Neshan dahid dar har Geraf namotenahi ,Ya binahayat raas vojod darad the har 2 ta az onha be ham vasl and Ya binahayat raas vojod darad the hich 2 ta az on ha be ham vast nistand .

avval sabet mikonim dar har majmooeye namotanahi az ras ha ($A$) ke hich zir majmooeye namotanahi az an daraye ras hayi ke hame baham mojaver bashand nist rasy vojood darad ke ba namotanahi ras az an majmooe mojaver ast.farz konid ingoone nabashad dar an soorat rasy vojood darad (b0) ke ba majmooei az namotanahi ras mojaver nist(majmooeye $B1$).badihi ast ke $B1$ khasiate majmooeye $A$ ra darad pas dar $B1$ niz rasy mesle $b1$ vojood darad ke ba hich kodam az ras haye majmooeye namotanahie $B2$ mojaver nist... mibinim ke mitavanim ozvi mesle $b0$ az $A$, $b1$ az $B1$ ... $bi$ az $Bi$ ... entekhab konim ke hich kodam ba ham mojaver nistand ke in ba farze ma motanaghez ast.pas dar har majmooei mesle $A$ rasy ($c0$) vojood  darad ke ba majmooei az namotanahi ras mojaver ast mesle C1.va mesle halate ghabl $c1$ ra az $C1$ , $c2$ ra az $C2$ ... entekhab mikonim.badihist ke majmooeie $\{c0,c1,c2,...\}$ majmooei namotahi az ras hast ke hame ba ham mojaverand . batavajjoh be tagharone masale tarafe digare hokm ham be hamin tarigh esbat mishavad.


How can we prove that all primes are of the form 6n +/- 1, where n is a whole number?

When you take numbers (mod6), they leave a residue of 0,1,2,3,4,5.  So every number can be written as 6n, 6n+1,..., 6n+5, n is a whole number.  6n, 6n+2, 6n+4 are all divisible by 2.  6n+3 is divisible by 3.  So all that's left is 6n+1, 6n+5, which can also be written as 6n-1.

With, of course, the exceptions of 2 and 3.


This may sound a little odd... but does anyone use any music to pump him/herself up for a math competition or just "mathing" in general?

I use "Rock and Roll All Night" by Kiss, "Ramble On" by Led Zepplin, "All My Life" by the Foo Fighters, and the theme music from "A Beautiful Mind"  ;)

Hmm, no, not really :P

I do listen to music to get pumped up...or relaxed...err, let's call it 'focused' for races.  I love The Division Bell by Pink Floyd.  Anything by Pink Floyd get's me focused.  I also like Boston for getting me pumped up.

Those are some good songs you picked there though.  Now that I have my mp3 player back, I'll probably be listening to different songs to get pumped to since I'll have a larger selection than I had with my CD's.

I like Megalomaniac by Incubus as a "pump up" song.  But that's not really necessary for math, lol.

The song that'll get anyone pumped up is "Lose yourself" by Eminem.

Rock songs get me hyperactive and far away from focussing :lol:


Let $ a,b,c,d>0$.Prove that:
$ \frac{1}{a^2\plus{}ab}\plus{}\frac{1}{b^2\plus{}bc}\plus{}\frac{1}{c^2\plus{}cd}\plus{}\frac{1}{d^2\plus{}da}\geq \frac{4}{ac\plus{}bd}$

BTW,
Try someone to solve [url]http://www.mathlinks.ro/viewtopic.php?t=213271[/url]question b)

[quote="Inequalities Master"]Let $ a,b,c,d > 0$.Prove that:
$ \frac {1}{a^2 \plus{} ab} \plus{} \frac {1}{b^2 \plus{} bc} \plus{} \frac {1}{c^2 \plus{} cd} \plus{} \frac {1}{d^2 \plus{} da}\geq \frac {4}{ac \plus{} bd}$
[/quote]
$ \frac {1}{a^2 \plus{} ab} \plus{} \frac {1}{b^2 \plus{} bc} \plus{} \frac {1}{c^2 \plus{} cd} \plus{} \frac {1}{d^2 \plus{} da}\geq \frac {2}{\sqrt{abcd}}$ is true too. :wink:

For you, Inequalities Master! :lol:
Let $ a,$ $ b,$ $ c,$ and $ d$ are positive numbers. Prove that:
\[ \frac {1}{a^2 \plus{} ab} \plus{} \frac {1}{b^2 \plus{} bc} \plus{} \frac {1}{c^2 \plus{} cd} \plus{} \frac {1}{d^2 \plus{} da}\geq \frac {2}{\sqrt {abcd}}
\]
Here is my proof:
Let $ abcd \equal{} 1,$  $ a \equal{} \frac {y}{x},$ $ b \equal{} \frac {z}{y},$ $ c \equal{} \frac {t}{z}$ and $ d \equal{} \frac {x}{t},$ where $ x,$ $ y,$ $ z$ and $ t$ are positive numbers.
Hence, $ \sum_{cyc}\frac {1}{a^2 \plus{} ab} \equal{} \sum_{cyc}\frac {x^2}{y^2 \plus{} xz} \equal{} \sum_{cyc}\frac {x^4}{x^2y^2 \plus{} x^3z}\geq$
$ \geq\frac {(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2)^2}{x^2y^2 \plus{} y^2z^2 \plus{} z^2t^2 \plus{} t^2x^2 \plus{} x^3z \plus{} z^3x \plus{} y^3t \plus{} t^3y}.$
Thus, it remains to prove that $ (x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2)^2\geq2(x^2y^2 \plus{} y^2z^2 \plus{} z^2t^2 \plus{} t^2x^2 \plus{} x^3z \plus{} z^3x \plus{} y^3t \plus{} t^3y),$
which is equivalent to $ (x^2 \plus{} z^2)(x \minus{} z)^2 \plus{} (y^2 \plus{} t^2)(y \minus{} t)^2\geq0.$
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81588

The following inequality is also true.
Let $ a,$ $ b,$ $ c,$ and $ d$ are positive numbers such that $a+b+c+d=1$. Prove that:
\[ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d}\leq \frac {1}{16abcd}
.\]


At room Temperature, the infra-red (iR) region of teh absorption spectrum for gas phase hydrogen choloride, H35Cl, consists of a series of absorption lines.  Starting with the lowest energy transitions, the first four of these are at energies of: 2885.98cm^-1, 5667.98cm^-1, 8646.78^-1, 10992.81cm^-1.
note: H35Cl = HCl, Chlorine with mass 35.

Q1) Assign the vibrational quantum number for the initial and final state for each of these four transitions.

Q2) Calculate zero-point evergy for H35Cl (units cm^-1)

Q3)  The difference in energy between these bands is NOT constant.  How does it change and why is this change observed.

Q5) Assuming H37Cl has the same force constant as H35Cl, predict the energy of the lowest energy iR absorption band in its spectrum (units: cm^-1).

thanks in advance!

Perhaps I am just ignorant, but what does the 35 represent in H35Cl?

Oh, wait I am ignorant.  Way for me to read the question.


Show that $x^{1992}-x^{1990}+(2n-1)x^3-x=y^2$ has no solutions in positive integers (x,y) when $n=1,2$

hmm i got for $n=1$

[hide]

factor it into $x(x-1)(x+1)(x^{1989} +1)$ which is obviously not a perfect square

the case of $n=2$ is trickier tho... cuz im not seeing how i could factor it

maybe we can bound it between two perfect squares to show that it always lies inbetween?

[/hide]

[quote="Hikaru79"][quote="gotztahbeazn"]hmm i got for $n=1$

factor it into $x(x-1)(x+1)(x^{1989} +1)$ which is obviously not a perfect square
[/quote]

I got that far too, but why is that "obviously not a perfect square"? How is it so obvious?  :blush:[/quote]

Well.  When $x$ isn't a square then there's a single factor of $x$ that isn't paired with any other factors, so the expression can't be a square.

When $x$ itself is a square then $x - 1$ doesn't share a factor with either of the other two so the expression still can't be square.

[hide="For n=1"]
$x^{1992} -x^{1990} +x^3-x=y^2$
$x^{1990} (x^2-1) + x (x^2 - 1) = y^2$
$(x^{1990} +x)(x+1)(x-1) = y^2$
$x (x^{1989} +1)(x+1)(x-1) = y^2$
$x (x+1)(x^{1988} - x^{1987} + x^{1986} - ... -x+1)(x+1)(x-1)=y^2$
$x (x+1)^2 (x^{1988} - x^{1987} + x^{1986} - ... -x+1)(x-1)=y^2$
Since $(x+1)^2$ is a perfect square, $(x^{1988} - x^{1987} + x^{1986} - ... -x+1)(x-1)x$ must be a perfect square
This means that $(x-1)x$ must be a factor of $(x^{1988} - x^{1987} + x^{1986} - ... -x+1)$ Since x clearly isn't a factor of $(x^{1988} - x^{1987} + x^{1986} - ... -x+1)$, x isn't a repeated factor and therefore the left hand side is not a perfect square unless $x$ itself is a perfect square..

By synthetic division, if we divide the 1988 degree polynomial by $x-1$, we notice the bottom row falls into a pattern of 2,1,2,1,2... etc. so the remainder is not 0, and the 1988 degree polynomial therefore does not divide $x-1$ either. Since $x$ and $x-1$ cannot both be a perfect square, (unless x = 1, in which case y is not positive), the left hand side is not a perfect square and there are no solutions.[/hide]

[hide="For n=2"]
The polynomial becomes:
$x^{1992}-x^{1990}+3x^3-x=y^2$
Which becomes:
$3x^{1992}-x^{1990}+3x^3-x -2x^{1992}=y^2$
Which factors to:
$x(3x^2-1)(x^{1989}+1) - 2x^{1992} = y^2$

For starters, notice that the left hand side must be even, so y cannot be odd.
Since y is even, we will represent it as $2z$, so the right hand side becomes $4z^2$, meaning the left hand side must be divisible by 4.

If x is odd, $x(3x^2-1)(x^{1989}+1)$ divides 4 but $- 2x^{1992}$ does not, so the left hand side does not divide 4 and x cannot be odd either.

If x is even, $x(3x^2-1)(x^{1989}+1)$ divides 4 only if x is a multiple of 4, and $- 2x^{1992}$ must divide 4, so x can only be a multiple of 4.

Let $x=4t$
$z^2 = t(48t^2-1)(2^{3978}t^{1989}+1) - 2^{3985}t^{1992}$

Again, with the same reasoning as used before, neither z nor t can be odd
Let $z = 2f$

$4f^2 = t(48t^2-1)(2^{3978}t^{1989}+1) - 2^{3985}t^{1992}$

$t$ must be a multiple of 4, and $x$ must be a multiple of 16.

By repeating this process indefinately, the lowest possible value of $x$ inflates to infinity because of the nature of the factors of the polynomial, therefore there are no possible integer solutions to this problem. Is there a sound way this can be shown using algebra?[/hide]


A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure 1. The cube is then cut in the same manner along the dashed lines shown in Figure 2. This creates nine pieces. What is the volume of the piece that contains vertex $ W$?
[asy]import three;
size(200);
defaultpen(linewidth(.8pt)+fontsize(10pt));

currentprojection=oblique;
path3 p1=(0,2,2)--(0,2,0)--(2,2,0)--(2,2,2)--(0,2,2)--(0,0,2)--(2,0,2)--(2,2,2);
path3 p2=(2,2,0)--(2,0,0)--(2,0,2);
path3 p3=(0,0,2)--(0,2,1)--(2,2,1)--(2,0,2);
path3 p4=(2,2,1)--(2,0,0);
pen finedashed=linetype("4 4");

draw(p1^^p2^^p3^^p4);
draw(shift((4,0,0))*p1);
draw(shift((4,0,0))*p2);
draw(shift((4,0,0))*p3);
draw(shift((4,0,0))*p4);
draw((4,0,2)--(5,2,2)--(6,0,2),finedashed);
draw((5,2,2)--(5,2,0)--(6,0,0),finedashed);

label("$W$",(3,0,2));
draw((2.7,.3,2)--(2.1,1.9,2),linewidth(.6pt));
draw((3.4,.3,2)--(5.9,1.9,2),linewidth(.6pt));
label("Figure 1",(1,-0.5,2));
label("Figure 2",(5,-0.5,2));[/asy]$ \textbf{(A)}\ \frac {1}{12}\qquad \textbf{(B)}\ \frac {1}{9}\qquad \textbf{(C)}\ \frac {1}{8}\qquad \textbf{(D)}\ \frac {1}{6}\qquad \textbf{(E)}\ \frac {1}{4}$

[hide="Hint"]What kind of solid is formed?[/hide]

[hide]I get 1/12:

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=324[/img]
[/hide]

[hide="Answer"]It is a square pyramid with base sides $s=\frac{1}{2}$ and $h=1$, so the area is $\displaystyle\left(\frac{1}{2}\right)^2\cdot 1\cdot \frac{1}{3}=\frac{1}{12}\Rightarrow \boxed{A}$.[/hide]

[hide]The piece is a triangular prism with a square base (side = $\frac{1}{2}$) and a height $=1$.

So the volume of that piece $V=\frac{1}{3}*1*(\frac{1}{2}*\frac{1}{2})=\frac{1}{12}$

$\boxed{A}$[/hide]

[hide]A. The base is $\frac{1}{4}$, and it's a pyramid, so it's multiplied by $\frac{1}{3}$ to get $\frac{1}{12}$.[/hide]


You do not need to reveal your location, but please participate in the poll.

I'll just answer your question: The majority of math people around the Bay Area are in Santa Clara County. Namely, San Jose, Cupertino, and Palo Alto.

Santa Clara, then likely Alameda (I don't think in SF they're so crazy about math)

...wow. My initial prediction:

50% Alameda
45% Santa Clara
2% San Francisco
2% San Mateo
1% Everywhere else

I guess people always tend to feel conceited.  :blush:

[quote="bookaholic"]I hadn't realized there were so many people from Alameda. Which schools do you Alamedans go to?[/quote]

MSJ/Hopkins.

And Harvest Park :)

Which was a school never noticed until next year.

[quote="dragon96"]I'd like to say I'm from Santa Clara County, so no one will know who was the one who got the crappy score at states.[/quote]

Theoretically you DID take the Santa Clara Valley chapter...

Hey, lookie! Someone is from Contra Costa County! Which city?

I'm from Santa Clara county like many others here.

errrrr what do u put if ur in the san jose/cupertino area?
we have like 15 AoPSers from our school lol.  :P

San Jose and Cupertino are both from Santa Clara County; just check it up.

We have about 15 AoPSers from our school as well.  :lol:

If I were to put up a similar poll in the Southern California ARML forum (which I won't, since I already know where everyone lives), Los Angeles County and Orange County would be pretty close to tied. There would be a small handful from Santa Barbara County and one person from Ventura County. I don't think I would get anyone from Riverside, San Bernardino, or Kern Counties. (San Diego County would not be included because that's a separate ARML area.)

But then there are 15 people in an ARML team, so therefore there would be two people  :P

[quote="DataBox"]Haha I guess... :blush:

In any case, you would normally expect more AoPSers than ARML team members[/quote]

what are you sure about this :O

I don't think that's right, I remember seeing lots more people at ARML practice than AoPS...although around here that might not necessarily be what you expect (crappy ARML so less people?)

I live in Virginia, but which County in California is the most EDUCATIONALEST county?

I know that in Virginia, Fairfax County is the county that is always first in education.

P.S. lol, EDUCATIONALEST, new word for AoPSers

How about just using "educated" instead of these weird new words?

Define "Educationalest". In fact, try to give a totally objective, fair, ranking approach for "best education".

Educationalest: the county that has the most smartest student and invests a lot of money in education.

The second definition is objective. The first is not.

Define "smartest" student.

Do we go by AMC scores, AIME scores, USAMO scores, IMO-ship, etc. And even then, "smartest" doesn't immediately imply mathematically smartest.

yeah... i think he means like best at standardized testing like STAR or something.

that doesnt count for actual smartness-tests create stupid mistakes

Thus there exists no objective way of measuring intelligence.

Actually "invests a lot of money in education" is objective, but in no way has much correlation to intelligence.

Any of you guys from the Harker School?  I saw their orchestra perfrom at the Lincoln Center (our HS also performed there)...and I also know that they are very good math school (had an IMO-or atleast IMO qualifier couple years ago?)

There are several Harker kids who used to post at least, but I'm not aware of any now.
Yes they are quite good. Christiano, Yi Sun, etc.

Harker had an IMO Silver Medalist last year as well as 4 8th grade USAMO qualifiers.  There are also USAPhO, USAChO (sp?), NACLO finalists that attend Harker.

[quote="DataBox"]13% of the Bay Area ARML A Team came from San Mateo County, however, I only think there are only two registered AoPS users from the county...[/quote]
Gotta love misleading ways of presenting information.
Like how our school had "36 times the national average" in USAMO qualifiers last year, and equalled/beat "21 states"? :P

I wonder who the third San Mateoan is.

By states do you include US territories/Canadian provinces?

[quote="r15s11z55y89w21"]I just find it hilarious how there are a few counties devoid of AoPSers and two counties teeming with AoPSers. I know 18 AoPSers from Alameda County (within contact via this site) and have met most of them in real life :) .[/quote]

Challenge: List them all.
Hint: I'm not one of them.

[quote="AIME15"]Am I counted in those 18? :)[/quote]

Yes.


let $a,b,c >0 , ab+bc+ca = 1$ find the greatest value of permutation :
$\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{b+c}}+\frac{1}\sqrt{a+c}}$

zhaobin : sorry

let $a \to \infty$ $y \to \frac 1 a$  $c \to 0$
then  $\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c} \to \infty$

[quote="black_puma"]let $a,b,c >0 , ab+bc+ca = 1$ find the greatest value of permutation :
$\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{b+c}}+\frac{1}\sqrt{a+c}}$
i'm soory[/quote]
If $b=c\rightarrow0+$ then $\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{a+c}}+\frac{1}{\sqrt{b+c}}\rightarrow+\infty.$ ;)

When $b=c \rightarrow 0$ condition isn't satisfied, I think?

[quote="black_puma"]let $a,b,c >0 , ab+bc+ca = 1$ find the least value of permutation :
$\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{b+c}}+\frac{1}\sqrt{a+c}}$

very very very soory  :rotfl:   :rotfl:[/quote]
Mabe $ab+bc+ca =210000$hahahahahahahahaha :rotfl:  :rotfl:


where can I find problems from this olympiad?

[quote="srulikbd"]thanks. this is a very good competition  :)
where can I find the MMO? moskov  mathematical olympiad
I found the LMO, here http://problemcorner.org/ thanks[/quote]
Is it also called the Leningrad Olympiad? If so there is a book on mathpropress.com, which I have. I wouldn't recommend it; the problems are shallow and not very hard. The AllUnion is much better.


Given a set of 100 different whole numbers, prove that one can choose either one number or several numbers whose sum is divisible 100.

Hint (in spoiler):  [hide]Let Sn be the sum of the first n numbers.  Show that two of the sums S0, S1, ..., S100 are equal mod 100.[/hide]

Spoiler..

[hide]Let S_n be the sum of the first n terms.  If some S_n is divisible by 100, then we are done.  Since we are dealing with divisibility by 100, let's look at this mod 100.  So, if some S_n=0 (mod 100), then we are done.  If not, this leaves 99 options for S_n (mod 100) (1, 2, ... 99).  However, we have 100 sums.  So, by the pigeonhole principle, at least two (call them S_k and S_m, with m>k) must be congruent mod 100.  So, the difference between S_k and S_m (not in mod 100) will be a multiple of 100.   Since every term in S_k is also a term in S_m, the difference between the two can be expressed as the sum of one or more of the terms in the sequence, and this is a multiple of 100.[/hide]

Hmm.. that wasn't very well written, sorry if it's confusing..

not sure if this works or makes sense but:

[hide]All numbers in hear refer to the number mod 100.

well, assume you have all ones.  You use 100 of them.  Imagine you replace a one with a two.  You use 99 of them. Imagine you make two a three. Now you use 98 of them.  Make two a four.  You use 97 of them and so on.  When your numbers are too big so that when you add the two smallest, their sum is greater than 100, you make negative modulos.[/hide][/hide]


1) Find an unbounded sequence with a convergent subsequence.
2) Find an unbounded sequence without a convergent subsequence.

A suggestion: think of an unbounded sequence. It will have either property (1) or property (2). You're halfway done!

[u]For number 1:[/u]

You can always interlace sequences:
$ a_n \equal{} 0$ for odd $ n$ and $ n$ for even $ n$.

If you don't like piecewise definition, you can always do something along the lines of:

$ a_n \equal{} n^n\plus{}(\minus{}n)^n$

for an unbounded sequence with a convergent sum.

[u]For number 2:[/u]

If it is monotonically increasing and does not converge, then no infinitely long subsequence can converge.

An example: $ a_n \equal{} n$: any two terms differ by at least $ 1$, so clearly, no subsequence can converge.


analytic geometry
integration
calculus
rotation
geometry
3D geometry
sphere

A one inch hole is drilled through a bowling ball with a two inch radius(small bowling ball!!!). How much volume is there left?

I'm having trouble with the book's explanation.

Using cylindrical coordinates (recall the Jacobian is $ r$),
\[ V \equal{} \int_{ \minus{} \sqrt {3}}^{\sqrt {3}}{\int_0^{2\pi}{\int_1^{\sqrt {4 \minus{} z^2}}{rdr}d\theta}dz} \equal{} \pi \int _{ \minus{} \sqrt {3}}^{\sqrt {3}}{3 \minus{} z^2dz} \equal{} 4\pi \sqrt {3}
\]

I'm pretty sure the original poster hasn't taken that course (third-semester calculus) just yet. We should rephrase the answer in terms of first/second semester calculus.

I do have one question for the original poster: what does "one inch hole" mean? That the hole is one inch long? That it is one inch in radius? That it is one inch in diameter? Each of those interpretations will lead to a different answer.

Good point, then I suppose the better way to go would be with volumes of revolution.  I did assume a hole of one inch radius was drilled passing through the center of the ball; if a one inch diameter was meant the answer becomes $ \frac {11\sqrt {3}\pi}{2}$.

the hole has a one-inch radius

And yes, I'm taking Calculus 2 and I just began finding volumes of solids

thanks in advance!

First, let's make a good drawing of the 2-dimensional object that we will rotate to get the volume. Start with a circle of radius 2 (equation $ x^2\plus{}y^2\equal{}4$) and draw the vertical line $ x\equal{}1.$ The D-shaped region to the right of this line and to the left of the curve will be what we want to rotate.

It may be useful to know the points of intersection; putting $ x\equal{}1$ into the other equation we get the points $ \left(1,\sqrt{3}\right)$ and $ \left(1,\minus{}\sqrt{3}\right).$

Now, decide which way we'd rather slice this D-shaped region. Vertical strips or horizontal strips? I'll add that when I teach this material, I emphasize making the choice here, choosing which slices seem more convenient, rather than planning up front to make "shells" or "washers." Decide which slices you want, then rotate them and see what you get. (I'll also add that if I were to ask this problem on a test, I wouldn't specify a method, and I assign the homework for the "shell' and "washer" sections of the chapter at the same time.)

If we slice vertically, both the top and bottom of the slice would be on the curve. If we slice horizontally, the right end of the slice is on the curve but he left end is on the straight line. It's a weak reason, but I'll use that as an excuse to slice horizontally. (When you make a choice for a weak reason like that, leave the door open to coming back to this point and trying the other way.)

Slice at height $ y.$ The width of the slice is $ dy.$ The left end of the slice is on $ x\equal{}1$ and the right end is on the curve. With $ y$ fixed, we solve for $ x$ to get $ x\equal{}\sqrt{4\minus{}y^2}.$ (We said right end - hence only the positive square root.)

We now rotate this slice about the $ y$ axis. You have to visualize this - draw all the pictures you need to. It rotates to a "washer": an annulus with inner radius $ 1$ and outer radius $ \sqrt{4\minus{}y^2}.$ The volume of this washer is the area of the annulus time $ dy.$ That is,

$ dV\equal{}\pi\left(\left(\sqrt{4\minus{}y^2}\right)^2\minus{}1^2\right)\,dy\equal{}\pi(3\minus{}y^2)\,dy.$

How far does $ y$ range? Look back on that two-dimensional diagram to see that $ y$ ranges between the two intersection points that we found.

Hence $ V\equal{}\pi\int_{\minus{}\sqrt{3}}^{\sqrt{3}}3\minus{}y^2\,dy$

(This is precisely the computation that JoeBlow arrived at. The Calculus III version may have different vocabulary but it isn't really a different calculation.)

The cute surprise is that we only need to know the length of the hole - in this case, $ 2\sqrt{3}$ - and not either the radius of the sphere or the radius of the hole to get this integral and this number.

That makes much more sense!  :idea:  Thanks!


you should type tittle of topic by English  :mad: But I think you are falling in love with a girl  :P .... For above inequality, I will try it  :roll:


geometry
cyclic quadrilateral
combinatorics proposed
combinatorics

If we assign every point in the plane 2 colours, does there necessarily exist a monochromatic square? What about a cyclic quadrilateral? What about a monochromatic regular n-gon? What about a cyclic n-gon? Does this generalise to k colours? I know the triangle case is true for any finite number of colours

Gallai's theorem states that for any finite colouring of the plane (or in finite dimension) and for any given finite set $ S$ of points, there exists a monochrome set homothetic to $ S$.
Thus, each of your questions have 'yes' for answer.

Pierre.

Is there a simple proof for this? I searched on google, but could not find any references to this theorem

Gallai's theorem is a generalization of Van der Waerden'theorem about monochromatic arithmetic progressions when colouring the positive integers. It can be proved with Hales-Jewett's theorem, which is more a well-known result (and that's the way I know the proof. But, it requires more times...).
In another hand, I think that, sometimes, Gallai is also known as Grünwald (why?).

Pierre.

Would you be able to give me a link to the proof, or to websites containing use of this theorem? I have seen one mention of it as Grunwald's theorem, but otherwise I cannot find it

You may find it (with Hales-Jewett's theorem) in the really nice "Ramsey Theory" by R.L. Graham, B.L. Rothschild, J.H. Spencer, published by Wiley.
Or, look at at book in Ramsey theory to get the Hales-Jewett theorem. Then, you will deduced Gallai's theorem.

Or, look at
http://www.ti.inf.ethz.ch/ew/courses/extremal04/razen.pdf

Pierre.

The proof defines a homothetic copy in only $ Z^m$, so does this still include non-lattice cases such as equilateral triangles, pentagons etc?


linear algebra
matrix
combinatorics unsolved
combinatorics

I really need help with this one!!!!!

Lets denote M the set of all matrices of the type (m,n) which have all elements 1 or -1 so that the product of all numbers on any colomn and line is 1. Compute the number of elements of M@!@!@!

bye bye!!!

pls help!!!

Are you sure it's 1 and not -1? I know a similar prb but the product on all rows and columns is -1. As I remember this one can be done recursively. Maybe there's a similar approach to your prb.

hey it's correct the way I posted!  :)  :)  :)  :)  :)  :)  :)

solution anyone? :?

bye bye

The solution has to be 2^((m-1)(n-1)), but you have to prove it!!! I don't have a solution! sorry!

cheers! :D :D

Mitzah! I can't understand your trouble!

The answer is 2<sup>(m-1)(n-1)</sup>, because we can fill elements in first (m-1) columns and (n-1) rows with arbitrary numbers:
[code]
1,1     1,2     1,3    ...     1,m-1     1,m
2,1     ...
...
n-1,1   n-1,2   ...            n-1,m-1  n-1,m
n,1     n,2      ...                     n,m
[/code]
then numbers in m-th column and n-th row will be determined uniquely.

Yeah, that's exactly how ypu solve the problem I told you abt, the onw with -1 instead of 1, so the same principle works.

I may sound too stupid, but here goes nothing! :D

can you post a combinatorical count?

bye bye!!  thank you very much ! :D

what about this problem?! please help! :D

Lets denote M the set of all matrices of the type (m,n) which have all elements 1 or -1 so that the product of all numbers on any colomn and line is -1. Compute the number of elements of M!

bye bye

[quote="Mitzah"]what about this problem?! please help! :D

Lets denote M the set of all matrices of the type (m,n) which have all elements 1 or -1 so that the product of all numbers on any colomn and line is -1. Compute the number of elements of M![/quote]
If $m$ and $n$ have the same parity then my previous solution work fine!
Otherwise, it is clear $M$ is empty, since product of all numbers in matrix should be $(-1)^m$, and on the other hand $(-1)^n$, but $(-1)^n\neq(-1)^m$.

Question is closed (?).


YEAST           23
WASH            25
SUMAC          57
SPREE           60
ARREST         64
LODGING        ?

fill in the question mark and explain why. :P

Can we have a hint? This is stumping me so far :oops:

the words are anagrams for someone famous... lol i know this is a really random puzzle

Is Wash Shaw? There was an air force pilot named Shaw that died at age 25, or else the playwright George Bernard Shaw?


A positive integer is equal to the sum of the squares of its four smallest positive divisors. What is the largest prime that divides this positive integer?

Source: Australian Math Competition, S30 (I30, J30), 2005

[hide="spolier"]
Notice that the number cannot be odd . So it can be written as

$n=1^2+2^2+a^2+b^2=5+a^2+b^2$ where $b>a$

if $n$ is divisible by $4$ , then taking $mod 4$ we see that RHS cannot be $0\mod 4$ . So $n$ only divisible by $2$.

clearly we must have $a$ be odd .And so this implies $b$ is even . And the only possiblity for that is $b=2a$ .Plug in we get $a=5$,$b=10$ and $n=130$ . So largest prime is $13$[/hide]


Express as a common fraction: $ 0.\overline5\plus{}0.\overline1\minus{}0.\overline3$

First we convert the decimals to fractions. $ 0.\overline5 \equal{} \frac{5}{9}$, $ 0.\overline1 \equal{} \frac{1}{9}$, and $ 0.\overline3 \equal{} \frac{1}{3}$. Now we can just compute: $ \frac{5}{9} \plus{} \frac{1}{9} \equal{} \frac{2}{3}$, which when added to $ \frac{1}{3}$ yields $ 1$.

Izzy, we're supposed to [b]subtract [/b]$ \frac {1}{3}$ from $ \frac {2}{3}$, instead of adding them together.  So our answer is actually $ \boxed{\frac{1}{3}}$


Fie A1 punctul unde simediana AK a unui triunghi oarecare ABC intalneste cercul circumscris triunghiului si A2, A3 picioarele bisectoarelor interioara si exterioara a unghiului A. Sa se arate ca dreptele A1A2 si A1A3 sunt dreptunghiulare.

Am pierdut ore bune să găsesc soluţia asta:

Fie $\mathcal C$ cercul circumscris triunghiului $ABC$ şi $\mathcal K$ cercul de diametru $A_{2}A_{3}$. Cele două cercuri se întâlnesc în exact două puncte, unul dintre ele fiind $A$. Îl notăm pe celălalt cu $D$.

Fie $T = AD \cap BC$. Din moment $AD$ este axa radicală a celor două cercuri, avem $\mathcal P_{\mathcal C}(T) = \mathcal P_{\mathcal K}(T)$, adică $\overrightarrow{TB}\cdot \overrightarrow{TC}= \overrightarrow{T A_{2}}\cdot \overrightarrow{T A_{3}}$.

Din [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=theorem+steiner&t=58872]topicul ăsta[/url] (teoremele 3 şi 4) deducem că $\overrightarrow{B A_{2}}= \frac{AB}{CA+AB}\cdot \overrightarrow{BC}$ şi că $\overrightarrow{B A_{3}}= \frac{AB}{AB-CA}\cdot \overrightarrow{BC}$.

Fie $t \in \mathbb R$ astfel încât $\overrightarrow{BT}= t \cdot \overrightarrow{BC}$.

Rezultă că \[\overrightarrow{TB}\cdot \overrightarrow{TC}= \left[-t \cdot \overrightarrow{BC}\right] \cdot \left[ (1-t) \cdot \overrightarrow{BC}\right] = \left( t^{2}-t \right) \cdot BC^{2}\] şi că
\begin{eqnarray*}\overrightarrow{T A_{2}}\cdot \overrightarrow{T A_{3}}&=& \left( \overrightarrow{B A_{2}}-\overrightarrow{BT}\right) \cdot \left( \overrightarrow{B A_{3}}-\overrightarrow{BT}\right) \\ \ &=& \overrightarrow{B A_{2}}\cdot \overrightarrow{B A_{3}}-\overrightarrow{BT}\cdot \left( \overrightarrow{B A_{2}}+\overrightarrow{B A_{3}}\right)+\overrightarrow{BK}^{2}\\ \ &=& \frac{AB^{2}}{AB^{2}-CA^{2}}\cdot BC^{2}-\frac{2t AB^{2}}{AB^{2}-CA^{2}}\cdot BC^{2}+t^{2}\cdot BC^{2}.\end{eqnarray*}

Egalăm cele două rezultate obţinute: \[\left( t^{2}-t \right) \cdot BC^{2}= \frac{AB^{2}}{AB^{2}-CA^{2}}\cdot BC^{2}-\frac{2t AB^{2}}{AB^{2}-CA^{2}}\cdot BC^{2}+t^{2}\cdot BC^{2}\]  \[\, \Longleftrightarrow \, \, t^{2}-t = \frac{AB^{2}}{AB^{2}-CA^{2}}-\frac{2t AB^{2}}{AB^{2}-CA^{2}}+t^{2}\]  \[\, \Longleftrightarrow \, \, \frac{2t AB^{2}}{AB^{2}-CA^{2}}-t = \frac{AB^{2}}{AB^{2}-CA^{2}}\]  \[\, \Longleftrightarrow \, \, t \cdot \frac{AB^{2}+CA^{2}}{AB^{2}-CA^{2}}= \frac{AB^{2}}{AB^{2}-CA^{2}}\]  \[\, \Longleftrightarrow \, \, t = \frac{AB^{2}}{AB^{2}+CA^{2}}.\] Astfel, $\overrightarrow{BT}= \frac{AB^{2}}{AB^{2}+CA^{2}}\cdot \overrightarrow{BC}$, deci $\frac{\overrightarrow{BT}}{\overrightarrow{TC}}= \frac{AB^{2}}{CA^{2}}$.

Din acelaşi [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=theorem+steiner&t=58872]topic[/url] (teorema 5), deducem că $AT$ (şi în consecinţă $AD$) este $A$-simediană în $\triangle ABC$.

Folosind $D \in \mathcal C, \, \mathcal K$, devine evident că $D \equiv A_{1}$ şi că $\measuredangle A_{2}A_{1}A_{3}= \measuredangle A_{2}D A_{3}= \frac{\pi}2$.

Am ataşat şi un desen:

[url=http://img59.imageshack.us/my.php?image=liliansimedianeny8.png][img]http://img59.imageshack.us/img59/4651/liliansimedianeny8.th.png[/img][/url]

E bună soluţia?

Se poate si mai usor totusi.Este suficient sa avem $AK\cdot KA_{1}=A_{2}K\cdot A_{3}K$ ca $A_{1}A_{2}A_{3}K$ sa fie inscriptibil.Avem $BK\cdot KC=AK\cdot KA_{1}$ si ramane sa avem $BK\cdot KC=A_{2}K\cdot A_{3}K$ si se face usor :lol:  verificarea.

In timp ce incercam sa rezolv problema asta, am descoperit o alta problema, care pare adevarata:

[i]Fie $S$ intersectia simedianei din $A$ a triunghiului $ABC$ cu cercul sau circumscris. Demonstrati ca simetricul lui $S$ fata de $BC$ se afla pe afla mediana $AM$, unde $M$ e mijlocul lui $BC$.[/i]


lol, these are the aops 1 questions on Exponents and Logarithims

[quote="math92"]Convert these equations into logarithmic equations.

1.  $3^{3}=27$

2.  $16^{1/4}=2$

3.  $x^{z}=y$[/quote]

#3 is sufficient to illustrate the rule for solving the other two:

[hide]

$z\log x =\log y$[/hide]

[quote="4everwise"]G-Unit is correct. They are. Anyway, they can still be beneficial to people who didn't do the problems in the book or don't have the books so I guess it's fine.[/quote]
Yes.
Hope you can post more questions from AoPS.

how is jimli correct for #3...its supposed to be log y base x is equal to z...just take the log base x of both sides...

I think the answers of Stan and chess make more sense.  But I have a feeling it was just a mistake for the original answerer.

I guess they can be benefitial but all you have to know is that logs are exponents. Whatever answer you get in logs is an exponent. Here's how I remember it, say you have Log4 of 64=x, that's the same thing as 4^x=64. It's really helpful when you start calculating big numbers. For others who don't know, the laws for logs are the same as laws for exponents. Same base, multiplication, add your exponents, etc. It's really easy if you know your exponents.


hello,

okay here is the problem.

I was wondering what equation I would use to calculate the center of gravity of a car. I don't need to know the height of the Center of Gravity, I only need the position laterally.

Then, I was wondering if there was another equation that I could use to find out the position and amount of added weight needed to move the center of gravity as close to the midpoint of the car as possible.

thanks to anyone that can help, it's greatly appreciated.

Maybe take a look at this link:
[url=http://en.wikipedia.org/wiki/Center_of_gravity]Center of Gravity[/url]


xy=27 and logxY + logyX =5/2
find x and y

[hide]Basically, convert $\log_{y}x$ into base x and get $\log_xy+\frac{1}{\log_xy}=\frac{5}{2}$. Let $w=\log_xy$ to get $w+\frac{1}{w}=\frac{5}{2}$. Solve this equation and get that $w=2, \frac{1}{2}=\log_xy$.

If $\log_xy=2$, then $x^2=y$, and since $xy=27$, substitute to come out with $x^2(x)=27$, $x=3$, and $y=9$. If $\log_xy=\frac{1}{2}$, $x^{\frac{1}{2}}=y$, and substitute into $xy=27$ to have that $x=9$, $y=3$. [/hide]

And also, this needs to move to getting started.


Solve the equation
$ sinx\plus{}sin2x\equal{}3cos^2x$
It seems simple,Yes :lol:

[quote="Allnames"]Solve the equation
$ sinx \plus{} sin2x \equal{} 3cos^2x$
It seems simple,Yes :lol:[/quote]
I think it isn't very hard, my friend :wink:

Condition : $ (sinx)( 3cos^2x \minus{} sin2x )\ge\ 0$

We have

$ sinx \plus{} sin2x \equal{} 3cos^2x$

$ \leftrightarrow sinx \equal{} 3cos^2x \minus{} sin2x$

$ \leftrightarrow sin^2x \equal{} (3cos^2x \minus{} 2sinx.cosx)^2$

$ \leftrightarrow sin^4x \plus{} sin^2x.cos^2x \equal{} 9cos^4x \minus{} 12sinxcos^3x \plus{} 4sin^2cos^2x$

$ \leftrightarrow sin^4x \minus{} 3sin^2x.cos^2x \plus{} 12sinxcos^3x \minus{} 9cos^4x \equal{} 0(1)$

Because $ cosx \equal{} 0 \leftrightarrow x \equal{} \frac {\pi}{2}$ isn't root of equality so we have

$ (1) \leftrightarrow tan^4x \minus{} 3tan^2x \plus{} 12tanx \minus{} 9 \equal{} 0$

Let $ t \equal{} tanx$ we have $ t^4 \minus{} 3t^2 \plus{} 12t \minus{} 9 \equal{} 0$

I think we can solve it by an easy way

[quote="onlylove_math"][quote="Allnames"]Solve the equation
$ sinx \plus{} sin2x \equal{} 3cos^2x$
It seems simple,Yes :lol:[/quote]
I
Let $ t \equal{} tanx$ we have $ t^4 \minus{} 3t^2 \plus{} 12t \minus{} 9 \equal{} 0$

[b]I think we can solve it by an easy way[/b][/quote]
Hic,I cant solve it :blush:

[color=darkblue]Other way for your problem:
Equation $ \Longrightarrow sinx \plus{} 2sinx.cosx \equal{} 3cos^2x$

Let $ t \equal{} tan\frac {x}{2}$ then we get equation: $ \frac {2t}{1 \plus{} t^2} \plus{} 2.\frac {2t}{1 \plus{} t^2}.\frac {1 \minus{} t^2}{1 \plus{} t^2} \equal{} 3.\frac {(1 \minus{} t^2)^2}{(1 \plus{} t^2)^2}$

$ \Longleftrightarrow 2t(1 \plus{} t^2) \plus{} 4t(1 \minus{} t^2) \equal{} 3(1 \minus{} t^2)^2$

$ \Longleftrightarrow 3t^4 \plus{} 2t^3 \minus{} 6t^2 \minus{} 6t \plus{} 3 \equal{} 0$

You can solve this equation by use [url=http://en.wikipedia.org/wiki/Galois_theory]Galois theory[/url].[/color]

hello,i think you need this formulas:
http://mathworld.wolfram.com/QuarticEquation.html
Sonnhard.

There is a negative root and a positive root $ t\approx{0.897}$.
$ t\equal{}\dfrac{\minus{}b^{3/2}\plus{}\sqrt{4ab\minus{}a^2b\plus{}11b\plus{}24a^{3/2}}}{2\sqrt{ab}}$ where
$ a\equal{}(35\plus{}6\sqrt{71})^{1/3},b\equal{}\sqrt{2a\plus{}a^2\minus{}11}$.
We can't construct $ t$ with compass and straightedge (because we can't construct $ a$).


Can square be partitioned in isosceles triangles with angles:$ 75,75,30$.

If it was possible , for the whole puzzle the number of 75° angles will be the double of the number of 30° angles . But on every vertex of the puzzle the number of 75° angles is at must the double of the number of 30° angles and on every vertex of the square there only 30° angles . Impossible !

I beg your pardon for my bad english  :blush:


function
calculus
search
algebra
polynomial
abstract algebra
Gauss

Is there a general formula that can be applied to any n-th degree function's coefficients to get the root, complex or real?  If so, please state it or post a link to a supporting website.
 :rotfl:  :rotfl:  :rotfl:

If I understand you correctly, there can't be a formula for degrees of higher than 4. :(

Masoud Zargar

There is no way to get exact roots for polys greater than degree 4, as boxedexe said, but using Newton's method roots can be approximated to however many decimal places of accuracy.
(Newton's method uses calculus, so I won't post it here, but feel free to search it, it's probably on the forum somewhere.)

It's a way to approach the zeros on the graph using derivtives.

People have proved that it is specifically impossible to have a general equation for the (quintic?) 5th degree. Now a better question is, how would one go about doing that??

Solving a fifth degree equation without using a fomrula.  ;)

EDIT: I forgot to write down the answer.
Using newton's method or factoring it should do the trick.

Is there anyway to know beforehand (a priori) whether it is possible to determine the exact values of the roots of a polynomial of degree five or greater?

All the mentioned stuff here is a result of Galois theory.
It's also possible to tell by Galois theory whether a polynomial can be solved by taking radicals (out of the coefficients/rationals) only.

But note: there are 'formulas' for the roots of polynomials of degree $\geq 5$, but they are not only using radicals; instead they need more complicated functions, e.g. the $\vartheta$-function.

I meant, how would you go about proving that a solution in general is impossible?

From Mathworld (Abel's Impossibility Theorem): "In general, polynomial equations higher than fourth degree are incapable of algebraic solution in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions."

I think Abel used some kind of precursor to Abstract Algebra (guess who Abelian groups are named after :)), like Gauss did with the Fundamental Theorem of Algebra.

Are there ways to know a priori whether a polynomial is "incapable of algebraic solution in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions" through Galois Theory?  Could someone lay out the basics, if possible?

[quote="mgao"]First of all. what is the $\vartheta$ function?
Second of all, what is the "formula"[/quote]
That's the Jacobi Theta function.  Also, there is a "quintic equation" of sorts using the Jacobi theta functions (which themselves are not solvable in a finite number of radicals, etc.), which if you really want to, you can check out here: http://mathworld.wolfram.com/QuinticEquation.html (look at equations 7 - 11).  I don't know about you, but the solutions aren't really meaningful to me.


Uita-te pe forumul de inegalitati. Sigur a fost discutata acolo. Si cred ca merge chiar forma mai tare, cu 2 in loc de 4.


Posted before here http://www.mathlinks.ro/Forum/viewtopic.php?t=56071


calculus
integration
calculus computations

Extremely useful calculators are at http://www.quickmath.com, for example definite/indefinite integrals can be done on [url=http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&section1=calculus&section2=integrate&section3=advanced]this page[/url].

How could I forget to post [url=http://integrals.wolfram.com/index.jsp]www.integrals.com[/url]? This site is from Wolfram Research who make Mathematica and host the [url=http://www.mathworld.com]online version of the CRC Concise Encyclopedia of Mathematics[/url].


How many combinations of pennies, nickels and/or dimes are
there with a total value of 25 cents?

Possibilities:

0 dimes:  0-5 nickels, so 6
1 dime: 0-3 nickels, so 4
2 dimes: 0-1 nickels, so 2.

summing yiels 12.

3+5x4/2=3+10=13-1(since this formula includes quarters)=12

I'd be safe and use AIME15's way, because this formula is unreliable, unproven, and does not work for all possibilities in the range from 10~50 cents.

It does work from 10-49. Find a counter-example.

21 cents.

EDIT: Oops, never mind. But generally, be wary of "formulas" that have never been proven.

[quote="ernie"]The formula which mewto talks about is this:

$ 3 \plus{} \frac {(\lfloor\frac {n}{5}\rfloor)(\lfloor\frac {n}{5}\rfloor \minus{} 1)}{2}$

This works for $ 10$ through $ 50$ cents, I believe.

$ n \equal{} \text{number of cents}$[/quote]

21:

3+(4*3)/2=3+6=9

Using my way:

0 dimes: 0-4 nickels, so 5
1: 0-2, so 3
2: 0, so 1

equals 9...

o_O

[hide="Proof"]
Lol my very first post on AoPS was about this. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=192642&start=20

To simplify our casework, we start with the fact that if $ 5n$ coins gives us $ k$ ways, then $ 5n\plus{}1,5n\plus{}2,...,5n\plus{}4$ each have $ k$ ways.
Proof: In each arrangement, say you have $ Q$ quarters, $ D$ dimes, $ N$ nickels, and $ P$ pennies. For every possible arrangement of $ Q,D,N$, you have $ 5n\minus{}(Q\plus{}D\plus{}N)$ pennies. So there is exactly 1 possible value of $ P$ for every possible arrangement of $ Q,D,N$. If we add any number from $ 0\minus{}4$ to $ 5n$, we will still have the same number of possible arrangements of $ Q,D,N$ (think about it, and it will make sense. What is we add $ 5$ to $ 5n$?). So that concludes our proof.
Then try out multiples of $ 5$ by casework...
(0-4)-1
(5-9)-2
(10-14)-4
(15-19)-6
(20-24)-9
(25-29)-13
(30-34)-18
(35-39)-24
(40-44)-31
(45-49)-39
Now subtract 3 from each of the cases above, and get $ 1,3,6,10,15,21,28,36$, which are triangular numbers!
Thus for $ 2\leq{n}\leq9$, we have $ k\equal{}3\plus{}\frac{n(n\minus{}1)}2$.
[/hide]

[quote="THECHAMP"]Wait the answer's 13![/quote]

The answer would be $ 13$ is quarters were included.

[quote="hunter34"]Actually the way to do this problem is generating functions.[/quote]

Yes, but keep in mind this is MC, and we need FORMULAS... :P  :wink:


The measures of the sides of an isosceles trapezoid are in the ratio of 3:4:3:6. The perimeter of the trapezoid is 48 cm. What is the number of square centimeters in the area of the trapezoid? Express your answer in simplest radical form.

The sides are $ 9$, $ 12$, $ 9$, and $ 18$.

From the top-left corner (assuming the lower base is $ 18$), you can form a perpendicular line to the base, creating a right triangle.

[asy]draw((0,0)--(3,0)--(3,sqrt(72))--cycle);
label("$9$",(1.5,sqrt(18)),NW);
label("$3$",(1.5,0),S);[/asy]

Thus, the height is $ 6\sqrt2$.

Then, the answer is $ 15 \times 6\sqrt2$ (note that $ 15$ is the median and the area of a trapezoid can be written as $ mh$) or $ \boxed{90\sqrt2}$.


AMC
AIME
geometry
3D geometry
tetrahedron
ratio
incenter

Here are a few random AIME problems:

1) Let N be the greatest integer multiple of 8, no two of whose digits are the same. What is N (mod 1000)?

I know the answer is simply 9876543120 (one of the largest integers, no two of whose digits are the same), but is there a more methodical way of doing this problem than just starting from 9876543210 and moving down?

2) In a regular tetrahedron, the centers of the faces are the vertices of a smaller tetrahedron. What is the ratio of the volume of the smaller tetrahedron to the volume of the larger one?

1) For a number to be divisible by 8, its last three digits must be divisible by 8. Find the largest three digit number divisible by 8 that uses the digits 0, 1, 2 exactly once. If it's not possible, try 0, 1, 3, etc. Then arrange the other digits in decreasing order.

2) The volume of a regular tetrahedron is proportional to the cube of its sidelength. Remember that in equilateral triangles, all the normal centers (incenter, circumcenter, orthocenter, centroid, etc.) are at the point, so we can use the properties of any of them! Let's use the centroid's property: a median is cut in a ratio of 2:1 by the centroid. Then form an equilateral triangle with two faces of the tetrahedron along the medians of the two faces. Since the length of each side is 1/3 as long as the median, it is 1/3*sqrt(3)/2=1/(2*sqrt(3)) as long as the side. Thus the ratio of the volumes is 1/(24*sqrt(3)).

Ok, I get the first one now, thanks. But for the second one, the problem actually asks for the sum of the numerator and denominator. Your answer would thus be 1 + 24*Sqrt(3), but the actual answer (of the AMC ppl) is 28, so...

By the way, a couple of side questions:

Does the "center of a triangle" (not necesasily an equilateral triangle, as in this case) mean its centroid?

I saw somewhere randomly (without justification) that the area of a regular tetrahedron with side s was Sqrt(2)/12 * s^3. Is this correct?

Thanks.

That means that the answer is 1/27, and I made some mistake somewhere. Maybe I'll check it some time.

Usually you don't refer to a center of a triangle unless it's equilateral. One reason is [url]http://faculty.evansville.edu/ck6/encyclopedia/[/url].

Yes, that is the volume of a tetrahedron. Draw a few triangles and use the Pythagorean Theorem a few times and the fact that the volume of a pyramid is 1/3 (area of base) (height).

the answer is indeed 1/27.

For any regular tetrahedron, the volume given the side, s, is s^3*sqrt[3] / 12 (you can derive this urself).

Then, s ^3*sqrt[3] / 12  is the volume of the large tetrahedron, and each side of the smaller tetrahedron is s/3, so apply the volume again, and the ratio is s/3

Hmm, I understand that the medians of a triangle divide one another into segments in the ratio 2 : 1, but I still don't see why the sides of the smaller tetrahedron are also 1/3 the length of the sides of the larger tetrahedron (even though it is somewhat intuitive)?

Oh, and are you sure that it is Sqrt(3), and not Sqrt(2), in your formula for the area of a regular tetrahedron?

And to convince you that a side of the tetrahedron is 1/3 the larger one...

Take one pair of parallel edges, one on the smaller tetrahedron and one on the larger tetrahedron. Then, form a triangle by drawing the medians of the two adjacent faces of the larger tetrahedron. Convince yourself that the points lie on the same plane, and then you will see that by similar triangles, the ratio of the sides is 1/3.

[quote]Take one pair of parallel edges, one on the smaller tetrahedron and one on the larger tetrahedron. [/quote]

Wait, how are any of their sides parallel? Let us imagine the larger tetrahedron ABCD as an upright pyramid-like object. Let the upper tip be A, and the points at the base be B, C, and D. The smaller tetrahedron WXYZ would thus be like the larger one, except scaled down and upside down. Let the bottom tip (on the same plane as B, C, and D) be W, and the other points be X, Y, and Z. Hopefully, you can imagine the same picture as I am imagining. Which two edges here are parallel?

Fierytycoon, see if this helps:

first, draw your tetrahedron in the following manner: first draw an equilateral triangle (labled ABC, clockwise with A at the top), and then from a point in the center (D), draw the lines connecting the vertices of the triangle with D. now, you have a regular tetrahedron. Imagine for a minute, that you are in 2-d space, and that ABC is merely a triangle and that D is just a point in the center of the triangle. Now, connect the centers of triangles ACD, ABD, and BCD. Let there centers be W, X, Y respectively. NOw, isn't WX parallel to BC. When you take this to 3-d, they're still parallel.

Hope this helps, or better yet, hope i'm rite

Oh, I see now. That image really helped. Hmm...would it be possible to quickly label the edges, etc.. in the following to correspond to the labelled diagram you described in your previous post?

[quote]Take one pair of parallel edges, one on the smaller tetrahedron and one on the larger tetrahedron. Then, form a triangle by drawing the medians of the two adjacent faces of the larger tetrahedron. Convince yourself that the points lie on the same plane, and then you will see that by similar triangles, the ratio of the sides is 1/3.[/quote]

I'm sorry, but I don't seem to understand exactly what you mean by the above. Thanks.

O sure.

Draw your diagram in the same manner, with tetrahedron ABCD as the big one, and tetrahedron WXYZ as the smaller one. Extend CW and BX to side AD, and they should meet a point P. Now, since P,W,X,B,C all lie on the same plane, TRiangle PWX ~ PBC, and the ratio of similitude is 1/3. Thus, the ratio of the sides is 1/3.

Hope this helps.

Oh, I get it now. That was a big help. Thanks!

The AIME problems you mentioned cannot be solved in the general case by your method because the program you wrote uses a non-standard library.

A better program is

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main (void)
{
  int i, k = 2;
  srandom(time(NULL));
  for (i = 0; i < k; i++)
    printf("%s\n", Problems[random() % nProblems]);
  return 0;
}

With this, the solutions should be obvious.

Ah, but you see, my statement was a single line of code. I had many #include statements prior to that line. You may often see me name threads with names such as "geometry_problem" and "AIME", but what you don't see is...

#define geometry_problem main(...)

This is no excuse for using some obscure 'int Rand (int, int);' function. How is anyone supposed to know what you mean? In particular, if you interpret the second argument as an inclusive upper bound, then your original statement is subject to an off-by-one error.

When you use non-standard features, you should give adequate documentation.

Well, I never ever mentioned anything about my Problems array, so why don't you pick on that? Its the same deal with my non-declared Rand function.


MATHCOUNTS
geometry
3D geometry
algebra
polynomial
Vieta
inequalities

Ok, on the Chapter Competition, I managed to get 1st place based on Sprint and Target, and also got 1st in our unnofficial Countdown round.

I have been to state once before in a team, but with a completely different team, I only qualified to go to state as an individual.

As of now, I'm roughly halfway through Saxon's Algebra 2. The majority of the questions I have struggled with in the past 3 years of competition have been primarily geometry, with the more algebraic type questions being easier.

My question is this: What should I study?

I plan on memorizing squares and cubes up through a certain number (undecided) and, of course, wherever Saxon takes me.

Thanks for your help! (In advance  :ninja: )

use the aops texts, they are a great resource, if you are short on time, i recommend volume 1, the basics. also there are many tests on [url] http://mnspe.org/mathcounts/coach.htm also practice ones from [url] http://www.yashar.biz/mathcounts i have done this, and i got first place in my chapter by a substantial margin, and i hope to go to denver. I am from Minnesota, and i wish you luck!

just as a sidenote, it's helpful to memorize cubes up to 12 and squares up to 31  :D (31^2<1000) for competitions, it saves a lot of time.

Also program a lot of program on your graphing calculator.
It will help for the targets and team rounds so you won't have to remember a lot of formulas real quick.

I do not recommend you use an Algebra II book to study for MATHCOUNTS; two reasons: first, much of the stuff in Algebra II isn't relevant to MATHCOUNTS, and secondly, much of the stuff in MATHCOUNTS is not covered in Algebra II.

Try doing lots of problems.

To study for MATHCOUNTS, the books that are most helpful are ones written for competitions, not for math knowledge.  Especially MATHCOUNTS, as now, more and more problems are logic-based rather than math-based.

Thanks for the advice!

And I agree with your comment about Algebra 2. There was very little that actually made a difference in the competition.

I'll check out some of the books you reccomended. And, of course, I'll be doing rounds from previous years as much as possible.

memorizing primes can be helpful (helpful for me at least a few times)
and also memorize pythagorean triples. it would be awfully embarassing to run out of time because you had 2 compute a leg to a 8-15-17 triangle.

Hmm... alright.

So:

-Cubes to 20
-Squares to 31
-Pythagorean Triples to (?)
-Extensive knowledge of 2007, 2008, and 2009
-Go through some Aops books
-(Erm... I'm sure I can find other things to put here...)

[quote="neurosurgeon1118"]I reccomend u program Herons formula[/quote]

It's really easy to remember it...

But if there is anything you [i]should[/i] program, it's the formula for the area of a regular polygon.

I also have Saxon Algebra 2. The work in there is just exercises. MathCounts involves Problems. Exercises are tasks that you know how to do instantly. Problems are tasks that you do not know how to do instantly. The best thing to do to prepare for MathCounts is to do past tests and read the AoPS texts.

What you should also know are:

Cubes up to 20
Squares up to 41
Common Pythagorean Triples(3-4-5, 5-12-13, 7-24-25, 8-15-17, 9-40-41, and 11-60-61)
Prime factorization of the current year($ 2^3\cdot 251$)
Primes up to 109
sums of powers($ 1\plus{}2\plus{}\cdots \plus{}n$, $ 1^2\plus{}2^2\plus{}\cdots \plus{}n^2$, and $ 1^3\plus{}2^3\plus{}3^3\plus{}\cdots\plus{}n^3$)
Knowledge of telescoping series

Doing some old AMC's are good too.

i would also memorize the formulas for the areas and volumes of 2D and 3D figures, respectively.

i know my cubes up to 30 and my squares up to 40.  :D and yet i fail at mathcounts.

Also, there is one thing Algebra II might help you with.  You should probably know many ways to factor polynomials.  You probably want to go beyond those covered in Algebra I, but most of what you need will not be in the Algebra II book.  Its much easier to find factoring tricks online.

[quote="1=2"][quote="neurosurgeon1118"]I reccomend u program Herons formula[/quote]

It's really easy to remember it...

But if there is anything you [i]should[/i] program, it's the formula for the area of a regular polygon.

I also have Saxon Algebra 2. The work in there is just exercises. MathCounts involves Problems. Exercises are tasks that you know how to do instantly. Problems are tasks that you do not know how to do instantly. The best thing to do to prepare for MathCounts is to do past tests and read the AoPS texts.

What you should also know are:

Cubes up to 20
Squares up to 41
Common Pythagorean Triples(3-4-5, 5-12-13, 7-24-25, 8-15-17, 9-40-41, and 11-60-61)
Prime factorization of the current year($ 2^3\cdot 251$)
Primes up to 109
sums of powers($ 1 \plus{} 2 \plus{} \cdots \plus{} n$, $ 1^2 \plus{} 2^2 \plus{} \cdots \plus{} n^2$, and $ 1^3 \plus{} 2^3 \plus{} 3^3 \plus{} \cdots \plus{} n^3$)
Knowledge of telescoping series

Doing some old AMC's are good too.[/quote]

I dunno, cubes up to $ 12^3$ and squares up to $ 32^2$ works really well for me already. :)

Updates:

-AoPS: The Basics

-Cubes to 20
-Squares to 31
-Pythagorean Triples
-Extensive knowledge of 2007, 2008, and 2009
-Go through some Aops books
-Herons Formula
-Have to remember 2^3 x 251
-Sums of powers to (?)
-Learn more about Telescoping Series
-Primes to 109
-Learn more ways to factor polynomials
-Learn alot more about geometry.

Speaking of geometry again... Any more thins that I definetly need to know or I'll get killed?

From the sounds of it, if I even make it to nationals I'll get slaughtered...  :|

Meh oh well  :P

Hahaha... Just took the Math Olympiad. That was fun.

Ok... the competition's Saturday. I'm gonna sleep all day Friday so I'll be awake  :sleeping:

If texas competition is so tough why don't the people that can't make nats move to a state with less competition. Like in alaska, the cutoff will probably be 10 less than the texas cutoff. Just an idea :lol:

[quote="wiseidiot"]If texas competition is so tough why don't the people that can't make nats move to a state with less competition. Like in alaska, the cutoff will probably be 10 less than the texas cutoff. Just an idea :Very Happy:[/quote]

You know me wiseidiot. From a firsthand point of view, I know that it isn't that simple to move. There are so many factors to consider when moving. Life is actually NOT about MathCounts :roll:  :D
And even though I would never have a chance of making it to nationals if I stayed in Texas..., well, I didn't even know about MC in 5th grade in my small town. :( Moving's hard.

True, true. I guess this is getting off topic though.

Who cares? This topic is going to be locked soon. Might as well spam as much as you want. jkjk :lol:

Anyone else aside from me think that the entire Middle School section is getting extremely spammy?

The best way to study for States is to review basic counting formulas (by deriving them), looking at the AoPS textbooks a little for the last week, and doing problems that focus on your weak points.

Am I the only person that wonders why California is so competitive (even more than Texas sometimes) yet we phail at Nationals.

[quote="gauss1181"][quote="SeventhSecret"]Good luck.

They're actually making me go to school tomorrow. Can you believe that? :P[/quote]
That's unbelievable. They made u skip the contest??? That's bad.

p.s. ryuk, stop posting so many times straight[/quote]

No, lol, not the entire contest. I mean the day before the contest.

My current programs are:
Distance formula (I took up a lot of space/time, making the simplification-of-square-root-thingy)
Heron area (same simplification-of-square-root-thingy)
Law of Sines/Cosines
Quadratic equation (need to put simplification-of-square-root-thingy here)
Solve a 2-variable equation (gaussian elimination)
Solve a 3-variable equation (gaussian elimination)
Solve a 4-variable equation (doesn't work)

I tend to make calculator programs WHILE doing the MC problems to brute force them, then I put them in action while I actually do the problems myself.

Go to Texas Instruments website and they have an application/program that solves equations with up to 10 variables. Downlaod it.

In case you can't find the application, here's the TI-84 Plus SE download.

If you don't have a SE, just go to google and type in "TI-(w/e you have) Simultaneous Equation Solver). The first link should have it.


advanced fields
advanced fields unsolved

Let $X$ be a path-connected space. Prove that the following two definitions for simply connected space are equivalent:
(1) any loop with basepoint at $p\in X$ is homotopic to the constant map (the constant value is $p$).
(2) any two paths with the same starting point and ending point are homotopic.
In addition, we require the homotopies be relative to $\{0,1\}$.

Clearly (2) implies (1) by taking any loop and the constant map as paths.

Converse:
Take any paths $\alpha , \beta$ from $A$ to $B$.
Now: $\alpha \simeq \alpha \cdot (\beta^{-1} \cdot \beta) \simeq (\alpha \cdot \beta^{-1}) \cdot \beta \simeq \beta$.


integration
calculus
function
real analysis
real analysis unsolved

Suppose that both f and g are of bounded variation on [a,b]. Prove that (g is Stieltjes-integrable with respect to f) <=> (f is Stieltjes-integrable with respect to g) <=> (f and g have no common points of left-discontinuity, and no common points of right-discontinuity), and in that case,

\[\int_{a}^{b}{fdg+}\int_{a}^{b}{gdf = f(b)g(b)-f(a)g(a)}\]

Supposing that g is stieltjes-integrable with respect to x, we have:

$L(m_{i},f(x_{i+1})-f(x_{i})) \leq \int g(f(x)d(f(x)) \leq U(M_{i},f(x_{i+1})-f(x_{i}))$ where $lim{i\to \infty}|L(m_{i},g(x))-U(M_{i},g(x)) |<e$ By definition of the riemman-stieltjes integral the deltas must be a monotonically increasing function and obviously f is stieltjes integrable. So, both functions are monotonically increasing .
The formula you posted  is taken  with integration by parts


group theory
abstract algebra
superior algebra
superior algebra unsolved

A really funny problem :rotfl:.

[i]Suppose $ |G|<\infty$ and $ p | (\text{Z}(G))$. Let $ K$ be Sylow $ p$-subgroup of $ \text{Z}(G)$ and $ P$ be Sylow $ p$-subgroup of $ G$. Prove that $ K\subseteq P$. [/i]

$ K$ is a $ p$-subgroup of $ G$, thus by one of Sylow's theorems contained in some $ x^{ \minus{} 1} Px$, thus $ K \equal{} xKx^{ \minus{} 1}$ is contained in $ P$. It suffices that $ K$ is normal, the center is not needed.

[quote="-oo-"]@sylow: and what's funny about that?[/quote]
Because its a one liner.
My solution was: note that $ \text{Z}(G) \subseteq \text{N}(P)$ and $ P \triangleleft \text{N}(P)$.
(A normal sylow p-subgroup contains every p-subgroup).


let ABCD be a convex quadricateral .let O be the intersection of diagnals and we know ^AOB=60.if CD=a & AB=a then find the maximum value of BC :?:

I got a maximum value, but it's a fairly dodgy solution  :blush:  I got a value of $ 2a$. Is that right?


modular arithmetic
number theory unsolved
number theory

x,y,z are non-negative integers
Solve
$ 2^x\plus{}3^y\equal{}z^2$
[hide="What I did"]
x,y >0
We see that both of them must be even.
x=2k
y=2m
$ 3^{2m}\equal{}(z\minus{}2^k)(z\plus{}2^k)$
Let $ z\minus{}2^k>1$ and $ z\plus{}2^k>1$, we easily get contradiction. So z-2^k must be 1. Here comes my problem. I am not able to get solutions for that.
We're left with x=0 or y=0 but that's trivial.
[/hide]

$ y \equal{} 0$  $ \implies$  $ 2^{x} \equal{} (z \minus{} 1)(z \plus{} 1)$ $ \implies$
$ z \plus{} 1 \equal{} 2^{m}$ ,$ z \plus{} 1 \equal{} 2^{k}$  $ \implies$  $ 2^{k} \minus{} 2^{m} \equal{} 2$  we know that only solution $ (2,1)$
$ k \equal{} 2,m \equal{} 1$  $ \implies$  $ x \equal{} 3$ $ \implies$  $ (3,0,3)$
$ y\geq 1$ $ \implies$  $ (z,3) \equal{} 1$  $ \implies$  $ z^{2}\equiv 1\pmod{3}$  $ \implies$
$ 2^{x}\equiv 1\pmod{3}$  $ \implies$  $ 2|x$  similarly $ 2|y$ $ (y \equal{} 2u)$
$ x \equal{} 2r$  $ \implies$  $ (z \minus{} 2^{r})(z \plus{} 2^{r}) \equal{} 3^{y}$
$ z \equal{} 3^{q} \plus{} 2^{r}$  $ \implies$  $ 3^{q} \plus{} 2^{r \plus{} 1} \equal{} 3^{p}$ $ \implies$
$ p > 0$  .if $ q > 0$  $ \implies$  $ 3|2^{r \plus{} 1}$  contradiction!
$ q \equal{} 0$  $ \implies$  $ z \equal{} 2^{r} \plus{} 1$  $ \implies$  $ 2^{r \plus{} 1} \plus{} 1 \equal{} 3^{2u}$
$ (3^{u} \minus{} 1)(3^{u} \plus{} 1) \equal{} 2^{r \plus{} 1}$  $ \implies$  only $ u \equal{} 1$
$ u \equal{} 1$  $ \implies$  $ y \equal{} 2$ , $ r \equal{} 2$ $ \implies$  $ x \equal{} 4$
$ \{(4,2,5),(3,0,3)\}$  :wink:

where do you take 2^{r+1} from
I dont understand that part

[quote="NikolayKaz"]$ x,y,z$ are non-negative integers.Solve the equation:

$ 2^x + 3^y = z^2$
[/quote]
If $ x = 0$,then $ 3^y = (z - 1)(z + 1)$ hence $ z - 1 = 3^a$ and $ z + 1 = 3^b$ with $ a + b = y$.We have $ 3^b - 3^a = 2 \Rightarrow b = 1$ and $ a = 0 \Rightarrow y = 1 \Rightarrow z = 2$.If $ x = 1$,then the equation becomes $ 3^y + 2 = z^2$ with $ y > 0$ which is impossible since a square is never congruent to $ 2(mod 3)$.If $ x > 1$,then looking at the equation $ (mod 4)$,It follows that $ y$ is even.We may rewrite the equation as $ 2^x = (3^y' - 1)(3^y' + 1)$ where $ y = 2y'$.Hence $ 3^y' - 1 = 2^a$ and $ 3^y' + 1 = 2^b$ with $ a + b = x$.Clearly $ 2^{b - 1} - 2^{a - 1} = 3^y' \Rightarrow a = 1$ and $ 2^{b - 1} - 1 = 3^y'$.For $ b = 2$,we obtain $ y' = 0$,It follows that $ x = 3,y = 0$ and $ z = 3$.For $ b = 3$,we easily deduce that $ x = 4,y = 2$ and $ z = 5$.If $ b\geq 4$,then $ 8|3^y' + 1$  which is a contradiction since $ 3^{y'} + 1\equiv 2,4 (mod 8)$.

All the solutions are $ (0,1,2);(3,0,3)$ and $ (4,2,5)$.


a)Determine the number of graphs having $ n$ vertices, so that the degree of every vertex is $ \geq1$

b)Which is the least and the greatest number of edges that a graph with property (a) may have?

c)Determine the number of graphs having $ n$ vertices, so that the degree of every vertex is $ \geq k$, where $ k<n$ is a fixed integer.

Labeled vertices ([url=http://www.research.att.com/~njas/sequences/A006129]A006129[/url]) or up to isomorphism ([url=http://www.research.att.com/~njas/sequences/A002494]A002494[/url])?  (The former is much easier to write down than the latter.)


A particular committee needs one $ 7^{\mathrm{th}}$-grader and three $ 8^{\mathrm{th}}$-graders. If there are four $ 7^{\mathrm{th}}$-graders and six $ 8^{\mathrm{th}}$-graders applying to the committee, in how many ways can the committee be filled?

the number of ways to fill the 7th grader are 4C1, which is 4.

and the number of ways to fill the 8th grades are 6C3, which is 20.

so total number of ways to fillthe commitee is 20*4.

answer : 80


function
real analysis
real analysis unsolved

Not an easy one:
   Prove that the linear functions are the only one that preserve the Cesaro-mean convergence: that is, if $x_n$ has Cesaro-mean convergence, then so does $f(x_n)$ (a priori, no relation between the two limits is imposed).

I am probably making terrible mistakes somewhere, but anyway, here are my ideas:
First, $f$ is continuous. (I don't think this part is necessary) Indeed, let $x_i$ converge (hence Cesaro converge) to $a$. Suppose $f(x_i)$ however Cesaro converges to something, but not $f(a)$. Then, by adding $a$'s to the sequence $x_i$ (and creating some sequence $y_i$) we can play around with the Cesaro convergence of the sequnce $f(y_i)$ without affecting the Cesaro convergence of $y_i$. (The idea would be that if we add lot's of $a$'s first, the Cesaro mean is pulled towards $f(a)$, but then, leaving enought $x_i$'s after this, the mean is pulled back to the limit different from $f(a)$)[A bit unclear, but I guess it is understandable]. So we got $f(x_i)$ Cesaro converges to $f(a)$. Now, to ensure it converges normally to $f(a)$: Suppose there exists an $\epsilon$ s.t. infinitely many of the $f(x_i)$'s are outside of the $\epsilon$-neighborhood of $f(a)$. But now, again, by increasing the "weight" (number of times they appear) of those $x_i$'s in the sequence, we can alter again the Cesaro convergence of $f(x_i)$, without affecting that of $x_i$, so indeed, $f$ is continuous...
Now, the argument for linearity follows in a similar way.
Take 2 different points, $a$ and $b$. Taking a sequence $a_i$ that converges to $a$ and shifting it, $b_i=b-a+a_i$, s.t. $b_i$ converges to $b$, and by playing in a similar way with the "weights", we can deduce that $\frac{f(a)+f(b)}{2}=f(\frac{a+b}{2})$. And then it's rather easy.....
The main idea here is to consider the new sequence $a_i,b_i$, that is the combination of our initial 2 sequences....
I hope it is rather clear...
Sorry if it's a waste of time...


So like how did other people from MA do? Sergei 0wned, and Andrew Geng won the Clay award thingy (what problem was it for?)

I got 6 due to my BS on #1

Where did you find out about the Clay Award? I do not see it anywhere on the AMC website.

Hmm, I got an 11 (from ABRHS). The other person from ABRHS didn't take it.

http://www.maa.org/news/052207usamoceremony.html

I didn't know they awarded it to non-HM/winners (everybody else on the list of awardees was a winner/HM).


geometry
geometric transformation
rotation
search
counting
derangement
combinatorics proposed

Given any natural no. n, find the no. of possible n^2Xn^2 sudoku?

[hide]Consider a $ k\times k$ board. There are $ k!$ possible arrangements for the first row. After the first row is arranged, there are $ (k\minus{}1)!$ possible arrangements for the second row. And so on. So there are $ 1!2!3!\ldots k!$ possible $ k\times k$ sudokus, where rotations are considered different. If rotations are considered the same, just divide by 4, because an arrangement has 4 possible rotations. So the answer is $ 1!2!3!\ldots(n^2)!$ (or its quarter).[/hide]

Johan Gunardi, you appear to be trying to count the number of [url=http://mathworld.wolfram.com/LatinSquare.html]Latin squares[/url], not [url=http://en.wikipedia.org/wiki/Sudoku#Mathematics_of_Sudoku]Sudoku boards[/url], of a given size, and you're not doing it right :)

Both problems are hard.  They are [url=http://www.research.att.com/~njas/sequences/A002860]A002860[/url] and [url=http://www.research.att.com/~njas/sequences/A107739]A107739[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

Oops. :blush:  Sorry, I forgot to consider the $ n\times n$ suboards.

I was trying to obtain the result by recursion for a 9X9 sudoku board, only to mess it up further. Some senior told me that the problem has been solved by Group Theory. Can anyone give me some idea about it? However, the problem still seems unsolvable to me.

For $ 9 \times 9$ no human has demonstrated a method to calculate it. The value known is due to a computer brute forcing the problem. Don't be surprised that you're messing it up.

As I mentioned before, I was successful in calculating the no. of ways the sudoku can be filled up by only one of the nine digits. It has been done by starting with any one end and then using the logic of filling up a sudoku.
The answer is (81X60X45X35X22X14X9X4)=8487309600!!!

My mind still tells that while filling up a sudoku, after filling up [b]certain nos[/b]. at appropriate places the rest nos. can be filled up in unique way, i.e. the sudoku becomes a fixed one. Can any body find out what is that "[b]certain number"[/b]??? :wacko:

[quote="Sunkern_sunflora"]As I mentioned before,[/quote]
Actually, I don't think you did mention it before, at least not on this thread.

I'm not sure what the last part of your post means.  Perhaps this paragraph answers your question?

[quote="[url=http://en.wikipedia.org/wiki/Sudoku#Mathematics_of_Sudoku]Wikipedia[/url]"]The maximum number of givens provided while still not rendering a unique solution is four short of a full grid; if two instances of two numbers each are missing and the cells they are to occupy form the corners of an orthogonal rectangle, and exactly two of these cells are within one region, there are two ways the numbers can be assigned. Since this applies to Latin squares in general, most variants of Sudoku have the same maximum. The inverse problem—the fewest givens that render a solution unique—is unsolved, although the lowest number yet found for the standard variation without a symmetry constraint is 17, a number of which have been found by Japanese puzzle enthusiasts, and 18 with the givens in rotationally symmetric cells. Over 47,000 examples of Sudokus with 17 givens resulting in a unique solution are known.[/quote]

[quote]JBL wrote:
Actually, I don't think you did mention it before, at least not on this thread.
[/quote]
I see! :blush:  Sorry!!
[quote]JBL wrote:
I'm not sure what the last part of your post means[/quote]
Have you ever played sudoku? If you do, then remember that some of the nos. are already given. You are to fill up the rest. Moreover, that fill-up is bound to be unique. The more the sudoku gets difficult, the less is the no. of initially given nos., yet the sudoku remains unique. I was trying to focus on this no. of initially given nos. i.e. [b]the least no. of nos. that should be provided in order to make the sudoku unique[/b](I'm talking about 9X9 sudoku). It might help our calculation.

Okay, and the question is answered by the paragraph I quoted:
[quote="[url=http://en.wikipedia.org/wiki/Sudoku#Mathematics_of_Sudoku]Wikipedia[/url]"]The inverse problem—the fewest givens that render a solution unique—is unsolved, although the lowest number yet found for the standard variation without a symmetry constraint is 17, a number of which have been found by Japanese puzzle enthusiasts, and 18 with the givens in rotationally symmetric cells. Over 47,000 examples of Sudokus with 17 givens resulting in a unique solution are known.[/quote]  (There are some references linked from the Wikipedia page that are not included in this quote, so go there for more information.)

My semi-knowledgeable advice is that you consider this question to be very hard and therefore not spend too much time on it -- there are many simple-to-describe enumeration problems that just seem to be completely intractable, and I think that in all likelyhood this is one of them.  I think the closely-related problem of Latin squares is similar (and has existed for longer): it's an elegant problem (with [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=918952861&t=202056]several equivalent phrasings[/url]) that just seems not to have a nice solution.

By the way, about your computation: I haven't tried to determine for myself if the approach you're thinking about actually works, but if it does there are still adjustments to be made.  At the very least, you need to divide by 9! for all the different orders in which you could choose the same nine positions to be the selected digit.  (This approach is definitely not likely to lead to a simple attack on the problem -- in the case of Latin squares, you get $ n!$ for the placement of the first entry, then the [[derangement]] number $ D_n$ for the placement of the second digit -- this is already quite messy!)

Thank you very much JBL for your kind advices.
Do there exist any complete solution for this problem??


I came upon a strange question that I do not quite understand the logic for. Just wondering if any of you can clarify it for me.

The problem is:

As indicated in the figure, a person on the deck which is $ h$ meters above the water level pulls a boat towards the deck with a rope. If the person pulls the rope with a constant velocity $ v_0$, find the velocity and acceleration of the boat while it has a horizontal distance $ S$ meters from the edge of the deck.

My work on it:

I split $ v_0$ into components, one parallel to the water surface and one perpendicular to it. The parallel component, $ v_0\cos\theta$, where $ \theta \equal{} \arctan\frac{h}{s}$ is the speed of the boat. However, I have the solution for the problem, and this is already wrong, apparently. The solution says that the speed of the boat parallel to the water is $ v$, which is split into a component parallel to the rope, and one perpendicular to the rope. This implies that $ v_0 < v$, or the speed of the rope being pulled is less than the speed of the boat!

If anyone can explain this, and solve the problem using this knowledge, please do. Thank you very much!

You could see, for instance, that the length between pulley and boat is $ \; l \equal{} \sqrt{s^2 \plus{} h^2}\;$ so  $ v_0 \equal{} \minus{} {{dl}\over{dt}} \equal{} \minus{} {s \over {\sqrt{s^2\plus{}h^2}}}.{{ds}\over{dt}} \;\;\;so\;\;\;v_0 \equal{} v. cos \theta \;\;\;$
 It could be done with the "trigo" functions also, of course.

I see where the formulas come from. Where does the intuition for this come from though? Also, there will be a component of the boat's speed that points into the water. Where does this come from (physically)?

Thanks!

You could think as the movement of the boat in relation to the pulley, with a radial velocity $ v_0 \equal{} v. cos \theta$ (actually, the ropes velocity) and a transversal ( kind of "a rotational one" in relation to the pulley) velocity,  $ v.sin\theta$ :that one that "points into the water".


Let $ T$ and $ U$ be subsets of a vector space $ V$ .

Prove that if $ T\subset U$ , then $ U^{\perp}\subset T^{\perp}$

Here is my idea to prove :

let $ x\in U^{\perp}$ then $ \forall y\in U$ , we have $ x\cdot y\equal{}0$ .

Since it is true for all $ y$ in $ U$ , it is true $ \forall y\in T$ (because $ T\subset U$ )
So $ x\in T^{\perp}$ .
Hence $ U^{\perp}\subset T^{\perp}$

Is my argument correct ? Thanks

[quote="shyong"] .......Is my argument correct ? Thanks[/quote]

[hide="only for shyong"]I bet one thousand won that yours proof is true :D[/hide]


quadratics
number theory proposed
number theory

Prove that if $a$ is a square modullo $p$ for all $p$ a prime, then $a$ is a square.

If $a$ is a particular square $\bmod p$ then $b = \sqrt{a}$ can take on one of two distinct values $\bmod p$.

Solving the system of infinite moduli over all primes we get two finite values of $b$ (additive inverses).  The others are infinite.  I don't want to work out the exact details  :P

Edit:  Actually, this problem is a lot easier than I thought.

Simply assume that $a$ is not square.  Then there exists $p > a$, in which case $a$ is not a quadratic residue $\bmod p$.

Ergo, if $a$ is a QR for all $p$ it follows that $a$ must be square.

(It seems to me that we only need an infinite subset of the natural numbers, not necessarily $\mathbb{P}$. )

[quote="t0rajir0u"][if] $a$ is not square.  Then there exists $p > a$, in which case $a$ is not a quadratic residue $\bmod p$.  [/quote]

It's not that easy.   This is a "local to global" problem where if the equation is solvable in its reductions modulo $n$ it is solvable without reduction.  I would be surprised if there is a solution not using quadratic reciprocity.

[quote="fleeting_guest"][quote="t0rajir0u"][if] $a$ is not square.  Then there exists $p > a$, in which case $a$ is not a quadratic residue $\bmod p$.  [/quote]

It's not that easy.   [/quote]

Could you explain why in more detail?  I'm afraid I don't quite follow  :|

[quote="t0rajir0u"]
Edit:  Actually, this problem is a lot easier than I thought.

Simply assume that $a$ is not square.  Then there exists $p > a$, in which case $a$ is not a quadratic residue $\bmod p$.

Ergo, if $a$ is a QR for all $p$ it follows that $a$ must be square.

(It seems to me that we only need an infinite subset of the natural numbers, not necessarily $\mathbb{P}$. )[/quote]

If $a$ is not a square, it does not mean it is not a quadratic residue $\mod{p}$ when $p > a$. For example, $2$ is a quadratic residue $\mod{7}$ even though $2$ is not a square.

[quote="probability1.01"]What? The username looks like Yichen, but I'm more inclined to believe it's someone else (Bohua?).[/quote]

Why's that?  :roll:
[color=white][size=59]You're wrong.[/size][/color]

If a is a quadratic residue of all primes < a it is a square modulo all primes > a. This is true, but how do you prove it?

If it is a quadratic residue of n primes the chances that it is not square is 1/2^n.

What you need to prove is that if it is not a q. res. of some prime <= a it is not square.

I once saw the solution to this problem. Basically, you have a on p, and you have to use the jacobi(?) thingy, then argue by contradiction.


[hide][quote="pascal_1623"]Bob is in 5th grade. Bob got 98% in science, 85% in writing, 87% in reading, 79% in social studies, and 99% in math. What is Bob's overall grade?[/quote]

98 + 85 + 87 + 79 + 99 = 448

448/5 = 89 and 3/5

-or-

89.6 %[/hide]

his grade is 89.6%.
98+85+87+79+99=448
448/5=89.6
not exactly a top-notch student.[hide][/hide]

all the numbers add up to 448, divide by 5 to get a percentage of 89.6 :idea: our answer


combinatorics proposed
combinatorics
Set systems
Subsets

$ S_{1},S_{2},...S_{n}$ are different subsets ($ n\geq 4$) of finite set $ \mathcal{S}$, $ |\mathcal{S}|=k \geq 4n$ such:
a) $ |S_{i}|=4$; for each $ i=1,...,n$
b) $ |S_{i}\cap S_{j}| = 1$; for each $ i,j$, $ i\ne j$

For witch minimal $ n$ following holds
$ {\Big|\bigcap_{i=1}^{n}S_{i}\Big|}= 1$?

[quote="Number1"]$ S_{1},S_{2},...S_{n}$ are different subsets ($ n\geq 4$) of finite set $ \mathcal{S}$, $ |\mathcal{S}|=k \geq 4n$ such:
a) $ |S_{i}|=4$; for each $ i=1,...,n$
b) $ |S_{i}\cap P_{j}| = 1$; for each $ i,j$, $ i\ne j$

For witch minimal $ n$ following holds
$ {\Big|\bigcap_{i=1}^{n}S_{i}\Big|}= 1$?[/quote]

b) $ |S_{i}\cap P_{j}| = 1$; for each $ i,j$, $ i\ne j$
ou want say
b) $ |S_{i}\cap S_{j}| = 1$; for each $ i,j$, $ i\ne j$


[quote="xxxxtt"]Which are all the 3-regular graphs with 6 vertices?
Is it only $K_{3,3}$?

This is connected with the following problem:

Determine all the 3-regular graphs G with minimum number of vertices such that $g(G)=4$ respectively $g(G)=5$ where $g(G)$ is the girth of the graph $G$ i.e. the lenghts of the shortest (elementary) cycle of G.

I have two example,for $g(G)=4$ the graph $K_{3,3}$ and for $g(G)=5$ the  Petersen graph.
Are there any others?
How do i prove these are the only ones?[/quote]

Sorry,the problem is trivial.


geometry
incenter
exterior angle
complex numbers
geometry solved

Let $ABCD$ be a circumscribed quadrilateral (i. e. a quadrilateral which has an incircle). The exterior angle bisectors of the angles $DAB$ and $ABC$ intersect each other at $K$; the exterior angle bisectors of the angles $ABC$ and $BCD$ intersect each other at $L$; the exterior angle bisectors of the angles $BCD$ and $CDA$ intersect each other at $M$; the exterior angle bisectors of the angles $CDA$ and $DAB$ intersect each other at $N$. Let $K_{1}$, $L_{1}$, $M_{1}$ and $N_{1}$ be the orthocenters of the triangles $ABK$, $BCL$, $CDM$ and $DAN$, respectively. Show that the quadrilateral $K_{1}L_{1}M_{1}N_{1}$ is a parallelogram.

This seems to be one of the easier Russian problems, so let's get it out of the way :).

Let $I$ be the incenter of $ABCD$. $I$ is also the intersection pt of the bisectors of $ABCD$, so $AI \perp AK \Rightarrow BK_1||AI$. In the same way we get $AK_1||BI$, so if we let $I$ be the origin of the plane and work with complex numbers, we get $k_1=a+b$.

After writing down the similar relations, we find $k_1+m_1=l_1+n_1 (=a+b+c+d)$, and this is the necessary and sufficient condition for $K_1L_1M_1N_1$ to be a parallelogram.

my problem give O is center of incircle ABCD then  (tex)AK1//BO//CL1;
   L1B//OC,K1B//AO =>BO=K1A=L1C
    similary AN1=M1C,AN1//M1C=>K1N1=L1M1; K1N1//M1L1(tex)

[hide = solution]
Observe that the incentre of $ABCD$ is the antipode of $K$ in $\odot(AKB)$ and similar results hold true for the other three triangles. It follows that after taking a homothety at $I$ with factor $1/2$ we're left to show that the midpoints of $AB, BC, CD, DA$ form a parallelogram, which is an easy and well known result.
[/hide]


Welcome, one and all, to the long-awaited Mafia Game 31.

19 people were on a private jet to Florida for a birthday, each bearing a gift for the party.  High altitude thunderstorms crashed the plane onto a small island off the coast of Georgia.  Fortunately, there is enough room for them all to sleep and a large bonfire burns in the middle, providing heat for all.  There's only one problem: every day they must kill one member to roast over the bonfire and eat for dinner.

Desperate to escape this dreadful island, the members turn to their items for assistance.  It is known that somebody's gift was a life raft... to fit four people.  This item is, of course, hunted down by many other members.  Some might even go to the point of murder to achieve this life raft.  Once these greedy anti-survival members are all dead or eaten, four members of the island will go out for help.  Too bad those anti-survival members are too self-centered to do that.

In addition to the life raft, many other members may find use of their items and may use them nightly for good reasons - to find evidence, to save others, etc.  [b]If you are told evidence that you did not ask for, this is not a mistake.[/b]  Use the information wisely.  Somebody may have given it to you for a reason.  Some unfortunate players seem to have items of no use to them.  A player like this is called a "townie" and will have a PM that looks like as follows:
[quote="Townie PM"][u]Role Info:[/u]
Your alignment is [b]pro-survival[/b].  You win if all anti-survival members of the game are dead.

[u]Item Info:[/u]
You item is a <Useless Item Here>.  Unfortunately there is no <Necessary Item> on this island, so your item will be of little use to you.[/quote]

[u]Player List[/u]
Hamster1800
mnmath
woser456
alanchou
ZhangPeijin
kimmystar94
zapi2007
1=2
perfect628
Temperal
jnub
CheeseIsGood
DonkeyKong
Berylliumnitride
math_explorer
Peter
Ars
erabel
xpmath

Good luck and have fun!  Please read the [b]entire[/b] post for clarity.

[quote][b]The Rules (PLEASE READ THESE, ESPECIALLY IF THIS IS YOUR FIRST GAME):
Votes must be in bold. If you do not bold your vote, it will not be counted.

Please be attentive and unvote, if necessary, before casting a new vote. This is not required, but I’d appreciate it.

Executions will require a simple majority of votes. Once a player has reached the necessary majority, his pleas are useless and any attempts to unvote will be unheeded. You may vote: no lynch – majority votes of this kind are necessary to end the day without a death.

The game is not to be discussed outside of the thread unless your role specifically states that you may do so.

Once your death scene has been posted, you’re dead. Stop typing.

Don’t edit/delete previously submitted posts

Don’t quote any PMs from me.

If you have a night choice to make, it is due by the posted deadline. I will not wait for you – if you do not submit a choice to me, tough! No choice will be made.

NO HIDE TAGS. Also, no changing font size, or posting in any other way that obscures something you're saying. Anything you say should be clearly visible to everyone without having to click a button or enlarge your text or anything like that.

If you anticipate being unavailable for more than a 48-hour period, please post a notice to that effect in the thread. Treat this game as a commitment. Be considerate – don’t leave us hanging.

Remember, it’s a game. It’s supposed to played. It’s supposed to be fun. Do your part.[/b][/quote]

Yay. First vote: [b]Vote: ZhangPeijin[/b].

P.S. can you post the player list?

Second vote: [b]Vote: woser456[/b]

Dude, all of his posts have been uber suspicious...man I think he's anti-survival.

Two notes: Please don't bandwagon and kill anyone without reason, and let's not turn this into miyomiyo's last game(8 pages lol).

[b]Vote: Temperal[/b] because I believe him =P

It doesn't say anything in the rules about not voting yourself, so [b]Vote:Temperal[/b].

Wait what?
Why did you vote for me first?
I didn't even check in yet.
 :wink:

Why do you care if someone randomly picks you from the list on the first vote?

I'm too lazy to vote yet.

Woah. Overreaction ZP...Maybe not the worst vote lol

Yay!  I'm on a desert island...
with 18 others... :(
well, i'm here

How was that overreacting?
Anyways, I'll go play some Halo.
Haha.

Ugh.  This game got messed up somehow...

technically, the mod screwed up the game (if you still don't trust me, I'll send it back!)

but i don't care by now. darn! i wanted this to be my first mafia game that i won.

[quote="kimmystar94"][quote="miyomiyo"]Ars is [b]pro-survival.[/b][/quote]

it was in my inbox. I'll resend it to you...
I am not lying. (why would i? it's the end of the game!)[/quote]
??!

This is very strange.  I carefully recorded every target, and I double-checked after you started claiming you had inspected Ars.  I see no way that this could have happened, I'm 99.9% sure you inspected zapi2007.

[quote="kimmystar94"]i wanted this to be my first mafia game that i won.[/quote]At the time we lynched woser456 we had everyone pro-survival. If that was everyone but Ars pro-survival, then Ars could dance the whoolahoop on his head, we'd have lynched him nonetheless. ;) And thus have won the game, even with you alive.

@Miyomiyo: could it be that kimmy asked for zapi, but you accidently typed Ars in return? (which she didn't worry about since you explicitly claimed that it's not a mistake to get info you didn't ask for)
I see no other explaining for this "strange" problem.

Anyway, beside this thing, I found it a great game. The theme was very coherent and original. :)

[quote="Peter"]@Miyomiyo: could it be that kimmy asked for zapi, but you accidently typed Ars in return? (which she didn't worry about since you explicitly claimed that it's not a mistake to get info you didn't ask for)
I see no other explaining for this "strange" problem.[/quote]This may be the case, and if it is, I hope you'll all be able to forgive me.

Though, had I given the correct results, would it have made a difference?  Would kimmystar have inspected Ars later on or just the others?

plus the innocent inspection is the reason kimmy was still alive at endgame
we had no reason to kill kimmystar

I asked for Ars. I remember. I never inspected zapi b/c the night I wanted to inspect zapi, he was lynched.

Hmm.
Weird fact.
Even over, Mafia 31 is more active than Mafia 32...

Well, it's night 5 in Mafia 32 now.  That's why it isn't so active now.

i think zhangpeijin means before night started

[quote="miyomiyo"]Though, had I given the correct results, would it have made a difference? Would kimmystar have inspected Ars later on or just the others?[/quote][quote="Peter"]At the time we lynched woser456 we had everyone pro-survival. If that was "everyone but Ars" pro-survival, then Ars could dance the whoolahoop on his head, we'd have lynched him nonetheless.[/quote]So even just not having any result for that night would have done it. But then again, mafia may have made other decisions when they really thought kimmy was inspector etc. etc...

[note that unlike you write, it was not "popular belief" that kimmy was naive, with the results she received we [b]knew[/b] something had to be wrong]

[quote="miyomiyo"]This may be the case, and if it is, I hope you'll all be able to forgive me.[/quote]Sure, it could happen to all of us. And I guess this will make mods double-check for the coming games when they see an accidental false claim.
And as I said before, it was one of the most enjoyable games nonetheless. :)

Well, though I can't see how this happened, I seem to have messed up.  Sorry, all.  Just to think: had I not made that mistake and had andersonw not killed non-stop, it could've easily been a town win.

Or if the hunting rifle worked on even nights, mnmath would've won!

On the last night I guessed it worked alternatively. I was going to submit Peter again too. It was fun.


I'm in 9th grade Accelearated Geometry AC, and this is what we got for homework. I've tried everything I know and I can't figure it out. Is it solvable?

Ignore the clue, it's basically asking "Is this sentence false?".

P.S. I know it is possible to get an answer, but is it possible to get one defeinte answer?

Firstly, i don't see what this has to do with geometry, but disregarding that:

If we ignore the clue as you instructed, we reaslise that room VIII is empty. Hence the sign is false, since there is no tiger in it. So Room IX is not empty. But again, reading sign IX, we get the "this sentence is false" paradox.  :huh:

[quote="tim1234133"]we reaslise that room VIII is empty.[/quote]

Why?  This is not at all obvious: it certainly can't have the lady, but it could have a tiger (so long as room IX has either a tiger or the lady).

oh, yes you are right.  :blush:  I had misinterpreted the question as being both parts of the statements of true/false together, rather than it being possible for one part of a statement to be true whilst the other is false.

When sign III says either sign V is right or sign VII is wrong, does it specifially mean one of the two (so if both sign V is right AND sign VII is wrong, then sign III would be wrong)? (similarly on sign V)

edit:
Assuming that it [color=red]doesn't[/color],
Sign VIII must be wrong (containing a tiger) and so Sign IX must also be wrong (containing a tiger). Therefore sign VI is right, and so sign III is wrong. So Sign VII is right, meaning that the lady is not in room 1. Also Sign V is wrong, meaning that both both Sign II and IV are wrong. Hence Sign I is right. We have already shown that signs III,V and IX are wrong, and that the lady is not in room I, so from this the lady is in room VII. I think if we have:
I: empty II: tiger III: tiger IV : tiger V: tiger VI: empty VII: lady VIII: tiger IX: tiger
then everything is consistent.

[quote="tim1234133"]

When sign III says either sign V is right or sign VII is wrong, does it specifially mean one of the two (so if both sign V is right AND sign VII is wrong, then sign III would be wrong)? (similarly on sign V)[/quote]

I think it means at least one of them. If both parts are true, then the sign is true because at least one of the statements is true. This is why this problem is so hard to solve, it is unclear when a statement is false and when it is true.


Alyssa, Juan, Myisha, Tracy, and Scott raised a total of $550.68 for the nearby homeless shelter by walking in a walk-a-thon. Before the walk, they each got promises (pledges) from a number of people to contribute money for each mile they walked. The average amount per mile which each person raised ranged from $7.10 a mile to $20.70 per mile, and the total distance each person walked ranged from 5 3/4 miles to 13 1/2 miles.
At the end of the walk-a-thon, they collected the exact amount that had been pledged to them for each mile they walked. All amounts are rounded to the nearest penny.
Using the information above and the clues below, how far did each person walk, and how much did they each earn to contribute to the shelter?
Alyssa walked the fewest miles, but her average pledge per mile was 50% more than Scott's.
Juan had an average of
$10.35 pledged per mile, and walked 3 1/4 more miles than Alyssa.
Myisha walked .25 mile more than Alyssa, but her average pledge per mile was twice Juan's.
Tracy had the least amount per mile pledged.
Scott walked twice as far as Alyssa and had an average of $1.45 more per mile pledged than Juan.

you might want to take off the latex there. i'm getting dizzy just trying to read it

I think she used a dollar sign for money and it trigered $\LaTeX$

Alyssa, Juan, Myisha, Tracy, and Scott raised a total of 550.68 for the nearby homeless shelter by walking in a "walk-a-thon." Before the walk, they each got promises (pledges) from a number of people to contribute money for each mile they walked. The average amount per mile which each person raised ranged from 7.10 a mile to 20.70 per mile, and the total distance each person walked ranged from 5 3/4 miles to 13 1/2 miles.

At the end of the walk-a-thon, they collected the exact amount that had been pledged to them for each mile they walked. All amounts are rounded to the nearest penny.

Using the information above and the clues below, how far did each person walk, and how much did they each earn to contribute to the shelter?

Alyssa walked the fewest miles, but her average pledge per mile was 50% more than Scott's.
Juan had an average of 10.35 pledged per mile, and walked 3 1/4 more miles than Alyssa.
Myisha walked .25 mile more than Alyssa, but her average pledge per mile was twice Juan's.
Tracy had the least amount per mile pledged.
Scott walked twice as far as Alyssa and had an average of 1.45 more per mile pledged than Juan.

The way I've understood to do this problem is to make a table (people versus their distances and pledges) and systematically use all the information you get.  Dont forget the ranges for least to most pledge/distance given in the problem statement!  If I didn't make any mistakes...:

(D=distance, P=pledge, C=overall contribution=D*P)
Alex: D=5.75, P=17.70, C=101.78
Juan: D=9.00, P=10.35, C=93.15
Myisha: D=6.00, P=20.7, C=124.20
Tracy: D=13.50, P=7.10, C=95.85
Scott: D=11.5, P=11.8, C=135.70

The sum of the indicated C's is 550.68 as the problem states.

Hey guys, sorry about the $\LaTeX$   :blush:

I forgot about this problem... was I right lol?


[quote="espark52"]what common fraction with a value between .81 and.91 has the least denominator?[/quote]
[hide="ohh"]$\frac{5}{6}$[/hide]
PS: Where are these problems from?

[quote="espark52"]that's correct, but show your work... :)[/quote]
My method was guess and check. I just did it uber fast cuz I saw a variation of this very problem earlier today...

The best way to go is to check fractions with a denominator of 2, then 3, then 4, then 5, then 6, then you get your answer.


algebra
polynomial
integration
real analysis
real analysis unsolved

Prove that the set of all linearly combinations (with real coefficients) of the system of polynomials $ \{ x^n\plus{}x^{n^2} \}_{n\equal{}0}^{\infty}$ is dense in $ C[0,1]$.

[i]J. Szabados[/i]

This one is surprisingly simple: Assume that there is a signed measure $ \mu$ annihilating all $ x^n\plus{}x^{n^2}$. Let $ a_m\equal{}\int_{[0,1]}x^m\,d\mu$. Then, on one hand, $ a_m\to \mu\{1\}$ by the dominated convergence theorem and, on the other hand, if $ a_m\equal{}a\ne 0$, then $ a$ and $ \minus{}a$ occur among $ a_m$ infinitely often, so we get a contradiction. Thus, our measure must annihilate every $ x^m$ and we have reduced everything to the classical Weierstrass theorem.

Let's upgrade the problem to the same question about the sequence $ \{x^n\plus{}x^{n^2}\plus{}x^{n^3}\}$ ;).

fedja, can you explain more clearly what does  it mean " annihilating" ?
[quote="fedja"] measure $ \mu$ annihilating all $ x^n \plus{} x^{n^2}$.  [/quote]


For positive reals $ a,b,c$, prove that
\[ ((a\plus{}b)(b\plus{}c)(c\plus{}a))^{\frac{2}{3}}\geq ((a\minus{}b)(b\minus{}c)(c\minus{}a))^{\frac{2}{3}}\plus{}4(abc)^{\frac{2}{3}}.\]


[quote="Fanatic"]A fly starts at one vertex of a cube with edge length 1 inch and randomly walks along edges of the cube. What is the number of inches in the longest distance that it could walk without retracing any edge?[/quote]
[hide]I just brute forced it and got 8. There may be more though.[/hide]

I found

[hide]at least nine. I posted to see if anyone could find more, and i wouldn't be surprised if they could.[/hide]

I found 9, go around the square, go down an edge and then go around a square again-4+1+4, so 9?

Does anyone have an answer key, or can anyone get 10 or more?

The answer key says 9.

[hide]Since no one posted a solution, I'll post one:

[img]http://i96.photobucket.com/albums/l167/adasarathy/cubesolution.png[/img]

[/hide]


calculus
integration
LaTeX
trigonometry
calculus computations

Can you use LaTeX? I can't understand what your formula stands for... especially the bold part in

'ntegral [b]'nt[/b] (s'na)^3/(cosa)^2.da

[quote="cosinerburc"]can you compute the 'ntegral  'nt (s'na)^3/(cosa)^2.da[/quote]
I'd assume it is $ \int\frac{\sin^{3}\alpha}{\cos^{2}\alpha}\,d\alpha$

If it so, then

\begin{eqnarray*}\int{\frac{{\sin^{3}\alpha }}{{\cos^{2}\alpha }}\,d\alpha }&=&\int{\sin\alpha (\sec^{2}\alpha-1)\,d\alpha }\\ &=&\int{\left({\sec\alpha\tan\alpha-\sin\alpha }\right)\,d\alpha }\\ &=&\sec\alpha+\cos\alpha+k\end{eqnarray*}

 :ninja: