Show that if s and t are odd then:

$ \frac {st \minus{} 1}{2}\equiv\frac {s \minus{} 1}{2} \plus{} \frac {t \minus{} 1}{2}\pmod{2}$

and

$ \frac {(st)^{2} \minus{} 1}{8}\equiv\frac {s^{2} \minus{} 1}{8} \plus{} \frac {t^{2} \minus{} 1}{8}\pmod{2}$

ps. Thanks to tonypr for the Latex help.

[quote="modularmarc101"][quote="tonypr"][quote="chetjan"](The equal signs should be congruences)

Show that if s and t are odd then:

(st-1)/2=(s-1)/2+(t-1)/2     mod 2
and
((st)^2-1)/8=(s^2-1)/8+(t^2-1)/8

ps. If anyone could tell me how write this in Latex that would be great.[/quote]

$ \dfrac{st - 1}{2}\equiv \dfrac{s - 1}{2} + \dfrac{t - 1}{2}\pmod{2}$

and

$ \dfrac{(st)^2 - 1}{8}\equiv \dfrac{s^2 - 1}{8} + \dfrac{t^2 - 1}{8}\pmod{2}$

... to use fractions is "\dfrac{a}{x}" and this gives $ \dfrac{a}{x}$, to use congruences is "\equiv" and that produces $ \equiv$. To put the mod at the end is "\pmod{n}" and this produces $ \pmod{n}$

Solution to the first one,

let $ s = 2k_1 + 1$, and $ t = 2k_2 + 1$, so that,

\begin{eqnarray*} \dfrac{st - 1}{2} & \equiv & \dfrac{s - 1}{2} + \dfrac{t - 1}{2}\pmod{2} \\
\dfrac{(2k_1 + 1)(2k_2 + 1) - 1}{2} & \equiv & k_1 + k_2\pmod{2} \\
\dfrac{4k_1k_2 + 2k_1 + 2k_2}{2} & \equiv & k_1 + k_2\pmod{2} \\
2k_1k_2 + k_1 + k_2 & \equiv & k_1 + k_2\pmod{2} \\
k_1 + k_2 & \equiv & k_1 + k_2\pmod{2} \end{eqnarray*}

Solved.

Edit: Your welcome.[/quote]

That isn't necessary.

Just notice that $ s,t \equiv 1 \mod 2$, and we can substitute.

$ \frac {(1)(1) - 1}{2} \equiv \frac {(1) - 1}{2} + \frac {(1) - 1}{2} \mod 2$

$ 0 \equiv 0 \mod 2$

Same for #2.[/quote]

How are you so sure that if $ k\equiv 0\pmod{2}$ then $ \dfrac{k}{2}\equiv 0\pmod{2}$ also? What if $ k=6$?

Oh, I didn't know that substitution with mods worked that way. Thanks.


Lola had 7 bags. Each bag contained 7 mice. Each mouse had
7 pieces of cheese. How many pieces of cheese were in all of
Lola's bags combined?


real analysis
calculus
integration
function
geometry
rectangle
topology

Let $\Omega$ be a bounded open subset of $\mathbb{R}^{n}$. We will call a function $\phi$ on $\Omega$ a special simple function if there is a finite number of open rectangles $R_{i}$ with real numbers $c_{i}$ such that $\phi = \sum c_{i}\chi_{R_{i}}$. Show that if $f$ is Lebesgue integrable on $\Omega$ then there is a sequence of special simple functions converging in the mean to $f$.

[hide="Discussion for $n=1$"]
I have done this problem before for the one dimensional case and $\Omega = [a,b]$. Then one could choose simple functions non-zero on [b]open[/b] sets approximating $f$. Because an open set is just a countable disjoint union of open intervals one can approximate the characteristic function of an open set with a continuous function, hence approximating $f$ with continuous functions. Then one can approximate these, and hence $f$, with special simple functions (step functions) by the usual Riemann integration technique. However, this argument does not seem to extend well to this more complicated case.
[/hide]

"converging in the mean" means converging with the $L^{1}$ norm ? If it is the case, it suffices to prove it for nice functions (they are dense for the $L^{1}$ norm), wich is easy ?

If $U\subset \mathbb R^{n}$ is an open set with finite measure, then the sequence of continuous functions $\phi_{n}(x)=\max\{1,n\,\mathrm{dist}\,(x,\mathbb R^{n}\setminus U)\}$ converges to $\chi_{U}$ in $L^{1}(U)$.

I didn't know continuous functions were dense in $L^{1}$, and I didn't think of mlok's nice way of showing it. I guess I thought it to be much more difficult than it is, so I didn't try for an explicit construction.


Έστω $a,b,c\geq 0$ με $a+b+c=1$ .Να βρείτε τη μέγιστη τιμή της παράστασης

$\frac{(ab)^n}{1-ab}+\frac{(bc)^n}{1-bc}+\frac{(ca)^n}{1-ca}$

Η ΑΣΚΗΣΗ ΜΕ ΔΥΣΚΟΛΕΨΕ ΠΟΛΥ,ΓΙΑΤΙ ΒΓΑΖΕΙ ΔΙΑΦΟΡΕΤΙΚΟ ΜΕΓΙΣΤΟ ΓΙΑ Ν>1 ΚΑΙ ΔΙΑΦΟΡΕΤΙΚΟ ΓΙΑ Ν=1 ΕΛΠΙΖΩ Ο Ν ΝΑ ΕΙΝΑΙ ΦΥΣΙΚΟΣ ΓΙΑΤΙ ΑΛΛΙΩΣ ………….. :o

ΓΙΑ Ν >1: K=$\frac{(ab)^n}{1-ab}+\frac{(bc)^n}{1-bc}+\frac{(ca)^n}{1-ca}$<=4/3(a^nb^n+b^nc^n+a^nc^n) γιατι α^νβ^ν/(1-αβ)<=4/3(α^νβ^ν)<=>αβ<=1/4 που ισχυει από ΑΜ-ΓΜ,κ.λ.π
Όμως αν mαx(α,β,γ)=α,εχουμε (1/4)^ν>=α^ν(1-α)^ν=α^ν(β+γ)^ν>=α^ν(β^ν+γ^ν+νβ^(ν-1)γ)>= a^nb^n+b^nc^n+a^nc^n αρα max=1/(3*4^(ν-1)) για α=β=1/2 και γ=0 .

Για Ν=1 θ.δ.ο Κ<=3/8 αρκει  λοιπον Κ+3=Σcyc(1/(1-ab))=[3-2(αβ+βγ+γα)+αβγ]/[1-(αβ+βγ+γα)+αβγ-(αβγ)^2]<=27/8 <=>(αβ+βγ+γα)<= (3+19αβγ-27(αβγ)^2)/11
Όμως αβ+βγ+γα<=(1+9αβγ)/4 από την προφανη αβγ>=Πcyc(α+β-γ) για α+β+γ=1
Αρκει λοιπον ν.δ.ο (1+9αβγ)/4<=(3+19αβγ-27(αβγ)^2)/11<=>-1/4<=αβγ<=1/27 που ισχυει απο αβγ>=0 και ΑΜ-GM.
Σορρυ για την Latex!?

ΟΥΦ!!!!! :D

                                 ΚΑΡΑΜΙΧΑΛΗΣ ΡΑΛΛΗΣ

Καλώς ήρθες Ράλλη .Εξαιρετικά  πολύ καλή η λύση  που δίνεις .  Τώρα θα παραθέσω ένα πρόβλημα εξαιρετικά δύσκολο το οποίο παραμένει για μένα άλυτο .Ελπίζω να με βοηθήσετε σας παρακαλώ .

Έστω $a,b,c\geq 0$  με $a^2+b^2+c^2=1$ .Βρείτε τη μέγιστη τιμή της παράστασης

$A=a^n(b+c)+b^n(c+a)+c^n(a+b)$  συναρτήσει του $n$ φυσικά . :D

Το διόρθωσα . c+a  είναι το σωστό .

Η ΑΣΚΗΣΗ ΜΟΥ ΦΑΙΝΕΤΑΙ ΣΧΕΔΟΝ Η ΙΔΙΑ!!!  :o ΙΣΧΥΕΙ:
ΓΙΑ Ν>1:
α^ν(β+γ)+β^ν(α+γ)+γ^ν(α+β)<=2^(1/2)(α^ν+β^ν+γ^ν) και(α^ν+β^ν+γ^ν)<=(1/(2^(1/2))^ν ΟΠΟΤΕ max=(1/(2^(1/2)))^(ν-1) ΓΙΑ α=β=1/(2^(1/2)) ΚΑΙ γ=0     (ΟΙ ΑΠΟΔΕΙΞΕΙΣ ΠΑΡΑΛΕΙΠΟΝΤΑΙ ΓΙΑΤΙ ΕΙΝΑΙ ΠΑΡΟΜΟΙΕΣ ΜΕ ΑΥΤΕΣ ΤΗΣ ΠΡΟΗΓΟΥΜΕΝΗΣ ΑΣΚΗΣΗΣ)
ΓΙΑ Ν=1:
ΜΕ ΠΡΑΞΕΙΣ.....max=2 ΓΙΑ α=β=γ=1/(3^(1/2))

ΜΗΠΩΣ ΟΙ ΑΣΚΗΣΕΙΣ ΑΥΤΟΥ ΤΟΥ ΤΥΠΟΥ ΓΕΝΙΚΕΥΟΝΤΑΙ ? ?
Α, ΚΑΙ ΜΙΑ ΖΟΡΙΚΗ......
ΑΝ α,β,γ ΕΙΝΑΙ ΠΛΕΥΡΕΣ ΤΡΙΓΩΝΟΥ ΜΕ α+β+γ=1 ν.δ.ο
Σcyc(α^ν+β^ν)^(1/ν)<1+(2^(ν/2))/2

ΡΑΛΛΗΣ

Ράλλη η λύση σου είναι απλά καταπληκτική .Πολύ απλή και κατανοητή σε ένα πρόβλημα που είναι άλυτο στο αγγλικό φόρουμ !!!! Θα δω το πρόβλημα που έθεσες.

[quote="RAKAR"]
 (α^ν+β^ν+γ^ν)<=(1/(2^(1/2))^ν [/quote]

Ράλλη νομίζω ότι αυτό εδώ είναι λάθος αφού δεν ισχύει για α=1 , β=γ=0

Θα είναι Σ(α^ν+β^ν)^(1/ν)<Σ(α+β)=2<=1+(2^(ν/2))/2 για ν>=2
Για ν=1 νομίζω πώς υπάρχει πρόβλημα αφού πρέπει 2<1+2^(-1/2) το οποίο δεν ισχύει.

Μπορεί κάποιος να με βοηθήσει με αυτήν εδώ ? Θέλω να βρω μια ωραία λύση γιατί αυτή που έχω με n-1EV και  SMV  είναι very ungly
Έστω $a,b,c\geq 0$  με $a^2+b^2+c^2=1$ .Βρείτε τη μέγιστη τιμή της παράστασης

$A=a^n(b+c)+b^n(c+a)+c^n(a+b)$  συναρτήσει του $n$ φυσικά .

Αυτη που προτεινεις στο τελευταιο ποστ μηπως τη λυνει ποιο πανω ο Ραλλης?

Ναι. Αρχικά και επειδή δε μποερούσα να τη διαβάσω νόμιζα ότι είναι σωστή όμως δεν είναι σωστή η λύση του...... ;) Οπότε περιμένω κάτι καινούργιο. ;)

Εχω βρει κατι αλλα δεν ειμαι σιγουρος.μηπως η λυση ειναι 2?

Όπως είπα και παραπάνω το μέγιστο είναι συναρτήσει του  $n$ . Πχ $\frac{n}{3n^2+1}$ .Δεν είναι τόσο και μάλιστα παίρνει τρεις διαφορετικές τιμές ανάλογα με τον $n$

Λοιπον μηπως αν παρεις την xy<=(x^2+y^2)/2 kai θεσεις χ= η n-οστη του α και y=b+c και κυκλικα .παιρνεις οτι   (b+c)^2<=2(b^2+c^2) kai oti a^2n+b^2n+c^2n<=a^2+b^2+c^2=1 briskeiw κατι. ειναι  λαθος? παρακαλω κοιταξτε το :!:

Λοιπόν παραθέτω τα αποτελέσματα σε μια εντελώς παρόμοια απλά με διαφορετική συνθήκη .Λοιπόν
Έστω $n,a,b,c>0$ με   $a+b+c=1$.
$F_{n}(a,b,c)=a^{n}(b+c)+b^{n}(c+a)+c^{n}(a+b)$
Τότε έχουμε
a)με $n \le 3$   $maxF_{n}(a,b,c)=max(\frac{2}{3^{n}},\ \frac{1}{2^{n}})$

b)με  $n >3$ τότε  $max F_{n}(a,b,c)=\frac{t^{n}+t}{(t+1)^{n+1}}$ με το  $t$ να είναι η ρίζα της εξίσωσης  $t^{n}-n\cdot t^{n-1}+n\cdot t-1=0$ που βρίσκεται στο $(0,1)$.

Πώς σας φαίνεται?? ;)

Σιλουανε οταν ειπα οτι η μεγιστη τιμη ειναι 2 εννοουσα τη την μεγιστη τιμη που μπορει να παρει γενικα(Θετωντας συγκεκριμενη τιμη για το n και τα a,b,c) αν δεν κανω λαθος πραγματι για ν=1 και a=b=c pairnei την τιμη 2.εσυ ζητας για καθε n να βρω μια καλυτερη προσεγγιση για το μεγιστο.λοιπον ειναι λιγο δυσκολο .μπορεις ισως να βρεις ενα τυπο με δοκιμες n=1,2,.. και μετα επαγωγικα.παντως αν εχεις λυση αν μπορεις πες την ή τουλαχιστον τη βασικη ιδεα :)

Δες το ποστ #16 .Τα αποτεσμάτα είναι παρόμοια .Όπως προανέφερα και παραπάνω έχω δύο διαφορετικες λύσεις αλλά είναι αρκετά μεγάλες .Η μία χρησιμοποιεί
την αρχή $(n-1)Equal\ Variable$  και η άλλη χρησιμοποιεί $SMV$ . ;)

Τι ειναι οι n-1 Equal Variable kai SMV?  :?:


[quote="neelnal"]How to design a website?
Is it hard? Does it require extensive knowledge?
Does it cost money?(Time is not money in this case)[/quote]

there are several websites that let you make a website for free but under their domain. if you want your own domain, then yeah, you need to pay money, don't know how much though

Come on. Just use some ASP, or PHP, and create an awesome website. Awardspace is a good space. Tripod and biz.ly have good templates.


[quote="mathnovice"]I don't understand how you can "compete" in biology. Who cares if you have memorized more than anyone else in the world? Whereas doing well in math competitions is a strong indicator of your ability in math, doing well in biology competitions doesn't necessarily imply that you will be discovering the cure for cancer.[/quote]

Testing your dissection skills with frogs and other animal?  :D

[quote="takuanitromars36"]How do I get into the USABO? What steps are there?[/quote]
Have you seen this site? [url=http://www.cee.org/usabo/index.shtml]CEE: USA Biology Olympiad[/url]  I think it should cover most of your questions.
[quote="takuanitromars36"]In addition, is there a good book equivalent in excellence to the AoPS book, but for biology instead?[/quote]
[url=http://www.cee.org/usabo/usabo_preparing.shtml]CEE: Preparing for Competition[/url]

It seems that the CEE recommends Campbell's, which is a decent book, but I wouldn't exactly call it equivalent in excellence to the AoPS books.

[quote="takuanitromars36"]True, but it always looks good for college. Besides, biology, as well as math, is fun fun fun! It's also an appropriate olympaid, since I want to work in medicine when I grow up.  :cool:[/quote]

I think biology research would be fun but just memorizing facts is boring.

[quote="mathnovice"]I think biology research would be fun but just memorizing facts is boring.[/quote]
Where do you draw the distinction? Research exists to find new facts.

The distinction is obvious: would you rather read about it or do it?

Put it another way, would you rather read about how to ride a bike or actually ride a bike? Just because memorizing facts about how to ride a bike  [i]may[/i] help you to ride a bike doesn't mean biking is all about memorizing facts about how to ride a bike. That is the distinction.

usabo? hmm.. at first i thought it was a typo

[quote="joml88"]IMO the other olympiads are not nearly as competitive as the math is.  I don't know of anything like AoPS for the other olympiads.  I think this link might help [url]http://www.cee.org/usabo/index.shtml[/url][/quote]

So untrue. The USACO has a big set of training pages for people to work on, and they are sufficient for getting into the camp. (Sorry its not biology, but I thought I'd bring it up.)

http://ace.delos.com/usacogate .

Where can I get practice exams for olympiads like biology and chemistry?


Each edge of a cube is colored either red or black.Every face of the cube has at least one black edge. The smallest possible number of black edges is

$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

[hide]The maximum number of faces that one edge van border with is $2$. Therefore, the answer should be $6/2=3$.

A quick drawing confirms this[/hide]

[hide]If you think about two faces, it can have one edge black, so since a cube has 6 faces, the minimum is $3$.[/hide]


An envelope contains eight bills: $ 2$ ones, $ 2$ fives, $ 2$ tens, and $ 2$ twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $ \$ 20$ or more?

$ \textbf{(A)}\ \frac {1}{4}\qquad
\textbf{(B)}\ \frac {2}{7}\qquad
\textbf{(C)}\ \frac {3}{7}\qquad
\textbf{(D)}\ \frac {1}{2}\qquad
\textbf{(E)}\ \frac {2}{3}$

[hide="Answer"]$\frac{1}{4}$ chance of a $1$, the next bill must be a $20$ to be greater than $20$ - $\frac{1}{4}\cdot \frac{2}{7}=\frac{1}{14}$
$\frac{1}{4}$ chance of a $5$, the next bill must be a $20$ to be greater than $20$ - $\frac{1}{4}\cdot \frac{2}{7}=\frac{1}{14}$
$\frac{1}{4}$ chance of a $10$, the next bill is either $10$ or $20$ to be greater than or equal to $20$ - $\frac{1}{4}\cdot \frac{3}{7}=\frac{3}{28}$
$\frac{1}{4}$ chance of a $20$, next bill can be anything - $\frac{1}{4}\cdot 1=\frac{1}{4}$
Total: $\frac{1}{14}+\frac{1}{14}+\frac{3}{28}+\frac{1}{4}=\frac{1}{2}\Rightarrow \boxed{D}$[/hide]

[quote="shake9991"]sorry for reviving, but there's an much easier way:

write out all of the possible sums:
2,6,10,11, 15, 20,21,25,30,40

clearly 1/2[/quote]

you got lucky...

there is a better chance to get a sum of 6 than something like a sum of 2..

it just happens that your way worked but it wont work in some other case, like say a sum of 21 dollars and greater

your way by listing, it would be 4/10=2/5.

However, the real answer would be 13/28

In total we can draw any 2 bills in 8*7/2=28 ways.If we to take bills from 2's, 5's and 10's we can get  20 dollars in  the only way,  otherwise we have less money. So,  6*5/2 - 1 = 14 ways to not receive 20 dollars. So, the answer is 14/28=0.5.

[hide="My Solution"]
Suppose you have drawn a $ \$20$. Clearly, no matter what the second bill is, you will obtain a sum greater than $ \$20$. The probably of you drawing a $ \$20$ is $ \frac {2}{8} \equal{} \frac {1}{4}.$

Suppose that you draw a $ \$20$ after drawing a non-$ \$20$. The probability of that occurring is $ \frac {6}{8}\cdot \frac {2}{7} \equal{} \frac {3}{4}\cdot \frac {2}{7} \equal{} \frac {6}{28}$.

The only other way to sum greater than $ \$20$ is to draw two $ \$10$'s. This is done with probability $ \frac {2}{8}\cdot \frac {1}{7} \equal{} \frac {1}{28}$.

Summing the probabilities, we obtain $ \boxed{\frac {1}{2}\text{ (A)}}$.
[/hide]

EDIT: Fixed Solution. Thanks, Ihatepie.

Not really, it seems more like luck. The probability of choosing two 10s is 1/28. Then, the probability of choosing a 20 and another one is 2*2/8*7/7-1/56*2.

Thanks. I just remembered that order matters because you are not replacing and therefore the cases I used don't encompass everything.

sorry for the revive, but TachyonPulse, $ \frac{1}{2}$ is actually the letter choice (D)

 :D


geometry
3D geometry
tetrahedron
summer program
Mathcamp
MIT
college

Have you guys gotten your new SAT scores yet? I got mines today, and all I can say is that I'm extremely disappointed.

Math-760
Critical Reading- 650
Writing-670

Total- 2080, which is 300 points lower than my old SAT I and my SAT II Writing combined.  :(

[quote="tetrahedr0n"]740 V 800 M 800 W (80/12)

So I'm done with this test.[/quote]

Wow, very impressive! Any tips on the essay? You seem to have it down.

if you people with 2000+ were disappointed, don't be.  I got:
740 M / 510 V / 610 W /

And i am extremely upset.

Chris.

jennifer lin from mc2004?
I'm not surprised.

Good job!!
How are you doing?

Wow, do I feel stupid or what?  :(

A lot of my friends say that if you write 5 paragraphs, then you get an 8/12 automatically. I'm not sure if the "automatically" is correct, but is there a better chance of getting a higher essay score if you write more? Considering that the essay is 20 minutes long, plus 5 minutes of outlining, that leaves you with about 3 minutes each paragraph if you're writing 5 paragraphs. Sigh, I got a 6/12.

Hi Geehoon :-) It's been a long time since MC...'ll PM you sometime.

About the essay -- mine was two and a half paragraphs and 1.5 pages long, and ended abruptly in mid-sentence. I recieved a score of 11.
I think the focus is mostly on organization of thought, development of a thesis... and I believe the graders like historical and literary examples [according to the review book, preferably at least one of each.] I'm not sure how much length is factored in.

Would anybody with 10 or above in the essay, like tetrahedron, mind letting me read your essay once it is posted on college board? I though I did well on the essay but I only got an 8. Like what 314159265358979323846 said, I put some historical facts and my essay was neat and organized but again I got an 8. And how did you guys study for the new test? How much do you guys think one will able to improve his score over a year?

Depends how hard you try. I only have 1 chance to improve my score, since May registration is over, and I'm taking SATII's in June, so I can only take it in October before I apply to college.

As long as you prepare a lot by taking practice exams, nothing is impossible.

[quote="anhnguyen"]Would anybody with 10 or above in the essay, like tetrahedron, mind letting me read your essay once it is posted on college board?[/quote]

Sure. My goal in writing my essay was to put a lot of varied arguments in it. As I was on the East Coast, I had the majority essay. Basically I had three paragraphs - one on the harm caused by dictatorial governments, one on Rand's Atlas Shrugged, and one on team basketball.

[quote="anhnguyen"]And how did you guys study for the new test? [/quote]

Not at all, and I imagine a lot of AoPSers didn't either considering it was right after the AIME.

Congrats, Jennifer! Hi Geehoon. This is Fred from MC '04.

Thanks a lot. I got 2000. I thought I could have done better. Extremely upset because of 740 M. English is not my first language so I don't blame myself that much. But still two more years to go  :) .

Congratulations Jennifer You are one of 107

http://wireservice.wired.com/wired/story.asp?section=Breaking&storyId=1017963&tw=wn_wire_story

And congrats everyone else, you all did great!

I got a perfect too  :lol: 12 on the essay  !! Very surprised and very happy and also VERY fortunate as well! BTW this is John Z. from mathcamp; good job jennifer and fred; watsup to geehoon!

Wow. How hard exactly is it to do well on the essay? Are you people very experienced writers and have you taken special preparation for the essay? Or did you just "wing it"? ;)

Its not graded very hard I think.

Congrats John. That makes 2 2400s I know of from MI. Cool!

[quote="juicybooty911"]any advice on how you can get a 800 on math?[/quote]

I did not, but I was told this by someone who did:

I assume that you are proficient in math, and that you have 10 minutes to spare before finishing math part. For each question:

1. Read question again.
2. Solve problem w/o looking at work.

Repeat twice.

For those here, the test is really not a math test, but a test in the ability to read a problem correctly.

My scores: 760 M  :( / 680 CR :( / 660 Wr  :(  :( . I dont mean to patronize others, but I expected better of myself, seeing as on the old SAT i got 770 M / 710 V, which I thought was okay. Needless to say, I will retake in June.

You got 20 points higher than me.  :(  Like you, I expected higher scores, since my old SAT scores were pretty good (770M, 770V) and my SAT II Writing was 800. But again, maybe it was just a bad day for me. I'm taking it again in October.  :(  What a waste of 4-5 hours.

800 M, 800 V, 760 W

I'm worried about the SAT2s though.. Chem >_<
How did people who took the May SAT do? Was the curve harder?

Math- 790
Critical Reading- 600
Writing- 670

I was among the March guinea pigs for the new SAT
800M/770V/740W (10 essay) had to stop the essay mid-sentence in my concluding paragraph, it was about 1.5 pgs

will be taking SAT II Math IIC, Physics, and US history october, and retaking SAT november (i really should be able to do better w/ grammar in the writing section). cutting it really close for MIT early action  :?

[quote="L_Li"]will be taking SAT II Math IIC, Physics, and US history october, and retaking SAT november (i really should be able to do better w/ grammar in the writing section). cutting it really close for MIT early action  :?[/quote]
I wouldnt do that if I were you.

I would take IIC and US history in October, Physics and US history in November(iff you did not like your October score), assuming math and physics are your strong subject (most likely to be true for most of AoPSers)

It is not worty to retake 2310 in my opinion, unless youre going for 2400 for scholarship or something and are positive that youre going to get it.

yeah i agree with gee hoon.  for admissions purposes 2310 is fine.  even for scholarships it should be fine, the exception being a presidential scholar, but i dont know what the standards are for that now, 2400?

770 M, 700 V/W. A kid at my school got 2330 and a 36 on the ACT. On the SAT II I got 800 math, (can barely remember these next two), about 760 chemistry (never want to spend another second on chemistry in my life), about 730 u.s. history.

[quote="AntonioMainenti"]about 760 chemistry (never want to spend another second on chemistry in my life)[/quote]that's how I felt after AP Chem test, but it is mendatory to take at least one class of chemistry for MIT students. :D


Source: John O'Bryan Math Competition 1997

A rectangle is divided into four smaller rectangles as shown.  The areas of three of the rectangles are given.  What is the area of the large rectangle?

Let the dimensions of the upper right rectangle be x by y, where x is the vertical side and y is the horizontal side. The lower right rectangle has dimensions 5/4x by y, because its area is 5/4 times the upper right rectangle, and they share a common base. Similarily the upper left rectangle has dimensions x by 3/8 y. So the total area is (x+5/4x)(y+3/8y)=99/32*xy. Since xy=8 (area of the upper right rectangle), the area of the big rectangle is 99/4.

I got

[hide]I got 15/4 its difficult to explain without a picture just label all the sides with variables and use ratios to describe their relationships.[/hide]

I got
[hide]$\frac {15}{4}$
If the width is split into $a$ and $b$ then $a = \frac 45 b$. If the length is $c$ and $d$, all we are looking for is $ad$. Since $\frac 45ad = 3$, then $ad = \frac {15}{4}$. Or you could just make the ratio $\frac {10}{8} = \frac {x}{3}$ and solve out for $x$.[/hide]

The answer is $\frac{99}{4}$.  The question asked for the area of the whole rectangle, not the one without an area given, but for that you two were correct.  The test could have put $\frac{15}{4}$ as a possible solution to trip people up, but it didn't.

I agree, these tests are worded weird very often.
Another question on this test has two points A and B on opposite bases of a cylinder, and AB intersects the axis of the cylinder.  It asks for the smallest distance between them but doesn't specify if it has to be along the surface or can go thru, but has the answers for both of them as choices...

Then the answer is probably whichever of the two is smallest.


It is asked to determine the type of the isolated singularity for:
$ f(z)\equal{}\frac{1\minus{}e^{z}}{1\plus{}e^{z}}$

We can simply say that the numerator is holomorphic in the points where the denominator has zeros, that is in the pints $ z_k\equal{}(2k\plus{}1)\pi{}i$. Then the function $ f(z)$ has the same singularity type as $ g(z)\equal{}\frac{1}{1\plus{}e^z}$. But the order of zero of the denominator of $ g(z)$ is 1, which can be seen form its power series expansion. This means that $ g(z)$ has a pole of order 1 in the relevant points. Is it correct?

But now I would like to use Riemann's theorem on removable singularities directly to arrive at the same conclusion. So I compute the limit:

$ \lim_{z\to{z_k}}(1\minus{}e^z)\frac{z\minus{}(2k\plus{}1)\pi{}i}{1\plus{}e^z}\equal{}\lim_{z\to{z_k}}(1\minus{}e^{z})(\frac{z\minus{}2k\pi{}i}{1\plus{}e^{z}}\minus{}\frac{\pi{}i}{1\plus{}e^{z}})\equal{}2\lim_{z\to{z_k}}(\frac{z\minus{}2k\pi{}i}{1\plus{}e^{z\minus{}2k\pi{}i\minus{}\pi{}i}}\minus{}\frac{\pi{}i}{1\plus{}e^{z\minus{}2k\pi{}i\minus{}\pi{}i}})\equal{}2\lim_{z\to{0}}(\frac{z}{1\plus{}e^{z\minus{}\pi{}i}}\minus{}\frac{\pi{}i}{1\plus{}e^{z\minus{}\pi{}i}})\equal{}2(\pi{}i\lim_{z\to{0}}\frac{1}{e^z\minus{}1}\minus{}\lim_{z\to{0}}\frac{z}{e^z\minus{}1})\equal{}2(\pi{}i\lim_{z\to{0}}\frac{1}{e^z\minus{}1}\minus{}1)$
since $ \lim_{z\to{0}}\frac{e^z\minus{}1}{z}\equal{}1$

Is there anything wrong I've done, because there is still one limit left which is not finite, isn't it? But I need something finite! Where is the mistake?

Oh my I see the mistake! It's time to have a rest!  :blush:


Q1. Find the sum to n terms of the series 1 + (1+2) + (1+2+3) + ..........

Q2. Find the sum of the series $ \frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}\dots\dots\ upto\ 16\ terms.$

Q3. On the ground n stones are placed.The distance between the first and the second is one metre, between second and third three metres, between third and fourth five metres and so on. How far will a person have to travel to bring them one by one to the basket placed at the first stone if he starts from the basket.

Q4. Find $ \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\dots\dots\ upto\ n\ terms.$

Q5. Find the sum of following series to infinity: $ \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}\dots\dots\infty$

P.S:- Here 1.2 means $ 1\times 2$

[hide="#1"]OK #1

1) Its straight forward. The $ r$-th term of the series is the sum of the first  natural numbers:

Hence, the sum upto the first terms of the series becomes:

$ S_n \equal{} \sum t_r \equal{} {\displaystyle \sum_{r \equal{} 1}^n \dfrac{r(r \plus{} 1)}{2}}$

Accordingly, we obtain

$ S_n\equal{}\frac{n(n\plus{}1)(n\plus{}2)}{6}$[/hide]

Q1: I think that's obvious:

$ S \equal{} \frac {{1.2}}{2} \plus{} \frac {{2.3}}{2} \plus{} \frac {{3.4}}{2} \plus{} ........ \plus{} \frac {{n(n \plus{} 1)}}{2} \equal{} \frac {{\left( {\sum\limits_{i \equal{} 1}^n i } \right) \plus{} \left( {\sum\limits_{i \equal{} 1}^n {i^2 } } \right)}}{2}$

$ \equal{} \frac {{\frac {{n(n \plus{} 1)}}{2} \plus{} \frac {{n(n \plus{} 1)(2n \plus{} 1)}}{6}}}{2}$

Q2: I think we use: $ 1^3 \plus{} 2^3 \plus{} 3^3 \plus{} ........n^3 \equal{} \left[ {\frac {{n(n \plus{} 1)}}{2}} \right]^2$.

Are you agree?

Q5 is difficult, I cannot find solution! :(

[hide]L_Euler is right about the formula, another formula to note is that the sum of the first n odd integers is $ n^2$

so each term reduces to $ \frac {(n \plus{} 1)^2}{4}$
and the sequence becomes $ \frac44 \plus{} \frac94 \plus{} \frac {16}{4} \plus{} \ldots \plus{} \frac {289}{4}$
or the sum of the first 17 squares, minus 1, all over 4

we use the identity $ 1^2 \plus{} 2^2 \plus{} 3^2 \ldots \plus{} n^2 \equal{} \frac {(n)(n \plus{} 1)(2n \plus{} 1)}{6}$ and we solve the problem $ \frac {(17)(18)(35)}{6} \equal{} 1785, 1785 \minus{} 1 \equal{} 1784, \frac {1784}{4} \equal{} 446$[/hide]
[hide]we again use the fact :the sum of the first n odd integers is $ n^2$
the distance from the basket to the kth stone is $ (k \minus{} 1)^2$, but to retrieve the nth stone,you must walk twice this distance (to and back)
we want the sum $ 2(0 \plus{} 1 \plus{} 4 \plus{} 9 \plus{} 16 \plus{} \ldots \plus{} (n \minus{} 1)^2)$
but the first term cancels out, and we again use the identity
$ 1^2 \plus{} 2^2 \plus{} 3^2 \ldots \plus{} n^2 \equal{} \frac {(n)(n \plus{} 1)(2n \plus{} 1)}{6}$
to find the distance to be
$ 2 \times \frac {(n \minus{} 1)(n)(2n \minus{} 1)}{6} \equal{} \boxed{\frac {(n \minus{} 1)(n)(2n \minus{} 1)}{3}}$[/hide]

[hide]rewrite the kth term in #4 as $ \frac {1}{k} \minus{} \frac {1}{k \plus{} 1}$
we now have the sum:
$ (\frac11 \minus{} \frac12) \plus{} (\frac12 \minus{} \frac13) \plus{} (\frac13 \minus{} \frac14) \ldots \plus{} (\frac {1}{n} \minus{} \frac {1}{n \plus{} 1})$

but many terms cancel leaving us with $ \frac11 \minus{} \frac {1}{n \plus{} 1} \equal{} \frac {n}{n \plus{} 1}$[/hide]

noone's perfect eh?

working on #5...
its probably an uber telescoper

my calculator tells me its $ \frac14$

[hide]time to verify
rewrite the kth term as $ \frac {1}{2k(k \plus{} 1)} \minus{} \frac {1}{2(k \plus{} 1)(k \plus{} 2)}$
this is valid since $ \frac {1}{2k(k \plus{} 1)} \minus{} \frac {1}{2(k \plus{} 1)(k \plus{} 2)} \equal{} \frac {k \plus{} 2}{2k(k \plus{} 1)(k \plus{} 2)} \minus{} \frac {k}{2k(k \plus{} 1)(k \plus{} 2)} \equal{} \frac {2}{2k(k \plus{} 1)(k \plus{} 2)} \equal{} \frac {1}{k(k \plus{} 1)(k \plus{} 2)}$

now we have our telescope
our sequence is now $ (\frac14 \minus{} \frac {1}{12}) \plus{} (\frac {1}{12} \minus{} \frac {1}{24}) \plus{} (\frac {1}{24} \minus{} \frac {1}{40}) \ldots \plus{} (\frac {1}{2n(n \plus{} 1)} \minus{} \frac {1}{2(n \plus{} 1)(n \plus{} 2)}) \ldots$
and the sequence continues forever so every term cancels out except $ \frac14$, which is our answer[/hide]

Will edit later:

[hide="Solution to #1"]1. Sum of triangles... answer is $ \binom{n + 2}{3} = boxed{\frac {n(n + 1)(n + 2)}{6}}$ by hockey-stick identity.
Alternately, $ \sum_{k = 1}^n{\frac {k(k + 1)}{2}} =$

$ \frac {1}{2}\sum_{k = 1}^n{(k^2 + k)} =$

${ \frac {1}{2}(\sum_{k = 1}^n{k^2}} + \sum_{k = 1}^n{k}) =$

$ \frac {1}{2}(\frac {(n(n + 1)(2n + 1)}{6} + \frac {n(n + 1)}{2}) =$

$ \boxed{\frac {n(n + 1)(n + 2)}{6}}$
[/hide]

[hide="Solution to #2"]
$ \frac {1^3}{1} + \frac {1^3 + 2^3}{1 + 3} + \frac {1^3 + 2^3 + 3^3}{1 + 3 + 5}\dots =$

$ \sum_{k = 1}^{16}{\frac {(\frac {k(k + 1)}{2})^2}{k^2}} =$

$ \sum_{k = 1}^{16}{\frac {(k + 1)^2}{4}} =$

$ \frac {1}{4}\sum_{k = 1}^{16}{(k + 1)^2} =$

At this point, it is much faster to sum up to $ 17^2$ and then subtract 1.

$ \frac {1}{4} (\frac {(17)(18)(35)}{6} - 1) =$

The answer is $ \boxed{446}$.
[/hide]
[hide="Solution to #3"]
Oh no, a word problem.

So basically, $ 2(1 + (1 + 3) + (1 + 3 + 5) + \dots + (1 + 3 + 5 + \dots + (2n - 1)))$.

That equals $ 2(1^2 + 2^2 + 3^2 + \dots + (n - 1)^2)$, written in sigma notation as $ 2\sum_{k = 1}^{n - 1}{k^2}$.

Using the sum of squares formula, this equals $ \frac {n(n - 1)(2n - 1)}{3}.$
[/hide]

[hide="Solution to #4"]
We have a telescoper. Using the identity $ \frac {1}{n} - \frac {1}{n + 1} = \frac {1}{n(n + 1)}$, we have
$ \frac {1}{1} - \frac {1}{2} + \frac {1}{2} - \frac {1}{3} + \frac {1}{3} - \frac {1}{4} + \frac {1}{4} - \dots - \frac {1}{n + 1}$.

The corresponding terms cancel very neatly.

$ 1 - \frac {1}{n + 1} = \boxed{\frac {n}{n + 1}}$.
[/hide]

[hide="UBER COOL BEAST #5 TELESCOPES (lyk yey I actually did it)"]
This problem was fun  :) .

We know $ \frac {1}{n(n + 1)} = \frac {1}{n} - \frac {1}{n + 1}$. We also know that, substituting $ n + 1$ for $ n$,
$ \frac {1}{(n + 1)(n + 2)} = \frac {1}{n + 1} - \frac {1}{n + 2}$.

Now we add the two equations.

$ \frac {1}{n(n + 1)} + \frac {1}{(n + 1)(n + 2)} = \frac {1}{n} - \frac {1}{n + 1} + \frac {1}{n + 1} - \frac {1}{n + 2}$

We have a nice cancellation on the right hand side. We simplify the left hand side.

$ 2(\frac {n + 1}{n(n + 1)(n + 2)}) = \frac {1}{n} - \frac {1}{n + 2}$

$ 2(\frac {n}{n(n + 1)(n + 2)} + \frac {1}{n(n + 1)(n + 2)}) = \frac {1}{n} - \frac {1}{n + 2}$

$ 2(\frac {1}{n + 1} - \frac {1}{n + 2} + \frac {1}{n(n + 1)(n + 2)}) = \frac {1}{n} - \frac {1}{n + 2}$

$ \frac {2}{n(n + 1)(n + 2)} = \frac {1}{n} - \frac {1}{n + 2} - \frac {2}{n + 1} + \frac {2}{n + 2}$

Now we cancel.

$ (\frac {1}{1} - \frac {1}{3} - \frac {2}{2} + \frac {2}{3}) + (\frac {1}{2} - \frac {1}{4} - \frac {2}{3} + \frac {2}{4}) + (\frac {1}{3} - \frac {1}{5} - \frac {2}{4} + \frac {2}{5}) + (\frac {1}{4} - \frac {1}{6} - \frac {2}{5} + \frac {2}{6})\cdots$

After some very exhaustive canceling, we get $ \sum_{k = 1}^{\infty}{\frac {2}{k(k + 1)(k + 2)}} = \frac {1}{1} - \frac {2}{2} + \frac {1}{2}$.

Almost there!

$ \sum_{k = 1}^{\infty}{\frac {2}{k(k + 1)(k + 2)}} = \frac {1}{2}$

Finally, divide both sides by 2.

$ \sum_{k = 1}^{\infty}{\frac {1}{k(k + 1)(k + 2)}} = \boxed{\frac {1}{4}}$.

 :)  :)  :)

Yey!
[/hide]

for #4, your current solution only goes up to n-1 terms.
add the last term and you get the same answer as mine

i suppose most of these questions were too simple for IITJEE that everyone from India works at.

We have got questions like $ \frac{1}{1.3.5.7.9}\plus{}\frac{1}{3.5.7.9}$ and so on in place of these questions

[quote="stevenmeow"]it seems a couple of people are unsure of how to make the dot be in the middle
use \cdots to do this
$ 1 \cdot 2$[/quote]

Sorry but it takes a lot of time to do that.

Not really, especially compared to matrices.

[quote="Textangle"][hide="UBER COOL BEAST #5 TELESCOPES (lyk yey I actually did it)"]
This problem was fun  :) .

We know $ \frac {1}{n(n \plus{} 1)} \equal{} \frac {1}{n} \minus{} \frac {1}{n \plus{} 1}$. We also know that, substituting $ n \plus{} 1$ for $ n$,
$ \frac {1}{(n \plus{} 1)(n \plus{} 2)} \equal{} \frac {1}{n \plus{} 1} \minus{} \frac {1}{n \plus{} 2}$.

Now we add the two equations.

$ \frac {1}{n(n \plus{} 1)} \plus{} \frac {1}{(n \plus{} 1)(n \plus{} 2)} \equal{} \frac {1}{n} \minus{} \frac {1}{n \plus{} 1} \plus{} \frac {1}{n \plus{} 1} \minus{} \frac {1}{n \plus{} 2}$

We have a nice cancellation on the right hand side. We simplify the left hand side.

$ 2(\frac {n \plus{} 1}{n(n \plus{} 1)(n \plus{} 2)}) \equal{} \frac {1}{n} \minus{} \frac {1}{n \plus{} 2}$

$ 2(\frac {n}{n(n \plus{} 1)(n \plus{} 2)} \plus{} \frac {1}{n(n \plus{} 1)(n \plus{} 2)}) \equal{} \frac {1}{n} \minus{} \frac {1}{n \plus{} 2}$

$ 2(\frac {1}{n \plus{} 1} \minus{} \frac {1}{n \plus{} 2} \plus{} \frac {1}{n(n \plus{} 1)(n \plus{} 2)}) \equal{} \frac {1}{n} \minus{} \frac {1}{n \plus{} 2}$

$ \frac {2}{n(n \plus{} 1)(n \plus{} 2)} \equal{} \frac {1}{n} \minus{} \frac {1}{n \plus{} 2} \minus{} \frac {2}{n \plus{} 1} \plus{} \frac {2}{n \plus{} 2}$

Now we cancel.

$ (\frac {1}{1} \minus{} \frac {1}{3} \minus{} \frac {2}{2} \plus{} \frac {2}{3}) \plus{} (\frac {1}{2} \minus{} \frac {1}{4} \minus{} \frac {2}{3} \plus{} \frac {2}{4}) \plus{} (\frac {1}{3} \minus{} \frac {1}{5} \minus{} \frac {2}{4} \plus{} \frac {2}{5}) \plus{} (\frac {1}{4} \minus{} \frac {1}{6} \minus{} \frac {2}{5} \plus{} \frac {2}{6})\cdots$

After some very exhaustive canceling, we get $ \sum_{k \equal{} 1}^{\infty}{\frac {2}{k(k \plus{} 1)(k \plus{} 2)}} \equal{} \frac {1}{1} \minus{} \frac {2}{2} \plus{} \frac {1}{2}$.

Almost there!

$ \sum_{k \equal{} 1}^{\infty}{\frac {2}{k(k \plus{} 1)(k \plus{} 2)}} \equal{} \frac {1}{2}$

Finally, divide both sides by 2.

$ \sum_{k \equal{} 1}^{\infty}{\frac {1}{k(k \plus{} 1)(k \plus{} 2)}} \equal{} \boxed{\frac {1}{4}}$.

 :)  :)  :)

Yey!
[/hide][/quote]
[hide="Much easier way"]We can reach $ \frac {2}{n(n \plus{} 1)(n \plus{} 2)} \equal{} \frac {1}{n} \minus{} \frac {1}{n \plus{} 2} \minus{} \frac {2}{n \plus{} 1} \plus{} \frac {2}{n \plus{} 2}$ much more quickly. We can do a [url=http://en.wikipedia.org/wiki/Partial_fraction_decomposition_over_the_reals]partial fraction decomposition[/url] as follows: let $ A,B,C$ be such that:

$ \frac{1}{n(n\plus{}1)(n\plus{}2)} \equal{} \frac{A}{n} \plus{} \frac{B}{n\plus{}1} \plus{} \frac{C}{n\plus{}2}$

Clearing denominators yields

$ 1 \equal{} A(n\plus{}1)(n\plus{}2) \plus{} Bn(n\plus{}2) \plus{} Cn(n\plus{}1)$

Now let $ n\equal{}0$. Then $ 1 \equal{} 2A \implies A \equal{} 1/2$ (I know I can't really divide by $ 0$, so if you really want to be picky we can get $ n$ arbitrarily close to $ 0$). Let $ n\equal{}\minus{}1$. Then $ B\equal{}\minus{}1$. Let $ n\equal{}\minus{}2$. Then $ C\equal{}1/2$.

So we get

$ \frac{1}{n(n\plus{}1)(n\plus{}2)} \equal{} \frac{1/2}{n} \plus{} \frac{\minus{}1}{n\plus{}1} \plus{} \frac{1/2}{n\plus{}2}$

This telescopes nicely (as you mentioned). See if you can generalize this (I haven't tried yet, but I'm sure it's possible).[/hide]

Oh, I just remembered.

Stevenmeow, \cdots gives you 3 of them. Use \cdot to get just 1.


induction
number theory proposed
number theory

Is there any validity of the statement that

$ 5F_{2n \minus{} 1}^2 \plus{} ( \minus{} 1)^n4$ is always a perfect square where $ n \geq 0$.
Assume $ F_{1} \equal{} 1$

Yes.

(Phrase your question better next time :P)

all the answers to $ x^2 \minus{} 5y^2 \equal{} \minus{} 4$ are obtained by...see PEN L8( http://www.mathlinks.ro/viewtopic.php?t=150786 ).
u can find that $ y_0 \equal{} 1,y_1 \equal{} 2,y_{n \plus{} 2} \equal{} 3y_{n \plus{} 1} \minus{} y_n$.
u can easily see that (by induction) $ y_n \equal{} F_{2n \plus{} 1}$
for example $ y_0 \equal{} F_1 \equal{} 1,y_1 \equal{} F_3 \equal{} 2,y_2 \equal{} F_5 \equal{} 5,...$.
so we have proven that $ 5F_{2n \minus{} 1}^2 \minus{} 4 \equal{} x^2$ for some integer $ x$.
which proves that your problem seems wrong for even $ n$.
i mean we must have both $ 5F_{4n \minus{} 1}^2 \plus{} 4$ and $ 5F_{4n \minus{} 1}^2 \minus{} 4$ be perfect squares.
so let's denote them by $ x^2$ and $ y^2$.
then u'll have $ x^2 \minus{} y^2 \equal{} 8$.
by this u can easily find what possible values can take $ x,y$.
am i missing sth? :oops:
for example $ F_7 \equal{} 13 \implies 5.13^2 \plus{} 4 \equal{} 849$ which is not a perfect square but $ 5.13^2 \minus{} 4 \equal{} 29^2$.

I never said that both of them must simultaneously be a square. Any one of them would suffice and in this case you can see that whether 4 is added or subtracted depends on n.

And where did you get $ 5F_{4n \minus{} 1}^2 \plus{} 1 and 5F_{4n \minus{} 1}^2 \minus{} 1 ?$

[quote="A K Bhattacharyya"]
And where did you get $ 5F_{4n \minus{} 1}^2 \plus{} 1 and 5F_{4n \minus{} 1}^2 \minus{} 1 ?$[/quote]
dear A K Bhattacharyya,
i don't see in my proof any of these terms: $ 5F_{4n \minus{} 1}^2 \plus{} 1$ and $ 5F_{4n \minus{} 1}^2 \minus{} 1$.
also u said that "it depends on $ n$" but i have proven that it 's not the fact since for all $ n  \in N$ we have $ 5.F_{2n\minus{}1}^2\minus{}4\equal{}x^2$.
and only for some small $ n$ we have $ 5.F_{2n\minus{}1}^2\plus{}4\equal{}x^2$.
plz tell me if i'm wrong
(sorry for my poor english :blush: )

[quote="manosij"]actually what i originally found was 5F<n>^2 + 4(-1)^n is a perfect square....
 :huh: [/i][/quote]
that's true,
we have proven it for the case $ n\equal{}2k\minus{}1$ and now it rests to prove it for even $ n$.
we can easily find all the answers(positive ones)  to the equation $ a^2\plus{}b^2\minus{}1\equal{}3ab$.
just see that by using the well-know idea we get all the answers which are :
$ (a,b)\equal{}(x_n,x_{n\plus{}1}),x_{n\plus{}1}\equal{}3x_{n}\minus{}x_{n\minus{}1},x_1\equal{}1,x_2\equal{}2$.
now one can easily prove by induction $ x_n\equal{}F_{2n}$.
now it rests to prove that $ 5.x_n^2\plus{}4\equal{}x^2$.
which is easy because by computing the discriminant we have
$ (3b)^2\minus{}4(b^2\minus{}1)\equal{}x^2 \implies x^2\equal{}5.b^2\plus{}4$.


Thought this was a very nice problem when I came across it.

You have trapezoid ABCD with base AB length 10 and base DC length 18. Point E is constructed on AD such that AE=3ED. Find the ratio of the area of BEC to the area of trapezoid ABCD.

[hide]
Pretend that AE is length 3 and ED is length 1. The altitude of AEB and the altitude of DEC are also in the ratio of 3:1. The area of triangle ABE is 10 * 3 / 2 = 15. The area of DEC is 18 * 1 / 2 = 9. The median of the trapezoid is 14, so the area of the trapezoid is 56. The area of BEC is 56 - 15 - 9 = 32. The ratio of the area of BEC to the area of ABCD is 32/56 which is $\frac{4}{7}$.[/hide]


One family has three children, at least one female, if having a male or female child is equally likely, what is the probability that at leat one of the three children is a male?

[hide]  There are 3 possibilties for how many boys and girls there are.
Case 1. Boy, Boy, Girl
Case 2. Boy, Girl, Girl
Case 3. Girl, Girl, Girl
The only way that there cannot be at least one boy is if Case 3 happens.  Now, the probabilty of Case 3 happening is (1/2)(1/2)=1/4 since the chance of at least 1 girl is 1 (since there is always 1 girl).  The chance of 2 girls is 1/2 (since there is an equal chance of getting a boy and a girl).  Finally, the chance of getting 3 girls is (1/2)(1/2) since the chance of each girl after the first one is 1/2.  Thus, the chance of at least one boy is $1-\frac{1}{4}=\boxed{\frac{3}{4}}$ [/hide]

[quote="The QuattoMaster 6000"][hide]Now, the probabilty of Case 3 happening is (1/2)(1/2)=1/4 since the chance of at least 1 girl is 1 (since there is always 1 girl).
[/hide][/quote]
[hide]That's not true. Case 3 has a 1/8 chance. The fact that there is 1 girl does not affect that.[/hide]

[hide]The possibilities are ggg, ggb, gbg, bgg, gbb, bgb, bbg, bbb. Of the 7 that at least one is female, 6 have at least on that is male. 6/7.[/hide]

[hide]This is conditional probability.  We have to find the probability something will happen if something else is true, so you have to be careful.

There are 7 possibilities that there will be 1 girl.  There is 1 possibility where there is not a boy.  So the answer is (7-1)/7=6/7

The problem with quattomaster's solution is that it finds the probability that there are not 3 girls or 3 boys, not the probability that there is at least 1 boy given there is 1 girl.[/hide]

Oh, yes you are right  :blush:

I wasn't thinking

I remember missing this problem when I was preparing for nationals last year, so there is nothing wrong.  I remember not understanding why I kept getting it wrong (I repeated tests).


Find all primes $p$ for which $(2^{p-1}-1)/p$ is a perfect square.

I remember having solved this before (right here on the forum).

Cool thanks.. I found it here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1055


probability
limit
calculus
calculus computations

I read that the probability of selecting 2 coprime integers is $\frac{6}{\pi^2}$, but I don't know why.  I posted this here because the identity $\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$ involves a fourier series.
Thanks.

More precisely, the statement that we want is:
Choose $a$ and $b$ independently from the set $\{1,2,\dots,N\}$ using a uniform distribution. Let $p_N$ be the probability that $a$ and $b$ are relatively prime.
We get $\lim_{N\to\infty}p_N=\frac{6}{\pi^2}.$

To prove this, ask the questions "Does $p$ divide both $a$ and $b$?" for each prime $p$. The probability that $p$ does not divide both $a$ and $b$ tends to $1-\frac1{p^2}$ as $N\to\infty$ and the answers to any pair of these questions tend to independence.
Therefore, the probability that the answer to all of these questions is no tends to
$\prod_{p\text{ is prime}}\left(1-\frac1{p^2}\right).$
By an Eulerian trick,
$\prod_{p\text{ is prime}}\frac1{1-\frac1{p^2}}=\prod_{p\text{ is prime}}\left(\sum_{k=0}^\infty\frac1{(p^2)^k}\right)= \sum_{j=1}^\infty\frac1{j^2}=\frac{\pi^2}{6}.$

The desired limiting probability is then the reciprocal $\frac6{\pi^2}.$

You are right that this comes from $\zeta(2)$, but there are many ways of evaluating that sum.

[quote="jmerry"]
The probability that $p$ does not divide both $a$ and $b$ tends to $1-\frac1{p^2}$ as $N\to\infty$ .[/quote]
Can you explain this?
Thanks

The probability "$p$ divides $a$" tends to $\frac{1}{p}$, since approximately $N/p$ numbers from $\{1,2,...,N\}$ are divisible by $p$. The events "$p$ doesn't divide both $a$ and $b$" and "$p$ divides both $a$ and $b$" complement each other, thus
P($p$ doesn't divide both $a$ and $b$)=1-P($p$ divides both $a$ and $b$)=1-P($p$ divides $a$)*P($p$ divides $b$)=$1-1/p^2$ (we used that events "$p$ divides $a$" and "$p$ divides $b$" are independent).


Are you sure you didn't mean $ {1991\choose 2n}$?

[hide="Solution if you did"] The sum (as I give it) can be written

$ \frac{(1\plus{}\sqrt{3}i)^{1991}\plus{}(1\minus{}\sqrt{3}i)^{1991}}{2^{1991}}\equal{} e^{i\frac{1991\pi}{6}}\plus{}e^{\minus{}i\frac{1991\pi}{6}}\equal{}\boxed{\minus{}\sqrt{3}}$

The given sum can be calculated by a similar method, but we must account for the missing last few terms, which is annoying. [/hide]

Sorry, it is actually $ \binom{1990}{2n}$.

Are all the 1991's in your solution actually 1990s?

Yes; [hide="it would be..."] $ e^{i\frac{1990\pi}{6}}\plus{}e^{\minus{}i\frac{1990\pi}{6}}\equal{}\minus{}1$ [/hide]


calculus
integration
trigonometry
calculus computations

$\int x \sin 2x \, dx = -\frac{x \cos 2x}{2} + \int \frac{\cos 2x}{2} \, dx = -\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} + C$ integrate by parts

[quote="pkothari13"]How would one evaluate the indefinite integral $\int (x \sin 2x)dx$

Apparently, this should use only elementary methods, because this appeared on a Calculus 1 Final.

The answer is [hide]$-\frac{1}{2}x \cos 2x + \frac{1}{4}\sin 2x +C$[/hide]

What's even more confusing to me is when I differentiate that answer, I don't get $x \sin 2x$.  :huh:

Rather, I get $x \sin 2x + \frac{1}{2}\cos 2x$[/quote]

yea use the product rule when you differentiate, but do it correctly (it helps, seriously)

[quote="MysticTerminator"]yea use the product rule when you differentiate, but do it correctly (it helps, seriously)[/quote]

Oh man, I totally forgot.  :blush:  Thanks MysticTerminator!

[quote="Kent Merryfield"]I have asked something very much like this on a test before we even studied integration by parts - as a multiple choice question, with about four possible answers. (It wouldn't be too hard for you to guess what a couple of the "distractors" or wrong answers might have been.) The point is simple: you can always check your result by differentiating, and faced with several possible answers (and not having learned the way to obtain them), you can differentiate them until you find the one that checks.[/quote]

I'm assuming that was the intent of the question, because I know integration by parts hadn't been taught.  Actually, I tried differentiating each answer choice, but because I forgot to use the product rule appropriately, so I was totally lost.  Therefore, I used a crude approximation of the integral with my own set bounds and I used Simpson's Rule.  I then checked using each of the answer choices, and I got the correct answer.

thats what my friend did last year (he was a Calc BC student then).

His advice to me was to try to avoid doing that. he says that because he did that he cant integrate very well. :rotfl: That and it is very time consuming. :P


hello, we get $ \sin(\frac{3\pi}{7})\cdot\sin(\frac{6\pi}{7})\cdot\sin(\frac{9\pi}{7})\equal{}\minus{}\frac{\sqrt{7}}{8}$.
Sonnhard.

I believe that the original poster was looking for solution, instead of just an answer...  :)

Sorry, but I've post this problem here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=284764[/url]

[quote="Dr Sonnhard Graubner"]hello, see also here
http://mathworld.wolfram.com/TrigonometryAnglesPi7.html
Sonnhard.[/quote]

Thanks! That's enlightening!

From the article:

How was the identity $ \sin(n\alpha) \equal{} 2\sin[(n \minus{} 1)\alpha]\cos\alpha \minus{} \sin[(n \minus{} 2)\alpha]$ and the complex exponentials obtained?


Here's the most recent [url=http://www.mathsci.appstate.edu/Events/HighSchoolMathContest/2007/registeredschools.html]list of schools[/url] participating at App State.  Chapel Hill usually dominates the event.  With a relatively small number of participating schools, there usually aren't many state qualifiers from App State.  Last year there were 2 state qualifier spots reserved for local students and 4 slots open to anyone.  Looks like it will be about the same this year.

Either Providence Day or Providence High won the 2007 UNCC overall competition.  Both schools fielded teams.  Those at Providence Day who competed in the Comprehensive Division at UNCC won't be at App State.

App State runs a good contest.  They have one section that requires work to be shown for credit.  It's amazing that they have time to grade the papers before the awards ceremony.

Yes I am. What test are you taking? Algebra I or Comprehensive?

can someone who went post the results? my school had to leave before the awards ceremony.

My sources are a bit sketchy on this.

App State Comprehensive state qualifiers:

1.  Asian name with F's (Chapel Hill)
2.  Alex Kritchevsky (Mt. Tabor)
3. ?
4.  Shannon Taylor (Mt. Tabor)

There were two local qualifiers in addition to the four above.  One of the locals plays flute.  I'll edit this post when I remember her name.

Comprehensive Team Placement:

1.  Chapel Hill
2.  Providence Day

There were 11 state qualifiers in Algebra 1.  I have no other information.

[url=http://www.mathsci.appstate.edu/Events/HighSchoolMathContest/]2007 App State Comprehensive and Algebra I results[/url] are now posted.

:rotfl:  haha apparently they miscored my test so i got 5th and qualified for states  :lol:

Congratulations!  It sounds like App State is sending the 2 locals and 5 non-local qualifiers this year.  In 2006, they only sent 4 non-locals along with the 2 local qualifiers to state.  There must have been more students participating at App State this year since only 7% are supposed to qualify.

Chapel Hill went to App State? Do they split up or something? 'Cause I saw a CH school at Wake Tech regional.

chapel hill usually goes to elon, unc charlotte, and app state. we split up to go to different contests but im pretty sure none of us went to wake tech.

I think Kimmy's talking about East Chapel Hill.

And Smith Middle... I guess Eastern CH goes to Wake Tech.


show that for positive reals $ a,b,c,d$,
$ \sum_{cyclic} \frac {a}{b\plus{}2c\plus{}d} \ge 1$ :wink:

$ \sum_{cyclic} \frac {a^2}{ab \plus{} 2ac \plus{} ad} \ge \frac {(a \plus{} b \plus{} c \plus{} d)^2}{2ab \plus{} 2bc \plus{} 2cd \plus{} 2da \plus{} 4ac \plus{} 4bd}$

$ \equal{}1 \plus{} \frac {(a \minus{} c)^2 \plus{} (b \minus{} d)^2}{2ab \plus{} 2bc \plus{} 2cd \plus{} 2da \plus{} 4ac \plus{} 4bd}\ge 1$

Where the first inequality is Titu's lemma.

[quote="peine"][quote="rofler"]$ \sum_{cyclic} \frac {a^2}{ab \plus{} 2ac \plus{} ad} \ge \frac {(a \plus{} b \plus{} c \plus{} d)^2}{2ab \plus{} 2bc \plus{} 2cd \plus{} 2da \plus{} 4ac \plus{} 4bd}$

$ \equal{} 1 \plus{} \frac {(a \minus{} c)^2 \plus{} (b \minus{} d)^2}{2ab \plus{} 2bc \plus{} 2cd \plus{} 2da \plus{} 4ac \plus{} 4bd}\ge 1$

Where the first inequality is Titu's lemma.[/quote]
It's also Cauchy-Schwarz,
what about this one, It's too easy,
let $ a,b,c,d$ be  positive reel numbers, prove that:
$ \sum \frac {a}{b \plus{} 2c \plus{} 3d} \geq \frac {2}{3}$[/quote]
If you multiply upsidedown by $ a,b,c,d$ respectively and then use the adreescu lema.....In the end you only have to prove that
$ a^2\plus{}b^2\plus{}c^2\plus{}d^2\geq ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd$ which is ubviosly true :wink:

[quote="Dimitris X"][quote="peine"][quote="rofler"]$ \sum_{cyclic} \frac {a^2}{ab \plus{} 2ac \plus{} ad} \ge \frac {(a \plus{} b \plus{} c \plus{} d)^2}{2ab \plus{} 2bc \plus{} 2cd \plus{} 2da \plus{} 4ac \plus{} 4bd}$

$ \equal{} 1 \plus{} \frac {(a \minus{} c)^2 \plus{} (b \minus{} d)^2}{2ab \plus{} 2bc \plus{} 2cd \plus{} 2da \plus{} 4ac \plus{} 4bd}\ge 1$

Where the first inequality is Titu's lemma.[/quote]
It's also Cauchy-Schwarz,
what about this one, It's too easy,
let $ a,b,c,d$ be  positive reel numbers, prove that:
$ \sum \frac {a}{b \plus{} 2c \plus{} 3d} \geq \frac {2}{3}$[/quote]
If you multiply upsidedown by $ a,b,c,d$ respectively and then use the adreescu lema.....In the end you only have to prove that
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2\geq ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd$ which is ubviosly true :wink:[/quote]
are sure?  :|  take ($ a\equal{}b\equal{}c\equal{}d\equal{}1$)
it's more easy than you think. :)

hello peine,
could u plz post ur solution ,
i did the sum this way,
write $ 1$ in the LHS as $ \frac {a \plus{} b \plus{} c \plus{} d}{a \plus{} b \plus{} c \plus{} d}$
then transposing to $ LHS$ with corresponding we get
to prove tht,
$ \sum_{cyclic} \frac {a(a \minus{} c)}{(a \plus{} b \plus{} c \plus{} d)(b \plus{} 2c \plus{} d)} \ge 0$
which can be finished of with AWLOG: or by
just taking out common terms which becomes square terms  :)
nice solution rolfer! titu's lemma got the nod alright
dimitris x ur statement isn't so obvious or rather false

[quote="peine"][/quote][quote="Dimitris X"][quote="peine"][quote="rofler"]$ \sum_{cyclic} \frac {a^2}{ab + 2ac + ad} \ge \frac {(a + b + c + d)^2}{2ab + 2bc + 2cd + 2da + 4ac + 4bd}$

$ = 1 + \frac {(a - c)^2 + (b - d)^2}{2ab + 2bc + 2cd + 2da + 4ac + 4bd}\ge 1$

Where the first inequality is Titu's lemma.[/quote]
It's also Cauchy-Schwarz,
what about this one, It's too easy,
let $ a,b,c,d$ be  positive reel numbers, prove that:
$ \sum \frac {a}{b + 2c + 3d} \geq \frac {2}{3}$[/quote]
If you multiply upsidedown by $ a,b,c,d$ respectively and then use the adreescu lema.....In the end you only have to prove that
$ a^2 + b^2 + c^2 + d^2\geq ab + ac + ad + bc + bd + cd$ which is ubviosly true :wink:[/quote][quote="peine"]
are sure?  :|  take ($ a = b = c = d = 1$)
it's more easy than you think. :)[/quote]
Oh yes....Now i see :(

let put $ S$ the left side, by Cauchy-Schwarz,
$ S(4(ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd))\geq (a \plus{} b \plus{} c \plus{} d)^2$
$ \Leftrightarrow S\geq \frac {(a \plus{} b \plus{} c \plus{} d)^2}{4(ab \plus{} bc \plus{} cd \plus{} da \plus{} bd \plus{} ca)}$
It's easy to prove that, $ (a \plus{} b \plus{} c \plus{} d)^2 \geq \frac {8}{3}(ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd)$
use AM-GM or $ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} d)^2 \plus{} (d \minus{} a)^2 \plus{} (c \minus{} a)^2 \plus{} (d \minus{} b)^2 \geq 0$
then $ S\geq \frac {2}{3}$
I think that was easy

er........... that looked easy thanks peine
i should have got it


If $ \log_{10}25=1.39794$, what is the number of digits in the complete expansion of $ 25^{100}$?

$ 25^{100}\text{ has }\lfloor1+\log25^{100}\rfloor=\lfloor1+100\log25\rfloor\approx\lfloor140.794\rfloor=140\text{ digits.}$


If $(x-a)(x-1)+1=(x+b)(x+c)$ is true for all $x$ and if $a,b,$ and $c$ are integers, find all possible values of $a$.

[hide]
$x^2 - (a+1)x + a + 1 = x^2 + (b+c)x + bc$
$-(a+1)x + (a+1) = (b+c)x + bc$
For this to be true for all $x$, the coefficients of x must be equal and the constants must be equal
$-(a+1) = b+c \implies a + b + c = -1 \implies a + 1 = -b - c$
$a + 1 = bc \implies bc = -b - c \implies (b + 1)(c + 1) = 1$
$(b,c) = (0,0),(-2,-2)$
$a = bc - 1 \implies a = -1, 3$
[/hide]

[hide]
substitute $x=-b$ both side we get

$(b+a)(b+1)+1=0$ or $(b+a)(b+1)=-1$

Since $a,b$ are integer , we have $b+a=1\land b+1=-1$ or $b+a=-1\land b+1=1$ . Hence we get $a=3,-1$
[/hide]


There are three liars. If one of them lie all the time, and the other lies half the time, and the third lies a quarter of the time, and they always go in the following order: (lie for L, truth for T)
1st one: LLLLL...
2nd one:LTLTLT...
3rd one:LTTTLTTT...
after how many lie/truth episodes after the first one, will they all lie together?

[hide]They all lie every 4 times...[/hide]

[hide]including the first one, five. Excluding the first one, four. This is because person one always lies, person two lies every odd episode, and person three lies where the first episode, n, he lies, and every other episode n (mod 4) he tells a lie. I just then found that since n can't equal 1, and it is (mod 4), the next smallest value is five. That is the fifth value in the sequence and the fourth value after the first.[/hide]

[hide]After the first one, they all lie every 4 times...[/hide]

[hid=solution]They lie every 4 times: 1, 5, 9, 13, 17.....[/hide]

They will lie together every 4 times because the first person always lies, the second person lies every 2 times, and the third lies every fourth.  the first, second, and third all go into 4, so every fouth time. :wallbash_red:


calculus
derivative
function
analytic geometry
graphing lines
slope
integration

i recently heard my math teacher talking about derivitives and their ability to give a person the general idea of what a graph looks like.  What are they and how can it be used to determine the shape of the graph with few calculations

a derivative of a function is a graph of the instantaneous slope of that function at each point. It can provide useful info about the function because if the derivative at a point is 0, then you know the function has a maximum of minimum at that point (only when there is a max or a min is the slope 0). If the derivative at a point is positive, then you know the function is increasing at that point. similarly, if the derivative at a point is negative, then the function is decreasing.

[quote="Yrael"]It can provide useful info about the function because if the derivative at a point is 0, then you know the function has a maximum of minimum at that point (only when there is a max or a min is the slope 0).[/quote]
Also note that the derivative could be undefined at a max or min.  For  example, the derivative of $f(x) = |x|$ with respect to x is undefined at x=0, but there is still a minimum at x=0.

The derivative of a function at a point means the slope of the line tangent to the function at that point.  If you know the equation for the derivative of the function you can find out a lot about the actual function.  In fact, let the equation of the derivative be f'(x).  We can find (with integrals) f(x).  Actually we would find that f(x)=g(x)+c where c is a constant.  So you can find a lot with the derivative.  Also, as noted you can find max's, min's, concavity etc.

Something that I find cool, although I don't know anything about are slope fields.  They give a basic idea of what a function looks like given it's derivative.  Wish I knew more about them  though!

The derivative= the limit as delta x approaches 0 of f(x+deltax)-f(x) divided by delta x.

[quote="demji"]how do you determine the derivitive of a point?[/quote]for a given function $f$ and a given point $x_0$ the derivative in that point is the limit
\[ \lim_{t\to 0} \frac { f(x_0+t) - f(x_0)}  t \] when it exists.

In general all known functions and their combinations are derivable, so we can use certain derivation formulae to help us compute the derivative in one point, faster and easier.

You need to understand basic calculus before understanding how derivatives can help you look and draw graphs.

[quote="ComplexZeta"]All functions are differentiable? Is this a joke?[/quote]

Who said that they were?

[quote="ComplexZeta"]All functions are differentiable? Is this a joke?[/quote]I said in general all known functions (and I was reffering to the initiator of this thread, which I assume that can't give me more than a couple of, if any, examples of non-derivable functions :) ;) since it seems that he hasn't yet taken any calculus classes).

These functions are called elementary functions, and all can be derived using a pack of rules (I think that he also asked if there is an easy way to compute the derivtative in a point).

Graphs with discontinuities or vertical tangents also don't have derivatives at that point.

It is wrong to think of the derivative as a slope, because very simply, the xy plane can be thought of as a number line with values associated with each point. This concept becomes increasingly important when you deal with complex calculus.

[quote="Treething"]Examples of graphs that are not differentiable:

y = x^(2/3)

x^2 + y^2 = r^2

y = 1/x

(sin(x))/x[/quote]
If you derivate each equation to $x$, you will get the following.
$\displaystyle y=x^{\frac{2}{3}}$ gives $y'=\frac{2}{3x^{\frac{1}{3}}}$
$\displaystyle x^2+y^2=r^2$ gives $y=\sqrt{r^2-x^2}$ gives $y'=\frac{-x}{\sqrt{r^2-x^2}}$
$\displaystyle y=\frac{1}{x}$ gives $y'=\frac{-1}{x^2}$
$\displaystyle y=\frac{\sin(x)}{x}$ gives $y'=\frac{\cos(x)}{x}-\frac{\sin(x)}{x^2}$

The easiest rule for derivitives is the polynomial rule.

Lets say we have this function: f(x)=2x^4-3x^2+x

The easiest way (and by far the fastest) is to just bring down the exponent and multiply it by the coefficient and then subtract the current exponent by 1:

f'(x)=(4)*(2)x^(4-1)-(2)*(3)x^(2-1)+(1)*x^(1-1)

that would simplify to:

f'(x)=8x^3-6x+1

other rules are harder (quotient rule, product rule, and chain rule).

Only other rule I'll take time to show now is the quotiet rule.
WARNING SMALL OFFENSIVE HUMOR
we have this function: h(x)/g(x)

the derivitive is: (g(x)*h'(x)-h(x)*g'(x))/(g(x)*g(x))

my math teacher showed us  this formula to help us remember:

HI/HO

so the derivitive is:

(HOdHI - HIdHO)/(HO*HO) and the best way to remember it is that there is a ho at every corner!


I'm creating Russian Roulette 2. For rules go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=222906]here[/url]

Post in thread to sign up. I'll accept sign-ups through October 1st, maybe not that long depending.

Sign up. I trust you will no rig the game.

I'm in.

However, I will win cuz I'm already dead.

I'm in.
Hopefully I won't die on a 3-3 split this time.  (looks at kimmystar94)

I need bullet options from anyone who has joined/will join. So far I only have Ernie's. Just a refresher, you get 2 guns and have to split 6 bullets between the 2 guns. For example, suppose I put 5 bullets in gun A. Then I would 1 bullet in gun B.

Need some peoples gun options. and bump.

/in

The message is too small. Please make the message longer before submitting.

/in

though I do not want to disgrace my awesome reputation in luck  :rotfl:

_________________
I will be modding stuff soon!

You made someone die with 1 bullet 2 rounds in a row...it's very suspicious, and interesting probability. I'm acusing it of not being fair enough, but I know that with random anything's possible...

But that means that I think random numbers are too important to leave to chance...

Since i don't know that amount of bullets in my guns

I give A to ernie

"C'mon... again?  How many friggin times do I have to do this," shouted ernie angrily.

"Why, are you scared," replied budi713, daringly.

"NOBODY DISSES THE PUPPET WITHOUT DYING!" shouted ernie.  "NOBODY!!!!!!!!!  And as for your gun, I will take it."

And then, it was night again.

My stories are very similar.

[hide="hmm..."]Budi713 decided to challenge ernie. Disgruntingly the puppet took the gun. Almost shocked crazychinesemaniac looked at the bullet options in the gun and saw that ernie was a perfect 3 for 3 in picking the right gun! This time he had picked the gun with only 1 bullet in it. Sniggering he put the gun to his head and pulled the trigger....Nothing happened and he went back to sleep.

Now it was budi713's turn, he slowly put the gun to his head trembling and pulled the trigger. BANG...but he had missed his head, instead he hit a wall and the bullet ricoched back and hit him in the heart. Squirming in pain he tried to crawl to the phone to call 911. However, ernie seeing what he was doing picked up the gun budi713 had just fired and emptied it on his latest victim. All bullets found their target and budi713 layed still.

[b][color=green]Ernie: Lives[/color], [color=red]Budi713: Dies[/color][/b]

[b]A mysterious voice then rang out on the speakers...I need gun options for round 2 from the remaining contestants!

Remaining Players:
Ernie
Brut3Forc3[/b][/hide]

I have a gut feeling ernie's going to win.

I have placed your 2 names into a generator to see who will go first.............

[hide="Ernie..."]is going second! Therefore [b]brut3forc3 [/b]gets to start off this round!

[u][b]Order:[/b][/u]
[b]Brut3Forc3
Ernie[/b][/hide]

I give Gun A, which I think has less bullets. Or more. Eh. I suppose I could check...

Is crazy away for a while? I'm bumping this up.

[hide="Sorry for my temporary absense..."]Guns at the ready men! One of you will leave here in eternal glory...the other will be dead! Ernie quickly and efficently said he would keep gun A. Again he picked the gun with less bullets! He laughingly pulled the trigger...Nothing happened. It was now brut3forc3's turn. He put the gun to his head and pulled the trigger...BANG! He was gone... Which means our winner is.....Ernie!

The puppet then appeared and his counterpart, Bert appeared as well. Wasting no time they both dived into the money and as my duty as the judge I asked Ernie and Bert, "What are you going to do with the money?" Confidently they responded, "We're going to move off seaseme street of course, with the housing market like it is, this is just the boost we need to get off that dumb street!" There we have it....our winners![/hide]

My gut feeling was correct.    :lol:   Now, if that could only translate to my gun decisions...

what the heck...i was killed with only 1 bullet in my gun...i shoulda put 0 in...

This game was in no way rigged...if you don't like the results take it up with. [url]http://www.random.org/lists[/url]

I know, it's funny if i stop looking at it from my own point of view :)


Given n real numbers.Let $k_{n}$ be "a best constant" with each n positive integer and satisfied:
$(S_1)^2+(S_2)^2+...+(S_n)^2$<=$k_{n}(x_{1}^2+...+x_{n}^2)$
S_i=(x_1+...+x_{i})/i for all i=1,n.
a)Prove that $k_{n}$<4
b)Find $k_{n}$.

Either I am very idiot now or your idea was already found by Hardy.


linear algebra
matrix
algebra proposed
algebra

hi,
it would be nice if somebody could post his solution or give me a hint.  :huh:

Create a new matrix $ B$ of dimension $ n$ by $ n$ as follows:
the first and the second rows are all $ 1$;
the rest of $ n\minus{}2$ rows are $ (a_{ij})$.

Obviously $ |B|\equal{}0$. If you expand $ |B|$ based on first row, you will get
$ |B|\equal{}A_1\minus{}A_2\plus{}A_3\minus{}A_4\plus{}\cdots.$.

Hopefully this is clear enough.


function
logarithms
limit
complex analysis
complex numbers
real analysis
complex analysis unsolved

A problem appeared on my complex analysis prelim problems:  Prove that a bounded harmonic function on the complex unit disk with a puncture at the origin is indeed harmonic on the whole disk.  I know there's a way to the problem using comparison principle, but is it possible to have a solution using complex?  I am not so sure if this harmonic function can be the real part of a complex function.

Let $ u$ be the function in question.

Write $ u(re^{i\theta})\equal{}\sum_{k\equal{}\minus{}\infty}^{\infty} \widehat{u}(r,k)e^{ik\theta}$ for $ 0<r<1.$

Since $ u$ is smooth, we have that $ \widehat{u}(r,k)$ is a smooth function of $ r,$ and also that for each $ r,$ $ \widehat{u}(r,k)$ tends to zero rapidly as $ k\to\infty.$

For each $ r,$ $ |\widehat{u}(r,k)|\le \sup_{\theta}|u(r,\theta)|\le M$ where $ M$ is a global bound on $ |u|,$ which exists by hypothesis.

Now use the fact that $ u$ is harmonic, differentiating under the sum to get that

$ \frac{d^2}{dr^2}\widehat{u}(r,k)\plus{}\frac1r\cdot\frac{d}{dr} \widehat{u}(r,k)\plus{}\frac{k^2}{r^2}\,\widehat{u}(r,k)\equal{}0.$

For $ k\ne0,$ this has solutions $ c_1r^k\plus{}c_2r^{\minus{}k},$ while for $ k\equal{}0$ it has solutions $ c_1\plus{}c_2\ln r.$ But we can only consider bounded solutions. Hence we have that

$ \widehat{u}(r,k)\equal{}a_kr^{|k|},$ so that

$ u(r,\theta)\equal{}\sum_{k\equal{}\minus{}\infty}^{\infty}a_kr^{|k|}e^{ik\theta}.$

This function is harmonic on the unit disk and takes on the value $ a_0$ at zero.

We can slightly weaken the condition that $ u$ is bounded. The condition $ \lim_{z\to 0}\frac{u(z)}{\ln|z|}\equal{}0$ would suffice for drawing the same conclusion.


linear algebra
matrix
invariant
geometry
geometric transformation
homothety
linear algebra unsolved

If $ A$ and $ B$ are in $ M_2(C)$ such that $ A^2\equal{}I$ and $ B^3\equal{}I$ but $ A$ and $ B$ is not equal to $ I$, prove that :
If $ D$ is a matrix commuting with both $ A$ and $ B$ (i.e $ DA\equal{}AD$ and $ DB\equal{}BD$), then $ D\equal{}kI$ for a scalar $ k$.
Thanks in advance.

Sorry, there are more criteria : $ ABA\equal{}{B}^{\minus{}1}$. Thanks in advance.

This is pretty much [url=http://en.wikipedia.org/wiki/Schur%27s_lemma]Schur's lemma[/url].
$ \langle A,B\rangle$ is a finite group ($ \cong S_3$) which acts on $ \mathbb{C}^2$.
This action is irreducible, because otherwise by [url=http://en.wikipedia.org/wiki/Maschke%27s_theorem]Maschke[/url] we have $ \mathbb{C}^2\equal{}\mathbb{C}v\oplus \mathbb{C}w$
with one-dimensional $ A,B$-invariant $ \mathbb{C}$-subspaces $ \mathbb{C}v, \mathbb{C}w$.
But then $ A,B$ would be diagonal matrices with respect to the basis $ \{v,w\}$.
So $ A,B$ would commute, which contradicts $ ABA^{\minus{}1}\equal{}ABA\equal{}B^{\minus{}1}\equal{}B^2\not\equal{}B$ for $ B\not\equal{}I$.

But now let $ C\in \mathcal{M}_2(\mathbb{C})$ commute with $ A,B$.
Then for some eigenvalue $ \lambda\in\mathbb{C}$ of $ C$ we have $ 0\not\equal{}\ker (C\minus{}\lambda I)$ $ A,B$-invariant.
This gives $ \ker (C\minus{}\lambda I) \equal{}\mathbb{C}^2$ and $ C\equal{}\lambda I$.

Beautiful proof Olorin but we can reason more simply.

We assume only $ B\not\equal{}I$;if $ A\equal{}\pm{I}$ then  $ B\equal{}B^{\minus{}1}$and $ B\equal{}I$;contradiction. $ A,B$ are diagonalizable.
Then we may assume $ A\equal{}\begin{pmatrix}1&0\\0&\minus{}1\end{pmatrix}$. $ spectrum(B)\subset\{1,j,j^2\}$; $ B$ and $ B^{\minus{}1}$ are similar then $ B\equal{}P\begin{pmatrix}j&0\\0&j^2\end{pmatrix}P^{\minus{}1}$. $ B$ is not diagonal because $ A,B$ do not commute. Then $ \{I,A,B\}$ is a free system.
Let $ C$ be a matrix which commutes with $ A,B$.     $ C\equal{}\alpha{I}\plus{}\beta{A}\equal{}\gamma{I}\plus{}\delta{B}$. Then $ \beta\equal{}\delta\equal{}0$ and $ \alpha\equal{}\gamma$.

More generally:
Let $ A,B\in\mathcal{M}_2(\mathbb{C})$ s.t. $ AB\not\equal{}BA$.
If $ CA\equal{}AC$ and $ CB\equal{}BC$ then $ C$ is a homothety.
Proof: Necessarily $ A,B$ are not homotheties; moreover the system $ \{I,A,B\}$ is free; else for example $ B\equal{}\alpha{I}\plus{}\beta{A}$ and $ AB\equal{}BA$. Then $ C\equal{}\alpha{I}\plus{}\beta{A}\equal{}\gamma{I}\plus{}\delta{B}$.............

[quote="loup blanc"]Then $ C \equal{} \alpha{I} \plus{} \beta{A} \equal{} \gamma{I} \plus{} \delta{B}$.............[/quote]

why is this true?

Edit: em I guess I figured it out - if $ C \in \mathrm{span}\{I,A,B\}$, by the commuting relations with $ A$ and $ B$ we are forced to conclude that $ C$ is a homothety. And if $ C$ is not within the span, then $ \{I,A,B,C\}$ is a basis of $ M_2(\mathbb{C})$. Then it is clear than $ C$ lies in the center of $ M_2(\mathbb{C})$, which means it is a homothety.

Hi Soarer,
If $ A\in\mathcal{M}_2(\mathbb{C})$ and if $ A$ is not a homothety then $ dim(\{C;CA\equal{}AC\})\equal{}2$. (even if $ A$ is not diagonalizable).

$ C$ has Jordan normal form $ \left(\begin{matrix}\lambda &1\\&\lambda\end{matrix}\right)$ or $ \left(\begin{matrix}\lambda &\\&\mu\end{matrix}\right)$ with $ \lambda\not\equal{}\mu$,
which have centralizers $ \mathbb{C}I_2\plus{}\left(\begin{matrix}&*\\&\end{matrix}\right)$ and $ \left(\begin{matrix}* &\\&*\end{matrix}\right)$ in $ \mathcal{M}_2(\mathbb{C})$,
both abelian.


[b]Problem[/b]: There are two different numbers.  The sum of the square of these numbers is a cube.  The sum of the cube of these numbers is a square.  Find the numbers.

[hide]Since it is not specified whether it has to be negative or positive 0 and 1 work.[/hide]

Yes, I think you need to work on your question statement a little bit.  What type of numbers are we dealing with?

The question doesn't mention any specific type of numbers but they should be real numbers.

You sure it doesn't mean positive (or non-negative) integers?

kool_dudy wrote:Problem: There are two different numbers.  The sum of the square of these numbers is a cube.  The sum of the cube of these numbers is a square.  Find the numbers.

Where did this problem come from kool_dudy?

It would probably be better stated: Find two distinct positive integers such that the sum of the squares of the numbers is a cube, and the sum of the cubes of the numbers is a square.

This is a pretty neat problem, but I'm pretty sure it doesn't belong in the getting started forum (though it's worthwhile for beginners to work on it). After a little searching I found a whole family of solutions - though there may still be other ones.

A hint on how I found my solutions: [hide]I noticed that 5 and 10 are "close" to being a solution.[/hide]

My actual solutions: [hide]54 and 2*54, 510 and 2*510, 516 and 2*516,... [/hide]

Okay, I found another one: [hide] 24*54*73 and 3*24*54*73. If you figured out how I found the ones up above you can figure out how I found this one. This method will give you an infinite family of infinitely many solutions. But there still may yet be even more.[/hide]

I'll promote this thread to Intermediate. I'm resolving all doubts in favor of promotion, because I want to leave plenty of room in Getting Started for really easy problems, and I hear Intermediate has been harder than people expected.

This is a problem that my teacher gave us from a book.

I can tell that the question wasn't stated correctly but I'm just copying the question from the book to this forum.

I also think that there are multiple answers to this question.


There is electrical circuit containing R,L and C.The oscilation is damping slowly.For destroying the damping,we use one way. When the current is maximum in the circuit we change the initial length of L($l_{0}$) by $\delta l$ and the new legth of inductor L is being $l_{0}+\delta l$.When the charge is maximum on the capacitor,we change the length into initial value.Find the ratio $\frac{\delta l}{l_{0}}$

I got
\[\frac{4R\pi}{\sqrt{4L/C-R^{2}}}.\]

Is that ok? :maybe:


function
limit
real analysis
real analysis unsolved

Let $f : \left[ 0, \infty \right) \to \mathbb R$ be a continuous function on $\left( 0 , \infty \right)$. Prove that $f$ has the property of Darboux iff there is $\left( a_n \right)_{n \geq 1} \subset \left( 0,\infty \right)$ s.t. $\lim_{n \to \infty} a_n =0$ and $\lim_{n \to \infty} f(a_n) = f(0)$.

Sorry, but what do you mean with Darboux property? :blush:

My question is : If $f$ is continous doesn`t it automatically have the Darboux property?

A function $f : I \to \mathbb R$ has the property of Darboux iff for all $a,b \in I$ and for any $f(a) \leq y \leq f(b)$ there are $a \leq x \leq b$ s.t. $f(x)=y$.

spix, $f$ is given continuous only on $\left( 0,\infty \right)$. There's one special case you must consider.

[quote="spix"]You are right. I misread. Anyway... if $\lim_{n\to\infty} a_n=0$ and $\lim_{n\to\infty} f(a_n)=f(0)$  then the function is continous in  $0$ which solves the problem.[/quote]

nope... we have just ONE sequence which fulfills this condition. this is not enough to conclude continuity.

Assume that there exists $b$ such that $f([0,b])$ is not an interval. Hence, there exists $\lambda\in [0,b]$ such that $f(x)\neq \lambda$. There can not be n such that  $a_n$ such that $f(a_n)\le \lambda \le f(a_{n+1})$ hence we can assume WLOG $f(a_n)\le \lambda$ for $n$. If $f(0)=\lim f(a_n)\le \lambda$. We know that there exist $x$ such that $f(x)\ge \lambda$.
$x>0$ is greater that some $a_n$. Hemce $f([a_n,x])$ is not an inerval. But $f$ is continuous on $[a_n,x]$.contradiction. Hence $f$ has mean value property.


When a positive three digit dividend is divided by a positive one digit divisor, the result is a whole number quotient with remainder 1. How many distinct values are possible for this three digit dividend?

I probably got it wrong, I'm really tired...

[hide]EDIT: lol, oops. I read the question wrong, or something.

 Prime numbers + 1  (so like 119+1 = 120) there are 21, and prime numbers * prime numbers (***These prime numbers must be 31 or below but 11 or greater, which is like 11*11 = 121.) There are 28.
21+28 = 59[/hide]

superwizard188, you seem to have counted the complement of the set in which we are interested.  :)


decimal representation
Digits
number theory
factorial
IMO Shortlist

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote="prasadb"]The solution posted seems too compilcated...(any number > 5  )! is divisible by 10 , so is modulo 0..that
means (n+k) <= 4 and there are only a finite few cases to verify k<=3 && n <= 3..
4!=4, 3!=6 and 2!=2 ..none of which can equal k[/quote]

The problem says the leftmost digit, not the unit digit.

[quote="Xell"]
By forming the quotient between two successive inequalities[/quote]
I don't think we can do this. a>b and c>d does not mean a/c > b/d. Take 4>3 and 100>1

[quote="dgreenb801"] I don't think we can do this. a>b and c>d does not mean a/c > b/d. Take 4>3 and 100>1 [/quote]

But he takes a/d > b/c. It's still valid since $ a,b,c,d>0$.

for every natural number $ m$,denote $ N(m)$ as:

$ N(m) \equal{} \frac {m}{10^{d(m) \minus{} 1}}$.

in which $ d(m)$ is the number of digits of $ m$.now notice that $ 1\leq N(m) < 10$.now it can be easily shown that for every two natural numbers $ m,\ell$:

$ N(m\ell)\leq N(m)\cdot N(\ell)$      (1)

now let $ n$ be a natural number such that for $ k \equal{} 0,1,2,\ldots ,9$,the leftmost digit of $ (n \plus{} k)!$ equals $ k$.if $ 2\leq k\leq 9$ then there exist a nonnegative integer $ r$,and a real number $ a$,for which $ (n \plus{} k)! \equal{} a\times 10^r$ and also $ k < a < k \plus{} 1$.with a similar argument there exist a real number $ b$ and a nonnegative integer $ s$,for which $ (n \plus{} k \minus{} 1)! \equal{} b\times 10^s$ such that $ k \minus{} 1 < b < k$.hence:

$ 1 < N(n \plus{} k) \equal{} N\left(\frac {(n \plus{} k)!}{(n \plus{} k \minus{} 1)!}\right) \equal{} \frac ab < \frac {k \plus{} 1}{k \minus{} 1}\leq 3$    (2)

on the other hand the inequality $ N(m \plus{} 1)\leq N(m)$ holds only if $ 9\leq N(m)$ and also it follows from (2) that:

$ 1 < N(n \plus{} 2) < N(n \plus{} 3) < \cdots < N(n \plus{} 9)\leq\frac 54$

and using (1) we get that:

$ N\left((n \plus{} 2)!\right)\leq N\left((n \plus{} 1)!\right)N(n \plus{} 2) < 20\times\frac 54$

$ N\left((n \plus{} 3)!\right)\leq N\left((n \plus{} 2)!\right) N(n \plus{} 3) < 20\times\left(\frac 54\right)^2$

$ N\left((n \plus{} 4)!\right)\leq N\left((n \plus{} 3)!\right)N(n \plus{} 4) < 20\times\left(\frac 54\right)^3 < 4$

but this contradicts the assumption that the leftmost digit of $ (n \plus{} 4)!$ equals $ 4$.

QED

[color=#444444]Here's another approach. Let $ \color[rgb]{.35,.35,.35} n \geq 4$ be a counterexample to the theorem. Logarithms are base $ \color[rgb]{.35,.35,.35} 10.$[/color]

Let $ \{n\}$ denote the fractional part of $ \log n,$ so that $ \{n\}$ increases until $ \log n$ [color=red]jumps[/color] an integer, at which point $ \{n\}$ decreases close to $ 0,$ starts increasing again, and so on. Prior to a jump, $ \{n\}$ is at least $ \tfrac 12,$ as $ \log n$ increments by at most $ \tfrac 2n.$ Divide $ [0,1]$ at the points $ \log 2, \ldots, \log 9,$ and see how $ \{(n+k)!\}$ belongs to the $ k-$th segment, for $ 1 \leq k \leq 9.$

No [color=red]jump[/color] occurs in $ \{n+2\}, \ldots, \{n+9\},$ as the lengths of the segments force them all to be $ < \tfrac 12,$ so this sequence is increasing. But $ \log 2 \geq \small size \{n+6\} + \cdots + \{n+9\} \geq \{n+2\} + \cdots + \{n+5\} \geq \log \tfrac 52$ is a contradiction.

[quote=orl]Prove that there is no positive integer $n$ such that, for $k = 1,2,\ldots,9$, the leftmost digit (in decimal notation) of $(n+k)!$ equals $k$.[/quote]

Isn't $n=4 , k=1$ a counterexample to this statement?
@below Oops sorry got the question wrong

This problem has a lot of different valid approaches.

[hide=Scientific Notation Solution]
First, we list out the factorials of the integers from $1$ to $11$, showing that there is no positive integer $n \le 11$ that works.

Suppose a larger positive integer $n$ does satisfy the problem conditions.

Let $\#(x)$ be equal to left part of $x$ when written in scientific notation (so excluding the power of $10$). If we want the leftmost digit of $(n+8)!$ to be $8$ and the leftmost digit of $(n+9)!$ to be $9$, we require multiplication by $n+9$ to take a value in $[8,9)$ into a value in $[9,10)$. Hence, $n+9 \in (1, 1.25)$. Notice that if $n+3 \le \sqrt{1.5}$, then even if we let $\#((n+1)!)$ be as close to $2$ as possible (so $1.999...$), multiplying by $n+2$ and then by $(n+3)$ will get us to at most $2.999...$, which is not in $[3,4)$, the required leading digit for $(n+3)!$. Hence, $\#(n+3) > \sqrt{1.5} > 11/9$. Because of our first observation, a 3-digit integer $n+3$ should be greater than $122$, a 4-digit integer $n+3$ should be greater than $1222$, and so forth. Remember that when we multiply by $n+3$, we take a value in $[2,3)$ to a value in $[3,4)$, so that $\#(n+3) \in (1, 2)$. Hence, $n+3$ has leading digit $1$, so there is no looping around.

On the other hand, we have the following. We know that $\#(n+9) > \#(n+8) > ... > \#(n+3)$ from our above observations (adding in a little intuition if I did not give complete details before). If we replace the strict inequalities with nonstrict inequalities, then we see that in the best case, $\#((n+9)!)$ is as close to $10$ as possible. Then $\#((n+7)!)$ is strictly bounded above by $10\cdot \frac{9}{11}\cdot \frac{9}{11} = 10\cdot \frac{81}{121} = \frac{810}{121} < \frac{847}{121} = 7$, contradicting the premise that $(n+7)!$ has leading digit $7$.

In conclusion, there are no positive integers $n$ satisfying the problem conditions.
[/hide]

We proceed by contradiction. Let the [i]reduced[/i] copy of a positive integer $n$ be $a$, where $n=a\cdot 10^b$ for a nonnegative integer $b$ and a terminating decimal $a$ such that $1\le a<10$. If the leading digit of $n$ is $k$ so that $k$ is from $1$ to $9$, then the reduced copy of $n$ is between $k$ and $k+1$. Since the reduced copy of $(n+1)!$ is between $1$ and $2$, and the reduced copy of $(n+2)!$ is between $2$ and $3$, the reduced copy of $n+2$ must be between $\frac{2}{2}=1$ and $\frac{3}{1}=3$. Doing this again, the reduced copy of $n+3$ must be between $\frac{3}{3}=1$ and $\frac{4}{2}=2$. The reduced copy of $n+9$ must be between $\frac{9}{9}=1$ and $\frac{10}{8}=1.25$. Note that the lower bound is always $1$ and the upper bound starts at $3$ and slowly reduces to $1.25$.

[b][color=#f00]Claim: [/color][/b] The reduced copy of all numbers from $n+2$ to $n+9$ now must be between $1$ and $1.25$.
[i]Proof.[/i] Suppose not, because the numbers from $n+2$ to $n+9$ are consecutive, then there will be a number from $n+2$ to $n+9$ with reduced copy between $9$ and $10$ which is obviously impossible because that does not fit into any of the earlier ranges (that had a lower bound of $1$ and an upper bound of at most $3$). $\blacksquare$

Note that the reduced copy of $(n+1)!$ is between $1$ and $2$. Since the reduced copies of $n+2, n+3, n+4$ are all between $1$ and $1.25$, the reduced copy of $(n+4)!$ is less than $2(1.25)^3=3.90625<4$, contradiction as it must be between $4$ and $5$.

ISL Marabot solve

Suppose for the sake of contradiction there does.

Let \[f(x)=\frac{x}{10^{\lfloor \log_{10} x\rfloor }}.\]

Note $1\le f((n+1)!)<2$ and $2\le f((n+2)!)<3$. Since $\frac{(n+2)!}{(n+1)!}=n+2$, we note $\frac{2}{2}=1<f(n+2)<\frac{3}{1}=3$.

Similarly, we can find the bounds $1\le f(n+k)< \frac{k+1}{k-1}\le 3$ for $k=2,3,4\ldots,9$ and $1\le f(n+9)<\frac{10}{8}=1.25$.

Claim: For all $k=2,3,4,\ldots,9$, $1<f(n+k)<1.25$.
Proof: Suppose not. Then since $n+2,n+3,\ldots,n+9$ are all consecutive, at least one of $\{f(n+2),f(n+3),\ldots,f(n+8)\}$ is between $9$ and $10$, a contradiction as that doesn't fit into the bounds we had found earlier.
$\blacksquare$

Now we bound $f((n+1)(n+2)(n+3))$. Note $4\le f((n+4)!)<5$ and $1\le f((n+1)!)<2$. So $\frac{4}{2}=2<f((n+1)(n+2)(n+3))<\frac{5}{1}=5$.

This is a contradiction as $f((n+1)(n+2)(n+3))=f(n+1)f(n+2)f(n+3)\le (1.25)^3=1.953125<2$

Let $f(n)$ be equal to $\frac{n}{10^{\lfloor\log_{10}{n}\rfloor}}.$ In other words, $f(n)$ is the result you get when you divide $n$ by $10$ until you get a number in the interval $[1,10).$

Note that $f$ is multiplicative, that is, $f(mn)=f(m)f(n).$ Assume for the sake of contradiction that $\lfloor f((n+k)!)\rfloor = k$ for $k=1,2,\dots,9.$ Note that $2\le f((n+2)!)< 3$ and $1\le f((n+1)!)< 2$ implies $1<f(n+2)<3.$ We also have from the same logic, $1<f(n+9)<\frac{5}{4}.$

Suppose $10^i<n+9<\frac{5}{4}\cdot 10^i$ for nonnegative integer $i.$ If $i=0$ then $n+9<1.25$ which doesn't make sense. If $n+9<12.5$ then $n<3.5.$ $n=1,2,3$ immediately fail the test that $(n+1)!$ begins with a $1.$

Thus, $i\ge 2$ so we can assume that $n+2,n+3,\dots,n+9$ have the same number of digits. Otherwise, $n+2$ will not be in the range $(10^i,3\cdot10^i)$

This implies that $f(n+2)<f(n+3)<f(n+4)<f(n+9)<\frac{5}{4}.$ Thus, $100\cdot 10^i<n+2<125\cdot 10^i$ for some $i$ and $100\cdot 10^i<n+3<125\cdot 10^i.$ Note that $1\le f((n+1)!)<2$ so $1<f((n+2)!)< 2.5.$ Finally, we have $1<f((n+4)!)<2\cdot 1.25^3=3.90625$ which is a contradiction!

We first note that: $$2\cdot10^{k_1}<\frac{(n+4)!}{(n+1)!}<5\cdot10^{k_1}$$
$$1.4\cdot10^{k_2}<\frac{(n+7)!}{(n+4)!}<2\cdot10^{k_2}$$
It suffices to show that for large enough $n$, $$1<\frac{(n+5)(n+6)(n+7)}{(n+2)(n+3)(n+4)}<\frac{14}{5},$$ and all $k$ below that large enough $n$ don't satisfy the wanted. The lower bound is obvious, and the upper bound happens when $n\ge6$:
\begin{align*}
&\frac{(n+5)(n+6)(n+7)}{(n+2)(n+3)(n+4)} \\
=&\frac{n+5}{n+2}\cdot\frac{n+6}{n+3}\cdot\frac{n+7}{n+4} \\
=&\left(1+\frac{3}{n+2}\right)\cdot\left(1+\frac{3}{n+3}\right)\cdot\left(1+\frac{3}{n+4}\right) \\
<&\left(1+\frac{3}{8}\right)^3<(1.4)^3<\frac{14}{5}.
\end{align*}
It can be easily verified that $n=1,2,3,4,5$ don't work, so we are done. QED.

For $1\leq k\leq 9$, let $$(n+k)!= a_k\textbf{e}b_k$$ in scientific notation. We have that $k\leq a_k<k+1$ for all $1\leq k\leq 9$.\\

Claim: $b_k$ is an arithmetic sequence. The idea here is that the mantissa doesn't "overflow" as we multiply here. So if at any point the arithmetic progression breaks, we would either have a multiplier that is about 10 times the previous multiplier, which obviously can't happen, or we would have a multiplier that is less than the previous multiplier (which also doesn't happen due to factorial).\\

Let the common difference be $d$. Then going from the first term to the third term, the multiplier is at least $(3/2)10^{2d}$. Going from the seventh term to the ninth term, the multiplier is at most $(10/7)10^{2d}$, which is smaller, contradiction since the multiplier is supposed to increase.

Solved with [b]GeNuAlCo_pi[/b]

FTSOC, let's assume that the question statement is wrong and there exists such $n$.
Now, we get, $10\cdot 10^{x+y} > (n+9)! \geq 9\cdot 10^{x+y}$
and, $9\cdot 10^{x} > (n+8)! \geq 8\cdot 10^{x}$ for some $(x,y) \in \mathbb N^2$.
Now by dividing these two inequalities, we get $\frac{10}8 \cdot 10^y > n+9 > 10^y$.
So, $n+9$ has first digit $1$.
So, $n+1$ must have first digit $1$ or $9$.

But, $3\cdot 10^{x+y} > (n+2)! \geq 2\cdot 10^{x+y}$
and, $2\cdot 10^{x} > (n+1)! \geq  10^{x}$ for some $(x,y) \in \mathbb N^2$
Then again dividing gives, $3 \cdot 10^y > n+2 > 10^y\implies 3 \cdot 10^y > n+1 \geq 10^y$
Therefore, $n+1$ also has first digit $1$.

So, $n+1, n+2, \dots, n+9$ all have same number of digits. let the number of digits be $b+1$. Then, $$\frac{10}8 \cdot 10^b > n+9 > n+1 \geq 10^b$$
$$\implies 1.25  > \frac{n+9}{10^b} > \frac{n+1}{10^b} \geq 1.00$$
From this we get, $$1.953125 = (1.25)^3 > \frac{(n+4)(n+3)(n+2)}{10^{3b}} > (1.00)^3 = 1$$
Therefore, first digit of $(n+4)(n+3)(n+2)$ is $1$.

But using the previous trick, $5\cdot 10^{x+y} > (n+4)! \geq 4\cdot 10^{x+y}$
and, $2\cdot 10^{x} > (n+1)! \geq 10^{x}$ for some $(x,y) \in \mathbb N^2$.
And then, $$ 5\cdot 10^y > (n+4)(n+3)(n+2) > 2\cdot 10^y$$
So we get, first digit of $(n+4)(n+3)(n+2)$ is $2, 3$ or $4$. But this is a contradiction. So our assumption was wrong and the problem statement is true. $\blacksquare$


dont get the title at all

i dont feel like posting the work if I am wrong and im not 100% sure, so i'll post my answer and maybe you could tell me if its right or not :)
[hide="answer"]
1.855
[/hide]

[quote="Ignite168"]Find the altitude of the longest side of a triangle with sides 3, 5, and 7.

Bonus point: Explain the topic title.[/quote]
[hide="bonus"]title=heron's anagram[/hide]


can someone tell me how the "if", "then", "else" things work when you want to program something on a TI-84?

for example:

If $0\leq R\leq 180$
$(\cos^{-1}R)\rightarrow \theta$
If $180<R<360$
$(\cos^{-1}(R-180))\rightarrow\theta$
Disp $\theta$

why doesn't this work?

[quote="iamagenius"]If $0\leq R\leq 180$
$(\cos^{-1}R)\rightarrow \theta$
If $180<R<360$
$(\cos^{-1}(R-180))\rightarrow\theta$
Disp $\theta$

why doesn't this work?[/quote]

because you don't have "then"

"then" comes after "if"

so IF in case 1 blah blah blah THEN this will happen...IF in case 2 blah blah blah THEN this will happen

here is the program fixed:
IF $0\leq R\leq 180$
THEN $(\cos^{-1}R)\rightarrow \theta$
IF $180<R<360$
THEN $(\cos^{-1}(R-180))\rightarrow\theta$
DISP $\theta$


trigonometry
inequalities unsolved
inequalities

Prove that if $x,y,z$ are real number and $x^{2}+xy+y^{2}=3;y^{2}+yz+z^{2}=16$ then $xy+yz+zx\leq 8$
This problem is a bit strange but it's not hard  :lol:

Construct a triangle $ABC$ such that $\angle ADB=\angle BDC=\angle CDA = \frac{2\pi}{3}$, $AD=x$, $BD=y$, $CD=z$, and $D$ is inside the triangle.
Since $x^{2}+xy+y^{2}=3$ and $y^{2}+yz+z^{2}=16$; $AB=\sqrt{3}$, $BC=4$.

$[ABC]=\frac{1}{2}AB \cdot BC \cdot \sin \angle ABC = 2\sqrt{3}\sin\angle ABC \leq 2\sqrt{3}$

$[ABC] = \frac{1}{2}(xy+yz+zx)\sin \frac{2\pi}{3}=\frac{\sqrt{3}}{4}(xy+yz+zx) \Rightarrow xy+yz+zx =\frac{4[ABC]}{\sqrt{3}}\leq 8$

$QED$

:|  I think it well-known :D
We have:
\[\begin{aligned}48&=\left(x^{2}+xy+y^{2}\right)\left(x^{2}+xy+y^{2}\right)\\ &=\left[\frac{3x^{2}}{4}+\left(y+\frac{x}{2}\right)^{2}\right]\cdot\left[\left(y+\frac{z}{2}\right)^{2}+\frac{3z^{2}}{4}\right]\\ &\ge\left[\frac{\sqrt{3}x}{2}\left(y+\frac{z}{2}\right)+\left(y+\frac{x}{2}\right)\frac{\sqrt{3}z}{2}\right]^{2}\\ &=\frac{3}{4}(xy+yz+zx)^{2}\\ \Longrightarrow&\, xy+yz+zx\le 8 \end{aligned}\]


geometry
circumcircle
power of a point
radical axis
geometry unsolved

Let $ (I)$ be the incircle of $ \triangle ABC$, which is tangent to $ BC,CA,AB$ at $ D,E,F$ respectively. $ D',E',F'$ be three midpoints of three altitudes of $ \triangle ABC$. Prove that $ DD',EE',FF'$ concurs on the line $ OI$ where $ (O)$ is the circumcircle of $ \triangle ABC$.

Dear Mashimura and Mathlinkers,
this problem has been posted in 2004 by
Döhring O., Midpoint of AH, BK, CL, Message Hyacinthos # 9539 du 13/03/2004.
http://tech.groups.yahoo.com/group/Hyacinthos/message/9539
Darij Grinberg for the proof: http://tech.groups.yahoo.com/group/Hyacinthos/message/9551
Sincerely
Jean-Louis

I'm sorry, I'm in a hurry now, so I'll only give a hint for this problem.  :)

First, call $ D_1$ the intersection of $ EF$ and $ BC$, and $ D_2$ the middle point of $ DD_1$. Points $ E_2$, $ F_2$ are defined similarly. Now, please prove that $ D_2$ is the pole of $ DD'$ with respect to $ (I)$. And similarly, $ E_2$ is the pole of $ EE'$ w.r.t. $ (I)$, $ F_2$ is the pole of $ FF'$ w.r.t. $ (I)$. And then, now we have to prove that $ D_2$, $ E_2$, $ F_2$ are collinear, and so $ DD'$, $ EE'$, $ FF'$ are concurrent at the pole of $ \overline{D_2E_2F_2}$ relative to $ (I)$. In fact, they lie on the radical axis of the circles $ (O)$ and $ (I)$, therefore $ \overline{D_2E_2F_2}\perp OI$, i.e., the pole of $ \overline{D_2E_2F_2}$ with respect to $ (I)$ lies on line $ OI$. The result follows.


The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability $ \dfrac{3}{5}$ and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent.

We use casework.

[hide="Cubs win 4-0"]

$ \left(\frac35\right)^4 \equal{} \frac {81}{625}$
[/hide]
[hide="Cubs win 4-1"]
The Red Sox will have to win either Game 1,2,3, or 4. If they win Game 5, they will have already lost 4-0. Therefore there are 4 orders it could happen in.

Thus it is $ \left(\frac35\right)^4\cdot\frac25\cdot{4} \equal{} \frac {648}{3125}$[/hide]
[hide="Cubs win 4-2"]
The Red Sox must win 2 of games 1,2,3,4,5. There are $ \binom52 \equal{} 10$ ways of choosing these games.

$ \left(\frac35\right)^4\cdot\left(\frac25\right)^2\cdot10 \equal{} \frac {648}{3125}$[/hide]
[hide="Cubs win 4-3"]
The Red Sox must win 3 of 1,2,3,4,5,6. There are $ \binom63 \equal{} 20$ ways of choosing these games.

$ \left(\frac35\right)^4\cdot\left(\frac25\right)^3\cdot20 \equal{} \frac {2592}{15625}$[/hide]

Adding these up we get $ \frac {11097}{15625}$ I probably made a mistake somewhere though.

Of course, if we are being serious, we know the Cubs will never win a World Series :rotfl:

Which, based on mewto's work, ends up at:

$ \dfrac{11097}{15625}\equal{}0.710208\approx 71 \%$

1000th post!!!

The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability $\dfrac{3}{5}$ and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent.
Your Answer: 71
Solution:
There are four possible cases for the Cubs winning the World Series, depending on the number of games that Red Sox win before the Cubs win their fourth game: the Red Sox can win no games, one game, two games, or three games. In general, if the Red Sox win exactly $k$ games before the Cubs win their 4th game, there will be a total of $3+k$ games played before the last one (which the Cubs must win), there will be a total of $\dbinom{3+k}{k}$ ways of selecting the games that the Red Sox win out of those, and for each of those arrangements the Cubs will win their 4 games with probability $\left(\dfrac{3}{5}\right)^4$ and the Red Sox will win the $k$ games that are selected for them with probability $\left(\dfrac{2}{5}\right)^k$, so we are left to evaluate the expression $\dbinom{3+k}{k}\left(\dfrac{3}{5}\right)^4\left(\dfrac{2}{5}\right)^k$ for $k = 0, 1, 2, 3$. This gives us our final probability of \begin{align*}
&\dbinom{3}{0}\left(\dfrac{3}{5}\right)^4\left(\dfrac{2}{5}\right)^0 + \dbinom{3+1}{1}\left(\dfrac{3}{5}\right)^4\left(\dfrac{2}{5}\right)^1 + \\
&\qquad\qquad\dbinom{3+2}{2}\left(\dfrac{3}{5}\right)^4\left(\dfrac{2}{5}\right)^2 + \dbinom{3+3}{3}\left(\dfrac{3}{5}\right)^4\left(\dfrac{2}{5}\right)^3
\end{align*} which simplifies to \begin{align*}
&\ \ \ \ 1\cdot(.1296)\cdot1+4\cdot(.1296)\cdot(.4)\\
&+10\cdot(.1296)\cdot(.16)+20\cdot(.1296)\cdot(.064)=.7102\ldots,
\end{align*} so our answer is $\boxed{71}$ percent.

[quote=Mewto55555]
Of course, if we are being serious, we know the Cubs will never win a World Series :rotfl:[/quote]

Hehe nope

Anyways,
the way I did this was so:
[hide=Not a Solution :/]
Case 1: Win 4 in a row. There's only 1 way this can happen, so the probability is $1\left(\frac{3}{5}\right)^4$.
Case 2: Win 4 but lose 1. There are $\binom{5}{1}=5$ ways this can happen, so the probability is $5\left(\left(\frac{3}{5}\right)^4\cdot \frac{2}{5}\right)$.
Case 3: Win 4 but lose 2. There are $\binom{6}{2}=15$ ways this can happen, so the probability is $15\left(\left(\frac{3}{5}\right)^4\cdot \left(\frac{2}{5}\right)^2\right)$.
Case 4: Win 4 but lose 3. There are $\binom{7}{3}=35$ ways this can happen, so the probability is $35\left(\left(\frac{3}{5}\right)^4\cdot \left(\frac{2}{5}\right)^3\right)$.

Adding up all of these cases gives $\sim 99\%$ of the cubs winning.[/hide]

Well, I figured out what I did wrong, but of course it was after I got the question wrong. :mad:

I honestly think Wolframming this is totally fine. I did the same as the official sol(casework, and probably the only sol), and even they did the massive bash shown at the bottom.


Of $n$ points in space, no four lie in a plane. Some of the points are connected by lines. We get a graph $G$ with $k$ edges. If $G$ does not contain a triangle, show that : \[ k \leq \Big[ \frac{n^2}{4} \Big]  \]

This is Turan's theorem for triangles. See

http://www.mathlinks.ro/Forum/viewtopic.php?t=1630
http://www.mathlinks.ro/Forum/viewtopic.php?t=21527
http://www.mathlinks.ro/Forum/viewtopic.php?t=5650
http://planetmath.org/?op=getobj&from=objects&id=3037

I have not read any of these proofs, since I am currently trying to find a proof on myself (I was successful in the triangle case, but the general case is keeping me busy for some months).

  Darij