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I recently recieved a copy of the 2008 National Sprint Round and Team Round.

Here is one of the problems that I couldn't figure out;

#9 2008 National Sprint Round;

Each of 'n' cats has 2n fleas. If two cats (and their fleas) are removed, and three fleas are removed from each remaining cat, the total number of fleas remaining would be half the original total number of fleas. What is the value of 'n'?

(By the way, Bobby Shen got 29/30 on this sprint round)

2n^2  - 4n - 3(n-2) = n^2
n^2 -7n +6 = 0
(n-1)(n-6) = 0

So there are six cats. (One cat doesn't work, since you have to take two cats to start with!)

good job eshen, thanks for helping me with this problem!

Here is another problem from National 2008 sprint, it is number 15;

15. In a soccer league with 8 teams, each team earns 2 points for a win, 1 point for a tie and 0 points for a loss. The league standings are based on the total points earned by each team. After a total of 40 games, each team has played the same number of games and the top four teams have earned point totals of 15, 13, 12, and 11 points. There is one team in fifth place and three teams are tied for sixth place. How many points has the fifth-place team earned?

Good luck with this problem!

As 2 points are awarded per game, after 40 games 80 points have been awarded. Subtracting the given numbers, there are 29 points for the other teams. 5th place must have 10 or less, so through trial and error we see that 5th place has 8 and 6th places have 7. Thus the answer is $ 8$.

thanks, I had no idea how to solve that problem,
here's number 19 on the same sprint round;

19. If 'm' workers can complete a job in 'd' days, how many days will it take 'n' workers, working at the same rate to complete one third of the job? Express your answer as a common fraction in terms of d, m, and n.

Good luck!

$ m$ workers do it in $ d$ days, so one worker does it in $ dm$ days.

$ n$ workers do it in $ \frac{dm}{n}$ days, so they do $ \frac{1}{3}$

of the job in $ \boxed{\frac{dm}{3n}}$ days.

[quote="abacadaea"]where do you find these?[/quote]

You have to get them from a state coach.

Nice solution BOGTRO, I didn't think of solving it that way, and yes 12markkram34, I do have a state coach, and he gave me all of these problems, I have probably solved about at least a thousand problems, I ranked 7th at state, but I'm just a sixth grader.

Here is another problem, it is number 22, it was a bit tricky for me to solve;

22. In a square grid of 8 dots along each side, how many distinct lines pass through the dot at the bottom left corner and at least one other dot of the grid?

and here's another too;

23. A right triangle had side lengths 3, 4 and 5 units. Another triangle has sides of length 4, 'x' and 'y' units and has the same area as teh right triangle. If x is less than or equal to y, what is the smallest possible value of 'y'? Express your answer in simp.rad.form.

There is already a discussion on #22. Look for it yourself. :)

[hide="Problem 23"]
First of all, the area of the right triangle is $ 6$, so the triangle with sides 4, $ x, y$ also has area 6. To minimize it, assume $ x \equal{} y$. Thus an isosceles triangle with sides $ 4, x, x$ has area 6, and by the Pythagorean Theorem, $ x^2 \minus{} 2^2 \equal{} 3^2$, or $ x^2 \minus{} 4 \equal{} 9$, or $ x^2 \equal{} 13$. Thus $ x \equal{} y \equal{} \boxed{\sqrt {13}}$. I'm not sure if I'm right though...[/hide]

Here's another problem;

26. Julian wants to label his fingers on his left hand with the integers one through 5, such that no two adjacent fingers have consecutive integers, and each integer is used exactly once. In how many ways can Julian label his fingers?  One such way is shown below;

                                              3  1  5  2  4

Can you find a way to solve this one without just listing all of the ways?

[quote="Yongyi781"]You can count the ways where the first number is smaller than the last number, then double it for reversals.
I'd just list the combinations for the first and last numbers. Then I'd fill them in. Note in most cases, there is only one way to fill them in, and for  those where the remaining three are consecutive (1____2, 1____5, 4____5), there is no way. There are 7, then multiply by 2 for an answer of $ \boxed {14}$ ways. (I made a mistake and used constructuve counting to get 72 ways, which was wrong. :| )

1____2
1____3 -> 14253
1____4 -> 13524
1____5
2____3 -> 24153
2____4 -> 25314
2____5 -> 24135
3____4 -> 31524
3____5 -> 31425
4____5[/quote]

I don't get this part:

You can count the ways where the first number is smaller than the last number, then double it for reversals.

Can someone please help me?  Thanks a lot!

That's ok.

Does anyone know how to do the last problem?

Gracie starts at her house and walks one block east. After each block she walks, she flips a fair coin. She walks one block east if the coin lands heads; otherwise one block west. She stops her walk the first time she lands back home. What is the probability that she stops her walk after walking exactly eight blocks? Express your answer as a common fraction.

That was the last problem, #30 of the 2008 National Sprint Round, right?[/hide]

[quote="Textangle"]That's ok.

[b]Does anyone know how to do the last problem?[/b]

Gracie starts at her house and walks one block east. After each block she walks, she flips a fair coin. She walks one block east if the coin lands heads; otherwise one block west. She stops her walk the first time she lands back home. What is the probability that she stops her walk after walking exactly eight blocks? Express your answer as a common fraction.[/quote]

I think he just said that.

We enumerate the possibilites. The next step can't be W, otherwise, she would have stopped before walking 8 blocks. So every possibility starts with EE.

EEEEWWWW
EEEWEWWW
EEEWWEWW
EEWEWEWW
EEWEEWWW

Those are the only five possibilities, each with eight steps, so the answer is $ \frac 5{256}$... wait! The problem stated that she ALREADY walked one block east, so the leading E actually had a probability of 1, making the answer actually $ \boxed {\frac 5{128}}$.

I remember seeing a problem similar to that on a state target round. Good Job isabella earlier. I know that what I'm saying is in random order.

That is a really good solution to the current problem Yongyi. You always make everything look so easy, while I try to find a difficult way of looking at it.

Hi Alex its Michael Chen!

Glad to see that you are still participating in MathCounts.

Hope to see you at Nats next year!!!

(btw, email me at michaelchen714@aol.com so i can get your email)

Oh hey Michael!!! I didn't know you had an account. I hope to see you at Nats. If we are aloud to go, then we will probably go, even if I don't make Nats', because most of my family is in Florida, and we would be able to see them there, and you and Mr. L.

We don't need to see a reunion on the MathCounts forum.  It's just spamming.

Hello, I am doing a project for my MathCounts class and therefore am getting problems from this post.

thanks!

oh dang....
bad target round memories...
 :P

[quote="mewto55555"]Oh yeah. I misread my tally marks on [b]both[/b] problems where it was simplest to count possibilities. I would've gotten top 57 easily and my team would have gotten in 5th or 6th  :mad:[/quote]

You are in the same boat as me. Poor me, I spent 5 minutes and 30 seconds on each pair thinking of a combinatorial way to approach those (evil) problems... blah...

[quote="unimpossible"]lol, well i forgot to divide by 4 at the end for number 7 and i don't know how i missed number 1. i did the problem again later (with the same type of solution) and got it.

Number 6...well i remember number 6 was a listing problem and i just forgot a case...not really a stupid mistake. just missed a case.[/quote]

For #1, you probably thought that LCM(2, 4) = 8 when you were doing PIE :P I did that too :(

I was trained to use combinatorial methods to solve problems. (AoPS Intro to Counting and Probability, previous Nationals rounds, FTW, etc). So when I saw problems like #6, #7, #8, I spent like almost the whole time looking for a combinatorial method to solve them. I totally did not even think about just listing them out (I thought it would a stupid question otherwise).


inequalities
function
geometry
inequalities solved

a,b,c,A,B,C > 0, a+A=b+B=c+C=k

Prove aB+bC+cA \leq k2.

Put A = k - a, B = k - b, C = k - c.
We have to prove that a(k - b) + b(k - c) + c(k - a) \leq k2.
Fix b and c in [0, k] and consider the left - handside as a function of a :
f(a) = a(k - b -c) + bk + ck - bc.
This is a linear function so the maximum is attained for a = k or a = 0.
Similarly, the maximum is attained for b = k or b = 0 and c = k or c = 0.
So we only have to check four cases :
(1) a = b = c = k -> LHS = 0.
(2) a = b = k, c = 0 -> LHS = k2.
(3) a = k, b = c = 0 -> LHS = k2.
(4) a = b = c = 0 -> LHS = 0.
So indeed, in all cases, LHS \leq k2.

Here's a cool idea for this one:

Take an equilateral triangle of side k ABC. Then take points M, N, P on (BC), (CA), (AB) respectively s.t. BM=a, CN=c, AP=b, and then we use the inequality [BMP]+[CMN]+[ANP]<=[ABC] where[XYZ] is the area of triangle XYZ. Pretty nice, isn't it? :D  :D

It is author's solution, grobber!
This problem was presented in all-russian math olympiad in 1987.


In $ \triangle ABC$, let $ O,H$ be the circumcenter and orthocenter respectively. Let $ A',B',C'$ be points on the circumcircles of triangles $ BHC,CHA,AHB$ respectively, such that each of the sets $ \{A',C,B'\},\{B',A,C'\},\{A',B,C'\}$ have points collinear. Let $ H'$ be the orthocenter of triangle $ A'B'C'$. Find $ \angle HOH'$, in terms of angles $ A,B,C$ of triangle $ ABC$.


If you qualify for the Chemistry Olympiad Study Camp, are you required to go?  I have almost no chance of qualifying, but I wanted to know in case I accidentally do better than I'm expecting.

Nope.  You can turn it down, although it think it's rather uncommon.  Actually, I do know some who did (albeit to attend MOP).

That being said, I believe that ostensibly one of the requirements for taking the nat'l exam is that you'll go to the camp if selected (please correct me if I'm wrong; I know this is true for selection of the international team)... but I don't think there'd be any bad blood if you turned it down.

OK, thanks.  I'm not sure yet if I would be available for that time, and I don't want to make them too angry.

Don't sweat it.  I would still take the exam, just for the experience at least.


quadratics
function
algebra
polynomial
system of equations

Find the remainder when you divide $x^{81}+x^{49}+x^{25}+x^{9}+x$ by $x^{3}-x$.

Since $x^{3}-x=x(x-1)(x+1)$, we'll use the remainders when $P(x)$ is divided by $x,x-1,x+1$, and those are $P(0)=0, P(1)=5, P(-1)=-5$ respectively.

Now

$P(x)=x(x-1)(x+1)Q(x)+ax^{2}+bx+c$

Substitute $x=0,x=1,x=-1$ to obtain the system of equations:

\begin{eqnarray*}c &=& 0\\ a+b+c &=& 5\\ a-b+c &=&-5\end{eqnarray*}

which has the solution $a=0,b=5,c=0$

Therefore the desired remainder is $5x$.

The remainder isn't a quadratic. It's a linear function, $5x$.

I mean when he solved for the remainder, he assumed it was the divisor function multiplied by the quotient polynomial plus an additional quadratic remainder.

Because whenever you divide some polynomial $P$ with another polynomial $Q$ with degree $n$, you always get a remainder of degree $n-1$.

[hide="Warning:  Extreme Elegance"] In other words, we want to find the residue class $\bmod (x^{3}-x)$.  Hence we operate $\in \mathbb{Z}[x]_{ x^{3}-x }$.

In this system, $x^{3}\equiv x$, so we simply reduce:

$x^{81}\equiv x^{27}\equiv x^{9}\equiv x^{3}\equiv x$
$x^{49}\equiv x^{17}\equiv x^{7}\equiv x^{3}\equiv x$
$x^{25}\equiv x^{7}\equiv x^{3}\equiv x$
$x^{9}\equiv x^{3}\equiv x$
$x \equiv x$

Hence our remainder is $5x$. ;) [/hide]

Edit:  [hide="An alternate explanation"] $(x^{2}-1) | (x^{2n}-1) \implies (x^{3}-x) | (x^{2n+1}-x)$.  Simply subtract $5x$ from our polynomial and the solution should be clear. [/hide]


Given are reals $a_1, \cdots a_n$.
Prove that $108^n \max_{x \in [0;1]} \prod_{i=1}^n |x-a_i| \geq \max_{x \in [0;2]} \prod_{i=1}^n |x-a_i|$.

Does the claim hold true for complex $a_i$?

[quote="harazi"]Yes, look at one of hxtung posts and you will see I proved this even with 12 replaced by 108![/quote]

where? :)


Hi,

I am in 7th grade in the Duke TIP, and will be taking the SAT in about a week.
I was wondering - what should my score be like in order to get recognition?...

Thanks for your help!

I am unfamiliar with Duke's program, but [url=http://cty.jhu.edu/ts/grades78.html]CTY[/url] requires 95th percentile.  Give or take, that's about a 700.

For grand recognition from Duke it should be about 670+ in any subject.  It's not that hard.

http://www.tip.duke.edu/talent_searches/recognition_ceremonies/

That's a list of requirements.


algebra
polynomial
linear algebra
matrix
function
linear algebra unsolved

Dear friends,

I have doubt like is finding the roots of a polynomial of order $ n$ for example $ f(x)\equal{}a_n x^n\plus{} a_{n\minus{}1} x^{n\minus{}1}\plus{} \cdots \plus{} a_1 x \plus{} a_0$ be the polynomial with roots $ \mathbf{r}\equal{}[r_0\, r_1\, \cdots\, r_n]^T$. My question is is there anyway to find a matrix $ X$ such that  \[ X \mathbf{a}\equal{}\mathbf{r}\] where $ \mathbf{a}\equal{}[a_n\, a_{n\minus{}1}\, \cdots\, a_0]^T$ is the polynomial coefficient vector.

Any help please.
bhupala

Er... why?  The roots of a polynomial don't depend linearly on its coefficients, so $ X$ is a function of $ f$ and I don't see it having any significance.

Actually I am working on constrained least squares $ \min_{a}\: \|a\|_2$ with the constraint that $ Ca\equal{}d$ and I also want to include constraint that roots of $ a$ satisfy $ |r_i|<1$. So I was trying to figure out if there is anyway to do this.
:)

Any help is very much appreciated.

bhupala.


I have a simple latex question that I can't find anywhere...

How do you do a bar in latex (for integrating an integral), for example x|-1 to 1

Here's an example:
$ \int_0^\infty \frac{\minus{}1}{x^2}dx\equal{}\frac{1}{x}$|0 to infty

I hope you understand what I mean... thanks a lot!

$ \left.\int_0^\infty \frac{\minus{}1}{x^2}dx\equal{}\frac{1}{x}\right|_0^{\infty}$
[code]\left.\int_0^\infty \frac{-1}{x^2}dx=\frac{1}{x}\right|_0^{\infty}[/code]It's a similar idea to getting resized (...) where you use \left(...\right). You don't want a left| but you need to tell LaTeX where to start to measure the height of the formula so you use \left[color=red][b][size=150].[/size][/b][/color] ... \right|

Or just use \Big| or \bigg| (for this case I'll use the first one):

$ \minus{}\int_0^\infty\frac1{x^2}\,dx\equal{}\frac1x\bigg|_0^\infty$

made it with

[code]$-\int_0^\infty\frac1{x^2}\,dx=\frac1x\bigg|_0^\infty$[/code]


trigonometry
function
absolute value
geometric series

For $ 0 < x < 2\pi$, compute the absolute value of the
difference between the solutions to the equation:

$ sec x \equal{} 1 \plus{} cosx \plus{} cos^2x \plus{} cos^3x \plus{}...$

$ \sec x \equal{} \frac{1}{1\minus{}\cos x}$

$ \sec x (1 \minus{} \cos x) \equal{} 1$

$ \sec x \minus{} 1 \equal{} 1$

$ \sec x \equal{} 2$

$ \cos x \equal{} \frac{1}{2}$

$ x \equal{} \{ \frac{\pi}{3} , \frac{5\pi}{3} \}$

So $ |\frac{5 \pi}{3} \minus{} \frac{\pi}{3}| \equal{} |\frac{4\pi}{3}| \equal{} \boxed{\frac{4\pi}{3}}$

I'm not sure if maybe the fact that $ \cos^k x$ changes signs affects the problem. I think it does so I think this only solves the problem for angles NOT in the interval $ (\frac{\pi}{2} , \frac{3\pi}{2} )$.

Wait...how did you get $ \sec{x} \equal{} \frac {1}{1 \minus{} \cos{x}}?$ Isn't the secant function by definition: $ \sec{x} \equal{} \frac {1}{\cos{x}}?$

EDIT: never mind, I overlooked that that was the simplification of  a geometric series :P

BY generating functions 1/1-x=1+x+x^2+x^3.....


geometry
Euler
radical axis
geometry solved

Let ABC be a triangle and let M be a point inside it. The lines AM, BM, CM cut the lines BC, CA, AB, respectively, at $A_o$, $B_o$, $C_o$. The circle with diameter $AA_o$ cuts the circle with diameter BC at $A_1$ and $A_2$. Similarly we define the points $B_1$, $B_2$, $C_1$, $C_2$. Prove that these six points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ lie on a circle.

[color=red][Moderator edit: Corrected some language mistakes and LaTeXed the text.][/color]

Here's a hint: try to prove that H (the orthocenter) is on $A_1A_2$ (as well as on $B_1B_2$ and $C_1C_2$, of course). Then the conclusion follows easily because H is the radical center of the circles (BC), (CA), (AB), so it has equal powers with respect to these 3 circles. On the other hand, if H is on $A_1A_2$, it has the same power with respect to the circles (BC) and $\left(AA_o\right)$, so its powers with respect to the 6 circles (BC), (CA), (AB), $\left(AA_o\right)$, $\left(BB_o\right)$, $\left(CC_o\right)$ are the same, so $HA_1\cdot HA_2=HB_1\cdot HB_2=HC_1\cdot HC_2$, so we get the conclusion.

It's easy to show that H is on $A_1A_2$, I just don't have the time to do it right now.

Prove that H lies on the line $A_1A_2$ (*) :
  AH cuts BC at A',
  BH cuts AC at B'
  consider these circles : (ABA') ; $\left(AA_o\right)$ ; (BAA') we will have
  AA' , BB', $A_1A_2$ are concurrent => (*)

What triangle center X(?) is the center of this circle?

    Pestich.

[quote="pestich"]What triangle center X(?) is the center of this circle?[/quote]

This actually depends on what triangle center is the point M. Initially M was an arbitrary point...

  Darij

The usual suspects for M:

   G,I,K,N ,   and incircle and Euler line.

  Pestich.

[quote="grobber"]Here's a hint: try to prove that H (the orthocenter) is on $A_1A_2$ (as well as on $B_1B_2$ and $C_1C_2$, of course). Then the conclusion follows easily because H is the radical center of the circles (BC), (CA), (AB), so it has equal powers with respect to these 3 circles. On the other hand, if H is on $A_1A_2$, it has the same power with respect to the circles (BC) and $\left(AA_o\right)$, so its powers with respect to the 6 circles (BC), (CA), (AB), $\left(AA_o\right)$, $\left(BB_o\right)$, $\left(CC_o\right)$ are the same, so $HA_1\cdot HA_2=HB_1\cdot HB_2=HC_1\cdot HC_2$, so we get the conclusion.[/quote]

Alas, this is wrong - from $HA_1\cdot HA_2=HB_1\cdot HB_2=HC_1\cdot HC_2$, we just get that the points $B_1$, $B_2$, $C_1$, $C_2$ lie on one circle, that the points $C_1$, $C_2$, $A_1$, $A_2$ lie on one circle, and that the points $A_1$, $A_2$, $B_1$, $B_2$ lie on one circle, but not the concyclicity of all six points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$. So the problem is not solved (I would be glad to solve it now, but I just don't have the time).

  Darij

I see. Sorry about that. Looks like the fact that $AA_0,BB_0,CC_0$ are concurrent is important. I wasn't using this at all :).

Grobber's proof actually can be corrected easily.
Let D,E,F,D',E',F' be the midpoints of BC,CA,AB,$AA_0,BB_0,CC_0$ respectively. We just have to prove DD',EE',FF' are concurrent. This can be done by applying Ceva to DEF and ABC.

[color=red][Moderator edit: This is actually the correct proof, but for those who want a more detailed writeup can look at http://www.mathlinks.ro/Forum/viewtopic.php?t=31399 .][/color]


area of trapezoid is average of basses times height

this height is 5
let's let the base with unknown length have length x
thus we have 5(12+x)/2=50
5(12+x)=100
12+x=20
x=8
since the trapezoid is isosceles the answer is (12-8)/2=2


geometry
circumcircle
trigonometry
geometry solved

Let $ABC$ be a triangle. Points $D$ and $E$ are taken on the line $BC$ such that $AD$ and $AE$ are parallel to the respective tangents to the circumcircle at $C$ and $B$. Prove that
\[\frac{BE}{CD}=\left(\frac{AB}{AC}\right)^2 \]

The problem is quite straightforward so I just point out main steps of solution:

1) Angle chasing [making use of those paralells and everything else you can find to get all the angles within the triangle ;) ]

2) Sine law applied to $ABC$, $AEB$ and $ACD$ - multiplying proper relations we get what we wanted

[quote="warut_suk"]$ABC$ is a triangle. $D$, $E$ are points on the line $BC$ such that $AD$, $AE$ are parallel to the tangents to the circumcircle at $C$, $B$. Show that $\frac{BE}{CD} = \left(\frac{AB}{AC}\right)^2$.[/quote]

The problem is almost trivial:

We will use directed angles modulo 180°.

Let y be the tangent to the circumcircle of triangle ABC at the point B. Then, by the tangent-chordal angle theorem, we have < (y; BC) = < BAC. On the other hand, by the conditions of the problem, the line AE is parallel to this tangent y to the circumcircle of triangle ABC at the point B; thus, < (AE; BC) = < (y; BC). Hence, < (AE; BC) = < BAC. In other words, < AEB = - < CAB. Also, trivially, < ABE = - < CBA. Thus, the triangles AEB and CAB are inversely similar; consequently, $\frac{BE}{AB}=\frac{BA}{CB}$, so that $BE=AB\cdot\frac{BA}{CB}=AB\cdot\frac{AB}{BC}$. Similarly, $CD=AC\cdot\frac{AC}{BC}$. Thus,

$\frac{BE}{CD}=\frac{AB\cdot\frac{AB}{BC}}{AC\cdot\frac{AC}{BC}}=\frac{AB^2}{AC^2}=\left(\frac{AB}{AC}\right)^2$,

and the problem is solved.

  Darij


Easy to prove that
$ a_{n}\equal{}n^{2}$

[quote="ashrafmod"]If the sequence $ a_{0}, a_{1}, a_{2}, ...$satisfies

            (1)           $ a_{1}\equal{} 1$, and

            (2)            for all integers m and n with m ≥ n ≥ 0,$ a_{2m}\plus{}a_{2n}\equal{} 2(a_{m\plus{}n }\plus{}a_{m\minus{}n})$,

determine $ a_{2005}$.[/quote]

A problem from Russia-1995 but in Russia -1995, we ditermine $ a_{1995}$ not $ a_{2005}$. And solution by inductive method same TTsphn 'solution  :)


vector
real analysis
real analysis unsolved

Prove that the line joining the midpoints of two sides of a triangle is parallel to and half as long as the third side.

Let $ ABC$ be a triangle with the mid points $ M,\ N$ of $ AB,\ AC$ respectively, we have $ \overrightarrow{AM} \equal{} \frac {1}{2}\overrightarrow{AB},\ \overrightarrow{AN} \equal{} \frac {1}{2}\overrightarrow{AC}$,

$ \therefore \overrightarrow{MN} \equal{} \overrightarrow{AN} \minus{} \overrightarrow{AM} \equal{} \frac {1}{2}(\overrightarrow{AB} \minus{} \overrightarrow{AC}) \equal{} \frac {1}{2}\overrightarrow{BC}\Longleftrightarrow MN\parallel{}BC\ \ \ Q.E.D.$


function
logarithms
calculus
integration
limit
real analysis
real analysis solved

find an equivalent of:
U_n=sum( ln(k)^p, k=2..n) where p is a fixed real...
(that is find a function f so that :  f(n)/U_n -> 1 )

The answer is not unique, but the simplest form of the answer is $f(n)=n(\ln n)^p$.

Note that the sum diverges, and hence any constants or bounded functions will be much smaller than $f(n)$ as $n\to\infty$ and can be safely ignored.

Since the terms of the sum are monotone, the sum may be approximated both above and below by an integral, with the difference between the upper and lower integral estimates being no more than either a constant or the last term, whichever is larger. In either case, that difference is much smaller than $f(n)$ as $n\to\infty$ and can be safely ignored.

So we consider $\displaystyle \int_a^x (\ln t)^p dt$ for some convenient $a$. Integrate by parts to get $\displaystyle x(\ln x)^p-a(\ln a)^p - \int_a^x \frac{pt}t (\ln t)^{p-1} dt$.

I now claim that that last integral is "small" and can be safely ignored. To show that, we calculate a limit by L'Hospital's rule:

$\displaystyle \lim_{x\to\infty} \frac{\int_a^x p(\ln t)^{p-1}dt}{\int_a^x (\ln t)^pdt} = \lim_{x\to\infty} \frac{p(\ln x)^{p-1}}{(\ln x)^p} = \lim_{x\to\infty} \frac p{\ln x} = 0$. This proves what we needed.


calculus
integration
trigonometry
logarithms
calculus computations

Help me to Evaluate:
\[ I \equal{} \int\limits_{0}^{ \plus{} \infty }{\frac {\sin x.dx}{{{x}^{n}}}}\]

\[ {{I}_{2}}\equal{}\int\limits_{0}^{\plus{}\infty }{\frac{\sin x.\log x.dx}{x}}\]

There exists $ \int_C e^{ \minus{} z}z^{\nu } d z \equal{} (e^{2\pi i \nu} \minus{} 1)\Gamma(\nu \minus{} 1)$in Arfken's Mathematical Methods for Physicists Page 505

The contour C is in the attachment .

I think take $ \minus{} z \equal{} ix$will solve this
\[ I \equal{} \int\limits_{0}^{ \plus{} \infty }{\frac {\sin x.dx}{{{x}^{n}}}}\]

[quote="Dr Sonnhard Graubner"]hello, for the first integral, what do we know about $ n$?
Sonnhard.[/quote]
n is a real number.


algebra
polynomial
algorithm
absolute value
Ring Theory
superior algebra
superior algebra unsolved

I've been trying to prove that the cyclotomic polynomials are irreducible over $ \mathbb{Q}$. I've shown that $ \Phi_{p^e}$ is irreducible via Eisenstein's criterion, but I'm having trouble in the general case. If we were trying to show the irreducibility of $ \Phi_n$ with $ n\equal{}p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, we could first adjoin $ \zeta_{p_1^{e_1}}$ and then $ \zeta_{p_2^{e_2}}$. We would need to show then that the degree of $ \zeta_{p_2^{e_2}}$ is still $ \phi(p_2^{e_2})$. This would require showing that we can still use Eisenstein's criterion for the prime $ p_2 \neq p_1$ in the integer ring of $ \mathbb{Q}(\zeta_{p_1^{e_1}})$. Is this true? I have a vague feeling it might have something to do with $ p_1$ being the only prime that ramifies in this extension, but I'm not that experienced with algebraic number theory.

If so, how would you prove that Eisenstein's criterion still works, and how would you prove that $ p_1$ is the only prime that ramifies? Is this even a natural way to show the irreducibility of cyclotomic polynomials?

There is a more elementary way, but it is sort of long. The basic outline is something like:

Let $ f$ be an irreducible monic factor, and $ \zeta$ a prim nth root of unity such that $ f(\zeta)\equal{}0$. By the division algorithm, for any m, $ f(x^m)\equal{}q_m(x)f(x)\plus{}r_m(x)$.

So $ f(\zeta^m)\equal{}r_m(\zeta)\Rightarrow r_{n\plus{}m}(\zeta)\equal{}r_m(\zeta)$. Thus there are only a finite number of distinct $ r_m$.

Thus $ r_{n\plus{}m}(x)\equal{}r_m(x)$ for all m.

Examine $ f(x^p)\equiv (f(x))^p \mod p$.

This gives $ r_p(x)\equiv (\minus{}q_p(x)\plus{}(f(x))^{p\minus{}1})f(x) \mod p$.

Since $ deg r_p(x)<deg f(x)$ and $ f$ monic $ r_p(x)\equiv 0\mod p$.

Let $ N$ be greater than the absolute value of all the coefficients on all the $ r_m$. Choose $ p>N$. Then $ r_p(x)\equal{}0$, so $ f(\zeta^p)\equal{}0$.

There are only finitely many primes less than $ N$. Let $ m$ be the product of them that do not divide $ n$. (If all divide $ n$, choose $ m\equal{}1$). Then $ (m,n)\equal{}1$. Let $ h$ be any integer such that $ (h, n)\equal{}1$.

By the Chinese Remainder Theorem, there is some $ h'$ such that $ h'\equiv h \mod n$ and $ h'\equiv 1 \mod m$.

Thus $ \zeta^{h'}\equal{}\zeta^h$ and all primes dividing $ h'$ are greater than $ N$. But for any prim nth root of unity such that $ f(\xi)\equal{}0$, we get $ f(\xi^p)\equal{}0$ for all primes greater than $ N$. Also, if $ p$ doesn't divide $ n$ , then $ \xi^p$ is a primitive nth root of unity.

Continue to use these facts to get $ f(\zeta^{h'})\equal{}0$ and $ f(\zeta^h)\equal{}0$. i.e. every primitive nth root of unity is a root of $ f(x)$. But this means it is the entire cyclotomic polynomial, and at we chose at the start $ f$ to be irreducible.

I think enough is there, I can fill in any details necessary.

Thanks. I'm still curious about Eisenstein's criterion in extensions, though.

Yes, you can still use Eisenstein's criterion (but for prime ideals, then) by considering a prime ideal factor of $ p$ in the extension.
Or: As $ p$ ramifies completely in $ \mathbb Q [\zeta_{p^s}] |  \mathbb Q$ and is unramified in $ \mathbb Q [\zeta_m] | \mathbb Q$ for $ \gcd(p,m)\equal{}1$, by looking at the degree of ramification,  the result follows immediatly. A similar way would be to use the discriminant formulas to get that the extensions are either disjoint (by coprimality) or must have the desired degree (by the formulas).


calculus
integration
trigonometry
function
calculus computations

The following integral once appeared in the "Landau Minimum" test:
Find:

\[ \displaystyle\int_{\minus{}\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{100}x \, \, dx \equal{}\, ?\]

[hide="hint"]Take into account that Landau was a physicist... :D [/hide]

[hide="Hint"]
Prove that $ \int_{0}^{\pi} \sin ^{2n} x dx \equal{} \frac {2n\minus{}1}{2n} \int_{0}^{\pi} \sin ^{2(n \minus{} 1)} x dx$
[/hide]

[hide="Answer"]One can prove that, for natural p,

$ \int_0^{\pi/2}(\cos \theta)^p\,d \theta \equal{} \frac {1 \cdot 3 \cdot 5 \cdot ... \cdot (p \minus{} 1)}{2 \cdot 4 \cdot 6 \cdot ... \cdot p} \cdot \frac {\pi}{2}$, for p even, and $ \equal{} \frac {2 \cdot 4 \cdot 6 \cdot ... \cdot (p \minus{} 1)}{1 \cdot 3 \cdot 5 \cdot ... \cdot p}$, for p odd.

Also, $ \int_{ \minus{} \pi/2}^{\pi/2} (\cos \theta)^p\,d \theta \equal{} 2 \int_0^{\pi/2}(\cos \theta)^p\,d \theta$.

So, in your case,

$ I \equal{} 2 \cdot \frac {1 \cdot 3 \cdot 5 \cdot ... \cdot 99}{2 \cdot 4 \cdot 6 \cdot ... \cdot 100} \cdot \frac {\pi}{2} \equal{} \frac {1 \cdot 3 \cdot 5 \cdot ... \cdot 99 \cdot \pi}{2 \cdot 4 \cdot 6 \cdot ... \cdot 100}$.[/hide]

CAn we use wallis formula????

That would be a better option...

this problem can also be solved using the [url=http://en.wikipedia.org/wiki/Beta_function]beta function[/url]

walli's formula would be a lot easier.. u know...

since it is an even function so the integral transforms to 2*integral between 0 to \pi/2... and then direct application of the wallis formula


Euler
geometry
analytic geometry
geometry proposed

Let triangle $ABC$ is given. $A_a$ is excenter of excircle that touches side $BC$. Proof that Nagel line of triangle $ABC$, Euler line of triangle $BCA_a$ and line $BC$ intersects in one point.

Use barycentric coordinates

$B=(0,1,0), C=(0,0,1), I=(a,b,c), G=(1,1,1), I_a=(-a,b,c)$

$O_{BCI_a}$ is midpoint of $II_a$, thus $O_{BCI_a}=(-a^2,b(b+c),c(b+c))$

$G_{BCI_a}=(-a,-a+2b+c,-a+b+2c)$

Equation for line $BC$ :  $(1,0,0)$ , for $IG$ : $(b-c,c-a,a-b)$ ,

for $O_{BCI_a}G_{BCI_a}$ : $((b+c)(b-c)(b+c-a),a(a-c) (b+c-a),a(b-a) (b+c-a))$

Hence it is enough to prove:
  $det\displaystyle \left( \begin{array}{ccc} 1 & 0 & 0\\ b-c & c-a & a-b \\ (b+c)(b-c) & a(a-c) & a(b-a) \\ \end {array}\right)= 0$

which is true.

I hope my calculations are corect.

The 3 concurrent lines: Nagel line of ABC, side of ABC and Euler line of triangle

   built on this side  must be radical lines of some 3 circles.

   What are these circles?

  T.Y.

  M.T.


:arrow:  [b]PROBLEM 1

The symbol 3! means 3 x 2 x 1 which = 6. Similarly 5! means 5 x 4 x 3 x 2 x 1, which = 120, and so on. suppose that N! ends in exactly 3 zeros after fully multiplying out. What is the smallest value that N can have?[/b]

There is already a marathon going.

[hide="Solution"]For there to be $ 3$ terminal zeros, there must

be at least $ 3$ factors of $ 5$ and $ 3$ factors of $ 2$ in the

prime factorization. The smallest such number will be $ 15!=
15 \cdot 14 \cdot 13 \cdot 12 \cdot \hdots \cdot 3 \cdot 2 \cdot
1$, which has the factors of $ 5$: $ 5$, $ 10$, and $ 15$

Therefore, the answer is $ \boxed{15}$[/hide]

New Problem:

Let $ S(n)=n(n+(n-1)+(n-2)+\hdots+3+2+1)$.

So, $ S(1)=1(1)=1$, $ S(2)=2(2+1)=6$, $ S(3)=3(3+2+1)=18$,

and so on. Compute $ S(100)$.

^^^^^

I made that one up so if it isn't MOEMS level tell me.

uh yeah that isn't moems level :cursing:  :cursing:  :cursing:

anyways it's just like the problem u posted in the other marathon. And we already have a running marathon. Please lock this mods.


geometry
circumcircle
similar triangles
geometry proposed

$ K \in BC, L \in CA, M \in AB$ are feet of the internal angle bisectors of a $ \triangle ABC$ and $ F$ its Feuerbach point - the tangency point of its 9-point circle and incircle. Show that the points $ K, L, M, F$ are concyclic.

Dear Yetti and mathlinkers,
this is a nice result.
You can see :  http://forumgeom.fau.edu/FG2001volume1/FG200117.pdf
Sincerely
Jean-Louis

Thank you very much Mr, jayme and Mr.Yetti for this nice concyclity result.

Thank you for the reference, I did not know this metric proof. But I have a very short synthetic proof, with no calculations:

Let $ (N)$ be the 9-point circle, $ (I)$ the incircle, $ (I_a), (I_b), (I_c)$ the excircles against $ A, B, C,$ and $ F_a, F_b, F_c$ are external Feuerbach points of the $ \triangle ABC.$ $ F$ is the external similarity center of $ (N), (I),$ $ F_a$ is the internal similarity center of $ (N), (I_a),$ and $ K$ is the internal similarity center of $ (I), (I_a).$ By Monge - d'Alembert theorem, $ F, F_a, K$ are collinear. Similarly, $ F, F_b, L$ and $ F, F_c, M$ are collinear. Therefore, $ \angle LFM \equal{} \angle F_bFF_c.$

According to [url=http://www.mathlinks.ro/viewtopic.php?t=276293]Two similar triangles[/url], the triangles $ \triangle KLM \sim \triangle F_aF_bF_c$ are similar $ \Longrightarrow$ $ \angle MKL \equal{} \angle F_cF_aF_b.$ Since $ F, F_a, F_b, F_c \in (N)$ are concyclic, so are $ F, K, L, M.$ And of course, I have a very short synthetic proof of $ \triangle KLM \sim \triangle F_aF_bF_c.$
______________________________

If $ X \in BC, Y \in CA, Z \in AB$ are feet of the external angle bisectors of the $ \triangle ABC,$ then $ (K, Y, Z, F_a), (L, Z, X, F_b), (M, X, Y, F_c)$ are concyclic in exactly the same way.

Dear Yetti,
your proof is the same as mine...
For more reference, the collinerity F, Fa, K has been proved by Naraniengar M. T., Progress Report of the India Mathematical Club (oct. 1908).
Sincerely
Jean-Louis

The following more general statement holds:

[b]Theorem[/b]. Let $ P$ be a point in plane of a triangle $ ABC$. Let $ A_{1}B_{1}C_{1}$, $ A_{2}B_{2}C_{2}$ be its pedal and cevian triangles, respectively. Prove that the circumcircles of $ A_{1}B_{1}C_{1}$, $ A_{2}B_{2}C_{2}$ meet on the nine-point circle of $ ABC$.

I remember Darij calls this concurrence point a Poncelet point (in an older thread). However, I can't find a synthetic proof. Any thoughts?

Dear Cosmin and Mathlinkers,
in my litterature, this common point is knowed under "Fontené's point". But, i think that there is other denomination.
What is new for me, is that the cevian circle passes also through this point...  Any reference?
Darij Grinberg has write a nice paper on his website concerning "A generalization of the Feuerbach's point".
Sincerely
Jean-Louis

Dear Jean-Louis,

That paper of Darij is not related to this result (that generalization involves Anti-Steiner points). I've taken the terminology from [url]http://www.mathlinks.ro/viewtopic.php?t=109112[/url].

Dear Mathlinkers,

[url=http://jl.ayme.pagesperso-orange.fr/Docs/31.%200.%20Le%20triangle%20I%20cevien.pdf] here [/url]  problem 24.

Sincerely
Jean-Louis

Does anyone know how to prove $ \triangle KLM \sim \triangle F_aF_bF_c$ synthetically?


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

here is what I have in mind:

Find out all X in M_2(R) so that we have [tex]X^{2004}-11X^{2003}+30X^{2002}=I_2. [/tex]

have fun!

This is not open question material.
Take p(x)=x^2004-...-1. Take m be the minimal polynomial of X. Then m|p.

All you have to do is to take the roots of p. Not an easy task, perhaps impossible, but it is the algorithm.

If deg(m)=1, then X is clear.
If deg(m)=2, then X is similar to
0 1
-a2 -a1
 where x^2+a1*x+a2 is m.


function
integration
calculus
limit
abstract algebra
advanced fields
advanced fields unsolved

Given the definition of dirac delta function as you can read it in a quantum mechanics book, i.e. $ \delta(x)=0 \ \text{for} \  x\neq 0, \delta(0)=\infty, \ \text{and} \int_a^b \delta(x) f(x) \ \mathrm{d} x=
\{
\begin{array}{rl}
	f(0) & \text{if} \ a<0<b \\
	0 & \text{otherwise}
\end{array} $, and using the fact that $ \delta(x)=\delta(-x)$ in the sense of distributions, "show" that $ \int_0^{\infty} \delta(x) f(x) \ \mathrm{d} x=\frac{1}{2} f(0)$.

Symbolic integral $ \int_0^\infty f(x)\,dx$ is fine for ordinary functions, since the two possible meanings of integrate on $ [0,\infty)$ and integrate on $ (0,\infty)$ yield the same answer.  But for delta functions and other distributions that are not functions, you need to specify which is your meaning.  Or make some other convention, such as "count the boundary point as half".

does it make any difference if i add that $ f$ is continuos at $ x\equal{}0$? or do you mean that without saying anything else than what i've said, which is exactly waht that book says, it's not possible to give an exact meaning to that formula?

The only difference continuity of $ f$ makes is that then $ f\cdot \delta$ is a meaningful distribution. The boundary convention issue still applies.

The "correct answer" is clearly $ \int_0^\infty\delta(x)f(x)dx\equal{}\frac{1}{2}\int_{\minus{}\infty}^\infty\delta(x)f(x)dx\equal{}\frac{1}{2}f(0)$, where the second integral uses the unique even extension of $ f$ to the entire real line.  Does this "answer" make any sense?  Absolutely not.  But neither does the question, since the question could only have been posed by a physicist.

But here's what the physicist meant:  If you think of "the delta function" as a limit of functions $ \phi_n$ with the property that the $ \phi_n$'s are even, then $ \int_0^\infty\delta(x)f(x)dx\equal{}\lim_{n\to\infty}\int_0^\infty \phi_n(x)f(x)dx\equal{}\frac{1}{2}f(0)$ using the argument given above for the $ \phi_n$'s in place of $ \delta$.  (I should point out that this is NOT how mathematicians think of the delta function.  In mathematics, you might refer to this as a specific "summability kernel" or some "particular model" of the delta function.)

Btw, I don't believe this has anything to do with "boundary conventions."  Even if you specify that you're working with $ [0,\infty)$ or $ (0,\infty)$, the question still makes no sense (and yields the same $ \frac{1}{2}f(0)$ answer using the physicist-type reasoning I described above).


I need help on the following problem:

In the figure below, PQ is a diameter of the circle, whose radius is 5 units. Line segment SR is perpendicular to line segment PQ. If PR = 9, find the area of triangle PSQ.
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=807[/img]

[hide="Answer"]The answer is 15 sq. units but there is no solution.[/hide]

Since $SR^2=PR\cdot QR=9\cdot (10-9)=9,$ so $SR=3$ units.
Thus the desired area is $\triangle{PSQ}=\frac{1}{2}\cdot 10\cdot 3=15$ sq. units.

[hide]Let a=PS, and h=SR.
$\triangle PRS\sim \triangle PSQ$
$\implies \frac{9}{a}=\frac{a}{10}\implies a^2=90$.

Using Pythagorean Theorem, $\sqrt{a^2-9^2}=\sqrt{90-81}=3=h$.
Area $=10*3/2=\boxed{15}$[/hide]

[hide]First of all, it's easy to see that $RQ=1$.
Because $\triangle PSQ\sim \triangle SRQ$, we know that $\frac{1}{SQ}=\frac{SQ}{10}$.  This gives $SQ^2=10 \Rightarrow SQ=\sqrt{10}$.
By the Pythagorean Theorem, $10-SR^2=1 \Rightarrow SR^2=9 \Rightarrow SR=3$.
So the area of $\triangle PSQ$ is $\frac{3\times10}{2}=\frac{30}{2}=\boxed{15}$[/hide]

[hide]
Because angles inscribed in semicircles are right angles, PSQ is a right triangle in S.
Because SR is the altitude to the hypotenuse, it is the geometric mean of 9 and 1 (we know RQ is 1 because the diameter is 10).
So the height is 3 and the base is 10, and the area is 15.
[/hide]


"Find a point (a,b) on the curbe y=x^2 such that the line through the point (a,b) and the point (1,1) has slope of 9/2."

Using slope formula we get 1-b=9/1-a=2. Base on the graph, you can't have a Y that is negative which would force b to be 10. If b = 10 then a has to be 3 to get a slope of -9/-2 which then converts to 9/2.

"Find the point above the X-axis on the circle of a radius 12 centered at (0,0) that is 10 units away from the point (12,0)."

I'm quite sure this one involves the distance formula, and i notice the triangle form by these points is a isosceles triangle.

Edit - Actually i think i found the solution to one of my problems but i don't think if it is right.

"Find a point (a,b) on the curbe y=x^2 such that the line through the point (a,b) and the point (1,1) has slope of 9/2."

ok we know y-1=2/x-1=5, ok we also know y=x^2, so i plug x^2 into the slope forumula.

X^2-1=2/x-1=5, then I multiple (x-1) to both sides to get 5x^2-1=(2/5x-2/5) and from furthur factoring I get x= -3/5,1 and since i know 1 is already taken i know that -3/5 is the number so i plug it into the graph of y=x^2 and i get y= 9/25. the coordinate is (-3/5,9/25) am i right?


The number 123 is shown on a computer monitor.

At the end of every minute the computer adds 102 to the number

on the  screen.

With his program, Smarty can interfere with this process by changing

the order of the digits of the number on the screen at any  time,

as many times as he wants to. Could Smarty write such a program that

would guarantee that we will see only 3-digit numbers coming up on the

screen forever?

Unless the three digits of the number on screen before him are $9,9,9$ or $8,9,9$ he just has to put the digits in increasing order to stay a 3-digit number.

We prove that he can alsways avoid being in this situation.
He need not ever get to 999, because if he is given 897 he can change it to 789. He need never get 998, because if he has 896, he can instead write 698. He need never get 989, because given 897, he can change it to 798. He need never get 899, because instead of getting 797 he can change it to 779. So he can always guarantee to not get $8,9,9$ or $9,9,9$


modular arithmetic
inequalities
quadratics
number theory
number theory proposed

$ p^{p \plus{} 1 } \plus{} (p \plus{} 1)^{p}$ can never be a perfect square for prime $ p$ .

Suppose the exist of primes $ p$ such that $ p^{(p\plus{}1)}\plus{}(p\plus{}1)^p$ is perfect square. Set:
$ p^{(p\plus{}1)}\plus{}(p\plus{}1)^p\equal{}k^2$.
If $ p\equal{}2,$ ta có $ p^{(p\plus{}1)}\plus{}(p\plus{}1)^p\equal{}...$, isn't perfect square.
If $ p\not\equal{}2$. We have:
$ (p\plus{}1)^p\equal{}(k\minus{}p^{\frac{p\plus{}1}{2}})(k\plus{}p^{\frac{p\plus{}1}{2}})$.
Set $ k\minus{}p^{\frac{p\plus{}1}{2}}\equal{}n, k\plus{}p^{\frac{p\plus{}1}{2}}\equal{}m.$ We have$ 2p^{\frac{p\plus{}1}{2}} \equal{}m\plus{}n$.
Therefore $ \gcd(m, n)\equal{}1$ or $ 2$.

[b]Cases 1[/b].$ \gcd(m, n)\equal{}1$. We have $ m\plus{}n$ is even number so $ m, n$ lodd.
On the other hand $ mn$ even.

[b]Cases 2[/b]. $ \gcd(m, n)\equal{}2$. Do $ mn\equal{}(p\plus{}1)^p$ and $ 2\parallel{}m\plus{}n$ assume $ m\equal{}2r^p$, $ n\equal{}2^{kp\minus{}1}s^p$.
We have $ p^{p\plus{}1}\equal{}r^p\plus{}2^{kp\minus{}2}s^p$ so $ p\mid r^p\plus{}2^{kp\minus{}2}s^p\Rightarrow p\mid r\plus{}2^{k\minus{}2}s\Rightarrow p\mid 1\plus{}(2^{k\minus{}1}s)^2$. Thus $ p$ without form $ 4z \plus{}3$. So $ k \equal{} 1$.
We have $ p^{p\plus{}1}\equal{}r^p\plus{}2^{p\minus{}2}s^p\Rightarrow p\mid 2r\plus{}s$.
Because $ 2rs\equal{}p$. Assume $ s<[\sqrt{p}]$, ta có $ s^2\plus{}1<2p$. Thus $ s^2\plus{}1\equal{}p$.
That $ 2rs\equal{}p\plus{}1$, so $ 2r\equal{}\frac{s^2\plus{}2}{s}$. Inferred $ s\in\{1, 2\}$.
Try both cases that do not such that.
So do not exist primes $ p$ satisfy.

the case$ p\equal{}2,3,5$ are easy to check.
we have $ (p\plus{}1)^{p}\equal{}(k\minus{}p^{\frac{p\plus{}1}{2}})(k\plus{}p^{\frac{p\plus{}1}{2}})$
assume $ s\equal{}\frac{p\plus{}1}{2}$
we have two cases:
1-$ k\minus{}p^s\equal{}2^{p\minus{}1}l^p \ , \ k\plus{}p^s\equal{}2m^p \ , \ 2ml\equal{}p\plus{}1$
by checking modulo $ p$ both of the first equation we get :$ l \equiv 2m \pmod{p}$
so it means that
a)$ p \le |2m\minus{}l|$ but we know that $ 2ml\equal{}p\plus{}1$ which implies $ |2m\minus{}l | \le p$ and the equality occurs if $ l\equal{}1,m\equal{}\frac{p\plus{}1}{2}\equal{}s$.but then , the equality must occur and we have $ p^s\equal{}m^p\minus{}2^{p\minus{}2}$ by putting $ m\equal{}s$ we get $ p^s\equal{}s^p\minus{}2^{p\minus{}2}$
mutiply each side by $ 2^p$ and we get:
$ 2^pp^s\equal{}(p\plus{}1)^p\minus{}2^{2(p\minus{}1)}$ or in other words $ 2^pp^s\equal{}p^p\plus{}p^p\plus{}(...)\minus{}2^{2(p\minus{}1)}$
but we have $ 2^pp^s\equal{}p^p\plus{}p^p\plus{}(...)\minus{}2^{2(p\minus{}1)}>p^p\plus{}(p^p\minus{}2^{p\minus{}1})>p^p$ or in other words we get:
$ 2^p>p^{\frac{p\minus{}1}{2}}$ or in other words we get $ p<7$.which has been checked.
b)$ 2m\equal{}l\equal{}d$ which directly implies $ m|p^s$ ad since $ m<p^2$ we get $ m\equal{}1,p$.if $ m\equal{}p$ then $ p|p\plus{}1$.contradiction.and if $ m\equal{}1$ then $ l\equal{}2$ and we have $ p\plus{}1\equal{}4$ or $ p\equal{}3$ which has been checked.or we can see that $ d^2\equal{}p\plus{}1$ and so we r done!!
2-$ k\minus{}p^s\equal{}2l^p \ , \ k\plus{}p^s\equal{}2^{p\minus{}1}m^p \ , \ 2ml\equal{}p\plus{}1$
proceeding similarly we get two cases:
a)$ p \le |2l\minus{}m|$ and also we have $ |2l\minus{}m|\le p$ which implies that $ l\equal{}s$ and $ m\equal{}1$ and then by using inequality we r done for this case.
so let's check the second case and finish it...
b)$ 2l\equal{}m\equal{}d$ and similarly it implies $ d^2\equal{}p\plus{}1$ and we r done!

If $ p = 2,3,5,7,11,13,17$ then it is easy to check that $ p^{p + 1} + (p + 1)^p$ is not a perfect square.
Now consider $ p \geq 19$ then $ p + 1 %Error. "divisible" is a bad command.
2$. Suppose that there exists such a positive integer $ a$ such that $ a^2 = (p + 1)^p + p^{p + 1}$ then $ \left(a - p^{\frac {p + 1}{2}}\right)\left(a + p^{\frac {p + 1}{2}}\right) = (p + 1)^p$. But $ \gcd{\left(a + p^{\frac {p + 1}{2}}, a - p^{\frac {p + 1}{2}}\right)} \in \{1,2\}$ and since $ (p + 1)^p %Error. "divisible" is a bad command.
2^p$ then $ \gcd{\left(a + p^{\frac {p + 1}{2}}, a - p^{\frac {p + 1}{2}}\right)} = 2$. Thus either $ a + p^{\frac {p + 1}{2}} = 2r^p$ and $ a - p^{\frac {p + 1}{2}} = 2^{p - 1}s^p$ or, $ a - p^{\frac {p + 1}{2}} = 2r^p$ and $ a + p^{\frac {p + 1}{2}} = 2^{p - 1}s^p$. Suppose that the latter case happen (we will use the same argument for the left case) then:
\[ 2p^{\frac {p + 1}{2}} = 2^{p - 1}s^p - 2r^p\]
But [i]Fermat's little theorem[/i] implies that $ 2^{p - 1}s^p - 2r^p \equiv s - 2r \pmod{p}$ so $ s - 2r %Error. "divisible" is a bad command.
p$. On the other hand $ 2rs = p + 1$ so:
\[ |s - 2r| \leq (p + 1) - 1 = p\]
This yields $ |s - 2r| = p$. Now [i]Viete's theorem [/i]implies that $ s$ and $ - 2r$ are the two integer roots of either one of the quadratic equations $ t^2 + pt - (p + 1) = 0$ or $ t^2 - pt - (p + 1) = 0$. Since $ - 2r$ is even, either $ - 2r = p + 1$ and $ s = - 1$ or $ - 2r = - (p + 1)$ and $ s = 1$. Therefore $ (r,s) \in \{\left(\dfrac{p + 1}{2},1\right), \left( - \dfrac{p + 1}{2}, - 1\right)\}$. Hence, either $ 2a = 2 \left( \dfrac{p + 1}{2} \right)^p + 2^{p - 1}$ or $ 2a = - 2 \left( \dfrac{p + 1}{2} \right)^p - 2^{p - 1}$. In both cases, notice that $ p \geq 19$, we have:
\[ a^2 = \left( \left( \dfrac{p + 1}{2} \right)^p + 2^{p - 2} \right)^2 \nonumber \\
\Rightarrow (p + 1)^p + p^{p + 1} = \left( \left( \dfrac{p + 1}{2} \right)^p + 2^{p - 2} \right)^2 \nonumber \\
\Rightarrow (p + 1)^p + p^{p + 1} = \left( \dfrac{p + 1}{2} \right)^{2p} + 2^{p - 1} \left( \dfrac{p + 1}{2} \right)^p + 2^{2p - 4} \nonumber \\
\Rightarrow \dfrac{(p + 1)^p}{2} + p^{p + 1} = \left( \dfrac{p + 1}{2} \right)^{2p} + 2^{2p - 4} \nonumber \\
\Rightarrow 2^{2p - 1} (p + 1)^p + 2^{2p} p^{p + 1} = (p + 1)^{2p} + 2^{4p - 4} \nonumber \\
\Rightarrow 2^{2p - 1} (p + 1)^p + 2^{2p} (p + 1)^{p + 1} > (p + 1)^{2p} + 16^{p - 1} \nonumber \\
\Rightarrow 2^{2p - 1} + 2^{2p}(p + 1) > (p + 1)^p + \dfrac{16}{p - 1} \nonumber \\
\Rightarrow 2^{2p - 1} + 2^{2p}(p + 1) + 1 \geq (p + 1)^p \nonumber \\
\Rightarrow 2^{2p - 1} (2p + 3) + 1 \geq (p + 1)^p \nonumber\]
Since $ 2p + 3 = o(2^p)$, there exists $ K$ such that $ (p + 1)^p < 2^{Kp}$. In fact, $ K$ is just about $ 4$ or $ 5$ then (easy to check). These rest cases obviously does not satisfies the condition, which ends the proof.

P/S: My $ 300^\text{th}$ post :lol: .


ratio
LaTeX
linear algebra
matrix
geometric sequence
geometric series
complex numbers

if $ z$=$ \frac{1+i}\sqrt{2}$ is a complex number the problem asked for the sum $ M=1+z+z^2+..+z{}^2{}^0$
my problem is that can i use the logic of sum of Geometric progression to solve this problem?
note that z is not real so anyone with a solution post it please

Question: does the proof of the formula for the sum of a geometric series depend on any special properties of the variable?
Answer: yes.  It depends on the fact that the basic laws of arithmetic apply to the variable, and that the variable is not equal to 2.
Follow-up question: Okay, but that answer is pretty irrelevant since the laws of arithmetic certainly apply in the complex numbers.  Does the proof depend on the fact that the variable is real?

(What's the point of these questions?  The point is that the logical structure in mathematics is all about [i]why[/i] things are true.  To memorize a mathematical result (like a formula) is not to understand it; to learn its proof (so that, for example, you could repeat it to someone who hadn't seen it before) is a vastly closer approximation.)

yes in the proof of the sum of gp it is said that if the variable/common ratio is greater than 1 you apply one formular
if the common ratio is between 1 and -1 you apply a second formular
also there is no way to say if a complex is greater than or less than one.
well the procedure i am thinking is converting to the trigonometric form and using demovires theorem
any additional suggestions?
any solutions?

[quote="21 7 15"]yes in the proof of the sum of gp it is said that if the variable/common ratio is greater than 1 you apply one formular
if the common ratio is between 1 and -1 you apply a second formular[/quote]  First of all, "formula."  Second of all, your response does not engender me with confidence that you know what the word "proof" means.  The proof is a demonstration that the formula is correct.  In this case, you're completely wrong that the proof depends on whether $ x$ is larger or smaller than $ 1$.  (This begins to matter when we consider the sum of an [i]infinite[/i] geometric series, but you have only finitely many terms written.)

By the way, your LaTeX code is broken in a coulple ways: you should have \frac{..}{\sqrt{2}}, the first equation should be a single pair of dollar signs (the first one before z, the last one after the end of the fraction), and for an elipsis you can use \ldots or \cdots.

all i am saying is that in the formular for sum of gp(both for finite and infinte series) a condition is given and that condition trys to compare the size of the common ratio and a given real number.
IN MY QUESTION,the common ratio is not real so cannot be compared(its size is not known).
i'm asking if it is possible to use the idea of gp.
i'll like to see any anwsers or opinion.
@JBL is $ i>1$ where $ i$ is a complex number.

[quote="21 7 15"]@JBL is $ i > 1$ where $ i$ is a complex number.[/quote]
excuse me? i think, complex numbers cant be compared with purely real numbers..

complex numbers cannot be compared with anything because they are not real but imaginary(imagination)
hence can a complex number be my common ratio?

Non-real numbers are not second class citizens.

You're adding terms that look like $ 1, z, z^2, z^3, \ldots$. The ratio between any two successive terms in that sequence is $ z$, so the common ratio is $ z$, whether $ z$ is real, or imaginary, or complex, or a matrix, or anything else that you can exponentiate and sum (so long as $ z \neq 0$).

If division by $ 1 \minus{} z$ is meaningful, then you can multiply a series of the form $ (1 \plus{} z \plus{} z^2 \plus{} z^3 \plus{} \ldots \plus{} z^N)$ by $ \frac {(1 \minus{} z)}{(1 \minus{} z)}$ (which is 1, so it doesn't change the series), multiply the numerators, and cancel terms to get the equivalent expression $ \frac {1 \minus{} z^{N \plus{} 1}}{1 \minus{} z}$. Try to do this yourself and see why the terms cancel.

Reflecting on the procedure, the only time in which you had to care in the least bit [i]what[/i] $ z$ is was the step when you had to divide by $ 1 \minus{} z$. When you're dealing with numbers, you can divide by $ 1 \minus{} z$ as long as $ z \neq 1$. And the value of $ z$ in your original post is certainly not equal to $ 1$.

Using an infinite series for $ \sum_{k\equal{}0}^{\infty} z^k$ and subtracting $ \sum_{k\equal{}21}^{\infty} {z^k}$ yields $ (1 \plus{} \sqrt2)i$.

@TZF i now see it but one more thing why not divide by z-1 ?
will it produce diffrent anwsers?
@goldenboy correct anwser did you use the exponetial rep. of complex numbers?

[b]@ 21 7 15:[/b] You're probably remembering the formula $ \frac {z^{N \plus{} 1} \minus{} 1}{z \minus{} 1}$, which is equal to $ \frac {1 \minus{} z^{N \plus{} 1}}{1 \minus{} z}$. The only difference is that the numerator and denominator both differ by a factor of -1, which therefore doesn't make a difference.

[b]@ goldenboy1.618:[/b] Careful! The two infinite series you used actually do not converge, so the values you computed for them are incorrect.

Note that the absolute value of $ \frac {1 \plus{} i}{\sqrt {2}}$ is $ 1$, and an infinite geometric series converges iff. the common ratio has absolute value [i]less than[/i] 1.

@TZF thanks for that
now all you ladies have had your chance to show that you know your math very well but you all forgot the anwser???

[quote="21 7 15"]all i am saying is that in the formular for sum of gp(both for finite and infinte series) a condition is given and that condition trys to compare the size of the common ratio and a given real number.[/quote]  I understand that you were saying this.  However, you are completely wrong: the proof (as now thoroughly demonstrated on this page; if you can't figure out the answer from what's here, you might be more incompetent than your well-deserved 1.5 rating suggests) only cares whether $ z \equal{} 1$; it doesn't matter in the slightest whether $ z > 1$ or $ z < 1$, and consequently the formula needs no such condition.

@JBL got it already.
then give me your solution/outlay
to much ARGUMENT about z has made me forget that the MODULUS of 1 is1
hence making complex i=2
@everybody just joking plz anybody with a solution/anwser?

do your own homework=i don't know how to do it/i wish i knew


algebra
function
domain
Ring Theory
superior algebra
superior algebra unsolved

Is it true that all ideals in ${\mathbb{Z}[\sqrt{-5}}]$ are of the form $(a,b)$? How would you prove?

(I'm trying to prove that its prime ideals are PID's, but I'm a bit stuck. Any help is welcome.)

See http://www.mathlinks.ro/Forum/viewtopic.php?t=75567 , giving that all ideals are even generated by two elements as $\mathbb{Z}$-module.  One can prove it only for this case or relevance with less algebra, but I don't want to "adapt" the proof now...

It's a rather interesting fact that all ideals in Dedekind domains are generated by (at most) two elements (but we don't totally have a Dedekind ring here), see http://www.mathlinks.ro/Forum/viewtopic.php?t=49426 .

Sorry for that amount of algebra.

PS: what do you mean with "prime ideals are PID"¿

prime ideals are PIDs: they are principal ideal domains, when considered as a standalone ring. :) Thanks for the quick reply, I'll try to understand.


algebra
polynomial
blogs
quadratics
number theory
greatest common divisor

Given the polynomial $x^3+ax^2+bx+c$ find the condition the polynomial must satisfy such that it has at least one nonreal root (based on $a,b,c$).

and generalize for $a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0$.

[hide="hint"][url=http://wangsblog.com/jeffrey/]See blog post[/url][/hide]

I have figured out this much:

[hide]First, if $a^2<3b$ there will be a nonreal root, for if the roots are denoted $r$, $s$, $t$, then $r+s+t=-a$, $rs+st+tr=b$, $rst=-c$.
Then $a^2<3b\Rightarrow (r+s+t)^2<3(rs+st+tr)$.
Expanding out the square, $r^2+s^2+t^2<rs+st+tr$.
Then subtract, divide by 2, and form squares to get $\left( \frac{r-s}{2} \right)^2+\left( \frac{s-t}{2} \right)^2+\left( \frac{t-r}{2} \right)^2<0$ so a root must be nonreal by trivial inequality.

Second, if $a^2=3b$ then if the roots are real, then $r=s=t$ (above sum of squares). Substituting, $a=-3r$, $b=3r^2$, $c=-r^3\Rightarrow a^3=27c$.
Therefore there will also be a nonreal root if $a^2=3b$ and $a^3\neq 27c$.[/hide]

[quote="breez"]One coefficient must be nonreal?

A cubic (odd power) polynomial with real coefficients always has at least 1 real root. If a cubic polynomial can have up to 3 complex roots, but has all real coefficients, then the complex roots must occur in conjugate pairs. There will be 1 pair left over, as only an even number of complex root pairs will form real coefficients. Thus all odd powered polynomials with all real number coefficients must have at least 1 real root. So if a, b, or c is nonreal, then it must have a nonreal root.[/quote]

Sorry, I meant with real coefficients.

scorpius is on the right track, i actually had intended the solution to be $a^2 < 3b$ but i suppose that other case works, too. for the generalization, find $k_1$ and $k_2$ if $k_1a_{n-1}^2 < k_2a_na_{n-2} \Rightarrow$ there is at least one nonreal root.

[hide]well, we know that the complex roots have to come in conjugate pairs, so we know that it factors into an irreducable quadratic

$(x-r)(x^2+px+q)$
$x^3+px^2+qx-rx^2-prx-qr$
$x^3+(p-r)x^2+(q-pr)x-qr$

so compare coefficients, and find p and q in terms of a, b and c then plug them into the inequality...

where $p^2-4q<0$,[/hide]

[quote="scorpius119"]For the generalization i get:

[hide]$k_1=d-1,\ k_2=2d$ where $d$ is degree. If the roots are denoted $r_y$ for $1\leq y\leq d$, then
\[ \sum_{1\leq y<z\leq d} \left( \frac{r_y-r_z}{2} \right)^2<0  \]
This is as strong as possible because there might not be nonreal roots if $(d-1)a_{n-1}^2=2da_na_{n-2}$ (if the polynomial has the form $a_n(x-r)^d$ for some real number $r$).[/hide][/quote]

That's what I get, too. Basically you find $gcd\left(n^2, \binom{n}{2}\right)$ so that you have an equal number of terms on both sides. Then you know it can be factored as the sum of the squares of the pairwise differences. From here $k_1$ and $k_2$ are evident.


Does someone know any function in matlab that generates random numbers in an arbitrary interval,
or a matlab code for a function that does that? Because the functions of matlab generate only in the interval of [0,1).

The numbers should be uniformly distributed.

To simulate random number U which has uniform distribution on (a,b):
x=rand;
u=(b-a)*x+a;


What is the units digit of the least whole number greater than 1000 whose digits are all different?

1234 is the least whole number after 1000 whose digits are all different so the units digit is 4

What about $ 1023$? That would give a units digit of $ \boxed{3}$.

I never thought of that.....
Yours is propbably right


Prove that if $ ab\plus{}bc\plus{}ca\equal{}\frac{1}{3}$ ,($ a,b,c$ are positive reals) then
$ \frac{a}{a^2\minus{}bc\plus{}1}\plus{}\frac {b}{b^2\minus{}ca\plus{}1}\plus{}\frac {c}{c^2\minus{}ab\plus{}1}\ge \frac {1}{a\plus{}b\plus{}c}$

[quote="ashrafmod"]Prove that if $ ab \plus{} bc \plus{} ca \equal{} \frac {1}{3}$ ,($ a,b,c$ are positive reals) then
\[ \frac {a}{a^2 \minus{} bc \plus{} 1} \plus{} \frac {b}{b^2 \minus{} ca \plus{} 1} \plus{} \frac {c}{c^2 \minus{} ab \plus{} 1}\ge \frac {1}{a \plus{} b \plus{} c}
\]
[/quote]

It is really easy one  :wink:
We have:
\[ \frac {a}{a^2 \minus{} bc \plus{} 1} \plus{} \frac {b}{b^2 \minus{} ca \plus{} 1} \plus{} \frac {c}{c^2 \minus{} ab \plus{} 1}
\]

\[ \equal{} \frac {a^2}{a^3 \minus{} abc \plus{} a} \plus{} \frac {b^2}{b^3 \minus{} abc \plus{} b} \plus{} \frac {c^2}{c^3 \minus{} abc \plus{} c}
\]

\[ \ge \frac {(a \plus{} b \plus{} c)^2}{a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \plus{} a \plus{} b \plus{} c}\ge \frac {1}{a \plus{} b \plus{} c}
\]
In fact:
\[ \frac {(a \plus{} b \plus{} c)^2}{a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \plus{} a \plus{} b \plus{} c}\ge \frac {1}{a \plus{} b \plus{} c}
\]

\[ \iff (a \plus{} b \plus{} c)^3\ge a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \plus{} a \plus{} b \plus{} c
\]

\[ \iff 3\sum ab(a \plus{} b) \plus{} 9abc\ge a \plus{} b \plus{} c
\]

\[ \iff 3\left(\frac{p}{3}\minus{} 3r\right)  \plus{} 9r\ge p
\]
It is always correct  :)

Almost the same:

$ \frac {a}{a^2 \minus{} bc \plus{} 1} \plus{} \frac {b}{b^2 \minus{} ca \plus{} 1} \plus{} \frac {c}{c^2 \minus{} ab \plus{} 1}\equal{}$

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

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

$ \frac{(a\plus{}b\plus{}c)^2}{(a\plus{}b\plus{}c)[(a\plus{}b\plus{}c)^2\minus{}3(ab\plus{}bc\plus{}ca)]\plus{}(a\plus{}b\plus{}c)}\equal{}$

$ \frac{(a\plus{}b\plus{}c)^2}{(a\plus{}b\plus{}c)[(a\plus{}b\plus{}c)^2\minus{}1]\plus{}(a\plus{}b\plus{}c)}\equal{}\frac{1}{a\plus{}b\plus{}c}$

 :)


geometry
geometric transformation
rotation

how do you rotate a point (x,y) by k degrees? also if you translate (x,y) first does that affect it?

First you need a centre of rotation. The easiest way to visualise the rotation is to draw a line segment from the centre of rotation to the point $ (x,y)$. Then rotate this segment the number of degrees that you wanted. The end of the rotated line segment is the new position of your point.

I think from this explanation it would be obvious that translating it first would affect it.

If you give your problem i will use that as a technique to actually assist you.
I know of a few tricks


geometry
analytic geometry
trapezoid
parallelogram
graphing lines
slope
angle bisector

I am weeks behind my work due to an illness, and I have no idea what I'm doing. Any help is appreciated. Thanks.

1. Given quadrilateral PQRS with vertices P (-2,3) Q (1,0) R (7,6) and S (0,5) prove that PQRS is a trapezoid.

2. Prove, using coordinate geometry, that the diagonals of isosceles trapezoid FISH with vertices F (2a, b) I (5a, b) S (7a,0) and H (0,0) are congruent

4. Given: Parallelogram ABCD
Line CE is perpendicular to Line AB
Line CF is perpendicular to Line AD
Prove: Triangle DCF ~ Triangle BCE

5. Given: Triangle ABC is isosceles with base Line AC
Line BD bisects angle ABC
Prove: Triangle ADB is congruent to Triangle CDB

6. Given: R is the midpoint of Line QS
QS bisects Line PT
Prove: Line PQ is congruent to Line TS

7. Given: Triangle BAT, line CBTD and Line AB are perpendicular to CD

Write an indirect proof to show that line AT is not perpendicular to Line CD

5. Here's a simple proof:
AB=CB (isosceles triangle)
BD=BD(Common side)
angle ABD= angle CBD (angle bisector)
Thus triangle ADB is congruent to triangle CDB by SAS congruency (Side-Angle-Side).

4. This is easier with a diagram:   :roll:
angle BEC=angle CFD (90 degrees)
angle CBE=angle CDF (opposite angles are equal in parallelograms)
Therefore triangle CBE=triangle CDF by AA similarity (Angle-Angle)
The last angle is equal too since the sum of the angles is 180 degrees.

1) a trapezoid has atleast 1 pair of opposite and parallel sides. Thus, if you calculate the slope of line segments that compose the quadrilateral, you should be able to find two opposite sides with the same slope.

2) Draw the trapezoid so that you can choose the right endpoints for the diagonals (choose values of a if it makes it easier). Then use the distance formula.


I am to calculate the number of states in a 3Dcubic potential well with impenetrable walls that have energy less than or equal to E

$ \ E_n\equal{}\frac{\hbar^2\pi^2}{\ 2 \ m \ a^2}\ (\ {n_x}^2 \plus{} \ {n_y}^2 \plus{} \ {n_z}^2)$

We may denote $ \ (\ {n_x}^2 \plus{} \ {n_y}^2 \plus{} \ {n_z}^2)\equal{}\ n^2$
and express n in terms of E_n

Then, we can evaluate the integral n(E')dE' for E'=0 to E'=E

I am not sure if this would give the correct answer.Can anyone please help?

[quote="kolahalb"]Then, we can evaluate the integral n(E')dE' for E'=0 to E'=E[/quote]
I didn't quite understand what you are trying to say.
A state is uniquely determined by $ n_x,n_y,n_z$, and so is its energy.

So the problem boils down to finding the number of solutions of
$ n_x^2\plus{}n_y^2\plus{}n_z^2\leq \frac{2ma^2 E}{\hbar^2\pi^2}$


What is the remainder when $5^{999999}$ is divided by $7$?

Kouichi,

You have provided an interesting solution.  Do you think you could discuss your solution a bit so that it is more easily accessible to middle school students?

Basically what he did was split the $5^{999999}$ into two parts, and then simplify each $\mod7$. Then, the number is small enough to find it $\mod7$.

[quote="indianamath"]5^3=-1(mod 7)[/quote]
And $-1\equiv6\mod7$.

[quote="Kouichi Nakagawa"]$5^{999999}=5^{6\times166666}\times5^{3}=\left(5^{6}\right)^{166666}\times5^{3}\equiv 1^{166666}\times 125 \equiv 6 \ \ (mod \ 7)$
($\because$ Fermat's Little Theorem)[/quote]
how did you go from $(5^{6})^{166666}$ to $1^{166666}$? sorry if this sounds naive.

[quote="diophantient"][quote="bpms"][quote="AstroPhys"]It is according to Fermat's Little Theorem [url]http://www.artofproblemsolving.com/Wiki/index.php/Fermat%27s_Little_Theorem[/url] ,
$x^{y-1}\equiv 1 \left( mod \ y \right)$
Since they are using $mod \ 7$, $7-1 = 6$.[/quote]
Note that that's only true where y is prime.[/quote]
That would be better phrased as that's only guaranateed when y is prime.
It's possible when y is composite.[/quote]
If its composite you can use the phi function.

$\phi(n)$ is the number of integers less than $n$ than have no common divisors with $n$.

[quote="Art of Owna"][/quote][quote="diophantient"][quote="bpms"][quote="AstroPhys"]It is according to Fermat's Little Theorem [url]http://www.artofproblemsolving.com/Wiki/index.php/Fermat%27s_Little_Theorem[/url] ,
$x^{y-1}\equiv 1 \left( mod \ y \right)$
Since they are using $mod \ 7$, $7-1 = 6$.[/quote]
Note that that's only true where y is prime.[/quote]
That would be better phrased as that's only guaranateed when y is prime.
It's possible when y is composite.[/quote][quote="Art of Owna"]
If its composite you can use the phi function.[/quote]
Well you can use totient when it's prime too :P

[quote="diophantient"][/quote][quote="Art of Owna"][/quote][quote="diophantient"][quote="bpms"][quote="AstroPhys"]It is according to Fermat's Little Theorem [url]http://www.artofproblemsolving.com/Wiki/index.php/Fermat%27s_Little_Theorem[/url] ,
$x^{y-1}\equiv 1 \left( mod \ y \right)$
Since they are using $mod \ 7$, $7-1 = 6$.[/quote]
Note that that's only true where y is prime.[/quote]
That would be better phrased as that's only guaranateed when y is prime.
It's possible when y is composite.[/quote][quote="Art of Owna"]
If its composite you can use the phi function.[/quote][quote="diophantient"]
Well you can use totient when it's prime too :P[/quote]

Yea, thats why its p-1

Enough discussion on the Totient function.  It is above the level of MATHCOUNTS.  The very first response, by vamatheletes, is a perfectly fine solution, and it is at the level of MATHCOUNTS.


algebra
function
domain
calculus
derivative
inequalities proposed
inequalities

[color=darkblue] Given are a real constant $k\in (0,1)$ and  the domain $D\subset (0,1)^{2}$ defined such :

$(x,y)\in D$ $\Longleftrightarrow$ $1<x+y=1+kxy\ .$ Ascertain ( [b][u]without derivatives ![/u][/b] ) $\max_{(x,y)\in D}\{\ x+y\ \}\ .$[/color]

Let $z : = x+y$. Then $z = 1+k \cdot xy \leq 1+\frac{k}{4}\cdot z^{2}$
$\Leftrightarrow \frac{k}{4}\cdot z^{2}-z+1 \geq 0$
$\Leftrightarrow z^{2}-\frac{4}{k}\cdot z+\frac{4}{k}\geq 0$
$\Leftrightarrow z \geq \frac{2(\sqrt{1-k}+1)}{k}(>2) \vee z \leq \frac{2(1-\sqrt{1-k})}{k}$
$\Leftrightarrow z \leq \frac{2(1-\sqrt{1-k})}{k}$

Hence $\max_{(x,y) \in D}\{ x+y \}\leq \frac{2(1-\sqrt{1-k})}{k}$. If $x=y= \frac{(1-\sqrt{1-k})}{k}$,

$1+k \cdot xy = 1+\frac{(1-\sqrt{1-k})^{2}}{k}= \frac{k+1+1-k+2\sqrt{1-k}}{k}= \frac{2(1-\sqrt{1-k})}{k}=x+y$

Hence $\max_{(x,y) \in D}{x+y}= \frac{2(1-\sqrt{1-k})}{k}$


conics
ellipse
hyperbola
parabola
complex numbers
complex analysis
complex analysis unsolved

Hi folks
Can any one help me to solve the following problem :
Let a be a real number and c is a positive real number (c>0) then  find all the complex numbers that satisfy :

|z-a|-|z+a|=2c

thanks so much

This is not complex analysis.  Look over the definitions of different kinds of conic sections; this is one of them.

hi t0rajir0u
Thanks for answer my question BUT can you explain more clearly for this problem?.

thanks

Recall the following definitions:

- A circle is the set of points with fixed distance from a given point.
- An ellipse is the set of points such that the sum of the distances from two given points is a fixed value.
- A hyperbola is the set of points such that the difference of the distances from two given points is a fixed value.
- A parabola is the set of points such that the difference of the distances from a given point and a given line are zero.

One of these definitions is encoded in the problem.  Which one?


inequalities
geometry
inequalities proposed

Prove that in an acute triangle we have \[ a^4+b^4+c^4\geq4P\left(\frac{bc}{a}AH+\frac{ca}{b}BH+\frac{ab}{c}CH\right)  \] where $a,b,c$ are the sides of the triangle, $P$ its area and $H$ its orthocenter.

I think we can also say that \[ a^4+b^4+c^4\geq4P\left(b\cdot AH+c\cdot BH+a\cdot CH\right) \] because if $a$ is the largest side, then $AH$ is the smallest and vice versa, so we can use rearangement.


floor function
logarithms
real analysis
arithmetic sequence
number theory proposed
number theory

The first to questions are quite simple, but maybe also nice :) These could have made fine olympiad problems :) The third question is loosely formulated, and at the moment I have no clue what an "answer" should be.

[b](i)[/b][Easy. ] Are there (infinitely many) $a \in (1,\infty)$ with the property that every prime divides some member of the sequence $\lfloor a^{n}\rfloor, n=1,2,\ldots$?

[b](ii)[/b][Somewhat easy. ] Are there (infinitely many) $a \in (1,\infty) \setminus \mathbb{Z}$ with the property that infinitely many primes do not divide any member of the sequence $\lfloor a^{n}\rfloor, n=1,2,\ldots$?

[I add "... but not trivial" in the description of both problems. ]

-----------------------------------------------------------------------

[b](III*)[/b] For "randomly" chosen $a \in (1,\infty)$, which of the following complementary events is "more probable?"

A. Every prime divides some member of the sequence $\lfloor a^{n}\rfloor, n=1,2,\ldots$.
B. There exists a prime which does not divide any member of the sequence $\lfloor a^{n}\rfloor, n=1,2,\ldots$.

Define the terms in the "$\cdot$"'s, and share your opinion. :)

[quote="Albanian Eagle"]$a=\frac{1+\sqrt{5}}{2}$
works in (i)
 :)  but we still have to find infitely many :([/quote]

$a,a^{2},a^{3},\dots$ :wink:

Exactly :) The number $\frac{1+\sqrt{5}}{2}$ is the first in an infinite series of multiplicatively independent solutions to (i). As I formulated (i) and (ii), it sufficed to give only one example in both cases, as pointed out by nayel.  :blush:  But now that you solved (i), let's ask for a nontrivial infinitude of the solutions:

[i] Construct an infinite sequence $a_{1},a_{2},\ldots \in (1,\infty)$ such that $\log{a_{i}}/\log{a_{j}}\notin \mathbb{Q}$ for $i \neq j$, and for all $n$ and all primes $p$ there is a $k = k(n;p)$ with $p \mid \lfloor a_{n}^{k}\rfloor$. [/i]

This is what I had in mind initially...  :blush:

[quote="Albanian Eagle"]eventhough I doubt it I'm trying
$\frac{a+\sqrt{a^{2}+4}}{2}$[/quote]

 :first:  Good job :) 6 points for that answer, only 1 point remains for the proof :)

[quote="Albanian Eagle"]well it's a linear recurrence...
we have to prove $\frac{F_{n+1}}{F_{n}}$ runs through all residues in $\mathbb{Z}_{p}$ right?
I think I'm blind now  :maybe:[/quote]

Well, it is THE recurrence $X_{n+2}-aX_{n+1}-X_{n}= 0$, with initial conditions $X_{0}= 1, X_{1}= a$. Extending this recurrence backwards, we have $X_{-1}= 0$. And it's well-known that a linear recurrence containing $0$ has a subrecurrence (that is, restriction of the indices to an arithmetic progression) which vanishes in any given modulus...

Why is this equivalent to $F_{n+1}/F_{n}$ taking all residues in $\mathbb{Z}/ p\mathbb{Z}$?  :maybe:

Next question on (i), still not difficult: construct a dense subset $S \subset (1,\infty)$ with the properties: (a) for every $a \in S$ and every prime $p$ the sequence $\lfloor a^{n}\rfloor$, $n = 0,1,2.\ldots$ has a member divisible by $p$; and (b) for any different $a,b \in S$, $\frac{\log{a}}{\log{b}}\notin \mathbb{Q}$.

[And the perhaps quite difficult problem: (**) is there such a set with positive inner Lebesgue measure? ]

Sorry vess, calculation mistakes. Of course I should have used modulo periodic sequences, like I did for the first case :oops: .

ANd, actually in (ii) we can take every $\sqrt{k}+1$ where k is not a square.
I was just going to ask the question with the dense sequence  :)


number theory
least common multiple
number theory unsolved

Prove that n^5 - n is divisble by 30.

I wrote it as (n - 1)n(n + 1)(n^2 + 1) and substituted values for n
on the form 5n, 5n - 1, 5n - 2, 5n - 3 and 5n - 4 which worked,
but isn`t there a faster way to solve it?

[quote="Antilopen"]Prove that n^5 - n is divisble by 30.[/quote]

Well, $n^5-n$ is divisible by 5 because of Fermat's little theorem (since 5 is prime ;) ).
Also, $n^5-n$ is divisible by 2, since $n^5-n=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)$, and at least one of the numbers n and n - 1 is divisible by 2.
Finally, $n^5-n$ is divisible by 3, since $n^5-n=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)$, and at least one of the numbers n, n - 1 and n + 1 is divisible by 3.

Since the number $n^5-n$ is divisible by the three integers 5, 2, 3, it must also be divisible by their least common multiple $\text{lcm}\left(5;\;2;\;3\right)=30$.

And we are done.

  darij

Thanks!

I am reading "An introduction to the theory of numbers" - fifth edition by I. Niven, H. S. Zuckerman and H. L. Montgomery. I started to work with it this week and it seem to be a very good introductionary book to the theory of numbers. But I am still in the first chapter. Fermat`s little theorem is mentioned in chapter 2...

Yes. Many problems exist of that form...

Bomb


Simplify the following expression:
$\sum_{k=1}^{N}\binom{2N}{2k}\binom{2k}{k}y^{k}$

hello, Maple says this here
$2\,N \left( 2\,N-1 \right) y{\3F2(1,1-N,-N+3/2;\,2,2;\,4\,y)}$
Sonnhard.


quadratics
number theory unsolved
number theory

Let number $x^4+x^{2}.y^{2}+y^4$ can be divisible with number 11 where x,y positive numbers.Show that this number also divisible with 14641.

Look at the quadratic residues $\bmod 11$ and note that $x \equiv y \equiv 0 \bmod 11$.


can you read and write fluently in korean?

너이거 일글수 있서?LOL I have an English/Korean keyboard :P :rotfl:

Even if you have a QWERTY keyboard, you can still type Korean. Depending on your computer's OS, you can pretty type any language (I think).

And I suck at speaking, writing, reading, etc. Korean...

Oh. I am too lazy to find what the typesetting language for Korean is so I will suffice with typing in English. But at least I am able to read them instead of getting ?????

[quote="hunter34"]I have a QWERTY keyboard, but I can read Korean somewhat.[/quote]

But what nationallity exactly are you??

[hide]I can read and write fluently but not well
[/hide]

hunter is Japanese, as you can see in the sidebar.

I read and write fluently, but my writing isn't exactly legible. I left Korea in the middle of 3rd grade, so my literary skills are kinda stuck there :P

@danny: why do you put everything inside hide tags? jw :roll:

I can say alot but can't write anything but my name...

I understand Korean OK, can't carry on much of a conversation, need to use the "guess and check" method to type in Korean, can read Korean (really really slowly) and write some Korean (although I misspell about everything).

Write?
I can sorta read... (I could read better when I was 3 and lived in Korea  :D )

I would always fail korean in korea, so that's one of the reason i came to Canda, so I wont fail anything :blush:

I'm korean and I can read korean but can't write so i voted somewhat.

um im the only one who filled out the poll lol

I can read, write, and speak, but typing = rip


Suppose you roll a pair of dice. What is the probability that the positive difference of the two numbers will be 2?

The possible pairs are:

1,3
2,4
3,5
4,6
5,3
6,4
3,1
4,2

which makes a total of 8 possibilities, out of $ 6 \times 6 \equal{} 36$ total. Thus, our answer is $ \frac{8}{36} \equal{} \frac{2}{9}$.


hello all
i think being 4th place in IMO 2005 is enough reason to have an Iranian Comminunity....

so could you please create an Iranian community?

also i surely think the number of ACTIVE iranians are way more than enough (you can take a look at Olympiad NT, or inequalities for example)...

thanks

list of iranian AoPSers in the first 100   out of 229 pages of Memberlist:

sam-n
Omid Hatami
lomos_lupin
ehsan2004
amir2
payman_pm
jensen
a_vakilian
hossein11652
ashegh
ali_ms2004
ALGEBRA
boxedexe
s.m.a
kvmedya
armanf
behzad
Naser_S
mathematica
jamshidi_f2004
dashmiz
Nima Ahmadi Pour
mahdi
Ali.Kh
shuresh
Ali_aomt
shinytick
ooiler
persian
Hamide_math
abbas_mehrabian
soroushafyouni
kharazmi
noshad68_ir
incompleteness
fesghelbahal
fasttwi1
gholamali_aminian
kasra_kh
ahmadmahmoody
sheillamahouti
shiva
ahmad p.k
ahmad
peivand
amin_mohammadpour
holakou_rahmanian
samangharib
masood
amirhtlusa

[quote="persian"]I second the motion.[/quote]

I also agree.

Masoud Zargar

Atlast ,Someone suggested this.
An Iranian Community can help those of mathlinks memebrs that can not use the forum with advanced English Translation(for Instance,Consider me:I shall ,put at least 5 minutes to find a way to translate my problem in to english which can be decreased to a short  time if i use my Native language)
Therefore, if we have a Community which we will be able to use our own language in it , we will be able to use the Mathlinks-AOPS to the best way possible and Help each other in learning Math(Which i think is the the Goal of Mathlinks-AOPS sites).
There are Iranian members of Mathlinks_Aops ,that can provide strong sources of learnig math in an Iranian community ,such as Iran IMO Traning courses Professors and IMO team memebrs .
Therefore I believe having an IRANIAN community would be a great help for Iranian members of Mathlinks-Aops site.

In any case i am very thankful from the Administration group of Mathlinks-Aops for providing such a possibility for us,To learn Math.
Thank you.

Hmmm, Actually I Asked this twice from Mr.Valentin Vornicu.But I don't know the reason Why we don't have Iran Community in the best MathSite of the world.

[quote="valentin vornicu"]why there's no iranian forum in AoPS ? Is it just because there are very few iranian people here?
Yup, that's the reason. As soon as you guys gather (3-4 persons at least) we can easily make one for you
[/quote]

I can assure you we are far more than 3-4 poeple .

[hide=" PS"]The real post was for japanies ,and i Adapted it in to this form, :) [/hide]

What are the needed conditions for having an Iranian community?

as far as i know,
when some people gather and want a community then after a while the admins will create one.
this is how, the African and the Japanese comunities were formed.

but non of the admins/mods have responded to this request yet.
i remmeber that the African comunity was made within a week, of request... ;) but this request is about 20 days old, yet none of the admins have responded. :huuh:

guess we just have to wait.

The forum has been created. Enjoy it!  :)

haha,thank you , :D  :D ,We have perfect admins in Aops-Mathlinks,Hmm or just kidding  :D .

Well guys lets speak persain ,i am tired of translating due it was good for my english learning

So where is this Iranian Community Forum?

Masoud Zargar

[quote="lomos_lupin"]Atlast ,Someone suggested this.
An Iranian Community can help those of mathlinks memebrs that can not use the forum with advanced English Translation(for Instance,Consider me:I shall ,put at least 5 minutes to find a way to translate my problem in to english which can be decreased to a short  time if i use my Native language)
[/quote]

I think I remember reading some of your questions pretty well, and had you posted them in persian, I wouldn't have been able to do so.
Also, remember one thing: this community is international because people make the effort to speak in English - even though many of us are not really fluent -; of course it would be much easier for me to speak in French, but if everybody did that, fewer people would benefit.
I think the communities should be there for regional information or for peope whose English is so bad that it would almost surely impair other people's comprehension of their posts. Using a national community beyond this scope is, in my sense, going against the purpose of the math community.

hmm...
no i don't think what you're saying is neccessarily true,

assume its true, then what do you is the purpose of english speaking communites? such as canada, U.S. States, Europe, etc.

Proof by cotradiction. :P


I found the 1999 state and sprint from klebian and biffandoc (you guys are my heroes for LIFE) but unfortunately, the answers were not provided :mad: , and yes i checked the 1999 sprint WITH ANSWERS file.  please help!!

Heh, whoops. I had uploaded the wrong file.
Now check, the right one's up.
By the way, some fraction bars are missing in mine, so I'd recommend printing the test from the other download and printing the answers only from this one.

Thanks klebian!!!!!!!!!!

(the resolution came out great)


function
topology
advanced fields
advanced fields unsolved

Let $ X$ be a metric space, and $ K$ a set dense in $ X$.  Find the least cardinality $ \alpha$ such that the following statement is true, independent of the choice of $ X$ and $ K$.

For every point $ \mathbf{k}\in K$, we have a set $ \mathfrak{N}_{\mathbf{k}}$ of open balls with center $ \mathbf{k}$, such that
[list]
[*] the set $ \bigcup_{\mathbf{k}\in K} \mathfrak{N}_{\mathbf{k}}$ is a base of $ X$;
[*] for all ${ \mathbf{k}\in K}$, the relation $ \lvert \mathfrak{N}_{\mathbf{k}} \rvert \le \alpha$ holds.[/list]

Axiom of choice allowed.

[hide="Some definitions"]
A set $ K$ is [i]dense[/i] in a metric space $ X$ if every element of $ X$ is either an element of $ K$ or a limit point of $ K$.

A [i]base[/i] of $ X$ is a collection $ \mathfrak{B}$ of open subsets of $ X$ satisfying the following condition :  for every point $ \mathbf{x}\in X$ and every open set $ G$ containing $ \mathbf{x}$, there exists some $ B \in \mathfrak{B}$ such that $ x \in \mathfrak{B} \subset G$.[/hide]

I must be missing something. Why isn't the answer $ \aleph_0$? Take the open balls of radius $ 1/n$ around the points of $ K$. This is a base: Take a point $ x \in X$ and an open set $ G$ containing $ x$. There exists an open ball $ B(x,r)$ around $ x$ contained in $ G$, with a certain radius $ r > 0$. Find a point $ k \in K$ whose distance from $ x$ is less than $ r/3$ (possible since $ K$ is dense), and take the ball of radius $ 1/n$ around this $ k$ for some $ n$ satisfying $ 1/n < r/3$. This ball belongs to our proposed base, and is contained in $ B(x,r)$ and therefore in $ G$.

EDIT: this was a bit sloppy; a simple fix is to take all balls of rational radius around points of $ k$. It's then easy to make sure that the open ball in the base contains $ x$ while being contained in $ B(x,r)$.

[quote="randomgraph"]I must be missing something. Why isn't the answer $ \aleph_0$?[/quote]
It is well-known that you can do this with $ \aleph_0$ balls centered around each element of $ K$.  But can you do it with only finitely many balls around each point?

[quote]take the ball of radius $ 1/n$ around this $ k$ for some $ n$ satisfying $ 1/n < r/3$.[/quote]
Note that the ball centered at $ k$ must also contain $ x$.  This doesn't happen the way you've written it; perhaps you made a typo or something.

(By the way, sorry, I should have said "open ball" wherever I said "neighborhood" in the original post.)

Hmm. For the case $ \mathbb{R}$ with dense subset $ \mathbb{Q}$, the answer is $ \alpha \equal{} 1$, right? Choose a ball of radius $ \frac{1}{2^n}$ about every dude of the form $ \frac{a}{2^n}$ where $ a$ is odd. I think this is true more generally (by choosing such balls around $ k$-tuples of odd integers all over $ 2^n$) for $ \mathbb{Q}^k \subset \mathbb{R}^k$ which is tasty since that makes it less likely $ \alpha$ is something like the topological dimension (I seem to almost always dislike dimension theory :( ).

Hm so I have some ideas of what to do when $ K$ is countable (perhaps put it in an isomorphism with $ \mathbb{Q}$ and construct the same balls) but I was worried about $ K$ uncountable. I think I can deal with this last part though by the following argument:

All metric spaces are certainly normal so Urysohn's lemma applies or whatever and we have that $ X$ is Tychonoff. We can hence Stone-Cech compactify so that $ X$ may be homeomorphically identified with a dense subset of this compactification, so by replacing $ X$ with the compactification $ \beta X$, we may assume that the original metric space is compact. We may now reduce to $ K$ countable by the following argument: for each $ n \in \mathbb{Z}$, consider the set of balls of radius $ \frac {1}{m}$ for all $ m > n$ about all points of $ K$. They cover the metric space by the density assumption and hence there exists a finite subcover. Taking the centers of the balls in this finite subcover and taking the union over all $ n$ gives a countable subset of $ K$ which is still dense.

So one may assume $ K$ is countable (and $ X$ is compact, although I'm not sure if I'll need to use this again).

edit: wait blargh if I do this X is probably no longer a metric space

sucks.

i think the answer is 1.
first, use the axiom of choic, we can prove that there exist such a open cover:$ \bigcup_{x_a\in K}B(x_a;r_i)$  $ \forall x_a,x_b$   $ d(x_a,x_b)>\frac{1}{2}r_i$
second, since"$ \forall x_a,x_b$   $ d(x_a,x_b)>\frac{1}{2}r_i$", K-{$ x_a$} is also dense.
third, make $ r_i\equal{}\frac{1}{2^i}$ and$ K_i$ the corresponding cover. than it's obvious that all balls in $ k_i$ i$ \in$N make up a base.

sorry, I didn't quite follow from the first step -- what is the cover here?

[quote="mathiq"]i think the answer is 1.
first, use the axiom of choic, we can prove that there exist such a open cover:$ \bigcup_{x_a\in K}B(x_a;r_i)$  $ \forall x_a,x_b$   $ d(x_a,x_b) > \frac {1}{2}r_i$
second, since"$ \forall x_a,x_b$   $ d(x_a,x_b) > \frac {1}{2}r_i$", K-{$ x_a$} is also dense.
third, make $ r_i \equal{} \frac {1}{2^i}$ and$ K_i$ the corresponding cover. than it's obvious that all balls in $ k_i$ i$ \in$N make up a base.[/quote]
The answer is not 1.  Let $ N$ denote the set of reals of the form $ 1/n$, for 1 an integer.  Consider the space
\[ N \cup \{0\} . \]
Then $ N$ is dense, but every element of $ N$ constitutes an open set, so if you limit yourself to 1 open set per element of $ N$, you won't get 0.

(I didn't really follow your first step, either, so I can't say what exactly is wrong with your proof.  Why do you need the axiom of choice, and what is the function of the string "$ \forall x_a, x_b d(x_a,x_b) > r_i/2$"?)

well :blush: , you are right. my prof holds only for a metric space that doesn't have an isolated point.  my idea was like MysticTerminator,find some open cover Ki and each open cover Ki consists of some open balls whose radiu is $ \frac {1}{2^i}$, and the distance of each two centers of open balls in Ki is biger than $ \frac {1}{2^i}$, and of course each center is in the given dense set. To get these open cover, i shall use the AC.  then it's obvious that any un-isolated point A and a open set U containing A, there exist an open ball in some certain Ki that is contained in U. (and don't foget every point apears to be a center in at most one Ki.)
as to those isolated points, each has to have a open ball that only contains himself. so the answer is 1+1=2

can you judge from my poor english speaking, where am i from? :roll:


function
logarithms
inequalities unsolved
inequalities

Find the maximum and the minimum value of $ \boxed{\large{x^{y^z}}}$ where

a) $ x \plus{} y \plus{} z \equal{} 20$ and $ \boxed{2 \leq x,y,z\in \mathbb{N}^*}$
[color=red]b)[/color] $ x \plus{} y \plus{} z \equal{} 20$ and $ \boxed{2\leq x,y,z \in\color{red}\mathbb{R}}$

Do you mean N* ={0,1,2,3,… } .If so :
Since y , z belongs to N*  y^z >=0   .So  x^(y^z) >= x^ 0 =1 (x>=2)  and   min(x^(y^z))=1.
If you mean  N* ={1,2,3,… } then
y^z >=1  thus    x^(y^z) >= x^ 1 =x >=2  and min(x^(y^z))=2 .
About max ,a  small program in MATLAB software such as following may be useful .
permutation :
function s=max(n)
s=0;
for i=2:n;
    x=i;
    for j=1:n-i;
        y=j;
        z=n-(i+j);
        t=x^((y^z)/10^10);
        if t>s
        s=t;
        end
end
end
>> permutation(20)

ans =

    1.162860658098168e+000
The intended max is  ans ^(10^10) .
If x,y and z belong to R there are not any min or max .Let’s try   x=2  , y= -101 ,z=119   and  x=100 ,  y= -100 ,z=20

[hide="Minimum in a)"] Let us fix $ x$. Let $ c \equal{} 20 \minus{} x$. Then $ y^z$ is minimum when its logarithm, $ zln y \equal{} (c \minus{} y)ln y \equal{} f(y)$, is minimum. Deriving it twice, we get $ f''(y) \equal{} (\frac {c \minus{} y}{y} \minus{} ln y)' \equal{} \minus{} \frac {c}{y^2} \minus{} \frac {1}{y} < 0$. Therefore $ f$ is concave $ \Longrightarrow$ our minumum occurs when $ y \equal{} 2$ or $ y \equal{} c \minus{} 2$, i.e., $ y \equal{} 2$ or $ z \equal{} 2$. But the only exception to $ a^2 \le 2^a$ is $ a \equal{} 3$, case in which we obtain $ x^{y^z} \equal{} 15^{2^3}$. This might be the minimum. If it is not, the minimum is when $ z \equal{} 2$. Now we want to minimize $ y^2lnx \equal{} (18 \minus{} x)^2lnx$. Its derivate is $ \frac {(18 \minus{} x)^2}{x} \minus{} 2(18 \minus{} x)lnx$.

$ \frac {(18 \minus{} x)^2}{x} > 2(18 \minus{} x)lnx$
$ \Longleftrightarrow 18 \minus{} x > 2xlnx$

For $ x \equal{} 2$ this is true. For $ x \equal{} 16$ this is false. Then $ (18 \minus{} x)^2lnx$ grows until some $ x$ and starts go become lower. The minimum can be when $ x \equal{} 2$ or when $ x \equal{} 16$. Of course $ 2^{16^2} \equal{} 2^{2^8} > 2^{16} \equal{} (2^4)^4 \equal{} 16^4 \equal{} 16^{2^2}$. Now, we just need to compare $ 15^{2^3} \equal{} 15^8$ and $ 16^{2^2} \equal{} 2^{16} \equal{} 4^8$. The second one is obviously smaller. So our minimum is when $ x \equal{} y \equal{} 2$.

This is very boring. I don't want to continue. Anyway, I think maximum in letter b) is not possible to compute. If I am correct, $ x, y$ and $ z$ must satisfy

$ x \plus{} y \plus{} z \equal{} 20$
$ z \equal{} ylny$
$ y \equal{} xzlnx$[/hide]

Well, it seems that there's no beautiful solution for these...  :huh:
Thank guys for your solutions  :wink:

Something is interesting in similar problems . When x, y and z are positive integers ,the max is where x=2  .I understood this  fact among solving several examples other than 20 as sum of x, y and z by MATLAB software.