Im not understanding a set of equations on my homework tonight:

[quote] Mr. Brock's class is planning a cookie sale to raise money for the local food bank. They took a survey to help them figure out how much to charge for each cookie. They found that the relationship between the price and the number of cookies they would sell is linear. According to the survey they will sell about 200 cookies if they charge 50 cents, and they will sell about 50 cookies if they charge $1.

A. Write an equation for the relationship between the cost and the number of cookies.

B. Find the slope and the y intercept for the equation, and explain what this information means in the context of this problem.

C. If they charge 70 cents for each cookie, about how many cookies will they sell?

D. If they want to sell 300 cookies how much should they charge?

[/quote]

Im think the answer to A is..

y = -300x but im not sure =/

Hi midnightz:

[quote]Mr. Brock's class is planning a cookie sale to raise money for the local food bank. They took a survey to help them figure out how much to charge for each cookie. They found that the relationship between the price and the number of cookies they would sell is linear. According to the survey they will sell about 200 cookies if they charge 50 cents, and they will sell about 50 cookies if they charge $1.[/quote][quote]A. Write an equation for the relationship between the cost and the number of cookies.
[/quote]
[quote]B. Find the slope and the y intercept for the equation, and explain what this information means in the context of this problem[/quote]

[hide="part A and part B"]You start by using the point-slope equation for a straight line: y=mx+b
let the x-coordinate represent the cost per cookie and let the y-coordinate represent the number of cookies sold.
Next, plot the two points that represent the data given in the problem.
One point would be(.50,200) and the other point would be (1,50).
Since the relationship is linear, draw a line between the two points you have plotted.
 The next thing is to find the slope of the line which is represented by "m" in the equation. That is found by dividing the difference of the two y-coordinates by the difference of the two x-coordinates. Your slope should come out to -300 for "m".
 To find "b" (your y-intercept)substitute the value of "m" into the formula and also the x and y values of one of your coordinates. The value of "b" should come out at 350.
So, the equation for your relationship should be y = -300x + 350.[/hide]

I'l leave it up to you to solve  parts C and D.


sabet konid agar $ x^k \plus{} \frac {1}{x^k}\in Q$ va $ x^{k \plus{} 1} \plus{} \frac {1}{x^{k \plus{} 1}}\in Q$
angah $ x \plus{} \frac {1}{x}\in Q$

it can be very easily proved through induction

kidding!easy induction!
i think you miss the problem.
can you send your solution in easy induction?

[quote="rt_08"]it can be easily proved[/quote]
would you mind sharing your solution with us?

then don't use $ \text{\LaTeX}$,write your solution as it is:

$ x+y=$x+y

$ x-y=$x-y

$ x*y=$x*y

$ \frac xy=$x/y

$ x^y=$x^y

and etc.


Define $ *$ so that $ A * B \equal{} A \plus{} 2(B \plus{} 3)$. Find the value(s) of $ x$ for which $ (x * x) * x \equal{} x * (x * x)$.

$ (x\plus{}2x\plus{}6)*x\equal{}x*(x\plus{}2x\plus{}6)$
$ x\plus{}2x\plus{}6\plus{}2x\plus{}6\equal{}x\plus{}2x\plus{}4x\plus{}12\plus{}6$
$ x\equal{}\minus{}3$


the problem is: given 1980 points in the plane, no two a distance less then 1 apart. show that there exist 220 points, no two a distance less than  \sqrt 3 apart.
I have found that among any 6 points, there are 2points with distance  \geq  \sqrt 3. also, if assume conversely, there are at most 1951245 such pairs according to Turan theorem.
However, I have not found its solution yet. :(

I was sure that this problem reminded me another one....
You may find a solution to your problem at :
http://www.mathlinks.ro/viewtopic.php?p=18282#18282

Wow! One solution and two problems.

Pierre.


The sequence $ a_1, a_2, a_3, ...$ of positive integers is determined by its first two members and the rule $ a_{n\plus{}2} \equal{} \frac{a_{n\plus{}1} \plus{} a_n}{gcd(a_n, a_{n\plus{}1})}$. For which values of $ a_1$ and $ a_2$ is it bounded?

The only solution for a bounded sequence is $ 2\equal{}a_1\equal{}a_2$. For the proof we will start with two properties for arbitrary sequences with the given recursive relation:

(i) If for some $ i$ the two numbers $ a_i,a_{i\plus{}1}$ are equal to $ 2$, then we have $ 2\equal{}a_1\equal{}a_2\equal{}\ldots$
(ii) If for some $ i$ the equation $ \gcd(a_i,a_{i\plus{}1})\equal{}1$ holds, we have for all $ k\equal{}1,2,\ldots$ the identity $ a_{i\plus{}k}\equal{}F_{k\minus{}1}a_i\plus{}F_ka_{i\plus{}1}$, with Fibonacci integers $ F_0\equal{}0,F_1\equal{}1,F_{n\plus{}2}\equal{}F_n\plus{}F_{n\plus{}1}$.

[i]Proof[/i]: (i),(ii) are straightforward inductions.

From (ii) we conclude the necessary criterion $ \forall i: \gcd(a_i,a_{i\plus{}1})\geq 2$ for a bounded sequence. The recursive definition yields then $ \forall n: a_{n\plus{}2}\leq \max\{a_n,a_{n\plus{}1}\}$ with equality iff $ 2\equal{}a_n\equal{}a_{n\equal{}1}$ holds, i.e. iff $ 2\equal{}a_1\equal{}a_2$ holds using (i). From now on we can therefore assume $ \forall n: a_{n\plus{}2}<\max\{a_n,a_{n\plus{}1}\}$, i.e. the sequence $ b_n\equal{}\max\{a_{2n\minus{}1},a_{2n}\}$ is strictly decreasing, a contradiction to $ b_n$ assuming values among the positive integers.


[quote="zheng"]Jeremy made a Venn diagram showing the number of students in his class who own types of pets. There are 32 students in his class. In addition to the information in the Venn diagram, Jeremy knows half of the students have a dog, 3/8 have a cat, six have some other pet and five have no pet at all. How many students have a cat, a dog and some other pet?

[asy]unitsize(50);
import graph;
pair A = (0,-1); pair B = (sqrt(3)/2,1/2); pair C = (-sqrt(3)/2,1/2);
draw(Circle(A,1.2) ^^ Circle(B,1.2) ^^ Circle(C,1.2));
label("10",A); label("2",B); label("9",C); label("$z$",(0,0)); label("$w$",(B+C)/2); label("$y$",(A+B)/2); label("$x$",(A+C)/2);
label("Cats",2.4C,C); label("Other Pets",2.4B,B); label("Dogs", 2.4A,A);[/asy][/quote]

You all fail for not answering this EASY(and easy) question.
 :furious:
you are going to  :wallbash_red:  when you see how easy it is.
Any way the answer is 1

Let's first look at the four things the question tells us: $ x\plus{}y\plus{}z\plus{}10\equal{}\frac{32}{2}$, so  $ x\plus{}y\plus{}z\equal{}6$
$ w\plus{}x\plus{}z\plus{}9\equal{} 32\times\frac{3}{8}$, so  $ w\plus{}x\plus{}z\equal{} 3$
$ w\plus{}y\plus{}z\equal{}4$
$ w\plus{}x\plus{}y\plus{}z\plus{}2\plus{}9\plus{}10\equal{}32\minus{}5$, so  $ w\plus{}x\plus{}y\plus{}z\equal{}6$
From here, we can subtract the third equation from the fourth, to find that $ x\equal{}2$. Plugging that into the first equation, we find $ x\equal{}2$. From the third equation, we can then see,$ w\equal{}0$ . Finally, from the second equation, we find $ z\equal{}1$. Thus, there is $ \boxed{1}$ student with cats, dogs, and other pets.

[quote="pinkmuskrat"]Don't post problems you know the answer for. That is not the purpose of the forum. Besides, please use the ask on forum button instead of posting it for yourself. :)

Oh BTW that prob is eezz.[/quote]

I did know the answer but only put this post here for friend of mine who is freakin bmud(read sdrawkcab)

BTW: if the prob was so eezz y did u not ans?

Um, do we need to answer every question we know? I'm lazy too. Okay, maybe it's not thaaaat easy, but still. I've memed most of these problems anyway. So sad I ran out.


let $a$ and $b$ be two integers. suppose that $\sqrt{7-4\sqrt{3}}$ is a root of the equation $x^{2}+ax+b=0$. find the value of b-a

[quote="p4fn2w"][quote="Kouichi Nakagawa"][hide="Hint"]$\sqrt{7-4\sqrt{3}}=2-\sqrt{3}$
[/hide][/quote]
Question -- can you please explain how you got that? I can see that if you square the RHS and put a radical over it, it becomes the LHS. But how did you see that the LHS is equal to the RHS when you first worked at simplifying this?[/quote]

[hide="to answer that question..."]

Well, let's just assume that we can express $\sqrt{7-4\sqrt{3}}$ as  $a+\sqrt{c}$

So...just set them equal, square, and solve the system of equations.

If there is supposed to be a coefficient in front of the $\sqrt{c}$, you will get a perfect square multiplied by some number when you solve for $c$, so just factor it out at the end[/hide]

[quote="p4fn2w"][quote="Kouichi Nakagawa"][hide="Hint"]$\sqrt{7-4\sqrt{3}}=2-\sqrt{3}$
[/hide][/quote]
Question -- can you please explain how you got that? I can see that if you square the RHS and put a radical over it, it becomes the LHS. But how did you see that the LHS is equal to the RHS when you first worked at simplifying this?[/quote]

Generally the following is known.

Positive integer $x, y$ satisfying $x+y=a, xy=b$
$\sqrt{a+2\sqrt{b}}=\sqrt{x}+\sqrt{y}$
$\sqrt{a-2\sqrt{b}}=\sqrt{x}-\sqrt{y}\ (x>y)$
$\left ( \because \sqrt{x}\pm\sqrt{y}= \sqrt{\left ( \sqrt{x}\pm\sqrt{y}\right )^{2}}= \sqrt{(x+y) \pm 2\sqrt{xy}}= \sqrt{a \pm \sqrt{b}}\ \right)$

Ah.. I get it now.

My first approach was to algebraically manipulate it, by either raising it to a power then taking the nth-root of that, or multiplying it by a fraction thats equivalent to 1.

Didnt really think of trying to express it in the form of $(\sqrt{a}+\sqrt{b})^{2}$

Thank you for answering my question, Zellex and Kouichi Nakagawa.

[hide]$\sqrt{7-4\sqrt{3}}=\sqrt{7-2\sqrt{12}}=2-\sqrt{3},2+\sqrt{3}$ is the other root. From veite, $a=-(sum)=-4$, $b=product=1$, $b-a=5$.[/hide]

The reason it might occur to you to try and re-express $\sqrt{7+4\sqrt{3}}$ is that the roots of the quadratic equation $x^{2}+bx+c=0$ for integers $b, c$ always have the form $\frac{m \pm \sqrt{n}}{2}$ for integers $m, n$.  If $\sqrt{7+4\sqrt{3}}$ is to be a root, it must be of that form.


Consider six points in the plane which form a convex figure of area $ S$.
Prove that we can always find a triangle whose area is at most $ S/6$.

I don't understand your problem. :huh:
Because we can obviously construct a triangle inside the convex figure having area greater than S/6,...... I might be wrong, but still the problem looks trivial to me.

See here: [url]http://www.mathlinks.ro/viewtopic.php?p=198932#198932[/url] (case $ n\equal{}1$).

Yes, I meant that triangle is formed by three vertices from the given six.

Thank you Cosmin (aka pohoatza),  I knew that this problem appeared somewhere,
but forgot where. Also, out of the hardest case with convex hexagon,
the small one with pentagon appears.


What is the sum of the even, positive integers less than 62?

Also, you could use the formula for an arithmetic series, which is $ n (a \plus{} z) \div 2$, where $ n$ is the number of terms, $ a$ is the first term, and $ z$ is the last term.

$ 30 (2\plus{}60) \div 2 \equal{} \boxed{930}$

2+4+6... + 60

= 1+3+5... + 59 + 30 (for the -1s that we subtracted)

Sum of n odd integers is just n^2

= 30^2 + 30

= 930

Using the sum for an arithmetic series is perfect but it remains that many people do not know the formula. Using the same method Gauss once did to compute $\sum_{i=1}^{100} i$, we notice that if we write the terms as $2, 4, 6, 8, \cdots 54, 56, 58, 60$ then we can see that $60+2=62$ and $58+4=62$, and so forth. There are $60/2=30$ values and so there are $30/2=15$ additions, thus the sum is $62 \cdot 15 = \boxed{930}$.

Thats too easy a question $2(\sum_{n=1}^{30}n)$


Harry Potter and the Half-Blooded Prince is coming out what are your predictions on who will die and who is the new Defense Against Dark Arts Teacher?

I think the real question might be "who will use the search button?"  ;)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35879

[quote="bleumoose"]don't jump the guy too much. i got yelled at a few months ago for bumping a thread that had fallen to the bottom of the page (i hadn't been on aops for a few months prior to that).

[quote]Please don't bump old threads, unless it's a math problem that hasn't had any responses. Otherwise, if a thread is more than a few days old since the last response, leave it alone.[/quote]

So what do you want us to do? Not bump old threads? Not start new threads on similar topics? Or just not talk about the topics that interest us, because we were too late to get into the discussion?[/quote]

Hm, well I didn't mean to jump on noneoftheabove, just to give him a friendly reminder for next time. And also, the thread wasn't exactly real old either, just less than three weeks. I was trying to point out there was already a thread about it and just reminding him that it's there. I didn't mean to sound mean or yelling at-ish, since I certainly did my share of bumping/starting topics about things already in place in the past.

[quote="noneoftheabove"]o sorry err ill try to use it next time because normally on math threads i dont really use them[/quote]

Generally for this board, we'll define "inactive" as a thread that hasn't had a response in a month or more. So in this case you could've posted on the other HP 6 thread and it would have been okay.

Past that, go ahead and start a new thread. Note though that if you start an identical thread too soon, most people will have already talked about it and there won't be much discussion and thus not much point to making a new thread.


geometry
geometric transformation
dilation

Hi
cud someone please answer this elementary question....what is NEUTRAL FeCl3 and why is it used to detect phenolic groups?? (i want to know wats the significance of NEUTRAL here)
Thank u..
Shyam

phenolic groups form a complex with the Fe3+ ion which gives the violet colour.....

Yes the phenol forms a complex $ \ce{[Fe(C6H5O)6]^{3\minus{}}}$

I thought his question was rather [u][b]why is it neutral [/u] $ \ce{FeCl3}$[/b] than the test with phenol itself  :huh:
I think it may be bcos side products may be formed and the complex may be formed only in a neutral soln and it might not form the complex in a non neutral soln  :maybe:  :maybe:

what is [b] neutral[/b] $ \ce{FeCl3}$ :maybe:

I think pH=7 that complex is formed and is violet in colour :maybe:  :maybe:

how can pH of a solution of phenol in water be 7 unles there is something to neutralize it's acidity :maybe:

Maybe its a soln of $ \ce{FeCl3}$ in sth else, bcos its acidic in water due to hydrolysis  :maybe:  :maybe:

Probably neutral meaning no interferring acid or basic radicals :huh:  :huh:

Start by noting that an aqueous solution of FeCl3 is slightly acidic. If you want a neutral solution of FeCl3 just add some base (NaHO for instance) until the pH is neutral.

but why do you want a neutral solution is the question? also fecl3 is stable in acidic medium, any fecl2 produced is immediately oxidised to fecl3 in acidic medium

IN basic medium, $ \ce{FeCl3}$ gets hydrolysed to $ \ce{Fe(OH)3}$

Probably in acidic medium the violet complex is not formed very easily :maybe:

probably in acidic medium $ \ce{C6H5O^\minus{}}$ IS not formed....

And probably in basic medium, phenol becomes soluble and hence a complex cant be formed  :maybe:  :maybe:

i have already given a reason why it is not possible in basic medium due to the hydrolysis of
$ \ce{FeCl3}$

are my reasons for acidic medium right??????

[quote="chemrock"]How illogical madness Phenol has to become soluble to form the complex :huh:[/quote]

Ithought that phenol's solubility in NaOH might interfere in the formation of the soluble complex itself  :huh:   :maybe:

u are forgetting something madness.....

u need only $ \ce{C6H5O^\minus{}}$ for the complex to be formed .........

IN basic medium it is easily formed....

i contacted madness............ he told me a valid reason..........
can u please explain madness........

don feel lazy this tiome

excuse me $ \ce{FeCl3}$ is inorganic and phenol is organic and the reaction to take place phase transfer of phenol should take place so that's why I think little amount of [b]NaOH[/b] is added.

well i think this is getting a bit long winded...cud someone please post a clear solution explaining as to WHY ONLY NEUTRAL FeCl3 is used???

THis is because in a basic medium $ \ce{FeCl3}$ gets hydrolysed to $ \ce{Fe(OH)3}$ .........
and in the acidic medium..$ \ce{C6H5O^\minus{}}$ is not formed for the complexation reaction to occur....

That's only in strong basic medium, Valerium.

ok, then why the reaction is not carried out in the basic medium....

In acidic medium, fecl3 is capable of getting reduced to fe2+, and carcul I think ranga is rite. In basic( even if it is weakly basic) medium, fe(oh)3 is formed. For eg, In qualitative analysis, even dilute naoh or nh4oh is used to ppt fe(oh)3. Even in presence of nh4+ salts, fe(oh)3 is ppted as its sol. Product is very low(3.8*10^-38)

In acidic medium, How will the $ {C6H5O^\minus{}}$ be formed varun....
and moreover why shloud the $ {Fe^3\plus{}}$ get reduced to $ {Fe^2\plus{}}$?? :maybe:

there is always an equilibrium btw fe3+ and fe2+ and in acidic medium fe2+ is more stable. and thats why i said acidic medium shud not be used too.

Does neutral ferric chloride give postitve test with alddehydes having at least one alpha hydrogen atoms?
Coz wiki says "ENOLS" give positive tests, so if a keto/aldo can form an enol which has higher content than the keto/aldo from, would it give positive test?
(Had a question from a paper, which says 2-methyl-4,4-dimethyl-3-keto pentanal gives postive test)

2,4,4-trimethyl-3-ketopentanal should be enolizable, so yes. Note that wikipedia's examples of oximes, hydroxamic acids, etc. imply that it is necessary for a functionality to have an $\ce{O-}$ group bonded directly to one atom participating in a double bond, so it would be inductively reasonable for the answer to your question to be $\boxed{\text{yes}}$.

Wait can somebody answer the original question though?


Yeah,I also want to know it.It had begun in the same time last year.

[quote="Hawk Tiger"]When will be the next edition?[/quote]It's been a little postponed due to some personal matters (and not only) that I have to attend to. I will make a General announcement when it will start (actually 1-2 weeks ahead of time).


[quote="jacob8888888"]In a magic square, numbers are arranged in an n x n array so that the sum of the nubers listed in each row and column is the same. THis sum is called the magic sum. Use the nine numbers -10, -8, -6, -4, -2, 0, 2, 4, 6 to form a 3 x 3 magic square. What is the magic sum for this magic square?

Thanks[/quote]

[hide][size=150]-2[/size]. I think it's just the averages of all the numbers...  :? [/hide]

The magic sum is the sum of the numbers in any row or column; you are close on the average, but not quite.

[hide]abc
def
ghj

magic sum=x

3x=a+b+c+d+e+f+g+h+j

3x=-10-8-6-4-2+0+2+4+6=-18

x=-6 magic sum[/hide]

The average of the numbers is -2. Therefore, the sum of three numbers is $-2 \times 3 = \boxed{-6}$


why is the c-c bond length in ethane and diamond different even when they both have only sp3 hybridized c atoms?
Answer:
Carbon in ethane has 3 bonds with H .so although the hibridizied are the same but the differance in other bonds causes different bond length{because of the differnce of the electronegativity of carbon in ethane and Diamond.
Saeed


conics
parabola
trigonometry
analytic geometry

The focus of a parabola is at the origin and its directrix is the line $ r\cos\theta\equal{}\minus{}4$. Find the polar equation of the parabola.

Also, what is the general formula for a polar parabola?

$ x \equal{} ay^2 \plus{} by \plus{}c \Leftrightarrow r \cos \theta \equal{} a r^2 \sin^2 \theta \plus{} b r \sin \theta \plus{} c$

I substituted and got 2 solutions

[hide]$ r \equal{} \frac {4}{1 \minus{} \cos\theta}$
and
$ r \equal{} \frac { \minus{} 4}{1 \plus{} \cos\theta}$[/hide]

...are they both right?

The equation of the parabola in rectangular coordinates is

$ x \equal{} \frac{1}{8}y^2 \minus{}2$

and so the equation in polar coordinates is

$ r \cos \theta \equal{} \frac{1}{8}r^2\sin^2 \theta \minus{}2$


function
geometry
geometric transformation
3D geometry
induction
calculus
integration

This problem was published in a book called "200 de ecuatii functionale pe N, Z, Q", in 2003 at GIL Publishing House ( http://www.gil.ro ). It can be found at page 27, problem nr. 139, author: Drimbe Onucu Mihai.

In French it's hard to understand. Can somebody translate (briefly)?

[quote="pigfly"]Find all functions $f: \mathbb{Z} \mapsto \mathbb{Z}$ satisfying the condition: $f(x^3 +y^3 +z^3 )=f(x)^3+f(y)^3+f(z)^3.$[/quote]if the added property:x,y,z are distinc,can you solve the problem so easily?

Who has the solution for the expanding: Given m,n integer numbers. Find out all integral functions $f$ satisfying:
\[ f(\sum_{k=1}^nx_k^m)=\sum_{k=1}^nf^m(x_k)  \]
For every integer numbers $x_1,x_2,...,x_n$

And for each integer m, what is the smallest n for which, there are finite function $f$.

This is a problem from American Mathematic Monthly, year 1997 (if i no wrong  :ninja: )  :D
[quote="pbornsztein"]This one has been used in our correspondance program last year.
See here for a solution (dossier 4 pb#2, in french but it doesn't matter here) :

http://www.animath.fr/tutorat/dossier_03044sol.pdf

Pierre.[/quote]

If $ x \equal{} y \equal{} z \equal{} 0$, we have: $ f(0) \equal{} 3f(0)^{3}$, so $ f(0) \equal{} 0$ is only integer solution.
If $ y \equal{} z \equal{} 0$, $ f(x^{3}) \equal{} f(x)^{3}$. So, $ f(y^{3}) \equal{} f(y)^{3}$.
If $ z \equal{} 0$, $ f(x^{3}\plus{}y^{3}) \equal{} f(x)^{3}\plus{}f(y)^{3}\equal{} f(x^{3})\plus{}f(y^{3})$. So, $ f(a\plus{}b) \equal{} f(a)\plus{}f(b)$, for any integers $ (a,b)$.
So, $ f(x) \equal{} f(1)\plus{}f(x\minus{}1) \equal{} 2f(1)\plus{}f(x\minus{}2) \equal{} ... \equal{} xf(1)$. By substituting $ f(1) \equal{} k$, where $ k$ is integer number, we get that $ f(x) \equal{} kx$ is solution for any integer number $ k$.

[quote="lasha"]...$ f(x^{3}\plus{}y^{3}) \equal{} f(x)^{3}\plus{}f(y)^{3}\equal{} f(x^{3})\plus{}f(y^{3})$. So, $ f(a\plus{}b) \equal{} f(a)\plus{}f(b)$, for any integers $ (a,b)$...[/quote]

I think this part is wrong. $ f(a\plus{}b)\equal{}f(a)\plus{}f(b)$ only for $ a,b$ perfect cubes. :(

[quote="mathmanman"]Well, now it's a bit late, but tomorrow I'll put the translation here ;)

EDIT : Ok, here goes :
First, they notice that $f(x) = 0$, $f(x) = x$ and $f(x) = -x$ are solutions, and, they want to show that these are the only ones.
Now, let $x=y=z=0$, we thus get : $f(0) = 3f(0)^3$., which leads to $f(0)=0$..
Then, if we let $y=-x$ and $z=0$, we get $f(-x)^3 = -f(x)^3$, which show that $f$ is an odd function.
With $(x,y,z) = (1,0,0)$, we have $f(1) = f(1)^3$, hence $f(1) = 0, 1 \ or -1$.
Moreover, with $(x,y,z)=(1,1,0)$ and $(x,y,z)=(1,1,1)$, we get that $f(2) = 2f(1)$ and $f(3) = 3f(1)$.
Thus, to conclude, we'd have to show that : $f(x) = xf(1)$ for all x.
To show this, we'll use the following lemma :
[b]Lemma[/b] : For all integer $x \ge 4$, $x^3$ can be written as a sum of five cubes of integers strictly less than $x$ in absolute value.
[b]Proof[/b] : I'll let this to the reader  :P  (if you can't prove it though, just ask and I'll translate that part too)

Then, thanks to this lemma, we can show via induction that $f(x) = xf(1)$, for all $x \in \mathbb{N}$ (and so, in $\mathbb{Z}$ too, since $f$ is odd).
And, eventually, we have that the only functions satisfying the asked relation are :
$f(x) = 0$, $f(x) = x$ and $f(x) = -x$.

:)[/quote]
can u translate the lemma for me, mathmanman?
thanks.

From AMM 1999
and appear in Venkatachala book in 2002 :lol:


integration
trigonometry
calculus
calculus computations

$ \int{\frac{\sin(x)}{1\plus{}\cos^2(x)}}$
Hint:
[hide]simple substiution[/hide]

[hide]$ (\cos x)' \equal{} \minus{} \sin x$, and so

$ \int \frac{\sin x}{1 \plus{} \cos ^2x}\,dx \equal{} \minus{} \int \frac{\minus{} \sin x}{1 \plus{} \cos^2x}\,dx \equal{} \minus{} \arctan(\cos x) \plus{} C$.[/hide]

1. $ \int{\frac {\sin(x)}{1 \plus{} \cos^2(x)}}$

2. $ \int{\frac {\cos(x)}{1 \plus{} \sin^2(x)}}$

3. $ \int{\frac {\sin(x)}{1 \plus{} \sin^2(x)}}$

4. $ \int{\frac {\cos(x)}{1 \plus{} \cos^2(x)}}$

1 and 2 can be solved with $ u \equal{} \cos(x)$ and $ u \equal{} \sin(x)$ respectively.
Using the identity $ \cos^2(x) \plus{} \sin^2(x) \equal{} 1$ and either of the substitutions for 1 and 2 one can solve 3 and 4.


Let be $P$ a point inside the equilateral $\triangle {ABC}$ such that $PA=5$, $PB=7$, $PC=8$.

Find the length of any side of $\triangle {ABC}$

Thanks shobber!!.

After that I got that, if $s$ is the side length, then $s^4=3(46s^2-537)$, but $s=?$

[hide]

using the identity in the post above, we know that:

$3(5^4 + 7^4 + 8^4 + s^4) = (5^2 + 7^2 + 8^2 + s^2)^2$

simplifying, we arrive at:

$s^4 - 138s^2 + 1161 = 0$

factoring, we get:

$(s^2 - 9)(s^2 - 129) = 0$

hence $s = \pm 3, \pm \sqrt{129}$, but since $s>0$ and $s$ obviously cannot be $3$ as the cevians to $P$ are even longer than that, $\boxed{s = \sqrt{129}}$

btw was this an AIME problem? and does anyone know a solution that doesnt make use of this identity?

[/hide]

Here's what I got...

Let a side be $s$.  Then there are circles centered at the three vertices, of radii 5, 7, 8, which intersect in a unique point.  They have equations \[ x^2+y^2=49 \] \[ (x-s)^2+y^2=64 \] \[ (x-\frac{s}{2})^2+(y-\frac{s\sqrt{3}}{2})^2=25 \]
Simplifying yields
\[ x^2+y^2=49 \] \[ x^2+y^2+s^2-2sx=64 \] \[ x^2+y^2+s^2-sx-sy\sqrt{3}=25 \]  From there, I used addition, and manipulation to get $s=\sqrt{15}$...

hmm check ur math... i did it ur way and ended up with the same quartic

but yea good solution ne ways.... pretty clever heh :)

Make a copy of $\triangle ABC$ and rotate it 60 degrees to get a traingle inside it with side lengths 5, 7, and 8.
$\\8^2=5^2+7^2-2(5)(7)\cos\theta\\\cos\theta=\frac{64-25-49}{-70}=\frac{1}{7}\\\theta=\arccos\frac{1}{7}$
$\\s^2=25+49-2(5)(7)\cos (60+\arccos\frac{1}{7})\\ =74-70[\cos 60\frac{1}{7}-\sin 60\sin\arccos\frac{1}{7}]\\ =74+55=129\\ \therefore s=\sqrt{129}$

Masoud Zargar

[quote="gotztahbeazn"]was this an AIME problem? [/quote]

No, it was a problem for an Iberoamerican Olympiad.

but this same problem with different numbers had been posted a million times


This is the most funny, cool, or just plain awesome stuff!

It's probably a bad ting, when a cat's got ur noses. :P

Don't u feeel bads 4 de bears? but it doesn't mean u can't laugh at dem!  :rotfl:


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[b] Some questions[/b]

A function $ f$ is defined on $ \mathbb R$* and  $ f^{ \minus{} 1} (x) \equal{} \frac {1}{f(x) } , \forall x$ which is not $ 0$. If $ f$ is strict  increasing on the interval of possitive numbres $ (0, \plus{} \infty ))$, can we find  a general form for this function. What more must we suppose , in order to determine all such functions , if possible? Can you find two such functions ?

 Thanks in advance - Babis

See:
Let $ f(x)\equal{}f$. Now:
\[ \frac{df^{\minus{}1}}{dx}\equal{}\frac{dx}{df}\equal{}\frac{1}{f'}\]
Using our condition we will have:
\[ \frac{1}{f'}\equal{}\minus{}\frac{f'}{f^2}\]
or
\[ {f'}^2\equal{}\minus{}f^2\]
This may be only if $ f\equal{}0$, but as we can see $ f\ne 0$. So does not exist such function $ f(x)$.

Your computation is not correct, Yuriy.

For a related problem, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1670930472&t=52068]here[/url].
Anyway, the main idea is that $ f(f(x)) \equal{} \frac1{x}$ (I'm assuming $ f$ to be bijective in your post). You can find a lot of solutions, I'm sure.

Yes, I see thanks. :) I was in math class in school, but in university we have high math only one year and I don't know what is bijection... :(

Definition of a bijection:
The function $ f$ from domain $ A$ to codomain $ B$ is bijective if for each $ y\in B$, there is exactly one $ x\in A$ with $ f(x)\equal{}y$.
This is the condition we need for an inverse function defined on $ B$ to exist.

There is actually some room here if we use the negative numbers. The function $ f(x)\equal{}\begin{cases}\minus{}x& x>0\\\minus{}\frac1x& x<0\end{cases}$, works. What happens if you put an arbitrary function on the positive side?

[quote="jmerry"]Definition of a bijection:
The function $ f$ from domain $ A$ to codomain $ B$ is bijective if for each $ y\in B$, there is exactly one $ x\in A$ with $ f(x) \equal{} y$.
This is the condition we need for an inverse function defined on $ B$ to exist.

There is actually some room here if we use the negative numbers. The function $ f(x) \equal{} \begin{cases} \minus{} x & x > 0 \\
\minus{} \frac1x & x < 0\end{cases}$, works. What happens if you put an arbitrary function on the positive side?[/quote]

 Dear Jmerry,

 Yes!This function with inverse signes has this property(both properties). A friend mathteacher in Greece tries to prove that this function is the only one. I think that this is not the case. This is the reason , I asked for two such functions. Any way , we will investigate the problem for a little more.
 Thanks,
 - Babis

Split into four regions; cut at $ 1$ and $ \minus{} 1$ as well (These map to each other). You can choose $ f$ almost arbitrarily on one of those regions, and fill in the others from it.

$ f(x)\equal{}\begin{cases}\minus{}e^{x\minus{}1}&x\ge 1\\ \frac1{\ln(\minus{}x)\plus{}1}&x\le \minus{}1\\\minus{}e^{1\minus{}\frac1x}&0<x<1\\1\minus{}\ln(\minus{}x)&\minus{}1<x<0\end{cases}$ works.


[hide="Solution"]

By the recurrence given, we have:

$x_{n}=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}$

and

$x_{n+1}=x_{n}-x_{n-1}+x_{n-2}-x_{n-3}$.

Adding these two equations and simplifying reveals:

$x_{n+1}=-x_{n-4}$

Using this equation, we can write the sum asked for as $x_{1}+x_{3}+x_{5}$. We can find $x_{5}$ using the recursion, showing that the sum is $211+420+167=\boxed{798}$[/hide]

[hide=Alternate Solution posted on wiki]
The recursive formula suggests telescoping. Indeed, if we add $x_n$ and $x_{n-1}$, we have $x_n + x_{n-1} = (x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4}) + (x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) = x_{n-1} - x_{n-5}$.

Subtracting $x_{n-1}$ yields $x_n = -x_{n-5} \implies x_n = -(-(x_{n-10})) = x_{n-10}$.

Thus,

$$x_{531} + x_{753} + x_{975} = x_1 + x_3 + x_5 = x_1 + x_3 + (x_4 - x_3 + x_2 - x_1) = x_2 + x_4 = 375 + 523 = \boxed{898}.$$

Notice that we didn't need to use the values of $x_1$ or $x_3$ at all.


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This is problem 2 of the 1st MathLinks contest (see http://www.mathlinks.ro/Forum/contest/1st%20Edition/1st%20round%20sol.pdf , where some proofs are presented). This is also inequality [b](1)[/b] in http://www.mathlinks.ro/Forum/viewtopic.php?t=5263 (where  I have given a proof without expanding).

  Darij

[quote="Armo"]Let $ a$, $ b$, $ c$ be positive reals. Prove that
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a \plus{} b}{a \plus{} c} \plus{} \frac {b \plus{} c}{b \plus{} a} \plus{} \frac {c \plus{} a}{c \plus{} b}$.[/quote]
Assume that $ a \ge b \ge c$.
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a \plus{} b}{a \plus{} c} \plus{} \frac {b \plus{} c}{a \plus{} b} \plus{} \frac {a \plus{} c}{b \plus{} c} \newline \iff \frac {c(a \minus{} b)}{b(b \plus{} c)} \plus{} \frac {b(c \minus{} a)}{a(a \plus{} b)} \plus{} \frac {a(b \minus{} c)}{c(c \plus{} a)} \ge 0 \newline \cdots
\]

You could also simplify and expand and bash this with AM-GM. ;)

[quote="Armo"]Let $ a$, $ b$, $ c$ be positive reals. Prove that
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a \plus{} b}{a \plus{} c} \plus{} \frac {b \plus{} c}{b \plus{} a} \plus{} \frac {c \plus{} a}{c \plus{} b}$.[/quote]

I think it isn't hard.
LHS-RHS
=$ \frac {abc[(\frac {ab}{c^2} \plus{} \frac {bc}{a^2} \plus{} \frac {ac}{b^2} \minus{} 3) \plus{} (\frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}) \minus{} (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})]}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

 You know how to do next! :)

[quote="darij grinberg"][quote="anthony19891019"]My SOLUTION~~~~~~[/quote]

Your last inequalities ($\frac{a-b}{b}\geq\frac{a-b}{b+c}$ et al.) need not be true.

  darij[/quote]

oh i seee. thank you

[quote="Lovasz"][quote="Armo"]Let $ a$, $ b$, $ c$ be positive reals. Prove that
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a \plus{} b}{a \plus{} c} \plus{} \frac {b \plus{} c}{b \plus{} a} \plus{} \frac {c \plus{} a}{c \plus{} b}$.[/quote]
Assume that $ a \ge b \ge c$.
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {a \plus{} b}{a \plus{} c} \plus{} \frac {b \plus{} c}{a \plus{} b} \plus{} \frac {a \plus{} c}{b \plus{} c}\newline \iff \frac {c(a \minus{} b)}{b(b \plus{} c)} \plus{} \frac {b(c \minus{} a)}{a(a \plus{} b)} \plus{} \frac {a(b \minus{} c)}{c(c \plus{} a)}\ge 0 \newline \cdots
\]
[/quote]
I think the inequality is cyclic, but not symmetric, so you can't assume that $ a\geq b\geq c$, you can only put $ a \equal{} \max(a,b,c)$. Am I right?

Yes, we need to put another assumption, $a \ge c \ge b$.

[quote=Armo]Let $ a$, $ b$, $ c$ be positive reals. Prove that
$$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a \plus{} b}{a \plus{} c} \plus{} \frac {b \plus{} c}{b \plus{} a} \plus{} \frac {c \plus{} a}{c \plus{} b}$$.[/quote]
[b] Belarus 1997 Test 4,problem 2 by I.Voronovich[/b]
[b]India tst 2002 p20[/b]
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge \frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2a}{b+c} $$
https://artofproblemsolving.com/community/c6h470229p2633711

Please, where I can find Problems from Belarus Math Olyimpiads ( for example 1996 ).
tnx

https://artofproblemsolving.com/community/u366128h1659644p10533555 [color=#f00]at post 8[/color]

Let $ a,b,c$ be positive reals. Prove that
$$\sqrt[n]{\frac {a}{b}}+\sqrt[n]{\frac {b}{c}}+\sqrt[n]{\frac {c}{a}}\geq \sqrt[n]{\frac {a + b}{a + c}}+\sqrt[n]{\frac {b + c}{b + a}}+\sqrt[n]{\frac {c + a}{c + b}}$$
Where $n\in N^+.$


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Is there a formula for $\sum_{k=0}^{\infty}\binom{3k}{k}x^{k}$?

For $\binom{2k}{k}$ there is a nice generating function:
$\sum_{k=0}^{\infty}\binom{2k}{k}x^{k}= \frac{1}{\sqrt{1-4x}}$

I am wondering if there is something similar for $\binom{3k}{k}$.

Mathematica says that sum is equal to

$\frac{2\cos(\frac{1}{6}\arccos(1-27x/2))}{\sqrt{4-27x)}}$

and I haven't given any thought at all as to why that should be.

Here is a few posts on information regarding it, and some generalization.
[url=http://www.mathlinks.ro/viewtopic.php?search_id=416999425&t=281653]Enjoy[/url]


[hide]
I'm not sure if i'm understanding the question right but I think each each edge has a one inch margin. $8\frac{1}{2}-2=6\frac{1}{2}$. THen we have the other side which is 11 inches. $11-2=9$ Then since the sides are $6\frac{1}{2}, 9$, we do $(6\frac{1}{2}\cdot2)+(9\cdot2)=31$
[/hide]

[hide="DarkKnight, read "] You're answer is right, but there is a tiny flaw in your solutions..

[quote=DarkKnight wrote]Then since the sides are 6\frac{1}{2}, 11, [/quote]
change that to...

6\frac{1}{2},9,[/hide]

[quote="MCrawford"]You are printing from your computer onto paper that measures $8 \frac{1}{2}$" by 11".  Each margin on the printed page is 1".  How many inches are in the perimeter of the rectangular region available for printing?[/quote]

[hide]
If each margin is 1", then you would have to subtract 2" from each dimension of the whole page to get the dimensions of the printed page. Therefore, the dimensions of the printed page are 6.5 by 9. To get the perimeter, do 2L+2W, which is 2*6.5 + 2*9, or 13+18, or 31. The perimeter of the rectangular region available for printing is 31".
[/hide]


Can someone explain to me why the Ramsey Theorem is considered a generalization of the Pigeonhole Principle? What is the connection between them?

First of all, we should not discuss Ramsey Theory in the intermediate forum. Now I'll answer the question.

The purpose of Ramsey Theory is to find the minimum number of points needed so that if we draw some colored lines from points to other points, then the colored lines must have some desired property (such as either a red triangle or a blue quadrilateral). The pigeonhole property is a very special case of Ramsey Theory since we want to know how many points we need so that when we draw colored lines with the additional property that if the line from A to B and the line from B to C are the same color, then the line from A to C is also that color for any points A, B, and C. Ramsey Theory is therefore much more general.

hey mathnovice, you should've went to the Combo class last summer but if you need any notes on Ramsey's/Pigeonhole, just pm me. Good Luck!

[quote="ComplexZeta"]The pigeonhole property is a very special case of Ramsey Theory since we want to know how many points we need so that when we draw colored lines with the additional property that if the line from A to B and the line from B to C are the same color, then the line from A to C is also that color for any points A, B, and C. Ramsey Theory is therefore much more general.[/quote]

How does this generalize to a statement like:

(The Strong Pigeonhole Principle)

Let q_1, q_2,...,q_n be positive integers. If

q_1 + q_2 + ... + q_n - n + 1

objects are put in n boxes, then either the first box contains at least q_1 objects, or the second box contains at least q_2 objects, ..., or the nth box contains at least q_n objects.

OR

(Pigeonhole Principle)

If n + 1 objects are put in n boxes, then at least one box contains two or more of the objects.

I don't understand how you go from something geometric like Ramsey Theorem to something algebraic like Pigeonhole Principle. How do you express Pigeonhole Principle in terms of the Ramsey Theorem?

I think you can draw a bunch of lines from each point to each other and use the same argument to get the strong (or intermediate, as Zeitz calls it) pigeonhole principle.

Here is an example of the relation of the two thereoms.

Think of 5 non-colineer points in a plane.

Then you have to draw all possible segments in color A or color B and not form any triangles with either color.

Now prove that each point must have exactly 2 segments of each color coming from it. (proof in spoiler)

[hide]If the points do not have 2 segments of each color coming from it then one color must be used on at least 3 of the segments. Let's say we draw 3 segments from a point in color A. Then the three points connected to the original point must eventually be connected to each other. None of these segments can be color A because then a triangle would be formed, but if you connect all of the three points with color B then a triangle would also be formed.[/hide]


Beginning with 2 and 7, the sequence 2, 7, 1, 4, 7, 4, 2, 8, ... is constructed by multiplying successive pairs of its members and adjoining the result as the next one or two members of the sequence, depending on whether the product is a one- or a two-digit number. Prove that the digit 6 appears an infinite number of times in the sequence.

I don't think I'm understanding the sequence completely.  The way you describe it makes it seem like it should be 2, 7, 1, 4, [b]4, 1, 6, 6[/b]

Am I missing something?

It's not the previous two terms. It's the first two terms you haven't used already.

[hide]

The sequence begins 27147428288161616..

It's obvious that any two digit number ab appearing n times, with a and b as digits, will produce (2n-1) of a*b, and any one digit number a will produce (2n-1) of a :^2:.  Because of the rapid growth of the number of elements in the subsequence, we will ignore this and show instead that the numbers begin cycling.

The repeating 16's will produce, in turn, a repetition of:

6

36

18

8

64

24

8

64

24

8

64

Therefore, the sequence will continue repeating with this part, each subsequence having more elements than the part before it.

[/hide]

Well, I'd like to say QED, but it always seems as though my proofs are somehow wrong.  So instead....

QED?

I know a sequence i did earlier thats kinda like that, but it isn't infinite because 1 digit numbers are expressed as "3" instead of "03".

6,2,1,2,2,4,8,3,2,4

Does anyone know a longer finite sequence?

I'm not sure exactly how you got that sequence (nor the thing about 03, in the sequence that was given before you didn't write 03 either.) If you worked it out in the same way as above, it would go 6,2,1,2,2,2,4,4,8,1,6.. and would go on forever. If instead you only multipled the last two digits at each point, you would get 6, 2, 1, 2, 2, 4, 8, 3, 2, 6, 1, 2, .. and that would go on forever too. Are you using some other rule?

No, its like when you multiply two of the digits, it gets moved over to the nearest open space, whether it be a one- or two- digit number. But it has to fit perfectly, you can't go on with 8,3,2,4 because 8*3=24 but 3*2=6, not 4. And you couldn't do anything before 6,2,1,2, because there is nothing times 6 that equals either 2 or 21.

I thought the "03" thing because to compensate for the products being mostly 2 digit numbers, you couldn't use 1 digit numbers or it would mess up the sequence, i dont know...


For  a constant number $ a$, let $ f(x) \equal{} x^2 \minus{} 4ax \plus{} a^2 \plus{} 5a \plus{} 2$.

(1) Let $ P$ be the vertex of $ f(x)$. Find the coordinate of $ P$.

(2) Find the value of $ a$ such that $ y$ coordinate of $ P$ is maximal.

(3) Find the range of $ a$ for which $ f(x)\geq 0$ for any real numbers $ x$.

(4) Find all values of $ a$ such that the minimum value of $ f(x)$ for $ x\geq 0$ is 2.

[hide="solution"]$ f(x)\equal{}(x\minus{}2a)^2\minus{}3a^2\plus{}5a\plus{}2$, so

[b](1) [/b]$ P\equal{}(2a,\minus{}3a^2\plus{}5a\plus{}2)$
[b](2)[/b]$ \minus{}3a^2\plus{}5a\plus{}2\equal{}\minus{}3(a\minus{}\frac{5}{6})^2\plus{}4 \frac{1}{12} \le 4\frac{1}{12}$
 $ \ Rightarrow a\equal{}\frac{5}{6}$
[b](3)[/b]Min. value $ \equal{}\minus{}3a^2\plus{}5a\plus{}2 \ge 0 \Leftrightarrow 3a^2\minus{}5a\minus{}2\equal{}(3a\plus{}1)(a\minus{}2) \le 0$
The range is $ \minus{}\frac{1}{3} \le a \le 2$
[b](4)[/b] min. value $ \equal{}\minus{}3a^2\plus{}5a\plus{}2\equal{}2 \Leftrightarrow a(\minus{}3a\plus{}5)\equal{}0 \Leftrightarrow a\equal{}0$ or $ 1\frac{2}{3}$[/hide]

Sorry, I made a typo in (4). I have just edited it.

kunny


The endpoints of a diagonal of a rectangle are $ (\minus{}4, \minus{}5)$ and $ (3,4)$, and the sides of the rectangle are parallel to the axes. How many square units are in the area of the rectangle?

The width of the rectangle is $ 3\minus{}(\minus{}4)\equal{}7$ and the length of the rectangle is $ 4\minus{}(\minus{}5)\equal{}9$, so the area is $ 7\cdot 9\equal{}\boxed{63}$


the complex number $ z$ is defined by $ r(\cos {\theta} \plus{} i \sin {\theta})$ and the complex number $ u$ is defined by $ r^2(\cos {2\theta} \plus{} i \sin {2\theta})$  Find $ \log_u{z}$

PS: show your work

By De Moire's Theorem, $ z^n\equal{}r^n(\cos n\theta \plus{}i\sin n\theta)$  This implies $ u\equal{}z^2$.  Thus, the desired answer is $ \dfrac{1}{2}$.

Sort-of: see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=207877 and the threads I linked to therein.


Let $ a\equal{}\overline{a_1a_2\ldots a_n}$ ($ a\in\mathbb{Z^\plus{}}$)
We know that $ \sum_{i\equal{}1}^na_i\equal{}\prod_{i\equal{}1}^na_i\equal{}9^{2007}$. Prove that $ n$ is a square number.

Take $ a$ such that the only digits are $ 1$s and $ 9$s; it has $ 9^{2007} \minus{} 2007(9)$ digits. Then replace a $ 9$ with the 5 digits $ 3,3,1,1,1$. So $ n$ can be both $ 9^{2007} \minus{} 2007(9)$ and $ 9^{2007} \minus{} 2007(9) \plus{} 4$. These can't both be squares... am I doing something wrong here?


If $p$ is a prime $\ge5$, prove that $p^{2}-1$ is divisible by $24$.

[hide="solution"]
Lets look at all the primes greater than or equal 5 in mod 24. They must be 1, 5, 7, 11, 13, 17, 19 or 23 mod 24 (they must be relatively prime to 24 to be prime). We can also see this as 1, 5, 7, 11, -11, -7, -5, -1 mod 24. 1^2-1 mod 24 = 0, 5^2-1 mod 24 = 0, 7^2-1 mod 24 = 0, 11^2-1 mod 24 = 0, and the same argument works with the additive inverses of these numbers. Thus, for any prime greater than or equal to 5, p^2-1 = 0 mod 24, or p^2-1 is divisible by 24. [/hide]

[hide] We know that $p^{2}-1=(p-1)(p+1)$.  Since p is greater than or equal to 5, we also know that p is odd.  Thus, p-1 and p+1 are consecutive even integers.  However, we also know that every consecutive even integers, one of them is also divisible by 4.  Therefore, one of them is divisible by only two and one of them is divisible by 4.  Hence, (p+1)(p-1) is divisible by 4*2=8.  Now, we also know that every 3 integers, exactly one of them is divisible by 3.  Take the integers p-1, p, and p+1.  This implies that one of them is divisible by 3.  p Cannot be divisible by 3 because it is prime, so p+1 or p-1 is.  This implies that (p+1)(p-1) is divisible by 8 and 3, so it is divisible by 8*3=24. [/hide]

My answer is pretty much the same as Quatto's.[/hide]

p=6k+1

 p^2-1=k^2+12k+1-1=12k(3K+1). similarly with the case p=6k-1

           And atleast one of the last 2 parantheses is even.
                         hence the problem!  :lol:

Curious.  Do there exist larger $d$ such that if $p \ge P$ for some prime $P$ then $d | p^{2}-1$?

Intuitively, no. (And if there is a similar identity, there are only finitely i.e. very few like 1 or 2 other numbers than 24). The only reason such tricks work for 24 is that it is entirely 2's and 3's. But the PHP type divisibility establishment isn't necessarily true for larger primes.

It's interesting, though.  Let $p_{n}$ be the $n^{th}$ prime, and consider the sequence $a_{n}= \frac{p_{n}^{2}-1}{24}$.

[b]Conjecture:[/b]  There does not exist $N$ such that for all $n > N$ the members of the sequence $a_{n}$ have a common factor.


define value of parameter $p$ for which those quadratic functions have exactly one common root:

$y=x^{2}+px+2$
$y=x^{2}+2x+p$

ehh... it must be trivial... :|

[hide="If they have a common root..."]
... then their difference is $0$.

$(p-2)x+(2-p)=0$
$x=1$

Hence $x^{2}+px+2=(x-1)(x-a)$
and $x^{2}+2x+p=(x-1)(x-b)$

We can see from the first one that $a=2$ and hence $p=-3$: in the second equation, if $b=-3$ then this works.

$\boxed{p=-3}$[/hide]


Define a $good~word$ as a sequence of letters that consists only of the letters $A,$ $B,$ and $C$ $-$ some of these letters may not appear in the sequence $-$ and in which $A$ is never immediately followed by $B,$ $B$ is never immediately followed by $C,$ and $C$ is never immediately followed by $A.$ How many seven-letter good words are there?

[hide="Solution"]WLOG, let $A$ be the first letter of the sequence. Then for each of the six remaining letters, there are two choices for a total of $2^{6}=64$ possibilities. Furthermore, since $A$, $B$, or $C$ can start a string, our answer is $64\cdot3=\boxed{192}$ seven letter words.[/hide]

[quote="Xaero"][hide="Solution"]WLOG, let $ A$ be the first letter of the sequence. Then for each of the six remaining letters, there are two choices for a total of $ 2^{6} \equal{} 64$ possibilities. Furthermore, since $ A$, $ B$, or $ C$ can start a string, our answer is $ 64\cdot3 \equal{} \boxed{192}$ seven letter words.[/hide][/quote]
I'm afraid your solution includes the possibility such as AAAAAAA. Does the question means that the good words should include A,B and C——all of them? :?:

[quote="4everwise"]some of these letters may not appear in the sequence[/quote] means that we are allowed to use only some of A, B, C.

Recursion also works (something I learned here).

Let $A(n)$ be the number of $n$ letter sequences beginning with an $A$ and respectively for $B(n), C(n)$.

Let $X(n) = A(n) + B(n) + C(n)$.

Quickly notice, $A(n) = A(n -1) + C(n-1)$ same for $B(n)$ and $C(n)$.

Then:

$$X(n) = 2(A(n-1) + B(n-1) + C(n-1))$$

$$\implies X(7) = 2(A(6) + B(6) + C(6))$$

$A(2) = 1 + 1, A(3) = 2 + 2 = 4$.

You see the pattern that: $A(n) = B(n) = C(n)$ so,

$A(4) =  4 + 4 = 8, A(5) = 8 + 8 = 16, A(6) = 16 + 16 = 32$.

$$X(7) = 2\cdot 3 \cdot A(6) = 192.$$

We first count the amount of two letter good words. These are $AA, BB, CC, AC, BA, CB$, for a total of $6$. Now note that we have two possible endings to add on to each of these, so the amount of three letter words is $6 \cdot 2^1$, the amount of four letter words is $6 \cdot 2^2$, so on, so the amount of seven letter words is $6 \cdot 2^5 = \boxed{192}$.


function
inequalities proposed
inequalities

Well Lets have something new about the convex functions:

Let $x_1<x_2<...<x_n$ and $f$ a convex function, then prove that :

$x_1f(x_2)+x_2f(x_3)+x_3f(x_4)+...+x_nf(x_1) \ge x_2f(x_1)+x_3f(x_2)+...+x_1f(x_n)$

I think it was discussed in http://www.mathlinks.ro/Forum/viewtopic.php?highlight=turkish&t=22923


$ A$ is a set of points on the plane, $ L$ is a line on the same plane. If $ L$ passes through one of the points in $ A$, then we call that $ L$ passes through $ A$.

(1) Prove that we can divide all the rational points into $ 100$ pairwisely non-intersecting point sets with infinity elements. If for any line on the plane, there are two rational points on it, then it passes through all the $ 100$ sets.

(2) Find the biggest integer $ r$, so that if we divide all the rational points on the plane into $ 100$ pairwisely non-intersecting point sets with infinity elements with any method, then there is at least one line that passes through $ r$ sets of the $ 100$ point sets.

Can anyone help with problem 2) ? :)
I've been stuck on it.

r=3
Consider diving a circle into 99 arc,and the rest to one set,then $r\leq 3$
And for all 4 points in different sets,there is at least one interestion point,consider this point,we can get a point with 3 points in different sets.So $3\leq r$