these are problems from a jump start on algebra book i have. I don't understand the different properties and am geting confused between associative,commutative and distributive properties. Please solve and post explanations so i can begin tackling some of these problems.

Use the distributive property to simply questions 1-4

1. 3 - (2+3x)
2. 2m+7-3(m+4)
3. 8d- (4d+2)
4. 5+4(5n-3) - 5n

In questions 5-11 tell if the property demonstrated in each equation is associative,commutative,distributive, or if neither side of the equation are equal to the other.

5. -1.7+15= 15+ (-1.7)
6. (20:4) : 2 =20 : (4:2)
7. 7-4+2 = 7+2-4
8. (5+n) - 2 = 5 + (n-2)
9. (5x3) x 7 = 5x (3x7)
10.  -(3n-5) = -3n+5
11. 2(4-2)= 2(4)-2(n)

[quote="DarkKnight"][quote="dotdomainz"]
Use the distributive property to simply questions 1-4

1. 3 - (2+3x)
2. 2m+7-3(m+4)
3. 8d- (4d+2)
4. 5+4(5n-3) - 5n
[/quote]

I can do these first four fast and I'll do the rest later.
[hide="1"]
$3-(2+3x)\\
\\
=3-2-3x\\
\\
=-3x+1$
[/hide]
[hide="2"]
$2m+7-3(m+4)\\
\\
=2m+7-3m-12\\
\\
-m-5$
[/hide]
[hide="3"]
$8d- (4d+2)\\
\\
=8d-4d-2\\
\\
=4d-2$
[/hide]
[hide="4"]
$5+4(5n-3) - 5n\\
\\
=5+20n-12-5n\\
\\
=15n-7$[/hide][/quote]

thank you, i am starting to understand it. The word "distributive was throwing me off. I still don't understand how to tell the property of an equation so if someone could help me with that i'd greatly appreciate it

[quote="dotdomainz"]
In questions 5-11 tell if the property demonstrated in each equation is associative,commutative,distributive, or if neither side of the equation are equal to the other.

5. -1.7+15= 15+ (-1.7)
6. (20:4) : 2 =20 : (4:2)
7. 7-4+2 = 7+2-4
8. (5+n) - 2 = 5 + (n-2)
9. (5x3) x 7 = 5x (3x7)
10.  -(3n-5) = -3n+5
11. 2(4-2)= 2(4)-2(n)[/quote]

5. It is commutative because you just flipped the -1.7 and the 15.
6. The two sides don't equal...
7. Commutative again because the -4 and the 2 are just flipped.
8. Associative property.
9. Associative property.
10. This is distributive. You multiplied 3n and -5 by -1.
11. Sides don't equal.


Find the set of reals $p$ for which $\sum_{n\ge 1}|q_{n+1}-q_n|^p$ diverges no matter what enumeration $(q_n)_{n\ge 1}$ of the rationals in $[0,1]$ we choose.

[quote="alekk"]call $a_i$ the number of $n$ such that $\frac{1}{i} > |q_{n+1}-q_n| \geq \frac{1}{i+1}$.
Then it is necessary that $\sum \frac{a_i}{i} = +\infty$ ($\sum |q_{n+1}-q_n|=\infty$ because $(q_n)$ is dense in $[0;1]$)

This shows that for $p=-1$, the series always diverges: $\sum |q_{n+1}-q_n| \geq \frac{a_i}{i+1} = +\infty$ because $\sum \frac{a_i}{i} = +\infty$
And for $p<-1$ it is easy to find examples where the series converges.[/quote]

Alekk ,i must have mixed something or anything else, but as you wrote we have
$\sum_{i=1}^{\infty} \frac{a_i}{(i+1)^p} \leq \sum |q_{n+1}-q_n| \leq \sum_{i=1}^{\infty} \frac{a_i}{i^p}$ so for $p \leq 1$ our series diverges no matter of the enumaration. And you probably meant that for $p>1$ we can find an example when it converges.
Am i right?

yep  :)  I was a bit tired this morning ..


ok, I am taking 10 people for this game. that means sign up. oh, and you might want to try some hidden contents. there is a prize at the bottom.

[hide][hide][hide]if you see this, pm me for 10000 coins[/hide][/hide][/hide][/hide]

I'll join.

FYI, people can just use the quote button to get it immeadiately.

/join

1. Butler
2. wrightboy10
3. westiepaw

so far every one who has pmed me, has joined up.

here is a hint:

W_ _ _ _
O_
T_ _ _
H_ _ _

the last one is similar to happy

joins :D

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

t does not mean the. here is a hint:

W_ _ _ _
Of
T_ _e
H_ _ _

how did you guess it james4l? anyway, for getting the answer right, you get 1000 special coins. the title is: World Of True Hope

thesignups are full. I will give you 24 hours to work out friends and rivals. and also, spaceguy has not foundthe hiddenmessage and neither has mister geek I think.

HEEEEEEEEHAAAAAAHAAAAAAA!!! I am by myself, don't even try to alliance yourself with me!!! You will pay the ultimate price!BWWAAAHAAAAA.lol  :P

ok then.
EVERYONE ALLIANCE UP TOGETHER AND GO AGAINST BUTLER!
no seriously do it  :D

hey butler want to be alliance with me? :rotfl:


[quote="mathomorphism"]I hope to see the official solution.[/quote]

[url=http://www.mathlinks.ro/viewtopic.php?t=161430]Topic[/url] appeared recently.


In a row are 23 boxes such that for $ 1\leq k\leq 23$ , there is a box containing exactly k balls.In one move we can double the number of balls in any box by taking balls from another box which has more. Is it always possible to end up with k balls in `k` th box   for $ 1\leq k\leq 23$ ?


Hi! Can you  slove this equation:
$ 8x^2\plus{}8x\plus{}1\equal{}\sqrt{x\plus{}5}$.

Square both sides and re-arrange:
$ 64x^{4}\plus{}128x^{3}\plus{}80x^{2}\plus{}15x\minus{}4\equal{}0$
Now find the roots using whatever methods you want.

[quote="JRav"]Square both sides and re-arrange:
$ 64x^{4} \plus{} 128x^{3} \plus{} 80x^{2} \plus{} 15x \minus{} 4 \equal{} 0$
Now find the roots using whatever methods you want.[/quote]

I think thanhnam2902's looking for a nice way to find the roots (similar to his other [url=http://www.mathlinks.ro/viewtopic.php?t=200717]thread[/url]). However, I can't find any nice solutions to this problem (the roots don't look nice at all).

That looks like completing the square.  Try it if you want.

Thank you ! It's a difficult equation. I tried to slove it, but I have'n got any way.


For each user, there is a comment followed by a series of stars. For example, as of today, I have the term "Hodge Conjecture" followed by a star. What is the full list of comments with their star number and the number of posts needed to achieve them? Also, where do these terms come from? I believe I have heard of the Reinman Hypothesis somewhere, but I am not sure about the others.

There are at least two threads discussing this question [i]on the first page of posts in this forum[/i].  There is also a [url=http://www.artofproblemsolving.com/Forum/search.php]search function[/url].

[quote="hm29168"]Also, where do these terms come from? I believe I have heard of the Reinman Hypothesis somewhere, but I am not sure about the others.[/quote]

They are the [url=http://en.wikipedia.org/wiki/Millennium_Problems]Millennium Problems.[/url]

Also, they're not ratings; ratings are something [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146659]entirely different.[/url]


inequalities
linear algebra
matrix
linear algebra unsolved

Given a matrix $ A\in \text{M}_n\left(\mathbb{R}\right)$. Prove that:
$ \det\left(I_n \plus{} A^{2}\right) \geq 0$.

Actualy a more general result is true :

If $ A,B\in \mathcal M_{n}(\mathbb R)$ , $ AB\equal{}BA$ , then $ \det(A^2\plus{}B^2)\geq 0$
[hide="Solution"]
$ A^2\plus{}B^2\equal{}(A\plus{}i\cdot B)(A\minus{}i\cdot B)$ $ \Rightarrow$

$ \det(A^2\plus{}B^2)\equal{}\det((A\plus{}i\cdot B)(A\minus{}i\cdot B))\equal{}\det(A\plus{}i\cdot B)(\overline{A\plus{}i\cdot B})\equal{}$

$ \equal{}\det(A\plus{}i\cdot B)\cdot \overline{\det(A\plus{}i\cdot B)}\equal{} |A\plus{}i\cdot B|^2\geq 0$
[/hide]


geometry
geometric transformation
homothety

[color=blue][i]Σε μια άσκηση από το αγγλικό φόρουμ βρήκα αυτό το λήμμα. Το κοίταξα πολύ προχειρα και δεν το κατάλαβα.
 Ας κάνουμε άνα σχήμα για να το μελετήσουμε προσεκτικά. Φαίνεται καλό συμπέρασμα.[/i][/color]

[color=blue][b]Lemma[/b]: Let $ ABC$ be a triangle and $ D$ be a point on $ CA$. $ X,Y$ are lie on $ BD,BC$ such $ AX,AY$ are isogonal line wrt $ \angle A$ and $ E,F$ be on $ BD,BC$ such $ XF\parallel YE$. Then $ AF,AE$ are isogonal line wrt $ \angle A$.[/color]
proof
Let $ P$ be a point of $ AB$ such triangle $ PXF$ and $ AEY$ are homothety with homothety center $ B$. It is obvious that $ P$ is well defined. From $ \angle PAX \equal{} \angle BAX \equal{} \angle YAC \equal{} \angle AYE \equal{} \angle PFX$,  $ A,P,X,F$ are cyclic. So $ \angle BAF \equal{} \angle PAF \equal{} \angle PXF \equal{} \angle AEY \equal{} \angle EAC$. So $ AF,AE$ are isogonal line wrt $ \angle A$.
I remember that this lemma is in crux.

[b]ΣΧΟΛΙΟ[/b]

 Αναρωτιέμαι για το εξής: Αν η ΑY τέμνει την ΒD στο Ε και φέρουμε ΧF // AEY , θα ισχύει πάλι το ζητούμενο σε αυτή την οριακή θέση ; Διότι το σχήμα άλλα λέει , αν και τα πράγματα είναι οριακά και ίσως σε αυτήν την περίπτωση να μην είναι απαραίτητο να ισχύει η πρόταση!
 Μπορεί όμως και να μου ξεφεύγει κάτι , ύστερα από τόση ώρα μάθημα.
 Μπάμπης

Δεν μπορώ να καταλάβω ακριβώς τι γίνεται αλλά σίγουρα το σχήμα δεν συμβαδίζει καθόλου με τη λύση!

Να εξηγήσω:

1)Λέει σε ένα σημείο $ \angle{YAC}\equal{}\angle{AYE}$ πράγμα το οποίο συμβαίνει μόνο όταν $ EY//AC$! (μήπως έχει παραλειφτεί αυτό στο λήμμα?...ίσως δεν ισχύει αλλιώς!)

2)Στο τέλος λέει $ \angle{AEY}\equal{}EAC$. Αυτό επίσης φανερά δεν ισχύει στο σχήμα!

Τέλος λέει ότι το εγγράψιμο είναι το $ APFX$ ενω θα μπορούσε να είναι και το $ APXF$ (milame gai kyrta tetrapleyra). Αυτό έχει επίσης σημασία γιατί χρησιμοποιείται η σχέση $ PAF\equal{}PXF$.

A btw να σημειώσω ότι το σχήμα θα ήταν αρκετά αλλιώς αν $ \angle{BFX}>90$ (νομίζω ότι αλλάζουν πολλά στις γωνίες που χρησιμοποιούνται σε αυτήν την απόδειξη)

Δημήτρη , δες όλο το σχετικό θέμα και εξέτασε τι ακριβώς γίνεται :

 http://www.mathlinks.ro/viewtopic.php?t=193904

 Ίσως έχει ενδιαφέρον.

 Μπάμπης


One hundred people reception where only coffee and doughnuts are served.

Sixty people drink coffee, 50 eat doughnuts and 10 neither drink coffee nor at doughnuts.

How many people had both items on the menu (coffee and doughnuts)?

 :cool:

[hide]
There are 100 people and 10 people not eating so there are 90 people eating.
$60+50-x=90$ where x is the number of people eating both.
$110-x=90$
$x=20$
[/hide]

[hide]twenty

100 minus the 10 people eating neither is 90. 50 + 60 is 110. 110-90 is 20![/hide]

[hide]The number of people eating or drinking something is 90.
60+50 is 110 things consumed, but only 90 people were there to have something, so
110-90 = [color=blue]20[/color][/hide]

[hide] 60-x   +   50-x   +   x   = 110-x  = 90.  Thus, x=20 and 20 people have both coffee and donuts on their list.[/hide]

[hide]20!  I'd have eaten 5 doughnuts  :P
[/hide]


Hi there!

If $ A$ and $ B$ are independent events ($ {\text{P}}\left({A \cap B}\right) = P\left( A \right) \cdot P\left( B \right)$) and the same may be said about events $ B$ and $ C$, are $ A$ and $ C$ independent?

[quote]
Suppose we are rolling two standard 6-sided dice. One is red, and one is blue. Let $ A$ be the event that an odd number comes up on the red die. Let $ B$ be the event that a 1 comes up on the blue die. Let $ C$ be the event that a multiple of three comes up on the red die. It's clear $ B$ is independent from both $ A$ and $ C$, but $ A$ and $ C$ certainly aren't independent. So independency is not transitive.
[/quote]

Thanks for the reply, MellowMelon, but I´m afraid $ P\left( C \right) = \frac{2}{6}= \frac{1}{3}= P\left({\left. C \right|A}\right)$ , therefore in your example $ A$ and $ C$ ARE independent!

A is independent of B
Then, B is independent of A
If independence were transitive, then A would be independent of A, which isn't true.

On the other hand, dependency IS transitive.

[quote="The Zuton Force"]A is independent of B
Then, B is independent of A
If independence were transitive, then A would be independent of A, which isn't true.[/quote]

A bit of clarification:  there are several properties that relations on two elements can satisfy.

[b]Reflexivity:[/b]  $ a \approx a$.  Independence is not reflexive.  Dependence is.
[b]Symmetry:[/b]  $ a \approx b \Leftrightarrow b \approx a$.  Independence and dependence are both symmetric.
[b]Transitivity:[/b]  $ a \approx b, b \approx c \implies a \approx c$.  Independence is not transitive (The Zuton Force has a nice demonstration of this), and dependence is.

A relation that satisfies all three properties is said to be an [b]equivalence relation.[/b]  Examples are equality and equality $ \mod m$.

Thanks for joining, "The Zuton Force".

Good argument, but a small detail:$ A$ is independent of $ A$ if (and only if) $ P\left( A \right)$ is zero or one.

Anyway (even after "case closed"), could you (or someone else) show me an "explicit" , "direct" , "concrete" (as MellowMelon´s) example of non-transitivity, please?

Suppose we have three people that are considering going to the mall.
Person A will go to the mall iff. person B will go to the mall. Person C does not know either A or B, and decides independently.

Then $ P(A+C)=P(A)P(C)$ and $ P(C+B)=P(C)P(B)$, but $ P(A+B)=P(A) > P(A)^{2}= P(A)P(B)$

You can make it even more definite by saying A and B will go iff. it is sunny (50%), and C will go iff. he has enough gas money (25%).

If you want a slightly less trivial example, consider the case when A will go iff. it is sunny (33%) and B will go iff. it is not raining (67%).

Excellent, The Zuton Force.

I really thing the whole thing is pretty clear to me now.

Thanks a lot.

P.S.: just one thing (as Ltn. Columbo used to say...) In your last (the less trivial) example, it seems to me that you violated the "smart choice" of $ A$ and $ B$ such that $ A$ occurs iff $ A+B$ occurs... in other words, is it still true that $ P(A+B)=P(A)$ if 33% is the "sunny" (A) probability and 67% is the "not raining" (B) one?
(There is no pun intended, I am only trying to understand your argument.)


logarithms
algebra
function
domain
complex analysis

I am not sure if this is an appropriate question. But I came up with the problem:

"Find the branch of $ f(z) = \ln(\ln z)$".

I understand this to means "what need to be removed" to make $ f(z)$ continous. But it really gets messy. And it doesnt seem like the answer is a single line.

Within your interpretation of the problem, all you need is to find a simply connected domain not containing $ 0$ and $ 1$. I would just make a cut from $ -\infty$ to $ 1$ but you have plenty of other options as well.

In fact, by "plenty of other options", you have infinitely many! Two other common cuts (although not nearly as convenient of this specific function -- but you may consider for others) would be from $ 0$ to $ \infty$ OR cut from $ -\infty$ to $ 0$ then also $ 1$ to $ \infty$. In fact, you'll see this last one a lot for making single-valued analytic continuations of functions where you must cut two points. Also note that you could do anything crazy such as cutting from $ 0$ to $ 1$, then cutting along $ g(y) = 1+y^{2}i$, $ y \in \mathbb{R}^{+}$ Maybe this helps?

This function has an interesting nested structure.  The plot below shows $ Im[Log[Log[z]]]$ fracturing (if that's a good enough term) into an infinite number (3 of which are shown) of sub-sheets for each sheet of Log[z].  If I define:

\begin{align*}LogPQ_{n,k}\equiv Log[Log[z]]&=ln\left[Abs[(ln(Abs(z))+i(Arg[z]+2k\pi)\right] \\ &+i\left[Arg[ln(Abs(z))+i(Arg[z]+2k\pi)]+2n\pi\right] \end{align*}

and select the principle sheet (n=0) and plot points on three sub-sheets (k=0,1,2), I get the plot below.

The red dot is $ Im[LogPQ_{0,0}[Log[3+3i]]]$, the green dot is $ Im[LogPQ_{0,1}[Log[3+3i]]]$, and the yellow, $ Im[LogPQ_{0,2}[Log[3+3i]]]$

Then I wonder if this infinite sum converges for all sheets of $ LogPQ$:

$ \sum_{k=0}^{\infty}\Delta Im\left(LogPQ_{0,k}[x+iy]\right)$

where the delta is just the difference between succeeding sub-sheets (red-green)+(green-yellow)+...

Here's the first 25 terms for $ z=3+3i$:

{0.871317, 0.0938531, 0.0343493, 0.0177683, 0.0108523, 0.00731484,
0.00526408, 0.00396951, 0.00310011, 0.00248809, 0.00204101, 0.00170448,
0.00144483, 0.00124031, 0.00107635, 0.000942876, 0.00083278, 0.000740903,
0.000663434, 0.000597512, 0.00054095, 0.000492056, 0.000449504, 0.000412243,
0.000379431, 0.000350386}

[img]http://img253.imageshack.us/img253/6745/loglogzfd4.jpg[/img]


I'm trying to see what the point of this problem is. Surely it's not just meant to be a calculation?  Is it meant to be done without a calculator?

[quote="shobber"][quote="Mathx"]Would you use LaTex please .I understood nothing.[/quote]
Isn't the graph clear enough!?
Find:
\[\sqrt[3]{\sqrt{\sqrt[3]{6860}-19}}\][/quote]

No, myth is right, there is nothing to understand/find here. There's no shorter way to write down those roots either.

So if it's supposed to be hard, the problem statement is wrong I guess...

[quote="ihsan_einstein"]I don't want about solution , I  want  certanly answers. I know answers. But I don't know solutions.
please answer the question.

ihsan[/quote] :?:

I can't follow you...

I think we should put  $\sqrt[3]{\sqrt{\sqrt[3]{6860}-19}}=x$

[quote="Peter VDD"][quote="Mathx"]Would you use LaTex please .I understood nothing.[/quote]
No, myth is right[/quote]

Is Mathx = Myth ? :huh:

Pierre.

Too bad, I would be pleased to see Myth again...

Pierre.

[quote="ihsan_einstein"]I don't want about solution , I  want  certanly answers. I know answers. But I don't know solutions.
please answer the question.

ihsan[/quote]
If you have either answer or solution, please post it.

BTW. Myth should have read some of the posts about his sudden disappearence, why he doesn't give us an answer?
BTW2. I thought he was Myth at first too! But not for his username, but his avatar!

That is probably what made me confuse too.... it is kind of a coincidence, "MaTHx" with the same avatar as "MyTH"? ;)

btw, a=1, x=24, y=25 --> this is too much for coincidence :P

[quote="shobber"][quote="Mathx"]Would you use LaTex please .I understood nothing.[/quote]
Isn't the graph clear enough!?
Find:
\[\sqrt[3]{\sqrt{\sqrt[3]{6860}-19}}\][/quote]

No it's not clear enough! You can't see it when not logged in, so I'm against such postings!

It is also quite strange to write $\sqrt[3]{\sqrt{\ }}$ instead of $\sqrt[6]{\ }$

What is the use of this topic ,Dairmuid find the best thing possible with out calculation.
[quote="Diarmuid"]If the objective is to do it without a caclulator, then we can at least make some progress:
$\sqrt[3]{\sqrt{\sqrt[3]{6860}-19}}=\sqrt[6]{\sqrt[3]{6860}-\sqrt[3]{6859}}$.

But $\sqrt[3]{6860}-\sqrt[3]{6859}=\frac{\sqrt[3]{6860}-\sqrt[3]{6859}}{6860-6859}=f'(c)$ where $f(x)=\sqrt[3]{x}$ and $6859\le c\le6860$.  This gives $\sqrt[3]{6860}-\sqrt[3]{6859}$ accurate to several significant figures as $\sqrt[3]{6860}-\sqrt[3]{6859}\approx\frac{1}{3\times6859^\frac{2}{3}}=\frac{1}{3\times19^2}=0.000923$.

I'm not sure how you'd go about finishing it off.[/quote]
Ehsan,What do you mean you want $\text{just answer}$,you mean just a number.
Well I think there is no use of just a number for answer is it?

[quote="NoName"]You can simplify it to
\[
\sqrt[3]{\sqrt{\sqrt[3]{6860} - 19}} = \frac{\sqrt[3]{20} - 1}{\sqrt[3]{12}}
\][/quote]that is not true... :?

The $1$ in the numerator should have been a $2$.

It seems like everybody started to enjoy this beautiful problem which is indeed a very strong result and has a huge impact on the whole algebra  :P  ;)

Ravi B is right, the result can be writen in a nice form
$\frac {\sqrt[3]{5}-\sqrt[3]{2}} {\sqrt[3]{3}}$.


Prove that the equation [tex]x^2+y^2+8z^2=2t^2[/tex] has infinitely many integer solutions.

Or t = x = 3y = 3z, and so on...

Pierre.

I would like to see as many solutions to this equation as possible. :D


There are nine unit triangles in the three rows of this
triangle. If you add three more rows to the bottom of this
triangle so that each row has 2 more unit triangles than
the previous row, how many unit triangles will there be?

[asy]draw((0,0)--(15,26)--(30,0)--cycle);
draw((5,8.7)--(10,0)--(20,17.3));
draw((20,0)--(25,8.7)--(5,8.7));
draw((20,0)--(10,17.3)--(20,17.3));[/asy]

Since there are 3 rows so far, adding 3 more rows will make 6 rows in total.
Summing the first 6 odd numbers (number of triangles in each row),
We have 1+3+5+7+9+11=[b]36[/b]

Sum of the first $ n$ odd numbers is $ n^2$.


hello,
is there any spcefic formula or relation 4 all positive integers $N$ which can b represented as a sum of the squares of 3 positive integers. eg 3,6,9...
pls help
its kinda urgent
thx

All number $N\not =(8k+7)4^{m}$ can be represent in form $N=a^{2}+b^{2}+c^{2}.$

This is not what was asked for ;)
Read the problem again!

can we find any relation between all natural nos $N$ st they cannot be represented as the sum of squares of 3 $non negative$ integers?

Those that are sum of three general squares, so with $0$ allowed, are those Rust mentioned.


calculus
integration
trigonometry
logarithms
calculus computations

Hey guys, I forgot to bring home the book for the course, so I don't have the exact problems that I assigned you. But attached below is a sheet with solutions to some of the odd problems from that page.

Feel free to post the problems here for discussion, or to ask questions about your work.

-----------------------------------------------------------------------------
Edit: Aja! I found the book. I had just left it in my car!

So here are the solutions to the problems that I assigned:

6.) $ \int \frac {10}{x^2\sqrt {25 \minus{} x^2}} dx$. Let $ x \equal{} 5 \sin \theta$ so $ dx \equal{} 5 \cos \theta d\theta$.

Then the integral becomes $ \int \frac {10}{25\sin^2 \theta \cdot 5\cos \theta}(5\cos \theta d\theta) \equal{} \frac {2}{5} \int \frac {d\theta}{\sin^2 \theta} \equal{} \frac {2}{5} \int \csc^2 \theta d\theta$

$ \equal{} \minus{} \frac {2}{5} \cot \theta \plus{} C \equal{} \minus{} \frac {2}{5} \cot \left( \sin^{ \minus{} 1} \left( \frac {x}{5} \right) \right) \plus{} C \equal{} \boxed{ \minus{} \frac {2}{5} \frac {\sqrt {25 \minus{} x^2}}{x} \plus{} C}$
-----------------------------------------------------------------------------

13.)$ \int x\sqrt {1 \plus{} x^2} dx$. Let $ x \equal{} \tan \theta$ and $ dx \equal{} \sec^2 \theta d\theta$.

Then the integral becomes $ \int \tan \theta \cdot (\sec \theta) \cdot (\sec^2 \theta d\theta) \equal{} \int \tan \theta \cdot \sec^3 \theta d\theta$.

Let $ u \equal{} \sec \theta$ so $ du \equal{} \sec \theta \tan \theta$, and the integral becomes $ \int u^2 du \equal{} \frac {1}{3}u^3 \plus{} C \equal{} \frac {1}{3}\sec^3 \theta \plus{} C \equal{} \boxed{\frac {1}{3}\left[\sqrt {x^2 \plus{} 1}\right]^3 \plus{} C}$

Note: This integral could, of course, be more easily solved by a u-substitution at the beginning.
-----------------------------------------------------------------------------

15.)$ \int \frac {1}{(1 \plus{} x^2)^2} dx$. Let $ x \equal{} \tan \theta$, $ dx \equal{} \sec^2 \theta d\theta$.

Then the integral becomes $ \int \frac {1}{\sec^4 \theta} \sec^2 \theta d\theta \equal{} \int \cos^2 \theta d\theta \equal{} \int \frac {1 \plus{} \cos(2\theta)}{2} d\theta \equal{} \frac {\theta}{2} \plus{} \frac {1}{4}\sin(2\theta) \plus{} C$

$ \frac {\theta}{2} \plus{} \frac {1}{4}(2\sin \theta \cos \theta) \plus{} C \equal{} \frac {1}{2}\tan^{ \minus{} 1}x \plus{} \frac {1}{2}\left(\sin(\tan^{ \minus{} 1}x) \cos(\tan^{ \minus{} 1}x)\right) \plus{} C$

$ \equal{} \boxed{\frac {1}{2}\tan^{ \minus{} 1}x \plus{} \frac {1}{2} \cdot \frac {x}{x^2 \plus{} 1} \plus{} C}$
-----------------------------------------------------------------------------

21.)$ \int \frac {x}{\sqrt {x^2 \plus{} 9}}dx$. Let $ x \equal{} 3\tan \theta$ and $ dx \equal{} 3\sec^2 \theta d\theta$.

Then the integral becomes $ \int \frac {3\tan \theta}{3\sec \theta} \cdot (3\sec^2 \theta d\theta) \equal{} 3\int \tan \theta \sec \theta d\theta \equal{} 3\sec \theta \plus{} C$

$ \equal{} 3\sec (\tan^{\minus{}1}x) \plus{} C \equal{} \boxed{\sqrt{x^2\plus{}9} \plus{} C}$

Note: Again, this could have been done more easily by a u-substitution at the beginning.
-----------------------------------------------------------------------------

27.)$ \int \frac {1}{\sqrt {x^2 \minus{} 9}}dx$. Let $ x \equal{} 3 \sec \theta$ and $ dx \equal{} 3\sec \theta \tan \theta d\theta$.

The integral becomes $ \int \frac {1}{3\tan \theta}(3\sec \theta \tan \theta d\theta) \equal{} \int \sec \theta d\theta \equal{} \ln|\sec \theta \plus{} \tan \theta| \plus{} C$

$ \equal{} \ln \left|\frac {x}{3} \plus{} \tan \left( \sec^{ \minus{} 1}\frac {x}{3} \right)\right| \plus{} C \equal{} \ln \left|\frac {x}{3} \plus{} \frac {\sqrt {x^2 \minus{} 9}}{3}\right| \plus{} C \equal{} \boxed{\ln |x \plus{} \sqrt {x^2 \minus{} 9}| \plus{} C_1}$
-----------------------------------------------------------------------------

47.)$ \int_0^\frac {\sqrt {3}}{2} \frac {t^2}{(1 \minus{} t^2)^{3/2}}dt$. Let $ t \equal{} \sin \theta$, $ dt \equal{} \cos \theta d\theta$.

The integral becomes $ \int_0^{\frac {\pi}{3}} \frac {\sin^2 \theta}{\cos^3 \theta}(\cos \theta d\theta) \equal{} \int_0^{\frac {\pi}{3}} \tan^2 \theta d\theta \equal{} \int_0^{\frac {\pi}{3}} (\sec^2 \theta \minus{} 1) d\theta$

$ \equal{} \tan \theta \minus{} \theta \Big|_0^{\frac {\pi}{3}} \equal{} \boxed{\sqrt {3} \minus{} \frac {\pi}{3}}$

The problems from the mixed review sheet that I gave today in class -- some of them are much more tedious/harder than the ones I will ask you to do. In particular, numbers 151, 153, 155, 159, 161, 165, 166, 168, 172, 174 (should be okay with this hint: let $ u \equal{} y^2$), 176, 177, 184, 185, 188, and 194.

You can try them for fun! But you will not be responsible for them in this class.


[quote="Tony2006"]prueba que:

$\sum_{k=1}^{n}(-1)^{k-1}\binom{k}{1}\binom{n-k}{k}2^{n-2k}= \binom{n+1}{3}$[/quote]
ESTAS SEGURO QUE ES ASI, EN LA SUMATORIA, POR Q MIRA KE PASA SI $n=k$, o si no prueba con valores pequeños y no cumple.
chau


geometry
3D geometry
pyramid
combinatorics unsolved
combinatorics

Is space is given a set S with 100 points such that no four points are coplanar. Prove that between the triangular pyramids with vertexes in the points of set S exist a maximum of 101x4x101 that has one or two vertexes in common.

So the triangular pyramids or have one vertex in common or one side in common!


algebra
polynomial
algorithm
Galois Theory
algebra unsolved

Find all real solution to the equation $x^5-5x+1=0$

The polynomial is not solvable, so it's not possible to express the zeros in terms of roots etc.
(there are three zeros, just use some algorithm to get them, e.g. Newtons)

What are general criterions to check quickly whether a polynomial is solvable or not? (I'm not asking for definition of solvability)

One criterion is (prove uses some Galois theory):
An irreducible polynomial of prime degree $p$ that has at least $2$ real and one complex solution is not solvable.
The prove works by showing (using some facts on permutations) that when the equation is solvable, the splitting field of such a polynomial is generated by any two of it's roots. But when taking two real roots, the splitting field would be purely real, contradicting the existence of a complex root.

I think the quickest here is Eisensteins criterion

Irreducibility alone helps nothing when trying to show solvability!
E.g. $x^2+1$ is irreducible, but also solvable.

We have showed it has no rational roots. That kinda helps!


Let $m$ be a natural number.[b] Prove that for each rational number $r, \; r>0\; ,$ there exists a positive integer  $s$  such that
\[ r= \frac{1}{m_1}+ \frac{1}{m_2}+ ... + \frac{1}{m_s}  \]

where $m_1,m_2,...,m_s$   are positive integers  with   $m<m_1<m_2 < ...< m_s \; .$[/b]

Even though the problem was a different one, a solution for this one can be read between the lines of post number $3$ in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Egyptian+fractions&t=16892]this thread[/url]. :)

I tried to solve it again without looking there, but couldn't :).


I'm sorry for posting so many problems- I only have 2 more to do, and I just need a little push! haha

I should know this one, I just can't figure out why I can't get it...
 Identify the kind of graph resulting from the equation:
        x^2 + 5y^2 + 16 = 0

I'm not sure what to do... is this correct?
        x^2 + 5y^2 = -16
   (then divide everything by -16 so the eq. = 1)

    I'm stuck after that.....

And here's another question:
 Solve graphically. 2^x = 2ln(x+3)   express your solution to the nearest tenth!

the first question doesnt seem to work on a regular cartesian coordinate... sum of square cannot be negative... so i would assume that you should use complex plane... :?:

question 2, graph both:

y=2^x
y=2ln(x+3)

then see where they meet. that is what you are looking for.

thanks sooo much!!

This is all review for me, and just the way they are worded tends to confuse the heck outta me!

Thanks!

[quote="frodo"][quote="Jenybeny27"]I should know this one, I just can't figure out why I can't get it...
 Identify the kind of graph resulting from the equation:
        x^2 + 5y^2 + 16 = 0
[/quote]

Doesn't the equation represent an ellipse?[/quote]

Not quite because there are no real coordinates that would satisfy it. It doesn't represent anything in the cartesian plane. However, if you change that to a $-16$ you get an ellipse.

[quote="Treething"]Is it meant to be graphed in the imaginary plane? With the a + bis and all that crap?[/quote]

Then you would need the real and imaginary parts of both x and y, resulting in 4 dimensions I think.

would that even be possible to coordinate??? I knew right away it had to do with imaginary, but i just wasnt sure after i said complex plane because it only contains y value of imaginary.


[hide="My sorution"]
well so the area of the whole circle is $1200$ im thinking. so if you let the radius of the circle be $r$ and do
\[\pi r^{2}=1200 \Rightarrow r = \frac{\sqrt{1200}{\pi}}\]
And if you find the area of the equilaterial triangle in terms of r you should get around $\frac{3\sqrt{3}}{4}r^{2}$ so the area of the equilateral triangle is
\[{\frac{3\sqrt{3}}{4}{(\frac{\sqrt{1200}}{\pi}})}=\boxed{\frac{45}{\pi}}\][/hide]

[quote="NeverOddOrEven"][hide="My sorution"]
well so the area of the whole circle is $1200$ im thinking. so if you let the radius of the circle be $r$ and do
\[\pi r^{2}=1200 \Rightarrow r = \frac{\sqrt{1200}{\pi}}\]
And if you find the area of the equilaterial triangle in terms of r you should get around $\frac{3\sqrt{3}}{4}r^{2}$ so the area of the equilateral triangle is
\[{\frac{3\sqrt{3}}{4}{(\frac{\sqrt{1200}}{\pi}})}=\boxed{\frac{45}{\pi}}\]
[/hide][/quote]

You didn't square r.

[quote="bpms"][quote="Go Around the Tree"]The area of a 30 degree sector of a circle is 100 square units. What is the area of an equilateral triangle inscribed in this circle?[/quote]
[hide]The area of the circle is 1200. the radius is $\frac{20 \sqrt{3 \pi}}{\pi}$ The radius is two thirds the side height of the triangle. SO the height is $\frac{30 \sqrt{3 \pi}}{\pi}$ So the side is $\frac{60 \sqrt{ \pi}}{\pi}$ . Which makes the area $\frac{900 \sqrt{3}}{\pi}$[/hide]
I'm 99% sure there is an elegant way, this went pretty well, I'll edit in the elgeant way if I find it.[/quote]

Shouldn't it be all over $\pi^{2}$?

Oh yeah never mind ur right. I said that because I assumed that NeverOddorEven's solution had the correct radius.

[hide="A solution, but to a different problem"]Actually, the correct answer is $A={150(2\sqrt{3}-3)\over\pi}$[/hide]

[quote="Farenhajt"]Actually, the correct answer is $A={150(2\sqrt{3}-3)\over\pi}$[/quote]

Um, why?

Oh, sorry :!:  :(  I misread the problem - what I've given is the area of the triangle inscribed in the [b]sector[/b], not in the circle. Disregard it. (${900\sqrt{3}\over\pi}$ is correct.)


search
inequalities unsolved
inequalities

a^4+b^4+c^4=3

We have to prove
1/(4-ac) + 1/(4-bc) + 1/(4-ab) <= 1

Can someone help me ?


Find the smallest number $\overline{xyz}$ for which:
$\overline{xyz}=4xyz$
By the way are these kind of problems still being given on the math contests !!?!?!? :huh:

[quote="Gohan"]
By the way are these kind of problems still being given on the math contests !!?!?!? :huh:[/quote]

Certainly not in the Bulgarian MO or something like that.

They do appear in "easy" olympiads though!

[quote="Arne"]
Certainly not in the Bulgarian MO or something like that.

They do appear in "easy" olympiads though![/quote]

Well it was given on a Bulgarian MO. Dont remember when exactly but it was between 1950-1970 :D .
Anyway ... does anybody has a solution for this problem because mine is very very very ugly :(

[quote="Gohan"] Well it was given on a Bulgarian MO. Dont remember when exactly but it was between 1950-1970 :D.Anyway ... does anybody has a solution for this problem because mine is very very very ugly :([/quote]

Well, no! First we learn proper topic naming and exact source specification!  :)

[quote="orl"]
Well, no! First we learn proper topic naming and exact source specification!  :)[/quote]
This could take a while..  :D

This is way too easy for an Advanced Section problem. Moving off to Pre-Olympiad :)


How many Februarys since 1900 have had five Saturdays?

(1) 4
(2) 5
(3) 6
(4) 7
(5) None of the above

Well the only way for there to be 5 Saturdays the Feburary must start on Saturday and be in the leap year.  Then we need to find a pattern for the first day of that particular Feburary.  However, i went through a calendar and it does not look like there is really any pattern.  Once it repeats a day you would think that it would repeat the pattern but it does not.  I just went through the calendar looking every leap year for a Feb starting with Saturday and found 4.  However i am not sure if the present calendar was in place at that time, meaning that all the leap years were in and other special cases.

[quote="demji"]Actually looking now i c that it repeats in the pattern sat, thurs, tues, sun, fri, wed, mon, sat.....[/quote]

so once every seven years, so 7*(2005-1900) then, correct?

[quote="demji"]Well the only way for there to be 5 Saturdays the Feburary must start on Saturday and be in the leap year.[/quote]

[hide]start from this year (2005), because we don't know what day feburary in 1900 started on
this year, Feburary started on a tuesday. 365/7 has a remainder of 1. 366/7 has a remainder of 2. so every year, the days move up one slot, except for leap years, when they move two.
so since 2004 was a leap year, tuesday was moved up from last year 2 slots, so Feburary started on a sunday. before that, 2003 started on saturday, 2002 started on friday, etc.
now, counting the number of years that start w/ a saturday, we have 2003, 1997, 1992, 1986, 1975, 1969, 1964, 1958, 1947...
there is a pattern of differences shown here
2003-1997 = 6
1997-1992 = 5
1992-1986 = 6
1986-1975 = 11
1975-1969 = 6
1969-1964 = 5
1964-1958 = 6
1958-1947 = 11
the pattern is repeated every 6+5+6+11 = 28 times

we also know that when a year is divided evenly by 4, it is a leap year. so far we have two answers, 1992 and 1964. we keep subtracting 28 to get more answers. 1936 and 1908

so the correct answer is (1) 4.

that was way too much work... but other methods i tried just didnt seem as justified[/hide]

OA is A.

But I am still not sure how to approach this problem given that the question does not state what day February started on 1900 or 2005 or 2004, unless it assumes that we know a handy trick.


calculus
integration
trigonometry
analytic geometry
calculus computations

Substitute $ x\equal{}\varphi\, ,\, y\equal{}r^{4}$ then $ I\equal{}\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\frac{4r^{3}\, dr\, d\varphi}{\sqrt{1\minus{}r^{2}\sin^{2}\varphi}}$ .

Shift from polar coordinates to Cartesian coordinates
$ x\equal{}r\cos\varphi$ , $ y\equal{}r\sin\varphi$ , $ x^{2}\plus{}y^{2}\equal{}r^{2}$ ,  $ r\, dr\, d\varphi\equal{}dx\, dy$

$ I\equal{}4\int_{0}^{1}\int_{0}^{\sqrt{1\minus{}y^{2}}}\frac{x^{2}\plus{}y^{2}}{\sqrt{1\minus{}y^{2}}}dx\, dy\equal{}4\int_{0}^{1}\frac{1}3\left(1\minus{}y^{2}\right)\plus{}y^{2}\, dy\equal{}\frac{20}{9}$

It is worth mentioning that, based on Misan's integral and Dr. Doe's calculations,  the following interesting series holds:

$ \sum_{n\equal{}0}^{\infty}\frac{((2n)!)^{2}}{16^{n}(n!)^{4}(n\plus{}2)}\equal{}\frac{20}{9\pi}$.

Let $ I$ be the value of the integral. Using the substitution $ y\equal{}t^{2}$ we get that

$ I\equal{}\int_{0}^{1}\int_{0}^{\pi/2}\frac{2tdtdx}{\sqrt{1\minus{}t\sin^{2}x}}\equal{}2\int_{0}^{1}tdt\int_{0}^{\pi/2}\sum_{n\equal{}0}^{\infty}\frac{\Gamma(n\plus{}1/2)}{\Gamma(1/2)n!}t^{n}\sin^{2n}xdx$

$ \equal{}\sum_{n\equal{}0}^{\infty}\frac{\Gamma(n\plus{}1/2)}{\Gamma(1/2)n!}\beta(n\plus{}1/2,1/2)$,

and we get after straightforward calculations that

$ I\equal{}\sum_{n\equal{}0}^{\infty}\frac{\Gamma^{2}(n\plus{}1/2)}{(n!)^{2}(n\plus{}2)}\equal{}\sum_{n\equal{}0}^{\infty}\frac{(2n!)^{2}\pi}{(n!)^{4}16^{n}(n\plus{}2)}\equal{}\frac{20}{9}$,

and the result follows.


Three positive integers have an arithmetic mean of 26 and a median of 27. If the median is 5 less than the largest integer, what is the smallest integer?

the mean is the numbers added together and divided by the number of numbers

so...

the mean is $ 26$, so now we find the sum

$ 26 \times 3 \equal{} 78$
now we subtract the median and the largest integer, which is $ 5$ more than the median

so $ 78 \minus{} 27 \minus{} 32 \equal{} 19$

ans is $ \boxed{19}$

Better yet, since you know 27 is one of the numbers, and 27 + 5 = 32 is the second number, and 26 is the average, you can do this:

32 is 6 above 26
27 is 1 above 26
So the third number must be (6+1) = 7 below the average to compensate.

26-7=19.

Dude nice solution iin77.

thats cool and clever


A certain number $ n$ has factors of 15 and 10.  What is the fewest total number of positive factors that $ n$ can have?

we basically want to find the number of factors in the lcm of 10 and 15. The lcm is 30 and their are 8 factors of 30.


rasti alaan khoondan ke hossien rezazadeh dobare ghahraman shode :first:  :clap:

heyf ke inja andazeye kafi emiticon nadare

[quote="mmahdi"]salam
man az aadam haye bozorg mitarsm(alaan shoma migin:"khob chikaret konam?" [/quote]

[quote]"khob chikaret konam?" [/quote]

 :P


Not so.  For example, the tower converges for, say, $x = 1.1$, to approximately $1.11178201104...$

[hide="The answer"] $x > e^{ \frac{1}{e}}$ [/hide]

[quote]Not so. For example, the tower converges for, say, x = 1.1, to approximately 1.11178201104...
[/quote]
 Or classic $\sqrt{2}^{{{\sqrt{2}}^{\sqrt{2}}}^{\dots}}= 2$

I was trying to remember that one, but had forgotten it  :P

it is also equal to four. in fact for all values less than e, it has exactly two values. try to prove that for an extra credit! :D

[quote="t0rajir0u"][quote="rishabh"]it has exactly two values.[/quote]

Nonsense.  We are considering the limit of the sequence

$x_{1}= x$
$x_{n+1}= x^{x_{n}}$

And limits have only one value.[/quote]

Exactly!  (One can easily show, for the $\sqrt 2$ case,  that for all $n$, $x_{n}\le 2 \implies x_{n+1}\le 2$)

 One "trick" question we  often asked (when we wanted to be  a little  mean to students)   was to ask them to find $a^{a^{a^{a^{\dots}}}}$ when $a=\sqrt[3] 3$, (  the obvious answer $3$ is, of course, a  wrong answer) .

Very similar to  that the value of
${x=1+\frac{1}{1+\frac{1}{1+\dots}}}$
is positive  (greater than 1) and [b]not the negative [/b]root of $x = 1+1/x$.

[quote="t0rajir0u"][hide="The answer"] $x > e^{ \frac{1}{e}}$ [/hide][/quote]

I agree with everything said above by both of you, though I'm still wondering how exactly to figure out that the limit tends to infinity when $x$ is greater than that value.

Thanks for your help!  :D

[quote="cincodemayo5590"]For what $x$ does $x^{x^{x^\cdots}}$ tend to infinity?[/quote]
What about complementation:
For what $x$ does $x^{x^{x^\cdots}}$ tend to finite value?
In other words - what is behavior of this limit if $x\in (0,1)$?  :P

Refer to the last 6 or so posts in:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=tower+power&t=77942[/url]
especially the second to last one (its by Gauss202).

There is, in fact, a lower limit on the power tower as well - numbers extremely close to zero will not cause the series to converge; rather, alternate terms will converge on different numbers. As $x$ approaches zero, the sequence $x, x^{x}, x^{x^{x}},$ etc... becomes more and more similar to $0, 1, 0, 1, 0, 1, 0, 1, ...$

[quote="gauss202"]
[hide]$[1/e^{e}, e^{1/e}]$[/hide][/quote]
Yes, exactly.
[hide]If $0 < x < e^{-e}$ then the sequence has 2 different limit points - one  for odd numbers and an other for even. If $x>e^{\frac{1}{e}}$ it tends to $\infty$[/hide]


calculus
derivative
logarithms
calculus computations

Does anyone know how to work out the differential of the following, could you tell me the identities you used in your solution??

[b]diff[/b](1/[b]ln[/b]2){(1-x)[b]ln[/b]((1-x)/(1-y)) + x[b]ln[/b](x/y)} [b]w.r.t y[/b]

Thankyou

hello,do you mean $ \frac{d}{dy}\left(\frac{1}{\ln(2)}(1\minus{}x)\ln\left(\frac{1\minus{}x}{1\minus{}y}\right)\plus{}x\ln\left(\frac{x}{y}\right)\right)$?
Sonnhard.


I am just starting these kinds of problems.  What is the best method to start to attack this problem?  How do we go from:

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

to

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


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

What is the group associated with the polynom $ t^3\minus{}3t\plus{}1$?

 :o

er, the group that consists of permutations

Permutations of ...¿ Well, probably really the Galois group is intended. Then you may start finding the group order.

That is not very precise. Permutations of what?

EDIT: If it's the Galois group, is it the Galois group associated with one of the roots, or the Galois group of the splitting field? And over what field? Q?

You can only associate Galois groups for splitting fields (well, in fact, to normal fields). There is for example no Galois group for $ \mathbb Q[\sqrt[3]2] | \mathbb Q$.

What kind of grouP? The group of the polynom over the field $ \mathbb{Q}$, which is the set of all permutations which preserve the validity of any equation over $ \mathbb{Q}$ of the form $ P(t_1,t_2,t_3,t_4)\equal{}0$ where the $ t_i$ are the zeros of the polynomial. Is this a Galois group?

Los

[quote="Los"] Is this a Galois group?[/quote]
Let $ K$ be a field and $ f(x)$ be a non-constant polynomial over $ K[x]$. Let $ F$ be the splitting field of $ f(x)$ over $ K$. Then $ \text{Gal}(F/K)$ is the [i]Galois group of $ F$ over $ K$ of the polynomial $ f(x)$[/i], it is defined to be the set of all automorphism of $ F$ leaving $ K$ fixed, i.e. if $ \sigma \in \text{G}(F/K)$ then $ \sigma: F\mapsto F$ is an isomorphism and $ \sigma (x) \equal{} x$ if $ x\in K$. It turns out that this set is always a group under functional composition.

So given $ f(x)\equal{}x^3 \minus{} 3x \plus{} 1$, it has three distinct real roots $ r_1,r_2,r_3$. Form $ F\equal{}\mathbb{Q}(r_1,r_2,r_3)$. In your case you are looking at the group $ G\equal{}\text{Gal}(F/\mathbb{Q})$. Now the question is to compute this group (i.e. what is it up to isomorphism) $ G$.

Abel preceeds Galois. I guess the group is a galois group.  but abel's paper on polynomials deg. >4 not generally admitting solutions doesn't use this language.

[quote="Los"]What kind of grouP? The group of the polynom over the field $ \mathbb{Q}$, which is the set of all permutations which preserve the validity of any equation over $ \mathbb{Q}$ of the form $ P(t_1,t_2,t_3,t_4) \equal{} 0$ where the $ t_i$ are the zeros of the polynomial. Is this a Galois group?[/quote]

yes. as you can see, this original definition of a galois group has been generalized to so called galois extensions.

[quote="ZetaX"]You can only associate Galois groups for splitting fields (well, in fact, to normal fields). There is for example no Galois group for $ \mathbb Q[\sqrt [3]2] | \mathbb Q$.[/quote]

I disagree. The Galois group is the trivial group in that case. You can associate a Galois group to any extension with the same definition as given in this thread, it is just that Galois' fundamental theorem only holds for Galois extensions. Even Galois' fundamental theorem itself deals with the possibility of non-Galois extensions -- an intermediate field is a normal extension if and only if it is the fixed field of a normal subgroup.

Nonetheless, and this is what I wanted to point to, it's not called Galois group, but only automorphism group then. The problem is that we have two different set's:
- the automorphisms (a group as pointed out).
- the homomorphisms to an algebraic closure (larger than the previous one).
Both differe for non-Galois extensions, so the question is what to sacrifice.


If we have 4 points $A_{i}(x_{i},y_{i},z_{i}), i=\overline{1,4}$ (with particulat coordinates) how can I prove that they are in the same plane?

My ideea(I don't know if it's good): Replacing the coordinates of all 4 points into the standard equation of a plane $ax+by+cz+d=0$ hence obtainig a linear system with 4 equations. Then we solve it(with Cramer's rule or something) and we get $a,b,c,d$ such our 4 points are verifying the equation of the plane. Any other methods?

Thanx :)

[hide="One way"]
you can find the equation of the plane and then satisfy the 4th point in that equation to prove.
[/hide]

[hide="Second Way"]
using vectors.
[/hide]


I have no idea to this problem

f (xy + f(x) ) = x f(y) + f(x)

how to solve?


Prove that sin1 and cos1 are irrational (1 radian not 1 degree!)

I'll do for sin1 as for cos1 is the same...

Lets consider that sin1 is rational. Then sin5=sin3cos2+sin2cos3=(3sin1-4(sin1)^3)(1-2(sin1)^2)+2sin1(4(cos1)^4-3(cos1)^2)=(3sin1-4(sin1)^3)(1-2(sin1)^2)+2sin1[4(1-(sin1)^2)^2-3(1-(sin1)^2)];
sin15=3sin5-4(sin5)^3
sin45=3sin15-4(sin15)^3
so we get that sin45 is rational which is false!
so sin1 is irrational!
cheers! :D :D

It is true I don't know much about degrees and radians, but sin 45 degreess is sqrt(2)/2, not sin 45 radians. I think you considered sin 1 degree.

oh yeah! not I see what he wrote in the brackets... thanks harazi!

suppose sin 1 is rational, there exists M large enough such that (M! sin 1) is an integer.

sin 1 = 1 - 1/3! + 1/5! - ... + (-1)^n / (2n+1)! + (lagrange's remainder).

where (lagrange's remainder) = sin(\theta x + (2n+3)\pi/2)/(2n+3)!

we can find M large enough such that

M! sin 1 = M!(1 - 1/3! + 1/5! - ... + (-1)^n / M!) + M!.(lagrange's remainder)

or

M! sin 1 = M!(1 - 1/3! + 1/5! - ... + (-1)^n / (M-1)!) + M!.(lagrange's remainder)

where M! sin 1 is an integer, and M!(1 - 1/3! + 1/5! - ... + (-1)^n / M!) is an integer too. this force that M!.(lagrange's remainder) is an integer.

0 < |M!.(lagrange's remainder)| < M!/(M+2)! = 1/(M+1)(M+2).

or

0 < |M!.(lagrange's remainder)| < M!/(M+1)! = 1/M.

how can M!.(lagrange's remainder) be an integer?

So sin 1 is irrational.

Similar arguments work for proving that cos 1 is irrational and e is irrational as well.

sin1 is irrational its continued fraction should not be periodic.
http://mathworld.wolfram.com/Sine.html

Does anyone know about Niven's proof for the irrationality of cos r for any rational number r?

http://www.briantaylor.com/Pi.htm

it includes Niven's proof of Pi's irrationality and transcendence.

maybe you can use the same idea for cosines.

Thx liyi, that's the technique I was looking for  :)


trigonometry
real analysis
real analysis unsolved

Let $ (c_n)$ be a decreasing sequence of positive numbers.If $ \sum c_n sin nx$ is uniformly convergent, show that $ lim (n c_n) \equal{} 0$.

Hint: Assume that $ c\sim \frac {1}{n}$ and show that the sum necessarily approaches something greater than $ 0$ for all small $ x$.  Show that this contradicts uniform convergence.

JoeBlow,can I know what do you mean by
$ c\sim \frac {1}{n}$ ?

Sketch: Let $ \epsilon > 0$. Then for large $ N$, $ |\sum_{n \equal{} N}^{2N} c_n \sin nx| < \epsilon$ for all $ x \in [0, 1]$, in particular for $ x \equal{} 1/(2N)$. This implies $ \sin (1/2)\sum_{n \equal{} N}^{2N} c_n < \epsilon$. Because the $ c_n$'s decrease, we have $ \sin (1/2)(N\plus{}1)c_{2N}< \epsilon$.


I want to know your favorite type of music because i am into music

Alternative rock. I can't stand heavy metal or rap.

EDIT: favorite bands: nirvana, REM

I listen to classical and rock.

Favorite Artists: Classical; Vivaldi, Bizet

Rock: Linkin Park, Green Day

 :)

-Hard rock, expect the new stuff
-Rap, particularly with lots of bass
-that "nerdcore/geeksta" posted a few days ago :D

rock, hip hop, soul, rap, and also musical songs like the phantom of the opera :) . basically anything except for country and folk.

Hard Rock, Rock, and some Metal. Rap is alright, as is some alternative. I have no favorite bands.
P.S. This topic has been done before.. i dunno where it is though.

I write my own music. (synthesized)

That way I don't have to spend any money.  :D

[quote="ToBe"]
P.S. This topic has been done before.. i dunno where it is though.[/quote]
this topic has been done numerous times. if you did a search, you should find the past topics. but since nobody uses the search function, this topic gets brought up every month or so.

Where is this search thing??

Anyway,i love rock,all types,maybe a little less metal(like korn,really bogus) and punk,except good charlotte and green day..

I love rap and hip-hop too,i just love eminem,no one less,for hip-hop,blackeyed peas...
Rockbands-rasmus,linkinpark,evenescence,creed,alterbrigde and many other bands i cant remember right now...i hate the old bands like gnr,nrvana,etc,no offence meant to their lovers :D

The best rock song i've heard so far is of therasmus-in the shadows..what a song,try downloading it...its awesome~!

[quote="shadysaysurspammed"]Where is this search thing??
[/quote]

in the box at the top of the page with all the tabs. its next to "memberlist" and above "logout".

[quote="noneoftheabove"]but these are peoples opinions not questions like where can i get (insert here) where they are looking for a answer then that is the purpose for the search button but i think with different people posting this not everyone would have posted in all the other topics maybe new members could voice there opinion without "reviving an old thread" is this logic wrong?[/quote]
a). your english is horrible
b). i never said it was bad. i just pointed out the the topics repeat themselves in fun&games.
c). and nothing's wrong w/ reviving an old thread either.

Rock
Hip Hop
Rap
...

Eminem, Rob Thomas, Kanye West, etc.

hey GUNS N ROSES for ever!!!!!

METALLICA
DEEP PURPLE

EAGLES - a bit on the classical side , but cool

I like 	Alkaline Trio, Bright Eyes, Crossfade, Death Cab for Cutie, Decemberists, Eve 6, Green Day, Hawthrone Heights, Keane, Lit, Offspring, Silverstein, Straylight Run, and Sugarcult.

Death cab for cutie!!!

Offspring??? :huuh: :noo:

[quote="eminem"]
DEEP PURPLE
[/quote]
Seconded.

yesssssssssssssssssssss!!! :lol:  :lol:  :lol:  At least some one agrees with me ! what else do you like, penguin ??


geometry
3D geometry
quadratics
rhombus
AMC
AMC 12
trigonometry

Here's the official problems (first 15) of the Fourtieth Annual Olympiad High School Prize Competition in Mathematics.

1.) Simplify (((1/64)+(1/128)+(1/256))/((1/1024)+(1/2048)+(1/4096)))/(((1+(1/2)+(1/4))/((1/8)+(1/16)+(1/32)))
A.) 1/2
B.) 1/8
C.) 4
D.) 2
E.) 8

2.) The expression (3:^2:):^3:-4 is equal to a perfect square times a prime number. What is the prime number?
A.) 25
B.) 19
C.) 13
D.) 29
E.) 7

3.) [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=1874]here[/url]
A.) 131
B.) 132
C.) 133
D.) 134
E.) 135

4.) [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=1874]here[/url]
A.) 9
B.) 10
C.) 11
D.) 12
E.) 13

5.) Amy works at a toy manufacturing company and her job is to make toy cars. After 5 hours, she made 4 toy cars. After 24 hours, she made 18 toy cars. Starting with the 5th hour, she makes the toy cars at a constant rate. During which hour will she make the 40th toy car?
A.) 40th
B.) 35th
C.) 52nd
D.) 53rd
E.) 50th

6.) What is the smallest four digit number that is the product of prime numbers A, BC, and DE where A, B, C, D, and E are distinct digits?
A.) 1234
B.) 1034
C.) 1222
D.) 1833
E.) 1066

7.) Three red cubes have the same weight as six blue cubes, two yellow cubes have the same weight as five blue cubes, and four white cubes have the same weight as six blue cubes. How many blue cubes are needed to match the total weight of four red, two yellow, and two white cubes?
A.) 10
B.) 14
C.) 20
D.) 16
E.) 5

8.) [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=1874]here[/url]
A.) 10
B.) 20
C.) 30
D.) 40
E.) 50

9.) The difference between the sum of 25 consecutive integers and the sum of the 10 consecutive integers that precede them is 1900, Find the largest of these numbers.
A.) 93
B.) 95
C.) 126
D.) 129
E.) 127

10.) MN and PQ are distinct non prime numbers each of which is divisible by a two digit prime number. What is the largest possible product of MN and PQ?
A.) 9405
B.) 8712
C.) 9504
D.) 9012
E.) 7402

11.) Consider the quadratic equation a:^2:+bx+c=0, where a, b, and, c are positive integers with b:^4:-16a:^2:c:^2:=65. The roots of this quadratic:
A.) are equal.
B.) are irrational.
C.) are imaginary.
D.) rational.
E.) cannot be determined.

12.) Each side of the square ABCD is 9 units long. The obtuse angle of rhombus BEDF measures 120o. What is the area o rhombus BEDF?
A.) 81
B.) 81:rt3:
C.) 18:rt2:
D.) 27:rt3:
E.) 27:rt2:

13.) A block of wood is in the shape of a cube. Each edge of the cube measures 10 inches. Two holes in the shape of right cylinders of height 10 inches are drilled such that the radius of the smaller hole is half the radius of the larger hole. The holes are drilled so that there is no overlap. To the nearest half inch, how many inches should the radius of the larger hole be so that the volume of the cube with the holes is 65% of the column of the original cube?
A.) 3
B.) 4
C.) 1.5
D.) 3.5
E.) 2.5

14.) The five-digit base 10 number babac is a multiple of 25. How many distinct numbers are possible if a, b, and, c can only take on the values 1, 2, 3, 4, 5, 6, or 7?
A.) 12
B.) 10
C.) 17
D.) 14
E.) 13

15.) How many three-digit numbers with exactly two repeating digits are a multiple of nine?
A.) 36
B.) 18
C.) 12
D.) 20
E.) 21

Tare you did it again, the record now holds at 140 lines :)

my answers

1. d
2 d
3 b
4 I put D, but i think it's E
5 didnt answer
6 I forget, southpaw's answer is right.
7 d
8 wasnt posted, i dont think.  I put D, but it's B i think. X
9 I forget what i put...
10 A
11 didnt answer, but B i think.
12 D
13 B X
14 D
15 E

[quote="cats..."]nice job.  can you tell me the lengths of the chords in the circle problem?  i want to check it.[/quote]
Er, problem 18? It's 2, 2, and 6.

How does it go though?  Give them like 6,2, x, 2 .  from up to down, then from left, to right.

x, 2, 2, and 6. I believe x is just 6 though.

yeah, you're right. thanks.  the answer's 5 sqrt 3, then.

Is it legal to post so many of these problems? If it isn't, will a moderator kindly delete this post please?

Often people who write math contests want to be able to sell copies of old contests. If the contests are on the web, no one will want to purchase them. Therefore, I believe many of these contests are copyrighted. Is that correct?

http://www.maml.org/wwwroot/sampleproblems/complete_sets/completesets.htm

this year is not up yet though.

It's free for peep's sake, didi you really think I'll break the law for some math problems? Use your common sense...</harshness in general>

Anyways what do you reckon the cut-off score is?

If you were to put all the AMC12 problems, I think some people might not be so pleased.

right.  These are not sold, and they even post the questions on the internet (link in my previous post).

I don't know what the cutoff will be, I have only seen one cutoff score so far and that was last year's.  If you can find others, that'd be good.  I think the questions were pretty conducive to making mistakes, so hopefully it'll be lower, but i dont know.  My guess is that the cutoff is pretty much the same each year, and if it does fluctuate, it's maybe by about a question.

What did you think of the questions?  I really didnt like them, there werent too many interesting ones (until the end).  And for many of them the fastest way to do it was not even to do math, but to just plug in the answers.

can you post how you did 13?  I got a different answer, but i might be interpreting wrong or something stupid.

So you have 1000 - 10:pi:r:^2:-2.5:pi:r:^2: =650

And pretty much just solve it.

There was dissent on #11:
a, b, c positive integers
b:^4: - 16a:^2:c:^2: = 65
(b:^2: - 4ac)(b:^2: + 4ac) = 65

Case 1:
b:^2: - 4ac = 1
b:^2: + 4ac = 65
2b:^2: = 66
b:^2: 33
Contradiction.

Case 2:
b:^2: - 4ac = 5
b:^2: + 4ac = 13
2b:^2: = 18
b:^2: = 9
b = 3
a = c = 1
Which gives the quadratic x:^2: + 3x + 1 = 0, which has only irrational real roots.
(B).

im not exactly sure what you're doing.  can you explain some more?

What i did was pi*r^2*10+1/4*r^2*pi*10=1000*.65

I thought that the sum of the volumes of the cylinders equals 65% of the volume, or .65*1000.

Anyone else have opinions on this one?

No no no, its the remaining volume of the cube thats 65% of the original, not the cyllinders.

For the other one, yeah, you're right, I'm wrong. darn.

yeah, alright so a's right.

For the trig one:[ sin(x/2)^2-2sin(x/2)*cos(x/2)-cos(x/2)^2]/ [i forget)  was it like 1-tanx or 1+tanx, i forget.  One of them was E, i put whatever was b or c that had a tan in it.  is that right?

hey guys who else is from mass here?
the answers i believe are correct:
1.) D
2.) D
3.) B
4.) E
5.) D
6.) C
7.) D
8.) B
9.) E
10.) A
11.) B
12.) D
13.) A
14.) D
15.) E
16.) D
17.) B
18.) E
19.) E
20.) C
21.) E
22.) C
23.) A
24.) D
25.) C
26.) D
27.) C
28.) D
29.) A
30.) D

anyways, i think i got 23 right, 2 wrong, and 5 blank....

how'd you guys do?

Um anyone who says they took the test would be from Mass, not to mention most people write MA or Massachustts for location (I'm also from Mass but I write "It's great to be here..." because of that commercial)

Pretty sure it was 1+tanx, so C. E was 1-tanx

If all of those answers are correct(which I'm pretty sure they are), then i got..

21 correct, 3 wrong, 6 blank.

Will that qualify???

yes.  that's a score of 111, last year's cutoff was 94 or 96 i forget, and i dont think it'll change much this year.

[quote="cats..."]right.  These are not sold, and they even post the questions on the internet (link in my previous post).[/quote]

Correction: there are a few years' worth of problems on the internet.

I agree with Simon.  The authors of this contest may be selling these, or planning on putting a book together.  I don't think they'd mind if we used their problems now and then (as I do in AMC - I'll use a few WMTS problems, a few from HMMT, a few from AIME, etc.), but one has to be careful about posting whole sets of problems.

The main rule of thumb is this: if it says on the sheet they gave you "copyright blah blah blah" or "do not reproduce these questions by e-mail, fax, copier, etc.", as the AMC booklets do, then I wouldn't post a mass of them on the internet.

But it's your call.  The evidence suggests that they don't mind, but it never hurts to play it safe.  I'm just highlighting the potential problems.

you're right; they only have 3 years of exams on there, and they're the 3 most recent exams.  What will happen is this year's exam will be posted there and then the 4th oldest exam will be removed.  So it's going to be posted on the internet anyways, i really dont see anything wrong with it.

Ah, I see.

Well, if you're sure that they're going to get posted, I guess I don't have a problem with it.  For that matter, I didn't have a problem with it in the first place, only a word of caution.  Eh, watch me ramble here.

Mmm, I'm going to leave that decision up to Mr. Rusczyk. If he honestly believe that this is copyright violation and that the MAML people get mad at me (or at AoPS) then I will agree with the decision to delete this.

...it took me like an hour and a half to type up just the first half, fyi.


Maybe this is just me, but where is the blog options button where I can actually make a blog?
I've been searching for a bit, it's probable that I overlooked it...

Try this link:

http://www.artofproblemsolving.com/Forum/weblog_config.php


modular arithmetic
number theory proposed
number theory

Prove that in the following sequence , there are infinity number that are not prime

$ t_n$ $ \equal{}$ $ n!(n! \minus{} 1) \plus{} 1$           ;         $ n\equal{}1,2,3,...$

[quote="behdad.math.math"]Prove that in the following sequence , there are infinity number that are not prime

$ t_n$ $ \equal{}$ $ n!(n! \minus{} 1) \plus{} 1$           ;         $ n \equal{} 1,2,3,...$[/quote]
Let odd $ n\geq 3$ and suppose $ t_n \equal{} p$ prime$ > 2$ and consider $ t_m$ with $ m \equal{} t_n \minus{} (n \plus{} 1)$

$ m > n$ (since $ n\geq 3$
We know (Wilson theorem) that $ (p \minus{} 1)! \equal{} \minus{} 1\pmod p$ and so $ m! \equal{} (p \minus{} (n \plus{} 1))! \equal{} \frac 1{n!}\pmod p$ (remember $ p$ is prime and $ n$ is odd)

Then $ t_m \equal{} \frac 1{n!}(\frac 1{n!} \minus{} 1) \plus{} 1\pmod p$ and so $ t_m \equal{} \frac {n!(n! \minus{} 1) \plus{} 1}{(n!)^2}\pmod p$ and, since $ n!(n! \minus{} 1) \plus{} 1 \equal{} p$ : $ t_m \equal{} 0\pmod p$ is not prime.

So :
Either there are infinitely many non primes $ t_n$ with odd $ n\geq 3$ and we got our result,
Either there are a finite number of non prime $ t_n$ with odd $ n\geq 3$ and so infinitely many prime $ t_n$ with odd $ n\geq 3$ but then, we can find for each of them a non prime $ t_m$.

Hence the result.


Prove that for any $a,b,c>0$ we have
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3k \ge (k+1)\frac{a+b+c}{\sqrt[3]{abc}}\]
for $k=\frac{3}{\sqrt[3]{4}}-1.$
:)

Dear Can,

This problem is pretty old. In "Sang Tao bat Dang thuc", published in 08-2006, I proposed the following inequality

If $x,y,z\ge 0$ and $k=\frac{3}{\sqrt[3]{4}}-1$ then
\[x^{3}+y^{3}+z^{3}+3kxyz \ge (k+1)\left(x^{2}y+y^{2}z+z^{2}x\right)\]

Although I created and proved it by myself, it seems to be too simple to be my own (so I did not write the author of the problem). WE can prove this one easy enough by EMV method.


quadratics
number theory
number theory proposed

Fix a prime $ p > 2$.An integer $ x$ is called a quadratic residue mod $ p$ if there exists an integer
$ y$ so that $ y^2 \minus{} x$ is a multiple of $ p$.
(a) Determine the number of integers $ x$ in {$ {0; 1; 2; : : : ; p\minus{}1}$}, for which both $ x$and $ x \minus{} 1$
are quadratic residues mod$ p.$
(b) Prove that $ \minus{} 1$ is a quadratic residue mod $ p$if and only if $ \frac {p \minus{} 1}{4}$ is an integer.

(b) It may be obvious from the fact $ \left( {\frac{{ \minus{} 1}}{p}} \right) \equal{} ( \minus{} 1)^{\frac{{p \minus{} 1}}{2}}$

(a) For $ 1 \le x \le p\minus{}2$, $ \frac{(\frac{x}{p})\plus{}1}{2}\frac{(\frac{x\plus{}1}{p})\plus{}1}{2}$ is $ 1$ if $ x, x\plus{}1$ are both quadratic residues, and 0 otherwise. Also, for $ x\equal{}0$, $ x$ and $ x\plus{}1$ are both quadratic residues. In addition, for $ x\equal{}p\minus{}1$, $ p\minus{}1$ and $ p$ are both quadratic residues iff $ (\frac{\minus{}1}{p})\equal{}1$. Hence, the answer is:
$ 1\plus{}\frac{(\frac{\minus{}1}{p})\plus{}1}{2}\plus{}\sum_{x\equal{}1}^{p\minus{}2} \frac{(\frac{x}{p})\plus{}1}{2}\frac{(\frac{x\plus{}1}{p})\plus{}1}{2}$
The last sum can be simplified. Since $ \sum_{x\equal{}0}^{p\minus{}1} (\frac{x}{p})\equal{}0$, we get:
$ \sum_{x\equal{}1}^{p\minus{}2} \frac{(\frac{x}{p})\plus{}1}{2}\frac{(\frac{x\plus{}1}{p})\plus{}1}{2}\equal{}\frac{\minus{}(\frac{\minus{}1}{p})\minus{}(\frac{1}{p})}{4}\plus{}\frac{p\minus{}2}{4}\plus{}\sum_{x\equal{}0}^{p\minus{}1} \frac{(\frac{x^2\plus{}x}{p})}{4}$
For $ p >2$:
$ \sum_{x\equal{}0}^{p\minus{}1}(\frac{x^2\plus{}x}{p})\equal{}\sum_{x\equal{}0}^{p\minus{}1}(\frac{(2x\plus{}1)^2\minus{}1}{p})\equal{}\sum_{x\equal{}0}^{p\minus{}1}(\frac{y^2\minus{}1}{p})$
It can be shown that the last expression is $ \minus{}1$ (I can elaborate if wanted), hence the final result is: $ \frac{p\plus{}2\plus{}(\frac{\minus{}1}{p})}{4}$.

The standard proof of b) uses Euler's criterion; $ \minus{}1$ is a quadratic residue if and only if $ (\minus{}1)^{ \frac{p\minus{}1}{2} } \equiv 1 \bmod p$ by Fermat's little theorem, and of course this occurs if and only if $ \frac{p\minus{}1}{2}$ is even.

[quote="bambaman"](a) For $ 1 \le x \le p \minus{} 2$, $ \frac {(\frac {x}{p}) \plus{} 1}{2}\frac {(\frac {x \plus{} 1}{p}) \plus{} 1}{2}$ is $ 1$ if $ x, x \plus{} 1$ are both quadratic residues, and 0 otherwise. Also, for $ x \equal{} 0$, $ x$ and $ x \plus{} 1$ are both quadratic residues. In addition, for $ x \equal{} p \minus{} 1$, $ p \minus{} 1$ and $ p$ are both quadratic residues iff $ (\frac { \minus{} 1}{p}) \equal{} 1$. Hence, the answer is:
$ 1 \plus{} \frac {(\frac { \minus{} 1}{p}) \plus{} 1}{2} \plus{} \sum_{x \equal{} 1}^{p \minus{} 2} \frac {(\frac {x}{p}) \plus{} 1}{2}\frac {(\frac {x \plus{} 1}{p}) \plus{} 1}{2}$
The last sum can be simplified. Since $ \sum_{x \equal{} 0}^{p \minus{} 1} (\frac {x}{p}) \equal{} 0$, we get:
$ \sum_{x \equal{} 1}^{p \minus{} 2} \frac {(\frac {x}{p}) \plus{} 1}{2}\frac {(\frac {x \plus{} 1}{p}) \plus{} 1}{2} \equal{} \frac { \minus{} (\frac { \minus{} 1}{p}) \minus{} (\frac {1}{p})}{4} \plus{} \frac {p \minus{} 2}{4} \plus{} \sum_{x \equal{} 0}^{p \minus{} 1} \frac {(\frac {x^2 \plus{} x}{p})}{4}$
For $ p > 2$:
$ \sum_{x \equal{} 0}^{p \minus{} 1}(\frac {x^2 \plus{} x}{p}) \equal{} \sum_{x \equal{} 0}^{p \minus{} 1}(\frac {(2x \plus{} 1)^2 \minus{} 1}{p}) \equal{} \sum_{x \equal{} 0}^{p \minus{} 1}(\frac {y^2 \minus{} 1}{p})$
It can be shown that the last expression is $ \minus{} 1$ (I can elaborate if wanted), hence the final result is: $ \frac {p \plus{} 2 \plus{} (\frac { \minus{} 1}{p})}{4}$.[/quote]
Good  :)


:welcomeani: :idea: [b]In a certain right triangle, the lengths of the legs have a sum of 50 and a product of 282. What is the length of the hypotenuse???[/b]
 :ddr:  :diablo:

[hide]\[\sqrt{1996}=2\sqrt{13*19*2}=2\sqrt{494}\][/hide]

I added something wrong... :blush: should be ....

[hide]\[\sqrt{1936}=44\][/hide]

[b]236factorial is correct.[/b]not that surprising...lol :D

[quote="JesusFreak197"]This is much simpler than that... :lol:

[hide="Answer"]$h^2=x^2+y^2$
$(x+y)^2=2500=x^2+2xy+y^2$
$xy=282 \Rightarrow x^2+y^2=2500-2(282)=1936$
$h^2=1936 \Rightarrow h=\boxed{44}$[/hide][/quote]
yeah...it is! :D $Great^2$


Is there a possibility to set a function which allow us to seek all threads about a specific member?

I mean, we can easily find all posts about a member (including threads) in the profile, but if we want to find all the threads (excluding posts, obviously)?, it's a little messy lookin' at post by post.

Check the search page carefully. There is an easy and obvious way to do that (author search, select view topics instead of posts).

I don't see how one can do that...
There is no way to view the threads one specific user has started, I think.

And by the way, the search seems to ignore some words, e.g. "value".

[quote="ZetaX"]There is no way to view the threads one specific user has started, I think.[/quote]
Exactly, that's what I'm talkin' about.

Actually I made a search yesterday and it doesn't show the specific member, appear many of them.

There's a "search post title only" option, but it only does something if you actually search for a word.
For example (all with "search post title only" and "Display results as topics"):
author=jmerry, empty search field: returns all 2395 threads (before this one) that I have posted in.
author=jmerry, search field=sum: returns the one thread I started with title "Sum".
author=jmerry, search field=*: returns nothing.

Modifying the behavior of this option would be the way to go.

It can't be very hard to apply this function; all you would need to do is scan the "By post" ones, and display each individual topic. Other features (i.e. last post in thread) would be trickier.


function calculus derivative algebra polynomial

A function is defined by $ f(0) \equal{} 0$ and $ f(n) \equal{} f(n \minus{} 1) \plus{} 2n$, for $ n > 0$. Find $ f(4)$.

$ 0,2,6,12,20$. We're just plugging and chugging. Alos not ethat the general fomrula is $ n(n\minus{}1)$, where n is the $ n$th term.

As an aside: the first derivative of f(x) would be a first degree polynomial (linear), so f(x) is second degree.

Which is not appropriate for this forum.

Umm...the last number BOGTRO posted which is $\boxed{20}$...again, it's a plug and chug.

Ohhhh, ok. I see how it works now. I didn't see how to do it at first.
Nvm, got it :)


algebra
polynomial
inequalities
search
AoPSwiki
geometry
3D geometry

[quote="usahw1011"]Explain everything throughly please.  thank you XDXD

[b]1. What number, when added to the numerator and the denominator of 5/8, results in a fraction whose value is 0.4?[/b]

[i](I didn't understand the question. if someone knows wht its talking about can u tell me?)[/i]

[b]2. For how many positive integers x is x^2 + 6x +9 between 20 and 40?[/b]

[i]Do we start off with this? 20< x^2+6x+9 <40 [/i]

[b]3. How many non-empty sets does a three-element set have?[/b]

[i]three-element sets? wouldn't it be like 7*8*9?? [/i]

[b]4.what is a composite number?[/b]

[b]5. We know the following to be true: 1. Z and K are integers with 500< Z < 1000 and K< 1  2. Z=K*K^2.
     What is the value of K for which Z is a perfect square?[/b]

[i]Do this without any plugging in[/i]

[b]6. What is the units digit of (20*21*22*23*24*25)/1000?[/b][/quote]
First, refrain from using chat textual smileys 'XDXD', it's unprofessional. However, you can do those in our Games & Fun Factory forum.

Second, I'll only guide you by giving you hints. You'll have to attempt to work on them. There is no sense of excitement if I show you the solution without you even trying at them.

1. Do what the problem says. What is numerator and denominator? What number (let that be $ x$), when added to the numerator and the denominator of 5/8 (numerator is 5 and denominator is 8) (now you do what the problem tells you: add $ x$ to 5 and add $ x$ to 8), results in a fraction (so the fraction is $ \frac{5\plus{}x}{8\plus{}x}$) whose value is 0.4 (set that fraction equal to 0.4...)

2. Yes start off with that inequality. A possible way is to look at $ x^2 \plus{} 6x \plus{}9$; this polynomial can be factored. Or you can just solve the inequality directly (the cumbersome way).

3. I think the question is "How many non-empty [b]subsets[/b] does a three-element set have?" Example of a 3-element set: $ \{a,b,c\}$. No, the answer is not 7*8*9. Write out each of the subsets of this 3-element set.

4. Google search or use AoPSWiki

5. There is a problem with the problem statement... in condition 1, it doesn't make sense if K<1 ... in condition 2, Z=K*K^2 ???

6. So what do you think? No comment from you... did you try to solve it?

[quote="usahw1011"][quote="10000th User"]
First, refrain from using chat textual smileys 'XDXD', it's unprofessional. However, you can do those in our Games & Fun Factory forum.

Second, I'll only guide you by giving you hints. You'll have to attempt to work on them. There is no sense of excitement if I show you the solution without you even trying at them.

1. Do what the problem says. What is numerator and denominator? What number (let that be $ x$), when added to the numerator and the denominator of 5/8 (numerator is 5 and denominator is 8) (now you do what the problem tells you: add $ x$ to 5 and add $ x$ to 8), results in a fraction (so the fraction is $ \frac {5 \plus{} x}{8 \plus{} x}$) whose value is 0.4 (set that fraction equal to 0.4...)

2. Yes start off with that inequality. A possible way is to look at $ x^2 \plus{} 6x \plus{} 9$; this polynomial can be factored. Or you can just solve the inequality directly (the cumbersome way).

3. I think the question is "How many non-empty [b]subsets[/b] does a three-element set have?" Example of a 3-element set: $ \{a,b,c\}$. No, the answer is not 7*8*9. Write out each of the subsets of this 3-element set.

4. Google search or use AoPSWiki

5. There is a problem with the problem statement... in condition 1, it doesn't make sense if K<1 ... in condition 2, Z=K*K^2 ???

6. So what do you think? No comment from you... did you try to solve it?[/quote]

I am sorry for using the textual smileys. Thank you for helping me.

1.  I understand the problem and I was able to solve it.

2. I still do not understand how I am suppose to do this problem. i understand it is 20< (x+3)^2 > 40. but wht then?

3. Is there any other way then just writing all the individual subsets down? Second, I could be having an misunderstanding of non-empty and empty subsets...just to clarify, what is the difference between a empty and non-empty subset?

5. Yes it is the other way around. its K>1. srry

6. Yes I have tried to solve it. At first I thought there must be some sort of pattern. first i believed maybe the fact that they're in consecutive order has something to do with the answer. of course artimetic or geometric series has nothing to do with this or i believe so... anyways I tried to mulitply the first and last but tht idea came out pretty fruitless. it was 500(first last), 504(second last, second first), and then 506(2 in the middle). i couldn't see any connection through that outlet....i can't really think of anything else. there is a connection i can't see here i think....or maybe im thinking too much.   :|...or too little...[/quote]

[hide]

For 2.

since 20 < x^2 + 6x + 9 < 40

as you said, 20 < (x+3)^2 < 40

so find all perfect squares in between 20 and 40.

They are 25, and 36. So set it equal to x.

(x+3)^2 = 25 and (x+3)^2 = 36

so solve it from there (don't forget to take the negative square roots.)

5. So K > 1 and     500 < Z < 1000 and  K*K^2=Z and they are all integers.

So first, you know that K*K^2=K^3. So find all perfect cubes in between 500 and 1000. so 512, 729 (you should have all of the cubes memorized)  so find which one is a perfect square (you should have those memorized also) (it is 729). the cube root of 729 is 9, so the answer is 9.[/hide]

Thx to Ihatepie, 2. and 5. are pretty much solved for you.

For 3., it's good to know that the total number of subsets of an $ n$-element set is $ 2^n$. To show where this formula comes from, there are many ways of proving it (like using Pascal's Triangle/Binomial Theorem) or the more classical combinatorial result: Either the element is in the subset or is not, so there are two choices. Now there $ n$ elements, so the total number of subsets is

$ 2\times2\times2\times\dots\times2 \equal{} 2^n$ (there are $ n$ 2's in the product on the LHS(left hand side))

Empty set is [u]the[/u] set with no elements in it: $ \{\ \}$. So if you are looking for non-empty subsets, you should take out the empty set from the total...

For 6., simplify the expression first by writing some of the factors into primes. Look at the prime factors of the denominator: $ 1000 \equal{} 2^3\times5^3$. Now, there are some factors in the numerator that would 'cancel' with the factors on the denominator. Once you simplify that, the rest is to multiply the units digit of the remaining factors to get your answer.

Easy Example for 6.: Find units digit of (12*13*14*15)/10
Notice that 10=2*5, now either 12 or 14 can give you a factor of 2 (it doesn't matter which one) and 15 gives you a factor of 5, (12*13*14*15)/10=6*13*14*3
We are looking for units digit so the tens digit are not important and we have
6*3*4*3=18*12=8*2=16=6 (OK I know my notation is unconventional but I hope you get the idea)

[quote="alanchou"]where did you get #2?  It seems too familiar...[/quote]

i found it while i was doing the FTW. actually all of these are from questions are from the FTW.

[hide="My Solutions"]
1)$ \frac{5 \plus{} x}{8 \plus{} x} \equal{} \frac {2}{5}$
Therefore,
$ 25 \plus{} 5x \equal{} 16 \plus{} 2x$
$ 5x \equal{} 2x \minus{} 9$
$ 3x \equal{} \minus{} 9$
$ x \equal{} \minus{} 3$
2)$ 20 < x^2 \plus{} 6x \plus{} 9 < 40$
$ 11 < x^2 \plus{} 6x < 31$
Using this, we can find that x could be equal to 2, 3, -8, -9
3) The number of subsets in a three element set is $ 2^3$
Therefore, since there is only one empty set, the number of non-empty sets would be $ 2^3 \minus{} 1 \equal{} 7$
4)A composite number is a number with more than two factors.
5)Since $ Z \equal{} K^3$ and Z has to be a perfect square, then $ Z \equal{} L^6$ The only perfect number to the 6th power between 500 and 1000 is [b]729[/b].
[/hide]

Well just to continue of ZhangPeijin, look at this if you want to know the answer to 6.

[hide]

so you have to find the units digit of (20*21*22*23*24*25)/1000

So first prime factorize 1000 so it is

(20*21*22*23*24*25)/2^3 * 5^3

So then start simplifying. Dividing two 5s from 25, and 5^3 you get

(20*21*22*23*24)/2^3 * 5

Dividing 5 from 20 and 5 gets

(4*21*22*23*24)/2^3

Dividing two twos off of 4 and 2^3 gives

(21*22*23*24)/2

Dividing two from  22 gives

(21*11*23*24*)

since you are only trying to find the units digit, multiply the unit digits of the numbers.

1*1*3*4 is 12, so the units digit is 2.[/hide]

Oh thanks, I didn't see number 6.
 :oops:
I feel blind.


The graphs of the equations \[ y=k, \] \[ y = \sqrt{3}x + 2k, \] \[ y = -\sqrt{3}x + 2k, \] are drawn in the coordinate plane for $k = -10, -9, -8, \dots, 9, 10$. These 63 lines cut part of the plane into equilateral triangles of side $\frac{2}{\sqrt{3}}$. How many such triangles are formed?

[quote="pkothari13"][quote="minsoens"][hide]
setting y=10, 0, and -10, we see that it's a regular hexagon with vertices at
$(\frac{5}{\sqrt{3}},10),(-\frac{5}{\sqrt{3}},10).(\frac{5}{\sqrt{3}},-10),(-\frac{5}{\sqrt{3}},-10),(\frac{10}{\sqrt{3}},0),(-\frac{10}{\sqrt{3}},0)$
side length$= \frac{10}{\sqrt{3}}$
dividing the hexagon into 6 eq. triangles of side length $\frac{10}{\sqrt{3}}$, we have that there are $1+3+5+7+9=25$ triangles of length $\frac{2}{\sqrt{3}}$ in each.
therefore, there are $6(25)=\boxed{150}$ little triangles

[/hide][/quote]

Whoa, that solution is awesome. :surf:

How did you decide to use -10, 0, and 10 though?[/quote]

But his solution is far from finished... (and it's not 150)  :(

[hide]Are you sure the side lengths of the individual triangles is $\frac{10}{\sqrt{3}}$?

That may be the mistake I think.[/hide]

:o Will anyone come to the rescue?? What was wrong with what minsoens did? Was it an arithmetical error? Or was there some concept that was overlooked?

oops jk
the sides of the individual triangles are supposed to be $\frac{20}{\sqrt{3}}$
then the coordinates of the hexagon ABCDEF are
$A=(\frac{-20}{\sqrt{3}},0); B=(\frac{-10}{\sqrt{3}},10); C=(\frac{10}{\sqrt{3}},10); D=(\frac{20}{\sqrt{3}},0); E=(\frac{10}{\sqrt{3}},-10); F=(\frac{-10}{\sqrt{3}},-10)$
so the total number is $\boxed{600}$(?) i hope i'm right this time

oh yeah and I used the values of -10, 0, and 10 because they produce the outermost intersection points of the hexagon.. for example, the intersection of the lines $y=\sqrt{3}x+2(10)$ and $y=-\sqrt{3}x+2(-10)$ produces the point $(\frac{-20}{\sqrt{3}},0)$

the subtle part to this problem is that, on the sides of the hexagon, there are more equilaterals on the outside, its almost like a hexagon with scales, theres 10 on each side, so 60 more, and you get 660


Anyone know where a list of flame tests can be found?  I realized I don't know any of them after seeing how my answer was completely wrong in the marathon thread... ahh well.

Actually I thought lead was greenish blue... but anyways:

http://chemistry.about.com/library/weekly/aa110401a.htm

[quote="g1sk8terz"]maybe try to google it[/quote]

hmm...did u guys not do labs on it? We usually keep a lab notebook with all the flame tests we did...but oh well. Usually, there are only 1 or 2 multiple choice tests which come up about the colors on the AP Exam anyways :P

we did do labs (took the class two years ago), however my teacher took the notebook at the end so that students wouldn't give them to next year's class.

yup we did do labs too...i meant google it if you didn't do labs and have no clue


Define the sequence {x_n}: a_1,a_2 ,... by a_1=1,a_2=b ,a_(n+2)=2a_(n+1) - a_(n) +2 where b,a are positive integer numbers .Prove that a_(n) a_(n+1) =a_m for sonme m is a positive integer. :roll:

This problem can be completed in two steps.  The first is noticing that $a_{n}=(n-1)b+(n-2)^{2}$ for all integers $n\ge 3$.  This can be easily proved by induction.

Secondly, through a good chunk of algebra, you can show that for all $n\ge 1$ we have that
$a_{k}a_{k+1}=a_{(k-1)(k-2+b)+2}$.
This completes the proof.


Find the minimum possible ratio of 5-digit number to the sum of its digits.

[hide]
We prove that the maximum comes for $ 10999$.  Note that $ \frac{10999}{28}\approx 393$.

First, say that we could get a smaller ratio for a number with leading digit $ 2$ or greater.  The smallest such number is $ 20000$.  The largest possible sum of digits is $ 9\plus{}9\plus{}9\plus{}9\plus{}9\equal{}45$.  So the smallest possible ratio is $ \frac{20000}{45}\approx 444$.  So the leading digit is $ 1$.

The maximum sum of the digits now is $ 1\plus{}9\plus{}9\plus{}9\plus{}9\equal{}37$.  For the ratio to be smaller than $ 393$, the number must be smaller than $ 393*37\equal{}14541$.  This reduces the maximal digit sum to $ 1\plus{}3\plus{}9\plus{}9\plus{}9\equal{}31$.  So the number must be smaller than $ 393*31\equal{}12183$.

This again reduces the maximum digit sum, to $ 1\plus{}1\plus{}9\plus{}9\plus{}9\equal{}29$.  So the number is smaller than $ 393*29\equal{}11397$.  Now the number with the greatest digital sum is $ 10999$.  So the desired number is no more than $ 10999$.

The smallest possible value for the $ 5$-digit number is $ 10000$.  So the digital sum must be at least $ \frac{10000}{393}\equal{}25$.  The smallest number with this digital sum is $ 10699$.  So we see that the digital sum must now be at least $ \frac{10699}{393}\equal{}27$.

The only numbers left with a digital sum of at least $ 27$ are $ 10899$ and $ 10999$.  We check that $ \frac{10899}{27}\approx 404>393$, so the minimum is indeed achieved for $ 10999$.
[/hide]


Determine a recurrence relation for the number h_n of ternary strings (made up of 0's, 1's and 2's) of length n that do not contain two consecutive 0's or two consecutive 1's. Then, find a formula for h_n.

I have a solution but need to verify its correctness. Can anyone help?

I think you made a mistake in Case 2, Subcase 2, because there aren't $ h_{n\minus{}3}$ sequences $ 011***$.

Here is the solution: Let $ h_{n,i}$ be the number of sequences which begins with $ i$ and has $ n$ elements $ i\in \{0,1,2\}$, and $ h_n\equal{}h_{n,0} \plus{} h_{n,1} \plus{} h_{n,2}$. Then it's not hard to see that $ h_{n,0}\equal{}h_{n\minus{}1,1} \plus{} h_{n\minus{}1,2}$, $ h_{n,1}\equal{}h_{n\minus{}1,0} \plus{} h_{n\minus{}1,2}$ and $ h_{n,2}\equal{}h_{n\minus{}1}$. Suming up this three, we get $ h_n\equal{}h_{n\minus{}1} \plus{} (h_{n\minus{}1,0} \plus{} h_{n\minus{}1,1} \plus{} h_{n\minus{}1,2}) \plus{} h_{n\minus{}1,2} \equal{} 2h_{n\minus{}1} \plus{} h_{n\minus{}2}$ and this is the recurence relation.

Bye


Between any two cities of a country there is only one one-way road. Show that there is a city from that every other city can be reached directly or by going over only one intermediate city.

[hide]
I'm sure it was posted before but couldn't find it.
[/hide]

Here's an quick, ugly, and brute-force induction solution:
[hide]
This is trivial for 1 and 2 cities. Assume this is true for n cities. Let's denote the starting city that satisfy the proposition for case n by A, the set of all cities that can be directly reached from A by R1, and the set of cities that can only be reached by taking 1 intermediate city as R2.

Notice that A, R1, R2 are mutually exclusive, and their union is all cities in case n. Also note that all cities in R2 are directly reachable from some cities in R1. Now denote the (n+1)th cities by B.

Consider picking city A. If at least one road from cities in A and/or R1 to B are toward B in direction, then we can go to that city B with at most 1 intermediate, and the rest also satisfy the proposition by assumption, and we're done with the induction.

Otherwise, all roads between B and any cities in A and R1 have direction away from B. Then we pick city B. Thus all cities in A and R1 are directly reachable, and all cities in R2 need at most 1 intermediate(because R2 are directly reachable from R1). This finish the induction.
[/hide]

I believe there is a variation of this problem in Engel, but I don't have access to the book at the current time; it's one of the examples in one of the first few chapters.

Masoud Zargar

Are we supposed to prove that there is a city that [i]can be reached by any other city[/i] in one or two moves or that there is a city that [i]can reach any other city[/i] in one or two moves?  Although I guess they would be very similar proofs...  :lol:


Let us define a "generalized diagonal" of a Tic-Tac-Toe board as any collection of 3 squares such that no two of the squares are in the same row or column.  You should be able to find 6 of these, 2 of which are the "normal" diagonals.  Generalized Tic-Tac-Toe is played just as normal Tic-Tac-Toe, except that if a player places marks on all 3 squares of a generalized diagonal, it is considered to be a win for that player.

1) (Hardest) Prove that ties are not possible in Generalized Tic-Tac-Toe.
2) (Easier) Using number 1, prove that the first player has a winning strategy.*
3) (Hard) Find such a winning strategy.
4) (Not hard) Does it matter what square the first player initially moves on?

*A winning strategy is a strategy that will guarantee a win, no matter what the opposing player does.  For example, no known winning strategy is known for chess.  However, in a simplified chess game where one player has a king and rook while the other has only a king, a winning strategy is known.  To prove the existence of a winning strategy, or to prove that a given strategy is a winning strategy, you much prove that it works, [i]regardless of the choice of moves[/i] of the other player.  This might also include a description of every move your player should make in respons to each possible move of the other player.

#1

[hide]Assume that there is a tie.  The first player fills in 5 squares.

Let (a, b, c) denote the number of squares in each column, in that order (WLOG).

In order to have a tie, no column may have three filled squares.

Therefore, we consider the only possible case, (2, 2, 1).  If not all rows contain a square, then we must have three filled squares in one row.  Hence, we can assume that each row also contains a filled square.

Let's first consider the rightmost column.  We will WLOG fill the square on the bottom.  We know that there are at least three filled squares in the other columns that are not on the same row, because if there weren't, the bottom row would make tic-tac-toe.

[][][]

[][][]

[][]()

[] - empty

() - filled

[] - range where three filled squares must be

We will consider only the red area.  Let us begin by filling all the squares.  We will then remove one filled square.

()()

()()

In this figure, both the following patterns exist.

()[]  []()

[]()  ()[]

Both of these patterns would result in a generalized diagonal.

Since the two patterns use different squares, removing a filled square will only destroy one of the two patterns.

QED[/hide]

#4

[hide]It does not matter where you start, there are exactly 4 (1 row, 1 column, 2 diagonals) tic-tac-toe possibilities for every square.

Let's assume this square:

[]()[]

[][][]

[][][]

Then, we can use its row and column.  It also has two diagonals, since there are two possible filled squares for the left column, and the filled square in the right column follows.

()()()  []()[]  []()[]  []()[]

[][][]  []()[]  ()[][]  [][]()

[][][]  []()[]  [][]()  ()[][][/hide]

More to come.  Are you sure you can easily use #1 to prove #2?  You can't just say that every position ends with player 1 winning...

beta wrote:The Winning Strategy(Tic-Tac-Toe isn't really a fair game, I don't really know why people would call it a strategic game)

[hide]For the first person to move,
place a X on one of the four corners
in the next move, put a X on one of the corners(at least two left) that would force the other player to block.

Next move: place a X on a corner(at least one left) and you win since the other player can only block one of the two lines.[/hide]

Not necessarily a win (for normal tic-tac-toe), and not a full explanation for generalized, since blue can gain the "initiative":

[][]()  [][]()  ()[]()  ()()()  ()()()

[][][]  []()[]  []()[]  []()[]  []()[]  It's tied from here on (normal tic-tac-toe).

[][][]  [][][]  [][][]  [][][]  []()[]

() -- player 1

() -- player 2

[quote="beta"][quote="JBL"]
*A winning strategy is a strategy that will guarantee a win, no matter what the opposing player does.  For example, no known winning strategy is known for chess.  However, in a simplified chess game where one player has a king and rook while the other has only a king, a winning strategy is known. [/quote]

That will be a good problem to do. Suppose the white king is at a1, black king is at h1, and the white rook is at a8. Find the winning strategy. It's black's turn.[/quote]

This will be obvious to most chess players.  For those of you who aren't, it will still most likely be obvious, but it's most effective to get a board and experiment a little (how would you try to control where the black king can go?)

Does anyone know if these same guiding principles/strategies for generalized tic-tac-toe hold for say a 4 x 4 grid or a 5 x 5 grid.  Now that I think about it there probably wouldn't be a winning strategy for an even x even case because both players go the same number of times.  What about an odd x odd grid?

Okay, your work looks pretty good for number 1 and 4, although there are better ways of stating your proof for number 4.  Specifically, I would like to point out that the fact that all squares have the same number of rows and diagonals is the same does not immediately imply that all squares are interchangable.

Yes, I am sure that number 2 follows easily from number 1.  And no, it is not because when 5 squares are filled in, they must contain a three-in-a-row, since the other player could have conceivably won before we get to that point.

Beta, your strategy ignores the case in which X's third move is required to block O.  That is, say X starts in the corner, O moves in the middle, and X goes in another corner.  On O's second move, he will create a 2-in-a-row.  If X moves in the corner, as you suggest, O could win on his move.  (This is roughly the same thing Probability said.)

amd -- I know that normal tictactoe on larger grids is always a tie, since O has ample opportunities to block anything X tries to do.  In fact, if a win is defined as 5 in a row, I don't think X can win even on an infinite board.  As for generalized tictactoe of larger sizes, I have no idea.  Higher-dimensional tic-tac-toe is also interesting.  A 3x3x3 is an easy win for the first player, but I think 4x4x4 is a draw.  It might be interesting to do some research on this.

[quote="Ravi B"]The 4x4x4 game, where you have to get 4 in a line, is a game called Qubic.  It has been shown by computer analysis to be a win for the first player.  (That surprised me, because I used to play the game during school, and I guessed that it was a draw too.)  See
[url]http://mathworld.wolfram.com/Qubic.html[/url].[/quote]

I too, used to play "3-D Tic-Tac-Toe" during school.  Beginning in high school, I wrote several different computer programs to play the game.  Most were easily beaten by any competent human.  My "best" version, which was written in IBM 370 assembly language, was available on our college mainframe.  (~1 million instructions per second, ~45-60 second limit per move.)

The program was completed about 1974, give or take.  Exclusive of playing against itself, in the years I was there, I only heard two claims that it had been beaten by a human, when the computer started the game.

As those who play the game know, 16 squares are of higher importance than the other 48 squares.  Watching the computer cream all humans, the advantage the first player had in choosing among those 16 squares was clear.  [Often by about the 5th pair of moves, the computer would already have a forced win.]

[I was unaware the game had been fully analyzed.  Thanks for the pointer, Ravi!]

beta wrote:The Winning Strategy(Tic-Tac-Toe isn't really a fair game, I don't really know why people would call it a strategic game)

[hide]For the first person to move,
place a X on one of the four corners
in the next move, put a X on one of the corners(at least two left) that would force the other player to block.

Next move: place a X on a corner(at least one left) and you win since the other player can only block one of the two lines.[/hide]

Sorry, dosn't work:

X goes first

X _ _

_ _ _

_ _ O

X O X

_ _ _

_ _ O

X O X

O _ _

X _ O

... and O has a generalized diagonal.

#2, #3: [hide]1. X goes in center. [Split into a and b] 2a. O goes in corner. 3a. X goes in a space horizontally next to the corner that O moved into. 4a. O is forced to block the vertical. 5a. X is forced to block the generalized diagonal. 6a. O has to block 2 squares but can't [O makes any move] 7a. X wins by horizontal or by generalized diagonal. [Other split] 2b. O goes next to the center. 3b. X goes on a corner next to where O last moved. 4b. O is forced to block the diagonal. 5b. X is forced to block the generalized diagonal. 6b. O moves anywhere since it can't block the 2 possible wins. 7b. X wins by vertical or horizontal.[/hide]

#2: [hide]Appling #1, there cannot be a tie so X or O must win.

If X keeps forcing O to block in a way that does not win, (except the first move) then O cannot attempt to win and X will win because he has the "initiative." [/hide]


Something is not right here since

$\sum_{n=1}^5 f(n) \: \approx \: 1,036 \: >\: 1$.

On the other hand, it is possible to prove that

$\sum_{n=1}^k f(n) \: < \: 2$

for all natural numbers $k$.

Yes, there was a mistake, I meant to start with f(1) = 1/3

how did you prove it?


How many ways are there to cover a 3 x 8 rectangle with 12 identical 1 x 2 dominoes?

Answer: very many ways:

Consider the 3 by 8:

|XX___________|
|ZYY__________|
|Z____________|

|XY___________|
|XY___________|
|AA___________|

Consider the leftmost 2 by 3. Either it is normal, i.e. contains only whole parts of dominos like the bottom figure.

Or, the only other option is to rotate the Y piece as in the first. This forces the following:

|XXBB_________|
|ZYY__________|
|ZAA__________|

Of this, there are two more cases:
either a normal 3 by 3 formed,

|XXBB_________|
|ZYYC_________|
|ZAAC_________|

or

|XXBBCC_______|
|ZYY__________|
|ZAA__________|

further top row extension.

The last resolves to a perfect 6 by 3

|XXBBCC__|
|ZYYDDF__|        (for example)
|ZAAEEF__|

or goes like

|XXBBCC_______|
|ZYYDEFFH|  (i.e., the top row seperation is forced, so it is like a forced 1 by 8 and a 2 by 8 tiled however)
|ZAADEGGH|

By case work, everything is resolved (of course, a perfect n by 3 implies reduction of rest of tiling to another m by 3, n+m=8)

Sorry for reviving an old thread.  The problem has been solved here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126738


If $ n > 3$ is odd, is it possible to place $ n$ black queens and $ n$ white queens on a infinite chessboard such that each queen is attacking four queens of the opposite color on each of it's four distinct lines of attack (one queen per line of attack).

[hide="What does line of attack mean?"]When I say line of attack I mean for instance the rook has two distinct lines of attack (horizontal and vertical). The bishop also has two lines of attack (left diagonal and right diagonal), and together these form the four distinct lines of attack of the queen.[/hide]

I think if we can do it for n=5,7,9 in m*m board then we can find another m*m board in first infinity board that isn't in attack line of old queens . then we can add new 5 black queen and 5 white queen and ... :oops:  :(


anyone have a solns. manual to "the basics of abstract algebra"
by paul e. bland....its a red and black book, it would really make this alot easier on me.

I don't know that book, but there is a lot of overlap between exercises in introductory textbooks on abstract algebra, so you may find what you want in book with many solved problems such as Schaums outlines, or a solution manual to one of the other texts.


Given $ \frac{2\minus{}x}{3\minus{}x}\equal{}4$ and $ \frac{3\minus{}y}{2\minus{}y}\equal{}4$, find the value of $ x\plus{}y$.

Solving the first equation,

$ \frac{2\minus{}x}{3\minus{}x}\equal{}4$
$ 4(3\minus{}x)\equal{}2\minus{}x$ (cross-multiply)
$ 12\minus{}4x\equal{}2\minus{}x$
$ 3x\equal{}10$
$ x\equal{}\frac{10}{3}$

Using a similar process for the second equation, $ y\equal{}\frac{5}{3}$.

Now, it's simple: just add our $ x$ and $ y$ values to get our final answer:
$ x\plus{}y\equal{}\frac{10}{3}\plus{}\frac{5}{3}\equal{}\frac{15}{3}\equal{}\fbox{5}$.