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$a_{n},(n=1,2,\dots)$ is a integeral sequence such that $a_{k+1}$ is the smallest postive integer which are different from $a_{1},a_{2},\dots,a_{k}$ and has the prperty $(k+1)|S_{k}+a_{k+1}$.
Is every positive integer a term of this sequence?
[hide]I think they are.[/hide]
Edit:I don't know if there is a solution.
Edit:$S_{k}=a_{1}+a_{2}+\dots+a_{k}$
Edit:more,for which $a_{1}$  every positive integer is a term of this sequence.

[quote="t0rajir0u"]I investigated the $a_{1}= 4$ case and I don't think $1$ appears anywhere in the sequence.  I'm too lazy to prove it, though.[/quote]
thank you.I have investigated the case $a_{1}=1$ and I found every postive integer $k\leq 30$ appear in the sequence.

If $a_{1}= 1$, then $a_{a_{n}}= n$ for all positive integer $n$. There are chances that this special case was already solved on the forum. See also http://mathforum.org/wagon/fall96/p818.html and http://www.research.att.com/~njas/sequences/?q=1%2C3%2C2%2C6%2C8%2C4&sort=0&fmt=0&language=english&go=Search

P.S: I will be unable to answer to any replies for the next two week.


geometry rectangle probability MATHCOUNTS

[hide="#14"]The letters A, B, C and D represent four distinct integers from 0 through 9. When A is added to B the result is C. When B is subtracted from A the result is D. How many possible ways are there to assign values to A, B, C and D?[/hide]
[hide="#24"]There is a 3 by 4 rectangle filled by 1 by 1 squares. The top row and rightmost column of the big rectangle has stars(only one star in each square) and the 3 squares on the rightmost column are shaded.(6 stars total w/3 shaded).
     Henri stands in the square on the bottom left corner marked "start." Tresa flips a fair coin. If the coin lands Heads, Henri moves one square upward: if it lands tails, henri moves one square to the right? This process is continued until henri moves into a square with a star, at which time the game ends. If this final square witht  he star is shaded, Teresa wins; if it is not shaded, Henri wins. What's the probablility that Teresa wins? Express as a common fraction. (i dont know how to draw a picture, so i described it best i can. sorry if the discription sucks-it'll help if you draw the thing first)[/hide]
[hide="#28"]There is a number line of integers from -1 to 12. Jeff will pick a card at random from ten cards numbered 1 through 10. The number on this card will indicate his starting point on the number line. He will then spin a fair spinner with 3 equal sectors, one of which says "move 1 space left", and two which says "move 1 space right", and follow the instruction indicated by his spin. From this new point he will spin the spinner again and follow the resulting instruction. What is the probability that he ends up at a multiple of 3 on the number line? express as a common fraction
<------------------------------------>
-1 0 1 2 3 4 5 6 7 8 9 10 11 12

[/hide]

#14 is easily brute forcable.. but
[hide]if a-b is greater than 0 then a is greater than or equal to b, so when a=0, there is 1 choice, a=1, there are 2 choices, when a=2, 3 choices.. all the way to a=4, 5 choices, but when a=5, a+b is less than 9 so there are again 5 choices, so when a=5, 5 choices, a=6, 4 choices..a=9, 1 choice so 1+2+3+4+5+5+4+3+2+1=30, however, if a b c and d are distint, then a cannont equal b or b cannot equal 0, there are 14 sceniros when this happens, s 30-14=16 [/hide]

remember that for #28 there's two types of numbers that can possibly result ina multiple of 3

-jorian

art of owna for #14, [hide]16[/hide] is correct

ii brute forced number 28.  i made it sorta like a casework problem.

i supposed that you started with each of them for example, if you started on -12, there are x ways.  if you start on -11, there are y ways. it sounds like it takes a long time, but it took me about 2 minutes and i got it right :D .

[quote="espark52"][quote="mathgeniuse^ln(x)"]16 isn't correct, it is 13 as written in teh Mathcounts Book.[/quote]

why dont YOU explain it then?-16 is what all my peers were telling me and what i got[/quote]

Okay

[hide]D is at least 1, since the integers have to be distinct.  Also, none of the integers can be 0, since then this would imply that two of the integers are the same.

So if D=1, then (A,B,C)=(3,2,5), (4,3,7), (5,4,9)
D=2, then (A,B,C)=(3,1,4),(5,3,8)
D=3, then (A,B,C)=(4,1,5), (5,2,7),
D=4, then (A,B,C)=(5,1,6), (6,2,8),
D=5, then (A,B,C)=(6,1,7), (7,2,9)
D=6, then (A,B,C)=(7,1,8)
D=7, then (A,B,C)=(8,1,9)
And that is all there is.

Count 'em 13.[/hide]

Mathgenius, you are right.. Not only can B not equal A or 0, B cannont be half of A because then B would equal D, so there are 17 cases-(for A and B)-(0,0 1,0 1,1 2,0 2,1 2,2 3,0 3,3 4,1 4,2 4,4 5,0 6,0 6,3 7,0 8,0 9,0) So 30-17=13..

Maybe I shouldve brute forced it..

[quote="vishalarul"][quote="espark52"][hide="#24"]There is a 3 by 4 rectangle filled by 1 by 1 squares. The top row and rightmost column of the big rectangle has stars(only one star in each square) and the 3 squares on the rightmost column are shaded.(6 stars total w/3 shaded).
     Henri stands in the square on the bottom left corner marked "start." Tresa flips a fair coin. If the coin lands Heads, Henri moves one square upward: if it lands tails, henri moves one square to the right? This process is continued until henri moves into a square with a star, at which time the game ends. If this final square witht  he star is shaded, Teresa wins; if it is not shaded, Henri wins. What's the probablility that Teresa wins? Express as a common fraction. (i dont know how to draw a picture, so i described it best i can. sorry if the discription sucks-it'll help if you draw the thing first)[/hide][/quote]

Is this what it looks like?[/quote]

sorry i took so long. yea that picture is correct.
(i was grounded from the computer for a couple weeks)

[quote="Zenox"][quote="archimedes1"][hide="24"]If you assign the number of ways to get to each square in the rectangle, you get

1   3   6   10
1   2   3   4
1   1   1   1

Each way is equally likely, so there are $1+3+6+10+4+1=25$ ways to play and finish the game.  Of these ways, $10+4+1=15$ ways result in Teresa's winning.  So, our probability is $\frac{15}{25}=\frac{3}{5}$.[/hide][/quote]
That doesn't work because...
[hide]Once you get to a star, you finish. So it looks like this

1  2  3  0
1  2  3  3
1  1  1  1

6 possible winning scenarios for Henri and 4 possible winning scenarios for Teresa scenarios, so it's 2/5
 [/hide][/quote]
Oops, you're right, I missed that part.

14
A>B
A+B<10
Using only these rules we get
b=1, a=2~8. 7 numbers
b=2, a=3~7, 5 numbers and etc so that the sum is 16
However, if A is twice B, B=C so we get rid of a=2,b=1
a=4,b=2 and a=6,b=4 (note that there is no more because a=8. b=4 is not possible)

24
1 2 3 0
1 2 3 3
1 1 1 1

ways to get to certain squares are denoted by the number on that respective square
Guys remember the formula for probability? number of stuff we want/total outcomes
(3+1)/(1+2+3+3+1)

4/10=2/5

28
first we add up the probabilities of landing on a multiple of 3 with the cards 1-10 and divide it over 10 since each card is equally likely.  ALso note that each card has exactly 1 way of getting onto a multiple of 3 --- either by going left 2 times (1/9) going right 2 times(4/9) and staying in the same place (going right then left or left then right 4/9)  so its (4+1+4+4+1+4+4+1+4+4)/9*(1/10)
31/90

EZZZZZZ

btw how do u do that hidden content thing (help out the new member)

On solution, they say answer is 5/16. Can someone pls explain????

The probability for each of the scenarios is different, so you have to calculate the probability of ending at each square, not just the number of paths.

@machack you see the eye crossed out? you click that and when it says the hide things put the hide=solution into the first one.

hey can someone send me the answer key? how do you find the answer key to these sprints


The inequality is trivial.  It's equivalent with the well known lnx<=x/e, x>=0.

A proof of that inequality is as follows.

$f(x)=\ln x$ is convex, and hence it lies below a given tangent to it at all points.  Thus at $x=e$, we get

$\ln x\leq \frac {1}{e}(x-e)+1=\frac {x}{e}$

Yes.  Don't worry, that was just a typo - the rest of my post reads correctly if you replace convex with concave.


function
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floor function
number theory
totient function
relatively prime

[quote="eryaman"]Whats the symbol represent?[/quote]

$\sigma (n)$ is the sum of the divisors of number n.

EDIT: I learnt this from the link in your signature, chess64. :lol:

[quote="eryaman"][quote="chess64"]$\sigma(n)$ is the number of [proper] divisors of $n$. I'm not sure about the "proper"; check mathworld.[/quote]That's phi, or the totient function I believe.

I think I figured out the problem.  Not the actual answer, but how to do it.
[hide="hint"]How many numbers from 1-64 does 1 divide into?  All of them, so add 64 1's=64
How many numbers from 1-64 does 2 divide into?  32 of them, so add 32 2's=64
How many numbers from 1-64 does 3 divide into? 21 of them, so ad 21 3's=63.

Hmmmm....So it seems that if $n|64$ then its sum will be 64, and if it doesn't....[/hide][/quote]
That's not totient function. You can use Search feature of ML/AoPS and look up what totient function means before posting a false claim.

Just to make things clear...
$d(n)$ represents the number of divisors of a number n.
$\sigma (n)$ represents the sum of the divisors of a number n.
$\phi (n)$ represents the number of positive intergers less than n relatively prime n.

[quote="chess64"]Ah, sorry I got them mixed up.

I also find that link very useful, so I put it in my sig. so people would be able to benefit from it. Unfortunately, as soon as I did, the mandelbrot people removed it from their site, and now I have to use google's cache :([/quote]

Thanks.  I printed it out for portable usage of the titles to excel. :lol:

well, i think that a proper divisor of n is every divisor except 1 and n, so we find all the numbers that are divided by 2,3,4,5,etc then subtract 63 from that because we overcount each divisor once because a proper divisor only includes divisors that are not n or 1

$-63+\sum_{k=2}^{64}\left\lfloor \dfrac{64}{k}\right\rfloor$

this can be simplified because we know every number above 32 will just give 1 so

$-63+32+\sum_{k=2}^{32}\left\lfloor \dfrac{64}{k}\right\rfloor$

i suppose a calc from here would work since we only have to do 31 numbers


function
algebra
polynomial
Rational Root Theorem

h(x) = x^3 - x^2 + 2x - 4

How would I find out what "x" is when h(x) = 0 ?
I tried factoring x^3 - x^2 + 2x - 4, but I couldn't get it simple enough that I could understand how to find what h(x) would be, if I knew what "x" was.
I sketched the graph, but I wanted to know how to solve for h(x) exactly when knowing what "x" is.
Please explain your work. Please be detailed. Thank you very much.

Another function, is g(x) = x^4 - 2x^3 + x - 3. I have the same problem with this one too. Any help would be much appreicated. If there is a certain method that is applied to equations like these, please tell me.  :)

[hide]for h(x) you can use Cubic Formula
http://mathworld.wolfram.com/CubicFormula.html
[/hide]

Wow, I must of spent like 10 mins trying to figure out what all that means and I still don't have a clue. Thank you felipesa for providing that information. Any help is appreciated, but I forgot to mention am not really a math wiz... if it can explain it in simpler terms (if possible) it would help a lot. As far as the Cubic Formula, I'll look more into that.

Avani?

Anyway, first you set the polynomial expression equal to zero by using the transitive property.  Then, it's best to try and factor it.  Use rational root theorem (look it up) to look for roots of the equation.  Since the equation is to the third degree, you should find 3 roots.


Find a positive solution to $x^{x^{2006}}=2006$

Please more experienced guys leave it for the less experienced.

(I define more experienced by: you solved this in less than 10 minutes!)

so i sorta simplified it to (not sure if it helps at all though)
[hide]
$x^{2006}ln(x)=ln(2006)$
[/hide]
but am not able to do anymore. could u provide a hint or something.

Hmmm...I got $\sqrt[2006]{2006}$...it checks out too...

[quote="Iversonfan2005"][quote="Treething"]It's x^x^2006, not x^2006 :P[/quote]

So, it still works out because we simplify the exponents first and so $\sqrt[2006]{2006}^{\sqrt[2006]{2006}^{2006}}=\sqrt[2006]{2006}^{2006}=2006$ ;)

EDIT: Oops..never mind..simplified the exponents wrong :blush: :fool:[/quote]

No, you did it right. If seamus meant (x^x)^2006, he would have written $(x^x)^{2006}$.
$x^{x^{2006}}$ means x^(x^2006), which is what you did.

Oh okay...for a second, I thought I needed to review exponents again :!:  :rotfl:

haha that seems like an obvious answer.  :rotfl:

i am wondering if there is a proof or some method you used to arrive at this solutoin.

Good one guys.

My first though was to sub the 2006 back into the tower of x's to obtain:

$x^{x^{x^{x^{2006}}}}=2006$

and continue doing this to get an infinate number of x's in the exponent.

This looked as if the solution to $x^{2006}=2006$ is the same as the solution to the one i posted. Subbing it in, it worked!


Let $x,y,z$ be positive real numbers with $x^{2}+y^{2}+z^{2}\geq 3$.
Show
\[\frac{x^{n}}{(1+y)(1+z)}+\frac{y^{n}}{(1+z)(1+x)}+\frac{z^{n}}{(1+x)(1+y)}\geq \frac{3}{4}\]
for a positive integer $n \geq 3$.

[quote="giancarlo"]Let $x,y,z$ be real numbers with $x^{2}+y^{2}+z^{2}\geq 3$.
Show
\[\frac{x^{n}}{(1+y)(1+z)}+\frac{y^{n}}{(1+z)(1+x)}+\frac{z^{n}}{(1+x)(1+y)}\geq \frac{3}{4}\]
for a positive integer $n \geq 3$.[/quote]

For $x,y,z$ real?

Counter example: $x=y=z=-2, n=3$ :D

indeed, in the rush i forgot to write 'positive'. :-)


It can be proved using Bertrand's Postulate

Using Bertrand is an overkill. :P Here's how I would proceed.  The proof of the hint is standard.  Now for a sufficiently large number $ n$ (eliminate the first couple values) take the largest prime $ p\leq n$.  Use the hint, and proceed from here.


Yay! correct (no surprise). Can you show your solution now?

[quote="math92"]
wow, nice job kunny, but i don't get your logic. :? [/quote]

The trick to this problem is converting mixed numbers to improper fractions, so we cancel out execessive numbers.

  Doing so yields:

$\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6} \cdots \frac{2005}{2004} \cdot \frac{2006}{2005}$

  Now we notice that every number cancel outs except for "2" and "2006" which gives us $\frac{2006}{2}=\boxed{1003}$

  Algebraicly this looks like:

$\frac{n+1}{n} \cdot \frac{n+2}{n+1} \cdot \frac{n+3}{n+2} \cdot \frac{n+4}{n+3} \cdots \frac{n+r-1}{n+r-2} \cdot \frac{n+r}{n+r-1}=\boxed{\frac{n+r}{n}}$


A basketball team has won 50 games of 75 played. The team still has 45 games to play. How many of the remaining games must the team win in order to win 60% of all games played during the season?

Essientially the same:
[hide] There are going to be 75+45=120 total games played.  They have to win .6*120=72 games.  Therefore, they have to win 72-50=$22$ more games.  [/hide]

Great replies and easy methods for solving this kind of question.

Thanks!


Would someone who has the Art of Problem Solving: volume 1 basics book be willing to post the problems and contents of page 46?

Thank you very much.

You know the book is copyrighted by AoPS so that wouldn't be right.

Isn't it cheap looking at the free excerpts and asking for the following pages after the excerpt ends?


Jane is 25 years old. Dick is older than Jane. In $ n$ years, where $ n$ is a positive integer, Dick's age and Jane's age will both be two-digit number and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let $ d$ be Dick's present age. How many ordered pairs of positive integers $ (d,n)$ are possible?

[hide="Answer"]We are looking for the number of 2 digit numbers greater than 25 such that the ones digit is greater than the tens digit.

For tens digits 3 through 8, when the tens digit is k, there are 9-k digits higher than k that can be the ones digits. When the tens digit is 2, the ones digit can be 6, 7, 8, or 9.

This makes a total of 4+6+5+4+3+2+1 = 25 ordered pairs.[/hide]


Given $ a<b$ prove that:

$ a^{3}\minus{}3a\minus{}2 \le b^{3}\minus{}3b\plus{}2$

thanx.

[quote="reason"]Given $ a < b$ prove that:

$ a^{3} \minus{} 3a \minus{} 2 \le b^{3} \minus{} 3b \plus{} 2$

thanx.[/quote]

let $ b\equal{}a\plus{}t,t>0$, ineq becomes

$ 3 a^2 t\plus{}3 a t^2\plus{}t^3\minus{}3 t\plus{}4 >0,$

and $ \Delta \equal{} (3t^2)^2\minus{}4(3t)(t^3\minus{}3 t\plus{}4)\equal{}\minus{}3 (t\minus{}2)^2 t (t\plus{}4) \le 0,$ so ineq holds.

[quote="kunny"]Why can we directly use Discriminant in this case :wink: ?[/quote]

it is a correct solution?

The inequality is equvialent to $ (b\minus{}a)(b^2\plus{}ab\plus{}a^2\minus{}3) \ge \minus{}4$. If $ b^2\plus{}ab\plus{}a^2 \ge 3$ we are done. Let $ b^2\plus{}ab\plus{}a^2 <3$. Then we have by AmGm $ (b\minus{}a)(3\minus{}a^2\minus{}ab\minus{}b^2) \le \frac{(3\minus{}a^2\minus{}ab\minus{}b^2\minus{}a\plus{}b)^2}{4}$ Hence it's left to prove that $ a^2\plus{}ab\plus{}b^2\plus{}a\plus{}1 \ge b$, which is equivalent to $ \frac{1}{2} \left( (a\plus{}b)^2\plus{}(a\plus{}1)^2\plus{}(b\minus{}1)^2 \right) \ge 0$, which is clearly true. Btw, equality does never occur.


The area of the rectangle is $ 3c \times x$.
The area of the triangle is $ \frac{2c \times x}{2}$.
$ \frac{\frac{2c \times x}{2}}{3c \times x} \equal{} \boxed{\frac{1}{3}}$.

hello, let $ x$ the side length of the given rectangle, then we have this area as
$ 3cx$ and the area of the triangle is $ \frac{1}{2}2cx$ and the ratio of both is $ \frac{1}{3}$.
Sonnhard.


algebra
polynomial
function
conics
parabola
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Solve the equation

[tex]\displaystyle\sqrt{5-x}=5-x^2[/tex]

without having to manipulate/factor an ugly quartic polynomial. (BTW, no ti89's or computer algebra systems either!)

EDIT --- solve the equation for REAL x.

[hide]x^2= 5- :sqrt: (5-x)

x= :pm: :sqrt:(5-:sqrt:(5-x))

Resubstitute it back into itself...

x= :pm:  :sqrt: ((5- :sqrt: (5- :sqrt: 5-.....)

Square it to get x^2=5-x

Solve.... x^2+x-5=0.... (1 :pm:  :sqrt: (21))/2[/hide]

Very interesting method, but you're not quite done. There could be other solutions than the ones you found (think about the quartic that forms when you  square the equation... there are 4 possible solutions, not 2) AND you have to check all 4 solutions to cast out the extraneous ones. That IS a very nice method, however both of mine are different.

This problem has one of the nicest, cleverest solutions of anything I have ever seen!  That is a nice method, by the way ... I wonder what happened to lose two roots.

I did it the not-so-ugly quartic polynomial way, using an interesting fact.

[hide]

The first thing I noted is that +/-LHS is the inverse function of RHS. I decided to find the solutions of +/-LHS=RHS and then toss out the extraneous ones. Roughly graphing +/-LHS and RHS, you see that you have 2 nice parabolas intersecting at 4 points. Interestingly, all 4 intersection points are in separate quadrants. Knowing that inverse functions are reflections across the line y=x, I realized that the intersection points in quadrant I and quadrant III must be on the line y=x. Solving for the x-values:

5-x^2=x

x^2+x-5=0

I won't use the quadratic formula yet, but I note that I want the positive root.

Now I find the ugly quartic polynomial to be x^4-10x^2+x+20. Then I use long division to divide it by x^2+x-5 to get x^2-x-4. This time I want the negative root.

Using the quadratic formula to find the positive root of x^2+x-5=0 and the negative root of x^2-x-4, I find that:

x=(-1+sqrt(21))/2 or x=(1-sqrt(17))/2

[/hide]

And then, because of my submission to the machine, I check my answer with a TI-89 .

Ragingg wrote:[hide]
x= :pm:  :sqrt: ((5- :sqrt: (5- :sqrt: 5-.....)

Square it to get x^2=5-x
[/hide]

You lost the two solutions in this step.

The solution Yasha gave --- noticing the inverse functions --- is one of my 2 cool solutions. Good job Yasha! But my other one is far cooler in my opinion. Can anyone come up with it?

[quote="beta"][quote="Yasha"]I did it the not-so-ugly quartic polynomial way, using an interesting fact....[/quote]

I don't see how that makes the problem easier, you can just square both side of the equation and factor it out, that might have been less complicated.

I am working on figuring the cool solution now.[/quote]

I don't know how you would go about factoring the quartic polynomial without knowing any more information or using a TI-89. None of the roots are rational and you wouldn't be able to factor by grouping either.

For this problem, an ugly way would be to factor the quartic polynomial with guess and check. BLEH, who would ever want to do that? Basically, a 'elegant' way would be to sidestep the necessity to factor by guessing. Yasha's way of noticing the inverse functions gave him a couple of solutions to the quartic WITHOUT having to guess at them, which is why I like that method. The other method is even cooler...

Square both side we have:

[tex]5-x=(5-x^2)^2[/tex]
[tex]5-x=25-10x^2+x^4[/tex]
[tex]x^4-10x^2+x+20=0[/tex]
Which factors into
[tex](x^2-x-4)(x^2+x-5)=0[/tex]

Solve each quadratic will give the solution, however that's not elegant and is brute-force factoring.

{EDIT}- Zabelman beats me by 2 minutes.

[quote="beta"]Which factors into... [/quote]

While this solution works, the reason I don't like it is... How'd you come up with that factorization? (I don't mind quadratic formula. At least the quadratic formula isn't pulled out of... er... you know where)

Beta, don't take this the wrong way. That solutions is perfectly fine, and there is nothing in the world wrong with it. However, there is an extremely elegant solution that I'm waiting for.

I am not sure if this is what you are looking for:

Add 20 to both side of the equation we have
[tex]20+\sqrt{5-x}=(5-x)(5+x)[/tex]
Let [tex]k=\sqrt{5-x}[/tex]

[tex]20+k=k^2(10-k^2)[/tex]
Simplify it we have
[tex]k^4-10k^2+k+20=0[/tex]
If we square the original equation and rearrange it
[tex]x^4-10x^2+x+20=0[/tex]

k=x is clearly a solution, which means [tex]x^2+x-5=0[/tex].
Synthetic division give us the other quadratic.

This surprisingly seems very similar to my solution. Here you solved x=LHS to get x^2+x-5=0, whereas I equated RHS=x to get the same quadratic.

My solution is not really "cool" since it's pure algebraic manipulation, yours is a bit cooler. I think the coolest solution is more geometric than algebraic.

For my solution... I noticed that when I resubstituted it back into itself, I only used the +... If I used the -, would the other two solutions come out of the strange new pattern?

Since no one has posted here in a few days, I think it's time to show ya'll the awesome solution...

Square the original equation to get

[tex]5^2+5(-2x^2-1)+(x^4+x)=0[/tex].

Then we use quadratic formula. You may think, "But it isn't quadratic in x!" That is correct. But it IS quadratic in 5. So, we can use the quadratic formula to solve for 5.

[tex]\displaystyle 5=\frac{2x^2+1\pm\sqrt{(-2x^2-1)^2-4(1)(x^4+x)}}{2}[/tex]

which simplifies to

[tex]\displaystyle 5=x^2+x[/tex] or [tex]5=x^2-x+1[/tex].

All that's left now is to solve these 2 quadratic equations for x and then to check the solutions.

Wow... that is really cool.  It makes the problem seem so much easier.  Where did you first see this problem?

[quote="zabelman"]
[tex]5^2+5(-2x^2-1)+(x^4+x)=0[/tex].

Then we use quadratic formula. You may think, "But it isn't quadratic in x!" That is correct. But it IS quadratic in 5. So, we can use the quadratic formula to solve for 5.
[/quote]

This is pretty cool!

If you can get a "clean" answer by using the quadratic formula, it is usually factorable in a clean way. So here goes,

Notice that [tex]x^4+x=x(x^3+1)=x(x+1)(x^2-x+1)[/tex]. If we let [tex]Y=5[/tex] in the above equation, then

[tex]Y^2+(-2x^2-1)Y+x(x+1)(x^2-x+1)=0[/tex]. Cross multipyling, the LHS factors as:

[tex](Y-x(x+1))(Y-(x^2-x+1))[/tex].

[quote]If you can get a "clean" answer by using the quadratic formula, it is usually factorable in a clean way.[/quote]

Exactly! Quadform is just a simple way to find the factors (if you are fortunate enough to see it).

Nice solution!

But what other further implications does that have exactly as a factoring tool? Does this mean that for any given expression which is nicely factorable into two factors, [b]there exists a constant[/b] for which if we define the constant to be the "variable," the discriminant (of the quadratic with the constant as the variable) will simplify nicely? Seems that the answer is a definite yes... but then how can we guarantee that this constant appears somewhere in the question? Is the constant the square root of any numerical (as opposed to algebraic) perfect square, like it was in this question (5, from 25)? How do we [b]know[/b]?


Try this one, found it in an old highschool textbook! :D

Use cos(3x) = cos(2x + x) to prove cos(3x) = 4cos^3x - 3cosx

Hint: Never forget the basics!

[hide="Proof"]\[ \begin{gathered} \cos 3x = \cos (2x + x) \hfill \\ = \cos 2x\cos x - sin2xsinx \hfill \\ = (\cos ^2 x - \sin ^2 x)\cos x - 2\sin x\cos x\sin x \hfill \\ = \cos ^3 x - \sin ^2 x\cos x - 2\sin ^2 x\cos x \hfill \\ = \cos ^3 x - (1 - \cos ^2 x)\cos x - 2(1 - \cos ^2 x)\cos x \hfill \\ = \cos ^3 x - (\cos x - \cos ^3 x) - 2(\cos x - \cos ^3 x) \hfill \\ = \cos ^3 x + (\cos x - \cos ^3 x)( - 1 - 2) \hfill \\ = \cos ^3 x - 3\cos x + 3\cos ^3 x \hfill \\ = 4\cos ^3 x - 3\cos x{\kern 1pt} \quad \hfill \\ \end{gathered}  \][/hide]

Congratulations, very nice work! Nice and neat too, I like that!  :lol:

Well, let's give it a Pre-Olympiad solution  :D

[hide="Solution"]
$\cos 3x = \Re \left( (\cos x + i \sin x)^3 \right) = \Re \left( \cos^3 x + 3i \cos^2 x \sin x - 3\cos x \sin^2 x - i \sin^3 x \right) = \cos^3 x - 3\cos x (1 - \cos^2 x) = 4 \cos^3 x - 3 \cos x$

I'd give a general solution for $\cos nx$, but it requires transforming the real terms with an even $\sin^k x$ power and it's rather ugly  :|
[/hide]


[quote="calc rulz"]I just noticed that the Math Jam isn't happening until next week. :o  Does this mean that we have to wait until then to discuss the problems?[/quote]

yes, see other topics

That's not necessarily true.  In the past they've allowed discussion before the Math Jam if it's later than normal.

Edit: nevermind, in this case it was confirmed for after the Math Jam; however, keep in mind that sometimes they do change this policy.


The sequence has $ a_1 \equal{} 0$ and $ a_{n \plus{} 1} \equal{}\pm(a_n \plus{} 1)$ for all $ n$. Show that the arithmetic mean of the first $ n$ terms is always at least $ \minus{} \frac12$.


A hyperbola whose foci are (0, 3 -sqrt{3}) and (0, 3 + sqrt{3}) has a conjugate axis of length 2.  Write an equation of the hyperbola in standard form.

The textbook tells me that the center (h, k) = (0, 3).

Where did that come from?

I need someone to please explain STEP BY STEP how to do questions like this one.

What about if the vertices were given and NOT the foci?

How would I solve something like that?

What is the transverse axis?

The textbook tells me that the center (h, k) = (0, 3).
Where did that come from?
The centre of the hyperbola is the midpt of the focii and the vertices.
What about if the vertices were given and NOT the foci?
Discussed above.
In genl focii are (+or-ae,0) or (0,+or-ae) depending on the major axis.
The hyperbola has 2 axes the major and the transverse axis .
 :)


Here are some more partial results

[code]SER 77 71 77 71 -1 61 (problems 4&5)
RUS 7 7 7 7 7 7 (problem 4)
SWE 7 - 7 - 7 7 (problem 4)
JAP - - 3 - 1 - (problem 3)
TPE 7 7 1 7 1 1 (problem 5)
NZL 7 4 7 7 7 3 (problem 4)[/code]

The minus sign represents either a 0 or an yet not coordinated paper.

[quote="Valentin Vornicu"]Some current partial scores that I can see on the tables here:

[code]BUL 7 7 3 3 7 7 on problem 1
MEX 7 7 7 6 7 7 on problem 4
PER 5 7 - - 1 6 on problem 5
HKG 7 7 7 7 7 7 on problem 2
HUN 6 7 - 7 7 1 on problem 2
EST 4 - 1 4 1 1 on problem 2
COL 1 2 1 1 1 - on problem 2[/code][/quote]

what's 'problem x' means?
and SER's score seems strange. :o

So based on these partial results was the marking scheme of problem 1  3+4 ??
to mathpk: The Serbians have two problems corrected :wink:

If you get results of Ukraine, tell please.

Can you tell me the results of [color=darkred]Kazakhstan[/color] team please?

Valentin,

 are the partial results you posted in order ,  i.e. the first score corresponds to the first contestant?

Could you post results of the Macedonian team, please?

Marko: yes the order is the one given by the IMO site to the contestants. I've uploaded the latest table with results, check the main page for that.

I mean http://www.mathlinks.ro/portal.php?t=160319

Or you can do

http://www.artofproblemsolving.com/Forum/files/result_2.htm
and
http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=9914

Wait, did no one from China get full marks!?

[quote="Nico0O"]Only 3 contestants resolved problem 6 up to now... :|

I think nobody will make a perfect score this year :o[/quote]

I think nobody will get more than 37 points

ITA 6 and RUS 4 can have 37 [or more]... (also PER 1, SRB 1, UNK 4, NOR 6 and AUS 2; but we haven't got their score on problem 3 and 6 yet...)

We'll see :P

Any new results from belgian team? Can someone check the scores for P2 and P3 for BEL4-6 please?

I think the cutoff for gold will be around 27 based on the partial results given...  :maybe:

[quote="Albanian Eagle"]Guys did any one calculate the cutoffs approximately?[/quote]

I did the following: I took the list of Joseph Myers and added for each pending result the average points for this problem. After sorting I got following estimated cutoffs:

Rank 43 (552/12) has 24 points.
Rank 131 (552/4) has 18 points.
Rank 261 (552/2) has 12,55 points.

I don't say this will be the cutoffs, just telling you the numbers. Especially the gold/silver cutoff may be higher as in my list from 44 to 48 there are 5 contestants (RUS1, RUS5, RUS6, POL6, PRK6) with 22 points and only problem 5 pending (estimated 23,86 points).

Gold : 28
Silver : 21-22 (In my statistics, 26.23% have >= 21 And 24.71% have >= 22)
Bronze : 13-14 (In my statistics, 50.95% have >= 13 And 48.48% have >= 14)

But it's just statistics  :D !

And the winner will certainly be RUS 4 or SRB 1 (if he answered correctly problem 6...)

But I haven't got results of China team... (except problem 6)

Xerus, there are 526 contestants.. not 552 (I think)

Xerus, your method will probably work to give an estimate for the bronze cutoff, however the people scoring around the Silver and Gold cutoffs are better than average and so will probably higher than what you predict. So not only those 5 but people you predict will be on 20 can easily score higher than 25 (and from the Chinese team only one result was known so you can easily assume you underestimate their scores). Moreover I count only 522 contestants as not all teams have 6 members.

[quote="Nico0O"]

And the winner will certainly be RUS 4 or SRB 1 (if he answered correctly problem 6...)

[/quote]

I think the winner will be Peter from Germany with 36 points.

Perhaps...
But if Konstantin Matveev (RUS 4) have 7 on problem 5, he will have 37 :| .

Can anyone send me the complete results for the macedonian team

[quote="Nico0O"]
And the winner will certainly be RUS 4 or SRB 1 [/quote]
SRB1 has 30 points,final score

[quote="Valentin Vornicu"]and
http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=9914[/quote]

What does P mean at problem 3 mostly?

[quote="KMN"]
What does P mean at problem 3 mostly?
[/quote]

I think it means [b]P[/b]ending

There is a rumour, The chinese team has 2 silver.

Can someone please post the final results of India? Eagerly awaiting some news here.

[quote="color"]Results of the russian team is out:

        Russia RUS4
	7 	7 	2 	7 	7 	7 	37 	Gold Medal

 	Russia 	RUS3
	7 	7 	5 	7 	7 	1 	34 	Gold Medal

	Russia 	RUS2
	7 	7 	2 	7 	7 	1 	31 	Gold Medal

	Russia 	RUS1
	7 	7 	1 	7 	7 	0 	29 	Gold Medal

 	Russia 	RUS5
	7 	7 	1 	7 	7 	0 	29 	Gold Medal

	Russia 	RUS6
	7 	7 	1 	7 	2 	0 	24 	Silver Medal[/quote]

Well, I see that [b]1 point from problem 3[/b] is really the gold :) A very good experience is that ... writing everything you can, for even problems you can not solve :D That sometimes give you the Gold medal, not the Silver  :lol:

Yes those cut-offs can be really cruel if you end up just below it  :(

TPE = 149 (total team scores) :)

Can someone please post the final results of TPE (each contestant's score)?
Thank you very much in advance.
 :)

Are the complete scores out as yet? If not could someone do us all a favour and put them up. Also, if that too is not possible - could you tell me the results of the India team?
Really awaiting results here - results that don't seem likely to come soon :(
Ashwath

final result sheet is available as below: :)
http://www.imo2007.edu.vn/statics/result/result%20imo2007.htm

Matching [url=http://www.imo2007.edu.vn/statics/result/result%20imo2007.htm]names[/url] with [url=http://www.mathlinks.ro/viewtopic.php?mode=attach&id=9914]results[/url] I am quite surprised, not only by the defeat of the Chinese team but especially the individual scores are partially unexpected. The previously mentioned fact that problem 3 and 6 are quite different in difficulty level than the rest of the problems played an important factor whether or not to get a gold medal, i.e. carefulness on problems 1/2/5/6 avoiding easy gaps/mistakes and hoping for partial credit on the hard problems. People successfully implementing this strategy got a gold medal but others who could have solved harder problems than 1/2/5/6 but not yet problem 3 or 6 might have missed out the gold medal due to levity.


in secter OAB, where O is the center of the circle and A and B are on the ends of the circle making it a quarter circle.  the radius of the circle is 3 and there is a circle inscribed in the sector tangent to all three sides. find the radius of the smaller circle.

i remeber this from a few years ago

i was stuck for days then BINGO i found out

here is the solution

[hide]you see a 1/4 circle draw the circle inside mark the center x lets have perpendicular lines drawn at x to the two straight sides. lets define r as the radius of circle. we the diangoal line through x that cuts the 1/4  circle in half is 3 (radius of big circle). and we know that trainlge formed from the lines of a base and the diagonal through x is a 45-45-90 triangle so 3=r+r*sqrt(2) so 3/(1+sqrt(2)) and you get -3+3sqrt(2) i believe[/hide]


Does anybody know the term for the moment when you say something that your brain is thinking, but you don't want to say out loud? I forgot it... I think it started with an F? and ended with Lapse?
like F------ Lapse?

Perhaps what you're looking for is "Freudian slip" - named for Sigmund Freud, of course. And pronounce that "eu" as it's pronounced in German, like the English "oy". But the second word of that phrase is always "slip," not "lapse."


[color=darkblue]Let $ x>1, y>1, z > 1$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 2$. Prove that $ \sqrt {x \plus{} y \plus{} z} \ge \sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1}$[/color]

$ \left(\sqrt{\frac{x\minus{}1}{x}\plus{}{\frac{y\minus{}1}{y}}\plus{}{\frac{z\minus{}1}{z}}}\right) (\sqrt{x\plus{}y\plus{}z} ) \ge (\sqrt{x\minus{}1}\plus{}\sqrt{y\minus{}1}\plus{}\sqrt{z\minus{}1})$
 but
$ \left({\frac{x\minus{}1}{x}\plus{}{\frac{y\minus{}1}{y}}\plus{}{\frac{z\minus{}1}{z}}}\right)\equal{}3\minus{}2\equal{}1$

I think this problem is in Iran Olimpiad.
It is killed by B.C.S

Yes, this is an inequality from Iran Mathematical National Olympiad 1998. It's has been solved similarly in my colection of inequality "Inequaltiy: elementary and advenced techniques"


i believe jane street has branches in chicago and tokyo.

There is at least one rather large investment bank here in Minnesota, too.

by the way, houston has a decent amount of quants working at energy companies that run var on energy trades and price out the optionality of power plants... although it is not as prestigious as wall st, etc they still pay pretty well and require a good head on your shoulders.

[quote="satyam"]From what I have heard, you can only find Quant jobs on Wall Street and in London. Is this true? Are there any jobs for quant available in other cities like Chicago(which has North America's second largest "Bank Market") and Toronto(which has North America's third largest "Bank Market")?

Also, if one cannot find a job as a quant, can he/she work as something else like a financial manger in a bank?
Thanks for your help.[/quote]

There are quant jobs all over, but most in the US are in NYC.

Financial manager in a bank is miles away from being a quant.  If you are trained for quantitative work in general, there are lots of fields for you to consider if being a quant in a financial firm doesn't work out for you.

[quote]Financial manager in a bank is miles away from being a quant. If you are trained for quantitative work in general, there are lots of fields for you to consider if being a quant in a financial firm doesn't work out for you.[/quote]

rrusczyk, can you please give me some examples of what kind of fields I will be able to work in financial industry if I am not hired as a quant.

Thanks.

The largest is consulting.  Others include government work (NASA, NSA, etc., all hire lots of quants), startups of various kinds (particularly if you know how to program), or starting your own company.  Many biotech companies now require serious modeling as well, though that may require more education.  Speaking of education, academia is also an option.

NYC & London - by far the most jobs, and funds
Greenwich - 2nd...but basically NYC
Chicago - the big C and a spin off, BofA took some of their quants back to NYC
Dallas - about 4 funds, one of them grew 200% this past year
Houston - as mentioned, tons of energy work there (NO State tax for Dallas and Houston...)
San Fran - decent amount of work, one of the super funds built a branch there, 5 other funds I know of
Minnesota - there is even a couple funds there!

[quote]Speaking of education, academia is also an option.[/quote]

Positions in academia are more selective and require more
time to pursue, than Wall St type quant.    I don't know what could possibly be meant by "academia is also an option" for a question like this.
If academia is an option then so is quant but not the other way around.

I work as a hedge fund quant in NYC, and despite the fact that I love my job, I am a bit frustrated by the lack of quant jobs outside the New York area.  They say NYC is the capital of the United States for finance jobs in general.  This holds even more true for quant jobs, IMO.


An complex # takes form of z= x+iy where x is Re and iy is Im.

This can be written in the form (cos theta + i sin theta) = z = 1 as polar coordinates.

my problem is how would you write z = 4+5i in polar coordinates. [/img]

[quote="kdog3682"]This can be written in the form (cos theta + i sin theta) = z[/quote]

Somewhat incomplete.  The form you actually want is

$z = r (\cos \theta+i \sin \theta)$

Where $r = |z|$, the "length" of $z$.  Plotting a complex number $z = x+yi$ as the point $(x, y)$, the complex numbers of the form $\cos \theta+i \sin \theta$ correspond only to the points on the unit circle $x^{2}+y^{2}= 1$.

Having said that,

[b]To convert from rectangular to polar:[/b]

$r = \sqrt{x^{2}+y^{2}}$
$\theta = \arctan \frac{y}{x}$ (be careful about quadrants!)

[b]To convert from polar to rectangular:[/b]

$x = r \cos \theta$
$y = r \sin \theta$

For $z = 4+5i$ we have that $x = 4$ and $y = 5$, and so

$r = \sqrt{4^{2}+5^{2}}= \sqrt{41}$ and $\theta = \arctan(\frac{5}{4}) = 51.3^{o}$.

So: $z = \sqrt{41}[\cos(51.3^{o})+i\sin(51.3^{o})]$.

Note that the angle $\theta$ must belong to the first quadrant because x > 0 and y > 0.


topology
vector
group theory
abstract algebra
invariant
advanced fields
advanced fields unsolved

Let G be a smooth manifold which is also a group.
Moreover suppose the group structure and the differentiable structure are compatible, that is, the multiplication and inversion maps are smooth.
Then G is called a smooth group.
Suppose p: G' -> G is a covering map.
Then the question is Under which conditions is G' also a smooth group?
If I am not wrong when G' is the universal covering of G then G' can be equipped with a smooth group structure.
However, what if G' is not the universal covering of G?
Thank you in advance for your help buddy...

The usual name for that structure is a Lie group. We should also assume that $ G$ is connected.

Let's use those tools, particularly the Lie algebra. Given any vector $ x$ at the identity of $ G$, we can move it to $ a$ by multiplying by $ a$; this gives a vector field $ X$ on $ G$, which we take to be an element of the Lie algebra $ H$. This vector field can be naturally pulled back to $ G'$.
This also gives us the exponential map $ \exp$ from $ H$ to $ G$. We can pull that back to a map $ \exp'$ from $ H$ to $ G'$: the path $ \exp(tX)$ in $ G$ can be pulled back to a path $ \exp'(tX)$ in $ G'$, and we just pick out the individual points on that path.
What can we do with that? We can follow the flow of a (complete) vector field, and use that to construct (left) multiplication by $ \exp(X)$ or $ \exp'(X)$. That isn't necessarily everything, but it does contain a neighborhood of the identity, and thus generates everything. This is automatically associative and includes inverses $ \exp(X)^{\minus{}1}\equal{}\exp(\minus{}X)$.

So, the answer is yes. $ G'$ is also a Lie group, and we can construct its multiplication from the multiplication on $ G$, together with the covering map.

[quote="jmerry"]The usual name for that structure is a Lie group. We should also assume that $ G$ is connected.

Let's use those tools, particularly the Lie algebra. Given any vector $ x$ at the identity of $ G$, we can move it to $ a$ by multiplying by $ a$; this gives a vector field $ X$ on $ G$, which we take to be an element of the Lie algebra $ H$. This vector field can be naturally pulled back to $ G'$.
This also gives us the exponential map $ \exp$ from $ H$ to $ G$. We can pull that back to a map $ \exp'$ from $ H$ to $ G'$: the path $ \exp(tX)$ in $ G$ can be pulled back to a path $ \exp'(tX)$ in $ G'$, and we just pick out the individual points on that path.
What can we do with that? We can follow the flow of a (complete) vector field, and use that to construct (left) multiplication by $ \exp(X)$ or $ \exp'(X)$. That isn't necessarily everything, but it does contain a neighborhood of the identity, and thus generates everything. This is automatically associative and includes inverses $ \exp(X)^{ \minus{} 1} \equal{} \exp( \minus{} X)$.

So, the answer is yes. $ G'$ is also a Lie group, and we can construct its multiplication from the multiplication on $ G$, together with the covering map.[/quote]

Thank you jmerry.
I have one more question Why do you need G to be connected?
Where does your argument fail if G is not connected?
Thank you in advance

The subgroup generated by the image of the exponential is the connected component of the identity. If that isn't everything, there's no canonical extension to other components by this method. Now, that's an open, closed, normal subgroup; if we quotient by it, we get a discrete group, which just gets duplicated some number of times. We can extend the group structure there, but how we do it isn't canonical.

I also note that your musings about universal coverings assume connectedness, as do all invocations of covering spaces I've ever seen.

Here is another approach using the universal covering and some more natural identifications around covering mappings in the category of connected pathwise connected topological groups which is larger than the category of Lie groups. If $ G$ is additionally assumed to carry a structure of a Lie group, then the mappings which will be derived in the larger category automatically descend to smooth mappings.

This approach has the advantage to globally define the topological group structure on $ G'$, whereas the method using the image of the exponential map uses the differential structure and the gap of being well-defined needs to be filled. The latter problem needs to be tackled if the image of exp is not surjective.

So let $ G$ be a connected topological group and $ p: G'\to G$ a covering map. Then let $ \pi': \widehat{G}\to G'$ be a universal covering of $ G'$. Note that $ \pi: \equal{} p\circ \pi': \widehat{G}\to G$ is a universal covering of $ G$. Picking an element $ e_{\widehat{G}}\in \pi^{ \minus{} 1}(\{e_G\})$ we can lift the group structure of $ G$ to a unique structure on $ \widehat{G}$ as a topological groups with identity element $ e_{\widehat{G}}$ such that $ \pi$ is a group homomorphism (monodromy theorem).

Now we want to push down the group structure of $ \widehat{G}$ to $ G'$ via $ \pi'$: For this we use the [b]natural[/b] group isomorphism $ \Phi_G: \pi_0(G)\cong ker(\pi)$ between the fundamental group of $ G$ and the kernel of $ \pi$:  where $ \Phi_G([\alpha]) \equal{} \widehat{\alpha}(1)$ with the lift $ \widehat{\alpha}$ w.r.t. $ \pi$ of a representing element of $ [\alpha]\in\pi_0(G)$. This gives the idea of how to construct the group structure of $ G'$ by stepping through the following road map:
[list]1 - $ p$ induces an injection $ \pi_0(G')\hookrightarrow\pi_0(G)$ providing a subgroup $ H$ of $ ker(\pi)$.

2 - $ H$ is a discrete subgroup of $ \widehat{G}$, hence it is a normal subgroup such that the abstract quotient group $ \widehat{G}/H$ carries a unique structure of a topological space such that the quotient map $ q_H: \widehat{G}\to \widehat{G}/H$ is a covering map.

3 - The group structure of $ \widehat{G}/H$ is automatically continuous such that $ \widehat{G}/H$ is a topological group and $ q_H$ is a group homomorphism of topological groups.

4 - $ \pi'$ is $ H$-invariant, i.e. $ \pi'\circ L_h \equal{} \pi'$ for all $ h\in H$, thus $ \pi'$ factors through $ \pi'': \widehat{G}/H\to G'$.

5 - $ \pi''$ is a covering mapping such that $ \pi_0(\pi''): \pi_0(\widehat{G}/H)\to \pi_0(G')$ is an isomorphism (use $ \pi_0(\widehat{G}/H)\cong H$). This implies that every fibre of $ \pi''$ has exactly one element, thus $ \pi''$ is a homeomorphism.

6 - Define the group structure on $ G'$ such that $ \pi''$ is an isomorphism of topological groups.

7 - Finally show that the group structures are compatible with the covering mappings, i.e. the covering mappings are group homomorphisms.[/list]


ratio
calculus
integration
number theory
prime factorization

(1)For the number 420, what is the ratio of the number of positive odd integral factors to the number of positive even integral integral.

I listed out all the factors, but it takes up a lot of time.  Is there a way to do this using the prime factorization?

That's basically what he did.

Hint: Odd times odd is even; even times even is even; odd times even is even.

Taking the prime factorization as nick42 did, $ 420 \equal{} 2^2\cdot{3}\cdot{5}\cdot{7}$. The even factors of 420 must have a two as one of there factors. So they either have to use $ 2^1$ or $ 2^2$(Thats two possibilities). The odd factors all use $ 2^0$ (1 possibilty). So the answer is $ \frac {1}{2}$.

Does that make sense?

positive even factors need to have at least one 2.

therefore, the number of positive even factors of 420 is number of factors of $ 2*3*5*7$ which is 16.

and the total number of factors of 420 is $ 2^2*3*5*7\equal{}24$

hence, the answer is $ \frac{8}{16}$

answer: $ \frac{1}{2}$

[quote="isabella2296"]That's basically what he did.

Hint: [b]Odd times odd is even[/b]; even times even is even; odd times even is even.[/quote]

Enough said.

[quote="FantasyLover"]positive even factors need to have at least one 2.

therefore, the number of positive even factors of 420 is number of factors of $ 2*3*5*7$ which is 16.

and the total number of factors of 420 is $ 2^2*3*5*7 \equal{} 24$

hence, the answer is $ \frac {8}{16}$

answer: $ \frac {1}{2}$[/quote]

When you find the even factors of 210, you still have to list them out, right?  What I was thinking is using the prime factorization $ 2^2\times3\times5\times7$ and multiply all the odds with odds and evens with evens and odds. So I did $ 3$, $ 5$, $ 7$ amongst themselves and found out there are 7 odds ($ 3, 5, 7, 15, 35, 21, 105$).  420 has $ 24$ factors, so the number of even factors must be $ 24 \minus{} 7 \equal{} 17$, and $ \frac {7}{17}$ doesn't equal $ \frac {1}{2}$.

EDIT!!!:::  Wait....for the odd factors I forgot $ 1$!!!, so that must mean: $ \frac {7 \plus{} 1}{17 \minus{} 1} \equal{} \frac {8}{16} \equal{} \boxed{\frac12}$!!!

I don't know id this is what you did FantasyLover....I kinda didn't get what you were saying.  Yay....

Yay is for me since I solved the problem...not because I didn't get your post.  yay...

[quote="FantasyLover"]420 has 8 odd factors: 1, 3, 5, 7, 15, 21, 35, 105

EDIT: sry im not that good a teacher...[/quote]

yeah....when I was doing it from the prime factorization, i forgot to include $ 1$, that was my mistake.  But thanks anyway!!

You don't even need to find the number of odd and even factors. The answer is always $ \frac{1}{n}$ where $ n$ is the power of the 2 in the number's prime factorization

[quote="tinytim"]You don't even need to find the number of odd and even factors. The answer is always $ \frac {1}{n}$ where $ n$ is the power of the 2 in the number's prime factorization[/quote]

Very helpful.  BTW: You copied Yongi's avatar, right?

Sorry to revive this, but tinytim why is that trick of yours true?

Because if it is 2^n, the factors are odd for 2^0, and even for 2^1,2^2...2^n. Thus for every one odd factor, there are n even ones.


[hide="Hint"]
Substitute values of $ x$ until you get $ P(4)$ on a side of the equation and then solve for it. Usually, $ 0$ is a good choice to plug in first.
[/hide]

Note that we must have $ \deg P\ge2$. Assume that $ \deg P\equal{}n>2$ such that $ P(x)\equal{}a_nx^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots$ so that in order for

$ \deg P(x)\plus{}\deg P(2x)\equal{}2$, we must have $ a_n\plus{}a_n\cdot2^n\equal{}0$, and since $ a_n\neq0$, $ 1\plus{}2^n\equal{}0$, which is clearly impossible.

Thus, we can only have $ \deg P\equal{}2\implies P(x) \equal{} ax^2 \plus{} bx \plus{} c$ for constants $ a,b,c$, so that

$ P(x) \plus{} P(2x) \equal{} 5ax^2 \plus{} 3bx \plus{} 2c \equal{} 5x^2 \minus{} 18$

$ \implies(a,b,c) \equal{} (1,0, \minus{} 9)\implies P(x) \equal{} x^2 \minus{} 9\implies P(4) \equal{} \boxed7$.


linear algebra
matrix
vector
linear algebra unsolved

Let $A \in R^{k \times n}$ be a matrix of full rank $k<n$.  Let the vector $b \in R^{k}$ and the symmetric positive definite matrix $M \in R^{n \times n}$ be given.

Find a matrix $B \in R^{(n-k) \times n}$ and a vector $d \in R^{n}$ so that $W=\left[ \begin{array}{c}A\\\noalign{\medskip}B\end{array}\right] \in R^{n \times n}$ is full rank and that the solution $x$ of $Wx=d$ solves the minimization problem:  $\min x^{T}Mx$ under the constraint $Ax=b$.

(The notation is lousy, but I did not write the problem).

By Lagrange's method if $x$ is the solution then there exists a $(1,k)$ row $\lambda$ s.t $\forall{h}\in\mathbb{R}^{n}: x^{T}Mh-\lambda{Ah}=0$. Thus $Mx=A^{T}\lambda^{T}$ or $Mx\in{Im(A^{T})}$.
Let $\{e_{1},\cdots,e_{n-k}\}$ a $ker(A)$-basis. $y\in{Im(A^{T})}$ is equivalent to $\forall{k}, e_{k}^{T}y=0$.
We choose \[B=\begin{pmatrix}e_{1}^{T}M\\\cdots\\e_{n-k}^{T}M\end{pmatrix},d=\begin{pmatrix}b\\0_{n-k}\end{pmatrix},\]

Remark: $y\rightarrow{y^{T}My}$ is an enclidean norm; thus $x$ is the orthogonal projection (modulo this norm) of the origin on the affine subspace of dimension $n-k$ defined by $Ax=b$.


analytic geometry
rotation
combinatorics unsolved
combinatorics

Players $ A,B$ alternately move a knight on a $ 1994\times 1994$ chessboard. Player $ A$ makes only horizontal moves, i.e. such that the knight is moved to a neighboring row, while $ B$ makes only vertical moves. Initally player $ A$ places the knight to an arbitrary square and makes the first move. The knight cannot be moved to a square that was already visited during the game. A player who cannot make a move loses. Prove that player $ A$ has a winning strategy.

Let's refer to places on the board by coordinates (x,y) where
x, y are both positive integers.  Let us assume that putting the knight on
(1,1) and then moving the knight to (3,2) puts A in a loosing position
(otherwise the proof is complete already).
Let us denote by (X,Y) the next move by B (such that it puts A into a
loosing position again).
Now, (just rotate the board) notice that putting the knight initially
at (2,3) then moving it to (Y,X) yields a winning strategy for A.
In fact, this situation differs from the former only by the fact that
the knight hadn't visited (1,1) prior, but this place is unreachable for
player B anyway so that the set of winning positions for A can only
be larger.

yours,
Wojtek


I asked this question. To solve it, I did this:

Let Great Aunt Minnie's age be $ m$.

Let Great Uncle Paul's age be $ p$.

$ m \plus{} p^2 \equal{} 7308$

$ p \plus{} m^2 \equal{} 6974$

We want to find $ p \plus{} m$.

Add our two equations together:

$ m \plus{} m^2 \plus{} p \plus{} p^2 \equal{} 14282$

$ m \plus{} p \equal{} 14282 \minus{} m^2 \plus{} p^2$

Write $ p$ in terms of $ m$.

$ m \plus{} 6974 \minus{} m^2 \equal{} 14282 \minus{} m^2 \plus{} (6974 \minus{} m^2)^2$

And then it starts to get messy. Is there an easier way?

$ p \equal{} 6974 \minus{} m^2$

Plug it into the other equation.

$ m \plus{} (6974 \minus{} m^2)^2 \equal{} 7308$

$ m \plus{} 48636676 \minus{} 13948m^2 \plus{} m^4 \equal{} 7308$

$ m^4 \minus{} 13948m^2 \plus{} m \equal{} \minus{}48629368$

Er... this is pretty messy...

Because the algebra would be too messy, Im not going to do it. But, I have a few suggestions:

1) Since the difference of the two equations is a bit far, then p might be greater than p by maybe a 19. Then you could start Brute Forcing, which I may attempt to do later.

2) Complete the square? that may be messy as well.

Brute Force.

First, I will make a table of the units digit for the possibitities.

$ m \plus{} p^2 \equal{} 7308$
if $ m\equal{}1$, $ p^2\equal{}7$ (impossible)
if $ m\equal{}2$, $ p^2\equal{}8$ (impossible)
if $ m\equal{}3$, $ p^2\equal{}9$ (p=3,7)
if $ m\equal{}4$, $ p^2\equal{}0$ (p=0)
if $ m\equal{}5$, $ p^2\equal{}1$ (p=1,9)
if $ m\equal{}6$, $ p^2\equal{}2$ (impossible)
if $ m\equal{}7$, $ p^2\equal{}3$ (impossible)
if $ m\equal{}8$, $ p^2\equal{}4$ (p=2,8)
if $ m\equal{}9$, $ p^2\equal{}5$ (p=5)
if $ m\equal{}0$, $ p^2\equal{}6$ (p=4,6)
Since a perfect square cannot end in 2, 3, 7, or 8 (or I think something like that), some of them are impossible

Then we try to see which one fits the other way around.
$ m^2 \plus{} p \equal{} 6974$
if m=3, p=5 (ruled out)
if m=4, p=8 (ruled out)
if m=5, p=9
if m=8, p=0 (ruled out)
if m=9, p=3 (ruled out)
if m=0, p=4

Thus, the only ordered pairs of (m,p), where m and p are the units digits of the Uncle's and Aunt's ages respectively are $ (5,9)$ and $ (0,4)$. Thats all I can get so far.

[quote="GameBot"]Great Aunt Minnie's age plus the square of Great Uncle Paul's age is $ 7308$. Great Uncle Paul's age plus the square of Great Aunt Minnie's age is $ 6974$. What is the sum of Great Aunt Minnie's age and Great Uncle Paul's age?[/quote]

Well

$ 84 < p \leq 85$
$ 82 < m \leq 83$

and the answer is 168.

There is no need to do any "algebra" since we are dealing with integers here.

Oh and dragon, please check your work.

Should be:

m=1 p^2=7
m=2 p^2=[b]6[/b]

then continues...

[quote="james4l"][quote="GameBot"]Great Aunt Minnie's age plus the square of Great Uncle Paul's age is $ 7308$. Great Uncle Paul's age plus the square of Great Aunt Minnie's age is $ 6974$. What is the sum of Great Aunt Minnie's age and Great Uncle Paul's age?[/quote]

Well

$ 84 < p \leq 85$
$ 82 < m \leq 83$

and the answer is 168.
[/quote]

And how did you get that?

LATEX:
Let Great Aunt Minnie's age be $ m$.
Let Great Uncle Paul's age be $ p$.
$$m + p^2 = 7308$$
$$m^2 + p = 6974$$
Subtracting the 2 equations gets:
$$m + p^2 - m^2 -p=334$$
$$m - p+(p-m)(p+m) = 334$$
$$(m-p)(p+m-1) =334 =167*2$$
so $m+p-1 = 167$, thus the answer is $\boxed {168}$


$ a_1\equal{}1$ and for every integer $ n\geq 2$, $ |a_n| \equal{} |a_{n\minus{}1}\plus{}2|$ is given. Find the minimum value of $ \sum_{i\equal{}1}^{2000}a_i$.

A) -4000 B) -3000 C) -2000 D) -1000 E) None

The way $ 1,3,\minus{}5,\minus{}3,\minus{}1,\minus{}1,...$ also gives the same result. So taking the negative is not a winning strategy. I think you need some proof.

I got
$ \sum_{i\equal{}1}^{n}a_{i} \ge\frac{\minus{}n}{2}$
so the correct answer is $ D$.
[hide="Hint"]Square and sum it till $ n$ terms[/hide]

[hide]
Let $ b_n \equal{} a_n \plus{} 1$. Then the sequence becomes $ b_1 \equal{} 2$ and for $ n\geq 2$, $ |b_n \minus{} 1| \equal{} |b_{n \minus{} 1} \plus{} 1|$. In particular, given $ b_k$, the possible values for $ b_{k \plus{} 1}$ are $ \{ \minus{} b_k,b_k \plus{} 2\}$.

[b]Lemma:[/b] $ \sum_{k \equal{} 1}^n b_k\geq 0$ for all integers $ n\geq 1$.
[b]Proof.[/b] The base case, $ n \equal{} 1$, is clear. Also, for $ n \equal{} 2$, the sum is either $ 2 \plus{} 4$ or $ 2 \plus{} ( \minus{} 2)$, both of which are nonnegative.

Now, for $ n\geq 3$, if all elements are nonnegative, we are done, so consider the first negative element $ b_k$. Then $ b_{k \minus{} 1}$ is positive, so $ b_k \equal{} \minus{} b_{k \minus{} 1}$ and $ b_{k \minus{} 1} \plus{} b_k \equal{} 0$. Therefore we can form a smaller sequence by excluding $ b_k$ and $ b_{k \minus{} 1}$ and the sum is nonnegative by the induction hypothesis. Also, $ b_{k \minus{} 2}\in\{ \minus{} b_k \minus{} 2,b_k\}$ and $ b_{k \plus{} 1}\in\{b_k \plus{} 2, \minus{} b_k\}$, so this new sequence still follows the rules defined by the sequence.

Now, $ \sum_{k \equal{} 1}^n a_k \equal{} \left(\sum_{k \equal{} 1}^n b_k\right) \minus{} n\geq \minus{} n$, and so for 2000 elements the lower bound is $ \minus{} 2000$. We can easily show that this can be achieved by either modularmarc101's or xeroxia's examples, hence the answer is $ \textbf {(C)}$.[/hide]

[quote="xeroxia"]$ a_1 \equal{} 1$ and for every integer $ n\geq 2$, $ |a_n| \equal{} |a_{n \minus{} 1} \plus{} 2|$ is given. Find the minimum value of $ \sum_{i \equal{} 1}^{2000}a_i$.

A) -4000 B) -3000 C) -2000 D) -1000 E) None[/quote]

Plz tell which is the correct answer,mine or Yongi's?

Two other users already showed how to get a sum of -2000, so the minimum value of the sum can't possibly be -1000  :roll:


Anyone know of an AoPS-like calculus book that starts at the introductory level? I'm starting calculus soon and it'd be really nice to have a good book, not just joe textbook.

I really liked Schaum's outline of Beginning Calculus for its clarity.  I didn't always get the notation part of it (like $\frac{d}{dx}$) at first, but besides that it's one of the best books i've used (probably next to AoPS and AaCoPS).  Preferably you should have someone you can at least ask for help but you'll be fine either way.

I personally like the book "A First Course in Calculus" by Serge Lang, Springer publications.

I like the books written by Anton or Stewart.
They are not like AoPS style, but Anton does a great job of providing proof on why stuff works and Stewart had great practice problems.


hi i'm new to this forum so i'm not quite sure if set problems belong to the intermediate section

it's not excatly an problem but more or less a concept question

is it possible to prove the principle of inclusion/exclusion for four intersecting sets?

i understand how it is proven with three sets and my txtbook says knowing the principle for three sets means you automatically should know the proof for the principle of 4 sets, without much explanation, therefore i am lost...  :(

any help is appreciated, thx :)

yes it is, but i forgot and i'm too lazy to derive it :|

You can just prove it for $n$ sets with a counting argument.

You can prove it for the general case if you can prove the identity \[ n-\binom{n}{2}+\binom{n}{3}-\binom{n}{4}+\cdots+(-1)^{r-1}\binom{n}{n}= 1\]

[quote="Fraxia"]You can prove it for the general case if you can prove the identity \[n-\binom{n}{2}+\binom{n}{3}-\binom{n}{4}+\cdots+(-1)^{r-1}\binom{n}{n}= 1\] [/quote]
Use Newton's binomial formula: $0=(1+(-1))^{n}=\sum_{i=0}^{n}\binom{n}{i}(-1)^{i}$.

[quote="princevegetam"][quote="Fraxia"]You can prove it for the general case if you can prove the identity \[n-\binom{n}{2}+\binom{n}{3}-\binom{n}{4}+\cdots+(-1)^{r-1}\binom{n}{n}= 1\] [/quote]

i'm sorry but what exactly does that mean?  :huh:

what does n stand for?

and what is happening to the n and the number inside the brackets? is that a fraction...[/quote]

In this case, $n$ is any positive integer. The thing in the parentheses, for example $\binom{n}{4}$ means the number of ways to choose 4 things out of a set of $n$. For example, if I have $n$ colored balls, each a different color, and I want to pick out 4 of the balls, $\binom{n}{4}$ is the number of different ways I can do this. There are mathematical ways to compute this.
As to your original question, you can prove the principle of inclusion-exclusion for 4 sets by using the one for 3 sets on the first 3 sets you have, then the one for 2 sets. Like this:
$a \cup b \cup c \cup d = (a \cup b \cup c) \cup d$
Hope that helps.

To find the total number of distinct elements among any number of sets, you alternate between adding and subtracting the intersections.

The total number of elements equals:

$(1)$ the sum of the number of elements in each individual set

$(2)$ minus the sum of the number of elements in each pair of sets

$(3)$ plus the sum of the number of elements in each set of three sets

etc.

To prove that this works, you want to show that each element, no matter how many sets its part of, is counted exactly once in the above expression.

For an element in $n$ sets, in $(1)$, it's counted $n$ times. In $(2)$, it's subtracted $ \binom{n}{2}$ times. In $(3)$, it's added back in $ \binom{n}{3}$ times etc... You can prove that this expression equals one by plugging in $x=1, y=-1$ in the expansion of $(x+y)^{n}$ with the Binomial Theorem.


[quote="hedeng123"][quote="vuthanhtu_hd"][b]Problem[/b]. Let $ a,b,c$ be non-negative real numbers and $ 0 \le k \le \frac {1}{2}$.Prove that

$ \prod (a \plus{} k\sqrt {bc}) \ge \prod ((k \minus{} 1)a \plus{} b \plus{} c)$

 :)

I think this problem is true and It's old problem but I donn't know author  :)[/quote]

$ a \equal{} x^2,b \equal{} y^2,c \equal{} z^2$

$ \prod (a \plus{} k\sqrt {bc}) \ge \prod ((k \minus{} 1)a \plus{} b \plus{} c)$

$ \Longleftrightarrow f(k) \equal{} ( \minus{} x^2y^4 \plus{} xzy^4 \plus{} 3y^2z^2x^2 \plus{} yzx^4 \minus{} z^4x^2 \minus{} x^4z^2 \minus{} y^4z^2$ $ \minus{} z^4y^2 \minus{} x^4y^2 \plus{} xyz^4)k^2 \plus{}$

$ (z^4y^2 \plus{} y^3x^3 \plus{} y^4z^2 \plus{} z^4x^2 \plus{} x^3z^3 \minus{} y^6 \plus{} x^4y^2 \plus{} x^4z^2 \minus{} 6y^2z^2x^2 \plus{} y^3z^3 \plus{}$ $ x^2y^4 \minus{} z^6 \minus{} x^6)k$

$ \minus{} z^4x^2 \plus{} x^6 \plus{} y^6 \plus{} z^6 \minus{} y^4z^2 \minus{} x^2y^4 \minus{} x^4y^2 \minus{} z^4y^2 \plus{} 3y^2z^2x^2 \minus{} x^4z^2\geq 0$

Note

$ \minus{} x^2y^4 \plus{} xzy^4 \plus{} 3y^2z^2x^2 \plus{} yzx^4 \minus{} z^4x^2 \minus{} x^4z^2 \minus{} y^4z^2$ $ \minus{} z^4y^2 \minus{} x^4y^2 \plus{} xyz^4$

$ \equal{} \minus{} (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \minus{} 2\sum(y^2z^2(x \minus{} y)(x \minus{} z))$ $ \minus{} xyz\sum(x(x \minus{} y)(x \minus{} z)) \minus{} xyz\sum((y \plus{} z)(x \minus{} y)(x \minus{} z))\le 0$

we only prove $ f(0)\geq 0$ and $ f(\frac {1}{2})\geq 0$

$ f(0) \equal{} \minus{} z^4x^2 \plus{} x^6 \plus{} y^6 \plus{} z^6 \minus{} y^4z^2 \minus{} x^2y^4 \minus{} x^4y^2 \minus{} y^2z^4 \plus{} 3x^2y^2z^2 \minus{} x^4z^2$

$ \equal{} \sum(x^4(x \minus{} y)(x \minus{} z)) \plus{} \sum(x^3(y \plus{} z)(x \minus{} y)(x \minus{} z)) \plus{} xyz\sum(x(x \minus{} y)(x \minus{} z))\geq 0$

$ f(\frac {1}{2}) \equal{} \frac {1}{4}(3x^2y^2z^2 \plus{} xy^4z \plus{} 2z^6 \plus{} x^4yz \minus{} 3x^4y^2 \minus{} 3x^4z^2 \plus{}$ $ 2x^3y^3 \minus{} 3x^2y^4 \minus{} 3y^4z^2 \plus{} 2y^6$ $ \plus{} 2y^3z^3 \minus{} 3y^2z^4 \plus{} 2x^6 \minus{} 3z^4x^2 \plus{} 2z^3x^3 \plus{} yz^4x)$
$ \equal{} \frac {1}{4}(2\sum(x^4(x \minus{} y)(x \minus{} z)) \plus{} 2\sum(x^3(y \plus{} z)(x \minus{} y)(x \minus{} z)) \minus{} (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2$ $ \plus{} xyz\sum(x(x \minus{} y)(x \minus{} z)) \plus{} xyz\sum((y \plus{} z)(x \minus{} y)(x \minus{} z)))\geq 0$[/quote]

Thanks hedeng123 :)


For matrices $ A \equal{} \left( \begin{array}{cc} 1 & 2 \\
\minus{} 3 & 4 \end{array} \right), \ B \equal{} \left( \begin{array}{cc} 4 & 1 \\
\minus{} 3 & 2 \end{array} \right),$ Find $ (2A \plus{} B)(2A \minus{} B) , 4A^2 \minus{} B^2$.

Question from kunny:

What should we learn from the problem?

$ 2A \equal{}\left(\begin{array}{cc}2 & 4\\ \minus{}6 & 8\end{array}\right)$

$ 2A\plus{}B\equal{}\left(\begin{array}{cc}6 & 5\\ \minus{}9 & 10\end{array}\right)$

$ 2A\minus{}B\equal{}\left(\begin{array}{cc}\minus{}2 & 3\\ \minus{}3 & 6\end{array}\right)$

$ (2A\plus{}B)(2A\minus{}B)\equal{}\left(\begin{array}{cc}6 & 5\\ \minus{}9 & 10\end{array}\right)\left(\begin{array}{cc}\minus{}2 & 3\\ \minus{}3 & 6\end{array}\right)$

$ \equal{}\left(\begin{array}{cc}\minus{}27 & 48\\ \minus{}12 & 33\end{array}\right)$

$ A^2\equal{}\left(\begin{array}{cc}1 & 2\\ \minus{}3 & 4\end{array}\right)\left(\begin{array}{cc}1 & 2\\ \minus{}3 & 4\end{array}\right)$

$ A^2\equal{}\left(\begin{array}{cc}\minus{}5 & 10\\ \minus{}15 & 10\end{array}\right)$

$ 4A^2\equal{}\left(\begin{array}{cc}\minus{}20 & 40\\ \minus{}60 & 40\end{array}\right)$

$ B^2\equal{}\left(\begin{array}{cc}13 & 6\\ \minus{}18 & 1\end{array}\right)$

$ 4A^2\minus{}B^2\equal{}\left(\begin{array}{cc}\minus{}43 & 34\\ \minus{}42 & 39\end{array}\right)$

$ \Longrightarrow (2A\plus{}B)(2A\minus{}B) \ne 4A^2\minus{}B^2$

[quote]
What should we learn from the problem?[/quote]

Algebraic identities aren't always true for matrices  :maybe:

You seemed to make a mistake in calculation of $ 4A^2\minus{}B^2$.

[quote]$ 4A^{2}\minus{}B^{2}\equal{}\left(\begin{array}{cc}\minus{}43 & 34\\ \minus{}42 & 39\end{array}\right)$[/quote]

It should have been $ \left(\begin{array}{cc}\minus{}33 & 34\\ \minus{}42 & 39\end{array}\right)$

[quote="picture74"][quote]$ 4A^{2} \minus{} B^{2} \equal{} \left(\begin{array}{cc} \minus{} 43 & 34 \\
\minus{} 42 & 39\end{array}\right)$[/quote]

It should have been $ \left(\begin{array}{cc} \minus{} 33 & 34 \\
\minus{} 42 & 39\end{array}\right)$[/quote]

Now that's okay. :lol:

[quote="picture74"]
[quote]
What should we learn from the problem?[/quote]

Algebraic identities aren't always true for matrices  :maybe:[/quote]

What would become the result of $ (2A \plus{} B)(2A \minus{} B)$ for matrices $ A,\ B?$

The issue is this: $ (2A\plus{}B)(2A\minus{}B)$ would equal $ 4A^2\minus{}B^2$ if $ A$ and $ B$ commuted - that is, if $ AB\equal{}BA.$ But these two matrices don't commute and the alleged identity doesn't work.

The lesson is this: algebra as you know it does work for matrices, mostly, but the non-commutativity of multiplication is a fact of life that you're just going to have to adjust to.

Non-commutativity is also why we never write $ \frac{A}{B}$ for $ A$ and $ B$ matrices; instead we write either $ B^{\minus{}1}A$ or $ AB^{\minus{}1},$ whichever of those two expression we mean (and we have to choose, since they may not be equal.)

[quote="picture74"][quote="kunny"]What would become the result of $ (2A \plus{} B)(2A \minus{} B)$ for matrices $ A,\ B?$[/quote]
I didn't really get what you meant  :([/quote]

The answer is $ 4A^2 \minus{} 2AB \plus{} 2BA \minus{} B^2$.


Show that $x=a+rcos\theta, y= b+rsin\theta$ defines a circle with $center (a,b)$ and radius $r$.


inequalities
Canada
am-gm inequality
AM-GM
algebra

By rearrangement: $a^3 \geq b^3 \geq c^3$ and $\frac{1}{bc} \geq \frac{1}{ac} \geq \frac{1}{ab}$.

Yes, the inequality is simple applications of $x^2+y^2+z^2\geq xy+yz+zx$ for $a^2, b^2, c^2$ and $ab, bc,ca,$.
We have
\[ a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2\geq abc(a+b+c). \]

Also you can prove the last one by the Muirheads Inequality...

Davron

Now sum cyclically using AM-GM that
$\frac{a^3}{2bc}+\frac{b^3}{4ac}+\frac{c^3}{4ab}\ge a$

By $Cauchy-Schwarz$ and $AM-GM$, we have:
\[\sum\frac{a^3}{bc}\ge \frac{(a^2+b^2+c^2)^2}{3abc}\ge \frac{(ab+bc+ca)^2}{3abc}=a+b+c\]

Very simple.
By AM-GM
\[\sum( \frac{a^3}{bc} +b+c)\ge 3\sum a\iff  \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c \]

Too easy.Using the fact that $\sum {x^2}\ge \sum{xy}$ we see that
$\sum{a^4} \ge \sum{a^2b^2} \ge \sum{a^2bc}$.
Result follows by dividing both sides by $abc$.

Bye...

It's trivial by chebychev, $\sum \frac{a^3}{bc} \ge\frac{a+b+c}{3} \sum \frac{a^2}{bc} \ge a+b+c$
I am learning inequalities and found chebychev very useful tool.

Inequality is equivalent to proving that $f(a,b,c) \geq 0$ where
\[
f(a,b,c)=a^4+b^4+c^4-a^2 b c-a b^2 c-a b c^2
\].
It is symmetric so we can take $a=c+x+y$ and $b=c+x$ for some $x,y \geq 0$ and then
\[
f(a,b,c)=f(c+x+y,c+x,c)=5 c^2 x^2+5 c^2 x y+5 c^2 y^2+6 c x^3+9 c x^2 y+11 c x y^2+4 c y^3+2 x^4+4 x^3 y+6 x^2 y^2+4 x y^3+y^4 \geq 0
\]
Done.

[quote=shobber]Prove that for all positive real numbers $a$, $b$, and $c$,
\[ \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c \]
and determine when equality occurs.[/quote]

The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers such that $a^{15}+b^{15}+c^{15}=3$. Prove that:
\[\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq3\]

[quote=shobber]Prove that for all positive real numbers $a$, $b$, and $c$,
\[ \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c \]
and determine when equality occurs.[/quote]

$\iff a^4+b^4+c^4\geq a^2b^2+c^2a^2+b^2c^2\geq abc(a+b+c) $ very simple

[quote=shobber]Prove that for all positive real numbers $a$, $b$, and $c$,
\[ \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c \]
and determine when equality occurs.[/quote]

alternatively by tchebytcheff we have
$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}\ge [\frac{(a+b+c)}{3}](\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab})$
but by cauchy
$[\frac{(a+b+c)}{3}](\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab})\ge a+b+c$

as $(a+b+c)^2\ge 3(ab+bc+ca)$

Here is another solution:
 Note that the inequality is homogeneous in $a,b,c$ so we let $abc=1$. The first observation is that, by AM-GM, $a+b+c \geq 3\sqrt[3]{abc} = 3$. Further, the given inequality reduces to $a^4 + b^4 + c^4 \geq a + b + c$. This is easy, as by the AM-QM (or the power mean) inequality, we have $a^4 + b^4 + c^4 \geq 4 \cdot (\frac{a+b+c}{4})^4 = (a+b+c) (\frac{a+b+c}{3})^3 \geq a+b+c$, and we are done.

[quote=shobber]Prove that for all positive real numbers $a$, $b$, and $c$,
\[ \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c \]
and determine when equality occurs.[/quote]
$$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}\geq\sqrt[15]{3^{14}(a^{15}+b^{15}+c^{15})}$$
is also true.
There is solution by hand.

[quote=shobber]Prove that for all positive real numbers $a$, $b$, and $c$,
\[ \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c \]
and determine when equality occurs.[/quote]

By am-gm: $$\sum_{\text{cyc}} \frac{a^3}{bc} +b+c \geq \sum_{\text{cyc}} 3a \geq \sum_{\text{cyc}} a$$ Equality when $a=b=c.$

Time $abc$ for both sides and we have to prove $a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab$ which is trivial by Muirhead

$\[ \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c \]$
Implies
$\[a^4+b^4+c^4\geq a^2bc+ab^2c+abc^2\]$
Since $(4,0,0)$ Majorizes $(2,1,1)$ This will always hold by muirhead.
Equality occurs when $a=b=c$

Trivial by Murihead Inequality:

Rearrange gives $$a^5bc+b^5ac+c^5ab\geq a^3b^2c^2+a^2b^3c^2+a^2b^2c^3$$

Which finishes since $(5, 1, 1)\succ \ (3, 2, 2)$. Equality holdes iff $a=b=c$.

[i]q.e.d.[/i]

Multiplying the inequality by $abc$ gives $$a^4+b^4+c^4 \geq a^2bc+ab^2c+abc^2$$
By AM-GM:
$$a^4+b^4+c^4 \geq a^2b^2+a^2c^2+b^2c^2$$
Also by AM-GM:
$$a^2b^2+a^2c^2+b^2c^2 \geq a^2bc+ab^2c+abc^2$$
Therefore,
$$a^4+b^4+c^4 \geq a^2bc+ab^2c+abc^2$$
Equality is when $a=b=c$. $\square$

[quote=Gibbenergy][quote="shobber"]Prove that for all positive real numbers $a$, $b$, and $c$,
\[ \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c  \]
and determine when equality occurs.[/quote]

  It isn't hard. :D
By AM-GM, we have
$\frac{a^3}{bc} +b+c \geq 3a$
Sum up  and done[/quote]

that's what I did lol

Given that $a, b, c$ are positive reals, multiplying both sides of the inequality by $abc$ attains
$$a^4+b^4+c^4 \geq a^2bc+ab^2c+abc^2$$
meaning that
$$\sum_{cyc}a^4\geq \sum_{cyc} a^2bc$$
since $(4,0,0)\succ (2,1,1)$ the inequality is true by Muirhead.

Multiply on $abc$ on both sides to get $a^4+b^4+c^4 \ge a^2bc+a^2bc+abc^2$
Because $(4,0,0) > (2,1,1)$, the inequality is true

[quote=CrazyInMath]One-Liner
\[\sum \frac{a^3}{bc}=\sum\frac{\frac{a^3}{bc}+\frac{c^3}{ab}}{2}\geq\sum\frac{ac}{b}=\sum\frac{\frac{ac}{b}+\frac{ab}{c}}{2}\geq\sum a\][/quote]

This is nice, but most solutions above are one-liner.
Here is a one0line rephrasement of the most commonly discussed solution.

By rearrangement $\frac{a^3}{bc} +\frac{b^3}{ca}+\frac{c^3}{ab} = \frac{a(a^2)}{bc}+\frac{b(b^2)}{ca}+\frac{c(c^2)}{ab} \ge \frac{a(bc)}{bc} + \frac{b(ca)}{ca}+\frac{c(ab)}{ab} = a+b+c.$

Multiply both sides by $abc$, so it suffices to prove $a^4+b^4+c^4\ge a^2bc+ab^2c+abc^2$.

Notice how $a^4+a^4+b^4+c^4\ge 4a^2bc$ by AM-GM. Summing this with its analogous inequalities proves the result, and equality occurs when $a=b=c$.

[quote=sqing]Prove that for all positive real numbers $a$, $b$, and $c$,
$$\frac{a^3}{bc} + \frac{b^2c^2}{a^2} \geq ab+c$$[/quote]

[b]Remark.[/b]  The claimed inequality $\frac{a^3}{bc} + \frac{b^2c^2}{a^2} \geq ab+c$ is [b]not[/b] true for all positive real numbers $a,b,c$. For instance,  if $\left(a,b,c\right)=\left(1,\frac{1}{2},3\right)$,  then
\begin{align*}\frac{a^3}{bc} + \frac{b^2c^2}{a^2}-( ab+c)
=&\frac{2}{3}+\frac{9}{4}-\frac{1}{2}-3\\
=&-\frac{7}{12}<0.
\end{align*}

Observe that $(4, 0, 0) \succ (2,1,1)$, so from Muirhead, $a^4 + b^4 + c^4 \ge a^2bc + b^2ca + c^2ab$, and dividing by $abc$ gives the desired result. Since Muirhead can be achieved from various applications of AM-GM, equality holds iff $a = b = c$.

Note that upon expansion we have,
\begin{align*}
\sum a^5bc \geq \sum a^3b^2c^2\\
\end{align*}
Then homogenize so that $abc = 1$. Then we wish to show,
\begin{align*}
\sum a^4 &\geq \sum a\\
\iff \sum a^4-a &\geq 0
\end{align*}
Now define $f(x) = x^4 - x$. We may check,
\begin{align*}
f''(x) = 12x^2
\end{align*}
and hence $f$ is convex everywhere. However then by Jensen's we may check,
\begin{align*}
f(a) + f(b) + f(c) \geq 3\cdot f\left(\frac{a+b+c}{3} \right)
\end{align*}
Note that by AM-GM
\begin{align*}
\frac{a+b+c}{3} \geq 1
\end{align*}
Finally we may check,
\begin{align*}
f([x \geq 1]) \geq f(1) = 0
\end{align*}
Indeed note that this holds as,
\begin{align*}
x^4 - x \geq 0 \iff x(x^3 - 1) \geq 0 \iff x \in (-\infty, 0) \cup (1, \infty)
\end{align*}
Thus we may finish as,
\begin{align*}
\sum f(a) \geq 3 \cdot f \left( \frac{a+b+c}{3} \right) \geq 3 \cdot f(1) \geq 0
\end{align*}


quadratics
arithmetic series
arithmetic sequence
number theory unsolved
number theory

[quote="hollandman"]Show there for any positive integer $ m$ there exist a positive integer $ N$ such that for all primes $ p > N$ there are $ m$ consecutive quadratic residues modulo $ p$.[/quote]

This is an amazingly difficult problem. It has appeared on one of the USA MOSP test and no one ever had ANY idea.

I can't recall the exact proof. But I can recall that it uses Van der Warden Theorem. Consider a number N so big that when we use red and blue to color numbers in range 1 to N-1, there'll always exist $ m^2$ numbers in an arithmatic sequence(like 2,7,12,17,...) that they are either all blue or all red. Now, let blue represent quadratic residue and red represent otherwise, the statement becomes that there exist $ m^2$ numbers(in arithmatic sequence) that either all quadratic residue or all otherwise.

I forget all the following things, and I can't really recall why $ m^2$ is needed.

[hide="a finish to the above"]
If $ 1,2,...,m$ are all quadratic residues, then we are done. Otherwise, suppose that there is at least one nonquadratic residue in that set. Of course, there is always at least one quadratic residue, such as $ 1$.

In the arithmetic series of $ m^2$ elements, let the common difference be $ d$. We can come up with arithmetic progression within this progression with $ m$ elements that have common difference of $ d,2d,...,md$. Among these possible differences, at least one must be a quadratic residue and at least one is a nonquadratic residue. Now, choose one of these differences $ ad$ such that $ \left(\frac{ad}{p}\right)$ is the same as the legendre of any element in this arithmetic progression.

Then, we can multiply each element in an arithmetic progression with difference $ ad$ by $ (ad)^{\minus{}1}$; this will give us an arithmetic progression of difference $ 1$, meaning they are consecutive integers. Also, since $ \left(\frac{ad}{p}\right)\equal{}\left(\frac{(ad)^{\minus{}1}}{p}\right)$ all elements in this new progression will all be quadratic residues.
[/hide]


Prove that any two numbers in the following sequence are prime.

$ 2 \plus{} 1, 2^{2^1} \plus{} 1, \dots , 2^{2^n} \plus{} 1$

Hints only, please, and hide them!

Thanks in advance.

EDIT: Oh, fixed.

[quote="hunter34"]Is $ n\in\mathbb{Z^ \plus{} }$[/quote]

i think $ n$ is just integers from 0-infinity, becuz it starts aas n=0

but some terms are not prime...
ehh
i didnt get the question

[quote="isabella2296"]Prove that any two numbers in the following sequence are [b]relatively[/b]prime.

$ 2 \plus{} 1, 2^{2^1} \plus{} 1, \dots , 2^{2^n} \plus{} 1$

Hints only, please, and hide them!

Thanks in advance.

[/quote]

Reverted back to original.

[hide]Fermat primes and look at the +1 part it makes a the problem trivial almost[/hide]

[quote="fishythefish"]I didn't check fully, but it seems that they are not just relatively prime, but completely prime.[/quote]

[hide]
No.  What about n=4, n=5, up to n=12?  Check wikipedia for "fermat numbers"
[/hide]

I figured it out. Thanks for your help, everyone!


algebra
polynomial
number theory
greatest common divisor
search
number theory unsolved

Let $P(x) \in Z[x]$ and $A=\{a_{1},a_{2},... a_{n}\}\subset N^{*}$ .
$\forall k \in Z, \ \ \exists a_{i}\in A$ such that $a_{i}| P(k)$
Prove that: $\exists a \in A$ such that $a | P(x) \forall x \in Z$

Rust, could you at least try to read the statement of these problems? Why on earth not taking $a=1$? After all, who said that $A$ contains the $gcd$ of its elements. By the way, the problem is taken from Leningrad olympiad and it is surely not an easy one. But it was discussed on the forum and Iura gave a solution. Use search button.

Yes, very obviously that $1 \in\{2,3\}$ if the polynomial is e.g. $2x$.

Ok, from now on I will just le you do whatever you want. It seems useless to try to say something, everything is too "obviously". Sleep in peace!


quadratics
number theory theorems
number theory

I was wondering: suppose you want to construct an equation to be solved in integers so that the sequence of solutions for one of the unknown $s_i$ has a property like: $2s_{i}+s_{i+1}=s_{i+2}$ (or any other recursive equation), does anyone know how to do this or maybe someone who can  show there isn't such an equtation?

You have $s_{i+2}=2s_{i}+s_{i+1}$
The charastic equation is $x^2=x+2 \rightarrow x^2-x-2=0$

$\text{With roots 2 and -1}$ therefore the general solution is:

$s_{n}=a({2}^n)+b({-1}^n)$

You get $a$ and $b$ from your starting values

ok I know, but you misunderstood the question: what I meant is can you find a diophantic equation (quadratic or whatever) like $3n^2+3n+1=m^2$ or $2^n+n^2=3^m$ which satisfies this: the sequence of solutions for $n$ satisfies an equation like the given one, so that it becomes easy to find the next solution.

I'll give an easy example: suppose you have to construct an equation so that the a sequence with all the solutions of one of the unknown satisfies $x_n=x_{n-1}-2$, then a good answer would be $x+2y=3$ because if $x_{n}$ provides a solution, then so does $x_{n}-2$ etc... This would give the sequence (...,-1,1,3,5,..) The corresponding sequence for $y_n$ would then satisfie $y_n=y_{n-1}+1$ and be (...,-2,-1,0,1,2,...).

I hope it is clear now although it is maybe a weird question

Ah, it is all clear now. And much more interesting!

does it have to give all the solutions, or just some of them?

if possible all of them, but even partial solutions may be usefull

Here i found a kinda answer for you.

Say were trying to find all $x$ such that the equation

$ax-2y=1$ has a solution, with $y$ an integer.

then if $x=s_i$ and $x=s_{i+1}$ are solutions, so is $x=s_{i+1}+2s_i$

I highly doubt its all of them, but im tired and want to go to bed...

indeed that is a nice equation (with solutions only if a is odd)
it will only give all (positive) solutions if you found out that $1$ and $3$ are solutions for $x$.

I think this might be generalized to all linear recursive equations, but I'm still thinking about it.


geometry
incenter
circumcircle
rectangle
angle bisector

In $ \Delta ABC$, $ M$ is the midpoint of $ II_{a}$. Show that $ M$ lies on the circumcircle of $ \Delta ABC$

[hide="hint"]$ MA\equal{}MB$[/hide]

[quote="CatalystOfNostalgia"]In $ \Delta ABC$, $ M$ is the midpoint of $ II_{a}$. Show that $ M$ lies on the circumcircle of $ \Delta ABC$

[hide="hint"]$ MA \equal{} MB$[/hide][/quote]
Is $ I_a$ the excenter of $ A$ in $ \triangle ABC$?  If so, then the hint should be:
[hide="Hint"]
$ MC \equal{} MB$.  [/hide]
Anyways, here is the solution.
[hide="Solution"]
Let the point of tangency with the $ A$-Excircle and $ BC$ be $ X$ and the point of tangency between the incircle and $ BC$ be $ Y$.  Furthermore, let the perpindicular to $ BC$ from $ M$ meet $ BC$ at $ Z$.
[b]Lemma: [/b]$ ZX \equal{} ZY$
[i]Proof:[/i] Let $ P$ be the point such that $ P$ is on $ I_aX$ extended and $ IP\perp I_aP$ and let $ Q$ be the point so that $ Q$ is on $ IY$ extended and $ IQ\perp I_aQ$. Since $ I_aX\perp XY$ and $ IY\perp XY$, we see that $ IP\parallel XY\parallel I_aQ$.  This means that $ \angle PXY \equal{} 180 \minus{} \angle I_aPI \equal{} 90$ and $ \angle PIQ \equal{} 180 \minus{} \angle IQI_a \equal{} 90$, so $ PIYX$ is a rectangle as are $ XYQI_a$ and $ PIQI_a$.  Thus, $ PQ$ bisects $ II_a$, so $ P$, $ M$, and $ Q$ are collinear.  This means that $ M$ is the center of rectangle $ PIQI_a$, so the altitude to $ PI$ from $ M$ meets $ PI$ at its midpoint.  Since $ XY\parallel PI$, we have that the altitude from $ M$ to $ XY$ meets $ XY$ at its midpoint.  Hence, $ XZ \equal{} YZ$.

Now, it is well-known that $ BX \equal{} YC$, so $ BX \plus{} XZ \equal{} YZ \plus{} YC\implies BZ \equal{} ZC$ and $ MZ\perp BC$, so $ MB \equal{} MC$.  Also, $ I$ and $ I_a$ are on the bisector of $ \angle A$, so $ II_a$ is the bisector of $ \angle A$.  Notice that $ II_a$ and the circumcircle of triangle $ ABC$ only meet at two points, which are $ A$, and say $ M'$.  Notice that $ \angle M'BC \equal{} \angle M'AC \equal{} \angle M'AB \equal{} \angle M'CB$, so $ M'B \equal{} M'C$, so $ M'$ is the intersection between the angle bisector of $ \angle A$ (or $ II_a$) and the perpindicular bisector of $ BC$.  This, however, is also $ M$, so $ M \equal{} M'$, which means that $ M$ is on the circumcircle of $ \triangle ABC$. [/hide]


For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added?

[hide]First, let $1abc$ represent each of the numbers in the set. Let its consecutive integer, $1abc+1$ be denoted as $1abc'$. Notice that if $a$, $b$, and $c$ are 0, 1, 2, 3, or 4, $1abc+1abc'$ requires no carrying. (125 possibilities) Additionally, there is no carrying when $c=9$and $a$ and $b$ are 0, 1, 2, 3, or 4. (25 possibilities) Finally, if $b=c=9$, then $a$ could be 0, 1, 2, 3, 4, or 9. (6 combinations) Thus, our answer is $125+25+6=\boxed{156}$[/hide]

[hide="Answer"]Looking at the first number of the pair only: for hundreds, only from $0$ to $4$ will work. For tens, we again have from $0$ to $4$. For units, we again have $0$ to $4$, but also $9$. The first condition removes $500$ numbers, and the second removes $250$. The third removes $100$, so we have $150$. However, we have oversimplified somewhat. $1999+2000=3999$, which involves no carrying, so we add one for that, and for the second condition, when the numbers ends with $99$, this will work. This adds $5$ numbers, so our answer is $\boxed{156}$.[/hide]

I understand where the $125$ comes from when considering $[0,5]$ and the $25$ when considering $(9,0)$ and the pair $(1999,2000)$, giving a total of $151$. Where do the last $5$ come from?

First of all if the last three digits are in $0,1,2,3,4$ we have $5$ ways for each digit, so $125$ total ways. The units digit could also be $9$, in which case the second and third digits are in $0,1,2,3,4$, for $25$ ways. Lastly, the final two digits could each be $9$ and the second digit could be $0,1,2,3,4,9$ for $6$ ways. Summing gives $\boxed{156}$.


algebra
polynomial
number theory proposed
number theory

Determine all non-constant polynomials $ f\in \mathbb{Z}[X]$ with the property that there exists $ k\in\mathbb{N}^*$ such that for any prime number $ p$, $ f(p)$ has at most $ k$ distinct prime divisors.

Nice problem.
[hide="Answer"]$ f(x)\equal{}nx^m$[/hide]
[hide="Proof"][b]Lemma[/b]
For each $ f\in\mathbb{Z}[x]$ let $ A(f)$ be the set of prime divisors of $ f(x)$ when $ x$ runs over the integers. If $ f$ has positive degree then $ A(f)$ is infinite.
[hide="Proof of lemma"]The result is clear for all $ f$ with $ f(0)\equal{}0$.
Suppose for the sake of contradiction that $ A(f)$ is finite for some $ f$ with nonzero independent term.

Let $ P\subseteq A(f)$ be the set of primes which divide $ f(x)$ for any integer $ x$ and let $ Q\equal{}A\setminus P$.

Let $ b$ and $ c$ be integers such that $ bc\equal{}f(0)$, $ (b,c)\equal{}1$, and $ p|b$ for any $ p\in P$.

For each $ q\in Q$ there exists an integer $ r_q$ such that $ q\not| f(r_q)$, because of the chinese remainder theorem there exists an integer $ d$ with $ b^2d\equiv r_q\mod q$ for each $ q\in Q$.

Now $ f(x)\equal{}bc\plus{}xg(x)$ with $ g(x)\in\mathbb{Z}[x]$ and $ f(b^2d)\equal{}b(c\plus{}bdg(b^2d))$.

Since $ f(b^2d)\equiv r_q\mod q$ we have that $ q\not| f(b^2d)$ for any $ q\in Q$. Also $ (p,c\plus{}bdg(b^2d))\equal{}1$ for any $ p\in P$, so $ c\plus{}bdg(b^2d)$ does not have prime factors in $ A(f)$.

As there exist inifinitely many such integers $ d$ it is sure that for some $ d$, $ c\plus{}bdg(b^2d)\not\equal{}\pm 1$ so it does have prime factors outside $ A(f)$ and this is the contradiction.[/hide]

(Returning to the problem...) Let $ f$ be one such polynomial ($ f(p)$ has at most $ k$ disctintc prime divisors) and rewrite it as $ f(x)\equal{}x^mg(x)$, with $ g\in\mathbb{Z}[x]$ and $ g(0)\not\equal{} 0$. Suppose for the sake of contradiction that $ g$ has positive degree. We can apply the lemma to $ g$ so $ A(g)$ is inifinite. In particular we can find $ k\plus{}1$ primes in $ A(g)$ not dividing $ g(0)$ and let $ s$ be their product. For each of these primes $ q$ there exist and integer $ r_q\not\equiv 0\mod q$ such that $ q|f(r_q)$. By the chinese remainder theorem there exists $ a\mod s$ such that $ a\equiv r_q\mod q$ for each of these $ q$. Since $ r_q\not\equal{} 0\mod q$ we have $ (a, s)\equal{}1$ then Dirichlet's theorem on primes in arithmetic progressions asures the existence of infinitely many primes $ p\equiv a\mod s$. Choose one of these $ p$ such that $ f(p)\not\equal{} 0$ and we have that $ s|f(p)$ so $ f(p)$ has more than $ k$ prime divisors, this is a contradiction.

So the only polynomials $ f\in\mathbb{Z}[x]$ with the mentioned property are $ f(x)\equal{}nx^m$[/hide]

:)


inequalities
calculus
integration
inequalities unsolved

Hi!

I'm not sure if this is the right place for asking this. My question is about the planar isoperimetric proof. Hopefully you can see the picture (equality.gif) at the end of this message.

Everything else is quite clear except for the last part of the equation. More precisely the part where dz^dz* (z* = the conjugate of z) "disappears" and integration over omega becomes somehow (with the help of Green's theorem?) integration along omega's bounding. How does this happen?

Can someone please help me?! If this isn't the right place to ask, does anyone know a better one?

Thank you!

I solved the problem by myself.  :first:


algebra
polynomial
induction
superior algebra
superior algebra solved

let f,g denote single variable poly. having coeff. in ring R.(commutative with identity)
 1.find all units in R[x]
 2.if f nonzero and fg=0,then show there exists constant c in R satisfying
    cg=0.

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=87441
2. use induction on the degree of $f$


Let array ${a_n}$ is defined by ${a_1} = \frac{27}{10}, {a_{n+1}^{3}} -3{a_{n+1}}({a_{n+1}-1})-{a_n}=1 \forall {n} > 1$
Prove that array has a limit and find that limit

[hide="almost complete solution"]just to start...looking for bounds on $x^3-3x^2+3x+c$, $a_{n+1} \le 3$ iff $9+c=8-a_n \ge 0$

or $a_n \le 8$ then $a_{n+1} \le 3$, since $a_1<8$, then by induction, $\forall n, a_n \le 3$

then by decartes', as long as -(a_n+1) is negative, there is no sign variation, and therefore, no negative roots, this similarly follows by induction, so we have $a_n \in (0,3)$

this does not proof it is convergent, but it is bounded...

and of course, when there is a limit, for large n, we can say that $a_n=a_{n+1}=L$, so we have a cubic...

$L^3-3L^2+2L-1=0$, by rrt, there are no rational roots, and since irrational/complex roots come in pairs, but there is only one limit, so it does not have a simple exact form...(you can use cardano's method), but using a calculator approximation, we get $L \approx 2.325$[/hide]

$a_{n + 1}^3 - 3a_{n + 1} \left( {a_{n + 1} - 1} \right) - a_n = 1 \Leftrightarrow$

$\Leftrightarrow a_{n + 1}^3 - 3a_{n + 1}^2 + 3a_{n + 1} - 1 = a_n \Leftrightarrow$

$\Leftrightarrow a_{n + 1} = 1 + \sqrt[3]{{a_n }}$

$a_{n + 1} = 1 + \sqrt[3]{{a_n }}\mathop \Leftrightarrow \limits_{\mathop {f\left( x \right) = 1 + \sqrt[3]{x}}\limits^ \uparrow } a_{n + 1} = f\left( {a_n } \right)$

then:
$\mathop {\lim }\limits_{n \to + \infty } a_n = l$

and:
$l - \sqrt[3]{l} - 1 = 0,l > 0$


A sumsquare is a 3 x 3 array of positive integers such that each row, each column and each of the two main diagonals has sum $ m$. Show that $ m$ must be a multiple of 3 and that the largest entry in the array is at most $ 2m/3\minus{}1$.

I'll prove that $ m\equal{}3x$, where $ x$ is the value in the central square.

Given three squares x, a, and b we can get the following:

[code]
   b    m-a-2b+x   a+b-x
   a        x      m-x-a
 m-a-b   whatever  m-x-b

[/code]

Looking at the right column, $ (a\plus{}b\minus{}x)\plus{}(m\minus{}x\minus{}a)\plus{}(m\minus{}x\minus{}b)\equal{}m$, so $ m\equal{}3x$.

Someone else can do the second part :p

Ok then, but you pretty much already did it.

Suppose the largest number is greater than or equal to $ 2m/3$. By the proof above, it can't be in the center. Pick $ b$ and $ a$ (as above) so that one of them is the largest number. If it is $ a$, then note we have $ m\minus{}x\minus{}a \equal{} 2m/3\minus{}a\leq 0$ in the square, contradiction. If it is $ b$, then note we have $ m\minus{}x\minus{}b \equal{} 2m/3\minus{}b\leq 0$ in the square, contradiction.

(EDIT: forgot the $ \equal{} 2m/3$ case)

I'll finish the [hide="second part"]
Lazarus proved that the center entry is one third of the sum.

Let $ a$ be the least entry, and $ x$ at the center. Then the entry opposite $ a$ is $ m\minus{}x\minus{}a\equal{}2x\minus{}a$. By definition $ a$ is positive, so $ a\geq1$.

We see that $ 2x\minus{}a\equal{}\frac{2m}{3}\minus{}a\leq\frac{2m}{3}\minus{}1$ since $ a\geq1$.

And we are done.
[/hide]