LaTeX
SFFT
MATHCOUNTS
articles
special factorizations

I'm just curious, how many of the "you are an aopser if..." apply to everyone on aops
personally, i'm qualified for 6

I'm hoping that you guys fulfill more than 2 because the poll doesn't fit...  :(

Four for me. I think, I don't remember what all 12 of the things were. I just think it was some where close to four. If not, then it was probably less.

9. You've already figured out this list is in base 12
6. You have a favorite number and it's not a whole number [color=blue](e)[/color]
2. You know Simon's Favorite Factoring Trick

B,9,3
Its in the Community on the left of the page you are on

http://www.artofproblemsolving.com/Community/AoPS_Y_Comm_TShirt.php

1/2 A (actually its asy and LaTeX)
9
1/2 5 (dunno anyone who can mem 1500 digits of pi)
3
2
1

I'm too lazy to write them out.

Oops, it is only 2 + 1/2 for me. I know Simons Favorite Factoring Trick, I figured out the list was in Base 12, and I know fifteen digits of pi.

3 apply to me, the computer can do >1500 digits of pi

I think that people are lying about being all twelve.

BTW, what does QED mean? I tried googling it, but it wouldn't tell me.

hmm... a more definitive answer for me would be

2 i know SFFT
5. i know 15 digits of pi, actually they never said how well we had to know the person, i could just pick the guy with the world record.  :|
6. favorite number is not a whole number, $ \pi$
9. base 12
B. I don't trust my calculator (I really don't)
10. I tune in to ESPN to watch a math contest (I have seen mathcounts nats on espn before, those kids were ridiculous of course this is before i actually did math competitively, i was just flipping channels

[quote="sooozyy"][quote="Wickedestjr"][quote="herefishyfishy1"]Quod Erat Demonstrandum (which was to be demonstrated)

Or sometimes it's Quantum electrodynamics. But not in this case.[/quote]

And that means ...[/quote]

you put a small square and QED at the end of a proof to say that
"this has been proved"[/quote]

Okay, thanks. Well, I do not know anybody that puts that at the end of their english paper.

7, 6, 5, 2

The person I know has about 2000 digits memorized, she was reciting them for a school contest, but messed up at about the 600th digit :P

Actually, all 12 apply to me. I know someone who knows 1600 digits.  :)

10. I satisify this.
B. I satisfy this.
A. I don't consider Latex my primary language. I consider math my primary language.
9. Yes.
8. no.
7. YESYESYESYESYESYESYESYESYESYESYES.
6. Yes. i.
5. I know infinitely many digits...in hexadecimal...so that counts as 1/2
4. haven't done it yet...
3. 1/2
2. yes
1. yes

I satisfy 8 of those...

I have a few...

10. We don't have ESPN.
A. I don't use my calculator. (This is probably yes then)
B. I consider my primary language math too.
9. This list has numbers?!! Oh yeah...
8. I've written my real name too many times. I'm more likely to type my real name accidentally.
7. If it's a 42-point test, yes.
6. Yes.
5. I know about 30 digits (I suck). The person that I know recited about 400 digits, but he was reading off a piece of paper.
4. Actually, I'm a pretty good English and History student too, but math is cooler.
3. I haven't taken any classes online. I don't have the time.
2. No, unless you count my friend Rahul, who is supposedly a "Simon"...
1. My parents would probably kill me.

How many is that?

[quote="Wickedestjr"]I think that people are lying about being all twelve.

BTW, what does QED mean? I tried googling it, [color=blue]but it wouldn't tell me.[/color][/quote]

How? If you google "qed" or "q.e.d.", with or without the quotes, the [u]first[/u] hit is the wikipedia article for Q.E.D.


algebra
polynomial
quadratics
Galois Theory
group theory
superior algebra
superior algebra unsolved

I need to find the Galois group of (x-1)(x+1)(x-2)=0 over the rationals Q.  And then what is the Galois group of x^3-1=0 over the rationals Q.
Please help :)

first try to find the splitting fields (smallest extension field of Q which contains all roots of that polynomial) and let us know what they are :)
(note that ur first poly is  reducible over Q (you just wrote it like that)

how do u find the splitting field of the second one

the splitting field needs to contain all roots:

obviously 1 is a root of x^3-1

I suggest u read http://en.wikipedia.org/wiki/Roots_of_unity to find out what the others are. However, it is not really required to know about roots of unity. After factoring (x-1) from x^3-1 you will get an irreducible (no roots in Q) polynomial of degree 2. Such  will give u always a quadratic extension (of degree 2), the galois group wil then be isomorphic to $Z_{2}$

anyway I would suggest you make yourself a bit comfortable in group theory first :) there will be some hardcore theorems about  connections between group theory and field theory. I'm afraid you won't survive them untill you review some basics about group theory

I'm not sure what the procedure is though for finding the Galois group

You just need to find the order of the group.


Each of the letters A, B, C and D represents a different odd integer between two and ten. What is the least possible value of $ \frac{A \times B \minus{} C}{D}$? Express your answer as a common fraction.

ANd just to clarify, that should be

(3x5-9)/7, including the parentheses.


I know that this website has practically every AIME problem ever on it:

http://www.kalva.demon.co.uk/

Does anyone know of another website that has a similarly comprehensive directory of AMC 12 problems?

I don't think there is a comprehensive one. Search the AMC forum and you'll find various websites that have a number of old tests.

However, buying the tests for $\$$1 a piece from the AMC would always work.

Some of the old AHSME/AMC tests (80's and '95-'01) are available in [url=http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm]this web site[/url] ;)

[quote="Valentin Vornicu"]We are just building one now on AoPS - MathLinks.[/quote]

Very cool!

[quote="Valentin Vornicu"]We are just building one now on AoPS - MathLinks.[/quote]
Awesome!

[quote="chess64"][quote="Valentin Vornicu"][quote="tetrahedr0n"]I don't think there is a comprehensive one.[/quote]We are just building one [b]now[/b] on AoPS - MathLinks. Check out http://www.mathlinks.ro/Forum/resources.php?c=182 for further details ( at http://www.mathlinks.ro/Forum/resources.php you can view the whole thing).[/quote]

Cool! But...isn't that illegal? :?[/quote]

There was a discussion about this in Moderator's forum and the fact is that AoPS received a "silent permission" to post the AMC problems in here. And even we didn't, most of the problems are here anyway.  ;)

It's very cool (I agree!!) but be patient on seeing the problems. It's going to take a lot of time for people to post all the problems.

I have to say this one thing though: Thanks to Valentin for making this happen with his computer skill.  :D  :)

[quote="Silverfalcon"]I have to say this one thing though: Thanks to Valentin for making this happen with his computer skill.  :D  :)[/quote]Hehe ...  :blush:


Let $OX$, $OY$ be two rays. Line $l$ meets $OX$, $OY$ at $A$ and $B$. $C$ and $D$ are on $l$ and on the different sides of $AB$ such that $AC=BD$.

A line through $C$ meets $OX$ at $K$ and $OY$ at $L$; a line through $D$ meets $OX$ at $M$ and $OY$ at $N$.

Line $NK$ meets $l$ at $E$, line $ML$ meets $l$ at $F$.

Prove: $AE=BF$.

Use the projective geometry: O :arrow: $\infty$ and the problem becomes rather simple.

My proof using Menalaus.

In $\triangle{AOB}$ and line $NKE$, by Menelaus theorem, we have:
\[ \frac{AK}{KO} \cdot \frac{ON}{NB} \cdot \frac{BE}{EA} = 1 \]
In $\triangle{AOB}$ and line $MLF$:
\[ \frac{AM}{MO} \cdot \frac{OL}{LB} \cdot \frac{BF}{FA} = 1 \]
In $\triangle{AOB}$ and line $LKC$:
\[ \frac{AK}{KO} \cdot \frac{OL}{LB} \cdot \frac{BC}{CA} = 1 \]
In $\triangle{AOB}$ and line $MND$:
\[ \frac{AM}{MO} \cdot \frac{ON}{NB} \cdot \frac{BC}{CA} = 1 \]
The last two equations tells us:
\[ \frac{AK}{KO} \cdot \frac{OL}{BL} \cdot \frac{AM}{MO} \cdot \frac{ON}{NB} = 1 \]
Multiply the first two equations togather and pludge the result above in, we have:
\[ \frac{BE}{AE}=\frac{AF}{BF} \]
\[ \frac{AB}{AE}=\frac{AB}{BF} \]
\[ AE=BF \]


A circle with centre I is the incircle of the convex quadrilateral ABCD. The diagonals AC and BD intersect at the point E. Prove that, if the midpoints of the segments AD,BC, and ID are collinear, then AB=CD.

ohh, sry. I got one thing wrong. Pls change to "AD,BC and IE are collinear,"


Solve the diopantine equation

$ \arctan(1/a) \plus{} \arctan(1/b) \plus{} \arctan(1/c) \equal{} \frac\pi 4$.

[hide="answer"]
Use the identity
\[ \arctan x \plus{} \arctan y \equal{} \begin{cases} \arctan \frac{x\plus{}y}{1\minus{}xy}\plus{}\pi & xy>1 \\ \arctan \frac{x\plus{}y}{1\minus{}xy} & xy<1 \end{cases}\]
to reduce the problem to
\[ 1\minus{}ab\minus{}ac\minus{}bc\equal{}a\plus{}b\plus{}c\minus{}abc\]
Use your favorite diophantine techniques to find the solutions (listed only those with $ a\leq b\leq c$)
\[ (a,b,c)\in \lbrace (\minus{}7,2,2),(\minus{}3,\minus{}2,0),(2,4,13),(2,5,8),(3,3,7)\rbrace\]
there are also infinitely many other solutions, though they are in some sense the trivial ones:
\[ (a,b,c)\in \lbrace (\minus{}k,1,k): k\in \mathbb{N}\rbrace\]
[/hide]


Because the remainder of the orchestra does not play string instruments, $ 100\% \minus{} 60\% \equal{} 40\%$ do not play string instruments.
$ 40\%\text{ of } 130 \text { members} \equal{} .4*130 \equal{} \boxed{52}\text{ musicians}$.

Multiply $.60$ by $130$ in order to find out how many musicians DO play string instruments - $78$ of them. Now subtract $78$ from $130$ to find how many musicians do NOT play string instruments. This leads us to $52$ musicians who do not play string instruments.

[quote=Juno]What is this problem doing in C&P?
[hide=Solution]$10\%$ of $130$ is $13$ and $13\cdot4=\boxed{52}$.[/hide][/quote]

Is it under Venn Diagrams?

[quote=titanium7]What's an instrument?[/quote]

Bruh

[quote=GameBot]In an orchestra, $ 60\%$ of the musicians play string instruments, and the remainder play woodwinds, brass, or percussion instruments. If the orchestra has 130 members, how many musicians would not play string instruments?[/quote]

Just do [hide=Solution]$\frac{2}{5}\cdot 130=52$[/hide]
Easier than complementary counting

[quote=titanium7]What's an instrument?[/quote]

A tool or implement, especially one for delicate or scientific work.

A measuring device used to gauge the level, position, speed, etc. of something, especially a motor vehicle or aircraft.

@above also something that you make music with: a piano, a clarinet, a saxophone, a violin, a drum, etc.

[color=#f00]REDACTED[/color] [size=50][color=#eee]But it was still a troll

[quote=happyhippos][quote=titanium7]What's an instrument?[/quote]

This was obviously a troll. :mad:[/quote]

Not necessarily, some people don't have the same experiences you do. Some people may not know what one is. Although this one might be a troll, not everyone knows what an instrument is.

[quote=Possible][hide=Sol]Use complementary counting. Subtract the invalid cases from the total. Ans = 130-.6*130 = .4*130 = 52[/hide][/quote]

Is it possible for possible to post solutions to every possible alcumus problem?

@above (S)he's trying their best. Probably possible will succeed


The area of a triangle is 600 ft$ ^2$. If the length of the base is 30 feet,
then what is the number of feet in the altitude?

The area of a triangle= 1/2bh.
Since 600=1/2bh,
bh=1200.
Since b=30,
30h=1200.
h=40.


Let $ a,$ $ b$ and $ c$ are non-negative numbers and
$ u=\frac{a+b+c}{3},$ $ v=\sqrt{\frac{ab+ac+bc}{3}},$ $ w=\sqrt[3]{abc}.$
Prove that $ 7u+w\geq8v.$

Yes, Yulia! My congratulations! :lol:
Try it:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156421
It's  little harder. :wink:

I wonder this inequality can be solved geometrically.

$ \frac{7u+w}{8}\geq v\Longleftrightarrow \frac{7u+w}{1+7}\geq v$.

Let $ t=\frac{7u+w}{8}$, which is the internally dividing point such that $ wt: tu=7: 1$


1) Let $ a,b,c \geq\ 0$ and $ ab \plus{} bc \plus{} ca > 0$ , $ k \geq\ 4$ . Prove that :

$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}} \plus{} k \sqrt {\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \plus{} c)^{3}}} \geq\ 2\sqrt {k}$

The equality occurs when :

$ (a,b,c) \equal{} (\frac {\sqrt {k(k \minus{} 4)} \plus{} (k \minus{} 2)}{2}x,x,0)$

$ (a,b,c) \equal{} (\frac {\sqrt {k(k \minus{} 4)} \minus{} (k \minus{} 2)}{2}x,x,0)$

Welcome  :)


probability
combinatorics unsolved
combinatorics

#1Suppose that you are playing balckjack against a dealer. In a
freshly shuffled deck, what is he probability that neiither you nor the
dealer is dealt a black jack?

#2An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3
balls is randomly selected, what is the probability that each of the
balls will be of the same color. (Repeat under the assumption that
whenever a ball is selected, its color is noted and it is then replaced in
the urn before the next selection).

For 1. wouldn't it just be the probability of (you not being dealt a black jack and not being dealt it again and dealer not being dealt one and dealer not being dealt one again)? So,

$\frac{50}{52}\cdot\frac{49}{51}\cdot\frac{48}{50}\cdot\frac{47}{49}=\frac{188}{221}\approx 0.85$.

according to the answer key, the correct answer should be .905

Sorry, I've never been too good at these, but since no one else is posting:

2. Find $P(R|RR)\cup P(B|BB)\cup P(G|GG)=\frac{5}{19}\cdot\frac{4}{18}\cdot\frac{3}{17}+\frac{6}{19}\cdot\frac{5}{18}\cdot\frac{4}{17}+\frac{8}{19}\cdot\frac{7}{18}\cdot\frac{6}{17}=0.0888$

And for the second condition $P=\left(\frac{5}{19}\right)^{3}+\left(\frac{6}{19}\right)^{3}+\left(\frac{8}{19}\right)^{3}=0.1244$.

Possible misinterpretation of "black jack"? Possibly the specific card combination considered a black jack?

This is a double post, and these problems are not at the Olympiad level.   Anyway, if you are not placing the marbles back in, the answer is
\[\frac{\binom{5}{3}+\binom{6}{3}+\binom{8}{3}}{\binom{19}{3}}.\]

If you place the a selected marble back in after each selection, the answer is
\[\frac{5^{3}+6^{3}+8^{3}}{19^{3}}.\]

The answers are self explanatory.


[size=150][i][u]The Circus Act [/u][/i][/size]

You may have seen or heard about the circus act in which someone dives off a high platform into a small tub of water. Well, the Interactive Circus Troupe has come up with a new wrinkle on this act.

They have attached the diver’s platform to one of the seats on a Ferris wheel, so that it sticks
out horizontally, perpendicular to the plane of
the Ferris wheel. The tub of water is on a moving cart that runs along a track, parallel to the plane of the Ferris wheel, and passes under the end of the platform.

As the Ferris wheel turns, an assistant holds the diver by the ankles. The assistant must let go at exactly the right
moment, so that the diver will land in the moving tub of water.
If you were the diver, would you want to trust your assistant’s on-the-spot judgment? A slight error and you could get a ‘splat!” instead of a”splash!”

Your Task
The diver has insisted that the circus owners hire your group to advise the assistant. You need to figure out exactly when the assistant should let go. (Your analysis will be tested carefully on a dummy before it is used with a human being.)

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

Although you solved a simplified version in Moving Cart, Turning Ferris Wheel, now you will take into account that the diver leaves the Ferris wheel with some initial velocity, with both horizontal and vertical components.

Again, here are the details about the setup for the act.
The Ferris wheel has a radius of 50 feet.
• The center of the Ferris wheel is 65 feet above the ground.
• The Ferris wheel turns counterclockwise at a constant
rate, making a complete turn every 40 seconds.
• When the cart starts moving, it is 240 feet to the left of  the base of the Ferris wheel.
• The cart moves to the right along the track at a constant speed of 15 feet per second.
• The water level in the cart is 8 feet above the ground.
• The cart starts moving as the platform passes the 3 o’clock position.

As before, let t = 0 represent the time when the Ferris wheel passes the 3 o’clock position, which is also when the cart begins moving. Let Wrepresent the number of seconds that elapse between t = 0 and the moment when the diver is released. Your assignment is to answer this question.

[b]For what choice of W will the diver land
in the tub of water? [/b]
---------------------------------------------------

you should probably ask a more specific question, like where you are stuck, how to get started, or whether you're on the right track, stuff like that.  cuz at my school this was used a unit long assessment, so...

I'm very lost in the unit, anyways, I got a long list of equations and the problem. I just don't know how to get started finding the time.

At the end of this page, these are the equations associated with the unit.
http://departments.cvuhs.org/allison/high_dive.htm

I'm really frustrated and need help!!


[hide] If the area is 64 the side must be 8 if the side is 8 the diagonal is $8\sqrt{2}$[/hide]

[hide=":-) click to reveal solution :-)"]THe area of a square can be written as:
$A=s^2$
were s is the length of a side. So plugging in 64 for A we get:
$64=s^2$
or
$\sqrt{64}=s$
We know that the square root of 64 is 8 so we can write:
$s=8$
The diagonal of a square can be written as
$s\sqrt{2}$
where s is the length of a side.  We know that 8 ist the length of the side so we have:
$\text{diagonal}=\boxed{8\sqrt{2}}$[/hide]

[hide]let a be the lenght of side of this squre and d be the lenght of diagonal of it we have $\rightarrow a^2=64 \rightarrow a=8$ and we have $2a^2=d^2 \rightarrow 128=d^2 \rightarrow d=8\sqrt{2}$and we are done.[/hide]

[hide]
If you cut the square in half, you have a 45-90-45 triangle; which means that the hypotenuse is $a\sqrt{2}$ where a is one of the legs.
So, the answer is $8\sqrt{2}$[/hide]

[hide]The side length of the square is 8. Since drawing a diagonal cuts the square into two 45-45-90 triangles with the side lengths as legs and the diagonal as the hypotenuse, the length of the diagonal is 8*rt2[/hide]

[quote="MCrawford"]A square has area 64 cm$^2$.  What is the length of a diagonal of the square?[/quote]

[hide]
Suppose a side of a square is x.
x^2=64
x=+- 8
-8 is not possible.

So the length of the square is 8

(8^2+8^2)^(1/2)=(64+64)^(1/2)=128^(1/2)

That is about [color=darkblue]8*2^(1/2)[/color] (or you can just use 45-45-90 triangles)

You know it is 45 degrees because 2 sides of the squares are always the same. Therefore it is isoceles.

The ratio of a 45-45-90 is $1$:$1$:$2^1/2$
If the smaller side is 8 (The hyp is always larger than any leg)
Then the hypotenuse would be 8 root 2 :D :P[/hide]

[hide]
The side length is 8
The diagonal is $8\sqrt{2}$[/hide]


modular arithmetic
number theory
relatively prime

What is the remainder when $ 2009^{2009}$ is divided by $ 11$?

I think you messed up a bit at the end (it should be $ 2009^9$)

$ 2009^{10}\equiv 1\bmod 11$ by FLT, and $ 2009\equiv 9\bmod 10$, so $ 2009^{2009}\equiv 2009^9\equiv (\minus{}3)^9\bmod 11$

$ 3^5\equiv 1\bmod 11$, so we have $ \minus{}1\cdot 4^4\equiv \minus{}3\equiv \boxed{8}\bmod 11$.

I haven't gone back to check it, but I probably messed up a remainder somewhere.

$ 2009 \equal{} 7 mod 11 \equal{} \minus{}4 mod 11$
4^5 also equals 1 mod 11 however I see you unawaredly fixed it up on the last step  :P
Jrav's error is he has $ 2009^{19} \equal{} 2009^{11}*2009^8 \equal{} 2009*2009^9 \equal{} 2009^9$


abstract algebra
superior algebra
superior algebra unsolved

In an exercise of my book, it defines "weak ring" (literally translated from Chinese) as follows.

$(R,+,\cdot)$ is a weak ring, if it satisfies:
(1) $(R,+)$ is an additive group; (it is abelian)
(2) $(R,\cdot)$ is a semigroup;
(3) There exist fixed elements $e_1,e_2$ such that
$x(y+z)=xy+xz-e_1$ and $(y+z)x=yx+zx-e_2$ for all $x,y,z\in R$.

I cannot find such a name on internet...
Is it a special ring? Or just an exercise?

I think this might be called a semiring...here's a link:

http://mathworld.wolfram.com/Semiring.html

It isn't. The distributivity is different.


For (almost?) any prime p, there exists a divisibility rule so that when a number's unit digit is multiplied by a constant k, and subtracted from the rest of the number and the result is divisible, then the number itself is too. (for example, the divisibility rule for 7 is k=2, so 49 would be $49\equiv 4-18\equiv 14\equiv 0\pmod{7}$, so it is divisible)

Try to do the following problems:

1) Is the almost at the top correct to add or not?  Prove your answer!
2) Now, try to develop a formula and/or an algorithm to find divisibility rules for any prime p.

1)
There is such k for all primes except 2 and 5.

If p is not 2 or 5:
Let k be an integer, which satisfies 10k=-1(mod p). This k always exists.
p|a-kb
-> p|10(a-kb)
-> p|10a-10kb
-> p|10a-10kb+(10k+1)b (because p divides 10k+1)
-> p|10a+b
so we are done.

If p is 2:
Suppose that there is such k. Consider the contrapositive of the statement. Then for 11 and 21, 1-k and 2-k do not divide 2. This is contradiction.

If p is 5:
Suppose that there is such k. Consider the contrapositive of the statement. Then for 11, 21, 31, 41, and 51, 1-k, 2-k, 3-k, 4-k, and 5-k do not divide 5. This is contradiction.

For 2), just repeat the divisibility rule(I think the number should get smaller), if p is not 2 or 5. If p is 2 or 5, look at the last digit.


Call a number "prime-looking" if it is composite but not divisible by 2, 3, or 5. The three smallest prime-looking numbers are 49, 77, and 91. There are 168 prime numbers less than 1000. How many prime-looking numbers are there less than 1000?

Is there any other way to do this other than trial and error?

Use the Principle of Inclusion-Exclusion, commonly referred to as PIE.


What significance do grades have in college? I assume they play a role in admission to graduate school -- probably more so if you want to be a lawyer or doctor? -- and maybe also in whether you could do a 5 year combined masters/b(s/a) program, but what else?

Thanks for the help.

College grades are often regarded in employment applications, at least for your first job or two out of college. That varies from industry to industry, and of course some colleges are more impressive with "average" grades than some other colleges are with high grades, but it is usually expedient to keep your grade average above about 3.5 for many purposes.


geometry
trapezoid
circumcircle
trigonometry
geometry proposed

Let $ABCD$ be a trapezium with $AB=a, CD=b$ $AC\perp BD$ and $AD$ cuts $BC$ at $F$ such that $\hat{DFC}= 45^\circ$. Caculate areas of $ABCD$

We have: $\angle AFB=\frac{1}{2}\angle AOB,$ so $O$ is the circumcenter of triangle $AFB$
Thus $OA=OB$ and then $ABCD$ is an isoceles trapezium.
And now, you can easy to show that: $S_{ABCD}=\frac{1}{4}ab(a+b)$

[quote="April"][color=darkred][b]The general problem.[/b] Let $ABCD$ be a trapezium $(AB\parallel CD)$ with $AB=a,\, CD=b,\,(a<b),\,AC\perp BD,\, AD\cap BC=E$ and $\widehat{CED}= \alpha.$
Caculate area of $ABCD$[/color][/quote]
[color=indigo][u][b]Proof.[/b][/u] Denote: $O=AC\cap BD;\,I,\,J$ be the midpoints of $AB,\,CD$ respectively.
easy to see that: $E,\,I,\,O,\,J$ are collinear and $IJ=\frac{a+b}{2}$
We have: $\frac{EI}{EJ}=\frac{AB}{CD}=\frac{a}{b}\Longrightarrow \frac{IJ}{EJ}=\frac{b-a}{b}\Longrightarrow EJ=\frac{b}{b-a}IJ=\frac{b(a+b)}{2(b-a)}$

On the otherhand, in $\triangle ECD,$ we have:
$\left\{\begin{array}{c}4EJ^{2}=2(EC^{2}+ED^{2})-CD^{2}\\ CD^{2}=EC^{2}+ED^{2}-2EC\cdot ED\cos\alpha\end{array}\right\|\Longrightarrow\left\{\begin{array}{c}\frac{b^{2}(a+b)^{2}}{(b-a)^{2}}=2(EC^{2}+ED^{2})-b^{2}\\ 2(EC^{2}+ED^{2})=2b^{2}+4EC\cdot ED\cos\alpha\end{array}\right\|\\ \Longrightarrow\frac{b^{2}(a+b)^{2}}{(b-a)^{2}}=b^{2}+4EC\cdot ED\cos\alpha\Longrightarrow\frac{1}{2}EC\cdot ED\sin\alpha=\frac{1}{8}\tan\alpha\cdot b^{2}\left(\frac{(a+b)^{2}}{(b-a)^{2}}-1\right)\\ \Longrightarrow S(ECD)=\frac{1}{2}\tan\alpha\frac{ab^{3}}{(b-a)^{2}}$ $\Longrightarrow S(ABCD)=\left(1-\frac{a^{2}}{b^{2}}\right)S(ECD)=\frac{b^{2}-a^{2}}{b^{2}}\frac{1}{2}\tan\alpha\frac{ab^{3}}{(b-a)^{2}}=\frac{1}{2}\tan\alpha\frac{ab(a+b)}{b-a}$[/color]


limit
trigonometry
calculus
calculus computations

Calculate: \[\lim_{x\to 0}\frac{1-\cos x\cdot\sqrt{\cos 2x}}{x^{2}}\]

Trivial by de L'Hôpital's.
Another solution is \[\begin{aligned}\lim_{x\to 0}\frac{1-\cos x\sqrt{\cos 2x}}{x^{2}}& = \left(\lim_{x\to 0}\frac{1}{1+\cos x\sqrt{\cos 2x}}\right)\left(\lim_{x\to 0}\frac{1-\cos^{2}x\cos 2x}{x^{2}}\right) \\ & = \frac{1}{2}\lim_{x\to 0}\frac{1-\cos^{2}x(1-2\sin^{2}x)}{x^{2}}\\ & = \frac{1}{2}\lim_{x\to 0}\left(\frac{\sin^{2}x}{x^{2}}+2\frac{\sin^{2}2x}{(2x)^{2}}\right) = \frac{1}{2}(1+2) = \frac{3}{2}\end{aligned}\]

Listen, this is really not so hard: we have forums specifically for calculus problems, and that includes limits.  Please put problems like this in those forums, where they belong.

[i]Mod edit: Problem originally appeared in the Pre-Olympiad forum and was subsequently moved.[/i]

hmmm I must be missing something... I don't understand how you got from line 2 to line 3?

[quote="JBL"]Listen, this is really not so hard: we have forums specifically for calculus problems, and that includes limits.  Please put problems like this in those forums, where they belong.[/quote]
Oh I'm sorry  :blush:
And here is another solution  :D
We have: \[\lim_{x\to 0}\frac{1-\cos x\sqrt{\cos 2x}}{x^{2}}=\lim_{x\to 0}\left(\frac{1-\cos x}{x^{2}}+\frac{\cos x(1-\sqrt{\cos 2x})}{x^{2}}\right)=\lim_{x\to 0}\left[\frac{2\sin^{2}\frac{x}{2}}{x^{2}}+\frac{\cos x(1-\cos 2x)}{x^{2}(1+\sqrt{\cos 2x})}\right]\\ =\lim_{x\to 0}\left[\frac{1}{2}\cdot\frac{\sin^{2}\frac{x}{2}}{\left(\frac{x}{2}\right)^{2}}+\frac{2\sin^{2}x}{x^{2}}\cdot\frac{\cos x}{1+\sqrt{\cos 2x}}\right]=\frac{1}{2}\cdot 1+2\cdot 1\cdot\frac{1}{2}=\frac{3}{2}\]


Find such digits x and y for which the 6-digit number xyxyxy (in base 10) has a 3-digit prime divisor.

Of course, you can take hours to attack it the hard way. How do you do it a little quicker?

[hide]Let the number $xyxyxy = a$. Then we have

$100000x + 1000x +10x + 10000y + 100y + y = a = 101010x + 10101y$

$10x \cdot 10101 + 10101y = a$

$(10x + y)(10101) = a$

$10101 = 3 \cdot 3367$ and $10x + y$ can't ever be a three-digit number, so it's impossible for the number $xyxyxy$ to have a three-digit prime divisor.[/hide]

There should be one, at least in the context of the problem.

tarquin:  Don't you still have to show whether or not your 4-digit number contains a 3-digit factor?

236factorial:  Why don't you list the source of your problems?  A problem quoted from a reputable source is more likely to have an answer than a problem your classmate made up.

[quote]Find such digits x and y for which the 6-digit number xyxyxy (in base 10) has a 3-digit prime divisor. [/quote]

$XYXYXY_{10}=(10x + y)(10101)$ as tarquin said, so

\[XYXYXY_{10}=(10x + y)(3 \cdot 7 \cdot 13 \cdot 37)\]
The prime factors of $XYXYXY_{10}$ are 3,7,13,37, and the prime factors of $10x+y$, none of which are 3 digits in length. Therefore, $XYXYXY_{10}$ has no prime 3-digit factors.

This problem is from the ABACUS International Math Challenge for 7th and 8th graders.

I thought 3367 was a prime number so I just left it as is. :o


ok these are all true or false. the program i use for this homework makes me get them all correct for the the problem to be right. Sadly it wont tell me which ones are wrong. I cant figure out which ones are so if someone would like to help me that would be great

7. The graph of a linear polynomial is a straight line.

8.The degree of a trinomial is at least 2.

9.The product of two polynomials is always a polynomial.

10. The quotient of two polynomials is always a polynomial.

11. The sum of two polynomials is always a polynomial.

12. The difference of two polynomials is always a polynomial.

[hide="All at once"] A linear polynomial is a straight line.  That's what "linear" means.

A trinomial in [i]one variable[/i] is at least degree two, since it must have at least three different terms.

The product of two polynomials is always a polynomial; one way to think about a polynomial is as an expression that can be obtained using a finite amount of addition, subtraction, and multiplication.

The quotient of two polynomials is [b]not[/b] always a polynomial.  For example, $\frac{1}{x}$ is not a polynomial.

The sum of two polynomials is always a polynomial; see above.

The difference of two polynomials is always a polynomial; see above. [/hide]


Let I be the incentre of the triangle XYZ, D the intersection point of the exterior bisector of the angle XZY with the interior bisector of the angle ZXY. Also $ E\in [ID]$ so that $ \angle YXZ \equal{} 2\angle YZE$. Prove that $ E$ is the midpoint of $ ID$ and $ YEZ$ is isosceles.

$ \widehat{YZE}\equal{}\widehat{YXE}\Rightarrow XYEZ$ is cyclic.
Then $ \widehat{EYZ}\equal{}\widehat{EXZ}\equal{}\widehat{ZYE}\Rightarrow\triangle YEZ$ is isosceles.
We have $ \widehat{EIZ}\equal{}\widehat{EZI}\equal{}\frac{A}{2}\plus{}\frac{B}{2}\Rightarrow IE\equal{}EZ$.
$ \triangle IZD$ is a right triangle $ \Rightarrow\widehat{EDZ}\equal{}\widehat{EZD}\Rightarrow EZ\equal{}ED$, so $ IE\equal{}ED$


inequalities
number theory unsolved
number theory

Let $ P(n)$ denote the product of all digits of a positive integer $ n$. Prove that for any positive integer $ a$, the sequence $ (a_n)$ defined by $ a_1\equal{}a, a_{k\plus{}1}\equal{}a_k\plus{}P(a_k)$ for $ k\geq 1$ is bounded.

[quote="easternlatincup"]Let $ P(n)$ denote the product of all digits of a positive integer $ n$. Prove that for any positive integer $ a$, the sequence $ (a_n)$ defined by $ a_1 = a, a_{k + 1} = a_k + P(a_k)$ for $ k\geq 1$ is bounded.[/quote]

$ a_n$ is a non-decreasing sequence.

$ 1)$ Assume $ a_n<10^{30}^$ $ \forall n$. Then $ a_n$ is bounded. Q.E.D

$ 2)$ Assume now exists $ m$ such that $ 10^{k+1}>a_m\geq 10^{k}\geq 10^{30}$
$ 2.1)$ Assume then $ a_n<10^{k+1}^$ $ \forall n$. Then $ a_n$ is bounded. Q.E.D
$ 2.2)$ Assume then exists $ p$ such that $ a_p\geq 10^{k+1}$
Then exists $ q$ such that $ a_{q+1}\geq 10^{k+1}>a_q\geq 10^k\geq 10^{30}$
So ${ P(a_q})\leq 9^{k+1}$
And, since $ k\geq 30$, $ 9^{k+1}<10^k$ (this inequality is true as soon as $ k>\frac{\log(9)}{\log(10)-\log(9)}=20.85\ldots$).
So $ P(a_q)<10^k$ and so $ a_{q+1}<a_q+10^k<10^{k+1}+10^{k}$

So $ 10^{k+1}\leq a_{q+1}<10^{k+1}+10^{k}$
And so the two leftmost digits of $ a_{q+1}$ are $ 10$ and $ P(a_{q+1})=0$ and so $ a_n=a_{q+1}$ $ \forall n>q$ and $ a_n$ is bounded. Q.E.D.


inequalities
geometry
perimeter
trigonometry
geometric transformation
inequalities unsolved

i have a big problem understanding the 2nd and 3rd relations on the
8th page in hojoo lee's paper on inequalities
his paper can be downloaded from here
http://www.mathlinks.ro/Forum/viewtopic.php?p=680991
(the second post)
thank you for any explanations to how one derives the 2nd and 3rd
formulas on the 8th page of that document.
Merry Christmas ! :)

What formulas are you referring to?

  Darij

i am sorry to see that the version of tin2006.pdf that i have printed
on paper is different from the one on the link that i have pointed out.
if i remember well,and i will correct myself if i am wrong,there was a
geometric construction like this
(it was problem 4 in the version of his paper that i have,
 it was exactly after erdos-mordell inequality)
let $\Delta ABC$ be an acute triangle
let D be the foot of the perpendicular from A to BC
analogous to E ,C,AB
analogous to F,B,AC
analogous to Q,A,EF
analogous to  P,C,ED
analogous to  R,B,FD

now one must proove that
$p(ABC)p(DEF)\geq p(PQR)^{2}$
where p represents perimeter

i am think that this was the statement of the problem,
i am at a netcafe,and i trust my memory that the statement i
wrote is correct,if not i promise to correct it in a day.

now the solution presented some equalities concenrning
the $\mathbf{R}$ of the triangle $ABC$ and
that was used to express some of the lines $DP,DR,DQ$
something like $DR=2\mathbf{R}cosA$

darij , if you have the version of the pdf where problem 4 is
similar to what i wrote here,then you can fill in the gaps,
if not i promise to post the problem exactly from there
in a day

Actually a question like this belongs into the geometry section...

Let ABC be an acute-angled triangle, and let D, E, F be the feet of its altitudes from the vertices A, B, C, respectively. Let R be the orthogonal projection of the point B on the line FD. Then, $DR=c\cos A\cos B$.

[i]Proof.[/i] Since < CFA = 90° and < CDA = 90°, the points F and D lie on the circle with diameter CA, so that the quadrilateral CDFA is cyclic. Thus, < CAF = 180° - < CDF. Hence, A = < CAF = 180° - < CDF = < BDR. Thus, in the right-angled triangle BRD, we have $DR=BD\cos\measuredangle BDR=BD\cos A$. Now, in the right-angled triangle ADB, we have $BD=AB\cos\measuredangle ABD=c\cos B$, and thus $DR=BD\cos A=c\cos B\cos A=c\cos A\cos B$, and the proof is complete.

I hope this was of help - I have proven $DR=c\cos A\cos B$ rather than your $DR=2R\cos A$ because $DR=2R\cos A$ is wrong, so it is possible that you meant something different, but the above should have given you an idea about how relations like this are proven. In fact, most trigonometric relationships in this configuration are pretty easy to prove. See also http://www.mathlinks.ro/Forum/viewtopic.php?t=24870 for a number of relationships with the orthocenter of an acute-angled triangle.

  Darij

i saw several times this "cyclic" term altough i never got to know what it means
could you explain?
also if possible give link or explain other common concepts used

[quote="spx2"]i saw several times this "cyclic" term altough i never got to know what it means[/quote]

A polygon is called [i]cyclic[/i] if there exists a circle passing through all of its vertices. It's that easy.

If you are trying to get used to geometry, check out
[url=http://www-math.mit.edu/~kedlaya/geometryunbound/]Kedlaya's Geometry Unbound[/url],
[url=http://cut-the-knot.com/geometry.shtml]Alexander Bogomolny's Cut The Knot[/url],
[url=http://michaj.home.staszic.waw.pl/prasolov.html]Victor Prasolov's geometry books in Dmitri Leites' translation[/url]
among other sources. A link list can be found on my homepage:
http://www.cip.ifi.lmu.de/~grinberg/Links.html

  Darij


As I promised, I posted the proof of this inequality. Here is the link: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=102983[/url].

Here is another solution with only AM-GM. We have to show that
$\sqrt[3]{(a+x)(b+y)(c+z)}\geq \sqrt[3]{abc}+\sqrt[3]{xyz}$. Get
this on third power so it should be $(a+x)(b+y)(c+z)\geq abc+xyz+3\sqrt[3]{a^{2}b^{2}c^{2}xyz}+3\sqrt[3]{abcx^{2}y^{2}z^{2}}$.

$(a+x)(b+y)(c+z)=abc+xyz+(abz+ayc+xbc)+(xyc+xbz+ayz)$,
but by AM-GM $abz+ayc+xbc\geq 3\sqrt[3]{a^{2}b^{2}c^{2}xyz}$ and $xyc+xbz+ayz\geq 3\sqrt[3]{abcx^{2}y^{2}z^{2}}$, and this is the end of the
proof.

Bye!

[quote="t0rajir0u"][quote="Rzeszut"]It is another form of particular case of some generalisation of Cauchy-Schwarz. This generalisation says that: \[\sum_{1\leq i\leq n}a_{i}^{(1)}\cdot a_{i}^{(2)}\cdot ... \cdot a_{i}^{(k)}\leq \sqrt[k]{\sum{{(a_{i}^{(1)})}^{k}}}\cdot...\cdot \sqrt[k]{\sum{{(a_{i}^{(k)})}^{k}}}.\] (we substitute $a=u^{3}$ etc). I will post the proof in Inequalities Theorems and Formulas.[/quote]

Correct me if I'm wrong, but that looks like Jensen's Extension to Holder's Inequality... ?[/quote]
Could you explain what is "Jensen's Extension to Holder's Inequality"? I can't find it on the forum.


Hi,

I came across this problem for long time but still don't know how to solve it. I'm not even sure how to graph this neither.

Find the area enclosed by the graph of $|x-60| + |y| = |\frac{x}{4}|$.

Like what equations do I have to set up?

basically, separate the $y$ from the $x$.  see if you can solve it from there :P

[hide="solution"]$|y| = |\frac x4 |-|x-60|$
$|\frac x4 | \ge |x-60|$
$\frac x4 \ge \pm (x-60)$
$\pm 60 \ge \pm x - \frac x4$
$48 \le x \le 80$ for a distance of $32$

each of the $|\frac x4 |$ and $|x-60|$ are linear terms, so they will form a triangle above and below the axis.  we need only to find the maximum of the expression (the height of the triangle), and multiply it by the base of $32$.  the maximum should occur when $|x-60|$ is lowest, since this increases much faster than $|\frac x4|$ thus $x=60$ and $|\frac x4 |-|x-60|=15$.  $15*32=\boxed{480}$[/hide]

[color=red]Edit: This solution is wrong.[/color]

First draw the graph of $y=|\frac{x}{4}|-|x-60|$. You can go step by step: $x \geq 60$, $0 \leq x \leq 60$, $x \leq 0$. Then transform it to $|y|$.

BTW: I think the area goes to infinity since the graph is NOT closed.

[hide]
You can get some equations by taking the absolute values off, one by one, and replacing it by + or - the contents inside (break it into 2 equations each time).
You end up with 4 linear equations for y.  Then you just have to find the vertices and then the area.  I got the vertices to be:
(48,0) (60,15) (80,0) (60,-15)
and the area to be 480.
[/hide]

I like alan's trick.

Anyway, so I'll get 4 equations (sorry I didn't have time to look at this thread until now.. and I'm eating :D )

I just did 1987 AIME. This is what I got for that problem.

[hide]
$|y| = |x/4| - |60-x|$

1) For $x \ge 60$, we have $y=\pm (-3x/4+60)$. We find $(x,y)=(60, \pm 15), (80, 0)$. Graph this...

2) For $0 \le x \le 60$, we have $y = \pm (5x/4-60)$. We find $(x,y)=(48,0),(60, \pm 15)$. Graph this....

3) If $x \le 0$, it's the same as $x \ge 60$.

Now we have two congruent triangles. The base is $80-48=32$, and the height is $15$ for both, so $32 \cdot 15 = 480$.
[/hide]


modular arithmetic
quadratics
algebra
polynomial
number theory
prime numbers
number theory unsolved

Find all primes p such that $ p\plus{}22$ is a perfect square.

[quote="Bugi"]Find all primes p such that $ p \plus{} 22$ is a perfect square.[/quote]

Hello!

Are you sure the problem is correct ?
A quick analysis gives $ p\equal{}3$ or $ p\equal{}2\pmod 3$ and many solutions :
$ 3\plus{}22\equal{}25\equal{}5^2$
$ 59\plus{}22\equal{}81\equal{}9^2$
$ 419\plus{}22\equal{}441\equal{}21^2$
$ 1499\plus{}22\equal{}1521\equal{}39^2$
$ 2003\plus{}22\equal{}2025\equal{}45^2$
$ 2579\plus{}22\equal{}2601\equal{}51^2$
$ 3947\plus{}22\equal{}3969\equal{}63^2$
$ 7547\plus{}22\equal{}7569\equal{}87^2$
$ 8627\plus{}22\equal{}8649\equal{}93^2$
ansd it seems to have a lot more ... .

I think we have to find the solutions in terms of congruences and cancel those that do not work, which seems difficult.  :( I have, however, deduced that if $ p\neq 3$, then $ p\equiv 59,131\pmod{144}$.

i agree with pco..

Your problem is equivalent to find all $ a \ge 0$ s.t. $ P(a): \equal{}(2a\plus{}5)^2\minus{}22$ is prime. As we must have $ 3 \mid a\minus{}2$ then (other p=3) it is equivalent to find all $ k \ge 0$ such that $ Q(K): \equal{}36k(k\plus{}3)\plus{}43$ is prime.
now i cannot see any kind of divisibility..(will it be still a coniecture? http://www.faculty.iu-bremen.de/sbaier/papers/primequadpoly.pdf )

[quote="Theorem1"]The problem can be reduced like this:
Find all natural primes in the form of : -13+36k²+36k.[/quote]

Hemmm, do you think this is simpler than "Find all natural primes in the form of : $ k^2\minus{}22$" ?

This is equivalent to proving that there is an infinite number of primes $ p$ such that $ p\plus{}22$ is a perfect square.

[quote="Theorem1"]there is infinetly many primes in the form of 5+6k such that:
p+22 is perfect square.[/quote]

[quote="boxedexe"]This is equivalent to proving that there is an infinite number of primes $ p$ such that $ p \plus{} 22$ is a perfect square.[/quote]

 :rotfl:

Attending news from Bugi..

Edit: @Boxedexe: i know that you are referring to the previous post, for that i was smiling!  :D

A clarification: I was referring to Theorem1's last post.  bboypa, your post is ambiguous ( :rotfl: makes it so). :)

Well, the problem is bad then.

I solved some problem, with integers, and came to that the discriminant of the quadratic is $ p\plus{}22$, so it must be a perfect square and so on...

Thanks all!

Without using dirichlet's theorem prove that there is infinitly many primes such that p+22 is perfect square.
Asir tqawad adak zamel "milan" qwad nkhrjou bra wnwrri mouk lqhba.  mathematitien nta ! zbbi fhmt wla walou!
Boxedexe hahahahahahaha! DHK A ZBI RWAH LMORROCO W NTFAHMOU!

Theorem 1, I understand moroccan language...

All these discussions lead to nowhere. The question of infinitely many pairs of primes $ (p, p \plus{} 2n)$, for fixed $ n$ is no more advanced than what we know for $ n\equal{}1$, i.e. for twin primes, and that is ... suspense ... zilch.

[quote="pco"][quote="Bugi"]Find all primes p such that $ p \plus{} 22$ is a perfect square.[/quote]

Hello!

Are you sure the problem is correct ?
A quick analysis gives $ p \equal{} 3$ or $ p \equal{} 2\pmod 3$ and many solutions :
$ 3 \plus{} 22 \equal{} 25 \equal{} 5^2$
$ 59 \plus{} 22 \equal{} 81 \equal{} 9^2$
...
ansd it seems to have a lot more ... .[/quote]
If so,it's same to prove that [b]there are infinitely many prime numbers of the form $ k^2 \minus{} 22,(k \ge 5,k \in Z)$[/b].Probably the solution of this one would be supposing the opposite, and to get a contradiction.If this do not work,so there are a finitely primes which fulfill the problem below,and there's a way to do it  :huh:

From wikipedia  [url]http://en.wikipedia.org/wiki/Formula_for_primes[/url]

"It is not known whether there exists a univariate polynomial of degree at least 2 that assumes an infinite number of values that are prime"

So if there are an infinite number of primes this is an unsolved problem.

I believe it is also true that no univariate polynomial of degree at least 2 is known to have finitely many primes unless it factorises

There is a conjecture of Hardy and Littlewood about the number of primes in a quadratic progression (that does not factorize) below a given bound, and it diverges. It is not known to be true or false for any given quadratic.

In other words, solving this problem would be a significant new breakthrough in number theory.

Hehe, I was solving some other problem, and came to this one, couldn't solve it (the original question had an other idea), now it can be a breakthrough in NT...

Maybe if there are infinitely many solutions, maybe we can find ll $ a$-s such that $ k^2\minus{}a$ are primes for infinitely many $ a$!


search
college
number theory
Berkeley Math Circle

I am searching  the solution of the"combinatorial number theory" -Gabriel D.Caroll,Berkeley Math Circle
Can you help me ?
[url]http://www.google.com/search?q=cache:HINtpNviAskJ:mathcircle.berkeley.edu/cnt.pdf+%22Combinatorial+number+theory%22&hl=vi&gl=vn&ct=clnk&cd=6[/url]

Contact me, Thangtonghop@yahoo.com, i 'll send to you


[quote="asinha"]Our school doesnt have a math club, but if it did, I belive it would be excellent practice for anything really.

Just my $0.02
^_^[/quote]
I think those $0.02 are mine. Can I have them back?

Yes I know about this bug. Please edit the problems where you can or use the $\$ command to create a dollar sign (or disable bbcode in that post).


[quote="blueray4"]okay, my psychology professor uses an online grading system, and i am not getting the grade it keeps telling me i have. i need some reassurance that i am right b4 i go into his office, or am i just screwing up somewhere.

Lab: 27/27 total points; worth 5% grade
Online: 6/6 points; worth 5%
Test1: 88/100 points worth 30%
Test2: 84/100 points worth 30%
Final: 94/100 points worth 30%
Extra Credit: 3/9 points worth 2%

 :wink:

tell me what you get[/quote]

your teacher does not have to do the scores by points...it is acceptable to alter grades based on effort, but i do calculate 90.47...

also don't post stuff like this anymore...

If you're making 90+ then why do you care what your grade is?

Because his grade most likely isn't showing as 90+?


[b]1.[/b] What are all the factors of a polynomial if two of the zeros of the polynomial are  $-17$ and $4+5i$?

[b]2.[/b] If $-4$ is a zero of $f(x) = 6x^{3}+23x^{2}-5x-4$, factor the expression $6x^{3}+23x^{2}-5x-4$.

[b]3.[/b]
[img]http://img249.imageshack.us/img249/8181/degreepolybi3.jpg[/img]

[hide="1"]
$(x+17)$, $(x-4-5i)$, and $(x-4+5i)$.

Complex roots come in conjugate pairs. [/hide]
[hide="2"]
Use synthetic division.

$(x+4)(6x^{2}-x-1)=(x+4)(2x-1)(3x+1)$.[/hide]
[hide="3"]
Number of turns = 6, so the lowest possible degree is 7.

Factors are determined from the x-intercepts:

(x+5), (x-1), and (x-2).

There are 7-3 = 4 complex roots (from the Fundamental Theorem of Algebra). [/hide]

Hey,

How're you doing rnwang2?

I will see these types of problems on my test, so this is really going to help me achieve that 100 %.  Thanks for your help!  :10:  :wow:  :clap:


hmm in HTML you could code the url to say "Click here!"

how do u do that with bbcode on ur signatures?

The bbcode for alternate text is (url=Address)text to display(/url), replacing ( with [. I know that works in posts; it should work in signatures (for legal links such as your weblog, anyway).


function
integration
calculus
algebra
polynomial
geometry
3D geometry

$ X_1,X_2,...$ i.i.d. uniform on (0,1)
$ S_n\equal{}X_1\plus{} ... \plus{} X_n$, $ T\equal{}\inf \{ n: S_n >1 \}$

Show $ P(T>n)\equal{}\frac{1}{n!}$

so the characteristic function of $ X_i$ is $ \frac{e^{it}\minus{}1}{it}$
so the char. func. of $ S_n$ is $ \phi (t)\equal{}\prod_{k\equal{}1}^n \frac{e^{it}\minus{}1}{it}\equal{}\left( \frac{e^{it}\minus{}1}{it}\right)^n$
so the density of $ S_n$ must  be
$ f_n(w)\equal{}\frac{1}{2\pi}\int e^{\minus{}itw}\phi (t) dt\equal{}\frac{1}{2\pi}\int e^{\minus{}itw} \left( \frac{e^{it}\minus{}1}{it}\right)^n dt$
$ \equal{}\frac{1}{2\pi} \sum_{k\equal{}0}^n(\minus{}1)^k \frac{n!}{k!(n\minus{}k)!}\int \frac{e^{it(k\minus{}w)} }{(it)^n}dt$
$ \equal{}\frac{1}{2(n\minus{}1)!}\sum_{k\equal{}0}^n(\minus{}1)^k \frac{n!}{k!(n\minus{}k)!}\frac{(w\minus{}k)^n}{|w\minus{}k|}$

hence $ P(T\le n) \equal{} \int_1^n f_n(w)dw$ and this integral is $ 1\minus{}\frac{1}{n!}$

this is the proof i found online.. i understand whats going on up until the integration for solving $ f_n$ maybe some substitution im not seeing
then im not sure why we can just say the "hence" statement, or how the integral works out to be so nice.

any help would be appreciated.

the equation to find $ f_n$ is just a Fourier inversion. And the conclusion follows from the fact that $ P[T \leq n] \equal{} P[S_n \geq 1]$

i know WHY we have the integral to solve for f_n, but the last step where the inttegral sign disappears is where my question is

also i still dont see how either of the things that $ P(T \le n)$ is can be changed into the integral from 1 to n..?

$ P[T \leq n] \equal{} P[S_n \geq 1] \equal{} \int_1^{\infty} f_n(t) dt \equal{} \int_1^{n} f_n(t) dt$

yeah okay i feel  dumb for not realizing the infinity can change to n,

but i still have issues when i try to calculate the integral in question.  i figure it would be easier to solve $ P(T>n)\equal{}\int_0^1 f_n(w)dw$
$ 2(n!P(T> n))\minus{}1 \equal{} 2\cdot n!\int_0^1 f_n(w)dw \minus{}1 \equal{}2\int_0^1 n\sum_{k\equal{}0}^n(\minus{}1)^k \frac{n!}{k!(n\minus{}k)!}\frac{(w\minus{}k)^n}{|w\minus{}k|}dw\minus{}1$
$ \equal{}n\sum_{k\equal{}0}^n(\minus{}1)^k \frac{n!}{k!(n\minus{}k)!} \int_0^1\frac{(w\minus{}k)^n}{|w\minus{}k|}dw\minus{}1$
$ \equal{}\sum_{k\equal{}1}^n(\minus{}1)^{k\plus{}1} \frac{n!}{k!(n\minus{}k)!} \int_0^1 n(w\minus{}k)^{n\minus{}1}dw$
$ \equal{}(\minus{}1)^{n\plus{}1}\sum_{k\equal{}1}^n(\minus{}1)^k \frac{n!}{k!(n\minus{}k)!} [k^n\minus{}(k\minus{}1)^n]$

but im not sure how to show this clears out to just 1.

In fact $ \sum_{k\equal{}0}^n(\minus{}1)^k \frac{n!}{k!(n\minus{}k)!} f(k) \equal{} 0$ whenever $ f$ is a polynomial of degree $ <n$.
If you use this for $ f(k) : \equal{} k^n \minus{} (k\minus{}1)^n$ you are done. To see that the above equation is true,
you only have to check for $ f(k) \equal{} \binom{k}{l}$ for $ 0 \le l <n$, but this is easy using the identity
$ \binom{n}{k} \binom{k}{l} \equal{} \binom{n}{l}\binom{n\minus{}l}{k\minus{}l}$.

If you don't want to use characteristic functions anyway, an easier way to see
$ P(T>n) \equal{} P(S_n < 1) \equal{} \frac{1}{n!}$ would be to show $ P(S_n < c) \equal{} \frac 1 {n!} c^n$,
which is true for all $ 0<c<1$ and $ n \ge 1$. This is inspired by the fact that $ \{S_n < c\}$ is an
$ n$-dimensional pyramid spanned by $ n$ vectors of length $ c$ that are mutually perpendicular,
and it can easily be shown by induction on $ n$ using $ P(S_n < c) \equal{} \int_0^c dx_n P(S_{n\minus{}1} < c\minus{}x_{n})$.


Bill, Phil, and Jenny are siblings.  Bill is twice as old as Phil.  Jenny is two years younger than Bill.  Currently, their dad is twice as old as the sum of their ages.  In nine years, Dad's new age will be equal to the sum of his three kids' new ages.  What is Jenny's current age?

Let $ b$= Bill's current age, $ p$= Phil's current age, $ j$= Jenny's current age, and $ d$= Dad's current age.

Thus, $ b\equal{}2p$ and $ j\equal{}2p\minus{}2$, so $ d\equal{}2(b\plus{}p\plus{}j)\equal{}2(2p\plus{}p\plus{}2p\minus{}2)\equal{}2(5p\minus{}2)\equal{}10p\minus{}4$

$ d\plus{}9\equal{}b\plus{}9\plus{}p\plus{}9\plus{}j\plus{}9$

$ 10p\minus{}4\plus{}9\equal{}5p\minus{}2\plus{}27$

$ 5p\equal{}20$

$ p\equal{}4$

So, $ j\equal{}2*4\minus{}2\equal{}6$

Let Bill's age be $ b$.

Let Phil's age be $ p$.

Let Jenny's age be $ j$.

Let their Dad's age be $ d$.

(1) $ b \equal{} 2p$
(2) $ j \equal{} 2p \minus{} 2$
(3) $ d \equal{} 2(2p \plus{} 2p \minus{} 2 \plus{} p)$ which equals $ d \equal{} 2(5p \minus{} 2)$ which equals $ d \equal{} 10p \minus{} 4$.

Now that we can write everything in terms of $ p$, we write our last equation.

$ 9 \plus{} 10p \minus{} 4 \equal{} 9 \plus{} 2p \plus{} 9 \plus{} 2p \minus{} 2 \plus{} 9 \plus{} p$

$ 5 \plus{} 10p \equal{} 25 \plus{} 5p$

$ 5p \equal{} 20$

$ p \equal{} 4$

So Paul is 4 years old. We want to find Jenny's age, so we plug this back  into (2).

$ j \equal{} 2p \minus{} 2$

$ j \equal{} 2(4) \minus{} 2$

$ j \equal{} 8 \minus{} 2$

$ j \equal{} \boxed{6}$

EDIT: Beaten to it...


A fair six-sided die is rolled until a sum greater than or equal to 700 is achieved. What is the expected number of rolls needed?

[quote="bertram"]Wild guess: 200

Motivation: On average you'll roll a 3.5 and $ 200\cdot 3.5 \equal{} 700.$[/quote]
if the sum required is 7
the expected no of turns isn't 2
so I think we have to calculate the probability in separate cases ( 117~700 turns)

[quote="madness"]Tenoreoz this sum is similar to the coin probability sum which you posted earlier  :huh:
So again the expected value E(x) is given by $ P(x).x$ where x is the no. of turns required to reach 700
(In this case the computing part is a little tedious but ......)  :wink:[/quote]

that's the point. try out your method and see what you get.

@ bertram

What if you end up rolling a sum greater than 700?

You might be able to bash this with generating functions. If $ e_n$ is the expected number of rolls to get a sum of $ n$, we have the recurrence

$ e_n \equal{} \frac{e_{n\minus{}1} \plus{} e_{n\minus{}2} \plus{} e_{n\minus{}3} \plus{} e_{n\minus{}4} \plus{} e_{n\minus{}5} \plus{} e_{n\minus{}6}}{6} \plus{} 1$
where $ e_{\minus{}5} \equal{} e_{\minus{}4} \equal{} e_{\minus{}3} \equal{} e_{\minus{}2} \equal{} e_{\minus{}1} \equal{} e_0 \equal{} 0$.

This is almost a question from PUMaC this year, however the PUMaC also said that answers within .0001 of the actual answer are correct, and it was this error allowance that made the problem feasible.

Their solution made use of the fact that certain probabilities approach certain values after a lot of rolls, and they are so close that you can assume they are those values without getting too much error (ie within 10^-30 let alone 10^-4).

This problem as it stands seems unlikely that there would be a solution without alot of cases and huge computations.

yea this was a question from PuMAC this year. I actually do have a quick solution but it's slightly off from the answer they give so I wanted to see what everyone else gets since I'm not sure if my answer is correct.


quadratics
algebra
polynomial
induction
number theory unsolved
number theory

$m$ is a natural number.
Given $a_1 = m+1$, $a_{n+1} = {a_n}^2-ma_n+m$
Show that $(a_p,a_q)=1$ for different $p,q$

Let $P(x) = x^2 -mx+m.$
For $m \neq 4$ this quadratic polynomial has no integer root. Since $P(a_n) = a_{n+1}$, it follows that $a_n \neq 0$ for all $n$.
For $m=4$, then $P(x) = (x-2)^2$. Since $a_1 = 5$, it easily follows by induction that $a_n \geq 5$ for all $n$ so that $a_n \neq 0$.

Now, let $n>0$ be given.
Clearly $a_{n+1} = m \mod [a_n]$.
Assume that $a_{k+n} = m \mod [a_n]$ for some $k >0.$ Thus $a_{n+k+1} = m^2-m^2+m = m \mod [a_n].$
It follows by induction that for all $p,q$ such that $p>q$ we have $a_p = m \mod[a_q].$

Assume that $p>q$ such that $(a_p,a_q) > 1$.
Wlog, we may assume that $q$ is minimal.
Therefore, if a prime $r$ divides both $a_p$ and $a_q$, it divides $r$. In that case, $r$ divides $a_n{q-1}^2$ too from the induction construction of the sequence, so that $a_q$ and $a_{q-1}$ are not coprime, which contradicts the minimality of $q$.
Thus, we must have $q=1$ but $r$ cannot divides $m$ and $a_1 = m+1$, a contradiction, and we are done.

Pierre.


what does it mean when a problem say {x} is the largest interger greater than or equal to x

$ \{x\}$

[b]It means the fractional part.[/b]

Yes $ \{x\}$ is the fractional part of the number.

Largest integer greater than or equal to $ x$? Not always defined!

$ \lfloor x \rfloor \equal{} [x]$ denotes the largest integer less than or equal to $ x$, and $ \lceil x \rceil$ denotes the smallest integer greater than or equal to $ x$.

[b]Contribute problems. See link below.[/b][/b]

And $ \{x\}\equal{}x\minus{}\lfloor x\rfloor$.

Example: $ x\equal{}3.1\rightarrow\lfloor x\rfloor\equal{}3, \{x\}\equal{}0.1$

Since its the "fraction part", wouldnt it be better to say that $ \{ 3.1 \} \equal{}\frac{1} {10}$ as opposed to $ 0.1$.

You can write $ \frac{1}{10}$ or $ 0.1$, nobody won't argue. I wrote $ 0.1$ because $ 3.1\equal{}3\plus{}0.1$ (it's easier to understand).

Well, you could say fractional or decimal, then.
Or you could say the non-integer part.

[quote="Bugi"]You can write $ \frac {1}{10}$ or $ 0.1$, nobody won't argue. I wrote $ 0.1$ because $ 3.1 \equal{} 3 \plus{} 0.1$ (it's easier to understand).[/quote]

But $ 3.1\equal{}3\plus{}\frac{1} {10}$ :P

[quote="abacadaea"][quote="Bugi"]You can write $ \frac {1}{10}$ or $ 0.1$, nobody won't argue. I wrote $ 0.1$ because $ 3.1 \equal{} 3 \plus{} 0.1$ (it's easier to understand).[/quote]

But $ 3.1 \equal{} 3 \plus{} \frac {1} {10}$ :P[/quote]

$ \frac {1}{10} \equal{} 0.1$ . No need to argue any more.


calculus
derivative
function
linear algebra
matrix
quadratics
vector

Let $ f:\mathbb R^{n}\rightarrow\mathbb R$ be a scalar field with continuous partial derivatives in the $ n$-ball $ B(\mathbf a)$, where $ \mathbf a$ is a stationary point of $ f$.  Prove that if $ D_{ii}f(\mathbf a )$ and $ D_{jj}f(\mathbf a)$ have different signs for some distinct $ i,j\in\{ 1,2,\ldots, n\}$, then $ f$ is a saddle point.

For a smooth function, Taylor's theorem tells us that

$ f(\mathbf{a}\plus{}\mathbf{h}) \equal{} f(\mathbf{a})\plus{}\nabla f(\mathbf{a})\cdot\mathbf{h}\plus{}\frac{1}{2}\mathbf{h}^{T}H\mathbf{h}\plus{}o(|\mathbf{h}|^{2}),$

where $ H$ is the Hessian matrix whose entries are $ D_{ij}(\mathbf{a}).$

The statement that $ \mathbf{a}$ is a stationary point is the statement that $ \nabla f(\mathbf{a})$ (or $ Df(\mathbf{a}),$ whatever notation you prefer) is zero, meaning that there are no first order terms in this expression. Hence whether $ f(\mathbf{a}\plus{}\mathbf{h})$ is greater or less than $ f(\mathbf{a})$ for sufficiently small $ \mathbf{h}$ is controlled by the quadratic form $ \mathbf{h}^{T}H\mathbf{h},$ except in directions in which that quadratic form is zero.

That means that a sufficient (but not necessary) condition for a saddle point would be that the quadratic form is indefinite. That is, we can find one $ \mathbf{h}$ for which the quadratic form is positive, and another $ \mathbf{h}$ for which the quadratic form is negative.

As an example to show that this condition is not necessary, consider the function $ f(x,y) \equal{} x^{4}\minus{}y^{4}.$ The origin is the only critical point (stationary point) and it is clearly a saddle point. Nonetheless, all second derivatives vanish as the origin, and the Hessian matrix (and hence the quadratic form that I am discussing) is identically zero, which is not indefinite.

Now, to the stated problem: If we let $ \mathbf{h}\equal{}\mathbf{e}_{j},$ the $ j$th standard basis vector, then $ \mathbf{h}^{T}H\mathbf{h}\equal{} D_{jj}(\mathbf{a}),$ which is one of the diagonal elements of the Hessian matrix. If one of those diagonal elements is positive and one negative, then the matrix is indefinite.

Hence, having diagonal elements of opposite sign is a sufficient (but very, very far from necessary) condition for having an indefinite matrix and hence a saddle point.

But it's not a very useful criterion.

Let $ f(x,y) \equal{}\frac{1}{2}x^{2}\plus{}2xy\plus{}\frac{1}{2}y^{2}.$ This has a unique critical point at the origin, and its Hessian matrix is $ H \equal{}\begin{bmatrix}1 & 2\\ 2 & 1\end{bmatrix}.$ Note that both diagonal elements are the same sign. Note also that for this quadratic function, the quadratic form is not merely an approximation of the function, it is the function, exactly. Note also that $ f(h,h) \equal{} 3h^{2}> 0$ but $ f(h,\minus{}h) \equal{}\minus{}h^{2}< 0.$ This is an indefinite quadratic form and a saddle point.

What we need to be able to identify about $ H$ is its [i]signature[/i]: how many positive eigenvalues does it have, how many negative eigenvalues, and how many zero eigenvalues. In other words, it's a question of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159717]congruence[/url] of matrices.


calculus
integration
geometry
calculus computations

Calculate $ \int_R \left(\sqrt{x} \minus{} y^2\right) dx dy$, with $ R$ being the bounded area between $ y \equal{} x^2$ and $ x \equal{} y^4$.

Looks like a straightforward problem. What have you tried; what are you having trouble with?

Thanks for your answer,
I don't know how to get started, how to use the given boundaries.

Probably you should start by drawing the region in question.  Then you can decide how to turn this double integral into an iterated integral.

Can I do it this way:
$ \int_R (\sqrt{x} \minus{} y^2) dx dy \equal{} \int_0^{x^2} \int_0^{y^4} (\sqrt{x} \minus{} y^2) dx dy$ ?

No. That integral is meaningless as stated. You wrote:

$ \int_0^{x^2}\int_0^{y^4}(\sim )\,dx\,dy.$

But the variable $ x$ only exists and has any meaning inside the inner integral. In that inner integral, $ x$ goes from $ 0$ to $ y^4,$ and in doing so, $ x$ is "used up". Once you're outside that integral sign, $ x$ no longer exists; there's only $ y.$

It would be legal for you to say $ \int_0^1\int_0^{y^4}(\sim )\,dx\,dy.$ Legal, in that it's allowed to have $ y$ go from $ 0$ to $ 1.$ (On the other hand, that's not the correct way to do your problem. Your problem has other limits.)

The outer integral's limits of integration must be constants. The inner integral's limits of integration may depend on the outer variable - in fact, they must depend on the outer variable if you're integrating over something other than a rectangle. The inner variable may not appear in any limits of integration.

Now, back to your problem. The boundary curves are $ y\equal{}x^2$ and $ x\equal{}y^4.$ These intersect at $ (0,0)$ and $ (1,1).$ Then there's a curve which is both the lower curve and the right hand curve; this can be written as either $ y\equal{}x^2$ or $ x\equal{}\sqrt{y}.$ And there's a curve which is both the upper curve and the left hand curve; this can be written as either $ x\equal{}y^4$ or $ y\equal{}\sqrt[4]{x}.$ How do I know which is lower right and which is upper left? I drew the picture. There's no substitute for drawing the picture. (Well, technically, I didn't personally draw it - I visualized what the drawing would look like. But it's the same principle.)

If you want to put $ dx$ inside, then you're first going from left to right, and you need both the left curve and the right curve solved for $ x$ in terms of $ y.$ That gives
\[ \int_0^1\int_{y^4}^{\sqrt{y}}(\sqrt{x}\minus{}y^2)\,dx\,dy\]
The alternative (which also works) would be to say
\[ \int_0^1\int_{x^2}^{\sqrt[4]{x}}(\sqrt{x}\minus{}y^2)\,dy\,dx\]


Define $ a\#b\equal{}ab\minus{}a\plus{}b\minus{}8$ for all real numbers $ a$ and $ b$.
$ x \# 3 \equal{} 37$. What is the value of $ x$?

Plug in to get $ 3x\minus{}x\plus{}3\minus{}8\equal{}37 \iff 2x\equal{}42 \iff x\equal{}21$.


quadratics
number theory solved
number theory

Let $T$ be the set of all rational numbers whose reduced form has
an odd denominator. Prove that the equations $y^2 = x^3 + 2x^2 +
2x + 3$ cannot be solved in $T$

i don't think it's a hard problem but i have no good ideas on this one. :maybe:

good luck and thank you

come on people!!! please....

  :maybe: any thoughts welcomed, even if not in connection with the problem... :P

hi!

y^2=x^3+2x^2+2x+3
y^2+1=(x^2+2)(x+2),

y=a/b with a and b coprime, b odd
x=p/q p and q coprime, q odd.

(a^2+b^2)/b^2=(p^2+2q^2)(p+2q)/q^3,

because of the coprimality thing between nominators and denominators in the two fractions, it can be seen that:

b^2=q^3
a^2+b^2=(p^2+2q^2)(p+2q)

since a and b are coprime a^2+b^2 can only have prime factors of the form 4k+1 a conclusion that follwos easily from quadratic residues considerations, so does the other side.
b is odd and so no matter what is a the left hand side is not divisible by 4, so p can't be even. Now it's clear that p^2+2q^2 has prime divisors of the form 4k+1 since itself is of this form, absurd.

psykyk

[quote="psykyk"] Now it's clear that p^2+2q^2 has prime divisors of the form 4k+1 since itself is of this form[/quote]

$p^2+2q^2$ has prime divisors of the form $4k+3$;

thank you for the solution!

Maybe you'll laugh, but I was actually starting to learn things about elliptic curves in order to tackle this one!! :D

Well...I hope elliptic curves will be more useful than to solve an olympiad-type problem  :D

Pierre.


algorithm
number theory
Euclidean algorithm
relatively prime
number theory proposed

Let $ n$ be a positive integer, $ n \ge 2$. Prove that $ n!$ is divisible with $ 1^2\plus{}2^2\plus{}...\plus{}n^2$ if and only if both $ n\plus{}1$ and $ 2n\plus{}1$ are compound.

[hide="Solution"]
$ \sum_{i\equal{}1}^n i^2\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}$
We want to show $ \frac{n(n\plus{}1)(2n\plus{}1)}{6}$ divides $ n!$, or $ m\frac{n(n\plus{}1)(2n\plus{}1)}{6}\equal{}n!$ which is $ (n\plus{}1)(2n\plus{}1)|6(n\minus{}1)!$.

Since both $ n\plus{}1$ and $ 2n\plus{}1$ are greater than $ n\minus{}1$, and $ n \geq 2$, if they are not composite then they cannot divide $ (n\minus{}1)!$.

By the Euclidean Algorithm,
$ gcd(2n\plus{}1,n\plus{}1)\equal{}gcd(2n\plus{}1\minus{}(n\plus{}1),n\plus{}1)\equal{}gcd(n,n\plus{}1)\equal{}1$
Thus, $ 2n\plus{}1$ and $ n\plus{}1$ are relatively prime, and they share no common factors.
$ 2n\plus{}1$ is strictly odd, so its factors must be less than or equal to $ \frac{2n\plus{}1}{3}$, which is less than $ n\minus{}1$. Thus it is clear that if $ n\plus{}1$ and $ 2n\plus{}1$ are not squares, then they can be expressed as the product of 2 pairs of distinct numbers less than $ n\minus{}1$, and so they divide $ (n\minus{}1)!$.

If $ n\plus{}1$ is a square, for $ n\plus{}1>4$, $ \sqrt{n\plus{}1}>2$ so $ n\plus{}1\equal{}\frac{n\plus{}1}{a}$ for some $ a\geq3$, thus $ n\plus{}1$ divides $ (\frac{n\plus{}1}{a})(2\frac{n\plus{}1}{a})$, and both are less than $ n\minus{}1$.

Similar reasoning for $ 2n\plus{}1$ being a square yields a similar result, but with $ 2n\plus{}1>16$, but the only odd square less than 16 is 9, and it is easy to check that for $ 2n\plus{}1\equal{}9$, $ n\equal{}4$ the result is true.

Thus $ (n\plus{}1)(2n\plus{}1)|(n\minus{}1)!$ for all $ n\geq2$, and so the statement is true, as we desired.[/hide]


For any positive integer $n \geq 1$ let $p(n)$ be the greatest prime divisor of $n$. Prove that there are infinitely many positive integers $n$ with
$p(n) < p(n+1) < P(n+2)$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=86080 , it gives some more of that problems.


Given $ a,b,c>0$.Prove that:
$ \frac{a}{\sqrt{a^2\plus{}8ab\plus{}b^2}}\plus{}\frac{b}{\sqrt{b^2\plus{}8bc\plus{}c^2}}\plus{}\frac{c}{\sqrt{c^2\plus{}8ca\plus{}a^2}} \ge \frac{3}{\sqrt{10}}$ :)

I am seeing Homogenisations everywhere.?!? :huh:
Replacing $ a, b, c$ by $ ka, kb, kc$ we see that the inequality remains unchanged.

So can we assume $ abc\equal{}1$?  Or shall we assume $ a\plus{}b\plus{}c\equal{}1$?
My idea is "foggy",......

*sigh*  Please help me, the experienced problem solvers.... :(

[quote="Potla"]I am seeing Homogenisations everywhere.?!? :huh:
Replacing $ a, b, c$ by $ ka, kb, kc$ we see that the inequality remains unchanged.

So can we assume $ abc \equal{} 1$?  Or shall we assume $ a \plus{} b \plus{} c \equal{} 1$?
My idea is "foggy",......

*sigh*  Please help me, the experienced problem solvers.... :([/quote]
Yes, you can assume $ a\plus{}b\plus{}c\equal{}1$ or $ abc\equal{}1$,But I think it can't solve this inequality, my friend :)

It's weaker and not nicer than the well-known:  :wink:
$ \sum\ \frac {a}{\sqrt{a^2 \plus{} 7ab \plus{} b^2}} \geq\ 1$

[quote="Potla"][quote="nguoivn"]It's weaker and not nicer than the well-known:  :wink:
$ \sum\ \frac {a}{\sqrt {a^2 \plus{} 7ab \plus{} b^2}} \geq\ 1$[/quote]

How did you compare these two inequalities, nguovin? Can you please elaborate a little? :blush:  :rotfl:[/quote]
use Holder inequality.This inequality by NGUYEN VAN THACH.I will post solution,my friend :)


topology
real analysis
real analysis unsolved

Let $X$ be a separable metric space(separable means that there is a countable dense set $A$ in $X$ ) and $C$ the set of all points such that all their neighbourhoods are uncountable.Show that $C$ is closed,it doesnt have any isolated point and $X-C$ is countable.

The metric space has a countable basis (in metric spaces separability is equivalent to this), so every point not in $C$ is a point of one of the countably many basis elements which contain only countably many points. It's clear from here that there can only be countably many such points, and the fact that $C$ is perfect (the rest of the problem) is obvious.


Anita attends a baseball game in Atlanta and estimates that there are 50,000 fans in attendance. Bob attends a baseball game in Boston and estimates that there are 60,000 fans in attendance. A league official who knows the actual numbers attending the two games note that:

i. The actual attendance in Atlanta is within $10 \%$ of Anita's estimate.
ii. Bob's estimate is within $10 \%$ of the actual attendance in Boston.

To the nearest 1,000, the largest possible difference between the numbers attending the two games is

$\textbf{(A)}\ 10,000 \qquad
\textbf{(B)}\ 11,000 \qquad
\textbf{(C)}\ 20,000 \qquad
\textbf{(D)}\ 21,000 \qquad
\textbf{(E)}\ 22,000$

[quote="chess64"][hide]If Anita estimates really low and Bob estimates really high, then the difference will be large.

$0.90 \cdot 50,000 = 45000$
$1.10 \cdot 60,000 = 66000$
Difference: $21000 \implies D$.
[/hide][/quote]

Read the question carefully.. To others, what will be the REAL answer? :)

Read the question carefully and note the difference between statement one and two.

Statement one says that[b]Atlanta's attendance is within[/b] $10%
$ of Anita's estimate.

Statement two says that [b]Bob's estimate is within[/b] $10%
$ of Boston's attendance.

[quote="Karth"]Sorry for reawakening this thread  :D But why isn't it [hide]$D$[/hide]?[/quote]
Because the actual attendance at Atlanta could be $66666$. :D And surge, you can't put % between the dollar signs. :D

[quote="surge"]Read the question carefully and note the difference between statement one and two.

Statement one says that[b]Atlanta's attendance is within[/b] $10\%$ of Anita's estimate.

Statement two says that [b]Bob's estimate is within[/b] $10\%$ of Boston's attendance.[/quote]

I think that will work.

[quote="lotrgreengrapes7926"][quote="Karth"]Sorry for reawakening this thread  :D But why isn't it [hide]$ D$[/hide]?[/quote]
Because the actual attendance at Atlanta could be $ 66666$. :D And surge, you can't put % between the dollar signs. :D[/quote]

I still don't get it.  Why can the actual attendance at Atlanta be $ 66666$?  Isn't it $ 1.1 \times 60000\equal{}66000$?

Thanks!

[quote="Silverfalcon"]
ii. Bob's estimate is within [b]$ 10 \%$ of the actual attendance in Boston[/b].
[/quote]

You are trying to go ten percent off of Bob's estimate, when the ten percent refers to "the actual attendance in Boston."  Thus, if $ B$ is the actual attendance, $ .9B < 60000 < 1.1B$, so $ 54545.45 < B < 66666.67$ (approximately).

Did we switch Atlanta and Boston somewhere?

[quote="Poincare"]shouldn't this be in the MC forum?[/quote]

Maybe they posted it in HSB because AHSME is a high school contest.


If, after two successive price reductions of 15% and 10%, an item is marked at 30.60 dollars , what was its original price in dollars?

[hide] Suppose that $ a$ was the amount of money before the $ 10\%$ reduction. Then, $ a(1\minus{}\frac{1}{10})\equal{}30.6 \implies a\equal{}30.6(\frac{10}{9}) \implies a\equal{}34$. Now let $ b$ be the amount before the $ 15\%$ reduction. Then we have $ b(1\minus{}\frac{15}{100})\equal{}34 \implies b\equal{}34(\frac{20}{17}) \implies b\equal{}\boxed{40}$.[/hide]


For every real $x \ne 1$ find the sum of the series $\ds\sum_{n=0}^\infty{\frac{{x^{3^{n}}+\left({x^{3^{n}}}\right)^{2}}}{{1-x^{3^{n+1}}}}}.$
(A. Kukush)

hint
[hide] use telescoping [/hide]

[hide="Solution"]As the hint says, it telescopes -- adding up the first two terms is sufficient to see what will happen, and one could induct from there.  Alternatively,
$\sum \frac{x^{3^{n}}+\left(x^{3^{n}}\right)^{2}}{1-x^{3^{n+1}}}= \sum \frac{1+x^{3^{n}}+\left(x^{3^{n}}\right)^{2}}{1-\left(x^{3^{n}}\right)^{3}}-\frac 1{1-x^{3^{n+1}}}= \sum \frac 1{1-x^{3^{n}}}-\frac 1{1-x^{3^{n+1}}}$.

This clearly telescopes, and the partial sums have value $\frac1{1-x}-\frac1{1-x^{3^{n+1}}}$.  If $|x| > 1$, the trailing term goes to zero and the sum goes to $\frac 1{1-x}$.  If $|x| < 1$, the trailing term goes to 1 and the sum goes to $\frac 1{1-x}-1 = \frac x{1-x}$.  Finally, if $|x| = 1$, then $x =-1$ and so the sum goes to 0.[/hide]


algebra
polynomial
Rational Root Theorem

Solve for x:

$ (x\plus{}4)^4\minus{}(x\plus{}4)^3\minus{}7(x\plus{}4)^2\plus{}x\plus{}10\equal{}0$

Okay, for this one, I made x+4 = y, for all the x+4's

Therefore, I got $ x^4\minus{}x^3\minus{}7x^2\plus{}x\minus{}4\plus{}10\equal{}0$ which simplifies to $ x^4\minus{}x^3\minus{}7x^2\plus{}x\plus{}6\equal{}0$

Now, I do not know how to find x. What should I do? I think there are several ways to do this, but I don't know one. =()

Use the [url=http://en.wikipedia.org/wiki/Rational_root_theorem]Rational Root Theorem[/url] to test for rational roots.

Or note that the coefficients add to 0 in the last polynomial (which should be in $ y$'s, btw), and factor out the $ y\minus{}1$. :)

[quote="now a ranger"]
Therefore, I got $ x^4 \minus{} x^3 \minus{} 7x^2 \plus{} x \minus{} 4 \plus{} 10 \equal{} 0$ which simplifies to $ x^4 \minus{} x^3 \minus{} 7x^2 \plus{} x \plus{} 6 \equal{} 0$

[/quote]

$ x^4 \minus{} x^3 \minus{} 7x^2 \plus{} x \plus{} 6 \equal{} 0$
$ x^4 \minus{} x^3 \minus{} (7x^2 \minus{} x \minus{} 6) \equal{} 0$
$ x^4 \minus{} x^3 \minus{} (7x \plus{} 6)(x \minus{} 1) \equal{} 0$
$ x^3(x \minus{} 1) \minus{} (7x \plus{} 6)(x \minus{} 1) \equal{} 0$
$ (x^3 \minus{} 7x \minus{} 6)(x \minus{} 1) \equal{} 0$
Rational root theorem gives us +/-6, 3, 2, 1
synthetic division gives 3 as a root leaving a quadratic...shouldnt be too hard from there


Given the polynomial P(x)=4x^4-8x^3-7x^2+11x-15

a) using the rational zero theorm (or any other you want to use) find all the rational zeros of the polynomial.

b) find [b]all[/b]zeros of the polynomial.

[quote="maniacman384"]Given the polynomial $P(x)=4x^4-8x^3-7x^2+11x-15$

a) using the rational zero theorm (or any other you want to use) find all the rational zeros of the polynomial.

b) find [b]all[/b]zeros of the polynomial.[/quote]

[hide="a"]possible roots are $\pm(\frac{15}{4}, \frac{15}{2}, \frac{15}{1}, \frac{5}{4}, \frac{5}{2}, \frac{5}{1}, \frac{3}{4}, \frac{3}{2}, \frac{3}{1}, \frac{1}{4}, \frac{1}{2}\frac{1}{1})$

I usually start from 1 and work my way up.  we find that $-\frac{3}{2}$ works.

Use synthetic division and you come up with the new equation: $2(2x^3-7x^2+7x-5)=0$

The possible roots are $\pm(\frac{5}{2}, \frac{5}{1}, \frac{1}{2}, \frac{1}{1})$

We find that $\frac{5}{2}$ works.

Use synthetic division and we get a quatratic with no rational roots.

So the aswer is $\boxed{\frac{-3}{2}, \frac{5}{2}}$

There is also a quartic equation but its a bit complicated.[/hide]

[hide="b"]From part a, now just solve the quadratic.

$x^2-x+1=0$

$\frac{1\pm\sqrt{1-4}}{2}=\boxed{\frac{1\pm i\sqrt{3}}{2}}$[/hide]

[quote="jli"]
So the aswer is $\boxed{\frac{-3}{2}, \frac{5}{2}}$
[/quote]
$\boxed{\frac{-3}{2}, \frac{5}{2}}$ :D


If $m$ and $n$ are positive integers prove that $3^m + 3^n +1$ cannot be a perfect square.

[hide="Edit; still a one-liner"]
$3^m + 3^n + 1 \equiv 3, 5, 7 \bmod 8$[/hide]

If $m$ and $n$ are of different parity then that is not true ^_^

[hide]$3^k\equiv 1,3 \pmod{8}$. Thus $3^m+3^n+1\equiv 3,5,7 \pmod{8}$. However, since a perfect square is $\equiv 0,1,4 \pmod{8}$, $3^m+3^n+1$ cannot be a perfect square.[/hide]

[quote="pieterminate"][quote="t0rajir0u"][hide="One-liner"]
$3^m + 3^n + 1 \equiv 3 \bmod 4$[/hide][/quote]

Umm, yeah.... counter example: $3+9+1 \equiv 1 \bmod 4$    :)[/quote]

Well thats 5 mod 8.


trigonometry
integration
calculus
calculus computations

A spherical tank contains $ 81.637$ gallons of water at time $ t\equal{}0$ minutes. For the next $ 6$ minutes, water flows out of the tank at the rate of $ 9\sin{\sqrt{t\plus{}1}}$ gallons per minute.  How many gallons of water are in the tank at the end of the $ 6$ minutes?

$ \frac {dV}{dt} = 9\sin\sqrt {t + 1} \\
 \\
dV = 9\sin\sqrt {t + 1}dt \\
 \\
\int\limits_{V_i}^{V_f} dV = \int\limits_0^{t_f} 9\sin\sqrt {t + 1}dt \\
 \\
V_i = 81.637,\;\;t_f = 6 \\
 \\
\text{Let, }t + 1 = x^2 \\
dt = 2xdx \\
 \\
\therefore \int 9\sin\sqrt {t + 1}dt = \int 2x\sin x dx \\
 \\
= 2x\{\cos x\} - 2\{ - \sin x\} \\
 \\
= \left[2\sqrt {t + 1}\cos\sqrt {t + 1} + 2\sin\sqrt {t + 1}\right]\limits_0^{6} \\
 \\
V_f - 81.637 = 2\sqrt {7}\cos\sqrt {7} + 2\sin\sqrt {7} - (2\cos\sqrt {1} + 2\sin\sqrt {1})$

You'll need a calculator there.


calculus
integration
function
trigonometry
calculus computations

[quote="N.T.TUAN"][hide]We only need evalue $\int \frac{\sin^{2}{x}}{\cos^{2}{x}}dx=\int \frac{1}{\cos^{2}{x}}dx-\int dx$

Now you see $(\tan{x})'=...$[/hide][/quote]

these are the answer choices:

a) ${\frac{\sec{3x}}3+C}$
b) $x-tan{3x}+C$
c) $x+tan{3x}+C$
d) ${x-\frac{\tan{3x}}3+C}$

are you supposed to multiply -1 by $\int\frac{1}{\cos^{2}{3x}}-\int dx$ even though $-tan^{2}{3x}=sec^{2}{3x}-1$

so which one would it be since the x from $\int\frac{1}{\cos^{2}{3x}}-\int dx$ would make the $etc.$-$\int dx = etc.-x$ so the x is negative.

[hide]$\int-\tan^{2}3x \,dx = \int-(\sec^{2}3x-1)\,dx = \int (1-\sec^{2}3x)\,dx$

You should know how to find the integral of 1 and $\sec^{2}3x$.
[/hide]


I'm curious: what version of phpbb is used to power this forum?

As for the Wiki, I've never seen that URL infrastructure for MediaWiki before. (e.g. .../Wiki/index.php/{YOUR PAGE}) Is it an edited version?

The wiki architecture is the default one. Changing it is what wikipedia did. See http://meta.wikimedia.org/wiki/Using_a_very_short_URL.

The version number of PHPBB upon which the forum is based is probably pretty much irrelevant by now considering how much of the code of the forum was written/modified by our beloved Valentin.

#H34N1: Really? When I downloaded and tried to use MediaWiki, I got the following URL:

[quote].../wiki/index.php?page={PAGE_NAME}[/quote]

as my only URL option.

Xevarion: Well, I know that a lot of modifications and extensions were obviously made, but I'm still interested in what version of phpbb was the base.

Base was like 2.0.2

If you seriously want an installation, I would go with phpbb3.

After a couple minutes, it will revert back to index.php/{PAGE_NAME}

It is in your LocalSettings.php and under $wgStringPath or something like that.

That didn't happen... I think it was because I was using 1.6.8 instead of the latest version.


algebra
polynomial
number theory unsolved
number theory

[u]A problem: [/u] Show that  $5(x-y)(y-z)(z-x) | (x-y)^5+(y-z)^5+(z-x)^5$. For all x,y,z pairwise distinct integers.

[u]In my book there is a solution there is written: [/u](*)
$(x-y)^5+(y-z)^5+(z-x)^5$ is zero for $x=y, y=z, z=x$. So the terms $x-y, y-z, z-x$ can be factored out.

How does the book came up with this?

Davron

I bet they are considering $(x-y)^5+(y-z)^5+(z-x)^5$ as a polynomial and applying Bezout's theorem to show this polynomial is divisible by $(x-y)(y-z)(z-x)$.

Please can you post here a Bezout's theorem, i don't about this theorem?

Davron

Look here (e.g. post #2):
http://www.mathlinks.ro/Forum/viewtopic.php?t=3751

Could you please write the application of the theorem to this problem.
But step by step please thanks.

Let $P(x-y, y-z)=(x-y)^5+(y-z)^5-((x-y)+(y-z))^5$. Fixing $y-z$ we may consider it as a polynomial with the variable $x-y$. Since $P(-(y-z), y-z)=0$ we can factor out $(x-y)+(y-z)=x-z$ and analogously for the other two factors.

Also cast a glance at http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%28x-y%29%5E5&t=55987


quadratics
number theory unsolved
number theory

Filnd all natural numbers $n$ such that $S(n^{2})=2006$ where $S(m)$ represents sum of digits of a natural number $m$.

There are none.
@Why? : http://www.mathlinks.ro/Forum/viewtopic.php?t=86721

LOOOL!!!!

I am so into some "hard" problems right now, I forgot that there are still some easy problems that can be solved just by taking quadratic residues mod $9$ :) :) :)
Bye


Please do not discuss the current problems in any fashion.


Prove that there aren't positive integers solutions to this equation:

$ 3^{m}\plus{}3^{n}\plus{}1\equal{}t^{2}$

I don't understand how to do this question. Please if anyone knows teach me more clearly possible  :blush:

Thanks.

Let $ m\ge n$, then $ 3^n(3^{m\minus{}n}\plus{}1)\equal{}(t\minus{}1)(t\plus{}1)$. Therefore $ t\equal{}a*3^n\pm 1$ and $ a*3^n\pm 3\equal{}\frac{3^{m\minus{}n}\plus{}1}{a}$ contradiction.

You could also look at it modulo 8 then

3,5,7=0,1,4 mod 8 -> Contradiction

[quote="Rust"]Let $ m\ge n$, then $ 3^n(3^{m \minus{} n} \plus{} 1) \equal{} (t \minus{} 1)(t \plus{} 1)$. Therefore $ t \equal{} a*3^n\pm 1$ and $ a*3^n\pm 3 \equal{} \frac {3^{m \minus{} n} \plus{} 1}{a}$ contradiction.[/quote]

Thanks Rust, but I don't understand this line:

$ t \equal{} a*3^n\pm 1$ and $ a*3^n\pm 3 \equal{} \frac {3^{m \minus{} n} \plus{} 1}{a}$

 :maybe:

It is my mystake.
Raficafi give good solution.

Basically if
$ a\equiv b$(mod m) if and only if $ m|(a\minus{}b)$


superior algebra
superior algebra unsolved

Is there a general rule for computing the conjugate tat^-1 of t using t (i.e. if you are given a and t, can we come up with a general rule for computing this) :)

In the world of permutations, conjugation is a permutation of the labels. For example, suppose $t$ is the permutation of $\{1,2,3,4,5,6\}$ which sends $1\to 2,2\to 4,3\to 3,4\to 1,5\to 6,6\to 5$, and $a$ has cycle decomposition $(1,2,3,4)(5)(6)$. Then $t^{-1}at$ has cycle decomposition $(2,4,3,1)(6)(5)$.

I don't think that really answers my question because the general rule is supposed to work for any 'a' and 't'

The example is typical; there's a reason I wrote the new permutation in that order. It's easier to give an example than to describe the exact procedure.


From the set {1, 2, 3, ..., 50}, we first write down all 50 of them, then we write down the product of the numbers taken two at the time, then three at a time, then four at a time, and so on until the product of all 50 of them.

What is the sum of the reciprocals of all the numbers written?

[hide="Hint 1"]Generalize the problem. [hide="Hint 1.1"]Let's take the set $ \{1,2,\dots,n\}$ and let's call $ s_{n}$ the sum of the reciprocals of all possible products of non-repeated elements.[/hide][/hide]
[hide="Hint 2"]Can you write $ s_{n}$ as a function of $ s_{n-1}$? [hide="Hint 2.1"]Separate the products with and without the element $ n$. [hide="Hint 2.2"]$ s_{n}=s_{n-1}+\frac{1}{n}+\frac{1}{n}s_{n-1}$.[/hide][/hide][/hide]
[hide="Hint 3"]Using Hint 2, find the first values of $ s_{n}$, guess the general term and prove it by induction. [hide="Solution"]$ s_{n}=n$.
In particular, $ s_{50}=50$.[/hide][/hide]

[hide="Solution"] I couldn't find a way to give a good hint for this, and the result is so nice that I need to give it.

The desired sum can be factored

$ \left( 1+\frac{1}{1}\right) \left( 1+\frac{1}{2}\right) ... \left( 1+\frac{1}{50}\right)-1 = \frac{2}{1}\cdot \frac{3}{2}\cdot ... \cdot \frac{51}{50}-1 = \boxed{50}$

The generalization is obvious. [/hide]

Im sorry t0rajir0u, but that solution isnt so obvious to me... can you explain a little more?

Why do you multiply all the $ (1+a_{i})$'s together (with $ a_{i}$ being all the elements of the reciprocals of the original set), and subtract 1 at the end?

I think it will be simplest to explain by way of example.

Suppose we simply have the set $ \{ 1, 2, 3 \}$.  We will write down the numbers $ 1, 2, 3, 1 \cdot 2, 2 \cdot 3, 1 \cdot 3, 1 \cdot 2 \cdot 3$ and then take the reciprocals.  This is

$ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{1 \cdot 3}+\frac{1}{1 \cdot 2 \cdot 3}$

Now, note that

$ \left( 1+\frac{1}{1}\right) \left( 1+\frac{1}{2}\right) \left( 1+\frac{1}{3}\right) = 1+\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{1 \cdot 2}+\frac{1}{1 \cdot 3}+\frac{1}{2 \cdot 3}+\frac{1}{1 \cdot 2 \cdot 3}$

Which is fairly trivial by expanding.  This is exactly the summation we want, but with an extra $ 1$ term, which we subtract.  Algebraically speaking,

$ (1+a)(1+b)(1+c) = 1+(a+b+c)+(ab+bc+ca)+(abc)$

Think of it as a choice: for each number $ a$, you choose whether to include the factor of $ a$ in the denominator (multiply by $ 1/a$) or whether not to (multiply by $ 1$).

You want to sum over all possible sets of choices, except for the case of having nothing in the denominator, so you have:

$ (\text{1 or not 1})\times (\text{2 or not 2}) \times ... \times (\text{50 or not 50})-(\text{nothing chosen})$

$ = \left(\frac{1}{1}+1\right) \left( \frac{1}{2}+1\right) ... \left( \frac{1}{50}+1\right)-1^{50}$

and t0rajir0u has shown how to simplify that. This is the underlying principle behind the Binomial Theorem.


All right,
  Using numbers from 0 to 9 (without repitition) create a ten digit number in which the first digit -from left- is divisible by 1, the first and second digits devisible by 2 , the first 3 digits deivisible by three, ... and the tendigits divisible by 10.

  For example imagine a number "ASDFGHJKLQ", A/1 and AS/2 and ASD/3 ,ASDF/4,..... ASDFGHJKLQ/10 will have no remainders.

(Optional) PROVE THAT THERE IS ONLY ONE SET OF NUMBERS WITH SUCH CHARACTERISTICS....

im a big believer in trial and error (genereally since i dont always see a pattern right away).  but you can use deduction such as the 5th number can only be a 5 or 0, but the last number has to be a 0 since its the only number that gives an interger when divided by 10.  Using deduction like that you could have a reaosnable stab at it.

THX guys,
but is there any method except trial and error(I know 5 and 0 are obvious, but how about the rest of the numbers?)

there was the exact problem posted by David,(djiminez) in the Puzzles and Brainteasers Forum. here is how you do it (proof by alan):
let the number be:$ABCDEFGHIJ$
obviously you know that $J=0$ and $E=5$, also $3|D+F+5$ and $D,F$ are even numbers so $D+F=4+6$ (not neccerly respectively).
$H=2$ because $8|FGH$, so $G$ is either $3$ or $7$
$B$ is 8 because it is the only unplaced even number
$D=6$ because $4|CD$ so the number so far looks like:$A8C654G2I0$
$7|A8C654G$ so $1000000A+10000C+G+806540\equiv 0 \mod{7}\rightarrow A+4C+G=14 \rightarrow C=1, I=9, $
$3|A+8+1$ so $A=3$ and $G=7, \Leftrightarrow$ the number is:$3816547290 $

if something is 0mod7, that means it is divisible by 7.

[quote="jamshidi_f2004"]Thanks for the solution but I did not understand couple of things, I would appreciate if you could help me ....
I. first of all, why should 3| D+F+5 and not 3| A+B+C+D+F+5.(since the first 6 digits are divisible by 6 then, ABCDEF must be divisible by 2 and 3 , so 3| A+B+C+D+F+5.)
II. Why does 8| FGH, (is it that only the last three digits can be considered for divisibilty rules?)
III. Same problem for 4| CD :?
IV. what does 0 mod 7 mean?

Thanks a lot :blush:[/quote]

I. Since $3|(A+B+C)$ and $3|(A+B+C+D+F+5)$, it immediately follows that $3|(D+F+5)$
II. The divisibility rule for $8$ is if the last three digits are divisible by $8$ then the whole number is.
II. Same with $4$, except the last two digits.

for divisibility tests take a look at here, :
http://mathworld.wolfram.com/DivisibilityTests.html

[quote="jamshidi_f2004"]Thanks a lot,
But do you know where can I find the proofs for divisibility rules of 8 and 4 as you mentioned?
 :)[/quote]

They are actually very simple. They are based on the fact that $1000 \equiv 0 \mod{8}$ and $100 \equiv 0 \mod{4}$... so clearly if you take the last $3$ digits you can tell if it's divisible by $8$ and the last $2$ for divisible by $4$.

[quote="paladin8"][quote="jamshidi_f2004"]OK thanks,
I got everything except one thing
7| 1000000A+10000C+G+806540 , I know 7| 806540, so 7| 1000000A+10000C+G,
but why A+4C+G= 14?

Are there any specific rules for divisibility of 7 used up here?[/quote]

Well, notice that $1000000A+10000C+g \equiv A+4C+G \mod{7}$ and I think it was logic that figured out that they sum to $14$. Not quite sure, though.[/quote]
yes Paladin8 is right. since
$ 1000000A+10000C+G+806540=0 \mod{7}=A+4C+G$
you know that $A+4C+G$ must be divisible by 7.
now if you look back you'll see why it has to equal $14$, because, your number till that point was:
$A8C654G2I0$, so based on the used and unused numbers, $A+4C+G=14$

because $999999A+9996C$ is divisible by 7.

remember, if $a, b \equiv 0\mod{c}$, then $a-b \equiv 0\mod{c}$.

yes, so if you divide the $ 1000000A+10000C+G$ by $7$, the remainder is the same as the remainder of when $ A+4C+G$ is devided by 7.

OK thanks I got it....
I don't know why did the same message appeared twice???????

[quote="jamshidi_f2004"]Iam sorry but I still so not get something..... :(  :blush:

If 7| 1000000A+10000C+G, then why should 7| A+4C+G?

I guess I have to study a little more in the field of number theory ;)[/quote]

didnt u just ask that question??

ok, here is the deal,
hopefully you know this:
that $999999$ is divisible by 7, (even if you don't, just use a calculator...) so $999999A$ must be divisible by 7 also.
also $9996$ is divisible by 7, so $9996C$ is divisble by 7 also.
so the remainder of $1000000A$ and $10000C$ by 7 must be equal to the remainder of  $A$ and $4C$ by 7 respectively.
therefore
$1000000A+10000C\equiv A+4C (mod 7)$
get it? ;)


I have a quadrilateral ABCD. I know that angle BAC is 2 times angle BDC, and angle CAD is 2 times angle CBD. Can I conclude that A is the center of circumscribed circle of BCD?

Yes.
 [geogebra]5ef11a1ba1f59af1276765fb4fd77be07d3068c9[/geogebra]

Let $ \angle CDB \equal{} \alpha$ and $ \angle CBD \equal{} \beta$
 $ \angle BCD \equal{} 180 \minus{} (\alpha \plus{} \beta) \Longrightarrow \angle BMD \equal{} \alpha \plus{} \beta$

Since $ \angle BAD \equal{} 2(\alpha \plus{}\beta) \Longrightarrow \angle BAD \equal{} 2\angle BMD$

Hence $ A$ is the center of the circle that circumscribes $ BCDM$

BTW does anybody know how to label the angles geogebra (i.e. $ \alpha$ and $ \beta$ and not their size in degree) :!:  :?:

Click on angle, and then on 3 points which form that angle. If you don't like the name, rename it, and you are done.


function
superior algebra
superior algebra unsolved

Hello. I am having a troublesome time showing that the localisation of $ \mathbb{Z} \times \mathbb{Z}$ about every pair $ (a,b)$ where neither are zero is isomorphic to $ \mathbb{Q} \times \mathbb{Q}$.
Whats troublesome is its so obvious that its difficult to know what to write. :(
Could someone post a solution?

Can I just define a homomorphism from $ \mathbb{Z} \times \mathbb{Z} \to \mathbb{Q} \times \mathbb{Q}$ by $ (x,y) \to \left(\frac xa , \frac yb \right)$?

If I understand you correctly, then you want to show that $ S^{\minus{}1}(\mathbb Z\times\mathbb Z)\cong\mathbb Q\times\mathbb Q$ with $ S\equal{}\mathbb Z^\times\times\mathbb Z^\times$.

That's correct. Try to show the following slightly general lemma:

For two commutative rings $ R_1$ and $ R_2$ with multiplicative subsets $ S_i$, $ i\equal{}1,2$, it holds \[ (S_1\times S_2)^{\minus{}1}(R_1\times R_2)\cong S_1^{\minus{}1}R_1\times S_2^{\minus{}1}R_2\,.\]
To see this it is sufficient to show that $ S_1^{\minus{}1}R_1\times S_2^{\minus{}1}R_2$ fulfils the universal property of $ (S_1\times S_2)^{\minus{}1}(R_1\times R_2)$.  See [url=http://www.mathlinks.ro/viewtopic.php?p=1668041#1668041]here[/url].

$ a,b$ aren't fixed; that's not the right homomorphism, because it's not the right structure on the left. You actually have an ordered pair of ordered pairs; either $ (x,y),(a,b))$ or $ ((x,a),(y,b))$, depending on how you look at it. Make that change, and it's the right function of those four variables. Then, you have to deal with the all the usual complication that goes into the definition of localization.