Can you solve the above question?

Have A Great Day!

I think the answer is:

$ \sum_{k\equal{}1}^{9n}\binom{3k\plus{}n\minus{}1}{n\minus{}1}$

[quote="binomial_4eva"]my answer: $ \infty$[/quote]
 :huh:
I think $ n$ is a given natural number... :maybe:

You're Persain?  :o  Please write in English so that everyone understands.  :wink:

Consider the following generating function:
\[ f(x) \equal{} (1 \plus{} x \plus{} \hdots \plus{} x^9)^n.
\]
Notice that the coefficients of the numbers give the number of n-digit integers composed of a specific set of digits.  The number of n-digit sequences that lead to numbers divisible by three, $ S_n$, is the sum of all coefficients of those $ x^k$ such that $ 3|k$.  This is clearly
\[ S_n \equal{} \frac {f(1) \plus{} f(\varepsilon) \plus{} f(\varepsilon^2)}{3} \equal{} \frac {10^n \plus{} 2}{3},
\]
where $ \varepsilon \equal{} e^{i\frac {2\pi}{3}}$.  However, this number counts the n-digit sequences of numbers that begin with 0 as an n-digit number.  Thus, We need to subtract $ S_{n \minus{} 1} \equal{} \frac {10^{n \minus{} 1} \plus{} 2}{3}$.  Equivalently, we have $ S_n \minus{} S_{n \minus{} 1} \equal{} \frac {10^n \minus{} 10^{n \minus{} 1}}{3} \equal{} 3\cdot10^{n \minus{} 1}$ as the required answer. $ \Box$

[u]Remark.[/u]  Notice that we can simply find the number of n-digit numbers, $ 10^n \minus{} 10^{n \minus{} 1}$ and divide them by 3 to get the desired answer.  However, the above gives a method that is useful in many cases.

EDIT: I think that the problem is poorly stated.  Please state it in a clear way.

I have another approach:

let $ S \equal{} \overline{x_1x_2\ldots x_n}$ be a n-digits number divisible by 3,so we must have:

$ x_1 \plus{} x_2 \plus{} \ldots \plus{} x_n \equal{} 3k$

where $ 1\leq k\leq 3n$,now note that $ S$ is a n-digits number,so we must have $ x_1\geq 1$ and $ x_i\geq 0$ for $ i \equal{} 2,\ldots ,n$.

now solving this equation gives us:

$ \sum_{k \equal{} 1}^{3n}\binom{3k \plus{} n \minus{} 2}{n \minus{} 1}$

is something wrong with this solution? :maybe:

[quote="boxedexe"]I believe that you cannot use the formula for the number of solutions with nonnegative $ x_i,\ i \equal{} 2,\hdots,n$ since you want the sum to be divisible by $ 3$.  you assume that the sum is disivible by three before you make the correct substitutions.  This is derived from my undestanding of the derivation for the formula for the number of nonnegative solutions of the sum being considered.  I hope that this satisfies your needs even though it is poorly written.  :oops:[/quote]
yeah you're probably right,cause it already fails for n=2.


Just what are the procees involved in the communication between a '[b] satellite[/b] Disc[b] '[/b] and '[b] another[/b] satellite[b] disc[/b]?[b] '[/b]
Aside from technical[b]/[/b] mechanical[b] work[/b], is there any mathematical[b] operations[/b] involved? What specific kind[b] of[/b] mathematics[b] is used so as to make such satellite disc work?

                                                        Respectfully,
                                                      A.Eddington[/b]

See, for example, the keywords [url=http://www.amazon.com/Satellite-Communication-Engineering-Signal-Processing/dp/082470777X]here[/url].  I don't know much about the subject, but a safe bet is that the mathematical techniques of [url=http://en.wikipedia.org/wiki/Signal_processing]signal processing[/url] are required ([url=http://en.wikipedia.org/wiki/Fourier_analysis#Applications_in_signal_processing]Fourier analysis[/url] and so forth).


conics
ellipse
geometry unsolved
geometry

i think i could solve this problem with Hire Theorem

i take perp to l1 from F1 and i have P1 point of large circle of ellipse, i take perp to l2 from F1 and i have P2 point of the large circle of ellipse so i need 3 third point P3 of large circle of ellipse to construct the circle and have the center of the ellipse. [size=150][b]How can i construct a P3 point of large circle from the point P of the ellipse? [/b][/size]  :(

The angle between the line through F1 and the point of intersection of the tangents(P) with one tangent is igual to the angle between the line F2P with the other tangent

[quote="jrrbc"][size=150][b]How can i construct a P3 point of large circle from the point P of the ellipse? [/b][/size]  :([/quote]
Polarity WRT any circle centered at $F_1$ takes the desired ellipse $\mathcal E$ to a circle $(E).$ That is, tangents of $\mathcal E$ go to points of $(E)$ and points of $\mathcal E$ go to tangents of $(E).$ To save construction of one polar, let $F_1P$ be radius of the polar circle $(F_1)$ $\Longrightarrow$ $p \perp F_1P$ at $P$ is polar of $P$ WRT $(F_1).$ Construct poles $T_1, T_2$ of the lines $t_1, t_2.$ WRT $(F_1),$ then the 2 circles $(E), (E')$ through $T_1, T_2$ tangent to $p.$ Inverting $(E), (E')$ in $(F_1)$ yields the pedal circles $(C), (C')$ of 2 possible ellipses $\mathcal E, \mathcal E'$ with given focus $F_1,$ point $P$ and tangents $t_1, t_2.$


conics
parabola
calculus
integration
analytic geometry
calculus computations

Find the centroid of the region bounded by the parabola $ y^{2}= x$ and the line $ y = x-2$ using integrals

and also..

Find the centroid of the region bounded by the curves y = square-root of x and $ y = x^{2}$

thanks  :D

The second one should be: $ x$-coordinate: $ \frac{\int_{0}^{1}x(\sqrt{x}-x^{2})dx}{\int_{0}^{1}(\sqrt{x}-x^{2})dx}=\frac{3/20}{1/3}=9/20$ and the $ y$-coordinate is $ 9/20$ due to symmetry.


Let a, b, and c be positive integers such that c+ba=30; and a+bc=18
How many values of the expression b+ca are possible?

this is just a sketch of the solution . first of all ,  usisng that $ a\plus{}bc\equal{}18$, you can limit the possibilities for b and c.using the other equation , yiu can limit it even more. Than just test the cases. (this is very sloppy. sorry).

[hide]$ a\plus{}bc\equal{}18, c\plus{}ba\equal{}30$
We must have $ b\leq17$.
Adding gives $ (a\plus{}c)(b\plus{}1)\equal{}48 \implies b\equal{}2,3,5,7,11,15$. We proceed by casework.
$ 1) b\equal{}15$ gives $ a\plus{}15c\equal{}18, c\plus{}15a\equal{}30$ which clearly has no solutions.
$ 2)b\equal{}11$ has no solutions either.
$ 3)b\equal{}7, a\plus{}c\equal{}6$ gives $ c\plus{}7a\equal{}30, a\plus{}7c\equal{}18 \implies (a,c)\equal{}(4,2)$
$ 4)b\equal{}5$ doesn't have integer solutions
$ 5)b\equal{}3, a\plus{}c\equal{}12$ gives $ c\plus{}3a\equal{}30, a\plus{}3c\equal{}18 \implies (a,c)\equal{}(9,3)$
$ 6)b\equal{}2, a\plus{}c\equal{}16$ gives $ c\plus{}2a\equal{}30, a\plus{}2c\equal{}18 \implies (a,c)\equal{}(14,2)$

Hence there are two possible values of $ b\plus{}ca$, namely $ 15$ and $ 30$.[/hide]

[hide="how I did it"]First I subtracted the second equation from the first.

$ ab \plus{} c \minus{} a \minus{} bc \equal{} 12 \implies a(b \minus{} 1) \minus{} c(b\minus{}1) \equal{} 12 \implies (b\minus{}1)(a\minus{}c) \equal{} 12$

Since a, b, and c are integers b - 1 and a - c are integers as well.  There are 6 pairs of numbers b-1 and a-c can be.  Those would be (1, 12), (2,6), (3,4), (4,3), (6,2), (12,1).

Next I added the two equations together getting $ a\plus{}c\plus{}ab\plus{}bc \equal{} 48 \implies (a\plus{}c) \plus{} b(a\plus{}c) \equal{} 48 \implies (a\plus{}c)(1\plus{}b) \equal{} 48$

Out of the 6 pairs we originally had we can test values of b in $ (a\plus{}c)(1\plus{}b) \equal{} 48$ to eliminate some pairs since a+c must be an integer.

Only b=2, b=3, b=5, and b=7 work.

b=2:
$ a\plus{}c \equal{} 16$
$ a\minus{}c \equal{} 12$
Adding the equations together we find that $ 2a \equal{} 28 \implies a \equal{} 14$.  This also means c = 2.

ac + b would be 28 + 2 = 30 in this case.

b=3:
$ a \plus{} c \equal{} 12$
$ a \minus{} c \equal{} 6$

We find that a = 9 and c = 3.  ac + b would equal 30 again in this case.

b=5:
$ a \plus{} c \equal{} 8$
$ a\minus{}c \equal{} 3$
2a = 11 so a = 11/2.  But this solution does not work since a and c must be integers.

b=7:
$ a\plus{} c \equal{} 6$
$ a \minus{} c \equal{} 2$
a = 4 and c = 2.  ac + b is 15.

Thus the only possible values of ac + b are 15 and 30.

This seemed kind of long writing it out, but it wasn't as long when I actually did it.  :P [/hide]


calculus
derivative
function
algebra
polynomial
calculus computations

Given the following information about a function f:

$x = 3.0, f(x) = 10 \\ \\ x = 3.5, f(x) = 8 \\ x = 4.0, f(x) =7 \\ x = 4.5, f(x) =4 \\ x = 5.0, f(x) =2 \\ x = 5.5, f(x) = 0 \\ x = 6.0, f(x) =-1$

Estimate the derivative of$f'$ at $x = 2$

Do you mean the derivative $f'$ or the derivative [b]of[/b] $f'$? Seeing that the function is almost affine, I'll go with $f'(2)\approx-4$ and $f''(2)\approx 0$. No calculations to back this up, sorry. I suppose you could write down a Lagrange polynomial, but I would not use it for [b]extrapolation[/b].


In Gazeta Matematica nr $\ 5$ din $\ 2002$, domnul profesor Gheorghe Szollosy propune urmatoarea problema:
  [i]Problema[/i]
   Fie $\ AA',BB',CC'$ ceviene concurente in triunghiul $\ ABC$, punctele $\ A',B',C'$ aflandu-se pe laturile $\ (BC),(CA),(AB)$.Daca notam masurile unghiurilor $\ CAA',ABB',BCC'$ cu $\ x,y,z$, demonstrati inegalitatea $\sin x\cdot\sin y\cdot\sin z\leq\frac{1}{8}$.Cand are loc egalitatea?

Prima oara cand am citit problema m-am gandit ca:

sinx * siny * sin z <= sin A/2 * sin B/2 * sin C/2

Iar:

Pentru a demonstra ca sin A/2 * sin B/2 * sin C/2 <= 1/8 folosim inegalitatea:

sin A/2 <= a/ b+c, apoi inegalitatea devenind imediata.

Nu am stat sa verific daka este corecta abordarea, dar cum am zis a fost primu lucru la care m-am gandit :)

notam cu $a_{i}$ unghiurile $x,y,z$ si cu $b_{i}$ celealte trei unghiuri formate. ca sa explic mai bine, in $A$ vom avea doua unghiuri : $a_{1}$ si $b_{1}$. $a_{1}$ este $x$, si $b_{1}$ este celalalt. etc... (astfel incat sa iasa teorema lu ceva frumos :) )
deci $\prod \sin a_{i}=\prod \sin b_{i}$. deci :
$\prod \sin a_{i}=\prod \sqrt{\sin a_{i}\sin b_{i}}$, iar
$\sqrt{\sin a_{i}\sin b_{i}}\leq\frac{\sin a_{i}+\sin b_{i}}{2}$.
apoi $\sin a_{i}+\sin b_{i}=2\sin\Bigg(\frac{a_{i}+b_{i}}{2}\Bigg)\cos\Bigg(\frac{a_{i}-b_{i}}{2}\Bigg)$. cum $a_{i}+b_{i}=A$, si cosinusul este $\leq 1$, obtinem ca $\sqrt{\sin a_{i}\sin b_{i}}\leq\sin A/2$, etc...

[quote="pohoatza"]Prima oara cand am citit problema m-am gandit ca:

sinx * siny * sin z <= sin A/2 * sin B/2 * sin C/2

Iar:

Pentru a demonstra ca sin A/2 * sin B/2 * sin C/2 <= 1/8 folosim inegalitatea:

[b]sin A/2 <= a/ b+c, apoi inegalitatea devenind imediata[/b].

Nu am stat sa verific daka este corecta abordarea, dar cum am zis a fost primu lucru la care m-am gandit :)[/quote]
   Da, asta e Inegalitatea lui Jean Balieu si e triviala demonstratia ei.Si eu m-am gandit la acelasi lucru la inceput, dar solutia lui Lucian e corecta si frumoasa.De fapt, el a folosit Teorema lui Ceva in forma trigonomterica, ce se deduce usor din Teorema Bisectoarei Generalizata. :lol:


[quote="Inequalities Master"]Prove that for any positive reals a,b,c the following inequality holds:
(4ab+bc+ca)/(a^2+b^2)+(ab+4bc+ca)/(b^2+c^2)+(ab+bc+4ac)/(c^2+a^2) is greater or equal than 4.[/quote]
$ \sum_{cyc}\frac{4ab\plus{}ac\plus{}bc}{a^2\plus{}b^2}\geq4\Leftrightarrow\sum_{cyc}ab(a\minus{}b)^4\plus{}\sum_{sym}\left(2a^4bc\plus{}6a^3b^2c\minus{}\frac{4a^2b^2c^2}{3}\right)\geq0,$
which obviously true.


First, arrange the cards in the following order:

RK, B6, BK, R8, R4, BA, BQ, BQ, R5, RQ, B7, B3, RK, RJ, B4, BJ, R6, R2, B10, R3, R10, B5, R9, B2, B9, B8, RA, R7, BA, R2, RJ, R3, R10, B10, R6, B2, R9, R4, B9, B8, RA, B3, R7, B5, BK, B6, R8, B7, R5, RQ, BJ, B4.

When putting the cards in order, you must place the red king down *face up* and then the black six on top of that (face up), and so on. When you're done you should see the black four. Next, play solitaire. I guarantee that you will never have won a game of solitaire with such ease in your life.

Yeah, I'm a dork. :D

I've won more easily:

Have someone else play the game up to the last card. Then put the last card in its respective place. Easy.

Can you [b]prove[/b] that that's the easiest way to win?

[quote="mathnerd314"]Can you [b]prove[/b] that that's the easiest way to win?[/quote]That would be a very silly thing to try to do.

[quote="noneoftheabove"]technically u could still not win like not finish the game or mess up the order[/quote]This is what I get for posting this on a math forum.  :D

of course but i dont think anyone else would care if it wasnt on a math forum unless it was like a card game forum then they would say what a stupid idea that would be because it would defeat the purpose of the game but i am not saying that am i  :?

[quote="AntonioMainenti"][quote="math92"]wow, you must have a lot of time on your hands. :D[/quote]Well it took me about 15 minutes to figure out the arrangement, and another 5 to post in on here. What do you have to show for from the last 20 minutes?  :P[/quote]

well...i found an SOJ, an IK helm, a Tal's belt, and a Thundergod's Vigor in 20 min before  :D

and wow...that's crazy lol
but i prefer hearts  :P

[quote="mwahahahaha"]
well...i found an SOJ, an IK helm, a Tal's belt, and a Thundergod's Vigor in 20 min before  :D
[/quote]

 :o Lucky! Do you still play D2? :P

hearts is more fun but playing in real life is way better

yeah spades and bridge but there is also president or vc which gets old though


Let $a_{n}$ be the  order term n of Sequence of Fibonacci. Prove that $a_{n}< (\frac{1+\sqrt{5}}{2})^{n}$ , for $n\ge2$

There is a stronger inequality.It says that  A^(n-2)<a_n<A^(n-1) Where A=[(1+sqr5)/2]^n
Iwill prove that a_n<A^(n-1).We suppose that the enequality is true for all k<=n .Because A^(n-2)+A^(n-1)=A^n =>A^n>a_(n-1)+a_n=a_(n+1)
so we proove it and for k=n+1 so it is true.Same we can proove that
A^(n-2)<a_n.    (sorry for LATEX :wink: )


For $ x > 1$, find the minimum value of $ 2x \plus{} \frac {1}{x^2}$.

By [b][i]AM-GM[/i][/b] we have
$ 2x\plus{}\frac{1}{x^2}\equal{}x\plus{}x\plus{}\frac{1}{x^2}\geq 3\sqrt[3]{x*x*\frac{1}{x^2}}\equal{}3$.
Why do u post so easy problems kunny? Any hidden messages here? :P

If $ x>1$, the minimum of $ f(x) \equal{} 2x\plus{}1/x^2$ doesn't exist (because $ f(x)>f(y)$ $ \forall x>y>1$).

LOL, your problems are always a bit [i]tricky[/i]! :P  :lol:


AMC
AIME
function
number theory
relatively prime

Let $S$ be the sum of all numbers of the form $a/b$, where $a$ and $b$ are relatively prime positive divisors of 1000.  What is the greatest integer that does not exceed $S/10$?

A solution (in spoiler of course)

[hide]The first thing you should notice is that 1000 = 2^3 * 5^3. Thus, m and n is some combination of 5^j and 2^k that remain separate (because of relative primeness) unless m or n = 1.

So, if you are familiar with generating functions, we should find some expression to GENERATE all these solutions. The most obvious one is this:

(1/8 + 1/4 + 1/2 + 1 + 2 + 4 + 8)(1/125 + 1/25 + ... + 125).

As you can see, the expansion of this covers all our possibilities, because it generates a combination of all powers of two and five between powers of -3 and 3. This, in fact, is our sum S.

Now add the individual sums in decimal form (fractions will get messy fast)

So S = (15.875)(156.248) :sim: 2480

So our answer is 248

[/hide]


Hello, welcome to the DMAT Round 2. If you're new, take a look at the pdf for all about the rules, prizes, problems, points, etc.

The goals of the competition:

(1) Create a non-middle math tournament geared for AIME preparation (and even some basic USAMO if possible), while also allowing people to have fun with math over the summer (and maybe into the school year).

(2) Hone problem creating and solving skills.

Here is the tentative schedule. The round has 10 problems for a total of 50 points. Get enough correct answers or points and you may win medals and ranks (see the pdf).

7/19: Easy problem (2 pts)
7/20: Word problem (4-6 pts)
7/21: Moderate problem (3-5 pts)
7/22: Moderate problem (3-5 pts)
7/23: Proof! (6-7 pts)
7/26: Word problem (4-6 pts)
7/27: Moderate problem (3-5 pts)
7/28: Moderate problem (4-6 pts)
7/29: Harder problem (5-7 pts)
7/30: Insane problem! (7 pts)

Please let gfour84 or me know if you want to participate (if you do, we expect you to try at least two problems!). I will primarily host this competition, as I am going away for vacation for all of August (may drop by the forums occasionally). A potential round 3 will be up to gfour84 and the success of this competition.

Do we have to sign up if we were in Round 1? If yes, I sign up.

Hey!  I don't want to be left out  :( I SIGN UP

though I'm new to this

Currently, 6(7?) people have signed up.

Note: You don't have to sign up, but you can only earn points from the round, not any of the 4 medals or the following awards:

n00b award: Don't get a single question correct. (If that happened, then your score on the AIME might be incredibly low).
1337 award: Score the highest out of all participants.
Haxxor award: Come up with an incredibly far-fetched but correct way to do a problem.

Maybe people are answering the questions incorrectly just so they can be dubbed noob.

The participant would have to consistently gives a plausible solution that just HAPPENS to give a wrong answer each time though.

My teacher at practice if you miss a really easy question will scowl and make a funny noise.  Sometimes I turn in a wrong answer to see that (you never know).

[quote="professordad"]I HAVE A QUESTION.

Where will the questions be posted up?  In the High School Intermediate Topics section?[/quote]

EDITED RESPONSE:
Since this thread gets repeatedly bumped down in the Intermediate Topics Forum, a link will be provided to the thread in the Middle School Tournament Forum. Round 2will begin sometime around Sunday with a trivial math question, as always.

Currently, there are 8 (9?) people signed up:
Brut3Forc3
Bugi
ghjk
mathwizarddude
narik
professordad
SnowEverywhere
sunehra
cyu2012?

P.S. Feel free to check out and solve the problems in Round 1 right now. You might want some easy points (well, depending on your definition of easy).

We will be opening up a new thread on the Intermediate forum sometime tomorrow. The questions will start posting there.

Note: if you do not specify, all people who submit solutions to #1 will be treated as Round 2 participants. Signups for earning medals will close after Question 1 submissions are closed, as the medals earned is based on the 10 questions.

When that thread opens up, please be sure to tell us you are committed and that you will participate.

Hello again! gfour84 and I have mostly determined the first four questions for DMAT Round 2, which will err on the easier side (equivalent to AIME #1-6) and are working on the proof question (which would be an AIME #10 probably). The questions will appear on the Middle School Tournament Forum, where they won't be bumped down.

The current sign-ups include the following. You are expected to be at least somewhat committed (there may be dire consequences if you do not submit any answers!):
AIME15
Brut3Forc3
Bugi
ghjk
mathwizarddude
narik
professordad
SnowEverywhere
sunehra
cyu2012(?)

The final participant sheet will consist of all people who submit solutions to question 2-A, worth three-points. It is set to be released tomorrow night.

Oh why not, I'll join. Need to practice my AIME skills.

Hello all, the problems will now start posting [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=1597]here[/url]. According to Sunehra, now it won't be pushed down.


geometry
circumcircle
parallelogram
geometric transformation
reflection
ratio
geometry proposed

H is the orthocenter of triangle ABC. The external bisector of angle C meets the circumcircle (ACH),(BCH) at D,E resp. Prove that three midpoints of DE,CH,AB are collinear.

$ (O)$ is circumcircle of the $ \triangle ABC$. $ A', K, M$ are midpoints of $ AB, CH, DE.$ $ X \in AB$ is foot of the internal bisector $ CX$ of the angle $ \angle C.$ $ CK \equal{} \frac {CH}{2} \equal{} OM$ and $ CK \parallel OM$ $ \Longrightarrow$ $ A'KCO$ is a parallelogram and $ A'K \parallel OC.$ The circumcircles $ (P), (Q)$ of the $ \triangle AHC, \triangle BHC$ are reflections of $ (O)$ in $ CA, BC$ $ \Longrightarrow$ they are congruent. Powers of $ M$ to $ (P), (Q)$ are in the constant ratio

$ \frac {\overline{MC} \cdot \overline{MD}}{\overline{MC} \cdot \overline{ME}} \equal{} \frac {\overline{MD}}{\overline{ME}} \equal{} \minus{} 1$

$ \Longrightarrow$ $ M$ is on a circle coaxal with $ (P), (Q).$ Since $ (P) \cong (Q),$ this circle is tangent to the parallels to $ PQ$ through $ C, H$ $ \Longrightarrow$ it is the circle $ (K)$ with diameter $ CH.$ Let $ MN$ be its diameter through $ K.$ Since $ MC \perp CX,$ it follows that $ N \in CX.$ $ CO, CH$ are isogonals with respect to the angle $ \angle A,$ therefore

$ \angle CNM \equal{} \angle CNK \equal{} \angle KCN \equal{} \angle HCN \equal{} \angle NCO$

and $ KM \parallel OC$ as well. Consequently, $ A', K, M$ are collinear.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $n\geq 3$ be an integer and $S_{n}$ the permutation group. $G$ is a subgroup of $S_{n}$, generated by $n-2$ transpositions. For all $k\in\{1,2,\ldots,n\}$, denote by $S(k)$ the set $\{\sigma(k) \ : \ \sigma\in G\}$.

Show that for any $k$, $|S(k)|\leq n-1$.

no solutions ?  :(
i enjoyed this problem in the contest because it was something different, and i think it`s a very nice problem.
[hide="my solution"]
consider the graph $T$ with $n$ vertices and $n-2$ edges, such that there exists edge $(i,j)$ if and only if the transposition $(i,j)$ is one of the $n-2$ transpositions from the hypothesis.
i claim now that the set $S(k)$ is the set of vertices $x$ such that there is a path from $k$ to $x$. this is easy to see, because any $\sigma \in G$ can be written as a finite composition of transpositions. now, if $\sigma(k)=x$, then there are transpositions that will map succesively $k \rightarrow k_{1}\rightarrow k_{2}\ldots \rightarrow x$, and that is the path in the graph. on the other side, i do the same thing in the other way, because a path can be seen as a composition of transpositions.
now, because $T$ has $n-2$ edges, is not connected (the minimal connected graph is a tree, that has $n-1$ edges), and now the answer is obvious.
[/hide]

I am sorry to revive this very old topic.

However, I am interested whether this problem can be solved using group actions. In particular, letting $G$ act on the set $\{1,2,...,n\}$, we have to show that the orbits of all elements have a length of most $n-1$.


AMC
AIME
search
USA(J)MO
USAMO
email
AMC 10

Perhaps this sounds like a stupid question, but how exactly does one go about acquiring their "official" AIME score? A brief search on the website didn't yield much besides a recommendation of sending in a letter/email and allow up to two weeks for it to process. Is there an alternative way that's simpler/faster, or would I have to just send the letter and wait?

(Yes, I know that the AIME score probably doesn't matter anymore since the USAMO is over, but it's just confusing to qualify for the USAMO and not even know what my AIME score is, since my proctor did not know what I received.)

Please do respond if you know the answer; thank you very much!

The contest manager at your school should receive an official score report...
But, if not, contacting the AMC will definitely get it to you (though it may take awhile)

Always check with the Contest Manager who registered for the AMC 10 or AMC 12 that you took that qualified you for the AIME.  We send an email report with  the AIME I results and the AIME II immediately after the AIME II.  We follow up with a postal mailed report.  Both go to the person who registered.

There is absolutely no reason that the Contest Manager should not have both of these reports and be able to tell you your results.

Work with your teacher or Contest Manager first, they DO have the results.  It's right there in your home town, in your school, just ask for it.

Steve Dunbar
MAA Director, American Mathematics Competitions


analytic geometry
quadratics
MATHCOUNTS
geometry
arithmetic series
Pythagorean Theorem

I tried this yesterday, and it was a little annoying and fun.

Try to prove the Diagonal Formula

$\frac{n(n-3)}{2}$

[hide="proof"]
For a n-gon, one point has n-3 diagonals (it can connect to all but itself and the two adjacent points). The next point has n-4 diagonals, a the way to the nth point which has 0. This forms an arithmetic series n-3, n-4, n-5, ... 1,0. Applying the formula sum=$\frac{\text{number of terms}}{2} \times$ (first term+last term). So we have $\frac{n}{2} \times (n-3)=\frac{n(n-3)}{2}$. So we have proven the theorem.[/hide]

[hide]Each point can connect to n-3 other points (can't connect to itself or the two adjacent points). That gives us n(n-3). Since we counted each diagonal twice, we must divide by two. [/hide]

[hide]
For a point in an n-gon, there is $n-3$ diagonals from that point.
The next one would have $n-4$, and so on to the one with 0 diagonals.
So, you do $n(n-3)$, but you have to divide by 2, because you counted each diagonal twice.
$\frac{n(n-3)}{2}$[/hide]

what I did is first put them as coordinates and using quadratic formulas and stuff then simplifing, I got the formula

I think what he meant was figuring how many diagonals there are in an n-gon.

Which is, n(n-3)/2

[quote="cj"][hide="my way (no annoying at all)"] C(n,2)-n. That is it.[/hide][/quote]

A little explanation to go with that?  :?

Oh. I just figured out that C(n,2) meant ${n\choose2}$. Now it's obvious.  :P

What do you mean by $a^2+b^2=c^2$?  That is the pythagorean theorem, which doesn't have anything to do with diagonals unless you put it on the coordinate plane...

[quote="Treething"]I think it's pretty self explanatory...

but anyways, there are n possibilities (vertices) and you have to choose 2 of them, can't doublecount so it's C(n,2) instead of P(n,2), minus n because you can't go from one vertex to the exact same vertex.[/quote]

Actually it's because you can't count the sides. Anyways I'm just a random former mathcountser who felt like disrupting the mathcounts forum :D.

[quote="mathcnt2005"]$a^2 + b^2 = c^2 ;$ that seems to work too.[/quote] which is the pythagorean theorum

Umm... we all know that. There was no point in stating it.

my 8th grade geometry book proves it nicely. to paraphrase...

[hide] each point on the polygon has n-3 diagonals to it- we get this from there are n points to draw diagonals to but you cant use the point itself or it's two neighbors so this rules out three. since each of the n points has this many diagonals able to be drawn from it, you multiply n(n-3). this counts each diagonal twice, however, so you must divide the whole thing by two giving (n(n-3))/2. It's not really that mysterious of an equation- it's easy to see just from the numbers right there.[/hide]


Can you guys give me a hint to this problem:

Prove that there are infinitely many positive integers n such that n^19 + n^99 can be written as a sum of two perfect squares in two different ways.

[hide="The relevant lemmas"] [b]Lemma 1:[/b]  A number can be written as the sum of two perfect squares if and only if the power of every prime congruent to $ 3 \bmod 4$ dividing it is even.

[b]Lemma 2:[/b]  A number can be written as the sum of two perfect squares in exactly one way if and only if it is of the form $ 2^a p$ where $ p$ is a prime congruent to $ 1 \bmod 4$. [/hide]


Suppose the digits of an integer N consist of the integers from 1 to 100 written in order from left to right, so that N = 123456789101112...9899100. What is the remainder when N is divided by 11

The divisibility rule for 11, if given a number such as abcdefg... is a-b+c-d+e-f+..etc. If this sum is divisible by 11, then the whole number is divisible by 11. If we apply this rule to your problem, we get 1-2+3-4+5-6+7-8+9-10+.... This is just a whole string of -1's. There are 50 pairs of these, so it is -50. -50 when divided by 11 is 4R6. So, the remainder is 6. If not, what am I missing? :maybe:

-rts2007

Consider the number $ a_n\cdot10^n\plus{}a_{n\minus{}1}\cdot10^{n\minus{}1}\plus{}\ldots\plus{}a_0$.

This number mod 11 is equivalent to $ \sum_{i\equal{}0}^{n} (\minus{}1)^i*a_i$. Same as rts2007's post above.

However, this is only for the digits, not the numbers themselves, so your method is incorrect.

Instead, take pairs of digits after reaching $ 10$.

So we have $ 1\minus{}2\plus{}3\minus{}4\plus{}5\minus{}6\plus{}7\minus{}8\plus{}9\equal{}5$ and then at $ 10$ onward, we take each number's first digit and subtract the second digit.

It's pretty much just pure calculation.

Is anything wrong with my "proof?"

-rts2007

Yes since you used $ 9\minus{}10$ and $ 11\minus{}12$. It should be $ 9\minus{}1\plus{}0\minus{}1\plus{}1\minus{}1\plus{}2\minus{}1\plus{}3 \ldots$ Do you see what I'm getting at?

You have to divide each two-digit number into single digits in order to use the divisibility rule.

Ah, yes. Now I see my mistake. Whatever, It's still ok. I'm pretty sure you can still solve the problem using my method or will it be sufficiently harder?

-rts2007

well no it's still just computation. not too bad and your method was correct except for that small error.

I got $ 35\minus{}360\minus{}1\plus{}5\equal{}319\equiv0\mod11$... is this correct?

hmm I got ${ -41\equiv\boxed{8}(\text{mod}11})$. It's just a bunch of computation, adding, subtracting (unless I can manage to add or subtract wrong :maybe:)

Write the number as

$ N\equal{}100\plus{}99\cdot 10^3\plus{}98\cdot 10^5\plus{}97\cdot 10^7\plus{}\dots\plus{}10\cdot 10^{181}\plus{}123456789\cdot 10^{183}$

Now note that all the powers of $ 10$ are odd, hence $ \equiv \minus{}1\pmod{11}$, and also $ 123456789\equiv \minus{}1\minus{}1\minus{}1\minus{}1\plus{}9\equiv 5\pmod{11}$. Therefore

$ N\equiv 1\minus{}(99\plus{}98\plus{}\dots \plus{}10)\minus{}5\equiv 1\minus{}{99\plus{}10\over 2}\cdot 90\minus{}5\equiv 1\minus{}109\cdot 45\minus{}5\pmod{11}$

Since $ 109\equiv \minus{}1\pmod{11}$, we get $ N\equiv 1\minus{}(\minus{}1)\cdot 45\minus{}5\equiv 41\equiv 8\pmod{11}$

[quote="mathwizarddude"]I got $ 35 \minus{} 360 \minus{} 1 \plus{} 5 \equal{} 319\equiv0\mod11$... is this correct?[/quote]
Ugh, it's perhaps not a wise choice to do math problems when you get tired...
It should have been $ 35*9 \minus{} 360 \minus{} 1 \plus{} 5 \equal{} \minus{} 41\equiv3\mod11$........  :oops:  :mad:
EDIT: DANG IT... there is an even number of digits so the very first power of 10 should be odd... therefore it's $ \plus{} 41$ or $ \minus{} 3\equiv8\pmod{11}$... man i stink today


Consider two number $ a,b>0$ such that
Suppose that $ a^p\minus{}b^p$ is a natural number for all p is prime .
Prove that $ a,b \in N$

It is true if $ a^2\minus{}b^2,a^3\minus{}b^3,a^5\minus{}b^5\in N$.

Can you prove it ? I don't think it is true.
For example $ a\equal{}\frac{x}{p},b\equal{}\frac{y}{p}$ and  $ p^5|x\minus{}y$
Easy to check that $ a^2\minus{}b^2,a^3\minus{}b^3,a^5\minus{}b^5$ is integer but $ a,b$ not neccesarry is integer. :wink:

Yes. From $ a^2\minus{}b^2\equal{}n_2,a^3\minus{}b^3\equal{}n^3,a^5\minus{}b^5\equal{}n_5$ we get
$ (b^2\plus{}n_2)^3\equal{}(b^3\plus{}n_3)^2,(b^2\plus{}n_2)^5\equal{}(b^5\plus{}n_5)^2$ and $ (a^2\minus{}n_2)^3\equal{}(a^3\minus{}n_3)^2,(a^2\minus{}n_2)^5\equal{}(a^5\minus{}n_5)^2$.
It give $ a,b$ are rational (not integer). Let $ a\equal{}\frac{x}{q},b\equal{}\frac yq, x,y,q\in N,gcd(x,y,q)\equal{}1$. Then $ (x\plus{}y,x^2\plus{}xy\plus{}y^2)\equal{}(x\plus{}y,xy)$, therefore $ (x\plus{}y,x^2\plus{}xy\plus{}y^2,q)\equal{}1$. It give $ q^m|x\minus{}y,q^{m\plus{}1}\not |x\minus{}y$. If $ p>m$ and $ q>1$ we get $ a^p\minus{}b^p\not \in N$.

[quote="Rust"]Yes. From $ a^2 \minus{} b^2 \equal{} n_2,a^3 \minus{} b^3 \equal{} n^3,a^5 \minus{} b^5 \equal{} n_5$ we get
$ (b^2 \plus{} n_2)^3 \equal{} (b^3 \plus{} n_3)^2,(b^2 \plus{} n_2)^5 \equal{} (b^5 \plus{} n_5)^2$ and $ (a^2 \minus{} n_2)^3 \equal{} (a^3 \minus{} n_3)^2,(a^2 \minus{} n_2)^5 \equal{} (a^5 \minus{} n_5)^2$.
It give $ a,b$ are rational (not integer). Let $ a \equal{} \frac {x}{q},b \equal{} \frac yq, x,y,q\in N,gcd(x,y,q) \equal{} 1$. Then $ (x \plus{} y,x^2 \plus{} xy \plus{} y^2) \equal{} (x \plus{} y,xy)$, therefore $ (x \plus{} y,x^2 \plus{} xy \plus{} y^2,q) \equal{} 1$. It give $ q^m|x \minus{} y,q^{m \plus{} 1}\not |x \minus{} y$. If $ p > m$ and $ q > 1$ we get $ a^p \minus{} b^p\not \in N$.[/quote]
Why from that $ a,b$ is rational ?  :wink:
2. Why we can resume $ \gcd(x,y,p)\equal{}1$

[quote="TTsphn"]Why from that $ a,b$ is rational ?  :wink:
2. Why we can resume $ \gcd(x,y,p) \equal{} 1$[/quote]
a and b are roots $ f(x)\equal{}0,g(x)\equal{}0, deg f\equal{}4,deg g\equal{}8$ polinoms with integer coefficints, suth that $ (f(x),g(x))\equal{}1$.
If $ a\equal{}\frac xq, b\equal{}\frac yq$ and $ (x,y,q)\equal{}d$, then $ a\equal{}\frac{x_1}{q_1},b\equal{}\frac{y_1,q_1},x\equal{}x_1d,y\equal{}y_1d,q\equal{}q_1d$ and $ (x_1,y_1,q_1)\equal{}1$.

I don't understand why from that $ a,b$ is rational .
You should post more clearly . It can be in the form $ a\plus{}b\sqrt{d}$


BoatsRUs built 7 canoes in January of this year and then each subsequent calendar month they built twice the number of canoes they had built the previous month. How many total canoes were built by BoatsRUs by the end of May of this year?

This is simply a geometric series with 5 terms, first term 7, and common ratio 2.

Total number of canoes built is $ 7\cdot \frac{2^5\minus{}1}{2\minus{}1}\equal{}7\cdot 31\equal{}\boxed{217}$ canoes.

So first there are 7, then 7*2, the so on.
Simply (1+2+4+8+16)*7= 31*7=217.
ergh beataen.


How many zeros does the number $ 75^{40} 8^{25}$ end in?

I had it down to $ (3*5^{2})^{40}((2^{3})^{40})$ but then got stuck.

$ (3\cdot5^2)^{40}((2^3)^{25})$

$ = 3^{40}\cdot5^{80}\cdot2^{75}$,

and since $ 2\cdot5 = 10$, this equals

$ 3^{40}\cdot5^5\cdot10^{75}\implies\boxed{75\text{ terminal zeroes}}$

$ \text{\LaTeX}$ made prettier, and yes, that's correct.


Probem 1:

If A^B^3 = B^A   and A is not equal with b and X

Find all  X  if :  3^444 = B^X
                            .They are real positive numbers .

Problem 2:

In a compotition, there are X televisions , Y mobiles and Z computers. There are N people who are winners and K losers.Every loser look at least one thing from thing above and every winner took at least 9 things. And:
5X+4Y+3Z=2N+K
Prove that: K is bigger than N is not true .

Problem 3:

Find all A in:
Sin A + Cos AB + tan ABC = 0 if XBC=A

Note: This is the final exam in Kuwait compotion year 1999.
Please post a complete solution.

for number 1, are you sure it's just "real positive numbers"? not "positive integers" or something? because  I would think there are infinite solutions...

for number 2, what is an "issue"?

Yes It is true that there are infinte answers.
For problem 2 : I meant with issue For loser : Every loser took at least one thing from above things . and so on with winers ( sorry for my weak English) :D

There is something wrong with number 2 as stated: if every winner took at least 9 things and every loser took at least 1, then the total number of things = X+Y+Z $\geq$ 9N + K, right?  So there is no way that 2N + K = 5X + 4Y + 3Z.

Also, for the last one, can't we simply choose A and B and then pick C to be whatever value we need, so long as A is not 0?

It is true 100% what you said and the solution that there is no solution for 2N+K=5X+4Y+3Z and the only solution that if N+K is bigger or equals 10
N=K=X=Y=Z=0 so there is no compotion so it is not true that K is bigger than N in any case.
( I knew that the trick very very easy ) and well done.

Problem 3  is very easy
Sin A + Cos AB + Tan ABC=0 if A=BCX

Sin BCX + Cos XC(B)^2 + Tan X(BC)^2 = 0
Now put it :

Sin BCX + Cos XC(B)^2 + (Sin X(BC)^2)/(Cos X(BC)^2) = 0
so : cos x(bc)^2 is not equal with 0  so : x(bc)^2 is not equal with
90 or 270..........

So now it is very easy to complete : What  Sin BCX + Cos BCX(B) + Tan BCX(BC) is equal with 0 ?                Just suggest any B  C  X  then find A

So infinite solutions we can find .  Well done JBL again


can someone help me on this prob? (and explain)

How many positive integers less than 10,000 have at most two different digits?

click to reaveal answer, but no explaination

[hide]answer is 927[/hide]

[hide]
we Have $\binom{10}{2}$ way to choose two different digits, and the following permutations:
$abbbb$
$aabbb$
$ababb$
$abbab$
$abbba$
$baabb$
$babab$
$babba$
$bbaab$
$bbaba$
$bbbaa$

the rest repeat themselves, so multiply by 2.  Take care of the 4, 3, and 2 digits case like this, and also the only 1 digit case.  Also include the case where only 1 kind of digit is present. That should be it
[/hide]

pkericchang: Shouldn't each number be only 4 digits, since 10,000 is the first 5 digit number?

you're right, I only glance at the thing so yeah...
Also, (I forgot to type this) substract any case with a 0 as the first digit.

Here is a slightly faster solution:

$\binom{10}{2}$ ways to choose 2 out of 10 digits, multiplied by $(2^{4}-2)$ because that at each of the 4 places in the number you can either choose the first or second digit, except in 2 cases all the digits are the same, and this would lead to double counting, so we will consider it at the end.

Now we account for leading zeroes:

As 0 can't be one of the digits, or we would be counting twice, we now have only $\binom{9}{2}$ options. You now have 3 digits left, so that is $(2^{3}-2)$ because we can't count the case where they are all the same.

We repeat this logic for two digits as well, and get $\binom{9}{2}$ and $2^{2}-2$.

Now add 10 due to 1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999, and 0.

You have:

$\binom{10}{2}(2^{4}-2)+\binom{9}{2}(2^{3}-2)+\binom{9}{2}(2^{2}-2)+37$

This evaluates to:

$45*14+36*6+36*2+10 = 928$

This is one more than you had, so I am going to assume that 0 is not supposed to be counted here.

0 isn't a positive integer  :wink:

Nice solution

[hide="I did it this way"]x = 9
xy or xx = 9 x 10
xxx = 9
xyx or xxy = 9x9x2
xyy = 9 x 9
xxxx = 9
xxxy, xxyx, xyxx = 9 x 9 x 3
xxyy or xyxy = 9 x 9 x 2
xyyx = 9 x 9
xyyy = 9 x 9
-----------------------
$\boxed{927}$[/hide]

Um...let's do some complimentary counting...I'd like to see if there's a PIE way to do this too >.<.

[hide]
Since we are looking for at most 2 distinct digits, we need to simply find how many numbers have at least 3 distinct digits, and subtract from the total.

There are $9999$ positive integers less than $10,000$.

$4$ Distinct Digits - $9*9*8*7\rightarrow 4536$
$3$ Distinct Digits (from $1000-9999$) -

There are 6 permutations...
aabc
abac
abca
abbc
abcb
abcc

Each permutation gives $9*9*8*1$. So there are $6*9*9*8 \rightarrow 3888$ numbers from $1000$ to $9999$ that have $3$ distinct digits.

$3$ Distinct Digits (from $100-999$) - $9*9*8\rightarrow 648$

Adding these numbers up, we get $4536+3888+648\rightarrow 9072$. We take the total and subtract to get the desired answer. $9999-9072\rightarrow \boxed{927}$[/hide]

[quote="JSteinhardt"][quote="pkerichang"]in your solution, how about 1, 2, 3, 11, 22, 33, ... Those numbers??? Also, what if the 2 out of 10 digits you choose for 4 digit has a zero in there??? won't you have a leading zero???[/quote]

I believe that you answered your own question, as the numbers that you are asking about are exactly the numbers that are counted with leading zeroes.

I apologize for not stating this more explicitly, but basically:

1 -> 0001
2 -> 0002
11 -> 0011

etc.

so these are all counted under the first case, where one of the two numbers you choose is a zero, and the other is 1, 2, etc., and then the first however many numbers is a 0. The reason I included 0 was because 0 -> 0000, although as Jason pointed out, 0 is not a positive integer,  :oops: .

Basically I am braking it up into the following cases:

1) No leading zeroes, or leading zeroes but zero is also one of the digits (or all of the other digits are the same), and not all [i]4[/i] numbers are the same.

2) 1 leading zero, and zero is not one of the digits, not all [i]3[/i] numbers are the same.

3) 2 leading zeroes, and zero is not one of the digits, not all [i]2[/i] numbers are the same.

4) All [i]4[/i] numbers are the same.

Potentially there is a case for 3 leading zeroes with not all [i]1[/i] numbers being the same, but it is easy to see that a number is the same as itself, so the numbers falling under this case are nonexistent.[/quote]

Nice solution, I did not understand it at first and had the same questions Pkerichang, but I did not realize that they were accounted for in the 10C2. I think that first step was great with the 10C2 to account for those extra cases.

Zellex I like your solution a lot! It is the cleanest one here I think. Maybe the resources section can change to include your proof instead.

I wonder if there is a counting argument for distinct permutations of ABC...all I can think of is PIE, which is not any nicer than listing out possibilities.

I like Zellex's solution a lot, also.

Change the list of permutations to $\frac{P(4,2)}{2!}$ and it's very clean!


When (2008!)/[(1004!)(1004!)] is expanded, how many terminal zeroes are there?

$ 2008!$ has $ 500$ factors of $ 5$ while $ 1004!$ has $ 249$ factors of $ 5$, so there are $ 2$ terminal zeroes.

Too see how many factors of $ p$ a number $ n!$ has, were $ p$ is prime, use the formula. $ e_p(n!) \equal{} \left[\frac{n}{p}\right] \plus{} \left[\frac{n}{p^2}\right] \plus{} \left[\frac{n}{p^3}\right] ...$ Were $ [x]$ denotes the smallest integer $ \le x$

Next time put this in the number theory forum:)


A cart moves with a constant speed along a horizontal circular path.From the cart,a particle is thrown up vertically with respect to the cart.The particle wil..
a)Land outside the path
b)On the path
c)follow an elleptical path
d)follow a parabolic path

[color=blue]I think that I 'll choose answer $d$[/color]

[color=blue]
Let $v$ denotes the velocity of the particle respect to the cart and $v_0$ is the cart's velocity.
The particle is thrown upward with the velocity $v'=\sqrt{v_0^2+v^2}$ make the angle $\theta$ to the horizontal where $\tan\theta=\frac{v}{v_0}$
Consequently, the particle will ffollow a parabolic part[/color]

It seems to me that any of the answers could be correct.  If the cart is moving pretty fast and the particle is thrown fairly high, then it will probably land outside the path.  However, if the particle isn't tossed very high, and if the cart is going slowly (and if the path is kind of wide) then the particle may land on the path.  Normally we would expect the path of the particle to be (approximately) a parabola.  But suppose the particle were thrown so high, and the cart were moving so fast, that the particle actually ended up in orbit around the earth.  Then it would follow (approximately) an ellipse....right?


a discussion we had tonight while playing pool:
there are $ 15$ balls on the table, arranged in a triangular form. $ 7$ are red, $ 7$ are yellow, $ 1$ is black. The fact of exchanging $ 2$ balls is called a 'move'. What is the minimal number of moves needed to go from any configuration to any other ?

Any two balls, or two adjacent balls?  If it's any two balls, then the answer is 8: we need 1 move to place the black ball where it belongs.  Then, in the worst case there are red balls in all the places we want yellow balls and vice-versa, so we must exchange 7 pairs in order to reverse the coloring.  (This also assumes that all red balls are alike, and likewise for yellow balls.)

oops .. I forgot to mention that the colors do not really matter for a configuration. It means that there are $ 2$ possibilities for a same configuration: one can exchange all the red balls with all the yellow ones, and one still call it the same configuation.

Well, we still clearly need one move for the black ball.  Then consider one of the two collections of seven spaces where we desire all balls of the same color.  It must contain at least 4 same-colored balls, so we need at most 3 switches to make it uniform in color, for a total of at most 4.

exactly  :lol:  so remember that next time you play pool ...


Given a,b,c>0 and $ k\geq\frac{7\plus{}3\sqrt{5}}{2}$or$ k\leq\frac{7\minus{}3\sqrt{5}}{2}$.Prove that
$ \sqrt{\frac{a}{kb\plus{}c}}\plus{}\sqrt{\frac{b}{kc\plus{}a}}\plus{}\sqrt{\frac{c}{ka\plus{}b}}\geq\frac{3}{\sqrt{k\plus{}1}}$

[quote="PHONGVAN"]Given a,b,c>0 and $ k\geq\frac {7 \plus{} 3\sqrt {5}}{2}$or$ k\leq\frac {7 \minus{} 3\sqrt {5}}{2}$.Prove that
$ \sqrt {\frac {a}{kb \plus{} c}} \plus{} \sqrt {\frac {b}{kc \plus{} a}} \plus{} \sqrt {\frac {c}{ka \plus{} b}}\geq\frac {3}{\sqrt {k \plus{} 1}}$[/quote]
The mine is stronger:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153325


Hey everybody, I'm not completely sure about this, but i read in a magazine that britney spears and Kevin Federline divorced after their second baby :( ,Is that true?

Who cares?  That's their business, not mine. :roll:

Hey man, you don't care cuz you ain't her fan :| , but it's important for her real fans.... I got really depressed and disappointed when i heard this news.... :(

Can we discuss about this without using google please :wallbash_red:
as you see I'm in Iran and i cant get these kind of news so to be sure that they are true or not....

As much as I could find out from a quick wikipediaing, she filed for divorce, but the battle for legal custody of their kids is still going on.


Let $ \zeta_{1}(O_{1},R_{1})$ and $ \zeta_{2}(O_{2},R_{2})$ be the circles that externally tangent to each other and $ l_{1}$ and $ l_{2}$ be the common external tangent lines to $ \zeta_{1}$ and $ \zeta_{2}$.Consider $ A,D\in l_{1}$ and $ B,C\in l_{2}$,such that $ ABCD$ is convex quadrilateral,$ AB$ is tangent to $ \zeta_{1}$ at point $ E$ and $ CD$ is tangent to $ \zeta_{2}$ at point $ F$.If $ AC\cap BD\equal{}O$ then prove that $ OE\plus{}OF\leq 2(R_{1}\plus{}R_{2})$. :)

Nobody?Does anybody want to know a solution? :wink:


The integers a and b have the property that for every nonnegative integer n the number of (2^n)*a + b is the square of an integer.  Show that a = 0.

(Problem may be familiar to many of you)

[hide]$a+b=s^{2}$
$2a+b=(s+k)^{2}$
so $\frac{2a+b}{a+b}=\frac{(s+k)^{2}}{s^{2}}$
After the 1 is subtracted, $\frac{a}{a+b}=\frac{2sk+k^{2}}{s^{2}}$
Taking reciprocals, $1+b/a=\frac{s^{2}}{2sk+k^{2}}$.
This implies that $s^{2}>2sk+k^{2}$ or $s^{2}-2sk-k^{2}>0$ or $s=k \pm \sqrt{k^{2}+1}$.  (Just use the quadratic formula.)
Plugging this in for s, $k^{2}+k^{2}+1\pm2\sqrt{k^{2}+1}-2k(k\pm\sqrt{k^{2}+1})-k^{2}>0$ or $1-k^{2}>0$, forcing k to be 0.  This means that a is 0.
There's two flaws with this though.  First, b/a must be positive to imply what I've shown.  However, b can be negative.  (If a is, eventually one will get negative numbers.)  Therefore, $\frac{a}{a+b}=1-\frac{b}{a+b}=\frac{2sk+k^{2}}{s^{2}}$, which implies that $2sk+k^{2}>s^{2}$ or $0>s^{2}-2sk-k^{2}$.  Solving for k this time, $k=-s\mp\sqrt{s^{2}+1}$.  After substituting, $0>s^{2}-1$.  In this case, s=0, so $a=k^{2}$ and b=$-k^{2}$.  But, 4a+b must also be a perfect square here, which is impossible since $3k^{2}$ is not a perfect square.
Unfortunately, there's a more troubling flaw.  b/a=b/0 in case 1, and  $\frac{b}{a+b}$ which is undefined.  Wouldn't that ruin the validity of my proof?
[/hide]
Also, let's expand the case: What if we were talking about perfect nth powers or if we were talking about $ak^{n}+b$ where k is an integer that isn't -1,1,0.

[hide]For the case $ak^{n}+b$ where $k\neq0, \pm1$,
Assume $a>0$, next let $ak^{n}+b=s^{2}$, then
$ak^{n}k^{2}+b=(ak^{n}+b-b)k^{2}+b=(s^{2}-b)k^{2}+b=(ks)^{2}-b(k^{2}-1)$
Since we can make $s$ arbitrary large by choosing a large $n$ and $b(k^{2}-1)$ is a constant, the difference of two consecutive squares will eventually exceed it. Hence, $ak^{n}k^{2}+b$ cannot be a perfect square.[/hide]


function
floor function
ceiling function
LaTeX

something is wrong with your solution   :?

oops, didn't see that the second one is ceil instead of floor :P , but then it makes the answer even neater :)   Will mine be right if both are floor, tho??

Yeah your solution would have been right  :lol:
Here's how I would have done it if it had two floor functions.
Look at the entire sum and pair terms from the $k$th term of the series with terms from the $2008-k$th term of the series. That yields a sum with $2007$ terms of the form
$\lfloor \frac{2007^{k}}{2} \rfloor +\lfloor \frac{-(2007^{k})}{2} \rfloor$.
Here's a nice identity for $\lfloor \frac{n}{2} \rfloor$ when $n$ is odd. Since $n$ is odd we can write $n=2j-1$ for some integer $j$. So $\lfloor \frac{n}{2} \rfloor=\lfloor \frac{2j-1}{2} \rfloor=j-1=\frac{n-1}{2}$.
So the general term of our new series becomes
$\lfloor \frac{2007^{k}}{2} \rfloor +\lfloor \frac{-(2007^{k})}{2} \rfloor=\frac{2007^k-1}{2}+\frac{-(2007^k)-1}{2}=-1$. Since we have $2007$ of these terms our final answer is $-2007*1=-2007$

[quote="srulikbd"][quote="pkerichang"][hide]
expand the sum we see that it is equivalent to $\sum^{2007}_{n=1}{\left\lfloor\frac{2007^n}{2}\right\rfloor + \left\lceil-\frac{2007^n}{2}\right\rceil}$.  Let $\left\lfloor\frac{2007^n}{2}\right\rfloor = p$, then $\left\lceil-\frac{2007^n}{2}\right\rceil = -p$.  Therefore, the answer is $0$.
[/hide]
edited: read the question wrong! and some latex errors[/quote]
I didn't understand it...how did u expand it?[/quote]

well, by expanding I meant write out all the terms (not literally).  But after you did that you'll see the power of both terms range from 1 to 2007, and therefore you can regroup them into my formula.
ex: writing:
$(1^1 + 2^3) + (1^2 + 2^2) + (1^3 + 2^1)$
is the same thing as:
$(1^1 + 2^1) + (1^2 + 2^2) + (1^3 + 2^3)$


When factoring, you have to put it in the form of $(x+a)(x+b)$ , with $a+b= $ the coefficient of $x$ and $ab$ is the constant. Using that you can try solving using simultaneous equations or any other method

Nice , thank you.

I kinda understand what you are saying but is there anyway you can do just one of those, step by step, so I can see what exactly it is I have to do..

Thank you very much. :blush:

Ok let's try #1

$x^2 - x - 12$

Now, notice that we want to get it into this form

$(x + a)(x + b)$

If we expand (FOIL) that form we will have

$x^2 + xb + xa + ab$

We can simplify this more by taking out a x from the middle two expressions by using reverse distribution --(Note that $xb + xa = (a + b)x$)

$x^2 + (b +a)x + ab$

Now let's compare this to our original equation

$x^2 - x - 12$

We can see that

$ab = -12$

and

$a + b = -1$

Now we've got to look for two numbers that multiply to -12 and add to -1. First we find all of the divisors of 12 -  $1\cdot12,2\cdot6,3\cdot4$
We can easily see that the first two choices won't work, because the numbers are way too distant to add up to one, but if we look at the last one, we can make that $-4$ and $3$ (remember we are trying to find two numbers that multiply to $-12$ and add to $-1$ So we can see that this pair works!

So we could now rewrite

$x^2 - x - 12$

into

$(x - 4)(x + 3)$

Now, if you look back, we can make a general rule for any equation that is $x^2 + (y)x + (z)$ (Note: the $y$ and the $z$ can be any numbers)
When we get it into that form we have to look for two numbers that muliply to get $z$ and add to get $y$

Well, hope I helped. If you are confused on any step, feel free to ask! If you think you understood it, feel free to try another problem from your list and post how you did and your answer, so we can see if you got it. (Don't worry about making it look good (like $x^2$ instead of x^2) doesn't really affect the way the math is read right now)

There's one tip: If the constant is positive, then $a$ and $b$ are both positive or negative, If the constant is negative, then either one of $a$ and $b$ is negative

WHOA windslicer you are so awesome man!!

That was a huge help!

Thank you so much! :)

Okay I pretty much understand it all except how a+b = -1 in the part wind mentioned...I know it is something stupid I am missing but any help would be appreciated..

THANKS GUYS!

No problem, anything I can do.

If you run into any trouble, feel free to post a question.

About the $(a+b) = -1$

Let's look at the model equation and the equation we have

Model:

$x^2 + (a + b)x + ab$    (remember comes from $(x+a)(x +b)$ which we foiled and simplified in my first post)

Our Equation:

$x^2 - 1x - 12$

If we look at the corresponding parts in both equations we see that

$x^2 = x^2$
$(a +b)x = -1x$
$ab = -12$

The first part is perfect. in the second part we can divide an x by both sides to get:

$\frac{(a+b)x}{x} = \frac{-1x}{x}$   which equals

$a + b = -1$

And the third part we can see that a multiplied by b is equal to -12.

Again, if any steps are unclear (like why $(x+a)(x+b) = x^2+ (a+b)x + ab$ feel free to ask (it's most important that you understand this - so feel free to ask any question)

ahhh, it says your offline now..Hopefully you saw the message right above that one you just posted :D

Since I editted my last post instead of posting a new one, I'll just post again so that if you are checking for new posts, you will see one.

Thank you very much WindSlicer. I totally understand it now :D

You explained it to me better than my math teacher could :roll:

[quote="TheCoug85"]Another user was already kind enough to tell me the answers to these problems, but I would appreciate it very much if somebody could show me the steps involved as well.

Help would be GREEEEEEEEEEATLY appreciated w/ factoring the following problems:

1.) X^2-X-12 (X^2 means squared just in case nobody knew that)

2.) X^2-4X-12

3.) X^2-X-6

4.) X^2-5X+6

5.) X^2-7X+12

Again Thanks a lot if you can help me.[/quote]

I guess you know how to do them but I'll just put the answers anyway: 1) (x+3)(x-4)
2)(x+2)(x-6), 3) (x+2)(x-3), 4) (x+1)(x-6), 5) (x-3)(x-4)

[quote="WindSlicer"]All of those are correct except for

4) $(x-3)(x-2)$

Since we are looking for numbers that multiply to 6 and add to -5, then we need two negative numbers, and the only ones that work are -3 and -2.[/quote]

Oops, my bad....allright


function
modular arithmetic
number theory
totient function

The answer key and solutions have different answers.

Short one is answer key and long one is solutions.

I believe the solutions is the correct one, because for the answer key they only put the tens digit, not the product of the tens and the units digit.

It's 8. I remember how I failed this problem (and #7) at state in sixth grade...bleh I would've made countdown if I had gotten those two right.

[hide="solution"] The product is $ 2^{(1\plus{}2\plus{}3\plus{}...\plus{}100)}\equal{}2^{5050}$. It's pretty well known that the last two digits of powers of 2 repeat every twenty (this could be a good thing to memorize, since [url=http://mathworld.wolfram.com/CarmichaelFunction.html]the Carmichael Function[/url] can be somewhat confusing). Alternately, use the Totient Function, which works just as well for this problem.

Thus, it reduces to $ 2^{10}\equal{}1024$, so our answer is $ 2\cdot{4}\equal{}8$.[/hide]

I believe Totient is only for the cases when the base is relatively prime to $ 100$, or whatever mod you're working in :wink:

It is. I just can never remember that. Brute forcing still works perfectly fine though :P .

I suppose you could use $ 2^{5050}\equiv24\pmod{25}$ and $ 2^{5050}\equiv0\pmod{4}$ and then solve the resulting linear congruences.


invariant
geometry
geometric transformation

You have $2006$ beads on a (closed) necklace in positions $0, 1, 2, 3, ... 2005$.  You are attempting to color the beads such that they satisfy the following property:

If beads in position $i$ and position $j$, $i > j$ have the same color, then neither the bead in position $i+(i-j)$ nor $j-(i-j)$ (taken $\bmod 2006$) can have that color.

What is the minimum number of colors needed?

Can you generalize?

when you say i+j and i-j can't have the same color, do you mean the same color as i and j have or just they can't have the same color as each other?

Neither can have the same color as the color of $i$ and $j$.  Sorry, it was unclear, I will edit.

[quote="t0rajir0u"]You have $2006$ beads on a (closed) necklace in positions $0, 1, 2, 3, ... 2005$.  You are attempting to color the beads such that they satisfy the following property:

If beads in position $i$ and position $j$, $i \neq j$ have the same color, then neither the bead in position $i+j$ nor $i-j$ (taken $\bmod 2006$) can have that color.

What is the minimum number of colors needed?

Can you generalize?[/quote]

Conjecture: For $n$ beads, the minimum number of colors needed is $n$. Also a rather elementary observation: the color of the bead at position $0$ is different from the color of the bead at any other position.

The conjecture is just to start the discussion to this interesting problem. Hope no one just writes down the complete solution. That would kill everything!

[quote="Kurt Gödel"][quote="FieryHydra"]That would kill everything![/quote]You're right!
Another observation is that bead 1 and bead 2 must have a different color. We allready have 3 colors!  :lol:[/quote]
My initial conjecture is wrong! Anyways, here is what we have for now. Let $a_{n}$ be the minimum of colors needed for $n$ beads. Then, $a_{1}=1, a_{2}=2, a_{3}=3, a_{4}=3,a_{5}=3,...$ It seems parity plays some role (don't know what.) Maybe $3$ different colors is all we need.

[quote="FieryHydra"]It seems parity plays some role (don't know what.)[/quote] What do you mean? That bead $k$ and $2k$ have a different color?

[quote="FieryHydra"]Maybe $3$ different colors is all we need.[/quote]According to my colorings $a_{6}=a_{7}=a_{8}=a_{9}=4$... 10 numbers to color for $a_{10}$ :weightlift:

[quote="FieryHydra"]Also a rather elementary observation: the color of the bead at position $0$ is different from the color of the bead at any other position.[/quote]

Could you explain this?  The solution should be invariant under translation of the positions of the beads, so I don't understand how you arrived at this conclusion  :huh:

Edit:  I am [b]so sorry.[/b]  The problem condition was written incorrectly.   :blush:  I will correct it now, and I'm very sorry for wasting your time.


Show that for positive reals $ a,b,c,d$
\[ \frac {1}{a^2 \plus{} ab} \plus{} \frac {1}{b^2 \plus{} bc} \plus{} \frac {1}{c^2\plus{}cd} \plus{} \frac {1}{d^2\plus{}da} \ge \frac {2}{\sqrt {abcd}}
\]

[quote="apollo"]Show that for positive reals $ a,b,c,d$
\[ \frac {1}{a^2 \plus{} ab} \plus{} \frac {1}{b^2 \plus{} bc} \plus{} \frac {1}{c^2 \plus{} cd} \plus{} \frac {1}{d^2 \plus{} da} \ge \frac {2}{\sqrt {abcd}}
\]
[/quote]
The hint see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81588

The solution see here: http://www.mathlinks.ro//viewtopic.php?t=102604 .

  darij


My problem is unique.

The problem is :

Why is Alcumus taking so long to come back online?

Sigh, don't ask..... It'll be done when they are ready.

Mods please move, but I did get a laugh, zheng.

idk!!
Maybe you should ask mr. rusczyk.
It's been  like more than three weeks.


group theory
abstract algebra
geometry
geometric transformation
rotation
reflection
superior algebra

List all subgroups of the dihedral group $D_4$, and divide them into conjugacy classes.

Any good idea?
I don't know brute-force..

Since all $x \in D_4$ are of the form $a^sb^t$ with $0\leq s \leq 3 , \leq t \leq 1$ where $a$ as a rotation about $90$ degree and $b$ as some reflection on a line. (Relations: $b^2=e,a^4=e,a^sb=ba^{-s}$...)
Then all subgroups are (prove for example by brute force or some simple geometric things):
$A=\{ e \}$
$B=\{ e,a^2\}$
$C=\{ e,b \}$
$D=\{ e,ab\}$
$E=\{ e,a^2b \}$
$F=\{ e,a^3b \}$
$G=\{ e,a,a^2,a^3 \}$
$H=\{ e,a^2,b,a^2b \}$
$I=\{ e,a^2,ab,a^3b \}$
$J=\{ e,a,a^2,a^3,b,ab,a^2b,a^3b \} = D_4$

Here the conjugation classes of the subgroups are (the ones with $1,4,8$ elements have to be normal, the four left can easily be separated):
$\{ A \}$
$\{ B \}$
$\{ C,E \}$
$\{ D,F \}$
$\{ G \}$
$\{ H \}$
$\{ I \}$
$\{ J \}$


I think that a fun idea for a math jam would be to discuss one of the Millennium Problems (like the Riemann Hypothesis).  We probably won't get very far on a solution, but it would be fun to discuss it.

I doubt many of us would be able to understand one of the millenium problems (like the Riemann Hypothesis), much less be able to discuss it.

I think that the Riemann Hypothesis isn't very hard to understand.  It's just the theory that the zeroes of

[tex]
\displaystyle \zeta(s) = \sum_{n = 1}^{\infty} \frac{1}{n^s}
[/tex]

have [tex]\Re\{s\} = \frac{1}{2}.[/tex]

P vs. NP, the Riemann Hypothsis, and the Poincar Conjecture can be understood by many people here. Perhaps the Navier-Stokes Equations can also be understood. I doubt the other three can.

I'm being a bit picky here, but nr1337's statement needs a bit of modification:

The zeros s of the analytic continuation of [tex]\displaystyle \zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}[/tex] to [tex]\mathbb{C}\setminus\{1\}[/tex] satisfy either [tex]\Im\{s\}=0[/tex] or [tex]\Re\{s\}=\frac{1}{2}[/tex].

[quote="zabelman"]P vs NP isnt too difficult to understand and discuss...[/quote]

I think it both is and isn't at the same time, since many people know what "NP-Complete" means (as a general idea) yet at the same time don't know what it <i> means</i> (as a technical term).

We were doing this sort of thing in my algorithms/data structures class... I really like the transformation from 3-satisfiability to vertex cover (probably only because I figured it out.... but its still cool anyway!)

[quote="kylekatarn"][quote="ComplexZeta"]The zeros s of the analytic continuation[/quote]

what's an "analytic continuation?"

Thanks in advance.[/quote]

You might notice that it is a little hard to plug in... say... negative numbers, into that form of the riemann zeta function and get back anything useful.

There is a way to take the riemann zeta function as is, where it makes sense (s > 1), and construct a function which is analytic in almost all of the complex plane (with the exception of the point s = 1, we're doomed there) that agrees with our old zeta function for Re(s) > 1.

Furthermore, this can be done in a unique way.

I think such discussions are better had on the message board than in a Math Jam.  Perhaps if there were a well-spoken professional mathematician who could conjure a one hour talk understandable by enough AoPS, we could do it.  However, it could take a lot of work to find that person.

However, now that you have me thinking, it might be good to have a well known mathematician give a Math Jam once in a while.  That's a hard schedule, but I'll keep in mind if I have the right opportunity to schedule such a Math Jam.  Perhaps they could even give the requested talk, though they might have more useful and perhaps more interesting (to AoPSers) subjects to talk about.

Keep in mind -- this is far from a promise.


does anyone know what "@n2" or something similar to that mean on the ti89 when using the "solve()" with a trig equation?  i'm guessing something with "+360n" or something but im not sure and can't find it in the manual

thanks

'@n[i]x[/i]' represents an arbitrary integer.

It numbers its arbitrary integers so it knows they are all different.  For example, the expression 3*@n2 + pi*@n3 means some integer multiple of 3 plus some (other, or not necessarily the same) integer multiple of pi.


This is a game we used to play during our english classes:We start with a (big) word and try to make as many words as we can using just the letters of the given word.For example I`ll start with the word ALPHABET and post 5 words:
ALPHA
BETA
PET
HAT
LAP
 You continue with other words.Have fun!

More words from ALPHABET (there might be some repeats):

ALPHABET (obviously)
ALPHA
BETA
BET
HAT
HATE
LATE
BATE
BAT
TAB
PATH
BATH
BATHE
LATHE
LAP
PAL
PALE
PETAL
to be continued...

howsabout... "uncopyrightable"
I bet you could get hundreds out of that :D

I was really bored one day and did one from

ARTOFPROBLEMSOLVING

ART
OF
PROBLEM
SOLVING
ARTS
PROBLEMS
A
RAT
RATS
ARTS
FOR
FORT
FORTS
ROB
ROBS
MORE
SOLVE
PROBE
PROBES
SOLVES
LOVES
LOVE
FALL
FALLS

...and so on

this is kinda like that game, Text Twist. That was a good game :)

[url=http://games.yahoo.com/games/downloads/tx.html]Play Text Twist HERE![/url]


Assume that you have a large box with 1,000,000 marbles.  On the surface of 940,000 marbles is the number 5, 50,000 marbles have the number 6, 5,000 marbles have the number 7 and 1,000 marbles have the number 8.  The marbles are extracted two at a time.  What is the probability of extracting two marbles with the number 8 (ie 500,000 total outcomes)?

Is it simply the product of the ratio of marbles with the number 8 to the total number of marbles or is it more involved (.001*.001 = 1 x 10-6)?

Thanks,

The probability is found when dividing the ways to choose two 8s divided by all the possibilities when drawing two marbles.  This is (1000C2)/(1000000C2).

This reduces to (1000)(999)/(1000000)(9999)

This reduces to 111/(1000)(1111).

Approx. equal to 9.991*10^-5

This is very close to yours actually...

Do you mean that the first two drawn are both 8 or that after 500,000 drawings there is at least one pair of 8s?

Thanks for the responses.  The latter.  What would be the total number of 8-8's drawn from the bucket?  I'm trying to apply this to a chemical reaction assuming equal reactivity.


Let ABC be a fixed triangle. D is a variable point on BC. The common tangent (other than BC) to the inscribed circle of △ADB and △ADC meets AD at K.
Show that the length of AK is a constant.

Pls explain to me with steps.. THZ A LOT!!

http://www.mathlinks.ro/Forum/viewtopic.php?t=28329

  darij


[quote="Tinfant1"]4^x= 5^2x-3[/quote]

In this mode, you *must* have grouping symbols around the exponent
on the right-hand side such as:

4^x = 5^(2x-3) if you mean $ 4^x \equal{} 5^{2x \minus{} 3}$.

-------------------------------------------------------------------------
As it is now, it is equivalent to $ 4^x \equal{} 25x \minus{} 3$.

As a lesser chance, it could mean the following with different placement:

4^x = 5^(2x)  - 3, which is $ 4^x \equal{} 5^{2x} \minus{} 3$, but this is not the expected,
typical type of exponential equation because of the two different bases
and a separate constant.  So, it would be ruled out.

[hide]
$ 4^x \equal{} 5^{2x\minus{}3}$

$ log(4^x) \equal{} log(5^{2x\minus{}3})$

$ xlog(4) \equal{} (2x\minus{}3)log5$

$ xlog(4) \equal{} 2xlog(5) \minus{} 3log(5)$

Subtract the $ xlog(4)$ term from each side, and add the $ 3log(5)$ term
to each side:

$ 3log(5) \equal{} 2xlog(5) \minus{} xlog(4)$

$ 3log(5) \equal{} x[2log(5) \minus{} log(4)]$

$ \frac{3log(5)}{2log(5) \minus{} log(4)} \equal{} x$

or

$ \frac{log(5^3)}{log(5^2) \minus{} log(4)} \equal{} x$

$ \frac{log(125)}{log(25) \minus{} log(4)} \equal{} x$

$ x \equal{} \frac{log{(125)}}{log(\frac{25}{4})}$

$ x \equal{} \frac{log(125)}{log(6.25)}$

[/hide]


function
induction
algebra proposed
algebra

Let $k$ be a natural number. For each function $f : \mathbb{N}\to \mathbb{N}$  define the sequence of functions $(f_{m})_{m\geq 1}$ by $f_{1}= f$ and $f_{m+1}= f \circ f_{m}$ for $m \geq 1$ . Function $f$ is called $k$-[i]nice[/i] if for each $n \in\mathbb{N}: f_{k}(n) = f (n)^{k}$.

(a) For which $k$ does there exist an injective $k$-nice function $f$ ?

(b) For which $k$ does there exist a surjective $k$-nice function $f$ ?

hint:first prove $b)$ and then prove $a)$
[hide]for $b)$
for $k=1$ the function $f(n)=n$ is trivialy a $1-nice$ surjective function...
so assume that$k>1$mthere doesnt exists any...
assume number $a$ s.t $a$ isnt the $k^{th}$ power of a natural number,then we know that there exists number $c$ s.t $f(c)=a$
and also we know that there exists $b$ s.t $f(b)=c$ with this method we can find a $x$ s.t $f_{k-1}(x)=c$
so we get that $f_{k}(x)=a$ now if $f$ were $k-$nice then we would have $f_{k}(x)=f(x)^{k}=t^{k}$ so we get that $a=t^{k}$ contradiction...
and for $a)$ there exists such finction,I`ll give the function later,there is a method of constructing this function some induction type of method...[/hide]

tnx [b]aviateurpilot[/b] I thought $k>1$ I have edited my post tnx...

[quote="aviateurpilot"][quote="lovejrz"]still dont understand[/quote]
for $b)$ [b]BaBaK Ghalebi[/b] say that if $f$ is a surjective $k$-nice function.
then we have $f_{k}$ is surjective and for exemple $\exists x\in\mathbb{N}: f_{k}(x)=2$
so $2=f_{k}(x)=f(x)^{k}$
then $\sqrt[k]{2}=f(x)\in\mathbb{N}$
then $k=1$
and we have $f(n)=n$ is a surjective $1$-nice function.

another solution.
all f surjective are a surjective $1$-nice function.
if $k>1$
$f_{k}$ is a surjective but $f^{k}$ is not surjective
then $f_{k}\neq f^{k}$.[/quote]
$ f $ must be surjective not $ f_{k} $.
the answer for both questions is yes.
the second is true because $ f_{k+1}(n)=n^{k} $ and we only have to prove the existence of such a function.
for the first we can construct some "chains".


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Find the Galois group of p(x)=X^3 - 2  over Z7

It's cyclic and of order $ \leq 3$. As the polynomial has no root, it has order $ 3$.

1.
$ 3^3 \equal{} \minus{} 1$ and not $ 2$ in your field. Check again.
2.
I said it is _cyclic_. Even without knowing this, there wouldn't be any other Galois groups as all groups of order $ 3$ are isomorphic.

[quote="romy"]What do you mean the polynomial has no root? How about x=3?

Also if it has order 3 there are many possibilities for the Galois group then?[/quote]

Well, $ 3$ is no root of this polynomial, because $ 3^{3} \minus{} 2 \equal{} 25$ is not divisible by $ 7$.
This polynomial is of degree $ 3$ and has no root, thus it is irreducible.
Let $ \phi$ a root in an extension of the field $ F_{7} \equal{} \mathbb{Z}/7\mathbb{Z}$. Since $ 1$, $ 2$ and $ 4$ are three distinct cubic roots of $ 1$ in the field $ F_{7}$, $ 1\phi$, $ 2\phi$ and $ 4\phi$ are three distinct roots of the polynomial $ X^{3} \minus{} 2$, thus the polynomial decomposes entirely in $ F_{7}(\phi)$. The three automorphisms of $ F_{7}(\phi)$ are defined by $ \phi \mapsto 1\phi$, $ \phi \mapsto 2\phi$ and $ \phi \mapsto 4\phi$. They form the Galois group.
Like each group of order $ 3$, this Galois group is cyclic.

Friendly,
P.

Edit : I had not seen the second answer of ZetaX. Sorry.

Well thank you both I have finally gotten there finding the splitting field to be Z7(cubrt(2))


calculus
geometry
integration
MIT
college
Harvard
analytic geometry

I am currently enrolled in a community college that is part of my state's program, which allows students to take college courses as part of our high school education. I am going to take the following math courses:

Calculus and Analytic Geometry 1 (Differential Calculus)
Calculus and Analytic Geometry 2 (Integral Calculus)
Calculus and Analytic Geometry 3
Calculus and Analytic Geometry 4 (Multivariable Calculus)
Differential Equations
Linear Algebra
Discrete Mathematics

Does anyone have an idea as to weather these math courses place me in either Math 25 or Math 55?

Also, does anyone know if MIT has courses that are as challenging as Math 25 or Math 55?

as far as i know, entering math 25 or 55 is not really about "placement."  math 55 is just a more rigorous, difficult, and homeowrk-intensive version of 25, but they are both about basically the same material.  MIT has 8.022, which is certainly at the honors level, but I'm not quite sure which of 23, 25, or 55 it is most similar to.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=55+math&t=24761]This[/url] thread should be useful.

[quote="JSteinhardt"]In addition if you are at either Harvard or MIT you have the oppurtunity to take all of the classes at both, due to some partnership they have.[/quote]

This is close to true, but not in fact true.  You can only cross-register when there is no equivalent course at your own school.  I'm not sure what that means as a practical matter, though (I've never tried to take any MIT courses).  I do know that a few MIT students do take math 55.

Via PM,
[quote="The Sigma"]Hello JBL,

I have a question about Math 25 and 55. Would you personally recommend starting in Math 25 and then moving to Math 55, or would you recommend starting directly in Math 55?

I am asking you this because I am taking advanced math such as differential equations, discrete mathematics, multivariable calculus, and linear algebra; and I was wondering if these courses at my community college would be enough preparation for Math 55.

*These community college courses are part of my high school curicullum, and I have to apply as a freshman.*[/quote]I'm not going to say, based on a list of courses you haven't taken yet and about which I know nothing but the course name, whether you should take math 55.  I would, however, direct your attention to some things I said in the earlier post joml linked to (12th post on that topic), with the most relevant excerpts being:
[quote="JBL"]For students deciding between math 25 and 55 (such as myself), there is a considerable grace period in which to make the decision. (In general, at Harvard, registration happens after one week of classes have passed. In this case, they give you at least 2 additional weeks to make the switch with not penalties whatsoever, and the classes are supposed to parallel each other somewhat for this time.)
...
I would encourage any student coming to Harvard and looking at majoring in math to sit in on {23, 25} or {25, 55}, depending on their strength.[/quote]

I should also amend something I said in that thread: it [i]is[/i] possible to major in math without taking 23 or above.  I have an acquaintance who's doing it (now entering his junior year).  He's a smart guy.  I have no idea why he didn't take 23, but he didn't, and he's doing fine.  However, I still think that it is a less-good idea, and the majority of math majors do take 23, 25 or 55 their freshman year.


hello, Maple gives $1/8\,\sqrt {\pi }
\text{MeijerG}\left(\left[[1],[\frac{3}{2},\frac{5}{2}]\right],\left[[1,1,\frac{1}{2}],[\hspace{0.2cm} ],\right],1\right)$ as the result.
Sonnhard.


trigonometry calculus geometry circumcircle inradius

Prove that triangle $ ABC$ is equilateral if it satisfies
$ cos^{2}\frac{A}{2}\plus{}cos^{2}\frac{B}{2}\plus{}cos^{2}\frac{C}{2}\minus{}2\equal{}\frac{1}{4}cos\frac{A\minus{}B}{2}cos\frac{B\minus{}C}{2}cos\frac{C\minus{}A}{2}$

[quote="stephencheng"]Necessity part: Substitute $ A \equal{} B \equal{} C \equal{} \frac {\pi}{3}$ and done.

Sufficiency part: We need to prove $ 4 \times L.H.S. \equal{} 4 \times R.H.S \le 1$ , i.e. $ cos A \plus{} cos B \plus{} cos C \le \frac {3}{2} \Leftrightarrow cos A \plus{} cos B \minus{} cos (A \plus{} B) \le \frac {3}{2}$ and equality holds only when $ A \equal{} B \equal{} C \equal{} \frac {\pi}{3}$ , which is true by using calculus .[/quote]

Are you sure? This doesn't seem right at all  :maybe:
I may agree if you can prove $ 4LHS<\equal{}1<\equal{}4RHS$
However, that is not true.

The identity $ cos\frac{A-B}_{2}$ =$ \frac{a+b}_{c}}.sin\frac{C}{2}$ and two similar identities obtained by cyclic interchange of variables gives :

$ \prod\frac{cos(A-B)}_{2}$ =$ \frac{(a+b)(b+c)(c+a)}_{abc}}.\frac{r}{4R}$.................(1)

R is the circumradius and r is the inradius of triangle ABC            (r=  4R$ \prod\sin\frac{A}{2}$ )

On the left hand side:

4($ \sum\cos^2\frac{A}{2}$ -2) =     4(1-$ \sum\sin^2\frac{A}{2}$) =      4(1-(1-$ \frac{r}{2R}$)) =            $ \frac{2r}{R}$

For a triangle with sides a,b and c (a+b)(b+c)(c+a)$ \geq$ 8abc, equality occurring when a=b=c , that is when the triangle is equilateral.
Therefore $ \frac{(a+b)(b+c)(c+a)}_{abc}}.\frac{r}{4R}$ $ \geq$ $ \frac{2r}{R}$ with equality only when the triangle is equilateral.

stephencheng how is it true by calculus?


ωραίο πρόβλημα


function
limit
integration
real analysis
real analysis unsolved

For each $ f: \mathbb R\longrightarrow \mathbb R$, let $ \Delta^nf \equal{} \Delta(\Delta^{n \minus{} 1}f)$, in which $ (\Delta f)(x) \equal{} f(x \plus{} 1) \minus{} f(x)$. Let $ f$ be a continuous function and $ \Delta^nf \equal{} 0$. Prove that there are functions $ a_0,\dots,a_{n \minus{} 1}: \mathbb R\longrightarrow\mathbb R$ such that
\[ f(x) \equal{} a_0(x) \plus{} a_1(x) \plus{} \dots \plus{} a_{n \minus{} 1}(x)x^{n \minus{} 1}
\]
and for each $ i,j$: $ \lim_{k\rightarrow \infty}\int_0^1a_i(x)a_j(kx) \equal{} \int_0^1a_i(x)dx\int_0^1a_j(x)dx$

[hide="Hint"]
Of course $ a_i$'s are periodic with period 1. You can use induction.[/hide]


logarithms
calculus
derivative
function
analytic geometry
graphing lines
slope

Consider the system of equations  $ y \equal{} a^x , x \equal{} a^y , x\ne y$ and $ a\in \mathbb R^{ \plus{} }$. Find the range of real number   "$ a$"  so that given system is consistent.

Are you sure you meant to say $ x\ne y$? How could these curves intersect other than on the line $ y = x?$

Anyway, to continue the thought -- drawing pictures leads us to the following idea:

The only possible intersections of these two curves occur in the first quadrant.

If $ 0<a\le 1,$ then the curves $ y=a^x$ and $ x=a^y$ intersect exactly once.

If $ 1<a<a_0$ (for an $ a_0$ yet to be determined), these curves intersect twice.

If $ a>a_0,$ then these curves never intersect. Note: since $ e^x>1+x$ and $ \ln x<1+x,$ we must have $ e>a_0.$

Finally, if $ a=a_0,$ the curves intersect once, at a point at which they are tangent to each other and to the line $ y=x.$

So what is this borderline case $ a_0?$ We must have a point $ x$ such that $ x=a^x$ and (by setting the derivative equal to that of $ y=x,$) $ a^x\ln a=1.$ Then, since $ x=a^x,$ we have $ x\ln a=1$ or $ x=\frac1{\ln a}.$ But then $ a^x=e^{x\ln a}=e^1=e,$ from which $ \ln \frac1e$ or $ a=e^{e^{-1}}.$ So that's our borderline case: $ a_0=e^{e^{-1}}\approxeq 1.447.$

If $ a<e^{e^{-1},}$ the curves intersect - once for $ a\le 1,$ twice for $ a>1.$

If $ a=e^{e^{-1}},$ the curves intersect once at a point of common tangency.

If $ a>e^{e^{-1},}$ the curves do not intersect.

[quote="Kent Merryfield"]Are you sure you meant to say $ x\ne y$? How could these curves intersect other than on the line $ y = x?$

Anyway, to continue the thought -- drawing pictures leads us to the following idea:

[hide]The only possible intersections of these two curves occur in the first quadrant.

If $ 0 < a\le 1,$ then the curves $ y = a^x$ and $ x = a^y$ intersect exactly once.

If $ 1 < a < a_0$ (for an $ a_0$ yet to be determined), these curves intersect twice.

If $ a > a_0,$ then these curves never intersect. Note: since $ e^x > 1 + x$ and $ \ln x < 1 + x,$ we must have $ e > a_0.$

Finally, if $ a = a_0,$ the curves intersect once, at a point at which they are tangent to each other and to the line $ y = x.$

So what is this borderline case $ a_0?$ We must have a point $ x$ such that $ x = a^x$ and (by setting the derivative equal to that of $ y = x,$) $ a^x\ln a = 1.$ Then, since $ x = a^x,$ we have $ x\ln a = 1$ or $ x = \frac1{\ln a}.$ But then $ a^x = e^{x\ln a} = e^1 = e,$ from which $ \ln \frac1e$ or $ a = e^{e^{ - 1}}.$ So that's our borderline case: $ a_0 = e^{e^{ - 1}}\approxeq 1.447.$

If $ a < e^{e^{ - 1},}$ the curves intersect - once for $ a\le 1,$ twice for $ a > 1.$

If $ a = e^{e^{ - 1}},$ the curves intersect once at a point of common tangency.

If $ a > e^{e^{ - 1},}$ the curves do not intersect.[/hide][/quote]

First i will give some background for the intersection of function with its inverse, I think it is a common misconception that they can  intersect at $ y = x$ only but to my understanding it is as follows:

If $ f(x)$ is an increasing function then solution set of $ f(x) = x\Longleftrightarrow$ solution set of $ f(x) = f^{ - 1}(x)$
But in case of $ f(x)$ is decreasing,$ f(x) = x\implies$ solution set of $ f(x) = f^{ - 1}(x)$ and converse may not be true.

Explanation:

Note regardless of $ f(x)$ increasing or decreasing we have
$ f(x) = x\implies f^{ - 1}(f(x)) = f^{ - 1}(x)\implies x = f^{ - 1}(x)\implies f^{ - 1}(x) = f(x)$

for converse,
Note $ f(x)$ is an increasing $ \Longleftrightarrow f^{ - 1}(x)$ is an increasing
Let $ x_{1}$ be a solution of $ f(x) = f^{ - 1}(x)$ and $ f(x_{1})\neq x_{1}$
suppose $ f(x_{1}) < x_{1}\implies f^{ - 1}(f(x_{1})) < f^{ - 1}(x_{1})$ (as $ f^{ - 1}(x)$ is an increasing)
$ \implies x_{1} < f^{ - 1}(x_{1}) = f(x_{1})$
Which is a contradiction
[size=117][b]
Hence in case of f is an increasing function we have "solution set of $ f(x) = x\Longleftrightarrow$ solution set of $ f(x) = f^{ - 1}(x)$"[/b][/size]

Now in case of a decreasing function if their is any point of intersection of the function with its inverse below the line $ y = x$ then its mirror image will be also a point of intersection.
e.g. consider $ f(x) = 1 - x^{3}$ and its inverse $ f^{ - 1}(x) = (1 - x)^{\frac {1}{3}}$; Clearly they are intersecting at $ (1,0)$ and at $ (0,1),$ both the points are not on the line $ y = x$
Observe why it is happening?
Because in our example function $ f(x)$ contains two points $ (1,0)$ and $ (0,1)$ which are mirror image of each other in the line mirror $ y = x$.

[size=117][b]Hence whenever curve $ y = f(x)$ contains two points symmetric to the line $ y = x,$ its inverse will pass through these points which are not the points of $ y = x$ [/b][/size]

[color=blue]Now for the problem before posting my solution i am giving an e.g.
Consider $ a = \frac {1}{16}$
Now see $ y = \left(\frac {1}{16}\right)^{x}$ and $ x = \left(\frac {1}{16}\right)^{y}$
Observe they are intersecting each other at $ \left(\frac {1}{2},\frac {1}{4}\right)$ and at $ \left(\frac {1}{4},\frac {1}{2}\right) .$ both are not on $ y = x .$[/color]
Dear Kent Merryfield please give your comment on my theoretical part.

Ah - I didn't draw enough pictures! I missed that configuration.

For the decreasing exponentials, the key issue is how steeply the curve is decreasing when it crosses the line $ y\equal{}x.$ The borderline case for this would be when the derivative equals $ \minus{}1$ at the point of crossing. Hence, for the borderline case:

$ a^x\ln a\equal{}\minus{}1$ when $ x\equal{}a^x.$ So $ x\equal{}\minus{}\frac1{\ln a},$ so $ \minus{}\frac1{\ln a}\equal{}e^{\minus{}\frac1{\ln a}}\equal{}e^{\minus{}1}.$ From this we get $ a\equal{}e^{\minus{}e}\approxeq .06599.$ Note that $ \frac1{16}$ is indeed less than this borderline case. So the revised statement:

If $ 0<a<e^{\minus{}e},$ the curves $ y\equal{}a^x$ and $ x\equal{}a^y$ intersect three times - once on the line $ y\equal{}x$ and twice off of that line.

If $ e^{\minus{}e}\le a\le 1,$ the curves $ y\equal{}a^x$ and $ x\equal{}a^y$ intersect once, on the line $ y\equal{}x.$

If $ 1<a<e^{e^{\minus{}1}},$ the curves $ y\equal{}a^x$ and $ x\equal{}a^y$ intersect twice, both on the line $ y\equal{}x.$

If $ a\equal{}e^{e^{\minus{}1}},$ the curves $ y\equal{}a^x$ and $ x\equal{}a^y$ intersect once at $ (e,e),$ which is a point of common tangency.

If $ a>e^{e^{\minus{}1}},$ the curves $ y\equal{}a^x$ and $ x\equal{}a^y$ do not intersect.

[quote="Kent Merryfield"]Ah - I didn't draw enough pictures! I missed that configuration.

For the decreasing exponentials, the key issue is how steeply the curve is decreasing when it crosses the line $ y \equal{} x.$ The borderline case for this would be when the derivative equals $ \minus{} 1$ at the point of crossing. Hence, for the borderline case:

$ a^x\ln a \equal{} \minus{} 1$ when $ x \equal{} a^x.$ So $ x \equal{} \minus{} \frac1{\ln a},$ so $ \minus{} \frac1{\ln a} \equal{} e^{ \minus{} \frac1{\ln a}} \equal{} e^{ \minus{} 1}.$ From this we get $ a \equal{} e^{ \minus{} e}\approxeq .06599.$ Note that $ \frac1{16}$ is indeed less than this borderline case. So the revised statement:

If $ 0 < a < e^{ \minus{} e},$ the curves $ y \equal{} a^x$ and $ x \equal{} a^y$ intersect three times - once on the line $ y \equal{} x$ and twice off of that line.

[hide]If $ e^{ \minus{} e}\le a\le 1,$ the curves $ y \equal{} a^x$ and $ x \equal{} a^y$ intersect once, on the line $ y \equal{} x.$

If $ 1 < a < e^{e^{ \minus{} 1}},$ the curves $ y \equal{} a^x$ and $ x \equal{} a^y$ intersect twice, both on the line $ y \equal{} x.$

If $ a \equal{} e^{e^{ \minus{} 1}},$ the curves $ y \equal{} a^x$ and $ x \equal{} a^y$ intersect once at $ (e,e),$ which is a point of common tangency.

If $ a > e^{e^{ \minus{} 1}},$ the curves $ y \equal{} a^x$ and $ x \equal{} a^y$ do not intersect[/hide].[/quote]

First very-very thanks for the complete and excellent solution. I also did this problem in same way.
Yes you are absolutely correct  :idea: "For the decreasing exponential functions, the key issue is how steeply the curve is decreasing when it crosses the line $ y \equal{} x.$  The borderline case for this would be when the derivative equals $ \minus{} 1$ at the point of crossing. It is also important to see for slope $ < \minus{} 1$ they will intersect thrice only; once above $ y \equal{} x$ and once below $ y \equal{} x$. :!:  I leave this part for readers to prove as an exercise (an easy exercise)
Answer to my question is: $ 0 < a < e^{ \minus{} e},$
I was planning to put each case as a separate question  but you already taken all cases in one go.Once again thanks for agreeing with my point of view. :lol:

Just for after thought:
I also tried this problem considering: $ y \equal{} a^x , x \equal{} a^y , x\ne y$  and  $ a\in \mathbb R^{ \plus{} }.$
$ \implies a \equal{} y^{\frac {1}{x}} \equal{} x^{\frac {1}{y}}$
$ \implies y^{y} \equal{} x^{x}$
Now Consider $ f(t) \equal{} t^{t},t\in \mathbb R^{ \plus{} }$
Notice the function $ f(t)$ is a strictly decreasing function for $ t \in \left(0,\frac {1}{e}\right]$ and increasing for $ t \in \left[\frac {1}{e},\infty \right)$. Also as $ x\rightarrow 0, t^{t}\rightarrow 1$ and $ f(1) \equal{} 1,$
Since we are interested in two values of $ t$ say $ x,y$ and  $ x\ne y$ ; we will consider $ f(t)$ in many to one region i.e.  $ t\in (0,1)$
our $ x \in\left(0,\frac {1}{e}\right)$ and $ y\in\left(\frac {1}{e},1\right)$
As  $ x$  increases from  $ 0$  to $ \frac {1}{e}$ , corresponding $ y$  decreases from  $ 1$  to  $ \frac {1}{e} .$
$ \implies$ range of $ x^{\frac {1}{y}}$ is $ \left(0,{\left(\frac {1}{e}\right)}^\frac {1}{\left(\frac {1}{e}\right)}\right) \equal{} \left(0,e^{ \minus{} e}\right)$
$ \implies$ desired $ a \in \left(0,e^{ \minus{} e}\right)$


Given triangle ABC with B<C which incribed in (O).AO intersects BC at D.E,F are circumcenters of triangle ABD and ACD respectively.Show that: the midpoints of AD,BE,CF is collinear iff C-B=60

A solution is basically based on the two following facts:

Lemma 1. Let ABC and A’B’C two similar triangles. If I = c(ABC) /\ c(A’B’C), then I lies on AA’ (and also lies on BB’).

Lemma 2. Let O be the circumcenter of ABC. If D = AO/\BC, then <EBC = <ACB, where E is the circumcenter od ABD.

I will later prove the lemmas and give all details of the proof.

[quote="sprmnt21"]I = c(ABC) /\ c(A’B’C), then I lies on AA’ (and also lies on BB’).
.[/quote]
  What does  c(ABC) mean here?Thank a lot!

with c(XYZ)  I denote the circumcircle of triangle XYZ.


geometry
geometric series
Pythagorean Theorem
similar triangles

In an iscoceles triangle, Circle A with radius $2$ is inscribed, which is tangent to all three sides of the triangle and is also tangent to another circle (Circle B which is above Circle A). Circle B has radius of $1$ and is tangent to Circle A and the two equal sides of the triangle. What is the area of the iscoceles triangle?

Source: USC Math Test #30

My reasoning so far:

[hide]Since Circle A has radius of $2$ and Circle B has radius of $1$ and since they are in the same triangle, you can draw a base and make Circle B inscribed in a triangle which is similar to the larger triangle. If this progression continues, you can have another circle, with radius of $1/2$ and another of radius $1/4$ and so on. So the diameters would be $4,2,1,1/2,1/4.........$ so the sum of all these are the sum of an infinite geometric series, so the sum is $8$, which is the height of the iscoceles triangle, however, I just can't find the base :(. [/hide]

I forgot which test I got this question from so I can't find the answer or solution so can someone help :roll:

This is how you find the base.

[hide="hint"] Remember that a line which is tangent to a circle is perpendicular to the radius of the circle which goes to the point of tangency.  Also remember the pythagorean theorem. :D [/hide]

[hide="solution"] As you said, the height is 8.  Now draw the radius from the center of $A$ (let that center be $A$) to one of the legs of the isoceles triangle.  Let the point of tangency be $C$.  You know that $AC=2$ and $AD=6$ (with $D$ being the vertex of the triangle).  Now use the pythagorean theorem to find that $CD=4\sqrt{2}$  Using the fact that triangles $ACD~EAD$ (where E is the point on the leg containing the point C such that AE is parallel to the base), we can find that $DE=\frac{9}{\sqrt{2}}$ and thus $AE=\frac{3}{\sqrt{2}}$.  Thus, using similar triangles, the base has length $4\sqrt{2}$, and thus the area of the triangle is $16\sqrt{2}$.[/hide]


We had to do some database maintenance this morning.  We think we've taken care of all the issues that came up, but if you get any database errors (or other new bizarre behavior), please post it here.

Okay....so that answers my old question.

New bizarre behavior:  Whenever i go to FTW, "View cart" appears at the top of the sidebar.  It's highlighted, like I had something in there.  WHen I click it, however, it says "Your shopping cart is empty".

FF 3.0 for OSX

Noted (I have the same problem).  We should have that fixed in the next day or so.  But I have confirmed that you can make the button go away by buying something :)

:D No thanks, I already have all the AoPS books except for Intermediate C&P.

Another bug:  My internet wasn't lagging at all (multiple tabs) but when I tried to post something, it said it couldn't connect to AoPS server.  However i was using aops in other tabs, so that shoudln't have been the problem.  When I pressed "try again", it said "You message has been posted" or that generic message it always says.

Well, not on my past 5 posts...but it happened twice.

Has the server gone down TWICE today, or is just my cmputer?

It had problems again an hour ago.  Let me know if anything is still broken.


Five people play a game where each player has a chance of ½ for winning any match (for example, a coin toss). All of the five players play against all other four. What is the probability that each player wins exactly two games?

[hide]I think the answer is $ \frac {3}{128}$. Call the teams 1,2,3,4, and 5. First consider the games 1-2, 1-3, 1-4, 1-5.
We can assume 1 beats 2 and 3 and loses to 4 and 5. There is a $ \frac {{4 \choose 2}}{16} \equal{} \frac {3}{8}$ chance 1 will lose exactly two games. Now consider the game 2-3. We can assume 2 beats 3. Now, the probability 2 will lose to exactly one of either 4 or 5 is $ \frac {1}{2}$, we can assume 2 beats 4, loses to 5. But then all of the rest of the games are determined. 3 and 4 must beat 5, as 5 has already one two games, and then 3 must beat 4. There is a $ \frac {1}{8}$ probability these final three games will occur like this. The final answer is $ \frac {3}{8} \cdot \frac {1}{2} \cdot \frac {1}{8} \equal{} \frac {3}{128}$.[/hide]


[quote="bjh7790"]there are non-negative numbers x, y, z  that satisfy $ x \plus{} y \plus{} z \equal{} 2$

show that $ (x^3 \plus{} y^3)(y^3 \plus{} z^3)(z^3 \plus{} x^3) \le 2$[/quote]
Let $ f(x,y,z)\equal{}2\minus{}(x^3 \plus{} y^3)(y^3 \plus{} z^3)(z^3 \plus{} x^3)$ and $ x\geq y\geq z.$
Hence, $ f(x,y,z)\minus{}f\left(x\plus{}\frac{z}{2},y\plus{}\frac{z}{2},0\right)\geq0$ because $ \left(x\plus{}\frac{z}{2}\right)^3\plus{}\left(y\plus{}\frac{z}{2}\right)^3\geq x^3\plus{}y^3,$
$ \left(x\plus{}\frac{z}{2}\right)^3\geq x^3\plus{}z^3$ and $ \left(y\plus{}\frac{z}{2}\right)^3\geq y^3\plus{}z^3.$
Hence, it remains to check that $ f(x,y,0)\geq0$ for $ x\plus{}y\equal{}2,$ which is obvious.


Find the integer numbers $ x,y,z$  such that  $ \frac {{x^2 \plus{} x \plus{} 1}}{{xy \minus{} 1}} \equal{} z$

Oh! That's old question!  :D  :D

[quote="Kouichi Nakagawa"]$ x, y \in \mathbb{Z}$?
or
$ x, y \in \mathbb{N}$?[/quote]

All $ x,y,z \in Z$. The solution is very cool!  :roll:  :roll:

[quote="Kouichi Nakagawa"]Easy Solution
$ (x, y, z) \equal{} (a, 0, a^2 \plus{} a \plus{} 1), (0, a, \minus{} 1), (a, \minus{} a \plus{} 1, \minus{} 1)$
But $ (x, y) \neq ( \minus{} 1, \minus{} 1), (1, 1)$

Another Soluction
$ (x, y, z) \equal{} ( \minus{} 2, \minus{} 2, 1), ( \minus{} 2, \minus{} 1, 3), ( \minus{} 1, \minus{} 2, 1), (1, 2, 3), (1, 4, 1), (2, 1, 7), (2, 4, 1), (4, 1, 7), (4, 2, 3)$[/quote]

Can you solve detailly!  :wink:  I need a detail solution! Thanks very much! :)


superior algebra
superior algebra unsolved

Suppose that $ F$ is a field of characteristic $ p$ and that $ \alpha$ is a separable element in some extension. Show that then $ F(\alpha^p)\equal{}F(\alpha)$.

here is a hint.
if $ \alpha^p$ satisfy $ f$, that is $ f(\alpha^p)\equal{}0$,
then since $ f(x^p)\equal{}f(x)^p$,
$ \alpha$ satisfy $ f$ too.


factorial
number theory proposed
number theory

Forgive me if this is a widely-known result, but I cannot find the answer.

Let S be a set of natural numbers such that the reciprocal sum of S diverges.
Is it necessarily true that there exists a fixed C such that every sufficiently large natural number can be expressed as a sum of C elements of S?

If not, can you give me a counterexample?

UPDATE: I just realized. Let S be the positive multiples of 5. All of their sums only add up to multiples of 5. So apparently, my statement as originally given is not true. But somehow this case where the set is an ideal seems like a trivial exception. Then again, we do have one kind of exception. Are there more interesting exceptions?

Here is a more concise way to ask the question.

Let S be a [url=http://en.wikipedia.org/wiki/Large_set_(combinatorics)]large set[/url] of natural numbers that is not a subset of a [url=http://mathworld.wolfram.com/ProperIdeal.html]proper ideal[/url].
Is S necessarily an [url=http://planetmath.org/encyclopedia/AsymptoticBasis.html]asymptotic basis[/url]?

Obviously not. Take a long interval, skip much longer interval, take an even longer interval, skip an even much longer interval, and so on. (If you want an explicit counterexample, $ S\equal{}\bigcup_{n\text{ odd}}[n!,(n\plus{}1)!]$ is one).

Yes, I see. Very nice example.

Divergence of the reciprocal sum can easily be shown by
sum > 1/(1+1) + 3/(3+1) + 5/(5+1) + ...

The fact it is not an asymptotic basis is immediate when you consider each number one less than the factorial of an odd number... they require more and more terms to be reached. Those are some mad gaps.

Thanks!


[quote="dragije"]On [url]http://jeka-math.blogspot.com[/url] I found this problem:
Prove that (2^2008 + 2042)^3 is divisible by 511, and I can`t solve it!
Any help would be appriciated.[/quote]

[hide="Hint"]Compute $ 2^9$ modulus $ 511$
Compute then $ 2^{9\times 223\plus{}1}$ modulus $ 511$
Compute then $ 2042$ modulus $ 511$[/hide]

Ty all.Helped me a lot.
I think I could try to solve harder problems on that site.


topology
advanced fields
advanced fields unsolved

given we say that a topological space $ X$ is a $ n+2$-quasimanifold if it's a co-countable subspace of a [i]connected[/i] $ n+2$-manifold $ M(X)$. ($ n\ge 0$).

prove that $ \pi_{k}(X) = \pi_{k}(M(X))$ for any $ 0\le k \le n$, and for any possible choice of $ M(X)$*.

just to give a shorter statement:
"low" homotopy groups of manifolds remain unaffected by removing countably many points.

* for example, $ X = \mathbb{R}^{2}\setminus\{0\}$ is a quasi manifold, and we can choose $ M(X) = X,\mathbb{R}^{2}, S^{2}$. in any case, $ X$ and $ M(X)$ are both connected, which means $ \pi_{0}(X) = \pi_{0}(M(X))$.

[color=darkred][size=75]EDIT: just a matter of english: from "countable points" to "countably many points"[/size][/color]

Vaguely speaking, the image of a homotopy between representatives of an element of $ \pi_{k}$ is "$ (k+1)$-dimensional", and since $ k+1<n+2$, a random perturbation of the homotopy will almost surely avoid the countable set.

However, this is too vague. For example, let $ X=\mathbb R^{3}\setminus \mathbb Q^{3}$ and let $ \gamma\colon S^{1}\to X$ be a closed curve of positive $ 3$-dimensional measure. How do we shrink $ \gamma$ into a point?  :maybe: