Find all the finite sequences with $ n$ consecutive natural numbers $ a_1, a_2,..., a_n$, with $ n\geq3$ such that $ a_1\plus{} a_2\plus{}...\plus{} a_n\equal{}2009$.

you gotta solve $ (2m\plus{}n\plus{}1)n\equal{}4018$ in non-negative integers, so just factor 4018...

[quote="pablo221092"]Find all the finite sequences with $ n$ consecutive natural numbers $ a_1, a_2,..., a_n$, with $ n\geq3$ such that $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} 2009$.[/quote]
I don't understand this. Do you mean that $ a_1, a_2,..., a_n$ are consecutive natural numbers? Or they are consecutive terms in the sequence?

[quote="The Nguyen"]I don't understand this. Do you mean that $ a_1, a_2,..., a_n$ are consecutive natural numbers? Or they are consecutive terms in the sequence?[/quote]

Sorry for the delay, $ a_1, a_2,..., a_n$ are consecutive natural numbers.
Many thanks.

you can see in this way a sum of consecutive naturals:
$ \sum_{n\equal{}1}^{n}{n}\minus{}\sum_{n\equal{}1}^{n\minus{}k}{n}$
so with the well known formula we have:
$ \frac{n(n\plus{}1)}{2}\minus{}\frac{(n\minus{}k)(n\minus{}k\plus{}1)}{2}\equal{} 2009\\
n^2\plus{}n\minus{}(n^2\minus{}2kn\plus{}n\minus{}k^2\minus{}k)\equal{}4018\\
k(2n\plus{}k\plus{}1)\equal{}41\cdot 7^2 \cdot 2$

now we have only to find n and k with systems of equations and after we must control when $ k>2$


At school I had to write an autobiography from the perspective of a lemon.  Can you guys look it over and give me tips please?

[hide="ESSAY"]
lol
[/hide]

I know it stinks.  That's why I need help.  Any help is appreciated.

I don't think it's that bad.  It's interesting.

[quote="JRav"]I don't think it's [b]that[/b] bad.  It's interesting.[/quote]

:rotfl:

OK, done laughing now.

Any tips?  Criticism?

You bolded "that," not me.

I think you're giving the lemon too many human qualities, like the ability to write and smile.  Does a lemon have hands and a face?  If you're going to endow this inanimate object with human qualities, elaborate.


A point $ (x,y)$ is a distance of 6 units from the $ x$-axis. It is a distance of 5 units from the point $ (8,3)$. It is a distance $ \sqrt{n}$ from the origin. Given that $ x<8$, what is $ n$?

Since the point is $ 6$ units from the $ x$-axis, we have that the $ y$ coordinate is $ \pm6$. Remembering Pythagorean triples (most importantly $ 3$, $ 4$, $ 5$) we find that the point is $ (4,6)$ (this also relies on the given condition that $ x<8$). So its distance from the origin is $ \sqrt{4^2\plus{}6^2}\equal{}\sqrt{52} \implies n\equal{}\boxed{52}$.

what do Pythagorean triples have to do with it????

[hide=My Sol]
Because(x,y) has a distance of $6$ from the $x$ axis, $y$ must have a value of $6$ or $-6$

If $y=6$
$5=\sqrt{(8-x)^2+(3-6)^2}$
$25=(8-x)^2+9$
$16=(8-x)^2$
Thus $x=4$ or $x=12$ but $x<8$ so the answer for this case is $x=4$

If $y=-6$
$5=\sqrt{(8-x)^2+(3-(-6))^2}$
$25=(8-x)^2+81$
$-56=(8-x)^2$, But the square of a real number is always positive, thus there are no solution for this case

From the 2 cases, we got that the only possible solution is $x=4$ and $y=6$
$\sqrt{n}=\sqrt{(4-0)^2+(6-0)^2}$
$n=16+36$
$\boxed{n=52}$
[/hide]

And @above, it just helps you to count this quicker, since this is a problem for FTW


1. if (x^2) + ax – 1 = 0, then find the probability that the quadratic always has one positive solution.(a belongs to the set of Integers)

2.A & B are playing a Game where either one of them wins the match or the match is drawn. The chance that A wins is 20% and that B wins is 30%. What is the probability that A wins the Game? - Edited - in the 5th round
2 - > Is this going to be the probability of all the possible ways the first 4 matches can end and with the fifth match being won by A?

2. Is there a fixed number of matches in the game?

[quote="fizsle"]1. if (x^2) + ax – 1 = 0, then find the probability that the quadratic always has one positive solution.(a belongs to the set of Integers)[/quote]

This is the same as asking what is $P(-a+\sqrt{a^{2}+4}> 0)$? The answer is obvious.

[quote="fizsle"]2.A & B are playing a Game where either one of them wins the match or the match is drawn. The chance that A wins is 20% and that B wins is 30%. What is the probability that A wins the Game?
2 - > Is this going to be the probability of all the possible ways the first 4 matches can end and with the fifth match being won by A?[/quote]If we consider $S$ as the sample space, and $E_{A}$ to be the event A wins the 5th match then the event that the first four matches end (somehow) $= S$ and the probability that A wins the game is $P(S\text{ AND }E_{A}) = P(E_{A}) = 0.20$.

[quote="fizsle"]Dr.No

Not clear with #2...  :huh: what is P(S and Ea)? can you explain a bit more?[/quote]

$S$ is the entire sample space, its subsets are the events that can occur. $P(S)\equiv 1$ since some event $\in S$ will certainly occur.

Likewise $S\cap E = E$, where $E$ is any event $\in S$.

$P(S\text{ AND }E_{A}) = P(S\cap E_{A}) = P(E_{A})$

The idea is that the joint probability of a completely certain event and a not so certain event is equal to the probability of the not so certain event all by itself.

[quote="fizsle"]thanks... i see it now...but the given answer choices are different...
a. 1.25%
b. 1.38%
c. 2.07%
d. 1.96%

??? :huh:[/quote]

Then I think I misunderstood the problem statement. There is a way, using a decision tree, to figure out the probability of A winning the game, but the rules of the game need to be clearly defined. The way the problem is currently stated I was under the impression that if A wins the 5th match, A wins the game.

Sorry everyone... for #2...the question is what is the probability of A winnning the 5th Game? i've edited the main post

Maybe the problem meant that A and B play until someone wins (no draw).
So, if A wins in the fifth round, it means that there have been four draws (each with prob. $50\%\equiv\frac{1}{2}$) and then A wins (with prob. $20\%\equiv\frac{1}{5}$).
If we consider each game statistically independent, the total probability is
$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{5}=\frac{1}{80}\equiv 1.25\%$.


[quote="vishalarul"]What does the $\sum_{sym}$ supposed to mean?[/quote]
$\sum_{\textrm{cyc}}$ means the sum of all cyclic permutations.

$\sum_{\textrm{sym}}$ means the sum of all permutations.

[quote="vishalarul"]What does the $\sum_{sym}$ supposed to mean?[/quote]
$\sum_{cyc}f(x,y,z) = f(x,y,z)+f(y,z,x)+f(z,x,y)$ and
$\sum_{sym}f(x,y,z) = f(x,y,z)+f(x,z,y)+f(y,x,z)+f(y,z,x)+f(z,x,y)+f(z,y,x)$

For example, $\sum_{cyc}xy = xy+yz+zx$ and
$\sum_{sym}xy = xy+xz+yx+yz+zx+zy = 2(xy+yz+zx)$

is it just a notation that simplifys your work or is there another reason?

also, are there common theorems dealing witht these notations?

well, it shortens the amount of stuff you need to write

theorems? its just notation.....

yes it shortens things out very much. Just compare:
$\sum_{sym}x^{4}y^{2}z$
with
$x^{4}y^{2}z+x^{4}yz^{2}+x^{2}y^{4}z+x^{2}yz^{4}+xy^{4}z^{2}+xy^{2}z^{4}$
 :wink:

can anybody prove this inequality by cauchy ??

[quote="parinaz.sa"]$(x,y,z>0)$then pove it:
$\frac{x^{3}}{x^{2}+xy+y^{2}}+\frac{y^{3}}{y^{2}+yz+z^{2}}+\frac{z^{3}}{z^{2}+zx+x^{2}}\ge\frac{x+y+z}{3}$[/quote]

I just wonder how this problem always shows up in Mathlinks. The easiest solution is that:
$\frac{x^{3}}{x^{2}+xy+y^{2}}\geq \frac{2x-y}{3}$

[quote="vishalarul"]How did you obtain that result? Which formula (or theorem) did you use?[/quote]

 You can use $\frac{x^{3}}{x^{2}+xy+y^{2}}\geq ax+by$
Assume that $x+y+z=1$
then change it to the function of x. We have two condition:
1> ${f(\frac{1}{3}})=0$
2> $f'(\frac{1}{3})=0$
solve for $a$ &$b$. Hope it helps!

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=107043&hilit=Lithuania]Lithuania1997[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=601089&p=3568666#p3568666]samc2013[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=54&t=3533&hilit=ineq]Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu:[/url]
Prove for all positive reals $x_1,x_2,\cdots, x_n $:\[\frac{x^3_1}{x^2_1+x_1x_2+x^2_2} +\frac{x^3_2}{x^2_2+x_2x_3+x^2_3} +\cdots+\frac{x^3_{n-1}}{x^2_{n-1}+x_{n-1}x_n+x^2_n} +\frac{x^3_n}{x^2_n+x_nx_1+x^2_1}\]\[ \geq \frac{x_1+x_2+\cdots +x_n}{3} .\]


Let $ P(x)$ and $ Q(x)$ be monic polynomials. Prove that the sum of the squares of the coeficients of the polynomial $ P(x)Q(x)$ is not smaller than the sum of the squares of the constant terms of $ P(x)$ and $ Q(x)$.

You will find the following lemma useful:  the sum of the squares of the coefficients of a polynomial $ R(x)$ is the constant term of $ R(x) R(x^{\minus{}1})$ (as a Laurent polynomial).

[quote="t0rajir0u"]You will find the following lemma useful:  the sum of the squares of the coefficients of a polynomial $ R(x)$ is the constant term of $ R(x) R(x^{ \minus{} 1})$ (as a Laurent polynomial).[/quote]

How do you prove it?  :maybe:


prove : $ tan\frac{3\pi}{11}\plus{}4sin\frac{2\pi}{11}\equal{}\sqrt{11}$

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 . Locked!


Hello guys!

I'm having a slight problem.  Basically I want the following:

a) A working reference linking to 'reference.bib'

b) Bibliography using the 'harvard' layout via 'bibliography.bib'

I tried the following:

in the same directory I have IEEEtran.bst, agsm.bst, harvard.sty

[code]
% preamble
\documentclass[a4page,12pt,twocolumn,twoside]{article}
\usepackage{harvard}
% end of preamble

\bibliographystyle{IEEEtran}
\bibliography{references}

\renewcommand\refname{Bibliography}
\begin{bibunit}[agsm]
\nocite{*}
\putbib[bibliography]
\end{bibunit}
[/code]

As it is an article the first bit comes up as references but I found as soon as I invoked the 'harvard' package, even though I use bibtex to process the aux file of the document and bu1 the links won't resolve for the references any more.  instead they stay as '?' but if I were to (even tho I do not want to) say cite a ref to the biblogs they resolve nicely.

Apart from this major glitch, I would like the refs to resolve citations the way the harvard package does i.e. (Last name of Author, year)...

I hope I'm making sense.

Any help would be greatly appreciated :)

--Mike

I seem to have fixed it with the following:

[code]

\bibliographystyle{agsm}
\bibliography{references}

\clearpage

\renewcommand\refname{Bibliography}
\begin{bibunit}[agsm]
\nocite{*}
\putbib[bibliography]
\end{bibunit}
[\code]

I.e. no longer using IEEEtran for the references and both using the harvard layout.

I think I'm gonna go with this :)

--Mike


inequalities
rearrangement inequality
inequalities proposed

If $ a,b,c>0$, then

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

[quote="Ligouras"]If $ a,b,c > 0$, then

$ \frac {a^3 \plus{} b^3}{b^2 \plus{} c^2} \plus{} \frac {b^3 \plus{} c^3}{c^2 \plus{} a^2} \plus{} \frac {c^3 \plus{} a^3}{a^2 \plus{} b^2}\geq a \plus{} b \plus{} c$[/quote]

assume  $ a\geq b \geq c$

the inequality is equivalent to

$ b^5c^2 \plus{} b^5a^2 \plus{} a^5b^2 \minus{} a^4b^2c \minus{} a^4bc^2 \minus{} b^4ca^2 \minus{} b^4c^2a \minus{} c^4ba^2 \minus{} c^4b^2a \plus{} c^5a^2 \plus{} c^5b^2 \plus{} a^5c^2\geq 0$

or
$ ((b^2 \plus{} c^2)a^3 \plus{} (b^3 \plus{} c^3)a^2 \plus{} (b^4 \plus{} c^4)a \plus{} 2b^5 \minus{} b^4c \minus{} c^4b \plus{} 2c^5)(a \minus{} b)(a \minus{} c) \plus{} 2(b \plus{} c)(b^2 \plus{} c^2)(ab \minus{} bc \plus{} ca)(b \minus{} c)^2\geq 0$

it easy to prove
$ 2b^5 \minus{} b^4c \minus{} c^4b \plus{} 2c^5\geq 0$
$ ab \minus{} bc \plus{} ca\geq 0$

First we show that
$ x^{3} \plus{} y^{3} \geq xy^{2} \plus{} x^{2}y$
For positive reals x and y
This is trivial (for example, it is equivalent to $ (x \minus{} y)^{2}(x \plus{} y) \geq 0$)
From this we get that
$ 2x^{3} \plus{} 2y^{3} \geq x^{3} \plus{} y^{3} \plus{} xy^{2} \plus{} x^{2}y \equal{} (x \plus{} y)(x^{2} \plus{} y^{2})$
And so
$ \frac {x^{3} \plus{} y^{3}}{x^{2} \plus{} y^{2}} \geq \frac {x \plus{} y}{2}$

Now note that
$ a^{3} \plus{} b^{3}, b^{3} \plus{} c^{3}, c^{3} \plus{} a^{3}$
and
$ \frac {1}{c^{2} \plus{} a^{2}}, \frac {1}{b^{2} \plus{} c^{2}}, \frac {1}{a^{2} \plus{} b^{2}}$
Are similarly sorted

So by the rearrangement inequality
$ \frac {a^{3} \plus{} b^{3}}{b^{2} \plus{} c^{2}} \plus{} \frac {b^{3} \plus{} c^{3}}{c^{2} \plus{} a^{2}} \plus{} \frac {c^{3} \plus{} a^{3}}{a^{2} \plus{} b^{2}} \geq \frac {a^{3} \plus{} b^{3}}{a^{2} \plus{} b^{2}} \plus{} \frac {b^{3} \plus{} c^{3}}{b^{2} \plus{} c^{2}} \plus{} \frac {c^{3} \plus{} a^{3}}{c^{2} \plus{} a^{2}}$
$ \geq \frac {a \plus{} b}{2} \plus{} \frac {b \plus{} c}{2} \plus{} \frac {c \plus{} a}{2}$
$ \equal{} a \plus{} b \plus{} c$

by Cauchy-Schwarz, $ LHS \ge \frac{4(\sum a^2)^{2}}{\sum a^{3} \plus{} 2\sum ab^{2} \plus{} \sum ac^{2}}$.

it suffices to show $ 4(\sum a^{2})^{2}\ge(a\plus{}b\plus{}c)( \sum a^{3} \plus{} 2\sum ab^{2} \plus{} \sum ac^{2})$, which follows from AM-GM.  :lol:


Prove, using mass points, that if the midpoints of consecutive sides of a quadrilateral are connected, the resulting quadrilateral is a parallelogram.

Sorry, no mass points here...

[hide="Hint"]Draw in the diagonals[/hide]

[hide="Solution..."]
Once we draw in the diagonals, we see that each diagonal divides the quadrilateral into two different triangles.

Now consider each triangle.  The segment connecting the midpoints of two of the sides of a triangle is parallel to the third side of the triangle (by similarity).

Since parallelism is transitive, we have that the opposite sides of the quadrilateral we create are parallel to eachother.  Thus, connecting the midpoints of consecutive sides of a quadrilateral creates a parallelogram.
[/hide]

Suffices to show that the diagonals of the midpoint quad bisect each other.

Assign 1 to each of the 4 vertices A,B,C,D.

A+B=2X
B+C=2Y
C+D=2Z
D+A=2W

Where X,Y,Z,Y are the midpoints of the sides.

The midpoint of 2X,2Z is A+B+C+D and has weight 4 by expanding X and Z, and same for the midpoint of 2Y and 2W.

Therefore, the diagonals bisect each other.


My school doesn't have a Mathcounts program so i have to participate individually. I was wondering how i would register and would i get the handbook or not. thanks :D

Please contact your chapter coordinator for information on this. You can search at http://mathcounts.org/NETCOMMUNITY/Page.aspx?pid=274&srcid=195


function
real analysis
real analysis unsolved

The function $f : \mathbb{R} \rightarrow \mathbb{R}$ is twice differentiable and satisfies the relations

$a)$  $|f(x)| \leq 1 \, \forall x \in \mathbb{R}$

$b)$  $f^{2}(x)+(f'(x))^{2} = 4$.

Show that there exists $c \in \mathbb{R}$ such that $f(c)+f''(c) = 0$.

I ommited the first question of the problem to make it more interesting.
I am almost sure that it is not necessary for the second part.  :D

differentiating the second relation we get $2f(x)f'(x) + 2f'(x)f''(x) = 0$, which gives $f'(x)(f(x)+f''(x)) =0$, so it suffices to show that $f'(x)$ is not everywhere 0. if it were, then $f^2(x) = 4$ everywhere, which contradicts $|f(x)|\leq 1$


geometry
perimeter
inequalities
triangle inequality

A triangle whose side lengths are whole numbers has one side which measures 25 inches and a perimeter of 80 inches. What is the fewest number of inches that can be the length of one of the remaining sides?

Well, there is $ 80\minus{}25\equal{}55$ left, and by the triangle inequality, the $ shorter side\plus{}25>55\minus{}shorter side$ so the $ shorter side>15$, meaning the minimum is 16


Can someone explain to me what it is? (Osculating Polynomial)

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

Is it something about the article that you don't understand?

No thank you. that was very helpful. I googled it but that did not come up.  :D


algebra
binomial theorem
number theory unsolved
number theory

Let $p$ be an odd prime number. For each $i=1,2,...,p-1$, we denote by $r_{i}$ to the remainder got when we divide $i^{p}$ by $p^{2}$. Compute the sum:

$r_{1}+r_{2}+...+r_{p-1}$

[quote]$r_{i}+r_{p-i}=i^{p}+(p-i)^{p}\equiv 0(\mod p^{2})$[/quote]
How do you get this? Surely $i^{p}+(p-i)^{p}\equiv p(\mod p^{2})$?

No, it's correct: use the binomial theorem for the second term; then the $i^{p}$ cancels and all remaining terms are divisible by $p^{2}$.

Ah, thanks. I was working mod p (using FLT), which explains the problem.


Suppose there is natural number 'n'. It can be the sum of various pairs of natural numbers. Prove that the product of the numbers in a pair having least positive difference between them is maximum as compared to other pairs's product
e.g. if, n=35, then pair(18,17) gives maximum product and it has least positive difference as compared to other possible pairs.

$ ab\equal{}\frac{(a\plus{}b)^2\minus{}(a\minus{}b)^2}{4}$
 :wink:


The real numbers $ \alpha$ and $ \beta$ satisfy the equations:

$ \alpha ^3\minus{}3 \alpha ^2\plus{}5 \alpha \minus{}17\equal{}0$
$ \beta ^3\minus{}3 \beta ^2\plus{} 5 \beta \plus{}11\equal{}0$

Compute $ \alpha\plus{}\beta$.

[quote="moldovan"]The real numbers $ \alpha$ and $ \beta$ satisfy the equations:

$ \alpha ^3 \minus{} 3 \alpha ^2 \plus{} 5 \alpha \minus{} 17 \equal{} 0$
$ \beta ^3 \minus{} 3 \beta ^2 \plus{} 5 \beta \plus{} 11 \equal{} 0$

Compute $ \alpha \plus{} \beta$.[/quote]

Put $ x \equal{} \alpha \minus{} 1$, $ y \equal{} \beta \minus{} 1$,  the original equations become

$ x^3 \plus{} 2x \minus{} 14 \equal{} 0........(1)$
$ y^3 \plus{} 2y \plus{} 14 \equal{} 0..........(2)$

$ (1) \plus{} (2)$, $ (x \plus{} y)(x^2 \minus{} xy \plus{} y^2 \plus{} 2) \equal{} 0$, for $ x,y$ real , we have

$ x^2 \minus{} xy \plus{} y^2 \plus{} 2 > 0$, so $ x \plus{} y \equal{} 0 \Leftrightarrow (\alpha \minus{} 1) \plus{} (\beta \minus{} 1) \equal{} 0$

$ \Rightarrow \alpha \plus{} \beta \equal{} 2$


Let ABCD  a convex quadrilateral.
I=intersection AC with BD
Take a points E and H on AB and BC respectively.
F= intersection EI with CD
G=intersection HI with AD.
M=intersection EG with AC
N=intersection FH with AC

Prove:

(AM*IN)/(IM*NC)=(AI/IC)

See, http://Forumgeom.fau.edu/FG2006volume6/FG200628index.html

Kostas Vittas.

Nice article, vittasko.

For a trig solution to the problem, notice that by Sine Law \[\frac{AI}{IC}=\frac{AB\cdot AD}{BC\cdot CD}\cdot\frac{\sin DAB}{\sin BCD}\]  \[\frac{AM}{MI}=\frac{AE\cdot AG}{EI\cdot IG}\cdot\frac{\sin GAE}{\sin EIG}\]  \[\frac{IN}{NC}=\frac{IH\cdot IF}{HC\cdot CF}\cdot\frac{\sin FIH}{\sin HCF}\] Thus, we are left to prove that \[\frac{AB\cdot AD}{BC\cdot CD}=\frac{AE\cdot AG}{EI\cdot IG}\cdot\frac{IH\cdot IF}{HC\cdot CF}\] which can also be easily proved by Sine Law.

Q.E.D.


linear algebra
matrix
limit
linear algebra unsolved

Denote by  $A_{n}$ the maximum of the determinant of an $n\times n$ matrix with entries $\pm1$. Does there exist a finite limit $\ds\lim_{n\to\infty}\sqrt[n]{A_{n}}$?

for $n=2^{k}$ there exists hadamard matrices. So there is no finite limit.

Yes, that's it.
By the way,don't you know, has this Hadamard's conjecture that Hadamard's matrices exist for all $n$ which are divisible by $4$ been already proved.


greatest common divisor
number theory
prime factorization

Why do you care if they're official or not as long as they work?

Here's a computationally effective method called the Euclidean Algorithm.  Suppose $a > b$.  Then $gcd(a, b) = gcd(b, a-b)$.

Example:  Suppose you wanted to find $gcd(84, 36)$.

$gcd(84, 36) = gcd(48, 36) = gcd(36, 12) = gcd(12, 24) = gcd(12, 12) = 12$

[hide="Proof"]
For two integers $a, b$, call $c = gcd(a, b)$.  Then we can write $a = cx, b = cy$ where $gcd(x, y) = 1$.  This means that $a-b = c(x-y)$ and $gcd(a, a-b) = gcd(b, a-b) = c$.
[/hide]

Another method useful for some competition problems:

$ab = gcd(a, b) \times lcm(a, b)$

thanks.  I came across the first method u showed me  today in class i just realized it always worked and my math teacher said he never heard of it but i knew that was impossible and so i just wanted to check.

[quote="Devendra Patel"]Thnaks for the 1st method ,I always thought finding the Factors was easier[/quote]

What if I asked you to find $gcd(21x + 4, 14x + 3)$?  :)  (IMO 1959 #1, remarkably easy)

[quote="Devendra Patel"]Thnaks for the 1st method ,I always thought finding the Factors was easier[/quote]
One obvious  thing to state here, that for large numbers finding factors may not be practical. Suppose you have two 20 digit numbers.. by the method given above, it won't take too long with even with  hand  while finding factors can take  years (or more ) even with  a superfast computer (assuming you are using simple factoring methods)

[quote="beta9"][quote="t0rajir0u"]
 This means that $a-b = c(x-y)$ and $gcd(a, a-b) = gcd(b, a-b) = c$.
$ab = gcd(a, b) \times lcm(a, b)$[/quote]

How do you get from

$a-b = c(x-y)$     to     $gcd(a, a-b) = gcd(b, a-b) = c$[/quote]

$gcd(a, a-b) = gcd(cx, c(x-y)) = c \times gcd(x, x-y) = c \times 1 = c$

(It should be obvious that $gcd(x, y) = 1 \implies gcd(x, x-y) = 1$.) [hide="Proof"]
Suppose that $gcd(x, x-y) = d > 1$.  But then $x - (x-y) = y$ would be divisible by $d$, and $x$ would also be divisible by $d$, contradicting the assumption that $gcd(x, y) = 1$.  [/hide]

[quote="pingy34"]Alright, I was wondering if anyone knew what all of the official recognized mathematical methods for finding gcf are.  Ne imformation whatsoever is greatly appreciated.  Thanks![/quote]

Well, as long as the numbers are not too huge, you can find the prime factorization.  All the common ones multiplied together give you the gcf.  But of course this doesn't work too well with super-huge numbers...
(Maybe somebody should solve that millenium problem  :D ).

[quote]Well, as long as the numbers are not too huge, you can find the prime factorization[/quote]
Exactly!, except that prime factorization for even the numbers which are not too huge but just huge :) is easier said then done.  (before the age of computers - even the  world class mathematicians did not know  factorization of say 2^32+1 -  and one of the facors here is mere 641)

The eculidian method is easier done than said (It is harder to describe than actually "do it") for random numbers.

So if numbers are small (or you know its factor) prime factorization method is straight forward (which is the case in almost all the tests :) ) If numbers are selected randomly and  not too small, the other method always works. (So it is nice to learn the method once for all)

Thanks t0rajir0u, that actually makes sense now.


Im planning to join my first Mathjam class today and was wondering how the space limit goes (and whether its actually hard to join a class)

It'll be no problem.  Just show up around 10 minutes before class starts.


I am a biginer in graph theory....so if you could help me with some material to study I would be very greatfull:)....I was searching but nothing seems to be good :(  .Please help me!

Thanks!


yesterday I was in class, going over the homework and I didn't quite get the solution, so could someone explain it to me?  The question was "If 'n set' contains n amount of elements how many subsets are there for each set?"
we decided that
if n={1} then there would be 2 subsets of the one-element set, {1} and the {nullset}
and if n={1,2} then there would be 4 subsets of the one-element set
n={1,2,3} then there would be 8 subsets
n={1,2,3,4} then there would be 16
we decided that the formula for this was 2[size=59]n[/size] but I don't understand how we can prove it. I understand that it's true, but is there another formula that proves this one?

please help! -mia

[hide="Informal Proof"]For each subset, we can choose to either include the element in the subset or not include. There are $ 2$ possibilities for each of $ n$ elements, and the choice for one element is independent of any other element, so there are $ 2^n$ subsets.[/hide]

[hide="Formal Proof"]We will prove by induction that there are $ 2^n$ subsets for a set of $ n$ elements.

Base cases: $ n \equal{} 0$; which trivially yields $ 1$ set, the empty set. Also, $ n \equal{} 1$, which yields $ 2$ sets, the set itself and the empty set.

Induction hypothesis: A set with $ k$ elements has $ 2^k$ subsets.

Inductive step: Consider a set with $ k\plus{}1$ elements. Single out one element; there remains a set of $ k$ elements, which has $ 2^k$ subsets by the induction hypothesis, each subset not having the removed element. Now, consider all the subsets with the removed element. There is now a set of $ k$ elements which can be partitioned in $ 2^k$ ways by the induction hypothesis, so there are $ 2^k \plus{} 2^k \equal{} 2^{k\plus{}1}$ subsets for a set with $ k\plus{}1$ elements, as desired.

Therefore, by induction, a set with $ n$ elements has $ 2^n$ subsets.[/hide]


I have a java applet that I want to host on the internet as to make it easily accessible publicly; does anyone know of a place where I can do so for free? (Google gave me some unhelpful hints such as "Buy your dedicated server for applet hosting for $****.**!")

[quote="kevmo314"]I would have to say your best bet is to either run your own server to host it or host it on a university's server if you have connections to one.  I doubt there's anywhere that'll host Java applets for free because there aren't very many, if any, decent places to host full websites for free.

If you'd like to get a domain name, that won't be free either, but you can get a free dynamic IP subdomain from [url=http://www.dyndns.org/]DynDns[/url].[/quote]

I guessed as much; I don't have computers without sensitive data on them that I can use as a web server (and most of the time they are rendering crap), and I'm still 2 years from college >_<. I suppose that I just won't be able to host stuff until I get my hands on an old PC that can run Apache. (I'll keep the free dynamic DNS in mind, though.)

You might be able to run virtualbox and set up a virtual host.  That way you shouldn't have to worry about your data.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose G is finite. Then G is:
(a). nilpotent $ \iff$ G has a normal subgroup of each order dividing |G|.
(b). cyclic $ \iff$ G has a unique subgroup of each order dividing |G|.

Note that $ G$ is nilpotent $ \iff$ $ G \equal{} P_1\times P_2\times\cdots\times P_n$, where $ P_i$ is a Sylow $ p_i$-subgroup.
(a). "<=",  obviously.
     "=>".  It is sufficent to prove the Sylow $ p_i$-subgroup $ P_i$ has a normal subgroup of each order of $ |P_i|$. But this is ture since $ p$-group is super-solvable.

(b).  "=>" is clear.
       "<=". Since the Sylow $ p$-subgroup $ P$ of $ G$ for every $ p$ is unique, we have $ G$ is nilpotent.  By the condition, $ P$
also has a unique subgroup of each order dividing $ |P|$.  So $ P$ has a unique subgroup of order $ p$.  Thus if $ p > 2$, then $ P$ is cyclic. If $ p \equal{} 2$, then $ P$ is cyclic or a generalized quaternion group. But a generalized quaternion group has more one subgroups of order 4. Then $ P$ is cyclic. Therefore, $ G$ is cyclic.


Hi, i have the following question and i'm just wanting to know if it's done correctly or not.
The question is:

The period P of a pendulum of length L is given by

P = 2(pi)sqrt(L/g)

where g is the acceleration of gravity.

(a) Assuming that L remains fixed, show that a 1% increase in g results in approximately a 0.5% decrease
in the period P.

And the worked solution to that is the following:

to find the change in t (delta t) we use the approximate result

(delta t )/(delta g) = dt/dg

so delta t = (dt/dg)*(delta g)

the % change in t is {(delta t)/t}*100

which is {(dt/dg)*(delta g)/t}*100

now we differentiate to find dt/dg...

dt/dg = 2*pi*L1/2*{-1/2*g-3/2}

and (dt/dg)*(delta g)/t is 2*pi*L1/2*{-1/2*g-3/2}*(delta g)*g1/2/(2*pi*L1/2)

which simplifies ( thank goodness ) to -1/2 * (delta g)/g

so {(dt/dg)*(delta g)/t}*100 = {-1/2 * (delta g)/g}*100

which is -1/2 * the % change in g. so if the % change in g is an increase of 1% it follows that the % change in t is a decrease of 1/2 %

Could someone tell me if that is correct?

Your calculation is correct. When you see a power law $y = Cx^a$, where C is some constant, then

$\frac{\text dy}{\text dx} = Cax^{a - 1},\ \ \text dy = Cax^{a - 1}\ \text dx$

$\frac{\text dy}{y} = \frac{Cax^{a - 1}\ \text dx}{Cx^a} = a\ \frac{\text dx}{x}$

In this case, $T = Cg^{-1/2},$ $a = -\frac 1 2$ (does not have to be positive or integer), so that

$\frac{\text dT}{T} = -\frac 1 2\ \frac{\text dg}{g}$

Ok now that i understand and have done that question (even if it did take me 3hrs :) ). Could someone help me solve this please?
It's obviously for the same question as above, but part b of it.

b) For fixed g, what percentage change in L will produce a 1% increase in P?

I just keep calling the period T. Now $T = CL^{1/2}$, so that $\frac{\text dT}{T} = \frac 1 2\ \frac{\text dL}{L}$. If $\frac{\text dT}{T} = +1 \% = +0.01$, then $\frac 1 2\ \frac{\text dL}{L} = +1 \% = +0.01$, so that $\frac{\text dL}{L} = +2 \% = +0.02$.

Sorry to be a nuisance, but what exactly is "C" in your definition? Can you interpret what you've said with the way it's done. Thanks mate

Also, where did the t come from?? I don't see a t in the equation :S

Why the calculus, surely that's over the top. Is that a question from a british olympiad paper, certainly remember one similar?
Why not just put in a value for g of 1.01g and work out the change in P ( 1/sqrt(1.01) ) then just work that out as a percentage of 1. I did the question that way and got full marks. You sit the second round of the BPho?

Kaiseraph, i'm not too sure what you're referring to. Can anyone help me please? Where did t come from? Or is that just a mistake for something else?


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be an arbitrary group, and let $ p$ be prime. If $ |G|\equal{}p$, then clearly $ G$ is cyclic, and thus Abelian. Now suppose $ |G|\equal{}p^2$. Does this imply that $ G$ is Abelian? This is the only question I couldn't get on my algebra 1 final exam, and the answer still escapes me.

Yes. Every group of order $ p^2$, where $ p$ is prime, is abelian.  :)

Here's how to prove it:

Notice that if any element of $ G$ has order $ p^2$, the group is cyclic and thus abelian and $ G \cong C_{p^2}$. Now suppose there is no such element, so every element has order $ p$. Then $ G$ is the union of $ p\plus{}1$ cyclic groups (all isomorphic to $ C_p$ of course) where every two of them have trivial intersection.

Now call $ \mathcal{P}$ the set of all $ p$-subgroups, so $ \left|\mathcal{P}\right| \equal{} p\plus{}1$. Consider the permutation representation $ (G,\mathcal{P})$, with conjugacy as the action. Clearly, there has to be an element of $ \mathcal{P}$ with orbit $ 1$. Call this element $ H$. Now consider the permutation representation $ (G,H)$ with conjugacy as action. Since $ e$ has an orbit of order $ 1$, every element must have an orbit of order $ 1$, since the only possible numbers of elements in an orbit are $ 1$ and $ p$. Therefor, every element of $ H$ commutes with every element of $ G$.

Now, pick an arbitrary $ K \in \mathcal{P}$ with $ K \neq H$. Consider the permutation representation $ (K, \mathcal{P})$. Clearly, $ H$ and $ K$ have an orbit of order $ 1$, so every element in $ \mathcal{P}$ must have an orbit of order $ 1$. Now, pick an arbitrary $ K' \in \mathcal{P}$. Now, consider $ (K,K')$, and observe that $ e$ has an orbit of order $ 1$. Therefor, every element must have an orbit of order $ 1$, hence every element of $ K$ commutes with every element of $ K'$. Since $ K'$ was an arbitrary element in $ \mathcal{P}$, it follows that $ G$ is abelian. Every element has order $ p$, so $ G \cong C_p \times C_p$.

Very nice and useful property!

Another route to this result: (though I haven't seen the approach with group actions before)

* Use the class equation to show that the center is nontrivial.

* Show that if $ x\not\in Z(G)$, $ G\equal{}Z(G)\times\langle x\rangle$.

[quote="coquitao"]Still another route:

1. Use the class equation to show that the center is nontrivial.

2. Hence, $ |\mathbf{Z}(G)| \equal{} p$ or $ |\mathbf{Z}(G)| \equal{} p^{2}$.

3. If  $ |\mathbf{Z}(G)| \equal{} p^{2}$ you are done... In the other case, you will have to resort to the fact that $ G/\mathbf{Z}(G)$ [b]cyclic[/b] implies G [b]abelian[/b].

That's all, folks!  :wink:[/quote]

This is what I tried on the exam, but I didn't know the class equation and so I merely reduced it to showing that the center is nontrivial. What is the class equation?

Take a look at the following [url=http://planetmath.org/encyclopedia/ConjugacyClassFormula.html]link[/url].  :wink:

So basically if the center was trivial, you would end up with an equation of the form:

$ p^2\equal{}1\plus{}\displaystyle\sum_i [G: C_G(x_i)]$

Then since $ |C_G(x_i)|\equal{}p$, $ [G: C_G(x_i)]\equal{}p$, so that the above equation becomes:

$ p^2\equal{}1\plus{}kp \;\;\;\; k\in \mathbb{N}$

Which clearly has no solutions, so that $ Z(G)$ must not be trivial. Is that how you do it?

[quote="Roosta"]Is that how you do it?[/quote]
It's how I do it.  :P

At the time of the exam, we had not learned about the class equation, so is there an easier way to prove that $ Z(G)$ must be nontrivial? A few members of my class are having trouble understanding the class equation, so any suggestions would be helpful. A hint rather than a proof is preferred (what can I say, I like proving things :D).

"Every element of $ G$ is either in $ Z(G)$, or it isn't."  The class equation is just a restatement of this fact.

Specifically, every element $ x$ lies in some conjugacy class $ C_x$.  The statement that $ x$ is not in $ Z(G)$ is exactly the statement that $ |C_x|>1$.  This gives us $ |G| \equal{} |Z(G)| \plus{} \sum_i |C_{x_i}|$, where the sum is over a set of representatives of the nontrivial conjugacy classes.

What is the order of the conjugacy class of $ x$?  It is simply the index of the centralizer: $ |C_x| \equal{} [G: C_G(x)]$, which gives us the class equation.

So really the only hard part is seeing that the index of the centralizer is the size of the conjugacy class.  But you've probably either seen this statement, or the more general theorem about group actions: the size of the orbit of any element under a group action is equal to the index of its stabilizer.

If you haven't seen this, it's not hard to prove.  Simply show that if $ y\equal{}g' \cdot x$, then the set $ \{g\in G | gx \equal{}y\}$ is just the coset $ g' \cdot Stab_G (x)$, giving a bijection between elements of the orbit of $ x$ and cosets of $ Stab_G (x)$. (note that $ \cdot$ here refers to conjugation--not multiplication--in the original problem)

[quote="coquitao"]Are you allowed to use [b]Cauchy's theorem[/b]?

If so, you can proceed as follows:

1. Use Cauchy's theorem to establish that G has an element of order p. Let us refer to that element as x.

2. Prove that <x> is a normal subgroup of G.

3. Prove that <x> $ \subset \mathbf{Z}(G)$.

4. Conclude now, from 3, that $ \mathbf{Z}(G)$ is a nontrivial subgroup of G.

Hope you  like this route better.

Best of luck.  :wink:[/quote]

When I did the problem, this is almost exactly what I tried to do, but the proof of 2 and 3 escaped me (I actually tried going from 1 to 3). I will try again and repost when I figure it out (or if I don't get it lol).

[quote="LydianRain"]"Every element of $ G$ is either in $ Z(G)$, or it isn't."  The class equation is just a restatement of this fact.

Specifically, every element $ x$ lies in some conjugacy class $ C_x$.  The statement that $ x$ is not in $ Z(G)$ is exactly the statement that $ |C_x| > 1$.  This gives us $ |G| \equal{} |Z(G)| \plus{} \sum_i |C_{x_i}|$, where the sum is over a set of representatives of the nontrivial conjugacy classes.

What is the order of the conjugacy class of $ x$?  It is simply the index of the centralizer: $ |C_x| \equal{} [G: C_G(x)]$, which gives us the class equation.

So really the only hard part is seeing that the index of the centralizer is the size of the conjugacy class.  But you've probably either seen this statement, or the more general theorem about group actions: the size of the orbit of any element under a group action is equal to the index of its stabilizer.

If you haven't seen this, it's not hard to prove.  Simply show that if $ y \equal{} g' \cdot x$, then the set $ \{g\in G | gx \equal{} y\}$ is just the coset $ g' \cdot Stab_G (x)$, giving a bijection between elements of the orbit of $ x$ and cosets of $ Stab_G (x)$. (note that $ \cdot$ here refers to conjugation--not multiplication--in the original problem)[/quote]

Are orbits and stabilizers algebra 1 level stuff? If so, we must have fallen behind in my class (after the Fundamental Theorem of Group Homomorphism we did decomposition of finite Abelian groups and then moved on to Rings).

[quote="Roosta"]Are orbits and stabilizers algebra 1 level stuff?[/quote]
Usually, but these designations are pretty arbitrary.  Anyway, you know it now, so it must be. :)

Hi there, [b]Roosta[/b]...

I wonder whether you've finished the alternative proof already.  :wink:

Regards,


topology
advanced fields
advanced fields unsolved

Given the topological space (M, {∅,M}), prove that there isn't a metric "d" on M, s.t. (M, {∅,M}) is the topological space arising from (M,d)...

Any suggestions?

In metric space, a single point is closed. This isn't true here if $ M$ has more than one element.

$ M$ should have at least two elements. metric spaces (resp. their associated topological spaces) are always hausdorff.

Say, M = {0,1} .. wouldn't that still be a closed set under all metrics? I thought automatically, sets of finite points were closed?

Also, I take it that (M, {∅,M}) is not a Hausdorff topological space?

Even if there was just one element, then (M,d) would be closed...and hence (M, {∅,M}) wouldn't be a topological space with d...since the topology has to consist of open sets? So why even mention that?

Please study your definitions again, you seem to have problems with fundamental definitions regarding topologies (topology itself, open/closed sets).
Roughly speaking, a topology consists of some space $ M$ and list of what is supposed to be "open", fulfilling some requirements. Closed sets are complements (in $ M$) of open sets by definition.
This is all.
Especially the topology mentioned in the first post shows that points are in general not closed (nor open). But if that topology is induced by some metric (open sets are defined to be those containing "neighboorhoods" of every elements), then every finite set of points is closed.

That the given topology is not Hausdorff is also a direct corollary from the definitions: take two distinct points $ a,b \in M$. Can you find an open set containing $ a$ but not containing $ b$¿ Why not¿

.. I understand everything you just said...and I never demonstrated a lack of understanding regarding open and closed sets...you might need to reread..

You said

[quote]In metric space, a single point is closed. This isn't true here if  M  has more than one element.[/quote]

I was just saying that this was questionable...because say M={0,1}, so M has more than one element...and is it closed under all metrics...? I have no problems with the fundamentals of open/closed sets...I know the def of Hausdorff...

I've got it now...(M, {∅,M})  isn't Hausdorff, cause under the topology of {∅,M}, you can't construct and open set that contains and elements of M without containing all other elements as well...

I am trying to learn topology out of a book. I just figure it is something I should learn before grad school, and was just wanting to use you peeps to bounce topics around with and help me answer the first couple of end chapter questions....I generally will come here when I don't grasp the book's explanation too well or am having trouble with a thought, so sayings similar to "Please resort back to your book" aren't really what I am looking for....

Zeta, thanks for the response, but I could have done perfectly fine without everything other than

[quote]Especially the topology mentioned in the first post shows that points are in general not closed (nor open). But if that topology is induced by some metric (open sets are defined to be those containing "neighboorhoods" of every elements), then every finite set of points is closed.

That the given topology is not Hausdorff is also a direct corollary from the definitions: take two distinct points a,b in M. Can you find an open set containing  a but not containing b ¿ Why not¿[/quote]


Don't know if this is off topic, still I am posting here.

Has anyone read or seen the Educative JEE Mathematics by KD Joshi and published by Universities Press India? Is it any good? BTW, I am not preparing for JEE but am just interested in JEE Math per se.

Thanks

I just got it --- great book if Math is really your interest and at the same time you prepare for IIT --- recommended to all!!  :D

IIT is a little advanced.  What level is this book written on?


This must be way too easy for you guys but how do you prove that $ (n\minus{}1)! \equiv 0 \bmod (n)$ using congruence theorems?

As stated this is wrong, for example, any prime number $ p$ doesn't divide $ (p\minus{}1)!$.

Maybe you want to prove Wilson Theorem which says that $ p$ divides $ (p\minus{}1)!\plus{}1$. Proofs on wiki

Yes, Crux, that's correct, for any prime number that is wrong per Wilson.
However, for any composite number larger than 5, the statement seems to hold. (The proof is also on Wikipedia, under Wilson's Theorem.)

Oops, sorry, I left out saying that n is a composite number and n>4. It is easy enought to prove intuitively, because the largest factor will be less than $ \frac{n}{2}$. Of course we will have to consider numbers of the form $ p^k$ where p is a prime separately.

Only I am looking for a proof that uses only the theorems in congruence. It is a problem from Niven and Zuckerman's book.

well, if n is not a square, it has trivially 2 factors belonging to (n-1)!

If n is a square of a number that is not prime, we can also easily solve (2 factors).

Otherwise, n is the square of a prime, by the condition this prime is $ \ge 3$

then p and 2p will be in (n-1)!

Yay.

Proof using congruenses only :D

$ n$ is composite so it is easy to see that we can write $ n \equal{} a \cdot b$ with $ a>1, b>1, (a,b) \equal{} 1$ unless $ n$ is on the form $ n \equal{} p^m$ where p is a prime.
Since $ (n\minus{}1)! \equiv 0 \bmod a$ and $ (n\minus{}1)! \equiv 0 \bmod b$ and since $ a$ and $ b$ is coprime, it yields $ \boxed{(n\minus{}1)! \equiv 0 \bmod n}$
Let's take a look on the case, where $ n \equal{} p^m$. We can easily find the number of times $ p$ divides(I mean the largest number $ e$ so $ p^e \equiv 0 \bmod (n\minus{}1)!$, but I don't know the english word..) $ (n\minus{}1)! \equal{} (p^m\minus{}1)!$. We have $ e \equal{} \left[\frac{p^m\minus{}1}{p}\right] \plus{} \left[\frac{p^m\minus{}1}{p^2}\right] \plus{} \left[\frac{p^m\minus{}1}{p^3}\right] \plus{} \cdots$
So if $ p^e \equiv 0 \bmod n$ we have $ (n\minus{}1)! \equiv 0 \bmod n$. Hence we only have to show that $ e \ge m$ And that is easy to see for $ n > 4$. Again we have $ \boxed{(n\minus{}1)! \equiv 0 \bmod n}$.

Take a look at the solution in your book and post it here:)

Thanks a lot. The book doesnt have solutions, but this forum is always there :)


Pack the entire plane with unit circles as tightly as possible. Now imagine each circle to be resistors, with resistance/length=1 unit/ohm. What is the equivalent resistance between two opposite points of a circle?

P.S.: I invented this question, so pardon me if it doesn't make sense!

This is the illustration for the question.


function
probability
calculus
derivative
logarithms
inequalities
real analysis

Firstly, I'd like to apologize if this thread is posted in a wrong section, but I didn't find any other suitable section in which I could post it.

Here is the problem:
Let $X$ be a random variable. Let $M_{X}$ be Moment Generating Function (MGF) of random variable $X$. We know that $M^{(k)}_{X}(0)=EX^{k}$, that is the k-th derivative of MGF in point 0 is the k-th moment of $X$. Let's define another function: $K_{X}=lnM_{X}$. This function is called Cumulant Generating Function. We know that for $k=1,2,3$  we have $(K_{X})^{(k)}(0)=E[X-EX]^{k}$, that is the k-th derivative of $K_{X}$ at point 0 is the k-th central moment of $X$. The problem is, however, that this formula is satisfied only for $k=1,2,3$. For bigger values of $k$ I found in Wikipedia that we use a special formula to obtain the k-th central moment of random variable $X$. However, the main problem I struggle with is how to prove that for $k>4$ contidion $(lnK_{X})^{(k)}(0)=E[X-EX]^{k}$ is not satisfied. I know that it is not satisfied, but how to show it?

If you have any ideas, I'll be very grateful.

Consider the case $EX=0$ (hence $M_{X}'(0)=0$). You can differentiate $K_{X}=\ln M_{X}$ (using the chain rule, product rule, and quotient rule) until you drop: $K_{X}'(0)=0$, $K_{X}''(0)=M_{X}''(0)$, $K_{X}'''(0)=M_{X}'''(0)$... but next comes $K_{X}''''(0)=M_{X}''''(0)-(M_{X}''(0))^{2}$. See, the fourth derivative is the first instance where you get a term with lower derivatives that are not of order $1$. So, unless $X=0$ a.s., you have the strict inequality $K_{X}''''(0)<M_{X}''''(0)$. (Under the assumption $EX=0$.)

Loosely speaking: If a calculus statement fails to be true, then it fails for "almost all" examples. That is, a simple example such as uniform distribution on $[-1,1]$ would suffice to disprove it. For this distribution $K_{X}(t)=\ln\sinh t-\ln t$, and you can use the Taylor series to find that $K_{X}''''(0)<0$. This example does not help for odd $k$, however.

For $k>4$ you can insert the Taylor expansion for $M_{X}$ into the Taylor expansion for $\ln$, and observe that powers $4$ and higher have messy coefficients.

Another vague notion: transforms in the Fourier/Laplace family are linear operators and thrive under linear formulas. The idea that taking a derivative on "one side" is equivalent to multiplying by a variable on the "other side" is a common notion across these transforms. And make no mistake: the moment generating function belongs to this family - specifically, it's a Laplace transform. Anyway, this very linear operator may not respond particularly well to having a nonlinear function like the logarithm applied to it.

Side note: If I'm calling $M_{X}(t)=E(e^{tX})$ a Laplace transform, what does that make the "characteristic function," defined by $\chi_{X}(t)=E(e^{itX})?$ That would be a Fourier transform - and it does what the moment-generating function does, with a few theoretical advantages, starting with the fact that $\chi_{X}(t)$ exists for all $t$ for any random variable $X,$ a guarantee I can't make with $M_{X}(t).$ But the practical uses are very similar.

If $X$ has finite variance, then $\chi_{X}''(0)=-E[X^{2}]$. Is there an interpretation for $\chi_{X}''$ when $X$ has infinite variance?

Well, for a classic example of that, let $X$ have density $\frac{1}{\pi(1+x^{2})}$ on $(-\infty,\infty).$ $\chi_{X}(t)=e^{-|t|}.$

(I think this is called the Cauchy distribution. Of course, we Fourier analysts tend to see it as the Poisson kernel.)

Note that $\chi_{X}''$ doesn't exist classically for all $t.$


analytic geometry
combinatorics proposed
combinatorics

There are given the integers $1 \le m < n$. Consider the set $M = \{ (x,y);x,y \in \mathbb{Z_{+}}, 1 \le x,y \le n \}$. Determine the least value $v(m,n)$ with the property that for every subset $P \subseteq M$ with $|P| = v(m,n)$ there exist $m+1$ elements $A_{i}= (x_{i},y_{i}) \in P, i = 1,2,...,m+1$, for which the $x_{i}$ are all distinct, and $y_{i}$ are also all distinct.


There are $n$ persons in a room where each person knows exactly three others. A "cangri" is defined the following way:
a) every "cangri" has 3 persons
b) In every "cangri" at least one of the persons knows the other two
c) Each person most belong to [i]at most [/i]one "cangri" .

Let $T(n)$ be the maximum number of cangris in a group of $n$ persons (where each persons knows exaclty 3 others)
Prove that $T(n) \geq \frac{2}{11}n$

Interesting cangri :)

We construct cangris in the following way: We choose A vertex such that at least $2$ of its neighboors are not in previous cangris. Then it is easy to see that if we have build $t$ cangris and we cannot build more, then we must have at most $\frac{5t}{2}$ vertex not choosen. So we get from here that $t$ is bigger than $\frac{2n}{11}$.

yes exactly. moreover i think the bound can be improved to  $\frac{n}{4}$.


inequalities proposed
inequalities
algebra

Let $ a,b,c$ be nonnegative real numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}1$. Prove that
\[ \frac{a^3}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{b^3\plus{}c^3}{a^2} \ge \sqrt{2}\]
:)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Prove that
\[ \frac {a^3}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {b^3 \plus{} c^3}{a^2} \ge \sqrt {2}
\]
:)[/quote]
No one like it? :(
I think it is very nice. :(

[quote="mitdac123"]Let me try :P
$ \frac { a^3}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {b^3 \plus{} c^3}{a^2} \ge \frac { (a^2 \plus{} b^2 \plus{} c^2)^2}{ a[ b^2 \minus{} bc \plus{} c^2 \plus{} a(b \plus{} c) ]} \ge \frac {1}{ 2.a[\frac {3 \minus{} 2a^2}{4}]} \equal{} \frac {1}{ 2\sqrt { a^2 ( \frac {\frac {3}{2} \minus{} a^2}{2} )^2} } \ge \sqrt {2}$
:)[/quote]
Nice you proof, :) I also used Cauchy Schwarz, but it is different from you. ;)


geometry
3D geometry
tetrahedron
vector
geometry solved

[quote="romano"]Let $ ABCD $ be a tetrahedron . $ O $ is the center of its circumsphere . $ H $ is the orthocenter of the tetrahedron $ ABCD $ . $ G $ is the centroid of the tetrahedron $ ABCD $ . Prove that :
 1)  vector $ OA $ + vector $ OB $ + vector $ OC $ + vector $ OD $ = 4. vector $ OG $
 2)  vector $ OA $ + vector $ OB $ + vector $ OC $ + vector $ OD $ = 2. vector $ OH $[/quote]

1) the equality is equivalent to vector $ GA $ + vector $ GB $ + vector $ GC $ + vector $ GD $ = the $ 0 $ vector.  and this is a characteristic property of the centroid, which isn't too hard to prove.  (let M and N be the midpoints of $ AB $ and $ CD $, respectively.  then $ G $ is the midpoint of $ MN $; vector $ GA $ + vector $ GB $ = 2 vector $ GM $ and vector $ GC $ + vector $ GD $ = 2 vector $ GN $, and since $GM $ = $ -GN $, the result follows)

2) this follows from part 1) and the property $ OH = 2OG $ (i.e. G is the midpoint of OH).


trigonometry
geometry
trig identities
Law of Cosines
Law of Sines
angle bisector

One angle of a triangle is twice another, and the sides opposite these angles have lengths 15 and 9. Compute the length of the third side of the triangle.

I don't think that's right... (if I remember correctly)

[hide="hint"]
Law of Sines
[/hide]

You're right.  I used a $\cos{x}$ in place of a $\cos{3x}$ by accident.

[quote="amcavoy"][hide]Use the Law of Cosines:

$c^{2}=306+270\cos{x}$

$81=225+c^{2}-30c\cos{x}$

So we have

$\frac{c^{2}-306}{270}=\frac{c^{2}+144}{30c}\implies c=\boxed{24}$[/hide][/quote]

I don't think that's right, since $15+9=24$ and that'd violate Triangle Inequality.

Also, isn't it $c^{2}=306-270\cos{x}$? (minus, not plus)

[hide]
Hopefully I didn't make any mistakes,  :blush:
Law of Sines
$\frac{\sin 2x}{15}=\frac{\sin x}{9}$
$18\sin x\cos x =15\sin x$
$\cos x=\frac{5}{6}$, $\sin x=\frac{\sqrt{11}}{6}$

Law of Sines, call the length of the third side $y$
$\frac{\sqrt{11}}{6\times9}=\frac{\sin (180-3x)}{y}$
\begin{eqnarray*} y&=&\displaystyle \frac{54}{\sqrt{11}}\times \displaystyle \left(\sin 2x\cos 2x+\sin x\cos x \right)\\ &=&\displaystyle \frac{54}{\sqrt{11}}\times \displaystyle \left(\frac{10\sqrt{11}}{36}\times(\frac{50}{36}-1) +\frac{5\sqrt{11}}{36}\right)\\ &=&\frac{54}{\sqrt{11}}\times\displaystyle \frac{41\sqrt{11}}{144}\\ &=&\frac{123}{8} \end{eqnarray*}
[/hide]

Here's a solution  ;)

Apply the law of sines: $\frac{9}{sin(x)}=\frac{15}{sin(2x)}$. Using the double angle formula on $sin(2x)$ and bringing it to the other side gives that $cosx=\frac{5}{6}$. Therefore $sinx=\frac{\sqrt{11}}{6}$.

Now we apply the law of sines again, but this time we include the unknow side length. $\frac{9*6}{\sqrt{11}}=\frac{y}{sin(180-3x)}$, where $y$ is the length of the unknown side. Note that $sin(180-3x)=sin(3x)= sin(2x)cosx+cos(2x)sinx= 2sinxcos^2 x+cos(2x)sinx= 2sinxcos^2 x+(cos^2 x - sin^2 x)sinx= 3sinxcos^2 x - sin^3 x$ .

Since we know $sinx$ and $cosx$, we can substitute to find $sin(3x)$. Thus we can return to our equation with $y$ and write $y=\frac{54}{\sqrt{11}}*sin(3x)=\frac{54}{\sqrt{11}}*(\frac{25*\sqrt{11}}{72} - \frac{11\sqrt{11}}{216})$, which you can verify simplifies to $y=16$.

adunokor, I was writing my solution when you posted. Looking back, I think you messed up when you rewrote $sin(3x)$. I hope there are no arithmetic errors in my answer

Yes, drunner's answer is correct. But, once you find $\cos x$, you can just law of cosines! :)

you're right, the law of cosines was quite a bit simpler. I guess right now my brain is wired towards thinking about the law of sines since I just looked through that section in "Geometry Revisted"

Oh right, $\sin 3x=\sin 2x\cosh+\sin x\cos2x$, not $\sin2x\cos 2x+\sin x\cos x$

And yeah, using Law of Cosines would have been much simpler  :P

why always resort to trig? there is an easy synthetic solution too!

[hide="big hint"]angle bisector[/hide]

Altheman, I got your hint! (I usually suck at pure geometry)

Label as triangle $ABC$ with $AB=15$ and $AC=9$. Let the angle bisector of angle $ACB$ intersect $AB$ at $D$. Notice that triangles $ABC$ and $ACD$ are similar (AA similarity). Before moving on let $a=BD$ and $b=DA$ and $y=BC$. From the similarity we have that $\frac{y}{a}=\frac{5}{3}$ so that $y=\frac{5a}{3}$. We also have that $\frac{y}{9}=\frac{a}{b}$, or $y=\frac{9a}{b}$. Setting this expression equal to our previous one for y, we find that $b=\frac{27}{5}$, which implies $a=\frac{48}{5}$. Now substitute into $y=\frac{5a}{3}$ to get $y=16$.


Calculate

Floor[ (n + 1) / 2 ] + Floor[ (n + 2) / 4 ] + Floor[ (n + 4) / 8 ] + ... + Floor[ (n + 2^k ) / 2^(k+1) ]

Somebody hints me to use binary expansion. I tried but haven't been successful.
The answer is [i]n[/i] as far as I recognize.

Anyone ?

Your conjecture is right, and this is IMO 1968 #6.

Pierre.

If you want the solution : http://www.mathlinks.ro/Forum/viewtopic.php?t=34668 :)


algebra
polynomial
quadratics
function
probability
functional equation
AMC

Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$, $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$

[quote="Altheman"]take $x=\pm i$, this suggests that $i$ is a root, then do $f(x)=(x^{2}+1)g(x)$...then you get the same identity for g(x)...then $f(x)=(x^{2}+1)^{n}$...the second thing gives $n=2$, so $f(5)=26^{2}=676$...i did not exclude other options...but this works out too well...[/quote]

Urg. I tried a bunch of values, but I never thought of trying $i$...

Dude that is a pretty tricky way Alex Anderson although if you do not guess the roots here's what you can do instead (pretty sim):

If x is a root, then so is 2x^3 + x = x(2x^2 + 1), so if x is a real root, we can create an infinite sequence of increasing real roots, which is contradiction (Unless x = 0 but the problem statement says NO). So whoa let's suppose x is the most magnificent complex root and it has the biggest magnitude ummm ok if the magnitude of x is greater than one, the magnitude of x(2x^2+1) is bigger and bleargh it sucks. Ummm also since f(0) = 1 the product of the roots has magnitude 1 so wtf all roots must be of magnitude 1 ok then we have x a root --> x(2x^2+1) a root omg 2x^2+1 must be of magnitude 1 omg x = i or -i omg ok I think you can take it from there...

i don't see what you mean by i is a root. how do you know

Bravo, Probability!  This is a beautiful solution.  I had figured out the part about how you can't have complex solutions with |z| > 1, but I didn't think about using the fact that the product of the roots is 1.

Actually, now that I think about it, I'm not sure I'm completely convinced of that part.  To conclude that the product of the roots is 1 (which you didn't say but I'm guessing that this is what you're using), you have to know that the lead coefficient of f is 1.  How do we know that?  Maybe we can get it just by inspecting the lead coefficient of f(x)f(2x^2) vs. f(2x^3 + x)?

Actually, now that I think about it, that should do perfectly.  Basically you get something to the effect of a^2 = a (where a is the lead coefficient), and so a = 1 or 0 (the latter doesn't fly since it's the lead coefficient).

Anyway, this is a very sophisticated approach.  I really do wonder whether this is what the author intended, because you really do have to use the kitchen sink to do it this way.  (Also, it seems like it takes a LOT more work to prove that your polynomial is the only possible one than it does to guess the polynomial, see where the 125 comes from, and get the answer.  And it always upsets me when you can get an answer in 1/4 the time using "bad mathematics" - i.e. come up with a polynomial and say, well, since the problem asks for f(5), the polynomial I came up with must give me the right value of f(5), without actually "knowing" that this is the case.)

Oh yeah.. I neglected the lead coefficient part  :blush:  although I had already established it at some point while solving the problem. I also suspect that the guessability might have been somewhat intended, since it seems like this problem ought to be generalizable (at least could have used something bigger than 125, if not tweaking the functional equation itself).

Edit--also, I just realized that the real root part in my solution is totally pointless... I guess I just did that before trying the complex case...

trying $f(i)$ is not that unreasonable...you want to have small numbers that come out of the function...like you try f(1), f(-1), f(0), etc...also the whole "real coefficients" deal leads you to try come complex numbers...in particular, i tried something like 2x^2=-(2x^3+x), x=-(2x^3+x), and solve those...that way you can conjugate both sides, and you get something like:

f(i)*f(-2)=f(-i)
f(-i)*f(-2)=f(i)

then eliminate f(-i)...to get that it can be 0...then -i is 0...blah

it is all just testing values/ equating the expressions so they cancel, etc

also this is pretty similar to an old mop problem...zhai used a similar method in his solution here...

the whole lead coefficient thing is standard for these types of problems [i left that out...but i was like ok a^2=a...0 no...1 yes...

[quote="Altheman"]

f(i)*f(-2)=f(-i)
f(-i)*f(-2)=f(i)

then eliminate f(-i)...to get that it can be 0...then -i is 0...blah
[/quote]

I don't see how it is [u]necessary [/u]that f(i) = f(-i) from these equations alone (it is reasonable, certainly, and more than a little hopeful, but what if $f(-2) = \pm 1$?)  Of course, it all works out in the end, but if one were being rigorous about it that might be a concern.

yea, i have the same problem with gregx007. Actually i thought of the $i$ substitution after some inspecting during the contest, but i cannot be sure if $f(i) =0$ or $f(2)=1$. In fact, it is easy to prove $f(x)$ is an even funtion, just substitue x and -x and divide each other. So $f(2)\not=1$. I set f(2)=1 and work out f(1)=1 and f(3)=124, and did some case working from there, setting the highest power of the function to 4, which does not work.

Anyhow, can anyone explain why $f(2)\not=1$.

i never said it was necessary, nor did i say it was a full solution...a full solution is in the mathjam, here: [url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=164[/url]

it is quite funny that my "solution" made it into the math jam...as mr. vornicu said, i pretty much cheated the problem

[quote="Phelpedo"]This problem isn't really original either...check problem 57on pg 257 of Engel. For those that don't have it, it is

Find all polynomials $ f(x)$ for which
\[ f(x)f(2x^2) \equal{} f(2x^3 \plus{} x)
\]
[/quote]
Do you have this book in PDF? Send me $ 1$ if you have. Thanks alot anyway

[quote="Phelpedo"]This problem isn't really original either...check problem 57on pg 257 of Engel. For those that don't have it, it is

Find all polynomials $ f(x)$ for which
\[ f(x)f(2x^2) \equal{} f(2x^3\plus{}x)
\][/quote]

By the solution of that book, how does it lead to $ f(x) = (x^{2}+1)^{n} $ ?

[quote=Aopser101]Does anyone have a real solution for this?

The one [url=https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_14]here[/url] assumes that $f(-2) \ne \pm 1$, which I don't understand why you can do.[/quote]

Yes. Can someone explain this as well? (Sorry for the incredibly large bump). Why can't this be true?

ur link isn't working.plz give a easy solution.

The link works for me. I think the Wiki has the best solution.

The solution on the Wiki gets that $f(x) = f(-x)$, and then says $f(-2) \neq 1$. Can someone explain why?

Well known trick  ;)
[hide=solution]Of course all the roots should have the same modulus since if it is not then $(r,p(r),p(p(r)), \cdots)$ ,where $p(x)=2x^3+x$, is an increasing set of roots of $f$ with increasing magnitude which is a contradiction.So let all the roots have modulus $k$.Since $f(0)=1$ we know $k=1$.

Let $r$ be a root of $f$.If we substitute $\tfrac{\sqrt{r}}{\sqrt{2}}$ we get that $\tfrac{\sqrt{r}}{\sqrt{2}}(r+1)$ is also a root of $f$.So $|\tfrac{\sqrt{r}}{\sqrt{2}}(r+1)|=1 \implies |r+1|=\sqrt{2}$ and since $|r|=1$ we get that $r=\pm i$.

So $f(x)=(x^2+1)^n$.Since $f(2)+f(3)=125 \implies n=2$.So $f(x)=(x^2+1)^2$.Now the requested answer is $(5^2+1)^2=26^2=\boxed{676}.$[/hide]

One of the easiest P14s ever

[hide = Solution]NOTE: This solution is an in contest solution and is not an actual proof.

We plug 1 into the relation to get $f(1)*f(2) = f(3)$ so $f(2)(f(1)+1)=125$.

Since this is an AIME problem, it is probably has a nice answer, so f(1)+1 is probably a factor of 125. So, we first try f(1) = 4. We get f(2)=25, f(3)=100. It's then easy to notice the sequence: $1^2,2^2,5^2,10^2$ for $f(0),f(1),f(2),f(3)$ respectively. So, we take the differences between the squared numbers to get that the next one will be 17^2 and $f(5) = 26^2 = 676$.[/hide]

No, 2014 I definitely had an easier #14. :)

[quote=NoSignOfTheta][quote=ilovepizza2020][b]One of the easiest[/b] P14s ever

[hide = Solution]NOTE: This solution is an in contest solution and is not an actual proof.

We plug 1 into the relation to get $f(1)*f(2) = f(3)$ so $f(2)(f(1)+1)=125$.

Since this is an AIME problem, it is probably has a nice answer, so f(1)+1 is probably a factor of 125. So, we first try f(1) = 4. We get f(2)=25, f(3)=100. It's then easy to notice the sequence: $1^2,2^2,5^2,10^2$ for $f(0),f(1),f(2),f(3)$ respectively. So, we take the differences between the squared numbers to get that the next one will be 17^2 and $f(5) = 26^2 = 676$.[/hide][/quote][/quote]

I edited it lol do you see the edit reason

[quote=NoSignOfTheta][quote=ilovepizza2020][b]One of the easiest[/b] P14s ever

[hide = Solution]NOTE: This solution is an in contest solution and is not an actual proof.

We plug 1 into the relation to get $f(1)*f(2) = f(3)$ so $f(2)(f(1)+1)=125$.

Since this is an AIME problem, it is probably has a nice answer, so f(1)+1 is probably a factor of 125. So, we first try f(1) = 4. We get f(2)=25, f(3)=100. It's then easy to notice the sequence: $1^2,2^2,5^2,10^2$ for $f(0),f(1),f(2),f(3)$ respectively. So, we take the differences between the squared numbers to get that the next one will be 17^2 and $f(5) = 26^2 = 676$.[/hide][/quote][/quote]

pizza edited it with "ok olyhero"

[quote=OlympusHero]Uh guys there is a problem I think I accidentally reported a post and when trying to click out I hit submit, will I be banned?[/quote]

No? I thought you knew that already

Ok, just checking. Does anyone have a solution with basic knowledge? Not a fakesolve is what I mean. I can't seem to find it and the solutions above (except for ilovepizza's but that's a fakesolve) do not make sense. Like $x=i$ gives $f(i)f(-2)=f(-i)$ and $x=-i$ gives $f(-i)f(-2)=f(i)$ so $f(-2)^2=1$ meaning $f(-2) = \pm 1$ but then what? I don't understand what Altheman said. And for rcv, how did he know to assume it was a quadratic? It could be any degree. I also don't know what probability1.01 means by create an infinite sequence of increasing real roots and what does he mean most 'magnificent' complex root? All really confusing.

Perhaps read the official solution in the "2007 II" file in this folder:

[url]https://drive.google.com/drive/u/1/folders/1i3oMxWIoK2gJt42BRbfIZt2DDoGmzWzP[/url]

Official solutions tend to be much easier to understand and follow through than AoPS solutions.

Given that we are only given 3 conditions, it makes sense that the polynomial is probably of degree $\geq 3$. Indeed, if it is a quartic, then we let it have leading coefficient 1. Then, if the polynomial is $x^4+ax^3+bx^2+cx+d$, we know by the first condition that $d=1$, and then we get
$f(2)=17+8a+4b+2c$
$f(3)=f(1)f(2)=82+27a+9b+3c$
Summing these gives $35a+13b+5c=26$. We get the intuition that $b=2$ and the rest are $0$, and using this gives us that the polynomial is $(x^2+1)^2$. Testing $5$ in this polynomial gives that the answer is $\boxed{676}$.


[quote="Pain rinnegan"]Let $ a,b,c > 0$. Prove that:
\[ (ab \plus{} bc \plus{} ca)\left(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\right)\le 2(a \plus{} b \plus{} c) \plus{} \frac {\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}{2}\]
[/quote]

$ LHS\equal{}a\plus{}b\plus{}c\plus{}\sum \frac{bc}{b\plus{}c} \le \frac{3}{2}(a\plus{}b\plus{}c) <2(a\plus{}b\plus{}c)<RHS$

[quote="caubetoanhoc94"][quote="Pain rinnegan"]Let $ a,b,c > 0$. Prove that:
\[ (ab \plus{} bc \plus{} ca)\left(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\right)\le 2(a \plus{} b \plus{} c) \plus{} \frac {\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}{2}\]
[/quote]

$ LHS \equal{} a \plus{} b \plus{} c \plus{} \sum \frac {bc}{b \plus{} c} \le \frac {3}{2}(a \plus{} b \plus{} c) < 2(a \plus{} b \plus{} c) < RHS$[/quote]

Yes, sorry i made some mistakes. I'll replace it with the following one:

Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 3$. Show that:
\[ \frac {a}{a^2 \plus{} bc} \plus{} \frac {b}{b^2 \plus{} ca} \plus{} \frac {c}{c^2 \plus{} ab}\ge \frac {18(2(a \plus{} b \plus{} c) \minus{} 3abc)}{(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)^2}\]


Tryouts for the 2007 Chicago ARML Teams will be held on Thursday, February 8, 2007, from 4 to 10 pm (with a break for dinner) at Evanston Township High School in Evanston, IL.

Tryouts are open to any student who has not yet completed high school and lives in the northern 2/3 or so of Illinois.

For more information, see [url]http://chicagoarml.org/[/url]

[quote="Altheman"][quote="Altheman"]woo that is in 3 months, i have to learn to work under time pressure :P[/quote]

woo that is in 3 [b]weeks[/b], i have to learn to work under time pressure :P[/quote]

woo that is in [b]1[/b] week, i have to learn to work under time pressure too.  :D

[quote="Altheman"]OH MAN 2 DAYS...

p.s. IL FORUM=DEAD

*tumble weed passes by*

looking for some good geo...lets just say that some recent contest [that is surely not the amc...] did not have good geo...except for one of them

also, we should totally get an illinois private forum for chicago arml...and we could do extra *pauses* MATH[/quote]

Rofl, one day now. And Idk, i basically have no idea as to where to get some good geo... errr... so, ya!

And we definitely should get a private Chicago ARML Forum; should we ask a mod or somethin? But nah, we wouldn't use that forum, for well... math or anything; just to chillax. (Wow, that's probably the only time I'll ever use that word.  :P )

Anyways, cy'all at the tryout tomorrow!  :D

[quote="Altheman"][quote="cincodemayo5590"] just to chillax. [/quote]

you pose a strong argument for a chicago arml forum...lol

as an actual argument, if we have a chicago arml forum, that forces people to signup for aops, exposing them to the site...which is beneficial for obvious reasons...not only can we use it [?], but it gives chicago arml kids a great resource

yea...

WOOO ARML TOMORROW>>>>>>I WISH I WOULD GET SLEEP>>>BUT 3 TESTS TOMORROW>>>STUDY>>>NO SLEEEP>>>>AWWWWW

P.S. SAM: TAKE MY AIME [in sig][/quote]

Ya, AoPS exposure is a VERY good thing for a math team. And heck yea, we can chillax.

P.P.S.S.P.S.P.S.P.S.P.S.P.S. ALEX: Meh. When I have time.

I need to avenge my dad's loss to andrew schults

must win before andrew schults Jr.


geometry
circumcircle
angle bisector
geometry proposed

Let $ABC$ be a triangle satisfying $BC < CA$. Let $P$ be an arbitrary point on the side $AB$ (different from $A$ and $B$), and let the line $CP$ meet the circumcircle of triangle $ABC$ at a point $S$ (apart from the point $C$).

Let the circumcircle of triangle $ASP$ meet the line $CA$ at a point $R$ (apart from $A$), and let the circumcircle of triangle $BPS$ meet the line $CB$ at a point $Q$ (apart from $B$).

Prove that the excircle of triangle $APR$ at the side $AP$ is identical with the excircle of triangle $PQB$ at the side $PQ$ if and only if the point $S$ is the midpoint of the arc $AB$ on the circumcircle of triangle $ABC$.

A very simple angle chase will show that $ABC,\ QRC$ are similar, and $P,Q,R$ are, in fact, collinear.

The condition expressed in excircles is equivalent, in this configuration, to the $c$-excircle of $ABC$ being tangent to $QR$. From the observation above, this can only happen in one situation, since $QR$ has a fixed direction and cuts $AC$ between $A$ and $C$. Since this obviously happens when $S$ is the midpoint of the arc $AB$, because that's when $A$ and $Q$, $B$ and $R$ are symmetric wrt the angle bisector $CS$ of $\angle BCA$, it means that this is the only time it happens.

I'm sure it lacks the rigor necessary for writing up a solution for a TST, but this is the main idea :).


100 pounds of chocolate is packaged into boxes each containing $1\frac{1}{4}$ pounds of chocolate.  Each box is then sold for $1.75.  What is the total selling price for all of the boxes of chocolate?

[hide]  100 pounds of chocolate/ 1.25 pound bags= 80 bags

80 * 1.75= [b]$140[/b] [/hide]

It's a pretty long time from May 21st [b]2005[/b] to March 23 [b]2007[/b] :P

[quote="redsmartie"][hide]100 pounds / 1.25 per box * $1.75 per box =$140[/hide]

[b]Easy as 1,2,3, ABC.[/b]
Any harder questions?[/quote]

If you want harder questions, you could go to a different forum.

100 / 1.25 = 80
80 x 1.75 = $ 140
* x means times
  i couldnt find a
  better symbol
ANSWER =$140

Welcom to AoPS, but before you post a lot, I recomend you read this:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11598

Although it is for the G&FF, it in general holds true to posting in the forum. Notice 8).

And think a moderator should lock this. How many times has this been revived? Like 3?

wouldnt the answer be 140??? tht is what i got


I am having difficulty with 19x + 5 = x - (-59). I can't figure out where I'm messing up.
Here are the steps that I am taking.
1) I subtract 5 from 5 and then subtract it from 59, leaving 19x = x - 54
2) I subtract the x - 54, leaving 19x = -53.
3) Then, I divided 19x by 19 and -53 by 19 and I get -2.73

I know that this is not right because my answer shouldn't have been a negative and it shouldn't have been a decimal. The answer is x=3. I just don't know how you get there, I'm overlooking something. If someone can help me out and show how you get there. I would appreciate it. Thanks

Your error is your first step.  You've messed up your signs-it should be $ \minus{}(\minus{}59)\minus{}5\equal{}59\minus{}5\equal{}54$, which positive.  So it turns into $ 19x\equal{}x\plus{}54\Rightarrow 18x\equal{}54\Rightarrow x\equal{}3$, which is what you know to be right.


[quote="epatjn"]A right triangle and an isosceles triangle have equal areas. The right triangle has sides of length 3, 4 and 5 units. The isosceles triangle has sides of 3, x and x units. What is the value of x ? Express your answer as a decimal to the nearest
thousandth.

Okay, so an isoceles triangle is really two right triangles, but how do I solve for x?
the sides are equal length, and an unknown
the 3 is actually 1.5 for the right triangle, and then how do I get the x?[/quote]

okay now let me explain:
[hide] we all can figure out that their area is 6 both right?
now who the hack cares about x right now let's figure out the isosceles' height
the area is 1/2h(3)
or 3/2(h)=6
so the height is 4
now Pythagorean theorem with that right triangle you talked about
16+2.25=18.25=x^2
take square root
and then your done, of course you need a calculator for this
.................................................
4.272
rounded

[/hide]

Like Funcia said,
[hide]The area of both triangles is 6.
The base of the isosceles is 3, so the height is 4. Using Pythagoreans therom, (a^2 + b^2 = c^2 where  a and b are side lengths and c is the hypotenuse)
[b]1.5^2 + 4^2 = x^2.
2.25 + 16 = x^2
x is about 4.2720018.[/b][/hide]

[quote="IntrepidFish"]I'm getting a waaaay different answer.[/quote]
How are you doing it?

[quote="IntrepidFish"]I'm getting a waaaay different answer.[/quote]

Don't worry about the isoceles triangle actually being two right triangles, like what funcia posted, worry more about finding the height.

If you focus on the length of the side of the isoceles, you'll brute force yourself until you get the answer.

If you focus on the height, Pythagorean Theorem will tell you the length of x. (especially since the height is easier to find).


Consider 6 positive numbers $ a,b,c,d,e,f$ satisfy: $ a\plus{}c\plus{}e\equal{}b\plus{}d\plus{}f$.
Prove that: $ \frac{af\plus{}be}{(a\plus{}b)(e\plus{}f)}\plus{}\frac{ad\plus{}bc}{(a\plus{}b)(c\plus{}d)}\plus{}\frac{cf\plus{}de}{(c\plus{}d)(e\plus{}f)} \geq \frac{3}{2}$

[quote="thesunflower"]Consider 6 positive numbers $ a,b,c,d,e,f$ satisfy: $ a \plus{} c \plus{} e \equal{} b \plus{} d \plus{} f$.
Prove that: $ \frac {af \plus{} be}{(a \plus{} b)(e \plus{} f)} \plus{} \frac {ad \plus{} bc}{(a \plus{} b)(c \plus{} d)} \plus{} \frac {cf \plus{} de}{(c \plus{} d)(e \plus{} f)} \geq \frac {3}{2}$[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=170558
Bad

Chien than, where's the influence between the 2 problem?

Please don't answer here and lock it, it's likely to be an attempt to cheat.


geometry
power of a point
radical axis
geometry proposed

Let $ABC$ be a triangle, and let $M$ and $M^{\prime}$ be two isogonally conjugate points with respect to triangle $ABC$.

Denote by $P$, $Q$, $R$ the feet of the perpendiculars from the point $M$ to the lines $BC$, $CA$, $AB$.
Denote by $P^{\prime}$, $Q^{\prime}$, $R^{\prime}$ the feet of the perpendiculars from the point $M^{\prime}$ to the lines $BC$, $CA$, $AB$.

Let the lines $QR$ and $Q^{\prime}R^{\prime}$ intersect at $E$.
Let the lines $RP$ and $R^{\prime}P^{\prime}$ intersect at $F$.
Let the lines $PQ$ and $P^{\prime}Q^{\prime}$ intersect at $G$.

Show that $AE\parallel BF\parallel CG$.

This looks hard but it's actually very simple:

We shall prove that the lines AE, BF and CG are all perpendicular to the line MM'.

In fact, since < AQM = 90 and < ARM = 90, the points Q and R lie on the circle with diameter AM. The center N of this circle is the midpoint of the segment AM.

Similarly, the points Q' and R' lie on the circle with diameter AM', and the center N' of this circle is the midpoint of the segment AM'.

Finally, it is well-known that since the points M and M' are isogonal conjugates, the points P, Q, R, P', Q', R' all lie on one circle. Call this circle u.

Now, the two common points of the circle with diameter AM and the circle u are the points Q and R; thus, the line QR is the radical axis of the circle with diameter AM and the circle u. Similarly, the line Q'R' is the radical axis of the circle with diameter AM' and the circle u. Since the three pairwise radical axes of three circles concur, the point E, defined as the point of intersection of the lines QR and Q'R', i. e. as the point of intersection of the radical axis of the circle with diameter AM and the circle u and the radical axis of the circle with diameter AM' and the circle u, must lie on the radical axis of the circles with diameters AM and AM'. This radical axis also passes through the point A (since the point A lies on both circles); hence, it follows that the line AE is the radical axis of the circles with diameters AM and AM'. Now, since the radical axis of two circles is perpendicular to the line joining their centers, this entails $AE \perp NN^{\prime}$ (in fact, the points N and N' are the centers of the circles with diameters AM and AM'). But since the points N and N' are the midpoints of the segments AM and AM', we have NN' || MM', and thus $AE \perp MM^{\prime}$. This shows that the line AE is perpendicular to the line MM'; similarly we proceed for the lines BF and CG, and the problem is solved.

  Darij

In general, we can show that if $AM,AM'$ are isogonally conjugate cevians, then $AE\perp MM'$. We can keep $M$ fixed and move $M'$ along the fixed line $AM'$. The transformation $MM'\leftrightarrow AE$ is projective, so by considering some particular positions of $M'$ we can determine it. We can first take $M'\to A$, and then put $M'$ on $MR,MQ$.

This even has a projective generalization, by considering $P\ell-$perpendicularity instead of perpendicularity. I won't state it, however, because there are too many sentences :).

[quote="grobber"]In general, we can show that if $AM,AM'$ are isogonally conjugate cevians, then $AE\perp MM'$.[/quote]

Indeed! This can be done in the same way as I proved it for two isogonal conjugate points M and M', since the points Q, R, Q', R' lie on one circle not only for isogonal conjugate points M and M', but also generally for any two points M and M' such that AM and AM' are isogonal conjugate cevians.

[quote="grobber"]This even has a projective generalization, by considering $P\ell-$perpendicularity instead of perpendicularity. I won't state it, however, because there are too many sentences :).[/quote]

Well, I usually consider projective generalizations only if they are particularly interesting. In this case, it seems to be just some straightforward replacement of perpendicularity with $P\ell$-perpendicularity and of isogonal conjugation with reciprocal conjugation...

  Darij

Let $PQ\cap P'Q'=Z$, $QR\cap Q'R'=X$ and $PR\cap P'R'=Y$, $O$ is center of circle $k$ (it's $ (PQRP'Q'R') $ - pedal circle).

We have that $X,Y,Z$ are collinear and this line passes through the points $M,O,M'$ by Pascal's theorem.
So the polar lines of the points $X,Y,Z$ with respect to $k$ are parallel. These polars are $AE,BF,CG$. :lol:

Dear Mathlinkers,

this is a generalization of a result of Schroeter...

You can see a synthetic proof on
http://perso.orange.fr/jl.ayme  vol. 2  Les deux points de Schroeter p. 4-7  where the commentar precises that the technic presented can be adapted to two isogonal points...
Sincerely
Jean-Louis


find all functions $f: \mathbb{R}\rightarrow\mathbb{R}$ such that $f\big(f(x)+y\big)=f\big(f(x)-y\big)+4f(x)y$, for all reals $x,y$.


do we have a formula to find $ \chi(X\bigvee Y)$ in terms of $ \chi(X)$ and $ \chi(Y)$ ? many thanks

The Euler characteristic is additive with respect to disjoint unions of complexes. Further, if $ X$ is a CW-complex and $ A$ a subcomplex, then by comparing the number of cells in $ X$, $ A$ and $ X/A$ we get that $ \chi(X/A) \equal{} \chi(X) \minus{} \chi(A) \plus{} 1$. Together, it follows that $ \chi(X\vee Y) \equal{} \chi(X) \plus{} \chi(Y) \minus{} 1$.

If you do not want to use CW-complexes but instead the homological definition of the Euler characteristic, then you may use the isomorphism $ H_k(X\vee Y)\cong H_k(X)\oplus H_k(Y)$ for $ k>0$ (this works at least for well-pointed spaces) and $ \text{rank} (H_0(X\vee Y)) \equal{} \text{rank}(H_0(X)) \plus{} \text{rank}(H_0(Y)) \minus{} 1$ (we glue together the path-components of $ X$ and $ Y$ containing their respective basepoints).

Since I abhor asymmetry, it can be noted $ \widetilde{H}_n(X \vee Y) \equal{} \widetilde{H}_n(X) \oplus \widetilde{H}_n(Y)$, since reduced homology is a more natural thing to consider for pointed spaces, though it still requires that $ X$ and $ Y$ are cofibrant; but that's more of a problem with the wedge sum operator than with the spaces themselves  :rotfl: .


conics ellipse geometry perimeter function calculus integration

What's the most exact approximate formule for the perimeter of ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

And the surface area of ellipsoid $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$

You can make an approximate formula arbitrarily exact by numerically integrating the appropriate elliptic integral.

you can learn how to google :)

http://mathcentral.uregina.ca/qq/database/QQ.09.99/proffit1.html

http://en.wikipedia.org/wiki/Ellipsoid

http://en.wikipedia.org/wiki/Ellipse

but those formulas are too advanced. one requeriment is know calculus.

a non-advances and poor approximation can be made using geometry

the area of this ellipse is approximatelly the mean of the area of a circle of radius a and a circle of radius b

the same thing for the ellipsoid

[quote="yagaron"]the area of this ellipse is approximatelly the mean of the area of a circle of radius a and a circle of radius b

the same thing for the ellipsoid[/quote]

The area of an ellipse with radii $a$ and $b$ is $\pi a b$, not $\pi \left(\frac{a^{2}+b^{2}}{2}\right)$.

Are you talking about surface area of an ellipsoid or volume?
I don't know about surface area, but the volume of an ellipsoid with radii $a, b, c$ is $\frac{4}{3}\pi a b c$.

I believe he meant that the perimeter of an ellipse is (really roughly) approximately $\pi (a+b)$.

And, vishalarul, not only is your volume equation wrong (it should be $\frac{4}{3}\pi abc$), but he specifically asked for surface area because, analogously to the perimeter of an ellipse, the surface area of an ellipse cannot be expressed as an elementary function of $a, b, c$.

like t0rajir0u. we are talking about perimeter, not area. i just write it wrongly. this is a little bit obviously...   :)

On what t0rajir0u said: it's actually very easy to approximate the elliptic integral numerically. If you're approximating $\int_{0}^{2\pi}\sqrt{a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta}\,d\theta$ for reasonable $a,b$, a left-endpoint Riemann sum with about 30 terms will give you accuracy to within the usual roundoff error of a standard floating-point system. The more sophisticated numerical integration methods don't actually work better in this case.


[size=200][color=green]
HAPPY NEW YEAR!! WELCOME TO 2010 ALL OF YOU :wow:
[/color][/size]

SAME TO EVERYONE!
I'm officially the first person to post in 2010 on aops India :D

I have the honour of being the second. Same to all.

[quote="Pocha"]How does New Year matter?,it's the same old thing newayz. :|[/quote]
NOOOOOOOO
I love New Years. I wish there were at least 3-4 new years EVERY year  :D

yes new year is a casual thing  but in this busy world v should celebrate such things to make ousrelf cheer up and lead the rest of our life happily   ........i think iam gng in a philosphical way   any ways here r my wishes

HAPPY NEW YEAR                                 Njyyyy it Yaaarrrr

[quote="abhitayal"]A VERY HAPPY NEW YEAR to all mathlinkers and AOPSians. :D
Sorry for being a bit late :rotfl:
though its still 31st as per US :rotfl:[/quote]
My career depends on that very fact!

Lol , Not finished the essays yet?  :P

Read somebody's else's message wishing u a happy new year, pick ur favourite & assume its from me. I wont even feel bad if u dont wish me back.

happy new year!!! :)
Well, I'm indian...too! It's sort of obvious, isn't it? :P


$\text{Find the sum of the variables if: }$

$8x+4z=25+5y$
$x=3+w$
$x+z=2-w+2y$

[hide]5 [/hide]

Are you sure?  Isn't the amount of equations needed equal to the amount of variables there are?  And in this case, there are four variables but three equations.  $w$ can be easily cancelled out, but that leaves two equations with three variables.  I'd like to see how this is solved... possibly you could add/subtract the equations in such a way that leaves $x+y+z+w=n$.

You cannot solve for w, x, y, and z, but you can probably solve for their sum. For example, if you know w+x=2 and y+z=3, then w+x+y+z=5, but you know nothing about the individual variables.

Yes, dk2 has the right idea.

[hide]Try rearranging the equations such that variables are one side and constants are on the other[/hide]

[hide]I also get 7.  Here is my solution if anyone wants to see it:

We have:

\begin{eqnarray}
8x-5y+4z &=& 25\\
x -w &=& 3\\
x-2y+w+z &=& 2
\end{eqnarray}

So now, we want $x+y+w+z$.  We can get this by some linear combination of (1), (2) and (3).   In other words, we will mulitiply (1) by some constant, say $k_1$, (2) by another constant, $k_2$, and (3) by $k_3$.  After we multiply each equation by its respective constant we will add all three equations together and get:

\[(8k_1+k_2+k_3)x+(-5k_1-2k_3)y+(-k_2+k_3)w+(4k_1+k_3)z=25k_1+3k_2+2k_3.\]

We set this equal to $x+y+w+z$ (since that's what we want) and then compare coefficients.  We end up with the system of equations:

\begin{eqnarray*}
8k_1+k_2+k_3 &=& 1\\
-5k_1-2k_3 &=& 1\\
-k_2+k_3 &=&1\\
4k_1+k_3 &=&1
\end{eqnarray*}

We solve this and get $k_1=1$, $k_2=-4$ and $k_3=-3$.

Thus $x+y+z+w=25(1)+3(-4)+2(-3)=7.$[/hide]

joml88 got 7. He just messed up his last number at the end.

Hmmm, here's my solution:

$3(x-2y+z+w = 2) = 3x-6y+3z+3w = 6$

$8x-5y+4z = 23$  - $3x-6y+3z+3w = 6$ will give you:
$5x+y+z-3w = 17$

Subtract 4 times the bottom equation (which gives you $4x-4w=12$ and you'll get $x+y+z+w = 5$

OH!!! OOPS! sorry!  :blush: the original problem had it with 23.. i just copied it down incorrectly.... yeah, so if it was 25 then it would be 7. my bad!


[hide="Solution"]Since $2n+1$ and $3n+1$ are perfect squares, both of them are $0,1 \text{ or } 4 \mod 5$ and $0,1 \text{ or } 4 \mod 8$. After checking all possible residues modulo $5$ and $8$, we see that $n= 0 \mod 5$ and $n=0 \mod 8$, thus $40|n$.[/hide]

[quote="Xixas"]A much more interesting question is to find all such $n$ that $2n+1$ and $3n+1$ are both perfect squares.[/quote]

Indeed! :)

This idea came up to me when I solved the problem : 2n+1 and 3n+1 perfect squares ==> 5n+3 composite.

It is not hard. All we have to do is to solve a generalized Pell's equation $3x^2-2y^2=1$.

[quote="abdurashidjon"]If n is element of natural number and 2n+1 3n+1 are squares, then n ios divisible by 40[/quote]

$ 2n+1 = x^2 $
$ 3n+1 = y^2 $

$ 5n+2 = x^2+y^2 $
when we check any square modulo $5$ we get residues as $0 , 1, 4 $
this means that $ x^2\equiv1 \text{mod}(5)  , y^2 \equiv1 \text{mod}(5) $

$ n = y^2 - x^2 $
$ n\equiv0 \text{mod}(5) $

since $ 2n+1 = x^2 $ it means that $ 4 | n  , n = 4k $
$ 12k+1 = y^2 $ this means that $ 2 | k $ gives $ 8|n $

since $ (5,8)=1 $
$ 40|n $

Hence done !


trigonometry
integration
real analysis
real analysis unsolved

Find solutions to the following boundary value problem
\[ \begin{cases}
\frac{\partial}{\partial t}u(t,x)\equal{}\frac{\partial^2}{\partial x^2} u(t,x) & \text{  in  } \mathbb{R}\times (0,T)
\\
u(x,0)\equal{}0 & \text{ on } \mathbb{R}\times \lbrace t\equal{}0 \rbrace
\end{cases}\]

I'm slightly confused on notation. You wrote in the first one $ u(t,x)$ and then in the second $ u(x, 0)$ on $ \mathbb{R}\times \{t=0\}$. I'm assuming that in keeping consistency with the first line you want $ u(0,x)$ on $ \{0\}\times\mathbb{R}$?

In that case, you should be able to do the standard separation of variables and get $ u(t,x)=\sum_{n=1}^\infty C_n\sin\left(\frac{n\pi x}{T}\right)e^{\frac{-n^2\pi^2 t}{T^2}}$ where $ C_n=\frac{2}{T}\int_0^T u(x,0)\sin\frac{n\pi x}{T}dx$.

My mistake, the second line should read $ u(0,x)$.  And in the first line $ (0,T)$ denotes the open interval $ \lbrace 0 < t < T\rbrace$, which I think is pretty standard notation.  So separation of variables only yields $ u\equiv 0$.


I have a lot of free time and have decided to start a game. I was thinking of an idea that has never been done before and all of a sudden I realized that no one has ever done Amazing Race. I have already made plans for all of the challenges and everything.

To Join:

You must find a partner that you trust and know will be using AoPS at around the same time as you. It would be a very good advantage to acctually know and be able to meet with this person in real life. If you would like to join simply say join and say your partners name in the game or if you don't care I'll put you with someone else who says they dont care. Posting in the acctually game thread will most likely be limited to eliminations as you do not want other people  to see your ideas. However, I will need a copy of the URL's to your quicktopic's. Finally, remember this is a game that will take you throughout cyberspace if you have filters on websites that are social websites please tell me so I will not have a location be their. Unlike the real Amazing Race I realize you and your partner may not be able to show up at the same place at the same time. So all I will require is that one of the two of you make it to the final location. Elimination is simply the last team to arrive at the location. If no more teams arrive in 5 days then any team that has not arrived will be eliminated.

I will accept 12 teams for this game.

I'll join.. I'm considering a certain partner.

[quote="perfect628"]Yeah I want to play, I do care about my partner, but I don't have one atm. I'll say soon, anyone who wants to be with me make a request.[/quote]

I'll be your partner.

OK [b]perfect628 and CatalystOfNostalgia[/b].

So like how exactly does this work?

I might have to drop out of this later...who knows what the workload will be like. I'm expecting 4-5 hours of homework on weekdays, and then hopefully around 2.5 hours of piano so I might not be able to do it. We'll see.

[quote="james4l"]I will join.[/quote]
Team?

The message contains only quote and code tags. Please add some text of your own.
Wait, no it doesn't...

[quote="CatalystOfNostalgia"]So like how exactly does this work?[/quote]

Well, its a bit like the acctual Amazing Race. At the beguinning of each competition I will give you your first clue. Then you do tasks to get the rest of your clues until you finally reach your final destination. The final destination maybe an acctual meet up with me or you may just have to PM me the URL of the destination website.

Umm yeah I don't really plan on leaving the state for an online game...

[quote="CatalystOfNostalgia"]Wait, like what? You want us to like go find you at your house? Are you paying for my plane ticket?[/quote]

Lol....you take things too literal. I mean meet up with me in a virtual reality setting. Ex. Second Life.

I hope what I said in the above post helps. If you have any more questions post them or PM me.

Id like to team up with pythag. But if he rejects Id like to team up with dragon.

[quote="crazychinesemaniac"][quote="pythag011"]I think dragon lives in San Francisco Bay Area....[/quote]

In that case I'll take that ticket and just get a cab. Thanks! You just saved me around 900 dollars flying you to where ever you wanted to go. :D

Almost forgot, as you all know Leg 3 is a reward leg. I may have one in round 4 or 5 but thats not totally settled. So, the reward for Leg 3 is...

[hide="*Drum Roll Please*"][b][color=green]Team who wins this will be allowed to receive their clue 1 up to 24 hours before the rest of you get the clue.[/color][/b][b]Please post that you want to use it at least 24-36 hours in advance before the next round will start for the other people.[/b][/hide]

[b]Also before I forget I will ask the team of Brut3Forc3 and Budi713 if they want to use their Exemption.[/b][/quote]

Actually.... Ummm...I ... kinda live on the opposite side of the world from SF...so you will have to buy me a plane ticket to SF please. heh heh heh Then I will take a cab home and sell the ticket on ebay when I get home and- whoops. Ignore the last sentence.[color=white]I actually do live in SF[/color]

Waiting on team of B and B if they want to use exemption. Probably will start tomorrow.

We decided not to use the exemption.

(i think)

I'll assume your team will [b]Not[/b] use your exemption. Clue 1 will be up tomorrow, if I'm on  :|

[size=200][color=green][u][i][b]Clue 1: Leg 3[/b][/i][/u][/color][/size]

[hide="Clue 1"]I hope you have been enjoying your travels. Now don’t forget, this competition is for the reward of receiving your first clue up to 24 hours before the other teams! You may want to get a leg on it (looks at team of CoN and perfect628). Anyway, here is your first clue. Please pick a plane and please do NOT make-out with your partner. You may do so after you win the race. Well good luck!

This city has 33 districts and has something named after Jekyll and Hyde (though it may have just been a coincidence). You could say this city is a bit [i]bloody[/i] and a book that is very popular among teenage boys takes place here.[/hide]

[hide="Teams Still In"][b][color=green]Cf249 and AIMEman
Pythag011 and James4l
Budi713 and Brut3Forc3
Ignite168 and Klebian
Perfect628 and CatalystofNostalgia[/color][/b][/hide]

dang why did i just read that book, notice a TV show copying them just yesterday, and still not see the connection?

Hmm...I haven't recieved a single answer in the last 3 days and no one has finished except 1 team.

[u][size=200][b][color=red]TEAMS HAVE 20 HOURS TO FINISH OR THE ONLY TEAM WHO HAS FINISHED WINS![/color][/b][/size][/u]

Clue for Clue4: Think of the periodic symbol for Oxygen.

With only 2 teams finishing we are down to our last 2 teams! The final teams are, Pythag011 and James4l and Ignite168 and Klebian!

[hide="The winner of the Reward..."][color=green][b]Ignite168 and Klebian[/b][/color]

I'll now ask the dumb question of whether or note Ignite168 and Klebian would like to use their reward. [/hide]

[hide="The Winners."][color=green][b]Taking full advantage of their reward the winners are.........Ignite168 and Klebian! Congratulations to those two![/b][/color]
I now open up the thread for discussion of the game. Oops...I just lost the game.[/hide]

can I see the last clue?

Well I didn't get the clue even at September 27 :(

You wouldnt have recieved it as it took ignite all of about 30 minutes to finish :/ however if you would like it I will post it. AH...nvm AoPS won't let me send .docx files sorry.

Clue 1: This city has a very popular name among girls. It is the most populous city of its country and has a magnificent view of the sea. It was established in 1788

Clue 2:  Welcome to Sydney, Australia. Here is your next clue: This next area is a bay and has a song named after it. In fact this song was created in the 1890’s and you’ve probably never heard it…Two runways of Sydney Airport extend into this bay and it is south of the Central Business district.

Clue 3: Welcome to Botany Bay! Here is your next clue: Not as famous as Sydney Opera House this is still a theatre of great history. This used to be a pier before in 1979 it was converted into a theatre.

Clue 4: I hope you remember Tower Bridge from our travels…Here is yet another bridge. This is probably the most important bridge in all of Sydney, Australia.

Actually it took under 6

Klebian solved a total of 0 clues on this race :(

klebian was told a total of like 1 clue on this race


From a)  and b), you know the number ends in 0. and from c) you know it starts with 987...

Thus there are 6*5*4 numbers, or 120 numbers.

well if a number is even and divisible by 5, we know it ends in 0.
Then we also know the first three digits.
Then doing 6*5*4=120.

EDIT: Beaten.

[quote="thetruthisfake"]Find number of different 7 digit phone numbers (not beginning with 0) that can be formed using digits 0 to 9 without repetition if :
a) The number is an even number
b) The number is a multiple of 5
c) The first 3 digits of the number is 987 in order

[EDIT]
Oops.. a bit of misunderstanding here.
It's actually,

a) Find number of different 7 digit phone numbers (not beginning with 0) that can be formed using digits 0 to 9 without repetition if the number is an even number
b) Find number of different 7 digit phone numbers (not beginning with 0) that can be formed using digits 0 to 9 without repetition if the number is a multiple of 5
c) Find number of different 7 digit phone numbers (not beginning with 0) that can be formed using digits 0 to 9 without repetition if the first 3 digits of the number is 987 in order[/quote]

Wait, so now there's 3 different problems??

[hide="SolutionA"]
There are 5 choices for the last number, and 8*8*7*6*5*4 choices for the other 6 numbers for a total of 268800 different phone numbers.
[/hide]
[hide="SolutionB"]
There is 2 choices for the last number, and 9*8*7*6*5*4 choices for the other 6 numbers for a total of 120960 different phone numbers.
[/hide]
[hide="SolutionC"]
There are 7*6*5*4 different choices for the last 4 digits, for a total of 840 different phone numbers
[/hide]
I hope these are correct..... :maybe:

Edit: Yeah, sorry about the double post. Can someone delete that?

I'm not very sure of the answer. It's based on my textbook, however, the answers from the textbook are slightly vague.

I did Q1 this way..
CASE #1 - Ends with 2,4,6,8

8x8x7x6x5x4x4 = 215040

CASE #2 - Ends with 0

9x8x7x6x5x4x1 = 60480

Answer : CASE #(1+2) = 275520

However, the answer from the textbook is only 215040, can anyone verify for me?


My nephew was excited as he came back from school and asked me this problem

A box has 10 balls numbered 1,2,....,10

A ball is picked at random and then a second ball is picked at random from the remaining nine balls. Find the probability that the numbers on the two selected balls differ by two or more.

i believe this is right
[hide]so the total number of selections is:

$\binom{10}{2}$ (since if you pick 3 and 5 its the same as picking 5 and 3)

So the number of ways the ball numbers differ by one is obviously $9$, subtract the two to get total number of ways the number on the balls differ by atleast two.

$S_{Total}=\binom{10}{2}-9=45-9=36\implies\frac{36}{45}=\boxed{\frac{4}{5}}$
[/hide]