limit
function
calculus
calculus computations

I am looking for a proof of the following limit theorems

$lim_{x\rightarrow a}[f(x)]^{n}= [lim_{x \rightarrow a}f(x)]^{n}$

[quote="FieryHydra"]I think this is a direct consequence of the following: $\lim_{x \rightarrow a}f(x) \cdot g(x) = \lim_{x \rightarrow a}f(x) \cdot \lim_{x \rightarrow a}g(x).$

Also, why did you post the same question three times?[/quote]

I don't know. I just posted once.

[quote="Xevarion"][quote="FieryHydra"]I think this is a direct consequence of the following: $\lim_{x \rightarrow a}f(x) \cdot g(x) = \lim_{x \rightarrow a}f(x) \cdot \lim_{x \rightarrow a}g(x).$[/quote]
What if $n$ is not an integer?[/quote] Then, of course, we need to do some more work.

But the function $g: \mathbb{R}\to \mathbb{R}, x \to g(x): =x^{n}$ is continuous! Hence $\lim_{x \to a}[f(x)]^{n}= \lim_{x \to a}g(f(x)) = g(\lim_{x \to a}f(x)) = [\lim_{x \to a}f(x)]^{n}$

Please, please, with questions of this sort include the words "provided all expressions in the question exist."  (Or some other condition to make the question actually true.)  As stated, the theorem (as well as fieryhydra's) are just wrong: let $f$ be any function which takes only the values 1 and -1 and takes each infinitely often near $a$, and let $n = 2$.  Then the left hand side is equal to 1, while the right hand side does not exist.

Once you have that $\lim f(x)$ exists, you can apply Pi-J's argument.  In the cases in which $n$ is not a rational number with odd denominator we need some conditions on the behavior of $f$ near $a$, as well.

JBL, thanks for pointing out the errors. I must admit I was very sloppy with my reasoning as well as writing, and that I need to be more careful from now on.


LaTeX
quadratics
algebra unsolved
algebra

Hello, I am new, so I'm still getting used to how this thing works... anyway I have this question that I have been trying to solve for quite some time, but I can't do it. I REAAAAAALLLY need help:

Romania TST 2001:

#3. Let n be an even integer greater than 2, and let "a" and "b" be real numbers with the relationship b^n = 3a +1. Prove that:

 p(x)=(x^2 +x +1)^n - x^n -a is divisible by q(x)= x^3 + x^2 + x + b iff b=1.

Can someone please help me?[/url]

I believe that something is wrong...
If $b=1$, then $a=0$.
Let $n=3$.  Then, $p(x)=x^6+3x^5+6x^4+6x^3+3x+1$
Note that $q(-1)=-1+1-1+1=0$, but $p(-1)=1-3+6-6+3+1=2$.
So, $p(x)$ is indivisible by $q(x)$.
[color=red]EDIT:Sorry, $3$ is not even.[/color]

$n$ is even, i think it's a hard problem.  The converse isn't bad i don't think, but I'm having problems with the forward direction.

Who can edit it by LATEX?
It is hard to read now.

[quote="lordWings"]Let $n$ be an [b]even[/b] integer greater than $2$,
and let $a$ and $b$ be real numbers with the relationship $b^n = 3a +1$.

Prove that:
$p(x)=(x^2 +x +1)^n - x^n -a$
is divisible by
$q(x)= x^3 + x^2 + x + b$
[b]iff[/b] $b=1$.[/quote]
Thanks.I am so stupy that I forgot we can post a new post to write the problem by LATEX. :blush:

[quote="sunchips"][quote="Hawk Tiger"][quote="sunchips"]THe "if" part is easy: by substituting roots of unity into $p(x)$. But I can't prove "only if". But If you let $z$ be a root of $q(x)$, and sub $z$ into $p(x)$, and let $z^2+z+1=-b/z$, then you get a quadratic in $z^n$. I don't know how to go from there  :([/quote]
Yes,you are right.My solution:
Firstly,$q(0)|p(0)$ then $b|1-a$ then $b|b-4$ so $b=1$ or$b=4$
if $b=4$then
we got $x_0^2n+\frac{4^n-1}{3}x_0^n-4^n=0$where $x_o$is the root of the equation$x^3+x^2+x+4=0$
let $t(x)=x^2n+\frac{4^n-1}{3}x^n-4^n=0$
then we can check that $q(x)$have a real root$x_1$ and two compelx roots$x_2,x_3$,and $-2<x_1<-1$
$|x_1|\not=|x_2|=|x_3|$
and $t(x)=(x^n-\alpha)(x^n-\beta)$where$\alpha<0<\beta$
so $|x_1|^n=\beta,|x_2|^n=|x_3|^n=|\alpha|$
we know $|\alpha \beta|=4^n$ and $(|x_1||x_2||x_3|)^n=4^n$
then we get$|\alpha|=1$ which is a contraduction.[/quote]

$a$ and $b$ may not be integers, thus we don't need to have only $b=1$ or $b=4$. However, I think that Hawk Tiger's solution can be generalized for any $a$ and $b$ with $b^n=3a+1$, so I think he has solved this problem. :)[/quote]
Thank you. I didn't  saw $a,b$be real numbers. :) what a coincidence! I didn't saw the condition clearly but use the correct method. :)


[quote="grafitti123"]pls delete. sorry[/quote] neverrrrrrrrrrrr


I seen this problem some time ago and I solved it  :ninja: , I thought it was really nice. (I hope I got the problem stated correctly  :maybe: ).

[i]Let $ f(x)$ be in $ \mathbb{Z}[x]$ with $ \deg f(x) \leq n$. Suppose that $ |f(x)| < n$ for all $ |x|\leq n^2$. Prove that $ f(x)$ is a constant polynomial.[/i]

[hide]
$ f(x)$ is an integer whenever $ x$ is an integer.  There are $ 2n^2\minus{}1$ integers $ x$ with $ |x|<n^2$, and $ 2n\minus{}1$ integers $ f(x)$ with $ |f(x)|<n$.  $ 2n^2\minus{}1>(2n\minus{}1)n$, so by Pigeonhole,  $ f(x)\equal{}k$, for some integer $ k$ with $ |k|<n$, for at least $ n\plus{}1$ integers $ x$ with $ |x|<n^2$.  Then $ f(x)\minus{}k$, at most an $ n$th degree polynomial, has at least $ n\plus{}1$ roots.  So $ f(x)\minus{}k$ is constant, and thus $ f(x)$ is constant.
[/hide]

in fact, $ f(x)\minus{}k\equiv 0$, so $ f(x)\equiv k$.


USAMTS AMC USA(J)MO USAMO LaTeX AIME inequalities

I think it's allowed.
http://usamts.org/Rules/U_RulesComplete.php
Rule 9 says:
Except where otherwise noted in problems, participants may use calculators and computers to solve problems. In this case, the participant must explain the program that was used to solve the problem, and must prove that the method used in the program does solve the problem. Symbolic manipulation calculators and mathematical programs such as Mathematica, Maple, and Matlab are considered programs for this purpose. Therefore, if a key step of a participant's solution uses these tools, the participant must demonstrate that the tool completed the step correctly.

(I wouldn't recommend it though.)

[quote="jhangil"]I think I heard that we could code programs to help us solve problems in USAMTS, but would it be considered a complete solution if we just have the code of the program and an explanation of why the programs solves the problem (or maybe even parts of the problem)?[/quote]

Since the USAMTS will post a solution after the round is done. We can be sure there're ways of solving any of them without using program.

More elegant solutions worth more points. (Would a solution with programming be more elegant than pure math computation or whatever you want to call it?)

Also recognize that can you fully justify you program, what if you make a mistake while making it, can you prove that the program is properly written?

lol, no offense, but it is quite ironic that a person with the username of Brut3Forc3 doesn't recommend using programming to brute-force a problem! :)

However, @jhangil, programming is completely fine as long as the solution is complete (unless if they changed the rules for this year), because when I did it last year, I got all five points for it (though my grader did say never to use Java ever again for problems like this...).  Back then, when they had commendations for solutions, programming wouldn't get you a commended solution, but if you're desperate for credit, go ahead.

[quote="moplam"]More elegant solutions worth more points. (Would a solution with programming be more elegant than pure math computation or whatever you want to call it?)[/quote]

There is a cap at 5, which is where you have a complete solution.

Thanks guys, I would personaly go for a mathematical probkemsolving approach, but some problems I cant think of a way to do it at the moment but seem possible with programming, so I was thinking I could answer using a program and later see if I can use a different approach when I know the answer.

[quote="dysfunctionalequations"][quote="moplam"]More elegant solutions worth more points.[/quote]

This is a false statement. Solutions are graded on mathematical correctness.[/quote]

I don't know about that, Sorry.

But USAMO mention that more elegant solution may serve as tiebreak so I mess the rules up.
It's my first USAMTS so I don't know the rule that well.

I just consider USAMTS a good practice for LaTeX

[quote="cyberspace"]There aren't any tiebreaks in USAMTS, they just rank everyone by score.

Also, I remember a problem maybe two years ago that I used a program on. It calculated exactly 858 cases to prove that none of them satisfied some condition, and I included the output table in my submission and explained what the program did. I ended up getting a 5 on it, so they definitely accept such solutions.[/quote]

I know there're no tiebreaker for USAMTS, I was talking about USAMO.

858 cases=.= seriously, I personally won't accept myself coming up with solutions for a question with so many cases when I'm so sure there's a nice solution.

If you happen to look a lot more in deep at the cases, I'm sure you can simplify the solution.

I don't like USAMTS anymore. People that use program to do the problem can get the same point as people that actually work out the problem=.=

[quote="Thunder365"][quote="moplam"]
I don't like USAMTS anymore. People that use program to do the problem can get the same point as people that actually work out the problem=.=[/quote]

Hmmmm... well I personally like USAMTS a lot.

See, Its all subjective. Some people may find the 1 month time limit a bit much, while I find it great. USAMTS doesnt stress time, and I like that.
Using your logic, I could say that competition with calculators allowed are dumb because you just punch in numbers and you get the answer. But the thing is, it's part of the rules. If you dont know how to use a calculator then its your loss, because the use of calculators is allowed. Similarly, if you arent great at programming- a valuable skill in my opinion- then its your loss.

But then again, we're all allowed our own opinions.[/quote]

USAMTS is a method of picking AIME qualifier but programmer not what the USA Math Team is looking for. I understand that programming is a valuable skill and I use programs to help me too, but I don't think people should stop at that point. They should go farther and find better solution.

[quote="NuncChaos"]Well, if you have IMO Us representative potential, you'd probably try to make it through AMC.[/quote]
You have the point.

[quote="cyberspace"][quote="moplam"]
858 cases=.= seriously, I personally won't accept myself coming up with solutions for a question with so many cases when I'm so sure there's a nice solution.

If you happen to look a lot more in deep at the cases, I'm sure you can simplify the solution.

I don't like USAMTS anymore. People that use program to do the problem can get the same point as people that actually work out the problem=.=[/quote]

Well, as it turned out, the day after I submitted that round I found out that if I just subtracted a 1 from both sides of an inequality it eliminated every single one of my 858 cases :P . I was just pressed for time though...[/quote]

lol, If you spend the time making the program on looking at the inequality, you maybe able to come up with the elegant solution. USAMTS gives enough time for you to investigate. A round usually only take me 2 hrs to solve. I have a lot extra time to come up with more elegant solutions.

Well yeah, now they take me only around 2 hours to solve. Remember I was only in 8th grade when I did the program so I wasn't exactly the greatest problem solver  :wink:  (I guess it was 3 years ago then not 2 but whatever...)

[quote="Thunder365"][quote="Differ"][quote="NuncChaos"]Well, if you have IMO Us representative potential, you'd probably try to make it through AMC.[/quote]

Except for those people who are good at proofs and not at solving short problems.[/quote]

But then you wouldnt get through AIME.

This is getting incredibly off topic  :rotfl:[/quote]

I agree.

PS: USAMO is just so much more difficult than USAMTS.

If you qualify for the AIME only through the USAMTS, it will be much more difficult to qualify for the USAMO. They use 100 as your AIME score to calculate your index, which means you are forced to score in the double digits to have a shot (which would probably address concerns about having to go through the AMC, because that route is definitely easier).

Also, about the computer solutions. I think that if it is completely correct, then you get a 5. However, if you make a mistake (not completely rigorous or wrong answer), it is much more difficult to get partial credit. What has worked in the past (for me) is using a program to brute force the answer, and then presenting a mathematical solution where you pull the answer out of your butt and show it's correct.


analytic geometry
algebra proposed
algebra

Let $a{}_i{}_j$, $i = 1, 2, 3; j = 1, 2, 3$ be real numbers such that $a{}_i{}_j$ is positive for $i = j$ and negative for $i \neq j$.
Prove that there exist positive real numbers $c_1, c_2, c_3$ such that the numbers
$  a{}_1{}_1c{}_1 + a{}_1{}_2c{}_2 + a{}_1{}_3c{}_3 \qquad a{}_2{}_1c{}_1 + a{}_2{}_2c{}_2 + a{}_2{}_3c{}_3  \qquad    a{}_3{}_1c{}_1 + a{}_3{}_2c{}_2 + a{}_3{}_3c{}_3 \end$
are all negative, all positive, or all zero.

I'm pretty sure you can find it around here somewhere.

We first find a solution for the system consisting of the following equations: $a_{11}x_1+a_{12}x_2+a_{13}x_3=0$ and $a_{21}x1+a_{22}x_2+a_{23}x_3=0$. We can give $x_3$ a positive value, and find $x_1,x_2$. We'l find that they are also positive.

Now, if $a_{31}x_1+\ldots=0$ as well, then we're done. Otherwise, if $a_{31}x_1+\ldots>0$, for example, then we can modify each $x_i$ a tiny bit so that $a_{11}x_1+\ldots>0, a_{21}x_1+\ldots>0, a_{31}x_1+\ldots>0$, and $x_i$ all remain positive.
It's not very rigorous, but I think it's it's pretty clear what I'm trying to say :).

Nice solution!  ive heard about another solution using O(0, 0, 0), P($a{}_1{}_1$, $a{}_2{}_1$, $a{}_3{}_1$), Q($a{}_1{}_2$, $a{}_2{}_2$, $a{}_3{}_2$), R($a{}_1{}_3$, $a{}_2{}_3$, $a{}_3{}_3$) in the three dimensional Euclidean space. then, its enough to find a point in the interior of the triangle PQR whose coordinates are all positive, all negative, or all zero. (but i didnt understand why its enough to find this point !!!!), can anyone help ?????!

[quote="metafisist"] ive heard about another solution using O(0, 0, 0), P($a{}_1{}_1$, $a{}_2{}_1$, $a{}_3{}_1$), Q($a{}_1{}_2$, $a{}_2{}_2$, $a{}_3{}_2$), R($a{}_1{}_3$, $a{}_2{}_3$, $a{}_3{}_3$) in the three dimensional Euclidean space. then, its enough to find a point in the interior of the triangle PQR whose coordinates are all positive, all negative, or all zero. (but i didnt understand why its enough to find this point !!!!)[/quote]

Hi man,  :lol:
I think if we found a point in the interior of triangle $PQR$, we are done.
just look at the projections of $O,P,Q,R$ on the $xy$ plane.


this inequality appeared in the Costarrican Math Olympiad (OLCOMA), and I think it's a shame since it's obvious that is was taken from another country olympiad or international competition shortlist. anyway it's very nice ... it seems hard but it's not  ;) i'm sorry if discussed before

Let $a,b,c$ be the sidelengths of the triangle $ABC$. Prove that
$\frac{3(a^4+b^4+c^4)}{(a^2+b^2+c^2)^2}+\frac{ab+ac+bc}{a^2+b^2+c^2}\geq 2$

solution:
[hide]
clearing the denominators it shows that the inequality is equivalent to proving that
$\sum_{cyc}a^4+a^3b+a^3c+a^2bc\geq 4\sum_{cyc}b^2c^2$, but by schur's inequality for $r=2$ and am-gm we have that
$\sum_{cyc}a^4+a^3b+a^3c+a^2bc\geq 2\sum_{cyc}b^3c+bc^3\geq 4\sum_{cyc}b^2c^2$, and we're done.
unfortunately this solution ocurred to me the next day after the test when going to take the breakfast...the day before i gave a 3 pages solution   :P
[/hide]

Good solutiopn but you don't use that $a,b,c$ are leghths of triangle?

$\frac{3(a^4+b^4+c^4)}{(a^2+b^2+c^2)^2}\geq 1$, but $\frac{ab+ac+bc}{a^2+b^2+c^2}\leq 1$

How could I procede without the use of sums?

 :?:

This is from Crux :
http://www.mathlinks.ro/Forum/viewtopic.php?t=14618 . :)


geometry
incenter
ratio
circumcircle
angle bisector
projective geometry
geometry proposed

Hello!

Let $ ABCD$ be a quadrilateral circumscribed around a circle $ \Gamma$, $ \{V\}=AC\cap BD$ and $ I$ is the incenter
of the triangle $ BCD$. If $ \{U\}=(IV\cap\Gamma$ then prove that $ UI$ is the bisector of the angle $ \angle BUD$.

Best regards, Petrisor Neagoe :)

Let $ (I), (J)$ be incircles of $ \triangle BCD, \triangle ABD$ and $ (I_c), (J_a)$ their excircles against $ C, A,$ respectively. Let $ (I), (I_c)$ touch $ BD$ at $ E, F.$ Since $ AB \plus{} CD \equal{} BC \plus{} DA$ $ \Longrightarrow$ $ FB \equal{} DE \equal{} CD \plus{} BD \minus{} BC \equal{} DA \plus{} BD \minus{} AB$ $ \Longrightarrow$ $ (J), (J_a)$ also touch $ BD$ at $ E, F.$ $ C$ is external similarity center of $ (I), \Gamma, (I_c)$ and $ A$ is external similarity center of $ (J), \Gamma, (J_a)$ $ \Longrightarrow$ $ CF, AE$ intersect at $ K \in \Gamma$ and $ CE, AF$ intersect at $ L \in \Gamma,$ so that $ KL \perp BD.$ $ C$ is external similarity center of $ (I), \Gamma$ and $ A$ is external similarity center of $ (J_a), \Gamma$ $ \Longrightarrow$ the external similarity center of $ (I), (J_a)$ is $ V \equiv AC \cap BD.$ $ A$ is external similarity center of $ (J), (J_a)$ and $ E$ is internal similarity center of $ (I), (J)$ $ \Longrightarrow$ the internal similarity center of $ (I), (J_a)$ is $ K \equiv CF \cap AE.$ Therefore, $ V, I, K, J_a$ are collinear and likewise, $ V, J, L, I_c$ are collinear. If $ X \equiv AB \cap CD, Y \equiv BC \cap DA,$ then $ P \equiv XY \cap BD$ is pole of the diagonal $ AC$ WRT $ \Gamma$ and the cross ratio $ (P, V, B, D) \equal{} \minus{} 1$ is harmonic. If $ PK, PL$ cut $ \Gamma$ again at $ T, U,$ the intersection $ KU \cap LT$ is on the polar $ AC$ of $ P$ and on $ BD$ as orthocenter of $ \triangle PKL,$ hence $ V \equiv KU \cap LT.$ The pencil $ U(P, V, B, D)$ is harmonic and $ UP \perp UV$ $ \Longrightarrow$ line $ UV \equiv UK \equiv UI$ bisects $ \angle BUD.$

I had another approach, but I have not in mind any proof of the below [b][size=100]Lemma 1[/size][/b], I thought as the key for the solution of this nice problem.

[b][size=100][color=DarkBlue]LEMMA 1. - Let $ ABCD$ be, an arbitrary harmonic quadrilateral with circumcircle $ (O)$ and we denote as $ P,\ Q,$ the points of intersections of the line segment $ BD,$ from the angle bisectors of $ \angle A$ $ ($ $ Q,$ between $ B,\ D$ and $ B,$ between $ P,\ Q$ $ ).$ Through the point $ P$ we draw an arbitrary line which intersects the circle $ (O),$ at points so be they $ E,\ F$ $ ($ in to the same side of $ BD,$ as vertex $ A$ $ ).$ Prove that the line segment $ DQ,$ passes through the incenter of the triangle formed by the tangent lines of $ (O)$ at points $ E,\ F$ and the line segment $ BD.$[/color][/size][/b]

Probably, this is a well known property in theory of harmonic quadrilaterals but, it is a new result to me.

I have now to read your solution dear [b][size=100]Vladimir[/size][/b], of this interesting problem of [b][size=100]Petry[/size][/b].

Thank both of you and best regards, Kostas Vittas.

Dear yetti and vittasko, thank you for your interest!

My solution:

$ O$ is the center of the circle $ \Gamma$, $ \left\{M\right\}\equal{}BC\cap\Gamma$ and $ \left\{N\right\}\equal{}CD\cap\Gamma$
$ ABCD$ is circumscribed quadrilateral $ \Rightarrow O\in CI$ and $ \frac{VB}{VD}\equal{}\frac{MB}{ND}$     (1)
The perpendicular to the line $ BD$ through the point $ O$ intersects the circle $ \Gamma$ at the point $ S$ ($ CS<CO$) and $ \left\{V'\right\}\equal{}SI\cap BD$.
From here [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=284806[/url], the problem 2 $ \Rightarrow\frac{V'B}{V'D}\equal{}\frac{MB}{ND}$     (2)
(1), (2) $ \Rightarrow\frac{VB}{VD}\equal{}\frac{V'B}{V'D}\Rightarrow V\equal{}V'$. So the points $ S,I,V$ are collinear.   (3)

$ U'\in\Gamma$ such that the points $ U'$, $ A$ are on the same side of $ BD$ and $ U'S$ is the bisector of the angle $ \angle BU'D$.
From here [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=284806[/url], the problem 5 $ \Rightarrow I\in U'S$.
So the points $ S,I,U'$ are collinear.     (4)

(3), (4) $ \Rightarrow$ the points $ S,I,V,U'$ are collinear $ \Rightarrow U'\equal{}U$.
So $ UI$ is the bisector of the angle $ \angle BUD$.

Best regards, Petrisor Neagoe :)

Let be the points $ E\equiv AB\cap CD,\ F\equiv AD\cap BC$ and $ Z\equiv EF\cap BD$ and it is easy to show that the points $ Z,\ B,\ X,\ D,$ are in harmonic conjugation ( from the complete quadrilateral $ ABCDEF$ ), where $ X\equiv AC\cap BD,$ instead of $ V.$

We denote sa $ K,\ L,\ M,\ N,$ the tangency points of the circumcircle $ (O)$ of $ ABCD,$ to the side-segments $ AB,\ AD,\ BC,\ CD,$ respectively and based on the [b][size=100]Newton theorem 1[/size][/b], we conclude that the cords $ KN,\ LM,$ pass through the point $ X.$

We draw the circle $ (V)$ with diameter the segment $ XZ$ and we denote as $ U,$ the point of intersection of $ (O),\ (V),$ in to the same side of $ BD$ as vertex $ A.$

Because of the pencil $ U.ZBXD$ is harmonic and $ UX\perp UZ$, we have that the line segment $ UX$ bisects the angle $ \angle BUD$ and so, it is enough to prove that the line segment $ UX,$ passes through the incenter $ I$ of the triangle $ \bigtriangleup BCD.$

$ \bullet$ Because of the points $ Z,\ B,\ X,\ D,$ are in harmonic conjugation, based on the  [b][size=100]Newton theorem 2[/size][/b], we have that $ (VB)\cdot (VD) \equal{} (VX)^{2} \equal{} (VU)^{2}$ $ ,(1)$

From $ (1)$ we conclude that the line segment $ VU,$ tangents to the circumcircle $ (O')$ of the triangle $ \bigtriangleup BUD,$ at point $ U.$

The line segments $ KL,\ MN,$ pass through the point $ Z$ as well ( the points of intersection of the opposite sidelines of the quadrilaterals $ ABCD$ and $ KLMN,$ are collinear ).

It is easy to show now, that the line segment $ AC,$ connecting the points $ A$ and $ X\equiv KN\cap LM,$ is the polar of the point $ Z$ with respect to the circle $ (O),$ because of the line segment $ KL$ as the polar of $ A$ with respect to $ (O),$ passes through the point $ Z.$

So, we conclude that the points $ Z,\ P,\ X,\ Q,$ are also in harmonic conjugation, where $ P,\ Q$ are the points of intersections of $ BD,$ from the circle $ (O)$ $ ($ $ P,$ between $ B,\ X$ $ ).$

Hence, aplying again the  [b][size=100]Newton theorem 2[/size][/b], we have that $ (VP)\cdot (VQ) \equal{} (VX)^{2} \equal{} (VU)^{2}$ $ ,(2)$

From $ (2)$ we conclude that the line segment $ VU,$ tangents to the circle $ (O)$ at point $ U$ and then, the circles $ (O),\ (O')$ tangent internally at point $ U.$

$ \bullet$ We have now, the configuration of the triangle $ \bigtriangleup BCD,$ with $ (O')$ as an arbitrary circle taken as cord the side-segment $ BC$ and the circle $ (O),$ as the mixtilinear incircle which tangents to $ CB,\ CD,$ at points $ M,\ N$ respectively and also tangents internally to $ (O')$ at point $ U.$

It is well known the result that the line segment $ UX,$ as the angle bisector of $ \angle BUD,$ passes through the incenter $ I$ of $ \bigtriangleup BCD$ and the proof is completed.

Kostas Vittas.

PS 1. Finaly, we don't need the [b][size=100]Lemma[/size][/b] I mentioned in my previous post but, it is equivalent.

PS 2. I think that [b][size=100]April[/size][/b] gave us a nice solution of the result I mentioned in the last words above ( as a generalization of a well known property in the mixtilinear incirles theory ), but I didn't find where it has been posted in the Forum. So, kindly please inform me if somebody knows a reference about it. Thanks in advance.

[quote="vittasko"]... We have now, the configuration of the triangle $ \bigtriangleup BCD,$ with $ (O')$ as an arbitrary circle taken as cord the side-segment $ BC$ and the circle $ (O),$ as the mixtilinear incircle which tangents to $ CB,\ CD,$ at points $ M,\ N$ respectively and also tangents internally to $ (O')$ at point $ U.$

It is well known the result that the line segment $ UX,$ as the angle bisector of $ \angle BUD,$ passes through the incenter $ I$ of $ \bigtriangleup BCD$ and the proof is completed.

... I think that [b][size=100]April[/size][/b] gave us a nice solution of the result I mentioned in the last words above ( as a generalization of a well known property in the mixtilinear incirles theory ), but I didn't find where it has been posted in the Forum. So, kindly please inform me if somebody knows a reference about it. Thanks in advance.[/quote]

Dear [b]vittasko[/b], I don't remember whether I proved the result you mentioned above or not. But it can be followed easily from the topic [url=http://www.mathlinks.ro/viewtopic.php?t=41667]Concyclic points with triangle incenter[/url] by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=5895]yetti[/url].

PS. Maybe the result above is known as a problem of Iran 1999? http://www.mathlinks.ro/Forum/viewtopic.php?t=6086 , http://www.mathlinks.ro/Forum/viewtopic.php?t=33

Thank you very much dear [b][size=100]Son[/size][/b] for all the useful references. For a moment I thought a problem of [b][size=100]Erken[/size][/b] of which you pointed out a solution and also its generalization but, probably I was confused ( naturally since I am sixty  :mad: ).

Best regards, Kostas Vittas.

[quote="vittasko"]...The line segments $ KL,\ MN,$ pass through the point $ Z$ as well ( the points of intersection of the opposite sidelines of the quadrilaterals $ ABCD$ and $ KLMN,$ are collinear ).

It is easy to show now, that the line segment $ AC,$ connecting the points $ A$ and $ X\equiv KN\cap LM,$ is the polar of the point $ Z$ with respect to the circle $ (O),$ because of the line segment $ KL$ as the polar of $ A$ with respect to $ (O),$ passes through the point $ Z.$

So, we conclude that the points $ Z,\ P,\ X,\ Q,$ are also in harmonic conjugation, where $ P,\ Q$ are the points of intersections of $ BD,$ from the circle $ (O)$ $ ($ $ P,$ between $ B,\ X$ $ ).$[/quote]
As a friend of mine [b][size=100]Andreas Varverakis[/size][/b] said, we can easier say that the line segment $ AC$ is the polar of $ Z$ with respect to $ (O),$ because it connects the point $ A,$ with the point $ Y\equiv KL\cap AX,$ as the harmonic conjugate of $ Z$ with respect to $ K,\ L$ $ ($ from the complete quadrilateral $ AKTLBD,$ where $ T\equiv BL\cap DK$ $ ).$

Many thanks to [b][size=100]Andreas[/size][/b], Kostas Vittas.

Let $O$ be the incenter of $ABCD$, let $P,Q,R,S$ be feet from $\overline{AB},\overline{BC},\overline{CD},\overline{DA}$, respectively.
Let $T=(BQI)\cap (DRI)$. Firstly, $T$ lies on the incircle of $ABCD$ as $$\measuredangle QTR=\measuredangle QBI+\measuredangle IDR=\tfrac{1}{2}\measuredangle QCR=\tfrac{1}{2}\measuredangle QOR.$$
Let $E,F$ be reflections of $R,Q$ over $\overline{DI},\overline{BI}$, respectively. Observe that $E$ lies on $(BQI)$ and $F$ lies on $(DRI)$ as by congruent triangles $IQ=IR=IE=IF$. Hence, $$\frac{VB}{VC}=\frac{\sin{\angle VQB}\cdot BQ}{\sin{\angle BVQ}}\cdot \frac{\sin{\angle DVS}}{\sin{\angle VSD}\cdot DS}=\frac{BQ}{DS}=\frac{BF}{DE}.$$
Furthermore, $VB\cdot VE=VD\cdot VF$, which means that $V$ lies on the radical axis of $(BQI)$ and $(DRI)$. We conclude that $T\equiv U$. Finally,
$$\measuredangle BUV=\measuredangle BQI=\measuredangle IRD=\measuredangle VUD,$$
as desired. $\blacksquare$


ratio
logarithms
limit
calculus
calculus computations

Start by considering the logarithm of that expression.

[quote="Altheman"]how do you compute:

$\lim_{n\rightarrow\infty }\frac{[\ln n]^{n}}{[\ln (n+1)]^{n}}$[/quote]
$\lim_{n\rightarrow\infty }\frac{[\ln n]^{n}}{[\ln (n+1)]^{n}}$~$\lim_{n\rightarrow\infty }\frac{[\ln n]^{n}}{[\ln (n)+1/n)]^{n}}$~$\frac{1}{(1+\frac{1}{nlnn})^{n}}$=$\frac{1}{e^{nln(1+\frac{1}{nlnn})}}$~$\frac{1}{e^{\frac{1}{lnn}}}$~$1$


ceiling function
analytic geometry
geometry
rectangle
graphing lines
slope
combinatorics unsolved

A flea jumps in a straight numbered line. It jumps first from point $0$ to point $1$. Afterwards, if its last jump was from $A$ to $B$, then the next jump is from $B$ to one of the points $B + (B - A) - 1$, $B + (B - A)$, $B + (B-A) + 1$.

Prove that if the flea arrived twice at the point $n$, $n$ positive integer, then it performed at least $\lceil 2\sqrt n\rceil$ jumps.

My solution:

We say that at step $i$, we move $d_i$ units, where $d_i$ is an integer.

From the condition we get that $|d_{i+1}-d{i}|$ is less than or equal to $1$

Suppose we fall two times on point $n$, there is a point to the right of $n$, or $n$ itself, such that the Flea makes a jump from it onto it (so it can go back).

Let $m$ be this point.

Take a coordinate system, where we will draw rectangles of base the segment between $i$ and $i+1$ and altitude $d_i$. Cleraly the sum (signed sum) of this rectangles from $0$ to $j$ is where the Flea is before the $j$ step.

So suppose our Flea arrives to $m$, in $x$ steps for the last time, $d_x$ is $1$ at most, because the next step has distance $0$. so the Flea takes at most $x+1$ steps on steping back in $n$.

Now draw the lines from $(0,1)$ with slope $1$, and from $(x,1)$ with slope $-1$, now traslade the altitude of the rectangles up, until they touch one of these lines with their leftmost vertex. The new area is bigger than or equal to the area of the rectangles, so in fact it is bigger than $m$ and $n$.

But it is easy to calculate this area and from here obtain the bound we want for $x$

I solved it in a very similar way.

If you want to pass through a point twice, you need to go there and go back (duh!). Since the length of the jumps stays the same or increases by $1$ or decreases by $1$, the flea must do a zero-jump (that is, a jump of lenght zero).

In order to make it easier to see, represent each jump by a rectangle of width $1$ and height $d$, where $d$ is the length of the jump. Put these side by side, in order of the jumps. For example, if the flea made jumps of lengths $1$, $2$, $3$, $2$, $2$, $3$, $2$, $1$, $0$, $-1$, draw these rectangles. Each X is a unit square and the last X is a "negative square".
[code]
  X  X
 XXXXXX
XXXXXXXX
         X
[/code]

This is exactly the same interpretation Pascual did, and as he said before, the point where the flea gets is equal to the sum of the signed areas (or the number of unit squares) of the rectangles. For example, the area of the previous rectangles is $1 + 2 + 3 + 2 + 2 + 3 + 2 + 1 + 0 + (-1)$, not $1 + 2 + 3 + 2 + 2 + 3 + 2 + 1 + 0 + 1$.

We are going to evaluate the further the flea can go with $k$ jumps, with the restriction that the flea is obliged to make a zero-jump. If $k = 2t$, the flea cannot make a jump bigger than $t$, since it takes $t+1$ jumps to make a $t+1$-lenght jump and more $t+1$ jumps to perform a zero jump. Thus the further the flea can go is to the point $1 + 2 + \cdots + t + (t-1) + (t-2) + \cdots + 1 + 0 = t^2$. Analogously, one can prove that if $k = 2t+1$, the further the flea can go is $t(t+1)$ (the same, but with another $t$-length jump).

Thus if $n > t^2$, the flea must do more than $2t$ jumps, that is, it must perform at least $2t+1$ jumps; and if $n > t(t+1)$, it must do at least $2t+2$ jumps.

It happens that if $t^2 < n \leq t(t+1)$ then $\lceil 2\sqrt n\rceil = 2t+1$ and it $t(t+1) < n \leq (t+1)^2$ then $\lceil 2\sqrt n\rceil = 2t+2$. Hence we are done.


algebra
difference of squares
special factorizations

Find all integer solutions to $ x^{2}+615=2^{n}$

[quote="mdk"]Find all integer solutions to $ x^{2}=615=2^{n}$[/quote]
 :blush: $ x^{2}=615$  :blush:
i think you mut edit your post


What is $\mu$?

The linear density of the spring?

[quote="perfect_radio"]An object of mass $m = 4 \, \textrm{kg}$ moves on a horizontal surface with a speed of $v_{0}= 4 \, \frac{\textrm m}{\textrm s}$. After some time, it hits head-on a fixed spring that deforms by $1 \, \textrm m$ if a force of $400 \, \textrm N$ acts on it. Considering $g = 10 \, \frac{\textrm m}{\textrm s^{2}}$, determine:
(a) the biggest deformation of the spring if $\mu = 0$;
(b) the biggest deformation of the spring if $\mu = 0.12$.

The answers in the book are $x_\textrm{max}= 0.4 \, \textrm m, \, x_\textrm{max}^\prime = 0.36 \, \textrm m$.

I can't manage to solve (b) :oops: (I keep getting the wrong answer.)[/quote]

perfect_radio:

$\mu N = m\ddot{x}+kx$

$\implies \text{Work}= \int_{0}^{x_{max}}m\ddot{x}dx = \int_{0}^{x_{max}}\mu N-kx dx = \mu N x_{max}-\frac{1}{2}k x_{max}^{2}$

$\text{Work}= \frac{1}{2}m\left( 0^{2}-v_{0}^{2}\right) = \mu N x_{max}-\frac{1}{2}k x_{max}^{2}$

$\implies k x_{max}^{2}-2\mu N x_{max}-m v_{0}^{2}= 0$

$\therefore x_{max}= \frac{\mu N \pm \sqrt{\left(\mu N \right)^{2}+k m v_{0}^{2}}}{k}$

For part a) $\mu = 0$ and $x_{max}= \frac{\pm v_{0}\sqrt{m k}}{k}= \frac{\pm 4\sqrt{4\cdot 400}}{400}= \pm 0.40$

For part b) $\mu = 0.12$ and $x_{max}= \frac{(.12)(4)(10) \pm \sqrt{\left((.12)(4)(10)\right)^{2}+(400)(4)(4)^{2}}}{400}= \frac{4.8 \pm \sqrt{23.04+25600}}{400}= \frac{4.8 \pm 160.072}{400}= 0.41 \text{ or }-0.39$

Since the spring is being compressed as this event takes place, the deformations correspond to the negative values. Hence,

$\boxed{ a) x_{max}=-.40 \text{ and }b) x_{max}=-.39.}$

[quote="perfect_radio"]I obtained the same result as you did, but, unlike you, I first solved the differential equation $m \left( x+\frac{\mu m g}{k}\right)^{\prime \prime}=-k \left( x+\frac{\mu m g}{k}\right)$, and only after I deduced the answer.

@PTynan89: $\mu$ must be the coefficient of friction. Why? Because this problem is in a ninth grade book, at the chapter about K.E., P.E., Mechanical Work and Friction.
Either Dr. No and I are wrong, or the book has some minor typos.[/quote]

In my own defense, the first equation I wrote, $\mu N = m\ddot{x}+k x$, [b]is[/b] the differential equation. $\ddot{x}\equiv \frac{d^{2}x}{d t^{2}}.$


modular arithmetic
algorithm
number theory
greatest common divisor
Euclidean algorithm

Find all the solutions of
$ 999x\minus{}49y\equal{}5000$

[hide]How to find the solutions when the coefficients of $ x$ and $ y$ are so large[/hide]

Write it as $ 999x\equiv \minus{}5000\pmod{49}$.  Let $ d\equal{}\gcd(999,49)\equal{}1$, which divides $ \minus{}5000$, so there is one unique solution up to a multiple of $ 49$.

Hint:You can use the Euclidean algorithm [url]http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm[/url] to find a solution (x,y) to ax+by=gcd(a,b), then you can use this initial solution to find infinitely mny more, then you can multiply both sides of the equation by whatever number you want.

There are two general approaches to diophantine equations.
1) See if you can solve it with modular arithmetic.
2) Experiment!!!!!!!!!!

With linear diophantine equations ($ ax + by = c$) it's best to use extended euclidean algorithm as dgreenb801 said and then express $ gcd$ as linear combination of $ a$ and $ b$ which will give us one particular solution to the equation $ (x_0,y_0)$. All solutions are given by:
\[ x = x_0 - bk, \ y = y_0 + ak, k \in Z\]
So, the [hide="solution:"]
$ \documentclass{article} \usepackage{amsmath} \begin{document} \begin{align*} 999 & = 20\cdot 49 + 19 \\
49 & = 2\cdot 19 + 11 \\
19 & = 11 + 8 \\
11 & = 8 + 3 \\
8 & = 2\cdot 3 + 2 \\
3 & = 2 + 1 \end{align*} \end{document}$

$ \implies gcd(999,49) = 1|5000 \implies \exists \ x_0,y_0 \in Z$

$ \documentclass{article} \usepackage{amsmath} \begin{document} \begin{align*} 1 & = 3 - 2 \\
& = 3 - (8 - 3\cdot 2) \\
& = 3\cdot 3 - 8 \\
& = 3(11 - 8) - 8 \\
& = 3\cdot 11 - 4\cdot 8 \\
& = 3\cdot 11 - 4(19 - 11) \\
& = 7\cdot 11 - 4\cdot 19 \\
& = 7(49 - 2\cdot 19) - 4\cdot 19 \\
& = 7\cdot 49 - 18\cdot 19 \\
& = 7\cdot 49 - 18(999 - 20\cdot 49) \\
& = 367\cdot 49 - 18\cdot 999 \end{align*} \end{document}$

$ \implies ( - 90000)\cdot 999 - ( - 1835000)\cdot 49 = 5000 \\
\implies x_0 = - 90000, y_0 = - 1835000 \\
\implies x = - 90000 + 49k, \ y = - 1835000 + 999k \\
\implies \boxed{x = 13 + 49k, \ y = 163 + 999k, k \in Z}$[/hide]

EDIT:

Also I'm pretty sure you could much simplify this by using modular arithmetic but I can't prove it strictly.
[hide]
$ c-ax \equiv 0 \pmod {b} \\
ax \equiv c \pmod {b} \\
999x \equiv 5000 \pmod {49} \\
980x+19x\equiv 2 \pmod {49} \\
19x \equiv 2 \pmod {49}$

Find one particular $ x$ from equation $ 19x-49y=2$ (same way we did up there) to get $ x_0=13$, then plug it in in the first equation to find $ y_0=163$.[/hide]


modular arithmetic
induction
number theory

I've been working on modular arithmetic, but I don't have many resources/problems. If anyone has any good modular arithmetic problems, please post them :)

If there are enough people, this might turn into a number theory marathon.

Prove that $ n^3\minus{}n$ is always divisible by 3, for all integers $ n$.

Prove that $ n^5 \minus{} n$ is always divisible by 5, for all integers $ n$.

Try these, they're not too hard.

Here's a hard problem (in other words, don't expect to get this immediately).

For any nonnegative integer $ n$, prove that $ 2^{5^n}\equiv 2^{5^{n \plus{} 1}}\equiv 2^{5^{n \plus{} 2}}\equiv \dots \pmod{10^{n \plus{} 1}}$.

In other words, prove that the $ n \plus{} 1$ digits from the right of the numbers $ 2^{5^n},2^{5^{n \plus{} 1}},2^{5^{n \plus{} 2}},\dots$ are all equal.

Also try this.

Each of the numbers $ a_1,\dots,a_n$ is either $ 1$ or $ \minus{} 1$, and:
\[ a_1a_2a_3a_4 \plus{} a_2a_3a_4a_5 \plus{} \dots \plus{} a_{n \minus{} 1}a_na_1a_2 \plus{} a_na_1a_2a_3 \equal{} 0.
\]
Prove that $ 4|n$.

[quote="modularmarc101"]Prove that $ n^3 \minus{} n$ is always divisible by 3, for all integers $ n$.

Prove that $ n^5 \minus{} n$ is always divisible by 5, for all integers $ n$.

Try these, they're not too hard.[/quote]

Hi, I want to practice modular arithmetic too. Below is how I proved it, but my methods doesn't seem very rigorous...
[hide]
1. $ n^3 \minus{} n\equal{}n(n\plus{}1)(n\minus{}1)$. Now we only need to consider 3 cases, since if x is a solution, so is x+3. Below, x is a multiple of 3:
a. $ x\equal{}n.$ Simple enough, x, a multiple of 3, was factored out.
b. $ x\equal{}n\plus{}1.$ Thus we have $ (x\plus{}1)(x)(x\plus{}2)$, and we see that x, a multiple of 3, appears in the factorization.
c. $ x\equal{}n\plus{}2.$ Thus we have $ (x\plus{}2)(x\plus{}1)(x\plus{}3)$, and since x is a multiple of 3, so is x+3.

2. $ n^5 \minus{} n\equal{}n(n^2\plus{}1)(n\plus{}1)(n\minus{}1)$. Likewise, we only need to consider 5 cases here, since if x is a solution so is x+5. Below, x is a multiple of 5:
a. $ x\equal{}n.$ Obvious again.
b. $ x\equal{}n\plus{}1.$ Obvious again (the n-1 factor is divisible by 5).
c. $ x\equal{}n\plus{}2.$ Thus we have $ (n\plus{}2)(n^2\plus{}4n\plus{}5)(n\plus{}3)(n\plus{}1).$ Note that $ (n^2\plus{}4n\plus{}5)$ is divisible by 5.
d. $ x\equal{}n\plus{}3.$ Thus we have $ (n\plus{}3)(n^2\plus{}6n\plus{}10)(n\plus{}4)(n\plus{}2).$ Note that $ (n^2\plus{}6n\plus{}10)$ is also divisible by 5.
e. $ x\equal{}n\plus{}4.$ Thus we have $ (n\plus{}4)(n^2\plus{}8n\plus{}17)(n\plus{}5)(n\plus{}3).$ A factor of n+5 appears, which is divisible by 5.

Are these results related to Fermat's Little Theorem?
[/hide]

[quote="modularmarc101"]Prove that $ n^3 \minus{} n$ is always divisible by 3, for all integers $ n$.

Prove that $ n^5 \minus{} n$ is always divisible by 5, for all integers $ n$.

Try these, they're not too hard.[/quote]

From factoring, $ n^3 \minus{} n \equal{} n(n\minus{}1)(n\plus{}1)$

Thus we have three consecutive numbers, and thus 3 must divide one of $ n, (n\minus{}1)$ or $ (n\plus{}1)$.

A similar logic applys to the other problem.

[quote="Yongyi781"]Each of the numbers $ a_1,\dots,a_n$ is either $ 1$ or $ \minus{} 1$, and:
\[ a_1a_2a_3a_4 \plus{} a_2a_3a_4a_5 \plus{} \dots \plus{} a_{n \minus{} 1}a_na_1a_2 \plus{} a_na_1a_2a_3 \equal{} 0.
\]
Prove that $ 4|n$.[/quote]

[hide="hint"]What happens when you change the sign of one of the $ a_i$?[/hide]

These are not hard, but are proofs (use mods in them).

Prove that the sum of three consecutive squares is never a perfect square.

Prove that if the sequence $ p_{1}, p_{2},...p_{k}$ is the first $ k$ primes, that $ p_{1}p_{2}...p_{k}\plus{}1$ cannot be a perfect square.

[quote="modularmarc101"]Prove that $ n^3 \minus{} n$ is always divisible by 3, for all integers $ n$.

Prove that $ n^5 \minus{} n$ is always divisible by 5, for all integers $ n$.

Try these, they're not too hard.[/quote]

I want to generalize, and state that:

$ n^k \minus{} n \equiv 0 \pmod {k}$ iff k is a prime. However, the only way I found to do this was by induction.

[hide="Induction Method"]Base Case: $ n \equal{} 1$

$ 1 \minus{} 1 \equal{} 0$ All integers k will divide 0, so this holds.

Assume the formula for $ n$, and show for $ n \plus{} 1$

$ (n \plus{} 1)^k \minus{} n \minus{} 1 \equiv 0 \pmod {k}$

Now, expand (n+1)^k using the binomial expansion theorem:

$ \binom{k}{0}n^k \plus{} \binom{k}{1}n^{k \minus{} 1} \plus{} ... \binom{k}{k \minus{} 2}n^2 \plus{} \binom{k}{k \minus{} 1}n \plus{} \binom{k}{k}1 \minus{} n \minus{} 1$

Now, we can simplify and rearrange to get:

$ n^k \minus{} n \plus{} \binom{k}{1}n^{k \minus{} 1} \plus{} ... \binom{k}{k \minus{} 2}n^2 \plus{} \binom{k}{k \minus{} 1}n$

We assumed that $ n^k \minus{} n$ was divisible by k, so it follows that:

$ \binom{k}{1}n^{k \minus{} 1} \plus{} ... \binom{k}{k \minus{} 2}n^2 \plus{} \binom{k}{k \minus{} 1}n \equiv x \pmod {k}$

----------------------------------------

Claim: Now, for all prime k, we can prove that x = 0, and thus proving the "if" in our proof as follows:

If every term in the expansion has a factor of k, then the congruence is proved to be true.

When k is a prime, the only factor on the denominator that can cancel it out (thus proving the congruence false) is k, by definition, because the numerator is $ k!$

Thus, only $ \binom {k}{k} \equal{} \dfrac{k!}{k!0!} \equal{} 1$ will cancel that factor. Because that term is not used in the simplified equation, every term in the equation is proved to have a factor of k.

-------------------------------------------

Now, we are guaranteed the congruence for all primes. Thus, it holds for k = 3 and k = 5, as the problems ask.

As for the other "only if" part, I'm sure it's true, and I have an idea on how to prove it, but I'll come back to that.[/hide]

Here is a simpler way to solve the original problems.

[hide="Cases"]

All positive integers can be expressed as 0, 1, or 2 mod 3. Thus, they are all in the form $ 3k$, $ 3k \minus{} 1$, or $ 3k \minus{} 2$.

Case 1: 3k

$ (3k)^3 \minus{} 3k \equal{} 27k^3 \minus{} 3k$ This is obviously divisible by 3.

Case 2: 3k-1

$ (3k \minus{} 1)^3 \minus{} 3k \plus{} 1 \equal{} 27k^3 \minus{} 27k^2 \plus{} 9k \minus{} 1 \minus{} 3k \plus{} 1 \equal{} 27k^3 \minus{} 27k^2 \plus{} 6k$ This is divisible by 3 as well.

Case 3: 3k - 2

$ (3k \minus{} 2)^3 \minus{} 3k \plus{} 2 \equal{} 27k^3 \minus{} 54k^2 \plus{} 36k \minus{} 8 \minus{} 3k \plus{} 2 \equal{} 27k^3 \minus{} 54k^2 \plus{} 33k \minus{} 6$. This is divisible by 3.

Similarly, we can prove it for n^5 - n with 5 cases.
[/hide]

I spent too long proving that generalization. I'll get to work on your problems Yongyi.

It's not true. Not only are there numbers for which there exist $ x$ satisfying that, the [url=http://en.wikipedia.org/wiki/Carmichael_numbers]Carmichael numbers[/url] are precisely those numbers so that all x within the conditions satisfy that.

[quote="Yongyi781"]

Each of the numbers $ a_1,\dots,a_n$ is either $ 1$ or $ \minus{} 1$, and:
\[ a_1a_2a_3a_4 \plus{} a_2a_3a_4a_5 \plus{} \dots \plus{} a_{n \minus{} 1}a_na_1a_2 \plus{} a_na_1a_2a_3 \equal{} 0.
\]
Prove that $ 4|n$.[/quote]

The first problem is tough (or maybe I'm just missing something). I think I have the solution for the second problem:

[hide="2nd Problem"]Firstly, we notice that the product of any of the terms must be either 1 or -1, and that there must be an even number of terms (as each term equalling 1 must have a corresponding term equalling -1).

Next, we notice that there are exactly n terms, as each term starts off with a number, followed by its three consecutive numbers in $ a_n$. Thus, there are terms $ 1 < \equal{} n < \equal{} n$, so there are n terms.
--------------------------------
Claim: Also, each number in $ a_n$ shows up in distinct products precisely 4 times.

Working in mod n (because looking at the next four terms, if the term $ a_n$ is already used, we simply start back over at the beginning), we see that $ a_i$ shows up in the products:

$ a_ia_{i \plus{} 1}a_{i \plus{} 2}a_{i \plus{} 3}$

$ a_{i \minus{} 1}a_{i}a_{i \plus{} 1}a_{i \plus{} 2}$

$ a_{i \minus{} 2}a_{i \minus{} 1}a_{i}a_{i \plus{} 1}$

$ a_{i \minus{} 3}a_{i \minus{} 2}a_{i \minus{} 1}a_{i}$

-----------------------------
Now, we can write the proof like this:

Start off with an arbitrary amount of $ a_n$ that are all equal to 1. The sum of the terms is thus n. The reason we choose this to start off with is that we can negate any number of terms to reach all possible combinations of the numbers in $ a_n$.

Negate precisely one $ a_n$, called $ a_x$. By doing so, we have negated the value of 4 of the terms, as $ a_x$ is present in four of the terms (as stated above).

Now, there are 5 cases for what can be the result of the total value after the negation of these 4 terms:
---------------------------------------
Case 1: All 4 terms are positive (1)

Call $ g$ the sum of all the other terms. Then:

$ g \plus{} 4 \equal{} n$

By negating the 4, which in turn negates the 4 terms, we see:

$ g \minus{} 4 \equal{} n$

Essentially, our total value $ n$ has been decreased by 8.

Proceeding in similar fashion, we find for the other cases that:

Case 2: 3 terms are positive (1), and 1 term is negative (-1)

The total value has been decreased by 4.

Case 3: 2 terms are positive (1) and 2 terms are negative (-1)

The total value has not been changed (decreased by 0)

Case 4: 1 term is positive, and 3 terms are negative (-1)

The total value has been increased by 4.

Case 5: All four terms are negative

The total value has been increased by 8.

--------------------------------------

Now, clearly any combination of adding or subtracting the numbers, 8, 4, 0, -4, and -8 will only result in a change divisible by 4.

Because the total value $ n$ is to be changed by only values divisible by 4, and to equal 0, it follows that $ 4|n$.[/hide]

I am not sure my proof at the end and I am wondering whether or not it is rigorous. I am still very bad with proofs, so can someone please tell me if my proof-writing in this case is rigorous?

[quote="SnowEverywhere"]

Prove that the sum of three consecutive squares is never a perfect square.

Prove that if the sequence $ p_{1}, p_{2},...p_{k}$ is the first $ k$ primes, that $ p_{1}p_{2}...p_{k} \plus{} 1$ cannot be a perfect square.[/quote]

[hide="1"]
Let's choose to work in mod 8 (which is quite convenient)

There are only 8 possibilities for numbers in mod 8. That is, we can only have numbers such that they are equal to either 0, 1, 2, 3, 4, 5, 6, or 7 mod 8.

For a number 0 mod 8, squaring it will result in:

$ 0 \equiv 0 \pmod {8}$

Similarly, squaring the modular congruences for the other possibilities yields:

$ 1 \equiv 1 \pmod {8}$

$ 2 \equiv 4 \pmod {8}$

$ 3 \equiv 1 \pmod {8}$

$ 4 \equiv 0 \pmod {8}$

$ 5 \equiv 1 \pmod {8}$

$ 6 \equiv 4 \pmod {8}$

$ 7 \equiv 1 \pmod {8}$

Thus, since these cases represent all positive integers, it holds that the modular values of perfect squares in mod 8 will have go in the order: 0, 1, 4, 1, 0, 1, 4, 1.

Now, the sum of any three consecutive perfect squares can only be:

0 + 1 + 4 = 5
1 + 4 + 1 = 6
4 + 1 + 0 = 5
1 + 0 + 1 = 2

Since none of these values are congruent to any of the possible values of a perfect square working in modulus 8, no 3 consecutive perfect squares can ever sum to a perfect square.
[/hide]

[hide="2"]

Note that the first prime number is 2. This will be key in the proof.

Next, we have that an even number can only be congruent to 0 or 2 mod 4.

Let's focus on just the product of the first k primes for now. 2 is the only even prime, and there are no other factors of two in the product. Thus, because of this, this product cannot be divisible by 4, and the only other possibility is the congruence 2 mod 4.

Therefore, the product plus 1 is 3 mod 4.
--------------------------------
Claim: Perfect squares can only equal 0 or 1 mod 4.

All positive integers can be divided up into either 0, 1, 2, or 3 mod 4.

Call a positive integer n. Let's go through the cases and square the modular congruence:

Case 1: $ n \equiv 0 \pmod {4}$

$ n^2 \equiv 0 \pmod {4}$

Case 2: $ n \equiv 1 \pmod {4}$

$ n^2 \equiv 1 \pmod {4}$

Case 3: $ n \equiv 2 \pmod {4}$

$ n^2 \equiv 0 \pmod {4}$

Case 4: $ n \equiv 3 \pmod {4}$

$ n^2 \equiv 1 \pmod {4}$

----------------------------

We have thus proved that the product of the first k primes plus 1 cannot be a perfect square by congruences in mod 4.
[/hide]

@veezbo: An easier solution to the first problem (sum of three consecutive squares) is to notice that exactly one square is divisible by 3; the rest are 1 mod 3. Their sum is hence 2 mod 3, and no squares are 2 mod 3.

Here's another problem from the Intermediate Modular Arithmetic section on AoPS.

Let $ D = d{}_1d{}_2d{}_3d{}_4d{}_5d{}_6d{}_7d{}_8d{}_9$

be a nine-digit positive integer (each digit not necessarily distinct). Consider

$ E = e{}_1e{}_2e{}_3e{}_4{}_e{}_5e{}_6e{}_7e{}_8e{}_9$,

another nine-digit positive integer with the property that each digit $ e_i$ when substituted for $ d_i$ makes the modified D divisible by 7. Let $ F = f{}_1f{}_2f{}_3f{}_4f{}_5f{}_6f{}_7f{}_8f{}_9$
be a third nine-digit positive integer with the same relation to E as E has to D.
Prove that every $ d_i - f_i$ is divisible by 7.
----------------------------------------
However, I think the posted proof is wrong. If anyone else tries this and successfully proves it, please post your solution.

[quote="SnowEverywhere"]These are not hard, but are proofs (use mods in them).

Prove that if the sequence $ p_{1}, p_{2},...p_{k}$ is the first $ k$ primes, that $ p_{1}p_{2}...p_{k} \plus{} 1$ cannot be a perfect square.[/quote]

sorry if i'm reviving an old thread, but i didn't feel like making a new one, so anyways..
i solved this without using modular arithmetic and wish someone could look at my solution and comment on it thx :)

[hide]We will prove this by contradiction.
So, we assume that there is an integer $ a$ that can make this sequence equal to $ a^2$.
Thus,$ p_{1}p_{2}...p_{k} \plus{} 1 \equal{} a^2 \minus{} 1 \equal{} (a \plus{} 1)(a \minus{} 1)$
We know that $ p_{1}$ is 2 and using this fact we can infer the following two facts.
First, $ a$ can't be an even number because the sequence is always even.
Second, $ a$ can't be an odd number either because we know there is only one even number that is a prime: $ 2$. If a were to be an odd number, both $ (a \plus{} 1)$ and $ (a \minus{} 1)$ would be even, which means the squence would have to be a multiple of four and it is not.
We find out that a can't be $ 0$ as well,because even there is no way to make the sequence equal to $ 0$.
Therefore, there is no integer $ a$ that can make the sequence a perfect square of an integer.[/hide]

As you can probably tell from my proof, I seem to be very wordy and feel like I'm proving it informally, syntax-wise.
If there is a thread you can point me to that can help me fix this that would be awesome. :)

[quote="veezbo"]

$ n^k \minus{} n \equiv 0 \pmod {k}$ iff k is a prime.

[/quote]

This is Fermat's little theorem ,see [url=http://en.wikipedia.org/wiki/Fermat's_little_theorem]here[/url]

Here's one more problem:
Show that every positive integer has a multiple whose decimal representation contains
only the digits 0 and 1.

(BTW does anyone know how to solve those unsolved problems in this thread?)


I guess this is the theorem or proposition, but I'd really like to know how the idea goes, please help.

Prove that "If $A$ is positive definite, then $det(A) > 0$"

what is your definition of such a matrix ?

Sorry I should state that $A$ is $n \times n$ matrix

I think alekk was getting at the following questions:

Is $A$ real or are you allowing $A$ to be complex?

Are you restricting this to symmetric (in the real case) or Hermitian (in the complex case) or are you allowing more general forms for $A?$

Are you distinguishing between positive definite and positive semidefinite?

$A$ is symmetric, in the real case with positive definite

so you must know that such a matrix is diagonalizable over $\mathbb{R}$, with strictly positive eigenvalues.

Could u explain more in detail? Thx in advance! :)

Theorem: A real symmetric matrix is orthogonally diagonalizable. A Hermitian matrix is unitarily diagonalizable, with real eigenvalues.

Proof: By Schur's Lemma, if $A$ is any $n\times n$ complex matrix, there is a unitary matrix $U$ such that $U^{-1}AU$ is upper triangular. If $A$ is Hermitian, then since $U^{-1}=U^*$, $U^{-1}AU$ is also Hermitian. A Hermitian upper triangular matrix is diagonal with real entries.

Now, we deal with the real symmetric case. By the above, the matrix has real eigenvalues, and is diagonalizable over $\mathbb{C}$. This forces $A$ to be diagonalizable over $\mathbb{R}$. It is easy to show that eigenvalues of different eigenvalues are orthogonal, and we can simply choose an orthonormal basis for each eigenspace.
Alternately, we go back to Schur's Lemma- if we know that the matrix is real and the eigenvalues are real, we can build our unitary matrix using real vectors. It works by induction.

Now, let $A$ be symmetric positive definite, and let $v$ be an eigenvalue with eigenvalue $\lambda$. We have $v^Tv>0$, and $v^TAv=\lambda v^Tv>0$, so $\lambda>0$.

Jmerry.. Thx for your exlaination. I guess I understand much more now.
As you used the Schur's Lemma and Hermitian, it was an interesting approach. Since I just learned those theorems days agao that I didnt know I should use them. Is that only or the best way to prove by using those theorems?

Thx again.  :)

There may be other ways, but this is the proof I remember. I got it from Kent- Schur's lemma is his favorite tool in linear algebra.


trigonometry
real analysis
real analysis unsolved

Consider the next ODE:
\[ y'' \plus{} (\tan x)y' \plus{} (\cos^2x) y \equal{} 0
\]
Using the variable $ t \equal{} \sin x$, transform this equation into a ODE with constant coefficients. And then, solve it.

And your question is?  :maybe:
Use the chain rule: $ \frac{dy}{dx}\equal{}\frac{dy}{dt}\frac{dt}{dx}$, etc.
To avoid ambiguity, write $ y''\plus{}(\tan x)y'\plus{}(\cos^2 x)y\equal{}0$.

I just can't solve the equation. Can you?

hello, the solution is
$ y(x)\equal{}C_1\sin(\sin(x))\plus{}C_2\cos(\sin(x))$.
Sonnhard.

Thank You ... I think that you solve this:
\[ z'' \plus{} az' \plus{} bz \equal{} 0
\]
with $ a,b,c$ constants.

And $ z \equal{} f(x)y$. How could you find $ f(x)$???

Now I'm stuck on this:
\[ \frac {d}{dx}\left( \frac {dy}{dt} \right) \plus{} y\cos x \equal{} 0
\]
where
\[ \frac {dy}{dx} \equal{} \frac {dy}{dt}\cdot \frac {dt}{dx}
\]

How did you arrive at that equation by using the substitution t = sin x?

This is what I did:
\[ t\equal{}\sin (x)\]
\[ \frac{dy}{dx}\equal{}\frac{dy}{dt}\frac{dt}{dx}\equal{}\frac{dy}{dt}\cos ( x)\]
\[ \frac{d}{dx}\left(\frac{dy}{dx} \right)\equal{}\frac{d^2y}{dx^2}\equal{}\frac{d}{dx}\left(\frac{dy}{dt} \right)\cos (x)\minus{}\frac{dy}{dt}\sin (x)\]

Substituting in the initial equation:
\[ \frac{d}{dx}\left(\frac{dy}{dt} \right)\cos (x)\minus{}\frac{dy}{dt}\sin (x)\plus{}\frac{\sin (x)}{\cos (x)}\left(
\frac{dy}{dt}\cos ( x)\right)\plus{}y\cos^2(x)\equal{}0\]

\[ \frac{d}{dx}\left(\frac{dy}{dt} \right)\plus{} y\cos (x)\equal{}0\]

am I right?

this is simple, jonysatie  :(

you are given, $ \frac{d^2y}{dx^2}\plus{}(\tan\,x)\frac{dy}{dx}\plus{}(\cos^2\,x)\;y\equal{}0\;\;\;\quad\quad\ldots(1)$

using $ t\equal{}\sin\,x\;\;\;\quad\quad\frac{dy}{dx}\equal{}\frac{dy}{dt}\frac{dt}{dx}\equal{}\frac{dy}{dt}\cos\,x$

and $ \frac{d^2y}{dx^2}\equal{}\frac{d}{dx}\left(\frac{dy}{dx} \right)\equal{}\frac{dt}{dx}\frac{d}{dt}\left(\frac{dy}{dt}\cos\,x\right)\equal{}\frac{dt}{dx}\left(\frac{d^2y}{dt^2}\cos\,x\minus{}\sin\,x\frac{dx}{dt}\frac{dy}{dt}\right)$
$ \equal{}\frac{d^2y}{dt^2}\cos^2\,x\minus{}\sin\,x\frac{dy}{dt}$

put these in $ (1)$ to get

$ \left(\frac{d^2y}{dt^2}\cos^2\,x\minus{}\sin\,x\frac{dy}{dt}\right)\plus{}(\tan\,x)\left(\frac{dy}{dt}\cos\,x\right)\plus{}(\cos^2\,x)\;y\equal{}0$
$ \Rightarrow\frac{d^2y}{dt^2}\cos^2\,x\plus{}(\cos^2\,x)\;y\equal{}0$

assuming $ x\neq\frac{(2n\plus{}1)\pi}{2}$ for $ n\in\mathbb Z$, we then only need to solve $ \frac{d^2y}{dt^2}\plus{}y\equal{}0$

we immediately see, $ y(t)\equal{}A\cos\,t\plus{}B\sin\,t$
$ \Rightarrow\;y(x)\equal{}A\cos\,(\sin\,x)\plus{}B\sin\,(\sin\,x)$

where $ A$ and $ B$ are arbitrary constants to be determined from initial value conditions.

[quote="misan"]this is simple, jonysatie  :(

and $ \frac {d^2y}{dx^2} \equal{} \frac {d}{dx}\left(\frac {dy}{dx} \right) \equal{} \frac {dt}{dx}\frac {d}{dt}\left(\frac {dy}{dt}\cos\,x\right) \equal{} \frac {dt}{dx}\left(\frac {d^2y}{dt^2}\cos\,x \minus{} \sin\,x\frac {dx}{dt}\frac {dy}{dt}\right)$
$ \equal{} \frac {d^2y}{dt^2}\cos^2\,x \minus{} \sin\,x\frac {dy}{dt}$
[/quote]

That was the step that I could not do.

Thanks


linear algebra
matrix
linear algebra unsolved

Let |A| be standard matrix norm (sup|Ax| : |x|=1). Is it true that |A|=|A*|, where * is conjugate transpose?

Yes, and it follows from $ \langle Ax,y\rangle \equal{} \langle x,A^*y\rangle$

Another question: what happens to the norm of a [url=http://en.wikipedia.org/wiki/Block_matrix]block matrix[/url] if you transpose each of its (square) blocks?


Also, the number cannot go over a limit.

Examples:

Base 2:
numbers 0,1 only

Base 5
numbers 0,1,2,3,4

Base 8
numbers 0,1,2,3,4,5,6,7

Base 10
numbers 0-9

Hmmm... I think he wants to know the process by which you can represent any number in base x in base y. :?

In base y,

find x

find what x- y :^n: is, where n is an integer.

This must be the biggest possible number.

Then, continue down, until you reach the ones digit, and that is the number.

Sorry!!!!

IT was supposed to be 0*2^0.

It's the binary 11110, not 11111.

This is for the b.8 too.

WWMIII

[quote="tmshu1"]OK. I've heard about bases in math competitions. I've asked people but they have all given me a really long and boring answer. Can someone explain to me what bases are (i know this is kinda a dumb question)? For instance "What is 14987 in base 8?"?[/quote]

Ok, here's my attempt at explaining it to someone in an interesting way!

Imagine a number line, where every number from 0 to "infinity" is represented by a charachter. This is basically to help you understand that without bases maths will still work. you can still multiply, add, and subtract (divide is a bit messy since it gives fractions!) numbers: 2+2 = 4, 4*4=, (where  is a charachter representing the sixteenth number in the line), and - =  (whatever that means...  and  can be on any scale and are single digits... they can represent a million minus seven and the answer would still be nine hundred thousand nine hundred and ninety three).

However,  when it comes to WRITING the numbers down and communicating them, we would find it rather hard to have an infinite amount of symbols as would be required! so when we get to 9, we use TWO DIGITS to represent the next number. in a sense, we were already using two digits with 01, 02, 03, or in a sence we were using an infinite amount of digits... 0000....01, 0000....02,... but we leave them out because they're stupid and irrelevant.

So when we write a number one then a number zero, which we ASSUME to be in base 10, we mean ten times one plus zero. when we write a three then a five, we mean ten times three plue five. carrying this on, when we write: 938, in base ten, we mean by it ten times ten times nine, plue ten times three, plue eight.

And 10293 means ten times itself four times, times 1, plus ten times itself three times, times 0, plus ten times itself twice times two plus ten times nine plus three.

Basically, you should by now have noticed that a number in base ten is in the form:

a*10^n + b*10^(n-1) + c*10^(n-2) + d*10^(n-3) + ... + e*10^3 + f*10^2 + g*10 + h

Where the number can be written abcd...efgh, and where a to h are the digits of the number! They of course cannot be larger than 9 because that would "spill" over onto the product of a larger power of ten and the higher digit (although we do it when we say "that car will cost you twelve hundred pounds"... but this is equivalent to one thousand two hundred pounds).

Now, what if we wanted to express this number (10293) in a different base? operations in different bases does not affect the actual result, since bases are simply a way of expressing the numbers. Therefore, this number squared, in base 5, 10, or no base in particular (what we did earlier where every number has its own charachter), will always be the same.

Say we use base three:

We are allowed to use 0, 1 and 2

So here we go: 0, 1, 2... errr... what next? well we want the equivalent to three now. Since three (like ten in base ten) is equal to three times one plus 0, we can write it as 10. The next number (four) is 11, since that is one times three plus one. And the next is 12 since that is one times three plus two.

The next number (six) is more exciting... it is 20, since six is equal to two times three plus nought. The seven is two times three plus one, (21), and so on.

When we get to expressing nine, we need more digits. Nine is equal to three times three times one plus nought times three plus nought, so we can write it 100.

And in general, we see the same pattern emerging: ANY integer can be written:

a*3^n + b*3^(n-1) + c*3^(n-2) + d*3^(n-3) + ... + e*3^3 + f*3^2 + g*3 + h

where the number in base three is abcd...efgh , where they are the digits and can equal 0, 1 or 2.

I hope this makes sense to you... it does to me but of course it would (I wrote it).
It's quite hard to elarn but you need to remember that a number is just an amount of something! A caveman can tap a stick on a rock any anout of times, and do addition with that! 2 taps plus 3 taps make 5 in total. fifteen taps plus one hundred taps make one hundred and fifteen in total. when we use digits and bases to describe the number, we are not changing the number at all! It is the same number of taps, and it's just a window we use to make the vast array of numbers more viewable.

ok. i think i got it. can someone give me another example (what is [i]anynumber[/i] in base [i]number[/i]) just so i can make sure?     :lol:

301 in base nine is 364 because it's 81*3 + 6*9 + 4*1.

counting in base n is like counting with only n fingers.

we have 10 fingers (most of us) so units are easily counted in bundles of 10. that's why normal math is done in base 10.

if some alien race had only 5 fingers, then they'd use base 5.

10(base 5) would = 5(base 10)

confusing enough?

ok. i still don't get it really. so... someone run it through for me again. explain every step. with a good number (not like 1111111). :(

If your dealing with base n than in number xyzw the place with the w is the units digit, the place with the z is the n spot every number there is worth n, the y spot is n^2 everything there is worth n^2, the z spot you should know now.

So the base ten number 102 in base seven is:
How many times does 343 fit in 102      0
How many times does 49 fit in 102        2
How many time does 7 fit in 102-98=4  0
How many times does 1 fit in 4              4
So the number is 0204 which can be simplifyed to 204.

[color=orange]HEY THAT HELPS ME ALOT!!!!!!!!!![/color] [color=green]THNX ALL U GUYS WHO HELPED ME!!!![/color]

[color=darkblue]BYE! [/color]

[quote="white_horse_king88"]Umm... Bases are Alkaline substances.

Haha... sorry. Couldn't help it.[/quote]

To clarify, it has a pH of more than seven and a hydroxide (OH) solution attached. chem...


ratio
number theory unsolved
number theory

Given a positive integer A with all digits being non-zero, call a number B a relative of A if B is fprmed by taking a cyclic permutations of the digits of A.
Find all such a such that all of A's relatives are divisible by A.

[quote="bomb"]Given a positive integer A with all digits being non-zero, call a number B a relative of A if B is formed by taking a cyclic permutations of the digits of A.
Find all such A such that all of A's relatives are divisible by A.[/quote]

Obviously all integers of the form d/9*(10^n-1), where d is a digit, satisfy the relation (these numbers consist of n copies of the same digit d).  In addition, somewhat less obviously, any number of copies of the string 142857 work.  These are the only solutions.

To show this, suppose there exist solutions not of this form.  Consider the smallest such solution S.  Obviously, it cannot be written as a number of repetitions of some other solution P (i.e., as PPPP...P).  Then S cannot have two digits the same, say, d and d', or else the two cyclic permutations starting with d and d' respectively would then be equal (since each is a multiple of S, and they start with the same digit).  It is then clear that S can be written as PPPP...P with P some (smaller) solution, contradiction.  Hence S contains no equal digits.  Obviously also S contains no digit 0.  In addition, S cannot have nine digits because then S would be at least 123456789, and require that 2S, 3S,..., 9S appear as relatives, but 9S is more than 10^9 and thus has ten digits.  Hence S has eight or fewer digits.

From here it's basically a slog to show there are no other solutions...
-> If there are two digits, putting S=ab with ba = k * ab gives (10k-1)a = (10-k)b which has no solutions with k not equal to 1. (19, 29, 39, 49 don't work, and if k is more than 5 then (10k-1)/(10-k) is greater than 10.
-> If there are three digits, then put S = abc with bca = k * abc so that (100k-1)a = (10-k)(10b+c), which also has no solutions with k not 1.
-> If there are six digits then putting S = abcdef with bcdefa = k * abcdef gives k * (10^5 k-1) * a = (10-k)*(bcdef).  So 10^5 * k - 1 has to have some factor less than 10.  But k = 2 gives 199999, prime; k = 3 gives 7 * 17 * 2521 so that 42857 a = bcdef so that abcdef = 142857 as given before or abcdef = 285714 which fails.  k = 4 gives 3*151*883, but then S is divisible by 151*883 = 133333 which makes at least four digits equal.  k = 5 gives 31 * 127^2, again a contradiction.  k 6 or greater makes the ratio (10^5*k-1)/(10-k) too large.

For 4, 5, 7, or 8 digits, sum over all cyclic permutations to yield the following formula:
k * S = (111...1) * (sum of digits),
where k is the sum of the "relative ratios" (e.g., for n=4, it is abcd/abcd + bcda/abcd + cdab/abcd + dabc/abcd) and is always at most 36, since there are at most eight relatives of S, and 9S cannot be one of them.  Also, the number of ones in the first factor is the number of digits in S.
-> For 4 digits this gives that k * S = 11 * 101 * (sum of digits), and since k < 37, S must be divisible by 101.  But then it can be written as abab for some a and b, so that ab is also a solution, impossible.
-> For 5 digits this gives that k * S = 41 * 271 * (sum of digits), and again k < 37 so that S is divisible by 41 and 271 and thus 11111, so all its digits are the same, impossible.
-> For 7 digits this gives k * S = 239 * 4649 * (sum of digits), so 1111111 divides S.
-> For 8 digits this gives k * S = 11 * 73 * 101 * 137 * (sum of digits), so 73, 101, and 137 divide S, so their product 1010101 does too.  But then S can be written as abababab for some a and b (since it has 8 digits), contradiction.

Hence there are no more solutions than those listed at the beginning.


find all of positve integers x,y,z that
2^x+3^y=z^2


hey

if f is differentiable at a, where a > 0, eveluate the following limit in terms of $\frac{dy}{dx}(a)$.  Here's the limit, $\lim_{x\to\a}$$\frac{f(x)-f(a)}{\sqrt{x}-\sqrt{a}}$


I'm planning to write mock AIME.

This would be close to my competition COP Championship Round but the rule of no calculator would follow. And rest of AIME rules. I'll probably derive the problems from old AIME and maybe other books and contests.

This would be perhaps even easier than AIME. But I'm doing for a) making resource for my competition COP, b) learning more about AIME.

If anyone's interested, please post here so I can get work to it.

Thanks. :)

^_^

i'll take it

anything for preperation will be appreciated

I'm up for it!

...I could certainly use the practice...

certainly. i would love to. just PM me with information about it. thanks

[quote="interesting_move"]When will it be administered?[/quote]

Good point.  I can't do it at all on Saturday.  That's the only day.

Science Bowl!

skcorlwobecneics

I'm in. But my schedule is very tight, if there is a conflict, could I possibly take it early?

I'll try to fit in this weekend. That's where I'm shooting at.

I can't guarantee you anything early.

Your score might not count but if you really want to take this before real AIME, you can always set your own time later and take it. :)

After all, mock AIMEs for doing good on real AIME, isn't it? :D

weekends are when events are jammed up :(

I'd like to do it... I'm doing the AIME in march...

Timezones being what they are though, i'll probably have to do it late or something...

Just to tell how far I am, I finished compiling 1-8.

Those aren't big deals.

Now, 9-15 are the harder ones. Because.... They have to be hard! :)  :D

i'm in
i feel like i'm joining some sort of brotherhood...

I'm interested too.  Hopefully, I'll be able to solve at least one problem  :D

In case anyone doesn't know, mock AIME is posted.

As I said, this is LOT easier than real AIME.  :D


It is possible to choose $ x > \frac {2}{3}$ in such a way that the value of $ \log_{10}(x^2 \plus{} 3) \minus{} 2 \log_{10}x$ is

$ \textbf{(A)}\ \text{negative} \qquad \textbf{(B)}\ \text{zero} \qquad \textbf{(C)}\ \text{one}$
$ \textbf{(D)}\ \text{smaller than any positive number that might be specified}$
$ \textbf{(E)}\ \text{greater than any positive number that might be specified}$

[hide="Solution"]$ \log(x^2\plus{}3)\minus{}\log(x^2)\equal{}\log\left(\frac{x^2\plus{}3}{x^2}\right)\equal{}\log\left(1\plus{}\frac{3}{x^2}\right)$.
$ \lim_{x\to\infty}\log\left(1\plus{}\frac{3}{x^2}\right)\equal{}\log 1\equal{}0$.

So, as $ x$ grows without bound, the function becomes infinitesimally small.
This is best characterized by answer choice $ \boxed{\textbf{(D)}}$.[/hide]


If the ratio of the legs of a right triangle is $ 1: 2$, then the ratio of the corresponding segments of the hypotenuse made by a perpendicular upon it from the vertex is:

$ \textbf{(A)}\ 1: 4 \qquad
\textbf{(B)}\ 1: \sqrt{2} \qquad
\textbf{(C)}\ 1: 2 \qquad
\textbf{(D)}\ 1: \sqrt{5} \qquad
\textbf{(E)}\ 1: 5$

[hide="Click for solution"]
[img]http://i43.tinypic.com/xfrlax.jpg[/img]
In any right triangle, we have $ x \cdot c\equal{}a^2$ and $ (c\minus{}x)c\equal{}b^2$.  Since $ \frac{a}{b}\equal{}\frac{1}{2}$ we have $ \frac{x}{c\minus{}x}\equal{}\frac{xc}{c(c\minus{}x)}\equal{}\frac{a^2}{b^2}\equal{}\frac{1}{4}$ or $ \boxed{\textbf{(A)}}$.
[/hide]


LaTeX
trigonometry
geometry
ratio
angle bisector
trig identities
Law of Sines

I'm not quite sure if everything I did was legal...<_< I feel like I'm consistently doing something incorrectly.
And I don't know Latex so this'll be kind of convoluted.

pg. 7 problem 3:

[hide]I drew the triangle, and since AD is an angle bisector, 7/BD = 5/(3-BD).  Solving that yields BD = 21/12 = 7/4, so DC = 5/4.

Then since BE is also an angle bisector, 3/EC = 7/(5-EC) --> EC= 15/10 = 3/2, so AE = 7/2.

If you assign C a mass of 7/2 and A 3/2, then B = 5/2.

E = (3/2)A + (7/2)C = 5   and   D = (5/2)B + (7/2)C=6.

Then checking to see if point P is the balancing point of BE:
(I don't remember if we were allowed to do this for cevians...I know we can't do this for lines inside quadrilaterals...>_> )

P=6D+(3/2)A=15/2
P=5E+(5/2)B=15/2, so P is the balancing point.

Since B=5/2 and E=5, x:y (which is BP:PE) = 5: (5/2) = [b]2:1[/b].[/hide]

p. 7 Problem 4:

[hide]Drew triangle as described...

I made AE=2x and AF=x, so then EB=12-2x and FC=16-x.

Mass of B=2x and since BM=MC, C=2x too.  Then from AB and AC, A= (12-2x)A + (32-2x)A.

F = 2xC + (32-2x)A = 32.

E = 2xB + (12-2x)A = 12.

M = 2xB + 2xC = 4x.

Again, not sure if I'm allowed to do this:

G = 4xM + ((12-2x)A + (32-2x)A) = 44
G = 12E + 32F = 44, so G is the balancing point of EF and AM.

EG:GF= 32:12 = [b]8:3[/b].
[/hide]

p. 7 Problem 5
[hide]
Drew triangle.

Since AT is an angle bisector , AB/BT = AC/TC, so 4/BT = 6/TC, so BT:TC= 2:3, so I assigned those as 2x and 3x.  Then B = 3 and C=2, so T=5.

From AB, B=3, so A=9.  From AC, C=2, so A=4.

D = 3B + 9A = 12.  E = 2C + 4A = 6.

AF is part of AT the angle bisector, so DF:FE = 1:2, so the masses of D and E work for F as the balancing point.

(and again the not-sure-if-this-is-allowed part)

F = 5T + (9A + 4A) = 18
F = 12D + 6E = 18.  F is the balancing point for AT and DE.

so AF:AT= 5: (5+9+4) = [b]5:18[/b].[/hide]

And can someone explain exercise 12 on page 6?

[quote="avcamethyst"]
pg. 7 problem 3:

[hide]Since B=5/2 and E=5, x:y (which is BP:PE) = 5: (7/4) = [b]20:7[/b].[/hide][/quote]

Everything is right until this last step. You used the wrong mass for B in your final calculation.

[quote="avcamethyst"]
p. 7 Problem 4:

Drew triangle as described...[/quote]

I think you drew your figure wrong. E is supposed to be on AC and F on AB. It appears you had them switched. Otherwise your calcuations and ideas are all correct! The answer should be EG:GF = 3:2.

p. 7 Problem 5 is totally correct.

[quote="avcamethyst"]And can someone explain exercise 12 on page 6?[/quote]

I think the answers to exercise 12 are incorrect. Let D have a mass of 5, and E have a mass of 1 (in order to make F the balancing point of DE). Now split the mass at B coming up the sides AB and CB. This should give point C a mass of 4, A a mass of 3/4, and B a mass of 1/4 + 1 = 5/4.

By splitting the mass at B it guarantees that F will be the balancing point of both BG and ED. Since B has a mass of 5/4 and F has a mass of 6, G must have a mass of 19/4.

You should check algebraically that G is actually the balancing point of AC - just because the masses add up doesn't gaurantee this. Once you know that it is, you know that AG:GC = 4:3/4 = 16:3 and BF:FG = 19/4:5/4 = 19:5 (which is different than what the handout says, but I believe these are right).

>.< I need to stop doing that...(plugging in wrong numbers and drawing incorrectly and other silly stuff like that. ick.) 3 is fixed, 4...not exactly.  It takes me forever to type that out. >.<

12:  Thanks!  No wonder the numbers didn't even work when I attempted to do it backwards haha.

*********************************

P. 3 exercise 3:[hide]
Can we look at figure 4b and see that 1A+2L=1B+2M=2N+1C=3G? Or am I supposed to do something more elaborate? @_@[/hide]

p. 4 exercise 5:
[hide]Draw picture. (I believe I did it right this time.  Hopefully, at least. XD )

I don't know if this is actually a proof or if I'm just being redundant.

from the picture...

A=2, B=1, so E=3.  --> 3E=2A+1B
A=2, D=1, so H=3.  --> 3H=2A+1D
D=1, C=2, so G=3.  --> 3G=1D+2C
C=2, B=1, so F=3.  --> 3F=1B+2C

3H+3F = (2A+1D)+(1B+2C) = (1B+2A)+(1D+2C) = 3E+3G

So K (intersection of EG and HF) is the balancing point, so HK:KF=1 and EK:KG=1,  which means the diagonals bisect each other, so EFGH is a parallelogram.[/hide]

p. 4 Exercise 6

[hide]Picture.

AH:HD=m:n
AE:EB=m:n
BF:FC=n:m
CG:GD=m:n

A=n, C=n, B=m, D=m.

(m+n)H=nA+mD
(m+n)G=mD+nC
(m+n)F=mB+nC
(m+n)E=nA+mB

(m+n)H + (m+n)F = (nA+mD) + (mB+nC) = (nA+mB) + (mD+nC) = (m+n)G + (m+n)E

same explanation as above only substitute in the variables.[/hide]

p. 4 exercise 7:

[hide]Angle bisectors of a triangle are concurrent (at I).

AD, BE, and CF are the angle bisectors.

since we have bB, cC, and aA,...

From both angle bisector theorem and mass point if we take D, E, and F as balancing points:
BF:FA=a:b
BD:DC=c:b
CE:EA=a:c

D=b+c
E=a+c
F=a+b

I=(a+b)F+cC=(b+c)D+aA=(a+c)E+bB=a+b+c.[/hide]

p. 4 exercise 9:
(a)[hide]
Picture.

Law of Sines:
sinC/AB=sinA/BC

Angle bisector theorem:
AB/AD=BC/DC

Multiply the two equations:
(sinC/AB)(AB/AD)=(sinA/BC)(BC/DC) --> sinC/AD=sinA/DC --> ADsinA=DCsinC.[/hide]

(b)
[hide]
Rearrange previous equation: DC/AD=sinA/sinC and plug in.

DC/AD=(4/5)(25/24)=5/6

Picture.

so AD=6 and DC=5, so A=5 and C=6, and since M is the midpoint of BC, B=6 and M=12.  D=5A+6C=11.

E=6B+11D=5A+12M=17, so E is the balancing point of BD and AM.

AE:EM=12:5
BE:ED=11:6[/hide]

And as a side note:  I did #1 and 6 on the power round last week, and the "official" solution for 1 was the same as mine (as expected), but their way of proving 6 was a lot more straight forward than my whole assigning variables to each segment thing.  @_@ Didn't even think of interpreting the lengths with semiperimeter-side.  >.<

Those all look good!

And on the Power Round a solution doesn't have to look exactly like the official solution to be correct. As long as it is logical and clearly written, it is fine. I thought as a team we did fairly well on the power round at Practice Session 2. All the solutions that were turned in were correct.

lovely. ^_^

Now on to the difficult ones.  wheeeeeee

Contest Problems on p. 7:

Problem 1:

[hide]Drew triangle.

A=3, B=1, so E=4.
B=1, C=2, so D=3.

4E+2C=6F, 3A+3D=6F, so F is the balancing point for EC and AD.

So then EF/FC=2/4=1/2, and AF:FD=3/3=1.

1+1/2=[b]3/2[/b].[/hide]

Problem 2:

[hide]Drew triangle.

A=4, C=3, so B'=7.
A=4, B=2, so C'=6.
B=2, C=3, so A'=5

4A+5A'=9P, 7B'+2B=9P, so P is the balancing point of AA' and BB' (but not necessarily of CC').

So AP:PA'=[b]5:4[/b].[/hide]

Problem 3:

[hide]Drew triangle.

By the angle bisector theorem, we get AE=7/2, EC=3/2, BD=7/4, and CD=5/4.

Then A=3/2, C=7/2, so E=5.
C=7/2, B=5/2, so D=6.

6D+(3/2)A=(15/2)P, and 5E+(5/2)B=(15/2)P, so P is the balancing point of AD and BE.

EP=y, PB=x.

x:y=5:5/2=[b]2[/b][/hide]

Problem 4: hint please? >.<

Problem 5:

[hide]Triangle.

Angle bisector theorem, BT:TC=3:2, so B=2, C=3, and T=5.

Since B=2, A=6 from side AB.  Since C=3, A=6 from side AC.  D=8 and E=9.

So A=12 from both sides AB and AC.

12A+5T=17F.  8D+9E=17F.  F is the balancing point of AT and DE, so AF:FT=[b]5:6[/b].
[/hide]

Problem 6:

[hide]Question: Is there extra information in this problem, or am I just doing it incorrectly?  I didn't use the 72 degrees at all.

Triangle.

Since E is the midpoint of AC, A=1, C=1, E=2.

CD:DB=2:1, so C=1, B=2, and D=3.

2B+2E=4F, 1A+3D=4F, so F is the balancing point of AD and BE, so EF:FB=1.

Then I labeled EF=FB=x and BD=y, and DC=2y.

[BDF]=(1/2)xysinB
[FDCE}=[BEC]-[BDF]= (1/2)(2x)(3y)sinB-(1/2)xysinB=(5/2)xysinB.

[BDF]/[FDCE]=((1/2)xysinB)/((5/2)xysinB))=(1/2)/(5/2)=[b]1/5[/b][/hide]

more to appear...

First, as a heads up, Problem 5 asks for AF:AT.

Problem 4
[hide="Hint: Method"]
For Problem 4, the ideal balancing point should be obvious.  After that, I assigned variables for the mass at B and C (R in the solution), the two masses at A (necessary for the transversal, S1 and S2 below) and the length of AF, to which FB, AE, and EC can be related. Then plow through the algebra (after determining what the masses of E and F would be in terms of R, and S1 and S2 respectively).
[/hide]

[hide="Solution"]
For Triangle ABC, with balancing point at G, set the masses of both B and C to $ R$, and the length of AF to $ x$.

Therefore:
$ \text{FB} \equal{} 12\minus{}x$
$ \text{AE} \equal{} 2x$
$ \text{EC} \equal{} 16\minus{}2x$

Let $ S_1$ and $ S_2$ be the masses assigned seperately to A from sides AC and AB respectively (so $ S_1$ relates to E and $ S_2$ relates to F).  For E and F to be balancing points of their repective sides, then:
$ (16\minus{}2x)R\equal{}(2x)S_1$
$ (12\minus{}x)R\equal{}(x)S_2$

Therefore:
$ \text{E's Mass}\equal{}R\plus{}S_1\equal{}\frac{8R}{x}$

$ \text{F's Mass}\equal{}R\plus{}S_2\equal{}\frac{12R}{x}$

Finally:
$ \frac{\text{EG}}{\text{GF}}\equal{}\frac{\text{F's Mass}}{\text{E's Mass}}\equal{}\frac{3}{2}$
[/hide]

Problem 7:

[hide]
Draw Triangle ABC with balancing point P.  PAF and PFB at 40 and 30, means that AF:FB = 4:3, as the two triangles share a height.  Let the masses at A, B, and C be 3, 4, and $ M$ respectively.  With D and E as balancing points of their respective sides, the area of PEA and PDC can be computed as $ 28M$ and $ 140\M$ because each triangle shares a height and base ratio (in terms of $ M$) with a known triangle, PEC (area 84) and PBD (area 35), and ratios of M:3 and 4:M from unknown:known.  As CFB and CFA share a common height, their areas are also in proportion:
$ 4(65+\frac{140}{M})=3(124+28M)$

which simplifies to:
$ 3M^2+4M-20=0$

The only positive solution for $ M$ is 2.
Plug 2 into $ 28M$ and $ 140/M$ (to give 56 and 70), and then add in the other 4 given triangles.  The total is the Area of ABC: [b]315[/b].
[/hide]

Problem 8:

[hide]
After drawing ABC, points E, D, and F, and rays AF and DF (given), draw a line to complete triangle AFB.  Then, let the masses of A, F, and B be such that E is the balancing point of ABF.

Therefore,
$ \text{Mass of A}: \text{Mass of B} \equal{} 2: 3$

Looking ahead, lets assign the masses at 4 and 6 respectively.

As:
$ \text{Mass of C}: \text{Mass of B}\equal{}\text{BE}: \text{EC}\equal{}3: 2$

the mass of C is 9.  The Mass of A + Mass of F = Mass of C, so the mass of F is 5.  The Mass of D = The Mass of A + Mass of B, so the mass of D is 10.

Finally:
$ \text{DE}: \text{EF}\equal{}\text{Mass of F}: \text{Mass of D}\equal{}1: 2$
[/hide]

By the way, these problems are found as the Contest Problems on the mass point geometry article/worksheet [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204362]here[/url].

Problem 9:

[hide]
First, let's give the triangle and its points labels.  Define the triangle as ABC, the trisection points DE opposite point B, with D closer to A, the F as the midpoint of BC defining the median as AF.  Also, let G be the intersection of AF and BD and H be the intersection of AF and BE.  Therefore:
$ x: y: z\equal{}\text{AG}: \text{GH}: \text{HF}$
or flipped around if it works out that AG is smaller that HF.

First, assume that G is the balancing point, making D and F balancing points of their respective sides. $ \text{AD}: \text{DC}\equal{}1: 2$, so the mass assigned to A must be twice that assigned to C.   $ \text{BF}: \text{FC}\equal{}1: 1$, so the masses assigned to B and C must be equal.  Therefore, let the mass of B and C be 1, and the mass of A be 2.  This means the mass of F will be 2, so:
$ \text{AG}: \text{GF}\equal{}\text{Mass of F}: \text{Mass of A}\equal{}1: 1$
$ \text{AG}: \text{AF}\equal{}1: 2$

Now, assume that H is the balancing point, making E and F balacing points of their respective sides.  Now, the mass of both C and B has to be twice that of A, so assign the mass of B and C to 2 and the mass of A to 1.  F is assigned a mass of 4.
$ \text{FH}: \text{HA}\equal{}\text{Mass of A}: \text{Mass of F}\equal{}1: 4$
$ \text{FH}: \text{FA}\equal{}1: 5$

Subtracting both of these ratios from 1 will give the ratio $ \text{GH}: \text{AF}\equal{}3: 10$.  Writing these ratios as fractions yields the result:

$ x: y: z\equal{}\text{AG}: \text{GH}: \text{HF}\equal{}\frac{1}{2}: \frac{3}{10}: \frac{1}{5}\equal{}5: 3: 2$

Note: if you worked from F to A, $ x$ would have been smaller than $ z$, so you would have been forced to flip the ratio around.
[/hide]


Let $ ABC$ be an acute-angled triangle, and $ D$ an arbitrary point of the edge $ BC$ different from $ B$ and $ C$. The outside common tangent to the incircles of $ \triangle ABD$ and $ \triangle ACD$ different from $ BC$ intersects the segment $ AD$ in the point $ E$. Prove that the length of $ AE$ doesn’t depend on the choice of $ D$.

Denote D,E be the tangency points of incricles of triangles ABD,ACD with BC, respectively.
$ DH \equal{} \frac {AD \plus{} BD \minus{} AB}{2},DK \equal{} \frac {AD \plus{} DC \minus{} AC}{2}\Rightarrow DE \equal{} HK \equal{} DH\plus{}DK\equal{}AD \minus{} \frac {AB \plus{} AC \minus{} BC}{2}$
$ \Rightarrow AE \equal{} AD \minus{} DE \equal{} \frac {AB \plus{} AC \minus{} BC}{2}$


What is the least integer value of x for which the ineequality $3x>2x+1$ is true?

[hide]3x > 2x + 1
x > 1
the answer is [b]2[/b][/hide]
ToBe did you even have the right problem?

He has the right problem. It looks weird because he put periods and they look like decimal points.

This is what he's trying to say:
[hide]
$3\cdot2=6\\ 2\cdot2+1=5\\ 6>5$[/hide]

the x looked like the variable so i didn't get it

Yup, looked like decimal points. Sry for confusion  :oops:

So he isn't solving it he is proving that the answer is right.

Click Mr. Green for the answer!
[hide=" :D "]3x>2x+1
3x-2x>1
x>1
2[/hide]


[quote="Ditdo"]convert to sequences to solve that function[/quote]
Yes,I know that...
But...How??? :?:

[quote="anhminh"]Find $f: R-\to R$ continues  satisfy:
Pro 1:
$f(f(x))=f(x)+2x$
Pro 2"
$f(f(x))=2f(x)-x$[/quote]
 Can [b]Ditdo[/b]post your solution?
 My solution not simple :lol:


algebra
polynomial
number theory unsolved
number theory

let $x_0=4,x_1=x_2=0,x_3=3$ and $x_{n+4}=x_n+x_{n+1}$.
Prove that $x_p$ is a multiple of $p$ if $p$ is a prime number.

I think it follows from writing $x_n=\alpha_1^n+\alpha_2^n+\alpha_3^n+\alpha_4^n$, where the $\alpha$'s are the roots of $x^4-x-1=0$. I'm not even sure that's true, because I haven't checked all the relations derived from Viete, but it seems Ok.. :).

Edit:

yes, actually, I think it does work. I think the computations work out fine, and we have $x_p=\sum \alpha_i^p=(\sum\alpha_i)^p=0$ in the splitting field of $x^4-x-1$ over $\mathbb Z_p$ (we also use $\sum\alpha_i=0$).

Anyone remembers the monster given in our TST this year. This problem is in AMM and a particular case of the monster.

I'm sure you're right grobber but i can't understand the end of your proof. What is the link between the spliting field and the original pb ?

We want to work in a field which contains $\alpha_i$ and in which $p=0$, so that we have $\sum\alpha_i^p=(\sum\alpha_i)^p$. We thus work in the splitting field of the polynomial $x^4-x-1$ over $\mathbb Z_p$.

[quote="kueh"]Let $a,b,c, m \in \mathbb{Z}$ with $b$ odd.

Define a sequence: $x_0=4,x_1=0, x_2=2c, x_3=3b, x_{n+4}=ax_n + bx_{n+1} + cx_{n+2}$.

Show that for $p$ prime, $p|x_{p^m}$.

PS: apparantly there's a nice proof of this more general thm? The first question is the case when: $a=b=m=1, c=0$[/quote]
This popped up on some correspondence work a few months back.. can't remember the full proof, but the gist was to use combinatorial considerations to build a ladder graph to make up an expression for $x_n$, or to be more specifically the term in $x_{4p+3q+2r}$ that contained $a^pb^qc^r$. It looked something like
$\frac{(p+q+r+1)!(4p+3q+2r)}{p!q!r!}$
but I probably haven't got it exactly right. Anyway, then you note that 4, 3 and 2 are arbitrary, and then I've forgotten. But it did work.
Strangely enough, I never used the fact that b was odd. I don't think it's needed.

I think this is the problem from the Romanian TST which was mentioned above by Harazi. Several people noticed then (when it was discussed, right after the TST's) that the condition that $b$ is odd appears to be unnecessary.

Clearly, $x_n=A\alpha_1+B\alpha_2+C\alpha_3+D\alpha_4$ where $\alpha_i$ are the roots of $x^4-x-1=0$, and we know too that $\sum_{i=1}^4 \alpha_i=0$, and $\sum_{1\le i<j\le 4} \alpha_i\alpha_j=0$ then by the problem
as $\sum_{i=1}^4\alpha_i^3=3(1)$ (See Newton's Formula) then $A=B=C=D=1$ by elementary algebra calculations.

Now, if $p$ is a prime number we now that \[\binom p n=0 \ mod \ p \quad {\rm if} \  1<n<p.\]
Then
\[
x_p=\sum_{i=1}^4 \alpha_i^p=\alpha_1^p+\alpha_2^p+\alpha_3^p-(\alpha_1+\alpha_2+\alpha_3)^p,
\]
and do the same with $\alpha_3$, $\alpha_2$ and $\alpha_1$ then adding everyone, we get that
\[
4x_n=\sum_{i=1}^{p-1} \binom p i Q_i(\alpha_1,\alpha_2,\alpha_3,\alpha_4).
\]
And its really easy to probe that $Q_i(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ is an integer for every $i=1,\dots,p-1$, [the general coefficient is different but binomial it's enought] then $4x_n=0 \ mod \ n$ so, if $p>2$ the result holds, and $x_2=0=0 \ mod 2$.

Any question?  :lol:  ;)

Best regards.

This is actually a restatement of what was done above :).

I know but some guys didn't understand your proof, only try to help a little.


function
invariant
geometry
geometric transformation
reflection
algebra
domain

how does the method of images work? I can see why it works (by going back to the form of the Green function and differentiating) but don't see how useful it is
to solve problems. At the moment I am basically memorising the different
images for each boundary condition which I am sure is not the way to go.
How do you think about it so that one can work out the solution logically from the method of images?

That's an excellent question.  I taught this topic last year, and I don't know if I have a good answer.  First, I assume that you're talking about Green functions for the Laplacian with Dirichlet boundary condition (rather than something more general).

Personally, I think of the method of images as an ad hoc trick.  The only "theory" behind it that I can describe is that harmonic functions are invariant under conformal maps, and reflection through a line and "reflection" through a circle are both conformal maps.  (Indeed, lines and circles are the same thing in conformal geometry.)

I suppose that a general principle might go something like this:  If your domain $\Omega$ can be "reflected" through some part (or all) of its boundary to obtain some new domain $\Omega'$ such that you already know the Green function for $\Omega'$, then you can deduce the Green function for $\Omega$ from it using the method of images.

Remember that a Green function on $\Omega$ at $x$ can be characterized as a function that is harmonic away from $x$, "looks like" the fundamental solution at $x$, and is zero at $\partial\Omega$.  By starting with the Green function on $\Omega'$ at $x$, you're halfway home.  You know that this function is harmonic away from $x$, "looks like" the fundamental solution at $x$, and is zero at the part of the boundary that you are *not* reflecting through.  All that's left is to subtract off some harmonic function that has the same boundary values at $\partial\Omega$.  That is where the the Green function on $\Omega'$ at *the reflection of $x$* comes in.  This function has the right boundary values at $\partial\Omega$ (after multiplication by a constant) and is harmonic in $\Omega$ (since the reflection of $x$ is not in $\Omega$).  That's it.

A physicist might say that you are looking for a configuration of point charges in the plane such that only one of these point charges lies in your domain $\Omega$, and so that your boundary $\partial\Omega$ has constant zero potential.  One can sometimes reason out what this configuration of point charges is using symmetry.  (This is just the math couched in physics words.)

As a pure mathematician, I don't consider the method of images to be very interesting.  To me, it's an elegant way to derive the Green function for the disk, no more.  Maybe someone else has another perspective on this, or can enlighten me about why the method of images is interesting.


Let $S$ be the sum of the base 10 logarithms of all the proper divisors of 1000000. What is the integer nearest to $S$?

[hide="Solution"]
$10^6 = 2^6 5^6$ has $7 \times 7 - 2 = 47$ proper divisors.  Note that, for any proper divisor of a number $N$ - call this divisor $a$ - , there is another proper divisor $b$ such that $ab = N$.

We can then add the logarithms symmetrically, adding $\log_{10} a + \log_{10} \frac{10^6}{a} = 6$ for each proper factor $a$.  Forty-six proper factors are paired in this way, giving us an initial sum $S = 23 \times 6 = 138$, and we must then add the unpaired factor, $\log_{10} 10^3 = 3$, giving us a final answer of $\boxed{141}$.
[/hide]

(6+1)(6+1)=49 divisors including 1 and $10^6$.  Subtract two to get all proper divisors: 47 of them that is.

Masoud Zargar

[quote="Chocoboom"]how do you know that (2^6)(5^6) has 7 x 7 - 2 proper divisors?[/quote]

It's actually a formula, if the integer $N$ can be factored into $a^x \cdot b^y$, the number of total divisors is $(x+1)(y+1)$ but a proper divisor does not include $1$ or $N$ so we subtract $2$ from that.

My solution:
[hide]
When you add logarithms, you multiply.

We find all of the positive integer divisors of $10^6$ by considering all terms of the expansion of $(1+ 2+ \ldots + 2^6)(1 + 5 + \ldots + 5^6)$. Since we are then going to multiply all these factors, notice that an expansion will yield $1(1 + 5 + \ldots + 5^6) + 2(1+5+\ldots+5^6)+\ldots+(2^6)(1+5+\ldots+5^6)$.

So there are respectively, $7$ multiples of $1$, $2$, $2^2$, . . . , and $2^6$. And multiplying them yields $2^{7(0 + 1 + \ldots + 6)} = 2^{147}$. The same reasoning applies for total multiples of $5$ in the final product. So the final product is $10^{147}$.

Since we want proper divisors, divide $10^{147}$ by $10^6$, so the answer is $\log {10^{141}} = 141$.[/hide]

o... i never knew that equations... thats interesting... :roll:

[hide]We start by noticing that $1000000=2^65^6$ has $(6+1)(6+1)-2=47$ proper divisors. Now, suppose that $d$ is a divisor of $10^6$. It follows that $\frac{10^6}{d}$ is also a divisor of $10^6$, and $d\cdot \frac{10^6}{d}=10^6$.

Now, each of our divisors except $10^3$ has another divisor to pair up with, giving us a total of $23$ pairs of $10^6$. Now, addition of logarithms is the equivalent of multiplication, so our answer is $26\log_{10} 10^6+\log_{10}10^3=26\cdot 6+3=\boxed{141}$.[/hide]

Why does it ask for the "integer nearest" when the answer is an integer? This confused me for a long time.

For any proper divisor $d$, we have $\frac{10^6}{d}$ is a proper divisor unless $d=1.$ So this is just, $26\log_{10} 10^6+\log_{10}10^3=26\cdot 6+3=\boxed{141}.$


1 liter = 0.001 meters cubed.

 :) ur welcome. :P  ;)

[quote="LuCy4EvA"]this is not a problem but we've been taught the relationship between inches and feet, meters and millimeters and stuff, but no one ever told me how to solve stuff like how many liters of water are in a box with the volume of 12 meters cubed....  :?[/quote]

That's a lot of water.

One cubic centimeter equals one milliliter. One milliliter of water weighs one gram.

If you have a TI-83, go to apps, then Science tools, and pick what the subject is.

[quote="Treething"]TI-83 Silver Edition comes with it, not the regular TI-83 :P .[/quote]

fine...TI-83 plus


A cylinder has radius $R=0.4m$ and height $h=1.6m$. Find the ratio between the mechanical work necessary to lift the body from horizontal to vertical and the mechanical work needed to do the opposite.

[hide]When lying on its side, the com of the cylinder will be at height h[size=75]1[/size] = R above the ground.
When the cylinder is raised up on its edge s.t. its com is passing through that edge. then after passing this point it will fall over to rest on its bottom face by virtue of its own weight and so no further effort is required.
Its com, when passing through the bottom edge, wil be d where d² = R² + (h/2)².
d² = (0.4)² + (0.8)²
d = 0.4√5
=======

The height difference is  Δh[size=75]1[/size] = d - R = 0.4√5 - 0.4 = 0.4(√5 - 1)
WD[size=75]1[/size] = mg*0.4(√5 - 1)
================

When the cylinder is upright, and is then pushed over s.t. its com is passing through that bottom edge again, then after passing this point it will fall over to rest on its side.
When upright, the cylinder's com is at h[size=75]2[/size] = (h/2) = 0.8
This time, difference  in height is Δh[size=75]2[/size] = d - h[size=75]2[/size]

Δh[size=75]2[/size] = 0.4√5 - 0.8
Δh[size=75]2[/size] = 0.4(√5 - 2)
WD[size=75]2[/size] = mg*0.4(√5 - 2)
================

Ratio of works done is
WD[size=75]1[/size] / WD[size=75]2[/size] = mg*0.4(√5 - 1) / mg*0.4(√5 - 2)
= (√5 - 1) /(√5 - 2)
==============[/hide]


Claim: Given $n$ balls, they are all of same color.

We use induction on $n.$
[b]Base case[/b]: For $n=1,$ this is trivially true.
[b]Induction case[/b]: Let the statement be true for $n=k, k > 1, k \in \mathbb{N}.$
Now, given $n+1$ balls, take the first $n$ balls (excluding the last one). These $n$ balls are all of the same color (by induction assumption).  Again, take the last $n$ balls (excluding the first ball.) These $n$ balls are of the same color (by induction assumption). But, these two sets of ball have some ball(s) in common. Therefore, all the $n+1$ balls are of the same color.

This completes our proof.

The inductive "step" falls short already on $n=2$ - the sets $\{\text{the first ball}\}$ and $\{\text{the second ball}\}$ have no common elements.


IS IT POSSIBLE TO GET AN INTEGER... AM I MISSING SOMETHING

No, I don't think there are any integer solutions. I agree with NSFY's calculations.

there definitley isn't

did you make it up?

[quote="FrankieBlueEyes"]Find three consecutive integers such that the sum of 2 times the first and 3 times the second equals 10 more than the third.[/quote]

So you have 3 integers: x, x+1, and x+2

Then you have 2x+3(x+1)=10+x+2, then you solve that out and get x=9/4, so pretty much this makes absolutely no sense  ;)

Did you mess up on the numbers or maybe I just didn't understand the problem?

[quote="ineedmathhelp"]Well, if the problem's really vague you could also think of the first number as x, the second as x-1, and the third as x-2...

But that gives an answer of 11/4 for the first integer, so I guess there are no solutions.[/quote]

You mean the third number in the sequence would be x, and the second x-1, and the third x-2, right?


If abc is a 3 digits number such $ 16*abc\equal{}49*cba$ find the digits $ a,b,c$

[hide="solution"]16(100a+10b+c)=49(100c+10b+a)

4884c+330b=1551a

444c+30b=141a

148c+10b=47a

Since they are digits, and the maximum value of the RHS is 47*9=423, c< 423/148=about 2.85.

Case 1: c=0

cba isn't 3 digits.

Case 2: c=1

Then b=4 and a=4.

abc=441, checking, is a solution.

Case 2: c=2

Then b=8 and a=8.

882 is indeed another solution.

Those are the only solutions.[/hide]

A variation of the solution posted by 1=2:
[hide]16(100a+10b+c)=49(100c+10b+a)
4884c+330b=1551a
Dividing by 33
148c+10b=47a
Rearranging,
7c+10b=47a-141c  and  7c+10b=47(a-3c)
Because b and c are between 0 and 9,
0<7c+10b=47(a-3c)<98
The choices are:
a-3c=1, so c=1, b=4, a=3c+1=4, abc=441
a-3c=2, so c=2, b=8, a=8, abc=882[/hide]
An alternate way to solve it, may be called “brute force”, but it does not take much longer:
[hide]16abc=49cba means that abc=49x, and cba=16x for some natural number x, so
(16)(49)x = 16abc = 49cba
Since cba>100 and abc<1000,
6<x<21, yielding 14 values of x to try, which yield the following values for abc=49x:
343, 392, 441, 490, 539, 588, 637, 686, 735, 784, 833, 882, 931, 980
Discarding the ones who would not yield a 3 digit cba divisible by 16, leaves only
441=9*49, 637=13*49, and 882=8*49 to be checked.
9*16=144 and 18*16=288 making abc=441, and abc=882 solutions
13*16=208 is not 637 reversed so 637 is not a solution[/hide]