United States brought home 2 gold medals and 2 silver medals. They couldn't repeat the quadruple gold medals brought home by the United States in 2004. Zhou Zhou from PDR China finished with the overall win, and China placed three out of four of their team members higher than the first place scorer of the U.S. team. I think the U.S. Biology Team wished they had Kay Aull back.

I don't believe any of our AoPS friends who went to the USABO made it to IBO, but I could be mistaken.

Hopefully I make it to the USABO and IBO in 2006  ;)   :D  lol

Good luck to everyone competing next year, hopefully the 2006 team can get four golds like the 2004 team!

My sister did the NBO last year.. was 8th... only 4 could come.. this year I'll try to beat her...:P

I guess the Chinese are good at everything. :D

Just wanted to express my annoyance at the attitude I've seen from Adidasty (particularly in other forums, not so much here). Eric, you understand why this kid annoys me. I didn't like the suggestion that my team hadn't fulfilled what was expected of us, or that we'd have been better off with one of us replaced by Kay.

joml88:
Wrong. There's no way a group of 100-some people all with different backgrounds and not sharing a common language could put together a comprehensive test in one day. The test is prepared by the host country. The jurors later remove a small number of questions they feel don't fit the requirements of the IBO guide. As only a small amount of the test gets edited, the whole 'feel' of the exam is dictated by the host country. As justdudxit said, the exam varies significantly from year to year, and this is why.

I made no accusation of cheating. I simply suggested that the Chinese used the openness of the writing process to write a test that emphasized areas they excel in, and ignored other aspects they find less important.

That's all.

- Chris
Team USA
IBO 2005

As an unbiased observer not knowing anything about adidasty, I did not pick up on any direct insults to the US team.

Your flame, Chris, was far worse than what adidasty said.  I find that it was totally uncalled for.  If you have an issue with another person on the forum, there are surely better ways to handle it.  [i]Insulting him publicly[/i] is not one of them.

The lowest point of your post, however, is this comment:

[quote]The Chinese prepared a test that followed their strengths, and we have no way to control that.[/quote]

First of all, this is a serious accusation.  In essence you are calling them cheaters.  Secondly, I am surprised that you can make such a comment without knowing how the problems are chosen.  Did you actually think that the problems for the competition are chosen completely by Chinese people?  That would raise a red flag to me, at least.

I looked [url=http://www.ibo-info.org/pdf/IBO_Guide15_e.pdf]it[/url] up just to be sure.

[quote]
1) The Ministry of Education or another representative institution of participating countries appoints one coordinator and one deputy for the IBO, both citizens of the country and informs the Coordinating Center about this representation and its duration by an official letter.

2) The task of coordinators is to decide at their annual meetings (cf § 2.1) about the principal questions of the IBO and to ensure requirements of the competition in accordance with recommendations for the theoretical and practical part (Appendix I and II).

Decisions of the Board of Coordinators are taken on the basis of majority votes, in the presence of at least 75 per cent of the coordinators. In case of equal votes, the chairperson of the coordinators takes the final decision.[/quote]

It's obvious why the 2004 US team with Kay Aull won 4 gold medals, but not as obvious why the 2005 team could not match that result.  There could have several reasons for that - one of which could be that the preparation regime in 2005 was not as perfectly aligned with the 2005 IBO as it was for 2004.  Biology Olympiad, more so than the other olympiads, varies wildly from year to year.  Look at the 2006 USABO vs. the 2005 one: there were several well-qualified non-seniors at the 2005 USABO who didn't qualify for the 2006 finals, for probably the same variety of reasons. The coaches probably have a better perspective of the results and could comment on that.

One thing I do know is that there was no lack of talent on the 2005 team.  The US team was downright amazing. You'd be hard-pressed to find someone who knew the ins and outs of biology as well as Robin, Chris, Grace, or Steven (near the end, I thought Robin was mutating into some sort of higher life form due to the sheer volume of facts in his head). They were a formidable force, they worked hard, they knew the material, and they did very well at IBO; 2 gold and 2 silvers isn't anything to scoff at. I was honored to meet them, even if they did always screw me over in Mafia :)

I agree with Joml88, but as someone who knows the spat, adid is rubbing it in a bit.  I see the humor in a public " f---  you" as much as the next guy, but flaming Adid in public isn't really all that helpful Chris. Come on Krueger, you're a smart guy.  You represented all of us (including Adid) at the IBO and won the precious Gold medal; the flaming is uncalled for, and you know that.


Let $n \ge 5$ be an integer. Show that $n$ is prime if and only if $n_{i} n_{j} \neq n_{p} n_{q}$ for every partition of $n$ into $4$ integers, $n=n_{1}+n_{2}+n_{3}+n_{4}$, and for each permutation $(i, j, p, q)$ of $(1, 2, 3, 4)$.

Actually the problem statement is false since it says "partition into integers" and we can just take n = n+0+0+0 so we proceed under the assumption that it means "positive integers."

First, I shall show the "if" direction.  Consider all composite positive integers n.  First of all, if n is even, the result is trivial since we can take n= 1 + 1 + (n/2 - 1) + (n/2 - 1).  Now, assume n is odd.  If we prove the result for n = pq where p and q are two not necessarily distinct odd primes then we clearly have the result for all odd composite n because if n = pqA for some positive integer A, then we can take the sum for n = pq and multiply all four summands by A.  Now, note that n = pq = (p-1)/2 + (p+1)/2 + ((q-1)/2 * (p-1)) + ((q-1)/2 * (p+1)) and clearly this sum satisfies the conditions.  Therefore, if n is composite, such a sum exists.

Now, we prove that if n is prime, such a sum does not exist.  Let n = a + b + c + (bc)/a where (bc)/a => c => b => a => 1 (these are not arrows but "equal to or greater than" signs).  Clearly a divides either b or c.  But then n = (b+a)(c+a)/a so it is clear that this sum is composite, contradicting the fact that n is prime.

So, the (very interesting) statement is correct.  As a side note, I do not know Latex and if anyone could rewrite my work nicely that would be highly appreciated.

Also, the part about the permutation is unnecessary.


[quote="anirudh"]Question 1: Somehow, my signature option disappeared in the profile tab, and i can't view other's signatures either. Why is this happening?

Question 2: I am currently in the MATHCOUNTS problem series. once it is over, would all the posts that i have posted over there be gone, andwould my post count go way down?[/quote]

I have no idea about Question 1; is it still a problem?

For Question 2, the topics will be moved to an Archive forum to which you will still have access.  Therefore, your post count won't change.

Well it is good that they will be moved to an archive forum. For problem 1, I just found out that if I logout, Everything is perfect again, and I can see signatures again. Also, the signature panel in the profile tab is not there.


Prove that for any $ a_1,...,a_n$ positive reals such that $ a_n \equal{} \min \left\{a_1,...,a_{n}\right\}$ we have $ a_1^n \plus{} ... \plus{} a_n^n \minus{} n a_1...a_n\geq (n \minus{} 1)(\frac {a_1 \plus{} ... \plus{} a_{n \minus{} 1}}{n \minus{} 1} \minus{} a_n)^n$.

Isn't this a direct application of Surany's inequality?

[quote="harazi"]May I please see your proof?  ;)[/quote]
have you proved it?

Of course, otherwise I wouldn't have said created by Vasc and Harazi. Anyway, I simply don't see how it is a simple consequence of Suranyi's inequality. Cezar, could you please enlighten us?  ;)

everyone who don't know before suranyi's inequality can see [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Surany&t=14008]here[/url]

I was just asking if this inequality is direct application of Surany. I'm not sure.
 Now,I'm thinking to give a solution and I will also try to solve it with Surany.

:wacko: Excuse me, but I do not find such questions very smart. Either you see a direct solution using Suranyi inequality and then it's supposed to be irony, either you don't see such a solution and then its non-sense to ask whether it is a direct consequence.

yeah,after thinking and thinking i come out the solution,if it is right i think it will be exciting although it is ugly ;)
$WLOG,a_1 \geq a_2 \geq a_3 \cdots \geq a_n$
let $f(a_1,a_2,\cdots,a_n)=a_1^n+...+a_n^n-n a_1...a_n- (n-1)(\frac{a_1+...+a_{n-1}}{n-1}-a_n)^n$
first i will prove  $f(a_1,a_2,\cdots,a_n) \geq f(a_2,a_2,\cdots,a_n)$
let $f_1(a_1)=a_1^n+...+a_n^n-n a_1...a_n- (n-1)(\frac{a_1+...+a_{n-1}}{n-1}-a_n)^n$
we have $f'_1(a_1)=na_1^{n-1}-na_2...a_n- n(\frac{a_1+...+a_{n-1}}{n-1}-a_n)^{n-1}$
we have  $f'_1(a_2)=na_2^{n-1}-na_2...a_n- n(\frac{2a_2+...+a_{n-1}}{n-1}-a_n)^{n-1} \geq na_2^{n-1}-na^{n-2}_2a_n- n(a_2-a_n)^{n-1} \geq 0$
and $f_1(a_1) \geq 0$
then $f(a_1,a_2,\cdots,a_n) \geq f(a_2,a_2,\cdots,a_n)$
with similar way we can prove $f(a_1,a_2,\cdots,a_n) \geq f(a_2,a_2,\cdots,a_n) \geq f(a_3,a_3,a_3,\cdots,a_n) \geq \cdots \geq f(a_{n-1},a_{n-1},\cdots,a_n)$
then the last inequality $ f(a_{n-1},a_{n-1},\cdots,a_n)\geq 0 $ will easy be proved.
i wonder how did you find it,what is the backgrond?
i guess the general one will still true:
$a_1^n+...+a_n^n-n a_1...a_n\geq k(\frac{a_1+...+a_{k}}{k}-\frac{a_{k+1}+\cdots+a_n}{n-k})^n$


[hide="10"]$ \frac {1} {2} (x\minus{}3) \plus{} (\frac {3} {2} \minus{}x)$ $ \equal{}$ $ 5x$ [hide="my answer"]$ \textbf{no solution, but I guess no solution too often so I am checking.}$[/hide][/hide]

[hide="14"]$ t\minus{}3(t\plus{}\frac {4} {3})$ $ \equal{}$ $ 2t\plus{}3$[hide="my answer"]\frac {-1} {4}[/hide][/hide]

[hide="17"]$ \frac {u} {5} \plus{} \frac {u} {10} \minus{} \frac {u} {6}$ $ \equal{}$ $ 1$ [hide="my answer"]$ \frac {15} {2}$[/hide][/hide]

[hide="10"]

$ \frac{1}{2}x\minus{}\frac{3}{2}\plus{}\frac{3}{2}\minus{}x\equal{}5x$

$ \minus{}\frac{11}{2}x\equal{}0$

$ x\equal{}0$

So there is a solution, but it is 0.[/hide]

[hide="14"]

$ t\minus{}3t\minus{}4\equal{}2t\plus{}3$

$ \minus{}4t\equal{}7$

$ t\equal{}\minus{}\frac{7}{4}$

You might have messed up your signs.[/hide]

[hide="17"]

$ \frac{6u}{30}\plus{}\frac{3u}{30}\minus{}\frac{5u}{30}\equal{}1$

$ \frac{4u}{30}\equal{}1$

$ u\equal{}\frac{30}{4}\equal{}\frac{15}{2}$

Your answer is correct.[/hide]

If you post your solutions, we can correct them.

Thank you so much! I realize my mistakes :) A lot of these topics may be coming in the future...

[quote="Eulers_Apprentice"]I have a question.
Are these really High school basics?  :o
It seems to me these are Algebra I problems.[/quote]
Algebra I=High school=basic high school stuff.

Oh. Algebra I is HSB? Nevermind...  :blush:

Its out of my Prentice Hall Algebra 2 workbook. Its only the third lesson but i thought still... Plus this forum is much more active than tho other classroom math at the moment...

[b]dynamo[/b], don't guess the solution of an equation without supporting statements. Do the rigorous work.[/b]

I agree. Otherwise, knowing how to solve equations is useless, and it's likely in most cases that you will not get the right answer.

Also, it probably does seem like Alg I, because the first couple chapters of most standard math textbooks review the last year's material. And though a lot of people do Alg I Honors in middle school through advanced placement, it's technically 9th grade material, and therefore is HSB.


So here is a really classic problem. If anyone knows the answer please dont buff it here. Let someone who didnt know this problem post the solution :)

We have two trains going with the same $ v$ velocity towards each other. The initial distance between the trains is $ l_0$. At the begining a small fly starts from one of the trains with velocity $ c\geq v$, it goes over to the other train then after it reaches it comes back to the first train and so on. During this the trains are moving constantly closer and closer to eachother.

Find how much distance did the fly fly altogether until the trains reached eachother?

[hide="hint"]"infinite geometric series?? huh? I have never heard about those"- says a seventh grader. still after five minutes he says-"ohh its really easy, even I can solve it"[/hide]

[quote="Thaakisfox"]infinite geometric series[/quote]

You don't need to say "infinite". A series is already infinite, by definition.

[quote="Carcul"]You don't need to say "infinite". A series is already infinite, by definition.[/quote]
A series is a sum of terms,
$ 1\plus{}2\plus{}3\plus{}4\plus{}\cdots\plus{}100$
Series are not necessarily infinite.

Hint
[hide]How is distance calculated in a simple problem?[/hide]

In my country, series refers to an infinite sum of terms. Anyway, nevermind, that's not the purpose of this thread.


Let $ a_n\equal{}1001001\cdots1001$, where $ n$ denotes the number of 1's in $ a_n$ and $ n\geq2$. Prove that all $ a_n$ are composite.

[quote="mountain.dew"]Let $ a_n \equal{} 1001001\cdots1001$, where $ n$ denotes the number of 1's in $ a_n$ and $ n\geq2$. Prove that all $ a_n$ are composite.[/quote]

[hide="Solution"]
If $ 3|n$ then $ 3|a_n$. Otherwise, since $ \omega^n \plus{} \omega^{n \minus{} 1} \plus{} \cdots \plus{} 1 \equal{} \omega^{3n} \plus{} \omega^{3n \minus{} 3} \plus{} \cdots \plus{} \omega^3 \plus{} 1 \equal{} 0$ (both sides have $ n$th roots of unity as roots), we have $ x^n \plus{} x^{n \minus{} 1} \plus{} \cdots \plus{} 1| x^{3n} \plus{} x^{3n \minus{} 3} \plus{} \cdots \plus{} x^3 \plus{} 1 \Longrightarrow \underbrace{111\cdots 11}_{n\ \text{times}} | 1001001 \cdots 001 \equal{} a_n$. Hence all $ a_n$ are composite.[/hide]


For which function does f(ab)= f(a) + f(b)

a. f(x)=x^2 b. f(x)= 1/x c. f(x)=4x+1 d. f(x)=4x

What is this a and b in the functions?!
Please help me. I do not understand this question at all!

Please do not double post. If you were lagging, please only click 'Submit' ONCE. Moderators can lock this topic, his question has been answered in here--->[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=132505[/url]


Solve for $ x$ and express as a common fraction: $ \frac12(4^{3x})\equal{}128$.

We rewrite this as $ 2^(6x) \equal{} 2^8$. Thus we have $ 6x \equal{} 8$, and $ x \equal{}\boxed{\frac {4}{3}}$.

4^3 = 64.
(64^x)/2 = 128.

64^x=256

2^6x=2^8.

Now we ask ourselves what times 6 = 8?

We quickly see that the[b]answer is 4/3[/b]


geometry
circumcircle
trigonometry
geometry unsolved

Find the locus of points M inside an equilateral triangle ABC such that $\angle MAB+\angle MBC+\angle MCA=\frac{\pi}{2}$.

I think that the altitudes $ AD,$ $ BE,$ $ CF$ of the given equilateral triangle $ \bigtriangleup ABC,$ is the wanted Locus of the point $ M,$ with the property as the problem states.

It is clear that the centroid $ G$ of $ \bigtriangleup ABC,$ has the property as the problem states

$ ($ $ \angle GAB \plus{} \angle GBC \plus{} \angle GCA \equal{} 30^{o} \plus{} 30^{o} \plus{} 30^{o} \equal{} 90^{o}$ $ ).$

We consider an arbitrary point $ M,$ on $ CF$ $ ($ $ M,$ between $ G,$ $ F$ $ )$ and it is easy to show that $ \angle MAB \plus{} \angle MBC \plus{} \angle MCA \equal{} 90^{o},$ because of $ \angle GAM \equal{} \angle GBM.$

$ \bullet$ WE denote as $ P,$ an arbitrary point between $ B,$ $ M$ $ ($ considered as an arbitrary point inwardly to $ \bigtriangleup ABC,$ doesn’t lie on one of the altitudes $ AD,$ $ BE,$ $ CF$ $ )$ and we will prove that $ \angle PAB \plus{} \angle PBC \plus{} \angle PCA \neq 90^{o}.$

We denote as $ A',$ the reflexion of $ A,$ with respect to the line segment $ BM$ and then, we have $ \angle PAM \equal{} \angle PA'M$ $ ,(1)$

We denote the point $ Q\equiv AD\cap BM$ and it is easy to show that the circumcircle of the triangle $ \bigtriangleup A'CM,$ passes through the point $ Q,$ because of $ \angle QA'M \equal{} \angle QAM \equal{} \angle QBG \equal{} \angle QCG\equiv QCM.$

So, the circumcircle of the triangle $ \bigtriangleup CMP,$ doesn’t pass through the point $ C$ and then, we have that $ \angle MCP\neq MA'P \equal{} \angle MAP.$

We conclude now, that $ \angle PAB \plus{} \angle PBC \plus{} \angle PCA \neq 90^{o}$ $ ($ In my schema, the point $ C$ lies inwardly to the circumcircle of $ \bigtriangleup CMP,$ because of the point $ P,$ has been considered between $ B,$ $ M$ and then, in this case of $ P,$ we have $ \angle MCP > \angle MA'P \equal{} \angle MAP$ $ \Longrightarrow$ $ \angle PAB \plus{} \angle PBC \plus{} \angle PCA > 90^{o}$ $ ).$

Hence, we conclude that any point $ M,$ inwardly to $ \bigtriangleup ABC,$ with the property as the problem states, lies on one of its altitudes $ AD,$ $ BE,$ $ CF,$ as the wanted Locus of $ M$ and the solution is completed.

Kostas Vittas.

Did anyone try solving the equation:

$ \sin{\alpha}\sin{\beta}\cos({\alpha\plus{}\beta})\equal{}\sin({\frac{\pi}{3}\minus{}\alpha})\sin({\frac{\pi}{3}\minus{}\beta})$

The following problem can be solved easily by use trigonometric, but in fact, it has a very nice synthetic solution.

Find the locus of point $ M$ inside the square $ ABCD$ such that $ \angle MAB \plus{} \angle MBC \plus{} \angle MCD \plus{} \angle MDA \equal{} 180^\circ$.


Find all natural numbers satisfying (y^2)(x-1)=x^5 -1

 :wallbash_red:

If $ x\equal{}1$, then $ y$ can be anything. Otherwise, $ y^2\equal{}x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1$. In this case, we have
\[ x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1>\left( x^2\plus{}\frac{x}{2}\right)^2\Rightarrow y>x^2\plus{}\frac{x}{2}\]
Let $ a\equal{}2y\minus{}2x^2\minus{}x$. Then $ a$ is a positive integer such that
\[ x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1\equal{}\left( x^2\plus{}\frac{x}{2}\plus{}\frac{a}{2}\right)^2\]
\[ \equal{}x^4\plus{}x^3\plus{}\left( a\plus{}\frac{1}{4}\right) x^2 \plus{}\frac{a}{2}x\plus{}\frac{a^2}{4}\]
\[ \Rightarrow 0\equal{} (4a\minus{}3)x^2\plus{}(2a\minus{}4)x\plus{}(a^2\minus{}4)\]
If $ a>2$, then all coefficients are positive, and if $ a\equal{}2$, then $ x$ must equal 0. So $ a\equal{}1$ giving
\[ 0\equal{}x^2\minus{}2x\minus{}3\equal{}(x\minus{}3)(x\plus{}1)\Rightarrow x\equal{}3\]
For $ x\equal{}3,a\equal{}1$, we get $ y\equal{}11$. So our solutions are $ \boxed{(1,y),(3,11)}$.


inequalities
quadratics
linear algebra
matrix
symmetry

$ \boxed{1}$ Let $ a, \ b,\ c,\ d,\ p,\ q$ are real constants.

(1) Find the condition for which the system equation in $ x,\ y\ : \left\{\begin{array}{ll}ax \plus{} by \equal{} p & \quad \\
cx \plus{} dy \equal{} q & \quad\end{array}\right.$ has no solution.

(2) Solve the inequality in $ x: ax^2 \plus{} bx \plus{} c > 0$.

Note that you may not use the formula of solution for a quadratic equation.

$ \boxed{2}$ In this problem, we consider the 3-digit number that doesn't have all same number in every digit such as 777 and we regard the number 4 as 004, the number 13 as 013. we define the procedure $ A$ as follows: We take the largest 3-digit number that can be formed by the 3-digits. For example, if we apply procedure $ A$ to the number 312, we get $ 321 \minus{} 123 \equal{} 198.$ If we apply procedure $ A$ again, we get $ 981 \minus{} 189 \equal{} 792$.  Consider we conduct procedure $ A$ repeatedly.

(1) If we apply procedure $ A$ to the number 413 five times, what number do you get?

(2) Suppose that we apply procedure to a 3-digit number to get the number $ M$, let denote $ x,\ y$ the third number from rightmost, the rightmost respectively of the number $ M$. Find all of pairs $ (x,\ y)$.

(3) Prove that we can apply the procedure $ A$ to 3-digit number many times.

(4) If we apply procedure $ A$ to any 3-digit number, the number will accord with the number obtained in (1).

[hide="one"]
Put it in matrix form; $ Ax = b$ then we have no solutions iff $ \det(A) = 0 \Longrightarrow ad - bc = 0$
[/hide]

[hide="two"]
$ ax^2 + bx + c > 0 \Longrightarrow a(x^2 + \frac {b}{a}x + \frac {c}{a}) > 0$

$ a( (x + \frac {b}{2a})^2 - \frac {(b^2 - 4ac)}{4a^2}) > 0$

$ (x + \frac {b}{2a})^2 > \frac {(b^2 - 4ac)}{4a^2}$    (assuming $ a > 0$)

$ {\frac { - b}{2a} - \frac {(b^2 - 4ac)}{4a^2} > x}$ and $ {x > \frac { - b}{2a} + \frac {(b^2 - 4ac)}{4a^2}}$ for $ a > 0$

$ {\frac { - b}{2a} - \frac {(b^2 - 4ac)}{4a^2} < x}$ and $ {x < \frac { - b}{2a} + \frac {(b^2 - 4ac)}{4a^2}}$ for $ a < 0$
[/hide]

[hide="three (a)"]
(I think this question only works if the digits of the number are distinct...)
We will call two numbers congruent if they use they same digits.

If we have three digits $ a > b > c$, then for any of the congruent numbers which uses those digits, $ A$ will make the transformation $ A: (100a+10b+c)-(100c+10b+c) = 100(a-c) - (a-c)$ Since$ a,b,c$ are digits and $ a>b>c$ we have$ 9\ge a-c \ge 2$

Therefore after one transformation $ A$ we will have one of $ 891 ( = 900 - 9), 792 ( = 800 - 8), 693,594,495,396,279$ or $ 198$ (note the symmetry)

Now consider the Transformations starting at $ 891(\equiv 198) \mapsto 792(\equiv279)\mapsto 693(\equiv 396) \mapsto 594 \mapsto 495 \mapsto 495...$ We move through all values and end at $ 495$ which then repeats itself

Hence after at most five transformations we arrive at $ 495$
[/hide]

I don't really understand what the other questions are asking...can you please clarify

[quote="ocha"][hide="one"]
Put it in matrix form; $ Ax = b$ then we have no solutions iff $ \det(A) = 0 \Longrightarrow ad - bc = 0$
[/hide][/quote]

Sorry, you get no point. We need to do casework.

[quote="ocha"]

[hide="two"]
$ ax^2 + bx + c > 0 \Longrightarrow a(x^2 + \frac {b}{a}x + \frac {c}{a}) > 0$

$ a( (x + \frac {b}{2a})^2 - \frac {(b^2 - 4ac)}{4a^2}) > 0$

$ (x + \frac {b}{2a})^2 > \frac {(b^2 - 4ac)}{4a^2}$    (assuming $ a > 0$)

$ {\frac { - b}{2a} - \frac {(b^2 - 4ac)}{4a^2} > x}$ and $ {x > \frac { - b}{2a} + \frac {(b^2 - 4ac)}{4a^2}}$ for $ a > 0$

$ {\frac { - b}{2a} - \frac {(b^2 - 4ac)}{4a^2} < x}$ and $ {x < \frac { - b}{2a} + \frac {(b^2 - 4ac)}{4a^2}}$ for $ a < 0$
[/hide][/quote]

We need to take into consideration the sign of Discriminant .

[quote="ocha"]

[hide="three (a)"]
(I think this question only works if the digits of the number are distinct...)
We will call two numbers congruent if they use they same digits.

If we have three digits $ a > b > c$, then for any of the congruent numbers which uses those digits, $ A$ will make the transformation $ A: (100a + 10b + c) - (100c + 10b + c) = 100(a - c) - (a - c)$ Since$ a,b,c$ are digits and $ a > b > c$ we have$ 9\ge a - c \ge 2$

Therefore after one transformation $ A$ we will have one of $ 891 ( = 900 - 9), 792 ( = 800 - 8), 693,594,495,396,279$ or $ 198$ (note the symmetry)

Now consider the Transformations starting at $ 891(\equiv 198) \mapsto 792(\equiv279)\mapsto 693(\equiv 396) \mapsto 594 \mapsto 495 \mapsto 495...$ We move through all values and end at $ 495$ which then repeats itself

Hence after at most five transformations we arrive at $ 495$
[/hide][/quote]

That's correct.  :)

[quote="ocha"]

I don't really understand what the other questions are asking...can you please clarify[/quote]

Sorry for my poor English, I edited the context, how about this now?

For number 1, consider it as two lines in cartesian graph
Since, they don't have solution, so the lines must be parallel
So, we must have $ \frac{a}{c}\equal{}\frac{b}{d}\equal{}\frac{p}{q}$

hope I'm not wrong

[quote="IW@IT"]For number 1, consider it as two lines in cartesian graph[/quote]

That's right.

[quote="IW@IT"]
Since, they don't have solution, so the lines must be parallel
So, we must have $ \frac {a}{c} \equal{} \frac {b}{d} \equal{} \frac {p}{q}$
hope I'm not wrong[/quote]

If so, where can we see the guarantee of $ cdq\neq 0$?

[color=blue] 1st place: ski   :winner_first:  :first:
  2nd place: skn  :winner_second:
3rd place: iijm :winner_third: [/color]

$ \displaystyle \displaystyle \displaystyle \begin{tabular}{|c|c|c|c|c|c\parallel{}c|}\hline Name & Problem 1 & Problem 2 & Total Score \\
\hline ski & 25 & 50 & 75 \\
\hline skn & 32 & 38 & 70 \\
\hline iijm & 9 & 45 & 54 \\
\hline Average & 9.1 & 19.1 & 28.2 \\
\hline\end{tabular}$

Entry 60


inequalities
calculus
real analysis
real analysis unsolved

If $(x_i)$ is a decreasing sequence of positive reals, then show that \[ \sqrt{ \sum_{i=0}^{n}{{x_i} ^2}}\leq\sum_{i=0}^{n}\frac{x_i}{\sqrt{i}} \]

Let's not put problems at this level into "Calculus Computations and Tutorials" - they look awfully scary to the inexperiences posters who frequent that forum. Come to think of it, don't we have a forum dedicated to the IMC?

[quote=Fermat -Euler]If $(x_i)$ is a decreasing sequence of positive reals, then show that \[ \sqrt{ \sum_{i=1}^{n}{{x_i} ^2}}\leq\sum_{i=1}^{n}\frac{x_i}{\sqrt{i}}\text{.} \][/quote]

Actually, we have \[\sqrt{\sum_{i=1}^n\mspace{-1mu}{x_i^2}}\mspace{2mu}\leq\sum_{i=1}^n\mspace{-4mu}{\left(\sqrt{i\mspace{1mu}}-\sqrt{i-1}\right)\mspace{-4mu}x_i}\textnormal{,}\] where \(\left\{x_i\right\}_{i\geq1}\) is still a decreasing sequence of positive numbers.


A hot air balloon is 10 feet off the ground and is rising at a constant rate of 2 feet per second. What is the height of the balloon, in feet, after 25 more seconds?

In 25 seconds, the hot air balloon will rise 2*25=50 more feet. So int 25 seconds, the hot-air balloon will be 10+50=60 feet above ground.


function
algebra
polynomial
topology
real analysis
real analysis unsolved

Prove that the metric space $ B[0,1]\equal{}${all bounded functions$ : [0,1]\to \mathbb{R}$} has an uncountable subset $ \{f_\alpha|\alpha\in A\}$ such that $ \rho(f_\alpha, f_\beta)\geq 1$ for all $ \alpha, \beta \in A$. Also, show that $ B[0,1]$ is not separable.

This is an especially neat result when you juxtapose it to the Stone-Weierstrass theorem that claims that polynomials with rational coefficients are dense in $ C[0,1]$. Now we have the coveted contradiction in mathematics and it is time to close the shop. (I wish this rule could be applied to philosophers and politicians :roll:). Well, I'm in a rather bad mood today, but the point should be clear: the problem statement is wrong (unless by $ C[0,1]$ you mean something other than the space of all continuous functions with uniform norm). To get the result you want, you need to consider the space of all bounded functions on $ [0,1]$ with $ \sup$-norm. Then you can just take the functions $ f_x$ that equal $ 1$ at $ x$ and $ 0$ everywhere else ($ x\in[0,1]$).

sorry about that. that was a typo. i restated the problem

It is not true.
E.g., consider the metric with
$ \rho(f,g) \equal{} \frac12$ if $ f\ne g$ and $ \rho(f,f) \equal{} 0$ for all $ f,g \in B[0,1]$.


$SOLUTION$.

[hide]
<=>  $15(\sum{a^{3}}-3abc+\sum{ab}+3abc)+9abc\geq 7$ (using the identity $\sum{a^{3}}-3abc=(\sum{a})(\sum{a^{2}}-\sum{ab})$), since  $\sum{a}=1$
<=> $15\sum{a^{2}}+54abc\geq 7=7\sum{a^{2}}+14\sum{ab}$
<=> $8\sum{a^{2}}+54abc\geq 14\sum{ab}$
<=> $8(\sum{a})^{2}+54abc\geq 30\sum{ab}$
<=> $4((\sum{a})^{3}+9abc)-9abc\geq 15\sum{ab}$ [/hide]
Our inequality is equivalent to:
$4(\sum{a}^{3}+9abc)\geq 15\sum{ab}+9abc$.
But using $Shur's$ inequality and $A(m)\geq G(m)$ we have:
$4((\sum{a})^{3}+9abc)\geq 16\sum{ab}\geq 15\sum{ab}+9abc$, which completes our proof  :wink: .


number theory
prime numbers
number theory proposed

Find all triples of prime numbers $(p, q, r)$ such that $p^{q}+p^{r}$ is a perfct square.

the firste observation is if we supose that $r\gneq{q}$ => $q=2$ and we have $1+p^{r-2}$ is perfect square.=>$p=3$ and $r=3$ :wink:

WLOG, $q \geq r$
$A^{2}= p^{q}+p^{r}= p^{r}(p^{q-r}+1) \Rightarrow r$ even
Furthermore, $p^{q-r}+1 = B^{2}\Rightarrow p^{q-r}= (B-1)(B+1)$.
Then, $B-1 = p^{k}, B+1 = p^{l}\Rightarrow p^{k}(p^{l-k}-1) = 2$

i) $k = 0 \Rightarrow p^{l}= 3 \Rightarrow p = 3,l=1, q-r = k+l =4$
This gives $(p,q,r) = (3,x+4,x), (3,x,x+4)$ with $x$ and $x+4$ prime numbers

ii $k = 1 \Rightarrow p = 2, l = 2, q-r = k+l = 3$
This gives $(p,q,r) = (2,5,2), (2,2,5)$

Remarks :
For the first case, $x$ can be $3$ or some odd prime $\equiv 1 (mod 6)$ so that $x+4$ is prime, too.


In general how does one prove a given set spans a space?

Probably to pick an arbitrary element $ x_0$ in the space, and see if you can form a linear combination of elements in your set to equal $ x_0$.


geometry
3D geometry
probability
logarithms
conics
parabola
calculus

MOCK AMC I-
CREATED BY: Lucky707 and Singthesorrow
DATE ADMINISTERED: January 30, 2005

Rules:
(Standard AMC Rules)
1. You have 75 minutes to complete the test.
2. Only non-programmable calculators are allowed.
3. 6 points for a correct answer, 2.5 points for a question left unanswered and 0 points for an incorrect answer so it is in your best interest NOT to guess.
4. All other standard AMC Rules apply... We've had plenty of Mock AMCs so you should know the rules by now.
5. At 5:15, you must submit the test. If you do not, it will not count. I will allow a few extra minutes for slow modems, etc. but please submit the test at 5:15.
6. My AOL is hezgotzbeenz if anyone has any questions during the test.

1. The I.Q. of a Martian varies directly with the square of the number of eyes it has. If a Martian with 5 eyes has an I.Q. of 75, find the IQ of a Martian with 8 eyes.
(A) 120
(B) 128
(C) 144
(D) 192
(E) 200

2. The number of cubic feet in the volume of a cube is the same as the number of square inches of its surface area. What is the length of an edge of it expressed in feet? (NOTE: 12 inches = 1 foot)
(A) 6
(B) 36
(C) 72
(D) 108
(E) 864

3. The first two terms of a geometric progression are $\sqrt{3}$ and the fourth root of three. What is the fifth term?
(A) $\frac{1}{\sqrt{3}}$
(B) $\sqrt{3}$
(C) fourth root of 3
(D) fourth root of $\frac{1}{3}$
(E) tenth root of 3

4. A leaky faucet drips at a steady rate. If we measure from the instant the first drop hits the bottom, to the instant the last drop falls, it takes 36 seconds for 9 drops to fall. How many seconds would it take for 12 drops to fall?
(A) 42 seconds
(B) 44 seconds
(C) 48 seconds
(D) 49.5 seconds
(E) 54 seconds

5. A triangle has sides of length 15, 20 and 25 units. It is inscribed in a circle. What is the circle's radius?
(A) $21/2$
(B) $43/3$
(C) $64/7$
(D) $25/2$
(E) $25$

6. How many zeroes are at the end of the product of the first fifty positive integers?
(A) 5
(B) 12
(C) 17
(D) 35
(E) 47

7. If $100! = (100)(99)(98)...(2)(1)$, then what is the largest value of $x$ such that $2^x$ divides $100!$ evenly?
(A) 50
(B) 97
(C) 98
(D) 99
(E) 100

8. I have a red shirt, a blue shirt and a green shirt. I also have blue jeans, black jeans and khakis. How many possible combinations can I make if I absolutely hate wearing the blue shirt with the blue jeans?
(A) 5
(B) 8
(C) 14
(D) 16
(E) 17

9. Let $x$ be the smallest perfect square that has 140 as a divisor. Then
(a) $0 < x < 1000$
(b) $1000 \leq x < 2000$
(c) $2000 \leq x < 3000$
(d) $3000 \leq x < 4000$
(e) $4000 \leq x < 5000$

10. What is the sum of the first twenty powers of 2 (2^1 + 2^2 + 2^3 ... + 2^20)?
(A) $2^{19} - 2$
(B) $2^{20} - 2$
(C) $2^{21} + 2$
(D) $2^{21} - 2$
(E) None of the above

11. $9ax+7y=12$ and $4x+2by=9$ represent the same line. $a$ and $b$ can both be expressed as fractions in lowest terms. What is the numerator of $a$ plus the denominator of $b$ equal to?
(A) 212
(B) 144
(C) 72
(D) 48
(E) 24

12. A square of side length 8 has its corners cut off to make a regular octagon. Find the area of the octagon to the nearest whole number.
(A) 32
(B) 48
(C) 53
(D) 55
(E) 59

13. A piece of string is cut in two at a random point. What is the probability that the larger piece is at least $x$ times as large as the shorter piece? (where $x \geq 1$)
(A) $\frac{2}{1+\sqrt{5}}$
(B) $\frac{x^2+x-1}{x}$
(C) $\frac{2}{x+1}$
(D) $\frac{3}{x+2}$
(E) None of the Above

14. Let $S$ denote the sum of all distinct four-digit numbers that contain only 1,2,3,4 or 5, each at most once. Then
(A) $S \leq 200,000$
(B) $200,000 < S \leq 300,000$
(C) $300,000 < S \leq 400,000$
(D) $400,000 < S \leq 500,000$
(E) $500,000 < S$

15. How many digits are in $a^b$ if $b=\sqrt{9530}$ and $\log a$ ~ $2.8356$
(A) 277
(B) 278
(C) 282
(D) 283
(E) 284

16. If $\frac{1}{(a)(b+1)} + \frac{1}{(b)(a+1)} = \frac{1}{(a+1)(b+1)}$, what is the value of $\frac{1}{a} + \frac{1}{b}$?
(A) $-1$
(B) $1$
(C) $2 \sqrt{5}$
(D) $-2 \sqrt{5}$
(E) Undefined

17. Let the parabola $x^2-11x+28$ meet the $x$-axis at two distinct points, $A$ and $B$. Let the vertex of said parabola be $C$. The area of triangle $ABC$ can be expressed as $\frac{f}{g}$, where $f$ and $g$ are relatively prime. $f$ and $g$ are both...
(A) Prime (but not in the form $2^n-1$)
(B) Perfect Squares
(C) Perfect Cubes
(D) In the form $2^n-1$
(E) None of the Above

18. Let $x$ be the number of positive integral factors of $90^9$. The sum of the digits of $x$ equals
(A) 7
(B) 10
(C) 13
(D) 16
(E) None of the Above

19. Bette visits her friend Keith and the nreturns home by the same route. She always walks 2 km/h when going uphill, 6 km/h when going downhill, and 3 km/h when on level ground. If her total walking time is 6 hours, then the total distance she walks, in km, is
(A) 9
(B) 12
(C) 18
(D) 22
(E) 36

20. If $JKLMN$ is a regular pentagon with side length $a$ and diagonal length $b$, what is the value of $\frac{b}{a} - \frac{a}{b}$?
(A) $\frac{3}{4}$
(B) 1
(C) $\frac{4}{3}$
(D) 2
(E) Depends on $a$

21. Let $P$ be a point outside circle $O$. Draw two tangents from $P$ such that they intersect $O$ at $A$ and $B$. If $AP=BP=7$ and the larger of the two arcs $AB$ is twice as large as the smaller of the two arcs, find the length of $AB$.
(A) $6$
(B) $7 \sqrt{3}$
(C) $\frac{11 \sqrt{3}}{2}$
(D) $7$
(E) None of the Above

22. The system of equations
$4y+gx=34$
$\frac{g^4x^2-16y^2}{g^2x+4y} = -10$
has how many solutions where $g$, $x$ and $y$ are all integers?
(A) $0$
(B) $1$
(C) $2$
(D) $4$
(E) None of the Above

23. How many arrangements of the letters of the word "INDEPENDENT" have no adjacent E's?
(A) $\frac{(2)(7!)}{5!}$
(B) $\frac{11!}{9!-2!}$
(C) $\frac{11!}{3!}$
(D) $\frac{11!}{3!2!}$
(E) None of the Above

24. If $f(x-y)=f(x)f(y)$ for all $x$ and $y$, and $f(x)$ never equals $0$, then $f(3)$ equals...
(A) $-3$
(B) $3$
(C) $9$
(D) :pm: 1
(E) None of the above

25. Pieces of paper numbered from $1$ to $100$ are each placed in one of three hats such that there is at least one piece of paper in each hat. In how many ways can this be done such that if two hats are selected and a card is taken from each, then the knowledge of their sum alone is always sufficient to identify the third hat?
(A) $6$
(B) $12$
(C) $18$
(D) $24$
(E) None of the Above

Is private message okay too or will you only accept IM?

Sorry I use those two interchangeably. You probably won't be able to IM me anyway.

The contest will begin in four hours. If I sent the test to you early, please try and have your answers in BEFORE 5:15 or you will not count as an official contestant.

EDIT: Contest will begin in 45 minutes.

The contest will commence in 20 minutes. I will edit this post every 5 minutes to keep you updated. When five minutes remain, I will edit this post every minute.

Good luck to all.

CURRENT TIME REMAINING: 2 minutes

P.S. If anyone sent me their Mock AMC I answers... my inbox was full, I just cleared it so send them again. If you see this after teh contest is done I can only list you as unofficial however, sorry, because I plan on releasing the results approximately 30-45 minutes after the end of the contest.

[quote="Lucky707"]DATE ADMINISTERED: January 30, 2005[/quote]

Isn't today the 29th?


function
calculus
derivative
inequalities unsolved
inequalities

Determine the max value of

$ 4(a^4+b^4+c^4+d^4) -(a^2bc + b^2cd + c^2da + d^2ab) -(a^2b+b^2c+c^2d+d^2a) $

for $ 0 \leq a,b,c,d \le 1 $

Okay... lets see if this works.

If we hold three of the variables as constant and view the expression as a function in the other variable, we see that the function is convex (the second derivative being always positive.)  So it remains to check the function at its endpoints, and with three variables 0 and one 1 we get 12-1-2 = 9, which is maximal.


An object in the plane moves from one lattice point to another.  At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down.  If the object starts at the origin and takes a ten-step path, how many different points could be the final point?

$ \textbf{(A) } 120 \qquad \textbf{(B) } 121 \qquad \textbf{(C) } 221 \qquad \textbf{(D) } 230 \qquad \textbf{(E) } 231$

[hide="Answer"]The sum of the coordinates of the endpoint must be an even number. Once we see that, it is obvious by plotting the possibilities that the number of possible endpoints is $(1+2+...+11)+(1+2+...+10)=\frac{11(12)}{2}+\frac{10(11)}{2}=121\Rightarrow \boxed{B}$.[/hide]

[hide=" answer"]An easy way is to show a pattern.

When n moves = $0$, there is $1$ coordinate.
When n moves = $1$, there are $4$ possible coordinates.
When n moves = $2$, there are $9$ possible coordinates.

We find that for $n$ moves, the number of possible endpoints is $(n+1)^2$

For $n = 10$, plugging in $n$ into our formula gives $\boxed{121}$.[/hide]

[quote="vvr1590"]wait how do you get that the endpoint must be an even number?
and how does that lead you to the other formula?[/quote]

The sum of the endpoints must be even. When you make a move, you are either adding or subtracting $1$ from one of the two coordinates. Therefore, after $10$ moves, you must have an even number between the two. I can't really explain beyond that; it's kind of intuitive.

As for the formula: by plotting the points, you have $1+2+3+...+10+11+10+...+3+2+1=\sum_{k=1}^{11}k+\sum_{k=1}^{10}k$, which is given by the formula I used.


linear algebra
matrix
geometry
3D geometry
superior algebra
superior algebra solved

hi guys,

if E is the set of matrix 3*3, find the set S={X=A^3 with A in E}  ..
any idea ??

bye..

In the open question section I've shown that if A is invertible then there exists B such that $B^3=A$. So $S\supset U(E)$ (U(E) are the invertible elements of E).
Also S contains no nilpotent matrices except for $0_3$.

On a second thought:
Let $J_n(a)=aI_n+E$
With $E=(e_{i,j})$
$e_{i,j}=\{\begin{array}{cc} 1,& \ if\ j=i+1; \\ 0,& \ if \ j\neq i+1\end{array}$.
All matrices whose Jordan canonical form doesn't have jordan cells $J_n(0)$ for n>1, are cubes.
----------------------------------------------------------------------------------------------
The converse (showing that cubes don't have jordan cells $J_n(0)$ for n>1) follows from the uniqueness of the Jordan canonical form.


algebra
polynomial
quadratics
Vieta
arithmetic sequence
arithmetic series
quadratic formula

1. Compute the infinite sum
1 /(1x2) + 1 /(1x2+2x3) + 1 /(1x2+2x3+3x4) +.................

2. Given f(x) = x^3-2x^2-ax+6 and g(x)= x^3+11x^2+bx-35, their highest common factor is x-c. Find the value of (a,b,c).

3. If f(x) is a third degree polynomial and f(1)=f(2)=0, and f(0)=2, f(-1)= -6, find f(3).

4. If x =(3+2sqrt2)/2, find 4x^4-8x^3-15x^2+13x.

5. If one root of x^4-4x^3-3x^2+ax+b is 1, and the other three roots are in arithmetic progression, find them.

6. If f(x)=|x-1|+3|x-2|+5|x-4| and g(x) = (x-1)^2+3(x-2)^2+5(x-4)^2, the smallest value of f(x) is ___. In g(x), at x =____, the value of g(x) is the smallest.

7. If f(x)=x^2 - (2m)x + (m+6), when x is between -1 and 3 inclusive, f(x) >0, find the value of A and B when A < m < B.

8. Find the remainder when f(x)=x^100 +2x+3 is divided by x^3+x^2+x+1.

9. If A, B, and C are the roots of x^3-3x^2+6x+2=0, find the equation with leading coefficient equal t0 1 whose roots are (A^3+2)/A-2 , (B^3+2)/B-2, and (C^3+2)/C-2 .

10. Let the arithmetic series 1, 4. 7, 10........ If I can only write 15 digits down in every page starting from page 1. In what page will 1234 be written?

11. a. Is x is the set of all real numbers, find the range of x in |3x-2|+|2x-5|=|x+3|.
b. Is x is the set of all real numbers, what is the smallest vaue of |x-1|+|x-4|. The smallest value occurs at what value of x?
c. Is x is the set of all real numbers, what is the smallest vaue of |x-1|+|x-2|+|x-4|. The smallest value occurs at what value of x?

[quote="orangefronted"]

3. If f(x) is a third degree polynomial and f(1)=f(2)=0, and f(0)=2, f(-1)= -6, find f(3).
[/quote]

[hide="My solution"]Since $ f(1)\equal{}f(2)\equal{}0$,we can say that 1 and 2 are two roots of f(x).

Let $ f(x)\equal{}(x\minus{}a)(x\minus{}1)(x\minus{}2)$ ---(1)

where $ a$ is the unknown root

Substitute $ x\equal{}0$ in (1)

$ f(0)\equal{}(\minus{}a)(\minus{}1)(\minus{}2)$

$ 2\equal{}\minus{}2a$

$ \Longrightarrow a\equal{}\minus{}1$

The polynomial is $ f(x)\equal{}(x\plus{}1)(x\minus{}1)(x\minus{}2)$

But if we substitute $ x\equal{}\minus{}1$ in f(x),we don't get $ \minus{}6$ which contradicts the question.

Now,Substitute $ x\equal{}\minus{}1$ in (1)

$ f(\minus{}1)\equal{}(\minus{}1\minus{}a)(\minus{}1\minus{}1)(\minus{}1\minus{}2)$

$ \minus{}6\equal{}(\minus{}1\minus{}a)(\minus{}2)(\minus{}3)$

$ \Longrightarrow a\equal{}0$

The polynomial is $ f(x)\equal{}x(x\minus{}1)(x\minus{}2)$

But if we substitute $ x\equal{}0$ here, we don't get 2 which again contradicts the question.

So,Please tell whether the mistake is in my solution or in the question![/hide]

[quote="picture74"][quote="orangefronted"]

3. If f(x) is a third degree polynomial and f(1)=f(2)=0, and f(0)=2, f(-1)= -6, find f(3).
[/quote]

[hide="My solution"]Since $ f(1) \equal{} f(2) \equal{} 0$,we can say that 1 and 2 are two roots of f(x).

Let $ f(x) \equal{} (x \minus{} a)(x \minus{} 1)(x \minus{} 2)$ ---(1)

where $ a$ is the unknown root

Substitute $ x \equal{} 0$ in (1)

$ f(0) \equal{} ( \minus{} a)( \minus{} 1)( \minus{} 2)$

$ 2 \equal{} \minus{} 2a$

$ \Longrightarrow a \equal{} \minus{} 1$

The polynomial is $ f(x) \equal{} (x \plus{} 1)(x \minus{} 1)(x \minus{} 2)$

But if we substitute $ x \equal{} \minus{} 1$ in f(x),we don't get $ \minus{} 6$ which contradicts the question.

Now,Substitute $ x \equal{} \minus{} 1$ in (1)

$ f( \minus{} 1) \equal{} ( \minus{} 1 \minus{} a)( \minus{} 1 \minus{} 1)( \minus{} 1 \minus{} 2)$

$ \minus{} 6 \equal{} ( \minus{} 1 \minus{} a)( \minus{} 2)( \minus{} 3)$

$ \Longrightarrow a \equal{} 0$

The polynomial is $ f(x) \equal{} x(x \minus{} 1)(x \minus{} 2)$

But if we substitute $ x \equal{} 0$ here, we don't get 2 which again contradicts the question.

So,Please tell whether the mistake is in my solution or in the question![/hide][/quote]

Perhaps its because you have not considered remainders?

There are 4 values they give ie f(1), f(2), f(0) and f(-1).

Also, because the remainder is less than the divisor, we could assume that R(x) is linear and hence $ R(x) \equal{} ax \plus{} b$

Then just proceed as you did in your work but you will have to solve a few simultaneous equations =)

[quote="xpmath"]The polynomial is not necessarily monic. The polynomial must be of the form $ f(x) \equal{} b(x \minus{} a)(x \minus{} 1)(x \minus{} 2)$, but $ b$ does not have to be $ 1$.

[hide="rest of solution"]Plugging in $ x \equal{} 0$, we have $ 2 \equal{} b( \minus{} a)( \minus{} 1)( \minus{} 2)$ so $ ab \equal{} \minus{} 1$.

Plugging in $ x \equal{} \minus{} 1$, we have $ \minus{} 6 \equal{} b( \minus{} 1 \minus{} a)( \minus{} 2)( \minus{} 3)$ or $ 1 \equal{} b(1 \plus{} a)$. This means $ ab \equal{} \minus{} b(1 \plus{} a)$ or $ a \equal{} \minus{} 1 \minus{} a$ so $ a \equal{} \minus{} \frac {1}{2}$, $ b \equal{} 2$.

Thus, the polynomial is $ f(x) \equal{} 2(x \plus{} \frac {1}{2})(x \minus{} 1)(x \minus{} 2) \equal{} (2x \plus{} 1)(x \minus{} 1)(x \minus{} 2)$, which satisfies the conditions.

From that, we find $ f(3) \equal{} 7\cdot2\cdot1 \equal{} 14$.[/hide]

Hm 2800 posts.[/quote]

ah, yes, of course *hits head*   :blush:

Congrats on 2800 posts  :lol:

Any takers for the other problems? :lol:

[hide="#1"]We need to evaluate $ S \equal{} \frac {1}{2} \plus{} \frac {1}{8} \plus{} \frac {1}{20} \plus{} \frac {1}{40} \plus{} \frac {1}{70} \plus{} \cdots$.

Multiplying every fraction by $ \frac {1}{3}$, we have $ S \equal{} 3\left(\frac {1}{6} \plus{} \frac {1}{24} \plus{} \frac {1}{60} \plus{} \frac {1}{120} \plus{} \cdots\right)$.

Note that $ 6 \equal{} 1\cdot 2\cdot 3, 24 \equal{} 2\cdot 3\cdot 4, 60 \equal{} 3 \cdot 4 \cdot 5, \cdots$.

Therefore, $ S \equal{} 3\left(\frac {1}{1\cdot 2\cdot 3} \plus{} \frac {1}{2\cdot 3 \cdot 4} \plus{} \cdots\right) \equal{} 3\displaystyle\sum_{x \equal{} 1}^{\infty}\frac {1}{x(x \plus{} 1)(x \plus{} 2)}$.

Noting that $ \frac {1}{x(x \plus{} 1)} \minus{} \frac {1}{(x \plus{} 1)(x \plus{} 2)} \equal{} \frac {2}{x(x \plus{} 1)(x \plus{} 2)}$, above summation simplifies to

$ \frac {3}{2}\displaystyle\sum_{x \equal{} 1}^{\infty}\left(\frac {1}{x(x \plus{} 1)} \minus{} \frac {1}{(x \plus{} 1)(x \plus{} 2)}\right) \equal{} \left(\frac {1}{2} \plus{} \frac {1}{6} \plus{} \frac {1}{12} \plus{} \cdots\right) \minus{} \left(\frac {1}{6} \plus{} \frac {1}{12} \plus{} \cdots\right)$
$ \equal{} \frac {1}{2}$.

Hence, the summation turns out to be $ \frac {1}{2}$, so the answer is $ \frac {3}{2}\cdot\frac {1}{2} \equal{} \frac {3}{4}$.

Answer: $ \frac {3}{4}$[/hide]

[hide="#4"]Let us construct a quadratic that has $ x$ as a solution.

By quadratic formula, we have $ \frac { \minus{} b \plus{} \sqrt {b^2 \minus{} 4ac}}{2a} \equal{} \frac {3 \plus{} \sqrt {8}}{2}$.

Comparing LHS and RHS, we have $ a \equal{} 1, b \equal{} \minus{} 3, c \equal{} \frac {1}{4}$.

Therefore, the quadratic with $ x$ as a root is $ x^2 \minus{} 3x \plus{} \frac {1}{4} \equal{} 0 \cdots \alpha$.

Denote $ 4x^4 \minus{} 8x^3 \minus{} 15x^2 \plus{} 13x \equal{} a$.

$ a \minus{} \alpha\cdot 4x^2 \equal{} 4x^4 \minus{} 8x^3 \minus{} 15x^2 \plus{} 13x \minus{} (4x^4 \minus{} 12x^3 \plus{} x^2) \equal{} 4x^3 \minus{} 16x^2 \plus{} 13x \equal{} a$.

$ a \minus{} \alpha\cdot 4x \equal{} 4x^3 \minus{} 16x^2 \plus{} 13x \minus{} (4x^3 \minus{} 12x^2 \plus{} x) \equal{} \minus{} 4x^2 \plus{} 12x \equal{} a$.

$ a \minus{} \alpha\cdot ( \minus{} 4) \equal{} \minus{} 4x^2 \plus{} 12x \minus{} ( \minus{} 4x^2 \plus{} 12x \minus{} 1) \equal{} 1$.

Answer: 1[/hide]

[hide="#5"]Let the rest of the roots be $ \alpha \minus{} \beta, \alpha, \alpha \plus{} \beta$.

By Vieta's, we have $ 4 \equal{} 1 \plus{} 3\alpha\implies \alpha \equal{} 1$.

Hence, the four roots are $ 1, 1 \minus{} \beta, 1, 1 \plus{} \beta$.

By Vieta's again, we have $ \minus{} 3 \equal{} 1 \minus{} \beta \plus{} 1 \plus{} 1 \plus{} \beta \plus{} 1 \minus{} \beta \plus{} 1 \plus{} \beta \plus{} 1 \minus{} \beta^2\implies \beta \equal{} 3 \ \text{or} \minus{} 3$

Both $ 3$ and $ \minus{} 3$ works, because either way, $ {1, 1 \minus{} \beta, 1, 1 \plus{} \beta} \equal{} {1, \minus{} 2, 1, 4}$.

Answer: $ \minus{} 2, 1, 4$
[/hide]


All you have to do is find the password. It's 8 letters.

muidmuitnornegyxotscinesraallap

[color=white][size=9]slobmys[/size][/color]
Hint: I've already given you two hints; this sentence is the third hint.

PM me if you need hints, or post help requests here. No spoilers please.
I'm working on more riddles at the moment, and will do something like mustafa.
PM me the answer if you get it.


Let $ a,b,c>0$ such that $ ab\plus{}bc\plus{}ca\equal{}a\plus{}b\plus{}c$
prove that:
$ \frac{3}{ab\plus{}bc\plus{}ca\plus{}1}\plus{}\frac{1}{(1\plus{}a)^2}\plus{}\frac{1}{(1\plus{}b)^2}\plus{}\frac{1}{(1\plus{}c)^2} \ge \frac{2(a\plus{}b\plus{}c\plus{}3)}{(a\plus{}1)(b\plus{}1)(c\plus{}1)}$

[quote="devil_maths"]Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} a \plus{} b \plus{} c$
prove that:
$ \frac {3}{ab \plus{} bc \plus{} ca \plus{} 1} \plus{} \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \ge \frac {2(a \plus{} b \plus{} c \plus{} 3)}{(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)}$[/quote]
Setting $ x \equal{} \frac {1}{1 \plus{} a},$ $ y \equal{} \frac {1}{1 \plus{} b},$ $ z \equal{} \frac {1}{1 \plus{} c}.$ We have
$ \begin{aligned} \frac {3}{ab \plus{} bc \plus{} ca \plus{} 1} & \equal{} \frac {9}{3(ab \plus{} bc \plus{} ca) \plus{} 3} \equal{} \frac {9}{2(a \plus{} b \plus{} c) \plus{} ab \plus{} bc \plus{} ca \plus{} 3} \\
& \equal{} \frac {9}{(a \plus{} 1)(b \plus{} 1) \plus{} (b \plus{} 1)(c \plus{} 1) \plus{} (c \plus{} 1)(a \plus{} 1)} \equal{} \frac {9xyz}{x \plus{} y \plus{} z},\end{aligned}$
and
$ \begin{aligned} \frac {a \plus{} b \plus{} c \plus{} 3}{(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)} & \equal{} \frac {1}{(a \plus{} 1)(b \plus{} 1)} \plus{} \frac {1}{(b \plus{} 1)(c \plus{} 1)} \plus{} \frac {1}{(c \plus{} 1)(a \plus{} 1)} \\
& \equal{} xy \plus{} yz \plus{} zx.\end{aligned}$
Therefore, the original inequality is equivalent to
$ x^2 \plus{} y^2 \plus{} z^2 \plus{} \frac{9xyz}{x \plus{} y \plus{} z} \ge 2(xy \plus{} yz \plus{} zx),$
which is Schur's Inequality.


Even better:

$\lambda(1000) = 100$
$\lambda(100) = 20$
$\lambda(20) = 4$

http://mathworld.wolfram.com/CarmichaelFunction.html


vector
superior algebra
superior algebra unsolved

Can a group have precisely 3 automorphisms?

Let $ G$ be the group, and let $ Z$ be its center. Because $ G/Z$ is isomorphic to the group of inner automorphisms, the index of $ Z$ in $ G$ is at most $ 3$. This means that $ G/Z$ is cyclic, so $ G$ is in fact abelian. Now $ x\mapsto\minus{}x$ (I'm using additive notation) is an automorphism of order $ 1$ or $ 2$, and this order divides $ 3$. It follows that $ x\mapsto\minus{}x$ is actually the identity, so $ G$ is a vector space over $ \mathbb F_2$. It's easy to see now that its group of automorphisms is either infinite or has even order, so it certainly can't have order $ 3$.


How do you obtain MATHCOUNTS problems before 1999 (my coach has a few)?  The MATHCOUNTS store doesn't sell them.

You can't buy them from MATHCOUNTS, you have to get them from your coach or convince them to let us share them :P

I once sent an email to Mathcounts.  Presumably, they thought I was a coach. :huh:

[quote]Thank you for contacting MATHCOUNTS.  Unfortunately, we no longer make those competition sets available for purchase.  I would suggest you post your question on our Coaches Forum at:  [url]http://forum.infosrc.com:8080/~mc_coaches[/url].  It is possible a number of coaches would be willing to send you a copy they have.  I apologize for any inconvenience.  Please do not hesitate to contact us should you have any further questions.

Chris Bright
State Support Manager

> [i]<myemail>[/i] Sent 11/12/05 01:36AM
> Dear Sir,

> I would like to inquire on whether it is possible to obtain national
> competitions prior to 1999. The only way to obtain them at this time is
> through a coach that has been to a national competition.
>
> Thank you,
> [i]<myname>[/i]
[/quote]
Maybe you should show that to your coach or something. :)

I think Im gonna write a letter to MATHCOUNTS about it


inequalities
function
inequalities unsolved

It is a Jensen's inequality for concave function $f(x)=x(S-x)$.

The inequality also follows directly from Chebyshev's Inequality.

The sequences $\left\{x_1,x_2,...,x_n\right\}$ and $\left\{(S-x_1),(S-x_2),...,(S-x_n)\right\}$ are obviously oppositely sorted. Chebyshev then implies \[\sum_{i=1}^n x_i(S-x_i) \leq \frac{\sum_{i=1}^n x_i \cdot \sum_{i=1}^n (S-x_i)}{n} = \frac{S \cdot \sum_{i=1}^n (S-x_i)}{n} = \frac{S \cdot (nS-S)}{n} = \frac{S^2 \cdot (n-1)}{n} \]


inequalities
trigonometry
real analysis
real analysis unsolved

[quote="Ottem"]Show that for all $0\leq x,y \pi/4$,
\[\sqrt{2}e^{\pi/4}-1 \leq \frac{e^{y}(y \cos x+x \sin x)-y}{x-e^{y}(x\cos x-y\sin x)}\]
[/quote]
How do you think: is it hard? I think no :wink:


I honestly have no clue how I even got into MC if I have trouble with this, but I'm honestly feeling brain-dead about the current challenge problem I'm doing...

[i] (x^2)+4(y^2)-4x-8y+4=0
(x^2)+4y-4=0

Solve the system, using ONLY algebra. Show all work. [/i]

Have no clue what the answer's even supposed to be, because I lost the TI's conics program...so if someone could point out the god-awfully obvious thing I'm missing, that would be very greatly appreciated.

ok i dont think that this is obscenely obvious and i dont think you need to know conics to solve this either. although the first is i beleive the equation of a circle.  But this is kind of useless

[hide="hint!"]
Ok i hate systems of simultanious equations with exponents longer than 1!!! so ill use substitution which i like better.  Solve one of the equations for y in terms of x and then plug that value for y into all of the y's in the other equations.  Solve for x and plug that value of x into any equation and sove for y.  Now i havent done this yet but there might be two answers for both x and y.
[/hide]

...yeah...about two seconds after I posted this, I figured it out...I know one solution is (0,1), by graphing it by hand and solving the first equation for y-1, but is there a secondary one?


inequalities
symmetry
inequalities unsolved

let $a,b,c\in R\ $ , $ x,y,z\in R\ $
prove that this inequality
$ax+by+cz+\sqrt{(a^2+b^2+c^2)(x^2+y^2+c^2)}\ \geq\ \frac{2}{3}\ (x+y+z)(a+b+c)$

Easy I think
WLOG $a\leq b\leq c$ and $x\leq y\leq z$ then using chebyshev on RHS
\[ax+by+cz+\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}\geq 2(ax+by+cz)\]
which becomes
\[(a^2+b^2+c^2)(x^2+y^2+z^2)\geq (ax+by+cz)^2\]
which is true by cauchy-schwarz

I think that you can't suppose that $a \leq b \leq c$ and $x \leq y \leq z$.

How is it symmetry ? What is the criteria for it to be symmetry ?

I think that this expression $ax+by+cz$ itself cant be symmetry   :?

you have right. i have not seeit this before


MIT
college
geometry
geometric transformation
LaTeX

denkt jemand, da MikTeX? so EINFACH ist?  Ich brachte es gerade (fhren Sie Satz, mit Winedt 5,4) durch an und wei bereits alles im System!!!  Ich knnte leicht answerefragen, die Sie der pplversuch zum Fragen mich :D  \emptyset

Translation:
does someone think that MikTeX?  like that SIMPLE is?  I attached it straight (you lead sentence, with Winedt 5.4) through and know already everything in the system!!!  I could to-who-ask easily, you pplversuch for asking me

Wrong forum topic. How is mikTex a Superior Algebra problem? Hehe, Peter can help us out here  ;)
PS Edit: So, is this another April Fool's joke too?


function
search
real analysis
real analysis unsolved

Find all functions $f: [0,\infty)\to[0,\infty)$, differentiable at $x=1$ and satisfying \[ f(x^3)+f(x^2)+f(x)=x^3+x^2+x \] for all $x\geq0$.
[i]proposed by Mihai Piticari[/i]

Hint: reduce it to finding function derivables in 0 and such that $f(x)+f(2x)+f(3x)=0$. If you still cannot do it, search on the forum a post by killer, where the problem is solved.


function
logarithms
search
Putnam
number theory
algebra unsolved
algebra

[quote="ywl"]a function f: N--N sastifies the following.
i)f(mn)=f(m)f(n) for all integers m,n
ii)f(1)=1
iii) f is monotonically increasing

what is f?

*I think it may be f(n)=n^k[/quote]
Let $a_{i}=f(p_{i})$ were $p_{i}$ - i-th prime. It give for $n=\prod_{i}p_{i}^{k_{i}}$ $f(n)=a_{i}^{k_{i}}$. By consider $n_{1}=p_{i}^{k_{i}},n_{2}=p_{j}^{k_{j}}$ we have $\frac{\ln a_{i}}{\ln p_{i}}=\frac{\ln a_{j}}{\ln p_{j}}$ (else we get contradition with monotonically increasing). It give exist $k\in N$, suth that $a_{i}=p_{i}^{k}$ and $f(n)=n^{k}$.

[quote="Rust"]we have $\frac{\ln a_{i}}{\ln p_{i}}=\frac{\ln a_{j}}{\ln p_{j}}$ (else we get contradition with monotonically increasing). It give exist $k\in N$, suth that $a_{i}=p_{i}^{k}$ and $f(n)=n^{k}$.[/quote]
We have some $k\in\mathbb{R}$ with $f(n)=n^{k}$ for all $n\in\mathbb{N}$, but why is $k\in\mathbb{N}$?
This seems to be non trivial, and I have no idea how to do it.

[quote="Kent Merryfield"]This would be called a "totally multiplicative" function in number theory; as this result shows, they're not very interesting.

In number theory, the phrase "multiplicative functions" refers to something else: $f : \mathbb{N}\mapsto\mathbb{Z}$ such that $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are relatively prime.[/quote]
My solution work for case exist $N$, suth that $f(mn)=f(m)f(n)$ if $(m,n)=1, \ m,n>N$.

Is this it?

http://www.mathlinks.ro/Forum/viewtopic.php?p=346551#p346551

[quote="harazi"]This has already been solved by Pierre Bornsztein on the forum. Try to use the search function with multiplicative increasing functions. It is a theorem of Erdos. That should be enough in order to locate the topic.[/quote]
I assume you mean [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=functions+increasing+multiplicative&t=28867]this[/url].
But there it is shown, that $k\in\mathbb{R}$ exists, such that $f(n)=n^{k}$ for all $n\in\mathbb{N}$.
This was not so hard, and I did it myself some time ago.

My question now is this: If $k\in\mathbb{R}$ with $n^{k}\in\mathbb{N}$ for all $n\in\mathbb{N}$, then $k\in\mathbb{N}_{0}$.

This seems very hard, and could need some deep transcendence theory.
(I don't know much about transcendental numbers :maybe: )

[quote="harazi"]Hmm, that's funny, I think if you take some random person who knows some mathematics, that person will surely say that proving Erdos's theorem is by far far more challenging than proving this thing you are speaking about. No, it does not need deep transcendence theorems. It needs just the mean value theorem. Any, this was discussed too on the forum, it is a Putnam problem and the topic surely was opened by vess or me. I forgot, but use again the search function.[/quote]
Sorry, maybe you are right. It's just that I can't solve this problem (no idea at all), while the proof of Erdos's theorem is technical and certainly challenging but doable.
Couldn't you just show me, or give me a hint, how to do that with the mean value theorem? Then I could sleep much better. :oops:

http://www.kalva.demon.co.uk/putnam/putn71.html  Unfortunately, I could not find the discussion on the forum and since the solution is identical, please follow the link.

Very nice. Thank you very much. :)
I didn't believe this would be so elementary.
I was thinking of $2^\alpha,3^\alpha\in\mathbb{N}\Rightarrow \alpha\in\mathbb{N}$. :oops:


Find all $f: R\rightarrow R$ such that $f(1)=1$ and

$f(f(x)y+\frac{x}{y})=xyf(x^{2}+y^{2})$ for all $x,y$ such that $y\neq 0$.

Please continuou here http://www.mathlinks.ro/Forum/viewtopic.php?t=148711  :wink:


linear algebra
matrix
linear algebra unsolved

Let $ A,B$ be 2-by-2 matrices with integer entries, such that $ AB\equal{}BA$ and $ detB\equal{}1$.
Prove that  if $ det(A^3\plus{}B^3)\equal{}1$ then $ A^2\equal{}I$

I assume that your problem is correct. Let $ A,B$ be 2 matrices over the integers which satisfy the requested conditions (hypothesis and conclusion).
case 1: $ A$ is similar (over $ \mathbb{R}$) to $ \begin{pmatrix}1&0\\0&-1\end{pmatrix}$; $ B=\lambda{A}+\mu{I}$ with $ 2\lambda\in\mathbb{Z}$ and $ \lambda^2-\mu^2=1$. $ det(A^3+B^3)=1$ implies that $ -2\mu^3-6\lambda^{2}\mu=1$ or $ 8\mu^3+6\mu+1=0$. This equation admits one real and 2 complex roots; in each case $ 4(1+\mu^2)$ isn't an integer; thus no solution in this case.
case 2:  then for example . $ spectrum(B)=\{\lambda,1/\lambda\}$ with $ \lambda+1/\lambda\in\mathbb{Z}$. $ det(A^3+B^3)=1$ implies that $ \lambda^6+\lambda^3+1=0$  then (for example) $ \lambda^3=exp(2i\pi/3)$, $ \lambda=exp(2i\pi/9)$ and $ \lambda+1/\lambda=2cos(2\pi/9)$ isn't an integer. Thus no solution in this case.
case 3: $ A=-I$. Here $ \lambda^6-\lambda^3+1=0$ then (for example) $ \lambda^3=exp(5i\pi/3)$, $ \lambda=exp(5i\pi/9)$ and $ \lambda+1/\lambda=2cos(5\pi/9)$ isn't an integer. Thus no solution in this case.
Conclusion: $ A,B$ don't exist!!

The conclusion is perhaps $ A^2\equal{}0$ ???


number theory
least common multiple
search
greatest common divisor

i found this question while searching on the net for questions...

The sum of 6 natural numbers (not necessarily distinct) is 23. Let M denote the positive difference of the maximum and minimum of the LCM of these 6 numbers. Then total number of divisors that divide M^2 but doesn't divide M are
(a) 22 (b) 33 (c) 18 (d) 26 (e) none of the foregoing

wat approach should one usually follow to solve questions of type which include 5 to 6 natural numbers sum as above..

[quote="ramanujam"]wat approach should one usually follow to solve questions of type which include 5 to 6 natural numbers sum as above..[/quote]

It depends quite a lot on what the problem is asking for.

[hide="Solution"] We want to find the maximum of the LCM of the $6$ numbers.  Now, $1+2+3+4+5+6 = 21$, so to maximize we probably want a $7$ in there and we'd like to repeat the smallest number, so I'm fairly sure that the maximum is given by the sum $2+2+3+4+5+7$, which gives the LCM $3 \cdot 4 \cdot 5 \cdot 7 = 420$.  This is likely the maximum because adding any terms larger than $7$ wouldn't increase the LCM significantly unless the new term were prime (say, $11$), and that would restrict the other terms too much because it is too large.

The minimum of the LCM appears when the numbers are all almost the same.  Well, $4 \cdot 6 = 24$, so a possibility is the sum $4+4+4+4+4+3$, which has the quite small LCM of $12$.  But in fact we can do better with the sum $1+2+4+4+4+8$.  This is almost definitely the maximum because among $a, b, c, d, e, f$ at least one must be greater than or equal to $4$, and the terms can't all be equal to $4$.

Hence $M = 412 = 2^{2}\cdot 103, M^{2}= 2^{4}\cdot 103^{2}$.  While $M^{2}$ has $5 \cdot 3 = 15$ divisors, $M$ has only $3 \cdot 2 = 6$ divisors, so I believe the answer is $\boxed{ 9, E }$.

Note that the conclusion is not solid; I haven't had time to really prove anything.  But on a contest, this would be acceptable since I am quite confident of the answer and I only need the answer.  [/hide]

hi t0rajir0u

i also worked on the problem in the same manner as u did,got a diff answer  :(  n i only got the right answer from where i got the question..this is the right answer...

Minimum occurs when the numbers are 6, 6, 6, 2, 2, 1. Maximum occurs when the numbers are 7, 5, 4, 3, 2, 2.  To achieve min LCM we have to search for max number of GCDs among the 6C2 pairs. To look for max LCM, we need the numbers as co-prime.

Thus, M = 414. Now, 414 can be written as 2*3^2*23 => M has 2*3*2 = 12 divisors => M^2 has 3*5*3 = 45 divisors.

Hence, choice (b) is the correct option

i did not get the right answer,but m working on more ques like these..actually somehow these ques are extremely dicey...the max is still ok to analyse but minimum creates a big problem..

[quote="t0rajir0u"]Tricky.  Well, let's do some closer casework on the minimal case.

[hide="Discussion"] Since one of the terms must be at least $4$, the minimum must be at least $4$.

[b]Case:[/b]  The minimum is $4$.  Then we can only use $1, 2, 4$.  But $4+4+4+4+4+4 > 23$ and $4+4+4+4+4+2 < 23$, absurd.

[b]Case:[/b]  The minimum is $5$.  Then we can only use $1, 5$.  But $5+5+5+5+1+1 < 23$ and $5+5+5+5+1 > 23$, also absurd.

[b]Case:[/b]  The minimum is $6$.  Then we can only use $1, 2, 3, 6$.  In fact as the given solution states this case is feasible. [/hide][/quote]

cool :cool: this method is good..


limit
logarithms
algebra unsolved
algebra

I agree with pigfly: [b]i[/b] is trivial. We can easily show that $|<s_n>|=|<s_{n-2}>|+|<s_{n-4}>|$. It leads to a Fibonacci-like sequence and we immediately obtain result.

I think, part [b]ii[/b] can be done in the following way.
Assume $0<\alpha<\frac{1}{2}\ln\frac{1+\sqrt{5}}{2}$.
Using [b]i[/b] we can find $t_n=(0,6,0,6,...,0,6)$ s.t. $\frac{\ln|<t_n>|}{n+2}>\alpha$.
Note that $|<(t_n,\underbrace{0,0,...,0}_{k},t_n>|=|<t_n>|^2$ (or something like that :?), where $k\geq 2$.
So we will construct $s$ as $(t_n,0,0,..,0,t_n,0,0,..,0,t_n,0,...)$ choosing number of zeros in each group in such way that $\lim_{n\to + \infty}\frac{\ln (|<s_n>|)}{n}=\alpha$.


Prove that    $1\cdot2\cdot3\cdots 2002<\left(\frac{2003}{2}\right)^{2002}.$

A similar ptoblem was in a Putnam contest.  It was:
[i]Prove that[/i]
\[ n!<\left(\frac{n+1}{2}\right)^n \]
for all $n>1$.
Simple application of AM-GM.

Well, I used induction [and proved a stronger hypothesis]

Let's take

$1\times 2\times \ldots \times k < \left( \frac{k+1}{2} \right)^{k}$

wich is valid for $k\geq 2$

now we have

$(k+1)!<\frac{(k+1)^{k+1}}{2^k}$

We only have to show that

$\frac{(k+1)^{k+1}}{2^k}<\left(\frac{k+2}{2} \right)^{k+1}$

wich could be written as

$2<\left(1+\frac{1}{k+1}\right)^{k+1}$

We know that for large k, the right side approaches to e...and then it's true =]

I hope there's no mistake =]

You can't use the behaviour of an inequality as k goes to infinity as a proof that it is true for all k. What you're saying is: given that the inequality works for small k, it works for large k. However, small k basically means finite k, so you've made no progress. But it's a good idea and very interesting.

By the way, what was the stronger hypothesis?

well, i think it is quite plausible to see that

$\left(1+\frac{1}{n}\right)^{n}< e$

for natural values is function always increasing, so, testing for n=1, we have 2, the worst case, for us, because we want to show that is greater then 2. I didn't mean infinity, you took that by yourself, I told that for large n the inequality is obvious...so the solution is right =]

We will adjust the inequality as such:

$2^{2002}* 2002! < 2003^{2002}$.

By a one to one correspondence, if we multiply each two on the left hand side with some other term in the $2002!$, we should have 2002 terms on each side. Then, if we can show that each term on the right hand side is greater than each term on the left hand side, the inequality should work out. So, we have:

$(2*2*...*2)*(1*2*...*2002) =$

$(2^{1}0)*(2*2^{9})*(3*2^{9})*(4*2^{8})*...$

This way, we get rid of:

$\sum_{n=0}^{9}2^{n}(10-n)$ 2's on the left hand side by increasing each factor of the $2002!$ but without exceeding the  value of 2003. If we evaluate the sum, we get $2036$, which is greater than $2002$. Thus, we can still add another $34$ 2's on the left hand side without having the inequality be false.


Let's have a positive integer n and we want to express it as the sum of several (at least two) positive integers. In how many different ways can we accomplish this task?
(note: for n=5 there is no difference between 3+2 and 2+3)

This is the number of partitions of $ n$ less one.  See [url=http://www.research.att.com/~njas/sequences/A000065]A000065[/url] and [url=http://www.research.att.com/~njas/sequences/A000041]A000041[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

It's like putting bars between two 1's in the sequence 1111...111 (with $ n$ 1's). For example, $ 111|11|1111$ corresponds with 3+2+4. In any space, we have two choices: to put a bar or not to put a bar. But we must put at least one bar. There are $ n\minus{}1$ spaces, so there are $ 2^{n\minus{}1}\minus{}1$ ways.

[quote="Johan Gunardi"]It's like putting bars between two 1's in the sequence 1111...111 (with $ n$ 1's). For example, $ 111|11|1111$ corresponds with 3+2+4. In any space, we have two choices: to put a bar or not to put a bar. But we must put at least one bar. There are $ n \minus{} 1$ spaces, so there are $ 2^{n \minus{} 1} \minus{} 1$ ways.[/quote]

But $ 111|11|1111$ as well as $ 11|111|1111$ corresponds to $ 2\plus{}3\plus{}4$, but you're counting it more than once.


Can somone list out all the pythagorean triplets here. As far as i know there are only 5 sets is it?

3,4,5
5,12,13
7,24,25
8,15,17
9,40,41

My teacher told me this 5 sets were the list of triplets left by pythagoras but after his death ppl found out few more , what are the rest?

what i meant is not like 6,8,10 that is twice the valueof the first triplets but other one.

there are infinite triples

btw here's a link to a small triple program I made:
[url=http://lacascadaerb.com/world/index.php?down=rightTRIG.zip]Triple program[/url]

The set of primitive Pythagorean triples is given by $ (2mn, m^2 \minus{} n^2, m^2 \plus{} n^2)$, where $ m, n$ have opposite parity and $ \gcd(m, n) \equal{} 1$.  In other words, we can generate [b]arbitrarily large[/b] Pythagorean triples.  For example, letting $ m \equal{} 28, n \equal{} 17$ gives

$ (952, 495, 1073)$.

A list of some triples can be found [url=http://www.tsm-resources.com/alists/trip.html]here[/url].


I am a new middle school math teacher, teaching from pre-algebra to algebra II. Many of my students are interested in taking AMC 8 on 11/18/2008. Where can I get or buy those materials? Your suggestions and assistance will be greatly appreciated. Thanks.

You can buy materials from the AMC website [url]http://www.unl.edu/amc[/url].  For a recent AMC 8 CD that is apparently not yet on the site, see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1245963#1245963[/url].

You can contact our office by any of the means listed in the bottom stripe on our webpage:
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inequalities
combinatorics proposed
combinatorics

Sets A0, A1, A2, ..., Ak are subsets, each of size 2n, of a set of 4n elements. Prove that there can be found two of them whose intersection has at least  $(1-\frac{1}{k})n$elements.

[u][i]Solution:[/i][/u]

Suppose that $ A_i \subset X$ where $ \mid A_i \mid \equal{}2n$ and $ \mid X \mid \equal{}4n$ ,and $ F$ be a set of $ A_i$. For an element $ x$ its [i]degree[/i] $ d(x)$ in a set $ F$ is the number of members of $ F$containig $ x$. Suppose that the assertion is not true then $ \forall i,j : \mid A_i \cap A_j \mid <(1\minus{} \frac 1k)n$. We know
\[ \sum_{x \in X} d(x)\equal{}\sum_{A_i \in F} \mid A_i \mid \equal{}2(k\plus{}1)n.\]
On the other hand, we have
\[ \sum_{x \in X} d(x)^2\equal{}\sum_{i\equal{}0}^k \sum_{j\equal{}0}^k  \mid A_i \cap A_j \mid \equal{}\sum_{i} \mid A_i \mid \plus{}\sum_{i \not\equal{} j} \mid A_i \cap A_j \mid< 2(k\plus{}1)n\plus{} \binom {k\plus{}1}{2}(1\minus{} \frac 1k)n.\]
By using the couchy-shwartz inequality we have ;
\[ \sum_{x \in X} d(x)^2 \geq \frac {( \sum_{x \in X} d(x))^2}{4n}\equal{}n(k\plus{}1)^2.\]
Then $ n(k\plus{}1)^2<2(k\plus{}1)n\plus{} \binom {k\plus{}1}{2}(1\minus{} \frac 1k)n \Longrightarrow k\plus{}1<2\plus{}\frac {k\minus{}1}{2} \Longrightarrow k<1$ and it's contradiction.


calculus
integration
trigonometry
linear algebra
matrix
calculus computations

Show    :o
$ \int_{\minus{}\infty}^{\plus{}\infty}\frac{dx}{(e^x\minus{}x\plus{}1)^2\plus{}\pi^2}\equal{}\frac{1}{2}$.

it seems you would want $ (e^{x} \minus{} x \plus{} 1)^{2} \equal{} \tan^{2}(\theta)$

Multiply the identity:

$ \tan^2(\theta) \plus{} 1 \equal{} \sec^2(\theta)$

by $ \pi^2$ and you get:

$ \pi^2\tan^2(\theta) \plus{} \pi^2 \equal{} \pi^2\sec^2(\theta)$

Then do a trig sub and the proper limits. It's Über clean.

Define $ f(z)\equal{}\frac{1}{\big(e^{\alpha z}\minus{}\beta z\plus{}1\big)^2\plus{}\big(\frac{\beta\pi}{\alpha}\big)^2}$ with $ \alpha,\beta>0$.

$ f$ has a pole iff $ e^{\alpha z}\minus{}\beta z\plus{}1\equal{}\pm \frac{i\beta \pi}{\alpha}$ and

$ \frac{1}{\beta} e^{\alpha z}\equal{}\minus{}\frac{1}{\beta}\pm \frac{i \pi}{\alpha}\plus{}z\equal{}: \mu \qquad \Big(z\equal{}\frac{1}{\beta}\mp \frac{i\pi}{\alpha}\plus{}\mu\Big)$

is equivalent to $ \frac{1}{\beta} e^{\frac{\alpha}{\beta}\mp i\pi\plus{}\alpha\mu}\equal{}\mu \iff \frac{\alpha}{\beta} e^{\frac{\alpha}{\beta}}\equal{}\minus{}\alpha\mu\, e^{\minus{}\alpha\mu}$

So $ \alpha \mu$ must have the form $ \minus{}W_k\left(\frac{\alpha}{\beta} \, e^{\frac{\alpha}{\beta}}\right)$

whereat $ k\in\Bbb{Z}$ stands for the branch number of the Lambert W-function.

Each pole of $ f$ can be written as $ z_k^{\pm}\equal{}\frac{1}{\beta}\mp \frac{i\pi}{\alpha}\minus{}\frac{1}{\alpha} W_k\left(\frac{\alpha}{\beta} \, e^{\frac{\alpha}{\beta}}\right)$ .

$ \forall t>0 \quad \text{Im}\left[W_k(t)\right]$ is $ \left\{\begin{matrix} > & (2k\minus{}1)\pi\ge \pi & \text{for} & k>0 \\ \\ < & (2k\plus{}1)\pi\le \minus{}\pi & \text{for} &  k<0 \end{matrix} \right.$

$ \Rightarrow \text{Im} \left(z_k^\pm\right)\equal{}\frac{1}{\alpha} \text{Im}\left[\mp i\pi\minus{}W_k\left(\frac{\alpha}{\beta} \, e^{\frac{\alpha}{\beta}}\right)\right]$ is $ \left\{ \begin{matrix} <0 & \text{for} & k>0 \\ \\ >0 & \text{for} & k<0 \end{matrix} \right.$

$ 2\pi i\, \text{res} (f,z_k^\pm)\equal{}\frac{2\pi i}{2 \, \underbrace{\left(e^{\alpha z_k^\pm}\minus{}\beta z_k^{\pm}\plus{}1\right)}_{\pm \frac{i\beta \pi}{\alpha}}\, \Big(\underbrace{\alpha\, e^{\alpha z_k^\pm}}_{\minus{}\beta \, W_k\left(\frac{\alpha}{\beta}\, e^{\frac{\alpha}{\beta}}\right)}\minus{}\beta\Big)}$

$ \minus{}\beta \, W_k\left(\frac{\alpha}{\beta} e^{\frac{\alpha}{\beta}}\right)\equal{}\alpha e^{\alpha z_k^\pm}$ because $ W_k\left(\frac{\alpha}{\beta}\, e^{\frac{\alpha}{\beta}}\right)\, e^{W_k\left(\frac{\alpha}{\beta}\, e^{\frac{\alpha}{\beta}}\right) }\equal{}\frac{\alpha}{\beta}\, e^{\frac{\alpha}{\beta}}$

$ \Rightarrow \minus{}\beta \, W_k\left(\frac{\alpha}{\beta}\, e^{\frac{\alpha}{\beta}}\right)\equal{}\minus{}\alpha \, e^{\frac{\alpha}{\beta}\minus{}W_k\left(\frac{\alpha}{\beta}\, e^{\frac{\alpha}{\beta}}\right) }\equal{}\alpha e^{\alpha \left(\frac{1}{\beta}\mp \frac{i\pi}{\alpha}\minus{}\frac{1}{\alpha} W_k\left(\frac{\alpha}{\beta}\, e^{\frac{\alpha}{\beta}}\right) \right)}$

Therefore $ 2\pi i\, \text{res} (f,z_k^\pm)\equal{}\mp \frac{1}{\frac{\beta}{\alpha} \left(\beta\, W_k\left(\frac{\alpha}{\beta}\, e^{\frac{\alpha}{\beta}}\right)\plus{}\beta\right) }$

All residues in the upper half plain cancel apart from $ \text{res} (f,z_0^\minus{})$ .

So $ \int_{\minus{}\infty}^\infty \frac{1}{\big(e^{\alpha z}\minus{}\beta z\plus{}1\big)^2\plus{}\big(\frac{\beta\pi}{\alpha}\big)^2}\, dz\equal{}\oint_{\begin{matrix} _\text{upper} \\ ^\text{semi\minus{}circle} \end{matrix}} f \, dz$

$ \equal{}2\pi i\, \text{res} (f,z_0^\minus{})\equal{}\frac{1}{\frac{\beta}{\alpha} \left(\beta \, W_0\left(\frac{\alpha}{\beta}\, e^{\frac{\alpha}{\beta}}\right)\plus{}\beta\right) }\equal{}\frac{\alpha}{\beta\, (\alpha\plus{}\beta)}$


Let a,b,c are length sides of triangel and x,y,z are real numbers such that $ ax \plus{} by \plus{} cz \equal{} 0$.
Prove that $ xy \plus{} yz \plus{} zx \leq 0$ :D

hi !!!!!!!
we supose  that a>b>c and x>y>z
then we have by Chebechev :
ax+by+cz=0 >=1/3(a+b+c)(x+y+z)
and we have ab,c>0 so x+y+z<=0
is it correct  :oops: ?

Sorry! I have edited the inequality.
 :D

easy problem  :D
we have $ ax\plus{}by\plus{}cz\equal{}0\Rightarrow x\equal{}\frac{\minus{}by\minus{}cz}{a}$
Then $ xy\plus{}yz\plus{}zx\leq 0\Leftrightarrow (y\plus{}z)(by\plus{}cz)\minus{}ayz\geq 0$
$ f(y)\equal{}by^2\plus{}z(c\plus{}b\minus{}a)y\plus{}cz^2$
with $ delta\equal{}z^2(a^2\plus{}b^2\plus{}c^2\minus{}2ab\minus{}2bc\minus{}2ca)\leq 0\Rightarrow f(y)\geq 0$
so $ xy\plus{}yz\plus{}zx\leq 0$


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate \[\int_{0}^{\frac{\pi}{4}}\left(\frac{\cos x}{\sin x+\cos x}\right)^{2}\ dx,\ \ \ \int_{0}^{\frac{\pi}{4}}\left(\frac{\sin x+\cos x}{\cos x}\right)^{2}\ dx.\]

[quote="kunny"]$\int_{0}^{\frac{\pi}{4}}\left(\frac{\sin x+\cos x}{\cos x}\right)^{2}\ dx$[/quote]
i first thought the two integrals were multiplied  :wallbash_red:
ill do the easier one:
[hide]
\begin{eqnarray*}\int_{0}^{\frac{\pi}{4}}\left(\frac{\sin x+\cos x}{\cos x}\right)^{2}\ dx &=&\int_{0}^{\frac{\pi}{4}}\frac{1+2\sin x \cos x}{\cos^{2}{x}}\ dx\\ &=&\int_{0}^{\frac{\pi}{4}}\sec^{2}{x}+2\tan{x}\ dx\\ &=&\tan{x}-2\ln(\cos{x})|^{\frac{\pi}{4}}_{0}\\ &=&1+\ln(2) \end{eqnarray*}
[/hide]

EDIT:
[quote="kunny"]$\int_{0}^{\frac{\pi}{4}}\left(\frac{\cos x}{\sin x+\cos x}\right)^{2}\ dx$[/quote]
[hide]
a friend gave some hints..  :lol:
we make the clever substitution $\sin{x}+\cos{x}=\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)$ and $\cos x=\cos\left(-\frac{\pi}{4}+\frac{\pi}{4}+x\right)=\frac{\sqrt{2}}{2}\left(\sin\left(\frac{\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}+x\right)\right)$
 and we proceed with the same method as for the other one:
\begin{eqnarray*}\int_{0}^{\frac{\pi}{4}}\left(\frac{\cos x}{\sin x+\cos x}\right)^{2}\ dx&=&\int_{0}^{\frac{\pi}{4}}\left(\frac{\frac{\sqrt{2}}{2}\left(\sin\left(\frac{\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}+x\right)\right)}{\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)}\right)^{2}\ dx\\ &=&\frac{1}{4}\int_{0}^{\frac{\pi}{4}}\left(\frac{\sin\left(\frac{\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}+x\right)}{\sin\left(\frac{\pi}{4}+x\right)}\right)^{2}\ dx\\ &=&\frac{1}{4}\int_{0}^{\frac{\pi}{4}}\csc^{2}{\left(\frac{\pi}{4}+x\right)}+2\cot{\left(\frac{\pi}{4}+x\right)}\ dx\\ &=&\frac{1}{4}\left(-\cot{\left(\frac{\pi}{4}+x\right)}+2\ln\left(\sin\left(\frac{\pi}{4}+x\right)\right)|^{\frac{\pi}{4}}_{0}\right)\\ &=&\frac{1}{4}+\frac{\ln 2}{4}\end{eqnarray*}
[/hide]

$\boxed{\ I=\int_{0}^{\frac{\pi}{4}}\left(\frac{\cos x}{\sin x+\cos x}\right)^{2}\ dx\ \ ,\ \ J=\int_{0}^{\frac{\pi}{4}}\left(\frac{\sin x+\cos x}{\cos x}\right)^{2}\ dx\Longrightarrow J=4\cdot I\ }$

[color=darkblue][b]Proof.[/b] I like this proposed problem, Kunny ! It is easy , but amazingly. I"ll use the substitution[/color] $t=\phi (x)=\frac{\pi}{4}-x\ :$
$J=2\int_{0}^{\frac{\pi}{4}}\frac{1+\sin 2x}{1+\cos 2x}\ dx=$ $2\int_{0}^{\frac{\pi}{4}}\frac{1+\cos 2x}{1+\sin 2x}\ dx=$ $4\int_{0}^{\frac{\pi}{4}}\frac{\cos^{2}x}{(\sin x+\cos x)^{2}}\ dx$ $\Longrightarrow \boxed{\ J=4\cdot I\ }\ .$

[color=darkblue][b]Another similar exercises[/b] from the my book [/color] [color=darkred][b][u]"Probleme de Analiza matematica, Ed. ALL EDUCATIONAL, Bucuresti, 2002".[/u][/b][/color]

$1.\blacktriangleright\ \int\frac{x^{2}+x}{(e^{x}+x+1)^{2}}\ dx=x-\ln (1+x+e^{x})+\frac{x+1}{1+x+e^{x}}+C\ .$
$2.\blacktriangleright\ \int_{0}^{\frac{\pi}{2}}\left(\frac{x}{x\sin x+\cos x}\right)^{2}\ dx=\frac{2}{\pi}\ .$
$3.\blacktriangleright\ \int_{0}^{2n\pi}\max\{\sin x\ ,\ \arcsin (\sin x)\}\ dx=\frac{n(\pi^{2}-8)}{4}\ .$
$4.\blacktriangleright\ \int_{0}^{\pi}x\left[\sin^{2}(\sin x)+\cos^{2}(\cos x)\right]\ dx=\frac{\pi^{2}}{2}\ .$
$5.\blacktriangleright\ \int_{0}^{\pi}x\sin(\cos^{2}x)\cos(\sin^{2}x)\ dx=\frac{\pi^{2}\sin 1}{4}\ .$
$6.\blacktriangleright\ a_{n}=\left(\int_{a}^{b}(x-a)^{n}(x-b)^{n}\right)^{\frac{1}{n}}\ dx\rightarrow\left(\frac{b-a}{2}\right)^{2}$, where $a<b\ .$
$7.\blacktriangleright\ a_{n}=\int_{0}^{1}\frac{1}{x^{n}+x+1}\ dx\rightarrow \ln 2\ ;\ b_{n}=\int_{0}^{1}\frac{1}{x^{2}-x+1}\ dx\rightarrow\infty\ .$
$8.\blacktriangleright\ x_{n+1}=\int_{0}^{1}\max\{x_{n},1-x\}\ dx\rightarrow 1\ .$
$9.\blacktriangleright\ \int_{0}^{\frac{\pi}{4}}\ln (1+\tan x)\ dx=\int_{0}^{1}\frac{\arctan x}{x+1}\ dx=\int_{0}^{1}\frac{\ln (x+1)}{x^{2}+1}\ dx=\frac{\pi\ln 2}{8}\ .$

[color=darkblue]I"ll solve all after few days ![/color]

Yes. This is the background of this problem.

Thank you for offering us problems, Virgil Nicula.

kunny

[quote="Virgil Nicula"][color=darkblue][b]Another similar exercises[/b] from the my book [/color] [color=darkred][b][u]"Probleme de Analiza matematica, Ed. ALL EDUCATIONAL, Bucuresti, 2002".[/u][/b][/color] $2.\blacktriangleright\ I=\int_{0}^{\frac{\pi}{2}}\left(\frac{x}{x\sin x+\cos x}\right)^{2}\ dx=\frac{2}{\pi}\ .$[/quote]$F\equiv\int \frac{x^{2}}{(x\sin x+\cos x)^{2}}\ dx=$ $\int\frac{x^{2}(\sin^{2}x+\cos^{2}x)+x\sin x\cos x-x\sin x\cos x}{(x\sin x+\cos x)^{2}}\ dx=\int\frac{x\sin x(x\sin x+\cos x )-x\cos x(\sin x-x\cos x )}{(x\sin x+\cos x)^{2}}\ dx=$
$\int\frac{x\sin x}{x\sin x+\cos x}\ dx-J$, where $J=\int \frac{(\sin x-x\cos x)(x\sin x+\cos x)'}{(x\sin x+\cos x)^{2}}\ dx\ .$ $\boxed{\ \begin{array}{ccc}u(x)=\sin x-x\cos x & \Longrightarrow & u'(x)=x\sin x\\ v'(x)=\frac{(x\sin x+\cos x)'}{(x\sin x+\cos x)^{2}}& \Longrightarrow & v(x)=-\frac{1}{x\sin x+\cos x}\end{array}\ }$ $\Longrightarrow$
$J=-\frac{\sin x-x\cos x}{x\sin x+\cos x}+\int \frac{x\sin x}{x\sin x+\cos x}\ dx$ $\Longrightarrow$ $\boxed{\ F=\frac{\sin x-x\cos x}{x\sin x+\cos x}+C\ }$ $\Longrightarrow$ $I=F\left(\frac{\pi}{2}\right)-F(0)$ $\Longrightarrow$ $I=\frac{2}{\pi}\ .$


I am homeschooled and I am going. What about you?

I know a group a couple of groups of homeschoolers in Florida that competed as a team.  There was a kid from one group who made top twenty...I think...not really sure though.  :?


According to [url]http://en.wikipedia.org/wiki/Floor_function[/url]

The floor function can be defined as

$ \lfloor x \rfloor = \textrm{max}\{n \in\mathbf{Z}| n \leq x \}$

Now I understand the concept behind floor and ceiling functions but this definition seems to make no sense. If I read it to myself, I get:

"The floor of x equals the maximum integer that divides itself less than or equal to x"

whereas I would intuitively think that it is "the floor of x equals the maximum integer less than or equal to x". This leads to 2 questions.

1) What is the purpose of using the $ |$ symbol, which usually means "divides". ex/ $ 3 | 6$ means "3 divides 6".

2) Why are the $ \{ \}$ delimiters used for the max? Usually, $ \{ \}$ denote the fractional part function, but I don't think the fractional part function is being used to define the floor function.

Can anyone help me with these notation issues?

The $ |$ should be read as "such that," not divides.

The $ \{$ and $ \}$ mean that there is a set of numbers that fit the condition, and we want the maximum one (hence, the max).

So for instance, we would have

$ \lfloor 2.3 \rfloor = \textrm{max}\{ \cdots,-3,-2,-1, 0, 1, 2 \}$

Therefore, the $ \{ \}$ indicate a set of solutions (correct me if I'm wrong).

So then, how do you differentiate between the two uses of $ |$ ?

Yes, the $ \{\}$ indicate a set of numbers.

Whenever $ |$ is used in a set, it means "such that," but when ever it is used in the form $ a|b$, then it means divides.

I guess it is just kind of intuitive as to which is meant (as you said, $ |$ clearly does not mean $ |$ in the given example) and look at the context.

Ok, final question.

Would this be an equivalent expression?

$ \lfloor x \rfloor = \textrm{max}\{n \in \mathbf{Z}\}\; \forall \; n \leq x$

On an unrelated note, is this correct? $ \forall a \in \mathbb{Z^{+}}\hspace{5pt}\land \hspace{5pt}\forall \{b,c\}\in \mathbb{R}$

Means "for all $ a$ belonging to the positive integers (including zero) and for all $ b$ and $ c$ belonging to all real numbers"

I had always thought that $ \mathbb{Z}^{+}$ meant positive integers, so 0 would not be included, but I could be mistaken.

Other than that, it's fine, but there I would not recommend writing up solutions that way because they are much less readable.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Show that the free group on $2$ elements is not solvable.

Assume that it is. Then we can exhibit a normal series $0=G_0\lhd G_1\lhd\ldots\lhd G_n=G$ where each $G_{i+1}/G_i$ is abelian (and non-trivial).

$G_1$ is an abelian free group, so it must be cyclic. Since $G_2/G_1$ is abelian, we have $[G_2,G_2]\le G_1$, so $[G_2,G_2]$ is cyclic, being a subgroup of a cyclic group. Now, we know from a problem previously posted here that the derived subgroup of a free group on more than one generator is not finitely generated. Since $G_2$ must be such a group (free but not cyclic), the fact that $[G_2,G_2]$ is cyclic contradicts the mentioned problem (which is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56412]this one[/url], by the way).

Hi Grobber,

I haven't been through your proof of the other problem yet but this doesn't seem elementary, that's why I think there must be some other easier solution somehow (especially considering that the problem was given in a prelim exam). Unfortunately I have no knowlegde about free groups to shed more light now.

Hi,

Eventually, I got something on another forum.
No free group (except that on one element) is solvable since you can find a surjection from that group onto $S_{5}$, which is not solvable (!!).

Why is there a surjection from the group onto $S_{5}$?  Because $S_{5}$ is finitely generated?

Take the surjective homomorphism uniquely defined by $f(a)=(1,\dots,n)$ and $f(b)=(1,2)$. Then the factor group $F_{2}/f^{-1}(S_{5})$ must be solvable, but it is isomorphic to $S_{5}$, which is not solvable, a contradiction. The proof is adaptable to any free group (except the one on one element of course). For a proof of the surjectivity see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=52287]www.mathlinks.ro/Forum/viewtopic.php?t=52287[/url]


linear algebra
matrix
combinatorics open
combinatorics

Given a simple graph $ G$ with adjacency matrix $ \mathbf{A}$, let $ G^n$ denote the graph with adjacency matrix $ \mathbf{A}^n$; in other words, $ (u, v)$ is an edge of $ G^n$ if and only if there is a path of length $ n$ from $ u$ to $ v$ in $ G$.  (Note that this is not the usual definition of a graph power, which involves paths of length less than or equal to $ n$.)

Find concise necessary and sufficient conditions on $ G$ such that $ G^n$ is connected for all $ n \ge 1$.

Well, $ G$ must clearly be connected itself.  Bipartiteness is then sufficient, since $ G^n$ will be disconnected for even $ n$, and in fact I'm fairly confident that this is necessary: if the graph has an odd cycle, I don't have any trouble convincing myself that I can march over to the odd cycle, spend some time walking around it (perhaps walking up and down the same edge many times if $ n$ is odd, or looping around repeatedly if $ n$ is even) and then march away from the odd cycle to my target vertex.  Actually writing this down formally is a little less pleasant, but not hugely horribly so -- it reduces to showing that for any integers $ m, n, a$ and any two vertices on a $ 2m \plus{} 1$-cycle, there is a path between these vertices whose length is congruent to $ a$ modulo $ n$.


yes we do specifically [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=32510]this[/url] guy :)

i request u guys to play more, i want my fellow desis involved in FTW!!!!!!!!!!!!!!

naangallam valkaiyulai munneranoomunu pakkarom unnakku yen da intha veen pratchanai ellariyum padathullendhu thavaravekkaruthu ?
P.S: I left that stupid passtime  :D

yes i did after the religious order  :P   :rotfl:


group theory
abstract algebra
inequalities
superior algebra
superior algebra unsolved

Let $ p \in N$ be a prime number and $ G$ a group with $ 2p^{2}$ elements.If $ H_{1}$ and $ H_{2}$ are subgroups of $ G$ with the property $ H_{1} \cap H_{2}\equal{}\{e\}$ and no matter how we chose $ p$ elements from $ G$ , at least $ 2$ of them are in $ H_{1} \cup H_{2}$,then prove that $ H_{1}\equal{}G$ or $ H_{2}\equal{}G$.

I'll write $ H$ and $ K$ instead of using subscripts.

We must have $ |H\cup K|\geq 2p^2\minus{}p\plus{}2$, otherwise choose $ p\minus{}1$ elements not in their union and the identity to get a contradiction. As $ |H\cap K|\equal{}1$, we obtain both of the following inequalities:
\[ |H|\plus{}|K|\equal{}|H\cup K|\plus{}|H\cap K|\geq 2p^2\minus{}p\plus{}3,|H\parallel{}K|\equal{}|HK\parallel{}H\cap K|\leq 2p^2\]
In particular, $ (|H|\minus{}1)(|K|\minus{}1)\leq p\minus{}2$. If neither $ H$ nor $ K$ is trivial, then $ |H|,|K|\leq p\minus{}1$ so that $ |H|\plus{}|K|\leq 2p\minus{}2<2p^2\minus{}p\plus{}3$, a contradiction.

So either $ H$ or $ K$ is trivial. WLOG let $ K$ be trivial, so $ |H|\geq 2p^2\minus{}p\plus{}2>p^2$ implying $ H\equal{}G$, as desired.


quadratics
calculus
integration
real analysis
number theory unsolved
number theory

Let $ 0 \equal{} s_1 < s_2 < ...$ be the set of integers of the form $ x^2\plus{}y^2$, for integral $ (x,y)$. Show that there exist arbitrarily large gaps between successive terms of $ s_n$, i.e. $ limsup_{n\minus{}>\infty} (s_n \minus{} s_{n\minus{}1}) \equal{} \plus{}\infty$

A positive integer x can be expressed as sum of two squares if and only if in its factorization $ \prod{p_i^{a_i}}$ all primes =3 mod 4 have even exponents, and it follows in a straghforward way by quadratic residues. So, if a prime q such that 4|q+1 and q||y then y is not expressibile as sum of two squares. But take q_1,q_2,....,q_k arbitrary primes =3 mod 4 and let z the solution of the system x+i=q_i mod (q_i ^2), and by TCR we have the solution. []
bye

What is TCR? Can you explain please? Is it Chinese Remainder Theorem? Thank you

Oh yes it mean CTR Chinese Theorem Remainder..
I hope it is clear
See you  :wink:


You probably have all played with spirographs before; you know, the little gears that spin around eachother? Anyway, here is a little math problem for you.

Let the number of teeth on the fixed spirograph be R, and let the number of teeth on the rotating spirograph be Q.

1.) Write an equation which relates the number of loops made by the gears to the number of teeth before it comes back to its starting point, using R and Q.

2.) Write an equation which shows how many times the rotating wheel will revolve around the fixed wheel before it comes back to its starting point, using R and Q.

3.) Write an equation that shows how many times the rotating wheel rotates around its axis before it comes back to its starting point, using R and Q.

Enjoy.


$ 1\blacktriangleright$ Let $ \{a,b,c\}\subset (1,\infty )$ so that $ c>\max\{\log_ab , \log_b a\}$.

Solve the equation : $ a^xb^{\frac 1x}+a^{\frac 1x}b^x=a^cb^{\frac 1c}+a^{\frac 1c}b^c\ .$

[b]Particular case.[/b] $ a=4$, $ b=9$, $ c=2$.

$ 2\blacktriangleright$ Solve the system (over $ \mathcal R$) : $ \{\begin{array}{c}
x^5+2y=3z^3\\\
y^5+2z=3x^3\\\
z^5+2x=3y^3\end{array}$.

May it be useful?
http://www.mathlinks.ro/Forum/viewtopic.php?t=163713

$ 2\blacktriangleright$ Solve the system (over $ \mathcal R$) : $ \{\begin{array}{c} x^5 + 2y = 3z^3 \\
\ y^5 + 2z = 3x^3 \\
\ z^5 + 2x = 3y^3\end{array}$.[/quote]
Use  method :$ x = \max\{x,y,z\}$ and consider to case:
[u]Case1:[/u]$ y\geq z$
[u]Case 2: [/u]$ z>y$


Suppose countrary

WLOG, we can assume $ (a,b)\equal{}1$ $ \rightarrow$ $ (a\minus{}b,a\plus{}b)\equal{}1$ or $ (a\minus{}b,a\plus{}b)\equal{}2$

If      $ (a\minus{}b,a\plus{}b)\equal{}1$    we have:
$ (a\minus{}b)|a^2\plus{}b^2$ $ \rightarrow$ $ (a\minus{}b)|2ab$

$ (a\plus{}b)|a^2\plus{}b^2$ $ \rightarrow$ $ (a\plus{}b)|2ab$

So   $ (a\plus{}b)(a\minus{}b)|2ab$, and with  respect to this fact that $ (a^2\minus{}b^2)|{(a\plus{}b)}^2\minus{}2ab$

  $ \rightarrow$  $ (a^2\minus{}b^2)|{(a\plus{}b)}^2$ $ \rightarrow$ $ (a\minus{}b)|(a\plus{}b)$ that is contradiction($ (a\minus{}b,a\plus{}b)\equal{}1$)

If   $ (a\minus{}b,a\plus{}b)\equal{}2$  with the same method way  we have

 $ (a\minus{}b)|2(a\plus{}b)$  $ \rightarrow$ $ (a\minus{}b)\equal{}2$ or $ (a\minus{}b)\equal{}4$

 $ \rightarrow$ $ a,b$ are odds

1)if $ a\equal{}b\plus{}4$ $ \rightarrow$ $ (4b\plus{}8)|b^2$ so $ b$ is even ,that is contradiction

2)If $ a\equal{}b\plus{}2$ $ \rightarrow$ $ 2b\plus{}4|b^2$ so $ b$ is even,that is contradiction


A balloon is ascending vertically up at 14m/s when a packet is drpped from it.The packet takes 6.1sec to reach the ground.What height was the packet dropped at?(g=10m/s^2)

[color=blue]
$\large y=h+v_0t-\frac{gt^2}{2}$

The packet hits the ground

$y=0\Longrightarrow h=\frac{gt^2}{2}-v_0t$[/color]

I agree.

"h" might be expressed as h sub 0, to be consistent with v sub 0, as they are both at the same time.

hello,wizard,haven't seen you for quite sometime :) well i don't really get what is written there but all i know is that we have to consider that the packets moves up a bit after it is dropped to bring its velocity to 0.(inertia)


Show by induction that $ n^2\plus{}n>\equal{}42$ when $ n>\equal{}6$ and $ n<\equal{}\minus{}7$

Reckon I've got it for $ n>\equal{}6$ but not sure. $ n<\equal{}\minus{}7$ is confusing as doesn't proof by induction rely on n being in the the set of natural numbers?

no, induction does not.  what we can do is show that if the statement is true for $ n$ then it is true for $ n \minus{} 1$ instead of the more commonplace $ n \plus{} 1$.

$ ( \minus{} 7)^2 \plus{} ( \minus{} 7) \equal{} 42$ so the statement is true for $ n \equal{} \minus{} 7$
Suppose $ n^2 \plus{} n \ge 42$ for some integer $ n \le \minus{} 7$.

Now $ (n \minus{} 1)^2 \plus{} (n \minus{} 1) \equal{} n^2 \minus{} n \ge n^2 \plus{} n \ge 42$ since $ n < 0$

So the statement $ n^2 \plus{} n \ge 42$ if $ n \le \minus{} 7$ is true for all INTEGER $ n$.

I really don't think that induction is necessary to prove this though.  Additionally, other proofs show that $ n$ need only be real.

For example
$ n^2 \plus{} n \minus{} 42 \ge 0$
$ (n \plus{} 7)(n \minus{} 6) \ge 0$

Which is clearly true if $ n \equal{} 6$ or $ n \equal{} \minus{} 7$.

The LHS is only positive if both factors are negative or both factors are positive.  The first condition occurs when $ n < \minus{} 7$; the second, when $ n > 6$.

Another alternate proof:
$ n^2 \plus{} n \plus{} 1/4 \ge 42 \plus{} 1/4$
$ (n \plus{} 1/2)^2 \ge 169/4$
$ n \plus{} 1/2 \ge 13/2$ or $ n \plus{} 1/2 \le \minus{} 13/2$
$ n \ge 6$ or $ n \le \minus{} 7$


[color=darkblue]Slove this equation:[/color] $ 5cos^{12}x\plus{}\frac{1}{46656}\equal{}cos^{10}x$

[quote="Ramanujan"][hide="hint"]$ 46656 \equal{} 6^{6}$ and now we can write $ \frac {1}{46656} \equal{} \frac {1}{6^{5}} \minus{} \frac {5}{6^{6}}$[/hide][/quote]

You give me the full of solution. Please!

We set $ \cos^{2}x \equal{} w$ and since $ \frac {1}{46656} \equal{} \frac {1}{6^{5}} \minus{} \frac {5}{6^{6}}$

we can rewrite the equation as follows :  $ 5w^{6} \minus{} \frac {5}{6^{6}} \minus{} \left( w^{5} \minus{} \frac {1}{6^{5}} \right) \equal{} 0$

But

$ 5w^{6} \minus{} \frac {5}{6^{6}} \minus{} \left( w^{5} \minus{} \frac {1}{6^{5}} \right) \equal{} \\
 \\
\left( w \minus{} \frac {1}{6} \right) \cdot \left(5w^{5} \minus{} \frac {1}{6}w^{4} \minus{} \frac {1}{6^{2}}w^{3} \minus{} \frac {1}{6^{3}}w^{2} \minus{} \frac {1}{6^{4}}w \minus{} \frac {1}{6^{5}} \right) \equal{} \\
 \\
\left( w \minus{} \frac {1}{6} \right)^{2} \cdot \underbrace{\left( 5w^{4} \plus{} \frac {4}{6}w^{3} \plus{} \frac {3}{6^{2}}w^{2} \plus{} \frac {2}{6^{3}}w \plus{} \frac {1}{6^{4}} \right)}_{p(w)} \equal{} \\
 \\
\left( w \minus{} \frac {1}{6} \right)^{2} \cdot p(w)$.

We have $ w \equal{} \cos^{2}x \geq 0$ therefore $ p(w) > 0$ and the equation

$ \left( w \minus{} \frac {1}{6} \right)^{2} \cdot p(w) \equal{} 0$ gives as solution $ w \equal{} \frac {1}{6}$

To solve the original equation we have to solve the equation $ \cos^{2}x \equal{} \frac {1}{6}$

The solutions are

 $ \left\{\arccos\left(\frac {1}{\sqrt {6}}\right), \minus{} \arccos\left(\frac {1}{\sqrt {6}}\right), \arccos\left( \minus{} \frac {1}{\sqrt {6}}\right), \minus{} \arccos\left( \minus{} \frac {1}{\sqrt {6}}\right) \right\}$

I cant check now my calculations, Sorry if something is wrong.  :)

[color=darkblue][u][b]Directive[/b][/u]

We use AG-AM inequality.[/color]


[quote="CdsdiverZX"]
1.Solution of -3x = -18
2.Standard form for (-1)8
3.Value of -1 / 3 x (-1 / 2)
4.Solution of 8x = -6
5.Value of 1 + (-2 + 4)5[/quote]
[hide="1"]
$-3x=-18$
Both sides are divisble by $-3$, so we have
$\frac{-3x}{-3}=\frac{-18}{-3}$
$x=6$
4 is very similar[/hide]
[hide="5"]$1+5(-2+4)=1+5(2)=1+10=11$[/hide]
hope that helps. :)

This should help with #3.
when multiplying two fractions, you multiply the numerators to get the new numerator and multiply the denominators to get the new denominator.
So, the answer to $\frac{-1}{2}\times\frac{-1}{3}$ is $\frac{-1^{2}}{2\times3}=\frac{1}{6}$

[hide="hi"]click [url=http://purplemath.com/modules/solvelin.htm]here[/url]
good explaination of linear equations[/hide]

[hide="2"] I'm not sure what you mean by standard form but,
$8(-1)=-8$[/hide]

[quote="CdsdiverZX"]I have to complete the following 5 problems but I want to understand and get the answer.

1.Solution of -3x = -18
2.Standard form for (-1)8
3.Value of -1 / 3 x (-1 / 2)
4.Solution of 8x = -6
5.Value of 1 + (-2 + 4)5

If someone could help me and or discuss this with me and show me how to get the answers I would be very grateful.I tried searching google but didn't find anything on these particular problems.

Thanks,

Cdsdiver[/quote]
[hide]
1. divide -18 by -3 = 6
2. multiply = -8
3. is that an x or times? if it is times it is 1/6
4. divide -6 by 8= -3/4
5. 1+ -10 +20=21-10=11[/hide]


Functional Analysis
linear algebra
linear algebra unsolved

Let $T$ be bounded linear operator on a Hilbet space $H$,and assume that $||T^n||\leq 1$ for some  natural number $n$.prove the existance of an invertible linear operator $A$ on $H$ such that  $||ATA^{-1}||\leq 1$.

Assume that $(x,y)$ is a scalar product of the Hilbert space $H$, and we have a bounded operator $T: H \rightarrow H$, such
that $|| T^{n} || \leqslant 1$. Define new scalar product $(x,y)_{1} : = (x,y) + (Tx,Ty) + (T^{2}x,T^{2}y) + \ldots + (T^{n-1}x,T^{n-1}y)$, for $x,y \in H$. Then $(x,y)_{1}$ is a
scalar product and we have that $(x,x) \leqslant (x,x)_{1} \leqslant C (x,x)$ for $C: = 1 + ||T||^{2} + ||T^{2}||^{2} + \ldots + ||T^{n-1}||^{2}$. Now, because of this and of the fact
that the space $(H,(\cdot,\cdot))$ is complete, we have that the
space $(H,(\cdot,\cdot)_{1})$ is complete, hence it is a Hilbert
space. Now, the spaces $(H,(\cdot,\cdot)), (H,(\cdot,\cdot)_{1})$ are isometric. It is clear when $H$ is finite-dimensional or
separable because if we have a dense countable subset of $(H,(\cdot,\cdot))$, then because of $(x,x) \leqslant (x,x)_{1} \leqslant C (x,x)$, this set is also dense in $(H,(\cdot,\cdot)_{1})$. In the general case one can also show
that they are isometric. So we have some isometry $A : (H,(\cdot,\cdot)) \rightarrow (H,(\cdot,\cdot)_{1})$, so $||Ax||_{1} = ||x||$ for every $x \in H$. Then because of $(x,x) \leqslant (x,x)_{1} \leqslant C (x,x)$, we have that $A$
is bounded and it's inverse is also bounded, when we consider it
as an operator $A : (H,(\cdot,\cdot)) \rightarrow (H,(\cdot,\cdot))$. On the other hand, if $x \in H$, then $(Tx,Tx)_{1} = (Tx,Tx) + (T^{2}x,T^{2}x) + \ldots + (T^{n-1}x,T^{n-1}x) + (T^{n}x,T^{n}x) \leqslant (Tx,Tx) + (T^{2}x,T^{2}x) + \ldots + (T^{n-1}x,T^{n-1}x) + (x,x) = (x,x)_{1}$. Therefore $||Tx||_{1} \leqslant ||x||_{1}$ for every $x \in H$. Now, if we have some $x \in H$, such that $||x|| \leqslant 1$, then $||Ax||_{1} = ||x|| \leqslant 1$, then $|| TAx ||_{1} \leqslant 1$, so $|| A^{-1}TA x || = || TAx ||_{1} \leqslant 1$. Therefore we have that $|| A^{-1}TA || \leqslant 1$.

I hope it is correct.