i am wondering if any one knows of a decent algorithm for finding the prime factors of a large (at least 12 digits) integer that can be programmed with c++.
so far the only one i can think of is seeing if $ t|n$ for $ t\leq{\sqrt n}$. this obviously is not very useful since for large numbers it will take way too long.

Efficient factoring is an open question (although much better algorithms exist than the one you are suggesting).  See http://en.wikipedia.org/wiki/Integer_factorization .


let $a,b,c$ the absissas of the vetrices of a triangle $ABC$ and a point $M$ in $x'x$ axis with absissa $k$ and we know that

$\dfrac{k(k-a)(k-b)}{c(c-a)(c-b)}+\dfrac{k(k-b)(k-c)}{a(a-b)(a-c)}+\dfrac{k(k-c)(k-a)}{b(b-a)(b-c)}=2$.

Find the min price of the sum of the distances of $M$ from the projections of $A,B,C$ at $x'x$

Now the problem is correct try to solve it

[quote="Arne"]Please clarify the problem. What is the $x'x$ axis? Etcetera.[/quote]
The axis $x'x$ is the axis of the absissas at an right-angled system of coordinates.
Do you want to clarify something else?


geometry
trapezoid
circumcircle
symmetry
rectangle
geometry proposed

A trapezium $ ABCD$ with $ AD\parallel BC$ is given and let $ P$ be, the intersection point of its diagonals $ AC,$ $ BD.$ We denote as $ E,$ $ F,$ the midpoints of the side segments $ BC,$ $ AD$ respectively and let $ E'$ be, the orthogonal projection of $ E,$ on $ AD.$ The circumcircles $ (O),$ $ (O'),$ of the triangles $ \bigtriangleup PAB,$ $ \bigtriangleup PCD$ respectively, re-intersect each other at a point so be it $ P'.$ Prove that the line segment $ E'P',$ bisects the segment $ EF.$

Kostas Vittas.

Let AB, CD meet at Q. By Ceva's theorem for centrally similar $ \triangle AQD \sim \triangle BQC,$ QP cuts BC, DA at their midpoints E, F. E' is the orthogonal projection of E to AD and similarly, let F' be the orthogonal projection of F to BC. Let U, V be the other intersections of (O) with AD, BC and let X, Y be the other intersections of (O') with BC, DA, forming another trapezoid UVXY.

$ \measuredangle AVB \mod \pi \equal{} \measuredangle APB \mod \pi \equal{} \measuredangle CPD \mod \pi \equal{} \measuredangle CXD \mod \pi$

$ \Longrightarrow$ DAVX is an isosceles trapezoid and similarly, BCYU is also an isosceles trapezoid. Their symmetry axes are the midpoint normals FF', EE' respectively. This means that E' is the midpoint of YU and F' is the midpoint of VX.

$ \measuredangle UP'P \mod \pi \plus{} \measuredangle PP'X \mod \pi \equal{} \measuredangle UAP \mod \pi \plus{} \measuredangle XCP \mod \pi \equal{} 0$

$ \Longrightarrow$ X, P', Y are collinear and similarly, U, P', V are collinear. Thus P' is the diagonal intersection of the trapezoid UVXY. Similarly as for the trapezoid ABCD, the line E'F' connecting the midpoints of its bases $ YU \parallel VX$ passes through its diagonal intersection P'. But EFF'E' is a rectangle, its diagonals cutting each other in half. Therefore, $ E'P' \equiv E'F'$ bisects the segment EF.

Thank you dear [b][size=100]yetti[/size][/b], for your interest.

For read easiness of the solution you presented (I have not in mind a different one), let me to post the relative schema and I think it is enough. Also, we can say that the angle chasing may be elementary.

Based on this result, we can solve the problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175434]Isogonality in a trapezium[/url].

Best regards, Kostas Vittas.


[quote="Ignite168"]Find all values of x such that: $9^{x+1}-3^{x+2}-3^{x}+log_{3}(3)=0$.[/quote]

[hide="hide your post"]UM

$3^{2x+2}-3^{x+2}-3^{x}=-1$
$10(3^{x})-3^{2x+2}= 1$
$\boxed{x=0}$[/hide]

M I RITE

no  [color=white]just kidding[/color] :D

[quote="#H34N1"][quote="Ignite168"]Find all values of x such that: $9^{x+1}-3^{x+2}-3^{x}+log_{3}(3)=0$.[/quote]

[hide]
That's $3^{2x+2}-3^{x+2}-3^{x}=-1\implies10(3^{x})-3^{2x+2}= 1$.
Therefore, $\fbox{x=0,-2}$. But it has to be an integer, so $x=0$.

I hope its not the same as everyone elses.[/hide][/hide][/quote]
The problem did not specify the equation has to be an integer.... and your solution looks pretty much like Treething's solution. The boxed answer is correct and I don't know why you said 'it has to be an integer, so...'


Find all positive integers $n\ge 3$ such that: $a_1a_2+a_2a_3+....+a_{n-1}a_n+a_na_1\le 0$ $\forall$ real numbers $a_1,a_2,...,a_n$ satisfy $a_1+a_2+...+a_n=0$

[quote="yptsoi"]$n=3,4$.[/quote]
Can you tell me why, yptsoi! Thank you!

easy,when $n=3,4$  It's easy to prove.
and when$n\geq 5$
let $a_1=1,a_2=1 a_4=-2$
$a_k=0,k=3,5,\cdots$
then the Inequality doesn't hold

Someone posted this problem at Algebra
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=89758]www.mathlinks.ro/Forum/viewtopic.php?t=89758[/url]


Quote:A sculpture for a math classroom is a figure composed of a 1" cube stacked upon a 3" cube
which in turn is stacked upon a 5" cube.
The entire (which includes the bottom :P ) sculpture is to be painted.
 Find the number of square inches of the surface that will be painted.[hide]

Assuming that the bottom surface will also be painted . . .

The 5" cube has six faces, each with 25 squares inches of surface.

It has 6 x 25 = 150, minus the 3x3 = 9 covered by the 3" cube,

or 141 squares inches.

The 3" cube will have five faces exposed, each with 9 square inches of surface.

It has 5 x 9 = 45, minus the 1x1 = 1 covered by the 1" cube,

or 44 squares inches.

The 1" cube will have face faces exposed, each with 1 square inch of surface.

So it has 5 square inches.

The total surface area to be painted is: 141 + 44 + 5 = 190 square inches.

[If the bottom surface is not to be painted, the answer is 165 square inches.][/hide]

A sculpture for a math classroom is a figure composed of a 1" cube stacked upon a 3" cube which in turn is stacked upon a 5" cube. The entire sculpture is to be painted. Find the number of square inches of the surface that will be painted.

(6*5^2-3^2)+(5*3^2-1)+(5)=141+44+5
                                              =190

[hide]Let's first find the surface area of all the cubes before we subtract.  The surface areas are 6,54, and 150.  That adds up to 210.  Because the 1" cube touches one squared inch of the 3" cube, one inch is subtracted from both cubes.  The 3" cube touches nine squared inches on the 5" cube so nine square inches are subtracted from both cubes.  Therefore 9+9+1+1=20 and 210-20=190sq. in.[/hide]


why 1 ^infinity is undetermined. if infinity is very large number then its answer must be one . where is the problem

[quote="mornik"]If by $ 1^{ \plus{} \infty}$ you mean $ \lim_{x\rightarrow \plus{} \infty}1^x$, this limit most certainly exists and isn't undetermined in any way: $ \lim_{x\rightarrow \plus{} \infty}1^x \equal{} 1$.

However, you're probably talking about something else, so it would be good if you made your question a bit clearer.

Also, what Null said is technically correct ($ \plus{} \infty\not\in\mathbb{R}$), that's why I'm interested in what you mean by $ 1^{ \plus{} \infty}$.[/quote]
can we say that  infinty is big number

No, of course not. Infinity is not a number. It is an element of the [url=http://en.wikipedia.org/wiki/Extended_real_number_line]extended real number line[/url], a notion useful in analysis, but it is not a number. Here's a thought: let's say that $ \infty$ is a "big number", as you say. What is, then, $ 0\cdot \infty$?

From the information $ \lim_{x \to a} f(x) \equal{} 1$ and $ \lim_{x \to a} g(x) \equal{} \plus{} \infty$ we cannot determine the value of $ \lim_{x \to a} f(x)^{g(x)}$  (it's an indeterminate form $ 1^{\plus{}\infty}$); the original poster doesn't understand why this should be the case (although it's not clear to me what the O.P. believes should be the alternative).


inequalities
advanced fields
advanced fields unsolved

Assume $ 0 < \beta < \gamma \leq 1$. Prove the interpolation inequality

\[ \parallel{}u\parallel{}_{C^{0,\gamma}(U)} \leq  \parallel{}u\parallel{}_{C^{0,\beta}(U)}^{\frac{1\minus{} \gamma}{1 \minus{} \beta}} \parallel{}u\parallel{}_{C^{0,1}(U)}^{\frac{\gamma \minus{} \beta}{1 \minus{} \beta}}\]

where \[ \parallel{}u\parallel{}_{C^{0,\gamma}} \equal{} \parallel{}u\parallel{}_{\infty} \plus{} sup_{\{x, y \; \in \; U, \;\; x \neq y\}} \frac{|u(x) \minus{}u(y)|}{|x\minus{}y|^{\gamma}}\]

First show that that result holds if you replace $ \|\cdot\|_\infty$ with $ [\cdot]$ , where $ [u]_{C^{0,t}} \equal{} \sup_{x\not \equal{} y} \frac {|u(x) \minus{} u(y)|}{|x \minus{} y|^t}$ (this is easy).  Call $ \alpha \equal{} (1 \minus{} \gamma)(1 \minus{} \beta)^{ \minus{} 1}$.

Let $ \phi_u(x) \triangleq (x \plus{} [u]_{C^{0,\beta}(U)})^\alpha(x \plus{} [u]_{C^{0,1}(U)})^{1 \minus{} \alpha} \minus{} (x \plus{} [u]_{C^{0,\gamma}(U)})$.

You've already shown that $ \phi_u(0)\geq 0$.  Differentiate $ \phi_u$, factor out $ (x \plus{} [u]_{C^{0,\beta}(U)})^\alpha(x \plus{} [u]_{C^{0,1}(U)})^{1 \minus{} \alpha}$, and what remains should be something that can be shown to be positive by Jensen's inequality.  Hence $ \phi_u(x)$ is increasing, and hence $ \phi_u(x)\geq 0$ for all $ x$.  But setting $ x$ equal to $ \|u\|_\infty$ then gives you exactly the inequality you wanted.

Using Young's inequality at the end works. Thanks :)


In the figure below, $ABCD$ and $AEFG$ are squares.

Find $\angle CPD$.

[quote="LoboSolitario"][hide]Let $l$ the length of side the square $AEFG$ and $L$ length of side the square $ABCD$

$\Delta FGP:$ Let $\angle GFP=\alpha$ and $\angle FGP=\beta$, so $\angle CPD=\alpha + \beta$.

Right triangle $FMC$ (where $DM =AG = l$ and $M \in DC$): $\tan \alpha = \frac{L-l}{L+l}$

Right triangle $DAG$: $\angle ADG = \angle FGP \Rightarrow \tan \beta = \frac{l}{L}$

Therefore

$\displaystyle{ \tan \angle CPD = \tan(\alpha + \beta) = \frac{\displaystyle \tan \alpha + \tan \beta}{ \displaystyle 1-\tan \alpha \cdot \tan \beta}= \frac{\displaystyle \frac{L-l}{L+l}+\frac{l}{L}}{1- \displaystyle \frac{L-l}{L+l}\cdot \frac{l}{L}}}$

$\tan \angle CPD = 1 \Rightarrow \angle CPD = 45^{\circ}$ .[/hide][/quote] What a clean resolution. Nice  :)


An airplane traveled from City A to City B and then from City B to City C. From City A, the airplane traveled east 700 miles and then north 800 miles to get to City B. From City B, the airplane traveled west 1,200 miles and then north 400 miles to get to city C. How many miles is City C from City A?

[hide="HINT"]
Break up the directions into component vectors with opposing directs having a change of sign and add them up, the magnitude resultant is the distance.
[/hide]
[hide="Solution"]
$AB_{x}=700 \ \ AB_{y}=800$
$BC_{x}=-1200 \ \ BC_{y}=400$
$\sqrt{ \left( AB_{x}+BC_{x}\right)^{2}+\left( AB_{y}+BC_{y}\right)^{2}}= 1300$
[/hide]


Problem 2 is very beautiful. :)
By the Cauchy-Schwartz inequality we have:
$ (a^2\plus{}\frac{1}{2}b^2\plus{}\frac{1}{3}c^2\plus{}\frac{1}{4}d^2)(1\plus{}2\plus{}3\plus{}4)\ge (a\plus{}b\plus{}c\plus{}d)^2$  (*)
Now, $ a^2\plus{}\frac{1}{2}b^2\plus{}\frac{1}{3}c^2\plus{}\frac{1}{4}d^2$
$ \equal{}\frac{1}{2}a^2\plus{}\frac{1}{6}(a^2\plus{}b^2)\plus{}\frac{1}{12}(a^2\plus{}b^2\plus{}c^2)\plus{}\frac{1}{4}(a^2\plus{}b^2\plus{}c^2\plus{}d^2)$
$ \le \frac{1}{2}*1\plus{}\frac{1}{6}*5\plus{}\frac{1}{12}*14\plus{}\frac{1}{4}*30$
$ \equal{}10$
Substituting this back into (*) obtain:
$ 10*10\ge (a\plus{}b\plus{}c\plus{}d)^2\Rightarrow a\plus{}b\plus{}c\plus{}d\le 10$ QED. :D

Mi first demonstration in english :D hope to be understood
Another way for the second:
By QM-AM we have $ 1 \equal{} \sqrt {\frac 55}\ge\sqrt {\frac {a_1^2 \plus{} \frac {a_2^2}{4} \plus{} \frac {a_2^2}{4} \plus{} \frac {a_2^2}{4} \plus{} \frac {a_2^2}{4}}{5}}$ $ \ge\frac {a_1 \plus{} \frac {a_2}{2} \plus{} \frac {a_2}{2} \plus{} \frac {a_2}{2} \plus{} \frac {a_2}{2}}{5}\rightarrow a_1 \plus{} 2a_2\le5$
By the same way we prove that $ a_1 \plus{} 2a_2 \plus{} 3a_3\le14$ and $ a_1 \plus{} 2a_2 \plus{} 3a_3 \plus{} 4a_4\le30$, and obiouvsly we have $ a_1\le1$.
Summing these four inequalities an oppurtune number of times, we have $ 12a_1 \plus{} 12a_2 \plus{} 12a_3 \plus{} 12a_4\le120$, QED


$\frac{1}{\sin 10^\circ}-\frac{\sqrt{3}}{\cos 10^\circ}= 4$
$\iff \cos 10^\circ-\sqrt{3}\sin 10^\circ = 2(2 \sin 10^\circ \cos 10^\circ)$
$\iff 2 \left ( \sin 10^\circ \times \left (-\frac{\sqrt{3}}{2}\right )+\cos 10^\circ \times \left ( \frac{1}{2}\right ) \right ) = 2(2 \sin 10^\circ \cos 10^\circ)$
$\iff 2 (\sin 10^\circ \cos 150^\circ+\cos 10^\circ \sin 150^\circ) = 2 \sin 20^\circ$
$\iff 2 \sin 160^\circ = 2 \sin 20^\circ$
$\iff 2 \sin 20^\circ = 2 \sin 20^\circ$

$\therefore \ \frac{1}{\sin 10^\circ}-\frac{\sqrt{3}}{\cos 10^\circ}= 4$


Let p a prime,let $ a_1,a_2,..,a_{p \plus{} 1}$ be distinct positive integer.Prove that there are indices $ i$ and $ j$ , $ 1\leq i < j\leq p \plus{} 1$
 such that:  $ \frac {max(a_i;a_j)}{gcd(a_i;a_j)}\geq p \plus{} 1$

"Nuber theory 1" [sic!] is not an appropriate title!

http://www.mathlinks.ro/viewtopic.php?t=135914


geometry
congruent triangles
geometry solved

Let $A,B,P_1,P_2,P_3,P_4,P_5$ and $P_6$ be eight point on the plane such that $P_i$ are on the same side of line $AB$ and triangle $ABP_i$ are all simnilar to each other. Prove that $P_1P_2P_3P_4P_5P_6$ is cyclic. For any given point $A$ and $B$ how does one construct points $P_i$.

Suppose the 3 angles of triangle $ABP_i$ are respectively x,y,z degrees.
Then there exists at least 2 equal angles in the 6 angles $AP_iB$, suppose $AP_1B$=$AP_2B$=x. We get that $AP_1B$ and $AP_2B$ are congruent. And it's easy to see that other angle at $P_i$ couldn't be equal to x, or we get 3 congruent triangles, then there is a situation of 2 $P_i$'s superposition,Contradiction. Similarly we get there are angles that $AP_3B$=$AP_4B$=y,$AP_5B$=$AP_6B$=z.
So the 6 angles $BAP_i$ satisfy that: 2 of them equal to x,y,z respectively.And for $AP_1B$=$AP_2B$=x, we have $BAP_1$ not equal to $BAP_2$,suppose $BAP_1$=y, $BAP_2$=z.Similarly suppose $BAP_3$=x,$BAP_4$=z,$BAP_5$=x,$BAP_6$=y.
Now draw a sketch by youself :blush:  the proof is obvious now.

P.S.  Sorry for my terrible English... :blush:

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=41712 .

  darij


calculus
integration
trigonometry
calculus computations

\begin{eqnarray*}
\int \sin^2 x \cos^2 x dx
& = & \int \left( \frac{\sin 2x}{2}\right)^2 dx \\
& = & \frac{1}{4} \int \frac{1- \cos 4x}{2} dx \\
& = & \frac{1}{8} \int (1- \cos 4x) dx \\
& = & \frac{1}{8} \left(x - \frac{\sin 4x}{4} \right) \\
& = & \frac{1}{32} (4x - \sin 4x)
\end{eqnarray*}

Ah, the class I'm teaching has a test on this chapter this week. This would be a perfectly fair test question, although I think I'd prefer to ask it as a definite integral - say

$ \int_0^{2\pi}\sin^2x\cos^2x\,dx.$

(In which case Kouichi Nakagawa should have no trouble producing the correct value of $ \frac{\pi}{4}.$)

There are many different methods that would work here; I don't see particularly strong reasons for preferring one of those methods over another.


Solve the equation :

$ (x^4\minus{}4x^2\plus{}5)(x^8\minus{}2x^4y^2\plus{}y^4\plus{}1)\equal{}4$

That's has an infinite amount of solutions, as it's
$ ((x^2\minus{}2)^2\plus{}1)((x^4\minus{}y^2)^2\plus{}1)\equal{}4$.

$ ((x^{2} \minus{} 2)^{2} \plus{} 1)((x^{4} \minus{} y^{2})^{2} \plus{} 1) \equal{} 4$

I must get faster at LaTeX. Most annoying.

Interesting.So can you find the solutions in set of $ Z$ ?

i guess just (x, y) = ( 1, 0) and ( -1, 0 )

LHS = product of two factors of 4: i.e $ \pm1, \pm2, \pm4$.

Go to go, but I guess someone will have beaten me to the post before this gets through.

This gives the same solutions as Nora.91's.


For $ a,b,c$ positive reals

$ \sum \frac{a}{2a^2\plus{}b^2\plus{}3} \leq \frac{1}{2}$

$ 2a^2\plus{}b^2\plus{}3\ge 2a\plus{}2ab\plus{}2\equal{}2(ab\plus{}a\plus{}1)$, the inequality turns into

$ \sum \frac{a}{ab\plus{}a\plus{}1}\le 1$, which is well known.

By Am-Gm and CS, we have:
$ LHS \leq\ \frac {1}{2}. \sum\ \frac {a}{2a \plus{} b} \leq\ \frac {1}{2}$  :)


inequalities
absolute value
algebra proposed
algebra

hi,
we consider the polynom with complex ceoficients
\[ P(x) \equal{} a_{0}x^{n} \plus{} a_{1}x^{n\minus{}1} \plus{} ... \plus{} a_{n}, a_{n} \not\equal{} 0,\]
with the property that there exists $ m \in \mathbb{N}$ s.t.
\[ |\frac{a_{m}}{a_{n}}| > \binom{n}{m}.\]
Prove that $ P(x)$ has at least a root $ \lambda$ with $ |\lambda| < 1$.

Suppose that the roots of $ P(x)$ are $ x_1,x_2,\cdots,x_n$ (maybe not distinct) and suppose that $ |x_i|\geq 1$ for all $ i\equal{}1,2,\cdots,n$
Therefore $ \left|\sum_{1\leq i_1<\cdots<i_{n\minus{}m}\leq n}\frac{1}{x_{i_1}\cdots x_{i_{n\minus{}m}}}\right|\leq 1^{n\minus{}m}\binom{n}{n\minus{}m}\equal{}\binom{n}{m}$
Since $ \displaystyle\binom{n}{m}<\left|\frac{a_m}{a_n}\right|\equal{}\left|\frac{\frac{a_m}{a_0}}{\frac{a_n}{a_0}}\right|\equal{}\left|\frac{(\minus{}1)^m\sum_{1\leq i_1<\cdots<i_m\leq n}x_{i_1}\cdots x_{i_m}}{(\minus{}1)^{n}x_1\cdots x_n}\right|$
$ \equal{}\left|(\minus{}1)^{m\minus{}n}\right|\left|\sum_{1\leq i_1<\cdots<i_{n\minus{}m}\leq n}\frac{1}{x_{i_1}\cdots x_{i_{n\minus{}m}}}\right|$
$ \leq\binom{n}{m}$ A contradiction occurs.
Therefore there must be a root that its absolute value is less than $ 1$.


Αλλιώς : Με αριθμητικό γεωμετρικό μέσο για τα πρώτα δύο .

Τότε $3^{\frac{ab}{c}}+3^{\frac{bc}{a}}\geq 2\cdot 3^{\frac{ab}{c}+\frac{bc}{a}}\geq 2\cdot 3^b$

Όμοια και για τα άλλα :D

Μπορείς επισης να θεσεις 3^ab/c=X^2, 3^bc/a=Y^2 κοκ. υπολογιζεις τα ανα δυο γινομενα τους και τοτε η ανισοτητα ειναι  προφανης.


algebra
polynomial
function
trigonometry
algebra unsolved

Let $ p_{1}(x) \equal{} p(x) \equal{} 4x^{3} \minus{} 3x$ and $ p_{n\plus{}1} \equal{} p(p_{n}(x))$ for each positive integer n. Denote by $ A(n)$ the set of all reals roots of the polynomial equation $ p_{n}(x) \equal{} x$. Show that $ A(n) \subset A(2n)$ and that each product of two elements of $ A(n)$ is the average of two elements of $ A(2n)$.

Consider $ x\geq 0$. If $ x>1$ then consider the function $ q(x)\equal{}p(x)\minus{}x\equal{}4x^3\minus{}4x$ which has $ q'(x)\equal{}4x^2\minus{}4>0$ then $ q'$ is increase on $ (1;\plus{}\infty)$ so $ q(x)>q(1)\equal{}0,\forall x>1$, or $ p(x)>x,\forall x>1$. Since, if $ x>1$, we have $ p_n(x)>p_{n\minus{}1}(x)>...>p_1(x)>x$ so $ p_n(x)\equal{}x$ has no solution on $ (1;\plus{}\infty)$.

Consider $ x<0$, similarly we obtain that $ p_n(x)$ has no solution on $ (\minus{}\infty;\minus{}1)$

For every $ x\in [\minus{}1;1]$ there exist an $ t$ such that $ x\equal{}\cos{t}$. Since, $ p_n(x)\equal{}\cos{3^n t}$, our equation is now:
$ \cos{3^n t}\equal{}\cos{t}\Leftrightarrow 3^nt\equal{}\pm t \plus{}k2\pi\Leftrightarrow t\equal{}\frac{k2\pi}{3^n\pm 1}$

So, $ A(n)\equal{}\{\cos{\frac{k2\pi}{3^n\pm 1}},k\in\mathbb{Z}\}$. Now it is so clearly that $ A(n)\subset A(2n)$.

I think the second request is not right :blush:

I can't see why do you think that the second statement is not true.....In that case something in my solution is wrong  :blush: , so, can you tell us?
Anyway, great solution for the first part!


inequalities calculus symmetry function algebra domain

Let $ x$,$ y$, and $ z$ be three positive real numbers whose sum is $ 1$. If no one of these numbers is more than twice any other, then the minimum possible value of the product $ xyz$ is

$ \textbf{(A)}\ \frac{1}{32}\qquad
\textbf{(B)}\ \frac{1}{36}\qquad
\textbf{(C)}\ \frac{4}{125}\qquad
\textbf{(D)}\ \frac{1}{127}\qquad
\textbf{(E)}\ \text{none of these}$

I'm not sure if this is correct. Feel free to tell me if not.

[hide]Let $ x \ge y \ge z$ and $ x \equal{} ay \equal{} abz, y \equal{} bz, z \equal{} z$ for $ a,b\in[1,2]$

$ x \plus{} y \plus{} z \equal{} 1 \\
z(ab \plus{} b \plus{} 1) \equal{} 1 \implies z \equal{} \frac {1}{ab \plus{} b \plus{} 1} \\
\implies x \equal{} \frac {ab}{ab \plus{} b \plus{} 1},\ y \equal{} \frac {b}{ab \plus{} b \plus{} 1} \\
\implies xyz \equal{} \frac {ab^2}{(ab \plus{} b \plus{} 1)^3}$
[/hide]

EDIT: It isn't so I'll just leave beginning so someone can continue from there.

[hide]
So we start with the equation.
$ x\plus{}y\plus{}z\equal{}1$
and also these equations, since no one number must be more than twice the other.
WLOG, we can just say x is less than or equal to y is less than or equal to z.

$ 2x\ge y$
$ 2x\ge z$

$ x\plus{}2x\plus{}2x \ge 1$
$ 5x \ge 1$
x is 1/5 when x+y+z equals one.

So $ 1/5\plus{}2/5\plus{}2/5$
and I multiplty to get the product $ 4/125$
C
[/hide]

You didn't prove that product is minimal.

[hide="Solution"]
Let $ x_my_mz_m$ denote the minimal product, and without loss of generality, let $ x_m\le y_m\le z_m$. Now we prove that $ 2x_m \equal{} z_m$ (we focus on this inequality, as the rest are weaker). Assume that $ 2x_m > z_m$. Then we can fix $ y_m$, and make new $ x_n \equal{} x_m \minus{} \epsilon$ and $ z_n \equal{} z_m \plus{} \epsilon$ that still satisfy $ 2x\ge z$ and $ x_n \plus{} y_m \plus{} z_n \equal{} 1$. The new product is $ x_ny_mz_n \equal{} (x_m \minus{} \epsilon)(z_m \plus{} \epsilon)y_m \equal{} y_m(x_mz_m \plus{} x_m\epsilon \minus{} z_m\epsilon \minus{} \epsilon^2)$
$ \le y_m(x_mz_m \plus{} z_m\epsilon \minus{} z_m\epsilon \minus{} \epsilon^2) \equal{} x_my_mz_m \minus{} y_m\epsilon^2 < x_my_mz_m$, a contradiction. Thus, $ 2x_m \equal{} z_m$.

Thus, we have $ y_m \equal{} 1 \minus{} x_m \minus{} z_m \equal{} 1 \minus{} 3x_m$, so $ x_m\le1 \minus{} 3x_m\le2x_m\implies\frac15\le x_m\le\frac14$ and $ x_my_mz_m \equal{} x_m\cdot(1 \minus{} 3x_m)\cdot2x_m \equal{} 2x_m^2 \minus{} 6x_m^3$.

I don't know how to proceed without calculus, though...with it, we can find that the relative minimum is at $ x_m \equal{} 0$ and the relative maximum is at $ x_m \equal{} \frac29$, so since $ \frac15\le\frac29\le\frac14$, the only two potential minima in the range $ \frac15\le x_m\le\frac14$ occur at $ x_m \equal{} \frac15\implies\frac4{125}$ and $ x_m \equal{} \frac14\implies\boxed{\textbf{(A)}\ \frac1{32}} > \frac4{125}$.

Edit: Eh, I wasn't really thinking. The general shape can easily be visualized.
[/hide]

[hide="Solution"]
Let $ m=x_0y_0z_0$ be the minimum value.  By symmetry, assume $ x_0 \le y_0 \le z_0$.  In fact $ z_0=2x_0$, for if $ z_0 < 2x_0$, then by decreasing $ x_0$ slightly, increasing $ z_0$ by the same amount and keeping $ y_0$ fixed, we would get new values which still meet the constraints but which have a smaller product - contradiction!  To show this contradiction formally, let $ x_1=x_0=h$ and $ z_1=z_0+h$, where $ h>0$ is so small that $ z_1 \le 2x_1$ also.  Then $ x_1$, $ y_0$, $ z_1$ also meet all the original constraints, and

\begin{align*}
x_1y_0z_1&=(x_0-h)y_0(z_0+h)\\
&=x_0y_0z_0+y_0  [ h (x_0-z_0 ) -h^2  ] < x_0y_0z_0
\end{align*}

So $ z_0=2x_0$, $ y_0=1-x_0-z_0=1-3x_0$, and $ m={2x_0}^2(1-3x_0)$.  Also, $ x_0 \le 1-3x_0 \le 2x_0$, or equivalently, $ \frac{1}{5} \le x_0 \le \frac{1}{4}$.  Thus $ m$ may be viewed as a value of the function $ f(x)=2x^2(1-3x)$ on the domain $ D=  \{ x  | \frac{1}{5} \le x \le \frac{1}{4}  \}$.  In fact, $ m$ is the [i]smallest[/i] value of $ f$ on $ D$, because minimizing $ f$ on $ D$ is just a restricted version of the original problem: for each $ x \in D$, setting $ y=1-3x$ and $ z=2x$ gives $ x,y,z$ meeting the original constraints, and makes $ f(x)=xyz$.  To minimize $ f$ on $ D$, first sketch $ f$ for all real $ x$.  Since $ f$ has a relative minimum at $ x=0$ ($ f(x)$ has the same sign as $ x^2$ for $ x<\frac{1}{3}$), and cubics have at most one relative minimum, the minimum of $ f$ on $ D$ must be at one of the endpoints.  In fact,

$ f  (\frac{1}{4}  )=\frac{1}{32} \le f  (\frac{1}{5}  )=\frac{4}{125}$, or $ \boxed{\textbf{(A)}}$.

(If $ f$ had another relative minimum between its two zeros, say at point $ x=a$, then the equation $ f(x)=f(a)$ would have at least $ 4$ roots -- draw a sketch.  But a cubic equation has at most three roots!)[/hide]

If we consider the function $ f(x,y)\equal{}xy(1\minus{}x\minus{}y)$ where the domain is the region in the $ xy$-plane that is the intersection of the graphs of

 $ y\ge x/2, \quad y\le 2x, \quad y\ge 1\minus{}3x, \quad y\le 1 \minus{} \frac{3x}2, \quad y\ge \frac{1\minus{}x}3, \quad y\le 2(1\minus{}x)/3$,

how would we find the global maximum then?

We first seek relative extrema within our region. Solve $ \left \{\begin{array}{c} \frac {\partial}{\partial x} f = 0 & \frac {\partial}{\partial y} f = 0 \end{array} \right \;$ to get $ (0,0)$, $ (0,1)$, $ (1,0)$, $ (1/3, \; 1/3)$.

Note that none of these but the last is within the desired region, and the last produces a local maximum. Thus, the minimum must lie on the boundary of the region. From here, it is a matter of minimizing the function on each of the individual line segments that make up the boundary of the region we are interested in.

[quote="math154"][hide="Solution"]
Let $ x_my_mz_m$ denote the minimal product, and without loss of generality, let $ x_m\le y_m\le z_m$. Now we prove that $ 2x_m \equal{} z_m$ (we focus on this inequality, as the rest are weaker). Assume that $ 2x_m > z_m$. Then we can fix $ y_m$, and make new $ x_n \equal{} x_m \minus{} \epsilon$ and $ z_n \equal{} z_m \plus{} \epsilon$ that still satisfy $ 2x\ge z$ and $ x_n \plus{} y_m \plus{} z_n \equal{} 1$. The new product is $ x_ny_mz_n \equal{} (x_m \minus{} \epsilon)(z_m \plus{} \epsilon)y_m \equal{} y_m(x_mz_m \plus{} x_m\epsilon \minus{} z_m\epsilon \minus{} \epsilon^2)$
$ \le y_m(x_mz_m \plus{} z_m\epsilon \minus{} z_m\epsilon \minus{} \epsilon^2) \equal{} x_my_mz_m \minus{} y_m\epsilon^2 < x_my_mz_m$, a contradiction. Thus, $ 2x_m \equal{} z_m$.[/hide][/quote]
Shouldn't we first prove it is impossible to have $ 2y_m\equal{}z_m$ since in setting $ z_n\equal{}z_m\plus{}\epsilon$, we'll have $ z_n>2y_m$? (Same as in AIME15's solution). Am I missing something?

[hide="Another Calculus Solution"]
Let $y = t_1x$ and $z = t_2x$, then $t_1,t_2 \in \left[\frac{1}{2},2 \right]$ and $x+y+z = 1 \implies x(1+t_1+t_2) = 1 \implies x = \frac{1}{1+t_1+t_2}$. Thus
\[xyz = t_1t_2x^3 = \frac{t_1t_2}{(1+t_1+t_2)^3}\]
Now, it suffices to find the minimum value of the function $f(t_1,t_2) = \frac{t_1t_2}{(1+t_1+t_2)^3}$ on the rectangle $\left[\frac{1}{2},2\right] \times \left[\frac{1}{2},2\right]$. Notice that
\[\frac{\partial f}{\partial t_1} = \frac{t_2(1+t_2-2t_1)}{(1+t_1+t_2)^3}\]
We see that for a fixed $t_2$, the function is increasing on $\left( \frac{1}{2}, \frac{1+t_2}{2} \right)$ and is decreasing on $\left( \frac{1+t_2}{2}, 2 \right)$. Therefore, it achieves it's minimum either when $t_1 = \frac{1}{2}$ or $t_1 = 2$.

[b]Case $t_1 = \frac{1}{2}$.[/b]
For $t_1 = \frac{1}{2}$, we have $\frac{\partial f}{\partial t_1} = \frac{t_2^2}{(\frac{3}{2}+t_2)^3} > 0$ and hence, it is increasing in $t_2$. Thus, it's minimum value is achieved when $t_2 = \frac{1}{2}$ and we have $f\left(\frac{1}{2},\frac{1}{2}\right) = \frac{1}{32}$.

[b]Case $t_1 = 2$[/b]
For $t_1 = 2$, we have $\frac{\partial f}{\partial t_1} = \frac{t_2(t_2-3)}{(3+t_2)^3} < 0$ for any $t_2 \in \left[\frac{1}{2},2\right]$ and hence, it is decreasing in $t_2$. Thus, it's minimum value is achieved $t_2 = 2$ and we have $f(2,2) = \frac{4}{125}$.

Since $\frac{4}{125} < \frac{1}{32}$, then the minimum value of $f$ on the rectangle $\left[\frac{1}{2},2\right] \times \left[\frac{1}{2},2\right]$ is $\boxed{\frac{4}{125}}$ which happens to be the answer of the initial problem too.
[/hide]


Assume that $a,b,c \in (0,\infty)$  and $X,Y,Z \in (0,\infty) .$
[b] Prove[/b] (perhaps well-known) [b] inequality [/b]
\[ \begin{array}{|c|} \hline \\ \displaystyle \frac{a}{b+c}X + \frac{b}{c+a}Y + \frac{c}{a+b}Z \ge \displaystyle X+Y+Z -\frac{(XY)^2 + (YZ)^2 +(ZX)^2}{2XYZ} \\ \\ \hline \end{array}\; .  \]

Any ideas to this quite nice inequality  ??

I think this yet unsolved inequality can be a way:
http://www.mathlinks.ro/Forum/viewtopic.php?p=496089#p496089


uhh where am i supposed to make the check out to?
i mean payable to?

   anddd for the early admission ones,
the payment is due on the 29th..as in Dr. Titu recieving it on the 29th through mail??

ookdsyf yeah kinda lost here..no info on where to pay the check payable to...
so a reply on this would be a LOT of help..considering the fact that i'll be gone for the entire week for nationals honor orchestra..and daddy left by himsef to figure that out;;
uh yeahh
thanks! :]

The application fee was supposed to be made out to AwesomeMath, so I see no reason why the payment for the program should be made out to anything else.

The address is, of course:

AwesomeMath
c/o Dr. Titu Andreescu
1721 Monaco Dr.
Allen, TX 75002 USA

and if you send it tomorrow you should be fine. You can send an email to Dr. Andreescu to tell him that you sent it, it might be late, and perhaps requesting him to tell you when he gets it, but I doubt he'll decide to un-admit you if he gets the check on March 1 and not February 29.

haha thanks :]
i was just unsure;

mmm yeahh excuse my stupidity please! lol


function
group theory
abstract algebra
geometry
geometric transformation
rotation
superior algebra

[b]Problem[/b]:
[i]determine total number K of essentially different colorings of a 5×5 square grid if each square can be colored black or white/left blank.
Treat two patterns as an identical one if the rotating of the grid by 90°,180°,or 270° generates identical view of the "board". [/i]
____________________________________________________
This more difficult variant of the problem posted here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=106467
_____________________________________________________

In order to solve the problem some definitions and classical theorems from group theory will be referred first:

[u]Definition 1[/u]:If $Z$ is a finite set and $G$ is a group ,we say that $G$ acts on $Z$ if there is a function $\pi : G\times Z\to Z$ marked $\pi (g,z)$ for $g\in G ; z\in Z$ such that :
1.) (gh)z = g(hz) for g,h$\in G$ and z$\in Z$
2.) 1z=z for all z$\in Z$ where 1 is the identity in $G$.

[u]Definition 2 [/u]:An orbit of $(C_{1},C_{2},...,C_{n})\in C^{n}$ we call a $(q,G)$-coloring of $Z$.

[u]Theorem 1[/u]:If group $G$ acts on set $Z$={$1,2,...,n$} and if $C$ is a set of $q$ colors,than $G$ acts on the set $C^{n}$ of all n-tuples of colors by:
$\pi (C_{1},...,C_{n})=(C_{\pi^{-1}1},...,C_{\pi^{-1}n})$ for all $\pi \in G$

[u]Theorem 2[/u]:Let $G$ be a subgroup of $S_{n}$ acting on set $Z$.
Then:
$K=\frac{1}{|G|}\sum_{\pi \in G}q^{t(\pi)}$ ,where $t(\pi)$ is the number of cycles in the complete factorization of $\pi$ as product of disjoint cycles.

______________________________________________________
Now the problem can be transferred in the framework of group theory like this:
Let $Z$ be the set containing 25 squares.Let $C$ contain 2 colors.
Since $Z$ is the set of squares one can consider the grid as a 2-coloring of $Z$.The orbit of $(C_{1},C_{2},...,C_{25})\in C^{25}$ corresponds to the 4 ways of viewing the board.Let $\pi$ be the rotation of the grid by 90°.This can be represented in cycle notation as:
$\pi =(1,5,25,21)(2,10,24,16)(3,15,23,11)(4,20,22,6)(7,9,19,17)(8,14,18,12)(13)$
Similary,the rotations by 180°,270° and 360° ,after the multiplycations are:
$\pi^{2}=(1,25)(5,21)(2,24)(10,16)(3,23)(15,11)(4,22)(20,6)(7,19)(9,17)(8,18)(14,12)(13)$
$\pi^{3}=(1,21,25,5)(2,16,24,10)(3,11,23,15)(4,6,22,20)(7,17,19,9)(8,12,18,14)(13)$
$\pi^{4}=\pi^{o}=(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14)(15)(16)(17)(18)(19)(20)(21)(22)(23)(24)(25)$
respectively.

$G=/\pi/$ acts on set $Z$ and by Theorem 1 $G$ acts on $C^{25}$-the set of all 25-tuples of colors.According to Theorem 2 the total number of grids is:$\frac{q^{t(\pi^{o})}+q^{t(\pi)}+q^{t(\pi^{2})}+q^{t(\pi^{3})}}{4}$
Since 25 cycles in the complete factorization of $\pi^{o}\Rightarrow q^{t(\pi^{o})}=2^{25}$
7 cycles in the complete factorization of $\pi\Rightarrow q^{t(\pi)}=2^{7}$
13 cycles in the complete factorization of $\pi^{2}\Rightarrow q^{t(\pi^{2})}=2^{13}$
and again 7 cycles in the complete factorization of $\pi^{3}\Rightarrow q^{t(\pi^{3})}=2^{7}$  we conclude that there are:
$K=\frac{2^{25}+2^{7}+2^{13}+2^{7}}{4}$ rotationally different patterns of a 5×5 black-white board.
Considerations of this kind can be extended to m×m grid with more than $q=2$ colors .

this is only an illustration of Polya enumeration theorem. The cycle index for this one is obviously $P(x_{1},x_{2}) = \frac{1}{4}(x_{1}^{25}+2x_{4}^{6}x_{1}+x_{2}^{12}x_{1})$, so the result falls out easily.
Using the cycle index, we can solve problems like
how many configurations have 20 black squares and 5 white squares?
How many configurations have at least 15 black squares?

If we consider white -> black, black -> white is the same configuration, then we need another group action. I think that's de brujin's theorem.

Edit: so for the problem in the link, we only need to use the cycle index polynomial. Invoking Polya's theorem, we substitute $x_{1}= 1+x$, $x_{2}= 1+x^{2}$, then substitute and read the coefficient of x^3 in the polynomial.

Except putting the emphasis on the clarity of presentation,what you see above is making use of Theorem 2.
And that one  is nothing else but a slight modification of the Theorem of Burnside.
OTOH,Polya theorem is more stronger and is generalization of Burnside theorem.
I agree that method of the cycle index of the rotation group is a powerful thing ,and cuts the process,particularly in counting problems of this sort when there's no placed restriction on the colors.


[color=darkred]
High schoolers and above, please try not to answer [img]http://www.artofproblemsolving.com/Forum/images/icons/icon3.gif[/img] problems.

EDIT: Don't answer them at all. This is an order from the mighty Mr. Ruscyk.

Also, please, please, please use the hide/spoiler tags. Otherwise I may have to ki... mffgrphgurgle... *cough* *splutter* contact an administrator to set your permission masks to one of a guest's for a while so you won't be able to post. Mwahahaha. Same with high schoolers posting here.

*innocently glances at TripleM and blahblahblah*
[/color]

[quote="rrusczyk"][b]Only middle school students should answer posts with [img]http://www.artofproblemsolving.com/Forum/images/icons/icon3.gif[/img] icons![/b][/quote]

However, I encourage elementary schoolers to try and answer the problems on this forum.

I would also add that middle schoolers who have been to State or Nationals in MATHCOUNTS, or think they have a very good chance of making it to State or Nationals this year, should leave the problems marked with [img]http://www.artofproblemsolving.com/Forum/images/icons/icon3.gif[/img] in this forum for less experienced students.  If you are one of these students, focus on those problems in The Basics, MATHCOUNTS, or Getting Started, instead.

shinwoo - make sure you don't have 'Disable Smilies in this post' marked in the options when you post.

[quote="eryaman"]Are high schoolers allowed to give hints if the posters are struggling with a specific problem?[/quote]
Please realize that multiple students will be working on challenge questions.  If one middle school student is having trouble, two middle school students will learn something when the 2nd middle school student provides the assistance.

Also, please realize that this forum was created, in part, because middle school students felt intimidated by the presence of higher level students in the forums.  To the extent possible, let's let the Middle School Forums belong to the middle school students.  OK?

We've decided to make this forum and The Basics 'middle schoolers and teachers/parents only', in the expectation that this will keep the conversation at the appropriate level.


algebra
polynomial
induction
Euler
quadratics
functional equation
binomial theorem

Let $ p(x)$ be a polynomial with real coefficients such that for all real $ x$,
\[ 2(1 \plus{} p(x)) \equal{} p(x \minus{} 1) \plus{} p(x \plus{} 1)
\] and $ p(0) \equal{} 8$, $ p(2) \equal{} 32$. Determine the value of $ p(40)$.

could someone tell me how to hide things?
a different solution is to notice that given functional equation is merely a recursion in terms of p(x). Then we could solve explicitly  for p(x) using characteristic equations.

For hiding stuff
[code]
[hide=Solution]
This is trivial.
[/hide]
[/code]
Gives

[hide="Solution"]
This is trivial.
[/hide]

[quote="Mat(h)s"][hide="solution"]We prove with induction that $ p(x) \equal{} x^2 \plus{} 10x \plus{} 8$:

$ p(1) \equal{} \frac {p(0) \plus{} p(2) \minus{} 2}{2} \equal{} 19$
So it is true for $ x \equal{} 0$, $ x \equal{} 1$ and $ x \equal{} 2$.

$ p(x \plus{} 1) \equal{} 2p(x) \minus{} p(x \minus{} 1) \plus{} 2 \\
\equal{} 2x^2 \plus{} 20x \plus{} 16 \minus{} (x \minus{} 1)^2 \minus{} 10(x \minus{} 1) \minus{} 8 \plus{} 2 \\
\equal{} x^2 \plus{} 12x \plus{} 19 \\
\equal{} (x \plus{} 1)^2 \plus{} 10(x \plus{} 1) \plus{} 8$

Indeed $ p(x) \equal{} x^2 \plus{} 10x \plus{} 8$.
So $ p(40) \equal{} 1600 \plus{} 400 \plus{} 8 \equal{} 2008$  :wink: [/hide][/quote]I don't like this solution as you haven't proven that there's only one possible $ p(x)$. You've proven that that one satisfies the conditiions, but not that it's equal to $ p(x)$. Is there any way to prove that you can't go above degree 2? Once you do that you have via the points the actual polynomial.

[hide="Edit: sketch of proof of that"]Write the polynomial in the general form, then just expand out by binomial theorem the RHS of the recurrence. Everything except the $ 1$ on the left cancels, leaving some terms on the right which are coefficients of the higher degree terms multiplied by $ x$, which must thus be $ 0$.[/hide]

I've proven that the value of $ p(x)$ equals $ x^2\plus{}10x\plus{}8$ for all positive integers $ x$. So indeed we need further argumentation if we want to prove that $ p(x)\equal{}x^2\plus{}10x\plus{}8$ for all $ x \in \mathbb{R}$. But since it is only asked to determine the value of $ p(40)$, I fail to see what's wrong with my solution (apart from not mentioning it is limited to positive integers).

Anyway, here's how I would prove the uniqueness of my p(x):
[hide]
If there would exist another polynomial $ q(x)$ which satisfies the conditions, consider the polynomial $ q(x)\minus{}p(x)$. Since for each positive integer $ p(x)\equal{}x^2\plus{}10x\plus{}8\equal{}q(x)$ we know that $ q(x)\minus{}p(x)$ is zero for each positive integer. A polynomial can't have infinitely many roots. Contradiction.
[/hide]

$ p(x\plus{}1)\minus{}2p(x)\plus{}p(x\minus{}1) \equal{}2$ which is constant. But that expression is the second difference of the polynomial, so it is a quadratic.

I tried solving for the characteristic equation, but I got something [hide="Weird"]$ p(n) = \left({4 + \frac{11}{2\sqrt{2}}\right)\left(1 + \sqrt{2}\right)^n + \left({4 - \frac{11}{2\sqrt{2}}\right)\left(1 - \sqrt{2}\right)^n}}$[/hide].

In this contest, I simply assumed the relation held after I noticed a pattern of the differences between consecutive terms (11, 13, 15, 17...) and got $ p(40) = 2008$.

Thats because you solved the characteristic equation incorrectly...

It should be $ x^2\minus{}2x\plus{}1\equal{}(x\minus{}1)^2\equal{}0$ which gives the quadratic.

Nice first post, euler_aficionado. That is the proper way to do it without guessing anything.

I assumed it was a quadratic, hence it is $ ax^2\plus{}bx\plus{}8$, subbing into the given easily gives a=1, since p(2)=32, b=10.

Let $ a_n\equal{}p(n)$ for integers $ n$. Then $ a_n$ satisfies that recurrence and we don't have to worry about whether or not that works for all x.

Of course we could let $ p(x)\equal{}a_nx^n\plus{}...$ where $ a_n\ne 0$. Then we can show that if $ n>2$, then the coefficients don't work out right.

[quote="Altheman"]Thats because you solved the characteristic equation incorrectly...

It should be $ x^2 \minus{} 2x \plus{} 1 \equal{} (x \minus{} 1)^2 \equal{} 0$ which gives the quadratic.[/quote]

Why isn't it $ x^2 \minus{} 2x \plus{} 1 \equal{} (x \minus{} 1)^2 \equal{} 2$?

I actually did it incorrectly. In this case, there is a constant term that is being added that cannot be eliminated by a substitution like $ a_n\equal{}b_n\plus{}k$ (a shift). Then the way to handle this is to say

$ 2\equal{}a_{n\plus{}1}\minus{}2a_n\plus{}a_{n\minus{}1}\equal{}a_n\minus{}2a_{n\minus{}1}\plus{}a_{n\minus{}2}$

Then the characteristic equation is $ (x\minus{}1)^3\equal{}0$, which gives you a quadratic...

$ (x\minus{}1)^2\equal{}0$ gives a linear function...


function
calculus
integration
algebra
polynomial
parameterization
real analysis

Let $f(x)$ be the unique differentiable real function satisfying \[f^{2n+1}(x)+f(x)-x = 0.\] Evaluate the integral $\int_{0}^{x}f(t)\,dt$

Try to substitute $u=f(t)$. This helps because $t=u^{2n+1}+u$; however, the answer I get is not completely explicit (it involves $f(x)$).

There's no hope of being explicit anyway; if we had an explicit answer, we could differentiate it and solve the polynomial equation $y^{2n+1}+y-x=0$ in terms of the parameter $x$. That's already very ugly for $n=1$, and impossible (with the usual functions) for $n>1$.


analytic geometry
function
algebra
polynomial
topology
Ring Theory
superior algebra

Hello,

once more I have a question trying to understand my algebra. :oops:

Let $k$ be an algebraically closed field.
Let $Y$ be an affine irreducible variety in $k^{n}$.

The coordinate ring is $A(Y)=A/I(Y)$, where $A$ is the ring of all polynomials in $n$ variables over $k$, and $I(Y)$ is the ring of all polynomials vanishing on $Y$.

Let $P$ be a point on $Y$
Define $m_{P}$ as the prime ideal within $A(Y)$ of all functions vanishing on $P$ , and let $A(Y)_{m_{P}}$ be the localization.

Let $O_{Y,P}$ be the ring of all germs : pairs $(U,f)$ where $O$ is an open subset within the topology of $Y$, containing $P$, and where $f$ is a regular function defined on $U$.

Are  $A(Y)_{m_{P}}$ and  $O_{Y,P}$ isomorphic ring? And if so, what is the isomorphism.

Sorry for the long question.  I have been trying for a long time.  I have a proof in front of me... the only problem is that it seems to give me a $U$ which explicitly excludes $P$ , rather than containing it.

So uhm, everyone with some experience is welcome to give me some hints about this, and will have my immense gratitude. :)

[quote="owk"]For any open subset $U\subseteq Y$, you have a restriction map $A(Y)\to A(U)$. This gives you a map $A(Y)\to\mathcal O_{Y,P}$. Now if some $g\in A(Y)$ does not vanish at $P$, it does not vanish in some neighbourhood $U$ of $P$, so $g|_{U}$ and, hence, $g_{P}$ is invertible. This induces a map $A(Y)_{\mathfrak m_{P}}\to\mathcal O_{Y,P}$. owk[/quote]
I'm sorry, this is too fast for me.  How is this a map from
$A(Y)_{\mathfrak m_{P}}$? :huh:

[quote="owk"]Universal property of localizations: as explained, all $g\notin\mathfrak m_{P}$ are mapped to units. owk[/quote]
No I'm sorry I still don't get it, you proceed quite fast.


[quote="orl"]Let $ a_1, a_2, \ldots, a_{2n}$ be $ 2n$ real numbers satisfying the condition $ \sum_{i \equal{} 1}^{2n \minus{} 1} (a_{i \plus{} 1} \minus{} a_i)^2 \equal{} 1$.  Find the greatest possible value of $ (a_{n \plus{} 1} \plus{} a_{n \plus{} 2} \plus{} \ldots \plus{} a_{2n}) \minus{} (a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n)$.[/quote]
I don't think this problem should be in this section.
Anyway, here is a brief hint.

[hide]$ (a_{n \plus{} 1} \plus{} a_{n \plus{} 2} \plus{} \ldots \plus{} a_{2n}) \minus{} (a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n)\equal{}\sum_{k\equal{}1}^{2n\minus{}1}{(2n\minus{}k)(a_{k\plus{}1}\minus{}a_{k})}\le \sqrt{\sum_{k\equal{}1}^{2n\minus{}1}(2n\minus{}k)^2}\sqrt{\sum_{k\equal{}1}^{2n\minus{}1}(a_{k\plus{}1}\minus{}a_{k})^2}$[/hide]

The equality you wrote in the hint doesn't agree about the coefficient of $ a_1$. In LH its -1, in in RH it's -(2n-1). I think you meant min(k,2n-k). Perhaps I'm wrong...

My solution: Denote $ x_n \equal{} a_{n \plus{} 1} \minus{} a_n$, and vector $ x \equal{} (x_n)$
The condition is $ \parallel{}x\parallel{}^2 \equal{} 1$
And the expression to maximize can be written as $ x\cdot(1,2,\cdots,n \minus{} 1,n,n \minus{} 1,\cdots,2,1)$
So, we need to take $ a \equal{} (1,2,\cdots,n \minus{} 1,n,n \minus{} 1,\cdots,2,1)$ and normalize it to 1:
After using sum of squares formula, $ \parallel{}a\parallel{}^2 \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1) \plus{} (n \minus{} 1)n(2n \minus{} 1)}{6} \equal{} \frac {n(2n^2 \plus{} 1)}{3}$

So, our vector will be $ x \equal{} \frac {a}{\parallel{}a\parallel{}}$, and $ x\cdot a \equal{} \parallel{}a\parallel{}$
I hope I didn't miscalculate anything important, but I think the idea is right.

Putting $ x_1 \equal{} a_1 ; x_2 \equal{} a_2 \minus{} a_1 ; x_3 \equal{} a_3\minus{}a_2 ; ... ; x_{2n}\equal{} a_{2n} \minus{} a_{2n\minus{}1}$
We have  $ x_2^2 \plus{} x_3^2 \plus{} .. \plus{} x_{2n}^2 \equal{} 1$

$ a_1 \equal{} x_1 ; a_2 \equal{} x_1 \plus{} x_2 ; a_3 \equal{} x_1 \plus{} x_2 \plus{} x_3 ; .. ; a_{2n} \equal{} x_1 \plus{} x_2 \plus{}..\plus{}x_{2n}$

So

$ S\equal{}(a_{n\plus{}1} \plus{} a_{n\plus{}2} \plus{}..\plus{} a_{2n} ) \minus{} ( a_1 \plus{} a_2 \plus{} ..\plus{} a_n ) \equal{} x_2 \plus{} 2x_3 \plus{} ... \plus{} (n\minus{}1)x_n \plus{} nx_{n\plus{}1} \plus{} (n\minus{}1)x_{n\plus{}2} \plus{} .. \plus{} x_{2n}$

Applying Cauchy-Swartz inequality :

$ S^2 \leq (  x_2^2 \plus{} x_3^2 \plus{} ...\plus{} x_{2n}^2 )(1^2 \plus{} 2^2 \plus{} ... \plus{} n^2 \plus{} ... \plus{} 1^2 )$

It's easy to check that the equality does hold :P

Synthesize a sequence $\{b_n\}$ by $a_1 = b_1$ and $b_i = a_i - a_{i-1}$ for $i \geq 2$. Let $S$ be the desired expression. We have $$\left(\sum b_i^2\right)(1^2+2^2+3^2+\cdots+n^2+(n-1)^2+\cdots+1^2) \geq S^2$$ by Cauchy-Schwarz, so $$S^2 \leq \frac 13(2n^3+n) \iff S \leq \sqrt{\frac 13(2n^3+n)}.$$ Equality obviously holds.


The "split selected posts" button just splits the first post that was selected, not all of them.

Bug confirmed. Will take steps to fix :)


function
trigonometry
algebra
domain
algebra unsolved

Find all functions $f: \mathbb R \to \mathbb R$ such that
\[f(x)\plus{}f(y)\plus{}f(z)\equal{}f(x)f(y)f(z)\]
happens for all reals $x,y$, and $z$.

[quote="reason"]Hi!!
find all functions:
$ f(x) \plus{} f(y) \plus{} f(z) \equal{} f(x)f(y)f(z)$

$ f: \mathbb{R} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} > \mathbb{R}$

thanx.[/quote]

$ f(z)(f(x)f(y)\minus{}1)\equal{}f(x)\plus{}f(y))$, and so :
either $ f(x)f(y)\equal{}1$ $ \forall x,y$ and so $ f(x)$ is constant
either $ \exists u,v$ such that $ f(u)f(v)\minus{}1\neq 0$ and so $ f(z)\equal{}\frac{f(u)\plus{}f(v)}{f(u)f(v)\minus{}1}$ is constant too.

So $ f(x)\equal{}a$ and $ 3a\equal{}a^3$ and so three solutions :

$ f(x)\equal{}0$
$ f(x)\equal{}\sqrt 3$
$ f(x)\equal{}\minus{}\sqrt 3$

Is it a real problem found in any contest / book / ... or just something you invented ?

this problem is for $ x\plus{}y\plus{}z\equal{}\pi$ but I edited it.

That restriction makes a very significant difference.

By removing that restriction, you allow $ z$ to be anything instead of $ \pi \minus{} (x\plus{}y)$.

With the restriction, $ f(x) \equal{} \tan(x)$ is a solution (and possibly others, I don't have time to investigate now).

BTW: use "f: \mathbb{R} \to \mathbb{R}" to produce "$ f: \mathbb{R} \to \mathbb{R}$" instead of $ \minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}>$.

I don't think it can be $ f(x) \equal{} \tan x$; the domain is all reals. We won't be able to plug in $ \frac{\pi}{2}$.

[quote="reason"]Hi!!
find all functions:
$ f(x) \plus{} f(y) \plus{} f(z) \equal{} f(x)f(y)f(z)$

$ f: \mathbb{R} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} > \mathbb{R}$

thanx.[/quote]
For $ x \equal{} y \equal{} z$ we will have: $ 3f(x) \equal{} (f(x))^3$, so $ D(f) \equal{} \{0,\sqrt{3}, \minus{} \sqrt{3}\}$.
Let $ f(\alpha ) \equal{} a$ and $ f(\beta ) \equal{} b$, and $ a\neq b$. For $ x \equal{} y \equal{} \alpha$ and $ z \equal{} \beta$:
$ 2a \plus{} b \equal{} a^2b$ (1).
And for $ x \equal{} y \equal{} \beta$ and $ z \equal{} \alpha$:
$ 2b \plus{} a \equal{} b^2a$ (2).
(1)-(2) gives $ ab \equal{} 1$. Contadiction (because $ a,b\in \{0,\sqrt{3}, \minus{} \sqrt{3}\}$).
So, $ f(x)$ - constant for all $ x$, and three constants available.


geometry
3D geometry
sphere
probability
algebra
binomial theorem

1. A cone is inscribed in a sphere of radius 2 in such a way that its base is one unit from the bottom of the sphere. What is the volume of the cone?

2. The contenders for the top positions in a basketball leagues are the Atlantics, the Buffaloes, and the Canards. It is certain that the Atlantics will end up in either first or second place, and the probability that the Atlantics will finish in a higher place than the Canards is 0.5. The probability that the Buffaloes will take second place is 0.1, and the probability that the Atlantics will take first place is 0.4. What is the probability that the Buffaloes will finish in third place?

3. Given a circle of radius 1 with an inscribed regular hexagon and a circumscribed regular hexagon, find the area between the two hexagons.

4. How many digits are there in the number $2003^5$

5. The four arcs are each a fourth of a circle of radius R. What is the enclosed area in terms of R?

(Number 5 image attached)

1.[hide]Draw a right triangle from the center of the sphere to one of the edges of the cone to the center of the base of the cone.  This triangle's hypotenuse is length 2, the leg perpidicular to the base has length 1.  This gives the radius of the cone to be $\sqrt 3$.  The height of the cone, extending the line segment perpindicular to the base of the cone, is 3 (length plus the radius).  Volume of a cone: $V=1/3\pi r^2h$, and thus $V=3\pi.$[/hide]

Rest o' the odds
[hide]3.  We see that the inner hexagon has side length 1 and that the outer has side length $2\sqrt3/3$.  Since each hexagon is composed of 6 equilateral triangles (A=s^2\sqrt3/4, we subtract the smaller hexagon's area from the larger hexagon's area.
$A=6\sqrt3/4((2\sqrt3/3)^2-1^2)=\sqrt3/2$

5.draw a square in which the circle is inscribed.  2r=s.  Therefore subtract the area of the four quarter circles from the total area of the square.
$A=(2r)^2-4(\pi r^2/4)=r^2(4-\pi)$
[/hide]

Here's my reasoning for #2:

[hide]There are only four possible situations based on the condition that Atlantic must be 1st or 2nd (ranked from 1st to 3rd, abbreviating Atlantic to 'A', etc.): ABC, ACB, BAC, CAB. It says that the probability for A to beat C is 50%; A beats be in 3 out of 4 situations, therefore there is a 50% one of these three, ABC, ACB, and BAC, will occur, and 50% chance that CAB will. Likewise, B's chance of coming in 2nd is 10%; thus, ABC has a probability of 10%. Also, A has a 40% chance of coming in 1st; Thus ACB and ABC have 40% between them; but since ABC has a 10% probability, ACB must therefore have a 30% probability. Since ACB and CAB are the only two situations in which B (Buffalo) can come in 3rd, their combined probability will be the probability that Buffalo will come in 3rd, and this is 80%. [/hide]

#4

[hide]$2003^5 = (2\cdot10^3 + 3)^5$. We can expand this using the binomial theorem, and the first term of the expansion is $2^5\cdot10^{15}$. It should be clear that the following terms will be two or three digit multiples of $10^{12}$, $10^9$, etc. and will therefore not affect the total number of digits in the number. The number of digits is the number of digits in $2^5\cdot10^{15}$, which is $16$ and our answer.   [/hide]

[hide="solution5"]
subtract the area of a full circle from square

${2R}^2 - R^2\pi$
$ = 4R^2 - R^2\pi$
$ = R^2 \left(4 - \pi \right)$[/hide]

i dont want to put 4-pi in a calculator... too lazy

I don't get the answer that is posted on #1.

 ;)

Can anyone show me a drawing?

For #4, what does probability 1.01 implying?

I'm not good with pictures, but a more detailed explanation of #1:

[hide]Basically, call the center of the sphere $O$. Let us call the bottom of the sphere $B$ and the top of the sphere $T$. Finally, let the center of the circle that is the base of the inscribed cone be $C$.

From the problem, we know that $CB=1$. Because the sphere has radius 2, we also know that $OB=2$. This means that $OC=1$. Now, we need to find the radius of the base of the cone. We draw a line from $C$ to the edge of the base of the cone that is tangent with the sphere, call this point $T$. This distance is $R$, the radius of the base, that we are looking for. Then, connect $O$ and $T$ so we have right triangle $OCT$. We know that $OC=1$ and also know that $OT=2$ because it's a radius of the sphere.

From Pythagorean Theorum, we find that $R=\sqrt{3}$. Finally, we can find the height $CT$ easily because it is merely $OC+OT$. $OC=1$ from above and $OT=2$ because it's a radius so the height $CT=3$. The volume of the cone is $\frac{1}{3}\pi r^2h=3\pi$.
[/hide]

As for #4, curl, you're right, but watch out, $10^{15}$ has 16 digits and $2^5>10$ :D

[hide="5"]Its just a square of side length 2R and 4 quarter circles of radius R

so A = R 2 (4-pi)[/hide]

[hide="Answer 1"] V = 1/3(pi)r 2 h
h = 3 , r =  :sqrt: (3)  So V = 3(pi)[/hide]

[hide="4"] The number of digits of any number can be calculated by int ( log n) + 1

In this case int ( 5log 2003) + 1 = 17[/hide]


geometry
abstract algebra
videos
calculus
linear algebra

i want:

- to dominate the november 28th competition
- to do halfway decently at the december 9th competition
- to catch up my online classes
- a russian doll (those wooden things that open up, then there's a smaller one inside, and that opens up too, etc.)
- ice cream
- someone in israel or palestine to realize that people's lives > some land claim
- PEACE ON EARTH

i want
-nothing
(holy crap that means i have no life at all)

Four Mathleague wins
Three perfect scores
Two A-plusses
And a calculator in a pear tree!

:D

[quote="13375P34K43V312"]I want...

-new cellphone, with BETTER service
-itunes gift card(s)
-Engel
-Challenging Problems in Geometry
-Xbox 360
-a life[/quote]

isn't that quite a bit to ask for?

:D

I think I already posted my wish list, but here it goes (i told my parents to get me whatever for everything else :) )

-3 intro books
-ti-89

-Jorian

-i already just got a razr
-i want money
-i want a calc
-just got that last algebra module
-my mum got me linear algebra by strang
-i want to place highly in MC(but people can't give you that!)
-i wish i'd not blanked out on the amc8(no one can give me that either)

in all i only really want two things wow i am almost with junggi
sure an itune card could be cool but i don't WANT it

-Some AMC problem books
-laptop
-a new bed my bed right now friggin pisses me off

i want money, lots lots of money from Santa.

Basically I'd be happy with a laptop. Basically.

I want ....

   - practice of AMC 10's and AIME's over break
   - Enjoy a Vacation
   - see my sisters I guess whom I haven't seen for an year maybe
   - eat food
   - enjoy the last winter break of middle school

  I never ask for anything for christmas anymore. I don't really think Christmas has to do with presents. Last year, I just enjoyed the break and the family time. This year, I don't really care what I get.
Kids think it's such an important part of Christmas, gifts that is.

there's much more to it

I wait...

Macintosh version of Ricochet
Macintosh version of Lego Star Wars the video game
Lego version of the Death Star in Star Wars

Greedy, aren't I?

[quote="joml88"]I want...

- to go back home to South Carolina  :)
-- see my friends, past teachers, and the town
- eat good, homecooked food
- a couple DVD's perhaps

Pretty much I'm just going to be glad going back home.  Hopefully I'll be able to see sunlight past 5 p.m. down there...[/quote]

That's life in Massachusetts buddy.

I want:

1.- To dominate the world
2.- A bed that makes itself :P

It's all :P

[quote="Dave Vather"]1.- To dominate the world
[/quote]
WHOA SO YOU ARE DARTH VADER AFTER ALL :o

[quote="Ignite168"]I don't celebrate Christmas.[/quote]

Let me guess.... Atheist?

Jewish

No, this message is not too small

Heh, if I could have one thing wished for in this thread, I'd choose Patricia's #4 :D.

But anyway, my wish list consists mostly of a 19-21" LCD monitor, to replace my current little 15" CRT...

my why wouldn't you it is the best holiday in the world if you are greedy and it isn't really Christian after all but if you are Christian you can celebrate it more as a Christian holiday instead of a greed holiday

I want
1. The Art and Craft of Problem Solving
2. New headphones for my ipod
3. A gift certificate to my local music dealership
4. To win.

[hide="s razhdestvom"]
Just to tell you, that means "merry christmas"in Russian

I would like:
An ipod
Less homework
a cell phone
A subscription to MAD magazine
Geometry, Tirginometry, and Calculus reference books
The Art of Problem Solving volumes 1 and 2
and the answer to life, the universe, and everything (42 aint gonna cut with me)
[/hide]

1. AoPS books
2. iPod
3. RAZR
4. laptop
5. Xbox LIVE!

my wishlist  :lol:
i might only get 1 of them for sure

I just want a TI-89 Titaniam calculater, but I am satisfied with just about anything.(It's the thought that counts!!!)

[quote="Piggypi314"][quote="Treething"]1. Rocket boosters custom-fitted for my shoes
2. Rocket boosters custom-fitted for my anus
3. Rocket boosters custom-fitted for the rest of my body

[quote="Piggypi314"]I want, in order of preference:
1. Straight A's and glowing comments
2. Global disarmament
3. A cure for AIDS
4. A week at Fry's alone with twenty thousand in cash
But who's giving?[/quote]

and you, sir/ma'am, need to get your priorities straight[/quote]

Rocket boosters custom-fitted for your anus?[/quote]

hey that was uncalled for.


IMO Shortlist
modular arithmetic
combinatorics
counting
triplets

Can two (or all three) of x,y,z be the same? I think that has to be true, otherwise, you can permute each working triple in 6 ways, but 6 doesn't divide 2007, which makes the problem trivial.

no x, y and z can't be the same.
The sets are ordered, so you can't permute them.

What do you mean by ordered triple?
Is triple $ 1,2,3$ considered to be the same as $ 3,2,1$ or different?

Another derivation of the equation $ 2007 \equal{} r^2 \minus{} rb \plus{} b^2$ using generating functions was just posted at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274884[/url].

[hide="Solution"]
We partition $\{1,2,...,n\}$ into two sets $R$ and $B$, of red numbers and blue numbers respectively.

Take $R(z) = \sum_{r\in R}z^r$ and $B(x)=\sum_{b\in B}z^b$. Then we have \[R(z)+B(z) = z+z^2+...+z^n.\]

Let $\zeta$ be a primitive $n$th root of unity, and note $\zeta^k$ is a root of $R(z)+B(z)$ for all $k$ not divisible by $n$.

The condition in the problem is equivalent to (by root of unity filter) \[\sum_{k=0}^{n-1}R(\zeta^k)^3+B(\zeta^k)^3 = 2007n.\]

All of the LHS vanishes except when we plug in $k=0$, since the $n$th roots of unity are roots of $R(z)+B(z)$, and $R(z)^3+B(z)^3 = (R(z)+B(z))(R(z)+\omega B(z))(R(z)+\omega^2B(z))$, where $\omega$ is a primitive third root of unity.

Now take $|R| = u$ and $|B| = v$. Then \[u^3+v^3 = 2007n\Longrightarrow u^2-uv+v^2=2007\] since $u+v=n$.

Looking modulo $3$ tells us that $3|u$ and $3|v$. Take $s=3u$ and $t=3v$. We now want to find nonnegative integer solutions to \[s^2-st+t^2=223.\]

The solutions turn out to be permutations of $(s,t) = (17,11)$ and $(s,t)=(17,6)$.

Then $\boxed{n = u+v=3(s+t) \in \{69,84\}.}$
[/hide]

Disappointing for a C3...

[hide]Suppose there are $b$ blue cells and $n-b$ red cells. We shall count the number of pairs $(x,y,z)$ where one of $x,y,z$ has a different color than the other two. WLOG let $x$ be red and $y,z$ be blue. Then there exist $b(n-b)$ such selections, since once we select $(x,y),$ $z$ is uniquely determined (as there exists a unique $1\leq z \leq n$ such that $x+y+z$ is a multiple of $n$). Accounting for the ordering, we get $3b(n-b)$ such selections. Now there are clearly $n^2$ pairs $(x,y,z)$ such that $n \mid x+y+z$ since for all $(x,y),$ there exists a unique $z \in [1,n]$ such that $n \mid x+y+z.$ Hence, we need $n^2 - 3b(n-b) = 2007.$

Note that $3 \mid n,$ so let $n= 3m.$ The equation transforms into $3m^2 - b(3m-b) = 669.$ Again, we get that $3 \mid b,$ so let $b= 3j.$ Our equation now becomes $m^2 - mj + j^2= 223,$ solving which is very standard. We eventually get the solutions $m \in \{23, 28\},$ so the only possibilities for $n$ are $69$ and $84.$
 [/hide]

Let $B,R$ partition $S$ such that $B$ contains only the blue elements and $R$ contains the red. Since $S$ forms a complete residue set $\mod n,$ for any ordered pair $(x,y)\in S^2$ there exists a unique $z\in S$ such that $n\mid x+y+z.$

Now we count the number of triples such that, WLOG $x\in B$ and $y,z\in R$ since the other cases are similar. If $|B|=x,$ trivially there are $x(n-x)$ such triples. We multiply by $3$ to account for permutations, and thus the problem reduces to solving $n^2-3x(n-x)=2007.$ Taking $\mod 3$ we have $3\mid n$ so let $n=3n'\implies 3n'^2-x(3n'-x)=669.$ $\mod 3$ again we have $x=3x'$ so $n'^2-x'n'+x'^2=223.$

Now since $n'\geq x,$ we have $n'^2-n'x'+x'^2=n'(n'-x')=223-x'^2$ so $x'\leq 14$ and checking cases we get $n'=23,28$ so $\boxed{n=69,84}$ are the only solutions. $\blacksquare$

tfw you can't do simple double counting but generating functions are OP :D

[quote=IMO ShortList 2007 C3] Find all positive integers $ n$ for which the numbers in the set $ S \equal{} \{1,2, \ldots,n \}$ can be colored red and blue, with the following condition being satisfied: The set $ S \times S \times S$ contains exactly $ 2007$ ordered triples $ \left(x, y, z\right)$ such that:

[b](i)[/b] the numbers $ x$, $ y$, $ z$ are of the same color,
and
[b](ii)[/b] the number $ x \plus{} y \plus{} z$ is divisible by $ n$.

[i]Author: Gerhard Wöginger, Netherlands[/i][/quote]

Answer: $\boxed{n \in \{69, 84\}}$.

Let $f$ be any formal series with complex coefficients and $g$ be an operator so that $$g(f)=\frac{1}{n} \left( \sum^{n-1}_{j=0} f(\varepsilon^j) \right),$$ where $\varepsilon$ is a primitive $n$-th root of unity. Let $R, B$ be formal series defined by $$R(x) \overset{\text{def}}{:=} \sum_{j \, \text{is red} } x^j$$ and $$B(x) \overset{\text{def}}{:=} \sum_{j \, \text{is blue}} x^j,$$ for any fixed red-blue coloring of $\{1, 2, \dots, n\}$. Set $F(x)=R(x)+B(x)=x \left(\tfrac{x^n-1}{x-1}\right)$.

Note that the desired number of triples is precisely equal to $$g\left(R^3+B^3\right)=g\left(F(F^2-3RB)\right)=g(F^3)-3g(FRB)=n^2-3r(n-r)=2007,$$ since for any $x $ red and $y$ blue, there is a unique $z \le n$ for which $n \mid x+y+z$. Solving this for $r \le n$ yields $n \in \{69, 84\}$ just as we remarked initially. $\blacksquare$

[quote=jgnr][hide]Answer: 84 and 69.

Let $ R$ and $ B$ be the sets of red and blue numbers in $ S$ respectively. We call an ordered triple $ (x,y,z)$ monochromatic $ x,y,z$ are of the same color, and bichromatic otherwise. We also call an ordered triple $ (x,y,z)$ divisible if $ x \plus{} y \plus{} z$ is divisible by $ n$. Let $ |R| \equal{} r,|B| \equal{} b$.

For all $ x,y\in S$, clearly there exists a unique $ z_{x,y}$ for which $ (x,y,z_{x,y})$ is divisible. So the number of divisible triples equals the number of pairs $ (x,y)$ where $ x,y\in S$, which is $ n^2$.

If a divisible triple $ (x,y,z)$ is bichromatic, exactly one of $ (x,y),(y,z),(z,x)$ belongs to $ R\times B$. On the other hand, if $ (x,y)\in R\times B$, then $ (x,y,z),(y,z,x),(z,x,y)$ are bichromatic triples which include $ (x,y)\in R\times B$. So the number of bichromatic divisible triples equals the number of elements in $ R\times B$, that is $ 3rb$. Therefore $ 2007 \equal{} n^2 \minus{} 3rb \equal{} (r \plus{} b)^2 \minus{} 3rb$.

Now we shall solve $ (r \plus{} b)^2 \minus{} 3rb \equal{} 2007$. Note that 3 divides $ r \plus{} b$, hence 9 divides $ (r \plus{} b)^2$, so 9 also divides $ 3rb$. One of $ r$ and $ b$ is divisible by 3. But $ r \plus{} b$ is divisible by 3, so both must be divisible by 3. Let $ r \equal{} 3m,b \equal{} 3k$. The equation becomes $ m^2 \minus{} mk \plus{} k^2 \equal{} 223$. Assume $ m\ge k$. So
\[ 892 \equal{} 4m^2 \minus{} 4mk \plus{} 4k^2 \equal{} (2k \minus{} m)^2 \plus{} 3m^2\ge 3m^2\ge 3m^2 \minus{} 3k(m \minus{} k) \equal{} 3(m^2 \minus{} mk \plus{} k^2) \equal{} 669.
\]
Thus $ 17\le m\le 15$. If $ m \equal{} 15$, $ 225 \minus{} 15k \plus{} k^2 \equal{} 223$ and $ k(k \minus{} 15) \equal{} 2$ which is impossible. If $ m \equal{} 16$, $ 256 \minus{} 16k \plus{} k^2 \equal{} 223$, which does not have integer solutions. Finally, if $ m \equal{} 17$, $ 289 \minus{} 17k \plus{} k^2 \equal{} 223$, whose solutions are $ k \equal{} 11$ and $ k \equal{} 6$.

So $ (m,k) \equal{} (17,11)$ or $ (17,6)$, and $ m \plus{} k \equal{} 28$ or $ 23$. Thus $ n \equal{} 3(m \plus{} k) \equal{} 84$ or $ 69$.[/hide][/quote]

Why isn't the number of bichromatic divisible triplets $6rb$ ?We should have six permutation of $(x,y,z)$ .

We claim that $\boxed{n \in \{69, 84\}}$. \\

Let the red numbers be $r_1, r_2, \ldots, r_k$, and the blue numbers be $b_1, b_2, \ldots, b_{n - k}$. Now, note that the coefficient of $x^i$ in the generating function
\begin{align*}
		P(x) &= \left(\sum_{i = 1}^k x^{r_i}\right)^3 + \left(\sum_{i = 1}^{n - k} x^{b_i}\right)^3
	\end{align*}
is the number of ordered triples $(x, y, z)$ such that $x + y + z = i$ and $x, y, z$ have the same color. By the roots of unity filter, the sum of the coefficients of $x^{kn}$, as $k$ ranges across the positive integers (note that $P$ has constant term $0$), is
	\begin{align*}
		S &= \frac{\sum_{i = 0}^{n - 1} P\left(\omega^n\right)}{n},
	\end{align*}
where $\omega = e^{\frac{2\pi i}{n}}$. We claim that $P\left(\omega^q\right) = 0$ for $q > 0$. Indeed, note that $x + x^2 + \cdots + x^n$ is a factor of $P(x)$. If $q > 0$, and $\gcd(q, n) = d$,
	\begin{align*}
		\omega + \omega^2 + \cdots + \omega^n &= \frac{n}{d}\left(\sum_{i = 0}^{d - 1} \omega^i\right) \\
		&= 0.
	\end{align*}
Hence, $0 \mid P\left(\omega^q\right)$ for $q > 0$, meaning that $P\left(\omega^q\right) = 0$. Therefore,
	\begin{align*}
		S &= \frac{P(1)}{n}.
	\end{align*}

We are given that $S = 2007$. Thus, we have
	\begin{align*}
		\frac{k^3 + (n - k)^3}{n} &= 2007 \\
		n^2 - 3kn + 3k^2 &= 2007.
	\end{align*}
Note that $k, n$ are positive integers. This is a quadratic in $n$. Since $n$ is an integer, its discriminant must be a perfect square, meaning that
	\begin{align*}
		4\cdot 2007 - 3k^2 &= t^2
	\end{align*}
for some nonnegative integer $t$. Hence, we have $t^2 + 3k^2 = 4 \cdot 2007$. By taking mod $3$, we find that both $t, k$ are divisible by $3$. Hence, write $t = 3u$ and $k = 3v$. The equation reduces to $u^2 + 3v^2 = 223 \cdot 4$. By exhaustive search, one may find that the positive integer pairs $(u, v)$ satisfying this equation are $(28, 6), (23, 11)$, implying that $k = 3v = 18, 33$. Solving the initial quadratic in $n$ then yields $n = 69, 84$ when $k = 18, 33$ respectively. Thus, $\boxed{n \in \{69, 84\}}$, as claimed. $\Box$

The answer is $69$ and $84$. Assume there are $k$ red numbers and thus $n-k$ blue numbers. Say a triple $(x,y,z)$ is [i]vanishing[/i] if $x+y+z$ is divisible by $n$.

[color=red][b]Claim:[/b][/color]     The number of vanishing $(x,y,z)$ containing both red and blue numbers is $3k(n-k)$.

[i]Proof.[/i]     The key is to count the number of vanishing triples with $x$ and $y$ different colors. Note that $x$ and $y$ uniquely determine $z$, so there are $2k(n-k)$ such triples. Similarly the number of vanishing triples with $y$ and $z$ different colors is $2k(n-k)$, and the number of vanishing triples with $z$ and $x$ different colors is $2k(n-k)$.

    For each such vanishing triple, among $(x,y)$, $(y,z)$, $(z,x)$ exactly two pairs have numbers with different colors, so double counting gives $3k(n-k)$ such vanishing triples. $\blacksquare$

Hence $2007=n^2-3k(n-k)$. Since $3\mid n$, let $n=3m$ for some $m$; then $669=3m^2-k(3m-k)$. Since $3\mid k$, let $k=3s$ for some $s$; then $223=m^2-3ms+3s^2$. The polynomial $3s^2-3m\cdot s+(m^2-223)$ in $s$ has perfect square discriminant $\Delta=2676-3m^2$. Since $3\mid\Delta$, let $\Delta=9t^2$. Then $892=m^2+3t^2$, and an exhaustive check gives $m\in\{23,28\}$, so $n\in\{69,84\}$, as desired.

The answers are $69$ and $84$. We claim that if there are $a$ red numbers and $b$ blue numbers, there are exactly $a^2-ab+b^2$ monochromatic triplets divisible by $n$. To do this, we use roots of unity filter. If we define $P(x)=\sum_{i\in R}x^i$, $Q(x)=\sum_{j\in B}x^j$, we want $\frac{1}{n}\sum_{i=0}^{n-1}P(\zeta^i)^3+Q(\zeta^i)^3$. However, $P+Q=\sum_{i=0}^{n-1}x^i$ divides $P(x)^3+Q(x)^3$, and it vanishes when $i\neq 0$. So, our equation becomes $\frac{1}{n}(P(1)^3+Q(1)^3)=\frac{a^3+b^3}{n}=a^2-ab+b^2$, as desired. To finish, we use quadratic formula to solve for $a$. Note that we need $b^2+4(2007-b^2)=8028-3b^2$ to be a perfect square, which is a finite case check. Checking, we see that only $b=18, 33, 51$ work, which yield $(a,b)=(18,51), (33, 51), (51, 18), (51,33)$.

[hide=solution]I claim that this is possible at $\boxed{n=69, 84}$, where you color $18$ and $33$ numbers blue, respectively. Let the generating function for blue be $F$ and $G=1+x+x^2+...+x^{n-1}$. Then the total generating function is $F^3+(G-F)^3$, and we can just sum this over the $n$th roots of unity. But notice that all the resulting terms have $G$ in it which is always zero except at $1$, thus this is actually equal to $\frac{k+(n-k)^3}{n}$ where $k$ is the number of blue numbers. (Surprisingly, this means that the number of triples does not depend on which numbers are colored blue, only how many!) We can simplify this down with mod $3$ to $k=3k_1$, $n=3n_1$, $n_1^2-3n_1k+3k^2=223$. By bounding we can see that $15\leq n_1<30$, and there are two triples that work-$(23,6), (28,11)$. Thus, multiplying back, the answers are $69$ and $84$, as desired.

Solved with [b]nukelauncher[/b].

Define generating functions for the red and blue numbers: polynomials $R(x),B(x)$ defined as $R(x)=\sum_{i \text{ red}}x^i$ and $B(x)=\sum_{i \text{ blue}}x^i$. So $R(x)+B(x)=x+\cdots+x^n=\tfrac{x(x^n-1)}{x-1}$. With $\omega$ being a primitive $n$th root of unity, roots of unity filter tells us that \[[R(\omega)^3 + \cdots + R(\omega^n)^3] + [B(\omega)^3+\cdots + B(\omega^n)^3] = 2007n.\] However, we also have \begin{align*}     \sum_{k=1}^n R(\omega^k)^3+\sum_{k=1}^n B(\omega^k)^3&=\sum_{k=1}^{n-1}\left[R(\omega^k)^3+B(\omega^k)^3\right]+(R(1)^3+B(1)^3)\\     &=\left[0+\dots+0\right] + R(1)^3+B(1)^3\\     &=2007n. \end{align*} Therefore, \[R(1)^3+B(1)^3=2007n, \quad R(1)+B(1)=n.\] This gives \[R(1)^2-nR(1)+\tfrac{n^2-2007}{3}=0.\] Since $R(1)$ is an integer, the discriminant is a perfect square. Thus, for some integer $k$, we have \[k^2=n^2-\tfrac43(n^2-2007)=2676-\tfrac13n^2.\] This finally becomes \[n^2+3k^2=8028.\]
We must have $3\mid n$, so let $n=3m$. Then $3m^2+k^2=2676$, which gives $3\mid k$ so let $k=3\ell$. This means $m^2+3\ell^2=892$. If suffices to test $\ell=1,\ldots,17$. We get $\ell=6,11,17$ which correspond to $n=84,69,15$. It remains to check if $0\le R(1)\le n$ for these choices of $n$. [list]      [*] For $n=15$, the solutions to the quadratic are $R(1)=-18,33$. These do not work.      [*] For $n=69$, the solutions are $R(1)=18,51$, which work.      [*] For $n=84$, the solutions are $R(1)=54,33$, which work. [/list] All cases checked, we conclude that the answer is $n=\boxed{69,84}$.

[hide=Remark]
The polynomials and roots of unity filter idea is very similar to [url=https://artofproblemsolving.com/community/c6h1664167p16875753]TSTST 2018/6[/url], albeit easier here.
[/hide]

[hide=Solution]
Let $f(x)=\sum_{i\text{ red}}x^i$ and let $g(x)=\sum_{i\text{ blue}}x^i$. Then, the number of ordered triples that satisfy the conditions is the sum of the $x^{ni}$ coefficients of $f(x)^3+g(x)^3$, which is equal to $\frac1n\sum_{i=0}^{n-1}(f(\omega^i)^3+g(\omega^i)^3)$. Since $f(\omega^i)+g(\omega^i)=\sum_{j=1}^n\omega^{ij}=0$ for $i\neq0$, this means that we must have $2007=\frac1n(f(1)^3+g(1)^3)$. Let $a$ be the number of red numbers and let $b$ be the number of blue numbers. Since $a+b=n$, we have $2007=\frac1n(a^3+b^3)=a^2-ab+b^2$. Therefore, we have $4a^2-4ab+4b^2=8028$. Assume without loss of generality $a\geq b$. Then, we get $(2a-b)^2+3b^2=8028$. Let $c=2a-b$. Then, $c^2+3b^2=8028$. Taking mod $3$, we find $3\mid c$, so let $c=3c_1$. Then, we get $3c_1^2+b^2=2676$. Taking mod $3$ again, we get $3\mid b$, so let $b=3b_1$. Then, we get $$c_1^2+3b_1^2=892.$$ Since $c>0$ and $a\geq b$, we must have $c_1\geq b_1$, so $4c_1^2\geq892$. Therefore, $c_1\geq15$. We must also have $c_1^2\leq892$, so $c_1\leq29$. Taking the equation mod $3$, we get $3\nmid c$. Taking mod $7$, we get $c_1^2-(2b_1)^2\equiv3\pmod7$. Since the quadratic residues mod $7$ are $0$, $1$, $4$, and $2$, this means that we must have $c_1^2\equiv0,4\pmod7$, so $c_1\equiv0,2,5\pmod7$. Therefore, we must have $c_1=16,19,23,26,28$. We have $3b_1^2=892-c_1^2$. If $c_1=16$, then $3b_1^2=892-256=636$, so $b_1^2=212$, which is impossible. If $c_1=19$, then $3b_1^2=892-361=531$, which means $b_1^2=177$, which is impossible. If $c_1=23$, then $3b_1^2=892-529=363$, so $b_1^2=121$, which means $b_1=11$. If $c_1=26$, then $3b_1^2=892-676=216$, which means $b_1^2=72$, which is impossible. If $c_1=28$, then $3b_1^2=892-784=108$, so $b_1^2=36$, which means $b_1=6$.

Therefore, we must have $b_1=11$ and $c_1=23$ or $b_1=6$ and $c_1=28$. In the first case, we have $b=33$ and $c=69$, so $n=a+b=\frac12c+\frac32b=84$. In the second case, we have $b=18$ and $c=84$, so $n=a+b=\frac12c+\frac32b=69$. Therefore, the only possible values of $n$ are $\boxed{69}$ and $\boxed{84}$.

Imagine forgetting roots of unity filter existed for the entirety of the problem. (Well now you have a solution without it.)

Let $C$ be a coloring of $S$, and let $N$ be the number of ordered triples $(x,y,z)$ satisfying the conditions in the problem. We would like all $n$ where $N=2007$.

First, notice that $3|N$ and all triples that aren't all the same number have a multiple of three number of permutations, so the number of working triples that are the same number is a multiple of 3. Thus $3|n$.

------
[b][color=#f00]Claim:[/color][/b] Suppose that a coloring $C$ has $x$ red and $y$ blue numbers, where $x\le y$. Then changing the color of a number $i$ from red to blue increases $N$ by $3(y-x+1)$.
[i]Proof.[/i] Label all blue numbers as $+3$ and label all red numbers other than $i$ as $-3$. We claim that $N$ will change by the sum of all labels, which would prove the claim. Notice that for all $j\in S$, there is exactly one $k\in S$ such that $i+j+k$ is a multiple of $n$. Now consider locally:

1. If $j,k$ are red and $n|(i+j+k)$, then we lose six ordered pairs (permutations). The sum of the two labels is $-6$ and corresponds with this. Similarly, if $j,k$ are blue and $n|(i+j+k)$, we gain six ordered pairs, and the sum of the labels of $j$ and $k$ correspond for that.
2. If $j$ is red and $n|(i+2j)$ or $n|(2i+j)|$, then we lose three ordered pairs (permutations). The label of $j$ (-3) corresponds with this. Similarly, if $j$ is blue and we gain three ordered pairs, the label for $j$ corresponds with this.
3. If one of $j,k$ is red and the other is blue, with $n|(i+j+k)$, then there is no change. The sum of the labels of $j$ and $k$ (0) correspond with this.

Since every $j\in S$ except for $i$ has exactly one $k\in S$ with $n|(i+j+k)$, every $j\in S$ except $i$ falls into exactly one of these three cases. Therefore the claim is proved. $\blacksquare$
------
Notice that there are a total of $n^2$ ordered triples (take the first two values arbitrarily and the third is fixed), when all numbers are the same color we have $N=n^2$. Therefore, by moving elements one by one to the other color we find that $N$ is only dependent on the number of numbers of each color. Moreover, one can check that:

1. When $n$ is a multiple of $6$, if the color with more numbers has $\frac{n}{2}+k$ numbers, $N=\frac{n^2}{4}+3k^2$;
2. When $n$ is $3$ modulo $6$, if the color with more numbers has $\frac{n+1}{2}+k$ numbers, $N=\frac{n^2+3}{4}+3k(k+1)$;

by showing that the claim holds in these cases and that when every number is the same color, $N=n^2$.

Now let's get the answer. When $n$ is a multiple of $6$, since our desired $N$ is a multiple of $9$, $k$ must be a multiple of $3$. Letting $n=6n'$ and $k=3k'$ and dividing both sides by $9$ gives us that we would like to solve $$n'^2+3k'^2= 223.$$ Checking gives the only solution in this case to be $n'=14, k'=3 \Rightarrow n=84$.

When $n$ is $3$ modulo $6$, letting $n=6n'+3$ and dividing both sides by $3$ gives us $$3n'^2+3n'+k(k+1)=668.$$ Checking gives the solutions to be $n'=11, k=16$, giving us $n=69$, and $n'=2, k=25$ (which gives us a contradiction as we must have $k\le \frac{n}{2})$.

Therefore the answer is $n=69, 84$. The construction is given by the $k$-value (giving us how many numbers should be red and how many should be blue).

The answer is $n=69,84$. First note that if $3 \nmid n$, there is exactly one valid pair—$(n, n, n)$—with all three values equal. If at least two values differ, then we can cyclically permute $(a,b,c)$ to $(b,c,a)$ and $(c,a,b)$. Thus if $3 \nmid n$, the number of working triples is $1 \pmod{3}$. Since $3 \mid 2007$, we have $3 \mid n$.
We now use a roots of unity filter to solve the problem. Let $R, B$ denote the red and blue colored elements of $S$ respectively; next let $R(x) = \sum_{i \in R} x^i$ and define $B(x)$ similarly. The number of admissible pairs is
$$\frac{1}{n}\sum_{i=0}^{n-1} (R(z^i)^3+B(z^i)^3),$$
where $z$ is a primitive $n$-th root of unity. Since $R(x) + B(x) \mid R(x)^3 + B(x)^3$, and $R(zi) + B(zi)$ vanishes if $i \neq 0$, it follows that the above expression is simply equal to $\tfrac{R(1)^3+B(1)^3}{n}=R(1)^2-R(1)B(1)+B(1)^2$. Since $R(1) = |R|$ and $B(1) = |B|$, the problem is reduced to finding all $n$ such that there exist $x,y$ with $x+y=n$ and $x^2-xy+y^2=2007$. WLOG, let $x>y$, and define $x-y=m>0$, so the equation becomes $3m^2+n^2=8028$, and we’re looking for possible values of $n$. We evidently require $3 \mid n$, so the equation becomes $m^2 + 3N^2 = 2676$ where $n=3N$; then $3 \mid m$, and the equation becomes $3M^2 + N^2 = 892$, where $m = 3M$. From here, a manual case check gives the solutions to be $(M,N) = (17,5),(11,23),(6,28)$. But since $N > M$, we must discard the first solution, which gives $N \in \{23, 28\} \implies n \in \{69, 84\}$ as desired. Construction is obvious. $\blacksquare$

Let $A$ be the set of red numbers, and $B$ be the set of blue numbers. We count the number of triples that are not of same color. Note that for each $(x,y)\in A\times B$, there is a unique value $z\in S$ such that $n\mid x+y+z$.

Therefore, there are $|A|\cdot |B|$ triples in the form $(A,B,A)$ or $(A,B,B)$. Similarly, there are $|A|\cdot |B|$ triples in the form $(A,A,B)$ and $(A,B,B)$ and the same amount of triples in the form $(A,A,B)$ and $(B,A,B)$. If we flip each of the colors in the previous result, we get that there are $6|A|\cdot |B|$ triples, but we count each case twice, so the complement of what we want is $3|A|\cdot |B|$. Thus, \[2007= n^2-3|A||B|\]
From here, it is easy to see that $n=69$ or $84$ are the only ones to work.

Let $R$ and $B$ be the integers colored red and blue respectively. Define \begin{align*} r(x) &= \sum_{r \in R} x^r \\ b(x) &= \sum_{b \in B} x^b \\ \end{align*} so $r(x) + b(x) = \frac{x^{n+1} - 1}{x - 1} - 1$.
Let $\omega$ be a primitive $n$th root of unity. Then it follows that \[ \sum_{i=1}^n r^3(\omega^i) + b^3(\omega^i) = 2007n. \]
Note that $r(\omega^i) + b(\omega^i)$ is $0$ whenever $i \ne n$.
As such, this just simplifies as \[ r^3(1) + b^3(1) = 2007n. \]
If $|R| = d$, then this is just \[ d^3 + (n - d)^3 = 2007n. \] which if we let $e = n - d$ simplifies as \[ d^2 - de + e^2 = 2007 \] of which we take $\pmod{3}$. By looking $\pmod{3}$ this is equivalent to the number of solutions to \[ d^2 - de + e^2 = 223 \]
It follows that the discriminant \[ 892 - 3e^2 = t^2 \] is a perfect square.
We can then exhaustively search to get that $e$ is either $6$ or $11$, which gives that $d$ is $17$, so the answers are $\boxed{69}$ and $\boxed{84}$ respectively.

Genfunc it. Let $g(x)$ be some genfunc thingy composed of the sum of some of the exponents $\{x^1, x^2, \cdots x^n\}$, representing the blue elements. WLOG that $x^n$ is not in $g(x)$.

Okay now we let function $f(x) = (g(x))^3 + ((x^1+x^2+\cdots x^n) - g(x))^3$, this creates a bunch of terms where the exponent of $x$ represents the sum of the three elements we choose in the triple. Now we use RoUF to find the terms with exponent divisible by $n$: (let $\omega$ be the primitive $n$th root of unity)
\begin{align*}
    \text{answer} &= \frac1n \sum_{i = 0}^{n-1} f(\omega^i)\\
    &= \frac1n \left(f(1) + \sum_{i = 1}^{n-1} f(\omega^i)\right)\\
    &= \frac1n \left(g(1)^3 + (n-g(1))^3 + \sum_{i = 1}^{n-1} (g(\omega^i)^3 + (0-g(\omega^i))^3)\right)\\
    &= \frac1n \left(g(1)^3 + (n-g(1))^3 + 0\right)\\
    &= 2007,
\end{align*}

so we want to find $n$ such that $x^3 + (n-x)^3 = 2007n$ has a solution $x\in [0, n-1]$. So
$$n^2 - 3nx + 3x^2 = 2007 \implies x = \frac{3n \pm \sqrt{(3n)^2 -12(n^2-2007)}}{6}$$

Thus we must have $2007\cdot 12 -3n^2 = m^2$ a perfect square. Noticing that $27 \mid 2007\cdot12$, we must have $3 \mid n, 9\mid m \implies n = 3k_1, m = 9k_2 \implies k_1^2 + 3k_2^2 = 892$. A simple trial and error now leaves us with only three possiblities: $n = 15, 69, 84$. It can be verified that $n=15$ does not give us an $x$ in the desired range, and furthermore that $69$ and $84$ both work, hence the answer is $\boxed{n \in\{69, 84\}}$. Done :)

[quote=Curious_Droid]Genfunc it[/quote]

You made my day :)

The answer is $69,84$.

Suppose for a certain coloring there are $b$ blue numbers and $r$ red numbers. Now we count the number of triples $(x,y,z)$ that aren't all of the same color (as in they are two of one color and one of the other) and $n\mid x + y + z$.  First we count the number of such triples when $x,y$ have different colors. There are in total $2 \cdot br$ ways to choose $x,y$. Notice that there is a unique $z$ with $n\mid x + y+ z$, and $(x,y,z)$ must have two of one color and one of the other. Therefore, there are $2br $ ways to choose $x,y,z$ in this case. If $y,z$ are of different colors, or if $x,z$ are of different colors, we similarly also have $2br$ ways. However, each triple has exactly two of $\{x,y\} , \{y,z\}, \{x,z\}$ having two different colors, which means that there are totally $\frac{2br + 2br + 2br}{2} = 3br $ ways to choose $(x,y,z)$ not all of the same color with $n\mid x+ y + z$.

There are $n^2$ ways to choose $x,y,z\in S$ with $n\mid x + y+ z$ because each $(x,y)$ admits exactly one possibility for $z$. Thus, there are $n^2 - 3br = r^2 - rb + b^2$ desired triples, so $r^2 - rb + b^2 = 2007$. The problem is equivalent to there existing positive integers $r+b= n$ with $r^2 - rb + b^2 = 2007$.  Considering $(r,b) = (51,18)$ and $(51,33)$, we see that $n = 69,84$ work. Now we prove that $n\in \{69,84\}$.

Now, we clearly have $3\mid n$, otherwise the number of triples couldn't be a multiple of $3$.

Claim: $3\mid r$ and $3\mid b$.
Proof: Suppose not. Then, $\frac{r^3 + b^3}{r + b} = 2007$, so $\nu_3(r^3 + b^3) - \nu_3(r + b) = 2 $. However, since $3\mid r + b$ but not $r$ or $b$, by LTE, we have $\nu_3(r^3 + b^3) = \nu_3(r + b) + 1$, contradiction. $\square$

Now, if $r = 3x$ and $b = 3y$, we have $x^2 - xy + y^2 = 223$, so $(x-y)^2 + xy = 223$. WLOG assume that $x \ge  y$.

So then $0\le x - y \le 14$. It is now casebash time. We claim that $x - y\in \{6,11\}$. If $x - y =0$, then $x^2 = 223$, absurd. If $x - y = 1$, then $xy = 222$, so $x - y \ge 37 - 6 = 31$. If $x - y = 2$, then $xy = 219$, so $x - y \ge 73 - 3  = 70$. If $x - y = 3$, then $xy = 214$, so $x - y \ge 107 - 2 \ge 105$. If $x - y = 4$, then $xy = 207$, so $x -  y \ge 23 - 9 = 14$. If $x - y = 5$, then $xy = 198$, so $ x - y \ge 18 - 11 = 7$. If $x - y = 7$, then $xy = 174$, so $x - y \ge 29 - 6 = 23$. If $x - y = 8$, then $xy = 159$, so $x - y \ge 53 - 3 = 50$. If $x - y = 9$, then $xy = 142$, so $x - y \ge 71 - 2 = 69$. If $x - y = 10$, then $xy = 123$, so $x - y \ge 41 - 3 = 38$. If $x - y = 12$, then $x y = 79$, so $x - y \ge 79 - 1 = 78$. If $x -  y = 13$, then $xy = 54$, so $x - y \in \{53,25,3\}$.  If $x - y = 14$, then $xy = 27$, so $x - y$ is equal to $26$ or $6$.

Therefore, we must have $x - y = 6$ or $x - y  = 11$. If $x - y = 6$, then $xy = 187\implies x = 17, y = 11$, so $n = 3(x + y) = 84$. If $x - y = 11$, then $xy = 102\implies x = 17, y = 6$, so $n = 3(x + y) = 69$. Thus, we must have $n\in \{69,84\}$, as desired.


Prove that there infinitely many numbers of the form $18^{m}+45^{m}+50^{m}+125^{m}$, divisible by 2006. $m\in N$

Sorry to revive old thread.

[quote="Rust"]2006=2*17*59. Expression:

$N(m)=18^{m}+45^{m}+50^{m}+125^{m}$ divisible to 2 for all m and divisible by 17 when m is odd, because

18=1(mod 17), 50=-1 (mod 17), 45=-6(mod 17) 125=6(mod 17).

$N(m)=(1+6^m)[1+(-1)^m]9mod 17)$, 50=-9(mod 59), 125=7(mod 59),45=-14 (mod 59). Therefore

$N(m)=[9^m+(-7)^m][2^m+(-1)^m]$. (7/59)=-(3/7)=1. Therefore, if m=29(mod 58) N(m) divisible by 2006.[/quote]
I don't understand the last line of your solution, from (7/59)=-(3/7) onwards.
Please could you explain it?
thanks
Marlin


In a rectangular coordinate system, the line $ 3y \equal{} x$ intersects the
line $ 2x \plus{} 5y \equal{} 11$ at point $ A$. What is the sum of the coordinates
of point $ A$?

Just substitute in for x. We have $ 11y\equal{}11$, so $ y\equal{}1$, and $ x\equal{}3$. The sum of x and y is $ \boxed{4}$.

[hide="my carefully explained answer"]this was really easy.

1.we know that there are 2 equations.

2.$ 3y = x $ and $ 2x+5y = 11 $

3.now we have 2 equations with 2 variables now we can just substitute for $ x=3y $ into the equation $ 2x+5y = 11 $

4.and we get an equation : $ 2(3y)+5y=11 $

5.simplifying this we get $ 11y=11 $  and $ y=1 $ and then we put that into $ x=3y $ and get $ x=3 $

6.[hide="this is my answer "]and we add together $ 3+1=4 $ and get 4 [/hide]
[/hide]

[hide="."][size=200][color=#0080FF][b][i][u]LPEASE RATE MY POST THANK YOU SO MUCH!!![/u][/i][/b][/color][/size][/hide]

:starwars:

may the force of AoPS rest within you!


Να βρεθούν οι ακέραιες λύσεις τις εξίσωσης

$ x^3\plus{}2y^2\plus{}4z^3\minus{}6xyz\equal{}1$

Έστω x = άρτιος τότε το αριστερό μέλλος θα είναι άρτιο ενώ το αριστερό περιττό, που είναι άτοπο
άρα πρέπει $ x \equal{} 2k \plus{} 1 < \equal{} >$
$ 8k^3 \plus{} 3k^2 \plus{} 3k \plus{} 1 \plus{} 2y^2 \plus{} 4z^3 \minus{} 12kyz \minus{} 6yz \equal{} 1 < \equal{} >$
$ 8k^3 \plus{} 2y^2 \plus{} 4z^3 \minus{} 12kyz \minus{} 6yz \equal{} \minus{} 3k^2 \minus{} 3k < \equal{} >$
$ 2(4k^3 \plus{} y^2 \plus{} 2z^3 \minus{} 6kyz \minus{} 3yz) \equal{} \minus{} 3(k^2 \plus{} k) < \equal{} >$ πολλ.2 = πολλ.3
άρα και τα δύο μέλη θα είναι πολλαπλάσια του 6
άρα θα είναι
$ 6|k^2 \plus{} k$
και
$ 6|4k^3 \plus{} y^2 \plus{} 2z^3 \minus{} 6kyz \minus{} 3yz$

[quote="maniopas"]
$ 2(4k^3 \plus{} y^2 \plus{} 2z^3 \minus{} 6kyz \minus{} 3yz) \equal{} \minus{} 3(k^2 \plus{} k) < \equal{} >$ πολλ.2 = πολλ.3
άρα και τα δύο μέλη θα ισούνται με 0
[/quote]

Μας βλέπουν ε .... :o

Sorry, το σκότωσα. Το διόρθωσα τώρα (σύλλαθο ήτανε). Αρκεί τώρα να λύσουμε το παραπάνω σύστημα.

Η αντικααταση ειναι λαθος αλλα και παλι η σκεψη σου δν με βοηθαει.....
Πολυ καλη ασκηση Σιλουαν....... :wink:

Για $ z \equal{}$ θετικός είναι:

$ x^3 \plus{} 2y^2 \plus{} 4z^3 \minus{} 6xyz \equal{} 1 < \equal{} > x^3 \minus{} 1 \equal{} 6xyz \minus{} 2y^2 \minus{} 4z^3 < \equal{} >$
$ (x \minus{} 1)(x^2 \plus{} x \plus{} 1) \equal{} 6xyz \minus{} 2y^2 \minus{} 4z^3$

αν $ x \equal{} 1$ τότε $ 6yz \minus{} 2y^2 \minus{} 4z^3 \equal{} 0$ παίρνποντας τριώνυμο ως προς $ y$ έχουμε
$ D >\equal{} 0 < \equal{} > 4z^2(9 \minus{} 8z) >\equal{} 0 < \equal{} > z <\equal{} \frac{9}{8}$ και επειδή $ z \equal{}$ θετικός
έχουμε $ z \equal{} 0$ ή $ z \equal{} 1$, δηλαδή τις λύσεις:
$ (x, y, z) \equal{} (1, 0, 0)$, $ (x, y, z) \equal{} (1, 1, 1)$, $ (x, y, z) \equal{} (1, 2, 1)$

αν δεν είναι $ x \equal{} 1$ τότε $ x^2 \plus{} x \plus{} 1 \equal{} \frac{6xyz \minus{} 2y^2 \minus{} 4z^3}{x \minus{} 1} < \equal{} >$
$ x^2 \plus{} x \plus{} 1 \equal{} 6yz \plus{} \frac{\minus{}2y^2 \minus{} 4z^3 \plus{} 6yz}{x \minus{} 1}$
και επειδή το αριστερό μέλος είναι πολυώνυμο με ακέραιους συντελεστές ως προς $ x$, πρέπει να είναι και το δεξί άρα:
$ \minus{}2y^2 \minus{} 4z^3 \plus{} 6yz \equal{} 0 < \equal{} >$ (όμοια με πάνω)
            έχουμε τις λύσεις $ (y, z) \equal{} (0, 0), (y, z) \equal{} (1, 1), (y, z) \equal{} (2, 1)$
            αν αντικαταστήσουμε τις λύσεις αυτές στην παρακάτω και επειδή το $ x \equal{}$ ακέραιος
            έχουμε ως δεκτές λύσεις μόνο αυτές που προκύπτουν για $ (y, z) \equal{} (2, 1)$, δηλαδή τις:
            $ (x, y, z) \equal{} (10, 2, 1), (x, y, z) \equal{} (\minus{}11, 2 ,1)$
και
$ 6yz \equal{} x^2 \plus{} x \plus{} 1$

Γενικά για ακέραιο z πώς βγαίνει; Πείτε μου αν έχω κάπου λάθος.

[quote="maniopas"] αν δεν είναι $ x \equal{} 1$ τότε $ x^2 \plus{} x \plus{} 1 \equal{} \frac {6xyz \minus{} 2y^2 \minus{} 4z^3}{x \minus{} 1} < \equal{} >$
$ x^2 \plus{} x \plus{} 1 \equal{} 6yz \plus{} \frac { \minus{} 2y^2 \minus{} 4z^3 \plus{} 6yz}{x \minus{} 1}$
και επειδή το αριστερό μέλος είναι πολυώνυμο με ακέραιους συντελεστές ως προς $ x$, πρέπει να είναι και το δεξί [/quote]

Αυτό το θεώρημα πρώτη φορά το βλέπω  :!:

Αν η άσκηση έβγαινε με μια διακρίνουσα ή ένα δύο αλγεβρικα τρικ δεν θα την είχαν αποκλείσει από την Shortlist με ***  :wink:

ΠΡΑΓΜΑΤΙΚΑ ΦΟΒΕΡΗ ΑΣΚΗΣΗ!!!Δεν την εχω ξαναδει και δεν ηξερα οτι ειναι shortlist.....Θα την κοιταξω πιο προσεκτικα....... :mad:

Κανεις δεν την ελυσε? :maybe:


[hide="Not sure"]So there are 4 slices of anchovies and 4 slices plain. The total is 12 dollars. So each slice is $\frac{12}{8}=\frac{4}{3}$. Jack eats 5 slices so $\frac{4}{3}\cdot5=\frac{20}{3}$. That means Billy Bob pays $\frac{12}{3}$. So Jack pays $\frac{8}{3}$ dollars more.[/hide]

Wait, I am not sure if that is correct
[hide] Since the pizza pie is 10 dollars with anchovies (and we know that the anchovies are 2 dollars), we have that a plain pizza pie is 8 dollars.  Therefore, each plain slice is worth 1 dollar.  Hence, Billy pays 3*1=3 dollars.  Take the anchovie half now.  Half of a plain pizza is 4 dollasr and he has to pay 2 dollars for the anchovies.  So, he has to pay 6 dollars for the anchovy half.  He also has to pay for 1 plain slice, which we already calculated as 1 dollar.  He must pay 6+1=7 dollars.  7 Dollars is 4 dollars more than 3 dollars.  So, 4 dollars is your answer. [/hide]

[hide="My answer (probably wrong though)."]10 / 8 = 0.8

0.8 per clean slice

2 / 4 = 0.5

0.5 more for anchovies

Jack eats:
4 anchovies slices:
4 * (0.8 + 0.5) = 5.2
1 clean:
0.8

5.2 + 0.8 = 6

Billy Bob eats 3 clean slices:
3 * 0.8 = 2.4

That means that Jack have to pay (6 - 2.4) dollars more which is 3.6 dollars.[/hide]

[quote="The QuattoMaster 6000"]Adam4444, $\frac{10}{8}=1.25\ne.8$.  Also, the plain slice is not 10/8 dollars.  10 Dollars is the pie including anchovies.  Since the anchovies cost 2 dollars, the plain pie costs 10-2=8 dollars, so each plain slice is 8/8=1 dollar.[/quote]Doh! I made some silly misstakes. I'm actually from Sweden and I think I read the question wrong too. :blush:
But thanks for explaining - I get it now.

[quote="Arvind_sn"]Captain Jack Sparrow and Billy Bob are sharing a Pizza Hut pizza. Billy Bob wants his part plain, and Jack wants his part with anchovies...now, the pizza is divided into eight equal-sized pieces. It costs two dollars to add anchovies to half the pizza, and the pizza costs ten dollars. Jack eats all of the anchovy covered pizza (half of the pizza) and one plain slice. Each pays for what he ate. How much more than Billy Bob does Jack have to pay?
 :D[/quote]
Is the pizza 10 dollars with or without the anchovies?
[hide="Assuming that the pizza is 10 dollars with the anchovies"]If you have a plain pizza, the whole pizza is worth 8 dollars. Then each piece is worth 1 dollars. Since Jack ate 5 pieces, he pays 5 dollars and Billy Bob pays 3 dollars. BUT, since we didn't add the anchovies, Jack also needs to add 2 dollars to get that Jack paid $\boxed{\$4}$ more[/hide]

[hide]The pizza costs ten dollars, and the extra anchovies two. Let's say there aren't any anchovies, so the pizza would be 8 dollars: Jack eats five slices, Billy Bob three. So Jack pays 5 dollars plus the two for the anchovies, and Billy Bob pays three: 7-3 = 4.
Jack pays $\$$4 more dollars for his share of pizza than Billy Bob.[/hide]

Is this right? What's the correct answer?

[quote="The QuattoMaster 6000"]xxxxx, the cost of the slices are different.  An anchovy slice is more than a plain slice; Bill and Jack aren't paying the average for each slice.[/quote]

woops. my fault

[hide="Clear Solution"]If the entire pizza was covered with cheese, then the cost of each [b]half-slice[/b] would be $\$4$. Since one half is covered with anchovies, the cost of the anchovy-half is $\$6$. So each [b]cheese-slice [/b]costs \[\$\frac{4}{4}=\$1\] So each [b]anchovy-slice[/b] costs \[\$\frac{6}{4}=\$1.50\] Jack eats one cheese and 4 anchovies, so he pays \[4\times\$1.50+1\times\$1=\$7\] Since the entire cost of the pizza is $\$10$, Bob pays $\$3$. So Jack pays $\$7-\$3=\$4$ dollars more.[/hide]


If a test asks you to show that there are $ x$ number of solutions to an equation or condition or something, could you simply list the $ x$ solutions? For one of the problems I know exactly how to show that there are a lot of solutions, but I don't know how to prove on paper that there are that many. Would it be acceptable to not show a rigorous proof for it, or would that look bad?

Listing the solutions shows that there are [i]at least[/i] $ x$ of them.  Your task is to show that there are [i]exactly[/i] $ x$ of them, which implies proving that there aren't any more.

the question states to show that at least $ x$ exist, and then asks if the student thinks there are more than that, and why or why not.

by the way, x is a specific number. For example, the problem could say to show that at least 4 exist.

Whichever way you answer, you need to prove that your answer is correct.