I am trying to format some text to match an equipment front panel ( for a user guide) and cannot seem to get the formatting to do what I want.

We are using \sffamily to get sans serif for the whole doc and I tried

...as indicated on the front panel \sffamily{\textsc{Test 1}} etc etc...

but the text comes out in serif font small caps - not sans serif.

Ideally, I would really like to be able to use Technics Bold TrueType font in Small Caps for this but have no idea how to do it. I have read the LateX Companion but am none the wiser. In general it would be great to know how to properly substitue fonts for LaTex.

I am a new convert to LaTex from Word and have only been going 1 week. A fragment is shown below (with irrelevant stuff stripped out)

all help gratefully received.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% HEADER
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\documentclass[a5paper,landscape,twoside,12pt]{book}
% a5paper is 148mm x 210mm

\usepackage[bindingoffset=8mm,inner=20mm,top=22mm,hmarginratio=1:1,vmarginratio=1:1]{geometry}
\usepackage{graphicx}
\usepackage{color}
\usepackage[sf,bf,compact,topmarks,calcwidth]{titlesec} %VT
\usepackage{nextpage}
\usepackage{multicol}
\usepackage{wrapfig}
\usepackage{fancyhdr}
\usepackage[font=sf]{caption}
\usepackage{amssymb}
\usepackage{makeidx}

\pagestyle{fancy}
\fancyhf{} %delete the current section for header and footer

\newcommand\FrontPanel[1]{\sffamily{\textsc{#1}}}

\begin{document}
\sffamily

All connections to the control console are made via the rear panel as shown in figure~\ref{fig:ConsoleRearPanel}. The principal connection to the rest of the RETB system is via the \FrontPanel{Control Rack} 37-way D connector.

\cleardoublepage
\printindex
\end{document}

If you look at the log (which was easy to produce thanks to the minimal example you gave) you will see
[quote]LaTeX Font Warning: Font shape `OT1/cmss/m/sc' in size <12> not available
(Font)              Font shape `OT1/cmr/m/sc' tried instead on input line 27.[/quote]which means small caps is not available in Computer Modern Sans Serif so it has been replaced by Roman (ie Serif).
You may be able to simulate small caps with \small \uppercase.

Of course choosing a font which has small caps would be best. Are you using MiKTeX? If so, there is a program called [url=http://mtfi.sourceforge.net/]MTFI - MiKTeX TTF Font Installer[/url] which makes installing true type fonts (such as Technics) for LaTeX easy but this may not have small caps either.

I see that [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=kpfonts]kpfonts[/url] says that:
[quote]Sans-serif fonts are available with upright, small caps and slanted shape, all with regular, bold and bold extended weight;[/quote]

Ok thanks once again!

I tried installing MTFI but it doesn't seem to work immediately with MikTex 2.6 ( there is a post about this on SourceForge).

I tried kpfonts and it worked immediately. It also changed my main font slightly but all-in-all good progress - thanks.

I also now get the idea of running a small example by ityself and checking the log - this may cut down on my 'newbie' questions!

[quote="pdjclark"]I tried installing MTFI but it doesn't seem to work immediately with MikTex 2.6 ( there is a post about this on SourceForge).[/quote]MTFI is just an automated version of [url=http://www.radamir.com/tex/ttf-tex.htm]Using TrueType fonts with TeX (LaTeX) and pdfTeX (pdfLaTeX)[/url] so you could still install Technics if you wanted to.

[url=http://groups.google.co.uk/group/comp.text.tex/browse_thread/thread/414c37d843954509/6d1b0a70f581353b]Report: How to install and to use TTF fonts in LaTeX (MiKTeX 2.4) - JUST IN 4 STEPS[/url] claims you must add another step (Step 2) to prevent rasterisation in pdfs and recommends [url=http://c.caignaert.free.fr/Install-ttf-Font.pdf]Step by step, installing a ttf-font for PdfTeX[/url] for its clarity.


for p=5,7,11,13,17,23,
C(2p,p)-2=0 (mod p^3)

investigate this "pattern"


What place are you in? I am in west coast.

So the west coast is the middle of nowhere.

I live in an area where many people would be offended by the second line of mathelete's signature.


Βρείτε όλες τις συναρτήσεις που ικανοποιούν τη σχέση

 $ f : \mathbb{R} \mapsto \mathbb{R}$ such that $ f(x \plus{} y)f(f(x) \minus{} y) \equal{} xf(x) \minus{} yf(y)$ for all $ \ x,y \in \mathbb{R}.$

Cute :)

[hide]
$ y\equal{}f(x)\minus{}x$
[/hide]

Cheerio,

Durandal 1707

για $ y\equal{}0$, $ f(x)f(f(x))\equal{}xf(x) \Leftrightarrow f(x)\equal{}0$ ή $ f(x)\equal{}x$

που ικανοποιούν τη δοθείσα σχέση

Βαζω αλλη μια πιο ευκολη.
Εστω $ f: R\to R$ μια παραγωγισιμη συναρτηση για την οποια ισχυει

$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\plus{}3xy(x\plus{}y)$ για καθε $ x,y\in R$

και $ f'(0)\equal{}0$

Να βρειτε τον τυπο της $ f$

Eυκολα φαινεται οτι [hide]$ f(x)\equal{}x^3$[/hide] ομως με ενδιαφερει πολυ να δω την λυση.

[quote="maniopas"]για $ y \equal{} 0$, $ f(x)f(f(x)) \equal{} xf(x) \Leftrightarrow f(x) \equal{} 0$ ή $ f(x) \equal{} x$

που ικανοποιούν τη δοθείσα σχέση[/quote]

Αν ήταν τόσο εύκολη θα την έβαζα στο δημοτικό  :P  Γενικά όταν έχεις $ f(x)g(x)\equal{}0$ τότε δεν συνεπάγεται ότι
$ f(x)\equal{}0$ για κάθε x ανήκει R ή $ g(x)\equal{}0$ για κάθε x ανήκει R

Μπορεί να έχεις ας πούμε $ f(x)\equal{}0$ για $ x>0$ και $ g(x)\equal{}0$ για $ x\leq 0$

@mailo : Καλό θα ήταν να την έβαζες σε ξεχωριστό τόπικ (δεν νομίζω να κολάει εδώ δεδομένου ότι δεν έχει αποδειχθεί ακόμα η προταθείσα :wink: )

$ \lim_{y\to0}\frac{f(x\plus{}y)\minus{}f(x)}{y}\equal{}\lim_{y\to0}\frac{f(y)}{y}\plus{}\lim_{y\to0}3x(x\plus{}y)$ ή
$ f'(x)\equal{}f'(0)\plus{}3x^2\equal{}3x^2$ άρα $ f(x)\equal{}x^3\plus{}c$ και επειδή $ f(0)\equal{}0$ που το έχουμε λάβει υπ' όψη μας και στο $ f'(0)$ προκύπτει ότι $ f(x)\equal{}x^3$
Δεν χρειάζεσαι την παραγωγισιμότητα παρά μόνον στο 0
Τα όρια "έσπασαν" επειδή και τα δυό τους υπάρχουν και είναι πραγματικοί αριθμοί

[quote="silouan"][quote="maniopas"]για $ y \equal{} 0$, $ f(x)f(f(x)) \equal{} xf(x) \Leftrightarrow f(x) \equal{} 0$ ή $ f(x) \equal{} x$

που ικανοποιούν τη δοθείσα σχέση[/quote]

Αν ήταν τόσο εύκολη θα την έβαζα στο δημοτικό  :P  Γενικά όταν έχεις $ f(x)g(x) \equal{} 0$ τότε δεν συνεπάγεται ότι
$ f(x) \equal{} 0$ για κάθε x ανήκει R ή $ g(x) \equal{} 0$ για κάθε x ανήκει R

Μπορεί να έχεις ας πούμε $ f(x) \equal{} 0$ για $ x > 0$ και $ g(x) \equal{} 0$ για $ x\leq 0$ [/quote]

Καταλαβαίνω αυτό που λες. Αλλά δεν λέω αυτό. Το έγραψα λάθος λόγω ταχύτητας  :blush:  ,  αλλά εννοούσα ότι $ f(x) \equal{}${$ 0$ or $ x$}.  Αλλά αυτό  θέλει συνλέχεια γιατί δεν τελειώνει εδώ το πρόβλημα...

[quote="Nick Rapanos"][quote="r_boris"]$ \lim_{y\to0}\frac {f(x \plus{} y) \minus{} f(x)}{y} \equal{} \lim_{y\to0}\frac {f(y)}{y} \plus{} \lim_{y\to0}3x(x \plus{} y)$ ή
$ f'(x) \equal{} f'(0) \plus{} 3x^2 \equal{} 3x^2$ άρα $ f(x) \equal{} x^3 \plus{} c$ και επειδή $ f(0) \equal{} 0$ που το έχουμε λάβει υπ' όψη μας και στο $ f'(0)$ προκύπτει ότι $ f(x) \equal{} x^3$
Δεν χρειάζεσαι την παραγωγισιμότητα παρά μόνον στο 0
Τα όρια "έσπασαν" επειδή και τα δυό τους υπάρχουν και είναι πραγματικοί αριθμοί[/quote]

Ναι αλλα έχετε αποδείξει ότι η συνάρτηση είναι πραγωγίσιμη στο 0 ?[/quote]
Aφού δίνει ότι $ f'(0)\equal{}0$ όχι μόνο είναι παραγωγίσιμη στο 0 αλλά και η τιμή της παραγώγου εκεί    δίνεται ίση με το 0

[quote="r_boris"][quote="Nick Rapanos"][quote="r_boris"]$ \lim_{y\to0}\frac {f(x \plus{} y) \minus{} f(x)}{y} \equal{} \lim_{y\to0}\frac {f(y)}{y} \plus{} \lim_{y\to0}3x(x \plus{} y)$ ή
$ f'(x) \equal{} f'(0) \plus{} 3x^2 \equal{} 3x^2$ άρα $ f(x) \equal{} x^3 \plus{} c$ και επειδή $ f(0) \equal{} 0$ που το έχουμε λάβει υπ' όψη μας και στο $ f'(0)$ προκύπτει ότι $ f(x) \equal{} x^3$
Δεν χρειάζεσαι την παραγωγισιμότητα παρά μόνον στο 0
Τα όρια "έσπασαν" επειδή και τα δυό τους υπάρχουν και είναι πραγματικοί αριθμοί[/quote]

Ναι αλλα έχετε αποδείξει ότι η συνάρτηση είναι πραγωγίσιμη στο 0 ?[/quote]
Aφού δίνει ότι $ f'(0) \equal{} 0$ όχι μόνο είναι παραγωγίσιμη στο 0 αλλά και η τιμή της παραγώγου εκεί    δίνεται ίση με το 0[/quote]

Ωχχχχ, χίλια συγγνώμη κ. Μπόρη. Νόμιζα ότι αναφέρεστε στην άσκηση του Σιλουανού... :blush:

Βάζω μια λύση στην άσκηση του silouan. Για λόγους ευκολίας τη γράφω στα αγγλικά.  :wink:

Set $ y \equal{} 0$ in the given equation and get $ f(x) \cdot f(f(x)) \equal{} x \cdot f(x) \ (*)$.

Therefore there exists at least one $ c \in R$ so that $ f(c) \equal{} 0$.

Now plug in $ y \equal{} f(x)$ to get according to (*) that $ f(0) \cdot f(f(x) \plus{} x) \equal{} 0$

and then $ y \equal{} \minus{} x$ to get $ f(0)f(f(x) \plus{} x) \equal{} x(f(x) \plus{} f( \minus{} x))$ and thus

$ f( \minus{} x) \equal{} \minus{} f(x)$, for every $ x \neq 0$.

Plugging in now $ x \equal{} c$ and $ y \neq 0$ we get $ f(y) \cdot (f(y \plus{} c) \minus{} y) \equal{} 0$.

As a result for every $ x \neq 0$ we have either $ f(x) \equal{} 0$ or $ f(x \plus{} c) \equal{} x$.

Let those $ x \neq 0$ satisfying the first equation form set $ A$ and the remaining $ x \neq 0$ form set $ B$.

We distinguish cases:

1. $ A$ is empty. Then we must have $ c \equal{} 0$ and therefore $ f(x) \equal{} x,\ \forall x \in R$.

2. $ A$ is non-empty. If $ B$ is empty then $ f$ is identically zero everywhere except for its value at $ 0$ which can be arbitrary.

Suppose now that $ B$ is non-empty. We will reach a contradiction.

Pick $ a \in A$ and $ b \in B$ and plug in $ x \equal{} a \plus{} b$ in (*) to get

$ f(b) \equal{} b\plus{}a$.  Therefore it is obvious that $ A$ has a single element $ a$. Then $ f(x)\equal{}x\plus{}a, \forall x \neq 0,a$

and hence as $ f$ is odd in $ R*$ we obtain $ a\equal{}0$ which is a contradiction.


Let $f(x)$ and $g(x)$ be polynomials with non-negative coefficients, and let $m$ be the largest coefficient of $f$. Suppose that there exist natural numbers $a<b$ such that $f(a)=g(a)$ and $f(b)=g(b)$. Show that if $b>m$ ,then $f=g$.

Let deg(f) = n and deg(g) = m, I can only show that n>=m, can anyone help me show that n=m?

oops, only now did I realize that we've already used m. So I will clarify the variables again:

deg(f)=n, deg(g)=m, the biggest coefficient of f is k. Positive integers a,b are still defined the same as above.

Suppose n<m, or, m>=n+1. Since every coefficient of f, c, satisfies 0 <= c <= k < b. Hence c<=b-1 for all coefficients of f. Then f(b) <= b^(n+1) - 1. However, g(b) >= b^m >= b^(n+1) > f(b). Contradiction.  :D


trigonometry
Trigonometric Equations
algebra
equation
IMO
IMO 1962

Solve the equation $\cos^2{x}+\cos^2{2x}+\cos^2{3x}=1$

I think we can use the fact that :
cos²x = (cos2x+1)/2

 :)

I will post my solution as soon as possible!

Some ugly trigonometric computation later :

$\cos^2{x}+\cos^2{2x}+\cos^2{3x} -1 = 2cos(x)^2 \times (2cos(x)^2-1) \times (4cos(x)^2-3)$

Is-it really IMO ?

Yes, this was an IMO question, the problems were trivial in the good old days :D

[quote="tµtµ"]Some ugly trigonometric computation
[/quote]

computation?

In fact, I was thinking about :
(cos2x)² = (cos4x + 1)/2
(cos3x)² = (cos6x + 1)/2

So, if we sum :
cos2x + cos4x + cos6x = -1
<==> -2cos2x + 2cos²2x + 4cos^3(2x) = 0
We can factor by 2cos2x, and this gives the solutions :)

Am I misreasoning?

By the double- and triple-angle formulas, $\cos2x=2\cos^2x-1$ and $\cos3x=4\cos^3-3\cos x$, so we have $\cos^2x+(2\cos^2x-1)^2+(4\cos^3x-3\cos x)^2=1$. Let $k=\cos x$. Then,
$$k^2+(2k^2-1)^2+(4k^3-3k)^2=k^2+(4k^4-4k^2+1)+(16k^6-24k^4+9k^2)=16k^6-20k^4+6k^2+1=1$$
implying
$$16k^6-20k^4+6k^2=2k^2(8k^4-10k^2+3)=2k^2(2k^2-1)(4k^2-3)=0.$$
The distinct solutions to this equation are $k=-\tfrac{\sqrt{3}}{2},-\tfrac{\sqrt{2}}{2},0,\tfrac{\sqrt{2}}{2},\tfrac{\sqrt{3}}{2}$. Hence, over the interval $x\in[0,\pi)$, the desired solutions to the original equation (noting $\cos x=k$) are $x=\boxed{\tfrac{\pi}{6},\tfrac{\pi}{4},\tfrac{\pi}{2},\tfrac{3\pi}{4},\tfrac{5\pi}{6}}$. If we do not restrict ourselves to $[0,\pi)$, we can add any multiple of $\pi$ to these angles. $\blacksquare$


Let be given $ n > 1$ $ n$ natural. Find number of all pemutations  $ ( a_1, a_2, a_3, ...,a_n)$ of the set
$ {1,2,...,n}$ such as

$ 1| a_1 \minus{} a_2$
$ 2| a_2 \minus{} a_3$
$ 3| a_3 \minus{} a_4$
...
$ n \minus{} 1|a_{n \minus{} 1} \minus{} a_n$

Let $ x_n$ (resp. $ y_n$) be the number of all 'good' permutations such that $ a_n \equal{} n$ (resp. $ a_n \equal{} 1$). Let $ p_n$ be the total number of 'good' permutations.
Clearly $ x_2\equal{}y_2\equal{}1$ and $ p_2\equal{}2.$

Now assume that $ (a_1, \cdots, a_n)$ is a good permutation. Then $ 1\minus{}n \leq a_{n\minus{}1}\minus{}a_n \leq n\minus{}1$ and $ n\minus{}1$ divides $ a_{n\minus{}1}\minus{}a_n$, so that $ (a_{n\minus{}1},a_n)\equal{}(n,1)$ or $ (a_{n\minus{}1},a_n)\equal{}(1,n)$ (we cannot have $ a_{n\minus{}1} \equal{} a_n$).
It follows that $ p_n \equal{} x_n\plus{}y_n.$

Moreover :
- If $ a_n \equal{} n$ and $ a_{n\minus{}1} \equal{}1$. Then, since there is no other divisibility condition on $ a_n$, we deduce that $ (a_1,\cdots, a_{n\minus{}1})$ is a good permutation with $ a_{n\minus{}1}\equal{}1$. Conversely, from a good permutation $ (a_1,\cdots, a_{n\minus{}1})$ with $ a_{n\minus{}1}\equal{}1$, we may form only one good permutation $ (a_1,\cdots, a_n)$ such that $ a_n \equal{} n$.
Thus $ x_n \equal{} y_{n\minus{}1}$.
- If $ a_n \equal{} 1$ and $ a_{n\minus{}1} \equal{}n$, then the other $ a_i$'s form a permutation of $ \{2, \cdots, n\minus{}1 \}$ satisfying the divisibilty conditions. Thus, for $ i\equal{}1,\cdots, n\minus{}1$, let $ b_i \equal{} a_i \minus{}1$. It's easy to verify that $ (b_1,\cdots, b_{n\minus{}1})$ is a good permutation with $ b_{n\minus{}1}\equal{}n\minus{}1$. Conversely, for any good permutation $ (b_1,\cdots, b_{n\minus{}1})$ with $ b_{n\minus{}1}\equal{}n\minus{}1$ we may form only one good permutation $ (a_1,\cdots, a_n)$ such that $ a_{n\minus{}1} \equal{} n\minus{}1$ and $ a_n \equal{} 1$ by adding $ 1$ to each $ b_i$ to get $ a_i$.
Thus, $ y_n \equal{} x_{n\minus{}1}$.

It follows that $ p_n \equal{} x_n\plus{}y_n \equal{} x_{n\minus{}1}\plus{}y_{n\minus{}1} \equal{} p_{n\minus{}1}.$
Thus, $ p_n \equal{} 2$ for all $ n>1$.

Pierre.

Your Proof is really great. Thanks for Your help.


Let be $ a,b,c,d\in \mathbb{N}^*$ . Show that :

$ \left\lfloor \frac{a^2\plus{}b^2}{2cd\plus{}1} \right\rfloor \plus{}\left\lfloor \frac{b^2\plus{}c^2}{2da\plus{}1} \right\rfloor \plus{} \left\lfloor \frac{c^2\plus{}d^2}{2ab\plus{}1} \right\rfloor\plus{}\left\lfloor \frac{d^2\plus{}a^2}{2bc\plus{}1} \right\rfloor\equal{}\left\lfloor \frac{2}{2abcd\plus{}1} \right\rfloor\Leftrightarrow a\equal{}b\equal{}c\equal{}d$

[hide]Since $ abcd\geqslant 1$, we get $ \left\lfloor{2\over 2abcd \plus{} 1}\right\rfloor \equal{} 0$, hence

$ a^2 \plus{} b^2 < 2cd \plus{} 1\iff a^2 \plus{} b^2\leqslant 2cd$
$ b^2 \plus{} c^2 < 2da \plus{} 1\iff b^2 \plus{} c^2\leqslant 2da$
$ c^2 \plus{} d^2 < 2ab \plus{} 1\iff c^2 \plus{} d^2\leqslant 2ab$
$ d^2 \plus{} a^2 < 2bc \plus{} 1\iff d^2 \plus{} a^2\leqslant 2bc$

Summing the right-hand inequalities up, we get

$ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} d)^2 \plus{} (d \minus{} a)^2\leqslant 0$

and the conclusion follows.[/hide]


calculus
integration
function
calculus computations

Let $f(x)$ the function such that $f(0)=0,\ |f'(x)|\leq \frac{1}{1+x}\ (x\geq 0).$ Prove that $\int_{0}^{e-1}\{f(x)\}^{2}dx\leq e-2.$

[quote="kunny"]Hello, liyi.

Where did the "2" of the left hand vanish? :maybe:[/quote]
I've fixed it. That seems to be quite a coincidence that I got the same answer...


algebra
polynomial
algebra unsolved
binomial coefficients

Prove that, for any sufficiently large integer n, there exists a polynomial p of degree at most $n-\frac{1}{10} \sqrt{n}$ such that:
$\mid p(0) \mid>\frac{1}{10}.{n\choose 1}.\mid p(1) \mid +{n\choose 2}.\mid p(2)\mid+...+{n\choose n}\mid p(n)\mid$

isn't it nice enough ?  I tried but without a success ,any help would be appreciated.


geometry
circumcircle
incenter
parallelogram
geometry proposed

Let a triangle ABC. Denote O,I,H,N be the circumcenter, incenter, orthocenter, Nagel point of triangle ABC. Prove that OI is parallel to HN.

Dear Mathlinkers,
let Be be the Bevan point of ABC (the circumcertee of the excentral triangle of ABC)
1. the quadrilateral HIBeN is a parallelogramm
2. O is the midpoint of IBe  (see http://perso.orange.fr/jl.ayme   vol. 3  Cinq théorèmes de Nagel p. 22)
Sincerely
Jean-Louis


A constant function and e^x are walking on Broadway. Then suddenly the constant function sees a differential operator approaching and runs away. So e^x follows him and asks why the hurry. "Well, you see, there's this differential operator coming this way, and when we meet, he'll differentiate me and nothing will be left of me...!" "Ah," says e^x, "he won't bother ME, I'm e to the x!" and he walks on. Of course he meets the differential operator after a short distance.
e^x: "Hi, I'm e^x"
diff.op.: "Hi, I'm d/dy"

*laughs*

Since I took calc, I get it  ;) .

$\text{Q: What is the difference between a psychotic and a neurotic?}$

$\text{A: A psychotic thinks that 2 + 2 = 5.}$

$.\quad\;\text{A neurotic know that 2 + 2 = 4 ... }but\;it\;worries\;him.$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\text{There are 10 kinds of people in the world:}$
$.\qquad\text{those who understand binary}$
$.\qquad\quad\text{and those who don't.}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\text{What is the difference between Data and Information?}$

$.\qquad\qquad\quad\text{382436: that's data.}$

$.\qquad\qquad\;\text{38-24-36: that's information.}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$.\qquad\quad\text{5 out of 4 people are mathematically challenged.}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$.\qquad\qquad\qquad\text{Two secrets for success}$

$.\qquad\text{1. Never tell anyone everything you know.}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

i don't get this at all. someone could explain?

u need calculus to comprehend the first
and e^x differentiated=e^x so that may shed some light onto the joke

usually $e^x$ differentiated is the same $e^x$. But if you consider this as a function of $y$, then it is constant also :)

I've heard one like it...

An insane mathemetician goes onto a bus and starts yelling, "I'll integrate you!  I'll diffrentiate you!"  All the passengers are freaked out and run off the bus, except for one lady.  The mathemetician goes up to her and says, "Aren't you afraid?  I'll integrate you!  I'll diffrentiate you!"  And the lady goes, "No, I'm not scared, I'm e^x."

[quote="white_horse_king88"]I... feel... lonely. Why haven't I taken calculus yet? I don't know. Sigh. I really wish I knew calculus. It sounds so fun... :(  :(  :([/quote]It will be extremely funny once you get some calculus (even the most basic one) :D

[quote="white_horse_king88"]I... feel... lonely. Why haven't I taken calculus yet? I don't know. Sigh. I really wish I knew calculus. It sounds so fun... :(  :(  :([/quote]

Cheer up. Most people don't take calculus until they're in 11th grade at my school. And at my old school, Calculus was standard 12th grade curriculum.

There's just a few weird, err I mean, special people who take it early.  ;)  :D  :P

Off Topic- I'm really digging these new smilies.  :lol:  :D

i haven't taken calculus yet either, so don't worry. ur not the only one who doesn't get it.  ;)

Same here...I just don't get it...

Of course, that could be equally applicable for a lot of things...

I did a quick google search and found this site:

http://www.math.utah.edu/~cherk/mathjokes.html

I had lots of fun laughing for an hour straight...

[quote=nwang_jd]Really, I think we should have 53 states in the US.
A nice prime number.
"One nation, under god, indivisible"[/quote]

yeah

yeah but why not kill off alaska, wyoming, and idaho and call it 47 states.

Or just conquer Mexico and Canada and split provinces/states into 2 until we have 101 states! The true alpha prime.

yo that's solid we control one of the 5 supersections of the world


function
calculus
derivative
algebra
polynomial
complex analysis
advanced fields

Let $f$ be an entire function. Prove that if $f$ is injective then $f(z)=az+b$, for some $a,b \in \mathbb{C}$, $a\neq 0$.

if $f$ is injective, $f'$ never vanishes. By Liouville, $f'$ is constant. thus $f'(z) = a, f(z) = az + b$.

hmm, it may be so, but I can think of an example of function that does not vanish at all, it is the derivative of an entire function, however, it is not constant. namely $f(z)=e^z$. maybe you can explain some more your argument?

my soln. is the following: the Great Picard theorem tells us that if an entire function is not a polynomial, then it assumes each value (besides at most one) an infinite number of times.

since our $f$ is injective, it must be a polynomial of degree $k$. But each polynomial of degree $k$, assumes each complex value $k$ times exactly. So $k=1$, and we are done.

OK, i see my "argument" was a bit too short. Let my try another way: $g(z) = f(\frac 1z)$ is injective from $\mathbb C^* \cup {\infty}$ to $\mathbb C$ and holomorphic except at $0$. It thus has a singularity at $0$, which is removable or a pole of order at most $1$ by the injectivity of $g$. $g$'s Laurent expansion at $0$ therefore is $g(z) = zR(z) + b + \frac az$, $R(z)$ being entire. Then $f(z) = g(\frac 1z) = \frac {R(\frac 1z)}z + b + az$. Since $f$ is analytic at $0$, $R(\infty) = 0$. $R \equiv 0$, because $R$ is entire and vanishes at infinity (Liouville finally!).

Ouf, I hope i got it right now.

[quote="Valentin Vornicu"]my soln. is the following: the Great Picard theorem tells us that if an entire function is not a polynomial, then it assumes each value (besides at most one) an infinite number of times. [/quote]

How does that latter follow?

An entire function that is not a polynomial has an essential singularity at $\infty$, so consider $g(z)=f(1/z)$ and apply Picard's Great Theorem.


In triangle ABC ,bisector of angles A,B,C meet the opposite sides at D,E,F ,respectively.
Prove that  $ \frac {1}{AD}\plus{}\frac {1}{BE}\plus{}\frac {1}{CF} \leq \frac {1}{r}$.

[quote="Dread Phantom"]In triangle ABC ,bisector of angles A,B,C meet the opposite sides at D,E,F ,respectively.
Prove that  $ \frac {1}{AD} \plus{} \frac {1}{BE} \plus{} \frac {1}{CF} \leq \frac {1}{r}$.[/quote]
Let $ h_a , h_b , h_c$ -altitudes of triangle ABC . Then $ h_a \le AD, h_b \le BE, h_c \le CF$ ;
$ \frac {1}{AD} \plus{} \frac {1}{BE} \plus{} \frac {1}{CF} \leq \frac {1}{h_a}\plus{}\frac {1}{h_b}\plus{}\frac {1}{h_c} \equal{} \frac {1}{r}$ ; :)

How is $ \frac{1}{h_a} \plus{} \frac{1}{h_b} \plus{} \frac{1}{h_c} \equal{} \frac{1}{r}$ ?
I only know that $ \frac{1}{r_a} \plus{} \frac{1}{r_b} \plus{} \frac{1}{r_c} \equal{} \frac{1}{r}$ (where $ r_a , r_b , r_c$ are the exradii of the triangle ABC)..

$ \frac{1}{r} \equal{} \frac{1}{h_a} \plus{} \frac{1}{h_b} \plus{} \frac{1}{h_c}$

$ 2A \equal{} ah_a \equal{} bh_b \equal{} ch_c\implies a \equal{} \frac{2A}{h_a} etc. As S \equal{} rp\implies p \equal{} \frac{A}{r}, (a \plus{} b \plus{} c) \equal{} \frac{2A}{r}\implies \frac{2A}{h_a} \plus{} \frac{2A}{h_b} \plus{} \frac{2A}{h_c} \equal{} \frac{2A}{r}, qed$


trigonometry
calculus
calculus computations

Today one of my classmates solved a problem that involved the following equality, which I think is not so obiouvsly. If anyone can post a proof it would be great :). My math teacher said that it can be prooven using elementary knowledge (such as 9-th grade trig) or calculus ones.

$\arcsin{x}-\arcsin{y}=\arcsin({x\sqrt{1-y^{2}}-y\sqrt{1-x^2}})$.

Take sine on both sides. Use the formula for $\sin(a\pm b)$ as well as that $\cos x = \sqrt{1-\sin^2x}$

note that $sin(cos^{-1}\theta)=\sqrt{1-\theta^2}$
and
$cos(sin^{-1}\theta)=\sqrt{1-\theta^2}$

and with this u just apply the normal sine addition formula.  :)


There are ten books, each with fewer than 100 pages.  Show that it is possible to arrange some or all of the ten books into two piles, with at least one book in each pile, so that the number of pages in each pile is the same.

Does anyone know how to go about doing this problem?? I am so stuck!!
Thanks!

[hide="hint"] try smaller cases: try for fewer number of books, and for each number of books, try to find the max number $ n$ where all the books have less than $ n$ pages.[/hide]

thanks for the tip!!

does less than 100 mean include 100?

No. If it less than 100 pages, then the maximum it can have is 99 pages.


The sum of four numbers $ w$, $ x$, $ y$, and $ z$ is $ 64$, and $ w<x<y<z$.  The sum of $ z$ and $ w$ is twice the arithmetic mean of the four numbers.  The number $ x$ is half the largest number.  What is the value of $ y$ if $ w\equal{}4$?

We see that $ z \plus{} w \equal{} 32$, and $ x \equal{} \frac {1}{2}z$. From $ w \equal{} 4$ we get $ z \equal{} 28$, now taking half of $ 28$, we get $ x \equal{} 14$. Adding up, and subtracting from 64, we get $ y \equal{} 18$.


quadratics
algebra
polynomial
geometry
geometric transformation
vector
abstract algebra

I have a question for which I am meant to use the following question to do. I am having problems generally with proving things to be isomorphic and so am finding the  following question hard to do and so I am unable to start on the main question. I would be grateful for anyhelp given. Thanks.

Let $ f \equal{} ax^2 \plus{}bx \plus{}c \in \mathbb{R}[x]$ be at most a quadratic polynomial with real coefficients and the discriminant D. Let $ A_{f} \equal{} \mathbb{R}[x]/(f)$.

Prove that if $ a\equal{}0$ then the ring $ A_{f}$ is isomorphic to one of the following three rings: $ \mathbb{R}, \mathbb{Z}_1, \mathbb{R}[x]$.

Prove that if $ a \neq 0$ and $ D > 0$ then the ring $ A_{f}$ is isomorhpic to the ring $ \mathbb{R}$ X $ \mathbb{R}$.

Prove that if $ a \neq 0$ and $ D < 0$ then the ring $ A_{f}$ is isomorhpic to the ring $ \mathbb{C}$.

Prove that if $ a \neq 0$ and $ D \equal{} 0$ then the ring $ A_{f}$ is isomorhpic to the ring $ \mathbb{R}[x]/(x^2)$.

Display the appropriate algebraic properties of these 6 rings $ (\mathbb{R}, \mathbb{Z}_1, \mathbb{R}[x], \mathbb{R}$ X $ \mathbb{R}, \mathbb{C}, \mathbb{R}[x]/(x^2))$ showing that they are not isomorphic.

I have recently done algebras and so I am pretty sure I will be able to do the last part. It is just showing things are isomorphic which i find difficult.

$ \bullet$ $ a \equal{} 0$. if $ b \equal{} 0$, the ideal $ (f)$ is $ 0$ or $ \mathbb{R}[x]$, thus $ A_f \equal{} \mathbb{R}[x]$ or $ A_f \equal{} 0$. otherwise $ x \mapsto bx \plus{} c$ defines an automorphism of $ \mathbb{R}[x]$ inducing $ \mathbb{R}[x]/(x) \cong A_f$. but $ \mathbb{R}[x]/(x) \cong \mathbb{R}$ which follows from the isomorphism theorem applied to the evaluation $ \mathbb{R}[x] \to \mathbb{R}, p \mapsto p(0)$.

$ \bullet$ $ a \neq 0$. multiplication with $ a^{ \minus{} 1}$ does not change the whole thing, so wlog $ a \equal{} 1$.
$ \bullet$ if $ D \geq 0$, factor $ f$ into two real linear factors. if $ D > 0$, they are coprime, and the chinese remainder theorem together with $ \mathbb{R}[x]/(x) \cong \mathbb{R}$ yields the desired isomorphism $ A_f \cong \mathbb{R} \times \mathbb{R}$. if $ D \equal{} 0$, translation with the unique root yields $ A_f \cong \mathbb{R}[x]/(x^2)$.

$ \bullet$ if $ D < 0$, $ f$ is irreducible, thus $ A_f$ a quadratic field extension of $ \mathbb{R}$. this must be $ \mathbb{C}$. to be explicit: if $ a$ is a complex root of $ f$, $ x \mapsto a$ induces a homomorphism $ A_f \to \mathbb{C}$. since both sides are fields, it is injective, and a comparison of vector space dimension over $ \mathbb{R}$ yields surjectivity.


[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \frac {a(b\plus{}c)\minus{}bc}{a^2 \plus{} 3b^2 \plus{} 3c^2} \plus{} \frac {b(c\plus{}a)\minus{}ca}{b^2 \plus{} 3c^2 \plus{} 3a^2} \plus{} \frac {c(a\plus{}b)\minus{}ab}{c^2 \plus{} 3a^2 \plus{} 3b^2} \le \frac {3}{7}$,

with equality for $ a \equal{} b \equal{} c$ and again when $ \frac a{4} \equal{} \frac b{3} \equal{} \frac c{3}$ or any cyclic permutation[/quote]
By a small trick (using our hands only), I found the following identity \[{\sum \frac{a(b+c)-bc}{a^2+3b^2+3c^2}-\frac{3}{7} =-\frac{1}{14} \sum \frac{(b-c)^2(3b+3c-7a)^2}{(b^2+3c^2+3a^2)(c^2+3a^2+3b^2)}-\frac{4\prod (a-b)^2}{\prod (a^2+3b^2+3c^2)},}\] from which the inequality immediately follows.

I used the following obvious identify to reduce to your result:
\[\sum_{cyc}{a^2(a-b)(a-c)(b-c)^2}=(a-b)^2(b-c)^2(c-a)^2\]
The idea to factorize such $(3c+3a-7b)^2$ is such a genius ideas, maybe notice that the equality cases $3\cdot 4+3\cdot 3-7\cdot 3=0$.


can some one tell me if this is true......and if yes please prove....

for 'a' being a real number ...0<a<2 is $a + 1/a >=2$ ?? I aint sure about this, cos my method's a bit rickety :D

Check out the last post in this thread:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49414

[quote="Carlyfosnizzle"]yeah its true
im not sure exactly how to phrase this mathematically but ok say

a is 1 then 1+1=2
a=1/2  then it would be 1/2+2=2.5>2
or if a was 3/2 then it would be 3/2+2/3=2.166>2

so no matter what the value is, its >= 2[/quote]A few things..

(a) just because you know it's true for those three particular values of $a$, you cannot say it's true for all. Your job is to prove that it is.

(b) $\frac{3}{2}+\frac{2}{3}=2+\frac{1}{6}>2$. Around here we generally don't put things into ugly decimal form when it's not necessary. The number $.166$ is only an approximation of $\frac{1}{6}$, so we wouldn't say they're equal. It's a technical detail you may not care about, but it bothers people like me.

Anyway, welcome to the forums.  :D


a,b integers , a>b
and

$a^{2}+b^{2}+ab = 961$
find a,b

An obvious solution is of course $a=31$, $b=-31$ :P

But how do you prove that it is also unique and there isn't any other solution :?:

The 'easiest' solution I found so far is to rewrite the equation as $(2a+b)^{2}+3b^{2}= 62^{2}$. Then remark that $62^{2}\geq 3b^{2}$, so $b^{2}\leq 35^{2}$. Then just find a systematic way to check al values for $b^{2}$ from 1 to $35^{2}$ to find the values $|b|\in\{11,24,31,35\}$. It's easy after that.


linear algebra
matrix
algebra
polynomial
IMC
college contests

The scores of this problem were:
one time 17/20 (by the runner-up)
one time 4/20 (by Andrei Negut)
one time 1/20 (by the winner)
the rest had zero... just to give an idea of the difficulty.

Let $A_{i},B_{i},S_{i}$ ($i=1,2,3$) be invertible real $2\times 2$ matrices such that [list][*]not all $A_{i}$ have a common real eigenvector, [*]$A_{i}=S_{i}^{-1}B_{i}S_{i}$ for $i=1,2,3$, [*]$A_{1}A_{2}A_{3}=B_{1}B_{2}B_{3}=I$.[/list] Prove that there is an invertible $2\times 2$ matrix $S$ such that $A_{i}=S^{-1}B_{i}S$ for all $i=1,2,3$.

We can replace $A_{i}, B_{i}$ by $xA_{i}, yB_{i}$ with $x,y$ non-zero real numbers; we can also replace $A_{i}, B_{i}, C_{i}$ by $X_{i}A_{i}X_{i}^{-1}, X_{i}B_{i}X_{i}^{-1}, X_{i}C_{i}X_{i}^{-1}$ with $X_{1},X_{2}\in GL_{2}(\mathbb{R}).$ We can thus assume $A_{1}=A_{2}.$ Of course, it is enough to find an $S$ such that $SA_{1}S^{-1}=A_{1}$ and $SB_{1}S^{-1}=B_{2}.$ We shall also assume that none of the matrices $A_{i}, B_{i},C_{i}$ are scalars.

Suppose first that the characteristic polynomial of $A_{1}$ is reducible over $\mathbb{R}$. Twisting $A_{1}$ by a scalar and conjugating appropriately, we need to consider the following two cases:

i) $A_{1}=A_{2}=\begin{pmatrix}x & 0\\ 0& 1\end{pmatrix}$ with $x\neq 1.$ Set $B_{i}=\begin{pmatrix}a_{i}& b_{i}\\ c_{i}& d_{i}\end{pmatrix}.$ The condition that $A_{1}B_{1}$ and $A_{1}B_{2}$ are conjugates forces $a_{1}=a_{2}, d_{1}=d_{2}, b_{1}c_{1}=b_{2}c_{2}.$ For an appropriate diagonal matrix $S$ we get $B_{2}=SB_{1}S^{-1}.$

ii) $A_{1}=A_{2}=\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}.$ Again, let
$B_{i}=\begin{pmatrix}a_{i}& b_{i}\\ c_{i}& d_{i}\end{pmatrix}.$ Comparing traces of $A_{1}B_{1}, A_{1}B_{2}$ gives $c_{1}=c_{2}.$ The condition on eigenvectors implies $c_{1}\neq 0.$ If we set $S=\begin{pmatrix}1 & x\\ 0 & 1 \end{pmatrix}$ where $x$ is chosen so that $SB_{1}S^{-1}=\begin{pmatrix}a_{2}& * \\ c_{1}& * \end{pmatrix},$ then $SA_{1}S^{-1}=A_{1}$ and $SB_{1}S^{-1}=B_{2}.$

Now suppose that $A_{1}=A_{2}$ has irreducible (over $\mathbb{R}$) characteristic polynomial. As in case i, we can find a $T\in GL_{2}(\mathbb{C})$ such that $TA_{1}T^{-1}=A_{1}$ and $TB_{1}T^{-1=B_{2}=S_{2}B_{1}S_{2}^{-1}.}$ Note the following: If $A\in GL_{2}(K)$ is a non-scalar ($K$ a field) and $MA=AM$ for some $2\times 2$-matrix over $K$, then $M=xA+yI$ with $x,y\in K.$ (To see this, conjugate $A$ to a matrix $\begin{pmatrix}0 & d\\ 1 & t\end{pmatrix}.$) We can thus find real numbers $x_{1},y_{1}, x_{2},y_{2}$ such that $S_{2}^{-1}(x_{1}A+y_{1}I)=x_{2}B_{1}+y_{2}I$ with $(x_{1},y_{1})\neq (0,0).$ (This can be seen after writing $T$ as a complex linear combination of $A$ and $I$,  and then separating real, imaginary parts.) Set $S=x_{1}A+y_{2}I.$ If this matrix $S$ is invertible, then we
are through. But this follows because the ring of matrices $\{ xA+yI: x,y\in \mathbb{R}\}$ is isomorphic to $\mathbb{C}$ by the map which sends $A$ to one of its eigenvalues.

(I might have posted the same solution in another forum. Apologies.)

Sounds pretty much like what I did though I am too lazy to go through every step of your proof...

:|

If you can solve this, you should really consider participating next year. ;)

(it is after the IMO in case you're still on IMO for a year)


$ \frac{7!6!}{5!4!3!}\equal{}\frac{7(6\cdot 5\cdot 4\cdot 3\cdot 2)^{2}}{5\cdot 4^{2}\cdot 3^{3}\cdot 2^{3}}$.
The 4's cancel completely.  We're left with a 7, 6, 6 and a 5 in the numerator, and a 3 and 2 in the denominator: $ \frac{7\cdot 36\cdot 5}{3\cdot 2}\equal{}7\cdot 6\cdot 5\equal{}210$


quadratics
number theory
prime numbers
number theory unsolved

Which prime number we can write as $ x^2\plus{}5y^2$?

[quote="Rust"]it is necessary to have $ (\frac p5)\equal{}1, (\frac{\minus{}5}{p})\equal{}1$ or $ p\equiv 1,9\mod 20$.[/quote]

[b]Proof of Rust's claim:[/b]
Obviously $ \gcd(x,y) \equal{} 1$ if $ p \equal{} x^2 \plus{} 5y^2$. $ p \mid x^2 \plus{} 5y^2 \iff x^2 \equiv \minus{} 5y^2 \bmod p$. Let $ y^{ \minus{} 1}$ be the inverse of $ y$: Then $ (xy^{ \minus{} 1})^2 \equiv \minus{} 5 \bmod p$. So $ \minus{} 5$ is a quadratic residue. Thus it is necessary that $ \left ( \frac { \minus{} 5}{p} \right ) \equal{} 1$ but $ \left ( \frac { \minus{} 5}{p} \right ) \equal{} \left ( \frac { \minus{} 1}{p} \right ) \cdot \left ( \frac {5}{p} \right ) \equal{} \left ( \frac { \minus{} 1}{p} \right ) \cdot \left ( \frac {p}{5} \right )$
But since $ x^2 \plus{} 5y^2 \equal{} p$ we get $ x^2 \equiv p \bmod 5$ so $ p$ is a quadratic residue $ \bmod 5$. Then $ \left ( \frac {p}{5} \right ) \equal{} 1$ $ \iff$ $ p \equiv \pm 1 \bmod 5$. And $ \left ( \frac { \minus{} 1}{p} \right ) \equal{} 1 \iff p \equiv 1 \bmod 4$. Thus: $ p \equiv 1,9 \bmod 20$.

See http://www.mathlinks.ro/viewtopic.php?t=232507 for a complete discussion of $ x^2\plus{}5y^2$ integers.

OK Mathias_DK You have shown that $ p$ must be congruent to 1 or 9  mod 20 but I think it doesn't have to be sufficient condition to express a prime number in the form $ x^2\plus{}5y^2$? Could You explain it futher?

[quote="matex (L)(L)(L)"]OK Mathias_DK You have shown that $ p$ must be congruent to 1 or 9  mod 20 but I think it doesn't have to be sufficient condition to express a prime number in the form $ x^2 \plus{} 5y^2$? Could You explain it futher?[/quote]
I haven't said it was a sufficient condition. I was merely explaining the post which Rust wrote.
It probably is though :P

Read my link, I would like to get corrections and additions  :)
Essentially, you have only to show lemma 1 to 5 if you are interested in primes only. The idea is:
We are interested in what primes are of type $ x^2\plus{}5y^2$. So we are interested in what primes are norms of numbers of type $ x\plus{}y\sqrt 5$. But instead of considering these numbers, we look at the much more complicated numbers of type $ a\plus{}bi\plus{}c \frac{1\plus{}\sqrt 5}{2}\plus{}d i \frac{1\plus{}\sqrt 5}{2}$. In fact, we now ask which prime numbers are of type $ p\equal{}(a^2\plus{}b^2\minus{}c^2\minus{}d^2\plus{}ac\plus{}bd)^2\plus{}5(ad\minus{}bc)^2$ [I hope I calculated that correctly, one doesn't need the exact term in the proof], the answer will be the same! So for every prime $ p\equiv 1,9 \mod 20$, we get that it is of that form, thus obviously of the much simpler form $ x^2\plus{}5y^2$ (with $ x\equal{}a^2\plus{}b^2\minus{}c^2\minus{}d^2\plus{}ac\plus{}bd,y\equal{}ad\minus{}bc$). To show that $ p$ can be written as required, one can use similar methods as for numbers of type $ x^2\plus{}y^2$ [which fail for $ x^2\plus{}5y^2$ but work for $ (a^2\plus{}b^2\minus{}c^2\minus{}d^2\plus{}ac\plus{}bd)^2\plus{}5(ad\minus{}bc)^2$  :D ].
This approach may sound very strange, but in fact there are good reasons for it.

This problem is very similar to the problem that tells us to prove that all prime of the form $ 4k\plus{}1$ can be expressed as sum of 2 squares
Thue's Lemma is very useful here.There is one book (that I see) that makes an approach of the numbers of the form $ x^2\plus{}dy^2$

wiosna  you hve not specified what x and y are.

Here goes an elementary proof (but boooooooooooring):

The only thing left to show is, that if a prime $ p$ fulfills $ p\equiv 1 \mod 4$ and $ \left( \frac p5 \right)$, then $ p$ is of type $ a^2 \plus{} 5b^2$.
Taking a $ c$ such that $ c^2 \equiv \minus{} 5 \mod p$, we use Thue's lemma (see  http://www.mathlinks.ro/Forum/viewtopic.php?p=293201#293201 ) to get $ 0<a,b < \sqrt p$ with $ a \equiv cb \mod p$, giving that $ p|a^2 \plus{} 5b^2$ and that $ a^2 \plus{} 5b^2 < p \plus{} 5p \equal{} 6p$. Thus $ a^2 \plus{} 5b^2$ is one of $ p,2p,3p,4p,5p$.
Case $ p$: done.
Case $ 2p,3p$: contradiction $ \mod 5$.
Case $ 4p$: $ \mod 4$ gives that $ a,b$ are even, thus $ \left(\frac a2\right)^2 \plus{} 5\left(\frac b2\right)^2 \equal{} p$ works.
Case $ 5p$: $ \mod 5$ gives that $ 5|a$, thus $ b^2 \plus{} 5\left(\frac a5\right)^2 \equal{} p$ works.


An ambulance travels at 40 mph and can follow a 20-mile route making no stops to get to the hospital. A helicopter travels at one mile per minute, and the air route is 15 miles to get to the same hospital. However, the helicopter takes three minutes for take off and three minutes for landing. How many fewer minutes does it take for the helicopter to complete its trip (take off, flight and landing) than for the ambulance to complete its trip?

The ambulance takes $ \frac{40}{20}(60)\equal{}30$ minutes while the helicopter takes $ (15\plus{}3\plus{}3)\equal{}21$. Our answer is $ 30\minus{}21\equal{}\boxed{9}$.


In ham yek soale mashhor va ziba ke omidvaram rahehal haye jadid va zibayi barash erae beshe

 Farz konid $ a,b,c$ adad hagigy namanfi hastand betory ke hich dotayi sefr nist sabet konid

\[ \frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(a\plus{}c)^2}\plus{}\frac{1}{(c\plus{}b)^2}\ge \frac{9}{4(ab\plus{}bc\plus{}ca)}\]

 Rasti kasi midone tarahe in soal kie ?
[hide]montazere hale zibaye man bashid[/hide]

[quote="M.A"]Rasti kasi midone tarahe in soal kie ? [/quote]
fek konam bedooni ke in too olampiade iranam oomade,vali khob tekrari boode,yeni tarrahesh irani nisto copysh kardan,ghablan yeki dge tarhesh karde boode... :)
hala esme tarrahesho mikhai hanooz? :maybe:
rahe halaye mokhtalefiam dare,be nazare man ma'ghooltarinesh ba SOSe,albate ta injaye kar,shayad rahe to behtar bashe... :wink:  :D

Mercy Babak
va amma rahe man
farz konid $ a \plus{} b \plus{} c \equal{} 1$ va $ ab \plus{} bc \plus{} ca \equal{} q$ va $ abc \equal{} r$ darim
\[ \sum \frac {1}{(a \plus{} b)^2} \equal{} \sum \frac {(1 \plus{} 6c)^2}{(1 \minus{} c)^2(1 \plus{} 6c)^2}\ge \frac {81}{\sum(1 \minus{} c)^2(1 \plus{} 6c)^2}
\]
Pas kafist sabet konim
\[ 13q\ge 1 \plus{} 36q^2 \minus{} 18r \Leftrightarrow 18r\ge 1 \plus{} 36q^2 \minus{} 13q
\]
Case1. q>=1/4 darim $ r\ge \frac {4q \minus{} 1}{9}$ pas kafie sabet konim
\[ 7q\ge 1 \plus{} 12q^2 \Leftrightarrow (1 \minus{} 3q)(4q \minus{} 1)\ge 0
\]
Cass2.$ \frac {1}{4} > q\ge \frac {1}{9}$ darim $ r\ge 0$ pas bayad sabet konim
\[ 0\ge 1 \plus{} 36q^2 \minus{} 13q \Leftrightarrow \frac {1}{4} > q\ge \frac {1}{9}
\]
Case3.$ q\le \frac{1}{9}$ darim
\[ \sum \frac {1}{(a \plus{} b)^2} \equal{} \sum \frac {c^2}{(1 \minus{} c)^2c^2}\ge \frac {1}{\sum (1 \minus{} c)^2c^2}
\]
Va edamash asoone
Rasti tarahe soal Mr.Ji Chen ast


function
Euler
logarithms
probability
probability and stats

take $ \zeta(z) \equal{} \sum_{n \geq 1} \frac{1}{n^z}$ the usual zeta function and take a real number $ u>1$. Show that $ \phi_u(y): \mathbb{R} \to \mathbb{C}$ defined by $ \phi_u(y) \equal{} \frac{\zeta(u\plus{}iy)}{\zeta(u)}$ is the characteristic function of a random variable $ X$  :)

This was proved(probably) by A. Khinchin. This characteristic function is infinitely divisible(proof requires some knowledge about infinite divisibility). Using the Euler formula for zeta function, taking the logarithm one will get the sum, with each term representing the characteristic function of Poisson-type random variable. Thus, the limiting distibution is infinitely divisible.

exactly  :)  I have read this in a book about Levy processes (I think so that it was first proved by Khinchin), but nothing more was said about a link between the Zeta function and probability theory. Do you know more about it ?
Thank you.

Wikipedia says that $ \frac{1}{\zeta(n)}$ is the probability that $ n$ randomly selected integers are relatively prime. I think you can find a lot of matches on google, for example this http://www.geocities.com/oshanker/RiemannZeta/RiemannZeta.pdf


algebra
polynomial
calculus
integration
algebra proposed

The sequence of real numbers $ a_1,a_2,...,a_n,...$ is defined by $ a_{n \plus{} 2} \equal{} a_{n \plus{} 1} \plus{} \dfrac{k}{n}a_n$,where k is a given real number.

 For which values of $ k$ there is a nonzero solution with $ a_n$ a polynomial in $ n$?

[hide="Solution"]$ k$ can be any non-negative integer:

It is given that $ a_{n+2} \&= a_{n+1} + \frac{k}{n}\cdot a_n$.

Notice that if $ \{b_n\}$ satisfies this definition, then so does $ \{k\cdot b_n\}$. Thus, we can normalize any polynomial solution for $ a_n$ to be monic:

\[ a_n \&= n^d + c_1 n^{d-1} + \cdots + c_d\]

Combining the two equations and subtracting, we have:

\[ ((n+2)^d-(n+1)^d) + c_1((n+2)^{d-1}-(n+1)^{d-1}) + \cdots + c_{d-1}((n+2)-(n+1)) + c_d - c_d\] \[ = k\cdot n^{d-1} + k\cdot c_1\cdot  n^{d-2} + \cdots + k\cdot c_{d-1} + \frac{k\cdot c_d}{n}\]

The L.H.S. is a polynomial of degree $ d-1$ with leading coefficient $ d$ after expanding, so $ c_d = 0$. Equating coefficients of $ n^{d-1}$, we have that $ d = k$, where $ d$ is the non-negative integral degree of the polynomial $ a_n$. That is, it is necessary that $ k$ is a non-negative integer for a solution to the recurrence that is a polynomial in $ n$.

We can show that it is also sufficient (proving that such a polynomial solution exists for every such $ k$) by writing the L.H.S. of the above equation fully as follows:

\[ d\cdot n^d + ((d-1)\cdot c_1 + h_1)n^{d-1} + ((d-2)\cdot c_2 + h_2\cdot c_1 + h_3)n^{d-2} + \cdots + (c_{d-1} + h_m \cdot c_{d-2} + \cdots + h_{m+d-2})\]

Here the $ h_j$ are constants that come from expansion of $ (n+2)^{\cdots} - (n+1)^{\cdots}$. Equating coefficients with $ k\cdot n^{d-1} + k\cdot c_1\cdot  n^{d-2} + \cdots + k\cdot c_{d-1} + \frac{k\cdot c_d}{n}$, we find that e.g. $ c_1 = h_1$, $ 2c_2 = h_2 \cdot c_1 + h_3$, ... We can solve this system by backsolving starting from $ c_1$ to find the unique (up to multiplication by a nonzero scalar) polynomial for each degree $ k=d$ that will satisfy the recurrence. For example, for $ k=0$ we can take $ a_n = 1$, for $ k=1$ we can take $ a_n = n$, and for $ k=2$ we can take $ a_n = n^2+3n$.[/hide]


Prove that if we select n + 1 numbers from the set S = {1, 2, 3, . . . , 2n}, among the numbers selected there are two such that one is a multiple of the other one.

$ \left\{1,2,...,2n\right\}$ has $ n$ subsets like:

$ \left\{1,2\right\},\left\{2,4\right\},\left\{3,6\right\},\cdots,\left\{n,2n\right\}$

so choosing $ n+1$ numbers we have to choose atleast $ 2$ numbers in one of these subsets and thats it... :)


Given the equation:

(1 - x^2)y"(x) - 2xy'(x) + 2y(x) = 0

How can you solve it using the Power Series?

hello, there exists an analytic solution of your equation.
$ y(x)\equal{}C_1x\plus{}C_2\left(\minus{}\frac{1}{2}x\ln(x\minus{}1)\plus{}\frac{1}{2}x\ln(x\plus{}1)\minus{}1\right)$.
Sonnhard.

Suppose your equation has a solution $ y(x)$ (defined in 0 for simpleness) that can be expressed as a power series, i.e.
$ y(x)\equal{}\sum_{n\equal{}0}^{\plus{}\infty}a_nx^n$ with $ y(0)\equal{}y_0$. Then you can write that $ y'(x)\equal{}\sum_{n\equal{}0}^{\plus{}\infty}(n\plus{}1)a_{n\plus{}1}x^n$ and $ y''(x)\equal{}\sum_{n\equal{}0}^{\plus{}\infty}(n\plus{}1)(n\plus{}2)a_{n\plus{}2}x^n$. Now substitute in your equation those expressions, you'll find an equation of the kind $ \sum_{n\equal{}0}^{\plus{}\infty}f(a_n,a_{n\plus{}1},a_{n\plus{}2})x^n\equal{}0$. A power series is equal to $ 0$ for all real $ x$ iff all its coefficients are $ 0$. Now you solve the equation $ f(a_n,a_{n\plus{}1},a_{n\plus{}2})\equal{}0$ finding the succession $ a_n$ of the coefficients of your solution.


For how many integers between 1 and 100 does $ x^{2}+x-n$ factor into the product of two linear factors with integer
coefficients?

I'm going to assume that the integers 1-100 are for the variable n (you were unclear)

[hide]
$ x^{2}+x-n$

$ (x-a)(x-b) = x^{2}-(a+b)x+ab$

$ a+b =-1$
$ ab =-n$

$ |a|$ and $ |b|$ must differ by 1, with one of $ a,b$ being negative and the other positive

this leaves the values of n
$ 1 \cdot 2$
$ 2 \cdot 3$
...
$ 9 \cdot 10$

or 9 integers
[/hide]


MATHCOUNTS
geometry
geometric transformation
reflection
AMC
AIME

I was wondering how many questions you need to correctly do to go to the Awesome Math Summer Program. And is there any other specific requirements(like mathcounts or amc score)?

i might be appling too and what i found was that they dont really ask about ur prior contest performance.

on the admission test, as long as u show your love for math ,ur fine
just try to get as much of them as you can and if you cant solve one of them, at least put as much as you can towards the solution and they will award partial credit.

AND MAKE SURE U SHOW ALLLLLLLLLLLLLL WORK.

they do ask about ur prior competition performance, however, it doesnt make a huge impact.
but iYOA is right. admission test are the important part. i answered.... 7 ? i think. and i got in. and i probably got a few wrong too :P yeah. don't worry bout it. just try your best :)

Don't worry about competition results that much... When I applied, I had NEVER taken a single AMC test, in practice or for real. At that time the only thing I ever did was MATHCOUNTS... if you call getting 22nd place doing something.

[quote="iYOA"]some of the competion results can be skewed and might not reflect how u are a mathmatician if you make too many careless mistakes,
also, the test might have been really easy that year and you might be the person who does better on harder tests.

  shelly32494: how old were u when you applied?

and what would you consider really bad competion results?[/quote]

definitely competition results can be skewed. i mean, when i was in 8th grade, i got a 15 on my state sprint [ somehow. -___-;; embarassing] anyways. and i was in 11th place [b/c of my state] and if i was in a good state, id be like.... 111th place or something. ><

i was 14. [rising sophomore ]
bad competition results? lol. it depends on the person . i really cant say, and its really not up to me.

yeah. well. cali is uber hard . >< everyone's amazing.

i dont know.

Well okay I'm sure that if it seems like you enjoy math and you don't solve very many problems, they won't reject you or anything. If you're thinking about spending that much time in a math camp for the summer I'm sure you have a good deal of interest already.

dont worry about how many questions you answered or your previous math experience

they mainly look for your interest in math.
you have to be able to solve at least some, but don't worry too much about scores.

remember: SCORES DONT MAKE FRIENDS >:{

[quote="sooozyy"]

remember: SCORES DONT MAKE FRIENDS >:{[/quote]

cant stress this one enough. ><
<3

[quote="AIME15"]Hm, my state sprint was double that, and I got 5th...

Anyway, I've heard that if you can solve about 4 problems on the admission test you're in; is that true?[/quote]
5th in California is still amazing man

[quote="AIME15"]
Anyway, I've heard that if you can solve about 4 problems on the admission test you're in; is that true?[/quote]

yeah
i did 4 cuz i forgot about it till the night before due

[quote="iYOA"]was that 22nd place at nats?
because thats really kindof good.[/quote]

State.

When I went there last year I felt like an idiot. :D

But this year I'm going to Mathcounts Nationals (Utah is an easy state) and I made AIME (which I know is nothing for most of you guys "I made Blue MOP in first grade. I feel like such an underachiever") so I hope I get in again this year and hopefully not feel like the stupidest one there.

I've solved like 6 problems and I'm going to put my work (not very much) on the four that I did not solve. Would I have a good chance of getting in?

Also, there isn't a specific part where they ask about previous competition math, but I put some scores under "extracurricular activities"...

I only submitted 6 because I forgot to write up the rest, and I got in. Don't worry about it.
Also if you have a complete solution don't show "work" that didn't have to do with it.

Well yeah, it makes sense that if you got a solution, then you don't show all the experimentation that you did, just the parts that lead to the solution.

But for the ones I couldn't solve, do I just put my experiments?

Got it. Nice job at states, by the way...

Earlier in here AIME 15 said that you only have to get like a 4 on the admission test to get in.

Really? How do you guys know that? Because I think I only got like a 5 or 6 on it. 7 at the max.  :(
And I really want to get in and am sort of nervous that I'll get rejected. :(

read korammath's post........

you'll be fine. so long as you display an interest for math. and 5-6 problems right is good.

Someone help me out here... but if I'm not mistaken, getting in is like.. NOT a problem?

5 or 6 or 7.. (wow. haha) is decent.. especially 7.

You probably won't get rejected, I mean, I didn't get rejected. (I'm not that smart.. like.. I epic-phailed the AIME twice)


geometry
3D geometry
algebra
difference of squares
special factorizations
number theory proposed
number theory

Given that a,b,c are positive integers, determine all triplets (a,b,c) such that they satisfy the equation
             a^2+b^3=c^4

Hello!!! Isn't anybody interested in posting a solution please.

My intution says that such triplets (a,b,c) is not possible, but I can't prove it. :(

No, there are infinitely many.
Start with $ (27,18,9)$ (found from $ 1 \plus{} 8 \equal{} 9$, then multiply by an appropriate common factor, namely $ 3^6$, to make appear the desired exponents).
Now, if $ (a,b,c)$ is a solution then so is $ (at^6,bt^4,ct^3)$ for all positive integers $ t$.

Pierre.

Good solution pbornsztein. Thanks.
But are you absolutely sure that for a solution (a,b,c) only solutions of that format exists? If so, then how is it justified??

[quote="Sunkern_sunflora"]
But are you absolutely sure that for a solution (a,b,c) only solutions of that format exists? [/quote]

I did never say that...

Pierre.

Write as $ b^3 \equal{} c^4 \minus{} a^2 \equal{} (c^2 \plus{} a)(c^2 \minus{} a)$ and stipulate that $ c^2 \plus{} a \equal{} b^2, c^2 \minus{} a \equal{} b$.  Then

$ 2c^2 \equal{} b^2 \plus{} b \implies$
$ 8c^2 \plus{} 1 \equal{} (2b \plus{} 1)^2 \implies$
$ (2b \plus{} 1)^2 \minus{} 2(2c)^2 \equal{} 1$.

(Note also that $ a \equal{} \frac {b^2 \minus{} b}{2}$, which is always an integer.)  This is a [url=http://en.wikipedia.org/wiki/Pell%27s_equation]Pell's equation[/url] whose solutions are generated by the powers of $ 3 \plus{} 2\sqrt {2}$.  $ 3 \plus{} 2 \sqrt {2}$ itself gives the trivial soution $ b \equal{} c \equal{} 1, a \equal{} 0$.  For example,

$ (3 \plus{} 2 \sqrt {2})^2 \equal{} 17 \plus{} 12 \sqrt {2}$

which gives $ b \equal{} 8, c \equal{} 6, a \equal{} 28$.  The next power is

$ (3 \plus{} 2 \sqrt {2})^3 \equal{} 99 \plus{} 70 \sqrt {2}$

which gives $ b \equal{} 49, c \equal{} 35, a \equal{} 1176$.  This gives us an infinite family of "primitive" solutions, but note that it does [b]not[/b] give us every such solution (for example, it doesn't give pbornsztein's solution).

[b]Edit:[/b]  Another infinite family along pbornsztein's solution's lines.  Pick any cube $ x^3$ and write it as a difference of squares $ y^2 \minus{} z^2$.  Then $ (zy^3, xy^2, y^2)$ is a solution.  pbornsztein's solution is given when $ x \equal{} 2, y \equal{} 3, z \equal{} 1$.  Note that unless $ y \equal{} 1$ we don't get solutions where $ c$ is squarefree unlike in the above family, although this infinite family is much "larger."


okay i'm currently studying for a ASVAB test for the military. and they have questions like the one below,
[b]
what number is 5% of 2000[/b]

how would i solve for something like that?

or they have

[b]35% of what number is 70[/b]

[b]if 45 is 120% of a number, what is 80% of the same number?[/b]

[b]if 6 is 24% of a number, what is 40% of the same number?[/b]

[b]25% of $900=[/b]

questions like those, please help

Use the is/of method: $ \frac{\%}{100}\equal{}\frac{\text{is}}{\text{of}}$.

So in your first example, it would be $ \frac{5}{100}\equal{}\frac{x}{2000}\Rightarrow x\equal{}100$

I always read these a little differently than what JRav wrote: "what number" means $ x$, "is" means =, "of" means $ \times$ and "$ a \%$" means $ \frac {a}{100}$.  So "what number is 5% of 2000" gets translated to $ x \equal{} \frac {5}{100} \times 2000$, and this agrees with what JRav wrote.  If you like, I suggest you try the other problems here now and someone will be able to tell you if you're correct.

Probably any algebra or pre-algebra textbook should cover this material, so you might go have a look at your local library, say.

When they say what number is x% of y (for example 5% of 2000) they basically mean 5%*2000. A percent is 1/100 or .01, so 5 percents are 5/100 or .05 So you then do .05*2000

Doing Vertical multiplication

 2000
 * .05
 ____
 10000
+00000
10,000
Now move the decimal point 2 spots over since there were two decimals in the multiplication. ( it is now invisible, right after the last 0 in 10,000)
So you get 100.00 or 100.

For your 2nd one, Instead of having what number is x% of y, you have the number and the percent. So You have 35% of y is 70. By doing the same thing as before, 35% of y is 35%*y, so .35*y=70.Since the opposite of multiplication is division, Divide both sides by .35.

So you have y=70÷.35  Use long division to now get the answer.

http://www.mathsisfun.com/long_division.html

You get the answer as 200.

The other questions are just combinations of these two properties.


How many non-integer solutions are there to $ a_{1}\plus{}a_{2}\plus{}a_{3}\equal{}n$ if $ 0\le a_{1},a_{2},a_{3} \le 2$?

If $ n \equal{} 0$ or $ n \equal{} 6$, the unique solution is integral.  If $ n < 0$ or $ n > 6$, zero solutions.  If $ 0 < n < 6$, infinitely many solutions.  I assume the question was meant to be different, though.

oh sorry what I meant to say was:

How many solutions are there, in terms of $ k$ and $ n$, to $ a_{1} \plus{} a_{2} \plus{} \cdots \plus{} a_{n} \equal{} k$ if $ a_{1},a_{2},\cdots,a_{n} \equal{} 0, 1,$ or $ 2$?

The $ x^k$ term in $ (x^2\plus{}x\plus{}1)^n\equal{}\frac{(1\minus{}x^3)^n}{(1\minus{}x)^n}\equal{}(1\minus{}x^3)^n\cdot \left( \sum_{j\equal{}0}^\infty \binom{n\plus{}j\minus{}1}{j}\cdot x^j\right)$ from here it is easy to get a summation. I don't know about a closed form..

Perhaps a recursive solution would work well here too.

Yes, easy to compute recursively.  In the [url=http://www.research.att.com/~njas/sequences/]OEIS[/url] it's sequence [url=http://www.research.att.com/~njas/sequences/A027907]A027907[/url].  No simple closed form.  (Sums of binomial coefficients, etc.)

hm according to http://mathworld.wolfram.com/TrinomialCoefficient.html, a closed form does exist, not very pretty though.


Each of the digits $1$, $3$, $7$ and $9$ occurs at least once in the decimal representation of some positive integers. Prove that one can permute the digits of this integer such that the resulting integer is divisible by $7$.

Number is greater than 10000, a permutation of it can be written $10^4*A + B$ where B is build with some permutation of 1,3,7,9, and $A \geq 0$

Mod 7, $r = 10^4*A$ is one of $0 ... 6$.

If r = 6 then take B = 1793
If r = 5 then take B = 3719
If r = 4 then take B = 1739
If r = 3 then take B = 1397
If r = 2 then take B = 1937
If r = 1 then take B = 1973
If r = 0 then take B = 1379

Is there a faster way rather than guess and check for the combinations of 1379 for each of the possible r (0,1,2,3,4,5,6) ?

$1, 3, 7, 9 \equiv 1, 3, 0, 2 \bmod 7$.  These numbers make up the first half of the residue class $\bmod 7$, and note also that

$10^0 \equiv 1 \bmod 7$
$10^1 \equiv 3 \bmod 7$
$10^2 \equiv 2 \bmod 7$
$10^3 \equiv 1 \bmod 7$

If I knew why it was so, I could say that a linear combination of these two residue classes should produce the entire residue class $\bmod 7$ (it makes sense to me), but I don't know why.  :(

[hide=My Solution]
There are two cases to consider based on the number of extra digits in the number we want to permute.

[b]Case 1: Extra digits (if any) are only zeroes[/b]

Essentially, we want a permutation of digits $1, 3, 9$ that is divisible by 7.  That way, we can "insert the zeroes" between said permutation and 7 to also get a number divisible by 7 because if $n \equiv 0 \pmod{7}$, then $10^k \cdot n \equiv 0 \pmod{7}$.  After trial and error, we find that $931$ is divisible by 7, so we can let the permutation be $9317$ and keep inserting zeroes between the $1$ and $7$.

[b]Case 2: Extra digits (if any) include at least one non-zero digit[/b]

Extra digits can be used to form a number greater than 0.  Then proceed like in the solution from tµtµ.
[/hide]


series summation
combinatorics
binomial coefficients
Summation
Combinatorial Identity
IMO Shortlist

We can rewrite it as $S=\sum_{k = 1}^{n/2} \frac{(-1)^k}{k} {n-k \choose k-1} = 0$, where $n=1990$.
Using identity $\binom{n-k}{k-1}=\binom{n-k}{n-2k+1}$, we see that \[S=\mathop{coef}_{u^{-1}}\sum_{k=1}^{n/2}\frac{(-1)^k}{k}(1+u)^{n-k}u^{-n+2k-2}=\] \[=\mathop{coef}_{u^{-1}}\frac{(1+u)^n}{u^{n+2}}\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\left(\frac{u^2}{1+u}\right)^k=\mathop{coef}_{u^{-1}}\frac{(1+u)^n}{u^{n+2}}\ln\left(1+\frac{u^2}{1+u}\right)=\] \[=\mathop{Res}_{u=0}\frac{(1+u)^n}{u^{n+2}}\ln\left(1+\frac{u^2}{1+u}\right).\]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=35810 for a continuation.

Though I don't think that ISL assumed such an approach, Peter.

forgive me for asking, 'cause i don't know, but what does Res refer to?

hey, and thanks for responding, Myth.

i think you're right, the proposers probably didn't have a solution like this in mind.  i haven't really worked on the problem yet, but it seems to me like it should have a combinatorial interpretation, probably having to do with inclusion-exclusion and probabilities.

Res stands for residue, e.g, the coefficient of $z^{-1}$ of the laurent series of $f(z)$

So does someone have a combinatorial proof ????

Let $S$ denote the sum then after some simplifications, I got
\[S=\sum_{k=0}^{995} (-1)^k \cdot \frac{(k+1)(k+2)\cdots (1989-2k)}{1991-2k}\]

Can anyone help me for further?

So there's no one have combinatorial proofs? :oops_sign:


search
number theory unsolved
number theory

[quote="sylow_theory"][quote="PenguinIntegral"]For any integer m, prove that $2m$ can be expressed in the form $a^{19}+b^{99}+k*2^{1999}$, where a and b are odd integers and k is a non-negative integer.

Sorry if this is a repost, but I could not find the problem with the search function.[/quote]

Okay, so how would you express 2m=2(2)=4 this way  :?:[/quote]
Make a and b negative.

[quote="PenguinIntegral"][quote="Baronov"]Ok you just need to prove that there exhist a and b such that :
$a^{19}+b^{99}\equiv 2m(mod 2^{1999})$ but since 19 and 99 are odd and $\phi(2^{1999})=2^{1998}$ it does not seem that hard.[/quote]
Try it. I think you will find it is not *that* easy.[/quote]

But why since 19 and 99 are odd if you take u and v be odd such that $u+v=2m$ you can find some x and y such that $x^{19}\equiv u(mod 2^{1999})$ and $y^{99}\equiv v(mod 2^{1999})$ now if $(x,y)$ is a solution to this then $(x-2^{1999},y-2^{1999})$ is also a solution so we can make x and y negative and so k will be positive.


algebra polynomial LaTeX articles search function trigonometry

The answer is any variant on the form:

[hide](x - y)(x - z)(x - w)(y - z)(y - w)(z - w).[/hide]

I'd post my solution, but I'm at MOP now and my solution is at home. I do remember that I did get the same polynomial as JBL.

ZennyK wrote:Since I have no idea when the solutions will be online, could someone please post their solution to problem 3 of round 4, or at least give the answer? Thank you in advance whoever replies to this.

I had written up a credits and quick answers page for round 4 months in advance. I don't know why it never got posted at the NSA web site.

I have the LaTeX file for it. So I will post it here. I will also enclose it in spoiler brackets even though Year 15 Round 4 is over. I hope you can read raw LaTeX.

Erin Schram

[hide]\documentclass[12pt]{article}

%\pagestyle{empty}

\setlength{\oddsidemargin}{0pt}

\setlength{\evensidemargin}{0pt}

\setlength{\topmargin}{-0.5in}

\setlength{\textwidth}{6.5in}

\setlength{\textheight}{9in}

\usepackage[pdftex]{graphicx}

\usepackage{amsmath,amssymb}

\begin{document}

\begin{center}\large\bfseries

USA Mathematical Talent Search\\

CREDITS and QUICK ANSWERS\\

Round 4 --- Year 15 --- Academic Year 2003--2004\\

\end{center}

\setlength{\parindent}{0pt}

\setlength{\parskip}{3pt}

\textbf{1/4/15.}

\parbox[t]{5.75in}{\footnotesize\sffamily

Find, with proof, the smallest positive integer  for which the sum of the digits of  is as small as possible.}

This problem was fashioned after Problem 3 of the 2nd Lithuanian Olympiad administered at Vilnius University on September 30, 2000.

We search for  rather than directly for . Numbers with only a single nonzero digit are that digit times a power of ten, which would not be divisible by . Of the numbers with two nonzero digits, the smallest digit sum occurs with both nonzero digits being~. The smallest such number must begin and end in~, because if it ended in zero, we could divide by . So we are looking for the smallest integer of the form  that is a multiple of .

The powers of  mod~ are , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . so . Therefore  is divisible by , and .

\bigskip

\textbf{2/4/15.}

\parbox[t]{5.75in}{\footnotesize\sffamily

For four integer values of  greater than six, there exist right triangles whose side lengths are integers equivalent to , , and  modulo~, in some order. Find those values. Prove that at most four such values exist. Also, for at least one of those values of , provide an example of such a triangle.}

This problem was devised by Dr.~Peter Anspach of the NSA, after a lunchtime conversation among mathematicians about trying to create a  right triangle.

Let , , and  be the sides of such a right triangle, with . If , then  so  is a factor of . If , then  is a factor of . If , then  is a factor of , but since , we can reject this case. Thus,  is a factor of  or , so  is , , , or .

Given an , a triple  that forms a right triangle can be found by trial and error, by applying number theory to an equation such as , or by the  method of finding Pythagorean triples after solving for , , and  that give the right values mod~. The first three solutions for each case are:

\begin{description}

\item for  and : , , ;

\item for  and : , , ;

\item for  and : , , ;

\item for  and : , , ;

\item for  and : , , .

\end{description} \vfil

\eject

\textbf{3/4/15.} \parbox[t]{5.75in}{\footnotesize\sffamily Find a nonzero polynomial  in the four indeterminates , , , and~ of minimum degree such that switching any two indeterminates in the polynomial gives the same polynomial except that its sign is reversed. For example, . Prove that the degree of the polynomial is as small as possible.}

This problem was invented by Dr.~David Grabiner, an NSA mathematician who helps with the evaluation and grading of problems for the USAMTS.

Consider any two of the indeterminants, say  and .

If the standard form of  has a term , then it must have a matching term . Grouping them together gives either  or , which both are divisible by . Since the

entire polynomial can be grouped this way, it is divisible by .

Thus, for every pair of indeterminants, the polynomial is divisible by their difference. This means  is divisible by . Letting  keeps the degree to a minimum, and it works because switching any two indeterminants in  gives the same polynomial except that

its sign is reversed.

\bigskip

\textbf{4/4/15.}

\parbox[t]{5.75in}{\footnotesize\sffamily

For each nonnegative integer  define the function  by

\begin{displaymath}

f_n(x) = \sin^n(x) + \sin^n(x + \frac{2\pi}{3}) + \sin^n(x + \frac{4\pi}{3})

\end{displaymath}

for all real numbers , where the sine functions use radians. The functions  can be also expressed as polynomials in  with rational coefficients. For example, \begin{eqnarray*}
&\displaystyle f_0(x) = 3,\qquad f_1(x) = 0,\qquad f_2(x) = \frac32,\qquad f_3(x) = -\frac34\sin(3x),\\
\noalign{\smallskip}
&\displaystyle f_4(x) = \frac98,\qquad f_5(x) = -\frac{15}{16}\sin(3x),\qquad f_6(x) = \frac{27}{32} + \frac{3}{16}\sin^2(3x),
\end{eqnarray*}

for all real numbers~. Find an expression for  as a polynomial in  with rational coefficients, and prove that it holds for all real numbers .}

This problem was devised by the distinguished Hungarian mathematician and poet Mih\'aly Bencze of Brasso, Transylvania.

The triple angle formula for sine is . But

since  also equals  and , it also expands to  and . Thus, for all real numbers~,

\begin{eqnarray*}
f_n(x)\sin(3x) &=& \sin^n(x)\sin(3x) + \sin^n(x + \frac{2\pi}{3})\sin(3x) + \sin^n(x + \frac{4\pi}{3})\sin(3x)\\
&=& \sin^n(x)\bigl(3\sin(x) - 4\sin^3(x)\bigr)\\
&& \qquad {}+ \sin^n(x + \frac{2\pi}{3})\bigl(3\sin(x + \frac{2\pi}{3}) - 4\sin^3(x + \frac{2\pi}{3})\bigr)\\
&& \qquad {}+ \sin^n(x + \frac{4\pi}{3})\bigl(3\sin(x + \frac{4\pi}{3}) - 4\sin^3(x + \frac{4\pi}{3})\bigr)\\
&=&3\bigl(\sin^{n+l}(x)+\sin^{n+l}(x + \frac{2\pi}{3})+\sin^{n+l}(x + \frac{4\pi}{3})\bigr)\\
&&\qquad {}-4\bigl(\sin^{n+3}(x) + \sin^{n+3}(x + \frac{2\pi}{3}) + \sin^{n+3}(x + \frac{4\pi}{3})\bigl)\\
&=&3f_{n+l}(x) - 4f_{n+3}(x).
\end{eqnarray*}

Setting  to 4 and rearranging gives

.

\bigskip

\textbf{5/4/15.} \parbox[t]{2.75in}{\footnotesize\sffamily

Triangle  is an obtuse isosceles triangle with the property that three squares of equal size can be inscribed in it as shown on the right. The ratio  is an irrational number that is the root of a cubic polynomial. Determine that polynomial.} \raisebox{-70pt}[0pt]{\setlength{\unitlength}{2pt}

\begin{picture}(95,40)

\put(10,6){\line(1,0){80}}

\put(10,6){\line(4,3){40}}

\put(90,6){\line(-4,3){40}}

\put (5,0){A}

\put (48,37){B}

\put(91,0){C}

\put(39.1,6){\line(0,1){21.8}}

\put(39.1,27.8){\line(1,0){21.8}}

\put(60.9,6){\line(0,1){21.8}}

\put(45.7,6){\line(-3,4){12.85}}

\put (45.7, 6){\line (4, 3){17.1}}

\put(50,36){\line(3,-4){12.85}}

\put(54.3,6){\line(3,4){12.85}}

\put(54.3,6){\line(-4,3){17.1}}

\put(50,36){\line(-3,-4){12.85}}

\put(29,24){D}

\put(36, 30){E}

\put(60,30){F}

\put(67,24){G}

\put(37,0){J}

\put(43,0){K}

\put (53,0){K}

\put(59,0){M}

\put (31,16){H}

\put(65.5,16){I}

\end{picture}}

\medskip

This triangle was apparently discovered by the Italian mathematician Eugenio Calabi. It is the only non-equilateral triangle into which we can fit three equal squares in this manner. The triangle is displayed on page 266 of {\em The Book of Numbers,} by John H. Conway and Richard K. Guy, published by Springer-Verlag in 1996.

\medskip

Let  be the length of a side of the squares. By symmetry, , so . . Triangle  is congruent to triangle , so . So , giving  and .

Let  be the midpoint of side . Triangle  is

similar to triangle . So . .

By the Pythagorean Theorem , which converts to \begin{eqnarray}
\left(2(AC-AB)\left(\frac{AB}{AC}\right)\right)^2
&=& (AC-AB)^2 + (2(AB)-AC)^2\\
\left(2(AC-AB)(AB)\right)^2
&=&(AC)^2(AC-AB)^2 + (AC)^2(2(AB)-AC)^2.
\end{eqnarray}

Let . Equation (2) converts to

\begin{eqnarray*}
4(x-1)^2 &=& x^2(x-1)^2 + x^2(2-x)^2\\
(4-x^2)(x-1)^2 - x^2(2-x)^2 &=& 0\\
(2-x)(x+2)(x-1)^2 - x^2(2-x)^2 &=& 0\\
(2-x)\bigl((x+2)(x-1)^2 - x^2(2-x)\bigr) &=& 0\\
(2-x)(2x^3-2x^2-3x+2) &=& 0
\end{eqnarray*}

By the triangle inequality, , so  and . Therefore,  is the root of the cubic polynomial .

\end{document}[/hide]

The End


algorithm
number theory
Euclidean algorithm

Find all possible 8 digit numbers n such that the last 8 digits of n^2 is n.  A great and beautiful solution would be greatly appreciated[/img][/code]

[hide="Solution"] We want to solve $n^{2}\equiv n \bmod 10^{8}$.

Well, let's solve $\bmod 2^{8}, \bmod 5^{8}$.  We have the solutions $n \equiv 0, 1$ in both cases, so by CRT two easy solutions are $n \equiv \boxed{ 0, 1 }$.

The other two solutions $n_{3}, n_{4}$ are when

$n_{3}\equiv 0 \bmod 2^{8}$
$n_{3}\equiv 1 \bmod 5^{8}$

And

$n_{4}\equiv 1 \bmod 2^{8}$
$n_{4}\equiv 0 \bmod 2^{8}$

By the Extended Euclidean Algorithm,

$390625 = 1525 \cdot 256+225$
$256 = 1 \cdot 225+31$
$225 = 7 \cdot 31+8$
$31 = 3 \cdot 8+7$
$8 = 1 \cdot 7+1$

This gives

$390625 \cdot 33-256 \cdot 50354 = 1 \Leftrightarrow$
$256 \cdot 340271-390625 \cdot 223 = 1$

So we conclude that

$n_{3}= 390625 \cdot 33 = \boxed{12890625}$
$n_{4}= 256 \cdot 340271 = \boxed{87109376}$ [/hide]

[quote="t0rajir0u"][hide="Solution (quoted)"]
By the Extended Euclidean Algorithm,

$390625 = 1525 \cdot 256+225$
$256 = 1 \cdot 225+31$
$225 = 7 \cdot 31+8$
$31 = 3 \cdot 8+7$
$8 = 1 \cdot 7+1$

This gives

$390625 \cdot 33-256 \cdot 50354 = 1 \Leftrightarrow$
$256 \cdot 340271-390625 \cdot 223 = 1$

So we conclude that

$n_{3}= 390625 \cdot 33 = \boxed{12890625}$
$n_{4}= 256 \cdot 340271 = \boxed{87109376}$ [/hide][/quote]

I'm sorry, but can you explain this further? How do you use the Euclidean Algorithm to find the unique solution provided by the CRT?  :)

[size=75](Oh, and I believe you have a typo at about half-way in your solution.)[/size]

[quote="t0rajir0u"][hide="quoted"]http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm

Given two positive integers $a, b, (a, b) = 1$ we can use the Extended Euclidean algorithm to find $x, y$ such that

$ax+by = 1$

Note that $ax$ satisfies

$ax \equiv 0 \bmod a$
$ax \equiv 1 \bmod b$

And that $by$ satisfies

$by \equiv 0 \bmod b$
$by \equiv 1 \bmod a$

I did some unnecessary extra work to find $by$.  Ignore that part.  Note that this allows us a much more general result:

Let $u = axn+bym$.  Then

$u \equiv m \bmod a$
$u \equiv n \bmod b$[/hide][/quote]

Oooh, alright. That's pretty tricky of you.  :P
Thanks for explaining it to me.


Solve equation:
$ 2x^2 \minus{} 6x \plus{} 1 \equal{} \sqrt{4x \plus{} 1}$

[hide]Squaring both sides and combining like terms, we have $ 4x^4\minus{}24x^3\plus{}40x^2\minus{}16x\equal{}0$. The roots of this polynomial are $ 0$, $ 2$, and $ 2\pm\sqrt{2}$. However, the only roots that work are $ \boxed{0, 2\plus{}\sqrt{2}}$.[/hide]


Hi everybody,
last Saturday i made the DC state team for the nationals and started studying.  Here is one of my questions if anyone could please help.

2004 National Sprint round # 26

The product 2^2 x 3^3 x 5^7 can be rewritten in the form a^2 x b^3 x c with positive integers a, b, and c.  What is the greatest possible value of a x b -- c?

Sorry about the weird equations.  i'm still tryin to figure out how to use the LaTex.  i meant for"--" to mean minus.

For this question the answer is 749 while the closest i got was 747 by grouping together 2, 3, 5, and 5 as "a" which would equal 150 and grouping the three 5's as "b" which would equal 5.  The product of 150 and 5 is 750 and the only number remaining for "c" would be 3.  When three is subtracted from 750, the answer is 747.  I am not sure as to how the book gets 749.

Thanks for the help

Well obviously you want to maximize a, then b. (You don't want anything in c)
You can do so with 50^2*15^3*1, for 749.

Hint: (2,5,5)^2*(3,5)^3


If we have a number and the sum of its digits is divisible on 3 then the number itself is divisible by 3.  Prove the statement.

consider the decimal representation of a number. since 10 is congruent to 1 modulo 3, the number is congruent to the sum of it's digits modulo 3. then it's divisible by 3 if and only if the sum of its digits is divisible by 3.

also, if a number is written in base b, the number is divisible by b-1 if and only if it's divisible by the sum of its digits (this also work for every divisor of b-1).

forgot to tell this only works for bases on the form 3k+1

[hide="formal solution"]A number is of the form $ a_n*10^n\plus{}a_{n\minus{}1}*10^{n\minus{}1}\plus{}\cdots \plus{}a_0$, where $ a_i$ are digits and $ a_n\neq 0$. Since $ 10^n\minus{}1\equal{}9999.....9999$, it's divisible by 3, so we subtract lots of 3's and we have the same remainder when divided by 3 as the original number: $ a_n\plus{}a_{n\minus{}1}\plus{}\cdots \plus{}a_0$. We can work backwards, so since that sum is divisible by 3, the original n+1 digit number is divisible by 3.[/hide]


superior algebra
superior algebra unsolved

For a commutative ring $ R$, find the two sided ideals of $ n \times n \minus{}$ matrices i.e. $ M_n(R)$.

Abstract algebra goes [url=http://www.artofproblemsolving.com/Forum/index.php?f=8]here[/url].


How do you find the mean of a set of numbers if one of the numbers is negative. For example, the set of numbers: 2,8,-7,3.

The mean of any set of real numbers $ a_1,a_2,\ldots,a_n$ is $ \frac {a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}n$. It doesn't matter if they're negative or not.

So $ \frac {2 \plus{} 8 \minus{} 7 \plus{} 3}4 \equal{} \frac64 \equal{} \boxed{\frac32}$.


quickly distributing, a,b, and c are equivalent; only (d) is not.


modular arithmetic
number theory open
number theory

Hi,

You probably already know the Lucas-Lehmer-Test (LLT) used for proving that a Mersenne number is prime or composite. (See: [url=http://mathworld.wolfram.com/Lucas-LehmerTest.html]MathWorld[/url] and a [url=http://www.mersennewiki.org/index.php/Lucas-Lehmer_Test#Proof_of_the_Lucas-Lehmer_test]proof[/url]).

The LLT is based on the properties of the Tree built by $x^2-2$ modulo a Mersenne number. And this Tree is only a part of the DiGraph under $x^2-2$ modulo a Mersenne prime. The other part is made of Cycles of lengths L such that L divides q-1. See: [url=http://www.cs.uwaterloo.ca/~tmjvasig/papers/newvasiga.pdf]Shallit and Vasiga[/url] paper. See: [url=http://www.mersenneforum.org/showthread.php?t=5777]GIMPS forum[/url] for finite formulae of the number of Cycles modulo a Mersenne prime.

Now, here is a [b]conjecture[/b] (checked up to M26 = $M_{23209}$) based on the properties of the Cycles of length (q-1) built by $x^2-2$ modulo a Mersenne number:

$\large M_q=2^q-1 \text{ is prime } \Longleftrightarrow \ S_{q-1} \equiv S_0 \ \pmod{M_q} \\ \\ \text{ where: } S_0=3^2+1/3^2 , \ S_{i+1}=S_i^2-2 \ .$

Can you prove it ?

Tony


modular arithmetic
search
number theory
prime numbers
number theory unsolved

prove that the equation :
$ 2^{x}\plus{}3^{y}\equal{}p$ , for positive integers $ x,y$ and the prime number $ p$ , has infinitely many solutions.
and find the conditions when is the result divisible by $ 5$ for some $ x,y$ .

Good luck :)

[quote="BadrGhamdi"]find the conditions when is the result divisible by $ 5$ for some $ x,y$ . [/quote]

what "result"?

the result obtaind from $ 2^{x}\plus{}3^{y}$ ..
i mean whe is $ p$ divisible by $ 5$ .. find the exact conditions for that .

[quote="pluristiq"]but you said $ p$ is prime.[/quote]

So $ p\equal{}5$
And $ 2^x\plus{}3^y\equal{}5$

And so $ x\equal{}y\equal{}1$

you misunderstood me ..  i'll rewrite the question

[b]problem-1 : [/b]
prove that the equation :
 $ 2^{x}\plus{}3^{y}\equal{}p$ , for positive integers $ x,y$ and the prime number $ p$  , has infinitely many solutions.

[b]problem-2 :[/b]
find the conditions where $ 2^{x}\plus{}3^{y}$ is divisible by $ 5$ . not necessary a prime.

[quote="BadrGhamdi"]you misunderstood me ..  i'll rewrite the question

[b]problem-1 : [/b]
prove that the equation :
 $ 2^{x} \plus{} 3^{y} \equal{} p$ , for positive integers $ x,y$ and the prime number $ p$  , has infinitely many solutions.

[b]problem-2 :[/b]
find the conditions where $ 2^{x} \plus{} 3^{y}$ is divisible by $ 5$ . not necessary a prime.[/quote]

Thanks for being more clear.
To be full precise, I think that, in problem 1, the variables are $ x,y$ and $ p$, and not only $ x,y$ (and $ p$ a parameter). Am I right ?

For second problem, you get $ 2^{x\plus{}y}\plus{}1\equal{}0\pmod 5$ $ \iff$ $ x\plus{}y\equal{}2\pmod 4$

[quote="pco"][quote="BadrGhamdi"]you misunderstood me ..  i'll rewrite the question

[b]problem-1 : [/b]
prove that the equation :
 $ 2^{x} \plus{} 3^{y} \equal{} p$ , for positive integers $ x,y$ and the prime number $ p$  , has infinitely many solutions.

[b]problem-2 :[/b]
find the conditions where $ 2^{x} \plus{} 3^{y}$ is divisible by $ 5$ . not necessary a prime.[/quote]

Thanks for being more clear.
To be full precise, I think that, in problem 1, the variables are $ x,y$ and $ p$, and not only $ x,y$ (and $ p$ a parameter). Am I right ?

For second problem, you get $ 2^{x \plus{} y} \plus{} 1 \equal{} 0\pmod 5$ $ \iff$ $ x \plus{} y \equal{} 2\pmod 4$[/quote]

yeah , you're right :) ..

the solution for the second problem is kind of  unclear ..
did you assume that we have no $ 3$ ?? will we get the same solution if we solve for $ 1 \plus{} 3^{x \plus{} y}\equal{}0 (mod5)$ ??

$ 3\equiv 2^{\minus{}1} \pmod{5}$, so $ 0\equiv 2^x\plus{}3^y\equiv 2^x\plus{}2^{\minus{}y} \pmod{5} \Leftrightarrow 2^{x\plus{}y}\plus{}1\pmod{5} \Leftrightarrow x\plus{}y\equiv 2 \pmod{4}$

[quote="pco"][quote="BadrGhamdi"][quote="pluristiq"]the first part seems difficult. additive results about primes are usually difficult to establish. can you give us a hint? :)[/quote]

i recommend an investigation .. plug some values for $ x,y$ .. i think you'll find a nice pattern[/quote]

I'm very surprised that a pattern exist on $ x,y$ which allow prime generation. It's a great great thing and I hope you'll soon give us the solution.

I thought that the problem was to show the existence of infinitely many solutions without general formulas.[/quote]

i didn't think you have that great sense of humor .. but seeing a pattern is always helpful in such a problem .. you can reach to properties which will lead you to the answer ..
thanks for your response

Give us a hint at least! Our primitive minds don't see the patterns you see.

i'll ask the problem proposer to give a hint ..

[quote="BadrGhamdi"] i didn't think you have that great sense of humor .. but seeing a pattern is always helpful in such a problem .. you can reach to properties which will lead you to the answer .. [/quote]

I'm sorry if my answer bothered you. Excuse me for that.

I saw no pattern at all which allow to be sure that $ 2^{\text{something}}\plus{}3^{\text{something else}}$ is prime.

To be honest, I was quite sure no such formula existed and I'm really very interested in your solution.

As Pluristiq said, could you just give us a hint (different from "hint : search for a pattern").

Thanks a lot.

Nota : it seems you are the problem proposer and, since you posted in "proposed and own" category, you have the solution.

[quote="pco"][quote="BadrGhamdi"]i was at an Olympiad training camp last week .. one of my friends their .. proposed this problem .. i was thinking a lot about it .. and when i asked him for a hint .. he said find a pattern :) , i tried to call him now  but he doesn't answer , so let's have some tries until he appears :) .. thanks for your replies ,[/quote]

Front page :

  Number Theory Unsolved Problems : Problems you couldn't solve and to which you know that there is a solution (i.e. a problem from a contest, etc.) but you don't know it.

  Number Theory Proposed & Own Problems : Problems[u][b] that you have already solved [/b][/u]and you are interested in second opinions or solutions.

  Number Theory Open Questions : An open question is a question that has no known solution up to this moment, and it is not known wheter the problem has one or not.[/quote]

sorry for that ,, i hope one of the mods will move it please

and please, for our mass peace of mind, ask your olympiad coach or whoever it was, what the solution to this problem is, if there really is one...

Let us fix $y$. How would we know how many (finite or infinite) prime numbers $p$ are of the form $2^x + 3^y$, if we can't even establish that for $y=0$, for which we get the Fermat numbers? It certainly is some hoax.


group theory
abstract algebra
superior algebra
superior algebra unsolved

let  H subgroup of G define N(H)={x/x belongs to G,s.t.(x)H( x inverse)=H}if  P  isp sylow subgroup of G  then N(N(P))=N(P).


If $ n$>$ 1$ and $ p_n$ is $ n^{th}$ prime number then $ p_n < 2^n$

[hide="induction"]
Trivial base case, for the inductive step, we have that $ 2p_{n}>p_{n\plus{}1}$ by bertrand's postulate.[/hide]

The priniciple of mathematical induction is basically that if you can prove something for a base case, (usually the smallest or simplest case) and then prove an "inductive step", (which is usually something like: "IF the statement is true for any particular case, THEN it's also true for the next case") then it is true for every case.    A simple example is to imagine you tip the first in a line of dominos, and you KNOW that every domino will knock over the next domino.  Then, you know that all the dominos will fall.

In this case, the base case would be $ n \equal{} 2$, in which case the inequality is obviously true.  diophantient is saying that if the inequality is true for any particular $ n$, then it's also true for $ n \plus{} 1$ by bertrand's postulate.  Now, since you know that the statement is true for $ n \equal{} 2$, and that if it is true for any particular $ n$ it is also true for $ n \plus{} 1$, by the principle of mathematical induction, you know that it is true for ALL numbers $ n$ greater than 1.

If you don't understand, I suggest you do some research on the internet on induction. (Note:  Induction has absolutly nothing to do with inductive reasoning.  Induction is in fact considered deductive reasoning.)

Yes ,I have some basic skills in mathematical induction , but there are many problems "by full induction " .I'm not sure what they mean by full induction  :wink:


linear algebra
matrix
algebra
polynomial

This was one question on an exam in a sophomore level Introduction to Linear Algebra course. It's my course and my question - and (I hope) it wasn't supposed to be all that difficult. The default assumption is that all matrices are over $\mathbb{R}.$

----

Mathematical true-false. For each of the following statements decide whether the statement is always true or sometimes false. If true, give a justification. If false, give a counterexample.

(a) If the characteristic equation for $A$ has real roots, then $A$ is diagonalizable.

(b) If the elements of each row of a square matrix $A$ add up to $137,$ then $137$ is an eigenvalue of $A.$

(c) If $A$ is a $9\times 7$ matrix of rank $7,$ then $Ax=0$ has only the trivial solution.

(d) If $A$ is $8\times 8$ and $A-9I$ has rank 5, then $(\lambda-9)^{3}$ is a factor of the characteristic polynomial for $A.$

(e) If all of the eigenvalues of $A$ are real, then $A^{2}+I$ is invertible.

(a) Sometimes false: $\left(\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}\right)$, could be slightly hard on the spot to think of, but definitely doable.

(b) Sometimes false: $\left(\begin{matrix}1 & 136\\ 136 & 1\end{matrix}\right)$, tons of examples, if people understand the concepts they should be fine.

I'll come back to the rest later, but these are my initial impressions.

Take another look at your answer to (b).

Wow, I feel like an idiot for making it sound like that was an obvious counterexample. Don't take that to be a hard question, though, in a test situation I wouldn't be listening to music and talking to people etc...excuses

redo (b) True: Let $A$ be an $n\times n$ matrix such that every row sums to $137$, then $Av=137v$ where $v=(1,1,...,1)$ with $n$ entries. So $137$ is an eigenvalue.

Might as well take a crack at the others:

(c) True: Suppose $A$ is $9\times 7$ and has rank $7$, then it rrefs down to 1's on the "diagonal" except for the last two rows which will all be zero (null space dimension 2). So $Ax=0$ means we get a set of equations $(x_{1}, 0, 0, 0, ..., 0)=0$ and $(0,x_{2}, 0, ..., 0)=0$ etc, so all the components must be zero, the trivial solution.

(d) I feel like this is true. Since $A-9I$ has rank $5$, the characteristic equation of degree $8$ should have a factor of $(\lambda-9)^{5}$ and thus has a factor of $(\lambda-9)^{3}$. My doubts come from remembering that the order of that factor has to do with raising $(A-9I)$ to a power and not the rank. Maybe I'm just remembering incorrectly or thinking of something else though.

(e) My comments are that if the eigenvalues of $A$ are real, then the eigenvals of $A^{2}$ are real. Not sure if this comes into play. I feel like this doesn't have to be true, but I don't have a counterexample.

On (d), we cannot guarantee a factor of $(\lambda-9)^{5}$, so I would not give you full credit for that answer.

Here's the real reason: Since $A-9I$ has rank $5,$ it has nullity $8-5=3.$ That means that $9$ is an eigenvalue of $A$ of geometric multiplicity $3.$ The algebraic multiplicity of an eigenvalue is greater than or equal to the geometric multiplicity - and saying that the algebraic multiplicity is at least $3$ is saying that $(\lambda-9)^{3}$ is a factor of the characteristic polynomial.

I would give you full credit for (c). For my own sake, I would phrase the answer this way: Since the matrix has rank $7,$ the dimension of the column space is $7.$ With just $7$ columns, the columns are independent. And the columns of $A$ being independent is precisely the same as saying that $Ax=0$ has only the trivial solution.

I won't comment on (e) just yet.

[quote="HilbertThm90"](e) My comments are that if the eigenvalues of $A$ are real, then the eigenvals of $A^{2}$ are real. Not sure if this comes into play. I feel like this doesn't have to be true,[/quote]

This is true (for any integer $k$, the eigenvalues of $A^{k}$ are the $k$-th powers of the eigenvalues of $A$, with the same algebraic multiplicities - but don't ask me how to prove this without some ugly polynomial arguments), but you don't need it. (e) is trivial if you work over $\mathbb{C}$: Since all eigenvalues of $A$ are real, in particular, it follows that neither $i$ nor $-i$ are eigenvalues of $A$. Hence, the matrices $A-iI$ and $A+iI$ are invertible. Therefore, their product, $\left(A-iI\right)\left(A+iI\right)=A^{2}-i^{2}I^{2}=A^{2}-\left(-1\right)I=A^{2}+I$, is invertible as well.

Here, we have silently used the fact that a real square matrix is invertible over $\mathbb{R}$ if and only if it is invertible over $\mathbb{C}$. This is trivial (e. g. using the criterion that a square matrix over a field is invertible if and only if its determinant is nonzero), but it is useful not to overlook such arguments.

  Darij

This is the first I have seen this problem.  For (e), couldn't you just argue that if $A^{2}+I=0$, then the minimal polynomial divides $x^{2}+1$. But if all the eigenvalues are real, you have a contradiction?

[quote="edoo"]This is the first I have seen this problem.  For (e), couldn't you just argue that if $ A^{2}+I=0$, then the minimal polynomial divides $ x^{2}+1$. But if all the eigenvalues are real, you have a contradiction?[/quote]

"Not invertible" and "equals 0" are pretty different things to say about matrices...

  Darij

For c, it's saying that the column space of A is 7 dimensional. Since the 7 columns of the matrix span the column space, they must be independent. Hence they only span 0 in one way.


Find all real [tex]\alpha[/tex] such that [tex]\{ j \cdot \alpha \}[/tex] rounded to the nearest integer is equal to [tex]1[/tex] when [tex]j[/tex] is even and equal to [tex]0[/tex] when [tex]j[/tex] is odd.

By $\{ j \cdot \alpha \}$ I think you are meaning the fractional part, am I wrong?

In that case, there is no such $\alpha$ :

- If $\alpha$ is irrational :
If $\theta$ is irrational it is a well-known result that $\{n \theta \}$ is dense in $[0;1]$. Using this for $\theta = 2 \alpha $, it follows that there are positive integers $j$ such that $\{2j \alpha \}$ is arbitrary close to $0$, which contradicts the desired property.

- If $\alpha = \frac a b$ for some integers $a,b$ then for $j=2b$ we have $\{j \alpha \} = \{2a \} = 0$, which also contradicts the desired property.

Pierre.

How well-known is that well-known result ? :) Could you give some link or reference, I'd be very grateful. :)

It's alomost the same as another well known theorem cited in http://www.artofproblemsolving.com/Forum/viewtopic.php?p=317796#p317796.

Or use your friend Google to look for Kroenecker's theorem.

Pierre.

Thank you very much pbornsztein, kondr :)

Here is a proof, which, since it's my own, is probably unnecessarily long.

$s_n=\{n\alpha\}$ is an infinite bounded sequence, so we may extract a convergent subsequence from it; for every $\epsilon>0$, there exist $i,j\in\mathbb{N}$ such that $|a(i-j)|<\epsilon$.   When we remove the absolute values, either we can extract a subsequence such that $s_{n_i}=\{n_i\alpha\}\to 0$ with $s_{n_i}$ monotonically decreasing $(*)$, or $s_{k_i}=\{k_i\alpha\}\to 1$ with $s_{k_i}$ monotonically increasing.  In the second case, fix $1/2>\epsilon>0$.  If $s_{k_i}=1-\epsilon$, then for the largest $m$ such that $m\epsilon<1$, we have $0<ms_{k_i}<\epsilon$, so we can extract a sequence with property $(*)$.

Now say that we want to come within $\epsilon$ of a given real $r$ in $[0,1]$.  We have already proven it for $0$, and for $1$, given $\epsilon>0$, there is an $i$ such that $0<s_{n_i}<\epsilon$, and $1-\lfloor 1/s_{n_i} \rfloor\cdot s_{n_i}<\epsilon$, so we are done there.  Otherwise, given $\epsilon>0$ and $r\in (0,1)$, there is an $i$ such that $r+s_{n_i}<1$.  Then for the $m$ such that $ms_{n_i}< r< (m+1)s_{n_i}$, we have $r-ms_{n_i}<\epsilon$ (if instead equality holds, it is trivial), so we are done.  This shows that $\{s_n\}$ is dense in $[0,1]$.


geometry
perpendicular bisector
geometry unsolved

Let $ ABC$ be an acute triangle, $ D$ be the foot from $ A$ to $ BC$ and $ E$ be the foot from $ B$ to $ AC$, $ H$ be the orthocenter of $ ABC$ and $ C_1$ is a midpoint of $ CH$.
$ P$ be the intersection point of $ AB$ and perpendicular bisector of $ DE$ and $ M$ be the intersection point of $ DE$ and $ PC_1$.
Show that $ M\hat{C}H\equal{}C_1\hat{P}C$.

[quote="HelloWorld"]Let $ ABC$ be an acute triangle, $ D$ be the foot from $ A$ to $ BC$ and $ E$ be the foot from $ B$ to $ AC$, $ H$ be the orthocenter of $ ABC$ and $ C_1$ is a midpoint of $ CH$.
$ P$ be the intersection point of $ AB$ and perpendicular bisector of $ DE$ and $ M$ be the intersection point of $ DE$ and $ PC_1$.
Show that $ M\hat{C}H \equal{} C_1\hat{P}C$.[/quote]
Because $ ED$ is the radical axes of $ (CEHD)$ and $ (AEDB)$ hence $ P$ will be the center of $ (AEDB)$, which implies that $ P$ is midpoint of $ AB$. Using this fact, by simple angle chasing, we have $ PE,PD$ are the tangents of $ (CEHD)$ or in other word, $ C_1E,CD$ are also tangents of $ (AEDB)\Longrightarrow C_1E^2 \equal{} C_1M.C_1P \equal{} C_1C^2$, which implies that $ C_1C$ is the tangent of $ (CMP)\Longrightarrow \angle C_1CM \equal{} \angle CPC_1$. The result is lead as follow.


geometry
power of a point
radical axis
geometry proposed

ABCD is an inscribed quadrilateral in a circle O. The extensions of sides AB and DC meet at point E, and the extensions of sides AD and BC meet at point F. EG and FH are tangent to the circle O. If EG = a, FH = b and EF = c, prove that $ a^{2}\plus{}b^{2}\equal{}c^{2}$

$ EG,EI,FH$ are the tangent segments from $E,F$ to the circle $ (O).$ $ IG$ is the radical axis of the circles $ (E,EG)$ and  $ (O).$ Since $ F$ lies on their radical axis (polar of E WRT (O)), then it has equal power with respect to $(O)$ and $(E,EG)$

$ FB \cdot FC \equal{} FH^2 \equal{} EF^2 \minus{} EG^2 \ \Longrightarrow EF^2 \equal{} EG^2 \plus{} FH^2$


geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

Consider a system of infinitely many spheres made of metal, with centers at points $(a, b, c) \in \mathbb Z^3$. We say that the system is stable if the temperature of each sphere equals the average temperature of the six closest spheres. Assuming that all spheres in a stable system have temperatures between $0^\circ C$ and $1^\circ C$, prove that all the spheres have the same temperature.

Discussed before, in a pretty long thread covering this problem in dimensions $ 2$ and $ 3$ and analyzing the need for the boundedness requirement. A good starting point is:

http://www.mathlinks.ro/viewtopic.php?p=159575#159575

it's a problem about converge progressions.
furthermore,the condition "between 0 and 1" is waste,for this can be reduced to "the temperature is bounded",and apparently temperature has bounds(at least the lower bound).


How do they pick the IMO team out of the MOPers?

i heard from someone that they come in weighted based on how they did on the USAMO, and then depending on how they do during the MOP they get more "points" and the highest "point" totals go to the IMO. i can't be sure though....

There are 12 Black MOPpers... the 12 winners of the USAMO. These people go through MOP training and 6 of them make IMO. There is some weight with regards to performance on the USAMO, but I think the TST plays a much more significant role.

Training actually occurs after TST for the most part, except for about a week of practice tests. The TST and USAMO are actually weighted equally, and in recent years we have gone down to like the top 15 in the U.S. instead of just top 12 for some reason.

[quote="JSteinhardt"]in recent years we have gone down to like the top 15 in the U.S. instead of just top 12 for some reason.[/quote]

ties

Well I could be wrong but I was under the impression that in the past they rounded down instead of up (i.e. if 11-15 are all tied, only take top 10).

Also remember, they don't take Canadians, so Black MOP does not necessarily consist of the top 12.