The score is positives 4, negatives 3, and we are at -34.
This round started at: 2006-Aug-31~18:22:14
This round will end at: 2006-Sep-07~18:22:14
Xantos C. Guin is responsible for keeping the time.
This post is $-35$

$-37$ well i did and i think i keep switching sides!

$-38$
uhhh
you need to pm bubka but he won't mind as long as you're hyperactive

read page1,2 on the original topic
$-43$

Team A increase the total by -1, team B by +1. Comments with each post are welcome, but not compulsory. Whoever is ahead after a week, wins. You can't post after yourself. Good luck.

The main tow thread is at -33. We will start at -33.

I make this -32.

(The first part of this tow was at 171 pages when deleted)

[quote="Xantos C. Guin"]This is $-54$ or $-31$, depending on what Klebian's post should have been.

BTW: How did the last thread get deleted?[/quote]

I deleted it. -30

KLEBIAN THIS ISN'T YOUR GAME, YOU CAN'T MAKE CRAP ESPECIALLY CHEATING FOR YOUR SIDE.

$-56$

 :idea:

$-57$
uhh w/e
ignite, happy almost navier-stokes day(uhh)!

WHAT A COINCIDENCE! I NOTICED TOO! SO... ARE YOU PLANNING ON POSTING IN IT?

$-60$

dunno
$-61$
I bet the random immunity is from a power role or something

[quote="Ignite168"]PLEASE POST!!!!!!!!!!!! NO ONE IS POSTING :(

AND "RANDOM" IMMUNITY IS DECIDED BY THE COMMITTEE OF BASURP, MAKY, AND ZAF007 SO YEAH, IT KINDA IS RANDOM.[/quote]
wait what?
darn I thought it was like a innocent power role or w/e(guilty maybe)
OK FINE i'll post in mafia21

ANYTHING SAID IN THIS THREAD IS NOT FDA APPROVED SO BELIEVE AT YOUR OWN RISK. QUOTING IT IN A MAFIA GAME = MODKILL BECAUSE OF DA RULEZ.

$-62$

LOL I WAS KIDDING ABOUT MAKY, BASURP, AND ZAF007, WHY DO YOU THINK THEY WERE ALL MODKILLED? I DECIDE WHO'S LYNCH IMMUNE

$-64$

THX FOR POSTING BTW.

BECUZ U POSTED, I'LL CONSIDER YOU GIVING YOU LYNCH IMMUNITY TOMMOROW, BUT THE RULES MIGHT SAY THAT'S NOT ALLOWED, I'LL HAVE TO CHECK


calculus
derivative
calculus computations

This is just a case of the pythagorean theorem.

$ x^{2}\plus{}y^{2}\equal{}z^{2}\Rightarrow x\frac{dx}{dt}\plus{}y\frac{dy}{dt}\equal{}z\frac{dz}{dt}$

Are you sure you mean [i]maximum[/i] distance? As $ t\to\infty,$ the distance increases without bound. The interesting question would be the minimum distance.

Did you not understand what Kent said?  Forget the numbers and just consider the scenario in real life.

Imagine that car B starts at the origin, and car A is now somewhere in the negative-x territory.  Eventually, the actual time is not important, car A will pass the origin and be in positive territory, while car B continues up in the positive y-direction.  Then by pythagoras, the two cars are getting farther and farther away by the instant, so unless there is some time frame or the cars move backwards, there is no maximum distance.

[quote="kim77"]3. Car is 125 km directly west of car B and begins moving east at 100 km/h. At the same time, car B begins moving north at 50 km/h. At what time t does the maximum distance occur?[/quote]

This is straightforward.  Suppose the minimum distance occurs at time $ t$, then we have $ z^2 \equal{} (125\minus{}100t)^2 \plus{} (50t)^2.$  And now it is a matter of minimize $ z^2.$

JRav: How would you justify that

$ x^{2}\plus{}y^{2}\equal{}z^{2}\Rightarrow x\frac{dx}{dt}\plus{}y\frac{dy}{dt}\equal{}z\frac{dz}{dt}$?

By implicit differentiation

Suppose that

$ x = x(t)$

$ y = y(t)$

$ z = z(t)$

Then

$ \frac {d[z^2]}{dt} = 2z\dot{z}$

Continue this way to get

$ 2x\dot{x} + 2y\dot{y} = 2z\dot{z}$

$ \Rightarrow{x\dot{x} + y\dot{y} = z\dot{z}}$

Yeah, but I don't see how that would solve this problem... perhaps I'm just missing some easy ways...

Well you have an expression for $ z^{2}$.  So differentiate it.


What is the measure, in degrees, of the acute angle formed by the minute hand and the hour hand on a standard clock when it indicates $ 9$:$ 40$?

Using the formula everyone knows, $ \frac{|60h\minus{}11m|}{2}$, 540-440/2=50.


[hide]In one hour, one man can paint $\frac{1}{4}$ of the house and the other man can paint $\frac{1}{3}$ of the house, so combined they can paint $\frac{7}{12}$ of the house. We can then set up the equation $\frac{7}{12}x=1$, and solving we find $x=\boxed{\frac{12}{7}}$ hours.[/hide]

[hide]
The first guy can paint a house in 4 hours, so in one hour he can paint $\frac{1}{4}$ of a house. Similarly, the second guy can paint $\frac{1}{3}$ of a house in an hour. So together, they would paint $\frac{1}{4}+\frac{1}{3}=\frac{7}{12}$ of a house in one hour; thus it would take them $\frac{12}{7}$ hours to paint the whole house.
[/hide]

The method the others have used is called harmonic mean.

[hide="solution"]The first man can do 1/4 of the job in an hour, and the second can do 1/3 of the job in an hour. So every hour, they finish 7/12 of the job.So 12/7 of an hour will bring them to the end of painting the house.[/hide]

[quote="xelngath"]Hello, can someone tell me how I would go about solving these kind of problems:

If a man can paint a house in 4 hours, and another can paint a house in 3 hours, how long will it take them to paint a house together?

I seem to have forgotten :/

Thanks a lot[/quote]

[hide="uh..."]
in 1 hr, man1 will paint $\frac{1}{4}$ of the house and man2 will paint $\frac{1}{3}$ of the house. $\frac{1}{4}+\frac{1}{3}= \frac{7}{12}$. Divide to get: $\frac{1}{\frac{7}{12}}= \frac{12}{7}$. It will take him $\frac{12}{7}$ hrs[/hide]

[quote="xelngath"]Hello, can someone tell me how I would go about solving these kind of problems:

If a man can paint a house in 4 hours, and another can paint a house in 3 hours, how long will it take them to paint a house together?

I seem to have forgotten :/

Thanks a lot[/quote]

x/3+x/4=1  I think


How many different three-digit odd numbers can be formed using the digits 3, 4, 5 and 6 if digits cannot be repeated?

For the three-digit-number to be odd, the last (units) digit must be odd. There are $ 2$ choices ($ 3$ or $ 5$) for this last digit. After this, there are $ 4\minus{}1\equal{}3$ choices left for the middle (tens) digit, since one of the $ 4$ choices for digits was already used in the units digit and repetition of digits is not allowed. Using the same logic with the tens digit, there are $ 3\minus{}1\equal{}2$ choices for the first (hundreds) digit, making a total of $ 2\times3\times2\equal{}\fbox{12}$ choices for the number.


http://en.wikipedia.org/wiki/Thue-Siegel-Roth_theorem

I don't get the part of taking infimum and getting the constant $ C(\epsilon)$. Also, shouldn't $ C(\epsilon)$ really be $ C(\epsilon, \alpha)$?

Did I say something stupid?  :P

Btw, how do you pronounce the name of that theorem?  :huh:

for any algebraic $ \alpha$ and any $ \epsilon > 0$ there exist $ q_{0}(\epsilon) > 0$ such as $ |\alpha \minus{} \frac {p}{q}|\geq q^{ \minus{} 2 \minus{} \epsilon}$ for any integers $ p,q$ with $ q > q_{0}(\epsilon)$
it's called Roth theorem .

$ q_0$ also depends on $ \alpha$, right? If this is true, I can follow the reasoning.


Let S be a set of 2×2 matrices with complex numbers as entries, and let
T be the subset of S consisting of matrices whose eigenvalues are ±1 (so
the eigenvalues for each matrix in T are {1,1} or {1,−1} or {−1,−1}).
Suppose there are exactly three matrices in T. Prove that there are matrices
A,B in S such that AB is not a matrix in S (A = B is allowed).


invariant
combinatorics proposed
combinatorics

Let $a_1, a_2, ..., a_{100}$ be an ordered set of numbers. At each move it is allowed to choose any two numbers $a_n,a_m$ and change them to
$\frac{a_n^2}{a_m}-\frac{n}{m}(\frac{a_m^2}{a_n}-a_m)$ and $\frac{a_m^2}{a_n}-\frac{m}{n}(\frac{a_n^2}{a_m}-a_n)$ respectively.
Determine if it is possible, starting with the set with $a_i = \frac{1}{5}$ for $i=20,40,60,80,100$ and $a_i=1$ otherwise, to obtain a set consisting of integers only.

Hint: $\frac{a_n}{n}+\frac{a_m}{m}$ is invariant.


integration
trigonometry
calculus
calculus computations

Find  $ \int \frac {x\plus{}sinx}{1 \plus{} cosx} \,dx$

$ \int \frac {x \plus{} \sin x}{1 \plus{} \cos x} \, \textbf d x \equal{} \int \left( \frac {x}{1 \plus{} \cos x} \plus{} \frac {\sin x}{1 \plus{} \cos x} \right) \, \textbf d x \equal{}$
$ \equal{} \int \left( \frac {x}{2 \cos^2 \frac {x}{2}} \plus{} \tan \frac {x}{2} \right) \, \textbf d x \equal{} \int \left(x \tan \frac {x}{2} \right)' \, \textbf d x \equal{} x \tan \frac {x}{2} \plus{} \textbf C$

This is my solution:

Use $ \sin x \equal{} 2\sin\frac {x}{2}\cos\frac {x}{2}$ and $ 1 \plus{} \cos x \equal{} 2 \cos ^2 \frac {x}{2}$. Then you can split it in two integrals:

$ \int \tan \frac {x}{2} ,dx$
$ \int \frac {x}{2 \cos ^2 \frac {x}{2}} \,dx$

First is straightforward, and the second is immediately solvable by partial integration. The sum yields

$ x \tan\frac {x}{2} \plus{} C$.

[quote="giannis"]Find  $ \int \frac {x \plus{} sinx}{1 \plus{} cosx} \,dx$[/quote]

$ \int \frac {x \plus{} \sin x}{1 \plus{} \cos x} \,dx
\equal{} \int \left(\frac{x}2 \sec^2 \frac{x}2 \plus{} \tan \frac{x}2 \right) \,dx
\equal{} (x\tan \frac{x}2 \minus{} \int \tan \frac{x}2 \, dx) \plus{} \int \tan \frac{x}2 \, dx
\equal{} x\tan \frac{x}2 \plus{} C$


calculus
integration
calculus computations

A couple of my friends are in 2nd quarter calculus (integration). I am only in first quarter calculus however (differentiation). Anyways, i know quite a bit of math beyond my current level and they were complaining about a tough exam problem they had and i took a crack at it just to see if i could do it. Anyways, here is the problem.

Given that the integral from 0 to 4 of f(x) with respect to x equals 2, find the integral from 0 to 2 of 2f(x) with respect to x.

I will not post my solution since it is very unorganized. I tried the problem several times and i mixed information i found in one aattempt and applied it to a different attempt and so on. The answer i came up with is that the integral that they are asking for is 8/3. Can anyone verify if this is correct (or incorrect)?

As written, the question is impossible to answer. The second integral needs to involve $f(2x)$.


(1) Prove that for $ m\in\mathbb{N}$, there exists $ \sqrt {2}$ between $ m$ and  $ \frac {m \plus{} 2}{m \plus{} 1}.$

(2) Which is closer to $ \sqrt {2},\ m$ or  $ \frac {m \plus{} 2}{m \plus{} 1}?$

I do believe...[hide="1"]$ 2>\sqrt{2}\Longrightarrow m\equal{}0,1$
$ 1<\sqrt{2}<\frac{3}{2}$ or $ 0<\sqrt{2}<2$[/hide][hide="2"]$ \sqrt{2}\approx 1.41$ or something, so $ \frac{m\plus{}2}{m\plus{}1}$ is closer in both cases[/hide]
Is that what you wanted?

1) is asking to prove that $ \sqrt{2}$ is between $ m$ and $ \frac{m\plus{}2}{m\plus{}1}$ (in either order) for every $ m \in \mathbb{N}$.

[quote="kunny"](1) For $ m\in\mathbb{N}$, there exists $ \sqrt {2}$ between $ m$ and $ \frac {m \plus{} 2}{m \plus{} 1}.$

(2) Which is closest to $ \sqrt {2},\ m$ or $ \frac {m \plus{} 2}{m \plus{} 1}?$[/quote]
[hide="1"]
Notice that for the case $ m\equal{}1$, we have that $ \frac{m\plus{}2}{m\plus{}1}\equal{}1.5$ and $ m\equal{}1$, so $ \sqrt{2}$ is between the two.  It suffices to show that $ \sqrt{2}$ is between the two numbers for $ m\ge 2$.  Now, if $ m\ge 2$, then $ \frac{m\plus{}2}{m\plus{}1}\le \frac{4}{3}<\sqrt{2}$ and $ m\ge 2>\sqrt{2}$.  Hence, $ \frac{m\plus{}2}{m\plus{}1}<\sqrt{2}<m$ for all $ m\ge 2$, and the proof is complete.  [/hide]

[hide="(1)"]If $ m \equal{} 1$ then this is equivalent to $ 1 < \sqrt 2 < \frac 32$, which is clearly true as $ 1 < 4$ and $ 9 > 8$. For $ m\ge 2$, we need to prove that $ \frac {m \plus{} 2}{m \plus{} 1} < \sqrt 2 < m$. The right side of the inequality is obvious as $ m\ge 2 > \sqrt 2$. And the left side is equivalent to $ m \plus{} 2 < \sqrt 2m \plus{} \sqrt 2\iff (m \minus{} \sqrt2)(\sqrt2 \minus{} 1) > 0$, which is again obviously true.[/hide]

[hide="(2)"]For $ m\equal{}1$ it's not difficult to see that $ \sqrt 2$ is closer to $ \frac{m\plus{}2}{m\plus{}1}$. This can be also checked for $ m\equal{}2$. Now for $ m\ge 3$ the difference between $ m$ and $ \sqrt 2$ increases rapidly. On the other hand as $ m$ approaches infinity $ \frac{m\plus{}2}{m\plus{}1}$ tends to $ 1$. Thus $ \sqrt 2$ is always closer to $ \frac{m\plus{}2}{m\plus{}1}$ than $ m$.[/hide]

[hide="1"]
$ x^2 \equal{} 2 \iff x \equal{} \frac {x \plus{} 2}{x \plus{} 1}$. Thus, $ \sqrt {2}$ is the unique positive number at which the two expressions are equal.

Also, $ m$ increases with $ m$ and $ 1 \plus{} \frac {1}{m \plus{} 1}$ decreases with $ m$

[i]Thus, $ \sqrt {2}$ must be strictly between the two expressions for all $ m > 0, \; m \neq \sqrt {2}$.[/i][/hide]

1
[hide]
If $ m \equal{} 1$, then $ 1 < \sqrt {2} < \frac {3}{2}$.
If $ m\geq 2 > \sqrt {2}$, then $ \frac {m \plus{} 2}{m \plus{} 1} \equal{} 1 \plus{} \frac {1}{m \plus{} 1}\leq 1 \plus{} \frac {1}{3} \equal{} \frac {4}{3} < \sqrt {2}$.[/hide]

2
[hide]
If $ m \equal{} 1$, then $ \frac {3}{2} \minus{} \sqrt {2} < \sqrt {2} \minus{} 1$
If $ m\geq 2$, in (1) we've proved there's always $ \frac {m \plus{} 2}{m \plus{} 1} < \sqrt {2} < m$, so take
$ f(m) \equal{} (m \minus{} \sqrt {2}) \minus{} (\sqrt {2} \minus{} \frac {m \plus{} 2}{m \plus{} 1})$
$ \equal{} m \plus{} 1 \plus{} \frac {1}{m \plus{} 1} \minus{} 2\sqrt {2}$
On interval $ [2, \plus{} \infty)$, $ f$ is increasing monotonically, and we have $ f(2) > 0$. Thus $ f(m) > 0$ for all $ m \geq 2$. So $ \frac {m \plus{} 2}{m \plus{} 1}$ is always closer to $ \sqrt {2}$.[/hide]

It suffices to compare the average of the functions with the root of 2.

[hide="oversight (maybe)"]
For both cases I notice that you leave out the case $ m\equal{}0$. This does not disprove either theory as you yield $ 0<\sqrt{2}<2$ and 2 is closer to $ \sqrt{2}$ than 0 is.
Or is $ 0\not\in\mathbb{N}$?
[/hide]

[quote="t0rajir0u"]The American convention is that $ \mathbb{N} \equal{} \{ 1, 2, ... \}$.  In Eastern Europe (and possibly some other places) $ \mathbb{N} \equal{} \{ 0, 1, 2, ... \}$.  To be unambiguous, we can use $ \mathbb{Z}_{\ge 1}$ and $ \mathbb{Z}_{\ge 0}$, respectively.[/quote]
Quite helpful, thank you, t0rajir0u. Although, are you sure you meant $ \mathbb{Z}$ as opposed to $ \mathbb{N}$? And it's interesting that I am of the former and agree with the latter.  :lol:

[quote="t0rajir0u"]$ \mathbb{Z}_{\ge 1}$ means "the subset of the integers that are greater than or equal to $ 1$."  Same with the other notation.[/quote]
Ah, okay then, thanks again. I'd always thought that $ \mathbb{Z}$ denoted all complex numbers, which was the source of my confusion. Wonderful clarification as always, t0rajir0u.


limit
function
calculus
derivative
real analysis
real analysis unsolved

Find $\lim_{n\to \infty}\sqrt{\frac{n^n}{n!}}$ ?

I think you mean $\lim_{n\to\infty} \sqrt[n]{n^n\over n!}$, since $\lim_{n\to \infty}\sqrt{\frac{n^n}{n!}} \to \infty$ obviously.

Hint:
If $a_n\to a$ then $\sqrt[n]{a_1a_2\cdots a_n}\to a$. Take $a_n = (1+\frac 1n)^n$.

Could someone just hide the answer instead of hinting at it?  I get close to nowhere on that limit.

try L'hopital's rule

the bottom of the fraction can be treated as the nth root of the gamma function and the top becomes the derivative of n which is one

To FMako:

Let $\displaystyle a_n = \frac{n^n}{n!}$.

We have $\displaystyle x_n = \frac{a_{n+1}}{a_n} = \left( 1 + \frac{1}{n} \right)^n$.

It is known that

(i) $x_n \to e$

(ii) if $\displaystyle \frac{a_{n+1}}{a_n} \to \ell$ then $\sqrt[n] {a_n} \to \ell$ (a consequence of Stolz-Cesaro)

Thus, the answer is $e$.

We also obtain a nice approximation, $n! \approx \left( \frac{n}{e} \right)^n$.


calculus
integration
limit
trigonometry
calculus computations

Find the following two limits:
1) $ \lim_{n\rightarrow\infty}\int_{0}^{\infty}\frac{\sin\left(\frac{x}{n}\right)}{\left(1\plus{}\frac{x}{n}\right)^{n}}\,dx$.
2) $ \lim_{n\rightarrow\infty}\int_{a}^{\infty}\frac{n}{1\plus{}n^{2}x^{2}}\,dx$.

I think the first one is either 0 or 1 but I'm not sure.
The second one obviously depends on $ a$: I got $ 0$ for $ a>0$, $ \frac{\pi}{2}$ for $ a\equal{}0$, and $ \pi$ for $ a<0$.  Yes, no?

for the second problem, we have $ \lim_{n \to \infty} \int_{a}^{\infty} \frac {n}{1 \plus{} n^{2} x^{2}} dx \equal{} \lim_{n \to \infty} \frac {\pi}{2} \minus{} \arctan {an} \equal{} L$.
This last limit, we have, for $ a <0, L \equal{} \pi$, for $ a \equal{}0, L \equal{} \frac{\pi}{2}$ and for $ a >0, L \equal{} 0$
Is this right?
EDITTED.

For the first one, we show that $ \lim_{n\to\infty} \int_{0}^{\infty} \frac{nx}{(1\plus{}x)^n} \, dx \equal{} 0$. But this easily follows from the direct calculation $ \int_{0}^{\infty} \frac{nx}{(1\plus{}x)^n} \, dx \equal{} \frac{n}{(n\minus{}1)(n\minus{}2)}$. Then we have

$ \left| \int_{0}^{\infty} \frac{\sin \left( \frac{x}{n} \right)}{\left( 1 \plus{} \frac{x}{n} \right)^n} \, dx \right| \equal{} \left| \int_{0}^{\infty} \frac{n \sin t}{\left( 1 \plus{} t \right)^n} \, dt \right| \leq \int_{0}^{\infty} \frac{n t}{\left( 1 \plus{} t \right)^n} \, dt$

and the rest follows from Squeezing Theorem.


Let $u_1=2<u_2<...$ be sequence of all natural square-free numbers. Prove that:

$\limsup _{n \longrightarrow \infty}(u_{n+1}-u_n)=\infty$

Check out [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48897]this topic[/url]. Ravi's proof can easily be applied to any $k$ to show that there is a square-free number $t=t(k)$ s.t. $t+1,t+2,\ldots,t+k$ are not square-free, and the desired result follows immediately.


floor function
Diophantine equation
number theory proposed
number theory

Do there exist $x,y>1$, both irrational, s.t. $\lfloor x^m\rfloor\ne\lfloor y^n\rfloor,\ \forall m,n\in\mathbb N^*$?

Yes, there exist. This is one of my favourite problems. Take x and y of the form $ a+\sqrt{a^2-1} $. After that it is easy. I know they gave in he official solution a surprising answer: $\sqrt{2}$, $\sqrt{6} $ if I remember corectly. Beautiful problem.

I remember giving an example of the form $a+b\sqrt c,\ d+e\sqrt f$ s.t. $a^2-cb^2=d^2-fe^2=1$, and then using some Pell equation, but I can't remember the details right now.


floor function
modular arithmetic
number theory unsolved
number theory

Hello,

Prove that if  $ n \in \mathbb{N}^{*}$  then  $ 6 \nmid \lfloor \big(\root3\of{28}\minus{}3\big)^{\minus{}n}\rfloor$

[quote="linda2005"]Hello,

Prove that if  $ n \in \mathbb{N}^{*}$  then  $ 6 \nmid \lfloor \big(\root3\of{28} - 3\big)^{ - n}\rfloor$[/quote]

Unfortunately, this is wrong, according to me.

Let $ a=\frac{1}{\root3\of{28}-3}$ We want to prove that $ [a^n]\neq 0\pmod 6$

We have $ (\frac{1}{a}+3)^3=28$ and so (developing this) $ a^3=27a^2+9a+1$

Then consider $ u_n=\{\frac{a^n}{6}\}$ (where $ \{x\}$ is the fractional part of $ x$)

We have $ \frac{a^n}{6}=27\frac{a^{n-1}}{6}+9\frac{a^{n-2}}{6}+\frac{a^{n-3}}{6}$

And so ${ u_n=\{27u_{n-1}}+9u_{n-2}+u_{n-3}\}$

And it's very easy to compute the first terms of this sequence with any basic spreadsheet (since all terms are in $ [0,1)$) :
$ u_{0}=0.1666666$
$ u_{1}=0.5551065$
$ u_{2}=0.4939759$
$ u_{3}=0.4999755$
$ u_{4}=0.5002285$
$ u_{5}=0.4999254$
$ u_{6}=0.5000190$
$ u_{7}=0.5000724$
$ u_{8}=0.5020522$
$ u_{9}=0.5560822$
$ u_{10}=0.0327628$

And this shows that $ \frac{a^{10}}{6}=k+0.03276...$ and so that $ a^{10}=6k + 0.1965...$ and so $ 6 | [a^{10}]$

Normally, this should need a demonstration including upper bounds and lower bounds estimation but for only 10 terms of a so simple serie, there is, IMHO, no doubt.

So, if anybodu has demonstrated the original question, I would be very interested in the demo.

[quote="pco"] Normally, this should need a demonstration including upper bounds and lower bounds estimation but for only 10 terms of a so simple serie, there is, IMHO, no doubt.[/quote]

Hemmm, maybe a doubt exist : with the 27 coefficient, $ u_{10}$ involves something about $ 27^7u_2$ and so, an imprecision of $ 27^{\minus{}7}$ on $ u_2$ may lead to a big error on $ u_{10}$ ...

To be continued.

Hemmmm  :blush:  . After some thoughts, the doubt is higher and higher. And here is a demo of the initial problem (which is ok, so ...).

Let $ a \equal{} \frac {1}{\sqrt [3]{28} \minus{} 3}$. We have $ (\frac {1}{a} \plus{} 3)^3 \equal{} 28$ and so $ a$ is a root of $ x^3 \minus{} 27x^2 \minus{} 9x \minus{} 1 \equal{} 0$

Notice that $ a \equal{} \frac {1}{\sqrt [3]{28} \minus{} 3}$ $ \equal{} \sqrt [3]{28^2} \plus{} \sqrt [3]{28*27} \plus{} \sqrt [3]{27^2}$ $ > 3\sqrt [3]{27^2}$ $ \equal{} 27$

Let $ z$ and $ \overline z$ the two other roots (both complex).
We have :
$ E1$ : $ a \plus{} z \plus{} \overline z \equal{} 27$
$ E2$ : $ az \plus{} a\overline z \plus{} z\overline z \equal{} \minus{} 9$
$ E3$ : $ az\overline z \equal{} 1$

Let the sequence $ u_n \equal{} a^n \plus{} z^n \plus{} \overline z^n$

$ u_0 \equal{} 3$
$ u_1 \equal{} a \plus{} z \plus{} \overline z \equal{} 27$ (see $ E1$ above)
$ u_2 \equal{} a^2 \plus{} z^2 \plus{} \overline z^2$ $ \equal{} (a \plus{} z \plus{} \overline z)^2$ $ \minus{} 2(az \plus{} a\overline z \plus{} z\overline z)$ $ \equal{} 27^2 \plus{} 18 \equal{} 747$ (see $ E1$ and $ E2$ above).

We also have $ u_{n \plus{} 3} \equal{} 27u_{n \plus{} 2} \plus{} 9u_{n \plus{} 1} \plus{} u_n$ $ \forall n\geq 0$

From this, we can easily conclude that $ u_n$ is always an integer and that $ u_n \equal{} 3\pmod 6$

We also have $ |z^n \plus{} \overline z^n|\leq 2|z|^n \equal{} 2a^{ \minus{} \frac {n}{2}}$ (since $ |z|^2 \equal{} z\overline z \equal{} \frac {1}{a}$, from $ E3$ above).

So we have $ a^n \equal{} u_n \plus{} \epsilon_n$ with $ u_n$ integer, $ u_n \equal{} 3\pmod 6$ and $ |\epsilon_n| < \frac {2}{27^{\frac {n}{2}}}$ $ < \frac {2}{5^n}$

Hence the result : $ \forall n\in\mathbb{N}^*$, either $ [a^n] \equal{} u_n \equal{} 3\pmod 6$, either $ [a^n] \equal{} u_n \minus{} 1 \equal{} 2\pmod 6$


probability
limit
logarithms
probability and stats

$ S$ is the set of $ n\times n$ matrices with entries $ 1,2, \ldots ,n^2$, each appearing only once. Taking an element $ M$ in $ S$ at random (uniform distribution), we denote by $ p_n$ the probabilty that $ \det M \neq 0$.
Prove or disprove $ \lim_{\infty} p_n \equal{} 0$

Actually, $ p_n\to 1$ :) Her is a related problem that may help a bit. Let $ x\equal{}(x_1,x_2,\dots x_n)$ be a random rearrangement of some $ n$-tuple of pairwise distinct positive real numbers. Let $ V$ be a $ n\minus{}k$-dimensional subspace of $ \mathbb R^n$. Prove that the probability that $ x\in V$ is at most $ \frac 1{k!}$. (This takes care of JC_math's problem almost up to the very end; all you need to do after that is to show that the same probability for $ n\minus{}1$-dimensional $ V$ can be also bounded by something tending to $ 0$ as $ n\to\infty$, which should not be too hard).

fedja,
Do you know how to estimate $ p_n$?

The elementary approach I outlined above should give something like $ p_n>1\minus{}\frac {\log n}n$, but it is a very weak bound. I expect the difference $ 1\minus{}p_n$ to decay at least exponentially. If you are interested in really good estimates, you may want to look at some recent papers by Tao and Wu, or, even better, by Rudelson and Vershynin concerning random matrices and see if their techniques can be modified to give the desired result. Actually, your problem seems to be simpler than the one they usually deal with (matrices consisting of randomly chosen $ \pm 1$), so I cannot exclude the possibility that there is a neat short solution yielding a really good estimate; I just don't see any at the moment :).

[quote="fedja"]If you are interested in really good estimates, you may want to look at some recent papers by Tao and Wu, or,[/quote]

Do you mean Tao and [url=http://www.math.rutgers.edu/~vanvu/]Vu[/url]?

Yes, of course. Sorry for misspelling  :blush:


group theory
abstract algebra
superior algebra
superior algebra unsolved

:)  A & B are normal subsets of some Sylow-p subgroup P of a finite group G.Prove that A & B are conjugate in G if and only if they are conjugate in N(P).

[quote="Subhajit"]I am eagerly waiting for a solution to the following problem.A & B are normal subsets of some Sylow-p subgroup P of a finite group G.Prove that A & B are conjugate in G if and only if they are conjugate in N(P).[/quote]

[quote="dphexen"]what is the definition of "normal subset" ?[/quote]
A subset $ A$ is normal in $ P$ if $ uau^{ \minus{} 1} \in A$ for all $ a \in A$ and $ u \in P$. For finite groups, this is the same as saying $ uAu^{ \minus{} 1} \equal{} A$ for all $ u \in P$.

[hide="Proof:"]Let $ g \in G$ such that $ gAg^{ \minus{} 1} \equal{} B$. As $ A$ is a normal subset of $ P$, $ B \equal{} gAg^{ \minus{} 1}$ is a normal subset of $ gPg^{ \minus{} 1}$. Therefore, $ P, gPg^{ \minus{} 1} \leq N_G(B)$. As $ P$ is a Sylow $ p$-subgroup of $ G$, both $ P$ and $ gPg^{ \minus{} 1}$ are Sylow $ p$-subgroups of $ N_G(B)$ and so there exists $ h \in N_G(B)$ such that $ P \equal{} hgPg^{ \minus{} 1}h^{ \minus{} 1}$, i.e., $ hg \in N_G(P)$. Finally, $ hgAg^{ \minus{} 1}h^{ \minus{} 1} \equal{} hBh^{ \minus{} 1} \equal{} B$, and so $ A$ and $ B$ are $ N_G(P)$-conjugate. [/hide]


Radius of the circle circumscribed on the triangle about angles $ \alpha, \beta, \gamma$ is equal 4, and the radius of the circle written down into this triangle is equal 2. Prove, that

$ \frac {1}{sin{\alpha} sin{\beta}} \plus{} \frac {1}{sin{\beta} sin{\gamma}} \plus{} \frac {1}{sin{\gamma} sin{\beta}} \equal{} 4$

Do you mean that $ R\equal{}4$ and $ r\equal{}2$?.
Well,
\[ \frac{1}{\sin{\alpha}\sin{\beta}}\plus{}\frac{1}{\sin{\beta}\sin{\gamma}}\plus{}\frac{1}{\sin{\gamma}\sin{\alpha}}\equal{}\frac{4R^2}{ab}\plus{}\frac{4R^2}{bc}\plus{}\frac{4R^2}{ca}\equal{}\\
\equal{}\frac{4R^2(a\plus{}b\plus{}c)}{abc}\equal{}\frac{8pR^2}{4prR}\equal{}\frac{2R}{r}\]
so it equals 4.
P.S. Anyway, because of $ R\equal{}2r$, the given triangle is equilateral :lol:


probability
MATHCOUNTS
geometry
LaTeX
percent
floor function
3D geometry

It's in all the other forums so why not bring it here?

PROBLEM 1
In math class, you have taken five tests. Your average grade on those tests is a
91.4%. Your first four test grades were 88%, 90%, 100%, and 89%. What was your fifth
test score?

cool, classroom marathon.

[hide]
91.4*5 - 88 - 90 - 100 - 89 = [b][u]90[/u][/b]
[/hide]

Problem II:

What is the next number in this sequence:

$1, 8, 27, 64, 125, ...$

[hide]the first term is 1 cubed, then 2 cubed, then 3 cubed, then 4 cubed, so the answer is 6^3 which is 216[/hide]
PROBLEM III
what is the next term in the sequence:
1,1,2,3,5,8,13...

[hide]friendly fibonacci! every term is the sum of the two preceding terms.  13+8 = 21[/hide].

[b][b]Problem 4[/b][/b]

Most ships that have traveled for long enough have barnacles attached to their bottoms. The captains of 4 such ships are talking about their barnacles; captain A has $a$ barnacles on his ship, captain B has $b$ barnacles, captain C has $c$ barnacles, and captain D has $d$ barnacles.  Research tells the captains that the number of barnacles on their ships satisfy the following equations:

$2a + 3b + 4c + d = 37$
$3a - 2b + 7c - d = 40$
$-a + c + 5d = 43$
$3a + 11c + 4d = 46$

What is the total number of barnacles on the ships?

You know, you could just make the problem just a little easier by not making 4 variables.

hint:  its very easy if you don't attempt to actually solve for the variables.

I want classroomers to break out of the habit of solving systems of equations when you are looking for things other than the individual variables (e.g. the sum of the variables).

try to manipulate the equations so that stuff cancels out to just the sum of the variables.  once you figure out the answer you'll realize that this problem is just as hard (or easy?) for classroomers as it is for anyone else  :)

Oh I got it!

Take the sum of the first three equations and subtract the fourth from it. that gives us an equation a + b + c + d = 74.

[size=200]Problem 5[/size]

If it is Saturday on January 1st, 2005, what would the day be on the second of January in 2009?

[hide]Friday[/hide]

[b]Problem #6[/b]

In how many different ways can 40 be written as a sum of two positive prime numbers? (a+b and b+a are the same)

[quote="star99"][hide]Friday[/hide][/quote]
Please show your work in the future.

[quote="star99"][b]Problem #6[/b]

In how many different ways can 40 be written as a sum of two positive prime numbers? (a+b and b+a are the same)[/quote]

[hide]3, 37
11, 29
17, 23

3 ways[/hide]

[size=200][b]Problem 7[/b][/size]

What is the sum of x and y satisfying the equations

3x - 4y = -19
2x + 8y = 94

[hide]
Solve the system of equations by solving for x in the second one.
$2x+8y=94$
$x=47-4y$
plug that in the the first equation.
$3(47-4y)-4y=-19$
$y=10$
Plug ten into the equation.
$2x+8(10)=94$
$x=7$
the sum of 7 and 10 is
$\boxed{17}$
[/hide]
PROBLEM 8
simplify:
(x + 1) + (x - 2) + (x + 3) + (x - 4) + … + (x + 99) + (x - 100)

[quote="math92"]
PROBLEM 8
simplify:
(x + 1) + (x - 2) + (x + 3) + (x - 4) + … + (x + 99) + (x - 100)[/quote]
[hide]
In this sequence, you can add together any two adjacent pairs of parentheses and get 2x-1. Since there are 50 pairs, multiply 2x-1 by 50 to get the answer, $100x-50$.[/hide]

$Problem$ $9$

Find the next three terms in this sequence

1, 1, 1, 2, 3, 7, 23, ..., ...

How many odd numbers with middle digit 5 are there between 40000 and 69999 inclusive, with no digits repeating?

If this is not considered 'classroom' math, someone else post the next problem.

[quote="me@home"][hide]There is one choice for the middle digit, times 3 choices for the first digit, times 8*7*6 digits for the rest
So we have: 1*3*8*7*6=168*6=960+48=[color=orange]1008[/color]
ooops... didn't read the odd part...
but since the range is from an even number inclusive to an odd, exactly half will be odd so its just:
1008/2=[color=green]504[/color] (real answer)
[/hide][/quote]
Hmmm..i got 336..
First we must consider the middle and last digits.
The middle digit is five. The last digit is odd. So we have
_*_*1*_*4
Since it is in bewteen 40000 and 69999, and 5 is already used, there are two choices. Then there are 7 choices for the second and 6 choices for the 4th. (we must count 0 too)
So we have
2*7*1*6*4=336

PROBLEM 258: What is the probability that if i roll a pair of dice and my friend also rolls a pair, we will get prime numbers on all of our dices, and I will have a greater product than him?

Still above Classroom, this is a Classroom level problem

In how many ways can at least one student from a group of four be selected for class honors if any combination of student(s) can be selected?

so what exactly would be classified as "classroom math"? quadratic formula/ polynomials/ binomial expansion? which?

Binomial expansion and the quadratic formula might be too hard (depends on how they are used in the problem). I'd guess that polynomials with no more than three terms are ok.

meh.. my problem isn't that hard. u could always list the possibilities, if its above classroom level.

[quote="mathcountsfreak"][hide=" PROBLEM 258"]For Ignite 168's, I did some case counting.
If one student is chosen, there are four possibilities (4 students)
If 2 are chosen, there are 6 (for $\binom{4}{2}$ (4 choose 2) because order (as I interpreted it) does not matter).
If 3 are chosen, $\binom{4}{3} = 4$.
If 4 are chosen, $\binom{4}{4} = 1$.
4+6+4+1=[color=red]15[/color][/hide][/quote]
They asked for probability. :)

Since Ignite said that that was a hard problem,

[color=indigo]PROBLEM 259[/color]

Solve-- $x^2+3x+2>0$

[quote="mbabbitt2003"]Prob' 260:

Mr. A, Mrs. B, Ms. C and Mas. D are standing in a line, wish respective positions. Mrs. B steals 5 dollars from Mr. A, but C steals 3 dollars from B. D steals one dollar from C. Let A, B, C and D, respetively, be the amount of money that Mr. A, Mrs. B, Ms. C and Mas. D have. If the following conditions are met:

$A+B+C+D=40$
$\frac{B}{D+C}=1$
$A-B+C-D=-8$
$D=10$
$2B-C=24$

find the amount of money that Mr. A used to have.[/quote]

The equations are after the stealing, right?
[hide] Adding the first and third equations we get $A+C=16$. From the second equation, we get $B=D+C$ or $B=C+10$. Substituting this in for $B$ in the first equation, we get $A+2C=20$. Subtracting $A+C=16$, we know that $C=4$. From the last equation, we know that $B=14$. Therefore, $A=12$. Since Mrs. B stole 5 dollars, Mr. A originally had $\boxed{\$17}$. :D [/hide]


I've already posted this in the high school basics forum.. but no answers yet..

"In the accompanying figure, the smaller circle is tangent to the larger circle. Ray [i]PQ[/i] is a common tangent and ray [i]PR[/i] passes through the centers of both circles. If the radius of the smaller circle is [i]a[/i] and the radius of the larger circle is [i]b[/i], show that $\sin{x}= \frac{b-a}{b+a}$"

O_O I have no clue..

[hide]
Let the small circle have center A and the large circle have center B. From A and B draw radii to the tangency points; the point of tangency in circle A is C and the tangency point in circle B is D.

draw a line from C parallel to PR and have it intersect BD at E. ABEC is a parallelogram. Opposite sides of parallelograms are congruent, so $CE=a+b$ and $BE=a\implies ED=b-a$. Since $CE\parallel AB$, $m\angle DCE=x$. Since $\angle CDE$ is right, the sine of x is $\frac{b-a}{b+a}$
[/hide]

Let $O,\ O'$ be the centers of the circles with radii $a,\ b$ respectively and similarlly $T,\ T'$ be the tangent points of the two circles and the ray $PQ$ respectively.Draw the  line passing through $O$ and parallel to the ray $PQ$.Denote the intersection point of $OH$ and $O'T'.$ Remark $\triangle{OO'H}$ is a right triangle with $\angle{OHO'}=90^\circ,$ we have $\sin x=\frac{OO'}{O'H}=\frac{b-a}{a+b}.$

tjhance, Sorry, you have posted the solution. I didn't look at your posting.

kunny


geometry
rectangle
analytic geometry
combinatorics proposed
combinatorics

We have an mxn table (m<n) we number the squares in the usual manner. the unit squares are crossed starting from number 1 in a spiral form in the clockwise direction.¿what are the values of m and n such that the las square crossed is the one with number 2005?

If I understood the conditions properly, the squares are $1, 2, ..., n$ in the first row, $n+1, ..., 2n$ in the second, ..., $(m-1)n+1, ..., mn$ in the mth, and the way we go over the squares is $1, 2, ..., n, 2n, ..., mn, mn-1, ..., (m-1)n+1, (m-2)n+1, ..., n+1, ..., 2005$, right?

It is clear that our path can be broken up into several rectangles, each containing the next (the first rectangle is formed by the squares from $1$ to $n+1$ as above, and then we're left with a $(m-2) \times (n-2)$ board, where we can get a rectangle in the same fashion, and so on). If $m=2k$, it is easy to see that the lengths of the rectangle sides are $(m, n)$, $(m-2, n-2)$, ..., $(2, n-m+2)$. The process will finish when we get to the lower-left corner of the $(2, n-m+2)$ rectangle. Now if we assign number $i+j$ to the number on the ith row and jth column, it's easy to see the lower-left corners of the rectangles are on a diagonal, e.g., their assigned numbers are equal. Also since the lower-left corner of the $(m, n)$ rectangle is on the mth row and first column, it has number $m+1$ assigned, so the coordinates of the finishing square satisfy $i+j=m+1$. Also, since in this case $m=2k$, it is easy to see that it is on the (k+1)th row, so $k+1+j=2k+1, j=k$, then the square where we finish is on the (k+1)th row and kth column.

By the method of numbering the squares, the square on the ith row and jth column had number $n(i-1)+j$. Then $n(k+1-1)+k=2005, nk+k=2005, k(n+1)=2005$. Since $n>m=2k$, and $2005=5 \times 401$, we have either $n+1=2005, k=1 \Rightarrow n=2004, m=2$ or $n+1=401, m=5 \Rightarrow n=400, m=10$.

If $m=2k+1$ it's similar. It's easy to see the ending square is the right-end of a $1 \times (m-n+1)$ rectangle. This one has the same coordinate sum as the upper-right corner of the board, that is $i+j=n+1$, and it is clearly on the k+1th row. So $k+1+j=n+1, j=n-k$. Then $n(k+1-1)+n-k=2005, nk+n-k=2005$. Here we rearrange and factor obtaining $(n-1)(k+1)=nk+n-k-1=2004=4 \times 501$. Since $n>2k+1$, we have either $n-1=501, k+1=4, n=502, m=7$ or $n-1=1002, k+1=2, n=1003, m=3$ or $n-1=2004, k+1=1, n=2005, m=1$.

very nice¡¡ your solution looks so elegant¡¡¿ it´s pretty much the same thing i did but mine has like 20 million diferent variables and very disorganised (all the solutions are scattered throughout 5 pages of calculations) but i got all the solutions or most of them at least  :blush:  ill get about 3 or 4 points on it.


function
ratio
calculus
integration
geometry
geometric transformation
dilation

Given an arithmetic sequence $ \{ a_i \}_{i\equal{}1}^{2008}$, and given that $ a_{1000} \plus{} a_{1008} \equal{} a_{2008}$, find the number of positive integer triples $ (p,q,r)$ such that $ a_p, a_q, a_r$ is an increasing geometric sequence.

Solution?

[hide]We rewrite the equation as $ a_1\plus{}999d \plus{}a_1\plus{}1007d\equal{} a_1\plus{}2007d$.  Solving yields that $ a_1\equal{}d$.  I'm guessing that we can assume all of them are positive and that $ a_1$ can equal 1.  In general, $ a_p, a_q$, and $ a_r$ can be written as $ k, kr$, and $ kr^2$.  If $ k\equal{}1$, then $ r^2 \le 2008$.  The greatest square that satisfies this is $ 44^2\equal{}1936$.  We have $ 44$ solutions already, but we can still change k.  If $ k\equal{}2$, then $ r^2 \le 1004$, and we have $ 31$ possibilities for r.  Continuing this sum yields a bunch of values.  Eventually, they sum up to $ 515$ (hopefully).

It looks like the floor function should be used, but I don't know how to apply it.[/hide]

[quote="azn6021023"]Solution?

[hide]We rewrite the equation as $ a_1 \plus{} 999d \plus{} a_1 \plus{} 1007d \equal{} a_1 \plus{} 2007d$.  Solving yields that $ a_1 \equal{} d$.  I'm guessing that we can assume all of them are positive and that $ a_1$ can equal 1.  In general, $ a_p, a_q$, and $ a_r$ can be written as $ k, kr$, and $ kr^2$.  If $ k \equal{} 1$, then $ r^2 \le 2008$.  The greatest square that satisfies this is $ 44^2 \equal{} 1936$.  We have $ 44$ solutions already, but we can still change k.  If $ k \equal{} 2$, then $ r^2 \le 1004$, and we have $ 31$ possibilities for r.  Continuing this sum yields a bunch of values.  Eventually, they sum up to $ 515$ (hopefully).

It looks like the floor function should be used, but I don't know how to apply it.[/hide][/quote]
You seem to be assuming that $ r$ is an integer.  Must this be the case?

[quote="Cronus"]We are looking for an integer r.So, why won't we be able to assume that r is an integer?[/quote]
Arg, naming issues.  azn6021023's solution involved the ratio between terms in the geometric sequence.  To avoid confusion, let's call this $ R$.

Azn demonstrated that $ a_{n} \equal{} nd$; we are looking, therefore for triplets $ \{j, jk, jk^{2}\}$, such that $ a_{j} \equal{} jd, a_{jk} \equal{} jkd, a_{jk^{2}} \equal{} jk^{2}d$,  where $ j,k$ are necessarily positive integers ($ d$, on the other hand, just has to be positive).

Oh OK.But I stil don't get it completely.I thought those were the same r.He was determining the number of "r"s through that R wasn't he?So, why aren't they the same? :huh:  :?:

8,12,18,27 is an increasing geometric sequence whose terms are integers, but the ratio is not an integer

As Altheman explained, the geometric series does not need to have an integral rate.  Thus, my solution was incomplete.

[hide]Notice that if $ k$ will be multiplied by a fractional growth rate, then it must be divisible by the denominator of $ r$.  Thus, we rewrite $ k\equal{}ab^2$ and $ d\equal{}c/b$ so that our three numbers are now $ ab^2, abc, and ac^2$.  Fortunately, we only have to check squares in this new sequence.  We realize that $ b\neq1$ because that case was covered in the previous portion of the solution.  I think the easiest way to count comes from the $ abc$ term with $ c\le sqrt{2008/a}$.  We also know that b and c are relatively prime, since they represent a fraction and could otherwise be reduced.  As stated earlier, $ c$ has a maximum of 44 different values, but we only need to count 42 values since $ 1/b$ and $ 2/b$ won't form an increasing geometric sequence for $ b\ge2$.  Counting all possible b values per c value...Okay, I have no good way of calculating the values that are relatively prime to  a given number.  I need help.  But from there, you basically then calculate possible dilations provided that a is relatively prime to b.  Sorry for the new incomplete solution.[/hide]


On each unit square of a $9 \times 9$ square, there is a bettle. Simultaneously, at the whistle, each bettle moves from its unit square to another one which has only a common vertex with the original one (thus in diagonal). Some bettles can go to the same unit square.
Determine the minimum number of empty unit squares after the moves.

Pierre.

I promise I won't spoil any more of these, but this problem is too cute :D:

Give the squares coordinates: from $(1,1)$ to $(9,9)$. The squares with both coordinates odd can only be occupied by beetles from squares with both coordinates even. There are $21$ squares with both coordinates odd and $12=21-9$ squares with both coordinates even, meaning that at least $9$ squares remain empty. It's very easy to find a set of movements leaving exactly $9$ squares empty, so the answer is $9$.

Here is a solution from a 12-year old student...
[hide]
Instead of counting the empty squares, let's count the squares that have two bugs.

Begin by coloring the 9x9 square like a checkerboard, with black squares in the corners.  Bugs on white squares have to stay on white squares.  Bugs on black squares have to stay on black squares.

The bugs on the white squares can find a place where they don't share a square.  (She demonstrated two proofs of this fact.)  Now, she considered the black squares.

First, the bugs at the corners have only one place to move.  Remember these four squares are occupied.  Now, look at the left columns of bugs.  There are still 3 bugs in the left column, but there are only 2 empty squares in the second column.  No matter what they do, two bugs will have to share a square.  Do the same for all four sides.  We have at least 4 squares occupied by two bugs.

Do the same thing with what's left [meaning the 5x5 square, since the second column from the left, second row from the top, second row from the right, and second row from the bottom are all full of bugs.]  The bugs in the corners [of the 5x5 square] have only one (empty) place to move (they move toward the center).  This leaves the other bug in the third column with no (empty) place to move.  Do the same for all four sides.  We have 4 more squares occupied by two bugs.

Next, look at the middle square.  This poor bug has to move to a square that's already full.

The answer is 9.[/hide]
This was a very nice little problem, Pierre!

Thanks you. I agree.
But, unfortunately, I'm not the creator of it. I just proposed it here from our pre-TST.

Pierre.

colouring in 4 colours makes the solution show up easily (its equivalent to all other solutions here actually)


How can i create a title for an article using TexnicCentre?
I can't find it in Latex Help...

To create a title put the command \title{ [Put title here] } before the \begin{document} command.  Then after the \begin{document} put \maketitle.

It will automatically insert a date so if you do not want the date put \date{} before the \begin{document} and it will leave it blank (you can also use that command to insert a date of your choice).

So, for example, your document might look like this:

[code]
\documentclass[11pt]{article}

\title{My Title}

\begin{document}
\maketitle

This is just a test document!
\end{document}

[/code]

Or you can place this piece of code wherever you want the title:
[code]
\begin{center}
\Large Title Here
\end{center}
[/code]

I don't recommend doing it the way Val suggested. In general, if you have a choice between using a built-in macro and doing the spacing commands on your own, take the built-in one. It probably knows how to do it better than you do.

[quote="shobber"]How can i create a title for an article using TexnicCenter?
I can't find it in Latex Help...[/quote]

TeXNicCenter is just an editor that has some built-in facilities, but most of the commands you would have to know them by yourself. Something you can do is to read the "Not so short introduction to Latex2e", available in the web as lshort.pdf or lshort.dvi, and also available in basically any distribution of latex. In my windows partition I have MikTeX installed, and this document is in texmf\sources compressed and texmf\docs\guides (In my case, the default, the folder texmf is in C:\) and I don't know why I have it in nine different languages! (I swear, I would like, but I don't speak that many)

It is a great reference, and I would say that every beginner should read it at least once almost complete. There are way better books in the market about Latex that work perfectly as reference, but not freely available. I really recomend it.

Best regards,


calculus
integration
derivative
logarithms
calculus computations

[hide="hint 1"]
Just use integration by parts, there's only really 2 choices for $ g(x)$ and the right one is obvious.
[/hide]

[hide="hint 2"]

Recall that the derivative of a^x is a^x ln(a). Integrate both sides, and factor out ln(a) (since it's a constant) to derive the integration rule you'll use to solve the
problem.
[/hide]

The $ 2^x$ may be distracting you- just write it as $ e^{\ln 2\cdot x}$, and note that we can integrate or differentiate it with equal ease. Have you seen $ \int xe^x\,dx$ before?

Thanks!
The reason I was stuck is I couldn't get the integral of (2^x), BTW how did you obtain the interal of (2^x)?

I explained it in the hint, but wtv.

$ (2^{x})' \equal{} 2^{x}\ln(2) \iff 2^{x} \equal{} \int 2^{x}\ln(2) \; dx$ now since $ \ln(2)$ is constant, we can factor it out.

$ 2^{x} \equal{} \ln(2)\int 2^{x} \; dx \iff \frac{2^{x}}{\ln(2)} \equal{} \int 2^{x} \; dx$

Replace $ 2$ by $ a$ for the general case and you're done.


geometry
parallelogram
geometry proposed

$AD, BE, CF$ are the altitudes of $\triangle ABC$. $K,M,N$ are the orthocenters of $\triangle AEF, \triangle BDF, \triangle CDE$ respectively. Prove that $\triangle KMN$ and $\triangle DEF$ are congruent.

Let $H$ be the orthocentre of $ABC$. It is obvious that $FM \parallel HD$ and $FH \parallel MD$, hence $DHFM$ is a parallelogram. Similarly, $DHEN$ is a parallelogram. Hence $FM = EN$ and $FM \parallel EN$, so $EFMN$ is a parallelogram and $EF = NM$. Similarly, $FD = KN$ and $DE = MK$ so we are done.


abstract algebra
Ring Theory
projective geometry
advanced fields
advanced fields unsolved

Hi,

if X is a scheme over Y, an invertible sheaf L on X is said to be very ample if there is an immersion $i: X\to P^{r}_{Y}$ for some r such that L is isomorphic to $i^{*}(O(1))$. Now let L be a very ample sheaf on X and $j: X\to P^{n}_{Y}$ an immersion for any n. Will L be isomorphic to $j^{*}(O(1))$?

My question arises from the proof of Theorem II/5.17 in Hartshorne. There, X is projective over a noetherian ring A, L is a very ample sheaf on X. He chooses a closed immersion $i: X\to P^{r}_{A}$ and says that L is isomorphic to $i^{*}(O(1))$. Why is he allowed to assume that the dimenions from the definition of a projective scheme and from the definition of a very ample sheaf are the same in this case?
Thanks.

Take $Y=P^{1}$, $L=\mathcal O(1)$, and choose the $n$-uple embedding $j: \mathbb P^{1}\to\mathbb P^{n}$. The invertible sheaf $j^{*}(\mathcal O_{\mathbb P^{n}}(1)$ is isomorphic to $\mathcal O_{Y}(n)=L^{\otimes n}$, so it is not $L$.

I don't have that proof you are refering to in front of me, but my guess is that he chooses $r$ and the immersion $i: X\to\mathbb P^{r}$ that [b]corresponds[/b] to the very ample sheaf $L$.

Hi, thanks for your response. Could you please give me the definition of the n-uple embedding?

My problem is: if he choses the immersion that corresponds to the very ample sheaf (maybe I don't understand what you mean by that) why is he allowed to assume that it is a closed immersion?
On the one hand I have the immersion from the definition of a very ample sheaf, on the other hand I have the closed immersion from the definition of a projective scheme, which may go to a projective space of a different dimension. Why is he allowed to assume that they are the same?

Any morphism from a proper scheme to a separated scheme (both over some base) is itself proper. In particular, any immersion from a projective scheme into a projective space is actually a closed immersion. The definition of "very ample" does not make any assumptions on $r$, it is just required that there exists an (non-unique) embedding into some projective space such that $\mathcal L=i^{*}\mathcal O(1)$. For the n-uple embedding, consult the index, letter "U". owk

Hi owk,
thanks for your response.
I've proven the following statement: Let f:X->S be universally closed, g:Y->S be seperated and h:X->Y be a morphism with g o h = f. Then h(X) is closed in Y.
So in the definition of a very ample sheaf Hartshorne actually means "an immersion of schemes over Y" and then I can apply the statement above to this situation, correct?

Yes. (If you insist, one could argue that X is declared to be a scheme over Y, hence "immersion" means "in the category of schemes over Y".) owk


calculus
integration
function
limit
real analysis
real analysis unsolved

Let $ f: \mathbb [a,b]\to\mathbb [0,1]$ a strictly increasing function. Prove that:

$ \begin{document}
\lim_{n \rightarrow \infty} \sqrt[n] { \int_a^b { {f}^{n}(x)dx }} = f(b-0)
\end{document}$

More generally:

Say that $ f\in L^p\cap L^{\infty}$ for some $ p\geq 1$.  Then
\[ \lim_{p\to\infty} \|f\|_p \equal{} \|f\|_{\infty}
\]


Find all strictly increasing functions $f: \mathbb{N}\to \mathbb{N}$ such that \[f(f(n))=3n.\]

first prove that $ f(3^{n}) \equal{} 2\times 3^{n}$ for $ n\geq 0$...
then we have:
$ f(f(3^{n})) \equal{} f(2\times 3^{n}) \equal{} 3^{n\plus{}1}$
thus:
$ \boxed{f(3^{n}) \equal{} 2\times 3^{n},f(2\times 3^{n}) \equal{} 3^{n\plus{}1}}$

now note that there are $ 3^{n}\minus{}1$ numbers $ k$ such that $ 3^{n}< k < 2\times 3^{n}$.
and also there are $ 3^{n}\minus{}1$ numbers $ t$ such that $ f(3^{n}) \equal{} 2\times 3^{n}< t < 3^{n\plus{}1}\equal{} f(2\times 3^{n})$.
so for $ k$ such that $ 3^{n}< k < 2\times 3^{k}$ and $ k \equal{} 3^{n}\plus{}m$ we have:
$ f(k) \equal{} 2\times 3^{n}\plus{}m$
and also if $ 2\times 3^{n}< k < 3^{n\plus{}1}$ and $ k \equal{} 2\times 3^{n}\plus{}m$ then:
$ f(k) \equal{} 3(3^{n}\plus{}m)$

so we have:

$ \boxed{f(3^{n}) \equal{} 2\times 3^{n},f(2\times 3^{n}) \equal{} 3^{n\plus{}1}}$

$ \boxed{0\leq m\leq 3^{n},k \equal{} 3^{n}\plus{}m: f(k) \equal{} 2\times 3^{n}\plus{}m}$

$ \boxed{0\leq m\leq 3^{n},k \equal{} 2\times 3^{n}\plus{}m: f(k) \equal{} 3(3^{n}\plus{}m)}$

[quote="BaBaK Ghalebi"]first prove that $ f(3^{n}) \equal{} 2\times 3^{n}$ for $ n\geq 0$...[/quote]Is it because I am sleeping, or is this not so obvious?

I don't think I'd call it obvious... the fastest proof I have is this.

First, since $ f$ is strictly increasing, we have $ f(1)\leq 3$. If $ f(1) \equal{} 3$, then $ f(f(1)) \equal{} f(3) > 3$, which is bad. But if $ f(1) \equal{} 1$ then $ f(f(1)) \equal{} 1$, which is bad. This leaves only $ f(1) \equal{} 2$.

By taking $ f$ of both sides of the original equation, we get that $ f(3n) \equal{} f(f(f(n))) \equal{} 3f(n)$. We conclude that $ f(3^{k}) \equal{} 3^{k}f(1) \equal{} 2\cdot 3^{k}$ for all nonnegative $ k$.

or you can just prove it usung induction on $ n$...
first note that $ f(1)\not \equal{}1$ because otherwise we would have:
$ f(f(1))\equal{}f(1)$
and on the other hand we have $ f(f(1))\equal{}3$ so we have $ 3\equal{}1$ contradiction...
so we have:
$ f(1)>1$ (*)
now $ f$ is strictly increasing so for $ x>y$ we have $ f(x)>f(y)$ so from (*) we get that $ f(f(1))>f(1)$ so $ f(1)<3$...
thus we have:
$ 1<f(1)<3$ so we must have $ f(1)\equal{}2$ as MellowMelon said...
so for $ n\equal{}0$ we have $ f(3^{0})\equal{}2\times 3^{0}$ as wanted...
now assume that the statement is true for $ n$ i.e. $ f(3^{n})\equal{}2\times 3^{n}$...
now we want to prove that $ f(3^{n\plus{}1})\equal{}2\times 3^{n\plus{}1}$...
note that we have:
$ f(3^{n})\equal{}2\times 3^{n}\Rightarrow f(f(3^{n}))\equal{}f(2\times 3^{n})\Rightarrow f(2\times 3^{n})\equal{}3\times 3^{n}\equal{}3^{n\plus{}1}$
$ \Rightarrow f(3^{n\plus{}1})\equal{}f(f(2\times 3^{n}))\equal{}3\times 2\times 3^{n}\equal{}2\times 3^{n\plus{}1}$
as wanted...
so by induction our claim is proved...

[quote="Peter"]Find all strictly increasing functions $ f: \mathbb{N}\to \mathbb{N}$ such that
\[ f(f(n)) \equal{} 3n.
\]
[/quote]
and more Find all strictly increasing functions $ f: \mathbb{N}\to \mathbb{N}$ such that

$ f(f(n)) \equal{} pn.$ wher $ p$ is a prime


Thad has an unlimited supply of 3-cent and 4-cent
stamps. If he has to put exactly 37 cents of postage on
a letter, how many different combinations of 3-cent
and/or 4-cent stamps could Thad use?

Alternatively, we want to put on enough 3s to make it a multiple of 4. We clearly need at least 3 so we have 28. Then we cn subtract multiples of 12 to get 16 and 4. So 4,16,28 so 3 ways.

Or, you can see that $37 \div 12 = 3\text{R}1$, where $12$ comes from $4\cdot3$. Since we get $3$, our answer is $\boxed{3}$.


trigonometry
inequalities
inequalities proposed

Prove that :
\[ \sum {x^4y} + \sum {x^2y^2z}\geq \sum{x^3yz} + \sum{x^3y^2} \]

Hi-hi-hi  :lol:
It is
\[(xy+yz+zx)(x^3+y^3+z^3+3xyz-\sum_{cyclic} x^2y)\geq 0.\]

dear Myth , What a nice solution ! Did you use Mapple or Mathematica ,.. ?

[quote="romano"]Did you use Mapple or Mathematica?[/quote]
You missed! Try again ;) :D

Wow, perfect! I start to feel that PNT is just a child while facing you, Myth.
Here's his proof:
Put $ xy+yz+zx=1$. Then put $ x=\tan{\frac{A}{2}},y=\tan{\frac{B}{2}},z=\tan{\frac{C}{2}}$. The inequality reduces to:
$ \sum \tan^2{\frac{A}{2}}+8\prod \sin{\frac{A}{2}} \geq 2$
which is famous

Actually, I'm not that good at inequalities but I got this one.

If it's homogeneous and Muirhead doesn't work, then Schur almost certainly will.  If Schur still doesn't work, start searching for a counterexample.  :lol:


I have a question about the TI-89:  When solving an equation, I got "false" as an answer.  Is this supposed to mean that there are no real solutions, and is there a way to change the mode or something so that it shows the complex solutions as well?

I believe the command CSolve will solve for complex roots.


function
integration
calculus
derivative
algebra
domain
algebra unsolved

[quote="Thales418"]

You write it wrong

${f(1990)=3979}$[/quote]

What is wrong?

Really sorry... the original problem:
Let $f: Z^{+}\to Z^{+}$ a function such:
$f(1)=1$
$f(2n)=2f(n)+1$
$f(f(n))=4n-3$
Calculate $f(1990)$

Thales418, it has no sense to take the derivative of f... Even if f:R->R, it's not always true that f has a derivative on his domain

[quote="Rust"]These conditions is correct. We have $f(1)=1=2*1-1,f(2)=3=2*2-1$.
Let $f(k)=2k-1 \ \forall k<n$. Then if n is even $f(n)=2f(\frac{n}{2})+1=2*(n-1)+1=2n-1$. If n is odd $f(n)=f(f(\frac{n+1}{2}))=4\frac{n+1}{2}-3=2n-1$. It prove $f(n)=2n-1 \ \forall n$.[/quote]
Thanks rust


geometry
geometric transformation
search

or is it a bug terrorizing the satellite camera?

[quote="perfect_radio"]Yes, of course.

From the producers of [i]Snakes on a Plane[/i]... Keanu Reeves... Morgan Freeman... [b]Bugs on a Satellite[/b].
One bug... one desire... Double the action! [i]Get down!!![/i]... Triple the excitement! [i]Get down again!!![/i][/quote]

Someone has been watching stand-up comedy :) (Pablo Francisco)

[quote="kstan013"]i infer that the giant super large man-eating bug is a common, yet deadly, household pet in rural German areas :wacko:[/quote]

1) How do you know it is man-eating?

2) If it is so common, why is there only 1 that big?

[quote="kstan013"]they could keep all the others in an opaque fish tank or inside their barns. maybe this one is crossbred with a giant squid

if i was that big, i would eat people[/quote]

Umm dinosaurs were even bigger and many of them were herbivores.

Also, giant squids don't get that big.

Do you see any opaque fish tanks or barns on the map?

It's sad but the phenomenon is not new at all.  Sometimes spiders terrorize entire meteors: :(

[img]http://www.clg.qc.ca/for/reg/dep/700/actint/2003/tintin/physique_files/image007.jpg[/img]

as can be seen in :

[img]http://i.biblio.com/m/97/2203001097.jpg[/img]

I should write a book on that called [i][u]Attack of the Super Evil Giant Creatures from Outer Space who Like to be Scary and Stuff[/u][/i]

[quote="Xevarion"][quote="fredbel6"]It's sad but the phenomenon is not new at all.  Sometimes spiders terrorize entire meteors: :(
[/quote]
rofl that's excellent. The sad thing is I knew what that was before I saw teh second picture...[/quote]
You read Tintin?

[quote="Xevarion"][quote="fredbel6"]You read Tintin?[/quote]
English translations  :blush:  :blush:  :blush:[/quote]
Well Tintin is Belgian (we Flemings (Dutchspeaking Belgians) call it Kuifje).
A few years ago, two Flemish comedians noted that there are numerous mysterious elements in that album, in fact, on the first pages! :P  :ninja:

[hide][img]http://www.bdnet.com/zi/bdweb/images/09/9782203001091_pg.jpg[/img]
In the second picture, Tintin points at a RISING falling star!
In the third picture, Tintin gets away with an unbelievably shameless act of animal abuse : he kicks Milou against that pole with his right foot (for the viewer) for no reason at all and acts like that's completely normal.

In the fourth picture, not only has Milou developed an extremely long hair protruding from his legs, the drawer proves one more time that he doesn't know anything about astronomy and draws stars around Milou's head.  Do note that the roofs have magically changed in the fifth picture.

The last two pictures are beyond comprehension?  What is going on here?  How can Milou's head be at the same height as Tintin's elbow?There are two theories :
1. Milou has developed an extremely long neck
2.Tintin's legs have disappeared.[/hide]

In the Engish translations they call the dog Snowy!

[quote="mustafa"]That might have been one of the funniest things that I have ever read :rotfl: Have you ever noticed how racist the author of tintin is?[/quote]
Tintin was written in the 1930s. I'm guessing that the audience at the time was significantly more racist and less cosmopolitan than today's.

[quote="mustafa"]That might have been one of the funniest things that I have ever read :rotfl: Have you ever noticed how racist the author of tintin is?[/quote]
Hmmm, please elaborate?  You do have a point  :( , but :
One thing you have to know is that Hergé didn't travel a lot.  Most of the information he used came from books or friends who visited those countries.  In his first works this is quite obvious, they are quite stereotypical about native Americans and black African tribes. When he was gonna make an album about China, students in Leuven urged him to do it right.  Zhang Chongren, a Chinese student, became a close friend, and was Hergé's inspiration for the Chang character, Tintin's Chinese friend in the resulting album : The Blue Lotus.  Later he lost contact with Zhang but found him back, this search inspired Hergé to make "Tintin in Tibet".  In fact, Tintin has a talk with Chang about prejudices about China  in Europe (and the converse too) after Tintin pulls Chang out of the water.  This shows that Hergé realizes his own mistake.
So he did become more accurate and less racist.
The interesting thing is that the first album The Blue Lotus mentioned the Mukden incident in 1931.  The album itself was published in 1936, so Hetgé was one of the few Europeans understanding its significance at that time.  Chang Kai-Shek was so pleased with it, he invited Hergé.
But the story about Tibet did create some controversy with the communist government.

I was talking about the stereotypicalness of the Africans.  :(

I think that Tintin is awesome anyways, I have the book where he goes to the moon in French, but I can read maybe one in three words :rotfl:

[quote="not_trig"]How did a bug get stuck on the satellite lens?[/quote]

a bug really can't get stuck on a satellite lens considering the satellite lenses are in space and the bug would not survive in space :roll:  :D

[quote="K81o7"]Some crazy people probably built that.

Oh yeah, Tintin, I have at least half the books.  Some of the best stuff there.  Dunno what I like more, that, or Asterix...used to read them a lot when i was young.[/quote]

They're not crazy, they're googlers!!! :rotfl:  :rotfl:  :rotfl:

It is pretty good quality, so probably an aerial pic (I think they only do that for populated areas.
Btw, how did you find that bug out of the whole world? (Still, even out of all of germany it is pretty lucky!)

I used to read tintin in english with my friends all the time when I was in elementary school, recently I checked one out of the library and read it for the memories... I think my mom read a version translated into turkish when she was young but I'm not sure :huh:

[quote="mustafa"]:ewpu:  :huh: ????? Googlers? What does that mean? I think I will start using that word in my conversations.

Somebody: Hi.
Me: You must be googlers!
Somebody: What are you talking about?
Me: (with eyes closed, hands over ears, yelling) GOOGLERS! GOOGLERS! GOOGLERS![/quote]

googlers as in "google" the search engine run by those 2 brothers in their garage

hmm 2 brothers in their garage? Google is a multi billion dollar company

Wehn I got the topic reply notification in mozilla thuderbird, it marked the message as possibly spam when it saw the words "Giant Bug Terrorizes Germany". :rotfl:

[quote="fredbel6"][quote="Klebian"]Tintin is quite awesome.

In case anyone was interested:

[hide][img]http://img149.imageshack.us/img149/4325/tintinpk1.png[/img][/hide][/quote]
Klebian, it's not working. :(

Satellite lens?  Don't forget that Google Earth is a combination of aerial pics and pics from space.  So it is possible.[/quote]

That's strange, it's working for me.


There is this problem I have :

Determine the value of the infinite series $ \frac{3^n\plus{}5^n}{8^n}$ from n=0 to $ \infty$.

Also, any sigma "properties" might help.

Thanks a lot.

We can easily show that it converges, as it is bounded above and below be geometric series. Now, we can re-order the summation:

$ \sum_{n\equal{}0}^{\infty} \frac{3^n\plus{}5^n}{8^n} \equal{} \sum_{n\equal{}0}^{\infty} \left( \frac{3}{8} \right)^n \plus{} \sum_{n\equal{}0}^{\infty} \left( \frac{5}{8} \right)^n$

And just to finish off... :D

Using the geometric series formula $ \sum_{n \equal{} 0}^\infty x^n \equal{} \frac 1{1 \minus{} x}$, the sum of the sums (no pun intended :D) yields
\[ \frac 1{1 \minus{} \frac 38} \plus{} \frac 1{1 \minus{} \frac 58} \equal{} \frac 85 \plus{} \frac 83 \equal{} \frac {64}{15}
\]


rotation
analytic geometry
geometry
geometric transformation

The triangle with vertices A (6, 1), B (4, 1) and C (4, 4) is
rotated 90 degrees counterclockwise about B. What are the
coordinates of the image of C (the point where C is located
after the rotation)? Express your answer as an ordered pair.

Well B and C look kinda like this:

                    . C

                    . B

So rotating C about B is basicaly moving it so that it is on the same horizontal line B is in AND keeping the same distance. Clearly the distance is 3 units so this is what it should look like.

                        . C

       C'.              . B

New point is just 3 units to the left of B or $ \boxed{(1,1)}$.


Let ABC be a triangle with mA=90 and BC=1. Let M, N on (BC) such that BM=MN=NC. If AM+AN= \sqrt 10/3, prove that AB=AC .

Let AB=x, AC=y. We have x2+y2=1.

From Stewart formula:
AN2=x2/3 + 2y2/3 - 2/9
AM2=2x2/3 + y2/3 - 2/9

So \sqrt (3x2+1) + \sqrt (3y2+1) =\sqrt 10.

Function \sqrt t is concave function ==> \sqrt (3x2+1) + \sqrt (3y2+1) \leq 2.\sqrt ((3x2+3y2+2)/2) = \sqrt 10 and identity occures iff x=y.


modular arithmetic
linear algebra
matrix
geometry
3D geometry
algebra
difference of squares

How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?

[hide="Solution"]Note that the numbers we choose would all have two factors in the form $(x+y)(x-y)$, and the difference between these two factors needs to be even, because $(x+y)-(x-y)=2y$.
So we will be able to get all odd numbers ($2n-1$), because two factors of any number is itself and $1$, and $2n-1-1=2n-2$, which is even. For example $519=(260+259)(260-259)$. The amount of odd numbers between $1$ and $1000$ is $500$.
Now to get an even number with factors that differ by $2y$, both factors need to be even. So it needs to be at least divisible by $4$, and the amount of these between $1$ and $1000$ is $250$.
Thus the total number of integers between $1$ and $1000$ that can be expressed as the difference between two squares is $\boxed{750}$.[/hide]

[hide]Squares are , so difference of squares is . So all integers that are  cannot be expressed as a difference of two squares, and that would leave  of the integers less than .[/hide]

It's small, but I think you still have to show that all numbers congruent to $0,1,3\pmod{4}$ can actually be expressed as the difference of two squares.

[hide]Only numbers that can't be expressed as differences of squares are those that are $2\mod4$. So that's 3 out of 4 numbers that can be expressed as differences of squares, which implies our answer is $.75*1000=\boxed{750}$[/hide]

towersfreak:

Any number that can be expressed in the form $(m+n)(m-n)$, where $m, n \in \mathbb{N}$, can be expressed as a difference of squares.  The only qualification this places on the number is that the sum of $(m+n) + (m-n) = 2m$ is even, and the difference $(m+n) - (m-n) = 2n$ is also even - that is, a number need only be expressible as the product of two integers with equal parity.

This is only not doable in numbers that have only one factor of two, such as $34 = 2 \times 17$ or $30 = 2 \times 3 \times 5$ - that is, numbers that are $\equiv 2 \bmod 4$, as has been stated previously.

The solution using mods wasn't immediately obviously to me because I didn't think to use mod 4 (If someone told me to consider the problem mod 4 then it would have been able to solve it). My question is what about this problems made you think of mod 4?
[Yes, I revived an old thread but it's linked to in the resources section]

Hello. Essentially, the idea is this. Suppose that $ x\equal{}(m\plus{}n)(m\minus{}n)$. Consider the factors $ (m\plus{}n)$ and $ (m\minus{}n)$ they must be both even or both odd (why). In those cases, $ x$ is [i]necessarily[/i] divisible by $ 4$ or $ \pm 1\mod 4$. So it just suffices to show which of those numbers can actually be expressed as a difference of squares. Considering mods is just a different way of looking at this argument.

Or I suppose you can be of the mind set to consider mod 4 whenever you see squares...

A common problem solving technique is to quantify the obstructions to a certain construction being possible.  In this case, the construction is, for a given integer $ x$, finding integers $ m, n$ such that $ x \equal{} (m \plus{} n)(m \minus{} n)$.  The obstruction is that $ m \plus{} n$ and $ m \minus{} n$ must have the same parity.  Then the work of the problem consists in showing that this is the only obstruction via the identities

$ m \equal{} \frac {(m \plus{} n) \plus{} (m \minus{} n)}{2}, n \equal{} \frac {(m \plus{} n) \minus{} (m \minus{} n)}{2}$.

In other words, it is necessary and sufficient that there exist a factorization of $ x$ into two factors of the same parity.  At no point do you have to "come up with" the idea to think about modular arithmetic.  In other words, the content of the problem is discussing the invertibility of the matrix

$ \left[ \begin{array}{cc} 1 & \minus{} 1 \\
\minus{} 1 & 1 \end{array} \right].$

The point is that its determinant is $ 2$.

On the other hand, it's always a good idea to look $ \bmod 4$ when you see squares, $ \bmod 7, 9$ when you see cubes, and so forth.  The reason is that the values of certain powers are strongly constrained modulo certain primes due to Fermat's little theorem (and more generally due to the totient theorem), and this is especially true of the powers of $ 2$:  squares, for example, are unusually constrained $ \bmod 8$.

[hide=Very quick sol] Easy to check that if the integer can be factored that way unless it's 2 mod 4. Pick an integer, there are 250 integers in the specified range equivalent to it mod 4, ans=750.[/hide]


$ a,b,c,d \in R,abcd\equal{}1$
Prove that \[ (\frac{1\plus{}ab}{1\plus{}a})^4\plus{}(\frac{1\plus{}bc}{1\plus{}b})^4\plus{}(\frac{1\plus{}cd}{1\plus{}c})^4\plus{}(\frac{1\plus{}da}{1\plus{}d})^4 \geq 4\]


Jake makes a list of all the four-digit positive integers with the property that the product of the four digits is 5. What is the arithmetic mean of the integers on his list?

1115, 1151, 1511, 5111 are the only four possible numbers.

add 'em up to get 8888.

divide by four, to get the answer of [b]2222[/b]


[url=http://story.news.yahoo.com/news?tmpl=story&u=/ap/20040415/ap_on_he_me/mcdonald_s_health]Yahoo News: McDonald's Launches Healthier Menu Items[/url]

McDonald's announced a few days ago that they would be offering happy meals for adults, which include bottled water, a salad, and a pedometer.  They're also doing some other minor stuff, like allowing people to exchange fries for apple slices (yeah, like anyone's gonna do that).  What do you think of these "changes"?

Woah, I'm sure they've been doing those sort of things in NZ for ages :D. Well OK, a few months.

Oops, just realised I didn't answer your question. Well, as the ads here say, its great. It means parents take you to McDonalds more. :D

Wow, I saw something about that and I assumed it was a joke ...

*grins* My Lit teacher told us about that. Class reaction: "[i]Pedometer?[/i] What do you need half a dozen pedometers for?!" and "[i]No toy?![/i] What further grave injustice is McDonald's trying to wreak upon the world?"

I was reading the paper the other day and the man waiting in line said, "I come to McDonalds for the fries.  If I want to eat healthy, I'll eat at home!"  I think thats a good point, which is why these healthy meals will flop.  It's a good effort, but I think that suing fast food stores for 'making one fat"  is so dumb it's unbelievable....

[quote="mathfanatic"][quote="rippledance"]"[i]Pedometer?[/i] What do you need half a dozen pedometers for?!"[/quote]

Yeah, I was wondering about that, too.  If you get one pedometer each time you go there, then you'll have...far too many...[/quote]
Perhaps it's designed to break by the time you go to McDonalds again...either that or it breaks so often you have to eat at McDonalds quite often :)

Did you know... McDonalds changed it's slogan and increased advertisement slots after the law suits and prior to the menu changes? Guess how they changed the slogan? :wink:

Who here thinks they should replace the sesame seeds on the bun to poppy seeds and replace Coke with coke and label it "The True Happy Meal, Shizzle my Nizzle yo!" :lol:

Maybe the pedometer has Disney charcters on it and also serves as a hypnotizer? So it is technically "a toy" :lol:

Even at home it may be impossible to eat healthy at a ranch/farm...unless you're on Atkins that is :lol:

Hey guys, I worked at McDonalds for 6 months back during freshman year.  Seariously -- I can tell you if the rumors that McDonalds does _________ are true.....

It actually wasnt a bad job.  $7.25/hour.  Thats pretty good for unskilled labor.  Plus you get 50% off on food.  Too bad I'm a vegeterian and only eat the fries.

Ha! The fries are fried in beef fat

[hide]They removed the beef fat a few years ago[/hide]

Not anymore. After Fast Food Nation came out there was apparently a huge uproar about the fries-in-beef-tallow so McDonald's uses vegetable oil now, with *ahem* certain [i]natural[/i] flavorings.

Still, it's arguable that while JS1527 was working at McDonalds, that change may not have yet been implemented.

This doesn't really surprise me... who knows what they put in anything they make?

When I was there they used vegetable oil for the fry-o-later.  However, I did learn that they take the fat from the burger grill and send it back to the main McDonalds HQ, where they sell it to farmers to feed to pigs. (Yuck!)  I worked there during my fresh/soph year (I am a senior now, so 2000/2001).

It was sealed big bottles that were labeled vegetable oil, so I am pretty sure its the real thing.  Want to know something else funny?  My manager went to Hamburger University or something like that.  LOL They like do drills and have contests to see who can serve customers the fastest.  However, its not a real University -- just a 2 week program.

[quote="JS1527"]They like do drills and have contests to see who can serve customers the fastest.[/quote]

Interesting.  Do they have accuracy drills (say, to make sure that the workers put the right food in the right bag)?

[quote="JS1527"]I did learn that they take the fat from the burger grill and send it back to the main McDonalds HQ, where they sell it to farmers to feed to pigs.[/quote]

Hmm.. all I can say is, at least they don't reuse it in the food.

The adult happy meal does seem like a silly idea though; I don't think anyone eats at McDonalds with the intention of eating healthy.

Well I don't know.  I was in the lower echelon of the McDonald's hierarchy.  I was not good enough to go to Hamburger University.  I didn't even apply because well to be honest I stood no chance.

 :)

jkjk.  Hmm... people can be pretty mean to McDonald's workers if they screw up on one thing.  Since they are there for a few weeks I suppose they do all sorts of drills....

[quote="Caroline"][quote="JS1527"]I did learn that they take the fat from the burger grill and send it back to the main McDonalds HQ, where they sell it to farmers to feed to pigs.[/quote]

Hmm.. all I can say is, at least they don't reuse it in the food.[/quote]

Amen.

[quote]The adult happy meal does seem like a silly idea though; I don't think anyone eats at McDonalds with the intention of eating healthy.[/quote]

Well, almost all of the "changes" McDonalds has made are just PR stunts.  They'll still sell unhealthy food, and they won't do anything about the fat content of their burgers, fries, triple shakes, or whatever they're famous for.  The option of substituting apple slices for fries, the removal of 7 oz. Super-Size fries (but 6 oz. Large fries still remain), and so on won't affect their customers at all.

I suppose it creates more options for healthy eaters, but maybe not.  I wouldn't eat at McDonalds if I wanted a salad; it would feel tainted somehow.  And interestingly enough, once you put the fried chicken and dressing on the salad, it has just as many calories as, say, a burger.

Still it is a fact chickens are "better" than cows since they have less saturated fat and whatnot and have less calorie per volume or weight or something like that...too lazy to go look it up...:) And what about the calorie? Would you rather consume protein, carbohydrate, and fat or consume dietary fiber, protein, and fat if they contain the same calorie? Heck I'm anti-Atkins but I'd still choose the salad, mainly because it's so much easier to grab food full of carbohydrate rather than something of dietary fiber.

[quote="Tare"]Still it is a fact chickens are "better" than cows since they have less saturated fat and whatnot and have less calorie per volume or weight or something like that...too lazy to go look it up...:) And what about the calorie? Would you rather consume protein, carbohydrate, and fat or consume dietary fiber, protein, and fat if they contain the same calorie?[/quote]

Heh.  Tare, chicken isn't any better than beef if you bread it, bathe it in grease, and fry it for a while.  Besides, lots of McDonalds food is ground up and stuff, so it's impossible to say whether the chicken contains less fat than the beef.

And my point about the salad was not that McDonalds salads are as unhealthy as the burgers; rather, my point is it is ironic that McDonalds' "healthy choice" still comes loaded with calories, fat, etc. thanks to the bacon, cheese, fried chicken, and salad dressing that they put on their salads.  I agree that salad is probably a better choice than a burger, because of the fiber, the vegetables, and so on.  But it's not a great choice compared to, say, eating a salad at home.

[quote="churchilljrhigh"]I love Mcdonalds. I can't believe there are those people who would sue Mcdonalds (overrepublicanism). If you don't want to eat it don't bother other's people's freedom. I especially love their $1 Big Mac days. However, honestly it isn't as good as any authentic chinese food.[/quote]

It's definitely not overrepublicanism. A lot of far-right republicans (Limbaugh, Hannity) agree that these lawsuits are stupid and frivolous; and they really are.

Besides, if republicans like capitalism, why would they support suing corporations?

Who thinks a greasy double quarter pounder (even though it tastes good) is good for you? It's not like McDonalds was marketing it as health food.

On a side note, I heard that their salads have more fat than a big mac. Don't know if its true, but if it were that would be hilarious.

According to the official website of McDonalds: Big mac has 600 calories. Grilled Chicken Bacon Ranch Salad has 250. Crispy Chicken Bacon Ranch Salad 350. Crispy Chicken California Cobb Salad has 370. Newman's Own Cobb Dressing
 has 120, Newman's Own Creamy Caesar Dressing has 190, Newman's Own Low Fat Balsamic Vinaigrette 40, Newman's Own Ranch Dressing 170. As you can see the most calorie you can get from a salad is 560, and that is if you use all of that huge dressing and eat the entire salad. However you could say there's a variation in quality, so maybe it could be over Big Mac...ah w/e I'm not eating at McDs so nmp :lol:

For your viewing pleasure: http://www.mcdonalds.com/usa/eat/willie.html

a big mac also fills you.

salads often dont. especially small salads.

Actually, salads can be quite filling.  I had a salad for lunch today, actually.  If I had had something with it (say, a bagel or a small portion of something meaty or dairy) it would have carried me all the way to dinner.

to me, choosing something "healthy" at McDonalds is like choosing which drug to take. it's just stupid for you to try to be healthy at greaseville, usa.

[quote="dogseatcheese"]Its been 2 years now and.... I have not seen an adult happy meal. So......where is it?[/quote]

In Argentina there is a sort of it, with water, salad and a peace of meat..... ahh!!! and an ice-cream.......

Diets are a fad in America today.  Just about every fast food chain/restaurant/etc. has began to offer healthier choices along with low carb options.

In response to the person who said that if they wanted to eat healthy, they would just go home...that would seem to be the case...but...the main reason people eat fast food is because they don't want to cook (or don't have enough time) not because they want to intentionally eat unhealthy.

It would also be interesting to note how much fast food has changed in the past two years.  If the people who posted in this thread were faced with similar news of a fast food chain coming out with a healthy option (say...KFC serving grilled chicken...) I imagine they would note be as skeptical.


The numbers 1 through 999, inclusive, are printed on a piece of
paper. How many digits are printed on the paper?


advanced fields
advanced fields unsolved

Let $ A\subset\mathbb{R}$ is measurable set, and let $ B\equal{}\{|x|: x\in A\}$. Show that $ B$ is also measurable set.

Thanks.

Hint: Write $ B\equal{}A_\plus{}\cup (\minus{}A_\minus{})$ with two measurable sets $ A_\plus{}, A_\minus{}$ derived from $ A$.


algebra
polynomial
superior algebra
superior algebra unsolved

let P be a polynomial with integer coeficients such that all his roots are integers and greter in modulus than 2.
P(0)+1 is prime that prove that P(X)+1 is ireduccible in Z[X].


(AIME 2002A/7)

What are the first three digits after the decimal point in (10^2002 + 1)^(10/7)?

[i]You may use the extended binomial theorem: (x + y)r = xr*(1 + 4 (y/x) + r(r-1)/2! (y/x)2 + r(r-1)(r-2)/3! (y/x)3 + ...) for r real and |x/y| < 1[/i]

[Edited to fix disabled HTML]

[hide](10^2002 + 1)^10/7 = (10^2002^10/7)(1+10^-2002)^10/7 = 10^2860*taylor series.

Notice that the taylor series is (1 + nx + n(n-1)x^2/2!....)

So in this case, we get 1 + (10/7)/10^2002 + (30/49)/2!/10^4004....We can omit terms past the third as they will have no effect on the first three decimal digits. In fact, the third term cannot have impact either...

Multiplying, we get the first two terms to be 10^2860 + 10^859/7. 10^859  :equiv: 3 mod 7. Since 1/7 = 0.142857142857...., we know 3/7 = 0.428571.... Thus our answer is 428.

[/hide]


There exists a 6-digit number $n$ such that $1n, 2n, 3n, 4n, 5n, \text{and }6n$ are rearrangements of the digits of $n$.  Find the remainder when $7n$ is divided by $1000$.

What question do you think this would be on an AIME?

[quote="miyomiyo"]There exists a 6-digit number $n$ such that $kn, k \in{1, 2, 3, 4, 5, 6}$ is a rearrangement of the digits of $n$.  Find the remainder when $7n$ is divided by $1000$.

What question do you think this would be on an AIME?[/quote]
I saw a very very similar problem on USAMTS
[hide]From prior knowledge, I know that if k=4 then n=102564 works.
So our answer is $3500+420+28=\boxed{948}$[/hide]

Sorry, the question wasn't worded clearly.  It's edited now.  If the question had been how you thought it was, it would be any number ($1n$ is always a "rearrangement" of $n$).

[hide="solution"]Recall that $\frac{1}{7}$ is a repeating decimal with a period of $6$ and that all multiples of it are rearrangements of its digits, still repeating. The latter is also true for $\frac{99999}{7}=\boxed{142857}$.[/hide]

You have the answer, but you didn't put it in the correct format.

Ouch. The night before the AIME and I still haven't learned how to read the question. [hide="The correct answer is this."]$143$[/hide]

[quote="chickendude"][hide]
no
wouldn't $n = 142857$
$7n = 999999$
$7n \mod 1000 = 999$
[/hide][/quote]

That's correct.


quadratics
geometry
3D geometry
function
tetrahedron
USAMTS
geometric transformation

my prediction

#1 wrong, but similar way... 1 or 2 pts
#2 perfect, 5 pts
#3 wrong, 0 or 1
#4 perfect, 5 pts
#5 perfect, 5 pts

blah... i hope i get around 20...

#1: but i have a nearly perfect solution, i just counted the one twice, i got 23 instead of 22. so maybe 4?
#2: perfect 5
#3: correct answer but not very rigorous solution 4 or 3
#4: correct answer: correct solution: horribly unslick solution maybe 4 or 5
#5 perfect solution 5

4 + 5 + 3/4 + 4/5 + 5 = anywhere from 21-23

hmm...
1. I dont know...i thought my solution was nice, i just missed 2 cases, so like 2
2. Probably not rigorous enough, but i got the right answers, so 4.
3. Pretty lame solution, very unelegant, but got the answer, so i guess i will get 5.
4. Got to 3s^4-28s^2+9 and then did the quadratic formula wrong, so ~3
5. Probably not rigorous enough...i dont know, ~3

so like 17, but i think thats pretty optimistic...

i phailed.

As said in the other thread;
1) Counted two extra cubes in one of my cases, however, I'm hoping for a 3-4.
2) Perfect, 5.
3) Good, but slightly unrigorous (is that a word), so 4.
4) Heh, not rigorous at all (I assumed that a line was a median) so 2-3..
5) Proved for all r besides odd primes. 3

So 17-19.

1. Brute forced it exactly like the provided solution. I am interested in the "nice" solution, though. Could anyone point me towards it?
2. I bashed w/ inequalities. Seems as rigorous or better than the provided solution.
3. A little wordier than the official solution, but demonstrated about the same things.
4. I actually proved that X, P, and Y were collinear instead of assuming it. From there, algebra.
5. I proved it slightly differently. I showed that the function $ \sqrt{q} \plus{} \sqrt{q\plus{}1}$ spanned the reals for a positive real $ q$, and then equated $ \sqrt{q} \plus{} \sqrt{q\plus{}1}$ and $ (\sqrt{m} \plus{} \sqrt{m\plus{}1})^r$ and showed that $ q$ was an integer.

1. PHAIL (just like me to fail, counting).
2. Hm, I don't know. 4 or 5.
3. 5
4. 5 (I embedded a tetrahedron inside a cube and solved it that way)
5. I used cosh??? I don't know...

Too early to tell. Oh yeah, and I did it at the last minute... long story...

Basically, I thought I wasn't going to do it because (1) the entry form seemed a hassle to complete and (2) nobody (I MEAN NOBODY) at my school did it either.

Then on October 13, I decided I was going to do it, sent the entry form in, blah blah, and did it at 9:00 PM to 10:00 PM.

Don't let this affect your grading though... I'll definitely work harder on Rounds 2-4.

hmm
1:attempted, but PHAIL
2:PHAIL
3:i attempted, but still, PHAIL
4. PHAIL
5: half right?

so about a 3.

oh my gawd
faulty algebra on problem 4, got wrong
miscounted on problem 1 got 21 instead of 22
completely did not get number 5
what do u think i could get on 1/4?
EDIT: perfect #2 might be my saving grace
#3 left a lot of things assumed, 3?

[quote="FantasyLover"][quote="abacadaea"][quote="pythag011"]Hmm...all correct but some of them I'm not sure if my proof was clear enough...hoping to get all 5s...[/quote]

im going to be shamed by a 7th grader :([/quote]

ur not the only one :([/quote]

I agree. Hmm, hopefully I can do better on Round 2.

[quote="xpmath"][/quote][quote="FantasyLover"][quote="abacadaea"][quote="pythag011"]Hmm...all correct but some of them I'm not sure if my proof was clear enough...hoping to get all 5s...[/quote]

im going to be shamed by a 7th grader :([/quote]

ur not the only one :([/quote][quote="xpmath"]

I agree. Hmm, hopefully I can do better on Round 2.[/quote]

same

FantasyLover-atleast you did better than me. Either that, or you are more optimistic than me. :P

You should've seen what I put for #1...

"It trivially follows that the answer is 30."

:rotfl: The result of my last-minute work and my inability to directly count combinations.

[quote="abacadaea"][/quote][quote="xpmath"][/quote][quote="FantasyLover"][quote="abacadaea"][quote="pythag011"]Hmm...all correct but some of them I'm not sure if my proof was clear enough...hoping to get all 5s...[/quote]

im going to be shamed by a 7th grader :([/quote]

ur not the only one :([/quote][quote="xpmath"]

I agree. Hmm, hopefully I can do better on Round 2.[/quote][quote="abacadaea"]

same

FantasyLover-atleast you did better than me. Either that, or you are more optimistic than me. :P[/quote]

*cough* i think its first one *cough*  :rotfl:

I think I perfect 2 to 5... all very neat solutions...
for 1... I counted fail.
I count 1 less for 1 case, 1 more for another... getting the same answer...-.-

1- probably 5
2- 5 (no problems at all)
3- 4-5 (got right answer, not sure about rigor of solution)
4- 2-3 (had right idea, but no answer)
5- 3-5 (I'm not sure about how my skewered induction process will be recieved, but in my view it is acceptable)

Likely score: 22

1: probably 4, i got the wrong answer but my solution was detailed
2: probably 2 or 3, i got the right answer but I had a horrible solution
3: 1 or 2...meh I got it wrong plus my solution seemed trivial....
4: 1 or 2...wrong answer, bad solution
5: probably 5, decent enough proof

1) Got 23 cubes... hopefully a 4
2) Correct and rigorous (5)
3) Correct, went in-depth (5)
4) Incorrect but went in depth as well (hopefully 3)
5) Correct, a lot of algebra (5)

therefore i'm hoping 22+/25 for first round... I really want to qualify for the AIME this year.  Is cutoff 69+/75? if so, I can't lose another 3 points =/

[quote="iironiic"]I really want to qualify for the AIME this year.  Is cutoff 69+/75?[/quote]

I don't think the USAMTS people set the cutoff score. At any rate, it's too early to say. Take the AMCs, too.

in my opinion, I was really successful with most the questions... it just takes a lot of thinking and many days of evaluation.  I will take the AMC 12, but I don't do too well on those tests usually, but let's hope this year is a great year, since this is my last year trying to qualify for the AIME.

Round 1 seems difficult in my opinion... especially since this is the first round...

is the cutoff score different each year?

# 1: Correct (I drew 3D pics for each one). (5)
# 2: Correct (5)
# 3: A lot of unnecessary proof, but still good. (5)
# 4: Good (5)
# 5: Totally crazy. Who knows, maybe 0, maybe 5 :lol: .(2~4)

So 22~24.

1) right answer at first i did their counting technique then i thought it want mathy enough so i made a totally absurd method that still got me the right answer
hoping for a 1-4
2)perfect  :lol: 5/5
3) totally misinterpreted the question
hoping for a 0 not a -1
4)perfect  :) 5/5
5) someone said that my soultion didnt include intgers just rational numbers? i think the Q was just rational numbers so
hoping for a 5/5 :D

[quote="gauss1181"][quote="1=2"]I tried doing the USAMTS, but I could only do two problems after two months. :(
And I've been doing very good on the AMC 12(near perfect) and am anticipating going to USAMO. Was I just being stupid on the test or are these questions harder than the AMC 12?[/quote]
USAMTS actually is harder than AMC 12 cuz you need to research. And you should plan your time wisely so you could finish every problem.[/quote]

I don't necessarily agree with this. The main aspect of the USAMTS that would make it harder than the AMC 12 is that you need a proof, while on some AMC 12 problems, you can just find a pattern and assume it works.

About research-I don't think much research is needed for any of the problems unless if there's a term you don't know, or a theorem you know exists and want to find (like in this year's round 1, there weren't any problems that needed research, unless if you didn't know what a tetrahedron was or something like that).

And planning your time wisely?! You have more than a whole month to do it! The only kind of planning that you need is not putting off the problems until the last days (which I admit, I did last year), which is more common sense than being wise.

maybe 10-14. I am not sure since I did not make a copy of my solutions before mailing them, and thus I am not really sure what I did for number 1.

1. don't remember what I got....(I had different cases for organization...I think I got it or was off by 1) 1 or 4?
2. 5, or 4, since I had no clue what bounds were (well I did not know that what I used was called bounds)
3. 1 or 2. right answer, but no proof. Well, I made up stuff for the proof
4. probably 4... or 3 or 5 - right answer, but horrible spacing and well, just bad presentation overall
5. 0 or 1 made stuff up with no proof

Hmm...

1:perfect, but i think i made some assumptions when trying to prove that 2 of the "2 edges" cases were distinct, so 4-5

2:perfect
3:OMG I AM A RETARD WHY DID I NOT JUST SEND IN MY ANSWER AND A BUNCH OF BS SO I COULD GET A 1-2 (meaning i got 0. guessed 300 and couldn't prove it so didn't send it in :mad: )

4:coordinate bash, 3-5
5:different solution 4-5
so about 16-20.
Which I'm pretty happy about, since I looked at the first round for like 5 min. last year and chickened out, so i got a 0

[quote="gauss1181"][quote="kyyuanmathcount"]I got a very unfortunate negative score :(.[/quote]
?????

I guess that reeks of sarcasm.. :huh:[/quote]

umm no, it is completely serious

he got a
[hide]NEGATIVE ZERO![/hide]

bah so I should not have sucked on usamts that round - I should try typing up solutions before 8 pm the day before next time...