A man stopped his car oopposite a hotel and immediately knew that he was bankrupt.   :cursing:  How?

Umm... maybe he saw a  :ninja: running out of the hotel with a large safe that has the guy's name on it??

To be very annoying, I think the proper term here is "broke" (def: having no money: without any money to spend), a condition of the bank account, not "bankrupt" (def: unable to pay debts: judged legally to be unable to pay off personal debts), a legal status that requires one to file for bankrupcy in court and which does not necessarily require that one has no monetary assets.

My answer would be, "He simultaneously looked in his wallet."

My answer would be, "He simultaneously looked at what was in his hand/in front of him."

This is tricky....I've heard of this before

[hide="DONT LOOK UNTILL YOU TRIED REALLY REALLY REALLY HARD"] He was playing Monopoly[/hide]

[quote="JBL"]Hmm, I guess I was wrong about the word choice, then  :oops:[/quote]
doesn't matter :lol:


inequalities
vector
geometry
real analysis
real analysis unsolved

Prove that in every prehilbert space(a vector space with inner product) the following inequality holds:
for every $a,b,c,d$ from the space $||a-b||||c-d|| + ||b-c||||a-d|| \geq ||a-c||*||b-d||$.

It reduces to the usual Ptolemy inequality in $\mathbb R^3$, which has been discussed several times on the forum (try the "Geometry" section) by considering the subspace spanned by $b-a,c-a,d-a$, which is isometrically isomorphic to $\mathbb R^3$.

Or you can directly suppose $a=0$ and work with $b/||b||^2$ and so on. But grobber's idea is more conceptual, I like it more.  :)


Η άσκηση μας λέει να βρούμε πόσοι πενταψήφιοι θετικοί ακέραιοι αριθμοί υπάρχουν που το καθένα απο τα ψηφία τους, εκτος του τελευταίου, είναι μεγαλύτερο ή ίσο του επομένου. Η λύση που δίνει η ΕΜΕ είναι ο συνδυασμός 14 ανα 5 πλιν της μονάδας(αφαιρεί τον αριθμό 00000). Δεν καταλαβαίνω όμως... Πως προκύπτει αυτό? :huh: γιατι 14?

Προσπάθησε να συσχετίσεις τέτοιους αριθμούς με ακολουθίες με 9 | και 5 *. Δίνω μία τέτοια συσχέτιση και το αφήνω σε εσένα να ανακαλύψεις γιατί δουλεύει.

23558 = ||*|*||**|||*|

[quote="Protonios"]Είναι οι συνδυασμοί [b]με επανάληψη[/b]. Ο τύπος είναι ν+κ-1 ανά κ. Οπότε είναι straightforward.[/quote]

now i get it. thanks :)


How could this cause an error in Java?

[code]
Integer.parseInt(get);
[/code]

Note that get is a String object.

The compiler complains that

[quote]method "parseInt" does not recieve ints as an argument.[/quote]

What's going on here?[/quote]

Are you sure $get$ is a String and not an integer?  What does the surrounding code look like?

I'm completely sure.

Here's a snippet of code directly above the above:
[code]
String get;
int num;
public class blahblahblah {
public void blahblahblah() throws IOException {
System.out.println("blahblahblah");
get = System.in.read();
[/code]
And then the shebang:
[code]
num = Integer.parseInt(get);
[/code]

Always believe the error message - if its telling you not to pass an int, you're passing an int.

System.in.read() reads a single byte - not a character, not a string. Thus it is indeed a number, not a String.

If you want to read input properly you will need to do this:

[code]BufferedReader br = new BufferedReader(new InputStreamReader(System.in));[/code]

and then use the br.readLine() method to get the next line of text as a String.


[quote="jensen"]$lim_{x \to 0} \frac{\sqrt{1+tanx}-\sqrt{1+sinx}}{x^3}=?$[/quote]
[hide="my solution "]$lim_{x \to 0} \frac{\sqrt{1+tanx}-\sqrt{1+sinx}}{x^3} =lim_{x \to 0} \frac{tanx-sinx}{{x^3}(\sqrt{1+tanx}+\sqrt{1+sinx})} =lim_{x \to 0} \frac{sinx(\frac{1}{cosx}-1)}{{x^3}(\sqrt{1+tanx}+\sqrt{1+sinx})} =lim_{x \to 0}\frac{1}{cosx}\frac{sinx}{x}\frac{1-cosx}{x^2}\frac{1}{\sqrt{1+tanx}+\sqrt{1+sinx}}$
$=\frac{1}{4}$ [/hide]


In a triangle $ ABC$ with $ AB \equal{} AC$ and $ \angle BAC \equal{} 100^o$. Let $ D$ lies on $ BC$ such that $ AC \equal{} DC$ and $ F$ lies on $ AB$ such that $ DF$ parallel to $ AC$. Find $ \angle ACF$.

Nice problem!
[hide]Rotate D 100 degrees about C away from the triangle to form point X. Then $ \triangle DCX$ is isoceles, and we find, F,D, and X are collinear. Let Y be the point on the same side of AC as X such that ACY is equilateral. Then $ \triangle BAY$ and $ \triangle XCY$ are isoceles. We find $ \angle BYA \plus{} \angle AYC \plus{} \angle CYX \equal{} 70 \plus{} 60 \plus{} 50 \equal{} 180$, so B, Y, and X are collinear. But then $ \angle FBC \equal{} \angle FXC \equal{} 40$, so FBXC is cyclic. So $ \angle CFX \equal{} \angle CBX \equal{} 30$. We now find $ \angle ACF \equal{} 30$.[/hide]
I left out a few steps that were just calculating the angles.

Take $E$ on $(BA$ so that $CE=AC$. Then $\triangle ACE$ is a [b]Langley triangle[/b], i.e. $80^\circ-20^\circ-80^\circ$. This makes $\triangle DEC$ equilateral, hence $DE=AB, \angle EDF=\angle ADF+\angle ADE=70^\circ+10^\circ=80^\circ=\angle AFD$, hence $\triangle DEF$ is a Langley triangle as well; we got $EF=CE, \angle FEC=80^\circ$, consequently $\angle ECF=\angle EFC=50^\circ$ and $\angle ACF=\angle ECF-\angle ACE=50^\circ-20^\circ=30^\circ$, done.

[b]Personal remark[/b]: Join this proof and the previous one to easily solve one of my few problems proposed on M.L.:

[size=150][i]For an isosceles triangle $ABC$ with $\hat A=100^\circ$ take $D$ on $(BC$, $E$ on $(AC)$, $F$ on $(BA$ so that: $\angle BAD=\angle CBE=10^\circ, \angle ACF=20^\circ$; then $AF=BD=CE$[/i][/size]

Best regards,
sunken rock


Prove, that if for positive integer $N$
$\exists a,b,c\in\mathbb{Z}$ are divisible by $3$ such, that $N=a^2+b^2+c^2$,
then $\exists x,y,z\in\mathbb{Z}$, which all not divisible by $3$ such, that $N=x^2+y^2+z^2$.

Three vectores $u=\{2,1,-2\},v=\{1,2,2\},t=\{-2,2,-1\}$ ortogonal and |u|=|v|=|t|=3. Therefore we have ortogonal matrice
A=(u/3.v/3,t/3). If a,b,c divisible 3 we have new solution (x,y,z)=A(a,b,c).

And why all $x,y,z$ are not divisible by $3$. It isn't true.

[quote="Rust"]It is not true. $11=9+1+1\not=x^2+y^2+z^2$ for $xyz\not=0$(mod 3),or
$19=9+9+1\not=x^2+y^2+z^2$ for $xyz\not=0$(mod 3).[/quote]
You didn't understand condition of problem: all of numbers $a,b,c$ are divisible by $3$ :arrow: $N$ is divisible by $3$, but neither $11$, nor $19$ isn't divisible by $3$

[quote="Rust"]If $3|N=a^2+b^2+c^2$, then or $3\not |a,3\not |b,3\not |c$ or $3|a.3|b,3|c$. If Aa divisible 3, then we can $A^ka$.
I don't undestand what is your question mean?[/quote]
Are you sure, that this prosess will end?

I undestand you. Let $n(a,b,c)=min(ord_3(a),ord_3(b),ord_3(c))$. We can prove that $n(x,y,z)<n(a,b,c)$ if (x,y,z)=A(a,b,c).
If $n(a,b,c)=ord_3(a)<ord_3(b)\le ord_3(c)$ it is obviosly. If $n(a,b,c)=ord_3(a)=ord_3(b)<ord_3(c)$ then
n(x,y,z)=-1+min(ord_3(2a+b),ord_3(a+2b),ord_3(-2a+2b)) can be equal n(a,b,c).
All nontrivial roots Aa=0(mod 3) are (1,1,0),(1,0,1),(0,1,1). Therefore for this cases we can use one of matrice $\sigma(A)$. Sufficiently A and $B=\sigma (A): A(a,b,c)=B(b,c,a)$.
It is easy to show, that $n(A(a,b,c))<n(a,b,c)$ or $n(B(a,b,c))<n(a,b,c)$.
This proces stopped.


1.) How do you solve this system

$ x \equiv 3 (mod 6)$
$ x \equiv 1 (mod 8)$
$ x \equiv 6 (mod 9)$

2.) How do you know if you can cancel in modulo? I know the rules before, but i already forgot.

For example

Does $ 2x \equiv 6 (mod 8)$ implies $ x \equiv 3 (mod 8)$?

2. No consider $ 2(7) \equiv 6 \pmod{8}$, but $ 7 \not\equiv 3 \pmod{8}$

1. $ x \equiv 3 \pmod{6} \rightarrow x \equal{} 6y \plus{} 3$, for some $ y$.
$ 6y \plus{} 3 \equiv 1 \pmod{8} \rightarrow 6y \equiv \minus{} 2 \pmod{8} \rightarrow 6y \equal{} 8z \minus{} 2$
$ 6y \plus{} 3 \equiv 6 \pmod{9} \rightarrow 6y \equiv 3 \pmod{9} \rightarrow 8z \minus{} 2 \equiv 3 \pmod{9} \rightarrow 8z \equiv 5 \pmod{9}$
Now just try to find the lowest $ z$ which satisfies, which happens to be $ 4$, and substitute.

$ 6y \equal{} 8z \minus{} 2$
$ 6y \equal{} 8(4) \minus{} 2 \equal{} 30$
$ 6y \plus{} 3 \equal{} 33$

Hope this makes sense.

2)  To be absolutely sure of anything you say in the world of modular arithmetic, rewrite in terms of variables.  In this case, $ 2x \equiv 6 \bmod 8$ means there exists an integer $ y$ such that

$ 2x \equal{} 6 \plus{} 8y \implies$
$ x \equal{} 3 \plus{} 4y$

hence it is true that $ x \equiv 3 \bmod 4$ but [b]not[/b] that $ x \equiv 3 \bmod 8$ because we can also have $ x \equiv 7 \bmod 8$.  In general, division by $ a \bmod m$ is always valid iff $ \gcd(a, m) \equal{} 1$.

This is pedantry, but where t0r wrote
[quote="t0rajir0u"]hence it is true that $ x \equiv 3 \bmod 4$ but [b]not[/b] that $ x \equiv 3 \bmod 8$[/quote]
I would have written, "hence it is necessarily true that $ x \equiv 3 \bmod 4$ but not necessarily the case that $ x \equiv 3 \bmod 8$."


Westpac is July 25 2007. Any old questions or tips on how to deliver on the contest and if possible, a Medal or a Prize?

Most old questions are in the Intermediate Forum. Pls put more.

Is there any website to obtain past papers for maths comp. ?

Damn bad - the questions are strictly copyrighted, and there are no sites currently having those questions. But look at [url]amt.edu.au[/url] and click the Activity button to see some past questions (eight per level).

I'm perfectly sure that teachers keep a few papers from previous years so the students are able to study it so they might stand a slight chance. All you need do is ask

The medals are up now........did you get one pda?
lol i sure didnt


There is a 5x5 chess board. Each square has a lightbulb.
There is a switch next to the first lightbulb of each row and there is another switch on top of the first lightbulb of each column (we have a total of 10 switches).
If you press a switch the state of every lightbulb in the corresponding row or column changes.
Prove that, no matter the first state, we can use the switches to make each row and column with more lightbulbs turned off than on.

If you know the solution, please don't write it. Kindly wait for others attempting to solve the problem. Any small hint would be good.

BTW, are we looking for an invariant or something?

[hide="Hint"]Keep track of the total number of lights that are off.[/hide]

Yay!  This is a pretty easy problem.  Consider[hide]the invariance principle.  Work in $\mathbb{Z}_{2}$.  No need for the extremal principle.[/hide]
Btw, it's a nice problem too. :)

Masoud Zargar


symmetry
graph theory
combinatorics unsolved
combinatorics

Six points A,B,C,D,E and F are given in the plane, no three of which are collinear.  Each segment XY where X and Y are any two of the six points is drawn using either red or blue ink.  Show that there must be four distinct points Q,R,S and T from among these six such that the segment QR, RS, ST and TQ are all the same colour.

I don't really get the question.  Could someone solve/prove/explain this question please?

Thanks

Oh... you should prove that there exists quadrilateral which segments have the same colour.  :wink:

[quote="me@home"]Isn't this the equivalent of saying: prove ramsey (4,2)?[/quote]

No, because it isn't a complete $K_{4}$ graph.

Also, it's not necessarily a quadrilateral.  The edges can cross.

It is a classical result that if you color all the edges between the six points with only two colors then there must be a monochromatic triangle.
Wlog, assume that $ABC$ is a red triangle.

Now, assume for a contradiction that no quadrilateral is monochromatic. Then, to avoid a red quadrilateral, there must be at least two blue edges between the three from $D$ to $A,B,C$. he same is true for $E$ and for $F$. Moreover, the three choices of these two blue edges have to be pairwise distinct to avoid a blue quadrilateral. This eans in particular that in each case the third edge has to be red.
By symmetry, we the may assume that $DB,DC$ are blue and $DA$ is red, that $FA,FB$ are blue and $FC$ is red, that $EA,EC$ are blue and $EB$ is red.
It follows that $ED$ is blue (otherwise $ABED$ would be a red quadrilateral) and $EF$ is blue (otherwise $EBCF$ would be a red quadrilateral). Therefore $EDBF$ is a blue quadrilateral. A contradiction.

It is a classical result that any simple graph with $n$ vertices and at least $\frac{n}{2}( 1+\sqrt{4n-3})$ edges contains a quadrilateral. The above result shows that this value is not optimal...

Pierre.

It is a classical result that any simple graph with $n$ vertices and at least ... contains a quadrilateral.
Do you mean that many colors??

I didn't talk about colors....
In another hand, you may understand that as follows. Color each edge of the complete graph $K_{n}$ with one from $p$ colors. If a color $c$ is used at least ... times (just see the bound in my previous post) then there is a quadrilateral with color $c$.

Pierre.

Okay, that's interesting!
My question was: you said "and at least ... contains a quadrilateral"
"..." what, is it edges, or (as i thought) colors, or something else?

Ok, sorry I didn't see the missing word :oops: . It's edited.

Pierre.

I believe the topic title should be "Snakes on a plane" instead :wink:.

http://www.imdb.com/title/tt0417148/

okay, i was a little curious pbornzstein...

@paladin8 haha funny... not. too many of those jokes lately

Thanks pbornsztein for your proof.
I was wondering if this is good enough?
I simplified it to:
As if the colours are interchanging from each new vertices, it is less likely to get a monochromatic quadrilateral;
AB,AC,AD,AE,AF are all red.
BC,BD,BE,BF are all blue.
CD,CE,CF are all red.
DE,DF are all blue, because here, if it is red, there is going to be a monochromatic quadrilateral.
Finally EF cannot be red or blue as both make monochromatic quadrilaterals.

Therefore as a contradiction, there exists such monochromatic quadrilaterals when joining six dots on a plane.

Could anyone help me prove that AB...AF, BC...BF, CD...CE are all the colours I have stated?(in turns;red,blue,red,blue... or vice versa)

Maybe this is not the BEST proof there is.  So could anyone help me improve on this? and I don't really get pbornsztein's graph explanation.  Is the N 6 or 4?  Please help me

Hey this is one of the NZIMO camp selection problems. It's suppose to be your own work.  :mad:

To the moderators please lock this topic

This isnt going to end until October 14th.

oh srry
didn't know tat
my friend gav me a sheet of paper with these questions for me to solve for him
Bu tstill any suggestions on my proof?

Oh yea also could someone explain briefly or tell me a good place where I can learn/find out about wat pbornsztein used with the graph for proofs and what ramsey(4,2) is ?

Thanks


Triangle ABC is an equilateral triangle and O is the center of its inscribed circle. If the area of the circle is 4 pi cm squared, what is the area, in square centimeters, of triangle ABC?

Also, if you know which year this question is from, can you please tell me?

Thank you.

I dont know the year, but I have the solution:

From the given, we determine that the radius of the circle is $ 2$. Thus, the altitude is $ 6$. Thus the area is $ \boxed{12\sqrt3}$.


If the repeating decimal .36 is expressed as a rational number a/b with a and b integers in reduced  form, then a + b is

(1) 15 (2) 16  (3) 17   (4) 18  (5) None of the above

[hide]the decimal is 36/99, but if some people dont know that, heres how u prove it
36.363636...  = x times 100
  0.363636...  =x
subrtacing both of those, you get 36=99x, so x=36/99, which can then be simplified to 4/11, so the answer would be 15[/hide]

Hi;

[hide].36363636... is the fraction 4 / 11. You only have to remember a fraction you must have seen many times 1 / 11 = .090909090909...

So a+b = 15, choice 1) is correct.[/hide]


geometry
Euler
incenter
circumcircle
geometric transformation
reflection
angle bisector

Prove that: in a triangle then the incircle and the Euler circle are tangent.

Dear Mathlinkers,
you can see my enterely synthetic proof on
http://perso.orange.fr/jl.ayme     vol. 1
or on "revistaoim" that you can find with google.
Sincerely
Jean-Louis

for a simple prof: call I the incenter, O the circumcenter, H the orthocenter and F the Nine-Point Center. Then using that IF is the I-median of IOH we calculate $ IF^2 \equal{} \left (\frac{R}{2}\minus{}r\right )^2$

[quote="mr.danh"]This problem is a well known theorem of Feuerbach. The most popular method is apply the properties of inversion.[/quote]
  Can you post this solution help me?

$ O$ is the circumcenter, $ M$ is the midpoint of $ BC,$ $ D$ foot of the A-altitude, $ X$ is foot of the internal bisector of the $ \angle A,$ $ U$ is the incircle tangency point with $ BC.$ WLOG, $ b > c.$ $ (M)$ is a circle with center $ M$ and radius $ MU \equal{} \frac {b \minus{} c}{2}.$ It is perpendicular to the incircle $ (I)$ $ \Longrightarrow$ inversion in $ (M)$ takes $ (I)$ to itself. The 9-point circle $ (N)$ passes through the inversion center $ M,$ its inversion image is a line $ n$ perpendicular to $ MN \parallel AO.$ The 9-point circle $ (N)$ also passes through $ D$ and $ MD \equal{} \frac {b^2 \minus{} c^2}{2a}.$ Inversion image of $ D$ is $ D'$ and

$ MD' \equal{} \frac {MU^2}{MD} \equal{} \frac {a(b \minus{} c)}{2(b \plus{} c)} \equal{} MX$

Therefore, $ X \equiv D'$ is the image of $ D.$ But the line perpendicular to $ AO$ through $ X$ is the reflection of $ BC$ in the angle bisector $ AX$ and this is tangent to $ (I)$ $ \Longrightarrow$ $ (N)$ is also tangent to $ (I).$


number theory
least common multiple
number theory proposed

Prove that in a subset of $ \{1,2,\cdots,2n\}$ which contains $ n$ elements, there always exists 2 numbers such that their smallest common multiple $ \le 3n \plus{} 6$

Are you sure?
what about $ n \equal{} 4$, the subset $ \equal{} \{5,6,7,8\}$?
the lcm of each pair in here is $ > 18 \equal{} 3n \plus{} 6$


This one is cute :

Let a and b be real numbers such that
a - 3a + 5a - 17 = 0 ; b - 3b + 5b + 11 = 0.
Find all possible values of a + b.

Enjoy :)

Usually when I find a solution to Arne's (nice   !) problems, he gives another solution 5 to 10 times shorter than mine  :evil: , this one will probably make no exception

[hide]

a+b = 2

[/hide]

You're right Belenos ... how about a proof :) ?
My solution is about four lines long :mrgreen:

[quote="Arne"]four lines long :mrgreen:[/quote]

I'm glad I didn't give my proof (:ge:  :^2:)  this time  :!::!:

Why wouldn't you post it ? I am curious to see your solution ... !

I'm always interested in other solutions.

This is my solution :

[hide]

Define f(x) = x - 3x + 5x = (x - 1) + 2(x - 1) + 3. If f(m) + f(n) = 6 then we have [(m - 1) + 2(m - 1)] + [(n - 1) + 2(n - 1)] = (m + n - 2)((m - 1) - (m - 1)(n - 1) + (n - 1) + 2) = 0. Since the second factor is always > 0, m + n = 2. Now notice that f(a) + f(b) = 17 - 11 = 6 so we indeed have a + b = 2.[/hide]

[hide]Let f (x) = x3 - 3x2 + 5x - 6. This function has only one real root x = 2.

Now substitute in x = a + b. f (a + b) = 3ab (a + b - 2), therefore f (a + b) = 0 only when a + b = 2.[/hide]


modular arithmetic
inequalities
geometric series
number theory proposed
number theory
residue

Let $ a, b, c$ be integers satisfying $ 0 < a < c \minus{} 1$ and $ 1 < b < c$. For each $ k$, $ 0\leq k \leq a$, Let $ r_k,0 \leq r_k < c$
be the remainder of $ kb$ when divided by $ c$. Prove that the two sets $ \{r_0, r_1, r_2, \cdots , r_a\}$ and $ \{0, 1, 2, \cdots , a\}$ are different.

If $ a\equal{}1$ the statement follows plainly since $ b>1$. Let now $ a>1$ and assume the contrary, namely, that $ \{0,1,\ldots,a\}\equal{}\{r_0,r_1,\ldots,r_a\}$. Extend the definition of $ r_i$ to be the remainder of $ bi$ when divided by $ c$. Since $ bj\equiv1\pmod c$ for some $ j$, we get $ \gcd(b,c)\equal{}1$. Then $ \{r_0,r_1,\ldots,r_{c\minus{}1}\}\equal{}\{0,1,\ldots,c\minus{}1\}$. From our initial assumption, we also obtain $ \{a\plus{}1,a\plus{}2,\ldots,c\minus{}1\}\equal{}\{r_{a\plus{}1},\ldots,r_{c\minus{}1}\}$.
Since either $ a$ or $ a\minus{}1$ equals $ r_k$ for some $ k<a$ (because $ r_{k\plus{}1}\in\{0,1,\ldots,a\}$) we have that $ a\ge r_{k\plus{}1}\equal{}(r_k\plus{}b)\pmod c\in\{(a\plus{}b)\pmod c,(a\plus{}b\minus{}1)\pmod c)\}$. From here $ c<a\plus{}b\minus{}1$, implying $ c\minus{}b<a\minus{}1$. However we have $ r_{c\minus{}1}\equal{}(c\minus{}1)b \pmod c\equal{}c\minus{}b\in\{a\plus{}1,a\plus{}2,\ldots,c\minus{}1\}$, hence $ a\plus{}1\le c\minus{}b\le c\minus{}1$, contradicting the pevious inequality $ c\minus{}b<a\minus{}1$.

I have some doubts as it seems too simple to be true...

We can always find  natural number 0<r<a such that   a< r+b <c

If k.b= m.a +r then the remainter of (k+1).b is not in the set {0,1,..,a}. We have contradiction if k<a

In the case k=a, we have a.b=m.a + r , hence a divides r ( contradiction )

Here's another solution, which is very unnecessary. Work $\pmod{x^c-1}$. Assume the contrary. Then,

$1+x^b+ x^{2b} + \ldots +x^{ab} = 1+x+ \ldots + x^a \pmod{x^c-1}$.
Easy manipulation with geometric series formula gives $\left(x^{b(a+1)}-1\right)\left(x-1\right) = \left(x^b-1\right)\left(x^{a+1}-1\right) \implies$ \[ x^{b(a+1)+1}-x^{b(a+1)}-x = x^{a+b+1}-x^{a+1}-x^b. \pmod{x^c-1} \]

Now since $0 < a < c-1,$ and $0 < b < c, a+b+1 \not= a+1, b$, so the RHS of the above is simplified $\pmod{x^c-1}$.
So, looking at the LHS, since we have the terms $x^{b(a+1)+1}-x^{b(a+1)}$, whose exponents differ by 1, we need two of  $(a+b+1, a+1)$ or $(a+b+1, b)$ to differ by exactly $1 \pmod{c}$. This contradicts $b \not= 1$ (since $a+b+1-(a+1) = b$) and $a+1 \not= 1 \implies a = 0$. (since $a+b+1-b = a+1$)

First, observe that it suffices to prove the problem for $a \le \frac{c}{2}$. This is because if $a \ge \frac{c}{2}$, we can take the last $c - 1 - a$ residues and reverse them, which yields the result. Note that $a = \frac{c}{2}$ is also trivial, so it suffices to show the problem for $a \le \frac{c}{2}$.

 We now go by contradiction. We first prove it for primes $c = p$. Observe that by summing the two set and setting them equal modulo $p$, we have \[ p \mid \frac{(k-1)(a)(a+1)}{2} \]. $k \not \equiv 1 \pmod{p}$, so $a$ or $a+1$ is a multiple of $p$, and at this point this is clearly contradiction, since it violates the bound $a < c-1$.

 Now we show that it is impossible for all integers $c$ using strong induction on the number of divisors of $c$. We have already proven the result for primes $p$, so that part is out of the way. For general $c$, we use casework:

 Case 1: $a < p$, where $p$ is the smallest prime divisor of $c$.

 I claim that, in this case, there is not a thing we need to do. Obviously, at least one of $r_k \neq kb$, or else the sums of the two sets would be different. Moreover, $kb - r_k$ is an increasing function in integers $k$, so $r_a \neq ab$. This implies that $ab \ge c$, so $b > \frac{c}{p}$. But $r_1 = b$, and now $b > a$, since $\frac{c}{p} \ge p$. Hence, we have contradiction

 Case 2: $a \ge p$, where $p$ is the smallest prime divisor of $c$.

 In this case, observe that $(k, c) = 1$, since there must be some element $r_x = 1$, and $(k, c) \neq 1$ would clearly contradict this. Hence, each multiple of $p$ in the first set corresponds to a multiple of $p$ in the second, and dividing by $p$, the non-existence of this is equivalent to the result for $\frac{c}{p}$, so we have contradiction.

A quicker way to finish after $x^c-1|(x^b-1)(x^{a+1}-1)$ would be noting that $\Phi_c(X)|x^c-1$ and $x^b-1=\prod_{d|b}\Phi_d(X)$ and as $\Phi(X)$ is irreducible over $\mathbb{Z}[X]$ we have $c$ divides one of $a,b$ which is impossible for size reasons.

Suppose that $\{r_0,r_1,r_2,\ldots,r_a\}\equiv\{0,1,2,\ldots,a\}\pmod c$. As there exists $k$ with $r_k=1$, $b$ is relatively prime to $c$. Hence, $\{r_{a+1},r_{a+2},\ldots,r_{c-1}\}\equiv\{a+1,a+2,\ldots,c-1\}\pmod c$ as $\{r_0,r_1,r_2,\ldots,r_{c-1}\}\equiv\{0,1,2,\ldots,c-1\}\pmod c$, so $\{r_1,r_2,\ldots,r_{c-a-1}\}\equiv\{1,2,\ldots,r_{c-a-1}\}\pmod c\implies\{r_0,r_1,r_2,\ldots,r_{c-a-1}\}\equiv\{0,1,2,\ldots,r_{c-a-1}\}\pmod c$. Hence, we may suppose that $a\le c-a-1$, or $a<\frac c2$.

Let $S\equiv\{0,1,2,\ldots,a\}\pmod c$ and $b^{-1}$ be the inverse of $b\pmod c$. We have that $bS=S$, so we also have $b^{-1}S=S$, which means that $b,b^{-1}\in S$. Now, note that $r_{b^{-1}-1}\equiv 1-b\pmod c$, so $r_{b^{-1}-1}=c-b+1$. But this means that $c-b+1\in S\implies c-b+1<\frac c2\implies b>\frac c2+1$, contradicting $b\in S$.

The shortest $``\textit{one-liners}"$ I've written (and one-liners couldn't be shorter, as the well-ordering principle on naturals say); especially one for a problem with a high spot on the contest. \\

$\color{green} \rule{25cm}{5pt}$ \\
$\color{blue} \textbf{Obvious Definition Change.}$ [color=magenta]Define $\color{green}r_i$ differently now: not only does range of the slider $\color{green}i$ is extended from $\color{green}(0,a)$ to $\color{green}(0,c-1)$, but also not by remainder: after defining $\color{green}r_0 = 0$, define $\color{green}r_{i+1} \equiv r_i+b \pmod c$.\\
$\color{blue} \rule{25cm}{0.4pt}$ \\
$\color{blue} \textbf{This Problem is So} \ \textbf{\textit{Easy.}}$ If $\color{green}(b,c) \ne 1$, it's impossible for $\color{green}1$ to be a part of $\color{green}r_i$, for 5 MOHS reasons. So, assuming that the set $\color{green}\{r_0,r_1,\ldots,r_{c-1}\}$ is complete, we get that $\color{green}\{r_{a+1},\ldots,r_{c-1}\}$ equals to $\color{green}\{a+1,a+2,\ldots\}$ ($\color{green}c-1$'s existence is not important here.) \\
$\color{blue} \rule{25cm}{0.4pt}$ \\
First, we will prove the obvious bound that $\color{green}b \leq a$. This is because $\color{green}r_1$ is $\color{green}\textit{exactly}$ $\color{green}b$, and if it exceeds $\color{green}a$, then, oh well, the world [/color] $\color{red} \boxed{\overline{\textbf{\textit{\underline{explodes}}}}}$ [color=magenta] (actually, this is because $\color{green}r_1 \in \{0,1,2,\ldots,a\}$).
$\color{blue} \rule{25cm}{0.4pt}$ \\
Then, since $\color{green}b \ne 1$, the two numbers $\color{green}a+1-b$ and $\color{green}a+2-b$ modulo $\color{green}c$ are greater than zero for $\color{green}b \leq a$ reasons; and are at most $\color{green}a$ due to $\color{green}b$'s positivity. Then, they should be among $\color{green}\{r_0,r_1,\ldots,r_a\}$ --- but only one of them (a.k.a. $\color{green}r_a$), when added with $\color{green}b$, should result in a member from the big set $\color{green}\{r_{a+1},\ldots,r_{c-1}\}$.  [/color]

We are done. (Forgive me for the absurdly colored Solution --- wanted to mess around a bit with posting options on AoPS) $\blacksquare$ $\blacksquare$ $\blacksquare$ \\

[hide = Properly written Solution.]
$\color{green} \textbf{Additive Remainders.}$ Define $r_i$ differently now: not only does range of the slider $i$ is extended from $(0,a)$ to $(0,c-1)$, but also not by remainder: after defining $r_0 = 0$, define $r_{i+1} \equiv r_i+b \pmod c$.\\
$\color{green} \rule{25cm}{0.4pt}$ \\
$\color{green} \textbf{In All Its Glory.}$ If $(b,c) \ne 1$, it's impossible for $1$ to be a part of $r_i$, as it will result in all $r_i$ being a multiple of $\text{gcd}(b,c)$. So, assuming that the set $\{r_0,r_1,\ldots,r_{c-1}\}$ is complete, we get that $\{r_{a+1},\ldots,r_{c-1}\}$ equals to $\{a+1,a+2,\ldots\}$ (I purposefully emitted the existence of $c-1$ there --- those two numbers are the one we will consider.) \\

$\color{green} \rule{25cm}{0.4pt}$ \\
First, we will prove the obvious bound that $b \leq a$. This is because $r_1$ is $\textit{exactly}$ $b$, and if it exceeds $a$, then $r_1 \not\in \{0,1,2,\ldots,a\}$. \\

$\color{green} \rule{25cm}{0.4pt}$ \\
Then, since $b \ne 1$, the two numbers $a+1-b$ and $a+2-b$ modulo $c$ are greater than zero for $b \leq a$ reasons; and are at most $a$ due to $b$'s positivity. Then, they should be among $\{r_0,r_1,\ldots,r_a\}$ --- but only one of them (a.k.a. $r_a$), when added with $b$, should result in a member from the big set $\{r_{a+1},\ldots,r_{c-1}\}$.
[/hide]
$\color{green} \rule{25cm}{5pt}$ \\

Note that $a+1$ and $a+2$ are members of $\{r_{a+1},\ldots,r_{c-1}\}$; else $a+1 = c-1$ and $r_{c-1} = c-1$ --- implying that $b = 1$. \\

(Guess they're still in four paragraphs, huh. A not easy one to find, and certainly not easy to write, but very easy to read, I hope ...) \\

[hide = Motivation: Quite the long journey: finding the tight bound from the given basic bound.]
Experimenting right of the bat, there are $c-2$ candidates for $a$ (from $1$ to $c-2$) and $c-2$ candidates for $b$ (from $2$ to $c-1$.) \\

$\color{cyan} \rule{25cm}{2pt}$ \\
$\color{green} \text{Sec 1: Preliminary Counting.}$ Excluding the blatant choices when $a = 0$ and $a = c-1$ (the problem is blatantly wrong if we pick only $r_0$ or all $r_0$ until $r_{c-1}$ --- and from here on I've set my eyes to consider all of $r_0$ until $r_{c-1}$ instead of ignoring the late $r_{a+1}$ until $r_{c-1}$), I tried to derive a contradiction from $a = 1$ or $a = c-2$ (the next bordered values.) \\

Surprisingly (or better worded: $\textit{obviously}$), those two can contradict the problem statement only when $r_1 = 1$ or $r_{c-1} = c-1$; when the multiplier $b$ is equal to $1$. This also makes spotting easy patterns to contradict on $a$ small to be useless --- any middle schooler would instantly recognize the fallacy of extracting a pattern with no intricacy at all. \\

Also, by the 5 MOHS argument of forcing gcd of $b$ and $c$ to be $1$, there's no point in analyzing the $``\text{degenerate}''$ cases when $b = 2, c= 4$; for example, so I directly skipped to $c = 5$ for less obvious examples. \\

$\color{green} \rule{25cm}{0.4pt}$ \\
For the still easy case of $c = 5$, there are two non-obvious $a$s to start: $a = 2$ and $a = 3$. Seeing that the condition
\[ \{r_0,r_1,r_2\} = \{0,1,2\} \Rightarrow \{r_1,r_2\} = \{1,2\} \]
quickly devolved into a senseless condition (after all, $r_0$ shouldn't be considered as a free variable/one thing to particularly zoom in, of course) I decided to stick to the remaining option: $a = 3$. Here I recalled choosing $a$ (the threshold) instead of $b$ (the multiplier) because the main problem seemed to be how big the threshold were, and I wanted the multiplier's weirdness to adapt to the threshold's (apparently fixed) size.

Well, if any of that makes sense for an $a = 3$ anyway; $(a,b) = (3,2)$ gave $\{r_1,r_2,r_3\} = \{2,4,1\}$ with $3$ disappearing --- and $b = 4$ made a $4$ in $r_1$. At this point I realized that the condition is equivalent to $4$ missing from the first four resides, and I was back to square one. \\

At least I got the impression that $1$ shouldn't be missing from any set, and that's apparentlya a big deal (from all attempts, I found that it's especially hard to make $1$ disappear unconditionally.) \\
$\color{cyan} \rule{25cm}{2pt}$ \\
$\color{green} \text{Sec 2: Some relatively useless progress.}$ Since small values of $c$'s behavior is easily manipulated and determined, I tried to jump towards high $c$, and a relatively large but comfortable prime to work on was $11$. Well, since this is (at least seems to be) a size problem, the instant loophole I spotted was that
\[\begin{tabular}{p{12cm}}
if 1 is in the front, then it's easy to find some small number which must lie at the back of the pack. \\
\end{tabular}\]
and so I fixed $1$'s position in the back, say at the seventh position $r_7$. By that, I obtained the decoy $b = 8$. \\

(By the way, the rigorous way to say the above observation was that if $b^{-1}$ is less than $c-a-1$, then we're done. Such an ugly statement but it usually works if $b^{-1}$ is indeed small.) \\

$\color{green} \rule{25cm}{0.4pt}$ \\
Now since I've represented the problem as following:
\[ \left\{ c \left\{\dfrac{0}{c}\right\}, c \left\{\dfrac{b}{c}\right\},\ldots, c\left\{\dfrac{ab}{c}\right\} \right\} = \{0,1,\ldots,a\} \]
I started to view $b$ not necessarily in the form of
\[ b \times \{0,1,\ldots,a\} = \{0,1,\ldots,a\} \]
as post $\#9$ views it; but in the additive sense in my Solution as they provide a means to bound $r_i$'s size. \\

Speaking of that, a problem which (also) immediately surfaced was the guaranteed existence of $b$, a.k.a. $8$, in the first remainder $r_1$. This is already a very big number, which kind of forced $a$ to be $9$ or more. So I cut it there; and got the very nice bound (which acutely narrows the possibility of $a$ in relation to a big summand $b$)
\[ a \geq b \]
and here I stopped at the configuration
\[ \{0,8,5,13 = 2,10,7,4,1,9,6\} = \{r_0,r_1,r_2,r_3,r_4,r_5,r_6,r_7,r_8,r_9\} \]
with apparently an (almost) perfect config with a misplaced $3$ in the back and a $10$ in the front. \\

(To add a bit of satisfaction: this kind of suits my initial goal of making a config $(a,b)$ looking kind of sus; apparently all almost-contradicting sets' misplaced numbers are very small or very big --- none in the middle.)\\

$\color{cyan} \rule{25cm}{2pt}$ \\
$\color{green} \text{Sec 3: The Lucky Break, Extended}$. Now I didn't really focus to the particular values of $r_0$ until $r_a$ (the method of precise output is usually put into a series/polynomial as in $\#4$ or $\#6$; I'll get to it towards the $\color{red} \text{end of this Motivation.}$ To begin, I picked another decoy prime $c = 17$ and set $a \geq b = 12 \geq 8$. \\

What can I infer from blindly picking these numbers? Clearly, I would want the extreme numbers $1$ and $16$ in the front/back of the whole pack, obviously, but what can I get from separating the remainders $r_0$ to $r_16$ somewhere in the middle (to be exact, after the $a^{th}$ remainder) and having them naturally arranged based on size? \\

Naturally, to contrast $\textbf{more}$ large/small numbers with each other! (The Lucky Break.)\\

Going away from the apparently must-be-big set $\{r_0,r_1,\ldots,r_a\}$, I went to look at the smaller
\[ \{r_{13},r_{14},r_{15},r_{16}\} = \{13,14,15,16\} \]
sized set. If you sniff $\textit{very}$ closely, it would seem that something is forced inside this set, is it not? Not only that, it didn't seem like a $14$ when added with a $8$ could also big; could it be that not some, but $\textit{all}$ terms don't actually have their identity to be $\textit{that}$ big? \\

Apparently, yes! While I initially considered terms where $r_i = c-1$ or $r_j = c-2$, since we're talking about the border of $r_a$ and $r_{a+1}$ here, it would seem best to make the gap insultingly small; just select $r_i$ and $r_j$ equal to $a+1$ to some $a+c$ ($c$ constant) since the transition can only happen once! \\

I was overjoyed, but my mind was still filled with a lot of other contradictions; I'd say what's most difficult in this problem is not about $\textit{finding a contradiction}$; rather; it's more to $\textit{finding the contradiction that finishes the problem.}$ \\
\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]
[hide = Bonus: why I think this problem is very beautiful.]
After writing the above Solution and before writing this Motivation, I skimmed the above Solutions --- and found out that they're $\textit{very much related}$ to one another. \\

$\color{red} \rule{25cm}{1pt}$ \\
$\color{red} \text{Adding from the back, or divisibilities from size issues.}$ The most similar ones out of all above seem to be post $\#2$;
[list = disc][*] Utilizing the transition from $\{r_0,r_1,\ldots,r_a\}$ to $\{r_{a+1},\ldots,r_{c-1}$ instead of mine, from $a+1$ and $a+2$ (elements of the back set) to the front; [*] ruling out the case $a = 1$ instead of $a = c-2$ in my part; [*] after pulling out the bound in $a+b-1$ and $c$, the finalization occured when comparing $r_{c-1}$ in relation to the set $\{a+1,\ldots,c-1\}$ instead of comparing $r_1 = b$ in relation to the set $\{0,1,\ldots,a\}$ [/list]
with the main idea being $c-b$, presumably with $b$ small, is still fairly large when compared with $a+1$, the transiting term towards $\{a,a-1,\ldots,1,0\}$. \\

$\color{red} \rule{25cm}{0.4pt}$ \\
And with that, reviewing the next post $\#3$: it has some very fatal typos which rendered the Solution wrong, but it has some interesting methodology to it (being a true impressive one-liner, at that.) \\

Now, the blatantly confusing pick of the number $r$ with
\[ 0 < r < a, a < r+b < c \]
was the groundbreaking idea: it highlights the $\color{green} \textbf{Additive Remainders}$ idea in an algebraic way, that is: assuming the set $\{r_0,r_1,\ldots,r_{a}\}$ is as the problem says, any remainder $r$ less than $a$ will happen, and here we'd like to specify one of them (say $r = r_k$) that will yield something greater than $a$, but not so big as to exceed the modulo class $c$, when added with the summand $b$. \\

However, $kb$ were not to be divided with $a$, but with $c$ --- so the correct expression would be
\[ kb = mc + [r] \Rightarrow (k+1)b = mc + [r+b]; \ \text{with} \ r+b > a \Rightarrow r \geq a-b+1\]
and modifying the bound a bit by considering $r_1$, we can be sure that $\textbf{that number is positive.}$ Since there's the problem of incidentally selecting $r_k = r_a$, the problem is not immediately over, since $k+1$ may as well be $a+1$, indicating that
\[ (a+1)b = mc + [r+b] \Rightarrow r_{a+1} = [r+b] = (a-b+1)+b=a+1 \]
if we turned around from this statement to selecting a number $a-b+2$ as the new $r$, we can proceed as in my Solution: but we'd have to verify that $(a-b+2)+b = a+2$ does not cycle back to zero (or reach $c$); i.e. $a+2$ shouldn't be equal to $c$ or $a \ne c-2$. \\

$\color{red} \rule{25cm}{0.4pt}$ \\
However, the equation
\[ r_{a+1} = a+1 \]
seemed to be very interesting to zoom into: and from here I was quite excited to see the link between this approach towards this approach to the Solutions in $\#5$ and $\#8$.

First of all, I deemed the divisibility
\[ (a+1)(b-1) = mc \Rightarrow c \mid (a+1)(b-1) \]
to be relatively useless on its own; as I didn't directly see a connection between $p \mid c$ implying $a \equiv -1 \pmod p$ and $b \equiv 1 \pmod p$. The latter even implies that when considering mod $p$ (prime) divisor of $c$, the remainders $r_1$ until $r_{c-1}$ rises in a way forbidden to what the problem dictates ($b \not\equiv 1$). \\

So, I took a step back and analyzed the terms $\textbf{surrounding}$ $r_{a+1}$ as follows:
\[ \begin{tabular}{c|c|c|c|c|c|c}
$k$&$\dots$&$a-1$&$a$&$a+1$&$a+2$&$\dots$ \\ \hline
$r_k$&$\dots$&$a-2b+1$&$a-b+1$&$a+1$&$a+b+1$&$\dots$ \\
\end{tabular} \]
and assuming we've already known that $a$ is greater than $b$, it would seem that before it reaches $k = 1$ (going down the ladder from $a$ to $1$), some small negative with magnitude less than $b$ will pop up and causing a term of magnitude at least $c-b+1$ (the term creating the negative would have to be $1$ or greater, for it cannot be zero) --- somewhat similar to the
\[ r_{b^{-1}-1} \equiv 1-b \pmod c = c-b+1 \]
argument in post $\#8$. But we're not done yet. \\

Since the identity $c \mid (a+1)(b-1)$ promises that if $c \nmid b-1$ (which should happen due to $b < c$, $a+1$ will have to share some prime divisor with $c$, rendering its gcd to $\textbf{not be}$ 1. Why should that be? If its gcd is equal to one, then $b = 1$, which results in the statement
\[ r_k = k \ \text{for every} \ k, \ 1 \leq k \leq c-1 \]
So what should happen if $\text{gcd}(a+1,c) = g$? Take $a = 5$ and $c = 15$, for example, yielding $\text{gcd}(6,15) = 3$. However, by Bezout (or Euclidean Algorithm in general), we get that adding $6$ to $6$ will yield
\[ r_{12} = 12, r_{18} \equiv 18 \ (\text{mod} \ 15), r_{9} = 9, r_{15} = 15 \ (\text{or} \ r_0 = 0) \]
which fulfilled our promise that $3$ should be the smallest $k$ so that $r_k = k$. Notice that by $r_{a+1} = a+1$ for any $a$ will create at least two blocks of similar sizes:
\[ \begin{tabular}{c|ccc|ccc|c}
$k$&0&1&2&$\color{red}3$&4&5&$\color{blue}6$ \\ \hline
$r_k$&0&&&$\color{red}3$&&&$\color{blue}6$ \\
\end{tabular}\]
$\color{red} \rule{25cm}{0.4pt}$ \\
From that, we can somewhat $\color{red} \textit{induct down}$ from the case $a = 5$ to the case $g = a' = 3$, being the common divisor of $6$ and $15$. This is because the placement of $1$ and $2$ in the second row (the $r_i$ row) $\color{red} \textit{must be in the first block}$, any later (for example $r_5 = 1$) would displace a $4$ in $r_8$, somewhere later than $a = 5$. \\

When I say that this is an inductive argument, however, this doesn't mean that I can use the term $a = g-1$ (here denoted by $2$) to lower it further; I'd have to resort to the similarities in block size (in contrast with what $\#5$ does.) \\

Now say that $1$ appears in $r_k$ where $k \ne 1$. Since it has to appear in the first block, we'd have $r_{lk} = l$ ($l \leq g-1$) as long as $lk$ does't exceed $c$. Then we can pick the first one to exceed $g$ and we are set. \\

$\color{red} \rule{25cm}{0.4pt}$ \\
A final remark: comparing with post $\#8$, after presumably establishing $a \geq b$ --- since we'd have to bound $b$ eventually if we were to use the (at least $c-b+1$) thing, the existence of similarly sized blocks may hint at the small size of $a$; as there are at least two of them. \\

Then, noticing the similar behavior occuring between the left set $\{r_0,r_1,\ldots,r_a\}$ and $\{r_{a+1},\ldots,r_{c-1}\}$ (i.e. they are additive mod $c$, and each term are $b$ more than its previous one), one may as well WLOG the one from $a+1$ to $c-1$ to be the desired left set and operate on a $c-a-1$-sized block (which has size at most $\dfrac{c}{2}$).

This will guarantee that the size of the left set ($a$) and the size of the summand $b$ which also is less than $a$ will not interfere with each other --- as $b \leq \dfrac{c}{2}$, a term as big as $c-b+1 \geq c-c\dfrac{c}{2}+1$ shouldn't appear in the small-sized left block $\{r_0,r_1,\ldots,r_a\}$.

Note that without specifying $b$'s small size, the term $1-b$ or $c-b+1$ before $r^{b^{-1}}$ wouldn't be sus at all, due to the sheer large negative it has --- see my first configuration on the end of $\color{green} \text{Sec 2}$. There, it'd seem like $b+1 = r^{b^{-1}+1}$ would be a better candidate to derive a contradiction, but the realization that
\[ b \leq a \Rightarrow b+1 \ \text{should be} \ \leq a \ \text{as well} \]
should be enough to rethink that approach.

$\color{red} \rule{25cm}{1pt}$ \\
$\color{red} \text{Exact values: Series mod a polynomial.}$ Speaking of determining the set $\{r_0,\ldots,r_a\}$ precisely, I'd like to share a problem I've done (a short while after doing this problem and before writing my post) which illustrates the idea wonderfully:
\[\begin{tabular}{p{12cm}}
Let $p$ prime, $k$ natural number, and $\mathcal{A}$ be a finite set of integers with at least $p^k$ elements. Denote $N_{\text{even}}$ to be the set of $\mathcal{A}$'s subsets with even cardinality, and sum of elements divisible by $p^k$ --- and $N_{\text{odd}}$ the same way (with odd cardinality.) \\
\\
Prove that $|N_{\text{even}}| \equiv |N_{\text{odd}}| \pmod p$. \\
\\
This was initially done with $k = 1$, but the author generalized it to $p^k$ due to Kummer. \\
\end{tabular} \\ \]
Since the best way to represent sum of elements is by manipulating power series, let $\mathcal{A} = \{a_1,\ldots,a_m\}$ with $m\geq p^k$ and construct the series
\[ S_{\mathcal{A}} = (x^{a_1}-1)(x^{a_2}-1)\ldots(x^{a_m}-1)\]
Why is this beautiful? Since the sum of ever subset of $\mathcal{A}$ is representable by multiplying $x^{\text{element included in that subset}}$ with each other (and discarding elements not included there by selecting the constant part), the problem is now equivalent to finding the total coefficient in powers divisible by $p^k$. \\

And the constant should also be negative as the number of non-included terms in $N_{\text{even}}$ are equal in parity, and each addition of a subset (as a product of $x^{a_i}$) should have one magnitude (whether the sign is positive/negative depends on the number of included terms.) \\

After absorbing those two observations by heart (sorry if that is too detailed, anyway) it should be easy by noting that
\[ (x-1)^{p^k} \mid S_{\mathcal{A}} \]
and that the coefficients of $(x-1)^{p^k}$ modulo $p$ (when grouped according to their powers mod $p^k$) is equal to $0 \pmod p$. Then the problem solves itself without needing to consider anything else (the construction of the series are the main difficulty.) \\

I decided to include this in the Motivation to motivate the construction of $x^{ab}+\ldots+x^a+1$ dividing $x^c-1$ --- as the above problem basically says that
\[ \text{some polynomial} \ ( \text{here it's} \ ) (x-1)^{p^k} \mid S_{\mathcal{A}}\]
solves the whole thing, so does determining what adds with each other (here, $r_k$ adds with each other; hence the definition of $r_{i+1} \equiv r_i +b \pmod c$ in my Solution), what congruence/divisibility is there ($x^c-1$ is due to its powers removing itself modulo $c$ to be included among $r_k$) and how to manipulate them (easy, by noting that the end result of the powers simplifying itself should be equal to $\{0,1,\ldots,a\}$ in series form.)
\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]
[/hide]
[/hide]


Compute

( 1 - 1/3)+( 1/2 - 1/4)+(1/3 - 1/5)+ ... +(1/123 - 1/125) . Express your answer to the nearest thousandths.

Can someone explain how its this answer?

[hide="answer"]1.483935 which 1.484 rounded to the nearest thousandths[/hide]

All of the terms cancel out except for $ 1 \plus{} \dfrac 12 \minus{} \dfrac 1{124} \minus{} \dfrac1{125}$, which rounds to your answer.


I read through the geometry chapters of Zeitz, but I still cannot solve the problems at the back of the sections ( i try them for 15 mins). For how long should I try them? I love geometry, but I'm not very good at it.

15 minutes seems kind of too short. You should try them for like at least half an hour to an hour, or more. Try every single method you can think of. (The most common feedback I got from Intro to Geometry was to not stare at a geometry problem, but to start labeling everything you know, and draw extra lines...if needed that is.) So maybe doing that would help?


integration
limit
logarithms
real analysis
real analysis unsolved

Let $a_n=\int_0^n \left[\frac x 1\right]+\left[\frac x 2\right]+...+\left[\frac x n\right]dx$ where $[.]$ is the biggest integer less than the number in brackets.  Find:
1) $\lim \frac{a_n}{n^2 \ln n}$
2) $\lim \sqrt[a_n] {1!2!3!...n!}$.

My solution :) :
(1)Because:
\[ a_n=\int_0^n \left[\frac x 1\right]+\left[\frac x 2\right]+...+\left[\frac x n\right]dx=a_n=\int_1^n \left[\frac x 1\right]+\left[\frac x 2\right]+...+\left[\frac x n\right]dx \]
Thus:\[ \int_1^n (\frac x 1+\frac x 2+...+ \frac x n-n)dx \le a_n \le \int_1^n(\frac x 1+\frac x 2+...+\frac x n )dx \]
Which means:
\[ \frac 12 \left(n^2+\frac{n^2}2+\cdots+\frac{n^2}n-\frac 1 1-\frac 1 2-\cdots-\frac 1 n \right)-n(n-1) \le a_n \le \frac 12 \left(n^2+\frac{n^2}2+\cdots+\frac{n^2}n-\frac 1 1-\frac 1 2-\cdots-\frac 1 n \right) \]
Thus easy to prove:$\lim_{n \to \infty} \frac{a_n}{n^2 \ln n}=\frac 1 2$.
(2)First I will find \[ L =\lim_{n \to \infty}\frac{2\left(\ln 2!+\ln 3!+\cdots+\ln n! \right)}{n^2 \ln n} \]
Use the Stolz Theorem:We have:
\[ L=\lim_{n \to \infty}\frac{2 \ln (n+1)!}{(n+1)^2\ln (n+1)-n^2 \ln n}=\lim_{n \to \infty} \frac{2 \ln (n+1)!}{(2n+1)\ln (n+1)} \]
\[ =\lim_{n \to \infty}\frac{\ln (n+1)!}{n\ln (n+1)}\underbrace{=}_{Stirling's formula}\lim_{n \to \infty}\frac{\ln (\frac{n+1}{e})^{n+1}\sqrt{2 \pi (n+1)}}{n \ln (n+1)}=1 \]
So combine 1) it is easy to find:$\lim_{n \to \infty}\sqrt[a_n]{1!2!\cdots n!}=e$.
I hope I am right. :)

I didn't tink that this aroximation is sharp enough.
\[ \int_1^n (\frac x 1+\frac x 2+...+ \frac x n-n)dx \le a_n \le \int_1^n(\frac x 1+\frac x 2+...+\frac x n )dx \]

Very nice solution... you are good!


trigonometry function calculus conics ellipse geometry perimeter

For All $x,y,z$ Reals

what is the maximum value of:

${|\sin (x-y)-\sin (x-z)+\sin (y-z)|}$

and what the values of $x,y,z$  for this maximum

$=|4\prod\cos{\frac{x-y}{2}}|\leq 4$

We need find $x,y,z$ such that $x-y=2m\pi,y-z=2n\pi,z-x=2p\pi(m,n,p\in\mathbb{Z})$, this is easy.

${\{x-y=2 m \pi ,y-z=2 n \pi ,z-x=2 p \pi \}}$

Solving

${m=-(n+p)}$
${y=2 \pi (n+p)+x}$
${z=2 \pi p+x}$

Substituing

${|\sin (2 n \pi )+\sin (2 p \pi )-\sin (2 \pi n+2 p \pi )|}$

Since

${\{n,p\}\in \mathbb{Z}}$

${|\sin (2 n \pi )+\sin (2 p \pi )-\sin (2 \pi n+2 p \pi )|=0}$

0 is a maximum?!!?

try mathematica

Reduce[{x - y == 2*m*Pi, y - z == 2*n*Pi, z - x == 2*p*Pi}, {x, y, z}]

this is not the solution guy :(

Sorry, I am wrong. It must is

$=4|\prod\sin{\frac{x-y}{2}}|\leq 4$, and you must slove $x-y=(2m+1)\pi,y-z=(2n+1)\pi,z-x=(2p+1)\pi(m,n,p\in\mathbb{Z})$.

Ok, I am wrong, again  :( . Now , I think we will use real analysis with function $f(a,b)=\sin{a}.\sin{b}.\sin{(a+b)}$ on $[0,2\pi]\times [0,2\pi].$. Do you agree?  :wink:

Forum index » College Playground » Calculus - Real Analysis » Calculus Computations and Tutorials

hehe

$f(a,b)=\sin{a}.\sin{b}.\sin{(a+b)}$

why this!?

need one more variable, no?!

like

${\sin \left(\frac{x-y}{2}\right) \sin \left(\frac{x-z}{2}\right) \sin \left(\frac{y-z}{2}\right)}$

$a=\frac{x-y}{2},b=\frac{y-z}{2},c=\frac{z-c}{2}$ then $a+b+c=0$ therefore $c=-a-b$. Also , we only need two variable :D

i did not realize this

hm... now become easy

hehe

congratulations

you won

another problem

this is wrong

the substituition is ok

but

when i solve for $a$ and $b$ i get a value

so , quem i get,  $x$  and $y$ in function of $z$  and use this on the expression  it give the result   0

i cannot think more

as son as possible i will show you what im talking about

Suppose a ellipse below

${\alpha >0\land \beta >0}$

${\phi (t)=\{\alpha \cos (t),\beta \sin (t)\}}$

For All

${(2 \pi \geq x\geq 0)\land (2 \pi \geq y\geq 0)\land (2 \pi \geq z\geq 0)}$
${x\neq y\land x\neq z\land y\neq z}$

We get $3$ points of a triangle

${A=\phi (x)\land B=\phi (y)\land C=\phi (z)}$

Lets length of the sides be

${\{\text{AB},\text{AC},\text{BC}\}}$

And the perimeter

${2 p=\text{AB}+\text{AC}+\text{BC}}$

By heron, the area of the triangle is

${\Delta =\sqrt{p (p-\text{AB}) (p-\text{AC}) (p-\text{BC})}}$

Simplifying

${\Delta =2 \alpha \beta \sqrt{\left(\sin \left(\frac{x-y}{2}\right) \sin \left(\frac{x-z}{2}\right) \sin \left(\frac{y-z}{2}\right)\right)^{2}}}$
${\Delta =\frac{1}{2}\alpha \beta \sqrt{(\sin (x-y)-\sin (x-z)+\sin (y-z))^{2}}}$

If we substitue

${(a=x-y)\land (b=x-z)\land (c=y-z)}$

${a-b+c=0}$

Simplifying and Creating a function

${f(a,b)=\sin (a)-\sin (a-b)-\sin (b)}$

Solving

${\frac{\partial f(a,b)}{\partial a}=0}$
${\frac{\partial f(a,b)}{\partial b}=0}$

${\left(c_{1}|c_{2}\right)\in \mathbb{Z}}$
${\left(a=2 \pi c_{1}\land b=-2 \pi c_{2}\right)\lor \left(a=2 \pi c_{1}-\frac{2 \pi }{3}\land b=\frac{2 \pi }{3}-2 \pi c_{2}\right)\lor \left(a=2 \pi c_{1}+\frac{2 \pi }{3}\land b=-2 \pi c_{2}-\frac{2 \pi }{3}\right)}$

Substituing

${f(2 \pi c_{1},-2 \pi c_{2})=0}$
${f\left(2 \pi c_{1}-\frac{2 \pi }{3},\frac{2 \pi }{3}-2 \pi c_{2}\right)=-\frac{3 \sqrt{3}}{2}}$
${f\left(2 \pi c_{1}+\frac{2 \pi }{3},-2 \pi c_{2}-\frac{2 \pi }{3}\right)=\frac{3 \sqrt{3}}{2}}$

So, the values of $a$ and $b$ are

${\left(a=2 \pi c_{1}\pm \frac{2 \pi }{3}\right)\land \left(\mp \frac{2 \pi }{3}-2 \pi c_{2}\right)}$

Since

${a=x-y\land b=x-z\land c=y-z}$
${c=b-a}$

${\left(x=z+\pi \left(\frac{2}{3}-2 c_{2}\right)\land y=z+\pi \left(-2 c_{1}-2 c_{2}+\frac{4}{3}\right)\right)\lor \left(x=z-\frac{2}{3}\pi \left(3 c_{2}+1\right)\land y=z-\frac{2}{3}\pi \left(3 c_{1}+3 c_{2}+2\right)\right)}$

conclusion

the angle between the vectors is 60 Degrees


geometry
geometric transformation
rotation
combinatorics proposed
combinatorics

We divide two concentric circles into $n$ equal sectors, and color the sectors of the large circle with different colors. For what $n$ does there exist a coloring of the sectors of the small circle with the same colors s.t. for any rotation of angle $\frac{2k\pi}n$ there are two overlapping sectors (one from each circle) with the same color?

Fix colors in the first circle and regard colors of the secound circle as a permutation of the colors. Then the question becomes: does there exist a permutation $p$ of $1,2,3...,n$ such that $p(i)-i$ are different modulo $n$?
suming $p(i)-i$ we get $1+2+3....+n=0$(mod n) so $n$ is odd, so for odd $n$ just consider $p(i)=2i$ mod n.


As our subforum is more and more active every day, it seems that we should have a moderator for it. That is why this topic is opened: who will we elect for the mod (and the question is, will the elected one accept the honour and responsibility?)

This topic is restricted to conversation in English, for the sake of legality. I didn't want to put a poll before we see your proposals... If there's a need for one, please say so.

I think that this person should be someone from college, like Monsieur Cochich or  somebody else who has time to deal with us.

Thank you for the nomination, Mademoiselle! (for Valentin: Kochich/Cochich=hsiljak... an intern nickname).

My opinion is also that the mod should be someone who is every day here, with a relatively high post count-proportional to ML experience. That's why I nominate [b]delegat[/b].

[quote="hsiljak"]
My opinion is also that the mod should be someone who is every day here, with a relatively high post count-proportional to ML experience. That's why I nominate [b]delegat[/b].[/quote]

I agree. I think a moderator should be [b]delegat[/b], [b]hsiljak[/b] or [b]vedran6[/b].

I would like to second the nomination of [b]hsiljak[/b].

[quote]Thank you for the nomination, Mademoiselle! (for Valentin: Kochich/Cochich=hsiljak... an intern nickname).

My opinion is also that the mod should be someone who is every day here, with a relatively high post count-proportional to ML experience. That's why I nominate delegat.[/quote]

Oh thanks for nomination but unfortunately I can not accept this because I am not sure about University (maybe I will not have 24 hour internet connection and some other problems ...)

I also nominate [b]hsiljak[/b]  (or possibly [b]vedran6[/b]) for this position because he is always active in these topics at our national forum and it was his idea to make it.

Valentin appointed me the new moderator. Thank you for the honour (and the responsibility which goes with the green nickname  :) )

Topic locked.


[b]1.3[/b] [i]In the lattice plane, $\mathbb{Z}\times\mathbb{Z}$, we assign to each point a real numbers.  Each of these is the arithmatic mean of the real numbers belonging to its four neighbours (above, below, to the right, and to the left).  Prove that all the labels are equal.[/i]

Masoud Zargar

This problem is mentioned in Arthur Engel's book.
Having dealt with the problem using positive integers, he goes on to say that the theorem is still valid if real numbers are used, though he does not know an elementary solution... so this might be well above Pre-Olympiad level.
Of course, I may be wrong.  :)

Yes.  However, it does not mention the level of difficulty, thus if it is not suitable for this forum, could a mod please move this to a higher level?  Thanks in advance.


geometry
incenter
circumcircle
geometry unsolved

Given a quadrilateral $ ABCD$ both circumscribed and inscribed, assume that none of its diagonals or sides is a diameter of the circumscribed circle. Let $ P$ be the intersection of the external angle bisectors at $ A,B$, and $ Q$ be the same of those at $ C,D$. If $ J,O$ are the incenter and the circumcenter of $ ABCD$, prove that $ JO$ is perpendicular to $ PQ$.

Problem is wrong.
we can name A,B,C,D in 2 different ways(clockwise or non-clockwise), but the result is different  for each case.


In a school there are $b$ teachers and $c$ students. Suppose that
(I) each teacher teaches exactly $k$ students and
(II) for each pair of distinct students, exactly $h$ teachers teaches both of them.
Show that $\frac{b}{h} = \frac{c(c-1)}{k(k-1)}$

Let $N$ be the number of triples formed by students $x,y$ and teacher $z$ such that $z$ teaches both $x$ and $y$.
There are exactly $\frac{c(c-1)} 2$ pairs of students. For each of them, there are $h$ teachers who teaches both of the students.
Thus $N = h\frac{c(c-1)} 2$.  (1)

In another hand, let's consider any teacher, say $z$. There are exactly $\frac{k(k-1)} 2$ pair of students who both are teached by $z$. Since there are $b$ teachers, we deduce that $N = b \frac{k(k-1)} 2$.  (2)

The desired equality follows from (1) and (2).

Pierre.


I alternately toss a fair coin and throw a fair die until I either toss a head or throw a 2. If I toss the coin first, what is the probability taht I throw a 2 before I toss a head?

I have no idea how to solve this problem. Could someone please explain it step by step? Thanks

The source is 5th ML, number 6, for this year.

The flipping a coin and tossing a die are independent, so you can multiply probabilities.
A success on the coin would be a tail, which has 1/2 probability; otherwise we would have flipped a head before rolling a 2, the failure.
After flipping a tail, a success would be rolling a 2, which has 1/6 probability. However, if you do not roll a 2, you go back to flipping the coin; that has 5/6 probability. When you go back to flipping the coin, it's as if you started over again.

Let p be the probability that you roll a 2 before flipping a head.
p=(1/2)(1/6)+(1/2)(5/6)(p)
The first fraction means that you flipped a tail and rolled a 2.
The second means that you flipped a tail but did not roll a two, so you continue flipping/rolling, as if you started over with the same probability of rolling a 2 before flipping a head.
Solve this to get 1/7.

This problem has already been discussed here

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30044

and here

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30496

this is a geometric sequence problem...

[hide="solution"]
let's say we toss a 2 on the first toss of the dice.
the probability of that happening is $\frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12}$

same thing, except this time it happens on the second toss of the dice:
$\frac{1}{2} \cdot \frac{5}{6} \cdot \frac{1}{2} \cdot \frac{1}{6} = \frac{5}{144}$

so each time, we have to multiplyu by another $\frac{5}{12}$ in order to find probability of 2 coming before a head on the next term. so this creates a geometric sequence with $F_1 = \frac{1}{12}$ and $r = \frac{5}{12}$

so finding sum of infinite geometric sequence: $\frac{F_1}{1 - r} = \frac{\frac{1}{12}}{1 - \frac{5}{12}} = \frac{1}{7}$

[/hide]


The function $ f: N\rightarrow N$ is such that $ f(1)\equal{}1,f(2n)\equal{}2f(n),$ and $ nf(2n\plus{}1)\equal{}(2n\plus{}1)(f(n)\plus{}n)$ for all $ n\geq 1$.

Prove that $ f(n)$ is always an integer, and for how many positive integers less than $ 2007$ is $ f(n)\equal{}2n$?

Call $ g(n) \equal{} \frac {f(n)}{n}$ then $ g(1) \equal{} 1,g(2n) \equal{} g(n),g(2n \plus{} 1) \equal{} g(n) \plus{} 1$.
Easy to check that $ g(n)\in Z^ \plus{} \forall n$ by induction!
(Assume that $ g(n)\in Z^ \plus{} \forall n\leq k$: if $ k \plus{} 1 \equal{} 2m$ then $ g(k \plus{} 1) \equal{} g(m)\in Z^ \plus{}$; if $ k \plus{} 1 \equal{} 2m \plus{} 1$ then $ g(k \plus{} 1) \equal{} g(m) \plus{} 1\in Z^ \plus{}$ (Because in all $ 2$ case we have $ m\leq k$))
We must find how many positive integers $ n$ less than $ 2007$ so that $ g(n) \equal{} 2$!
With $ n \equal{} 2^k.l: k\geq 0,l$ odd we have $ g(n) \equal{} g(l)$ then if $ g(n) \equal{} 1$ it mean $ g(l) \equal{} 1\Leftrightarrow l \equal{} 1$ (If $ l \equal{} 2m \plus{} 1\geq 3$ then $ g(l) \equal{} g(m) \plus{} 1\geq 2$)
Then $ g(n) \equal{} 1\Leftrightarrow n \equal{} 2^k$!
Now with $ n \equal{} 2^k.(2m \plus{} 1),k\geq 0,m\geq 1$ we have if $ g(n) \equal{} 2$ it mean $ g(2m \plus{} 1) \equal{} 2\Leftrightarrow g(m) \equal{} 1\Leftrightarrow m \equal{} 2^h,h\geq 0\Leftrightarrow n \equal{} 2^{k \plus{} h \plus{} 1} \plus{} 2^k,k,h\geq 0$
$ \Leftrightarrow n \equal{} 2^p \plus{} 2^q,p > q\geq 0$
Finally, $ n\leq 2007\Leftrightarrow 0\leq q < p\leq 10$, it have $ C^2_{11} \equal{} 55$ numbers $ n$ so that $ f(n) \equal{} 2n$!

For the record, this was BrMO 1 2008 Question 6


calculus
integration
function
real analysis
real analysis unsolved

Let  f be a nonnegative function define in [0,00) such that
1) f(0)=0
2) f is derivable in [0,00)
3) f is increasing
 4) lim{x-00}f(x)=00.
Fet F be a primitive of f such that F(0)=0. Show that F(x)/f(x) has a positive limit (i.e. >0) when x goes to infinity.

source: unknown

I don't think that's true.

Take f(x)=ee<supx-e. Then f'(x)=ex*ee<supx. It's clear that f(x)/f'(x)->0 as x->00 so, from L'Hospital's rule, F(x)/f(x)->0 as x->00, so the limit isn't >0, but 0.

Maybe it's supposed to be >=0? (that's obvious, so maybe we just have to show that the limit exists).


inequalities
LaTeX
Gauss
search
geometry
geometric transformation

sorry for don`t using LATEX :(

1) Let $x,y,z$ be positive numbers such that $x+y+z=1$. Prove that
\[(1+x)(1+y)(1+z) \geq 8(1-x)(1-y)(1-z).\]

2) Let $x,y,z$ be positive numbers such that $xyz=1$. Prove that
\[\frac{x}{y}+\frac{y}{z}+\frac{1}{x}\geq x+y+1.\]

3) Let $x,y,z$ be positive numbers such that $x^{2}+y^{2}+z^{2}=1$. Prove that
  \[x+y+z+1/xyz \geq 4\sqrt3.\]
4) Let $x,y,z$ be positive real numbers, prove that
 \[\frac x{\sqrt{x+y}}+\frac y{\sqrt{y+z}}+\frac z{\sqrt{z+x}}\leq \frac54 \sqrt{x+y+z}.\]
5) Let $0<x\leq y\leq z$. Prove that
\[(x+y+z)(xy+yz+zx) \geq 9xyz+(y-x)(z-x)^{2}.\]
sorry if you already met this problems.

You can learn $\text{\LaTeX}{}$ in this forum by looking at post sources in the forum.

pvthuan

ok, I understood.
I found problem 4. is here: http://www.mathlinks.ro/Forum/viewtopic.php?t=1459

1. The inequality is equivalent to $(2x+y+z)(x+2y+z)(x+y+2z) \geq 8(x+y)(y+z)(z+x)$
By letting $a = y+z, b = z+x, c = x+y$, then we have $(a+b)(b+c)(c+a) \geq 8abc$ which is clear by $AM-GM$.

$(x+y+z)(xy+yz+zx)-xyz=(x+y)(y+z)(z+x)$
so given inequality equivalent to
\[(x+y)(y+z)(z+x)\ge 8xyz+(y-x)(z-x)^{2}\]
maybe this will help. at least the inequality looks nicer. :D

2. We apply the substitution $x=\frac{a}{b}, y=\frac{c}{a}, z=\frac{b}{c}$

Multiplying through by $abc$ gives $a^{3}+c^{3}+b^{2}c \ge a^{2}c+bc^{2}+abc$

Which is true in light of the 3 AM-GMs:

$\frac{c^{3}+b^{2}c}{2}\ge c^{2}b$,

$\frac{a^{3}+\frac{c^{3}}{2}}{3}\ge \frac{a^{2}c}{2}$,

$\frac{a^{2}c}{2}+\frac{b^{2}c}{2}\ge abc$

[quote="Little Gauss"]by EMV method[/quote]

You are right, nice and simple proof  :lol:

I'm sorry, but can you explain how the method works in that problem ? I don't too familiar with EMV method.. Thanks a lot..

[quote="RDeepMath91"][quote="Little Gauss"]#5.
$(x+y+z)(xy+yz+zx)\ge 9xyz+(y-x)(z-x)^{2}$
$\iff x(y-z)^{2}+y(z-x)^{2}+z(x-y)^{2}\ge (y-x)(z-x)^{2}$
so we can assume that $x=0$.
->$(y+z)yz\ge yz^{2}\iff y+z \ge z$. this is trivial.[/quote]

Why can we assume that $x = 0$ ?[/quote]

Well, this one is pretty simple. So I can explain for you.

EMV="Entirely Mixing variable" method, that means you change $x,y,z$ to $x-t,y-t,z-t$ (mix all variable at one). If we choose $x'=x-t,y'=y-t,z'=z-t (t\le z)$ then
\[x(y-z)^{2}+y(z-x)^{2}+z(x-y)^{2}\ge x'(y'-z')^{2}+y'(z'-x')^{2}+z'(x'-y')^{2}.\]
\[(y-x)(z-x)^{2}=(y'-x')(z'-x')^{2}.\]

That means, one side is unchanged but one side is decreased. So if we choose $t=c$, or in other word, let one of three variable $a,b,c$ equal to $0$, we can continue as Little Gauss do.

About this method, you can search my article in Mathematics Reflections (issue 2006) to see more examples. Hope that this explanation satisfy you.

Actually, EMV method is a particular case of the substitution method: for $0 \le x\le y \le z$ we make $y=x+a,z=x+b, a\ge 0, b\ge 0$.

In some situation only. My improvement of EMV can solve Suranji, VASC's inequality and Turkevici inequality very easily in just some line. Furthermore, the original EMV is very easy to use and simple. If letting $x=z+a,y=z+b$, we must use a lot of computing.

Anyway, problem 4 is very nice by EMV but not the direcly expanding way as  $x=z+a,y=z+b$ (If we do that, it is much complicated).  :lol:  I just love simple method only.


don't open it. I copy-pasted it here and only three symbols showed up when I previewed.

[quote="mathelete"]The message is too small. Please make the message longer before submitting.[/quote]

..........tell us what this is and we might acctually dl it....

[quote="AIME15"]Seriously mathelete...

If you want to spam a forum, go to some other forum, not AoPS.[/quote]

I don't think mathelete is as bad as some other people on this forum. *cough*[size=34]you should know who im thinking of[/size]
He's actually pretty smart, I know him in real life.

No he means the person above him...

lol its been downloaded 11 times.

Whoops, I forgot to tell you to press the enter key multiple times. The numbers will change.

[quote="AIME15"]No thanks, I am NOT opening a .exe file until you tell me what it is.[/quote]

You think I am going to crash your computer?

It works in only microsoft-based operating systems


[quote="story"]Three people claim that they are psychic. In order to test there skills, the villagers held a contest.
2 Gamecubes and a X Box was put in three different boxes and was distributed to the people claiming to be psychic. None of them saw what they got, but they knew of the fact that there were 2 Gamecubes and a X Box in the boxes.
There's an ironic end to this, though. Person A looked at what Person B got, and Person B saw what Person C had. Person C was an honest person, so he didn't cheat, but he knew that person A and person B cheated. When the contest started, neither person A nor person B guessed an answer. Person C noticed that neither one of them guessed an answer, so he came to a logical conclusion as to what's inside his box, and he was right.
The moral of the story is: Don't lie.[/quote]
In the story, what did each person get?

(The original story is from the movie "Trick," but was translated and modernized.)

[hide]

The reason person A and B didn't say what they had was because they didn't know what they had I'm assuming. Thus, person B and person C had the gamecubes and A had the XBOX.

[/hide]

Correct.

...all this typing for 5 minutes... :(

yup, since if either person A or B saw an X-Box, then they would immediately know that they had a Gamecube themselves, since there was only one X-box

Can we generalize this? (For example, is there other cases where the reader can determine what the psychics had? Can we make a generic case?)

person C has xboc because person B couldn't hazard a guess


function
algebra
rational function
algebra unsolved

It is some very close approximation of $4e - 1$ -- what kind of closed-form are you expecting?

Looks a lot like Canada 1994: [url=http://www.kalva.demon.co.uk/canada/can94.html]www.kalva.demon.co.uk/canada/can94.html[/url]

[quote="warut_suk"]Find the sum of $\frac{k^2+k+1}{k!}$ from k=1 to 1994[/quote]

    [hide="Hint"]$k^2+k+1$ = $k(k-1) + 2(k)+1$[/hide]

For example, $\Sigma \frac {k}{k!} = \Sigma \frac{1}{(k-1)!}$, so the sum is expressible in terms of the partial sums of $1/{n!}$.

i don't think the sum can be expressed by a simple form

I'm sure fleeting_guest thought you wanted to compute the entire series, i.e. from $k=0$ to $\infty$ (that's also what I thought :)).

The idea was to express the sum to $n$ terms (here $n=1994$) in terms of $e_n = (1 + \dots 1/{n!})$.

Now i don't think it is not going to be so easy and even i think impossible :D

But well, warut what is the problem source? can u look over it again whether ur typing was correct or not! :P

By the way, I think that in the cases where the sum to $n = \infty$ does not involve $e$, the sums for finite $n$ always are telescopic and can be expressed as a rational function of $n$ and $n!$.
e.g.  $\sum \frac{k^2 + k - 3}{k!}$.

[quote="mahbub"]the aim to evalute sum is not express as function in olympiad. right? Yes in these types of probs telescoping is the general idea. so i want to be sure if this prob is correct at all![/quote]

The given sum equals $\sum \frac {k^2 + k - 3}{k!} + 4 \sum \frac {1}{k!}$.  The first one is telescopic, so the  problem  is equivalent to computing finite sums of $1/n!$.  In other words, there is no hope of further simplification.

[quote="mahbub"]but i don't think there is any hope to solve the later sum :P[/quote]

Yes, that is a repetition of what was explained above.


inequalities
function
inequalities proposed

A real number r is said to be attainable if there is a triple of positive real numbers a,b,c such that a,b and c are not the lengths of the sides of a triangle and satisfy

rabc>(a^2)b +(b^2)c + (c^2)a

1.Determine if 7/2 is attainable

2.Find all positive integers n such that n is attainable

To 1:

7/2 is attainable. The triple a=1/2^(1/3), b=1, c=a+b shows it.

To 2:

I don't have a complete solution but I think you should begin like that:
Let c>a,b. I rewrite the inequality as follows: n>a/c+b/a+c/b. The right side is a strictly increasing function in c so that n>a/(a+b)+b/a+(a+b)/b is the necessary condition. I made a lot of calculations after that but I didn't become a pretty inequality yet :( .

Misha

from Cauchy's inequality we have:$a^2b +b^2c + c^2a \geq 3(a^3b^3c^3)^{\frac{1}{3}}=3abc$, so the only integers which can be attainnable are $1,2$. Now you just have to find an example

Zorro, the problem is we have the inequality the other way round. From your post it follows n>3 and not n<3.

Misha

Now I see that we already have solved the problem. We have shown that n must be greater than 3 and for all n>3 we can take the same a,b,c as for 7/2.


limit
trigonometry
integration
calculus
calculus computations

Let f (x) be continuously differentiable on (0,∞) and suppose lim
x→∞ f '(x)=0.
Prove that lim x→∞ f (x)/x = 0.

Do you want an $ \varepsilon-\delta$ proof, or something less formal? Since $ \lim_{x\to\infty}f'(x)=0$ and $ f$ is continuously differentiable on $ (0,\infty)$ we know that $ \lim_{x\to\infty}f(x)=C$ tends to some constant. Therefore, $ \lim_{x\to\infty}\frac{f(x)}{x}=0$ since the numerator tends to a constant while the denominator tends to infinity.

[quote="HilbertThm90"]Since $ \lim_{x\to\infty}f'(x)=0$ and $ f$ is continuously differentiable on $ (0,\infty)$ we know that $ \lim_{x\to\infty}f(x)=C$ tends to some constant.[/quote]
That is absolutely false. Just consider $ f(x)=\sqrt{x}.$ Or $ f(x)=\sqrt[4]{x}\sin(\sqrt{x}).$

As for correctly proving it (from scratch, rather than claiming that it follows from Cesaro-Stoltz):

Choose $ \epsilon>0.$ Find $ M$ such that if $ x>M,$ then $ |f'(x)|<\frac{\epsilon}2.$

For $ x>M$ we have $ f(x)=f(M)+\int_{M}^{x}f'(t)\,dt.$

But $ \left|\int_{M}^{x}f'(t)\,dt\right|<\frac{\epsilon}2(x-M).$

Then $ \left|\frac{f(x)}x\right|<\frac{|f(M)+\frac{\epsilon}2(x-M)}x< \frac{|f(M)|}x+\frac{\epsilon}2.$

Choose $ N$ so that for $ x>N,\ \frac{|f(M)|}x<\frac{\epsilon}2.$ Then for $ x>N,$ we have $ \left|\frac{f(x)}x\right|<\epsilon.$

Since we can do this for any $ \epsilon>0,$ we are done.


geometry
projective geometry
geometry solved

this is venkat from india.
is euclidean geometry enough for solving imo geometry problems, or is it important to know concepts like projective geometry etc  :)  ?

Euclidean geometry should be sufficient, however, some knowledge of projective and inversive geometry can be useful and won't hurt you :)


LaTeX
superior algebra
superior algebra unsolved

Let G be a group, and x; y have order m and n, respectively, where
(m; n) = 1. If xy=yx^a for some positive integer a. Prove that m is a divisor of a^n-1.

just to make sure,

is that $ yx^a$ or $ (yx)^a$?

is it $ a^n\minus{}1$ or $ a^{n\minus{}1}$?

The automorphism $ x \mapsto y^{\minus{}1} xy$ has order $ n$.

I just thought we can consider the group action
   (y^i)(x^j) ---> X^(ja^i)
Then consider its stabalizer.

  Thank for your hint !!!
  (Sorry, I cannot play Latex, but I will learn)


[hide="Hint"]For a number to end in 0, it must have a multiple of 10.   To create a multiple of 10, a number must have factors of 2 and 5.[/hide]

[quote="bpms"][quote="Ignite168"]How many 0s does $\frac{127!}{64!*32!*16!*8!*4!*2!*1!}$ end in?[/quote]
Is it [hide]0?[/hide][/quote]
Please provide a solution.

I get 6, but I did it im my head, so perhaps I divided wrong...

EDIT: Typo, I really did get 6.

[quote="Ignite168"]How many 0s does $\frac{127!}{64!*32!*16!*8!*4!*2!*1!}$ end in?[/quote]

[hide]127! has 31 zeros
64! has 14 zeros
32! has 7 zeros
16! has 3 zeros
8! has 1 zero
4!,2! and 1! have none

Therefore, the answer is $\boxed{6}$[/hide]

There was another discussion like this. $\frac{5!}{4!}$ ends in 0 zeroes, even though $5!$ has 1 zero and $4!$ has 0 zeroes.
[hide] In this case, we are not worried about the factors of 5, but the factors of 2. $127!$ has $120$ factors of 2. $64!$ has $63$ factors of 2, $32!$ has $31$ factors of 2, $16!$ has $15$ factors of 2, etc. Adding $63+31+15+7+3+1=120$. Therefore, it ends in $120-120=\boxed{0}$ zeroes. [/hide]

6 is incorrect. lotrgreengrapes is the only one that is correct. Tricky, eh?

[quote="mysmartmouth"]Haha, I've seen a similar problem in the SC forum, and I still made the same mistake.

Here's the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84288]problem[/url].[/quote]

I was hoping someone would catch that as the bottom of the fraction I made was all powers of 2 will the top was 1 less than one.


calculus
derivative
trigonometry
calculus computations

$ f: \mathbb{R}x\mathbb{R} \rightarrow R: \ f(x,y) \equal{} \sin(x) \sqrt {|y|}$

I think that the total derivate of $ f$ only exist on the set ([b]R[/b]x[b]R[/b]) \ ([b]R[/b]\(2$ \pi \mathbb{Z}$)x{[b]0[/b]}), is this correct ?

[quote="fedja"]What you wrote is a bit ambiguous and formally incorrect, but if you really meant $ \mathbb R^2\setminus[(\mathbb R\setminus\pi\mathbb Z)\times\{0\}]$, then yes. :) The idea is right: either $ y$ should be different from $ 0$ or $ \sin x$ should be equal $ 0$.[/quote]

Thanks, I had worked it out, so probably I did it good :) Except I forgot that $ sin(\pi)\equal{}0$ (the $ 2$ comes from there)   :blush:


Determine each real root of
\[x^{4}-(2*10^{10}+1)x^{2}+10^{20}+10^{10}-1 = 0 \]

[hide]
Rewrite this as
\[10^{20}-10^{10}(2x^{2}-1)+x^{4}-x^{2}-1 = 0 \]
Now we will apply the quadratic equation to $10^{10}$:
\[10^{10}= \frac{2x^{2}-1 \pm \sqrt{4x^{4}-4x^{2}+1-4(x^{4}-x^{2}-1)}}{2}\]

\[10^{10}= \frac{2x^{2}-1 \pm \sqrt{5}}{2}\]

\[x= \sqrt{\frac{2*10^{10}\pm \sqrt{5}+1}{2}}\]
And then we can take the square root. Unfortunately, this method wouldn't be feasible on the actual test because calculators are not allowed, but I think it's much simpler than other solutions I've seen.
[/hide]


Putnam
floor function
ratio
IMO
IMO 2007

What if you ranked the countries at the IMO by the same scheme used to rank colleges on the Putnam? Would it come out differently? The idea: first, convert all scores into ranks, averaging the ties. Thus, Konsantin Matveev gets a rank of 1, Caili Shen and Peter Scholze are both 2.5, Alex Zhai and Sherry Gong are among five people with rank 10, and so on. Then, determine the team score by adding the individual ranks. For the Putnam, that's the rank of three people, chosen beforehand. The only way I could see to do it for IMO countries was to limit it to countries that fielded full 6-member teams, and all all six ranks. It turns out that this would make very little difference. Here's the top 20 in this system.

$ \begin{array}{cccc}\text{No.}&\text{Country}&\text{Rank sum}&\text{Point sum}\\ 1.&\text{Russia}&164&184\\ 2.&\text{China}&176.5&181\\ 3.&\text{S. Korea}&267&168\\ 4.&\text{Vietnam}&273.5&168\\ 5.&\text{Japan}&413&154\\ 6.&\text{USA}&418.5&155\\ 7.&\text{Ukraine}&441.5&154\\ 8.&\text{Bulgaria}&445&149\\ 9.&\text{Taiwan}&462&149\\ 10.&\text{Romania}&492.5&146\\ 11.&\text{Hong Kong}&518&143\\ 12.&\text{N. Korea}&541.5&151\\ 13.&\text{Iran}&546.5&143\\ 14.&\text{Thailand}&629.5&133\\ 15.&\text{Hungary}&736&129\\ 16.&\text{Germany}&776.5&132\\ 17.&\text{Turkey}&807.5&124\\ 18.&\text{Belarus}&827&119\\ 19.&\text{Poland}&832&122\\ 20.&\text{Moldova}&855.5&118\end{array}$

Vie got 3 gold medals, Kor got 2 gold medals, therefore the rank of Vie is 3 and Kor is 4th...  :|

bookworm_vn: you are missing the whole point of the post.

I did not know that Putnam teams were ranked this way. And I can't say that I like it. Counting medals and such is counting positive contributions to the team. The worst a team member can do is to contribute nothing (which is not something unusual). Adding up individual ranks means counting negative contributions of team members, i.e. their failure to reach the top. One unsuccessful participant can bring down the entire team by a lot. I would not want to be that participant. :|

Of course, the psychological difference between $ \min\sum_{i}x_{i}$ and $ \max\sum_{i}\left(n-x_{i}\right)$ is immense...

The difference between comparing the sum of ranks and the number of medals is actually more a difference between $ \min\sum_{i}x_{i}$ and $ \max\sum_{i}\left\lfloor n-x_{i}\right\rfloor$, since medals are basically the same as reversed ranks after a kind of rounding operator (this is the way medals are defined on the IMO - the cutoffs have to be set in a way to make the ratio of gold:silver:bronze:nothing as near as possible to 1:2:3:6). Still, both ways to rank the quality of a team are pretty much of the same kind. It's just that one is "positively formulated", with a good result contributing to the score, and the other one is "negatively formulated", with a bad result mutilating the score of the team, but this is a purely emotional difference.

  Darij


Ένα ενδιαφέρον θέμα που άκουσα (και θέλω να πιστεύω πως έδειξα  :oops: )

Αν από την αρμονική σειρά ($ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}...$) αφαιρέσουμε όλους τους όρους που περιέχουν το ψηφίο 9 μέσα τότε η σειρά θα συγκλίνει.

Hmm , nice ! Λοιπόν κάθε τέτοιος k-ψηφιος αριθμός έχει τη μορφή $ 10^{k \minus{} 1}a_1 \plus{} ... \plus{} a_k$ όπου $ 0\leq a_i < 9$
Για δοσμένο $ k$ έχουμε $ 8\cdot 9^{k \minus{} 1}$ τέτοιους αριθμούς , κανείς από τους οποίους δεν ξεπερνά το $ 10^{k \minus{} 1}$

Άρα το άθροισμα της σειράς που προκύπτει είναι $ < \sum_{k \equal{} 1}^{\infty} 8\frac {9^{k \minus{} 1}}{10^{k \minus{} 1}} \equal{} 80$


Determine (algebraically) whether each function is even, odd, or neither.

(a) f(x) = 2x^4 - x^2

(b) F(x) = x/|x|

NOBODY GIVE HIM THE SOLUTION PLEASE!!!

[hide="hints only"]It is odd when $f(x)=-f(-x)$

Its even when $f(x)=f(-x)$

its neither when its neither[/hide]

[quote="abacadaea"]b can be any of them[/quote]

a function can't be a combination of even, odd, and neither. just like a number.

i said i [b]could[/b] be. it depends on what x is

abcdabcd, even and odd aren't referring to parities of #s-they are charecteristics of functons. An even function $ f$ is such that $ f(x)\equal{}f(\minus{}x)$, and an odd is such that $ f(x)\equal{}\minus{}f(\minus{}x)$.

Yeah, it's true. A function can't be BOTH even and odd. Get it now, abacadaea?

oooh i see what you mean. yeah i know but i meant for different values of x. so if we have like $ f(x)\equal{}x\plus{}2$, f(x) is even when x is even and odd when x is odd. so it is possible, for different values of x

You're thinking of "even" and "odd" as divisibility by 2. When working with functions, the terms "even" and "odd" refer to the characteristics of a function, which were described in perfect628's post.

The example you provided, $ f(x) \equal{} x\plus{}2$, is neither because $ f(x)\ne f(\minus{}x)$ and $ \minus{}f(x)\ne f(\minus{}x)$.


Given that $a,b,c>0$ and $abc=1$ prove that $a^{2}+b^{2}+c^{2}\ge a+b+c$

can someone please give me a hint about where to start. I am thinking Cauchy but its not getting me anywhere.

[hide="hint"]
find reasonable constants $x,y,z$ such that $xa^{2}+yb^{2}+zc^{2}\geq a$
[/hide]

best thing for this kind of inequality(where you have the condition $abc=1$) is to try and homogenize it with a couple of nice substitutions(so that every summand is of the same degree). in example
$a^{2}$ is degree $2$, $ab$ is degree $2$,$a^{7}b^{4}c^{6}$ is degree $17$, $a$ is degree one,constant is degree $0$...
here is a way to homogenize it and a start to the solution.
[hide]
$a=x^{3},b=y^{3},c=z^{3}$
then it becomes $x^{6}+y^{6}+z^{6}\geq x^{3}+y^{3}+z^{3}$ and now using that $xyz=(abc)^{\frac{1}{3}}=1$
$x^{6}+y^{6}+z^{6}\geq x^{4}yz+xy^{4}z+xyz^{4}$
now try to find the constants that scorpious asked for
$px^{6}+qy^{6}+rz^{6}\geq (p+q+r)x^{4}yz$[/hide]

[hide="Cauchy-Schwartz and AM-GM Solution:"]
\[a^{2}+b^{2}+c^{2}\geq \frac{(a+b+c)^{2}}{3}\geq (a+b+c)\sqrt[3]{abc}=a+b+c \]
[/hide]


Two players A and B take stones one after the from a heap with n>=2 stones. A begins the game and takes at least 1 stone but no more then n-1 stones.Each player on his turn must take at least 1 stone but no more than the other player has taken before him.The player who takes the last stone is the winner. Find who of the players has a winning strategy.

What I had posted before was totally wrong :). I hope it's not the case now.

If $n$ is odd, then $A$ takes one stone, and it's clear that he wins. We can generalize this to all the cases when $n$ i not a power of $2$. Suppose $n=2^{a_0}+\ldots+2^{a_k},a_0<a_1<\ldots<a_k$. $A$ can take $2^{a_0}$ stones. After $B$ makes a move, the number that's left has, in its binary expansion, the smallest power of $2$ equal to at most the maximal power of $2$ which divides the number of stones taken by $B$, so $A$ can repeat the procedure.

If $n$ is a power of $2$, After $A$ makes a move, we're left with some number $2^{a_0}+\ldots+2^{a_k},k\ge 1$, just as above. Then $B$ can take $2^{a_0}$ stones, and it continues just as above, except that $B$ wins.

In conclusion, $B$ has a winning strategy iff $n$ is a power of $2$.

Is it correct?

Aleks I couldn't check  your previous solution...But this's very nice solution.


Among any 79 consecutive positive integers, show that there is at least one whose sum of digits is divisibly by 13.

[quote="neelnal"][hide]
In any 79 consecutive integers, the sum of the digits always vary from n, some number, to n+(7+9-1)=n+15.

So, n+15, since n is a postive integer, will always contain a multiple of 13.[/hide][/quote]

Eh?

What if the sequence contains something like ...,998,999,1000,1001,...

It doesn't vary from just n to n+15

[quote="neelnal"][hide]
In any 79 consecutive integers, the sum of the digits always vary from n, some number, to n+(7+9-1)=n+15.

So, n+15, since n is a postive integer, will always contain a multiple of 13.[/hide][/quote][color=blue][b]neelnal[/b][/color], how can we be sure about that?
Is it a theorem?
If yes, where can we find it?

Anyway, [color=blue][b]Jumbler[/b][/color] is right.

Take for example the 79 consecutive numbers $980, ..., 1058$

We get two groups of numbers $1,...,14$ and $17,...,27$

We don't have any group $n, ..., n+15$

[hide="hint"]$79=6(13)+1$...hm this smells like pigeons [/hide]


For anyone who's going to MOP this year, could you please help? I seem to not have gotten the email from AMC about MOP and I never got the MOP homework problems either. I'd like to at least look at them before going . . . thanks!

PM me your email adress and they shall be sent.

{EDIT} Issue has been resolved. This probably should be deleted. or you can leave it here as a memoir of uh. . . something.


Let a triangle ABC, O is the circumcenter, K is the nine-point center of ABC . BO meets AC at D, CO meets AB at E. Denote M,N,P be the midpoints of BC,BD,CE. Prove that the bisector of angle NKP passes through M.

Hello!

     Let $ H$ be the orthocenter of the triangle $ ABC$, {$ B'$}${ = BH\cap CA}$, {$ C'$}${ = CH\cap AB}$, $ A"$ is the midpoint of $ AH$,
$ B"$ is the midpoint of $ BH$ and $ X$ is the midpoint of $ AC$.

It's well known that $ OX = \frac {BH}{2}$ and the points $ B"$, $ K$, $ X$ are collinear.
It's easy to see that the quadrilateral $ BOXB"$ is parallelogram. So $ B"X\parallel{}BD$. But $ K$ is the midpoint of $ B"X$ and
$ N$ is the midpoint of $ BD$, so the points $ B'$, $ K$, $ N$ are collinear    (1).
Similarly we have that the points $ C'$, $ K$, $ P$ are collinear     (2).
It's well known that the points $ A"$, $ K$, $ M$ are collinear     (3).
The point $ A"$ is the center of the circle through the points $ A$, $ C'$, $ H$, $ B'$.
So $ A"B' = A"C'\Rightarrow KA"$ is the bisector of the angle $ \angle B'KC'$     (4).
(1), (2), (3), (4) $ \Rightarrow KM$ is the bisector of the angle $ \angle NKP$.   :)


Prove that if V  and W are finite-dimensional vector spaces such that dimW <dimV , then there is no one-one linear transformation T:V→W.

Obviously if V is R^m and W is R^n, with n < m, then it could be solved by the very fact that the nullity of T isn't zero.

Is there a general solution for the general vector spaces?

Yes, it's exactly the same argument: the nullity of $ T$ can't be zero.


hello, i think your product $ \prod_{i\equal{}1}^m 2\minus{}\frac{2i\minus{}1}{i}\equal{}\frac{1}{m!}$.
Sonnhard.

[hide]the left hand side is $ \frac{1}{m} \cdot \frac{3}{m} \cdot \ldots \cdot \frac{2m\minus{}3}{m} \cdot \frac{2m\minus{}1}{m}$
The right hand side is $ 1 \cdot 2 \cdot \ldots \cdot m\minus{}1 \cdot m$
BOth sides have $ m$ terms. Furthermore,
$ 1 \ge \frac{1}{m}$
$ 2 > \frac{3}{m}$ (given that $ m>1$)
$ \vdots$
$ m\minus{}1 > \frac{2m\minus{}3}{m}$ (given that $ m>1$)
$ m > \frac{2m\minus{}1}{m}$ (given that $ m>1$)
All inequalities except for the first can be seen because each term on the Left hand side is less than 2.
The first inequality is true only for $ m\equal{}1$, which is the equality case for the original inequality.
For all other m, each term on the right hand side is greater than its corresponding term on the left hand side, so the inequality holds.[/hide]

$ (2m \minus{} 1) \cdot (2m \minus{} 3) \cdots 3 \cdot 1 \neq (2m \minus{} 1) \plus{} (2m \minus{} 3) \plus{} \ldots \plus{} 3 \plus{} 1$  :roll:

Given that the stated product is less than 1 for all $ m > 1$, I'm inclined to think the OP made some error in stating the problem.

[hide=" my solution"]
$ (2 \minus{} x) < 2$ for all positive $ x$

Therefore $ (2 \minus{} \frac {1}{m}) . (2 \minus{} \frac {3}{m}).(2 \minus{} \frac {5}{m})...............(2 \minus{} \frac {2m \minus{} 1}{m}) < 2^{m}$ since $ m$ is positive.

Also $ 2^{m} < m!$ for all $ m \ge 4$

Therefore $ (2 \minus{} \frac {1}{m}) . (2 \minus{} \frac {3}{m}).(2 \minus{} \frac {5}{m})...............(2 \minus{} \frac {2m \minus{} 1}{m})\le m!$ for all $ m \ge 4$

The cases when $ m < 4$ can be done manually.

 [/hide]

Equality occurs when $ m \equal{} 1$


Prove that for every natural n the  following equation has infinitely many solutions in natural numbers: $x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}=y^{2}$.

If $a$ is odd. Then:
$a+(\frac{a-1}{2})^2=(\frac{a+1}{2})^2$.
So we can choose $x_1,x_2,\ldots,x_{n-1}$ in such way that $x_1^2+x_2^2+\ldots+x_{n-1}^2$ is odd. We can do it in infinitely many ways. Then we choose $x_n$ and $y$ and we're done.


What is the eccentricity of the conic defined by $x^{2}-4y+6x=0$?

[quote="AstroPhys"]HINT
Put it in form $\frac{\left(x-h\right)^{2}}{a^{2}}+\frac{\left(y-k\right)^{2}}{b^{2}}= 1$
Then the eccentricity is
$\epsilon = \sqrt{1-\frac{b^{2}}{a^{2}}}$.[/quote]

Please hide your "hint". I'm pretty sure that's only for an ellipse.  :wink:

[hide="Anyway, doing it,"]$4y = x^{2}+6x$

$y = \frac{x^{2}}{4}+\frac{3x}{2}$

I don't think the eccentricity is defined for parabolas; if you meant the "shrink", that would be the coefficient of the $x^{2}$ term, $\frac{1}{4}$.[/hide]

The eccentricity of a parabola is $1$.  Think of a parabola as an ellipse with an infinitely long semimajor axis.

eccentricity is also defined as the ratio distance(point on locus to focus) : distance (point on locus to directrix)

which is why t0rajir0u is correct: a parabola is the locus of all points equidistant from a given point (the focus) and a given line (the directrix)

The eccentricty of a parabola is one. Even though most people don't use this fact as something important in solving math problems, it is simply an ellipse stretched out. (I remember learning all of this in Algebra II. Yay!)

You can simply change the equation into:

(x+3)= 4(y+9/4),

which is the standard form of a parabola. From this we know that the eccentricity of this conic is ONE!


The function $ f$ is defined by $ f(n) \equal{} f(n \minus{} 1) \plus{} f(n \minus{} 2)$. It is also
true that $ f(1) \equal{} 3$ and $ f(3) \equal{} 10$. What is the value of $ f(6)$?

We're not sure how to find $ f(6)$, so let's find out stuff and then see if that helps us. Since $ f(3) \equal{} 10$ and $ f(1) \equal{} 3$, that means $ f(2) \equal{} 10 \minus{} 3 \equal{} 7$. From that we can deduce that $ f(4) \equal{} 10 \plus{} 7 \equal{} 17$ and $ f(5) \equal{} 7 \plus{} 17 \equal{} 24$. Aha! $ f(6)$ must be $ 17 \plus{} 24 \equal{} \boxed{41}$.

[quote="isabella2296"]We're not sure how to find $ f(6)$, so let's find out stuff and then see if that helps us. Since $ f(3) \equal{} 10$ and $ f(1) \equal{} 3$, that means $ f(2) \equal{} 10 \minus{} 3 \equal{} 7$. From that we can deduce that $ f(4) \equal{} 10 \plus{} 7 \equal{} 17$ and [b]$ f(5) \equal{} 7 \plus{} 17 \equal{} 24$[/b]. Aha! $ f(6)$ must be $ 17 \plus{} 24 \equal{} \boxed{41}$.[/quote]
Actually, $ f(5)\equal{}10\plus{}17$, since we have $ f(4)\plus{}f(3)$.  So the answer is actually $ 17\plus{}27\equal{}\boxed{44}$


Να βρεθουν ολα τα συνολα τεσσαρων φυσικων αριθων, διαφορετικων ανα δυο, τετοια που καθε στοιχειο να διαιρει το αθροισμα των αλλων!!

 :)

Αρχική ιδέα: το πρόβλημα είναι ισοδύναμο με: έχουμε τον αριθμό $N$ και θεωρούμε το σύνολο των διαιρετών του: $D_{N}= \{ d \geq 1 \wedge d \: | \: N \}$. Αν υπάρχουν 4 στοιχεία στο $D_{N}$ που έχουν άθροισμα ίσο με  $N$ είναι λύση στο πρόβλημά μας, και εύκολα παρατηρούμε ότι όλες οι λύσεις του προβλήματος ανάγονται στην παραπάνω ιδέα.

Οπότε ψάχνουμε $a > b > c > d$ θετικούς διαιρέτες του Ν ώστε:

$a+b+c+d = N \Leftrightarrow \frac{N}{x}+\frac{N}{y}+\frac{N}{z}+\frac{N}{t}= N \Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}= 1$.

Άρα το πρόβλημα είναι: ψάχνουμε τους αριθμούς $x < y < z < t$ με $(x, y, z, t) \in N^{4}$ ώστε $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}= 1$.

Η εξίσωση λύνεται εύκολα (αλλά επώδυνα!!) γι' αυτό και δεν κάθομαι να τη λύσω (ή μάλλον γιατί βαριέμαι).

Απο τα $x, y, z, t$ μπορούμε να γυρίσουμε πίσω σε άπειρες τετράδες $(a, b, c, d)$ καθώς μπορούμε να επιλέξουμε τυχαίο $N$ πολλαπλάσιο του $LCM(x, y, z, t)$ (LCM = ΕΚΠ).

~

πολυ ωραια σκεψη!!!

 :lol:

Οντως ωραια ιδεα.....
Απλα μια διευκρινηση.Λεει να βρεθουν ολα τα συνολα.....
Μ αυτο τον τροπο κατασκευαζεις απειρα τετοια συνολα,αλλα πως αποδεικνυεται
οτι ολα τα συνολα με αυτη την ιδιοτητα προερχονται μονο απο αυτο τον τυπο+ και οχι και απο καποιον αλλο?...(με τον οποιο ενδεχωμενως να εχουν καποια συνολα <κοινα>).Θελω να πω μηπως θα ηταν καλυτερα να ζητουσε <νδο υπαρχουν απειρα συνολα τεσσαρων...>?

ΡΑΛΛΗΣ :D

Κάθε σύνολο που είναι λύση του αρχικού προβλήματος παράγεται με τον τρόπο αυτό.

Γιατί; Έστω ένα σύνολο-λύση είναι το: $\{a, b, c, d\}$. Αν το $a$ διαιρεί το $b+c+d$ τότε διαιρεί και το $a+b+c+d = N$ και αντίστροφα. Ομοίως για τα $b, c, d$. Άρα τα $a, b, c, d$ είναι διαιρέτες του $N$, κλπ.

Τωρα ενταξει καταλαβα,παω πασο....

ΡΑΛΛΗΣ :D

εγω εβγαλα (θα σας πώ μόλις κατορθώσω να φτιάξω τις σημειώσεις μου τις λύσεις
(α,β,γ,δ)=(π(μλ-1),πμκ,πμ,π) όπου π,μ,κ,λ φυσικοί
κι (νομίζω δεν ειναι απολυτο) κ/λ+1
παρακαλώ ελεξτε αν ειναι σωστό
σε λιγο κανα δυο μερες θα πώ αναλυτικά πώς το βρήκα

Θα ηθελα Αμβροσιε οταν μπορεις να διευκρινησεις αυτο που ειπες γιατι φαινεται ενδιαφερον. :wink:

Παρακαλώ πολύ να δείξετε κατανόηση καθότι έχω χαθεί στις σημειώσεις μου βοηθήστε με ελέγχοντας τις λύσεις μου ξανά
δ=π
γ=πμ
β=πμκ
α=π(μλ-1)
π,μ,κ,λ φυσικοί κι(έχω αμφιβολία αν ειναι απαραίτητο) κ/λ+1
εγγυόμαι οτι πριν την αρχή της επόμενης εβδομάδας θα το δημοσιεύσω

αρα φίλε cmad δεν θα ισχυει κ/λ+1 κι γενικά αυτό Πρέπει να ισχύει

Προχθες το βραδυ,δεν ειχα τι να κανω και ξαναειδα αυτη την ασκηση....
Με προβληματισε το γεγονος οτι στη λυση του cmad,αναφεροταν οτο προκυπτουν απειρες τριαδες,αλλα δεν αναφεροταν η μορφη τους....
Απο περιερφεια ,λοιπον, βρηκα οτι:Θεωρωντας τους 4 αυτους αριθμους σε διαταξη a<b<c<d,afoυ a+b+c<3d kai d|a+b+c,pairnoyme a+b+c=d i 2d
Estw a+b+c=d.AFOU a|b+c+d=2d-a,a|2d kai ομοιωςb|2d,c|2d,αρα θετντας
2d=ax=by=cz me x>y>z pairnoyme 1/x+1/y+1/z=1/2
Gia  z>=6 den yparxi lisi
z=5,oute
z=4 (a,b,c,d)=(k,4k,5k,10k) kai (k.2k,3k,6k)
z=3  (k,6k,14k,21k)  (k,3k,8k,12k)  (k,2k,6k,9k) (2k,3k,10k,15k)

Gia a+b+c=2d exoyme a,b,c|3d.Entelws omoia 3d=ax=by=cz me x>y>z>3 1/x+1/y+1/z=2/3 opοte z>=4.y>=5,x>=6και 1/x+1/y+1/z<=37/60<2/3 και αρα δεν υπαρχει λυση.

ΡΑΛΛΗΣ :D

ωραια :) ειναι και οι μοναδικες?

Εγώ δεν κατάλαβα ποιο είναι το πρόβλημα rakar :?:. Στη λύση μου λύνεις την $\sum_{\mbox{cyc}}\frac{1}{x}= 1$ (που έχει πεπερασμένο πλήθος λύσεων) και παράγεις τη μορφή των $a, b, c, d$, απλώς δεν έκατσα να τη λύσω.

Δεν ειπα οτι υπαρχει κανενα προβλημα.... :!: Η λυση σου ειναι ολοσωστη...
Απλως ελεγες <λυνουμε την εξισωση αυτη και βρισκουμε απειρες τετραδες..>
Ε,εγω εκατσα και την ελυσα,επειδη δεν εγραφες την μορφη τους....,για να βρω την μορφη αυτων των τετραδων....
Ναι  kosta ειναι οι μοναδικες μορφες αυτων των απειρων τετραδων,γιατι πολυ απλα αυτες ειναι οι μοναδικες λυσεις της εξισωσης με x,y,z που προκυπτει...

ΡΑΛΛΗΣ :D


[quote="ismee"]in the form of y=mx+c[/quote]

since it has a natural logarithm plus an exponent doesn't that mean it's not linear?

I think ismee means linear programming where non-linear equations are rearranged to a 'linear' form so that certain constants (in this case a and b) can be calculated by plotting a straight-line graph:

y=ln(ax^2+b)
so
e^y=ax^2+b

Y=e^y, X=x^2, m=a, c=b


Programming game! K, there will be a bunch of teams. One team challenges another team to solve a computational problem with a computer. Please no open problems. Also, no using stuff like MatLab or Mathematica.  And, no problems that by the judgment of players need to much computing power.  Also, no problems that require networks. I don't like sign ups but we have to have one for this game.

i know about project euler 1=2, but i thought it'd be a lot more fun if there were teams involved.

[quote="1=2"]projecteuler.net[/quote]
This.

And what's wrong with Mathematica?

[quote="Poincare"]I mean, some people can afford mathematica others can't. Also, Mathematica gives features not found in normal programming languages which would not be fair.[/quote]
I don't have mathematica, but it would seem that most needed features should be easily programmable.
What kind of "features" are you talking about?

i don't have mathematica either but i have heard that it can solve equations, graph, integrate, all kinds of stuff.

How big will the teams be? Is it possible to make a one-man team?


vector
invariant
linear algebra
matrix
complex numbers

Seems to be a standard result, but i've been trying a lot and can't
just figure it out. Can anyone can give a detailed proof?

Let V be a (real or complex) finite dimensional (inner product or hermitian)
vector space, and A: V -> V a normal linear operator.

* Prove that if F is a subspace of V invariant under A, then F is
invariant under A* (the adjoint of A), and the same holds for the
orthogonal complement of A.

Thanks.

Extend to $ \mathbb{C}$. A matrix is normal if and only if it is unitarily diagonalizable. It's easy to characterize the invariant subspaces of a diagonal matrix.

But the hard part is really the real case.
And I'm not quite sure I understand how the diagonal form
would help, since it assumes an orthonormal basis specifically
chosen to diagonalize, while none of the elements of this basis
has to be part of a basis for the subspace F.

The point is that if you start with a diagonal matrix, the basis elements of an invariant subspace [i]must[/i] be original basis elements. The only exception comes if we have multiple copies of the same element on the diagonal; in that case, we can draw elements from anywhere in the eigenspace.

What are all invariant subspaces under $ \begin{bmatrix}1&0\\0&2\end{bmatrix}$? They're the lines $ \{ce_1\}$ and $ \{ce_2\}$, along with the trivial subspace and the full space.

Also, there's no problem with going to complex numbers from the reals. Our result gives that the complexification $ F\plus{}iF$ is invariant under $ A^*\equal{}A^T$; since $ A^T$ is real, $ F$ and $ iF$ are each invariant and we have what we want.

Sorry jmerry, but it seems I'm still missing your point here.  :maybe:

It is now clear to me that for a diagonal matrix, its only invariant nontrivial
subspaces are those made of eigenvectors, but I fail to see that 'taking the
problem to the complex numbers' part. If:
$ A : V \rightarrow V$ is normal and $ V$ is a real vector space, isn't the
complexified of $ V$ just an 'embedding' of real scalars of $ V$ into complex scalars?
I don't know what you mean by the complexified $ F \plus{} iF$.
Also, the only thing that changes about $ A$ is that you could now split it into
a form $ A_{1} \plus{} iA_{2}$, where each of the operators $ A_{1}, A_{2}$ is real diagonal.
Then, if $ A(F) \subseteq F$, we have $ (A_{1} \plus{} iA_{2})(F) \subseteq F$, but its
not necessarily true that $ F$ is invariant under $ A_{1}$ or $ A_{2}$.
Could you please clarify that part?
Thanks.

When we move to complex scalars, we have to change the underlying vector space. The complex entries in the matrix aren't an issue, since we can make the matrix real again by a change of basis. The change to the vector space is; I'm choosing to represent the new vector space as the sum of two copies $ V\plus{}iV$ of the original space, and the complex span of the invariant subspace we started with is $ F\plus{}iF$.