function
induction
number theory
prime factorization

If $ f$ is a multiplicative function and $ n = p_{1}^{\alpha_{1}} .......p_{k}^{\alpha_{k}}$
Prove that $ F(n) = \displaystyle\prod_{i = 1}^k\ ( 1 + f(p_{i}) + ...........f(p_{i}^{\alpha_{i}}))$

[quote="ThinkFlow"]Is $ F(n) \equal{} \sum^{n}_{k \equal{} 1}f(k)$?[/quote]
For a multiplicative function $ f$ ,its summation function $ F$ is defined as $ F(n) \equal{} \displaystyle\sum_{d \mid n}\ f(d)$

Every divisor of n is of the form $ p_1^{\beta_1} p_2^{\beta_2} p_3^{\beta_3}...p_k^{\beta_k}$ where $ 0 \le \beta_i \le \alpha_i$

$ \sum_{d|N} f(d) \equal{} \sum f(p_1^{\beta_1} p_2^{\beta_2} p_3^{\beta_3}...p_k^{\beta_k})$ (with summation over all possible values of the $ \beta_i$'s )

$ \equal{} \sum f \left(p_1^{\beta_1} \right) f \left(p_2^{\beta_2}\right)...f \left(p_k^{\beta_k} \right)$

$ \equal{} \sum f^{\beta_1} (p_1) f^{\beta_2} (p_2)...f^{\beta_k} (p_k)$

$ \equal{} \prod_{i \equal{} 1}^k \left(1 \plus{} f(p_i) \plus{} f^2(p_i) \plus{} ... \plus{} f^{\alpha_i} (p_i) \right)$

[quote="hsbhatt"]

$ \equal{} \sum f^{\beta_1} (p_1) f^{\beta_2} (p_2)...f^{\beta_k} (p_k)$

$ \equal{} \prod_{i \equal{} 1}^k \left(1 \plus{} f(p_i) \plus{} f^2(p_i) \plus{} ... \plus{} f^{\alpha_i} (p_i) \right)$[/quote]
 Can you  explain this .....I dont understand ?

Ignore that bit -- hsbhatt was doing a computation with strongly multiplicative functions, which is not what you wanted.

Write out a few small examples: do $ n \equal{} 6$, $ n \equal{} 12$, $ n \equal{} 24$ and $ n \equal{} 30$ and you'll see what's going on.

strongly multplicative functions are same as completely multiplicative functions.?
That is a function $ f$ is said to be completely multiplicative if $ f(mn) \equal{} f(m)f(n)$ for all positive integers $ m$ and $ n$.
Is this the definition for strongly multiplicative functions also

Yes.

If you haven't already, I really do suggest you write out both the sum and product for $ n \equal{} 12$ and $ n \equal{} 30$.

Well.......for $ n = 12$
I  get for Sum
$ F(12) = \displaystyle\sum_{d \mid 12}f(d)$
$ F(12) = f(1) + f(2) + f(3) + f(4) + f(6) + f(12)$
$ F(12) = f(1) + f(2) + f(3) + f(4) + f(3)f(2) + f(3)f(4)$
The product
$ F(12) = \displaystyle\prod\ (1 + f(2) + f(4))$
$ F(12) = ( 1 + f(2) + f(4))(1 + f(3))$
$ F(12) = f(1) + f(2) + f(3) + f(4) + f(3)f(2) + f(3)f(4)$
Sorry if I am acting stupid  but is this what you wanted me to try?.I am sorry,I cannot understand what you mean by telling " both the sum and product for $ n = 12$"
This is what you meant right?

Yes.  Do you see how the terms of the sum and product correspond based on the prime factorization of $ d$?  This is the entire result.  (To do it more rigorously is an induction on the number and multiplicity of the primes in the factorization, but being able to write the proof is equivalent to understanding how the factorization works, and to understand it for a few small examples is probably enough to understand it in general.)


Quadrilateral ABCD is inscribed in circle O. Diagonal BD bisects AC. If AB=10, AD=12, and CD=11. Compute BC.

[hide]Let the intersction of AC and BD be E, and let AE = CE = m. Triangles ABE and DCE are similar, so BE/10 = m/11 <=> BE = (10/11)m. Also, triangles AED and BEC are similar, so 12/m = BC/((10/11)m) <=> BC = 120/11.[/hide]


Let be three circles exteriors tangents in $ A,B,C$ of centers $ O_1,O_2,O_3$ such that $ O_1A\equal{}a\sqrt{3}\ ,\ O_2A\equal{}a$ and $ O_3B\equal{}2a\plus{}a\sqrt{3}$ .

$ l(\widetilde{AB})\plus{}l(\widetilde{BC})\plus{}l(\widetilde{CA})$=?

$ l(\widetilde{AB})$= the length of the AB arc

Sorry if my english isn't very good . Can somebody help me?

Is this what you mean:

$ O_1$, $ O_2$, and $ O_3$ are externally tangent circles, with $ O_1$ and $ O_2$, $ O_2$ and $ O_3$, and $ O_3$ and $ O_1$ tangent at $ A$, $ B$, and $ C$, respectively. Given that $ O_1 A \equal{} a \sqrt{3}$, $ O_2 A \equal{} a$, and $ O_3 B \equal{} 2a \plus{} a \sqrt{3}$, find the sum of the lengths of arcs $ AB$, $ B$, and $ CA$?

[quote="Zhero"]Is this what you mean:

$ O_1$, $ O_2$, and $ O_3$ are externally tangent circles, with $ O_1$ and $ O_2$, $ O_2$ and $ O_3$, and $ O_3$ and $ O_1$ tangent at $ A$, $ B$, and $ C$, respectively. Given that $ O_1 A \equal{} a \sqrt {3}$, $ O_2 A \equal{} a$, and $ O_3 B \equal{} 2a \plus{} a \sqrt {3}$, find the sum of the lengths of arcs $ AB$, $ BC$, and $ CA$?[/quote]

Yes , $ O_1,O_2,O_3$ are the centers of those circles .


Since it seems to be somewhat of a tradition, I will start a topic for the next competition--Mercer.

Anyway, who's coming? (the "traditional first question", because it makes sense to ask)

I won't being going this year as I will be at PUMaC (which I have to wake up at 4 AM tomorrow to leave for :( ). Last year I got 5th place - my first time to ever place in a varsity tournament. I do not particularly like their written tests because most of the problems are unoriginal and/or can easily be solved with a calculator (which you are allowed). Also, I don't think they have any team rounds (correct me if I'm wrong) which seems kind of weird. Nevertheless, it is usually a well-run tournament, so I would probably go if I was in town.

I am, as stated in the Kennesaw post, sick, so I will not be going (my team will be, though).


$   \frac{200!}{100!100!}$
$= \frac{200 \cdot 199 ... \cdot 101}{100!}$
$= 2 \cdot 199 \cdot 2 \cdot 197 ... \cdot 101 \cdot 50!$
the biggest two digit prime is 191/3 (because 191 is in the numerator) = 67

by the way, this is my wild guess, kind of... correct me if i'm wrong

Next time, I'm going to move posts like this.

This problem, if my memory is correct, 1983 #9 AIME problem. Although 1983 AIMEs are famous for their easy problems (all contests in first year are easy usually), this is stil #9..

[hide="Solution"]

Since expansion gives $\frac{200!}{100!100!}$, we want the largest prime factor such that it appears 3 times on the numerator (because two of them will cancel out on the denominator). To do that, we note that the largest prime factor such that $p \cdot 3 < 200$ is $\fbox{61}$. [/hide]

You guys can still solve it.. I don't mind that but if I see an AIME problem that is over this level, I'll move it. And please note the level of this forum.

[quote="Silverfalcon"]
[hide="Solution"]

Since expansion gives $\frac{200!}{100!100!}$, we want the largest prime factor such that it appears 3 times on the numerator (because two of them will cancel out on the denominator). To do that, we note that the largest prime factor such that $p \cdot 3 < 200$ is $\fbox{61}$. [/hide]

[/quote]

To add to the solution above, see if you can see all  the two  digit prime factors  are in this order ..
[hide] $(2)^3(3)(5)(11)(13)^2(17){37)(53)(59)(61)}$  (and of course all three-digit prime factors) [/hide]
( Nice exercise to see if you can figure this  out  with the logic suggested in the above post.)

[hide="Solution"]Expanding the binomial, we get ${200 \choose 100}=\frac{200!}{100!100!}$.

Therefore, our two digit prime $p$ must satisfy $3p<200$. The largest such prime is $61$, which is our answer. [/hide]

[hide="Answer"]$\binom{200}{100}=\frac{200!}{100!100!}$
All prime factors of $\binom{200}{100}$ are those where, if $n$ appear in $100!$, then at least $2n+1$ appear in $200!$. Since we are looking for two-digit prime factors, all of those will be found in $100!$, but all above $50$ will be found only once in each, so we must find the first prime that appears three times in $200!$, which is $\boxed{61}$.[/hide]

[hide]A simpler way is to see that two-digit prime x will cancel numerator/denominator so must be seen one more time numerator. Therefore a multiple of 3. Building off of knowledge factorials, the problem basically asking highest prime less than floor(200/3). This be 61.[/hide]

[quote="PenguinIntegral"]What is the largest 2-digit prime factor of the integer $n = \binom{200}{100}$?[/quote]

[hide="Solution"]
Notice that $\dbinom{200}{100}=\dfrac{200!}{100!100!}$. $200!$ and $100!100!$ each contain two factors each of $97$, $89$, and all primes $\ge \dfrac{200}{3} \approx 66.6$. But for $61$, $200!$ contains $3$ factors of $61$, while $100!100!$ only contains $2$, making the answer \boxed{061}$.
[/hide]

If u got enough time then check cases
$p\in\mathbb{11,13,17,........,97}$

We want the largest prime that appears twice in the numbers 101-200 and only once in the numbers 1-100, which is clearly $\boxed{61}$.

No, this is P18 of Lehigh year 2018 [url=https://www.lehigh.edu/~dmd1/hstest18.pdf], here.[/url]

You guys are all wrong about the answer, it is $\boxed{061}.$

[hide]Note that for large 2-digit primes, the factorial on the top of the fraction is divisible by two copies of the prime (like how $200!$ is divisible by at most $97^2$ because it is divisible by $97 \cdot 194$), and the bottom product of factorials is also divisible by two copies of the prime (like how $100! \cdot 100!$ is divisible by at most $97^2$ because each of the factorials is divisible by $97$). Therefore, we want to find a prime such that the numerator factorial is divisible by three copies of the prime while keeping the denominator product divisible by $2$. We note that the prime must be less than $\frac{200}{3}$ in order to be divisible by three copies of the prime (there has to be 3 multiples of the prime from $1$ to $200$). The first such prime is $61$, and we see that this indeed works.[/hide]


[quote="Moonmathpi496"]Today 5th Bangladesh math olympiad has started.I am sending another problem for under 10th grade students.

f(x)=x^6+x^5+...+x+1.Find the remainder when dividing f(x^7) by f(x).

try to solve
>moon[/quote]
[hide]i just realised...umm this competition is over right?

i don't understand the rules on the webpage anyway[/hide]
[hide]
well i'll actually write out tho no need to
x^7((x^5+x^4+x^3+x^2+x^1+1/x))
x(x^5+x^4+x^3+x^2+x^1+1/x)

so the answer is x^6 right?[/hide]

hey function

you are wrong....

http://en.wikipedia.org/wiki/Polynomial_remainder_theorem
http://en.wikipedia.org/wiki/Polynomial_long_division

use this for the solution

the solution is 7

:)

agh tall guy you are right i should think before posting lousy solutions


Το 1955 ο Furstenberg όρισε ένα τοπολογικό χώρο στο $ \mathbb{Z}$ και τον χρησιμοποίησε για να αποδείξει ότι υπάρχουν άπειροι πρώτοι αριθμοί. Για όσους δεν την έχετε ξαναδεί προσπαθήστε να ανασκευάσετε την απόδειξή του. (Λίγο ζόρικο το πρόβλημα, αλλά αν σας δώσω τον τοπολογία γίνεται σχετικά εύκολο)

Φοβερή σύμπτωση! Μόλις πριν λίγες μέρες έτυχε να την πετύχω την εν λόγω απόδειξη online, και τολμώ να πω ότι εκνευρίστηκα πολύ γιατί ο author του post πήγαινε βήμα βήμα στην κατασκευή χωρίς να προϊδεάσει που θέλει να καταλήξει, μέχρι που έφτασε στο αποτέλεσμα, and it blew my mind.

Προσπαθείστε την, αξίζει όσο λίγα πράγματα.

Επειδή δεν βλέπω προσπάθειες, δίνω σε hidden την τοπολογία

[hide]
Στην τοπολογία αυτή, όλα τα σύνολα της μορφής $ a\mathbb{Z} \plus{} b$ θα είναι ανοιχτά
[/hide]

Για να το κλείνουμε και αυτό:

Στην τοπολογία του hidden παρατηρούμε ότι τα σύνολα $ \alpha\mathbb{Z} \plus{} \beta$, $ \alpha \neq 0$, είναι closed (γιατί?), και άρα αν είχαμε το πολύ πεπερασμένους πρώτους, το σύνολο $ \{\pm 1\} \equal{} \bigcap_{p \in \mathbb{P}} \mathbb{Z} \setminus p\mathbb{Z}$ θα ήταν open, κάτι που είναι αδύνατο.


limit
logarithms
real analysis
real analysis unsolved

Find the following limit value.

\[\lim_{n\to\infty} \left(1-\frac{1}{1+\sqrt{2}}\right)\left(1-\frac{1}{\sqrt{2}+\sqrt{3}}\right)\cdots\cdots\left(1-\frac{1}{\sqrt{n-1}+\sqrt{n}}\right)\]

Consider $a_n = \ln\left(1 - \frac{1}{\sqrt{n}+\sqrt{n+1}}\right)$. It is easy to show that $a_n = -\frac{1}{2\sqrt{n}}+ O(1/\sqrt{n})$ and therefore $\sum a_n = -\infty$. So the limit in the problem is 0.

$c(n) =\left(1-\frac{1}{1+\sqrt{2}}\right)\left(1-\frac{1}{\sqrt{2}+\sqrt{3}}\right)\cdots\cdots\left(1-\frac{1}{\sqrt{n-1}+\sqrt{n}}\right)$

Give an asymptotic expansion of c(n)

In using $e^{-x}>1-x$, for $k\geqq2, 0<1-\frac{1}{\sqrt{k-1}+\sqrt{k}}=1-(\sqrt{k-1}+\sqrt{k})<e^{\sqrt{k-1}-\sqrt{k}}$,yielding

$0<c(n)<e^{({1-\sqrt{2})+\cdots\cdots+(\sqrt{n-1}-\sqrt{n})=e^{1-\sqrt{n}}}}$

Therefore the desired limit value is $\lim_{n\to\infty} c(n)=0$


[quote="laughlaughalot"]Kind of slow today. :(
Prove: that the real part of z, Re(z), equals: (z+conjugate of z)/2; that modulus of z= (zxz(conjugate))^1/2; Im(iz)=Re(z)[/quote]

Let $ z\equal{}a\plus{}bi$. Thus, $ Re(z)\equal{}a\equal{}\frac{a\plus{}bi\plus{}a\minus{}bi}{2}$.

$ Im(iz)\equal{}Im(ai\minus{}b)\equal{}a\equal{}Re(z)$.

$ \sqrt{(a\minus{}bi)(a\plus{}bi)}\equal{}\sqrt{a^2\minus{}b^2}$.

[quote="BOGTRO"]
$ \sqrt {(a \minus{} bi)(a \plus{} bi)} \equal{} \sqrt {a^2 \minus{} b^2}$.[/quote]
Small typo: $ \sqrt {(a \minus{} bi)(a \plus{} bi)} \equal{} \sqrt {a^2 \plus{} b^2}$


n \geq 2 , x1,x2,...,xn  are reals  such  that  x1+x2+...+xn \geq 0  and
x1 2 + x2 2 +...+xn 2 =1. Let
M=max{x1,x2,...,xn} . Prove that  M \geq 1/(sqrt[n(n-1)])
thanks

now ican be  stupid
we can easily find n*(M) 2  \geq  \sum (x1) 2 =1 <=> M   \geq  1/ \sqrt n \geq 1/ \sqrt n*(n-1)

The x_i's are not supposed to be positive....

Pierre.


Yes, the one in Lawrence, Kansas... will it take place on Friday May 13 or Saturday May 14?

Could you please post it here when the date is finalized?

Hello,
I am going to the regionals as well, and the web site says that it will be held on Friday the 13th.


How would you factor the polynomial f(x)= x^3 + x + 1 ? Is is possible to factor it without resorting to numerical methods?

The polynomial $x^{3}+x+1$ is irreducible over $\mathbb{Z}[x]$. Indeed, if it were a product of two (nontrivial) polynomials, then one would be a first degree polynomial, the other would be a second degree polynomial. Therefore $x^{3}+x+1$ would have a rational root. But if $p/q$ is a root, then $q$ divides the leading coefficient, $p$ divides the constant coefficient. Hence $p/q$ is either $1$ or $-1$. But neither of these numbers is a root. We conclude that the polynomial is irreducible over $\mathbb{Z}[x]$.

The fundamental theorem of algebra states that every polynomial can be factored into linear factors over $\mathbb{C}[x]$:
\[f(x) = (x-r_{1}) \times \dots \times (x-r_{n}),\]
where $r_{1},\dots, r_{n}$ are the roots of $f(x)$. One might use Cardano's Formula to find all the roots of $f(x)$ - but I think that the roots are not "nice".


I was wondering where I could find the winning problem from the 2002 National Countdown round? I am currently teaching a MathCounts class, and we went over exponents and getting stuff down in terms of powers of 2, so I thought it would be neat to use this problem as a reference. However, I don't actually remember what the problem was! I'm at college, so I can't just pop in the videotape of the round.

I do know the answer was $(-2)^{34}$, at least.

I googled it and got [b]very[/b] lucky. It just so happened that the problem I found was the winning 2002 nats countdown problem (and there was only 1 problem!). Albert Ni won that year I think.

Which is the greatest: $4^{16}$, $2^{33}$, $(\frac{1}{8})^{-10}$, $(-2)^{34}$ or $16^8$?

Isn't too bad  :P

Guess what? This is REALLY funny. I mean seriously, really really really funny.

Darktreb's real name is ALBERT NI! :D

(btw that was me signed on to my sister's account--i was using it after her :D)

[quote="diophantine27"]Darktreb's real name is ALBERT NI! :D[/quote]

You're kidding right?  :o

and chess, never play a mean trick like that again :D

[quote="chess64"]Actually, I'm not kidding. Didn't you know that Darktreb was on the US IMO team last year? ;)[/quote]

he was? why didn't I see his name?  :?

Oh sorry, actually he didn't make IMO but he won USAMO. http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/05usamowinners.html

I don't get it though. Did you skip 12th grade, darktreb?

I went to only three years of high shcool, as I went to the Illinois Math and Science Academy, a three year (sophomore, junior, senior) boarding school; only I went straight out of 8th grade. I considered attending a 4th year at Exeter or something like that, as I might have made IMO that way (in other words I would have spent doing geometry :)). However, I decided not to, so now I'm at MIT.

Thanks 236factorial for that. I realize it's kind of funny that I'm asking for my own winning question, but at least I remembered the answer. It took me about 3 seconds from the start of problem to ring in (rarely does a National Countdown problem take more than 5 seconds I'd say, especially once someone gets on a roll. Many of my answers came within 2 seconds of seeing the problem), so I agree, it wasn't too bad.

Oh yeah, I should note that I'm listed as a 11th grader since it was just my third year of high school, and Zuming told me to keep writing down a grade that corresponded what year of HS I was in, since he knew I might attend four years in the end.


geometry
3D geometry
sphere
LaTeX
AMC
AIME

Okay, so we have a cylinder with height $ 24$ and radius $ 4$, and we must subtract $ 3$ spheres each with radius $ 4$.  The area of the cylinder is $ 16\pi \times 24$, or $ 384\pi$.  We must subtract $ 3 \times \frac{4}{3} \pi \times 64$, or $ 256\pi$.  This yields $ \boxed{128\pi}$.

[quote="MCNational!"]Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can. If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls. Express your answer in terms of $ \pi$.[/quote]

ezpz......Ok...so lets first find the volume of the cylindrical can. To do that we need to find the radius and height of the can. The radius is obviously 4, and the height is just 3 x 8,(because there are 3 tennis balls with a diameter of 8), so $ /pi*r*r*h$ = $ /pi * 4*4*24$ = $ 384/pi$. Yay....we still have half of the problem left..lol.
Ok..now lets find the volume of the three balls: The volume of one ball is $ 4/3*/pi*4*4*4$ = $ 256/pi/3$ To find the volume of all three...multiply by 3! Then you get......$ 256/pi$. Now subtract! $ 384/pi \minus{} 256/pi$ = $ 128/pi$. So the answer is $ 128/pi$!!!!
meh...correct me if i'm wrong...i did this in a rush..lol

[quote="AIME"][quote="MCNational!"]Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can. If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls. Express your answer in terms of $ \pi$.[/quote]

ezpz......Ok...so lets first find the volume of the cylindrical can. To do that we need to find the radius and height of the can. The radius is obviously 4, and the height is just 3 x 8,(because there are 3 tennis balls with a diameter of 8), so $ \pi\cdot r\cdot r\cdot h$ = $ \pi \cdot 4 \cdot4\cdot24$ = $ 384\pi$. Yay....we still have half of the problem left..lol.
Ok..now lets find the volume of the three balls: The volume of one ball is $ \frac43\cdot\pi\cdot4\cdot4\cdot4$ = $ \frac{256\pi}3$ To find the volume of all three...multiply by 3! Then you get......$ 256\pi$. Now subtract! $ 384\pi - 256\pi$ = $ 128\pi$. So the answer is $ \boxed{128\pi}$!!!!
meh...correct me if i'm wrong...i did this in a rush..lol[/quote]

Fixed your $ \text{\LaTeX}$ :)

Ugh AIME

ill teach u lol...

$ \text{\LaTeX}$


A subset of $S=\{1,2,\ldots,n\}$ is said to be [i]alternating[/i] if its elements, when arranged in increasing order, follow the pattern odd,even,odd,even,$\ldots$.
For example, if $S=\{1,2,3,4,5\}$ then $\{1,4,5\}$ and $\{3,4\}$ are alternating but $\{1,3,4\}$ and $\{2,3,4,5\}$ are not. The empty set is considered alternating.
Let $a_n$ denote the number of alternating subsets of $S$
Define $a_n$ recursively
Find an explicit formula for $a_n$

Let $b_n \; (c_n)$ be the number of alternating subsets of $S_n = \{1,2,\ldots, n\}$ for which the largest integer in the subset and $n$ has same (different) parity. Then we have the following recursive formulas:

$(1) \; a_n = a_{n-1} + b_{n-1},$

$(2) \; b_n = b_{n-1} + c_{n-1},$

$(3) \; c_n = 2b_{n-1}.$

According to (3) $c_{n-1} = 2b_{n-2}$, hence

$(4) \; b_n = b_{n-1} + 2b_{n-2}$

by (2). Futhermore (1) gives

$(5) \; b_n = a_{n+1} - a_n.$

Using formula (5) in (4), the result is

$a_{n+1} - a_n = a_n - a_{n-1} - 2(a_{n-1} - a_{n-2}),$

i.e.

$(6) \; a_{n+1} = 2a_n + a_{n-1} - 2a_{n-2}.$

The characteristic equation of (6) is

$r^3 - 2r^2 - r + 2 = 0,$

which is equivalent to

$(r - 2)(r - 1)(r + 1) = 0$.

Thus the roots are $r =-1,\, 1,\, 2$. This means that there exist constants $c_2$, $c_1$ and $c_0$ such that

$(7) \; a_n = c_2 \cdot 2^n + c_1 \cdot (-1)^n + c_0$

Now it is easy to determine that $a_1 = 2$, $a_2 = 4$ and $a_3 = 7$. Inserting these three values into (7), the resulting system of linear equations has the solution $c_2 = 5/6$, $c_1 = 1/6$ and $c_0 = 1/2$. Consequently

$a_n = \frac{5 \cdot 2^n + (-1)^n + 3}{6}$.


When will the round 3 LaTeX form be released?  I want to start typesetting my solutions using the template, but obviously I can't, as I don't have the template.  Also, as a note to the USAMTS site administrators, the round 15 problem and solution files (on the "past years" page) seem to be misisng.

I believe that you use the same form to typeset each round of questions.  So just go and find round 1's form.

Just remember to change the argument of the \usamtsround{ } command which should be a few lines before the \begin{document} in the template.

We've uploaded the new form.  The change mentioned above is the only one you'll have to make.


Barney gives three great big hugs each to two kids. Both of those two kids each give four great big hugs to both their parents (assume that they both have two parents). How many hugs are given out?

[hide]Barney gives $3 \times 2$ or 6 hugs.  The two kids each give out 4 hugs, so $4 \times 2$ is 8.  6+8=$14$.[/hide]

OK. This is changed too.

fix'd, I feel like it's chapter level :P .

You're missing something in your answer.

Umm... I get a different answer.

[hide] Barney gives 2 kids 3 hugs, for a total of [b]6[/b], then the two kids give four hugs to four parents, for a total of 2x4x4 = [b] 16 [/b] hugs given by kids. [b] 16 + 6 = 22 [/b][/hide]

[quote="Treething"]Both of those two kids each give four great big hugs to both their parents (assume that they both have two parents). [/quote]

I though it meant the 2 kids each give 4 hugs to both their parents combined, not each.  So, you wouldn't multiply by 2.  :? It said 4 hugs to both their parents not each of their parents.  Am I missing something?  :?

I don't think it's worded correctly. I understood it to mean that each kid gives four hugs to each parent, meaning $2(4)(2)=16$.

(Fanatic, you wrote 2x4x4. I'm guessing you either meant 2x4x2 or 4x4. ;))

The answer is

[hide]
22
[/hide]

barney's hugs count too ;-;

[quote="easyas3.14159..."]I got...

[hide=":)"]22 hugs[/hide]

click the smiley for the answer[/quote]

after he gives everyone the answer.  :D


Do you know the answers to 4, and 5?

http://www.math.okstate.edu/~hsc/2007/20072.pdf

[hide="#5"]\begin{eqnarray*}\angle EAD=80^\circ\implies\angle EAC=60^\circ\implies EA=AC=CE\implies\angle CED=40^\circ\end{eqnarray*}

Hence $ \angle EDC=70^\circ\implies\angle BDC=30^\circ$[/hide]

Here's my solution for problem 5( I think u should check it carefully):
Applying Cosine's theorem we have: DC^2=AD^2+AC^2-2AC.AD.cos20
                                                        = BC^2+AC^2-2BC.AC.cos20
                                                        = AC^2(sin^2(20)/sin^2(80) +1- sin40/sin80) (1)
Then applying Sine's theorem into triangle ABC I have:
BC/AC=sin20/sin80 (2)
Then replacing (2) into (1) I have:sin^2(20)+sin^2(80)-sin40.sin80/(sin^2(80).sin^2(20))=1/sin^2 (BDC)
After substituting many times I got: 2(1+cos20)/sin^2(20)=1/sin^2(BDC)
Then I got: 1/sin^2(10)=1/sin^2(BDC).Therefore,BDC= 10(since BDC<110)(This makes me suspect since the figure shows me that it maybe bigger than 10) :maybe:

[hide="#4"]Extend $ EC$ over $ C$ to the intersection $ C'$ with the left circle. Similarly, extend $ ED$ over $ D$ to the intersection $ D'$ with the right circle.

Then $ \tan\alpha \equal{} {AC'\over C'E}$, but by the triangle congruence, namely $ \triangle EDC\cong \triangle C'AC$, we get $ AC' \equal{} ED\land C'E \equal{} 2CE$, hence $ \tan\alpha \equal{} {ED\over 2CE}$

Similarly, $ \tan\beta \equal{} {BD'\over D'E} \equal{} {CE\over 2ED}$

Therefore $ \tan\alpha\tan\beta \equal{} {ED\over 2CE}\cdot{CE\over 2ED} \equal{} {1\over 4}$[/hide]

[quote="ghjk"][quote="Farenhajt"][hide="#5"]\begin{eqnarray*}\angle EAD = 80^\circ\implies\angle EAC = 60^\circ\implies EA = AC = CE\implies\angle CED = 40^\circ\end{eqnarray*}

Hence $ \angle EDC = 70^\circ\implies\angle BDC = 30^\circ$[/hide][/quote]

How can you havve EDC=70 while you don't calculate ECD??[/quote]

$ \triangle CED$ is isosceles with the apex angle of $ 40^\circ$.

OK! Now i see your solution. But can you help me find my mistake? I try using it without using the hint below! :blush:.It's very nice solution if it's true!

I have:BC/AC=sin20/sin80. Then I easily find: BC^2=AC^2.sin^2(20)/sin^2(80) :)

I see. I don't have time to delve into it right now... retrace your steps, check your math and try to find it yourself.


topology
vector
Functional Analysis
advanced fields
advanced fields unsolved

Let $X,Y$ be a Banach spaces, $(T_{n}) \subset B(X,Y)$. Let $T \colon X \to Y$ be a linear map such that $T_{n}x \rightharpoonup Tx$ weakly in $Y$ for every $x \in X$.

Must $T$ be bounded?

let $B$ be a Banach space, $\{e_{1}, e_{2}, .... \}$ a basis where $\|e_{k}\|=1$. Define $T_{n}(e_{k})=ke_{k}$ for $k \leq n$ and $T_{n}(x)=0$ for $x \not \in Span(e_{1}, e_{2}, \ldots, e_{n})$. Then $T_{n}(x)$ converges strongly towards a vector $T(x)$. But $T(e_{n})=n e_{n}$.

uhm..
alekk, are you sure $T_{n}x$ strongly converges somewhere?
i mean, if you consider the banach space $\ell^\infty$ with the usual basis, this doesn't seem to converge for $\sum e_{k}$...  :huh:

More simply. If $X$ is a separable Hilbert space with orthonormal basis $(e_{k})$, then take $x = \sum_{k=1}^\infty k^{-2}e_{k}$. It is easily seen that $T_{n}x = \sum_{k=1}^{n}k^{-1}e_{k}$ isn't a Cauchy sequence in $X$. So alekk's definition of $T$ doesn't fit.

I was talking about algebraic basis (every $x \in H$ is a Finite combination of elements of the basis ). Sorry not to have been clearer about that ..

It follows from Baire Category Theorem that infinite-dimensional Banach space cannot have countable Hamel basis.   :wink:

Sorry for the possible nonsense I am going to write. Already, for all $x$, the family $T_{n}(x)$ is bounded because it converges weakly (Banach-Steinhauss). Thus again by Banach-Steinhauss one can find $C$ such that $\parallel T_{n}(x)\parallel \leq C\parallel x\parallel $ for all $n$ and all $x$. Now, consider the set $F=T(B)$ where $B$ is the unit ball of $X$. I claim that it is bounded. By Banach-Steinhauss again, it suffices to show that the filtration of this set by any functional of $Y$ is a bounded set. Fix $f$ a functional on $Y$. Then for all $x\in B$ we know that $f(T_{n}x)$ converges to $f(Tx)$ but on the other hand $\parallel f(T_{n}x)\parallel \leq \parallel f\parallel C$ for all $n$, thus $\parallel f(Tx)\parallel \leq \parallel f\parallel C$ and so $f(F)$ is bounded. Therefore, $T$ is bounded. WHere is the mistake? [/hide]

seems right to me. But I still don't see where I did a mistake (Volodja: I just suppose that $e_{1}, e_{2}, ..$ are algebraicly independant, and complete it in an Hamel basis)  :blush:

[i]To Alekk:[/i]

How do you define $T$? I think the problem with your example is exactly in the definition of $T$.  :maybe:  E.g., modernizing my example, take $x = \sum_{k=1}^\infty k^{-1}e_{k}$ then $T_{n}x = \sum_{k=1}^{n}e_{k}$. From here one sees that $(T_{n}x)$ doesn't converge even weakly because the sequence $\|T_{n}x\|$ is unbounded.

[i]To Harazi:[/i]

Personally, I don't see any mistake. I also obtained that $T \in B(X,Y)$, and my solution nearly the same as yours. The only difference is that after proving that $fT \in X^{*}$ for every $f \in Y^{*}$, I used Closed Graph Theorem instead of Banach - Steinhauss.

[b]P.S.[/b] This is the problem number 5 from [url]http://www.sissa.it/fa/download/phd/year1990.ps[/url].

Dear Volodja, where are these examinations taken from? The problems seem nice, maybe you could give some links to some other such exams. Thanks in advance.

While searching for web page of Antonio Ambrosetti, I faced the site of S.I.S.S.A. ([url]http://www.sissa.it/[/url]) where he works.

I took problems from here [url]http://www.sissa.it/fa/phd/entrance_examinations/[/url].

[quote="alekk"]But I still don't see where I did a mistake [/quote]
Your operators $T_{n}$ are unbounded!

yep  :D  I realized that yesterday evening ..


Think of modular arithmetic as "remainder algebra".  It deals with the remainder when one integer is divided by another.  In general, mods are written like $ a\equiv b \pmod{m}$.  Let's look at the equation: $ 12\equiv 2 \pmod{5}$.  This means that the remainder when 12 is divided by 5 is 2.  Let's look at another equation: $ 17\equiv 12 \pmod{5}$.  This just means that the remainder when 17 is divided by 5 is the same as the remainder when 12 is divided by 5.  This is true because both remainders are 2.  (Also, the operator "$ \equiv$" had to used only in the second example because 12 is obviously not the remainder when 17 is divided by 5.  The equal sign can be used only when the answer is the remainder.  For example the first example could be written: $ 2 \equal{} 12\pmod{5}$. When in doubt use "$ \equiv$".)
Try the following problems to see if you have a good grasp on mods.

1.  What is the remainder when 17 is divided by 7?
[hide]Answer: Since 17=7*2+3, 17 is 3 more than a multiple of 7, so the answer is [b]3[/b]. (This is the definition of remainder).[/hide]

2.  If $ x \equal{} 32 \pmod{6}$ , find $ x$.
[hide]The remainder when 32 is divided by 6 is 2.  Hence the answer is [b]2[/b].[/hide]

3.  If $ x \equal{} 2011 \pmod{1000}$ , find $ x$.
[hide]The remainder when 2011 is divided by 1000 is 11.  Hence the answer is [b]11[/b].[/hide]

4.  Find an integer $ 40 < a < 50$ such that $ a\equiv 29 \pmod{9}$.
[hide]The remainder when 29 is divided by 9 is 2.  We are trying to find a number between 40 and 50 that also has a remainder of 2 when divided by 9.  The only such number is 9*5+2 = [b]47[/b][/hide]

**5.  Find the remainder when $ 2^{2009}$ is divided by 7.
[hide]
Start listing out the remainders when the numbers $ 2^1$, $ 2^2$, $ 2^3$, $ 2^4$,... are divided divided by 7:  2,4,1,2,4,1,2,4,1,...  -notice a pattern?  The pattern repeats every 3 terms, so find the remainder when 2009 is divided by 3, which is 2.  The second number of our pattern stem is 4, so the answer is [b]4[/b].
Be sure you understand how mods are used in the solution and why the pattern must repeat.
[/hide]

For #2, #3 $ x$ can represent an infinite # of values

Do not confuse people with =
Standard symbol is $ \equiv$, and I have never seen = in equations in my life.

Another definition of mod:

$ a\equiv b\pmod m$ if and only if $ m\mid a\minus{}b$.

For 5, if the answer is to be explained, that can't be the solution.

No, you have to use $ a\equiv b\pmod m <\equal{}> ac\equiv bc\pmod m$ for integers, and $ a\equiv b\pmod m <\equal{}> a^n\equiv b^n\pmod m$.
Then, as $ 2^3\equiv 1\pmod 7$, we have $ 2^{2007}\equal{}(2^3)^{669}\equiv 1^{669}\equiv 1\pmod 7$
Now, $ 2^{2009}\equal{}4\cdot 2^{2007}\equiv 4\cdot 1\equiv 4\pmod 7$

But, if just answer is needed, you can do the pattern way.

[quote="AMC 14"]okay, so I am attempting to self teach my self the subject of Number Theory. I have purchased the Art of Problem Solving addition for this subject, but the first chapter starts pretty fast. I have never done anything with Number Theory, so would somebody please explain the concept of 'modulo' and basic number theory...[/quote]

Chapter 1 in Introduction to Number Theory just talks about multiples, divisors, and divisibility - are you talking about Volume 1 and not Introduction to Number Theory?

If you really want to learn Number Theory, and you haven't been exposed to any of it, you should try Introduction to Number Theory.  If you still don't understand the material after several read-throughs, feel free to private message/facebook me and I'll try to help.

The [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=2]corresponding AoPS class[/url] also should be helpful.

[quote="Bugi"]Do not confuse people with =
Standard symbol is $ \equiv$, and I have never seen = in equations in my life.
[/quote]
I do agree that it is always more clear to use "$ \equiv$", however "=" is also correct in some cases.  From wikipedia: "The notion of modular arithmetic is related to that of the remainder in division. The operation of finding the remainder is sometimes referred to as the modulo operation and we may see 2 = 14 (mod 12). The difference is in the use of congruency, indicated by ≡, and equality indicated by =. "

I also do agree that my answer for #5 was not the best method, however it is the most elementary.

If, for some reason, you do not see that $ 2^3 \equiv 1 \pmod{7}$, you can use FLT or Euler's Theorem to get $ 2^6 \equiv 1 \pmod{7} \implies 2^{2009} \equiv 2^5 \equal{} 32 \equiv 4 \pmod{7}$.

i think he's doing number theory, and hasn't read intro?

Okay modulo arithmetic is all about finding the reminders of things.
The purpose of modulo arithmetic is to find number that have the same reminder when divided by the modulo. 2 mod 10 , and 12 mod 10 are congruent with each other because they have the same reminder when divide by ten. Is this more clear? if not please reply.


limit
real analysis
real analysis unsolved

just for laugh :rotfl:
let $p_n = e^{\pi^{{p_{n-1}}}}$ and $r_n = \pi^{e^{{r_{n-1}}}}$ for $n=1,2,...$
with $p_0=1$ and $r_o=1$
what can you say $\displaystyle\lim_{n\to\infty}\frac{p_n}{q_n}$
 :huh:  :huh:  :P


number theory
greatest common divisor
number theory proposed

Find all natural number $ n$ so that $ 2^n\plus{}1\vdots n^2$

This is IMO 1990 problem 3.  Here's one solution:

[hide]
We claim $ n\equal{}1, 3$ are the only solutions.

Clearly $ n$ is odd.  Suppose that $ n\equal{}3^kr$, where $ 3 \nmid r$.  Then
\[ 2^n\plus{}1\equal{}2^{3^kr}\plus{}1\equal{}(2^r\plus{}1)h\]
Because $ r$ is odd, $ 3|2^r\plus{}1$.  So let $ 3^m$ be the greatest power of $ 3$ dividing into $ 2^r\plus{}1$.  Then $ 2^r \equiv \minus{}1\mod{3^m} \iff 2^{3^kr} \equiv \minus{}1\mod{3^{m\plus{}k}}$, so it follows that $ 3^k$ is the greatest power of $ 3$ dividing $ h$.

Because $ n^2|2^n\plus{}1$, we have $ 3^{2k}|(2^r\plus{}1)h$, and so $ 3^k|(2^r\plus{}1)$.  But because $ 3 \nmid r$, we have $ 9 \nmid 2^r\plus{}1$: hence, $ k \in \{0, 1\}$.

If $ r \neq 1$, let $ p$ be its smallest prime factor.  Then $ \text{gcd}(r, p\minus{}1)\equal{}1$.  So we have
$ 2^n \equiv \minus{}1\mod{p}$.  Let $ \text{gcd}(n, p\minus{}1)\equal{}j$.  Then $ j\equal{}nx\plus{}(p\minus{}1)y$ for some integers $ x, y$.  So \[ 2^j\equal{}(2^n)^x(2^{p\minus{}1})^y \equiv (2^n)^x \equiv \pm1 \mod{p}\]
according as $ x$ is even or odd.  But
\[ j\equal{}\text{gcd}(n, p\minus{}1)\equal{}\text{gcd}(3^kr,p\minus{}1)\equal{}\text{gcd}(3^k, p\minus{}1)|\text{gcd}(3, p\minus{}1)|3\]
so because $ p|2^j \mp 1$, we have $ p$ divides one of $ 1, 3, 7$ or $ 9$.  Because $ p>3$, we have $ p\equal{}7$.  But $ 7 \nmid 2^m\plus{}1$ for any integer $ m$, because $ 2$ has odd order mod $ 7$.  Hence, $ 7 \nmid 2^n\plus{}1$, and so $ n \nmid 2^n\plus{}1$.  Therefore, $ r\equal{}1$, and so $ n\equal{}1$ or $ 3$, as desired.[/hide]

There exists another solution...i'll see if i can find it.  The key lies in realizing that if $ p$ is the smallest prime divisor of a number $ m$, then $ \text{gcd}(m, p\minus{}1)\equal{}1$, and so $ p|2^{am}\plus{}1 \implies p|2^a\plus{}1$...

[quote="ZetaX"][quote="isomorphism"][quote="kunny"][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55436]This is IMO 1990[/url][/quote]

Hey [b]kunny[/b],
Nobody solved it there  :([/quote]

It _is_ solved there![/quote]

Well since Arne kept on reminding people their solution is not right, I did not read the proof that was criticized by him. Unfortunately for me I missed [b]pluricomplex[/b]'s solution.

[quote]There exists another solution...i'll see if i can find it. The key lies in realizing that if p is the smallest prime divisor of a number m, then $ \text{gcd}(m, p\minus{}1)\equal{}1$, and so $ p|2^{am}\plus{}1 \implies p|2^a\plus{}1$...[/quote]

Ah thanks [b]K81o7[/b] :D
I believe the other solution you were mentioning is akin to [b]pluricomplex[/b]'s solution in the other thread :)


logarithms
integration
probability
calculus
expected value
special factorizations
probability and stats

The question is: Find $ E(max(e^X - a,0))$, knowing that $ X$ is standard normal. I wasn't entirely sure how to approach this, but this is what I tried:
Break it up into two cases, one where $ e^X - a > 0$ and the second where $ e^X - a \leq 0$. Then:
Note that $ e^X - a > 0 \rightarrow e^X > a \rightarrow x > \ln a$   Thus,
Expected Value of Case 1 = ${ \int_{ln a}^{\infty} \frac {x}{\sqrt {2 \pi}} e^\frac { - x^2}{2}} = \frac {1}{\sqrt {2 \pi e^f}}$ where $ f = (\ln a)^2$
 Expected Value of Case 2 = 0 (since it's negative, so when taking the max it will always be zero)
Probability of Case 1 =${ \int_{\ln a}^{\infty} \frac {1}{\sqrt {2 \pi}} e^\frac { - x^2}{2}}$
Probability of Case 2 = ${ \int_{ - \infty}^{\ln a} \frac {1}{\sqrt {2 \pi}} e^\frac { - x^2}{2}}$

Then just do (Expected Value of Case 1) (Probability of Case 1) + (Expected Value of Case 2)(Probability of Case 2) = (Expected Value of Case 1) (Probability of Case 1). My only problem is figuring out the Prob. of Case 1. Is there any easier way to do this? I feel like you could use mgfs and this might be easier. Help please! :)

In general, $ E[g(x)] = \displaystyle\int_{-\infty}^{\infty} g(x)\dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\, dx$ so what you want is:

\begin{eqnarray*}
E[\max\{e^X - a, 0\}] &=& \displaystyle\int_{-\infty}^{\infty}
\max\{e^X - a, 0\} \dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\, dx
\\&=&
\displaystyle\int_{-\infty}^{\ln a}
\max\{e^X - a, 0\} \dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\, dx
+ \displaystyle\int_{\ln a}^{\infty}
\max\{e^X - a, 0\} \dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\, dx
\\&=&
\displaystyle\int_{\ln a}^{\infty}
(e^X - a)\dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\, dx
\\&=&
\displaystyle\int_{\ln a}^{\infty}
e^X\dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\, dx
-a\displaystyle\int_{\ln a}^{\infty}
\dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\, dx
 \end{eqnarray*}

The 'answers' will all be in terms of the CDF of the normal distribution. The first integral can be evaluated by completing the square in the exponential and noting that it's still normal but now with a non-zero mean.
The second integral is
$ \displaystyle\int_{\ln a}^{\infty}
\dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\, dx = \displaystyle\int_{-\infty}^{-\ln a}
\dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\, dx
= \Phi(-\ln a)$


Consider all the 'connected' configurations of 4 unit squares, where each square shares a side with at least one other.  There are seven of these, the seven Tetris pieces.  Can we fit all of them without overlap into a  7 x 4 rectangle?  (you may assume that the pieces have to be placed parallel to the sides of the rectangle, in the obvious sense)

I don't think so. Color the $7 \times 4$ rectangle alternating with black and white. Every piece covers two blacks and two whites except the T-shaped piece, which covers three of one color and one of the other. So the number of white spaces covered is not equal to the number of black spaces covered, which clearly cannot be the case if the entire $7 \times 4$ rectangle is covered.


inequalities
number theory
prime factorization

Dude.
It's either 0, 1, 2, 3, 4... 8 or 1, 2, 3... 9.

Come on.

But now we check the sums:
$ 1+2+3 ..+9 = 45$.
$ 1 * 2 * 3 * (2^{2}) * 5 * (3*2) * 7 * (2^{3}) * (3^{2}) = 2^{7}* 3^{4}* 5 * 7 = 128 * 81 * 5 * 7 = 362880$

But of course they do.
If it was the other set, you'd be multiplying by zero.

(and none can be negative, because the sum wouldn't be CLOSE to 45 if they were).

[quote="vishalarul"][quote="xxxyyyy"]I have thought of nine single digit integers whose sum is $ 45$ and their product is $ 362880$.

What are these nine integers?[/quote]

[hide="Hint"]$ 45 = \sum_{i=1}^{9}i$
$ 362880 = 9!$[/hide]

[hide="Solution"]From the hint, we immediately see that the nine integers are
$ 1,2,3,4,5,6,7,8,9$[/hide][/quote]

what if some numbers are positive and some are negative?

[quote="DiscreetFourierTransform"]
(and none can be negative, because the sum wouldn't be CLOSE to 45 if they were).
[/quote]

[quote="t0rajir0u"][quote="xxxyyyy"]I have thought of nine single digit integers whose sum is $ 45$[/quote]

This is enough to do the problem, since $ 45$ is the maximum sum of nine single-digit integers.[/quote]

Well, some of the integers can be repeated... It does'nt say that they are distinct.

[quote="t0rajir0u"]True.   :oops:  Now the solution can be proven to be unique either by considering prime factorizations or by standard methods in inequalities.[/quote]

How to go about the proof?

The prime factorization solution is just casework (we cannot, for example, have two $ 7$s because $ 362880$ is only divisible by $ 7$ once), and the shortest inequality solution uses majorization  :|

Does it mean that there is no other solution?

Can anyone provide the proof of the same?

Well, there are only a finite number of sets of $ 9$ single-digit integers with sum $ 45$, so a "proof" could consist of all possible products of the elements of such sets.

$ 1 \times 2 \times 4 \times 4 \times 4 \times 5 \times 7 \times 9 \times 9 = 362880$
$ 1+2+4+4+4+5+7+9+9 = 45$

how about it?


Let $ A_1A_2...A_n$ be a convex polygon. Prove that the sum of distances from an interior point to its sides does not depend on the position of that point if and only if the following equality holds:
$ \frac{\overrightarrow{A_1A_2}}{A_1A_2}\plus{}\frac{\overrightarrow{A_2A_3}}{A_2A_3}\plus{}...\plus{}\frac{\overrightarrow{A_nA_1}}{A_nA_1}\equal{}0$

This problem was already posted... see [url]http://www.mathlinks.ro/viewtopic.php?t=119086[/url]


algebra
polynomial
trigonometry
linear algebra
complex numbers

Okay, this is one question that I kind of didn't really get when I studied Taiwanese math.  So I might not use all the proper terms because I studied this in Chinese.  Sorry in advance.

I'll use the Fibonacci Sequence to illustrate:
The sequence can be given as: $a_{n+2} = a_{n+1} +a_n$
Dividing by $a_{n+1}$ gives: $\frac {a_{n+2}} {a_{n+1}}$ $=$ $1 +$ $\frac {a_n} {a_{n+1}}$.
This can be given by the characteristic $x = 1 +\frac {1} {x}$ -- (equation 1)
Wherethen we solve and obtain $x = \frac {1+\sqrt{5}} {2} or \frac {1-\sqrt{5}} {2}$ -- (equation 2)
Using the solved x, it can be concluded that:

$a_n = a(\frac {1 + \sqrt{5}} {2})^n + b(\frac {1 - \sqrt{5}} {2})^n$ -- (equation 3)

And through solving $a$ and $b$ using the given conditions (usually $a_0$ and $a_1$), we can solve for any $a_n$ in the series.

The question is, [i][u]why[/u][/i] does equation 3 relate to equation 2 (the solution of equation 1)?  Math texts usually gives equation 3 as a natural follow-up to equation 2 but don't give the reason.

Also, as a follow-up, when the characteristic equation is, for example, $(x-2)^3 = 0$, which yields 3 identical roots $x = 2, 2, 2$, $a_n$ becomes $a_n = a(2)^n + bn(2)^n + cn^2(2)^n$.  [i][u]How/why is[/u][/i] this so?

What if the characteristic equation yields $i$ as solutions?  For example, if the characteristic equation is $x^2 = -1$?  What would $a_n$'s equation look like then?

Sorry for this extended post, but thanks for all help in advance.

I don't understand how you get equation one.

It is only true when you take the limit on the ratio. Fibonacci is not Geometrical.

Well, the terms $r_0, r_1$, and the relation $r_{n+2}=r_{n+1}+r_n$ define exactly ONE sequence, right?

And it's clear that if $\zeta_1,\zeta_2,\ldots,\zeta_n$ are the roots of the polynomial $\sum_i a_iz^i$, then

$r_n=\sum_i b_i\zeta_i^n$ satisfies the recurrence relation $\sum_i a_ir_i=0$

Now we just need some linear algebra to show that there exist unique $b_i$'s such that the initial conditions for our recurrence relation are satisfied.

The point of adding terms of the form $na^n,n^2a^n,n^3a^n,\ldots$ when there are repeated roots in the characteristic equation is that these terms satisfy the recurrence relation, and are linearly independent, so we can use them to satisfy the initial conditions.

if the roots of characteristic polynomial are complex numbers $a \pm bi$, then the solution will looks like
$(\sqrt{a^2+b^2})^n ((K-M)i \sin \theta +(K+M) \cos \theta)$ where $\tan \theta = \frac{b}{a}$
to derive this you'll need De' movrie formula
$(\cos \theta + i \sin \theta )^n = (\cos n\theta + i \sin n\theta )$ which can easily proved by induction.


limit
function
algebra
polynomial
calculus
calculus computations

How do you prove the Taylor's Theorem? My professor prove the case that

$ f(x)\equal{}f(x_0)\plus{}f'(x_0)(x\minus{}x_0)\plus{}\frac12f'(\xi)(x\minus{}x_0)^2.$

First he suppose $ k$ is a number such that

$ f(x)\equal{}f(x_0)\plus{}f'(x_0)(x\minus{}x_0)\plus{}\frac k2(x\minus{}x_0)^2$

and let

$ \Phi(t)\equal{}f(x)\minus{}f(t)\minus{}f'(t)(x\minus{}t)\minus{}\frac k2(x\minus{}t)^2.$

But how do we know the equation has a solution for any $ x$, $ x_0$, $ f$, etc.? And how can we spread it to the general case?

You mean the second equation? Given $ f$, $ x_0$, and $ x\ne x_0$, you can easily solve it for $ k$.

what about something like $ f(x) \equal{} \lim_{h \to x}e^{\minus{}h^{ \minus{} 2}}$?
it's taylor series about 0 is just 0

[quote="0714446459923"]what about something like $ f(x) \equal{} \lim_{h \to x}e^{ \minus{} h^{ \minus{} 2}}$?
it's taylor series about 0 is just 0[/quote]

This function does not satisfy all the hypotheses of Taylor's theorem. It is possible to find a Laurent series for the function, where you allow the series to have negative indices.

No, that function [i]does[/i] satisfy Taylor's theorem.

By that, I mean the following theorem: suppose for a given $ n,$ that $ f\in C^{n\plus{}1}$ on some interval. (That's overkill, but I'm not currently interested in minimizing the hypotheses.) Then for any $ a$ and $ x$ in that interval, there exists $ \xi$ in between $ a$ and $ x$ (that means either $ \xi\in (a,x)$ or $ \xi\in (x,a),$ whichever is correct), such that

$ f(x)\equal{}\sum_{k\equal{}0}^n\frac{f^{[k]}(a)(x\minus{}a)^k}{k!}\plus{}
\frac{f^{[n\plus{}1]}(\xi)(x\minus{}a)^{n\plus{}1}}{(n\plus{}1)!}.$

(That's the Lagrange or mean value form of the remainder; there are other variants of the theorem with other forms of the remainder.)

Taylor's theorem is about polynomials, not series. The truth of Taylor's theorem is only loosely connected to the notion of Taylor series. Note that Taylor's theorem does not require that the function be $ C^{\infty}$ - just that it have "enough" derivatives.

So Taylor's theorem is that for a fixed $ n,$ if we write

$ f(x)\equal{}\sum_{k\equal{}0}^n\frac{f^{[k]}(a)(x\minus{}a)^k}{k!}\plus{}R_n(x),$

then we have a specific estimate on the size of $ R_n(x).$

Now, what if $ f$ is infinitely differentiable and we try to take the limit as $ n\to\infty?$ in that last expression. Then we find ourselves facing the the question of whether or not $ \lim_{n\to\infty}R_n(x)$ equals zero.

In some cases it does; in some cases it doesn't. Taylor's theorem may provide tools for showing that limit (although they're not always easy to use), but Taylor's theorem does not by itself assert that that that limit exists.

If $ \lim_{n\to\infty}R_n(x)\equal{}0$ for all $ x$ in some neighborhood of $ a,$ then we have that $ f(x)\equal{}\sum_{k\equal{}0}^{\infty}\frac{f^{[k]}(a)(x\minus{}a)^k}{k!}.$ That's a Taylor series; such functions are called [i]real analytic[/i]. But not all $ C^{\infty}$ functions are real analytic. (In fact, a $ C^{\infty}$ function need not be analytic [i]anywhere[/i].)

---

So what about $ f(x)\equal{}\begin{cases}e^{\minus{}x^{\minus{}2}},x\ne0\\0&x\equal{}0\end{cases}$ ?

This function is in $ C^{\infty}.$ For each $ n,\ f^{[n]}(x)$ is bounded and $ f^{[n]}(0)\equal{}0.$ Taylor's theorem is true for this. If $ M_n\equal{}\max|f^{[n]}(x)|,$ then for each $ n$, Taylor's theorem (centered at zero) asserts that

$ |f(x)|\le \frac{M_{n\plus{}1}|x|^{n\plus{}1}}{(n\plus{}1)!}.$

However, we cannot estimate $ M_n$ in such as way as to be able to take a limit of that as $ n\to\infty.$ The Taylor series for $ f$ centered at zero is identically zero and does not equal $ f.$

[quote]Taylor's theorem is about polynomials, not series. The truth of Taylor's theorem is only loosely connected to the notion of Taylor series.[/quote]
Which makes me wonder why my calculus textbook has Taylor polynomials buried under a pile of convergence tests for series.


probability
geometry
3D geometry
absolute value

[quote="lkryptonicl"]How many positive integers can be written as the sum of two cubes less than 500?[/quote]

$c=a^3+b^3 ; 500 > a ; 500 > b ; c > 0$

Then, the answer is $\boxed {500^2}$

[quote="fireemblem13"][hide]I get 21.
The problem with ur solution is that a^3 < 500, not a < 500. same with b. There are 7 positive perfect cubes less than 500, the cubes of the numbers 1 -> 7. 7 choose 2 = 21[/hide][/quote] Ummmmmm..... I got $\boxed{23}$.

[hide]
5 possibilities for 7.
6 possibilities for 6.
7 possibilities for 5.
7 possibilities for 4.
7 possibilities for 3.
7 possibilities for 2.
7 possibilities for 1.

So the answer I think is: $\frac{5+6+7*5}{2}=\boxed{23}$.[/hide]

The answer is the 7th triangular number minus 2 = $\boxed{26}$ (for 343+343 and 216+343)

@ Kryptonic: the numbers in the form $n^3+n^3=2n^3$ have no repeats. There are 6 of them, and by dividing those by 2 also you loose 3.

Here's a visual:

Does it mean the sum is less than 500 or each of the 2 cubes is less than 500?

[hide="What I think"]I think the latter, and then since there are 7 cubes, then $28$?[/hide]

P.S How is this probability?  :?  :?  :?

[quote="lkryptonicl"]How many positive integers can be written as the sum of two cubes less than 500?[/quote]

Maybe I was wrong, but I thought by reading it that the numbers where lees than 500, not their cubes...

It means that the cubes are less than 500, but I'm pretty sure that it should be 21, like fireemblem said.  [hide]There are $7$ cubes, less than 500, and you can choose any 2 of those to add together to get distinct numbers.  Therefore, it is$7\choose2$, or $\boxed{21}$.[/hide]
Edit:
[hide]Actually, let's make it 42.  Because they are cubes, there are negatives, so you can add any of the negative ones to a perfect cube that has a greater absolute value than them and still have a postive integer.   Therefore, you can add 6 negative cubes to 343 ($7^3$) and still have an answer that fits the question.  Likewise, you can add 5 to 216 ($6^3$), etc.  Therefore, we have $6+5+4+3+2+1=21$ more combinations that we need to add to the 21 we got from $7\choose2$.  Thus, $21+21=\boxed{42}$[/hide]

The largest cube less than 500 is ${7^3}$. Thus we have
$\binom{7}{2} \implies 21$ ways. However, We can also use negative cubes and 0, as long as the result is positive. Thus we have an Additional ${7+6+5+4+3+2+1} = 28$ ways. Thus our answer is $21 + 28 = \boxed{49}$

Darn it, can't believe I forgot 0...  Normally I'm the person at our school who remembers things like that...

[quote="lkryptonicl"]I know its somewhere in the 20's. Or maybe not....Can someone confirm it?

[hide="zuton force was right"]

For 1---7 possibilities
For 2---6 possibilities
For 3---5 possibilities
For 4---4 possibilities
For 5---3 possibilites
For 6---1 possibility
For 7---0 possibilities

So $7+6+5+4+3+1=26$.[/hide]

PS: 0 isnt positive.[/quote]
But 0 is a cube, and we only care about whether the end result is positive or not.

Yeah, any of the cubes can be negative or 0, as long as the end result is positive, so we get the extra 28 values from that...

[hide]Consider the set of numbers {0, 1, 2, 3, 4, 5, 6}. We can choose any two of these numbers (except 0 and 0), add their cubes up, and get a number under 500. that's ${7\choose2}-1=20$ possibilities. Now consider what can be added to 7 to get a number under 500. $7^3+5^3=468<500$, but $7^3+6^3>500$, so we can have $7^3$ plus any other number from 0-5 cubed, which is 6 numbers.

So 20+6=26. [/hide]


Καλημερα σας, σημερα γυρισα απ τις διακοπες (βασικα ειμαι ακομα στο πλοιο και γραφω απο εκει), συγχαριτηρια στους 6 μαθητες γι αυτη την επιτυχια και μακαρι να τα πανε τοσο καλα και σε λιγα χρονια οταν θα δινουν στην ΙΜΟ και στη ΒΜΟ.

Με τα προβληματα θα ασχοληθω το βραδυ που θα βρησκομαι σπιτι μου, με μια πρωτη γρηγορη ματια παντος τα 3 πρωτα μου φαινονται σαν θεματα μετριας δυσκολιας, ενω το 4ο αρκετα δυσκολο (Ισος επηδη ακομα δεν εχω ασχοληθει με συνδιαστικη :P)

Αρχικά θα ευλογήσω τα γένια μου ...  :)  (στην πραγματικότητα τον συμπολίτη και φίλο μου  :) )
Δεν είναι άλλωστε και μικρή υπόθεση να είσαι Α Γυμνασίου και να γράφεις 28 στην JBMO .
Μπράβο ρε Παναγιώτη , πολλά πολλά συγχαρητήρια . :lol:  Βλέπω στον ορίζοντα το πρώτο gold σε ΙΜΟ . :wink:
Μετά να συγχαρώ όλα τα παιδιά για την πολύ καλή προσπάθεια και τα καλά αποτελέσματα .

Από διαγωνσιτικής απόψεως ήταν μια εύκολη JBMO με κλασσικά θέματα και πολύ πιο βατή από πέρσυ  :P

Συγχαρητηρια σε ολους τους μικρους  που μετειχαν στην JBMO. Τα πηγατε παρα πολυ καλα.
Παναγιωτη εκπληκτικοςςςςς.....τα ειπε ολα ο SIL . Μετεφερε στον $ BigL$ οτι η επιστημη που υπηρετει ειναι χρησιμοτατη αλλα δεν δινει συγκινησεις .Ευτυχως που το καταλαβε και εκανε στροφη....και η αποδειξη εισαι εσυ.

[quote="cretanman"]Συγχαρητήρια σε όλη την Ελληνική Ομάδα των μαθητών για την κατάκτηση 2 αργυρών [Παντέλα (34), Τσαμπασίδης (30)] και 3 χάλκινων μεταλλίων [Λώλας (28), Καλαντζής (27), Λιάμπης (21)] στη 12 Βαλκανική Ολυμπιάδα Νέων που έγινε στην Αυλώνα της Αλβανίας από 23 έως 28 Ιουνίου.

Τα cut offs για τα μετάλλια ήταν

Χρυσό 38 (υπήρχαν 8 μαθητές με βαθμολογία 40)
Αργυρό 30
Χάλκινο 13

Απ'ότι είδα τα προβλήματα έχουν συζητηθεί στις σελίδες

Πρόβλημα 1
http://www.mathlinks.ro/Forum/viewtopic.php?t=211771

Πρόβλημα 2
http://www.mathlinks.ro/Forum/viewtopic.php?t=211772

Πρόβλημα 3
http://www.mathlinks.ro/Forum/viewtopic.php?t=211774

Πρόβλημα 4
http://www.mathlinks.ro/Forum/viewtopic.php?t=211773

Για το 4ο πρόβλημα να σημειώσω ότι πράγματι οι λύσεις είναι ΟΛΟΙ οι άρτιοι αριθμοί οι μεγαλύτεροι ή ίσοι του 4 . Κάποιος όμως οφείλει να δείξει ότι με περιττού πλήθους κινήσεις δεν μπορούμε να επανέλθουμε σε κατάσταση όπου όλα τα τετράγωνα θα έχουν μετατραπεί σε μαύρα (και είναι πολύ όμορφος ο τρόπος απόδειξης αυτού του βήματος). Σύμφωνα με το πλάνο βαθμολόγησης κάποιος έχανε τη μισή άσκηση εάν δε το έκανε αυτό. Οπότε οι περισσότεροι μαθητές έχαναν πολλές μονάδες στο συγκεκριμένο πρόβλημα.

[color=red][size=150][b]Και πάλι συγχαρητήρια στους μαθητές[/b][/size][/color]

Αλέξανδρος[/quote]

 [color=red][b]Συγχαρητήρια  και από μένα σε όλα τα παιδιά που συμμετείχαν  !!![/b][/color]

 Η επιτυχία αυτή είναι μόνο η αρχή . Χαρείτε όμως τα αποτελέσματα και τα ξαναλέμε. Εννοείται ότι από το Σεπτέμβρη σας θέλουμε ενεργά μέλη στο φόρουμ!

[color=red][b] Καλό καλοκαίρι ! [/b][/color]

Sigxaritiria k apo mena stin elliniki Omada! Mipos 3erete pos ta pige i kipriaki?
Chrystalla Kipros


search
combinatorics unsolved
combinatorics

There are $ n$ circles of radius $ 1$ on the plane, no two tangent. No circles is such that it intersects no other circles. Prove that the number of intersection points is not less than $ n$. Find all cases of equality.

Posted before.
See http://www.mathlinks.ro/Forum/viewtopic.php?search_id=92193884&t=79046

Pierre.

Actually I did realize that solution. But it seems so difficult to find the equality case. That was why I posted it.

Maybe you can help.

By the way, I think the condition "No circles is such that it intersects no other circles" is enough. We don't need "no two of them are tangent and the subset of the plane formed by the union of them is connected. Prove that the number of intersection points is at least n."

Or I'm wrong?


SQL requests not achieved

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SQL Request: DELETE FROM phpbb_weblog_contributors WHERE weblog_id = 34 AND user_id IN ( )
Line : 256
File : class_blogs.php

THis occurs anytime I try to remove a contributor to my blog.

Billy

This happens to me too. I'll have to think twice now before letting anyone contribute. :(

I use XPSP2, also tried with VistaRCL1, didn't work

Tried on Firefox, IE6, IE7, Opera, Mozilla, and Netscape. No luck

why can't we remove contributors? I get the same message!

I get the same problem, and my browser is Safari.

Err... guys, it doesn't matter what your browser is :rotfl:

You'll have to get Mr. Vornicu to fix it.


modular arithmetic
Miscellaneous Problems

Unfortunately bboypa wolfram alpha is giving me an answer of 8, not 6.  Your mistake is failing to account for what happens to the numbers that are divisible by 5 or 2 after you divide the 5's and 2's out.

[quote=Peter]Counting from the right end, what is the $2500$th digit of $10000!$?[/quote]

We have to get $x\equiv \frac{10000!}{2^{2499}\cdot 5^{2499}}$ $mod 10$
Since $ord_{2}10000!>ord_{5}10000!=2499$,$x$ is even, namely $x\equiv 0,2,4,6,8$ $mod 10$.

On the other hand, $\frac{10000!}{5^{2499}} \equiv (1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdot 8\cdot 9)^{100}\equiv 6^{1000}\equiv 6$ $mod 10$

Then $2^{2499}\cdot x\equiv 6$ $mod 10$ $\Leftrightarrow$ $x\equiv 2,7$ $mod 10$

Therefore $x\equiv 2$ $mod10$.

The answer is $\boxed 2$. $\blacksquare$


function
trigonometry
complex analysis
complex analysis unsolved

We know that if $ f$ has an essential singularity at $ z_{0}$ then by Picard's theorem, given $ \delta > 0$ then $ \{f(z): 0 < |z \minus{} z_{0}| < \delta\}$ is either $ \mathbb{C}$ or $ \mathbb{C} \minus{} z'$ for  $ z'\in\mathbb{C}$

Now fixed $ n > 0$, let $ f(z) \equal{} z^{n}sin(\frac {1}{z})$, $ \delta > 0$ then[b] find[/b] $ \{f(z): 0 < |z \minus{} 0| < \delta\}$?

thanks for your hints

Suppose that $ z$ is close to $ 0$ and $ f(z)\approx a\ne 0$ ($ 0$ is, clearly, in the range). Note that then $ |f'(z)|\approx |a|\cdot|z|^{ - 2}$ and $ f''(z)$ approx $ |a|\cdot |z|^{ - 4}$. Hence, the inverse function theorem works in a neighborhood of size $ %Error. "const" is a bad command.
|z|^2$ (well, to be precise, you need to estimate $ |f''|$ from above in that neighborhood rather than at the point, but it is not hard either and the estimate is the same up to a constant factor), so the image contains a disc of some fixed radius around $ f(z)$ (this radius may depend on $ a$ but not on $ z$) and you are done (Sohotski theorem is enough for this argument, Picard is a big overkill here).

based on what you wrote I cannot figure out what is the image :blush: . Do you mean it is some disc ?

In this case we have $ z \equal{} 0$ is an essential singular so the image must be $ \mathbb{C}$ or $ \mathbb{C} \minus{} z_{0}$ for some $ z_{0}$

for $ e^{\frac {1}{z}}$ , its image near $ 0$ is $ \mathbb{C} \minus{} 0$
 since $ \sin(1/z) \equal{} \frac {e^{\frac {i}{z}} \minus{} e^{\frac { \minus{} i}{z}}}{2i}$ and clearly $ 0$ is in its image so its image near $ 0$ must be $ \mathbb{C}$
so the image of $ z^{n}\sin(\frac{1}{z})$  near $ 0$ must be $ \mathbb{C}$. I guess it is the whole complex plain but I have not had it done yet

It is the whole plane and what I wrote was assumed to be a proof that every $ a\ne 0$ is in the image. The idea is to choose a small $ z$ with $ f(z)$ almost equal to $ a$ and show that then in a small neighborhood of $ z$ there exists $ w$ with $ f(w)\equal{}a$ using the (quantitative version of the) inverse function theorem. Since $ 0$ is also in the image, that is the end of the story.


Let $ O$ in triangle $ ABC$ satisfy $ OA\equal{}OB\plus{}OC$. Let $ B',C'$ respectively be the middle point of triangle
$ COA,AOB$. Prove that two circle $ COC',BOB'$ is tagent

I didn't knew that Iran had 5 (or more) rounds.

not exactly...
after two rounds there remains about 40 contestants,among these students,12 of them will be rewarded by gold medal and finally between these 12 students,they choose 6 persons to participate in IMO...
so somehow the mathematical olympiad in Iran consists of 4 rounds...
this problem is from the third round,the round where 40 students are being challenged and 12 of them get the gold medal...

[quote="nktp"]Let $ O$ in triangle $ ABC$ satisfy $ OA \equal{} OB\plus{}OC$. Let $ B',C'$ respectively be the middle point of triangle
$ COA,AOB$. Prove that two circle $ COC',BOB'$ is tagent[/quote]
what is definition of midpoint of triangle?

$ G$ is the middle point of the triangle $ ABC$ mean $ \overrightarrow{GA}\plus{}\overrightarrow{GB}\plus{}\overrightarrow{GC}\equal{}\overrightarrow{0}$

no nktp youre wrong,as I sent the link before,here is the right form of the question:
[quote="Omid Hatami"]$ O$ is a point inside triangle $ ABC$ such that $ OA = OB+OC$. Suppose $ B',C'$ be midpoints of arcs $ \overarc{AOC}$ and $ AOB$. Prove that circumcircles $ COC'$ and $ BOB'$ are tangent to each other.[/quote]


This is a thread dedicated for people who ask advice on how to do well on AMC/AIME/USAMO. Normally, I would discourage such action because the ONLY way you can really get better at ANYTHING is to keep practicing and there isn't really anything that other people can do for you. But since people are asking many, many, many, times (idea of making the thread dedicated for this came from my friend who used to be a moderator; it will be better, in my opinion, to have this kind of thread so we don't have to answer same type of question everytime), so if you have something to ask, ask HERE.

Other advice threads will now be locked from now on.

To give some links to those threads to begin with (just don't post in them, ask any further questions here):

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=69293]Clicky #1[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=69982]Clicky #2[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=69262]Clicky #3[/url]

Good idea.  I was also thinking of making a thread like this, and asking to get it stickied.  Maybe you should also have people who have questions about rules, USAMO qualification, etc.etc. to ask here.

I think the best advice thread was started by FieryTycoon in this forum last year, I think it was called "Advice..."

EDIT: fixed the thread name

Putting all the threads together is a great idea, but just saying "practice" isn't all of it.

Often, one feels that what one is practicing is inefficient or useless, so it's good to know what others feel is best way to practice. Obviously no-one can get better without practicing, but for example, suppose you were working towards USAMO qualifying (ALOT of people), it's hard to tell whether and how to practice AIME problems, harder Olympiad problems, or some percentage of both.

Other times, people just want to make sure they're not missing something when they feel like they're never improving (it happens to everyone). So too many of the same thread is generally bad, but not all advice is bad really.


function
geometry
limit
greatest common divisor
trigonometry
algebra
polynomial

(All of the following problems were posed by me.)

1. Suppose there exists a function $ A(n), \forall n \in \mathbb{N}\backslash\{0,1,2\}$ such that it returns the area of a regular $ n$-sided polygon that is inscribed inside a circle of radius $ R$. Must the following statement be true?

$ \lim_{n \to \infty}{A(n)}=2\pi R$

(It's stating that as the number of sides of the $ n$-sided polygon reaches infinity, the area of the polygon will approach the area of the circle.)

2. Show mathematically why you can group digits in a base-$ n$ number into $ k$ digits and have the groups convert to a digit in an equivalent base-$ n^k$ number.

[hide="Clarification"]
Suppose we have the number $ 101011110101_2$.
Grouping the digits into 4 groups, we obtain:
$ 1010 1111 0101$

Converting each of the groups into decimal, we obtain:
$ 10.15.5$

Now replacing the double digit numbers with letters, we obtain $ \text{AF}5_{16}$.
[/hide]
[hide="My Solution"]
I do not know whether this is good enough to answer the problem, so please let me know if it is good enough.

Suppose each digit in a base-$ n$ number $ N$ is represented by $ d_i$ ($ i$ representing the number of digits from the right). So the actual representation of the number is

$ N = d_i \cdot n^i + d_{i-1} \cdot n^{i-1} + d_{i-2} \cdot n^{i-2} + \dots + d_2 \cdot n^2 + d_1 \cdot n^1 + d_0 \cdot n^0$

Now let's try plugging in a certain value for $ k$, such as $ 2$. (Let's assume that $ i$ is odd to keep it simple.)

$ N = (d_i \cdot n^i + d_{i-1} \cdot n^{i-1}) + (d_{i-2} \cdot n^{i-2} + \dots + d_2 \cdot n^2) + (d_1 \cdot n^1 + d_0 \cdot n^0)$

We notice that we can factor in those groups. (All but the first group from the right.)

$ N = n^{i-1}(d_i \cdot n + d_{i-1}) + (d_{i-2} \cdot n^{i-2} + \dots + n^2 (d_3\cdot n + d_2) + (d_1 \cdot n^1 + d_0 \cdot n^0)$

Now from factoring the greatest common factor out, we notice that we obtain a representation in another base system - in that of $ n^2$. That's because dividing the digits into groups of two leaves a representation of the form $ c_0 \cdot n^0 +  c_1 \cdot n^2 + c_2 \cdot n^4 + \dots + c_i \cdot n^{2i}$. We also notice that the digits are being converted (because they are being multiplied by $ n$). If we substitute $ m = n^2$ into the new representation, we just obtained a base conversion from base-$ n$ to base-$ n^2$($ m$). Now we know it works for $ k=2$. We need to prove that it works for all natural values of $ k$. To do so, let's take a look at the original representation of base-$ n$.

$ N = d_i \cdot n^i + d_{i-1} \cdot n^{i-1} + d_{i-2} \cdot n^{i-2} + \dots + d_2 \cdot n^2 + d_1 \cdot n^1 + d_0 \cdot n^0$

When we grouped for $ k=2$, it is noticeable that we found that we factored out $ n^{2i}$. This is because the first 2 digits from the right were multipled by $ n^0$ and $ n^1$ and everything afterward had a group of the next two consecutive numbers. The GCF was always a multiple of $ 2$ because that was the lowest power of $ n$ that was common to both terms. And because the terms were grouped into 2's, the GCF's power increased by 2 for every group. So if we grouped the terms in a representation into groups of $ k$, we obtain a representation in which the powers of $ n$ being multiplied by are multiples of $ k$. Also, if the number of terms is not a multiple of $ k$, then you can add 0's multiplied with powers of $ n$ to keep the method valid.

Thus, the conversion method mentioned in the problem statement can work for all $ k$ and all $ n$ because when the terms of the representation of the number are broken into groups of $ k$, they can be factored such that after factoring, the representation is turned into a base-$ n^k$ number. (Of course, they have to be natural numbers.)
[/hide]

3. Find an expression that gives you the maximum base-$ 10$ value that can be stored in a base-$ n$ number containing $ d$ digits which requires [i]the shortest number of calculations[/i].

[hide="My Solution"]
We can determine the maximum base-$ 10$ value that can be stored in a base-$ n$ number containing $ d$ digits by making all the digits the highest digits they can be. The highest digit in any base is $ n-1$ (if $ n-1 \ge 10$ then a letter, symbol, or some special character is used). By a summation, we can come up with the following:

$ \sum_{k=0}^{d-1}{(n-1)\cdot n^k}$

The above expression does give what the problem is looking for, but it does not have the shortest number of calculations. We need to think a little deeper to find a better solution.

Now there is something special about the expression above. We can see it once we try it out with some values for variables. Suppose we want to find the maximum value in base-$ 5$ with $ 2$ digits.

$ \sum_{k=0}^{1}{4 \cdot 5^k}=4\cdot 5^0 + 4\cdot 5^1 = 24$

We note that the the number shown above is $ 44_5$. What is special about that number? What number is that number close to? The answer is that the number is just $ 1$ below the first $ 3$-digit base-$ 5$ number. And this applies to every base with any number of digits. Instead of using the summation above, why not convert the number that is 1 above the maximum and subtract 1? The reason why we would do that is because it is [i]much[/i] easier to convert the first $ m$-digit number in a base rather than the last $ (m-1)$-digit number. The first $ m$-digit number of any base starts with a 1 and the rest of the digits are zeros. So the solution is $ \boxed{n^{d}-1}$. This is because the next power of a base multiplied with the first digit in the next order of magnitude in a base is $ d$. (The power you multiply for the first digit is always the number of digits minus 1.)
[/hide]

4. Why is it that we cannot have a base-$ 1$ number system? (I somewhat know the answer to this, but I want to see if you guys can come up with a real good one.)

Remark on base one:

http://en.wikipedia.org/wiki/Unary_numeral_system

http://www.research.att.com/~njas/sequences/A000042

Remark on the polygon limit:

It follows from $ \lim_{x\to 0}\frac{\sin x}{x}\equal{}1$

http://www.maa.org/joma/Volume7/Aktumen/Polygon.html

Generally, unrelated problems are posted and discussed in separate threads.  Otherwise, an individual thread can become cluttered.  Some comments.

[hide="1"] It is nice to see that you are stating this problem in a very formal way, but the function $ A(n)$ obviously exists.  Usually problems of this form are stated with the implication that there may either exist none or may such functions, but the function you have described is unique. [/hide]
[hide="2"] One way to think about a base-$ b$ representation is as a sequence of remainders upon division by $ b$:  divide by $ b$, subtract the remainder and remove the factor of $ b$, divide by $ b$ again, subtract the remainder and remove the factor of $ b$... the set of remainders gives you the set of digits in base $ b$.

If you start with a number's digits in base $ n$, its remainders upon division by $ n^k$ are clearly expressible in terms of those digits and the above process precisely groups and removes $ k$ digits at a time.

(Just because I didn't write any summation notation or digits does not make this explanation "non-mathematical": here I find that the details get in the way of what's really going on.  If you think of numbers as being polynomials in $ n$, this should make enough sense that I do not have to resort to actually writing out what's going on.) [/hide]
[hide="3"] The expression you've got is the sum of a geometric series.  I'm also not sure why you specified "base-$ 10$" value; the answer is a number and numbers exist independently of their base representations. [/hide]
[hide="4"] That depends on what you mean by a number system.  hsiljak has given you an example of what could be considered a base-$ 1$ number system; instead of each place value representing $ b^0, b^1, b^2, ...$, each place value represents $ 1^0, 1^1, 1^2, ...$.

On the other hand, the remainder construction I described fails miserably when we try to divide by $ 1$, so in that sense it is not particularly meaningful to talk about base-$ 1$ representation.  For further investigation it might help to be familiar with [url=http://en.wikipedia.org/wiki/Factoradic]some[/url] [url=http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibrep.html]other systems[/url]. [/hide]

On #1, the limit of $ A(n),$ assuming that $ A(n)$ is the area of the polygon, is $ \pi R^2.$

If you want something whose limit is $ 2\pi R,$ that would be $ P(n),$ the perimeter of the polygon.

=========================

Something I like to do in our "History of Mathematics" class:

WLOG, we may choose $ R$ to be any convenient number, so let $ R\equal{}1.$

Let $ p_n$ be the half of the perimeter of of the regular $ n$-gon inscribed in the unit circle.

Let $ s_n$ be the length of one side of that regular $ n$-gon. Note that $ p_n\equal{}\frac{n}2\,s_n.$

Derive a recursive formula for $ s_{2n}$ in terms of $ s_n.$ The only operations required should be ordinary arithmetic operations and square roots - in particular, there is no need to name any trigonometric functions. In terms of geometry, you really only need to know the Pythagorean Theorem.

Starting with an easily-described base case (either $ n\equal{}4$ or $ n\equal{}6,$ your choice), use the recursion to write $ s_n$ and $ p_n$ for successive doublings for a modest number of iterations. That is, find $ s_n$ and $ p_n$ either for $ n\equal{}4,8,16,32,64,128$ or $ n\equal{}6,12,24,48,96,192.$

[That much is well-known in history, both from Hellenistic (e.g., Archimedes) sources or Chinese sources.]

[And then the part that I don't give to the history class because not everyone has had a numerical analysis class.]

Use a Richardson extrapolation table (even exponents only) to accelerate the convergence of your sequence to $ \pi.$

Ok thanks everyone. My bad about the incorrect formula  :blush:.


geometry
circumcircle
absolute value
complex numbers

A circle that passes through the origin has a rightmost point of $(2,8)$.  Compute its radius.

[hide]
if the rightmost point is at (2,8) then the center will be at a point (x,8), which will be equidistant from (2,8) and the origin.  therefore $x+2=\sqrt(x^2+8^2)$, which solves to x=15, so the radius is 17.

That works, I guess... but you should hide it.

Here's what I was thinking:
[hide]Use the complex plane.
The fact that $(2,8)$ represents the rightmost point would show that we have a complex number, plus its absolute value, giving us $2+8i$.  So we have $z+|z|=2+8i$.  Let $a+bi=z$.  So now we have $a+bi+\sqrt{a^2+b^2}=2+8i$.  So $b=8$.  Now we have $a+\sqrt{a^2+64}=2$.  Solving gets us $a=-15$.  So the radius would be $|z|$, or $\boxed{17}$.

In fact, that's how I got the problem: $z+|z|=2+8i$, compute $|z|$.  I just fooled around with it a bit.  Apparently, I just made it a heck of a lot easier.[/hide]

wow that's really creative lingomaniac. Aside from roots of unity, I've always wondered how complex numbers can relate to geometry. I used a geometric way, but it seems a little involved when I compare it to blackwellbc's solution.

[hide]Since the point $(2,8)$ is the rightmost point, the radius to $(2,8)$ is perpendicular to the $y$ axis. Therefore the chord lying on the $y$ axis is bisected by the radius, so another point on the circle is $(0,16)$. Now realize that the circle is the circumcircle of the triangle with vertices at $(0,0)$, $(2,8)$ and $(0,16)$. Knowing that the area of the triangle, $A$, relates to the circum radius $R$ by $A=\frac{abc}{4R}$ where $a$, $b$, and $c$ are the side lengths, we have $R=\frac{abc}{4A}=\frac{16\sqrt{68}\sqrt{68}}{4(16)}=17$.[/hide]

since you know that $x=2$ is tangent to the circle, you can also use power of a point

Here's a really simple one (less than 30 seconds) using Thale's
[hide]
since $(2, 8)$ is the right-most point, the left-most point will be $(a, 8)$, and these two points along with the origin must form a right triangle.  So we have $2^2 + 8^2 + a^2 + 8^2 = (2-a)^2$, solving we have $a = -32$, therefore the diameter is $34$ and the radius is $17$.
[/hide]
btw, some exercises in Art and Craft of Problem Solving the author wants you to solve WITHOUT algebra, meaning that you have to think of the geometric meaning behind complex numbers.  I always think those kind of problems are nice.


Let P is an arbitrary point in a quadrilateral $ ABCD$ such that  $ \widehat{PAB},\widehat{PBA},\widehat{PBC},\widehat{PCB},\widehat{PCD},\widehat{PDC},\widehat{PAD},\widehat{PDA}$ are all acute angles. Let $ M,N,K,L$ are feets of the altitude from $ P$ on $ AB, BC, CD, DA$. Find the smallest value of $ \frac{AB}{PM}\plus{}\frac{BC}{PN}\plus{}\frac{CD}{PK}\plus{}\frac{DA}{PL}$

This is a double-post of a question that is likely a current contest (?) problem (see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164085 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164102
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=914160#914160 )

Moderator, please lock or delete (as well as this poster's other recent threads, all of which are both double-posted and likely fall into the same category).


Let $x,y,z$ be positive integers, and let $h$ denote their greatest common divisor.  If $1/x-1/y=1/z$, prove that both $hxyz$ and $h(y-x)$ are perfect squares.

Let $x=ha, y=hb, z=hc$, so that $(a, b, c)=1$. We have $(y-x)z=xy$, so dividing by $h^2$, $(b-a)c=ab$. $hxyz=h^4abc, h(y-x)=h^2(b-a)$, so we want to show that $abc, b-a$ are perfect squares.

Suppose that some prime $p$ divides $c$, and let $d$ be such that $p^d \| c$. Since $(b-a)c=ab$, we have $c \mid ab$; but $p$ can't divide both $a, b$ because $(a, b, c)=1$. Then it divides just one of them, which means it doesn't divide $b-a$. So we have $p^d \| LHS, p^d \| RHS$, then $p^d \| ab$. We can do this for all primes that divide $c$. Then the highest power of $p$ dividing $abc$ is $2d$, which is even, and the highest power of $p$ dividing $b-a$ is $0$, which is also even.

Now suppose some prime $q$ divides $a$ but not $c$. Then it must also divide $b-a$; so it divides $b$. Similarly any prime which divides $b$ but not $c$ divides $b-a$ and thus $a$. Let $e, f$ be such that $q^e \| a, q^f \| b$. If $e \neq f$, $q^{e+f} \| RHS$ whereas $q^{min(e,f)} \| LHS$, contradiction. So $e=f$. Then the highest power of $q$ dividing $abc$ is $e+f=2e$, which is even. We've characterized all prime divisors of $abc$ as having even exponents, so it is a perfect square. Then as $ab=c(b-a), abc=c^2(b-a)$. Thus as $abc$ is a square, $c^2(b-a)$ is a square and $b-a$ is a square.


geometry
geometric transformation
rotation
trigonometry

A particular Ferris Wheel has a radius of 60 feet and its center is 75 feet off the ground. It takes this Ferris Wheel 120 seconds to make one full rotation. Assume that a rotation starts at the 3:00 position of the Ferris Wheel and that it rotates counterclockwise.

a) write an equation to show a rider's position at a given time t, to the left or right of an imaginary vertical line drawn down the center of the Ferris Wheel.

I started out with f(t)=75+60cos3t and when I started doing b, I knew mine is not correct.

b) Use your equation from part a to find at what time(S) a rider would be 10 feet to the left or center during the first rotation.

Please help!

the required displacement is to the right or
 left of the centre of the Ferris Wheel. So the height is immaterial.
 The required equation is $x = 60 \cos(\omega*t)$
 i.e basically $x = 60 \cos \displaystyle\left(\frac{\pi t}{60}\right)$
 where $x$ is the displacement off centre.
 It is negative on the left side of the wheel and
 positive on the right.


D and E are two points on AB side of ABC triangle and AD/DB . AE/EB=(AC/CB)^2.
prove that <ACD=<BCE.

Dear "behdad...",
this problem has been very recently send to this site...
Make a link with the concept of symmedian and you will find the Steiner' relation.
Sincerely
Jean-Louis


[quote="aleph0"]I got (6+1)3=21. The 3 is for 3 directions and the 6 is for small rhombi and the 1 for large.[/quote]

Is there a formula for these types of problems rather than trying to count all of them?

nah
each question is different and every diagram is different, so you just have to go with the flow :)
Seriously though, casework is most effective in these kind of problems


Euler
function
floor function
induction
AwesomeMath
summer program
number theory

Prove that
\[ \sum_{i\equal{}1}^n \phi(i)\left\lfloor\frac{n}{i}\right\rfloor\equal{}\frac{n(n\plus{}1)}{2}\]

I have an inductive proof but I was wondering if anyone had a more intuitive approach.

My proof also uses induction, so I won't post it.

Algebraically, we have
\begin{align*} \frac {n(n + 1)}{2} & = \sum_{i = 1}^n i \\
& = \sum_{i = 1}^n \sum_{d | i} \phi\left(\frac {i}{d}\right) \\
& = \sum_{i = 1}^n \sum_{b | i} \phi(b) \\
& = \sum_{b = 1}^n \sum_{k = 1}^{\left\lfloor \frac {n}{b} \right\rfloor} \phi(b) \\
& = \sum_{b = 1}^n \phi(b) \cdot \left\lfloor \frac {n}{b} \right\rfloor. \end{align*}
We can also express this combinatorially, as follows:  $ \frac {n(n + 1)}{2}$ is the number of pairs $ (a, i)$ with $ 1 \leq a \leq i \leq n$.  To get to the second line, we replace each pair with the ordered triple $ (d, q, i) = \left( \gcd(a, i), \frac {a}{\gcd(a, i)}, i\right)$, and this is a bijection between the pairs and triples satisfying $ \gcd\left(\frac {i}{d}, q\right) = 1$, $ d | i$ and $ d \cdot q \leq i \leq n$.  To get to the third line, we replace the triple $ (d, q, i)$ with the triple $ (b, q, i) = \left( \frac {i}{d}, q, i\right)$ which satisfies $ \gcd(b, q) = 1$, $ b | i$, $ q \leq b$ and $ i \leq n$.  Finally, for each $ b$, the number of triples in which it appears is the number of ways to choose a number $ q$ no larger than $ b$ and relatively prime to it, together with a multiple $ i = kb$ of $ b$ no larger than $ n$, and this is $ \phi(b) \cdot \left\lfloor \frac {n}{b} \right\rfloor$.

@panos_lo: what was the point of your post?  (This is a rhetorical question, i.e., there is no reason to respond to it.)

Oh man this was on the Awesomemath Team contest this year :O


videos
geometry
geometric transformation
blogs
MIT
college

[url=http://www.spiegel.de/schulspiegel/wissen/0,1518,636316,00.html]Here[/url] is a "Spiegel" artikel about the German team with some pictures, or translated into [url=http://209.85.227.132/translate_c?hl=en&ie=UTF-8&sl=de&tl=en&u=http://www.spiegel.de/schulspiegel/wissen/0,1518,636316,00.html&rurl=translate.google.com&usg=ALkJrhiWh1S46Vq9YojF_GtbUG520hfCIA]English by google translator.[/url]

- German interview (.mp3-file) titled [url=http://www.kakadu.de/blob/article/6352/audio.mp3]"Perfekt in Mathematik: Lisa Sauermann"[/url]

[b]Some short videoclips from the German news show "Tagesschau.de (dail show)" in German language:[/b]

- [url=http://www.tagesthemen.de/multimedia/video/video532442.html]Internationale Mathematik-Olympiade in Bremen, 14.07.2009[/url]

- [url=http://www.tagesschau.de/multimedia/video/video532606.html]Mathe-Olympiade: Bremen Gastgeber der Mathematik-Olympiade, 15.07.2009[/url]

- [url=http://www.tagesschau.de/multimedia/video/sendungsbeitrag1382.html]Mathematik-Olympiade: Deutsche Schüler auf Platz Neun bei Internationaler Mathematik-Olympiade, 21.07.2009[/url]

Here are two German articles about the IMO, including a German video clip with Christian Reiher who explains [url=http://www.mathlinks.ro/viewtopic.php?p=1561571#1561571]problem 1 of the IMO 2009.[/url] Unfortunately user [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=26547]gjw[/url] deleted his very interesting comment. I do not see how any IMO Shortlist related comments about problems that made it to the IMO 2009 paper would be disagreed by anybody (unless the description contains comparisions/descriptions to other shortlisted problems.)

- [b]FIRST ARTICLE:[/b] [url=http://www.spiegel.de/wissenschaft/mensch/0,1518,636670,00.html]The first article[/url] is called "MATHE-OLYMPIADE 2009: Universal-Formel gegen peinliche Geschenke" by Spiegel Online. The article discusses exam conditions, estimates by the German team regarding number of solved problems, Elias Wegert discusses [url=http://www.mathlinks.ro/viewtopic.php?p=366562#366562]IMO problem 1986/3,[/url] Christian Reiher explains [url=http://www.mathlinks.ro/viewtopic.php?p=1561571#1561571]problem 1 of the IMO 2009[/url] in a video clip and finally they talk about soccer and politcs. [url=http://translate.google.ca/translate?u=http%3A%2F%2Fwww.spiegel.de%2Fwissenschaft%2Fmensch%2F0%2C1518%2Cdruck-636670%2C00.html&sl=de&tl=en&hl=en&ie=UTF-8]Here[/url] is a google translation of the first article.

- [b]SECOND ARTICLE:[/b] [url=http://www.mz-web.de/servlet/ContentServer?pagename=ksta/page&atype=ksArtikel&aid=1246046512417&openMenu=1013016724285&calledPageId=1013016724285&listid=1018881578312]Here[/url] you can find an article about Bertram Arnold regarding A-levels, musical talent, IMO preparation and family. [url=http://translate.google.ca/translate?u=http%3A%2F%2Fwww.mz-web.de%2Fservlet%2FContentServer%3Fpagename%3Dksta%2FXPage%26atype%3DksArtikel%26aid%3D1246046512417%26template%3DDruckfassung%26sp%3D1022083369355&sl=de&tl=en&hl=en&ie=UTF-8]Here[/url] is a google translation of the second article.

[quote="orl"] Unfortunately user [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=26547]gjw[/url] deleted his very interesting comment. I do not see how any IMO Shortlist related comments about problems that made it to the IMO 2009 paper would be disagreed by anybody (unless the description contains comparisions/descriptions to other shortlisted problems.)
[/quote]

I asked him not post that because I thought it was not ok to do that during the competition.
 If I was wrong or I was rude I'm deeply sorry. :)

Thank you for the nice articles!

PS: The comment was reposted by  gjw, so we still have the post after all!

[quote="Teodoros"][quote="orl"]Some pictures:

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=62&userid=64]50th IMO 2009 Exams[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=63&userid=64]50th IMO Anniversary celebration[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=65&userid=64]50th IMO 2009 Wangerooge[/url][/quote]

thank you for your kind contributions, since this year IMO's site is simply pathetic (it's hard to find a euphemism).
if it wasn't for mathlinks, ... but, there are at least five posts where i could see the german team!!![/quote]

I completely agree with the previous poster...

Not a single picture has been uploaded yet to the IMO 2009 website... some of us who can't attend IMO want to see pictures of our friends there and we can't :(

Thank you very much orl!

[quote="loose_tongued"]Does anyone know where can I find hymn of IMO?[/quote]

Yes, since the 38th IMO 1997 in Argentinia IMOs have an own hymn whose extract can be found [url=http://www.mathematik-olympiaden.de/IMOs/dokumente/hymne.wav]here (0.8 MB).[/url]

There are some more news but I post the links later. [url=http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=30&Itemid=39&lang=en]Here[/url] you can find the links to the daily news, or directly here:

- [url=http://www.imo2009.de/imo/downloads/day1.pdf]Daily News, Day 1[/url]
- [url=http://www.imo2009.de/imo/downloads/day2.pdf]Daily News, Day 2[/url]
- [url=http://www.imo2009.de/imo/downloads/day3.pdf]Daily News, Day 3[/url]
- [url=http://www.imo2009.de/imo/downloads/day4.pdf]Daily News, Day 4[/url]
- [url=http://www.imo2009.de/imo/downloads/day5.pdf]Daily News, Day 5[/url]
- [url=http://www.imo2009.de/imo/downloads/day6.pdf]Daily News, Day 6[/url]
- [url=http://www.imo2009.de/imo/downloads/day7.pdf]Daily News, Day 7[/url]
- [url=http://www.imo2009.de/imo/downloads/day8.pdf]Daily News, Day 8[/url]
- [url=http://www.imo2009.de/imo/downloads/day9.pdf]Daily News, Day 9[/url]

Terence Tao uploaded his talk from the 50th IMO anniversary celebration about "Structure and randomness in the prime numbers" on 19th July [url=http://terrytao.wordpress.com/2009/07/20/structure-and-randomness-in-the-prime-numbers-2/]here.[/url] He is also doing an experiment on crowdsourcing research/olympiad problems. In some ways it is a bit similar to the [url=http://www.mathlinks.ro/weblog_entry.php?t=285839]Netflix Challenge[/url] except that it is intended to be much more collaborative. Also look at this article on [url=http://www.scientificamerican.com/article.cfm?id=mindless-collectives-rational-decision-making]"Mindless Collectives Better at Rational Decision-Making Than Brainy Individuals".[/url] You can see the respective blog posts by Terence Tao [url=http://terrytao.wordpress.com/2009/07/20/imo-2009-q6-as-a-mini-polymath-project/]there[/url] and [url=http://terrytao.wordpress.com/2009/07/21/imo-2009-q6-mini-polymath-project-cont/]here continued.[/url] The problem he is crowdsourcing is [url=http://www.mathlinks.ro/viewtopic.php?t=289051]IMO Problem 2009/6.[/url]

yea at the end of the video me and my team . .
thanks  for the video orl :lol:
and for the daily news i was searching of it !

[url=http://www.mathetalente.de/community/fotos.html]Here[/url] you can find an overview of IMO picture sets, or directly below:

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=62&userid=64]50th IMO 2009 Exams, 24 pictures by Stephanie Schiemann[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=63&userid=64]50th IMO 2009 Anniversary celebration, 47 pictures by Stephanie Schiemann[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=65&userid=64]50th IMO 2009 Wangerooge, 27 pictures by Stephanie Schiemann[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=68&userid=64]50th IMO 2009 Opening-trial 1, 60 pictures by Stephanie Schiemann[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=69&userid=64]50th IMO 2009 Opening-trial 1, 68 pictures by Stephanie Schiemann[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=67&userid=64]50th IMO 2009 Arrival, 16 pictures by Stephanie Schiemann[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=71&userid=64]50th IMO 2009 Siegerehrung / Closing Ceremony, 63 pictures by Stephanie Schiemann[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=76&userid=64]50th IMO 2009 End-of-IMO-Party, 19 pictures by Stephanie Schiemann[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=70&userid=121]50th IMO 2009, Team Netherlands (NLD), 15 pictures by Marie van Amelsvoort[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=72&userid=126]50th IMO 2009, Picture Set 1, 154 pictures by Jan-Hendrik Bruns[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=73&userid=126]50th IMO 2009, Picture Set 2, 116 pictures by Jan-Hendrik Bruns[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=75&userid=128]50th IMO 2009, IMO Teams, 100 pictures by Anke Allner[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=74&userid=129]50th IMO 2009, Wangerooge, 22 pictures by Mats Vermeeren[/url]

- [url=http://www.mathetalente.de/community/fotos/album.html?albumid=77&userid=125]50th IMO 2009, Picture Set 1, 84 pictures by David Bauer[/url]

[b]Some news articles about the IMO 2009:[/b]

- [url=http://translate.google.com/translate?u=http%3A%2F%2Fwww.shz.de%2Fschleswig-holstein%2Fpanorama%2Fartikeldetail%2Farticle%2F529%2Fabiturient-rechnet-der-welt-etwas-vor.html&sl=de&tl=en&hl=en&ie=UTF-8]Article about Malte Lackmann (Germany)[/url]

- [url=http://translate.google.com/translate?u=http%3A%2F%2Fwww.welt.de%2Fdie-welt%2Farticle4142716%2FJunge-Talente-im-Wettstreit.html&sl=de&tl=en&hl=en&ie=UTF-8]Young talents in competition[/url]

- [url=http://translate.google.com/translate?u=http%3A%2F%2Fwww.kreiszeitung.de%2F%2Fbremenalles%2F00_20090627010029_Viele_Wege_fuehren_zu_Ruhm.html&sl=de&tl=en&hl=en&ie=UTF-8]Many roads lead to glory[/url]

- [url=http://translate.google.com/translate?u=http%3A%2F%2Fwww.kreiszeitung.de%2F%2Fbremenalles%2F00_20090714010044_Mathematik_in_fuenfneun_Sprachen.html&sl=de&tl=en&hl=en&ie=UTF-8]Mathematics in 59 languages[/url]

- [url=http://translate.google.com/translate?u=http%3A%2F%2Fwww.kreiszeitung.de%2F%2Fbremenalles%2F00_20090715010439_Aufgaben_von_besonderer_Qualitaet.html&sl=de&tl=en&hl=en&ie=UTF-8]Tasks of particular quality[/url]

- [url=http://translate.google.com/translate?u=http%3A%2F%2Fwww.kreiszeitung.de%2F%2Fbremenalles%2F00_20090720010124_Professoren_wie_Popstars_Jugendliche_wollen_Au.html&sl=de&tl=en&hl=en&ie=UTF-8]Professors like popstars[/url]

- [url=http://translate.google.com/translate?u=http%3A%2F%2Fwww.kreiszeitung.de%2F%2Fbremenalles%2F00_20090721010331_Mit_einem_hoffnungsvollen_Blick_ins_naechste_J.html&sl=de&tl=en&hl=en&ie=UTF-8]With a hopeful look towards next year[/url]

- [url=http://translate.google.com/translate?u=http%3A%2F%2Fwww.kreiszeitung.de%2F%2Fbremenalles%2F00_20090722010254_Von_Grashuepfern_und_Medaillen.html&sl=de&tl=en&hl=en&ie=UTF-8]Of grasshoppers and medals[/url]

- [url=http://news.xinhuanet.com/english/2009-07/23/content_11761579.htm]HK students win five maths medals[/url]

- [url=http://www.tehrantimes.com/index_View.asp?code=199384]Iranian students win 6 medals in IMO 2009[/url]

- [url=http://www.hindu.com/thehindu/holnus/001200907220325.htm]India bags five medals at International Math Olympaid[/url]

- [url=http://www.abc.net.au/news/stories/2009/07/21/2631673.htm?section=world]Aussie students win Maths Olympiad golds[/url]

- [url=http://www.abc.net.au/news/stories/2009/07/21/2631673.htm?section=world]Philippine Team Garners Bronze at the International Mathematical Olympiad in Bremen, Germany[/url]

- [url=http://www.cpv.org.vn/cpv/Modules/News_English/News_Detail_E.aspx?CN_ID=351469&CO_ID=30293]Vietnam’s Nung ethnic student won golden medal at International Math Olympiad[/url]

- [url=http://www.livinginperu.com/news/9641]Peru won four silver and two bronze medals in International Math Olympiad[/url]

- [url=http://www.newswire.ca/en/releases/archive/July2009/22/c7678.html]Canadian Math Students Win Olympiad Gold, Silver, And Bronze Medals[/url]

- [url=http://services.inquirer.net/mobile/09/07/26/html_output/xmlhtml/20090726-217240-xml.html]RP student wins bronze in math tilt[/url]

- [url=http://www.cogito.org/Articles/ArticleDetail.aspx?ContentID=17802]Cogito, IMO 2009[/url]

[quote="epsilonist"][quote="orl"][quote="loose_tongued"]Does anyone know where can I find hymn of IMO?[/quote]

Yes, since the 38th IMO 1997 in Argentinia IMOs have an own hymn whose extract can be found [url=http://www.mathematik-olympiaden.de/IMOs/dokumente/hymne.wav]here (0.8 MB).[/url]
[/quote]
What about its meaning (also, which language is it?) and its proposer?[/quote]

It's Spanish, the lyrics and translation can be found [url=http://www.oma.org.ar/imo/song.html]here[/url]

[quote="Mat(h)s"][/quote][quote="epsilonist"][quote="orl"][quote="loose_tongued"]Does anyone know where can I find hymn of IMO?[/quote]

Yes, since the 38th IMO 1997 in Argentinia IMOs have an own hymn whose extract can be found [url=http://www.mathematik-olympiaden.de/IMOs/dokumente/hymne.wav]here (0.8 MB).[/url]
[/quote]
What about its meaning (also, which language is it?) and its proposer?[/quote][quote="Mat(h)s"]

It's Spanish, the lyrics and translation can be found [url=http://www.oma.org.ar/imo/song.html]here[/url][/quote]

[url=http://www.oma.org.ar/imo97/index_en.html]Here[/url] is a link to the "38th IMO official song" but unfortunately the link has expired. I did not quite understand whether this song was only meant for the 38th IMO, i.e. each IMO has its own song, or at the IMO 1997 in Argentinia this song was established as the hymn for all coming IMOs?

I have edited postings 1, 10 and 11 in this thread.

[quote="epsilonist"][quote="orl"][quote="loose_tongued"]Does anyone know where can I find hymn of IMO?[/quote]

Yes, since the 38th IMO 1997 in Argentinia IMOs have an own hymn whose extract can be found [url=http://www.mathematik-olympiaden.de/IMOs/dokumente/hymne.wav]here (0.8 MB).[/url]
[/quote]

So that downloadable song was just the extracted version and was cut, right?
Then where can I find the full version, please?
I'm really into it.  :)[/quote]

I may suggest you email [url=http://www.math.uni-rostock.de/~gronau/]Dieter Gronau[/url] or some people who were involved into the 38th IMO 1997 in Argentinia and tell us your findings!  :)

- The guy who is "really into it" is apparently to contact those suggested people to find out more about the IMO hymn.

- Anyway [url=http://www.imo-register.org.uk/2009-report.html]here[/url] is the "50th IMO, Bremen, Germany, 10–22 July 2009, report by Geoff Smith".

- At the [url=http://www.mathlinks.ro/viewtopic.php?p=1573049#1573049]end of posting 10[/url] I added a link to a new picture set.

[quote="epsilonist"][quote="orl"]- The guy who is "really into it" is apparently to contact those suggested people to find out more about the IMO hymn.
[/quote]

Yep, ...
Thanks to orl really much. I've got the file now.
The full song is very good![/quote]

Good job! How about uploading it here?

Help me  :(  ,
I somehow cannot upload it!
Firstly, I did upload the .mp3 file of it but then an error occured.
Later, I packed the file in .rar and then [b]"General error : tried to upload empty file"[/b] appeared!