Is it possible to combine >> with getline?

For example, I want to read in the following graph, I use
fin>>H>>W;
fin.getline(str, 100, '\n');

but i can't read the graph. I know there is something with the delimiter and getline's way to reading it, but I don't know how to solve it. Please lemme know how would I be able to read the follow, with 5 and 3 stored in int varibales.

5 3
+-+-+-+-+-+
|         |
+-+ +-+ + +
|     | | |
+ +-+-+ + +
| |     |
+-+ +-+-+-+

u can just use;

2 for loops with scanf("%c",&ch)

 :D

or fgets is fine. :P

Your problem is when getline is called, it automatically calls the carriage return (i.e "\r").  This can be fixed like so.

[code]#include <iostream>
using namespace std;

string temp;
int H, W;
fin>>H>>W;
fin.ignore(numeric_limits<streamsize>::max(), '\n');
getline(fin, named);[/code]

The cin.ignore command causes the current position in the stream to skip to the next line instead of returning to the beginning of the current line and placing the entire line or the first 100 chars of it in your string.  It should be used between any traditional >> calls and getline calls.


compare the numbers $ \pi^{e} , e^{\pi}$

$ e^{\pi} > \pi^{e}$ iff $ (e^{\pi})^{\frac{1}{\pi e}} > (\pi ^{e})^{\frac{1}{\pi e}}$

iff $ e^{\frac{1}{e}} > \pi ^ {\frac{1}{\pi}}$

Now consider the graph of $ y\equal{}x^{\frac{1}{x}}$.

We could differentiate to find the maximum at $ x\equal{}e$ and conclude that $ e^{\pi}$ is indeed gretaer that $ \pi^{e}$, but differentiation is not appropriate for this forum so I'm not sure what to suggest.

My calculator gives $ e^{\pi}$ as $ 23.14069263$ and $ \pi ^{e}$ as $ 22.45915772$, which confirms that the above result is correct.

Haven't thought of another approach yet.

We could take logs to get $ e^{\pi} > \pi^{e}$

iff

$ \pi \ln e > e \ln \pi$

iff

$ \pi > e \ln \pi$

iff

$ \frac{\pi}{\ ln \pi} > e$,

so we need to show that the minimum of $ y\equal{} \frac{x}{\ln x}$ occurs for $ x\equal{}e$, which is true (again) by differentiation.


I have nine coins which have a total worth of 36¢. How many dimes do I have?

[hide="my answer"] 3? It seems too simple, but I just thought 6 pennies and 3 dimes.... [/hide]

could you possibly... [i]prove[/i] it???

[hide="Proof"]First, there obviously needs to be a penny.
But if there is only one penny then we cannot cover 35 cents within 8 non-pennies. [8 nickels = 40 cents > goal]
So there are six pennies.

Now we need to cover 30 cents with 3 coins.
Obviously no quarter here - then we would cover 5 cents with 2 non-pennies
since the average coin in these 3 coins is 10 cents, there are 3 dimes[/hide]

nice proof james4l!
btw, doesn't mod arithmetic help here also?

james4l put in "modular arithmetic", he just didn't type it out.


The base three representation of $x$ is \[ 12112211122211112222. \]The first digit (on the left) of the base nine representation of $x$ is

$\text{(A)} \ 1 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 4 \qquad \text{(E)} \ 5$

[hide]Notice that $3^{2n}=9^n$. Therefore, we can group terms of $x$ to find the base $9$ representation of $x$.

Only the first two digits of our number are large enough to have significance. We want the $x^9$th term.

Therefore, $1^{19}_3+2^{18}_3=(3\cdot 1+2)3^{18}=(5)9^8$, and our answer is 5.[/hide]

Two digits in base 3 represent one digit in base 9. There are an even number of digits, so the leftmost 2 make a pair. Converting, we have ${3^1} + 2({3^0}) = 5 \implies\boxed{E}$


$ a^{4}\plus{}b^{4}\plus{}c^{4}\equal{}3,a,b,c>0$
prove that $ \sum_{cyclic}\frac{1}{4\minus{}ab}\leq 1$

[quote="apollo"]$ a^{4}\plus{}b^{4}\plus{}c^{4}\equal{} 3,a,b,c > 0$
prove that $ \sum_{cyclic}\frac{1}{4\minus{}ab}\leq 1$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30764


parameterization
function
algebra
functional equation

Solve for $ x$: $ x\equal{}\sqrt{a\minus{}\sqrt{a\plus{}x}}$, where $ a>0$ is a parameter.

The thing that I'm not clear with is the argument of the restraints on $ x$.

Thanks for any help.

Answer
[hide]$ x\equal{}\frac{\minus{}1\plus{}\sqrt{4a\minus{}3}}{2}$ when and only when $ a\geq 1$[/hide]

Mmm...unluckily, this is not a one that can be rewritten as composite functional equation: the signs under two radical are not the same.

My solution was straightforwad: square, rearrange, square again, and factor the resulted polynomial. However, after that, it becomes hard to tell which of the four roots are to be chosen, and for what values of the parameter $ a$:

$ x_{1,2} \equal{} \frac { \minus{} 1\pm \sqrt {4a \minus{} 3}}{2} \text{ and } x_{3,4} \equal{} \frac {1\pm \sqrt {4a \plus{} 1}}{2}$

[hide="Variance in a"]
$ \sqrt{a-\sqrt{a+x}}$ is nonnegative, so squaring twice gives $ a^2-a \geq x \iff  \(a - \frac{1}{2}  \)^2 - \frac{1}{4} \geq x$.

Now, we know that $ x \geq 0$, so $ ( a - \frac{1}{2} ) ^2 \geq  \frac{1}{4} \iff | a - \frac{1}{2} | \geq \frac{1}{2}$ $ \iff a \leq 0 \; or \; a \geq 1$.

And since $ a>0$, this means $ a \geq 1$.
[/hide]

[hide="Picking your solution"]We know that $ x \geq 0$. Consider the graphs of $ x$ and $ \sqrt{a-\sqrt{a+x}}$. At $ x=0$, $ \sqrt{a-\sqrt{a+x}} > x$. however, $ \sqrt{a-\sqrt{a+x}}$ is decreasing while $ x$ is increasing, and we achieve one unique solution. It is now our task to figure out which of the four solutions to your quartic is the actual solution.

Because of the nature of the problem (continuity of functions and etc etc) we can assume that this solution takes the same form in terms of $ a$ each time.

So you know $ x$ is positive since you're working in the reals, so that eliminates $ x_2$ and $ x_4$ (consider sufficiently large $ a$).

Then plug in the known solution, such as $ (x,a) = (1,3)$, to find that $ x_1$ is the solution.[/hide]


probability
number theory unsolved
number theory

There are 9 cards in a box, labelled from 1 to 9.
How many cards should be taken so that the probability of getting a card whose label is multiple of 4 will be greater than $ \frac{5}{6}$


Let n be a positive integer and let $x_1,x_2,...,x_n,y_1,y_2,...,y_n,z_2,z_3...,z_{2n}$ be positive reals satisfying that
$x_iy_j\geq z_{i+j}^2$ for all $1\leq i,j\leq n$
And set $Z=max(z_2,z_3,...,z_{2n})$. Prove that
$
(\frac{x_1+...+x_n}{n})(\frac{y_1+...+y_n}{n})\leq (\frac{Z+z_2+z_3+...+z_{2n}}{2n})^2
$

Do you think this wonderful inequality hadn't been discussed at MathLinks?

http://www.mathlinks.ro/Forum/viewtopic.php?t=6062&start=0

  darij

Thanks. I searched for it before I posted but I failed.


Let $a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that $\left(\frac{a}{b+c}\right)^{2}+\left(\frac{b}{c+a}\right)^{2}+\left(\frac{c}{a+b}\right)^{2}+\frac{2abc}{(a+b)(b+c)(c+a)}\ge 1$

[quote="goc"][quote="behemont"]think it can be easily done with setting $p=a+b+c, q=ab+bc+ca, r=abc$, expanding and using
$p^{3}+9r\geq 4pq$ and $p^{2}\geq 3q$..[/quote]
learning fast i see ;) all though i didn't check to see if your solution is correct, i remember that you got this exact hint earlier in the day...:)[/quote]

first of all, all though is spelled although, if you mean "iako", and i suppose you do...
and for learning fast, that's correct..everyone says me so... :P

[quote="goc"][quote="behemont"]says me so... :P[/quote]
seljak... tells me so... :lol:[/quote]

we're even now... :D


In a triangle $ABC$ we have $\angle ABC = 2 \angle ACB$. Point $D$ divides segment $BC$ in a ratio $2: 1$. $E$ is chosen so that $D$ is midpoint of segment $AE$. Prove that:

$2\angle ECB = \angle EBC + \pi$

http://www.mathlinks.ro/Forum/viewtopic.php?t=18153

  darij


One color printer produces 300 dots per linear inch while another can produce 600 dots per linear inch.  If the two printers both print the same 8-inch by 10-inch photograph, what is the ratio of the total number of dots produced by the first printer to the total number of dots produced by the second printer?  This figure shows what a printer using four dots per linear inch would produce.  Express your answer as a common fraction.

[asy]filldraw(Circle((0,0),0.5));
filldraw(Circle((1,0),0.5));
filldraw(Circle((2,0),0.5));
filldraw(Circle((3,0),0.5));

filldraw(Circle((1,1),0.5));
filldraw(Circle((2,1),0.5));
filldraw(Circle((3,1),0.5));
filldraw(Circle((0,1),0.5));

filldraw(Circle((0,2),0.5));
filldraw(Circle((1,2),0.5));
filldraw(Circle((2,2),0.5));
filldraw(Circle((3,2),0.5));

filldraw(Circle((0,3),0.5));
filldraw(Circle((1,3),0.5));
filldraw(Circle((2,3),0.5));
filldraw(Circle((3,3),0.5));

draw((-0.8,-0.5)--(-0.8,3.5),Bars);
draw((-0.5,3.8)--(3.5,3.8),Bars);

label(rotate(90)*"1 inch",(-0.8,1.5),W);
label("1 inch",(1.5,3.8),N);[/asy]

The ratio of the dots per linear inch is $ \frac{1}{2}$, thus the ratio of the dots per square inch is $ \frac{1}{2}^2\equal{}\boxed{\frac{1}{4}}$.


Lately i've been getting tasks where i have to make the definition set for a function (sorry, i don't know what you call it in english, but plz tell me if you don't understand what i mean). The book i'm studying doesn't really make it clear (if you ask me) what this definition set is - i would have liked it more described.

Like this one:
g(x)=3/x

How do i find the definition set for this function?

i think you mean the set of all numbers for which the function gives returns a value, it is called the domain of a function. for y=3/x there is only one x-value that will not result in a nice y-value, and that is x=0 since 3/0 is undefined.

Yeah, that sounds about right. But what i really think is odd is that there are several tasks where i have to set the domain of a function. At the beginning of the chapter they are talking about a domain of a function for a barrel where 'x' is indicate radius of the barrel and thereby sets 'x' to be[0.3;1]. But then you've got to know what the function indicates in the real world, and that was what i found odd.

And what about these 3 functions? Do you just remove the 'x' value that makes 'y'=0?

f(x)=x^2-1
g(x)=sqrd(x)-3
h(x)=(2/x)+(3/2x-10)

No,  you just remove the 'x' values for which you can not calculate a value of the function.
eg:  g(x) cannot be calculated for x=-4 since there is no such thing as sqrt(-4) (working with reals, i mean). So the domain of g is the set of nonnegative reals.

[quote="Sporally"]Sorry i was away for a bit, hope you are still in on this thread.

What would you say about f and h then?[/quote]

For $f(x)=x^{2}-1$ since $x^{2}\geq 0$, $x^{2}-1 \geq -1$
 so the domain of $f$ is the set of reals $\geq$ -1.

For h, do you mean
$\frac{2}{x}+\frac{3}{2x-10}$
or
$\frac{2}{x}+\frac{3}{2x}-10$
?

The one where -10 is below the equation line.

For f, why can't x^2 be negative?

Because, when you work with the real numbers, a square is always positive..

I recall i have heard that before. Not sure that was the rule here aswell though. But thx, think i know almost everything there is to know about the domain of a function, or else expect me to drop in for a cup of wine ;)

Alright :)
So, now, as an exercice, give the domain of h ;)

Now this feels like in was in the old days in school, and i'm suppose to practise my math skills only :D

How about R (R is what we use for all real numbers) but 0 and 5?

Yes, that's right, congrats :D

So, to sum it up, the main rules when dealing with domains of definition are :
-you cannot have a negative number under a square root (when working with reals)
-you cannot divide by zero

Great. Thx for all the help on this one. Might not have to drop in for a cup of wine on this thread, but expect me to be back somewhat later - still have math for 2,5 years to sum up on :roll:

Hehe ok :)
Actually, that's cool you could understand, and, if you ever need help, that's why this forum is for.. ;)
So, you're welcome!

Well there are some cases of domain of real functions:

[b]1)[/b] $y= \frac{f(x)}{g(x)}$

To get domain you must do $g(x)\not= 0$, so domain is $D=\{x \in \mathbb{R}\, |\, g(x) \not= 0 \}$

[b]2)[/b] $y= {\sqrt[n]{f(x)}}$

If $n$ is even, $D=\{x \in \mathbb{R}\, |\, f(x) \geq 0 \}$

If $n$ is odd, $D=\mathbb{R}$

[b]3)[/b] $y=\frac{f(x)}{\sqrt[n]{g(x)}}$

If $n$ is even, by (1) you have $g(x) \not= 0$ and by (2) you have  $g(x) \geq 0$, hence $D=\{x \in \mathbb{R}\, |\, g(x) > 0 \}$

If $n$ is odd, by (1) you have $g(x) \not= 0$ and by (2) you have $\mathbb{R}$, hence $D=\{x \in \mathbb{R}\, |\, g(x) \not= 0 \}$

And you can have a few more cases, but knowing those above you can get the rest :)

Well, that's what i call an explanation. Thx a lot again ;)


function
limit
real analysis
real analysis unsolved

let $ f_k$ be a sequence of functions converging pointwise to $ f$ on $ [a,b]$. Assume that each $ f_k$ is of bounded variation and that $ V(f_k)\leq M < \infty$ for all k . ( $ V(f_k)$ denotes the total variation of f_k on $ [a,b]$ ) . prove that f is also of bounded variation and $ V(f)\leq M$

Let $ P$ be a fixed partition of $ [a,b].$ Then

$ V(P;f)\equal{}\sum_{j\equal{}1}^m|f(x_j)\minus{}f(x_{j\minus{}1})|\equal{}$

$ \equal{}\lim_{n\to\infty}\sum_{j\equal{}1}^m|f_n(x_j)\minus{}f_n(x_{j\minus{}1})|\equal{}
\lim_{n\to\infty}V(P;f_n)\le V(f_n)\le M.$

Now take the supremum over all partitions $ P.$

---

I'm teaching a course this semester, starting next week, for which I am using Wheeden and Zygmund as textbook. That book draws on my mathematical ancestry.

Ah, I see Antoni Zygmund is your mathematical grandparent: http://www.genealogy.ams.org/id.php?id=9336 :)

[quote]I'm teaching a course this semester, starting next week, for which I am using Wheeden and Zygmund as textbook. That book draws on my mathematical ancestry[/quote]

oh you are? we follow the same book too in my class. If i should come across some difficulties in the text book, may i ask you? thanks


ı dont know answer. but is seems as a binomial intergrate.
I am very curious me too

[hide="A Possible Start"]Let $y=\sqrt{x+\sqrt{x+\ldots}}$.
$y=\sqrt{x+y}$
$y^{2}=x+y$
$y^{2}-y-x=0$
$y=\frac{1\pm\sqrt{4x+1}}{2}$
$sqrt{x+\sqrt{x+\ldots}}=\frac{1\pm\sqrt{4x+1}}{2}$
I think that we should go with the positive root.
$\int \frac{1+\sqrt{4x+1}}{2}\,dx$[/hide]

I interpreted it as ending after 4 x's.

However, if it was intended to go on indefinitely, then i suppose that's what it would be.

Oh.  For some reason, I thought that it was supposed to go on indefinitely.  But apparently, that's not the case.

Hey, I was tired.

[quote]The Zuton Force AstroPhys, I don't understand your hint. Where would the chain rule apply? It's an integral. Please explicate.

Personally, I have strong doubts about this function having an elementary anti-derivative. I see no approach or no substitution that would make it work. Mathematica agrees with me also.

luciferit, do you know the solution/answer to this problem? If so, I am very curious to see it. If it is a problem you made up out of curiosity, it likely won't have an answer (it's been said jokingly that the only anti-differentiable functions are the ones that appear in the textbook). [/quote]
 i agree with the zuton force, this function does not have an elementary anti-derivative.

It turns out that the function
\[\int_{0}^{x}\sqrt{t+\sqrt{t}}\ dt\]
is elementary, even though the original problem does not result in one. And of course this topic should be moved to calculus section.
[hide="hint with new problem"]
The expression under the integral is
\[t^\frac{1}{4}\cdot \sqrt{1+\sqrt{t}}\]
[/hide]


Consider the geometric sequence $ \frac{16}{9}, \frac{8}{3}, 4, 6, 9, \ldots$. What is the eighth term of the sequence? Express your answer as a common fraction.

EDIT: Sorry, I thought I had found a shortcut.

We see that the common ratio is $ \frac {3}{2}$. So we have $ \frac {16}{9}*(\frac {3}{2})^7 \equal{} \frac {243}{8}$.

Edit: Uh modularmarc, I think your answer is incorrect.


1. Let ABC be a triangle, and let D and E be points on sides AB and AC, respectively, such that DE is parallel to BS. Prove that AD/AE=DB/CE

2. Let ABC be a triangle, and let D and E be points on AB and AC, respectively, such that AD/AE=BD/EC. Prove that DE is parallel to BC.

hmm... I'm not sure if these questions really belong in the Mathcounts forum, but here's how you can prove them.
[hide="#1"]Because DE||BC, corresponding angles are congruent, so triangle ADE~ABC. So AD/AB=AE/AC.
since AB=AD+DB and AC=AE+AC, plug in these equations to AB and AC, simplify it, and you'll get AD/AE=DB/CE.[/hide]
[hide="#2"]It's the converse of #1. You know AD/AE=DB/CE, so you have ADE~ABC. Thus, angle ADE=angle ABC. Since they are corresponding angles, DE||BC.[/hide]

*On #1, I'm sure when you said BS, you meant BC.

[quote="AIME15"]It's his homework problems.[/quote]
AIME15 says it nicely to him...but basely to me... :(  :furious:  :cursing:


function
complex analysis
algebra
domain
complex numbers
real analysis
complex analysis unsolved

Let $ f(z) = \frac 1 {\sqrt {z^2 - 1}}$, but for $ \sqrt {\sqr z^2 - 1}$ is chosen a branch (regular) wich is positive for $ |z| > 1$. Find a series of function $ f(z)$ in a neighborhood of $ a = \infty$ and determine radius convergence for that series.

The function is not analytic at a neighborhood of $ \infty$, so the problem doesn't make sense as written.

[quote="Solunac"]... for $ {\sqrt z} \minus{} 1$ is chosen a branch (regular) wich is positive for $ |z| > 1$. [/quote]How can that be if positive means real and positive?

[quote="Solunac"]... but for $ \sqrt {\sqr z^2 - 1}$ is chosen a branch (regular) wich is positive for $ |z| > 1$.[/quote]Do you mean that the square root has [b]positive real part[/b] for $ |z| > 1$?

No. Now it is apparent that the OP really meant "positive for real $ z>1$".

Well, that is the text. I have seen that it can be real and positive only for certain arguments of $ z$, but I don't see what $ |z| > 1$ has to do with it. Maybe it is a mistake in text, but what kind of a mistake? It could be what you suggested, but still what does it has with $ |z| > 1$? Sorry if I am not exact as I would have to be.

By, the way thanks for suggestions.

...or it could be what fedja suggested. Thanks to you too fedja.

Supposing that $ g(z)\equal{}\sqrt{z^2\minus{}1}$ is a square root with real positive values for $ z>1$, do you assume that $ f$ is holomorphic at $ \infty$ such that $ f$ [b]can[/b] be expanded at $ \infty$?
I am not so sure about this since this strongly depends on the chosen domain of $ g$. ;) E.g., what happens if the plane is sliced by a straight line starting from $ 1$?

Transformed like
$ f(z)\equal{}{1\over z}\left(1\minus{}{1\over z^2}\right)^{\minus{}1/2}\;;$
it could be used
$ (1\plus{}w)^{\alpha}$
(for small $ w$)

... and maybe it ment "wich is if z "real" pozitive for |z|>1" (for regulating difficulties with domain).

Note that you urgently need to be very careful with your symbols. First you might want to tell us the domain and the definition of your functions before you use the formalism only - hoping that the relations applied really work. Since we are dealing with complex analysis here, the intuition gathered in real analysis does not work or has to be subtly refined.

Please define $ f$ is rigorous terms, e.g., by using a branch of a logarithm. Then proceed - and we will see.

At the end the question invokes that $ f$ is holomorphic at $ \infty$ which - as I already said - strongly depends on the domain of $ f$.


[quote="A smileys notice"]Too many smilies in your post: 12.
You can have at most 10 smilies in this message. Please remove some smilies before submitting this message.[/quote]

[quote="Another smileys notice"]Too many smilies in your post: 10
You can have at most 9 smilies in this message. Please remove some smilies before submitting this message.[/quote]

[quote="Another smileys notice"]Too many smilies in your post: 9
You can have at most 8 smilies in this message. Please remove some smilies before submitting this message.[/quote]

[quote="Another smileys notice"]Too many smilies in your post: 8
You can have at most 7 smilies in this message. Please remove some smilies before submitting this message.[/quote]

[quote="Another smileys notice"]Too many smilies in your post: 6
You can have at most 3 smilies in this message. Please remove some smilies before submitting this message.[/quote]

What's with the constantly changing number of smileys allowed?

[quote="vallon22"]it is supposed to be 3

hmm... i tested a bit (using this post, deleted now because i really spammed the post up) and the smilies count restriction didn't seem to apply

i posted at least 20 smilies using 5 different ones[/quote]

Maybe that's because you're a moderator?  :P

Apparently, in the test forum, this thing doesn't happen, but it did in the FTW forums.

Maybe a mod type thing, you people with extra powers.

Sorry to revive but what is a smileys notice?

:-D  :-D  :-D  :-D :-D  :-D  :-D  :-D  :-D  If you post more than 5 smileys, (like I just did) apparently it used to give you a notice saying you have too many smileys. It is gone with new AOPS. Now it just does not render the extras.

If you put a line between it works:

:D :D :D :D :D

:D :D :D :D :D

Therefore, the rule has been removed.

[hide]:D :D :D :D :D

:D :D :D :D :D

:D :D :D :D :D

:D :D :D :D :D

:D :D :D :D :D

:D :D :D :D :D
:D :D :D :D :D

:D :D :D :D :D[/hide][/quote]
So maybe they just removed ti completley.

this is turning into spam after a 8-year revive. [s]Mods, please lock[/s] Reporting.


Let $ n\ge 5$ be a positive integer. Prove that the set $ \{1,2,\ldots,n\}$ can be partitioned into two non-zero subsets $ S_n$ and $ P_n$ such that the sum of elements in $ S_n$ is equal to the product of elements in $ P_n$.

Perhaps you mean that $ \{1,2, \ldots, n\}$ [b][color=red]can [/color][/b]be partitioned into two such non-empty subsets $ P_n$ and $ S_n$? For example,

$ 1 \times 2 \times 4 \equal{} 3 \plus{} 5$ is a partitioning of $ \{1,2,3,4,5\}$

$ 1 \times 2 \times 6 \equal{} 3 \plus{} 4 \plus{} 5$ is a partitioning of $ \{1,2,3,4,5,6\}$

$ 1 \times 3 \times 6 \equal{} 2 \plus{} 4 \plus{} 5 \plus{} 7$ is a partitioning of $ \{1,2,3,4,5,6,7\}$

I can check if this holds for all $ n \geq 5$ but I'd rather first make sure this is actually the problem.

EDIT: Yes, this works for all $ n \geq 5$. For $ n\equal{}2k$ take $ P_{2k} \equal{} \{1, k\minus{}1, 2k\}$ and for $ n\equal{}2k\plus{}1$ take $ P_{2k\plus{}1} \equal{} \{1, k, 2k\}$.

http://www.mathlinks.ro/viewtopic.php?p=131398
I just wrote it here.
I don't know are there mistakes in the problem.

Well, so now you do know  :roll: (and that link doesn't work, but no matter).

Sorry for the wrong link:
http://www.mathlinks.ro/viewtopic.php?t=212129


logarithms
integration
calculus
derivative
function
real analysis
calculus computations

More general $ g'(x) \equal{} g(x)h(x)$, with  $ h(x)$ continuous, $ g(0)\neq 0$ . Then $ \forall x, g(x)\neq 0$ .
Let's supose that there is   $ z$ such  $ g(z) \equal{} 0$ , so there is $ x_0$ such that $ g(x_0) \equal{} 0$   and $ \forall _{|x| < |x_0|},g(x)\neq 0$
For all $ \ |x| < |x_0|$ we  can write $ \frac {g'(x)}{g(x)} \equal{} h(x)$ so $ \ln |g(x)| \minus{} \ln |g(0)| \equal{} \int_0^x h(t)dt$  , and for $ x\rightarrow x_0$ we obtain $ \int_0^{x_0} h(t)dt \equal{} \minus{} \infty$ (manifestly false !)
Finally $ \forall x, g(x)\neq 0$ ... :cool:

And here is a proof that if $ g(c) \equal{} 0$ for some $ c > 0$ and $ g(0) \neq 0$, then there is some $ x_0 > 0$ such that $ g(x) \neq 0$ for all $ 0 < x < x_0$.

Let $ S \equal{} \{c | g(c) \equal{} 0 , c > 0\}$. Let $ x_0 \equal{} inf(S)$.

Now there is a sequence $ \{c_n, c_n \in S\}$ such that $ g(c_n) \equal{} 0$ and $ c_n \to x_0$. Since $ g$ is continuous, we see that $ g(x_0) \equal{} 0$, thus $ inf(S) \in S$, as $ x_0 \neq 0$.

Another proof that probably works(I might be handwaving) for the original problem is that:

if $ g(c) \equal{} 0$ then for all $ n \ge 1$, $ g^n(c) \equal{} 0$ (the nth derivative)

(Note that $ g(x)$ is infinitely differentiable).

Using Taylor's formula then gives us the result.

[quote="Aryabhatta"] ....

Another proof that probably works(I might be handwaving) for the original problem is that:

if $ g(c) \equal{} 0$ then for all $ n \ge 1$, $ g^n(c) \equal{} 0$ (the nth derivative)

(Note that $ g(x)$ is infinitely differentiable).

Using Taylor's formula then gives us the result.[/quote]

Warning , you can't use simply Taylor  , because you dont know if $ g$ is analytical   :D

(remember  the non-analytic smooth function : $ f(x) \equal{} e^{\minus{}1/(x\minus{}c)^2}$ if $ x \neq c$ and $ f(c) \equal{} 0$,  $ f^{(n)}(c)\equal{}0$ ).
 :cool:

Yes. I did say I was handwaving  :)

btw, thanks for pointing that out, I had forgotten about that.

Diogene thx for your proof though is exactly the same with the one i already know. Hoping to see a different way

[quote="mathxl"]If $ f: R \to R$,$ f\left( 0 \right) \equal{} 1$ and $ f'\left( x \right) \equal{} \left( {2 \minus{} f\left( x \right)} \right){e^{f\left( x \right)}}$,$ \forall x \in R$

Prove that $ f\left( x \right) \ne 2,\forall x \in R$[/quote]

Something wrong ???  :maybe:

For $ x\equal{}0$ , we have : $ f^{'}(0)\equal{}(2\minus{}1)e^1\equal{}e ,\forall x \in R$ (*)

If $ f(x)\equal{}2 , \forall x \in R$ , we have : $ f^{'}(x)\equal{}0 , \forall x \in R$

It's unreasonable with (*) .

P/S : am i right ???  :maybe:

[quote="Materazzi"]If $ f(x) \equal{} 2 , \forall x \in R$ , we have : $ f^{'}(x) \equal{} 0 , \forall x \in R$
[/quote]

The negation of $ f(x) \neq 2, \forall x \in R$ isn't what you have.

It is: there is some $ x \in R$ such that $ f(x) \equal{} 2$.


floor function
combinatorics proposed
combinatorics

Prove that the sum:

\[ S_n=\binom{n}{1}+\binom{n}{3}\cdot 2005+\binom{n}{5}\cdot 2005^2+...=\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\binom{n}{2k+1}\cdot 2005^k \]

is divisible by $2^{n-1}$ for any positive integer $n$.

I'm not certain about it, but i think that, if $A_n$ is the expression, $A_{n+2} = 2(A_{n+1} + 1002A_{n})$.

so, the problem follows by induction.  :)

(is easy to see that $A_n = \frac{1}{2\sqrt{2005}}((1+ \sqrt{2005})^n - (1 - \sqrt{2005})^n)$)

It's already quite old I know... but I want to post a sketch of my solution

consider $ \frac{(\sqrt{2005}\plus{}1)^n}{\sqrt{2005}}\equal{}a_n\plus{}b_n*\sqrt{2005}$
so $ S_n \equal{} a_n$
$ \frac{(\sqrt{2005}\plus{}1)^n(\sqrt{2005}\plus{}1)}{\sqrt{2005}}\equal{}(a_n\plus{}b_n*\sqrt{2005})(\sqrt{2005}\plus{}1)$
so we get $ a_{n\plus{}1}\equal{}2005b_n\plus{}a_n$ and $ b_{n\plus{}1}\equal{}a_n\plus{}b_n$
now use indeuction

How do you reach to conclusion of that recurrence relations. Pls elaborate.
I am unable to get that thing.
Thanks in advance.


A zealous geologist is sponsoring a contest in which entrants have to guess the age of a shiny rock. He offers these clues: the age of the rock is formed from the six digits 2, 2, 2, 3, 7, and 9, and the rock's age begins with an odd digit.

How many possibilities are there for the rock's age?

Choose first digit: 3, 7, or 9, so 3 options.

The other 5 digits can be rearranged in 5*4 ways, so the answer is 3*5*4=[b]60[/b].

Shouldn't it be 360, because it was never specified that all digits should be used, or am I misterpreting the question?


geometry
geometric transformation
rotation
calculus
derivative
complex analysis

What angle of rotation is produced by the transformation $w= \frac{1}{z}$ at the point

1. $z=1$

2. $z=i$

I got the answers of part 1 is $\pi$ and part 2 is $0$, need some explaination... Thx

Just look at the derivative of $\frac 1z$, which is $-\frac 1{z^2}$. At $1$ it is $-1$ and at $i$ it is $1$, hence the answers (in general, the angle of rotation is just the argument of the derivative).


$ Let A_{k}$ [b] be a real symmetric nxn matrices that are positive and [/b] $ x_{k}> 0, (x_{k} \in R ) , \forall k \in {1,2,..,n}$ .[b] Prove that [/b] :
        $ det(\sum_{k\equal{}1}^{n}{x_{k}.A_{k}}) \geq  \sum_{k\equal{}1}^{n}{x^{n}_{k}.det(A_{k})}$ .
[b] "...  If I feel unhappy , I do mathematics to become happy , If I feel happy , I do mathematics to keep happy .''
                                                                                (Alfred Renyi)[/b]
 :P  :P  :P

Remark that if A, B are two symmetric matrices, where A is positive. Then there exists a matrice T such that

C=T.A.T^(-1)  and D=T.B.T^(-1) are diagonal matrices

If both A and B are positive, the elements on C,D 's diagonals are all positive

det(A+B)=det(C+D) > det(C) + det(D) = det(A) + det(B)

We have

det(x1.A1+...+xn.An) > det(x1.A1)+...+det(xn.An)=x1^n.det(A1)+...+ xn^n.det(An)


$ \left\{ \begin{array}{l}
 x \equal{} y^2  \plus{} 1 \\
 y \equal{} z^2  \plus{} 1 \\
 z \equal{} x^2  \plus{} 1 \\
 \end{array} \right.$

I don't think this has real roots...

By using AM-GM over and over,
$ x\equal{}y^2\plus{}1\ge 2y \equal{} 2z^2\plus{}2 \ge 4z \equal{} 4z^2\plus{}4 \ge 8z$ and $ x\equal{}y^2\plus{}1\ge 1$ which is a contradiction.

Define $ f(x)\equal{}x^2\plus{}1$. By substitution, we find $ x\equal{}f^3(x)$
But $ f(x)\minus{}x\equal{}x^2\minus{}x\plus{}1\equal{}(x\minus{}\frac{1}{2})^2\plus{}\frac{3}{4}>0$
Therefore, $ f^3(x)>f^2(x)>f(x)>x$. No real solution.

uh...How can we solve for the complex solutions? I only find a trivial one: let $ f(x)\equal{}x$, then $ x\equal{}y\equal{}z$ turn to be the one of the primitive cubic roots of $ \minus{}1$.

Have $ x,y,z > 0$
Put $ f(x) \equal{} x^2 \plus{} 1$

$ x \ge y \ge z \Rightarrow f(x) \ge f(y) \ge f(z) \Leftrightarrow z \ge x \ge y$
$ \Rightarrow x \equal{} y \equal{} z$

[quote="timwu"]Define $ f(x) \equal{} x^2 \plus{} 1$. By substitution, we find $ x \equal{} f^3(x)$
But $ f(x) \minus{} x \equal{} x^2 \minus{} x \plus{} 1 \equal{} (x \minus{} \frac {1}{2})^2 \plus{} \frac {3}{4} > 0$
Therefore, $ f^3(x) > f^2(x) > f(x) > x$. No real solution.

uh...How can we solve for the complex solutions? I only find a trivial one: let $ f(x) \equal{} x$, then $ x \equal{} y \equal{} z$ turn to be the one of the primitive cubic roots of $ \minus{} 1$.[/quote]

The general solutions are given by $ (f^2(z),f(z),z)$ where $ f^3(z)\equal{}z$. This is a polynomial of degree 8, which can be reduced to one of degree 6 by dividing by $ x^2\minus{}x\plus{}1$. After that, one has to find some solution... there can, however, only be a maximum of 8 (nonordered)  triples satisfying the equation.


inequalities
complex numbers
triangle inequality
algebra unsolved
algebra

let,
$ z_{i}$ be complex numbers with, $ |z_{1}|\equal{}\cdots\equal{}|z_{n}|\equal{}1$
$ z\equal{}(\sum_{k\equal{}1}^{n}z_{k}).(\sum_{k\equal{}1}^{n}\frac{1}{z_{k}})$

Prove that, $ z$ is a real number and $ 0\leq z\leq n^{2}$

[hide]
\[ |z_{k}|\equal{}1\Rightarrow\frac{1}{z_{k}}\equal{}\overline{z_{k}}\]
\[ \sum_{k\equal{}1}^{n}\frac{1}{z_{k}}\equal{}\sum_{k\equal{}1}^{n}\overline{z_{k}}\equal{}\overline{\sum_{k\equal{}1}^{n}z_{k}}\]
\[ z\equal{}\left|\sum_{k\equal{}1}^{n}z_{k}\right|^{2}\]
This shows that $ z$ is a nonnegative real number. The fact that $ z\leq n^{2}$ follows from the triangle inequality
\[ \left|\sum_{k\equal{}1}^{n}z_{k}\right|\leq\sum_{k\equal{}1}^{n}|z_{k}|\equal{}n\]
[/hide]

generaly fi $ |z_{1}| \equal{} |z_{2}| \equal{} .... \equal{} |z_{n}|$
we can suppose for $ k\in\{1,2,..,n\}$ that $ z_{k}\equal{} re^{ia_{k}}$ for $ a_{k},r\in\mathbb{R}$.
$ z \equal{}\sum_{0\le i,j\le n}\frac{r}{r}e^{i(a_{i}\minus{}a_{j})}\equal{}\sum_{0\le i,j\le n}\frac{r}{r}e^{i(a_{j}\minus{}a_{i})}\equal{}\overline{z}$ so $ z\in\mathbb{R}$

Very nice solution,both!
What I can remember about the problem is that, It is Romanian MO 1987 problem.
(anyway I thought that the problem may be harder...But it is not so much hard)

$ \ z_{i}\equal{} cis a_{i}$
$ \ 1/z_{i}\equal{}cis (\minus{}a)_{i}$
$ \ 1/z_{i}* z_{i}\equal{} 1$
$ \ cis a_{i}* cis (\minus{}a)_{j}\plus{}cis a_{j}* cis (\minus{}a)_{i}\equal{} 2*cos(a_{i}\minus{}a_{j})\to $
$ \ z\equal{}(\sum_{i\equal{}1}^{n}cis a_{i}).(\sum_{i\equal{}1}^{n}cis (\minus{}a)_{i}) \equal{} n\plus{}2*\sum_{i>j}cos(a_{i}\minus{}a_{j})\leq  n\plus{}2*1*n*(n\minus{}1)/2\equal{}n^{2}$
Can anyone prove that $ \ z$ is not negative using this approach?


Just seeing if my scores are good enough

I have no idea about the average, but the [i]best[/i]? Well, I haven't been to nationals yet (as of this May!) but I would guess like a 27/7 or 27/8. Maybe even higher in the sprint round. But don't quote me.

[quote="Magixmaniak"][quote="tarquin"]I think the best national scores are perfect... because I remember last year a Texas kid named Chen got 2nd and I think he got a 45.[/quote]

No, last year the best score was a 44(28/8) by Gregory Gauthier (is that how you spell his name?) from Illinois. Jeffrey Chen tied for second and won the tiebreaker (because he had a higher sprint score) with a 40(28/6).[/quote]

Phew. That makes my 41 sound better.

I'd never taken last years. I wish I had :(

are we talking national or state scores? because there's a difference

Mr.Ocax

[quote="Magixmaniak"]What year did you take??
Because in my opinion, last year's test was really hard compared to the previous ones.[/quote]
2003

[quote="Phelpedo"]Jeffrey Chen got 28/6? Target problems are supposed to be freebees.... Yeah, I think 2004 set was much harder than usual. Take 2003 for example. Greg Gauthier also got a 44, but he was sixth place.

Biffandoc, are you saying that you got a 41 in 2003? What did you get in 2004?[/quote]

Yea, thats what Jeffrey got. Last year's target was really hard (if you've ever taken it).

I've been through last year's sprint round, and phew! I hope this year is at least a [i]little[/i] bit easier, to say the least. :|

I agree. Not many people could finish last year's sprint test at Nationals except for the top students.

[quote="Phelpedo"]Jeffrey Chen got 28/6? Target problems are supposed to be freebees.... Yeah, I think 2004 set was much harder than usual. Take 2003 for example. Greg Gauthier also got a 44, but he was sixth place.

Biffandoc, are you saying that you got a 41 in 2003? What did you get in 2004?[/quote]I didnt go to nationals 2003 or 2004 just took the test for 2003 this year. Son I hope to take the 2004 test (if I can get my hands on it)

[quote="jb05"]I don't know who Kevin Chen is, but it appears that all of you are like that. I know I am. I don't know about the apple sauce though; maybe granny smith apple sauce. And what's wrong with walking into the room knowing everything???[/quote]

Actually, I think what's wrong is walking into the room [b]thinking [/b]you know everything.

In my opinion, I would say that the best scores are either perfect or almost perfect. However, I haven't been to nationals yet. I'm going in May because I came in 1st in Nevada. :)

When was the last time anyone got a perfect?

[quote="bob123"][quote="probability1.01"][quote="nat mc"] I bet someone will get perfect at nationals[/quote]

Not necessarily.  Several points here:

1. Top scores at nats are usually about 30% luck.
2. Top ten or so nats scores vary from 36+ to 40+.
3. It is very hard to get a perfect score consistently on national competitions, and it is quite possible that no one gets a perfect.

The key to doing well (at least for me) is to practice a lot under time pressure, and learn to be fast and accurate, since that will be the major factor.  Well, sadly for me, I was not very fast or accurate in general :P.[/quote]

I saw you in the countdown round last year (on TV)... I thought you did pretty well.[/quote]

Who is robability 1.01

Did Gauthier get a perfect? I know Hesterberg didn't (45), and I don't have much info from before that.

Yeah and noone else got above 40. Probability 1.01 is Alex Zhai.

Ahhh!!! How did you know? :P

Well I was pretty bad at countdown, actually.  And it wasn't a fluke--I normally am anyway.


I think Bergstrom/ engel etc helps here :wink:

Here is my solution .
By Cauchy-Swartz in Engel form we have that
$LHS\geq \frac{(x+y+z)^{2}}{x\sqrt{y}+y\sqrt{z}+z\sqrt{x}}$

so it remains to prove that
$\sqrt{8(x+y+z)^{3}+27xyz}\geq 3\sqrt{3}(x\sqrt{y}+y\sqrt{z}+z\sqrt{x})$

Now put x=a^2 etc and by squaring we have to prove that
$8(a^{6}+b^{6}+c^{6})+24(a^{2}b^{4}+b^{2}c^{4}+c^{2}a^{4})+75a^{2}b^{2}c^{2}\geq 3(a^{4}b^{2}+b^{4}c^{2}+c^{4}a^{2})+54abc(ab^{2}+bc^{2}+ca^{2})$

Now to prove this use AM-Gm or rearrengement with AM-GM to QED :wink:

Actually the following also holds:
\[\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{z}}+\frac{z}{\sqrt{x}}\geq \sqrt{\frac{(x+y+z)^{3}}{xy+yz+xz}}\geq \frac{3\sqrt{3}(x+y+z)^{2}}{\sqrt{8(x+y+z)^{3}+27xyz}}\]

In this form I found it easier (I don't know why :) )

By Cauchy-Swartz in Engel-form we have that
$LHS\geq \frac{(x+y+z)^{2}}{x\sqrt{y}+y\sqrt{z}+z\sqrt{x}}$ so we have to prove that

$\frac{(x+y+z)^{2}}{x\sqrt{y}+y\sqrt{z}+z\sqrt{x}}\geq\sqrt{\frac{(x+y+z)^{3}}{xy+yz+zx}}$

or $(x+y+z)(xy+yz+zx)\geq (x\sqrt{y}+y\sqrt{z}+z\sqrt{x})^{2}$

which is true by Cauchy -Swartz


algebra
polynomial
absolute value
number theory proposed
number theory

[quote="ThangTongHopK41"]Let $ n$ be a natural number and $ f_{1}, f_{2},..., f_{n}$ be polynomials with integer coefficients. Show that there exists a polynomial $ g(x)$ which can be factored (with at least two terms of degree at least 1) over the integers such that $ f_{i}(x) \plus{} g(x)$ cannot be factored ( which at least 2 terms of degree at least 1) over the integers for every $ i$[/quote]
The first we have an well know lemma (Eisentein) :
[i]Let $ p(x)\equal{}a_nx^n\plus{}...\plus{}a_0$ is an integer polynomial coefficient and
i)$ p\not|a_n$
ii) $ p|a_j,\forall j\equal{}0,...,n\minus{}1$
iii) $ p^2 \not | a_0$
Then P(x) is irreducible on $ \mathbb{Z[x]}$
 [/i]

Now consider problem . Let $ m$ is the greatest degree of polynomials $ f_1,..,f_n$ . We can rewrite all $ f_j(x)$ in form :
$ f_k(x)\equal{}\sum_{i\equal{}0}^m a_{i,k}x^i,\forall k\equal{}1,...,m$ (although degree of some polynomials are smaller than $ m$ )
Let M is the greatest sum of absolute value of each coefficient with the same exponent . (i.e the sum of  absolute  of $ a_{j,i},\forall j\equal{}1,..,n$
Consider the sequence of prime $ M<p_1<p_2<...<p_n$ and
Now for each $ p_j$ consider the system of congruence :
$ b_0\equiv p_j\minus{}a_{j,0} (\mod p_j^2),\forall j\equal{}1,...,n$
$ b_i\equiv \minus{}a_{j,i} (mod p_j),\forall i\equal{}1,..,m ,j\equal{}1,...,n$
From chinese remainder theorem each of system has a solution on $ Z[x]$ mod p.
Call root of each system equation are $ (b_0,b_1,...,b_m)$
We have for each $ p_j$
$ \sum_{i\equal{}0}^m b_i\equiv \minus{}\sum_{i\equal{}0}^m a_{j,i} (\mod p_j)$
Because $ p_j>M$ so  $ \sum_{i\equal{}0}^m b_i$ can't divide by any $ p_j,\forall j\equal{}1,..,n$
We chose $ a\equal{}\minus{}(b_0\plus{}...\plus{}b_m)$ and consider polynomial
$ g(x)\equal{}ax^{m\plus{}1}\plus{}b_mx^m\plus{}...\plus{}b_0$
This polynomial has at least solution x=1 so it reducible . Apply the Eisentein's lemma we also have $ g(x)\plus{}f_i(x)$ are irreducible for all over i ,proof completes .

A Good solution
Another problem:
Let $ q$ be a positive integer
Prove that there exists integer x and prime p such that
$ x^{q}\minus{}1 \vdots p$ and $ x^{q\minus{}1}\minus{}x\plus{}1 \vdots p$

[quote="ThangTongHopK41"]A Good solution
Another problem:
Let $ q$ be a positive integer
Prove that there exists integer x and prime p such that
$ x^{q} \minus{} 1 \vdots p$ and $ x^{q \minus{} 1} \minus{} x \plus{} 1 \vdots p$[/quote]
It seems not true  :)  . For example $ q\equal{}3$ .If exist p and x such that $ p|x^3\minus{}1$ and $ p|x^2\minus{}x\plus{}1$
Then we have $ p|x^3\plus{}1$. Therefore $ p|2$ ,but it gives contradiction because $ x^2\minus{}x\plus{}1$ is always an odd number .

Sorry i miss the condition $ q \geq 7$, now we can solve the problem


ratio
trigonometry
geometry
Pythagorean Theorem

[hide]$AB=12$
$sin(30)=\frac{1}{2}$ We know angle B is 30 degrees. So, $\frac{AC}{AB}=\frac{1}{2}$. AB is twice AC or 12. You can simply use the Pythagorean Theorem to get the third side.[/hide]

[quote="krnboy32"]Suppose the triangle ABC has C = 90degrees, A = 60degrees, and AC = 6. Find AB and BC[/quote]

[hide]It's a 30-60-90 degree triangle, where the hypotenuse is AB, the big side is BC, and the small side is AC=6.

Since the sides are in a ratio $1: \sqrt{3}: 2$ (by the Pythagorean Theorem), the big side (BC) is $6\sqrt{3}$ and the hypotenuse is $6\cdot2=12$.[/hide]

i still dont know trigonometry but I know kinda the ratio of sides and simple ones lol im taking geometry :(


conics
ellipse
geometry
geometric transformation
reflection
hyperbola
parabola

i thought about this when i saw an isl problem (2000) i think, which said
"prove that there exist points $D,E,F$ on sides $BC,CA,AB$ of triangle $\triangle ABC$ such that $AD,BE,CF$ are concurrent and $OD+HD=OE+HE=OF+HF$".
well, there exists an ellipse with foci $O$ and $H$ that is tangent to $BC,CA,AB$. let the tangency points be $D,E,F$, the hexagon (degenerate) formed by $A,B,C,D,E,F$ has a circle inscribed in it, so by brianchon theorem $AD,BE,CF$ are concurrent.

this was more the "introduction". $O$ and $H$ are isogonal conjugates...my question is : is the question still good for $P$ and $Q$ isogonal conjugates ? that is, does there exists an ellipse with foci $P$ and $Q$, that is tangent to the sides of the triangle ?
(another interesting case is $I$, which is it`s own isogonal conjugate, and the ellipse is a circle in that case - the inscribed circle).

the statement is true for every two isogonal conjugate points $P$ and $Q$.

[quote="maky"]my question is: is the question still good for $P$ and $Q$ isogonal conjugates? that is, does there exists an ellipse with foci $P$ and $Q,$ that is tangent to the sides of the triangle ?[/quote]
According to the location of $P$ in $\triangle ABC,$ the "shape" of the inconic varies.

We assume that point $P$ lies inside $ \triangle ABC,$ so is its isogonal conjugate $Q.$ The remaining case is treated analogously. Let $ P_1,P_2,P_3$ and $ Q_1,Q_2,Q_3$  be the perpendicular feet of $ P,Q$ on sidelines $BC,CA,AB.$ It's well-known that $ P_1,P_2,P_3 ,Q_1,Q_2,Q_3$  lie on a circle centered at the midpoint $K$ of $PQ,$  thus let $ \varrho$ be its radius. Let $ Q_0$ be the reflection of $ Q$ across $ BC$ and $ A_0 \equiv PQ_0 \cap BC.$ Then $ BC$ is external bisector of $ \angle PA_0Q$ and since $ K,Q_1$ are midpoints of $ PQ,QQ_0$ we get $ A_0P \plus{} A_0Q \equal{} A_0P \plus{} A_0Q \equal{} PQ_0 \equal{} 2\varrho.$ Similarly, there exists $ B_0,C_0$ on $ CA,AB$ such that $ CA,AB$ are external bisectors of $ \angle PB_0Q , \angle PC_0Q,$ respectively and whose sum of distances to $ P,Q$ equals $ 2\varrho$ $ \Longrightarrow$ there exists an inellipse $ \mathcal{K}$ with foci $ P,Q$ and pedal circle $ (K,\varrho).$ When $ P$ is outside $ \triangle ABC,$ $\mathcal{K}$ becomes hyperbola, or parabola if $P$ lies on $\odot(ABC),$ since $Q$ is at infinity. Conversely, foci of a conic $ \mathcal{K}$ are isogonal conjugates with respect to any triangle bounded by three tangent lines of $ \mathcal{K}.$


Look, I've been trying to do this question for over 5 days; I'm almost ready to give up. I'd appreciate any input you can give me. Here it goes:

Given: (P^Q)=>R
Prove:  (P=>R)v(Q=>R)

Thanks!!

What exactly do you mean, natural deduction? Do you not use truth tables? If so---

(P^Q)=> R
~R => ~(P^Q)  - contrapositive
~R => (~Pv~Q) (DeMorgans law)
(~R=>~P) v (~R => ~Q) (I guess here'd be the 'natural deduction' step; it's pretty simple distributive)
(P=> R) v (Q => R) (contrapositive)

Nice solution previous.

Using truth tables would be quite bulky on this one. -bleah- Using natural deduction is easier.

is ~PvQvR logically equivalent to ~PvQvRvR?

Your question is ambiguous due to lack of parentheses/spacing.  (R v R) <=> R is a true statement, if that answers your question.


[quote="cellular_mem"]A box contains over 100 marbles.  The marbles can be divided into equal shares among 6, 7 or 8 children with 1 marble left over each time.  What is the least number of marbles that the box can contain?

-MOEMS[/quote]
[hide]$x\equiv 1\mod 6$
$x\equiv 1\mod 7$
$x\equiv 1\mod 8$
$x-1\equiv 0\mod 6$
$x-1\equiv 0\mod 7$
$x-1\equiv 0\mod 8$
$lcm(6,7,8)=168$
$x=\boxed{169}$[/hide]

[hide]Since we want the least amount, we want the LCM of 6,7 and 8. We find that it is 168. then, we simply add 1. 169 is our answer.[/hide]

[hide="solution"]
Notice that having 168 marbles will create no remaining marbles for each of the divisions. (See somebody else's solution for why.) If you add one marble, there will be one marble left over for each of the divisions. This means that the answer is [b]169[/b].
[/hide]

The answer is obviously the Ignite igniting into a +1.

Okay; so am I allowed to answer this? if yes; then continue; if not then just ignore this-

[hide]
So; if there was over 100 marbles in a box, and they could be devided between 6, 7 , or 8 children with one left over each time; then we need to find the LCM of 6, 7, and 8. I don't know what you got-but I get 168. Now then; because we need one left over; we add 1.

  168
[u]+   1[/u]
  169

[quote]169 marbles in a box![/quote]
[/hide]

[hide="solution"]We are asked to find a number greater than 100 and it leaves a remainder of 1 if it's divided by 6, 7 or 8. The lcm of 6 7 and 8 is 168. The answer is 169.[/hide]

[hide]So basically the problem is asking for the number one higher than the LCM of 6, 7, and 8. We find that number is [b]169[/b].[/hide]

[quote="cellular_mem"]A box contains over 100 marbles.  The marbles can be divided into equal shares among 6, 7 or 8 children with 1 marble left over each time.  What is the least number of marbles that the box can contain?

-MOEMS[/quote]

[hide=" :) "]All we have to do is find the LCM of 6 7 and 8 and add 1. This is 168+1=169 [/hide]

If the lcm of the numbers is greater than one hundred why do we need to know that it has over one hundred marbles.

[quote="someperson01"]The LCM of 6,7,8 is 168. 168+1=169, the answer.[/quote]

actually it is 84, but that is under 100 so doesn't count  :wink:

[quote="bpms"][quote="SplashD"]That's just a piece of useless info that if people read fast, they might think the number of marbles is under 100[/quote]
Actually I think I figured it out. It prevents the answer from being one marble.  :D[/quote]
it also stops it from being 85

[quote="daermon"][quote="someperson01"]The LCM of 6,7,8 is 168. 168+1=169, the answer.[/quote]

actually it is 84, but that is under 100 so doesn't count  :wink:[/quote]
No, $[6,7,8]=3\cdot7\cdot8=168$.

[quote="daermon"][quote="someperson01"]The LCM of 6,7,8 is 168. 168+1=169, the answer.[/quote][/quote]

actually it is 84, but that is under 100 so doesn't count  :wink:
84/8 =10.5; which isnt an integer


vector
algebra
polynomial
calculus
integration
product of roots
geometry unsolved

There exists a plane f(a).x+f(0).y+f(c).z=1 which is passing through the point (0,1,1) where y=f(x) is a second degree polynomial which cuts the line x+y=1 at 2 distinct points x=1,x=c where c=rational number.It is also given that there exists a positive integral value of a for which the product of roots of f(x)=0 becomes -1

Q1 Find value of f(a)+f(0)+f(c)
Q2 Find hte equation of plane parallel to given plane for negative value of c and passing through (1,1,-1)
Q3 Find hte eqn of line on 3D space such that line os perpendicular to a plane for c>1 and passing through (1,1,1)
Q4 If there exists two such planes ,then find hte nagle between them.


modular arithmetic
number theory
greatest common divisor

61.  What is the greatest integer that divides $ p^4 \minus{} 1$ for every prime number $ p$ greater than $ 5$?

(A) 12
(B) 30
(C) 48
(D) 120
(E) 240

I know what the answer is, and I guessed it correctly (by writing out the prime factorizations of $ p^4\minus{}1$ for 7, 11, and 13, and looking at the common factor.  However, I would like to get a more definitive answer.  Thanks.

$ p^4\minus{}1\equal{}(p^2\minus{}1)(p^2\plus{}1)\equal{}(p\minus{}1)(p\plus{}1)(p^2\plus{}1)$
Now clearer?

That's what I already worked out.  It doesn't make it clear why it has to be one of those though.

All primes, except $ 2$ and $ 3$, can be expressed as $ \pm 1 \pmod {6}$.

[hide]It suffices to find the highest power of $ 2,3,5$ that divides it. No other prime factor $ q$ appears in all numbers of the form $ p^4 \minus{} 1$ because it won't appear in $ q^4 \minus{} 1$.

For $ 2$: Using the factorization provided above $ (p\minus{}1)(p\plus{}1)(p^2\plus{}1)$. $ p$ is odd, so all of these are divisible by $ 2$. One of $ p\minus{}1$ and $ p\plus{}1$ is additionally divisible by $ 4$ (two consecutive even numbers). So there's at least a factor of $ 16$.

For $ 3$: $ p \neq 0$, so by Fermat's Little Theorem, $ p^4 \equiv p^2 \equiv 1 \pmod{3}$. So $ p^4 \minus{} 1$ is divisible by $ 3$.

For $ 5$: $ p \neq 0$, so by Fermat's Little Theorem, $ p^4 \equiv 1 \pmod{5}$. So $ p^4 \minus{} 1$ is divisible by $ 5$.

Ordinarily you'd have to verify that the next power doesn't always divide the result (not hard, construct counterexamples in mods), but this is a multiple choice, so we know the answer is $ 240$ now.[/hide]

[hide]$ \blacktriangleright \; p$ is odd, so $ p \plus{} 1$, $ p \minus{} 1$, and $ p^2 \plus{} 1$ must all be even, contributing three factors of $ 2$.

$ \blacktriangleright$ One of $ p \minus{} 1$ and $ p \plus{} 1$ must be divisible by $ 4$, contributing a fourth factor of $ 2$

$ \blacktriangleright$ For any three consecutive numbers, one must be divisible by $ 3$. Because $ p$ is not divisible by $ 3$, either $ p \minus{} 1$ or $ p \plus{} 1$ must be divisible by $ 3$

$ \blacktriangleright \; p \equiv \pm 1, \pm 2 \pmod{5}$. If $ p \equiv \pm 1$, then $ p \mp 1 \equiv 0 \pmod{5}$, and is thus divisible by $ 5$. If $ p \equiv \pm 2$, then $ p^2 \equiv 4 \equiv \minus{} 1 \implies p^2 \plus{} 1 \equiv 0$.

So, we've guaranteed that $ p^4 \minus{} 1$ is divisible by $ 2^4 \cdot 3 \cdot 5 \equal{} 240$. From here, we can do one of two things:

$ [1] \blacktriangleright$ Looking at our list of choices, the only multiple of $ 240$ is $ 240$ itself, so this is our answer.

$ [2] \blacktriangleright$ Without looking at the list of choices, we know the answer must be some multiple of $ 240$. Note that $ \text{GCD} [7^4 \minus{} 1 , \; 11^4 \minus{} 1 ] \equal{} 240$, so the answer must be $ 240$ itself.[/hide]

[b]Edit:[/b] See above, he beat me to it.


A quadratic equation $ ax^2\minus{}2ax\plus{}b\equal{}0$ has two real solutions. What is the average of the solutions?

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ \frac{b}{a} \qquad
\textbf{(D)}\ \frac{2b}{a} \qquad
\textbf{(E)}\ \sqrt{2b\minus{}a}$

[hide]the sum of the solutions is $ \frac{\minus{}b}{a}$ so if we apply this we get the sum is $ \frac{\minus{}2a}{\minus{}a} \equal{} 2$ and since the sum is $ 2$, the average of the 2 solutions is $ \frac{2}{2} \equal{} 1 \longrightarrow \fbox{\mbox{A}}$[/hide]

[hide="other way"]
The average of the solutions is the x-value of the vertex. You find this by using the formula $ \frac{\minus{}b}{2a}$. Plugging in $ 2a$ for b and $ a$ for a, you get $ \frac{2a}{2a} \equal{} 1$
[/hide]
Pretty similar to cognos's way, but combines everything into one step. Be sure to remember to divide by $ 2a$, not $ a$, or you'll miss it like I did  :blush:

not easiest way soltixnj lol... Use Vieta's like cognos.

[quote="soltixnj"]Easiest way to solve (how I did it)

Put as (rx + c)(sx + d) = 0.

-c/r =0,                         -d/s=0.
 -(cs+dr) /2rs is the average.

We have cs + dr = -2a.
Also, rs = a.
Thus the answer is just -(-2a) / 2a = 1.[/quote]
plugging in numbers also creates a greater chance of making a careless mistake, especially if theres a lot of arithmetic involved. if you do it cognos's way, it is a simple mental math problem (unless you need pencil and paper to do 2/2).

[hide=Solution]
We use Vieta's to get that the average of the roots is $\frac{-\frac{-2a}{a}}{2}=\boxed{1}$.
[/hide]


Find all finite sequences $(x_{0},x_{1},...,x_{n})$ such that for every $j$, with $0\leq j\leq n$, $x_{j}$ equals the number of times $j$ appears in the sequence.

$x_{0}+x_{1}+x_{2}+...+x_{n}=$ sum of the numbers of each number in the sequence $=$ the length of the sequence $=n+1$
$0x_{0}+1x_{1}+2x_{2}+...+nx_{n}$ equals to the sum of the numbers in the sequence because $ix_{i}$ is a sum of $i$ in the sequence.
Therefore, $0x_{0}+1x_{1}+2x_{2}+...+nx_{n}=x_{0}+x_{1}+x_{2}+...+x_{n}=n+1$.


function
number theory
relatively prime
number theory proposed

(x,y) is called relatively prime divisor pair of A, if x is relatively prime to y and x,y are both divisors of A. Show that number of relatively prime divisor pairs of A is d(A^2) where d(x): Number of positive divisors of x.

It is not true.
The number of relative prime divisor of n must be $ 2^{d(n)}$
Proof quite easy .

Suppose, for example take A=7.
relative prime divisor pairs=(1,7),(7,1) and (1,1)
So, d(A^2)=3 whereas 2^d(n)=4

If x,y are relatively prime and divisors of A, then LCM(A/x,A/y)=A
No. of relatively prime divisor pairs of A = No. of solutions of LCM(x,y)=A
Let this function be L(A)
If A is power of prime(=p^k), then L(A)=2k+1
L(A) is multiplicative
=> L(A) = (2k_1+1)*(2k_2+1)*.... = d(A^2)


function
calculus
integration
real analysis
real analysis unsolved

Find a sequence of functions $q_{n}$ and a function $q$ on $[0,1]$ such that $q_{n}(x)\to q(x)$ for all $x \in [0,1]$, and $q_{n}$ is integrable but $q$ is not integrable.

Since in your other question, you meant "Riemann integrable" when you said "integrable," I'll assume that that's what you mean here as well.

The classic example: Let $\{r_{k}\}$ be an enumeration of all of the rational numbers in $[0,1].$ Define $q_{n}(x)$ to take on the value $1$ on the points $\{r_{1},r_{2},\dots,r_{n}\}$ and the value $0$ everywhere else. Each $q_{n}$ is zero except at finitely many points, hence Riemann integrable with integral zero. The pointwise limit if this sequence of functions is the characteristic function of the rationals, which is not Riemann integrable.


induction
combinatorics unsolved
combinatorics

Let $ n,k$ be two positive integers satisfies $ n > 2^k$ and $ A \equal{} \{1,2,...,n\}$. Prove that there exists a subset $ S \subset A$ such that:

1. $ |S| \equal{} k \plus{} 2$.
2. For each $ m$ such that $ 1 \leq m \leq k \plus{} 2$, every subset of $ S$ with $ m$ elements has pairwise distinct sum of elements, i.e. if $ S_1, S_2 \subset S$ and $ |S_1| \equal{} |S_2| \equal{} m$ where $ 1 \leq m \leq k\plus{}2$ then
\[ \sum_{x \in S_1} x \neq \sum_{x \in S_2} x\]

I think that the set $ 2^0, 2^1, 2^2 \cdots 2^k, 3$ works.

[quote="Mashimaru"]Let $ n,k$ be two positive integers satisfies $ n > 2^k$ and $ A \equal{} \{1,2,...,n\}$. Prove that there exists a subset $ S \subset A$ such that:

1. $ |S| \equal{} k \plus{} 2$.
2. For each $ m$ such that $ 1 \leq m \leq k \plus{} 2$, every subset of $ S$ with $ m$ elements has pairwise distinct sum of elements, i.e. if $ S_1, S_2 \subset S$ and $ |S_1| \equal{} |S_2| \equal{} m$ where $ 1 \leq m \leq k \plus{} 2$ then
\[ \sum_{x \in S_1} x \neq \sum_{x \in S_2} x\]
[/quote]
I think that the set: $ S\equal{}\{1,4,7,8,2^3\plus{}1,2^4\plus{}1,....,2^k\plus{}1\}$ such that condition problem.

[quote="math10"][quote="Mashimaru"]Let $ n,k$ be two positive integers satisfies $ n > 2^k$ and $ A \equal{} \{1,2,...,n\}$. Prove that there exists a subset $ S \subset A$ such that:

1. $ |S| \equal{} k \plus{} 2$.
2. For each $ m$ such that $ 1 \leq m \leq k \plus{} 2$, every subset of $ S$ with $ m$ elements has pairwise distinct sum of elements, i.e. if $ S_1, S_2 \subset S$ and $ |S_1| \equal{} |S_2| \equal{} m$ where $ 1 \leq m \leq k \plus{} 2$ then
\[ \sum_{x \in S_1} x \neq \sum_{x \in S_2} x\]
[/quote]
I think that the set: $ S \equal{} \{1,4,7,8,2^3 \plus{} 1,2^4 \plus{} 1,....,2^k \plus{} 1\}$ such that condition problem.[/quote]

Why does that set satisfy the problem?

This is a fantastic solution I have ever known.

We will call a subset $ S$ of $ A$ satisfy the condition that for $ m \leq |S|$ and $ S_1, S_2 \subset S$ such that $ |S_1| \equal{} |S_2| \equal{} m$ then
\[ \sum_{x \in S_1} x \neq \sum_{x \in S_2} x\]
is a [i]good[/i] subset.

We will prove the problem for $ A \equal{} \{a_1, a_2, ..., a_n\} \subset \mathbb{N}^*$. With induction on $ k$, the base cases $ k \equal{} 1$ is trivially. Now suppose that for $ n \geq 2^k \plus{} 1$, there exists a good subset $ S \subset \{a_1, a_2, ..., a_n\}$ and $ |S| \equal{} k\plus{}2$. Consider $ k\plus{}1$. Let $ A' \equal{} \{b_1, b_2, ..., b_{n'}\}$ be the set of $ n' > 2^{k\plus{}1}$ elements. Base on the Principle of Pigeon's Hole, there exists $ B_1 \equal{} \{b_{i_1}, b_{i_2}, ..., b_{i_{2^{k}\plus{}1}} \} \subset A'$ such that $ b_{i_j}$ have the same parity. By the induction hypothesis, there exists a good subset $ S^*$ with $ k\plus{}2$ elements of $ B_1$. Now set $ B_2 \equal{} A' \minus{} B_1$. If $ B_2 \neq \emptyset$ then it is easy to see that $ S \equal{} S^* \cup \{x\}$ for $ x \in B_2$ is a good subset with $ k\plus{}3$ elements of $ A'$ (this fact can be proved using the parity and the property [i]good[/i] of $ S^*$). Thus in this case we are done. If $ B_2 \equal{} \emptyset$ then $ A'$ consists of just even numbers or just odd numbers. If even, let $ k$ is the minumum power of $ 2$ of every elements in $ A'$ then consider $ A'_1 \equal{} \dfrac{A'}{2^k}$, then apply the above for $ A'_1$, notice that whenever $ S$ is good then $ kS \plus{} r$ with $ k, r \in \mathbb{R}$ is also good. Finally, if $ A'$ only consists of odd numbers then consider $ A'_2 \equal{} A' \plus{} 1$ then $ A'_2$ only consists of even numbers and it has been shifted to the above situation. Thus the problem is solved.


Let's have
$A=(2^{n})\cdot(2^{n}+2^{m-1})+(2^{m})\cdot(2^{m}+2^{n-1})$ objects and we want to split them into $2^{n}+2^{m-1}$ groups containing $2^{n}$ objects and $2^{m}+2^{n-1}$ groups containing
$2^{m}$ objects. Then $\frac{B}{A}$ is the number of ways we can do it and thus a positive integer.


linear algebra
matrix
superior algebra
superior algebra unsolved

[quote="maff"]Show that the trace of a nilpotent matrix with entries in an arbitrary field(not necessarily algebraically closed) is zero without invoking a field extension.[/quote]

Even more general: http://www.mathlinks.ro/viewtopic.php?t=233169 .

But in the field case, you can just use $ 0 \equal{} \mathrm{Im}A^m\subseteq\mathrm{Im}A^{m \minus{} 1}\subseteq\mathrm{Im}A^{m \minus{} 2}\subseteq ...\subseteq\mathrm{Im}A^2\subseteq\mathrm{Im}A^1\subseteq\mathrm{Im}A^0 \equal{} V$ (where $ A$ is the nilpotent matrix, and $ m$ is an integer such that $ A^m\equal{}0$) to find a basis $ \left(e_1,e_2,...,e_n\right)$ of $ V$ which satisfies $ Ae_i\in\left < e_1,e_2,...,e_{i \minus{} 1}\right >$ for every $ i$. In this basis, the trace of $ A$ is obvious (the whole diagonal of $ A$ is zero in this basis).

  darij


You can read the first few pages of my note "The Neuberg-Mineur circle" ( downloadable from http://reflections.awesomemath.org/2007_3.html or http://www.cip.ifi.lmu.de/~grinberg/NeubergMineur.pdf ).

  Darij


ratio
inequalities
function
inequalities open

Here's what I was thinking: what if we write the sum as $\sum\frac {(x_ix_j)^2}{x_ix_j(1-x_i)(1-x_j)}\ge E=\frac {(\sum x_ix_j)^2}{\sum x_ix_j(1-x_i)(1-x_j)}$?.

I'm not sure right now, but I think that $E$ can be shown to be $\ge \frac n{2(n-1)}$ by considering the $\binom n2$ numbers $a_{ij}=x_ix_j$ as our variables. Intuitively that's Ok because at the numerator we have $\sum a_{ij}^2+2\sum a_{ij}a_{i'j'}$, and at the denominator we have $\sum a_{ij}^2+k\sum a_{ij}a_{i'j'}$, where $k>2$. What we have to do now is settle the details.

Could you please post those details? I don't find them so details...

Well, the fact that that ratio is written as $\frac {\sum a_{ij}^2+2\sum a_{ij}a_{i'j'}}{\sum a_{ij}^2+k\sum a_{ij}a_{i'j'}}$ with $k>2$ is obvious because the summands from the denominator are of the type $\displaystyle x_ix_j(\sum_{t=1}^nx_t-x_i)(\sum_{t=1}^nx_t-x_j)$ and then we expand $(\sum_{t=1}^nx_t-x_i)(\sum_{t=1}^nx_t-x_j)$. We find an $x_ix_j=a_{ij}$ (we're now working with a particular pair $(i,j)$) and some other $a_{i'j'}$s. This means that each $a_{ij}^2$ appears once, and since the sum is symmetric in $a_{ij}$, we find the denominator to be $\sum a_{ij}^2+k\sum a_{ij}a_{i'j'}$ and $k>2$ because there are more terms in the denominator than in the numerator.

So what we should actually prove is this (a change of notation):

$\frac {\sum x_i^2+2\sum x_ix_j}{\sum x_i^2+k\sum x_ix_j}\ge \frac {n^2}{n^2+(k-2)\binom n2}$ ($i,j$ are between $1,n$), and this is easily done by using the fact that $\sum x_i^2\ge \frac 2{n-1}\sum x_ix_j$.

Note that in this last paragraph I have changed the notation, so $n$ from this last paragraph is not the same as the $n$ from the initial problem. Here $n$ denotes the number of $x_i$s, which have replaced the $a_{ij}$s from the initial problem.

I hope it's clear (and correct :)).

Dear grobber, I apologise, but I still dont understand something. Im sure Im wrong and that is why I ask you to be a little bit more explicit. Here is what I understood: ok, take $\frac{n(n-1)}{2} $ variables $ a_{ij} $ corresponding to all products $ x_i\cdot x_j $ with $ i<j $. Let these new variables be called $ s_k$, with $ k=1,2...,\frac{n(n-1)}{2}=p $ . Of course, in the numerator we will have the square of the sum of these new variables. I agree. Now, you say that in the numerator we will have something like $ s_1^2+...+s_p^2+q( s_1s_2+s_1s_3+...+s_{p-1}{s_p}) $. Of course, in thise case q must be an integer. Now, let us count how many terms we have in the numerator and in the last expression. In the numerator we clearly have $ p\cdot (n-1)^2 $ terms. In the last expression we have $ p+q\cdot\frac{p(p-1)}{2} $. From the equation I find $ q=\frac{4n}{n+1} $ which is not an integer. Could you please explain to me where I am wrong? Im sure Im wrong, but again I dont see some obvious things. Thanks

Yes, I made make a mistake :(.

Here's what's going on (assume we work with ordered pairs, $i<j$): each product $a_{ij}a_{i'j'}$ where the pairs $(i,j),(i',j')$ have one common element appears $3$ times, and each such product for which $(i,j),(i',j')$ have no common elements appears $4$ times, so the expression in the denominator wasn't symmetric in my $a_{ij}$ (or your $s_i$) as I had initially rushed to assume :(.

Sorry I did this, but the question is: can this be repaired? In the denominator we get something of the form $\displaystyle \sum a_{ij}^2+3\sum a_{ij}a_{i'j'}+(S=\sum_{(i-i')(j-j')(i-j')(j-i')\ne 0} a_{ij}a_{i'j'})$. Could we find an inequality of the type $\sum a_{ij}^2\ge tS$ for some $t>0$? (we could easily find what $t$ should be, but I'm too lazy :))

What you say about this
$\displaystyle \sum \frac{x_ix_j}{(1-x_i)(1-x_j}\leq 1$ ,
where  $\displaystyle x_i$ are positive numbers such that $\displaystyle x_1+...+x_n=1$ ?
It is nice and I hope it isn't triviale.

This is so so nice and beautiful. Thanks, Vasc, for this masterpiece. Let us suppose that $ x_1\geq x_2\geq ...\geq x_n $. Then the left-hand side is at most $\frac{\sum_{i<j} x_ix_j}{(1-x_1)(1-x_2)}$. Now, I will prove the last quantity is at most 1. But ${\sum_{i<j} x_ix_j}=\frac{1-x_1^2-...-x_n^2}{2}\leq\frac{1-x_1^2-x_2^2}{2} $ and so we are left with $\frac{1-x_1^2-x_2^2}{(1-x_1)(1-x_2)}\leq 2$ which is the same as $ (x_1+x_2-1)^2\geq 0 $. Hope there arent mistakes, Im so excited now that I cant verify the solution. Even if its wrong, the problem remains superb. But what about my open question?

Beautiful solution !

Your open inequality is valid.
My solution uses the inequality
$F(x_1,x_2,...,x_n)\geq F(t,t,x_3,...,x_n)$
where $t=(x_1+x_2)/2$. We have to prove that
$\frac{2x_3}{1-x_3}+...+\frac{2x_n}{1-x_n}-\frac{1-2t}{1-t} \geq 0$.
It suffices to show that
$2(x_3+...+x_n)-\frac{1-2t}{1-t} \geq 0$,
that is
$2(1-2t)-\frac{1-2t}{1-t} \geq 0$,
or
$(1-2t)^2\geq0$.
I hope this solution is good.

Remark.
What you say about this very hard inequality:
$\sum\frac{x_ix_j}{(1-x_i)(1-x_j)}\leq\frac{n(n-1)}{2(2n-1)^2}$,
where $x_i$ are non-negative numbers such that $x_1+...+x_n=\frac{1}{2}$ ?

Super-nice solution, Vasc. I think Ive done your inequality, but Im not sure, since Im in a hurry now. The inequality is $ (\sum_{i=1}^n \frac{a_i}{2-a_i})^2\leq\sum_{i=1}^n \frac{a_i^2}{(2-a_i)^2}+\frac{n(n-1)}{(2n-1)^2} $ where $\sum_{i=1}^n a_i=1 $. Using Cauchy-Schwarz (how I like this inequality!) we find that the left-hand side is at most $\sum_{i=1}^n\frac{a_i}{(2-a_i)^2} $ and so we are left with $\sum_{i=1}^n\frac{a_i(1-a_i)}{(2-a_i)^2}\leq\frac{n(n-1}{(2n-1)^2} $. But the function $ f(x)=\frac{x(1-x)}{(2-x)^2} $ seems to be concave.

Harazi, you are the The Big-Master of the Special Cases !
Your last proofs are ingenious.
But that inequality has still other special cases.
	Case 1. $\sum\frac{x_ix_j}{(1-x_i)(1-x_j)}\leq\frac{1}{4}$,
where $x_i$ are non-negative numbers such that $x_1+...+x_n=\frac{2}{3}$ .
	Case 2. $\sum\frac{x_ix_j}{(1-x_i)(1-x_j)}\leq\frac{3}{16}$,
where $x_i$ are non-negative numbers such that $x_1+...+x_n=\frac{3}{5}$ .
	Case 3. $\sum\frac{x_ix_j}{(1-x_i)(1-x_j)}\leq\frac{n-1}{8(n-2)}$,
where $x_i$ are non-negative numbers such that $x_1+...+x_n=\frac{1}{2}+\frac{1}{2(2n-3)}$ .
I think these inequalities are more and more difficult.
Can someone prove one of them?


An equilateral triangle and a regular hexagon share a side. The perimeter of the triangle is 20 cm. How many centimeters are in the perimeter of the hexagon?

A triangle has half as many sides as a hexagon.  Since they are both regular and have the same side length, the perimeter of the triangle is half the perimeter of the hexagon.  Thus, the answer is $ 20 \times 2$ or $ \boxed{40}$.


inequalities
logarithms
limit
real analysis
real analysis unsolved

Let  f(x)=max{n: n!≤ x} and x>1     Show that  f(x)/(lnx/(lnlnx))→ 1 (as n→ ∞)

you have that for $n!\le x\le (n+1)!$ the limit respects the following inequalities:

$\frac{n \ln(\ln n!)} {\ln (n+1)!} \le \frac{f(x) \ln(\ln x)} {\ln x} \le \frac{(n+1) \ln(\ln (n+1)!)} {\ln n!}$

So, it's enough to show that the two limits expresed in terms of $n$ tend to $1$ as $n$ tends to infinity. By Cesaro-Stolz theorem we have that:

\[ L=\lim \frac{n \ln(\ln n!)} {\ln (n+1)!}= \lim \frac{(n+1) \ln(\ln(n+1)!)-n \ln(\ln n!)}{\ln (n+1) }= \lim \frac{n( \ln (\frac{\ln (n+1)! }{\ln n!})+\ln(\ln(n+1)!)}{\ln (n+1)}= \lim \frac{n \ln (1+ \frac{\ln (n+1)}{\ln n!} )}{\ln (n+1)}+\lim \frac{\ln(\ln(n+1)!)}{\ln (n+1)}= \lim \ln\left(\left ( 1+ \frac{\ln (n+1)}{\ln n!}\right)^{\frac{\ln n!}{\ln (n+1)}}\right)^{\frac{n \ln (n+1)}{\ln (n+1) \cdot \ln n!}}+\lim \frac{\ln(\ln (n+1)!)-\ln(\ln n!)}{\ln (n+1) -\ln n}= \ln e^{\lim \frac n {\ln n!}}+\lim \frac{n \ln(\frac{\ln (n+1)!}{\ln n!})}{n \ln (1+\frac 1 n)}=\ln e^{\lim \frac n {n!}} + \frac{ \ln \lim \left (\left(1+\frac{\ln (n+1)}{\ln n!}\right)^{\frac{\ln n!}{\ln (n+1)}}\right)^{\frac{n \ln (n+1)}{\ln n!}}} {1} = \ln e^{\lim \frac n {n!}}+ \ln e^{\lim \frac{n \ln (n+1)} {\ln n!}}. \]

From Cesaro-Stolz theorem we get that \[ \lim \frac{n} {\ln n!}=\lim \frac{1}{\ln (n+1)}=0 \] and \[ \lim \frac{n \ln (n+1)} {\ln n!}=\lim \frac{n (\ln (n+2)-\ln (n+1))+\ln (n+2)}{\ln (n+1)}=\lim \frac{\ln \left(1+\frac 1 {n+2}\right)^n}{\ln n}+\lim \frac{\ln (n+2)}{\ln n}=\lim \frac 1 {\ln n}+1=1 \]
Hence L=1. Doing the same with RHS of the double ineq we reach the conclusion.

Note: I used the fact that $\lim (1+x_n)^{\frac{1}{x_n}}=e$ when $x_n \to 0$. All limits in proof are when $n \to \infty$.


trigonometry
function
trig identities
sine double angle

1) find the numerical value of sin10sin50sin70.( all angles are in degrees).

2) show that sin pi/10 * cos pi/5  = 1/4

Just Remember  sin (x) = cos (90 -x ) if  angles are in degrees and
  sin 2x = 2 sin x cos x... So

If  All angles are in degrees and ..

A = sin 10 sin 50 sin 70

[hide](cos 10 ) * A =  cos 10 sin 10 sin 50  sin 70
               = (1/2)(2 cos 10 sin 10) cos 40 cos 20
               = (1/2) sin 20  cos  40 cos 20  = (1/2) sin 20 cos 20 cos 40
                      = (1/4) sin 40 cos 40
                      = (1/8)  sin 80
                       = (1/8) cos 10 [/hide]

so  A = (1/8)

Similarly if   (pi/10 = 18 degrees)
B =  sin (18) cos (36)
[hide]Then  (cos 18) B = cos 18  sin 18 cos 36
                     = (1/2) sin 36 cos 36
                     = (1/4) sin 72
                     =  (1/4) cos 18[/hide]
so B = 1/4

its a beautiful proof. the book too had a similar hint but i couldn't get it. is this a famous problem or something very common.

This sort of trig manipulation is something you should get comfortable with (i.e. you shouldn't think of this as a trick that just crushes this problem).

We see the strings of sines multiplied, and we start to think

[hide]If some of those were cosines, and were the same angle as one of the sines, we could use sine double angle to combine a pair of the trig functions into one trig function.  This would make the problem simpler, so we try to find a way to do it.

We use sin x = cos(90-x) to write sin 70 = cos 20 and sin 50 = cos 40, so we have sin 10 cos 20 cos 40 (we pick the sin 70 and sin 50 to convert since we know it will get us angles 10, 20, 40, which fits nicely with the double angle formula).  Then we just bash away as Gyan suggested:

sin 10 = (sin 20)/(2 cos 10), so

sin 10 cos 20 cos 40 = (sin 20)(cos 20)(cos 40) / (2 cos 10)
= (sin 40)(cos 40)/(4 cos 10)
= sin 80 / (8 cos 10)
= 1/8

Once we've done that one, the other one seems easy.  (Though admittedly, this one isn't obvious unless you're very experienced with trig identities)
[/hide]


algebra
polynomial
function
real analysis
real analysis unsolved

Hi
I was wondering if the following is true :
let $ \left( a_{i}\right) _{i\in \mathbb{N}}$ be a strictly monotonic sequence with $ a_{i}\in \left[ 0,1\right]$ for all $ i\in \mathbb{N}$
and let an arbitrary real sequence $ \left( y_{i}\right) _{i\in \mathbb{N}}$
do there exist a continuous function $ f: \left[ 0,1\right] \rightarrow R$ such that : $ \forall i\in \mathbb{N}: f\left( a_{i}\right) \equal{}y_{i}$ ?

Not always.

E.g. define $ a_i \equal{} 1 \minus{} \frac {1}{i}$ and $ y_i \equal{} ( \minus{} 1)^i$ for $ i \equal{} 1,2,3,...$

If there is a continuous function $ f: [0,1]\to \Bbb{R}$ with $ f(a_i) \equal{} y_i$ then $ f(a_i)$ has to converge to $ f(1)$ .

But $ ( \minus{} 1)^i$ doesn't seem to converge.

We have existence if and only if the $ y_i$ converge. If they do, just construct the piecewise linear function with the $ (a_i,y_i)$ as vertices, and extend to the limit by continuity. For any remaining parts of the interval, fill in with a flat line.


Solve the equations
\[ 4x\plus{}2\sqrt{9\minus{}2x^2}\minus{}x^2\sqrt{9\minus{}2x^2}\equal{}10.\]
\[ 4x\minus{}2\sqrt{9\minus{}2x^2}\plus{}x^2\sqrt{9\minus{}2x^2}\equal{}10.\]

I don't understand.  Clearly $ (2 \minus{} x^2) \sqrt{9 \minus{} 2x^2} \equal{} 0$ and $ 4x \equal{} 10$ so there are no solutions...

[quote="t0rajir0u"]I don't understand.  Clearly $ (2 \minus{} x^2) \sqrt {9 \minus{} 2x^2} \equal{} 0$ and $ 4x \equal{} 10$ so there are no solutions...[/quote]
i think it's not a system


LaTeX
AMC
AIME
geometry
3D geometry
trigonometry
complex numbers

Let x1 = 97, and for n > 1 let xn = n/(xn−1)
. Calculate the product x1x2x3x4x5x6x7x8.
 :D

~More problems coming~

Sure, it's a great problem, but $ \text{\LaTeX}$ would make it easier to read. For subscripts like these, enter:

[code]$x_{n+4}$[/code]
for: $ x_{n+4}$

So I assume you meant:
[quote="jkl381"]Let $ x_1 = 97$, and for $ n > 1$ let $ x_n = \frac{n}{x_{n-1}}$. Calculate the product $ x_1x_2x_3x_4x_5x_6x_7x_8$.[/quote]

[hide="Solution"]
What we are trying to look for simplifies to:

$ x_1\left(\frac2{x_1}\right)x_3\left(\frac4{x_3}\right)x_5\left(\frac6{x_5}\right)x_7\left(\frac8{x_7}\right)\\
=2\cdot4\cdot6\cdot8\\
=384$[/hide]

This problem is from the [url=http://www.artofproblemsolving.com/Wiki/index.php/1985_AIME_Problems/Problem_1]1985 AIME[/url]. Just thought it was worth mentioning.

:D

USA
AIME
1983
1 Let x,y, and z all exceed 1, and let w be a positive number such that logx w = 24, logy w = 40,
and logxyz w = 12. Find logz w.
2 Let f(x) = |x−p|+|x−15|+|x−p−15|, where p  x  15. Determine the minimum value
taken by f(x) by x in the interval 0 < p < 15.
3 What is the product of the real roots of the equation x2 + 18x + 30 = 2
p
x2 + 18x + 45?
4 A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the
circle is p50cm, the length of AB is 6cm, and that of BC is 2cm. The angle ABC is a right
angle. Find the square of the distance (in centimeters) from B to the center of the circle.
[img]6435[/img]
5 Suppose that the sum of the squares of two complex numbers x and y is 7 and the sum of
the cubes is 10. What is the largest real value of x + y can have?
6 Let an equal 6n + 8n. Determine the remainder upon dividing a83 by 49.
7 Twenty five of King Arthur’s knights are seated at their customary round table. Three of
them are chosen - all choices being equally likely - and are sent of to slay a troublesome
dragon. Let P be the brobability that at least two of the three had been sitting next to each
other. If P is written as a fraction in lowest terms, what is the sum of the numerator and the
denominator?
8 What is the largest 2-digit prime factor of the integer

200
100

?
9 Find the minimum value of
9x2 sin2 x + 4
x sin x
for 0 < x < .
10 The numbers 1447, 1005, and 1231 have something in common. Each is a four-digit number
beginning with 1 that has exactly two identical digits. How many such numbers are there?
11 The solid shown has a square base of side length s. The upper edge is parallel to the base and
has length 2s. All edges have length s. Given that s = 6p2, what is the volume of the
solid? [img]http://www.artofproblemsolving.com/Forum/albumpic.php?picid = 791sid =
cfd5dae222dd7b8944719b56de7b8bf7[/img]Inagivencircle, thelengthofdiameterAB is a two
digit integer. Reversing the digits of AB gives the length of a perpendicular chord CD.
The distance from their intersection point H to the center of the circle is a positive rational
number. Determine the length of AB.
123 For {1, 2, 3, . . . , n} and each of its non-empty subsets, an alternating sum is defined as follows.
Arrange the number in the subset in decreasing order and then, beginning with the
http://www.artofproblemsolving.com/
This file was downloaded from the AoPS − MathLinks Math Olympiad Resources Page
Page 1 http://www.mathlinks.ro/
USA
AIME
1983
largest, alternately add and subtract succesive numbers. For example, the alternating sum
for {1, 2, 3, 6, 9} is 9 − 6 + 3 − 2 + 1 = 6 and for {5} it is simply 5. Find the sum of all such
alternating sums for n = 7.
14 In the adjoining figure, two circles with radii 6 and 8 are drawn with their centers 12 units
apart. At P, one of the points of intersection, a line is drawn in such a way that the chords
QP and PR have equal length. Find the square of the length of QP. [img]6437[/img]
15 The adjoining figure shows two intersecting chords in a circle, with B on minor arc AD.
Suppose that the radius of the circle is 5, that BC = 6, and that AD is bisected by BC.
Suppose further that AD is the only chord starting at A which is bisected by BC. It follows
that the sine of the minor arc AB is a rational number. If this fraction is expressed as a
fraction m
n
in lowest terms, what is the product mn? [img]6438[/img]

Use the Resources/Contests section.

http://www.mathlinks.ro/resources.php?c=182&cid=45&year=1983


$ x^2\plus{}y^2\plus{}z^2\equal{}2^m$ To prove that there are  No desicions in natural numbers.THANS A LOT.

Since the LHS is greater than or equal to $ 3$, $ m\ge2$, and because $ k^2\equiv\{0,1\}\pmod4$, $ x^2 \plus{} y^2 \plus{} z^2\equiv\{0,1,2,3\}\equiv2^m\equiv0\pmod4\implies x^2,y^2,z^2\equiv0\pmod4$
$ \implies x,y,z\equiv0\pmod2\implies(x,y,z) \equal{} (2a,2b,2c)$ for some integers $ a,b,c$. But this means that $ 4a^2 \plus{} 4b^2 \plus{} 4c^2 \equal{} 2^m\implies a^2 \plus{} b^2 \plus{} c^2 \equal{} 2^{m \minus{} 2}$, with the same conditions that we started with. This means that for any solution, there is a smaller solution that can be found. However, this is impossible, since all solutions must be in the natural numbers. Thus, there are no solutions.

[quote="math154"] k^2\equiv\{0,1\}\[/quote]What means this

I mean that any perfect square of an integer leaves a remainder of $ 0$ or $ 1$ when divided by $ 4$.


floor function
number theory unsolved
number theory

[quote="keyree10"]Find all natural numbers $ n > 1$ such that $ n^{2}$ does $ \text{not}$ divide $ (n \minus{} 2)!$.[/quote]

It suffices to Find all natural numbers $ n > 1$ for which $ n^2\mid (n \minus{} 2)!$ since $ \gcd(n^2,n \minus{} 1) \equal{} 1$ it's equivalent to $ n^2 \mid (n \minus{} 1)!$ or $ n^3 \mid n!$.  Let $ p$ be a prime divisor of $ n$ and let $ \alpha$ such that $ p^{\alpha} \parallel{} n$ Acording to Legendre Formula we have:
 $ v_p(n!) \equal{} \sum_{k \equal{} 1}^{\infty} \left \lfloor \frac {n}{p^k} \right \rfloor$

if $ n$ has more than two prime divisors and $ v_2(n) \ge 2$ or $ v_2(n) \equal{} 0$. Then for all $ k \le \alpha$ we have $ \frac {n}{p^k} \ge 3$
and hence $ v_p(n!) \ge \sum_{k \equal{} 1}^{\alpha} \frac {n}{p^k} \ge 3 \alpha \equal{} v_p(n^3)$

$ \bullet$ if $ n \equal{} 2p^{\alpha}$ and $ p\ge 3$ and $ \alpha \ge 2$ Then $ v_p(n!) \equal{} \sum_{k \equal{} 1}^{\alpha} \frac {n}{p^k} \equal{} 2\left(\frac {p^{\alpha} \minus{} 1}{p \minus{} 1}\right)\ge 3^{\alpha} \minus{} 1 \ge 3 \alpha$

$ \bullet$ if $ n \equal{} 2p$ we have $ v_p(n!) \equal{} 2 \le 3$ hence $ \{2p \mid p\in \mathbb{P} \}$ are a solution

$ \bullet$ if $ n \equal{} 2^{\beta}$ Then $ v_2(n!) \equal{} 2^{\beta} \minus{} 1 \le 3\beta$   if $ \beta \in \{1,2,3\}$  Thus $ 2,4,8$ are solution

$ \bullet$ if $ n$ has one prime divisor greater than $ 3$ Then $ v_p(n!) \equal{} \frac {p^{\alpha} \minus{} 1}{p \minus{} 1}$

$ \bullet$ if $ \alpha \ge 3$ Then $ \frac {p^{\alpha} \minus{} 1}{p \minus{} 1} \ge \frac {3^{\alpha} \minus{} 1}{2} \ge 3\alpha$

$ \bullet$ if $ \alpha \equal{} 2$ Then $ v_p(n!) \equal{} p \plus{} 1 < 6$ if $ p \equal{} 3$ Hence $ 9$ is a solution

$ \bullet$ if $ \alpha \equal{} 1$ it's easy to show that $ n^2 \not | (n \minus{} 2)!$ hence $ \mathbb{P}$ is a solution

Hence the solution are $ S \equal{} \{8,9\} \cup \mathbb{P} \cup \{2p \mid p\in \mathbb{P} \}$

[quote="Rijul saini"]I'm not sure about your solution as $ 9$ is also a solution.[/quote]

Yes you're right I just forgot to include $ 9$ in the case when $ \alpha\equal{}2$  :)

By the way guys, how many questions do you think one needs to do or the minimum marks required to get through the INMO?

dear viewers solutions to the given set are updated by HBCSE

http://olympiads.hbcse.tifr.res.in/uploads/inmo-2010

I don't want to spam
http://olympiads.hbcse.tifr.res.in/uploads/inmo-2010

http://olympiads.hbcse.tifr.res.in/subjects/mathematics/previous-question-papers-and-solutions
This is the official link

Here is AoPS link
http://www.artofproblemsolving.com/Forum/resources.php?c=78&cid=46&sid=839a08f83416b59b2c30ede458e49e9b

And no need to be so professional with me atleast...
I am still not worth that much respect !

Could someone explain me?
[quote=Abdek]

if $ n$ has more than two prime divisors and $ v_2(n) \ge 2$ or $ v_2(n) \equal{} 0$. Then for all $ k \le \alpha$ we have $ \frac {n}{p^k} \ge 3$[/quote]

See [url=http://www.artofproblemsolving.com/community/c6h1229295]here[/url] also :)

[quote=UzbekMathematician][quote=Abdek]Hence the solution are $ S \equal{} \{8,9\} \cup \mathbb{P} \cup \{2p \mid p\in \mathbb{P} \}$[/quote]
$n=4$ and $n=6$  ????? :maybe:[/quote]

Oh!! Surry. I unterstand  :wallbash:

The moment when you realize you made a seemingly insignificant mistake but later miss about 3 cases because of that and realize that you have just wasted 30 minutes :wallbash_red:

So no. of factors of such number n is 2,3,4,5.
So if it's 2 then n=p,
If it's 3 then n=p^2,
If it's 4 then pq or p^3,
If it's 5 then p^4.

n=p is a solution.
If n=p^4  by comparing the exponent of p in our original problem we get no solutions.
If n=pq by comparing exponent we get q=2(W.L.O.G p>q).
If n=p^3 we get p=2.
If n=p^2 we get p=3.

Here is a simpler solution.
First observe that by wilson theorem $(n-1)(n-2)!$ is divisible by $n$ for all composite except $4,$ implying
$-(n-2)!$ is divisible by $n$.
Now obviously if $n$ is a prime then it does not divide $(n-2)!$
If $n$ is a power of prime then it will not divide $(n-2)!$ if and only if $p^{\frac{n}{2}}-1<n$
Which only happen when $p=2, n=3$ and $p=3, n=2$
Now if $(n-2)!<n^{2}$ then it will obviously not divide $(n-2)!$ which happen for $n<7$
Now if $n=2p$ where $p$ is a prime then in order for denominator to completely divide the numerator, $2p-2$ should contain 2 $p$ a contradiction.
Hence $n^{2}$ does not divide $(n-2)!$ when $n=2p$.
Now we will prove that if $n=kp$ where $ k$ is an integer greater than $2$ then,$ n^{2}$  divide $(n-2)!$
The following argument is true if we prove that we find an square $n$ between $n$ and $kn$
Note that if we are able to prove this for $k=3$ then it is obivously true.
Which is equivalent to proving
$\sqrt{3n}-\sqrt{n}>1$
Squaring, we get the following the quadratic in $n$:
$4n^{2}-8n+1$ which is true for $n>1$. Therefore,
$n=p, n=2p, n=8,9$ where $p$ is a prime are the only  $n$ whose square does not divide $(n-2)!$

[quote=nathantareep][quote=pkrmath2004]If a number has alteast 6 factors then n^2|(n-2)![/quote]

How do you say that?[/quote]

im not sure how pkrmath2004 did it but i found the same thing
$n|(n-2)!$ except when $n$ is a prime, here only two divisors are enough
$n^2|(n-2)!$ will happen when $n$ has at least 4 factors as we can write $\frac{(n-2)!}{n.n}$
i think he meant 6 divisors including 1 and $n$ itself but well we dont have them in $(n-2)!$


Let $ m$ be the number of five-element subsets that can be chosen from the set of the first $ 14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $ m$ is divided by $ 1000$.

Forgot to minus off the 252 at the end (was suspicious that there was no need for mod 1000 but I didn't check.)

[quote="MCNational!"]i brute forced this one  :P[/quote]

As did I, but I only found 240 consecutive ones.

[quote="MCNational!"]i brute forced this one  :P[/quote]

me too sorta. case work... :P  lol i came up with like 2200. somehow.

I just found my error. I missed one case out of 18.

[quote="funtwo"]i brute forced the 252 part... then i was like w8 coincidence?[/quote]
Same here. It was just a lot of adding up.

[quote="dgreenb801"]This is almost exactly like the example given in Aops volume 2 on one-to-one correspondences.
[hide]Let the O's be the selected objects.
XXOXXOXOXXXOXO is an example of a sequence with 5 selected numbers out of 14, where none are consecutive. Since none are consecutive, we can take away an X from in between each consecutive O (4 of them), so this bijects to (forms a one-to-one correspondence with)
XXOXOOXXOO
But the total possibilities for this is $ {10 \choose 5}$, so we have to subtract it from the original $ {14 \choose 5}$.[/hide][/quote]

I haven't studied combinatorics much, and I dont understand bijections very much. Could someone please explain to me why/how this is a bijection?

Edit: Ohhhh, I get it now. If you arrange the final sequence in any order you wish, by adding those four x's back, we will still ensure that no 5 of the numbers is consecutive. That's pretty sweet.
On the real test, I figured out that I needed to find the ways they weren't consecutive, and I knew I needed to find a nice bijection for it, I just didn't know how to approach it.

[quote=FibonacciFan][hide]
We use complementary counting. There are $ \binom{14}{5} \equal{} 2002$ total subsets we could choose. The subsets with no consecutive numbers can be counted as follows.

Bin1 Elem1 Bin2 Elem2 Bin3 Elem3 Bin4 Elem4 Bin5 Elem5 Bin6

We have five elements and six bins. The bins between two elements represent the difference between those elements and Bin1 represents the number of elements below the first element and Bin6 represents the number of elements above the last element. There must be at least one numbers in Bins 1-4, so there are 14 - 5 - 4 = 5 numbers left to place. By balls and urns, there are $ \binom{6 \plus{} 5 \minus{} 1}{5} \equal{} \binom{10}{5} \equal{} 252$ ways to do this. Therefore, the answer is $ 2002 \minus{} 252 \pmod{1000} \equal{} \boxed{750}$.
[/hide][/quote]

I do not understand this. What do you mean the "bins between two elements represent the difference between those elements?"

Can somebody post a simpler solution?

Okay, so I think I found a simpler way in the book, A Gentle Introduction to AIME.

Lets say that X|XX|XXX|XX|X| is what we are trying to find. The Xs are the numbers that are not chosen, and the |s are numbers that are chosen. There are 14c5 ways we can choose the numbers, so there are 2002 ways we can choose the numbers.

However, there is a problem. Some of them are not consecutive! So, we have to do complementary counting.

We have to count the number of ways they are not consecutive. How will we do that? We know that all the X's will add up to 9. So we count them, by saying how many there Xs there are, before the |. So, x[size=50]1[/size] + x[size=50]2[/size] + x[size=50]3[/size] + x[size=50]4[/size] + x[size=50]5[/size] + x[size=50]6[/size] = 9. (x[size=50]6[/size] is how many Xs there are after the final |. X[size=50]1[/size] is how many Xs there are before the first |.)

Now, there do not have to be any Xs before X[size=50]6[/size] and X[size=50]1[/size]. But for every other x, it has to be greater than, or equal to 1.
Now, we subtract 4 from 9. So there are five stars, and 5 chosen ones(bars). 10c5 is 252, and 2002 - 252 is 1750, and the remainder is 750.

Why do we subtract 4? If it is not consecutive, the Xs can fill up the space before or after the x[size=50]1[/size], or x[size=50]2[/size], and so on and so forth.

Are there any holes in my solution? Please tell me if there are.

[hide=@Calculus123]In this problem, the easiest approach is by using complementary counting and the one-to-one correspondences mentioned by dgreenb801 above. As you said, there are $\binom{14}{5}$ total ways to choose a 5-element subset. Now, we have to find the total number of consecutive 5-element subsets. To do this we subtract 4 from the total because each of the 5 chosen numbers must have at least one space between them (for a total of 4). We can arrange the remaining numbers in any order, so that's how we get $\binom{10}{5}$. As a result the answer is $\binom{14}{5} - \binom{10}{5} = 750$ mod $1000$

This is a problem which can be solved with one-to-one correspondences.
[hide=solution]Let the five numbers selected be $a<b<c<d<e$. We will count the complement; i.e. no two of the five numbers are consecutive. Consider each of the numbers $a, b^\prime=b-1, c^\prime=c-2, d^\prime=d-3,$ and $e^\prime=e-4$. Then $1\leq a<b^\prime<c^\prime<d^\prime<e^\prime\leq 10,$ with no further restrictions. There are $\binom{10}{5}=252$ ways to do this, since there's only one way to arrange a set of five numbers to be in increasing order. However, this is not the answer since we counted the complement. The answer is $\binom{14}{5}-252=2002-252=1\boxed{750}$.[/hide]

What if there are several consecutive sets?

For example consider the set {1, 2, 5, 7, 9, 10}

[quote=OlympusHero]I understand the complementary counting solution, but why is this solution invalid?

We will use PIE. If two numbers are consecutive, there are $13 \cdot \binom{12}{3}$ ways. If three numbers are consecutive, there are $12 \cdot \binom{12}{2}$ ways. If four numbers are consecutive, there are $11 \cdot \binom{12}{1}$ ways. If five numbers are consecutive, there are $10$ ways. Therefore, the total amount of ways is $13 \cdot \binom{12}{3} - 12 \cdot \binom{12}{2} + 11 \cdot \binom{12}{1} - 10 \equiv 190 \pmod {1000}$.[/quote]

this is why you don't memorize PIE - try to understand it. if you count how many times two numbers are consecutive, you are actually counting the case where three numbers are consecutive $\binom{3}{2}=3$ times.

Thanks - I literally read that in the C&P book and completely forgot about it :maybe: I'll solve tomorrow.

To amopuri: Don't I count those in the case with two consecutive numbers? There are two sets of consecutive numbers, sure, but it's still counted in that case. I think the problem is I didn't subtract off the appropriate number of times.


induction
greatest common divisor
number theory
relatively prime

Let $a_{1} < a_{2} < a_{3} < \cdots $ be an infinite increasing sequence of positive integers in which the number of prime factors of each term, counting repeated factors, is never more than $1987$. Prove that it is always possible to extract from $A$ an infinite subsequence $b_{1} < b_{2} < b_{3} < \cdots $ such that the greatest common divisor $(b_i, b_j)$ is the same number for every pair of its terms.

If I grasp the problem correctly,the condition turns to" for every positive integer $ i$,$ a_i\equal{}p_1^{\alpha_1}*...*p_k^{\alpha_k},then,\sum \alpha_i \le 1987$
We replace $ 1987$ by $ n$ and consider an induction on $ n$,the foundation is trivial,we assume it is true for all positive integer $ <n$,now consider the case $ n$

Note that if there are infinite terms share a common prime divisor $ p$,we can just divide these terms by $ p$ and use induction hypothesis and reach our
goal

So we can simply assume that there exsists no prime $ p$such that we can find infinitely many terms which is a multiple of $ p$    $ ......(*)$

Thus,we pick out $ a_1$,by $ (*)$ we can find infinitely many terms which is [b]relatively prime [/b]to $ a_1$,denote the infinite sequence $ \{a_i'\}$

then we pick out $ a_2$ as what we did with $ a_1$...Finally we can pick out an infinite subsquence $ \{b_i\}$such that $ (b_i,b_j)\equal{}1$ complete our induction.

QED


geometry
trapezoid
email
geometry solved

Hello  :D ,
I have a question in relation to a flower bed:

[i]A landscape gardener uses 36 paving bricks 22cm by 11cm to form a 'circular' flower bed.[/i] (See attachment below for image * note: not to scale)
[b]a.[/b] What is the radius of the circle?
[b]b.[/b] There is a V-shape between each brick. What is the angle of the V?
[b]c.[/b] The gardener wishes to eliminate this V by cutting a portion for the corners of each brick so that the face of each brick is in the shape of an isosceles trapezium ABCD. What is the length of the side, AB (see image for representation)?

Can you please help?

Thank-you!

Jpm911

Because we are using 36 bricks, i believe that means a 36-gon with sides of 22 is inscribed in the circle. Each angle of a 36-gon is 170. Using trig. i found the radius to be 126.21.  ( $\frac {11}{Cos 85}$ )  Before i attempt the other parts of this problem ~ did anyone else get this?

Oh, thanks xenon_nightmare152.
It is actually a question that is meant to involve trigonometry.
Could you please go into a little more detail and explain  to me how you go to (11/cos 85)?
Thanks.

Jpm911

:blush: i don't know how to add an attachment/pic , but i can try to explain. Each side of the 36-gon is a chord of the circle. Hence, the apothem of the polygon and what would be part of the radius of the circle will be perpendicular to the chord and bisect the chord as well, resulting in a right angle with one legs length equivalent to $\frac {22}{2} = 11$ We figured that each angle was 170 degrees. The hypotenuse of the right triangle (also the radius of the circle!) is bisecting this angle, so one angle of the right triangle will be 85 degrees. $Cos85 = \frac {11}{x}$ where x is the length of the radius.  :huh:  That explanation is confusing without a diagram...try sketching it out

Oh, I am kind of understanding where you're coming from.

It's quite easy to post an attachment.....

Once you've made a "Post a reply" you will notice two options (which I'm positive you're familiar with "Preview" and "Submit"). Directly below these options is a box titiled "Attachment Posting Control Panel".
Scroll down a little more and you will notice that there is an options section... here you will see the button "Add an Attachment".  :lol:

If you could add one that would be fantastic!

Thank-you!

Can anyone else please help answer my question...
It's starting to become an urgent priority of mine!  :(

Thankyou  :P

Jpm911

Here you go....I had to do the same thing for school...hope its not too late...and I dont guarantee that its all correct either...

See attachment.

P.S. What school are your from??

 :arrow: Email mitchell.cain@gmail.com

Oh, thanks a lot!

I'm from a school in Palm Beach, Florida.

But, I know this question is from an Australian text book.

I guess you're Australian.


induction
linear algebra
matrix
vector
linear algebra unsolved

This looks familiar. Has it appeared in this forum recently? If so, I can't find it.

Not exactly induction: The matrix is $ I\plus{}aa^T$ for some column vector $ a.$ If we subtract the copy of $ I$ back off of it, we have a rank-1 matrix. That means that $ 1$ appears as an eigenvalue with geometric multiplicity (hence also algebraic multiplicity) $ n\minus{}1.$

Meanwhile the trace of the original matrix is $ n\plus{}a_1^2\plus{}a_2^2\plus{}\cdots a_n^2.$ Since the sum of the eigenvalues is the trace, that means that the one eigenvalue we haven't accounted for yet is $ 1\plus{}a_1^2\plus{}a_2^2\plus{}\cdots a_n^2.$

The determinant is the product of the eigenvalues, which is to say the product of $ n\minus{}1$ copies of $ 1$ with $ 1\plus{}a_1^2\plus{}a_2^2\plus{}\cdots a_n^2.$ We're done.

Thank you.
It is a very decent solution.
But I still want to know how to solve this problem by induction. My friend told me that he had seen a solution using induction before. Unfortunately he doesn't remember now.