Scores came out online for the January SAT. What'd y'all get?

I got:

CR: 520 :(
M: 790 :( (DANG I MADE A STUPID MISTAKE ON JUST ONE QUESTION)
W: 710 :D

With a 10/12 on the essay.

CR: 570
M: 790 (Same boat as you.)
W: 640

Got a 7/12 on my essay.

Nice writing score!

Wait, CR = Critical Reading, W = ?

I'm going to be taking it soon and I have to drive an hour to the testing location because the one that was actually close to us was full.  :ninja:

W = Writing

garr, for me from two years ago
CR: 680 (  :mad: )
W: 760
M: 790 (  :blush: )

[quote="Stargirl240"]*Sigh*... I feel stupid, lol.
CR=680
Math=720 (*cries*...)  :o  :(
Writing=610
And from the other thread, this thread wasn't up when I asked![/quote]
LOL YOU SHOULD HAVE JUST CLICKED, "NEW TOPIC" and made the thread yourself! :P

xD Also, what did y'all get on teh essay pplz?

[quote="Stargirl240"]You can see the essay score? Lol, I probably wrote at a... 3rd grade level? Meh. Above 2000, all I was aiming for lol. Did you take ACT's?[/quote]
I didn't do the ACT's. But when you go online to see your scores, it shows your multiple choice raw score and your essay score.

In 8th grade I got
CR - 690
Maths - 790 (we all love getting pwned by one question don't we)
Writing - 690 (essay 9)
I'm not taking it this year. Next time I take it, I'm gonna be aiming for a >2300...

Wow it's only been a year since I took my last SAT (the retired version), yet it feels like 2-3 years. (Hanging around this forum makes me seem too old :p.)

My score when I was in 8th grade on retired:

CR: 540
Math: 590 (Epic fail. This is the first time I ever used that phrase)

[quote="avec_une_h"]I got perfect sentence correction...  I'm not great at writing in general, just grammar.[/quote]
Same here. But I had a 10 on my essay.

And hello people here, how do I stop making stupid mistakes in the "easy" math tests? It's like I always tend to do so in SAT Math, IL Math League, MATHCOUNTS, AMC 10, and like a bunch of random crap.

Oh, wait til you take the AIME. Then you'll know how badly careless mistakes hurt.

Uh, yeah...I *MUST* get greater than an 8 on this year's AIME...or else I won't make USAMO :maybe:  :o  :|

don't. make. dumb. mistakes.

read the problems really carefully. walking next to a moving walkway is totally different from walking on it.
although, i didn't fall for that. i actually kept making stupid mistakes and got non-integer answers for that problem, so i read the problem very, very carefully...

[quote="ManHatTaNatiVe94"]Lol Rose weren't you like National Spelling Bee contestant?  :P

Also, randomly, I saw you at that one scholastic bowl invitational at Wheaton North a month ago I believe, because I remember Oak Park was there.[/quote]

Yeah, I was in the National Spelling Bee.

Was it the fresh/soph Scholastic Bowl meet?  If so, we were there.  Our team came in second.


calculus
integration
algorithm
LaTeX
conics
parabola
group theory

find all $x,y\in\mathbb{Z}$ such that$x^{2}+x+2=y^{3}$

If $(x,y)$ is a solution, then $(-1-x,y)$ is also a solution. Moreover, $RHS = LHS > 0$. Thus, we can just consider $x,y \geq 0$.

$x = 0 \Rightarrow y^{3}= 2$
$x = 1 \Rightarrow y^{3}= 4$
$x = 2 \Rightarrow y^{3}= 8 \Rightarrow y=2$
Since $(2,2)$ is a solution, then $(-1-2,2) = (-3,2)$ is also a solution.

*) For $x \geq 2$,
$x^{3}-(x^{2}+x+2) = (x-2)(x^{2}+x+1) > 0 \Rightarrow x^{3}> x^{2}+x+2$ for $x > 2$. Then, let $x^{2}+x+2 = (x-2k)^{3}$ for a positive integer $k$.
We have $P(x) = x^{3}-x^{2}(6k+1)+x(12k^{2}-1)+(8k^{3}-2) = 0$
By AM-GM, $x^{3}+x(12k^{2}-1) > 2x^{2}\sqrt{12k^{2}-1}> x^{2}(6k+1)$ for $k \geq 2$ and yields $P(x) > 0$.
$k=1$ gives $x^{3}-7x^{2}+11x+6=0$ which gives no integer solutions.

Hence, the solutions are $(-3,2)$ and $(2,2)$

EDIT : @Albanian Eagle : Have I corrected the mistake ? Thx..  :)

your solution is wrong as it is !
check your calculations :wink:

er....
you want to prove $p(x)=(x-2k)^{3}-(x^{2}+x+2)>0$ when k>=2?
I can't understand you.
I think no matter how you modify your answer ,you can't prove this :P

[quote="Noobiel"]Trivial (elliptic curve)[/quote]
I don't know anything about elliptic curve,can you say the proposition or everything that make this problem trivial?is there infinitely many solution?

RDeep, could you please explain why if $ (x,y) is a solution   then so is (\minus{}1\minus{}x,y)$ ? Thanks :)
Btw I did a preview, and know there are no spaces (I could have just did two sets of cash marks for the brackets) but to learn, how do I get spaces in there? :S

This follows from the calculation of the automorphism group.

[quote="Euclid's Turtle"]RDeep, could you please explain why if $ (x,y) is a solution then so is ( - 1 - x,y)$ ? Thanks :)
Btw I did a preview, and know there are no spaces (I could have just did two sets of cash marks for the brackets) but to learn, how do I get spaces in there? :S[/quote]
For the math question: if $ x,y$ is a solution, put $ x^\prime=-1-x,y^\prime=y$ into the equation. Simple calculation gives that it is again a solution.
For the $ \text{\LaTeX}$ question: If you want to insert text, use \text{insert text here}, this gives $ \text{insert text here}$. But it's often better to close the TeXcode and write normally (like in: $ abc$ text here $ def$).

Thanks ZetaX :)
And Noobiel, calculation of the automorphism group? Woosh over my head :)

Also, yes I see that it gives the same, $ x' \equal{} \minus{}1\minus{}x$ but ... is that just skill and hindsight?

[quote="Noobiel"]An elliptic curve has finitely many integral points, so they're torsion, and there is an algorithm for calculating the torsion group. As our curve is defined over the rationals, there is even an upper bound: http://en.wikipedia.org/wiki/Mazur's_torsion_theorem Probably there is some implementation in the internet. You can read Silverman, Arithmetic of elliptic curves.[/quote]
Thanks Noobiel :)
what about$ x^2\plus{}x\plus{}1\equal{}y^3$?or $ x^2\equal{}y^3$(obviusly it has infainitely many solution)

[quote="Euclid's Turtle"]Also, yes I see that it gives the same, $ x' \equal{} \minus{} 1 \minus{} x$ but ... is that just skill and hindsight?[/quote]
Any parabola has a line of symmetry.

[quote="ali666"]what about$ x^2 \plus{} x \plus{} 1 \equal{} y^3$?or $ x^2 \equal{} y^3$(obviusly it has infainitely many solution)[/quote]
This is a [b]singular[/b] elliptic curve; it has a singular point at $ (0, 0)$.  The interesting theorems about elliptic curves apply to nonsingular elliptic curves.  By the way, Noobiel, I don't think it's true that every integral point is in the torsion subgroup (you're probably thinking of Nagell-Lutz, which is not an if-and-only-if).  The problem of finding all of the integer points is algorithmic.

is it true that every nonsingular elliptic curve has finitely many integral points?and therefore$ x^2\plus{}b\equal{}4y^3$(which $ b$ is nanzero) have finitely many solutions?
thanks... :D


topology
geometry
geometric transformation
reflection
advanced fields
advanced fields unsolved

Let $M$ be a closed (i.e. compact, boundaryless) connected smooth manifold. Also, let $f: M\to\mathbb R$ be a non-constant smooth function. Prove that we can find diffeomorphisms $h_{i}: M\to M,\ i=\overline{1,N}$ such that the map sending $p\in M$ to $(f\circ h_{1}(p),\ldots,f\circ h_{N}(p))$ is an embedding of $M$ in $\mathbb R^{N}$.

(here, smooth means $C^\infty$, and embedding means one-to-one immersion)

uhm..
maybe there's something to work out, but this idea seems to solve the question:
let $n$ be the dimension of our manifold, and consider $n$ smooth maps $g_{i}$ from $\mathbb{R}^{n}$ to itself such that $g_{i}$ restricted to $B_{n}$ (the unitary ball) is a reflection through $H_{i}= e_{i}^{\perp}$, and outside $2B_{n}$ is the identity.
now cover your manifolds with charts $\psi_{p}: \mathbb{R}^{n}\to M$ such that $\psi_{p}(0)=p$, one for each $p\in M$.
apply $g_{i}$ to each of these charts, and paste each automorphism $g_{p,i}= \psi_{p}\circ g_{i}\psi_{p}^{-1}$ to the identity on $g_{i}(\mathbb{R}^{n})^{c}$. this gives a locally injective map $g_{p}= (g_{p,1},\dots,g_{p,n})$ (around p).
finally, fix a point $%Error. "widebar" is a bad command.
{p}$ such that $df_{p}\neq 0$, and move this point around with automorphisms of $M$ (which act transitively on $M$ itself): you obtain a cover of $M$ made of open subsets which $g_{p}$ are injective on (you can choose them around $p$). since $M$ is compact, you choose a subcover (indexed by points $p_{1},\dots,p_{m}$).
now, let $f_{i}= g_{p_{i}}$: i guess that if you call $F = (f_{1},\dots,f_{m}): M\to\mathbb{R}^{mn}$ you have an [i]imbedding[/i] of $M$, and $M$ should be a covering of its image.
but, since the map is locally injective, it should suffice to separate points adding some more $f\circ h$, where $h$ is such that separates properly those points.. it shouldn't be difficult (i hope :p)

uhm.. does anyone have a nice solution, anyway?


advanced fields
advanced fields unsolved

Some simple thoughts lead me to the following conjecture I have not yet been able to verify:

Let $ X,Y$ be Banach spaces (or, more generally, Fréchet spaces) and $ \sigma, \tau$ the respective weak topologies. Let $ T: X \to Y$ be a linear mapping.
Then $ T$ is continuous if and only if $ T: (X,\sigma)\to (Y,\tau)$ is continuous.

Sorry for bothering you, I think I have a proof now and it's actually pretty simple.


Hi can someone help me on this problem for my combinatorics class?  I need a COMBINATORIAL PROOF for this problem to get some extra credit for this class.  I don't really need the extra credit, but it would be really nice, plus I want to know how to do this problem.

Prove:

  n
SUM [(2k choose k) * (2n-2k choose n-k)]  =  4^n
k=0

See

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=4%5En&t=40150

This problem is not that out of place, but we usually put problems like these in the Olympiad combinatorics forum.

Yeah, I checked that.  The combinatorial proof is a bit vague, though.  I don't get the part about the "easy bijection" between paths from (0,0) to (2k, 0) and random walking 2k steps from (0,0) without crossing the x-axis.  Could someone please elaborate on that point?  Thanks.


[quote="Raja Oktovin"]Let $ C_1$ be a circle and $ P$ be a fixed point outside the circle $ C_1$. Quadrilateral $ ABCD$ lies on the circle $ C_1$ such that rays $ AB$ and $ CD$ intersect at $ P$. Let $ E$ be the intersection of $ AC$ and $ BD$.
(a) Prove that the circumcircle of triangle $ ADE$ and the circumcircle of triangle $ BEC$ pass through a fixed point.
(b) Find the the locus of point $ E$.[/quote]
a, Let $ PK,PQ$ are two tangents from P to $ (C_1), PO\cap KQ \equal{} \{H\}, CH\cap (C_1) \equal{} \{J\}, BH\cap (C_1) \equal{} \{F\}$
We have $ HA.HO \equal{} HK.HQ \equal{} HC.HJ$
so PJOC is cyclic we get $ \angle OJH \equal{} \angle OPC (1)$
On the other hand, $ OH.OP \equal{} R^2 \equal{} OD^2$ then $ \frac {OH}{OD} \equal{} \frac {OD}{OA}$
therefore $ \Delta HOD$~$ \Delta DOP$ then $ \angle OPC \equal{} \angle ODH (2)$
From (1) and (2) we have $ \angle HJO \equal{} \angle HDO$ so $ HO\perp JD$
therefore $ JD//KQ$
Similarly, $ AF//KQ//JD$ then $ arcAD \equal{} arcJF$
We get $ \angle BHC \equal{} \angle BEC$ then BHEC is cyclic, similarly AHED is cyclic
So the circumcircle of triangle $ ADE$ and the circumcircle of triangle $ BEC$ pass through a fixed point H
b, We have $ 180^o \minus{} \angle BHE \equal{} \angle BCE \equal{} 1/2(arc BK \plus{} arc KA) \equal{} 1/2(arc BK \plus{} arc QF)$
$ \equal{} \angle BHK$
Then K,H,E is collinear
So $ E\in [KQ]$

Let $O$ be the center of $C_1$ and $\tau$ be the polar of $P$ WRT $C_1$ (Locus of E), $F \equiv OP \cap \tau.$ Denote $\omega_1,\omega_2$  the circumcircles of $\triangle EBC ,\triangle PBC,$ respectively. Line $AD \equiv a$ is both image of $\omega_1$ and $\omega_2$ through the inversions with poles $E,P$ and powers $p(E,C_1),$  $p(P,C_1),$ respectively. Hence, by conformity $\angle (a,C_1)=\angle (\omega_1,C_1)=\angle (\omega_2,C_1).$  Since $B,C$ are double points in the inversion $\mathcal{I}$ about $C_1(O),$ then  $\omega_1$ and $\omega_2$ are homologous under $\mathcal{I}.$ Keeping in mind that $\mathcal{I}: F \mapsto P,$  it follows that $F \in \odot(BEC).$ By analogous reasoning we get that $F \in \odot(AED).$


Compute all (x,y) such that
$xy+11=x^{2}$
$xy-7=y^{2}$
my bad, yes I changed the problem  sorry

[hide]
This isn't going to be very detailed:
Subtract the 1st equation from the 2nd to get $(x-y)(x+y)=18$. Call this equation 3.
Add the 1st equation to the 2nd to get $(x-y)^{2}=4$
So $x-y=2~\text{or}-2$
Substitute back into equation 3 and $x+y=9~\text{or}~x+y=-9$
So the possibilities for $(x,y)$ are $(\frac{11}{2},\frac{7}{2})$ and $(\frac{-11}{2},\frac{-7}{2})$
[/hide]

I think I missed something  :maybe:

CircleSquared, your solution looks rights.
It does not restrict it to integers though.

Ummmm... my answer isn't integers, if that's what you mean.
I'm pretty sure my solution covers all reals. :maybe:

[quote="CircleSquared"]
Add the 1st equation to the 2nd to get $(x-y)^{2}=4$
So $x-y=2~\text{or}-2$
[/quote]

 :maybe: Like the part of the solution, but I could be wrong. :wink:

[quote="CircleSquared"]Umm, how can there be non-integer real roots of 4?[/quote]

what are you talkign about, I think your answer is right, I just made up this problem to help me with these types, plugging your answers back in we get it right.


I have 6 shirts, 4 pairs of pants, and 6 hats. The pants come in tan, black, blue, and gray. The shirts and hats come in those colors, and also white and yellow. I refuse to wear an outfit in which all 3 items are the same color. How many choices for outfits, consisting of one shirt, one hat, and one pair of pants, do I have?

If there were no restrictions, he would have $ 6\times{4}\times{6}\equal{}144$ possible choices. The restriction stated implies that there are $ 4$ outfits we must reject (all tan, black, blue or gray). So the answer is $ 144\minus{}4\equal{}\boxed{140}$.

[quote="mathyoga"]If there were no restrictions there would be 4*6*4 or 144 ways. But wouldn't the 4 other restrictions come from the shirts and hats 2*2.[/quote]

Actually, not, because the question states that the person wouldn't wear an outfit where all [u]3[/u] items are the same color, not just 2.


Professor I has 240 students, professor II has 120 students, porfessor III has 40 students. The avg. are 2.75 for professor I, 2.00 for professor II and, 1.3 for professor III. Use matrix multiplication to determine how many A's, B's, C's, D's and F's there are in total.
 :maybe:

I started it like this: 400 students in total, 2.016667 avg. for all of the students. But, how will I go about putting that into a matrix. Thanks in advance for help.

Not enough information. Professor II could have given $ C$s to the entire class, or he could have given half $ B$s and half $ D$s.

Sorry about that.

I didnt think you need the first part if a problem b/c I solved it already. I guess I have to use part one as well regardless here it is.

           A       B         C        D       F
Prof I    25     35       30       10      0
Prof II   10     20       40       20      10
Prof III  5       10       20       40      25

Sorry about that. By the way I dont know how to write it in the code sorry about that. Please put up with me for a while.

So there are $ 25+10+5=40$ $ A$s in total, etc. If you really want to interpret addition as matrix multiplication, write
\[ \begin{pmatrix}25 & 10 & 5 \\ 25 & 20 & 10 \\ 30 & 40 & 20\\ 10 & 20 & 40 \\ 0 & 10 & 25\end{pmatrix}\begin{pmatrix}1 \\ 1 \\ 1 \end{pmatrix}=\begin{pmatrix}40 \\ 55 \\ 90 \\ 70 \\ 35 \end{pmatrix}\]
However, I don't see what is gained by using matrices here.

But it states there are 240 students in professor I class, 120 in professor II class, and 40 in professor III class. That is why I am confused.

Same here. Maybe you should carefully state the problem in its entirety.

Hey thanks. Actually you guys helped me out. I re-read the problem and it stated the % not the people in class. Wow, I feel retarded sorry for that. Either way here is the problem and my solution. I believe I got it right, but you can check it either way, again thanks.

The % of each grade given by the professor is summarized in the table.
*Rows are Professor I, Professor II and Professor III, respectfully.
And Columns are A, B, C, D, F respectfully.

\[ \begin{pmatrix}25 & 35 & 30 & 10 & 0\\ 10 & 20 & 40 & 20 & 10\\ 5 & 10 & 20 & 40 & 25\\ \end{pmatrix}\]

Problem: Professor I has 240 students, Professor II has 120 students, professor III has 40 students. Use matrix multiplication to determine the number of students who received A’s, B’s, C’s, D’s and F’s in total.

This is relatively easy problem, IF YOU READ IT CORRECTLY. Stupid me.

\[ \begin{pmatrix}240 & 120 & 40 \\ \end{pmatrix}\begin{pmatrix}.25 & .35 & .30 & .10 & 0\\ .10 & .20 & .40 & .20 & .10 \\ .05 & .10 & .20 & .40 & .25 \end{pmatrix}=\begin{pmatrix}75 & 112 & 128 & 64 & 22 \end{pmatrix}\]

That is my answer.
75 A’s, 112 B’s, 128 C’s, 64 D’s and 22 F’s.

Looks right. I like the way Professor II grades.  :thumbup:

[quote="rateesquad"]*Rows are Professor I, Professor II and Professor III, respectfully.
And Columns are A, B, C, D, F respectfully.[/quote]

"Respectfully" means "in a respectful manner" or "with respect."  The word you want is "respectively," which is used when matching two lists of the same length and means that you match the first member of the first list to the first member of the second list, and so on.


Determine with justification whether there exist functions f and g such that f(x)+g(y)=xy.

Do you mean without any appearance of $ y$ in $ f(x)$, and none of $ x$ in $ g(y)$? Because we could just have $ f(x)=xy$ and $ g(y)=0$...  :P

[quote="cincodemayo5590"]Do you mean without any appearance of $ y$ in $ f(x)$, and none of $ x$ in $ g(y)$? [/quote]

A function of one variable, by definition, should only vary according to one variable.  $ xy$ is a function of both $ x$ and $ y$.

[hide="Solution"] $ f(0)+g(y) = f(x)+g(0) = 0 \implies f(x)+g(y) =-f(0)-g(0) = xy$, contradiction. [/hide]


what are the essentials, besides Blackjack, Slot?

[quote="crazychinesemaniac"]Don't forget Texas Hold Em. And is this really appropriate for "Games Forum".[/quote]

why wouldnt it be?

is gambling really good for the middle school/high school/college student? that comment was rhetorical :P

Otherwise, for the smaller kid, "Go Fish" is a nice game.  So is "Old Maid".


By Lagrangian method we get $ x_1x_2...x_6\equal{}x_i(\mu_1 \plus{}\mu_2x_i), i\equal{}1,2,...,6$.
Therefore for exremal mean $ x_1x_2...x_6\minus{}\mu_1(x_1\plus{}x_2\plus{}..\plus{}x_6)\minus{}\frac{\mu_2}{2}(x_1^2\plus{}x_2^2\plus{}...\plus{}x_6^2\minus{}6)$ we get that $ x_i$ had only 2 means. We can consider only $ x_1\equal{}x_2\equal{}..\equal{}x_k\equal{}x,x_{k\plus{}1}\equal{}...\equal{}x_6\equal{}y, k\equal{}1,2,3.$
For $ k\equal{}1$ extremal mean $ f_1\equal{}\minus{}\frac{1}{25}$.
For $ k\equal{}2$ extremal mean $ f_2\equal{}\frac 12$.
For $ k\equal{}3$ extemal mean is $ f_3\equal{}\minus{}1$.
Therefore $ \minus{}1\le x_1x_2...x_6\le \frac 12$.

This is the problem for 9 class. I think here exist more easier solution :)

If we have a odd number of $ \{x_i\}\in R^ \minus{}$ or at least one of them s.t. $ x_i \equal{} 0$ than is obviously.
so, we assume a even number of $ \{x_i\} < 0$, 2 or 4: wlog we can assume $ x_1,x_2 < 0$ and $ x_3,x_4,x_5,x_6 > 0$.
so we can get the problem all in $ R^ \plus{}$ : "if $ \{x_i\}_6\in R^ \plus{}$, and $ \sum_{i \equal{} 1}^{6}{x_i^2} \equal{} 6$ and $ x_1 \plus{} x_2 \equal{} x_3 \plus{} x_4 \plus{} x_5 \plus{} x_6$ then $ \prod_{i \equal{} 1}^{6}{x_i} \le \frac {1}{2}$ ".

we have:
$ 6 \equal{} (x_1^2 \plus{} x_2^2) \plus{} (x_3^2 \plus{} x_4^2 \plus{} x5^2 \plus{} x_6^2) \ge {(\frac {x_1 \plus{} x_2}{\sqrt2})}^2 \plus{} {(\frac {x_3 \plus{} x_4 \plus{} x_5 \plus{} x_6}{2})}^2 \equal{} \frac {3}{4}{(x_1 \plus{} x_2)}^2 \implies {(x_1 \plus{} x_2)}^2 \le 8$,it holds iff $ x_1 \equal{} x_2 \equal{} \sqrt2$ and $ x_3 \equal{} x_4 \equal{} x_5 \equal{} x_6 \equal{} \frac {x_1}{2}$.

and in fact:
$ (x_1x_2)(x_3x_4x_5x_6) \le {(\frac {x_1 \plus{} x_2}{2})}^2{(\frac {x_3 \plus{} x_4 \plus{} x_5 \plus{} x_6}{2^2})}^4 \equal{} \frac {{(x_1 \plus{} x_2)}^6}{2^{10}}\le \frac {1}{2} \leftrightarrow {(x_1 \plus{} x_2)}^2 \le 8$.

bye
paolo


calculus
integration
logarithms
calculus computations

[quote="*210*"]It 'seasy to prove that

 $ \forall x\in [\sqrt {2008},\sqrt {2009}] \frac {1}{\sqrt {x^2 \plus{} 1}} < \frac {\sqrt {1 \minus{} e^{ \minus{} x^{2}}}}{x} < x$[/quote]

You have almost solved the problem, but you seemed to make a typo.

It is easy to prove that !!! My english is so bad   :blush:   Sorry for that .

But I think that is not what do you mean by a typo ! is not it ???

Please take a look [url=http://www.mathlinks.ro/viewtopic.php?p=1728007#1728007]this[/url]. :)

Thank's  Kunny I have  corrected  my Typo  :)


inequalities
three variable inequality
Cauchy Inequality

Well,easy one:
[tex]a+\sqrt{ab}+\sqrt[3]{abc} \leq a+ \sqrt[3]{ab\frac{a+b}{2}}+\sqrt[3]{abc} [/tex]
Now I will prove that:
[tex] a+ \sqrt[3]{ab\frac{a+b}{2}}+\sqrt[3]{abc} \leq \sqrt[3]{a.\frac{a+b}{2}.\frac{a+b+c}{3}}[/tex]
By Cauchy,we have:
[tex] 3\sqrt[3]{1.\frac{2a}{a+b}.\frac{3a}{a+b+c}} \leq \frac{1+\frac{2a}{a+b}+\frac{3a}{a+b+c}}{3}[/tex]
[tex] 3\sqrt[3]{1.1.\frac{3b}{a+b+c}} \leq \frac{2+\frac{3b}{a+b+c}}{3}[/tex]
[tex] 3\sqrt[3]{1.\frac{2b}{a+b}.\frac{3c}{a+b+c}} \leq \frac{1+\frac{2b}{a+b}+\frac{3c}{a+b+c}}{3} [/tex]
Now,just add it up and we will have what we need :)
The equality occurs when $a=b=c$

Excellent problem and proof!

Anh Cuong! Try to use \cdot next time instead of "."
Compare $a.b$ and $a\cdot b$

[quote="Anh Cuong"]By Cauchy,we have:
[tex] 3\sqrt[3]{1.\frac{2a}{a+b}.\frac{3a}{a+b+c}} \leq \frac{1+\frac{2a}{a+b}+\frac{3a}{a+b+c}}{3}[/tex]
[tex] 3\sqrt[3]{1.1.\frac{3b}{a+b+c}} \leq \frac{2+\frac{3b}{a+b+c}}{3}[/tex]
[tex] 3\sqrt[3]{1.\frac{2b}{a+b}.\frac{3c}{a+b+c}} \leq \frac{1+\frac{2b}{a+b}+\frac{3c}{a+b+c}}{3} [/tex]
[/quote]

errr... don't you mean:
(by AM-GM or Cauchy (i see you like Cauchy a lot ;)) --  this is not important)

\[ \displaystyle \sqrt[3]{1 \cdot \frac{2a}{a+b}
\cdot \frac{3a}{a+b+c}} \leq
\frac{1+\frac{2a}{a+b}+\frac{3a}{a+b+c}}{3},\]
\[ \displaystyle \sqrt[3]{1 \cdot 1 \cdot \frac{3b}{a+b+c}} \leq
\frac{2+\frac{3b}{a+b+c}}{3},\]
\[\displaystyle  \sqrt[3]{1 \cdot \frac{2b}{a+b} \cdot \frac{3c}{a+b+c}} \leq
\frac{1+\frac{2b}{a+b}+\frac{3c}{a+b+c}}{3}\]

Just because in Viet Nam, we usually call it Cauchy inequality :)
I really felt strange before with the name AM-GM,but now it is o.k for me to understand,but not for me to use. :D
And sorry about that crazy mistake too :blush:

[quote="iura"]I know the proof for general case uses generalised Holder Inequality for n*n matrix, but I can't remember it even for n=3 (this problem)![/quote]

Yeah, Holder works for 3 or 4 variables, but I have no idea how to make n variables work. Now you may ask why there are two copies of exact the same thing below, it's obviously a bug in the LaTeX processor which I have no idea again :-)

[tex]\begin{eqnarray*}
a+\sqrt{a b}+\sqrt[3]{a b c}
&=& a^{\frac{1}{3}}a^{\frac{1}{ 3}}a^{\frac{1}{3}} + a^{\frac{1}{3}}(\sqrt{a b})^{\frac{1}{3}}b^{\frac{1}{3}}
	+ a^{\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}} \\
&\leq & (a+a+ a)^{\frac{1}{3}} (a+ \sqrt{a b} + b)^{\frac{1}{3}} (a+b+c)^{\frac{1}{3}} \\
&\leq & (a+a +a)^{\frac{1}{3}} (a+ \frac{a + b}{2} + b)^{\frac{1}{3}} (a+b+c)^{\frac{1}{3}} \\
&=&3\sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}.
\end{eqnarray*}
[/tex]

[tex]\begin{eqnarray*}
a+\sqrt{ ab}+\sqrt[3]{a b c}+ \sqrt[4]{abcd}
&=& a^{\frac{1}{4}}a^{\frac{1}{4}}a^{\frac{1}{4}}a^{\frac{1}{4}}
	+ a^{\frac{1}{4}}a^{\frac{1}{4}}b^{\frac{1}{4}}b^{\frac{1}{4}}
	+	a^{\frac{1}{4}}b^{\frac{1}{4}}(\sqrt[3]{abc})^{\frac{1}{4}}c^{\frac{1}{4}}
	+ a^{\frac{1}{4}}b^{\frac{1}{4}}c^{\frac{1}{4}}d^{\frac{1}{4}}  \\
&\leq & (a+a+a+a)^{\frac{1}{4}} (a+a+b+b)^{\frac{1}{4}}(a+ b+\sqrt[3]{abc} + c)^{\frac{1}{4}} (a+b+c+d)^{\frac{1}{4}} \\
&\leq & (a+a+a+a)^{\frac{1}{4}} (a+a+b+b)^{\frac{1}{4}}(a+ b+\frac{a+b+c}{3} + c)^{\frac{1}{4}} (a+b+c+d)^{\frac{1}{4}} \\
&=&4\sqrt[4]{a \cdot \frac{a +b}{2} \cdot \frac{a+b+c}{3}\cdot \frac{a+b+c+ d}{4}}.
\end{eqnarray*}
[/tex]

Very nice, the LaTeX double display problem has been fixed. Cheers!  :D

[quote="Lagrangia"]And from a quick view at Anh Cuong's solution, I think it can solve this one too, following the same pattern's, right? :D[/quote]

Try it, I would like to see how it works out. Thanks!

[quote="Valiowk"]From what I remember, Kedlaya proved the more general result
\[\frac{x_1 + \sqrt{x_1x_2} + \sqrt[3]{x_1x_2x_3} + \cdots + \sqrt[n]{x_1x_2 \cdots x_n}}{n} \leq \sqrt[n]{x_1 \cdot \frac{x_1 + x_2}{2} \cdot \frac{x_1 + x_2 + x_3}{3} \cdots \frac{x_1 + x_2 + \cdots + x_n}{n}}.\][/quote]

Finally, I found where it came from.

It was a Conjecture by F. Holland and proved by Kiran Kedlaya in the paper:

Proof of a mixed arithmetic-mean, geometric-mean inequality, [i]American Mathematical Monthly[/i] 101 (1994), 355-357.

Here is the URL for the LaTeX file of his paper:

http://www-math.mit.edu/~kedlaya/papers/mixed_am-gm.tex

[quote=Lagrangia]Let a,b,c be positive reals. Prove that:
\[\displaystyle \frac{a+\sqrt{ab}+\sqrt[3]{abc}}{3} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}.\] When does the equality occur?
cheers! :D[/quote]
Let $a,b,c$ be positive reals. Prove that:
$$a + \sqrt {ab}  + \sqrt[3]{{abc}}\le \frac{4}{3}(a + b + c)$$ When does the equality occur?

I managed to get a partial solution. If I find the rest I'll update this post.

[hide="Case where c is at least as big as a"]
We see by AM-GM that

$$\frac{4}{3}(a + b + c) = \frac{a + b + c}{3} + \frac{a + b}{2} + \frac{b + c}{2} + \frac{c + a}{2} \geq$$
$$\sqrt[3]{abc} + \sqrt{ab} + \sqrt{bc} + \sqrt{ca} \geq \sqrt[3]{abc} + \sqrt{ab} + a$$

Note this bound is fairly weak, particularly in the last step.
[/hide]

[hide="Remark"]
If $c \geq b \geq a$ then we actually get the stronger bound

$$a + b + c \geq a + \sqrt{ab} + \sqrt[3]{abc}$$

Here's how: by AM-GM we have

$$a + b + c \geq 3\sqrt[3]{abc}$$

Then using the order of the variables we rewrite it as

$$3\sqrt[3]{abc} = \sqrt[3]{abc} + \sqrt[3]{abc} + \sqrt[3]{abc} \geq \sqrt[3]{abc} + \sqrt{ab} + a.$$
[/hide]

[quote=Diamondhead][quote=sqing]Let $a,b,c$ be positive reals. Prove that:
$$a + \sqrt {ab}  + \sqrt[3]{{abc}}\le \frac{4}{3}(a + b + c)$$ When does the equality occur?[/quote]
Wow. Nice problem.
From A.M-G.M inequality, $$3a+12b \geq 2\sqrt{(3a)(12b)}=12 \sqrt{ab}$$ and $$a+4b+16c \geq 3\sqrt[3]{(a)(4b)(16c)}=12 \sqrt[3]{abc}.$$
After summing up, therefore, $4a+16b+16c \geq 12(\sqrt {ab}+\sqrt[3]{abc})$, so $a+\sqrt{ab}+\sqrt[3]{{abc}} \leq \frac{4}{3}(a+b+c)$ as desired.

Note. The equality holds when $a=4b=16c$, i.e, $(a,b,c)=(16k,4k,k)$ for some positive real $k$.[/quote]
Very nice.
Thanks.
$\sqrt{ab}=\frac{1}{6}\sqrt{3a\cdot 12b}\le \frac{1}{4}a+b ,$ $\sqrt[3]{abc}=\frac{1}{4}\sqrt[3]{a\cdot 4b\cdot 16c} \leq\frac{1}{12}a+\frac{1}{3}b+\frac{4}{3}c ,$
so $a+\sqrt{ab}+\sqrt[3]{{abc}} \leq \frac{4}{3}(a+b+c)$ .
Equality holds when $(a,b,c)=(16k,4k,k)$ for $k >0$.

[quote=Lagrangia]Let a,b,c be positive reals. Prove that:
\[\displaystyle \frac{a+\sqrt{ab}+\sqrt[3]{abc}}{3} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}.\] When does the equality occur?
cheers! :D[/quote]
[url=http://www.artofproblemsolving.com/community/c6h1273221p6663760]Bulgaria 2016,P 3:[/url]
For $a,b,c,d>0$ prove that $$\frac {a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4} \leq \sqrt[4]{a.\frac{a+b}{2}.\frac{a+b+c}{3}.\frac{a+b+c+d}{4}}$$

[quote=sqing][hide=Lagrangia][quote=Lagrangia]Let a,b,c be positive reals. Prove that:
\[\displaystyle \frac{a+\sqrt{ab}+\sqrt[3]{abc}}{3} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}.\] When does the equality occur?
cheers! :D[/quote][/hide]
Let $a,b,c$ be positive reals. Prove that:
$$a + \sqrt {ab}  + \sqrt[3]{{abc}}\le \frac{4}{3}(a + b + c)$$ When does the equality occur?[/quote]
Let $0\leq a<b<c<d$ . Prove that:
$$\sqrt[6] {ab}  + \sqrt[12]{{abc}} + \sqrt[20]{{abcd}}<  3 \sqrt{{a}} + 3\sqrt[4]{{b}} + 2\sqrt[6]{{c}} + \sqrt[8]{{d}}$$
( Daniel Sitaru)

[quote=Lagrangia]Let a,b,c be positive reals. Prove that:

\[\displaystyle \frac{a+\sqrt{ab}+\sqrt[3]{abc}}{3} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}.\] When does the equality occur?

cheers! :D[/quote]
Solution by Soumitra Mandal-Chandar Nagore-India:

[quote=ytChen]$\boxed{\text{Prove that if }x,y,z>0\text{, then }\sqrt{\frac{x}{y}} +2\sqrt{\frac{y}{z}}+3\sqrt{\frac{z}{x}}\le\sqrt{6\left(\frac{x}{y}+\frac{2y}{z}+\frac{3z}{x}\right)}.\text{ --- Daniel Sitaru}}$

[b]Solution[/b]. Since $x,y,z>0$, from the Cauchy-Schwarz's inequality it follows that
\begin{align*}6\left(\frac{x}{y}+\frac{2y}{z}+\frac{3z}{x}\right)=(1+2+3)\left(\frac{x}{y}+\frac{2y}{z}+\frac{3z}{x}\right)\ge\left(\sqrt{\frac{x}{y}} +2\sqrt{\frac{y}{z}}+3\sqrt{\frac{z}{x}}\right)^2,
\end{align*}which yields the desired inequality. $\blacksquare$[/quote]
[url=https://artofproblemsolving.com/community/c6h108180p10648645]Yeah.[/url]
Thanks.

[quote=Valiowk]From what I remember, Kedlaya proved the more general result
\[\frac{x_1 + \sqrt{x_1x_2} + \sqrt[3]{x_1x_2x_3} + \cdots + \sqrt[n]{x_1x_2 \cdots x_n}}{n} \leq \sqrt[n]{x_1 \cdot \frac{x_1 + x_2}{2} \cdot \frac{x_1 + x_2 + x_3}{3} \cdots \frac{x_1 + x_2 + \cdots + x_n}{n}}.\][/quote]
 https://artofproblemsolving.com/community/c6h305822p1653937

[hide=Thanks.][quote=Jason99][quote=sqing]Let $a,b,c>0,a+b+c=1 .$ Prove that$$(a+b)(b+c)(c+a)+ab+bc+ca\leq a^2+b^2+c^2+8abc.$$[/quote]
Setting $c = 1-a-b$, we have
$$(a+b)(1-a)(1-b) + ab+b(1-a-b)+(1-a-b)a \leq a^2 + b^2 + (1-a-b)^2 + 8ab(1-a-b)$$
This is equivalent to
$$a(9b-4)(a+b-1) \leq (1-2b)^2 \implies ac(4-9b)\leq (1-2b)^2$$
But this is true since by AM-GM,
$$ac(4-9b) \leq \frac{(a+c)^2}{4} (4-9b) = \frac{(1-b)^2}{4} (4-9b)$$
and trivially
$$\frac{(1-b)^2}{4} (4-9b) \leq (1-2b)^2 \iff 0 \leq b(3b-1)^2$$

Equality occurs when $(a,b,c)=\left(\frac{1}{3},\frac{1}{3}, \frac{1}{3} \right)$.[/quote]

[/hide]
[quote=sqing]Let $a,b,c$ be positive reals. Prove that$$a + \sqrt {ab}  + \sqrt[3]{{abc}}\le \frac{4}{3}(a + b + c)$$ When does the equality occur?[/quote]
Let $a,b,c$ be positive reals such that $a + b + c=21.$ [url=https://math.stackexchange.com/questions/477283/a-b-c0-abc-21-prove-that-a-sqrtab-sqrt3abc-leq-28]Prove that[/url]$$a + \sqrt {ab}  + \sqrt[3]{{abc}}\le28$$
[url=https://math.stackexchange.com/questions/627543/inequality-fraca-sqrtab-sqrt3abc3-leq-sqrt3a-cdot-frac]here[/url]


trigonometry
limit
real analysis
real analysis unsolved

I know that I should make a change of variable,t=x-pi/2 = 0, to get sinx/x but if I use trigonometry identities the problem gets too long.

btw, "a" is a constant.

[quote="kunny"]What does $ sen$  mean?, maybe the context seems to be Spanish.
Intuitively by judging from the form of the problem, we could use $ \boxed{\lim_{\theta \rightarrow 0} \frac {1 \minus{} \cos \theta}{\theta ^2} \equal{} \frac {1}{2}}$.[/quote]

sen is sine.

but, how you make $ \boxed{\lim_{\theta \rightarrow 0} \frac {1 \minus{} \cos \theta}{\theta ^2} \equal{} \frac {1}{2}}$. appear

Here is the solution for Japanese High School students who will have the entrance examination of Japanese University.

$ \sin (a + 2x - \pi) - 2\sin (a + x - \frac {\pi}{2}) + \sin a$

$ = - \sin (2x + a) + 2\cos (x + a) + \sin a$

$ = \sin a - \sin (2x + a) + 2\cos (a + x)$

$ = - 2\cos (x + a)\sin x + 2\cos (x + a)$

$ = 2\cos (x + a)(\sin x - 1)$.

$ \therefore \lim_{x\rightarrow \frac {\pi}{2}} \frac {\sin (a + 2x - \pi) - 2\sin (a + x - \frac {\pi}{2}) + \sin a}{(x - \frac {\pi}{2})^2}$

$ = \lim_{x\rightarrow \frac {\pi}{2}} \frac {2\cos (x + a)(\sin x - 1)}{(x - \frac {\pi}{2})^2}\ Let\ x - \frac {\pi}{2} = \theta$

$ = \lim_{\theta \rightarrow 0} \frac {2\cos (\frac {\pi}{2} + \theta + a){\sin (\frac {\pi}{2} + \theta )} - 1}{\theta ^2}$

$ = 2\lim_{\theta \rightarrow 0} \sin (\theta + a)\cdot \frac {1 - \cos \theta }{\theta ^2}$

$ = 2\sin a\cdot \frac {1}{2} = \boxed{\sin a}$.

Rider:

$ \lim_{\theta \rightarrow 0} \frac {1 - \cos \theta }{\theta ^2} = \lim_{\theta \rightarrow 0} \frac {(1 - \cos \theta )(1 + \cos \theta)}{\theta ^2 (1 + \cos \theta)}$

$ = \lim_{\theta \rightarrow 0} (\frac {\sin \theta}{\theta})^2 \cdot \frac {1}{1 + \cos \theta}$

$ = 1^2\cdot \frac {1}{2} = \frac {1}{2}$.

Maybe someone would expalin the solution by L'Hopital's theorem.


Let the incircle of triangle $ABC$ have the center $I$ and touch the sides $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. Now let $AI \cap EF=O$, $BI \cap FD=P$, $CI \cap DE=Q$. Let the lines $AQ$ and $AP$ intersect the line $EF$ at $R$ and $S$; the lines $BO$ and $BQ$ intersect the line $FD$ at $T$ and $U$; the lines $CP$ and $CO$ intersect the line $DE$ at $V$ and $W$.

[b]1)[/b] Show that $RU\parallel DE$, $SV\parallel FD$ and $TW\parallel EF$.
[b]2)[/b] Show that the lines $RU$, $SV$ and $TW$ are concurrent.

Did you just copy the first post from http://www.mathlinks.ro/Forum/viewtopic.php?t=15765 ? In fact, it looks so, with exactly the same problem wording...

  Darij


ABCDEDF is a regular  octahedron (vertices such that ABCD is a square). Draw the "normal common line"* to AB and CE and find d(AB,CE).

*normal common line = a line perpendicular to both lines at the same time :maybe:

[hide="Hint"]Consider the projection of the octahedron on a plane that is perpendicular to the lines AB and CD.[/hide]


I can't solve this problem. I try to draw a picture. Please help me.  :(

from the picture ABC is a triangle , AB = 12 , AC = 15, BC = 20, AD = 10
BE and CE are tangent line of two circles. Find AE

You can show that $ABE$ and $ADC$ are similar, so $\frac{AE}{AC}=\frac{AB}{AD}$, and since you know $AB,AC,AD$, you're done.

To show that similarity, all you need is a bit of angle chasing.

thanks $grobber$  :D
I " m so stupid. ! :P


The sides $ a, b, c$ and $ u, v, w$ of two triangles $ ABC$ and $ UVW$ respectively are related by the equations

$ u(v\plus{}w\minus{}u) \equal{} a^{2}$

$ v(w\plus{}u\minus{}v)\equal{}b^{2}$

$ w(u\plus{}v\minus{}w)\equal{}c^{2}$.

Prove that triangle $ ABC$ is acute-angled and express the angles $ U, V, W$ in terms of $ A, B, C$.

hello, solving your eqaution system for $ u,v,w$ i have got
$ u\equal{}\frac{a^2}{(a^2\plus{}c^2\minus{}b^2)\sqrt{\frac{a^2\plus{}b^2\minus{}c^2}{c^4\minus{}(b^2\minus{}a^2)^2}}}$
$ v\equal{}\frac{b^2}{(a^2\plus{}b^2\minus{}c^2)\sqrt{\frac{a^2\plus{}b^2\minus{}c^2}{c^4\minus{}(b^2\minus{}a^2)^2}}}$
$ w\equal{}\sqrt{\frac{a^2\plus{}b^2\minus{}c^2}{c^4\minus{}(b^2\minus{}a^2)^2}}$
therefore the triangle $ ABC$ is acute-angled.
Sonnhard.

[hide="Solution"]

Since $ \cos A \equal{} \frac{a^2\plus{}b^2\minus{}c^2}{2bc}$ etc. we only need to verify that WLOG $ a^2\plus{}b^2>c^2$

Substituting, we see that this is equivalent to proving that $ w>|u\minus{}v|$ which is true as u,v,w are themselves sides of a triangle

[/hide]

Hello,

[quote]express the angles $ U, V, W $ in terms of$ A, B, C $[/quote]

How do you do this bit?


Sa se gaseasca expresia termenului general al sirului $\ (x_{n})_{n\in\mathbb{N}}$ dat prin relatiile \[x_{0}=1,x_{1}=2\]  \[\textrm{si}\]  \[x_{n}=4x_{n-1}-3x_{n-2}+4n-8,\forall n=2,3..\]

[hide="Romana"]E corecta solutia ta, Kunny.Asa mi-a dat si mie.Insa am facut-o in alt mod.Am inceput prin a nota un sir $\ (y_{n})_{n\in\mathbb{N}}$, cu proprietatea $\ y_{n}=x_{n}+n^{2}$ si ajungeam la o ecuatie de tipul $\ y_{n}=ay_{n-1}+by_{n-2}$.De aici e totul bine.[/hide]
[hide="English"]Your solution is correct, Kunny.That is also my result.But I solved this problem in another way.I started by setting the sequence $\ (y_{n})_{n\in\mathbb{N}}$ whith $\ y_{n}=x_{n}+n^{2}$ and the equation becomes something like $\ y_{n}=ay_{n-1}+by_{n-2}$ .From here everithing is ok. [/hide]

Sorine, e tare rezolvarea ta, dar daca iti dadeai seama de forma termenului general asa din intuitzie mai simplu apelai la inductie de tipu 2, dar eu unul nu cred ca este solutia ta cea cu $...+n^{2}$ :oops:


Prove that $ \sqrt{p}$ is irrational where $ p$ is prime.  :D

[hide="The popular proof"]Let $ \sqrt {p} \equal{} \frac {m}{n}$, for $ m, n \in \mathbb{Z^{ \plus{} }}$, and $ (m, n) \equal{} 1$. Squaring, we have $ p \equal{} \frac {m^{2}}{n^{2}} \Longrightarrow pn^{2} \equal{} m^{2} \Longrightarrow p \mid m$. Making the substitution $ m \equal{} px$, we have $ pn^{2} \equal{} p^{2}x^{2} \Longrightarrow n^{2} \equal{} px^{2}$, so $ p \mid n$, which makes $ p$ a common factor of $ m, n$, but one of the conditions is that the gcf of $ m,n$ is $ 1$.[/hide]


function
group theory
superior algebra
superior algebra unsolved

Assuming that $ G$ is a finite group with $ n$ elements, and $ m$ is a positive integer, we define $ \Gamma(G) = \sum_{g\in G}o\left(g\right)^{-m}$. Here, for any $ g\in G$, the notation $ o\left(g\right)$ means the order of $ g$. Prove that $ \Gamma\left(G\right)\geq\Gamma\left(\mathbb{Z}_{n}\right)$, where $\mathbb Z_n$ is the cyclic group with $n$ elements. When does equality occur?

I haven't found a proof yet, but I believe that the orders of the elements of a cyclic group majorize those of any other group with the same number of elements; if $k$ divides the order of the group, there are at least $k$ elements of order dividing $k$. This would be enough to prove your problem, and we could replace $k^{-m}$ with any other decreasing function of $k$. If this is true, equality would only occur for cyclic groups.

Now I've got it. In fact, if $k|n$, the number of elements with order dividing $k$ is divisible by $k$. In all that follows, let $n$ be the order of $G$ and let $k$ be a fixed divisor of $n$.

Define $f(m)=$ (#of elements with order dividing $m$). Let $S$ be the set of ordered $k$-tuples of elements of $G$ $(a_1,a_2,\dots,a_k)$ with $a_1a_2\cdots a_k=e$. Let $R$ be the permutation of $S$ with $R(a_1,a_2,\dots,a_k)=(a_2,a_3,\dots,a_k,a_1)$

We consider the orbits of $S$ under the action of $R$. Let $g(m)$ be the number of elements of $S$ which are contained in orbits of size $m$. Clearly, $m|g(m)$

For the size of an orbit to be a divisor of $m$, we must have $a_i=a_{k/m+i}$ and $(a_1a_2\cdots a_{k/m})^m=e$. Counting these, we have
$\sum_{j|m}g(j)=f(m)n^{k/m-1}$ (we can choose the block freely until its last element)

Now, by the Mobius inversion formula, $g(m)=\sum_{j|m}f(j)n^{k/j-1}\mu(\frac kj)$.

Evaluating at $k$ and reducing mod $n$, we have
$g(k)=\sum_{j|k}f(j)n^{k/j-1}\mu(\frac kj)\equiv f(k) \mod n$
Since $k|n$ and $k|g(k)$, $k|f(k)$. Since $f(k)\ge 1$, $f(k)\ge k$.

By my previous post, your problem is an easy corollary.

I was readin this nice proof of this powerful theorem, and seem not to understand this:

[quote="jmerry"]Now I've got it. In fact, if $k|n$, the number of elements with order dividing $k$ is divisible by $k$. In all that follows, let $n$ be the order of $G$ and let $k$ be a fixed divisor of $n$.

Define $f(m)=$ (#of elements with order dividing $m$). Let $S$ be the set of ordered $k$-tuples of elements of $G$ $(a_1,a_2,\dots,a_k)$ with $a_1a_2\cdots a_k=e$. Let $R$ be the permutation of $S$ with $R(a_1,a_2,\dots,a_k)=(a_2,a_3,\dots,a_k,a_1)$

We consider the orbits of $S$ under the action of $R$. Let $g(m)$ be the number of elements of $S$ which are contained in orbits of size $m$. Clearly, $m|g(m)$

For the size of an orbit to be a divisor of $m$, we must have $a_i=a_{k/m+i}$ and $(a_1a_2\cdots a_{k/m})^m=e$. [/quote]

Jmerry, could you please explain how exactly you got this last sentence.

Thanks a lot in advance,

It's a mistake- I got the roles of $m$ and $\frac km$ mixed up. Here's a revised version of the proof:

In all that follows, let $n$ be the order of $G$ and let $k$ be a fixed divisor of $n$.

Define $f(m)=$ (#of elements with order dividing m). Let $S$ be the set of ordered $k$-tuples of elements of $G$ $(a_1,a_2,\dots,a_k)$ with $a_1a_2\cdots a_k=e$. Let $R$ be the permutation of $S$ with $R(a_1,a_2,\dots,a_k)=(a_2,a_3,\dots,a_k,a_1)$.

We consider the orbits of $S$ under the action of $R$. Let $g(m)$ be the number of elements of $S$ which are contained in orbits of size $m$. Clearly, $m|g(m)$.

For the size of an orbit to be a divisor of $m$, we must have $a_i=a_{m+i}$ and $(a_1a_2\cdots a_m)^{k/m}=e$. Counting these, we have
$\sum_{j|m}g(j)=f(\frac km)n^{m-1}$ (we can choose the block freely until its last element)

Now, by the Mobius inversion formula, $g(m)=\sum_{j|m}f(\frac kj)n^{j-1}\mu(\frac mj)$.

Evaluating at $k$ and reducing mod $n$, we have
$g(k)=\sum_{j|k}f(\frac kj)n^{j-1}\mu(\frac kj)\equiv f(k)\mu(k) \mod n$.
Since $k|n$ and $k|g(k)$, $k|f(k)$ if $k$ is squarefree.

This isn't enough- I'm not sure if I can fix it now.


abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Let $ p$ be an odd prime and $ a$ an integer not a perfect $ p$-th power. If $ \xi$ denotes a $ p^k$-th root of unity and $ \zeta\equal{}\sqrt[p]a$, could the extension $ \mathbb{Q}(\xi,\zeta): \mathbb{Q}(\xi)$ be trivial?

No: the Galois group of $ \mathbb{Q}(\xi)/\mathbb{Q}$ is abelian, so any subextension must be normal, which $ \mathbb{Q}(\sqrt[p]{a})/\mathbb{Q}$ is not for $ p>2$.

This seems to work for an arbitrary root of unity $ \xi$.


[quote="akimatsu"]i'm doing divisibility proofs, and i need to mention each digit.

i use let: $D_x$ be a digit of a number, where the x is the power of 10 of the digit place
Ex: $D_0=10^0=\text{one's place}$      $D_1=10^1=\text{ten's place}$ etc.

like for example, when i'm proving divisibility of 5 i need to say:

any number can be expressed as the sum of 10(all the digits until the 10s digit (inclusive)) + its one's digit
like 1234=10(123)+4 or 45678=10(4567)+8

so i do that with variables like $\text{for any number }n=10(D_xD_{x-1}....D_2D_1)+D_0$

but i think it's really hard to understand. do you have idea of how to do this?[/quote]

[size=150]D[/size] as in what? A decimal? A binary number? I don't know what you are trying to simplify.
 :?:

D as in digit. $D_x$ simply represents a digit in a number.

In many problems, you can represent a two digit number as $10a+b$ or a three digit number as $100a+10b+c$.

the easiest way for me to represent the numbers is any integer can be represented in the form

$\sum_{i=1}^{\infty}d_i*10^{i-1}$ where $d_i$ is the digit corresponding to the $i$th place value and $d_i$ is an integer $\in [0,9]$

for example if you wanted to do the divisibility for $9$, then you would look in $\mod 9$, and notice that the 10's cancel, and it becomes
$\sum_{i=1}^{\infty}d_i$

or if the sum of the digits is divisible by 9, then the number is divisible by 9 (make sure when you do the proof, you show going in the other direction, because i just showed the converse)


A fair 6-sided die is rolled once. If I roll $ n$, then I win $ 6\minus{}n$ dollars. What is the expected value of my win, in dollars?

This is just a typical expected value problem, is there any faster way to do it besides just plugging in numbers?

$ 1/6*(5)\plus{}1/6*(4)\plus{}1/6*(3)\plus{}1/6*(2)\plus{}1/6*(1)$
Which is:
$ 15/6\equal{}5/2\equal{}2.5$
SO your expected winnings is:
$ 2.5$ dollars.


I got this wrong about a year ago. (by rushing through it and misunderstanding the problem) I did the problem again and got a different answer. I'm not sure what the answer is, so I figure that the people here ought to get it right.

[i]When 1000 is divided by some positive integer x, the remainder is 16. What is the sum of the smallest and largest value of x?[/i]

[hide]Is the answer 1008? I got the largest value of x as 984 because 1000/984 has a remainder of 16. And then isn't the smallest value of x equal to 24?[/hide]

Thank you in advance.

[hide="Hint"]See if you can do this using modular arithmetic.[/hide]

Yeah, I basically got the same solution. Is it wrong?

[hide]Well, $ x < 1000$. The largest possible value is 984. The smallest possible value is the smallest divisor of 984 such that $ x > 16$. $ 984 \equal{} 2^3\cdot3\cdot43$, so $ x \equal{} 24$. So the answer is:

$ 984 \plus{} 24 \equal{} 1008?$[/hide]

Am I missing something or is the problem asking for the solutions to the problem:

$ 1000\equiv 16\pmod{x}$ where $ x\in\mathbb{R}^\plus{}$

If this is the case, then I can't see why there would be a possible sum.

$ x \in \mathbb{N}$.  And that condition is weaker; in order for $ 16$ to be a remainder upon division we also require that $ x > 16$.


calculus
integration
algebra
function
domain
real analysis
real analysis unsolved

is this integral  at all doable ?   :(

$ \int_0^1\int_0^1\int_0^1\int_0^1\dfrac{dw\;dx\;dy\;dz}{(wxy\minus{}1)(zwx\minus{}1)(yzw\minus{}1)(xyz\minus{}1)}$

Partial fractions doesn't work? I try later...

$ \int_0^1\int_0^1\int_0^1\int_0^1 \dfrac{dw\;dx\;dy\;dz}{(1 \minus{} wxy)(1 \minus{} zwx)(1 \minus{} yzw)(1 \minus{} xyz)} \equal{} \sum_{k,i,j,l\equal{}0}^{\plus{}\infty} \int_0^1\int_0^1\int_0^1\int_0^1 (wxy)^k (zwx)^i (yzw)^j (xyz)^l \, \mathrm{d}(w,x,y,z) \equal{} \sum_{k,i,j,l\equal{}0}^{\plus{}\infty} \frac{1}{(1 \plus{} i \plus{} j \plus{} k)(1 \plus{} i \plus{} j \plus{} l)(1 \plus{} i \plus{} k \plus{} l)(1 \plus{} j \plus{} k \plus{}l) }$
:huh:  :?:

Just a Sketch!

$ \int_0^1 \ldots \int_0^1 \left(\prod _{j \equal{} 1}^n \left(\frac {1}{t_j}\prod _{i \equal{} 1}^n t_i \minus{} 1\right)\right)^{ \minus{} 1}dt_1\ldots dt_n$

Misan is the case $ n \equal{} 4$
Making the substituition:

$ t_s \equal{} \frac {1}{\alpha _s^{n \minus{} 1}}\prod _{i \equal{} 1}^n \alpha _i$

This gives:

$ \left(\prod _{j \equal{} 1}^n \left(\frac {1}{t_j}\prod _{i \equal{} 1}^n t_i \minus{} 1\right)\right)^{ \minus{} 1} \equal{} \left(\prod _{j \equal{} 1}^n \left(\alpha _j^{n \minus{} 1} \minus{} 1\right)\right)^{ \minus{} 1}$

To evaluate The Jacobian we need a little trick, I will leave this as a exercise:

$ J \equal{} (1 \minus{} n)^{n \minus{} 1}$

So the Integral

$ J . \int \prod _{j \equal{} 1}^n \frac {1}{\alpha _j^{n \minus{} 1} \minus{} 1}d\alpha _1\ldots d\alpha _n$

I was thinking in split this n-Integral in a simple integral power n, since there is some simetry, but I was unable do determine the domain of this function. and how to do this and even if this is possible.

Maybe this leads to nothing and there is a easy way.


[quote="stvs_f"]let a,b,c   are positive  that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1 .$prove that :
\[ \sum(\frac {a^5 \plus{} b^5}{ab(a \plus{} b)}) \geq 3(ab \plus{} bc \plus{} ca) \minus{} 2
\]
[/quote]
Apply AMGM we have $ a^5\plus{}b^5\ge a^2b^2(a\plus{}b)$ and $ \sum ab\le \sum a^2\equal{}1$ we have done

Nice proof Allnames.

You can also say how you got $ a^5 \plus{} b^5 \ge a^2b^2(a \plus{} b)$ by power mean inequality that
$ (a^5 \plus{} b^5)/2 \ge {(a \plus{} b)^5/2}^5$ to make it more comprehendable.

[quote="stvs_f"]let a,b,c   are positive  that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1 .$prove that :
\[ \sum(\frac {a^5 \plus{} b^5}{ab(a \plus{} b)}) \geq 3(ab \plus{} bc \plus{} ca) \minus{} 2
\]
[/quote]
$ \iff$
$ \sum(\frac {a^5 \plus{} b^5}{ab(a \plus{} b)} \minus{} 3ab \plus{} a^2 \plus{}b^2) \geq 0$ $ \iff$

$ \sum(\frac {a^5 \plus{} b^5 \minus{} 2a^3b^2 \minus{} 2a^2b^3 \plus{}  a^4b \plus{} ab^4}{ab(a \plus{} b)}) \geq 0$ $ \iff$

$ \sum(\frac {(a^2\minus{}b^2)^2}{ab}) \geq 0$. Done :)


geometry
probability
calculus
integration
3D geometry
trapezoid
analytic geometry

$Question 1$ We have a triangle whose vertices are at $(0,0)$, $(4,4)$, $(10,0)$. A Square is drawn in it such that its base
is on x-axis and its two corners are on the $2$ sides of the triangle. What is the area of the square?

$Question 2$ Two circles of radius $1 m$ each intersect, such that the center of one is on the circumference of the other.
What is the area of the region common to both the circles?

$Question 3$ You are given a $28$digit number : 5-383-8-2-936-5-8-203-9-3-76 , of which you can only see $18$ of the digits and the rest are blank. These blanks are filled using the digits $0$ to $9$ once. This can of course be done in $10!$ ways, i.e. more than 3 million ways. What is the probability that the number so formed is divisible by $396$?

$Question 4$ We know that $5^{3}= 125$ and $6^{3}= 216$. Given that $148877$ has an integral cube root, can u find this cube root without using a calculator ?

$Question 5$ A circle is inscribed in an isosceles trapezium whose parallel sides measure $8cm$ and $18 cm$. Would it be possible to determine the area of the circle ?

[quote="xxxyyyy"]$Question 2$ Two circles of radius $1 m$ each intersect, such that the center of one is on the circumference of the other.
What is the area of the region common to both the circles? [/quote]
[hide="2"] Let's call the 2 circles $A$ and $B$, with their respective centers. And the 2 intersection points $C$ and $D$. The area of the overlap is equal to sector $ACD$ plus sector $BCD$ minus the quadralateral $ACBD$. $\frac{120}{360}\cdot\pi\cdot1^{2}+\frac{120}{360}\cdot\pi\cdot1^{2}-2\cdot\frac{1^{2}\sqrt3}{4}=\boxed{\frac{2}{3}\pi-\frac{\sqrt3}{2}}$[/hide]

[quote="xxxyyyy"]
$Question 4$ We know that $5^{3}= 125$ and $6^{3}= 216$. Given that $148877$ has an integral cube root, can u find this cube root without using a calculator ?
[/quote]
[hide]
The only units digit the number can have is 3. Since $50^{3}=125000$ and $60^{3}=216000$ the number is between 50 and 60 making the number $\boxed{53}$[/hide]

[hide="1"]
Find equations of lines containing sides of triangle other than one on x-axis:
Line passing $(0,0)$ and $(4,4)$ has equation $y=x$
Line passing $(4,4)$ and $(10,0)$ has equation $3y+2x=20$
One verticle of square lies on first line and has coordinates $(a,a)$
Second one's coordinates $(2a,a)$, and since it lies on second line
$6a+2a=20 \Rightarrow a=\frac{5}{2}$
$S=\frac{25}{4}$
[/hide]

[hide="3"]
Is answer $\frac{6}{11}$?
[/hide]

[hide="4"]
Try some number first, for example $50^{3}=125000$, so it must be higher.$148877$ ends at $7$, so its cube root must end at $3$, we try $53$ and win.
[/hide]

[hide="5"]
Since trapezium has an inscribed circle sums of its opposite sides are equal $2x=8+18=26 \Rightarrow x=13$
Draw altitude from verticle on smaller side to other side
$4r^{2}=13^{2}-5^{2}=144 \Rightarrow r=6$
Area is $36\pi$

[/hide]

[quote="xxxyyyy"]$Question 3$ You are given a $28$digit number : 5-383-8-2-936-5-8-203-9-3-76 , of which you can only see $18$ of the digits and the rest are blank. These blanks are filled using the digits $0$ to $9$ once. This can of course be done in $10!$ ways, i.e. more than 3 million ways. What is the probability that the number so formed is divisible by $396$?[/quote]

[hide="Solution"] $396 = 2^{2}\cdot 3^{2}\cdot 11$, so we need to investigate when \[A = \overline{5a_{1}383a_{2}8a_{3}2a_{4}936a_{5}5a_{6}8a_{7}203a_{8}9a_{9}3a_{10}76},\] where $\left(a_{1}, a_{2}, \cdots, a_{10}\right)$ is a permutation of $\left(0, 1, \cdots, 9\right)$, is divisible by 4, 9 and 11. But $A$ ends in 76, so it's always divisible by 4. For $a$ to be divisible by 9, the sum of the numbers should be congruent to 0 modulo 9. But \[5+3+8+3+8+2+9+3+6+5+8+2+0+3+9+3+7+6+\sum_{i=1}^{10}a_{i}\equiv 90+(0+1+2+\cdots+9) \equiv 135 \equiv 0 \mod{9},\] so $A$ is always divisble by 9. Now, because $10^{n}\equiv (-1)^{n}\mod{11}$, we know that \[A \equiv-5+a_{1}-3+8-3+a_{2}-8+a_{3}-2+a_{4}-9+3-6+a_{5}-5+a_{6}-8+a_{7}-2+0-3+a_{8}-9+a_{9}-3+a_{10}-7+6 \equiv-56+\sum_{i=1}^{10}a_{i}\equiv-56+45 \equiv-11 \equiv 0 \mod{11}.\] Therefore, 11 does always divide $A$. We conclude that $A$ is always divisible by 396.[/hide]

[quote="xxxyyyy"]$Question 1$ We have a triangle whose vertices are at $(0,0)$, $(4,4)$, $(10,0)$. A Square is drawn in it such that its base
is on x-axis and its two corners are on the $2$ sides of the triangle. What is the area of the square?

[/quote]

I extend the above question....

In the same triangle $ABC$ draw a square $PQRS$ ( in that
order) with diagonally opposite vertices $P$ and $R$ on
$AC$, $Q$ on $AB$, $S$ on $BC$ ( both sides produced if
necessary). Give the construction of this square and
its area.

??

[quote="midorya"][hide="1"]
Find equations of lines containing sides of triangle other than one on x-axis:
Line passing $(0,0)$ and $(4,4)$ has equation $y=x$
Line passing $(4,4)$ and $(10,0)$ has equation $3y+2x=20$
One verticle of square lies on first line and has coordinates $(a,a)$
Second one's coordinates $(2a,a)$, and since it lies on second line
$6a+2a=20 \Rightarrow a=\frac{5}{2}$
$S=\frac{25}{4}$
[/hide][/quote]

[hide="Correction"]Your equation for $a$ should have been $3a+4a=20$.  :wink: [/hide]

[quote="matt276eagles"][hide="Correction"]Your equation for $a$ should have been $3a+4a=20$.  :wink: [/hide][/quote]
Yes, you are right  :oops:

[hide="4"]Note that $148877\equiv 2\pmod{3}, 1\pmod{4}, 2\pmod{5}$.  After a little testing, we find that its cube root must equivalent to $2\pmod{3}$, $1\pmod{4}$, and $3\pmod{5}$.  Therefore, it is equivalent to $53\pmod{60}$.  Clearly, the cube root can't be greater than $100$, so the cube root is $53$.[/hide]


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=13981[/img]

 :(

Call r the perpendicolar line to OMa on Ma, call E the feet of the perpendicolar from I to r. So $ IF \equal{} r$. Now we have $ OI^2 \equal{} R(R\minus{}2r)$ and so $ R \equal{} r \plus{} \sqrt{OI^2 \plus{} r^2}$. But $ \sqrt{OI^2 \plus{} r^2}$ can be easy costruct with pytagora, so we can costruct R, then The circle with center O and radius R meet r on B and C and the tangent from B to the incircle meet the tangent from C to the incircle in A.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=13996[/img]

 :thumbup:


induction
number theory proposed
number theory

Show that for any positive integer n, we can find a positive integer t such that 19^t - 97 is divisible by 2^n.
comeup with sumthing nice,guys.mine is tortuous :D

I hope the following is correct:
  Proof by induction. First cases are easy. Suppose we have found t such that 19^t-97 is divisible by 2^n. Suppose 19^t-97=k*2^n. If k is even, then the induction is finished: take t for the step n+1.So, assume k is odd. Let f(x) be the exponent of 2 in x. It is easy to see that f(19^t-1)+f(19^t+1) \leq n. Indeed, suppose it is not so. Between f(19^t-1) and f(19^t+1) at least one is 1. It would follow that the other one is at least n. So we would have 2^n| 19^t-1 or 2^n|19^t+1 and also 2^n|19^t-97. It follows that n \leq 5, but for this values examples can be found ( we can start the induction with n=6 bla bla bla...). So, let us take p=n+1-f(19^t-1)-f(19^t+1)>0. Let us choose M=t(1+2^p). We show that 2^(n+1)| 19^M-97. But 19^M-97= 19^t(19^(M-t)-1)+k*2^n. So, it suffices to prove that f( 19^(M-t)-1)=n. But this follows easily from the choice of p and from the formula 19^(M-t)-1=(19^t-1)(19^t+1)(19^2t+1)...(19^(2^(p-1)t-+1) and the fact that f(x^2+1)=1 for odd x.

Well , using induction method is a briliant idea! I saw the official proof! It used Dirichlet principt and some simple remarks!

my proof also uses induction just like harazi's but Ha Duy Hung,could u post the official solution .i'm interested in knowing how neat the official solution is :D

I'll be very happy if someone answer to these questions:
1. What does it means: 'Let $ f(x)$ be the exponent in $ x$'?
2. Why: 'Between $ f(19^{t}\minus{}1)$ and $ f(19^{t}\plus{}1)$ at least one is $ 1$?
If you don't want to answer these questions - please write some more detailed and easy for understanding solution.


If Eric can paint $ 3$ cars in $ 4$ hours and $ 2$ trucks in $ 5$ hours, then how long, in hours, would it take him to paint $ 4$ cars and a truck?

We simply set up equations, 3/4=4/x, whereas x=the number of hours it takes eric to paint 4 cars. 2/5=1/y, whereas y=the number of ours it takes eric to paint one truck. Therefore x=16/3 and y = 5/2 and when we add these fractions together, we get 47/6 hours.


Find a pattern for

2^1 + 2^2 + 2^3 + 2^4 + 2^5 +... 2^n

Go to highschool basics for a harder version.

To get to each number, multiply by 2. For instance, to get from $ 2^n$ to $ 2^{n\plus{}1}$, multiply by 2.

What, do you mean $ 1\plus{}2\plus{}2^2\plus{}2^3\plus{}\cdots\plus{}2^n\equal{}2^{n\plus{}1}\minus{}1$ as a "pattern"?

ROFL. So you want a recursion: $ 2S_{n\minus{}1}\plus{}2\equal{}S_n$.

I was actually looking for 2^(n+1) - 2, but that'll work 2.

[quote="math154"]What, do you mean $ 1 \plus{} 2 \plus{} 2^2 \plus{} 2^3 \plus{} \cdots \plus{} 2^n \equal{} 2^{n \plus{} 1} \minus{} 1$ as a "pattern"?[/quote]

$ 1$ is not the first number in the sequence, that's why you were wrong. :P

[quote="Wickedestjr"]The sums are 2, 6, 14, 30, 62, 126,....

 What is the pattern?[/quote]

Do you mean like... the next number equals the previous number doubles and plus 2?  :|

Well if you add 2 to each number, then if becomes clearer.

cherryblossomkiwi add 2 to each number.

This is probably an odd way to look at it, but when I glanced at it, I actually saw a different equation than the simplified version. So yeah, I had to do extra work. :P

This is the relationship from one sum to the next, not finding the sum for the $ n$th integer...

[hide="My Solution"]
$ p \equal{}$ Previous Sum
$ s \equal{}$ New Sum

Therefor,

$ s \equal{} 2(p \plus{} 2) \minus{} 2$
$ s \equal{} 2p \plus{} 4 \minus{} 2$
$ s \equal{} 2p \plus{} 2$
[/hide]

That would basically say that:

for $ p \ge 1 \in \mathbb{Z}, T_p \equal{} 2T_{p\minus{}1} \plus{} 2$

I hope I stated that correctly.  :maybe:

Also, it's interesting to note that any number $ 2^n$ equals the sum of all $ 2^a$ where $ n>a\ge0$ and $ \{a,n\}\in\mathbb{Z}$, minus $ 1$. This is the basis of binary.
Therefore, the sum of the first $ n$ powers of $ 2$ is $ 2^{n\plus{}1}\minus{}1$.

But because this sequence does not have $ 2^0$, or $ 1$, we subtract $ 1$ to get $ 2^{n\plus{}1}\minus{}2$.

[quote="ernie"]Didn't funkymunk just prove that?[/quote]

No he didn't, he just showed where he got the equation from.


2^100 digit's are added to get another number whose digits are then added and so on until a sigle digit is gotten.  what is the digit

[quote="chess64"]Some number from 1 to 9... :P

Seriously though, doesn't it depend on what the first number is?[/quote]

What do you mean?

BornForMath, please give solutions with your answers.

[hide="hint1"]Look for an invariant[/hide]

[hide="hint 2"]We are adding up the digits of a number...what does this remind you of?  When else would you add the digits of a number up?[/hide]

It's no problem.  That's actually how I intrepreted the first time I read the problem also :)

Btw, I just wanted to remind everyone that you shouldn't be using a calculator/CAS for this problem ;)

[quote="joml88"]
[hide="hint 2"]We are adding up the digits of a number...what does this remind you of?  When else would you add the digits of a number up?[/hide][/quote]

When we're doing divisibility, okay, so...

Right, now be a little more specific and then find a way to apply it to the problem at hand.  You're almost there so just think about it some more.

[quote="chess64"][quote="joml88"][quote="chess64"]Some number from 1 to 9... :P

Seriously though, doesn't it depend on what the first number is?[/quote]

What do you mean?
[/quote]

OH, nvm, I thought it said $2^{100}$ digits, not the digits of $2^{100}$...[/quote]

Yeah that's what I thought too.  And rereading it, that is what it says.  If it is supposed to be the digits of $2^{100}$, it should say "2^100's digits are added..." but instead it says "2^100 digits are added..."  Confusing.

[quote="deej21"][/quote][quote="chess64"][quote="joml88"][quote="chess64"]Some number from 1 to 9... :P

Seriously though, doesn't it depend on what the first number is?[/quote]

What do you mean?
[/quote]

OH, nvm, I thought it said $2^{100}$ digits, not the digits of $2^{100}$...[/quote][quote="deej21"]

Yeah that's what I thought too.  And rereading it, that is what it says.  If it is supposed to be the digits of $2^{100}$, it should say "2^100's digits are added..." but instead it says "2^100 digits are added..."  Confusing.[/quote]
i was just about to say that :)

[quote="236factorial"]Yeah... $2^{100}$, when it's divided by 9, has a remainder.... of 7.  :!:   :idea:
That means its sum of its digits also is 7 (mod 9). So the last digit left also must be 7 (mod 9).
I see now  :D[/quote]

 :yup:

$2^3 \equiv -1 \text{(Mod 9)}$
$2^{100} = (2^3)^{33} \cdot 2 \equiv  (-1)^{33} \cdot 2 \equiv -2 \equiv  7 \text{(Mod 9)}$

[quote="math92"][/quote][quote="deej21"][/quote][quote="chess64"][quote="joml88"][quote="chess64"]Some number from 1 to 9... :P

Seriously though, doesn't it depend on what the first number is?[/quote]

What do you mean?
[/quote]

OH, nvm, I thought it said $2^{100}$ digits, not the digits of $2^{100}$...[/quote][quote="deej21"]

Yeah that's what I thought too.  And rereading it, that is what it says.  If it is supposed to be the digits of $2^{100}$, it should say "2^100's digits are added..." but instead it says "2^100 digits are added..."  Confusing.[/quote][quote="math92"]
i was just about to say that :)[/quote]

PLEASE DO NOT SPAM...FOR THE LAST TIME!!!

Next time, you could jsut delete some of those quotes. They are clogging up the page...  ;)


[quote="Nora.91"][b]p[/b] and [b]q[/b] $ \in$ [b]N[/b] and

$ \frac {p}{q} \equal{} 1 \minus{} \frac {1}{2} \plus{} \frac {1}{3} \minus{} \frac {1}{4} \plus{} ... \minus{} \frac {1}{1318} \plus{} \frac {1}{1319}$.

Prove that [b]p[/b] is divisible by 1979.[/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367332#p367332]IMO first problem 1979[/url]


[quote="stevo2007"]Could anybody help with solving this.

[u]3x^3 2x^2 - 5x +4[/u]
             x+2[/quote]

$\frac{3x^{3}\cdot 2x^{2}-5x+4}{x+2}$

is this what you mean?

@Geómetra,  :spam: ....

It won't happen again.

Sorry, I didn't see he was new here.

[quote="ckck"]I think stevo2007 means $\frac{3x^{3}+2x^{2}-5x+4}{x+2}$. Is that right?

And to Geómetra thats not nice. If someone doesn't know how to use latex at least translate the problem for them.[/quote]

Yes that is the right equation as for latex what program is that.

[quote="stevo2007"]as for latex what program is that.[/quote][code]$\frac{3x^{3}+2x^{2}-5x+4}{x+2}$.[/code]You can go to the LaTeX forum(under site and general information) to learn most of this, that's where I learned most of my TeX. Don't be afraid to ask questions there. They will help you a lot.

$\frac{3x^{3}+2x^{2}-5x+4}{x+2}$
I don't think this is factorable so the easiest method would probably be long division.
[hide]$x+2)\overline{3x^{3}+2x^{2}-5x+4}=3x^{2}-4x+3-\frac{2}{x+2}$
Note: the ) thing is for the long division sign... :P [/hide]


A simple but very elegant problem.

  Let $x$ be any number less than 1, and let $y$ be any number greater than 1.  Let $S$ be the sum of $x$ and $y$, and let $P$ be the product of $x$ and $y$.  Prove that the difference between $S$ and $P$ must be greater than 1.

Heh, I liked this problem. Nice and quick.

Just for fun, I decided to prove why x + y > xy as well.

[hide="Solution"]
First, you have to prove that x + y > xy.

There are three cases: x is greater than 0, x is 0, and x is less than 0.

If x is greater than 0 but less than 1, xy < y. x + y > y (since x is greater than 0), so x + y > y > xy. Holds true in the first case.

The second case is very simple; plugging in 0 for x gets y > 0. Correct.

The third case is a little bit trickier. Subtract y from both sides of the initial equation to get  x > xy - y. Factor to get x > y(x - 1). Now, x > x - 1, and as you know, both sides are negative. Multiplying a negative number by a positive number greater than 1 gets an even smaller number, so x > y(x - 1) holds true as well.

Now for the fun part.

Since we proved that x + y > xy, and we need to prove that the positive difference of the two is greater than 1, we do this.

x + y - xy > 1
-xy + x + y - 1 > 0
xy - x - y + 1 < 0
(x - 1)(y - 1) < 0

x - 1 is negative, as x < 1.
y - 1 is positive, as y > 1.

A negative number times a positive number is always negative. Tada.
[/hide]

I haven't completely looked over your answer Treething but I can tell you theres a much more elegant solution.  Try and see if you can find it.

[quote="chess64"][hide]
$x<1, y>1$.
$x+y-xy>0$. Multiply by $-1$.
$xy-x-y<0$. Add 1 and factor.
$(x-1)(y-1)<1$.

Because $x<1$ and $y>1$, we have $\text {negative } \cdot \text { positive } = \text { negative } <1 \ \ \ \ \ \ \ \ \blacksquare$
[/hide]

edit: oops :blush: proved the wrong thing :D[/quote]

  Somethings need changing, but other than that you quite close! :)

[quote="AntonioMainenti"][hide="solution that is partially borrowed from chess64"]We want to show that \[x<1\text{ and }y>1\Longrightarrow|x+y-xy|>1.\]Assume otherwise. We have $x-1<0$ and $y-1>0$, and so $(x-1)(y-1)<0<1$. Then $(x-1)(y-1)<1\Longrightarrow{x}y-x-y+1<1\Longrightarrow{x}+y-xy>0$. By our previous assumption we have $|x+y-xy|=x+y-xy\le1$. But this implies $x+y-xy-1\le0\Longrightarrow(x-1)(1-y)\le0\Longrightarrow(x-1)(y-1)\ge0$, a contradiction.

This is a proof by contradiction, and not a very elegant one at that. Is there a nice direct proof you had in mind G-UNIT?[/hide] :D[/quote]

  A nice solution at that, but yes, a nice direct proof exists.

[quote="G-UNIT"]A nice solution at that, but yes, a nice direct proof exists.[/quote]Oh yeah. Of course: [hide="were you thinking something other than this?"]We want to show that \[x<1\text{ and }y>1\Longrightarrow|x+y-xy|>1.\]$x-1<0\text{ and }1-y<0\Longrightarrow(x-1)(1-y)>0\Longrightarrow{x}+y-xy-1>0\Longrightarrow{x}+y-xy>1\Longrightarrow|x+y-xy|>1$

And now that we have a direct proof, $\text{QED}$ :D
[/hide]

Pth, the last six lines is the actual proof. That's quick enough for me >_> . I added the first part because I originally proved the wrong thing, and I wanted to cover my butt.

look at this....
[hide="solution"]
to prove: $x+y-xy>1$
So, $x(1-y)+y>1$
$(1-y)$ is always negative and the only case for x to be positive is $0$ so $x(1-y)$ is always positive  and so is y so it is always greater than $1$
[/hide]
Correct?

I just wanted to ask if x could be like 0.5 and stuff...because in his solution,riddler's used the value of x as 0 or less... :?:


Let $ ABC$ be a triangle. Prove that
$ \sqrt{\frac a{bc}}\plus{}\sqrt{\frac b{ca}}\plus{}\sqrt{\frac c{ab}} \ge \frac1{\sqrt{b\plus{}c\minus{}a}}\plus{}\frac1{\sqrt{c\plus{}a\minus{}b}}\plus{}\frac1{\sqrt{a\plus{}b\minus{}c}}$

[quote="dduclam"]Let $ ABC$ be a triangle. Prove that
$ \sqrt {\frac a{bc}} \plus{} \sqrt {\frac b{ca}} \plus{} \sqrt {\frac c{ab}} \ge \frac1{\sqrt {b \plus{} c \minus{} a}} \plus{} \frac1{\sqrt {c \plus{} a \minus{} b}} \plus{} \frac1{\sqrt {a \plus{} b \minus{} c}}$[/quote]
Try $ a\equal{}1\plus{}\sqrt{2},b\equal{}\sqrt{2}\plus{}\sqrt{3},c\equal{}\sqrt{3}\plus{}1$,
$ LHS\minus{}RHS\equal{}\minus{}0.018...<0$

Yes,NguyenDungTN,i made mistake  :( But i think the following inequality is true

$ \frac1{3}(\sqrt [4] {\frac a{bc}} \plus{} \sqrt [4] {\frac b{ca}} \plus{} \sqrt [4] {\frac c{ab}})^2 \ge \frac1{\sqrt {b \plus{} c \minus{} a}} \plus{} \frac1{\sqrt {c \plus{} a \minus{} b}} \plus{} \frac1{\sqrt {a \plus{} b \minus{} c}}$


Could someone please show me the derivation of the formula for the sum of squares?
(The formula is $1^2+2^2+3^2...+(n-1)^2+n^2=\frac{n(n+1)(2n+1)}{6}$)

well you can see it by inspection after some work and prove it with induction fairly easily.

How would you do that though?

I was trying some stuff using triangular numbers, since $S_n=T_n+T_{n-1}$, but I wasn't getting anything too useful...

After looking at a few sums, you can notice that the sum varies with $n^{3}$.  This being the case, the formula must be of the form $S_{n}=an^{3}+bn^{2}+cn+d$.  Now notice that

\[ S_{n}-S_{n-1}=n^{2} \]

\[ \left(an^{3}+bn^{2}+cn+d\right)-\left(a\left(n-1\right)^{3}+b\left(n-1\right)^{2}+c\left(n-1\right)+d\right)=n^{3} \]

From here, you can solve for $a$, $b$, $c$, and $d$, getting the desired result.

As stated above, these are much easier to prove by induction.  This method I posted can be quite tedious when working with higher degree polynomials.

Thanks.  Now for one more question about one of the explanations...

What are telescoping sums?

A telescoping sum is a sum where some terms cancel each other out, leaving you with something easier to deal with.  Usually this means you are left with only the first and last terms.


Hello all,
I am a COMPLETE newbie and am truly not all that great with computers. I have downloaded all the necessary components to installing Texnic and Miktex. When I first open Texnic, there is a configuration wizard and I, being the fool that I am with computers, do not know what to put for anything.

Any help will be greatly appreciated.
Thank you very much.

Normally you just need to agree to everything it suggests as a default (at least, that's what I did and it worked just fine). If something malfunctions after the installation, or if you need more options, you'll always be able to take care of that later. Just be sure that you install MikTeX first and Technic Center afterwards :)

[quote="iamhe"]When I first open Texnic, there is a configuration wizard and I, being the fool that I am with computers, do not know what to put for anything.[/quote]See [[LaTeX:Downloads#TeXnicCenter]] item 7. If you have lost the Wizard you can get it back again by going to Build, Define Output Profiles, Wizard button in bottom left corner.


Find the integer solutions x, y of the equation
$x^{6}-y^{2}=648.$

[hide]$(x^{3}-y)(x^{3}+y)=648$

By factorization and taking into account that $(x^{3}-y)+(x^{3}+y)=2x^{3}$, we find only two possibilites:

$x^{3}-y=18\land x^{3}+y=36\implies x=3\land y=9$

$x^{3}-y=36\land x^{3}+y=18\implies x=3\land y=-9$

Hence all solutions are $(3,\pm 9)$[/hide]

$(3\alpha ,9\beta)$, where $\{\alpha ,\beta\}\subset \{-1,1\}$ (four solutions).


AMC
AIME
trigonometry
modular arithmetic

9. For how many positive integers $n$ less than or equal to 1000 is  $(\sin t + i \cos t)^n=\sin nt + i \cos nt$ true for all real $t$?

I got the answer, but it was rather guess and check....

I made $t$=90 deg. and found that the equation is true when n=1,5,9,...,993,997. So total 250 values....

Is there any other method??

[hide]

$\sin{(t)} = \cos{(90-t)}$ and $\cos{(t)} = \sin{(90-t)}$

So $(\sin{(t)}+i\cos{(t)})^n = (\cos{(90-t)}+i\sin{(90-t)})^n = \cos{(n(90-t))}+i\sin{(n(90-t))}$.

Equating real and imaginary parts, we have

$\cos{(n(90-t))} = \sin{(nt)}$ and $\sin{(n(90-t))} = \cos{(nt)}$

Analysis of $n \pmod{4}$ is probably the quickest way here, since after that it only repeats itself.

$n \equiv 1 \pmod{4}$ works and the others don't, so we should have $250$ solutions.[/hide]

Yea, the major step is noting the striking resemblance between the given expression and Demoivre's Theorem.

This is a rather intuitively simple problem for a #9 on an AIME right?


I was wonder how you all think the test was in difficulty compared to last year. Also, the blank value decreased to 1.5.

Do you think it is possible?

From my quick glance, it seemed easier than last year. It's rare for the cutoff to be lowered; it only happened once, in 2004.

Do you think a 112.5 with a stupid mistake on one problem is good for a 9th grader?

I either made 115.5 or 114, maybe 115.5.

I personally highly highly doubt the cutoff will be lowered this year considering almost everyone I talked to thought the AMC10 was pretty easy compared to past years.

I'm okay with that. I'll just do my best on the 10B.

I just want to take AIME as a learning experience mostly.

Well judging by your score it means you're pretty close and that the problems that came up were just not your forte. Aim for the B I guess.


logarithms
number theory unsolved
number theory

Let $k\geq 2$ be a natural number.Prove that there exists a power of $2$ such that the number of $9$'s in its last $k$ digits is not less than half.


For any given positive integer n, determine (as a function of n) the number of ordered pairs (x,y) of positive integers such that x^2-y^2 = 10^2*30^(2n). Prove also that the number of such pairs is never a perfect square.

The number of ways to represent $M$ is a difference of two squares is a number of ways to expand $M$ into product of two non-equal multipliers of the same parity.

Direct application to our problem gives $\frac{(2n+1)^{2}(2n+3)-1}{2}$. Who will prove it is not a perfect square? It is easy ;)


[hide="better LaTeX"]$ 2^{x\plus{}2}(1\minus{}2\minus{}4)>5^{x\plus{}1}(1\minus{}5)$

Which is just $ 2^{x\plus{}2}*\minus{}5>5^{x\plus{}1}(\minus{}4)$

We divide by -1 to get

$ 2^{x\plus{}2}*5<5^{x\plus{}1}*4$

$ 2^x<5^x$

which is true for all $ x>0$.[/hide]

By the way, instead of using asterisks to indicate multiplication, you can use these two bits of code:

[code]$5 \cdot 6 = 30[/code]
creates $ 5 \cdot 6 \equal{} 30$ and
[code]$5 \times 6 = 30$[/code]
creates $ 5 \times 6 \equal{} 30$.


Hey guys.  I just wanted to post a clarification on what Max has said.  ANYBODY can play Cody Bowl, regardless of age, education, and experience.  You can sign up as an individual, or form your own team.  If you are interested in creating a team, please send me a message, either on the forum or on AOL Instant Messenger (mathjoker).

I won't claim that the problems are necessarily AIME level for individual rounds and USAMO for vertical (team) rounds - in many cases they are significantly easier.  Still, if you want practice for these two competitions, it is hard to go wrong with Cody Bowl.

Hope to hear from you soon!

- Cody Patterson, founder and commissioner

[quote="mathjoker"]Hey guys.  I just wanted to post a clarification on what Max has said.  [b]ANYBODY [/b]can play Cody Bowl, regardless of [b]age, education, and experience[/b].  You can sign up as an individual, or form your own team.  If you are interested in creating a team, please send me a message, either on the forum or on AOL Instant Messenger (mathjoker).

I won't claim that the problems are necessarily AIME level for individual rounds and USAMO for vertical (team) rounds - in many cases they are significantly easier.  Still, if you want practice for these two competitions, it is hard to go wrong with Cody Bowl.

Hope to hear from you soon!

- Cody Patterson, founder and commissioner[/quote]

So wait a second, I could go ask Mr. Crawford or my math teacher to be on my team?!

lol so danny has realized I am not the smartest person around anymore ....  ;)

Danny, that is correct - anyone is allowed to play.  Of course, one big question is whether the people you would ideally want on your team will be willing to play.  Cody Bowl is designed as an amateur math league, so if you shoot too high (let's say you get a former IMO-er on your team), the contest may not be challenging enough to be interesting for that person.  With no cash or other tangible prizes on the line, it would come down to how much that person wants to play.

The reason I allow this is because I want to experiment with a completely open, completely free math league.  Last season, I had some undergraduate students involved and I found that if anything, high school students had the advantage because that's the kind of math my contest focuses on.  People who are closer to high school math competitions will enjoy the greater advantage.

Hope this clears up your question.


function
Functional Analysis
real analysis
real analysis unsolved

(X,M,u) is a measure space, E in M, indentify L^2(E,u) with
the subspace of L^2(X,u) consisting of functions that
vanish outside E. If {En} is disjoint and X=\/En. Then
{L^2(E,u)} is mutually orthogonal subspaces.
Prove: every f in L^2(X,u) can be uniquely written as
f=sum(fn). (series converging in norm) where fn in
L^2(En,u).
(I know charicteristic functions X_En is an orthonormal basis
in L^2(X,u), so we can take fn = <f, X_En >. But why this
decomposition is unique?)

easy again, Trick is to think in abstract Hilbert space, don't
use anything L^2. show fn is unique for all n, by using orthogonal
projection to 1-dim (closed) subspace.


This will be directed toward beginners in problem solving, so if you are an expert, maybe you'll want to do it for fun. But, the problems will be pretty easy, so don't be discouraged from this one based on previous results.

The test will likely be the weekend after Christmas break, so sure I am posting this pretty early, but as a heads up also.

is there an exact date going to be set soon? if so, i'll take the test (for fun... my AMC8 days are over)

Ok, are people still interested? I have it done, so count this as the Mock AMC K, which I will post a little after the Mock AMC I, which is tomorrow.

It will be about AMC 10 level, so it won't be as challenging as some of the previous AMC's.

When is this gonna be?
cause i want to take this before AMC 10 A.....

Hmm, I don't know if enough people will take it if it's today because short notice.

I'll take it.  It'd be better to post it today than tomorrow.

How about saving it for the AMC B? We'll probably need practice then; because no one will be motivated enough to post another AMC.

Ok, I am offering it to anyone who wants to take it early now.

My apologies to Lucky707, who I kept waiting for a long time, since the MOCK AMC K wasn't perfected when he requested it.

The date will probably be either Sunday, 6:00 PM Eastern, or Monday 5:30 eastern, but I think today suits people better. Just ask.

So that's 3:00 PST...right after I come home from Chinese school...today (Sunday) works for me I guess. If I don't forget.

[quote="white_horse_king88"]Ok, I am offering it to anyone who wants to take it early now.

My apologies to yifman12, who I kept waiting for a long time, since the MOCK AMC K wasn't perfected when he requested it.

The date will probably be either Sunday, 6:00 PM Eastern, or Monday 5:30 eastern, but I think today suits people better. Just ask.[/quote]

Umm, what'd I do?

could u send the mock amc k test to me? i'm pretty busy at both scheduled times but i really, really, really, (really x5) need to practice for the amc on tues. thanks :D

[quote="white_horse_king88"]Sorry Yifman12, got my usernames mixed up.  :blush:

Hehe, I sent it to you accidentally, so please ignore it thanks!

All other people, do you have a prefered time? Or should I post it at the set time of 6:00 PM eastern?[/quote]
As I said...6:00 PM Eastern is fine. If I were more selfish I'd want it a little later, like 6:30 or 7:00, so I'm less likely to be late...but I'm not sure if people on the east coast would want it so late (for them). Because I would probably object if it were 7 PM PST. So yeah.

if we take it early, are you going to time us or something? i don't know when i could have 75 min to myself so i can't exactly give u a specific time. if possible, could u just send it to me the moment you see this message?

I have a class later today. Please send it to me now.
thanks

Could you send it to me early as well?
Thanks.

What time is good for me to send it NightFlarer?

Right now is best.  I may not take it right away, but I will take it in 75 minutes and send you my answers when I'm done (should finish today)

Pardon the errors... there seem to be endless things. I may have sent multiple PM's to some people because or various reasons. Please excuse the discrepancies. Thanks.

Correct answer banks:

2: (A)$-4$  (B)$0$  (C)$4$  (D)$8$  (E)$12$
3: (A)$85$ (B)$86$ (C)$87$ (D)$88$ (E)$89$
16: A) 219 B) 220 C)228  D) 256 E) 288

15: The fourth sentence, replace Bobby with Adam.
22: I'm asking for radius not area sorry.

6 is good
and seeing it is 6 right now can i have it?


linear algebra
matrix
linear algebra unsolved

Given $ V\equal{}M_n(K)$ and $ W$ is the set of $ C\in V$ such that there exists $ A,B \in V$ such that $ C\equal{}AB\minus{}BA$ .Prove that $ W$ is a subspace of $ V$.

$ W\equal{}\mbox{Ker(Tr)}$, so it's a subspace of $ V$ of dimension $ n^2\minus{}1$.

$ x \in W \Rightarrow tr(x)\equal{}0 \Rightarrow x \in Ker(tr)$.So $ W$ is a subset of $ \mbox{Ker(tr)}$.I can show that there are $ n^{2}\minus{}1$ matrices, each can be written as the form $ \mbox{AB\minus{}BA}$,which are linear independent.So that is a base of $ \mbox{Ker(tr)}$.
For $ x \in W$,$ x$ is a linear combination of the base above.But how to prove if $ x$ is a linear combination of that base,then $ x$ can be written as the form $ \mbox{AB\minus{}BA}$?

that's not what it means to be spanned by a basis. it means if the trace of A is 0 then A can be written as a linear combination of matrices of the form CD-DC; not necessarily that A itself is of that form.

then $ A \in \mbox{Ker(Tr)}$ but it doesnt mean $ A \in W$

Yes it does.

W is the [i]span [/i]of matrices of the form AB - BA. Not (in general; n = 1, for example) the set of matrices of the form AB - BA. There's a difference.

[quote="§outh§tar"]W is the [i]span [/i]of matrices of the form AB - BA.[/quote]
$ W$ is defined in the original post as follows.
[quote]$ W$ is the set of $ C\in V$ such that there exists $ A,B \in V$ such that $ C\equal{}AB\minus{}BA$[/quote]
If you want to consider a different object, it would help to give it another name.

I believe that it has been discussed many times on the forum, but here is an approach. I will take $ \mathbb{K} \equal{} \mathbb{R}$ to keep things simple.
1/ show that if $ Tr(C)\equal{}0$, then $ C$ is similar to a matrix whose diagonal is null
2/ now suppose that the diagonal of $ C$ is null, and take for example $ B \equal{} diag(1,2,3, \ldots, n)$. Find $ A$ such that $ C\equal{}AB\minus{}BA$
3/ conclude

That argument doesn't work across different fields- in particular, it doesn't work in finite characteristic, where the identity may be in $ W$.


function
Euler
search
logarithms
number theory
totient function
prime numbers

Let $\mathbb{N}$ be the set of positive integers. Let $n\in \mathbb{N}$ and let $d(n)$ be the number of divisors of $n$. Let $\varphi(n)$ be the Euler-totient function (the number of co-prime positive integers with $n$, smaller than $n$).

Find all non-negative integers $c$ such that there exists $n\in\mathbb{N}$ such that \[ d(n) + \varphi(n) = n+c , \] and for such $c$ find all values of $n$ satisfying the above relationship.

If $n=1$ then $\varphi(n)+d(n)=2=n+1$.
Suppose $n>1$. In this case we have all $d(n)$ divisors of $n$ except 1 are not coprime with $n$. It follows that
\[\varphi(n)+d(n)\leq n+1.\]
Moreover,
1) $\varphi(n)+d(n)=n+1$ $\Leftrightarrow$ if $m<n$ and $(m,n)>1$ then $m\,|\,n$    [b](*)[/b].
2) $\varphi(n)+d(n)=n$ $\Leftrightarrow$ for all $m$ s.t. $m<n$ and $(m,n)>1$ we have $m\,|\,n$ and this rule has exactly 1 exception    [b](**)[/b].

Now we consider these cases.

1) If $n$ is prime or $n=4$ then $\varphi(n)+d(n)=n+1$. Suppose $n$ is composite number and $n\neq 4$. It implies that $n=ab$, $a,b\in\mathbb{N}$, $a\geq 2$ $b\geq 3$. Consider $m=a(b-1)$. We have $(m,n)=a>1$ and $n$ isn't divisible by $m$ -- contradiction with [b](*)[/b].

2) Suppose $n$ isn't prime and $n\neq 4$. If $n=6$, $n=8$, $n=9$ or $n=12$ then $\varphi(n)+d(n)=n$. Suppose $n\neq 6$. There are some subcases:
  2.1. $n=p^k$, where $p$ is prime number, $k\geq 2$. In this case $\varphi(n)=p^k-p^{k-1}$ and $d(n)=k+1$, so $\varphi(n)+d(n)=n-(p^{k-1}-(k+1))$, hence we search for equality $p^{k-1}-(k+1)=0$.
    2.1.1. If $p=2$ then $k\geq 4$ (since $n\neq 4,8$) and using unduction it is easy to show that $2^{k-1}-(k+1)>0$;
    2.1.2. If $p=3$ then $k\geq 3$ (since $n\neq 9$) and using unduction we show $3^{k-1}-(k+1)>0$.
    2.1.3. If $p\geq 5$ then $p^{k-1}-(k+1)>0$.
    Conclusion: case 2.1 doesn't give new values for $n$.

  2.2. $n=p^kq^ls$, where $p<q$ are maximal distinct prime divisors, $s\geq 1$.
    2.2.1. $p,q\geq 3$. Consider $m_1=(p^k-1)q^ls$ and $m_2=p^k(q^l-1)s$. We have $m_1$ and $m_2$ are two exception from [b](**)[/b] -- contradiction.
    2.2.2. $p=2$ $\Rightarrow$ $s=1$, so $n=2^kq^l$.
      2.2.2.1. $q=3$ and $l=1$. In this case $k\geq 3$ (since $n\neq 6,12$). Consider $m_1=3\cdot 3$ and $m_2=5\cdot 3$ -- contradiction with [b](**)[/b].
      2.2.2.2. $q=3$ and $l\geq 2$. Consider $m_1=2\cdot 5$ and $m_2=2\cdot 7$ -- contradiction with [b](**)[/b].
      2.2.2.3. $q\geq 5$. Consider $m_1=2\cdot 3$ and $m_2=2^k\cdot (q-1)$ -- contradiction  with [b](**)[/b].
      Conclusion: case 2.2 doesn't give new values for $n$.

valentin where do you find cwmo problems?

[quote="Myth"]zhaoli posted CWMO 2001, 2002, 2003 and 2004
http://www.mathlinks.ro/Forum/viewtopic.php?t=15596
http://www.mathlinks.ro/Forum/viewtopic.php?p=124375
http://www.mathlinks.ro/Forum/viewtopic.php?p=124142[/quote]

ah ic, thx myth
Do you think that the papers are becoming easier ??
Personally i think cwmo 2004 looks easier than 2003 one.

[quote="Myth"]Actually, I didn't compare them, because I have no time to carefully read CWMO 2003.[/quote]

ic.

Spell it to hear it....you will see...

Pierre.

For all non-negative integers $c$ , $c\in\mathbb{N}$ what can we say about the solutions of \[ n + 1 - d(n) - \varphi(n) = c , \] and for each such $c$ we may have solutions (a finite set always?)
Does exist a subsequence of non-negative integers such that $ \frac{d(n) + \varphi(n)}{n}$ tends to 0 - yes or we can get in terms of some elementary function also a lower bound: \[ L(n) \leq d(n) + \varphi(n) \leq n+1 \]

Consider $n_k=p_1p_2...p_k$, where $p_k$ is a sequnce of all prime numbers.
Then $d(n_k)=2^k$ and $\varphi(n)=(p_1-1)...(p_k-1)$. And we have $\frac{d(n_k)}{n_k}\to 0$ and $\frac{\varphi(n_k)}{n_k}\to 0$ (last one is not so obvious, but it is equaivalent to $\frac{1}{p_1}+...+\frac{1}{p_k}\to +\infty$ as $k\to +\infty$).

I knew these and i wrote because it seems related, but i wonder for the 1st part which is the answer and for the lower bound $ L(n) $ if we can express using only a combination of rational, logarithmic and exponential.

I think I can prove something like $\frac{n}{\ln^2(n+1)}<\varphi(n)$. In the same time $d(p)=2$ for all prime numbers $p$, so there are no estimates.

[quote="heartwork"][quote="Myth"]Consider $n_k=p_1p_2...p_k$, where $p_k$ is a sequnce of all prime numbers... $\varphi(n)=(p_1-1)...(p_k-1)$... $\frac{\varphi(n_k)}{n_k}\to 0$ (last one is not so obvious, but it is equaivalent to $\frac{1}{p_1}+...+\frac{1}{p_k}\to +\infty$ as $k\to +\infty$).[/quote]

$n_k=p_1p_2...p_k$ is the primordial, while the fact $\frac{\varphi(n_k)}{n_k}\to 0$ has an elementary proof, by considering for example a set of integers from $1$ to some given value $a$ , $A=${$1,2,...,a$} and the subset of $A$ , $B$ that contains only integers from $A$ that does not divide with any of the primes $p_i$. It follows that $\frac{(p_1-1)(p_2-1)...(p_k-1)}{p_1p_2...p_k}\to 0$ by making $a$ large enough as $k\to +\infty$.[/quote]
No, it is not clear for me. :maybe:
Maybe I am missing something.

Fix an integer $a$ , $A=${$1,2,...,a$}. Construct $B$ by eliminating step by step all integers which are multiple of $p_1$, $B$ will have at most $\frac{(p_1-1)}{p_1}B$ elements, then eliminate from this new set all integers which are multiple of $p_2$, then $B$ will have at most $\frac{(p_2-1)}{p_2}\frac{(p_1-1)}{p_1}B$ elements and so on, for $p_k$ large enough $B$ will contain only $1$, so $\frac{|B|}{|A|}\to 0$ as both $k\to +\infty$ and $a\to +\infty$. I hope it is correct.

[quote="heartwork"]Construct $B$ by eliminating step by step all integers which are multiple of $p_1$, $B$ will have at most $\frac{(p_1-1)}{p_1}B$ elements, then eliminate from this new set all integers which are multiple of $p_2$, then $B$ will have at most $\frac{(p_2-1)}{p_2}\frac{(p_1-1)}{p_1}B$ elements[/quote]
It is not true, because in your esimate you exlude some numbers twice.

What i wrote is only a sketch of proof and i think it works since between the remaining numbers after such step this is the closer percentage from $B$  (it's about distribution of all rests modulo $p_i p_j$ which are not multiple of $p_j$ - how many of them are not multiple of  $p_i$). Do you think that the only proof for this fact it comes from the Euler product and Riemann-Zeta for $s=1$ ?

[quote="heartwork"]Do you think that the only proof for this fact it comes from the Euler product and Riemann-Zeta for s=1 ?[/quote]
Actually, it is not that difficult. Just not mention Euler product and Riemann-Zeta ;) :D

I'll post the solution that Gabriel (silver medal in IMO2004) and me got to the original CWMO problem. It's a nice (and short!!) solution.

Well, it's fairly easy to prove that $\phi(n) + d(n) \leq n+1$ by a counting argument: $\phi(n)$ is the number of positive integers less than $n$ and coprime with $n$. Let $A$ be the set of such numbers. On the other hand, $d(n)$ is the number of positive divisors of $n$. Let $B$ be the set of such numbers.

It's easy to see that $A \cap B = \{1\}$, so since $A$ and $B$ are subsets of $[n] = \{1,2,\dots,n\}$, $n\geq |A\cup B| = |A| + |B| - |A\cap B| = \phi(n) + d(n) - 1$ and the result follows.

So $c \leq 1$. If $c = 1$, by the previous counting argument, every number less or equal to $n$ must be a divisor of $n$ or a number coprime with $n$, which is "kinda hard to happen". OK, not so hard: if $n$ is a prime, $\phi(n)=n-1$ and $d(n) = 2$, so all primes $n$ are solutions if $c=1$. Let's find the other solutions in this case. Suppose $n$ is not prime and let $p$ be the smaller prime divisor of $n$. Since $\gcd(n,n-p)=p\neq 1$, $n-p$ divides $n$ and thus $n-p$ divides $p$. Hence $n-p \leq p \iff n\leq 2p$. But $n\geq p^2$, so $p=2$ and $n\leq 4$. $n=4$ is the only other solution.

If $c=0$, let $p$ be as before and consider $n-p$ and $n-2p$. If $n-p$ divides $n$ the problem reduces to the former case. So $n-p$ is not a divisor of $n$ nor coprime to $n$ and it must be the only one. Thus $n-2p$ divides $n$ and consequently $2p$ as well. Hence $n-2p \leq 2p \iff n \leq 4p$ and since $n \geq p^2$, $p \leq 3$ and $n \leq 12$. Testing, we get the other solutions: $6$, $8$, $9$ and $12$.

Well, I didn't think of a lower bound yet (I didn't read all the other messages) but I know that the average number of divisors of $n$ is close to $\log n$. More precisely: let $\overline{d}(n)$ be the average of the number of divisors of all numbers less or equal to $n$, that is,
\[\overline{d}(n) = \frac{1}{n}\sum_{1\leq k\leq n}d(k)\]

Thus $|\overline{d}(n) - \log n|\leq 1$.

To manage to prove this, count in two ways the number of pairs $(x,y)\in Z^2$ such that $1\leq x, y\leq n$ and $x$ divides $y$. Or look it in "Proofs from The Book".

Oh, $\log n = \log_e n$, by the way. You can replace it by $H(n) = \sum_{1\leq k\leq n}\frac{1}{k}$ (OK, the Euler-Mascheroni constant is missing, but it is indeed a good estimate).

Actually, what I did is so lengthy and not so rigorous, that I will write it.
But here is the sketch.

We exploit this fact $p^{k-1}>k+1$,for all $p>=5,k>=2$,
then we just have to check  the following cases to search where $n-d(n)-\varphi(n)$ is actually non negative.

1)$n$ is a product of primes
2) $n$ is a a prime power (also includes one and primes here)

Then with  a good deal of intuitional bounding and a bit heavy casework, we get that only these choices of $n$ give a non-negative $c$
They are just $n=1,p,4,8,6$ and all in all $c$ can be just $0,1$ and $n$ is already found,just assign them to their respective values of $c$.

My non-rigorous solution can be made rigorous but it will be extremely long.
To make it rigorous,I would have to replace my intuitional bounding with some analytic Number theory as  used in the above solutions.

An example of the 'intuitional bounding' that I was referring to is as follows,
When I had to verify that $pqr....t-(p-1)(q-1).....(r-)$ is greater than $$2^{no. of primes-1}$$ for primes $p,q,r,...t$ and also # of primes>2, I actually took $2,3,5$ to verify it and then used an 'intuitional bounding'to see that this value should now only increase.This seems true obviously but I dont want to prove,since it seems tedious and maybe uses analytic Number theory ,which I am not aware of .
But I think I solved the problem, with only such obvious looking bounds to be proved.


geometry
geometric transformation
reflection

Does anyone know how to play Chinese musical instruments (or just know what is Chinese music)?

I am not able to play, but I love listening to music performed on traditional Chinese instruments such as the pipa, erhu, or zheng. That has long been one of my favorite forms of music.

Interestingly, ensemble pieces for those instruments, in which you have a pipa, zheng, erhu, maybe a yangqin, and other instruments performing together are often very Western-influenced. Much of the ensemble music of that kind was only composed in the last century, and the composers were aware of Western compositional techniques for ensemble pieces. Nonetheless, those pieces sound very exotic to a Western ear, because of the unfamiliar instruments used in the performance.

My favorite Chinese piece is a pipa piece often called "Yizu Wuqu," which is probably familiar to most people who listen to a lot of Chinese music.

I listen to Chinese music extensively, both lyrical and intrumental.

I don't really know artist names of the singers, but my favorite Chinese instrument is the erhu.

I can't really play it that well, but I learned how to play some basic songs on it.

I can play loads of cool Chinese songs on the violin. The key is to use a lot of glissando.

[quote="Drunken_Math"]my favorite Chinese instrument is the erhu. [/quote]
Why do you like the erhu?
I am learning it just now.

[quote="Drunken_Math"]I can play loads of cool Chinese songs on the violin. The key is to use a lot of glissando.[/quote]

How do you do glissando on the violin?  :?

When I first saw Memoirs of a Geisha, The house of the flying daggers, etc I decided that I love chinese and japanese music. Unfortunatelly I do not know any artists; could someone help?

There's some Chinese music that was composed for western instruments that aren't bad to listen to. The Butterfly Lovers Violin Concerto and The Yellow River Piano Concerto come to mind. My piano teacher once gave me a short piece to play, called 平湖秋月 (Autumn Moon Over The Calm Lake), which sounded quite nice.

i only know that $KOTO$, is a chinese instrument,

which has really nice sound,, :?

Lots of chinese music use pentatonic scale. It's like a normal scale without the subdominant and the leading note. There are heaps of chinese music written for the piano but it would sound better if you play it on a real chinese instrument. And the pipa has four strings like the violin but it doesn't stay in one key. It can be tuned into all keys according to the music.

Yeah, there are many types of Chinese Musical Instruments available, but I think only five items are best loved by Chinese people, i.e.Chinese Pipa, Erhu, Guqin and Guzheng, and Bamboo Flute.

On my website you would find more about such chinese musical isntruments.

[url]http://www.shoppingchinanow.com/[/url]

ive seen them. they look hard to play...

I was in New York this weekend.  When walking in Central Park, a Chinese man was sitting there playing the erhu.  Everything that comes out of it is so depressing, but I like it lots!  :D

I love the yangqin.

Once, I was at a subway station and this lady was playing it and I just kinda stood there until she finished her piece. Beautiful instrument.

Having lived in China for about a year already, I definitely hear Chinese music !  :D I think it's more relaxing and musical than the modern crap we have in France.

I'm into western music, but my dad (being Chinese) is really into it. He arranged (for me) the "Reflections of Moon" piece (er4 chuan2 ing4 yue4 in bad pinyin) from erhu, including lots of the "special chinese string techniques." He also brought to my attention this Chinese violin concerto called the "Butterfly Lover."

my piano teacher teaches me various
chinese songs. they are great for 3 reasons
1). they sound good
2). they captivate the audience at recitals
3). because judges at auditions are nube to Chinese music, chances are that you will get a higher score by playing a chinese piece rather than some well known Sonata or Nocturne.

my friend knows how to play some weird and not too common chinese instrument, which i forgot what it was called....

i made a shakuhashi by myself, seeing pictures on internet

and using some papers of arthur h benade and others i make the fingerholes.

the sound is good, but have less octaves.

if you want i can take some photos of this. and the others flutes i made.

i had make a dizi too. but i dont have this anymore...

it's quite simple play chinese music, the most difficult in chinese music is not the encore, is the technique.

I personally play the guzheng, along with classical piano.  They are very different, but they are similar.  Guzheng tends to be more expressive for me.  Playing Guzheng is more fun, but sometimes the technical stuff can be rather annoying.

and, btw, to a poster above, the koto is the japanese version of the guzheng, with fewer strings

chinese instruments are amazing!!  :jump: i am learning the erhu and it's sooooo fun i luv it i wish i had more time to play it, but violin comes first.....

I had my elementary school years in China, and I was a percussionist (still one now, except it's drumline), and we had to play a variety of Chinese percussion instruments, my favorite being what I call a Chinese Drumset (I don't know the english for it), which basically is around 5 drums arranged according to size small on the left and big on the right, and you play it like marching tenors (not really because we don't get much rebound from the drum and we rarely do doubles or french rolls because the stick is too short).  I like it cuz I often got solos if I play the instrument.  And our school actually won the national championship 5 times in a row, I believe (can't really remember properly, it's been many years and I'm getting old).
Anyways, this is an image of the drum
http://netdrive.puiying.edu.hk/~club_hkpyco/hkpyco/image/m8.gif

I play 古筝guzheng
it resembles a harp, but it's a little more complex since it has bridges across for altering the sound when playing. it has 21 strings.
i play it since i was 6 XD

I like to add another note to the pentatonic scale: if you're using C-D-E-G-A, add B and if you do it right it can still sound Chinese-ish.