15  phones have just been received at an authorized service center. 5 of these phones re cellular, 5 are cordless, and the other 5 are corded phones. Suppose that these components are randomly allocated the numbers 1,2,...,15 to establish the order in which they will be serviced.

1) What is probability that all the cordless phones are among the first ten to be serviced?
2) What is probability that after servicing 10 of these phones, phones of only 2 of the 3 types remain to be serviced?
3) What is the probability that two phones of each type are among the first 6 serviced?

Disclaimer: I'm bad at combinatorics in any shape or form...  :|

[hide="Solution to #1"]
There are $ 15\choose10$ ways to pick the first 10 phones.
If all the cordless phones are put into the first 10, there are 10 phones left to choose 5 from.
So, $ {10\choose{5}}/{15\choose{10}} = \frac {252}{3003}$
[/hide]

[hide="Incomplete Solution to #2"]
Say we choose 5 phones out of one kind and $ n$ phones out of another. Then we must choose $ 5 - n$ phones out of the last kind.
Let $ P(n) = {10\choose5}{5\choose n}{5\choose{5 - n}}$ (This is wrong...)
Now, $ n$ can be 1, 2, 3, or 4. But 3 and 4 would be superfluous because you'd be repeating stuff.
So the probability is given by $ \frac {P(1)*P(2)}{15\choose10} = \ldots{?}$
I'll try to finish this later...
[/hide]

[hide="Solution to #3"]
There are $ 15\choose6$ ways to pick the first 6 phones.
The number of combinations that include 2 of each kind of phone is given (I think) by $ {5\choose2}^3$
So... $ \frac {1000}{5005}$
which simplifies to $ \frac {200}{1001}$
[/hide]

yeah, that's what I got for part a. I hope it's right. Parts b and c i'm not sure.

for part 3, $ \frac{1000}{5005}$ simplifies to $ \frac{200}{1001}$. For part 2, the answer is 3 times the answer in part 1 (because it could be any type) minus twice the probability of getting two types of phone completely serviced (one for overlap and one because having two types serviced would leave only one type, not two). This gives $ \frac{250}{1001}$.

Yes, I meant $ \frac{200}{1001}$, not $ \frac{500}{1001}$. I'm tired  :lol:
I'll think about #2 a little more.

the answer to part b is .24975. Can someone give me a better explanation?
Also, why is C() cubed in part c. A clear explantion would be appreciated

In #3, there are 3 sets of different kinds of phones. Each set contains 5 phones. Out of each set of 5, you want to choose 2, so that the total is 6 phones, 2 from each set.
The number of ways to choose 2 from a set of 5 is $ 5\choose{2}$. The number of combinations of 6 that can be made with 2 of 3 sets of 5 is $ {5\choose{2}}{5\choose{2}}{5\choose{2}}\equal{}{5\choose{2}}^3$


__   __
RS//UT, WX is a median. Find each indicated measure.
 [geogebra]ef5cb112fb00527b3edcafac04858e14dc339861[/geogebra]

TX=______
WX=______
m<U=_____
m<1=______
m<2=_______

Angle U=60 degrees because of consecutive interior angle theorem
I don’t know what median is, so im not sure on the next part
Angle 2=60 degrees from consecutive interior angle theorem
Angle 1=106 from consecutive angles postulate,
And I think the WX is 18 if median means the connection between the middle of  RU and ST

http://regentsprep.org/Regents/math/geometry/GP9/LTrapezoid.htm

It's called the midsegment theorem.


inequalities proposed
inequalities
algebra

$ a,b>0, ab\ge 1$, prove
$ \frac{1}{(2a\plus{}3)^2}\plus{}\frac{1}{(2b\plus{}3)^2}\ge \frac {2}{5(2ab\plus{}3)}$
P.S. I have ugly solution  [hide] After expanding we get $ 20(a^3b\plus{}b^3a)\plus{}12(a^2b\plus{}b^2a)\plus{}54\ge 16a^2b^2\plus{}6(a^2\plus{}b^2)\plus{}54ab\plus{}18(a\plus{}b)$ but by AM-GM
1) $ 8(a^3b\plus{}b^3a)\ge16a^2b^2$
2) $ 12(a^3b\plus{}b^3a)\minus{}12(a^2\plus{}b^2)\ge 24(ab\minus{}1)$
3) $ 12(a^2b\plus{}b^2a)\minus{}12(a\plus{}b)\ge 24(ab\minus{}1)$
4) And CS $ 6(a^2\plus{}b^2)\ge 6(a\plus{}b)\plus{}6(ab\minus{}1)$ after summing we get our result :wink: [/hide]

[quote="dima ukraine"]$ a,b > 0, ab\ge 1$, prove
$ \frac {1}{(2a \plus{} 3)^2} \plus{} \frac {1}{(2b \plus{} 3)^2}\ge \frac {2}{5(2ab \plus{} 3)}$
P.S. I have ugly solution  [hide] After expanding we get $ 20(a^3b \plus{} b^3a) \plus{} 12(a^2b \plus{} b^2a) \plus{} 54\ge 16a^2b^2 \plus{} 6(a^2 \plus{} b^2) \plus{} 54ab \plus{} 18(a \plus{} b)$ but by AM-GM
1) $ 8(a^3b \plus{} b^3a)\ge16a^2b^2$
2) $ 12(a^3b \plus{} b^3a) \minus{} 12(a^2 \plus{} b^2)\ge 24(ab \minus{} 1)$
3) $ 12(a^2b \plus{} b^2a) \minus{} 12(a \plus{} b)\ge 24(ab \minus{} 1)$
4) And CS $ 6(a^2 \plus{} b^2)\ge 6(a \plus{} b) \plus{} 6(ab \minus{} 1)$ after summing we get our result :wink: [/hide][/quote]
Yes, you are right, Loc. Sorry, I will think on it again.

[quote="can_hang2007"]
It suffices to prove that
\[ 5(a \minus{} b)(2ab \plus{} 3) \ge 3(2a \plus{} 3)(1 \minus{} b)
\]
[/quote]
I think it should be $ 5\left( {a \minus{} b} \right)\left( {2ab \plus{} 3} \right) \ge 3\left( {2a \plus{} 3} \right)\left( {1 \minus{} b} \right)\left( {2b \plus{} 3} \right)$


In how many ways is it possible to arrange the digits of 1150 to get a four-digit multiple of 5?

I don't see what I did wrong.

So there are 3 ways to pick the first digit, 3 for the second, 2 for the third, and 1 for the fourth IF THERE ARE NO RESTRICTIONS.  Multiplying, we get 18.

It has to end in a 0 or a 5 to be divisible by 5, so half of those, so we have 18/2=9.

The official solution said the answer was 5.

Pick the last digit first, it has to be $ 0$ or $ 5$.

Then, you must arrange the first three digits.  This is $ \frac{3!}{2!}$ (because the $ 1$'s are indistinguishable).  This yields $ 3$.  But we have counted $ 0115$.  So the answer is $ 2 \times 3 \minus{} 1$, or $ \boxed{5}$.

I don't understand why my Solution is wrong, either:
[hide=Sol] In order for the number to be a multiple of $5$, it has to end in $0$ or $5$. Thus, we can create the number: "11X", where the X represents the block of $0$ and $5$.There is $\frac{3!}{2!}$ ways, or $3$ ways, to order "11X". We have to multiply by $2!$, however, since we have to compensate for the fact, that X represents the "05" block. Thus, the answer is $\boxed{6}$.[/hide]

You subtract $1$ from your answer, as stated by ernie, because you overcount $0115.$

I first got 6 and i was confused why i got it wrong

and then i realized i overcounted and put 5


Is there a positive real number "closest" to 0?

[quote="randomdragoon"][hide]The real numbers are not well-ordered, so no.

That is,

Let $a$ be the positive real number closest to $0$. But then $a/2$ is closer to $0$. But we said $a$ is the closest one, a contradiction. Therefore $a$ doesn't exist.[/hide][/quote]
Just like how there is no "smallest" number or "greatest" number.


Gonna watch the new Harry Potter movie? When?

I'll see it soon after it comes in theaters.  It looks good.

I'll see it soon after it comes out.
I probably won't be able to see it the day it comes out, but I think I will on the weekend.

As soon as possible is my answer.  I can't wait!!! :D

The movie is only released on the 2nd of December in South Africa :( . Incidentally, I'm going on a maths camp from the 4th, and I don't want to wait till after it to see the movie. Also, it's the end of our school term the 2nd, so to celebrate, my friends and I am going to watch the movie.

out here in India too its goin to hit the theaters pretty late. I'll watch it within a week of it being released, even though i aint obsessed with harry potter but the trailer looks cool :)

I'm definately going to see it within the first week of it being released!!! :)  Hope it's better than the 3rd one! Can't wait!! :)

Heh, it has to be better than the third one: just look at the trailer!

I'll be watching it on 11/19  :D
(comes out on the 18th)

When I saw "Harry potter IV", I immediately thought "What, is this a rehab device for HP addicts?  :P

Oh, and please, guys, don't post any "spoilers" about the movie here, make a seperate thread for it :)  There are some people here who still want to watch the movie but can't do so yet because it hasn't been released in their country.

I just watched it. Pretty good but they skipped a lot from the book of course.

boy i saw it today and was it a big bummer!!, they missed out lots from the book and even if that was expected , i dunn know , i didnt like it...

Volde was looking DAMN DAMN DAMN stupid , they made his voice absloutely boring , opposite of the expected volde voice

sorry if this is a "spoiler" :D

another thing. the only thing that gave credit to the movie was the effects, they were nice!! and boy do daniel radcliffe and ron, hermione( i dont know their actual names :D )   look BIG now!!

How is it possible that there will be spoilers? The spoilers are in the book... :huh: and everyone seems to get disappointed with the movies. It is only because they leave stuff off the book which is expected because, otherwise, it would be a 5 hour long movie

[quote="eminem"]boy i saw it today and was it a big bummer!!, they missed out lots from the book and even if that was expected , i dunn know , i didnt like it...

Volde was looking - - - stupid , they made his voice absloutely boring , opposite of the expected volde voice
[/quote]

Wow, I thought it was the best movie so far.  The skipped parts bugged me, and they should have shown more of the game at the beginning, but I thought it was really well done.

The actors are getting old, but we've all seen like 25 year olds play 16 year olds before.  Besides, changing the actors would bug me more than them looking too old for the part.

Some people I know actually went and saw it at 12 midnight, the very first show, the first minute of that morning.

[quote="noneoftheabove"] How is it possible that there will be spoilers? [/quote]
Well I just thought maybe there is something in the movie that is really shocking or different than in the book and then some people would spoil it for those that didn't see it yet.

The movie is horrible.  Save your money, and rent it from NetFlix when it comes out on DVD.

Is it horrible compared to the book, or can you judge it objectively, solely as a movie? As a movie, I thought it moved fast enough and kept you interested. It skipped over many elements, but I can see why some of those were omitted to better the flow and to cut down on time. I thought the new director did an amazing job with this one. He introduced a new style that was bright and provocative, and he even had a little fun with some elements - some of it was pretty odd I'll admit. Still, I wouldn't be surprised if this movie becomes this year's second highest net profit movie, after Star Wars.

i totally agree on the "new director did a much better job" part ,like, this movie was totally different from the other potter movies  but dont you think cutting off loads of the book, even if it is necessary ,doesnt "entertain" you much, especially if you have read the book and know all about it and all the stuff that has been cut.

you talked about bood and movie and i remembered this:
in a book there is something that can't be in the movie and of course movie has something that a book can't.but both of them are nice and has a lot of fun. :D

[quote="yif man12"]I didn't like it at all. Wasted 8 dollars on it. The ordering of events was screwy but that didn't bother me too much. I hated the animations though. Dunno why, but to me, the animations looked video-game-ish, and that shouldn't happen in a movie. There was way too much of an appearance difference between actors and special effects people.[/quote]
i think you are right but the animations are not too bad.some movies like SHREK,BRIDE GROUP & ... are good.of course the y are not as good as the movies that has real actors but they are good. :D

I watched the movie today and I liked it :)  I don't think the animations look too bad. Voldemort doesn't look too good, though, and his voice doesn't sound right.

I didn't like the movie either .... I prefer the book ( same thing with other Harry Potter movies )

didnt like the movies.... ><

way tooo choppy....

and crouch jr freaked me out...

just watched it, my first film of 06  :D , surprisingly the theatre is still 90%full, so i had to sit in the seconed row, o man..my neck, the movie looked weird from that angle too.
i must say great movie, better that the first 3, very different feeling to this one. i like it, a new change, always good.  however, the movie cut way too much stuff from the book (as expected) but only 2 and half hour, that's  kinda too short for book4.  i'm kinda disappointed that they didn't do more with the tri wizard tournament.  it's a great event and a great chance to make this movie magnifiencent, but they didn't do enough with it, (just some pretty nice special effects and sound, not enough real thing, especially the third event, maze, was way too short).  i don't understand why they spent so much time on the bath and moaning myrtle part.  and no world cup(they build my excitment up with all that noise and the huge stadium but then boom nothing!)

and last thing, don't know if anyone also felt this: i felt like the movie is trying to evoke our emotion, especially with the Ball.  The emotion betwenn Ron harry and hermione kinda felt little romantic.  however, the feelling is only half way there.  anyone know what i'm taking about? can't discribe it well.


geometry
incenter
ratio
trigonometry
geometry unsolved

Quadrilateral $ABCD$ is inscribed in circle $\omega$. The bisector of $\angle ADC$ passes through the incenter of triangle $ABC$. Pick point $M$ on arc $ADC$ of $\omega$. Let points $P$ and $Q$ be the incenters of triangles $ABM$ and $BCM$, respectively.

a) Prove that all triangles $DPQ$ are similar, regardless of which point $M$ is picked.
b) Express $\angle PDQ$ and the ratio $DP/DQ$ in terms of $\alpha$ and $\beta$, where $\alpha = \angle BAC$ and $\beta = \angle BCA$.

I drew a picture and "saw" the solution for (a), but it's hard not to miss some details, so I'll just write some obsrvations about the configuration.

- Let $c$ be a circle tangent at $AB,BC$ in $R,S$ respectively and internally tangent to $\omega$. Then the point of tangency is $D$.

- Show that $DR,DS$ are the angle bisectors of $\angle BDA,\angle BDC$ respectively. Assume they intersect $\omega$ at $X,Y$ respectively. Then $DXY,DRS$ are similar.

- Show that triangles $DPX,DQY$ are similar, which implies (a). we show two things: $\angle DXP=\angle DYQ$ (obvious) and $\frac {XP}{YQ}=\frac {XD}{YD}$. This last equality can be derived as follows:

$XP=XB,YQ=YB$. We must thus show $\frac {XB}{YB}=\frac {XD}{YD}$. This is true because $DB$ is a symmedian in $DXY$.

As for (b), all the observations above lead to the conclusion that $\angle PDQ=\angle XDY=\frac {\alpha+\beta}2$. $\frac {DP}{DQ}=\frac {DX}{DY}=\frac {BX}{BY}=\frac {\sin \frac {\beta}2}{\sin \frac{\alpha}2}$.


I don't think that many countdown rounds are official in chapter anyway.
In my chapter (which is xpmath's chapter as well) they let us choose our bracket position, so it would be slightly unfair if it was official (because you place your name in the bracket based upon if you were called up first, and they call you up in a random order).

But if the ladder was completely random, then yes, that would be the only way to have an official countdown.

To be an official countdown, the countdown [b]must[/b] be done in a ladder-format.

[quote="jjx1"]thats not true. Ours is official but its not ladder format. its 10th vs 9th. then the winner vs 8th and so on.[/quote]

Um, I thought that was the ladder format.

Then, can this be the reason to protest the results?

[quote="jjx1"]thats not true. Ours is official but its not ladder format. its 10th vs 9th. then the winner vs 8th and so on.[/quote]
That's ladder, like Fantasy said.

Yeah, you should definitely protest this. Also, you should probably protest the disqualification of your team.

Good for her! I was planning on doing the same thing. (Note: this is BOGTRO,sorry)

[quote="FantasyLover"][quote="AIME15"]I heard that isabella2296 has already protested.  She has emailed the chapter coordinator and the state coordinator.[/quote]

Yeah, she pm-ed me.

Lol, I never knew that a cd round is not official unless it's in ladder format until xpmath told me...[/quote]
What are you talking about? I told you when we were leaving. fail.

And yes, from the MC website:

An official Countdown Round is defined as one that determines an individual’s final overall rank in the competition. If the Countdown Round is used officially, the official procedures as established by the MATHCOUNTS Foundation [b]must be followed[/b].

If a Countdown Round is conducted unofficially, the official procedures do not have to be followed. Chapters and states choosing not to conduct the round officially must determine individual winners on the sole basis of students’ scores in the Sprint and Target Rounds of the competition.

In an official Countdown Round, the top 25% of students, up to a maximum of 10, are selected to compete. These students are chosen based on their individual scores. The two lowest‑ranked students are paired, a question is projected and students are given 45 seconds to solve the problem. A student may buzz in at any time, and if s/he answers correctly, a point is scored; if a student answers incorrectly, the other student has the remainder of the 45 seconds to answer. Three questions are read to each pair of students, one question at a time, and the student who scores the most points (not necessarily 2 out of 3) captures the place, progresses to the next round and challenges the next highest‑ranked student. (If students are tied after three questions (at 1‑1 or 0‑0), questions continue to be read until one is successfully answered.) This procedure continues until the fourth‑ranked Mathlete and her/his opponent compete. For the final four rounds, the first student to correctly answer three questions advances. The Countdown Round proceeds until a first‑place individual is identified. (More detailed rules regarding the Countdown Round procedure are identified in the "Instructions" section of the School Competition booklet.)

Add the disqualfication, the throwing out my brohter's team scores for no good reason (in fact, no reason),
and Chapter was a total fiasco. The only good thing is that I made state.

I have a nice good solid protest out of all of this. I might send it.

fantasy. if you team was really dqed for being too good that should be protested. I doubt that is though. Please tell me the real reason

p.s You do have a reason to protest this fantasy. But, lol, if you would have made state or been 1st i'm sure you wouldn't be protesting .

Ok, there were two reasons:

First, as I said, we were too strong to compete with the other teams.

Second, our team was consisted of students from different schools, but this was unavoidable.

Two kids (including me) didn't have MATHCOUNTS Team at their school, one was homeschooled, and one did have MATHCOUNTS Team, but his team's practice time didn't match with his free time.

[quote="jjx1"]p.s You do have a reason to protest this fantasy. But, lol, if you would have made state or been 1st i'm sure you wouldn't be protesting .[/quote]

No, I didn't make states because of the BS Countdown round.

Sadly, I must agree with the "P.S." in jjx1's post.

@Fantasy: So your team was disqualified for not being from the same school, not because you were "too good"?

Why is izzy protesting this?  Either way, she's not in top 4 :maybe:

Wow, AIME...

As I was keep saying, coming from different school was UNAVOIDABLE, and our team was also TOO GOOD.

For your last question, it's a bit complicated to answer...

Okay first of all if you do win this protest what will happen. Do you have to cd again? or what. ANd also do you have the rule of top two or top three teams make state. THen the 4 people or so people not in the top teams make state? if so. None of the top four people were in a top team???

Hm, your chapter is [size=200]really[/size] screwed up.

Those that made state because of this "bs" countdown round will be mad if this happens.

I have already protested (for your sake) and will now exit this discussion.

[quote="jjx1"]Okay first of all if you do win this protest what will happen. Do you have to cd again? or what. [/quote]
At this point, I do not know.

[quote="jjx1"]ANd also do you have the rule of top two or top three teams make state. THen the 4 people or so people not in the top teams make state? if so. None of the top four people were in a top team???[/quote]

Nope. None of the top four people after CD round were in either first or second team.

i thought the ladder was for the old format...


email
videos
geometry
geometric transformation
reflection

Take an average person.

In America.

Not on AoPS.

Middle Class.

Here's a standard workday:
Get up
Get ready
Go to work
Have lunch
Do more work
Come home
Watch TV
Eat Dinner
Watch some more TV
Get Ready For Bed
Watch more TV
Go to bed
Watch TV
Go to sleep

Disclosure: I have no idea if this is correct, but i'm pretty sure that this is close to it.

And when people go to work they turn on the radio because there's nothing else to do.

And when they check their email at work they might throw in 5 minutes of news websites.

And the children of this guy go to school, do school, come back, watch TV, have dinner, more TV, then go to bed.

All in all, someone is bombarded with advertising, media, and entertainment for a lot of the day(at least 3 hours give or take). What is your opinion on this?

[size=0]I am prepared for lots of scrutiny, but please don't bash at me, I'm just asking for your opinion. :maybe: [/size]

Yeah, I might be biased b/c I don't watch TV (except some comedy) or play video games, but I think that's pretty pathetic. It's not even fun to rot in front of the TV -- it's just pseudo-engaging. If people took some time for self-reflection and evaluated what they spent their time on, they'd probably realize that watching so much TV isn't really worth it.

It's also funny how everyone's trying to get entertained all the time. Theres phone and movies on your ipod, and music and movies on your phone, and radio on your TV, and generally, every form of entertainment on every device. They're all the same, and all being advertised by shamelessly frequent usage of the catch-word "entertainment".

Why can't people spend 2 seconds without being "entertained" by media?

Also, I noticed you didn't include breakfast in your list of things to do. Yeah, people really need to start eating breakfast regularly.

What is the meaning of ¨an average person¨? I propose that such a thing is not well-defined.

An average person is one who doesn't engage himself in intellectual endeavors/ pursuits rather he/she is more concerned with worldly things like T.V., computer games. An average person is one who might waste time on mundane things.

I myself am not sure of what traits constitute an 'average person' but people maintain that some of those I had cited are some general characteristics that are exhibited.  :)

Maybe this?

http://en.wikipedia.org/wiki/American_middle_class#Working_class_majority


geometry
incenter
analytic geometry
geometry proposed

Let $\triangle ABC$ be a triangle, denote $G,I,S$ the centroid, the incenter and the Spieker center, respectively. Show that :

(1) $G$ is contained in the incircle $(I)$ if and only if $5(a^2+b^2+c^2) \leq 6(ab+bc+ca)$

(2) $S$ is contained in the incircle $(I)$ if and only if $ab(a+b)+bc(b+c)+ca(c+a) \leq 7abc$

P.S. The [i]Spieker point [/i]of a triangle $\triangle ABC$ is the incenter of its medial triangle.

Let M a point with the baricentrical coordinates (x,y,z), x+y+z=1. Then

$M\in Circle(I,r)\Longleftrightarrow MI^2=r^2\Longleftrightarrow \sum xbc-\sum yza^2=r(4R+r)
\Longleftrightarrow$

$\sum xbc=\sum yza^2+\sum (s-b)(s-c)\Longleftrightarrow\sum 4xbc=\sum 4yza^2-\sum a^2+\sum 2bc$

$\Longleftrightarrow\sum (1-4yz)a^2=2\sum (1-2x)bc.$ In conclusion,
\[M\in Circle(I,r)\Longleftrightarrow\sum (1-4yz)a^2=\sum (1-2x)bc.\ \ (*)\] [u]The first point of the problem.[/u]
Particularly, $M:=G\left(\frac  13,\frac 13,\frac 13\right)$ we have $M\in Circle(I,r)\Longleftrightarrow 5\sum a^2=6\sum bc$.....a.s

[u]The second point of the problem.[/u]
Because $G\in (IS),IG=2GS,\ I\left(\frac{a}{2s},\frac{b}{2s},\frac{c}{2s}\right)$ results 2S+I=3G, i.e. $S\left(\frac{b+c}{4s},\frac{c+a}{4s},\frac{a+b}{4s}\right).$
From the relation (*) results that

$S\in Circle(I,r)\Longleftrightarrow\sum \left[1-\frac{4(a+b)(a+c)}{16s^2}\right]\cdot a^2=2\cdot\sum\left[1-\frac{2(b+c)}{4s}\right]$ $\cdot bc\Longleftrightarrow$

$\sum\left[(a+b+c)^2-(a+b)(a+c)\right]\cdot a^2=4s\sum abc\Longleftrightarrow\sum\left[2s(b+c)-bc\right]\cdot a^2$ $=6abc(a+b+c)$

$\Longleftrightarrow\sum (b+c)a^2=7abc\Longleftrightarrow\sum ab(a+b)=7abc$...a.s.o.

These problems are nicely because they are equivalencies with inequalities.
(See http://www.gil.ro)
.

[quote="levi"]Let M a point with the baricentrical coordinates (x,y,z), x+y+z=1. Then

$M\in Circle(I,r)\Longleftrightarrow MI=r\Longleftrightarrow \sum xbc-\sum yza^2=r(4R+r)
\Longleftrightarrow$

[/quote]

Can you explain this in more detailed way ?

Leon

Let $M_1(x_1,y_1,z_1),M_2(x_2,y_2,z_2)$ be two points with baricentric coordinates with respect to the triangle ABC.
Then there is the relation:
\[4Rr+\overline{IM_1}\cdot\overline{IM_2}=\frac 12 [(z_1+z_2)ab+(x_1+x_2)bc+(y_1+y_2)ca]-\] \[-\frac 12 [(y_1z_2+y_2z_1)a^2+(z_1x_2+z_2x_1)b^2+(x_1y_2+x_2y_1)c^2]\] For $M_1=M_2:=M(x,y,z)$ we obtain the relation (*).


function
induction
algebra unsolved
algebra

Find all functions f such that :f is from N to N
in this form :
f(2f(x)+f(y))=2x+y
for all natural x & y

Suppose there exists some $ m\not \equal{}n$ such that $ f(m)\equal{}f(n)$
Then $ 2m\plus{}m\equal{}f(2f(m)\plus{}f(m))\equal{}f(2f(m)\plus{}f(n))\equal{}2m\plus{}n$. Contradiction. Thus $ f$ is injective.
[u][b]Case 1:[/b][/u]
Suppose $ f(a)<a$
there exists $ [\frac{a\minus{}1}{2}]$ pairs $ (x,y)$ such that $ 2x\plus{}y\equal{}a$ where $ [q]$ denotes greatest integer function.
We will have $ 2f(x)\plus{}f(y)<a$ for all such $ x,y$.
$ \implies a\minus{}2\geq 2f(x)$
$ \implies [\frac{a\minus{}2}{2}]\geq f(x)$
but for $ [\frac{a\minus{}1}{2}]$ distinct $ x,$ $ f(x)$ are distinct.
For odd $ a$ it is impossible.
If $ a$ was even then $ f(1)\equal{}1,f(2)\equal{}2,...f(\frac{a\minus{}2}{2})\equal{}\frac{a\minus{}2}{2}$
But in that case $ a\equal{}f(2f(\frac{a\minus{}2}{2})\plus{}f(2))\equal{}f(a)$. Contradiction.
[u][b]Case 2:[/b][/u]
Suppose $ f(b)>b$
$ 3b\equal{}f(3f(b))$
Putting $ 3f(b)\equal{}a$ we get Case 1.

So $ \boxed{f(x)\equal{}x}$ for all $ x$

[hide]Step 1: $ f$ is injective.
If $ f(a) \equal{} f(b)$, then $ 3a \equal{} f(3f(a)) \equal{} f(3(f(b)) \equal{} 3b$, so $ a \equal{} b$.

Step 2: $ f(k\plus{}1)\minus{}f(k) \equal{} f(m\plus{}1)\minus{}f(m)$ for all naturals $ k,m$.
We show by induction that $ f(k\plus{}1)\minus{}f(k) \equal{} f(2)\minus{}f(1)$ which is sufficient. Base case $ k \equal{} 1$ is pretty evident.
Assume true for $ k$. Notice that $ f(2f(1)\plus{}f(k\plus{}2)) \equal{} k\plus{}4 \equal{} f(2f(2) \plus{} f(k))$.
By $ f$ injective, $ 2f(1) \plus{} f(k\plus{}2) \equal{} 2f(2) \plus{} f(k)$, implying that $ f(k\plus{}2) \equal{} 2f(2) \minus{} 2f(1) \plus{} f(k)$.
By the inductive hypothesis, $ f(k\plus{}2) \equal{} 2f(k\plus{}1) \minus{} 2f(k) \plus{} f(k)$, implying that $ f(k\plus{}2) \minus{} f(k\plus{}1) \equal{} f(k\plus{}1) \minus{} f(k)$.
Another application of the inductive hypothesis gets that $ f(k\plus{}2)\minus{}f(k\plus{}1) \equal{} f(2)\minus{}f(1)$ as wanted.

Step 3: $ f(x) \geq x$.
We show $ f(2) \minus{} f(1)$ is greater than $ 0$. It can't be negative by the well-ordering principle. $ f(2) \minus{} f(1) \equal{} 0$ implies $ f$ is constant, which also doesn't work. This means $ f$ is strictly increasing by step 2's conclusion, and $ f(x) \geq x$ follows.

Step 4: $ f(x) \equal{} x$.
If $ f(n) > n$, then $ f(3f(n)) \geq 3f(n) > 3n$, contradiction. So $ f(x) \equal{} x$ for all $ x$.[/hide]

$ f(x) = f(y)$
$ \Longrightarrow$ for any $ "a"$ $ f(2f(a) + f(y)) = f(2f(a) + f(x))$$ \Longrightarrow 2a + x = 2a + y$  $ \Longrightarrow x = y$ therefore $ f$ is injective ,now we note that $ f(2f(a) + f(b)) = 2a + b = f(2f(b) + f(2a - b))$ for $ 2a > b$
$ \Longrightarrow 2f(a) + f(b) = 2f(b) + f(2a - b)$
$ \Longrightarrow 2f(a) = f(b) + f(2a - b) ______*$
now $ 3f(3f(1)) = 3*3$ $ \Longrightarrow f(9) = 9f(1)$ substiuting  $ a = 5$ and $ b = 9$ in $ *$
we get $ f(5) = 5f(1)$  now substiuting  $ a = 3$ and $ b = 5$ in $ *$we get $ f(3) = 3f(1)$
also $ f(2f(2) + f(1)) = 5 = f(2f(1) + f(3))$ gives $ f(2) = 2f(1)$ let $ f(1) = c$
now $ f(2) = 2c$ and $ f(3) = 3c$
assume till $ f(i) = ic$ for $ i = 1$ till $ k - 1$
now $ f(2f(1) + f(k)) = k + 2 = f(2f(2) + f(k - 2))$
applying injectivity we get $ 2f(1) + f(k) = 2f(2) + f(k - 2)$
$ \Longrightarrow f(k) = f(k - 2) + 2c = (k - 2)c + 2c = kc$
therefore $ f(x) = xc$ substituting in the equation we get $ c = 1$ as $ c$ is positive]
therefore $ f(x) = x$ for all $ x$ in $ naturals$

[color=green]1. $ f$ is injective.
2. $ f(2f(n)\plus{}f(n\minus{}1))\equal{}3n\minus{}1\equal{}f(2f(n\minus{}1)\plus{}f(n\plus{}1))$ then $ f(n\minus{}1)\plus{}f(n\plus{}1)\equal{}2f(n)$.
3. $ f(n)\equal{}n$.[/color]

[quote="quangpbc"][color=green]1. $ f$ is injective.
2. $ f(2f(n) \plus{} f(n \minus{} 1)) \equal{} 3n \minus{} 1 \equal{} f(2f(n \minus{} 1) \plus{} f(n \plus{} 1))$ then $ f(n \minus{} 1) \plus{} f(n \plus{} 1) \equal{} 2f(n)$.
3. $ f(n) \equal{} n$.[/color][/quote]
 :clap2:


number theory
prime numbers
combinatorics unsolved
combinatorics

S is a set of 9 prime numbers ,we consider 1985 natural numbers which all prime divisor in S .
prove that there exists 4 number such that their product equals $ n^4$ ( n is some natural  number)

1985 IMO, #4
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60788[/url]


well the series was 1,1,2,3,6,8,13...
I know the rule is the after number equals the sum of two number before ... ..I think there's an easier way for us to find the nth number without using calculator to add the numbers..please tell me thank you..

This is very famous sequence in mathematics. Hint: You can find this sequence in pascal's triangle.

I know pascal's triangel however I dont get that you can find it using pascal's triangel.....can you tell me more clearly plz

At first it hast to be 1,2,3,[b]5[/b],8,13,...

The name of this sequence is 'Fibonacci sequence', I'am sure it was discussed before. There is no good formula that doesn't need a calculator, but look for 'Binet's formula'

Although there is a theorem to find the mth term in the nth row of Pascal's triangle, so if you know the row and term, you could apply the formula (I think it's nCm, but correct me if i'm wrong)

Well, these numbers (yes, your formula, that I would take as definition, is correct ) only occure as 'srew' sum of some numbers in pascal's triangle, so it doesn't help much (I think it are even more calculations)

to find the general term of a homogeneous sequence, you can use the method of characteristic equation or generating function

We have $a_{n+2}-a_{n+1}-a_{n}=0, \forall n \in \mathbb{N}$.

We first calculate the roots of the equation: $x^2-x-1=0$, which are $x_{1}=\frac{\sqrt{5}+1}{2}$ and $x_{2}=\frac{-\sqrt{5}+1}{2}$.

It is known that the general term is given by the formula: $a_{n}=C \cdot x_{1}^n+D \cdot x_{2}^n, \forall n \in \mathbb{N}$.

Now we construct a system:
$a_{1}=C \cdot x_{1}+D \cdot x_{2}=1$
and
$a_{2}=C \cdot x_{1}^2+D \cdot x_{2}^2=1$

from which we find constants $C$ and $D$.

This is obviosuly the Fibonacci Sequence which has a formula:
\[ F_n=\frac{\phi^n-\left(\frac{-1}{\phi}\right)^n}{\sqrt{5}}  \]
where $\phi$ is the golden mean which equals $\frac{1+\sqrt{5}}{2}$.

EDIT: This does require a calculator.
Masoud Zargar


search
function
complex analysis
complex analysis unsolved

Suppose that f:G -> C is analytic except poles. Show that the poles of f can not have a limit in G

Use search:
[url=http://www.mathlinks.ro/viewtopic.php?t=294931]found by searching "poles have limit"[/url]

A pole is, by definition, an isolated singularity. A limit point of poles is a non-isolated singularity.

Invert the function, the poles become zero. If they accumulate in $ G$, then by the rigidity theorem, $ 1/f\equal{}0$ over $ G$, which is a problem.

[quote="LkNsngth"]Invert the function, the poles become zero. If they accumulate in $ G$, then by the rigidity theorem, $ 1/f \equal{} 0$ over $ G$, which is a problem.[/quote]
$ f$ might be zero occasionally. Of course, these zeroes are apart from poles, but can accumulate at the same point.


According to a textbook, the atmosphere, hydrosphere, biosphere, and solid Earth are all related to the death of a star occuring billions of years ago, but I don't really understand this concept.

Could someone give me a little bit of insight on this?

If you're talking about what I think you're about, then consider this:  at some point in the past, the universe was composed solely of hydrogen.  Yet the universe (and Earth) now contains many other elements; where did they come from?

The answer is fusion.  The fusion that occurs in stars is (believed to be) the source of most of the universe's elements heavier than hydrogen.  The heavier the star, the more fusion it undergoes before it dies.  Now, stars can die in a variety of ways, but on the rare occasion that a star goes nova, it spits out all the heavy elements that have been fused in its core, and these elements scatter across the cosmos.  It is from these elements that planets are formed.

This process is known as [url=http://en.wikipedia.org/wiki/Stellar_nucleosynthesis]stellar nucleosynthesis.[/url]  As an example of how this process directly affects the mixture of elements on planets such as ours,

[quote="Wikipedia"]Iron and nickel are notable for being the final elements produced by stellar nucleosynthesis, and thus are the heaviest elements which do not require a supernova or similarly cataclysmic event for formation. Iron and nickel are therefore the most abundant metals in metallic meteorites and in the dense-metal cores of planets such as Earth.[/quote]

Wow very nice post t0rajir0u. :)

Just like to add something, to elaborate.  How our solar system was created, was when a star went super nova and the gas and debris from the explosion formed a cloud.  From this cloud a center star formed of condensed matter and the surrounding matter formed clumps that got larger and larger to form the planets.  Even today the remnants of the cloud of debris is seen in the Kuiper Belt, Asteroid Belt, and Oort Cloud.

Another way to think of it is, since the Earth is here, you assume it was made of the stuff in the universe which are stars and clouds of gas.

the theory for the origin of our solar system seems plausible.
but i always wonder where did our solar system's parent star go?

it should be a neutron star somewhere still around, right?

Well when a star goes supernova, the star does not always remain afterwards.  According to the theory it went "completely" supernova so the core neutron star did not last.  Where the star went is what now makes up the present Sun and planets.

A supernova is triggered by a shock wave from the core in degenerate state. So a star can't go "completely" supernova, i.e. the core cannot participate in the explosion. The solar neighbourhood up to few tens of light years shows the structure of an explosion product ("the local fluff"). As yet the remnant has not been found.
*All this is to the best of my knowledge, heard in lectures etc.

A recent issue (May 21, 2007) of TIME magazine has an article talking about the supernova from a massive blue supergiant (around 100 to 200 times as massive as the Sun).

According to that article, this type of supernovas completely destroys the core, leaving nothing to form a black hole or neutron star.

I am not sure if it's just a hypothesis or has some solid theoretical basis.

The article james4l was talking about it is [url=http://www.time.com/time/magazine/article/0,9171,1619560,00.html]here[/url].

Reagarding shreyas_patankar 's post, the defintions in Astrophysics change over time; our present definitions are just from what we have observed or from what we found to be theoretically possible.
In that article they believe they have found another type because they have seen something they have not seen before.
[url=http://en.wikipedia.org/wiki/Supernova]Wikipedia Supernovae Article[/url].

For this reason, I am never completely confident when posting astronomy.
[quote="shreyas_patankar"]*All this is to the best of my knowledge, heard in lectures etc.[/quote]


inequalities
trigonometry
function
geometry unsolved
geometry

In a triangle $ ABC$ right-angled at $ C$ , the median through $ B$ bisects the angle between $ BA$ and the bisector of $ \angle B$. Prove that
\[ \frac{5}{2} < \frac{AB}{BC} < 3\]

[quote="castigioni"]The solution is a=2.65896...b
So a/b=2.65896... that is between 3/2 and 3[/quote]
Please write your full proof...I would really like to see it...

Hey can't anyone give a full proof of this...
The 2nd post is definitely done by calculator....That is NOT ALLOWED in out national olympiads....

[quote="Rijul saini"]In a triangle $ ABC$ right-angled at $ C$ , the median through $ B$ bisects the angle between $ BA$ and the bisector of $ \angle B$. Prove that
\[ \frac {5}{2} < \frac {AB}{BC} < 3\]
[/quote]
$ \angle\frac {B}{4}$ $ \equal{}$ $ \alpha$ and $ M$ is the midpoint of $ AC$
We can compute that $ \sin3\alpha$ $ \cos4\alpha$ $ \equal{}$ $ \sin\alpha$ ,
We know that $ \frac {c}{sin90}$ $ \equal{}$ $ \frac {a}{sin(90 \minus{} 4\alpha)}$, Hence $ \frac {c}{a}$ = $ \frac {1}{\cos4\alpha}$ = $ \frac {\sin3\alpha}{\sin\alpha}$ = $ 3 \minus{} (\sin\alpha)^2$.
$ \Longrightarrow$ $ \frac {c}{a}$ $ <$ $ 3$.
$ \cos2\alpha$ $ \equal{}$ $ \frac {(\frac {c}{a} \minus{} 1)}{2}$ $ \Longrightarrow$ $ \frac {a}{c}$ $ \equal{}$ $ \cos4\alpha$ $ \equal{}$ $ \frac {1}{2}$$ (\frac {c}{a} \minus{} 1)^2 \minus{} 1$.

Consider $ \frac {c}{a}$ $ \leq$ $ \frac {5}{2}$ and this is  contradict with the initial consideration, namely, $ \frac {c}{a}$ $ >$ $ \frac {5}{2}$

Since $E$ is the midpoint of $AC$, we have $AE=EC=\frac{b}{2}$. Since $BD$ bisects $\angle ABC$, we also have $CD=\frac{ab}{a+c}$. Since $BE$ bisects $\angle ABD$, we have \[\frac{BD^2}{BA^2}=\frac{DE^2}{EA^2}.\] However we have \[\begin{cases}BD^2=BC^2+CD^2=a^2\frac{a^2b^2}{(a+c)^2} \\ DE^2=\left(\frac{b}{2}-\frac{ab}{a+c}\right)^2\end{cases}\] Using these in the above expression and simplifying yields \[a\left[(a+c)^2+b^2\right] = c^2(c-a)^2.\] Using $c^2=a^2+b^2$ and eliminating $b$, we obtain \[c^3-2ac^2-a^2c-2a^3=0.\] Substituting $t=\frac{c}{a}$, we reduce this to a cubic: \[t^3-2t^2-t-2=0.\] Now, consider the function $f(t)=t^3-2t^2-t-2$ for $t>0$, since $c/a\in\mathbb{Z}^{+}$. For $\<t\le 2$, we see that $f(t)=t^2(t-20-t-2<0$. We also observe that $f(t)=(t-2)(t^2-1)-4$ is strictly increasing on the interval $(2, \infty)$. It is easy to compute $f(5/2)=-\frac{11}{8}<0$ and $f(3)=4>0$. Hence there is a unique value of $t$ in the interval $(\frac{5}{2}, 3)$ such that $f(t)=0$. Hence \[\frac{5}{2}<\frac{c}{a}<3\] as desired. $\Box$

my solution

solution =

$\angle\frac {B}{4} = \alpha$ and$ M$ is the midpoint of $AC $
using sine law in triangle $ABE$ and$BEC $, we obtain
that $\sin3\alpha \cos4\alpha = \sin\alpha , $

We know that$ \frac {c}{sin90} = \frac {a}{sin(90 - 4\alpha)}$, Hence
$\frac {c}{a} = \frac {1}{\cos4\alpha} = \frac {\sin3\alpha}{\sin\alpha} = 3 - (\sin\alpha)^2.$
trivially,
$\Longrightarrow \frac {c}{a} < 3.$

now , to prove that $(5/2) < (c/a) = 3- 4sin^2\alpha$
this is equivalent to $sin\alpha< (1/2^{3/2} )$
now we know
$sin\alpha < sin (22.5) < (3^{1/2} - 1)/(2^{3/2}) < 1/(2^{3/2} )$

and we are done :P :D


function
real analysis
calculus
integration
trigonometry
real analysis unsolved

If $f\in L[c,d]$(Means Lebesgue intergal),and $c<a<b<d$,such that:
$\int_{a}^{b}|f(x+h)-f(x)|dx=O(|h|)$
Prove that there exist a function $g$ such that:
(1)$f(x)=g(x)$,for almost everywhere.
(2)g is a bounded variation function.

Notice:
A bound variation function means that:for every $a=x_{0}<x_{1}<\cdots<x_{n}=b$
then $\sum_{i=1}^{n}|g(x_{i})-g(x_{i-1})|$ is bounded.(I don't know how to say it in English,so I just write out its defination :P )
Thanks for help.

Edit by the advice below.

It is a function of bounded variation. :)

A counterexample: $f(x)=\sin\frac{1}{x}$ on the interval $(0,1)$.
You should assume that $O(|h|)$ it uniform with respect to $x$. Formally, \[\sup_{x\in [a,b-h]}\int_{0}^{h}|f(x+t)-f(x)|dx =O(h)\] if $h$ is positive, and similarly for negative $h$.

Sorry.  :blush:
I mistake the problem,and wrote a nonsence condition(Because $\int_{0}^{h}|f(x+t)-f(x)|dt=O(|h|)$ is always true.)
Now you can do it,Thank you.


Given a continuous function $\textsl{f} : \Re \rightarrow \Re$ satisfying the conditions:
$\textsl{f}(1000) = 999$,
$\textsl{f}(\textsl{x})\bullet \textsl{f}(\textsl{f}(\textsl{x})) = 1$ for all $\textsl{x} \in \Re$.
Determine $\textsl{f}(500)$.

[hide="solution"]We have $f(1000)=999\Longrightarrow999\cdot{f}(999)=1\Longrightarrow{f}(999)=\frac{1}{999}.$ Then by the Intermediate Value Theorem, there exists $\alpha\in(999,1000)$ such that $f(\alpha)=500$. Thus $f(\alpha)\cdot{f}(f(\alpha))=500\cdot{f}(500)=1\Longrightarrow{f}(500)=\frac{1}{500}$. I'm not sure, but this might be too easy for intermediate.

I'm interested in what this function would look like. Clearly for $x\in[\frac{1}{999},999]$ we have $f(x)=\frac{1}{x}$, but then it jumps up to $999$ at $x=1000$. What is the explicit formula?[/hide]


logarithms
real analysis
real analysis unsolved

i) Prove that

$ \prod_{n\equal{}1}^{\infty}\left(1\plus{}\frac{1}{n^{2}}\right)\equal{}\frac{\sinh\pi}{\pi}$

ii) Using i), evaluate

$ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n}}{n}\zeta(2n)$

After integrating the formula $ \pi\coth\pi z\minus{}\frac{1}{z}\equal{}\sum_{n \equal{} 1}^{\infty}\frac{2z}{n^{2}\plus{}z^{2}}$ from $ 0$ to $ x$ you obtain
$ \ln\frac{\sinh\pi x}{\pi x}\equal{}\sum_{n \equal{} 1}^{\infty}\ln\left(1\plus{}\frac{x^{2}}{n^{2}}\right)$ . Use $ \exp$ on both sides and set $ x \equal{} 1$ .

$ \sum_{n \equal{} 1}^{\infty}\frac{(\minus{}1)^{n}}{n}\sum_{k \equal{} 1}^{\infty}\frac{1}{k^{2n}}\equal{}\sum_{k \equal{} 1}^{\infty}\sum_{n \equal{} 1}^{\infty}\frac{(\minus{}1)^{n}}{n}\left(\frac{1}{k^{2}}\right)^{n}\equal{}\sum_{k \equal{} 1}^{\infty}\minus{}\ln\left(1\plus{}\frac{1}{k^{2}}\right) \equal{}\minus{}\ln\prod_{k \equal{} 1}^{\infty}\left(1\plus{}\frac{1}{k^{2}}\right)$


calculus
integration
trigonometry
calculus computations

Alright I tried but I was unable to get this one: Origanaly It is integrate: 2sin(x)/(6sin(x)+4cos(x)) I've got it to 2(integrate: sin(x)[3sin(x)+2cos(x)]^-1) I can't figure out where to go from here, any help?

Notice that $(3\,sinx+2\,cosx)'=-2\,sinx+3\,cosx$.
Now try to write $sinx$ as a linear combination of $(-2\,sinx+3\,cosx)$ and $(3\,sinx+2\,cosx)$.
  :)


vector
calculus
derivative
function
algebra
domain
linear algebra

are linear mappings defined for all elements of a vector space? so can't you have an "undefined" mapping for an element? so if a linear mapping is not onto, that would mean some vectors map to zero (null space is not just {0}), but all vectors DO have a mapping?

thank you

Yes; there's no reason for this not to be true for finite-dimensional vector spaces.  For Hilbert spaces one sometimes permits operators such as differentiation to be defined on some dense subset instead.

That's a very confused statement (I'm referring to the post by mche.). Let's back up, and not talk about vector spaces or linear mappings, but just return to basic "function-speak."

When we say $ f: A\to B,$ for which we use words like "$ f$ is a function from $ A$ to $ B$" or "$ f$ maps $ A$ into $ B,$" we do imply that $ f$ is defined for every element of $ A.$ (In the specialized situation that t0rajir0u referred to, we'd still cover ourselves with wording like "the operator is densely defined" or "the operator is defined on such-and-such subspace." I'd suggest not paying too much attention to that just yet.)

In $ f: A\to B,$ $ A$ is the [i]domain[/i] of $ f.$ $ B$ is the [i]codomain[/i] or "target space." The question of whether a function is onto (surjective) depends on what we have specified the codomain to be.

The set $ \{y\in B\,|\,\text{there exists }x\in A\text{ such that }f(x) \equal{} y\}$ is the [i]range[/i] of $ f.$ $ f$ is onto if the range of $ f$ is all of $ B,$ and it is not onto if the range of $ f$ is something less than all of $ B.$ A function not being onto has nothing whatever to do with it not being defined at any point.

Here's an example: $ A \equal{} B \equal{} \{1,2,3\}.$ Define $ f: A\to B$ by $ f(1) \equal{} 1,f(2) \equal{} 3,f(3) \equal{} 1.$ Note that this is defined for all $ x\in A.$ It's not onto, because there is no $ x$ such that $ f(x) \equal{} 2.$ It's not one-to-one (injective), because $ f(1) \equal{} f(3).$  For both of those reasons, it doesn't have an inverse function.

okay, I think I see what you're saying.. so you can only say f: R -> R is the function is defined for all R?  but what I said initially was wrong as well, if a mapping is not onto, it doesn't mean that the "excluded" elements are the null space or anything like that.. thank you for your paitence


Suppose that $ a,b,c$  are the side-lengths of a triangle such that $ a\plus{}b\plus{}c\le 1$ . Prove that
$ \sqrt[3]{2(b\plus{}c\minus{}a)}\plus{}\sqrt[3]{2(c\plus{}a\minus{}b)}\plus{}\sqrt[3]{2(a\plus{}b\minus{}c)}\ge 64abc$

Equality holds for $ a\equal{}b\equal{}c\equal{}0$ or $ a\equal{}b\equal{}\frac1{4},c\equal{}\frac1{2}$ up to permutation!  :)

your inequality is very nice!!
I don't have solution at the moment!!
but I don't believe that it is true!!

[quote="can_hang2007"][quote="dduclam"]Suppose that $ a,b,c$  are the side-lengths of a triangle such that $ a \plus{} b \plus{} c\le 1$ . Prove that
$ \sqrt [3]{2(b \plus{} c \minus{} a)} \plus{} \sqrt [3]{2(c \plus{} a \minus{} b)} \plus{} \sqrt [3]{2(a \plus{} b \minus{} c)}\ge 64abc$

Equality holds for $ a \equal{} b \equal{} c \equal{} 0$ or $ a \equal{} b \equal{} \frac1{4},c \equal{} \frac1{2}$ up to permutation!  :)[/quote]
My solution :)[/quote]

Thank you,Can. This's a nice but easy problem. My solution using AM-GM inequality,only one line  :lol:


Perfect Squares
number theory
IMO Shortlist

An integer $n$ is said to be [i]good[/i] if $|n|$ is not the square of an integer. Determine all integers $m$ with the following property: $m$ can be represented, in infinitely many ways, as a sum of three distinct good integers whose product is the square of an odd integer.

[i]Proposed by Hojoo Lee, Korea[/i]

This problems had appeared here early and I solved it. But I don't remember its previous subject.

This problem was proposed in 2003 IMO Shortilist...
It was also proposed for the first Moldavian TST this year....
It is a very hard one! I admit it :):):)

[quote="vetali"]Shi bai eshti krutoi. Vrei sa ai mai multi posturi. (palosh)

Problema data e intr-adevar nu una din cele mai ushoare.
se arata ca nr de forma 4k+1 sunt cele dorite.[/quote] please use english language. if you do not know english, please ask your fellow users (both Romanian and Moldovian) to guide you. Thanks.

The source of the problem is ISL 2003, proposed by Hojoo Lee, and also MathLinks Contest, 3rd Edition, Round 3, problem 3.

I knew that I had solved this problem early.

Consider following representations
n=3(n+6k2)-2(n+6k2)-6k2
n=4(n+6k2)-3(n+6k2)-3k2
n=9(n+6k2)-8(n+6k2)-2k2
where k is arbitrary integer.
It is easy to prove that one of these representations satisfies all conditions.

Valentin! What is official solution for this problem?

Myth, could you please post a complete solution, because I don't get why is easy to solve the problem from ther  :?  :(

Leonardo!
I know you is in Venezuela's IMO team. Just try to complete solution yourself.
OK?

Best regards

But Myth, think about those who are not qualified for IMO.... :(

Pierre.

You want to say that you cannot complete solution? :)

[quote="darij grinberg"]Let me just speak out my greatest thanks to all posting ISL 2003 problems on this wonderful forum. I hope they will be of much use for me on my two remaining TSTs in the end of May.

  Darij[/quote] note that only problems previously proposed in other TSTs have been revealed on the forum. no other ISL problems are revealed.

for example, this problem has been used in the Moldovian TSTs 2004.
the other problems were either used in German TSTs, or Singaporean TSTs.

Well, I think this is mostly true, but - for instance - [url=http://mathlinks.ro/viewtopic.php?t=5642]this geometry problem[/url] was used in the Singapore [i]and[/i] in the German TSTs (and in the Belgian ones, if I remember correctly). But I guess this is not often the case.

  Darij

Excuse me. My previous solution is incorrect, because n<>4(n+6k2 )-3(n+6k2)-3k2.

New solution.
Consider following representations:
(1) n=2(n+2k2)-(n+2k2)-2k2
(2) n=4(n+3k2)-3(n+3k2)-3k2
Let k be sufficiently large integer. It is clear 2k2 and 3k2 are good numbers.
a) n=4l+1 then we use (1) where k is odd ==> (n+2k2)=3 (mod 4) and 2(n+2k2)=2 (mod 4). So addends are good numbers.
b) n=4l+2 then we use (2) where k is even ==> (n+3k2)=2 (mod 4) and 3(n+3k2)=2 (mod 4).
c) n=4l+3 then we use (1) where k is even ==> (n+2k2)=3 (mod 4) and 2(n+2k2)=2 (mod 4)
d) n=4rt then we apply 1), 2) or 3) for t and multiply addends by 4r.

[hide=Disclaimer] I don't know Quadratic Reciprocity, so I can't prove my assmption, but it is based on computational evidences, and it's very probably likely to be true. [/hide]

For mod $4$ reasons, we see that the integer must be $3 \mod 4$. For any integer $i \equiv 3 \mod 4$, let $S$ denote the set of primes $p_i$ such that $i$ is not a quadratic residue modulo $p_i$. If I assume $|S|$ is infinite, then the question is trivial, because (for $p_i \in S$, $p_i > |i|$) $$ \displaystyle (a,b,c) = \{-p( \frac{ i + p^2}{2} + p), -p(2+p), ( \frac{ i + p^2}{2} + p)(2+p) \} $$ satisfies the condition of the question, so all such $i$ works.

[hide=Concrete examples in python] Click on Pop Out to have a  better experience, then click run. No need to have any programming knowledge - just run it. $i$ denote $i$, in above notation (it may be negative), and $k$ denote the maximum searching limit, i.e the programme checks the members of $S$ until $k$th prime, so it must be positive and big.

 [pywindow]
def prime(K):
	t = [2]
	n = 1
	j = 3
	while (n < K):
		for i in range(n):
			if (j % t[i])== 0:
				j = j+1
				break
			else:
				if (j < t[i]**2):
					t.append(j)
					j = j+1
					n = n+1
					break
	return t

def qr(i, p):
	i = i%p
	for j in range(p):
		if j**2 % p== i:
			return 1
	return 0

def s(i, k):
	a = prime(k)
	a = a[2:]
	b = []
	for j in a:
		if qr(i, j)-1:
			b.append(j)
	return b

def corr(b, i):
	i = -i
	if b[-1]**2 < i:
		return -1
	else:
		for j in range(len(b)):
			if b[j]**2 > i:
				return j

def trip(p, d):
	print("----------------------")
	r = int(((d+p**2)/2 + p))
	s = int((2+p))
	print( "a=-"+str(p)+"*"+str(r), "b=-"+str(p)+"*"+str(s), "c="+str(r)+"*"+str(s))
	print (-p*r, -p*s, r*s)

def main():
	i = int(input("i -->Please input a number 3 modulo 4"))
	k = int(input("k---> Tests upto this number of primes. Make this large, if you typed previous one negative"))
	a = s(i, k)
	c = corr(a, i)
	if c == -1:
		print("Please restart the programme - make the k larger")
	else:
		a = a[c:]
	print("**********************ACTION BEGINS :P************************************")
	print("Note that a,b,c are always not square, and c*a*b is always square, and c+a+b is always i")
	for j in a:
		trip(j, i)

main()

 [/pywindow] [/hide]

First, note the even integers obviously don't work since they can't be the sum of three odd integers. In addition, numbers that are $1 \mod 4$ don't work either because if three numbers $a,b,c$ multiply to a perfect square, we must have $(a,b,c) \equiv (1,1,1),(1,3,3) \mod 4$, up to permutation, both sets which have sums that are $3 \mod 4$. Now we will prove that all numbers $3 \mod 4$ work. Write $s=(p_1)^{e_1}(p_2)^{e_2}...(p_k)^{e_k}$. Fixing $s$, we choose a prime $p \equiv 3 \mod 4$, $p \equiv 1 \mod s$, $p \equiv -2-r \mod r^2$, where $r$ is a large prime not dividing $s$ by CRT and Dirichlet.
We claim that the construction $(a,b,c) = (-p( \frac{s + p^2}{2} + p), -p(2+p), ( \frac{s + p^2}{2} + p)(2+p))$ satisfies the condition of the question for sufficiently large $p$. To show this, note that if the first one were a perfect square absolute value, $p|s$, impossible for large p. Therefore, the third one must be a perfect square. Let $q_1,q_2,...q_k$ denote the primes with odd exponent in the factorization of $p+2$. Then $q_i|s$ for all $1 \leq i \leq k$. But $p+2 \equiv 3 \mod s$, so the only possibility is the $q_i=3$. However, $q_i=r$ for some $i$ and since $r$ was arbitrarily large, we get a contradiction.

Clearly, no even integer can be expressed as the sum of three odd integers.

If $n\equiv 1\bmod{4}$ and $n=a+b+c,$ since $a\cdot b\cdot c$ is an odd perfect square, then $(a,b,c)$ reduces to $(1,1,1)$ or $(1,3,3)$ mod $4$ which both add up to some $m\equiv 3\bmod{4},$ so we get a contradiction.

If $n\equiv 3\bmod{4},$ we can, indeed, write it in infinitely many ways!

Consider the triple $(x,y,z)=(p+2f(n),p+2g(n),-p)$ for some prime $p>2.$ Then, let $(a,b,c)=(xy,yz,zx).$ Clearly, $abc$ is indeed the square of an odd integer.

It boils down to finding some $f$ and $g$ such that $a,b$ and $c$ are good and the following identity holds\[(p+2f(n))(p+2g(n))-p(p+2f(n))-p(p+2g(n))=n.\]Notice that the latter gives us $4f(n)g(n)=n+p^2.$

By taking $p$ to be greater than $n, \ p\nmid n+p^2$ so $p\nmid f(n)$ or $g(n)$ so $-p(p+2g(n))$ and $-p(p+2f(n))$ are good, since the exponents of $p$ in their prime decompositions is $1.$

It remains to find $f$ and $g$ such that: \[f(n)\cdot g(n)=\frac{n+p^2}{4}\text{ and }(p+2f(n))(p+2g(n))\neq \pm k^2.\]Without complicating ourselves too much, we can choose $g(x)= 1$ and $f(x)=(x+p^2)/4.$ Simply choose some prime $p$ such that $p+2\equiv q\bmod{q^2}$ for some other prime $q>x.$ It then follows that $\nu_q((p+2f(n))(p+2g(n)))=1$ so that number is good as well, finishing the proof.

If $m=a+b+c$ where $a$, $b$, $c$ multiply to a number $\equiv 1\pmod 4$ then $(a,b,c)\pmod 4$ is $(1,1,1)$ or $(1,3,3)$ and therefore $m\equiv 3\pmod 4$. Let $m=4k-1$ then consider the following construction:

Let $4k-1=xy+yz+zx$, where $x$, $y$, $z$ are odd numbers. It just remains to satisfy the equality and make sure that $xy$, $yz$, $zx$ are not perfect squares or the negations of them.

Let $x=2p+1$, $y=-2p+1$, and $z=2p^2+2k-1$. Then \[xy+yz+zx=2z+xy=4k-1\]
Note that $xy=1-4p^2$ is always good except at $0$. We need to show that infinitely many times, $(2p^2+2k-1)(2p+1)$ and $(2p^2+2k-1)(2p-1)$ are good. Choose some large prime $q$ and there are infinitely many $p$ such that $\nu_q(2p-1)=1$. In this case, $q\mid 4p^2+4k-2$ so $q\mid 4k-1$, contradicting largeness. We can impose a similar restriction to $2p+1$ by CLT and we are done.


Let $ABC$ be an equilateral triangle and $D$, $E$ points on the sides $[AB]$ and $[AC]$ respectively. If $DF$, $EF$ (with $F\in AE$, $G\in AD$) are the interior angle bisectors of the angles of the triangle $ADE$, prove that the sum of the areas of the triangles $DEF$ and $DEG$ is at most equal with the area of the triangle $ABC$. When does the equality hold?

[i]Greece[/i]

We have \[\frac{[DEF]}{[DEA]} = \frac{EF}{EA} = \frac{DE}{DE+AD}\] Similarly \[\frac{[DEG]}{[DEA]} = \frac{DG}{DA} = \frac{DE}{DE+AE}\] Thus $[DEF] + [DEG] = DE[DEA](1/(DE+AD) + 1/(DE+AE))$. We have $[DEA] = 1/2 AD\cdot AE\sin A$, and $[ABC] = 1/2 AB\cdot AC \sin A$, so \[\frac{[DEF] + [DEG]}{[ABC]} = \frac{DE\cdot AD\cdot AE}{AB^2}\cdot\frac{1}{(DE+AD) + 1/(DE+AE)}\]
Obviously $DE \le AB$, so it is sufficient to prove that $AD\cdot AE(1/(DE+AD) + 1/(DE+AE)) \le DE (*)$.
The cosine rule applied to $ADE$ gives $DE^2 = AD^2 + AE^2 - AD\cdot AE$. Hence also $DE^2 - AD\cdot AE = (AD - AE)^2\ge 0$. Thus we have $DE(AD - AE)^2 + DE^2(AD + AE) \ge AD\cdot AE(AD + AE)$.

So \[AD\cdot AE(AD + AE + 2DE) \le DE(AD^2 + AE^2 + DE(AD + AE)) = DE(DE + AD)(DE + AE)\], which is $(*)$.

[quote=jakarta962]can we use trigonometry in JBMO?[/quote]

of course

Bonus question:

In the above problem, prove that $DF/EG = AD/AE$.


In a modified game of football you can score 21 or 42 or 43 points.

21-field
42-touchdown
43-touchdown with extra point

what is the largest number not achievable?

[hide]Because 42 is a multiple of 21, we can just use the McNugget formula with 21 and 43.

$43 \cdot 21 - 43 - 21 = 839$[/hide]

Just if case people dont know what the McNugget Therem is...

the formula is $mn-m-n$

EXAMPLE:

If Mc Donalds sells their chicken nuggets in boxes of 5 and 7,

the largest nuber of chicken nuggets not able to make is 5*7-5-7=23


Let (an)n>=0 be a sequence of natural numbers so that a_(n+1)>a_n and:
x_n=1/a0!+1/a1!+..+1/an!, for any n natural!
Prove that the limit of xn is an irational number!!!

bye bye!

Hehehe  :D You went to that contest in Vaslui, didn't you?

This wasn't proposed for my grade, but I can remember there were some helping points in the problem:

a) Prove that x_n is convergent (easy to do: x_n increasing and x_n <= 1+1/1!+..+1/n! -> e)

b) Let l be the limit of x_n. Prove that l-x_n<1/an*an!

This is equivalent to l-xn=Sn=1/a(n+1)!+1/a(n+2)!+... < 1/an*an!. It's obvious that Sn <= 1/(an+1)!+1/(an+2)!+..., so it would be enough to prove this:

For any natural n we have 1/(n+1)!+1/(n+2)!+.. < 1/n*n!. We multiply this with n! to get An = 1/(n+1)+1/(n+1)(n+2)+.. < 1/n. We can see that An < 1/(n+1)+1/(n+1)^2+... = 1/(1-1/(n+1)) - 1 = (n+1)/n - 1 = 1/n, so we've proven b).

c) Prove that l is irrational

I don't know if I can solve it completely, but here's the main thought:
We assume l=p/q.
We have 1/a0!+..+1/an! < p/q < 1/a0!+..+1/an!+1/an*an! for all n. We can bring it all to the same denominator an*an! and we increase an until q|an!. We then multiply it all with q and we get p between 2 numbers which aren't integers (I think). We use the same idea as that for proving that e is irrational. Try this for an=n and see what you get. I can't remember it all right now, but this is the idea. Mayeb some of the things I've said so far are wrong though :D

btw congrats grobber, I heard you turned out first at that contest. Iandrei was second, right?

Thanks! I think Iandrei was second, although Dinu Razvan was the second one on the list. I guess that's because D is before I in the alphabet  :D (they had equal scores from the first 3 problems, but there was this fourth problem for making the difference between contestants with equal scores and I think Iandrei got more on that one).

First of all congratulations grobber and iandrei!  and to all the contestants!

2nd this problem is not very hard, especially if you had solved the problem given some years ago, 2 or 3 years ago,( i don't remember exactly) at the "Bacalaureat" in Romania;
this is the generalization of that problem!
(there we had that a0=1; a1=1; a2=2; ...;an=n) etc.

cheers! :D :D


Um, isn't this almost the equivalent of "My Fourth Practice Senior Meet #2"?

So [hide]11^10=...01

01-9:equiv:92 (mod 100)

92[/hide]

lols
yea
but
this is from the junior meets
so i guess they duplicate


Similar to the other one, except you have to calculate:

$ 1 \plus{} \frac{1}{4} \plus{} \frac{1}{9} \plus{} \frac{1}{16} \plus{} \frac{1}{25} \plus{} \frac{1}{36} \plus{} \frac{1}{49} \plus{} ...$

$ \frac{\pi^2}{6}$, although you should watch what forum you're posting in, Yongyi

[quote="alanchou"]$ \frac {\pi^2}{6}$, although you should watch what forum you're posting in, Yongyi[/quote]

Although I forgot to mention that calculators were allowed... hence the "Let's wee if you're thinking" part :D

[quote="micromath"]sorry to be the idiot among the crowd, but how did u guys just do that, i realize that this may be high school level but ive seen problems that resemble these in some amc tests, so without further ado a solution is requested.[/quote]

the reason I didn't post a solution is because it uses material beyond mathcounts, and thus doesn't go in this forum unless moved

Historically, this has been called the [url=http://en.wikipedia.org/wiki/Basel_problem]Basel problem[/url]. The link shows Euler's nonrigorous solution as well as a more rigorous one, and also gives some history.


[quote="tuandokim"]Prove that:if x+y+z=1 then
$ \sum xy \plus{} 12xyz\sum(xy) \geq xyz \plus{} 4(\sum xy)^2$[/quote]
Try $ x\equal{}2,$ $ y\equal{}\minus{}1$ and $ z\equal{}0.$

[quote="arqady"][quote="tuandokim"]Prove that:if x+y+z=1 then
$ \sum xy \plus{} 12xyz\sum(xy) \geq xyz \plus{} 4(\sum xy)^2$[/quote]
Try $ x \equal{} 2,$ $ y \equal{} \minus{} 1$ and $ z \equal{} 0.$[/quote]
Sorry but x,y,z>0.Try again!

Let $ xy \plus{} yz \plus{} zx \equal{} p$ and $ abc \equal{} t$.
We have to prove

$ 4p^2 \minus{} (12t \plus{} 1)p \plus{} t\leq 0$
$ \Leftrightarrow \frac {(12t \plus{} 1) \minus{} \sqrt {144t^2 \plus{} 8t \plus{} 1}}{8}\leq p \leq \frac {(12t \plus{} 1) \plus{} \sqrt {144t^2 \plus{} 8t \plus{} 1}}{8}$

From the inequalities
$ \sqrt {3xyz}\leq xy \plus{} yz \plus{} zx \leq \frac {1 \plus{} 9xyz}{4}$
$ \Leftrightarrow (xy\plus{}yz\plus{}zx)^2 \equal{} (xy)^2\plus{}(yz)^2\plus{}(zx)^2\plus{}2xyz(x\plus{}y\plus{}z)\geq 3xyz(x\plus{}y\plus{}z)$
and $ xy\plus{}yz\plus{}xz\leq\frac{(x\plus{}y\plus{}z)^3\plus{}9xyz}{4(x\plus{}y\plus{}z)}$
we know $ \sqrt {3t}\leq p\leq \frac {1 \plus{} 9t}{4}$ and from AM-GM we have $ t\leq \frac {1}{27}$ too.
That may conclude the solution but I haven't done it yet because I've got another work :roll: .

[quote="tuandokim"]Prove that:if x+y+z=1 and x,y,z>0 then
$ \sum xy \plus{} 12xyz\sum(xy) \geq xyz \plus{} 4(\sum xy)^2$[/quote]
It's equivalent to $ \sum_{cyc}(x\minus{}y)^2xy(x\plus{}y\minus{}z)\geq0,$ which is obvious.


probability
geometry
3D geometry
function
analytic geometry
graphing lines
slope

Can anyone help me with these problems please?

1. Three red balls, three green balls, and three blue balls are placed in a row in random order. Find the probability that no two blue balls are adjacent.

2. Two A's and two B's were randomly put inside a 3x3 grid. What is the probability that the line connecting the two B's intersect? (Note: A line is extended in both directions.)

3. A student is taking a 20=item quiz having 4 choices. The probability that he knows the answer given that he is correct is 6/7. How many questions was he prepared?

4. 27 unit cubes were painted orange such that exactly 4 of its faces are painted and the unpainted faces share a common side. A 3x3x3 cube is formed using the 27 unit cubes. What is the probability that each face of the resulting 3x3x3 cube is entirely orange?

5. An ant is situated at the vertex of triangle ABC. EVery m0vement of the ant consists of moving to one of the other two adjacent vertices from the previous vertex where it is situated. The probability of going to any of the other two adjacent vertices of the triangle is equal. If the ant will move 7 times, find the probability that he will back at vertex A at the end of the journey.

6. Among the 10 people seated at a round table, there is a group of 3 friends. In how many ways is it possible to seat 10 people at a round table if no two members of the group of 3 friends are seated next to each other?

7. Urn 1 has 5 red balls and 1 blue ball. Urn 2 has 2 red balls and 4 blue balls. A fair die is tossed. If it results in an even number, balls are repeatedly drawn one at a time with replacement from Urn 1. If it is an odd number, balls are repeatedly drawn one at a time with replacement from Urn 2. If the first two draws resulted in a blue ball being drawn, what is the probability of getting a red ball in the third draw?

8. A number is called "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?

9. A drawer contains several pairs of socks. Not wanting to count, Melissa asked Krissa, "How many socks are there in the drawer?" Krissa, not wanting to give away the answer. replied, "Well, each sock has exactly one matching sock, and the probability that two socks simultaneously drawn form a matching pair is 1/15." What is the answer to Melissa's question?

10. How many 4-digit positive integers have the product of their digits divisible by 3?

Thank you!

I can try.

[hide="1"]Place the red and green balls all together.  There are $ \frac {6!}{3!3!} \equal{} 20$ ways of doing this.  Now, if we want the blue balls to be separated, we can place them in between the other balls present.  There are $ 7$ places where the blue ball will not border another blue ball, and we need to choose three of them.  $ \frac {7!}{3!4!} \equal{} 35$ ways to do this last step.  The final answer becomes $ 20*35 \equal{} 700$.[/hide]

[hide="10"]

The product will be divisible by 3 if atleast one digit is 0, 3, 6 or 9
hence we find all numbers without these 4 digits
which are 6x6x6x6=1296

Total numbers is 9x10x10x10=9000
Hence required number is 9000-1296=7704[/hide]

[hide="9"]Let $ a$ be the number of socks in the drawer.  There are $ \frac {a}{2}$ total pairs in the drawer.  The number of ways to choose two socks simultaneously is $ \binom {a}{2}$.  Probability is "win" conditions divided by total.  Thus, $ \frac {1}{15} \equal{} \frac {\frac {a}{2}}{\frac {a(a \minus{} 1)}{2}}$.  Cancellation (since $ a$ shouldn't be zero) yields $ a \equal{} 16$.[/hide]

[hide="I think this is 8"]
Use Euler's totient function $ \phi{n}$ to find the number of relatively prime integers less than $ n$.  It's important to realize that if $ n$ is prime, $ \phi{n}\equal{}n\minus{}1$.  Furthermore, $ \phi{abc}\equal{}\phi{a}*\phi{b}*\phi{c}$ if those three are relatively prime.  Thus $ \phi{30}\equal{}\phi{2}*\phi{3}*\phi{5}\equal{}8$.  Since $ 990\equal{}30*33$ and Euler's totient function cycles for the given interval ($ \phi{60}\equal{}2*\phi{30}$), we have $ 33*8\equal{}264$ numbers.  In the interval $ 990\rightarrow 1000$, we can verify quickly that $ 991, 997$ are the only two numbers we aren't sure are prime.  This gives $ 266$ total numbers that could be prime-looking.  $ 168$ are prime, so $ 98$ are prime-looking.[/hide]

Thank you to those who responded.

1. 5/12

8. 100

9. 16

10. 7704

The other numbers please  :idea:

I get 98 for 8 also. Where did we go wrong? I used inclusion exclusion. Let div n be the amount of numbers less than 1000 divisible by n.
Prime looking=Composite-div2-div 3-div 5+div 6+div15+div10-div30=(999-168)-499-333-199+166+66+99-32=98.

[hide="8"] First, we count all the numbers divisible by 2, 3, or 5. Using PIE, that is 499+333+199-99-166-66+33=733. There are then 999-733=266 numbers not divisible by 2, 3, or 5. Subtracting the other 165 prime numbers besides 2, 3, or 5 gives 101. Then, subtracting 1, which is neither prime nor composite, we have 100.[/hide]

EDIT: Thanks, math154.

Shouldn't it be $ 101$ then? You used PIE counting $ 1000$, but then you subtracted that from $ 999$, which would not include $ 1000$.

Edit: Never mind--forgot about $ 1$--it's neither prime nor composite, so we have to subtract that too.

The answer is $ 999 \minus{} 733 \minus{} 165 \minus{} 1 \equal{} \boxed{100}$.

[hide="2)"]subtract the complement+casework
There are $ \binom{9}{2}\binom{7}{2}\equal{}756$ ways to put the A's and B's
for the lines to be parallel there are 7 cases
horizontal vertical diagonal(slope of 1) diagonal(slope of -1)slope of 2 slope of -2 slope of 1/2 slope of -1/2
for horizntal there are $ \binom{3}{2}^3\equal{}27$ arrangements. Same for vertical. For diagonal there are $ \binom{3}{2}\plus{}\binom{3}{2}\plus{}1\equal{}7$ arrangements. For each of slope 2, slope -2, slope 1/2, and slope -1/2 there are 2 arrangements. So there are $ 27\plus{}27\plus{}7\plus{}2\plus{}2\plus{}2\plus{}2\equal{}69$ ways to make parallel lines and the probability of intersecting lines is $ \boxed{\dfrac{229}{252}}$[/hide]
[hide="3)"]Say the student gets x questions correct. Then he got x/7 questions correct out of luck. Since there are 20 questions on the test, he got 20-x questions wrong. Because there are 4 choices, the chance of getting a question wrong is 3 times more than getting a question right from luck. So we have
20-x=3x/7
20=10x/7
x=14
and only 6/7 of these 14 he got correct were solved by luck, so the student was prepared for $ \boxed{12}$ questions.[/hide][/hide]

Anyone else who can help me out with the other problems?

6. In  order to not have to do a lot of casework, I did this using complementary counting.  Therefore, you have two cases:
[hide="1. All three friends are together"]
You can arrange these friends in 3! ways, and the other 7 people in 7! ways.  Thus, there are 3!*7!*10/10=3!*7! ways to arrange them such that all three friends are together.[/hide]
[hide="2. Two friends are together"]
There are $ {_3}C_2$=3 ways to choose 2 friends.  You can arrange these people in 2 ways.  In the remaining 8 spots, there are 8! ways to arrange the remaining people.  However, we have to subtract those combinations which involve the remaining friend next to the other two.  tHIS IS JUST 3!*7!, which we got at first. So, we get that the number of ways is:
6*8!-6*7!=7!(42)=7!(42)=6*7![/hide]
The total number of possible combinations is 10!/10=9!, so the number of combinations where all three friends are separated are:
9!-3!*7!-42*7!=
7!(72-6-42)=24*7!

EDIT:arghhhhhhhhhhhhhhh in my second case i made a mistake; it should be 36*7!making the answer[b] 30*7![/b]

Ok, I get a different answer for #6. What did I do wrong?

[hide="6"] Place the first friend in a seat to account for rotations. Now we have two cases:

[u]Case 1: The second friend and the first friend have only one seat in between[/u]
There are two ways to decide where the second friend sits now. Then, there are 5 spots for the third friend to sit. The other 7 people can then sit in 7! ways. The total here is $ 2\cdot5\cdot7!$.

[u]Case 2: The second friend and the first have more than one seat in between.[/u]
There are 5 ways to place the second friend, and then 4 ways to place the other friend. Again, 7! ways to sit the other 7 people. The total here is $ 5\cdot4\cdot7!$ here.

In all, there are $ 2\cdot5\cdot7! \plus{} 5\cdot4\cdot7! \equal{} 6\cdot5\cdot7! \equal{} 30\cdot7!$ ways.[/hide]

EDIT: Oh apparently someone told me that I didn't distinguish between the second and third friends. Darn.

I, too, have a different answer. =O

[hide="6"]
Let's call the three friends A, B, and C, and let's number the seats 1-10. Without loss of generality, since this is a circular table, let A be seated at seat 1. Since the people next to A cannot be B or C, there are $ 7 \choose 2$ ways to choose people to sit next to A (the 7 people who aren't A, B, or C.) We now have 7 remaining people (we may think of them as being arranged on a line.) Here, we use complementary counting. The number of ways to seat the two together is 2*6! = 1440. The total number of ways to seat the people is 7! = 5040, so the total number of ways is 3600. Thus, the answer is $ {7 \choose 2} 3600 \equal{} 21 * 3600 \equal{} 93,600$.
[/hide]

Hm... input would be appreciated.

[quote="Attila the Hun"]5)

[hide]you can represent the sequence of steps by
(ABCABCA), etc.

we are given that the first and last positions are A. so, we have to find the no. of ways to fill in the remaining positions. each of the five remaining positions can be filled in 2 ways. so, the number of such permutations is 2^5.
total number of permutations is 2^7. so the probability that the ant will end up at A ( given all permutations have equal probability ) is

2^5 / 2^7 = 1/4[/hide][/quote]

I believe the ant starts on A and then makes 7 steps. Thus you have A-1-2-3-4-5-6-7 and you want to find the probability that 7 is A. It's correct that the total number of possible paths is 2^7. However you can't just say that 1, 2, 3, 4, 5, and 6 each have 2 possibilities; e.g. if 5 was B, then 6 can only be C.

[hide="my solution"]
Let $ P_n$ denote the probability that the ant ends up at A after n steps.

After the (n-1)th step, the ant can only be at B or C, and for both of these, it has a 1/2 chance of moving to A.

Thus $ P_n \equal{} 1/2 \cdot (1\minus{}P_{n\minus{}1})$

For $ n\equal{}1$ the probability is 0 so we can recursively compute $ P_7$ given that $ P_1 \equal{} 0$.

We end up with $ P_7 \equal{} 21/64$
 [/hide]

In number 6 its unclear to me whether rotationally identitical seatings should be counted as ¨different seatings¨. But if so,

[hide]There are 10 places to seat person A. There are 7 remaining seats in which to seat persons B and C. There are $ (7)(6)\equal{}42$ total ways to seat two distinguishible people in those 7 seats, but $ (2)(6) \equal{} 12$ of them are adjacent. So for each seating of person A, there are $ 30$ ways to seat persons B and C so that none of the three are adjacent. Then for the remaining 7 persons there are 7! ways to seat them in the remaining seats.  Thus there are $ (10)(30)(7!) \equal{} 1,512,000$ ways to seat all of them so that A, B, and C are not adjacent.[/hide]

Zhero, there aren't C(7,2) ways to put the first 2 people next to person A.  There are 7*6 ways. Order matters because you're only taking a part of the circle, not the whole circle.

6. 151,200

Anyone for 1, 2, 4, and 7? Thanks!

For #1, azn already found there are 700 possibilities, but this is out of a total $ \frac {9!}{3!3!3!}$, thus the answer is $ \frac {5}{12}$.
[hide="4"]There are 6 center cubes, each with a $ \frac {4}{6}$ chance of being orange. For the 12 edge cubes, 5 out of the 12 edges on the cube can be used. Also, for the eight corner cubes, only two out of the eight corners on the cube may be used. The answer is (I think)
$ (\frac {4}{6})^6 \cdot (\frac {5}{12})^{12} \cdot (\frac {2}{8})^8 \equal{} \frac {5^{12}}{2^{34} \cdot 3^{18}}$[/hide]
[hide="7"]If two blue balls were drawn already, then they must have been drawn from Urn 2, which means there are now two of each color in Urn 2 and thus a $ \frac {1}{2}$ chance of drawing red. Is this what the question was asking?[/hide]
[hide="2"]The lines between A and A and B and B must be parallel if they don't intersect, so we find this and subtract it from 1. Assume first they are both vertical. There are 9 possible vertical pairs for A and A, and 6 for B and B after they are inserted, resulting in 54 possibilities. There are another 54 possibilities if the pairs are horizontal. If the pairs are diagonal with a slope of 1, then  there are two possibilities if the pairs cover a diagonal of length 2, and if one pair covers a digonal of length 3 and the other of length 2, then there are $ 2 \cdot {3 \choose 2} \cdot 2 \equal{} 12$ possibilities, so there are a total of 14 possibilities when the slopes are either 1 or -1. Now, if the slopes are 2, then it is easy to see there are only 2 possibilities, and this is the same as when the slopes are 1/2, -2, and -1/2. Thus, I think the answer is
$ 1 \minus{} \frac {54 \plus{} 54 \plus{} 14 \plus{} 14 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 2}{\frac {9!}{2!2!5!}} \equal{} \frac {17}{21}$[/hide]

dgreen, I think you missed the "with replacement" part.

We use conditional probability, $ P(A|B)\equal{}\frac{P(A \cup B)}{P(B)}$.  $ A$ is the probability of being from a given urn, and $ B$ is the probability of choosing two blue balls.

Notice that $ B$ is the sum of two probabilities.  If Urn 1 is chosen, there is a $ \frac{1}{2}*(\frac{1}{6})^2$ probability two blue balls are chosen.  If Urn 2 is chosen, there is a $ \frac{1}{2}*(\frac{1}{3})^2$ probability two blue balls are chosen.  The total probability is $ \frac{1}{72}\plus{}\frac{1}{18}\equal{}\frac{5}{72}$.

Now, for Urn 1, $ P(A \cup B)\equal{} \frac{1}{72}$.  This makes $ P(A|B)\equal{}\frac{\frac{1}{72}}{\frac{5}{72}}\equal{}\frac{1}{5}$.

For Urn 2, $ P(A \cup B)\equal{}\frac{1}{18}$.  This makes $ P(A|B)\equal{} \frac{\frac{1}{18}}{\frac{5}{72}}\equal{}\frac{4}{5}$.

If Urn 1 was chosen, there is a $ \frac{5}{6}$ chance the next ball will be red.  If Urn 2 was chosen, there is a $ \frac{2}{3}$ chance the next ball will be red.  Summing probabilities yields $ \frac{5}{6}*\frac{1}{5}\plus{}\frac{2}{3}*\frac{4}{5}\equal{}\frac{7}{10}$.

O whoops, I really wanted to use conditional probability, I even looked it up on wikipedia, but yes, I did miss the with replacement part and thought it couldn't be used  :( .

Do you mean, for number two, that the line connecting the two A's and the one connecting the two B's intersect?  The answer is $ \frac{1}{4}$, will explain at lunchtime.

[hide="#7"]If [b]a blue[/b] ball was drawn, there's 2 possibilities: either it was drawn from Urn $ 1$ or Urn $ 2$.  We thus have 2 cases:
[hide="Blue Ball Drawn from Urn 1"]
We have two possibilities, either the blue ball is drawn first or it is drawn second.  Both have the same probability:
$ \frac{1}{6} \bullet \frac{5}{6} \bullet \frac{5}{6}\equal{}\frac{25}{216}$.  The probability of getting an even number on the die is $ \frac{1}{2}$, so the probability for this case is given by: $ \frac{1}{2} \bullet 2 \bullet \frac{25}{216}\equal{}\boxed {\frac{25}{216}}$.[/hide]
[hide="Blue Ball Drawn from Urn 2."]
This case is slightly harder.  If a blue ball was drawn, either:
[hide="Case 1: 2 Blue Balls were drawn"]
The probability that $ 2$ blue balls were drawn is: $ \frac{2}{3} \bullet \frac{2}{3}$.  The probability that, then, a red ball is drawn is $ \frac{1}{3}$.The probability of this case is thus $ \frac{4}{27}$.[/hide]
[hide="Case 2: 1 Blue Ball is drawn"]
This blue ball can be drawn in two ways, first, or second.  If it is drawn first, a red ball must follow, and then another red ball, for a total of $ \frac{2}{27}$.  If the blue ball is drawn SECOND, then a red ball must be drawn first and third, for a probability of $ \frac{2}{27}$.  Hence, the total probability for this case is $ \frac{4}{27}$[/hide]
The sum of these two cases is $ \frac{8}{27}$.  The probability of getting an odd number on a roll of a dice is $ \frac{1}{2}$.  The probability for this case is thus $ \boxed {\frac{4}{27}}$[/hide]

The total probability for these two big cases is thus $ \boxed {\frac{25}{216}\plus{}\frac{4}{27}}$.
There's a possibility I made a mistake because I'm doing all of this in my head..[/hide]

Would some kind person please tell me what's wrong with my solution to #6?

[hide="Solution"]Let's take the 7 people who aren't 'friends' and organize them around the circle. There are $ 7!$ ways to do this. Then, we can partition the 'friends' between the seated people. Since there are $ 7$ 'partitions', we choose three to make$ 7\choose3$ and multiply by $ 3!$ to distinguish each 'friend'.
Then, our answer comes out to $ 7!*3!*{7\choose3}\equal{}1058400$[/hide]

[hide="#6"]First let's count "bad" ways (the ones where at least 2 of friends seat next to each other). We will count possible seats first, then distinguish friends later.
Let's place 2 of the seats where friends will sit together. There are $ 10$ ways to do this. Now let's place the third seat so it's not next to seats we already placed. We have $ 6$ possibilities to do so. If we distinguish friends and count other people we get total of $ 10*6*3!*7!$ ways.
Now lets put all the 3 seats together. There are $ 10$ ways to place them and if we distinguish friends and count other people, we get total of $ 10*3!*7!$ ways which makes total of $ 420*7!$ "bad" ways to place friends.
There are $ 10!$ ways to put people on their seats of which there are $ 10!\minus{}420*7!\equal{}300*7!\equal{}\boxed{1512000}$ "good" ways (where no friend sits next to another).[/hide]

[hide="@ aguoman"]I think your solution is good if we count "rotations" as same solution. All of us distinguished rotations as unique solution and that's why we get bigger number of possible ways.
Somebody correct me if I'm wrong[/hide]


calculus
derivative
trigonometry
real analysis
real analysis unsolved

[quote="Virgil Nicula"][color=darkred]Prove that ([u]with or without derivatives[/u]) for any $ x\ne 0$ exists the relation

$ \boxed {\ \arctan x \plus{} \arctan \left(x^3\right) \plus{} \arctan\frac {1 \minus{} x^2}{x} \equal{} \frac {\pi}{2}\cdot\frac {|x|}{x}\ }\ .$[/color][/quote]

use
$ \tan^{\minus{}1}\,A\plus{}\tan^{\minus{}1}\,B\plus{}\tan^{\minus{}1}\,C\equal{}\tan^{\minus{}1}\left(\dfrac{A\plus{}B\plus{}C\minus{}ABC}{1\minus{}AB\minus{}BC\minus{}CA}\right)$

so, for $ x>0$, we will get $ \frac {\pi}{2}$, and for $ x<0$, we will get $ \minus{}\frac {\pi}{2}$


Does anyone know where I can get a book with the AIME (questions + solutions) from 1988+?  (or if it even exists).

[quote="t0rajir0u"]Save yourself money and use Kalva.  :)

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

(Unless you specifically want a book for classroom purposes or something...)[/quote]

Hey! Don't direct them to kalva :o , use the contest resources right here! :D

EDIT: Then again, some AIMEs are not posted

Is kalva legal?
Probably as legal as Linewire.

I don't think there's any copyright on the problems, so Kalva is legal. But whether or not it should be legal is a different question...

Anyway, the Contests link near the top of your screen will take you to a section of the site that includes many AIME questions from the past (and we are working on adding more...). I urge you to consider ordering a paper copy from the AMC office though. The exams are not terribly expensive and working the problems under contest conditions from the original document is different from doing them off of a AoPS pdf. At least, I think so. *shrugs*

[quote="MithsApprentice"]I urge you to consider ordering a paper copy from the AMC office though. The exams are not terribly expensive and working the problems under contest conditions from the original document is different from doing them off of a AoPS pdf. At least, I think so. *shrugs*[/quote]

Lol yeah I tend to get one more problem right (one of the last six usually) when I do it with a contest paper than the PDF... idk why....

I meant like a book with AIME problems like how Contest Problem Book V has AIMEs 1983-1988 as well as the AHSMEs.  Is there a reason why they stopped making AIME books?  Contest Problem book VI only has AHSMES.


[hide]Wouldn't it be 320 5/8 lbs?  If the iron wrecking ball has a diameter of 8", the volume  of the wrecking ball would be (4/3)  :pi:  r :^3: , which would be 256 :pi: /3. The volume of the second wrecking ball would be (4/3) :pi: 6 :^3: = 228 :pi: . The amount of times the volume of the second wrecking ball is multiplied would be 228 :pi:/(256 :pi: /3)= 171/64. To find the new weight of the wrecking ball, take 171/64 * 120 = 2565/8, which equals 320 5/8 lbs.  :)  :)  :)  [/hide]

[hide]Form the ratio $\frac {2^3}{3^3} = \frac {120}{x}$  So it must be $405$.[/hide]

[hide] 180    8=120 and 12=x solve the proportions u get 180[/hide]

[quote="MCrawford"]A wrecker's iron ball eight inches in diameter weighs 120 pounds.  How many pounds would a similar iron wrecking ball twelve inches in diameter weigh?[/quote]

[hide="Solution"]
Each dimension is increased by 3/2.

120(3/2)(3/2)(3/2)
120(27/8)
15(27)
300+105
[b]405 pounds[/b]

*looks at biffanddoc strangely* :P
[/hide]

biffanddoc, and mathprincess09
check your answers again ;)

and

elementsofmathkid, your method was correct, but the calculation was wrong

228 should be 288

which the new answer would be

[hide="elementsofmathkid only"]405[/hide]

Ok. Thanks. (Stupid mistakes  :blush: )

 :)  :)  :)

[quote="mathprincess09"]i dont get wat u mean
arent you supposed to use proportions? :blush:[/quote]

You do use proportions on lengths, but let's see how proportions of lengths relate to proportions of volume.

If $r$ is the radius of a sphere (a length measurement), the volume of the sphere is
\[ \frac{4}{3}r^3 \pi. \]
If we multiply that radius by a proportion $x$, the new radius is $xr$ and the volume is
\[ \frac{4}{3}(xr)^3 \pi = \frac{4}{3} x^3 r^3 \pi = x^3\left( \frac{4}{3} r^3 \pi\right). \]
and the ratio between the volumes is
\[ \frac{x^3\left( \frac{4}{3} r^3 \pi\right)}{\frac{4}{3}r^3 \pi} = x^3. \]

Because a radius increase is not simply in one dimension of the sphere, but three, we see that the ratios of the volume is $x^3$ -- the cube of the proportion between the radii of the spheres.

[quote="mathprincess09"]I dont get what you mean.
Aren't you supposed to use proportions? :blush:[/quote]
There is one thing that nobody has explicitly mentioned.

The mass of an object of uniform density is proportional to the volume of the object.

Once you think about it, this is probably "obvious" to most people.  But if you haven't thought about it before, in terms of a math problem, it might not be obvious.

I'll bet this concept (mass ~ volume) trips up a lot of students.

[quote]You do use proportions on lengths, but let's see how proportions of lengths relate to proportions of volume.

If  is the radius of a sphere (a length measurement), the volume of the sphere is

If we multiply that radius by a proportion , the new radius is  and the volume is

and the ratio between the volumes is

Because a radius increase is not simply in one dimension of the sphere, but three, we see that the ratios of the volume is  -- the cube of the proportion between the radii of the spheres.[/quote] i thought u would use proportions too, thanks for explaining it!

I know this is a state question but what is the year?


Out of all 6 harry potter books, which do you think was the best and which was the worst?

Personally, I liked the third one the best.

I think that the worst one was the forth one.(no quiditch with harry)[sob] [sob] :(

You should make a poll.

I hated the fourth one, and the first one was pretty good.

I liked the third one the best too.  But I think that the 5th one was the worst.  What's weird about that one though, is that it seems the least darkest to me, even though the series is supossed to gradually get darker and darker.

[quote="236factorial"][quote="shobber"]Definitely.
Because of Lupin. He is nearly a god![/quote]

Who are you talking to?[/quote]
I mean: I love book 3 because Lupin was a main charactor in that book.

I forgot what happened in all the books. I guess that's what you get when you don't read Harry Potter for 3 years!

the 4th was the best by far in my opinion.  i loved the whole mad-eye moody thing and the tournament between schools.

i've read all of them and they all seem fairly equal. in terms of the quality of the books, i'm not exactly thrilled, but, they're not bad either.

The sixth one was the best.

The fourth one was the worst.

[quote="solafidefarms"]Out of all the Harry Potter books, I think they are all bad. :mad: They promote witchcraft.[/quote]

haha. We have an anti-witchcraft person here.  :ninja:

Anyway, I think the 4th was the best and the second was the worst. The second was boring me to death.

[quote="math92"]i like the 4th one the best becuase of the tournament and i hate the 2nd the most becuase it was very boring in my opinion. :)[/quote]

I agree.  The Tournament did make the 4th one interesting.  What part of the Tournament did everyone like the most?

The 2nd one was definitely the worst because it was so boring. :sleeping:

[quote="solafidefarms"]Out of all the Harry Potter books, I think they are all bad. :mad: They promote witchcraft.[/quote]

 lol.....I cant believe you said that..........one of the saddest things ive heard today

The 6th is obviously the worst. Argh I was so dissappointed with Rowlings. She spent, what, 650 pages?, talking about absolutely nothing! There were a few interesting subjects...Dumbledore dying (although there's speculation that he's actually not "dead"...odd stuff) and Snape's betrayal for instance, not to mention the introduction of Horcruxes...but overall, it was a terrible book. Almost no plot.

I'd have to say the 3rd and 4th are the best, the 4th being slightly better. The introductions of Lupin and Mad-Eye were simply ingenius on Rowling's part. *nods* And the storyline in those were excellent.

None of the 6 were good (other than the first few). They got worse and worse like pokemon.

[quote="236factorial"]None of the 6 were good (other than the first few). [/quote]

That's...uh...contradictory? Although, I know what you mean. ;-)

[quote]They got worse and worse like pokemon.[/quote]

Oh. I still stand by how the 4th was the best. But 5 and 6 were definitely dissappointing. Remember the excitement of finding out how Quirrell was the bad guy in The Sorcerer's Stone back in elementary school? Yea...no more of that.

[quote="236factorial"]Lock this topic. Harry Potter just sucks  :mad:   :rotfl:
I guess I'm not into witchcraft, or whatever that's called.[/quote]

Unfortunately, closing a topic because one individual is not interested in it may be considered abuse of power.  ;)

I, myself, do not in any way like witchcraft. But nonetheless, the books are good for light reading if you appreciate fiction for its...well...fictitiousness.

I was kidding about the locked thing.

the first book took me 1 week, so did the second book, the third one took 10 days, the fourth one took over a month  :(  then I gave up and didn't read the fifth, which not only was longer, but had smaller font (and bigger pages?!).

I don't really read anymore  :|

harry potter is not a good book series.  I dont read it :mad:

sixth was super-dissapointing cuz all the hype that mass media gave it made it so good in theory, that it kinda slacked in real

but otherwise...second was boring and the rest are yummy =D

and reading is a good thing...you should start reading again lol

i think the worst one is 1,2,3,4,5,and 6 all tie for a horrible book series called harry potter :P

i totally totally totally totally totally totallytotally totally totallytotally totally totallytotally totally totally
absolutley agree wiht you ....er i forgot your username someone.
that wasn't sarcasm btw.
this thread is just disgusting. :mad:
how could problemsolving math people even  think about hairy poter being a redable book.
i read all seven and now i think jk rowlings just some idiot who didn't have a job with no creativity at all.
she probably just wanted money.
mega complaints:[hide]book1: what? :huuh:  really lame why couldn't just dumbledorejust do it. book 2 really really lame. why THE HELL COULDN"T uh ya harry just bomb the stupid basilik  instead using his''magical sword? couldn't he just shot tom? why oculdn't dumble dore just come in the chamber and do the job?
oh ya and also harryy pooter is lame since its world is a forest and castle. it becomes way ot serial at the end[/hide]
i will continue my stronger complaints later since i am bored. :D  :weightlift:
like
fine.
i
was
a
little
confused.
not the record though.
once did a 8 year one.
EPIC NECROPOST BATMAN!
ok some mod or amin or manager or cot can lockit.

Seriously? This hasn't been posted on in 6 years.

[quote="Canton"]I think someone should have noticed that the 7th book isn't in the poll...

MODS PLEASE LOCK![/quote]
Uh...when this thread was made and was active, the 7th book did not even come out. :P

Edit: yeah...this should be locked

Yeah, someone should edit the poll cause the 7th book has already came out and i would put that as my fav book in the series i probably don't like the 3rd the most. But someone should definitely change the poll asap cause i want to vote!!!  I cant wait till the movie for Deathly Hallow part 2 comes out !!!!!! XD


It is a cool trick that when a+b<10 ab x 11=a(a+b)b, so for example, 45x11=4(4+5)5=495.

Can you expand on this for multiple digits (like 326x11)?

From about.com

The 11 Times Rule #1

Take any number to 10 and multiply it by 11.
Multiply 11 by 3 to get 33, multiply 11 by 4 to get 44. Each number to 10 is just duplicated.
The 11 Times Rule #2

Use this strategy for two digit numbers only.
Multiply 11 by 18. Jot down 1 and 8 with a space between it. 1   8.
Add the 8 and the 1 and put that number in the middle: 198

using my technique (gotten from my coach), it is [u]1(carry)[/u] [u]9+1(carry)(=0)[/u] [u]9+9(=8)[/u] [u]9[/u]=1089
where the 1's are carries and the (=_) means you only write the unit digit since you carried

$ \overline{a_1a_2a_3\ldots a_n}\cdot1 \equal{} \overline{a_1a_2a_3\ldots a_n}$, before carrying.
$ \overline{a_1a_2a_3\ldots a_n}\cdot11 \equal{} \overline{a_1(a_1 \plus{} a_2)(a_2 \plus{} a_3)\ldots(a_{n \minus{} 1} \plus{} a_n)a_n}$, before carrying.
$ \overline{a_1a_2a_3\ldots a_n}\cdot111 \equal{} \overline{a_1(a_1 \plus{} a_2)(a_2 \plus{} a_3\plus{}a_4)(a_3\plus{}a_4\plus{}a_5)\ldots(a_{n\minus{}3}\plus{}a_{n \minus{} 2} \plus{} a_{n\minus{}1})(a_{n\minus{}1}\plus{}a_n)a_n}$, before carrying.
Notice the pattern. :roll:

[quote="mewto55555"]That's pretty sketchy. 99x11=9189? And also I posted this exact thing in FTWPD [b]when it was enitrly relevant to the problem[/b] and it got deleted. Some one up there hates me.[/quote]
I think it depends on the number. Like if it's below a certain number or something. (I don't agree entirely with that, either)

Im not sure if this has been posted... I can't like understand smart people.

Its just:
_abcde
[u]abcde_[/u]

Where there are spaces and you just add.

actually, all multiplying by 11 is doing that way is

10 * whatever number that is + whatever number that is= 11* whatever blah...

Thats another one.
I'm sure you could just google it and get one bazillion results.

$ \begin{eqnarray*}(11)\cdot x&=&(10+1)x\\&=&10x+x$

[quote="stevenmeow"]only i consecutive digits sum to less than 10
or 326*11 its
[u]3[/u] [u]3+2[/u] [u]2+6[/u] [u]6[/u]=3586
i any pair of consecutive digits add to 10 or more, then you can 'carry' to the next place value[/quote]

lol I learned this trick in 3rd grade. 11 makes my days! :D


for all reals x and y prove that $ x^y \plus{} y^x >1$

you can view my solution by downloading it from http://www.filehosting.org/file/details/69692/RAHUL_DULGGAL.doc

can you please post it here.some people would not like downloading it


Find all natural solutions of $ x(x\plus{}1) \equal{} y(y\plus{}2)$

[hide]No solutions (as the title implies), since LHS has different parities while the RHS has the same parities, thus LHS is even while RHS could be even or odd.

edit: Oops... I answered too quickly!  :blush:

This question is harder than I thought it was... I apologize for posting so rashly

[/hide]

[quote="p4fn2w"][hide]No solutions (as the title implies), since LHS has different parities while the RHS has the same parities, thus LHS is even while RHS could be even or odd.[/hide][/quote]
I'm not sure I follow your logic.  For example, take the equation
$ 2n\equal{}m$ in natural numbers.
The LHS is obviously even and the RHS could be even or odd, but that doesn't mean there are no solutions.

[quote="archimedes1"][quote="p4fn2w"][hide]No solutions (as the title implies), since LHS has different parities while the RHS has the same parities, thus LHS is even while RHS could be even or odd.[/hide][/quote]
I'm not sure I follow your logic.  For example, take the equation
$ 2n \equal{} m$ in natural numbers.
The LHS is obviously even and the RHS could be even or odd, but that doesn't mean there are no solutions.[/quote]

I think RHS must be even

[hide]
The right hand side is one less than $ (y\plus{}1)^{2}$ and is greater than $ y^{2}$, while the right hand side is $ x\plus{}1$ less than $ (x\plus{}1)^{2}$ and is greater than $ x^{2}$.  Therefore, $ x\equal{}0$, so there are no natural solutions.
[/hide]

[quote="K81o7"][hide]
The right hand side is one less than $ (y\plus{}1)^{2}$ and is greater than $ y^{2}$, while the right hand side is $ x\plus{}1$ less than $ (x\plus{}1)^{2}$ and is greater than $ x^{2}$.  Therefore, $ x \equal{} 0$, so there are no natural solutions.
[/hide][/quote]

I'm sorry, I don't understand...

[hide="My solution"]
Obviously, $ x\neq y$.

Now,
\[ x(x\plus{}1)\equal{}y(y\plus{}2)\Leftrightarrow x^{2}\minus{}y^{2}\equal{} 2y\minus{}x\Leftrightarrow x\plus{}y \equal{}\frac{y}{x\minus{}y}\minus{}1\le y\minus{}1\implies x\le\minus{}1\]
Contradiction.
[/hide]

[quote="not_trig"][quote="K81o7"][hide]
The right hand side is one less than $ (y\plus{}1)^{2}$ and is greater than $ y^{2}$, while the right hand side is $ x\plus{}1$ less than $ (x\plus{}1)^{2}$ and is greater than $ x^{2}$.  Therefore, $ x \equal{} 0$, so there are no natural solutions.
[/hide][/quote]

I'm sorry, I don't understand...

[/quote]

$ x(x\plus{}2) \equal{} x^{2}\plus{}2x \equal{} y^{2}\plus{}y$ so $ (x\plus{}1)^{2}\equal{} y^{2}\plus{}y\plus{}1$

But this is impossible because $ (y\plus{}1)^{2}> y^{2}\plus{}y\plus{}1 > y^{2}$, ($ y > 0$).

ps: your solution is nice :)

The object of solving this problem was to determine  that there's no perfect square of the form \[ n\equal{}p^2\plus{}p\plus{}1\]

[quote="jonysatie"]Find all natural solutions of $ x(x \plus{} 1) \equal{} y(y \plus{} 2)$[/quote]
[hide]
Substitute $ z\equal{}y\plus{}1$ so $ x^2\plus{}x\equal{}z^2\minus{}1\iff (x\minus{}z)(x\plus{}z)\equal{}\minus{}(x\plus{}1)$.  Then, the magnitude of the LHS is always greater or equal to that of the RHS with equality at $ x\equal{}z\minus{}1$ or $ x\equal{}\minus{}(z\minus{}1)$, meaning $ x\equal{}y$ or $ x\equal{}\minus{}y$, so $ x\equal{}y\equal{}0$ is the only solution.
[/hide]

[quote="jonysatie"]The object of solving this problem was to determine  that there's no perfect square of the form
\[ n \equal{} p^2 \plus{} p \plus{} 1
\]
[/quote]

If I understand you correctly, $ p^2\plus{}p\plus{}1$ can never be a square anyways because it's strictly between $ p^2$ and $ (p\plus{}1)^2$ for naturals $ p$.

Okay...here is something that I thought of...

Let $ n$ be a positive integer. Prove that $ n^4\plus{}2n^3\plus{}3n^2\plus{}4n$ is never a perfect square.

Looks like it's squeezed between $ (n^2 \plus{} n \plus{} 1)^2$ and $ (n^2 \plus{} n \plus{} 2)^2$.

[quote="Altheman"]Okay...here is something that I thought of...

Let $ n$ be a positive integer. Prove that $ n^4 \plus{} 2n^3 \plus{} 3n^2 \plus{} 4n$ is never a perfect square.[/quote]

$ (n^2\plus{}n\plus{}1)^2<n^4\plus{}2n^3\plus{}3n^2\plus{}4n\iff 1<2n$

$ n^4\plus{}2n^3\plus{}3n^2\plus{}4n<(n^2\plus{}n\plus{}2)^2\iff 0<2n^2\plus{}4$

Hence the expression is squeezed by consecutive squares.

Cool technique! First time I've thought about using that to prove something.
It's as nifty as the Squeeze theorem in calculus.