induction
modular arithmetic
number theory unsolved
number theory

Let $ p$ be a prime number and let $ a_1,a_2,\ldots,a_{p \minus{} 2}$ be positive integers such that $ p$ doesn't $ a_k$ or $ {a_k}^k \minus{} 1$ for any $ k$. Prove that the product of some of the $ a_i$'s is congruent to $ 2$ modulo $ p$.

I got some ideas about the problem.
[hide="take a look at them"]Let $ w$ be a primitive root of $ p$. $ a_i\equal{}w^{k_i} (mod p)$ and $ a^s\equal{}2 (mod p)$. Hence the problem turns to this.
$ k_1,\ldots,k_{p\minus{}2}$ are positive integers such that $ p\minus{}1$ doesn't divide $ i.k_i$ for any $ i$. Prove that there are $ k_i$'s such that their sum is congruent to $ s$ modulo $ p$.
I think also this more general statement is true.
$ k_1,\ldots,k_{n\minus{}1}$ and $ n$ be positive integers such that $ n$ doesn't divide $ i.k_i$ for any $ i$. Prove that for every $ 0<r<n$ ($ r$ is a positive integer) there are $ k_i$'s such that their sum is congruent to $ r$ modulo $ n$.[/hide]

I think it is not weird BaBaK Ghalabi. Thanks for the solution.

[quote=BaBaK Ghalebi]a weird solution:

we prove by induction on $ k = 2,\ldots , p - 1$ that there exists a set of integers $ \{b_{k,1},b_{k,2},\ldots ,b_{k,k}\}$ such that:

(I)each $ b_{k,i}$ either equals $ 1$ or is the product of some terms of the sequence $ a_1,a_2,\ldots ,a_{p - 2}$.

(II)$ b_{k,i}\not\equiv b_{k,j}\pmod p$ for $ 1\leq i < j\leq k$

suppose that we have chosen the set $ \{b_{k,1},b_{k,2},\ldots ,b_{k,k}\}$ for some integer $ 2\leq k\leq p - 2$. because $ a_k\not\equiv 0\pmod p$, no two of the numbers $ a_kb_{k,1},\ldots ,a_kb_{k,k}$ are congurent modulo $ p$. also,because $ a_k^k\not\equiv 1\pmod p$, we have:

\begin{eqnarray*}\left(a_kb_{k,1}\right)\left(a_kb_{k,2}\right)\ldots\left(a_kb_{k,k}\right)\not\equiv b_{k,1}b_{k,2}\ldots b_{k,k}\pmod p\end{eqnarray*}

hence, we cannot permute $ \left(a_kb_{k,1},\ldots ,a_kb_{k,k}\right)$ so that each term is congurent modulo $ p$ to the corresponding term in $ \left(b_{k,1},\ldots , b_{k,k}\right)$. because the $ a_kb_{k,i}$ are distinct modulo $ p$, there must exist $ j$ such that no two elements of the set $ \{b_{k,1},\ldots ,b_{k,k},a_kb_{k,j}\}$ are congurent modulo $ p$. we relabel this set of numbers as $ \{b_{k + 1,1},b_{k + 1,2},\ldots ,b_{k + 1,k + 1}\}$. each of these $ k + 1$ numbers equals $ 1$ or is the product of some terms of the sequence $ a_1,\ldots ,a_{p - 2}$, and the induction is complete.

now consider the resulting list $ b_{p - 1,1},\ldots ,b_{p - 1,p - 1}$. exactly one of these numbers is congurent to $ 2$ modulo $ p$; because this number is not equal to $ 1$, it is congurent to the product of some of the $ a_k$, as desired.[/quote]

Can I ask what motivates this solution. Considering the sequence in that way is quite random I suppose


summer program
Mathcamp
geometry
MathPath
HMMT
Duke
college

Hey, y'all, congrats to those who made it!!! I can't wait to see you all!!!  :D

Anyway, if you're going, take the poll and reply with your name, age, where you live, and anything else you want.  :)

Yeah, so I'm Alina, I live in Illinois, and I'm 15 (barely :P And Damn that poll makes me feel old...). I like to do math, laugh ( :rotfl: ) and make jokes. I don't know what else to say, so yeah...

Oh yeah, do you guys have facebooks? I should make a group :P

53 days till MathCamp!!!!!!!!!!!!

Hmm should I still post if I'm a returning alumnus? Anyway hi, I'm David, 16 years old from Massachusetts. I have a lot of interests (not including math, of course  :wink:  jk), and you can check out my facebook profile to see them all. Hmm just to name a few, languages (french, latin, hebrew, more), linguistics, philosophy, classical music, politics (strong leftist), judaism, etc. Mathcamp will be fun this year =)

Yes, everyone who's going, please post! And if you took the poll, you should post too. :)

Yaaaayyyyy

I'm totally copying this from my introduction last year... (with some parts changed)

I'm Laura Zehender, from near Richmond, VA (I have lived here my whole life). I am 16 and will be a junior next year, and have been homeschooled since 2nd grade. My interests include math (especially combinatorics [even more so now from taking classes at MC] and geometry), dance (ballet and tap, jazz and pointe previously, and have done some other kinds of dancing), Latin (my only AP test left now), piano (especially Tschaikowsky), listening to country music, ancient Greek literature and myths, and art (especially modeling clay and drawing with pencil). I also am a bookworm and chocoholic. This is my second year at Mathcamp (went in '07), and I went to MathPath in '06, and my older brother went to Mathcamp twice. Some other things that came up in last year's thread... I'm a Libertarian and a ENFJ (Myers-Briggs type).

Also, I now know some cubing, but only a very little bit. I joined a 4-H shooting club this year, so I've learned shotgun and archery, besides continuing rifle shooting which I've done for awhile. This year I've joined a lot of math teams, so I've been to PUMaC, HMMT, and Duke math meets, and I'm going to ARML now ([i]finally[/i] there is a team for the rest of VA)... anyone else been to those, or is going?

Hey, I'm Alex, a homeschooled 17 year old junior from Maryland.  I have been to Mathcamp in 2006 and 2007 (and MathPath in 2005).  I'm interested in math ( :D ), politics, philosophy, chemistry, games and puzzles of all sorts, and most other things.  If you want to find me at camp, locate the nearest Mao game and check there.  :lol:

In response to LadyKn1ght's question, I was at HMMT.  Unfortunately due to extenuating circumstances I couldn't go to PUMaC or Duke.  I'm also going to ARML (this should not come as a surprise to you . . .  :P )

[quote="E^(pi*i)=-1"]I'm also going to ARML (this should not come as a surprise to you . . .  :P )[/quote]

Hmmm, I wonder why. Maybe because you're my teammate...  :wink:

Hi, I'm Tom.  I am 17.  I live in New Jersey.  I will be a senior next year.  I am homeschooled.  I went to Mathcamp last year.  This will be my second year.

I like math.  I fence.  I play Bridge.  I can cube.  I shoot rifles.  I am not and do not want to be in the possession of nuclear weapons.  I do not like short sentences.

I have attended PUMaC and HMMT once each.

My name is Michele and I'm a 16-year-old rising senior from New York (the State not the City). This is my first time going to Mathcamp and I'm really excited  :D

I like to play tennis, classical guitar, and a plethora of board games including (but not limited to) Risk, Stratego, Scrabble, and Blokus. My fastest cube time was 2 minutes (and I feel like a wimp because of it  :P ).

I was at HMMT and PUMaC this year and will be at the Penn State ARML.

Hi, I'm Billy, another homeschooled junior, and my favorite word is 'poecilonym'. I do absolutely love COUNTRY MUSIC! and singers thereof (particularly Sara and Brad and Danielle and Martina and Eric, in no order). I'd like to imagine I can carry on an intelligent although divagating conversation about politics, religion, philosophy, history, vocabulary, competition, &c. I've been to HMMT and Pumac and DMM and ARML; the latter 2 2 times each.

I would LOVE to possess nuclear weaponry. It'd help in [url=http://rubberband-girl.blogspot.com/2008/04/everybody-want-to-rule-world-part-1.html]world domination[/url], which is a side interest of mine.

Oh yes, I do math, and computer science, and sesquipedaliophilia. (I also talk archaicly when I am either uncomfortable with what I'm saying or am trying to be humorous) I shotgun also, I enjoy Attack! (risk++), hearts, stratego, and I run slowly. I've never solved a cube. I can play bridge, very badly as Tom can attest to. And I'm sure I forgot at least something else about me. Oh yes, I enjoy country music greatly. :)

Hmm, copying the intro from the previous thread sounds like a decent idea. Needs editing though.

Palmer Mebane, from near Richmond, Virginia (born there also), graduating this year, second time going. Originally I said I'm probably the person most disconnected from immediate reality you'll ever meet, and then I went to Mathcamp last year; that claim is apparently not true. My biggest interests are math and logic-oriented puzzles: I did fairly well on the USAMO this year and last year, and Mystery Hunt 2008 (at MIT) was the most fun I've had in my life. When not doing math or puzzles I play lots of classic and old-school electronic games and program some. Since we're on a hosted math competition name dropping spree, I went to PUMAC, remotely went to HMMT (both on Quagga), and will be going to ARML on a team with Laura and Alex. I hadn't been to any of those before this year.

Hi,
I'm Kyumin Kim, (although many of my friends call me Joe),
I'm 15, a rising Junior in Leland high in California.

My hobby includes playing the violin and trying to learn to program
in assembly and opcode (for my Ti84), and never having solved a cube.
Oh, and I like math too.

Congratulations to all the admits  :)

I'm Tom, I'm 18 and I'll be a college freshman in a few short months. MC '07 was top-notch and this summer is going to be great as well  :D

I'm Leon Zhou, a 15 year-old sophomore in San Diego. First year at Mathcamp, though I did go to Mathpath in '06 (with Ladyknight, whom I sort of remember).

Math, A-Team, video games, board games, books, all the standard stuff. I'll be going to ARML on the San Diego A team; if you see a bunch of people playing SSBM, I'll be one of them.

See you all there.

I'm Jacob Steinhardt. I think liking math is obvious, so here are some non-math things that I intend to do at MathCamp, and hope others will do as well:

(1) Play FOOTBALL (known as soccer to some in the United States)
(2) Play ultimate frisbee
(3) Play bridge
(4) Play bughouse
(5) Play the piano/sing

If you are interested in any of the first four, please let me know. (The last one is pretty spontaneous so no use planning it, but if you have good sheet music bring it in; I'll be bringing Phantom of the Opera, although I'd guess that other people have it as well.)

EDIT: Like Laura and Palmer, I live in Virginia. And I also like to program computers, although that is more or less math so I'm not sure that counts as a non-math interest.

I'm James Koppel (although my friends call me "Jimmy"), and I'm from St. Louis. I am 16, and will be a senior come fall.

I am very into computers, both programming them and using them (primarily for games). My favorite programming languages are JavaScript, Ruby, and a new dialect of Lisp called Arc; the latter was created by entrepeneur, programmer, and essayist Paul Graham, whom I admire. On a small tangent, I must reccommend that everyone read some of Paul Graham's essays, especially [url=http://www.paulgraham.com/nerds.html]this one[/url], [url=http://www.paulgraham.com/say.html]this one[/url], and [url=http://www.paulgraham.com/lies.html]this one[/url].

In gaming, I like pretty much all genres (except, for the most part, sports and racing games [save Mario Kart and the like]), but I prefer strategy games, adventure games, and RPGs. My two favorite series are Legend of Zelda and Heroes of Might and Magic. I also have the skills of a low-level SSBM tournament player.

In non-computer entertainment, I'm a fencer (my fencing club is classical-style, French school; we fence with a variety of modern and historical weapons, but the instructor won't let me progress beyond foil), a tournament chess-player (yes, I'll play bughouse with you Jacob), and a 1st-degree black belt in Tae Kwon Do.

In books, I mainly read high fantasy such as Dragonlance and R.A. Salvatore, although I enjoy authors in other fantasy subgenres and other genres altogether.

Any, since many people are listing their favorite words, I'll list mine: contretemps (mainly because of the pronunciation), kakorrhaphiophobia, quomodocunquize, cosmarchy, and scacchic. Sorry, I don't have any particlar attatchment to the hippopotomonstrosesquipedalian that would cause me to seek them out over their shorter brethren unless I want to intimidate someone.

Ok, since everyone's putting a lot of stuff in theirs, maybe I should say a little more about myself. :)

I love science almost as much as I love math, and I have had a recent interest in biology (hence I want to check out the mathematical biology classes I saw on the MathCamp brochure thing). I go to IMSA (Illinois Math and Science Academy - horrible school, never go here if you value your health and/or sanity at all - I have seriously gotten an average of 3.75 hours of sleep each night this week, not even exaggerating) and I skipped 9th grade (pssh freshmen... :P). I play clarinet and piano, Bach is my favorite composer for piano. Umm, I can solve a Rubik's Cube, but I'm not too good. I think my best time is like a minute and a half or something. Then again my cube has never seen a lube. Uh... I was on the volleyball team at my school (only because we suck at sports) and might bring my ball to camp. I'm also on the robotic's team on my school (we do First, if anyone knows about it). Hmm, what else to say... I'm doing ARML and I'll probably be on C team since I'm not really that good at timed things and I've been failing my homework. And I was also at HMMT, it was really fun. Yeah, I think that's about all that's interesting about me... I think...

51 days!!!!!!!!

By the way, our new facebook group can be found here: [url]http://www.facebook.com/group.php?gid=27473818432[/url]
Join!

There's already a big group for Mathcamp here: http://www.facebook.com/group.php?gid=2207038462

And a group for Mathcamp 2007: http://www.facebook.com/group.php?gid=2386374244

What we need is a Mathcamp 2008 group.

I SWEAR that wasn't there when I searched Mathcamp...

Hi! My name is Victor Luo, and I am a 14 year-old sophomore (don't worry, I'm mature  :P ). I live in Virginia and go to school at W. T. Woodson (close to TJ). I went to HCSSiM last year and had a lot of fun learning mathematics, so I look forward to the Canada/USA Mathcamp. I play cello (Bach is my favorite composer) and I take Latin (currently learning about Cicero's In Catilinam speeches). I also play football (soccer in US) and enjoy any type of sports. If anyone is taking a plane from Dulles International Airport or Reagan National Airport, please let me know, so we can maybe catch the same plane. You can find me on facebook.

Adrian Chan. Essentially Los Angeles, CA. Rising college freshman. I was here last year.

If you

- love French (I swear... I'm at least a LITTLE better this year)
- love playing cliché chord progressions and improvising lyrics/melodies over it
- love singing cheesy alternative rock music
- listen to progressive rock (Dream Theater, Rush, etc.)
- play some sort of fingerstyle guitar
- play some sort of electric guitar
- play some sort of bass
- play some sort of drums
- play any instrument
- Ultimate

then we can get along, especially the guitar... I see there's a classical fingerstyle guitarist?? Awesome.

I really want to form a band, but I failed last year. Bummer. Anyone want to bring their drum set and cymbals over to Reed? Maybe a nice Mesa Boogie double stack guitar amp?

[quote="JSteinhardt"]laughinghead, do you know Tony Liu?[/quote]

 :maybe: I don't think so... I am I supposed to?
After a quick facebook search I see that you might be referring to kid who graduated from IMSA last year, in which case, no, I don't know him because this is my first year at IMSA. But a lot of people I know know him.  :D

[quote="VDLmath"]Hi! My name is Victor Luo, and I am a 14 year-old sophomore (don't worry, I'm mature Razz ). I live in Virginia and go to school at W. T. Woodson (close to TJ). I went to HCSSiM last year and had a lot of fun learning mathematics, so I look forward to the Canada/USA Mathcamp. I play cello (Bach is my favorite composer) and I take Latin (currently learning about Cicero's In Catilinam speeches). I also play football (soccer in US) and enjoy any type of sports. If anyone is taking a plane from Dulles International Airport or Reagan National Airport, please let me know, so we can maybe catch the same plane. You can find me on facebook.[/quote]

Yeah Bach!!!
Yeah soccer!!
 :P

@post above that I'm too lazy to quote:

Yeah alternative!!

Hi, my name is Leonid Grinberg; I'm 16 and a rising Junior. I went to Mathcamp in 2007.

Besides mathematics, I'm into computers (programming, web design, networking, you name it), physics, linguistics, grammar, philosophy, psychology, politics, religious discussions, debating, and, to a lesser extent, history.

I know English and Russian equally well and have been studying Latin for five years. I am also fluent in Foodtongue -- for those that are first-years, you'll find out what that is when you get to camp (or, if someone less lazy than I feels like explaining).

Is ultimate frisbee popular? (I personally think it's second only to tennis) Also, does anyone know where I can get a regulation frisbee?

Ah Food Tongue! Apple cherry oyster Food Tongue, maize apple spinach ham-sandwich. Apple cauliflower Sweet Tea. :)

(I met Expresso at ISEF this past week; I am pretty sure the two of us are the only Food Tongue speakers who haven't been to Mathcamp yet.)

Hmm just to reply to a few people's interests, I also love French, so I think Eternica and I will get along. Bach is also my favorite composer. As well, I've been taking Latin (I hope we can have conversational Latin at MC this year   :D  ...)

[quote="solafidefarms"](I met Expresso at ISEF this past week; I am pretty sure the two of us are the only Food Tongue speakers who haven't been to Mathcamp yet.)[/quote]

You are wrong. A friend of mine knows a bit, and he's not a Mathcamper (nor is he going this year).

Haha, conversational Latin. I can become completely confused again! I'm okay at reading Latin, but speaking... fun :P

Cool! Mango grass mint pear cauliflower?

Anybody else know Food Tongue speakers who learned pre-Mathcamp?

Grass Lion-Meat. Pear grass cherry Foodtongue, cucumber grape?

You know, we probably shouldn't continue having a conversation... it'd be more than somewhat rude. No need to scare people :-)

Yeah, save the Food Tongue and certain other traditions for week two  :lol:

Ohhh Adrian I eagerly await hearing your new and improved French  :D

I firmly believe in only learning one language unless one has a very practical use for others. Learning foreign languages is, in my opinion, generally not worth the trouble; therefore, I will most likely refuse to learn "Food Tongue" (also, I am a picky eater, so I would only be able to use a few words :wink: ).

I'd hardly consider Food Tongue foreign. :P

Hi, I'm Benjamin, 14, from Singapore, and other than math I also play the piano and tennis. I'm keeping my fingers crossed that my school will allow me to go (our holidays have already ended by the time mathcamp starts).

I have a rather large collection of math ebooks, most of which I have not read, and I like some games you've probably never heard of before.

ok thats all.

[quote="Eternica"]Ultimate is not THAT crazy (well... I'm comparing this to HCSSiM '06 where people would ALWAYS play ultimate after lunch), but I'm pretty sure it's crazy enough for people to find a group of people to play it occasionally/frequently. You could just post what and where on the schedule board and people will know what's going on.

and David Corwin, I knew you put French as one of your interests, but I never heard you speak it last year. The French dudes were making fun of me because I kept butchering the language. They even tricked me into saying bad words. Ai-yah.[/quote]

Yeah I spoke French some with Justin last year, but most of the French people just wanted to speak really quickly. Also, I was pretty rusty in French last summer after a not-so-great French teacher, but this year I've improved a LOT in French. Hopefully this summer I get to practice my French at MC =)

Hi! I'm Grant Mathews, from Maryland. This is my first time so I have no idea what to expect but oh well :D I'm a big programmer guy; it's tied with math on interest level for me.

Other than that, I like... (no particular order)

ping pong
weird music
good books
funny things
arrested development (i try not to burn too many hours with TV but this is one show that I just can't pass up)
piano (dunno how that one's gonna work)
writing
video games
a good puzzle
board games

stuff like that!

Oops, I just realized that I haven't introduced myself yet. Hehe :blush:
My name is Andrew Yang Guo, 14 years-old, Chinese and rising freshman in New York. I'm taking Pre-Calc and aspiring to take part in a lot of math competitions next year (I'm relatively new to these things, so I'm hoping to catch up fast.)

Other than math, I have an assortment of hobbies including (and not limited to):
      Learning Chinese (Needs much improvement)
      Playing the cello (Does anyone know if I can bring mine to Mathcamp?)
      Tennis, cross-country and track
      Reading, a LOT (You can put raiding the library under here :D)
      Chess
      Science (It was either Mathcamp of CTY's Physics. Guess which I chose?)
      Debate (Nothing like a good [i]argument[/i] every so often. That always irritates my speech teacher :lol:)
      Listening to classical music (Rock music? Pfft)

Oh, and I'll put in a good word for Spanish, just because it's feeling left out. ¡Te sabes que estoy diciendo?

Obviously, I am not a Mathcamp staff member, so nothing I say is official, but I am fairly certain that you can bring a cello, but you're responsible for it.

My suggestion, though, is that you won't want it -- there are plenty of other things to do.

I know I'll be extremely busy with all the classes and activities going on at Mathcamp, but frankly, I'm afraid of what would happen to my cello ability during 5 weeks of in activity. Also, towards the beginning of September, I have some important auditions that I want to practice for. So I would definitely like to take some time to practice during Mathcamp.  Rest assured, it won't have any negative effects. I'll probably be spending my practice time wishing I could be doing other stuff.

No, I was just warning you. Furthermore, having no musical abilities of my own, I tend to find it difficult to imagine someone wanting to play :-).

Aguoman, are you who I think you are? This is David Ding, from the Shanghai summer camp several years ago and the cruise last year...good times  :roll: . I only remember that you played a string instrument and enjoyed a good puzzle.

Unfortunately, I'm on Mathcamp's waitlist and still haven't received any word from them in several weeks, so I shouldn't be posting here.

:w00t: Woah!!! :w00t:
David, is that you??
Wow, what a coincidence! How's life in California going?
I hope you get accepted. It'd be great to see you at Mathcamp!

Send me a pm or email me!

You're welcome to bring your cello (and I'd find that cool, because I used to play the cello myself --- no doubt much more poorly than you do!).  Lots of people have brought their violins to past camps.

However: you are responsible for what happens to it!  In particular, you'll have a roommate, so you won't have total control over locking your room.  We've never had anything go wrong, but if something does happen, we can't be responsible for it either.

I'm Justin Brereton from Michigan. I am 17 and will be a senior next year. I went to MAthcamp last year and it was AWESOME.
My interests include

Chess - (sometimes I play at 2000 strength, sometimes at 1500 strength, current rating 1850)
I am definately looking forward to some hardcore Bughouse, but dreading the Steinhardt defense.

Math, duh

Soccer, don't play competitively anymore, but I will be playing some at mathcamp

Tennis, I will bring my racket to Mathcamp, to anyone else who wants to play tennis bring yours too.

Frisbee, ultimate is always fun, Campers need to dominate campers v staff

Congratulations to all who were accepted to mathcamp, it will be the best five weeks of your summer.


probability
quadratics
number theory
relatively prime
algebra
quadratic formula
AMC

When rolling a certain unfair six-sided die with faces numbered $1, 2, 3, 4, 5$, and $6$, the probability of obtaining face $F$ is greater than $\frac{1}{6}$, the probability of obtaining the face opposite is less than $\frac{1}{6}$, the probability of obtaining any one of the other four faces is $\frac{1}{6}$, and the sum of the numbers on opposite faces is $7$. When two such dice are rolled, the probability of obtaining a sum of $7$ is $\frac{47}{288}$. Given that the probability of obtaining face $F$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.

Stupid question: Does that mean that all the other faces are fair?

[hide]
Let the probability of geting each of $(1,2,3,4,5,6)$ be $(a,b,c,d,e,f)$.  Let the face $F$ be $1$ and the face opposite $F$ be $6$, since it doesn't matter...

You can get $7$ from $af$, $fa$, $be$, $eb$, $cd$ or $dc$, so $2(af+be+cd)=\frac{47}{288}$ or $af+be+cd=\frac{47}{576}$.

Assuming all the other faces are fair, $af+\frac{1}{18}=\frac{47}{576}\implies af=\frac{15}{576}=\frac{3\cdot5}{24\cdot24}$.  So $a=\frac{5}{24}$, since it is larger than $\frac16$ and $m+n=29$.
[/hide]

It's supposed to say "the probability of obtaining each of the other faces is 1/6" somewhere in there.

A slightly simpler way.
[hide]
Let the probability of getting the unfair face be $\frac16+x$ and on the other side it is $\frac16-x$.  All the other probabilities are $\frac16$.  So the probability of getting a $7$ is $\frac23\cdot\frac16+2\cdot\left(\frac16+x\right)\cdot\left(\frac16-x\right)=\frac16-2x^2=\frac{47}{288}$. So $x=\frac1{24}$. $\frac16+x=\frac5{24}$. So the answer is $\boxed{029}$[/hide]

The probability of obtaining a sum of $7$ is equal to the probability of obtaining any one of the four fair sides and then its opposite, plus the probability of obtaining side $F$ and then the side opposite of $F$, plus the probability of obtaining the side opposite of $F$ and then side $F$. Let the probability of getting $F$ be equal to $x$, or $x = \frac{m}{n}$. Since the probability of getting  a sum of $7$ is $\frac{47}{288}$, $\left(\frac{4}{6}\right)\left(\frac{1}{6}\right)+x\left(\frac{1}{3}-x\right)+x\left(\frac{1}{3}-x\right) = \frac{47}{288}$. Simplifying the equation yields $18x^{2}-6x+\frac{15}{32}= 0$. By the quadratic formula, $x = \frac{5}{24}$ or $\frac{1}{8}$. Since $\frac{1}{8}< \frac{1}{6}, \frac{m}{n}= x = \frac{5}{24}$, so $m+n = \boxed{029}$.

Let the probability of obtaining face $F$ be $x$. Then $2 \cdot x\left(\frac{1}{3}-x\right) + 4 \cdot \frac{1}{6} \cdot \frac{1}{6}=\frac{47}{288}$, so $x=\frac{5}{24}, \frac{1}{8}$. We must have $x>\frac{1}{6}$, so $x=\frac{5}{24} \implies \boxed{29}$.


algebra
polynomial
Vieta
AMC
USA(J)MO
USAMO
system of equations

The product of two of the four zeros of the quartic equation
\[x^{4}-18x^{3}+kx^{2}+200x-1984=0 \]
is $-32$. Find $k$.

[hide]Let the roots be $a,b,c,d$ and WLOG let $ab=-32.$
By Vieta's,
$abcd=-1984\implies cd=62$
Also by Vieta's,
$a+b+c+d=18$
and
$abc+abd+acd+bcd=ab(c+d)+cd(a+b)$
Letting $a+b=x,c+d=y,$ we have a system of equations in $x,y$
$x+y=18$
$62x-32y=-200$
Solving, we get
$x=4,y=14.$
Now by Vieta's again,
$k=ab+ac+ad+bc+bd+cd=-32+62+a(c+d)+b(c+d)=30+xy=86$
So $\boxed{86}$ is the answer
[/hide]

[hide="Partial Solution"]
let $a,b,c,d$ be the roots of the equation

$a+b+c+d=18$
$ab+ac+ad+bc+bd+cd=k$
$abc+abd+acd+bcd=-200$
$abcd=-1984$

so if we let $ab=-32\text{ then }cd=62$ Manipulating the second equation we get..
$(a+b)(c+d)+ab+cd=k$
$(a+b)(c+d)+30=k$
$(a+b)(18-a-b)+30=k$
$18a+18b-2ab-a^{2}-b^{2}+30=k$
$18a+18b-a^{2}-b^{2}+94=k$

substituting for a..just makes it real ugly..so not sure what to do...
[/hide]
Forgive me if i have any sign errors..i probably do..
EDIT: junggi beat me but i know what to do now

[hide="Solution"]

By Vieta's Formulas we know that

$a+b+c+d=18$

$ab+cd+bc+ad+ac+bd=k$

$abc+bcd+acd+abd=-200$

$abcd=-1984$

where a,b,c, and d are the roots of the polynomial. WLOG let $ad=-32$. By the last formula we know that $bc=62$. We can plug this information in to the third equation and use the first and third equations to find that $a+d=4$ and that $b+c=14$. We can substitute the values for $ad$ and $bc$ into the second equation. Note that $ab+cd+ac+bd=(a+d)(b+c)$, so $k=30+(14)(4)=\boxed{86}$[/hide]

Correct  :) The problems from earlier years are definitely easier than the more recent problems.

Yay, another USAMO problem I can claim to have solved:
[hide]$abc+abd+bcd+acd=-200$
$31a+31b-16c-16d=-100$

$a+b+c+d=18$

$a+b=4$
$ab=-32$
a and b are the roots of
$x^{2}-4x-32$
a=-4, b=8

$c+d=14$
$cd=62$
$c=7+i\sqrt{13}$
$d=7-i\sqrt{13}$
So plugging this into $ab+ac+ad+bc+bd+ad$ we get $86$ .[/hide]
PS: Where did you get this problem?


what is the measure of the smaller angle made by the minute hand and the hour hand at 2:48?

** i know you can draw it out, but there is a formula for these kind of probs....does anyone know it?  it is something like \[ |
\frac{11h\minus{}60m}{2}|\], but i dont know.. :huh:

[quote="DaReaper"]what is the measure of the smaller angle made by the minute hand and the hour hand at 2:48?

** i know you can draw it out, but there is a formula for these kind of probs....does anyone know it?  it is something like
\[ | \frac {11h \minus{} 60m}{2}|
\]
, but i dont know.. :huh:[/quote]
[hide="Generalization"]
Yeah, we can generalize it.  Say that the time is currently x:y.  Notice that for each minute, the minute hand moves 6 degrees.  So, the minute hand has moved 6y degrees.  The hour hand moves 30 degrees per hour.  There has been $ x \plus{} \frac {y}{60}$ hours by x:y.  So, the hour hand has moved $ 30x \plus{} \frac {y}{2}$.  Thus, the degree measure between the hands is $ |30x \plus{} \frac {y}{2} \minus{} 6y| \equal{} \boxed{|\frac {60x \minus{} 11y}{2}|}$

Plug in y=48 and x=2 to get that the degree measure is $ |\frac{120\minus{}528}{2}|\equal{}\frac{528\minus{}120}{2}\equal{}\frac{408}{2}\equal{}208$.  We want this angle to be below below 180, so it is $ \boxed{152}$.  [/hide]

[quote="DaReaper"]where did you get the 152 from?[/quote]
$ 360 \minus{} 208 \equal{} 152$

Thats because we want the smaller angle.

[quote="DaReaper"]what is the measure of the smaller angle made by the minute hand and the hour hand at 2:48?

** i know you can draw it out, but there is a formula for these kind of probs....does anyone know it?  it is something like
\[ | \frac {11h \minus{} 60m}{2}|
\]
, but i dont know.. :huh:[/quote]

[hide="I Like Pie s formula"]
For the Non-Reflex Angle between the hour and minute hands (in degrees): $ 180 \minus{} \left|180 \minus{} 30h \plus{} \frac {11m}{2} \plus{} \frac {11s}{120}\right|$
[/hide]

[quote="DaReaper"]WHAT ABOUT THE REFLEX ANGLE? WHAT IS THE FORMULA FOR THAT?[/quote]
The reflex angle of x is simply 360-x (normally it is above 180, so people often like avoiding it).  So, i_like_pie's formula for the reflex angle is $ 180\plus{}|180\minus{}30h\plus{}\frac{11m}{2}\plus{}\frac{11s}{120}|$.  [/hide]

[quote="DaReaper"]NEVER MIND, THATS 208 RIGHT?[/quote]
Yes, that would be the 208.  The non-reflex angle would be the 152.  I chose the non-reflex angle so we don't have to deal with numbers above 180.

Try to turn off all caps. It doesn't look nice.

sorry, i was working on a project that involved all caps... :blush:

my bad... :blush:


geometry
geometric transformation
reflection
rotation
dilation
trapezoid
ratio

Let ABC be a triangle and let ABD be an "external" (D on the opposite halfplane defined by AB wrt C) rectangular triangle at B (<ABD=90). If ACE is similar to ABD and "external" too wrt ABC, then it holds that the perpendicular bisector of BC bisects DE too.

The subject line of your post gives a big hint...  :D

  darij

Could someone please post the solution?  :maybe:

Posted before.
My solution was like this:
Take $M, N, P$ the projections of $A, D, E$ respectively on $BC$. See that $\triangle BND\sim \triangle AMB$ and $\triangle EPC\sim \triangle CMA$, so $\frac{AM}{NB}=\frac{AB}{BD}=\frac{AC}{CE}=\frac{AM}{CP}$, hence $NB=CP$, that is, the perpendicular bisector of $BC$ is the perpendicular bisector of $NP$ as well, and it is midline of the trapezoid $DNPE$, i.e. passes through the midpoint of $DE$, done.

Best regards,
sunken rock

Nice solution, sunken rock  :)
I'd found the same solution, but I want the solution using just reflection, rotation and dilation.
Perhaps Darij could post it  :maybe:

Dear Mathlinkers,
in a particular case, the synthetic proof of this result of O. Bottema can be seen on
http://perso.orange.fr/jl.ayme   vol. 5  La figure de Vecten p. 72
Sincerely
Jean-Louis

[quote="jatin"]I want the solution using just reflection, rotation and dilation.
Perhaps Darij could post it  :maybe:[/quote]

Oh, right, so I posted a one-liner almost 7 years ago... Time to explain it a bit:

We work with directed angles modulo $360^{\circ}$ (not modulo $180^{\circ}$). We WLOG assume that the angles $\measuredangle CAE$ and $\measuredangle DAB$ are $45^{\circ}$ (and not $-45^{\circ}$). (In fact, if they are $-45^{\circ}$, we can simply reflect everything in a line.)

Let $F$ be the point satisfying $\measuredangle FCB = \measuredangle CBF = 45^{\circ}$. Then, clearly, triangle $BFC$ is a 45-45-90 triangle (this means a triangle with two $45^{\circ}$ angles and one $90^{\circ}$ angle), so that $\measuredangle BFC = 90^{\circ}$ and $BF=CF$, so that the point $F$ lies on the perpendicular bisector of the segment $BC$. Thus, in order to solve the problem, it is enough to show that $F$ is the midpoint of $DE$.

Let $x$ be the spiral similitude with center $A$ mapping $E$ to $C$. Clearly, the ratio of this similitude is $\frac{1}{\sqrt 2}$ (since $\frac{AC}{AE}=\frac{1}{\sqrt 2}$ (because $ACE$ is a 45-45-90 triangle)) and the angle of this similitude is $-45^{\circ}$ (because $\measuredangle EAC = -\measuredangle CAE = -45^{\circ}$).

Let $y$ be the spiral similitude with center $F$ which maps $C$ to $B$. Clearly, the ratio of this similitude is $1$ (since $BF=CF$ yields $\frac{CF}{BF}=1$) and the angle of this similitude is $\measuredangle CFB = -\measuredangle BFC = -90^{\circ}$. In other words, $y$ is a rotation with angle $-90^{\circ}$.

Let $z$ be the spiral similitude with center $A$ mapping $B$ to $D$. Clearly, the ratio of this similitude is $\sqrt 2$ (since $\frac{AD}{AB}=\sqrt 2$ (because $ABD$ is a 45-45-90 triangle)) and the angle of this similitude is $-45^{\circ}$ (because $\measuredangle BAD = -\measuredangle DAB = -45^{\circ}$).

Now, it is known that the composition of spiral similitudes with ratios $u$, $v$, $w$ and angles $p$, $q$, $r$ is a spiral similitude with ratio $uvw$ and angle $p+q+r$, unless $p+q+r=0^{\circ}$ (in the case when $p+q+r=0^{\circ}$, it is a parallel translation). Applying this to our spiral similitudes $x$, $y$, $z$, we see that the composition $z\circ y\circ x$ is a spiral similitude with ratio $\frac{1}{\sqrt 2}\cdot 1\cdot \sqrt 2=1$ and angle $\left(-45^{\circ}\right)+\left(-90^{\circ}\right)+\left(-45^{\circ}\right)=-180^{\circ}=180^{\circ}$ (because $\left(-45^{\circ}\right)+\left(-90^{\circ}\right)+\left(-45^{\circ}\right)\neq 0^{\circ}$). In other words, this composition $z\circ y\circ x$ is a central reflection. Since this reflection maps $E$ to $D$ (because

$\left(z\circ y\circ x\right)\left(E\right)=z\left(y\left(\underbrace{x\left(E\right)}_{=C}\right)\right)=z\left(\underbrace{y\left(C\right)}_{=B}\right)=z\left(B\right)=D$),

the center of this reflection must be the midpoint of $DE$. Now let us prove that this reflection leaves $F$ invariant.

In fact, let $X$ be the point satisfying $\measuredangle FAX = \measuredangle XFA = 45^{\circ}$. Then, triangle $AXF$ is a 45-45-90 triangle, so that $\frac{AF}{AX}=\sqrt 2$ and $\measuredangle XAF = -45^{\circ}$. Thus, the map $z$ maps $X$ to $F$ (since $z$ is the spiral similitude with center $A$, ratio $\sqrt 2$ and angle $-45^{\circ}$). In other words, $z\left(X\right) = F$.

Further, let $Y$ be the point satisfying $\measuredangle AFY = \measuredangle YAF = 45^{\circ}$. Then, triangle $AYF$ is a 45-45-90 triangle, so that $\frac{AY}{AF}=\frac{1}{\sqrt 2}$ and $\measuredangle FAY = -45^{\circ}$. Thus, the map $x$ maps $F$ to $Y$ (since $x$ is the spiral similitude with center $A$, ratio $\frac{1}{\sqrt 2}$ and angle $-45^{\circ}$). In other words, $x\left(F\right) = Y$.

Since $AXF$ and $AYF$ are 45-45-90 triangles, it is clear that $AXFY$ is a square, so that $XFY$ is a 45-45-90 triangle. Thus, the rotation around $F$ with angle $-90^{\circ}$ sends $Y$ to $X$. Since the rotation around $F$ with angle $-90^{\circ}$ is $y$, this rewrites as follows: The map $y$ sends $Y$ to $X$. In other words, $y\left(Y\right)=X$. Hence,

$\left(z\circ y\circ x\right)\left(F\right)=z\left(y\left(\underbrace{x\left(F\right)}_{=Y}\right)\right)=z\left(\underbrace{y\left(Y\right)}_{=X}\right)=z\left(X\right)=F$.

Thus, $F$ is a fixed point of $z\circ y\circ x$. Since $z\circ y\circ x$ is a central reflection, and the only fixed point of a central reflection is its center, this yields that $F$ is the center of the reflection $z\circ y\circ x$. But we know that the center of the reflection $z\circ y\circ x$ must be the midpoint of $DE$. Thus, $F$ is the midpoint of $DE$, qed.


2b-6a+8c+a+2b-5ab= -396

8bc+2b^2-3a^2+4ac= 887

Compute a+b+c

This question is from a Mock AMC12

hello, is
$ 8bc\plus{}2b^2\minus{}3a^2\plus{}4ac\equal{}887$
$ 4b\minus{}5a\plus{}8c\plus{}5ab\equal{}\minus{}396$
an equation system?
Sonnhard.

I have been given the 2 above equations
and I suppose that were suppose to find out what does a+b+c equals to by solving the above equations


In the figure below, the large circles are congruent and tangent to each other. Their radii are equal to 6 inches. The smaller circle in the middle is tangent to the larger circles. What is the radius of the small circle?

[img]http://www.spirit-lake.k12.ia.us/new/HS/Morris/p096.jpg[/img]

Math is a subset of life.  :P

Hint: you can find the solution if you use a right triangle.
________________________________________________
Math is a subset of life  :P

[quote="Kevin K."]Hint: you can find the solution if you use a right triangle.
[/quote]

Um...That is what I did. Why don't you check my solution first.

[hide]There is an isosceles right triangle with legs 12, 12, 12{2^(1/2)}. Since the diameter of the small circle is part of the hypotenuse, the diameter of the circle is 12{2^(1/2)} - 12. Divide the diameter in two, and the radius of the small circle is [b]6(2^(1/2))-6[/b].[/hide]

(I just realized that your answer was right)

[hide]You can make a 45-45-90 right triangle with any three of the four centers of the circles. Since the radii of the circles are 6 inches, then the two legs of the triangle are 12 inches and the hypotenuse is 12*sqrt.(2). We also see that the hypoteneuse of this triangle is equal to the sum of 2 times the radii of the big circles (6 in.) and the diameter of the small circle. Therefore, the diameter of the little circle is 12*sqrt(2)-12 and the radius of the small circle is half of that, which equals [b]6(sqrt(2)-1)[/b][/hide]


1. If $ z$ is an n-th root of unity, then find the value of the sum $ 1 \plus{} z \plus{} z^2 \plus{} z^3 \plus{} ... \plus{} z^{n\minus{}1}$.

2. If the coefficients of the polynom $ z^2 \plus{} az \plus{} b$ are real and its zeros are $ w$ and $ \bar{w}$, show that $ a \equal{} \minus{}2 Re(w)$ and $ b \equal{} |w|^2$.

3. For a > 0, what does the equation $ \left|\frac{z\plus{}1}{z\minus{}1}\right| \equal{} a$ represents

(a) for $ a \equal{} 1$?

(b) for $ a \neq 1$?

4. Show the following properties of complex numbers:

(a) $ \bar{\bar{z}} \equal{} z$

(b) $ |\bar{z}| \equal{} |z|$

(c) $ z \bar{z} \equal{} |z|^2$

(d) $ \bar{z \plus{} w} \equal{} \bar{z} \plus{} \bar{w}$

(e) $ \bar{zw} \equal{} \bar{z}\bar{w}$

(f) $ |zw| \equal{} |z| \cdot |w|$

(g) $ |z\plus{}w| \leq |z| \plus{} |w|$

5. Solve, in $ \mathbb{C}$, each of the following equations:

(a) $ z^3 \minus{} 1 \equal{} 0$

(b) $ z^4 \plus{} 1 \equal{} 0$

(c) $ z^6 \plus{} i \minus{} 1 \equal{} 0$

(d) $ z^3 \equal{} z$

(e) $ z \equal{} (\bar{z})^2$

(f) $ \bar{z} \equal{} z^4$

(g) $ (\bar{z})^5 \equal{} z$

(h) $ z^3 \bar{z} \minus{} 1 \plus{} 1 \equal{} 0$

(i) $ z^3 \equal{} (2 \minus{} 2i)^3$

(j) $ |z|z^2 \equal{} (\sqrt {3} \minus{} i)^3$

(k) $ z \bar{z} \minus{} i \bar{z} \minus{} 1 \equal{} i$

Some problems are trivial if vieta's formulas are used like 1,5(a,d) and maybe more(didn't notice that much.
Anyway ans. to#1 is[hide]0[/hide]

Your answer is correct but show your reasoning.

[quote="Carcul"]1. If $ z$ is an n-th root of unity, then find the value of the sum $ 1 \plus{} z \plus{} z^2 \plus{} z^3 \plus{} ... \plus{} z^{n \minus{} 1}$.

2. If the coefficients of the polynom $ z^2 \plus{} az \plus{} b$ are real and its zeros are $ w$ and $ \bar{w}$, show that $ a \equal{} \minus{} 2 Re(w)$ and $ b \equal{} |w|^2$.

3. For a > 0, what does the equation $ \left|\frac {z \plus{} 1}{z \minus{} 1}\right| \equal{} a$ represents

(a) for $ a \equal{} 1$?

(b) for $ a \neq 1$?[/quote]
hi ,i see your preparing for the complex seriously !!! good luck !
1) $ \sum z^k  \equal{}\frac {z^n \minus{} 1}{z \minus{} 1}  .......................$
2) $ z^2\plus{}az\plus{}b \equal{} (z\minus{}w)(z\minus{}\bar{w})$ and we know that $ w\plus{}\bar{w}\equal{}2Re(w)$ and $ |w|\equal{}(w*\bar{w})^(1/2)$
3) !!!!!! waht does the equation represent !!!!!  :maybe:  $ |z\plus{}1|\equal{}a|z\minus{}1|$ ==>$ AM\equal{}AB*a$
for $ a\equal{}1:  AM\equal{}AB$  for $ a \neq 1$ transaltion i think


I'm bit unsure about what successive and consecutive mean regarding the sides of the polygon. Like, I thought successive meant in that order so:

Successive sides of pentagon is are 2000,1999,1998,1997,1996. Then if I draw it, shouldn't 2000 always be next to 1996 and 1999 (that is, I can't have 2000,1998,1997,1996,1999 or something like differently).

Now, how would that change or be same if it said "consecutive sides..."?

Thanks.

As far as I can tell, it would be the same up to orientation, and the answer to most problems shouldn't change based on orientation.  If the orientation affects the answer, it should be given (i.e. "in counterclockwise order").

I think...

"successive" implies that something follows another (ie, "next"). This word is better used for things like iterations, recursions, repetitive processes, and such.

"consecutive" implies that they are next to each other, but does not have as much of a forward-directional implication

"adjacent" is best used when things are physically touching but order is not implied.

You could probably just say "in that order", or say something like "going counterclockwise, the side-lengths are blah" / "the side lengths are blah in counterclockwise order".

Or, you can say that going counterclockwise, the sidelengths form an AP, or something like that.


algebra
polynomial
superior algebra
superior algebra solved

Hello All,
I am pondering the following problem:

Let $\mathbb F = \mathbb F_2$ be the field of two elements, $\mathbb K=\mathbb F(x)$ and $\mathbb L = \mathbb K(\sqrt x).$  Show the trace map is identically zero.

(I assume they mean the trace of $\mathbb L/\mathbb K.$ )

Now, my thoughts on this are that since we are initially extending by an indeterminate, we essentially have a transcendental extension.  This eventially works down to the idea that L is an inseperable extension of K and hence the trace is zero.  Am I thinking correctly?

-JNG

$\mathbb{L}/\mathbb{K}$ is an extension of degree $2$, and it can be seen as the splitting field of the polynomial $X^{2}-x=0$ over $K$. In particular, $|Aut(\mathbb{L}/\mathbb{K})| \leq 2$, whence $|Aut(\mathbb{L}/\mathbb{K})|=1$ since $X^{2}-x=0$ is not separable (since $\mathbb{L}$ has characteristic $2$; in particular, this also says our extension is not Galois). The group of the $\mathbb{K}$-isomorphisms is thus trivial, and the trace of any element $u$ is (taking into account the multiplicity) $2u=0$.


Yesterday I read in the schaum’s solved problem series "there is no node for an s orbital .thus an electron can reside in the nucleus too"
Isn’t this absurd regarding the fact that by the Heisenberg’s uncertainty principle an electron cannot reside in the nucleus: if it were to do so, it would have velocities greater than the velocity of light. Is there any advertent error in my book or this is a fact? If so, please explain

[quote="_anant"]by the Heisenberg’s uncertainty principle an electron cannot reside in the nucleus: if it were to do so, it would have velocities greater than the velocity of light.[/quote]

Hmm.  Could you explain your logic here?  I don't see the contradiction, considering it just means that part of the electron's probability distribution across space is in the nucleus.

by the aforementioned principle, del(x) * m((del v))>= h/4pie .keeping the values of del (x) =approx. diameter of the nucleus. will give del V of the order of c. also it cannot be digested by me how there will be even 1% probability of the standing wave of the electron to reside inside the nucleus. A nuclear reaction will start as it will be captured by a proton.

From http://www.newton.dep.anl.gov/askasci/chem99/chem99283.htm :

[quote="Christopher Grayce"]What about electrons in s orbitals?  Well, they can and do hit the
nucleus.  Not often.  Although the probability per unit volume of an s
electron being at the nucleus is very high, the volume of the nucleus is
only about a trillionth the volume occupied by the typical electron.  But
occasionally one strays in, and then (if certain factors relating to the
physics of the nucleus are right) it gets absorbed by a proton, which turns
into a neutron and emits a neutrino.  This, as you've guessed, is a form of
radioactive decay of elements, called "K-shell capture" because it normally
only happens for electrons in the lowest (K) shell.  The atom loses 1 in
atomic number (e.g. a chlorine atom turn into a sulfur atom) but stays
neutral.  The neutrino is hard to detect, but the hole the captured
electron leaves is filled by an electron "falling down" from a higher
energy level in the atom, which results in an X-ray being emitted that you
can detect.
     Incidentally the s electrons that come very close to the nucleus and
don't get captured are the source of "spin-spin coupling" in nuclear
magnetic resonance (NMR) spectra.  This effect basically allows you to
count the number of hydrogens bonded to each carbon atom in a molecule.
Pretty helpful if you're trying to figure out what the molecule looks like.[/quote]

So it looks like you are right; it will be captured by a proton, and such phenomena are known in nature.  No contradiction!   :D


algorithm
vector
linear algebra
matrix
number theory
superior algebra
superior algebra solved

What's the order of $SL_2(\mathbb{Z} / n\mathbb{Z})$ ??

[quote="jmerry"]Let $f_k(n)$ be the number of $k\times k$ matrices with determinant 1 over $\mathbb{Z}_n$, the ring of integers mod $n.$
We clearly have that $f_k$ is multiplicative in the usual number-theory sense (use the Chinese Remainder theorem),
This leaves only the question of finding $f_k(p^j)$ for primes $p$ and positive integers $k$.

A universal algorithm for constructing and counting these matrices:
Step 1: Choose vectors $v_0,v_1,\dots,v_{k-1}$ such that their reductions mod $p$ are linearly independent. There are $\left(p^k-p^i\right)\cdot p^{k(j-1)}$ ways to choose $v_i$, for a total of $(p-1)(p^2-1)\cdots(p^k-1)\cdot p^{k^2(j-1)+k^2/2-k/2}$ ways to do this.
Step 2: Divide $v_{k-1}$ by the determinant of the matrix with columns $v_0,v_1,\dots,v_{k-1}$ (mod $p^j$) to obtain $w_{k-1}.$ This reduces the number of possibilities by a factor of $p^{j-1}\cdot(p-1),$ since different choices for the last column lead to the same final matrix.
Step 3: Build the matrix with columns $v_0,v_1,\dots,v_{k-2},w_{k-1}.$

Thus $SL_k(\mathbb{Z}_{p^j})$ has $(p^2-1)(p^3-1)\cdots(p^k-1)\cdot p^{(k^2-1)(j-1)+k^2/2-k/2}.$ For $k=2$, this is $(p^2-1)p^{3j-2}.$
To generate other values, multiply together what you get at prime power values.[/quote]

 Hi jmerry. This is really helpful. I get stuck because I didn't think of the multiplicative property :blush:


function
integration
limit
logarithms
real analysis
real analysis solved

A function $f(x)$ is defined at $x>0$ as follows.

\[f(x)=\int^x_0 \frac{t}{(t+1)(t+a)}dt\ (a>0)\ \]
Let $g(a)=\lim_{x\to\infty}\{\ln (x+1)-f(x)\}$,examine the continuity of the function $g(a)$  at  $a=1$.

Since $\ln(x+1)=\int_0^x \frac1{t+1} dt$,
$g(a)=\lim_{x\to\infty}\int_0^x \left(\frac1{t+1}-\frac{t}{(t+1)(t+a)}\right) dt =\int_0^{\infty}\frac{a}{(t+1)(t+a)} dt$

In this form, $g$ is obviously smooth at 1 and at all other $a>0$. Its value at 1 is 1.


LaTeX
SFFT
special factorizations
\/closed

A-O-P-S is the Georgia pronunciation as far as I know, with few exceptions. However, I have been told that most non-Georgians pronounce it A-Ops.

[quote="Ovrlndnsea"]How do I pronounce "AoPS" if I don't want to say "Art of Problem Solving?"[/quote]

i say A-ops

We say "A-Ops" here around the office when referring to ourselves.  It's not a big deal though.

More challenging: pronounce "Rusczyk" :)

I seem to remember Amber saying Rus-sic...

[quote="b-flat"]I always thought it was ruz-e-ik or something like that.

While we are at it, how is $\LaTeX$ pronounced?[/quote]
LaTeX should be pronounced 'lay-tech' where tech is said as in technology.

Mm, I pronounce it "A-O-P-S" and "Rus-ik".  :P
And I always just read LaTeX just as it's spelled.

I say Ay-ops.
Isn' it Ruh-chick? Or is it Rush-ik?
Mr. Frost says the former.

And isn't $\LaTeX$ Lah-Teck? I say Lay-tecks anyway.

[quote="mysmartmouth"]I always say A-O-P-S, but that's probably just me.

Rusczyk...I would say Roo-Shick, but if that's wrong, I would appreciate a correction.[/quote]

I say A-O-P-S, as I haven't heard of anyone else who has said it before, and nothing else seems right to say.

A-ops sounds more like a robot.. :blush:

I would say Ruh-sick, as it sounds more like english.

Lah-tecs

Ruhk-zic

I have no need to say Art of Problem Solving in Public, but if I did, I would probably say A-Ops

[quote="[url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=latexpronounce]the UK List of TeX Frequently Asked Questions[/url]"]How should I pronounce “LaTeX(2e)”?

Lamport never recommended how one should pronounce LaTeX, but a lot of people pronounce it ‘Lay TeX’ or perhaps ‘Lah TeX’ (with TeX pronounced as the program itself; see the rules for TeX). It is definitely not to be pronounced in the same way as the rubber-tree gum. [/quote]

[quote="[url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=TeXpronounce]the UK List of TeX Frequently Asked Questions[/url]"]How should I pronounce “TeX”?

The ‘X’ is “really” the Greek letter Chi, and is pronounced by English-speakers either a bit like the ‘ch’ in the Scots word ‘loch’ ([x] in the IPA) or (at a pinch, if you can’t do the Greek sound) like ‘k’. It definitely is not pronounced ‘ks’ (the Greek letter with that sound doesn’t look remotely like the Latin alphabet ‘X’).

This curious usage derives from Knuth’s explanation in the TeXbook that the name comes from the Greek word for ‘art’ or ‘craft’ (‘techni’), which is the root of the English word ‘technology’. Knuth’s logo for TeX is merely the uppercase version of the first three (Greek) letters of the word, jiggled about a bit....[/quote]

Is this any different than bpms? ruh-sic

well Rusczyk pronounced his name on the Simon's FAvorite factoring trick vid so i knwo how to prounounce that...

as for AoPS i pronounce it "a-ops"

I would pronounce A-Ops and Rush-chick.

If I was talking to someone who didnn't know it at all, I would say "Art of Problem Solving".


Let $ ABC$ be a triangle with three bisector $ AA';BB';CC'(A' \in BC;B' \in AC;C' \in AB)$.
The circumcircle of triangle $ A'B'C'$ cuts $ BC;CA;CA$ at $ X;Y;Z$
Prove that three sides $ A'X;B'Y;C'Z$ exist a sides is sum of two sides remaining.

it sounds like some old Kurshak competition problem.


#1. DarkKnight (graphic)

[img]http://img179.echo.cx/img179/7674/mostlyharmless9df.jpg[/img]

#2. sky9073 (graphic)

[img]http://artofproblemsolving.com/Forum/files/thumbs/t_mharmless_195.gif[/img]

#3. LynnelleYe (graphic)

[img]http://img.photobucket.com/albums/v189/bigsister/mostlyharmless3.png[/img]

#4. Dnas (graphic)

[img]http://dnas.outer-heaven.net/vogon.png[/img]

#5. ToBe (animation)

[url]http://www.artofproblemsolving.com/Forum/files/dogggyanimmostharmless_225.swf[/url]

You have one week to vote. Next Monday we will find out the winner! I will vote in case of a tie.

NOTE: The competitors CAN vote for themselves.

Of course I voted for myself, but if I weren't biased I'd probably vote for Dnas's pic. (No offense to the others, I love all of them, but Dnas's is just...cool.)
But the link to ToBe's animation doesn't work.

Dnas's is definitely the coolest pic, but I voted for sky because I thought hers fit in more with "fits best with the topic"

Oh, and in case I win, which I won't, the topic for the next round is 'time'.


Maybe if you make them a foe? If not, you can just delete the PMs.

[quote="kstan013"]I'd go with the adding to the list of foes :D That's the whole point of a foes list anyway, right?[/quote]
I don't believe that works yet

[quote="chess64"]Is it possible to ignore all PM's from certain users? For example, there is a certain user who PMs me several times every day, but I am not interested in reading his messages.[/quote]

of course if it is spam you can just tell an admin and they can either pinpoint the user's real account and hopefully ban their IP address or something  :lol:

[quote="chess64"]Good point, can someone ban anirudh's ip address? j/k. i guess i'll just keep deleting any pm from #H34NI :roll:[/quote]

You probably didn't read like them.

It was mainly about asking what I did wrong. :|

And I wanted a reply so I could stop it but you never replied so I sent it again, thinking that it never went through. :|

Whoa haha I guessed who you were talking about WOOT.  Just waiting for the foes list, eh?


Does there exist a set of 11 integers such that no 6 of these integers have their sum divisible by 6 ?

(Is the question clear ?)

That shouldn't matter:  only the mod-sixiness of them is relevant.  That is, if you think you need a solution that has the number 4 twice, just change one of them to a 10 and it will still work.

This was a clever problem.  There exists no such set; solution follows:

[hide]Per Joel's comment, we need only consider sets (I suppose that they would be multisets, since the elements are not all distinct, but the terminology is irrelevant) such that each integer is in the range 0:le:k<6 (ALL sets considered hereafter will be such sets).   Furthermore (and this is important), we need only consider sets such that 0 is a mode.  This is because subtracting a constant from each of the elements in the set does not change whether the sum of a given 6 elements is divisible by 6.  By pigeonhole, then, there are at least 2 0s.  We can now do a (slightly messy) case analysis based on how many 0s there are in the set.

Case 1: 6 or more zeros:

We're already done.

Case 2: 4 or 5 zeros:

It suffices to show that in a set of 6 numbers, none of which is zero, there must exist some subset of the set the sum of whose members is divisible by 6.  Now we will build a set of size 6, trying to avoid such a subset, and show it is impossible.  First, such a set cannot contain both 1 and 5, both 2 and 4, or two 3s.  Hence it can have at most one 3 and only two other distinct numbers.  Suppose these numbers are 1 and 2 (the argument is analogous for the cases of 1 and 4, 2 and 5, and 4 and 5).  Then we have five elements fitting into these two classes, and thus by pigeonhole some one of these classes must contain at least three elements of the set.  If there are three 2s, we are done, so suppose instead that there are three or more 1s.  Then we cannot have a 3, so we must then have four or more 1s.  We cannot have six 1s (0 was a mode!); if we have five or four 1s, we must have a 2, and four 1s and a 2 sum to 6.  This concludes case 2.

Case 3: 3 zeros:

We now need to show that any set of 8 elements, none of which is zero and with no more than 3 elements taking any given value (0 is a mode!), has some subset of size 3 or more the sum of whose members is divisible by 6.   Now 1+2+3 = 6, and 3+4+5=12, so either two of {1, 2, 4, 5} are not in our set, or 3 is not in our set.  In the former case we can use an argument similar to the second part of case 2:  There are 8 elements and at most three 3s, so if the other two numbers used are 1 and 4 or 2 and 5, we can use the exact same argument as in case 2.  If the other two numbers used are 2 and 4, we are done, since there must be three of at least one of them.  If they are 1 and 5, there are again three of at least one of them, and so we are done if there are any 3s at all.  In the case that there are then no threes, we must have at least 4 of one of the others, which is a contradiction (there are at most three elements with any given value).  In either case we are done.

In the latter case (no 3s), we are done if there are {two 1s and two 5s}, {two 2s and two 4s}, or {one each of 1, 2, 4, 5}.  Suppose, then, that none of these occurs.  Then we must have at most 1 of one of these and none of some other.  But then we have seven elements to distribute among two numbers, and by pigeonhole, there are thus four of some number.  But this is a contradiction, since there are at most three elements with any given value.

Case 4: 2 zeros:

We must now show that in a set of 9 elements, none of whose elements are zero and that contains no more than 2 elements with any given value, we have some subset of at least four elements whose members' sum is divisible by 6.

Since we have nine elements and five possible values for each to take, and no value is taken more than twice, four of these values are taken twice and one once.  But then we either have {two 1s and two 5s} or {two 2s and two 4s}, and we are done.

This completes the proof.

I'm sure that this proof could be considerably shortened by careful arrangement, but I was wondering if there was a different, nicer way to do it.

[/hide]

My original solution was a bit shorter. Have a look at

http://www.kalva.demon.co.uk/bmo/bsoln/bsol004.html ;

the solution is kindly submitted by myself :wink:.

Yours is definitely a better solution.  I'm kind of surprised I didn't see that...   My case analysis was way too nasty.

Ah well.

Does anyone still have this solution? It seems vanished from Kalva.

Of course I do, since I submitted the solution.

Lemma 1: among any three integers, two integers can be found having an even sum.

Lemma 2: among any five integers, two integers can be found having a sum which is divisible by 3.

(Proof: almost trivial, pigeonhole principe.)

Now, given $a_{1},a_{2},a_{3},\,\cdots,a_{11}$, we can assume without loss of generality that $b_{1}= a_{1}+a_{2}$ is even, by Lemma 1. Also, we can suppose that $b_{2}= a_{3}+a_{4}$ is even, ..., and that $b_{5}= a_{9}+a_{10}$ is even. Among the five integers $b_{1},b_{2},\,\cdots,b_{5}$, three can be found having a sum which is a multiple of 3, by Lemma 2. But the sum of these $b_{i}$ is equal to the sum of six of the numbers $a_{i}$, so we have found six numbers among the given ones which have their sum divisible by 6.

We can conclude that such a set cannot exist.


1) find the vqlue of m, expressed as an improper fraction, so that this relation is not a function.
{(-2m+1, -4), (-6m+8, 0)}.

2) find all integer values of x for which:
(x-3)(x+1)(x-5) is smaller than zero.

[hide="1."]
If it is not a function, $-2m+1=6m-8$
Solving for m, we get
$m=\frac{7}{4}$
[/hide]
[hide="2."]
Wouldn't any negative smaller than -1 work? A negative times a negative = a positive
A positive times a negative=a negative
[/hide]

2. [hide]If you look at the intervals in which (x-3), (x+1), and (x-5), are negative/positive.  we see that if X<3, (x-3)<0, but so is (x-5).  If  X< -1, then X+1 is negative, so the first set of solutions is (-infinite, -1).  Then, we see that if 3<X<5, then only X-5 is negative, so we have (3,5). Thus, the answer is
(-infinite, -1)U(3,5)[/hide]

2.[hide]one way that you can find the sign of a polynomial functions is drawing a number line with all asymptotes(rational functions) and zeroes.  Determine the change of sign as each term switches its sign.  eventually, multiplying the signs of each term will give a final sign.  use the intervals given to get the integers[/hide]


analytic geometry
geometry
ratio
Pythagorean Theorem

Two walls are 10 ft. apart. Two ladders, one 15 ft. long and one 20 ft. long, are placed at the bottoms of the walls leaning against the opposite walls. How far from the ground is the point of intersection?

[hide]Using the Pythagorean Theorem, we find the heights of the walls: $\sqrt{125}$ and $\sqrt{300}$.

The formula for the height of the intersection is $\frac{ab}{a+b}$.

$\frac{\sqrt{125}\sqrt{300}}{\sqrt{125}+\sqrt{300}}\approx\boxed{6.8\text{ feet}}$[/hide]

How to derive the formula:

Consider right intersecting triangles $ABD$ and $CDB$, sharing side $BD$.

$BC$ and $AD$ intersect at $E$

We know that $\frac{CD}{EF}= \frac{BE+EC}{BE}$
and $\frac{CD}{AB}= \frac{EC}{BE}$

so combining the top two ratios,
$\frac{CD}{EF}= 1+\frac{CD}{AB}$ so
$EF = \frac{(CD)(AB)}{AB+CD}$.

It turns out that in questions like these, the height of the intersection is the product of the heights over the sum of the height, or in other words, half the harmonic mean of the heights


Find the value of $ n$ if $ 5000 \minus{} n \equal{} 991 \plus{} 993 \plus{} 995 \plus{} 997 \plus{} 999$?

hello, we have $ 5000\minus{}n\equal{}4975$ so $ n$ is $ 25$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we have $ 5000 \minus{} n \equal{} 4975$ so $ n$ is $ 25$.
Sonnhard.[/quote]
I don't think your solution to have good sense.

[b]My solution[/b]
$ 991\equal{}1000\minus{}9$
$ 993\equal{}1000\minus{}7$
$ 995\equal{}1000\minus{}5$
$ 997\equal{}1000\minus{}3$
$ 999\equal{}1000\minus{}1$
So, $ 991 \plus{} 993 \plus{} 995 \plus{} 997 \plus{} 999 \equal{} 5000\minus{}(9\plus{}7\plus{}5\plus{}3\plus{}1)\equal{}5000\minus{}25$.
Thus, $ n\equal{}\boxed{25}$.


What is the solution you had n mind, April?
I'm sure arqady's stronger ineq. is just s.o.s. brute force proof;
the given conditions are hard to work with, I need ideas.

Here is what it is equivalent to:\[3\left(\sum a^{2}bc+\sum_{sym}a^{3}b^{2}+\sum a^{4}bc\right)\geq  \sum a^{2}b^{2}c+\sum_{sym}a^{3}b^{2}c+\sum_{cyc}a^{3}b^{2}c^{2}\]

[hide="the original one"]
The condition $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ is equivalent to $abc=ab+bc+ca$:
\[\frac{a^{2}}{a+bc}=\frac{a^{3}}{a^{2}+abc}=\frac{a^{3}}{a^{2}+ab+bc+ca}=\frac{a^{3}}{(a+b)(a+c)}\]
WLOG let $a\geq b\geq c$; then we also have
\[\frac{a^{2}}{(a+b)(a+c)}\geq \frac{b^{2}}{(a+b)(b+c)}\geq \frac{c^{2}}{(a+c)(b+c)}\]
because $\frac{a^{2}}{(a+b)(a+c)}=\frac{1}{1+\frac{ab+bc+ca}{a^{2}}}$ etc. Applying Chebyshev,
\[\sum_\text{cyc}\frac{a^{2}}{a+bc}=\sum_\text{cyc}\frac{a^{3}}{(a+b)(a+c)}\geq \frac{a+b+c}{3}\cdot \sum_\text{cyc}\frac{a^{2}}{(a+b)(a+c)}\]
Now we only need to prove
\[\sum_\text{cyc}\frac{a^{2}}{(a+b)(a+c)}\geq \frac{3}{4}\]
After clearing denominators, it reduces to
\[a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b\geq 6abc \]
which is just AM-GM.
[/hide]

[hide="the stronger one given by arqady"]
Again we simplify $\frac{a^{2}}{a+bc}$ by writing it as $\frac{a^{3}}{(a+b)(a+c)}$. Now use Cauchy:
\[\left( \sum_\text{cyc}a(a+b)(a+c) \right) \left( \sum_\text{cyc}\frac{a^{3}}{(a+b)(a+c)}\right) \geq (a^{2}+b^{2}+c^{2})^{2}\]
So it only remains to prove that
\[4(a^{2}+b^{2}+c^{2})(a+b+c)\geq 3(a(a+b)(a+c)+b(a+b)(b+c)+c(a+c)(b+c)) \]
But we have
\[a(a+b)(a+c)+b(a+b)(b+c)+c(a+c)(b+c)=(a^{2}+b^{2}+c^{2})(a+b+c)+3abc\]
so that the above is equivalent to
\[(a^{2}+b^{2}+c^{2})(a+b+c)\geq 9abc \]
Using AM-GM on each of the factors on the left proves this.
[/hide]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=128442&highlight=xy+sqrt]here[/url]
is a similar one, see if there is one of your nice methods for this one, scorpius :)

[quote="scorpius119"]
The condition $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ is equivalent to $abc=ab+bc+ca$:
\[\frac{a^{2}}{a+bc}=\frac{a^{3}}{a^{2}+abc}=\frac{a^{3}}{a^{2}+ab+bc+ca}=\frac{a^{3}}{(a+b)(a+c)}\]
[/quote]
Apply the [b]AM-GM Inequality[/b] we have:
\[\frac{a^{3}}{(a+b)(a+c)}+\frac{a+b}{8}+\frac{a+c}{8}\ge\frac{3}{4}a \]

\[\Longrightarrow\sum\frac{a^{2}}{a+bc}+\frac{\sum a}{2}\ge\frac{3}{4}\sum a \]

\[\Longrightarrow\sum\frac{a^{2}}{a+bc}\ge\frac{\sum a}{4}\]

Yes, scorpius119.
But the following inequality is not true:
$\frac{a^{2}}{a+bc}+\frac{b^{2}}{b+ca}+\frac{c^{2}}{c+ab}\ge\frac{3(a^{3}+b^{3}+c^{3})}{4(a^{2}+b^{2}+c^{2})}.$ :wink:

[quote="April"]Let $a,\,b,\,c$ be positive real numbers such that: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$. Prove that:
\[\frac{a^{2}}{a+bc}++\frac{b^{2}}{b+ca}+\frac{c^{2}}{c+ab}\ge\frac{a+b+c}{4}\]
[/quote]
[hide="hint"]App chebychev
$\frac{a^{2}}{a+bc}+\frac{b^{2}}{b+ca}+\frac{c^{2}}{c+ba}\geq\frac{1}{3}(a+b+c)(\sum\frac{a}{a+bc})$
Then app schwarz
$\sum_{cyc}\frac{a}{a+bc}\geq\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}+3ab+3ac+3bc}\geq 3/4$[/hide]


[color=darkblue]Let $ABC$ be a triangle. Denote  for a point $D\in (BC)$ the intersection $E$ between the $B$- bisector and the line $AD\ .$ Denote the intersection $F\in BE\cap AC\ .$

Prove that if two of the following sentencies are truly, then and the third sentence is truly :

$\begin{array}{ccccccc}\blacktriangleright DB+BA=AC\blacktriangleleft & \ \maltese\ & \blacktriangleright BF^{2}=CB\cdot CF\blacktriangleleft & \ \maltese\ & \blacktriangleright\mathrm{the\ quadrilateral\ }DEFC\mathrm{\ is\ cyclically}\blacktriangleleft\end{array}$[/color]

i'm curious, how was it i inspired this? :)


I have seen a similar post about a message board problem, but it has been a while and I still can't access my message board. Does anyone have any suggestions?

P.S. I have Olympiad Geometry and Intermidiate Algebra.

The class message boards are located in the Online School section of the Forum (the same section that this board is located in).  I've checked and you should have access.


Yesterday Han drove $ 1$ hour longer than Ian at an average speed $ 5$ miles per hour faster than Ian. Jan drove $ 2$ hours longer than Ian at an average speed $ 10$ miles per hour faster than Ian. Han drove $ 70$ miles more than Ian. How many more miles did Jan drive than Ian?

$ \textbf{(A)}\ 120 \qquad \textbf{(B)}\ 130 \qquad \textbf{(C)}\ 140 \qquad \textbf{(D)}\ 150 \qquad \textbf{(E)}\ 160$

Let Ian's speed be $ x$ and his time driving be $ y$

Therefore, Ian drove $ xy$ total miles

Han drove $ (x\plus{}1)(y\plus{}5)\equal{}xy\plus{}5x\plus{}y\plus{}5$

Jan Drove $ (x\plus{}2)(y\plus{}10)\equal{}xy\plus{}10x\plus{}2y\plus{}20$

The problem says that Han drove 70 miles more than Ian so:

$ xy\plus{}5x\plus{}y\plus{}5\minus{}xy\equal{}5x\plus{}y\plus{}5\equal{}70$

$ 5x\plus{}y\equal{}65$

Jan drove $ xy\plus{}10x\plus{}2y\plus{}20\minus{}xy\equal{}2\times65\plus{}20\equal{}130\plus{}20\equal{}150\ (D)$ miles more than Ian.

Or... Let Ian drive for 0 hours.  This implies Han drove for one hour at $ 70$ miles per hour.  Then Jan drove for 2 hours at 75 miles per hour. $ 2*75\equal{}150\equal{}\boxed{D}$


algorithm
number theory
Euclidean algorithm

Find the least positive integer $ n$ for which $ \frac{n\minus{}13}{5n\plus{}6}$ is non-zero reducible fraction.

$ \textbf{(A)}\ 45 \qquad \textbf{(B)}\ 68 \qquad \textbf{(C)}\ 155 \qquad \textbf{(D)}\ 226 \qquad \textbf{(E)}\ \text{none of these}$

[hide="Solution"]
Using the Euclidean Algorithm, $ \gcd(n\minus{}13,5n\plus{}6)\equal{}\gcd(n\minus{}13,71)>1\implies n\minus{}13\equal{}71k$
$ \implies n\ge84\implies\boxed{\textbf{(E)}\ \text{none of these}}$.
[/hide]

[quote="math154"][hide="Solution"]
Using the Euclidean Algorithm, $ \gcd(n \minus{} 13,5n \plus{} 6) \equal{} \gcd(n \minus{} 13,71) > 1\implies n \minus{} 13 \equal{} 71k$
$ \implies n\ge84\implies\boxed{\textbf{(E)}\ \text{none of these}}$.
[/hide][/quote]

Your solution is correct, except 155-13= 71k, when k = 2, so the answer is [b]155 C[/b].   This question was way too brute-forcable for multiple-choice.

That doesn't make sense. The question asks for the [b]least[/b] integer $ n$ such that the fraction given is reducible. math154 clearly showed that the least $ n$ is $ 84$ so his answer was correct. $ 155$ does work, but it is not the [b]least[/b] integer.

For that reason, it's not at all brute-forceable. The MAA isn't stupid :lol:

Putting $ n = 84$ gives ${ \frac\{71}{426} = \frac {1}{6}$.


hello, i have found $ x\equal{}16777216$ as a solution.
Sonnhard.

[hide="Extremely elegant solution"]Noticing all the divisors of $ 24$ and the power of $ 2$, we substitute $ 2^{24}$ in and it happens to work. Hurray![/hide]


what is the unit digit of $ \frac{2^{1993}\plus{}3^{1993}}{5}$?

[hide][b]3[/b].
 What i did was try to factor x^1993+y^1993 and then try to find modular patterns for 2 and 3.[/hide]

[hide]use patterns. the patterns of unit digits for powers of 2 are: 2,4,8,6,2,4,8,6...there are 4 #s in that pattern. 1993=1 mod 4 so its the first # in the pattern, or 2.

do the same for 3: pattern=3,9,7,1...1993=1 mod 4, so its the 1st # in the pattern, or 3.

thus, you add up the 2 units digits to get 5, divide by 5, and the units digit is:

[hide="boy, and to think you scrolled all the way down for no reason but to see one digit"]1[/hide][/hide]

[quote="abacadaea"]you cant factor $ 2^{1993}\plus{}3^{1993}$ :ninja: my friend made me put that[/quote]
you can factor x^m+y^m if m is odd

[quote="mz94"][hide]use patterns. the patterns of unit digits for powers of 2 are: 2,4,8,6,2,4,8,6...there are 4 #s in that pattern. 1993=1 mod 4 so its the first # in the pattern, or 2.

do the same for 3: pattern=3,9,7,1...1993=1 mod 4, so its the 1st # in the pattern, or 3.

thus, you add up the 2 units digits to get 5, divide by 5, and the units digit is 1[/hide][/quote]

no 25/5=5. the units digit is not 1...

We want the residue of $ \frac{2^{1993}\plus{}3^{1993}}{5}$ modulo 10, equivalently we can find the residue of $ 2^{1993}\plus{}3^{1993}$ modulo 50. By Fermat-Euler, $ 3^{\phi50}\equal{} 3^{20}\equiv1\pmod{50}$, so $ 3^{1993}\equiv3^{13}\equiv81\cdot81\cdot81\cdot3\equiv31^{2}\cdot\minus{}7\equiv11\cdot\minus{}7\equiv23\pmod{50}$. By Fermat-Euler, $ 2^{\phi25}\equal{} 2^{20}\equiv1\pmod{25}$, so $ 2^{1993}\equiv2^{13}\equal{} 8192\equiv17\pmod{25}$, and since $ 2^{1993}\equiv0\pmod{2}$, we get $ 2^{1993}\equiv 42\pmod{50}$. Now, $ 2^{1993}\plus{}3^{1993}\equiv 23\plus{}42\equiv15\pmod{50}$, so $ \frac{2^{1993}\plus{}3^{1993}}{5}\equiv 3\pmod{10}$, and the rightmost digit is $ \boxed{3}$.

For factoring $ x^{2k\minus{}1}\plus{}y^{2k\minus{}1}$, note that $ x\plus{}y$ is always a factor because $ x\plus{}y\equal{}0$ implies $ x^{2k\minus{}1}\minus{}x^{2k\minus{}1}\equal{}0$, so by the factor theorem the claim holds.

yeah that solution is much better than mine  :P

I think mz 94 explained it quite well and comprehensively.

It's easy to understand.

But it's not the way you do it..

(I thought of it as that way too, but other guys made a brilliant point on that one..)

thats what i thought was wrong with it. it could be 1,3 or 5.

d**n it. i was about to say that 2. the my brain died. whatever

the answer is 3 and cannot be anything else  :P


[hide="Solution 1"]Clearly, one of the roots is 0 . WLOG, let $ r_4 \equal{} 0$.

By Vieta's formulas, we know that

$ r_1 \plus{} r_2 \plus{} r_3 \equal{} 0$

Expanding the sum, we have that

$ \sum_{cyc} \frac {r_1 \plus{} r_2}{r_3 \plus{} r_4} \equal{} \frac {r_1 \plus{} r_2}{r_3 \plus{} r_4} \plus{} \frac {r_4 \plus{} r_1}{r_2 \plus{} r_3} \plus{} \frac {r_3 \plus{} r_4}{r_1 \plus{} r_2} \plus{} \frac {r_2 \plus{} r_3}{r_4 \plus{} r_1} \equal{} \frac {r_1 \plus{} r_2}{r_3} \plus{} \frac { r_1}{r_2 \plus{} r_3} \plus{} \frac {r_3 }{r_1 \plus{} r_2} \plus{} \frac {r_2 \plus{} r_3}{ r_1} \equal{} \frac { \minus{} r_3}{r_3} \plus{} \frac { r_1}{ \minus{} r_1} \plus{} \frac {r_3 }{ \minus{} r_3} \plus{} \frac { \minus{} r_1}{ r_1} \equal{} 4( \minus{} 1) \equal{} \boxed{ \minus{} 4}$[/hide]

[hide="Solution 2 (same as first but skipping a step)"]By Vieta's formulas,

$ r_1 \plus{} r_2 \plus{} r_3 \plus{} r_4 \equal{} 0$.

So,

$ \sum_{cyc} \frac {r_1 \plus{} r_2}{r_3 \plus{} r_4} \equal{} \frac {r_1 \plus{} r_2}{r_3 \plus{} r_4} \plus{} \frac {r_4 \plus{} r_1}{r_2 \plus{} r_3} \plus{} \frac {r_3 \plus{} r_4}{r_1 \plus{} r_2} \plus{} \frac {r_2 \plus{} r_3}{r_4 \plus{} r_1} \equal{}\frac {\minus{}(r_3 \plus{} r_4)}{r_3 \plus{} r_4} \plus{} \frac {\minus{}(r_2 \plus{} r_3)}{r_2 \plus{} r_3} \plus{} \frac {\minus{}(r_1 \plus{} r_2)}{r_1 \plus{} r_2} \plus{} \frac {\minus{}(r_4 \plus{} r_1)}{r_4 \plus{} r_1} \equal{} 4(\minus{}1) \equal{} \boxed{\minus{}4}$[/hide]


calculus
derivative
function
real analysis
real analysis unsolved

Show that there are no functions f(x), g(x) R-->R such that

d/dx (f(x)/g(x)) = d/dx (g(x)/f(x))

And how about: $ f(x) \equal{} K$ and $ g(x) \equal{} C$ where $ C \in \mathbb{R}$ and $ K \in \mathbb{R}$? Actually, there is an infinitive number of functions that are good for this. Three simple conditions that should be true are:

$ h: \mathbb{R} \rightarrow \mathbb{R}$ (even this isn't necessary)

$ C \in \mathbb{R}$

$ K \in \mathbb{R}$

Then,

$ f(x) \equal{} K h(x)$

and

$ h(x) \equal{} C h(x)$

or simpler:

$ \frac {f(x)}{g(x)} \equal{} M\in \mathbb{R}$


Let $ABCDEF$ be a cyclic hexagon with $AB=CD=EF$. Prove that the intersections of $\overline{AC}$ with $\overline{BD}$, of $\overline{CE}$ with $\overline{DF}$, and of $\overline{EA}$ with $\overline{FB}$ form a triangle similar to $\triangle BDF$.

Lovely!
Maybe it's just the fact that I haven't been able to do problem-solving for the past 14 days, and that I am experiencing an example in real life of the proverb "absence makes the heart grow fonder", but I am experiencing a great adrenaline rush solving this problem. It's very nice...

For the names of the points, look at the figure (Click it for a better resolution)
Here is my solution:

Because $\left|AB\right|=\left|CD\right|$, we know that $\left|BI\right|=\left|CI\right|$, and hence $GI$ is the perpendicular bisector of $\left[BC\right]$.

Hence, we see that $\angle GID = \frac{\pi}{2}-\angle CBD$
Similar, $\angle GHF = \frac{\pi}{2}-\angle FDE$ and $\angle GJB = \frac{\pi}{2}-\angle BFA$.

Now we can easiley see that $\angle GID = \angle GHF = \angle GJB$.
This implies that $\Delta JIH$ and the pedal triangle of $G$ wrt $\Delta BDF$ are similar. (This is well known, and if you don't know it, it's easy to prove)

This pedal triangle is the medial triangle of $\Delta BDF$ because $G$ is the circumcenter. Hence we are done!

Are there other solutions?


Find all functions $ f: \mathbb{R} \longrightarrow \mathbb{R}$ that satisfy the following equation for every two real numbers x,y:

$ f[(x \minus{} y)^2] \equal{} [f(x)]^2 \minus{} 2xf(y) \plus{} y^2$.

[hide]
Putting $ x\equal{}y$ we get

$ f \left( 0 \right) \equal{} \left( f \left( x \right) \minus{} x \right)^2$

Putting $ x \equal{} 0$ we obtain, that $ f \left( 0 \right) \equal{} 0$ or $ f \left( 0 \right) \equal{} 1$.

Thus, there are two cases to consider:

$ \left( f \left( x \right) \minus{} x \right)^2 \equal{} 0$ or $ \left( f \left( x \right) \minus{} x \right)^2 \equal{} 1$.
[/hide]

[b]Solution:[/b]
1.Substitute $ x \equal{} y \equal{} 0$:
\[ \boxed{f(0) \equal{} f(0)^2}
\]
From boxed relation we conclude that either $ f(0) \equal{} 0$ or $ f(0) \equal{} 1$.Consider these two cases:
[b]1.[/b]If $ f(0) \equal{} 0$:
Substitute $ x \equal{} 0$ in the main equation:
\[ f(y^2) \equal{} y^2
\]
Therefore,$ f(x) \equal{} x$ for all $ x\geq 0$.Now let $ x$ be some nonnegative real,hence $ f(x)^2 \equal{} x^2$:
\[ (x \minus{} y)^2 \equal{} x^2 \minus{} 2xf(y) \plus{} y^2
\]
Therefore $ f(y) \equal{} y$ for all $ y\in\mathbb{R}$.
[b]2.[/b]If $ f(0) \equal{} 1$:
Once again we substitute $ x \equal{} 0$ in the main equation:
$ f(y^2) \equal{} y^2 \plus{} 1$.Therefore,
\[ f(x) \equal{} x \plus{} 1
\]
for all $ x\geq 0$.
Now let $ x\geq 0$ therefore:
$ (x \minus{} y)^2 \plus{} 1 \equal{} (x \plus{} 1)^2 \minus{} 2x\cdot f(y) \plus{} y^2$.
Therefore,
\[ f(y) \equal{} y \plus{} 1
\]
for all $ y\in\mathbb{R}$.
It is pretty easy to check that both answers satisfy to equation,but  they can not satisfy simultaneously.
$ \blacksquare$.

(2) Find all functions f: R ---> R such that $ f(x) f(y) \equal{} f(x \minus{} y)$, for all $ x,y \in \mathbb{R}$.

$ f^2(x)\equal{}f(0) \Longrightarrow f(x)\equal{}c \Longrightarrow c^2\equal{}c \Longrightarrow c\equal{}0 or 1$
Is there any typo in question?

There is no typo. Another problem:

(3) Find all functions $ f: \, \mathbb{R} \longrightarrow \mathbb{R}$ such that $ f[x^2 \plus{} f(y)] \equal{} y \plus{} [f(x)]^2$ for all $ x,y \in \mathbb{R}$.

[quote="Carcul"] Find all functions $ f: \, \mathbb{R} \longrightarrow \mathbb{R}$ such that $ f[x^2 \plus{} f(y)] \equal{} y \plus{} [f(x)]^2$ for all $ x,y \in \mathbb{R}$.[/quote]
[b]Solution:[/b]
Substitute $ x \equal{} 0$ in the main equation:
\[ \boxed{f(f(y)) \equal{} y \plus{} f(0)^2}
\]
From boxed relation we conclude that $ f$ is bijective function,therefore,there exist $ x_0$,such that $ f(x_0) \equal{} 0$,(in fact $ x_0 \equal{} f( \minus{} f(0)^2)$).
Substitute $ x\rightarrow x_0$ and $ y \equal{} 0$:
\[ f(x_0^2 \plus{} f(0)) \equal{} 0 \equal{} f(x_0)
\]
But as we know $ f$ is injective therefore:
\[ x_0^2 \plus{} f(0) \equal{} x_0
\]
On the other hand $ f(0) \equal{} f(f(x_0)) \equal{} x_0 \plus{} f(0)^2$,combining it with last equation:
\[ x_0^2 \plus{} x_0 \plus{} f(0)^2 \equal{} x_0
\]
Therefore $ x_0 \equal{} f(0) \equal{} 0$.
First boxed equation implies that $ f(f(y)) \equal{} y$ and let's substitute $ y \equal{} 0$ into the main equation:
\[ \boxed{f(x^2) \equal{} (f(x))^2}
\]
It means that we can rewrite the main equation as:

$ f(x^2 \plus{} f(y)) \equal{} y \plus{} f(x^2)$.

Now substitute $ f(y)$ instead of $ y$ in the main equation and use relation $ f(f(y)) \equal{} y$:

$ f(x \plus{} y) \equal{} f(y) \plus{} f(x)$,for all $ y\in\mathbb{R}$ and $ x\in\mathbb{R}_{ \plus{} }$

Second boxed relation implies that $ f(x) > 0$ if $ x > 0$,therefore,if $ x > y$ then $ f(x) > f(y)$.Here we've just used previous condition of $ f$.Therefore,our function is strictly increasing.
Now suppose to the contrary that $ f(x)\neq x$,for some $ x$,but then it follows that either $ f(x) > x$ or $ f(x) < x$.
Consider both cases:
[b]1.[/b]Suppose that $ f(x) > x$,then $ f(f(x)) > f(x) > x$,contradiction.
[b]2.[/b]Suppose that $ f(x) < x$,then $ f(f(x)) < f(x) < x$,contradiction.
It means that the only possible solution is $ f(x) \equal{} x$.

An other method .
Easy to prove that $ f(0)\equal{}0$ .
Take $ x\to 0$ then
\[ f(f(y))\equal{}y,\forall y\in \mathbb{R}
\]
It gives f is an inject function .
Take $ y\to 0$ then
\[ f(x^2)\equal{}f(x)^2
\]
Take $ x\to \minus{}x$ then
\[ f(x)^2\equal{}f(\minus{}x)^2
\]
Because f is an inject function so $ f(x)\plus{}f(\minus{}x)\equal{}0$
From above condition we have
$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ and $ f(x^2)\equal{}f(x)^2$
Easy to check that $ f(2x)\equal{}2f(x)$
We carculus in two ways :
$ f((x\plus{}1)^2)\equal{}f(x^2)\plus{}2f(x)\plus{}1\equal{}f(x\plus{}1)^2\equal{}(f(x)\plus{}1)^2$
$ \Rightarrow f(x)\equal{}x$
 The function have only solution : $ f(x)\equiv x$

[quote="TTsphn"]
We carculus in two ways :
$ f((x \plus{} 1)^2) \equal{} f(x^2) \plus{} 2f(x) \plus{} 1 \equal{} f(x \plus{} 1)^2 \equal{} (f(x) \plus{} 1)^2$
$ \Rightarrow f(x) \equal{} x$
 The function have only solution : $ f(x)\equiv x$[/quote]
I don't understand this part.
What did you calculate in two ways?
How did you conclude that $ f(x) \equal{} x$?
In my humble opinion,determination of $ f(0)$ is the most important and rather hard part of the solution,of course,only if there doesn't exist any other easier way to find $ f(0)$,than I did it in my solution...


algebra
polynomial
number theory proposed
number theory

Let $ f$ be a polynomial with integer coefficients. Assume that there exists integers $ a$ and $ b$ such that $ f(a)\equal{}41$ and $ f(b)\equal{}49$. Prove that there exists an integer $ c$ such that $ 2009$ divides $ f(c)$.
[i]Nanang Susyanto, Jogjakarta[/i]

[hide]Clearly for all $ m,n \in \mathbb{Z}$, $ 41|f(a\plus{}41m)\minus{}f(a)$ and $ 49|f(b\plus{}49n)\minus{}f(b)$ but since $ 41|f(a)$ and $ 49|f(b)$, then $ 41|f(a\plus{}41m)$ and $ 49|f(b\plus{}49n)$ for all integers $ m$ and $ n$. Since $ (41,49)\equal{}1$, then there exists $ m,n$ such that $ a\plus{}41m\equal{}b\plus{}49n\equal{}c$, and thus $ 2009|f(c)$ in this case.[/hide]

Since the polynomial has integer coefficient , then if $ x,y$ are integers

$ f(a\plus{}41x)\equiv f(a) \equiv 0 \ (\text{mod} \ 41)$
$ f(b\plus{}49y)\equiv f(b) \equiv 0 \ (\text{mod} \ 49)$

There exist integer $ c$ and $ (x_0,y_0)$ as a solution for $ (x,y)$ such that $ c\equal{}a\plus{}41x_0\equal{}b\plus{}49y_0$ , thus $ 2009|f(c)$

 :P


How did you guys do?
Here's my embarassing report-
 	Total  	 53
 	All India Rank  	 734
Maths  	15(I couldn't do more than 10-20 coz the examiner robbed me of a full 25 mins. at my centre. I somehow stopped short of calling up unified council and lodging a complaint.  :ninja: )
Physics  	 14
Chemistry  16
Gen. Quest 8

I only know of 3 friends in Lucknow in top 150 now. Will find out more at school/coaching.
Right now-
Mayank Srivastava- 83(KVPY+NTSE+RMO+NSEA and all that)
Ritwik Yadaw- 120 odd?
Piyush Kapoor- 115

mine is worse than urs  :o  :mad:

i got rank 1300 something..........

got 22 in maths out of 25....(was only able to attempt 22 due to shortage of time....... :rotfl: )...........

not same but the case was also not easy at our centre

at my centre invigilator was a fool i guess :P

Me p. mittal class 11
Rank 181
Marks 64  :) So far so good

My friend KARN VOHRA
Rank 104
Marks 68
Depressing indeed :(

Whats more depressing he got 6 Marks more than me in chemistry :sigh:  :wallbash:

me rank 4 in class IX! :lol:
maths - 25/25

phy - 25/25

chem - 17/20(lost marks here)

bio- 16/20(lost even more here!) :(

g.k. - 8/10

total - 91/100

what's really depressing is that 3rd ranker also got 91!!!!!!!!!!!!!!! :mad:

got 2 in gk and around 10 in bio........

i didnt write it this year but last year my rank was really cool...
1234

i got a real depressing result.....
11th after last yr's 1st :(

what a great deal of cheating in my centre.......... :o

see all of brigth lands school in the list.......

http://www.unifiedcouncil.com/achievers.aspx

they were asking the answers to questions during the examination tym from my coaching teachers...........
thankfully i had filled the form but not given it........

all the children from my centre are in top100....... :rotfl:  :rotfl:

Still I dont understand  :huh:
why it is written that Karn topped ddun with rank 104 :o

it happened to me also//......
i got 82 rank in 6th(got highest rank frm ddun.......) and it showed the name of someone at 238 rank....
same last year....
i got 79th but it showed someone at 183.......

well state achievers means that ppl who r not in top 100 but toppers in state.........
top 100 r national achievers......... :rotfl:
 :rotfl:

congrats to fbaggins and rituraj
I am in class XIth
Got 70 marks and 63rd rank :)

congrats to you all  :)

esp ritu and fbaggins

i was thinking of giving it but later cudn't coz of mine laziness  :rotfl:

well done

there's nothing  good abt my performance :(  :cry:
it nearly sent me in depression.... :(  :mad:

well it happens after 1st rank

but except 4 AIR 1(92),most were quite close to u(many 82)

its not abt that but following facts.....

1.AIR 1 is from Narayana(i m in Aakash) and DPS (the skool i dislik most).and he got 40 more marks than me in NSEB :mad:
neways he is a good frnd of mine :first:  but sm bloody competition
Moreover he got 25 in Phy,25 in Chem....my favs

2.So many Kota guys crushed me.

3.The girl at No.8--she is in my batch...and hadnt got a better rank than me in ne test (excpt 1 ):P

4.My 12th has already been toooo depressing...and if INBO's gone,that will be toooooooooooo bad :(

You will still have JEE to leave your mark :omighty:

i did really badly i got all india rank 782

[quote="Arnav...!!!"]ritu is not givin jee :rotfl:  :rotfl:[/quote]
He would have cleared jee with rank 1 in bio stream as well.  :rotfl:

@ ritu:the exam counts much more than anything else

and abt competition,let him read these  :rotfl:  :rotfl:

Wat abt ur hopes in INBO(98% or even more???????)

[quote="rituraj007"]me and JEE!! god gracious!!!!!!! :rotfl:  :rotfl:

AND MY HOPES IN inbo ARE 49% :o[/quote]
He meant IBO gold medal.

Finely carved joke :P

[quote="skand"]and abt competition,let him read these  :rotfl:  :rotfl:
[/quote]
Well he is a really good frnd of mine..so i hope he doesnt feel bad after reading this
And frankly speaking,i wont mind him getting the first rank (as long as i m second :P)........................[i]virtually[/i]

[hide="PS"]actually we r competing 4 more than just a first rank.....
[hide]
[hide]

u kno how to read this.....

a gal!!!! :blush:  :rotfl:

[/hide][/hide]

[/hide]

[quote="rituraj007"]Finely carved joke :P

[quote="skand"]and abt competition,let him read these  :rotfl:  :rotfl:
[/quote]
Well he is a really good frnd of mine..so i hope he doesnt feel bad after reading this
And frankly speaking,i wont mind him getting the first rank (as long as i m second :P)........................[i]virtually[/i]

[hide=PS][/quote]

I never knew the 'quote' button was so useful  :rotfl:  :rotfl:

[quote="skand"]rank 6???????????????[/quote]

no but u werent too far.....rank 8 :P


The radius of the first circle is 1 inch, that of the second $\frac{1}{2}$ inch, that of the third $\frac{1}{4}$ inch and so on indefinitely.  The sum of the areas of the circles is:
\[ \text{(A)}\ \frac{3\pi}{4} \qquad \text{(B)}\ 1.3\pi \qquad \text{(C)}\ 2\pi \qquad \text{(D)}\ \frac{4\pi}{3} \qquad \text{(E) none of these} \]

It's a geometric sequence with the first term being $1, 1/4, 1/16, ....$ and so on because you're looking for the area, therefore, using the formula for geometric sequence $a/(1-r)$, you get $1/(1-1/4)$, therefore the answer is $4/3pi$  :winner_first:  :flex:

A geometric sequence. Apply the formula $\frac{t_1}{1-r}$ to get $\frac{4}{3}\pi$


ARE YOU READY FOR A CHALLENGE

The demand equation for x camcorder is

p = 5000 [ 1 - (4 / 4 +  [i]e[/i]^-0.002x)]

Find the demands x for prices of (a) p = $600 and (b) p =$400

[quote="kryms3n"]ARE YOU READY FOR A CHALLENGE[/quote]

No.

The message contains only quote and code tags. Please add some text of your own.

Do you mean

$p=5000\left(1-\left(\frac{4}{4-e^{-0.002x}}\right)\right)$?

[hide="solution"]

We solve for x.

$p=5000-\frac{20000}{4-e^{-0.002x}}$

$p-5000=-\frac{20000}{4-e^{-0.002x}}$

$\frac{1}{5000-p}=\frac{4-e^{-0.002x}}{20000}$

$\frac{20000}{5000-p}=4-e^{-0.002x}$

$4-\frac{20000}{5000-p}=e^{-0.002x}\implies \log_{e}\left(4-\frac{20000}{5000-p}\right)=-\frac{x}{500}$

So, in terms of p, we have x as

$-500 \log_{e}\left(4-\frac{20000}{5000-p}\right)$

You just plug in the requested values of p and you have your answers.[/hide]

Am I doing something wrong here? Maybe it's because my hands are freezing...


linear algebra
matrix
superior algebra
superior algebra unsolved

Find all norm $N$ on $M_n(C)$ such that:
$N(A)^2=N(A^{*}A)$ for all matrix $A \in M_n(C)$.

Assuming the norm is positive definite, we have that $N(I)=1$, and hence $N(U)=1$ when $N$ is a unitary matrix. We also have that $N(AU)=N(A)$ for unitary $U$ and any $A$. Assuming continuity of the norm, the norm of a diagonal matrix must be the largest diagonal element. This, along with everything else, is enough to specify $N$ as the 2-norm; $N(A)$ is the largest singular value of $A$.

yes it is  :)
(Moubi, your example is not a norm)


The real numbers x,y and z are not equal and satisfy

x+1/y=y+1/z=z+1/x=k

Determine all possible values of k.

[url=http://www.mathlinks.ro/viewtopic.php?search_id=1106793479&t=120984&sid=9d69d25d2a73e589b54d1f4b7d224851]You can find the solution here.[/url]


calculus
derivative
logarithms
calculus computations

x^y=y^x

find y'?

This is what I have done so far:

yln(x)=xln(y)
y= (xln(y))/ln(x)

Then I apply the quotient rule, Is that right? But that seems pretty complicated.  I have a feeling I am doing something wrong.  Can anyone give me any clues?  Thanks.

The problem I can't figure out is there is y on both side.  How do I simplify the equation to make y only appears on one side?

Have you learned implicit differentiation yet?

If so, this is how you do it:

\[ x^y=y^x \Rightarrow y\ln x = x \ln y  \]
\[ \frac{dy}{dx} \ln x + \frac{y}{x} = \frac{x}{y}\frac{dy}{dx} + \ln y  \]
\[ \frac{dy}{dx} = \frac{xy\ln y - y^2}{xy\ln x -x^2}  \]


Prove that among the numbers $ n,n\plus{}1,\ldots,2n$ with $ n\in\mathbb{N}^*$ there is at least $ 1$ perfect square.

$ (k\plus{}1)^2 \minus{} k^2 \equal{} 2k\plus{}1$

So the diference between 2 squares is strictly increasing.
Then our best chance, if you know what i mean, would be that $ n\minus{}1$ would be a square, but than $ n\minus{}1 \equal{} k^2 \Rightarrow n \leq(k\plus{}1)^2 \leq 2n$

You mean that among the above numbers there are no perfect square?  :wink:

I mean that if there were no squares in $ \{n,...2n\}$ you could take the largest integer $ \leq n$ that is a square, lets call him $ M^2$

Well, $ (M \plus{} 1)^2$ is largest than $ n$ because $ M^2$ is the largest square below $ n$, but it is also lower than $ 2n$ (for the fact i mentioned earlier) because $ M^2$ is at the most $ n \minus{} 1$


[color=blue][size=150] APMO 2006[/size]

Como saben el lunes se realizo el examen Asia Pacifico ...
quisiera que las personas que cuentan con informacion de las preguntas
del examen..
las posten aki...

como tambien la discusion de estos.

Gracias,

Carlos Bravo ;)[/color]

Carlos, no se si sea prudente poner eso ya, aca en Colombia nos dijeron cuando terminamos que no discutieramos la prueba hasta que nos avisaran, y todavia no nos han dicho nada.

JAS

[color=blue] OK ...
dejemos esto pendiente ...

Gracias por el consejo...

Carlos Bravo ;)[/color]

Aqui,
cuelgo el enunciado del ultimo examen APMO,
esta en version en espaoñl.

Buenísimo!!!!!!!!!  Alguien tiene las respuestas????

Esta ha sido la apmo mas facil de los ultimos 5 años....

espero quienesparticiparon posteen sus soluciones.

[color=blue] Bueno, parece ser
que ya tenemos las soluciones oficiales. .

Gracias por el link ....

esto fue publicado recien esta semana ;)

PD_ SOLO QUEDA ESPERAR LOS RESULTADOS

Carlos Bravo :)
Lima Peru[/color]

¿en cuánto tiempo, aproximadamente, llegan los resultados?

[quote="José"]http://www.kms.or.kr/competitions/apmo/[/quote]

[color=blue]No lo tengo muy claro...
pero de seguro que en las reglamentaciones del evento
podras encontrar la informacion ...

Carlos Bravo ;)[/color]

Las soluciones serán enviadas a los países participantes el 2/5 (a esperar amigos, ejejeje)...

bueno  yo  presente el examen la semana  pasada ,  desafortunadamente no me fue muy  bien  solo  saque  7  puntos en el  problema 3 ,  realmente los problemas eran muy  feos a mi  no me gustaron mas que el 3 jejejeje  y  los criterios de calificación eran muy  raros  daban  muchos puntos por cosas triviales que cualquiera  hubieRA   hecho .  aunque eso  me enseña que siempre se deben  intentar  todos los problemas.

  marko  avila

Si, por lo que hablé con algunos que participaron, el 3 fue el que más les gustó... igual, 7 puntos en un solo problema está muy bien, ya tenés la mención asegurada!!!!!!

Marko, como eran los criterios de calificación. Cuantos puntos daban si se probaba que $p^4$ dividia??
Los  problemas de esta APMO estaba considerablemente mas faciles que los del año pasado. Yo crei que si me iba bien iba a hacer 1 problema, pero pude con el 1 y con el 5.

A mí me salió el 1 y me sentí decepcionado del 5. Yo creo en ese me dan 5 puntos pero como no escribí muy bien no es seguro. En otros problemas no hice nada interesante excepto en el 4, que pude rasgar 2 puntos

Ya llegaron los resultados a colombia. Tuvimos 1 Plata (yo!!!) y 4 menciones.
 :D


Would it be possible to include bbcode commands in the blog custom block? Thanks a lot.

Yeah, I'll do that in the near future :)

Um...is this going to be done sometime soon?  :lol:

*bump* will this be done soon?

Thanks.

Sorry about appearing kinda impatient

Well better late than never :D BBCode support enabled in the Custom Block (this includes of course $ \text{\LaTeX}$ and Asymptote.)
You can now use this Custom Block field to create, say, a Blog Roll (using the Url tag.)


[hide]$x+y=$? :) [/hide]

P.S.  Please do not double post.

[quote="jsteves18"]at a sporting event tickets are 4.50 for the first person and 2.50 for all remaining family members. in a section there is 480 people who have in total paid 1550. how many 4.50 tickets were sold?

Equation 1: 4.5x + 2.5y = 1550

Equation 2: ??????????????

Please HELP! Thanks! :D[/quote]

You already post this question in another topic, so please don't do so.

Hint: You have to use all the data in the question, so what number have you not used here?


Okay - I have a genetics problem that I am doing for practice for AP Biology next year...but I can't seem to set them up. Here they are:

1.) Michelle can't roll her tongue and has brown eyes. Her mother is unable to roll her tongue and has blue eyes. Michelle marries James, who can roll his tongue and has blue eyes. James' father could not roll his tongue and had blue eyes. Give the phenotypes and genotypes of any offspring likely to be produced by James and Michelle.

Okay - well this is what I have so far...

[hide]In humans, the ability to roll tongues is dominant ($T$) over the inability to roll tongues ($t$), so we let the genotype for a person who can't roll their tongue to be $tt$. Also, brown eyes are dominant ($B$) over blue eyes ($b$), so we let the genotype for a person with brown eyes to be $BB$, and a person with blue eyes to be $bb$.[/hide]

So what do we do from here? Can anyone provide a solution to this problem?

Since James' father was homozygous recessive for both traits, you know James is heterozygous (TtBb).  From the question, you know the mother is homozygous recessive for tongue rolling and heterozygous for eye color (ttBb).  Do the punnett square and you have your answer.

Attached is a visual aid of the Punnett square I utilized.

Genotype:
TtBB- 2/16- 12.5%
TtBb- 4/16- 25%
Ttbb- 2/16- 12.5%
ttBB- 2/16- 12.5%
ttBb- 4/16- 25%
ttbb- 2/16- 12.5%

Phenotype:
Able to roll tongue and brown eyes- 12.5%+25%- 37.5%
Able to roll tongue and blue eyes- 12.5%
Not able to roll tongue and brown eyes- 12.5%+25%- 37.5%
Not able to roll tongue and blue eyes- 12.5%

-arb95


I managed to find out, before I'm getting to eat breakfast, that for x > 4, $y^{2}= 1 (\mod{10})$. But I can't find specific solutions.
For x <= 4 we have x = 4 and y = 5.

Check cases and there are no solutions until $(x,y)=(4,5)$, so now assume $x\geq 5$.

$y$ is obviously odd, so let $y=2a+1$, giving $x! = 4a(a+1)$

So we are essentially demanding that there exists $a(a+1)=6(5\times 6\times 7\times...x)$ for some $x$. This makes the solutions $(x,y)=(5,11)$ and $(7,71)$ obvious.

Now, since a,a+1 are coprime, 36 must divide either one or the other. That's as far as I got before getting stuck and getting coffee :).

$x\in(4,5,7)$ is all solutions, but whether it is possible it to prove?

[quote="juna"]Solve in positive integers $x!+1=y^{2}$[/quote]
why it isn't
$x!+2=y^{2}$
?  :lol:

[quote="juna"]If $x^{2}+k=y^{2}$, $k>1$ - all very simple. There is only final set of the solutions.[/quote]

You meant  $x!+k=y^{2}$, didn't u?

Oooh, I think the proof can be done with.

(a, a+1)=1, so 36 divides only one of them, into which all multiples of 2 and 3 in the whole product must also go.

Intuition tells me that the the product of the powers of 2 and 3 in x! will be more than 1 greater than the product of the powers of all other primes in x! - can somebody show this completing the proof? (or am I just wrong?)

Oooops, 36 isn't a prime power :blush:. Another approach needed methinks.

I shall add $\forall p|x(p\not | y)$, $p$ - prime.

[quote="juna"]If $x!+k=y^{2}$, $k>1$ - all very simple. There is only final set of the solutions.[/quote]can you prove it please?

[quote="cocoowner"]if $x >3$, $y^{2}$ can't be mod 2 so, $x<4$
And you can check x=1,2,3 There's no solution.[/quote]
What do you mean mod 2. You are wrong. You only have that $y$ is odd.

That's famous open problem. See http://mathworld.wolfram.com/BrocardsProblem.html


How can i derive avogadro's number using these values;
NACL
Distance = 2.89 x 10 ^ -8
Molecular mass = 58g
Unit Cells = 4

Can anyone help or tell me what i need to do
Thanks in adavance

I'm not sure why, but the numbers seem to work in this formula.

Since the distance of the particles is $2,89 \cdot 10^{-8} mm$ (I don't know, is this mm?), so in $1mm$ there are about $\dfrac{1}{2,89 \cdot 10^{-8}} = 3,46 \cdot 10^7$ particles

So, in $1mm^3$ there are $(3,46 \cdot 10^7)^3 = 4,143 \cdot 10^{22}$ particles

Now, if you multiply by $58$ and divide by $4$, you obtain the Avogadro's number, but I don't know why

(so $\dfrac{4,143 \cdot 10^{22}}{4} = \dfrac{6,023 \cdot 10^{23}}{58}$, which is how many particles there are in $1gr$, but I still don't know why)

And what are these "unit cells" ?