[color=blue][size=150]NEW GAME: QuasiJeopardy[/size][/color]

The way you play it is that one person asks a question. Another person tells an answer off the top of their head. A second person checks the answer on google, and verifies it by giving a link to the correct answer. Then, the second person asks a question. If the first person gets it right, then the second person recognizes teh first person with the amount of points given. The amount of points for a question may vary from 1-5 and is specified during the problem's posting. If the first person gets it wrong, then teh second person answers it correctly from thier head. A third person then verifies. If the 1st gets it wrong, and the 2nd person gets it right, then the value of the question doubles, as the 2nd person takes his quota of points as well as the first person's quote.

[quote="Me-[i][size=200]THIS IS AN EXAMPLE, NOT THE REAL QUESTION[/size][/i]-as an example just"]Question 1 [5 points]: Who is John Kerry

If the first person answers, "A candidate in the democratic party for the 2004 elections", he takes 5 points."

If he gets it wrong by saying "the president of the us now".

Second guy writes: No. He's some dude who is in bush's cabinet.

Third guy writes: Both wrong. He is A candidate in the democratic party for the 2004 elections.

Fourth guy gives proof, and the third guy gets 5 from 1st guy+5 from 2nd guy+5 from 3rd guy, and the 3rd guy takes home 15 points.

The 4th guy gets the amount of points offered for showing proof.

Scoreboards will be posted every page. Someone post a scoreboard and get 3 points for doing so.[/quote]

OK. Now the real thing:

[size=200]Question 1 [2 points]:[/size]
[size=150]Raisa Gorbachev, a prominent lady, had a large amount of power in which country?[/size]

If watching Jeopardy has done me any good, I would have thought that you're given an "answer" and you have to state the question.

Ah but this is quasi jeopardy.
Like my dad has some quasi judicial functions, and some bodies are quasistellar! :rotfl:

[quote="anirudh"]So Koda gets 2 Points.

Scoreboard:

Koda--2 points

In what year did Iran capture 79 people from the US Embassy and cut diplomatic ties with the US?[/quote]
when chuck norris created god

LOL. NO seriously, LOL!!!.  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:
Next person who gets it correct gets 10 points. (5+5)

No, that's wrong--Chuck Norris didn't create himself.

when chuck norris created Billybob the carpenter?
ccy:fyi, chuck norris IS god

Yes, you win: you get 15 points.

Scoreboard:

junggi--15 points
Koda--2 points

New question: Orange juice is the official beverage of which US state?

junggi please stop saying things about chuck norris like that or i will have to ...just stop it please

[quote="anirudh"]And the link: http://en.wikipedia.org/wiki/Orange

So the scoreboard:

Anirudh-15
Junggi-15
Koda-2

QUESTION 3:

The world's first atomic bomb was dropped on which city? [4 points][/quote]
1) how did you get 15 pts? should be moer like 5
2) wikipedia is uncredible(according to some of my uncredible sources)
3) so you shouldnt get any pts
4) the answer to question 3 is MTV.

[quote="junggi"][quote="anirudh"]And the link: http://en.wikipedia.org/wiki/Orange

So the scoreboard:

Anirudh-15
Junggi-15
Koda-2

QUESTION 3:

The world's first atomic bomb was dropped on which city? [4 points][/quote]
1) how did you get 15 pts? should be moer like 5
2) wikipedia is uncredible(according to some of my uncredible sources)
3) so you shouldt get any pts
4) the answer to question 3 is MTV.[/quote]
1. That was a hard question.
2. No, wikipedia is incredible.
3. Ok, he gets 5.
4. Yes, you get 10 points, so...

Scoreboard:

Junggi-25
Anirudh-5
Koda-2

New question (1 point):
Who is Donald Cox?

[quote="jackdillon"]A person( just try to prove me wrong!!!)[/quote]
HAHA WRONG
he's an ANIMAL

Remove me from teh scoreboard. I'm not playing.  :P

Scoreboard:

Junggi-35
Koda-2

and the answer to q3 is wrong.

Going again:

QUESTION 5:

The world's first atomic bomb was dropped on which city? [10 points]

The world's first atomic bomb was tested at Trinity, and the world's second atomic bomb was dropped at Hiroshima.

No, it was dropped at "Hero of Shima" first.

[url=http://uncyclopedia.org/wiki/Atom_bomb]proof[/url]

WRONG!!!

Since that is unlawful  :rotfl: , you just got slapped with a 50 point fine~ :rotfl:

[quote]At 5:29:45 (Mountain War Time) on July 16, 1945, in a white blaze that stretched from the basin of the Jemez Mountains in northern New Mexico to the still-dark skies, "The Gadget" ushered in the Atomic Age. The light of the explosion then turned orange as the atomic fireball began shooting upwards at 360 feet per second, reddening and pulsing as it cooled. The characteristic mushroom cloud of radioactive vapor materialized at 30,000 feet. Beneath the cloud, all that remained of the soil at the blast site were fragments of jade green radioactive glass created by the heat of the reaction.[/quote]

from http://inventors.about.com/library/weekly/aa050300a.htm opposes you. So, I just won myself 60 points, and donate it to junggi!

[url]http://en.wikipedia.org/wiki/Trinity_site[/url]

I think I should get it - the test was called Trinity but no the place!  :oops:

what's wrong with my website proof  :mad:  :mad:
[url=http://uncyclopedia.org/wiki/Atom_bomb]proof[/url]

[quote="Koda"][quote="junggi"][quote="Koda"]Nothing much, except that it is wrong.[/quote]
why  :( [/quote]

because it is not right  :D[/quote]
but it's uncyclopedia  :(

ugh, here is the next problem:

Fill in the blank:

"Winken, Blinken, and ___, one night,"

[10 points]

Nod.  :roll:

cool

donating points to junggi

SCOREBOARD:

junggi-55
koda-2

next question: who is my brother? (54 points)

[quote="mustafa"]Art Garfunkel. I'll donate my 54 points to junggi.[/quote]
????????????/
WHOA IM COMPLETE PWNING THIS GAME

[quote="mustafa"]I'm amazed by your incredible skill junggi![/quote]
Haha, yes I am a genius at QuasiJeopardy(whatever that is)

:rotfl:  :rotfl:  :rotfl:

ok, anirudh is right, 10 points for junggi.

Next question:

What is the current name of the Persian language?

[20 points]

You may NOT donate these 20 points to anyone!

[quote="nutz_for2.718281828"]:rotfl:  :rotfl:  :rotfl:

ok, anirudh is right, 10 points for junggi.

Next question:

What is the current name of the Persian language?
[/quote]
persian, duh

next question[$\pi^{e!}$ points]
What alternative rock band consists of 4 guys who served as the 43rd-46th presidents(with spongebob squarepants as vice pres), that other guy that nobody cares about, and oscar wilde?

[quote="junggi"][quote="nutz_for2.718281828"]:rotfl:  :rotfl:  :rotfl:

ok, anirudh is right, 10 points for junggi.

Next question:

What is the current name of the Persian language?
[/quote]
persian, duh[/quote]

uhh... FALSE.

My questions answer was false.
It will be Kenai, and I do NOT donate my 54 points to junggi.

So I am on 56 and in the lead! :lol:


[quote="brady00n"]The probability of getting a two of a kind in a five card poker hand is:

$\displaystyle{13\choose 1}\displaystyle{4\choose 2}\displaystyle{12\choose 1}\displaystyle{4\choose 1}\displaystyle{4\choose 1}\displaystyle{4\choose 1} all over \displaystyle{52\choose 5}.$

We must choose which card is the pair, and there are 13 choose 1, we must choose two of the four cards of that number, which is 4 choose 2, then there are 12 choose 1 ranks left.  But here's where I'm lost.  Why is it four choose 1 here, and not four choose 1 alternating with 11 choose 1, and 10 choose 1, since the cards chosen have to be different (otherwise, it's a full house...)

Thanks![/quote]

Maybe you wrote it down wrong.

It should be
$\frac{\displaystyle{13\choose 1}\displaystyle{4\choose 2}\displaystyle{12\choose 3}\displaystyle{4\choose 1}\displaystyle{4\choose 1}\displaystyle{4\choose 1}}{ \displaystyle{52\choose 5}}$

You choose which rank is the pair, then which suits of that rank, then 3 of the ranks that are not the pair one, hence the 12. Then (4C1)^3 to choose the suits of each.

By the way, use \frac{num}{den} to make fractions.


calculus
integration
calculus computations

Could not solve it
 $ \int \frac{\sqrt{x}dx}{(1\plus{}x^3)}$

then,actually,I must integrate
$ \int \frac{2t^2dt}{1\plus{}t^6}$


In the figure below, points are equally spaced horizontally and vertically one centimeter apart. By what factor is the area of the polygon multiplied if its dimensions are doubled?
[asy]dot((0,0));
dot((0,1));
dot((2,0));
dot((3,0));
dot((1,0));
dot((1,1));
dot((2,1));
dot((3,1));
draw((0,0)--(1,0)--(3,1)--(1,1)--cycle);[/asy]

Since area is directly proportional to the square of a linear measurement, the factor is $ 2^2\equal{}\fbox{4}$.


If two circles intersect, we can apply an inversion having an intersection pt as pole, and that would turn the problem into showing that there is a circle tangent to two lines and a circle. If, on the other hand, the circles are pairwisely disjoint, after inversion with a point on one of the circles as pole, we have to show that there is a circle tangent to two circles and a line.

Both of these can be proven by using continuity, for example. Take the first case. The two lines divide the plane in four regions. Consider a region which the circle (call it $\mathcal C$) intersects. Now start with a circle inside the region, tangent to the two lines, and large enough so that it does not intersect $\mathcal C$, and then make it smaller. If we make its radius $\inf r$ for $r$ s.t. whenever the radius is $R>r$, it does not intersect $\mathcal C$, then it will be tangent to $\mathcal C$.

Stuff like that.. :); of course, this is only part inversion.

[quote="Leonardo"]Prove, using inversion, that given 3 circles $k_1, k_2, k_3$ there is a circle $k$ that is tangent to all of them.[/quote]
This proposition, as stated, is [b]false[/b]. For example, if the circle $k_1$ is inside the circle $k_3$ and the circle $k_2$ outside the the circle $k_3$, a circle tangent to all 3 given circles does not exist. There is the following theorem for inversion:

A pair of circles $(C_1), (C_2)$ can be transformed by an inversion into 2 intersecting lines (if the 2 circles intersect), into 2 parallel lines (if the 2 circles touch), or into 2 concentric circles (if the 2 circles do not have a common point).

A given circle can always be transformed by an inversion into a line - it is sufficient to choose the inversion center on the given circle.

[b][color=blue](1)[/color][/b] If the 2 circles intersect at points $A, B$, selecting the inversion center at one of their intersection points, say $A$ and the power of inversion equal to $AB^2$ (i.e., the radius of the inversion circle is $AB$) carries them into 2 lines intersecting at the point $B$ (this point, being on the inversion circle, stays in place).

[b][color=blue](2)[/color][/b] If the 2 circles touch at a point $A$, selecting the inversion center at the tangency point and the power of inversion arbitrary (i.e., the radius of the inversion circle is arbitrary) carries them into 2 parallel lines.

[b][color=blue](3)[/color][/b] If the 2 circles do not intersect, a circle $(D)$ exists, with center $D$ on the line $c = C_1C_2$ connecting their centers, perpendicular to both the circle $(C_1)$ and $(C_2)$. Let $P, Q$ be the intersections of the circle $(C_2)$ with the line $c$. Perform an inversion with center at one of these intersection points, say $P$. In this process, the circle $(C_2)$ is carried into a line line $c_2 \perp c$ intersecting the line $c$ at a point $Q'$ and the circle $C_1$ into a circle $(C_1')$ centered on the line $c$ and not intersecting the line $c_2$. Let $(Q')$ be the circle centered at the point $Q'$ and with radius equal to the tangent length from the point $Q'$ to the circle $(C_1')$. Obviously, the circle $(Q')$ is perpendicular to both the circle $(C_1')$ and to the line $c_2$. Since an inversion is an involution (a transformation that is its own inverse), performing the same inversion again carries the circle $(C_1')$ back to the circle $(C_1)$, the line $c_2$ back to the circle $(C_2)$ and the circle $(Q')$ into a circle $(D)$ centered on the line $c$ and perpendicular to the circles $(C_1), (C_2)$, because inversion preserves angles between curves.

Let $A, B$ be the intersections of the circle $(D)$ with the line $c$. Perform an inversion with center at one of these intersection points, say $A$. The line $c$ passing through the inversion center $A$ is carried into itself, the circle $(D)$ into a line $d \perp c$ perpendicular to the line $c$ and the circles $(C_1), (C_2)$ into circles $(C_1''), (C_2'')$ centered on the line $c$ and perpendicular both to the line $c$ and to the line $d$. But if a circle is perpendicular to a line, the center of this circle lies on this line. Consequently, the circles $(C_1''), (C_2'')$ are both centered at the intersection of the perpendicular lines $d \perp c$, i.e., they are concentric.

Returning to the proposition of the problem. Assume that none of the 3 circles $k_1, k_2, k_3$ intersect, the circles $k_1, k_2$ are either both outside of the circle $k_3$ or both inside of this circle, the circle $k_1$ is not inside of the circle $k_2$ and the circle $k_2$ is not inside of the circle $k_1$. Transform the circles $k_1, k_2$ by an inversion into 2 concentric circles $k_1', k_2'$. The circle $k_3$ is then carried into a circle $k_3'$ between the concentric circles $k_1', k_2'$. A circle $k'$ touching the 2 concentric circles $k_1', k_2'$ clearly exists, centered either on the circle $m$ concentric with $k_1', k_2'$ and with radius $\frac{r_1' + r_2'}{2}$ or on the circle $n$ also concentric with with $k_1', k_2'$ and with radius $\frac{|r_1' - r_2'|}{2}$. Using continuity, it can be moved to touch the circle $k_3'$. Performing the same inversion again carries the circles $k_1', k_2', k_3'$ back to the circles $k_1, k_2, k_3$ and the circle $k'$ into a circle $k$ tangent to all 3 original circles.

Of course, I disregarded circles wich lie inside one another without mentioning it. Stupid me! :)


Find an expression for the sum of the following series:

$  1(1!) + 2(2!) + 3(3!) + 4(4!) ................ + n(n!)$

Hint:

Sum of the first n couting numbers is n(n+1)/2.

That's interesting. I've never thought about that before.

It appears the pattern goes
$1^2(0!) + 2^2(1!) + 3^2(2!) + ... + n^2((n-1)!)$

I dont' see how that helps, and guile's hint is not meaning anything to me.


For what value of $ n$ will $ (7!)(5!)(3!)(1!) \equal{} n!$ ?

$ 7! \times 5! \times 3! \equal{} 7! \times 5 \times 4 \times 3 \times 2 \times 3 \times 2 \equal{} 7! \times 8 \times 9 \times 10 \implies n\equal{}\boxed{10}$.


[hide="Possibly Wrong"]
Infinitely many, such that $a = \frac{-b-1}{2}$, where $a < 0$  and $b > 0$ or $a> 0$ and $b<0$.
[/hide]

[quote="unknown_sequence18"]find a,b, $4a+2b+2=0$
how many solution does this have ?[/quote]

[hide="Integers"]Dividing by 2,

$2a+b+1 = 0$

Let $x$ be any integer. It is easy to check that:

$(a,b) = (x,-2x-1)$

produces an integral solution.[/hide]
[hide="If not"]Let $x$ be any real. Then
$(a,b) = (x,-2x-1)$

produces a real solution.[/hide]
[hide="Answer in both cases"]This should be obvious, there are infinitely many solutions.[/hide]


I think most of us were severely trying to help as we do with others, we just don't like it when others use criticism when asking for tests and to the people that don't help at all.

Thank you and I hope your team does wonderful,
Jorian Hoover

[quote="life1"]Hello all -- to those who've been supportive of me, thank you -- and to those I've helped (many of you have pm'd me privately), you're welcome.  I will be severely limiting my input on this forum from now on (you're welcome) - and may leave altogether, but in my absence I want to let those who have lodged public and on-forum criticisms of me to kindly reread the agreement we all signed when registering, especially letter d:  "d. not to upload, post, e-mail, transmit or otherwise make available any Content that is unlawful, harmful, threatening, abusive, harassing, tortuous, defamatory, vulgar, obscene, libelous, invasive of another's privacy, hateful, or racially, ethnically or otherwise objectionable;" -- without citing specific legal verse, suffice it to say that often "defamatory, libelous, hateful (someone actually used the word "Hate" when referencing me in a recent post!!),  :huh: objectionable, etc, are VERY subjective terms, often open to interpretation", so to avoid any possible Misinterpretation I am asking you to kindly STOP posting highly critical, and sometimes personal comments, whether it be aimed at me or anyone else.  You may not have liked how I worded a post, but I tried to limit my posts to objective questions about math, problems, contests, etc, and when discussing anything of a personal nature, talked only about myself or my child, not any users on this site!  The only time I referenced others was to respond to a criticism or question, or to thank someone for their help. Thank you for your consideration here. NOTE: don't bother asking me to cite specific examples, or to name names -- I won't do that on this forum, and I will not be responding to [u][b]any[/b][/u] further comments regarding this post. You have a right to respond, but please don't ask for any further follow-up from me - as far as I'm concerned this topic is now "locked".[/quote]
Bleh, sorry if I was overly critical, you were frustrating me


Oops!...first example in NT packet  :blush:

The surest way to get people to look at something is to tell them to ignore it.


Out of curiosity, can the books be bulk-ordered?

Yes - we offer a discount of 10% on orders of 10 texts or more (solutions don't count as texts).


algebra
polynomial
inequalities open
inequalities

And the most interesting one:
Prove or disporove: for all positive $a,b,c\ge0$, every integer $n$ such that $n\ge w_{1}\ge w_{2}$, $n\ge v_{1}\ge v_{2}$, $n\ge t_{1}\ge t_{2}$, $n\ge u_{1}\ge u_{2}$, (for some condition with these numbers) we have

$(n-w_{1},w_{1}-w_{2},w_{2})+(n-v_{1},v_{1}-v_{2},v_{2})\ge(n-t_{1},t_{1}-t_{2},t_{2})+(n-u_{1},u_{1}-u_{2},u_{2})$

Notation of my brackets can be found here:

http://www.mathlinks.ro/Forum/viewtopic.php?t=56369

Wow. Of course you need to add at least something like $w_{1}+v_{1}\le u_{1}+t_{1}.$ Otherwise you may swap both sides :)

And once again :whistling:
[quote="conceive"]
Notation of my brackets can be found here:

http://www.mathlinks.ro/Forum/viewtopic.php?t=56369[/quote] I quote from there:[quote="conceive"][size=150]I suggest this notation:[/size]
$[r,s,t]$ will denote SYMMETRIC sum while
$(r,s,t)$ will denote CYCLIC sum[/quote] So please stick to that and use [b]square brackets[/b]!! I'm sure you don't mean cyclical sums here. :o


Prove that there are infinite number of n s.t $ p | 2^{n}-n$

If p is prime work $ n=(p-1)(pk-1), k\in N$


Has anyone got a proof that the nine points described on the nine point circle are concyclic? Any problems concerning this circle?

It is easy if we have got  some figures, and here

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

[color=red]There are  aplications to this beautifull circle:[/color]
See one here...
[b]1)[/b] [url]http://www.mathlinks.ro/viewtopic.php?t=197473[/url]

Dear Mathlinkers,
a nice proof of the nine points circle based on the eight points circle of Louis Brand is presented (in French with many figures) on my website
http://perso.orange.fr/jl.ayme  vol. 4 : Du cercle des huit points au cercle des neuf points.
Sincerely
Jean-Louis


Could someone attach it. I need it really bad for school.  :( I wanted to look over it during the weekend, as well as other tests and aops vol 1

I need it lol. I'm running out of time. oh well  :(


It's just my wondering.. From two different kinds of people in the world, one kind of person who is really smart and other kind of person who's pretty smart but has really quick speed of playing with numbers, which kind of person would do better on their competition and school? It's just my wondering you know...

I think intelligency is more than speed but..

[quote="Syntax Error"]It really depends on the competition.  In mathcounts the faster guy would benefit because it's more of a guess and check competition dare I say.  It also depends on how smart and fast they both are.  On competitions like AIME and IMO speed doesn't matter as much.[/quote]

Don't forget last year's National Countdown Round finals, though.

That was pretty incredible.

[quote]Don't forget last year's National Countdown Round finals, though.

That was pretty incredible.[/quote]

Huh? What happened last year at countdown?

You mean how they both constantly missed questions even when there was time left over to think?  Hm.

Objection! For all of the questions that I (Adam) missed, I had buzzed in first - I never "missed questions even when there was time left over to think," much less "constantly."

Fierytycoon, to summarize: Andrew Chien (2nd place written) went up 3-0, then I came back to win 4-3.

i love countdown problems! u dont even have to read half the question.  scan real quick, look for the numbers, then do somethin with the numbers.  i got first in countdown in chapter last year and this year and first in countdown at state this year.  i wish i could go to nationals just so i could do the countdown .  :(

Well... I'm suck at countdown round but what countdown requires it is not just scann the problem.. You need to have good determination that your answer is absolutely correct.  This year, I was third but went down to fourth on chapter because of stupid countdown round.. I got one question right, though.. My favorite one:

(3/4)^-1 or something ilke that...

I have horribly slow reactions.. most of the time my opponent person would have the answer before i even finished reading the problem.. *sigh* =(

Free practices and discussions
Visit Notice Topic on Getting Started Forum

personally, i'm the type that is fast, and pretty smart, but not the smartest, and i got to nationals that way, but in taking previous tests (nationals) i find that i'm not getting the best scores shall i say.  So it all depends on where you are competing, but in countdown round, it's speed, but also depends if you can quickly find the key parts in the problem.
For example: What is the probability of flipping a coin three times and getting a Heads-Tails-Heads in that order?  You just really have to see "Heads-Tails-Heads" to be able to get the problem, and that is just 1/8 (1/2^3) so you should get that problem quickly if you can scan well, and are fast.

I think in the overall scheme of things, speed does not matter at all.  Sure, on a couple of competitions, like Mathcounts countdown round, or even the AMCs sometimes, speed is a major factor, but when it comes to devising proofs and stuff, who cares about speed?   If you don't have the mathematical talent, you can't do it, regardless of how fast you work on simple problems.

Though I must admit, speed math can be fun - countdown round is fun, as well as games like 24 =)


number theory
modular arithmetic
number theory unsolved

Okay, I recently learned a few tricks for doing modular arithmetic problems, but I still do not understand how to solve a problem like this;

12^23 (mod 34)

It seems like there is no way to simplify, unless you actually evaluate it, or look for a pattern. Please Help!!!

Ohh, I get it, so you just factor it into pieces, solve for the pieces, and then multiply them all. That is extremely helpful, but how did you know that 12^11 = 6 (mod 34) and how did you know that
12^12 = 4 (mod 34)?

Yeah, but you can't calculate it if the numbers are much larger. Solve this one;

73^15 (mod 97)

hello, you must not calculate the whole number, make it like the following example
$ 73\equiv 73 \mod 97$
$ 73^2 \equal{} 5329\equiv 91 \mod 97$
$ 73^3\equiv 73\cdot91 \equal{} 6643\equiv 47 \mod 97$
and so on.
Sonnhard.

Oh, now I get it. Okay, I thought that there was a faster solution than that.

hello, i have $ 73^{15}\equiv 50 \mod 97$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, it is your last problem $ 73^{15} \mod 97$.
Sonnhard.[/quote]

No, it is supposed to be 73^(15), if it were 73^1 times 5, then I would have put 5(73^1)

hello, $ 73^{15}\equal{}73^{\left(15\right)}$.
Sonnhard.

yes, i know. i dont understand wut ur trying to say. in last step , i just broje that down by a exponential identity. we have conflicting answers.
$ \bullet$ see if u can spot any mistake in my solution.

$ \bullet$ then please spot the mistake in my solution.

I don't think analytic is wrong, but I am not so good with modular arithmetic. I think that there could be a mistake in Dr Sonnhard Graubner's solution.

hello, the mistake is on your first line
$ 73^2\equiv (\minus{}24)^2 \mod 97\equiv91\equiv\minus{}6 \mod 97$.
Sonnhard.

How did you even get the exact number for 73^15?

Oh, well that is helpful. Anybody that thinks that they know the answer to this problem please say so, because analytic and the other person with the really long name, can't agree on the answer.

Your original question is made much simpler by the fact that $ 34$ is not prime.  The standard (paper-and-pencil) technique is to compute the remainder $ \bmod 17$ and $ \bmod 2$ and then apply the [url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem]Chinese remainder theorem[/url].  Here, this gets you

$ 12^{23} \equiv (\minus{}5)^{7} \equiv \minus{}5 \cdot 25^3 \equiv \minus{}5 \cdot 8^3 \equiv \minus{}5 \cdot 512 \equiv \minus{}5 \cdot 2 \equiv 7 \bmod 17$

and $ 12^{23} \equiv 0 \bmod 2$, hence $ 12^{23} \equiv 24 \bmod 34$.  On the first equivalence I have used [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url].

As for direct calculation without theory, you can still speed up the exponentiation a lot using [url=http://en.wikipedia.org/wiki/Exponentiation_by_squaring]exponentiation by squaring[/url].

wickedestjr dont say that doctors s.g and i dont agree. he spotted the mistake and i think the answer would be 50 although i i didnt calc. it.

@torajirou: thanks for the last link

[quote="t0rajir0u"]Your original question is made much simpler by the fact that $ 34$ is not prime.  The standard (paper-and-pencil) technique is to compute the remainder $ \bmod 17$ and $ \bmod 2$ and then apply the [url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem]Chinese remainder theorem[/url].  Here, this gets you

$ 12^{23} \equiv ( \minus{} 5)^{7} \equiv \minus{} 5 \cdot 25^3 \equiv \minus{} 5 \cdot 8^3 \equiv \minus{} 5 \cdot 512 \equiv \minus{} 5 \cdot 2 \equiv 7 \bmod 17$

and $ 12^{23} \equiv 0 \bmod 2$, hence $ 12^{23} \equiv 24 \bmod 34$.  On the first equivalence I have used [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url].

As for direct calculation without theory, you can still speed up the exponentiation a lot using [url=http://en.wikipedia.org/wiki/Exponentiation_by_squaring]exponentiation by squaring[/url].[/quote]

That question has already been solved, but it turns out that Dr Sonnhard Graubner was correct, but what did analytic do wrong?

Dr Grubner already identified analytic's mistake.  In any case, I was demonstrating a solution using the two standard techniques (the Chinese remainder theorem and Fermat's little theorem) that generally make problems like this very straightforward.  The wording of your first post implies that you weren't aware of these theorems.

I've heard of those two theorums before, but I do not understand the meaning of the Chinese Remainder Theorum. Fermat's Little Theorum is the most straightforward theorum I have ever seen, but the Chinese Remainder Theorum is like the opposite.

as dr. spotted it should be -6 in first step in place of 6.
$ 73\equiv\minus{}24 and \minus{}24^2\equiv\minus{}6 and \minus{}6^3\equiv\minus{}22 and \minus{}22^2\equiv\minus{}1mod{97}$ from these,$ \minus{}24^{12}\equiv\minus{}1 mod{97}$
so, $ 73^{15}\equiv\minus{}24^{12}\times\minus{}24^2\times\minus{}24\equiv\minus{}1\times\minus{}6\times\minus{}24\equiv\minus{}144\equiv\minus{}47\equiv50 mod{97}$

[quote="t0rajir0u"][url=http://www.cut-the-knot.org/blue/chinese.shtml]This[/url] explanation might be a little clearer.[/quote]

That did make it a bit easier, but I still don't understand the formula.

[quote="t0rajir0u"]Don't worry about the formula.  The important thing is that the answer exists, and usually when computing it you don't use the formula.  For example, once I knew that the remainder we wanted $ r$ satisfied

$ r \equiv 7 \bmod 17$
$ r \equiv 0 \bmod 2$

there were only two choices:  either $ r \equiv 7 \bmod 34$ or $ r \equiv 24 \bmod 34$ (why?).  And only the second one is even.[/quote]

Because if r = 7 (mod 17), then r = 24 (mod 17), and that's where you get the 24, and then it could be 41 (mod 17) which leaves a 7. Is that right?

EDIT: I got some practice with these kind of problems, and I tried a problem similar to this one. Tell me if this is a good solution;

[hide]96^85 (mod 74)
22^85 (mod 74)
22^(1 + 11) (mod 74)
22^12 (mod 74)
484^6 (mod 74)
40^6 (mod 74)
8^6 x 5^6 (mod 74) - just to make it a little easier
8^6 x (5^3)^2 (mod 74)
8^6 x (125)^2 (mod 74)
8^6 x 51^2 (mod 74)
8^6 x 3^2 x 17^2 (mod 74)
(64)^3 x 289 x 9 (mod 74)
64 x (64)^2 x 67 x 9 (mod 74)
(-10) x 4096 x (-7) x 9 (mod 74)
70 x 4096 x 9 (mod 74)
(-4) x 4096 x 9 (mod 74)
9 x (-16,384) (mod 74)
-(2^14) x 9 (mod 74)
-(2^7) x 2^7 x 9 (mod 74)
-(2^7) x 9 x 128 (mod 74)
-(2^7) x 9 x (-20) (mod 74)
2^7 x 9 x 20 (mod 74) - the negatives cancel out
128 x 180 (mod 74)
54 x 32 (mod 74)
27 x 64 (mod 74) - just multiplied one by 2, and divided one by 2.
27 x (-10) (mod 74)
-270 (mod 74)
26 (mod 74)
  [/hide]


The below problem is a nice and beautiful property that [b]Mr. Dung Tran Nam[/b] (nickname on mathlinks: [b]namdung[/b]) found independently and has been proved by himself in 1983, which is also the year that he attended the IMO competition with Vietnam team:

[b]PROBLEM-[color=darkblue] Let $ ABC$ be a triangle with $ G$ is its centroid. Let $ A_0,B_0,C_0$ respectively are the midpoints of $ BC,CA,AB$. Prove that the circumcenters of triangles $ AGC_0,AGB_0,BGA_0,BGC_0,CGA_0,CGB_0$, respectively are concyclic[/color][/b].

Dear Mathlinkers,
for an entirely synthetic proof see also

http://perso.orange.fr/jl.ayme   vol. 2  Le cercle de van Lamoen

Next, I will put an additionnal historic note concerning Mr. Dung Tran Nam.
Sincerely
Jean-Louis

For arbitrary P, those six circumcenters lie on a same conic.


AMC
AIME
modular arithmetic
algebra
polynomial
trigonometry
geometry

Answers:

1.	120 (6 correct solvers)
2.	019 (3 correct solvers)
3.	017 (0 correct solvers)
4.	648 (2 correct solvers)
5.	289 (3 correct solvers)
6.	070 (1 correct solver)
7.	669 (4  correct solvers)
8.	648 (4  correct solvers)
9.	481 (2  correct solvers)
10.	032 (0 correct solvers)
11.	533 (2  correct solvers)
12.	465 (1 correct solver)
13.	301 (0 correct solvers)
14.	025 (0 correct solvers)
15.	166 (1 correct solver)

Total Number of Participants: 11
Average Score: 2.64
Mode Score: 4, 0

High Scores:

scorpius119:  7   ( 2, 4, 7, 8, 9, 12, 15)

issacchao:      5   (1, 2, 4,  7, 8)

Thanks to people who participated !

Wow, I had the highest score with a SEVEN!!! That shows how impossible it was...

Anyway, an outline of a solution to one of the problems i was the only one to solve...

[hide="15"]If $P(x)=\sum_{k=0}^n a_nx^n$ and $P(-2)=P(-5)=2006$, simplify the eqautions mod 2 and mod 5. Then $a_0\equiv 0\pmod{2}$ and $a_0\equiv 1\pmod{5}$. Combining, $a_0\equiv 6\pmod{10}$. But $a_0$ is between 0 and 9 so $a_0$ must be 6.

To find the coefficient, consider the polynomial $P_1(x)=\sum_{k=0}^{n-1} a_{n+1}x^n=\frac{P(x)-6}{x}$. Then $P_1(-2)=-1000$ and $P_1(-5)=-400$. This gives $a_1\equiv 0\pmod{2}$ and $a_1\equiv 0\pmod{5}\Rightarrow a_1=0$.

Continue with the polynomials $\frac{P(x)-6}{x^2}$ (which shows $a_2$ is also 0), $\frac{P(x)-6}{x^3}$, etc. until you get $P_{n+1} (-2)=P_{n+1} (-5)=0$.

After some computation, find that $P(x)=x^{11}+6x^{10}+6x^9+6x^8+5x^7+x^5+9x^4+4x^3+6\Rightarrow \boxed{166}$.[/hide]

Ooh ooh I can do 14 it's $\sin \angle BAC = \frac{12}{13}$! Actually I'm not 100% sure as I was too lazy to redraw the diagram but it's something like that.

Since $\triangle ABE$ and $\triangle ADC$ share the same altitude, $\frac{[ABE]}{[ADC]}=\frac{BE}{CD}=\frac{59}{61}$. Note that $\angle BAE = \angle DAC$, so $\sin BAE = \sin DAC$

$[ABE]= \frac{AB(AE)\sin BAE}{2}$

$[ADC]= \frac{AD(AC)\sin DAC}{2}$

So $\frac{[ABE]}{[ADC]}=\frac{AB(AE)}{AD(AC)}=\frac{59}{61}$. $m+n=120$.
\

[quote="1234567890"]Ha, i got 2, 6 and 8 (right, I think) before deciding this AIME was impossible and I didn't want to do it. I should've done it, though. Plus, I had my English paper due. :(

Could someone post number one in a way that doesn't coordinatize? I think it is fairly straightforward through coordinazation.[/quote]

Sorry, I meant #3. :(

# 2 is 4/15, right? I think I got the answer through the wrong way, so I'd appreciate it if someone posted the solution for that, too.

[quote="1234567890"][quote="1234567890"]Ha, i got 2, 6 and 8 (right, I think) before deciding this AIME was impossible and I didn't want to do it. I should've done it, though. Plus, I had my English paper due. :(

Could someone post number one in a way that doesn't coordinatize? I think it is fairly straightforward through coordinazation.[/quote]

Sorry, I meant #4. :([/quote]

You mean #6?

No, I meant 3. Wow, I am off today. :huh:

I just let A be on (0,0), B be on (14,0), C be on (5,12) and applied the angle bisector thm to get D as another point. Then I had some messy circles (realizing each pt. of an circumcenter was equidistant to points on a triangle to get center of circles. Then I just found equation of the lines BO and CP, then found the intersection. Very nasty. :mad:

[hide="solution for number 13"]we know that it is an oblique ellipse, so when we rotate it the correct amount, we will have no xy term

we know that $b^2-4ac$ is the same under rotation, and the constant term is the same under rotation

make the constant term 1 by dividing out 2304,  so if we rotate it so the $xy$ term is $0$, and the constant term is $1$, so the coefficients of $x^2$ and $y^2$ are $\frac{1}{a^2}$ and $\frac{1}{b^2}$, also note that the area is $ab\pi$

$\frac{45}{2304}x^2-\frac{36}{2304}xy+\frac{20}{2304}y^2=1$

so under the rotation, we have that $B^2-4AC=B'^2-4A'C'$
$\frac{36^2-4*45*20}{2304^2} =-\frac{4}{a^2b^2} \implies ab= \sqrt{9216}=96$, so $\lfloor A \rfloor = \lfloor 96\pi \rfloor = \boxed{301}$[/hide]

#2

sample space = 6!/6 = 5!

# of legal arrangements:
5!  ( all arangements)
- 2*3C1*4!   (one of the couples together *2 for switching place, 3C1 for there are 3 groups 4! becasue 5!/5)
+ 4*3C2*3!   (add back overlaps when two couples are adjacent. 2*2 for switching, 3C2 for picking 2 couples, 4!/4 for ring arranging)
- 8*3C3*2!   (subtract overlap when all couples are adjacent. 8 for switching, 2! for ring arrangments)

P = (5! - 2*3C1*4! + 4*3C2*3! - 8*3C3*2! )/5! = 4/15

i also noticed that number 13 could be solved (very directly) with integration...just use the quadratic formula to isolate y, then integrate between the 2 functions that you obtain from the formula in the range of numbers that satisfy it (look at the square root produced by the quad formula to find this range)

[quote="beta"][quote="1234567890"]No, I meant 3. Wow, I am off today. :huh:
[/quote]

Okay, surprisingly no one got this:

Outline Solution:

Draw OD and PD, angle BOD = 2(angle BAD) = BAC=DPC. Since triangle BOD and DPC are both isosceles so they are similar so angle OBD = PDC => BQ is parallel to DP => angle DPC = angle BQC = angle BAC. Hence AQCB is cyclic. Also triangle BOD similar to BQC, hence BQ=QC.

Now $\cos BQC = \cos BAC = \frac{5}{13}$.  we apply LoC on BQC, say $BQ=x$

$x^2 + x^2 - 2x^2(\frac{5}{13}) = 15^2$, solve that,

$x = \frac{15\sqrt{13}}{4}$. Now Apply Ptolemy's Theorem on AQBC,

$AQ(15) +13(\frac{15\sqrt{13}}{4}) = 14(\frac{15\sqrt{13}}{4})$

Solve $AQ=\frac{\sqrt{13}}{4}$.

[quote="Silverfalcon"]beta, do you plan to post a solution to this test? It will be very nice to see them especially the geometry ones.[/quote]

I wanted to but it's gonna take forever :([/quote]

Ptolemy's for a #3? Now that's hardcore.

[quote="isaacchao"]i actually started doing that during the exam. I stopped because i thought it would turn out ugly.

i think there is a generalized formula for area of rotated ellipse. something like 2pi/sqrt(b^2 - 4ac)
where a b c are coefficients of x^2, xy, y^2, respectively[/quote]

Rotating an ellipse is simple enough if you forgot the formula.

[quote="paladin8"]

Ptolemy's for a #3? Now that's hardcore.[/quote]

No Ptolemy is not the only approach to calculate AQ. You can do this with Law of Cosines as well, once you see the cyclic quadrilateral. We have calculated above(before Ptolemy)
$QC=\frac{15\sqrt{13}}{4}$, now we can calculate $\cos ABC = \frac{33}{65}$(you have to apply $\cos (\alpha + \beta)$ formula).
So $\cos AQC = -\frac{33}{65}$.

Now apply Law of Cosines on AQC we have

$AQ^2 + (\frac{15\sqrt{13}}{4})^2 - 2 AQ(\frac{15\sqrt{13}}{4}) = 14^2$. Solve for $AQ$ you should get the same answer.

Well, there are probably a few more methods using the cyclic quadrilateral AQCB(like Law of Sines, you can calculate $\sin ABQ$ using $\sin(\alpha - \beta)$ formula), but in my opinion Ptolemy uses the least amount of computation. But the key is to see AQCB is cylic.

sketches of solutions

1. trivial
2. Fix the first guy at the top; either his spouse is sitting directly opposite and you can seat the others in 4*2*2 (4 for where the second guy goes, 2 to decide which one of the third couple is sitting next to the second guy, 2 to switch around the last two people) or his spouse is sitting not-directly opposite from him and you can seat everybody in 2*4*2 ways (2 to decide which not-opposite his spouse is in, 4 to decide who's sitting between him and his spouse, 2 to decide which way the last couple can be seated) for a total of 32 ways out of 120.
3. Calculate angles to find that Q is at the top of the circumcircle; then use Ptolemy to get sqrt(13)/4.
4. It'll be the total number of 4-digit numbers with distinct nonzero digits (9*8*7*6) plus the number of such numbers with the sum of the first+fourth and the second+third digits equal all over two; find that latter amount by noting that if there are x ways for two nonzero distinct digits to add up to n, then there will be 4*x(x-1) ways total in which you can form a four digit number with etc etc whose first+fourth and second+third digit sums are both n, do this as n runs from 5 to 15 to get that this is 272 or so and plug it back into the original formula.
5. 17^2^k = 17^2^k + 1 - 1, break it up now (factor the bottom).
6. E is the excenter opposite C. I didn't feel like being clever so I just dropped the perpendicular from E to AB and found the legs of the right triangle (one side is the exradius K/(s-c) and the other side is s - a).
7. Write everything in ternary (2006 = 2202022) and note that the expression is the sum of the last four digits. Since 2006 ends with a 2, we can group everything into triplets wherein only the last digit changes; there'll be 2007/3 of these. Obviously, for each one, exactly one of the triplets will have the sum of the last four digits divisible by 3.
8. Use binomial formula and wonder how the heck everything cancelled so nicely. Use Fermat-Euler at the end.
9. Basically, the odd numbers come in equivalence classes mod 6 - the first one that we can get congruent to 3 is 27, the first congruent to 1 is 43, and the first congruent to 5 is 53. After we get the first in any equivalence class, we can get all the subsequent ones (the crux here is that it's REALLY EASY to get odd composite numbers divisible by 3 and less so to get ones that aren't).
10. I have a way but it's incredibly brute-force; I didn't feel like doing it during the contest. I'll either work it out or look for a better way and come back later.
11. We have a tangent to an ellipse. I originally wanted to use calculus but decided it might be easier to use that clever fact about light from a focus reflecting off a tangent to meet the other focus - if you let theta be the angle whose sine squared we are looking for and alpha the incident angle = reflected angle, you do some trig to get cos theta = 3 cos alpha and later 3 sin theta = 4 sin alpha; square and add these.
12. I had a method but got the wrong answer. Eh. Possibly an arithmetic error, possibly some crucial flaw in the method. Will check.
13. Use the transformation
(x')      (    1         k  ) (x)
(   ) =  (                   ) (  )
(y')      (    0         1  ) (y)

where those are two column vectors and a matrix to transform the equation. Note this leaves the area constant as the determinant of the matrix is 1 (this is known as a shear, I think). Solving this in terms of k to get rid of that xy-term, we get k = -2/5 or something; now go through the equations to get an ellipse in (thankfully) standard form, and we know how to get the area of that.
14. Didn't even have time to draw the diagram. I think those perpendiculars pass through the circumcenter though, and the incenter of the orthic triangle is the orthocenter.
15. I had a method but got the wrong answer - pretty sure this one was an arithmetical error as my method was pretty arithmetic heavy.

Wow, I am really stupid.

I put $\frac{8}{30}$---->>>>>38  for #2   :blush:


a= -1
b= 3
c= -3
d= -2

Problem:
[u]ab - bc + cd[/u]
(a - c)(b - d)

I read in the book it relays to me that I should go from left to right using addition first and then subtraction.. What bothers me about this is that to get 6/5 from this problem I would have to subtract first and then add. Was the book wrong? What's going on? I don't understand..

The answer in the book says it's 6/5

I'm pretty sure I'm suppose to simplify 12/10 into 6/5 but i'm not sure.. I redid the problem to find a way to get 6/5 but I don't understand why the book told me a person is suppose to go from left to right adding and then subtracting..

if I did addition first and then subtraction I come out with

0/10 I think..

What is the correct process when it comes to addition and subtraction?

[quote="Cyberman"]What is the correct process when it comes to addition and subtraction?[/quote]

order of operation states that for addition and subtraction it does not matter, except when there is a parenthesis

e.g. 3+4-3 = 3+1 =4(subtraction first)

or

e.g. 3+4-3=7-3=4(addition first)

EX) 3-(4+3) = 3-7=-4(parenthesis)
3-(4+3)=-3+1=-2(wrong answer)

How is that related to Algebra ?????
P.S: Could someone move this to a middle school section ?

:ninja:  It was in my Algebra book.  :ninja:

What bothered me here was this:

(-1 x 3) - (3 x -3) + (-3 x -2) =

-3 - -9 + 6

-9 + 6 = -3
-3 - -3=0

 :|  But yet if i do it a different way i get 12...

So, order of operation does not matter for subtraction or addition...meaning it does not matter which sign is there, as long as i go from left to right.
This means it is most important that I make sure I go from left to right though.

 :?: Something like that?

If a person must always go from left to right, then I would obtain 12.. ok then..

just watch your signs when you plug in the values... after plugging in everything and doing the operations in the parenthesis you should get:

$\frac{-3-(-9)+6}{2*5}$, which equals $\frac{12}{10}$, then $\frac{6}{5}$


function
calculus
derivative
complex analysis
complex numbers
complex analysis theorems

f - complex function of one variable. What is the difference between R-differentiability and C-differentiability? Can you please give some examples?

 Thanks in advance.

if its real differentiable that means its two partial derivatives exist, if its complex differentiable, it means its two partial derivatives exist, AND satisfy the Cauchy-Riemann equation. Basic example:

$ f(x \plus{} iy) \equal{} x \minus{} iy$. This is the simple "complex conjugation map. As a real function in two variables, it looks like $ f(x,y) \equal{} (u(x,y),v(x,y))$ where $ u(x,y) \equal{} x$ and $ v(x,y) \equal{} \minus{} y$. These functions are real differentiable (hence you can write down the Jacobian etc.) Now, for complex differentiability, the key is to see if it satisfies the Cauchy Riemann equations (which basically means taking the limit as $ h\rightarrow 0$ is the same no matter which direction $ h$ goes to $ 0$). Lets see if the Cauchy-Riemann conditions are satisfied:

They read $ \partial_x u \equal{} \partial_y v$ and $ \partial_y u \equal{} \minus{} \partial_x v$. Ok, so lets see if this works:

$ 1 \equal{} \minus{} 1$ and $ 0 \equal{} 0$. So the first equation isn't satisfied. There you have it, the function is real differentiable and not holomorphic. In fact, holomorphicity puts a HUGE restraint on the niceness of the function. It turns out holomorphic functions are: smooth, analytic and determined by how they look on an arbitrary small neighborhood of $ \mathbb{C}$. That is, if there exists open sets $ U$ and $ V$ such that two functions satisfy $ f|_U \equal{} g|_V$ then $ f \equal{} g$

Thanks LkNsngth! Trying to understand your last statement :).


LaTeX
induction
number theory
relatively prime

let n be an integer $ > 2$,

P.T.    $ 2^2^^^n ^+^ 1$ +$ 2^2^^^n^$  + $ 1$  is a product of three integers always $ > 1$

When n=4, we have $ 2^8\plus{}1\plus{}2^8\plus{}1\equal{}514\equal{}2\cdot257$ which is a counterexample.

wouldn't it be 2^10+2^8+1? (I'm assuming he meant 2^(2(n+1))+2^(2n)+1

That would give 3*61*7.

I don't have time now, I'll solve it later.

Oh sura edited it.

Ok I will give it another go:

$ 2^{2(n\plus{}1)}\plus{}2^{2n}\plus{}1\equal{}5\cdot2^{2n}\plus{}1$

Taking it mod 3, we have $ 5\cdot 1 \plus{}1\equal{}6$ so it is divisible by 3.

Oh hey wait, if that is a correct interpretations, then n=3 gives $ 2^8\plus{}2^6\plus{}1\equal{}321\equal{}3\cdot107$ so it is still untrue.

[quote="Mewto55555"]Oh sura edited it.

Ok I will give it another go:

$ 2^{2(n \plus{} 1)} \plus{} 2^{2n} \plus{} 1 \equal{} 5\cdot2^{2n} \plus{} 1$

Taking it mod 3, we have $ 5\cdot 1 \plus{} 1 \equal{} 6$ so it is divisible by 3.

Oh hey wait, if that is a correct interpretations, then n=3 gives $ 2^8 \plus{} 2^6 \plus{} 1 \equal{} 321 \equal{} 3\cdot107$ so it is still untrue.[/quote]

SORRY Mewto55555.about another edision as i am a new learner of latex[/hide]

I believe a generalization of this is in Engel. We can prove that it has at least $ n\plus{}1$ different prime factors through induction.

[hide="solution outline"] Because I don't feel like typing more than I need to, let $ a\equal{}2^{2^{n\minus{}1}}$.

Then $ 2^{2^{n\plus{}1}}\plus{}2^{2^{n}}\plus{}1\equal{}a^4\plus{}a^2\plus{}1\equal{}(a^2\plus{}1)^2\minus{}a^2\equal{}(a^2\minus{}a\plus{}1)(a^2\plus{}a\plus{}1)$.

Through induction, you can prove that if $ a^2\plus{}a\plus{}1\equal{}2^{2^{n}}\plus{}2^{2^{n\minus{}1}}\plus{}1$ has $ n$ different prime factors, then $ 2^{2^{n\plus{}1}}\plus{}2^{2^{n}}\plus{}1$ has at least $ n\plus{}1$, as the two factors are relatively prime, from the Euclidean Algorithm.[/hide]

Bleh, if we only want to prove that  this thing always has at least 3 factors greater than 1 when $ n > 2$, then all we need to do is let $ a \equal{} 2^{2^{n \minus{} 2}}$. Then
\[ 2^{2^{n \plus{} 1}} \plus{} 2^{2^n} \plus{} 1 \equal{} a^8 \plus{} a^4 \plus{} 1 \equal{} (a^2 \minus{} a \plus{} 1)(a^2 \plus{} a \plus{} 1)(a^4 \minus{} a^2 \plus{} 1).\]
Since $ n > 2$, it follows that $ 2^{n \minus{} 2} > 1$ and $ a\equal{}2^{2^{n \minus{} 2}} > 2$, and is a positive integer, and so all factors are positive integers greater than 1, QED.


[quote="bgbgbgbg"]segma (n^n) from 1 to 2001[/quote]

No closed formula for this sum.

The guy who invented this problem and asked it to you is quite idiot. Forget him.

i want to know the first digit in the sum  :1^1+2^2+3^3+4^4+....+2001^2001
i.e (he digit of ones) for example: 1234567 the first digit here is 7

Just add all those 21 equalities written (not just the last one, you refer at). I used the obvious fact that $ \ell(n \plus{} m) \equal{} \ell(\ell(n) \plus{} \ell(m))$, and the periodicity (easy to check) of the last digit of a power of a digit (the orbits are $ (0), (1), (2,4,8,6), (3,9,7,1), (4,6), (5), (6), (7,9,3,1), (8,4,2,6), (9,1)$).

EDIT. In answer to a post that has meanwhile been deleted ...


[quote="math92"]The lengths of a right triangle are 6, 100, and x.  Find x using the pygortheaguem(how do you spell it?) thereom.[/hide][/quote]

Would 6 and 100 be the two legs, or is 6 a leg and 100 the hypoteneuse?

[hide="The First Case"]$6^2+100^2=10036=2\sqrt{2509}$
[/hide]
[hide="The Second Case"]$100^2-6^2=9964=2\sqrt{2491}$[/hide]

First of all, it is called the "Pythagorean theorem"

Concerning the solution to your problem:

[hide]  Consulting the "triangle inequantity" will tell you that the two smaller sides of a triangle must have a sum greater than the longest side of the triangle.  Evidently the hypotenuse of this triangle cannot be 6, because 100 is bigger and the hypotenuse must always be the largest.  Thus it can be either 100 or x.  Now we must evaluate both cases and see whether each case satisfies both the "pythagorean theorem" and the "triangle inequantity".

Case 1:
a=6, b= x, c=100

6^2 + x^2 = 100^2

36 + x^2= 10000

x= 99.819837 apporxamitely

The two smallest sides: 99.819 + 6 > 100

Case 2:

a= 6, b= 100, c=x

6^2 + 100^2 = x^2

36 + 10000 = x^2

x=100.17983

The two smallest sides:  6+100 > 100.17983

  Since both cases work x can be either 9964sqrt or 99.81   or   10036 sqrt or 100.17.  ALthough x is only a tenth sometimes a hundreth of of being the same value of one of teh other legs and ultimately making the answer invalid, the fact remains it is still a teeny bit smaller, and the answer remains the same.[/hide][/hide]

sorry guys, i meant that 6 and 100 were the measurements of the legs.

[quote="math92"]what's lmao[/quote]
i don't think i want to spell it out. as for the answer, i got $x\approx100.17983$

This is getting a little out of hand. I know the moderators are itching to get the posts out of this topic.

[hide="anyway"]$6^2 + 100^2 = c^2$
$10036 = c^2$
$c = \sqrt{10036}$
$c = 2\sqrt{2509}$[/hide]

thanks everyone for telling me how to spell it! :lol:

It's Pythagorean theorem. It is kind of hard to spell it. :). According to Pythagorean theorem, $6^2$+$100^2$=$x^2$. Therefore, $x^2=10036$. So $x=\sqrt{10036}=100.179$ ;) .

u sound like a EXCITED kid @ math 92
lol i love excitin ppl nd ppl who talk a lot!

[quote="math92"]The lengths of a right triangle are 6, 100, and x.  Find x using the pygortheaguem(how do you spell it?) thereom.[/hide][/quote]

[hide]you have to assume 2 possibilities.

One is the legs are 6 and 100. then the hyp would be 10036
The other is that 100 is the hyp. then the other leg would be 9964[/hide]


Suppose that $ r\equal{}\frac{1}{n}\sum_{i\equal{}1}^n \lambda _i$ is an algebraic integer where all $ \lambda _i$ are roots of unity.
Show that either $ r\equal{}0$ or all $ \lambda_i$'s are equal to each other.

Generally, the ring of integers in $ \mathbb{Q}[\zeta_m]$ is $ \mathbb{Z}[\zeta_m]$.  I'm not sure how to prove it, though.

the hint that was given together with the problem
Consider the cyclotomic extension containing these roots of unity, and look at the norm.

It is also a direct consequence of a theorem of Kronecker that we discussed on the forum: the only algebraic integers all of whose conjugates are on the unit circle are the roots of unity.


in hungkhtn's post about SOS, the document mentioned somethig about the Basic-form of S.O.S method and some analysis techniques do you know where i could get it, could someone explain it to me (like how would you rewrite an inequality in terms of $\sum (x-y)^{23}$ etc...

thanks in advance

As I wrote in my book "Sang tao bat dang thuc" published in Vietnam, this skill (how to analyze an expression to square) is very simple. There is only one theorem proving the existance of this representation. If you can wait, see the book "SI Inequalities, A story of Math" which will be published in Gil this summer. There is a lot of SOS-kind problem and solution there to help you skillful at this method.

thanx for your reply

unfortuantely i cnat get that book for GIL :(

care to enlighten me with this simple skill

As to answer your question, the SOS-representation of your expression is very easy,

\[\sum (x-y)^{23}=\sum (x-y)^{2}(x-y)^{21}.\]

that was a typo :blush:  is was meant to be $\sum (a-b)^{2}$

my question is are there any certain type of analyzing technique which could convert an inequality interms of $(a-b)^{2}, (b-c)^{2}, (c-a)^{2}$

if so could you post it

thanks

where cn i find ‘Squares analysis method s.o.s’？

[quote="hungkhtn"]Actually, it is a quite simple but effective method posted here

http://www.mathlinks.ro/Forum/viewtopic.php?t=80127[/quote]

but it is just a part

Every part is found in Sang Tao bat dang Thuc, published in 2006 in Vietnam, written by me. In the second volume which is published in Gil this summer, SOS is considered much meticulously and carefully.


algebra
polynomial
inequalities
Cauchy Inequality
algebra unsolved

let $P(x) =x^{2004}-x^{2002}+3$
  if a,b,c positive real  then prove that :
 $P(a).P(b)P(c) \geq 9(ab+bc+ca)$

oh, hien's solutiom is very nice .
  thank you, Hien .


inequalities
calculus
integration
function
calculus computations

Prove that every continuous functions $ f : [\minus{}1,1] \to \mathbb{R}$ takes place the inequality :
    $ \int_{\minus{}1}^{1}{f^2(x)dx} \geq \frac{1}{2}(\int_{\minus{}1}^{1}{f(x)dx})^2\plus{}\frac{3}{2}(\int_{\minus{}1}^{1}{xf(x)dx})^2$


[code]\documentclass{minimal}
\usepackage[svgnames]{xcolor}
\usepackage{pst-plot,pstricks-add}

\begin{document}
  \begin{pspicture}[showgrid=true](-4,-4)(9,6)
    \psaxes[tickstyle=bottom,ticksize=3pt]{<->}(0,0)(-4,6)(9,-4)
    \pscustom[fillstyle=solid,fillcolor=yellow,linestyle=none]{%
      \psplot{0}{3}{x x add}
      \psplot{0}{3}{x}
      \psline[liftpen=1](6,6)(3,6)
    }
    \rput(2.3,3){A}
    \psplot{-1}{3}{x x add}
    \psplot{-1}{6}{x}
    \pscircle(2,0){2}
    \pscircle(4,0){4}
  \end{pspicture}
\end{document} [/code]

I must make a color amongst big,small circle and y=x,y=2x which is A region.I help you.

I want to make like this:
http://www.mathematica-pro.com/math/images/graphic.jpg

psclip would seem to be the answer. See section 28 Clipping in the [url=http://www.dante.de/CTAN/graphics/pstricks/base/doc/pstricks-doc.pdf]PSTricks manual[/url] and the examples at [url=http://tug.org/PSTricks/main.cgi?file=Examples/clip]Clipping[/url] and further examples at [url=http://tug.org/PSTricks/main.cgi?file=pst-plot/FunctionExamples#reciprocal]Reciprocal function[/url]


[color=blue]Solve in real numbers the follwing equation?????[/color]

[size=200](x-1)^1/2 +(2-x)^1/2  =x^2 -3x+3[/size]

[color=blue]hide the solution [/color]

[hide="Solution"]

We can see that two solutions are $ x\equal{}1$ and $ x\equal{}2$.

Squaring both sides,

$ 1 \plus{} 2\sqrt{(x\minus{}1)(2\minus{}x)} \equal{} x^{4} \minus{}6x^{3} \plus{}15x^{2} \minus{}18x \plus{} 9$

$ 2\sqrt{(x\minus{}1)(2\minus{}x)} \equal{} (x\minus{}1)(x\minus{}2)(x^{2} \minus{}3x\plus{}4)$

Again we can see that $ x\equal{}1$ and $ x\equal{}2$.

Squaring both sides again,

$ 4(x\minus{}1)(2\minus{}x) \equal{} (x\minus{}1)^{2}(x\minus{}2)^{2}(x^{4}\minus{}6x^{3} \plus{}17x^{2} \minus{}24x\plus{}16)$

$ 4\equal{} (2\minus{}x)(x\minus{}1)(x^{4}\minus{}6x^{3}\plus{}17x^{2}\minus{}24x\plus{}16)$

Now it's getting a bit messy. I suspect that there aren't any other real solutions.

[/hide]

yes there aren't any other real solutions

    but there are another way to solution

Hi salhcool.

What is the other way to the solution?


inequalities
induction
topology
triangle inequality
real analysis
real analysis unsolved

I can't get this problem at all. I've read through compactness, connectness, and Cauchy sequences in Analysis, but this seems particularly difficult. Can anyone help me with it?

Let $E$ be a compact metric space, $\{U_{i}\}_{i \in I}$ a collection of open subsets of $E$ whose union is $E$. Show that there exists a real number $\epsilon > 0$ such that any closed ball in $E$ of radius $\epsilon$ is entirely contained in at least one set $U_{i}$.

The hint the textbook gives (Introduction to Analysis, Max Rosenlicht) is to use contradiction: "take bad balls of radii $1, \frac{1}{2}, \frac{1}{3}, \ldots$ and a cluster point of their centers. But my problem is... I have no idea what this means.

I do know a cluster point is any point $p$ in a subset $S$ of a metric space $E$ such that any open ball with center $p$ contains an infinite number of points of $S$.

Here's my take on the problem: I was kind of leaning towards using $Contradiction$ to prove this. So assume this $\epsilon$ does not exist. So any closed ball has to be contained in the union of at least 2 different sets, $U_{1}, U_{2}$. I somehow want to lead this up to finally contradict the statement That "$E$ is a compact metric space.

can anyone give any useful pointers? Also, I appreciate the vast knowledge of many posters here, but please explain in elementary terms of compactness, cauchy sequences, connectness, etc... if possible b/c this is an introductory analysis course problem and all the problems should be solvable using basic properties.

Thanks a bunch.

Please help me
I have a problem with Abel
Where can i find something about Abel?
              Thanks

[quote]Show that there exists a real number $\epsilon > 0$ such that any closed ball in $E$ of radius $\epsilon$ is entirely contained in at least one set $U_{i}$.[/quote]
In a proof by contradiction, you start by assuming the negation of this statement, which is:
[quote]For every real number $\epsilon > 0$ there exists a closed ball in $E$ of radius $\epsilon$ that is not entirely contained in any set $U_{i}$.[/quote]
(Note that you formed the negation incorrectly.)

This assumption can be used with any $\epsilon>0$, in particular with $\epsilon=1$, $\epsilon=1/2$, etc. Thus, for each positive integer $n$ there exists a closed ball $B_{n}$ of radius $1/n$ such that $B_{n}$ is not contained in any $U_{i}$.

The centers of these balls form an infinite sequence in $E$. Any infinite sequence in a compact space has at least one cluster point. Let $p$ be a cluster point for this sequence. Since $p\in E$ and $E$ is covered by the sets $U_{i}$, there is $i_{0}$ such that $U_{i_{0}}$ contains $p$. Since $U_{i_{0}}$ is open, it also contains an open ball centered at $p$. Let $r$ be the radius of this ball. It's customary to use notation $B(p,r)$ for open balls and $\overline{B}(p,r)$ for closed balls.

It remains to prove that $B_{n}\subset U_{i_{0}}$ when $n$ is sufficiently large, since this will contradict our assumption. By the definition of a cluster point, the center of $B_{n}$ belongs to $B(p,r/2)$ for infinitely many values of $n$. Choose $n>2/r$ for which this happens. By the triangle inequality, $B_{n}$ is contained in $B(p,r)$ and therefore in $U_{i_{0}}$, a contradiction.

-------
Abel is dead, [b]angelinheart_147[/b]. Abel is dead.

If you have a problem with Abel, you have a problem with me.  :mad:

What you need is called Abel's lemma, or [url=http://en.wikipedia.org/wiki/Abel_transformation]Abel's transformation[/url], or summation by parts. Whatever. Of course, we want to use the arithmetic-geometric mean inequality. However, the numbers $a_{k}$ are probably not all equal in the extremal case, which would make the AGM inequality not so sharp. Let's adjust for this by switching to $b_{1}=1\cdot 2\cdot a_{1}$, $b_{2}=3\cdot 4\cdot a_{2}$, ... $b_{k}=(2k-1)(2k)a_{k}$ for $k=1,\dots,n$. Now we have $b_{1}\ge 1$, $b_{1}b_{2}\ge 1$, $b_{1}b_{2}b_{3}\ge 1$, etc. The AGM inequality implies $b_{1}\ge 1$, $b_{1}+b_{2}\ge 2$, $b_{1}+b_{2}+b_{3}\ge 3$, etc. To estimate $\sum_{k=1}^{n}\frac{b_{k}}{(2k-1)2k}$ from below, apply Abel's transformation with $B_{k}=b_{1}+\dots+b_{k}$, which is at least $k$. Hmm... the sum becomes messy, but since the coefficients of $B_{k}$ are positive, the sum is minimized when $B_{k}=k$ for all $k$. This happens precisely when $b_{k}=1$ for all $k$, and it remains to prove that \[\sum_{k=1}^{n}\frac{1}{(2k-1)2k}=\sum_{k=1}^{n}\frac{1}{n+k},\] by induction in $n$.


Alice and Barbara play a game with a pack of 2n cards, on each of which is written a positive integer. The pack is shuffled and the cards laid out in a row, with the numbers facing upwards. Alice starts, and the girls take turns to remove one card from either end of the row, until Barbara picks uo the final card. Each girl's score is the sum of the numbers on her chosen cards at the end of the game.
Prove that Alice can always obtain a score at least as great as Barbara's.

[hide]let the cards be coloured alternatively, black and white.

so that it is like this

B W B W B W ........  B W B W

now Alice finds the sum of all the integers on black cards, let this be sumB and she finds the sum of all integers on white cards, let this be sumW.

just suppose sumB>=sumW

Alice picks the B on the left hand side and this leaves Barbara to choose a white card, because the leftmost and rightmost card will be White.

This allows Barbara to pick another black card, forcing Alice to pick a white card and so on.

So in the end, Alice's sum will be sumB which is >= sumW (Barbara)[/hide]


absolute value
number theory proposed
number theory

if $c$ is a positive  integer and $p$ is an ood prime ,what is the smallest residue (in absolute value) of $\sum_{n=0}^{\frac{p-1}{2}}{2n\choose n}c^n (mod p)$.


geometry
circumcircle
incenter
geometry proposed

Let M be a variable point on the side BC of the equilateral triangle ABC, I and J the incenters of the triangles ABM and AMC respectively, while K being the circumcenter of the triangle AIJ.  Then MK and BC are perpendicular.

Best regards,
sunken rock

Since $ \angle AJM \equal{} \angle AIM \equal{} 120^{\circ}$ and $ \triangle KIJ$ is equilateral, it follows that the pair of lines $ JK,IK$ are the isogonals of $ JA,IA$ with respect to $ \triangle MIJ,$ since they form equal and oppositely directed angles WRT $ IJ,MJ$  and $ IJ,MI,$ respectively $\Longrightarrow$ $A,K$ are isogonal conjugates in $ \triangle MIJ$ $\Longrightarrow$ $ MK, MA$ are isogonal rays $\Longrightarrow$ $ \angle IMK \equal{} \angle AMJ.$  But $ \angle AMJ \equal{} \angle JMC$ (MJ is bisector of <AMC) and $ \angle IMK \plus{} \angle KMJ \equal{} 90^{\circ}$ $ \Longrightarrow$ $ \angle KMJ \plus{} \angle JMC \equal{} 90^{\circ}$ $ \Longrightarrow$ $ KM \perp BC,$ as desired.

Well, Luis, I got an easier way to solve this: if B' and C' are circumcenters of triangles BMI and CJM respectively, then B' is the midpoint of arc BM of the circle (ABMJ) i.e. A, I and B' are collinear,  C' the midpoint of arc CM of (AIMC), thence MK, B'J and C'I are concurrent and make 60 degs angles among them, that is, they are perpendicular on BC, CA and AB respectively.

Note: I hope the same, this to assist in solving that locus but, I do not see how, as yet!

Best regards,
sunken rock

sunken rock, why are $ MK$, $ B'J$, and $ C'I$ concurrent?

Edit: I see it now.  Triangles $ IJK$, $ IB'M$, and $ JC'N$ are all equilateral, so $ MK$, $ B'J$ and $ C'I$ concur at the Fermat point of triangle $ MIJ$.


Hi. Just a quick question to ask...

I have submitted solution to the round 2 problems; right next to round 2, I have green check mark to assure me that my solution has been logged in. However, I am wondering why my score is not up yet? Is it that scores of small fraction of people have not been yet posted? I am not sure what happened here! Please suggest me what I need to do in order to find out my score..

Thank you very much-

P.S. Happy Holidays- Happy New Year!!

If your results didn't get posted, make a post in the grading disputes forum, and an admin will check it out for you.


trigonometry
inequalities
inequalities proposed

$ (\forall)a,b,c,d,x,y\in \mathbb{R}$ show that :

$ \frac{(a\plus{}b\sin\ x)^2\plus{}(c\plus{}d\sin\ y)^2\plus{}(a\plus{}b\cos\ x)^2\plus{}(c\plus{}d\cos\ y)^2}{a^2\plus{}b^2\plus{}c^2\plus{}d^2}\le 3$

[quote="alex2008"]$ (\forall)a,b,c,d,x,y\in \mathbb{R}$ show that :

$ \frac {(a \plus{} b\sin\ x)^2 \plus{} (c \plus{} d\sin\ y)^2 \plus{} (a \plus{} b\cos\ x)^2 \plus{} (c \plus{} d\cos\ y)^2}{a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}\le 3$[/quote]
Thank you, Alex :lol:
$ \frac {(a \plus{} b\sin\ x)^2 \plus{} (c \plus{} d\sin\ y)^2 \plus{} (a \plus{} b\cos\ x)^2 \plus{} (c \plus{} d\cos\ y)^2}{a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}\le 3$

$ \Leftrightarrow 2(a^2 \plus{} c^2) \plus{} b^2(\cos^2x \plus{} \sin^2x) \plus{} d^2(\cos^2y \plus{} \sin^2y) \plus{} 2(ab\sin x \plus{} cd\sin y \plus{} ab\cos x \plus{} cd\cos y)\le 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)$

$ \Leftrightarrow 2(a^2 \plus{} c^2) \plus{} b^2 \plus{} d^2 \plus{} 2(ab\sin x \plus{} cd\sin y \plus{} ab\cos x \plus{} cd\cos y)\le 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)$

$ \Leftrightarrow a^2 \plus{} 2b^2 \plus{} c^2 \plus{} 2d^2 \minus{} 2ab\sin x \minus{} 2cd\sin y \minus{} 2ab\cos x \minus{} 2cd\cos y \ge 0$

$ \Leftrightarrow a^2(\cos^2 x \plus{} \sin^2 x) \plus{} 2b^2 \plus{} c^2(\cos^2 y \plus{} \sin ^2y) \plus{} 2d^2 \minus{} 2ab\sin x \minus{} 2cd\sin y \minus{} 2ab\cos x \minus{} 2cd\cos y\ge 0$

$ \Leftrightarrow (a\cos x \minus{} b)^2 \plus{} (a\sin x \minus{} b)^2 \plus{} (c\cos y \minus{} d)^2 \plus{} (c\sin y \minus{} d)^2\ge 0$
Which is always true  :)

Nice , Khanh . My solution use Cauchy's inequality :

$ LHS\le 2\plus{}\frac{(a^2\plus{}b^2)(\sin^2\ x\plus{}\cos^2\ x)\plus{}(c^2\plus{}d^2)(\sin^2\ y\plus{}\cos^2\ y)}{a^2\plus{}b^2\plus{}c^2\plus{}d^2}\equal{}3$ , so we are done .


prove the product of n consecutive integer number is divisible by n!
sounds right, but... how??

[quote="NapoleonXIV"]pruduct of n consecutive integers in $ n!$[/quote]
He means any string of n consecutive integers I think.
Proof (I think this works...):
We know that C(n+k,n) is a natural number if k is a natural number, because C(n,k) is a natural number for all natural n and k, with n>=k.
C(n+k,n)=(n+k)(n+k-1)(n+k-2)...(k+1)/n!. Therefore n! must divide any string of n natural numbers.


[color=darkblue]$ \lim_{n \to \infty} \left [ cos(\pi n \sqrt[3]{n^3\plus{}3n^2\plus{}n\plus{}1})\plus{} sin(\pi n \sqrt[3]{n^3\plus{}3n^2\plus{}n\plus{}1})  \right ]$ , $ n \in N$.[/color]

Let $ x_n\equal{}\sqrt[3]{n^3\plus{}3n^2\plus{}n\plus{}1}\equal{}\sqrt[3]{(n\plus{}1)^3\minus{}2n}\equal{}(n\plus{}1)[1\minus{}\frac{2n}{3(n\plus{}1)^3}\plus{}O(n^{\minus{}4})]$.
Therefore $ y_n\equal{}\pi nx_n\equal{}\pi n(n\plus{}1)\minus{}\frac{2\pi}{3}\plus{}O(n^{\minus{}1})$ and
$ cosy_n\equal{}\minus{}\frac 12 \plus{}O(n^{\minus{}1}),siny_n\equal{}\minus{}\frac{\sqrt 3}{2}\plus{}O(n^{\minus{}1})$.


geometry
perimeter
trigonometry
trig identities
Law of Sines

Given an obtuse triangle ABC, where c>90. If all sides are integers, and angle A is twice angle B, find the minimum perimeter of ABC.

Is the answer 77 by any chance??

[color=cyan]I hope whatever answer you get doesn't show up on your homework . . .[/color]

wat do u mean?

i did the problem today, and i got 77 as my answer.

is it correct?

[color=cyan]Ha, like I know the answers to the questions that end up on the board in that room . . .  I'm too tired to work on it now.  Maybe later.  How did you do it?[/color]

I think we can apply the law of sines to this problem.  Also we need to know some values such as sin 15, sin 75...and others that are multiples of 15 (except for obvious ones such as 30, 45, etc...).

-interesting_move

[color=cyan]Why do you suppose the angles will work out to such nice numbers?[/color]

Well, those numbers must have some common divisor (not necessarily rational) in order for the sides to be integers right?

-interesting_move

basically, wat i did was i drew two heights, one from A and one from B, and then using the definition of cosine, i manipulated the expressions so that the lengths of the sides were x, 2xcosA, and 2x(4cosA*cosA-1). Since we want x to be minimized, we cosA to have the smallest possible denominator, which comes out to be 8. cosA=7/8. Thus, the sides are 16, 28, and 33.

[color=cyan]When you say, "those numbers," what numbers are you refering to?  Also, all numbers share "not necessarily rational" divisors -- you can factor a real number into real numbers in very many ways.[/color]

yes but, the perimeter is definitely the smallest when cosA=7/8. Try it out.

[color=cyan]Well, your answer definitely works . . . I'm approaching it differently, though, wondering if I can get at it without using trig.  From what you've written, I'm not convinced that it must be the smallest.[/color]

well, think bout it like dis.

cosA=rational number, in m/n, where (m,n)=1. Therefore, since 2xcosA is an integer, 2x must divide n. Hence, if n is increased, x is also increased, which is not good. Therefore, u r looking for the smallest possible n, which is 8.

[color=cyan]I understand everything you're saying now, but I don't follow at all how you found 7/8 to be the best . . . what about, say, 2/3?[/color]

well, u kno that 3A<90, since ABC is obtuse, which means A<30. it follows tthat sqrt[3]/2<cosA<1. sqrt[3] is app. 0.866. So i try to find the fraction with the lowest denominator that is bigger than 0.866. ANd that is 7/8, which is .875. 6/7, for exmaple, will not work since it is .857, and 2/3 obviously is less than .866.

BTW, if i hand this solution in to u guys, won't it be so obvious who i am?? And wats so bad about this username? Its just not as conventional as ur initials. LoL.

[color=cyan]I don't check the homework from that class, thankfully.  I have enough with the geometry.  I am mostly convinced you are correct, partly because it doesn't seem likely that there is a smaller solution.  Actually, I could probably get mathematica to check all cases in a couple of minutes.  But at the very least, the next time you explain something, explain all of it at once, don't wait until the 6th post to give a fundamental justification for one of your statements.  It does seem like you're right, though.[/color]

[quote="interesting_move"]Also we need to know some values such as sin 15, sin 75...and others that are multiples of 15.[/quote]

If anybody didn't know, you can often use the laws of sines/cosines of sums and differences. sin 15 = sin(45-30). sin 75 = sin(45+30) and so on and so forth.

-Isaac


linear algebra
matrix
number theory
greatest common divisor

Can anyone help me with this question?

For $n \geq 1$ let

$B_{n}=\left(\begin{array}{cc}a_{n}&b_{n}\\ c_{n}&d_{n}\end{array}\right) = \left(\begin{array}{cc}3 &2\\ 4&3 \end{array}\right)^{n}-\left(\begin{array}{cc}1 &0\\ 0&1 \end{array}\right),$

Denote $g_{n}$ the greatest common divisor of $a_{n}, b_{n}, c_{n}, d_{n}$. Show that $lim_{n \to \infty}g_{n}= \infty$

[quote="violetcraze"]Can anyone help me with this question?

For $n \geq 1$ let

$B_{n}=\left(\begin{array}{cc}a_{n}&b_{n}\\ c_{n}&d_{n}\end{array}\right) = \left(\begin{array}{cc}3 &2\\ 4&3 \end{array}\right)^{n}-\left(\begin{array}{cc}1 &0\\ 0&1 \end{array}\right),$

Denote $g_{n}$ the greatest common divisor of $a_{n}, b_{n}, c_{n}, d_{n}$. Show that $lim_{n \to \infty}= \infty$[/quote]

Try figuring out the odd-even parity of the elements in the matrix taken to the nth power.

yeah, i noticed that. but how is that going to help me out with figuring out a conclusion for the gcd?    :maybe:

[quote="violetcraze"]I know that the matrix $B_{n}$ is in the form of $\left(\begin{array}{cc}a&b\\ 2b&a\end{array}\right)$
but how is that going to help?[/quote]
With $(a+1)^{2}-2b^{2}=1,$ which you can rewrite as $a(a+2)=2b^{2}.$

The gcd you're looking for is $\text{gcd}(a,b).$ What can you get from that last equation in $a$ and $b$?


Prove that for any prime $p>5$ the equation $x^{4}+4^{x}=p$ does not have any whole number solution.

[quote="Karth"][hide="Lazy Outlined Solution"]
[b]Case 1[/b]
If $x \equiv 0\bmod{2}$, obviously $p$ has no solutions.

[b]Case 2[/b]
If $x \equiv 1\bmod{2}$, let $p = x^{4}+4^{2x+1}= x^{4}+4\cdot 4^{2x}.$ The result follows from Sophie Germain. Oh yea, and you have to prove that none of the resulting factors are one. I'll do that later...
[/hide][/quote]

 :)  That's pretty much all you need. Yep, that's how I did it.

[quote="Karth"][hide="Lazy Outlined Solution"]
[b]Case 1[/b]
If $x \equiv 0\bmod{2}$, obviously $p$ has no solutions.

[b]Case 2[/b]
If $x \equiv 1\bmod{2}$, let $p = x^{4}+4^{2x+1}= x^{4}+4\cdot 4^{2x}.$ The result follows from Sophie Germain. Oh yea, and you have to prove that none of the resulting factors are one. I'll do that later...
[/hide][/quote]
You should put $x=2k+1$.

for n odd
$n^{4}+4^{n}=(n^{2}+2^{n}+2^{k+1}n)(n^{2}+2^{n}-2^{k+1}n)$
its quite obvious that none of the factors is one.

[quote="white_horse_king88"]A $\pmod{5}$ argument works very well here.[/quote]
i would like to see that. :wink:

well $x^{4}\equiv 1 \mod 5$ (unless $x\equiv 0 \mod 5$)

so $1+(-1)^{x}\equiv p \mod 5$

if $p=1,3$ or $4 \mod 5$
$(-1)^{x}\equiv 0,2$ or $3 \mod 5$

which is impossible.

p can obviously not equal 0 $\mod p$

as for the $p\equiv 2 \mod 5$ case, i can't think of anything... not sure about the $x\equiv 0 \mod 5$ either

Well, I clearly miss-stated. My bad. That what you get for doing quick stuff in your head that's wrong.

I might be thinking of another problem (esp. after white knights comments) but I remember having a proof using only mod 5 (this was the WOOT problem, right?) because there was some WOOT problem where they pulled out Sofie Germain that I got without using it... :maybe:
perhaps we made the same mistake? (but I was pretty confident with mine I remember)

I also used $\bmod 5$ the first time I saw this problem, and it doesn't work.


Looks like a hat in the middle of a triangle. I don't know about the shade of blue that you chose, are there any other shades that you could change it to?

P.S. I like it :D

Actually the two last questions are importand... but thanks!

It reminds me of a Texan cult symbol :rotfl:

It's a cool cult symbol though. I like it :D

[quote="thatin"]So what so you think about this?
Does it remind you something?
And if yes what?[/quote]

To me, it looks like a bell inside the star of david.  Does it symbolize a synagogue?

A star of David is like this [img]http://www.isymbolz.com/spiritual/judaism/jd008-Star%20of%20David-04.jpg[/img]

[quote="thatin"]right! well both have 2 triangles but... they are different![/quote]

Yeah... true.  My mind just has a tendency to make random associations.

anyway I think that I shall will delete it... too many associations with signs of religions

how about this?

I like it, but the thingy in the middle reminds me of a mix between omega and a cowboy hat.

i dont like it sorry
it reminds me of texas and it kinda scares me about texan diamond sellers

no problem...!
at least I know that you are honest!

are you trying to find out what the logo is or sumthin

do u really think that thingy looked like the star of david?? cuz i thot it was a star...
ps its my star...im david!!!!!

I think that it looks like a top view of a triangular pyramid, with a platform on top. Maybe the platform is a helicopter landing pad? I dunno.


linear algebra
matrix
induction
superior algebra
superior algebra theorems

Prove that unitary matrix $A$ is representable in form $A=XYX^{-1}Y^{-1}$, where $X,Y$-unitary matrices  iff  $detA=1$

I think the condition that the matrix is unitary is useless. In fact, this is true for any matrix with determinant 1 in any infinite field of zero characteristic (and I think even in more general cases). On the other hand, in a post Moubi presented some indications I think. If you do not find it, I will post some steps of the proof, which is anything but not simple.

Could you please specify what kind of the Moubinool's post i should look for or  maybe you could give some steps of the proof if it is not difficult for you, beacuse i can't find this post(there are so many Moubinool's posts!!!)

I do not know where it is. Ok, let's give a solution.
Consider the matrices $ A(i,j,b)=I_n+bE_{i,j}$ where $ E_{i,j}$ is the matrix that has 0 everywhere except in the place $ (i,j)$.
It is not difficult to prove by induction and using elementary transformations that any matrix of determinant 1 is the product of such matrices. Thus, it suffices to prove that each such matrix is of the form $ XYX^{-1}Y^{-1}$. To prove this, assume first that $n$ is not 2 and take such a matrix $ A(i,j,b)$ and a number $k$ different from i and j. Then prove that $A(i,j,b)=A(i,k,b)A(k,j,1)A^{-1}(i,k,b)A^{-1}(k,j,1)$.
If $n=2$ then the result is false if the field has less than 4 elements and true if the field has at least 4 elements. To see this, use the identity $ (d,0,0,1/d)\cdot (1,c,0,1)\cdot (1/d, 0,0,d)\cdot (1, -c, 0,1)=(1, c(d^2-1),0,1)$. Here I put for a matrix $A=(a_{ij})$  the representation $ A=(a_{11}, a_{12},a_{21}, a_{22})$.


Stanford
college
geometry
Berkeley Math Circle

Any one belong to a math circle?

I'm in the SJMC

Stanford Math Circle
cool stuff(omg Mr. Sam Vandervelde is amazingly cool)

yeah.  he came to speak last wednesday, about the fourth dimension.

um there is a club at my school but no im not in it

San Francisco math circle!  :D
I went to the new peninsula math circle for mid schoolers but its kinda easy so Im gonna try the stanford one.  it was too hard last year...  :maybe:

[quote="lena_siz"]Sorry I am reposting in such an old thread, but I have a question right on this topic. I'm moving to Fremont, CA, next school year, and I really wanted to join a math circle. I heard that the Berkeley Math Circle is really good, but it seems it is a long way from Fremont. What are the good math circles in Fremont Area, or it is worth going to Berkeley Math Circle?

Thank you for your help.[/quote]
what grade are you in?
im assuming you're in high school or an advanced middle schooler, or you wouldn't be asking about berkeley
i'm in seventh grade, and I'm gonna commute with the guy that won 8th and lower BAMO from san francisco to berkeley next year
its gonna be awesome
i think they probably have BART in fremont, its a really good public transit system
i wouldn't try to drive to berkeley, there is a lot of traffic
i dont know exactly where fremont is (bleh) so i might mention some that are too far away
stanford math circle is awesome, but i dunno if they're gonna have it again next year (sam is leaving)
the parents are trying to put something together
i went to the san francisco math circle this year, but its probably not advanced enough
its focus is on people that havent been exposed to problem solving
i know theres a san jose one, but i think it might be for middle schoolers (star99, care to back me up on this?)
hope this helps!
welcome to SFBA :)

the san jose math circle isn't like the other math circles, from what i know.  I havn't been there for a while, but  there used to be a speaker (usually different each time) who would talk about some topic, and they would usually bring worksheets, etc.  the last few times i was there, they decided to change it up a bit, since they didnt have enough speakers, and they started this russian math circle format, where they have some pretty hard questions, and you would work on them there and at home, and then the next time there would be people there to help you and talk you through them.

also, sjmc doesnt do any math competitions as a group, which is the key difference, i think.

for the age group, there are mostly middle schoolers, a few high schoolers who dont do much, and a few elementary school kids whose parents make them come.

@Daermon & star99:
Thanks for the explanation! I will be in gr. 10 when I would move to CA. I think that Berkeley would be the best choice, because I assume I can take a train, (or BART bus) to Berkeley from Fremont (which is in Alameda County), and from what I read it's a really good math circle. Who here attends/attended the Berkeley Math Circle, can you please talk about your experience as a part of the group?


algebra
polynomial
vector
function
linear algebra

I am having trouble with a homework problem. I have to find a basis for the space of polynomials p(x) of degree less than or equal to 3 and find a basis for the subspace with p(1)=0. Our book didn't cover much information on vectors as functions so I'm not sure how to begin. Any help would be greatly appreciated.

What does a polynomial with degree $ \le 3$ look like?

How about $ p(x)\equal{}a_0\plus{}a_1x\plus{}a_2x^2\plus{}a_3x^3.$

Any restrictions on those coefficients? No, it looks like all of $ a_0,a_1,a_2,a_3$ should be "free."

Does it make sense to add two such polynomials? Does it make sense to multiply a polynomial by a constant? In both cases, do you get another polynomial of the same form? Since the answer to all of these questions is "yes," we have a vector space. With that said, the representation above shows such polynomials as linear combinations of $ \{1,x,x^2,x^3\}.$ That's a spanning set; to show that it is a basis, you would need to show that it is independent. I'll leave that to you. They are independent, and this is the "natural" basis for this space.

Now, we have the subspace such that $ p(1)\equal{}0.$ You can consider that to be the condition $ a_0\plus{}a_1\plus{}a_2\plus{}a_3\equal{}0.$ How to find a basis?

Well, let me ask this: suppose the question were to find a basis for the subspace of $ \mathbb{R}^4$ given by $ \{(x_1,x_2,x_3,x_4)\in\mathbb{R}^4: x_1\plus{}x_2\plus{}x_3\plus{}x_4\equal{}0\}.$ How would you proceed in that case?

[Note that the question "find a basis" never has a unique answer. There are many, many possible different correct answers.]


I have an exercise where I must find convergence radius of series $\sum a_n^2z^n$, $\sum a_n z^{2n}$, $\sum k^na_nz^n$ with convergence radius of $\sum a_n z^n$ is R
I want use D'alembert but I don't succeed, is there a way for using it ?
Thanks

I think, it is impossible, because D'alembert doesn't concern convergence radius in general case.
Why do you not want to use Cauchy?

do you know Hadamard Theorem: $1/R=limsup (a_n)^{1/n} $

Or more exactly Cauchy-Hadamard theorem, though all people call it [i]radical Cauchy's formula[/i].  :)

I didn't know formula $\displaystyle 1/R=limsup (a_n)^{1/n}$
Thanks

Now you know it, so your problem becomes absolutely easy.  :)

[quote="alekk"]$1/R=limsup (a_n)^{1/n} $[/quote]

Do not forget $|...|$

$\frac{1}{R}=\limsup |a_n|^{\frac{1}{n}}$

[quote="Myth"]Or more exactly Cauchy-Hadamard theorem, though all people call it :)[/quote]
Textbooks in China call it exactly Cauchy-Hadamard  :D


44. A fair coin is to be tossed 100 times with each toss resulting in a head or a tail.  If H is the total number of heads and T is the total number of tails, which of the following events has the greatest probability?

A) $ H\equal{}50$
B) $ T \ge 60$
C) $ 51 \le H \le 55$
D) $ H \ge 48$ and $ T \ge 48$
E) $ H \le 5$ or $ H \ge 95$

Thoughts:

Obviously no calculator/charts allowed, and I need a method that I can use to get the solution within 3 minutes.  I almost immediately process of eliminated all but B or D through just thinking about the normal curve.  However, I have no easy way of really comparing B and D, and the people on this forum always have much better reasoning/ideas than mine, so I thought I would see if anyone has a nice, clear, definite solution (with any accompanying theory behind the madness).

Cheers.

I think it is D. B is much less than $ \frac {40 \cdot {100 \choose 60}}{2^{100}}$. D is around $ \frac {5 \cdot {100 \choose 50}}{2^{100}}$. But I'm not sure.

Hmm I think it's A.

The expected number of heads is $ \frac {1}{2}\times 100 \equal{} 50$, so it seems logical that $ 50$ heads would have the greatest probability.

Edit: Nvm, I forgot it's $ \ge 60$...

RisingFlame, 50 alone has the greatest probability, but it may not have a higher probability than the number of heads falling in a certain range. In fact, it can't be A because D includes A.

it's 'clearly' D.

the variance is $ npq\equal{}25$, so the standard deviation is $ 5$.  $ P(T\geq 60)$ is roughly $ P(X>2)$ where $ X\sim N(0,1)$, whereas $ P(H,T\leq 48)$ is roughly $ P(|X|<1/2)$.  Normals have fast decaying tails, so $ P(T\geq 60)$ is about 1 percent if you remember the rules of thumb for the normal distribution.


# of children:  0        1        2        3        4        5
Probability:   .15      .25      .30     .15      .10     .05

a) find prob. that a randomly chosen couple has odd number of children?
b) find the mean?
c) What is the expected number of children?
d) What is the standard deviation?

My answers:

a) I just added the prob. of 1,3, and 5 together = 0.45?

b) I multiplied the number of children times the probability of all 6 options and added them together all over 1.
Which gave me a mean of 1.95 which makes sense to me.

c) My text has me confused a bit on expected numbers, it refers to it as though it is the mean... so I'm not sure if that means that I should round up the mean and say 2 children is what is expected???

d) I found the variation by going (0-1.95)^2 * (0.15) + (1-1.95)^2 * (0.25) +... =1.8475

I then took the square root of this = 1.3592277 ?


Does anyone know what the score was to get into countdown at Pennsylvania's MathCounts States?

The cutoff's usually in the mid 30's.  In my year (2007) the top two had 41, and I'm pretty sure it went down to around 33 or 34 for the cutoff.

Also, last year they excluded one set of target questions from the final score.  So any statistics from last year are considering that the max possible score was 4 less than usual

[quote="ged3.14"]Yeah, so heres what i no:
the top 4 people written each had 37
I was fifth written & I had 36

So im guessing u would need, like, maybe 31 to make the top ten :?:[/quote]

from 2009?

Yeah, those numbers would be from 09

Oh and 1 more thing: this year they had to skip the last 2 questions on target, idk y, so on a normal year the minimum score 2 make the top 10 would be more like 33/34

The top 5 people ALL scored 37, which is a really low score. However, what happened was that there was a problem leak. Like, some guy accidentally sent 7/8 on the target to some coaches, so they canceled problems 7/8(which i was pretty mad about). Im going to guess the cutoff would have been 39. Usually its like 30-32 to make the countdown, around 38 to make nats...


geometry
geometric transformation
reflection
linear algebra
matrix
analytic geometry
graphing lines

[quote="gotztahbeazn"]The graph $xy = 1$ is reflected in $y = 2x$ to give the graph $12x^2 + rxy + sy^2 + t = 0$. Find $rs$.

anyone have a slick solution to this?[/quote]

Or just start graphing $xy = 1$ and $y = 2x$ and then see, or use a matrix.

[quote="Altheman"]the reflection matrix gives you a direct answer[/quote]

Er... what do you mean?  :huh:

yea alex can u explain the reflection matrix? ive nvr heard of it heh

suppose you reflect $(x,y)$ over $y=mx$, to $(x',y')$ then

$\left[ \begin{array}{c} x'\\ y'\\ \end{array}\right] = \frac{1}{m^2+1}* \left[ \begin{array}{cc} 1-m^2&2m\\ 2m&m^2-1\\ \end{array}\right] * \left[ \begin{array}{c} x\\ y\\ \end{array}\right]$

This can be proven using matricies, the rest is pretty trivial.

[quote="gotztahbeazn"]but this will only get u another poitn right? from there you'd have to do the same thing i did, find 3 points or w/e put them into the formula and solve a system for $r,s,$ and $t$[/quote]

The point is...instead of a value into $x$ and $y$, put $\frac{1}{x}$ into $y$.

[quote="Virgil Nicula"]Here is the your teacher ! I will help you with all one's heart.[hide="A lemma and the proof."][color=darkred][b][u]Lemma.[/u][/b] If the point $S(x,y)$ is the reflection of the point $A(x_1,y_1)$ w.r.t. the line $d$ with the equation $D(x,y)\equiv ax+by+c=0$, $a^2+b^2\ne 0$, then\[ \boxed {\ x=x_1-\frac{2a}{a^2+b^2}\cdot D\left(x_1,y_1\right)\ ,\ y=y_1-\frac{2b}{a^2+b^2}\cdot D\left(x_1,y_1\right)\ }  \][u] Proof.[/u] Define the middlepoint $M$ of the segment $[AS]$. Then
$M\in d$ and $AS\perp d\Longleftrightarrow D\left( \frac{x+x_1}{2},\frac{y+y_1}{2}\right)=0$ and $\frac{x-x_1}{a}=\frac{y-y_1}{b}\Longleftrightarrow$
$D(x,y)+D(x_1,y_1)=0$ and $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{D(x,y)-D(x_1,y_1)}{a^2+b^2}\Longleftrightarrow$
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=-\frac{2}{a^2+b^2}\cdot D(x_1,y_1)$ a.s.o.[/color]

[color=blue][u]Proof of the proposed problem.[/u] In the our particular case we have: $D(x,y)\equiv 2x-y=0$, i.e. $a=2$, $b=-1$, $c=0$ and $A\left(m,\frac 1m\right)$, where $m\in R^*$, i.e. $x_1=m$ and $y_1=\frac 1m$. Thus, $D(x_1,y_1)=\frac{2m^2-1}{m}$ and $x=m-\frac{4(2m^2-1)}{5m}=\frac{-3m^2+4}{5m}$, $y=\frac 1m +\frac{2(2m^2-1)}{5m}=\frac{4m^2+3}{5m}$. Therefore,
$3m^2+5x\cdot m-4=0\ ,\ 4m^2-5y\cdot m+3=0\Longrightarrow$$\left|\begin{array}{cc}3 & -4\\ 4 & 3\ \end{array}\right| ^2 =\left|\begin{array}{cc} 3 & 5x\\ 4 & -5y\ \end{array}\right|\cdot \left|\begin{array}{cc} 5x & -4\\ -5y & 3\ \end{array}\right|\Longleftrightarrow$
$\boxed {\ 12x^2-7xy-12y^2+25=0\ }$, i.e. $r=-7$, $s=-12$, $t=25$.[/color]

[i]Remark.[/i] I hope I didn't make a mistake.[/hide][/quote]

I can't understand your solution....but i have another one
Contruct 2 points in xy=1 and the reflection line: (x,y),(x'y') ,connect the two points, then we can see: this line is perpendicular to y=2x and be divided into 2 equal parts.the coordinate of the intersection point is (X,Y)
so the slope of this line is -1/2, then we have the general equation: (Y-y)=-1/2(X-x). Cuz (X,Y) is on y=2x, so Y=2X, sub it in to the equation (Y-y)=-1/2(X-x),
2X-y=-1/2(X-x), we can get: X=2/5 y+1/5 x,     so Y=4/5 y+ 2/5 x. Because (X,Y) is the midpoint of the connection line, so X=(x+x')/2, Y=(y+y')/2.
2/5 y+ 1/5 x=(x+x')/2, 4/5 y+ 2/5 x=(y+y')/2,  then we get: x'=4/5 y- 3/5 x, y'=3/5 y+4/5 x
3x'=-9/5 x+12/5 y, 4y'=16/5 x+12/5 y, subtract the 2 equations we have: 3x'-4y'=-5x, x=-(3x'-4y')/5.
Because (x,y) is on xy=1, so x=1/y, y=-5/(3x'-4y'). Sub x=-(3x'-4y')/5 and y=-5/(3x'-4y') into y'=3/5 y+4/5 x. Then simplify the large equation, finally we get the equation we want: 12x'^2-7x'y'-12y'^2+25=0, so that r=-7, s=-12, t=25

By the way, will we learn matrix in high school??

I am 64 years old ([u]a retired teacher with a pension 230 Euro/month[/u]) and I learned in a week from the [b]LaTeX Help[/b] on mathlinks.ro (see up in the second line of commands; the first is with [b]Home[/b]).

haha then i have 47 years to learn it  :lol: , i am 17 now

Haha, you well-know the subtraction ! It is a stupid joke. You didn't understand what I wrote upper: [b]I learned in a week, the my dear joker ![/b] [u]A happy new year to you ![/u]


What is the sum of all the two-digit integers?

[hide="this is my answer"]the 1st # in a 2 digit # is = 10
the last # in a 2 digit # is = 99

there are 90 #'s so we put them into a arithmatic series

and get the equation

(99+10)90/2=4905

so the answer is 4905 or $\boxed{4905}$[/hide]

[b][i][u]please rate my post thanks![/u][/i][/b]

please be fair though!!

:heli: :heli: :heli: :heli:

I don't understand, because integers are negative too. Therefore, shouldn't the negative counterparts offset their positive counterparts? I put 0 as an answer and I was wrong.

I think it's implied that they are positive.

FTW says the answer is 0, however, so I think the question refers to all integers, positive and negative.

Yes, they have changed the answer to 0 because the question only says integers, which includes negatives, zero, and positives.


topology
abstract algebra
group theory
symmetry
advanced fields
advanced fields unsolved

As [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=639144#639144]before[/url], a monothetic group is a locally compact Hausdorff abelian group having a dense cyclic subgroup.

The title says it all :):

Prove that a monothetic group is either compact or discrete.

(edited: I've replaced "compact" with "locally compact" in the definition; thanks everyone for noticing the mistake, and I apologize for the inconvenience)

I don't think that's what he meant...
but i may be wrong

Yes, I've made the necessary correction. Thank you for noticing.

Couldn't you please change "compact" in the conclusion to "locally compact" too?   :)

OK, here's my proof.

Let A be the locally compact Hausdorff abelian group with a dense cyclic subgroup generated by a in A.

Let C be any symmetric compact neighbourhood of 0 in A. Symmetry means that x in C implies -x in C. It is always possible to find one such, because if U is any neighbourhood of 0 then the intersection of U and -U is symmetric. Put Cn = na + C, for n in Z. The interiors of these sets cover A, so they cover C+C, so a finite number of them cover the compact set C+C.

Notice that C(-n) = -na + C = -na - C = -(na + C) = -Cn. Let n1 be an index such that the union of the Cn from 0 up to and including n1, together with their negations, cover C+C.

If C is finite, the topology on A is discrete, and we are done. If C is infinite, then the sequence na (n = 0,1,2,...) must come back to C infinitely many times. So we can find an index n2 such that n1 <= n2 and (n2)a is in C. Then C(n2) = (n2)a + C < C+C. Also, C(-n2) = -C(n2) < -C-C = C+C.

Put A' = union of Cn and -Cn when n runs from 0 up to and including n2. For any subset S of A' we can split it into pieces, one in each Cn, and travel (add a in each step) each piece separately until we get to C(n2). When we get there, we split whatever we have into new pieces, one in each Cn, and travel on. We see that we can travel as far as we want, always staying within A'. We can also travel in the negative direction by adding -a with the same result.

In particular, C < C+C < A' so we can travel C infinitely far in both directions, always remaining within A'. Therefore the union of all the Cn is contained in A'. But this union is all of A, so A' = A. A' is a finite union of compact sets, so A is compact.


modular arithmetic
algebra proposed
algebra
number theory

[quote="pco"][quote="Leon"]Find the least natural number whose last digit is 7 such that it becomes 5 times larger when this last digit is carried to the beginning of the number.[/quote]
The required number is $ 10n\plus{}7$ with $ n\in[10^{p\minus{}1},10^p)$

The equation is $ 7\times 10^p\plus{}n\equal{}5(10n\plus{}7)$ $ \iff$ $ n\equal{}\frac{10^p\minus{}5}{7}$

The first value of $ p$ such that $ 10^p\equal{}5\pmod 7$ is $ p\equal{}5$, and so $ n\equal{}14285$ and so the required number is $ \boxed{142857}$[/quote]
 gr8 sir :)

Or we can simply go digit by digit:

$\overline{pq\dots abcde7}\cdot 5\stackrel{\mathrm{multiplication\ by\ 5}}{=}\overline{rs\dots xyzwt5}\stackrel{\mathrm{given\ condition}}{=}\overline{7pq\dots abcde}$

Hence $e=5$

$\overline{pq\dots abcd57}\cdot 5=\overline{rs\dots xyzw85}=\overline{7pq\dots abcd5}$

Hence $d=8$

$\overline{pq\dots abc857}\cdot 5=\overline{rs\dots xyz285}=\overline{7pq\dots abc85}$

Hence $c=2$

$\overline{pq\dots b2857}\cdot 5=\overline{rs\dots xy4285}=\overline{7pq\dots ab285}$

Hence $b=4$

$\overline{pq\dots a42857}\cdot 5=\overline{rs\dots x14285}=\overline{7pq\dots a4285}$

Hence $a=1$

$\overline{pq\dots 142857}\cdot 5=\overline{rs\dots 714285}=\overline{7pq\dots 14285}$

Since we completed the cycle, we stop here as we're looking for the smallest such number, which is $142857$.

[quote="geniusramanujan"]the no is 142857[/quote]
Great contribution !
Thanks a lot !
Your post will be quite useful for a lot of readers.


Good luck to everyone on the MAML this Tuesday.

I believe you can use anything that doesn't do symbolic algebra.

None of the questions will require a calculator. There were 2 questions last year that asked for decimal approximation, so having a calculator will gain you maybe a 30 seconds on those depending how fast you can do mental arithmetic.

im aiming 4 about 120 if its about the same difficulty as last year...et vous?

Keep in mind that the contest is 25 questions this year, not 30, so if the difficulty of the questions is the same, scores should be higher (since the duration is still 90 minutes).  The scoring is 6-2-0.

Any calculator without a CAS is allowed.  Also, you are explicitly allowed graph paper, a ruler, a compass, and a protractor.  If you want any of those things, you should probably bring them yourself.

Will the problem types change drastically from previous years?

If there are only 25 problems this time, they should be harder - too many people will get perfect scores, I'd think.

This is the MAML?!?!? I thought this was the MMO. I'm confused now. got a 100 last year, don't know what to expect for this year.

I hope the problems will be harder; but less computational, as in some cases in previous years.

[quote="rjyanco"]Keep in mind that the contest is 25 questions this year, not 30, so if the difficulty of the questions is the same, scores should be higher (since the duration is still 90 minutes).  The scoring is 6-2-0.
[/quote]

I didn't know that.  :huh:

I felt that last year's problems were somewhat less mature in that most involved some little manipulation and then were easily finished.


function
inequalities proposed
inequalities

Prove that for positive $a,b,c$ is $2(a^3+b^3+c^3)+3abc \geq 3(a^2b+b^2c+c^2a)$.

If $f(a,b,c)=2(a^3+b^3+c^3)+3abc-3(a^2b+b^2c+c^2a)$, by computations,
$f(a+d,b+d,c+d) - f(a,b,c) = \frac{3}{2}d (\sum (a-b)^2)$
That means for any set of $a,b,c$, WLOG assume $c$ is the smallest, then $f(a,b,c) \ge f(a-c,b-c,0)$

Thus, we can assume one of the variable to be 0. The rest is trivial.

If we consider the expression
\[ 2(x^3+y^3+z^3)+3xyz-3(x^2y+y^2z+z^2x)  \]
as a function of any one of the variables, say $x$. Then we use properties of monotone fuctions.

We suppose that $x\geq y\geq z$, and then prove that $f''(x)\geq0$, $f'(x)\geq0$, and $f(y)\geq0$. The last one reduces to $2y^3+z^3-3y^2z\geq0$, just AM-GM.

Actually, we can prove that
\[ 2(x^3+y^3+z^3)+3xyz\geq 3\max\{x^2y+y^2z+z^2x, xy^2+yz^2+zx^2\}.  \]