Utilizand doar trei cifre, se pot gasi 16 numere de trei cifre ast incat sa nu exista doua dintre ele care sa dea acelasi rest la impartirea cu 16?

Astept rezolvari pls :)

Pai cam asta era si problema mea :) ca bineintzeles puteam sa ma uit la caut rezolvarea sau ceva in genul asta :) asta e enuntzul dat la OBMJ 1998:|

uita`te la resources, acolo enuntu e bun. sigur sunt si rezolvari, dar nu m`am uitat si acolo... dar nu prea imi place problema :(

pai ma uitam acu pe subiectele astea si erau neasteptat de usoare adika asa si era asta problema 4 de ...nu prea intzelegeam :| mai era o problema misto

Fie ABCDE pentagon, astfel incat AB=AE=CD=1 si m(ABC) = m(DEA)=90 de grade si BC + DE = 1 . Aflatzi aria pentagonului :)

mi`aduc aminte asa vag de problemele astea...le`am facut acu vreo 3 ani si ceva...
oricum, astea erau totusi greute... la obmj in 2003 am avut ceva noroc de subiecte usoare.. am iesit din sala dupa o ora jumate :)

bleah...
[hide="solutie"]presupunem ca exista 16 numere $z_{j}\equiv j (\bmod ~ 16)$ cu $j=\overline{0,15}$, si fie $a_{i}$ cele trei cifre. ($i=1,2,3$).
atunci exista cel putin o cifra para si cel putin o cifra impara. presupunem $a_{1}$ para si $a_{2}$ impara.
daca $a_{3}$ este para, atunci ne uitam la cele 8 numere impare. acestea sunt neaparat cele care se termina in $a_{2}$. cum maxim 3 din aceste numere pot avea cifra zecilor tot $a_{2}$, rezulta ca celelalte 5 vor avea cifra zecilor para, deci vor fi congruente cu $a_{2}(\bmod ~ 4)$.
daca $a_{3}$ este impara, atunci ne uitam la cele 8 numere pare. acestea se termina neaparat in $a_{1}$. cum maxim 3 din aceste numere pot avea cifra zecilor tot $a_{1}$, rezulta ca celelalte 5 vor avea cifra zecilor impara, deci vor fi congruente cu $a_{1}+2(\bmod ~ 4)$.
in ambele cazuri am obtinut 5 numere congruente modulo $4$.
contradictie, deoarece, pentru $r\in\left\{0,1,2,3\right\}$, exact patru din numerele $z_{i}$ dau restul $r$ modulo $4$, si anume $z_{4s+r}$, cu $s=\overline{0,3}$.[/hide]
bleaaah...  bine ca au iesit repede cazurile, ca altfel nu cred ca aveam rabdare

da, mijto :) tot asa am inceput si eu presupunand si ajungand la contradictzie, dar rezolvarea din barem era ciudata :(

da, la inegalitatea aia de la OBMJ 2003 de care ai zis mai sus, asa am inceput si eu. astept sa vad finalizarea :)

[quote="pohoatza"]moama ...felicitari :) eu acu ma uitam pe net la subiecte si n-am rabdare sa citesc de inegalitatea si crek ca am mai facut ceva asemanator ma gandeam la faptu ca (1+x^2) / (1+y+z^2) poate fi trat in :

Cum 2y <= y^ 2 + 1 rezulta ca y < y^2 + 1 /2

deci fractia aia e >= 1+ x^2 / 1 + z^2 + (y^2+1)/2 si se ajunge la

fractia aia :) >= 2 ( 1+ x^2 ) / 2(1+z^2) + (y^2+1) ...asta prin amplificare cu 2...procedam analog pt fiecare si daici...:|:|:|:| maine :)[/quote]
   Scrie mai baiatule in $\text{LaTex}$!Ca eu nu-nteleg nimik.

problema este 4/OBMJ 2003
enuntul suna cam asa
$x,y,z>-1$. aratati ca
$\sum\frac{1+x^{2}}{1+y+z^{2}}\geq 2$.
si
[quote]
Cum $2y\leq y^{2} + 1$ rezulta ca $y\leq\frac{y^{2} + 1}{2}$

deci $\frac{1+x^{2}}{1+y+z^{2}} \geq \frac{1+ x^{2}}{1 + z^{2}+ \frac{y^2+1}{2}}$ si se ajunge la

$\frac{1+x^{2}}{1+y+z^{2}}\geq \frac{2(1+ x^{2})}{2(1+z^{2})+ (1+y^{2})}$ ...asta prin amplificare cu 2...procedam analog pt fiecare si daici... maine [/quote]
am transcris eu in latex

ms maky ca ai scris in latex aia :) O sa postez continuarea pe la 10 ...:) caci acu trebuie sa plec pana la mall :)...M-am mai gandit azi la ea :)

totusi a fost cea mai grea problema din concurs asta :)
nici un bulgar nu a facut`o. hehehe

[quote="PhilAndrew"]OBMJ 2003 - Nu mai e nevoie sa postezi nimic rezolvarea e evidentå  :)
[hide="Evident"]
Facem substit. $a=1+x^2$...
Trebuie aråtat cå $\sum \frac{2(1+x^2)}{1+y^2+2(1+z^2)} \ge 2 \Leftrightarrow S=\sum \frac{a}{b+2c} \ge 1$.
Din Cauchy $\sum a(b+2c) \cdot \sum \left( \frac{a}{b+2c} \right) \ge (a+b+c)^2 \Rightarrow S \ge \frac{(a+b+c)^2}{3(ab+ac+bc)} \ge 1$.[/hide][/quote]
   Se putea face mai frumos, mai Andrei.Oricum e frumoasa solutia ta.Uite cum m-am gandit eu:
  Dupa ce folosim inegalitatea evidenta $\ x^2+1\geq\ 2x$ si analoagele e suficient sa aratam ca $\sum\frac{x}{1+y+z^2}\geq\ 1$.Doar folosind Cauchy in Engel form ajungem ca trebuie sa demonstram inegalitatea
$\ x(x-1)(1-y)+y(y-1)(1-z)+z(z-1)(1-x)\geq 0$, care e evident adevarata, facand substitutiile $\ x=a-b+1,y=c-a+1,z=b-c+1$ si obtinem exact Inegalitatea lui Schur.We are done. ;)

Da...Exact asa facusem problema ...cum a facut-o si Phil :) dar nam putut s-o postez atunci din lipsa de timp si din nestiintza de Latex :)

[quote="maky"]n`am inteles cum vine asta...scrie si tu mai detaliat...[/quote]
   Bine,..imi pare rau, dar era tarziu cand am postat solutia si n-am mai stat sa explic.Poate acum vei intelege mai bine.
    $\text{Solutie}$
  Folosind inegalitatea evidenta $\ x^2+1\geq\ 2x$ si analoagele, problema revine la a demonstra ca $\ 2\left(\sum\frac{x}{1+y+z^2}\right)\geq\ 2$, sau $\sum\frac{x}{1+y+z^2}\geq\ 1$.Acum folosim Inegalitatea lui Cauchy in forma[i] Engel[/i] si ramane sa aratam ca avem $\ (x+y+z)^2\geq\ x+y+z+xy+yz+zx+xz^2+yx^2+zy^2$.Dupa ce grupam termenii in mod convenabil inegalitatea de mai sus e echivalenta cu $\ x(x-1)(1-y)+y(y-1)(1-z)+z(z-1)(1-x)\geq\ 0$, care dupa ce substituim $\ x=a-b+1,y=c-a+1,z=b-c+1$ devine
$\ (a-b+1)(a-b)(a-c)+(c-a+1)(b-c)(b-a)+(b-c+1)(c-a)(c-b)\geq\ 0$, care e [i]Inegalitatea lui Schur[/i].
   E ceva alternativ..si eu am avut ideea lui Kosmin, dar cand am vrut s-o postez era deja acolo la locul ei, asa ca m-am mai gandit la una. :)


Let $ f: A \to B$ and let $ C,D$ be two subsets of $ B$.

Are the following true, and under which hp?
1. $ f^{-1}(C \cap D)=f^{-1}(C) \cap f^{-1}(D)$
2. $ f^{-1}(C \cup D)=f^{-1}(C) \cup f^{-1}(D)$

$ 1.$true
$ f^{-1}(C\cap D)=\{a\in A|\exists c\in C\cap D: f(a)=c\}$
$ =\{a\in A|\exists c\in C: f(a)=c\}\cap \{a\in A|\exists c\in D: f(a)=c\}$
$ =f^{-1}(C)\cap f^{-1}(D)$

$ 2.$true
$ f^{-1}(C\cup D)=\{a\in A|\exists c\in C\cup D: f(a)=c\}$
$ =\{a\in A|\exists c\in C: f(a)=c\}\cup \{a\in A|\exists c\in D: f(a)=c\}$
$ =f^{-1}(C)\cup f^{-1}(D)$


inequalities
inequalities unsolved
algebra

[quote="radio"]Let $ a,b,c > 0$. Prove that

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq 3\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}}$[/quote]
We can suppose that $ a \geq b \geq c$ then we have $ \sum_{cyc} \frac {b}{a} \geq \sum_{cyc} \frac {a}{b}$
Note that we can prove:
$ \sum_{cyc} \frac {a^2}{b^2} \plus{} 6 \geq \frac {3}{2}\sum_{cyc} \frac {a^2 \plus{} b^2}{ab}$
So $ (\sum_{cyc} \frac {a}{b})^2 \equal{} \sum_{cyc} \frac {a^2}{b^2} \plus{} 2\sum_{cyc} \frac {b}{a} \geq \frac {3}{2}\sum_{cyc} \frac {a^2 \plus{} b^2}{ab} \minus{} 6 \plus{} \sum_{cyc} \frac {a}{b} \plus{} \sum_{cyc} \frac {b}{a} \equal{} \frac {5}{2}\sum_{cyc} \frac {a^2 \plus{} b^2}{ab} \minus{} 6$
We need prove that:
$ \frac {5}{2}\sum_{cyc} \frac {a^2 \plus{} b^2}{ab} \minus{} 6 \geq \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}$
or $ \sum_{cyc} (a \minus{} b)^2(\frac {5c}{a} \plus{} \frac {5c}{b} \minus{} 4) \geq 0$
Note $ a \geq b \geq c$ so $ \frac {5a}{b} \plus{} \frac {5a}{c} \minus{} 4 \geq \frac{5b}{c} \plus{} \frac {5b}{a} \minus{} 4 \geq \frac {5c}{a} \plus{} \frac {5c}{b} \minus{} 4$
Done!

[quote="chien than"][quote="radio"]Let $ a,b,c > 0$. Prove that

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq 3\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}}$[/quote]
We can suppose that $ a \geq b \geq c$ then we have $ \sum_{cyc} \frac {b}{a} \geq \sum_{cyc} \frac {a}{b}$
Note that we can prove:
$ \sum_{cyc} \frac {a^2}{b^2} \plus{} 6 \geq \frac {3}{2}\sum_{cyc} \frac {a^2 \plus{} b^2}{ab}$
So $ (\sum_{cyc} \frac {a}{b})^2 \equal{} \sum_{cyc} \frac {a^2}{b^2} \plus{} 2\sum_{cyc} \frac {b}{a} \geq \frac {3}{2}\sum_{cyc} \frac {a^2 \plus{} b^2}{ab} \minus{} 6 \plus{} \sum_{cyc} \frac {a}{b} \plus{} \sum_{cyc} \frac {b}{a} \equal{} \frac {5}{2}\sum_{cyc} \frac {a^2 \plus{} b^2}{ab} \minus{} 6$
We need prove that:
$ \frac {5}{2}\sum_{cyc} \frac {a^2 \plus{} b^2}{ab} \minus{} 6 \geq \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}$
or $ \sum_{cyc} (a \minus{} b)^2(\frac {5c}{a} \plus{} \frac {5c}{b} \minus{} 4) \geq 0$
Note $ a \geq b \geq c$ so $ \frac {5a}{b} \plus{} \frac {5a}{c} \minus{} 4 \geq \frac {5b}{c} \plus{} \frac {5b}{a} \minus{} 4 \geq \frac {5c}{a} \plus{} \frac {5c}{b} \minus{} 4$
Done![/quote]

you can't suppose $ a \geq b \geq c$ because the initial inequality isn't symmetric..

[quote="radio"]Let $ a,b,c > 0$. Prove that

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq 3\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}}$[/quote]
after clearing denominators,It's equivalent to prove that
\[ \sum_{cyc}a^5bc^2\plus{}\sum_{cyc}a^5c^3\plus{}\sum_{cyc}a^4bc^3\plus{}2\sum_{cyc}a^4b^3c\plus{}2\sum_{cyc}a^3b^3c^2 \ge 7\sum_{cyc}a^4b^2c^2\]
Which is obviously true by AM-GM;\[ a^5bc^2\plus{}a^5c^3\plus{}a^4bc^3\plus{}a^4b^3c\plus{}a^4b^3c\plus{}a^3b^3c^2\plus{}a^3b^3c^2
\ge 7a^4b^2c^2\]

[quote=radio]Let $ a,b,c>0$. Prove that

$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq 3\sqrt{\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}}$[/quote]
[url=https://artofproblemsolving.com/community/c6h2520123p21377185]Mongolian 2007[/url]


Let $x\geq 1\geq y\geq 0$. Prove or disprove that

$x^{2}+y^{2}+2xy+4\geq 4x+4y$.

$\{x,y\}\subset\mathrm R\Longrightarrow (x+y-2)^{2}\ge 0\Longleftrightarrow x^{2}+y^{2}+2xy+4\ge 4x+4y$ with equality iff $x+y=2$.
In the our case it is truly and we have equality iff $x=y=1$.

Not iff $x=y=1$, but for any $x>1, y\in[0,1)$ (check the conditions) such that $x+y=2$ - e.g. $x=1.2; y=0.8$

Yes, Farenhajt, you are right ! I thaught that $x\ge 1$ and $y\ge 1$.

yep, this follows from the inequality (x+y-2)²>= 0 as the responses above noted.


AMC
USA(J)MO
USAMO
geometry
3D geometry
sphere

Suppose your worst enemy gives you n red points and n blue points in the plane, no three of them collinear. (Your enemy picked them, so they are not nicely arranged.) Show that it is possible to draw n line segments, each segment joining a red point to a blue point (so that each red point is paired with a unique blue point, and vice versa) in such a way that none of the line segments intersect.

if this problem is for "People Practicing for NAts", then why does it require proof? You know that this is far greater than MC Level as I do, especially if you're going to nats.

[quote="akhil0422"]You totally copied nikeballa in his avatar and his "define location" answer for location.[/quote]

Actually one of my friends did this for me....gave me the avatar..

[quote="USAMO13"]Its funny how the "super-smart" nike says this problem is hard...[/quote]

if you find something comical, feel free to PM me, just don't spam the forums with useless posts. i think you realize that this problem is far too hard for MC Level. i'm sure you made USAMO though, by your username. i never said i was super-smart, i'm actually quite the contrary.

and if your friend gave you the avie, why does it have nike signs, and why is it entitled "NIKEBALLA96" in Dojo's album of spinning spheres?

[quote="USAMO13"]No one has posted a solution yet..
@nike..lets stop fighting and make a peace treaty..[/quote]
uh. because you've stolen my avie, my location, and not even asked me, so i should make a peace treaty, right?
i'm still wondering if yuo made usamo.

fine nike...ill change my avatar if u can find me a good one...
SOMEONE POST A SOLUTION!!!

This is [b]not[/b] a Nats practice problem.  This is another problem stolen from:

http://www.math.oakland.edu/ousmi02/prob.html

HOW MANY TIMES DO I HAVE TO I TELL YOU???
MY COACH RECCOMENDED THOSE PROBLEMS!!!

How many times does everyone on this forum have to tell you?

[b]POST ALL OF YOUR PROBLEMS IN THE SAME POST.[/b]


Show that:\[\alpha = \pm \frac{\pi}{12} + k\cdot \frac{\pi}2
(k\in \mathbb{Z}) \Longleftrightarrow\  |{\tan \alpha}| + |{\cot
\alpha}|  = 4\]

i just prove $\alpha=\pi/12$. The other cases can be deduced easily by the basic properties of $\tan$ and $\cot$.

we are going to prove $\tan \pi/12 + \cot \pi/12 = 4$

$\tan \pi/12 = \tan \frac{\pi/6}{2} = \frac{1-\cos \pi/6}{\sin \pi/6} = 2-\sqrt3$

thus $\cot\pi/12=1/(2-\sqrt3)=2+\sqrt3$ and $\tan \pi/12+\cot \pi/12=4$

heh :) original thoughts..

but if you notice that considering only x=pi/12 means that the absolute value is innecessary, just put them on the same numerator and you get $\frac{\sin^2x+\cos^2x}{\sin x \cos x}$, which is equivalent to 1/(1/4) = 4


Let $ D$ , $ E$ , $ F$ be the projections of the orthocenter $ H$ of $ \triangle ABC$ on the sidelines $ BC$ , $ CA$ , $ AB$ respectively. Define :

 the midpoints $ M$ , $ S$ of $ [BC]$ , $ [AH]$ respectively ; $ P\in MF\cap AD$ ; $ R\in ME$ for which $ FR\parallel BC$ . Prove that $ PE\perp SR$ .

[b]Indication.[/b] See [url=http://www.mathlinks.ro/viewtopic.php?t=284253][color=red][b]here[/b][/color][/url].

Choose $ \{S, F, E, R, P\}$ with $ \angle MFS \equal{} \angle MES \equal{} 90^o, SP\perp FR$ then applying [url=http://www.mathlinks.ro/viewtopic.php?t=284253]this lemma[/url] $ \rightarrow$ QED


let a is  an integer. prove that the number solution of the equaltion :
          x^2 + y^2 + z^2 = 2axyz  ( mod p )  , p is a prime , is ( p+(-1)^((p-1)/2))^2

Your equation equavalent to $(z-axy)^{2}=a^{2}x^{2}y^{2}-x^{2}-y^{2}(mod \ p)$.
Let $N_{0}$ number of solutions $a^{2}x^{2}y^{2}=x^{2}+y^{2}(mod \ p)$, $N_{1}$ number of solutions $f(x,y)=a^{2}x^{2}y^{2}-x^{2}-y^{2}\in (Z_{p})^{*}^{2}$.
Then answer is $N_{0}+2N_{1}$.
If $p|a$ and $p=-1(mod 4)$, then $N_{0}=1$, therefore $N_{0}+2N_{0}$ is odd, but $(p-1)^{2}$ is even.
Therefore your problem is not correct.

Maybe the formulation of the problem should be:

Let $p$ be an odd prime and $a$ an integer not divisible by $p$. Show that the number of triples of integers $(x,y,z)$, $0\leq x,y,z\leq p-1$, satisfying $(x+y+z)^{2}\equiv axyz\ (\mbox{mod }p)$ is $p^{2}+1$.

This is problem from India IMO training camp 2002.

sorry, i don't understand. can you post your solution, GOBLIN?
             thanks.


number theory
greatest common divisor
number theory unsolved

[quote="Johann Peter Dirichlet"]n is a positive integer such that n(n+1)/3 is a square. Show that n is a multiple of 3, and n+1 and n/3 are squares.[/quote]
Too easy problem.
If $3|n$=>$gcd(\frac{n}{3},n+1)=1$=>
$\frac{n}{3}=x^2$ and $n+1=y^2$,
now let that $n+1=3k$=>$k(3k-1)=t^2$=>
$k=p^2$ and $3k-1=q^2$=>$3p^2=q^2+1$=>$3|(q^2+1)$
but for any integer $x$ we have that $3|(x^2-1)$ or$3|x^2$ so contradiction.


(A.Zaslavsky) Given two triangles $ ABC$, $ A'B'C'$. Denote by $ \alpha$ the angle between the altitude and the median from vertex $ A$ of triangle $ ABC$. Angles $ \beta$, $ \gamma$, $ \alpha'$, $ \beta'$, $ \gamma'$ are defined similarly. It is known that $ \alpha \equal{} \alpha'$, $ \beta \equal{} \beta'$, $ \gamma \equal{} \gamma'$. Can we conclude that the triangles are similar?

I have a negative answer. The angles $ \alpha \equal{}\alpha'$,$ \beta \equal{}\beta'$, $ \gamma \equal{}\gamma'$ are equal if and only if the corresponding least angles ($ <90^o$) between the medians and the respective side are equals. But the last condition its true for the following triangles. Take any triangle $ ABC$, and the triangle $ A'B'C'$ such that its sides are the medians of $ ABC$  (we can construct it by an easy way), it is not hard to prove that for this two triangles the conditions of the problem holds, but they doesn't have to be similar.


algorithm
LaTeX
induction
geometry
3D geometry
calculus
function

Find an easy algorithm for:
1)The sum of the first $n$ positive integers.
2)The sum of the first $n$ positive odd integers.
3)The sum of the first $n$ positive perfect cubes.
4)The sum of the first $n$ positive perfect squares.
5)The sum of the first $n$ positive even integers.

Remember to show work!

[hide="1."]
First n terms of arithmetic sequence: $S_{n}=n(\frac{a_{n}+a_{1}}{2})$. We know that $a_{n}=n$ and $a_{1}=1$, so $\boxed{S_{n}=\frac{(n)(n+1)}{2}}$.[/hide]
[hide="2."]Using formula for nth term we find that $a_{n}=2n-1$. So $S_{n}=n(\frac{2n-1+1}{2})=n(\frac{2n}{2})=n(n)=n^{2}$. So $\boxed{S_{n}=n^{2}}$[/hide]

What i'm basically asking for here is a proof, not just a definition.

Umm... is anybody gonna post the proofs? :?:

You can count in pairs.  If you have $n$ odd numbers, and you pair them least-greatest and then backwards, each pair sums to $2n$. So the sum is

$2n(n)/2=2n^{2}/2=n^{2}$

Use a similar approach for the sum of the first n integers and even integers.

[hide="Proof for 5."]
Use nth term formula to find that $a_{n}=2n$. Plug this in to sum of n terms formula: $S_{n}=n(\frac{2n+2}{2})=n(\frac{2(n+1)}{2})=n(n+1)$ Thus, $\boxed{S_{n}=n(n+1)}$[/hide]

[quote="max_tm"]$i^{2}$ and $i^{3}$ are a bit more involved... (I couldn't do it without the aid of the book personally, but there are probably many people here who could if they want to latex it all) they can be found in AoPS vol. 2 but I also think they're featured in the proofs without words animations on the left side of your screen.[/quote]
lol. Well, I can't afford AoPS volumes, so can someone post it? Thanks.

Do you want the proofs by induction?  They are valid proofs of these identities. However, you could also think of the sum of cubes as the square of the sum of the integers, i.e.

$(n(n+1)/2)^{2}=n^{2}(n+1)^{2}/4$, which is the sum of cubes.  So what kind of proofs are you looking for?

[quote="bigboy82892"]Proof by any method except guess and check. :D  :lol:[/quote]
Alright, so I believe that 1, 2, and 5 were explained quite well.
Here is the proof for 4:
[hide="4"] Say that $0^{2}+1^{2}+2^{2}+3^{2}...+x^{2}=ax^{3}+bx^{2}+cx+d$.

Step 1. Now, say that x=0.  We have that $0=a(0)+b(0)=c(0)+d\Rightarrow d=0$.

Step 2. Now, say that x=1.  We have that $0+1=a(1)+b(1)+c(1)+d(1)=a+b+c\Rightarrow a+b+c=1.$  Thus, we have that $2(a+b+c)=2\Rightarrow 2a+2b+2c=2$ and $3(a+b+c)=3\Rightarrow 3a+3b+3c=3$

Step 3. Now, say that x=2.  We have that $0+1+4=a(8)+b(4)+c(2)+d=8a+4b+2c\Rightarrow 5=8a+4b+2c$

Step 4. Now, subtract the second-to-last equation of Step 2 from the final equation of Step 3.  We get that $8a+4b+2c-(2a+2b+2c)=5-2\Rightarrow 6a+2b=3\Rightarrow 3(6a+2b)=9\Rightarrow 18a+6b=9$.

Step 5. Now, say that x=3.  We have that $1+4+9=a(27)+b(9)+c(3)+d=27a+9b+3c\Rightarrow 14=27a+9b+3c.$

Step 6. Now, subtract the final equation of Step 2 from the final equation of Step 5 to get that $27a+9b+3c-(3a+3b+3c)=14-3\Rightarrow 24a+6b=11$

Step 7.  Now, subtract the final equation from Step 4 from the final equation of Step 6.  We get that $24a+6b-(18a+6b)=11-9\Rightarrow 6a=2\Rightarrow a=\frac{1}{3}.$  Now, from our 24a+6b=11, we can plug a in to get that $24(\frac{1}{3})+6b=11\Rightarrow 6b=3\Rightarrow b=\frac{1}{2}$.  Finally, from our a+b+c=1, we can say that $\frac{1}{3}+\frac{1}{2}+c=1\Rightarrow c=\frac{1}{6}$

Step 8.  Now, we know that $a=\frac{1}{3}, b=\frac{1}{2}, c={1}{6}, d=0$, so \[0^{2}+1^{2}+2^{2}+3^{2}...+x^{2}=ax^{3}+bx^{2}+cx+d=\frac{x^{3}}{3}+\frac{x^{2}}{2}+\frac{x}{6}+0=\frac{2x^{3}}{6}+\frac{3x^{2}}{6}+\frac{x}{6}=\frac{2x^{3}+3x^{2}+x}{6}=\frac{(x)(2x^{2}+3x+1)}{6}=\frac{x(x+1)(2x+1)}{6}\] which is our formula.
 [/hide]

Now, from this idea, I will give you a hint for 3.
[hide="Hint"] We can say that $0^{3}+1^{3}+2^{3}...+x^{3}=ax^{4}+bx^{3}+cx^{2}+dx+e$, and plug in values to try and figure out what a, b, c, d, and e are.  [/hide]

Second-to-worst way to prove those identities other than induction.   :|

[hide="A more fun method"] $\sum_{k=1}^{n}{k \choose 2}={n+1 \choose 3}$ by Hockeystick.  But ${k \choose 2}= \frac{k(k-1)}{2}$, and the result follows.

Similarly $\sum_{k=1}^{n}{k \choose 3}={n+1 \choose 4}$.  :) [/hide]

[quote="mathmaster46"][hide="1"](n*(n+1))/2[/hide]
[hide="2"]n^2[/hide]
[hide="3"](1+2+3+...+n)^2[/hide]
[hide="4"]n(n+1)(2n+1)/6[/hide]
[hide="5"]n(n+1)[/hide][/quote]
I think bigboy82892 wanted the proofs, not what they are.

[hide="hopefully the worst way"]
No seriously, this probably isn't good for eyes that want HSB stuff
[hide="calculus and generating functions contained within"]

Ok so here's how you can do this with generating functions. I will demonstrate with sum of squares. I will also just give the output of my TI-89 because doing this stuff by hand is a pain.

$x^{n}+x^{n+1}+x^{n+2}+x^{n+3}+\ldots = \frac{x^{n}}{1-x}$

Differentiating both sides with respect to x and multiplying by x.

$nx^{n}+(n+1)x^{n+1}+(n+2)x^{n+2}+\ldots = \frac{-(x(n-1)-n)x^{n}}{(x-1)^{2}}$

Repeating:

$n^{2}x^{n}+(n+1)^{2}x^{n+1}+... = \frac{-(x^{2}(n-1)^{2}-x(2n^{2}-2n-1)+n^{2})x^{n}}{(x-1)^{3}}$

Now we plug in 0 for n to get the infinite sum

$\sum_{n=0}^{\infty}n^{2}x^{n}= \frac{-x(x+1)}{(x-1)^{3}}$

Subtracting:

$\sum_{i=0}^{n-1}i^{2}x^{i}=\frac{(x^{2}(n-1)^{2}-x(2n^{2}-2n-1)+n^{2}) x^{n}-x^{2}-x}{(x-1)^{3}}$

Now we need to take the limit as $x$ approaches $1$.

$\sum_{i=0}^{n-1}i^{2}= \lim_{x\to1}\frac{(x^{2}(n-1)^{2}-x(2n^{2}-2n-1)+n^{2}) x^{n}-x^{2}-x}{(x-1)^{3}}= \frac{n(n-1)(2n-1)}{6}$.

Since this is $n-1$ squares, just plug in $n+1$ to get $n$ squares which is

$\frac{(n)(n+1)(2n+1)}{6}$ as desired.

This can be trivially generalized to any power, but you have to take the derivative $n$ times for $i^{n}$.
[/hide]
[/hide]

If t0rajir0u was thinking of a different worst way, I'd kind of like to see it. The method I outlined is extremely nasty, but works.

I'd suggest using The QuattoMaster6000's method for any serious proofs, or induction if it just says to prove it.

Hamster1800, do you know if the generalization has a name or a link to it? I'm a big fan of generalization  :P (you can still use it on a non-calculus based contest if there's calculus involved in the solution, right?)

Hamster1800: the worst way to which I referred was induction because induction assumes you already know the formula and says nothing about how one would go about [i]discovering[/i] such formulas.

max_tm: http://mathworld.wolfram.com/FaulhabersFormula.html

[hide="one way to generalize the advanced method outlined above"]
This is a general thing that generates the formulas so kindly written out in t0rajir0u's link.

$\sum_{i=1}^{n}i^{k}= \lim_{x\to1}\left[\left(x\frac{d}{dx}\right)^{k}\left(\frac{1-x^{n+1}}{1-x}\right)\right]$.

Where $\left(x\frac{d}{dx}\right)$ means to differentiate and multiply by $x$.

If I made a mistake in writing this, please tell me.
[/hide]


Let k be a positive integer. A positive integer n is said to be k-flip if the digits of n are reversed in order when it is multiplied by k. For example, 1089 is a 9-flip because 1089*9 = 9801, and 21978 is a 4-flip because 21978*4 = 87912. Explain why there is no 7-flip integer.


BU is my first choice school and I want to know my chances.  I see posts on the College Confidential Forum with people having 1400+ SATs, extremely high GPAs [in the 3.8-4.0+ range], massive EC's and leadership positions [atleast 10?], and VERY high SATII scores. And yet, they are asking what their chances are at BU?!?!?!

OK. Back to my question. I will be in my Junior Year this September. I plan to major in mathematics/actuarial science after High School. I am really a math person. If I receive extremely high grades in all my math courses [96-100 range], get 750+ SAT math, 750+ SAT 1C, 750+ SAT 2C, will this make up for my extremely LOW SAT critical reading score [possibly in the high 400s and low 500s] and a downward trend in my english classes [90,87,83,70]. But I plan to get the english grade up into the 90's my Junior Year. So am I screwed?

I'm part of my schools math team. I rank in top 15% if not 10%. My unweighted GPA so far is 3.8 and weighted GPA so far is 4.0. I did some volunteering at a public middle school through an internship in my school. I also worked over the summer at a small public elementary school. I don't like to read and my vocabulary is very low. So do I still have a chance? Is there anything I can possibly do to make up for those low grades in english and SAT critical reading scores? Please help. Thanks. Any chance at all?

BU is not an extremely competitive and difficult school to get into, so as far as I know you shouldn't have a problem getting in as long as you keep to it and do as you have planned.  Obviously they want more than a 70-80 range in English, but if you get that up than you probably shouldn't have a problem.  Try working on vocab and reading as much as you hate it so that you can push up your SAT scores and do better in English.  It'll all be worth it in the end.

Thanks a lot. I'll try to read as much as I can even though I hate it.

Yes, be sure to read about something you are interested in, and mention your independent reading somehow in the application process. (Does BU do interviews?) Some of the people posting on the College Confidential boards are more than a little obsessive, and of course they will get into some colleges, and maybe end up not even enrolling at BU. Show your interest in BU, and show your interest in learning about the subjects related to your planned career, and you should have a decent chance of getting in. "Show your interest" means mention those things in your application essays, as well as an interview if you have one.

What is your plan B? You do have to apply to more than one school if your favorite school is not a SURE thing for getting in.

Good luck.

Yes I have a few other schools in mind. They are George Washington University, University of Maryland - College Park, University of Maryland - Baltimore County, and SUNY - Stonybrook so far.

just out of curiosity...isn't Washington University significantly more difficult to get into than Boston University?

Umm.. I guess so. I just got back my scores from the Princeton Review Practice Test I took on August 20, 2005.  My score was a 1560 (750M, 350V, 460W, 6E) Is it possible to raise that 1560 to a 2000+ by the March/May test? The March/May test will be my first time taking it for real. How much do I need to study? Is reading 3+ hours daily enough to boost that verbal score up atleast 200 points?

i'm sure it's possible to raise it to a 2000.  but in order to do this, you certainly have a lot of work to do.  reading 3 hours a day will certainly improve your score, but as you read, make sure you are actually learning the vocabulary you encounter.  what kind of things do the princeton review tell you to work on?

The Princeton Review gave me a few comments on each section. I'll just name a few.

Math:
1. Plug in at least twice on MUST be questions.
2. Be careful, you chose a partial answer.

Verbal:
1.Read CR questions carefully
2. Read five lines above and below on line ref/lead word questions.
3.Prove your answers with evidence in the passage.

Writing:
1. Approximately 1/5 of ID questions have no error.
2. For content questions, always go back to the passage.
3. Watch out for idiom errors.

The verbal is really killing me. When I read the passages, I don't seem to understand it's meaning.

if it's more the comprehension than the vocabulary that's killing you, then reading a lot will be very helpful.  as you read, go back every few paragraphs and think to yourself, "what did i just read?".  make sure you could explain it to yourself so that you will be ready for SAT passages.  read some complicated stuff to practice with.

BU is a party school with a population full of affluent internationals (moreso pre-2001), affluent US Asians, and (you can guess the pattern) affluent suburban Boston whitebread.  The lack of visible US lower-class population is striking in comparison to more "diverse"/PC schools.  It unapologetically admits moneyed applicants, for the money, and recruits heavily outside the US.  If you want to date the children of the South American upper classes, sent abroad for safety, BU is a good place. The selectivity level is moderate and a strong math background would probably get you in.  Their math department is better than the selectivity level would indicate, largely because of the proximity to MIT and Harvard (top-level PhD's who stay in town and work at BU).  I believe they also run a large summer math program for high-school students.

[quote="fleeting_guest"]I believe they also run a large summer math program for high-school students.[/quote]

Yay for PROMYS.

[quote="paladin8"] Yay for PROMYS.[/quote]

That's the one.   It was/is administrated by Glenn Stevens, who I think was one of Barry Mazur's PhD students working on modular forms.  That is, he is one of the MIT/Harvard PhDs and postdocs who ended up at BU and contribute to the strength of the math department.

The summer program is conceivably a pleasant form of employment available to BU math undergrads, if that is a consideration for the original questioner.


Tunning our PCs is an art too , so ..  my pc :)

[img]http://static.flickr.com/27/70050460_bbde3d6881_o.jpg[/img]

This is mine:

[img]http://img55.imageshack.us/img55/1606/dsktp7vw.jpg[/img]

how does one get the style of your computer and your background?!!?!?!?!?!?!??!?!

For the second screenshot, you can get themes [url=http://www.stardock.com/]here.[/url]

I think the first one is Linux.

digitalblasphemy.com has some nice ones. :)

[quote="MithsApprentice"]Andrew's desktop doesn't look like it was something from WindowsBlinds, but I could be very wrong. O.o

Anyway, since we're on this topic...anyone know of where/how I could get a hold of some nice desktop wallpapers?[/quote]

I actually do use WindowsBlinds. :P But anyway, for nice desktop wallpapers, go to http://www.deviantart.com. It's got nice abstract backgrounds and much more.

Andrew which is your wallpaper  :?:
I found many on Devian Art but i couldn't find yours.
Just give me a link or the name of that wallpaper.

[quote="MithsApprentice"]He's got this software that does Windows themes, and that wallpaper is part of it. I'd assume it's not on devantart.

I've looked through some of the devantart stuff and didn't find anything too appealing. Anyone want to share some links to good ones?[/quote]

Here's a couple ones I have saved on my computer from Deviant Art.

http://img220.imageshack.us/img220/3189/amped8xr.jpg
http://img220.imageshack.us/img220/3796/beyondthegraceofgod0pz.jpg
http://img146.imageshack.us/img146/8366/bg8sm.jpg
http://img220.imageshack.us/img220/1093/colossal8tc.jpg
http://img503.imageshack.us/img503/3007/cryogenics7137lc.jpg
http://img503.imageshack.us/img503/9175/datascape9fr.jpg
http://img220.imageshack.us/img220/8458/destinysending3aa.jpg

Go Ubuntu!
(still using Windows XP, but trying out the ubuntu livecd)

go Linux+Enlightenment+MacOSX theme - the perfect clean desktop and wm combination. :)

wow where did you get the background?  :lol:

Comes with Linux.

I think the second one is from Windows Vista.

My background is fractal.
I made it myself by Ultra fractal 4.0

dual booting windows vista, xp, gonna try out linux ;)

[url=http://img524.imageshack.us/my.php?image=prntscrngf8.jpg][img]http://img524.imageshack.us/img524/9752/prntscrngf8.th.jpg[/img][/url]

[quote="Eli"]On windows it's something like Ctrl+Prnt Scrn for the whole desktop and Alt+Prnt Scrn for the active window.  On Linux, I believe there are screenshot programs.[/quote]
On Kubuntu you can press Prnt Scrn (One button), which takes a snapshot of the whole desktop and opens KSnapshot.

Strange. I just press the print scrn button on every computer that I've ever used.

My laptop :D

[url=http://truongton.net/forum/picture.php?albumid=56447&pictureid=1004647]Link[/url]

[url=http://img191.imageshack.us/img191/6380/desktoplae.png]Here is my desktop.[/url]

[quote="funcia"]how do you screenshot :blush:[/quote]
Press Ctr+Print Screen

alt+prntscrn takes a screenshot of the current active window, very useful.


function
algebra
binomial theorem
combinatorics proposed
combinatorics

Let $E=\{1,2, \cdots, n \}$.
Determine the number of triples $(A,B,C)$ of subsets of $E$ such that $A \cap B \neq \emptyset,B \cap C \neq \emptyset,C \cap A \neq \emptyset$ and $A \cap B \cap C = \emptyset .$

Pierre.

We can use generating functions. (I love generating functions!)

Let A and B contain x elements together, B and C have y elements together, and C and A have z elements together. Furthermore, A has p other elements, B has q others, and C has r others. And s of them are unused. Then x+y+z+p+q+r+s=n, and we want the sum of $\displaystyle\frac{n!}{x!y!z!p!q!r!s!}$ for all partitions (x,y,z,p,q,r,s) of n with x,y,z>=1 and p,q,r,s>=0.

Well, this is exactly the coefficient of $\frac{x^n}{n!}$ in the expansion of

$\left(\frac{x^1}{1!}+\frac{x^2}{2!}+\cdots\right)^3\left(\frac{x^0}{0!}+\frac{x^1}{1!}+\cdots\right)^4=(e^x-1)^3\cdot e^{4x}=e^{7x}-3e^{6x}+3e^{5x}-3^{4x}$.

This coefficient is

$7^n-3\cdot6^n+3\cdot5^n-4^n$.

My approach was more computational...
Using the notations and results here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=20934

we deduce that th number of ways to select two disjoints and non-enmpty subsets of a set with $k \geq 0$ elements is $N_k = 3^k - 2^{k+1} + 1.$
Thus, we construct all the good triples as follws :
- First we select $B$ with $k\geq 2$ elements : $C_n^k$ choices.
- For each of them, we select two disjoint non-empty subsets from $B$, which will be $A \cap B$ and $B \cap C$. This leads to $N_k$ choices.
- Then we select any two non disjoint (and then non-empty) subsets from $B' = \{1, \cdots, n \} \setminus B$, which will be $A \setminus (B \cap A)$ and $C \setminus (B \cap C)$. There are $4^{n-k}$ ways to select two subsets from $B'$, for $2 \times 2^{n-k}$ among them at least one of the subset is empty, and $1$ one of them correspond to tha case where both are empty.
Thus there are $4^{n-k} - 2\times 2^{n-k} + 1 - N_{n-k} = 4^{n-k} - 3^{n-k}$ ways to complete $A$ and $C$ as desired.
Thus the total number of good triples is :
$\dsp T = \sum_{k=0}^n [C_n^k(3^k - 2^{k+1} +1)(4^{n-k} - 3^{n-k})]$
Using the binomial theorem, it leads easily to $T = 7^n - 3 \times 6^n + 3\times 5^n - 4^n$, as above  :)

Pierre.


In my computer science class, we are supposed to make a program that simulates a game:

Player A and Player B flip a coin at the same time.  (Comes up either heads or tails.)  They flip simultaneously until one or both players has flipped 4 heads in a row.  Then the game ends.

We are supposed to run the game 100+ times to calculate the expected number of flips one player will go through each game.

This is easier with a program, but is there a mathematical way to find the expected number of flips?

[hide]I think it's somewhere between 10-11.
If there's less than 4 turns, the game doesn't end.
Since one must get 4 in a row to end the game, the last 4 turns in the game must be heads on at least one player.
If the game ends in 4 turns, one or both players got 4 heads.  If there's one person playing, his chances of getting 4 heads is 1/16 so if there's two people playing, the chances is 1/8 though it overcounts the probability that both of them get it by 1.  So the probability is 1/8-1/256.
If the game ends in 5 turns, then the first flip must be tails.  The chances of that happening is 1/2 and the chances of getting 4 heads in a row after that is 1/16 so the combined probability is 1/32.  Two people reduce it to 1/16 though we still overcount this entire thing.  1/16-1/1024
Even if there's 6,7,8 turns, the idea's the same.  The only thing that has to be a tails is the one before the 4 in a row.  The other 1-3 flips can be anything since they can't form four heads in a row.  So the probability of each is 1/16-1/1024.
However, when we reach nine turns, we run into some problems.  There must be an additional restriction on the first four flips because the first four could result in 4 heads.  Since that's the only possiblity that's voided, we get rid of it so the restriction is (15/16)*1/2*1/16 for one person.  That's 15/512 and the overcount is 225/262144, while the probability for two people is 15/256.  Altogether, that's 15/256-225/262144.
When we reach 10 turns, a similar thing happens.   The first 5 numbers could have HHHHH HHHHT or THHHH.  Out of the 32 possible flip results only those three are forbidden.  The probability for one person is (29/32)(1/2)(1/16).  That's 29/1024 and the overcount is 841/1048576.  For two people, probability is 29/512.  Overall, that's 29/512-841/1048576.  (This is why a computer or calculator is needed.)
Before we go any further, let's add it up.  $1/8-1/256+4(1/16-1/1024)+15/256-225/262144+29/512-841/1048576.$  That's $1/4+1/8+15/256+29/512-2/256-225/262144-841/1048576.$  That's 0.4807 so we probably only need 1 more.  Let's show that anyway.
For 11, the first 6 numbers are of concern.  The possibilites are HHHHHH, HHHHHT, HHHHTH, THHHHH, THHHHT, TTHHHH.  THose are the only 6 possibilites that create 4 in a row.  There's 64 total so we have 58/64 allowed or 29/32 allowed.  Since that's the same as last, the probability is the same as last.  29/512-841/1048576=0.0558 which does cause the probability to exceed 50%.  [/hide]

well, that is a reasonably difficult problem for just 1 person, let f(n) be the expected number of flips to get n tails in a row, if we start our sequence with a head

f(n+1)=f(n)+1/2(1)+1/2*f(n+1)

this is because if we have n tails in a row, then we have gone f(n) flips, then the next flip could either be tails, which would end the sequence with one flip, or we get a heads, and start over

g(n) is n tails in a row if we start with a tail

$g(n)=1+1/2*f(n)+1/4*(1+f(n))+1/8(2+f(n))+...$
$+1/2^{n-1}*(n-2+f(n))+1/2^{n-1}*(n-1)$

$g(n)=1+\frac{1}{2^{n-1}}+ \sum_{k}^{n-2} \frac{k}{2^{k+1}}+f(n)*\sum_{i=1}^{n-1}\frac{1}{2^i}$

this just goes through all the possibilities sequences, TH..., 2TH..., 3TH..., ... (n-1)TH, (n)T, notice that g(n) and f(n) include their first roll in thier own function, this explains the confusing terms that are like 1/2^{n-1}(n-2+f(n)), to explain the one term, after we get the first tail, we then get n-1 more flips plus the initial flip, giving a total of n-1 tails then 1 head, that adds only n-2 flips to the total because the 1 head is counted in the f(n)

now that we have those 2 functions, f(4)+g(4) is our answer, start with f(1)=1+1/2+1/2*f(1)

since we get a head, then 1/2 chance ending, 1/2 chance of getting a head again and starting over, f(1)=3, i don't feel like doing the rest, this has taken much longer than i expected...

if it is this difficult for 1 person, 2 people even more difficult, either i am doing this wrong, or it should go in the olympad section


algebra
function
domain
vector
algebra proposed

[quote="Mr.Doe"]This doesnt make sense to me...

Is it maybe ;
$f(xyz)=f(\frac{xf(y)}{z})f(\frac{yf(z)}{x})f(\frac{zf(x)}{y}), f(0): =0$,or something like that?

Am I right  :wink:  ,or not???[/quote]

Right, right, what??

[quote="cckek"]$f(xyz)=f\left(\frac{xf(y)}{z}\right)\left(\frac{yf(z)}{x}\right)\left(\frac{zf(x)}{y}\right),f(0)=0$.[/quote]

May I ask what are the domain and the range of the function?

[quote="cckek"]$f: \mathbb{R}\to \mathbb{R},\quad f(xyz)=f\left(\frac{xf(y)}{z}\right)f\left(\frac{yf(z)}{x}\right)f\left(\frac{zf(x)}{y}\right),f(0)=0$[/quote]

Ofcourse $f(x)=x$ and $f(x)=1,x\ne 0$ and $f(x)=-1, x\ne 0$ and $f\equiv 0$ are solutions:)

[quote="cckek"]Ofcourse $f(x)=x$ and $f(x)=1,x\ne 0$ and $f(x)=-1, x\ne 0$ and $f\equiv 0$ are solutions:)[/quote]
And for every $\lambda : \mathbb{R}\rightarrow \mathbb{R}$ satisfying
$\lambda(x+y)=\lambda(x)+\lambda(y)$ and $\lambda(\lambda(x))=x$ for $x,y\in\mathbb{R}$
we get a solution
$f(x)=\pm e^{\lambda(\ln|x|)}$ for $x\not =0$.
And for every decomposition $\mathbb{R}=V\oplus W$ in $\mathbb{Q}$-vector spaces we can take $\lambda : \mathbb{R}\rightarrow \mathbb{R}, \lambda(v+w)=v-w$ for $v\in V,w\in W$.


linear algebra
matrix
vector
algebra
polynomial
linear algebra unsolved

Let $ K$ be a field. What are the conditions for the permutation matrix $ P_{\sigma}$ to be diagonalizable in $ M_n(K)$? In that case, find a basis of proper vectors.

The permutation $ \sigma$ of the set $ X$ can be wrote as $ \sigma_{k}\ldots\sigma_{1}$, where $ \sigma_{1},\ldots,\sigma_{k}$ are cyclic permuations of some sets $ X_{1},\ldots,X_{k}$ such that $ X_{1}\cup\ldots\cup X_{k} \equal{} X$ and $ X_{i}\cap X_{j} \equal{} \emptyset$ for $ i\neq j$ and the order of $ \sigma_{i}$ as an element of the permuation group is $ \# X_{i} \equal{} : s_{i}$. Thus, after some change of basis by permutation of basis vectors we get that the matrix $ P_{\sigma}$ is similar to a the matrix of the form
$ \mathrm{diag}\left(\ldots,\begin{bmatrix} 0 & 1 & & 0 \\
& & \ddots & \\
& & & 1 \\
1 & & & 0\end{bmatrix},\ldots\right)$,
where the $ i$-th block is of size $ s_{i}$; we will denote it by $ P_{s_{i}}$. We can diagonalize the whole matrix $ P_{\sigma}$ if and only if we can diagonalize each of these blocks. It is easy to see that the eigenvectors of $ P_{s}$ are $ \left(\omega^{0\cdot j}, \ldots, \omega^{\left(s \minus{} 1\right)\cdot j}\right)$ for $ j \equal{} 0,\ldots,s \minus{} 1$, where $ \omega$ is one of the primitive roots of unity of order $ s$ and the $ j$-th eigenvector corresponds to the eigenvalue $ \omega^{j}$. This set of vectors is a basis since they form a nonsingular van der Monde matrix. By extending eigenbases of the matrices $ P_{s_{i}}$ with some zeroes we get the eigenbasis of the matrix $ P_{\sigma}$, which is diagonalizable iff the polynomials $ x^{s_{i}} \minus{} 1$ split into linear factors over $ K$.

A narrower version of the question: for what $ n$ and what $ p$ is an $ n$-cycle diagonalizable over $ \mathbb{F}_p\equal{}\mathbb{Z}/p\mathbb{Z}\ ?$

For $ n\equal{}p,$ the answer is clearly no, but when is it yes?

For the narrow question: precisely when $ n|p\minus{}1$.

If we make that the field with $ q\equal{}p^k$ elements, precisely when $ n|q\minus{}1$.

In my previous post, I simplified the matter too much. The permutations $ \sigma_{i}$ are cyclic, but they are not necessairly of order $ n$. The block corresponding to $ \sigma_{i}$ should be the circulant matrix whose first row is $ \left(\underbrace{0,\ldots,0}_{r_{i}},1,0,\ldots,0\right)$ for some $ r_{i}$. Its eigenvectors are the same I wrote, but the eigenvalue corresponding to the $ j$-th vector should be $ \omega^{j\cdot r_{i}}$.


geometry
trigonometry
trig identities
Law of Sines

The altitudes through the vertices $A,B,C$ of an acute-angled triangle $ABC$ meet the opposite sides at $D,E,F$, respectively, and we have $AB>AC$. The line $EF$ meets $BC$ at $P$, and the line through $D$ parallel to $EF$ meets the lines $AC$ and $AB$ at $Q$ and $R$, respectively. $N$ is the midpoint of the segment $BC$. Prove or disprove that $\angle{NQP} + \angle{NRP} \geq 180$. [edit] The original problem was to prove that if  $\angle{NQP} + \angle{NRP} < 180$ than $BN > NC$.

Could anyone please just give me a hint!

Try to prove that $RNQP$ is cyclic knowing that $RBQC$ is cyclic(uses the power of the point), and the conclusion follows! :)

I guess your hint was helpfull :D
$\angle(CQN) = 180-\angle(ACB)$
$\angle(QRB) = \angle(PFR) = 180- \angle(AFE) = 180 - \angle(ACB)$ (because $EF$ and $RD$ are parallel) so the quadrilateral $QCRB$ is cyclic indeed :D
This of course implies that $CD*DB = QD*DR  = CN^2-DN^2$
If we can prove that $PD*DN = QD*DR$ that the quadrilateral will also be cyclic but this is equivalent to showing that $PD*DN = CN^2-DN^2$ or to show that $NC^2 = ND*NP$
One can easily prove that $\angle(FPC) = \angle(ACB) - \angle(ABC)$ and $\angle(PFN) = \angle(CAB)$ so we can write now $PN = \frac{FN*sin(\angle(CAB))}{sin(\angle(ACB) - \angle(ABC))}$
and also $DN = CN - CD = \frac{2CN*sin(\angle(ABC))cos(\angle(ACB))}{sin(\angle(CAB))}$
So $ND*NP = CN^2(1-\frac{2sin(\angle(ABC))cos(\angle(BCA))}{sin(\angle(CAB))}*\frac{sin(\angle(CAB))}{sin(\angle(ACB) - \angle(ABC))}$ which is equivalent to proving that $sin(\angle(ACB) - \angle(ABC)) = sin(\angle(CAB)) - 2sin(\angle(ABC))sin(\angle(ACB))$ which in turn is equivalent to
$sin(\angle(ACB) + \angle(ABC)) = sin (\angle(CAB))$ which is surely true  :D  :P
Maybe I have a mistake somewhere but I am sure that this is the right solution!
SAILOR could you please post your solution?!

The problem is equivalent to $BD\cdot CD=ND\cdot DP$.
By Ceva and Menelaus, $\frac{BD}{CD}=\frac{BP}{CP}$. Let ND=a,DC=b, then $BD\cdot CD=ND\cdot DP$ and $\frac{BD}{CD}=\frac{BP}{CP}$ are both equivalent to $CP=\frac{b(a+b)}{a}$. Thus the problem is solved.

Just for the case somebody is interested in a fully synthetic solution:

[color=blue][b]Problem.[/b] Let D, E, F be the feet of the altitudes of a triangle ABC. Let the line EF meet the line BC at a point P, and let the parallel to the line EF through the point D meet the lines CA and AB at the points Q and R, respectively. Finally, let N be the midpoint of the segment BC. Prove that the points N, P, Q and R lie on one circle.[/color]

[i]Solution.[/i] We will use directed angles modulo 180°. First we show a basic fact :

[b]Lemma 1.[/b] If H is the orthocenter of a triangle ABC, and D is the foot of the altitude from the vertex A, then $DB\cdot DC=-DA\cdot DH$, where the segments are directed.

[hide="Proof of Lemma 1"]
[i]Proof of Lemma 1.[/i] Since the point H is the orthocenter of triangle ABC, the lines AH and BH are altitudes in triangle ABC, and thus perpendicular to the lines BC and CA, respectively. Hence, < (BC; AH) = 90° and < (BH; CA) = 90°. Consequently, < (BH; CA) = < (BC; AH), and thus

  < HBD = < (BH; BC) = < (BH; CA) + < (CA; BC) = < (BC; AH) + < (CA; BC) = < (CA; AH) = < CAH = < CAD.

Also, trivially, < HDB = < CDA (since both angles < HDB and < CDA equal 90°, as $AH\perp BC$).

Now, from < HBD = < CAD and < HDB = < CDA, it follows that the triangles HBD and CAD are directly similar. Hence, using non-directed segments at first, we get $\frac{DB}{DH}=\frac{DA}{DC}$. With directed segments, this becomes $\frac{DB}{DH}=-\frac{DA}{DC}$, and thus we get $DB\cdot DC=-DA\cdot DH$, and Lemma 1 is proven.[/hide]

A less trivial fact we will use is the following one:

[b]Lemma 2.[/b] The line PH is perpendicular to the median AN of triangle ABC.

For the proof of this lemma, I refer to http://www.mathlinks.ro/Forum/viewtopic.php?t=552 .

Finally, a very simple lemma:

[b]Lemma 3.[/b] The points B, C, Q and R lie on one circle.

[hide="Proof of Lemma 3"]
[i]Proof of Lemma 3.[/i] Since < BEC = 90° and < BFC = 90°, the points E and F lie on the circle with diameter BC. Thus, the points B, C, E, F are concyclic, and it follows that < CEF = < CBF. In other words, < (CA; EF) = < (BC; AB). On the other hand, < (CA; EF) = < (CA; QR), since the line QR is parallel to the line EF. Thus, < (CA; QR) = < (BC; AB); in other words, < CQR = < CBR, and it follows that the points B, C, Q and R lie on one circle. This proves Lemma 3.[/hide]

Now we come to the solution of the problem:

Let H be the orthocenter of triangle ABC. Then, by Lemma 1, applied to the triangle ABC, we have $DB\cdot DC=-DA\cdot DH$.

On the other hand, by Lemma 2, the line PH is perpendicular to the line AN. In other words, the point H lies on the perpendicular from the point P to the line AN. On the other hand, the point H, being the orthocenter of triangle ABC, must lie on the altitude of triangle ABC issuing from the vertex A; in other words, it lies on the perpendicular from the point A to the line BC. Since the line BC is the same as the line PN, we can thus say that the point H lies on the perpendicular from the point A to the line PN. So the point H lies on the perpendicular from the point P to the line AN and on the perpendicular from the point A to the line PN. In other words, the point H lies on two altitudes of the triangle APN. Thus, this point H is the orthocenter of the triangle APN. The point D is, clearly, the foot of the altitude of this triangle from the vertex A. Hence, we can apply Lemma 1 to the triangle APN, and conclude that $DP\cdot DN=-DA\cdot DH$.

Comparing this with the equation $DB\cdot DC=-DA\cdot DH$, we get $DP\cdot DN=DB\cdot DC$. Since - after Lemma 3 - the points B, C, Q and R lie on one circle, the intersecting chords theorem yields $DB\cdot DC=DQ\cdot DR$. Thus, $DP\cdot DN=DQ\cdot DR$. And after the converse of the intersecting chords theorem, this equation shows that the points N, P, Q and R lie on one circle. This solves the problem.

[b]EDIT:[/b] I just noted this problem was also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=54020 .

  darij

[color=#000080]Without loss of generality we let $\angle{B}>\angle{C}$.Applying the law of sines in $\triangle{PBF}$ we get $\boxed{PB=\frac{acosBsinC}{sin(B-C)}}$Similarly application of law of sines in $\triangle{RDB},\triangle{DQC}$ we get $\boxed{RD=\frac{ccosBsinB}{sinC}}$ and $\boxed{DQ=\frac{bsinCcosC}{sinB}}$.Also $\boxed{DM=\frac{a}{2}-ccosB}$Thus points $P,R,M,Q$ are concyclic

$\Leftrightarrow DM*PD=RD*DQ$

$\Leftrightarrow DM*(PB+BD)=RD*DQ$

$\Leftrightarrow (\frac{a}{2}-ccosB)(\frac{acosBsinC}{sin(B-C)}+ccosB)=bccosBcosC$

$\Leftrightarrow (sinA-2sinCcosB)(\frac{2sinAcosBsinC+2sinCcosBsin(B-C)}{sin(B-C)})=4sinBcosBsinCcosC$

$\Leftrightarrow (2sinAcosBsinC+2sinCcosBsin(B-C)=4sinBcosBsinCcosC$

$\Leftrightarrow sinA+sin(B-C)=2sinBcosC$

$\Leftrightarrow 2sinBcosC=2sinBcosC$

and we are done![/color]

[u][b]Note[/b][/u]:It should be stated in the problem that $\triangle{ABC}$ is such that $AB \neq AC$ (otherwise $FE$ shall not meet $BC$).


Suppose that $ 4^a \equal{} 5$, $ 5^b \equal{} 6$, $ 6^c \equal{} 7$, and $ 7^d \equal{} 8$. What is $ a\cdot b\cdot c\cdot d$?

$ \textbf{(A)}\ 1\qquad
\textbf{(B)}\ \frac{3}{2}\qquad
\textbf{(C)}\ 2\qquad
\textbf{(D)}\ \frac{5}{2}\qquad
\textbf{(E)}\ 3$

[hide="Solution 1"]
Continually substituting, we find that $ 8\equal{}7^d\equal{}6^{cd}\equal{}5^{bcd}\equal{}4^{abcd}\implies abcd\equal{}\boxed{\textbf{(B)}\ \frac32}$.
[/hide]
[hide="Solution 2"]
From the given, $ a\equal{}\log_45$, $ b\equal{}\log_56$, $ c\equal{}\log_67$, and $ d\equal{}\log_78$, so multiplying this and using the property $ \log_mn\log_nk\equal{}\log_mk$, $ abcd\equal{}\log_45\log_56\log_67\log_78\equal{}\log_48\equal{}\log_{2^2}2^3\equal{}\boxed{\textbf{(B)}\ \frac32}$.
[/hide]

[hide="Click for solution"]
We have:

\[ 4^{abcd}\equal{}\left ( \left ( \left ( 4^a \right )^b \right )^c \right )^d\equal{} \left ( \left (5^b \right ) ^c \right )^d\equal{} \left (6^c \right )^d\equal{}7^d\equal{}8\equal{}4^{\frac{3}{2}}
\]

so $ abcd\equal{}\frac{3}{2}$ or $ \boxed{\textbf{(B)}}$.
[/hide]


$2$ circles are tangent internaly at point $A$.a secant intersects the circles in$M,N,P,Q$.prove that $\angle MAP$=$\angle NAQ$

i thinkwith homogeneous this problem is  so easy.

Well , consider the common tangent $KAL$. Use the theorem for the exterior angle in $\triangle APN$ ( points P , N belong to small circle) and the << Chord - tangent  >> theorem.(for example $\angle N = \angle KAM + \angle MAP$ and $\angle Q = \angle KAM$). But $\angle N = \angle Q + \angle NAQ$, so .....

 If you still have problem , please write.

Babis


geometry rectangle Alcumus algebra function domain \/closed

The animation on the left-hand side of the screen with the different-colored dots that illustrate various formulas is really distracting when I try to write long proofs in posts. I would recommend getting rid of it.

Hmmm... I absolutely love it and would like it to stay. Well, in such matters you get as many opinions as you have people, you know :roll:. Nothing satisfies everyone...

I used to like it too...but at some point (perhaps after the two-thousandth time I saw the same formula being derived) I stopped liking it.

Maybe there should just be a way to turn it off.

That seems like a good idea. Maybe include an option in the preferences to turn off the proof animation?

Well, you can right-click on it and choose "Rewind" if it starts distracting you too much when you are typing a long post. Then you'll see just a blue rectangle until you finish typing and go to another page view. This doesn't solve the problem completely but may help a bit. Or you can disable ActiveX when browsing AoPS. That'll kill it too. Of course, adding extra settings in the profile preferences is a better idea but it seems like the site developers have a much bigger fish to fry right now.

Ahh, thanks fedja, that will work. I should have known...everything is clickable :)

What is the "bigger fish"? Alcumus? :maybe:

See #13 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=255835]this[/url] thread. Since Valentin put it in the "public domain" himself, I do not think I'm disclosing anything "top secret" here.


Compute the number of $ z$, such that  $ 0\leq  Arg(z)\leq 2\pi$ and the absolute value of $ z =1$, satisfying the following.

$ z^{29}-z^{7}-1=0$


Ok people for those of you that havent seen 1 vs 100 on TV, its pretty much what is means, 1 VS 100! However a) were using only math problems and b) I don't know if 101 people are willing to preticipate. So here's how it's going to work we will have multiple choice questions you have 3 days to enter the answer. If you get the question right you get to keep going. HOWEVER, if one if the 100 misses the question there out! The question's get harder through every round. If at anytime the "1" misses the question they automatically lose and the "100" win and if all of the "100" are eliminated then the "1" wins! If you are interested in playing send me a private message by Saturday, March 8th with the reason why you should be the "1". If you're confused watch the video.
[youtube]D4McSpANs4c[/youtube]
Also the "1" can not select to just take the money and walk away.
And it isn't battle of the sexes style.

EDIT: If no winner is clear by the 10th question then the "1" automatically wins.
EDIT2: I have decided to tone the number of people down to 20 instead of 100 because 100 did not seem plausible.

I'll be part of the mob.

And there is NO way that there will be 100 people.

[i]And[/i], what's the use of winning?

[quote="MustItAlwaysBeMe"]
[i]And[/i], what's the use of winning?[/quote]

glory. be happy with that! :D

I think there might be a higher chance for success if you lower the number of mob members.

That was my point, but then again, it will be better if it doesn't last as long.  Who thinks we should start an AoPS bank and keep track of prize money won on this forum?

I would like to be the "1", but I'll be part of the mob if there already is a "1".

I want to try and be part of the mob, if I am active enough.  :)

MOB!!!!!!!!

And how hard are these questions gonna be?

inky pinky pon- oh wait the I was chosen...

Mob.

When is this going to take place, and how is this going to take place?

1. The questions should be MOEMS/MATHCOUNTS/HSB difficulty questions.

2. We need a particular number of Mob members (I think a good number would be 10).

The I: lotsofmath

[u]Mob Members:[/u]
MustItAlwaysBeME
D3m0n Shad0w
zapi2007
mathcountsfreak
1=2

So these questions are math questions?

*evil laughter*

i'll be in the mob.. :D

wait.. instead of math, can we do random trivia facts?? they're more fun...

[quote="kimmystar94"]wait.. instead of math, can we do random trivia facts?? they're more fun...[/quote]

Yeah, with math some people will be real pwnage...

I will suck at this, though.

[quote="7h3.D3m0n.117"]Since answers are due within 24 hours, I won't lock this topic for every question. Just don't post your answers in the thread until the deadline is over.

And how is that a trivia question? In the regular 1 vs 100 it was a question about modern facts, like movies or Hollywood stars, not specific questions that Googling will help solve.[/quote]

Well there are many factors that change this from the regular 1 vs 100. For instance we can't control people googling the facts, however it would be more difficult to google something so specific like I have put. Also what do you mean "Answers are due within 24 hours, I won't lock this topic for every question. Just don't post your answers in the thread until the deadline is over." That was slightly confusing because you have close to 3 days and are you encouraging people to drop out by saying answer after the deadline?

Whoa, my bad, I thought today was the 9th for some reason  :blush: . In this case, I will lock this topic temporarily until the deadline is over. Wouldn't want anyone to accidently post the answer and spoil it for the others now would we? I think the solution would be just to PM crazychinesemaniac the solution and wait til the deadline is over.

Hi just incase you are looking on this post instead of the new one the question is how many NBA teams have won 19 straight games or more in its history?

A)9
B)12
C)4

Ok answer was _ and really no one responded except a few people so I think I'll start a new 1vs15 unless you email me answers in like 24hours....

Err can you post the results of the question before moving on? I think you're supposed to knock out a few mob members before going onto the next question.

I'm not going on to the next question...yet..I got about 5 responses out of 11..

1.  You are supposed to eliminate everyone who got the 1st question wrong.
2.  The second question should be a math question, as should the 3rd, so I did not answer the 2nd question.

The "1"
Lotsofmath

The MOB...
1.Mustitalwaysbeme-Out

3.Zapi2007 -Safe
4.Mathcountsfreak -Safe
5.1=2-Out
6.Kimmystar94-Out
7.CircleSquared -Out
8.Xpmath -Safe
9.Andersonw -Out

12.Brut3Forc3 -Out
13.cf249 -Safe
14.Shaggy75-Out

There you have it people we are down to 4 in the mob! The answer was "C". People left are number 3,4,8,13 and the one!

Question 2! Math word problem
A play ground that is a perfect rectangle is being built near a daycare. The daycare wall is one of the long sides of the playground boundaries. You have 125 meters of fencing and the area of the playground is 1500 sq meters. How long is the side of the daycare?Answers rounded to nearest thousandths

A)62.532
B)92.607
C)82.992

Its OVER! The one has one since the answer to the last question was B! I think I will do this again when I am less busy so it can be more sucessful.

yeah... it has to be organized WAY better.... you changed question #1, you should've at least notified the ones who were in the mob and the one...

there were a lot of loopholes..

Also, I don't think most people know exactly how many notes are in a piano piece.  Make the trivia something that will come to mind instantly.  Also, I didn't know you ACTUALLY decided to change the first question.  So yeah.

umm... i searched for about 15 minutes trying to find a FREE pdf file... and then i got out my sheet of paper and started counting.. :D

lol thanks for the advice also question one was on wikipedia :)

oh it was? lol i found a pdf and counted the notes...took forever too.


??only a formula,what do you mean,which problem do you want to post? :)   ;)


I have not visited this forum in a very long time. In fact, I have no idea I have to register a USAMT ID # and submit some other paper. But my solution to the first round has already been sent a long time ago, will my solution still be valid? I hope it is not too late.  :?

Send an email to solutions@usamts.org with your username and real name.  Go to the USAMTS pages by clicking the USAMTS on the sidebar and complete the registration there next time you log in.

We received about 2 dozen papers without any username or ID #; chances are your is one of them.  It will be scored officially.  Please help us attach your paper to your ID by sending your information.


anyone majoring in pre-med or know anything about how med school and everything works?

thanks for the help!

Most of the better undergraduate schools don't have a major labeled "pre-med." Instead, they have students taking a variety of different majors who intend to apply to medical school. Medical schools have certain course requirements that they expect all applicants to fulfill while taking their undergraduate courses. For example, I'm pretty sure that all medical school applicants are required to have so many college credits of specified courses in biology, chemistry, and a few other subjects. But if you meet the prerequisite course requirements, you can take any other undergraduate courses you please for your college major subject. I have heard of math majors, English majors, physics majors, biology majors, history majors, engineering majors, and students with various other undergraduate majors all getting into medical school.

I'm told that a really crucial thing to do if you desire to go to medical school is to get really good grades in your undergraduate courses. And at the school where I got my undergraduate degree, the courses that were prerequisite courses for medical school application were HARD courses in which it was not easy to get good grades. So learn your study skills while you are young, and be prepared to work hard to ensure a successful application to medical school.

One physician I know locally is a neurosurgeon who was an undergraduate math major. She was told by her undergraduate advisor, a woman mathematician, that she was wasting her life by going into medicine.   :P  She is one of the best neurosurgeons in our town, so I don't think she wasted her life by going to medical school.

The specific required courses for pre-med (or pre-dental, or pre-veterinary, or a couple of other things) tend to match up well with the lower division and mid-level requirements for a major in the Biology/Biochemistry/Chemistry cluster of majors. That cluster of majors is divided up in many different ways and goes by many different names at various colleges and universities. Because of that close match, many pre-med students adopt majors from within this grouping. But the overlap is just the lower division and mid-level courses. The upper division courses for those majors are not required for med school - hence what tokenadult said about students from many other majors. Note, however, that you can't just finish your history major, or math major, or whatever following just the requirements of that major, and then decide to apply to med school - you do need those science classes. And you should start taking the science classes right away, as a freshman. Most schools have some sort of pre-health-professions advisor or advising center, or at least some source of information from which you can see what courses you need to take.

Here's the question you should ask: if you wind up not going to med school, what would you do then? That's part of what should determine what major you choose. Many things can happen. Medical schools are quite hard to get into, and many students who want to apply simply don't have the college grades (or MCAT scores) to be competitive applicants. Some students, once they get a good look at what the profession entails, lose their enthusiasm for it for themselves. I've also see students attracted so strongly by the science courses that they wind up as research scientists rather than physicians.

The short version: have a Plan B.

The most common major of students going to med school from Princeton is History. So it's possible to get into med school with just about any major. The important thing is to get strong grades in your undergraduate classes (especially the sciences), to score well on the MCAT, and to have a strong personal statement and interview.

The [url=http://www.aamc.org/students/mcat/start.htm]MCAT[/url] in particular means much more for medical school admission than does, say, the SAT for undergraduate admission, and it is a much harder test. The MCAT covers Physical Sciences (Physics and Chemistry), Biological Sciences (Biology, Chemistiry, and Organic Chemistry), Verbal Reasoning, and Writing. All of these topics are covered at the basic undergraduate level. No advanced coursework in these areas is really necessary.

A lot of pre-med students major in history because it gives them a nice break from the science prerequisites (Calculus, Physics, Chemistry, Organic Chemistry), while at the same time helps them with their verbal reasoning and writing skills. History and humanities classes also tend to be easier to get higher grades than science classes so it pads the GPA.

Biology is also a common major of students who go on to med-school, as are psychology, mathematics, and english.

[quote="juicybooty911"]what's premed, prelaw, etc?  what are those part of? undergradute or something else?[/quote]

A program that prepares you for medical school in the United States is an undergraduate program, because United States medical schools expect their applicants to be completing an undergraduate degree program (there are a few exceptions). Similarly, a program that comes before applying to law school could be called a "prelaw" program, but in fact there are no course requirements AT ALL to apply to law school--any kind of undergraduate degree, with any combination of courses taken or not taken, is valid for applying to law school. (I know about law school requirements, because I have a law degree that I gained ten years after my undergraduate degree.) If a school says it has a "premed" program, it may mean nothing more than that it offers the courses that are prerequisites for medical school admission--but nearly any university and most liberal arts colleges offer those courses. If a school says it has a "prelaw" program, it is fooling you, because there is no such thing.

Many undergraduate schools have student clubs such as the Prelaw Society or some such name, and it may be valuable to join a club like that just to become acquainted with other students who are thinking about going to law school or to medical school. But joining such a club while an undergraduate student, which may be a very good idea, is not a requirement for applying to medical school or to law school. Some undergraduate schools also have special premed or prelaw advisors, who may give you useful advice about how to prepare for your next school. But, again, you can get the same advice in many places (for example, here
 :)  ) and it is not necessary to consult a premed advisor before applying to medical school.

Usually when someone says they have a pre-med or pre-law program that means they have a special advisor or some sort of guidance system to help students apply to medical school or law school. For example, Washington University in St. Louis has a "pre-med" program in that they have very good courses for the basic requirements for med school and they have an advising system especially for those students in the pre-med program. I believe they may also offer some kind of combined undergrad/medical degree program to a select few students, but don't take my word for it.

I'm going to be a freshman in the fall and I'm a Moleular/Cell Biology major, and I'm on a pre-med pathway. There are set courses you must take before applyin to med school. My major overlaps with many of these courses.

Another factor to consider if you want to go to med school is what your major will be. So many applicans are science majors, that some med schools value other majors, for they want diversity in their entering class. I chose not to do something such as some Humanities major as mine because if I don't ge into med school, I still want to do something science related.

That's a good policy: decide what you would really like to do even if you don't get into med school, and pursue that while being aware of med school requirements.

[quote="Cyberman"]M.D.s get sued so often, it may have been better to be a lawyer..  :P[/quote]
Lol. Get a phD then  :D That way you won't get sued :P

Depends on where you work

If you own your own practice, then you are responsible for your own malpractice insurance and what not

That's opposed to a doctor that works for a company like Kaiser, who insures all their workers...so they are protected against frivolous lawsuits


The largest whole number such that seven times the number is less than 100 is

$ \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 16$

Let $ x$ be the number.  We have that $ 7x<100$, or $ x<14.2$.  Our answer is thus $ [14.2]\equal{}\boxed{14}$

[quote="AIME15"]Do you realize that the 1980 AHSME has already been posted??  If you are going to post problems for the Resources Section, make sure they haven't been posted yet...[/quote]

Sigh yeah, I sort of realized that... I meant to post 1985 AHSME...


vector
geometry
geometric transformation
reflection
trigonometry

an object O is just about to strike a perfectly reflecting inclined  plane of inclination 37 degrees .its velocity is 5m/s.find velocity of its image.

i also thought that.but the options are

(A)3i + 4j
(B)4i + 3j
(C)4.8i + 1.4j
(D)1.4i + 4.8j

[quote="Rajiv"]Simple! If velocity of object is $ \minus{} 5\hat{j}$, then that of image is $ 5\hat{i}$.

If you mean relative velocity of image with respect to object, that too is not tough, just simple vector addition work which, I am sure, you can do.[/quote]

i think htats true when the incline is 45...

draw the normal to the plane and take the reflection

answer will be 5i only wen angle is 45

otherwise ans as in this cae wil be (C) :D

just simple vector addition and trigonometry :oops:

:oops:

Right! The velocities will be same only when angle is $ 45$...didn't notice the $ 37$...


calculus
integration
trigonometry
calculus computations

Show that, for any integer m,

$ \int_{0}^{2\pi}\, e^{x}\cos mx\, dx \equal{} \frac{1}{m^{2} \plus{} 1} (e^{2\pi}\minus{}1).$

(i) Expand $ \cos (A\plus{}B) \plus{} \cos (A\minus{}B)$. Hence show that

$ \int_{0}^{2\pi}\,e^{x}\cos x\cos 6x\,dx \equal{} \frac{19}{650}(e^{2\pi} \minus{}1).$

(ii) Evaluate $ \int_ {0}^{2\pi}\, e^{x} \sin 2x \sin 4x\cos x\,dx$.

[quote="AndrewTom"]Show that, for any integer m,

$ \int_{0}^{2\pi}\, e^{x}\cos mx\, dx \equal{} \frac {1}{m^{2} \plus{} 1} (e^{2\pi} \minus{} 1).$[/quote]

$ (e^{x}\cos mx)'\equal{}e^{x}\cos mx\minus{}me^{x}\sin mx\ \cdots [1]$

$ (e^{x}\sin mx)'\equal{}e^{x}\sin mx\plus{}me^{x}\cos mx\ \cdots [2]$

$ [1]\plus{}2*[2]: (1\plus{}m^2)e^{x}\cos mx\equal{}(e^{x}\cos mx)'\plus{}m(e^{x}\sin mx)'$

$ \therefore \int_0^{2\pi} e^{x}\cos mx\ dx\equal{}\left[\frac{e^{x}}{1\plus{}m^2}(\cos mx\plus{}m\sin mx)\right]_0^{2\pi}\equal{}\frac{1}{1\plus{}m^2}(e^{2\pi}\minus{}1)$

You are welcome! :lol:

Repeat Interal by parts two times,

$ I_m\equal{}\int_0^{2\pi} e^{x}\cos mx\ dx\equal{}[e^{x}\cos mx]_0^{2\pi}\plus{}m\int_0^{2\pi} e^x\sin mx\ dx$

$ \equal{}e^{2\pi}\minus{}1\plus{}m\left([e^{x}\sin mx]_0^{2\pi}\minus{}m\int_0^{2\pi} e^{x}\cos mx\ dx\right)$

$ \equal{}e^{2\pi}\minus{}1\minus{}m^2I_m\equal{}I_m$, yielding the result.

Yea there are many ways to the result. I switch between doing such integrals with the method I used here and the ones you displayed :).


In acute $\triangle{ABC}$ the lengths of a,b, and c are x+1,x, and x-1. BD is perpendicular to AC. Find CD-DA.

[hide]Let $AD=e$ and $CD=f$, so $e+f=x$. Then, by the Pythagorean Theorem, $(x-1)^2-e^2=(x+1)^2-f^2 \Rightarrow -2x-e^2=2x-f^2 \Rightarrow e^2-f^2=-4x \Rightarrow (e+f)(e-f)=-4x \Rightarrow x(e-f)=-4x \Rightarrow e-f=-4 \Rightarrow f-e=4 \Rightarrow CD-AD=\framebox{4}$.[/hide]

Assuming that a=BC, b=AC, c=AB,
[hide]We know that CD+DA=AC.  Since triangles ADB and CDB are right with right angles at D, $CD^2+BD^2=(x+1)^2$ and $(x-CD)^2+BD^2=x^2-2xCD+CD^2+BD^2=(x-1)^2$.  Subtracting the second equation from the first, we get that $2xCD-x^2=4x$ and $2CD-x=2CD-(CD+AD)=CD-AD=4$.[/hide]

Hmm..can anyone see where I messed up my signs?

arghh... your work is all in 3 lines, not pretty, im not gonna look at it :p
but i think 4 is right

Yeah I found the mistake. Should i put linebreaks? What's the convention?

Every time you have that $\Rightarrow$ symbol, replace it with a line break and then start the line with the $\therefore$ (therefore) sign.

It's a easy problem, I think. And the way to get the solution for this problem is a bit like the way to proof Heron formula.
(My English is too bad, so may be you cant understand what I wrote)
$S= \sqrt{p(p-a)(p-b)(p-c)}$


search
induction
Gauss
number theory unsolved
number theory

Apart from 3^2-1, prove that a^n-1 always has a prime factor of the form nk+1.

Thx

Bomb

It's been shown before on the forum (by peter Scholtze, I believe) that any prime factor of $\varphi_n(a)$ which does not divide $n$ is of the form $nk+1$, where $\varphi_n$ is the $n$'th cyclotomic polynomial. Since $\varphi_n|x^n-1$, this may help.

I'm sure you have enough keywords now to perform a search :).

However,  one needsa to prove first that such a prime p exists that divides a^n-1 but does not divide into n.

Thx

Bomb

Yes, I know that problem doesn't exactly solve this one, but, like I said, it may help.

Sorry but my level of knowledge is too low to figure out how it helps.

Bomb

For the little Fermat's theorem, except for Cheamicheal's numbers, $a^{p-1}\equiv 1$ mod $p$, so let's take the prime factors of $a^n-1$ and call them $q_i$: since $q_i|a^n-1$, or rather $a^n\equiv 1$ mod $q_i$, we'll have (because of the Gauss' demostration) that $n$ divides $q_i-1$, or rather $kn+1=q_i$, as we want to demostrate

But how do u prove the existence of such a prime p.

Bomb

Do you mean how can I proof that $a^n-1$ isn't a prime? Well, it's simple, it's $(a-1)(\sum^{n-1}_{k=1} a^k)$

You must shows there exist a prime p s.t. p divides a^n-1 but does not divide a^d-1 where d is a factor of n.
Then and only then is p of form nk+1.

Bomb

$p-1$ is even for each prime $>2$. So $2|p-1$. $GCD(a^k-1;a^{p-1}-1)=a^k-1$ because $a^{p-1}-1=(a^k-1)t$, where $t>a^k-1$ and $k|p-1$. But $a^k-1$ hasn't all the primes of $t$, except for some cases $((3+1)(3-1))$: you can demostrate it by the distinction among even numbers, odd nos. and powers of two. Sorry if I made stupid mistakes, but i'm exausted now

But  hasn't all the primes of , except for some cases : you can demostrate it by the distinction among even numbers, odd nos. and powers of two.

But, how do you prove that a^k-1 has not all the primes of t, using your notation, JohnConstantine.

Bomb


If  $m, n$  are integers and $m^2 + 28mn + n^2$ is a multiple of  $13$.   , prove that $m^3 + n^3$ is also a multiple of  $13$.

[u] Babis[/u]

26mn is a multiple of 13,so m*m+28mn+n*n-26mn is a multiple of 13.That is to say
(m+n)*(m+n)is a multiple of 13,so m+n is a multiple of 13.m*m*m+n*n*n=(m+n)(m*m-mn+n*n),of course it is a multiple of 13.
   By the way,which software do you use to write m*m and so on? :P

[quote="qmdtc"]   By the way,which software do you use to write m*m and so on? :P[/quote]

Latex
see more on [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5261[/url]

$LaTeX$

[quote="socrates"]Which class was this for?

About the solution notice that $m^2+n^2+28mn=(m+n)^2+26mn$
so $(m+n)^2$ is a mulpiple of 13.
since 13 is a prime number we conclude that $m+n$ is divisible by 13 as well
so the prolem is solved by just factoring $m^3+n^3$=$(m+n)k$
where $k \in \mathbb{Z}$

Nice problem but based in a very common idea. :)  :D[/quote]

Is it even necessary to write $m^3+n^3 = (m+n)k$ where $k \in \mathbb{Z}$? I mean, if you know $13|(m+n)$, it's obvious that $13|(m^3+n^3)$ since $m^3+n^3 = (m+n)(m^2-mn+n^2)$ and obviously it's true..

 :D

Very good! This was  my solution ,too.

[u] To Socrates[/u]

The problem was: No 2 , Classe 12 !!!. Of course very easy.But this is the first round. In the second round the problems are more interesting.[b]Thanks for answering  QMDTC in his post.[/b]

[u]Babis[/u]


calculus
integration
real analysis
function
probability
limit
real analysis solved

$x\in R$, $x=E(x)+\sum \epsilon_{k}(x).2^{-k}$

$\epsilon_{k}(x) = 0 \;or \; 1$

Compute

1/

$\int_{0}^{1}\epsilon_{k}(x)dx$

2/
$\int_{0}^{1}\epsilon_{k}(x)\epsilon_{q}(x)dx$ where k,q positive integer $k \neq q$

3/ $\int_{0}^{1}|S_{n}(x)-\frac{1}{2}|^2dx$ where

$S_{n}(x)=\frac{1}{n}\sum_{1\leq k \leq n}\epsilon_{k}(x)$

I assume you mean $E(x)$ on the first line to be the integer part of $x.$ Since everything in this problem concerns the interval $[0,1],$ we might as well assume $E(x)=0$ and ignore it.

There are countably many points (the dyadic rationals) for which $e_k(x)$ is not well defined. That's a set of measure zero, hence harmless as long as we interpret your integals as Lebesgue integrals.

Of course, $\int_0^1e_k(x)\,dx=\frac12.$

Furthermore, it is easy to show (just by looking at the sets on which the functions take on each possible value) that

(deleted - ack, what was I thinking!)

Try this: $\int_0^1\left(e_k(x)-\frac12\right)\left(e_q(x)-\frac12\right)dx=0$

That says that the $e_k-\frac12$ are orthogonal. In fact, they're much more than that - they are probabilistically independent.

Choosing $x\in[0,1]$ is choosing a member of a probability sample space. Consider this the space of flipping a fair coin infinitely many times. $e_k(x)$ is 1 if the $k$th coin is heads, and 0 of it is tails.

That identifies $S_n(x)$ as the average fraction of flips so far that are heads. The Strong Law of Large Numbers says that $\lim_{n\to\infty}S_n(x)=\frac12$ for almost every $x.$

The actual result Moubinool asked for is far weaker than that, and has several reasonably easily accessible proofs. I'll back away and let others provide those.

The answers to Moubinool's #1 and #2  are that $\int e_k=\frac12$ and $\int e_ke_q=\frac14$ for $k\ne q.$ As for #3:

Suppose $(e_k)$ is a sequence of square-integrable ($L^2$) functions on a measure space of measure 1 such that there are constants $\mu$ and $\sigma$ such that for all $k\ne q,$ $\int e_k=\mu,\,\int e_ke_q=\mu^2,$ and $\int e_k^2=\mu^2+\sigma^2.$

Let $S_n=\frac1n\sum_{k=1}^ne_k.$ We wish to compute

$\int(S_n-\mu)^2.$

This equals $\int S_n^2-2\mu\int S_n +\mu^2.$

$\int S_n = \frac1n\sum_{k=1}^n\mu = \mu$, so $-2\mu\int S_n=-2\mu^2.$

$\int S_n^2=\frac1{n^2}\sum_{k=1}^n\sum_{q=1}^ne_ke_q$

$=\frac1{n^2}\left(\sum_{k=1}^n\sum_{q=1}^n\mu^2+ \sum_{k=1}^n(\mu^2+\sigma^2)\right)$

$=\frac1{n^2}\left(n^2\mu^2+n\sigma^2\right)=\mu^2+\frac1n\.\sigma^2.$

So, $\int(S_n-\mu)^2=\mu^2+\frac1n\,\sigma^2-2\mu^2+\mu^2= \frac1n\,\sigma^2.$

In our case, $\sigma^2 = \int e_k^2-\left(\int e_k\right)^2=\frac12-\frac14=\frac14.$

Thus, $\int(S_n-\mu)^2=\frac1{4n}.$

Of course, my entire language and notation in this post is quite silly. These are random variables. They are a sequence of uncorrelated random variables with the same mean and same variance, and everything we said can be phrased in the language of variances.

These random variables are in fact independent and identically distributed, but for the result in #3 we can weaken the hypothesis of independent to uncorrelated, and we can weaken the hypothesis of identically distributed to having the same mean and variance.


How many permutations of 1,2,3,4,5,6,7 are not derangements?

thanks~

[quote="AndrewTom"]See here http://en.wikipedia.org/wiki/Derangement[/quote]

Using the list for $ d_n$ in this link provided by Andrew Tom:
[hide]
The number of derangements for $ n \equal{} 7$ is $ 1854$.

The number of permutations is $ 7! \equal{} 5040$.

The number of permutations without derangements is

$ 5040 \minus{} 1854 \equal{} 3186$.

Note:  The number of derangements for seven elements, here,
divided by the number of permutations for seven elements is

1854/5040 ~ 0.367857 and 1/e ~ 0.367879
 (its limit as the number of derangements increases
 without bound).[/hide]


[i]Second Test, May 16[/i]

Let $a$, $b$, and $c$ be positive real numbers such that $abc = 1$. Prove that
$\frac{ab}{a^{5}+b^{5}+ab}+\frac{bc}{b^{5}+c^{5}+bc}+\frac{ca}{c^{5}+a^{5}+ca}\le 1$ .
When does equality hold?

What about \[a = \frac{x}{y}\] etc. and then Muirhead? (with previous basic transormations to "clear" form)

From what I calculated fast in my notebook seems to work, tho it's way more brutalforce and less elegant solution than proposed by Megus.

(and takes 5 more minutes)

[quote="Beat"]$a^{5}+b^{5}\ge a^{2}b^{2}(a+b)$ ? Prove it[/quote]

Monotonic crap and its obvious.

btw. Xevarion your AM-GM proof is boring. Freaking boring. :D

Erm, from what i calculated in my mind multiplying everything by everything and then using abc = 1 also works.

Also and perhaps simpler $a^{5}+b^{5}\geq a^{3}b^{2}+a^{2}b^{3}$ by rearrangement on $a^{3},b^{3}: a^{2},b^{2}$.

Use the identity $a^{5}+b^{5}=(a+b)(a^{4}-a^{3}b+a^{2}b^{2}-ab^{3}+b^{4})$ and upon simplifying you need to prove $a^{4}+b^{4}\geq a^{3}b+b^{4}a$,which,by putting all the terms in the left side,is equivalent to: \[(a^{3}-b^{3})(a-b)\geq0\] .I think it is clear how to finish this last one..

about $a^{5}+b^{5}\geq a^{2}b^{2}(a+b)$ we can say that it is true according to [b]Muirhead[/b]:

      $(5,0)\succ(3,2)$

[quote="Megus"]$\frac{ab}{a^{5}+b^{5}+ab}= \frac{1}{c(a^{5}+b^{5})+a^{2}b^{2}c^{2}}\leq \frac{1}{c(a^{2}b^{2}(a+b))+a^{2}b^{2}c^{2}}= \frac{c}{a+b+c}$

similarly for other fractions, adding these up you get the inequality[/quote] :lol: nice puzzle proof!  :lol:


AMC
AIME
LaTeX
USAMTS
email
calculus
linear algebra

Who's coming?

Ironically, I don't know if I'll be on my school's team yet, but even if I won't be, I'll be there as an individual.

Me too.

Our Varsity team has a better chance of winning, but the JV team was invited instead.

How many people total are competing?

Circa 50 or a little fewer schools with teams of four, plus 15 individuals, but many of them are probably coming with school teams anyhow, so that's about 200.

I'm coming.

Does anyone have a copy of last year's (or any other year's) Varsity tests?  I'm wondering where to get them...

Most teachers in GA who have been to varsity state before have copies of old tests, I think.  But you're in Canada, aren't you, lowest_reynolds?  I really don't know.  It would take a lot of time to make an electronic file (there are 50 problems, at least 40 of which are multiple-choice), and I don't know if we would be allowed to anyhow.  Do you know a teacher in GA who could send you copies?

I don't know of a teacher in GA that could send me copies.  Booo...

Couldn't you just scan me a copy?

I'm in Georgia right now btw.  But I do mostly Math stuff in Canada and whatnot.  I guess I got invited to this thing on the basis of my AMC + AIME composite.

My email's jmpham89@yahoo.ca
...

Psssh...send me some tests there...it would realllllly help  ;)

I don't have a scanner.  Maybe my parents have one at work, but that would be a great deal of trouble.  I'm working on a $\text{\TeX}$ file, though.

Alright, thanks.  :)

Post it or send it when you're done.

I would also like a copy.

Thanks alot Boy Soprano II.

All right, I've finished.  I won't post them here, though, since it is a "public" forum (although it's pretty obscure at the moment).

Could you e-mail it to me?
[hide]xantoscguin@gmail.com[/hide]

Methinks I already did.  Oh, well, I'll send it again, since $\text{\TeX}$ files are small.

Got it. Thanks.

Now I must install $\LaTeX$ on my comp.

BTW: What is the time limit on this test? And are calculators allowed?

The rules change over time, but in 2003, it was 90 minutes, and calculators were allowed.  Scoring was $50+4x-y$, where $x$ was the number of questions correct and $y$ was the number of questions incorrect.

amy is the arml chick with the blonde hair?

lol. how many of you georgia arml people are going to be returning? i'll be on the team this coming year. dont know too many people still on the team though....

[quote="Xantos C. Guin"]
I added JV State since that is the tournament where non-ARML members have the highest chance to beat ARML members since it is somewhat fluky.[/quote]

!!! haha thats how i first got invited to arml

just wondering, xantos, where do you teach?

Wes?

I'm pretty sure I'll be returning. So will Boy Soprano II (the other 9th grader who got on the A team.)

BTW: Why did you move to New Hampshire?

[quote="captcha000"]oops lol you sound like a teacher  :P

haha didn't move. went to a boarding school there. i'm coming back this year. going to lakeside, which doesnt have a great math team, but oh well. were you on the team two years ago, or just this past year?[/quote]

No, just this year. I might have made the team last year had I taken the AMC10/AIME but I barely knew about it.

[quote="drunner2007"]wes brown right?
so you were at exeter?
i thought mr. sigur said you were headed to mit without finishing high school.

hey Xantos, every time i saw penguinintegral i kept thinking it was u. its not right?[/quote]

captcha000 is Howard Tong.

I was asking him if Wes was also there when he was in GA ARML. (He made a statement about Linda and Harrison being the only ones still there, but that statement isn't there anymore.)

penguinintegral is not me, just another penguin on AoPS.

lol yeah i did know wes. he's not on the team anymore though, didnt he already graduate? he joined my third year.

lol you read quickly, i found the thing about harrison and linda kinda irrelevant so i took it out. there are other people i know on the team (peter, brayden, ruby were with me at mathcounts) but no one else from ARML. you might know livia, she was at varsity state, but she told me she didnt feel like doing ARML this year and steve got really angry at her =\.

you might have said so before, but what school do you go to, xantos?

He goes to Wheeler High School, a magnet school in Cobb County... for the time being.  (I go to Lassiter.)  Hmm.  Steve seemed kind of philosophical about Livia's non-participation on the bus.  The coaches were happy with how the team seemed to be turning out, anyhow.  But then we got eighth.  Well, I'm looking forward to next year.  Especially since our B team beat all the other B teams, and we'll still have three of our five high scorers from this year.

By the way, good job on the latest mock AIMEs, Santhosh.

Oh, does anybody know where what seemed to be the only non-.xls ARML scoresheet in existance (which used to be posted on the GA ARML website) went?  Or does anybody who has Excel (or some other such program [unlike me]) want to make some other file type out of the .xls version? :D

Yeah, Wes is going to MIT without having had a senior year.  It said on the website that he didn't graduate, but that's really a rather trivial detail when you consider it. :D

Wes says he might attend some summer school in GA just so he can be on the ARML team next year. Atleast that's what he told me on the bus.

[quote="Boy Soprano II"]Oh, does anybody know where what seemed to be the only non-.xls ARML scoresheet in existance (which used to be posted on the GA ARML website) went? Or does anybody who has Excel (or some other such program [unlike me]) want to make some other file type out of the .xls version?  [/quote]

What other file type do you have in mind?

oh my, wheeler and lassiter...oh yeah phillip mote is from lassiter too isnt he? i dunno if you know the "fab four" from a few years ago: adam tart, brooks andrews, andy vesper, johnny rodean? adam was in the class of 04, the rest were 03, so they were from quite a few years back.

xantos, maybe you remember oliver khamky? he also graduated in 04.

Yes, Philip goes to Lassiter, so do 3 other GA ARML 2006 team members.

I never heard of the "fab four". I was in 7th grade in 04. (That was when if I heard "AMC", I would automaticaly think American Movie Channel.)

I never knew Oliver since I entered Wheeler for the 05-06 year. But I do know that he was the only USAMO qualifier and ARML member ever from Wheeler. Our Varsity team is doing alot worse without him.

Does Miles speak of PDF or HTM? Or perhaps Miles would like to know about openoffice.org? For OO.Org opens and edits xls files, and does run on Windows, and does everything else that M$\$$ Office does. Sure, it is sometimes a little clunky, but before you go blaming it for a bad UI, look at Office 2007. YEEEEEEEEEEECCCCCHHHHHHHH!!!!!!!!!!!!

Mrs. Poss sometimes talks about them.  Haha, if Adam had still been here this year, we would have had more than a bit of trouble determining whom to put on our team for Varsity State. :D   (We had four people on the team this year.)

I would really like almost any type of file except .xls.  PDF, html, htm, .txt, .tex, .dvi, or even .jpg or .bmp or (ew) .doc.  But not .ppt (although I seriously doubt that anybody would do that).  But I think I might install Open Office one of these days.  (Too late at night to install at the moment.)  (After our hard-drive crashed in December, our new one had practically nothing on it.  Fortunately, we have a fair amount of CDs.)  Okay, if I install Open Office before anybody else converts the file, I'll convert it myself, for the benefit of humanity.

(I wonder if Mathematica can read .xls (we got student Mathematicas at MOP) :rotfl: (I have Mathematica but I can't process .xls or .ppt files))

We should tell Wes's story (perhaps slightly altered) to a stranger, and see what the reaction would be.  "Wes dropped out of high school so he could go to MIT.  But then he had to go to summer school in Georgia so he could go to Pennsylvania for a math tournament."

Okay, I have been too lazy to install OpenOffice recently, but I have discovered that my cheapo program was capable of reading .xls documents.  So anyhow, I converted it to a .tex document, which I shall now attach.  Plus, in case something funny happens, I shall also link to it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=5115]here[/url].


a current $ I$ flows down a wire of radius $ a$ it is distributed uniformly.then find the surface current  density

I think the surface current density is $ 0$.

@golduck no it flows on the surface of wire

@shreyas
$ K=\frac{d \vec I}{d \vec l}$

where $ K$ is the surface current density
and $ l$ is width perpendicular to current flow

The volume current density is $ j=\frac{I}{\pi a^{2}}$. The current flowing between radius $ x$ and $ x+dx$ is $ I(x)=j\cdot 2\pi xdx$.
As the current is distributed uniformly, the surface current density is equal in every point on the surface of the wire, so we can write:
$ K=\frac{I(a)}{2\pi a}=\frac{I}{2\pi a^{2}}dx$. As we are talking about surface charge density than $ dx\rightarrow 0$, so $ K\rightarrow 0$.
Or in other way: if we have got a surface current density, then the component of the magnetic induction vector perpendicular to this surface changes by $ \mu_{0}K$. But for this wire we have got (from Ampere's law) $ B_{1}(x)=\frac{\mu_{0}Ix}{2\pi a^{2}}$ inside the wire and $ B_{2}(x)=\frac{\mu_{0}I}{2\pi x}$ outside the wire (of course $ B$ vector is always perpendicular to the radius) so $ B_{1}(a)=B_{2}(a)$ and $ K$ must be equal to $ 0$

[quote="pardesi"]a current $ I$ flows down a wire of radius $ a$ it is distributed uniformly[/quote]
I think it means it is distributed uniformly in the volume of the wire.

[quote="pardesi"]golduck it is given to u that current flows uniformly only on the surface[/quote]
so is it $ K=\frac{I}{2\pi a}$ :?:

sorry if my statement was confusing.but since the question demanded surface density i thought it would be pretty clear
yes of course the ans is right...but i had posted this to double check my understanding of the surface current density because it is not that clear (at least to me) in the book

Consider the same problem without the condition of uniform flow. Current I flows through wire of radius a, resistivity rho. Wire obeys Ohm's law. Find volume current density [b]j(r)[/b]
I have read (Feynman Lectures II) that [b]j(r)[/b] is such that power dissipated in the wire is minimum. Thus we have 2 integrals of [b]j[/b], one saying total current in the wire is I and the other giving the amount of power dissipated(to be minimised). I can't take it any further than that.

well first most importantly when u say current $ I$ flows through the wire then does that mean the entire wire (Assuming it to be a part of another wire) or through a cross section....

Current [b]I[/b] flows through a long wire and we are considering a small part of length l


I was looking at prime number theorem and suddenly got interested in this:

we know for $ n > 2$, $ p_1*p_2*...*p_n \minus{} 1$ always has some primes before it that is greater than $ p_n$, but I was wondering when will $ p_1*p_2*...*p_n \minus{} 1$ be a prime?   I tried for the first few, 2*3-1=5 is a prime, so is 2*3*5-1 = 29, and then 219 is not a prime,  I don't know about 2309 yet.  Is there a pattern?  I am sure it's been done to the death already, but just curious since this is very known from Euclid's proof of the infinite primes.  Anyone?

Let $ p_n$ be the nth prime, does anyone knows this?  Hello?

You're more likely to get a legitimate answer if you post your problem [url=http://www.artofproblemsolving.com/Forum/index.php?f=58]here[/url], the Number Theory Open Problems forum.

Either ask a moderator to move this one, or just post it in there and ask a moderator to delete this one (you can just append "[DELETE]" to the title of this particular thread). I doubt this subforum gets much traffic.

[Moderator says: or just wait until a moderator sees that topic ;) ]

Short answer is No, there is no pattern.

[quote="randomgraph"]Letting $ q_n \equal{} p_1p_2\cdots p_n \minus{} 1$ where $ p_1 \equal{} 2$, $ p_2 \equal{} 3$, $ p_2 \equal{} 5$ etc. is the sequence of primes, we find that $ q_n$ is prime for $ n \equal{} 2,3,5,6,13,24,66$ and 68 but no other $ n$ below 160.

EDIT: This is on [url=http://www.research.att.com/~njas/sequences/?q=2%2C3%2C5%2C6%2C13%2C24%2C66%2C68&language=english&go=Search]OEIS[/url], of course, and continues 167, 287, 310, ..., 1849. Checking $ q_{1849}$ must have been a significant effort. This is pretty sparse, but who knows if it goes on forever.

BTW, I used this [url=http://www.alpertron.com.ar/ECM.HTM]ECM-based online calculator[/url]to make sure that $ q_{66}$ and $ q_{68}$ are indeed prime.[/quote]

that was a really cool site.  the OEIS.  thanks!  let me know if you ever get anywhere.


Prove that there is no integer solution to the equation
\[x_{1}^{4}+x_{2}^{4}+x_{3}^{4}+\cdots+x_{14}^{4}= 1599. \]

$x^{4}\equiv_{16}0,1\implies x_{1}^{4}+\hdots+x_{14}^{4}\not\equiv 15\equiv 1599\pmod{16}\implies\text{No Solution}.$

yeah, hat's pretty much how to do it...

But please hide your soutions from now on.


linear algebra
matrix
superior algebra
superior algebra solved

for any $A\in M_n(\mathbb C)$ we denote with $m(A)$ the numbers of its nonzero minors.
a)prove that $m(I_n)=2^n-1$
b)prove that for any $A\in M_n(\mathbb C)$ notsingular matrix we have that $m(A)\geq 2^n-1$.

A minor of a square matrix is the determinant obtained by eliminated p lines and p columns.

The of minors
a)
size 1 : |1| obtained by deleting n-1 rows and n-1 columns
there are C(n,n-1)

size 2 :
|1 0|
|0 1| by deleting n-2 rows and n-2 columns there are C(n,n-2)

etc
size n : |I_n| just C(n,0)

There are

C(n,0)+C(n,1)+C(n,2)+...+C(n,n-1)=  \sum_(k=0 to n) C(n,k) -1 =
2^n - 1

moubinool,

i think you wanted to say 'size'. 'seize' is another word with different meaning.

And the problem says "m(A) the numbers of its nonzero minors"...
How do we know the determinants are non-zero?...

for a)
we can obtain a nonzero minor from I_n if we eliminate the lines i_1,i_2,...,i_k and the same columns... (otherweise we will obtain a null line or column)...
k\in {0,1,2,..,n-1}
for a certain k we can choose those lines (and columns) in C(n,k) ways... so for all k in {0,1,..,n-1} we can choose them in C(n,0)+C(n,1)+..+C(n,n-1)=2^n-1

[quote="Kuba"]b)prove that for any A\in M_n(C) notsingular matrix we have that m(A)\geq 2^n-1.
[/quote]

Chose k lines from A. The Binet Cauchy formula for $\det(A)$ in terms of the k-order minors on these k lines looks like:
 $\displaystyle \det A= \sum_{i=1}^{\left( \begin{array}{cc}  n\\ k \end{array}\right)} d_i\cdot c_i$
,where $d_i$ are all the minors formed on the k-lines of A.

If A is non-singular, then one of those $d_i$'s is not zero.
This shows that we have at least $\left( \begin{array}{cc}  n\\ k \end{array}\right)$ k-order non-zero minors, so at least $2^n-1$ non-zero minors as k ranges from 1 to n.


[quote="paradox102"]what happens if you mail the entry and permission forms, and you email the solutions?  Meaning, what happens if the person recieves the solutions first and then your entry/permission forms?  Is that okay?[/quote]

I don't work in USAMTS but I believe that they ONLY grade solutions that have entry form. In another word, it will be fine since they won't grade your solution until the entry/permission form is received.

As for mailing the entry form and permission forms and emailing the solutions, this is the method the USAMTS people recommend, especially if you used LaTeX in your solution.

Nope. They grade them without entry forms too. I accidentally forgot to send my entry form last year (round 4) but they still graded my solutions. ;)

[quote="DPatrick"]Right.  We will grade them.  But if you don't send us an entry form, we're not going to send you any prize that you might win![/quote]

Any update on when scores will be posted?

And after that, you still have to put them into the computers, right? How long does that usually take?

Once we find out our scores, can we ask questions about why we were graded a certain way, if we have any?

[quote="DPatrick"]We're waiting to get the graded papers back from the NSA.[/quote]

When do you expect to get them? :D Sorry, I can't wait.

We expect to have the scores up at the end of week if all goes as planned.

Do we need to send in entry and permission forms for each round or just the first one we participate in.

you guys still haven't answered my one question!  Let me rephrase it:

I have mailed the entry and permission forms.  They were sent at October 24.  I have finished my solutions, and I am ready to hand them in.

My question is:

can i send my solutions now via email?

How would I know if you guys got my entry/permission forms?

This is my first year doing USAMTS, and I want to know ASAP!

 :P

[quote="paradox102"]can i send my solutions now via email?[/quote]Yes.  As it says on the USAMTS website:
[quote]You may submit the Entry Form and the Under 13 Permission Form with your first set of solutions. The USAMTS Permission Form may be submitted at any time, and is not required for USAMTS participation. If you are emailing your solutions to us, you can send the Entry and Permission forms separately via snail mail.[/quote]

[quote="paradox102"]How would I know if you guys got my entry/permission forms? [/quote]You can check your "Profile" on the "My USAMTS" page on the USAMTS website.

Snail mail is normal mail.

[url=http://www.google.com/search?q=define%3Asnail+mail&sourceid=mozilla-search&start=0&start=0&ie=utf-8&oe=utf-8&client=firefox-a&rls=org.mozilla:en-US:official]Click here for more info...[/url]

You can, but since they aren't due for another few weeks, I would recommend taking some time to review them to make sure that you didn't make any dumb mistakes and that the solutions are clearly written.

[quote="paradox102"]so can i send in my solutions now?[/quote]

You can send them in now, but you can't re-submit a new set of solutions once you have sent solutions in.