[hide]1+2+3+4+5...+1100 wow this will take ages...lol

ok the formula to find the sum of the first n positive integers is n(n+1)/2
so 1100(1101)/2= [b]605550[/b][/hide]

If you know how to add up to 100, it would be fairly easy to do it...

[hide="looky Siverfalcon"]1100 + 1 = 1101
1101 * 1100 = 1211100
1211100 / 2 = 605550

that's the answer 605550

by the way this is the concept of how you solve the problem....

since it's labeled like this,

1      + 2      + 3       + 4 + 5 + 6 +...........  1100
1100 + 1099 + 1098 +...............................+   1

which is

1101 + 1101 + 1101 +......

and since we labeled 1100 times, we multiply it by 1100

and since we doubled them, we divide the total number by 2

or

you could have just multiplied 550

it's so inconveinient typing all these without LaTeX.. :( [/hide]

[hide="solution"]
0.

You didn't specify that you needed to add integers =o =o =o
[/hide]

See the difference now, Treething.

shinwoo, that's really good thinking.

That's what I used to do when I had no knowledge of formulas.

From there, we can actually derive this formula. This might be again little bit hard but...

[hide="Proof_2"]

Our goal is to get $\frac {n \cdot (n+1)}{2}$

Use 1+2+3...n.

Using what shinwoo said, it goes like 1+n, 2+(n-1) which is also n+1. Now, how many pairs like that are there?

Well, if you set 1+2+3...8 then 8/2 = 4 pairs.

Which means that if it's n, that's n/2.

So, n+1 for n/2 pairs.

\[(n+1) \cdot \frac {n}{2} = \frac {n(n+1)}{2}\]

Maybe this is not exactly way you do it but that's what I did for the first time. [/hide]

Notice that all I did was develop from what shinwoo said.

You can use the 'sum of arithmetic series" equation:
sum=n/2(1st term+last term)
where n is the number of terms.

So you would solve it like this:
1100/2 (1+1100)
(550)(1101)
605550

That's basically what some others are saying, but that is a formula that can be used for all numbers no matter what the amount being added is...

[quote="Silverfalcon"]Nice job.

Can you prove the formula though?

This may not be so basic but...

[hide="Proof"]

Assume k works. Then k+1 must work. If we prove this, then it means it works for all integers.

\[\frac {k \cdot (k+1)}{2} + (k+1) = \frac {(k+1) \cdot (k+2)}{2}\]

This then equals to:

\[\frac {k^2+3k+2}{2}\] and the second part is:

\[\frac {k^2+3k+2}{2}\]

So, this works and we just proved our formula.[/hide][/quote]
Silverfalcon, don't forget the BASE case. It is essential that you prove it, even if it is obvious.

[quote="shinwoo"]hm..

I don't get silverfalcon's first proof...

 :(  :?[/quote]
[hide="mathematical induction"]
Basically, Silverfalcon used a method called [b]mathematical induction[/b] or simply induction. There are many things that can be said about the induction principle, but as a starter, let me give you a simple way of using it for now until you are more comfortable on how to work with it at a higher level.

Let me start off by saying what is this mathematical induction: it is a proof technique that associates the statement that you want to prove with the natural numbers. The induction consists of two parts:

1) Prove that the statement is true for the base case. Usually the base case is $n=1$.

2) Prove that if the statement is true for $k$, then it follows that the statement is also true for $k+1$.

Mathematical induction is simply making a matching game between the set of natural numbers and the values of a given formula.

[b]Part 1)[/b] First prove that it holds for the base case, which is $n=1$. This case you can easily see that it is obviously true because $1=1$.

[b]Part 2)[/b] Next step is done by Silverfalcon's proof: assume that this statement is true--> "the sum of the first $k$ natural numbers is $\frac{k(k+1)}{2}$." Then this part is tricky, you have to prove "the sum of the first $k+1$ natural numbers is $\frac{(k+1)((k+1)+1)}{2}$" from your induction hypothesis, which is "the sum of the first $k$ natural numbers is $\frac{k(k+1)}{2}$."[/hide]

Sorry.

1(1+1)/2 = 1

Which is just n=1.

Yup. :)  :D

OOO!!!!

thanks alot 3X.lich!!

I really appreciate your help :D

[quote="Silverfalcon"]Find the sum of first 1100 natural numbers.

 :)[/quote]

[hide="click to reveal hidden contest"]$\frac{(1+1100)1100}{2}=605550$[/hide]


Given $f: X \to Y$, there exists $g: Y\to X$ such that $g(f(x))=x$ and $f(g(y))=y$ for all $x\in X$ and $y\in Y$. Prove (as rigorously as possible) that $f$ is a bijection.

[b]A hint.[/b]

 Let  $f(a) = f(b)$ . Then $g(f(a)) = g(f(b))$ , that is $a = b$.Thus f is injection. (In the same way g is injection).The second of the relations says that f is  <<onto>> B, so  $f$ is injection and << onto>> , that is a bijection.

   Babis


Support
vector
geometry proposed
geometry

$ \Delta_1,\ldots,\Delta_n$ are $ n$ concurrent segments (their lines concur) in the real plane. Prove that if for every three of them there is a line intersecting these three segments, then there is a line that intersects all of the segments.

The concurrency may even be taken in projective way (the case when the support lines of the segments are parallel, and for that particular case the problem is a well-known application of Helly's theorem (mentioned in the Encyclopaedia Britannica), as sketched below).

Consider the (dual) $2$-dimensional vector space $V$ of the lines of the plane. Define by $L_k$ the set of lines intersecting the segment $\Delta_k$, for $1\leq k\leq n$. The sets $L_k$ are convex in $V$, and by the givens, any three have a non-empty meet. By Helly's theorem, them all $n$  have a non-empty meet, and so any line in that meet intersects all $n$ segments.

Rembetophile :-)

Do you mean Wikipedia? Anyway, Wikipedia contains it :-)

http://en.wikipedia.org/wiki/Helly%27s_theorem

No, phile  :)  I mean the problem with segments on parallel support lines (vertical) is given as example of an application to Helly's theorem in the Encyclopaedia Britannica.


[quote="amir2"]  may be hard [/quote]
Is it a joke or a mistake ?!
 :cool:


Here are two cool trig questions:

1. $\text{Compute } \sin 12^\circ(\sin 12^\circ+\sin 36^\circ+\sin 60^\circ+\sin 84^\circ+\sin 108^\circ)$

2. $\text{Calculate } \cos 36^\circ - \cos 72^\circ$

[hide="Answers"]
1. 3/4
2. 1/2[/hide]

Any suggestions as to how to solve them?

Sum to product, then use some induced formula, then multiply the least cosine value to it, the product to sum and double angle, then simplify...

lol do you mind showing your exact process?

for the second one
$\displaystyle cos36^\circ-cos72^\circ=cos\frac{\pi}{5}-cos\frac{2\pi}{5}$
$\displaystyle =\frac{(cos\frac{\pi}{5}-cos\frac{2\pi}{5})cos\frac{\pi}{10}}{cos\frac{\pi}{10}}$
$\displaystyle =\frac{cos\frac{\pi}{5}cos\frac{\pi}{10}-cos\frac{2\pi}{5}cos\frac{\pi}{10}}{cos\frac{\pi}{10}}$
Use the formula  $cosacosb=\frac{1}{2}[cos(a-b)+cos(a+b)]$
We obtain $\displaystyle cos36^\circ-cos72^\circ=\frac{1}{2}\frac{cos\frac{\pi}{10}+cos\frac{3\pi}{10}-cos\frac{3\pi}{10}-cos\frac{\pi}{2}}{cos\frac{\pi}{10}}$
simplify this we have $\displaystyle cos36^\circ-cos72^\circ=\frac{1}{2}$

hmm yea pkothari13, the order of the answers doesnt match up with the order of problems *yawns* im tired  :sleeping:  dreaming of  :yoda:   :sleep2:

[quote="towersfreak2006"][quote="pkothari13"]
1. $\text{Compute } \sin 12^\circ(\sin 12^\circ+\sin 36^\circ+\sin 60^\circ+\sin 84^\circ+\sin 108^\circ)$
[/quote]
[hide="First problem...finally"]
\[\sin{\alpha}+\sin{\beta}=2\sin{\frac{\alpha+\beta}{2}}\cos{\frac{\alpha-\beta}{2}}\]\[\sin{36^\circ}+\sin{12^\circ}=2\cos{12^\circ}\sin{24^\circ}\]\[\sin{108^\circ}+\sin{84^\circ}=2\cos{12^\circ}\sin{96^\circ}\]So our expression becomes: \[\sin{12^\circ}(2\cos{12^\circ}(\sin{24^\circ}+\sin{96^\circ})+\sin{60^\circ})\]\[2\sin{12^\circ}\cos{12^\circ}(\sin{24^\circ}+\sin{96^\circ})+\sin{12^\circ}\sin{60^\circ}\]Again by sine sum-to-product:\[\sin{24^\circ}+\sin{96^\circ}=2\sin{54^\circ}\sin{60^\circ}\]Also, from sine double angle:\[2\sin{12^\circ}\cos{12^\circ}=\sin{24^\circ}\]Then we want the value of:\[2\sin{24^\circ}\sin{54^\circ}\sin{60^\circ}+\sin{12^\circ}\sin{60^\circ}\]\[\sin{60^\circ}(2\sin{24^\circ}\sin{54^\circ}+\sin{12^\circ})\]Sine product to sum:\[2\sin{24^\circ}\sin{54^\circ}=\cos{30^\circ}-\cos{78^\circ}=\cos{30^\circ}-\sin{12^\circ}\]Finally, we have:\[\sin{60^\circ}(\cos{30^\circ}-\sin{12^\circ}+\sin{12^\circ})\]\[\sin{60^\circ}\cos{30^\circ}=\boxed{\frac{3}{4}}\]
And happiness ensues...:):):)
[/hide][/quote]

wowowow! thank you so much towersfreak! that must have been a latex nightmare! but you came through  :D now we can say we finally figured this stupid problem out.

and yeah, sorry for putting the answers in the wrong order... i must have been daydreaming or something


calculus
trigonometry
geometry
geometric transformation
rotation
integration

An ant is walking on the second hand of a clock at a rate of 6 cm per minute. The second hand measures 6 cm. He starts out from the tip of the second hand and walks towards the center. How long (a distance, not time) does the ant travel?

Note: the clock is ticking while the ant walks

What do you mean by "He starts out from the middle and walks towards the center"?

should this be here? or in the physics forum maybe

I don't know how to use spoilers, otherwise I would, I would appreciate it if someone would tell me.

In my solution below, I'm assuming "in the middle", means in the middle of the hand, that is 3 cm away from the center of the clock.  I'm also assuming that when you say second hand you mean the hand that ticks 60 times per minute.

I can't be totally sure of the problem you are asking because if you want the distance relative to the clock hand, which is unbelievably simple look here:

The clock is six (6) cm. long.  Half of that is three (3) cm. long.  Therefore the ant travels 3 cm.

If you want the distance the ant travels relative to the surface of the clock (the piece of paper with no. 1 - 12 on it) the problem gets a lot more complicated and calculus becomes involved.

Since the mathematics involving polar graphs usually behave where $\theta$ increaces as the angle moves counterclockwise, I will assume the clock is moving counterclockwise and $\theta_0 = 0$, (the second hand starts at the 3 o'clock position).   Sorry to disapoint you, but the answer should be the same.

The distance from the center of the clock over time is:
 $r(t) = 3 - \frac{t}{10}$,
where $t$ is the time in seconds and $r$ is the radius.

$t$ and $\theta$ are related by the equation $t = \frac{30\theta}{\pi}$.  Substituting this into the previous formula:  $r(\theta) = 3 - \frac{3\theta}{\pi}$.

Paramaterizing this:
$x = (3 - \frac{3\theta}{\pi})\cos{\theta}$
$y = (3 - \frac{3\theta}{\pi})\sin{\theta}$

Differentiating:
$x' = \frac{-3\cos{\theta}}{\pi} - (3-\frac{3\theta}{\pi})\sin{\theta}$
$y' = \frac{-3\sin{\theta}}{\pi} + (3-\frac{3\theta}{\pi})\cos{\theta}$

The ant goes through $\pi$ radians of rotation before it meets the center of the hand on the clock, so the total distance will be the arclength formula for $r(\theta)$ where $\theta$ is on the interval $[0,\pi]$.

$L = \displaystyle\int_0^{\pi}{\sqrt{(x')^2 + (y')^2}}$

or, written out longer,

$L = \displaystyle\int_0^{\pi}{\sqrt{(\frac{-3\cos{\theta}}{\pi} - (3-\frac{3\theta}{\pi})\sin{\theta})^2 + (\frac{-3\sin{\theta}}{\pi} + (3-\frac{3\theta}{\pi})\cos{\theta})^2}}$

I don't feel like doing this by hand, and since I'm lazy I'm getting the calculator to do it for me.
...... After pressing clear 50 times.....

$L$ = 5.8345432 cm.

unless im misunderstanding what you are doing, your answer should be wrong since your answer is less than 6
-edit-
nevermind, as i misread bob the first time around

hey wouldnt the clock being a wall clock or a floor clock(couldnt think of a word) make a difference to the sum?

Does the clock tick or does the hand travel continuously?  Does the sudden movement of the hand at all affect the ant?

For math's sake, i think you should assume the answer is no, for the both of you

I made a typo earlier, where it said middle, it was supposed to be the tip of the second hand. Sorry about that... I haven't been getting enough sleep lately.

And yes, the clock travels continuously (otherwise, this problem would be a lot easier)


Help me to draw the graph of the curve $ r\equal{}a(2cos \theta\plus{}cos 3\theta)$.

Hello
Here is the graph.
[hide="one interesting thing"]
any equation of this form $ r \equal{} acos(N\theta)$ is an equation of N leaf rose.[/hide]
Thank u.


Let $ T$ be the centroid of triangle $ ABC$ and let $ TM,TN,TP$ be the altitudes in the triangles $ ATB,BTC,TCA$ respectively.
Show that $ {{[ABC]} \over {[MNP]}} \ge 4$

Without inequalities there is a interesting result.

${ \frac {[ABC]^3}{[MNP]}} = \frac {(6 AB AC BC)^2}{AB^2 + AC^2 + BC^2}$

[quote="HelloWorld"][quote="mr.danh"]MNP is the pedal triangle of P wrt triangle ABC, so the result of this problem is obvious because of Euler's theorem.[/quote]
Why? How to do that ? . Please post your solution  :)[/quote]
[quote="mr.danh"][b]The Euler's theorem for the pedal triangle:[/b] $ S[MNP] \equal{} \frac {1}{4}S[ABC]|1 \minus{} \frac {OT^2}{ R^2}|$, whre O is the circumcenter and R is the circumradius of triangle ABC, and MNP is the pedal triangle of T wrt ABC.[/quote]

[quote="mr.danh"][b]The Euler's theorem for the pedal triangle:[/b] $ S[MNP] \equal{} \frac {1}{4}S[ABC]|1 \minus{} \frac {OT^2}{ R^2}|$, whre O is the circumcenter and R is the circumradius of triangle ABC, and MNP is the pedal triangle of T wrt ABC.[/quote]
How do we proof this formular? $ S[MNP] \equal{} \frac {1}{4}S[ABC]|1 \minus{} \frac {OT^2}{ R^2}|$ Help me please. :10:


MIT
college
Stanford
Princeton
geometry
3D geometry

mine would be IISER,Kolkatta  > IISER,Pune $ \approx$ NISER,Bhubaneswar > IISER,Mohali > CBS-DAE,Mumbai > CMI > ISI,Bangalore > >>>>>>>> IIT's ,BITS NIT Trichy/Warangal/Surathkal >>>>>> other NIT's

i wanna do integrated 5 year MS physics :)

[b]DO add what course u would  be pursuing[/b]

Mine: IIT > BITS > NIT > CMI  :)

Lets see what i wanna do later ok  :P

My preferences will be IIT Chennai> CMI> IIT powai> all iits. thats all.. Im not even plannin ny other college.. And you might wonder why im preferin iit m over iit b.. Its coz i wanna b in chennai.. And cmi stands over iit b for the same reason..

IIT-chennai>any other IIT>BITS>any of the NIT's
we'll see abt the course :P  :P  :P

MIT, Stanford, Princeton > IIT > Univ of Cal, Drexel, Northwestern, NUS, NTU > BITS, NIT

As for what I wanna do, well... If I decide to go abroad, then I want to take up a double major in Bio and Environmental Engineering with minors in Neuroscience and Robotics (If you didnt get that, its ok.. I am pretty confused myself)

And if I am in India, I have no idea whatsoever...

P.S - God save the queen.... Hail Rasputin

Wth the order changed? I thought iit was lesser than NUS and NTU or wasnt it?  :huh: And ofc NUS was clearly $ \geq$ Drexel wasnt it  :huh:

Hmmm... It really doesnt matter you know. I am just praying, it wouldnt get down to that level actually. Preferably, one of my top choices should click. Well, if it doesnt, then  :D ~Ladelilolou~

Hmm..Interesting order for Varun.

Anyway, I have IIT>BITS...

Dont want to join NIT... Not applied for anything else but I think I have direct admission to CMI.

I want to do some eng course csc or mech or electronic (However I would avoid this just to get away from Semiconductors :D)

And then do Robotics course in Kharagpur or the States later.

well i guess its IIT - M > any other IIT's > bits>nit trichy > ntu> dumeelkuppam university for only high intellectuals :D

and as for the course i wanna take up, csc is first choice, followed by aeronautical perhaps :)

speaking of which, does any1 know of any college that offers a good course in aeronautics(except for iit and the like ofc)??
ive heard hindustan in chennai is pretty good.. tho not sure..

anything goes for me
for me iit=nit =bits
its same for me
i have as such no preferences :D

[quote="hash_include"]well i guess its IIT - M > any other IIT's > bits>nit trichy > ntu> dumeelkuppam university for only high intellectuals :D

.[/quote]
sorry da forgot dumeelkuppam university.....and Dumeelkuppam Gaja the Dean needs a special mention da :D  :D  :rotfl:  :rotfl:  :P

Actually MIT is my worst resort  :D
[hide]
Its Mariamman Institute of Technology  :P   :rotfl:
[/hide]

[quote="pardesi"]

sorry da forgot dumeelkuppam university.....and Dumeelkuppam Gaja the Dean needs a special mention da[/quote]

ofcourse!! how silly of me!  :rotfl:  :rotfl:  :rotfl:

what an idiotic question! :huh:  Infinity is not a real number!   :lol:

oops,i forgot this very point

right,leonardo

U people r making this forum very boring :cursing:  :wallbash_red:

skand's answer is rite

and talking abt stupidity, i can do the ques correctly in rough and rite the wrong answer :rotfl:

[quote="rituraj007"]skand's answer is rite

[/quote]

huh?

dont tell me infty multiplied by zero is zero! :huh:  :huh:

actually its bein funny and borin' also nothin is infinity i suppose its limit just tends to infinite

hey one of my answers was right! infinite is not to be messed with! :D

well
can't we define infinity sometimes in future?/

my frnds always say 4 proofs
multiply LHS and RHS by inf and u can write "Hence proved" :rotfl:  :rotfl:

man its becomng ........................
well now i am starting to spam

rituraj- the future holds the unexpected but we cannot mess with it untill t's value i found and the will take infinity years itself.....scared yet? :D

[quote="Da Vinci"]
huh?

dont tell me infty multiplied by zero is zero! :huh:  :huh:[/quote]
 infty multiplied by zero is zero! :rotfl:  :rotfl:  :rotfl:  :D

i was trying to prove that :rotfl:  :rotfl:  :rotfl:

[quote="rt_08"]well now i am starting to spam[/quote]
Sorry, but i was (am) under the impression that you were (are) doing that most of the time :huh: :huuh:


logarithms
limit
induction
ratio
real analysis
real analysis unsolved

Let $ a_n$ be a sequence with $ a_0=2009$ and $ a_{n+1}=\ln (1+a_n)$. Evaluate the limit
\[ \lim_{n\rightarrow \infty}\sqrt[n]{a_n}.\]
if it exists.

note that sequance $ a_{n}$ is decreasing and nonegative (for example by induction). thus it has limit.  note that therefore limit will be equal to zero. Hence for sufficiently large $ n$ , sequance $ a_{n}$ behaves like zero  $ a_{n} \approx 0$
thus  $ a_{n\plus{}1} \equal{} \ln(1\plus{}a_{n}) \approx a_{n} \minus{}\frac{a_{n}^{2}}{2}$  roughly speaking we have $ a'_{n} \approx \minus{}\frac{a_{n}^{2}}{2}$ or $ a_{n} \approx   \frac{C}{n}$ where $ C$ is some positive constant (in this case $ C\equal{}4$) thus  limit is equal to $ 1$

$ \{ a_n \}$ is a strictly decreasing positive sequence (property of log) and thus in particular the desired limit exists: $ a_n ^{{\textstyle{1 \over n}}}$ is therefore strictly decreasing, I believe (though this is not necessary for what follows).

Simple induction shows $ 1 \le a_n ^{{\textstyle{1 \over n}}}$ and since $ \{ a_n \}$is strictly decreasing we have $ a_n ^{{\textstyle{1 \over n}}}  \le 2009^{{\textstyle{1 \over n}}}$

Squeeze  theorem.

Note that $ \lim\sqrt[n]{a_n}\equal{}\lim\frac{a_{n\plus{}1}}{a_n},$ provided the latter limit exists. (See the proofs of the root test and ratio test for the convergence of series.)

As has already been noted, $ \lim a_n\equal{}0.$ It's decreasing and bounded below, so it must have a limit; that limit must be a root of $ x\equal{}\ln(1\plus{}x),$ and the unique root of that equation is $ x\equal{}0.$

Hence, $ \lim\frac{a_{n\plus{}1}}{a_n}\equal{}\lim\frac{\ln(1\plus{}a_n)}{a_n}\equal{}\lim_{x\to 0}\frac{\ln(1\plus{}x)}x\equal{}1.$

Hence our limit is $ 1.$

Prove that $ \lim_{n\to\infty}\ na_n\ \equal{} \ 2$ .

Extremal says that $ \lim na_n\equal{}4.$ Virgil Nicula says that $ \lim na_n\equal{}1.$

Based on numerical evidence, I say they're both wrong. That limit exists, but it's not $ 4$ and it's not $ 1.$

I have written $ a'_{n} \equal{} \minus{} \frac {a_{n}^{2}}{2}$ and hence $ a_{n} \approx \frac {C}{n}$
 it's not hard to find $ C$, for example via the substitution in diferential equation
$ \left( \frac {C}{n} \right)' \equal{} \minus{} \frac {C^{2}}{2n^{2}}$ from this follows that $ C \equal{} 2$ (I'm sorry in my first post there was just mechanical mistake, I have written  $ C \equal{} 4$), anyway you must yourself determine or get correct constant, since I have already written simplest differential equation  :wink:


calculus
integration
induction
Euler
real analysis
real analysis unsolved

Can someone help me on these challenge questions please ?

For n = 0, 1, . . ., let In = S(Intergral) (pi/2)(upper bound) and 0 (Lower Bound) sin^n(x) dx.

(a) Show that I0 > I1 > I2 > · · · > 0.

(b) Use integration by parts to show that

In+2 = ((n + 1)/(n + 2))(In) if n is greater or equal to 0.

(c) Use part (b) and the fact that In+2 < In+1 < In, by part (a), to show that ((In+1)/(In)) ->1 as n -> infinity.

(d) Use (b) and induction to show that for all n greater or equal to 0,

I2n =(((2n)!(pi))/((2^(2n))((n!)^(2))(2))) and I2n+1 =((2^(2n))((n!)^2)/((2n + 1)(2n)!))

(e) From the Euler-Maclaurin Formula, for some constant C,

n! = Ce^(-n)n^(n+1/2)an

where an -> 1 as n -> infinity. Use (c) and (d) to show that C = root(2pi)

The following will help you!

[url]http://mathlinks.ro/Forum/viewtopic.php?t=80515[/url]

First of all, thanks for the hints kunny, but is it possible for you or anyone else to give me more hints or perhaps do some parts of the problems for me so I can understand it please ?

I know I am not as smart as some of the people in this forum, but I really need help on this because I will have an exam on this soon. :(


Stiti cumva subiecte,rezultate de la barajul de seniori(din 12 mai)?

[b]Domnul [color=blue]Gologan[/color][/b] a zis sa le tinem "secrete" pana saptamana viitoare.
(Adica sa nu le putem pe net ca unele din probleme erau din diverse shortlisturi si vor fi folosite probabil si in alte tari.)

[EDIT] None taken...

"[spam]

Se zice Gologan (eventual cu d-l. in fata), nu Gogolan.

[/spam]"

PS: mi-am luat revansa, no offense.  :P

1. zahariuc - 99p (din $7 \times 16$ posibile - zicea domnul Gologan ca-l ia la corectura la ultimele baraje :D )
2. ilie livia - 70 si ceva
3. turea lucian - cam tot ata
4. persu madalina
6. dumitrescu mihai
9. ionescu ciprian - aprox 60p
12. deneanu andrei - 54p
prodescu, munteanu sinziana - pe undeva prin primii 12
parca mai era si unu' de-a noua care a fost la jbmo

apar si eu pe undeva pe acolo (printre cei necalificati), locul 16 - 46p

PS@pohoatza: I-am nimerit numele. Cred ca greseala de ieri are legatura cu putinele ore dormite in ultimul timp...

care va sa zica, Papara din Sibiu a fost intamplator in primii 6 la baraj la nationala?adica ce loc a obtinut acuma, stii?

De la a 12-a au trecut mai departe doar 3 :!: Papara n-a fost printre ei... Locul exact nu-l stiu. Poate ne ajuta cineva rezultatele in format electronic :cool:

Cine e pe 5? Si care sunt punctajele primilor 12 (exacte)? :)

vad ca deja sunt discutate rezultatele, si cred ca e ok sa le postez.
totusi, subiectele sa le tinem secrete inca vreo saptamana doua.

aici sunt rezultatele
[url=http://img520.imageshack.us/my.php?image=rezultateseniorihs0.jpg]rezultate seniori[/url]

am fost io mai retard putin...si am prins si o zi proasta cred (nu erau foarte grele 2 si 3)
sper sa nu se mai intample :)

si acum cei 12 cand vor sustine urmatorul ( urmatoarele ) baraje ? si toate acestea au avut rolul doar de a injumatati lotul ?

impresionant cel de-a 9-a ... cam putini de a 12-a , toti sub linie ... ciudat ca papara nu a reusit

[quote="pureblack2003"]impresionant cel de-a 9-a ... [/quote]Nu e chiar atat de surprinzator (avand in vedere performantele lui de la juniori).

madalina persu n-a avut rezultate deosebite la juniori ? totusi , nu s-a impus cum s-a impus baiatul asta ... si totusi , are 3-4 ani in minus de pregatire fata de ceilalti , d-asta e asa deosebit

[quote="pureblack2003"]madalina persu n-a avut rezultate deosebite la juniori ? totusi , nu s-a impus cum s-a impus baiatul asta .[/quote]Eu vad ca se impune destul de bine si ea :) In general, incepand de la mine  :P toti care au fost la JBMO si au luat aur au ajuns si la IMO (nu neaparat in a 9-a).

Stie cineva rezultatele de la juniori de anul asta?

Rezultate partiale:

1. Chindea Filip
2. Ciolan Alex
7. Anghel Patricia (49 pct)
8. Onul Darius (48 pct)
9.  Filip Laurian :D (47 pct)
11.Ivanovici Stefan (44 pct)

Daca aveti si restul rezultatelor, va rog sa le postati aici.

[quote="H3llShadow"]Loku 9 EU :D (47 pct)

[/quote]
Fii un pic mai explicit :P
Stii cva de cei din Vianu?

Ye Kay s-a situat pe locul 18, iar Milu Alexandru a fost pe 10. Nu am informatii in legatura cu ceilalti.

Cred ca te intreba, printre altele, cum te cheama...

Eu stiu de Bumbacea Radu, ca a intrat in lot, pe locul 3 (apropo felicitari daca citeste acest mesaj), in rest as fi curios daca stie cineva de Iordache Bianca Teodora ce a facut.

Apropo, felicitari si tie HellShadow.

Multumesc :P . Eu sunt Filip Laurian din Arad.

Cu toate ca Bianca e in primii 6 nu va putea participa la balcaniada din cauza varstei.

Cate puncte avea radu?

Poate sa puna cineva ultimele baraje de la seniori la [color=blue]Resources[/color]? As fi foarte recunoscator.

la resources nu pot, dar
[url]http://www.mathlinks.ro/viewtopic.php?t=153375[/url]


When I do a practice state test (sprint round), I usually don't get to the last 5 problems or so. How can I improve at speed so that I'll have time to do them?

Do LOTS AND LOTS OF PROBLEMS. Remember, practice makes perfect! Being good at Countdown also helps to improve speed. You can also practice your speed on FTW.

Thanks! Just one question.

Should I do mostly hard Problems? Easy Problems? Or some of both?

You should probably do harder problems, like national or other stuff  :D

do mathcounts state-level problems because that's what u're stuck on


posted before. I give solution for $a_{n+2}=aa_{n+1}+\frac{b}{2}a_{n}$, were b is odd.


Why does it say "Hodge Conjecture" under my Avatar that I set?
Also, what does the 1/5 stars under that mean -- am I being rated somehow?

There's a reason we have a search function.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159956[/url]

That still doesnt really help. It just measures how many posts you have. Dont worry about it.


[quote="Interval"]Julio’s wages vary directly as the number of hours that he works. If his
wages for 5 hours are 29.75 dollars, how much will he earn for 30 hours?[/quote]
[hide="Definition of direct variation - y=kx"]$w=\text{wage}$
$h=\text{hours}$
$w=kh$ <- definition of direct variation ($k$ is a constant)
$29.75=k\times5$
$k=5.95$
$w=(5.95)30$
$w=178.5$[/hide]

[size=150]Different types of variations[/size]

[u]$k$ is a constant for all[/u]
Direct variation = $y=kx$ In words: y varies directly with x.

Indirect variation = $y=\frac{k}{x}$ In words: y varies indirectly with x.

Joint: $z=kxy$ In words: z varies jointly with x and y.

$400^{th}$ post!

In order to solve this problem, you'll have to use th variation equation  y= kx.  "x" will be the number of hours he works.  "k" will be the constant. And "y" will be the number of hours.

[quote="Interval"]What is the difference between direct variation and indirect variation?[/quote]
Look at my post

You could also note that in 30 hours, there are 6 groups of 5 hours, and in each group of 5 hours, he earns $\$29.75$, so in 30 hours he earns $6\times29.75=\$178.50$.


[hide]$\displaystyle \sin\frac{\pi}{9}\sin\frac{2\pi}{9}\sin\frac{4\pi}{9}$

$\displaystyle =\frac{1}{2}(\cos\frac{\pi}{9}-\cos\frac{\pi}{3})\sin\frac{4\pi}{9}$

$\displaystyle =\frac{1}{2}\cos\frac{\pi}{9}\sin\frac{4\pi}{9}-\frac{1}{4}\sin\frac{4\pi}{9}$

$\displaystyle =\frac{1}{4}(\sin\frac{5\pi}{9}-\sin\frac{-\pi}{3})-\frac{1}{4}sin\frac{4\pi}{9}$

$\displaystyle =\frac{1}{4}(\sin\frac{5\pi}{9}-sin\frac{4\pi}{9})+\frac{\sqrt3}{8}$

$\displaystyle =\frac{\sqrt3}{8}$
[/hide]


If you analyze the number "def" and the number "abc", you'll get an interesting relationship that bears fruit.  E.g. [hide]7000abc + 7def=[/hide].

darkquantum wrote:7(a b c d e f)=6(d e f a b c), where a,b,c,d,e,f are digits, not necessarily distinct.
Find a b c d e f.

[hide]Let 100a+10b+c=x and let 100d+10e+f=y. Then this problem gives us 7000x+7y=6000y+6x => 6994x=5993y.  => 538x=461y. Therefore x=461 and y=538, since (538,461)=1 and x and y must be 3 digit numbers. The big number we get by smushing together x and y, so we get a b c d e f =461538. Correct?[/hide]

you're right, that has nothing to do with the problem. simon i'm ashamed of you

1/7

[quote="ComplexZeta"]This doesn't have that much to do with the problem, but what is interesting about the number 142857?[/quote]
I have no clue as to why you're bringing this up, but 0.142857... is indeed 1/7. Not only that, but you have to "shift" the decimals for the other 5 fractions. For example, 2/7=0.285714... 6/7=0.857142...

Hehe actually it has a fair amount to do with the problem. Divide 6 by 13, and maybe (please?) you'll see what I'm talking about.

Oh I see, if you rearrange the digits of the answer around you get 1/7...so how did you get that idea? :)

not exactly

hmmm...why is that? just coincidence? because i can't think of a reason it should hold true in general...

Here's what I figured out about this. Since I came up with it on my own, it may or may not have any relationship to the truth.

Consider 1/n, where (n, 10)=1. Suppose we multiply by 10^k and then keep only the fractional part. We're going to get a bunch of fractions of the form m/n. But since all we did was multiply by a bunch of factors of 10 and then remove the integer parts, the new ones should be the same digits as 1/n but moved around cyclically. Unfortunately, sometimes the period is shorter than n-1. (For example, if n=13, the period is only 6.) That means we can start with various things /n and do the same thing and still get stuff moved around cyclically.

I explained that poorly, didn't I? Maybe someone else will understand what I'm saying and can explain it better.

well technically,
i think 000000 works fine

Does it say ``Find all solutions...'' by any chance?

It was a sophomore meet, and this is one of the easier problems. What are you talking about?

As long as we're talking about 142857... Examine the following (geometric) series:

.14 + .0028 + .000056 + .00000112 + .0000000224 + ...

The key here is that while the smallest $10^n-1$ that 7 divides is $10^6-1$, it does divide $10^2-2$.

[quote="Rodolphe2005"]14+28+57=99
142+857=999

I have study those numbers, maybe we can talk about them ?

142857=(10^(7-1)-1)/7

142857 is the first "nombre ttu" in french (i don't the english name) and it is linked to the prime number 7 by this formula:
n = (10^(p-1) - 1)/p if p is a long prime number

a long prime number verify: min(10^k = 1 (p), k in IN*) = p-1 and p!=2 and 5

try this formula for p=17 and multiply this number by an other in |[2;16]| ...
An other propertie of those numbers is that if you "cut" it in two and sum the two new numbers you have 10^((p-1)/2) - 1[/quote]

And how can we use that in terms of being practical?  Does it have any application?

I think "nombre te^tu" means "stubborn number."
Just guessing.  I don't speak french.

I don't know applications for this but it's a problem unsolved to find if 10 is a primitive root of p.


hello, the solution is $ a\equal{}b\equal{}c\equal{}0$.
Sonnhard.

Guten Tag, Dr Sonnhard Graubner. :)

Please justify your answer.

hello kunny, it is
$ x^4\minus{}2x^3\minus{}x^2\plus{}2x \ge 0 \Leftrightarrow$
$ \minus{}\infty<x \le \minus{}1,0 \le x\le1,2\le x< \infty$.
And $ 1\minus{}x^2 \ge 0 \Leftrightarrow \minus{}1 \le x \le 1$.
So the solution is $ a\equal{}b\equal{}c\equal{}0$.
Sonnhard.

O.K. :lol:

Sorry, the problem should be posted in Intermidate Forum.

kunny

Are you sure that the solution is complete? We just get, that $ x\in[0;1]$ Consider the function $ f(x)\equal{}\sqrt{x^4\minus{}2x^3\minus{}x^2\plus{}2x}\plus{}\sqrt{1\minus{}x^2}$. Than, because of$ f(0.8)\equal{}1.18787754$, and $ f(0.9)\equal{}0.869594863$, there must be such $ x_0\in(0.8,0.9)$, that $ f(x_0)\equal{}1$. It means that we have at least one another solution :$ a\equal{}b\equal{}c\equal{}x_0$. Am I right?

hello, you should solve the equation $ f(x)\equal{}1$ with respect to x.
Sonnhard.

greenvert has a valid point here, By looking at the graph (or using the intermediate value theorem) we see that there exists a point $ 0<x<1$ such that $ \sqrt{x^4\minus{}2x^3\minus{}x^2\plus{}2x} \plus{}\sqrt{1\minus{}x^2}\equal{}1$, and there exists other solutions as well. And as far as I see it your solution is not complete: all you have done is point out that $ \minus{}1\le x \le 1$ (infact it should be $ 0\le x \le 1$) which is neccesary for the radical to be defined.

hello, yes, you have right, the other solution is
$ 0.8646876844181..$ which you can calculate for example with Maple.
Sonnhard.

Actually there seems to be infinitely many solutions Sonnhard, just vary a, holding $ b,c$ constant.


看到一道题是“……圆C与圆I直交……”什么叫“直交”？

正交, 叫垂直也可以...
曲线和曲线的交角就是看交点处切线的夹角

谢谢liyi!
我记得以前我也问过什么叫两圆垂直,结果yetti 扯了一大堆......原来就是这么回事呀!  :D

哦，原来如此。thanks.


knowing that:  $ a^3\plus{}\frac{1}{a^3}\equal{}18$

calculate this:  $ a^4\plus{}\frac{1}{a^4}\equal{}18$

What exactly are we supposed to calculate?

ah!I'm sorry.
knowing that:
$ a^3\plus{}\frac{1}{a^3}\equal{}18$
calculate this:
$ a^4\plus{}\frac{1}{a^4}$

let $ a\plus{}\frac{1}{a}\equal{}k$
we will got that $ a^3\plus{}\frac{1}{a^3}\equal{}k^3\minus{}3k\equal{}18$
so $ k\equal{}3$
so that
$ a^2\plus{}\frac{1}{a^2}\equal{}7$
then
$ a^4\plus{}\frac{1}{a^4}\equal{}47$

[quote="Rose_joker"]we will got that $ a^3 \plus{} \frac {1}{a^3} \equal{} k^3 \minus{} 3k \equal{} 18$
so $ k \equal{} 3$[/quote]

Um, can't $ k$ be a root of $ k^2 \plus{} 3k \plus{} 6$? ($ k^3 \minus{} 3k \minus{} 18 \equal{} (k \minus{} 3)(k^2 \plus{} 3k \plus{} 6)$). I know that this implies that $ k$ is complex, but why can't it be? (it wasn't specifically mentioned in the problem) Anyway, my calculator-bash reveals the answer is something different in that case(not a nice number).


Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$, and $PQ$ is parallel to $AB$.  Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm.

[asy]
size(250);
pair A=origin, B=(96,0), C=(96,22), D=(0,22), W=(16,22), X=(20,0), Y=(80,0), Z=(76,22), P=(24,11), Q=(72,11);
draw(P--X--A--D--W--P--Q--Y--B--C--Z--Q^^W--Z^^X--Y);
dot(A^^B^^C^^D^^P^^Q^^W^^X^^Y^^Z);
pair point=(48,11);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NE);
label("$D$", D, NW);
label("$P$", P, dir(point--P));
label("$Q$", Q, dir(point--Q));
label("$W$", W, N);
label("$X$", X, S);
label("$Y$", Y, S);
label("$Z$", Z, N);[/asy]

[hide]We have that $[WZQP] = [XYQP]$ and $PQ || XY$, $PQ || WZ$. Also $XY = WZ$.
So the altitude of $XYQP$ equals the altitude of $WZQP$ and it's equal to $\frac{19}{2}$.
So $[ABCD] = 19AB = 19(XY + 87)$. So $AB = 87 + XY$.
On the other hand $2AB = AX + WD + YB + CZ + 2XY = XY - 19 + XY - 19 + 2XY = 4XY - 38$. So $AB = 2XY - 19$.
Solving the system we get $AB = 193$.[/hide]

Sorry for bumping this up, but for the $\left[ABCD\right] = 19AB = 19(XY+87)$ part, how do we know that $19AB = 19(XY+87)$? How did we know $AB$ is $XY+87$?

This reply is in response to a question made by ark_xm in [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=771699#771699[/url]

[quote="ark_xm"]Here's my question: for the $\left[ABCD\right] = 19AB = 19(XY+87)$ part, how do we know that $19AB = 19(XY+87)$? How did we know $AB$ is $XY+87$?[/quote]

Andreas expressed the area of the rectangle ABCD in two ways: one way is base times height obviously, and the second way he used the fact that the rectangle is divided into four parts of equal area and 4 times the area of trapezoid XYQP, in other words,

$[ABCD]=BC\cdot AB=4\cdot\frac12\left(\frac12BC\right)(XY+PQ)$.

Extend PQ to meet AD and BC at R and S. Then $ [AXWD] \equal{} 19/2 (\frac {DW \plus{} PR}{2} \plus{} \frac {AX \plus{} PR}{2}) \equal{} 19/2 (\frac {XY \minus{} 19}{2} \plus{} PR) \equal{} [XYPQ] \equal{} 19/2 \cdot \frac {XY \plus{} 87}{2}$
XY cancels, and we get PR=53. By the same logic, QS=53, so the answer is 53+53+87=193.


MATHCOUNTS
geometry
3D geometry
AMC
AIME
videos
inequalities

So Bluestalk claimed that a discussion topic would still be helpful despite everyone apparently abandoning AoPS, but that we only need one for now. So, until/unless this thread grows overly long, which I doubt it will, all discussion from week 8 onward goes here. If there's any.

P.S. Junggi and moogra, either contribute in a mathematical way or shut up. Start your own thread if you think whatever you were arguing over is still worth arguing over.

suggestion:
maybe we should announce a mathcounts aim chat in a chatroom (have you used this feature?). Since there are many users of AIM, this might attract more members to discuss the hw problems. Also, just remind everyone about AoPS at mathcounts. I hope you are doing a good job on them.

btw the cube in my avatar is impossible. where the green side is it should be blue :)

[quote="bookaholic"]
P.S. Junggi and moogra, either contribute in a mathematical way or shut up. [/quote]
you might have noticed that it stopped when ccy deleted all spam....(not saying moogra\ne idiot.)
anyway, I suggest you have some kids take the new mock amc(seperately at home, probably.)(any mock amc, for that matter). Yea it might be a bit screwy, it was made by a guy getting about 5-8(I think?) on AIME, and it's pretty calc dependent, but it could be good practice(well but I don't know what good mc practice is). You could look at it first, and have them take it if it looks okay. and, heh, I pdfed/edited it lol. and this could make some of them actually get aops accounts.

edit:better yet, have them go to mathcounts forum, and try out some mock mathcounts/competitions. that might get them to stay on more

And for countdown (if anyone needs practice for that useless round), the countdown rounds that were done over aim REALLY helped.  Not sure if they are still going on... :roll:

Actually another unsuggested way is to make accounts for people FOR THEM. Then they might come on here, and since they are willing to go on mathcounts, they may go here. It might fail because this isn't nice, and many people don't like forums (i.e. Brian Wang, Jim, Janice, Boris, and many more)

-edit-
instead of just giving suggestions, heres a good site for mathcounts preparation:
http://mathcounts.saab.org/mc.cgi
As for countdown, we can the old countdown round from ms beddoes. Those were really fun, and everyone was happy about the round. This might encourage more members

BOOKAHOLIC, when are you posting solutioNS???

2006 amc 8 school honor roll
[quote="AMC site"]
Black Pine Circle School  	Berkeley  	CA  	US
Carmel Valley Middle School 	San Diego 	CA 	US
Chadwick School 	Palos Verdes Penn 	CA 	US
Cupertino Middle School 	Sunnyvale 	CA 	US
De Portola Middle School 	Tarzana 	CA 	US
EPGY-Stanford University 	Stanford 	CA 	US
Harvard Westlake School 	Los Angeles 	CA 	US
Harvest Park Middle School 	Pleasanton 	CA 	US
Hopkins Junior High School 	Fremont 	CA 	US
Horner (John M ) Junior High School 	Fremont 	CA 	US
Hyde Junior High School 	Cupertino 	CA 	US
[b]Jordan (David Starr) Middle School 	Palo Alto 	CA 	US[/b]
Miller Middle School 	San Jose 	CA 	US
Palos Verdes Intermediate School 	Palos Verdes Estates 	CA 	US
Redwood Middle School 	Saratoga 	CA 	US
South Pasadena Middle School 	South Pasadena 	CA 	US
[b]Terman Middle School 	Palo Alto 	CA 	US[/b]
The Harker Sch Blackford Cps Middle 	San Jose 	CA 	US
Warren (Earl) Middle School 	Solana Beach 	CA 	US[/quote]
Jordan's there....Terman's there.....WHERE'S JLS
(or is it just not added there yet)
Do you guys know if you guys got 66>?

[b]JLS ISN'T ON THE HONOR ROLL??[/b] [i]I THOUGHT WE HAD AT LEAST THREE 22's.....[/i]
Well, I doubt that we got into the honor roll since it isn't listed. I mean, if they have Terman AND Jordan, why wouldn't they have JLS? It's probably because we didn't make it...(yeah I know I have so much school spirit =P), but as of last week, I thought there were at least three 22's-- [b]AT LEAST[/b]. I guess I'm wrong. People probably just bubbled in the wrong bubble even though they had the correct answer on their booklet thing.

All I know for sure is that Jordan got two 25's (John and Nathan, DUH), 584320503475 people got 23's, and the rest of the people (in my math class at least) got 22's, 19's, 17's, 13's. And Terman...I have no idea.

Yay! We're on there. We only had one 25, two 24's, and one 23 (me). :ninja:

I have a feeling(just a feeling) that jls might not make state  :ninja:
(yea that's supposed to be a provocative comment thing)

sad... jls is going to fail this year... who got the 3 22's? It could just be that the AMC people haven't updated yet. It certainly updated from last time I checked

Welllll...
http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2006-8a/2006-amc8schHonorRoll.shtml
[quote]This is a [b]preliminary[/b] list of the AMC 8 Participating Schools who have qualified for the AMC 8 Honor Roll. This Roll is achieved by attaining a score of 66 through 75 when adding together the the top three student scores (within a school). [As answer forms are run and checked, additions will be made to this list - last update: 12/5/06 ][/quote]

So there's still a chance of a comeback there..

well good job ccy. Still a chance. What happened to JLS anyway? It was like the best last year... Still isn't that what I said? It could just be that the AMC people haven't updated yet. It certainly updated from last time I checked. :)

[quote="moogra"]well good job ccy. Still a chance. What happened to JLS anyway? It was like the best last year... [/quote]
yea no thanks to you

But now really, 3 22's? cmon you guys can do better than that.....(hopefully)
We're pretty much expecting you guys to make state, so if you think you can't, you know what to do

p.s. hey bookaholic, how's w00t pset going?

[quote="bookaholic"]Okay, the party will be next Tuesday, 5/22. You guys at Gunn should come, but only if you bring food. Bluestalk, are you going to be busy? If so, will you at least have time to fill out the survey?[/quote]
lol
sorry thats when we have our debate party

Darn. You guys are ending already? Our debate buffet dinner won't be until June 8.

[quote="bookaholic"]Darn. You guys are ending already? Our debate [b]buffet[/b] dinner won't be until June 8.[/quote]
WTF?
OMG
DLGKJADSGLKAJDSGLKASDGASLDKGJADSG
wait you're in debate?

Apparently one of the debaters has an insanely generous parent...

I didn't debate this year, but Paly will have a policy team for the first time next year, and decided that people who want to debate policy next year should start meeting this year. So I'm technically part of the team, even though I haven't done anything except go to meetings and listen to the coach talk about S[size=59]ignificance[/size]H[size=59]arms[/size]I[size=59]nherency[/size]T[size=59]ext[/size]S[size=59]olvency[/size].

[quote="ccy"]POLICY??? ARE YOU OUT OF YOUR MIND??? haha :o  My god, that's scarier than insanely speedreading circuit-LD. :ninja:

Haha Gunn doesn't have a policy team at all due to "lack of interest".[/quote]
except you and pamudh
but hoon had to ruin it all for some reason i forgot
dang it wouldve been SOOOOOOO FUNNY to watch you 2 in policy...
=(

Wait...why is policy scary? Other than the fact that I have no idea how I will ever begin to actually be able to debate it?

there's nothing wrong with it
except for the fact that its competitive debate, which by definition sucks. [i]a lot[/i]
apparently the gunn debate team doesn't like policy bc they seem to think everyone in policy reads 10000000 words per second
which i dont quite want to believe
but eh ok

I don't get it. LD isn't competitive?
It's true that our coach will go "THEPLANWILLLEADTOGLOBALNUCLEARWARTHEIMPACTISEXTINCTION" and call that "slow" style, leaving what "fast" style is supposed to be up to the imagination...
But the neat thing about policy is that you spend a LOT of time thinking up new ways to wipe out the human species. Apparently writing policy cases consists of coming up with a plausible causal chain between the opponent's position and extinction. Um. It's great.

LD is competitive
which is why it also sucks
[i]a lot[/i]

hmm policy sounds like a lot of fun

i'm gonna try to join a debate team in high school, it sounds fun
i like arguing :P

[quote="Ubemaya"]I can't stand doing anything in front of an audience.[/quote]
the "audience" consists of a judge, your opponent, and several empty desks(depending on what room youre in)

@daermon:ohhh youre in for a treat
competitive debate is not about arguing
there's only 2 things important in debate
1)how fast you can read
2)how much of your life youre willing to sacrifice
and thats called circuit style debating
the league style debating or whatever is dependent on only one factor and that is
1)how well you speak
as you can see it has nothing to do with actual arguing

[quote="Bluestalk"]q/a about clubs? Because...
1. Is debate club worthwhile *to you guys*?
2. WHY IS THERE A TIC TAC TOE CLUB???
3. Are most clubs time consuming?[/quote]
1. uh, if it's just doing debate, then its a complete waste of time, but the club itself is really fun
like the presidents this year, awesome dudes, funny and such
yea
2. some people have no lives =(
3. no. see: tic tac toe club
most clubs arent time consuming at all.
its just the ones that involve contests that are time consuming. like math, debate, quiz kids, etc.


It is really hard to read chatspeak, like ppl, nd, u, cuz, etc. It obscures what you are trying to say and makes you seem like you are immature and a newbie, since experienced users write out everything. Please spell out your words in the future.

Thanks!!

Billy

[quote="sugarwergoingdown"]:( dang i use shthand alla time....y cant we use cuz nd etc? ppl unnerstand them...
nd hows it immature its jus writing faster....saves time!  ;)[/quote]

Ok, what does "alla" mean? "All the?" Most people on the forums don't feel like translating words into proper grammar and it's immature as that's what you'd expect a third grader typing with his elbow to write.

Please don't use chatspeak... I don't know many of the abbreviations, and it makes it hard to read people's posts. Since this is a forum, you do not need to type fast to say something like in an IM or chatroom. It doesn't take that much longer to write words out ;) .

By the way, what does 13375p33|< mean? I've looked at it sideweays and upside-down and still haven't figured it out.

Why does it matter if there's a list of definitions? It still makes the poster, and AoPS in general, seem puerile and jejune. Spell out your words - it helps develop your typing skills.

Locked.

Billy


inequalities
calculus
calculus computations

They're all of the form $\sum_{n}c_{n}w^{n}$, where $w=w(x)$ is some expression involving $x.$ They're power series in $w,$ so we can bring to bear all of our usual methods for dealing with power series to find a radius of convergence (for $w$) and even convergence on the boundary. Then we simply replace $w$ by whatever it is in terms of $x.$ This is exactly what diogene has done, although in cases 2) and 3) he has not taken the last step of "solving" those inequalities for $x.$


Let $ \{f_k\}$ be a bounded sequence in $ L^p(\mathbb R^n)$, $ 1<p<\infty$. Is it always true that there exists $ f \in L^p(\mathbb R^n)$ and a subsequence $ \{f_{n_i}\}_i$ that converges to $ f$ almost everywhere?

Any ideas?  :(

[quote="Yustas"]No: choose something like $ f_n(t) \equal{} e^{int}, \ t\in[0,T]$ and use Riemann-Lebesgue lemma. If it were true, it would imply sequential compactness of the unit ball in $ L_p$, which is not true.[/quote]

Here we may prove by contraction using the unit ball in $ L_p$ is not sequential compact and Bounded Convergence Theorem is enough. I don't know what the sequence $ f_n(t) \equal{} e^{int}, \ t\in[0,T]$ play the role in here?  :)


function
logarithms
analytic geometry
graphing lines
slope
calculus
calculus computations

I have two problems which I am having a hard time solving. I don't know if this is the right forum, but please move if it is not.

Problem 1:

The table below shows lead emissions "L" in millions of metric tons from 1970 to 1990. Here "t" is years since 1970.

t - L
0 -  199.1
5 - 143.8
10 - 68.0
15 - 18.3
20 - 5.1

A) Use exponential regression to find an exponential model for the data.
B) What is the yearly decay factor and percentage decay rate?

Problem 2:

The table below shows the relationship between the length "L" in centimeters and weight "W" in grams of a certain species of fish.

L - W
31 - 250
33 - 308
35 - 363
37 - 420
39 - 520

A) Find a formula that models W as a power function of L.
B) If one fish is twice as long as another, how do their weights compare?
C) If one fish is twice as heavy as another, how do their lengths compare?

[quote="Starz4eva2"]
A) Find a formula that models W as a power function of L.
[/quote]

What is a power function? Is it simply a function of the form y = a*x^z, where a is a constant and x is the independent variable raised to the power z? In that case, and for your first question too I think, a good first step would be to take the logarithm of the dependent variable. Then plot the data on graphs. Hopefully the graphs will indicate whether the relationship is linear or quadratic. If it is, then the job is pretty much done. If not, you could do a linear least-squares fit to model the data; the slope of that line would give you the power coefficient.


geometry
geometric transformation
reflection
LaTeX
ARML
MATHCOUNTS
email
\/closed

First, there's a VIP status at the bottom (on the "who's on" list). What's that for?

Second, on the old forum (on the main forum index) you could click anywhere inside the box for that forum - not necessarily the name - and it would go there. Basically, the entire box was a link. Could we have that back?

Thirdly, the "quote" button is now at the bottom next to the contact info for the poster. I personally think it's better where it used to be, along with the other options concerning that post. Just a suggestion though.

And one more thing...would it be possible to enable poll-making for all users on all forums? I can definitely understand disabling them for Site and General Info boards, but otherwise I'd like to request getting them back if none of the admin have anything against doing so.

Sorry if it sounds like I'm complaining. The new forum looks and feels great. :-) Thanks for your efforts

Another important thing: The "SOURCE" fields of the old messages are not shown when I browse the subforum (e. g. "Geometry Proposed&Own Problems").

  darij

[quote="MithsApprentice"]Sorry if it sounds like [u]I'm complaining[/u]. The new forum looks and feels great. :-) Thanks for your efforts[/quote]It sure does  :lol:

VIP status will probably be given to some of the people on the board that are not moderators, however they are earned merits. It will not have any privileges other than a different color from the all users.

All the other viewable things will be back. Those are minor things to be added. For now more important is the overall functionality of the board :) in both styles.

[quote="MithsApprentice"]Thirdly, the "quote" button is now at the bottom next to the contact info for the poster. I personally think it's better where it used to be, along with the other options concerning that post. Just a suggestion though.[/quote]

I couldn't agree more with this.  I notice that the "quote" button occurs [i]before[/i] the ICQ button for the users who display their ICQ number, and I'm pretty sure that is certainly not supposed to be the case.  Furthermore, it doesn't make much sense to put the "quote" button down there.

[quote="Valentin Vornicu"]VIP status will probably be given to some of the people on the board that are not moderators, however they are earned merits. It will not have any privileges other than a different color from the all users.[/quote]

Speaking from my experience at another forum where VIP status was also available, there's not much purpose in giving special colours to some users if they're not on the admin team.  There's a tendency for people to wonder exactly what one has to do to get the "colours", which kind of misses the entire point.  I'm sure the people who have contributed a lot to this forum won't mind sticking with the normal colours.

I notice that the username/details column is quite a bit wider than before.  Is it possible to reduce it back to the previous size, which I think looked aesthetically better?  The flag seems rather squashed below the "location" field, so perhaps you could insert another line in between.

Once again, the MathLinks section has been reduced to two small sections on the index page.  Can we have the more detailed section of before, and is it possible to put back the "+" and "-" that we could click on each category to either show or hide it?  I think the Config option in the previous version worked quite well.

I realise that I seem to be complaining a lot too :blush: .  Sorry!

A small request for now: Is it possible to remove the section which lists all the users in the past 24 hours?  For one, it's certainly unnecessary, and it's also quite a nuisance to look at. :(  I think the one which reflects online users is sufficient.

The [b]visual[/b] appearance of both Styles is not yet finished. Please stop complaining about it untill it's finished :lol:

At this point you cannot shut the viewonline box, but it may become an option for users in the future. I will consider that, but right now you'll just have to live with it. Just don't scroll down there ;) ... btw the portal is working now, so you can see there only the current users online.

The + and - have to be re-thinked programming-wise because of the (completely) new structure of the forum.

"Mark all topics read" appears under Forum Index - just go up and after first two subforums there's a line called Forum Index and "Mark all topics read" on it  :)

[quote="billzhao"]Whatever happened to the "Mark all topics read" on the main forum index?

And as for LaTeX, how come the $\backslash [$ $\backslash ]$ tag nolonger centers the equations? Is this change on purpose?[/quote]The Mark all topics read is on  right of the upper navigation bar.

They will be centered, I'll fix that.

[quote="TripleM"]I'm not sure if this used to be true on the mathlinks style board before, but previously on the AOPS one when clicking what forum you wanted to go to, you could click anywhere within the box instead of just on the link - this doesn't seem to happen any more. Any chance of turning this back on, it was very useful.[/quote]First of all it's the same board, just the styles are different (ie images and colors).

Second of all I asked everyone to stop complaining about visual things, untill we haven't announced them finished. They will be fixed. For now the main issue are safe resuming of all posting activities, and security ;)

[quote="RC-7th"]Just to be clear, I'm [u]not[/u] complaining. When the symbol besides a thread has a red dot on it, you've posted in it. What does a greyish dot mean?[/quote]It means the topic is also hot (ie there are at least 15 replies on that particular topic).

[quote="Valentin Vornicu"]
Second of all I asked everyone to stop complaining about visual things, untill we haven't announced them finished. They will be fixed. For now the main issue are safe resuming of all posting activities, and security ;)[/quote]

Complaining? It was a question, not a complaint  :? I think the new look is great..

[quote="TripleM"]I think the new look is great..[/quote]That's the spirit :P :P

[quote="zmli"]Since the forums list is getting longer and most of us will be only interested in few of the forums, it there a way to have a My Forums option so that when clicked, only the forums that I am interested in regularily participate will be listed.  Certainly will eliminate frustrations in hunting for my favorite forums.[/quote]The forum remembers your settings in categories collapsed or not. You can just close everything that you are not interested in, and you don't see them.

[quote="zmli"]Well, that will not work.   For example, here might be the only forum that I will be interested in and want it displayed as a default when I get into the forum:

My Forums:
- San Diego Math Circle
- Southern California ARML
- MATHCOUNT
- Online Classes

I can then quickly check each of the forums to see new messages.[/quote]Subscribe to each of the forums. You'll get emails when a new post is made, with a link to it. We do not have the time to even try to do what you're suggesting at this moment (it's definitely not an easy thing to do).


LaTeX
floor function
function
algorithm
continued fraction
algebra

Can somebody please tell me what continued fractions are? I heard about them somewhere, but I don't know what they are.  :maybe:

continued fractions are like

$ \cfrac{2}{1 \plus{} \cfrac{2}{1 \plus{} \cfrac{2}{1 \plus{} \cfrac{2}{...}}}}$

if you wanna solve example, let $ \cfrac{2}{1 \plus{} \cfrac{2}{1 \plus{} \cfrac{2}{1 \plus{} \cfrac{2}{...}}}}$ be x.

then it becomes $ x \equal{} \frac {2}{1 \plus{} x}$, which is $ x \plus{} x^2 \equal{} 2.$

so its $ x^2 \plus{} x \minus{} 2 \equal{} 0.$ $ (x \plus{} 2)(x \minus{} 1) \equal{} 0$, and x=-2 or 1.

it can't be negative, so the value is 1.

Oh, I see! What would you call something like this *LaTeX challenge, wish me luck*:

$ \sqrt{12\minus{}\sqrt{{12}\minus{}\sqrt{12...}}}$

*Yay! I did it!*

hmm i don't know about that...

anyway if you want to solve that example, let $ \sqrt{12\minus{}\sqrt{{12}\minus{}\sqrt{12...}}}\equal{}x$

then its $ x\equal{}\sqrt{12\minus{}x}$

so $ x^2\equal{}12\minus{}x$.

and $ x^2\plus{}x\minus{}12\equal{}0$.

if you factor it, it becomes $ (x\plus{}4)(x\minus{}3)\equal{}0$.

it can't be negative, so it's 3.

[quote="isabella2296"]Oh, I see! What would you call something like this *LaTeX challenge, wish me luck*:

$ \sqrt {12 \minus{} \sqrt {{12} \minus{} \sqrt {12...}}}$

*Yay! I did it!*[/quote]

It is called a [url=http://mathworld.wolfram.com/NestedRadical.html]nested radical[/url] ;) But you can call it continued square root-it is also a common name for it.

Another term you might come across is complex fractions. This just means that there is a fraction in the numerator or the denominator. (No imaginary numbers involved.)

Complex fractions? Oh, like this:

$ \frac {\frac {\frac {1}{11}}{111}}{1111}$

which simplifies to this:

$ \frac {1}{1356531}$

Right?

You should know how to write expansions of fractions such as $ \frac{127}{22}$ as well.
For example, you can write $ \frac{127}{22}\equal{}5\plus{}\frac{1}{1\plus{}\frac{1}{3\plus{}\frac{1}{2\plus{}\frac{1}{2}}}}$

A continued fraction in which numerators of all fractions are 1's are called proper.
You can also write continued fractions for irrational numbers as well, like $ \sqrt{23}$ or $ \pi$, etc.

[quote="xpmath"]You should know how to write expansions of fractions such as $ \frac {127}{22}$ as well.
For example, you can write $ \frac {127}{22} \equal{} 5 \plus{} \frac {1}{1 \plus{} \frac {1}{3 \plus{} \frac {1}{2 \plus{} \frac {1}{2}}}}$
[/quote]

How'd you get that?  :huh:

The basic idea is that an expression in the form $ 5 \plus{} \frac{1}{\text{stuff}}$ has the property that the second term is less than $ 1$.  So if you know that the above expression has value $ \frac{127}{22}$, you know that

$ 0 < \frac{127}{22} \minus{} 5 < 1$;

in other words, $ 5 \equal{} \lfloor \frac{127}{22} \rfloor$ (the [url=http://en.wikipedia.org/wiki/Floor_function]floor function[/url]), which is true.  Then

$ \frac{127}{22} \minus{} 5 \equal{} \frac{127 \minus{} 110}{22} \equal{} \frac{17}{22} \equal{} \frac{1}{\text{stuff}} \implies$
$ \frac{22}{17} \equal{} \text{stuff} \equal{} 1 \plus{} \frac{1}{\text{more stuff}}$

and again since the second term is less than $ 1$ the only way this can work is if $ 1 \equal{} \lfloor \frac{22}{17} \rfloor$, which is true.  This repeated taking of the floor, subtracting, and taking the reciprocal eventually stops if the original number you're working with is rational.

i think there is a algorithm
but i think its like this:

$ \frac{127}{22}\equal{}\frac{110}{22}\plus{}\frac{17}{22}\equal{}5\plus{}\frac{17}{22}\equal{}5\plus{}\frac{1}{\frac{22}{17}}\equal{}5\plus{}\frac{1}{\frac{17}{17}\plus{}\frac{5}{17}}\equal{}\ldots$

Oh, I think i get it...Thanks, everyone, for the great help!


real analysis
calculus
derivative
real analysis unsolved

This problem is from Rudin, Chapter 8, Problem 22.  This is one of Rudin's hinted steps for proving Newton's Binomial Theorem.

Let $ f(x) \equal{} \sum_{1}^{\infty} \frac {\alpha(\alpha \minus{} 1)\cdots (\alpha \minus{} n \plus{} 1)}{n!}x^{n}$

Prove that $ (1 \plus{} x)f'(x) \equal{} \alpha f(x)$

I arrived at $ (1 \plus{} \alpha)s_{n} \equal{} (1 \plus{} x)f'(x)$, where $ s_{n}$ is the sum.

However, I am stuck at this point.

Where $ s_n$ is the sum of what?

There are certain things you need to do to prove that differentiation term-by-term is valid; prove them.  What's left to do after this?

We need $ f'$ to converge uniformly for $ \minus{} 1 < x < 1$ (This is one of the conditions given; my mistake in not stating this in the original post.)

But we can just use Weierstrass M-Test to show this.


[b]exponential/logarithmic function[/b]

solve for x:
1. 2 * 9^(x-1) = ((1/4)^x) / 27

2. 2log[size=75]3[/size] (x + 3) - log[size=75]3[/size] (x + 4) = 0

thanks. is there anyone who has an answer to no. 2?

#2.
$2\log_{3}{(x+3)}-\log_{3}{(x+4)}=0$
$\Leftrightarrow\log_{3}{(x+3)^{2}}=\log_{3}{(x+4)}$
$\Leftrightarrow (x+3)^{2}=x+4$

We have now a 2nd degree equation $x^2+5x+5=0$ with roots $x_{1}=\frac{-5+\sqrt{5}}{2}$ and $x_{2}=\frac{-5-\sqrt{5}}{2}$

I don't think the first is correct solved. :)

[quote="Slizzel"]I don't think the first is correct solved. :)[/quote]

I think it's solved correctly  :? I checked his answer.

Yeah. Sorry I though that $2^1=4$ :rotfl:. Apologize :D!


calculus
integration
logarithms
calculus computations

show :)

$ \int_0^1\int_0^1\frac{1}{(\phi\minus{}xy)\cdot\ln\,(xy)}\;dx\;dy\;\equal{}\;\boxed{\ln\,\left(\frac{1}{\phi^2}\right)}$

where $ \phi\equal{}\frac{1}{2}(\sqrt 5 \plus{} 1)$

Claim: If $ a > 1$, then $ I \equal{} \int_{0}^{1} \int_{0}^{1} \frac{dx dy}{(a \minus{} xy)\ln (xy)} \equal{} \ln\left( 1 \minus{} \frac{1}{a}\right).$

Use the substitution $ (x, y) \equal{} (e^{\minus{}u}, e^{\minus{}v})$, we have $ I \equal{} \minus{} \int_{0}^{\infty} \int_{0}^{\infty} \frac{e^{\minus{}(u\plus{}v)} \, du dv}{(a \minus{} e^{\minus{}(u\plus{}v)})(u\plus{}v)}$. Again applying the substitution $ (z, w) \equal{} (u\plus{}v, u\minus{}v)$,

$ I \equal{} \minus{} \int_{0}^{\infty} \int_{\minus{}z}^{z} \frac{e^{\minus{}z}}{(a \minus{} e^{\minus{}z})z} \, \frac{dw dz}{2} \equal{} \minus{} \int_{0}^{\infty} \int_{\minus{}z}^{z} \frac{e^{\minus{}z}}{(a \minus{} e^{\minus{}z})} \, dz \equal{} \ln \left( 1 \minus{} \frac{1}{a}\right).$

Applying this,

$ \int_{0}^{1} \int_{0}^{1} \frac{dx dy}{(\phi \minus{} xy)\ln (xy)} \equal{} \ln\left( 1 \minus{} \frac{1}{\phi}\right) \equal{} \ln \left( \frac{\phi^2 \minus{} \phi}{\phi^2} \right) \equal{} \ln \left( \frac{1}{\phi^2}\right)$

since $ \phi^2 \minus{} \phi \minus{} 1 \equal{} 0$.


let $ABC$ be a triangle.and $G$ be the centroid of it.line $s$ passes through $G$ and intersects $AB$ and $AC$ in points $M$ and $N$ respectively.prove that \[ AN \times MB +AM \times NC=AM \times AN \] (if $AN=-NA,...$) :D

[quote="ashegh"]i think by using petalemy theorem we can solve it... :)[/quote]
yes i think it will be useful. :D

:huh: Petalemy? do you mean Ptolemy's Theorem?

[quote="Arne"]Yes. Always remember that some (Iranian) people use different names here:

Fishagoures = Pythagoras
Ptolemy = Petalemy = Batlamious

:D Amusing[/quote]
yes sometimes all of us even you Arne will make mistakes. :D


termination of transcription

"In eukaryotes, however, the pre-mRNA is cleaved from the growing RNA chain while RNA polymerase II continues to transcribe the DNA"

this comes from pg 317 of BIOLOGY Campbell, Reece 7th edition

why does RNA pol II continue to transcribe the DNA even though the pre-mRNA is already completed and what happens to this extra RNA transcription when the RNA polymerase eventually falls off (termination of transcription)?

Past the quoted section, it reads, "Then at a point 10 to 35 nucleotides downstream from the AAUAAA signal, proteins associated with with the growing RNA transcript cut it free from the polymerase, releasing the RNA."  And then it transcribes hundreds of nucleotides past the release site.

I don't know for sure but I would imagine that this is just do to some intrinsic property of the polymerase and does not have a function.  This is supported by the fact that prokaryotes immediately stop transcription when the RNA is released.

Are you talking about the pre-mRNA being released, cut free, from the RNA polymerase after it transcribes the polyadenylation signal? If so, as you said, the RNA polymerase continues to transcribe (I don't know what the product of this transcription is though. Do you?) hundreds of nucleotides until it falls of the template DNA by some unknown mechanism - that is when transcription stops, not when the pre-mRNA is released...if that is what you are saying.

oh yea nvm...I see what you are saying....what i just described is what eukaryotes do (the whole post pre-mRNA transcription thing) but prokaryotic transcription does end when the polyadenylation signal is transcribed.

so yea the extra eukaryotic transcription probably has no function.

I'm pretty sure that at this moment, the origin and functions of this excessive transcription are still not well-understood.


A box contains 5 white balls and 6 black balls.  Five balls are drawn out of the box at random.  What is the probability that they all are white?

[hide]This is simply $ \frac{5}{11} \times \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} \times \frac{1}{7}\equal{}\frac{1}{462}$.[/hide]

[hide]Alternatively, there are 5c5=1 ways to choose the white balls and11c5=462 ways of picking all balls. 1/462.[/hide]

[hide]the first ball has 5/5+6 chance of being white.        5/11
the second has 4/4+6 chance of being white           2/5
the third has 3/3+6 chance of being white.           1/3
the fourth has 2/2+6 chance of being white.          1/4
the last has 1/1+6 chance of being white.            1/7

5/11*2/5*1/3*1/4*1/7 simplifies down to 1/11*2*3*7, which is 1/462   !

 :surf:  :starwars:  :first:  :ddr:[/hide]


What does a catalyst in a chemical reaction? I read wikipedia and the information is not very detailed to satisfy me. It basically says:

[quote] This means that catalysts reduce the amount of energy needed to start a chemical reaction. Molecules that would not have had the energy to react or that have such low energies that they probably would have taken a long time to react are able to react in the presence of a catalyst. Thus, more molecules that need to gain less energy to react will go through the chemical reaction.[/quote]

[quote="g1sk8terz"]catalysts make the reaction faster...because more molecules are available for the reaction..[/quote]

not a good defenition...

most catalysts speed up reactions by lowering the Activation Energy neccessary ($E_a$),for the reaction.

some other catalysts speed up reactions by providing a surface in which the reaction occurs.. (Pt for example).

But how do they lower the activation energy?

A catalyst does not lower the activation energy of a reaction.  It provides an alternate way for the reaction to happen which has a lower activation energy.

How that happens though.. I have no idea.

[quote="SnowStorm"]A catalyst does not lower the activation energy of a reaction.  It provides an alternate way for the reaction to happen which has a lower activation  .[/quote]

your first statement is absolutly wrong! ;)

[quote="PPonte"]No, SnowStorm is completly right.

You may read this:
http://www.chemguide.co.uk/physical/catalysis/introduction.html#top[/quote]

no! not quite completely...
remmeber i said the first statement:
[quote]A catalyst does not lower the activation energy of a reaction[/quote]
is false!

the following equation(which is commonly known as Arrhenius equation ) can be used to determine how the catalyst effects the activation energy of a reaction

$\ln{(\frac{k_1}{k_2})}=\frac{E_0}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

I think you guys are just misunderstanding each other.  SnowStorm was correct in saying that a catalyst doesn't truly reduce the activation energy of a reaction.  What it does is provide an alternate pathway which has a lower activation energy so that in essence, the reaction is able to proceed with a lower activation energy, but the activation of the original reaction is not altered.

and in response to the original question, catalysts can catalyze by acting as a surface to react on as already stated.  They can also work via the lock and key mechanism (mostly in biological pathways) in which a reactant binds to an enzyme at its active site.  The enzyme then is able to strain certain bonds so that they are easier to break, thus allowing the chemical reaction to happen.  Another way catalysts work are by actually participating in the reaction (very typical in organic chem).  According to the def of catalysts, they can't be consumed in reactions, and they're not b/c they are restored to their original form at the end of the reaction.


We all seem to have problems with not reading the questions carefully enough.  What can we do to stop making these mistakes!  I'll start this topic with 2 suggestions, then you can add your own as well.

1.  Read [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]Stop Making Stupid Mistakes[/url]

2.  Underline important information as you read the question.

What suggestions do you have?

eh, I missed a perfect on a practice chapter cause I missed one word: "even"


Let ABC be a triangle with bisector CD and height AH. AH and CD intersects at point E. If AD=5 and DC= 25 then DE= ?

you're given too little information. under the given info, the $ DE$ isn't uniquely determined.


Hello peoples:
I need help on a problem. If you know the solution, could you please post it, and tell me how to do it? Here is the problem:
"The product (66)(9)(22)(39) has a prime factorization of the form (2A)(3B)(11C)(13D). What is the value of ac-bd? "

  {Note: 2a actually represents "2 to the power of a", same goes for 3b,11c, & 13D}

PLEASE HELP ME!   :?:  :!:  :!:  :!:

Hint: Write out the Prime factorizations of each of the multiples

[hide="Solution"]
So we have

$ 66 \equal{} 11 \cdot 3 \cdot 2$
$ 9 \equal{} 3^{2}$
$ 22 \equal{} 2 \cdot 11$
$ 39 \equal{} 13 \cdot 3$

So it can be written as $ 2^{2}3^{4}11^{2}13$

So $ a \equal{} 2$
$ b \equal{} 4$
$ c \equal{} 2$
$ d \equal{} 1$
and $ ac \minus{} bd \equal{} 0$
[/hide]

Thanks sniped you, I really appreciate it. Your really smart !


number theory
prime numbers
number theory open

Hi!

I'm struggling with the following two problems:

1)
Do four different prime numbers $ a$, $ b$, $ c$ and $ d$ exist, such that $ a^3\plus{}b^3\equal{}c^3\plus{}d^3$?

2)
Do four different prime numbers $ a$, $ b$, $ c$ and $ d$ exist, such that $ a^4\plus{}b^4\equal{}c^4\plus{}d^4$?

Any help is appreciated!

/Kon


Se considera cercurile $\ C(O_1,r_1),\ C(O_2,r_2),\ C(O_3,r_3)$, avand raze diferite, care sunt tangente unei drepte $\pi$.O dreapta $\psi$, paralela cu $\pi$, determina pe cele 3 cercuri segmente de lungimi nenule
$\ l_1,\ l_2,\ l_3$.
$\ a)$ Sa se determine pozitia $\psi_0$ a dreptei $\psi$ astfel incat suma $\ l_1^2+\ l_2^2+\ l_3^2$ sa fie maxima
$\ b)$ Sa se arate ca $\psi_0$ trece prin centrul de greutate al triunghiului $\ O_1O_2O_3$.
                                        $\ Mihaela\ Sterpu,\ Craiova$

Asta e nerezolvata de mult...are cineva vreo solutie?

[hide="Generalizare."]Se considera cercurile $\ C(O_{1},r_{1}),\ C(O_{2},r_{2}), ...,\ C(O_{n},r_{n})$, $n\in N, n>1$, care sunt tangente unei drepte $\pi$. O dreapta $\psi$, paralela cu $\pi$, determina pe cele $n$ cercuri segmente de lungimi $\ l_{1},\ l_{2}, ..., \ l_{n}$.
$\ a)$ Sa se determine pozitia $\psi_{0}$ a dreptei $\psi$ astfel incat suma $\ l_{1}^{2}+\ l_{2}^{2}+...+\ l_{n}^{2}$ sa fie maxima.
$\ b)$ Sa se arate ca $\psi_{0}$ trece prin baricentrul sistemului de puncte $\ O_{1}, O_{2}, ...,O_{n}$.[/hide]

[hide]ciudat..cum sa se dea baricentre la clasa a VII-a??[/hide]

Generalizarea nu este neaparat la nivel de clasa a 7a si de obicei unele generalizari presupun cunostinte mai avansate decat cazul paricular.

Rezultatul mai general pe care l-am postat se trateaza asemanator cazului particular. Sorine, daca stii sa lucrezi cu trinomul de gradul al doilea, poti rezolva cu usurinta ambele probleme.


quadratics
algebra
polynomial
Vieta
binomial theorem
complex numbers

Hmm this a a good problem that tests some clever factoring skills but everything cancels out and you are left with $ 1^{5}=1.$

How do you get 1?  the roots are irrational.

[quote="13375P34K43V312"][quote="vikas"]the roots are irrational.[/quote]

so? it doesn't matter what the roots are, it only matters what the sum of the roots is. actually, come to think of it, the roots aren't even real, but the answer remains unchanged.[/quote]

i mean unreal.  so is 1 correct? i dont know how you get that.

$ (r^{5}+s^{5})+5rs(r^{3}+s^{3})+10r^{2}s^{2}(r+s)=(r+s)^{5}$ if we expand and apply the binomial theorem, and by Vieta $ r+s=1$, so our answer is 1.

oh...yeah, 'cause the sum of the roots is -b/a...so its 1337/1337=1.. genius!!!

Does Vietas therom work on polynomials with unreal coefficiants?  or only with real?

What the heck who's rating all my posts 1?

And yes, Vieta's Theorem is valid for polynomials with non-real coefficients as well, because if you look at the deriviation of it (or intuitive deriviation), you'll see that you don't have to assume anything is real, because in the complex numbers, addition and multiplication are defined just as they are in the real number system.

http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_formulas
I think that $ \frac{-b}{a}$ is one case of it.


geometry
circumcircle
ratio
invariant
perpendicular bisector
geometric transformation
geometry proposed

For an arbitrary point $ D$ on side $ BC$ of an acute-angled triangle $ ABC$, let $ O_1$ and $ O_2$ be the the circumcenters of the triangles $ ABD$ and $ ACD$, and $ O$ be the circumcenter of the triangle $ AO_1O_2$. Find the locus of $ O$ when $ D$ runs over the side $ BC$.

[hide="hint"] it is a line passinge through $ Q$ where $ Q$ is the cicumcenter of $ AMN$ ( $ M,N$ are the midpoints of $ AB,AC$ )and also this line constructs an angle equal with $ 90^\circ\minus{}\angle B$ with $ OM'$ where $ M'$ is the midpoint of $ AM$. [/hide]

Let $ O'$ be, the circumcemter of the triangle $ \bigtriangleup AO_{1}O_{2}$ ( instead of $ O$ ) and also let $ O$ be, the one of the given triangle $ \bigtriangleup ABC.$

We denote the points $ M\equiv AB\cap OO_{1}$ and $ N\equiv AC\cap OO_{2},$ as the midponts of the side-segments $ AB,\ AC$ respectively and let be the point $ K\equiv AD\cap MN,$ as the midpoint of $ AD$ ( easy to prove applying the [b]Thales theorem[/b] ).

Because of now, $ O_{1}O_{2}\perp AD$ and $ K\equiv AD\cap O_{1}O_{2},$ it is easy to show that $ \angle KO_{2}A \equal{} \angle KNA \equal{} \angle C$ $ ,(1)$ and $ \angle KO_{1}A \equal{} \angle KMA \equal{} \angle B$ $ ,(2)$ from the cyclic quadrilaterals $ AKO_{2}N$ and $ AKMO_{1}.$

From $ (1),\ (2)$ $ \Longrightarrow$ $ \angle O_{1}AO_{2} \equal{} \angle A$ and so, the quadrilateral $ AO_{1}OO_{2}$ is cyclic.

Hence, the Locus of the circumcenter $ O'$ of the triangle $ \bigtriangleup AO_{1}O_{2}$ as the problem states, is the midperpendicular of the segment $ AO,$ where $ O$ is the circumcenter of the given triangle $ \bigtriangleup ABC$ and the proof is completed.

Kostas Vittas.

A more general result:

In $ \triangle ABC,$ let $ D$ be a variable point on $ BC$ and $ O_1,O_2$ the circumcenters of $ \triangle ABD$ and $ \triangle ACD.$ $ U$ and $ U'$ are two points equally characterized WRT $ \triangle ABC$ and $ \triangle AO_1O_2.$ Show that $ U'$ moves on the perpendicular bisector of $ AU.$

Let $ U''$ be the midpoint of $ AU$.
obviously $ U,U''$ are two corresponding points of $ \Delta AMN$ and $ \Delta ABC$ ($ M,N$ are the midpoints of $ AB,AC$ as in my last post).
now it sufficies to prove that $ \angle U'U''A\equal{}90^\circ$.
draw the perpendicular from $ U''$ to $ AM,AN$ and let them intersects these two segements at $ X,Y$.
also these two lines intersects $ AO_1,AO_2$ at $ X',Y'$ respectively.
since $ XX' \parallel{} O_1M \perp AM$ we have $ \frac{AX}{XM}\equal{}\frac{AX'}{X'O_1}$ and so $ X,X'$ are two corresponding points therefore we have $ U'X' \perp AO_1$ and similarly we have $ U'Y' \perp AO_2$.
so we have $ \angle XU''Y\equal{}\angle X'U''Y'\equal{}180^\circ\minus{}\angle A\equal{}\angle X'U'Y'$.
so $ X'U''U'Y'$ is cyclic.we have
$ \angle U'U''A\equal{}\angle AU''Y\plus{}\angle YU''U'\equal{}\angle AU''Y\plus{}\angle Y'U''U'\equal{}\angle AU''Y\plus{}\angle Y'X'U'\equal{}\angle AU''Y\plus{}\angle U'AY'\equal{}
\angle AU''Y\plus{}\angle U''AY\equal{}90^\circ$
so we r done.

[quote]In $ \triangle ABC,$ let $ D$ be a variable point on $ BC$ and $ O_1,O_2$ the circumcenters of $ \triangle ABD$ and $ \triangle ACD.$ $ U$ and $ U'$ are two points equally characterized WRT $ \triangle ABC$ and $ \triangle AO_1O_2.$ Show that $ U'$ moves on the perpendicular bisector of $ AU.$[/quote]
It's easy to see that the isosceles $ \triangle AO_1B$ and $ \triangle AO_2D$ are directly similar due to $ \angle AO_1B \equal{} 2\angle ADB \equal{} \angle AO_2D.$ Thus, $ \angle O_1AO_2 \equal{} \angle BAC$ and $ \frac {_{AB}}{^{AC}} \equal{} \frac {_{AO_1}}{^{AO_2}}$ $ \Longrightarrow$ $ \triangle ABC \sim \triangle AO_1O_2 \sim \triangle AMN,$ where $ M,N$ denote the midpoints of $ AB,AC.$ If $ U$ and $ U'$ are equally characterized with respect to $ \triangle ABC$ and $ \triangle AO_1O_2,$ we claim that $ \angle BAU \equal{} \angle O_1AU'$ and  $ \angle ABU \equal{} \angle AO_1U'$ (magnitud and orientation).

Ratio $ k \equal{} \frac {_{AO_1}}{^{AU'}}$ remains invariant and since $ O_1$ moves on the perpendicular bisector $ \mathcal{L}_c$ of $ AB,$ then $ U'$ moves on another fixed line $ \mathcal{L}$ which is the image of $ \mathcal{L}_c$ through the spiral similarity centered at $ A$ with coefficient $ k$ and rotational angle $ \angle BAU$ (clockwise). Clearly, $ \mathcal{L} \perp AU$  contains the homologous point of $ U$ WRT $ \triangle AMN$ $ \Longrightarrow$ $ \mathcal{L}$ is the perpendicular bisector of $ AU.$

The subject circle passes through $O$, the circumcenter of $\triangle ABC$, hence $A$ and $O$ are fixed points, consequently its circumcenter lies on the perpendicular bisector of $OA$.

Best regards,
sunken rock

another solution for the generalization:
just see that $\angle O_1AB=\angle U'AU=90^\circ-\angle ADB$ and we have obviously $\frac{AU'}{AU}=\frac{O_1A}{AB}$ and so $AO_1B \sim AU'U$ which implies that AU=UU'.


function
set theory
real analysis
real analysis unsolved

This is probably pretty simple, but I am bad with this sort of thing, so I'll just ask here.

If $ X$ and $ Y$ are two infinite sets with strictly different cardinalities, can the sets of functions $ \{f: X \to \mathbb{Z}\}$ and $ \{f: Y \to \mathbb{Z}\}$ have the same cardinality?

Please? I need to know this for something... A simple yes or no is ok.

Try to look up Easton's theorem on Wikipedia and you'll see that "a simple yes or no" may be impossible here. :) (of course, $ \aleph_0^{|X|}\equal{}2^{|X|}$ for all infinite $ X$)

I would be less generic and restrict the hatred to set theory :P Also isn't it just beautiful that we cannot say boring "yes" or "no" but have to say something like "we can prove that it is impossible to prove or disprove that it is impossible to prove or disprove that the statement is true" instead?  :wink:

[quote="fedja"]Try to look up Easton's theorem on Wikipedia and you'll see that "a simple yes or no" may be impossible here. :) (of course, $ \aleph_0^{|X|} \equal{} 2^{|X|}$ for all infinite $ X$)[/quote]

But that defies common sense! If $ |X| > |Y|$, one would think that $ |X \times \mathbb{Z}| > |Y \times \mathbb{Z}|$. Please enlighten me.

For every infinite set $ |X\times \mathbb Z|\equal{}|X|$, so $ 2^{|X|}\equal{}2^{|\mathbb Z\times X|}\equal{}(2^{\aleph_0})^{|X|}\equal{}c^{|X|}\ge \aleph_0^{|X|}\ge 2^{|X|}$.


calculus
derivative
real analysis
real analysis unsolved

do I take the norm of it?

[img]http://farm4.static.flickr.com/3078/3117153021_4256662c0f_o.png[/img]

What's the derivative of $ \mathbf{c}'(t) \cdot \mathbf{c}'(t) \equal{} \parallel{} \mathbf{c}'(t) \parallel{}^2$?


Mia orea askisi apo tis shortlist tu 1984...
[b][u]askisi 11[/u][/b]
Vrite olus tus 8etikus akereus $n$ oste
   $n=d_{6}^{2}+d_{7}^{2}-1$
opu $1=d_{1}<d_{2}<...<d_{k}=n$ ine oli i dieretes tu $n$.

 :D

Katapliktiko provlima!! Aporw pws den mpike tote ws thema ston diagwnismo!!

Loipon ap'oti eida iparxei ki edw mesa se kapoio post alla den exei dwthei i lisi (den milaw gia to elliniko subforum), opote mporeite na anatreksete sto IMO Compendium (iparxei stin gnwsti istoselida me onoma Vivlos twn Diethnwn Mathimatikwn Olimpiadwn, kai to evala prin apo kamposo kairo). Ekei exei analitika tin lisi... Htan pragmatika mageia auti i askisi tote pou mpike! Otan deite tin teliki lisi tis tha katalavete giati... (Made in USSR. Ouden sxolion).

Alexandros

Θα προτεινα να μην ανατρεξει κανεις να δει την λυση παρα μονο εαν νομιζει οτι την εχει λυσει.......Εγω παντος εγω παντος τιποτε απο τα δυο εχω κανει ακομα. :)

φαντάζει κάτι πολύ μαγευτικό αλλά κι πολυπλοκο όποις βάλει λύσει ας είναι αναλυτικός

ASKHSH No 12

για ποια $a,b$ η παρασταση
$A=a^{3}+b^{3}+3a^{2}b$  ειναι τελειο τετραγωνο ?
....

φίλε κώστα γιατί βιάζεσαι ?? ας βάλει κάποιος την λύση στην προηγούμενη άσκηση εν ανάγκει ας την αντιγράψει άν μπορει
ευχαριστώ πολύ

ok. αλλα στα αλλα τοπικ μαζευονται δυο τρεις αλυτες και μετα λυνονται, με ποιο τροπο θα δουλεψουμε ?!?

[b][color=red]ΑΣΚΗΣΗ 13 [/color][/b]

[color=blue] Μια ασκησούλα για Junior ! [/color]

 Να αποδείξετε ότι στο σύνολο  $\{ 11, 111, 1111, 111...1, ... \}$ δεν υπάρχουν  τέλεια τετράγωνα !

 [u]Μπάμπης[/u]

[color=red]ΑΣΚΗΣΗ 13 [/color]

[b]Hint[/b]

[hide]Καθένας από αυτούς τους αριθμούς μπορεί να γραφεί $4k+3$. Επίσης $k^{2}\equiv0\pmod4$ ή $k^{2}\equiv1\pmod4$.[/hide]

Και μία άσκηση για τον επόμενο Αρχιμήδη  :P

[color=red][b]ΑΣΚΗΣΗ 14[/b][/color]

Να βρεθούν όλα τα ζευγάρια θετικών ακεραίων $(a,n)$ που είναι τέτοια ώστε ο $\frac{(a+1)^{n}-a^{n}}{n}$ να είναι ακέραιος.

[quote="stergiu"][b][color=red]ΑΣΚΗΣΗ 13 [/color][/b]

[color=blue] Μια ασκησούλα για Junior ! [/color]

 Να αποδείξετε ότι στο σύνολο  $\{ 11, 111, 1111, 111...1, ... \}$ δεν υπάρχουν  τέλεια τετράγωνα !
[/quote]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=779875#p779875]εδώ[/url]


Let $ a,$ $ b,$ and $ c$ are non-negative numbers such that $ a\plus{}b\plus{}c\equal{}3.$ Prove that:
\[ a^5\plus{}b^5\plus{}c^5\plus{}6\geq3(a^3\plus{}b^3\plus{}c^3)\]

Let $ f(x) \equal{} x^5 \minus{} 3x^3 \plus{} 4x \minus{} 2\ (x\geq 0)$, we have $ f'(x) \equal{} 5x^4 \minus{} 9x^2 \plus{} 4 \equal{} (x \plus{} 1)(\sqrt {5}x \plus{} 2)(x \minus{} 1)(\sqrt {5}x \minus{} 2)$.
Thus the sign of $ f'(x)$ is equal to that of $ (x \minus{} 1)(\sqrt {5}x \minus{} 2)$ for $ x\geq 0$. By this $ f(x)$ is local maximum at $ x \equal{} \frac {2}{\sqrt {5}}$ and local minimum at $ x \equal{} 1$,
yielding $ f(x)\geq f(1)$. Since $ f(1) \equal{} 0$, we have $ f(x)\geq 0$. For $ a,\ b,\ c\geq 0$, $ \therefore f(a) \plus{} f(b) \plus{} f(c)\geq 0\Longleftrightarrow a^5 \plus{} b^5 \plus{} c^5 \plus{} 4(a \plus{} b \plus{} c) \minus{} 2\cdot 3\geq 3(a^3 \plus{} b^3 \plus{} c^3)$.

By $ a \plus{} b \plus{} c \equal{} 3$, we have $ a^5 \plus{} b^5 \plus{} c^5 \plus{} 6\geq 3(a^3 \plus{} b^3 \plus{} c^3)$. $ Q.E.D.$

I think Mixing Variable Method killed it easily.
We have:
\[ f(a,b,c) \equal{} 5(a^5 \plus{} b^5 \plus{} c^5) \plus{} 6 \minus{} 3(a^3 \plus{} b^3 \plus{} c^3)
\]
It is easy to see that:
\[ f(a,b,c) \minus{} f\left(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}\right) \equal{} \frac {(b \plus{} c)(b \minus{} c)^2(15b^2 \plus{} 10bc \plus{} 15c^2 \minus{} 36}{16}\ge 0
\]
In fact, without loss of generally we can assume that $ a \equal{} min\{a,b,c\}$ so :
\[ 15b^2 \plus{} 10bc \plus{} 15c^2 \minus{} 36\equal{}15b^2 \plus{} 10bc \plus{} 15c^2 \minus{} 4(a\plus{}b\plus{}c)^2\]
\[ \equal{}11(b^2\plus{}c^2)\plus{}2bc\minus{}4a^2\minus{}8ab\minus{}8bc\ge 0\]

And we also have:
\[ f\left(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}\right) \equal{} \frac {3(a \minus{} 1)^2(5a^3 \plus{} 15a^2 \minus{} 17a \plus{} 5)}{16}\ge 0
\]
because:
\[ f(a) \equal{} 5a^3 \plus{} 15a^2 \minus{} 17a \plus{} 5\ge f\left(\frac {4\sqrt {30}}{15} \minus{} 1\right) > 0
\]
We are done  :wink:

I don't think your solution is correct, kunny. Let $ x\equal{}0.2\to f(x)<0$  :wink:

[quote="kunny"]
yielding $ f(x)\geq f(1)$. Since $ f(1) \equal{} 0$, we have $ f(x)\geq 0$. For $ a,\ b,\ c\geq 0$[/quote]
Try $ x\equal{}0.$ :wink:

Oh, I see. This problem is made in artificial way, arqady. :wink:

[quote="zaizai-hoang"][quote="kunny"]How can you prove for $ c \equal{} 0$?[/quote]

What do you mean? I think your solution is wrong clearly  :wink: See the tittle of the topic "Anti tangent line"[/quote]

You have changed your paper. Where is $ x\equal{}\frac{a}{c}$? :wink:

This problem is made to let us know the limitation of using "tangent line".

[quote="kunny"][quote="zaizai-hoang"][quote="kunny"]How can you prove for $ c \equal{} 0$?[/quote]

What do you mean? I think your solution is wrong clearly  :wink: See the tittle of the topic "Anti tangent line"[/quote]

You have changed your paper. Where is $ x \equal{} \frac {a}{c}$? :wink:

This problem is made to let us know the limitation of using "tangent line".[/quote]
yeah, I have edited something in my post before I saw some mistakes  :wink: Of course, tangent line technique doesn't work in this case  :wink: Have any proof for this?!

Nice work, zaizai-hoang!  :)  What about the following inequality?
Let $ a,$ $ b,$ $ c$ and $ d$ are non-negative numbers such that $ a\plus{}b\plus{}c\plus{}d\equal{}4.$ Prove that:
\[ 2(a^5\plus{}b^5\plus{}c^5\plus{}d^5)\plus{}12\geq5(a^3\plus{}b^3\plus{}c^3\plus{}d^3)\]

[quote="arqady"]Nice work, zaizai-hoang!  :)  What about the following inequality?
Let $ a,b,c$ and $ d$ are non-negative numbers such that $ a\plus{}b\plus{}c\plus{}d\equal{}4$. Prove that:
$ 2(a^5\plus{}b^5\plus{}c^5\plus{}d^5)\plus{}12\geq5(a^3\plus{}b^3\plus{}c^3\plus{}d^3)$[/quote]
This problem has been posted earlier,.... I don't remember exactly the link...... probably at unsolved problems. I think it was an unique $ Chebyshev$ solution.


Hi all,

This problem, the one below, is really tiring(phew, i am tired)..

It asks you to come up with an expression which will tell u the maximum lcm of numbers you can obtain which in turn are derived by partition of a natural number n.

That is, you are given a number n....You gotta partition it into m natural numbers adding upto n, of course. (where m can be anything)

You have to come up with a technique which maximises the lcm of these m natural numbers obtained from partition of n..

is the problem clear enough?

Well, then thanx in advance for solving it...(and posting the solution)

This is quite a classic problem.. its easy to prove that you only ever need prime powers (or 1). Then it just becomes a 2 dimensional DP where you're allowed to use the first i primes to make up a sum of at most j.
Of course, you'll probably need BigIntegers (or equivalent) if the input is anything more than tiny.

sorry to say but u sound a bit greek to me :D ...of course thats me being silly or fool or whatever u think :( ...
But i would ask u (if u can find time) to go into some more details...

Thanks and regards
-Akash Kumar

Alright.

Step 1: prove that all numbers in the partition should be 1 or a prime power.

Suppose we use n as one of the numbers, which isn't 1 or a prime power. Then we can take out some prime p to get $n = p^{a}b$. Then it definitely won't be worse to replace n by the two numbers p^a and b, since the LCM stays the same but the sum gets smaller, leaving us more 'room' to get a bigger LCM.

Now, lets define f[i][j] to be the biggest LCM we can get amongst prime powers which add to <= j and only use the first i primes. (Since we're saying <= j, that means we can now ignore 1s altogether). For example, f[2][10] = 12, since if we restrict ourselves to the first 2 primes, the best we can do is 2^2 + 3 (plus three ones) which gives LCM 12. f[3][10] = 30, since we can now have 2 + 3 + 5, with LCM 30. And the answer to the main problem for input 10 is f[4][10], since there are only 4 primes under 10. (That turns out to be 30 too).

Now, how do we calculate f[i][j]. If i = 0, then we aren't allowed any primes and the answer is just 1. So lets assume i>=1.
There are two cases - either we don't use the ith prime at all, or we do. The first case, where we don't use the ith prime, just reduces to the best answer for f[i-1][j].
For the second case, lets suppose we use $p_{i}^{k}$. The rest of the sum must be made up of powers of the first i-1 primes, whose sum is $j-p_{i}^{k}$. So we take the LCM of $f[i-1][j-p_{i}^{k}]$ and $p_{i}^{k}$, and that just involves multiplying them together since the first one has smaller prime factors only.

Thus to calculate f[i][j] we check through all possibilities for the prime power, and take the biggest answer.

The following code will print out the first N answers (for input 1 through N). Takes a couple of seconds to work it out up to 3000.

[code]import java.util.*;
import java.math.*;

public class BigLCM {

	static final int N=3000;

	public static void main(String[] args) {
		// calculate primes

		boolean[] prime = new boolean[N+1];

		int primecount=0;
		Arrays.fill(prime,true);

		for (int i=2; i<=N; i++) {
			if (prime[i]) {
				primecount++;
				for (int j=i*i; j<=N; j+=i) prime[j]=false;
			}
		}
		int[] primes = new int[primecount];
		primecount = 0;
		for (int i=2; i<=N; i++) {
			if (prime[i]) {
				primes[primecount++]=i;
			}
		}

		BigInteger[][] f = new BigInteger[N+1][N+1];

		// base case
		for (int j=0; j<=N; j++) f[0][j] = BigInteger.ONE;

		for (int i=1; i<=primecount; i++) {
			for (int j=0; j<=N; j++) {
				// don't use ith prime at all
				f[i][j] = f[i-1][j];
				for (int primepow = 1; primepow<=j; primepow*=primes[i-1]) {
					// use the prime power 'primepow'. note the loop stops when the prime power is too big.
					BigInteger poss = new BigInteger(""+primepow).multiply(f[i-1][j-primepow]);
					if (f[i][j].compareTo(poss)<0) f[i][j]=poss;
				}
			}
		}
		for (int j=1; j<=N; j++) System.out.println(j+": "+f[primecount][j]);

	}

}[/code]

Nice!!!! Thank You..frnd. really nice of u to explain in such depths..and thax for the code too

Gr8 job :D .....


trigonometry
Pythagorean Theorem
geometry

[quote="Ignite168"]Treething is on a hot air balloon 2500 ft above sea level. If the radius of the earth is 3960 miles and there are no obstructions, how far on Earth could Treething see?
Use 5280ft. = 1 mile.[/quote]

What do you mean by how far? Do you calculate the distance as going on the circumference or the direct distance?

How far can treething see in any direction (in a line) anyway?

Can you be a little more specific?

When you say see, do you mean by the farthest distance possible that Treething can see from where he is to a point on the Earth.  If it is, then I have an answer.

[hide="Solution"]1.  The radius of the earth is $3960*5280=20908800$ feet.  So the distance from the center of earth to where Treething is is $20911300$ feet.

Now we draw a picture showing that the farthest distance that Treething can see is perpendicular to a radius of the Earth.

So by using the Pythagorean Theorem, the answer is

$\sqrt{(20911300)^{2}-(20908800)^{2}}$ feet which is equal to 323342.3109 feet approximately.

So converting it to miles the answer is approximately 61.23907403 miles

$61.23907403$

So Treething can see any distance on the earth less than or equal to 61.23907403 miles.[/hide]

If you mean the distance between Treething and the farthest point on earth from him that he can see, then here is the solution:
[hide]The radius of the earth is $3960\times5280=20908800$ feet, so Treething is $3960\times 5280+2500=20911300$ feet from the center of the earth.
Let $T$ be the point where Treething is, $C$ be the center of the earth, and $P$ be the point farthest away from Treething that he can see. Then $\angle CPT=90^o$. The distance we are looking for is $TP$. By the Pythagorean Theorem,
$TP=\sqrt{TC^2-CP^2}=\sqrt{104550250000}\approx\boxed{323342.31}$ feet.[/hide]

[quote="SplashD"]That's the distance from Treething, not the distance on Earth[/quote]

The distance  on earth would have to involve trig, so I am going to assume that it is the distance from Treething, and I should be correct.

I blame Ignite for the unclearity of the problem.

[quote="mathgeniuse^ln(x)"][quote="SplashD"]unclearity[/quote]

[hide=" :rotfl: "]I think you mean "bad wording".[/hide][/quote]

No comment.

The correct answer is "any point on Earth less than or equal to approx. 61.24mi. away from Treething"

Okay, I pretty much got that question right except for the less than part.


Compute the number of positive integers less than 512 whose rightmost two digits are both 1 when expressed in base 2.

[hide]512= 2^9 = 1000000000 in base 2 . All nonnegative integers smaller than 512 can be expressed in the form _ _ _ _ _ _ _ _ _  , where a 0 or 1 goes in each blank. To satisfy the condition given in the problem, the rightmost two digits have to be 1s; the integers less than 512 which satisfy this condition take the form _ _ _ _ _ _ _ 11. Once again, each of the blanks is occupied by either a 0 or 1. Since there are 7 blanks, there are 2^7= 128 ways to fill those blanks, or 128 numbers that meet the condition.  [/hide]

OO, nice way of thinking hooyoung...

Ur both correct!

Here's the answer in the spoiler:

[hide]The maximum binary number that is less than 512 in base 10 and ends with two 1s is 111111111.  Try for yourself to see.  There are 128 numbers with  _ _ _ _ _ _ _ 11 in binary form since it's 2^7.[/hide]

[quote="teratomato"]Those numbers are 3 mod 4, it's one of four residue classes, the answer is 512/4 = 128.[/quote]
Can you explain this solution? I don't understand the residue classes but a general explanation would be nice.

Residue classes are basically the remainders when divided. Thus, in a (modn), there are n-1 possible residue classes, 0 to n-1.

NOw, a number in base 2 with both rightmost digits 1's, will translate into 3(mod4) since if we were to add up all the "digits", they would look someting like this:

...+64*(0 or 1)+32*(0 or 1)+16*(0 or 1)+8*(0 or 1)+4*(0 or 1) +2*1+1*1.

Thus, we can see in mod4 that it will always equal 3.

close...there are n possible residue classes.

yea my bad

0 to n-1 is n.

Hmm, still need for the bus-lag (is there such a thing??) to wear off.

teratomato wrote:I believe that the "method" I used is slightly superior since it can be used with little modification for limits that are not powers of two.

Hmph.  Superiority, eh?  Well, I say "all methods are created equal."  [hide]Not.[/hide]

Well, i have to admit, some methods ARE superior to others. Most notably in polynomials: elagant manipulation vs. brute force. I tend to think the former is superior


modular arithmetic
number theory unsolved
number theory

Prove that equation $ y^2\equal{}x^3\plus{}7$ has no solution in $ Z$

its obvious that $ x$ is odd.
$ (x \plus{} 2)((x \minus{} 1)^2 \plus{} 3)) \equal{} y^2 \plus{} 1$
$ ((x \minus{} 1)^2 \plus{} 3)\equiv 3 \pmod 4$ so it has a prime number of form$ 4t \plus{} 3$.but $ y^2\plus{}1$ dont have such prime.


Find all primes $p$ such that the number of distinct positive factors of $p^2+2543$ is less than 16.

For all primes except $2$, $p^2\equiv 1$ mod $8\Rightarrow p^2+2453\equiv 0$ mod $8$.
For all primes except $3$, $p^2\equiv 1$ mod $3\Rightarrow p^2+2453\equiv 0$ mod $3$.

But now, suppose that some prime $p$ other than $2$ and $3$ divides $p^2+2453$.  Then $2^3.3.p|p^2+2453$, and hence $p^2+2453$ has at least $16$ factors, so only $2$ and $3$ can be prime factors of $p^2+2453$.

But the highest power of $3$ dividing $p^2+2453$ must be $2$, since if $2^33^3|p^2+2453$, we get at least $16$ factors.  Similarly, the highest power of two dividing must be $6$, since $2^7.3|p^2+2453$ also gives at least $16$ factors.

Combining these gives $p^2+2453\le2^63^2=576$ for $p\not\in \{2,3\}$, which is an obvious contradiction.

It is easy to check that $p=3$ gives exactly $16$ factors, but $p=2$ gives $6$ factors, and hence $p=2$ is the unique valid solution.

This problem was used in a Thai competition two years ago. I think another solution develops from substituting $p = 6k+1, 6k-1$ and checking $p = 2,3$