Prove that there are infinitely many ordered quadruples of positive integers $(x,y,u,v)$ such that \[xy+1,xu+1,xv+1,yu+1,yv+1,uv+1\] are all perfect squares.

[quote="N.T.TUAN"]Prove that there are infinitely many ordered quadruples of positive integers $(x,y,u,v)$ such that
\[xy+1,xu+1,xv+1,yu+1,yv+1,uv+1 \]
are all perfect squares.[/quote]

It is famous Diophant's problem. I heard it isn't solved yet, is it true?

[quote="A. S. K."][quote="N.T.TUAN"]My friend said that, this is one of problems from MOSP. And I think it is solved.  :wink:[/quote]

Yes, it is  :)
http://web.math.hr/~duje/pdf/dioeul2.pdf[/quote]
Thanks for your help but this link is not works. Please fixed it!  :wink:

[quote="N.T.TUAN"]
Thanks for your help but this link is not works. Please fixed it!  :wink:[/quote]

Now it works.


Find all values of $ a$ such that the system \begin{align*}ax+2y+3z&=1\\2x+ay-3z&=4\\3x-y+2z&=-5\end{align*} cannot be solved for a unique solution $ (x,y,z).$

Find the determinant of the coefficient matrix. If it is equal to zero, then the equation has no unique solutions.

if x = Dx/D, and then D = 0. Do you mean that the determinant D will be evaluated and equated to zero? Then what?

[b]Edit:[/b] After reading Dr Graubner's post, I realized I neglected the case in which the three planes had pairwise intersections, but these pairwise intersections are three parallel lines; ie, when the three normal vectors are co-planar, though no two are parallel.

So actually, the fastest way of solving the problem [i]would[/i] be to use the determinant. Still, my note in my original post still stands, and I think it is a good idea to realize [i]why[/i] the determinant of the coefficient matrix has anything to do with the number of solutions. Hint: The coefficient matrix is the transpose of the matrix whose columns are the normal vectors of the three planes.

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

[color=red]Geometrically, you won't get a unique solution when two planes are parallel. Two planes are parallel if the $ x,y,z$ coefficients are proportional. Using this, work pairwise -- it's pretty straightforward, and you don't need to compute the determinant.[/color]

[b]Note:[/b] [i]In general, I'm not a huge fan of using the determinant without an understanding of what it is, what it does, and how it does it. If a person doesn't understand the determinant, it has the potential of coming across as a random and arbitrary computational technique. Of course, if it's the first approach you see, it may or may not be worth using in a timed environment.[/i]

hello, adding the first and the second equation we get
$ (a\plus{}2)x\plus{}(a\plus{}2)y\equal{}5$
multiplying the second equation by $ 2$ and the third by $ 3$ we get
$ 13x\plus{}(2a\minus{}3)y\equal{}\minus{}7$
from the last equation we have
$ x\equal{}\minus{}\frac{1}{13}\left(7\plus{}(2a\minus{}3)y\right)$
inserting this in the first equation above and solving by $ y$ we have
$ y\equal{}\minus{}\frac{1}{2}\frac{7a\plus{}79}{(a\plus{}2)(a\minus{}8)}$
and analogously we get
$ x\equal{}\frac{1}{2}\frac{17a\minus{}1}{(a\plus{}2)(a\minus{}8)}$
$ z\equal{}\minus{}\frac{1}{2}\frac{(5a\plus{}14)(a\minus{}3)}{(a\plus{}2)(a\minus{}8)}$.
for all $ a$ with $ a\ne\minus{}2$ and $ a\ne8$.
Sonnhard.


[quote="NextPeace"]Let $ x_1,x_2,\cdots,x_n \geq 0$
$ \sum\frac {1 \plus{} x_1x_2 \plus{} x_1x_3 \plus{} \cdots \plus{} x_1x_n}{(1 \plus{} x_2 \plus{} x_3 \plus{} \cdots \plus{} x_n)^2}\geq1$[/quote]
By Cauchy Schwarz Inequality, we have
\[ (1\plus{}x_2\plus{}\cdots \plus{}x_n)^2 \le \left( 1\plus{}x_1(x_2\plus{}\cdots \plus{}x_n)\right) \left( 1\plus{}\frac{x_2\plus{}\cdots \plus{}x_n}{x_1}\right) \equal{}\frac{(1\plus{}x_1(x_2\plus{}\cdots \plus{}x_n))(x_1\plus{}x_2\plus{}\cdots \plus{}x_n)}{x_1}\]
Hence
\[ \frac{1\plus{}x_1(x_2\plus{}\cdots \plus{}x_n)}{(1\plus{}x_2\plus{}\cdots \plus{}x_n)^2} \ge \frac{x_1}{x_1\plus{}x_2\plus{}\cdots \plus{}x_n}\]
Therefore
\[ LHS \ge \frac{x_1\plus{}x_2\plus{}\cdots \plus{}x_n}{x_1\plus{}x_2\plus{}\cdots \plus{}x_n} \equal{}1\]
Done!
:)


Find :
[color=darkblue]
[b]  2/3! + 11/4! + ....... + S-1/n! + .....  = ?[/b][/color]

where S is the sum of the products of first [i]n[/i] natural numbers taken two at  a time.

 :D

could you please explain the question?
the first two terms do not seem to satisfy the expression for generic term [(S-1)/n!]
that is the generic expression right? not S - (1/n!)?(which also does not generate the first two terms)

I think the generic term should be $ S_{n}/(n\plus{}1)!$.

forgive me if I am wrong.

Do you mean $ \frac{S_{n\minus{}1}}{n!}$  ?

I got the $ n^{th}$ term as $ 3n\plus{}2 /24(n\minus{}2)!$

But i am not able to make this telescoping.

I could write $ (3n\plus{}2)/24(n\minus{}2)!$ as $ 14/24(n\minus{}2)!  \minus{} 6/24(n\minus{}1)!$

Ugh but the series is not telescoping.

Its $ S_{n\minus{}1}/n!$ arjun. The answer is 11e/24. But i do not know how to arrive at the answer.

sums of this type are done by expressing numerator as A(n-2)!+B(n-3)!+...
then splitting the given series
in this case the genaral term is 1/8(n-2)! + 1/3(n-1)! which sums to1/8e+1/3e

[quote="befuddlers"]I could write $ (3n \plus{} 2)/24(n \minus{} 2)!$ as $ 14/24(n \minus{} 2)! \minus{} 6/24(n \minus{} 1)!$

Ugh but the series is not telescoping.[/quote]

Oh well i did not know that we had to write the series as sum of two terms (I wrote it as a difference of two terms) :oops:

And i did not know that the sum of the factorial series converged to e.

Nice solution basil.


Let x, y, and z be the set of all real numbers such that
6x^2+y^2+z^2=2x(y+2z). What is the value of x+y+z?

[hide="Solution"]Rearrange the given expression as $ x^2\minus{}2xy\plus{}y^2\plus{}4x^2\minus{}4zx\plus{}z^2\equal{}\minus{}x^2$.

Therefore we have $ (x\minus{}y)^2\plus{}(2x\minus{}z)^2\equal{}\minus{}x^2$.

If $ x$ is not 0, then LHS is always nonnegative but RHS becomes negative, which is impossible.

Hence, the only possible value for $ x$ is $ 0$.

$ x\minus{}y$ and $ 2x\minus{}z$ are $ 0$ also.

Thus, $ x\equal{}0, y\equal{}0, z\equal{}0$, and $ x\plus{}y\plus{}z\equal{}0$.

Answer: $ 0$[/hide]


algebra
polynomial
quadratics
inequalities
geometric inequality
algebra unsolved

let $ m$ be a finite set of non-zero real numbers. Prove that there exists a constant $ C$ with the folowing property: if $ P(x)$ is a polynomial of degree $ n$ with $ n$ real (possibly repeated) roots, all of whose coefficients belong to $ M$, then $ n \leq C$

We can even have $ 0\in M$ if we assume $ P(0)\not \equal{} 0$!

$ T: \equal{} \max \{ |\left ({y\over x}\right)^2 \minus{} 2{z\over x}|: x,y,z\in M ,x\not \equal{} 0\}$
$ t: \equal{} \min \{ {1\over 2}, |{x\over y}|: x,y\in M, x,y\not \equal{} 0 \}$
Then $ T\ge 0$ and $ 0 < t\le {1\over 2}$.

Assume for some $ n\in\mathbb{N},n > {T\over t},n\ge 2$
we have some polynomial
$ P(x) \equal{} a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \cdots \plus{} a_0$
of degree $ n$ with only real roots and all its coefficients $ a_0,\ldots,a_n\in M$ and $ P(0)\not \equal{} 0$.

Then $ a_0,a_n\not \equal{} 0$ and if $ \lambda_1,\ldots ,\lambda_n\in\mathbb{R}$ are the roots of $ P(x)$ with multiplicities,
we get by [url=http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formulas]Viète[/url]
$ \lambda_1 \plus{} \cdots \plus{} \lambda_n \equal{} \minus{} {a_{n \minus{} 1}\over a_n}$
$ \sum_{1\le i < j\le n}\lambda_i\lambda_j \equal{} {a_{n \minus{} 2}\over a_n}$
$ \lambda_1\cdots \lambda_n \equal{} \pm {a_{0}\over a_n}$.

Then
$ |\lambda_1|^2 \plus{} \cdots \plus{} |\lambda_n|^2 \equal{} (\lambda_1 \plus{} \cdots \plus{} \lambda_n)^2 \minus{} 2\sum_{1\le i < j\le n}\lambda_i\lambda_j \equal{} \left ({a_{n \minus{} 1}\over a_{n}}\right)^2 \minus{} 2{a_{n \minus{} 2}\over a_{n}}\le T$
and
$ |\lambda_1|\cdots |\lambda_n| \equal{} |{a_0\over a_n}|\ge t$
and by the [url=http://en.wikipedia.org/wiki/Generalized_mean]quadratic-geometric inequality[/url] we get using $ n > {T\over t}$
$ \sqrt {t} > \sqrt {T\over n}\ge \sqrt {|\lambda_1|^2 \plus{} \cdots \plus{} |\lambda_n|^2\over n}\ge \sqrt [n]{|\lambda_1|\cdots |\lambda_n|}\ge \sqrt [n]{t}$
and $ t^n > t^2$, which contradicts $ 0 < t\le {1\over 2}$ and $ n\ge 2$.

So we can take $ C \equal{} \max \{{T\over t},1\}$ to solve the problem.

P.S. We also only need the first, second, third and last coefficient of $ P(x)$ to be in $ M$. :D


calculus
integration
logarithms
trigonometry
real analysis
real analysis solved

Evaluate of the following definite integrals.

(1)$\int^a_0 \ln (a^2+x^2) dx$    (a:positive constant number)

(2)$\int^{\frac{\pi}{2}}_0 \frac{\sin \theta}{\sin\theta+\cos \theta} d\theta$

(1) integrating by parts we obtain:
$\displaystyle\int_0^a\ln(x^2+a^2)dx=a\ln(2a^2)-2\int_0^a\frac{x^2+a^2-a^2}{x^2+a^2}dx=a\ln(2a^2)-2a+2a^2\int_0^a\frac1{x^2+a^2}dx=a\ln(2a^2)-2a+a\frac\pi2$

(2)
$I=\displaystyle\int_0^{\frac\pi2}\frac{\sin x}{\cos x+\sin
x}dx=\int_0^{\frac\pi2}\frac{\tan x}{1+\tan
x}dx=\frac\pi2-\int_0^{\frac\pi2}\frac1{1+\tan x}dx$

$\tan x=t\Leftrightarrow t\in(0,\infty)$.

$I=\displaystyle\frac\pi2-\int_0^\infty\frac{dt}{(1+t)(1+t^2)}=\frac\pi2-\frac12\int_0^\infty\left(\frac1{t+1}-\frac{t-1}{t^2+1}\right)dt
=
\frac\pi2-\frac12\int_0^\infty\left(\frac1{t^2+1}+\frac1{t+1}-\frac12\frac{2t}{t^2+1}\right)dt
=
\frac\pi2-\frac\pi4+\frac\pi8-\frac12\left(\ln(t+1)-\frac12\ln(t^2+1)\right)|_0^\infty=\frac{3\pi}8-\frac12\ln\frac{t+1}{\sqrt{t^2+1}}|_0^\infty=\frac{3\pi}8$

For the first one,you are right,but the second one,wrong.

kunny

maybe I misscalculated smth... another approach:
Denote by $I=\int_0^{\frac\pi2}\frac{\sin x}{\sin x+\cos x}dx$ and by $J=\int_0^{\frac\pi2}\frac{\cos x}{\sin x+\cos x}dx$
Now $I+J=\frac\pi2$
$J-I=\ln(\sin x+\cos x)|_0^{\frac\pi2}=0$
So $I=\frac\pi4$
is it correct now?


there is one bad part in middle but rest is good.

http://www.collegehumor.com/video:1727961

wtf.  that was the wierdest video ive ever seen

That is amazing...how long would that have taken?

i dunno but I guess a week maybe a couple.


For how many non-negative real values of $ x$ is $ \sqrt{144\minus{}\sqrt[3]{x}}$ an integer?

We can see that $ 0\leq\sqrt{144 \minus{} \sqrt [3]x}\leq12$, (given the domain), and since there is a distinct integer value for $ x \equal{} \{0^6,1^6,2^6,\ldots,12^6\}$, our answer is $ \boxed{13}$.

oh oops, when I read nonnegative, I excluded 0 for some odd reason....

Eh???
You put x=1^6, you get root of 143, that's not an integer!!!


hello, your given equation is equivalent with
$ \frac{\cos(\frac{\pi}{6}\plus{}x)\csc(x)^2\sec(x)\sin(\frac{\pi}{4}\minus{}x)\sin(\frac{\pi}{4}\plus{}x)}{1\minus{}\cos(2x)}\equal{}0$
Sonnhard.

Dr Sonnhard Graubner, how did you think of that factorization?

i think he just made the stuff into cosines and sines then used periods...

[color=darkblue]Hi ! Dr Sonnhard Graubner.
Can you give me the full of solution? [/color]


symmetry
geometry
parallelogram
function
3D geometry
tetrahedron
probability

1.  The real numbers $x_1, x_2, ... , x_n$ belong to the interval $(0,1)$ and satisfy $x_1 + x_2 + ... + x_n = m + r$, where $m$ is an integer and $r \in [0,1)$. Show that $x_1 ^2 + x_2 ^2 + ... + x_n ^2 \leq m + r^2$.

2.  For a permutation $P = (p_1, p_2, ... , p_n)$ of $(1, 2, ... , n)$ define $X(P)$ as the number of $j$ such that $p_i < p_j$ for every $i < j$. What is the expected value of $X(P)$ if each permutation is equally likely?

3.  $W$ is a polygon which has a center of symmetry $S$ such that if $P$ belongs to $W$, then so does $P'$, where $S$ is the midpoint of $PP'$. Show that there is a parallelogram $V$ containing $W$ such that the midpoint of each side of $V$ lies on the border of $W$.

4.  $d$ is a positive integer and $f : [0,d] \rightarrow \mathbb{R}$ is a continuous function with $f(0) = f(d)$. Show that there exists $x \in [0,d-1]$ such that $f(x) = f(x+1)$.

5.  The sequence $a_1, a_2, a_3, ...$ is defined by $a_1 = a_2 = a_3 = 1$, $a_{n+3} = a_{n+2}a_{n+1} + a_n$. Show that for any positive integer $r$ we can find $s$ such that $a_s$ is a multiple of $r$.

6.  Find the largest possible volume for a tetrahedron which lies inside a hemisphere of radius $1$.

1. http://www.mathlinks.ro/Forum/viewtopic.php?p=348855#p348855
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=348856#p348856
3. http://www.mathlinks.ro/Forum/viewtopic.php?p=348857#p348857
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=348859#p348859
5. http://www.mathlinks.ro/Forum/viewtopic.php?p=348862#p348862
6. http://www.mathlinks.ro/Forum/viewtopic.php?p=348863#p348863


geometry
3D geometry
calculus
integration
function

[quote="M4RI0"]Find all positive integers n such that $ n^3 \minus{} 18n^2 \plus{} 115n \minus{} 391$ is the cube
of a positive integer.[/quote]

Usually, when you copy a problem from somewhere else, the $ \minus{}$s change after you paste, so make sure you check.

I finally solved it :D . But I'll wait to see your solutions.

Well, just like the well known method of eliminating integers by placing integral functions between squares, we place $ n^3 \minus{} 18n^2 \plus{} 115n \minus{} 391$ between two cubes to eliminate some cases.

[u]Solution.[/u] Notice that $ (n \minus{} 5)^3 > n^3 \minus{} 18n^2 \plus{} 115n \minus{} 391 > (n \minus{} 6)^3$ for all $ n > 25$, and $ (n \minus{} 6)^3 > n^3 \minus{} 18n^2 \plus{} 115n \minus{} 391 > (n \minus{} 7)^3$ for $ 25 > n > 12$ (Why?).  Notice that for $ n \equal{} 12,25$, we obtain cubes.  Checking the values of the function for $ n < 12$-though a bit tedious-we find that $ n \equal{} 11$ also leads to a cube.  As a result, $ n\in\{11,12,25\}$ are the only possible values that satisfy the conditions of the problem. $ \Box$

Just one mistake, for n=7 you don't get the cube of a positive integer.

I've got a question:

I found that $ u_n\equal{}n^3\minus{}18n^2\plus{}115n\minus{}391$ is given by the recurrence relation $ u_{n}\equal{}4u_{n\minus{}1}\minus{}6u_{n\minus{}2}\plus{}4u_{n\minus{}3}\minus{}u_{n\minus{}4}$;$ n\geq 5$;$ u_1\equal{}\minus{}293,u_2\equal{}\minus{}225,u_3\equal{}\minus{}181,u_4\equal{}\minus{}155$

and  for $ a_n\equal{}n^3$, we have the same recurrence relation

Could this have helped?

I thought you asked for all positive integers that lead to cubes. I will edit my post.


there is a 3 tuples set of number which will make all numbers's total is a square root.
for example: (6,19,30) 6+19=25;6+30=36;19+30=49

there is infinite solutions for that..
as  x+y=49
x+(y+32)=81
y+(y+32)=64
y=16
y+40=48
x=33 so (16,33,48)

Is there a solution for 4-tuples? I cannot find it... and cannot prove there is none.

don't u know something about it? if you don't i'll go to next level' students ;)

When you are asking for a solution in 4-tuples, do you mean that you get a perfect square if you add any 2 numbers, or do you get a perfect square if you add any 3 numbers.

Well..
One is (0, 0, 0, 0)
I know its not what your looking for, but one exists at least :).

So basically the question is:  if we have three positive integers that satisfy

a + b  \
b + c   |  perfect squares
a + c  /

is there any reason that a, b, and c can't all be the same distance d below various perfect squares such that

a + d
b + d
c + d

are also all perfect squares?  In other words, is there some reason that a, b, and c can't be distributed as perfect squares (that is, with gaps of 1, 3, 5, 7, ... or sums of these odd prime integers between them)?

I don't actually have a solution, but mebbe that'll help someone.  (I'm working on a Java program to find some concrete example, which would eliminate the issue of proving that one doesn't exist.)

The short answer:  yes, it's possible.  I have lots of examples (through a Java program); here's one:

482
3362
359
2

Er... I don't know any 'equation' for these, although logically it's possible.  My code checks likely possibilities, it doesn't use an equation to generate results.

Here's the code I used (unfortunately uncommented so far..)  It's pretty inefficient, and it lists different permutations of the same fourtuple separately (yeah, easily fixable, but this gets the job done.)  the forum won't let me post a .txt file, so they're in word.

thank you but i don't understand codes and programming languages

Oh - sorry.  Basically, what my program does is

1) Find a list of all triples that work, with a, b, c  <= 250.  It does this by choosing a value for a, then picking some b such that a + b is a perfect square.  Then it chooses c such that a + c is a perfect square; if b + c is also a perfect square it "keeps" the triple.

2) Once it has a good-sized list of triples that work, it searches for values of d that it could add.  It tries all the values of d that make a + d a perfect square, testing to see that b + d and c + d are also perfect squares.

It's inefficient, I know, but at least it doesn't just try every combination of four integers a, b, c, d <= 1000.

From here I suppose the interesting question is:  Is there ANY n for which it is impossible to choose n integers that form a set in which the sum of any two distinct members is a perfect square?  I smell an interesting research topic.

This is really rambling -- it doesn't much answer the question, I'm afraid.

Well, for three variables, the general solution is $\{x,y,z\} = \{\frac{a^2+b^2-c^2}{2},\frac{a^2-b^2+c^2}{2},\frac{-a^2+b^2+c^2}{2}\}$ provided those three things are all integers, so in other words we have either 2 or 0 of a, b, c being odd.  So then we want to try and find integral solutions for $w + x = d^2, w + y = e^2, w + z = f^2$ which would require us to find solutions to $d^2 - e^2 = b^2 - c^2, e^2 - f^2 = a^2 - b^2, f^2 - d^2 = c^2 - a^2$.  So, having picked our three perfect squares, we need to find another set of 3 squares with the same common differences.   (Actually, I could have gotten to that much more easily.  Oh, well.)  There are lots and lots of such sets of squares -- as long as we are careful about picking the differences between them (no primes, for example) we can get a solution.

Let's say we have d, e, f larger than a, b, c (respectively) and d < e < f.  $f^2 - d^2 = (2d + 1) + (2d + 3) + \ldots + (2f - 1) = (2d + 1) + \ldots + (2e - 1) + (2e + 1) + \ldots + (2f - 1)$ and we need to find another sequence of odd numbers that has total sum the same and can be broken into two pieces with the same sums as those.


linear algebra
matrix
linear algebra unsolved

Let $ R$ be a ring and $ A$ be a $ n$ dimensional matrix over $ R$. If $ det(A)$ is a zero-divisor in $ R$, prove that $ A$ is not injective. Any help is appreciated.

More generally: http://www.mathlinks.ro/viewtopic.php?t=124137

  darij


[url=http://h1.ripway.com/gabriel240587/IMO_1959-2006_Problems.pdf]IMO Problems, 1959 - 2006[/url]

[url=http://h1.ripway.com/gabriel240587/KumpulanSoalOlimpiadeMatematika.pdf]Indonesian Mathematical Olympiad Problems[/url]

can someone possibly translate the second one?

i'm so sory, my english is bad, my toefl just 470... but i'll try to translate that file to english... but, there're few problem in english... oh yeah, if you want some e-books and files, please visit : [url=http://olimpiadematematika.net/forum]here[/url] and search click "gudang arsip".. in that forum you can find some e-books and files in english.... please try...  :D

oh, i'm so sorry , this the true link [url=http://olimpiadematematika.net/Forum]here[/url] .... i'm very2 sorry.....  :blush:


Anyways, it's quite straightforward if you write $1=k+1-k$ a couple of times. I guess this works for the other two.


[hide]
2^16 is 65536.[/hide]

What happened to your wall then?  :huh:

[hide]$65536$....... on the wall?? maybe a good hole.[/hide]

How are you able to hold that thing?  The cubes aren't all the same size right?  Either way, my answer is

[hide]65536[/hide]

[hide="Answer"]65536[/hide]

45.02 is pretty darn good  :D
I still can't figure out how to do it in 10000000  :lol:

any advice at solving it with two layers done and an X on top?

[quote]After solving his Rubik's cube in EXACTLY 45.02 seconds (he's trying to break 45 seconds), Treething throws his Rubik's cube against the wall, and it breaks into 2^(2^(2^2)) pieces. How many pieces is this?[/quote]

Dang...you must be superman or something...  :showoff: Anyway, I think the answer is
[hide]2^(2^(4))=2^16=65536[/hide]

[quote="Treething"]After solving his Rubik's cube in EXACTLY 45.02 seconds (he's trying to break 45 seconds), Treething throws his Rubik's cube against the wall, and it breaks into 2^(2^(2^2)) pieces. How many pieces is this?[/quote]

You are fast with you rubiks cube.  Mine took about 75 seconds.

I started with it at the beginging of this year and got my time down to near 30 seconds.  I can do it one handed in 3 minutes. :D
[hide]65536 was the answer by the way and treething is really strong. :) .[/hide]


Let $ p(x)=x^{4014}+13*x^{2007}+1$ and let $ q(x)$ be the monic polynomial of minimum degree such that if $ p(r)=0$, then $ q\left(r+\frac{1}{r}\right)=0$ for all roots, $ r$. It can be shown that $ p(x)$ has no double roots. Find $ q(1)$.

Assume the root is $\alpha$ instead of $r$ we get, $$P(\alpha)=0\implies Q\left(\alpha+\frac{1}{\alpha}\right)=0.$$ By factor theorem we get, $$P(x)=x^{4014}+13^{2007}+1=(x-\alpha)$$ Therefore $$Q(x)=\left(x-\alpha-\frac{1}{\alpha}\right)=\left(x-\frac{\alpha^2+1}{\alpha}\right)$$ Now Plug in $x=1$ we get, $$Q(1)=\frac{\prod_{cyc}{}(\alpha^2-\alpha+1)}{\prod_{cyc}{}\alpha}\implies\boxed{11.}$$


[color=darkblue]Let $ \triangle ABC$. Prove that: $ \frac{1}{3}cosA\plus{}\frac{1}{4}cosB\plus{}\frac{1}{5}cosC\le\frac{5}{12}$[/color]

hello, by the theorem of cosine we have
$ {\displaystyle \frac{5}{12}\minus{}\frac{b^2\plus{}c^2\minus{}a^2}{6bc}\minus{}\frac{a^2\plus{}c^2\minus{}b^2}{8ac}\minus{}\frac{a^2\plus{}b^2\minus{}c^2}{10ab}}\equal{}$
$ {\displaystyle \frac{50abc \minus{}20ab^2\minus{}20ac^2\plus{}20a^3\minus{}15a^2b\minus{}15bc^2\plus{}15b^3\minus{}12a^2c\minus{}12b^2c\plus{}12c^3}{120abc}}$
By substituting
$ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y$, we have to prove that
$ 36z^2y\plus{}16z^2x\plus{}16zx^2\plus{}36y^2z\plus{}4yx^2\plus{}4yx^2\geq88xyz$ this is equivalent
with
$ 36\frac{z}{x}\plus{}16\frac{z}{y}\plus{}16\frac{x}{y}\plus{}4\frac{x}{z}\plus{}36\frac{y}{x}\plus{}4\frac{y}{z}\geq88$
This is true because
$ 36\frac{z}{x}\plus{}16\frac{z}{y}\plus{}16\frac{x}{y}\plus{}4\frac{x}{z}\plus{}36\frac{y}{x}\plus{}4\frac{y}{z}\geq2\sqrt{144}\plus{}2\sqrt{16\cdot4}\plus{}2\sqrt{16\cdot36}\equal{}88$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, by the theorem of cosine we have
$ {\displaystyle \frac {5}{12} \minus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{6bc} \minus{} \frac {a^2 \plus{} c^2 \minus{} b^2}{8ac} \minus{} \frac {a^2 \plus{} b^2 \minus{} c^2}{10ab}} \equal{}$
$ {\displaystyle \frac {50abc \minus{} 20ab^2 \minus{} 20ac^2 \plus{} 20a^3 \minus{} 15a^2b \minus{} 15bc^2 \plus{} 15b^3 \minus{} 12a^2c \minus{} 12b^2c \plus{} 12c^3}{120abc}}$
By substituting
$ a \equal{} y \plus{} z,b \equal{} x \plus{} z,c \equal{} x \plus{} y$, we have to prove that
$ 36z^2y \plus{} 16z^2x \plus{} 16zx^2 \plus{} 36y^2z \plus{} 4yx^2 \plus{} 4yx^2\geq88xyz$ this is equivalent
with
$ 36\frac {z}{x} \plus{} 16\frac {z}{y} \plus{} 16\frac {x}{y} \plus{} 4\frac {x}{z} \plus{} 36\frac {y}{x} \plus{} 4\frac {y}{z}\geq88$
This is true because
$ 36\frac {z}{x} \plus{} 16\frac {z}{y} \plus{} 16\frac {x}{y} \plus{} 4\frac {x}{z} \plus{} 36\frac {y}{x} \plus{} 4\frac {y}{z}\geq2\sqrt {144} \plus{} 2\sqrt {16\cdot4} \plus{} 2\sqrt {16\cdot36} \equal{} 88$.
Sonnhard.[/quote]
Thank you very much! :lol:


number theory
Diophantine equation
relatively prime
prime numbers

prove that there's no integer solution for the following equation (x is not 0)

2x^4 + 2x^2y^2 + y^4 = z^2

   [hide] try to use pythagoras' substitution  and then try to show that if x is minimal then there's one more solution which is less than x[/hide]

i've solved it without mods, just a little bit mod   the main idea is pythagoras triples
  in the original solution they solve it by using those triples 2 times, but i've solved it as in the hidden part

[quote="xxreddevilzxx"]is this necessarily diophantine?

i see mods.[/quote]

If an equation is a [url=http://mathworld.wolfram.com/DiophantineEquation.html]diophantine equation[/url] that just means that you are only looking for integer solutions to that equation.

isn't it a diophantine equation??? x,y,z are integers    it's only  given that x is not 0

Yeah, I was agreeing that it is a diophantine equation.

so no one's gonna solve it? i've got a lot of pleasure when i solved it

is it so hard?) i've solved it for 1.5-2 hours

maybe this would help:
your equation would be:
$x^2+(x^2+y^2)^2=z^2$
then you can define two relatively prime numbers such as $u,v (u>v)$ where:
$(u^2-v^2)^2+(2uv)^2=(u^2+v^2)^2$ ;)

it is obvious, the main part is after this
 there are 2 wawys i kow to do it   the first one to use pythagor triples one more or to take x minimal(positive) and to  show that there is a solution (x',y',z') with x' less than x

That proof method is called infinite descent and the proof of the fact that there are no integer solutions to the equation $\ x^4 + y^4= z^2 $ uses that common proof method.

yeah   x^4 - y^4= z^2  the same method too[/list]


LaTeX
modular arithmetic
number theory unsolved
number theory

how can i go on solving $p^q(2^{q-1}-1)=p^7\cdot q$ if $(p,q)\in Z^2$

[quote="Arne"]Well, there are only a few possible values for $q$, why don't you check these seperately? :)[/quote]I think you should read his question as "why are there only a few cases" ;)

That's pretty obvious, since \[p^{7 - q} = 2^{q - 1} - 1,\] so if $q > 7$ the LHS is not an integer but the RHS is. And we can't really have $q < 0$ either since then the LHS is an integer but the RHS isn't.

Of course you can also say that it's trivial by the Catalan conjecture/theorem :D

Yeah in deed is as easy as arne says
Well lets see that $p^q(2^{q-1)+1)=p^7}$ then for
$q=1$ then  $p^6=2^0+1$ then p is not an integrer
$q=2$ then  $p^5=2^1+1$ then p is not an integrer
$q=3$ theb  $p^4=2^2+1$ then p is not an integrer
$q=4$ then  $p^3=2^3+1$ then p is not an integrer
$q=5$ then  $p^2=2^4+1$ then p is not an integrer
$q=6$ then  $p^1=2^5+1$ then p=33
$n=7$ then  $p^0=2^6+1$ the p=any number

now you would probably take this as a joke,i have verry seriously and stupidly made a mistake in the
hypothesis namely i forgot a $q$.
the equation is $\displaystyle p^q(2^{q-1}-1)=p^7q$
the big problem was that i wrote in latex this : \displaystyle p^q(2^{q-1}-1)=p^7\cdotq and i thought it would
print also a q but it didnt.excuse me please for beeing stupid

The problem is still quite obvious as it just reduces to $p^{q-7}(2^{q-1}-1) = q$. If $q \geq 7$, then L.H.S. $\geq (2^{q-1}-1) > 2^{q-2} > q$ (Contradiction). So we can rearrange the eqn to $\frac{2^{q-1}-1}{q} = p^{7-q}$, which is an integer, thus we have $2^{q-1} \equiv 1 \pmod{q}$, and consequently we have $q$ must be odd. (This is actually totally unnecessary, but at least with this observation, we reduce the number of cases to check for $q$ later on) Looking back at the eqn $\frac{2^{q-1}-1}{q} = p^{7-q}$, we find that if $q<0$, then L.H.S. $\not\in \mathbb{Z}$, but R.H.S. $\in \mathbb{Z}$. (Contradiction) Thus we have $1 \leq q \leq 6$. Checking for the odd values of $q$ in this range, we get the solutions $(p, q) = (1, 0), (3, 1)$.

Now, we must also be careful, since in our algebraic manipulation, we have assumed $p, q \neq 0$. So checking this case also in our original eqn, we get the solution $(p, q) = (0, k)$, where $k$ is any arbitrary integer.

Done.


trigonometry
geometry
Pythagorean Theorem

hi there
h r u ?
i want to prove this
tan(a)+tan(b)+tan(c)=tan(a)tan(b)tan(b)
a,b,c<>90 :blush:
sorry i miss this
a,b,c
is a 3 angles in a a triangle
sorry for bad english
 :(

[quote="rnwang2"][hide="Sketch"]
$\tan{A}+\tan{B}=[\tan{(A+B)}](1-\tan{A}\tan{B})$.

$\tan{A}+\tan{B}+\tan{C}$

$=[\tan{(A+B)}](1-\tan{A}\tan{B})+\tan{C}$.

Now, notice

$A+B=\pi-C\implies \tan{(A+B)}=-\tan{C}$. [/hide][/quote]
thanx alto man
i never think this is easy as this

[quote="The QuattoMaster 6000"][quote="BlackIce"][quote="vishalarul"]Impossible:

Let $a=b=c=\frac{\pi}{4}$

$\tan(a)+\tan(b)+\tan(c) = 3 \neq 1 = \tan(a) \cdot \tan(b) \cdot \tan(b)$[/quote]
sorry i miss somthing
a,b,c is a 3 angle in a triangule[/quote]

[hide]
Take the triangle ABC and drop a perpindicular from A to BC.  Call the point where this perpindicular intersects BC to be D.  Say that $AD=h$, $BD=a$, and $CD=b$.  Notice that $\tan B=\frac{h}{a}$ and $\tan C=\frac{h}{b}$.  We also have that $BC=a+b$, so $[ABC]=\frac{(BC)(AD)}{2}=\frac{h(a+b)}{2}$ (if you don't know it already, then $[ABC]$ means the area of Triangle ABC).  Now, we have that by the Pythagorean Theorem, $AC=\sqrt{h^{2}+b^{2}}$.

Now, draw perpindicular from B to AC.  Say that it hits AC at E.  We have that
\[\frac{h(a+b)}{2}=[ABC]=\frac{BE(AC)}{2}=\frac{BE(\sqrt{h^{2}+b^{2}})}{2}\Rightarrow BE=\frac{h(a+b)}{\sqrt{h^{2}+b^{2}}}\]
Now, we can say that, by the Pythagorean Theorem, $BE^{2}+EA^{2}=AB^{2}$.  We have that by the Pythagorean Theorem, $AB^{2}=h^{2}+a^{2}$.  Thus,
\[h^{2}+a^{2}=(\frac{h(a+b)}{\sqrt{h^{2}+b^{2}}})^{2}+EA^{2}\]
Here, we can do some simplifications and get
\[\frac{(h^{2}+a^{2})(h^{2}+b^{2})}{h^{2}+b^{2}}=\frac{h^{2}a^{2}+2abh^{2}+h^{2}b^{2}}{h^{2}+b^{2}}+EA^{2}\]
Thus, we have that
\[EA^{2}=AB^{2}-BE^{2}=\frac{h^{4}+a^{2}h^{2}+h^{2}b^{2}+a^{2}b^{2}-(h^{2}a^{2}+2abh^{2}+h^{2}b^{2})}{h^{2}+b^{2}}\]
We can simplify a bit to get that
\[EA^{2}=\frac{a^{2}b^{2}-2abh^{2}+h^{4}}{b^{2}+h^{2}}=\frac{(h^{2}-ab)}{b^{2}+h^{2}}\]
Squarerooting the left and right sides, we get that $EA=\frac{h^{2}-ab}{\sqrt{b^{2}+h^{2}}}$.  Thus, we have that
\[\tan A=\frac{BE}{EA}=\frac{\frac{h(a+b)}{\sqrt{b^{2}+h^{2}}}}{\frac{h^{2}-ab}{\sqrt{b^{2}+h^{2}}}}=\frac{h(a+b)}{h^{2}-ab}\]
Now, going back a bit, we found that
\[\tan A=\frac{h(a+b)}{h^{2}-ab}\]

\[\tan B=\frac{h}{a}\]

\[\tan C=\frac{h}{a}\]
Now, we must prove that tan A+tan B+tan C=(tan A)(tan B)(tan C)

Plugging in our values, we must prove that
\[\frac{h}{a}+\frac{h}{b}+\frac{h(a+b)}{h^{2}-ab}=(\frac{h}{a})(\frac{h}{b})(\frac{h(a+b)}{h^{2}-ab}) \]
We can simplify the left hand side to obtain
\[\frac{(a+b)(h)(h^{2})}{(h^{2}-ab)(ab)}\]
and we can simplify the right hand side to obtain
\[\frac{h^{3}(a+b)}{(h^{2}-ab)(ab)}\]
Setting them equal, we get that
\[\frac{(a+b)(h)(h^{2})}{(h^{2}-ab)(ab)}=\frac{h^{3}(a+b)}{(h^{2}-ab)(ab)}\]
Simplifying, we obtain that
\[1=1 \]
which is always true.  [/hide][/quote]
thanx alot man i try this and do it my self
nice work

[quote="rnwang2"][hide="Sketch"]
$\tan{A}+\tan{B}=[\tan{(A+B)}](1-\tan{A}\tan{B})$.

$\tan{A}+\tan{B}+\tan{C}$

$=[\tan{(A+B)}](1-\tan{A}\tan{B})+\tan{C}$.

Now, notice

$A+B=\pi-C\implies \tan{(A+B)}=-\tan{C}$. [/hide][/quote]
Just as a note, I actually ended up proving that $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$ by accident.

Solution:
We must prove that $\tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}$ if $A+B+C=\pi$

Then:
$A+B+C=\pi$
$A+B=\pi-C$
$\tan{(A+B)}=\tan{(\pi-C)}$
$\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=-\tan{C}$
$\tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}$


How to figure out if 993!^2 is congruent to 1 modulo 1987?

Show that $ (993!)^2 \equiv (\minus{}1)^{993} 1986! \bmod 1987$.  Are you familiar with Wilson's theorem?

I know that 1986! is congruent to 11 mod 1987, but ~~` still confused

Note that $ 1986 \cdot 1985 \cdot ... \cdot 994 \equiv (\minus{}1) \cdot (\minus{}2) \cdot ... (\minus{}993) \bmod 1987$.  And you mean $ 1986! \equiv \minus{}1 \bmod 1987$.


I'm just wondering what the motivation behind this problem was, the equation isn't really beautiful it seems pretty random lol :lol:


I don't quite understand your problem. Do you mean this:
"Take a chord passing through $F_{2}$, and let the intersection points with the ellipse be $A$ and $B$. In how many ways can the tangent lines to the ellipse at $A$ and $B$ intersect?"

If so, I think the answer is $1$.

I believe the locus is taken over all chords through one of the foci...otherwise it's trivially a point

[quote="shobber"]Given an ellipse with foci $F_{1}$ and $F_{2}$. If the chord passing though $F_{2}$ intersect the ellipse at point $A$ and $B$, then find the locus of the intersection of the tangents to the ellipse at $A$ and $B$.[/quote]
OK, I shall rewrite my problem.

Given an ellipse with foci $F_{1}$ and $F_{2}$. Chord $AB$ of the ellipse passes through $F_{2}$. The tangent lines to the ellipse at $A$ and $B$ intersect at point $L$. Find the locus of $L$ when $A$ and $B$ moves on the ellipse.


If x is a real number,find the minimum value of  $ |{sinx\plus{}cosx\plus{}tanx\plus{}cotx\plus{}secx\plus{}cosecx}|$

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=242211[/url]


calculus integration trigonometry function derivative Euler logarithms

Show that $ \int_0^1 (\cos y(x)) \sqrt {1 \plus{} y'(x)^2} \, dx$ has a strong local minimum under the conditions $ y(0) \equal{} 0$, $ y(1) \equal{} \pi/4$

I haven't taken a course in differential equations, so I have problems motivating the existence of a field to embed the sought extremal in. What is an existence theorem I may use?

I don't understand the phrase "existence of a field to embed a sought extremal in" but what is meant, if I remember it right, is that you need to show the existence of a $ C^1$-function $ f$ and $ \varepsilon>0$ such that for any $ C^1$-function $ g$ with $ \|f\minus{}g\|_{C}<\varepsilon$, the value of the functional on $ g$ is at least as large as on $ f$. ($ C$ instead of $ C^1$ is not a typo).

If you mean that you got a differential equation for the extremal and just want to show that it has a solution, the main problem is that it is a boundary problem and not an initial value problem, so, perhaps, you'll have to do 2 things: first show that the solution of the initial value problem is unique and depends continuously on the initial conditions at $ 0$ (the standard Lipschitz criterion will probably suffice here) and, secondly, show that you can get $ y(1)$ both $ >\frac\pi 4$ and $ <\frac\pi 4$ choosing the derivative at $ 0$ appropriately.

One more option would be to take a minimizing sequence $ y_n$ and try to show that it has a limit. First, you restrict yourself to the functions whose values are between $ \minus{}\pi/4$ and $ \pi/3$ to guarantee some a-priori bounds on $ \cos$ . Then you say that if your functional is less than some constant, then you know the bound for $ \|y'\|_{L^1}$. Then I don't really know what to do :(

I'm afraid that I'm trying to answer questions you didn't ask though. So, can you ask your question in a more understandable way?

Ok, I am reading a swedish text, so maybe "fält" does not translate to field. What I mean by a field is basically a set of of solutions to the corresponding euler equation for different initial values. Essentially it just boils down to showing that the euler equation has a solution with the right boundary values (I also need the existence of surrounding solutions of an initial value problem though). Then I want to use a theorem about the Weierstrass E-function to show that indeed this is a strong minimum. In the statement of that theorem it uses the lingo I used above. I haven't realized that terminology differs so much in math.

I will try the first approach you gave to solve it, I think I can do it like that. Sorry about the confusion. I don't really know too well what I am doing - I basically am trying to take the exam in a course I haven't taken because I need university credits, so I am hastily studying things. (This course has "methods", so I think I can do it. Maybe seems ethically wrong, but at least I am learning a little).

So, are there no general theorems about solutions to boundary value problems? If not, why is that?

Ok, I think this is right:

The euler equation is
$ \frac {y'y''}{1 \plus{} (y')^2} \equal{} \frac { \minus{} y'\sin y}{\cos y}$
It's integrated to
$ y' \equal{} \pm \sqrt {A\cos^2 y \minus{} 1}$ where $ A \geq 1$.
We choose the plus sign to solve the problem.

By the Lipschitz condition you refer to, plus some extra squabble about continuing solutions, for every $ A > 1$ this equation with the plus has a unique solution in $ [0,1]$ for every initial value in $ ( \minus{} \arccos \frac {1}{\sqrt {A}}, \arccos \frac {1}{\sqrt {A}})$. Let $ y_A$ be the solution with $ y_A(0) \equal{} 0$.

Calculation gives $ y_A'(x) \leq \tan \arccos \frac {1}{\sqrt {A}}$ (maybe I should have chosen better notation ...), which is less than $ \pi/4$ for some $ A$. Also, calculation shows that I can get $ A$ so that $ y'_A(x)$ bigger than $ pi/4$. I realized it's too much notation to write out though. So there is $ A$ so that $ y_A(1) \equal{} \pi/4$.

I would say that for this $ A$, the integrated equation with the plus defines a field with the wanted extremal in it embedded. It's now easy to see that Weierstrass E-function for this equation is $ \geq 0$ for $ x\in [0,1]$ and $ |y| < \pi/2$ so it's a strong local minimum.

I guess kind of pointless to write it down, especially when I didn't finish the details, but I did it mostly for myself. On all other problems I tried, it was possible to solve the equation exactly so I didn't need to argue existence of solutions. Thanks for the hints. Do you know how to finish your second suggested approach?

[quote="Kalle"]
So, are there no general theorems about solutions to boundary value problems? If not, why is that?[/quote]
The reasons are many. Try to solve something like $ y''\plus{}y\equal{}0$ with conditions $ y(0)\equal{}0,y(\pi)\equal{}1$ and you'll see that you are in a big trouble. And this is just for the linear equation! If you get some non-linearities involved, then the things get even worse. Try $ y''\equal{}\minus{}y'^3$ and you'll discover that starting from $ y(0)\equal{}0$, you'll never have any chance to reach $ y(1)\equal{}1000$ though the boundary value problems with small $ y(1)$ are perfectly well-posed for this equation. Any general existence theorem should include the full description of such phenomena. Now go figure :P. Of course, you can state [b]some[/b] theorems but their scope will be fairly limited.

[quote]
for every $ A > 1$ this equation with the plus has a unique solution in $ [0,1]$ for every initial value in $ ( \minus{} \arccos \frac {1}{\sqrt {A}}, \arccos \frac {1}{\sqrt {A}})$.
[/quote]
Hmmm... I'm not that sure: you have trouble when approaching $ Y\equal{}\arccos\frac{1}{\sqrt {A}}$ because the equation essentially becomes $ y'\equal{}\sqrt{Y\minus{}y}$, which allows reaching the level in finite time.

Also, I do not see how you integrated your Euler equation to what you wrote. If I reduce by $ y'$, I'll get $ \arctan y'\equal{}\log \cos y\plus{}C$ after taking the antiderivatives, which doesn't quite agree with what you claimed.

[quote][quote]
for every $ A > 1$ this equation with the plus has a unique solution in $ [0,1]$ for every initial value in $ ( \minus{} \arccos \frac {1}{\sqrt {A}}, \arccos \frac {1}{\sqrt {A}})$.
[/quote]
Hmmm... I'm not that sure: you have trouble when approaching $ Y \equal{} \arccos\frac {1}{\sqrt {A}}$ because the equation essentially becomes $ y' \equal{} \sqrt {Y \minus{} y}$, which allows reaching the level in finite time.[/quote]
For fixed A, there is through every point in the area $ \{x, y : 0 < x < 1, |y| < \arccos \frac {1}{\sqrt {A}}\}$ exactly one solution curve. Every solution $ y$ exists as long as $ |y| < \arccos \frac {1}{\sqrt {A}}$ and is growing where it exists. If $ y(x) \to \arccos \frac {1}{\sqrt {A}}$ as $ x\to x_0$, then we can extend uniquely the solution by setting $ y(x) \equal{} \arccos \frac {1}{\sqrt {A}}$ for $ x \geq x_0$. I don't think there is a hole now, but indeed I did not consider this originally.

[quote]Also, I do not see how you integrated your Euler equation to what you wrote. If I reduce by $ y'$, I'll get $ \arctan y' \equal{} \log \cos y \plus{} C$ after taking the antiderivatives, which doesn't quite agree with what you claimed.[/quote]
I made a mistake in writing down the equation

OK, if you are just after the existence, you are in a good shape. I was primarily worried about uniqueness that is lost, but if you do not need it, there is no problem. Also, it is clear now that your particular solution indeed gives $ y(1)$ continuously depending on $ y'(0)$, so you should be fine here too.

As to the terminology, even if it is standard, it is always a good idea to remind what it is unless you see it used in every other post. I can say about myself that my knowledge of mathematics can be best compared to the knowledge of the surroundings a person travelling with a weak flashlight over an unfamiliar terrain possesses. I see a small spot I am directing the beam at now more or less clearly, I also have some vague memories of what I've seen before, but the rest is completely black and uncertain. So, in many cases I have no idea of what the standard words mean if they do not directly relate to the field I'm working in now. You may say that a good mathematician should definitely know more than that, but I have never pretended to be any good, just somewhat lucky :P.

I do not need the uniqueness, but I think that the extension is unique. How could it be anything else when the function is bounded by $ \arccos \frac {1}{\sqrt {A}}$ and has derivative $ \geq 0$?

Every mathematician says they're not good, don't be so modest. I will be thrilled if I can become a PhD student; you've accomplished much more than that

With positive square root and real valued solutions, you are right. I just meant that the general theorem does not guarantee you anything once you reach the level. By the way, for the mere existence, you do not even need Lipschitz condition, just continuity.

Anyway, I think that you may safely consider the problem solved and go to the next one. The only thing I want to add is that if you ever want a reasonably complete and really well-written book on Variational calculus, try Gelfand and Fomin. It is also extremely cheap on Amazon (about 9 bucks for 200+ pages in paperback). I think I should reread it some day too :)

I found this question to be pretty darn difficult, and interesting, too.  For one thing, I didn't even know how to integrate the Euler-Lagrange equation until I read Kalle's post.   :blush:

You may very well have the complete solution, but I don't see it yet.  The problem has two parts: existence, and showing that the solution is a local minimum.

For existence, how did you compute that there is some $ A$ such that $ y_A(1)\geq \frac{\pi}{4}$?  This is non-obvious to me.

For the local minimum, I don't know what the Weierstrass E-function is.  Is that the same as the second variation, aka the Hessian, of the functional?  If so, then it is not obvious to me why this should be positive.  I'd be interested in seeing the proof.  (My best guess is that you can compare it to some constant coefficient linear problem that is known to have positive eigenvalues.)

To finish fedja's thought, for existence, you can bound the Sobolev norm of a minimizing sequence and then use a compactness theorem to extract a limit in a bigger space.  Then you prove regularity of the solution.  This is a standard PDE technique with the one hitch being that you have to rule out solutions whose $ y$ values are too large (say, bigger than $ \frac{\pi}{3}$).

Here's another perspective on this problem (that of a geometer):  The functional is computing the arclength of a curve with respect to the metric $ g_{ij}\equal{}(\cos^2 y) \delta_{ij}$, using the $ x$-coordinate as the parameter for the curve (as long as $ y<\frac{\pi}{2})$.  You can then appeal to the Hopf-Rinow Theorem to show that there exists a minimizing geodesic.  (Since we're not working on a *complete* Riemannian manifold, you again have to show that you don't have to worry about large values of $ y$ to make this kosher.)  Then the geodesic equation will tell you that that it's okay to parametrize your solution by the $ x$-coordinate.

Note that although this proof tells you that your solution is a local minimum, it doesn't tell you that it "passes" the 2nd Derivative Test (which is what I assume you mean by "strong" local min).  Again, as I mentioned above, this seems nontrivial to me.  Again, you probably have to do some comparing to a constant coefficient linear problem.

[quote="yenlee"]
For existence, how did you compute that there is some $ A$ such that $ y_A(1)\geq \frac {\pi}{4}$?  This is non-obvious to me.
[/quote]
As long as $ y_A(x) \leq \pi/3$ we have
${ y_A'(x) \geq \sqrt {\frac {A}{4} - 1}}$
And this is bigger than $ \pi/4$ for big $ A$, so we can find such an $ y_A$.
(Of course, implicitly I am also using $ y(1) = \int_0^1 y'(x) \, dx$).

The Weierstrass excess function is defined in this paper for instance [url=http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pjm/1102968298]here[/url]
I tried to find something better, but google mostly gives papers that you have to pay for.
Maple can compute it, and in the manual it also states the sufficiency I am using here.
It could be the same as the Hessian, but I don't want to think about this. I am not well versed in this subject. Anyway it's non-negativity is a sufficient condition, and in this particular case it's positivity follows from the convexity of $ z \to \sqrt {1 + z^2}$, although at first I did it by looking at derivatives and stuff. Does this argument also work for the Hessian?

For clarity of difficulty level, this is the hardest of five problems given in five hours, so it's not that bad. Usually the last problem is almost impossible to solve. I've only done it on one exam on higher level. That's sad to think about. In my mind I tell myself that I sometimes solve them. Oh well, I can usually solve them at home given unlimited time

Actually the Weierstrass function is a very beautiful idea. I'll outline it in the one-dimensional setting making a few shortcuts, so we'll have to rely on one miracle at one place. Good textbooks explain why it is not a miracle, but it normally takes 10 to 15 pages :P. It all starts with the so called "Hilbert invariant integral". Suppose that we have an extremal curve $ \gamma_0$. It may happen (and really happens most of the time) that it can be included into a family of extremal curves $ \gamma_t$ ($ t\in(-\delta,\delta)$ such that these curves sweep an open simply connected region $ R$ around $ \gamma_0$ without overlaps. In our case, what Kalle did was first to integrate the Euler equation to get the equation $ y'=\sqrt{A\cos^2y-1}$ for the extremals, then he found an appropriate $ A$ to have the curve joining the required endpoints and then using the solution curves $ \gamma_t$ for this equation starting at $ y(0)=t$.

Now, at each point of the region $ R$ swept by $ \gamma_t$, consider the function $ \psi(x,y)$ such that $ (1,\psi)$ is the tangent vector to the extremal passing through $ (x,y)$. The next thing is to try to set up a closed differential form $ \omega$ such that
\[ \int_{\gamma_t} \omega=\int_0^1 F(x,y_t,y'_t)\,dx    \qquad(1)\]
for every our extremal curve $ \gamma_t$ given by $ y=y_t(x)$. To this end, notice that since $ y'_t(x)=\psi(x,y_t(x))$ by the definition of $ \psi$, the best chance to find our form is to search for a form of the kind
\[ \omega=[F(x,y,\psi(x,y))-H(x,y)\psi(x,y)]\,dx+H(x,y)\,dy.\]
Then on each $ \gamma_t$ the $ H$ terms will cancel out when integrating. Actually it is the only possible kind of the form if we insist that the integrals over any part of $ \gamma_t$ coincide in (1).

To make $ \omega$ closed, we must have
\[ \frac{\partial F}{\partial y}+\frac{\partial F}{\partial y'}\frac{\partial \psi}{\partial y}-\frac{\partial H}{\partial y}\psi-H\frac{\partial \psi}{\partial y}=\frac{\partial H}{\partial x}\]
Now (here is the miraculous part), put
\[ H(x,y)=\frac{\partial F}{\partial y'}(x,y,\psi(x,y))\]
Then the terms with $ \frac{\partial \psi}{\partial y}$ disappear and what is left is
\[ \frac{\partial F}{\partial y}=\frac{\partial H}{\partial x}+\frac{\partial H}{\partial y}\psi\]
Note also that $ H(x,y)=\frac{\partial F}{\partial y'}(x,y,\psi(x,y))$ so $ H(x,y_t(x))=\frac{\partial F}{\partial y'}(x,y_t(x),y'_t(x))$.
Now, differentiating the last identity with respect to $ x$, we get
\[{ \frac{\partial H}{\partial x}}+\frac{\partial H}{\partial y}\psi
=\frac{d}{dx}\frac{\partial F}{\partial y'}(x,y_t(x),y'_t(x))\]
and we see that the identity we need is just the Euler equation!

The rest is no miracle. We just write the difference between the functional on any curve $ \gamma$ given by $ y=y(x)$ with given endpoints and the integral over $ \gamma_0$ as $ \int_0^1 F(x,y,y')-\int_\gamma\omega$ and, using our parametrization, get
\[ \int_0^1 E(x,y,\psi,y'-\psi)\,dx\]
where
\[ E(x,y,z,w)=F(x,y,z+w)-F(x,y,z)- \frac{\partial F}{\partial y'}(x,y,z)\cdot w\]
is the Weierstrass E-function. Its non-negativity is a simple consequence of convexity of $ F$ in $ y'$ in our case. That's it.


Hi all, I have a couple of rather strange questions.

Firstly, I would like to know-- How many people on AoPS identify themselves (or believe) themselves to be introverts or extroverts?

Secondly, Which is more better or has more advantages and less disadvantages: Being an introvert or an extrovert?

I would like to say that, in the present context an introvert doesn't necessarily mean a person who is bashful, shy or a person who doesn't have many friends. In the present context, an introvert may also mean-- a person who engages himself/herself in frequent introspection or deliberation.

These days, I'm inclined to think (it maybe untrue too) that only charming* people are better off in today's society-- despite the fact that the most significant contributions made to society were by highly brilliant scientists-- who were at most times introverted.

I also began to observe that people who are highly extroverted tend to make highly over-simplified arguments, frequently indulge in making sweeping generalizations, etc. They tend to display unsound reasoning. Again, I'm postulating based upon observation-- I don't have enough concrete evidence(link, etc) to substantiate this particular argument--which is based upon empirical evidence.

Personally, I, myself am neither an introvert or an extrovert. I possess qualities of both an introvert and an extrovert.

* Charming people (who're also good orators) win Nobel peace prize(s) :P  and get elected to high positions of power. :D

I am an introvert in the sense that I am shy.

I also believe that being slightly extroverted would have advantages.  I'm slightly introverted, and accordingly have a distaste of communication with strangers (but I'm working on overcoming that :) ).  The only advantage to being introverted is that I only 'run off at the mouth' to people I know really well (so I don't offend anyone accidentally).

I am and hate how I am an introvert,too .Yes,I am fraid of being among a crowd ,especially strangers I don't know.
And I suppose that being an introvert gets more disadvantages than an extrovert.Give me an specific example.When giving presentations in front of others people on classes,I always lost my confidence,remble at the thought of that someone will laugh at me.As consequence of that ,I forgot many things that  I prepared carefully at home and receive a lower grade then.
I think I should change myself to become more active,talkative,..but just don't know how   so I really nead some good advice from smart and nice friends here  :D

I would rather describe myself as an introvert in the sense that I  remain confined to myself.However,I can actually turn myself to an extrovert if required in some situations.In fact,in my opinion it is better to be a mixture of both.It helps in having a look at the flip side of the coin too.

You might want to look at this thread. It is on Myers-Briggs personality types. "I" stands for introverted and "E" stands for extroverted.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88374[/url]

I took the test on that thread and I got 66% E, which is about what I expected.  I'm certainly extroverted; I can make friends with strangers easily, share my thoughts easily, and give public speeches easily.  I also agree that extroverts have the advantage over introverts that more things will happen to them.  What I mean is by being extroverted, you get your name out there much more easily than if you keep your thoughts to yourself.

Wow, I took the test on the other forum and got:

INTJ (100	75	62	78)

It looks like I am 100% introverted.

I would guess that most people on AoPS are deeply involved in math, so they spend most of their time doing math. Then, they don't have time for socializing.

I spend almost of my free time doing math, but I still spend alot of time socializing.  I don't think you should have either one at the expense of the other, but I think you should have both at the expense of sleep.

[quote="Ihatepie"]I would agree with nobelium that it's best to be a mix of both. You need to be extroverted, but know when to be introverted or vice versa.[/quote]

I don't think you can just switch back and forth like you're suggesting.  Your personality doesn't change minute-to-minute.

Also, introversion doesn't imply a lack of social skills.  I'm very introverted in the sense that I enjoy spending time alone, and being around other people for too long drives me crazy - the personality test just told me that I'm 78% introverted.  But I think most people who know me would agree that I'm pretty easy to be around, and I've spoken in public a handful of times, and all of them went well.  Don't get me wrong - I'm not trying to brag, and I don't really have anything to brag about anyway - but I'm saying that introverted people can also be pretty sociable people.  I know plenty of other people who are this way too.

I guess I need to clarify what I said. It's good to be sociable and be good at meeting people, but it's also good to feel comfortable while alone and to be quiet when you need to.

When I'm with people I know well, I can be talkative (usually way too talkative :P ) and in conversations with another person, I'm usually talking about 65-75% of the time. I enjoy hanging out with my friends, but when I socialize with more than one person for longer than an hour or so, I get tired and want to be by myself, and the last time I went to a party was in third grade (it was a birthday party, and I didn't even like those.) Generally, I like being alone, and I absolutely hate talking to strangers or people I don't know well. It took me about five minutes to get up the courage to call a classmate about some homework I missed when I was sick once. In this sense, I suppose I'm introverted.

The obvious disadvantages of this are that I feel self-conscious when I have to converse with others (like in the homework example I gave), and sometimes this stops me from doing simple things, such as calling the library about a book I reserved or something, and I often avoid these situations.

On the other hand, as Ihatepie said, it's good to not mind being alone. I think I'd prefer being a little more extroverted, but being an introvert isn't completely bad.


Please choose an answer. If you like one over the other choose that one even if you like the other one OK also. If you like them both equally choose that or if neither are satisfactory choose neither.

[hide]Did you know that if you put spoiler around your text like this and have BBcode enabled then it will go into hide-a-text?[/hide]

[quote="h_s_potter2002"]Hide-a-text is better mainly becuase it hides LaTeX, which spolier does not.[/quote]
And images. Very helpful when you're answering a geometry question and a diagram is important but would give it away.

[quote="Super Naterul"][quote="Valentin Vornicu"]hmm nobody asks us the admins :D[/quote]

 What does "nobody asks you" supposed to mean? :?[/quote]just a joke ;)

I don't think the two are really comparable, hide does exactly what spoilers did and more :) The only difference being that you had to highlight the text instead of clicking on a link, which takes up a lot more room on the page.. its far better this new way.

Yeah, I like the

[hide]

hide-a-texts

[/hide]

better, but I miscasted my vote...

[quote="MysticTerminator"][hide=" "]blah[/hide][/quote]right :) .. let's not forget though that this is not the test forum, ok? ;)

i think it meant a vote for spoilers cuz arnav has had a history of adding in random comment that blend into the background with using spoilers

[quote="Super Naterul"][quote="theone853"]vote for spoilers cuz [u]arnav[/u] has had a history [/quote]

What is arnav???[/quote]

I think we can assume that he's MysticTerminator.

haha thats great

i definitely prefer hide. with spoilers, you couldnt hide LaTeX and you couldnt tell if there was actually something there or not. I remember trying to scroll down by highlighting stuff and accidentally catching the spoiler

[quote="theone853"][hide][hide][hide]There is such a thing as quoting, you know?[/hide][/hide][/hide][/quote]

[hide][hide][hide]Say what?[/hide][/hide][/hide]

Hm, this is probably getting a little annoying. Maybe we should stop before the mods come. Who knows what they'll do to us...  :D  ;)

[quote="yif man12"][quote="theone853"][quote="beta"][hide][hide][hide]I think I really like hide-a-text, because I can make people who wants to know what I'm saying keep clicking[/hide][/hide][/hide][/quote]

[hide][hide][hide]There is such a thing as quoting, you know?[/hide][/hide][/hide][/quote]

[hide][hide][hide]Say what?[/hide][/hide][/hide]

Hm, this is probably getting a little annoying. Maybe we should stop before the mods come. Who knows what they'll do to us...  :D  ;)[/quote]

[hide][hide][hide]Say what?[/hide][/hide][/hide]

[quote="KBabe"][/quote][quote="yif man12"][quote="theone853"][quote="beta"][hide][hide][hide]I think I really like hide-a-text, because I can make people who wants to know what I'm saying keep clicking[/hide][/hide][/hide][/quote]

[hide][hide][hide]There is such a thing as quoting, you know?[/hide][/hide][/hide][/quote]

[hide][hide][hide]Say what?[/hide][/hide][/hide]

Hm, this is probably getting a little annoying. Maybe we should stop before the mods come. Who knows what they'll do to us...  :D  ;)[/quote][quote="KBabe"]

[hide][hide][hide]Say what?[/hide][/hide][/hide][/quote]

[hide][hide][hide][/hide][/hide][/hide] :)

[hide][hide][hide]mine has moremwahahahahahahahahahahahahahahahaha :rotfl: [/hide][/hide][/hide]

[color=red]Modified by Admin: next time, the topic gets deleted![/color]

You guys really need to stop now.  This page is all screwed up at the bottom now...at least for my computer.

Hm, that post seems awfully familiar. (The weird one)

And yeah, my computer screen viewing is really funky too.

i guess you remember the ziggurat thread where it screwed up everyone's computers with at least 100 embedded quotes

To quote Valentin: ALWAYS hit quote, before starting to click  :P

You have a point.  [color=white]But white text still blends in! ; - )[/color]

[quote="Valentin Vornicu"][quote="cardwacko9999"]although the above trick is funny,  I still think that spoilers are better: you cant hide hide-a-texts from the unwary user.[/quote]What exactly do you mean?!  :?[/quote]

Before, you could have a couple of lines of seemingly empty space, that actually had hidden text in it... You could spoiler text in a non-obvious place and see if anyone would think to select it, as davidcyl did in [url=http://www.artofproblemsolving.com/Forum/topic-16485-15.html]this old thread[/url].

Yeah, it sure would be nice to write text that didn't seem to be there  :(  [size=67][color=white]It apears that noone has really read my last post..heh!!![/color][/size]

i saw it...i saw it. i just wanted to see if anyone else would b4 i ruined it.

[quote=barry.yyc]this is a 15 year old revival[/quote]

plz do not bump 15 year old threads


Two strings [b]I[/b] and [b]II[/b] of lengths $ l$ and $ 2l$ respectively are arranged as shown in the figure. The ends are connected to point masses two of mass $ m$ and one with mass $ 2m$. The separation between the balls is $ l$ as shown in figure. String [b]I[/b] is on the verge of getting taut. The entire arrangement is placed on a smooth horizontal table. The lower mass is given a horizontal velocity $ v \perp$ to the line joining it and the mass above it.

_____________________

$ 1)$ The impulse experienced by point mass of $ 2m$ due to jerking of string [b]II[/b] is:

$ A)$ $ \frac {mu}{\sqrt3} \plus{} \frac {mu\cos^27^o}{\sqrt3(6 \plus{} 3\sin^27^o \plus{} 2\cos^27^o)}$

$ B)$ $ \frac {\sqrt3mu\cos^27^o}{6 \plus{} 3\sin^27^o \plus{} 2\cos^27^o}$

$ C)$ $ \frac {\sqrt3mu}{8}\sin 14^o$

$ D)$ $ 0$

_____________________

$ 2)$ the velocity of $ m$ connected to string [b]I[/b] just after the string becomes taut is:

$ A)$ $ \frac {\sqrt3u}{8}\sin 14^o$

$ B)$ $ \frac {\sqrt3u\cos^27^o}{6 \plus{} 3\sin^27^o \plus{} 2\cos^27^o}$

$ C)$ $ \frac {\sqrt3mu\sin7^o\cos^27^o}{2\{6 \plus{} 3\sin^27^o \plus{} 2\cos^27^o\}}$

$ D)$ $ 0$

cmon this qn is not meant to be in the physics forum it is to be in the maths one as it involves just solving 9 linear eqns (if you do it using impulse momentum thm that is)i shall solve that hopefully and tell the answer :blush:

[quote="UNEARTHLY"]
...just solving 9 linear eqns...

[/quote]
9? :wink:

I did all that. But my answer does not match any of the options.

I understand this may be easy for you, but for general enlightenment where did you get those equations from? Conservation of linear momentum gives me only

$ 0 \equal{} v_{2I} \plus{} v_1$
$ 0 \equal{} v_{2II} \plus{} v_3$

where we're looking for $ mv_{2II}$. I know it must have something to do with $ l$, as that's information I'm clearly wasting. Also, my first thoughts were that $ v_1 \equal{} v_2 \equal{} v_3$, but that wouldn't lead to a very interesting answer.

P.S: Is $ u$ in the answers supposed to be $ v$?

man apply impulse=change in momentum for every one of these balls you willend up having two impulses one of these balls and then try resolving the velocities like in x and y directions then you must get some answer Rajiv bhaiya but i shall solve it and send in a reply as soon as i am free.. :oops:  :oops:

[quote="ffao"]Is $ u$ in the answers supposed to be $ v$?[/quote]

Yes. Sorry!

(I'm using this as scratch, hope you don't mind ;))
I'm still missing one equation. 8 variables, 7 equations... A and B are both balls of mass m.

By conservation of horizontal momentum:

$ v{}_A{}_x = 2v{}_I{}_x$
$ v{}_B{}_x = v - 2v{}_{II}{}_x$

Due to equilibrium, the horizontal velocities must be the same:

$ v{}_B{}_x = v{}_I{}_x + v{}_{II}{}_x$
$ [v{}_A{}_y - (v{}_{II}{}_y - v{}_I{}_y)] \tan 7^\circ + v{}_A{}_x = v{}_B{}_x$

By conservation of vertical momentum:

$ v{}_A{}_y = 2v{}_I{}_y$
$ v{}_B{}_y = 2v{}_{II}{}_y$

By equilibrium, we have that:

$ v{}_B{}_y + (v{}_{II}{}_y - v{}_I{}_y) = \sqrt2 [\frac {v{}_A{}_y - (v{}_{II}{}_y - v{}_I{}_y)}{\cos 7^\circ}]$


When is MOSP?
WHere is MOSP?
what do they do in MOSP?
do they select ppl by AIME score or grade or USAMO score???

http://www.unl.edu/amc <-- go browse this site is the best answer I think

[quote="white_horse_king88"]OR:[url]http://moppers.kaseorg.com/[/url]

This place is sweet.[/quote]

Thx. It is SWEET:
Error: HTTP 404 Who sent you here?

At MOP, you do math. 6 to 9 or 10 hours of it every day. *shrugs* After all, it is a math camp.

NICE!!!!!! to tell the truth ive never been to any math camps.....
and 8 to 10 hours a day??? sounds gr8!

Wow, that's a lot of math. Don't think I'd be able to take all of that...

They do stuff like ultimate frisbee there, right?

[quote="yif man12"]They do stuff like ultimate frisbee there, right?[/quote]

When you have free time in the afternoons, there's an excellent field across the street from the dorm...

[quote] What do they really do in MOSP??????? [/quote]

I'm sorry, but this topic just tempts it so much.

What do they really do in MOP?  Why they stuff you into a cell and force you to take math medicine.  It turns you into a good problem-solver, but they say side effects may include no life, obsession with math, and varying degrees of brain fry.  Other rare but serious side effects may occur.

hahahahahaha......that's pretty good! Even though it might have those side effects, I wouldn't mind qualifying! :lol:

Shoot. I must be immune to the part about "good problem-solver" and "obsessed with math".

It's incredibly intense- it won't all sink in until months later. Keep looking at problems afterward; you never know what you'll find out you learned. If you're not already obsessed about the problems, you didn't belong there anyway

Side effects may include:
dramatic improvement in Scrabble and other word games
ego deflation; you know who's better
inexplicable inside jokes ($\infty=$ Walter)
thinking you're really not all that weird, by comparison (SPUI?)
inventing new problems you don't know how to solve (Tetris against an evil computer)
trying to come up with an end-of-proof symbol
searching for that elusive 1.0 style score

I wandered around that dorm in Lincoln dazed for a month, and I came back a different person. The second time, I could actually keep up.


Find the greatest positive integer $n$ for which there exist $n$ nonnegative integers $x_1, x_2,\ldots , x_n$, not all zero, such that for any $\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n$ from the set $\{-1, 0, 1\}$, not all zero, $\varepsilon_1 x_1 + \varepsilon_2 x_2 + \cdots + \varepsilon_n x_n$ is not divisible by $n^3$.

Valentin: you use a  nice $-$ sign  :D

Claim: $n=9$.
When taking the $\epsilon_i$ from the set $\{0,1\}$ only, all numbers that are reached by this have to be different $\mod n^3$ since otherwise  subtracting them gives a number that can be written with $\epsilon_i \in \{-1,0,1\}$ and is divisible by $n^3$.
Thus $2^n \leq n^3$, which is just true for $2 \leq n \leq 9$.
But now we can take $x_1=1 , x_2 = 2 , x_3=4 , ... , x_9=2^8=256$ and it's easy to see that the moduli of the possible sums are not $0$ (use that the $2$-adic valuation is not $\infty$ for all sums that are allowed) and smaller than $9^3>511=2^8+2^7+...+4+2+1$, so not divisible by $9^3$.

@above

I think that you missed the case that, if we take $\varepsilon_3, \varepsilon_4,\cdots,\varepsilon_n=0$ and $\varepsilon_1=-1, \varepsilon_2=-1, \varepsilon_3=1$, then the sum $\varepsilon_1 x_1 + \varepsilon_2 x_2 + \cdots + \varepsilon_n x_n=0$, which is divisible by $n^3$.

[quote=Severus]I don't think I understand this  problem. If we take $x_i=i$, then the maximum value of $|\varepsilon_1x_1+\dots+\varepsilon_nx_n|=1+2+\dots+n=n(n+1)/2<n^3$, for $n\geq 2$. Then $n^3\nmid \varepsilon_1 x_1 + \varepsilon_2 x_2 + \cdots + \varepsilon_n x_n$. Thus shouldn't all $n$ work? I am surely misunderstanding the problem statement. Please help :)[/quote]

$\varepsilon_1x_1+\dots+\varepsilon_nx_n$ can be equal to 0, and $n^3 | 0$

Ah...Thanks@sansae and MessingWithMath!
That clears my confusion :)

My solution from WOOT:

Assume that $n\ge10$. There are $2^n$ expressions $\sum_{i=1}^n\varepsilon_ix_i$ with $\{\varepsilon_i\}\subseteq\{0,1\}$. Since $2^n>n^3$, the Pigeonhole Principle shows that at least two must have the same remainder$\pmod{n^3}$. Subtracting these, we have an expression in the form $\sum_{i=1}^n\varepsilon_ix_i$ with $\{\varepsilon_i\}\subseteq\{-1,0,1\}$ and $\exists i:\varepsilon_i\ne0$.

So $n\le9$. Now we claim that $n=\boxed9$ works. Choose $x_i=2^{i-1}$, and assume that $n^3\mid\sum_{i=1}^n\varepsilon_ix_i$.
The maximum of $\sum_{i=1}^n\varepsilon_ix_i$ is $\sum_{i=1}^nx_i=2^n-1<n^3$ while the minimum is $1-2^n>-n^3$. The final case is $\sum_{i=1}^n\varepsilon_ix_i=0$, which we assume.
We will prove (by induction) that $\varepsilon_i=0$ for all $i$. First, taking$\pmod2$ gives $\varepsilon_1$ even, so $\varepsilon_1=0$. Now assume that $\varepsilon_i=0$ for all $i<k$ for some $k$. Dividing our equality by $x_{k-1}$ and reading$\pmod2$ gives that $\varepsilon_{k+1}=0$.
Since all $\varepsilon_i=0$ after induction, we have a contradiction.

This is a slightly variation of the ZetaX´s solution
[quote]Claim: $n=9$.
When taking the $\epsilon_i$ from the set $\{0,1\}$ only, all numbers that are reached by this have to be different $\mod n^3$ since otherwise subtracting them gives a number that can be written with $\epsilon_i \in \{-1,0,1\}$ and is divisible by $n^3$.
Thus $2^n \leq n^3$, which is just true for $2 \leq n \leq 9$[/quote]
Taking  $x_1=1 , x_2 = 2 , x_3=4 , ... , x_9=2^8=256$, as $n^3\mid \varepsilon_1 x_1 + \varepsilon_2 x_2 + \cdots + \varepsilon_n x_n$ only if theres a non trivial combination such that $ x_1 + \varepsilon_2 x_2 + \cdots + \varepsilon_9 x_9 = 0$, but this is impossible, because this wold imply the existance of two distinct binary representation of a same number.

ISL 2002/N5 trick strikes again

The problem condition is equivalent to $a_1x_1+\cdots+a_nx_n$ being distinct modulo $n^3$ for all $n$-tuples $(a_1,\ldots,a_n)$ whose entries are either $0$ or $1$. Since there are $2^n$ such tuples, we need $2^n \leq n^3 \implies n \leq 9$. For $n=9$, take $(x_1,\ldots,x_9)=(1,2,4,\ldots,256)$. Every combination $a_1x_1+\cdots+a_nx_n$ is a distinct integer between $0$ and $511$, hence all distinct modulo $9^3=729$. $\blacksquare$


vector
trigonometry
geometry proposed
geometry

Assume that x, y, z are vectors on plane such that x + y + z = 0 and ||x|| = ||y|| = ||z|| = 1.
Prove that angle between any two such vectors is equal to 120 degrees.

This looks like it can be done with the dot product formula: $ \cos{\theta}\equal{}\frac{\textbf{a}\cdot\textbf{b}}{\parallel{}\textbf{a}\parallel{}\parallel{}\textbf{b}\parallel{}}$

$ x\cdot (x\plus{}y\plus{}z)\equal{}0\equal{}x\cdot x\plus{}x\cdot y\plus{}x\cdot z\equal{}1\plus{}x\cdot y\plus{}x\cdot z$

Then $ \minus{}1\equal{}x\cdot y\plus{}y\cdot z$ and cyclical relations...it is obvious then that $ x\cdot y\equal{}\minus{}\frac{1}{2}\equal{}|x\parallel{}y|\cos\theta\equal{}\cos \theta$

I feel so stupid asking this question.
I guess I need to come up with tougher ones.
Thanks.


modular arithmetic
calculus
integration
algebra
function
domain
polynomial

show that if p is prime,then (p-1)!=-1(mod p).

[quote="t0rajir0u"]Every number except $ 1,-1$ has a unique inverse $ \bmod p$ which is not itself.[/quote]
For all $ x\in \{1,2,...,p-1\}$ there is $ x'\in\{1,2,...,p-1\}$ (unique) such that $ xx'\equiv 1\pmod p$ because $ \gcd (x,p)=1$. We have $ x=x'$ iff $ x^{2}-1\equiv 0\pmod p$ iff $ x=1$ or $ p-1$.

[quote="t0rajir0u"][quote="pengyijie"]yes.but why?I hope you can give me more details.I am a greenhand.[/quote]

Suppose there is some $ a \not \equiv 0 \bmod p$ with two inverses $ b, b'$.  Then

$ ab \equiv ab' \equiv 1 \bmod p \implies a(b-b') \equiv 0 \bmod p \implies b \equiv b' \bmod p$

So the inverse is unique $ \bmod p$.  Here we have used the fact that $ \mathbb{Z}_{p}$ is an integral domain; in other words, for prime $ p$ we have

$ p | ab \implies p | a \vee p | b$[/quote]
yes.I know what you said.The uniqueness of inverse can be got directly by the fundamental theorem of group,because (Zp,*)forms a group.But what I really want to know is why x^2=1(mod p) holds only when x=1 and x=p-1.

Ok.Thanks a lot!
A friend of mine tell me how to solve it.p|x^2-1,so p|x-1 or p|x+1.

Generally, [b]Lagrange's Theorem[/b] states that a polynomial of degree $ n$ has at most $ n$ distinct solutions $ \bmod p$.


I can do this with case work,  but missed it in practice, (thought the sets had to be distinct). Anyone have a better way?

[quote]S has 6 elements. How many ways can we select two (possibly identical) subsets of S whose union is S? [/quote]

[hide="QEFD"]Every element is either only in setA, only in set B, or in both set A and B.

So that's 3^6 ways with overcounting(each selection is counted twice except A=B=S). identical sets happens when A=B=S, so (3^6-1)/2 gives you the ways to choose two non-identical sets, +1 gives you the answer, that's like 365 or something. [/hide]

I still don't understand how you get 3^6 ways of overcounting.......

I don't understand why its even $\ 3^n$ aren't there are $\ n$ options in each case making it $\ n^3$ not $\ 3^n$.  I don't know, I could be really ignorant right now but I'm not seeing how its $\ 3^n$ .

OKay,
Every element is either only in set A, only in set B, or in both set A and set B. Thus you have three options for each element, hence 3^6 (The idea is similar to that the proof of 2^n subsets, where the proof is each element is either in the subset or not in the subset).

As for overcounting, just think about it. Say our set is {1, 2, 3, 4, 5, 6}.
 Set A {1, 2, 3, 5}, Set B {4, 5, 6}.  Now this selection is same as:
 Set A {4, 5, 6}, Set B {1, 2, 3, 5}. Now both selections are counted in our 3^6.


[quote="white_horse_king88"]If $3^6*5^4=a^b$, where both $a$ and $b$ are positive integers, find the smallest possible value for $a + b$.[/quote]

[hide="solution"]$3^6\cdot 5^4=27^2\cdot 5^4=27^2\cdot 25^2=675^2$

$675+2=677$

i'm not sure if this is the smallest[/hide]

[quote="jli"]

[hide="solution"]$3^6\cdot 5^4=27^4\cdot 5^4=135^4$

$135+4=139$

i'm not sure if this is the smallest[/hide][/quote]

Hmm... $3^6=27^4$? You're on the right track, but exponential error.

[quote="Silverfalcon"]There is actually easy way.

[hide="Sol"]

Use calculators to get 3^6*5^4 is a number that is pretty high. Notice that taking square root of it simplify it to 675^2 but 675 doesn't simplify from here anymore.

So, 675+2 = 677 is final answer.[/hide]

Was this question like #17 or something? I can't remember exactly but I think it was somewhere around it.[/quote]

How does that demonstrate that it is the LOWEST value? 677 is pretty high. How do you know there isn't a smaller?

[quote="Silverfalcon"]Perhaps this might help:

[hide="Prime_Factorization"]

3^6*5^4 is what we have.

We can rewrite 3^6 as 3^2 and 3^4. Now, we can pair up 3^4 and 5^4, having 15^4. That leaves us 3^2. To have them in one exponent, we have to have them in same power. Now, we can't simplify 3^2 but we can simplify 15^4 into 225^2. Multiply them. You'll get 675^2.[/hide]

Did that answer your question?[/quote]

Perfectly, thank you =) I'm loving this place more every day!


number theory
superior algebra
superior algebra unsolved

Maybe it has been posted before,but I couldn't find it.
Find the product of the non null elements of $Z_{n}$.

Basically what you are asking is...
What is (as Gockel puts it): $1\cdot 2\cdot 3\cdot ... \cdot (n-1) = (n-1)! \equiv ? (\bmod n)$

Now if $n$ is prime then [u]Wilson's theorem[/u] implies $(n-1)! \equiv-1 (\bmod n)$.

What happens if $n$ is composite?
If $n=1$ then $(n-1)! \equiv 0 (\bmod 1)$.

If $n>1$? Since $n$ is composite we can write $n=ab$ non-trivially, that is, $2\leq a,b \leq n-1$. Now if $n$ is [b]not[/b] a square then $a\not = b$ and so we can find its factors among $1,2,...,(n-1)$ so then $(n-1)! \equiv 0 (\bmod n)$.

So the only question remains, what happens if $n$ is a square?
Say $n=c^{2}=cc$ and $c$ [b]is not[/b] prime, then $n=cab$ and the same argument implies as above. That is $(n-1)! \equiv 0 (\bmod n)$.

Finally what if $n=p^{2}$ and $p$ is prime. Well, if $p=2$ then $(n-1)! \equiv 2 (\bmod n)$. But if $n>4$ then and $n=p^{2}$. I claim that $2p$ is among the number $1,2,...,n$ because $p=\sqrt{n}\leq (n-1)/2$ so $2p$ is in the factors. And hence $(n-1)! \equiv 0 (\bmod n)$.

So, to summarize,
$n$ is composite and [b]not[/b] $=4$ then the product of its elements is 0.
$n$ is prime then the product of its elements is -1.
$n$ is composite and $n=4$ then the product of its elements is 2.

[quote="Gockel"]If n is composite, there are zerodivisors in $\mathbb{Z}_{n}$, so the product is 0.
If n is prime, every nonzero element is invertible. So most of the factors cancel out, since their inverse is also a factor of this product. The only factors that don't cancel out are the self-inverse elements. These are the solutions of $x^{2}=1$ which are exactly +1 and -1 (remember that $\mathbb{Z}_{n}$ is a field if n is prime). So the product is -1, if n is prime.[/quote]
This problem was in a contest earlier this day, and at first look this was also my solution too, later though I realized that $Z_{4}$ was an exception.

The general problem is: Let $A$ be a finite, commutative, unitary ring. Find the product of all non-zero elements of $A$ (it has been solved on the forum as part of a local olympiad problem).


$ lim_{n\to\infty}\sum_{i \equal{} 1}^n\ln(1\pm a^{ \minus{} i}) \equal{} 0$

is it true for all $ a > 1 \plus{} \sqrt {2}$ ?

I need for $ a > 1 \plus{} \sqrt {2}$, probably the answer is 'yes' and for $ a > 1$, not onty $ a > 1 \plus{} \sqrt {2}$ (I think so) but I may be wrong.......

any ideas? :wink:

For $ a > 1$ we may write
$ \lim_{n\to\infty}\sum_{i \equal{} 1}^{n}\ln (1 \plus{} a^{ \minus{} i}) \equal{} \sum_{i \equal{} 1}^{\infty}\ln \left (1 \plus{} \frac {1}{a^i}\right ) \equal{} \ln\prod_{i \equal{} 1}^{\infty}\left (1 \plus{} \frac {1}{a^i}\right ) > \ln 1 \equal{} 0$.
Similarly for $ \lim_{n\to\infty}\sum_{i \equal{} 1}^{n}\ln (1 \minus{} a^{ \minus{} i})$. But I'm not sure what you mean by the $ \pm$ sign. Hence
$ \lim_{n\to\infty}\sum_{i \equal{} 1}^{n}\ln (1\pm a^{ \minus{} i})\neq 0$ for $ a > 1$.


[quote="hoainamss"]Let $ x,y,z$ be positive real numbers.
Prove that: $ \frac {5x^3 \minus{} y^3}{xy \plus{} 3x^2} \plus{} \frac {5y^3 \minus{} z^3}{yz \plus{} 3y^2} \plus{} \frac {5z^3 \minus{} x^3}{zx \plus{} 3z^2}\leq x \plus{} y \plus{} z$[/quote]

without loss of generality, we may assume that $ x \ge y \ge z$ or $ x \le y \le z$.

$ \frac {5x^3 \minus{} y^3}{xy \plus{} 3x^2} \plus{} \frac {5y^3 \minus{} z^3}{yz \plus{} 3y^2} \plus{} \frac {5z^3 \minus{} x^3}{zx \plus{} 3z^2}\leq x \plus{} y \plus{} z$
$ \Longleftrightarrow \frac {(x \minus{} y) \left(x^2 \minus{} y z\right) \left(3 x^2 \plus{} x y \plus{} x z \plus{} 3 y z\right)}{x (3 x \plus{} y) z (x \plus{} 3 z)} \plus{}$ $ \frac {(y \minus{} z) \left(x y \minus{} z^2\right) \left(3 x y \plus{} x z \plus{} y z \plus{} 3 z^2\right)}{y z (3 y \plus{} z) (x \plus{} 3 z)} \ge 0$

which is ture by $ x \ge y \ge z$ or $ x \le y \le z$.

[quote="hoainamss"]Let $ x,y,z$ be positive real numbers.
Prove that: $ \frac {5x^3 \minus{}y^3}{xy\plus{}3x^2}\plus{}\frac {5y^3 \minus{}z^3}{yz\plus{}3y^2}\plus{}\frac {5z^3 \minus{}x^3}{zx\plus{}3z^2}\leq x\plus{}y\plus{}z$[/quote]
There is another solution for this type of inequality
Firstly , we prove that
$\frac{5x^3-y^3}{xy+3x^2} \le 2x-y (1)$
$ \leftrightarrow 5x^3 -y^3 \le -x^2y+6x^3-xy^2$
$\leftrightarrow x^3+y^3 \ge xy(x+y)$
$\leftrightarrow (x+y)(x-y)^2 \ge 0 $
The last inequality is true for all positive value of $x,y$

Similarly,we have $\frac{5y^3-z^3}{yz+3y^2} \le 2y-z (2)$

and $\frac{5z^3-x^3}{zx+3z^2} \le 2z-x (3)$

From three inequalities above , we have the conclusion

 $\frac{5x^3-y^3}{xy+3x^2} +\frac{5y^3-z^3}{yz+3y^2} +\frac{5z^3-x^3}{zx+3z^2} \le x+y+z$


I've seen this problem in 2 or 3 different places:

Prove that the units digit of n^5 is the same as the units digit of n.

The only thing that comes to my mind is this:

[hide]

Prove it for the numbers 0 to 9 by hand.

Let n=k (mod 10), 0<=k<=9

Therefore n^5=k^5 (mod 10).

We have already proved it for all numbers 0 to 9.Hence Proved.[/hide]

Is there a better(elegant) solution, i mean one which doesnt have the tedious stuff?

Hi,

[hide]If we consider factoring (n^5-n), we see that for all n, the term is divisible by 10, and hence the units digit of n^5 and n are the same.[/hide]

Sambit

n^5 - n = n (n^2+1)(n+1)(n-1)
then how do u see that all term are divisible by 10... can give a more detail solution..
Andy[/tex]

Thanks Sambit for your solution.

Quote:n^5 - n = n (n^2+1)(n+1)(n-1)
then how do u see that all term are divisible by 10... can give a more detail solution..
Andy

In spoiler:

[hide]Its divisible by 2 as either n or n+1 is divisible by 2

Also, n^5 - n is divisible by 5 by Fermat's Little Theorum[/hide]

I cannot see how factoring of n^5 - n can prove the question.. btw, u are using mod 5 and not mod 10.. can give a more detail solution..
thanks..

n^5-n is divisible by 10 if and only if it is divisible by 5 and  2, as GCD(2,5)=1. From Misha's factorization it is obvious, that n^5-n is divisible by both 5 and 2, because among five consecutive integers there will allways be a multiple of 5 and a multiple of 2. Hence n^5-n must be divisible by 10.

[quote="Misha123"]I think what Sambit meant is follows:

$n^5-n=n(n^2+1)(n-1)(n+1)\equiv n(n^2-4)(n-1)(n+1) mod 5$ and

$n(n^2-4)(n-1)(n+1)=(n-2)(n-1)n(n+1)(n+2)$.

Now the statement is obvious :D .

Misha[/quote]

Ok, hehe, I guess its sometimes better to stick to the basics.

Thanks..


Yek varag rast gooshe ba abaade $10\times 12$ ba geychi be do bakhsh ba masahathaye mosavi tagsim shode ast. Neshan dahid ke toole masire boresh hadeaghal $10$ boode ast.

[quote]Pas hallesh chi shod[/quote]

 :P

az hendeseye tahlili estefadeh kon;

masalan A (0,0) ast va B(x,0).....
hala mitoni begi ke do sare khatti ke mostatil ro be do ghesmate ham masahat taghseem kone bar azlae rooberoo be ham gharar daran.
Az onja ham formul faseleye do noghtaro benevis halle!!!! ;)

faghat tanha nokteye ke hast ine ke aya vaghean lazemeh ke do sar khat taghseem konandeh bar azlae roobero bashand?

halat bandi mikonim 6 halat badihi darad 1 halat ham(do zele mojaver) ba estefade az nabarabari va chand halat bandi digar hal mishavad.

rahe ason,kotah age kasi dare lotfan bege


*Taken from Mcrawford's lesson on sequences*

How many possible sequences of 8 coin flips are there so that no 2 sequential flips are heads?

Hint: [hide][hide]Use the Fibonnaci Sequence[/hide][/hide]

This is similar to a probability question from an AIME test... but the hint makes the answer come out rather quickly.

Would you like to post your answer along with the solution?

I basically gave away the answer with the hint. Would someone like to explain how they got the answer? It's rather interesting.

Would you like to post your idea so that other poeple can understand as well? It's a really cool problem.

hm... I don't get it  :blush: can someone please explain?


Prove that in every triangle $ABC$:

 $4 \cos{{A \over 2}} \cos{{B \over 2}} \cos{{C \over 2}}\geq \sin{A}+ \sin{B}+\sin{C}$.

Always we have equality : $\sum \sin A=4\prod \cos\frac A2\ .$ Indeed, $\sum \sin A=\sum \frac{a}{2R}=\frac pR\ ;$ $4\prod \cos\frac A2=4\prod \sqrt{\frac{p(p-a)}{bc}}=4\frac{pS}{abc}=\frac{4pS}{4RS}=\frac{p}{R}\ .$


The width of a rectangle is $ \frac23$ of its length and its perimeter is 4 units. Express as a common fraction the number of square units in the area of the rectangle.

Let the perimeter be $ l$. We then have $ \frac{2}{3}l \plus{} l \equal{} (4 \div 2) \implies 1\frac{2}{3}l \equal{} 2 \implies l \equal{} 1.2$. Thus, the area is $ 1.2 \times \frac{2}{3} \cdot 1.2 \equal{} \frac{24}{25}$.

I believe isabella2296 means that $l$ is the length of the rectangle because then you would be saying that $5/3$ of the perimeter is equal to $1/2$ of the perimeter, which is obviously false. The rest of her steps follow from the assumption that $l$ is the length, though.


For a point $ M$ inside an equilateral triangle $ ABC$, let $ D,E,F$ be the feet of the perpendiculars from $ M$ onto $ BC,CA,AB$, respectively. Find the locus of all such points $ M$ for which $ \angle FDE$ is a right angle.

The quadrilateral MDBF is cyclic, so $ \angle MDF \equal{} \angle MBF \equal{} 60^o \minus{} \angle MBC$.
Similarly, $ \angle MDE \equal{} 60^o \minus{} \angle MCB$. Hence $ \angle FDE \equal{} 120^o \minus{} (\angle MCB \plus{} \angle MBC)$.
Thus $ (\angle MCB \plus{} \angle MBC) \equal{} 30^o$. Hence $ \angle BMC \equal{} 150^o$. So the locus is just the arc of the circle through B and C such that $ \angle BMC \equal{} 150^o$.


function
trigonometry
calculus
calculus computations

I am having another problem with an inverse problem. I do not feel if I know for sure if I am doing it correctly.

The problem asks for the (f^1)'(a) for the function, f(x) = sin x, and real number a. a is .5.

Thank you. [/code]

We do this the same way as your last problem:

$f(x) = \sin x = .5 \Rightarrow x = \frac{\pi}{6}$

$f^{-1}'(.5) = \frac{1}{f'\left(\frac{\pi}{6}\right)} = \frac{2\sqrt{3}}{3}$

We could also solve this problem more explicitly:

$f(x) = \sin x \Rightarrow f^{-1} (x) = \sin^{-1} x \Rightarrow \frac{d}{dx} f^{-1} (x) = \frac{1}{\sqrt{1-x^2}} \Rightarrow f^{-1} (.5) = \frac{2\sqrt{3}}{3}$


induction
modular arithmetic
number theory

Prove, if y,z,w are odd integers, and x is an interger, then 17 divides

   $x^{y^{z^w}}-x^{y^z}$

Hint: use induction at you peril...

[hide]assume $x$ isn't divisible by 17 (otherwise the result follows immediately).  it then suffices, by fermat's little theorem, to show $y^{z^w} \equiv y^z \pmod{16}$.  since $y$ is odd, it then suffices to prove that $z^w \equiv z \pmod{8}$ (since $\phi(16) = 8$), or, since $z$ is odd, to prove that $z^{w-1} \equiv 1 \pmod{8}$, or $(z^{\frac{w-1}{2}}-1)(z^{\frac{w-1}{2}}+1) \equiv 0 \pmod{8}$, which is obvious because one of the terms on the left is $0 \pmod {4}$ while the other is $2 \pmod{4}$ (as they're even numbers that differ by 2).[/hide]

Very good! Our exam was today, thats how i did it after trying induction (which was a warning!) except a bit more laboriousally...

heh, i'm curious how you approached it by induction


Let $ a_1,a_2,\dots$ be a sequence of integers determined by the rule $ a_n\equal{}a_{n\minus{}1}/2$ if $ a_{n\minus{}1}$ is even and $ a_n\equal{}3a_{n\minus{}1}\plus{}1$ if $ a_{n\minus{}1}$ is odd. For how many positive integers $ a_1 \le 2008$ is it true that $ a_1$ is less than each of $ a_2$, $ a_3$, and $ a_4$?

$ \textbf{(A)}\ 250 \qquad
\textbf{(B)}\ 251 \qquad
\textbf{(C)}\ 501 \qquad
\textbf{(D)}\ 502 \qquad
\textbf{(E)}\ 1004$

[hide="Solution"]
Clearly, $ a_1$ is odd or else $ a_2 \equal{} a_1/2 < a_1$.

Then, $ a_2 \equal{} 3a_1 \plus{} 1 > a_1$, and $ a_2$ is even.

This forces $ a_3$ to be $ \frac{3a_1 \plus{} 1}{2} > \frac{3}{2} a_1 > a_1$.

We now note that $ 2 | a_3 \implies a_4 \equal{} \frac{3a_1\plus{}1}{4} \equal{} \frac{3}{4} a_1 \plus{} \frac{1}{4} \leq a_1$, hence $ a_3$ must be odd.

Thus, $ \frac{3a_1 \plus{} 1}{2} \equal{} 2k\plus{}1 \implies 3a_1 \plus{} 1 \equal{} 4k\plus{}2 \implies 3a_1 \equal{} 4k\plus{}1$

Notice that:
$ a_1 \equiv 3 \pmod{4} \implies 3a_1 \equiv 9 \equiv 1 \pmod{4}$, as desired.
$ a_1 \equiv 1 \pmod{4} \implies 3a_1 \equiv 3 \not \equiv 1 \pmod{4}$.

Hence, we want the set of $ a_1 \in \mathbb{N} ^{\leq 2008} \; : \; a_1 \equiv 3 \pmod{4}$

Partition 2008 into the sets: (1,2,3,4), (5,6,7,8), ... (2005, 2006, 2007, 2008), and exactly one element of each works.

Hence the answer is $ \frac{2008}{4} \equal{} 502 \implies \boxed{\text{D}}$
[/hide]

After getting $ 3a\equal{}1\plus{}4k\equiv 1 \mod 4$, we just multiply by the multiplicative inverse 3 mod 4 which turns out to be 3 by inspection, and we get the result $ a \equiv 3\mod 4$ easily.

In the official solution, they jump straight into representing numbers as 4n, 4n+1, etc. This makes sense, but how did they know to do this? How can I instantly see methods such as this after reading through the question? (I may have recognized this had I worked at the problem for a few minutes)

[hide = official solution 1]
All positive integers can be expressed as $4n$, $4n+1$, $4n+2$, or $4n+3$, where $n$ is a nonnegative integer.

If $a_1=4n$, then $a_2=\frac{4n}{2}=2n<a_1$.
If $a_1=4n+1$, then $a_2=3(4n+1)+1=12n+4$, $a_3=\frac{12n+4}{2}=6n+2$, and $a_4=\frac{6n+2}{2}=3n+1<a_1$.
If $a_1=4n+2$, then $a_2=2n+1<a_1$.
If $a_1=4n+3$, then $a_2=3(4n+3)+1=12n+10$, $a_3=\frac{12n+10}{2}=6n+5$, and $a_4=3(6n+5)+1=18n+16$.
Since $12n+10, 6n+5, 18n+16 > 4n+3$, every positive integer $a_1=4n+3$ will satisfy $a_1<a_2,a_3,a_4$.

Since one fourth of the positive integers $a_1 \le 2008$ can be expressed as $4n+3$, where $n$ is a nonnegative integer, the answer is $\frac{1}{4}\cdot 2008 = 502 \Rightarrow D$.

This is the same as the above solution, except they explicitly start off by saying we can partition the numbers into four classes. In general, the solution is not how you should attempt to solve it. Write backwards, solve forwards.

clearly $a_1$ has to be odd so $a_2=3a_1+1$

since $a_1$ is odd, $a_2$ is even, so $a_3=\frac32a_1+\frac12$

$a_3$ must be odd so $a_1$ must be $3\pmod4$

so our answer is 2008/4=502


[quote="hmoonster"]how long does the personal statement have to be?[/quote]

They vary quite a bit.  Anywhere from a few paragraphs to a page or two is typical.   It should be long enough to say what you want to say, and doesn't need to be any longer than that!

[quote="xxreddevilzxx"]Can i write about my Math competition "career" in the personal statement section??[/quote]

Absolutely.

It is also reasonable to write about this in your math background section.


Αποδείξτε ότι για κάθε πρώτο $ p$ υπάρχουν άπειροι αριθμοί της μορφής $ 2^n\minus{}n$ που διαιρούνται απο τον $ p$.

[quote="Γιώργος"]Αποδείξτε ότι για κάθε πρώτο $ p$ υπάρχουν άπειροι αριθμοί της μορφής $ 2^n \minus{} n$ που διαιρούνται απο τον $ p$.[/quote]

Σχετικα ευκολο αλλα ωραιο
Για $ p\equal{}2$ καθε αρτιος αριθμος $ >0$ εχει αυτη την ιδιοτητα. Για $ p>2$ επιλεγουμε τους αριθμους $ (p\minus{}1)^{2k}$ για $ k\equal{}1,2,3,...$, και απο το μικρο θεωρημα του Fermat ειναι: $ 2^{(p\minus{}1)^{2k}} \minus{} (p\minus{}1)^{2k} \equal{} (2^{(p\minus{}1)})^{(p\minus{}1)^{2k\minus{}1}} \minus{} (p\minus{}1)^{2k} \equiv 1 \minus{} 1 \equiv 0(modp)$.

[quote="Athinaios"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=243741[/quote]

kala dn paizomai :P  :P  :P


Would someone be so kind as to direct me to the thread that discussed what the optimal number of questions to answer for the newest scoring rules was on the AMC 10 and 12 given that one is on the borderline for qualifying for the AIME?

If there was none, how many questions would you answer on the AMC 10 if you weren't sure that you could qualify for the AIME?

I think I remember it.
AMC-10: Answer 21, or 25. For 21, you can get 2 wrong and still make it. With 25, you can get 5 wrong and still make it.
AMC 12:14,18,22, or 25. With 14, you need all of them correct for 100.5, with 18, you can get 3 wrong and make it, with 22, you can get 6 wrong and still make it, and with 25, you can get 8 wrong and make it.

Although, I think 17 is good on the 12, you can get 2 wrong and make it.

Last year someone at my school thought they needed 120 to get in.

So they had like 18 correct and 7 blank.  That was a 118.5.  So he thought he needed to randomly guess atleast one correct inorder to get over 120.  So he guessed on three of the questions (becuase 19 correct and 4 blank and 2 wrong is still 120 and qualifying)-hoping he guessed atleast 1 of the 3 questions right to qualify.  The person guessed all 3 wrong, leaving himwith 18 correct and 4 blank and 3 incorrect.

That was a 114.  Too bad last year his first score of 18 correct and 7 of 118.5 wouldve gotten him in (becuase less than 1% got 120), but since he was playing the "numbers game" you are trying to play right now, he ended up not getting in with a 114. (the cutoff was 117).

Infact if he had just tried to guess 1 randomly (not 3), he wouldve at worst ended up with 18 C/6B/1 I which is still 117 and qualifying.  However, since he didnt think that guessing 1 or 3 would make any difference (in both cases if you guessed 1 right your score would be over 120 and would qualify-and obviously you have a better chance at getting 1/3 than 1/1), he guessed all 3.

So basiclaly, do as many as you can.  Dont try and estimate how many you need right to get in and try to calculate what  stategy you need to use.

First, thank you xpmath, MellowMelon, and thefuture. All three of you are absolutely right. We're not going to try to answer only 21 if we can answer more, and we won't guess on problems.

I guess that I'm really more interested in your timing and strategies while taking the test rather than answering the least number possible to qualify. Is it rude for me to ask the below hidden questions in this thread that was originally about trying to answer questions optimally?[hide]How important is it to check your work and when do you decide to do so if you do?

Suppose you answered twenty-one questions and had five minutes left.
Would you go back and check your work or try to answer the rest of the questions 22-25?

What if you had ten minutes or fifteen minutes left in the above scenario?

Am I missing the point or are my questions valid?[/hide]

I'll hazard a guess that you're practicing past AMC tests. If timing and such is really something you're worried about, you should try to figure out how long on average you take to do the harder problems. That should help you make decisions in cases like the hypothetical one you gave.

I'm guessing that for most people checking work is going to be more important, now that calculators are not allowed.

I suggest you also read the advice that is part of the instructions read by the contest adminstrator at the time of the contest:
Teachers' Manual, page 5, Section II, paragraph 5.
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/1012TM/2008AMC1012A-TM.pdf

See also Frequently Asked Questions AIME #7 and AIME #8
http://www.unl.edu/amc/f-miscellaneous/faq.shtml#aime

Steve Dunbar
MAA Director, American Mathematics Competitions

I would suggest on at least #1-15 you should do them as fast as you can and not check your answer because it's multiple choice.  Most of the time, you check your answer and find that it's correct, which wastes precious time.  Ideally, less than a minute per problem for 1-15, so then the last ten should be easily doable with about 6 minutes per problem.

last year i didn't check any at all, answered all 25, and got 22 right (23, but i had a bubbling error).  EDIT: i guessed one one of them

as mentioned, the multiple-choice format helps with avoiding stupid mistakes.  In your scenario, with 5 minutes left, i would rather think about problems i havn't answered yet. I find that my "checking-work" quality when i rush is extremely poor, and i sometimes even change a right answer to a wrong one.

The ones that i would check with ample time remaining are:

1. Counting and probability problems
2. Algebraic simplifications
3. Problems that rely on arithmetic

I would advise against checking the easy ones because it's simply a waste of time you could be using to check the tougher ones that are more easily gotten wrong.

Well, congratulations for never making stupid mistakes on the easy ones, but I personally have caught several mistakes by spending a minute or so looking back over the first 10 problems. :|

Especially reading mistakes.

I always make those.

The cutoff is not necessarily 120/100, especially after the scoring change (blanks worth 1.5 instead of 2.5) and now this year the no calculator rule. It will be hard to tell, but if the AMC "feels harder than average" to you I would probably throw any point counting strategy out the window and go for the 1%/5% cutoff.

Everyone's tendencies to make silly errors is different. Judge for yourself by taking practice tests whether checking over the easy problems is worthwhile or not.

i take the past AMC's without a calculator in preparation for the upcoming AMC's..

i usually manage to finish the first 15 in about 20 minutes, the next five in another 7, leaving me about 8 minutes per each of the five problems left with time left to spare at the end. however, instead of checking all problems at the end, i find it more efficient to make sure of my answers [at least for the first 12 or so] as i go, so i don't have to waste time in reconsidering the problems at the end. true, this is kinda risky, but i think it works once you get faster at solving the problems.

[quote="MellowMelon"]As many as you can solve, really. I would worry more about doing the math well than trying to beat the test.[/quote]

I'm with MellowMelon on this one


Dear Mathlinkers
let ABCD a convex quadrilateral
The conjecture :
(sin <BAC / sin <BDC). (sin <ACD / sin < ABD) = (sin <ACB / sin <ADB) . (sin <CAD / sin < DBC)
I don't know if it is written in the best esthetic form?
Sincerely
Jean-Louis

$ \frac{\sin \angle BAC}{\sin \angle BDC}\frac{\sin \angle ACD }{ \sin \angle ABD} \equal{} \frac{\sin \angle ACB}{\sin \angle ADB} \frac {\sin \angle CAD }{ \sin \angle DBC} \leftrightarrow$

$ \frac{\sin \angle BAC}{\sin \angle ACB}\frac{\sin \angle ACD }{ \sin \angle CAD} \equal{} \frac{\sin \angle ABD}{\sin \angle ADB} \frac {\sin \angle BDC }{ \sin \angle DBC} \leftrightarrow$

$ \frac{BC}{AB} \frac{AD}{CD} \equal{} \frac{AD}{AB} \frac{BC}{DC}$

It's definite!  :D

Dear Mathlinkers,
an article concerning the ‘’Euler-Poncelet point'' has been put on my website
http://perso.orange.fr/jl.ayme    vol. 8 Le point d’Euler-Poncelet d’un quadrilatère  p. 100.
You can use Google translator
Sincerely
Jean-Louis