How many perfect squares are factors of \\ $ 2 \times 4 \times 6 \times 8 \times 10 \times 12$?

[hide]
Each 2 can be written as $ 2^x\cdot y$ where $ n$ and $ y$ are a positive integer. This leads to:
\begin{align*} 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 & = 2\cdot2^2\cdot(2\cdot3)\cdot2^3\cdot(2\cdot5)\cdot(2^2\cdot3) \\
& = 2^{10}\cdot3^2\cdot5 \end{align*}
There are 6 spots for the powers of 2 and 2 spots for the powers of 3. Thus the answer is $ 2\cdot6 = \boxed{12}$
[/hide]

I agree with violin and have a rigorous [hide="solution."]Each square can be written as $ p_1^{2e_1}\cdot p_2^{2e_2}\cdots p_x^{2e_x}$, where $ p_i$ are primes and $ e_i$ are non-negative integers. The value of $ 2\cdot 4\cdot 6\cdot 8\cdot 10\cdot 12$ is $ 2^{10}\cdot 3^2\cdot 5$. The 5 is ruled out since it is not a perfect square and has no perfect square factors other than 1. Thus we are left with $ 2^{10}\cdot 3^2$, and we pluck out 2's and 3's from that to get a perfect square number $ 2^{2e_1}\cdot 3^{2e_2}$, where $ 0\leq e_1\leq 5$ and $ 0\leq e_2\leq 1$. There are 6 possible values $ e_1$ could be and 2 possible values $ e_2$ can be. Thus there are 12 perfect square factors.[/hide]

Easy way to find number of square factors:

Prime factorize it, then round each of the exponents in the factorization to the next highest even number. Divide each of the new exponents by 2 and multiply them together.


what kind of attractive forces dominate in the following systems:

a) liquid xenon
b) an aqueous solution of pottassium bromide
c) frozen water

I'm pretty sure the answer to c is hydrogen bonds...

what kind of attractive force must be overcome to

a) boil water
b) melt KCL
c) sublime diiodine (I2)
d) boil dihydrogen sulfur

I'm pretty sure the answer to a and d is hydrogen bonds but im not completely sure.

Please explain your answers :)

Xenon atoms are non-polar but easily polarizable: therefore, in liquid Xenon London forces dominate.

In an aqueous solution of KBr we have molecules of water and solvated K+ and Br- ions. Therefore, we have hydrogen bonds between water molecules and electrostatic atractions between each ion and molecules of water.


calculus
integration
algebra
function
domain
polynomial
abstract algebra

Let $ A$ is an integral domain and let $ i,j\in N$ and $ (i,j) \equal{} 1$ show that $ < x^i \minus{} y^j >$ in the polynomial ring $ A[x,y]$ is prime.

consider the homomorphism $ A[x,y] / (x^i \minus{} y^j) \to A[z] , x \mapsto z^j , y \mapsto z^i$. we show that it is injective; in particular, $ (x^i \minus{} y^j)$ is prime.

using polynomial division (which works in general for polynomials whose leading coefficient is a unit), we see that every element in $ A[x,y]/(x^i \minus{} y^j)$ has the form $ \sum_{u<i,v<j} a_{uv} x^u y^v$ with (unique) coefficients $ a_{uv} \in A$. if it lies in the kernel, we have $ \sum_{u<i,v<j} a_{uv} z^{ju \plus{} iv} \equal{} 0$. now the $ ju \plus{} iv$ are pairwise different: if $ ju \plus{} iv \equal{} ju' \plus{} iv'$, we have $ j(u\minus{}u')\equal{}i(v'\minus{}v)$. since $ i$ and $ j$ are coprime, we get $ j | v' \minus{} v$. since $ \minus{}j < v' \minus{} v < j$, this shows $ v' \equal{} v$. in the same way we get $ u' \equal{} u$. thus, by comparing coefficients, we get $ a_{uv}\equal{}0$ for all $ u<i, v<j$, as desired.

That doesn't quite work; we can't bound the degree in both variables simultaneously. After all, $ x^{2i}$ maps to $ z^{2ij}$, and you can't get that with combinations of $ x^uy^v$ with $ u<i$ and $ v<j$.

The appropriate reduced form is actually $ \sum_{0\le u<i,v\ge 0}a_{uv}x^uy^v$ or the counterpart restricting $ v$. Running your argument with this form, we get $ j(u\minus{}u')\equal{}i(v\minus{}v')$, so $ i|u\minus{}u'$. Since $ \minus{}i<u\minus{}u'<i$, $ u\equal{}u'$. From the equation, $ v\equal{}v'$ as well and the rest follows.


calculus
integration
logarithms
search
calculus computations

Splendid! I love these sort of problems... post many, many more of these. Check out Kunny's Today's Calculation of Integral #xxx series of posts in this forum (search for it).

@Fitim : What you said is true only if $ n\to \infty$


a ball was being dropped from a height of 3m. It always bounces back to 1/10 of the previous height. what will be the total height the ball has travelled when it stops bouncing.

* I am not sure abt the numbers but is it possible to solve such a question?

Ok im confused.  If the ball ALWAYS bounced back to 1/10 of the previous distance how could it posibly stop bouncing?  Maybe im overthinking the question but idk

[hide="solution"]
you can use infinite geometric series.  the sum is going to be

$ 3 \plus{} 2\times3\times \frac {1}{10} \plus{} 2\times3\times\frac {1}{10^2} \plus{} ...$
$ 3 \plus{} 2\left(\frac {3}{10} \plus{} \frac {3}{100} \plus{} \frac {3}{1000} \plus{} ...\right)$.
Call the sum in the parentheses $ x$, then
$ x \equal{} \frac {3}{10} \plus{} \frac {3}{100} \plus{} \frac {3}{1000} \plus{} ...$
Now, divide both sides by 10
$ \frac {x}{10} \equal{} \frac {3}{100} \plus{} \frac {3}{1000} \plus{} \frac {3}{10000} \plus{} ...$

So $ \frac {x}{10}$ is the exact same as x, except that it's missing a 3/10, so we subtract the second equation from the first

$ \frac {9x}{10} \equal{} \frac {3}{10}$, so $ x \equal{} \frac {1}{3}$.  Plugging this into our original expression, the ball has bounced $ 3\frac {2}{3}$ meters.
[/hide]

I am assuming that you're talking about the height it bounces up and not falling down.

[hide]At first it bounces up 3 x 1/10, then 3 x (1/10)^2, then 3 x (1/10)^3.....
So if you add this infinite geometric sequence by using the formula:
first term/1-the common ratio.
(3/10)/(1-1/10)
Solving you would get 1/3 of a meter.  [/hide]

Sorry if my explaination is hard to understand

[quote="zolojetto"]but doesnt it say that it will always bounce back up no matter how small of a distance between it and the ground.  Maybe im confused but doesnt the question ask how far the ball has traveled when it has stopped bouncing and the way i interpret the question is that the ball never stops bouncing.[/quote]

yes, the idea of infinity is hard to grasp at first.  I suppose the question could have been better worded.

i am sorry for the ambiguous question but wat i meant was it was dropped from 3m, 1/10 of it would be 0.3m so it bounces up 0.3m and drops again, bounces back to 1/10 of 0.3m, that would be 0.03 etc.

sry though and ty for response

According to the question the ball would bounce infinetly. However Newtons laws state something in motion will stay in motion unless acted upon by an outside force. That force is gravity....so....physically not possible


geometry
trigonometry
cyclic quadrilateral
geometry proposed

Given an acute triangle ABC such that $\measuredangle{A}= 60^{0}$. Let $H, I$ be respectively the orthocenter and incenter. Then

$\frac{BH-CH}{BI-CI}= \frac{HO}{HI}$

Since $\angle{A}=60^{o}$, so we have $B$, $O$, $I$, $H$, $C$ concyclic. Thus in cyclic quadrilateral $BOIH$, by Ptolemy theorem, we have: \[OI \cdot BH+HI \cdot BO=OH \cdot BI\] In cyclic quadrilateral $COIH$, by Ptolemy theorem, we have: \[OI \cdot CH+HI \cdot CO = OH \cdot CI\] Substract the two equalities, we have: \[\frac{BH-CH}{BI-CI}=\frac{OH}{OI}\] Thus it only remains to show that $OI=HI$. Since $AH=2R |\cos{A}|=R=AO$, and $\angle{OAI}=\angle{HAI}$, so we have $\triangle{OAI}\cong \triangle{HAI}$, thus $OI=HI$, so we finally have: \[\frac{BH-CH}{BI-CI}=\frac{OH}{HI}\]

Very nice! Your solution is the same as mine :)


Triangle ABC and Triangle DCE are on the same side of Line ACE
AB=BC=AC,    DC=CE=DE
BE and AD intersect at R
Prove that RC is angle bisector of  angle ARE
Also prove that Quadrilateral RDEC is cyclic

After exploring the problem by drawing a figure with the geometer sketchpad, I realized some constant angles, namely angle ARC = angle ERC = angle ARB = angle EDR = 60 degrees.
Spending more than an hour last night, I didn't come up with any relevant method to prove this simple relation.
Finally I joined the line BD and realized a pattern.
This figure is a special case for the proof of the existence of Fermat point!
Here is a link to Fermat point for your reference.

[url=http://mathworld.wolfram.com/FermatPoints.html]mathworld.wolfram.com/FermatPoints.html[/url]

This problem was tripple-posted. The [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36474]solution[/url] is in the Olympiad section, Geometry, Unsolved problems.


algebra
difference of squares
special factorizations

given a natural number could you find the number of ways in which you can write it as the difference of two squares of natural numbers

$ a^2 \minus{} b^2 \equal{} c \implies (a\plus{}b)(a\minus{}b) \equal{} c$ Now factorise c into 2 factors d and e so that d > e and d = a+b and e = a-b such that a and b are natural numbers.
Or are you asking for a general solution? :|

[quote="befuddlers"]I asked for the general solution :D[/quote]

You can represent all numbers of the form

4*k+1 and 4*k-1 and 4*k as difference of two squares but  you can never represent natural numbers of the form

4*k+2 as difference of squares if you want msg me on manup86@gmaill.com i will mail you the solution

A detailed account on this could be found in Burton, whr der s a complete chapter devoted 2 sum and difference of squares. In the meanwhile you can catch up with a number of other indians at http://joigroup.freezoka.com/


Let n \geq 3 be odd.
Given numbers ai, bi in [0, 1], where 1 \leq i \leq n, prove that there exist i, j such that

0 \leq aibj - ajbi \leq 2/(n - 1).

Well, I may be wrong but I do not see what the hypothesis 'n odd' is made for...

For i = 1,...,n, let A_i(x_i,y_i). Then A_i belongs to the unit square C with vertives O(0,0), (1,0),(1,1) and (0,1).

Wlog, we may suppose that the points A_i have been labeled so that <A_iOA_(i+1) are in not decrasing order.
Then if i<j the number a_ib_j - a_jb_i is equal to 2*[OA_iA_j], twice the area of the (possibly degenerate) triangle OA_iA_j.

In another hand, we have :
(n-1)*min[OA_iA_(i+1)]  \leq sum [OA_iA_(i+1)]  \leq [C] = 1.
And the result follows.

Pierre.


calculus
integration
calculus computations

Evaluate the integral:
$ \int_{0}^{2} \sqrt{1\plus{}x^3} dx$
I used my calculator to evaluate it, and the result is approximately(?) $ 3.421$. I'm curious whether there exists a non-calculator solution or not.I couldn't find one though even though I used integration by part and other techniques that I know. So, pleased show me your solution, if possible.

The second integral there is (as far as I can tell) an [url=http://en.wikipedia.org/wiki/Elliptic_integral]elliptic integral[/url].

[quote="kunny"]$ I \equal{} \int_0^2 \sqrt {1 \plus{} x^3}dx \equal{} [x\sqrt {1 \plus{} x^3}]_0^2 \minus{} \int_0^2 x\cdot \frac {3x^2}{2\sqrt {1 \plus{} x^3}}dx$

$ \equal{} 6 \minus{} \frac {3}{2}\int_0^2 \left(\sqrt {1 \plus{} x^3} \minus{} \frac {1}{\sqrt {1 \plus{} x^3}}\right)dx$

$ \therefore \frac {5}{2}I \equal{} 6 \plus{} \frac {3}{2}\int_0^2 \frac {1}{\sqrt {1 \plus{} x^3}}dx$

$ \Longleftrightarrow I \equal{} \frac {12}{5} \plus{} \frac {3}{5}\int_0^2 \frac {1}{\sqrt {1 \plus{} x^3}}dx$.[/quote]
A very nice solution. Thank you very much Mr. Kunny! :)
To finish it, I think we need to use $ \int_0^2 \frac {1}{\sqrt {1 \plus{} x^3}}dx \equal{} ln/x \plus{} \sqrt {x^{3/2} \plus{} 1}/$

hello, the equation $ \int\frac{1}{\sqrt{1\plus{}x^3}}\,dx\equal{}\ln|x\plus{}\sqrt{x^{\frac{3}{2}}\plus{}1}|$ is wrong.
Sonnhard.

There is no elementary antiderivative. You will have to satisfy yourself with some numerical approximation.

$ \int_{0}^{2} \sqrt {1 \plus{} x^3} dx$ means the arc length of $ y\equal{}\frac{2}{5}x^2\sqrt{x}\ (0\leq x\leq 2)$,

[quote="kunny"]$ I \equal{} \int_0^2 \sqrt {1 \plus{} x^3}dx \equal{} [x\sqrt {1 \plus{} x^3}]_0^2 \minus{} \int_0^2 x\cdot \frac {3x^2}{2\sqrt {1 \plus{} x^3}}dx$

$ \equal{} 6 \minus{} \frac {3}{2}\int_0^2 \left(\sqrt {1 \plus{} x^3} \minus{} \frac {1}{\sqrt {1 \plus{} x^3}}\right)dx$

$ \therefore \frac {5}{2}I \equal{} 6 \plus{} \frac {3}{2}\int_0^2 \frac {1}{\sqrt {1 \plus{} x^3}}dx$

$ \Longleftrightarrow I \equal{} \frac {12}{5} \plus{} \frac {3}{5}\int_0^2 \frac {1}{\sqrt {1 \plus{} x^3}}dx$.[/quote]

thank you extermely much kunny

very nice solution

hello, but you can't solve the integral $ \int_{0}^2\frac{1}{\sqrt{1+x^3}},\,dx$ expressed  by the known elemetary functions.
Sonnhard.


\documentclass{article}
\usepackage{latexsym}

\begin{document}

Este adevarata propozitia de mai jos?

``Fie $f: I\to R$ o functie continua pe \textit{intervalul I.} $I$
este un interval oarecare, de oricare tip (inchis sau deschis la
stanga sau la dreapta, sau in ambele parti, marginit sau nemarginit,
sau combinatii de acestea). Mai presupunem ca functia este strict
monotona pe $I$. Fie a si b capetele lui $I.$ Ele pot face parte sau
nu din interval si pot fi reale sau infinite. Atunci:

Functia are limita in a si in b ? Indiferent daca limitele, sa zicem
l si respective L sunt finite sau infinite, exista ele?

Multimea valorilor functiei, adica f(I) = intervalul cu capetele l
si L (in ordinea in care sunt l si L?

3. Daca de exemplu I = [a, b) si f este strict descrescatoare,
atunci f(I) = (L, f(a)]?

\end{document}

Desi vad ca topicul e de 2 ani,raspunsul este afirmativ la toate intrebarile!

Mai exact,orice functie monotona are limita in orice punct de acumulare al domeniului.(Deci la prima intrebare,nici macar nu-i nevoie ca functia sa fie continua).
O functie continua si strict monotona,da,transforma intervalul dat in intervalul ale carui capete sunt limitele functiei la extremitatile intervalului de definitie(acelasi tip de interval,desigur,cand e descrescatoare se schimba corespunzator capetele inchise)


Hi, I've been learning Java for a few weeks in a technology class, so I am not good and don't know very much.  However, I have a major project due tomorrow, and am having major problems.  Here is one problem I am having.(I cut out a lot of unnecessary stuff)
[quote]import java.io.*;

public class digital
{
	public static void main(String[]args)
	throws Exception
	{
		digital();
	}
	public static void digital()
	throws Exception
	{
		int mpixels;
		int pixels;

		mpixels=(int)System.in.read();
		pixels=mpixels*1000000;

		//Output
		System.out.print("This is equal to "+pixels);
		System.out.println(" pixels");
}
}[/quote]
The program is supposed to have th user enter a value, then multiply it by 1 milliion then display the result.  However, I get weird results.  For example when I enter 5, it displays 53 milion, which is obviously not correct.  Please help, this is urgent   :(

The issue here has to do with how input/output works in java.
System.in is of type InputStream, and read() in InputStream reads the next character
so what you're doing is reading the first character, which gets it's ASCII value, and then you multiply that by a million.

since what you probably want to do is read the whole line, and not just the first character, here's one way to do it (though i'm not great at java, so it's probably not the best way):

replace:
mpixels=(int)System.in.read();

with:
BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
String s=in.readLine();
mpixels=Integer.parseInt(s);

the three lines do this: first set up a "BufferedReader," which can read a whole line at a time.  then read the whole line into a string.  Integer.parseInt(s) converts the String s into an integer.

I think that that should do it.

-Kevin

what if I want to use decimal?  (using double)  would I do mpixles=Double.parseDouble(s);

yeah, that should work.
the online Java API at java.sun.com is a life saver

Cool, thanks.  I got a 100 on my program.  :D


probability
function
integration
logarithms
real analysis
calculus
derivative

This is the same question which I asked in the probability section. I am putting it here in a different way.
Suppose that $ f_n$ is a sequence of (measurable) functions such that $ \int f_n \log f_n d\mu \le k < \infty$ where $ \mu$ is a finite measure. Does there exists a convergent subsequence $ \{f_{n_i}\}$ in $ L^1$?

Perhaps you need the notion: "uniformly integrable"

Some notational comments:

It looks like you're tacitly assuming that $ f_n\ge 0.$ If not, you really should have some absolute values somewhere.

The other thing is that the logarithm has problems near zero that aren't really what you want to be important here. I'd probably write the $ L\log L$ condition this way:
\[ \int |f_n|\log(1\plus{}|f_n|)\,d\mu\le k<\infty.\]
But that doesn't really matter all that much, as long as we interpret $ 0\log 0\equal{}0.$

Now, just to see why there's something to do here: what if you relaxed the $ L\log L$ condition to being uniformly bounded in $ L^1?$ That is, assume only that $ \int|f_n|\le k<\infty?$

That gets shot down by this example. Let the measure space be $ [0,1]$ with Lebesgue measure and let $ f_n\equal{}n\mathbf{1}_{[0,1/n]}.$ Then $ \|f_n\|_1\equal{}1$ but there is no convergent subsequence in $ L_1.$

Note that the $ L\log L$ condition rules out this example.

$ \int f_n\log(f_n)\equal{}\log n$ and is not bounded.

You still have to answer the original question, of course - I haven't answered that for you yet.

[quote="Kent Merryfield"]It looks like you're tacitly assuming that $ f_n\ge 0.$ If not, you really should have some absolute values somewhere.[/quote]
Yes $ f_n\ge 0$.

[quote="Kent Merryfield"]The other thing is that the logarithm has problems near zero that aren't really what you want to be important here. But that doesn't really matter all that much, as long as we interpret $ 0\log 0 \equal{} 0.$[/quote]
Yes absolutely!

Moreover one can assume $ \|f_{n}\|_{1}\equal{}1$ for all $ n$. After all these $ f_n$'s are Radan-Nykhodym derivatives of prob. measures $ P_n$ with respect to $ \mu$, i.e. $ f_n\equal{}\frac{dP_n}{d\mu}$.

[quote]The family ${ \{f_{\alpha}\}}$ is '''uniformly integrable''' iff there exists a nonnegative increasing function $ G(t)$ such that $ \lim_t \frac{G(t)}{t} = \infty$ and $ \sup_{\alpha} E(G(|f_{\alpha}|)) < \infty$[/quote]

By the above result we can conclude that $ \{f_n\}$ is uniformly integrable. From this is it possible to extract a convergent subsequence?


geometry
combinatorial geometry
combinatorics unsolved
combinatorics

Hello ,

I am trying to find the triangle with maximum area , from a given set of 'n' integer co-ordinates of the form x,y .
Only problem is 'n' is huge ( of the order of 10^6) so i have to use an efficient method.

So i constructed a convex hull from the given n points. Now i have the vertices of the convex polygon and i need to calculate the area of the maximum triangle.

Is it correct that if i choose any two vertices of the convex hull and then loop over the remaining vertices one by one , every time calculating the area and marking the one with the maximum area , i will get the right answer?
If not please correct me .

thanks

Don't double post. http://www.mathlinks.ro/viewtopic.php?p=1555764#1555764

yes sorry , just i wasnt sure if i am posting in the right category .


I know that the MC directors are probably grading like crazy right now to look at all applicants entries. Any idea when we'll know yes, no, or waiting list?

Thanks. :)

I thought the [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=156]math jam about MathCamp[/url] suggested that there would be two weeks of grading quizzes after the application deadline, and then applicants would be informed of results of their applications. The whole math jam transcript is well worth reading.

Thank you for the link! I seem to have missed the jam.

I got waitlisted for the 2006 mathcamp :(

How much chance do I have of going then?

Some people who get accepted will not be going as they will be attending other programs. (You will see if you read the Admissions subforum). Depending on how well you think you did on your quiz, you can probably judge whether you'll get in or not. Put it this way, at least you weren't rejected ;)


I have a LaTeX document that is predominantly two-columned (using \documentclass[12pt, twocolumn]{article}), but I want certain sections of the document to be one-columned (without starting a new page). How would I do this?

Nevermind; I found a solution. Is multicol the best package to use?

[quote="123456789"]Is multicol the best package to use?[/quote]Yes.


logarithms
trigonometry
calculus
integration
real analysis
real analysis unsolved

Prove the following equality. \[\ln \frac{1+a}{1-a}=2\sum_{n=0}^{\infty}\frac{a^{2n+1}}{2n+1}\ (-1<a<1).\]

Note the similarity to the arctangent series:

$\tan^{-1}a=\sum_{n=0}^{\infty}\frac{(-1)^{n}a^{2n+1}}{2n+1}$

Could you show, without going through the series, that

$\tanh^{-1}a=\frac12\ln\left(\frac{1+a}{1-a}\right)\ ?$

[b]Solution [/b]to the problem posted by Kunny.
For $|a|<1,$ we have $\ln (1+a) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}a^{n}}{n}$ and $\ln (1-a) =-\sum_{n=1}^{\infty}\frac{a^{n}}{n}.$

Subtracting the second from the first, we obtain

$\ln \frac{1+a}{1-a}=2\sum_{n=0}^{\infty}\frac{a^{2n+1}}{2n+1}$

Kunny, this problem is too easy. But, it's good to see you back on this part of the forum after a long time. Hope you will keep those integrals and sums coming just as you did earlier! I certainly miss those problems :wink:
-------------------------------------------------
And, here is the [b]solution[/b] to the problem posed by Prof Merryfield.

We know that $\tanh a = \frac{\sinh a}{\cosh a}= \frac{e^{a}-e^{-a}}{e^{a}+e^{-a}}= y$ (say.)

Then, solving for $a,$ we have
$\frac{e^{2a}-1}{e^{2a}+1}= y$

$\Rightarrow \frac1{y}= \frac{e^{2a}+1}{e^{2a}-1}$

$\Rightarrow \frac{1+y}{1-y}= e^{2a}$ ... [Applying componendo and dividendo.]

$\Rightarrow a =\frac12 \ln \left( \frac{1+y}{1-y}\right)$

$\Rightarrow \tanh^{-1}a=\frac12\ln\left(\frac{1+a}{1-a}\right)$


Let be $ n\ge 4\ ,\ n\in \mathbb{N}$ such that $ \left\lfloor \frac{2^n}{n}\right\rfloor \equal{}2^k\ ,\ k\in \mathbb{N}$ . Prove that $ n\equal{}2^m\ ,\ m\in \mathbb{N}$

Assume $ n \ne 2^m$.

First, the consequence of the floor function is such that if $ \left\lfloor\frac {2^{n}}{n}\right\rfloor \equal{} 2^{k}\ ,\ k\in\mathbb{N}$, then $ \frac {2^n}{n} \equal{} 2^k \plus{} p, \mbox{where } p < 1$.

Then, we have $ 2^n \equal{} n2^k \plus{} np$.

Here $ np$ is an integer, which is the remainder when $ 2^n$ is divided by $ n$.

We divide by $ 2^k$ to get $ 2^{n \minus{} k} \equal{} n \plus{} \frac {np}{2^k}$

Since the LHS is an integer and $ n$ is an integer, $ \frac {np}{2^k}$ must also be an integer. But that means $ np$ must either be a multiple of $ 2^k$, or equal 0. If $ np$ is zero then the problem is solved. So we assume the other case.

If $ np$ is a multiple of $ 2^k$, then we have $ np \equal{} q(2^k)$, q being a positive integer.

Therefore, we have:
$ 2^n \equal{} n2^k \plus{} q2^k$.
$ 2^n \equal{} (n \plus{} q)2^k$.
$ 2^{n \minus{} k} \equal{} n \plus{} q$.

I don't know where to go from here; sorry!

The question $ \Leftrightarrow 2^n \le 2^k n<2^n\plus{}n$
$ \Leftrightarrow 2^{n\minus{}k} \le n <2^{n\minus{}k}\plus{}\frac{n}{2^k}$

Now assume that $ n \neq 2^m$, then $ \frac{n}{2^k} \ge 1$

$ \Rightarrow 2^n \le 2^k n \le n^2$, which is impossible for $ n>4$, and if $ n\equal{}4$ then $ n\equal{}2^2$, contradiction, done.


Does anyone else here do Math League?

If you have, post your scores.

Mine (Middle School):
 6th grade: 40
 7th grade (forced by MC coach to take it w/ 8th grade): 39
 8th grade: 40.

If you ever look it up to see whether I'm lying, know that Qian and Cathy are both me. I just didn't wanna be called "Queen Dog"...

What the hell are the Math Leagues? The only middle school contest I know of is MathCounts. Then again, I might be out of touch w/ current developments, being in 10th grade and all.

I did a Maryland Math League in 6th grade and I did terrible (about 37?) because I wasn't good at math back then  :oops: .

6th grade i won in indiana, idk either 39 or  40
7th grade i failed and was like 3rd, i think 37
8th grade i FAILED and came 9th with a 38.

Uh. I don't remember. It was really easy....I think I got first in sprint and target.
It was disguised as a "MATHCOUNTS" competition--generic version.
Don't ask. :lol:

Anyways, it wasn't official, in a way. It was a school hosted math competition. I don't think it really counted for anything besides that.

....I see that it's not very popular.

That's not what I'm talking about though...
It's where you do 40 fairly easy questions in 30 min. The only way MC people can miss a question is by stupid mistakes. So, if one of the MC 1st team members miss a question or 2, they're like... just kill me now.
Every state/region has its own math league contest, but all the problems are the same. People just aren't being compared to others not in their own region/state.
For example, MA does NEML. They had a billion people get 40.
GA does GML. This year there's one person with a 40, and that person averaged 38 in 8th grade practices AND failed at state mathcounts.
I think in math league GA isn't the worst, but... it's not the best either.
EDIT: Yeah, I shouldn't feel too bad. There are states with no perfects.

I got 40 last year in 6th grade, and my school doesn't offer now.

In 5th and 6th grade i had it and got perfect scores.

My school area has something similar to what you guys do, except it's much,much,much easier. They basically give you a question, you get 6 minutes to answer it. The problem is probably around the difficulty of a mid-level Mathcounts states question. They give us 6 questions in this manner. So each school has a team of 6, and we each do these 6 questions individually. Then we add up our scores, and 36 is the highest possible.

   Amazingly, my school only gets around 25 averages. (I haven't gotten a question wrong since I started this in 6th grade.)

   I know how undeveloped math competition is at my school. One thing is that this was the first year my school had an mc team, and the school didn't host it. The students had to take the initiative. In addition, none of the kids are allowed to advance past geometry in 8th grade, It doesn't matter if we know all of the topic. (or Algebra in 7th, or Pre-Algebra in 6th)


Ok, the challenge is to solve this question with geometry. it is a somewhat easy problem with a few precalc formulas, but where is the fun in that?
Evaluate: $ \displaystyle\sum^\infty_{n\equal{}1}\dfrac{n}{(n\plus{}1)!}$

hello, prove by induction that
$ \sum_{n\equal{}1}^{k}\frac{n}{(n\plus{}1)!}\equal{}1\minus{}\frac{1}{(k\plus{}1)!}$ and then you can calculate your searched sum.
Sonnhard.

That is how my friend did it when I posed the problem to him.
I solved it by breaking it into $ \frac {n \plus{} 1}{(n \plus{} 1)!} \minus{} \frac {1}{(n \plus{} 1)!}$ and then $ \frac {1}{n!} \minus{} \frac {1}{(n \plus{} 1)!}$ and it then telescopes.
However I found a geometry way to solve it and want to see if anyone gets it.

[edit: I am bad at latex]


28) Cobalt oxide used in glazing pottery contains 71.06% by mass cobalt and 28.94% oxygen. What is the empirical formula of cobalt oxide?

1)CoO

2)Co3O4

3)Co2O

4)CoO3

5)Co2O3


A box of mass $ m_1$ sits on a frictionless plane, and a box of mass $ m_2$ sits on top of $ m_1$, with coefficient of friction $ \mu$ between the two of them. With how much force can you push $ m_1$ without $ m_2$ sliding on top of it? I found this to be pretty straightforward; if you push with any more force than $ F \equal{} \mu g m_2$, then the box on top will slip.

But: if you push on $ m_2$ instead, how much force can you use without $ m_2$ sliding on top of $ m_1$? My immediate answer was that it would be the same amount of force, but the next problem, which is already answered for you, suggests that it's different. The forces acting on $ m_2$ are its own weight $ W_2$, the normal force $ N$ that $ m_1$ exerts on $ m_2$, the applied force $ F$, and the friction force $ f \equal{} \mu \cdot N \equal{} \mu \cdot m_2 g$. The friction force is in the opposite direction to the applied force. The forces acting on $ m_1$ are its weight $ W_1$, the normal force of the surface on $ m_1$ which we don't really care about, and the reaction to the friction force $ f$ which will point in the same direction as the force applied to $ m_2$ with magnitude $ \mu g m_2$. So, the friction force acting on $ m_1$ will give a total force acting on it in the $ x$ direction of $ \mu g m_2$. This means that the frame of reference for $ m_2$ will be non-inertial, because it will be in the accelerating frame of sitting on top of $ m_1$. So all told the force acting on $ m_2$ will be $ F$. In order for $ m_2$ to not slip on top of $ m_1$, $ F \le \mu g m_2$. Again, I get the same answer as I did before after more careful though, which still contradicts the book. What am I doing wrong?

[quote="JRav"]Just stick with the inertial frame of reference.  It makes everything a lot easier.  Then just draw your free-body diagram and you should be able to do it.[/quote]

Even when I do this, I inexorably wind up concluding the same thing, that the total force you can apply to the top box to make them move without slippage is the same as if you were pushing on the bottom box.

You have to take into account that there is a coefficient of friction between $ m_{1}$ and $ m_{2}$ and we have no reason to believe it is the same between $ m_{1}$ and the table.


algorithm
number theory
greatest common divisor
Euclidean algorithm

For how many integers n between 1 and 100, do the numbers $n^2 +1$ and $n+1$ have a common factor greater 1.

[hide]

By the Euclidean Algorithm:

So any odd value of  works, and the answer is .

[/hide]

I believe that is right, except that I typed the problem wrong. It was supposed to be $n^2 + 2$ . Try it now.
So your use of the euclidean algorithm implies that adding any multiple of $n+1$ does not change the gcd of the two numbers. How could you rigorously prove this?


Let $ a,b,c$ be positive real numbers, such that $ a\cdot b\cdot c\equal{}1$.
Prove that the following inequality holds
$ \frac{1}{3a^2b^2\plus{}1}\plus{}\frac{1}{3b^2c^2\plus{}1}\plus{}\frac{1}{3c^2a^2\plus{}1}\plus{}\frac{3}{4}(a\plus{}b\plus{}c)\geq3$

[quote="marin.bancos"]Let $ a,b,c$ be positive real numbers, such that $ a\cdot b\cdot c \equal{} 1$.
Prove that the following inequality holds
$ \frac {1}{3a^2b^2 \plus{} 1} \plus{} \frac {1}{3b^2c^2 \plus{} 1} \plus{} \frac {1}{3c^2a^2 \plus{} 1} \plus{} \frac {3}{4}(a \plus{} b \plus{} c)\geq3$[/quote]

Nice your inequalities marin bancos  :)

[i]Solution:[/i]

$ \sum_{cyc}\frac {1}{3a^2b^2 \plus{} 1} \equal{} \sum_{cyc}\left(1 \minus{} \frac {3a^2b^2}{1 \plus{} 3a^2b^2}\right) \equal{} \sum_{cyc}\left(1 \minus{} \frac {3a^2b^2}{a^2b^2(3 \plus{} c^2)}\right) \equal{} \sum_{cyc}\left(1 \minus{} \frac {3}{3 \plus{} c^2}\right)$

Hence it suffices to show that :

$ a \plus{} b \plus{} c \ge \frac {4}{3 \plus{} a^2} \plus{} \frac {4}{3 \plus{} b^2} \plus{} \frac {4}{3 \plus{} c^2}$

Or,

Acording to Am_Gm inequality :

$ a \plus{} b \plus{} c \ge \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \equal{} \sum_{cyc}\frac {4}{4\sqrt [4]{a^2}} \ge \sum \frac {4}{3 \plus{} a^2}$

And the result follows  :P

This is the solution to the problem! Congratulations, [b]Abdek[/b]! [b]You are goo...od![/b]

[quote="Abdek"][quote="marin.bancos"]Let $ a,b,c$ be positive real numbers, such that $ a\cdot b\cdot c \equal{} 1$.
Prove that the following inequality holds
$ \frac {1}{3a^2b^2 \plus{} 1} \plus{} \frac {1}{3b^2c^2 \plus{} 1} \plus{} \frac {1}{3c^2a^2 \plus{} 1} \plus{} \frac {3}{4}(a \plus{} b \plus{} c)\geq3$[/quote]

Nice your inequalities marin bancos  :)

[i]Solution:[/i]

$ \sum_{cyc}\frac {1}{3a^2b^2 \plus{} 1} \equal{} \sum_{cyc}\left(1 \minus{} \frac {3a^2b^2}{1 \plus{} 3a^2b^2}\right) \equal{} \sum_{cyc}\left(1 \minus{} \frac {3a^2b^2}{a^2b^2(3 \plus{} c^2)}\right) \equal{} \sum_{cyc}\left(1 \minus{} \frac {3}{3 \plus{} c^2}\right)$

Hence it suffices to show that :

$ a \plus{} b \plus{} c \ge \frac {4}{3 \plus{} a^2} \plus{} \frac {4}{3 \plus{} b^2} \plus{} \frac {4}{3 \plus{} c^2}$
[/quote]
my solution is identical to Abdek's one to this moment.
now, 2 cases:
1.$ \sum_{}^{} ab \ge  \sum_{}^{} a$
it's enough to prove:
$ 3 \ge  \sum_{}^{}  \frac{4}{2\plus{}2a}$
which is trivial
2. $ \sum_{}^{} ab <  \sum_{}^{} a$
inequality
$ \frac{1}{2}  \sum_{}^{} a \ge  \sum_{}^{}  \frac{1}{1\plus{}a}$
is equivalent to:
$ 2 \sum_{}^{} a\plus{}( \sum_{}^{} a)^2 \plus{} \sum_{}^{} a \sum_{}^{} ab \ge 2 \sum_{}^{} ab\plus{}4 \sum_{}^{} a\plus{}6$
which is true because
$ 2 \sum_{}^{} a \ge 6$
$ 2 \sum_{}^{} ab\plus{} 4\sum_{}^{}a<6 \sum_{}^{} a \le  ( \sum_{}^{} a)^2 \plus{} \sum_{}^{} a \sum_{}^{} ab$
last ineq is equivalent to
$ \sum_{}^{} (a\plus{} \frac{1}{a})  \ge 6$
done

[quote="marin.bancos"]Let $ a,b,c$ be positive real numbers, such that $ a\cdot b\cdot c = 1$.
Prove that the following inequality holds
$ \frac {1}{3a^2b^2 + 1} + \frac {1}{3b^2c^2 + 1} + \frac {1}{3c^2a^2 + 1} + \frac {3}{4}(a + b + c)\geq3$[/quote]
Let $ a = \sqrt {\frac {x}{y}}, b = \sqrt {\frac {y}{z}}, c = \sqrt {\frac {z}{x}}$. Then by cauchy-schwartz:
$ \sum_{cyc} \frac {1}{3a^2b^2 + 1} = \sum_{cyc} \frac {z}{3x + z}  ) \ge \frac { ( \sum_{cyc} z  )^2}{ ( \sum_{cyc} 3xz + z^2  ) } \ge \frac {3}{4}$.
And by AM-GM we have: $ \frac {3}{4}(a + b + c) \ge \frac {9}{4}$. Adding up we get the desired :)

Thank you, [b]Dumel[/b]! Thank you, [b]Mathias_DK[/b]! Interesting solutions!
Dear [b]Mathias_DK[/b]! Very nice your way of "breaking" this inequality in two independent inequalities. I wouldn't think it works. Congrats!
My solution is, in fact, the same as [b]Abdek[/b]'s solution.


if just someone please can help me??? it's really important and i can't colve these things!
thank you very much
maria

[img]http://img356.imageshack.us/img356/9079/111cm7.jpg[/img]

[img]http://img529.imageshack.us/img529/4198/77680984si0.jpg[/img]

use the value of det A to compute:
[img]http://img204.imageshack.us/img204/3139/73697062xs1.jpg[/img]

A $ 4\times 5$ matrix doesn't have a determinant.

(It appears you've cropped that image badly)

Let $ A, B \in M_{n \times n}$ and $ \alpha$ be a scalar. Then:

1) $ det(\alpha \, A) \equal{} \alpha^n \, det(A)$

2) $ det(A^T) \equal{} det(A)$

3) $ det(AB) \equal{} det(A) det(B)$

4) It follows from 3) that $ det(A^n) \equal{} [det(A)]^n$


[quote="M@re"]In triangle $ABC$ bisector of $\angle BAC$-$AA'$ is constructed ($A'\in BC$). Let $X$ be the point on $AA'$. Poitns $B'$ and $C'$ are s.t. $BX \cap AC=B'$ and $CX\cap AB=C'$. Let $A'B'\cap CX=P$ and $A'C'\cap BX=Q$.
Prove that $\angle PAC=\angle QAB$.[/quote]

According to http://www.mathlinks.ro/Forum/viewtopic.php?t=21297 , since < XAB = < XAC (as the point X lies on the angle bisector of the angle BAC), we have < XAP = < XAQ. Thus, < PAC = < XAC - < XAP = < XAB - < XAQ = < QAB.

  Darij


factorial
limit
calculus
calculus computations

Find $ \lim_{n\rightarrow \infty}{\sqrt[n]{\dfrac{1!\plus{}2!\plus{}\ldots\plus{}n!}{n^{n\plus{}1}}}}$.

Let $ a_n \equal{} \sqrt [n]{\dfrac{1! \plus{} 2! \plus{} \ldots \plus{} n!}{n^{n \plus{} 1}}}$

$ n! < 1! \plus{} 2! \plus{} ... \plus{} n! < n n!$

So $ \frac 1e(2\pi n)^{\frac {1}{2n}}n^{ \minus{} \frac 1n}$ $ (\frac {n!}{\sqrt {2\pi n}(\frac ne)^n})^{\frac 1n}$ $ < a_n$ $ < \frac 1e(2\pi n)^{\frac {1}{2n}}(\frac {n!}{\sqrt {2\pi n}(\frac ne)^n})^{\frac 1n}$

So the required limit is $ \frac 1e$

Thank you very much for your answer and interest, pco.
Stealing your first line of the proof, we can come out with this proof too.
Make use of the following lemma:
If $ \lim_{n\to\infty}x_{n}^{\frac {1}{n}} \equal{} l$, then $ \lim_{n\to\infty}\frac {x_{n \plus{} 1}}{x_{n}} \equal{} l$.

Take  $ a_n \equal{} \sqrt [n]{\dfrac{1! \plus{} 2! \plus{} \ldots \plus{} n!}{n^{n \plus{} 1}}}$

Then $ \sqrt [n]{\dfrac{n!}{n^{n \plus{} 1}}} \equal{} b_n\le a_n\le c_n \equal{} \sqrt [n]{\dfrac{n!}{n^n}}$

By our lemma
 $ \lim_{n\to \infty} {b_n} \equal{} \lim_{n\to \infty}{\dfrac{(n \plus{} 1)!}{(n \plus{} 1)^{n \plus{} 2}}\cdot \dfrac{n^{n \plus{} 1}}{n!}} \equal{} \lim_{n\to \infty}{\left(\dfrac{n}{n \plus{} 1}\right)^{n \plus{} 1}} \equal{} \lim_{n\to \infty}{\left(\dfrac{n}{n \plus{} 1}\right)^{n}}\cdot \lim_{n\to \infty}{\dfrac{n}{n \plus{} 1}} \equal{} \dfrac{1}{e}$

and also similarly $ \lim_{n\to \infty}{c_n} \equal{} \lim_{n\to \infty}{\left(\dfrac{n \plus{} 1}{n}\right)^n} \equal{} \dfrac{1}{e}$
So the required answer is $ \boxed{\lim_{n\to \infty}{a_n} \equal{} \frac {1}{e}}$.

I don't think the above solutions make use of the hint requirement.

Basically it is known that ((n!)^(1/n))/n tends to 1/e (stolz-cesaro). Applying the hint gives you the result.


Show that  $ \frac{a}{b\plus{}2c\plus{}3d}\plus{}\frac{b}{c\plus{}2d\plus{}3a}\plus{}\frac{c}{d\plus{}2a\plus{}3b}\plus{}\frac{d}{a\plus{}2b\plus{}3c}\ge\frac{2}{3}$   for any positive reals

[quote="super_math"]Show that  $ \frac {a}{b \plus{} 2c \plus{} 3d} \plus{} \frac {b}{c \plus{} 2d \plus{} 3a} \plus{} \frac {c}{d \plus{} 2a \plus{} 3b} \plus{} \frac {d}{a \plus{} 2b \plus{} 3c}\ge\frac {2}{3}$   for any positive reals[/quote]
We have:
$ \frac {a}{b \plus{} 2c \plus{} 3d} \plus{} \frac {b}{c \plus{} 2d \plus{} 3a} \plus{} \frac {c}{d \plus{} 2a \plus{} 3b} \plus{} \frac {d}{a \plus{} 2b \plus{} 3c}\ge \frac {(a \plus{} b \plus{} c \plus{} d)^2}{4(ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd)}$
Then $ (a\plus{}b\plus{}c\plus{}d)^2\equal{}2(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd) \plus{}a^2\plus{}b^2\plus{}c^2\plus{}d^2 \geq \frac{8}{3}(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd)$
Done!

You can see at:
http://www.mathlinks.ro/Forum/viewtopic.php?p=354552&sid=ec2ea982593b5d02f0a55978f601c47a#354552


Solve the equation:
$ (\frac{2}{3})^{x}\plus{}\frac{4}{3}\equal{}2^{x}$

*$ x>1\Rightarrow(\frac{2}{3})^{x}<\frac{2}{3}$ (because$ 0<\frac{2}{3}<1$)
$ \Rightarrow LHS<\frac{2}{3}\plus{}\frac{4}{3}\equal{}2$
But $ RHS>2$ (because$ x>1$)$ \Rightarrow$ reasonless.

*$ x<1\Rightarrow(\frac{2}{3})^{x}>\frac{2}{3}$ (because$ 0<\frac{2}{3}<1$)
$ \Rightarrow LHS>\frac{2}{3}\plus{}\frac{4}{3}\equal{}2$
But $ RHS<2$ (because$ x<1$)$ \Rightarrow$ reasonless.

*So $ x\equal{}1$.DONE :)

[quote="ADN"]*$ x > 1\Rightarrow(\frac{2}{3})^{x}<\frac{2}{3}$ (because$ 0 <\frac{2}{3}< 1$)
$ \Rightarrow LHS <\frac{2}{3}\plus{}\frac{4}{3}\equal{} 2$
But $ RHS > 2$ (because$ x > 1$)$ \Rightarrow$ reasonless.

*$ x < 1\Rightarrow(\frac{2}{3})^{x}>\frac{2}{3}$ (because$ 0 <\frac{2}{3}< 1$)
$ \Rightarrow LHS >\frac{2}{3}\plus{}\frac{4}{3}\equal{} 2$
But $ RHS < 2$ (because$ x < 1$)$ \Rightarrow$ reasonless.

*So $ x \equal{} 1$.DONE :)[/quote]
You are correct  :)


Suppose that the letters O and T are drawn with infinitesimally thin lines. Show that uncountable infinitely many copies (in diverse sizes) of the letter O can be placed in the plane without overlapping, but only countably infinitely many copies of T can be placed.

O's are obvious; consider the set of circles centered at the origin.

Let the point where the two lines of a T meet be called its center and let its "radius" denote the minimal distance from its center to one of its three extreme points.  Two T's of radius at least $ r$ have centers at least $ r$ apart.  Therefore we can replace a T of radius $ r$ with a [b]solid[/b] circle of radius $ \frac{r}{2}$.  (An example of a countable tiling here, then, is the set of Ford circles.)

At this point I am not aware of the technical machinery required to prove that the set of centers must be countable.  I can only give an intuition:  the sum of the areas of the circles contained in any region $ R$ of finite area is finite, so the sum of the squares of the radii of all the circles in $ R$ converges, which appears to show that the (multi)set of radii is countable.  The plane is clearly a countable union of such regions.

[quote="t0rajir0u"]Two T's of radius at least $ r$ have centers at least $ r$ apart.[/quote]
Uhm... are you really sure?  :huh:

Correction:  let the center denote the point one-fourth of the way from the intersection of the two lines to the bottom of the T.  Let the distance between that intersection and the center be $ a$ and let the distance between the intersection and one side of the T be $ b$.  Define the radius of a T as $ \min(2a, c)$ where $ c$ is the altitude to the hypotenuse of a right triangle with legs $ 4a, b$.


Problem 1
In an obtuse triangle $AB\Gamma$ $(\angle A\Gamma B=120{\ensuremath{^\circ})}$, $A\Gamma=6$ and $B\Gamma=2$. The interior bisector of the angle $\angle A\Gamma B$ intersects the side $AB$ at the point $\Delta$. Calculate the length of the line segment $\Gamma\Delta$.

Problem 2
The sides of an acute triangle are succesive integers $\nu$, $\nu+1$, $\nu+2$ $(\nu>3)$. If the altitude to the side whose length is $\nu+1$ divide the side into two line segments prove that their difference is 4.

Problem 3
Find all the values of the natural number $\nu$ such that

$2^{8}+2^{11}+2^\nu$

is a perfect square (ie square of a natural number).

Problem 4
Let $\alpha,\beta,\gamma >0$, $\alpha<\beta\gamma$ and $1+a^{3}=\beta^{3}\gamma^{3}$. Prove that $1+\alpha<\beta+\gamma$.

In the first the side opposite to the obtuse angle should be the greatest.

[hide="2"]Let the two line segments be $x$ and $y$. Then $x+y = \nu+1$.
Also $(\nu+2)^{2}= x^{2}+h^{2}$ and $\nu^{2}= y^{2}+h^{2}$. Hence $4(\nu+1) = (x-y)(\nu+1)$. [/hide]

[hide="3"]$2^{8}+2^{11}+2^{\nu}= (2^{4})^{2}+2.2^{4}.2^{6}+2^{\nu}$. Hence $\nu = 12$.[/hide]

[quote="Andreas"]In the first the side opposite to the obtuse angle should be the greatest.[/quote]

I am sorry. There was a typo in latex. I 've correct it.

[quote="Andreas"][hide="3"]$2^{8}+2^{11}+2^{\nu}= (2^{4})^{2}+2.2^{4}.2^{6}+2^{\nu}$. Hence $\nu = 12$.[/hide][/quote]

What are the other solutions or how can you prove that this is the only one solution?


LaTeX
number theory unsolved
number theory

Find all non-negative integers $ x,y,$ such that $ 3^x\plus{}5^y\equal{}7^z\plus{}1$

Represent $ 7$ as $ 5\plus{}2$ and $ 3$ as $ 5\minus{}2$ then rearrange the equality as $ 5^y \equal{} (5 \plus{} 2)^z \minus{} (5 \minus{} 2)^x \plus{} 1$.
By lookig at this $ (mod 4)$ we see that $ x$ and $ y$ have to be of the same parity. The look at it $ (mod 5)$ and you are done. Sorry for the shortness, but i think it's pretty straightforward.

:rotfl:  Do you say that there aren't any solutions, cause... (x,y,z)=(1,1,1). If you look mod 5 only thing you get is that x and z are in form $ 4k\plus{}1$.  :wink:

[quote="10maths_tp0609"]Find all non-negative integers $ x,y,$ such that $ 3^x \plus{} 5^y \equal{} 7^z \plus{} 1$[/quote]

we take the equation mod. 3 and have $ y\equal{}2k\plus{}1$
taking mod. 4 we obtain: $ x$ and $ z$ are both even or odd.

1) when $ x\equal{}2m$, $ z\equal{}2n$, then $ 9^m\plus{}5\cdot25^k\equal{}49^n\plus{}1$, but it follow $ 9^m \equiv \minus{}3 (mod. 24)$, then $ 3 \equiv \minus{}1 (mod. 8)$, a contradiction.
2) when $ x\equal{}2m\plus{}1$, $ z\equal{}2n\plus{}1$, then $ 3\cdot9^m\plus{}5\cdot25^k\equal{}7\cdot49^n\plus{}1$
and i dont know more..., somewhere must be a contradiction for $ m\geq1, n\geq1, k\geq1$

Spongebob i  forgot to mention the trivial solution. But the thing is if I was was good at working with LaTeX i'd write the solution out - and i'm not. My point is that only you do the exapansion of $ (5\minus{}2)^x$ and $ (5\plus{}2)^z$, because x and z are of the same parity we get certain terms divisible by 5 and in one case the RHS=5k+1 and in other it follows that $ 2^z\minus{}2^x\plus{}1$ is divisible by 5 and that proves wrong - or at least I think that's what I had written.

Yes you did, and I tell you $ 2^5 \minus{} 2^3 \plus{} 1\equal{}25$ is divisible with $ 5$ and $ 3$ and $ 5$ are of the same parity. If $ 5|2^z \minus{} 2^x \plus{} 1$ and $ x\equiv_2 z$ the solutions are $ (x,z)\equal{}(4k\plus{}1, 4l\plus{}3)$ because $ 5|2^{4k\plus{}1} \minus{} 2^{4l\plus{}3} \plus{} 1$.
Bye

it's an ok idea but i don't think it leads to a solution just that easy

Can somebody give a full solution, because i spent many time on this problem, and i am just not able to solve it.

[quote="10maths_tp0609"]Find all non-negative integers $ x,y,$ such that $ 3^x \plus{} 5^y \equal{} 7^z \plus{} 1$[/quote]
We see that $ ( \minus{} 1)^y \equiv \minus{} 1 \bmod 3 \iff 2 \nmid y$.
We see that $ ( \minus{} 1)^x \plus{} 1^y \equiv ( \minus{} 1)^z \plus{} 1 \bmod 4$, So $ x$ and $ z$ has same parity.
Also notice $ 3^x \equiv 2^z \plus{} 1 \bmod 5$ then $ x \equiv z \equiv 1 \bmod 4$
Yet again $ 5^y \equiv 5 \bmod 16$ so $ x \equiv y \equiv z \equiv 1 \bmod 4$

My plan at this point is to prove that $ 2^n \mid x \minus{} 1, y \minus{} 1, z \minus{} 1 \forall n \in \mathbb{N}$ And thus implying $ x \equal{} y \equal{} z \equal{} 1$ is the only solution.

We know that $ 2^n \mid x \minus{} 1, y \minus{} 1, z \minus{} 1$. We check $ \bmod 2^{n\plus{}3}$
We have $ (4\minus{}1)^x \plus{} (4\plus{}1)^y \equiv (8\minus{}1)^z \plus{} 1 \bmod 2^{n\plus{}3}$
It's easy to see that $ (8\minus{}1)^{2^n} \equiv 1 \bmod 2^{n\plus{}3}$
On the other hand $ (4 \pm 1)^{2^n} \equiv 1 \plus{} 2^{n\plus{}2} \bmod 2^{n\plus{}3}$
Thus we have $ 2^{n\plus{}1} \mid x\minus{}1, y\minus{}1$, and we "just" have to prove $ 2^{n\plus{}1} \mid z\minus{}1$.

I dont get your proof.  how did you come up with the last 2 lines etc.

Oh my guys,The result is $ (0,0,0,(1,1,1)$
Our work is only checking $ mod 3$,and $ mod 8$, I think all of you can complete,it is not too hard

Could please somebody give a full solution, because i still cant solve this problem.


First, a short explanation about this article. It is intended for participants of the [b]Getting Started[/b] forum. A short introduction of number systems is presented, followed by some simple exercise questions intended to pique interest and reinforce techniques. The article finishes with some challenging questions.

[b]Number Systems[/b]

[b]Introduction - Decimal System[/b]

In everyday life, we use the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. There are 10 digits in total, and we call this the "base-10" or "decimal" system.

The number 15, in base 10, has 2 digits; however, the numerical values expressed by the digits 1 and 5 are different. Since 1 is located to the left of 5, its value is greater than 5; in particular, the 1 actually represents 1 * 101. That is,

15 = 1 * 101 + 5 * 100.

(This is called the base-10 expansion of 15.)

In the same vein,

234 = 2 * 102 + 3 * 101 + 4 * 100.

Here, "2" is called the most significant digit, for its value is the greatest; 4 is called the least significant digit, for its value is the smallest.

Notice that as the digits progress from least significant to most significant, the exponent to which 10 is raised increases. Notice that we are expressing this number in base-[b]10[/b], and we multiply each digit by a power of [b]10[/b] to arrive at the final value of the number.

[b]Other Number Systems[/b]

There are other number systems in which different amounts of digits are used. For example,

the base-2 (binary) system uses only 2 digits, 0 and 1;
the base-8 (octal) system uses 8 digits, 0 through 7;
the base-16 (hexadecimal) system uses 16 digits, 0 through 9 and A through F.

In each of these bases, the numerical digits have the same values as their decimal counterparts. In base-16, A represents the number 10, B represents the number 11, and so on; F represents the number 15.

[b]Problem 1.[/b] How is the number 1610 represented in base-16?

These three number systems are commonly used in computers; other number systems exist as well. For example, the analog clock is a combination of a base-12 system (hour) and a base-60 system (minute and second).

We represent a number n of base r (r stands for radix, another name for base) as (n)r.

[b]Problem 2.[/b] Write the binary number, (1001 0010)2, as its base-2 expansion.

[b]Problem 3.[/b] Solve for x: (x)10 = (1001 0010)2.

[b]Problem 4.[/b] Solve for x: (x)10 = (F6A)16.

[b]Problem 5.[/b] Solve for x: (x)10 = (125)8 + (16)16.

[b]Problem 6.[/b] Solve for x: (x)10 = (25)8 * (A3)16. In how many ways can you solve this?

[b]Conversion Between Bases[/b]

In problems 3 through 6, you are asked to convert numbers of different bases to base-10. How, then, do you convert base-10 numbers to other bases?

[b]Example.[/b] Convert (125)10 to base-2.

We divide 125 by 2, then divide the quotient by 2, and so on. We note the remainders:

125 / 2 = 62, remainder 1
62 / 2 = 31, remainder 0
31 / 2 = 15, remainder 1
15 / 2 = 7, remainder 1
7 / 2 = 3, remainder 1
3 / 2 = 1, remainder 1
1 / 2 = 0, remainder 1

The division ends when the quotient reaches 0. Then, we read off the quotients, starting at the bottom and working our way up. We arrive at

(0111 1101)2 = (125)10.

(Note that binary numbers are often written in groups of 4 digits.)

[b]Problem 7.[/b] Solve for x: (x)16 = (1155)10.

One may often wish to convert between base-2 and base-8 or base-16. This can be done in a simple manner.

[b]Example.[/b] Convert (A63)16 to base-2.

We utilise the following chart:

[code] base-10 | base-2 | base-16 || base-10 | base-2 | base-16
---------+--------+---------++---------+--------+---------
  0      | 0000   | 0       ||  8      | 1000   | 8
  1      | 0001   | 1       ||  9      | 1001   | 9
  2      | 0010   | 2       || 10      | 1010   | A
  3      | 0011   | 3       || 11      | 1011   | B
  4      | 0100   | 4       || 12      | 1100   | C
  5      | 0101   | 5       || 13      | 1101   | D
  6      | 0110   | 6       || 14      | 1110   | E
  7      | 0111   | 7       || 15      | 1111   | F[/code]

(Can you see how the chart is constructed? Can you construct a similar chart for base-8?)

We simply look up A, 6, and 3 in the chart. A corresponds to (1010). 6 corresponds to (0110). 3 corresponds to (0011). Therefore, (A63)16 = (1010 0110 0011)2.

Note that conversion between base-10 and base-2 can [b]NOT[/b] done in this manner.

[b]Problem 8.[/b] Express (1101 0110 1000)2 in base-16.

[b]Problem 9.[/b] Express (235)8 in base-16.

[b]Arithmetic in Other Bases[/b]

Problem 6 can be solved by converting the right side to base-10; however that is usually cumbersome.

[b]Example.[/b] Calculate (2A)16 * (C9)16 and express the answer in base-16.

We begin by converting the individual digits to base-10 (with the help of the chart). See the calculations below.

[code]        ( 2) (10)
*       (12) ( 9)
-----------------
        (18) (90)
  (24) (120)
-----------------
  (24) (138) (90)
[/code]

Now, 90 / 16 = 5 with remainder 10; 138 + 5 = 143, 143 / 16 = 8 with remainder 15; 24 + 8 = 32, 32 / 16 = 2 with remainder 0; 2 + 0 = 2, 2 / 16 = 0 with remainder 2. Therefore, the final answer is (2) (0) (15) (10). We convert this back into base-16 and obtain (20FA)16.

[b]Problem 10[/b]. Redo problem 6, but express the answer in base-8.

[b]Challenges[/b]

[b]Challenge 1.[/b] For a number (a1a2a3...an)r with each a representing a digit (0 :le: a < r), find its base-r expansion.

[b]Challenge 2.[/b] Calculate (1001 0101.11)2 + (2A.9)16, and express the answer in base-10.

I'll post the answers at a later date.

Hope you've enjoyed this article! Please don't hesitate to contact me with comments and/or suggestions. There'll be more of these in the future, so stay tuned! :)

Wow! That's great what you did Osiris.  I was going to post something like that but never got around to it.  You explained the topic well.   If we have more posts as informative as this one then I believe that the users of this forum will learn more(unless they already know the material).

That's a great expository article for this level of mathematics.

Osiris wrote:Challenge 1. For a number (a1a2a3...an)r with each a representing a digit (0 :le: a :le: r), find its base-r expansion.

Challenge 2. Calculate (1001 0101.11)2 + (2A.9)16, and express the answer in base-10

For the first one shouldn't it be  0 :le: a<r

but anyways heres my answer

[hide][tex]a_1*r^{n-1}+a_2*r^{n-2}+...+a_n[/tex][/hide]

The second one:  [hide]Convert both to base 10.  10010101.11_2=128+16+4+1+1/2+1/4=149 3/4

2A.9_16=32+10+9/16=42 9/16

Adding them gets 192 5/16[/hide]

[edited]

Yes, thank you, corrected.

(you messed up the calculation a TINY bit for challenge 2 - but the procedure is correct.)

O whoops a slight adding error. :oops: I fixed it.


Κοιτάξτε την άσκση 3 σελίδας 2. Έχουμε ότι

$ (x \plus{} y)^{4} \equal{} x^{4} \plus{} y^{4}\Leftrightarrow x^{4} \plus{} y^{4} \plus{} 6x^{2}y^{2} \plus{} 4x^{3}y \plus{} 4xy^{3} \equal{} x^{4} \plus{} y^{4}\Leftrightarrow3x^{2}y^{2} \plus{} 2x^{3}y \plus{} 2xy^{3} \equal{} 0\Leftrightarrow xy[(x \plus{} y)^{2} \plus{} (x \plus{} \frac {1}{2}y)^{2} \plus{} \frac {3}{4}y^{2}] \equal{} 0$

Άρα πρέπει $ (x \plus{} y)^{2} \plus{} (x \plus{} \frac {1}{2}y)^{2} \plus{} \frac {3}{4}y^{2} \equal{} 0$ το οποίο ισχύει αν και μόνο αν $ x \equal{} y \equal{} 0$ δηλαδή άτοπο αφού $ x,y\neq 0$

Έτσι δεν λύνεται, διότι μου είπαν πως είναι λάθος... :o

Φιλε Γιωργο προφανος  εκανε λαθος αυτος που το διαβασε...Η λυση σου ειναι σωστη. :wink:


For positive integers n, denote by D(n) the number pairs of different adjacent digits in the binary (base two) representation of n. For example, D(3)=D($11_{2}$)=0, D(21)=D($10101_{2}$)=4, and D(97)=D($1100001_{2}$)=2. For how many positive integers n less than or equal to 97 does D(n)=2?

You've got a 1 at the beginning and a 1 at the end. Combos in the middle are obviously triangle numbers, so just find number of combos for up to 97.

yifman, you're right about the triangles, but I have no idea how it was "obvious" to you as I reached this conclusion only after lots of work. Care to elaborate?

[hide]The number must start with one, and then switch from a string of ones to a string of zeros, then back to a string of ones, where "string" is one or more bits in a row. The smallest number with D(n)=2 is then n=101 with 3 bits. All the other numbers that qualify, then, must have between 3 and 7 bits.

For a number with n bits, the switch from 1 to 0 may occur at any bit between bit 2 and bit (n-1), then the switch back to 1 may occur anywhere between bit 3 and bit n.

If the first switch is at bit 2, the second may be at 3, 4, 5, ..., n -- n-2 possibilities
If the first switch is at bit 3, the second may be at 4, 5, 6 ..., n -- n-3 possibilities
...
If the first switch is at bit n-1, the second may be at n -- only 1 possibility

The total number of ways to make this work is then, (n-2)+(n-3)+...+2+1 = (n-2)(n-1)/2.

Plugging in n=3, 4, 5, 6, 7 gives 1, 3, 6, 10, 15, which are, as yifman said, the first few triangular numbers.

Adding these together, there are 35 solution numbers with 7 bits or less. However, several of them are >97, so we observe that the only ones lower than 97 are:

1000001
1000011
1000111
1001111
1011111

So the last term is 10 instead of 15, and the final answer is 30. Unless I screwed up somewhere.[/hide]

I brute forced this and got 26. I used only logic, because I'm not freakeskly intelligent. Evidently my logic failed. Could someone list the combinations of digits/permutations of digits please. Also, could someone explain triangle numbers and how they apply.

The list, in increasing order:
[hide="List"]101
1001
1011
1101
10001
10011
10111
11001
11011
11101
100001
100011
100111
101111
110001
110011
110111
111001
111011
111101
1000001
1000011
1000111
1001111
1011111
1100001=97[/hide]
There are 26. MatrixFrog counted the last term wrong.

The triangular numbers come in counting how many there are with $ k$ bits- there are 1 with 3 bits, 3 with 4 bits, 6 with 5 bits, and so on. In this order, it's fairly easy to see the representation of triangular numbers as sume of consecutive integers.

This was from one of my years- I know I answered it because I answered everything, and I have no idea whether it was one of the ones I got wrong.

[quote="MatrixFrog"]yifman, you're right about the triangles, but I have no idea how it was "obvious" to you as I reached this conclusion only after lots of work. Care to elaborate?[/quote]

I wouldn't go as far as calling it "obvious", but basically, you have some 1's, some 0's, and then some more 1's.

The first term must be a 1. Of the remaining n-1 positions, we pick two of them. There are $ \binom{n \minus{} 1}{2} \equal{} \frac {(n \minus{} 1)(n \minus{} 2)}{2}$ ways to do this.

Let the one to the left be the leftmost 0, and the one to the right be the leftmost of the second set of 1's. Together, they determine the number.


$ ABCD$ is a rhombus. $ C_A$ is the circle through $ B, C, D$; $ C_B$ is the circle through $ A, C, D$; $ C_C$ is the circle through $ A, B, D$; and $ C_D$ is the circle through $ A, B, C$. Show that the angle between the tangents to $ C_A$ and $ C_C$ at $ B$ equals the angle between the tangents to $ C_B$ and $ C_D$ at $ A$.

could someone please explain?

Just a note it is from 1903, but that is besides the point.

[hide]HINT:
Try to find some equalities of various angles, and use many properties of cyclic quadrilaterals.[/hide]

I'll post solution if necessary.

Greetings, and welcome to the internet. Here on the internet there are websites called "search engines." They are very useful for finding a wide variety of information. I suggest you choose your favorite search engine and type in the words "Eotvos competition" to see what it gives you. (On the first page of results, my favorite search engine gives several links that look very promising.)

I'll help him out.
[url=http://www.justfuckinggoogleit.com]http://www.google.com/search?q=Eotvos+competition[/url]


limit
logarithms
real analysis
real analysis unsolved

$ \lim_{t\to 0\plus{}}\frac{e^{\minus{}\log^2 t}}{t}$

Thanks in advance!

I assume that's the natural logarithm.

The obvious step is algebraic manipulation: $ e^{\minus{}\ln t\cdot\ln t}\equal{}t^{\minus{}\ln t}$, so we have $ \lim_{t\to 0^\plus{}}t^{\minus{}1\minus{}\ln t}$. That's not an indeterminate form.


[quote="socrates"]$(a^2-bc)^2+(b^2-ca)^2+(c^2-ab)^2=(a^2+b^2+c^2+ab+bc+ca)(a^2+b^2+c^2-ab-bc-ca)$

Is this the answer??

 :)[/quote]
You can continue to factorize it:
${(a^2+b^2+c^2)}^2-{(ab+bc+ac)}^2$
 :)


Let $ R$ be a ring and $ a_1,\dots, a_n\in R$. Define $ [a_1, a_2,\dots, a_n]\equal{}\sum_{\pi\in S_n}\text{sgn}(\pi)a_{\pi1}a_{\pi2}\dots a_{\pi n}$.

Suppose $ k$ is a natural number and that $ A_1,A_2,\dots, A_{2k}$ are $ k\times k$ matrices. Prove or disprove that $ [A_1, A_2,\dots, A_{2k}]\equal{}0$.

The bracket is an $ R$-multilinear alternating form (just like the wedge product for vectors). It suffices to consider the behavior on elements of a basis, such as matrices with a $ 1$ in one position and zeros elsewhere.

Let $ M_{i,j}$ have a $ 1$ in the $ (i,j)$ position and zeros otherwise. Let $ A_k\equal{}M_{i(k),j(k)}$. The ordered pairs $ (i(k),j(k))$ form a directed graph on $ k$ vertices, and each nonzero term in the sum corresponds to an Eulerian path (using all edges).


Find the least value of $\frac{(x^{2}-x+1)^{2}}{x^{6}-x^{3}+1}$, where $x\in R$

Let $f(x)=\frac{(x^{2}-x+1)^{2}}{x^{6}-x^{3}+1}$. We have $f(x)>0\forall x\in\mathbb{R}$ and $\lim_{n\to\infty}f(x)=0$ ,therefore NOT exist least value!

This problem is more interesting than the first.

$f(x)\equiv\frac{(x^{2}-x+1)^{3}}{x^{6}-x^{3}+1}=\{\begin{array}{ccc}1 & \mathrm{if}& x=0\\\\ h(x+\frac{1}{x}) & \mathrm{if}& x\ne 0\end{array}\mathrm{\ ,}$

where $h: (-\infty ,-2]\cup [2,\infty )\rightarrow \mathbb R\ ,\ h(t)=\frac{(t-1)^{3}}{t^{3}-3t-1}$ a.s.o.


logarithms
calculus
calculus computations

Just apply $\ln$ to both sides and you will get a geometric progression. It is to much to write.

really simpl: I think everyone can find this sum $\sum\ln^{2k}{\frac{2}{e}}$ and this $\sum\ln^{2k+1}{\frac{2}{e}}$


superior algebra
superior algebra theorems

hi everyone,

here is a well known fact but funny one:

Let Sn be the group of permutation of n elements.Prove
that all bijective morphism f of Sn are of the form:
  f(A)=SoAo(S^-1)

except for n=6 ...  :D

Here's a partial result: If we could show that $f$ takes transpositions to transpositions, then we would be done, and here's why:

Clearly, the automorphism is determined by the images of transpositions. Let, in general, $f((x,y))=(x',y')$. Then $f((a,b)(a,c)),\ f((a,c)(a,d)),\ f((a,d)(a,b))$ are all $3$-cycles, so the pairs [$f((a,b)),f((a,c))$], [$f((a,c)),f((a,d))$], [$f((a,d)),f((a,b))$] each have one component in common. However, since $f((a,b)(a,c)(a,d))$ is a $4$-cycle, it means that that common component is one and the same for all three pairs, so $f((a,b))=(a',b'),\ f((a,c))=(a',c'),\ f((a,d))=(a',d')$. This means that there is a bijection $g:\overline{1,n}\to\overline{1,n}$ (a permutation) s.t. $f((a,b))=(g(a),g(b))$. From here we get $f((a,b))=g^{-1}\circ (a,b)\circ g$ (there might be something wrong here, but this is the main idea), and the conclusion follows.

Now I guess we should prove that, except for $n=6$, an automorphism of $S_n$ turns transpositions to transpositions. In any case, the image of a transposition has order $2$, so it must be a product of disjoint transpositions (an odd number, because it should also be an odd permutation).

Moreover, the fact that $f$ takes transpositions to transpositions is also necessary (it's not just something I'd like to be true :)), because it's well-known that a permutation is the conjugate of another iff they have the same type of cycle decomposition (in the unique decomposition in disjoint cycles, they have the same number of cycles of length $1,2,\ldots$).

I must have been really tired last night if I couldn’t see the rest of the solution.

Since any two transpositions are in the same conjugacy class, it means that their images are also in the same conjugacy class, so if the image of a transposition is a product of $2k+1$ disjoint transpositions, then this is true for all transpositions. Vice-versa, the preimage of any product of $2k+1$ disjoint transpositions is a transposition. This means that for this particular $n$ and this particular $k$ the number of transpositions is the same as the number of products of $2k+1$ different transpositions.

If $n<6$ there are no products of $\ge 3$ transpositions in $S_n$, and if $n>6$ a counting argument will show that there are more products of $2k+1>1$ disjoint transpositions than there are transpositions.

so for n=6 .. what's the remaining automorphism ??

You know, I think there's no such an automorphism for $n=6$ either. I've tried to figure out what's wrong with thism but failed.

The elements of order $3$ which are products of $2$ transpositions are the $3$-cycles, while the elements of order $3$ which are products of $2$ products of $3$ disjoint transpositions are the products of $2$ disjoint $3$-cycles, and these are twice as many as the $3$-cycles. However, an automorphism as the one we're looking for, would yield a bijection between the $3$-cycles and the products of $2$ disjoint $3$-cycles.

What am I doing wrong?

Hi Alek

I see you are interested in math problem Oral Examination ENS
you will a solution of this one in RMS Revue Mathmatiques Spciales
Mai/Juin 1994 page 685.

Here's a construction of one, although I haven't gone through all the dirty work to prove that it works. We define the automorphism $F$ by its action on the transpositions:
$F(1,2)=(1,2)(3,4)(5,6)$
$F(1,3)=(1,3)(2,5)(4,6)$
$F(1,4)=(1,4)(2,6)(3,5)$
$F(1,5)=(1,5)(2,4)(3,6)$
$F(1,6)=(1,6)(2,3)(4,5)$
$F(2,3)=(1,6)(2,4)(3,5)$
$F(2,4)=(1,5)(2,3)(4,6)$
$F(2,5)=(1,3)(2,6)(4,5)$
$F(2,6)=(1,4)(2,5)(3,6)$
$F(3,4)=(1,2)(3,6)(4,5)$
$F(3,5)=(1,4)(2,3)(5,6)$
$F(3,6)=(1,5)(3,4)(2,6)$
$F(4,5)=(1,6)(3,4)(2,5)$
$F(4,6)=(1,3)(2,4)(5,6)$
$F(5,6)=(1,2)(3,5)(4,6)$
Every transposition is mapped to a product of three disjoint transpositions, and in fact $F$ is an involution. Two transpositions are disjoint if and only if their images share a pair.
An example of the relations that need to be tested:
$F(1,2)*F(1,3)*F(1,2)=F(2,3)$
$1 \to 2 \to 5 \to 6$
$2 \to 1 \to 3 \to 4$
$3 \to 4 \to 6 \to 5$
$4 \to 3 \to 1 \to 2$
$5 \to 6 \to 4 \to 3$
$6 \to 5 \to 2 \to 1$
$F(2,3)=(1,6)(2,4)(3,5)$

By the way, grobber- your counting was wrong; there are exactly as many three-cycles as products of two disjoint three-cycles. The choice of direction for the second cycle is balanced by the double-counting that comes from not knowing which is first.

Why resurrect a very old thread to post a completely new question? Something like that deserves its own thread, [b]if[/b] you actually know what you're trying to ask.


USAMTS is over...
When does round 2 score come out? I mean... for those who submitted solutions by mail.
Can anyone clarify the "pro-rated score"? is it (R1 + R2 + R3) / 3?

"What we will do is use your scores from Rounds 1, 3 and 4 to give you a pro-rated score for Round 2."

I thought it said not to ask about this.

It probably won't be the average, since this doesn't account for difficulty factors. Since Rd. 2 was the easiest, I expect pro-rated scores wil be relatively high. For example, 13/?/18/13 are my scores, so I expect my rd. 2 to be 17 or something.

If my round 2 had been "lost" I would have gotten 100 >_<

round 2 was the easiest..... if usamts did not lose my solutions, i could have received silver....now i am expecting bronze..... kinda upset....

I pretty much agree that Round 2 was the easiest (part. b/c it had more geometry, which is my best subject, and the counting/probability problem wasn't too evil), since I had the highest on that. :)

@barkashdanbar: aww...that'd really suck. :(  I recommend that you should have saved an extra copy of your USAMTS solutions (like I did, fortunately) so that they aren't lost if USAMTS has this issue. :)

Well, if they go by just strictly pro rated  based strictly on averages, my score will be lower than what I would have gotten had I estimated by score for round 2.  I estimated, right after I turned it in, that I would get a 22, if its average, I would get a 21...shouldnt make a difference in medals but still a lower score.

If Round 2 was the easiest round and they dont do averages, then I might get as low as 19 or something.

I think they said they will make announcement first week of April.  Dont worry, there are multiple people, they wont forget.

round 2 was definitely the hardest.i managed to get as low as a 13 on that round lol.

Argh I slacked off on round 4 a lot and only answered 3 problems (one of which I misinterpreted). Thus I only have a 44 for rounds 1, 3, 4. I was expecting ~18 for round 2 and I doubt I'll get bronze through a pro rated score...bleh I'll just have to wait.

[quote="Temperal"][quote="abacadaea"][quote="Temperal"]Round 2 was most definitely not the easiest - in my opinion it was by far the hardest, but judging by the statistics, difficulty was nonincreasing, hence round 2 and round 3 were about the same with round 1 harder and round 4 easier.[/quote]

Those statistics were as they are because some people did not submit.[/quote]

All right, then, I can safely say that round 2 was by far the hardest.  :wink:[/quote]

Ummm...if people DID submit, more people would have higher scores...making it "easier"

[quote="abacadaea"][/quote][quote="Temperal"][quote="abacadaea"][quote="Temperal"]Round 2 was most definitely not the easiest - in my opinion it was by far the hardest, but judging by the statistics, difficulty was nonincreasing, hence round 2 and round 3 were about the same with round 1 harder and round 4 easier.[/quote]

Those statistics were as they are because some people did not submit.[/quote]

All right, then, I can safely say that round 2 was by far the hardest.  :wink:[/quote][quote="abacadaea"]

Ummm...if people DID submit, more people would have higher scores...making it "easier"[/quote]

...how would you know that the unsubmitted scores were higher? For all you know, they could be on average much lower, making my claim true. Or they could be all 20+, for all we know, which is the point of my last post.

[quote="Temperal"][/quote][quote="abacadaea"][/quote][quote="Temperal"][quote="abacadaea"][quote="Temperal"]Round 2 was most definitely not the easiest - in my opinion it was by far the hardest, but judging by the statistics, difficulty was nonincreasing, hence round 2 and round 3 were about the same with round 1 harder and round 4 easier.[/quote]

Those statistics were as they are because some people did not submit.[/quote]

All right, then, I can safely say that round 2 was by far the hardest.  :wink:[/quote][quote="abacadaea"]

Ummm...if people DID submit, more people would have higher scores...making it "easier"[/quote][quote="Temperal"]

...how would you know that the unsubmitted scores were higher? For all you know, they could be on average much lower, making my claim true. Or they could be all 20+, for all we know, which is the point of my last post.[/quote]

Well, atleast there would be more scores >20 on the leaderboard, so it wouldnt look as hard.

we're talking about actual difficulty, not "look as hard" here.


search
inequalities
function
inequalities unsolved

$Let$ $a, b, c$ $be$ $positive$ $real$ $numbers$  $prove$ $that$

$\frac{a+b}{b+c}\cdot\frac{a}{2a+b+c}+\frac{b+c}{c+a}\cdot\frac{b}{2b+c+a}+\frac{c+a}{a+b}\cdot\frac{c}{2c+a+b}\geq\frac{3}{4}$

hello

well i don't quite understand the generalisation of this ineq that Akhrif sent right now but i do hv a proof of this one
but there is one problem

normalise \[ a+b+c = 1 \] to obtain

\[ \sum \frac{a(a+b)}{1-a^2}\geq \frac{3}{4} \]

now \[ \sum \frac{a(a+b)}{1-a^2} = \sum \frac{a(1-c)}{1-a^2} \]

hence we r left to prove \[ \sum \frac{a}{1-a} + \sum \frac{a(b-c)}{1-a^2} = -3 + \sum \frac{1}{1-a} + \sum \frac{a(b-c)}{1-a^2} \geq \frac{3}{2} \]

 \[ \sum \frac{1}{1-a} \geq \frac{9}{2} \] by am hm

hence we r left to prove
\[ \sum \frac{a(b-c)}{1-a^2} \geq 0 \]
ie \[ \sum ab(\frac{a^2 - b^2}{(1-a^2)(1-b^2)}) \geq 0 \]

ie. \[ \sum ab(a-b)(1+c)(1-c)^2\geq 0 \]
which reduces to

\[ \sum a^2b \geq \sum ab^2 \]

ie \[ (a-b)(a-c)(b-c) \geq 0 \] if we had assumed in the beginning that \[ a\geq b\geq c \] wlog then this was definitely true

but now we can't assume such a thing though in the beginning the func was symmetric
i tried many times but always reached this same pt.

there must be some other good method whats wrong here :o  :o

plz send a reply

akhrif ,will you participate in IMO ? :D

now just the national but i hope the next year :D

thanks for your answer  ;) .

I asked you because i don't know a lot  about your

country concerning IOM.Other thing is that i don't find  morrocan

tests of selecting team(the same thing happens when i try to

search for tunisia and algeria ... :blush: ).Finally ,i would like to

tell us a little about the level of education in your country  :huh:  .

advanced thanks for any new answer.

Since the inequality is homogenize , we can WLOG let $a+b+c=1$

Consider the function $f(x)=\frac{x}{1-x^2}$ over $I=(0,1)$ where $x\in I$

$f''(x)=\frac{2x(3-x^2)}{(1-x^2)^3}>0$  (hopefully no mistake  :P )

Then we apply weight Jensen

$\frac{(a+b)f(a)+(b+c)f(b)+(c+a)f(a)}{2(a+b+c)}\geq f\left(\frac{a(a+b)+b(b+c)+c(c+a)}{2(a+b+c)}\right)\\=f\left(\frac{a(1-c)+b(1-a)+c(1-b)}{2}\right)\\=f\left(\frac{1-(ab+bc+ca)}{2}\right)$

But from $(a+b+c)^2\geq 3(ab+bc+ca)$ we have $ab+bc+ca\leq \frac{1}{3}$ , hence

$\frac{1-(ab+bc+ca)}{2}\geq \frac{1}{3}$

So $f\left(\frac{1-(ab+bc+ca)}{2}\right)\geq f(\frac{1}{3})=\frac{3}{8}$  ($\because f'(x)>0$ so it is increasing )

Hence

$(a+b)f(a)+(b+c)f(b)+(c+a)f(a)\geq 2\cdot \frac{3}{8}=\frac{3}{4}$

 :)

[quote="Lovasz"]Both shyong's solution and arpit goel's one are quite complicated.
Here’s my solution. Hope it's new!

[quote] Prove that:
\[ \frac{a+b}{b+c}\cdot \frac{a}{2a+b+c}+\frac{b+c}{c+a}\cdot \frac{b}{2b+c+a}+\frac{c+a}{a+b}\cdot \frac{c}{2c+a+b} \ge \frac34  \]
Where a, b, c are positive reals. [/quote]
Normalize choice by letting $a+b+c=1$. The ineq above become:
\[ \frac{a(a+b)}{1-a^2}+\frac{b(b+c)}{1-b^2}+\frac{c(c+a))}{1-c^2} \ge \frac34 (*)  \]
Easy to see that:
\[ \frac{a}{1-a^2} \ge \frac38 \\ \iff (a-\frac13)(a+3) \ge 0  \]
And
\[ \begin{eqnarray*} LHS_{(*)} &\ge& \sum_{cyc} \frac38 \cdot (a+b) \\ &=& \frac34(a+b+c) \\ &=& \frac34  \]
Done.[/quote]

but $a,b,c$ not neccesary $\leq \frac{1}3$  :|

Oh my mind,  :huh: .
The solution is wrong. I'll try to find another one.


is there $ S\subset R^2$ such that $ S$ and $ S^c$ are not contain a imge of continuse curve?

As a guess, Let $ S \equal{} \{(x,y)\in\mathbb{R}^2: \mbox{x or y are rational}\}$, though I can't prove it right now (busy).

(Edit: I guessed this with only a moment's thought, and it isn't quite right, since only $ S^c$ is unable to contain the image of a continuous curve (with more than one point, of course).)

This is too nice problem to let it slide into oblivion.

One way to approach it is to play the cardinality game. There are only continuum continuous curves, each of them of cardinality continuum. Let's ask this. Given a set $ A$ of cardinality $ \alpha$ and a family $ F$ of subsets of cardinality $ \alpha$ that is itself of cardinality $ \alpha$, is it always possible to construct a subset $ B\subset A$ such that neither $ B$, nor $ A\setminus B$ has any element of $ F$ as its subset?

When $ \alpha\equal{}\aleph_0$, the answer is "Yes": just take the first set in $ F$ and put one its element to $ B$ and the other one to $ A\setminus B$. Then take the second set in $ F$ and do the same, and so on (you'll always have some elements that haven't been used during previous steps). The proof of the general result should be the same, only using the transfinite induction instead of the ordinary one.

Of course, the answer here should not depend on AC, so many other approaches are possible.

Thanks a lot for the solution. I was also finding answer for the similar problem. I was just tired from the problem and got stuck after searching too many forums for the answer but finally I got the answer. It's feel so relief because I have to submit assignment tomorrow..


Prove that for all $ x, y \in \Re^ \plus{}$,

$ 2x\sqrt {x} \plus{} y^3 \geq 3xy$

[hide]By AM-GM, \[ \frac{x\sqrt{x}\plus{}x\sqrt{x}\plus{}y^3}{3}\ge\sqrt[3]{(x\sqrt{x})^2y^3}\] \[ 2x\sqrt{x}\plus{}y^3\ge3xy\][/hide]

[hide]$ \frac {x\sqrt {x} \plus{} x\sqrt {x} \plus{} y^3}{3} \geq \sqrt {x\sqrt {x}\cdot x\sqrt {x} \cdot y^3} \\
\frac {x\sqrt {x} \plus{} x\sqrt {x} \plus{} y^3}{3}\geq xy \\
x\sqrt {x} \plus{} x\sqrt {x} \plus{} y^3 \geq 3xy \\
2x\sqrt {x} \plus{} y^3 \geq 3xy$[/hide]
Edit: Darn, got beat by towers. :P

Let $ u\equal{}\sqrt{x}$, then it is $ 2u^3\plus{}y^3\ge 3u^2y\iff (u\minus{}y)^2(2u\plus{}y)\ge 0$

Or if you want to be slightly more tricky, you could just directly say:

$ 2x\sqrt{x}\plus{}y^3\minus{}3xy\equal{}(\sqrt{x}\minus{}y)^2(2\sqrt{x}\plus{}y)\ge 0$


(Note: This is a spin-off of a Warmup from my school that I tweaked to make a bit harder) Let there be a # AB such that 98>AB>23 and A is the tens digit and B is the ones digit.  When you subtract A+B from AB, you get a new #, UV.  Find the sum U+V, [u]AND PROVE YOUR ANSWER[/u].

[hide]10a+b-(a+b)=9a
So whenever you perform the process with a two digit number the result is always 9a. All multiples of nine under 99, have a sum of 9. Answer:9[/hide]

http://www.bored.com/mysticalball/index.htm

same thing.


[quote="Anonymous52111"]Please show your step-by-step work. Thank you.

Solve: 1/5 (10x - 15) = 6 (1/3x - 1/2) + 10[/quote]
Do you mean
\[\frac{1}{5}(10x-15) = 6\left(\frac{x}{3}-\frac{1}{2}\right)+10\]

or do you mean
\[\frac{1}{5}(10x-15) = 6\left(\frac{1}{3x}-\frac{1}{2}\right)+10\]

[quote="ckck"][quote="vishalarul"][quote="Anonymous52111"]Please show your step-by-step work. Thank you.

Solve: 1/5 (10x - 15) = 6 (1/3x - 1/2) + 10[/quote]
Do you mean
\[\frac{1}{5}(10x-15) = 6\left(\frac{x}{3}-\frac{1}{2}\right)+10 \]
or do you mean
\[\frac{1}{5}(10x-15) = 6\left(\frac{1}{3x}-\frac{1}{2}\right)+10 \]
[/quote]
I think he means the second one because it gets an answer.[/quote]
Then
[hide]$\frac{1}{5}(10x-15) = 6\left(\frac{1}{3x}-\frac{1}{2}\right)+10$
$2x-3=2/x-3+10$
$\frac{2x^{2}-2}{x}=10$
$2x^{2}-2=10x$
$x^{2}-5x-1=0$
$\frac{5\pm\sqrt{25+4}}{2}=\frac{5\pm\sqrt{34}}{2}$


Let $m$ and $n$ be integers greater than 1. Consider an $m$ by $n$ rectangler grid of points in the plane. Some $k$ of these points are coloured red in such a way that no three red points are vertices of a right-angled triangle two of whose sides are parallel to the grid.
Determine the greatest possible value of $k$.

I might be wrong

[hide]Call a horizontal line of points a row and a vertical line of points a column

Let a point that cannot be colored red without making a right angled triangle be called voided.

When there are two points on the same row, two columns are "voided"

When there are two poitns on the same column, two rows are voided

The intersections of the rows and the lines of any two colored points are also voided

We want to find the most amount of points we can color until the whole rectangle is voided.

The the rectangle be n long and m high

The most columns and rows we can take off and still have stuff left over is m-1+n-1

However, that calls for n-1 colored points on one row and m-1 points on one column that is not on the same column as one of the points on the row.

that gives m+n-2 and I can't write the rest right now because I have to leave[/hide]


easy.
let $ S\equal{}c/{(a \plus{} 2b)} \plus{} d/{(b \plus{} 2c)} \plus{} a/{(c \plus{} 2d)} \plus{} b/{(d \plus{} 2a)}$
with caushy:
$ \left(c/{(a \plus{} 2b)} \plus{} d/{(b \plus{} 2c)} \plus{} a/{(c \plus{} 2d)} \plus{} b/{(d \plus{} 2a)}\right)(2(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd))\geq(a\plus{}b\plus{}c\plus{}d)^2$
with caushy:
$ 2ab \leq a^2\plus{}b^2$
$ 2bc \leq c^2\plus{}b^2$
$ 2cd \leq c^2\plus{}d^2$
$ 2da \leq a^2\plus{}d^2$
$ 2ac \leq a^2\plus{}c^2$
$ 2bd \leq d^2\plus{}b^2$
summation:
$ 2(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd)\leq 3(a^2\plus{}b^2\plus{}c^2\plus{}d^2)$
$ \Leftrightarrow 8(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd)\leq3(a\plus{}b\plus{}c\plus{}d)^2$
therefore:
$ \frac{3}{4}(a\plus{}b\plus{}c\plus{}d)^2S \geq S(2(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd))\geq (a\plus{}b\plus{}c\plus{}d)^2$
then: $ S\geq \frac{4}{3}$

[quote="orl"]For the positive real numbers $ a,b,c$ prove that $ c/{(a \plus{} 2b)} \plus{} d/{(b \plus{} 2c)} \plus{} a/{(c \plus{} 2d)} \plus{} b/{(d \plus{} 2a)} > \equal{} 4/3$[/quote]
by Jensen:
$ LHS \ge (a\plus{}b\plus{}c\plus{}d)^2\frac{1}{2(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd)} \ge \frac{4}{3}$

So $\frac c{a \plus{} 2b} \plus{} \frac d{b \plus{} 2c} \plus{} \frac a{c \plus{} 2d} \plus{} \frac b{d \plus{} 2a} = \frac {c^2}{ac \plus{} 2bc} \plus{} \frac {d^2}{bd \plus{} 2cd} \plus{} \frac {a^2}{ca \plus{} 2da} \plus{} \frac {b^2}{db \plus{} 2ab} \geq$ $ \dfrac {(\sum a)^2} {2\sum ab} \geq \dfrac {4}{3}$, since $\left ( \dfrac {\sum a}{4} \right )^2 \geq \dfrac {\sum ab}{6}$, by the Maclaurin inequality.

making the form into  $\sum_{cyc}  \frac{1}{ \frac{a}{c}+\frac{2b}{c} } $ then use T2+Cauchy

[quote=sqing][quote=orl]For the positive real numbers $ a,b,c$ prove that \[ \frac c{a \plus{} 2b} \plus{} \frac d{b \plus{} 2c} \plus{} \frac a{c \plus{} 2d} \plus{} \frac b{d \plus{} 2a} \geq \frac  43.\][/quote]
See also here
http://www.artofproblemsolving.com/community/c6h246429p1353592[/quote]

??? The link is here

[quote=adityaguharoy][quote=Skravin]or using the WLOG condition between a,b,c,d then using the rearrangement theorem and Cauchy[/quote]

For assuming a condition of comparing you must see if a cyclic transformation holds good.[/quote]

oh... I missed it


$ x^2\plus{}y^2\minus{}13(x\minus{}y)\equal{}0$
$ x,y\in Z\plus{}$

Very weak, since the discriminant of the equation in $ x$ is $ \Delta \equal{} 338 \minus{} (2y\plus{}13)^2$.

Solutions $ (3,2)$, $ (10,2)$ and $ (0,0)$, $ (13,0)$.

[quote="wall e"]$ x^2 \plus{} y^2 \minus{} 13(x \minus{} y) \equal{} 0$
$ x,y\in Z \plus{}$[/quote]

mavropnevma,

you included too many many solutions.  You note the restrictions
on x and y.  So each of them must be positive.

The solutions are:  $ (3,2)$ and $ (10,2)$.

It is a matter of notation. For most mathematicians $ \mathbb{Z}_\plus{}$ includes zero; if we want to exclude it, either use $ \mathbb{Z}_\plus{}^*$ or use plain English and say "positive integers".

[quote]It is a matter of notation. For most mathematicians  includes zero;[/quote]
how is this ?the definition of a positive no. is that it should be>0how can you include 0

Discussions about ambiguous notations are ridiculous: if you want people to know what you mean, use an unambiguous notation like $ \mathbb{Z}_{> 0}$ or $ \mathbb{Z}_{\geq 0}$ or just define what you mean in words.  Of course, in this case [i]it doesn't matter[/i] since if you personally don't think that 0 is in whatever the set in question is, you can simply throw out those solutions that include it.


calculus derivative inequalities limit function

Prove that $e^x\ge x+1$ without derivatives.

\[ \mathrm {\ Without\ derivatives\ !\ } \]

[quote="ThAzN1"]I guess it's kinda easy with Maclaurin series... (assuming $x \geq 0$)[/quote]

In fact, the inequality holds even for $x < 0$, and the whole point is to solve it without calculus :)

Oh? I didn't know that counts... (it said without derivatives ;) ) I've seen $e^x$ defined as $1 + x + \frac{x^2}{2!} + \cdots$ in some contexts.  Anyways, $e > 2$ so $e^x \geq (1+1)^x \geq 1 + 1\cdot x$ by Bernoulli if $x \geq 1$ or $x \leq 0$.  Not sure about $0 \leq x \leq 1$ off the top of my head.

Well, I think the case $x \in (-1, 1)$ can be proved with calculus (Lagrange error bound for the negative case?  Correct me if I'm wrong), but I expect proving it without calculus will be quite a challenge.  :(

[quote="Virgil Nicula"][b]ThAzN1[/b], the Bernoully's inequality is $(1+x)^n\ge 1+nx$, where $\boxed {n\in N},\ x>-1 !$
[quote][b]T0rajir0u wrote :[/b]
In fact, the inequality holds even for $x < 0$, and the whole point is to solve it without calculus :)[/quote]
Why $x<0\Longrightarrow e^x\ge x+1\ ?$
Maybe $x\le -1\Longrightarrow e^x\ge x+1\ !\ (x\le -1\Longrightarrow x+1\le 0<e^x)$
Remain $x\in (-1,1)\ ,$ in close vecinity of $0$.[/quote]

No, there's a generalization  :)

http://mathworld.wolfram.com/BernoulliInequality.html

I am not sure whether this proof is acceptable or not, but hopefully it is.

$e^x\neq x$
$\\\because\lim_{x\rightarrow -\infty^+}e^x=0>x\\ \lim_{x\rightarrow\infty^+}e^x=\infty>x\\ e^0=1> 0$

No matter what you do, there is no solution to $e^x=x\implies e^x>x$.

$\\\lim_{x\rightarrow\infty^-}e^x-1=\infty>x\\ \lim_{x\rightarrow\-infty^+}e^x-1=-1>x\\ e^0-1=0=x$

This shows that $e^x-1\geq x\implies e^x\geq 1+x\ \forall x\in\mathbb{R}$.  Equality holds when $x=0$.

$\mathbb{QED}$

Masoud Zargar

Again, it's not a proof.  You need more than three values of a function (yes, even a continuous one) to conclude that one is greater than the other everywhere.

The easiest calculus proof is as follows.  For any $x\geq 0$, there exists $c\in [0,x]$ such that $\exp(x)-\exp(0)=\exp(c)(x)$ by the mean value theorem.  We have $\exp(c)\geq 1$ on $[0,x]$; hence $\exp(x)\geq x+1$ as desired.  If $x<0$, there exists $c\in[x,0]$ such that $\exp(0)-\exp(x)=\exp(c)(-x)$.  Since $\exp(c)<1$, we get that $1+x\leq \exp(x)$, which is what we want again.  More generally, it follows from the fact that $\exp$ has a convergent power series for all $x\in\mathbb{R}$, and it follows from the Cauchy estimate on the error term.

To blah....

  Good , but the mean value theorem is ... derivativative !!!
 I think Virgil asks for  a pure classic proof, based on inequalities and the limit (of function or sequence) OR NOT ?
  It can be derived of course by $lnx \leq x - 1$ , but then we 'll have a faulos circle (remember the : chicken and the egg !)

  babis

What about drawing the grahic represantation of the functions $e^x$ and $x+1$?

[quote="stergiu"]...but then we 'll have a [color=red]faulos circle[/color] (remember the : chicken and the egg !)[/quote]"[color=red]faulos circle[/color]" ="[color=blue]vicious circle[/color]"

I am sorry, it is enough ! I see that you didn't understand what means " [b]without derivatives[/b] ".[hide="Solution."][quote][color=darkred]\[ \mathrm {A\ classical\ proof\ only\ with\ limit\ of\ a\ certain\ sequence\ and\ the\ \underline {Bernoulli's\ inequality}\ .}  \] For a certain real number $x$ and for any $n\in N^*$, $n> -[x]$ we have $\frac xn\ge \frac{[x]}{n}>-1$, i.e. $\boxed {\ \frac xn>-1\ }$. In the [u]Bernoulli's inequality[/u] $\left(1+\frac xn\right)^n\ge1+n\cdot \frac xn=1+x$, for $n\to\infty$
we obtain $\boxed {\ e^x\ge 1+x\ }\ \ !\ .$ Indeed, $L=\lim_{n\to\infty} \left(1+\frac xn\right)^n=e^l$, where $l=\lim_{n\to\infty}n\cdot \left[ \left( 1+\frac xn\right)-1\right]=x$ $\Longrightarrow$  $l=x$ $\Longrightarrow$ $L=e^x\ .$[/color][/quote][/hide]

No, I understood perfectly well that my solution wasn't what virgil was looking for.  I just included it for completeness.

Well, I see that nobody wants to give a proof without derivatives. here is one:
By Bernoulli's inequality, we have \[ (1+\frac{x}{n})^{n}\geq 1+x. \]
by passing to the limit when $n\to\infty$ the conclusion follows. :)

very nice. i was about to suggest using  Bernoulli's inequality as well. i had just learned it, i however would not have thought of the limit part though.  :blush:

That was Virgil Nicula's solution :) :
[quote="Virgil Nicula"][color=darkred]\[ \mathrm {A\ classical\ proof\ only\ with\ limit\ of\ a\ certain\ sequence\ and\ the\ \underline {Bernoulli's\ inequality}\ .}  \] For a certain real number $x$ and for any $n\in N^*$, $n> -[x]$ we have $\frac xn\ge \frac{[x]}{n}>-1$, i.e. $\boxed {\ \frac xn>-1\ }$. In the [u]Bernoulli's inequality[/u] $\left(1+\frac xn\right)^n\ge1+n\cdot \frac xn=1+x$, for $n\to\infty$
we obtain $\boxed {\ e^x\ge 1+x\ }\ \ !\ .$ Indeed, $L=\lim_{n\to\infty} \left(1+\frac xn\right)^n=e^l$, where $l=\lim_{n\to\infty}n\cdot \left[ \left( 1+\frac xn\right)-1\right]=x$ $\Longrightarrow$  $l=x$ $\Longrightarrow$ $L=e^x\ .$[/color][/quote]


What is the value of $ abc$ is $ ab \equal{} 4, bc \equal{} 2, ac \equal{} 18$ and $ a>0$?

We have $ ab\equal{}4$, $ bc\equal{}2$, and $ ac\equal{}18$.  Dividing the first equation by the second, we get $ \frac{a}{c}\equal{}2$, or $ a\equal{}2c$.  Substituting into the third equation, we get $ 2c^2\equal{}18 \implies c\equal{}\pm 3$.  However, we know that $ a>0$, so all the numbers are positive, so $ c\equal{}3$.  $ a\equal{}6$ and $ b\equal{}\frac{2}{3}$ quickly follow for our solution set of $ (a,b,c)\equal{}\boxed{(6,\frac{2}{3},3)}$.

However, we wish to find the product, so our answer is $ \boxed{12}$.

$ab \times bc \times ac=a^2b^2c^2=18*2*4=144$

$\sqrt{144}=12$

[size=50][color=#FF0000]Moderator edit: LaTeXed it to make it easier to read.[/color][/size]


number theory
relatively prime
number theory unsolved

Find all positive integers $ n$ such that $ (n\plus{}1)(n\plus{}2)|2C(2n,n)$

Does $ C(2n,n)$ mean $ \binom{2n}{n}$? Assuming this is the case, the $ (n\plus{}1)$ part doesn't make any difference: $ n\plus{}1$ is relatively prime to $ n\plus{}2$, and $ \binom{2n}{n}$ is always divisible by $ n\plus{}1$ (see "Catalan Numbers"). Therefore the question is equivalent to: Find all positive integers $ n$ such that $ n\plus{}2$ divides $ 2\binom{2n}{n}$.

Having said all this, I don't think I'm able to find any useful pattern in those numbers.

Yeah! I mean $ 2\binom{2n}{n}$.
No, you're wrong. It doesn't mean "Find all positive integer $ n$ such that $ n\plus{}2$ divides $ 2\binom{2n}{n}$ , but it means "Find all positive integer $ n$ such that $ \frac{2\binom{2n}{n}}{(n\plus{}1)(n\plus{}2)} \in \mathbb{N}$

[quote="t0rajir0u"]Did you or did you not actually read the argument in that post?  It's valid.  The two statements are equivalent.[/quote]
Sorry! I'm wrong. I forget $ (n\plus{}1,n\plus{}2)\equal{}1$. :(

Do you have reason to believe that these numbers have some alternative useful characterization? (For that matter, if this problem is taken from some contest then that's probably reason enough).

The OEIS has an entry for those $ n$ such that $ n\plus{}2$ divides $ \binom{2n}{n}$. It's not exactly your question (although they coincide for odd $ n$), but the fact that OEIS has nothing interesting to say about these numbers worries me. It doesn't [i]prove[/i] anything, of course.

So - what is the motivation for this problem?


Just for the sake, please do provide a solution.

I can look in the answer key too :)

Do you mean,indianamath [hide] 4sqr3?[/hide]

[hide="hint"]Solve two equations. The intersection of the parabola and the line can be solved using two variables and two equations.[/hide]

[hide="Solution"] Say that the y-axis hits AB at M.  Notice that OM is the perpindicular bisector of AB, so AOM is a 30-60-90 triangle.  Now, say that point A is at $(x,\frac{-x^{2}}{2})$, so $OM=\frac{-x^{2}}{2}$ and $OA=x$.  Thus, since $OA\sqrt{3}=OM, \frac{-x^{2}}{2}=x\sqrt{3}\Rightarrow x^{2}+2x\sqrt{3}=0\Rightarrow x=0, x=-2\sqrt{3}\Rightarrow AO=2\sqrt{3}\Rightarrow AB=\boxed{4\sqrt{3}}$ [/hide]

Dude, you've already posted the problem, like what's up with that?
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=analytic&t=120939

[hide]
let $y = \sqrt{3}*x$

then

$\sqrt{3}*x =-\frac{1}{2}x^{2}$

$-2\sqrt{3}= x$

since it's a length, just make it positive and multiply by $2$ to get side $AB = 4\sqrt{3}$.
[/hide]


quadratics
modular arithmetic
number theory proposed
number theory

Let $p$ be a prime number such that $p=a^{2}+5b^{2}$, where $a,b$ are positive integers and $a$ is odd. Then:
$a$ is a quadratic residue $\pmod{p}$ if and only if $p \equiv 1 \pmod{5}$

Let $(\frac{a}{p})=(\frac{p_{1}}{p})(\frac{p_{2}}{p})...(\frac{p_{n}}{p})$ (we consider prime factors of $a$ which are dividing it in the odd power). Since $p_{i}\not = 2$ we have:
$(\frac{p_{i}}{p})(\frac{p}{p_{i}})=(-1)^{\frac{p-1}{2}\frac{p_{i}-2}{2}}=1$ (it's easy to verify that $p \equiv 1 \mod{4}$) from the The Law of Quadratic Reciprocity. But:
$(\frac{p}{p_{i}})=(\frac{5b^{2}}{p_{i}})=(\frac{5}{p_{1}})(\frac{b^{2}}{p_{i}})=(\frac{5}{p_{i}})=(\frac{p_{i}}{5})$
where the last equality follows again from The Law of Quadratic Reciprocity. Now it's easy to see that $p \equiv 1 \mod{5}\leftrightarrow a \equiv 1 \mod{5}\leftrightarrow$ there is even number of $i$'s such that $p_{i}\equiv \pm 2 \mod{5}\leftrightarrow a$ is a quadratic residue $\mod{p}$.


Find all solutions of the equation $x^y+y=y^x+x$ in the positive integers.

It was done fast, so I hope that's true:

[tex]x^y + y = y^x + x \Leftrightarrow x^y - y^x = x - y.[/tex]
Let's look at three cases:
If [tex]x - y > 0[/tex] then for [tex]x > 3[/tex] and [tex]y \geqslant 2[/tex] we have [tex]x^y - y^x
\leqslant 0[/tex]. For [tex]x > 3[/tex] and [tex]y = 1[/tex] we have always a solution (because then [tex]x^1-1=x-1[/tex])
It suffice now to enumerate cases for [tex]x = 3[/tex] and [tex]y = 2, 1[/tex] and [tex]x = 2[/tex] and [tex]y = 1[/tex]. We get the solutions [tex]( x, 1 ) ; ( 3, 1 ) ; ( 3, 2 ) ; ( 2, 1 )[/tex]
For the case [tex]x - y < 0 \Leftrightarrow y - x > 0[/tex] it suffice to substitute [tex]x[/tex] by [tex]y[/tex] and [tex]y[/tex] by [tex]x[/tex] in the previous case. And we obtain solutions: [tex]( y, 1 ) ; ( 1, 3 ) ; ( 2, 3 ) ; ( 1, 2 )[/tex]
If [tex]x = y[/tex] then [tex]y^x - x^y = 0[/tex], so for any [tex]x[/tex] being integer [tex]( x, x )[/tex] is a solution


hey there can you change your signature?

I guess nobody would like that sort of language on the site! :D :D

so ... change it!!!

cheers!

i know my opinion on this topic isnt reqd but neway,wot  does the message say,i was just curiou.i think "merde" in french was "s**t",if i remember correctly.but vialli's italian.but i think i see the connection
.and i i'm thinking correctly,vialli SHOULD  change the signature

People, stop reviving old threads please.


function
topology
advanced fields
advanced fields unsolved

Let $X$ be a compact Hausdorff topological space, and let $C(X,\mathbb R)$ be the linear space of continuous real functions on $X$ with the supremum norm.

If $E$ is a closed linear subspace of $C(X,\mathbb R)$ such that for every $f\in E$ the set $f(X)$ is finite, then $E$ is finite dimensional.

[hide="Step 1"]Let's say that two points $x_{1}, x_{2}\in X$ are separated by $E$ if there is $f\in E$ such that $f(x_{1})\ne f(x_{2})$. The relation "not separated by $E$" is an equivalence on $X$. It suffices to prove that there are only finitely many equivalence classes.[/hide]
[hide="Step 2"]Now suppose that $\{A_{n}\colon n\ge 1\}$ is an infinite family of distinct equivalence classes. Choose $a_{n}\in A_{n}$ for each $n$. Choose a set of real numbers $\alpha_{m,n}$ that is linearly independent over $\mathbb{Q}$. For each pair of indices $1\le m<n<\infty$ choose a function $f_{m,n}\in E$ such that $f_{m,n}(a_{m})-f_{m,n}(a_{n})=\alpha_{m,n}$. Finally, let $f=\sum_{m,n}c_{m,n}f_{m,n}$, where $c_{m,n}$ are rational coefficients chosen so that the sum converges uniformly. We have $f(a_{m})\ne f(a_{n})$ for any $1\le m<n<\infty$, a contradiction.[/hide]


Rectangle $ PQRS$ is inscribed in rectangle $ ABCD$, as shown. If $ DR \equal{} 3, RP \equal{} 13,$ and $ PA \equal{} 8,$ compute the area of rectangle $ ABCD$.

That's what I got as well. Care to show you got that?

[hide="solution"]

Draw a perpendicular line from $ R$ to $ AP$, and call the intersection point $ X$.

Now, $ RXAD$ is a rectangle, so $ AX \equal{} 8$.

Thus, $ XP \equal{} 5$, and by Pythagorean theorem, we have that $ RX\equal{}12\equal{}AD$

Uh...$ BP\equal{}RD\equal{}3$, so the area is $ 12\times 11 \equal{} 132$[/hide]

:) Same way as I've done. I don't think there's any other way to go about this problem.


prove that the sum of two odd squares cannot be a square

modulo 4 the sum of two odd squares is 2, whereas a square has remainders 0 (even case) or 1 (odd case)

[quote="1=2"]This is from Engel's book.

[hide="solution"]When an odd number is squared, the result is a number that has a remainder of 1 when divided by 4. When we sum two odd squares, we get a number that has a remainder of 2 when divided by 4. But no square can have a remainder of 2 when divided by 4.
[/hide][/quote]
er this is a really basic/well known problem ie engel wasnt the first to put it in his book/the OP probably didnt find it in engel....
also the same solution was already posted


[quote="ElChapin"]A] Let a straight line $ m$, intersect the sides $ BC$, $ CA$ and $ AB$ of a triangle at $ D$, $ E$ and $ F$ respectively. Prove that the cicumcirlcles of triangles $ AEF$, $ BDF$ and $ CDE$ have a common point.

B] Find the locus of this intersection point.

 :)[/quote]
[hide="A"]
Allow the intersections between the circumcircles of $ \triangle AEF$ and $ \triangle BDF$ to be $ P$.  Now, drop perpindiculars from $ P$ to $ AC$, $ AB$, $ BC$, and $ EF$.  Since $ P$ lies on the circumcircle of $ \triangle AEF$, the feet of the perpindiculars from $ P$ to $ AC$, $ EF$, and $ AB$ must be collinear by Simson's Theorem.  Similarly, we notice that since $ P$ is on the circumcircle of $ \triangle DFB$, the feet of the perpindiculars from $ P$ to $ AB$, $ EF$, and $ BC$ are collinear.  Thus, we notice that all of these feet from $ P$ are collinear; in particular, the feet from $ P$ perpindicular to $ BC$, $ AC$, and $ EF$ are collinear.  By the converse of Simson's Theorem, $ P$ must lie on the circumcircle of $ \triangle CDE$, so all of the circumcirlces share point $ P$ and hence they intersect at one point. [/hide]

[hide="B"]
We have previously shown that the feet of the perpindiculars from $ P$ to $ AB$, $ AC$, and $ BC$ are collinear, and hence $ P$ must lie on the circumcircle of $ \triangle ABC$.  Clearly, we can take any point on the Simson Line of $ P$ (which is on the circumcircle of $ \triangle ABC$), call it $ Q$, and draw line $ DEF$ through $ Q$ that is perpindicular to $ PQ$.  Hence, any point on the circumcircle of $ \triangle ABC$ can be $ P$.  Thus, the locus of $ P$ is the circumcircle of $ \triangle ABC$. [/hide]

[quote="CatalystOfNostalgia"][quote="The QuattoMaster 6000"] Clearly, we can take any point on the Simson Line of $ P$ (which is on the circumcircle of $ \triangle ABC$), call it $ Q$, and draw line $ DEF$ through $ Q$ that is perpindicular to $ PQ$.  Hence, any point on the circumcircle of $ \triangle ABC$ can be $ P$.  Thus, the locus of $ P$ is the circumcircle of $ \triangle ABC$. [/hide][/quote]

I don't quite understand this. How can you "draw line DEF through Q?" Q isn't on line DEF.[/quote]
Sorry, what I am trying to do is show that any point $ P'$ on the circumcircle of $ \triangle ABC$ can be $ P$ (i.e. I will show that $ P$ can be any point on the circumcircle of $ \triangle ABC$; meaning the locus of $ P$ is the entire circumcircle instead of random points on the circumcircle.)  Now, I will show that there is a line $ D'E'F'$ such that $ P'$ is the intersection between the circumcircles of $ \triangle AE'F'$, $ \triangle BD'F'$, and $ \triangle CD'E'$.  We construct line $ D'E'F'$ by considering any point $ Q$ on the Simson Line of $ P'$ with respect to $ \triangle ABC$ and then drawing $ D'E'F'$ through $ Q$ such that $ P'Q\perp D'E'F'$.  Hence, we can find line $ D'E'F'$ and we are done.  I hope that helps  :)

Again, I apologize for the lack of clarity.


The graph of the function $ f(x) \equal{} 2x^3 \minus{} 7$ goes:

$ \textbf{(A)}\ \text{up to the right and down to the left} \\
\textbf{(B)}\ \text{down to the right and up to the left} \\
\textbf{(C)}\ \text{up to the right and up to the left} \\
\textbf{(D)}\ \text{down to the right and down to the left} \\
\textbf{(E)}\ \text{none of these ways.}$

[hide="Click for solution"]
Graphing:

    [asy]import graph;
    size(200);
    Label f;
    f.p=fontsize(6);
    xaxis(-16,16,Ticks(f, 8, 0.25));
    yaxis(-16,16,Ticks(f, 8, 0.25));
    real f(real x)
    {
    return 2*x^3-8;
    }
    draw(graph(f,-1.6,2.3));[/asy]

The answer is thus $ \boxed{\textbf{(A)}}$.
[/hide]

Alternatively, one could say any function with an odd leading exponent, and a positive leading coefficient is up to the right and down to the left.


There is a solid matter with unknown mass.We want to heat it by amount $\delta Q$.Then its volume will be $\delta V$ greater than initial volume.$c_{p}$,$\mu$ molar mass and young modulus E of this matter is known.Find the ratio $\frac{\delta Q}{\delta V}$.

[b]Here is the solution.

$\delta V=V_{0}3\alpha \delta T$

$\delta l=l_{0}\alpha \delta T$

$\frac{\delta V}{\delta l}=\frac{3V_{0}}{l_{0}}$

$\delta Q=mc_{p}\delta T$

$PV_\mu=\frac{m}{\mu}R\delta T$

$P=E\frac{\delta l}{l_{0}}$

Using these equations we get the following answer:

$\frac{\delta Q}{\delta V}=\frac{\mu c_{p}E}{3R}$