be glad to see
any your ideas/solutions/comments/emotions or so on concerned with it.

Is there any intelligent solution for this problem? I mean, after 40 minutes I also found this horrible thing, but I could not believe that a nicer solution doesn't exist for this problem (whose statement is, by the way, very nice). So, does the proposer have something smart?

Here is almost the same solution but presented a bit differently:
[b]Lemma[/b] $(a-d)(a-e)\ge 0$
[b]Proof[/b] \[\begin{aligned}0\le 2b^{2}c^{2}&=(b^{2}+c^{2})^{2}-(b^{4}+c^{4})\\ &=(d^{2}+e^{2}-a^{2})^{2}-(d^{4}+e^{4}-a^{4})=2(a^{2}-d^{2})(a^{2}-e^{2})\\ &=2(a-d)(a-e)(a+d)(a+e)\,. \end{aligned}\] Similar inequalities hold for $b$ and $c$.
Let now $P(x)=(x-d)(x-e)=x^{2}-\alpha x+\beta$ ($\alpha,\beta>0$). Write \[a^{2}P(a)+b^{2}P(b)+c^{2}P(c)\ge 0=d^{2}P(d)+e^{2}P(e)\] and cancel the sums of the fourth and of the second powers (with coefficients $1$ and $\beta$ respectively). You'll be left with the inequality for the third powers with the coefficient $\alpha$.
The end :).

Nice solution, Zhaobin!  :lol: Great solution, Fedja!  :D

Fedja's solution is easier then mine. On the other hand, my solutions works also for the following problem:

Let $s>u>t>0$ and $a, b, c, d, e$ be strictly positive real numbers such that
$a^{s}+b^{s}+c^{s}=d^{s}+e^{s},$ $a^{t}+b^{t}+c^{t}=d^{t}+e^{t}.$
Compare $a^{u}+b^{u}+c^{u}$ with $d^{u}+e^{u}.$

Try to solve it! :wink:

I'll post my solution later and now is only key lemma:
[hide="Hint (Key lemma)"]Let $s>t>0$ and $a\ge b\ge c>0,$ $d\ge e>0$ be strictly positive real numbers such that
$a^{s}+b^{s}+c^{s}=d^{s}+e^{s},$ $a^{t}+b^{t}+c^{t}=d^{t}+e^{t}.$
Then $a>d\ge e>b\ge c.$

[hide="proof for any $s,t$"]
If $d\ge a$ then $(d^{t},e^{t},0)\succ(a^{t},b^{t},c^{t})$ and by Karamata inequality for convex function $x^{s/t}$ we have $d^{s}+e^{s}>a^{s}+b^{s}+c^{s}.$ Hence $a>d,$ $a^{t}>d^{t},$ $b^{t}+c^{t}<e^{t}$ and $b<e.$ Therefore $a>d\ge e>b\ge c.$ [/hide]
[hide="one more easy proof for $t=2,s=4$"]
Let $c^{2}=e^{2}+\Delta,\, \Delta\in \mathbb{R}.$ Then $d^{2}=a^{2}+b^{2}+\Delta$ and the second equation implies $a^{4}+b^{4}+(e^{2}+\Delta)^{2}=(a^{2}+b^{2}+\Delta)^{2}+e^{4},$ $\Delta=-\frac{a^{2}b^{2}}{a^{2}+b^{2}-e^{2}}.$
(Here $a^{2}+b^{2}-e^{2}\ge\frac{2}{3}(a^{2}+b^{2}+c^{2})-\frac{1}{2}(d^{2}+e^{2})=\frac{1}{6}(d^{2}+e^{2})>0.$)
Since $c^{2}=e^{2}-\frac{a^{2}b^{2}}{a^{2}+b^{2}-e^{2}}=\frac{(a^{2}-e^{2})(e^{2}-b^{2})}{a^{2}+b^{2}-e^{2}}>0$ then $a>e>b.$ Therefore $d^{2}=a^{2}+b^{2}-\frac{a^{2}b^{2}}{a^{2}+b^{2}-e^{2}}<a^{2}$ and $a>d\ge e>b\ge c.$
[/hide][/hide]

I have talked to the bulgarian guy from Princeton University about this problem and he said that he had solved this problem using Lagrange's multipliers. :maybe:
I tried that, too, but I didn't have the courage to finish it. :(

I did it also with Lagrange Multipliers but they didn't understand my solution.  :( Here is a brief explanation of what I did :
We prove that $d^{3}+e^{3}\geq a^{3}+b^{3}+c^{3}$ if $1\geq a,b,c\geq0$.
If one of $a,b,c$ (suppose $a$) are zero then you could prove the equality of $d^{3}+e^{3}$ and $c^{3}+b^{3}$. If one of $d,e$ is zero also the inequality will be proved easily. WLOG suppose that $c=\max\{a,b,c\}=1$. You see the all of $a,b,d,e$ with these conditions make a compact space. Suppose at the point $a,b,d,e$ the function $f(a,b,d,e)=a^{3}+b^{3}+1-d^{3}-e^{3}$ takes its maximum. You could see that there exist $\lambda_{1},\lambda_{1}$ which $a,b,d,e$ are root of quadratic equation $x=\lambda_{1}x^{2}+\lambda_{2}$. If $\{a,b\}\cap\{d,e\}\neq\emptyset$ then ineqaulity is obvious. So $a=b$ and  $d=e$. Now you can calculate $f(a,b,d,e)$ and show that it's not positive.

I also solved it like Zhaobin, but I like neither this solution nor the official one. [quote="zhaobin"]$2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})\sqrt{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}\ge 3a^{2}b^{2}c^{2}+\sum_{sym}a^{3}b^{3}$[/quote] Somehow I cancelled the $3a^{2}b^{2}c^{2}$ term on my paper, thinking it was negative. :roll: but apparently they only deducted 2 points as the solution doesn't change.

Is there in fact a way to show that inequality easily without Muirhead? If there is a nice way, I will probably change my mind on this solution. (and considering the elegant problem statement, on the problem as a whole)

My team leader told me it could be done with AM-GM, but I don't see how to do this without squaring the whole pack out...

Rogue, could you send your solution for generalization?

Let $s>u>t$ and $a,b,c,d,e$ be strictly positive real numbers such that $a^{s}+b^{s}+c^{s}=d^{s}+e^{s}$ and $a^{t}+b^{t}+c^{t}=d^{t}+e^{t}.$ Compare the numbers $a^{u}+b^{u}+c^{u}$ and $d^{u}+e^{u}.$
[b]Solution.[/b] Without loss of generality $a\ge b\ge c,\, d\ge e.$
[b]Lemma.[/b] Let $s>t$ and $a\ge b\ge c,$ $d\ge e$ be strictly positive real numbers such that $a^{s}+b^{s}+c^{s}=d^{s}+e^{s},$ $a^{t}+b^{t}+c^{t}=d^{t}+e^{t}.$ Then $a>d\ge e>b\ge c.$
[i]Proof.[/i] If $d\ge a$ then $(d^{t},e^{t},0)\succ(a^{t},b^{t},c^{t})$ and by Karamata
inequality for convex function $x^{s/t}$ we have $a^{s}+b^{s}+c^{s}>d^{s}+e^{s},$ a contradiction. Hence $a>d,$ $a^{s}>d^{s},$
$b^{s}+c^{s}<e^{s}$ and $b<e.$ Therefore $a>d\ge e>b\ge c.$

Consider a function $f(x)=a^{x}+b^{x}+c^{x}-d^{x}-e^{x},\, x\in \mathbb{R}.$ We shall prove that $f(x)$ has only two zeroes $x=2$ and $x=4$ and changes the sign at these points. Suppose the contrary. Then Rolle's theorem implies that $f'(x)$ has at least two distinct zeroes. Without loss of generality $a=1.$ Then $f'(x)=\ln b\cdot b^{x}+\ln c\cdot c^{x}-\ln d\cdot d^{x}-\ln e\cdot e^{x},\, x\in \mathbb{R}.$ If $f'(x_{1})=f'(x_{2})=0,\, x_{1}<x_{2},$ then
$\ln b\cdot b^{x_{i}}+\ln c\cdot c^{x_{i}}=\ln d\cdot d^{x_{i}}+\ln e\cdot e^{x_{i}},\quad i=1,2,$
but since $1>d\ge e>b\ge c$ (See lemma) we have
$\frac{(-\ln b)\cdot b^{x_{2}}+(-\ln c)\cdot c^{x_{2}}}{(-\ln b)\cdot b^{x_{1}}+(-\ln c)\cdot c^{x_{1}}}\le b^{x_{2}-x_{1}}<e^{x_{2}-x_{1}}\le\frac{(-\ln d)\cdot d^{x_{2}}+(-\ln e)\cdot e^{x_{2}}}{(-\ln d)\cdot d^{x_{1}}+(-\ln e)\cdot e^{x_{1}}},$
a contradiction. Therefore $f(x)$ has a constant sign at each of the intervals $(-\infty,t),$ $(t,s)$ and $(s,\infty).$ Since $f(0)=1$ then $f(x)>0,$ $x\in(-\infty,t)\bigcup(s,\infty)$ and $f(x)<0,$ $x\in(t,s).$ In particular, $f(u)=a^{u}+b^{u}+c^{u}-d^{u}-e^{u}<0.$

[quote="Cezar Lupu"]Let $a, b, c, d$ three strictly positive real numbers such that \[a^{2}+b^{2}+c^{2}=d^{2}+e^{2},\]  \[a^{4}+b^{4}+c^{4}=d^{4}+e^{4}.\] Compare \[a^{3}+b^{3}+c^{3}\] with \[d^{3}+e^{3},\][/quote]
Let $\epsilon = \frac{2}{3}$. Let $f(a,b,c) = \left ( \begin{array}{c} f_1(a,b,c) \\ f_2(a,b,c) \end{array} \right ) = \left ( \begin{array}{c} a^{1-\epsilon}+b^{1-\epsilon}+c^{1-\epsilon} \\ a^{1+\epsilon} + b^{1+\epsilon} + c^{1+\epsilon}\end{array} \right )$.
Assume there exists $m,n,p,q,r>0$ s.t. $f(m,n,p) = f(q,r,0) = \left ( \begin{array}{c} x \\ y\end{array} \right )$ for some $x,y > 0$.
Let $V = \left \{ (a,b,c) \in [0,\infty) \mid f(a,b,c) = \left ( \begin{array}{c} x \\ y\end{array} \right ) \right \}$. Then it is obvious that if $V \subseteq [0,\sqrt[1-\epsilon]{x}]^3$, and that $V$ is closed, hence $V$ is compact.
Let $g(a,b,c) = a+b+c$. Since $g$ is continous it has a maximum on $V$. Assume for the sake of contradiction that a maximum is $(a,b,c) \in V$, with $a,b,c > 0$.
Hence $Df(a,b,c)$ has rank 2. Now it is pretty easy to see (Lagrange?) that $\text{grad} f_1(a,b,c)$ and $\text{grad} f_2(a,b,c)$ are linearly dependent, i.e. $(a^{-\epsilon},b^{-\epsilon},c^{-\epsilon})$ and $(a^{\epsilon},b^{\epsilon},c^{\epsilon})$ are linearly dependent, so $a=b=c$. But this is impossible cause then there couldn't exist $q,r > 0$ s.t. $f(q,r,0) = f(a,b,c)$. Contradiction.
Let $(a,b,c)$ be a maximum of $g$ on $V$. Then at least one of $a,b,c$ are zero, wlog because of symmetry let $c=0$. Then since $a,b \ge 0$ they are decided by $a^{1-\epsilon} + b^{1-\epsilon} = x, a^{1+\epsilon} + b^{1+\epsilon} = y$, so $\{a,b\} = \{q,r\}$. Hence $g(q,r,0)$ is a maximum, and hence $g(q,r,0) \ge g(m,n,p)$.

Changing to $(a^3,b^3,c^3,d^3,e^3) = (m,n,p,q,r)$ we get the desired.


A teacher has made ten statements for a True-False test. Four statements are true and six are false. How many distinct answer keys could there be for the test?

There are $ \dbinom{10}4\equal{}\dbinom{10}6\equal{}\boxed{210}$ ways to choose $ 4$ true statements and $ 6$ false statements.

[quote=ishankhare][quote=RadioActive]Yea, I just did $10$ factorial divided by $6$ factorial times $4$ factorial.[/quote]
That's also what I did. However, I think $\binom{10}{4}$ is much slicker and faster.[/quote]

They actually are the same thing.

[quote=stitch04]There are $\binom{10}{4}=\boxed{210\text{ ways}}$ to choose four of the answers to be the true answers.[/quote]

If you want to bump a thread from a year ago, post a new solution to go with it.

[quote=ishankhare][quote=RadioActive]Yea, I just did $10$ factorial divided by $6$ factorial times $4$ factorial.[/quote]
That's also what I did. However, I think $\binom{10}{4}$ is much slicker and faster.[/quote]

I was thinking about doing that but instead I did 2^10, but that just gives you the total number of ways, not what the question is asking.

I know that you are a new user. Read the rules about posting (don't bump posts)

It's not bad to bump post if you post a new viewpoint of the problem.

[quote=ishankhare][quote=RadioActive]Yea, I just did $10$ factorial divided by $6$ factorial times $4$ factorial.[/quote]
That's also what I did. However, I think $\binom{10}{4}$ is much slicker and faster.[/quote]

They are both the same because choosing 6 objects is the same as not choosing 4 objects


Bobby is climbing Mount LeLo in La La Land.
He is going through 2 time zones to get to the summit, one time zone is an hour; Patonic, Lobotomy, then Gigabyte.
He starts out in Patonic, goes through Lobotomy, then Gigabyte.
He starts out at 12:00 P.M. and hikes at an average speed of 5 miles per hour, if he needs to climb 25 miles; what's the time in Gigabyte when he climbed up the mountain, and then back down?

For clarification: In La La Land, there is 24  hours in a day.

When he goes up the mountain and change time zones, does he add or subtract an hour?


calculus MATHCOUNTS inequalities quadratics

[hide]If you think about it, you'll notice that as $n$ comes closer to $m$, the expression gives $1+1=2$.  If n gets far apart from m, then the answer gets really big with a tiny fraction added to it.  Clearly, we want the biggest fraction possible ; the small fraction doesn't make a difference.  So, we choose the maximum $n$ and the minimum $m$.  So, $\frac{80}{20}+\frac{20}{80}=4+\frac{1}{4}$.[/hide]

Yes, but we don't need to formally prove it.  We just need to convince ourselves that it makes sense.

[quote="me@home"]To truly know this requires calculus, which isn't spoken about here...

is there a way a mod can move this (or split it) to another forum where such comments can be made?[/quote]

but this is a MATHCOUNTS problem... they shouldnt have a problem on there that requires calc...

archimedes1's answer is correct. the only time you would use calculus would be if m and n were dependent on each other, in which case you might use related rates.

Just a tip: don't give up on a mathcounts problem just because it seems like a calculus problem, because its not. this is just a question where you need to know how to make the biggest possible number with the given limitations. getting intimidated by a problem that looks like it requires something that is out of the mathcounts level will just lose you a point.

You don't need to prove it with Calculus.  Using inequalities is enough.  Or you can even use quadratics:

Let $\frac{n}{m}=x$.  Let $x+\frac{1}{x}=c$.  Then:
$x^{2}+1=cx$
$x^{2}-cx+1=0$
$x=\frac{c\pm\sqrt{c^{2}-4}}{2}$

Let's use the solution $\frac{c+\sqrt{c^{2}-4}}{2}$ since one is the reciprocal of the other.

You can see as c increases, x increases or as x increases, c increases.  So in order to get the maximum value of c, you need the maximum value for x.  And that is when $\frac{n}{m}$ or $\frac{m}{n}$ is maximized.  Since the maximum value for either of them is $\frac{80}{20}=4$, the answer is 4+1/4

But there is no need to do this.  You can just try a few numbers and make the guess you want to maximize $\frac{n}{m}$.  You don't need to prove anything.

There are already some good answers.

[hide]Number sense might tell you that $a+\frac{1}{a}$ is always $\geq 2$ when $a$ is positive.  [The smallest possible value occurs when $a=1$.]  If you knew that, it might lead you to try the largest possible value for $a$ and the largest possible value for $\frac1a$.

Otherwise, I might build a guess-and-check table.  I'd put in the minimum, median, and maximum value for $m$ and for $n$, and work out each of the 9 answers:

$\begin{array}[b]{l|c|c|c|}m\downarrow \qquad n\rightarrow&40&60&80\\ &&&\\ \hline &&& \\ 20 & 2+\frac12 =2\frac12& 3+\frac13=3\frac13 & 4+\frac14=4\frac14 \\ &&&\\ \hline &&& \\ 30 & \frac43+\frac34=2+& 2+\frac12=2\frac12 & \frac83+\frac38=3+\\ &&&\\ \hline &&& \\ 40 & 1+1=2 & \frac32+\frac23 =2+& 2+\frac12=2\frac12 \\ &&& \\ \hline \end{array}$

This doesn't exactly "prove" the answer, but it sure looks like the number in the upper right is the biggest possible.  I'd probably go with this and move on.[/hide]


In a set of $ n$ people, each of its $ 2^n\minus{}1$ nonempty subsets is called a company. Each company should elect a leader, according to the following rule: If a company $ C$ is the union $ A\cup B$ of two companies $ A,B$, then the leader of $ C$ is also the leader of at least one of the companies $ A,B$. Find the possible number of choices of leaders.

This is from the first round of the Polish MO -- please cite your sources when posting problems!  There was at least one poster who tried to post this problem before the deadline; see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=173273 and that poster's other posts.  Since the deadline is passed, I've returned that thread to the forum -- it contains a complete solution.


linear algebra
matrix
vector
algebra
polynomial
linear algebra unsolved

This is a problem about linear dependence, basis and nullspace.

Let $A$ be a non-zero $3\times3$ matrix such that $A^2 = 0$.

(i) Prove that there exists some $v \in \mathbb{R}^3$ such that $\{v, Av\}$ is a linearly independent set.
(ii) Let $v$ and $Av$ be given as in (i). Prove that $\mathbb{R}^3$ possesses a basis $\{v, Av, w\}$ such that $Aw = 0$.

For the existence, should we provide a general vector $v$?

This is a hard problem, at least to me.

OK, can I do it this way?

Let $A$ be as given in the problem. We then test linear independence using the equation $xv + y(Av) = 0$. If we can show that $x=0$ and $y=0$, we are done for the first part.

Since $A$ is non-zero, we may select a vector $v \in \mathbb{R}^3$ such that neither $v$ nor $Av$ is zero. (Because if either one is zero, then the coeffienct of the zero matrix may not be zero.) Then premultiplying $A$ to both sides gives

   $A[xv + y(Av)] = A0 = 0$, that is,

   $x(Av) + y(A^2v) = 0$

But $A^2v = 0v = 0$, we further have $x(Av) = 0$. As $Av$ is not zero, we may conclude that $x=0$. It should then follow from $y(Av) = 0$ and the fact that $Av$ is not zero that $y=0$.

My only problem here is I am not sure if I am at all justified in making the statement 'we may select a vector $v$ from $\mathbb{R}^3$ such that neither $v$ nor $Av$ is zero.'

I suppose the argument is as follows: Suppose the nonzero entry (any one) in A is in the $i$-th column, we just select any vector $v \in \mathbb{R}^3$ which has a non-zero entry in the $i$-th row. Then $Av$ is not zero. To avoid the unpleasant possibility that the product may be subtracted to become zero, let us make all other entries of $v$ zero.

Any comments?

OK, somebody please check...

It is evident from the fact that $A^2 = 0$ that $A$ is not invertible. Then there exists a non-zero vector $w$ such that $Aw = 0$.

Consider the equation $xv+y(Av)+zw = 0$. Then as in part (i) we premultiply both sides of the equation to get

$A[xv+y(Av)+zw] = A0 = 0$. or equivalently,

$x(Av)+y(A^2v)+z(Aw) = 0$. But we know that $A^2v$ and $Aw$ are both zero. Therefore $x = 0$. Suppose for the sake of contradiction that $w = \lambda Av$ for some $\lambda \in \mathbb{R}$ which is not zero. But I am stuck here.

Perhaps: Suppose $y(Av) + zw = 0$. Suppose $z \neq 0$. Then $y(Av) = -zw \neq 0$, therefore $y\neq 0$. So $Av = (-\frac{z}{y})w$...

Here's your way out of this puzzle:

If $Av=0$ for all $v,$ then $A$ is the zero matrix. One way to see this: multiplying $A$ by the elements of the standard basis gets us the columns of $A.$ If $Av=0$ for all $v,$ that includes the standard basis elements, hence all the columns of $A$ are zero, hence $A=0.$

So if $A\ne0,\,\exists\,v$ such that $Av\ne0.$ Automatically, that would mean $v\ne 0.$ From there, you have a proof. Another way of looking at it: the only way for $\{v,Av\}$ to be linearly dependent would be for $Av=\lambda v$ for some $\lambda.$ But that would make $\lambda$ an eigenvalue of $A,$ and all eigenvalues of $A$ satisfy the same polynomial equation that $A$ satisfies: $\lambda^2=0.$ $\lambda=0$ would violate $v\ne0.$

Your proof of the last part is basically correct. You said:
[quote]Suppose for the sake of contradiction that $w = \lambda Av$ for some $\lambda \in \mathbb{R}$ which is not zero. But I am stuck here. [/quote]
But, as I've mentioned above, the only eigenvalues of $A$ are zero, so this can't happen.

This amounts to a proof of a very specific case of Jordan canonical form. Basically, you are finding a basis in which the matrix $A$ can be represented as
$\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}.$

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59105]this topic.[/url]

But my problem is with $Av$ and $w$ and not $Av$ and $v$ (this was settled long time ago).


Find a polynomial with integer coefficents which has the roots:
\[\sqrt{\frac{3+\sqrt7}{5}}\]
\[-\sqrt{\frac{3+\sqrt7}{5}}\]
\[\sqrt{\frac{3-\sqrt7}{5}}\]
\[-\sqrt{\frac{3-\sqrt7}{5}}\]

Solution in spoiler:

[hide]

Since the four things are the roots, P(x)=(x^2-(3+sqrt(7)/5))(x^2-(3-sqrt(7)/5) after multiplying the conjugate roots.   It simplifies to x^4-(6/5)x^2+2/25 => P(x) = 25x^4-30x^2+2.

[/hide]

-interesting_move

[hide]

Notice that the solution to a polynomial of the form

ax^4+bx^2+c

is

x = :pm: sqrt((-b :pm: sqrt( b^4-4ac))/2a)

which is the form of all the solutions given.

Thus one such polynomial, though it does not have integer coefficients, would be

2.5x^4 - 3x^2 + .2

which, when we multiply by 10, we get

25x^4 - 30x^2 + 2

[/hide]

Obviously both solutions posted are correct, but I like nr's better; it doesn't require a calculator or anything too complicated.

nr1337 wrote:[hide]
x = :pm: sqrt((-b :pm: sqrt( b^4-4ac))/2a)
[/hide]

Shouldn't it be[hide] x = :pm: sqrt((-b :pm: sqrt( b^2-4ac))/2a)[/hide]

[quote="JBL"]Neither solution requires a calculator, and nr's solution requires you to effectively figure out the answer before doing the problem.  Also, he picks his solution out of a hat -- how did he get values for a, b and c?[/quote]

Haha, I guess I didn't read it too carefully.  I thought that he paired up the conjugate roots and guessed at values for a,b,c and then multiplied them.  Neither solution requires a calculator, but the first one requires a lot more tedious multiplication of radicals.


function
limit
algebra
domain
Functional Analysis
real analysis
real analysis unsolved

Does anyone know an example about Closed linear operator which is not Continuous .

I have differential operator .....I need anyone else ??

On $ L^2(\mathbb{R}),$ let $ Tf(x)\equal{}xf(x).$

Of course, seen through the Fourier transform, this is a differential operator in disguise. But you could replace $ x$ by some other unbounded function to get a more general example.

Note that any such operator cannot be defined on the whole Banach space. The "nice" examples are defined on a dense subspace of the Banach space in question - as is true of this example, which is defined for (at least) compactly supported $ f.$

How to prove it's closed ?

if $ f_n \rightarrow f$ and $ T(f_n)\rightarrow F$ , then we need to prove that $ F\equal{}T(f)$

Let's see you try that. This is one of those "you do what you have to do" proofs.

Since , $ f_n \rightarrow f, \Longrightarrow xf_n \rightarrow xf$

I need to prove that :

$ \parallel{} xf \minus{} F\parallel{} \equal{} 0$

But $ \parallel{}xf \minus{} F\parallel{} \equal{} \lim_{ n\to \infty} \parallel{}xf_n \minus{} Tf_n\parallel{} \equal{} 0$ , So $ xf \equal{} F$

Does that the true way ?

Give me a hint if it's not true .I don't know how to do the technique.

I just think about this example :

let $ X$ be any banach space of $ \ell^p$ , then if we assumed the domain of  linear operator $ T$ to be :

$ D(T)\equal{}\{ x : a.x \in X \}$ , where $ a$ is a fixed sequence in $ \mathbb R$ .

$ T(x) \equal{} a.x$ .

This is  closed and it's bounded iff $ a$ is .

Does this example true ? Or have I missed something?


Let $H$ and $O$ be the orthocenter and the circuncenter of triangle $ABC$ respectively,$N$ is the midpoint of $BC$, $D$ is the feet of the altitude on side $BC$.It's given that $HOND$ is a rectangle and $HO=11$ and $ON=5$ find the lenght of $BC$

Let E, F be the feet of the B-, C-altitudes of the triangle $\triangle ABC$. Denote a = BC, h = AD the side and A-altitude of the triangle $\triangle ABC$. Also, denote p = HO = DN, q = HD = ON the sides of the rectangle HOND. The right angle triangles $\triangle BHD \sim \triangle BCE$ are similar, hence, $\angle BHD = \angle C$ and

$\tan \widehat C = \frac{BD}{HD} = \frac{BD}{q}$

From the right angle triangles $\triangle ABD, \triangle ACD$ with the common side h = AD,

$\tan \widehat B = \frac{h}{BD},\ \ \tan \widehat C = \frac{h}{CD}\ \Longrightarrow\ BD \tan \widehat B = CD \tan \widehat C$

The angle $\angle BON = \frac{\angle BOC}{2} = \angle A$ and then from the right angle triangle $\triangle BON$,

$\tan \widehat A = \frac{BN}{ON} = \frac{a}{2q}$

Substituting $BD = BN - DN = \frac a 2 - p,$ $CD = CN + ND = \frac a 2 + p$ into the above 2 equations for $\tan \widehat B,\ \tan \widehat C,$

$\tan \widehat C = \frac{a - 2p}{2q}$

$(a - 2p) \tan \widehat B = (a + 2p) \tan \widehat C\ \Longrightarrow \tan \widehat B = \frac{a + 2p}{2q}$

For any 3 angles, such that $\widehat A + \widehat B + \widehat C = 180^\circ$, we have

$\tan \widehat A + \tan \widehat B + \tan \widehat C =$

$= \tan (\widehat A + \widehat B)\ (1 - \tan \widehat A \tan \widehat B) + \tan \widehat C =$

$= \tan (180^\circ - \widehat C)\ (1 - \tan \widehat A \tan \widehat B) + \tan \widehat C =$

$= -\tan \widehat C\ (1 - \tan \widehat A \tan \widehat B) + \tan \widehat C =$

$= \tan \widehat A \tan \widehat B \tan \widehat C$

Substituting for $\tan \widehat A,\ \tan \widehat B,\ \tan \widehat C$ leads to

$\frac{a}{2q} + \frac{a + 2p}{2q} + \frac{a - 2p}{2q} = \frac{a}{2q} \cdot \frac{a + 2p}{2q} \cdot \frac{a - 2p}{2q}$

$\frac{3a}{2q} = \frac{a}{2q} \cdot \frac{a^2 - 4p^2}{4q^2}\ \Longrightarrow a^2 - 4p^2 = 12q^2$

$a = 2 \sqrt{3q^2 + p^2} = 2 \sqrt{3 \cdot 25 + 121} = 2 \sqrt{196} = 28$


analytic geometry
superior algebra
superior algebra unsolved

find an ideal from a commutative ring with unity.so that every member of ideal be nilpotent but the ideal will not be nilpotent.

Hmmm. How about $ R \equal{} \prod_{i > 1} \mathbb{Z}/(p^i)$, and take the ideal $ I$ generated by the element $ p$ in each coordinate (and 0 elsewhere). Each element in $ I$ is nilpotent, but there is no $ n$ such that $ I^n \equal{} 0$.

@nukular: p is not nilpotent. in general: a finitely generated ideal of nilpotent elements is nilpotent.
@mj: you just introduce unbounded exponents of nilpotency. $ k[x_2,x_3,...] / (x_2^2 , x_3^3,...)$

It's not finitely generated. It's generated by many elements $ x_i$ where $ x_i$ is $ p$ in the $ i$th coordinate and zero elsewhere. Then each $ x_i$ is nilpotent and the whole ideal is not.

ah ok. I thought you mean I is generated by (p,p,...).


HI! I am new here. I am a teacher at a scool with classes 9-12.
A student asked me how to demonstrate that (a+b) all at square is a(at square) + 2ab + b(at square). I didnt know. how can i demonstrate?

Proving that $ (a\plus{}b)^2\equal{}a^2\plus{}2ab\plus{}b^2$:
Note that $ (a\plus{}b)^2\equal{}(a\plus{}b)(a\plus{}b)$ (definition of squaring a number)
Multiply this using FOIL (First, Outside, Inside, Last)
$ (a\plus{}b)(a\plus{}b)$
$ a(a)\plus{}a(b)\plus{}b(a)\plus{}b(b)$
Rewrite: $ a^2\plus{}ab\plus{}ab\plus{}b^2$
Combine like terms:
$ a^2\plus{}2ab\plus{}b^2$

You have just to express $ 2ab \equal{} ab \plus{} ab$ than :

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

$ a(a \plus{} b) \plus{} b(a \plus{} b)\Leftrightarrow(a \plus{} b)(a \plus{} b) \equal{} (a \plus{} b)^{2}$

If this is what you mean

You can also take a square with area $ a^2$, 2 squares with area $ ab$, and a square with area $ b^2$ to make a square with side length $ a\plus{}b$ and thus area $ (a\plus{}b)^2$.

Not only should you be familiar with the FOIL method, distribution is also good in case you deal with powers that are greater than 2.


Solve for all $ x,y$:

$ \sqrt{8x\minus{}3y\plus{}1}\plus{}\sqrt{8x\minus{}3y}\equal{}y$

$ \sqrt{8x\minus{}3y\plus{}1}\minus{}\sqrt{8x\minus{}3y}\equal{}\frac{2}{y^2\plus{}1}$

Multiply the two equations to get $ 1\equal{}\frac{2y}{y^2\plus{}1}$ and $ y\equal{}1$. Then add the two equations to solve for $ x$.


function
limit
integration
real analysis
real analysis unsolved

Find all differentiable functions $f:[0,\infty)\to\mathbb{R}$ such that exists $\lim_{x->\infty} f'(x)$ and

                             $\int_{2x}^{2x+1} (f(t)+f(\frac{1}{2})\) dx=f(x)+f(x+1)$ for every $x\geq 0$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46587]It's right here[/url], although there's no answer there either. I guess it just doesn't look that nice :).

Well, I'm not sure but I think that this problem works without the first condition. Hope I haven't made a mistake in my calculatuions. :?

But as usual solutions are not presented. What should we understand from that, Cezar? I mean, I can also say that I can solve all the problems on the forum, but who will believe me?


function
inequalities
analytic geometry
graphing lines
slope
real analysis
real analysis unsolved

Let g: R -> R be a bounded convex function.

Prove that g must be a constant function.

If possible, can you please explain your methods in the proof.

Draw a secant line through two points on the graph. The graph passes above the secant to the left and to the right of the two points. If the secant line is not horizontal, the graph is unbounded.

Consider some $a<b.$

Suppose $x>b.$ Then since we can write

$b=\frac{x-b}{x-a}\cdot a+\frac{b-a}{x-a}\cdot x$ it follow that

$g(b)\le\frac{x-b}{x-a}\cdot g(a)+\frac{b-a}{x-a}\cdot g(x)$

We can rearrange this inequality to

$g(x)>\left(\frac{g(b)-g(a)}{b-a}\right)x+\frac{bg(a)-ag(b)}{b-a}$

If $g$ is bounded above as $x\to\infty,$ then we must have that $\frac{g(b)-g(a)}{b-a}\le 0.$

A very similar calculation for $x<a$ will result in the exact same ineqality:

$g(x)>\left(\frac{g(b)-g(a)}{b-a}\right)x+\frac{bg(a)-ag(b)}{b-a}$

This time, boundedness of $g(x)$ as $x\to-\infty$ will give us that $\frac{g(b)-g(a)}{b-a}\ge 0.$

Geometrically, all we've done is draw the chord line through two points on the curve and argue that outside of the interval between the two points, the curve must lie above or on that chord line. If the chord line had a nonzero slope, then $g(x)$ would have to be unbounded above. Boundedness therefore implies that the chord line has zero slope. That is, $g(b)=g(a).$ Since $a$ and $b$ were arbitrary, this proves that the function is constant.

[color=green][Once again, I'm slow. Consider this the explanation of mlok's post.][/color]


conics
parabola
geometry proposed
geometry

Here's another fun question involving envelopes

Prove geometrically that if the corner of a rectangular piece of paper is folded down so that the sum of the edges left unfolded is constant then the creases so formed will all touch a parabola.

This is obtained by taking a point $A\in (Ox$ and a point $B\in (Oy$ s.t. $OA+OB=c$ is constant. In the given conditions, we can show that the midpoint of $AB$ lies on the line $x+y=\frac c2$, and it's now easy to show that the parabola in question is the one having the line $x=-y$ as its directrice, and $(\frac c2,\frac c2)$ as its focus, since the lines $AB$ are the perpendicular bisectors of segments connecting $(\frac c2,\frac c2)$ with arbitrary points on the line $x=-y$.


[quote="SBHD"]It's enough if I show that
$b^{2}+2005a+2006 \geq 2007a+2005$ and $a^{2}+2005b+2006 \geq 2007b+2005$.
After solving this inequalities, we get $b^{2}-2a+1 \geq 0$ and $a^{2}-2b+1 \geq 0$. Adding these two inequalities we get $a^{2}+b^{2}-2a-2b+2 \geq 0$ --> $(a-1)^{2}+(b-1)^{2}\geq 0$
q.e.d.[/quote]

I don't think you can do that  :o

$x+y \geq 0$ does not imply $x\geq 0$ and $y\geq 0$!

Just to show you: you thought you proved $b^{2}+2005a+2006 \geq 2007a+2005$ for all non-negative numbers $a,b$. Fill $a=2$ and $b=1$ in and you'll get that $6017 \geq 6019$.

You are right, made a mistake, sorry, but x,y => 0 (First post).

So, trying to get another prove. Thanks

[quote="arqady"]Let $a$ and $b$ are non-negative real numbers. Prove that:
$(a^{2}+2005b+2006)(b^{2}+2005a+2006)\geq(2007a+2005)(2007b+2005).$[/quote]
Generalization 1
Let $a,b\geq 0$ and $\alpha >0,\beta >1$.Prove that
$\left( {{a}^{2}}+\alpha b+\beta  \right)\left( {{b}^{2}}+\alpha a+\beta  \right)\ge \left( \left( \alpha +2 \right)a+\beta -1 \right)\left( \left( \alpha +2 \right)a+\beta -1 \right)$
Generalization 2
Let $a,b\geq 0$ and $\alpha >0,\beta \ge n-1,n>1$.Prove that
$\left( {{a}^{n}}+\alpha b+\beta  \right)\left( {{b}^{n}}+\alpha a+\beta  \right)\ge \left( \left( \alpha +n \right)a+\beta +1-n \right)\left( \left( \alpha +n \right)a+\beta +1-n \right)$


This problem is quite easy to solve using algebra. However, someone told me that there is a more unique and original solution that makes use of a simple geometric identity(or proof, if you want to call it). After scrunging around, I still can't think of it. Anyone welcome to try?

4 cars are traveling on a road in the same direction. Car A passes car B at 8 AM, car C at 9 AM, car D at 10 AM. Car B passes Car D at 12 PM, and car C passes Car D at 2 PM. At what time did car B meet car C?

Try solving it using geometry instead of algebra.

[hide]Answer: The time that car B meets car C is 10:40. [/hide][/hide]

Here's how you do it:

[hide]

Graph it! The graph is going to have time along the x-axis and distance along the y-axis. You don't really care about distance though. Now label your x-axis from 8 to 14 (use military time). It often helps if you draw vertical lines at each hour.

Since each car is going at a constant speed, their graphs are lines. The way I drew it is I let the distance at the intersection of A and B be 0. But since you don't really care about distance, it doesn't matter. Draw two lines from the origin, (which is at 8 hours, 0 distance). Line A should have a greater slope. Now draw a point where A intersects 9 o'clock and another one where A intersects 10 o'clock. Make another point where B intersects 12 o'clock. Draw a line through that point and the point where A intersects 10. Label this line D. Draw a point where D intersects 2. Make a line through this point and the point where A intersects 9. Call this line C. Label the intersection of B and C. This is the point you're looking for.

If you draw a line between the origin and the inersection of C and D, you see that you have a triangle. The sides are A, D, and the new line you just drew. Hopefully you can see that B and C are the medians of the triangle. There are lots of ways to find the coordinates of the intersection of the medians of a triangle. I'll use the fact that the coordinates are the arithmetic mean of the coordinates of the vertices of the triangle. Since you only care about the x-coordinates, you average 8, 10, and 14 to get 32/3=10 2/3=10:40.

[/hide]

This is a very cool problem.