Latexopoiimeno  :P

[quote="xp"]Να αποδειχθεί οτι $ (2^2\plus{}1)(2^3\plus{}1)...(2^{n}\plus{}1)<2^{\frac{n(n\plus{}1)}{2}}$ για κάθε n>1[/quote]

Να υποθέσω ότι η διατύπωση του Silouan είναι η σωστή διατύπωση της άσκησης?

Ρωτάω επειδή με βάση τη διατύπωση του xp (είναι εκείνο το $ 7$ που χαλάει το pattern) φαίνεται η LHS να είναι το γινόμενο όλων των περιττών μέχρι το $ 2^n\plus{}1$, και η ανισότητα δεν ισχύει...

Cheerio,

Durandal 1707

A ναι, να βάλω επίσης τη λύση μου (πολύ επιγραμματικά) με βάση τη διατύπωση του Silouan. Καλά, βγαίνει και πολύ πιο απλά, εύκολη είναι η άσκηση, αλλά αυτό ήταν που μου΄ρθε με την πρώτη ματιά...

Λογαριθμίζοντας, αρκεί νδο: $ \frac {n(n \plus{} 1)}{2} \minus{} \sum_{k \equal{} 2}^n \log_2 (1 \plus{} 2^k) > 0$ ή, ισοδύναμα: $ \sum_{k \equal{} 2}^n \log_2 (1 \plus{} 2^{ \minus{} k}) < 1$.

Έστω $ f(x) \equal{} \log_2 (1 \plus{} 2^{ \minus{} x})$, οπότε η $ f$ θα είναι φθίνουσα και κυρτή. Έτσι, αρκεί να έχουμε $ \int_1^\infty f(x) dx < 1$; ωστόσο, εύκολα βρίσκουμε ότι $ f(x) \leq \frac {2^{ \minus{} x}}{\log 2}$, ενώ $ \frac {1}{\log 2}\int_1^\infty 2^{ \minus{} x} dx \equal{} \frac {1}{\log 2 \log 4}$.

Αν δεν κάνω λάθος, το παραπάνω κλάσμα είναι μικρότερο του $ 1$ και τελειώσαμε, δεν έχω όμως την υπομονή να το ελέγξω. Αν δεν ισχύει αυτό, τότε ένα υψηλότερο ανάπτυγμα λογικά θα αρκεί (εννοώ δηλ. να χρησιμοποιήσουμε την ανισότητα $ \log_2 (1 \plus{} y) \leq \frac {1}{\log 2} (y \minus{} y^2/2 \plus{} y^3/3)$).

Αν η λύση δουλεύει, αφιερωμένη στον r_boris ;)

Cheerio,

Durandal 1707

PS: By the way, είναι ενδιαφέρον ότι με την παραπάνω προσέγγιση αποδεικνύουμε το πιο ισχυρό αποτέλεσμα ότι η εν λόγω ανισότητα ισχύει ακόμα και με την RHS πολλαπλασιασμένη με μία σταθερά $ A < 1$. Δεν το έχω κοιτάξει καλά, αλλά ίσως θα ήταν ενδιαφέρον πρόβλημα να βρεθεί η ελάχιστη σταθερά $ A$ για την οποία η ανίσωση παραμένει αληθής (λογικά θα ανάγεται στον υπολογισμό μιας σειράς, με το μάτι όμως δε βλέπω αν αυτή θα υπολογίζεται η όχι...).

Εντυπωσιακή προσέγγιση (το κλάσμα είναι πράγματι μικρότερο της μονάδας).

Αλλά αμφιβάλλω πάλι για τη διατύπωση.

Παίρνοντας υπ' όψιν το 7 της εκφώνησης,
μάλλον ο γενικός όρος είναι $ 2k \plus{} 1$ με $ 2 \le k \le 2^{n \minus{} 1}$
(μου φαίνεται πιο ισχυρή από τη διατύπωση του Silouan).

Δε νομίζω ότι η αρχική διατύπωση είναι σωστή: $ 5\cdot 7\cdot 9 \equal{} 315 > 64 \equal{} 2^{3\cdot 4/2}$...

Cheerio,

Durandal 1707

Πάλι καλά που έδωσες το αντιπαράδειγμα γιατί προσπαθούσα να τη βγάλω κάμποση ώρα με την αρχική διατύπωση. (Δε σκέφτηκα ότι μπορεί είναι λάθος.  :blush: )

Μια πιο κλασική λύση:

από B.C.S. είναι $ LHS^2 \le (2^{2\cdot2} + 2^{2\cdot3} + ... + 2^{2\cdot n})(1 + 1 + ... + 1) =$

(επειδή η πρώτη παρένθεση είναι άθροισμα n - 1 όρων γεωμετρικής προόδου με αρχικό όρο $ 2^4$ και λόγο $ 4$)

$ (n - 1)(\frac {2^4(2^{2\cdot(n - 1)} - 1)}{4 - 1}) = \frac {n - 1}{3}(2^{2n + 2} - 2^4) <$

(προφανέστατα) $ 2^{n(n + 1)} \Leftrightarrow$

$ LHS < 2^{\frac {n(n + 1)}{2}}$

Τελικά ήταν πολύ χαλαρή. :lol:

$ f(n)\equal{}2^2\plus{}1)(2^3\plus{}1)..(2^n\plus{}1)<(2^2\plus{}2)(2^3\plus{}2)...(2^n\plus{}2)\equal{}3.2^{n\minus{}1}f(n\minus{}1)\equal{}>f(n)/f(n\minus{}1)<3.2^{n\minus{}1}$και πολλαπλασιαζοντας απο το 2 εως το n προκυπτει το ζητουμενο. :wink:

Hμουν εκτός δυκτίου για τεχνικούς λόγους μια εβδομάδα .Οντως το 7 είναι typo για αυτό είπα στον Σιλουανο να γράψει το σωστό .Η δική μου λύση : πολ/ζω με 3 και τα δύο μέλη και αφού διαιρέσω kαι τα δύο μέλη με το β΄ μέλος λογαριθμίζω και με την βοήθεια της ln(1+a)<=a.....

[quote="Durandal"]A ναι, να βάλω επίσης τη λύση μου (πολύ επιγραμματικά) με βάση τη διατύπωση του Silouan. Καλά, βγαίνει και πολύ πιο απλά, εύκολη είναι η άσκηση, αλλά αυτό ήταν που μου΄ρθε με την πρώτη ματιά...

Λογαριθμίζοντας, αρκεί νδο: $ \frac {n(n \plus{} 1)}{2} \minus{} \sum_{k \equal{} 2}^n \log_2 (1 \plus{} 2^k) > 0$ ή, ισοδύναμα: $ \sum_{k \equal{} 2}^n \log_2 (1 \plus{} 2^{ \minus{} k}) < 1$.

Έστω $ f(x) \equal{} \log_2 (1 \plus{} 2^{ \minus{} x})$, οπότε η $ f$ θα είναι φθίνουσα και κυρτή. Έτσι, αρκεί να έχουμε $ \int_1^\infty f(x) dx < 1$; ωστόσο, εύκολα βρίσκουμε ότι $ f(x) \leq \frac {2^{ \minus{} x}}{\log 2}$, ενώ $ \frac {1}{\log 2}\int_1^\infty 2^{ \minus{} x} dx \equal{} \frac {1}{\log 2 \log 4}$.

Αν δεν κάνω λάθος, το παραπάνω κλάσμα είναι μικρότερο του $ 1$ και τελειώσαμε, δεν έχω όμως την υπομονή να το ελέγξω. Αν δεν ισχύει αυτό, τότε ένα υψηλότερο ανάπτυγμα λογικά θα αρκεί (εννοώ δηλ. να χρησιμοποιήσουμε την ανισότητα $ \log_2 (1 \plus{} y) \leq \frac {1}{\log 2} (y \minus{} y^2/2 \plus{} y^3/3)$).
Ευχαριστώ για την αφιέρωση Παναγιώτη . Πράγματι εντυπωσιακός τρόπος

Αν η λύση δουλεύει, αφιερωμένη στον r_boris ;)

Cheerio,

Durandal 1707

PS: By the way, είναι ενδιαφέρον ότι με την παραπάνω προσέγγιση αποδεικνύουμε το πιο ισχυρό αποτέλεσμα ότι η εν λόγω ανισότητα ισχύει ακόμα και με την RHS πολλαπλασιασμένη με μία σταθερά $ A < 1$. Δεν το έχω κοιτάξει καλά, αλλά ίσως θα ήταν ενδιαφέρον πρόβλημα να βρεθεί η ελάχιστη σταθερά $ A$ για την οποία η ανίσωση παραμένει αληθής (λογικά θα ανάγεται στον υπολογισμό μιας σειράς, με το μάτι όμως δε βλέπω αν αυτή θα υπολογίζεται η όχι...).[/quote]


function algebra polynomial conics parabola quadratics geometry

Three years ago, I had a factoring program where it would list the prime factorization of a number that you entered. Unfortunately, I did not back the program up and it got deleted when my RAM cleared. Does anyone know the general form of the program?

Also, is it possible to display answers (possibly off of the factoring source) in simplest radical form (as a program not as a feature like in the TI 89)?

For example, $ \sqrt {18}$ as $ 3 \sqrt {2}$ instead of $ 4.2426...$.

There's (at least) one on the TI site for each of these things.

http://www.ticalc.org/archives/files/fileinfo/226/22626.html

In addition to having both functions you mention, this has many other cool features.

There are probably more programs as well.

[quote="7h3.D3m0n.117"]Ah. Before I download it, I assume the source code can be viewed (because I have an 83+ not an 84, hence I do not have the cable that attaches to my computer) by itself?[/quote]

Actually its an application... I don't know where you could view the source code.

Sorry, I didn't notice you were looking for a BASIC program (my bad).

[quote="tjhance"]Sorry, I didn't notice you were looking for a BASIC program (my bad).[/quote]

That's okay. An ASM program would be helpful, but I'm more interested in implementing that [clever] manipulation of BASIC onto other math programs.

Flash applications $ \ne$ ASM programs.

I guess you could do the highly inefficient method of using SoE to check primes up to half the number.

[quote="Temperal"]I guess you could do the highly inefficient method of using SoE to check primes up to half the number.[/quote]

But how would I do that?

[code]
Input " ",X
\sqrt{X} -> A

...
[/code]

As you can see once you get "A" you're stuck  :( . Also, how can you program the calculator to distinguish between the integers and decimals?

[quote="7h3.D3m0n.117"]As you can see once you get "A" you're stuck  :( . Also, how can you program the calculator to distinguish between the integers and decimals?[/quote]

Hit "Math" and go over to "NUM". the function iPart returns the integer part of a number, while fPart, returns the decimal part. So

[code]fPart(A)=0[/code]

would test if a value is an integer.[/code]

Why can't you just use a FOR() loop to cycle through integers until you're up to your square root, and check each one individually?

Get AFACTOR2 for prime factorization, and FACTOR9. The latter does radical simplifications, polynomial factorizations, lists the factors of a number, gives tons of info on parabolas of quadratics, and shows cubic divisions(w/long division).

[quote="abacadaea"]i have a formula that converts radicals to simplest radical form but not decimals.
its
[hide]
input A
sqrt(A)->b
0->c
1->d
while c<b
c+1->c
if b/c^2=int(b/c^2)
cd->d
end
clrhome
output 1,1, d
output 1,4,A/d
[/hide]
something like that[/quote]

I tried on a TI-84 Plus Silver, but it just output 1 and A every time.

I have a program on the area of a regular polygon with side length S if you want it. N=number of sides.

[hide="in here"]Prompt N,S
(NS^2)/(4tan(180/N)) -> A
Disp "AREA:",A[/hide]

Very simple.

I really don't know how to program very well, but I have had experience with loops.

For the prime factorization, maybe the GCD function could help?[hide="Something like"]Prompt A
1->U
Lbl 1
If GCD(A,U)=1
Then
U+1->U
GoTo 1
Else
Disp "",U
If (A/U)=1
Then
Goto 2
Else
U+1->U
Goto 1
Lbl 2
End[/hide]

Could someone check that plz? EDIT: It just checks the numbers not relatively prime to A. :(

[quote="1=2"]
I tried on a TI-84 Plus Silver, but it just output 1 and A every time.[/quote]

Yeah if I had an 84, I'd just download the ASM app.

[quote="1=2"]
I have a program on the area of a regular polygon with side length S if you want it. N=number of sides.

[hide="in here"]Prompt N,S
(NS^2)/(4tan(180/N)) -> A
Disp "AREA:",A[/hide]
[/quote]

Nice! Too bad my perimeter/area test in geometry already passed  :( .

my brother won an 84 silver but he already had an 89 titanium so he gave it to me.  :lol:


There is a pipe situated vertically in water with an end in the water and the other end in the air.The length of the pipe $in$ the water is $H$ and the the section of the pipe(horizontally) is  surface $S$.Somebody blows at the end of the pipe wanting to make air come out from the end in the water.How much work is the ''somebody'' doing?We also know the atmospheric pressure$P_{0}$.

[hide="solution"]Does it depend on the atmospheric pressure $P_{0}$ ?
To be able to blow air out of the lower end in water one has
to create an excess pressure in the tube that corresponds to the
pressure at the depth $z$ (see figure), in order to lower the surface to that
level.
Now the work becomes ($W=F\cdot s$):
$dW=Sp(z)dz$
Now the function $p(z)=\rho g z$ is the excess pressure. (The total pressure at a depth
is given by Pascal's law $p(z)=P_{0}+\rho g z$). Integrating yields:
$W=S\rho g\int_{0}^{H}zdz=\frac{1}{2}\rho gSH^{2}$[/hide]

your question has it's answer in my question :) .you must push the water in the pipe against the atmospheric pressure too.

So you mean that the answer is
$W=\frac{1}{2}\rho gS(H^{2}+H)$ ?

Yes sorry thats what I meant  :oops:

Also its correct to consider the total pressure as you did ($P_{0}+p(z)$), I made a mistake sorry.


hmm should we go

actually I don't know if anyone uses this forum whom I haven't already asked

but heck doesn't hurt to try

!?!?!?!?!?

hilton!?

apparently you haven't told me

Eh, it's just Jeremy and I, and you've probably already talked to us.

Do you know of anyone who's actually going?

Berman, please tell me what this is

I tried asking Arnav and he refused to answer

He refused to answer to me too, but I was very insightful and figured out what he meant.

The awards ceremony for winning ARML last year, which happens to be two days after the state contest.


Good after noon

I need to select an arc by clicking with the mouse. For that, a rectangle is used. I already did the same to select a line and it works.

How can I calculate it?

I will appreciate your prompt help
Best regards

Pedro Ferrer

Is it necessary to use a rectangle?
It will be easier if you use a little circle around the mouse pointer instead of a rectangle.

Let $x,y$ the coordinates of the mouse $M$

Let $(M,r)$ the circle around the mouse and $(O,R)$ the given circle.

You need to calculate the distance $MO$.
If $R-r\leq MO\leq R+r$ then you are too close to the arc (another expression is $-r\leq MO-R\leq r$)

This will tell you if you intersect the circle.
If it is important to know whether you intersect the arc, then you sould also find the direction of the line $OM$.

I think that the case with the rectangle is more complicated.

By the way, what do you want to do with this?
A geometric sketchpad maybe?

The rectangle isn't actually worse- it's equivalent to a circle with radius equal to half the diagonal.


does anyone know a website about the [color=blue][i]silk road mathematical competition[/i][/color]??

Containing info, problems etc...

thanks!
 :)

I am almost sure there is no site about it ,but if you want some informations or problems PM.

Omid Hatami ,He is one of our coordinators and he is related somehow with these SIlk road things

look out the web site daryn.kz i think you can find nfo about silkroad oly.
if you cant i can help you (i try:)  i think

I hope someone can translate into English and post them
http://matol.kz/nodes/93
Have Fun!
Note: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=365714&hilit=Silk+Road


For i:
[hide] (3n)! is the product of the terms 1 thru 3n, and every third term is a multiple of 3, so there is atleast 3^n in the numerator. Also, every other term is a multiple of 2, so there is atleast 2^(3n/2) in the numerator.[/hide]

For ii:
[hide] In (6n)!, every fifth term is a multiple of 5, so there is atleast 5^(6n/5) in the numerator. Since every third term is a multiple of 3, there is atleast 3^(2n) in the numerator. Every other term is a multiple of 2, yielding 2^(3n), and every fourth term is a multiple of four, yielding an additional 2^(3n/2). The product is 2^(4.5n) which is greater than the exponent in the denominator.[/hide]

For iii:
[hide]This problem requires some analytical thinking. Separate (n^2)! into n!, (2n)Pn, (3n)P(2n), ... ,(n^2-n)P(n^2). There are n of these sets. Each set aPb is divisible by n!, (this is the same as aCb since a-b=n) so n! cancels out n times. This is the same as (n!)^n, which is the denominator.[/hide]


function
number theory proposed
number theory

Ha-I think i find this problem really interesting ,Really interesting:

Here $f^n=f(f(f( N times)$
Find all of the functions like:

$1. F: \mathbb{N} \longrightarrow \mathbb{N}$
$F^2(n)=2n$

$2.F: \mathbb{N} \longrightarrow \mathbb{N}$
$F^p (N)=pn$ where $p$ is a prime.

Similar method can be used to solve f(f(n)=2n to that of f(f(n))=3n.

Bomb

For p=2, we have F is defined uniquely for odd numbers from which even follow, we define
f(x)=y and f(y)=2x and so forth a unique pairing between each odd integers and we are done since f(2n)=2f(n).

For p>2, we have f(np)=pf(n) and so on...

Bomb


symmetry
LaTeX
combinatorics proposed
combinatorics

2n +1 objects from the selected objects n, meeting with the situation that  for any integer k, 0<k<n+1, 2k +1 former objects have at least k objects.
Ask for a different total number of election.
f(1)=3,f(2)=9,f(3)=28.......

[quote="FSheva"]bad English[/quote]
sorry for disturbing. The question has been solved with the help of my friend.
symmetry is used in the solution to it. f(n)=C(2n+1,n)-C(2n+1,n-2)
Now I have another question when I went deep through it.

I want more ideas about it.

what do you mean by $ C^{n}_{2n\plus{}1}$? :huh:  :huh:

$ C_{m}^n \equal{} \binom{m}{n} \equal{} \frac{m!}{n!(m\minus{}n)!}$

The terms of that series are asymptotic to $ \frac{3}{2\sqrt{\pi n}}$, so it diverges.  Please read about using [[LaTeX]] (see also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 ) and in the future, please put unrelated problems in distinct threads.

Also, can anyone tell me what the first problem actually is supposed to say?

I think he is asking for the number of all $ n$-element subsets $ A$ of $ \left\{1,2,...,2n \plus{} 1\right\}$ such that $ \left|A\cap \left\{1,2,...,2k \plus{} 1\right\}\right|\geq k$ for all $ k\in\left\{1,2,...,n\right\}$.
This is a shade of a famous question to find the number of all $ n$-element subsets $ A$ of $ \left\{1,2,...,2n\right\}$ such that $ \left|A\cap \left\{1,2,...,2k\right\}\right|\geq k$ for all $ k\in\left\{1,2,...,n\right\}$ (the answer is the $ n$-th Catalan number, since the $ n$-th Catalan number is the number of Dyck paths of length $ 2n$).

  darij


inequalities
induction
inequalities unsolved

[quote="selmanfarisi"]if $ n > 2$ prove that
 $ (n!)^2 > n^n$[/quote]

We can easy to check following inequality :
$ 1< k<n\minus{}1\rightarrow k(n\minus{}k\plus{}1)> n \Leftrightarrow (n\minus{}k)(k\minus{}1)>0$
So we have
$ (n!)^2\equal{}n^2((n\minus{}1)\cdot 2)\cdots(2\cdot(n\minus{}1)) >n^n$


abstract algebra
Ring Theory
superior algebra
superior algebra solved

Let $A$ be a noetherian ring. Prove that it is reduced iff it does not have embedded primes and every localisation at a minimal prime is a field.

Reduced=having no nilpotents
embedded primes= non-minimal prime ideals associated to $A$ or equivalently non-minimal prime ideals that appear as radicals of primary components of a reduced primary decomposition of $0$.

The only place where the fact that the ring is noetherian is used is in drawing the conclusion that every ideal, (and, in particular, the zero ideal) has a primary decomposition.

$\Rightarrow$

Suppose the ring is reduced. $(0)$ is then the intersection of the minimal prime ideals $\frak p_{i},\ i=\overline{1,t}$, and this is a minimal primary decomposition. Since the isolated components of a primary decomposition do not depend on the decomposition, every primary decomposition contains the minimal prime ideals, and hence can contain no other ideals; the conclusion is that $(0)=\bigcap_{i=1}^{t}\frak p_{i}$ is the only primary decomposition of $0$, so there are indeed no embedded primes.

The localization $A_{\frak p_{i}}$ is a ring with precisely one prime ideal, so it's a field iff it has no non-zero nilpotent elements. If $x\in A$ becomes nilpotent after localization, then $x^{n}$ must be annihilated by some element $y$ outside $\frak p_{i}$ and some $n$. But this means that $(xy)^{n}=0$ (in $A$), so $xy=0$, i.e. $x$ is annihilated by an element $y$ outside $\frak p_{i}$ and is thus zero after localization. This proves that $A_{\frak p_{i}}$ is a field.

$\Leftarrow$

Assume now that $(0)$ has no embedded primes and each localization $A_{\frak p_{i}}$ is a field. Assume also that we can find a non-zero nilpotent $x$. The annihilator of $x$ consists of zero divisors, so it's contained in the union of all primes belonging to $0$, i.e. the union of all minimal primes. An ideal contained in the union of finitely many prime ideals is actually contained in one of those ideals, so there is a $\frak p_{i}$ which contains $\mbox{Ann}~x$. On the other hand, $x$ must be zero after localization at $\frak p_{i}$, so it must be annihilated by an element outside $\frak p_{i}$. We have a contradiction.


I'll start off by saying "Hi", since this is my first post on this forum. :)

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

First here's some examples:

37789 x 3256 = 123040984
4326 x 456765347 = 1975966891122
5327 x 467 = 2487709

Ok, now here's the problem I am facing...  I would like to do the reverse of what I have shown above.  I start off with a large number and I would like to find 2 INTEGER numbers, which when multiplied together gives you the results of the starting number. (ie.  210950 = 8438 x 25)

Is there a way I can do this?  Btw, please explain in novice terms since I am new to learning math. :)

Or if anyone is kind enough to write a simple Windows, DOS, or Flash based program which would do that for me, that would be even better. :)  That way I could type in a large number and then press enter and then it lists many different combinations to choose from.  This would save me ALOT of time in the long run, since I need to do this with hundreds, maybe even thousands of numbers. :)

There is a BIG reason why I need to do this, but I wont go into the reasons why I need it since it would take too long to explain, and I would probably bore you all to death with my story, and my fingers would hurt badly from all the typing.  hehe :D

[quote="Saul"]This should be useful:
http://www.cryptographic.co.uk/factorise.html
It's a java applet. Instructions are on the page.[/quote]

Sorry, that is not very useful to me.  Thanks for sharing it anyway. :)

When I typed '543543' into that java applet and pressed the button it came up with this result:

543543 = 3.7.11.13.181
Which translates into:  3x7x11x13x181 = 543543

This is far too many numbers for me, since I only want multiplication by 2 integar numbers (ie. 33 x 16471 = 543543).  Btw, I completely fluked that answer and I didn't suddenly figure out how to do it.  lol :D

one way to get around that, if you have a calculator, is to just put all those numbers into two sets and multiply each set together. So for 543543 you could get (3.7.11) x (13.181) = 231 * 2353 = 543543.

Factoring an arbitrarily selected integer is a difficult problem that increases in difficulty RAPIDLY as the integer to be factored grows larger. Most computer algebra systems (e.g., Mathematica) have integer-factoring capabilities that do well up to a fairly large number of digits. BUT most public-key encryption schemes, such as the secure online connections used in online commerce, count on factoring large integers being so difficult that it is practically impossible in reasonable amounts of time. If someone finds a handy-dandy way to factor an integer of any moderate size rapidly, he will cause billions of dollars of changes in the world economy.

If you want a good list of factors.
1) Find the prime factorization of the number. (ie the prime factorization of 543543 is $3*7*11*13*181$)
2) Any combination of these numbers multiplied into 2 numbers will give you what you want, for example:
$3 * (7*11*13*181) = 3 * 181181 = 543543$
$(3 * 7) * (11 * 13 * 181) = 21 * 25883 = 543543$
$(3 * 7 * 11) * (13 * 181) = 231 * 2353 = 543543$
There are a lot more, but this gives you the general idea.

[quote]If someone finds a handy-dandy way to factor an integer of any moderate size rapidly, he will cause billions of dollars of changes in the world economy.[/quote]

Tokenadult,

About how large is 'moderate size'?

Well public key-private key uses some pretty large primes... a couple hundred digits long I think


Is it acceptable to take the AMC B in a state other than your own.  I'm going to be on vacation in Florida, but fortunately I'll be near one of the testing sites.  I took the AMC A in New York, by the way.

You can definitely take it at different schools, so I don't see why you couldn't in different states.

The main issue is that you generally have to take the AIME at the same place as where you took your qualifying AMC.  So if you didn't qualify for the AIME on the A date at your school, then take the AMC B elsewhere and qualify for the AIME, you'd have to take the AIME elsewhere too, which might not be so convenient if it's out-of-state.  I am definitely not sure about this policy, so I would advise contacting the AMC if you need further clarification.

Ok well I qualified for the AIME at my school (which is where I would be taking the AIME).  However, I did not do as well as I would have liked.  So I'm hoping to do better in Florida, in which case I would use that score as my AMC score.


A trapezium is inscribed in a circle of radius 3 units so that one base is a diameter of the circle and the other base has length 3 units. What is the perimeter of the trapezoid

A trapezium is a trapezoid, right???

If so, then the bases are $ 3$ and $ 6$.

The height (I think) must be half the radius, or $ 1.5$.

Then, use the Pythagorean Theorem.

$ 1.5^2 \plus{} 1.5^2 \equal{} c^2$

$ 4.5 \equal{} c^2$

$ \frac{45}{10} \equal{} c^2$

$ \frac{3\sqrt{5}}{\sqrt{10}} \equal{} c$

$ \frac{3\sqrt{50}}{10} \equal{} c$

$ \frac{15\sqrt{2}}{10} \equal{} c$

$ \frac{3}{2} \sqrt{2} \equal{} c$

Then, find the perimeter.

$ 3 \plus{} 6 \plus{} 2 (\frac{3}{2} \sqrt{2}) \equal{} p$

$ 3 \plus{} 6 \plus{} 3\sqrt{2} \equal{} p$

$ 9 \plus{} 3\sqrt{2} \equal{} p$

I think the height must be more than 1.5, because then three points of the circle would have to be collinear. (The place where one end of the diameter intersects the circle, the place where the corresponding end of the smaller base intersects the circle, and the "top" of the circle if the two bases were horizontal, and the diameter was "lower" than the other base.

 [geogebra]975b55e0aaca4a6d92bea5425120b5bc943637d2[/geogebra]

Uh... Repetam por favor :blush:   :blush:

Repeat what? And could you not speak in Spanish?


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topology
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I feel that most American's are not informed enough to vote smartly. I think that the government should propose that voters have certain IQ or higher. My 23 year old friend proposed that you earn the right to vote when you earn your college degree.

The reasons I feel most American's are not informed enough to vote is because so many people have the most essentric reasons for voting for a candidate. One person I asked said they are voting McCain because they think Obama is a terrorist. That is flat out the most obsurd thing I've ever heard. Another voting for Obama said that they want a president who won't die in office. I find that insensitive and just an excuse for "I don't have a reason".

I want to find out what others think about this.

[quote]I feel that most Americans are not informed enough to vote intelligently. The government should mandate that voters have certain IQ or higher. My friend proposed that you earn the right to vote when you earn your college degree.

The reason I feel most Americans are not informed enough to vote is because so many people have the most eccentric reasons for voting for a candidate. One person I asked said they are voting McCain because they think Obama is a terrorist. That is flat out the most absurd thing I've ever heard. Another voting for Obama said that they want a president who won't die in office. I find that insensitive and just an excuse for "I don't have a reason".

I want to find out what others think about this.[/quote]

bleh I felt like fixing up your post since if you are advocating a position which entails only smart people having a say in a democracy, it is usually prudent not to have your opening lines be tattered by various grammatical and rhetorical errors.

IQ is certainly not a reasonable standard for intelligence; nobody whom I consider intelligent or smart even knows what their IQ is. The college degree thing is less flawed.

I, for now, will refrain from giving any opinion on whether or not the fundamentals of this idea are sound; it is a tricky issue and a slight misstatement could destroy your position.

this is one of the worst ideas I have ever heard of and would create 1) corruption 2) ruin the middle class 3) a society in which your part in the country is decided the moment you are born (as in, you can't really change iq level)

horrible idea, once again

A counterargument: the [url=http://en.wikipedia.org/wiki/Voting_Rights_Act]Voting Rights Act[/url] of 1965. A college degree is in indicator of wealth more than anything else, and this would be blatantly discriminatory. I would much rather further strengthen the right to vote- there are far too many schemes aimed at preventing voting right now, particularly targeted at the poor and minorities.

Yes, some voters don't know much about the candidates or issues. In high-profile races, this is because they don't particularly [i]want[/i] to know; the campaigns and independent sources are easily available, but they've already formed an opinion and don't care to examine it. You can't stop that sort of thing.

What if people, before voting, had to take a short quiz on the candidate's stands on various issues? I think we could all agree that this is not discriminatory, as one should know what their candidate stands for before voting.

[quote="jmerry"]

Yes, some voters don't know much about the candidates or issues. In high-profile races, this is because they don't particularly [i]want[/i] to know; the campaigns and independent sources are easily available, but they've already formed an opinion and don't care to examine it. You can't stop that sort of thing.[/quote]

I think this is skirting the issue -- the whole idea is to expressly stop this sort of thing by considering alternative voting schemes if need be; to note that it can't be stopped according the current political picture merely reinforces the need to consider alternatives.

@travis this was my second idea, after I came up with the first idea of restricting voting to persons of suitable intelligence. It's prone to run into issues in implementation though, I think -- who will decide what questions go on there? The questions could very well be biased towards a particular political leaning or (worse) towards a particularly ethnic or socioeconomic group. It seems difficult to make up this quiz without intense political debate.

I think this is a terrible, terrible idea. This is America, land of the free. Trying telling George Washington that there's a law saying you can't vote if some guy decides that you're too dumb.

hey, turns out he would have agreed

see, for example, direct election of senators only happening with the progressive movement in the twentieth century (seventeenth amendment)

in fact, I think the original purpose of the electoral college was to ensure that those actually voting for the highest positions in the land (namely, the electors) would be respectable members of the community instead of just any old dude. this idea of letting the common man have a vote only began with jacksonian democracy (jackson is pretty much my least favorite president ever), well after the washington administration.

IRT OP: I think it's a good balance right now. Everyone under 18 are not allowed to vote, and people 18 or over can register to vote.

Most of the time, the ignorance is mainly from people under 18, which can't vote anyways. People 18 or older are mature enough to decide who to vote for. Even though you might not consider those valid reasons to not to vote for someone, they consider them valid reasons. Their vote counts just as much as yours does, because to them their reason is as valid as your reasons for voting is to you.

If we wanted absolute equal votes regardless of maturity, then we'd allow every single person to vote, including those younger than 18. Eventually we'd find day cares where they train little kids to vote for the "right candidate".

If we wanted votes weighted based on value, then we'd give the vote value to each person based on the share of the US they own. In other words, the more US gov stocks you buy, the more your vote counts. If anyone can get a hold of more than 50% of all the US stocks, then that person basically decides all the major decisions of the country (including who's president, etc). If that person can buy more than 2/3 of all the US shares, then he can have absolute authority. In addition to this, the poor would be selling all their shares because many of them would value money more than their share in the government.

If we were more theocratic, we could make it so that one has to acknowledge the authority of the almighty God before being allowed to vote. This can prevent all the self-righteous people from influencing the vote, but would also encourage corruption and mixture of religion into politics.

I think this 18 years or older limit is a good balance in the middle between equal value and preventing stupidity.

eh... we all know that politics is mostly first impressions and charisma, and not actual issues... I'm surprised no one mentioned that...

EDIT: Voting for people, that is. Voting for laws - now that's a different story.

[quote]
Most of the time, the ignorance is mainly from people under 18, which can't vote anyways.[/quote]

wtf

[quote]People 18 or older are mature enough to decide who to vote for. Even though you might not consider those valid reasons to not to vote for someone, they consider them valid reasons. Their vote counts just as much as yours does, because to them their reason is as valid as your reasons for voting is to you.[/quote]

so you accept "obama is a terrorist" as a valid reason for voting for a particular candidate? it's nice to think that everything is equal and whatnot but seriously, this is not the case. someone who wants to walk around saying 1 + 1 = 3 is being an idiot, regardless of his personal beliefs of why he feels that it is justified to him.

[quote="ProtestanT"]If we were more theocratic, we could make it so that one has to acknowledge ... God before being allowed to vote. This can prevent all the self-righteous people from influencing the vote[/quote]  Can someone explain to me what this is supposed to mean?

eh, all the random examples at the end of the post don't seem terribly relevant to the initial point; it seems better to let them go rather than distracting from the question at hand

Okay, fair enough.  Here's a question for the O.P.: suppose that we agree to your plan, but under the condition that [i]I[/i] get to choose what the relevant credentials for voting are.  How good of an idea do you still think this is?  (Thing to know about me: I'm vindictive and so I will almost certainly design a rule to exclude you and people like you.)*

Really, I think this question should be broken into two parts:
1) What properties must a person possess in order to make an informed voting decision?
2) Should voting be restricted to people who possess said properties?

So far in this thread, the answers "high IQ," "a college degree," "18 years of age" and "ability to pass a test about the positions of various candidates" have been proposed as answers to question 1.

* What's the point of this comment?  To help tease out the difference between, "I wish only us smart people got to have a say," and "I wish only them smart people got to have a say."

the obvious answer to 2 seems to be yes (in fact, ideally, we would restrict exactly to the set of people who make an informed voting decision, no more and no less), with the difficulty lying in answering question 1 (and suitably implementing this answer)

your question above is terrible. assuming a priori that the OP (or anyone posing the question) should not be in the group of people eligible to vote is as flawed as assuming the opposite; the answer to such a flawed question is going to be "no" simply because the question is flawed, not because you have cleverly managed to tease out the difference which you indicate.

On what standard can you justify that one reason is more valid than another? And if a standard is chosen, how can you justify that standard?

We don't always have to choose the person that is really the best for the job. We choose the person that [b]appears[/b] to be the best for the job. Rumors matter in the appearance whether you want them to or not.

If we want the real best candidate for the job, then it is way too important for a majority to decide (Why should we decide the best leader based on what the majority choose? The majority can be rebellious, greedy, selfish, etc. In fact, the best we can settle with is one that the majority likes the most, but isn't necessarily the best.). In other words, I don't trust the American people in picking the right president. However, I feel it's better if they pick a president that they want than if a right president is picked that the population doesn't want, and it turns out that it's a "wrong" president.

Take the Republican nomination for example. The majority (well...it's more of a plurality) chose John McCain. However, McCain is obviously not the best candidate from the Republicans. Probably many people can name at least 5 Republicans that are better than John McCain.
But we can agree that McCain is the candidate that the most Republicans support. In other words, if he's the president, he can be the president that the most Republicans are happy with. That's what's important here, because it's impossible for a majority to decide the best candidate.

So you sacrifice "future" for "happiness"...
I don't think this is a good idea (as it isn't in most other real cases, too).

And I wouldn't make let money decide anything, by this hard workers have much less potential than rich snobs ;) Additionally, a very amall group would probably hold 90% of the power without really doing the state any good.
The requirement should be something that costs at most 1 hour a week so that practically everyone who wants can afford it (necessary materials should be given by the state itself). Maybe reading some papers and doing some tests about their content...

[quote="Singularitarian"]What if all the people who are allowed to vote decide to enslave the people who aren't allowed to vote?

Have you ever heard that nightmare word, that word so sickening it should make you vomit----that word, "aristocracy"?[/quote]
It's already happening. Many people between ages 5 and 18 are already forced to go to school against their will. If they could vote, obviously a large amount of them would make it a right to choose not to go to school.

[quote="ProtestanT"][quote="ZetaX"]So you sacrifice "future" for "happiness"...
I don't think this is a good idea (as it isn't in most other real cases, too).[/quote]
No, because if the future is bad then in most cases the happiness is bad. The majority considers its future if it feels that it must be happy in the future.[/quote]
No they don't think about future; they just think about stupid promisses and speeches they heard; not more.
School is a good example why not everyone should be able to decide on everything: school _is_ necessary (believe it or not), but a high amount of those visiting it doesn't like school. They are just to "stupid" torealise that it is necessary. Now replace school by a general political decision and schoolars by uninformed voters...

Singularitarian: I don't get your point. Aristocracy was at no point based on any votings, nor would modern laws allow enslaving others (at least here as there are unchangeable fundamental laws). And at least my proposal would avoid this: practically everyone can get his right to vote, they just need to do something for it (note that rich people need to do the same amount of work and can't let it be done by minors!).

After centuries, even millenia, of gruesome tyranny, political thinkers finally discovered the idea of "equality before the law".  It was a radical idea at the time.  Revolutions were fought, tyrants were overthrown.  It was slow to catch on---African-Americans and women were not granted equality under the law in the U.S. until the 20th century---but civil rights activists worked tirelessly and dedicated their lives to this principle.

Was that all much ado about nothing?  Or have you failed to learn one of the primary lessons of history.

Humans are so bad at thinking about politics.  Someone can be a genius at math, literally a genius, and then they turn around and say they want to toss out equality before the law, introduce a large class of second-class citizens, and bring back aristocracy.

So the people who are allowed to vote won't be allowed to enslave the people who aren't allowed to vote?  And they won't be allowed to oppress them in any other repulsive ways?  What's going to stop them?  A constitution?  Is it the example of the United States which makes you think the constitution will be adhered to?  Because the United States government tossed the constitution out the window a long time ago.

Is the intelligence of the aristocrats supposed to imply that they won't choose to oppress the second-class citizens?  Please.  Don't project your own benevolence onto others.  History is filled with examples of smart people doing terrible things.

Let's say the test you have to pass to join the aristocracy is that you have to graduate from college.  Then the government is forcing people to go to college.  (Or else live as a second-class citizen.)  This could only be acceptable to someone who is really locked into the idea that every smart person has to go to college.  What if I don't want to go to college?  What if I think it's a waste of money and time?  What if I want to teach myself instead?  Would you point a gun at me and force me to go to college, or to jump through whatever other hoops pass for an "intelligence test"?  Would you have the government point the gun for you?

Let's say that someone isn't that smart, doesn't like to learn and isn't any good at it, but he loves his friends and family and so his plan for his life is to work at a gas station and when he's not working spend the rest of his time with his friends and family.  And let's say his common-sense political views are that he wants to be left alone, he doesn't want to be forced to do anything, he doesn't want the country to go to war, he doesn't want high taxes.  You don't have to be a genius to feel that way.  Well, here we have a wonderful person whose political views are more reasonable than those of most "intelligent" people, and yet under this aristocratic system he would be a second-class citizen, denied the right to vote.  Are his taxes about to be raised?  Nothing he can do about it.  Is his country about to initiate a war?  Nothing he can do about it.  Is there about to be a draft?  Nothing he can do about it.  There are a million ways he could be oppressed, and there is nothing he could do about it.

Once I am admitted into the aristocracy, I'm going to push for the return of Droit de seigneur.  Sure, most people think something as barbaric as Droit de seigneur could never return.  But hey, most people thought something as barbaric as aristocracy could never return either.

I don't understand what all this talk about an aristocracy is...  Anyone at any time could be come elligible to vote.  Any such test would have to be very well made.

Someone might understand they don't want to pay as much in taxes, but do they know what taxes are for?

I do agree with your point below, however:
[quote]A constitution? Is it the example of the United States which makes you think the constitution will be adhered to? Because the United States government tossed the constitution out the window a long time ago.[/quote]

If any such test was made, it and it's regulations would have to be made very carefully.  As you have said, it could easily backfire, although you added on that the idea was a barbaric one.

But surely you agree that the ultimate idea of preventing those who know ABSOLUTELY nothing about politics and aren't educated from voting?

while I'm not happy with some of the decisions that the current administration has made, I am skeptical as to how you conclude that we "threw out the constitution a long time ago". anyway, that is still largely off-topic.

your main assumption seems to lie with this "large class of second-class citizens". nobody's saying it has to be large. the expectation is that all people will be able to and hopefully will eventually obtain the privilege of voting.

okay then you blather on about the lessons of history; that's fine if you can relate it more directly to the current situation, which you do only somewhat tangentially. finally, isn't the government all about making people do things they don't normally want to do? or do all people just really want to go to school, and later in life, pay taxes? now, for the particular example you give, I do not personally find it necessary that all people have a college education, but whatever the ultimate "civic intelligence test" may prove to be, then yes, I am arguing that I would be in favor of a system which mandates this knowledge in order to vote.

[quote="alexhhmun"]But surely you agree that the ultimate idea of preventing those who know ABSOLUTELY nothing about politics and aren't educated from voting?[/quote]Why?  I asked this question before, as have others, but it's been either ignored or branded 'obvious'.  Why do you feel that someone needs to know about some range of political issues to have a valid reason to prefer one candidate over the other?

A few random examples that I feel illustrate this notion is flawed:
- A farmer without a high school education who knows nothing about politics except that one of the candidates supports farm subsidies and the other does not.  Though the farmer is, by any measure, uninformed about the political system in general, he knows exactly who he wants to vote for.  Should he be excluded?
- A single mother working two jobs to support her kids, who has no means or time to inform herself about the political system or the candidates in particular, and who only knows one thing about politics and the government that she heard from a local grassroots campaign: one candidate wants to raise the minimum wage and expand federal funds to send underprivileged kids to college, the other does not.  Should she be excluded?
- A strongly religious person without much education and little knowledge of politics is vehemently against abortion, as they feel it is murder.  They know one candidate supports federal enforcement of abortion rights and the other opposes it.  Should they be excluded?
- A 90 year old billionaire with Alzheimer's and no idea of current events may only want to know which of the candidates supports reducing the estate tax, even if he knows absolutely nothing else about them.  Should he be excluded?

I bet if you tried you can come up with many more profiles of such uninformed voters.  I assert furthermore that such examples could account for at least several million people (mostly from among the poor).  What rationale is there for excluding such people?  This brings me to what is, perhaps, the more significant question (which has also already been raised and ignored as if it were obvious):

By what specific mechanism do you feel excluding some people from the process will result in an improved outcome in the first place?  If this question cannot be answered concretely, the entire discussion seems moot.

If I think how plain stupid some people acted about 80 years ago when voting for the NSDAP (so practically for Hitler), but also for more modern kinds of radical nationalism...
They vote(d) for them, beacuse of simple speeches that told them that everything will get better with them (less foreigners, no unemployment, a good "german" (or french, polish, ...) nation, ...). If they would thought further or would take other sources into consideration (thinking about all claims and promisses is a part of consideration!), they would surely be more aware of the truth.

At the moment, the system supports those can make great speeches, can lie very good, make good-looking promisses, or just pick up on single themes that wouldn't be so important in general but can decide the election. Is this really how it's supposed to be¿ I think not! And this is where excluding uninformed people will cause an improvement. Note that in the way I define "informed", this is a state practically everyone could archieve; and note that thinking is a necessary step to get really informed, reading some stuff isn't.

@joe

It's a legitimate point as to whether or not these voters should be included or excluded. Certainly, at this point, I feel they should be included, but I could imagine a working political system that goes even further than I would and demands knowledge on at least a wider range of issues. (I think Hilbert made the point in another thread about the current election that even if one, say, opposes abortion rights, it would still be selfish folly to support the McCain/Palin ticket for solely that reason as there's a greater picture that needs to be considered. While such a political system, where everyone is forced to consider the grand effect of their political choices, seems ideal, I think that would be far, far more difficult to implement than the small step we're proposing here, so let's not deal with that at this point.)

So anyway, now that I've granted that a working knowledge of at least one point is sufficient to earn the privilege of voting, do you seriously propose to tell me that all voters in the upcoming election have this working knowledge of even one point? I very much doubt it as yes, I do mean a moderately complete knowledge of at least this one issue. For example, voting against Obama because you think he is going to raise taxes would be an example of something I would wish to eliminate -- the voter is legitimately concerned about the issue at hand, but nonetheless perpetuates, through his or her voting decision, a very basic misunderstanding of where exactly the candidates stand on the issues. Again, I'm not saying that a majority of voters are completely uneducated or anything of the sort as I still do believe that a large portion of the populace should and would retain the privilege of voting, but I nonetheless think certain restrictions are simply appropriate.

I have mixed feelings about how to express myself in a discussion like this.  On one hand, I have strong feelings about certain issues, and I want to make my perspective clear.  On the other hand, I recognize that I'm by no means an expert on politics, and I think it's important for me to admit that.

When I say humans are bad at thinking about politics, I might be the best example!  So, young AoPSers, I think you are excellent thinkers, and I hope one day you'll solve our world's political problems.  (Goodness knows they need to be solved.)

[quote="MysticTerminator"]So anyway, now that I've granted that a working knowledge of at least one point is sufficient to earn the privilege of voting, do you seriously propose to tell me that all voters in the upcoming election have this working knowledge of even one point?[/quote]Yes, I do believe that every person that voted had at least one well-formed reason to vote, otherwise I imagine they wouldn't have voted.

But let's say you're right; let's say there is a group of people who for some reason decide to vote even though they have no compelling reason to do so, and no knowledge of either candidate or any political issue whatsoever.  Let's also imagine that we've decided that taking away their vote will, somehow, improve the outcome of the election.  How do you propose to differentiate these people from those who do have a legitimate reason to vote, but little knowledge of other issues?  Surely a test with one or two dozen pre-defined questions cannot differentiate these two groups.

dude what, singularitarian ? you made a post saying you don't know what to post ? that almost seems a bit superfluous.

joe, yes. choosing the test would be difficult. I have never made any attempt to deny this. my primary purpose in the thread so far was to argue that this proposition is not as ludicrous and politically incorrect as it may in fact seem. my main reservation in discussing this particular idea was that it seems fundamentally contrary to american principles, and I anticipated spending far too long fending off such attacks and reexplaining why this is in fact not the case. anyway, I seem to have succeeded on some level -- you at least seem to be willing to grant that the proposal, at some weak level, may be valid. (I will admit also that in my own thoughts, I've spent far more time trying to figure out the basic question of validity and not quite that much to the next logical question of to what degree this principle may be valid, how to implement it, and so forth.)


calculus
integration
logarithms
calculus computations

Can anyone please help me asap in simplifying or estimating the following integral, as i require it in some physics problem:
$ \int_{0}^{a} \frac {t^{3}}{e^{t} \minus{} 1} dt$
where $ a$ is a constant, $ a \in \mathbb{R^{\plus{}}}$

I'm fairly sure that it can't be done using only elementary functions.

If you allow polylog then it might be doable in closed form.

Maybe the integral $ \int_0^{\infty} \frac {t^3}{e^t \minus{} 1}\ dt$ is related to statistical physics.
Can anyone have Landau Liphshiz's statistical physics?

hello, Maple 12 gives
$ \displaystyle{\minus{}\frac{1}{15}\pi^4\minus{}\frac{1}{4}a^4\plus{}a^3\ln(\minus{}e^a\plus{}1)\plus{}3a^2\text{polylog}(2,e^a)\minus{}6a\text{polylog}(3,e^a)\plus{}6\text{polylog}(4,e^a)}$.
Sonnhard.

[quote="Carcul"]That integral arises in the derivation of Stefan-Boltzmann's law.[/quote]

See this Appendix: [url]http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law[/url]

Yes indeed Stefan's Law was the context in which posed a problem. the actual problem was the following:
Prove that :
$ \frac{\int_{0}^{\lambda_{max}}\rho_{T}(\nu)d\nu}{\int_{0}^{\infty} \rho_{T}(\nu)d\nu} \approx 0.25$
where $ \rho$ is the  energy density of stnding waves in cavity radiation in frequency $ \nu$ and $ \nu \equal{} d\nu$

albeit a physics problem, i need help in the integration, which requires that integral.

hello, we have $ \int_{0}^\infty \frac{t^3}{exp(t)\minus{}1}dt\equal{}\frac{\pi^4}{15}$.
Sonnhard.

[quote="Euclidean Geometer"]Can someone please give me a numerical approximation?[/quote]

Did you see the link I showed?

[quote]Did you see the link I showed?
[/quote]
Yes sir i did. thanks.  :)


A subset of the integers 1,2,...,100 has the property that none of its members is 3 times another. What is the largest number of members such a subset can have?

$\text{(A)} \ 50 \qquad \text{(B)} \ 66 \qquad \text{(C)} \ 67 \qquad \text{(D)} \ 76 \qquad \text{(E)} \ 78$

Edited to make answer correct

--

Suppose $12 \le x \le 33$ is in a subset S that is largest.  Then $3x \notin S$.  We can construct $S'$ with $|S'| = |S|$ by deleting $x$ and adding $3x$.

After this procedure, we can have $12, \dots, 33$ empty.  A similar procedure with $x \in \{2,3\}$ shows that we can have $2,3$ empty.  Total answer 76.

[quote="Singular"]Suppose $1 \le x \le 33$ is in a subset S that is largest.  Then $3x \notin S$.  We can construct $S'$ with $|S'| = |S|$ by deleting $x$ and adding $3x$.

After this procedure, we necessarily have $1, \dots, 33$ empty.  Thus the largest is 67.[/quote]

Not quite. We can have $\{34, 35, \ldots, 100\} \cup \{1,2,4,5,7,8,9,10,11\}$ which gives $76$ and I believe that's the largest.

[hide]

Number of multiples of $3$ from $1-100$ is $33$.

Number of multiples of $9$ from $1-100$ is $11$.

Number of multiples of $27$ from $1-100$ is $3$.

Number of multiples of $81$ from $1-100$ is $1$.

Basically, we have to subtract all multiples of $3$ because $3=1*3$ and so on. So, we have $100-33=67$.

However, we then realize that multiples of $9$'s are okay because they are $3$ times the multiples of $3$. Since we took out the multiples of $3$'s, we can include multiples of $9$. So, we have $67+11=78$.

Then we realize that the multiples of $27$'s are not okay because they are $3$ times a multiple of $9$. Therefore, we have to subtract them out. So, we have $78-3=75$.

Then, we realize that we have accounted for everything except for $81$, which is $3$ times a multiple of $27$. Since we just ruled out all the multiples of $27$, we can safely add that back in to get $75+1=76\rightarrow \boxed{D}$[/hide]

[quote="paladin8"][quote="pkerichang"]but it's not in the answer choice...(I got 75, too)[/quote]

Uhh... mine was just a typo actually. If you add $67+9 = 76$ that should be right...[/quote]

how can I get $67 + 9 = 75$??!!?? :wallbash_red:  :wallbash:

[hide]
Therefore are a total of 67 numbers without the factor 3.  to add on more numbers, notice that if a number has more than two factors of 3 (or has a factor of 9), then it'll be three times a multiple of 3, which is not counted yet.  These numbers are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, and 99.  Notice that 27 and 54 are invalid because they are multiple of some numbers in the list mentioned.  Therefore adding 9 numbers to 67, we get $76$
[/hide]


geometry
3D geometry
tetrahedron
algebra
polynomial
congruent triangles

I'm at MOP, and I came up with this wonderful proof. Some people enjoyed it here, so I decided to post it. I hope that's okay...

I'm not sure if I'm allowed to post the problem, but it involves making a tetrahedron and deciding whether or not all the faces have to be congruent.

Here's my answer:

All four triangles must be congruent. We will prove this using a well-known children's poem.

First, we will label the edges
[img]http://img14.photobucket.com/albums/v43/picounts/mopdiagram.jpg[/img]

Six edges, all tame
were really bored, and feeling lame.
One said "Hey, I have a game;
let's go around and become the same."

Edge P was about to die,
so he said with a sigh,
"Who to choose? My oh my!
I'll be congruent to edge Y."

Next came the edge L.
And when he looked, as far as he could tell,
being equal to E would be pretty swell.
E said, "No! Go to... Nebraska!"

L kept looking, as one might guess.
He wanted the edge that would be the best.
And so the edge looked to the west.
"I'll be congruent to the edge S!"

The next edge was as ready as can be.
Edge E was shrieking wth glee.
And with a smile, he said to T,
"Come, let's have congruency."

All the edges looked around.
They were so happy, they almost frowned
"We have broken new ground!
Four congruent triangles! This will astound!"

This proof hdn't taken a while.
It's not a real proof, not within a mile.
But hopefully, it got a smile.
What can I say? I'm just going for style.

um, tim? the answer was "no, the triangles do not all need to be congruent"

I know, but that wouldn't have made as good a poem, and I didn't know that at the time.

ah...i wonder if anyone actually solved that one...stupid tests

ahh...we're all getting pwned...

if you changed the y to an a, then the tetrahedron would be named "staple"! sorry for being off topic, i just always notice weird stuff

[quote="picounts"]Stuff.  You should all be glad I'm not actually putting it in the quote.

-Magnara[/quote]

That was beautiful.

[quote=picounts]Here's one I came up with on the last test. Po-Ru gave me a .9 style for it. The problem was asking if there existed a polynomial with certain properties. Here's what I came up with: (And as a note: Josh is putting together a database of all the fake proofs submitted to him at MOP '04)

No such polynomial exists. We will begin with a lemma.

LEMMA: There exists a Babel fish such that when one sticks it in one's ear, one can immeadiately understand and speak any language in the universe (yes, Douglas Adams did come up with this)

PROOF: Assume such a fish does not exist. Our assumption implies that anyone who understands a language must have bothered to take time out of their schedule to learn it. But this implies that people would actually WANT to speak French, a contradiction.
[]

Assume such a polynomial exists. Such a polynomial could only be created by
 A. God, or
 B. Po-Ru

However, such a polynomial could only have been created by a supergenious, but the school newspaper, "The Insolent," said the following on Page 8 on April 1, 2003:
[img]http://img14.photobucket.com/albums/v43/picounts/mostlikelynot2.jpg[/img]
(For those of you who can''t see the picture, it says: "Most likely NOT to be a supergenious: Po-Ru Loh. Explanation: If he's been guessing all along, his luck has got to run out sometime.")

Therefore, such a polynomial could only have been created by God. Therefore, if we can prove that God does not exist, we are done. We will use a method similar to the one popularized by Douglas Adams in <u>The Hitchhiker's Guide to the Galaxy</u>. There's only one good way to prove the non-existance of God; we have to go have a talk with him. Since Po-Ru's not a supergenious anymore, he has lots of spare time, so he might as well come along. And since Misha likes to follow people, he'll come along too. So pack your bags, it's gonna be a long journey to the land of God. [Graders note: "Oh Boy!"]
>
>>
>>>
>>>>
>>>>>
>>>>
>>>
>>
>
Ah, here we are. O Our journey was four centimeters (I told you it would be a long trip) (yes, even God uses the metric system). Unfortunately, Misha didn't survive the journey. I'm not going to provide the gruesome details, but thw short story is that Misha learned the hard way that pi does not equal four. [Grader's note: "Uh-oh"] Luckily, Po-Ru made it, and he is talking to God right now. Actually, they're using AIM, but that's not the point. Let's view their conversation:
[img]http://img14.photobucket.com/albums/v43/picounts/aimwithgod.jpg[/img]
[b][color=red]p_o_r_u:[/color][/b] from the lemma, there exists a babel fish
[b][color=blue]iamgod2004:[/color][/b] yea
[b][color=red]p_o_r_u:[/color][/b] such a creature could not have been created coincidentally
[b][color=blue]iamgod2004:[/color][/b] tru
[b][color=red]p_o_r_u:[/color][/b] therefore, only u could make n e thing like that
[b][color=blue]iamgod2004:[/color][/b] wtf?
[b][color=red]p_o_r_u:[/color][/b] if god duznt make it, who duz?
[b][color=blue]iamgod2004:[/color][/b] o
[b][color=red]p_o_r_u:[/color][/b] b4 i go on, tell me, would u ever let humans kno u existed?
[b][color=blue]iamgod2004:[/color][/b] well, i'm talking to u, rn't i?
[b][color=red]p_o_r_u:[/color][/b] I'm not human; I'm poru
[b][color=blue]iamgod2004:[/color][/b] o yea
[b][color=red]p_o_r_u:[/color][/b] so tell me, y don't u let ppl kno of ur existance?
[b][color=blue]iamgod2004:[/color][/b] cuz proof denies faith, n w/o faith, i m nuthing
[b][color=red]p_o_r_u:[/color][/b] but teh babel fish is a dead giveaway of ur existance. Therefore, by ur own logic, u dont exist.
[b][color=red]p_o_r_u:[/color][/b] QED
[b][color=blue]iamgod2004:[/color][/b] omg
[b][color=blue]iamgod2004:[/color][/b] I hadn't thought of that
[b][color=blue]iamgod2004:[/color][/b] *disappears in a puff of logic*
[b][color=blue]iamgod2004 has signed off... permanently[/color][/b]
[b][color=red]p_o_r_u:[/color][/b] MWAHAHAHAHAHAHA
[b][color=blue]iamgod2004 appears to be offline and may not reply.[/color][/b]
--------------------------------------------

Now we know that Po-Ru is not a supergenious, and God doesn't exist. Therefore, both are incapable of creating the desired polynomial. A contradiction. Therefore, no such polynomial exists.
Q.E.D.

EDIT: Sorry about making the page scroll horizontally; I don't know how to fix that :([/quote]

I think this is from Hitchhikers Guide to the Galaxy

"RT-2PM2 Topol-M Silo-based intercontinental ballistic missile hits streak"


A sum is formed by alternately adding and subtracting consecutive odd integers starting with 1 and ending with 1997 as indicated. What is the sum?\\
$ 1 \minus{} 3 \plus{} 5 \minus{} 7 \plus{} 9 \minus{} 11 \plus{} \ldots \plus{} 1993 \minus{} 1995 \plus{} 1997$

We notice that the sum of the first two consecutive terms is -2.
1995 is the $ \frac {1996}{2} \equal{} 998^{th}$ odd number.
$ \frac {998}{2} \equal{} 499^{th}$ -2.
So the sum of the terms up to 1995 is $ 499 * \minus{} 2 \equal{} \minus{} 998$.
Adding this to 1997 gives $ \boxed{999}$.


The symbol $ 25_b$ represents a two-digit number in the base $ b$.  If the number $ 52_b$ is double the number $ 25_b$, then $ b$ is:

$ \textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12$

[hide="Solution"]
$ 25_b \equal{} 2b \plus{} 5\\
52_b \equal{} 5b \plus{} 2$

Using the information from the problem, we have:

$ 4b \plus{} 10 \equal{} 5b \plus{} 2\\
b \equal{} 8$

Thus, $ b \equal{} \fbox{8 (B)}$.
[/hide]

[hide="Solution"]$ 2(2b\plus{}5)\equal{}5b\plus{}2\implies 4b\plus{}10\equal{}5b\plus{}2\implies b\equal{}\boxed{\textbf{(B)}\ 8}$.[/hide]


algebra
polynomial
quadratics
Rational Root Theorem
complex numbers

Factor the following [b][i]WITHOUT[/i][/b] "Guess and Check".  Show your method.
 $\displaystyle{3x^2+14x+8}$

P.S.  I can do this!  :D
        The problems can get a lot harder...if you want.

if the discriminant for this problem is positive then it can be factored

you could also try the quadratic formula

um, factors of $24(=3x8)$ that add up to $14$ are $2,12$ so
$3x^2+14x+8=3x^2+12x+2x+8+3x(x+4)+2(x+4)=(3x+2)(x+4)$
There are much harder eqtns to factor

[quote="spx2"]now ask yourself why cant you factor $x^2+1$ ?[/quote]
It has no real roots.  :)

Hoshattack, you said you had harder problems? :?  :D

more carefull lotrgreengrapes7926
i asked why doesnt it have real roots(why cant you factor it?) over the field of reals
(not complex).

By rational root theorem the only possible real roots are $\pm1$.  Since $\pm1$ obviously don't work, $x^2+1$ has no rational roots.

[quote="nayrb"]The discriminant of x^2 + 1 is negative. Thus, there will be a complex number and its conjugate as the roots.[/quote]

Actually, it's a double root of $i$.

[quote="Jhonny Ruiz"]No. Nayrb is right.

[hide]$x^2+1$ is not a perfect square binomial for complex numbers.
I think I already solved this before...
$x^2+1=$
$(x+i)(x-1)=$
$x= \pm i$

or of course...
$x=\frac{\pm \sqrt{-4}}{2}=$
$x=\frac{\pm 2i}{2}$
$x= \pm i$[/hide]

[hide] $i$ is the double root of...
$x^2+2ix-1$[/hide][/quote]

Sorry, I wasn't thinking straight....  :oops:
Also, I believe it should be
$(x+i)(x-i)$, not $(x+i)(x-1)$.

Well because b=0 in the trinomial x^2 + 1, and the discriminant is negative, that means that both roots will be pure imaginary numbers, each the opposite of each other.

[quote="nayrb"]Well because b=0 in the trinomial x^2 + 1[/quote]...

That should not be trinomial, but quadratic.

[quote]
Sorry, I wasn't thinking straight.... Embarassed
Also, I believe it should be
(x+i)(x-i), not (x+i)(x-1).[/quote]

I'm sorry... really dumb mistake...

$x^2+1=(x+i)(x-i)$
$x= \pm i$

In general, if we want to factor something in the form
$a^2+b^2=c$ we can use complex numbers to say...

$a^2+b^2=c$
$(a+bi)(a-bi)=c$

just to clarify, when you say "can factor" any polynomial [i]can[/i] be factored when allowing complex numbers (and knowing the roots), however you might have learned in class that things like $x^2+1$ cannot be factored because you are only considering real numbers, i find it is good to clarify if you are supposed to factor over rational, integer, real, complex, etc

100% agree with that altheman
Thanks for the correction


trigonometry
trig identities
tangent half angle

How do you derive the $\tan$ half angle formula from the  $\sin$ and $\cos$ half angle formulas?

What do you mean? Using $\sin^2\frac12\theta=\frac12(1-\cos\theta)$ and $\cos^2\frac12\theta=\frac12(1+\cos\theta)$ to find $\tan\frac12\theta$?

But the answer is right there by my post #2 in this topic... Just divide $\sin^2\frac12\theta$ by $\cos^2\frac12\theta$ and then take square root and you get your proof for post #3

Better yet, if you don't want square roots, you can do the following: after you get tangent formula in post #3, multiple both top and bottom by $1-\cos x$. Top part is a square so you can take it out of the square root. Bottom is easy, $1-\cos^2x=\sin^2x$ and you can take it out of square root too.

Another formula is if you multiply both top and bottom by $1+\cos x$.

Thanks for the proof (the other formulas were in my book too. Somehow I could derive the other formulas from the tangent half angle formula, but I couldn't derive the tangent half angle formula from the sin and cos half angle formula.)


Let $ ABC$ be an acute triangle. Prove that: \[ \sum_{cyc} \left(\frac{\sin(A)\sin(B)}{\sin(C)}\right)^2 \ge \frac{9}{4}\]

[quote="night^wish"]Let $ ABC$ be an acute triangle. Prove that:
\[ \sum_{cyc} \left(\frac {\sin(A)\sin(B)}{\sin(C)}\right)^2 \ge \frac {9}{4}
\]
[/quote]
You can see my attach file:

If $ A\plus{}B\plus{}C\equal{}90^\circ$, prove that \[ \frac{\sin A.\sin B}{\cos C}\plus{}\frac{\sin B.\sin C}{\cos A}\plus{}\frac{\sin C.\sin A}{\cos B}<1\]


As it says above, I'm looking for a good self-study modern algebra text.  I've heard that Artin's book is the standard on this particular subject, but I've also heard that it is not the clearest when it comes to SELF-study.

Does anyone have other suggestions?

Artin's book is good, but I personally dont think it's a good book for an initial introduction. It doesn't spend enough time with some of the topics that will give you basic grounding in the subject, like Sylow's Theorem. I would recommend [u]Topics in Algebra[/u] by I.N. Herstein for a good basic introduction, or [u]Abstract Algebra[/u] by Dummit and Foote for a more comprehensive (and modern) view.

My undergrad book was:

Basic Abstract Algebra by P. B. Bhattacharya, S. K. Jain, and S. R. Nagpaul

which I enjoyed immensely.  However, I did take algebra with Jain himself, so I'm not sure how this book goes when you're completely on your own.

I found [url=http://www.math.uiuc.edu/~r-ash/Algebra.html]this book[/url] online which I thought was quite readable, especially with its "Enrichment" section, but I don't know how it compares to other texts.

i'm using topics in algebra by herstein, i like it a lot

I read a large portion of Topics in Algebra (2nd edition) and Algebra (3rd edition); however, I did not like the fact that it does not cover enough topics.  So I switched to Abstract Algebra (Dummit and Foote).  The only problem with "Abstract Algebra" is that the authors are sometimes prone to verbiage.

Dummit and Foote is very good, and it's a good text to own to keep as a reference (what doesn't it have in there?)


If two standard 6 sided dice are rolled indefinitely and the top faces are summed each time what is the probability of rolling two 7's before a 6 and an 8?

Wow, seriously? :|

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202582

[color=blue]Mods, please lock this.[/color]


AMC
AIME
geometry
trapezoid
3D geometry
tetrahedron
prism

anyone know a good solution for 1983 #11?

i dunno how to post the diagrams so ill just link everyone

http://www.kalva.demon.co.uk/aime/aime83.html

Here's how I'd do it:

Split it into 3 chunks, like this: take two planes, perpendicular to the square and going through BC and AD. Figure out the distance from EF to the square by making another plane, this time through EF and perpendicular to the square. It should be a isoseceles trapezoid, that you know all 4 sides of.

I got that the volume of the middle "chunk" is 216, gonna work on the tetrahedrons after dinner.

Tetrahedrons should be 1/3rd of the triangular prism that they're in. Volume of the prism is 108, so they should be 36 each.

216+36+36=288.

[quote="Iversonfan2005"][quote="d343seven"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46266

This is the same problem posted a while ago.

All you have to do is "complete" the figure into a triangular prism by adding a pyramid onto both sides.[/quote]

Surprisingly this is number 11  ;)[/quote]

Look at the year.... it was 1983, wthe time when the AIME problems were actually considered one of the easiest of all time ;)


induction
function
algebra
functional equation
number theory
relatively prime
algebra unsolved

Solve this functional equation

-$ f : N$---->$ N$
-$ f(2) \equal{} 2$
-$ f(mn) \equal{} f(m).f(n)$   for all m and n [b]relatively prime[/b]
-f is increasing

Claim: $ \forall n: f(n)\equal{}n$

Proof: Let $ n$ be minimal with $ f(n)\not\equal{} n$. Clearly, $ n>2$ and $ \forall N\geq n: f(N)\geq N\plus{}1$ holds, becaue of the "increasing"-property. If $ n$ is even, then it must be a power of $ 2$, otherwise we can represent it as a product of proper co-prime factors smaller than $ n$, yielding the contradiction $ f(n)\equal{}n$. Since the even number $ n\plus{}2$ is not divisible by $ n,n\plus{}1$, it must be a power of $ 2$ as well, but this gives the contradiction $ n\equal{}2$. So, $ n$ must be odd and the even number $ n\plus{}1$ must be a power of $ 2$. Since $ n\plus{}3$ cannot be a power of $ 2$, we can write it as $ n\plus{}3\equal{}2^tm$ with $ m\geq 3$, odd. Since $ 2^t<n$, the other factor must fullfill $ m\geq n\geq 2m\minus{}3$ to avoid a contradiction, i.e. $ n\equal{}3$.

However, $ 4f(5)\equal{}2f(10)>2f(9)\equal{}f(18)>f(15)\equal{}f(3)f(5)$, implying $ f(3)<4$, a contradiction.

[quote="hjbrasch"]implying $ f(3) < 4$, a contradiction.[/quote]

$ f(3) < 4$ is not a contradiction,

Actually I've proved that: $ f(3)\equal{}3$

I have an incomplete solution

I proved:
$ f(3)\equal{}3$
$ f(2^n\plus{}1)\equal{}2^n\plus{}1$ for every $ n$

Now we should prove $ f(n)\equal{}n$ by induction
Wich is possible but I dont know how to do it
Some one help please.

see the link
Hint
First prove that
$ f(n)\equal{}n$ for all compose number.
From $ f(n)$ is monotone then
$ f(p\minus{}1)<f(p)<f(p\plus{}1)$

[quote="TTsphn"]http://www.mathlinks.ro/Forum/viewtopic.php?t=167412
see here .[/quote]

Its not it.
m and n must be prime to say $ f(mn)\equal{}f(m)f(n)$
and its a big difference  :wink:

[quote="TTsphn"]compose number.
[/quote]

Do you mean the opposite of prime?

I recorder you result $ f(2^n \plus{} 1) \equal{} 2^{n }\plus{} 1$
$ f(3) \equal{} 3,f(5) \equal{} 5,f(4) \equal{} 4$
Now we induction
$ f(n) \equal{} n,\forall n$
Suppose chose for n-1 we prove for $ n$
Case 1: $ n \equal{} pq$ where $ \gcd(p,q) \equal{} 1$
Then $ f(n) \equal{} f(p)f(q) \equal{} n$
Case 2 :$ n \equal{} p^a$ ($ a\in N$)
If $ p \equal{} 2$ then $ f(n \plus{} 1) \equal{} n \plus{} 1$ so $ f(n) \equal{} n$
 If n not is a power of 2 then
$ n \plus{} 1 \equal{} p^a \plus{} 1 \equal{} 2^{k}q$ where $ q\equiv 1(mod2)$
then easy to check $ f(n) \equal{} f(2^k)f(q) \equal{} n$
So $ f(n) \equal{} n$

I have an other method.
If $ m>n$
$ f(m)\equal{}m,f(n)\equal{}n$ then $ f(k)\equal{}k,\forall k\in [n,m]$
Proof:
Consider $ k\in (n,m)$ then $ n<f(k)<m$ so
$ f(n\plus{}1),..,f(m\minus{}1)$ is a permutation of $ (n\plus{}1,..,m\minus{}1)$
Because $ f(n)$ is monotone then $ f(k)\equal{}k,\forall k\in[n,m]$

try to solve a more general problem : find all increasing multiplicative functions $ f: \mathbb{N} \to \mathbb{R}$ .it has been posted befor but i want other solutions  :D

short answer to FOURRIER's claim on my post

$ f(3)<4$ is a contradiction, since the assumption of a counterexample to $ f\not\equal{}id_{\mathbb{N}}$ has led to the necessary condition $ f(3)\geq 4$

Two remarks after reading some of the other proofs:

1. It was sometimes overlooked, that in the induction step/minimal counterexample choice $ f(n)\not\equal{}n$ the integer $ n$ might have the form $ p^m$ and its successor the form $ q^k$ for primes $ p,q$ making it not immediately accessible to a reduction/induction argument.

2. The wrong assumption was made at some stage that one can always find co-prime values $ a,b<n$ with $ ab>n$ (this argument was used to conclude $ f(n\minus{}1)\equal{}n\minus{}1,f(ab)\equal{}ab\Longrightarrow f(n)\equal{}n$ due to the monotonic behavior). This is only true for $ n>3$ (actually, showing $ f(3)\equal{}3$ is the critical step in the proof)

let $ f(3) \equal{} a$,we claim that $ a \equal{} 3$

let $ f(3) \equal{} a$ thus:

$ f(5)\geq a \plus{} 2$

now we know that $ f(15) \equal{} f(5).f(3) \equal{} a.f(5)\geq a^2 \plus{} 2a$

and we know that $ f(18)\geq f(15) \plus{} 3 \equal{} a^2 \plus{} 2a \plus{} 3$ (I)

in the other hand $ f(6) \equal{} f(2).f(3) \equal{} 2a$ thus:

$ f(5) < 2a\Rightarrow f(5)\leq 2a \minus{} 1$

$ \Rightarrow f(10) \equal{} f(5)f(2) \equal{} 2f(5)\leq 4a \minus{} 2$

$ \Rightarrow f(9)\leq 4a \minus{} 3$

$ \Rightarrow f(18) \equal{} f(9)f(2) \equal{} 2f(9)\leq 8a \minus{} 3$ (II)

now from (I),(II) we get that:

$ a^2 \plus{} 2a \plus{} 3\leq 8a \minus{} 6\Rightarrow a^2 \minus{} 6a \plus{} 9\leq 0\Rightarrow (a \minus{} 3)^2\leq 0\Rightarrow a \equal{} 3$

so our claim is proved and $ f(3) \equal{} 3$

now let $ p_k$ be the $ k^{th}$ prime number,and assume that $ f(p_k) \equal{} p_k$ we want to show that $ f(p_{k \plus{} 1}) \equal{} p_{k \plus{} 1}$...

first of all we know that $ p_{k \plus{} 1} < 2p_k$,and hence $ f$ is increasing we get that $ f(2p_k)\geq f(p_{k \plus{} 1}) \plus{} 2p_k \minus{} p_{k \plus{} 1}$ (*)...

also we know that:

$ f(2p_k) \equal{} f(2)f(p_k) \equal{} 2p_k$

so according to (*):

$ f(p_{k \plus{} 1}) \plus{} 2p_k \minus{} p_{k \plus{} 1}\leq f(2p_k) \equal{} 2p_k$

$ \Rightarrow f(p_{k \plus{} 1}) \minus{} p_{k \plus{} 1}\leq 0\Rightarrow f(p_{k \plus{} 1}) \equal{} p_{k \plus{} 1}$

as wanted...

so for every prime number $ p$ we have $ f(p) \equal{} p$

now we know that for natural number $ n$ there exists a prime number $ p$ s.t $ p > n$ now according to the fact that $ f$ is increasing we get that:

$ f(p)\geq f(n) \plus{} p \minus{} n$

$ \Rightarrow p\geq f(n) \plus{} p \minus{} n$

$ \Rightarrow f(n) \equal{} n$


algebra
difference of squares
special factorizations

The factors of the expression $ x^2\plus{}y^2$ are:
$ \textbf{(A)}\ (x\plus{}y)(x\minus{}y) \qquad\textbf{(B)}\ (x\plus{}y)^2 \qquad\textbf{(C)}\ (x^{\frac{2}{3}}\plus{}y^{\frac{2}{3}})(x^{\frac{4}{3}}\plus{}y^{\frac{4}{3}}) \\
\textbf{(D)}\  (x\plus{}iy)(x\minus{}iy) \qquad\textbf{(E)}\ \text{none of these}$

$ x^2\plus{}y\plus{}2$ can't be factored without imaginary roots, which rules out choices $ A$, $ B$, and $ C$. Choice $ \fbox{D}$ is the correct answer since the difference of squares theorem gives $ (x\plus{}iy)(x\minus{}iy)\equal{}x^2\minus{}(\sqrt{\minus{}1}^2y)\equal{}x^2\plus{}y^2$.

I agree , I think the answer is $ \boxed D$


real analysis
induction
real analysis unsolved

Is the Lebesgue sigma-algebra (on [0,1], say) countably generated ?
(the sigma-algebra of Lebesgue-measurable sets is meant)

The answer is supposed to be "no" , but i can't understand how to prove it.
A possible way is to consider the Cantor set (which is Lebesgue-measurable, but is not countable).
It supposed to be a "counterexample". But how?

Also, there should be another way to prove it. Does anyone have an idea?

Thanks in advance for any help/ideas.  :blush:

The cantor set has cardinality $ \mathfrak{c}$, so that there are (at least, and hence exactly) $ 2^{\mathfrak{c}}$ lebesgue measurable sets.  Can you do something with this?

Hmm.. Then we need to prove that  if  a sigma-algebra is countably generated, then we should have only countably many measurable sets.

Say A is a countably generated sigma-algebra with S - a countable set that generates A.
Then A consists of
sets in S (countable),
sets like A\x (where x is in S) (countable),
finite union of sets of S and of the described above (countable),
countable (or finite) union or intersection (and "A\x") of sets described above (countable),
countable (or finite) union or intersection (and "A\x") of sets described above (countable),
...
, right?

Should we apply induction here?  :juggle:
(or maybe it's much easier?)

Thank you very much!

Not quite -- for example, the Borel sigma-algebra is countably generated (I think, I'm not 100% sure what your definition of countably generated is!), but has cardinality $ \mathfrak{c}$!

That's a pretty big hint, I guess.

Ah! I've managed confuse a couple of things there..
Would the following work?

C (Cantor set) is Lebesgue measurable (also Borel, right?), with measure zero, and is of cardinality $ 2^{\aleph _0}$. Any subset of C is Lebesgue measurable (with measure zero), hence there are at least $ 2^{2^{\aleph _0}}$ Lebesgue measurable sets (and hence exactly). (As you said before, and i didn't get that)

Now, if Lebesgue sigma-algebra was countably generated, there could be only $ 2^{\aleph_0}$ in it.
Which is a contradiction.

---
The definition in use: A sigma-algebra is called countably generated, if there exists a finite or countable set that generates it.
---

[quote="kar"]C (Cantor set) is Lebesgue measurable (also Borel, right?), with measure zero, and is of cardinality $ 2^{\aleph _0}$. Any subset of C is Lebesgue measurable (with measure zero), hence there are at least $ 2^{2^{\aleph _0}}$ Lebesgue measurable sets (and hence exactly). (As you said before, and i didn't get that)

Now, if Lebesgue sigma-algebra was countably generated, there could be only $ 2^{\aleph_0}$ in it.[/quote]

Yeah this is right, although if this is an assignment, I would prove the last statement.  The Cantor set is indeed Borel measurable (it's closed), but one can find subsets of it which are not Borel measurable (recall that the Borel measure is not complete).

Ok, I'll try to think of a proof for this statement. (It's not an assignment, but it doesn't matter)

Thanks for your help !


[hide]
$ \text{Im}((9 \plus{} bi)^{3}) \equal{} \text{Im}((9 \plus{} bi)^{2})$
$ \text{Im}(729 \plus{} 243bi \minus{} 81b^{2} \minus{} b^{3}) \equal{} \text{Im}(81 \minus{} b^{2} \plus{} 18bi)$
$ 243b \minus{} b^{3} \equal{} 18b$
$ 243 \minus{} b^{2} \equal{} 18$
$ 225 \equal{} b^{2}$
$ b \equal{} \pm 15$
Since we are told that b is positive, our answer is $ \boxed{15}$[/hide]

woah only one reply? wow.

[hide=Solution]
We use binomial theorem to calculate the following.

$\mathrm{Im}((9+bi)^2)=\binom{2}{1}9b=18b$.

$\mathrm{Im}((9+bi)^3)=\binom{3}{1}9^2b-\binom{3}{3}b^3=243b-b^3$.

We equate these to get $18b=243b-b^3$.

We know that $b>0$ so we can divide by $b$ on both sides to get $18=243-b^2$.

We simplify to get $b^2=225$.

We know that $b>0$ so $b=\boxed{015}$ is the only solution.
[/hide]

[hide= Sol]We know that $(9+bi)^2 = 81 - b^2 + 18bi.$
Then $(9+bi)^3$ is equal to $729 -27b^{2} + 243bi - b^3i$.  If we set the imaginary parts equal to each other, we get $18bi = 243bi - b^3i$.
Divide everything by $bi$ and we have $18 = 243 - b^2$.
So we have $b^2 = 225$ and we want the positive root so $b = 15$[/hide]

I wish 2021 AIME I was as easy as this test...

Also I think this was an Alcumus problem which I read wrong and answered -15 too :(

[hide=Solution]
If the imaginary parts of $z^2$ and $z^3$ are equal, then we simply apply the operations on $9+bi=z$. Squaring gets $(9+bi)(9+bi)=(81-b^2)+(18b)i$. We also get that cubing gives $(729-27b^2)+(-b^3+243b)i$. Therefore, we see that $18b=-b^3+243b$. This simplifies to $b^3-243b+18b=0$. This gets us that $b(b^2-225)=b(b-15)(b+15)=0$. We get that $b=15$ as $b$ is a positive number.

This was AIME? a bit surprised at first. Remembered doing something like this in IA alcumus and it was like level 5 IA level 2 precalc lol

Sweet and short AIME Problem:

[hide = Solution]$\text{Im}(z^2) = \text{Im}({z^3}) \Rightarrow \text{Im}[(9 + bi)^2] = \text{Im}[(9 + bi)^3] \Rightarrow \text{Im}[(81 - b^2) + 18bi] = \text{Im}[(729 - 27b^2) + (-b^3 + 243b)i] \Rightarrow 18b = -b^3 + 243b \Rightarrow b^3 - 225b = 0 \Rightarrow b(b^2 - 225) = 0 \Rightarrow b = 0, b = \pm 15.$ But, since $b$ is a positive real number, the only possible option is $b = \boxed{15}.$[/hide]

$(9+bi)^2=18bi$, so the imgainiry, part is $18b.$ Now, $(9+bi)^3$, has imigniary part 243b-b^3, hence, $18b=243b-b^3 \implies 18=243-b^2 \implies b^2=225 \implies b=\pm 15$, since $b,$ is positive, hence, $b=\boxed{15}.$

[quote=mathmax12][quote=lpieleanu]@above, $(9+bi)^2=18bi$ is not true.

[hide=Solution]
Note that $$(9+bi)^2=(91-b^2)+18bi,$$ $$(9+bi)^3 = (729-27b^2)+(243b-b^3)i,$$ so we have
\begin{align*}
18b=243b-b^3 &\implies b^3-225b = 0 \\
&\implies b(b-15)(b+15) =0\\
&\implies b=-15, 0, \text{or } 15.
\end{align*}
Since we are given that $b$ is positive, the answer is $$\boxed{015}.$$ $\square$[/hide][/quote]

i mean that the imiginary part is 18bi[/quote]

Okay, but still, the equation isn't true, so I just wanted to point that out.

[quote=ChromeRaptor777][quote=mysmartmouth]The complex number $z$ is equal to $9+bi$, where $b$ is a positive real number and $i^{2}=-1$.  Given that the imaginary parts of $z^{2}$ and $z^{3}$ are equal, find $b$.[/quote]

you can pretty much just bash it out[/quote]

[quote=v_Enhance]
Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a [color=#f00]full solution[/color] to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. [color=#f00]We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.[/color]
[/quote]


function
algorithm
absolute value
algebra proposed
algebra

Prove that every bijective function $ f: \mathbb{Z}\rightarrow\mathbb{Z}$ can be written in the way $ f\equal{}u\plus{}v$ where $ u,v: \mathbb{Z}\rightarrow\mathbb{Z}$ are bijective functions.

I vaguely remember having seen this problem somewhere else, but not vividly enough to prevent me spending a while finding the solution.

We can wlog let $ f(n)\equal{}n$ (by permuting the arguments of u and v accordingly).

I claim the following algorithm generates bijections u and v summing to identity, where I make both odd functions (u(-x)=-u(x) etc) and u(0)=v(0)=0.

For k, starting at k=0 and continuing upwards, let $ u(2k\plus{}1)\equal{}\minus{}m, v(2k\plus{}2)\equal{}\minus{}l$ where $ m$ is the least positive integer such that $ u(x)$ does not already contain $ m$ in its image between $ \minus{}2k$ and $ 2k$, and similar for $ v,l$, and correspondingly let $ v(2k\plus{}1)\equal{}2k\plus{}1\plus{}m, u(2k\plus{}2)\equal{}2k\plus{}2\plus{}l$

Suppose this algorithm fails to generate a bijection. It clearly can exclude no integer from its image, because it specifically demands inclusion of all integers not yet in its image, so it will only break if injectivity fails. Suppose k is the least integer for which we obtain u(x)=u(y) or v(x)=v(y) for |x|,|y|<=2(k+1).

If u(x)=u(y) then 2k+2+l = u(z) for |z|<2k+2. z must be odd, for obvious reasons (if it's even, l hasn't been chosen correctly). This would imply that the image of u(x) as x varies on -(|z|-1),|z|-1 must contain all the positive integers between 1 and 2k+2+l. This is clearly absurd, since |z|-1<2k+1<2k+2+l.

We get similar contradictions in the v(x)=v(y) case, and thus conclude that u,v are both bijections, as required.

I think we can do even less work.
Draw a table containing three rows.  The first row contains the integers (in, say, the order 0,1,-1,2,-2,...), while the second and third rows shall contain the values of u(n) and v(n) respectively (n being the value in the first row).
We repeat the following process to fill in all the columns:

i)look at the first column which we have not yet filled, with first row n.  Choose u(n) and v(n) to have absolute value larger than anything we have yet chosen so that they sum to f(n).

ii)look at the first number, k, which we have not yet chosen as a value u(i).  Choose n such that f(n)-k has not yet been chosen as a value for v, and fill in the n-column accordingly.

iii)repeat ii with u and v reversed.

Cycling through i, ii and iii generates bijections u, v with u+v=f

Jack

Here is a slightly more explicit construction:
Note that we only need to consider $f$ as the identity function by a suitable permutation
Now consider, $$U=[-2,1,-3,0,3,-1,2]^n$$ $$V=[-1,-3,2,0,-2,3,1]^n$$
(By $[\dots]^n$ we mean placing copies of $[\dots]^{n-1}$ each shifted by a corresponding entry of $[\dots]\cdot|[\dots]^{n-1}|$. For example, $$[-2,1,-3,0,3,-1,2]^2=[-16,-13,-17,-14,-11,-15,-12,5,8,4,7,10,6,9,-23,-20,-24,-21,-18,-22,-19,-2,1,-3,0,3,-1,2,19,22,18,21,24,20,23,-9,-6,-10,-7,-4,-8,-5,12,15,11,14,17,13,16]$$ Centering both $U$ and $V$ at $0$ will result in two bijections that sum to the identity.

[quote=vjdjmathaddict][size=100]is this trivial :maybe:  simply define v=f(f)+f and u=-f(f).[/size][/quote]

Why is $f(f)+f$ bijective?


integration
trigonometry
calculus
calculus computations

1. $\int \tan^{-1}\left(\frac1x\right) dx$

2. $\int e^{\sqrt{x}} dx$

These antiderivatives shouldn't be hard to find, but somehow I got stuck :(

Can anyone help?
(Sorry if these are not appropriate for this level.)

For 1:$\int \tan ^ {-1} \left(\frac{1}{x}\right) dx=\int (x)' \tan ^ {-1}\left(\frac{1}{x}\right) dx$

$=x\tan ^ {-1} \left(\frac{1}{x}\right) - \int x\left(\tan ^ {-1}\left(\frac{1}{x}\right)\right)' dx.$

For 2:Let $\sqrt{x}=t,$ we have $x=t^2\Longrightarrow dx=2tdt.$ Thus

$\int e^{\sqrt{x}}dx=\int e^t\cdot 2tdt=2\int te^t dt.$ :)

Alternate viewpoint on #1: $\tan^{-1}\left(\frac1x\right)=\cot^{-1}(x)=\frac{\pi}2 -\tan^{-1}(x).$

You still have to do an integration by parts similar to what kunny showed you; this use of the trigonometric identities is meant mostly to clean things up so you can see and think more clearly.


search
modular arithmetic
number theory
Diophantine equation

A confused bank teller transposed the dollars and cents when he cashed a check for Ms Smith, giving her dollars instead of cents and cents instead of dollars.  After buying a newspaper for 50 cents, Ms Smith noticed that she had left exactly three times as much as the original check.  What was the amount of the check?

There is no way this is Pre-Olympiad.   :huh:

Unless I'm misunderstanding the question or something.

Let $ d$ be the amount of dollars and $ c$ be the amount of cents.

$ 100c\plus{}d\minus{}50\equal{}3(100d\plus{}c)$.
$ 97c\minus{}299d\minus{}50\equal{}0$

The solutions to the above Diophantine equation are:
$ c \equal{} 56 \plus{} 299 t
d \equal{} 18 \plus{} 97 t$

However, we note that $ c$ must be less than 100.  $ d$ has no limit.

Now I'm a bit stuck.   :huh:

[quote="AdithyaGanesh"]$ 97c \minus{} 299d \minus{} 50 \equal{} 0$[/quote]
$ d\equal{}\frac{97c\minus{}50}{299}$ where $ 0\le c<100$.
Search for an integer, c, in $ 97c\minus{}50\equal{}0\pmod{299}$.

[hide="Solution"]We have that $ 100c\plus{}d\minus{}50\equal{}3(100d\plus{}c)$, so $ 97c \minus{} 299d \equal{} 50$. We have that $ c < 100$ and $ d < 100$. $ c \equal{} 56$, $ d \equal{} 18$. So, the original check had a value of $ \boxed{\$18.56}$.

[/hide]


Εστω οι τυχαίοι πραγματικοί $ a_1,a_2,..., a_n$. Να αποδείξετε ότι υπάρχει πραγματικός $ x$ ώστε καθένας απ' τους αριθμούς: $ x \plus{} a_1, x \plus{} a_2,..., x \plus{} a_n$ να έιναι άρρητος.

[hide]Pigeonhole principle![/hide]

ΥΓ. Νομίζω ότι το αποτέλεσμα γενικεύεται και για άπειρο πλήθος όρων, αλλά η απόδειξή του ξεφεύγει απ' τα πλαίσια των Ολυμπιάδων

[quote="NickNafplio"]Λοιπον. Επηδη μεχρι στιγμης εχω διαβασει ελαχιστα πραγματα απο Συνδιαστικη. Θα προσπαθησω να βαλω μια λυση που μπορει και να ξεφευγει απο το επιπεδο των (σχολικων) ολυμπιαδων και η οποια βασιζεται σε... Θεωρια Συνολων

[hide]Θεωρω το συνολο $ A_i, i \equal{} 1,2,...n$ το οποιο περιεχει της $ n \minus{}$αδες $ \{(x \plus{} a_1, x \plus{} a_2, ..., x \plus{} a_n), x \in R\}$ για τις οποιες το στοιχειο $ x \plus{} a_i$ ειναι ρητος, προφανος υπαρχει μια 1-1 και επι αντιστοιχια απο το υποσυνολο των ρητων αριθμων $ K_i \equal{} \{a_i \plus{} x | a_i \plus{} x \in Q\}$ στο $ A_i$, αρα το $ A_i$ ειναι ισοδυναμο με ενα υποσυνολο των ρητων αριθμων και επομενος ειναι αριθμισημο συνολο. Επηδη ομως το x παιρνει ολες τις πραγματικες τιμες, το συνολο ολων των $ n \minus{}$αδων $ X_n \equal{} \{(x \plus{} a_1, x \plus{} a_2, ..., x \plus{} a_n), x \in R \}$ ειναι ισοδυναμο με το συνολο των πραγματικων αριθμων το οποιο ειναι υπεραριθμισημο. Αν ισχυει το ζητουμενο τοτε θα ειχαμε: $ \cup_{i \equal{} 1}^{n}{A_i} \equal{} X_n$ το οποιο ειναι ατοπο αφου το $ X_n$ ειναι υπεραριθμισημο συνολο ενω η ενωση πεπερασμενου (και γενικα αριθμισημου) πληθους αριθμισημων συνολων ειναι παντα αριθμισημο συνολο.

Παρατηρηση: Απο αυτη τη λυση φαινεται οτι η ασκηση μπορει να γενικευθει και για απειρο πληθος αριθμων $ a_i$, αρκει το πληθος τους να ειναι αριθμισημο.[/hide]

ΥΓ: Επηδη δεν βλεπω πως μπορει αυτη η ασκηση να λυθει με αρχη Dirichlet, mim οταν μπορεσεις βαλε και τη δικη σου λυση...[/quote]

Νίκο ωραία η λύση σου. Με αρχή Dirichlet, πάρε ένα άρρητο x, και δες τι γίνεται αν προσθέτεις τα x,2x,3x,...

Αυτο που παρατηρησα αμεσως ειναι οτι υπαρχει μοναδικο μη μηδενικο πολλαπλασιο $ kx$ του $ x$ ωστε το $ a_i \plus{} kx$ να ειναι ρητος, απο οπου προκειπτει αμεσως το ζητουμενο :p

Να προσθεσω κατι ακομα για τη λυση που εβαλα πιο πανω, αν εχουμε απειρους ορους οι οποιοι δινονται απο μια ακολουθια $ a_1, a_2, ...$ τοτε το πληθος τους ειναι προφανος αριθμισημο αρα γι αυτους ισχυει το ζητουμενο.

[quote="NickNafplio"][quote="Demetres"][quote="NickNafplio"]Αυτο που παρατηρησα αμεσως ειναι οτι υπαρχει μοναδικο μη μηδενικο πολλαπλασιο $ kx$ του $ x$ ωστε το $ a_i \plus{} kx$ να ειναι ρητος, απο οπου προκειπτει αμεσως το ζητουμενο :p
[/quote]

Νίκο, μπορείς να εξηγήσεις λίγο περισσότερο την λύση σου;[/quote]

Αν εχεις 2 ρητους, $ a_k \plus{} nx$, $ a_k \plus{} mx$, $ m,n \in N$... τοτε η διαφορα τους $ (m \minus{} n)x$ ειναι ρητος. Αρα και ο $ x$ ειναι ρητος, ατοπο
Αρα για πεπερασμενο πληθος αριθμων $ a_i$ υπαρχει πεπερασμενο πληθος ακεραιων πολλαπλασιων του $ x$ τα οποια οταν τα προσθεσουμε στους $ a_i$ παιρνουμε τουλαχιστον εναν ρητο αριθμο.
Αυτη δεν ηταν και η δικη σου λυση?[/quote]

Ναι έτσι το έκανα. Επειδή όμως στο προηγούμενο σου post έγραψες $ a_i \plus{} kx$ αντί $ kx$ ήθελα να σιγουρευτώ ότι αυτή ήταν και η δική σου λύση.

[quote="bilot4"]
ΥΓ. Νομίζω ότι το αποτέλεσμα γενικεύεται και για άπειρο πλήθος όρων, αλλά η απόδειξή του ξεφεύγει απ' τα πλαίσια των Ολυμπιάδων[/quote]

Σωστός για αριθμήσιμο πλήθος όρων (βασικά αφού είναι σε ακολουθία είναι και αριθμήσιμο το πλήθος τους):

πρόχειρα βγαίνει, αφού (πληθικοί αριθμοί) $ |N|<|R|$ και άρα Dirichlet.

Note: Αριθμήσιμο σύνολο είναι αυτό που έχει 'ίδιο' πλήθος στοιχείων με αυτό των φυσικών, δηλαδή μπορεί να απεικονιστεί αμφιμονοσήμαντα επί των φυσικών.


Να βρεθει μια ακολουθια (εκτος απο την τετριμμενη) που η γραφικη της παρασταση,δηλαδη τα σημεια $ (n,a_n)$,
να ειναι συμμετρικη ως προς την ευθεια $ y\equal{}x$.

Για να δούμε

$ a_n\equal{}n\plus{}1$,n άρτιος και
$ a_n\equal{}n\minus{}1$,n περιττός.
Και νομίζω ότι δουλεύει

Σωστος :!:
Δεν ειμαι σιγουρος οτι δεν υπαρχουν αλλες. Μπορουμε να το δειξουμε?

@mast:
[hide]τελικα εισαι must[/hide]

[quote="AlexandrosG"]Σωστος :!:
Δεν ειμαι σιγουρος οτι δεν υπαρχουν αλλες. Μπορουμε να το δειξουμε?
[/quote]

Σίγουρα υπάρχουν και άλλες. Π.χ. $ a_1 \equal{} 3, a_3 \equal{} 1$ και $ a_n \equal{} n$ για $ n \neq 1,3$.

thx ρε φίλε. Και εσύ είσαι πολύ καλός μαθηματικός. Τα λέμε την ΣΕΜΦΕ


Anyone know how to prove the Kepler' second law (the one about the planetary motion)?

[hide="hint"]
since the force due to sun is central the angular momentun of the planets about the sun is conserved .also make use of the relation that the angular momentum equals the areal velocity of an object about a pont[/hide]

The involves the use of the Principle of Conservation of Angular Momentum.

Angular momentum $ \equal{} L \equal{} mvr \equal{} mr^{2}w \equal{} mr^{2}\frac{ d\theta}{dt}$ where ($ w \equal{}$ angular velocity)

Area swept out in time $ dt \equal{}\frac{1}{2}r^{2}d\theta \equal{}\frac{1}{2}\frac{L}{m}dt$

So, if m is constant (which it is for a particuar planet), then for constant time intervals ($ dt$), the area swept out must be the same since L is constant.


Let us call a sum of integers cool if the ﬁrst and last terms are $ 1$ and each term
differs from its neighbours by at most $ 1$. For example, the sum $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 3 \plus{}
2 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 2 \plus{} 3 \plus{} 3 \plus{} 2 \plus{} 1$ is cool.

How many terms does it take to write $ 2008$ as a cool sum if we use no more terms
than necessary?

[hide="Solution"]

I think a symmetric increasing sum would work best. The biggest such sum under 2008 is $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} ... \plus{} 43 \plus{} 44 \plus{} 44\plus{} 43 \plus{} ... \plus{} 1 \equal{} 1980$. Add a 28 in there somewhere, and I get a minimum of 89 terms.
[/hide]

Really? For $ 1\plus{}2\plus{}3\plus{}...\plus{}63\equal{}2016$ decrease the last 8 terms in the arithmetic series by 1 to get a sum of 2008 with 63 terms... (not sure how to prove if it is the minimum)

[quote="kurt.math"]Let us call a sum of integers cool if the [b]ﬁrst and last terms are $ 1$[/b] and each term
differs from its neighbours by at most $ 1$.[/quote]


search
parameterization
algebra
polynomial
algebra theorems

[i][b]What are the discriminants?[/b][/i]They were mentioned(well,actually they were the weapon for solving the problem) in the solution 3 of the problem 4,round 14,International Mathematical talent search,but i cant write it here cuz i don't know the LaTex.

If I am not mistaking the the discriminant is $\vartriangle=b^{2}-4ac$ for a second degree ecuation... $ax^{2}+bx+c=0$. But maybe I didn't understand the question..  :blush:

You're not mistaken :) .By looking at the solution in view of your reply,i realised you were right,but there's sth i don't understand.In the solution of my problem it's stated:

"[i]a(ax + y) + 1 = a2x + ay + 1 must be a perfect square, hence its discriminant, y2  4x, should vanish[/i]"

[b]Why should the disriminant vanish??[/b] :maybe:

Because you would be left with a root of a non-square, making the whole thing be irrational.

Why y 2 -4x is a non square?if it's a square(a perfect one)then it's not necessarey to have the discriminant vanished.

There should be something else in the problem. Could you post the solution or at least the problem so that we can know exactly what x,y are and what parameters they meet?

it's about de whole ecuation being a perfect square.....not about the discriminant!that's why the discriminant should be 0!

What they meant is that if you want any cuadratic
$ax^2+bx+c=0$
To be a perfect square, both roots must be equal, or what is the same, the discriminant $b^2-4ac$ must vanish.

To first guy, discriminant of a polynomial is defined by products of its roots, it is used to determine realness of the roots of a polynomail.

Bomb