How many of the National MATHCOUNTS Participants are also going to compete in the National Middle School Science Bowl held in Denver, Colorado, on June 21-24?  Respond and talk about how MSSB and MATHCOUNTS kind of go together.

i don't think our school does science bowl. is it like the science fair?

No, MSSB is totally different.  I didnt know about it till this year, but i did find the site, look below

[url]http://www.scied.science.doe.gov/nmsb/[/url]

Never heard of this...my middle school only does MC (no spelling bee, science fair, and so on...)

that is so weird. my math teacher has one of those posters with einstein on it in her room. i wonder why i haven't heard about this from my school *suspicious look* jk? when does it take place and is it too late to sign up? i can't find any info on that on the website.

sorry, regionals were in february and march.  i had mine on 2/17.  i posted this because i wanted to see if any of the mathcounts people competed in this since theres a math category and math and science r so closely related, but the next part of the competition is the national one, and thats in late june in denver, co

nope, i'm not going. i asked my principal and he said that we didin't do it. :(

[quote="erdogankerem123"]No, I seem to have heard of it, but other than that I don't know anything about it. How does it work?[/quote]
It's like a Countdown round with buzzers and everything, but Science-related questions.

[quote="Fanatic"]We're having state round this saturday.

While practicing for Nationals, we randomly got pulled over by Mr. Fischer and his colleague Dr. Chi, who runs the state science bowl, and they said we should do it.

And we did. With one week's notice, haha.

I've been doing all the sample problems, but we didn't pick our school's best science students [not even close], so i'm  :mad: because we won't be competitive.[/quote]

hey good luck, i hope u make it, so then we'll be facing each other.  make sure u have a solid background in all of the scientific areas, like it sez on the site, and u might want to break it up into areas of expertise.  thats how my school divided up- im good at math and chem, my friend's good at anatomy, my other friend's good at astronomy and earth sci, and my other friend's good at biology and animal and plant sci.  divide it up so u can conquer, oh and dont buzz in early unless u $KNOW$ the answer.

EDIT: our school didnt pick the kids until the monday before the comp, and then we had that tues, wed, thurs, and fri off because of snow days.  oh well, we still won, and all of the kids ranked 2nd,3rd,4th, and 6th in MATHCOUNTS for the school (i was first, like 60 points ahead :lol:) respectively.

I was extremely annoyed by the people that got picked... I know mathcrazed will read this eventually. Our school didn't pick the teams, the math team, while at Terre Haute, "picked" the team.

Doesn't seem like anyone else is preparing. We didn't choose the strongest science students (not even close, as I said earlier), so I'm not feeling so good about our chances.

I'm on a team of four 7th graders. Seems to me like they're thinking "let's feel this out so we can know what's coming next year" and not really looking to compete this year. I'm especially pissed because one team member just randomly picked her best friend as the 5th person and didn't consult with us at all. There are a couple science gems in our school that weren't asked to be on the team.

/rant.

Well there's Science Bowl for our high school, but for me being active in: it's ONLY science olympiad, scholar's bowl, and math team. We do everything, but were not the best at anything.

[quote="Fanatic"]I was extremely annoyed by the people that got picked... I know mathcrazed will read this eventually. Our school didn't pick the teams, the math team, while at Terre Haute, "picked" the team.

Doesn't seem like anyone else is preparing. We didn't choose the strongest science students (not even close, as I said earlier), so I'm not feeling so good about our chances.

I'm on a team of four 7th graders. Seems to me like they're thinking "let's feel this out so we can know what's coming next year" and not really looking to compete this year. I'm especially (mad)  (Edited for content)   because one team member just randomly picked her best friend as the 5th person and didn't consult with us at all. There are a couple science gems in our school that weren't asked to be on the team.

/rant.[/quote]

of course terre haute picked their team.  watch out for them.  they have won the competition like the last two years.  and theyve been dominating indianas mathcounts team too havent they.  up until this year that is, they onle have 1 kid :lol:  is thats an off year, then id hate to see what a good year would be.  im the only that gone to nats from my school since 1983.  and for terre haute, that kid will be the only kid from his school to go to nats since last year.  :lol: just the thought of how sad my school is compared to others is funny :lol:


In this Exercise, solve the exponential equation algebraically. Round your result to three decimal places. Check if your answer if correct

525 / (1 + (e^-x)) = 275

CAN YOU SOLVE IT?

[hide="hint"]Keep in mind that $ln(e^{n})=n$[/hide]


NOTE: The solutions must be nonnegative

1) Find the number of possible solutions to $x+y+z=20$

2) Find the number of possible solutions to $x+y+z+a+b=75$ where $x>=20$

3) Same problem as number $2$ but with also $b<10$.

Natural numbers, and zero only?  Then you can make them into generating sequences...
Then the first one is just the $x^{20}$ term in $(1+x+x^2+...)^3$...which comes out to
$(1+x+x^2+x^3...)(1+2x+3x^2+4x^3+...)$
$=(1+3x+6x^2+10x^3+...)$
which makes the answer equal to 210.  There's a Math Jam on generating functions which occurred which helps in solving these kinds of problems, and there's a transcript for it...there are all sorts of nifty things in there...

Nice work agolsme!

Also note that instead of using cases, you can subtract the number of solutions to $x+y+z+a+b=75$ with the constraints $a\ge20, b\ge10$ (in addition to $x,y,z,a,b$ being nonnegative integers) from your answer on part (b).

[quote="agolsme"][hide]
a) Encode it same as $\bullet\bullet\bullet \ldots | \bullet\bullet\bullet\ldots | \bullet\bullet\bullet\ldots$ there are 20 $\bullet$s and 2 $|$s. So thats: $\displaystyle{22 \choose 2}$
b) this is the same thing as $x + y + z + a + b = 55$ where $x,y,z,a,b$ are all $\ge 0$. So, do the same thing as above: $\displaystyle{59 \choose 4}$
c) well you could case work it, b can be $9,8,7,6,5,4,3,2,1,0$ so $x + y + z + a = 46,47,48,49,50,51,52,53,54,55$
Thus: \[\displaystyle{49 \choose 3} + \displaystyle{50 \choose 3} + \displaystyle{51 \choose 3} + \displaystyle{52 \choose 3} + \displaystyle{53 \choose 3} + \displaystyle{54 \choose 3} + \displaystyle{55 \choose 3} + \displaystyle{56 \choose 3} + \displaystyle{57 \choose 3} + \displaystyle{58 \choose 3}\]

[/hide][/quote]

Thank you so much, this really helps me with Combinatorics


Prove that if $p>3$ is prime then $p^{2}$ divides the numerator of $1+1/2+...+1/(p-1)$

Isn't this Wolstenholme's Theorem? :maybe:

Edit: Yes, see http://www.mathlinks.ro/Forum/viewtopic.php?highlight=wolstenholme&t=302


function
integration
real analysis
real analysis unsolved

Let $ (X, \mathcal{B}, \mu)$ be a finite measure space and $ f$ be a nonnegative measurable function of $ X$. For each $ A \in \mathcal{B}$, set
$ \nu(A)\equal{}\int_Af d\mu$.

Verify that it is a finite measure if and only if $ f$ is integrable.

I do not see how this if and only if follows. Any hints on this would be great. Thanks in advance.

[quote="eskimo343"]Let $ (X, \mathcal{B}, \mu)$ be a finite measure space and $ f$ be a nonnegative measurable function of $ X$. For each $ A \in \mathcal{B}$, set
$ \nu(A) \equal{} \int_Af d\mu$.

Verify that it is a finite measure if and only if $ f$ is integrable.

I do not see how this if and only if follows. Any hints on this would be great. Thanks in advance.[/quote]

The "only if" is trivial. What are you having trouble with on the "if" part?

What I have done so far is:

It is clearly a measure, and $ \int_X f d \mu \leq \int_X |f| d \mu < \infty$.


1/Let ABC be the triangle . (I) is the incircle of ABC touches the sides
BC,CA,AB respectively at D,E,F . AD \cap  (I)=L,LB  \cap (I) =M ,LC \cap  (I)=N.
Prove that ME=NF if AL=DL

Hmmm... No one liked this problem?

I'll try to show that NE is parallel to AD. In the same way we can show FM || AD, so ENMF is a cyclic trapezoid, thus it's an isosceles trapezoid=>ME=NF.

Proof for NE || AD:

AL*AD=AE2=2AL2=>AE=\sqrt2*AL.
We want to prove CNE similar to CLA, iff CLA similar to CEL (in this order of the vertices) iff CL2=CE*CA.

Since CL is the median in CAD, we can compute Cl2=(CA2+CD2)/2-AL2. Replace here CA=CE+EA and EA=\sqrt2*AL and you'll get the conclusion CL2=CE*CA, which is what we wanted.

I'm not surte if you meant "if" or "iff", so I only tried to solve this part of the problem.


trigonometry
inequalities
inequalities unsolved

Let $ABC$ an acute triangle. Prove that:

$(\sin{2B}+\sin{2C})^{2}\sin{A}+(\sin{2C}+\sin{2A})^{2}\sin{B}+(\sin{2A}+\sin{2B})^{2}\sin{C}\leq 12\sin{A}\sin{B}\sin{C}$

It qppears somewhat tight, because after applying the easy inequality $\sin 2b+\sin 2c\leq 2\sin a$ it results in AM-GM flipped...


Let $ a,b,c>0 ,abc\equal{}1$ Show that

$ \sqrt {a^3  \plus{} b^3  \plus{} 2abc}  \plus{} \sqrt {b^3  \plus{} c^3  \plus{} 2abc}  \plus{} \sqrt {a^3  \plus{} c^3  \plus{} 2abc}  \ge 6$ :)

From QM-AM inequality we have
\[ \sum_{cyc}\sqrt{a^3\plus{}b^3\plus{}1\plus{}1}\ge\sum_{cyc}\frac{\sqrt{a^3}\plus{}\sqrt{b^3}\plus{}1\plus{}1}{2}\equal{}3\plus{}\sum\sqrt{a^3}\ge 3\plus{}3\equal{}6.\]


I want to see your solutions

Joe is playing a type of poker called texas hold' em.  There are two cards given to him and five on the table.  His "hand" is the best combination of five cards in poker.  If he uses a standard deck, what is the probability that he gets a straight (aces can be high or low)?

Let's see if you agree with me

[hide]768/238901

10 numerical combos to make a straight
4*4*4*4*4 ways to change the suit
42 ways to chose 5 of the 7 cards
430080 7 card combos to make a straight

52chose7=133784560 ways to get 7 cards

P=430080/133784560
P=768/238901[/hide]

[hide="this is what i did"]
$\displaystyle \frac{{4}\cdot{4}\cdot{4}\cdot{4}\cdot{4}\cdot{47}\cdot{46}\cdot{10}\cdot{5}\cdot{4}\cdot{3}\cdot{2}\cdot{1}\cdot{nCr(7,5)}}{{52}\cdot{51}\cdot{50}\cdot{49}\cdot{48}\cdot{47}\cdot{46}}$

which equals

$\displaystyle \frac{128}{1547}$
[/hide]

Basically, you are asking the probability of getting a 5-card straight from a 7-card hand.

This problem is too hard for MATHCOUNTS.  You can find very good calculations (and some very good explanations) for poker probabilities at either of these two sites:

1.  [url]http://www.math.sfu.ca/~alspach/comp20[/url]
2.  [url]http://www.wizardofodds.com/games/pokerodd.html[/url]


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[quote="Gyan"]What's new there?[/quote]

Look at the release notes.

[quote="A+MATH"]I got it yesterday.[/quote]
so did i!  lets get more people to get it!

[url=http://www.spreadfirefox.com/?q=affiliates&id=14949&t=86]Get Firefox[/url] and  [url=http://www.spreadfirefox.com/?q=user/register&r=14949]spread the word![/url]


modular arithmetic
combinatorics unsolved
combinatorics

Let N be a positive integer greater than 2. We number the vertices
of a regular 2n-gon clockwise with the numbers 1, 2, . . . ,N,−N,−N +
1, . . . ,−2,−1. Then we proceed to mark the vertices in the following way.
In the first step we mark the vertex 1. If ni is the vertex marked in the
i-th step, in the i+1-th step we mark the vertex that is |ni| vertices away
from vertex ni, counting clockwise if ni is positive and counter-clockwise
if ni is negative. This procedure is repeated till we reach a vertex that has
already been marked. Let $f(N)$ be the number of non-marked vertices.
(a) If $f(N) = 0$, prove that 2N + 1 is a prime number.
(b) Compute $f(1997)$.

It looks like a very nice problem :).

We modify the game as follows: relabel $-t$ with $2N+1-t$ for $t=\overline{1,N}$, and insert a $0$ between $1$ and $2N$. This way, we get a circle with the residues modulo $2N+1$ arranged starting from $1$ in a clockwise direction.

It's clear that when we reach a vertex which was initially labeled $-t,t\in\overline{1,N}$, instead of thinking that we go $t$ steps in a counter-clockwise direction on the old circle, we can think that we go $2N+1-t$ steps in a clockwise direction on the new circle (with the added $0$). This way, the rules of the game can be restated as follows: whenever we reach the residue $t$, we go to the residue $2t\pmod{2N+1}$.

The fact that $f(N)=0$ translates to $2^{2N}\equiv 1\pmod{2N+1}$, and $2N$ is the smallest exponent for which this happens. Since the order of $2$ in the multiplicative group of invertible residues modulo $2N+1$ is a divisor of $\varphi(2N+1)$, it means that $\varphi(2N+1)\ge 2N$, which implies that $2N+1$ is a prime.

The approach above gives an idea about how to find $f(N)$ for some general $N$: $f(N)$ equals $2N$ minus the smallest $k$ for which $2^k\equiv 1\pmod {2N+1}$.

In our case, with $N=1997$, we must find the smallest $k$ for which $2^k\equiv 1\pmod{2\cdot 1997+1=5\cdot 17\cdot 47}$. It's easy to obtain this value: it's $8\cdot 23=184$. I guess the answer should be $2\cdot 1997-184=3810$.

I hope there are no computation mistakes in the last part.


[hide="Hint"]How far is the end of the tunnel from the back of the train?[/hide]

[hide="I think this is the answer"]I think it's 1/10 of a mile.

The train travels 30 miles per hour, so that's one mile per 2 minutes.
Assuming the problem means that the edge of the train is touching the edge of the tunnel when the train reaches the tunnel and completely gets out, there is a distance of 1 mile between the back edge of the tunnel and front edge of the train when it gets out.
Think about several equal sized blocks: Let's say one block represents the train. Since the tunnel is 9 times as long as the train, 9 blocks placed next to each other would represent the tunnel. There is a distance of 1 mile between the front edge of the train and the back edge of the tunnel, so if all ten blocks were aligned next to each other, it would represent one mile. That's 10 blocks. So the train is 1/10 mile long.[/hide]

[quote="bryantacademy"]A train traveling 30 miles an hour reaches a tunnel that is 9 times as long as the train. If the train takes 2 minutes to completely clear the tunnel, how long is the train.

I have seen this before but I do not remember the solution.[/quote]

[hide]It takes 2 minutes to clear the tunnel and so, you get 2 minutes to travel 1 mile. The tunnell is 9 times the train. It takes 9 times the tunnel to clear the front of the train. And a 1 times the tunnel so that the back is out. THat equals 10 times the tunnel for 1 mile. Thus the train is 1/10 mile.[/hide]

I believe that this is right.

[hide][i]Supposing x is the length of the train, the tunnel is 9x[/i]
The train takes 2 min to clear the tunnel at a constant rate of 30 (mph)
 :arrow: the tunnel is 1 (mile); 2/60 *30
Since the tunnel is 9x + x long and that would equal 1mile;
[b]The train is 1/10 miles long
[/b][/hide]

[hide]
From the back of the train to the end of the tunnel is $10x$ and it takes the train $\frac{1}{30}$ of an hour to go through the tunnel at 30 mph. If the train is $\frac{1}{2}$ miles long, it takes it 5 minutes. If it is $\frac{1}{10}$ of a mile long, it takes 2 minutes. The answer is $\frac{1}{10}$ of a mile.[/hide]

[hide]wasn't this on a 150 greatest problem book?

1/10?[/hide]


Yeah, you are right  :blush:

Do you have any prove?

In fact it was posted again see here for another solution
http://www.mathlinks.ro/Forum/viewtopic.php?t=5825

[quote="shyong"]
isnt that 1/99 istead of 1/98 which eventually gives us a nice number $|a_k|\leq \sqrt{100}=10$[/quote]
You are right. Thank you.


$1).\ \ f: [a,b]\rightarrow R,\ f\in C^1,\ f'(a)=f'(b)\Longrightarrow \left(\exists\right) c\in (a,b)$ so that

$(b-c)\left[f'(c)-\frac{f(c)-f(a)}{c-a}\right]=(c-a)\left[f'(c)-\frac{f(b)-f(c)}{b-c}\right].$

$2).$ Let $f: [a,b]\rightarrow R$ be a function $f\in C^1$ for which exists $m\ne n$ so that

$\frac{mf'(a)-nf'(b)}{m-n}=\frac{f(a)-f(b)}{a-b}$. Prove that exists $c\in (a,b)$ so that

$m(b-c)\left[f'(c)-\frac{f(c)-f(a)}{c-a}\right]=n(c-a)\left[f'(c)-\frac{f(b)-f(c)}{b-c}\right].$

1) Consider $g: [a,b]\to \mathbb{R}$ such that $g(x)=(b-x)\left[f'(x)-\frac{f(x)-f(a)}{x-a}\right]-(x-a)\left[f'(x)-\frac{f(b)-f(x)}{b-x}\right]$ extended by continuity in $a$ and $b$ We have that g is continuous and we have that $g(a)=(b-a)(f'(b)-f'(a))$ and $g(b)=-g(a)$. From mean value theorem someone can find $c$ such that $g(c)$=0. The consequence follows.

[quote="Virgil Nicula"]

$2).$ Let $f: [a,b]\rightarrow R$ be a function $f\in C^1$ for which exists $m\ne n$ so that

$\frac{mf'(a)-nf'(b)}{m-n}=\frac{f(a)-f(b)}{a-b}$. Prove that exists $c\in (a,b)$ so that

$m(b-c)\left[f'(c)-\frac{f(c)-f(a)}{c-a}\right]=n(c-a)\left[f'(c)-\frac{f(b)-f(c)}{b-c}\right].$[/quote]

Apply Rolle theorem to

$g(x)=m\frac{f(x)-f(a)}{x-a}+n\frac{f(b)-f(x)}{b-x}$

[quote="xirti"]1) Consider $g: [a,b]\to \mathbb{R}$ such that $g(x)=(b-x)\left[f'(x)-\frac{f(x)-f(a)}{x-a}\right]-(x-a)\left[f'(x)-\frac{f(b)-f(x)}{b-x}\right]$ extended by continuity in $a$ and $b$ We have that g is continuous and we have that $g(a)=(b-a)(f'(b)-f'(a))$ and $g(b)=-g(a)$[/quote]

xirti  g(a)=g(b)=0


Follows directly from the fact that $\forall A,B,C\subset \mathbb{N}: \min(A,B,C) = \min(A,\min(B,C))$.


According to J.H. Plumb, "History is now strictly organized, powerfully disciplined, but it possesses only a modest educational value and even less conscious social purpose." According to George Santayana, "Those who cannot remember the past are condemned to repeat it." Discuss your point of view.

also,

"We might say that we were looking for global schemas, symmetries, universal and unchanging laws--and what we have discovered is the mutable, the ephemeral, the complex." Support or challenge Nobel Prize winner Ilya Prigogine's assertion.

[quote="mdk"]According to J.H. Plumb, "History is now strictly organized, powerfully disciplined, but it possesses only a modest educational value and even less conscious social purpose." According to George Santayana, "Those who cannot remember the past are condemned to repeat it." Discuss your point of view.[/quote]

In my opinion there are many places where you can see history repeating itself.

For instance, if you read the bible, there are many stories there about what happened then, and what happened say in the last 100-200 years
there are similarities.

Also a fun fact... Germany was one of the first nations to embrace democracy. Yet was the cruelest dictatorship during the 2nd world war.

As a student of statistics, I can't throw away history, or past data, on which statistical conclusions are fully dependent. :roll:

It only says that the science of history is too complex to eb defiend in terms of rigid laws.


Let the sequence $\{a_{n}\}$ be defined as $a_{1}=1$ and $a_{n}$ is the greatest positive integer such that $a_{n}\not\in\{a_{1},a_{2},\ldots,a_{n-1}\}$ and $n\mid a_{1}+a_{2}+\ldots+a_{n}$. Prove that  $a_{a_{n}}=n$ for all $n\in\mathbb{N}$.


[hide="Sorry to revive, but . . ."]
Note that if $ A$ and $ B$ meet, their paths can be connected to form one path from $ (0,0)$ to $ (5,7)$.  There are $ \binom{12}{5}\equal{}792$ of those, which gets us that figure much more easily.
[/hide]

[quote="ch1n353ch3s54a1l"][quote="worthawholebean"]Objects $ A$ and $ B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $ A$ starts at $ (0, 0)$ and each of its steps is either right or up, both equally likely. Object B starts at $ (5, 7)$ and each of its steps is either left or down, both equally likely. Which of the following is closest to the probability that the objects meet?

$ \textbf{(A)}\ 0.10 \qquad \textbf{(B)}\ 0.15 \qquad \textbf{(C)}\ 0.20 \qquad \textbf{(D)}\ 0.25 \qquad \textbf{(E)}\ 0.30$[/quote]

[hide]
Each object has to make 6 steps so there are only a few possible meeting points: $ (5,1), (4,2), (3,3), (2,4), (1,5), (0,6)$. Then just take each point separately and add 'em up...the number of ways they can meet is $ \binom{6}{1}\binom{6}{0} \plus{} \binom{6}{2}\binom{6}{1} \plus{} \ldots \plus{} \binom{6}{5}\binom{6}{4} \plus{} \binom{6}{6}\binom{6}{5} \equal{} 792$.

The number of ways either object A or B can move six steps is $ 2^6$ so the probability is $ \frac {796}{2^6\cdot 2^6}\approx \boxed{\textbf{(C)}\ 0.20}$[/hide][/quote]

how do we know for sure that these points can meet only at  $ (5,1), (4,2), (3,3), (2,4), (1,5), (0,6)$?

[quote="talkinaway"]Let $A$ be the sum of the coordinates of $A$, and $B$ be the sum of the coordinates of $B$.

After $t$ moves, $A$ is $t$, while $B$ is $12 - t$.  If $A$ and $B$ are at the same point, then the sum of their coordinates must be the same.  Therefore $t = 12 - t$, or $t = 6$.  That means that they meet after $6$ moves, when the sum of both of their coordinates is $6$.  Hence the list of coordinates given.  The only "catch" is that $(6,0)$ is not possible since you can't get from $(5,7)$ to $(6,0)$ without moving $B$ to the right, and $B$ only moves to the left.

This explanation uses the property of an "invariant" - while the coordinates may vary, the sum of the coordinates will always be the same after $t$ seconds.

That said, if you're reviewing solutions, check out the other solution.  It's very slick.  It turns this "unfamiliar" problem into the rather familiar problem of moving from $(0,0)$ to $(x,y)$ along lattice points - a problem that's familiar to many math teamers.  The solution is $\binom{x+y}{x}$, because you have $x+y$ moves to make, and $x$ of those moves are right, while the other ones are up.[/quote]

Ok let me try to understand $\binom{x+y}{x}$... So this is my understanding, please correct me if I'm wrong:

So out of the total possible moves we are selecting the possible moves to right or moves to up (depending on which one you define x as). So its basically asking: out of x+y moves, how many ways can we select x right or x up moves....? What I'm not getting is that wouldn't order matter? Like right right up and right up right would be totally different so why use combinations? Could you give me like the intuition behind (x+y choose x) ?

Thanks

Each time you rearrange the order of, for example, "up up right up right up", you're taking a distinct path.  If you ONLY change the order of "up" and "right", and you don't change the total number of ups and rights, you wind up at the same target point $(x,y)$ (presuming you start at the origin).

Thus, all you need are the total number of "instructions", which is $x+y$, of that total number of slots, you pick which ones go "right" (traditionally denoted as $x$).  That's how you get $\binom{x+y}{x}$.  All the remaining slots become "up", but you have no choices once you pick the "right"s.

(Of course, you can pick the ups instead of the rights, giving $\binom{x+y}{y}$.  That just means $\binom{x+y}{x} = \binom{x+y}{y}$, which is to be expected.)

@above write it as:
$$ \binom{6}{5}\binom{6}{0} + \binom{6}{4}\binom{6}{1} + \ldots + \binom{6}{1}\binom{6}{4} + \binom{6}{0}\binom{6}{5} $$

By Vandermonde's Identity, this becomes $\binom{12}{5}$, which is $792$.

[quote=candyofthesun]but how do you write it like that? what's the reasoning?

Edit: i'm not talking about vandermondes, but the thing before that[/quote]

${n\choose k}={n\choose n-k}$ as in the later you are choosing $k$ to not be in your set.

sorry I just realized my question was unclear, I meant to ask how post #2 got this part only: $ \binom{6}{1}\binom{6}{0} + \binom{6}{2}\binom{6}{1} + \ldots + \binom{6}{5}\binom{6}{4} + \binom{6}{6}\binom{6}{5}$.

I believe that each $\binom{6}{a}\binom{6}{b}$ represents choosing how many vertical steps or something.

In case anyone is curious, this is [url=https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_22]2003 AMC 12A Problem #22[/url]

[quote=worthawholebean]Objects $A$ and $B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $A$ starts at $(0,0)$ and each of its steps is either right or up, both equally likely. Object $B$ starts at $(5,7)$ and each of its steps is either left or down, both equally likely. Which of the following is closest to the probability that the objects meet?

$ \textbf{(A)}\ 0.10 \qquad
\textbf{(B)}\ 0.15 \qquad
\textbf{(C)}\ 0.20 \qquad
\textbf{(D)}\ 0.25 \qquad
\textbf{(E)}\ 0.30$[/quote]

note the original coordinates are opposite parity.  you can use parity arguments about which moves correspond.

What I have done was to first let if there are x right steps and y up steps then accordingly there are 5-x left steps and 7-y down steps
Adding them up gives exactly 12 steps
Hence the total steps is 2^12
But each right and up steps gives us an unique left and down acc. To the previous definition
For example if there is RU
THEN AN UNIQUE STEP WOULD BE LDD thus
The problem remains to solve for how many ways we can put five R's and 7 U's that would we 792 by vandermondes identity we can easily calculate it
Hence the answer is 792/2^12 is approx 0.20

[quote=worthawholebean]Objects $A$ and $B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $A$ starts at $(0,0)$ and each of its steps is either right or up, both equally likely. Object $B$ starts at $(5,7)$ and each of its steps is either left or down, both equally likely. Which of the following is closest to the probability that the objects meet?

$ \textbf{(A)}\ 0.10 \qquad
\textbf{(B)}\ 0.15 \qquad
\textbf{(C)}\ 0.20 \qquad
\textbf{(D)}\ 0.25 \qquad
\textbf{(E)}\ 0.30$[/quote]
they should connect in 6 rounds, where each round has each object going once. so denominator is $2^{12}$ If we just connect the lines we find that its the same as how many ways to move from 0,0 to 5,7 which is obviously 12C5 so our answer is 12C5/2^12


They will almost certainly not get done today.

We are shooting for Wednesday (11/8).

Oh... I've been checking for the last 10 days or so, thinking "Where is my score?"......

Thank you.

Does that mean you filed all the forms too?  :maybe:

[quote="DPatrick"]They will almost certainly [b]not[/b] get done today.[/quote]
That means they haven't logged the forms yet.

LOL I like how you guys keep building the suspense.

Thanks for running this, and working so hard to grade everything in such a short period of time  :)

How can we tell if they have received our forms?  :blush:

EDIT: Oh, my name's in the leaders thing, so they've gotten it. :D

[quote="h_s_potter2002"]Are scores up yet?  Because they seem to be up for everyone else, but I haven't gotten anything yet.[/quote]

i haven't gotten my scores either :/


function
trigonometry
logarithms
complex numbers
complex analysis
complex analysis unsolved

find an analytic function satisfying $ |f(x\plus{}iy)|\equal{}e^{xy}$

can take for example $ f(z)\equal{}e^{Re(z)Im(z)}$ where is the problem here.
i am sure that i don't understand the probleme, because i am very bad in english

[quote="kouboy"]find an analytic function satisfying $ |f(x \plus{} iy)| \equal{} e^{xy}$[/quote]
\[ f(z) \equal{} e^{(z^2)/2i} \equal{} e^{xy}\left(\cos \alpha(z) \plus{} i\sin(\alpha(z))\right)\]
with $ \alpha(z) \equal{} \frac{y^2 \minus{} x^2}{2}$ ought to do it.

The general method for finding $ f$ given $ |f|\equal{}g$ is: (1) check that $ \log g$ is harmonic apart from its singularities (otherwise no $ f$ exists); (2) find a conjugate harmonic function $ h$ (related to $ \log g$ by Cauchy-Riemann equations); (3) set $ f\equal{}ge^{ih}$.

Here $ \log g\equal{}xy$, so C-R equations are $ h_x\equal{}\minus{}x$ and $ h_y\equal{}y$.

To avaiateurpilot: the key word is [url=http://en.wikipedia.org/wiki/Analytic_function]analytic[/url]. Your proposed example is differentiable in the $ \mathbb{R}^2$ sense, but not in the complex sense. The real and imaginary parts are not analytic functions.

thanks mlok and hpe.

log g should be harmonic because its the real part of the analytic function log f right? ( of course where f doesnt vanish). a direct way of checking that $ log|f|$ is harmonic seems a lot of work and may be clumsy too

One way for $ Log[g(x,y)]\in\mathbb{H}$ is for $ g(x,y)$ to satisfy the following coupled set of PDEs:

$ \frac {\partial^2 g}{\partial x^2} \equal{} \frac {1}{g}\left(\frac {\partial g}{\partial x}\right)^2$

$ \frac {\partial^2 g}{\partial x^2} \equal{} \frac {1}{g}\left(\frac {\partial g}{\partial x}\right)^2$

The most general solution to this is:

$ g(x,y) \equal{} c_1e^{c_2 x y}$

[edit]  The most general expression $ g(x,y)$ must satisfy seems to be solutions to:

$ \frac {\partial^2 g}{\partial x^2} \plus{} \frac {\partial^2 g}{\partial y^2} \minus{} \frac {1}{g}\left[\left(\frac {\partial g}{\partial x}\right)^2 \plus{} \left(\frac {\partial g}{\partial y}\right)^2\right] \equal{} 0$

What are some other solutions?


probability
calculus
calculus computations

Laplacian is defined as the gradient of the gradient. The Laplacian operator is one that widely used in Computer Graphics and Computer Vision.

My question is: Why people use Laplacian, but not the gradient or the gradient of Laplacian? What's so special about Laplacian? How did the first researchers discover that Laplacian is something that's going to be important in today's research?

The laplacian is the [i]divergence[/i] of the gradient.  It comes up naturally in (say) studying diffusion, or fluid dynamics, or electromagnetism.  (see any of: laplace's equation, heat equation, wave equation).  It is the simplest example of an elliptic operator and motivates many of the results one proves on elliptic PDE in a typical graduate course on the subject.  It is the infinitesimal generator of Brownian motion and therefore of great importance in probability theory.  It satisfies various other nice properties as well.

It's important because of its applications and because of the theory which has been built around it.


Find all continuous functions$ f: \mathbb R \to \mathbb R$ which satisfies
\[ f(x \plus{} y \plus{} axy) \equal{} f(x)f(y)
\]
for all $x, y \in \mathbb R$.

[quote="Xaenir"]Find all continuous functions$ f: R \to R$ which satisfy:
\[ f(x \plus{} y \plus{} axy) \equal{} f(x)f(y)
\]

\[ \forall x,y
\]
[/quote]

We have the obvious constant solutions $ f(x)\equal{}0$ and $ f(x)\equal{}1$.  Let us now look for non constant solutions :
If $ a\equal{}0$, we get (since $ f(x)$ is continuous) the standard non constant solution  $ f(x)\equal{}e^{bx}$

If $ a\neq 0$, let $ g(x)\equal{}f(\frac{x\minus{}1}{a})$

So $ f(x)\equal{}g(ax\plus{}1)$ and the initial equation becomes $ g((ax\plus{}1)(ay\plus{}1))\equal{}g(ax\plus{}1)g(ay\plus{}1)$, so $ g(xy)\equal{}g(x)g(y)$ and so $ g(x)\equal{}x$ and so $ f(x)\equal{}ax\plus{}1$

As a synthesis :
Constant solutions are $ f(x)\equal{}0$ and $ f(x)\equal{}1$

Non constant solutions are :
If $ a\equal{}0$, $ f(x)\equal{}e^{bx}$, for any real $ b\neq 0$

If $ a\neq 0$, $ f(x)\equal{}ax\plus{}1$


Look at these!

[img]http://photos1.blogger.com/blogger/5639/2020/1600/original.jpg[/img]
Is it WTF or THE Mac Store
[img]http://photos1.blogger.com/blogger/5639/2020/1600/funny%20fart.jpg[/img]
LET IT RIP!!!!

I checked the site out. Nice site! I am a fan of Escher.


If $ x\equal{}\frac{10}{3^n}$ and $ y\equal{}\frac{1}{2^n}$, what is the smallest positive integer $ n$ such that $ x < y$?

$ x\equal{}\frac{10}{3^n}<y\equal{}\frac1{2^n}\implies10\cdot2^n<3^n\implies n\ge\boxed6$. Bad problem in my opinion.


rotation
geometry
parallelogram
geometric transformation
vector
geometry solved

I don't think the conclusion of this question is true, since D (or E) can move along the perpendicular bisector of AB (or AC).

ah...i've modified the problem..
i forget to give a condition.

Rotate $D$ around $E$ with an angle of $-\frac {\pi}2$. Let the image be $D$. Then $ED'\perp ED$ and $ED'=ED$. It's easy to find the angle $\angle FCD'$. It's precisely $\angle ABC$. On the other hand, $\frac {BC}{CF}=\frac {BA}{CD'}=\frac {BA}{AD}=\sqrt 2$, so the triangles $ABC,D'CF$ are similar. This makes it easy to show that $FD'||AE,FD'=AE$, so $AFD'E$ is a parallelogram, so $ED'||AF,ED'=AF\Rightarrow ED\perp AF,ED=AF$.

[quote="liyi"]$ABC$ is an triangle. $D,E,F$ are lying outside the triangle and $\triangle DBA,\triangle AEC,\triangle BFC$ are isoceles right triangles with $DB=DA, EA=EC, FC=FB$. Join $DE$ and $AF$. Show that $DE$ is perpendicular to $AF$[/quote]

I don't really like these notations, since they are asymmetric. Hence, I will rename the points F, E, D as X, Y, Z, respectively. Then, we have the isosceles right triangles XBC, YCA and ZAB. In other words, the points X, Y, Z are the centers of the squares constructed on the segments BC, CA, AB and pointing into the exterior of triangle ABC. So we arrive at the following equivalent formulation of your problem:

[i]Let ABC be a triangle, and let X, Y, Z be the centers of the squares BCTU, CAVW, ABSR constructed on the segments BC, CA, AB and pointing into the exterior of triangle ABC. Show that the line YZ is perpendicular to the line AX.[/i]

Here is a solution:

I will prove that the three lines YZ, ZX, XY are perpendicular to the lines AX, BY, CZ, respectively.

The spiral dilatation with center B, rotation angle 45 and dilatation factor $\sqrt{2}$ maps C to T and Z to A. Hence, the segment TA is the image of the segment CZ, so that

  < (CZ; TA) = 45

(where < (k; l) denotes the directed angle between two lines k and l).

The spiral dilatation with center C, rotation angle 45 and dilatation factor $\displaystyle \frac{1}{\sqrt{2}}$ maps T to X and A to Y. Hence, the segment XY is the image of the segment TA, so that

  < (TA; XY) = 45,

and < (CZ; XY) = < (CZ; TA) + < (TA; XY) = 45 + 45 = 90, so that CZ is perpendicular to XY. Similarly, AX is perpendicular to YZ and BY is perpendicular to ZX, and we are done.

[By the way, the same approach using spiral dilatations shows that AX = YZ, BY = ZX and CZ = XY.]

  Darij

[quote]Let ABC be a triangle, and let X, Y, Z be the centers of the squares BCTU, CAVW, ABSR constructed on the segments BC, CA, AB and pointing into the exterior of triangle ABC. Show that the line YZ is perpendicular to the line AX. [/quote]

This may seem trivial but you can also solve it using vectors: AX*YZ=0. Then represent AX using AB, BC, and YZ using AB, BC and CA. After some cancellations, you get an expression of ab*sinC or ab*cosC and some cyclic parts, but ab*sinC=2S, where S is the area of ABC and ab*cosC=a^2+b^2-c^2. So eventually everything dies and you get 0=0.  :cool:

The details are as straightforward as the idea, so I don't think it's necessary to show them here. Btw, the rotation and spiral dilatation solutions are cool.  :lol:


Prove that for positive numbers $a_1$, ..., $a_n$ the following inequalities hold
a) $\frac{a_1}{a_1}+\frac{a_2}{\sqrt{a_2}}+\frac{a_3}{\sqrt[3]{a_3}}+...+\frac{a_n}{\sqrt[n]{a_n}}\leq (1+\sqrt{S})^2$;
b) $\frac{a_1}{a_1}+\frac{a_2}{\sqrt{a_2}}+\frac{a_3}{\sqrt[3]{a_3}}+...+\frac{a_n}{\sqrt[n]{a_n}}\leq \left(\frac{1}{2}+\sqrt{S}\right)^2$,
where $S=a_1+a_2+...+a_n$.

Is there some typo in b)? Take $n=1$ and $a_1\to 0$. Then the LHS is $1$ but the RHS is only about $\frac14$.   :?

There is no any comments about that in the source. I guess, $n>1$ for b).

Sorry, I said nonsense.

The ineq a) was proposed at Kolmogorov Cup, and b) is written in Khrabrov's selected problems, where he writes that RHS can be replaced with $\left(\frac{1}{2}+\sqrt{S}\right)^2$. There is a typo, I think. Guess a right version.

[quote="Myth"]Prove that for positive numbers $a_1$, ..., $a_n$ the following inequalities hold
a) $\frac{a_1}{a_1}+\frac{a_2}{\sqrt{a_2}}+\frac{a_3}{\sqrt[3]{a_3}}+...+\frac{a_n}{\sqrt[n]{a_n}}\leq (1+\sqrt{S})^2$.[/quote]
Well, I will write my solution for it. I don't think my approach is nice, but...

The inequality is equivalent to $\frac{a_2}{\sqrt{a_2}}+\frac{a_3}{\sqrt[3]{a_3}}+...+\frac{a_n}{\sqrt[n]{a_n}}\leq 2\sqrt{S}+S$. So it is clear we may assume $a_1=0$.

Let $f(a_2,...,a_n)=RHS-LHS$ an $a$ is a minima point of $f$. We have $f'_{a_k}(a)=1+\frac{1}{\sqrt{S}}-\frac{k-1}{ka_k^{1/k}}$. We see that $f'_{a_k}<0$, hence $a_k>0$ for all $k$, moreover it is clear that all $a_k\leq 1$. It follows that we necessarily have $1+\frac{1}{\sqrt{S}}-\frac{k-1}{ka_k^{1/k}}=0$, i.e. $a_k=\left(\frac{k-1}{k}\cdot c\right)^k$ for all $k$, $c<1$ and $\sqrt{S}=\frac{c}{1-c}=c+c^2+c^3+...>c+c^2+...+c^n$ [b](*)[/b].

Therefore, we are to prove $\sum_{k=2}^n \left(\frac{k-1}{k}\cdot c\right)^{k-1}\leq 2\sqrt{S}+\sum_{k=2}^n \left(\frac{k-1}{k}\cdot c\right)^k$.
But $\left(\frac{k-1}{k}\cdot c\right)^{k-1}\leq \frac{1}{2}c^{k-1}$, so use [b](*)[/b] and we are done  :?

P.S. We see that in this way the more strong inequaity is true, namely $\frac{a_1}{a_1}+\frac{a_2}{\sqrt{a_2}}+\frac{a_3}{\sqrt[3]{a_3}}+...+\frac{a_n}{\sqrt[n]{a_n}}\leq \left(\frac{1}{2}+\sqrt{S}\right)^2+\frac{3}{4}$.
So maybe it is the right variant for b).

Hope it is correct, though I am sure fedja will find at least 5 errors here. :(

I posted the part a) as a problem from AMM and the solution presented was very nice and with no computations. I will try to find it. The great part in your solution is that it is stronger inequality.

[quote="harazi"]I posted the part a) as a problem from AMM and the solution presented was very nice and with no computations.[/quote] Indeed, my solution couldn't appear in Kolmogorov Cup.

Ok, I cannot find it, but I remember the very nice solution: for any $x<1$ we have $\sum_{k=1,n}{a_k^{1-1/k}}=\sum_{a_k\geq x^k}{a_k^{1-1/k}}+\sum_{a_k<x^k}{a_k^{1-1/k}}\leq\frac{\sum_{k=1,n}{a_k}}{x}+\frac{x}{1-x}$. Next, choose $x=\frac{\sqrt{S}}{1+\sqrt{S}}$ to minimize the last expression and it's done.


Why can't they both be less than zero?

If $ x \equal{} (\minus{}1)$ and $ y \equal{} (\minus{}1)$, then this gives you an answer of $ 5$ which is less than $ 12$.

Please do not put large blocks of text in LaTeX, buzzer.  Also, I moved this to HSB; this is not middle school classroom math.

[quote="buzzer11"]Mickey cannot afford to buy Minnie a diamond, so he draws her a picture of one instead. He graphs the set $ D$ defined by $ |3x| \plus{} |2y| \le 12$  What is the area of $ D$?

$ \text{a. 24 b. 36 c. 48 d. 96 e. 144}$[/quote]

In general, if you don't know what a graph with absolute value signs looks like, just graph all possible cases: $ 3x \plus{} 2y\le12$, $ 3x \minus{} 2y\le12$, $ \minus{} 3x \plus{} 2y\le12$, and $ \minus{} 3x \minus{} 2y\le12$, which forms a rhombus with vertices $ (\pm4,\pm6)$, so the area is $ \frac12\cdot8\cdot12 \equal{} 48\implies\boxed C$.

Edit: nick42, you're basically correct; you just forgot to divide by $ 2$.

Generalization: Find the area bounded by the graph $ |ax\plus{}b|\plus{}|cx\plus{}d|\le e$.

Sorry about the latex.

Yes, the answer is C. The area of a orthodiagonal quadrilateral is half of the product of its diagonals.


Hi! I have this problem that I cannot solve myself. Pls help me!
Let be given a circle with center[b] C[/b] and a point [b]A[/b] outside the circle. [b]B,C[/b] are points on the circle such that [b]AB[/b] and [b]AC[/b] are touching with the circle. A line from [b]A[/b] cuts the circle at 2 points [b]M, N[/b] but not passing through [b]C[/b]. The tangents to the circle from [b]M, N[/b] intersect at[b] K[/b]. Prove that [b]B, C, K[/b] are collinear.

It was posted many times before, but its faster to post solution.

You should change your notation first: you use C as the center of the circle and as a tangency point. Anyway let's call center O. A belongs to MN - polar of K, so K belongs to polar of A which is BC. Hence B,C,K are collinear.


This was part of a problem that I was using generating functions for. I reduced it down to finding the coefficient in $x^{60}$ in the expansion of $(1+x+x^{2}+x^{5})(1+x+x^{2}+x^{5})...$.

Help?

Well, what's the ...  How many times is that expression multiplied by itself.

[quote="jclarkemathL314159"]This was part of a problem that I was using generating functions for. I reduced it down to finding the coefficient in $x^{6}0$ in the expansion of $(x+x^{2}+x^{5})(x+x^{2}+x^{5})...$.

Help?[/quote]

I think he want us to find the coefficient of $x^{60}$ in expression $(x+x^{2}+x^{5})^{n}$ where $n$ goes to infinitely.

[quote="jclarkemathL314159"]goes on infinitely.[/quote]
Wait, are you sure about that?  :huh:
Because if it does go to infinity, then every term (by that I mean the $x+x^{2}+x^{5}$) will contribute a power of $x$ that is at least 1, so there's no term that has a finite exponent, which means that the coefficient of $x^{60}$ is simply 0.  :wink:

[quote="jclarkemathL314159"]Oh sorry, you're right. I messed up my generating function. here's what it's supposed to be.

Find the coefficient of $x^{60}$ in the expansion of $(1+x+x^{2}+x^{5})(1+x+x^{2}+x^{5})...$[/quote]

[hide="Uhm..."]I counted the number of non-negative integer solutions to $A+2B+5C=60$ and got $\boxed{205}$...[/hide]

wait, that can't be right...

the expansion $(1+x+x^{2}+x^{5})^{n}$ depends on $n$.  If $n$ is arbitrarly large, then the answer is infinity...

jclarkemathL314159, could you post the original problem?  It seems that you have made a mistake in writing the generating functions - the coefficient of $x^{60}$ is infinite for the generating function you gave.

[quote="Treething"]eh he fixed it to $(1+x+x^{2}+x^{5})$[/quote]

If you have it infinitely couldn't you just keep multiplying the $x^{60}$ factor by 1 or something...then it's still infinite i think..

[quote="jclarkemathL314159"]OK here is the original problem:

So to keep it short, find the number of solutions to A+2B+5C=60. Positive integer solutions that is.[/quote]

 :D

[hide]I got 259... I'm not completely sure if it's right so I won't post my solution yet..[/hide]

[hide]This is quite easily brute forceable... C can be anything from 1 to 11. If C=1, then A+2B=55, and A has to be odd, so there are 27 values there, If C=2, A+2B=50, so A is even, and there are 24 values. If C=3, A+2B=45, there are 22 values. C=4, 19 values. C=5, 17 values...... Final answer:157 unless I did my math wrong.[/hide]

I am quite sure that there is a beautiful way to do this, and I will probably think about this question all day tomorrow.

I think if you leave the one out of your original generating function you'll get the answer. Putting in the one is the same as letting you add on as many 0's as you want (i think, my generatingfunctionology isnt to great)

I haven't worked through it completely, but if you instead write three generating functions for $A$, $B$, and $C$ as $(x+x^{2}+x^{3}+\cdots)$, $(x^{2}+x^{4}+x^{6}+\cdots)$, and $(x^{5}+x^{10}+x^{15}+\cdots)$, respectively, I think it then becomes finding the coefficient of $x^{60}$ in the product of those three functions which is \[ \frac{x^{8}}{(1-x)(1-x^{2})(1-x^{5})}\]


From: http://www.xxhh.net

1、钞票不是万能的，有时还需要信用卡；

2、每个人都应该热爱动物，尤其是煮熟的；

3、要节约用水，尽量和女友一起洗澡；

4、要用心去爱你的邻居，不过不要让她的老公知道

5、每个成功男人的背后，都有一个女人。 每个不成功男人的背后，都有两个；

6、再快乐的单身汉迟早也会结婚，幸福不是永久的嘛；

7、聪明人都是未婚的，结婚的人很难再聪明起来；

8、成功是一个相关名词，他会给你带来很多不相关的亲戚；

9、不要等明天交不上差再找借口，今天就要找好；

10、爱情就象照片，需要大量的暗房时间来培养；

11、后排座位上的小孩会生出意外，后排座位上的意外会生出小孩；

12、现在的梦想决定着你的将来，还是再睡一会吧；

13、应该有更好的方式开始新一天，而不是千篇一律的在每个上午都醒来；

14、努力工作不会导致死亡，不过我不会用自己去证明；

15、工作好有意思，尤其是看着别人工作；

16、神决定了谁是你的亲戚，幸运的是在选择朋友方面他给了你留了余地；

17、两个人的状态是不稳定的，三个人才是！

18、服饰就象铁丝网，它阻止你冒然行动但并不妨碍你尽情的观看；

19、学的越多，知道的越多。知道的越多，忘记的越多。忘记的越多，知道的越少。为什么学来着？

[quote="shobber"]From: http://www.xxhh.net
5、每个成功男人的背后，都有一个女人。 每个不成功男人的背后，都有两个；
[/quote]
which 2?


For a sequence ${a_{n}}$, $a_{1}=1$ and $a_{n+1}=3^{a_{n}}+1$ for all positive integer $n$. Let the remainder when sequence $a_{n}$ is divided by 100 be $b_{n}$. Find the value of $b_{2006}+b_{2007}+\cdots+b_{2015}$.

[quote="Sepp"][hide="Solution"]Since $\phi(100) = 40$ and $(3, 100) = 1$, $3^{40}\equiv 1 \pmod{100}$. Also $\phi(40) = 16$ and $\mathrm{ord}_{40}(3) = 4$. Then $a_{n}\equiv 2 \pmod{40}$ with $n \geq 3$.
Hence $b_{n}= 3^{2}+1$ for $n \geq 4$. And so the desired sum is equal to $100$.[/hide][/quote]

Why is it $a_{n}\equiv 2\pmod{40}$ for all $n\ge 3$? :blush:

Sorry, it should be $a_{n}\equiv 10 \pmod{40}$ with $n > 3$ and $b_{n}\equiv 3^{10}+1 \equiv 50 \pmod{100}$ with $n > 4$.


ANYH0W, TH1S F0RUM IS G01NG T0 D1E! LARRY GET ()N M()RE 1NSTEAD 0F P|_AY1NG WITH PUPPeTS.


modular arithmetic
number theory unsolved
number theory

suppose that [b]a[/b] is a natural number and [b]a >1[/b] .[b]p[/b] is a prime number and [b]a 2 n +1[/b] divisible by [b]p[/b],prove that:
[b]p=2 n+1k+1[/b].

Hmm.. This actually looks rather straightforward: the order of $a\pmod p$ is a divisor of $2^{k+1}$, because $a^{2^{k+1}}\equiv 1\pmod p$. Hiwever, this order cannot be $<2^{k+1}$, because we have $a^{2^k}\equiv -1\pmod p$, so the order is, in fact, $2^{k+1}$, and the order of an element must divide $p-1$, hence the result: $2^{k+1}|p-1$, Q.E.D.

[by order of $a\pmod p$ I mean the order of the residue of $a\pmod p$ in the multiplicative group of non-null elements in $\mathbb Z_p$]

[Edit: isn't it $a^{2^n}+1$? It kind of looks like $a^{2n}+1$]

oh if p is prime and p is divisor a^{2^n}+b^{2^n} we have that p form 2^{(n+1)}*k+1;
it's great.

You prove it in the same way, actually, but you need the extra condition that $p$ doesn't divide either $a$ or $b$.


inequalities
search
inequalities proposed

1.Let $ a,b,c \in \mathbb{R}^\plus{}$ Prove that
\[ \sum_{cyclic}a\sqrt{7a^2\plus{}9b^2} \geq \frac{4}{3}(a\plus{}b\plus{}c)^2\]
2.Let $ a,b,c \in \mathbb{R}^\plus{}$ and $ ab\plus{}bc\plus{}ca \equal{} abc$
Prove that \[ \sum_{cyclic}\frac{1}{\sqrt{ab}\minus{}1} \leq \frac{3}{2}\]  :lol:

[quote="apollo"]2.Let $ a,b,c \in \mathbb{R}^ \plus{}$ and $ ab \plus{} bc \plus{} ca \equal{} abc$
Prove that
\[ \sum_{cyclic}\frac {1}{\sqrt {ab} \minus{} 1} \leq \frac {3}{2}
\]
:lol:[/quote]

Setting $ a: \equal{} \sqrt {ab} ,b \equal{} \sqrt {bc} ,c: \equal{} \sqrt {ca}$ then $ a^2  \plus{} b^2  \plus{} c^2  \equal{} abc$

And the inequality becomes

$ \frac{1}{{a \minus{} 1}} \plus{} \frac{1}{{b \minus{} 1}} \plus{} \frac{1}{{c \minus{} 1}} \le \frac{3}{2}$

$ \sum {\frac{1}{{a \minus{} 1}}}  \equal{} \frac{{\sum {\left( {b \minus{} 1} \right)\left( {c \minus{} 1} \right)} }}{{\left( {a \minus{} 1} \right)\left( {b \minus{} 1} \right)\left( {c \minus{} 1} \right)}} \equal{} \frac{{\left( {ab \plus{} bc \plus{} ca} \right) \minus{} 2\left( {a \plus{} b \plus{} c} \right) \plus{} 3}}{{abc \minus{} \left( {ab \plus{} bc \plus{} ca} \right) \plus{} a \plus{} b \plus{} c \minus{} 1}}$

Setting $ p \equal{} a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$ then we need to show that

$ 3p^2  \plus{} 7p \minus{} 11q \minus{} 9 \ge 0$

By Schur ineq we have $ q \le \frac{{p^3  \plus{} 9p^2 }}{{18 \plus{} 4p}}$

It suffices us to show that

$ 3p^2  \plus{} 7p \minus{} 11\left( {\frac{{p^3  \plus{} 9p^2 }}{{18 \plus{} 4p}}} \right) \minus{} 9 \ge 0$

$ \Leftrightarrow p^3  \minus{} 17p^2  \plus{} 90p \minus{} 162 \ge 0$

$ \Leftrightarrow \left( {p \minus{} 9} \right)\left( {p^2  \minus{} 8p \plus{} 18} \right) \ge 0$

Which is true by $ p^2  \minus{} 8p \plus{} 18 \ge 0$ and $ p \minus{} 9 \ge 0$

We have done :).

[quote="apollo"]1.Let $ a,b,c \in \mathbb{R}^ \plus{}$ Prove that
\[ \sum_{cyclic}a\sqrt {7a^2 \plus{} 9b^2} \geq \frac {4}{3}(a \plus{} b \plus{} c)^2
\]
[/quote]
By the Cauchy Schwarz Inequality, we have
\[ \sum a\sqrt{7a^2\plus{}9b^2} \ge \frac{1}{4}\sum a(7a\plus{}9b) \equal{}\frac{1}{4}(7\sum a^2 \plus{}9\sum ab)\]
It suffices to prove that
\[ \frac{1}{4}(7\sum a^2 \plus{}9\sum ab) \ge \frac{4}{3}(\sum a)^2\]
\[ \Leftrightarrow 21\sum a^2 \plus{}27\sum ab \ge 16\sum a^2 \plus{}32 \sum ab\]
\[ \Leftrightarrow 5\sum a^2 \ge 5\sum ab\]
which is true. :)

[quote="can_hang2007"][quote="apollo"]1.Let $ a,b,c \in \mathbb{R}^ +$ Prove that
\[ \sum_{cyclic}a\sqrt {7a^2 + 9b^2} \geq \frac {4}{3}(a + b + c)^2
\]
[/quote]
By the Cauchy Schwarz Inequality, we have
\[ \sum a\sqrt {7a^2 + 9b^2} \ge \frac {1}{4}\sum a(7a + 9b) = \frac {1}{4}(7\sum a^2 + 9\sum ab)
\]
It suffices to prove that
\[ \frac {1}{4}(7\sum a^2 + 9\sum ab) \ge \frac {4}{3}(\sum a)^2
\]

\[ \Leftrightarrow 21\sum a^2 + 27\sum ab \ge 16\sum a^2 + 32 \sum ab
\]

\[ \Leftrightarrow 5\sum a^2 \ge 5\sum ab
\]
which is true. :)[/quote]
Nice solution  :lol:
My solution is similar
But Problem 2 this is my solution
[u]sol[/u] substitute $ a=\frac{1}{x^2},b=\frac{1}{y^2},c=\frac{1}{z^2}$
therefore $ x^2+y^2+z^2 =1$
$ \sum_{cyc}\frac{1}{\sqrt{xy}-1} = \frac{xy}{1-xy}$
It's remain to prove $ \frac{xy}{1-xy} \leq \frac{3}{2}$
but It's very old and well known.

[quote="great math"]Apollo, can you solve it clearly?[/quote]
See here : http://www.mathlinks.ro/viewtopic.php?search_id=567651858&t=36873

[quote="cyclic"][quote="can_hang2007"]
By the Cauchy Schwarz Inequality, we have
\[
\sum a\sqrt {7a^2 + 9b^2} \ge \frac {1}{4}\sum a(7a + 9b) [/quote]
Why? Can you explain more clearly, can_hang2007?[/quote]
note that $ 16(7a^2\plus{}9b^2)\geq (7a\plus{}9b)^2$ :wink:

[quote="cyclic"][quote="can_hang2007"]
By the Cauchy Schwarz Inequality, we have
\[
\sum a\sqrt {7a^2 + 9b^2} \ge \frac {1}{4}\sum a(7a + 9b) [/quote]
Why? Can you explain more clearly, can_hang2007?[/quote]
Using Cauchy Schwarz
$ \sqrt {a^2  \plus{} a^2  \plus{} ... \plus{} a^2  \plus{} b^2  \plus{} b^2  \plus{} ... \plus{} b^2 }  \ge (a \plus{} a \plus{} .. \plus{} a \plus{} b \plus{} .. \plus{} b) \equal{} (7a \plus{} 9b)$( 7a and 9b)
I think
$ a{\kern 1pt} _1  \plus{} a_2  \plus{} .. \plus{} a_n  \le \sqrt {n(a_1 ^2  \plus{} a_2 ^2  \plus{} ... \plus{} a_n ^2 )}$  :maybe:  :)

[quote="tronghieu"][quote="cyclic"][quote="can_hang2007"]
By the Cauchy Schwarz Inequality, we have
\[
\sum a\sqrt {7a^2 + 9b^2} \ge \frac {1}{4}\sum a(7a + 9b) [/quote]
Why? Can you explain more clearly, can_hang2007?[/quote]
Using Cauchy Schwarz
$ \sqrt {a^2 \plus{} a^2 \plus{} ... \plus{} a^2 \plus{} b^2 \plus{} b^2 \plus{} ... \plus{} b^2 } \ge (a \plus{} a \plus{} .. \plus{} a \plus{} b \plus{} .. \plus{} b) \equal{} (7a \plus{} 9b)$( 7a and 9b)
[/quote]
Thanks!


I'm not sure about this problem so I ask for help.

If x = 4 is a solution for the equation $a(x-1)^4 + b(x-1)^2 + c = 0$, then a second solution is:

[hide="My answer"]

$x = 1$.

If I put 4 for x, I get:

$81a + 9b + c = 0$ and if a = 1, b = -9 and c = 0, the equation is true.

Back into original equation, then I have:

$(x-1)^4 - 9(x-1)^2 = 0\\
(x-1)^2((x-1)^2 - 9) = 0$

Right side's answer is 4 and the left side is 1.[/hide]

Is this right? Also, is there any sure way to solve this?

hmm ;)  i don't think your proof is 100% right, but your answers may be...

no, $C$ does not have to be $0$. as i said before, your answer is correct but your proof is not. your proof is to risky because the $81a+9b+c=0$ can hold for many $(a,b,c)$ such as $(1,-8,-9)$
so you need a better way to solve this problem and here is how you do it.

let $(x-1)^2$ be $K$. then your equation will be
$a.K^2+bK+c=0$, if $x=4$, then $K=9$. so $K=9$ is one answer to the equation $a(K^2)+bK+c=0$. :arrow: $81a+9b+c=0$
lets $Z$ be the other root of $a(K^2)+bK+c=0$. then
$-b/a=Z+9.$
and $-c/a=9Z$
from the first one you get $Z= -(b/a)+9$
and from the second one you get: $Z= -c/(9a)$
then $-(b/a)-9-(c/(9a))=2Z$, if you get the common denominator you get $Z=(-(8b+81a+c))/18a$. and since you know that $8b+81a+c$ is equal to zero, then $Z$ will be equal to zero.
and if $Z$ is equal to zero then $(x-1)^2$ will be zero which you get [b]$x=1.$ [/b]

I think djshowdown2 is correct... -2 will give the same thing as 4.

[quote="djshowdown2"]Hmm, I don't think so since that holds true if and only if c=0.

[hide]I'm not sure on this, but I was thinking let $k=(x-1)^2$ so you get $ak^2+bk+c$, which one could solve for two solutions k, then use $k=(x-1)^2$ to find all for solutions of the original equation, but since we know x=4 is a solution, then say k=(4-1)^2 = 9 so 9 = (x-1)^2 also has the solution $\fbox{-2}$?[/hide][/quote]

I think what djshowdown2 said makes sense because since the $(x-1)\text{'s}$ are both raised to even powers, it is easy to find another $x$ that will give the same result, ie, $x=-2$--->because $x-1=3 \ \text{or} \ -3$. it's probably what the problem intended. at least that's what i think

EDIT-sorry i see someone already posted the exact same thing

yeah, i made a mistake where is said:
$-b/a=9+Z$.

Silverfalcon, the question wants you to find a second solution, regardless of the values of a, b and c.  That is, you know that 4 is a solution, but you don't know about any other relationships between a, b and c.  So while your choice of a, b and c might make 1 another solution, you don't know for sure that your choice of a, b and c is right.


Functional Analysis
advanced fields
advanced fields unsolved

Let H be complex Hilbert space and T$\in BL(H)$.
If $\langle Tx,x\rangle$ is real for all $x\in H$ then show that T
is self adjoint.

First we prove that if in a complex Hilbert space we have $(Tx,x)=0,\ \forall x$ for some linear operator $T$, then $T=0$.

By expanding $(T(x+y),x+y)=0,\ \forall x,y$ we get $(Tx,y)+(Ty,x)=0,\ \forall x,y\ (*)$. Now apply $(*)$ to $x,iy$ instead of $x,y$. We get $-i(Tx,y)+i(Ty,x)=0,\ \forall x,y$. This, together with $(*)$, shows that $(Tx,y)=0,\ \forall x,y$, which clearly implies that $T\equiv 0$, as desired.

Now for our problem: $(T^*x,x)=(x,Tx)=(Tx,x),\ \forall x$, so we can apply the result above to $T-T^*$ and conclude that it's trivial, i.e. $T$ coincides with $T^*$.


geometry
geometric transformation
reflection
function
analytic geometry
floor function
number theory

Three mirrors are arranged in the shape of an equilateral triangle, with their reflective surfaces pointing inwards.
There is an infinitesimal gap at each vertex of the triangle through which a laser beam may pass.

Label the vertices A, B and C. There are 2 ways in which a laser beam may enter vertex C,
bounce off 11 surfaces, then exit through the same vertex: one way is shown below;
the other is the reverse of that.

[img]http://mathfun84.googlepages.com/laserbeam.gif[/img]

There are 80840 ways in which a laser beam may enter vertex C, bounce off 1000001 surfaces, then exit through the same vertex.

In how many ways can a laser beam enter at vertex C, bounce off n surfaces, then exit through the same vertex?
Or how is it computable in a fast way, if say C is about 1 billion?

Could you tell us the source of the problem?
I think it is very nice if a nice solution exists.
If you made the problem: How did you find out the picture? How did you find out--->
"There are 80840 ways in which a laser beam may enter vertex C, bounce off 1000001 surfaces, then exit through the same vertex."

[quote="Akashnil"]Great problem. I've solved it :)[/quote]

And what is your method?
I've tried to solve it by say calculating the positions of the bounces by simple geometry and seperating two consecutive starting laser hits that after n bounces it returns to the starting position, but this is really inefficient, and we have to check also that we haven't returned to the starting position earlier than n bounces.

Thanks for the hints, I see your method.

[quote="kevinatcausa"]Two suggestions:

1.  A general hint for these sorts of problems is that it's usually easier to work with the diagram formed by repeatedly reflecting the triangle across edges (with all of the reflections of vertex $ A$ also labeled $ A$, etc.).   This has the nifty effect of changing ugly geometric paths into pretty straight lines.

2.  Once you've done this, the question of "Which vertices correspond to paths returning only once to the starting position" and of how to count those vertices becomes a question of number theory rather than geometry or combinatorics.[/quote]

Hi, I'm not sure I understand point 1 of your suggestion. Could you please explain it?

[quote="kevinatcausa"]Two suggestions:

1.  A general hint for these sorts of problems is that it's usually easier to work with the diagram formed by repeatedly reflecting the triangle across edges (with all of the reflections of vertex $ A$ also labeled $ A$, etc.).   This has the nifty effect of changing ugly geometric paths into pretty straight lines.

2.  Once you've done this, the question of "Which vertices correspond to paths returning only once to the starting position" and of how to count those vertices becomes a question of number theory rather than geometry or combinatorics.[/quote]
This is exactly what I did. I'll post a solution later.
And 2 requires the Euler's phi function of $ n\plus{}1$ and greatest integer function etc.

[quote="Akashnil"]
This is exactly what I did. I'll post a solution later.
And 2 requires the Euler's phi function of $ n \plus{} 1$ and greatest integer function etc.[/quote]

I'm really curious to see that!

Yama,

To get a sense of what's going on, let's look at what happens on a square group of mirrors as opposed to an equilateral triangle.  We can view the square as being embedded in the Euclidean plane, with vertex $ A$ at $ (0,0)$, $ B$ at $ (0,1)$, $ C$ at $ (1,1)$ and $ D$ at $ (1,0)$.  If we were to reflect the square across the line $ BC$, point $ A$ would end up at $ (0,2)$ and point $ D$ at $ (1,2)$.

If we fire a laser from $ A$ towards $ (\frac {1}{2},1)$ (the middle of the opposite side).  It's pretty easy to see geometrically that it will bounce off and hit $ D$.  Now suppose that we fired the laser, but we removed all the mirrors first.  The laser beam's path would just be a straight line, and the first integer point it would hit would be $ (1,2)$.  Note that that point is labeled $ D$ (the same point it hit when the mirrors were still around).

This is not a coincidence!  If you work it out geometrically, the path the mirrorless laser beam takes in the reflected square one it passes through $ (\frac {1}{2},1)$ is the exact reflection of the path it took before we removed the mirror, so the final vertex will be an exact reflection of the final vertex before we removed the mirror.  An alternative "physics-y" explanation: From our point of view the laser is bouncing off the mirrors.  By reflecting the square instead of the beam we're effectively moving to the laser beam's frame of reference, where the mirrors are bouncing off of it instead.  It should exit at the same vertex either way.

This works just as well if the laser is bouncing off of several walls in succession; you just have to reflect the square several times.

Here's an example of a problem that can be solved by other techniques, but is particularly natural this way.  The solution is also relevant to the triangular problem in the original post:  [b]Prove that in the square grid the laser beam never exits at $ A$[/b]

Hint:

[hide]The reflections of $ A$ all have both their coordinates even.  [/hide]

@kevinatcausa:

Thank you, for all this explanation. I realized I read "vertex" instead of "edge" and therefore I was making reflections on vertexes and not edges and it is a bit hard like that, hehe.
I get the point now. You just draw reflections of that triangle and when the "snake" you have drawn lets you draw a straight line within itself that connects vertex C of original triangle to vertex C of the last reflections, you got a correct path.

What I don't understand still, is how to get to the problem's solution, and what the totient function has to do with it...

Any help will be very appreciated.
Cheers

Nevermind, I solved it.

Thank you for your help.
Cheers

Does my answer match with yours?
Number of reflections=$ n$
$ n\equiv 1\ (\mod 6)\implies 2\lfloor \frac{\phi (\frac{n\plus{}3}{2})}{6}\rfloor$
$ n\equiv \minus{}1\ (\mod 6)\implies \minus{}2\lfloor \minus{}\frac{\phi (\frac{n\plus{}3}{2})}{6}\rfloor$
$ 0$ in all other cases.

Honestly I don't know. I solved without the totient function, because I didn't figure out how to use it.
I just made a grid of triangles and got the coordinates of the desired reflections of C.
Checked if there are points with integer coordinates on the line that connects C to one of its reflections and that's all.
Your solution seems to be much faster and elegant, but honestly I don't know how to get there.

Anyway, for n=12017639147 I got a different value from the one your formula gets. It's "almost" the same, but mine ends in ...624. Your is a little more evilish and ends in ...666. (o;

Cheers

[b]Please post your formula.[/b]
Does your formula tell that there's nothing if $ n\not \equiv 1$ or $ \minus{} 1\ (\mod 6)$?
I also made a lattice of equilateral triangles. The totient function comes up because there must not be another lattice point on the line connecting the origin and the reflection point. And thus the coordinates of the reflection-point must be relatively prime. And how was your method? And there's no reason to believe my method is more elegant.

And one more thing, if my formula is correct $ n \equal{} 1$ is an exceptional case.

Well, when I told you that your solution was more elegant I meant you have found a closed formula for the problem, while I didn't.
I made these considerations: The initial angle of the laser must be between 60 and 90 degrees (60, 90, excluded of course).
Now, if you mirror the original triangle up, then right, then up, then right, and so on, you find that when r is the number of reflections, you get x=(r-1)/2 and y=(r+3)/2, coordinates of the upper-right reflected point. Since when y is odd you find C points at x=3 mod 6, you just check in a loop every number of the form 6n+3. You check if it doesn't have common factors with y. If it doesn't, then they are coprime, and therefore you've found a good path. Stop the loop when 6n+3>x=(r-1)/2.
Double the result to consider upper left part of the cartesian plane and you're done.

I'm sorry I can't answer your questions about special cases, since I don't have a closed formula to solve this one.

Best regards.
Yama

The [url=http://projecteuler.net/index.php?section=problems&id=202]problem[/url] appeared in projecteuler.net last week
It is really **RIDICULOUS** to post and discuss the solution of the problem here.
It spoils the fun and the point of the [url=http://projecteuler.net/index.php?section=scores]score table[/url]


Triangle ABC is inscribed in a circle with center O.  A circle with center I is inscribed in trangle ABC. AI is drawn, and extended to intersect the larger circle at D.  Prove that D is the circumcenter of triangle CIB.

Proofs don't belong here. Someone move this to Intermediate Topics.

Wow, I solved this problem!

Download the picture for an image (it will pop up as a new tab).

[hide="Click here for the Solution"]I will prove that $ DB = DI$. The proof that $ DI = DC$ is analogous. Together, these statements prove that $ D$ is the circumcenter of $ \triangle CIB$.

I is the intersection of the angle bisectors of $ \triangle ABC$.

Let $ m\angle BCA = 2\theta$ and $ m\angle BAC = 2\alpha$. Let $ E$ be the intersection between segment $ BI$ and the angle bisector of $ \angle BDI$.

Since they intercept the same arc,
$ 2\theta = m\angle BCA = m\angle BDA$

$ \boxed{\theta = m\angle BDE}$

Since they intercept the same arc,
$ \alpha = m\angle ADC = m\angle DBC$
$ \boxed{\alpha = m\angle DBC}$

Now, our goal is to show that $ m\angle EBD = 90^{\circ}-\theta$.
[hide="If we showed that, then the result will follow because"]
If we showed that,
$ m\angle BED = 90^{\circ}$
$ m\angle IED = 90^{\circ}$

So by ASA triangle congruency, $ \triangle BED \cong \triangle IED$, proving that $ DB = DI$.[/hide]

Since ABC is a triangle (duh),
$ m\angle ABC = 180^{\circ}-2\alpha-2\theta$

Since $ AI$ is the angle bisector of $ m\angle ABC$,
$ m\angle IBC = (1/2)m\angle ABC = 90^{\circ}-\alpha-\theta$
$ \boxed{m\angle EBC = 90^{\circ}-\alpha-\theta}$

Adding the last two boxed statements,
$ m\angle EBD = 90^{\circ}-\theta$.

That's what we needed to show. $ \blacksquare$[/hide]


trigonometry probability MATHCOUNTS AMC USA(J)MO USAMO AIME \/closed

My school is competing in the GBML (Greater Boston Mathematical League) and I need help finding a book for it. Its covers algebra 2, trig, number theory, and a lot of factoring. Does AoPS provide a book that covers factoring and number thoery while touching the other two subjects?

thanks

You're looking for the AoPS Classic books (Volume 1 and Volume 2).

Sry but would the classic books be:

The Art of Problem Solving, Volume 1: the Basic
The Art of Problem Solving, Volume 2: and Beyond

Yes.

However, I would say the AoPS book 1 is plenty. It also covers counting and probability, I believe.

i have volume 1 and it does not go in dept to what i need to know. It requires algebra II factoring which the basics does not cover..does "beyond" cover this?

like past difference of squares and stuff like that

Beyond covers that... and much more beyond. If you are just looking for an algebra book, the intermideate algebra book should be good too.

Aops Volume 1 Description: A classic problem solving textbook for students in grades 7-10. Used by many successful MATHCOUNTS and AMC participants.

Aops Volume 2 Description: A classic problem solving textbook for students in grades 9-12. Used by many successful AMC participants, USAMO qualifiers and Math Olympiad Program invitees.

Intermediate Algebra Description: A comprehensive textbook covering Algebra 2 and topics in PreCalculus. This book is the follow-up to the acclaimed Introduction to Algebra textbook. In addition to offering standard Algebra 2 and PreCalculus curriculum, the text includes advanced topics such as those problem solving strategies required for success on the AMC and AIME competitions.

Intermediate Counting and Probability Description: An intermediate textbook in counting and probability for students in grades 9-12, containing topics such as inclusion-exclusion, recursion, conditional probability, generating functions, graph theory, and more.

You can find all of these books (plus more) and their descriptions at:

http://www.artofproblemsolving.com/Books/AoPS_B_Texts_FAQ.php


The average (mean) of $ 20$ numbers is $ 30$, and the average of $ 30$ other numbers is $ 20$. What is the average of all $ 50$ numbers?

$ \textbf{(A)}\ 23 \qquad
\textbf{(B)}\ 24 \qquad
\textbf{(C)}\ 25 \qquad
\textbf{(D)}\ 26 \qquad
\textbf{(E)}\ 27$

[hide="Click for solution"]
The sum of the numbers is $ 20 \times 30\plus{}30 \times 20\equal{}1200$, so the average is $ \frac{1200}{50}\equal{}24$, or $ \boxed{\textbf{(B)}}$.
[/hide]


Let $ f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that for all $ x,y \in \mathbb{R}$, $ f(x^3\plus{}y^3)\equal{}(x\plus{}y)(f(x)^2\minus{}f(x)f(y)\plus{}f(y)^2).$ Prove that for all $ x \in \mathbb{R}, f(1996x)\equal{}1996f(x)$.

[quote="moldovan"]Let $ f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that for all $ x,y \in \mathbb{R}$, $ f(x^3 \plus{} y^3) \equal{} (x \plus{} y)(f(x)^2 \minus{} f(x)f(y) \plus{} f(y)^2).$ Prove that for all $ x \in \mathbb{R}, f(1996x) \equal{} 1996f(x)$.[/quote]

Let $ P(x,y)$ be the assertion $ f(x^3 \plus{} y^3) \equal{} (x \plus{} y)(f(x)^2 \minus{} f(x)f(y) \plus{} f(y)^2)$

$ P(1, \minus{} 1)$ $ \implies$ $ f(0) \equal{} 0$
$ P(\sqrt [3]x,0)$ $ \implies$ $ f(x) \equal{} \sqrt [3]xf(x)^2$. So $ f(x)$ has the same sign as $ x$

Suppose then $ f(ax) \equal{} af(x)$ $ \forall x$ for some real $ a > 0$.
$ P(\sqrt [3]ax,0)$ $ \implies$ $ f(ax^3) \equal{} \sqrt [3]axf(\sqrt [3]ax)^2$ $ \implies$ $ af(x^3) \equal{} \sqrt [3]axf(\sqrt [3]ax)^2$
$ P(x,0)$ $ \implies$ $ f(x^3) \equal{} xf(x)^2$ and so, with line above : $ axf(x)^2 \equal{} \sqrt [3]axf(\sqrt [3]ax)^2$
So $ f(\sqrt [3]ax)^2 \equal{} \sqrt [3]{a^2}f(x)^2$ and, since $ a > 0$ and we know that $ f(x)$ has the same sign as $ x$ : $ f(\sqrt [3]ax) \equal{} \sqrt [3]{a}f(x)$

$ P(\sqrt [3]ax,x)$ $ \implies$ $ f((a \plus{} 1)x^3) \equal{} (\sqrt [3]a \plus{} 1)x(\sqrt [3]{a^2}f(x)^2 \minus{} \sqrt [3]af(x)^2 \plus{} f(x)^2)$ $ \equal{} (a \plus{} 1)xf(x)^2$
And, since $ P(x,0)$ implies $ f(x^3) \equal{} xf(x)^2$ we get $ f((a \plus{} 1)x^3) \equal{} (a \plus{} 1)f(x^3)$

So, if $ f(ax) \equal{} af(x)$ $ \forall x$, then $ f((a \plus{} 1)x) \equal{} (a \plus{} 1)f(x)$ $ \forall x$

And so, since $ f(1\cdot x) \equal{} 1\cdot f(x)$, we get $ f(nx) \equal{} nf(x)$ $ \forall x$, $ \forall n\in\mathbb N$

Hence the result.


geometry
circumcircle
algebra
polynomial
geometry proposed

Let $\mathcal{P}=P_1P_2 \ldots P_n$ be a convex $n$-gon, and let $\mathcal{I}$ be its (closed) interior. By the Weierstrass principle, the minimum
\[
d(\mathcal{P}) :=  \min_{X \in \mathcal{I}} \max_{1 \leq i \neq j \leq n} \frac{P_iX}{P_jX}
\]
is attained at some point of $\mathcal{I}$. Call every point $X_0$ at which this minimum is attained a [i]pseudo-circumcenter[/i] of the convex $n$-gon $\mathcal{P}$.

[b]Problem. [/b] (i) Show that every convex $n$-gon has at least one and at most finitely many pseudo-circumcenters;
(ii) For each $n \geq 3$, determine the maximal possible number of pseudo-circumcenters that a convex $n$-gon can have;
(iii)* Foe each $n \geq 3$, determine
\[
\sup \big\{ d(\mathcal{P}) \big\},
\]
where the supremum is over all convex $n$-gons $\mathcal{P}$.

Enjoy! (Yes, I know that (i) is trivial, but I wanted to make this problem interesting to beginners as well. )

--Vesselin

[* means that I haven't (yet) solved this part. ]

Just the trivial, as you call it, (i), up until now.. :(

We can write $d(\mathcal P)=\min \frac{\max XP_i}{\min XP_j}$. Consider now the set of those pseudo-circumcenters $X$ for which $\frac{\max XP_i}{\min XP_j}=\frac{XP_{i_0}}{XP_{j_0}}$. If we assume that $XP_i<XP_{i_0},\ \forall i\ne i_0$ and $XP_j>XP_{j_0},\ \forall j\ne j_0$, then th same is true when we replace $X$ with $Y$ in a small region around $X$, and we can find such a $Y$ in the region (consider the region a small disk, for example) s.t. $\frac{YP_{i_0}}{YP_{j_0}}<\frac{XP_{i_0}}{XP_{j_0}}$, meaning that $X$ was not a pseudo-circumcenter in the first place, contradiction.

We have shown that either $XP_{i_0}$ or $XP_{j_0}$ is achieved twice (i.e. for two vertices of the polygon), and from here it's easy to show that for the fixed $i_0,j_0$, there are only finitely many pseudo-circumcenters (they are intersections between an Apollonian circle of triangle $XP_{i_0}P_{j_0}$ and finitely many other Apollonian circles, corresponding to a few other triangles formed by $X$ and two vertices of the polygon).

Did you conjecture anything for the third problem, vess? I think there are constructions which show that for a fixed $n$, we have $\sup[d(\mathcal P)]\ge n$, but I couldn't go further up until now.

Let $d_n$ denote the supremum. As you remarked, we obviously have $d_n \geq n$, with equality when $n=3$. I couldn't invent better examples for the case $n=5$; however, I have an example showing that
\[
d_4 \geq \frac{1}{\sqrt{5}-2} = 4.236\ldots.
\]
I think I can show that this bound gives the value of $d_4$. On the other hand, I think that $d_n < cn$ for some constant $c$, though I don't know how to show this. It also seems probable that $d_5=5$, and that $d_n>n$ strictly when $n$ is even.

Well, I also had some problems with the case $n$ even, and my example sort of "forced" $d(\mathcal P)=n$ for those specific polygons $\mathcal P$, so I'm not surprised to hear that there are better examples. Actually, all my examples were degenerate polynomials (a segment with some points on it :)). This degenerate case is not covered by the problem, but it's just as good :).

Thanks for the answer.


function
floor function
inequalities
algebra
combinatorics

[quote="1234567890"][quote="Rushil"]Find the number of ordered triples $(x,y,z)$ of non-negative integers satisfying
(i) $x \leq y \leq z$
(ii) $x + y + z \leq 100.$[/quote]

I am bad at counting, but let me try:
[hide]Without (i), there are
$6(\binom{2}{2} + \binom{3}{2} + \binom{4}{2}.......\binom{99}{2})$ ways to arrange the numbers. Dividing by the 6 permutations because of the restrictions of (i), you have:
 $\binom{2}{2} + \binom{3}{2} + \binom{4}{2}.......\binom{99}{2}$ ways to arrange the $x,y,z$.

Now using the hockey-stick identity, we have: ${\binom{100}{3}}$ ways to arrange the triples.[/hide][/quote]

I think that this solution is wrong !
In fact a computer calculation show that the number of ordered triples $(x,y,z)$ satisfyng (i), (ii) is 30787 .
Can anyone justify this result ?

Leon

[quote="1234567890"]Oops, I also checked my work and found the error. Haven't solved it, though. Any help?[/quote]

Since $0 \le x \le y \le z$ , there are integer numbers $a,b,c \ge 0$ such that :

$x = a{\rm{ }},{\rm{ }}y = a + b{\rm{ }},{\rm{ }}z = a + b + c$

and the relation $x + y + z \le 100$  becomes:

$3a + 2b + c \le 100$ with $a,b,c \ge 0$           (*)

To find the number of solutions $\left( {a,b,c} \right)$ of (*) I will use the following

[color=blue][b]Lemma.[/b][/color] The number  $a_n$  of solutions $\left( {a,b,c} \right)$ of  the Diophantine equation $3a + 2b + c = n$ ,
with $a,b,c \ge 0$  is given by

$a_0 = 1$

$a_n = \frac{1}{{72}}\left[ {6n^2 + 36n + 47 + 9\left( { - 1} \right)^n + 8\theta _n } \right]$        (1)

where $\theta _n = 2$ if $n \equiv 0\left( {\bmod {\rm{ }}3} \right)$ and $\theta _n = -1$ if $n\not \equiv 0\left( {\bmod {\rm{ }}3} \right)$.

[color=blue][b]Proof.[/b][/color] The generating function of the sequence $a_n$ is

$\left( {1 + x^3 + x^6 } \right)\left( {1 + x^2 + x^4 } \right)\left( {1 + x + x^2 } \right) = \frac{1}{{1 - x^3 }} \cdot \frac{1}{{1 - x^2 }} \cdot \frac{1}{{1 - x}}$      (2)

The right side of (2) can be written as sum of partial fractions

$\frac{{{\textstyle{{17} \over {72}}}}}{{1 - x}} + \frac{{{\textstyle{1 \over 4}}}}{{\left( {1 - x} \right)^2 }} + \frac{{{\textstyle{1 \over 6}}}}{{\left( {1 - x} \right)^3 }} + \frac{{{\textstyle{1 \over 8}}}}{{1 + x}} + \frac{{{\textstyle{1 \over 9}}}}{{1 - \omega x}} + \frac{{{\textstyle{1 \over 9}}}}{{1 - \omega ^2 x}}$

where $\omega = e^{{{2\pi i} \mathord{\left/ {\vphantom {{2\pi i} 3}} \right. \kern-\nulldelimiterspace} 3}}$ is a complex cube root of 1 , satisfying $\omega + \omega ^2 = - 1$ .

From the well known relations

$\begin{array}{l} \frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {x^n } \\ \frac{1}{{\left( {1 - x} \right)^2 }} = \sum\limits_{n = 0}^\infty {\left( {n + 1} \right)x^n } \\ \frac{1}{{\left( {1 - x} \right)^3 }} = \sum\limits_{n = 0}^\infty {\frac{{\left( {n + 1} \right)\left( {n + 2} \right)}}{2}x^n } \\ \frac{1}{{1 + x}} = \sum\limits_{n = 0}^\infty {\left( { - 1} \right)^n x^n } \\ \frac{1}{{1 - \omega x}} = \sum\limits_{n = 0}^\infty {\omega ^n x^n } \\ \frac{1}{{1 - \omega ^2 x}} = \sum\limits_{n = 0}^\infty {\omega ^{2n} x^n } \\ \end{array}$

we have :

\[ \begin{array}{l} \sum\limits_{n = 0}^\infty {a_n x^n = } \\ = \frac{{17}}{{72}}\left( {\sum\limits_{n = 0}^\infty {x^n } } \right) + \frac{1}{4}\left( {\sum\limits_{n = 0}^\infty {\left( {n + 1} \right)x^n } } \right) + \frac{1}{6}\left( {\sum\limits_{n = 0}^\infty {\frac{{\left( {n + 1} \right)\left( {n + 2} \right)}}{2}x^n } } \right) + \frac{1}{8}\left( {\sum\limits_{n = 0}^\infty {\left( { - 1} \right)^n x^n } } \right) + \frac{1}{9}\left( {\sum\limits_{n = 0}^\infty {\left( {\omega ^n + \omega ^{2n} } \right)x^n } } \right) \\ \end{array}  \]

Collecting all terms and equating the coefficients we get the formula (1).  $\blacksquare$

[color=blue][b]Corollary.[/b][/color] The number S of solutions $\left( {a,b,c} \right)$ of  $3a + 2b + c \le n$ with $a,b,c \ge 0$ is

\[ \begin{array}{l} S = 1 + \sum\limits_{k = 1}^n {a_k } = 1 + \frac{1}{{72}}\sum\limits_{k = 1}^n {\left( {6k^2 + 36k + 47 + 9\left( { - 1} \right)^k + 8\theta _k } \right)} = \\ = 1 + \frac{1}{{72}}\sum\limits_{k = 1}^n {\left[ {6\frac{{k\left( {k + 1} \right)\left( {2k + 1} \right)}}{6} + 36\frac{{k\left( {k + 1} \right)}}{2} + 47k} \right] + \frac{1}{{72}}} \sum\limits_{k = 1}^n {\left[ {9\left( { - 1} \right)^k + 8\theta _k } \right]} = \\ = 1 + \frac{1}{{72}}\left( {2n^3 + 21n^2 + 66n} \right) + \frac{1}{{72}}\sum\limits_{k = 1}^n {\left[ {9\left( { - 1} \right)^k + 8\theta _k } \right]} \\ \end{array}  \]
For $n = 100$ we find the number of solution of (*) : $S=30787$

[quote="scorpius119"]I get a completely different answer...   :huh:

[hide]From $0\leq x\leq y\leq z$, $0\leq y-x\leq z-x$. The equation can be written $(y-x)+(z-x)=100-3x$. Then if $N(k)$ denotes the number of nonnegative integer solutions $(a,b)$ with $a\leq b$ and $a+b=k$, the number of ordered triples is given by the sum $\sum_{x=0}^{33} N(100-3x)$.

Now finding the formula for $N(k)$: $a\leq \frac{a+b}{2}=\frac{k}{2}\leq \left\lfloor \frac{k}{2}\right\rfloor$. Since each value of $a$ between $0$ and $\left\lfloor\frac{k}{2}\right\rfloor$ inclusive has exactly one solution for $b$, $N(k)=1+\left\lfloor\frac{k}{2}\right\rfloor$.

The result sum becomes $\sum_{x=0}^{33} 1+\left\lfloor \frac{100-3x}{2}\right\rfloor =\sum_{x=0}^{33} 51+\left\lfloor \frac{-3x}{2}\right\rfloor=(51*34)+\sum_{x=0}^{33} \left\lfloor \frac{-3x}{2}\right\rfloor$. Separating into even and odd cases, this is $(51*34)+\left( \sum_{x=0}^{16} \left\lfloor \frac{-3(2x)}{2}\right\rfloor +\left\lfloor \frac{-3(2x+1)}{2}\right\rfloor \right)=(51*34)-\left( \sum_{x=0}^{16} 6x+2\right)=(50*34)-(3)(16)(17)=(50*34)-(24*34)=26*34=884$

This method gives an answer of 884, which is much less than the other proposed answers...[/hide][/quote]

Dear Scorpius 119, your calculation is wrong. I have verified by computer that the solution is 30787 (see post #3).
On the other hand your method applied to the analogous problem with $n=10$

\[ x + y + z \le 10  \]

with $0 \le x \le y \le z$ gives:

\[ \sum\limits_{x = 0}^3 {\left( {1 + \left[ {\frac{{10 - 3x}}{2}} \right]} \right)} = 4 + \left[ 5 \right] + \left[ {\frac{7}{2}} \right] + \left[ 2 \right] + \left[ {\frac{1}{2}} \right] = 14  \]

but we can easily see that the number of solutions is greater than 14.

Leon

oops i did $x+y+z=100$. To apply my solution to the inequality, I could get the formula for sum of three variables for general $n$ and sum from $n=0$ to $100$. This produces an extremely confusing double sum with floors all over the place, though...

[quote="Leon"][quote="1234567890"]Oops, I also checked my work and found the error. Haven't solved it, though. Any help?[/quote]

Since $0 \le x \le y \le z$ , there are integer numbers $a,b,c \ge 0$ such that :

$x = a{\rm{ }},{\rm{ }}y = a+b{\rm{ }},{\rm{ }}z = a+b+c$

and the relation $x+y+z \le 100$  becomes:

$3a+2b+c \le 100$ with $a,b,c \ge 0$           (*)

To find the number of solutions $\left({a,b,c}\right)$ of (*) I will use the following

[color=blue][b]Lemma.[/b][/color] The number  $a_{n}$  of solutions $\left({a,b,c}\right)$ of  the Diophantine equation $3a+2b+c = n$ ,
with $a,b,c \ge 0$  is given by

$a_{0}= 1$

$a_{n}= \frac{1}{{72}}\left[{6n^{2}+36n+47+9\left({-1}\right)^{n}+8\theta_{n}}\right]$        (1)

where $\theta_{n}= 2$ if $n \equiv 0\left({\bmod{\rm{ }}3}\right)$ and $\theta_{n}=-1$ if $n\not \equiv 0\left({\bmod{\rm{ }}3}\right)$.

[color=blue][b]Proof.[/b][/color] The generating function of the sequence $a_{n}$ is

$\left({1+x^{3}+x^{6}}\right)\left({1+x^{2}+x^{4}}\right)\left({1+x+x^{2}}\right) = \frac{1}{{1-x^{3}}}\cdot \frac{1}{{1-x^{2}}}\cdot \frac{1}{{1-x}}$      (2)

The right side of (2) can be written as sum of partial fractions

$\frac{{{\textstyle{{17}\over{72}}}}}{{1-x}}+\frac{{{\textstyle{1 \over 4}}}}{{\left({1-x}\right)^{2}}}+\frac{{{\textstyle{1 \over 6}}}}{{\left({1-x}\right)^{3}}}+\frac{{{\textstyle{1 \over 8}}}}{{1+x}}+\frac{{{\textstyle{1 \over 9}}}}{{1-\omega x}}+\frac{{{\textstyle{1 \over 9}}}}{{1-\omega^{2}x}}$

where $\omega = e^{{{2\pi i}\mathord{\left/{\vphantom{{2\pi i}3}}\right. \kern-\nulldelimiterspace}3}}$ is a complex cube root of 1 , satisfying $\omega+\omega^{2}=-1$ .

From the well known relations

$\begin{array}{l}\frac{1}{{1-x}}= \sum_{n = 0}^\infty{x^{n}}\\ \frac{1}{{\left({1-x}\right)^{2}}}= \sum_{n = 0}^\infty{\left({n+1}\right)x^{n}}\\ \frac{1}{{\left({1-x}\right)^{3}}}= \sum_{n = 0}^\infty{\frac{{\left({n+1}\right)\left({n+2}\right)}}{2}x^{n}}\\ \frac{1}{{1+x}}= \sum_{n = 0}^\infty{\left({-1}\right)^{n}x^{n}}\\ \frac{1}{{1-\omega x}}= \sum_{n = 0}^\infty{\omega^{n}x^{n}}\\ \frac{1}{{1-\omega^{2}x}}= \sum_{n = 0}^\infty{\omega^{2n}x^{n}}\\ \end{array}$

we have : \[\begin{array}{l}\sum_{n = 0}^\infty{a_{n}x^{n}= }\\ = \frac{{17}}{{72}}\left({\sum_{n = 0}^\infty{x^{n}}}\right)+\frac{1}{4}\left({\sum_{n = 0}^\infty{\left({n+1}\right)x^{n}}}\right)+\frac{1}{6}\left({\sum_{n = 0}^\infty{\frac{{\left({n+1}\right)\left({n+2}\right)}}{2}x^{n}}}\right)+\frac{1}{8}\left({\sum_{n = 0}^\infty{\left({-1}\right)^{n}x^{n}}}\right)+\frac{1}{9}\left({\sum_{n = 0}^\infty{\left({\omega^{n}+\omega^{2n}}\right)x^{n}}}\right) \\ \end{array}\] Collecting all terms and equating the coefficients we get the formula (1).  $\blacksquare$

[color=blue][b]Corollary.[/b][/color] The number S of solutions $\left({a,b,c}\right)$ of  $3a+2b+c \le n$ with $a,b,c \ge 0$ is \[\begin{array}{l}S = 1+\sum_{k = 1}^{n}{a_{k}}= 1+\frac{1}{{72}}\sum_{k = 1}^{n}{\left({6k^{2}+36k+47+9\left({-1}\right)^{k}+8\theta_{k}}\right)}= \\ = 1+\frac{1}{{72}}\sum_{k = 1}^{n}{\left[{6\frac{{k\left({k+1}\right)\left({2k+1}\right)}}{6}+36\frac{{k\left({k+1}\right)}}{2}+47k}\right]+\frac{1}{{72}}}\sum_{k = 1}^{n}{\left[{9\left({-1}\right)^{k}+8\theta_{k}}\right]}= \\ = 1+\frac{1}{{72}}\left({2n^{3}+21n^{2}+66n}\right)+\frac{1}{{72}}\sum_{k = 1}^{n}{\left[{9\left({-1}\right)^{k}+8\theta_{k}}\right]}\\ \end{array}\] For $n = 100$ we find the number of solution of (*) : $S=30787$[/quote]
wow! is there no simpler solution?  :huh:

[quote="Rushil"]Find the number of ordered triples $(x,y,z)$ of non-negative integers satisfying
(i) $x \leq y \leq z$
(ii) $x+y+z \leq 100.$[/quote]
I should mention this is a much simpler problem than one might think it is. Also, please keep in mind that this is a Regional Mathematical Olympiad (RMO) problem; so, it shouldn't be too hard to do.

As a point of departure, for people who might not know this, for the purpose of olympiad contest, India is divided into several regions and each region conducts its own RMO. This means that the question sets differ from one region to another even for the same year. Around top 30 students from each region then proceed to the next level: the Indian National Math Olympiad (INMO). The top 30 contestants based on the INMO are then invited to a summer camp where they are provided instruction and training for a month. Finally, 6 participants are then chosen to represent India in the IMO.

Sorry for the small digression.

Here is a simple solution, which I am pretty sure is correct.

$x=0:$ Possible solutions for $(y,z)$ are $(0,100), (1,99), (2,98),...,(50,50),$i.e. 51 solutions.
$x=1:$ Possible solutions for $(y,z)$ are $(1,98),(2,97),(3,96),...,(49,51), i.e$ 49 solutions.
$x=2:$ Possible solutions for $(y,z)$ are $(2,96),(3,95),(4,94),...,(49,49), i.e.$ 48 solutions.
$x=3:$ Possible solutions for $(y,z)$ are $(3,94),(4,93),...,(48,49),i.e.$ 46 solutions.
$x=4:$ Possible solutions for $(y,z)$ are $(4,92),(5,91),...,(48,48),i.e.$ 45 solutions.
$...$
$...$
$x=33:$ The only solution for $(y,z)$ is $(33,34).$
If $x \geq 34,$ then there are no solutions for $(y,z)$ because then at least one of $y$ or $z$ will be less than $x,$ contradicting condition $(ii).$

So, the number of solutions for $(x,y,z)$ satisfying the two given conditions
$=(51+49)+(48+46)+(45+44)+(43+41)...+(3+1)$ (34 terms in all)

Someone, please compute this. (Or, I will do this later.)

[quote="FieryHydra"][quote="Rushil"]Find the number of ordered triples $ (x,y,z)$ of non-negative integers satisfying
(i) $ x \leq y \leq z$
(ii) $ x \plus{} y \plus{} z \leq 100.$[/quote]
[hide]I should mention this is a much simpler problem than one might think it is. Also, please keep in mind that this is a Regional Mathematical Olympiad (RMO) problem; so, it shouldn't be too hard to do.

As a point of departure, for people who might not know this, for the purpose of olympiad contest, India is divided into several regions and each region conducts its own RMO. This means that the question sets differ from one region to another even for the same year. Around top 30 students from each region then proceed to the next level: the Indian National Math Olympiad (INMO). The top 30 contestants based on the INMO are then invited to a summer camp where they are provided instruction and training for a month. Finally, 6 participants are then chosen to represent India in the IMO.

Sorry for the small digression.

Here is a simple solution, which I am pretty sure is correct.

$ x \equal{} 0:$ Possible solutions for $ (y,z)$ are $ (0,100), (1,99), (2,98),...,(50,50),$i.e. 51 solutions.
$ x \equal{} 1:$ Possible solutions for $ (y,z)$ are $ (1,98),(2,97),(3,96),...,(49,51), i.e$ 49 solutions.
$ x \equal{} 2:$ Possible solutions for $ (y,z)$ are $ (2,96),(3,95),(4,94),...,(49,49), i.e.$ 48 solutions.
$ x \equal{} 3:$ Possible solutions for $ (y,z)$ are $ (3,94),(4,93),...,(48,49),i.e.$ 46 solutions.
$ x \equal{} 4:$ Possible solutions for $ (y,z)$ are $ (4,92),(5,91),...,(48,48),i.e.$ 45 solutions.
$ ...$
$ ...$
$ x \equal{} 33:$ The only solution for $ (y,z)$ is $ (33,34).$
If $ x \geq 34,$ then there are no solutions for $ (y,z)$ because then at least one of $ y$ or $ z$ will be less than $ x,$ contradicting condition $ (ii).$

So, the number of solutions for $ (x,y,z)$ satisfying the two given conditions
$ \equal{} (51 \plus{} 49) \plus{} (48 \plus{} 46) \plus{} (45 \plus{} 44) \plus{} (43 \plus{} 41)... \plus{} (3 \plus{} 1)$ (34 terms in all)

Someone, please compute this. (Or, I will do this later.)[/quote][/hide]

ur answer is wrong
 :(  :maybe:
i think u have made the mistake of assuming $ x\plus{}y\plus{}z\equal{}100$

[quote="10000th user"]Is anyone able to provide a simpler and definite solution to settle this problem from the Resources?[/quote]
this problem can be done in a very nice way

on checking we get that if $ x\equal{}0$ $ \rightarrow z$ has (i.e total no. of solutions)$ 51 \cdot 51$ solutions
then if $ x\equal{}1$ $ \rightarrow z$ has (i.e total no. of solutions)$ 49 \cdot 48$ solutions
if $ x\equal{}2$ $ \rightarrow z$ has $ 48 \cdot 48$ solutions
if $ x\equal{}3$ $ \rightarrow z$ has (i.e total no. of solutions)$ 46 \cdot 47$ solutions
$ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots$
if $ x\equal{}32$ $ \rightarrow z$ has $ 3 \cdot 3$ solutions
if $ x\equal{}33$ $ \rightarrow z$ has $ 1 \cdot 2$ solutions
If $ x \geq 34,$ then there are no solutions for $ (y,z)$ because then at least one of $ y$ or $ z$ will be less than $ x,$ contradicting condition $ (ii).$
therefore the pattern we find is that for even $ x$ $ \rightarrow$ total no. of solutions =$ 51^2 \plus{} 48^2 \plus{} \cdots \plus{} 3^2$(i.e $ 51^2 \plus{}(51\minus{}3)^2 \plus{} \cdots \plus{} (51\minus{} 16 \cdot 3)^2$
and for odd $ x$ $ \rightarrow$ total no. of solutions = $ 49 \cdot 48 \plus{} 47 \cdot 46 \plus{} 43 \cdot 44 \plus{}\cdots \plus{} 1 \cdot 2$(i.e the product of the consecutive nos. after the square of the previous one)

Summing up....... :) we are done
(Phew..finally this question has got a good answer)

The solutions to $ a \plus{} 2b \plus{} 3c \le 100$ are the lattice points of a tetrahedron with side lengths $ 100, 50, \frac{100}{3}$, so it should in principle be possible to obtain an answer using [url=http://en.wikipedia.org/wiki/Ehrhart_polynomial]Ehrhardt polynomials[/url].  In any case, the theory of counting such points is much more interesting than the generating functions computation alone would suggest :)

[quote=shivangjindal]Its , $\frac{\dbinom{103}{3}-3\cdot 51^2+2(34)}{6}+3\cdot 51^2+34 = 30787$. I found this expression by calculating , when all of $a,b,c$ are unequal(where , $a<b<c$), when two are equal and all three are equal. The first part, i did by finding , number of solutions of $a+b+c+d=100$ where $a \neq b \neq c$ , then dividing it by $3!=6$ since we want , $a<b<c$ . Rest can be done similarly. This is much smaller method, but it will not be practical idea to use this method when there are more variables.[/quote]

What do you mean by "rest can be done similarly"? please elaborate.. why were you adding $3.51^2+34$


Prove that if $ x^2 \plus{}y^2\equal{}z^2$ then $ xyz \equiv 0 (mod60)$

Please explain it very clear and easy. Thanks :)  :lol:

Note that $ 60\equal{}2^2 \cdot 3 \cdot 5$.

Consider quadratic residues mod 3. Clearly, they are 0 and 1. Thus, we must have one of the squares a multiple of 3, as $ 1\plus{}1 \neq 1$.

Thus, one of x,y,z, is a multiple of 3.

Consider quadratic residues mod 4. Clearly, they are $ 0,1$ again. Thus, by the same reasoning, one of x,y,z is a multiple of 4.

Consider quadratic residues mod 5. The resiudes are $ 0,1,4$. If none of the numbers is a multiple of 5, we either have

1+1=4 or 4+1=1, neither of which are true.

Thus, we have x,y,z multiples of 3,4,5, so xyz is a multiple of 60.

i don't understand why did you write $ 1\plus{}1\neq 1$  i think you are supposing that z,x,y are congruent 1 in modulo 3, but why did you simply substitude the three numbers by 1?

He said this because $ x^2$ and $ y^2$ will both be either $ 0$ or $ 1$ mod 3. If either of them were $ 0$ mod 3, it would be a multiple of 3. The only way for none of the three to be a multiple of three is if they were all $ 1$ mod 3, which would lead to $ 1\plus{}1\equal{}1$, which clearly isn't true.

That might've just created more confusion...

yes...i undestand that, but my question is about the notation, how can you prove that it is impossible that x,y,z  are all congruent 1 modulo 3 just by saying that 1+1=1 is not true? As far as i know substituding  by 1 doesn't lead us to prove it

[quote="BOGTRO"]Note that $ 60 \equal{} 2^2 \cdot 3 \cdot 5$.

Consider quadratic residues mod 3. Clearly, they are 0 and 1. Thus, we must have one of the squares a multiple of 3, as $ 1 \plus{} 1 \neq 1$.

Thus, one of x,y,z, is a multiple of 3.

Consider quadratic residues mod 4. Clearly, they are $ 0,1$ again. Thus, by the same reasoning, one of x,y,z is a multiple of 4.

Consider quadratic residues mod 5. The resiudes are $ 0,1,4$. If none of the numbers is a multiple of 5, we either have

1+1=4 or 4+1=1, neither of which are true.

Thus, we have x,y,z multiples of 3,4,5, so xyz is a multiple of 60.[/quote]
Take quadratic residues of what? How are they clearly 0 and 1?

Ok, suppose $ (3p\plus{}1)^2\plus{}(3q\plus{}1)^2\equal{}(3r\plus{}1)^2$, then if you take the remainder of both sides when divided by 3 you get $ 1\plus{}1\equal{}1$, a contradiction.

Explained more:
When you expand the three squares, you get
$ 9p^2\plus{}6p\plus{}1\plus{}9q^2\plus{}6q\plus{}1\equal{}9r^2\plus{}6r\plus{}1$
Moving all terms that have multiples of 3 on one side, we get
$ 9p^2\plus{}6p\plus{}9q^2\plus{}6q\minus{}9r^2\minus{}6r\equal{}1\minus{}1\minus{}1$
$ 3(3p^2\plus{}2p\plus{}3q^2\plus{}2q\minus{}3r^2\minus{}2r)\equal{}\minus{}1$
We assumed $ p$, $ q$, and $ r$ are integers, so $ 3p^2\plus{}2p\plus{}3q^2\plus{}2q\minus{}3r^2\minus{}2r$ is an integer. Therefore $ \minus{}1$ is a multiple of 3, which is wrong. We have reached a false fact. Therefore one of our assumptions was wrong. We assumed that there existed integers $ p$, $ q$, and $ r$ such that $ (3p\plus{}1)^2\plus{}(3q\equal{}1)^2\equal{}(3r\plus{}1)^2$. Therefore there do not exist integers such that $ (3p\plus{}1)^2\plus{}(3q\equal{}1)^2\equal{}(3r\plus{}1)^2$.


group theory
abstract algebra
superior algebra
superior algebra solved

Need some help with the following, thx alot.

Let $p>2$ be a prime

1. Prove that there is just one element of order $2$ in $Z_p^{\times}$  (consider when $p | k^2 - 1$ )

2. Write down the automorphism of order $2$ of a cyclic group of order p

3. Prove there are only two groups of order $2p$ (up to isomorphism)

1.: please don't post into the Solved Problems section.
2.: exactly the same questions (except the third) occured with exactly the same wording in http://www.mathlinks.ro/Forum/viewtopic.php?t=62068
3.: use the Sylow theorems and let $\mathbb{Z}/2\mathbb{Z}$ act on the subgroup with order $p$ for the third problem

could u write down the automorphism of order 2 of cyclic group of order p

Edit: just send every element to it's inverse.


function
modular arithmetic
induction
algebra
polynomial

Given that $ F_n$ is the nth Fibonacci number. Prove that:$ F_{4n \plus{} 2}$ - $ (2n \plus{} 1)$ is divisible by 5 with every positive integers n.

The sum of two integer-periodic functions is periodic, so this reduces to a finite calculation.  So this computation is only interesting if it can be generalized appropriately.

Edit:  Ah, I forgot; $ 5$ is special with respect to $ F_n$.  I claim that $ F_n \equiv 2n \cdot 3^n \bmod 5$; see a sketch of the results leading up to the proof [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=229442]here[/url].

I know i am an idiot, but still does the given eq satisfy for n=1 :( .

[hide="Details of what t0rajir0u said"]
The Fibbonacci numbers mod 5 repeat every 20 numbers:
11230331404432022410
And then they begin to repeat.
Case 1: $ n \equiv 0 \pmod 5$
Then $ 2n \plus{} 1 \equiv 1 \pmod 5$ and $ F_{4n \plus{} 2} \equal{} F_{20k \plus{} 2} \equiv 1 \pmod 5$, so $ 5| F_{4n \plus{} 2} \minus{} (2n \plus{} 1)$
Case 2: $ n \equiv 1 \pmod 5$
Then $ 2n \plus{} 1 \equiv 3 \pmod 5$ and $ F_{4n \plus{} 2} \equal{} F_{20k \plus{} 6} \equiv 3 \pmod 5$
Case 3: $ n \equiv 2 \pmod 5$
Then $ 2n \plus{} 1 \equiv 0 \pmod 5$ and $ F_{4n \plus{} 2} \equal{} F_{20k \plus{} 10} \equiv 0 \pmod 5$
Case 4: $ n \equiv 3 \pmod 5$
Then $ 2n \plus{} 1 \equiv 2 \pmod 5$ and $ F_{4n \plus{} 2} \equal{} F_{20k \plus{} 14} \equiv 2 \pmod 5$
Case 5: $ n \equiv 4 \pmod 5$
Then $ 2n \plus{} 1 \equiv 4 \pmod 5$ and $ F_{4n \plus{} 2} \equal{} F_{20k \plus{} 18} \equiv 4 \pmod 5$
[/hide]

[quote="mittal"]does the given eq satisfy for n=1 :( .[/quote]
You may be indexing incorrectly; the standard indexing is $ F_0 \equal{} 0, F_1 \equal{} 1$, which gives $ F_6 \equal{} 8 \equiv 3 \bmod 5$.  This is also a consequence of Binet's formula after expansion, but the disadvantage of this approach is that it does not generalize (as it stands) to primes other than $ 5$, at which $ F_n$ behaves differently.

[quote="dgreenb801"][hide=Details of what t0rajir0u said]
The Fibbonacci numbers mod 5 repeat every 20 numbers:
11230331404432022410
And then they begin to repeat.
Case 1: $ n \equiv 0 \pmod 5$
Then $ 2n \plus{} 1 \equiv 1 \pmod 5$ and $ F_{4n \plus{} 2} \equal{} F_{20k \plus{} 2} \equiv 1 \pmod 5$, so $ 5| F_{4n \plus{} 2} \minus{} (2n \plus{} 1)$
[/quote]
How can you prove this property using induction?? I tried but I didn't prove it, so I assumed that the remainder is not periodical for mod 5 :(

[quote="t0rajir0u"]Two consecutive values of the sequence $ \bmod 5$ determine the rest of the sequence, so once you see any two consecutive values that are the same as what they were before, that's enough to prove periodicity.  In this case, after the terms listed the next two terms are $ 1, 1$ - and you're done.[/quote]
Is this a proof for periodicity? Since I try to find out the mathematical proof for this periodicity, not just list all of the terms to see where the terms occur again(we can do it on AMC but I think we need to have a more rigorous proof). Is this the way we prove for the periodicity of the function? I think we have to prove: f(x)=f(x+p) with some p, so f(x) is periodical after p numbers??

[b]Lemma:[/b]  Let $ s_n$ be a sequence described by a linear recurrence relation of order $ k$ with constant integer coefficients.  Then $ s_n \bmod p$ is periodic with period at most $ p^k$.

[i]Proof.[/i]  $ k$ consecutive values $ (s_{n\plus{}1}, s_{n\plus{}2}, ... s_{n\plus{}k}) \bmod p$ determine the rest of the sequence, and there are $ p^k$ such $ k$-tuplets, so the result follows by Pigeonhole.

In the post that I linked to I describe a more systematic way to determine the exact period (at least when $ p$ is prime).

There is a small subtlety here: what you've written allows for a sequence to be "eventually periodic," in the same way that $ \frac{5}{18}$ has an "eventually periodic" decimal expansion.  In the case of the Fibonacci numbers, any $ k$-tuple defines the sequence going both backwards and forwards in any mod and so the sequence must be truly periodic, but this is not true of all linear recurrences.

True.  For this proof to work, one requires the additional condition that the last coefficient in the recurrence is invertible $ \bmod p$, which is true for any prime $ p$ if the last coefficient is $ \pm 1$ as it is in the Fibonacci case.  But I believe actual periodicity only requires that the extension generated by adjoining the roots of the characteristic polynomial to $ \mathbb{F}_p$ is separable.


limit
logarithms
calculus
calculus computations

Evaluate

   lim     (Sqrt.(9 + 17lnX + 9ln^2X) - 3*Sqrt.(1 + lnX + ln^2X))    if the limit exists.
                x-> 0+

Where Sqrt. stands for the square root.

$ \lim_{x\to 0^\plus{}} \sqrt{ 9 \plus{}17 \ln x \plus{} 9 \ln ^2 x } \minus{} 3\sqrt{1\plus{}\ln x \plus{} \ln^2 x}\equal{}$

$ \lim_{x\to 0^\plus{}} 3\sqrt{ 1 \plus{}\frac{17}{9} \ln x \plus{}  \ln ^2 x } \minus{} 3\sqrt{1\plus{}\ln x \plus{} \ln^2 x}\equal{}$

$ \lim_{x\to 0^\plus{}} 3\frac{(1 \plus{}\frac{17}{9} \ln x \plus{}  \ln ^2 x) \minus{}(1\plus{}\ln x \plus{} \ln^2 x)}{\sqrt{ 1 \plus{}\frac{17}{9} \ln x \plus{}  \ln ^2 x } \plus{} \sqrt{1\plus{}\ln x \plus{} \ln^2 x}}\equal{}$

$ \lim_{x\to 0^\plus{}} 3\frac{\frac{8}{9} \ln x }{\sqrt{ 1 \plus{}\frac{17}{9} \ln x \plus{}  \ln ^2 x } \plus{} \sqrt{1\plus{}\ln x \plus{} \ln^2 x}}\equal{}$

$ \lim_{y\to \minus{}\infty} \frac{\frac{8}{3} y }{\sqrt{ 1 \plus{}\frac{17}{9}y \plus{}  y^2} \plus{} \sqrt{1\plus{}y \plus{} y^2}}\equal{}$

$ \lim_{y\to \minus{}\infty} \frac{8}{3}\frac{1 }{\sqrt{ y^{\minus{}2} \plus{}\frac{17}{9}y^{\minus{}1} \plus{}  1} \plus{} \sqrt{1\plus{}y^{\minus{}1} \plus{} y^{\minus{}2}}}\equal{}\frac{4}{3}$


inequalities
number theory unsolved
number theory

And the second problem is from A. E. Perlin
$a,b,x,y$ are natural numbers, such that $ax+by$ is devisible by $a^2+b^2$. Prove that
$gcd(x^2+y^2,a^2+b^2)>2$.

[quote="Beat"]$a,b$ are different natural numbers, such that $ab(a+b)$ is devisible by $a^2+ab+b^2$. Prove that $|a-b|>\sqrt[3]{ab}$. (S. Berlov)

Problem N6 from:
http://www.zaba.ru/cgi-bin/tasks.cgi?tour=russia.final.XXVII.11klass&solution=1

So many beautiful problems from Russia.
 :lol:[/quote]

Look at this :[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58754]www.mathlinks.ro/Forum/viewtopic.php?t=58754[/url]

Thank you ! And sorry i didn`t know for the post  :(