calculus
integration
limit
algebra
binomial theorem
real analysis
real analysis unsolved

Find

$\lim_{n\rightarrow \infty}\frac{n}{2^{n}}\int_{0}^{1}x^{2n}(x^{2}+1)^{n}dx$.

Didi

Well expanding using the binomial theorem we can get

$\int_{0}^{1}\sum_{k=0}^{n}\binom{n}{k}x^{2k+2n}dx = \sum_{k=0}^{n}\binom{n}{k}\frac{1}{2n+2k+1}$

so

$\frac{n}{2^{n}}\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2n+2k+1}= \frac{1}{2^{n}}\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2+\frac{2k+1}{n}}$.

Since the sum of the coefficients is $2^{n}$, it's a weighted average of all those $\frac{1}{2+\frac{2k+1}{n}}$, which range from $\frac{1}{2}$ to $\frac{1}{4}$ so it's probably around $\frac{1}{3}$, but I don't know how to evaluate it.

I think it's not hard to prove that
\[\sum_{|k-\frac{n}{2}|>n^{\tfrac23}}\binom nk=o(2^{n})\]
That's why the lim is equal to $\frac13$

Let's try it without expanding out the binomial. We'll start with a sequence of substitutions:

$\frac{n}{2^{n}}\int_{0}^{1}x^{2n}(1+x^{2})^{n}\,dx$ Let $y=x^{2}.$

$=\frac{n}{2^{n+1}}\int_{0}^{1}y^{n-1/2}(1+y)^{n}\,dt$ Let $u=1-y.$

$=\frac{n}{2}\int_{0}^{1}(1-u)^{n-1/2}\left(1-\frac{u}2\right)^{n}\,du$ Let $v=nu.$

$=\frac12\int_{0}^{n}\left(1-\frac{v}{n}\right)^{n-1/2}\left(1-\frac{v}{2n}\right)^{n}\,dv$

Now, take the limit as $n\to\infty,$ interchanging limit and integral. We can find a Dominated Convergence Theorem justification for the exchange. The limit will be

$\frac12\int_{0}^{\infty}e^{-v}e^{-v/2}\,dv= \frac12\int_{0}^{\infty}e^{-3v/2}\,dv=\frac12\cdot\frac23= \frac13.$


This is false.  If $ \omega$ denotes the first uncountable ordinal, then $ [0, \omega)$ in the order topology is limit point compact but not compact (it's not even Lindelof).  Did you intend for $ X$ to be a metric space?

[quote="sondhtn"]Prove: If every infinite subset of X always have limit point $ \Rightarrow$ X is compact space[/quote]

Yes, X is a metric space

Do you know that a metric space $ X$ is compact if and only if it is complete and for every $ \varepsilon>0$, one can find a finite $ \varepsilon$-net in $ X$?

I can proof $ X$ is complete but I can't find a finite $ \varepsilon$-net in $ X$. Can you help me !

Just take any maximal $ \varepsilon$-separated set. It is a $ \varepsilon$-net and has no limit points.


快開學了, 先祝願各位學業更進幾步,
畢業班的能考進理想的大學, 參加數學競賽的能奪得優異成績...
不知大家有何"新學年"願望?
對於我來說, 能取得保送資格已經是不錯的了 :P

[quote="ifai"]取得保送資格 :P[/quote]
Which university?

浙大吧! 不過成功機會不大...
據我所知, 我校曾成功保送(獲錄取)最好的大學只是中山大學. 而北大, 清華等的名額都被別的學校佔了. 我校有一位曾代表澳門參加cmo, 並獲全澳校際數學比賽冠軍, imo 銅獎, 而且多才多藝的校友, 他當時保送去清華, 但沒有被錄取. 最後他還是自己考了上去. 現在他升大四, 物理系(基礎科學班).

[quote="shobber"]我们这里情况不一样. 由于清华北大都是北京的学校, 所以只要在北京赛区拿一等奖就行了.[/quote]
羨慕...不過我們的高考比你們簡單得多

Shobber,你太爽了,我们这边进了CMO都不一定能上北大啊.......

為何這個 Chinese Communities 總是那麼冷清?
這裡view的人不少, 但post的人(數)實在...

[quote="dingdongdog"]Of course I'm in China.
My goal in this term is to be in CMO.[/quote]
你在廣東何市? 今年幾年級?
(不便回答的話可以不答)

我住在深圳,我家窗前是深圳河,河对面是新界.现在上高三.

[quote="dingdongdog"]我住在深圳,我家窗前是深圳河,河对面是新界.现在上高三.[/quote]
深圳? 不錯啊.  我也是上高三, 不過奧數一定跟你相差很遠. 你(們)懂微積分嗎? 國內在高中階段會否教初等微積分?

[quote="dingdongdog"]微积分是自学的.[/quote]
現處於哪個程度? 正在學甚麼內容?

高三课本应该有极限和初步积分吧? 不过高考涉及很少, 所以没人把它们当回事.

另一件想提的是, 澳門和去年一樣, 都不搞CMO預賽, 直接由之前"全澳數學比賽"的優勝者參賽. 雖然搞預賽我也一定不會入選, 但總算可以學點東西. 今年又不搞, 真是有點失望...

[quote="Lilian"]Would you write it in English, please![/quote]
This is a forum for CHINESE, don't you know that!

[quote="Lilian"]Would you write it in English, please![/quote]
Just like asking you to post in Chinese.

[quote="Mathsniper"][quote="Soarer"][quote="Mathsniper"]hello, everyone, 我是來自澳門的 :) 就讀濠江中學, 請問ifai是就讀那間學校的? 本人就讀高二, 都係研究數學的, 甚麼內容也可(Real analysis, Complex analysis, Graph theory, Combination...), 多多指教  :)[/quote]

sniper, do you remember me?
還有，你讀得咁快？complex analysis??[/quote]
yaaaa, 我當然記得你 :P 我現在嘗試了解將觀點放高一點去看數學, 到一定時間時在中途深入了解所學的知識, 好似是Real and Complex anaylsis, 當然, 現在我只是懂, 一般的還能做, 難的還有待研究, 因為兩個理論已是多年來通過多位數學家奠立的, 以我角度, 要細究則必然要了解其一切(包括理論創立的背景及其意義). 所以要投放更多的時間下去, 但現在我還不得不快研究圖論... 因為其應用廣泛, 本人是有學習程序設計的背景(從初一, 現在高二), 在比賽中使用率高(但我要指明... 我對比賽並不看重, 我只在乎, 學懂, 精), 而且Graph theory亦是重要理論!!! 所以我亦都想此為首要學習內容.[/quote]

太快了... complex analysis =.="

[quote="Soarer"][/quote][quote="Mathsniper"][quote="Soarer"][quote="Mathsniper"]hello, everyone, 我是來自澳門的 :) 就讀濠江中學, 請問ifai是就讀那間學校的? 本人就讀高二, 都係研究數學的, 甚麼內容也可(Real analysis, Complex analysis, Graph theory, Combination...), 多多指教  :)[/quote]

sniper, do you remember me?
還有，你讀得咁快？complex analysis??[/quote]
yaaaa, 我當然記得你 :P 我現在嘗試了解將觀點放高一點去看數學, 到一定時間時在中途深入了解所學的知識, 好似是Real and Complex anaylsis, 當然, 現在我只是懂, 一般的還能做, 難的還有待研究, 因為兩個理論已是多年來通過多位數學家奠立的, 以我角度, 要細究則必然要了解其一切(包括理論創立的背景及其意義). 所以要投放更多的時間下去, 但現在我還不得不快研究圖論... 因為其應用廣泛, 本人是有學習程序設計的背景(從初一, 現在高二), 在比賽中使用率高(但我要指明... 我對比賽並不看重, 我只在乎, 學懂, 精), 而且Graph theory亦是重要理論!!! 所以我亦都想此為首要學習內容.[/quote][quote="Soarer"]

太快了... complex analysis =.="[/quote]
keke~ 你當佢興趣書咁睇吧 :P Complex analysis我只是映射那part沒有看, 但問過阿sir話映射個度好鬼多理論, 我阿sir以前係學函數論, so得閒都搵佢教下我 :D

[quote="shobber"][quote="Mathsniper"]keke~ 你當佢興趣書咁睇吧 $\cdots$得閒都搵佢教下我 :D[/quote]
看不懂...[/quote]
你当它兴趣书那样看吧
有时间都找他教一下我

那些都是广东那边的自创词汇么? 还有, "得闲" 二字, 大有古风...

是广东话（或香港话）的口语。

[quote="shobber"]那些都是广东那边的自创词汇么? 还有, "得闲" 二字, 大有古风...[/quote]
Not exactly 自创词, just the difference between Cantonese and Putonghua.


Mike paid $ \$1.25$ for a stamp three years ago. He was just offered double that amount for the stamp. Assuming the stamp's offer price doubles every three years, how many dollars will he be offered in 12 more years?

He will be offered $ \$1.25\cdot2^{1\plus{}\frac{12}3}\equal{}\$\frac54\cdot2^5\equal{}\$5\cdot2^3\equal{}\boxed{\$40}$.


inequalities
geometry
triangle inequality

Find smallest real number $p$ for which is this statement true:

For every positive integer $n$ this inequality holds:
$\sqrt{1^2+1}+\sqrt{2^2+1}+\ldots\sqrt{n^2+1}\leq \frac{1}{2}n(n+p)$

I think $p=2\sqrt{2}-1$.I'll post my solution if answer is right.

It would seem to be, theres just one part of my proof i cant do! Because im to lazy to multiply out two hugh brackets.... :)

Yup your right, a completely brief proof follows... its not even a proof, it just outlines the method, ask if you want more!

 [hide]You can easily prove that when you add an extra term to the sequence that the RHS grows more than the LHS so the min value is indeed 2 (:sqrt:2)-1, corresponding to the first term in the sequence.[/hide]

[quote="ondrob"]Yes, it's right.
Also you can find an geometric interpretation (after you find $p$)...[/quote]
Yes, I've just found very nice geometric solution for it. I'll post it soon.   :)

Ok, here is my solution. We have that: $p\geq{\frac{2\sum_{i=1}^{n}{\sqrt{i^{2}+1}}}{n}}-n.$Now I'll prove (using geometry) that RHS is decreasing ,since $p=2\sqrt{2}-1$ when $n=1$.So we have to prove that $\sum_{i=1}^{n}{\sqrt{i^{2}+1}}+\frac{n(n+1)}{2}\geq{n\sqrt{(n+1)^{2}+1}}.$
Let $P=(0;1)$ and $A_{i}=(i;0)$  $(i=1,2,..,n)$.Using triangle inequality we obtain $\sqrt{i^{2}+1}+(n-i+1)>\sqrt{(n+1)^{2}+1}$  $(i=1,2,..,n)$, adding these up we get the result , since $\frac{n(n+1)}{2}=\sum_{i=1}^{n}{i}.$
Very nice!!  :)

Yes, congratulations Armo, you've found very nice geometrical solution. I'll post another inequality that can be done with geometrical approach soon...


hi    :lol:

Im Rachel...Just joined this forum out of my interest in educational forums.

I like to say hi to all of ya n would like to give a small intro abt me..

well...im just a high school student..like browsing the net...i do chat all the time.(in the forums usually)

so thought to register in this wonderful interesting forum talking much abt ma interests

ok guys...so can i know all of ya :roll:  :roll:  :roll:

Hi there.
I'm *points to my username* nutz_for2.718281828.
I'm officially in middle school, but I sometimes surf around the highschool forums for something to do, instead of the usual beginners stuff.

You can take classes here:

Introductory classes, intermediate classes and olympiad classes. I reccomend introductory if you domn't have a firm grip on it. I then suggest intermediate. After you have mastered that, then the olympiad courses would be just right for you.

Also, you can browse around the forums for questions out of the blue. like Billy Bob questions. I'm just entering intermediate, but the problems are too hard for me.

Well, for your entertainment, there is also a games and fun factory forum.

I hope you enjoy AoPS, I do.

Hello,

welcome to Mathlinks.  I'm fredbel6, university maths student.

In "Games and Fun Factory" there is a "sticky" topic where people can introduce themselves (I have done so , and maybe you want to do it too).  Click on this link to go there :
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=15747[/url]

Hi   :lol:   ,

Thanx to all the members for ur warm heart welcome.

Bye  :)

Hello rachel!

Welcome to AoPS! It is a great place to be if you are interested in maths - both serious (olympiad type) and for fun (bubka type :D ).

Explore games and fun factory - you will find lots of interesting things there. Enjoy and keep writing about your experiences.

Cheers,
bubka

Thanx for u all fun loving guyz.

byeeeeeeeeee!


[quote="N.T.TUAN"]Let $P(x)=x+x^{2}+...+x^{n}=x.\frac{x^{n}-1}{x-1}(x\not = 1)$.

We have $P'(x)=1+2x+...+kx^{k-1}+...+nx^{n-1}$ therefore sum is $P'(2)$.[/quote]
is there any problem if the sum is a not continous function?


limit
logarithms
calculus
calculus computations

Find the limit:

$\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{i=1}^{n}\frac{1}{k}$

It is equal to $0$ because $\sum_{i=1}^{n}\frac{1}{i}-\ln{n}$ is convegers.


HAPPY NEW YEAR!!!!

I'm SURE 2007 will be quite a special year!  :lol:


How many times does the digit 9 appear in the list of all integers from 1 to 500?

FTW says the answer is 100... however I checked by computer and the solution is indeed 95.

Actually, there are TWENTY 9's for each set of 100 numbers starting at _01 and going to _99, (except for the 900's, which are not included here), so  5*20=100.

Maybe the confusion comes from the fact that 99 has 2 9's in it.

yeah.  i messed up on that the first time.

an alternate way of explaining the "there are 20 9's from 00 to 99" is as follows
there are 100 total numbers, each number is two-digit.
Each digit has the same frequency in this interval, and there are $100\cdot2 = 200$ digits.
So there are $\frac{200}{10}=20$ 9s.
and since 9 doesn't appear in the hundreds digit from 0 to 500, we can multiply $20$ by $\frac{500-0}{100}=5$ to obtain our answer, 100.


AMC
USA(J)MO
USAMO
inequalities
geometry
function
number theory

What does everyone think about the topics of the USAMO questions are gonna be?  I looked online, and there seems to be no general pattern that they follow.

Also, what do you think will be the cutoffs for MO(S)P?

There's no way to tell what the USAMO problems will be; topic distribution is pretty random, except that usually the first question each day is not very technical, so as to give more people a chance.

The name of the residential math program to which you referred is MOP.  The cutoff for black (the top twelve or so) is usually in the 20s; blue (next 15-18 non-seniors) is usually in the high teens or low 20s, and red (9th grade) is more variable.

Last year the cutoffs for blue and red were 17 and 9. I'm not sure what the cutoff for black was.

I think generally you should need 2.5 problems to make blue and 1.5 to make red. 3.5-4 is probably good for black.

[quote="randomdragoon"]All I'm going to say is, I think pure inequalities have gone the way of the dodo.

Don't take my word for it though, but I would bet money on this assertion.[/quote]

They come up when you least expect them: STUDY INEQUALITIES!!! :D

(of course, I'm telling you this to throw you off so you study inequalities while I prepare for the 6 geometry problems on the 2008 USAMO  :P )

Hmmm one thing I noticed was there's a good few problems like "find with justification, the largest number of...". Maybe I just haven't seen all that many problems.
Wait what's .5 of a question? Usually they give very little credit for BS, right, so you need to get a ton of good BS and collect a few points.

No...the way to get partial credit is if you have significant parts of a solution.

Most olympiad problems at least have some sort of bounding aspect to them. I doubt that you will see a three variable symmetric inequality or something...those are pretty much 'solved'. But you will most likely need to use inequalities as parts of solutions.

Actually usually MOSP is somewhere around 3-3.5 problems for blue and 4 for black. Last year was really weird, because the test was... bad. I hated that test.

What will be the fate of this new era of combo problem favoring rigor over creativity (last year's #2, #4)? As long as those don't show up en masse, I will be happy.

[quote="randomdragoon"]All I'm going to say is, I think pure inequalities have gone the way of the dodo.

Don't take my word for it though, but I would bet money on this assertion.[/quote]

There were pure inequalities on the IMO just two and three years ago. And last year's USAMO #6 may not have been pure, but it definitely used the same ideas as your standard inequality. They might be afraid of putting them on the USAMO because the level of inequalities these days that would challenge really good people might be impossibly difficult for the neophytes (?).

Yeah, actually I've noticed a general decline in quality in AMC/AIME/USAMO problems of late. Maybe I need to submit some problems after I graduate  :o

[quote="jb05"]There were pure inequalities on the IMO just two and three years ago. And last year's USAMO #6 may not have been pure, but it definitely used the same ideas as your standard inequality. They might be afraid of putting them on the USAMO because the level of inequalities these days that would challenge really good people might be impossibly difficult for the neophytes (?).[/quote]

Yeah, I was referring to the USAMO about this claim, for the reasons you outlined above.

Last year, with the question on Tyrannosaurus Rexes (dinosaur problem), if you had simply found the answer with little justification how many points should you get?

[quote]actually I've noticed a general decline in quality in AMC/AIME/USAMO problems of late[/quote]

Can you quantify that, or make your statement precise?  Exactly how, and in what degree is there a general decline on quality?
Can you compare to the quality over time of the CMO, of the BMO, the Australian MO and the corresponding preliminary contests?

I just finished looking through 60 years of the AMC contests in preparation for posters and brochures for our 2008-2009 contest year.  My perception is that the problems now have greater variety (more number theory, more interesting counting, more sequences, more meaningful questions about functions) and less routine computation.  I think the problem statements are now more precise.

You probably have too great a familiarity with the problems, you have studied them too much and so they seem routine to you now.  What once was fresh and new, now is ordinary for you.  But objectively, I believe the quality is bettr, and certainly no worse.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="Hamster1800"]1, possibly 0 or 2. 0 corresponds to them deciding that the answer is easy enough to see and not enough work was done. 2 corresponds to the answer being slightly nontrivial and your making nontrivial progress towards the solution.[/quote]
So would "nontrivial progress" be equivalent to writing down the first half of a full solution, or writing down a "full" solution but with nontrivial holes?

Nontrivial progress is any nontrivial work that will lead to a correct solution, or a solution with a major flaw in it that could (with difficulty) be fixed. (Depending on how big the flaw is, the latter may receive points comparable to the former, so for this reason many people overestimate their scores on the USAMO.)

Note the "nontrivial" qualifier on the work; you can't write down something mildly obvious and expect to get points from it. Nor will you get points for an interesting observation that doesn't get you anywhere.

As an example of nontrivial progress, on #1 last year I had the correct idea that when you get a term equal to the average of all the terms before it, the sequence becomes constant (this is pretty much trivial, actually, if you ask me), but my proof for the fact that there must exist such a term was complete BS, basically i did a whole bunch of random stuff and ended up where i started with completely different notation (pathetically, I had no idea that it was wrong until like the next day).

I believe I managed to get a 3 for that, and thinking back, I probably should have gotten a 1. But I mean, I did get a 6 total, and I had the standard BS on #4 and #5 and that's it, so I probably didn't get any lower than a 3.

I guess that counts as nontrivial, hmmm

Hmmm... I got 11 last year, which included #1 and "standard BS" on #4, #5.  I am inclined to believe that I got 2 on each of those. (7,2,2)

This would make your distribution 2,2,2, which I think is more feasible than anything involving a 3 or a 4 on one problem, which is always rare.

Still, if you got points, then that's a good example of "nontrivial" nonetheless.


How many shells of elecrons orbit the nucleus of an atom?

Er.. it depends on the atom (how many electrons it has)?
H has one shell.. while Na has 3 (energy levels)

Yeah...but are you (ln(dx/dy)) referring to orbitals of a certain sublevel?

that is a completely vague question. it can definitely range depending on principal quantum number or energy level, excited states, etc. and theyre not all really shells per se, the s orbitals resemble a shell of sorts but the p-orbitals are dumbbell shaped with x, y, or z orientation. the d and f-orbitals are still more complexly contructed because they can contain more electrons.

it can be much more!!!
with new elements being discovered everyday,nothing is impossible :rotfl:  :D  :rotfl:

my post edited:  :rotfl: [size=150]uptil now seven[/size]


Questions for people who've been to AMSP.

     Is it possible to take an Intermediate Aops class during AMP without skipping classes? Would it be a good idea to bring an instrument to practice? Does the university wash your laundry for you? Sorry I haven't been to AMSP before.

The AMSP classes are earlier in the day. I think that somebody tried to do some AoPS midterm assignment or something but then forgot about it (I'll ask him). I think that you'll be pretty busy (especially if want to work on extra problems after class is over).

There some free time after dinner that you can do assorted activities. Some people did sports like frisbee or soccer. Some did more math, some just hung out, talked, went swimming, played things like cards, etc. There was also ping-pong/DDR. I know there were some people who practiced violin daily so you can definitely do that if you want.

The way that laundry worked last year was that you sort of live in an apartment that has a washer and a dryer (and soap). So you do have to wash your own clothes, but it is pretty easy.

whewww
its good that we can bring our instrumentss!
yayy ;]

so it's pretty much self laundry...and soap's provided so we dont have to bring any or buy any there?
thankss

[quote="CatalystOfNostalgia"]There is a place where you can play piano: there are two practice rooms with upright pianos that aren't too bad, as well as a concert hall that may or may not be locked and may or may not have no lights on and may or may not have the lighting room locked.

diophantient was taking an AoPS class that ran during AMP, I forget what he did though. Maybe that's the same person that Altheman is referring to.[/quote]
Yeah, I ended up just missing 3 classes and screwing the midterm.

Midterm? AoPS classes have midterms? Does that mean finals too? Are they hard?

some i guess. the ones i did didnt have mdtrms, but i guess some do...
good luck on acceptance!
i feel stupid saying this as i havent been accepted yet either.
wish me luck 2!

[quote="diophantient"][quote="sooozyy"]DAI! did you get your plane tickets yet?

or even any other peoplee?

oh and quickk question,
when we arrive at the airport, how do we get to the campuss?[/quote]
There are AwesomeMath people there who have AwesomeMath shirts and AwesomeMath signs who lead you to an AwesomeMath shuttle which takes you to AwesomeMath.[/quote]

Airport? Plane tickets? Where do you find  the instructions to go onto AwesomeMath??

TPBM is the (in-)famous David, Suzy.
Anyways, I'm probably considering WOOT next year instead. And no, I didn't buy my tickets yet. My parents will probably wait til later. Did you?

next year..i'm not doing any academic camps.
probably focus on my early college apps and intership or whatnot.
nope didn't buy my ticket yet.
i should though.

Speaking of planes,

I already set up the flight to Dallas in Flight Simulator X and I'm gonna fly it tomorrow. I'll try an ILS appraoch, which I hate to do. Anybody else like flying or flight simulators?

:pilot:

DAI
hopefully you got my pm.

i reserved my flight todayy.
infact like ten minutes ago.
checkk your inbox!


Let $(F_{n})$ be a decreasing sequence (i.e. $F_{1}\supseteq F_{2}\supseteq...$) of closed bounded nonempty sets in $R^{k}$. Prove that $F=\bigcap_{n=1}^{\infty}F_{n}$ is also closed, bounded, and nonempty.

proving that $F$ is closed and bounded it is easy but i having trouble in proving that $F$ is non empty.

Note that while the first and second property can be proved by themselves, the third is wrong if you omit the closed and bounded hypotheses (aka compactness). owk

Choose a $f_{j}\in F_{j}$ for each $j$.  The sequence $(f_{j})_{j=1}^\infty$ is contained in $F_{1}$.  $F_{1}$ is compact, so according to Bolzano-Weiestrass there is a convergent subsequence with limit in $f$ in $F_{1}$. But this subsequence, except possibly the first element, is also in $F_{2}$. So since $F_{2}$ is closed, $f$ is also in $F_{2}$. Similarly we have $f \in F_{j}$ for all $j$.


Let $ABC$ be a triangle
$M$ is a variable point on the side $BC$ and $D$ is a point on the line $AB$ such that $MA=MD$.
The parallel line from $B$ to $MD$ intersects the line $AM$ at $P$.
Find the locus of $P$

pontios, can you please explain that part please????.

[quote="Farenhajt"][quote="MatrixFrog"]"The parallel line from B to MD intersects the line AM at P."

I assume you mean perpendicular line. I can't figure out what it would be parallel to.[/quote]

My guess is "parallel to MD and passing through B".[/quote]

Yes.  $BP \parallel MD, \ P \in AM$
I'm sorry

[quote="DanK"]And while he's doing that, can someone refresh my memory of what a locus is.  I've used them before, but it's been quite a while...[/quote]

You have to find where the point $P$ is moving on (a line or a curve).

In other words the locus is the set of all the positions of $P$

For example, a circle $(O,R)$ is the locus of the points $M$ with the property $MO=R$

Hmmm, all I can figure out so far is that it will be an expontential equation...

[hide]
Since $DM\parallel BP$, we have $\angle ADM=ABP$.
But we also know that $\angle ADM=\angle MAD$ because $AM=MD$, so $\angle MAD=\angle ABP$ (or $\angle PAB=\angle PBA$), which implies $PA=PB$.
Hence the locus of $P$ is simply the perpendicular bisector of $AB$, but only to the direction of $C$.

[hide="Strictly speaking"]If $P$ is on the other side of $C$ with respect to $AB$, then $M$ will not be on $\overline{BC}$, so the perpendicular bisector of $AB$ to the opposite direction from $C$ cannot be included.

If $P$ is on the same side as $C$ but inside the triangle, then $M$ will be still on $\overline{BC}$, while $D$ will not be on $\overline{AB}$. However, the problem states that $D$ could be anywhere on $\overleftrightarrow{AB}$, so this case is still fine.

Finally, if $P$ is on the same side as $C$ and outside the triangle, the property obviously holds, so this case is fine as well.[/hide]
[/hide]

Oops.  I forgot to accomodate for the fact that $MA=MD$.  :blush:


geometry
3D geometry
octahedron
real analysis
real analysis unsolved

Let $H$ be a hexagone symmetrical with respect to the origin and having non void interiour. There exists a plane $P$ in $\mathbb{R}^{3}$ and a cube $C$ centered at the origin such that the intersection between $C$ and $P$ is $H$.

In dual situation we must proof that arbitrary symmetrical hexagone is a projection of octahedron, but it's obvious.


If we have $v_0$, and $\forall n \;v_{n+1}=\frac{1}{a+bv_n}$. Can we evaluate $\prod_{k=0}^{n}v_k$?


Determine all positive integers n with the following property: Every natural number which is written in base 10 has n digits, in particular (n-1) Ones and one prime !

I can't exactly understand what you're asking. Could you explain?

Determine all positive integers n with the following property: Every natural number whose decimal expansion contains n digits, one digit is 7 and the other (n-1) digits are 1s, is a prime !

I hope the question is clear now !  :)

I hope I interpreted the qn correctly.

Firstly, the qn is equivalent to finding all n satisfying the condition that (10^n - 1)/9 + 6*10^k is prime for all k \in {0, 1, ..., n-1}. Obviously n is \neq 0 (mod 3), since otherwise the sum of digits of the number is = 0 (mod 3), implying the number = 0 mod 3 and so cannot be prime.

Consider n \geq 6. Note that 6*10^k are all distinct (mod 7) for k = 0, 1, ..., 5, corresponding to 1, 2, ..., 6 (mod 7) in some order, so if (10^n - 1)/9 \neq  0 (mod 7), then there exist some k \in {0, 1, ..., 5} s.t. (10^n - 1)/9 + 6*10^k = 0 (mod 7) and therefore not prime.

Thus, for all n \geq 6 satisfying the above condition, we must have (10^n - 1)/9 = 0 (mod 7), implying n = 0 (mod 6), which is a contradiction since we have shown earlier that n \neq 0 (mod 3).

It then suffices to check the cases for n < 6. Note that 3 divides 711, 13 divides 7111, & 17 divides 71111. Note that 7, 17 & 71 are all prime. Therefore the only possible values for n are 1 & 2.


I'm a beginner in C, and I got some troubles to understand this code, after running it, it shows some values of N P Q, but I don't understand how can we guess them from the code, for example, how can the first part of the code give us this ==>   C : N=6 P=2 Q=1  ?   normally if N=5, then after executing the program, we should have N=6 and Q=5 (since we have ++ after the N) !! I am confused,  can someone explain at least 1 part of the program  :help:
This is what u see after running the program :
C : N=6 P=2 Q=1
D : N=6 P=3 Q=1
E : N=6 P=2 Q=0
F : N=6 P=3 Q=1
G : S S
H : 83 83
I : 53 53

[code]#include <stdio.h>
#include <conio.h>
main()
{
   int N=10, P=5, Q=10, R;
   char C='S';

   N = 5; P = 2;
   Q = N++ > P || P++ != 3;
   printf ("C : N=%d P=%d Q=%d\n", N, P, Q);

   N = 5; P = 2;
   Q = N++ < P || P++ != 3;
   printf ("D : N=%d P=%d Q=%d\n", N, P, Q);

   N = 5; P = 2;
   Q = ++N == 3 && ++P == 3;
   printf ("E : N=%d P=%d Q=%d\n", N, P, Q);

   N=5; P=2;
   Q = ++N == 6 && ++P == 3;
   printf ("F : N=%d P=%d Q=%d\n", N, P, Q);

   N=C;
   printf ("G : %c %c\n", C, N);
   printf ("H : %d %d\n", C, N);
   printf ("I : %x %x\n", C, N);
   getchar();
   return 0;
}[/code]
[/hide]

[code]Q = N++ > P || P++ != 3;[/code]
Let me add brackets to show you how the compiler understands it:
[code]Q = (N++ > P) || (P++ != 3)[/code]

In other words, its not setting Q = N++; its setting Q to be the result of OR-ing those two bits together. First it compares N>P which is true, then it does N++ and sets N to 11. Since it already knows true || anything is true, it doesn't need to look at the second bracket, and therefore never does P++.
So we get N=6, P is still 2, and Q is true, ie Q=1.

[code]Q = N++ < P || P++ != 3;[/code]

This time N<P, so the first bit returns false - so now it does look at the second part. P isn't equal to 3, so that returns true, so Q is true. Both N and P have been incremented by 1 in the process, so we get N=6, P=3, and Q=1.

[code]Q = ++N == 3 && ++P == 3;[/code]

First it increments N to 6, and this isn't equal to 3. So we already know the entire statement is false, and don't look at the second part. Q=0, N=6, P=2.

[code]Q = ++N == 6 && ++P == 3;[/code]
First it increments N to 6, and so N==6 is true. Then it increments P to 3, and P==3 is true, so Q is true. Q=1, N=6, P=3.

The last bit is just printing a number as a character, integer, and hexadecimal.

[quote="TripleM"]
The last bit is just printing a number as a character, integer, and hexadecimal.[/quote]
َH : 83 83
I : 53 53
Is the integers in H and I standard ? why not  H: 95 95 and I : 24 24 for example ?
what's the relation between "int N=10" that we have declared and that 83 and 53 ?

I hope I understood your question in the proper manner.

There's NO relation between int N=10 and the last results. Note

[code]N=C[/code]

That's where it came from. N is given the same value (and that value is 'S', ASCII equivalently, dec 83 /oct 123 hex53/ ). That's why you got 83 83 (53 53).

(a joke: why not 95 95? Because 'S'!='_')


modular arithmetic
Euler
number theory
IMO Shortlist
number theory solved
Hi

Determine all positive integers $ n\geq 2$ that satisfy the following condition: for all $ a$ and $ b$ relatively prime to $ n$ we have \[a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.\]

Indeed, it's quite easy: It says that the inverse of any $a$ in $\mathbb Z_n$ is actually $a$. This means that in the group $U(\mathbb Z_n)$ each element has order $2$. It's easy to see that unless $n$ is $1,2,3,4$, we will be able to find a number which is coprime with $n$, $>1$ and $<\sqrt n$. That's because if $n$ is prime we take $2$ and if $n$ isn't prime we take $p-1$, where $p$ is the smallest prime factor of $n$.

Edit:

Looks like I keep missing cases :): if $2|n$ then this might not work; Anyway, we know that $\phi(n)$ must be a power of $2$, which tells us that $n$ has form $2^tp_1p_2\ldots p_k$, with $p_i$ odd primes of the form $2^a+1$. Morevoer, we know that there are no numbers $a$ s.t. $(a,n)=1$ and $1<a\le\sqrt n$. I'll see what I can do.

I may be wrong, but I think that grobber is wrong. I don't this are the answers. And I wouldnt call it exactly  very easy number theory. As far as I remember this problem was given in a TST in Hungary, it also appeared on IMO Shortlist and finally a more general case was given in our TST this year (and contrary to most peoples expectations, it didnt turn out to be easy). It isnt difficult, but not very easy.

Do you by any chance know what the answer is?

Huh, sorry, I must have been wrong...

  darij

I'm terribly sorry I rushed into it like that. Ok, so after the above observations, the proof isn't hard to come by: if $n\ge 49$ then we take $a=7$. $7$ doesn't divide $n$ because it's not a Fermat prime, and $1<7\le\sqrt n$. On the other hand, if $n\ge 25$, then $30=2\cdot 3\cdot 5|n$. This means that the only number $>25$ that we have to check is $30$, and it doesn't work because we take $a=7$ again.

The solutions can now be found manually.

I think Harazi means that one :
http://www.kalva.demon.co.uk/short/soln/sh00n1.html

Pierre.

Yes, I rushed a little bit, but it's practically the same problem. Anyway, what's with these complicated solutions? The solution given by Sasha to my problem for the TST is just perfect: smart and very short.

Really? Well, Peter, when I said they are practically the same I wasn't wrong. Think a little bit: they both ask  the numbers n for which any relatively prime a with n has order 2 in Z_n? Isn't this the same?

[quote="harazi"]they both ask the numbers n for which any relatively prime a with n has order 2 in Z_n? Isn't this the same?[/quote]

Sorry hazari, I just learnt the euler phi function yesterday, orders are planned for in a couple of days, currently I got even no clue what they are :D

[quote="Peter VDD"]ab+1 divides n isn't the same as ab = 1 mod n... :D[/quote]

He replaced b by -b. Then, of course, the condition "ab + 1 is divisible by n" becomes "ab = 1 mod n", and the condition "a + b is divisible by n" becomes "a = b mod n".

  Darij

Yes, sorry

didn't think of that since both problems state natural numbers :blush:

Sorry, of course I meant: replace b by any number equivalent to -b modulo n. This number needs not to be negative.

  Darij

[color=#00f]Hi, this is my first post! :welcomeani:
I usually try to be very formal in my proofs, I hope you find it useful.
Any flaws or misstypes that you find you can tell me, I'd like to know. Any kind of constructive feedback is welcomed![/color]

[color=#f00][size=150][b][u]Solution:[/u][/b][/size][/color]

Let $a \equiv b \pmod{n}$ and write $a = ka' + r$ and $b = kb' + r$, so $gcd(a,n)=gcd(b,n)=gcd(r,n)$. So we simplify the problem to be that if $r$ is coprime with $n$ then $r^2 \equiv 1 \pmod{n}$.

Now we are going to bound the prime factors of $n$. So, take $p$ to be the greatest prime that divides $n$, suppose that $p \geq 5 \implies \phi (p) = p-1 \geq 4$. It means there exists three distinct positive integers $x, y, z < p$ such that $gcd(x,p)=gcd(y,p)=gcd(z,p)=1$. Now, consider $x+px'$, we know that $gcd(x+px',p)=gcd(x,p)=1$ and let $n_x = \dfrac{n}{p^{V_p(n)}}$. By definition we have that $gcd(n_x,p)=1$ so there exists a multiplicative inverse of $p$ modulo $n_x$, call that $p^{-1}$. Now take $x' \equiv p^{-1}(1-x) \pmod{n_x} \implies x+px' \equiv 1 \pmod{n_x}$ so $gcd(x+px',n_x)=1$ and because we have $gcd(x+px',p)=1 \implies gcd(x+px', p^{V_p(n)})=1 \implies gcd(x+px',n)=1$ so we have that $(x+px')^2 \equiv 1 \pmod{n} \implies (x+px')^2 \equiv x^2 \equiv 1 \pmod{p}$. Analogously we get that $y^2 \equiv z^2 \equiv 1 \pmod{p}$, but if we look at $k^2 \equiv 1 \pmod{p} \iff p|k^2-1=(k-1)(k+1)$ so $p|k-1$ or $p|k+1$, resulting in only two solutions modulo $p$, so by the [i]Pigeonhole Principle[/i] we obtain that two of $x, y, z$ are congruent modulo $p$. Let them be $x$ and $y$, then $x \equiv y \pmod{p} \implies p|x-y \implies x=y$ since $0<x,y<p$, contradiction! Thus we cannot have a prime factor greater or equal than 5 dividing $n$.

Write $n=2^{\alpha} \cdot 3^{\beta}$. Note that $n=2,4$ or $8$ are solutions. But $n=16$ isn't, that is because $5^2 = 25 \not\equiv 1 \pmod{16}$ so, suppose $\alpha \geq 4 \implies 16|n$. Take $r=5$, note that $5 \nmid n \implies gcd(5,n)=1 \implies 5^2 \equiv 1 \pmod{n} \implies 5^2 \equiv 1 \pmod{16}$ because $16|n$, contradiction! So $\alpha \leq 3$. Doing the same for $\beta$, note that $n=3$ works but $n=9$ doesn't, because $5^2 = 25 \not\equiv 1 \pmod{9}$. So suppose $\beta \geq 2 \implies 9|n$ and take $r=5$ and note that $gcd(5,n)=1$ because $5 \nmid n$ so it implies that $5^2 \equiv 1 \pmod{9}$ because $9|n$, contradiction! So the only solutions are of the form $n=2^{\alpha} \cdot 3^{\beta}$ where $0 \leq \alpha \leq 3$ and $0 \leq \beta \leq 1$, that is $\boxed{n \in \{2, 4, 8, 3, 6, 12, 24\}}$ because $n > 1$. And in fact all of those work! :D

We claim that the answer is $n = 2,4,8,3,6,12,24.$ These clearly work. Now we will show that these are the only ones that work.

The given condition is equivalent to $a^2 \equiv 1 \pmod{n}$ for all $a$ such that $\gcd(a,n) = 1.$ We will split things up on which primes do not divide $n$ or not.

[b]Case 1:[/b] $2 \nmid n.$ Plugging in $a = 2$ yields $4 \equiv 1 \pmod{n},$ yielding $n = 3.$

[b]Case 2:[/b] $2 \mid n$ but $3 \nmid n.$ Plugging in $a = 3$ yields $9 \equiv 1 \pmod{n},$ implying that $n = 2,4,8.$

[b]Case 3:[/b] $2 \mid n$ and $3 \mid n,$ but $5 \nmid n.$ Plugging in $a = 5$ yields $25 \equiv 1 \pmod{n},$ implying that $n = 6,12,24.$

[b]Case 4:[/b] $2,3,5 \mid n$ but $7 \nmid n.$ Plugging in $a = 7$ yields $49 \equiv 1 \pmod{n},$ implying that $5 \nmid n,$ contradiction.

[b]Case 5:[/b] $2,3,5,7 \mid n,$ but $11 \nmid n.$ Plugging in $a = 11$ yields $121 \equiv 1 \pmod{n},$ implying that $7 \nmid n,$ contradiction.

In general, if $p_1, p_2, \dots, p_k$ are the first $k$ primes in order, then

[b]Case $n$:[/b] $p_1, p_2, \dots, p_{k-1} | n$ but $p_k \nmid n.$ Plugging in $a = p_k$ yields $p_k^2 \equiv 1 \pmod{n},$ or $(p_k+1)(p_k-1) \equiv 0 \pmod{n}.$ The second-largest prime factor of $p_k - 1$ is $\frac{p_k - 1}{2},$ and this is less than $p_{k-1}$ by Bertrand's Postulate. Hence $p_{k-1} \nmid p_k - 1.$ Again by Bertrand's Postulate, we can deduce that $p_{k-1} \nmid p_k + 1$. This is a contradiction.

Therefore, the only $n$ that work are $n = 2,3,4,6,8,12,24.$

Not sure what I was doing during June that gave me such a weird solution.. came across this again in otis/otis excerpts somewhere

    Let $n=\prod_{i=1}^{k}p_i^{e_i}:a^2\equiv1\pmod{p_i^{e_i}}\implies a\equiv\{1,-1\}\pmod{p_i^{e_i}}$, which implies p is 2 or 3. Now, setting a=5 implies $25\equiv1\pmod n\implies n\mid24$, so all factors of 24 (except 1) are our answers. It's easy to check that these work because the square of any integer relatively prime to 8 or 3 is 1 mod 3,8. $\blacksquare$

fixed latex

[hide=Sol]

If $a \equiv b \pmod{n}$, and $ab \equiv 1 \pmod{n}$, for all $a$, $b$ relatively prime to $n$ then we must have for any residue $r$ relatively prime to $n$, $r^2 \equiv 1 \pmod{n}$, implying the order is at most 2. Then the only primes dividing $n$ can be 2 or 3.

Claim: Any divisor of 24 is a valid $n$.
Proof. Let $\nu_3(n) = k$, and $\nu_2(n) = \ell$. It is well known that the maximal order modulo $n$, is the lcm of the maximal orders modulo its prime factors. Thus $k \leq 3$, and $\ell \leq 1$, as desired.
[/hide]

Note that the condition is equivalent to saying that for all mods $k$ relatively prime to $n$, $k^2$ is $1$ mod $n$. \\\\
I claim that no prime $p\ge5$ can divide into $n$. FTSOC, assume that such a $p$ divides into $n$. Now note that for any $p\geq 5$, $2^2=4$ is not $1$ mod $p$, but rather $4$ mod $p$ since $p>4$. So if we find a number $x$ mod $n$ that is $1$ mod all other prime divisors and $2$ mod $p$ (this is always possible by CRT), then $x^2$ cannot be $1$ mod $n$ since $x^2$ is not $1$ mod $p$. \\\\
From this, we have that $n$ is in the form of $2^a3^b$ for some nonnegative integers $a$ and $b$. Now I also claim that $9$ also cannot divide into $n$. This is because from a similar argument used before for $p\ge 5$, we have that $2^2$ is not $1$ mod $9$. \\\\
Finally, I also claim that $16$ cannot divide $n$. This is because from an argument similar to the argument used before, $3^2$ is not $1$ mod $16$, meaning that $16$ cannot divide $n$. \\\\
Therefore this gives us that the only possible values for $n$ are $2$, $4$, $8$, $3$, $6$, $12$, and $24$. Through trial and error, we find that all of these work, therefore our solutions for $n$ are $2$, $4$, $8$, $3$, $6$, $12$, and $24$, finishing the problem.

This is the same as finding all $n\ge 2$ such that $a^2\equiv 1\pmod{n}$ for all $a$ such that $\gcd(a, n)=1$. We only need to focus on when $n$ is a power of a prime because of CRT. Plugging in $a=2$ means that all odd $x>4$ can't divide $n$. Plugging in $a=3$ gives that all $3\not|x>9$ can't divide $n$. This means the solutions for $n$ are $2, 3, 4, 6, 8, 12, 24$.

This equivalent to showing that $a^2 \equiv 1 \pmod n$ for all $a$ relatively prime to $n$. By CRT, we can split this into a system of congruences of the form $a^2-1 \equiv 0 \pmod{p^k}$ for $a$ relaitvely prime to $p$.

If $p \ne 2$, we need $2^2-1 \equiv 0 \pmod{p^k} \implies p=3, k=0,1$.

If $p=2$, we need $3^2-1 \equiv 0 \pmod{2^k} \implies k=0,1,2,3$.

Thus our solutions for $n$ are $\boxed{2,3,4,6,8,12,24}$, which we can test to ensure they all work. $\blacksquare$

Sketch:

Fundamentally, this is stating that when you square a relatively prime residue of $n$, we get $1 \pmod{n}$. This reminds of us the very well-known fact that $a^2 - 1 \equiv 0 \pmod{24}$ if $\gcd(a, 24) = 1$. Thus, we can easily deduce that all factors of $24$ suffice (besides $1$ as $n \geq 2$). Now if a prime $p > 3$ existed such that $p  | n$, then we know that any relatively prime to $n$ residue that is $2 \pmod {p}$ will become $4 \pmod {p}$ when squared which is not $1$. A similar argument holds if $9 | n$. Moreover, if $16 | n$, then we choose some residue relatively prime to $n$ that is $3 \pmod{16}$. Squaring this would yield $9 \pmod{16}$ which is not $1 \pmod{16}$.

This question is essentially finding all positive integers $n$ such that all $a$ satisfying $\gcd(n,a) = 1$ also satisfies $a^2 \equiv 1 \pmod{n}$. Because of CRT, only need to find the possible prime factors of $n$. Plugging in $a=2$ shows that every odd prime greater than $3$ fails, so $n$ consists of only factors of $2$ and $3$. Then, plugging in $a=5$ shows that all factors of $24$ work except $1$.

The answer is $n = \boxed{2,3,4,6,8,12,24}$.

This problem is equivalent to finding all $n$ so that all $a$ relatively prime to $n$ satisfy $a^2 \equiv 1\pmod{n}$.
By CRT we can just look at the separate prime divisors $p^m$ of $n$, so if $a^2 \equiv 1 \pmod{n}$, then $a^2 \equiv 1 \pmod{p^m}$ also holds.

If $p > 3$, then $a = 2$ immediately fails, so the prime divisors of $n$ are $2$ and $3$.
$\newline$
\[a^2 \equiv 1 \pmod{2^m}\]
It's easy to check that all $m \leq 3$ work, as plugging in $a = 3$ for $m > 3$ fails.
$\newline$
\[a^2 \equiv 1 \pmod{3^m}\]
All $m \leq 1$ work, as plugging in $a = 2$ for $m > 1$ fails.
$\newline$
So our answers are $n = 2, 3, 6, 8, 12, 24$(as $1$ is not included due to the problem statement.)

[hide=Sketch]
Let $p$ be a prime such that it divides n. Then ,$a^2 \equiv 1 (mod p), GCD(a,p ) =1$. From this it is clear that $p$ can take the values of  $2,3$ only. Let $n=2^x3^y$, now we just have to check the numbers of the form of $n$ upto $25$. Therefore we get $1,2,3,4,6,8,12,24.$ :D

[b][color=#f00][size=150]A nice and standard one.[/size][/color][/b]
Notice that suppose we have $a \equiv b (mod \ n)$ and $ab \equiv 1 (mod \ n)$, it is very tempting to consider $a^2 \equiv 1 (mod \ n)$ [W.L.O.G]. Infact, reverse engineering we get that having this condition imposes both the conditions starting from either, given that $gcd(p,a)=gcd(p,b)=1$.
We take cases, either n is odd or it is even.
[color=#0f0][b]Case 1: n is odd[/b][/color], $gcd(n,2)=1$ thus $2^2 \equiv 1 (mod \ n)$ $\implies$ $3 \equiv 0 (mod \ n)$. We get $\boxed{\text n=1,3}$ which satisfy our conditions.
[color=#0f0][b]Case 2: n is even[/b][/color], now we take subcase where $gcd(n,3)=1$ then $3^2-1 \equiv 8 \equiv 0 (mod \ n)$ $\implies$ $\boxed{\text n=2,4,8}$
Else, $3 \mid n$ and we take further cases $gcd(n,5)=1$ wherein we achieve, $5^2-1 \equiv 24 \equiv 0 (mod \ n)$ $\implies$ $\boxed{\text n=6,12,24.}$
Again we can assume $5|n$ which would lead to further cases for the prime being $7$ but manual checking gives no further solution.
[b][size=150][color=#f00]Claim: The maximum prime dividing n must either be 2 or 3.[/color][/size][/b]
[b]Proof:[/b] Consider $n=p_1^{z_1}p_2^{z_2}\ldots p_k^{z_k}$ wherein $k \in \mathbb{N}$.
[color=#f00]By FTA (Fundamental Theorem Of Arithmetic)[/color], we know that the form is finite and representable. Combined with euler's proof of existence of infinite primes. Let $P$ be the minimum prime not occuring in n greater than the maximal prime occuring in it. Thus we get $P \nmid n$ and $P>p_k$ assuming that $p_k>p_{k-1}>\cdots>p_1$. We consider the equation, $P^2 \equiv 1 (mod \ p_k)$ $\implies$ $P \equiv 1,(-1) (mod  \ p_k)$. Notice that we can extend the argument to $P_k$ representing the $k-th$ smallest prime greater than the maximal prime dividing n such that $P_k \nmid n$.
This would imply that every prime not dividing n must be in this sequence but more importantly, all primes $> p_k$ are of the form $P_k=p_k*t \pm 1$. Now, since the form of $n$ is finite it cannot just contain something like [color=#0f0][b]"all primes of a dirichlet form"[/b][/color]. We use the general statement of [b]Dirichlet's Theorem[/b] which tells us that there are infinite primes in the arithmetic progression (let us consider a particular form) say, $a_n=p_k*(n-1)+2$ and obviously all such primes cannot be in n. The only other possibility is when atleast this form corresponds to $P_k=p_k*t \pm 1$ (we can extend to $a_n=p_k*(n-1)+3$ and so on) thus we get $2 \equiv \pm 1 (mod \ p_k)$ $\implies$ $3,1 \equiv 0 (mod \ p_k)$ $\implies$ $p_k=1,3$.Notice that, $p_k=1$ doesn't make any sense whereas $p_k=3$ is our original maximum through manual work. Infact, the reason for $p_k=2$ not occuring is because we need to also take into account that $p_k*t+2 \equiv 0 (mod \ p_k)$ which then forces $p_k=2$ and is our edge caese.  This completes the proof of claim.
[b]Now, we use this claim combined with our original cases thus limiting us to the fact that primes starting from 5 and so on can never divide n and thus the answers we found are the only ones.[/b]
[size=100][b]Q.E.D[/b][/size] $\blacksquare$

Everything is trivial by CRT :) [INMOTC Karnataka insider joke]
Consider the case where n is prime. Clearly as $a^2 \equiv 1$ mod $n$, $a \equiv 1, -1$ mod $n$ $\implies n = 2$ or $3$.
Now by CRT, every such number has to be of the form $2^x3^y$.
We look for the highest exponent of 2, which by trial and error is $2^3 = 8$. Similarly, the highest power of 3 is $3^1 = 3$. Therefore, the only possibilities are all factors of $2^3 \cdot 3^1 = 24$ (except $1$ as it is excluded). Clearly all these numbers work, so we are done. $\square$

Note: for higher powers of $2$, consider the case $a = 3$, and for higher powers of $3$, consider the case $a = 2$ for contradictions.

Why is the problem equivalent to $(A)$ $x^2 \equiv 1 \pmod n$ for all $\gcd(x,n) = 1$?
We are given that $a \equiv b \pmod n \Leftrightarrow ab \equiv 1 \pmod n$ for $a$ and $b$ relatively prime to $n$.
But we used $a \equiv b \pmod n \Rightarrow ab \equiv 1 \pmod n$ to get $(A)$?

Only if case:
From the first $12$ values, get that $2,3,4,6,8$, and $12$ work.
Then, if $n$ does not divide $2$, then $a=b=2$ results in $4 \equiv 1 \pmod n$, which is false.
Similarly, if $n$ does not divide $3$, then $a=b=3$ results in $9 \equiv 1 \pmod n$, which is false.
So $n$ has to divide $6$
Get that $24$ also works but $18$ does not.
If $n \ge 25$, then if $n$ does not divide $5$, then $a=b=5$ results in $25 \equiv 1 \pmod n$, which is false.
So $n$ has to divide $30$.
Get that $30$ does not work and for $n \ge 60$, $a=7$ results in it not working
So $n$ has to divide $420$, but then $a=11$ does not work, so it has to divide $420*11$, and then it repeats, so it does not work since $p_1*p_2*...*p_n > (p_{n+1})^2$ for large $n$ where $p_n$ is the nth prime. This can be proved by induction because it is true for $n=4$ and if $n$ works, then the LHS is multiplied by $p_{n+1}$, but the RHS is multiplied by $(\frac{p_{n+2}}{p_{n+1}})^2 \ge 2^2 = 4$, so the LHS will remain larger. So the only numbers that work are $\boxed{2,3,4,6,8,12,24}$
Testing, all of those satisfy the if case.


What's the greatest possible value of $ a^3b\minus{}ab^3$ if $ a^2\plus{}b^2\equal{}1$?

You can set $ a \equal{} \cos \theta ,\ b \equal{} \sin \theta$, so $ a^3b\minus{}ab^3 \equal{} ab(a^2 \minus{} b^2) \equal{} \frac {1}{4}\sin 2\theta \leq \frac{1}{4}$.

[quote="kunny"]You can set $ a \equal{} \cos \theta ,\ b \equal{} \sin \theta$, so $ a^3b \minus{} ab^3 \equal{} ab(a^2 \minus{} b^2) \equal{} \frac {1}{4}\sin 2\theta \leq \frac {1}{4}$.[/quote]

Is any other method without using trigonometry?  :maybe:

[hide="Solution Without Trig"]$ S = ab(a^2 - b^2) \implies S = ab(a + b)(a - b)$

$ S^2 = a^2 b^2 (a + b)^2 (a - b)^2$
[size=59][color=white]-[/color][/size]
$ = a^2 b^2 (a^2 + b^2 + 2ab)(a^2 + b^2 - 2ab)$
[size=59][color=white]-[/color][/size]
$ = a^2 b^2 (1 + 2ab)(1 - 2ab)$
[size=59][color=white]-[/color][/size]
$ = a^2 b^2 (1 - 4a^2b^2)$
[size=59][color=white]-[/color][/size]
$ = \frac {1}{4} (4a^2b^2)(1 - 4a^2b^2)$

Let $ x = 4 a^2 b^2$:
[size=59][color=white]-[/color][/size]
$ \implies 4 S^2 = x(1 - x)$

This is a parabola, and achieves a maximum value of $ 1/4$ when $ x = 1/2$.

Thus, $ 4S^2 \leq \frac {1}{4} \implies S^2 \leq \frac {1}{16} \implies S \leq \frac {1}{4}$

Now we have to show that this is attainable. By GM-QM, we have that:
[size=59][color=white]-[/color][/size]
$ \sqrt{ab} \leq \sqrt{\frac{a^2+b^2}{2}} = \sqrt{\frac{1}{2}}$
[size=59][color=white]-[/color][/size]
$ \implies ab \leq \frac{1}{2}$
[size=59][color=white]-[/color][/size]
$ \implies 4 a^2 b^2 \leq 1$
[size=59][color=white]-[/color][/size]
$ \implies x \leq 1$

Therefore, $ x=\frac{1}{2}$ is attainable, the maximum of $ S$ is indeed $ \frac{1}{4}$
[/hide]

[b]Notice:[/b] Kunny had a typo in his solution; the last two steps should read $ = \frac{1}{4} \sin 4 \theta \leq \frac{1}{4}$

[hide="Explanation without parabolas"]

I used TZF's solution, just if you don't understand parabolas:

$ x(1\minus{}x)\equal{}x\minus{}x^2\equal{}\minus{}(x^2\minus{}x\plus{}\frac{1}{4})\plus{}\frac{1}{4}\equal{}\minus{}(x\minus{}\frac{1}{2})^2\plus{}\frac{1}{4}$
It's maximum for $ \minus{}(x\minus{}\frac{1}{2})^2\equal{}0$ or $ x\minus{}\frac{1}{2}$[/hide]

[quote="Bugi"][hide="Explanation without parabolas"]

I used TZF's solution, just if you don't understand parabolas:

$ x(1 \minus{} x) \equal{} x \minus{} x^2 \equal{} \minus{} (x^2 \minus{} x \plus{} \frac {1}{4}) \plus{} \frac {1}{4} \equal{} \minus{} (x \minus{} \frac {1}{2})^2 \plus{} \frac {1}{4}$
It's maximum for $ \minus{} (x \minus{} \frac {1}{2})^2 \equal{} 0$ or $ x \minus{} \frac {1}{2}$[/hide][/quote]

I understand parabolas and quadratic equations  :)

Thanks for all solution you've posted :)


What does $\mod 7$ help¿
It's an open problem to solve $x^{3}+y^{3}+z^{3}=3$. Note that $x=4=y, z=-5$ is a solution.

[quote="fermat3"]sorry zetax   we can prove that x^3 +y^3 +z^3 +t^3=3 has ifinitly solutions in Z :ninja:  YOU CAN TRY AGAIN[/quote]
zetax didn't say the opposite  :ninja: .

hi mathlinkers i waited a   another solution for my problem  then i  ll poste mine   let x^3 +y^3 +z^3 +t^3=3 .  we find infinitly many solutions in Z  let a=4-24n^3  b=4+24n^3  c=-24n^2  and t=-5 and done .azbi :ninja:


Interesting question. I am pretty new, so I can't type in Latex (yet). So please bare with my lack of professionalism  :blush:

F{net} = rho g V - mg= ma

v = 4/3 pi r^3

so

rho g 4/3 pi r^3 - mg = m r"  (??)
(then solve the differential equation, and I think you can go from here..I think)

I think that's how I would do it. Not sure if I am right.  :(

[quote="Immanuel Bonfils"]Seems the ball holds the air inside the bottle (no air in the neck) in order to make sens the "pressure-volume variations of air to be isothermal".
In this case $ \rho$ varies.[/quote]

Interesting observation. I completely miss that line.  :surrender:

I don't see how this would contribute to the problem solving though.

You have to fix your differential equation quite a bit!

The way you put it's not linear ($ r^3$ !)

Also r, the sphere radius, is a constant; so r" =0 !

[quote="MNrule"]An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction. For figure: http://www.flickr.com/photos/30014811@N06/3249114189/

Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for  the time period of oscillations assuming the pressure-volume variations of air to be isothermal.[/quote]
[hide="My thoughts"]

The differential equation describing SHM is well known:
\[ \ddot{z} \plus{}\omega^2z \equal{} 0\]

where, in this case, $ z$ is the vertical coordinate.

From the Lagrangian for this system:
\[ m\ddot{z} \plus{} mg \equal{} F_z\]

One now has to determine $ F_z$.
\[ F_z \equal{} F_{\text{buoyant}} \plus{} F_{\text{press}} \]

If the ball is neutrally buoyant then $ F_{\text{buoyant}} \equal{} mg$.

Under this assumption, $ m\ddot{z} \equal{} F_{\text{press}}$.

However there is no indication in the problem statement as to how $ F_{press}$ ought to be modeled.

But if one further lets $ F_{\text{press}} \equal{} p_{acoustic}A \equal{}\rho_{air}A^2v^2$, where $ A$ is the cross section of the ball, and $ v$ is the speed of sound in the air column. In order to solve the problem, one needs to set this equal to some spring force, i.e. $ F_{\text{press}} \equal{} \minus{}kz$, and then solve for $ k$.

Then $ \omega \equal{}\sqrt{\frac{k}{m}} \equal{}\frac{2\pi}{T}$, where $ T$ is the oscillation period to be found.

At this point, however, I am uncertain how to make the expression $ \rho_{air}A^2v^2\propto z$.[/hide]

[hide="With the assumptions I posted earlier,"]
the body's form doesn't matter; it could be a cilinder (piston) as well..
\[ T \equal{} {{2\pi}\over a} \sqrt {{{mV} \over p_0}}\]

where  $ p_0$ is the equilibrium pressure, when the volume of air in the bottle is V.
 [/hide]


[quote="linzk3"]A 12 in. by 16 in. painting is surrounded by a frame of uniform width.  The area of the frame equals the area of the painting.  Find the width of the frame.[/quote]
[hide]$(12+2w)(16+2w)=2(12)(16)$

$4w^{2}+56w+192=384$

$w^{2}+14w-48=0$

$w=\frac{-14\pm\sqrt{196+192}}2$

$w=\frac{-14\pm2\sqrt{97}}2$

$w=-7+\sqrt{97}\approx\boxed{2.85\text{ inches}}$[/hide]

[quote="i_like_pie"][quote="ProtestanT"]$(12+x)(16+x)=12(16)(2)$[/quote]
Sould be $(12+2x)(16+2x)=2\cdot12\cdot16$ because the frame is on both sides.

[quote="ProtestanT"]$x^{2}+28x+192=384$
$x^{2}+28x-192=384$[/quote]
Umm...[/quote]

It doesnt matter if its x or 2x because you can just divide by 2 in the end, which is what he did.

how did ProtestanT get two equations from 1?

the one with +192 is the right one.

[quote="Castor496"][quote="i_like_pie"][quote="Castor496"]I'll try to learn how to use Latex soon.[/quote]

Welcome to AoPS! Here are some sites that might help with $\LaTeX$:

http://tobi.oetiker.ch/lshort/lshort.pdf

http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/quote]

$Thanks,$ $those$ $links$ $were/are$ $helpful.$  :)[/quote]
don't put text in latex, some people have slow computers/internet

[quote="i_like_pie"][quote="ProtestanT"]$(12+x)(16+x)=12(16)(2)$[/quote]
Sould be $(12+2x)(16+2x)=2\cdot12\cdot16$ because the frame is on both sides.

[quote="ProtestanT"]$x^{2}+28x+192=384$
$x^{2}+28x-192=384$[/quote]
Umm...[/quote]
actually, the second one is supposed to be $x^{2}+28x-192=0$

i subtracted 384 from both sides

sry bout the typo


[quote="asdf2345"]That actually was pretty good...but the people needed to stop going "heh eheh heheh" every five seconds.[/quote]

yeah, if they didn't keep saying it, it would be a lot better

I thought the guy on the left was really freaky.


integration
limit
real analysis
real analysis unsolved

$\displaystyle\int_{R_+}{f(x) dx}<\infty$, $\displaystyle f(x)\searrow 0$ as $x\to\infty$. Prove that $\displaystyle \lim_{x\to\infty} xf(x)=0$.

It is really well known:

We have that $lim _{x \to \infty} \int_x^{2x}f(t)dt=0$, but $xf(2x)\leq \int_x^{2x}f(t)dt$, so $\lim_{x \to \infty}xf(x)=0$


algebra
function
domain
number theory proposed
number theory

Let $f(z) : = \sum_{n \geq 0}z^{2^{n}}$ on $|z| < 1$. Prove that $f(2/3)$ is irrational.

[The general result, due to Mahler, is that $f(\alpha)$ is transcendental for every algebraic number $\alpha \neq 0$ in the domain $|z| < 1$ of $f(z)$; this is not much harder to prove than the special case presented above. ]

This is my approching method.

If f(2/3) is rational, Then there exist coprime integer a, b such that $f(\frac{2}{3})=a/b$. so $a f(\frac{2}{3})=b$.
Then there exist integer n such that $2^{n}>a$
But in base-3, $f(\frac{2}{3})=0.22020002000000020 ...._{(3)}$
and $2^{n+1}>2a=a_{k}a_{k-1}...a_{0}_{(3)}$
Therefore, $b=a f(\frac{2}{3})=0. blahblahblah... a_{k}a_{k-1}...a_{0}0a_{k}...a_{0}00a_{k}..._{(3)}$ can't be integer! contradiction!

So $f(\frac{2}{3})$ is irrational.

I think that we can apply to z is rational, than f(z) is transcendental by this method. Is it true?

I'm sorry, my English is poor.

It's not this simple: of course the problem is equivalent to showing that the the base-3 expansion of $f(2/3)$ is not periodic, but this expansion is not what you claim it to be! Notice that, in order to obtain the expansion, you must write each $2^{2^{n}}$ in base $3$, and then proceed to add all terms $2^{2^{n}}/3^{2^{n}}$ and make all base-$3$ carries --- and you'll see it's not at all that simple. In fact, you don't have any obvious pattern for the base-3 expansion of $f(2/3)$, so don't try to look for that expansion directly; try some other method.

If the problem were, instead, to show that $f(1/3)$ is irrational, then this would be trivial just by looking at the base-2 expansion of that number. However, for $f(2/3)$, things are not as straightforward. This is  why I gave the problem with $2/3$ --- it is the simplest non-trivial case.


The Round 3 scores are now available on the USAMTS web site http://www.usamts.org.

I'm just curious, do you have an estimate as to approximately how long it will take for them to grade the Round 4 submissions?

We're hoping second week of April.  Shouldn't be too much later than that.


An alternative approach.. though joml88's is faster.

For the equation of a line, y = mx +b.
substitute m=1, and (2, 3) to find b.
Then substitute (5, y) in the equation.

[quote="cheezymoogle"][quote="joml88"]First the equation for slope (denoted $m$) is:

\[m=\frac{y_2-y_1}{x_2-x_1}\]

if you are given two points $(x_1,y_1)$ and $(x_2,y_2)$.  So in your example $(x_1,y_1)$ is $(5,y)$ and $(x_2,y_2)$ is $(2,3)$.

Can you solve it from here?[/quote]

Nope. This is the other problem of the same type that I'm having problems with.

Find x if the slope of a line passing through points A (-2,x) and B (4,1) is equal to -2/3.

The book I'm using doesn't explain how to do it, it just gives a solution to it.[/quote]

We can plug the numbers into the formula: \[\frac{(x)-(1)}{(-2)-4}=-\frac{2}{3}\]

\[\frac{x-1}{-6}=-\frac{2}{3}=-\frac{4}{6}\]

\[x-1=-4\]

\[x=-3\]

[quote="cheezymoogle"]when the slope is given?

As in "Find y if the slope of a line passing through points A (5,y) and B (2,3) is equal to 1."

How do you do it? Any help would be appreciated.[/quote]

with the definition of slope, you know that the slope 1=y/x; meaning that for every change in y, the x changes the exact same amount, and vice versa.  in this problem, the x goes from 5->2, a change of -3.  so y->3 must be a change in -3.
setting up the equation, you get:
y-3=3
y=6

[quote="cheezymoogle"]when the slope is given?

As in "Find y if the slope of a line passing through points A (5,y) and B (2,3) is equal to 1."

How do you do it? Any help would be appreciated.[/quote]

if the slope is 1, y=x+b

3=2+b

b=1

y=x+1

y=5+1=[b]6[/b]

y=mx+b(standard form), where m is the slope
b is the yintercept


ARML quadratics conics ellipse geometry trapezoid AMC

I made so many careless mistakes. Like for number 7, I set an equation and got sin(75+x)=-1, so 75+x=270, but then I forgot to subtract the 75 and get 195.

[quote="nat mc"]I got a 4. (1,2,3,4).  Should have gotten 123456 and maybe 7, but oh well I have four more years.[/quote]

I got a 4 also. (1, 2, 5, 7)  Definitely should have had 3 and maybe 4(i was using weighted averages backwards), but I don't blame myself for missing 6 or 8.  I started to work on 6 after taking about 7 minutes to convince myself of my answer to number 5 (a fairly guessable question in my opinion).  After looking at the solution, I saw how far I was away from the answer.  I had just barely subtracted to cancel y from the equations when time was up and put down a guess.  Then number 8, well, I can't say that I personally woiuld have guessed at a root in that equation, but I also didn't think to rewrite as quadratic in a which is a nice solution.  Anyway, I'm not too unsatisfied and I've got a couple more years.  Congrats to all. :)

I also got 4, (1,3,4, and 7).
I was a little miffed cuz for 2 I had the right answer, then I was using pythagorean theorem and checking and I messed it up and got a idfferent answer and put it down and it was wrong, of course.  then later I saw how you didn't even need to pythag it, oh well.  And #5 too  :( but whats past is past, I guess.

yeah, i got a 4 also. i realized i put the amount of positive numbers in that 5 member set (3) and so i got that wrong, and i just had no clue how to do that ellipse problem.  i realized the similar triangles in the trapezoid problem with only about 30 seconds left and i put a<0 for that last problem because of descarte's rule of signs. ohh welllll! it's all good i suppose

I got a 3 (1,3,5), but I really should have had at least a five, since I made extremely dumb mistakes on #2 and #7, and #4 and #8 were actually extremely easy also...

I got 4 (1,2,5,7). Really stupid; misread #3 and spent the whole 10 minutes on it, but could've got #4 pretty easily too... *sigh* well maybe next year... I should've spent more time doing #8, too, but I kept checking #7 to make sure I was right.

almost identical situation to paladin8.  1,2,5,7.  misread 3, spent most of the time on 7

I got a 4 also - 1,3,4,5. I had absolutely no idea on the circle and ellipse one. Can someone explain how to get the solution to that?

hmm I wonder who that 1 vote of 7 is... maybe Zachary Abel?

Anyhow, I got a 5 (1,2,3,4,5) yes,yes, #7 was easy.... And I realized that by using the equation that I had, I could have pulled out #8, had I not stopped and just guessed  :mad: which makes me very mad. Could have been up there for Site competition had I put 195 instead of 255, and colud have been up there for National competition had I put $\frac{3}{4}$ instead of $\frac{1}{2}$....

but I have 2 more yrs... and I tied a USAMO HM(hehe).

I could have been up for site if I hadn't made any stupid mistakes and had gotten the incredibly easy #4, and I could have been up for national - in 9th grade - if I had gotten the easy #8. But I didn't. So, don't worry about this year, just practice for next year, and get it then. ;)

Yeah, 7 was a very interesting problem. A lot of people got 255 instead because when you look at the coordinates, you realize that a line going from the two points would pass through the origin, so it would seem like you could just add 180 to 75, but they switched the sin and the cos for one of the coordinates, so instead of calculating 180+75 it would be 180+15.

I used the distance formula instead (surprisingly it was quite easy to simplify). I didn't have to worry about the trap, but I ended up making an arithmetic error. Sheesh.

I got 5 (don't remember which ones). For two of them I ran out of time and correctly guessed them :) . Anyway, 5 is good and I still have four more years so I'm happy. And nat mc: 4 is great for an 8th grader. You said you should have gotten 6 or 7, but very, very few middle schoolers are at the top of the [b]nation[/b] in high school competitions. Perhaps Gabriel, Reed, or Tiankai, but they were in middle school too long ago so I wouldn't know.

3 here. (1,3,5) 7 tricked me cuz i fell into the sin-cos trap cuz i didn't read the problem closely. and suprisingly enough, i got #1 right by guessing a random prime. i think that used up all my luck for the year.  :lol:

I got 1. One problem out of the whole set. And it wasn't 1 - 4 either. One problem. All your scores trump mine. Oh well, I still have a few more years... :(

[quote="joml88"][quote="Elemennop"]I didn't see what was so hard about #7, either, so long as you stayed in the algebraic mode instead of going geometrically.[/quote]

What do you mean?  Almost instantly after I decided to go geometric I saw the solution...[/quote]

Really? I went algebraic and didn't hit a single snag, but later tried to do it geometrically and couldn't get it.

Maybe that's just cause I'm really bad at geometry.

I think joml might mean in terms of a circle of radius 1 and a circle of radius 2, which, if they are at opposite degree measurements, are obviously 3 units apart. That's what I did.

[quote="white_horse_king88"]My score is too embarrasing to talk about... but I must say the problems often do appear harder than they turned out to be. Any tips for ARML?

I mean, I'm just weak at ARML. That's weird. I always never end up finishing some problem or other because I split my time wrong.[/quote]

Same. This year I missed both 3 and 4 because I misread 3 and spent the whole 10 minutes on it. Blah.

Does anyone know who was the highest scoring freshmen at ARML (and the score)?

Well, I got a 3. :P

I think one on the Southern California A team got a 5.

We (South Carolina) had a freshman with a 5 and also an 8th grader with a 5.

[quote="JesusFreak197"]I think one on the Southern California A team got a 5.[/quote]
We had two freshmen with 5's - Macky and Yolanda. I'm sure if you look nationally, you'll turn up a several 6's - there are a lot of teams, after all.

We (LV!!!) had a freshman with a 5.

Me, I only managed to get 3 (1,2,5); messed up on 3 and 7 and took too long trying to do 6. (If only I could have seen the cheap calculus trick then...)

I (Central Jersey) got 4 {1, 2, 3, 5}.  Could you tell me how to do number 6 w/ calculus?

6 works like this: [hide]u write ${\bf f}(t)=<5\cos t,4\sin t,0>$ where $<x,y,z>$ are the components. Then use the formula for curvature:
$\frac{||{\bf f}^\prime (t) \times {\bf f}^{\prime \prime} (t) ||}{||{\bf f}^\prime (t)||^3}=\frac{20}{(16+9\sin^2 t)^\frac{3}{2}}$

At $t=0$ this is $\frac{5}{16}$. Take the reciprocal to get $\boxed{\frac{16}{5}}$. (The circle fits here because curvature is at a maximum at $t=0$; it wouldn't work for $t=\frac{\pi}{2}$.)[/hide]

i love how the score distribution (for the mathlinkers/AoPSers who voted) is sort of like a bell curve!

I'm pretty sure a few did, but whoever did probably made a dumb mistake on another problem, as those were definitely the two hardest.


[quote="Nam Luu"]Find all positive integers that cannot be written as the sum of two composite numbers.
Can anyone give me the solution of this problem? Thanks a lot. :lol:[/quote]

I suppose we are speaking about sum of [u][b]positive [/b][/u]composite numbers.

Let $ n\ge 12$ such a number. This means that $ n \minus{} 4$, $ n \minus{} 6$ and $ n \minus{} 8$ are primes(else $ n \equal{} 4 \plus{} (n \minus{} 4)$, or $ n \equal{} 6 \plus{} (n \minus{} 6)$ or $ n \equal{} 8 \plus{} (n \minus{} 8)$ would be a decomposition in sum of two positive composite numbers)
But exactly one of the three numbers $ n \minus{} 4$, $ n \minus{} 6$ or $ n \minus{} 8$ is divisible by $ 3$ and greater than $ 3$, so non prime.

So $ n < 12$ and then :
$ 8 \equal{} 4 \plus{} 4$
$ 10 \equal{} 6 \plus{} 4$

Hence the result : $ \{1,2,3,4,5,6,7,9,11\}$

[quote="pco"]
I suppose we are speaking about sum of [u][b]positive [/b][/u]composite numbers.

Let $ n\ge 12$ such a number. This means that $ n \minus{} 4$, $ n \minus{} 6$ and $ n \minus{} 8$ are primes(else $ n \equal{} 4 \plus{} (n \minus{} 4)$, or $ n \equal{} 6 \plus{} (n \minus{} 6)$ or $ n \equal{} 8 \plus{} (n \minus{} 8)$ would be a decomposition in sum of two positive composite numbers)
But exactly one of the three numbers $ n \minus{} 4$, $ n \minus{} 6$ or $ n \minus{} 8$ is divisible by $ 3$ and greater than $ 3$, so non prime.

So $ n < 12$ and then :
$ 8 \equal{} 4 \plus{} 4$
$ 10 \equal{} 6 \plus{} 4$

Hence the result : $ \{1,2,3,4,5,6,7,9,11\}$[/quote]
Of course we are talking about positive number.
Thanks again for the solution.
This problem can't be called "very difficult" :D  :D  :D


linear algebra
matrix
linear algebra unsolved

Consider a matrix $A\in\mathcal M_{n}(R)$ such that $A^{n}=\alpha\cdot A$ with $\alpha\neq\{\pm 1\}$. Proove that $B=A+I_{n}$ is inversable.

The condition $A+I_n$ is invertible is equivalent to "$A$ hasn't eigenvalue -1". But from the condition $A^n=\alpha A$ we get $\lambda^n=\alpha \lambda$ which means that $\lambda \neq -1$ under the hypothesis.

yes beacause $\det(X)=\prod x_i$ where $x_i$ are the eigenvalues of $X$.


induction
combinatorics proposed
combinatorics

[quote="Arne"]A nice third problem: :D

The edges of a regular 2k-gon are coloured in red and blue. At each step every edge whose adjacent edges have the same colour, turns into red and the other edges turn into blue. Prove that after 2k - 1 steps all edges will be red, and that 2k - 1 is the best possible.[/quote]

 Your problem remind me one that I solved last week( a bit easier).
 Let A1,..An be a regular n-gon. At each of the vertices of the n-gon we have a number(real) a_1,a_2,..a_n.at each step we replace the number at Ai by a_i*a_(i+1)
 1)Prove that if n=2  k then after some steps all the number
at each vertice are positive.
 2)prove that the statement is not always true if n<>2  k
Answer:
 1)by induction we can prove that
 after m step the number at Ai will be
   a_i^(0,m)*a_(i+1)^(1,m)*...*(a_i+m)^(m,m)
  with (a,b)=b!/a!/(b-a)!
  and consider a_(n+1)=a1...
  if n=2 k then after 2k steps the number at Ai will be
  ai^2*a_(i+1)*(1,n)+...
  If n=2 k it's easy to prove that (1,n),(2,n)...(n-1,n) are even so the number at Ai is positive.
  I strongly believe that this technique can also be used to prove your problem. :D  :D  :D
  I also think that my question2 is also right for your problem
  :D  :D  :D

Something is wrong here.
Consider k=2 and a square with 2^k=4 edges BBBR.

After one step, this becomes RRBB.
After two steps, this becomes RBRB.
After three steps, this becomes BBBB.
After four steps, this becomes RRRR.

Four steps are not 2^(k-1) steps.

$BBBR\rightarrow BRBR\rightarrow RRRR$ and that is two steps.


Check it out: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81064

That's a great idea.  The only problem in Latex in AIM.  I know you can use latex in GAIM, but it's hard to set up, or so I am told.

that is a good idea, here's the website that allows you to download it, after you have downloaded GAIM http://linux.softpedia.com/get/Communications/Chat/Gaim-LaTeX-3544.shtml

Although if this doesn't work, I still think it's worthwhile doing the AIM practices anyway since not all questions require latex anyway.

Yes, but it's not easy to compile latex in GAIM.  You also need to download other stuff.  Scroll down on that website for details.


limit
real analysis
real analysis unsolved

Please try the following problem and post the solution please.

a(n)=a(n)+[1+{a(n)}^2]^(1/2),a(0)=0.Find Lim a(n) as n---> infinity.
Thanks in advance

I think you wanted to write that $ x_{n \plus{} 1} \equal{} x_n \plus{} \sqrt {1 \plus{} x_n^2}$ , right ? If so then we have that $ x_{n \plus{} 1} \minus{} x_n > 0$ so $ x_n$ is strictly increasing therefore $ \exists \lim_{n\to \infty} x_n \equal{} l$ ($ l > 0$) .If $ l \in \mathbb{R}$ taking the initial relation to limit we obtain $ l \equal{} l \plus{} \sqrt {1 \plus{} l^2}$ contradiction.So the only posibility is that $ \lim_{n \to \infty} x_n \equal{} \infty$

Alternately, $ x_1\equal{}1$ and each term is at least twice the previous term.

An other one : $ x_{n\plus{}1} \geq x_n\plus{}1 \geq x_{n\minus{}1}\plus{}2 \geq ... \geq x_1\plus{}n$ . Manifestly $ x_n \rightarrow \plus{}\infty$
 :cool:


the given is

        tan(tan┘2 + tan┘3 + .... tan┘1995)


geometry
incenter
circumcircle
vector
geometry unsolved

In triangle ABC , <C=30 degrees.
From AC and BC , AE and BD are cut off respectively equal to AB ,
Prove that DE is perpendicular to IO as well as equal to IO, where I and O are respectively the incenter and circumcenter of triangle ABC .
[ i have a synthetic geometric solution, i am posting this as a challenge to solve the problem by complex-number or vector method ]

Dear apratimdefermat,
I don't understand your problem. Perhaps a figure will be welcome or something else...
Thank in advance
Sincerely
Jean-Louis

Dear Apratimdefermat,
thank for your figure.
This interesting problem coming from Laisant has been posted on this site.(see : http://www.mathlinks.ro/Forum/viewtopic.php?t=232923)
This problem is true with  any <A.
Any precise reference?Sincerely
Jean-Louis

[size=150]i think the problem u linked to is different from mine

and in that problem the equality doesn't  occur[/size]


I was looking over AoPS Volume 1, and it said that sqrt(xy) is inadmissible as part of a factorization.  I was wondering why we don't include sqrt(xy) as part of a factorization.

This depends upon what we're factoring over.  If we are factoring with integer exponents, then this would be knocked out.  If we are factoring to do some problem solving, then you can use any type of term you like.


My friends and I are planning to make model rockets.
We're hoping to enter competitions so we're not so into rocket model kits.

Any advice on basic design (esp. legal stuff(I happen to live in California btw)), where to launch, etc. would be greatly appreciated.

I am very into rockets, PM me if you want some advice.

it was actually 3 3/4 years.

just sayin'

So you guys are saying this person already did this rocket and moved on.

Sometimes I feel like doing, math, ok?

I'm not entirely in my skater world...
Not yet...

Why do you people keep reviving threads?
It's spamming u know. :wink:


John borrows $ \$1,000,000$ from the bank. If a bank charges $ 10\%$ interest (compounded monthly). What will John pay back after $ 20$ years.

[hide="wtf?"]20 years is equivalent to 240 months. Each month, we multiply his previous amount by 1.1.
That is, after the first month, he pays $ 1000000\cdot 1.1$
On the second he pays $ 1000000\cdot (1.1)^2$, and so on.

Our answer is then $ 1000000\cdot (1.1)^{240} \equal{} \boxed{\$8594971441069317}$
[/hide]

Wow. That can't be right. I must not be thinking clearly.

Edit: lol, that was stupid.

Lol, it's $ 1\plus{}\frac1{120}$ each month, not $ 1\plus{}\frac1{10}$. So you get the slightly nicer amount of $ \boxed{\$7328073.63}$. :P

:D LOL grn_trtle
I solve it like this
$ \begin{array}{l}
 A\left( t \right) \equal{} 1\left( {1 \pm r} \right)^t  \\
 A \equal{} 1000000\left( {1 \plus{} \frac{{0.1}}{{12}}} \right)^{240}  \\
 \end{array}$
And I come up with $ \boxed{ \approx 7328074}$ But I just couldn't believe it so I think I made a mistake but now math154 is like my answer so I think its correct now.

Thanks a lot  :)


[hide="Solution"]The total miles he rides after x days looks like this: 10+20+30+...+x*10
We can factor out 10 to get 10(1+2+3+...+x).
The expression in the parenthesis is equal to x(x+1)/2.
So 10+20+30+...+x*10=5x(x+1)
Setting it equal to 500:
500=5x(x+1)
100=x(x+1)
Trying 10 and 9 for x:
10(11)>100
9(10)<100
So we see that 9<x<10, so Lance Armstrong arrives on September 10th.[/hide]

I think, the easiest way is to draw a graph.

[quote="rrusczyk"]Lance Armstrong wants to ride to his friend's house 500 miles away. He rides his bike 10 miles on September 1.  He then rides another 20 miles on September 2.  Each subsequent day, he rides 10 miles more than he rode the day before.  On what day will he show up at his friend's house?[/quote]

[hide]instead of working with big numbers, if you divide everything by 10 then the problem becomes 1+2+3... +n = 50 and the sum of consecutive numbers are n (n + 1)/2 and so  n (n + 1) is equal to 100 and when you solve, the closet you can get is 90 because  9(9 + 1)  = 90, but the answer would be 10 because by the 9th day, he wouldn't be there yet. So it would be September 10[/hide]

[hide][color=indigo]He will arive on September 10[/color][/hide]

[hide]a_1=10
 a_2=20

 a_n=10n.
The sum of the distances is equal to 10*n*(n+1)/2=5*n*(n+1)=500 or n^2+n-100=0. 9<n<10 .So on September 9 he will show up at his friends house.[/hide]

[quote="rrusczyk"]Lance Armstrong wants to ride to his friend's house 500 miles away. He rides his bike 10 miles on September 1.  He then rides another 20 miles on September 2.  Each subsequent day, he rides 10 miles more than he rode the day before.  On what day will he show up at his friend's house?[/quote]

[hide="solution"]it is an arithmetic sequence (or series, i forgot?)

$\frac{10+10n}{2}\cdot n=500$ where n is the number of days

$5n+5n^2=100$

$n^2+n-100=0$

so n is 9.5 which means he will arrive on the [hide="10th"]10th[/hide][/hide][/hide]


Due to budget cuts, there are seventeen prisoners confined to the solitary confinement block. If each solitary confinement cell has, on average, 3.4 prisoners and prisoners are randomly assigned to cells, what is the probability that two prisoners, Arnav and Damien, are assigned to the same cell?

walkaroundtheriver, that was a typo.
17/3.4 = 5.

[quote="Walk Around The River"][quote="13375P34K43V312"]There are $ \frac{17}{3.5}=5$ cells
so it's just 1/5[/quote]

nevermind this.[/quote]

uh sorry i'm not perfect and type 5's instead of 4's sometimes
as i said in my blog AoPS hates me and rates my good posts very low

well hmmmm i wonder what this post will be rated

rating posts? why do they do that?

hmm. that was an easy problem. Too bad more of them aren't like that  :D


$ 8\%$ of $ (n \plus{} 2)$ equals 12. What is $ n$?

Set up an equation. \[ \frac{8}{100}(n\plus{}2)\equal{}12\implies n\plus{}2\equal{}150\implies n\equal{}\boxed{148}\]


Let $a,b,c,d$ be real numbers such that $|a|+|b|+|c|+|d|=8,abcd\neq 0$.

Prove that $\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}+\frac{1}{d^4}\geqq \frac{1}{4}$.

Let $p,q,r,s$ be the absolute values of $a,b,c,d$ so $p+q+r+s = 8$ and $p,q,r,s >0$.  By Holder,

\[\left(\sum_{\text{cyc}} \frac{1}{p^4}\right)^{1/5} \left(\sum_{\text{cyc}} p\right)^{4/5} \geq 4,\]

so we have

\[\sum_{\text{cyc}} \frac{1}{p^4} \geq \frac{4^5}{8^4} = \frac{1}{4}.\]

Thank you for your replies, ThAzN1,Urs Schoenenberger.

Here is the back ground of making this problem.

$\frac{1}{x^4}\geqq -\frac{1}{2}|x|+\frac{17}{16}\ (x\neq 0)$


complex numbers
complex analysis
complex analysis unsolved

Let z_1, z_2, z_3 be complex numbers such that abs(z_1)=abs(z_2)=abs(z_3). Prove that z_1, z_2, z_3 are vertices of an equilateral triangle if and only if z_1 + z_2 + z_3 = 0.

The points $z_{i}$ divide the unit circle into three arcs. The triangle is equilateral iff all three arcs are equal (to $2\pi/3$).


Rafael Nadal FTW!!!

That's why.  Cuz I said so.

I I voted voted for for Rafael nadal Rafael Nadal. I have trouble with polls too. :D

I'm always rooting for Roddick, just because he's fun to watch. And Blake, too.

As for who will win, I really don't know. Federer seems to have lost it - I mean he got eliminated from the Olympics AND Wimbledon, but he was always pretty good on the solid court.


vector
geometry
3D geometry
integration
calculus
calculus computations

Consider the vector field  $ \vec{F} \equal{} 10\vec{i} \plus{} y\vec{j} \plus{} 5\vec{k}$ .

Find the flux of the right face of the cube in the first octant of the given vector field. The cube edge length is c.

The flux is
$ \Phi_f \equal{} \int_S \mathbf{F} \cdot \mathbf{dA}$ where $ S$ is the surface you're integrating over and $ dA$ is the surface normal.

In our case, I think the surface normal is c^2 in the positive x direction.

The cube face we're looking at has $ x \equal{} c$, $ 0<y<c$, $ 0<z<c$
So we want
$ \Phi_f \equal{} (10\vec{i}\plus{}y\vec{j}\plus{}5\vec{k}) \cdot (c^2, 0, 0)$

I threw this out pretty fast.. it's probably wrong. :P


Inside acute triangle ABC we take a point D. Let K,L,M be projections  of D on AB, BC, CA consecutively. Prove that ABC and KLM are similar to one another if and only if D is the center of circumscribed circle on the triangle ABC.

its easy problem.i give you a hint
let the intersection of $ AD,BD,CD$ with circumcircle $ ABC$ be $ A_1,B_1,C_1$ and show that $ KLM$ is similir to $ C_1A_1B_1$and.....

and how do i show this? really can't figure it out;/

its really easy.just use that $ LDMC$ and $ BKDL$ and $ AKDM$ and $ AC_1BA_1CB_1$ is cyclic. :P


trigonometry
inequalities proposed
inequalities

[size=200]Let ABC be a triangle,prove that:
$ \sum\frac{r_a}{r_b} \geq \sum\frac{h_bh_c}{w_a^2} \geq \frac {\sum\cot\frac {A}{2}}{\sum\tan\frac {A}{2}} \geq \sum\frac {a}{b} \geq \sqrt {3}\sqrt {\frac {\sum\cot\frac {A}{2}}{\sum\tan\frac {A}{2}}} \geq \sum\frac {a \plus{} b}{b \plus{} c}$
BQ[/size]


[quote="Kent Merryfield"]¿Cuántos enteros en el conjunto 100, 101, 102, 103, ... , 999 no conitenen los dígitos 2, 5, 7, u 8?[/quote]

Quien se anima a hacer el calculito de inclusion exclusion?

Creo q una forma mas facil es:
Como no se van a utilizar los numeros 2, 5, 7, 8 quedarian los numeros 0, 1, 3, 4, 6, 9
ahora hacemos una permutación con repetición de tres cifras
resulta 6^3=216...pero como el cero en algunas ocaciones entro en las centenas sacamos estos hacinedo permutacion con los digitos 1-3-4-6-9 resultando 20
Ahora el resultado sería 196
(para ahorarse la fatiga de exclusion e inclusion) :D

No entendí bien tu solución americoperu...  bueno pero creo también que no es necesario usar inclusión-exclusión.

[hide="Mi Solución "]
La cifra de las centenas tiene 5 posibilidades (1, 3, 4, 6, 9), la de las decenas 6 (0, 1, 3, 4, 6, 9) y la de las unidades 6 también (0, 1, 3, 4, 6, 9); en total $5 \times 6 \times 6 = 180$ posibilidades.
[/hide]

$Tipe$


Flip a fair coin repeatedly until the sequence HTH occurs. What is the probability that
the sequence THTH has not yet occurred? (For example, the event whose probability
we seek includes the sequence HHTH, but not the sequence TTHTH.)

[hide]
There is a $ \frac 18$ chance that the first three throws are HTH; otherwise there is a $ \frac 12$ chance that the throw preceding the last three throws (HTH) is a head. The answer is $ \frac 18 \plus{} \frac 12\left(1 \minus{} \frac 18\right) \equal{} \boxed{\frac 9{16}}$.[/hide]

Nice thought, but it's not enough.

[hide="Here is why"]You made the excellent observation that the only question is whether the prefix preceding the final HTH ends in T or not.  Unfortunately, you decided that (when this string is nonempty) this happens with probability $ \frac{1}{2}$.  This would be true if the prefix could be [i]any[/i] string, but it can't be: it can only be a string that contains no HTH block.  Thus, you need to compute the probability that a string with no HTH block ends in H, and this might not be $ \frac{1}{2}$.[/hide]

Actually... you can have any random string before the HTH. You don't have to consider that no HTH is in it 'cause... well, if an HTH was in it, then you wouldn't be considering that part of the string.

But the fact that there is no HTH in it means that it is not a [b]random[/b] string. There are conditions on it.

Define $ P(xyz)$ as the chance of getting a winning string from that initial configuration (where $ 1$ represents heads and $ 0$ represents tails).  Thus, you have the following possibilities.

$ P(000) \equal{} \frac{1}{2}(P(000) \plus{} P(001) )$

$ P(001) \equal{} \frac{1}{2}(P(010) \plus{} P(011) )$

$ P(010) \equal{} \frac{1}{2}(P(100))$

$ P(011) \equal{} \frac{1}{2}(P(110) \plus{} P(111) )$

$ P(100) \equal{} \frac{1}{2}(P(000) \plus{} P(001) )$

$ P(101) \equal{} 1$

$ P(110) \equal{} \frac{1}{2}(P(100) \plus{} 1 )$

$ P(111) \equal{} \frac{1}{2}(P(110) \plus{} P(111) )$

From the last equation, $ P(111) \equal{} P(110)$.  Plugging that into the fourth equation, $ P(111) \equal{} P(011)$.  Similarly, from the first and fifth equations, $ P(000) \equal{} P(001) \equal{} P(100)$.  To simplify things, let $ P(111) \equal{} x$ and $ P(000) \equal{} y$.  Also, $ P(010) \equal{} \frac{1}{2}y$.  From the second and seventh equations, $ y \equal{} \frac{y}{4} \plus{} \frac{x}{2}$ and $ x \equal{} \frac{y}{2} \plus{} \frac{1}{2}$.  Thus, $ 3y \equal{} 2x$ and $ 3y \equal{} 6x \minus{} 3$.  Solving for $ x$ yields a value of $ \frac{3}{4}$ and solving for $ y$ yields $ \frac{1}{2}$.  Adding together all the possibilities and dividing by $ 8$ yields $ \frac{3x \plus{} 3.5y \plus{} 1}{8}$ $ \equal{}$ $ \boxed{\frac{5}{8}}$.


Can it be said that the center (center of gravity, like a median of a triangle) of any polygon can be found by putting the figure on a graph, taking the average of the x-coordinates, then taking the average of the y-coordinates?
Like if a quadrilateral has vertices $ (x_1,y_1), (x_2,y_2), (x_3,y_3)$, and $ (x_4,y_4)$ then its center is at $ (\frac {x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4}{4}, \frac {y_1 \plus{} y_2 \plus{} y_3 \plus{} y_4}{4})$

And is this true in general?
If so, why?

Well, the center of mass in general is just the average of masses in each dimension. And since the points don't really have a mass, then you just need to take the average of all the distances from a given point, in this case, the origin.
So, it'll be true no matter what type of polygon, no matter how many dimensions it has (although I can't really imagine the center of mass in more the 3 dimensions).


group theory
abstract algebra
superior algebra
superior algebra unsolved

I need to prove that the group $ G$ of order $ 255$ is cyclic .

Method (1) :

I proved $ Z(G)\equal{}G$ and so the Sylow p-subgroups are coprime with order.therefore $ G$ is isomorphic to external product of its Sylow cyclic subgroup .

In fact I need someone to complete the second Method :

$ G\equal{}3\times 5\times 17$ ,

Since (say) $ H$ is the only Sylow 17-subgroup it will be normal , and (say) $ K$ is the only Sylow 5-subgroup it will be normal .

So, |H\cap K|=1, since $ gcd(|H|,|K|)\equal{}1$ , then $ P\equal{}HK$ is a subgroup of order $ 85$ , and In fact it's normal ( since its index of the smallest  prime divide $ G$ , from my post here [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1440742#1440742]Here[/url] ) , also $ P$ is cyclic , since $ 5 \not | {17\minus{}1}$

I stuck here  :o  , Could someone help me to complete my second solution  :maybe:

I found this link :

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cyclic&t=39870

which is very nice post .

[quote="t0rajir0u"]If the Sylow $ 3$-subgroup $ L$ isn't normal then there are $ 85$ of them, which takes up $ 168$ non-identity elements.  There are $ 87$ remaining elements, and $ P$ takes up $ 85$ more of them, so there are $ 2$ non-identity elements left.  What can their orders be?

(Hence, if I'm not mistaken, the Sylow $ 3$-subgroup is also normal.)[/quote]

Let me count with u , we assume $ L$ is not normal .So, there $ 85$ of L Sylow , and which they are distinct , then the number of non-identity elements $ \#= 2\times 85 =170$ .Thus , we left $ 85$ elements to $ P$ with identity for sure .

how did u conclude they are two elements left ? the two elements must be of order $ 3$ as I see .

That doesn't mean we get L is normal .

Sorry, I don't understand your counting , and for what u point ?

This was posted in Ask an algebraist.

Let $ H$ be the unique Sylow 17-subgroup of $ G$; then $ N_G(H)/C_G(H)$ is isomorphic to a subgroup of Aut$ (H)$, so $ |N_G(H)/C_G(H)|$ divides both $ 255$ and $ 16$. Hence $ N_G(H) \equal{} C_(H) \equal{} G$, so $ H$ is central, in which case $ |Z(G)|$ has order a multiple of $ 17$ and is a divisor of $ 225$. It follows $ |G/Z(G)| \equal{} 1, 3, 5, 15$, hence $ G$ must abelian.

Finally, since the order of $ G$ is a product of distinct primes, it follows from the fundamental theorem of finite abelian groups that $ G$ is cyclic.

Or also: since there's a unique Sylow $ 17$-subgroup $ P$ it is normal. There can't be both $ 51$ Sylow $ 5$-sbgroups and $ 85$ Sylow $ 3$-subgroups.(why?), so either there's unique and thus normal Sylow $ 5$-subgroup $ F$ or $ 3$-sbgp T ==> anyhow, we have a subgroup $ FT$ of $ G$ of order $ 15$, and since clearly $ P \cap FT \equal{} 1$, we get $ G \equal{} P*FT$ ==> $ G$ is the semidirect product of $ P$ and $ FT$ ==> there exists a homomorphism $ FT \to Aut(P)$, but this homomorphism has to be the trivial one (why?) ==> the above product is actually direct, and since both $ P$ and $ FT$ are abelian you're done.

[quote="puuhikki"]This was posted in Ask an algebraist.

Let H be the unique Sylow 17-subgroup of G; then N_G(H)/C_G(H) is isomorphic to a subgroup of Aut(H), so |N_G(H)/C_G(H)| divides both 255 and 16. Hence N_G(H) = C_(H) = G, so H is central, in which case |Z(G)| has order a multiple of 17 and is a divisor of 225. It follows |G/Z(G)| = 1, 3, 5, 15, hence G must abelian.

Finally, since the order of G is a product of distinct primes, it follows from the fundamental theorem of finite abelian groups

Or also: since there's a unique Sylow 17-sbgp P it is normal. There can't be both 51 Sylow 5-sbgps. and 85 Sylow 3-sbgps. (why?), so either there's unique and thus normal Sylow 5-sbgp F or 3-sbgp T ==> anyhow, we have a sbgp. FT of G of order 15, and since clearly P /\ FT = 1, we get G = P*FT ==> G is the semidirect product of P and FT ==> there exists a homom. FT --> Aut(P), but this homom. HAS to be the trivial one (why?) ==> the above product is actually direct, and since both P and FT are abelian you're done.[/quote]

The 1st I have it .

let me check if I got the second one

"There can't be both 51 Sylow 5-sbgps. and 85 Sylow 3-sbgps. (why?)" we will exceed the elements of $ G$.

"HAS to be the trivial one (why?) ==> the above product is actually direct"

since |Aut(P)| is coprime with |FT| .

It seem similar to my thinking for second solution .

Are my explanation true ???

 :)


How many integers from 100 through 999, inclusive, do not
contain any of the digits 2, 3, 4 or 5?

There are $ 5$ choices for the first digit and $ 6$ for each of the other two.  Thus, the answer is $ 5 \times 6 \times 6\equal{}\boxed{180}$.


A set of circles in the plane with the sum of diameters equal to 41 can be completely covered with a circle of diameter 10. Prove that there exists a line which intersects more than 4 of the circles from the given set.

[i]Croatia 1993, 11th grade problem[/i]

This is a well-known idea (I would have posted this earlier, but this is the first time I see your post).

We project everything on a line, so the problem can be reduced to this: a set of collinear segments with the sum of their lengths 41 can be covered by a segment of length 10. Show that there is a point which belongs to more than 4 of the segments. This is obviously true:

Let's assume every pt on the segm of length 10 belongs to at most 4 segms. We add up all the lengths of the segments, so every point appears at most 4 times, so the length can't be more than 4*10=40, but the length we get is 41, so we have a contradiction.


abstract algebra
superior algebra
superior algebra unsolved

Hello,

I have to proof that:

$ \begin{gathered}
  {\bigwedge ^m}\left( \mathbb{Q} \right) \equal{} 0{\text{  (with }}m > 1) \hfill \\
  \mathbb{Q} {\text{ should be regarded as a }}\mathbb{Z}{\text{ \minus{} module}} \hfill \\
\end{gathered}$

Has anybody a hint for me?

Well, you have to show that any element $ \frac{p_1}{q_1} \wedge \frac{p_2}{q_2}$ for $ \frac{p_1}{q_1}, \frac{p_2}{q_2} \in \mathbb{Q}$ is actually zero. How can you show that this is zero?

[hide="Hint"] Remember that any element of the form $ r \wedge r$ is zero. Use the multilinearity over $ \mathbb{Z}$ of the wedge product.[/hide]

Hi,

I know this calculation rules of the wedge prodcut:

$ \frac{{{p_1}}}
{{{q_1}}},\frac{{{p_2}}}
{{{q_2}}} \in \mathbb{Q} \hfill \\
  \frac{{{p_1}}}
{{{q_1}}} \wedge \frac{{{p_2}}}
{{{q_2}}} \equal{} \left( {\frac{{{p_1}}}
{{{q_1}}} \plus{} \frac{{{p_2}}}
{{{q_2}}}} \right) \wedge \frac{{{p_2}}}
{{{q_2}}} \equal{} \frac{{{p_1}}}
{{{q_1}}} \wedge \left( {\frac{{{p_2}}}
{{{q_2}}} \plus{} \frac{{{p_1}}}
{{{q_1}}}} \right) \hfill \\
  r \in \mathbb{Z} \hfill \\
  r\left( {\frac{{{p_1}}}
{{{q_1}}} \wedge \frac{{{p_2}}}
{{{q_2}}}} \right) \equal{} \frac{{r \cdot {p_1}}}
{{{q_1}}} \wedge \frac{{{p_2}}}
{{{q_2}}} \equal{} \frac{{{p_1}}}
{{{q_1}}} \wedge \frac{{r \cdot {p_2}}}
{{{q_2}}} \hfill $

But unfortunately I still couldn't show that
$ {\bigwedge ^m}(\mathbb{Q}) \equal{} 0$

It would be nice, if anybody would help me again  :)

Bye
Edgar

[quote="edgar01"]Hi,

I know this calculation rules of the wedge prodcut:

$ \frac {{{p_1}}} {{{q_1}}},\frac {{{p_2}}} {{{q_2}}} \in \mathbb{Q} \hfill \\
\frac {{{p_1}}} {{{q_1}}} \wedge \frac {{{p_2}}} {{{q_2}}} \equal{} \left( {\frac {{{p_1}}} {{{q_1}}} \plus{} \frac {{{p_2}}} {{{q_2}}}} \right) \wedge \frac {{{p_2}}} {{{q_2}}} \equal{} \frac {{{p_1}}} {{{q_1}}} \wedge \left( {\frac {{{p_2}}} {{{q_2}}} \plus{} \frac {{{p_1}}} {{{q_1}}}} \right) \hfill \\
r \in \mathbb{Z} \hfill \\
r\left( {\frac {{{p_1}}} {{{q_1}}} \wedge \frac {{{p_2}}} {{{q_2}}}} \right) \equal{} \frac {{r \cdot {p_1}}} {{{q_1}}} \wedge \frac {{{p_2}}} {{{q_2}}} \equal{} \frac {{{p_1}}} {{{q_1}}} \wedge \frac {{r \cdot {p_2}}} {{{q_2}}} \hfill$

But unfortunately I still couldn't show that
$ {\bigwedge ^m}(\mathbb{Q}) \equal{} 0$

It would be nice, if anybody would help me again  :)

Bye
Edgar[/quote]

$ \frac{p_1}{q_1} \wedge \frac{p_2}{q_2} \equal{} p_1p_2q_1q_2 \left(\frac{1}{q_1q_2} \wedge \frac{1}{q_1q_2} \right).$

Oh, thank you for your great help, ysharifi.


Let $n=2^{31}3^{19}.$  How many positive integer divisors of $n^2$ are less than $n$ but do not divide $n$?

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our final answer is the total number of divisors of $n^2$ which are less than $n$, minues the factors of $n$(which are obviously factors of $n^2$).
$n^2=(2^{31}3^{19})^2=2^{62}3^{38}$ thus $n^2$ has 63*39 factors, since we have a perfect square it will have an odd number of divisors, thus the product of pairs of factors taken from the from the first and last elements of factors, will result in $n^2$, looking at symmetry since each pair of factors we see that there are $\frac{63*39-1}{2}=1228$ factors of $n^2$ less than $n$.
$n$ itself has 639 factors (other than itself) so the desired number is: $1228-639=\boxed{589}$ [/hide]

Sorry to revive, but I'm trying to study for AIMEs and I'm wondering: how did you find out how many factors of $n^2$ are less than $n$? I didn't get the explanation.

If $d$ is a divisor of $n^2$, then so is $\frac{n^2}{d}$. That means that after we subtract $1$ (to account for $n$), half of the remaining divisors are less than $n$.

First we will find the amount of factors of $n^2$ that are less than $n$. $n^2$ will have $63 \cdot 39$ total factors so out of those $\frac{63 \cdot 39-1}{2}$ are less than $n$, which is equal to $1228$. Out of these we subtract $32 \cdot 20 = 640$ to get $588$, but we add back one for $n$ itself to give $\boxed{589}$.

n^2 has $63\cdot39=2457$, now since we have an odd number of factors, $\frac{2457-1}{2}$ are less than n so we have $1228$, now we subtract the number of factors of $n$ which is 640, so 1228-640+1=$\boxed{589}$

$n^2$ has $63 \cdot 39$ total factors. Out of those, $\frac{63 \cdot 39-1}{2}$ are less than $n$ and after subtracting the number of factors of $n,$ which is $32 \cdot 20$ and adding back one for $n,$ we get our answer of $\boxed{589}.$