factorial
number theory unsolved
number theory

[quote="hsbhatt"]Find numbers a,b and c such that a!+1, b!+1, c!+1 and a!+b!+c! are all perfect squares[/quote]

There are only three positive integers $ n$ known, for which $ n!\plus{}1$ is a square: $ n\equal{}4,5,7$.
Brocard has conjectured that there are no others; see for instance
http://mathworld.wolfram.com/BrocardsProblem.html

This makes your problem very easy.

Thank you very much. I was thinking along the same lines and was vainly trying to prove that n!+1 is not a perfect square for n>7.

Your post saves me a lot of effort!!  :)

In fact, I was trying on these lines. Can you tell me whether there is a mistake in my logic

Now $ 7!\plus{}1 \equal{} 71^2$

Suppose there exists k>7 for which $ k!\plus{}1 \equal{} m^2$

Then $ k! \equal{} 7!.n$

Thus $ (n\minus{}1).7! \equal{} 5040.(n\minus{}1) \equal{} (m^2\minus{}71^2) \equal{} (m\minus{}71)(m\plus{}71)$

Now since we know that the result is not true for k=8,9,10 and 11, $ n \ge 7920 \Rightarrow m \ge 5000)$

Since both m-71 and m+71 are even numbers, 5040.(n-1) should be factorised as 2520. (2n-2)

Since m-71<m+71, this means m is at most 2591 which does not meet the requirement that $ m \ge 5000$

[quote="hsbhatt"]
Since both m-71 and m+71 are even numbers, 5040.(n-1) should be factorised as 2520. (2n-2)
[/quote]
this step is wrong!for example maybe $ n\minus{}1$ is composite and  $ n \minus{} 1 \equal{} a b$ and $ 5040.(n \minus{} 1)$ factorised as $ (2520a)(2b)$

yeah, ok. that had to be wrong somewhere.


geometry
3D geometry
algebra
polynomial
modular arithmetic
algebra unsolved

Does exist $ Q_{1},...,Q_{n}$ polynomilas in $ Z[X]$

such that $ 2X\equal{}{Q_{1}(X)}^3\plus{}...\plus{}{Q_{n}(X)}^3$ ?

:huh:
If $ 2X \equal{} Q_1(X)^3 \plus{} \cdots \plus{} Q_n(X)^3\equiv Q_1(X^3) \plus{} \cdots \plus{} Q_n(X^3)\pmod{3}$,
then looking at the coefficient of $ X$ we get $ 2\equiv 0\plus{}\cdots\plus{}0\pmod{3}$. Contradiction!

You are wrong

2X=2.X^3 (mod 3), but there is no contradiction here when u compare the coefficients

I'm not wrong.
$ 2x\not\equiv 2x^3\pmod{3}$, as $ 3\nmid 2x^3\minus{}2x$. (meaning $ 3$ doesn't divide all coefficients)


calculus
integration
logarithms
calculus computations

How to solve this integral
$ \int \frac {1}{\sqrt {1 \plus{} e^x} \plus{} \sqrt {1 \minus{} e^x}}dx$

mhmh..$ \begin{array}{l}
 {\left( {\sqrt {1 \plus{} {e^x}}  \plus{} \sqrt {1 \minus{} {e^x}} } \right)^2} \equal{} 2\left( {1 \plus{} \sqrt {1 \minus{} {e^{2x}}} } \right) \\
  \Leftrightarrow \sqrt {1 \plus{} {e^x}}  \plus{} \sqrt {1 \minus{} {e^x}}  \equal{} \sqrt 2 \sqrt {1 \plus{} \sqrt {1 \minus{} {e^{2x}}} }  \\
 \end{array}$
so \[ \begin{array}{l}
 I \equal{} \frac{1}{{\sqrt 2 }}\int {\frac{1}{{\sqrt {1 \plus{} \sqrt {1 \minus{} {e^{2x}}} } }}dx\mathop  \equal{} \limits_{dx \equal{} \frac{{2\left( {{u^2} \minus{} 1} \right)}}{{\left( {{u^2} \minus{} 2} \right)u}}du}^{\sqrt {1 \plus{} \sqrt {1 \minus{} {e^{2x}}} }  \equal{} u} }  \equal{} \frac{1}{{\sqrt 2 }}\int {\frac{1}{u} \cdot \frac{{2\left( {{u^2} \minus{} 1} \right)}}{{\left( {{u^2} \minus{} 2} \right)u}}du \equal{} }  \\
  \equal{} \frac{1}{{\sqrt 2 }}\int {\left( {\frac{1}{{{u^2}}} \plus{} \frac{1}{{{u^2} \minus{} 1}}} \right)} du \equal{}  \minus{} \frac{1}{{u\sqrt 2 }} \plus{} \frac{1}{{2\sqrt 2 }}\int {\left( {\frac{1}{{u \minus{} 1}} \minus{} \frac{1}{{u \plus{} 1}}} \right)du \equal{} }  \\
  \equal{}  \minus{} \frac{1}{{u\sqrt 2 }} \plus{} \frac{1}{{2\sqrt 2 }}\ln \frac{{u \minus{} 1}}{{u \plus{} 1}} \plus{} c\mathop  \equal{} \limits^{u \equal{} \sqrt {1 \plus{} \sqrt {1 \minus{} {e^{2x}}} } } ... \\
 \end{array}\]


trigonometry
geometry
circumcircle
perpendicular bisector

In the figure below, $\angle CAD=90^\circ$, $AC=AD$, $\angle ABC=15^\circ$ and $\angle ACB=30^\circ$.

Find $\angle ADB$.

[hide="Is this correct?"]$\angle ADB=105^\circ$[/hide]

[hide="Solution"]Let $M=AD\cap BC$ and $a=AC=AD$
We'll also need:
\begin{eqnarray*}\sin 15^\circ=\sqrt{{1-\cos 30^\circ}\over 2}=\sqrt{{1-{\sqrt{3}\over 2}}\over 2}=\sqrt{{2-\sqrt{3}}\over 4}=\sqrt{{4-2\sqrt{3}}\over 8}={{\sqrt{3}-1}\over{2\sqrt{2}}}\end{eqnarray*}.

$\triangle ADC$ is 90-45-45 $\Longrightarrow \angle BCD=45^\circ-30^\circ=15^\circ$

\begin{eqnarray*}\angle ABC=\angle BCD\Longrightarrow AB\parallel CD\Longrightarrow\angle BAD=\angle ADC=45^\circ\Longrightarrow\angle AMB=120^\circ\Longrightarrow \angle AMC=60^\circ\end{eqnarray*}

$\triangle AMC$ is 30-60-90 $\Longrightarrow AM=a{\sqrt{3}\over 3}\Longrightarrow MD=a(1-{\sqrt{3}\over 3})$

Sine Law for $\triangle AMB$:
\begin{eqnarray*}BM=AM{{\sin 45^\circ}\over{\sin 15^\circ}}={a\sqrt{3}\over 3}{{1\over\sqrt{2}}\over{{\sqrt{3}-1}\over{2\sqrt{2}}}}=a(1+{\sqrt{3}\over 3})\end{eqnarray*}

Cosine Law for $\triangle BMD$:
\begin{eqnarray*} BD^2 &=& MD^2+MB^2-2MD\cdot MB\cos 60^\circ\\ &=& {a^2}({4 \over 3}-{{2\sqrt{3}} \over 3})+{a^2}({4 \over 3}+{{2\sqrt{3}} \over 3})-2{a^2}\cdot{2\over 3}\cdot{1\over 2}\\ &=& 2a^2\Longrightarrow BD=a\sqrt{2}\end{eqnarray*}

Sine Law for $\triangle BMD$:
\begin{eqnarray*}\sin\angle MBD={MD\over BD}\sin 60^\circ={{1-{\sqrt{3}\over 3}}\over{\sqrt{2}}}{\sqrt{3}\over 2}={{\sqrt{3}-1}\over{2\sqrt{2}}}=\sin 15^\circ\Longrightarrow\angle MBD=15^\circ\end{eqnarray*}

Hence from $\triangle MBD$ we get $\angle ADB=105^\circ$[/hide]

[hide="Here are a interesting lemma and a classical solution."]
[b]Lemma.[/b] Let $ABCD$ be a convex quadrilateral. I note:
$m(\widehat {BAC})=a_1,\ m(\widehat {CBD})=b_1,\ m(\widehat {DCA})=c_1,\ m(\widehat {ADB})=d_1;$
$m(\widehat {DAC})=a_2,\ m(\widehat {ABD})=b_2,\ m(\widehat {BCA})=c_2,\ m(\widehat {CDB})=d_2.$
Then there is the following relation:
$\boxed {\ \sin a_1\cdot \sin b_1\cdot\sin c_1\cdot\sin d_1=\sin a_2\cdot\sin b_2\cdot\sin c_2\cdot\sin d_2\ }\ \ (1).$

[u]The solution of the proposed problem.[/u]

I note $m(\widehat {ADB})=x$. Then the relation $(1)$ becomes:
$\sin 15\sin (60+x)\sin 45\sin 30=\sin 90\sin 15\sin x\sin 15\Longleftrightarrow$
$\sin (60+x)=4\sin x\sin^215\Longleftrightarrow \sin (60+x)=2\sin x(1-\cos 30)$
$\Longleftrightarrow\sqrt 3\cos x=(3-2\sqrt 3)\sin x\Longleftrightarrow \cos x=-(2-\sqrt 3)\sin x$
$\Longleftrightarrow\tan x=-(2+\sqrt 3)\Longleftrightarrow \boxed {\ x=105\ }.$

[u]Remark.[/u] $\tan 75=\tan (45+30)=\ldots =2+\sqrt 3.$
[/hide]

I offer a similar problem (exercise) for using the lemma.

 $\triangle ABC$, $A=50$, $P\in (ABC)$, $PB=PC$; $E\in PB\cap AC$, $F\in PC\cap AB$, $m(\widehat {PEF})=50$, $m(\widehat {PFE})=30$ $\Longrightarrow$ $m(\widehat {ABE})=\boxed {\ ?\ }.$

$\boxed {\ 03=rewsnA\ }$

A synthetic solutioon:
Construct $E$ the circumcenter of $\triangle ABC$, see that $\triangle ABE$ is equilateral, while $\triangle AEC$ has angles $75^\circ-75^\circ-30^\circ$. Take $F$ the circumcenter of this triangle, hence $\triangle AFC$ is equilateral, consequently we shall get $AF=AC=AD$ and $\angle DAE=\angle FAE=15^\circ$. Due to $EF=AF$ we find that $D$ lies onto the perpendicular bisector of $AE$, i.e. $\angle ABD=\angle EBD=30^\circ\implies\angle ABD=105^\circ$.

Note: some steps of the proof have been skipped, but you may easily make your own angle chase.

Best regards,
sunken rock


Does anyone know of any piano competitions for intermediate to advanced players?

If you're not really really good look into local competitions. They probably don't have very strict requirements. (that's what I'm gonna do :))

There are tons of piano concerto competitions out there. Depends on what state you're in.

There is also the MTNA Junior Performance Competition that is very competitive.


function
linear algebra
linear algebra unsolved

I'll start out simple to introduce the basic idea.

Let's say you have a function f(s) that diverges to positive infinity as s approaches 1 from above. And let's say you know for certain that for all s > 1, f(s) x g(s) = 1. Then it is obvious that g(s) tends to 0 as s approaches 1 from above.

But what if you had a more complicated system?
Let's jump into four variables, for example.

For all s > 1, you know that:

A(s) x a(s) + B(s) x c(s) + C(s) x b(s) + D(s) x d(s) = 1
A(s) x b(s) + B(s) x a(s) + C(s) x d(s) + D(s) x c(s) = 0
A(s) x c(s) + B(s) x d(s) + C(s) x a(s) + D(s) x b(s) = 0
A(s) x d(s) + B(s) x b(s) + C(s) x c(s) + D(s) x a(s) = 0

Let's say you know that A(s), B(s), C(s), and D(s) all diverge to positive infinity as s approaches 1 from above, and you know they all stay very close together as this happens (all staying within some constant amount of each other).

From this information, how would you go about proving that the limits of a(s), b(s), c(s), and d(s) are finite as s approaches 1 from above?

What about a situation where the number of variables and equations are both h?

Sorry to double post, but I think I figured out my own question!
I want to see if someone else comes up with the same technique.
You might need one more hint, though...


superior algebra
superior algebra unsolved

If $ L$ is a lattice we say that an element $ a\in L$ is [i]join irreducible[/i] if $ a \equal{} b\vee c$ implies $ a \equal{} b$ or $ a \equal{} c$. If $ L$ is a finite lattice show that every element is of the form $ a_1\vee  a_2\vee  ...\vee  a_n$ where each $ a_i$ is join irreducible.

if the given $ x \in L$ is irreducible, we're done. otherwise we may write $ x \equal{} a \vee b$ with $ x \neq a, x \neq b$. now continue with $ a,b$, and so on. the process has to terminate since in every step all  appearing elements are different.


Find all pairs $ (a, b)$ of real numbers such that whenever $ \alpha$ is a root of $ x^{2} \plus{} ax \plus{} b \equal{} 0$, $ \alpha^{2} \minus{} 2$ is also a root of the equation.
[b][Weightage 17/100][/b]

[quote="makar"]Find all pairs $ (a, b)$ of real numbers such that whenever $ \alpha$ is a root of $ x^{2} + ax + b = 0$ (1), $ \alpha^{2} - 2$ is also a root of the equation.[b]
[Weightage 17/100][/b][/quote]

If $ a$ and $ a^2-2$ are roots then from (1) we get $ a^2+a^2+b=0$ (2) and $ (a^2-2)^{2}+a(a^2-2)+b=0$ (3)
(2)=(3) so $ a^4+a^3-6a^2-2a+4=0\Longleftrightarrow (a-2)(a+1)(a-\frac{1+\sqrt{3}}{2})(a+\frac{1+\sqrt{3}}{2})=0$... we examine then for the different values of $ a$ which values of $ b$ we get

For $\alpha$ is a root, then $\alpha^2- 2$ is a root. We have following cases here.
Case 1: $\alpha = \alpha^2 -1$. This gives $\alpha = -1, 2.$ From these values of $\alpha$, Obtain the values of $a$ and $b.$ Here $(a,b) =\{ (-4,4), (2,1)\}.$
Case 2: $\alpha \neq \alpha^2 -1.$ Quite Complicated(think later).

For $\alpha ^ 2 -2$ as a root, then $(\alpha^2 -2)^2 -2$ is also a root and $\alpha$ may or may not be a root. We have following cases.
Case 3: $\alpha ^ 2 -2 =(\alpha^2 -2)^2 -2 \implies \alpha = \pm 1, \pm 2 $. Thus the pairs of $(a, b) = \{ (0,-4), (0,-1)\}. $
Case 4: $(\alpha^2 - 2)^2 - 2 = \alpha \implies \alpha = \dfrac{-1\pm \sqrt{5}}{2}.$ Thus the pairs of $(a,b) = \{ (1,-1)\}.$

[quote=guptaamitu1]file:///C:/Users/arjun/Documents/IMG_20200923_0001.pdf
This is my solution (the link for the solution is given above). I think it is complete and a correct solution, though it may contain some small mistakes.
If you find some mistake, please tell me so that i can improve upon that.[/quote]

Umm.... That is a local file, only accessible on your computer. Try making a new post, and click on the little paperclip button, then upload that file.

[quote=smo]Wouldn't Vieta lead to infinite solutions?[/quote]

Are you crazy?
It says $a, b \in \mathbb{R}$.
[quote=makar]Find all pairs $ (a, b)$ of [b]real[/b] numbers such that whenever $ \alpha$ is a root of $ x^{2} \plus{} ax \plus{} b \equal{} 0$, $ \alpha^{2} \minus{} 2$ is also a root of the equation.
[b][Weightage 17/100][/b][/quote]

[quote=Brut3Forc3]Let $ f(x)\equal{}x^2\minus{}2$. If $ r$ is a root of the equation, then $ f(r)$ must be as well. One case this is possible when $ f(r)\equal{}r \iff (r\minus{}2)(r\plus{}1)\equal{}0$, or $ r\equal{}2,\minus{}1$. This corresponds to the solutions $ (4,4), (\minus{}2,1)$.

Now let $ u$ be a root of the equation such that $ f(u)\ne u$. We must either have $ f(f(u))\equal{}u$, or $ f(f(u))\equal{}f(u)$. The second case is simply $ f(u)\equal{}2,\minus{}1$, or $ u\equal{}\pm2,\pm1$. We have the possible set of roots $ \{2,2\},\{\minus{}2,2\},\{1,1\},\{\minus{}1,1\}$. This gives us two new solutions: $ (0,\minus{}4),(0,\minus{}1)$.

The final case gives us $ (u^2\minus{}2)^2\minus{}2\equal{}u \iff u^4\minus{}4u^2\minus{}u\plus{}2\equal{}(u\minus{}2)(u\plus{}1)(u^2\plus{}u\minus{}1)\equal{}0$. We can discard $ u\equal{}2,\minus{}1$, since it is already in the previous cases, and the final solution is $ (1,\minus{}1)$.

All solutions are $ (4,4),(\minus{}2,1),(0,\minus{}4),(0,\minus{}1),(1,\minus{}1)$.[/quote]

Just the exact solution from the internet....  :ninja:


The incircle $ C(I)$ of the triangle $ ABC$ touches the its sides in

the points $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ . Define the points

$ \{\begin{array}{c}
M\in DE\ ,\ AM\parallel BC\\\
N\in DF\ ,\ AN\parallel BC\end{array}$ and $ \{\begin{array}{c}
X\in (DM)\ ,\ XD=XM\\\
Y\in (DN)\ ,\ YD=YN\end{array}$ .

Prove that the points $ A$ , $ X$ , $ Y$ and $ I$ are concyclically.

$ \angle DMA \equal{} 90 \minus{} \frac{\gamma}{2}$, $ \angle DNA \equal{} 90 \minus{} \frac{\beta}{2}$, $ NE \parallel CI \perp ED$, $ MF \parallel BI \perp DF$, $ H: MF \cap NE$.

So H in on the incircle and I is the midpoint of DH. MNFE in cyclyc so A is the midpoint of MN. So X, Y, A and I are point on the feuerbach circle of DMN.

[quote="Virgil Nicula"]The incircle $ C(I)$ of the triangle $ ABC$ touches the its sides in

the points $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ . Define the points

$ \{\begin{array}{c} M\in DE\ ,\ AM\parallel BC \\
\ N\in DF\ ,\ AN\parallel BC\end{array}$ and $ \{\begin{array}{c} X\in (DM)\ ,\ XD = XM \\
\ Y\in (DN)\ ,\ YD = YN\end{array}$ .

Prove that the points $ A$ , $ X$ , $ Y$ and $ I$ are concyclically.[/quote]
Nice problem!
[hide="Alternate Solution"]
We see that $ \angle MFA = \angle DFB = \angle FDB = \angle FMA\implies MA = FA$.  Similarly, $ AE = AN$.  Since $ AF$ and $ AE$ are tangents to the incircle, $ AF = AE$.  Thus, $ MA = AF = AE = AN$, so $ A$ is the midpoint of $ MN$.  Also, $ Y$ is the midpoint of $ ND$, hence $ AY\parallel MD$.  Now, note that $ AE = AF\implies \angle AIE = \frac {\angle EIF}{2} = \angle MDN = \angle MXA$.  Thus, we obtain that $ AIXE$ is cyclic.  Similarly, $ AFYI$ is cyclic.  Since $ \angle IEA = 90 = \angle IFA$, we have that $ AEIF$ is cyclic.  Therefore, $ A,E,X,I,Y,F$ are concyclic and we are done. [/hide]


abstract algebra
superior algebra
superior algebra unsolved

Let $ A$ be a commutative ring with finite number of divisors of zero and at least one divisor of zero different from the null one. Prove that $ A$ is finite and if $ |A|\equal{}n$ then $ |U(A)|<n\minus{}[\sqrt{n}]$.

Consider the (group)homomorphism $ R\to R: x\mapsto ax$. If a is a nontrivial zero divisor, both kernel and image are contained in the set of zero divisors. So they are finite and because of $ |R| \equal{} |ker(f)|\cdot|im(f)|$ R is finite too.

If $ |R| \equal{} n$, then $ |ker(f)|$ and $ |im(f)|$ are divisors of n. One of them is $ \geq\sqrt {n}$, so there are at least $ \sqrt {n}$ zero divisors. The set of units is the complement of the set of zero divisors (the ring is finite!) and so $ |R^\times|\leq n \minus{} \sqrt {n}$.

Nice solution, thanks, but I have a question, can you please post a demonstration of the following:
[quote="Gockel"] $ |R| \equal{} |ker(f)|\cdot|im(f)|$ .[/quote]?(I didn't see it until now though I guess it's somethinkg well known :( )

$ R/ker(f) \equal{} im(f)$ as quotient group


I want to make something like:

[img]http://img74.imageshack.us/img74/7550/dfsdhf.png[/img]

which means, centered horizontal and vertical.

$ \vspace*{\fill} \centerline{TEXT} \vspace*{\fill}$

It doesn't work...

It works perfectly. Try
[code]\documentclass{article}
\begin{document}
\vspace*{\fill}
\centerline{Text}
\vspace*{\fill}
\newpage
more text
\end{document}[/code]and you'll see that the document has 'Text' in the centre of the first page and 'more text' starts on page 2. If that doesn't happen then you need to post the log errors here.


pigeonhole principle
algorithm
number theory
greatest common divisor
least common multiple
probability
Euclidean algorithm

Using the pigeonhole principle to prove the following version of the Chinese Remainder Theorem:
[quote]If [tex]m[/tex] and [tex]n[/tex] are relatively prime positive integers and [tex]a[/tex] and [tex]b[/tex] are any integers, there exists an integer [tex]x[/tex] such that [tex]x\equiv a\mod m \mbox{ and } x\equiv b\mod n[/tex][/quote]
By the way, most textbooks provide algorithmic proof to the Chinese Remainder Theorem using (extended) Euclidean Algorithm for GCD of two integers.

The fact that you haven't used pigeonhole principle in your solution doesn't mean that there isn't such a solution.

[quote="probability1.01"]WLOG, a < m and b < n.  LCM(m,n)=mn, so for every x such that 1 :le: x :le: mn there must be a distinct pair (a,b) where a = x mod m and b = x mod n.  If this is not true, then it would contradict the fact that LCM (m,n)=mn.  Since 0 :le: a < m and 0 :le: b < n, there are only m*n possible pairs of a and b.

This isn't really the pigeonhole principle.  The pigeonhole principle is generally used for showing that at least one case must satisfy something, not that all cases satisfy something.[/quote]

Try to be a little more rigorous... its a relatively simple theorem, so rigor becomes important.

[quote="zscool"]a try: consider the n numbers a, a + m, a + 2m, ..., a + (n-1)m, each is congruent to a mod m.  Suppose one of the n numbers is not congruent to b mod n, then there are n - 1 possible remainders for the n numbers, so by the pigeon-hole principle there exists i, j with i [tex]\neq[/tex] j and i,j < n s.t. a + im :equiv: a + jm mod n => m(i - j) :equiv: 0 mod n => i :equiv: j mod n (since (m,n) = 1) and because i, j < n, i = j...contradiction.

Therefore there exists a t s.t. a + tm :equiv: b mod n, and a + tm :equiv: a mod m.[/quote]

I know there are a lot proofs. This is the one I am looking for. The one sketched by probablity 1.01 is pretty nice too.


vote what place you got at state. like the 2007. wow no one has posted in a year. I KNOW WILLIAM. hes my friend=)

useless spamming  :D

EDIT: Oh, wait crap... did I answer the poll. I never mad nats in 7th grade, sigh  :(


number theory
prime numbers
number theory open

[quote="energizersss"]can someone give me the solution of:

for which prime numbers x and y (if any) that 5yx^2+16xy^2 is a perfect square.

if possible can you provide the proof of the theorm used in solving this problem (if any)[/quote]
If x and y are distinct primes then x, y and 5x+16y are all coprime. So the prime factors x and y can only occur once in xy(5x+16y), which cannot therefore be a square. [b]Edit.[/b] Corrrection: I was overlooking the possibility that x or y could be 2 or 5. There is a solution (x,y) = (2,5) !

If x=y then $ xy(5x \plus{} 16y) \equal{} 21x^3$, which again cannot be a square.

Therefore: no prime solutions [except (2,5).]

It's an interesting problem to find (nonprime but nonzero) integer solutions. If (x,y) is a solution then obviously so is $ (xk^2,yk^2)$ for any natural number k. So we look for solutions in which x and y do not share a squared factor. One such solution is (4,1). There is also an infinite family (x,y) = (9,5), (25,45), (49,180), ..., given by $ x \equal{} (2n \plus{} 1)^2$, $ y \equal{} \textstyle\frac54n^2(n \plus{} 1)^2$. I don't know if there are others.


How many positive even perfect cubes are less than 2008?

The cube root of 2008 is roughly 12. Half of these are even, so our answer is $ 12 \div 2$ or $ \boxed{6}$.


Θα ήθελα μέσω αυτού του τόπικ να συζητηθούν τα παρακάτω θέματα:
1)Κατά πόσο η κατάληψη υποβαθμίζει τα ελληνικά πανεπιστήμια
2)Κατά πόσο η κατάληψη είναι μέσο πίεσης
3)Αν είναι όντως μέσο πίεσης, τι διαφορά έχει από τα μέσα πίεσης που λέγετε ότι εφαρμόζουν ενώσεις (lobby) και είναι κατά πολλούς καταδικαστέα
4)Το κατά πόσο είναι θεμιτό μια ομάδα να αποφασίζει για τα δικαιώματά της αδιαφορώντας για την κοινή γνώμη και παραβλέποντας τη νόμιμη αρμόδια αρχή
σε όσους αναρρωτιούνται τι δουλειά έχει αυτό το τόπικ σε αυτό το φόρουμ θα ήθελα να απαντήσω ότι με ενδιαφέρει μια συζήτηση με επιχειρήματα που να προσεγγίζουν τη μαθηματική απόδειξη και όχι τα σοφίσματα πολλών ατόμων σε άλλα φόρουμ

Πολύ ενδιαφέρον και επίκαιρο θέμα thatin. Αν και δεν είμαι φοιτητής, θεωρώ ότι δεν είναι τόσο τα ελληνικά πανεπιστήμια που υποβαθμίζονται (πολλά τμήματα ούτως ή άλλως δε θεωρούνται και ό,τι καλύτερο, ας μην κρυβόμαστε πίσω από το δάχτυλό μας), αλλά το όλο εκπαιδευτικό σύστημα.
Είναι μπούμεραγκ για τους φοιτητές. Αργά ή γρήγορα θα το καταλάβουν. Τι γνώμη έχεις εσύ?

Σεβαστό Mostel αλλά προσωπικά δε θα θυσίαζα τέτοιες σπουδές για το ελληνικό σουβλάκι :lol: (ούτε εσύ θέλω να πιστεύω).

Άκουσες ποτέ για προβλήματα σε Αμερικάνικα πανεπιστήμια? Ποτέ!!! Εδώ στην Ελλάδα πότε ο Γιάννης δεν μπορεί πότε ο...ξέρεις τη συνέχεια φαντάζομαι.

Σοβαρά τώρα στην Αμερική οι σπουδές είναι επένδυση, άσχετο αν βασανίζεσαι. Γι' αυτό βγαίνουν όλα τα μυαλά από κει. Έχω θείο καθηγητή engineering εκεί και ξέρω τι σου λέω.

Κοίταξε , τα παραπάνω τα είπα χάριν αστεϊσμού.

Φυσικά, δεν πιστεύω κάποιος να του δινόταν η ευκαιρία να πάει για undergraduate στο εξωτερικό (σε ενα αξιοπρεπές πανεπιστήμιο) και να μην πήγαινε.

Απλά, θέλω να πιστεύω πως τα ελληνικά πανεπιστημία ΜΠΟΡΟΥΝ και ΘΑ ΠΑΝΕ μπροστά.

 :)

Έχουν τα φόντα και το δυναμικό. Υποδομή δεν έχουν γιατί δεν πέφτει χρήμα. Μακάρι πάντως...

kata ti gnomi mou dn einai to xrima o logos

alla i politiki pou akolou8eitai kai i prosoxi pou dinetai se ayta...

se ligo kairo eidika an (opos ola deixnoun) epitrapei i dimiourgia idiotikon panepistimion ta dimosia 8a pane ap'to kako sto xeirotero
...oti egine diladi kai me ta antistoixa gymnasia kai lykeia...

Δε συμφωνώ. Ιδιωτικά λένε και ιδιωτικά δε φαίνονται. Κάτι για ΜΙΤ έλεγαν, χρήμα δεν έπεσε, άφαντο. Το χρήμα φίλε Δημήτρη παίζει καθοριστικό ρόλο και αφού ζούμε στην Ελλάδα που δεν της τρέχει απ τα μπατζάκια, δε χρειάζονται ιδιωτικά πανεπιστήμια για να πάνε απ το κακό στο χειρότερο τα δημόσια. Πάνε από μόνα τους αν με πιάνεις... :)

Όντας ένας μαθητής δημόσιου λυκείου, εν μέρει έχεις δίκιο για την κατάσταση που αναφέρεις. Ωστόσο, προσωπικά, αν και η κατάσταση είναι έτσι, δεν έχω παράπονο από την παιδεία που μου παρέχεται. Βελτιώσεις μπορούν σίγουρα να γίνουν ! Βλέπεις, δεν φταίνε τόσο τα χρήματα που παρέχονται στην παιδεία όσο η νοοτροπία του Έλληνα, του οχαδελφισμού. Ο ρόλος του μαθητή πρέπει να είναι διττός, πρώτον σωστός πολίτης και δεύτερον άξιος μαθητής. Αλλά και τα δύο βαυκαλίζονται από εξωγενείς παράγοντες στους οποίους εμείς οι μαθητές είμαστε επιρρεπείς (π.χ. καταπάτηση δικαιωμάτων, καταστροφή ιδιοκτησίας κ.λπ.). Βλέπεις, ο Έλληνας εκτιμάει κάτι μόνο όταν πληρώνει. Όταν είναι δημόσιο το κατασπαράζει ανελέητα. Ίσως είναι υπερβολικά αυτά που γράφω, αλλά αυτή είναι η πραγματικότητα που ταλανίζει αδίσταχτα την ελληνική μαθητική κοινωνία. Ο Μουζέλης γράφει: [i]"Από τη μια, οι πολίτες υποστηρίζουν ανεπιφύλακτα και με απόλυτα πατριωτικό πνεύμα τα εθνικά ιδεώδη, μη διστάζοντας ακόμη και να δώσουν τη ζωή τους σε περιπτώσεις πολέμων. Από την άλλη μεριά, όμως, οι ίδιοι αυτοί πολίτες δεν έχουν κανέναν ενδοιασμό να κλέβουν το κράτος (είτε έμμεσα, μέσω της φοροδιαφυγής, είτε άμεσα, ξοδεύοντας αφειδώς ή οικειοποιούμενοι δημόσιο χρήμα), να καταστρέφουν το περιβάλλον ή να προωθούν με φανατισμό τα προσωπικά τους συμφέροντα εις βάρος των συμφερόντων της πλειοψηφίας των πολιτών."[/i]

Συνέχιζω ακράδαντα να πιστεύω πως η κατάσταση στο προσεχές μέλλον θα παρουσιάσει σταδιακά σημάδια βελτίωσης.

i pragmatiki omos aitia,kosta, dn einai i elleipsi xrimaton alla i askopi dapani toys(se polemikous skopous k.t.l)

ayto einai pou 8elo na diloso :wink:

file mostel,

to 8ema einai pos dn ginontai vimata pros ayti tin katey8insi kai etsi dn yparxoun simadia pou na me kanoun na eyelpisto gia tin tyxi ton ellinikon panepistimion... :(

Mostel συμφωνώ όμως δεν είναι πάντα έτσι. Μια μεγάλη μερίδα μαθητών-φοιτητών έχει μάθει να σέβεται και να εκτιμά και είναι αρκετά συνειδητοποιημένοι, ίσως η πλειοψηφία. Δε γίνονται προσπάθειες από το κράτος όμως και αυτό τα λέει όλα γιατί κακά τα ψέματα από το κράτος εξαρτάται. Οι άνθρωποι δεν αλλάζουν από μόνοι τους (αν το δεις φιλοσοφικά). Εύχομαι να μην διαψευστούν οι ελπίδες σου μόνο και μόνο για να αλλάξει το image της χώρας μας στο εξωτερικό...

Τι μου λες τώρα ρε Δημήτρη για αλόγιστη δαπάνη? Η Ελλάδα είναι χρεωμένη από δω κι από κει. Τα πανεπιστήμια θα σωθούν? Εδώ το έθνος πάει για χρεωκοπία. Και με όλα αυτά που γίνονται με τους γείτονες είναι αλόγιστη δαπάνη οι πολεμικοί σκοποί? Anyway ξεφεύγουμε νομίζω αλλά κάθε τι που λέγεται καλό είναι μήπως γίνει κάτι.

Δεν θέλω να έρθω σε αντιπαράθεση μαζί σου, αλλά δυστυχώς, τα παραδείγματα που αναφέρεις ανήκουν στη μειοψηφία. Στο σχολείο που φοιτώ εγώ, το 90% των μαθητών φέρουν την "ιδεολογία" που ανέφερα ανωτέρω.

ΔΕΝ ξέρω τι γίνεται στα ιδιωτικά λύκεια ! Εκεί η κατάσταση ίσως είναι αρκετά καλύτερη.

Άλλωστε αν ήταν η πλειοψηφία συνειδητοποιημένοι, δεν θα είχαμε ούτε καταλήψεις, ούτε.. ! Η κατάληψη δεν είναι τρόπος διεκδήκησης (κατ' εμέ), τουλάχιστον στην έκταση που έχει λάβει αυτούς τους μήνες και στον τρόπο με τον οποίο γίνεται.

Όσον αφορά το image της χώρας στο εξωτερικό, είναι το τελευταίο που πρέπει να σε νοιάζει. Τουλάχιστον, η κατάσταση θέλω να βελτιωθεί ΠΡΩΤΑ για 'μένα.

Επίσης, αναφέρεις ότι το κράτος είναι υπεύθυνο και πως θα πρέπει να νουθετήσει κατάλληλα τους πολίτες. Οι άνθρωποι ΜΠΟΡΟΥΝ να αλλάξουν από μόνοι τους. ΑΝ τα περιμέναμε όλα από το κράτος... Ας αρχίσουν οι προσπάθειες πρώτα από εμάς , και σιγά σιγά, μέρα με τη μέρα, μαζί με τη βοήθεια του κράτους, τα πράγματα θα βελτιωθούν (θα το λέω συνέχεια για να το εμπεδώσω).

Όλα αυτά τα λέω πάντα σε φιλικό επίπεδο ώστα να γίνει και μια εποικοδομητική συζήτηση. Σέβομαι την άποψή σου, απλώς παρέθεσα ένα-δύο σημεία στα οποία δεν συμφώνησα. Χαίρομαι ιδιαίτερα που υπάρχουν άτομα με παρόμοιες ανησυχίες!

Στέλιος
 :)

Για την κατάσταση στα Πανεπιστήμια σίγουρα δεν φταίει μόνο ένας.

Φταίνε οι πολιτικοί (κανένα σημάδι επιθυμίας για βελτίωση της κατάστασης).

Φταίνε οι Πανεπιστημιακοί (ακόμα και να υπάρχουν κάποιοι που θέλουν ν ακάνουν σωστά τη δουλεία τους είναι δαχτυλοδεικτούμενοι αό τους υπόλοιπους)

Φταίνε οι φοιτητές (απόδειξη: η κατάληψη χωρίς κανένα πλέον νόημα)

Αλλά κυρίως φταίει ολόκληρη η κοινωνία. Τέτοια πανεπιστήμια ζητάει και τέτοια παίρνει. Γι΄αυτό και πιστεύω ότι, παρ' όλο που θα θέλαμε να υπάρξει βελτίωση της κατάστασης στο μέλλον, δύσκολα θα γίνει κάτι ουσιαστικό.

Χάρηκα για τη συζήτηση Στέλιο. Ίσως τελικά να έχεις και δίκιο. Φαίνεται πως υποστηρίζεις ένθερμα όσα λες. Ας ελπίσουμε ότι η κατάσταση θα βελτιωθεί. Πραγματικά θέλω να πιστέψω ότι είναι στο χέρι μας απλά δυσκολεύομαι ακόμα...Ο Αλέξανδρος τοποθετήθηκε ωραία και διπλωματικά :)

[quote="mostel"]Λόγοι για να μείνει κάποιος στο ΕΜΠ:

Τις προάλλες μιλούσα με τον ξάδερφο μου που είναι στο Stanford και του λέω:

"Να έρθω εγώ USA και συ στο Greece?"

Και μου απαντάει:

"ΕΤΟΙΜΑΣΟΥ, ΕΓΩ ΕΧΩ ΠΑΚΕΤΑΡΕΙ ΤΑ ΠΡΑΓΜΑΤΑ ΜΟΥ ΗΔΗ."

Όπως είναι φυσικό, μου φάνηκε περίεργη η αντίδρασή του. Όταν τον ρώτησα γιατί το είπε αυτό μου απήντησε:

"Εδώ δεν έχει ούτε σουβλάκια, ούτε πολλή ζωή, ούτε..."

ΣΥΜΠΕΡΑΣΜΑ:

Ναι, οκ, η παιδεία ίσως είναι καλύτερη αλλά... τι προτίματε; Τα σουβλάκια ή τον Riemann ?  :P

----------

Τα ξένα π[b]αινε[/b]πιστήμια ΔΕΝ είναι πανάκια.[/quote]
μπορεις να παρεις μερικα σουβλακια μαζι...μαλιστα τωρα που το σκεφτομαι καλο θα ηταν ενα σουβλατζιδικο με διεθνες delivery... :rotfl:  :rotfl:
παντως περα απο την πλακα ειδα σημερα την αποψη του κυριου κανακη στην τηλεοραση και με εκνε να αναλογιστω την ακρεα θεση μου(δλδ οτι οι καταληψεις ειναι παρανομες και πρεπει να μπουν μεσα να συλλαβουν καμια δεκαρια μπασ και παμε καθολου μπροστα)...ειπε λοιπον ο αντωνης(επιτρψτε μου ημουν και ειμαι φανατικος του ΑΜΑΝ..το νιωθω σα κοντινο φιλο) οτι μια τετοια κατασταση ειναι υγειης και ειναι καλυτερη απο το να καθονται στα φραπεδοκαρεκλα και να τα ξυνουν(να με συμπαθατε για την εκφραση αλλα ετσι ειναι)...επισης ειπε οτι τον στεναχορησε ιδιετερα το γεγονος οτι οι φοιτητες δερνονταν μεταξυ τους...πραγματι καλα να δερνονται με τα ΜΑΤ αλλα ειναι λυπιρο να βλεπει κανεις μια τετοια δυχονοια...

Να σας πω την αποψή μου:

Έχω προσωπική εμπειρία και με το προαναφερθέν ΜΙΤ  και με το καλύτερο ίσως πολυτεχνέιο της χώρας μας το ΕΜΠ στο οποίο κάνω διδακτορικό στον τομέα τη ρομποτικής και του αυτομάτου ελέγχου (αυτό το τονίζω όχι για επείδηξη αλλά για να τονίσω ότι κάνω phd σε έναν τομέα αιχμής τεχνολογίας).
Λοιπόν θεωρώ ότι το ΜΙΤ είναι λίγο υπερτιμημένο και η θεωρία ότι εμείς εδώ κτυπάμε μύγες είναι λάθος.[b]Στο ΜΙΤ γίνεται καλύτερη δουλειά γιατί[/b]:
[b]α)[/b] έχουν άπειρα κονδύλια (DARPA κτλ.) [b]β)[/b] οι αγορές γίνονται άμεσα.Παραγγέλνεις κάτι το βράδυ και έίναι στο γραφέιο σου το μεσημέρι.Εδώ θέλεις 10 ημέρες, να λαδώσεις το τελωνείο, και μπορεί να μην έρθει και όλας λόγω γραφειοκρατίας.  [b]γ)[/b]οι μηχανικοί (που έχω προσωπικές επαφές) στο ΜΙΤ έχουν μεγάλη έμφαση στο design και τις βασικές αρχές λειτουργίας της φύσης.Από θεωρητικές γνώσεις και μαθηματικά είναι  αιώνες πίσω.Ακόμα θυμάμαι τον μορφασμό ενός μηχανικού από το ΜΙΤ όταν του μίλησα για partial differential equations!
Απεναντίας εμείς μπορεί να ξέρουμε τα απίστευτα αλλά να πνιγόμαστε σε μια κουταλιά νερό σε βασικά πράγματα (του στυλ ποιές δυνάμεις ασκούνται πάνω σε ένα ψάρι κτλ).Μας λείπουν λίγο οι πρακτικές εμπειρίες και τα πρότζεκτ.[b]δ)[/b]Στο ΜΙΤ υπάρχει φοβερός ανταγωνισμός και αξιοκρατία
αλλά στην Ελλάδα όλα επισκιάζονται από τις φοιτητικές παρατάξεις.Οι εκλογές, οι αναδείξεις καθηγητών, η ανάθεση διδακτορικών,διανομή κονδυλίων γίνονται τις περισσότερες φορές με συνθήκες μη ηθικού εμπορίου με αυτές.
Στην Ελλάδα όλοι οι φοιτητες αντιγράφουν από παλιά θέματα ή ασκήσεις. Τα μαθήματα που γίνονται π.χ σε φοιτητές ηλεκτρολόγους μηχανικόύς εδώ, δεν είναι παρά επιλογής σε μεταπτυχιακό
στο ΜΙΤ! Δεν είναι τυχαίο ότι καθηγητές Έλληνες διαπρέπουν. Υπάρχουν περισσότεροι καθηγητές στο ΜΙΤ που αποφοίτησαν από το ΕΜΠ παρά από το Caltech (Γιαννάς,Μπαρμπαστάθης,Τριανταφύλλου κτλ.). Και ακόμα περισσότερoι που ήταν εκεί και ήρθαν εδώ για οικογενειακούς λόγους και λόγω καιρού.Ξέρετε τι διαολοκρυο τρώς χειμώνα στην Βοστόνη?Πάρτε μία γευση από την κάτωθι φωτογραφία.Εγώ προτιμώ τα σουβλάκια του Mostel. Άλλωστε είναι τυχαίο ότι το εργαστήριο μου κάνει 5-10 journal και 10-15 paper στα καλύτερα συνέδρια το χρόνο?Και από τους καλύτερους φοιτητές που θα μπορούσαμε όλοι να πάμε έξω μόνο ένας έφυγε για ΜΙΤ.Οι άλλοι προτίμησαν ελλάδα! Παρά τα προβλήματα κάνουν και οι Έλληνες καλή δουλειά! Φανταστείτε να είχαμε και κονδύλια! ΟΙ γνώστες θα δουν και τον τρούλο του ΜΙΤ  που αρχίζει ο πιο τρελός διάδρομος σε campus, που υπάρχει...

πάντως ωραίο είναι το κρύο...
και όσον αφορά το θόλο... το περιπολικό που είναι? :rotfl:

τι εννοείς <Στην Ελλάδα όλοι οι φοιτητες αντιγράφουν από παλιά θέματα ή ασκήσεις>??

Θα συμφωνήσω με τον mimmis. Επαφή κι εμπειρία από ξένα Πανεπιστήμια δεν έχω. Αλλά έχω μία (ατομική κι ίσως εκτός πραγματικότητας) ιδέα για το δικό μου.

Δεν ξέρω τί γίνεται με τις παρατάξεις και τους φοιτητές. Έχω ακούσει για αντιγραφές και λυμένα θέματα, έχω ακούσει για ομαδικό "κόψιμο" από καθηγητή γι' αυτόν ακριβώς το λόγο, έχω ακούσει για αδικίες και μεθόδους "ανεμιστήρα" στη διόρθωση τελικών γραπτών.

Από προσωπική εμπειρία, πάντα πήρα το βαθμό που περίμενα. Υπάρχει μία μόνο εξαίρεση μέχρι στιγμής, αλλά αυτό είναι πονεμένη ιστορία. Τέλος πάντων. Οι περισσότεροι καθηγητές στους οποίους παρακολούθησα εγώ, είχαν όρεξη. Και είχαν πολλή όρεξη!

Δεν έχω ιδέα τί γίνεται με κονδύλια και χρήματα και παροχές και όλα τα σχετικά. Η πραγματικότητά μου διαφέρει πολύ από αυτή των περισσότερων. Αλλά όποτε χρειάστηκα κάτι, σε όποιον καθηγητή και να απευθύνθηκα, ανταποκρίθηκε άμεσα και επαρκώς.

Επίπεδο σπουδών δε ξέρω συγκριτικά, αλλά θεωρώ πως το δικό μας είναι καλό. Αν ασχοληθεί ο φοιτητής όπως πρέπει, είναι πολύ καλό. Ξέρετε τί δεν είναι καλό; Οι φοιτητές...

Κι εδώ ερχόμαστε στο αρχικό θέμα. Η κατάσταση είναι τραγική, κατά την άποψή μου. Καθότι δεν εξετάζουμε αν είναι καλό η όχι η δημιουργία ιδιωτικών Πανεπιστημίων, θα εστιάσω απευθείας στο θέμα των καταλήψεων. Δεν είναι μέτρο πίεσης. Είναι μέτρο εκτόνωσης των φοιτητών σε βραδινές ή ανάλογες εξόδους.

Θα προκαλούσαν πολύ περισσότερη ζημιά στην κυβέρνηση αν ήθελαν να αντιδράσουν οργανώνοντας 3 πορείες την εβδομάδα στο κέντρο της Αθήνας, παρά 3 μήνες κατάληψη. Όπως και να το κάνουμε, η κατάσταση αυτή ζημιώνει τους φοιτητές περισσότερο από οποιονδήποτε άλλο. Αλλά όπως ήδη αναφέρθηκε, δεν έχουμε στην πλειοψηφία μας την ωριμότητα και τη θέληση να το καταλάβουμε.

Αντιδρούμε για να αντιδρούμε, γιατί κάποιος μας είπε πως οι φοιτητές είναι το αύριο του τόπου και το έχουμε δει όλοι επαναστάτες άγνωστης αιτίας. Δε λέω πως οι αλλαγές είναι σωστές (ούτε και πως δεν είναι). Λέω όμως ότι κανένας δεν αναρωτήθηκε για το τί θα ήταν σωστό. Κανένας δεν πρότεινε, δεν ενημερώθηκε, δεν σκέφτηκε.

Όλοι θέλησαν να αντιδράσουν...
- Γιατί ρε παιδιά κάνουμε κατάληψη;
- Έλα μωρέ τώρα, έχουμε χρόνια μπροστά μας. Θες εσύ ιδιωτικά Πανεπιστήμια;
- Και τί αλλαγές θα φέρουν τα ιδιωτικά Πανεπιστήμια ρε παιδιά;
- Ε;... Δεν ξέρω, αλλά καλές δε θα είναι.

Και το χειρότερο όλων είναι πως κανένας δε συμμετέχει ατομικά, αλλά όλοι συμμετέχουν τυφλά και κομματικά.

Ξέφυγα λίγο από το αρχικό θέμα.  :oops:

[quote="VickiL"]
Δεν ξέρω τί γίνεται με τις παρατάξεις και τους φοιτητές. Έχω ακούσει για αντιγραφές και λυμένα θέματα, έχω ακούσει για ομαδικό "κόψιμο" από καθηγητή γι' αυτόν ακριβώς το λόγο, έχω ακούσει για αδικίες και μεθόδους "ανεμιστήρα" στη διόρθωση τελικών γραπτών.

Από προσωπική εμπειρία, πάντα πήρα το βαθμό που περίμενα. Υπάρχει μία μόνο εξαίρεση μέχρι στιγμής, αλλά αυτό είναι πονεμένη ιστορία. Τέλος πάντων. Οι περισσότεροι καθηγητές στους οποίους παρακολούθησα εγώ, είχαν όρεξη. Και είχαν πολλή όρεξη!

Δεν έχω ιδέα τί γίνεται με κονδύλια και χρήματα και παροχές και όλα τα σχετικά. Η πραγματικότητά μου διαφέρει πολύ από αυτή των περισσότερων. Αλλά όποτε χρειάστηκα κάτι, σε όποιον καθηγητή και να απευθύνθηκα, ανταποκρίθηκε άμεσα και επαρκώς.

Επίπεδο σπουδών δε ξέρω συγκριτικά, αλλά θεωρώ πως το δικό μας είναι καλό. [/quote]

Ζηλεύω :mad: Θέλω να 'ρθω κι εγώ!

μιλοντας για ΕΜΠ σκεφτειτε οτι τη φημη του την εχτισε απο συγκεκριμενα μυαλλα που εβγαζε κατα καιρους ,και οχι απο αυτους που τρεχουν για τσαμπουκαδες στις συνελευσεις κτλ,θελω να πω σε αρκετα πανεπιστημια υπαρχουν φοβεροι επιστημονες και βοηθοι μαθητες τους.παντα θα υπαρχει ποιοτητα εστω και λιγη


Find all real values of $ x$ that satisfy $ x^{\minus{}\frac{4}{5}}\equal{}2^9\minus{}2^8$.

[quote="123456789"]Find all real values of $ x$ that satisfy $ x^{ \minus{} \frac {4}{5}} \equal{} 2^9 \minus{} 2^8$.[/quote]

$ 2^9 \minus{} 2^8 \equal{} 2^8$

So we are trying to solve

$ x^{ \minus{} \frac {4}{5}} \equal{} 2^8$

$ 2^{ \minus{} 8} \equal{} x^{\frac {4}{5}}$

Take logs base 2

$ \minus{} 8 \equal{} \frac {2}{5} \log_2 x^2$

So $ \minus{} 20 \equal{} \log_2 x^2$

Then $ 2^{ \minus{} 20} \equal{} x^2$

So $ x \equal{} \pm 2^{ \minus{} 10} \equal{} \pm \frac {1}{2^{10}}$


function
calculus
derivative
analytic geometry
graphing lines
slope
real analysis

Can anyone explain why l'hopitals works.

Look up the proof in mathematical analysis textbook. I think most books contain it.

L'Hopital's rule is really two different rules in one, one describing what happens to a quotient of two functions as you approach a point, the other describing what happens as you head towards infinity.

For the first one (approaching a point), keep in mind that functions (when they are differentiable) behave very similarly to linear functions over very small ranges.  You can see this happening if you plot a function on a graphing calculator and keep on zooming in at a point--the function will look more and more like a straight line as you zoom closer in, and the derivative is the slope of that straight line.

Normally in L'Hopital's rule you'll have something that looks like $\frac{f(x)}{g(x)}$ with $f(a)=g(a)=0$.   Keeping in mind what I said above, what straight line would $f$ behave like?  What straight line would $g$ behave like?

For the second one (when $x$ is heading towards infinity), it helps to think of the in terms of rate of growth.  Of the functions $y=1.01x-10000$ and $y=0.99x+10000$, which one will become larger for very large $x$?  The point is that if $x$ is large enough, those "little" 10,000's really don't make much of a difference.  What matters is the 1.01 and the 0.99 in front of the $x$.  In the same way, if $f'(x)$ is approaching 2 as $x$ approaches infinity, then $f(x)$ will in the long run behave like $2x$, even if $f$ does strange things for small $x$.

Keeping this in mind, what should happen to $\frac{f(x)}{g(x)}$ for very large $x$?

The above doesn't constitute a proof in any proper sense of the world, and indeed it ignores many important cases (for example, what happens when $f'(x)$ and $g'(x)$ both are tending to 0?), but hopefully it gives a rough idea of what's going on.

i dont know how to do the syntax for equations, but i'll do the best i can

f prime = f(x+h) - f(x)
                     h

as h approaches 0.
but just say you make x + h = c
then h = c - x
so as h approaches 0, x approaches c.
you can either one minus either one

f prime = [u]f(x) - f(c)[/u]
                   x - c

as x approaches c.

if f(c) equals 0 and
g(c) = 0, then:

[u]f prime[/u]   =  [u]( f(x) - 0 ) / ( x - c )[/u]
g prime  =  ( g(x) - 0 ) / ( x - c )

or:

[u]f prime[/u] = [u]f(x)[/u]
g prime = g(x)
one of the indeterminate forms is 0/0
another +-oo/oo, can be expressed as:
+ [u]1/oo[/u]
-  1/oo
where f(c) and g(c) equals zero

i'm not as clear on the rest of the indeterminate forms 1^oo involves taking the natural logarith.  i think wikipedia might have the rest

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55058]More[/url] on L'Hôpital's rule.


factorial
floor function
number theory
least common multiple
relatively prime

Not really all that hard, you only have to know what to do
[hide]
2003 state
1. You just have to realize that $10^{100}$ has 101 digits and subtracting $9^{100}$ gets rid of one of them. The answer is 100
2. Notice that for 5! and up the last digit is 0. So for 5!+5 to 9!+9, the last digit is 9. For the rest, the factorials are easy.
1,2,6,24 so adding 1,2,3,4, we get 2,4,9,8, and adding, we get a units digit of 3. For the others, it is 5. 5+3=8
3. Use the formula $\frac{n(n+1)(2n+1)}{6}$ Put in 10.
$\frac{10(11)(21)}{6}$ and compute the answer. 385

2002 state
1. Each pair (1and 2,3 and 4...) give -1 to the total -1*50=-50
2. You have to take two powers of 2 off of each in the numerator . So now it is $2^2+2^3+2^4+2^5$. Easy enought to compute. 4+8+16+32=60.
3. For a number to be divisible by 3,4,5 it has to be divisible by their lcm. $3*4*5=60$ $500/60=50/6\approx8$ so the answer is 8[/hide]

Most of those are pretty easy, you just need to know how to approach them.

[hide="1"]$9^{100}$ is too small to remove any more than one digit from $10^{100}$, so the answer is $\boxed{100}$. This can be proved in a few different ways. For example, $a^x-b^x=(a-b)(a^{x-1}+a^{x-2}b+...+b^{x-1})$. Therefore, $a^x-b^x=a^{x-1}+a^{x-2}b+...+b^{x-1}$, which will fairly obviously have 100 digits.[/hide]

[hide="2"]This question isn't worded properly, but I'm assuming the units digit of the sum of the terms. The first four are straightforward, as they're rather small, and you don't need to calculate the factorial for the rest, as they will all have a factor of $10$ in them. Therefore, the units digits of each one is just the added number, so our answer will be $((1!+1)\equiv 2\bmod{10})+((2!+2)\equiv 4\bmod{10})+((3!+3)\equiv 9\bmod{10})+((4!+4)\equiv 8\bmod{10})+5+6+7+8+9\equiv \boxed{8}\bmod{10}$.[/hide]

[hide="3"]There are two ways to approach this. The sum of the first $n$ squares is $\frac{n(n+1)(2n+1)}{6}$, so the given expression will be equivalent to $\frac{10(11)(21)}{6}=\boxed{385}$. Or, just work them out and add them up, which isn't too difficult. $1+4+9+16+25+36+49+64+81+100=385$.[/hide]

[hide="4"]Each pair sums to $-1$, so the answer is $\boxed{-50}$.[/hide]

[hide="5"]Factor out $4=2^2$ from each term in the numerator to see $2^2+2^3+2^4+2^5$. From there, you can either just sum them: $4+8+16+32=60$. Or, you can use the formula $(1+2+2^2+2^3+...+2^n)=\frac{1-2^{n+1}}{1-2}=2^{n+1}-1$. Since you're adding up to $2^5$, your answer is $2^6-1$, except you also added $1+2$ at the beginning, so subtract that to find $60$.[/hide]

[hide="6"]Separately or together? If separately, use Principle of Inclusion-Exclusion: $\left\lfloor \frac{500}{3}\right\rfloor +\left\lfloor \frac{500}{4}\right\rfloor +\left\lfloor \frac{500}{5}\right\rfloor -\left\lfloor \frac{500}{3\cdot 4}\right\rfloor -\left\lfloor \frac{500}{3\cdot 5}\right\rfloor -\left\lfloor \frac{500}{4\cdot 5}\right\rfloor +\left\lfloor \frac{500}{3\cdot 4\cdot 5}\right\rfloor =\boxed{300}$. If together, they're relatively prime, so it must be divisible by $60$. $\left\lfloor \frac{500}{60}\right\rfloor =\boxed{8}$.[/hide]

[quote="goldendomer"]I looked over a few old countdown round questions and some just seems impossible to do in 45 seconds. Ex.

2003 state
-How many digits are in the whole-number representation of (10^100)-(9^100)?
-What is the units digit of the sequence 1!+1, 2!+2, 3!+3, ......, 9!+9?
-What is the sum of 1^2+2^2+3^2+......10^2?

2002 state
-Compute 1-2+3-4+5-...+99-100
-Compute (2^4+2^5+2^6+2^7)/4
-How many of the first 500 positive integers are divisible by 3, 4, and 5?

Any tricks to the answer?[/quote]

[hide]
No there are tricks to those. Lemme take a look at them.

Ill comment on #6

[hide]This is a great example of PIE, Principle of inclusion and exclusion (i hope).

When you have 3,4, and 5, you find all the multiples of each by dividing them in turn into 500, then rounding down.

Then, you have to realize you double counted those that are multiples of both 3 and 4, 4 and 5, or 3 and 5.  So, you take aways multiples of 12, 20 and 15.

Then, you have again taken away too many of LCM(3,4,5), so you need to add back those that are multuiples os 3*4*5=60. [/hide]

[hide="1"]
100, because $10^{100}$ has 101 digits and $9^{100}$ takes away one digit.[/hide]
[hide="2"]
$5!$ and up end in zeros.
The first 4 end in 1, 2, 6, 4; $1+2+3+4+5+6+7+8+9=45$.
$5+3=\boxed{8}$[/hide]
[hide="3"]
I just added them up because I don't know the formula:
$1+4+9+16+25+36+49+64+81+100=\boxed{385}$[/hide]
[hide="4"]
Each pair is $-1$ and there are 50 pairs: $\boxed{-50}$[/hide]
[hide="5"]
Since you divide by 4, you take 2 off of the exponent: $4+8+16+32=\boxed{60}$[/hide]
[hide="6"]
The LCM of 3, 4, and 5 is 60:
$\left\lfloor\frac{500}{60}\right\rfloor=\boxed{8}$[/hide]

[quote="ccy"]I disagree.  At chapters, I buzzed in too early and I got the wrong answer 3 times (At our chapter, 10 people do the questions at the same time.  The first 3 to answer correctly proceed to the next round).

Well maybe that was because I sucked...[/quote]

I got first doing the same thing! :P

You have to measure you're opponent's ability.

If they are "Scary Fast," then you should buzz in a little earlier.

If they are average, just get the problem right.

[quote="xxm0rg07hxx"]Another trick to countdown is to buzz in like 2 seconds before you get the answer.
They give you 3 seconds to answer.
1. People will think you're a bit faster than you really are.
2. You might beat the other person.[/quote]

My mind, without exception, froze up when I did that. Don't do it unless you're REALLY good under stress.

[quote="X_wing1redleader"]problem: you never had time to check
cuz i got booted out of the countdown round because i didn't check my answer... cuz i thought he would buzz in in the time i was checking :([/quote]

That's why they're so easy- so you don't need to check them.
I won mine.  :)

When I buzz a few seconds early, not only does my mouth screw up, I can't do so much as 13+17=30.

Under the pressure, I would suggest not only not buzzing in early, but writing down the answer, and staring at it while saying it, it helps reduce careless errors in speech.

If you remember at nats, Neal Wu didn't use his pencil in the countdown.  I usually don't use it too but consider that he did win nats without using it much on the countdown.

I hate the countdown round! Basically, what I learned from chapters is to read fast, or not read the entire thing, and don't be nervous. Nerves during the countdown round make it really hard to win. There was a girl who went from 10th to 1st because of the countdown round.

Occasionally, I see the guy who won the 2003 nationals, Adam Hesterberg (he also helped out at our chapter) so I'm going to ask him for advice. (I do know of 2 people who used to go to my school who have beaten Adam at the countdown round though.)


Does there exist an enumeration (q_n) of the rational numbers in [0,1] such that
\[\sum_{n = 1}^\infty \frac{q_n}{n}\]
converges?

Of course!

Let $a_n$ be the $n$'th number which is not a square. Then let the $a_n$'th rational in our enumeration be $\frac 1{2^n}$, and enumerate the rest of the rationals however you want. We get that sum to be $\le \sum_{n\ge 1}\frac 1{2^n}+\sum_{n\ge 1}\frac 1{n^2}$.

We can even make it as small as we want.

hmm...if i'm not wrong it is possible. first, you see that if it is possible for some series containing all rational numbers, but possibly some elements twice, then it is possible for a sequence containing every rational number once(just take, if some element occurs the second time, an arbritary smaller rational number and continue that way). now, we easily see that
$\sum_{k=0}^{\infty} \sum_{m=1,m\ \text{odd}}^{\infty} \frac{1}{1+\frac{m+1}{2(k+1)}}\frac{1}{2^km}$
converges, since the first term factor is smaller than $\frac{2(k+1)}{m}$, so the sum is smaller than $\sum_{k=0}^{\infty} \frac{\pi^2}{8}\frac{2(k+1)}{2^k}$, which converges. now, set $n=2^km$ and $q_n=\frac{1}{1+\frac{m+1}{2(k+1)}}$ where $m\geq 1$ is odd, $k\geq 0$. it is quite easy to show that every rational number of $(0; 1)$(if you want it to be closed, take 0 and 1 at the beginning of the sequence) occurs in the sequence $q_n$.

i just saw that grobber already gave a much nicer solution, but now i wrote down my proof.

Peter


analytic geometry
geometry
graphing lines
slope

Hello

How do you find the distance between 2 points on a line when the coordinates of both lines are given? If I know this I can find the equation of the straight line 'L' passing through these points, expressed in the form ax+by+c=0

Thanks.

I'm not really sure what you're asking, but if you want to find the equation of a line that passes through 2 given points, first you must find the slope:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Then you can use the general point-slope equation

$y=m(x-h)+k$ where $m=$ slope, and $(h,k)$ is a point on the line.  Plug in your slope and one of the two points to find the equation of the line.  Then you can rearrange to whatever form you want.

I hope this is what you're asking... or did you want to know how to find the distance between two points or two lines?


Sorry if this was posted before. It's not very difficult but I like it  :lol:
Find all the sets $X$ of positive integers different from $0$ which has at least $2$ elements and satisfy the following proprety : For all $m,n \in X$ such that $m<n$ there exist $k \in X$ such that $n=mk^2$

this seems really easy, unless i made some stupid mistake

let $x=min X$

case 1: $x=1$
let $y= min (X-\{ x \})$
then we have $y=x \cdot k^2$, with $k \in X$, i.e. $k=1$ or $k=y$, impossible

case 2: $x>1$
let $y= min (X-\{ x \})$
since $y=x \cdot k^2$, with $k \in X$, it is clear that $k=x$, so $y=x^3$
Suppose $X$ has more than 2 elements.
Then $z=min(X- \{ x,y \})$.
$z=x \cdot p^2$, $z=y \cdot q^2$, $p,q \in X$
From the first one we have either $p=x$ or $p=x^3$. but $z>y$, so $p=x^3$ and $z=x^7$.
therefore $x^7=x^3 \cdot q^2$, i.e. $x^4=q^2$, $q=x^2$, so $x^2 \in X$, contradiction.

therefore $X= \{ x,x^3 \}$, with $x \in \mathbb{N} - \{ 1 \}$


geometry
circumcircle
angle bisector
geometry proposed

Let $ABC$ be a triangle and $D$ be the foot of $A$ bisector of $ABC$. Let $K$ be the midpoint of $AD$. Suppose $B', C'$ be the intersections of $BK$ and $AC$, $CK$ and $AB$. Let $C_1$ be the second intersection of the circumcircle of $BC'C$ and $AB$. Let $B_1$ be the second intersection of the circumcircle of $BB'C$ and $AC$. Let $M, N$ be the intersections of $BB_1$ and $CC_1$ with $AD$. Prove that $AK^2 = AM.AN$

The claim is valid for a generic point K of angle bisector AD.

Let H be the simmetric of C wrt AD. Of course, H lies on AB. Is easy to see that <MBK = <NHK and that <DKC - <DKB = <NKC so we have that BM//HK and BK//HN. From here by Thales is easy to derive the claim.

Thank you sprmnt21. That is true.
 Just another solution
 From $K$, we draw a parallel line to $BC$ that cuts $AB$ and $AC$ at $U$ and $V$. Let $A'$ be the second intersection of $AI$ and the circumcircle $(ABC)$. Then we can prove that $A'BUM$ and $A'CVN$ are cyclic. From this, we can easily obtain the equality by applying $AD.AA' = AB.AC$.


Find the ordered triple $ (a,b,c)$ where $ a$, $ b$ and $ c$ are integers, and $ c$ is prime, if
$ \sqrt{4\plus{}\sqrt{8\minus{}\sqrt{33\minus{}8\sqrt{2}}}} \equal{} a \plus{} b\sqrt{c}$.

Letting $ \sqrt{m_1}\minus{}\sqrt{n_1}\equal{}\sqrt{33\minus{}8\sqrt2}$, we find $ m_1\plus{}n_1\equal{}33$ and $ m_1n_1\equal{}32$, so this simplifies to $ \sqrt{32}\minus{}\sqrt1\equal{}4\sqrt2\minus{}1$.

Now, we want to simplify $ \sqrt{9\minus{}4\sqrt2}\equal{}\sqrt{m_2}\minus{}\sqrt{n_2}$, so $ m_2\plus{}n_2\equal{}9$ and $ m_2n_2\equal{}8$ gives $ \sqrt8\minus{}\sqrt1\equal{}2\sqrt2\minus{}1$.

Finally, we want $ \sqrt{3\plus{}2\sqrt2}\equal{}\sqrt{m_3}\plus{}\sqrt{n_3}$, which gives $ m_3\plus{}n_3\equal{}3$ and $ m_3n_3\equal{}2$ or $ \sqrt2\plus{}\sqrt1\equal{}1\plus{}\sqrt2\implies\boxed{(1,1,2)}$.

hello we have
$ 33\minus{}8\sqrt{2}\equal{}32\plus{}1\minus{}8\sqrt{2}$
$ 8\minus{}\sqrt{33\minus{}8\sqrt{2}}\equal{}8\plus{}1\minus{}4\sqrt{2}$
$ \sqrt{8\minus{}\sqrt{33\minus{}8\sqrt{2}}}\equal{}2\sqrt{2}\minus{}1$
$ 4\plus{}\sqrt{8\minus{}\sqrt{33\minus{}8\sqrt{2}}}\equal{}1\plus{}2\plus{}2\sqrt{2}$
$ \sqrt{4\plus{}\sqrt{8\minus{}\sqrt{33\minus{}8\sqrt{2}}}}\equal{}1\plus{}\sqrt{2}$.
Sonnhard.


parameterization
combinatorics proposed
combinatorics

Let $0<\alpha<\frac{1}{2}$ be a real and $C$ be a circle with parameter $1$. The distance between two points on this circle is defined by the lenght of the smallest arc between them. Let $T=I_{1}\cup\cdots\cup I_{m}$ where $I_{j}$s are some distinct arcs on the circle. If the distance between each two points in $T$ is smaller or equal to $\alpha$, Prove that:

                     $\sum^{m}_{j=1}len(I_{j})\leq\alpha$
(where $len(I_{j})$ is the lenght of $I_{j}$...!)

What do you mean by parameter? Is that radius or diameter?
   AR

well it means that the radius is $\frac{1}{2\pi}$

I think you mean ‘perimeter’ instead of ‘parameter’.

Maybe this is a solution...
Every arc in T of lenght l has its shadow of length l+1-2alpha which
cannot contain points from T. If two consecutive arcs have a hole
of legth <1-2alpha between them, then their shadows intersect mutually,
which means that we can freely fill this hole to get one greater arc.
Now there are m-1 arcs in T, but the condition still holds. So we can
suppose that we have come to k arcs and now all the holes are greter than
1-2alpha. Now all the shadows are disjoint, so if S is summary legth of
T, then we have 2S+k(1-2alpha)<=1.  From here its easy to get
S<=alpha.
Is this correct?


Να δείξετε ότι $ (1\plus{}\frac{1}{x^{2}})xsin(\frac{1}{x})>1$ για $ x\ge\frac{1}{\sqrt5}$
(εξ αρχής η παραγώγιση της $ (1\plus{}\frac{1}{x^{2}})xsin(\frac{1}{x})\minus{}1$ δεν συμφέρει)

η λύση είναι στο attachment


algebra
polynomial
function
system of equations

Polynomial $ P(x)$ contains only terms of odd degree.  When $ P(x)$ is divided by $ x\minus{}3$, the remainder is $ 6$.  What is the remainder when $ P(x)$ is divided by $ x^2\minus{}9$?

[hide="Answer"]$ 2x$[/hide]

I say that $ x^2\minus{}9\equal{}(x\plus{}3)(x\minus{}3)$ so it's basically the beginning of the second sentence of the problem divided by $ x\plus{}3$.  However, I don't see what this tells me about the remainder. :(

Thanks in advance!

[hide="Solution"]Since $ P(x)$ has only odd degree terms, it is an odd function: $ P( \minus{} x) \equal{} \minus{} P(x)$. So $ P(3) \equal{} 6$ and $ P( \minus{} 3) \equal{} \minus{} 6$.
Now let $ P(x) \equal{} Q(x)(x \minus{} 3)(x \plus{} 3) \plus{} ax \plus{} b$. Substituting the above gives a system of equations:
$ 6 \equal{} 3a \plus{} b$
$ \minus{} 6 \equal{} \minus{} 3a \plus{} b$
$ 0 \equal{} 2b \Rightarrow b \equal{} 0, a \equal{} 2$.
Hence the remainder is $ 2x$.[/hide]

Why does the remainder have to be $ ax\plus{}b$?

$ (x \minus{} 3)(x \plus{} 3)$ is of degree $ 2$, so the remainder is at most degree $ 1$. Generally, if you're dividing $ P(x)$ by a polynomial of degree $ n$, the remainder is at most degree $ n \minus{} 1$.


function
real analysis
calculus
calculus computations

Assume that $ f(x)\in C(\minus{}\infty,\plus{}\infty)$ s.t. $ f(x)$ is a periodic function while $ T$ is its period.

Then prove: $ f(x)$ is uniformly continuous when $ x\in(\minus{}\infty,\plus{}\infty)$.

[quote="pleurestique"]since $ f$ is continuous, it is uniformly continuous on any closed bounded interval, in particular any interval of the form $ [a,a \plus{} T]$. the conclusion follows easily from here...[/quote]

Thank you for your answer, but I think it is not so simple. If the number of intervals is limited, of course it will be true, but here the number is infinite. For example, $ f(x) \equal{} x^2$ is not uniformly continuous though in every closed bounded interval it's uniformly continuous. So could please show me more details about this proof? :blush:  :blush:  :blush:

uniform continuity guarantees a universal $ \delta$ for every $ \epsilon$, etc... yes, there are infinitely many intervals, but by *periodicity* the $ \delta$ will be the same on all intervals of the form, say, $ [(k \minus{} 1)T,kT)]$, whose union is all of $ \mathbb{R}$ ...


geometry
3D geometry
pigeonhole principle
sphere
area of a triangle
combinatorics unsolved
combinatorics

Given $75$ points in a unit cube, no three collinear, show that we can choose three points which form a triangle with area at most $\frac{7}{72}$.

Divide the cube into 36 $\frac13 \times \frac13 \times \frac14$ rectangular boxes.  Some such box must contain 3 points, by pigeonhole principle.  Each such box fits within a sphere of radius $R = \frac{1}{2}\sqrt{\left(\frac13\right)^{2}+\left(\frac13\right)^{2}+\left(\frac14\right)^{2}}= \frac{\sqrt{41}}{24}$.  The area of the largest triangle which fits in such a sphere, and thus an upper bound on the area of any triangle in such a box, is $A = \frac{3R^{2}\sqrt{3}}{4}= \frac{41}{256\sqrt{3}}$ (for an equilateral triangle).

$\frac{41}{256\sqrt{3}}/ \frac{7}{72}= \frac{123\sqrt{3}}{224}< 1$ (we can estimate $\sqrt{3}< \frac95$, for instance), so the triangle formed by the three vertices in this box has area less than $\frac{7}{72}$, and we're done.

This is an ugly method -- probably there is something at least a little more elegant.  I'm curious if dividing the box into $\frac12 \times \frac13 \times \frac16$ solids was the "intended" path of solution (with an argument that the largest area of a triangle in a rectangular box is at most $\frac{d^{2}}4$, where $d$ is the diagonal), since in that case the diagonal is exactly $2\sqrt{\frac{7}{72}}$.  In fact, this is almost certainly what was desired.

my solution is the same
use $S^{2}={S_{x}}^{2}+{S_{y}}^{2}+{S_{z}}^{2}$ with S_x;S_y;S_z is the correspond area in Oyz;Ozx;Oxy(i don't know how to describe in E)
then divide the cube as you said...


Well, for one thing, they'll never consider anything from current contestants. That's just common sense.

You might consider sending proposals to problem solving journals such as [i]Crux Mathematicorum with Mathematical Mayhem[/i].

[quote="towbomb"]Does anyone know how to send a proposed problem to the AMC committee for consideration?  I have a pretty good problem that I think might work well on the USAMO...[/quote]

They ask for problems from a wide variety of sources I believe. I know of several people who write problems for the CAMC. You can probably send yours in next year (given, of course, that your problem is of appropriate difficulty, you are not a contestant, and you are an adult) for consideration.

so if we are nto participating for USAMO, technically we CAN send in problems for consideration???

Yes. but i think somehow u need to prove that.

Not only do you need to prove that, but by the time you know you are not participating in USAMO, I bet it is already written =) (assuming you are not an adult).

With my hobby of problem writing, I find this to be an intriguing thread. I'd never heard of such procedures before. A quick question: Do they only want olympiad problems or can we submit material for the other rounds?

I believe that you can submit problems for any round.

[quote="MithsApprentice"][quote="Mildorf"]Do they only want olympiad problems or can we submit material for the other rounds?[/quote]

Well they obviously have committes of people who submit and select problems every year. But I'm not sure if they'd let just anyone submit a problem. And I'm almost positive a student submitting questions is unthinkable.[/quote]

By that comment I mean next year, when I am no longer in the group that actually takes the contests.

Right. :-) I was wondering if they'd let anyone who wants to submit a problem submit one. Perhaps they have requirements of some sort or they only take problems from a certain group of people.


calculus
derivative
integration
real analysis
real analysis unsolved

Let f(x)=\[ \int_{1}^\infty\frac{e^{\minus{}xy}}{y^3}dy\].
Show the f(x) is differentiable on (0, infinity) and find the formula f'(x)

$ \frac{e^{\minus{}xy}}{y}\equal{}\int_x^{\infty}e^{\minus{}ty}\,dt$

Thus,
$ f(x)\equal{}\int_1^{\infty}\frac{e^{\minus{}xy}}{y^3}\,dy$
$ f(x)\equal{}\int_1^{\infty}\frac1{y^2}\int_x^{\infty} e^{\minus{}ty}\,dt\,dy$
$ f(x)\equal{}\int_1^{\infty}\int_x^{\infty} \frac{e^{\minus{}ty}}{y^2}\,dt\,dy$
$ f(x)\equal{}\int_x^{\infty}\int_1^{\infty} \frac{e^{\minus{}ty}}{y^2}\,dy\,dt$ (Tonelli's theorem)

Then we just differentiate $ f$ by the Fundamental Theorem:
$ f'(x)\equal{}\minus{}\left(\int_1^{\infty} \frac{e^{\minus{}ty}}{y^2}\,dy\right)_{t\equal{}x}$
$ f'(x)\equal{}\minus{}\int_1^{\infty}\frac{e^{\minus{}xy}}{y^2}\,dy$

That integral (and the integral that defines $ f$) converge for all $ x>0$. That's the best you'll get on the formula; that integral can't be evaluated in terms of elementary functions.

[quote="santug"]Let f(x)=
\[ \int_{1}^\infty\frac {e^{ \minus{} xy}}{y^3}dy\]
.
Show the f(x) is differentiable on (0, infinity) and find the formula f'(x)[/quote]

Try letting $ y\equal{}z/x$.

[quote="fedja"]Not a really good idea because you'll get an integral with variable lower limit. ;)[/quote]

Setting $ y \equal{} z/x$ gives $ f(x)\equal{} x^2\int_{x}^\infty\frac {e^{ \minus{} z}}{z^3}dz$. Use the product rule and the fundamental theorem to get $ f'(x)$ very quickly. What's the problem?

I just do not see how it is advantageous to saying that the derivative of the integrand has an integrable majorant ($ \frac 1{y^2}$) and differentiating under the integral sign, that's all. The point is not that what you are saying is wrong but merely that it is not any better than what has been done already. :) Sorry if this was unclear. :blush:


[quote="santosguzella"]$Z$ and $w$ are complex numbers and $u=\sqrt{zw}$. Prove that: $|z|+|w|=|\frac{z+w}{2}-u|+|\frac{z+w}{2}+u|$[/quote]
$u\in \sqrt{zw}=\left\{ -\alpha ,\alpha  \right\},{{\alpha }^{2}}=zw$


The deadline for the enrollment for this course is Sunday, October 26.  The class will cover those topics of algebraic insight and approaches that are most useful for algebraic problem solving of the variety found on the AMC & AIME.  These also, of course, form the basic foundations needed to approach even more difficult problems.  If you have any questions about the course, please post them here or email us at classes@artofproblemsolving.com.

Um, are you refering to the "Algebraic Problem Solving" or "AMC/AIME Problem Series"? I believe that you're talking about the first, but just to make sure.

I'm talking about the Algebraic Problem Solving.


How do you make the "for all" (upside-down A) sign in $\LaTeX$?


X is a binomial random variable where the number of trials is 5 and the probability of success in each trial is p. Find the value of p if P(X = 4) = 0.12

P(X =4) = 5 (p^4)(1-p) = 0.12

So,

5p^4 - 5p^5 = 0.12

How to solve for p??

If you have $5p^4 - 5p^5 = 0.12$, then the only two positive value of $p$ are approximately $.973$ and $.459$.

The answer is that you solve numerically, for example using a graphing calculator or mathematics software.  There is no nice way to get an exact value for such an equation.  Being a quintic, its solutions may not even be expressible in radicals.


$\sqrt{8}+\sqrt{18}=$

$\textbf{(A)}\ \sqrt{20} \qquad \textbf{(B)}\ 2(\sqrt{2}+\sqrt{3}) \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 5\sqrt{2} \qquad \textbf{(E)}\ 2\sqrt{13}$

A rather straight-forward solution

[hide]$\sqrt8 + \sqrt{18} = 2\sqrt2 + 3\sqrt2 = 5\sqrt2$ = D[/hide]

[quote="joseluis"]\sqrt{8}+\sqrt{18}=

[hide]2\sqrt{2}+3\sqrt{2}=5\sqrt{2}[/hide][/quote]
add the dollar signs around the code   ;)

[hide]Everyone already knows, but i got D. :) [/hide]

[hide]sqrt 8 -> 2(sqrt2) and sqrt 18 ->3(sqrt 2). 2(sqrt 2)+3(sqrt 2)=5(sqrt 2), so answer D.[/hide]

[hide]um, that equals 2sqrt2 + 3 sqrt2... factor out the sqrt 2 and you get sqrt2(2+3) = 5 sqrt 2, or D[/hide]

[quote="4everwise"]$\sqrt{8}+\sqrt{18}=$

\[ \text{(A)}\ \sqrt{20} \qquad \text{(B)}\ 2(\sqrt{2}+\sqrt{3}) \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 5\sqrt{2} \qquad \text{(E)}\ 2\sqrt{13}  \][/quote]

[hide]root 8 = 2 root 2
root 18 =3 root 2

add them up up it would be

2+3=5 last time I checked
so it would be 5 root 2[/hide]

[quote="José"][hide] D is the correct one[/hide][/quote]

Please stop writing solutions that have absolutely no content except for an answer.

[quote="José"]√8+√18=2√2+3√2=5√2 ------>>>> (D).[/quote]
[quote="José"][quote="4everwise"]$\sqrt{8}+\sqrt{18}=$

\[ \text{(A)}\ \sqrt{20} \qquad \text{(B)}\ 2(\sqrt{2}+\sqrt{3}) \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 5\sqrt{2} \qquad \text{(E)}\ 2\sqrt{13}  \][/quote]

[hide] D is the correct one[/hide][/quote]He answered this problem twice. I wonder what he is up to...

5root2 . any1 want a british math/physics penpal? im lonely here lol every1's an idiot. email me!

i bet D is the most common letter on this page. :rotfl:
but seriously do not post if u don't have a solution.
i've seen it in many other threads.
DON'T SPAM


This is my first problem on number theory :)
Here it goes:
Let $ a,b \in Z$ Find all integral solutions of the equation:
$ (a^2 \plus{} b^2)(a^2b^2 \plus{} 1) \plus{} 2ab(a \minus{} b)^2 \equal{} 2(ab \plus{} 1)(a^2b^2 \minus{} ab \plus{} 2)$
I hope you will like it :wink:

One solution is $ a\equal{}0,b\equal{}4$

The point is to find all of the solutions...

[hide]By expanding the second term on the LHS and regrouping, we get

$ (a^2 \plus{} b^2)(ab \plus{} 1)^2 \minus{} 4a^2b^2 \equal{} 2(ab \plus{} 1)(a^2b^2 \minus{} ab \plus{} 2)$

Hence $ ab \plus{} 1\mid 4a^2b^2$, but as $ (ab \plus{} 1,ab) \equal{} 1$, we get $ ab \plus{} 1\mid 4\iff ab\in\{ \minus{} 5, \minus{} 3, \minus{} 2,0,1,3\}$

From $ a^2 \plus{} b^2 \equal{} \frac {2(ab \plus{} 1)(a^2b^2 \minus{} ab \plus{} 2) \plus{} 4a^2b^2}{(ab \plus{} 1)^2}$ we find non-negative integer values:

$ ab \equal{} \minus{} 2\implies a^2 \plus{} b^2 \equal{} 0\implies a \equal{} b \equal{} 0\implies ab\neq \minus{} 2$, hence no solution;
$ ab \equal{} 0\implies a^2 \plus{} b^2 \equal{} 4$;
$ ab \equal{} 1\implies a^2 \plus{} b^2 \equal{} 3\implies (a \plus{} b)^2 \equal{} 5$, which has no integer solution.

Therefore $ ab \equal{} 0\land a^2 \plus{} b^2 \equal{} 4\iff (a,b)\in\left\{(0,\pm 2),(\pm 2,0)\right\}$[/hide]

[hide="solution"]We let $ a+b=s$, $ ab=p$.

We get the equation $ (s^2-2p)(p^2+1)+2p(s^2-4p)=2(p+1)(p^2-p+2)$

We can expand and bring the $ s$'s to one side, and everything else to the other side:

$ s^2p^2+s^2+2ps^2=4p^3+8p^2+4p+4$

We divide by 4 and factor to get $ (s(p+1)/2)^2-1=p(p+1)^2$.

Dividing by $ (p+1)^2/4$ and moving stuff around, we get

$ (2/(p+1))^2=s^2-4p$

The right side is an integer, so the left side must be too.  Thus $ 2/(p+1)$ can only be $ -2, -1, 1, 2$.

For \begin{tabular}{c|c|c|c|c}\\
2/(p+1)=&-2&-1&1&2\\
p=&0&1&-3&-2\\
4p=&0&4&-12&-8\\
s^2=(2/(p+1))^2+4p=&4&5&-11&-4\end{tabular}

Thus the only time $ s$ is an integer is when $ p=0$, and then $ s=\pm 2$.  Thus the solutions $ a$ and $ b$ are the solutions to the equation $ x^2\pm 2x$ which are $ (2, 0), (-2, 0), (0, 2), (0, -2)$.

Now what if $ p=-1$? Then we can't divide by $ p+1$ like we did above.  No worries, though.  The equation $ (s(p+1)/2)^2-1=p(p+1)^2$ would turn into $ -1=0$, impossible.[/hide]

Very well....
I creat the equation as follow:
$ a^4b^2 \plus{} a^2b^4 \plus{} a^2 \plus{} b^2 \plus{} 2ab(a \minus{} b)^2 \equal{} 2ab \plus{} 2a^3b^3 \plus{} 4 \Leftrightarrow(a \minus{} b)^2 \plus{} a^2b^2(a \minus{} b)^2 \plus{} 2ab(a \minus{} b)^2 \equal{} 4 \Leftrightarrow(a \minus{} b)^2(a^2b^2 \plus{} 2ab \plus{} 1) \equal{} 4 \Leftrightarrow (a \minus{} b)(ab \plus{} 1) \equal{} 2,$ or $ (a \minus{} b)(ab \plus{} 1) \equal{} \minus{} 2$...
We solve the systems and we take the solutions..
I hope you enjoyed it....

BARCA IS BACK :)  :P


Παρακατω προτεινω μια δικη μου ασκηση για την οποια δεν ξερω ουτε εχω ψαξει ουτε θα ηθελα να βρω τωρα καποιες πληροφοριες απτο ιντερ. κι ετσι σας την παρουσιαζω ωστε να πειτε τις ιδεες σας και τις λυσεις σας.
Για λογους συντομιας ονομαζω την παρασταση:
D(n)=$x^{n}+y^{n}-z^{n}$ οπου x,y,z,n θετικοι ακεραιοι.
Zητω να αποδειξεται
α)καθε ακεραιος παριστανεται με την μορφη D(2) με απειρους τροπους.
β)Δεν παριστανονται ολοι οι ακεραιοι υπο την μοπφη D(3)
γ)Εξεταστε το προβλημα για οσους περισοτερους φυσικους n μπορειτε και κοιταξτε τις ειδικες περιπτωσεις οπου n πρωτος.

Για το α) έχω αποδείξει  ότι κάθε ακέραιος μπορεί να παρασταθεί με τη μορφή D(2)... άλλα δεν έχω αποδείξη ότι γίνεται με άπειρους τρόπους...Βασικά εξέταζω τους αριθμούς με βάση τα υπόλοιπά που δίνουν με το $4$ και έτσι πχ για $n=4k$ έχω βρει την ταυτότητα
$4k = (k+1)^{2}-(k-1)^{2}+0^{2}$. Αντίστοιχες ταυτότητες έχω βρει και για $n=4k+1$ κτλ...

Επίσης απέδειξα μια πρόταση που διάβασα ότι κάθε ακέραιος μπορεί να εκφρασθεί ως διαφορά τετραγώνων αν και μόνο αν δεν είναι της μορφής $n=4k+2$

Για το β) νομίζω  ότι υπάρχουν άπειροι ακέραιοι που δεν μπορούν να εκφραστούν ως D(3) και αυτοί είναι οι $n=9k+4$ αφού $x^{3}\equiv 0 \bmod{9}$
ή $x^{3}\equiv-1 \bmod{9}$  ή   $x^{3}\equiv 1 \bmod{9}$

Βασικά παρατήρησα και κάτι άλλο.Για να μελετήσουμε τέτοιες παραστάσεις πχ την D(x) συμφέρει να εξετάζουμε τους ακεραίους mod x^2....

 :D

Εχεις κανει ενα λαθος ομως foti.Γραφω θετικοι ακεραιοι.Καλη δουλεια παντος.Εχεις δρομο ακομα.......συνεχισε...... :wink:

Ναι το ξέρω....Απλά γράφω τι βρήκα.Δεν είπα ότι έλυσα την άσκηση... :)  :)

Πρσπαθησε παντος να λυσεις πληρως το α ερωτημα.Νομιζω πως μπορεις....για δες λιγο καλυτερα..... :)

Βρήκα μία λύση για το πρώτο ερώτημα Δημήτρη, αλλά είναι αρκετά μεγάλη! Έχει να κάνει με σχετικά πρώτους αριθμούς κ.λπ.

Ας πω εν ολίγοις τα βασικά βήματα:

Υποθέτουμε ότι $x,y,z$ είναι μεταξύ τους σχετικά πρώτοι αριθμοί.

$x^{2}+y^{2}=z^{2}$

Εύκολα βρίσκουμε ότι ο $z$ θα είναι περιττός. Επίσης, εύκολα προκύπτει ότι $x$ π.χ. άρτιος και $y$ περιττός.

Ακόμη

$x^{2}=y^{2}-z^{2}=(y-z)(y+z)$

Θα πρέπει $y-z$, $y+z$ να είναι άρτιοι (αφού υπεθέσαμε ότι $x$ είναι άρτιος)

Στη συνέχεια θέτουμε $x=2m$ $z+y=2k$, $z-y=2n$

Θα πάρουμε ότι $k,n$ είναι σχετικά πρώτοι. Άρα θα υπάρχουν αριθμοί τέτοιοι ώστε $k=p^{2}$ και $n=q^{2}$ κ.λπ. κ.λπ.

Υ.Γ. Πρόκειται για το περίφημο πρόβλημα των Πυθαγόρειων Τριάδων που λύνεται με Διόφαντο.

EDIT:

Μία αναλυτικότερη και διαφορετική λύση μπορείτε να δείτε εδώ:

Δεν καταλαβα πως ακριβως προκυπτει λυση της ασκησης αλλα σε πιστευω.Ελπιζω οποτε εχεις χρονο να δωσεις πληρως την λυση σου.

Η δικη μου λυση βασιζεται στο απλο λημμα που λεει οτι καθε ακεραιος εκτος αυτων που ειναι της μορφης $4k+2$ παριστανεται ως διαφορα τετραγωνων.

Αυτό το απέδειξα και εγώ χωρίς να  ξέρω ότι είναι λήμμα... :lol:

Ας το παμε σιγα-σιγα.Προσπαθηστε να αποδειξεται οτι δεν ειναι δυνατον ολοι οι θετικοι ακεραιοι αριθμοι μπορουν να παραστηθουν με την μορφη $D(f(p^{x}))$ οπου $p>2$. (οπου $f$ η συναρτηση Οιλερ).Aποδειξτε επισης οτι το ιδιο ισχυει για την μορφη $D(3k)$

μηπως μπορουμε να το δειξουμε για $D(p^{n})$  οπου p πρωτος ? το κοιταξες?

Αυτην την περιπτωση δεν την εχω κοιταξει Κωστα αλλα για την περιπτωση $D((p-1)k)$ απεδειξα οτι δεν παριστανονται ολοι οι θετικοι ακεριοι.Θα κοιταξω παντος το ερωτημα σου.

Εικαζω οτι ειναι αδυνατον να βρουμε φυσικο $n>2$ τετοιον ωστε ολοι οι ακεραιοι να μπορουν να παραστηθουν με την μορφη $D(n)$ και αυτο γιατι εχω κανει αρκετες δοκιμες αριθμων $n$ αλλα αυτο πρεπει να αποδειχθει........καμια ιδεα καποιος? :)

Αν και ξεχαστηκε σιγουρα, θα γραψω την κεντρικη ιδεα μου και θα αποδειξω αυτο που ειχα γραψει στο προηγουμενο μηνυμα μου....Ειχα πει οτι μονο ο αριθμος $ n=2$ ειναι αυτος ωστε καθε ακεραιoς να παριστανεται υπο την μορφη $ D(n)$,Συγκεκριμενα για $ n>2$ υπαρχουν απειροι ακεραιοι οι οποιοι δν παριστανονται υπο την μορφη $ D(n)$.
[url]LEMMA[/url]
Αν $ a=bmod{p}$ τοτε $ a^{p}=b^{p}modp^{2}$.Με την βοηθεια αυτου του λημματος αποδεικνυω οτι τα δυνατα υπολοιπα του $ D(n)modp^{2}$ ειναι $ 0$,$ +-1$,$ +-a_{1}$,$ +-a_{2}$,......,$ +-a_{(p-3)/2}$ οπου $ o\leq a_{i}<p^{2}$ που ειναι αδυνατο αυτο το συστημα υπολοιπων $ x+y-z$ με $ x$,$ y$,$ z$ να παιρνουν τις παραπανω τιμες $ 0$.$ +-1$,$ +-a_{1}$.....να ισουται με το πληρες συστημα υπολοιπων $ modp^{2}$ .Δες π.χ για $ p=5$ ειναι αδυνατο το συστημα υπολοιπων $ x+y-z$ με τις μεταβλητες να παιρνουν τις τιμες $ 0$,$ +-1$,$ +-a_{1}$ να ισουται με το πληρες συστημα υπολοιποιπων $ mod25$ :wink:


Let $a,b,c$ be positive real numbers.
Prove that:$\displaystyle \sum \frac {a}{\sqrt{a^2+2(b+c)^2}} \geq 1$.

Stronger than IMO 2001,right ?However,it can be done by Cauchy...
We have:
[tex] (a\sqrt{a^2+2(b+c)^2}+b\sqrt{b^2+2(a+c)^2}+c\sqrt{c^2+2(a+b)^2}) \cdot[/tex]
[tex](\frac{a}{\sqrt{a^2+2(b+c)^2}}+\frac{b}{\sqrt{b^2+2(a+c)^2}}+\frac{c}{\sqrt{c^2+2(a+b)^2}}) \geq (a+b+c)^2 [/tex].
Cauchy again:
[tex] (\sqrt{a}\sqrt{a^3+2a(b+c)^2}+\sqrt{b}\sqrt{b^3+2b(a+c)^2}+\sqrt{c}\sqrt{c^3+2c(a+b)^2})^2 \leq (a+b+c)(a^3+b^3+c^3+2a(b+c)^2+2b(a+c)^2+2c(a+b)^2 ) \leq (a+b+c)^4 [/tex]
So we have done. :)
Hm,I wonder if Cauchy can be a verb,because sometime we say: Cauchy it... :D

Instead of "Cauchy it", let's "Jensen it", :)

WLOG we may assume $a+b+c=1$. Since $\displaystyle f(x)=\frac{1}{\sqrt{x}}$ is convex on $(0,\,\infty)$, we have

$\displaystyle
\frac{a}{\sqrt{a^2+2(b+c)^2}}+\frac{b}{\sqrt{b^2+2(a+c)^2}}+\frac{c}{\sqrt{c^2+2(a+b)^2}} $
$\displaystyle
\geq \frac{1}{\sqrt{a^3+2a(b+c)^2+b^3+2b(a+c)^2+c^3+2c(a+b)^2 }}
$
$\displaystyle
\geq \frac{1}{\sqrt{(a+b+c)^3}} = 1$

since $a^3+b^3+c^3+2a(b+c)^2+2b(a+c)^2+2c(a+b)^2  \leq (a+b+c)^3$ as pointed out by [b]Anh Cuong[/b].

Cheers!  :D

I have the expansion for this inequality .

Let  a,b,c  be possitive real numbers , k be an possitive integer number .

Prove that :
\[ \displaystyle\sum\frac{a}{\sqrt{a^2+k(b+c)^2}} \geq \frac{3}{\sqrt{4k+1}} \]

Can somebody solve this inequality ?

Your inequality is only true for [tex] k\geq 2 [/tex].
For [tex] k<2 [/tex],o.k,take [tex] k=1,a=1000,b=1,c=1 [/tex] then it will be wrong.

Yes , you are right Anh Cuong.
The inequality is only true for $ k \geq 2 $
Can you solve this inequality ? Try !

Yes,since we have:
[tex] a^3+b^3+c^3+ka(b+c)^2+kb(a+c)^2+kc(a+b)^2 \leq \frac{12k+3}{27}(a+b+c)^3 [/tex] where [tex] k\geq 2 [/tex] :)

An other soln:
Prove that: $\displaystyle \frac {a}{\sqrt{a^2+2(b+c)^2}} \geq \frac {a^{4/3}}{a^{4/3}+b^{4/3}+c^{4/3}}$.

[quote="hxtung"]Let $x,y,z$ be positive real numbers. Prove that $\frac{x}{\sqrt{x^{2}+2(y+z)^{2}}}+\frac{y}{\sqrt{y^{2}+2(z+x)^{2}}}+\frac{z}{\sqrt{z^{2}+2(x+y)^{2}}}\geq 1$.[/quote]
Using the arithmetic mean - geometric mean inequality and the Holder inequality, we have

$\left(\sum{x}\right)^{3}\left[\sum\frac{x}{\sqrt{x^{2}+2(y+z)^{2}}}\right]^{2}\geq\left\{\sum{x[x^{2}+2(y+z)^{2}]}\right\}\left[\sum\frac{x}{\sqrt{x^{2}+2(y+z)^{2}}}\right]^{2}$

$\geq\left(\sum{x}\right)^{3}$.


Let $a,b,c,d$ be non-negative numbers such that.
$a^2+b^2+c^2+d^2=2$.
Prove that
$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \geq 2$.

A marvellous inequality :!:

That is to say that hungkhtn's result and this one are so inspiring. I was also inspired by this but my result was weaker.


function
algebra
functional equation
algebra proposed

Let   [b]f: R -> R[/b]   f(x) be a function such that
   [i](a)[/i] f(x+y)=f(x)+f(y) for all real number x, y
   [i](b)[/i] f(xy)=f(x)f(y) for all real numbers x, y
      f(x)=? :?:

$ f(x^2)\equal{}f(x)^2\ge 0$.
If $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ and $ f(x)\ge 0$ when $ x\ge 0$, then $ f(x)\equal{}ax$. In these case $ a\equal{}0$ or $ a\equal{}1$.

[quote="Rust"]$ f(x^2) \equal{} f(x)^2\ge 0$.
If $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ and $ f(x)\ge 0$ when $ x\ge 0$, then $ f(x) \equal{} ax$. In these case $ a \equal{} 0$ or $ a \equal{} 1$.[/quote]

 can you prove it  if  f(x+y)=f(x)+f(y) and f(x)>=0 when x>=0 then f(x)=ax

[quote="hsiljak"][quote="ABRORBEK"]can you prove it  if  f(x+y)=f(x)+f(y) and f(x)>=0 when x>=0 then f(x)=ax[/quote]

Doesn't it follow easily from [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]well-known facts about Cauchy's functional equation[/url]?[/quote]

what do you mean? you want say that if f(x+y)=f(x)+f(y) then f(x)=cx where c - is constant?

One of its solutions, yes (with an additional assumption of continuity, monotonicity, boundedness, etc). As Engel calls it, it is the "tamed" solution (there are also some "wild" solutions, check out the link in my original post).

One more thing:

[quote="MathWorld"]It was proved by Cauchy in 1821 that the only continuous solutions of this functional equation from R into R are those of the form f(x)=kx for some real number k. In 1875, Darboux showed that the continuity hypothesis could be replaced by continuity at a single point and, five years later, proved that it would be enough to assume that f(x) is nonnegative (or nonpositive) for sufficiently small positive x.[/quote]

The last sentence is approval of Rust's approach.

You don't need any well-known facts. Since $ f(x) \geq 0$ if $ x \geq 0$, then if $ a > b$ put in $ x \equal{} b, y \equal{} a \minus{} b$ and you get $ f(a) \minus{} f(b) \equal{} f(a\minus{}b)$, which is nonnegative, so $ f$ is nondecreasing. It's easy to show from $ f(x\plus{}y) \equal{} f(x) \plus{} f(y)$ that $ f(x) \equal{} x$ for all rationals, and since the rationals are dense in the reals you're done.


geometry
3D geometry
combinatorics unsolved
combinatorics

Prove that exists an integer $N_d$ such that for all integers greater than it $N \geq N_d$, we can always fill exactly  a square with $N$ smaller squares. Determine explicitely the values for which this is not possible - quite easy.
(IMC London 2000)
For each valid $N$ we may define a sequence $a_N$ by number of different ways - up to diiferent sizes for the $N$ squares - this can be done, what can say about it?!?
I put the same questions (and which is maybe also a known problem) - try the same for a cube - there is such $N_c$.

and it is also true for nth dimensional space.... :)


One of the questions in Larson's [i]Problem Solving Through Problems[/i] reads as follows:

"Let S denote an n-by-n lattice square, n>=3. Show that it is possible to draw a polygonal path consisting of
2n-2 segments which will pass through all n^2 lattice points of S."

Does 'polygonal' imply that it is closed? If so, I haven't been able to establish the n=3 case. Any ideas?

BTW, it is problem 2.4.5 on page 74 in case you want to look it up.

I don't think it makes any difference whether it's closed or not: if you have a closed path you can stop right before you reach the point where you began, thus making it open.

The case [tex]n=3[/tex] is a well-known puzzle. I don't know anyone who solved it the first time they saw it :). Here's the picture:

Ya, I found that solution (the first time I saw it :)) but I didnt think it met the condition since it wasnt a closed polygon. I guess I'm just reading too much into the wording of the question.

It may just be that "polygonal" is there to say that all of the line segments are straight.

Ok, here's a proof for open paths:

[hide]Start with the n=3 solution and proceed by induction.  Add an extra row of points on the right and bottom of the figure to make it a 4x4 square; two more line segments can be drawn through them from the diagonal line jutting out of the top-right of the square.  For 5x5, draw two more rows on the top and left side of the figure.  When the end line from the path for the 4x4 is extended, it can connect to two more line segments passing through these two new rows of points.  By alternating between bottom & right rows and top & left rows, this method can result in a path for any nxn square.  Since only 2 new line segments are needed each time, every solution thus generated has exactly 2n-2 line segments.

So, for all nxn squares with solutions, the (n+1)x(n+1) square has a solution, and the proof is complete.[/hide]

(sorry I can't spoilerize the picture)


is it the beta one? i don't like that one firefox is still better

[quote="funcia"]they now have one that is tabbish so i can have i thik 5 sires in one screen and i can change my default search engine easily but it is confusing!  :juggle:
it is like i think 7 edition it has the little "e" with a yellow thing over it....it was really shocking to me that i can google with it and wiki with it and alot of other things like ebay with it!!!
but it looks like people don't even use internet explorer anyway![/quote]

Welcome to 2 years ago with Firefox.

[quote="funcia"]OK I woke up at 5 am to research something and what was i to find but my internet had a fourth place ribbon on it!!!!!! and i got to put the wiki and google as my default search!!! And i can have more than 1 screen up!!!! What is microsoft thinking? this is so cool!!!! Do you guys like it???[/quote]
$\mathfrak{firefox}>IE$
$\mathbb{QED}$

@david: get out of here

It's always for download... [url=http://portableapps.com/apps/internet/firefox_portable]portablefirefox[/url]
and [url=http://www.mozilla.com/en-US/firefox/]normal firefox[/url]

If you want the cutting edge build of Firefox (2.0 rc3) get it from here:

http://www.mozilla.org/projects/bonecho/all-rc.html

It's stable enough for everyday use, and adds a bunch of features to the current 1.5 version.  (2.0 final should be released in the next couple of weeks)

(post count=500 now)

The 2.0 one is way better. :D I have been using FF for like 3 years.  :)

Firefox hasn't even been out yet for 2 years..

[quote="funcia"]they now have one that is tabbish so i can have i thik 5 sires in one screen and i can change my default search engine easily but it is confusing!  :juggle:
it is like i think 7 edition it has the little "e" with a yellow thing over it....it was really shocking to me that i can google with it and wiki with it and alot of other things like ebay with it!!!
but it looks like people don't even use internet explorer anyway![/quote]

yeah i have it too. its not that good. i only use it for fun. it is way harder to use than the regular one (menus ar ebad), but it has tabs


is it just my browser, or am I seeing $ signs instead of the nice math symbols for all the recent posts?  if it's just my computer, how do I fix it?

thanks

This should be in the support forum. :wink:
My browser shows the $ \text{\LaTeX}$.


By looking at $x^{2}-y^{2}= 16.$ we know that the answer is $25$ and $9$. Is there an algebraic way of doing it?

Please remember when posting in high school basics to use language and symbols that beginning high school students will understand.

It's Norwegian for "and".

http://www.freedict.com/onldict/nor.html

[quote="guile"]But where's the answer?[/quote]
I just answered Tiago Guerreiro's question. If you'd really like an answer, read Solar Plexsus' solution or just see that $(4,y,x)$ should be a Pythagorean triple, which is only possible if $(x,y)=(5,3)$.


Let $ a,$ $ b$ and $ c$ are real numbers such that a sum all two of them is positive. Prove that:
\[ \frac{a^3}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{b^3}{a^2\minus{}ac\plus{}c^2}\plus{}\frac{c^3}{a^2\minus{}ab\plus{}b^2}\geq a\plus{}b\plus{}c\]

[quote="aidan"]Do you mean "a sum of two of them is positive"?[/quote]
I have meant that $ a \plus{} b > 0,$ $ a \plus{} c > 0$ and $ b \plus{} c > 0.$
My inequality is wrong! I am sorry.
I change it. The Darij's nice problem:
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ \frac {a^2(b \plus{} c)}{b^2 \plus{} c^2} \plus{} \frac {b^2(a \plus{} c)}{a^2 \plus{} c^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} b^2}\geq a \plus{} b \plus{} c
\]


vector
linear algebra
linear algebra unsolved

For any vectors a,b,c and d in R3, prove that

(a x b).(c x d) = (a.c)(b.d) - (a.d)(b.c)

Any help would be greatly appreciated. Many thanks,
Mike

LHS we have
$(a \times b) \cdot (c \times d)$
can be written as
$[(c \times d)] \cdot (a \times b) \equiv a \cdot (b \times (c \times d))$ by cyclic interchange.
now looking at the RHS
$(a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c)$
$= (a \cdot [c(b \cdot d)] - a \cdot [d(b \cdot c)]$
$= a \cdot (c(b\cdot d) - d(b \cdot c))$

so if the LHS = RHS ie

$a \cdot (b \times (c \times d)) = a \cdot (c(b\cdot d) - d(b \cdot c))$
then we have shown what we were meant to show.

you can see that as
$b \times (c \times d) = c(b\cdot d) - d(b \cdot c)$
this is the vector triple product
so therefore the LHS = RHS

you can show the vector triple product to be true by making vectors and multipling out each side and showing they are the same, but that would mean way too much typing.
you can show the original problem to be true by making vectors and multipling them all out also, however it is faster to show the vector triple product.


This is :wow:

Simplyfy the fraction $y=\frac{(x^2+3x+3)^2-1}{(x^2+5x+5)^2-1}$. Find the minum of $y$ and it's extreme values. Enjoy :)

[quote="JBL"]You do not have all the local extrema.  Also, I can't for the life of me figure out why you left the polynomial in the form you did.  It can be simplified further![/quote]

...How?  Unless you want complex numbers...it would be (x^2+3x+4)/(x+3)(x+4)

He simplified it more after my post.  You can see by the fact that his post was edited at a time after mine was posted.  There is still one local extremum waiting to be found ... :)

This is an also 9th grade problem. Please stop using calculus  :)

i did simplify it the first time, but i did not have enought time to post it, i edited later.
there is a min at $x=-.585$ and there is a max at $x=-3.4142$

I think we can change it into a quadratic equation and use $b^2-4ac$ to find $y$.


Let be given a table nxn.Write $a_{j,1},a_{j,2},a_{j,3}(j=1,2,3,...,n)$ in $n$ columns and $b_{i,1},b_{i,2},b_{i,3}(i=1,2,3,...,n)$ in $n$ lines such that:
$a_{j,1}+a_{j,2}+a_{j,3}=b_{i,1}+b_{i,2}+b_{i,3}(\forall i,j=\overline{1,n})$.
Prove that:
\[\sum_{i=1}^n a_{j,1}a_{j,2}a_{j,3}=\sum_{i=1}^n b_{i,1}b_{i,2}b_{i,3}\]
If you solve it then you give me a special gift for my birthday(today)! :)  :D

[quote="jhaussmann5"]Happy birthday!  Now, about your problem, I don't think it's true.

For example, take $n = 1$, $a_{1,1} = a_{1,2} = -1$, $a_{1,3} = 2$, $b_{1,1} = b_{1,2} = 1$, $b_{1,3} = -2$.  Then $a_{1,1} + a_{1,2} + a_{1,3} = b_{1,1} + b_{1,2} + b_{1,3} = 0$, but $a_{1,1} a_{1,2} a_{1,3} = 2$ and $b_{1,1} b_{1,2} b_{1,3} = -2$.[/quote]
Thanks!:D
But I think you had a mistake,from we write 3 number in the column or the line so $n\ge 3$.Right?

I guess I'm confused... can you give an example, say for $n = 4$?

[quote="pigfly"]Ok,here is the example:
  :)      2        3      - 5
  -6      4         2        :)
   2      :)        -5       3
   4      -6       :)       2[/quote]

I'm sorry, I still don't see it.  What are the values of $a$ and $b$ here?


So, uh, who will be at PUMaC, and on what team?

I'm going with the Upstate New York Math Cabal.

Central MD

(I'm the only one from Pakistan, so it should be easy to find me...

and also I will probably be wearing a UMD sweatshirt like the whole time :D)

Carolinas will be there

I'll be the 7 foot, red-haired (with dreadlocks), caucasian on the team. I'll be wearing a pink sweatshirt with "I'm sexy" written on the front.

[quote="Iversonfan2007"]I'll be the 7 foot, red-haired (with dreadlocks), caucasian on the team. I'll be wearing a pink sweatshirt with "I'm sexy" written on the front.[/quote]

OK I"LL TRY TO FIND YOU MAN

I will be there. ;)
I'll be the tall dude with curly hair who is in charge of everything. :D

[quote="joml88"]I might be there.  I'm a skinny asian with glasses.  Basically just another asian so good luck finding me.  I like yellow chives, if that helps.  If towersfreak goes with me then I'll probably be hanging out with him.  He's caucasian with a light Italian skin tone, about 145 pounds, 5'9.[/quote]
wait HUH I thought you were caucasian and towersfreak was asian :maybe:

I'll be there.

I'll be struttin in with the uber Quagga team and the Florida B team.

And yes, I'll be keeping my pimp hand strong.

I'll be there.   :D
Short Asian girl, running around, probably looking five times as stressed as Nathan, even though I'll probably have one fifth of his responsibility.  I'm not good with stress... :P

i'll be there, with the really awesome team otherwise known as upstate new york.

[quote="E^(pi*i)=-1"][quote="eyefragment"]I'll be there, getting absolutely demolished as part of the Lawrenceville School team.  I'll be sporting an overgrown beard and a Canada/USA MathCamp shirt.[/quote]

Hey, cool, I didn't know you were going.

I'll be the other one in a Mathcamp shirt (Maroon, if that helps), on the Central MD team.[/quote]

Yay!  On a related note, Yasha will be there as part of the NC team.  Look forward to seeing you next week.

[quote="eyefragment"]I'll be sporting an overgrown beard and a Canada/USA MathCamp shirt.[/quote]

It's possible that so will Professor Conway.   :P

I will be there, along with the rest of the Livingston High School team =)

hey. I'm indian, 5' 7'', black hair.

will be coming with the south brunswick high school team A and team B from my school is coming too.

I'll be there.  I'll be making sure people get destroyed from lack of rigor on the power round :)  I'm semikidding.

OH YEAH WELL IT IS ON MY RIGOR BRINGS ALL THE BOYS TO THE YARD


inequalities
graph theory
combinatorics unsolved
combinatorics

$ G$ is a tree. $ x, y$ are vertex of $ G$, $ e(x)$ is eccentricity of $ G$.
Let  $ L$ be a walk of length  $ e(x)$  which starts  at  $ x$.
If  $ L$ does not contain $ y$ , then $ e(y)\equal{}e(x)\plus{}1$.

why?

Is that true? according with this definition
[quote]Let $ G$ be a graph and $ v$ be a vertex of $ G$. The eccentricity of the vertex $ v$ is the maximum distance from $ v$ to any vertex. That is, $ e(v) \equal{} \max\{\textrm{d}(v,w) \ : \ w \in V(G)\}$.[/quote]
let us consider the tree made of a path $ v_1v_2\ldots v_n$, with $ n > 5$. Then $ e(v_3) \equal{} n \minus{} 3$, $ L \equal{} v_3v_4\ldots v_n$ is a walk of length $ e(v_3)$, starting at $ v_3$ and not containing $ v_1$, while $ e(v_1) \equal{} n \minus{} 1 \equal{} e(v_3) \plus{} 2$.

All we can say, under your notations, is that $ e(y) \geq e(x) \plus{} 1$, which is clear by the definition (the tree is connected, so there is a path $ P$, of length at least $ 1$, connecting $ y$ to $ x$; then consider the walk $ PL$).

Sorry... $ y$ and $ z$ are adjacent to $ x$ ...

1) If $ L$ does not contain $ y$, then $ e(y)\equal{}e(x)\plus{}1$, and

2) if $ L$ contain $ y$, then $ e(y)>e(x)$

Is it right??

Thanks to mavropnevma. :D

Who is $ z$ that just appeared?

Almost right. Yes, if $ x,y$ adjacent and $ L$ does not contain $ y$, then $ e(y) \equal{} e(x) \plus{} 1$.
If $ L$ contains $ y$, I would say $ e(y) \equal{} e(x) \minus{} 1$, not $ e(y) > e(x)$ ...

oh..I see  :)

one more qusetion to mavropnevma

but sorry for my bad english

qusetion:
For connect graph $ G$,
$ 2f(G) \ge d(G)$

$ f(G)$ is radius and $ d(G)$ is diameter of $ G$

why?

Thanks to mavropnevma  :)

One should use standard notations (see R. Diestel: Graph Theory), $ \textrm{diam}(G)$ for the diameter, and $ \textrm{rad}(G)$ for the radius.

The following is true for any graph: $ \textrm{rad}(G) \leq \textrm{diam}(G) \leq 2\textrm{rad}(G)$.

The first inequality is indeed trivial; the second one (you asked about) is true since if we take $ v$ a central vertex (with largest distance to any other vertex equal to $ \textrm{rad}(G)$), and $ x,y$ two vertices realizing $ \textrm{diam}(G)$, then $ \textrm{diam}(G) \equal{}  \textrm{d}(x,y) \leq \textrm{d}(x,v) \plus{}   \textrm{d}(v,y) \leq 2\textrm{rad}(G)$.

Thank you very much for your reply.

I am studying about tree but very difficult

when I can't understand it , please help  :D


[quote="hien"]if $a,b,c$ are sides length of triagle ABC. It is true or false:
$\frac{a}{\sqrt{bc}}+\frac{b}{\sqrt{ca}}+\frac{c}{\sqrt{ab}}\leq 3$[/quote]
false
take
$a=4$
$b=3$
$c=5$
 :)


In trapezoid $ABCD$, sides $AB$ and $CD$ are parallel, and diagonal $BD$ and side $AD$ have equal length. If $m\angle DBC=110^\circ$ and $m\angle CBD =30^\circ$, then $m \angle ADB=$

$\text{(A)}\ 80^\circ \qquad \text{(B)}\ 90^\circ \qquad \text{(C)}\ 100^\circ \qquad \text{(D)}\ 110^\circ \qquad \text{(E)}\ 120^\circ$

Doesn't make sense. How can $m \angle DBC$ and $m \angle CBD$ be different in measure?

I think it should be $\angle DCB=110^\circ$.

Probably. It is the only angle that looks obtuse.

[hide="solution if it is true"]$m \angle CDB = 180 - m \angle CBD - m \angle DCB$. So, $m \angle CDB = 180 - 30 - 110 = 40$. $m \angle CDB = m \angle DBA = 40$ because $DC || AB$. Because $DA = BD$, $\triangle BDA$ is isosceles. So $m \angle DAB = m \angle DBA = 40$. $m \angle ADB = 180 - m \angle DAB - m \angle DBA = 180 - 40 - 40 - \boxed{100}$[/hide]

$\angle ADB$ looks obtuse to me, but that's the one you're solving for, so... :P


inequalities
vector
limit
triangle inequality
calculus
calculus computations

If $ \sum_{n\equal{}1}^{\infty}x_{n}$ is a convergent series in a normed vector space $ X$, show that $ \left\|\sum_{n\equal{}1}^{\infty}x_{n}\right\|\leq\sum_{n\equal{}1}^{\infty}\left\|x_{n}\right\|$.

We can write
$ \left\|\sum_{n\equal{}1}^{\infty}x_{n}\right\|\equal{}\lim_{m\rightarrow\infty}\left\|\sum_{n\equal{}1}^{m}x_{n}\right\|$
can't we?  Does that necessarily mean that
$ \lim_{m\rightarrow\infty}\left\|\sum_{n\equal{}1}^{m}x_{n}\right\|\leq\lim_{m\rightarrow\infty}\sum_{n\equal{}1}^{m}\left\|x_{n}\right\|$?
Or am I going about this the wrong way entirely?

You haven't mentioned triangle inequality.  :maybe:

If one (okay, me) does a proof by contradiction, then we ultimately get a violation of the triangle inequality so I've got a proof.  Thanks anyway.


25. In triangle JSR, <R is a right angle and the altitude from R meets JS at Z. The lengths of JSR are integers, SZ=1331, and cos(S)=$\frac{p}{q}$ where p and q are relatively prime. Find p+q.

i get more than 1 answer, so could you list the choices?


function
group theory
abstract algebra
set theory
algebra unsolved
algebra

Find all function $f: \mathbb{R}\rightarrow \mathbb{R}^{+}$ that is onto and statisfies $f(x+y)=f(x)f(y)$ for all $x,y \in \mathbb{R}$.

[quote="Rust"]Let $g(x)=ln(f(x)): R\to R$, then g(x) additive: $g(x+y)=g(x)+g(y)$.[/quote]

NOT sufficient for solution of this problem! You have to prove $g$ is monotone, or...

All additive functions, satisfyed $g(x+y)=g(x)+g(y)$ can be represented by Hamel's basis or by subgroup A (annulator), belongs R, suth that $R/A=Q$ and $a\in R\A$ (if $g(x)\not =0 \ \forall x$), suth that $g(a)=c\not =0, c\in R$.
Then for all $x\in R$ exist $r\in Q$, suth that $x-ra\in A$ and $g(x)=cr$.

Some assumption on the function is needed, in order to force it to be the standard solution.   This follows from the theorem that Zermelo-Fraenkel set theory with a countable form of the axiom of choice, is consistent with the statement "all real functions are measurable".


[quote="nealth"]What is the value of cos20*cos40*cos80?[/quote]
Trig isn't supposed to be posted in the Mathcounts forum...

really easy, just use the double angle formula to get [hide]1/8[/hide]

$S=\cos 20 \cos 40 \cos 80= \cos 20 \cos 40 \sin 10$
$S \cos 10=\frac18 \sin 10 \implies S=\frac18 \tan 10$

[quote="me@home"]$S \cos 10=\frac18 \sin 10 \implies S=\frac18 \tan 10$[/quote]
This should be $S \cos 10=\frac{1}{8}\sin 80 \implies S=\frac{1}{8}$


I enjoyed this funny little problem:

Let $ABC$ be a given triangle and let $k$ be its incircle. Three tangents to $k$ are drawn, parallel to the sides of the triangle. These tangents divide the triangle into a hexagon and three small triangles (similar to $ABC$). Let $r$ be the radius of $k$ and let $r_1$, $r_2$ and $r_3$ be the radii of the incircles of each of the small triangles.

Show that $r_1 + r_2 + r_3 = r$.

$ABC$ must be equilateral, by calling each side of the biggest equilateral triangle equal to $x$, we have by $rs=A$, so $r=\frac{x\sqrt{3}}{6}$ now the other three smaller circles are inscribed in equilateral triangles also which are the same size, and their sides are $\frac{1}{3}$ of the bigger triangle which can be proven because the hexagon the circle with radius $r$ is inscribed in is regular with side length equal to a third of the side of the equilateral triangle, thus each of the smaller circles have a radius of, using the $rs=A$ formula, of $\frac{x\sqrt{3}}{18}$, so $3 \cdot \frac{x\sqrt{3}}{18}=\frac{x\sqrt{3}}{6}$.

Correct me if this is not a good enough proof since this is the first proof I have done.

This can be proven by just proving that the opposite sides of the hexagon are congruent, because the triangles are similar anyways.  There, I'm still thinking...

I'm thinking the opposite sides actually might not be congruent, but the sum of the 3 sets of opposite sides is actually equal, because of tangents

The opposite sides of the hexagon are actually equal in length, but you need to prove that :)

Well, here's how I did it, and you don't need to prove that :D

So the ratio of r1 to r is equal to smaller triangle "1"'s perimeter to the whole triangle.

Do that for the other 2.

However, you can easily see from tangents that the 3 little triangle's perimeters added together are the same as the big triangles, so (r1+r2+r3)/r=1, and we're done


For all integers $n$ such that $n\in[0,N]$ where $N$ is also an integer we are given a sequence such that $F_{n}=(p)F_{n+1}+(1-p)F_{n-1}$ where $p$ is a constant that satisfies $0<p<1$.

Also $F_{0}=0$ and $F_{N}=1$
Find $F_{n}$ in terms of $n$.

Use the characteristic equation:

$pF_{n+1}-F_{n}+(1-p)F_{n-1}=0$ gives

$pt^{2}-t+(1-p)=0$, and the solutions are

$t_{1}=\frac{1-p}{p}, t_{2}=1$.

By the method of undetermined coefficients we get

$F_{n}=\frac{\left(\frac{1-p}{p}\right)^{n}-1}{\left(\frac{1-p}{p}\right)^{N}-1}$

two comments:
1) how do you go from $t_{1}=\frac{1-p}{p}$ and $t_{2}=1$ to $F_{n}=\frac{\left(\frac{1-p}{p}\right)^{n}-1}{\left(\frac{1-p}{p}\right)^{N}-1}$?

and

2) what about when $p=\frac{1}{2}$. [hide="answer to this case"] $F_{n}=\frac{n}{N}$ but i would like to see somone else prove it [/hide]

other than that very nice.

Into the general formula $F_{n}=At_{1}^{n}+Bt_{2}^{n}=A\left(\frac{1-p}{p}\right)^{n}+B$, where $A, B$ are constants, we plug $n=0$ and $n=N$ and solve the system for $A, B$. The solution is $A=\frac{1}{\left(\frac{1-p}{p}\right)^{N}-1}, B=-\frac{1}{\left(\frac{1-p}{p}\right)^{N}-1}$

And yes, I missed the case $p={1\over 2}$, sorry about that.


Ana and Bill are playing a game in which they take turns. In each turn, they put one bead into one of ten boxes (randomly, with each box having equal probability). Each box can hold an indefinite number of beads. Once all of the boxes are filled, the person who put the last bead wins.

Question 1: What is the probability that the game lasts more than 12 turns?
Question 2: What is the expected value of the number of turns needed to finish the game?
Question 3: Does the person who gets the first turn have an advantage, disadvantage, or no advantage?

[quote="PhireKaLk6781"]Each box can hold an indefinite number of beads. Once all of the boxes are filled[/quote]  :huh: This is incomprehensible.

Whoops. Sorry for the lack of clearness. "Once all of the boxes are filled" should be "Once each box as at least one bead".  :P

Would someone be willing to answer and provide a solution?   :|

I hope this isn't getting missed. Would someone please tell me how this problem is solved?

Here are some hints:

Question 1 is a counting problem: count the total number of possible ways the first 12 moves can go, then count the number of ways that result in all 10 boxes being filled, then subtract.  (The answer is larger than 99%, i.e., it's extremely unlikely that the game ends in 10, 11 or 12 turns.)

For question 2, perhaps the easiest thing to do is a state diagram approach: let $ e(n)$ be the number of expected moves required when there are $ n$ boxes left unfilled.  Thus, we're trying to compute $ e(10)$, and we have (for example) that $ e(6) \equal{} 1 \plus{} \frac {4}{10} e(6) \plus{} \frac {6}{10} e(5)$.

For question 3, a different but similar state diagram approach seems in order.  Let $ p(n)$ be the probability that the first player wins when starting with $ n$ unfilled boxes.  We have $ p(0) \equal{} 0$ (the first player has just lost) and we want to compute $ p(10)$.

These questions are one form of the coupon-collector problem: part 2 is the most common version, I think.  Question 1 can be answered in more generality using inclusion-exclusion, another common form.  A search for the word "coupon" on the forum brings up several recent discussions.