I don't know if this goes here...
but i think it does.

CAN ANYONE HELP ME IN MODULAR MATH?
more specifically,
what is modular math, and what are mods?

Thanks!

thanks for the help, pascal12.

However, I was looking for basic modular math.
Can anyone explain in a simple-ish way?
thanks!

Are you looking for a definition? I'm not sure how basic you want.

Mods are like remainders, and you would express that when 12 is divided by 5, there is a remainder of 2 as:
$ 12 \equal{} 2\mod5$ (Not sure how to the congruence sign.)
So with the general form $ a \equal{} b\mod{c}$, a is what you start out with, c is what you divide by, and b is your remainder.
So $ 8 \equal{} 3\mod5$ and $ 39 \equal{} 3\mod6$

There's also rules of mods like addition/subtraction and multiplication, where you do these operations to both a and b.

So you can say $ 12 \equal{} 2\mod5$
$ 12 \plus{} 1 \equal{} 2 \plus{} 1\mod5$
$ 13 \equal{} 3\mod5$

[quote="pascal12"](Not sure how to the congruence sign.)
[/quote]
You can do \ equiv,
$ 3^2 \plus{} 4^2\equiv0\pmod{25}$


Two dice are rolled. If one of the dice rolled an odd number, what is the probability that their sum is 10?

Problem needs to specify whether exactly one die is odd or at least one die is odd.  I assumed that there was one odd and one even.

You can use the "Report error with problem or solution" button at the bottom of the solution page.

If the problem specified one odd and one even, I think it would be worded "If [b]exactly[/b] one of the dice rolled an odd number".  Still, I agree that it is confusing.


[quote]شرکت در (مسابقه فقط برای کسانی که به ادامه زندگی علاقه‌مندند مجاز است[/quote]
 :rotfl:  :rotfl:  :rotfl:

nemidonam chera fek mikonam dar har hal ehtemal terekidanemon 1/10  ;)  :D

bayad nafare akhar raft age didi kasi oonja pokide bedone inke kasi befahme mipichooni (albate age did ham khiali nist)

[quote="nodet"]در اتاقی ۱۰ صندوق وجود دارد. در ۹ صندوق پول و در یک صندوق بمبی است که با بازشدن صندوق منفجر می‌شود. شما و نه نفر از دوستانتان در صفی ایستاده‌اید، به نوبت وارد اتاق می‌شوید و یکی از صندوق‌ها را باز می‌کنید.  شما ترجیح می‌دهید چندمین نفر صف باشید؟ چرا؟ (شرکت در (مسابقه فقط برای کسانی که به ادامه زندگی علاقه‌مندند مجاز است![/quote]
avalin nafr chon ehtemale zende moondan maximum(90%)hast.
albate agar har kasi pas az baz kardan jabash az otagh bere biroon ke fekr konam manzoore masale hamine. :maybe:

Forget the bomb for now and think that there are $9$ boxes with money and $1$ empty box. Now without loss of generality we may suppose that the persons pick up the boxes in order $1,2,\ldots,10$.

The probability that the empty box is the $i$-th box is obviously $\frac 1{10}$. So the answer is, everyone has the same chance to survive.

دروود

مسلما نفر آخر چون احتمال زندگیش خیلی زیاده
دفعه ی اول احتمال یک دهم هست و اگه اتفاق نیفتاد میشه یک نهم.
و باز اگه نشد میشه یک هستم و.... یعنی 9 بار قبل از ما احتمال های مختلف اتفاق میافته  و به  احتمال خیلی زیاد زنده میمونیم ما و پولدار میشیم.

من نمیدونم که حساب کردن ریاضیش چجوری میشه ولی جواب همینیه که گفتم!

سلامت باشین دوستان :)

[quote="ArmanBM"]
مسلما نفر آخر چون احتمال زندگیش خیلی زیاده
دفعه ی اول احتمال یک دهم هست و اگه اتفاق نیفتاد میشه یک نهم.
و باز اگه نشد میشه یک هستم و.... یعنی 9 بار قبل از ما احتمال های مختلف اتفاق میافته  و به  احتمال خیلی زیاد زنده میمونیم ما و پولدار میشیم.
من نمیدونم که حساب کردن ریاضیش چجوری میشه ولی جواب همینیه که گفتم!
[/quote]

na, eshtebah nakon! baraye inke nafare dovvom bemire, bayad nafare avval namire (ke ehtemalesh $\frac{9}{10}$ ast) va az $9$ ja'beye mande bomb tu ja'beye avval bashe (ke ehtemalesh $\frac{1}{9}$ ast) ke mishe $\frac{9}{10}\times \frac{1}{9}= \frac{1}{10}$ va hamintor baraye baghie, ehtemal $\frac{1}{10}$ ast.


I have to write graduating paper on subject of diophantine equations for my final year in high school. I would really appreciate if someone knows where to find some good material on the subject (online of course). Thank you in advance.

Exactly what kind of information are you looking for? On how to solve them?

Yes, I can solve them, but I don't know how to systematize it and write a paper on it.

Don't know if this will help, but try [url=http://en.wikipedia.org/wiki/Diophantine_equation]this[/url].

You may want to look at the following:
[url]http://209.85.173.132/search?q=cache:UMX8DlxDXVUJ:matwbn.icm.edu.pl/ksiazki/aa/aa83/aa8316.pdf+Diophantine+equations+papers&hl=en&ct=clnk&cd=8&gl=us&client=firefox-a[/url] and [url]http://www.astro.virginia.edu/~eww6n/math/DiophantineEquation.html[/url].

Flame , I would not prefer to you to copy and paste from wikipedia, at least try to edit and make some self-contained research

Thank you.

Of course I won't copy paste but the problem is that I have to write some literature I used. I could write everything by myself almost without any external source. But the rules are the rules, and i have to follow them.

As you are from Croatia, this might help (Serbian site):
http://www.diofant.org/


linear algebra
matrix
linear algebra unsolved

Prove that  there exist $(x_n,y_n,z_n,t_n)\in Z^4-((0,0,0,0))$
such that $t_nx_n^2-4x_ny_n^2+4z_ny_n^2-z_n^2t_n=(-8)^n$ for any $n \geq 1$

If the problem is as stated, there is a trivial solution $x_n=t_n=(-2)^n$, $y_n=z_n=0$. Seems like Moubinool meant something else. I wonder what it might be :?

And that's not a solution in $Z^4-\{0,0,0,0\}$?

Probably, Moubinool means that none of $x_n,y_n,z_n,t_n$ should be zero. But even then, we have a simple solution $x=t=1$, $y=z=-1$ for $n=1$, from which we can get a solution for any $n$ by multiplying each number by $(-2)^{n-1}$ :?


Prove that for each $n$ there are $n$ distincts numbers $a_1, a_2, ...,a_n$ such that for each $a_i$ the product of the other $n-1$ numbers is congruent with 1 mod $a_i$.

Consider this sequence:$a_1=2,a_{n+1}=a_n^2-a_n+1$
Easy to see $a_{n+1}=a_1a_2\cdots a_n+1$
We can prove that $a_1,a_2,\cdots,a_{n-1},a_n-2$ satisfied the condition.


We have :

$ 2(ab \plus{} bc \plus{} ca \plus{} 1) \equal{} (a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) > 0$

 :P


Let $ a_1, a_2,\ldots  ,a_8$ be $8$ distinct points on the circumference of a circle such that no three chords, each joining a pair of the points, are concurrent. Every $4$ of the $8$ points form a quadrilateral which is called a [i]quad[/i]. If two chords, each joining a pair of the $8$ points, intersect, the point of intersection is called a [i]bullet[/i]. Suppose some of the bullets are coloured red. For each pair $(i j)$, with $ 1 \le i < j \le 8$, let $r(i,j)$ be the number of quads, each containing $ a_i, a_j$ as vertices, whose diagonals intersect at a red bullet. Determine the smallest positive integer $n$ such that it is possible to colour $n$ of the bullets red so that $r(i,j)$ is a constant for all pairs $(i,j)$.

Note that coloring $1$ bullet red, say $A_j A_k \cap A_xA_y$, increments each of the $\binom42$ $r(s, t)$'s with $s, t \in \{j, k, x, y\}$ by one. Therefore, the problem is equivalent to finding the minimal $n$ such that we can pick $n$ quadruples of $A_i$'s such that every edge $A_iA_j$ is contained in the same number $z$ of quadruples. Observe that by an easy double-counting argument, we have that $\binom82 \cdot z = \binom42 \cdot n$, and so we clearly obtain that $3 | z$ by mod $3$. This means that $n \ge \frac{\binom82 \cdot 3}{\binom42} = 14.$

All that remains is to construct an example with $14$ quadruples. To do so, simply use $\{1, 2, 3, 4\}, \{1, 2, 5, 6\}, \{1, 2, 7, 8\}, \{3,4,5,6\}, \{3,4,7,8\}, \{5,6,7,8\},$ and all of the $\frac{2^4}{2} = 8$ quadruples in $\{1, 2\} \times \{3, 4\} \times \{5, 6\} \times \{7, 8\}$ which have the property that the sum of all four elements is even.


function
integration
calculus
real analysis
real analysis unsolved

Is there a function $f : \mathbb R \to \mathbb R$ which is three times differentiable in only one point $x_0 \in \mathbb R$?

First note that if $f: \mathbb{R}\to \mathbb{R}$ is bounded in a neighborhood of $0$ then $g(x)=x^2f(x)$ is differentiable at $0$.

Now take  for $f$ a continuous non differentiable at all point, then $g$ must be differentiable only at $0$, so for  $G(x)=\int_0^x g(t)dt$, $G'=g$ is two times diff only at 0 and a second integral will work.

Something I was wondering is if here exists a continuous solution not diff at all point but 0, but 3 times diff at 0. Does anybody has any idea?

Good idea.
but there is no continuous function that is not differentiable anywhere but 0 and is 3 times differentiable at 0 for otherwise it should be differentiable at a neighborhood of 0.
But there does exist a continous function not differentiable any where.
It is the so called Werestrass function.
Denote this function by $f$. Let $g(x)=x^2f(x)$.
Then $g$ is continuous and differentiable only at 0.
Put $h(x)=\int_{0}^{x}g(t)dt$, then $h'(x)=g(x)$.
Now $F(x)=\int_{0}^{x}h(t)dt$ is the one we want.

well using Baire's theorem in the space of continuous functions  of $\mathbb{R}$ you can prove that almost everyone is not differentiable in every point. It can be found in Dujunji's Book.


During the monsoon season, torrential rainstorms occur. In how many distinct ways can the letters of MONSOON be arranged?

The 8 letters can be arranged 8! ways. You must then divide by 3! for the indistinct O's and by 2! for the indistinct N's. this gives you $ \frac{8!}{3! \cdot 2!}\equal{}\boxed{3360}$

[quote="Ebob"]The 8 letters can be arranged 8! ways[/quote]

Count them. There is actually 7 letters. Therefore:

$ \frac{7!}{3! \cdot 2!}\equal{}\boxed{420}$


Let f:N→N be a function satisfying $ kf(n)\le f(kn)\le kf(n) \plus{} k \minus{} 1$ for all $ k, n\in N$.

(a)Prove that $ f(a) \plus{} f(b)\le f(a \plus{} b)\le f(a) \plus{} f(b) \plus{} 1$ for all $ a, b\in N$.

(b)If $ f$ satisfies $ f(2007n)\le 2007f(n) \plus{} 200$ for every $ n\in N$, show that there exists $ c\in N$ such that $ f(2007c) \equal{} 2007f(c)$.

Suppose there exist $a,b$ such that $f(a+b)\leq f(a)+f(b)-1$
Now $abf(a+b)\leq f(ab(a+b))\leq abf(a+b)+ab-1$
And $(a+b)f(ab)\leq f(ab(a+b))\leq (a+b)f(ab)+a+b-1$
Now from assumption we've $ab(f(a)+f(b)-1)+ab-1\geq (a+b)f(ab)$
But then we get $0>a(f(ab)-bf(a))+b(f(ab)-af(b))>0$
so it's a contra and $f(a+b)\geq f(a)+f(b)$ for all $a,b\in\mathbb N$
Now similarly let exist $a,b$ such that $f(a+b)\geq f(a)+f(b)+2$
So we've now $(a+b)f(ab)+a+b-1\geq ab(f(a)+f(b)+2)$
So we've now $a(b-1)+b(a-1)\geq a(f(ab)-bf(a))+b(f(ab)-af(b))\geq 2ab-(a+b)+1$
so again it's a contradiction and hence,we get
$f(a)+f(b)+1\geq f(a+b)\geq f(a)+f(b)$ for all $a,b\in\mathbb N$
Now for second part we'll use the fact $af(2007n)\leq 2007f(an)+2006$
Assume there is no such $c$ so we've now $f(2007n)\geq 2007f(n)+1$ for all $n\in\mathbb N$
Now take $a=(2007)^x,n=(2007)^y$ so we have
${LHS\geq (2007)^{x+1}f((2007)^y}+(2007)^x$
and so $LHS \geq (2007)^{x}+(2007)^{x+1}+(2007)^{x+2}f((2007)^{y-1})\geq .......$
and going on $RHS\geq (2007)^x+(2007)^{x+1}+......+(2007)^{x+y}f(2007)$
Now $RHS\leq (2007)^2f((2007)^{x+y-1})+403436\leq......\leq (2007)^{x+y}f(2007)+m$
where $m$ is a number independent from $x$ dependent on choosing of $y$
so can choosing such $x$ such that $m$ doesn't satisfy that property, and hence done.

I think we can prove an even stronger statement which trivializes both (a) and (b).

(a) We claim that there exists some positive real $r \ge 1$ for which $f(n) = \left \lfloor nr \right \rfloor$. This would clearly imply the desired equation since $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor \le \left \lfloor x + y \right \rfloor \le \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor  + 1, \forall x, y \in \mathbb{R}^{+}.$ Observe that the condition given is equivalent to $\left \lfloor \frac{f(kn)}{k} \right \rfloor = f(n).$

Let's define $I_n$ as the closed (possibly empty) interval of positive real numbers $r$ satisfying the condition that $f(i) = \left \lfloor ir \right \rfloor$ for all $1 \le i \le n.$ Then, notice that $I_{n+1} \subseteq I_n$ for all $n \in \mathbb{N},$ and that $I_1$ is nonempty. Now, we are going to claim that $I_n$ is nonempty for all $n \in \mathbb{N}.$ Indeed, simply observe that $\frac{f(n!)}{n!}$ is in this interval, due to the fact that $ \left \lfloor \frac{f(n!) i}{n!} \right \rfloor = f(i)$ for all $i = 1, 2, \cdots, n$ from out condition above. Therefore, since $I_1 \supseteq I_2 \supseteq I_2 \supseteq \cdots$ is a descending chain of nonempty subsets, we know by Cantor's intersection theorem that there exists a real number $r$ such that $r \in I_n, \forall n \in \mathbb{N}.$ However, this would then imply that $f(n) = \left \lfloor nr \right \rfloor, \forall n \in \mathbb{N}.$ Note that $f(n) \ge nf(1)$ implies that $r \ge 1$ as we claimed above.

(b) Let $r$ be the real number that we found above. The condition is then equivalent to $\left \lfloor 2007nr \right \rfloor \le 2007 \left \lfloor nr \right \rfloor + 200,$ which is equivalent to $\left \{ \frac{nr}{2007} \right \} < \frac{201}{2007}.$ However, this clearly implies that $\left \{\frac{nr}{2007} \right \} = 0, \forall n \in \mathbb{N},$ which hence means that $r$ is actually a positive integer divisible by $2007.$ As a result, we can conclude that $f(2007c) = 2007cr = 2007 * cr = 2007 f(c)$ for $\mathbf{all}$ $c \in \mathbb{N}.$

$I_n$ is not closed, and I'm unsure if this can fixed. So far I have only found that $f(n) = kn - 1$ is a solution for $k \in \mathbb N$, but there may be more.

However, this is sufficient to solve the problem as given, I think, as for every $n \in \mathbb N$ there is $r_n \in \mathbb R$ such that $f(k) = \lfloor r_nk \rfloor$ for all $k \leq n$.


Does anyone know of a good book on the consequences of the Riemann Hypothesis with lots of good proofs? Or else, would someone mind explaining to me why the following statement is equivalent to the Riemann Hypothesis:

sigma(n) <= H_n+e^H_n*ln(H_n) for all n>=1, with equality only if n=1.

Thanks very much!

The proof you want can be found here - Lagarias, J. C. "An Elementary Problem Equivalent to the Riemann Hypothesis" 22 Aug 2000: http://xxx.lanl.gov/abs/math.NT/0008177/

There is also a pretty good bibliography on the Riemann Hypothesis here: http://mathworld.wolfram.com/RiemannHypothesis.html

And also at the back of the book The Music of the Primes by Marcus du Sautoy.


[quote="dragon96"]How do I shift something like 2 units down? I think theres a shift() command, but how do I use it?[/quote]

[code]shift(a_pair)*a_path;[/code]
OR
[code]shift(a_triple)*a_path3;[/code]

[asy]size(5cm);
path thesquare=((0,0)--(0,1)--(1,1)--(1,0)--cycle);
filldraw(thesquare, black, 2mm+red);
filldraw(shift(0.5,0.5)*thesquare, red, 2mm+black);[/asy]
;-)

[quote="fedja"]Except shift(triple)*picture didn't work in one of the recent versions. I do not know if they fixed this bug by now :?.[/quote]
Yes, it is fixed in the incoming version 1.44.
The Asymptote version 1.44 allows [b]to produce and embed[/b] PRC code.
The code...
[code]import graph3;
import solids;
size(0,150);
currentprojection=perspective(0,0,10,up=X);
currentlight=(3,2,-1);
pen color=green;
real alpha=250;

real f(real x) {return x^2;}
pair F(real x) {return (x,f(x));}
triple F3(real x) {return (x,f(x),0);}

pair H(real x) {return (x,x);}

path p=graph(F,0,1,n=10,Spline)--graph(H,1,0,n=10,Spline)--cycle;
surface s=surface(bezulate(p));
path3 p3=path3(p);

revolution a=revolution(p3,X,-alpha,0);
draw(surface(a),color);
draw(s,color);
draw(rotate(-alpha,X)*s,color);

xaxis3(Label("$x$",1,align=-2X+Y),dashed,Arrow3);
yaxis3(Label("$y$",1,align=X),Arrow3);
dot(Label("$(1,1)$"),(1,1,0),Y);
arrow("$y=x$",(0.7,0.7,0),Y,0.75cm,red);
arrow("$y=x^2$",F3(0.7),X,0.75cm,red);
draw(arc(1.25X,0.3,90,90,3,-90),Arrow3);[/code]
... compiled with the command [i]asy -f pdf -prc file.asy[/i] gives this interactive pdf file:


limit
calculus
derivative
logarithms
calculus computations

Since the limit is indeterminant, you can use L'Hopital's Rule.  The derivative of the denominator is 1, but the top is pretty messy.  I velieve it is
$ \frac{\frac{1}{1+\ln(x)}-\ln(1+\ln(x))}{x^{2}}e^{\frac{\ln(1+\ln(x))}{x}}$
So now we just have to plug in $ e^{-1}$.  Unfortunately, that too is an indeterminate.  I believe  you need to do it again to get rid of the indeterminants but I'm completely sure.

[quote="JRav"]Since the limit is indeterminant, you can use L'Hopital's Rule.  The derivative of the denominator is 1, but the top is pretty messy.  I velieve it is
$ \frac {\frac {1}{1 + \ln(x)} - \ln(1 + \ln(x))}{x^{2}}e^{\frac {\ln(1 + \ln(x))}{x}}$
So now we just have to plug in $ e^{ - 1}$.  Unfortunately, that too is an indeterminate.  I believe  you need to do it again to get rid of the indeterminants but I'm completely sure.[/quote]

No, The numerator is not differentiable at $ e^{-1}+$

The numerator is approximately $ (1\plus{}\ln x)^e$, which is approximately a constant multiple of $ (x\minus{}\frac1e)^e$. Once you do that, the answer is obvious.

[quote="jmerry"]The numerator is approximately $ (1 \plus{} \ln x)^e$[/quote]

Well, I think you are using here Taylor Series, I haven't studied these things yet but it doesn't matter if you post such things, because Ill check them later...

but there's another way

[hide="Hint:"]Multiply by $ ln(x)\minus{}ln(e^{\minus{}1})$[/hide]

[quote="jmerry"]No, I'm not; I'm approximating the $ \frac1x$ power with a $ \frac1c$ power, where $ c$ is the limit of $ x$. Simplifying things that aren't indeterminate can be very useful.[/quote]

Do you mean $ e^x$ is approximatly $ (1\plus{}lnx)^{\epsilon}$, please detail...

[quote="Virgil Nicula"][quote="jmerry"]The numerator is approx. $ (1 \plus{} \ln x)^e$, which is approx. a constant multiple of $ (x \minus{} \frac1e)^e$.
Once you do that, the answer is obvious.[/quote]
Yes, the answer is evidently !

$ f(x)\equiv\frac {e^{\frac {ln(1 \plus{} ln(x))}{x}}}{x \minus{} e^{ \minus{} 1}}$ $ \implies f(x) \equal{} e\cdot\frac {\ln\ (ex)}{xe \minus{} 1}\cdot (1 \plus{} \ln x )^{\frac {1 \minus{} x}{x}}$ $ \implies$ $ \lim_{x\searrow\frac 1e}f(x) \equal{} 0\ .$[/quote]

I didn't understand this too  :(

[quote="FOURRIER"][quote="Virgil Nicula"]$ f(x)\equiv\frac {e^{\frac {ln(1 \plus{} ln(x))}{x}}}{x \minus{} e^{ \minus{} 1}}$ $ \implies f(x) \equal{} e\cdot\frac {\ln\ (ex)}{xe \minus{} 1}\cdot (1 \plus{} \ln x )^{\frac {1 \minus{} x}{x}}$ $ \implies$ $ \lim_{x\searrow\frac 1e}f(x) \equal{} 0\ .$[/quote]
I didn't understand this too  :([/quote]
$ 1\blacktriangleright\frac {e}{ex \minus{} 1} \equal{} \frac {1}{x \minus{} e^{ \minus{} 1}}$ and $ A \equal{} e^{\ln A}\implies$  $ \left\{\begin{array}{c}
\ln (ex) \equal{} 1 \plus{} \ln x \equal{} e^{\ln (1 \plus{} \ln x)} \\
 \\
(1 \plus{} \ln x )^{\frac {1 \minus{} x}{x}} \equal{} e^{\frac {1 \minus{} x}{x}\cdot \ln (1 \plus{} \ln x )} \end{array}\right\|$ $ \implies$

$ \ln (ex)\cdot (1 \plus{} \ln x )^{\frac {1 \minus{} x}{x}} \equal{}$ $ e^{\ln (1 \plus{} \ln x) \plus{} \frac {1 \minus{} x}{x}\cdot \ln (1 \plus{} \ln x )} \equal{}$ $ e^{\frac {\ln (1 \plus{} \ln x)}{x}}\ .$

$ 2\blacktriangleright\lim_{x\to 1}\frac {\ln x}{x \minus{} 1} \equal{} 1$ $ \implies$ $ \lim_{x\searrow\frac 1e}\frac {\ln ex}{ex \minus{} 1} \equal{} 1\ .$

$ \lim_{x\searrow\frac 1e}(1 \plus{} \ln x )^{\frac {1 \minus{} x}{x}} \equal{} ( \plus{} 0)^{e \minus{} 1} \equal{} 0\ .$

[quote="jmerry"]The numerator is approximately $ (1 \plus{} \ln x)^e$, which is approximately a constant multiple of $ (x \minus{} \frac1e)^e$. Once you do that, the answer is obvious.[/quote]

Could you explain that and justify approximating the power?


FTW
geometry
search
Support
HMMT
factorial
3D geometry

CHAPTER 1: THE HATCHING OF THE EGG

A long, long time ago, when the earth was still young, there was a forgotten land called Arthera.
It was conquered by 5 brothers, each very talented in their own unique way.
When they died, they became gods.  The island was still guarded by them but it was empty.  Everything was dying and it was just too boring.  The gods decided to make creatures on the island.  Together, they created an egg.  Long it lay,  there in the volcano, Quecan.  Years turned into decades, decades turn into centuries, and soon the egg was forgotten.  Suddenly, on the 18th of October on some forgotten year, it hatched!  But it didn't create 1 baby, it created thousands of babies!  They were split into 5 different groups by the gods.  The orcs, the elves, the goblins, the wizard/witches, and the monsters.  Soon, they were different clans.  Over the course of 3141 years, they started hating each other.  Each group believed that they were the best of the land.  The gods were deeply saddened by this discovery and had no choice but to leave the foolish creatures.  The land was no longer immune to danger.

You are the leaders of your clan.  I will give you guys the options once the game begins.  Good luck and have fun!

Orcs (Extremely strong, good melee, average arching, slow and stupid, bad attitude and temper.)
1.
2.
3.
4.
Elves (Kind, great archers, bad melee, really fast and average at knowledge, get along with everyone well.)
1.
2.
3.
4.
Goblins (Super strong melee, bad arching, slow and stupid, lots of people, kinda strong.)
1.
2.
3.
4.
Witches/Wizards (Great long-ranged attacks and really really powerful, slow, very smart, weak, and little people, really bad melee.)
1.
2.
3.
4.
Monsters:  Ghost(invisibility), Earthshaker(can control trees), Spedsnake(very very fast), Fistflood(Can do special move of making hundreds of fists launch at the enemy but it takes very long to reload.)
1.
2.
3.
4.

Chose your clans players!  Lets see who will be the strongest clan of all!

I'm in!

Orcs (Extremely strong, good melee, average arching, slow and stupid, bad attitude and temper.)
1.
2.
3.
4.
Elves (Kind, great archers, bad melee, really fast and average at knowledge, get along with everyone well.)
1. footballer
2.
3.
4.
Goblins (Super strong melee, bad arching, slow and stupid, lots of people, kinda strong.)
1.
2.
3.
4.
Witches/Wizards (Great long-ranged attacks and really really powerful, slow, very smart, weak, and little people, really bad melee.)
1.
2.
3.
4.
Monsters: Ghost(invisibility), Earthshaker(can control trees), Spedsnake(very very fast), Fistflood(Can do special move of making hundreds of fists launch at the enemy but it takes very long to reload.)
1.
2.
3.
4.

[hide]I'm in!

Orcs (Extremely strong, good melee, average arching, slow and stupid, bad attitude and temper.)
1.
2.
3.
4.
Elves (Kind, great archers, bad melee, really fast and average at knowledge, get along with everyone well.)
1. footballer
2. bluecarneal
3.
4.
Goblins (Super strong melee, bad arching, slow and stupid, lots of people, kinda strong.)
1.
2.
3.
4.
Witches/Wizards (Great long-ranged attacks and really really powerful, slow, very smart, weak, and little people, really bad melee.)
1.
2.
3.
4.
Monsters: Ghost(invisibility), Earthshaker(can control trees), Spedsnake(very very fast), Fistflood(Can do special move of making hundreds of fists launch at the enemy but it takes very long to reload.)
1.
2.
3.
4.[/hide]

Orcs (Extremely strong, good melee, average arching, slow and stupid, bad attitude and temper.)
1.
2.
3.
4.
Elves (Kind, great archers, bad melee, really fast and average at knowledge, get along with everyone well.)
1. footballer
2. bluecarneal
3. andrewjjiang97
4.
Goblins (Super strong melee, bad arching, slow and stupid, lots of people, kinda strong.)
1.
2.
3.
4.
Witches/Wizards (Great long-ranged attacks and really really powerful, slow, very smart, weak, and little people, really bad melee.)
1.
2.
3.
4.
Monsters: Ghost(invisibility), Earthshaker(can control trees), Spedsnake(very very fast), Fistflood(Can do special move of making hundreds of fists launch at the enemy but it takes very long to reload.)
1.
2.
3.
4.

Orcs (Extremely strong, good melee, average arching, slow and stupid, bad attitude and temper.)
1.
2.
3.
4.
Elves (Kind, great archers, bad melee, really fast and average at knowledge, get along with everyone well.)
1. footballer
2. bluecarneal
3. andrewjjiang97
4. ksun48
Goblins (Super strong melee, bad arching, slow and stupid, lots of people, kinda strong.)
1.
2.
3.
4.
Witches/Wizards (Great long-ranged attacks and really really powerful, slow, very smart, weak, and little people, really bad melee.)
1.
2.
3.
4.
Monsters: Ghost(invisibility), Earthshaker(can control trees), Spedsnake(very very fast), Fistflood(Can do special move of making hundreds of fists launch at the enemy but it takes very long to reload.)
1.
2.
3.
4.

hmmm..
People seem to like elves...
well, no one can choose elves anymore haha.

Speedsnake
Orcs (Extremely strong, good melee, average arching, slow and stupid, bad attitude and temper.)
1.
2.
3.
4.
Elves (Kind, great archers, bad melee, really fast and average at knowledge, get along with everyone well.)
1. footballer
2. bluecarneal
3. andrewjjiang97
4. ksun48
Goblins (Super strong melee, bad arching, slow and stupid, lots of people, kinda strong.)
1.
2.
3.
4.
Witches/Wizards (Great long-ranged attacks and really really powerful, slow, very smart, weak, and little people, really bad melee.)
1.
2.
3.
4.
Monsters: Ghost(invisibility), Earthshaker(can control trees), Spedsnake(very very fast), Fistflood(Can do special move of making hundreds of fists launch at the enemy but it takes very long to reload.)
1. Richard_Min (Speedsnake)
2.
3.
4.

Orcs (Extremely strong, good melee, average arching, slow and stupid, bad attitude and temper.)
1.
2.
3.
4.
Elves (Kind, great archers, bad melee, really fast and average at knowledge, get along with everyone well.)
1. footballer
2. bluecarneal
3. andrewjjiang97
4. ksun48
Goblins (Super strong melee, bad arching, slow and stupid, lots of people, kinda strong.)
1.
2.
3.
4.
Witches/Wizards (Great long-ranged attacks and really really powerful, slow, very smart, weak, and little people, really bad melee.)
1.
2.
3.
4.
Monsters: Ghost(invisibility), Earthshaker(can control trees), Spedsnake(very very fast), Fistflood(Can do special move of making hundreds of fists launch at the enemy but it takes very long to reload.)
1. Richard_Min (Speedsnake)
2. pi31415926 (Fistflood)
3.
4.

Orcs (Extremely strong, good melee, average arching, slow and stupid, bad attitude and temper.)
1.
2.
3.
4.
Elves (Kind, great archers, bad melee, really fast and average at knowledge, get along with everyone well.)
1. footballer
2. bluecarneal
3. andrewjjiang97
4. ksun48
Goblins (Super strong melee, bad arching, slow and stupid, lots of people, kinda strong.)
1.
2.
3.
4.
Witches/Wizards (Great long-ranged attacks and really really powerful, slow, very smart, weak, and little people, really bad melee.)
1.
2.
3.
4.
Monsters: Ghost(invisibility), Earthshaker(can control trees), Spedsnake(very very fast), Fistflood(Can do special move of making hundreds of fists launch at the enemy but it takes very long to reload.)
1. Richard_Min (Speedsnake)
2. pi31415926 (Fistflood)
3. spaceguy524 (Ghost)
4.

heh people are teaming up like crazy.
You know, you will end up teaming up with 3 other people anyways so whats the point ? :O

what do yo u think?
of course ill make it a little different but..

Orcs (Extremely strong, good melee, average arching, slow and stupid, bad attitude and temper.)
1.
2.
3.
4.
Elves (Kind, great archers, bad melee, really fast and average at knowledge, get along with everyone well.)
1. footballer
2. bluecarneal
3. andrewjjiang97
4. ksun48
Goblins (Super strong melee, bad arching, slow and stupid, lots of people, kinda strong.)
1.
2.
3.
4.
Witches/Wizards (Great long-ranged attacks and really really powerful, slow, very smart, weak, and little people, really bad melee.)
1. musicnmath
2.
3.
4.
Monsters: Ghost(invisibility), Earthshaker(can control trees), Spedsnake(very very fast), Fistflood(Can do special move of making hundreds of fists launch at the enemy but it takes very long to reload.)
1. Richard_Min (Speedsnake)
2. pi31415926 (Fistflood)
3. spaceguy524 (Ghost)
4. Meewgee009 (Fistflood)

Orcs (Extremely strong, good melee, average arching, slow and stupid, bad attitude and temper.)
1. [b]eqjj168[/b]
2.
3.
4.
Elves (Kind, great archers, bad melee, really fast and average at knowledge, get along with everyone well.)
1. footballer
2. bluecarneal
3. andrewjjiang97
4. ksun48
Goblins (Super strong melee, bad arching, slow and stupid, lots of people, kinda strong.)
1.
2.
3.
4.
Witches/Wizards (Great long-ranged attacks and really really powerful, slow, very smart, weak, and little people, really bad melee.)
1. musicnmath
2.
3.
4.
Monsters: Ghost(invisibility), Earthshaker(can control trees), Spedsnake(very very fast), Fistflood(Can do special move of making hundreds of fists launch at the enemy but it takes very long to reload.)
1. Richard_Min (Speedsnake)
2. pi31415926 (Fistflood)
3. spaceguy524 (Ghost)
4. Meewgee009 (Fistflood)

nice ending and great game!!!

 :clap:  :clap2:  :10:  :thumbup:

loved it 10!!!!!!!!!!!!!!!!!!
if you are playing my game sorrry for delay.

Thank you all!
I (sadly) and not making a sequel to this game.  If it were as successful as this one, it would take a while :D.
You guys could make one.
I really want to thank all of you for your support to help this game run so smoothly!
I'm probably not making another game in a long time.

If I could rate your performance in the game, it's a 10!

:D :D

I wasn't included...

oh well

I rated it 6 stars

from 1-10, rate it 20

gratz everyone.  Yes this is abcak not pie.

How high did your post rating increase now, piegraph?

So what everyone died?
You guys succeeded your goal :P
But only andrew survived...

[quote="piegraph"]But only andrew survived...[/quote]

W00t!

Awesome

I was in almost all of the stories

Hooray


quadratics
vector
superior algebra
superior algebra unsolved

Hello,

I have been struggling with this problem for quite a while.

Let $K=GF(q^{2})$ and $F$ be the unique subfield of order $q$, where $q$ is an odd prime power.

Consider this (nonsingular) quadratic form : $f=X_{0}^{2}+\cdots+X_{2 n}^{2}=0$ over $K$.

It can be verified that the number of vectors in this $2 n+1$-dimensional space over $K$ satisfying $f=0$ is $q^{4 n}$.

Now consider : $g=a f+a^{q}f^{q}$, with $a$ some element of $K$, different from zero.
It can be checked quite easily that this will be a nonsingular quadratic form in a $4 n+2$ dimensional space over $F$.

But of what type? It can be "hyperbolic" or "elliptic", meaning : it can have $q^{ 4 n+1}+q^{2 n+1}-q^{2 n}$ or $q^{ 4 n+1}-q^{2 n+1}+q^{2 n}$  vectors satisfying  $g=0$.

I don't have a lot of experience with this sort of manipulation of fields. :huh:
Obviously the choice of $a$ will be crucial.

Thanks for every thought you can give,
fredbel6

I have been messing around and I think the following thinks might be relevant :  :maybe:

whether or not $q\equiv 1 \bmod{4}$
whether or not $a$ is square in the bigger field $K$
whether or not $n$ is odd


Hi.

Im watching a solution of a test of how to know if 17 divides a number n.

$ abcde \equal{} 10 * [\frac {abcde}{10}] \plus{} e \\
\equal{} 10 * [\frac {abcde}{10}] \minus{} 109e \plus{} 110e \\
\equal{} 10 * [\frac {abcde}{10} \plus{} 11e] \minus{} 109e$

Last it says 109e divides 17 (= 7e) .

Can someone explain what was done here ?

Thanks.

Maybe 17|119...
Suppose we formally write $ n\equal{}d_0d_1...d_m$, with $ 0\leq d_i \leq 9$ for $ n\equal{}\sum_{i\equal{}0}^md_{m\minus{}i} 10^i$. Then, 17 divides $ n$ if and only if 17 divides $ d_0d_1...d_{m\minus{}1}\plus{}12d_m$.


If I am able to solve pretty much all the problems in Aops vol 1 then approximately what scores do you think I can get on the AMC 12 and the AIME(if I make make it).

It varies, but I'll definitely bet that you could get a 105+ on AMC12 and at least a 5 on the AIME.

Yknow, you can just try some though to see where you are in the resources section.


group theory
abstract algebra
superior algebra
superior algebra unsolved

How can we determine all groups of order 12 using group presentations, without using semidirect products.

It just seems to split into so many cases, and I have trouble identifying the presentations with 3 generators. Is there a smart shortcut to reduce the amount of cases?

Smart would be to use semidirect products. :wink:

There are only 5 possible groups of order 12
$ C_4\times C_3\cong C_{12}$
$ C_2\times C_2\times C_3\cong C_2\times C_6$
$ C_4\ltimes C_3\cong \langle a,b : a^4 \equal{} b^3 \equal{} 1, a^{ \minus{} 1}ba \equal{} b^{ \minus{} 1}\rangle$
$ (C_2\times C_2)\ltimes C_3\cong C_2\times D_6$
$ (C_2\times C_2)\rtimes C_3\cong A_4$

I know, but this is an old exam problem (I have exam next week), and we don't know about semidirect products.

That doesn't mean you can't use it, at least indirect.
Giving a group presentation is only the final step after you know the structure of the group.
(group presentations aren't very useful anyway)

Ok, let me ask for more concrete advice.

Suppose there is only one Sylow 4-subgroup so that it's normal, and suppose it is the Klein 4-group, with generators $ a,b$. Then we take an element $ c$ of order 3. Now, when I conjugate $ a$ and $ b$ with $ c$ loads of things can happen. How can I determine which things are possible?

I mean, I suppose I could do it brute force, but then I have to do it for a cyclic Sylow 4-subgroup (ok, this is easy). Then I have to repeat the procedure with a normal Sylow 3-subgroup, which together with a Sylow 4-subgroup generate the group, and then I have many cases to check again.

Conjugation with $ c$ permutes $ \langle a,b\rangle\setminus \{1\}\equal{}\{a,b,ab\}$ and has order $ 3$.
So it must be a 3-cycle and after renaming the generators of $ \langle a,b\rangle$ we get $ c^{\minus{}1}ac\equal{}b$ and $ c^{\minus{}1}bc\equal{}ab$ and $ c^{\minus{}1}abc\equal{}a$.
This gives only one non-abelian group of order 12 with normal Sylow 2-subgroup, i.e. $ A_4$.

I don't see your problem. There are $ 5$ different groups, so brute casework is necessary.
But all cases are pretty trivial. And it should be allowed to skip some details.

My problem is that I am not good at this. Your help has taught me several things I did not think of.  Very good idea to view conjugation as a permutation of order 3!

Apart from my conjugation issues, what I was doing was basically to write presentations for $ A_4$ and $ C_2 \times D_6$ without recognizing them, and trying to show that the presentations had order $ 12$ without referring to these concrete realizations. This makes it a lot of work. To establish that  $ \langle a,b : a^4 \equal{} b^3 \equal{} 1, a^{ \minus{} 1}ba \equal{} b^{ \minus{} 1}\rangle$ really has order 12, my best idea is to realize it by finding suitable permutations in $ S_{12}$ ( a = (1,2,6,3)(4,8,11,10)(5,7,12,9) and b = (1,4,5)(2,7,8)(3,9,10)(6,11,12) work, but it's a bitch to find them). But perhaps this is also a known group like $ A_4$ or $ C_2 \times D_6$ that is easy to directly establish the isomorphism with?

If I want to try to realize it in $ GL_n(\mathbb{Z})$ instead, what $ n$ should I use?

Let $ \zeta_3\equal{}e^{2\pi i/3}\equal{}{\minus{}1\plus{}\sqrt{3}i\over 2}\in \mathbb{C}^\times$ of order $ 3$.
Then $ \langle \zeta_3,j \rangle\le \mathbb{H}^\times$ would be a realization.
Or the subgroup of $ C_4\times S_3$ of index $ 2$ given by $ \langle x(12),(123)\rangle$ for $ C_4\equal{}\langle x\rangle$.


Let $ABCD$ be a paralelogram. Let $O$ be a point inside $ABCD$ such that $\angle AOB + \angle COD = 180$. Prove that $\angle OBC = \angle ODC$.

[hide]I have a solution using Translations, i would like to see a different solution. [/hide]

[quote="jasamper88"][quote="shobber"]When you say translation, do you mean $T(\overrightarrow{AD})$?[/quote]

Yes, and tranlating $\triangle AOB$. So you have a cyclic quadrilateral.[/quote]

i think the most nice solution is doing this fact.


Let I denote the interval [-pi/4,pi/4]. Determine the function f defined on the interval [-1.1] such that f(sin2x) = sinx +cosx and simplify f(tan^2(x)) for x in the interval I

Hey buddy

here is your solution

f(sin2x)={(sinx+cosx)^2}^(1/2)=(1+sin2x)^1/2

therefore the function is f(x)=(1+x)^1/2
similarly f{tan^(2)x}={1+tan^(2)x}^1/2=secx :|


anybody would help me to post some others problems similar this problem

I'm just going to put this into a format that's easier to read:

[quote="sigma"]Let $\left \{b_n \right \}_{n \ge 0}$ be a sequence of positive integers given by $b_0=1$, $b_1=4$, and $b_n=4b_{n-1}+b_{n-2}$ for integers $n \ge 2$.  Determine the maximum of $m \in \mathbb {R}^+$ such that for all nonnegative integers $y<b_{n+1}$, there is $\left |x-\sqrt{5}y\right | \ge \frac{m}{b_n}$ for every positive integer x.[/quote]

Now only if sigma could delete the picture to make it readable in one screen (wink, wink, nudge nudge)...

On a more relevant note, I actually have no idea what the problem is asking, let alone how to solve it.


Prove that $ x^n\plus{}y^n\equal{}(x\plus{}y)(x^{n\minus{}1}\minus{}x^{n\minus{}2}y\plus{}...\minus{}xy^{n\minus{}2} \plus{}y^{n\minus{}1}$) holds for all odd integers n.

Also, is there a similar formula for even integers n?

Proof: expand.  :wink:

No, there is not a similar factorization for even $ n$. However, if $ n\plus{}1$ is composite, then you can clearly factor out $ n^{d\minus{}1} \plus{} n^{d\minus{}2} \plus{} \cdots \plus{} 1$, where $ d$ divides $ n\plus{}1$.

Ahh, thank you. I was trying to find a way using induction :P

I don't understand your second part though. How do you factor $ n^{d\minus{}1} \plus{} n^{d\minus{}2} \plus{} \cdots \plus{} 1$ from $ x^n \plus{} y^n$ ?

OMG I AM SO STUPID.  :oops_sign:

So sorry. Nevermind about that.

[size=59]I was thinking of something else.[/size]

[b]Inductive Step:[/b] We assume $ x^n \plus{} y^n \equal{} (x \plus{} y)(x^{n \minus{} 1} \minus{} x^{n \minus{} 2}y \plus{} ... \minus{} xy^{n \minus{} 2} \plus{} y^{n \minus{} 1})$

Multiply by $ y^2$ to get $ x^n y^2 \plus{} y^{n \plus{} 2} \equal{} (x \plus{} y)(x^{n \minus{} 1}y^2 \minus{} x^{n \minus{} 2}y^3 \plus{} ... \minus{} xy^{n} \plus{} y^{n \plus{} 1})$

What to add in: $ \underbrace{x^{n \plus{} 2} \minus{} x^n y^2}_{\text{add this to LHS}} \equal{} x^n (x^2 \minus{} y^2) \equal{} (x \plus{} y)(x \minus{} y)x^n \equal{} \underbrace{(x \plus{} y) (x^{n \plus{} 1} \minus{} x^n y)}_{\text{add this to RHS}}$

Cancel on left, factor out the common $ (x \plus{} y)$ on right to get:

$ x^{n \plus{} 2} \plus{} y^{n \plus{} 2} \equal{} (x \plus{} y)(x^{n \plus{} 1} \minus{} x^{n} y \plus{} ... \minus{} xy^{n} \plus{} y^{n \plus{} 1})$

If n is not a power of 2 (if it is, there is no factorization)
then rewrite n as $ k \cdot l$ where l is odd and k is a power of 2.
then the factorization is $ \left(x^k \plus{} y^k\right)\left(x^{k(l \minus{} 1)} \minus{} x^{k(l \minus{} 2)}y^{k} \plus{} x^{k(l \minus{} 3)}y^{k(2)} \minus{} x^{k(l \minus{} 4)}y^{k(3)} \ldots \minus{} x^{k(3)}y^{k(l \minus{} 4)} \plus{} x^{k(2)}y^{k(l \minus{} 3)} \minus{} x^{k}y^{k(l \minus{} 2)} \plus{} y^{k(l \minus{} 1)}\right)$
and proof is similar to TZF/leoxnlin's


What's the unit vector that parallel to vector V=(2,-3,6)?  I totally confused :oops:

Parallel vectors are constant multiples of each other. How long is the vector you started with, and how much do you have to scale it to reach length 1?


ALGEBRA data la examenul de grad II Bucuresti

2005:
Fie polinomul f = X4 +X3+X2+X+1
a)	Descompuneti polinomul f in factori ireductibili in R[X] si in C[X]
b)	Aratati ca f este polinom ireductibil in Q[X]
c)	Daca α Є C este o radacina a lui f, notam F α ={ a α3+b α2+c α+d  ; a;b;c;d Є Q} Aratati ca F α   este un subcorp al lui C si un Q-spatiu vectorial de dimensiune 4
d)	 Determinati inversul numarului x = α3 + α in F α
2004:
1)	Sa se arate ca daca f Є Z[X] are trei radacini intregi care au resturi distincte la impartirea cu 3, atunci pentru orice n din Z avem 3 | f (n)
2)	Fie K un corp comutativ, K[X] un K-spatiu vectorial cu operatiile uzuale si Vn = { f Є K[Z] , grad f ≤ n , n Є N }. Sa se arate ca :
a)	Daca f1 , f2 ,…, fm Є K[X] sunt polinoame nenule astfel incat grad fi ≠ grad fj pentru orice i≠j atunci f1 , f2 ,…, fm sunt liniar independente peste K
b)	 Vn este un subspatiu vectorial al lui K[X] de dimensiune n+1
c)	 B = { 1; X-a; ….., (X-a)n } este o baza a lui Vn , unde aЄ K

Pentru 2005:

(a) Fie $\omega = \exp \left( \frac{2 \pi i}5 \right)$. Avem $f = \left( X-\omega \right) \left( X-\overline{\omega}\right) \left( X-\omega^{2}\right) \left( X-\overline{\omega}^{2}\right)$. Pentru descompunerea in $\mathbb R[X]$, grupam primii doi termeni si ultimii doi.

(b) Criteriul lui Eisenstein pentru $p=5$ si $f(X+1)$.

(c) Prima parte: Multe calcule. E importanta observatia $\alpha^{4}=-\alpha^{3}-\alpha^{2}-\alpha-1$.

A doua parte: Fie $p,q,r,s$ astfel incat $p \alpha^{3}+q \alpha^{2}+r \alpha+s = 0$, i.e. \[(p-1) \alpha^{3}+(q-1) \alpha^{2}+(r-1) \alpha+(s-1) =-\left( \alpha^{3}+\alpha^{2}+\alpha+1 \right) = \alpha^{4}.\] Rezulta ca $\alpha$ e radacina a polinomului $g \in \mathbb Q[X]$, unde $g = (1-s)+(1-r) X+(1-q) X^{2}+(1-p) X^{3}+X^{4}$. Dar $f$ este polinom ireductibil, deci este polinomul minimal al lui $\alpha$. De aici deducem ca $1-s = 1-r = 1-q = 1-p = 1$, de unde $p=q=r=s=0$, ceea ce voiam.

(d) $\left( \alpha^{3}+\alpha \right) \left(-\alpha-1 \right) = 1$.

Pentru 200[b]4[/b]:

1. Fie $\alpha_{i}\equiv i \left( \bmod \, 3 \right)$ trei dintre radacinile lui $f$. Daca $\alpha \in \mathbb Z$, atunci $\alpha-\alpha_{i}| f \left( \alpha \right)-f \left( \alpha_{i}\right)$, unde $\alpha \equiv \alpha_{i}\left( \bmod \, 3 \right)$, deci $3 | f \left( \alpha \right)$.

2. (a) Presupunem ca $\deg f_{1}> \deg f_{2}> \ldots > \deg f_{m}$, dupa care e evident.

(b,c) Fie $f_{0}, f_{1}, \ldots, f_{n}$ polinoame din $K[X]$ astfel incat $\deg f_{i}= i$. Demonstram prin inductie (finita) dupa gradul lui $g$ ca exista $a_{0},\ldots,a_{n}$ astfel incat $g = \sum a_{i}f_{i}$ (pentru pasul $\left( 0,1,2,\ldots,k \right) \mapsto k+1$ folosim $g = \gamma f_{k+1}+h$, unde $\deg h \leq k$).

2004

[hide="solutie 1"]fie $k,p,q$ cele trei radacini.
atunci $(x-k)(x-p)(x-q)|f$.
deci $(n-k)(n-p)(n-q)|f(n)$ pentru orice $n\in\mathbb{Z}$. cum $\left\{k,p,q\right\}\equiv\left\{0,1,2\right\}(\bmod\ 3)$, rezulta ca una din parantezele din stanga este divizibila cu $3$. deci $3|f(n)$.[/hide]

[hide="solutie 2"]a)
presupunem prin absurd ca exista scalarii $a_{i}\in K$ astfel incat $\sum a_{i}f_{i}=0$. (*)
fie $F$ polinomul de grad maxim dintre $f_{i}$. acesta exista si este unic (deoarece gradele sunt distincte doua cate doua). sa notam cu $\alpha$ coeficientul sau dominant si cu $s$ gradul sau. dupa ce exprimam toate $f_{i}$`urile in functie de baza canonica a lui $K[X]$, egalitatea (*) devine :
$\alpha X^{s}+\sum_{k=0}^{s-1}\beta_{k}X^{k}=0$ (**) , unde $\beta_{k}$ sunt scalari. cum $1,X,X^{2},\ldots$ este baza canonica a lui $K[X]$, rezulta ca $1,X,X^{2},\ldots,X^{s}$ sunt liniar independenti, deci egalitatea (**) poate avea loc decat daca toti scalarii sunt nuli, si deci $\alpha=0$, contradictie cu alegerea lui $\alpha$.
b)
$V_{n}\subset K[X]$ si este generat de $1,X,X^{2},\ldots,X^{n}$. (trecand peste toate detaliile)
c)
elementele lui $B$ sunt vectori din $V_{n}$.
elementele lui $B$ sunt liniar independente : daca notam $f_{k}=(X-a)^{k}$, observam ca $\deg f_{i}\neq \deg f_{j}$ $\forall\ i\neq j$. deci conform a), $f_{i}$`urile sunt liniar independente.
fie $W$ subspatiul lui $V_{n}$ generat de $B$. atunci $\dim W=n+1$, deci $W=V_{n}$, adica $B$ este baza a lui $V_{n}$.[/hide]

edit : iar mi`a luat`o inainte perfect_radio :)


superior algebra
superior algebra unsolved

Prove that $ x^4\plus{}1$ is reducible over $ Z_{p}$ ,where $ p$ is a prime number.

Show that it is reducible if one of $ \minus{}1,2,\minus{}2$ is a square; conclude.


[quote="maggot"]Let $M,N$ be points in triangle $ABC$ such that

$\angle BAM=\angle CAN, \angle ABM=\angle CBN$

Show that $\angle BCM=\angle ACN$.[/quote]

Anybody?


2 circles lie in a plane. The smaller circle has a radius of 7 and the distance between their centers is 14. The area of the intersection of the circles is $ \frac{245\pi}{6} - 49\sqrt{3} $
What is the radius of the larger circle?

I don't think you needed to know the area of intersection.
[hide]Draw the radii of the smaller circle that are tangent to the larger circle. Draw the radii of the larger circle that hit those points of tangencies. Two right triangles are formed that share a hypotenuse of 14 and have legs of 7. $\sqrt {196 - 49} = \sqrt {147}$.[/hide]

[quote]Draw the radii of the larger circle that hit those points of tangencies. Two right triangles are formed that share a hypotenuse of 14 and have legs of 7[/quote]
can you prove that the triangles are right angled?

When a radius hits the tangent line then the angle formed is always ninety degrees.

you are right, i feel so stupid  :blush:

But a radius of the smaller circle is not neccesarily tangent to the the larger circle.

well there has to be one that is
and plus, the answer $7\sqrt{3}$ that he gave also gives the correct area :)

The formula for the interesection of circles with radii $p$ and $q$ with distance between the centers $d$ is:

set $\displaystyle \frac{p+q+d}{2}$ to be $s$

set $\displaystyle \frac{\sqrt{s(s-p)(s-q)(s-d)}}{d}$ to be $m$

set ${2}\sin^{-1}{(m/p)}$ to be $\alpha$

set ${2}\sin^{-1}{(m/q)}$ to be $\beta$

Then the area is $\displaystyle \frac{(\alpha{p}^2+\beta{q}^2)\pi}{360}-\displaystyle \frac{p^2\sin{\alpha}+q^2\sin{\beta}}{2}$

Tarquin, how would u know that the tangents from the center of the smaller circle to the the larger circle are even radii?  You would need to know the area of the intersection in order to know that.  Think about it: if you aren't given the area of intersection, the radius of the larger circle could be anything.

Yeah you're right.
Since the area of intersection is pretty small compared to the smaller circle's area, then the tangents probably are radii...


A travel agency offers an organization an all-inclusive tour for Y8000 per person if not more than 100 persons join the tour. However, the cost per person will be reduced by Y50 for each person in excess of 100. Find the largest possible gross sales revenue for the travel agency. How many persons should join the tour and what is the cost per person so that this maximum revenue is achieved?

[hide]Let $ x$ be the number of people in excess of 100.

Then the revenue is $ (100 \plus{} x)\cdot(8000 \minus{} 50x) \equal{} 800000 \plus{} 3000x \minus{} 50x^2$.

We set the derivative to zero: $ \minus{} 100x \plus{} 3000 \equal{} 0 \Rightarrow x \equal{} 30$.

So the number of people is 130.
The revenue is $ 130 \cdot 6500 \equal{} 845000$.[/hide]

[hide="Non-Calculus"]$ ax^2\plus{}bx\plus{}c$ takes on a minimum/maximum value at $ x\equal{}\minus{}\frac{b}{2a}$.[/hide]

yeah...i used $ \frac{\minus{}b}{2a}$ because calculus is so operational, no offense, i only use it when i'm in a hurry..hehe thanks to you guys anyways:)


Prove or disprove
Let $ A,B,P\in C^{n\times n}$ and $ rankP\equal{}r$, where $ 0<r<n$. If $ PA\equal{}BP$, then $ A$ and $ B$ have $ r$ eigenvalues the same.

[quote="miwalin"]
Let $ A,B,P\in C^{n\times n}$ and $ rankP \equal{} r$, where $ 0 < r < n$. If $ PA \equal{} BP$, then $ A$ and $ B$ have $ r$ eigenvalues the same.[/quote]
 It is true !  :wink:
  $ rank(P) \equal{} r$ then the exist $ R,S \in GL_n{\mathbb{(C)}}$ such that : $ P \equal{} R\left[\begin{array}{cc}I_r & O \\
O & O\end{array}\right]S$ ,$ I_r$ is $ r$x$ r$ identity matrix .then we have :
 $ R\left[\begin{array}{cc}I_r & O \\
O & O\end{array}\right]SA \equal{} BR\left[\begin{array}{cc}I_r & O \\
O & O\end{array}\right]S$
 $ \left[\begin{array}{cc}I_r & O \\
O & O\end{array}\right]SAS^{ \minus{} 1} \equal{} R^{ \minus{} 1}BR\left[\begin{array}{cc}I_r & O \\
O & O\end{array}\right]$
Put  $ SAS^{ \minus{} 1} \equal{} \left[\begin{array}{cc}U_1 & U_2 \\
U_3 & U_4\end{array}\right]$,$ R^{ \minus{} 1}BR \equal{} \left[\begin{array}{cc}V_1 & V_2 \\
V_3 & V_4\end{array}\right]$
 $ U_1,V_1 \in M_r{\mathbb(C)}$ ,$ U_4,V_4 \in M_{n \minus{} r}{\mathbb{(C)}}$,$ U_2,V_2 \in M_{rx(n \minus{} r)}{\mathbb{(C)}}$,$ U_3,V_3 \in M_{(n \minus{} r)xr}{\mathbb{(C)}}$
Then we have :
$ \left[\begin{array}{cc}I_r & O \\
O & O\end{array}\right]\left[\begin{array}{cc}U_1 & U_2 \\
U_3 & U_4\end{array}\right] \equal{} \left[\begin{array}{cc}V_1 & V_2 \\
V_3 & V_4\end{array}\right]\left[\begin{array}{cc}I_r & O \\
O & O\end{array}\right]$
$ \left[\begin{array}{cc}U_1 & U_2 \\
O & O\end{array}\right] \equal{} \left[\begin{array}{cc}V_1 & O \\
V_3 & O\end{array}\right]$
 Hence we have $ U_1 \equal{} V_1,U_2 \equal{} O,V_3 \equal{} O$ as :
$ SAS^{ \minus{} 1} \equal{} \left[\begin{array}{cc}V_1 & O \\
U_3 & U_4\end{array}\right]$,$ R^{ \minus{} 1}BR \equal{} \left[\begin{array}{cc}V_1 & V_2 \\
O & V_4\end{array}\right]$
 From $ \det(A \minus{} \lambda I_n) \equal{} \det(SAS^{ \minus{} 1} \minus{} \lambda I_n) \equal{} \det(V_1 \minus{} \lambda I_r)\det(U_4 \minus{} \lambda I_{n \minus{} r}) \equal{} O$
          $ \det(B \minus{} \lambda I_n) \equal{} \det(R^{ \minus{} 1}BR \minus{} \lambda I_n) \equal{} \det(V_1 \minus{} \lambda I_r)\det(V_4 \minus{} \lambda I_{n \minus{} r}) \equal{} O$
then $ A$ and $ B$ have $ r$ eigenvalues the same.


So, do you have to formally register in order to participate?

Yes.  See http://www.usamts.org for details.


Χαιρετώ όλους τους Έλληνες mathlinkers .

Πολύ κίνηση βλέπω τελευταία στο φόρουμ .Νέα μέλη και δραστήρια . Πολύ καλό αυτό .
Η πρόταση του nickolas προς την ΕΜΕ (για ξένο trainer) είναι πολύ καλή αλλά δεν πιστεύω ότι θα γίνει δεκτή .

Προτείνω και μία άσκηση Γεωμετρίας :

"Έστω κυρτό πεντάγωνο ABCDE ,τέτοιο ώστε Β = Ε = 90 και BAC = EAD . Aν Ο είναι το σημείο τομής των διαγωνίων του BD και CE , να δείξετε ότι οι ευθείες ΑΟ και ΒΕ είναι κάθετες ."

Νίκο γεια και πάλι . Λοιπόν χθες που μίλησα με το Γιώργο το Δέσκο μου είπε ότι την έλυσε πολύ εύκολα με διανύσματα .Αλλά δεν το κοίταξα ακόμα. Κοίτα το αν θέλεις.
Από ποια χρονιά είναι η άσκηση ????

Σιλουανέ , στο βιβλίο (Γεωμετρία Β Λυκείου - Γ. Μπαιλάκης ) που την είδα δεν ανέφερε χρονιά .

H άσκηση παραμένει άλυτη και στο αγγλικό forum πάνω απο 10 ημέρες !!
  Πρέπει να είναι αρκετά δύσκολη .

Για να είναι από τη Ρωσσία....... :lol:
Δοκίμασες καθόλου με γεωμετρία διανυσματικη και εσωτερικό γινόμενο.?

Ποια άσκηση στο αγγλικό φόρουμ είναι;

  Μπάμπης.

Ας δούμε μία προσέγγιση με κλασσική γεωμετρία ( στο σχήμα μου είναι, AC = 7.5, AD = 7.8,
CD = 6.3, AB = 4.5 ).

Τα ορθογώνια τρίγωνα ABC ( <ABC = 90º ) και AED ( <AED = 90º ), είναι όμοια και άρα έχουμε
<ACB = <ADE, (1).

Δια των κορυφών C, D, του τριγώνου ACD, φέρνουμε τις ισογώνιες ευθείες των CB, DE, ως προς τις γωνίες <ACD, <ADC, αντιστοίχως και έστω Α' το σημείο τομής των. Λόγω της (1), έχουμε ότι <CDA' = <DCA' και άρα το τρίγωνο A'CD είναι ισοσκελές ( A'C = A'D ).

Στο τρίγωνο ACD, οι πλευρές των τριγώνων A'CD, BAC, EAD, που συντρέχουν στις κορυφές A, C, D, είναι ισογώνιες ως προς τις αντίστοιχες γωνίες. Επομένως σύμφωνα με το θεώρημα του Jacobi, οι ευθείες AA', BD, CE, τέμνονται στο ίδιο σημείο και άρα η ευθεία ΑΑ', διέρχεται δια του σημείου Ο, τομής των BD, CE.

Έστω F, G, τα μέσα αντιστοίχως των AC, AD και B', E', τα σημεία τομής των μεσοκαθέτων των AC, AD, από τις ευθείες AB, AE. Αποδεικνύεται εύκολα ότι B'E' // BE και επομένως αρκεί να αποδείξουμε ότι η ευθεία AA' ≡ ΟΑ, είναι κάθετη επί την ευθεία B'E'.

Έστω K, L, τα σημεία τομής της ευθείας FG, από τις ευθείες A'C, A'D, αντιστοίχως.

Από CD // KL, => <KLA' = <CDA' = <DCA' = <LKA', (2) και άρα το τρίγωνο A'KL είναι ισοσκελές (A'K = A'L ). Εάν K', L', είναι αντιστοίχως τα μέσα των A'C, A'D, αποδεικνύεται εύκολα ότι KK' = LL', (3), ( επειδή είναι A'C, A'D ).

Από την καθετότητα των ευθειών AB, BC και AC, B'F, συμπεραίνουμε ότι <CB'F = <BB'F = <ACB, (4) (BB'CF εγγράψιμο τετράπλευρο).

Αλλά <ACB = <DCA', (5)  ως ισογώνιες ως προς τη γωνία <ACD.

Από (2), (4), (5), => <CB'F = <LKA', => <CB'F = <CKF, (6).

Από (6), συμπεραίνουμε ότι το τετράπλευρο B'KCF είναι εγγράψιμο και άρα έχουμε <B'KC = <B'FC = 90º. Δηλαδή η ευθεία B'K, είναι κάθετη επί την ευθεία A'K. Ομοίως αποδεικνύεται ότι και η ευθεία E'L, είναι κάθετη επί την ευθεία A'L.

Στο τρίγωνο A'B'C σύμφωνα με το δεύτερο θεώρημα των διαμέσων,
έχουμε ότι (A'B')² - (CB')² = 2(A'C)·(KK'), (7).

Ομοίως από τρίγωνο A'E'D, => (A'E')² - (DE')² = 2(A'D)·(LL'), (8).

Από (7), (8), =>  (A'B')² - (AE')² =  (CB')² - (DE')², (9) ( επειδή A'C = A'D και KK' = LL' ).

Από (9), =>  (A'B')² - (AE')² =  (ΑB')² - (ΑE')², (10) ( επειδή CB' = AB' και DE' = AE' ).

Από (10), συμπεραίνουμε ότι η ευθεία ΑΑ' είναι κάθετη επί την ευθεία Β'Ε' και η πρόταση έχει αποδειχθεί.

Από τα παραπάνω είναι φανερό ότι η πρόταση αληθεύει και όταν το δοσμένο πεντάγωνο, είναι μη κυρτό.

Κώστας Βήττας.

ΥΓ – Περιμένω ανυπόμονα τη δημοσίευση και άλλων ( απλούστερων ) αποδείξεων. Αυτήν εδώ, πρωτοεμφανιζόμενη νομίζω στην Ελληνική βιβλιογραφία, την αφιερώνω σε ένδειξη τιμής, σε όλα τα μέλη του Ελληνικού mathlinks Forum.

:coolspeak: Κωστα ωραια λυση


For any $ 4$ points in space $ A,B,C,D$ , Prove that $ AC^2 \plus{} BD ^2 \plus{} AD^2 \plus{} BC^2  \ge  AB^2 \plus{} CD^2$.

Let $ \overrightarrow{AB} \equal{} \overrightarrow{b},\ \overrightarrow{AC} \equal{} \overrightarrow{c},\ \overrightarrow{AD} \equal{} \overrightarrow{d},$

$ AC^2 \plus{} BD ^2 \plus{} AD^2 \plus{} BC^2 \equal{} |\overrightarrow{c}|^2 \plus{} |\overrightarrow{d} \minus{} \overrightarrow{b}|^2 \plus{} |\overrightarrow{d} |^2 \plus{} |\overrightarrow{c} \minus{} \overrightarrow{b}|^2$

$ \equal{} 2(|\overrightarrow{b}|^2 \plus{} |\overrightarrow{c}|^2 \plus{} |\overrightarrow{d} |^2) \minus{} 2\overrightarrow{b}\cdot (\overrightarrow{c} \plus{} \overrightarrow{d})$

$ AB^2 \plus{} CD^2 \equal{} |\overrightarrow{b}|^2 \plus{} |\overrightarrow{c}|^2 \plus{} |\overrightarrow{d} |^2 \minus{} 2\overrightarrow{c}\cdot \overrightarrow{d}$

$ \therefore AC^2 \plus{} BD ^2 \plus{} AD^2 \plus{} BC^2 \minus{} (AB^2 \plus{} CD^2)$

$ \equal{} |\overrightarrow{c}|^2 \minus{} 2(\overrightarrow{b} \minus{} \overrightarrow{d})\cdot \overrightarrow{c} \plus{} |\overrightarrow{b} \minus{} \overrightarrow{d} |^2$

$ \equal{} |\overrightarrow{b} \minus{} \overrightarrow{c} \minus{} \overrightarrow{d}|^2\geq 0\ Q.E.D.$

If $ M,N$ are the midpoints of $ AB$ and $ DC,$ according to a [url=http://mathworld.wolfram.com/Quadrilateral.html]Casey's theorem[/url] we have

$AC^2 \plus{} BD^2 \plus{} BC^2 \plus{} AD^2 \equal{} AB^2 \plus{} CD^2 \plus{} 4MN^2$

The inequality obviously follows from $MN \ge 0$

luisgeometria: Shouldn't it be $ M,N$ midpoints of $ AC,BD$?
In the website, it is given that the points are the midpoints of the diagnols of the polygon.?

Yes, MN is the segment connecting the midpoints of the diagonals, but indeed the relation holds for both complex and simple quadrilaterals. We could state that AB,DC are diagonals of the complex quadrilateral  ADBC or AC,BD are diagonals of the convex/concave quadrilateral ABCD.


geometry
geometric transformation
homothety
angle bisector
geometry proposed

Dear Mathlinkers,
reading a paper on Forum Geometricorum,
see: http://forumgeom.fau.edu/FG2009volume9/FG200915.pdf
I propose to find another proof of theorem 1 by considering the orthocenter...
Sincerely
Jean-Louis

Another proof: We continue with the notations from the article. Let $ M_{b}$, $ M_{c}$ be the midpoints of sides $ CA$, $ AB$, respectively, and let $ \Gamma$ be the circle centered at $ E_{a}$ passing through $ M_{a}$. Obviously, the nine-point center $ M_{a}M_{b}M_{c}$ is tangent to $ \Gamma$ at $ M_{a}$. Now, consider the homothety $ \mathcal{H}$ mapping $ \Gamma$ into the nine-point circle. Let $ Q$ be the other intersection of the $ A$-altitude with $ \Gamma$. The image $ P'Q'$ of $ PQ$ is the diameter of the nine-point circle, which is perpendicular to $ BC \| M_{b}M_{c}$. Thus, $ P$ is the midpoint of the arc $ M_{b}M_{c}$, i.e. $ M_{a}P$ is the $ M$-internal angle bisector of the medial triangle. Since $ M_{a}M_{b}M_{c}$ and $ ABC$ are clearly homothetic, we therefore have that $ M_{a}P \| AT_{a}$.

Dear Cosmin,
my idea is based on this nice result:
the internal bissector of <BHC where H is the orthocenter of ABC, is parallel to the internal bissector of <BAC...
Sincerely
Jean-Louis


Let $p>2$ be a prime number and let $ N=\frac{(\frac{3p-3}{2})!}{[(\frac{p-1}{2})!]^3} $. Find $ N (mod p^2) $.

Simplify with ((p-1)/2)!, then take a $p$ from the numerator. We should calculate the result of what's left (mod p). Break the product forming the numerator in (p+1)/2*(p+3)/2*..*(p-1) and (p+1)*(p+2)*..*(3p+3)/2, and check the residue of each of these (mod p) when divided with ((p-1)/2)!. The first one is (-1)(p-1)/2, while the other one is 1/((p-1)/2) (mod p), which is the same as -2, so the product is (-1)(p+1)/22 (mod p), and after multiplying it with p we get (-1)(p+1)/22p (mod p2). Is it Ok? (the cases p=3,5 work :)).


calculus
integration
function
real analysis
real analysis unsolved

Let f , g be two nonnegative measurable functions on [0, 1] such that $ \int_{0}^{1} f(x)dx \equal{} 1$, $ \int_{0}^{1} g(x)dx \equal{} 2$, and $ \int_{0}^{1} g(x)^{2}dx \equal{} 5$. Let $ A \equal{} \{x: g(x) \geq f(x)\}$. Show that $ m(A) \geq \frac {1}{5}$, where m is Lebesgue measure.

I haven't solved this yet - I've tried some things that haven't worked. But I immediately noticed that the only role $ [0,1]$ seems to be playing is that of a set of measure one. Hence, I've been thinking of this under the following rephrasing:

$ X$ and $ Y$ are nonnegative random variables with $ E(X)\equal{}1,\ E(Y)\equal{}2$ and $ E(Y^2)\equal{}5.$ Prove that $ P(Y\ge X)\ge\frac15.$

One side note: we have $ \text{Var}(Y)\equal{}1.$

Actually, one has very little freedom as to what to try here :). Just write
$ 2\equal{}\int_{[0,1]}g\equal{}\int_A g\plus{}\int_{A^c}g\le \int_A g\plus{}\int_{A^c}f$
$ \le
m(A)^{1/2}\left(\int_{[0,1]} g^2\right)^{1/2}\plus{}1\equal{}\sqrt{5m(A)}\plus{}1\,,$
and the result follows immediately.

The sharp version here: $ m(A)>\frac{5\minus{}\sqrt{5}}{10}\approx 0.264$

The "equality" case: $ (f(x),g(x))\equal{}\begin{cases}(s,s)&0\le x\le \frac1s\\(0,\frac{s}{s\minus{}1})&\frac1s< x\le 1\end{cases}$, with $ s\equal{}\frac{5\minus{}\sqrt{5}}{2}$


Let A a ring such that x^3=x for every x in A.Than A is comutativ.

I think it has been solved before on the board.

Let's use [b]lemma 1[/b]: in a ring where the implication $x^2=0\Rightarrow x=0$ holds, every idempotent element is in the center (i.e. commutes with every other element).

[b]Proof[/b]: Let $x^2=x$ ($x$ is an idempotent element) and let $y$ be any element in the ring. We have $(xyx-xy)^2=xyxyx-xyxyx-xyxy+xyxy=0\Rightarrow xy=xyx$. In exactly the same manner we show $yx=xyx$, so $xy=yx$, Q.E.D.

[b]Lemma 2[/b]: If in a ring we have $x^n=x,\ \forall x$, then $x^{n-1}$ is in the center ($\forall x$).

[b]Proof[/b]: Assume we have $x^2=0$. Then $x^n=0\Rightarrow x=0$, so we can use Lemma 1: we have $(x^{n-1})^2=x^{2n-2}=x^nx^{n-2}=x^{n-1}$, so $x^{n-1}$ is idempotent, so, by Lemma 1, $x^{n-1}$ is in the center of the ring ($\forall x$).

We apply all of this to our problem, and we find that $x^2$ is in the center of the ring, $\forall x$.

We also have $(x+1)^3=x+1=x^3+1$, so $3x^2+3x=0$, meaning that $3x$ is in the center of the ring, $\forall x$. At the same time, $2x=(x+1)^2-x^2-1$ is in the center of the ring, so $x=3x-2x$ is in the center of the ring, $\forall x$, Q.E.D.

Yes, this is the solution I also gave during the magnificient kholle last week. Nice problem, anyway!


how can i ascend to spread the picture? :(

i type the question and save it.how can i copy these Inequality ascend to spread to
forum :?

Do you mean how to upload a picture attachment? To do so, just click the " Add an attachment link " below the typing area , next to [b]Options[/b]

if x,y,z>0,and x+y+z=3.proof:x/(x2+y+z)+y/(y2+z+x)+z/(z2+x+y)<=1
thank u for ur help :oops:

Proofs do not belong in the Getting Started forum, the go into Intermediate at least ;)  I've found a very weird way to do this inequality and I don't think it's correct. I used rearrangement. If somebody can confirm that it is possible to use rearrangement hen I will post my solution :)

You can use rearrangement, and you're still hanging around in getting started? I really think you should consider moving to a higher forum then. :P

Anyway, to know if it is possible with rearrangement, the easiest way will be to post your solution. :)
I think it is possible but it is not easy.

Anyway, here's how I did it:[hide="my sol"]$\sum \frac{x}{x^2+y+z}=\sum \frac{x}{x^2+3-x}\le\sum \frac{x}{2x+2-x} = \sum \frac{x}{x+2}$ (where the middle step is AM-GM in the denominator)
Now by Jensen on the concave function $f(x)=\frac{x}{x+2}$ we get that this is $\le 3 f\left(\frac{x+y+z}{3}\right)=3f(1)=1$[/hide]
PS: this is at the very least pre-olympiad level :P

I'm going to lock this topic because of two reasons:

A. Your first post asked about pictures so a good place will be Questions,Suggestions forum.
B. But then you changed to ask about the question involving proof (which according to Peter VDD, it can be solved by Rearrangement  :D I didn't take a look at it though).

So, because you're asking for two way different things, the topic is locked. For pictures, ask in the right forum. For proofs, most algebraic ones go to Intermediate but the advance ones, like ones that use Holder's or Schur's, will likely to be on Pre-Olympiad or higher.


algebra
polynomial
superior algebra
superior algebra unsolved

Find a splitting field of the polynomial $X^{p^{m}}-1$ over $\mathbb{F}_{p}$.

The polynomial is the same like $(x-1)^{p^{m}}$ if $p$ is odd ;)

@ZetaX: And it's the same like $(x-1)^{p^{m}}$ if p is even ;-)

Oh I forgot $1=-1$ for the even case  :blush:  :D

[quote="Rodolphe2005"]eugene it's more interesting to see the splitting field of $X^{p^{m}}-X$ ... Maybe an error in your post ?[/quote]

It is $F_{p^{m}}$. No,it wasn't a mistake i was just blind at the moment of posting it,though right after having posted realized that it was trivial.


How long does it usually take to be notified of your USAMTS score?  Haven't seen this asked at all so sorry if it's been posted.

We hope this year in general to get the papers graded within 3-4 weeks of the entry deadlines.  Please note that these are postmark deadlines, which means we generally have to wait for 7-10 days after that before we have all the papers.

Results will be available online within a couple days of grading.  (Note: you will not be mailed any results - everything, including both your scores and grader comments, will be online.)


Here are some of the sums I was unable to solve . Please reply if you know the way :

Q.1:One or more of five siblings have eaten some cakes without their
mother’s permission.When questioned , they gave the following answers
–
1st child: Only one of us ate the cake 2nd child:Exactly two of us ate
the cake 3rd child:Exactly two of us did not eat the cake. 4th child :
Only one of us did not eat the cake. 5th child :All of us ate the cake.
Their mother knew from past behaviour that the guilty children lied
while te others told the truth.The number of children who ate the cakes
were ?

Q.2 : Two and a half work crews take two and a half days to clear
two and ahalf hectares of forest . How many work crews are required
to clear four hectares of forest in two days ??

Q.3 : If I were to start writing positive integers in the order 1 , 2 ,3,
....with the condition that each digit (0-9) can only be used 100 times
, which is the first integer that cannot be written ?

Q.4:Let x = p7 + p3 and y = p7 − p3.Find the value of x6 + y6.

Q.5 : Let x , y , z and w represent four distinct positive integers such
that
x2 − y2 = z2 − w2 = 81
Find the values of xz + yw + xw + yz.

Q.6 : Some unit cubes are assembled to form a larger cube and then
some of the faces of the larger cube are painted.After the paint dries ,
the larger cube is disassembled into the unit cubes and it is found that
96 of these have no paint on any of their faces.How many faces of the
larger cube were painted ?

Q.7 : The number A = 200420052006...2040 is formed by putting the
consecutive integers from 2004 to 2040 together.What is the remainder
when A is divided by 9 ?

Q.8 : Let n = 2004 ^2 +  2005^2.Find the value of square root 2n − 1

Q.9 : It is known that n = 10^100 −1 has 100 digit 9’s.How many digit
9’s are there in n^3?

Q.10: Let a and b be positive integers such that 2
3 < a
b < 5
7.Find the
value of a+b when b has the minimum value.

Q.11 : Matt , who is often late for appointments ,walks up a moving
escalator at the rate of one step at a time.When he moves quickly ,at
the rate of two steps per second , he reaches the top after taking 24
steps . When he is tired ,he climbs at the rate of 1 step per second
and reaches the top after taking 15 steps.How long would Matt take to
reach the top , if on a rare day , he just stood on the escalator ?

Q.12 :A set S of positive integers <= 2004 satisfies the condition that
no number in S is double in value of another number in S.What is the
largest possible number of integers in S ??

Q.13 :Find the number of digits inN where N is the product of all the
the positive divisors of 100, 000, 000.

Q:14 :Numbers like 541 and 93210 are said to be strictly descending.
Find the number of strictly descending integers which are less than
10^5

Vignesh

Here are hints, partial solutions or full solutions to some of the questions (in arbitrary order, and not chosen by difficulty):

Q.5: a^2 - b^2 = (a - b)(a + b) = 81 = 3^4. Since a - b < a + b, a - b is 1 or 3 and a + b is 81 or 27. We now have four candidates for x, y, z and w, and xz + yw + xw + yz = (x + y)(z + w) = 3^4 * 3^3 = 3^7.

Q.4: (a - b)^6 + (a + b)^6 = 2(a^6 + 15a^4*b^2 + 15a^2*b^4 + b^6).

Q.3: We are looking for the first positive integer in which the same digit is repeted 101 times, that is 10^101.

Q.8: n = 2004^2 + (2004 + 1)^2 = 2(2004^2 + 2004) + 1, so 2n - 1 = 4(2004^2 + 2004) + 1 = (2*2004 + 1)^2.

Q.13: Let m = 10^8 = 2^8*5^8. Now, for a divisor k = 2^x*3^y of m we have 9 candidates for x and 9 for y. If we are going to multiply all 81 divisors, we would get n = 2^(9*(1 + 2 + 3 + ... + 8))*5^(9^(1 + 2 + ... + 8)) = 10^324, so the number of digits is 325. (The exponent 9*(1 + 2 + ... + 8) follows because there are 9 divisors for each candidate x and y, respectively).

Q.14: We have the sequence 9876543210, and shall pick out at most 5 numbers to make a smaller sequence, for instance (98)7(6)5(43)2(10), where the numbers in paranthesis are taken away; 762. This can be done in 1 + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153 different ways.

Q.12: The idea is to divide the set {1,2,...,2004} into some smaller and more easily understandable sets. For instance, we can make subsets {a,2a,4a,...} where a is some odd number and the set stops with the last number 2^k*a <= 2004. What we then remark is that the restriction that both s and 2s cannot both be in S is nothing more than the same restriction made on the subset in which both s and 2s belongs. In this subset, we easily see that the maximum number of integers chosen to be in S is the number we get by chosing alternating numbers; a, 4a, 16a, ..., the number simply determined by k. That is, we have to find k for the different a:

2004/4 = 501, so 2004 > a >= 501 gives k = 1 or 2, choosing just one number to be in S.
2004/16 = 125.25, so for 501 > a > 125 we choose two numbers (of three or four candidates).
2004/64 = 31.x, so for 125 >= a > 31 we choose three numbers.
2004/256 = 7.x, so for 31 >= a > 7 we choose four numbers.
2004/1024 = 1.x, so for 7 >= a > 1 we choose five numbers.
For 1, we choose six numbers.

Now, 6 + 3*5 + ... = (the answer)

Q.9: n^3 = (10^100 - 1)^3 = 10^300 - 3*10^200 + 3*10^100 - 1 = 999....9997000.....0003000....0002, having one less 9's than 10^300 - 10^200 = 10^200(10^100 - 1), having 100 9's. The answer follows easily.

Q.7: The remainder of A when divided by 9 is the remainder of the digitsum of A.

Sorry if some answer is wrong or misleading; I did this in a rush and has to go on now.

sorry for the late reply here. just joined a couple of days ago.

i am not an olympiad of my country so this is the first time i am looking at these questions.

i gtg in 10 min so i just pick a qn and start trying to solve.

qn3. (using my brains to show the way i think though.. haha.. kinda long)

i guess the qn is trying to say that u can only write 1 integer once in that series. so for example, writing 1 to 10 has got to use 1 twice, 2 once, 3 once, and so on.

so for sure, the first integer that cant be written is  in the 100 series.

frm 1 to 100, 1 is being used 21 times.
1 to 10 - 2
11 to 20 - 10
21 to 30 - 1
31 to 40 - 1
41 to 50 - 1
51 to 60 - 1
61 to 70 - 1
71 to 80
81 to 90
91 to 100 - 2

so frm
101 to 110 1 is being used another 11 times which adds up to 32
111 to 120 1 is being used another 20 times which adds up to 52
121 to 130 1 is being used another 11 times which adds up to 63
131 to 140 1 is being used another 11 times which adds up to 74
141 to 150 1 is being used another 11 times which adds up to 85
151 to 160 1 is being used another 11 times which adds up to 96

so 164 is the integer =)

check me if i am wrong.

gtg srry rushing to sch


Solve in positive integers:
$5^{x}= z^{2}-12$

[hide="Solution"]$z^{2}\equiv 0,1,4\pmod{5}$, hence RHS is never divisible by $5$, so there's no solution.[/hide]


Let $z\in\mathbb{C}^{*}$ such that $\left |z^{3}+\frac{1}{z^{3}}\right |\leq 2$. Prove that $\left |z+\frac{1}{z}\right |\leq 2$.

$z^{3}+\frac{1}{z^{3}}=\left(z+\frac{1}{z}\right)^{3}-3\left(z+\frac{1}{z}\right)$, use the triangle inequality.

$2\geq \left|\left(z+\frac{1}{z}\right)^{3}-3\left(z+\frac{1}{z}\right)\right|\geq \left|\left (z+\frac{1}{z}\right)^{3}\right|-\left|3\left(z+\frac{1}{z}\right)\right|$.

Let $\left|z+\frac{1}{z}\right|=r$, we have $2\geq r^{3}-3r\Longleftrightarrow (r-2)(r+1)^{2}\leq 0$. Done!

WOW. That works out nicely. I have been looking at this problem for a while. I did not see that factorization.

I know the generalizations are sometimes pointless, but an induction argument clearly gives:

If $x\in \mathbb{C}$, then $\left|x^{3^{j}}+\frac{1}{x^{3^{j}}}\right|\le 2\implies \left|x+\frac{1}{x}\right|\le 2$ for all positive integers, $j$.

This statement is counterintuitive in my opinion. I think it would make a better problem.

This is the first problem for me,  but I solved it in one minute. :P

[quote="boxedexe"]This is a very well-known problem dealing with complex numbers.  For example, you can find this problem in the first chapter of Complex Numbers from A to...Z by Dorin Andrica and Titu Andreescu, and in Mathematical Olympiad Challenges by Razvan Gelca and Titu Andreescu.  I am pretty sure that I have seen this problem appear in other books as well.[/quote]

I think this problem or something close to it appearead on of the latter problems of an aime.

To further clarify my comment, I think that it would be less obvious where one should start [and hence it would be a better problem] if my formulation of the question were asked.

@boxedexe

Yes. I first saw this problem in Titu's book.


I know I posted about this before in the other forum but several days ago I got the aim virus, today in college they deleted it but they said to be safe from people that could have accessed the thing before they deleted the virus a reformat of the hard drive is necessary, is that true? I dont know what to do.

Well,

1) Not everyone is online 24/7
2) Not everyone knows every detail about the AIM virus

My suggestion is that you just reformat the hard drive like they said--details on how I would not know.

I just don't really want to reformatt because I am bad in that stuff and I am nto sure if I have these special cd's for refromatting.

I think I also caught that virus, but my virus blocker (i think actually the firewall) stopped it from working. and later I scanned my harddrive and cleaned it is yours called mgs.exe?

I dont know what it is called but the word this was underlined.

http://securityresponse.symantec.com/avcenter/venc/data/w32.allim.b.html

This would probably help...


Pine Lane Middle School's student council has 10 boys and 8 girls serving as representatives. A subcommittee containing 3 boys and 3 girls is formed to organize a school dance. How many different subcommittees can be formed?

[hide]how many ways to pick the three girls?
8 pick 3 which is 56
how many ways to pick the 3 boys?
10 pick 3 which is 120
56*120=6720
that's a lot of ways.
[/hide]

[hide]
$\binom{10}{3}=120$ ways to choose the boys.
$\binom{8}{3}=56$ ways to choose the girls.
Multiply to get 6720[/hide]

[hide]Just do 10 choose 3*8 choose 3 and you get 6720.[/hide]


probability
geometry
rectangle
Alcumus
videos
floor function

What is the probability that an integer in the set $ \{1,2,3,\ldots,100\}$ is divisible by $ 2$ and not divisible by $ 3$?

$ \textbf{(A)}\ \frac{1}{6} \qquad
\textbf{(B)}\ \frac{33}{100} \qquad
\textbf{(C)}\ \frac{17}{50} \qquad
\textbf{(D)}\ \frac{1}{2} \qquad
\textbf{(E)}\ \frac{18}{25}$

There are 50 numbers divisible by 2, 33 by 3, and 16 are divisible by both. Thus 34 numbers are divisible by 2 and not 3, giving us a probability of $ \frac{34}{100} \implies \frac{17}{50}$. Answer choice C is the answer.

If we were to draw a venn diagram to represent this figure, then we would draw a rectangle to represent the 100 integers. We would draw one circle to represent the even numbers, then we would draw another circle to represent the numbers that are divisible by three. The numbers that are divisible by two, but not three is the region in the circle representing the even numbers that is outside of the intersection of the two circles. The amount of even numbers is 50, because 100/2 is 50. The intersection between the two circles is the numbers divisible by (2)(3), or 6. There are 16 numbers between 1 and 100 that are divisible by 6. So the answer is 50 - 16, or 34. 34/100 is 17/50

The answer is C.

Do we need a venn diagram or such a little long way to do this question or there is an easier method?

Well, no, but that is the easiest way to explain. You should watch the Alcumus videos for venn diagrams.

I have a solution that I think is a little more understandable than the one isabella2296 provided; it is nearly the same, except it is faster.

[hide="My Solution"]
There are $ \left \lfloor \frac {100}{2}\right \rfloor \equal{} 50$ integers that are divisible by $ 2$.  The numbers divisible by both $ 2$ and $ 3$ need to be subtracted now; these are simply the ones divisible by $ 6$.  We perform the same operation that we performed earlier to find that this number is $ \left \lfloor \frac {100}{6} \right \rfloor \equal{} 16$.  So there are $ 50 \minus{} 16 \equal{} 34$ numbers that are divisible by $ 2$ but not by $ 3$.  There are $ 100$ numbers total in the set, so our probability is $ \frac {34}{100} \equal{}\frac {17}{50}$, which is equivalent to answer choice $ \boxed{\textbf{C}}$.
[/hide]

I hope I helped!

There are $50$ numbers divisible by $2$ in the set. And since the lcm of $2$ and $3$ is $6$, we overcounted the multiples of $6$, and the largest multiple of $6$ less than $100$ is $96$, which gives a quotient of $16$, so there are $16$ multiples of $6$. So we have $50-16 \rightarrow 34$. Since there are a total of $100$ numbers in the set $\frac{34}{100} \rightarrow \frac{17}{50}$.


Let $ X$ be a quasi-compact scheme, $ A = O_{X}(X)$. Consider the morphism $ f: X \to \mathrm{Spec}A$ induced by the identity on $ A$. Show that $ f(X)$ is dense in $ \mathrm{Spec}A$.

Start with some $ a\in A$ such that $ D_{\mathrm{Spec}\,A}(a)$ is non-empty and disjoint to $ f(X)$. owk

disjoint to $ f(X)$ means that $ a(x) \neq 0$ for some $ x \in X$, but $ D(a)$ nonempty means that $ a(x) = 0$ for all $ x \in X$, contradiction

is this correct?

No. $ D_{\mathrm{Spec}\,A}(a)$ disjoint to $ f(X)$ implies $ a(x)=0$ for all $ x\in X$. owk

EDIT: clarified

No, see clarification in my previous post. owk

I guess this has been proven in http://www.math.umn.edu/~bahra004/alg-geo/week8.pdf

Actually this is easier and you don't need reduced schemes...

By quasi-compactness, one can cover $X$ by finitely many spectras of rings, i.e by $U_i $ such that $U_i \cong \text{Spec} A_i$, for $1 \le i \le n$. Let $A := \Gamma( X, \mathcal O _X )$.

Now the map  $f: X \to \text{Spec}A$ induced by identity restricts on the $U_i$'s as

$f|_{U_i}: \text{Spec}A_i \to \text{Spec} A$, where $f|_{U_i}$ arises from the restriction map $\varphi_i : A \to A_i$ in the sheaf $\mathcal {O} _X$.

So a point $\mathfrak p \in \text{Spec} A_i$ goes to $\varphi_i ^{-1} (\mathfrak p)$.

Let $g \in A$ be such that $V((g))$ contains $f(X)$.

This means that for all $\mathfrak p \in \text{Spec} A_i$, we have $g \in \varphi_i ^{-1} (\mathfrak p )$ i.e $\varphi_i(g) \in \mathfrak p$.

i.e restriction of $g$ to $U_i$, which we denote by $g_i$, is a nilpotent in $A_i$, so $g_i ^{k_i} = 0 \in A_i$.

Now let $m := \text{sup}_{1 \le i \le n}(k_i)$. So, $g^m$ is zero on the open cover we have taken, hence $g^m = 0$ in $A$. i.e $g$ is a nilpotent of $A$.

Which means $V((g)) = \text{Spec} A$.

This implies that if the basic open sets $D(f)$ are non-empty, then $D(f)$ must intersect $f(X)$, hence $f(X)$ is dense!


Find the fundamental period of the following functions:

(i) f(x) = sin(2m(pi)(x-[x]))

(ii) f(x) = sin((2m+1)(pi)(x-[x]))

where 'm' is an integer in both the functions and [x] denotes the greatest integer less than or equal to x

On plotting the graph I found that period of (i) is 2/m and that of (ii) is 1.

But I am not able to get these results analytically i.e. by use of algebra and trigonometry.


Let $a,b,c$ be nonnegative real numbers such that $a+b+c=1$, prove that
\[1\le \sum\frac{a}{\sqrt{a+2b}}\le \sqrt{\frac{3}{2}}\]
:)

@:I have a nice solution for both of 2 sides, hope someone will also have a nice solution for them. :)

For the left side we may consider $f: (0,\infty)\to  \mathbb R$ $f(x)=\frac{1}{\sqrt{x}}$ f convex.So using Jensen we have:
\[\sum af(a+2b) \geq f((a+b+c)^{2})=1\]

For the left, we can use Cauchy-Schwarz. Notice that
\[(\sum a\sqrt{a+2b})^{2}\le (a+b+c)(\sum a(a+2b))=1.\]

Yes, the left sides is trivial, how about the right?

Can you post your solution to the right side ?

I've read a solution of the author by entirely mixing variable method (shown in Math Ref) but I believe that an easier solution exists.

[quote=toanhocmuonmau]Let $a,b,c$ be nonnegative real numbers such that $a+b+c=1$, prove that\[1\le \sum\frac{a}{\sqrt{a+2b}}\][/quote]

[quote=toanhocmuonmau]Let $a,b,c$ be nonnegative real numbers such that $a+b+c=1$, prove that\[\sum\frac{a}{\sqrt{a+2b}}\le \sqrt{\frac{3}{2}}\][/quote]


calculus
integration
modular arithmetic
number theory unsolved
number theory

Find all Possible positive integral solutions of the equation
$x^{3}+y^{3}$ $=$ $3z^{3}$

[hide] just consider equation mod 9[/hide]

I don't see how "consider it mod 9" kills the problem. Surely 0+0=0 and 1+(-1)=0 are still potential solutions which need to be eliminated.

If there exist solutions, there exists a solution $(x_{0},y_{0},z_{0})$ with the smallest possible $z$-value.
A quick check mod 9 reveals that $3|z_{0}$, so $x_{0}^{3}+y_{0}^{3}=81z_{1}^{3}$, for some positive integer $z_{1}$

If $3|x_{0},y_{0}$ also, dividing through 27 we get a new solution $(x_{1}, y_{1}, z_{1})$ smaller than $(x_{0},y_{0},z_{0})$. Contradiction.

Therefore WLOG $x_{0}=3x_{1}+1, y_{0}=3y_{1}-1$, and I should really be sleeping now (it's getting close to midnight UK time).

[quote="kouboy"] just consider equation mod 9[/quote]

I don't think its working

Please send complete solution if you have.

I think this is not so easy
For the proof use the following lemmas( using some theory about the field $\mathbb{Z}[\sqrt{-3}]$)
[b]Lemma 1:[/b]
All solutions of the equation \[x^{2}+3y^{2}=z^{3}\] in $\mathbb{Z}^{3},$ where $z$ is odd are given by the formula: \[x=\frac{p^{3}-9pq^{2}}{8},\ y=\frac{3p^{2}q-3q^{3}}{8},\ z=\frac{p^{2}+3q^{2}}{4},\] where $(p,q)\in\mathbb{Z},$ $p=q= 1\pmod{2},$ $(p,q)=1$ or $(p,q)=2$ as $p=2,q=0\pmod{4}$ or $p=0,q=2\pmod{4}.$

[b]Lemma 2:[/b]
All solutions of the equation \[x^{2}+3y^{2}=4z^{3}\] in $\mathbb{Z}^{3},$ where $(x,y)=1$ and $x,y$ are odd, are given by the formula: \[x=\frac{x_{0}-3y_{0}}{2},\ y=\frac{x_{0}+y_{0}}{2}\] or \[x=\frac{x_{0}+3y_{0}}{2},\ y=\frac{x_{0}-y_{0}}{2},\] where $x_{0}=\frac{p(p^{2}-9q^{2})}{4},y_{0}=\frac{3q(p^{2}-q^{2})}{4},\ (p,q)\in\mathbb{Z}^{2}$ and $(p,q)=1$ or $(p,q)=2$ and $8/pq.$
For the proof we use the Fermat Method of Infinite Descent and consider two cases:
[b]Case 1:[/b] $x,y$ are odd. Therefore we can find $(u,v)\in\mathbb{Z}^{2}$ such that $x=u+v,\ y=u-v.$ Then use Lemma 1.
[b]Case 2:[/b] $x,y$ are from different parity. Then we consider $u=x+y,\ v=x-y.$ Then some calculations and use Lemma 2.
 :P


Ok, so I found this problem on the Mathcounts forums, but I don't believe it to be a Mathcounts question. I need help in doing it, and I think this is the best place to post it....so here it is:

Connie wants to find the number of ways to rearrange the letters in HAPPY TOESDAY, with two conditions. The arrangements have to be distinct & the first three letters in each arrangement have to be in alphabetical order. In how many ways can she do this?

I got very lost in understanding this problem, especially with the second condition(I didn't post this problem). If somebody can come up with a solution/answer, that would be extremely helpful. Thanks.

[hide=" click to reveal hidden content"]

I'm not sure, but I get $ 635040$:

The $ 8$ possibilities for starting letters seem to be:

$ AAD, DEH, EHP, HPP, PPO, OST, STY, TYY$.

Each of these leaves $ 9$ letters to arrange; in two cases there are three pairs (AA, PP, YY) and in the other six cases there are two pairs.

So this gives $ \frac{6\cdot 9!}{2! \cdot 2!} \plus{} \frac{2 \cdot 9!}{2! \cdot 2! \cdot 2!} \equal{} \frac{7 \cdot 9!}{4} \equal{} 635040$.

Hope this is of some help to you. I also hope it's right.

[/hide]

[hide]
This is what I got:
There are 9 choose 3 ways to pick the first 3 letters, without any of them being the same. With that, there are 3*8 ways to pick the first 3 letters, with one pair of some letter being in the first 3 (there are 3 pairs, AA, PP, YY). When the first 3 letters are all distinct, then there are 9!/(2*2*2) ways to arrange the last 9 letters, while if there is one pair in the first 3 letters, then there are 9!/(2*2) ways to arrange the last 9 letters, so I got:
$ \binom{9}{3} * \frac{9!}{2*2*2} \plus{} 8*3*\frac{9!}{2*2} \equal{} 5987520$[/hide]

three people, three different answers
i'm a little worried =)

[quote="AndrewTom"][hide=" click to reveal hidden content"]

I'm not sure, but I get $ 635040$:

[b]The [/b]$ 8$ [b]possibilities for starting letters seem to be[/b]:

$ AAD, DEH, EHP, HPP, PPO, OST, STY, TYY$.

Each of these leaves $ 9$ letters to arrange; in two cases there are three pairs (AA, PP, YY) and in the other six cases there are two pairs.

So this gives $ \frac {6\cdot 9!}{2! \cdot 2!} \plus{} \frac {2 \cdot 9!}{2! \cdot 2! \cdot 2!} \equal{} \frac {7 \cdot 9!}{4} \equal{} 635040$.

Hope this is of some help to you. I also hope it's right.

[/hide][/quote]
[hide="analysis"]
why are there only 8 possibilities?
the starting letters could be APY
[/hide]
[quote="parukia911"][hide]
This is what I got:
There are 9 choose 3 ways to pick the first 3 letters, without any of them being the same. With that, there are 3*8 ways to pick the first 3 letters, with one pair of some letter being in the first 3 (there are 3 pairs, AA, PP, YY). [b]When the first 3 letters are all distinct, then there are 9!/(2*2*2) ways to arrange the last 9 letters, while if there is one pair in the first 3 letters, then there are 9!/(2*2) ways to arrange the last 9 letters[/b], so I got:
$ \binom{9}{3} * \frac {9!}{2*2*2} \plus{} 8*3*\frac {9!}{2*2} \equal{} 5987520$[/hide][/quote]
[hide="analysis"]
this was the way I was doing it, until I noticed something that made me make a face and conclude it was going to take a lot more casework than I originally thought
everything seemed fine (to me at least) until we get to the bolded part

when the three letters are distinct, the number of ways to arrange the last letters is unfortunately not $ \frac {9!}{2\times2\times2}$
consider the following case:
say the first 3 letters are AEH
in this case, the number of ways to arrange the last letters would be $ \frac {9!}{2\times2}$ because we already used an A

same idea for the case where one pair is in the first 3 letters[/hide]

[quote="LincolnMS nerd herd"][hide]I'm going to start off by apologizing at my lack of LaTeX. I'm not that skilled yet. I'll try to learn. :P

Ok, well this was my solution.

1)I thought, in total, there would be 12! ways to arrange letters.

2)However, in each arrangement, there are 6 ways to arrange the front three letters, and only one way to have it be alphabetical, so we would have to divide by 6.

3)Now, we have 3 pairs...so then we would have duplicates 3 times, so it's (2!)^3=8 which we would also have to divide by.

4)So, now we have 12!/48=9,979,200 :huh:

I have no idea whether that's right or not. I personally don't like the problem, so you guys don't have to reply. Once again, sorry for the lack of LaTeX, and I hope it's still readable. :oops: [/hide][/quote]
[hide="analysis"]
this is an interesting solution
it is logical, but i don't know if it is right[/hide]

[hide="here is my solution"]
we have many cases to consider

[u]Case 1 - one pair and one letter not in a pair in first three letters[/u]
there are 3 possible pairs to choose from and 6 other letters for the last one
there are $ \frac {9!}{2\times2}$ ways to arrange the last letters

so the number of combinations in this case is $ \binom{3}{1}\binom{6}{1}\frac {9!}{2\times2} \equal{} 3\times6\times90720 \equal{} 1632960$

[u]Case 2 - one pair and one letters in a pair in the first three letters[/u]
there are 3 possible pairs to choose from for the pair and 2 pairs left for the last one
there are $ \frac {9!}{2}$ ways to arrange the last letters

so the number of combinations in this case is $ \binom{3}{1}\binom{2}{1}\frac {9!}{2} \equal{} 3\times2\times181440 \equal{} 1088640$

[u]Case 3 - three distinct letters in first three letters with none of them from the pairs[/u]
there are 6 possible letters to choose from for the first three letters
there are $ \frac {9!}{2\times2\times2}$ ways to arrange the last letters

so the number of combinations in this case is $ \binom{6}{3}\frac {9!}{2\times2\times2} \equal{} 20\times45360 \equal{} 907200$

[u]Case 4 - three distinct letters in first three letters with one of them from a pair and the other two not[/u]
there are 3 pairs to choose from and 6 other letters to choose the rest from
there are $ \frac {9!}{2\times2}$ ways to arrange the last letters

so the number of combinations in this case is $ \binom{3}{1}\binom{6}{2}\frac {9!}{2\times2} \equal{} 3\times15\times90720 \equal{} 4082400$

[u]Case 5 - three distinct letters in first three letters with two of them from a pair and the other one not[/u]
there are 3 pairs to choose from and 6 other letters to choose the last one from
there are $ \frac {9!}{2}$ ways to arrange the last letters

so the number of combinations in this case is $ \binom{3}{2}\binom{6}{1}\frac {9!}{2} \equal{} 3\times6\times181440 \equal{} 3265920$

[u]Case 6 - three distinct letters in first three letters, but all of them from pairs[/u]

there are 3 pairs to choose from
there are $ 9!$ ways to arrange the last letters

so the number of combinations in this case is $ \binom{3}{3}9! \equal{} 9! \equal{} 362880$

now we add up all the numbers we've gotten:

$ 1632960 \plus{} 1088640 \plus{} 907200 \plus{} 4082400 \plus{} 3265920 \plus{} 362880 \equal{} \boxed{11,340,000}$

hmm.... another different answer  :(
[/hide]

Thats why i love combinatorics, you can even make combination out of answers itself.

First, are you sure it isnt TUESDAY?
I agree with AndrewTom's method as it is most clean and presice. And easy to understand for stupid people like me  :D
Although he missed some combinations
Letters
AADEHOPPSTYY
Arrangements
I can group them into sets
1 ( x, y, z) 2 (a, x, y) 3 (a, b ,x) 4 (a, a, x) 5 (a, a, b) (a ,b, c)including permutations
in which x y z are letters appearing once, and a, b are letters appearing twice in the list

G1: DEH DEO DES DET DHO... EHO EHS EHT ... HOS HOT .... OST (20)                      20
G2: ADE ... ADT, AEH ... AST (15), DEP DHP ... DPS (15) , DEY ... STY (15)               45
G3: a = 3 OPTIONS b = 2 OPTIONS divide by 2 repeating, X = 6                               18

Hey, isnt this the first three combination, and the last few permutation?!

G4: 3x6=                                                                                18
G5: 3x2=                                                                                6
G6:1

Now calculating the last few letter's permutation in easy
G1: $ \frac {9!}{2*2*2}*20$
G2: $ \frac {9!}2*245$
G3: $ \frac {9!}*2*18$
G4: $ \frac {9!}2*2*18$
G5: $ \frac {9!}*2*6$
G6: $ \frac {9!}*1$
Total = \frac {9!(10+45+36+18+12+4)}4= 11 340 000

I got the same as Vallon 12, and Im pretty sure we used the same method...  :|
Note to self: Read posts before posting...

[quote="LincolnMS nerd herd"]Hmm, your casework could work. Like what you said about me, it is interesting, but different still. :P

When talking with my friend, he said that there was something wrong about mine. He says, what if the first three were like AAY? Then there would be only 3 different arrangements of the first, not $ 3!$.

So then he suggested, like you, to use casework. So I proceed like he suggested, and got...(I think)

[hide]1) There can be none of the 6 repeating in the first 3 letters. Then there would be $ \frac {\binom{6}{3}9!}{2!*2!*2!} \equal{} 907,200$ The combination comes from how to pick the first 3, cause each "pick" gives one distinct alphabetical arrangement. The rest is just how to arrange the 9 letters left

2) There is 1 repeating letter in the first 3 letters. Then there's $ \frac {(6*6*5)/2*9!}{2!*2!*2!} \equal{} 4,082,400$ The reason we divide by 2 in the numerator is because for each repeating letter, the remaining pairs are counted twice, once for say ABC and another for ACB.

3) There are 2 repeating letters, so now it's $ \frac {(6*5*6)/2*9!}{2!*2!*2!} \equal{} 4,082,400$ This is much like step 2, because it's just how to have 1 non-repeating letter in the first 3.

4)There are 3 repeating letters in the first 3 letters. Again, it should be like step one, just different wording. So...$ \frac {\binom{6}{3}9!}{2!*2!*2!} \equal{} 907,200$

5) Add those up to get...$ 9,979,200$ Which happens to be what I got with my formula...either we're both wrong, or we're both right. This is all very confusing... :([/hide][/quote]

wait, i don't get it
does "repeating letter" mean either A, P, or Y? or does it mean letters that repeat in within the first three letters?

Ah, yes. Very sorry. The Repeated letter means letters A, P, or Y. I think yours might be right though, mainly cause you seem smarter than me. :P

i'm kinda disappointed
two people that wrote my name, and neither of them spelled it right =P

anyway, could you please explain how you got $ 6\times6\times5$ in the numerator of number 2?

Umm, would you have 6 choices for the repeating letter... and 6 choices for the first non-repeating letter, 5 choices for the 2nd non-repeating letter, and now you've counted them twice, so you divide by 2...

Ex. Like how EU is the same as UE, but only EU is alphabetical.

Sorry, VALLON.


[quote="riddler"]1. $\text{I have ten pairs of socks in my drawer,}$ $\text{but all of the socks are separate. Each pair is a different colour.}$ $\text{I shut my eyes and pull out some socks.}$ $\text{How many must I take out to guarantee that I get a pair?}$ $\text{How many more do I need in order to get a second pair?}$

2. Let $x$, $y$ and $z$ be positive real numbers such that $xyz = 1$. Show that \[\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}= 1\] 3. Show that a $3$-digit number $abc$ is divisible by $7$ if and only if $2a+3b+c$ is divisible by $7$.[/quote]

Please try not to type regular text in LaTeX, for us dial-up folks.

[hide="First one"]Worst case scenario: You pick out ten socks and they're all different colors. Okay, pick out another one. By pigeonhole, it has to match one of them. So you need to pick out at least eleven socks to guarantee one pair. And you need to pick out twelve to guarantee two pairs.[/hide]

[hide="Hint on 2"] Let $x = \frac{a}{b}, y = \frac{b}{c}, z = \frac{c}{a}$. [/hide]

[hide="Hint on 3"] The number in question is $100a+10b+c$.  What is this $\bmod 7$? [/hide]

[quote="riddler"]
2. Let $x$, $y$ and $z$ be positive real numbers such that $xyz = 1$. Show that \[\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}= 1\] [/quote]

[hide="2"]Use $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$.

Then $\sum_{cyc}\left(\frac{a}{1+\frac{a}{b}+\frac{a}{c}}\right)=1$,
$\sum_{cyc}\left(\frac{abc}{a(ab+ac+bc)}\right)=1$
$\sum_{cyc}\left(\frac{bc}{ab+ac+bc}\right)=1$

Since the denominator is the same in each term you get $\frac{ab+ac+bc}{ab+ac+bc}=1$[/hide]

[hide="Another method for 2"]Substitute $z={1\over xy}$ and simplify[/hide]

[hide="3"]The number is equal to 100a+10b+c, which in mod7 is 2a+3b+c[/hide]


AMC
AMC 10
AIME
search
probability
expected value

I heard it was 21 or 25 for the AMC 10 what about amc 12

I believe it's 19 for the AMC10 and 14 for the AMC12.

If you take the 12, and are aiming to qualify for AIME, your best strategies are these:
Answer 14 correct, 11 blank for 100.5 pts
Answer 18 correct, 7 blank for 118.5 pts
Answer 22 correct, 3 blank for 136.5 pts

If you answer 18, you can afford to miss three and still qualify (getting a 100.5)

If you answer 22, you can afford to miss six (getting a 100.5)

That applies if the qualifying standard is 100. It might be lower- see the announcement for this year's A test.

Yup, it was 97.5. Shouldn't make too much of a difference, though.

The Search button is rather easy to use, and it gives http://www.artofproblemsolving.com/Forum/viewtopic.php?t=130853 quite quickly

[quote="now a ranger"]typically, trying to answer them all is best since the blank score is only 1.5 as to 2.5 in 2006. I am going to try to answer all of them, even if it is blind guessing, whereas I can at least have a chance, since no one usually gets 19 right and the exact number of blank for a 120+.  :huh:[/quote]

Yes, it is a good bet to answer all of the questions... But last year, I actually got 19 right and left exactly enough blank to qualify for the AIME (121.5) :lol:  :P

[quote="Boy Soprano II"][quote="now a ranger"]typically, trying to answer them all is best since the blank score is only 1.5 as to 2.5 in 2006. I am going to try to answer all of them, even if it is blind guessing, whereas I can at least have a chance, since no one usually gets 19 right and the exact number of blank for a 120+.  :huh:[/quote]
You are joking?

Anyhow, to repeat something Mr. Patrick said, it's kind of futile to try to answer a fixed number of questions, because the qualifying score might fluctuate, in which event the reasons behind these schemes are absurd.  How about answering exactly the problems you can solve?  Why don't we hear more of that strategy?[/quote]

maybe because novice borderline test takers might not be able to do well under pressure and get all the problems they do right and/or can't do enough of them; this leads to need for guessing, educated and non-educated ( preferably educated ). How else can they make AIME then?

[quote="Boy Soprano II"]How about answering exactly the problems you can solve?[/quote]
Some educated guesses are probably a good idea- if you can eliminate all but two answers, guessing increases the expected value of the score. If you can eliminate two, leaving three, guessing still helps a little on average (with the new scoring system).

Unless your goal is to get at least $n$ points, where $n$ is some fixed value (100,120,150), don't guess problems you don't know anything about. If your goal is the AIME this year, your target is not a fixed number of points.

[quote="now a ranger"]maybe because novice borderline test takers might not be able to ... get all the problems they do right[/quote]
I had very high scores every year I took the AHSME/AMC 12, but I answered at least one question wrong every time. These were not guesses- they were mistakes I made trying to solve the problems. Drop "novice/borderline" from that statement. Anyone can make mistakes.
My lowest score came in the year that I left several problems blank- I did better as the foolhardy 9th grader answering every question than as the cautious 10th grader leaving most of the back page blank.

You should probably try to do all the problems you can, plus guess on two or so that you are quite confident of the answer, but perhaps a bit doubtful on. I remember I did this on one of the AMC 12's like two years ago... I wasn't that great at math then, but I remember this question about a spider, socks, and shoes. I got the question right simply by eliminating 3 answer choices and guessing from the remaining two. That sort of helped me go to the aime.

[quote="now a ranger"]...maybe because novice borderline test takers might not be able to do well under pressure and get all the problems they do right and/or can't do enough of them; this leads to need for guessing, educated and non-educated ( preferably educated ). How else can they make AIME then?[/quote]My feeling is that "novice borderline test takers" shouldn't even be thinking about the AIME until they can remove those adjectives from their self-description.  The AIME is for more experienced contest problem solvers (that's why it's got that "I" in its name).  If someone finds the AMC sufficiently challenging, why rush to get to the AIME?  Learn problem solving at an AMC-level first.

You have to learn to walk before you can run.