quadratics
modular arithmetic
number theory proposed
number theory

$X,Y,Z$ E N .If $(x,y)=1$ and $x^{2}+y^{2}=z^{4}$ show that $7$ divides $xy$ :)

     edited; :wink:

[quote="Maruf"]$X,Y,Z$ E N .If $(x,y)=1$ and $x^{2}+Y^{2}=z^$ show that $7$ divides $xy$ :)[/quote]
maybe the problem is:
$X,Y,Z$ E N .If $(x,y)=1$ and $x^{2}+Y^{2}=z^{2}$ show that $7$ divides $xy$ :)

maybe the problem is $x^{2}+y^{2}=z^{4}$ then $7|xy$ :P

We may suppose  y is even. Then
\[x=m^{2}-n^{2}, y=2mn,z^{2}=m^{2}+n^{2}\]
Quadratic residues $\pmod7$ are 0,1,2,4...

[quote="RIP"]We may suppose $(x,y,z)=1$ and y is even. Then
\[x=m^{2}-n^{2}, y=2mn,z^{2}=m^{2}+n^{2}\]
Quadratic residues $\pmod7$ are 0,1,2,4...[/quote]

 can you explain  your solution,thanks!!

That's wrong solution.
No, that's right.


[quote="neelnal"]1. What is mod, how to find it easily, and give me some applications/problems with it.

2. How to find a number in base of another number easily and give me some applications/problems.

3. What is the thing with the parantheses with one number on top of the other? What is the concept in words, formula, and what are applications at high levels? A problem using it would be nice too.

Thanks.[/quote]

Okay, one at a time.

"Mod" is a shorthand way to write "modulo". The best example of modular arithmetic is the clock. For example, if it's midnight right now and we want to find what time it will be 100 hours from now, we would take $100 \mod 12$. This is the same thing as saying "the remainder that you get when you divide 100 by 12".

The thing with parentheses is the "choose" function. For example, if you want to find the number of ways you can "choose" 2 objects out of a bag with 5 objects in it, you would take $5 \choose 2$. This assumes that the order doesn't matter. It is equivalent to saying $\frac{5!}{2!3!}$.

And I don't know much about converting one base to another.

[quote="neelnal"]3. What is the thing with the parantheses with one number on top of the other? What is the concept in words, formula, and what are applications at high levels? A problem using it would be nice too.[/quote]
$\binom{n}{k}\equiv_{n}C_{k}\equiv\frac{n!}{k!(n-k)!}$
It is how many ways there are to choose, without replacement and without order, $k$ objects from $n$ objects.

So, does anyone know a quick way to do the base thingy?

Like what is 500 in base 17?

Is there a quick way? Or do I just have to divide and stuf?

In bases, each digit signifies $x*y^{z}$ where x is the digit, y is the number base that you are in, and z is the number of digits to the left of the units digit that you are. For instance, in base 10, the number 4239 is equal to $9*10^{0}+3*10^{1}+2*10^{2}+4*10^{3}$.

The base 2 number 1001 in base 10 would be equal to $1*2^{0}+1*2^{3}$ which is equal to 9.

The base 10 number 283 in base 5 would be equal to $2*5^{3}+1*5^{2}+1*5^{1}=3*5^{0}$.

500 in base 17 is equal to $2*17^{2}+B*17^{1}+7*17^{0}$

Note that when there are bases higher than base 10, letters are used to represent higher numbers, so that $A=11, B=12, C=13,$ and so on. The most common number base higher than 10 is 16.

To get a number from base x to base y where neither x nor y are equal to 10, I normally change the number in base x to base 10, and then I change that number to base y.

I think you mean A = 10, B = 11, C = 12 etc

[quote="mustafa"]The base 10 number 283 in base 5 would be equal to $2*5^{3}+1*5^{2}+1*5^{1}=3*5^{0}$.[/quote]
The equal sign is supposed to be a plus.

$283=2\cdot5^{3}+1\cdot5^{2}+1\cdot5^{1}+3\cdot5^{0}=2113_{5}$

Some info...

The reason base ten is the one we use is that we have ten fingers and count by tens...

The reason base sixteen is the most common base above ten is that it is used in computer programming...

$\binom{n}{k}$ is almost always read as "n choose k" but in some cases it is read as "n take k"...

You probably know this, but $n!=n(n-1)(n-2)...(3)(2)(1)$...

IMPORTANT
The reason $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ is because think about it. If order doesn't matter then you have to pick k things out of a set of n. You have n choices for the first thing, n-1 for the second, n-2 for the third, ..., n-k+1 for the kth item. So the number of choices is $(n)(n-1)(n-2)...(n-k+1)$. This quantity can be rewritten as $\frac{n!}{(n-k)!}$. The thing is, we made this formula not counting the fact that order matters. So we need to factor in that order doesn't matter. There are k numbers you multiplied in making our answer. There are $k!$ ways to make the combination, when we really only have 1. $\frac{n!}{(n-k)!k!}$.


[quote="carlosbr"]Two players, A and B, play the following game: they retire coins of a pile which contains initially 2006 coins. The players play removing alternatingly, in each move, from 1 to 7 coins, each player keeps the coins that retires. If a player wishes he can pass(he doesn't retire any coin), but to do that he must pay 7 coins from the ones he retired from the pile in past moves. These 7 coins are taken to a separated box and don't interfere in the game any more. The winner is the one who retires  the last coin, and A starts the game. Determine which player can win for sure, it doesn't matter how the other one plays. Show the winning strategy and explain why it works.[/quote]

If they won't may pass at that way:
$A$ starts with taking $6$ coins and after $B$ he does $8-i$ or if $B$ and he can win by take the last after $251$ moves.

At the other way:
$A$ takes always $7$ coins until they reach a number, after at least $143$ moves each, he has to bring the number on $8$, or B did, but $A$ can pass at least once more, (142*14<1998), (143*14>1998), so B has to take sth and then A do $8-i$ and win.
If B did earlier less then $8$, A win very simply.

[hide=My solution]$A$ removes always $7$ coins. If $B$ removes always $7$ coins (without passing) then after $143$ moves each one, there are $2006-143\times 14=2006-2002=4$ coins, so $A$ wins. If $B$ removes always $7$ coins (without passing), in his last move he can pass and then $A$ removes $3$ coins, or remove at most $3$ coins, and $A$ removes the remaining coins (may be $0$); but now $A$ can pass more times than $B$, so $B$ will have to remove at least one coin, then $A$ removes the remaining coins and win.
If $B$ pass or remove less than $7$ coins in some turn, then we can get again to $8$ coins and $A$ can pass more times than $B$, so $B$ will have to remove at least one coin, then $A$ removes the remaining coins and win.[/hide]

Let $n(X)$ be the total of coins of player $X$. Because every number is a lose number or a win number, the better strategy to $A$ is to maximize $n(A)$. Indeed, if $n(A)=7q+r$ and $n(B)= 7l+r$ and $q>l$,  $A$ will win because he can pass in a lose number and win in a win point.
Thus, the better strategy to $A$ is always choose $7$. Because $B$ is not interesting in lose (=0), by symmetry he has to choose always 7.
Then, after several time, the pile will have $18$ coins and $n(A)=n(B)$. Thus,  $A$ can choose $7$ again, and $11$ is a lose number, because if $B$ chooses $t>3$, $A$ can easily win and if $B$ choose $x$, $A$ can choose $3-x$ putting $B$ in number $8$, and because in this time $n(A)>n(B)$,  $B$ will have to choose a number $v$ and of course $8-v$ less equal to $7$. Thus $A$ always wins.


A teacher wants to arrange 3 copies of $ \emph{Introduction to Geometry}$ and 4 copies of $ \emph{Introduction to Number Theory}$ on a bookshelf. In how many ways can he do that?

hello
no. of ways=$ \frac{(3\plus{}4)!}{3!4!}\equal{}35$
thank u

Additionally, we can pick $ \binom{7}{3}\equal{}\boxed{35}$ spots to put the $ \emph{Introduction to Geometry}$ books.

Or $ \binom{7}{4}=\boxed{35} $ for $ Introduction $  $ to $ $ Number Theory $

I don't think so if they are the same kind...

They should have specified distinguishable ways or not

All copies of both books are indistinguishable, so I would assume that's why they didn't specify that.

What they said all copies are identical so they are indistinguishable. (or maybe that was just added later)


a) Prove that we can assign one of the numbers $1$ or $-1$ to the vertices of a cube such that the product of the numbers assigned to the vertices of any face is equal to $-1$.

b) Prove that for a hexagonal prism such a mapping is not possible.

For (a): we put $-1$ in two opposite corners and $1$ in the other corners.
I'll think later about (b).

Valentin, will you post grades 9/10? I'm interested in those...

[hide="Major Hint"]For b, consider the products on the opposite hexagons and prove they are different.[/hide]


As I've heard, IMC 2004 will take place in Skopje, Macedonia. They have announced last year that Bulgaria will organise the 2004 IMC, but I guess they have changed their mind.
For us, the romanians, this only means higher transport cost, since Bulgaria is our neighbour wile Macedonia is not.


calculus
integration
search
real analysis
real analysis theorems

I'm sorry if this already has been asked. But my search on this forum did not pay off...

Anyways, I was wondering if there is a book out there containing challenging integrals (problems, solutions and hopefully some hints on how to reason)? So I'm basically want integrals, nothing else. A book with the same philosophy as Problems and Theorems in Analysis by George Polya, Gabor Szegö.

Thank you.

You might try [url=https://www.cambridge.org/us/catalogue/catalogue.asp?isbn=0521796369]Irresistible Integrals[/url] (or try [url=http://www.amazon.com/Irresistible-Integrals-Symbolics-Experiments-Evaluation/dp/0521796369]Amazon[/url].)


function
real analysis
real analysis unsolved

Let $ f$ and $ g$ be measurable functions on $ E$.
a) If $ f$ and $ g$ are finite a.e., show that $ f\plus{}g$ is measurable no matter how we define it at the points when it has the form $ \infty\minus{}\infty$ or $ \minus{}\infty\plus{}\infty$.
b) Show that $ fg$ is measurable without restriction on the finiteness of $ f$ and $ g$.  Show that $ f\plus{}g$ is measurable if it is defined to have the same value at every point where it has the form $ \infty\minus{}\infty$ or $ \minus{}\infty\plus{}\infty$.

For a), I am thinking that since $ m(\left|f\right|\equal{}\infty)\equal{}m(\left|g\right|\equal{}0)$, we can write $ \{f\plus{}g>\alpha\}\equal{}\bigcup_{r\in\mathbb{Q}}\left(\{f\equal{}\infty\}\cup\{\infty>f>r\}\right)\cap\left(\{g\equal{}\infty\})\cup\{\infty>g>\alpha\minus{}r\}\right)$, but the sets where $ f,g\equal{}\pm\infty$ are null sets, they are measurable so the entire union\intersection above is measurable.

Not quite as sure about b).

I am sure you mean this: ;)
[quote="JRav"]$ m(\left|f\right| \equal{} \infty) \equal{} m(\left|g\right| \equal{} 0)$[/quote]
$ m(\left|f\right| \equal{} \infty) \equal{} m(\left|g\right| \equal{} \infty)\equal{}0$
[quote]$ \{f \plus{} g > \alpha\} \equal{} \bigcup_{r\in\mathbb{Q}}\left(\{f \equal{} \infty\}\cup\{\infty > f > r\}\right)\cap\left(\{g \equal{} \infty\})\cup\{\infty > g > \alpha \minus{} r\}\right)$[/quote]$ \{f \plus{} g > \alpha\} \equal{} \bigcup_{r\in\mathbb{Q}}\left(\{f \equal{} \infty\}\cup\{\infty > f > r\}\right)\cap\left(\{g \equal{} \infty\}\cup\{\infty > g > \alpha \minus{} r\}\right)$ - here clearing one right bracket.

For (b) use the decomposition \[ f\equal{}f 1_{\{0<f<\infty\}}\plus{}f 1_{\{\minus{}\infty<f<0\}}\plus{}\infty1_{\{f\equal{}\infty\}}\minus{}\infty1_{\{f\equal{}\minus{}\infty\}}\] - and analogously for $ g$ - to reduce to the case that $ f$ and $ g$ are finite using the convention that $ 0\cdot \infty\equal{}0$. Then use the composition of measurable mappings $ f\cdot g\equal{}M\circ(f,g)$ with the multiplication map $ M: \mathbb R^2\to\mathbb R, (x,y)\mapsto xy$.

[quote]Show that $ f\plus{}g$ is measurable if it is defined to have the same value at every point where it has the form $ \infty\minus{}\infty$ or $ \minus{}\infty\plus{}\infty$.[/quote]This is a special case of (a).

Indeed, that is what I meant. :)

The second part that you quoted, (a), is that correct?

For (b), we would get that $ g=g\chi_{\{0<g<\infty}\}+g\chi_{\{-\infty<g<0\}}+\infty\chi_{\{g=\infty\}}-\infty\chi_{\{g=-\infty\}}$.  Perhaps I'm just being stupid (probably  :wink: )but where are we using the fact that $ 0\cdot\om\infty=0$? (I agree with the convention, I'm just not seeing where we're multiplying something that's 0 by something that's infinite).

Your proof of (a) looks good to me.

$ 0\cdot\infty\equal{}0$: What is $ \infty\cdot 1_A$ outside the measurable set $ A$? ;)

Yes, I realized that a little bit later, when I wrote it out on a chalkboard. :)   And I think its easier to write $ \{f \plus{} g > a\} \equal{} \{<\infty>f \plus{} g > a\}\cup\{f\plus{}g\equal{}\infty\}\cup\{f \plus{} g \equal{} \infty \minus{} \infty\}\cup\{f \plus{} g \equal{} \minus{} \infty \plus{} \infty\}$, where the last two sets are null sets, so the whole thing is measurable.  Correct?

Hmm, do you mean $ \{f\plus{}g\equal{}\infty\minus{}\infty\}\equal{}\{f\equal{}\infty\}\cup\{g\equal{}\minus{}\infty\}$? I would reduce to the finite case by restricting to the complement of the null-set $ \{f\equal{}\pm\infty\}\cup\{g\equal{}\pm\infty\}$. Then everything is fine. Otherwise you need to run through too many cases what I usually try to avoid.  :o

I'm not sure if that is what I "mean," but it seems that $ \{f\plus{}g\equal{}\infty\minus{}\infty\}$ and $ \{f\equal{}\infty\}\cap\{g\equal{}\minus{}\infty\}$ are the same thing.  It seems like I just need to consider that case, and $ \{f\plus{}g\equal{}\minus{}\infty\plus{}\infty\}\equal{}\{f\equal{}\minus{}\infty\}\cap\{g\equal{}\infty\}$, in which case I just get one big union and intersection.

Apology for the late reply. Of course there was a typo in my last posting - and you have corrected it in your reply. But with that everything seems to be fine with the solutions to your questions. :)

Just one more question: you said that my final question (part (c)) was just a special case of part (a).  Why is that true since I am not necessarily assuming $ f$ and $ g$ to be finite a.e. in the third part?

Right, got me! ;) I think we need to amend your proof of (a) a bit - since, e.g. what happens to \[ \{f+g >\alpha\}=\bigcup_{r\in\mathbb{Q}}\left(\{f =\infty\}\cup\{\infty > f > r\}\right)\cap\left(\{g =\infty\})\cup\{\infty > g >\alpha-r\}\right)\] when we set $ \infty-\infty=\infty$? It is still correct? I missed this nasty detail, to be honest. And there will be another as we will see below.

While writing the first reply I had my proof of (a) in mind which I will briefly sketch: Let $ A$ be the set of nasty points, i.e. $ A=A_1\cup A_2$ with $ A_1=\{f=+\infty\}\cap\{g=-\infty\}$ and $ A_2=\{f=-\infty\}\cap\{g=+\infty\}$. We know that $ A_1$ and $ A_2$ are measurable sets of measure zero. The function $ f+g$ restricted to $ E\setminus A$ is measurable; we could use your proof for this. Here I assume that $ \infty+\infty=\infty$ and  $ -\infty+(-\infty)=-\infty$.

Now the problem comes in when we have a look at the measurability of $ f+g$ restricted to $ A_i$, $ i=1,2$. If $ A_i$ does not contain any proper measurable subset, then $ f+g$ must be constant on $ A_i$. This does not happen if the $ \sigma$-algebra is complete with respect to the measure on $ E$, i.e. that every subset of a null-set is measurable. If every subset of $ A$ is measurable we can certainly define $ f+g$ on $ A$ by any functions with values in $ \overline{\mathbb R}$.

Now you might follow my comment on (c) which can be solved [i]like[/i] the addition of (a). Hope that this makes sense to you. I think with this you can amend your proof.


[quote]Let $ P$ be the common point of three circles $ \|\begin{array}{cccc}C_1(O_1,r_1)\\
\\
C_2(O_2,r_2)\\
\\
C_3(O_3,r_3)\end{array}$ and let $ \{\begin{array}{cccc} P_1\in C_1\cap C_2\\
\\
P_2\in C_2\cap C_3\\
\\
P_3 \in C_3\cap C_1\end{array}$
The tangent in $ P$ at $ C_1$ meets $ C_2$ in $ N_1$ and $ C_3$ in $ M_3$,the tangent in $ P$ at $ C_2$ meets $ C_3$ in $ N_2$and $ C_1$ in $ M_1$,and the tangent in $ P$ at $ C_3$ meets $ C_1$ in $ N_3$ and $ C_2$ in $ M_2$.Prove that \[ P_1M_1 \cdot P_2M_2 \cdot P_3M_3=P_1N_1 \cdot P_2N_2 \cdot P_3N_3\][/quote]

here is a proof [url]http://www.mathlinks.ro/weblog_entry.php?p=1126659#1126659[/url][/url]


What is the final temperature in degrees centigrade of 100.0 g of water at 30.0 degrees C if it is mixed with 50.0g of water at 0.0 degrees C?

First, we must assume that:

I. The system in study is isolated.

II. There is transfer of energy between the two parts of the system only under the form of heat.

With these assumptions made, we can write

$ Q_1 \plus{} Q_2 \equal{} 0$

where $ Q_1$ is the heat transfered by the water at 30.0ºC and $ Q_2$ is the heat transfered to the water at 0.0ºC.

Now we are going to make two more assumptions:

III. The water at 0.0ºC is in the liquid state.

IV. The heat capacity of water is constant (in the temperature range of this problem, this assumption is valid).

Therefore:

$ m_1\,\,c \,\, (T \minus{} 30.0) \plus{} m_2 \,\, c \,\, (T \minus{} 0.0) \equal{} 0$,

where T is the final temperature of the system. Now just substitute $ m_1 \equal{} 100\,\,g$, $ m_2 \equal{} 50.0\,\,g$, and $ c \equal{} 4.18\,\,J/g/K$.


In my country people use max flow or Hungarian method for Bipartite Matching(BPM). But I know hoffman algorithm is the best for BPM. But I don’t know its implementation. And as far I know in abroad Hoffman is popular. So if any one have the implementation of Hoffman Algo can u post it here or send me by email?

Mahbub

Really? What's the time bounds for it?

Hopcroft-Karp is usually the one people refer to the fastest BM algo... Time: O(E sqrt(V))

well what do u mean by time bound?

as far i know, BPM is better than maxflow, and Hopman (or known as hopkroft Karp) is better than BPM.

do u have implimentation of HK? can u plz send that?

Mahbub

Nope, I've never written any HK implementation... Never understood it really, and never had the need for it either. ;)

What I do know about it, is that it's basicly the same as the normal augmenting path algorithm, but finds multiple augmenting paths each "cycle", instead of just one...

Hopcroft-Karp is actually almost the same as the "normal" augmenting-path Bipartite Matching algorithm, but you search for many paths at once. I attach a paper on it. I have coded it once, but I think on a competition like IOI it's better to know how to code normal BPM quickly and to make it efficient, the improvement of sqrt(N) is not very impressing in practice.

Great!

Well, HK is a great algorithm to have for competitions where "prewritten code" is allowed. :)

[quote="kubus"]to code normal BPM quickly[/quote]

What's the normal algorithm?

normal algo is Hungerian Algo or u can also use Maxflow-BPM

Well, if you use the maxflow and bear in mind that what you are trying to do is just a matching, where capacities are all 1 and so on, you can make it quite simple - just keep finding an augmenting path using simple dfs.

actually the method (DFS) kubus is statis is nothing but hungerian tree algortihm. but th difference is that we mainly understand that hangerian is BFS algorithm... but kubus is just using DFS instead of DFS.


I would like to know (using some illusionary example) the following:

$A$ is said to be complete [b]if[/b] it contains $B$.

Is it the same as saying

$A$ is said to be complete [b]if and only if[/b] it contains $B$?

I know the usage of [b]if[/b] and [b]if and only if[/b] have different meaning in general, but in the case of the example above, are they different?

Here is how I see the first statement is equivalent to the second statement:
If $A$ is complete, then it must then contain $B$. Suppose $B$ is not contained in $A$, then $A$ wouldn't have been complete.

Yes, the definitional meaning of "if" is different from the logical meaning of "if". I prefer to use other words to avoid this confusion.
On the other hand, I don't like using "if and only if" in a definition either. That phrase is much better suited to a theorem.

Also, be careful with your examples- you've given a statement and its contrapositive, not the statement and its converse.

[quote="jmerry"]Also, be careful with your examples- you've given a statement and its contrapositive, not the statement and its converse.[/quote]

 :huh: I am not entire clear. Examples would be appreciated.

I personally would see the top statement as misleading, because it still implies (albeit only to the pedant) that there could exist an entity $A$ that does not contain $B$, but could still be said to be complete. It could be equivalently rephrased as "any $A$ that contains $B$ must be called complete".

Personally I would use the "[b]if and only if[/b]" definition, hence eliminating this ambiguity entirely while maintaining the integrity of the original definition. It could be rephrased as above as "any $A$ that contains $B$ must be called complete and any $A$ that does not contain $B$ cannot be called complete".

I see no reason for this situation to be different to that of theorems. Why not have a consistent usage of "if" and "iff"?

To make myself clear, my interest is about knowing proper usage of [b]if[/b] and not to make the original statement unambiguous. Here is the reason why I raise this question. Many times I encounter statements using only [b]if[/b] to describe definitions. Although it is one directional, the author seems to assume the other direction also to be valid in subsequent theorems (Yes, the author could have strengthen his definition like what you just gave to avoid ambiguity). It has led me to believe that [b]if[/b] and [b]if and only if[/b] can be used interchangebly in some cases.

[quote="Ilthigore"]there could exist an entity $A$ that does not contain $B$, but could still be said to be complete.[/quote]
I think you have a point there. The statement

$A$ is complete [b]if[/b] it contains $B$

leaves out some [i]possibility[/i] that $A$ can be complete if it contains $C$ (with or without $B$). However the other statement

$A$ is complete [b]if and only if[/b] it contains $B$

does not leave out such possibility.


IN THE NAME OF GOD

hiiiiiiii

If a man can walk 4km in 1 hour and his wife can walk 5km in 1 hour, how many minutes will it take them to meet if they begin walking at the same time at opposite ends of a 12km road, and have agreed to meet at the half way point?

May God help   you to all the best

SALAH

EDIT: Sorry I didnt see it said "halfway there" .

This is not the correct solution
is there another solution??????

regards

salah

[hide]The man will make it to the halfway point in $ \frac {6 km}{4 \frac {km}{hr}} \equal{} 1.5 hr \equal{} 90 minutes.$
His wife will make it in $ \frac {6 km}{5 \frac {km}{hr}} \equal{} 1.2 hr \equal{} 72 minutes.$

The wife will make it first, and spend 18 minutes waiting for her husband. The answer is thus $ 90 minutes$. Is that correct?[/hide]

Silly miscalculation, I edited.

Yes, I agree with -Elixer-'s answer of $ 90$ minutes (except that the wife makes it in $ 72$ minutes (your $ 1.2$ hours) and waits for $ 18$ minutes).

I did it like this:
[hide]
First I found the rates that each person can walk 1 km at.
Wife: 5 km in an hour. 60 minutes ÷ 5 km = 12 minutes to walk 1 km.
Man: 4 km in an hour. 60 minutes ÷ 4 km = 15 min to walk 1 km.
The wife will get to the halfway point in 12*6 = 72 min, and the man will get there in 15*6 = 90 minutes.
Therefore, they will meet at the halfway point in [b]90[/b] minutes, and the wife will just have to wait for 18 minutes.  :no: [/hide]


Τα σημεια $ K,L,M,N,J,O$ κινουνται εσωτερικα των τμηματων $ AB,BC,CD,DE,EZ$ ενος κανονικου εξαγωνου $ ABCDEZ$ με περιμετρο $ 2$.
Nα βρεθει το ελαχιστο της περιμέτρου του εξαγώνου $ KLMNJO$

Ωραίο πρόβλημα!

Η απάντηση είναι $ \sqrt{3}$, σωστά?

Η γνωστή τεχνική του ότι η ευθεία είναι μικρότερη απ' την τεθλασμένη πάντα πιάνει σε τέτοιου είδους προβλήματα :wink:

σωστος αν εχεις χρονο καντο αναλυτικα

Μια περιγραφικη λυση για οποιον την θελει.
Απο Ν.συνιμητονων ....εχουμε οτι για την πλευρα του νεου εξαγωνου ισχυει $ \sqrt{x^2+y^2+xy}\geq{\sqrt{3}/2(x+y)}$ .
Εφαρμογη για ολες τις πλευρες , προσθεση κατα μελη.....και παιρνουμε $ \sqrt{3}$ οπως αναφερθηκε απ τον bilot4

Συγγνώμη που δεν απάντησα νωρίτερα. Είναι πολύ ωραία η λύση σας:wink:
Εγώ το έκανα τελείως αλλίως (καθαρά γεωμετρική λύση)... Εχω συνάψει το σχήμα και νομίζω ότι δεν θέλει πολύ περαιτέρω εξήγηση... Απλά έχω πάρει $ CD \equal{} CD', CM \equal{} CM', \angle{BCD'} \equal{} 120$ και αντίστοιχα για τα άλλα. Αυτό μας δίνει εύκολα ότι $ LM \equal{} LM'$ κτλ. το οποίο σημαίνει ότι η περίμετρος ισούται με το $ KL \plus{} LM' \plus{} M'N' \plus{} N'O' \plus{} O'P' \plus{} P'K'$. Το τελευταίο όμως είναι μεγαλύτερο ή ίσο του $ KK' \equal{} AA'$ (ως παραλληλόγραμμο). Τελικώς εύκολα προκύπτει ότι $ AA' \equal{} 3AB \equal{} \sqrt {3}$.

Y.Γ. Θα έβαζα τη λύση νωρίτερα αλλά είχα ένα πρόβλημα με το συνάψω το σχήμα. Ευχαριστώ το Σιλουανό για τη βοήθειά του σε αυτό :)

Απλα εκπληκτικη λυση ............

[quote="xp"]Απλα εκπληκτικη λυση ............[/quote]

Ναι, πραγματικά υπέροχη λύση! :coolspeak:

Cheerio,

Durandal 1707

Σας ευχαριστώ πολύ  :)

Πράγματι πολύ καλή λύση  :)  Τώρα προσπαθείστε το γενικό πρόβλημα για ν-γωνο  :wink:

Eισαι ανησυχο πνευμα .Μια κουβέντα ειπαμε και το ψαξες.....Αν στο διαλλειμα 9-9.10 δεν εκανα λαθος εχουμε:
Aν $ S$ ειναι η περίμετρος το $ n$ γωνου ειναι $ S'_{min}\equal{}Scos(\frac{\pi}{n})$


Determine the number of perfect squares that there are between $ 40000$ and $ 640000$ that are multiples of $ 3, 4$ and $ 5$, simultaneously.

[hide]
$ 3|n^{2}\implies 3|n$
$ 4|n^{2}\implies 2|n$
$ 5|n^{2}\implies 5|n$
$ 3,4,5 | n^{2}\implies 3 \times 2 \times 5 | n \iff 30 | n$

Assuming between means exclusive, we want numbers between $ \sqrt{40000}=200$ and $ \sqrt{640000}=800$ that are multiples of $ 30$. We can transform our range by shifting it by multiples of $ 30$:

$ \#(200,800) \equiv \#(200-180, 800-180) \equiv \#(20, 620) \equiv \#(20, 20 \times 30+20)$

$ 20 < 30 \times 1 < 30 \times 20 < 30 \times 20+20$

Thus, there are $ 20$ such numbers.

[/hide]


From the mailing I received today from Tom Kilkelly:

1. 40 Sofia Math Circle (Bulgaria)
2T. 37 Lynbrook High School
3T. 37 Phillips Exeter Academy
4. 36 Illinois Math and Science Academy
5T. 35 Thomas Jefferson HS
5T. 35 Western Washington
7T. 34 Georgia ARML
7T. 34 Mongomery Blair HS
9T. 33 Colorado ARML
9T. 33 Long Beach Math Circle
9T. 33 AAST
12T. 32 New Mexico Math League
12T. 32 Phillips Academy
12T. 32 Western MASS A
12T. 32 Iowa City West HS
16T. 31 Naperville North HS
16T. 31 Wayzata HS
16T. 31 Daderyds Gymnasium (Sweeden)
16T. 31 Stuyvesant HS

35 is an [b]amazing[/b] score for us considering we didn't seriously answer the part worth 5 points  :D

[quote="Kent Merryfield"]There were two 5-point parts; #3 and #4e. Our 33 comes from getting a 1 out of 5 on #3 and 2 out of 5 on #4e, and all of the other points.

["Long Beach Math Circle" is a large fraction of the Southern California ARML team - as I assume that "Western Washington" is a large fraction of the Washington ARML team - right?][/quote]

Western Washington is a large fraction, but there are also a number of strong students who couldn't make it (3 hour drives and such). I think we lost a majority of points on #3; basically, nobody could do it and in the last 3 minutes we scrambled to write random numbers down, and we didn't even finish that.

[quote="t0rajir0u"][quote="paladin8"]basically, nobody could do it[/quote]

Probably because only two people were working on it until the last few minutes  :|

We should probably divvy up the work more efficiently next time  :)[/quote]
You and us both. That's our story too.

man...i wish my school would do these types of things...sadly, there is not enough interest

p.s. is there anywhere i can see the contest?

Grrrrr, I was almost sure we had closer to a perfect than that... But we counted the later parts more important than the early ones and we didn't have a great proof for 4e, so I guess that's why we got but a 34. :(

[quote="Kent Merryfield"]You live in Illinois, but I take it you don't go to either Naperville North or IMSA, right?[/quote]

yes...although my school is well-rounded, the math program only teaches basic proficiency, (my school is new trier...)

I would like to bring the ARML Power Contest and/or the Mandelbrot to my school by next year, but there might not be enough interest.


Hi,

You are given an undirected graph.Now the problem is to find number of loop less paths between two given nodes in this graph.

Are there any algorithms for solving this problem.

Are there any special cases where it can be solved ??

Thank You.
Abhilash I.

For small n (up to 20 something-ish) a simple 2^n*n DP would work. I can't think of anything for larger n though.

This paper might be of interest: http://maths.uq.edu.au/~kroese/ps/robkro_rev.pdf

Well, we would need to define some more specific rules for the program before we can create the algorithm. Can links be traveled only once? Once in each direction? An arbitrary number of passes (would frequently result in infinite paths)?

But regardless, the algorithm should be fairly simple. You create a stack or queue (same result) containing all the nodes in the graph, and begin by adding the first of the two source nodes. Then for each node in the stack, you add all the linked nodes to the stack/queue (or all the linked nodes that have not already been added, if you want each node to only be usable once). If you want to avoid back-and-forth (you probably do, otherwise there would almost always be infinite paths) then give the new node a parent (there may be more than one) which is the identifier of the old node which led to that node, and add the additional condition that new nodes must not be the parent of the node currently being processed. Finally, when the ending node is reached, increment a paths_found variable by one. (Depending on the rules of the system, you probably don't want to add the ending node to the path, but it would be neither correct nor incorrect to do so.) It's a lot like A* pathfinding - store nodes in a stack/queue, and when each node is processed add each adjacent node to the stack/queue if certain conditions are met.

You must have missed the 'loopless' criterion he mentioned ;)

Anyway, of course a simple traversal would give you the right answer, but thats easily going to be exponential. I'm pretty sure we're looking for an algorithm that would actually run in a limited amount of time (the brute force method won't work as high as the DP method will).

[quote="TripleM"]You must have missed the 'loopless' criterion he mentioned ;)[/quote]
Sorry - I did.

[quote="TripleM"]Anyway, of course a simple traversal would give you the right answer, but thats easily going to be exponential. I'm pretty sure we're looking for an algorithm that would actually run in a limited amount of time (the brute force method won't work as high as the DP method will).[/quote]
Well, the problem there is that there is no such algorithm. Given an arbitrary number of nodes and an arbitrary number of connections of arbitrary complexity, there is no method of finding the number of paths between two nodes without actually traversing the path. There are certain optimizations that can be made to avoid redundant calculations, but attempting the problem without traversal is akin to searching for an object before identifying it - it's simply a necessary part of processing the given input.

In any case, it wasn't stated or implied in the original post that a DP is what we're looking for. If someone asks how to find the shortest path between two points with arbitrary obstacles, it seems reasonable to point to A*, given the lack of equivalent alternatives. I don't see the difference here.

Can you actually prove its NP complete? There are very similar problems in graphs which do have polynomial-time solutions, so I wouldn't go saying that there is no algorithm without being very sure.
Of course, the link I gave above probably does prove it is NP complete. Then again, some exponential algorithms are far faster than others.

Anyway, he asked if there were special cases the problem could be solved on. The DP gives a way of solving it for small n which a brute force approach does not.

(edit - by the way, you said a shortest path algorithm with arbitrary obstacles should use A* - but of course, Dijkstra is a simple polynomial algorithm for that - you must have meant something else).

[quote](edit - by the way, you said a shortest path algorithm with arbitrary obstacles should use A* - but of course, Dijkstra is a simple polynomial algorithm for that - you must have meant something else).[/quote]
I said arbitrary obstacles, not arbitrary nodes and links. Sorry if I wasn't clear.

Hi,
Problem Statement:
Given an undirected graph find the number of edge disjoint paths between
two given nodes from that graph ?

Is there any polynomial algorithm  or is it NP-complete??

Thank You.
Abhilash I.


AMC
AMC 10
AMC 12
AIME
calculus
trigonometry
logarithms

From the AMC12 Discussion Thread:
[quote="DPopov"]AMC 10 is harder to qualify with in my opinion. There is greater stress on not making stupid mistakes.[/quote]
And I agree.

Is it time to turn AMC10/12 back into one, again?

Or, just phase out AMC10!

I may be influenced by the skill level of the AoPS members, but it seems like AMC12 is now appropriate for many middle schoolers, that is, as long as they are guided correctly. Also, by keeping some problems sufficiently challenging (certainly the 20s), it will remain appropriate for high schoolers as well.

No, no, no. Obviously, AoPS is nowhere near an accurate sample of AMC-takers. Most of my friends who took it did 10-15 problems on the 10 and would only have answered 8-10 on the 12. The 12 is far too advanced for most freshmen and below. With that few questions answered, you just get bored with the remaining time.

[quote="mysmartmouth"][quote="MathChauffeur"]
I may be influenced by the skill level of the AoPS members, but it seems like AMC12 is now appropriate for many middle schoolers, that is, as long as they are guided correctly. Also, by keeping some problems sufficiently challenging (certainly the 20s), it will remain appropriate for high schoolers as well.[/quote]

When I was in middle school, I could barely qualify for AIME via the AMC 10 (120.5).  So the AMC 12 would have been IMPOSSIBLE.[/quote]

Yeah. I qualified last year (8th grade) with a 124.

9th grade, took the 12, and did respectably. 118.5.

Comments, scores, and opinions on this forum are not representative of the AMC 10.

About 4000 teachers believe the AMC 10 is appropriate for at least 100,000 students.  Scores, averages,  number of problems answered, and the number of students participating, along with teacher comments we receive all convince the Committee on the American Mathematics Competitions that the AMC 10 has a place in our contest sequence.

Read the letter from the AMC 10 Chair, Doug Faires, in the 2006 Summary of Results and Awards, page 12 and 13.  (All contest managers from last year received this book in June or July 2006.)

Steve Dunbar
AMC Director

At our school, we've given only the AMC-12, to all grades 9-12 for the past five years.

Last year our top three freshman scores were 102, 101.5, and 95.  Our top three sophomore scores last year were 122, 114.5, and 103.

We will probably give the AMC-10 and AMC-12 next year due to the change in scoring.  Under the old system, a freshman could find eleven problem she could do correctly, do them, and qualify for the AIME.  Now, she'd have to find fourteen, which may be next to impossible for a student who hasn't been practicing on AMC tests to know the style of questions.

Ideally, I'd still give the AMC-12 exclusively, as I believe it sets the bar higher for the underclassmen.  I don't like the idea of giving an intentionally easier test to sophomores (many of them are in pre-calculus at our school) just because they're a year younger than the juniors.  But, I want to get them exposed to the AIME under contest environments (even when freshmen get one or two right, it's still an accomplishment to be proud of.)

TDP.

Which contest (AMC 10 or AMC 12) to give to a student is a question best left to the discretion of the teacher who knows the student well.    If you have sophomores in pre-calculus, then maybe the AMC 12 is the proper contest to administer to those students.  Your experience and judgment is important in helping students.  That's wh we often refer to the teachers and contest managers as coaches.

I don't think the word "easier" is the right word to distinguish between the AMC 10 and the AMC 12.  The AMC 10 contest does not contain trigonometry, logarithms, complex numbers, some advanced algebra of polynomials and their roots, and does not use function notation.  As we say in the brochures, the "AMC 10 covers material normally associated with grades 9 and 10."  For a student who has covered only that material, the AMC 10 may not be easier but it certainly is less advanced.

I have written comments on the Certification Form from many teachers who appreciate the distinction between the AMC 10 and the AMC 12.  The number who have written those comments is many more than have written in this forum.  Again, I don't think you can take the comments of this forum as representative.  Remember, your experience may be different, and our rules accommodate both experiences.

Steve Dunbar
AMC Director

I guess it depends on the purpose of the AMC-10 test.

If the purpose of the test is to determine participation on the AIME or USAMO, where trigonometry, logarithms, complex numbers, advanced algebra of polynomials and their roots, and functional notation all appear, then no, I don't think it's fulfilling it's purpose. If a student designs to take the AIME or USAMO, I think they should take the AMC-12 test, even if they are a freshman, and even if they haven't had the requisite mathematics.  Exposure to the AMC-12 will be more of help experience-wise than the AMC-10 for those level of exams.

The AIME floor value last year was eight.  I find it difficult to believe that a student who could get eight right on the AIME would not be able to get at least eleven right on the AMC-12 last year.  (Per http://www.unl.edu/amc/b-registration/b1-archive/2003-2004/faq2003-04.html, there are about a dozen problems in common between AMC-10 and AMC-12)

If, on the other hand, the purpose of the AMC-10 test is to challenge students with "outside the box" mathematics at a 9th or 10th grade "appropriateness level" (read, Algebra I, II, and Geometry) then, yes it is fulfilling it's purpose.

I'm not saying these goals need be mutually exclusive, but it should be noted that they are not very similar.  Obviously, students participating in this forum are going to be more desirous of an AIME qualification, or even an USAMO qualification, and so make comments to that effect.

And my choice of "easier" in the prior post may not have been appropriate.  Less rigorous wouldn't be appropriate either, but I do want an exam that will push beyond currently known content.  If one is able to reach the stars, one will be content with that.

TDP.


geometry algebra polynomial conics inequalities function trigonometry

Since I finished summer school Geometry, I would now like to know what material will be covered in Algebra 2......Anything will be helpful :)

Matrices
Polynomial Factorization (Just a little harder then algebra I?)
Binomial Theorem
Non-Linear Systems of Equations
Conics
Synthetic Division/Algebraic Long Division (Think that's what it is called)
Inequalities/Systems of Inequalities
Exponential Function
Rational Functions
Logarithms/Natural Log
A little basic trig

Those are just a few of the topics...


take $x=y=z=\frac{1}{2}$ and u get a contradiction

Sorry,I posted carelessly.Let $x$,$y$,$z$>0 and $\sum x$$\leq$ $\frac{3}{2}$.Prove that
$2\sum x-\sum \frac{(4y^2z^2-1)x}{y^2+z^2-1}$ $\leq$ $\frac{3}{4}$


inequalities
calculus
inequalities proposed

After applying Rolle's theorem to $(x-a)(x-b)(x-c)(x-d)$ we get an inequality in $u,v,w$ s.t. $u+v+w=0$. (technique similar with the one used for proving Schur for several variables) I solved that through $w=-(u+v)$; then I considered $LHS-RHS$ as $f(u)$; $f' \left( - \frac{v}{2} \right)=0$; the rest is easy. :D

Can anyone find a nice solution for the inequality with $u+v+w=0$:

\[ (uv+vw+wu)^2 + 3 \geq 6uvw \ ?  \]

Perfect_Radio, can you post your solution in detail? :)  ;)

We have $P(x) = x^4+(ab+ac+ad+bc+bd+cd)x^2-x(abc+abd+acd+bcd)+abcd$, so $P'(x) = 4x^3 + 2(ab+ac+ad+bc+bd+cd)x-(abc+abd+acd+bcd)$.

By Rolle's Theorem, there are $u,v,w \in \mathbb{R}$ s.t. $P'(x) = 4(x-u)(x-v)(x-w)$. (in case we have $a=b$ for example, then $a$ is also a root $P'(x)$; else we apply the Rolle argument)

Now equate the two formulas and substitute in the inequality. As I said in my previous post, my solution for the $u+v+w=0$ inequality is pretty ugly (you can do it yourself). Try to solve this inequality. Maybe you'll find a nice solution.

[quote="perfect_radio"]

Can anyone find a nice solution for the inequality with $u+v+w=0$:

\[ (uv+vw+wu)^2 + 3 \geq 6uvw \ ?  \][/quote]
You can easy see that we only need to prove this ineq.
 $x^2y^2+(x^2+y^2)(x+y)^2+3\geq 6xy(x+y)$wher $x,y\geq 0$. :)
Let's use Caush's ineq. for 12 numbers
$x^2y^2+8*\frac{(x^2+y^2)(x+y)^2}{8}+1+1+1\geq \sqrt[12]{x^2y^2*(\frac{(x^2+y^2)(x+y)^2}{8})^8}\geq 6xy(x+y)$
Mybe not wery nice :(  but not wery ugly too :D .

[quote="nttu"][quote="perfect_radio"]Can anyone find a nice solution for the inequality with $u+v+w=0$:

\[ (uv+vw+wu)^2 + 3 \geq 6uvw \ ?  \][/quote]
  Suppose $u \geq 0 \geq v \geq w$
 Put $m=-v$ ; $n=-w$
 So $u =m+n$ (1)
 Replace (1) into the ineq , we need to prove :
 $(m^n+n^2+mn)^2 +3 \geq 6mn(m+n)$ , where $n \geq m \geq 0$
  ;)  Maybe this ineq can be solved by AM-GM , Cauchy ,... but I used calculus to solve this ineq . It's not nice but easy .[/quote]
 :rotfl:  :blush: I think you have a typo ther must be $(m^2+n^2+mn)^2 +3 \geq 6mn(m+n)$  :P  :D ,and  for this ineq. see my laste post.(change m,n->x,y :rotfl:  :oops: )

:oops_sign:
$ d\equal{}\minus{}a\minus{}b\minus{}c$
$ abc\plus{}abd\plus{}acd\plus{}cbd\equal{}abc\minus{}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ac)\equal{}\minus{}(\minus{}abc\plus{}\sum_{sym}^{}a^2b\plus{}3abc)\equal{}\minus{}(\sum_{sym}^{}a^2b\plus{}2abc)\equal{}\minus{}(a\plus{}b)(a\plus{}c)(b\plus{}c)$
$ ab\plus{}ac\plus{}bc\plus{}ad\plus{}cd\plus{}bd\equal{}ab\plus{}bc\plus{}ac\minus{}(a\plus{}b\plus{}c)^2\equal{}\minus{}\frac{1}{2}((a\plus{}b)^2\plus{}(a\plus{}c)^2\plus{}(b\plus{}c)^2)$
Let  $ x\equal{}a\plus{}b$
       $ y\equal{}a\plus{}c$
        $ z\equal{}b\plus{}c$
We need $ \frac{1}{4}(x^2\plus{}y^2\plus{}z^2)^2\plus{}12\ge \minus{}6xyz$
$ x^2\plus{}y^2\plus{}z^2\ge 3|xyz|^{\frac{2}{3}}$
So, $ \frac{1}{4}(x^2\plus{}y^2\plus{}z^2)^2\plus{}12\ge$ $ \frac{9}{4}|xyz|^{\frac{4}{3}}\plus{}12\equal{}\frac{3}{4}|xyz|^{\frac{4}{3}}\plus{}\frac{3}{4}|xyz|^{\frac{4}{3}}\plus{}\frac{3}{4}|xyz|^{\frac{4}{3}}\plus{}12\ge$ $ 4\times(\frac{12\times3^3}{4^3})^{\frac{1}{4}}|xyz|\equal{}6|xyz|\ge\minus{}6xyz$

As a three-year late side note, we can prove the inequality in $ u,v,w$ by using the fact $ x^3 \plus{} ax \plus{} b$ has three real roots iff
\[ 4 a^3 \plus{} 27 b^2 \leq 0 .
\]

[quote="litbon"][quote="Valentin Vornicu"]Let $ a,b,c,d$ be real numbers with sum 0. Prove the inequality:
\[ (ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd)^2 \plus{} 12\geq 6(abc \plus{} abd \plus{} acd \plus{} bcd).\]
[/quote]
My soloution is here [url]http://vimf.freeforums.org/m-t-s-bdt-olympiad-t162-20.html[/url][/quote]

Next time just post the solution here. it seems your forum is out of service now..

Can anyone find a nice solution for the inequality with $u+v+w=0$:

\[ (uv+vw+wu)^2 + 3 \geq 6uvw \ ?  \][/quote]

Substituting $w=-(u+v)$, we have the inequality
         $(u^2+uv+v^2)^2+3\geq6uv(u+v)$
After writing $u=z+t, v=z-t$, the inequality becomes
$(3z^2+t^2)^2+ 3 \geq 12z(z^2-t^2)$
As $t^2 \geq 0$ , it is sufficient to prove that
$(3z^2)^2+3 \geq 12z^3$, which is obvious from AM-GM

Does anyone find the a,b,c,d when LHS and RHS are equal?

hi can any one help me with this problem i ve seen seen the aops solution but didnt understand
we have 0<=x,y,z<=1 prove that
(x/(y+z+1))+(y/(x+z+1))+(z/(x+y+1))+(1-x)(1-y)(1-z)<=1

[quote=Valentin Vornicu]Let $ a,b,c,d$ be real numbers with sum 0. Prove the inequality:
\[ (ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd)^2 \plus{} 12\geq 6(abc \plus{} abd \plus{} acd \plus{} bcd).
\][/quote]

Let $4u=a+b+c+d$, $6v^2=ab+cd+ac+bd+ad+bc$ ($v^2$ can be negative), $4w^3=abc+bcd+cda+dab$ and $P(x)=(x-a)(x-b)(x-c)(x-d)=x^4-4ux^3+6v^2x^2-4w^3x+abcd$.
Thus by Rolle's theorem $P'(x)=4(x^3-3ux^2+3v^2x-w^3)$ has a three real roots, which gives $$\Delta=6912(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=-6912(4v^6+w^6)\geq 0\text{.}$$
So we need to prove
$$3v^4+1\geq 2w^3$$
,after squaring it's equivalent to
$$9v^8+6v^4+1\geq 4w^6$$
or
$$9v^8+6v^4-4w^6+1\geq 9v^8+16v^6+6v^4+1=(3v^2+1)^2(9v^4-2v^2+1)\geq 0$$
, which is true.


What are some of the scholarship competitions in MAth??
(LIke essays/researches)

Try searching on a website such as Fastweb.com or Collegeboard's scholarship search. I doubt if there's math scholarship competitions out there, but it's worth a try.


number theory
prime numbers
number theory unsolved

Find all prime numbers $p$ which satisfy the following condition: For any prime $q < p$, if $p = kq + r, 0 \leq r < q$, there does not exist an integer $q > 1$ such that $a^{2} \mid r$.

I guess this is supposed to be "...there does not exist an integer $ a>1$ such that $ a^2 | r$"?

I'm still new to this writing a proof thing, so I'd appreciate it if people could comment on how I could improve, thx

[hide]The only such p are $ 2$, $ 3$, $ 5$, $ 7$, and $ 13$.
Proof:
I will divide up our search in to two cases.

Case 1: p is a prime NOT in the form $ 3^k \plus{} 4$.
In this case, we are guaranteed that as long as $ p$ is greater than $ 5$, $ p\minus{}4$ will be divisible by a prime larger than $ 3$.
Thus we see that we can have $ p \equal{} kq \plus{} 2^2$, and $ q > 2^2$. This rules out any prime NOT of the form $ 3^k \plus{} 4$ that is greater than 5. By inspection we see that every prime less than or equal to 5 works, thus 2, 3, and 5 are numbers that we are looking for.

Case 2: p is a prime of the form $ 3^k \plus{} 4$.
In this case, let us consider $ k \ge 4$. Then $ p \minus{} 7^2 \equal{} 3^k \minus{} 45 \equal{} 9(3^{k\minus{}2} \minus{} 5)$. Since $ k \ge 4$, the expression is positive, and is divisible by the square of a prime. Trying cases for $ k \equal{} 1$ to $ k \equal{} 3$, we see that $ 7$ and $ 13$ are also numbers we are looking for.

Thus the numbers we are looking for are $ 2$, $ 3$, $ 5$, $ 7$, and $ 13$.
[/hide]

This is not a problem too difficult.
I will try to make a resolution in the shortest time(by the busy work :lol: )

math10 may have given up, so here's my solution.

We prove that if $ p > 13$ is prime, there is a prime $ q < p$ such that the least residue of $ p$ modulo $ q$ is not square-free, which is the property described in the problem. In fact, we prove a stronger statement: if $ p>13$ is odd (not necessarily prime), there is a prime $ q$ such that the least residue of $ p$ modulo $ q$ is either $ 4$, $ 8$ or $ 9$.

Suppose that some odd $ p>13$ is a counterexample: there's no prime $ q < p$ for which the remainder upon dividing $ p$ by $ q$ belongs to $ \{4,8,9\}$.

If there is a prime $ q > 3$ diving $ p\minus{}4$, then $ 4$ is the least residue of $ p$ modulo $ q$, which contradicts our assumption on $ p$. Since $ p$ is odd, $ p\minus{}4$ is also not divisible by $ 2$, so the only prime divisor of $ p\minus{}4$ must be $ 3$. In other words, $ p\minus{}4 \equal{} 3^k$ for some natural number $ k$.

Consider now $ p\minus{}8$. If it has any prime factor greater than $ 8$, we once again contradict our assumption on $ p$. Thus we may assume that all of its prime factors belong to $ \{2,3,5,7\}$. But $ 2$ is impossible since $ p\minus{}8$ is odd, and $ 3$ is impossible since $ p\minus{}4$ is divisible by $ 3$. Thus $ p\minus{}8$ is divisible only by $ 5$, $ 7$ or both.

Next consider $ p\minus{}9$. For the same reason as above, it too can only have prime factors among $ \{2,3,5,7\}$. Since $ p\minus{}4$ is divisible by $ 3$ and not by $ 5$, $ p\minus{}9$ is divisible by neither $ 3$ nor $ 5$. It may or may not be divisible by $ 7$, but we claim that in either case our previous friend $ p\minus{}8$ cannot be divisible by $ 7$. Here's why: If $ p\minus{}9$ is divisible by $ 7$, then $ p\minus{}8$ certainly isn't. If $ p\minus{}9$ is not divisible by $ 7$, then it must be a power of $ 2$ (since no other prime factors are available). But powers of $ 2$ can only leave remainders of $ 1, 2, 4$ upon division by $ 7$, so $ p\minus{}8$ which is one greater than a power of $ 2$ cannot be divisible by $ 7$.

Recall that the only possible prime factors of $ p\minus{}8$ were $ 5$ and $ 7$, and we just ruled out $ 7$. We may therefore conclude that $ p\minus{}8$ is a power of $ 5$: $ p\minus{}8 \equal{} 5^m$. Comparing this with $ p\minus{}4 \equal{} 3^k$ we obtain:

(*) $ 5^m \plus{} 4 \equal{} 3^k$

We conclude the proof by showing that (*) only has the trivial solution $ 5\plus{}4 \equal{} 3^2$, which corresponds to $ p\equal{}13$. Reducing mod $ 4$ we see that $ k$ must be even, so $ k \equal{} 2t$ for some $ t$. Now $ 5^m \equal{} 3^{2t} \minus{} 4 \equal{} (3^t\minus{}2)(3^t\plus{}2)$. But these factors cannot both be divisible by $ 5$ (they differ by $ 4$), so the only way their product can be a power of $ 5$ is if one of them is $ 1$ and the other is a power of $ 5$. This forces $ t\equal{}1$, $ k\equal{}2$ as required. QED

[quote=randomgraph]Case 1 seems fine, however:
[quote="Math4tots"]
Case 2: p is a prime of the form $ 3^k \plus{} 4$.
In this case, let us consider $ k \ge 4$. Then $ p \minus{} 7^2 \equal{} 3^k \minus{} 45 \equal{} 9(3^{k \minus{} 2} \minus{} 5)$. Since $ k \ge 4$, the expression is positive, and is divisible by the square of a prime. Trying cases for $ k \equal{} 1$ to $ k \equal{} 3$, we see that $ 7$ and $ 13$ are also numbers we are looking for.
[/quote]
This is not very clear to me. What do you take $ q$ and $ r$ to be in this case? If not the proof itself, at least the exposition seems to be flawed.[/quote]

This finish is somewhat similar.
Take $r=5^2$, $q =$ a prime that divides $3^{k \minus{} 1} \minus{} 7 = \frac{p-5^2}{3}$ that is greater than $5$. Since $k \geq 4$ the number $3^{k-1}-7 >1$ and $3 \not |5$, so it is not a power of 3, and if $5|3^{k-1}-7$ then $5|p$, contradiction. Thus if $q$ does not exist, then $3^{k-1}-7$ is a power of 2. However $3^{k-1}-7 \equiv 4,2 (mod 8),$ so $3^{k-1}-7 \leq 4 \implies k<4$, contradiction. Now finish as in the quote by considering cases $k\in [1,3]\cup \mathbb{N}$.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G \equal{} \langle g_1, \ldots, g_r \rangle$ be a finitely generated (not necessarily finite) group, en let $ N$ be an abelian normal subgroup of $ G$. Show that every element of $ [G,N]$ can be written in the form

\[ g \equal{} [g_1,n_1]\ldots [g_r,n_r]\]

for certain elements $ n_1, \ldots, n_r \in N$. Note that $ g_1, \ldots, g_r$ are fixed.

[quote="Jan"]Let $ G \equal{} \langle g_1, \ldots, g_r \rangle$ be a finitely generated (not necessarily finite) group, en let $ N$ be an abelian normal subgroup of $ G$. Show that every element of $ [G,N]$ can be written in the form
\[ g \equal{} [g_1,n_1]\ldots [g_r,n_r]
\]
for certain elements $ n_1, \ldots, n_r \in N$. Note that $ g_1, \ldots, g_r$ are fixed.[/quote]

Let $ [a,b] \equal{} aba^{ \minus{} 1}b^{ \minus{} 1}.$ Forget for now that G is finitely generated. Show that $ [a_1a_2,n_1] \equal{} [a_1,a_2n_1a_2^{ \minus{} 1}][a_2,n_1],$ and $ [a_1, n_1][a_1,n_2] \equal{} [a_1,n_1n_2],$ for all $ a_i \in G, n_i \in N.$

Now your problem should be easy.


Gauss
induction
modular arithmetic
algebra
polynomial
pen
number theory

Didn't you mean $x^2+y^2+1 \equiv 0$ $mod$ $n$ ?

[quote="amir2"]prove that for every natural $n$

$x^2+y^2+1 \equiv 0$

has solution.[/quote]

Well it's a very nice question. I think the result should be

Problem: Let $n$ be a positive integer. Prove that $4\nmid n$ if and  only if there exist $x,y$ integers so that $n\mid x^2+y^2+1$

Here's my very very ugly solution:

If $4\mid n$, as a comment of al. M.V , no such pair $(x,y)$ exist!

Now let me consider the case $4\nmid n$ we will so that $n|x^2+y^2+1$ for some $x,y$.

1) $n=p$ where $p$ is an odd prime : then $p$ or $2p$ is  not in the form $4^k(8r+7)$. So by Gauss Theorem $p$ or $2p$ can be writen in the form $x^2+y^2+z^2$. Of cause there one of $x,y,z$; namely $z$ is relative to $p$ ie... $\gcd(p,z)=1$. By muliple the conveser of $z$ modulo $p$ we get the problem is true in that case.

2) $n=p^a$ for any $a$ is positive integer. Now use induction on $a$; just put $x_{a+1}=x_a+tp^a$ for $t$ is integer we get: the problem is true in the case

3) if $n=2$ then chose $x=0$ and $y=z=1$ so problem is true in that case

4) At last use Chines Remainder Theorem  so if $4\nmid n$ then such pair $(x,y)$ exists.

It's ugly but i hope it's clear  :D

It's not ugly, but you can avoid aplying Gauss' Theorem. If we want to show that we can find $x,y$ s.t. $p|x^2+y^2+1$, it suffices to notice that the sets $\{x^2\pmod p|x\in\overline{0,p-1}\}$ and $\{-1-y^2\pmod p|y\in\overline{0,p-1}\}$ together have $p+1$ elements, so they must have an element in common.

In LIV Polish Math Olympiad we had a problem:
Prove that for every prime $p>3$ there exist integers $x,y,k$ with $0<2k<p$ and $kp+3=x^2+y^2$.

[quote="bomb"]All you need is Chevalley's theorem and you are done.

Bomb[/quote]

Can somebody supply the details of this? We have a degree 2 polynomial in 2 variables, so we can't directly apply Chevalley's theorem. Also, how do you get rid of prime powers? (doesn't Chevalley only hold over finite fields with characteristic p?)

An other problem
Prove that for all $ (a,p)\equal{}(b,p)\equal{}1$  then the equation
$ ax^{2}\plus{}by^{2}\plus{}c (mod p)\equal{}0$ has solution.

(This answers one of my questions above)

If you allow infinite solutions, Chevalley's theorem will now kill it $ g(y_{0},y_{1},y_{2})\equal{}ay_{1}^{2}\plus{}by_{2}^{2}\plus{}cy_{0}^{2}\equal{}0$ has more variables than its degree so by Chevalley there exists a notrivial zero $ (p,q,r)\not\equal{} (0,0,0)$. However, if we wish to only consider finite solutions, we have $ p\not\equal{}0$ which we cannot get by Chevalley.


$ ABCD$ is  circumscribed quadrilateral .$ [AS]\perp [AB]$ and $ [AS\cap[BO \equal{} {M}$
$ [AQ]\perp[AD]$ and $ [AQ\cap[DO \equal{} {N}$
$ \implies$ $ [MN]\perp[AC]$ show that

See in [b]Darij Grinberg[/b]'s article-theorem 4:
http://www.cip.ifi.lmu.de/~grinberg/CircumRev.pdf

(Sorry, but I was not able to see the link, I have some problems with adobe).

Take $ NE \perp CD, MF \perp BC $, with E and F on CD, respectively F.
By symmetry, $ AN = NE, AM = MF $ ( 1 ) and ( 2 ). Next, call X, Y, Z, W the contacts with the incircle of the sides AB, BC, CD, AD.
Again by symmetry, AX = FY, ZE = AW; from AX = AW we get ZE = FY, which makes CE = CF.
But, to prove the perpendicularity, we have to prove, by Carnot lemma, $ NC^2 - AN^2 = MC^2 - AM^2 $ ; from the relations (1) & (2) we must show $ CF = CE $, which is true.

Best regards,
sunken rock


geometry
circumcircle
conics
projective geometry
power of a point
radical axis
geometry proposed

This is a small beautiful problem, which was created by treegoner and me a few monthes ago:

Consider a triangle $ABC$ and two points  $P, Q$ in it's plane. Let $A_{1},B_{1},C_{1}$ and $A_{2},B_{2},C_{2}$ be cevians in our triangle (intersections of $AP,BP,CP,AQ,BQ,CQ$ with opposite sides). Denote $U,V,W$ second intersections of circles $(AA_{1}A_{2}), (BB_{1}B_{2}), (CC_{1}C_{2})$ with circumcircle $(ABC)$. Let $X$ be point of the intersection of $AU$ with $BC$, similarly $Y$ for $BV$ and $AC$ and $Z$ for $CW$ and $AB$. Prove that $X,Y,Z$ are collinear.

A short note to this problem.

Actually this problem is still true if you replace the condition $A_{1}, A_{2}, B_{1}, B_{2}, C_{1},C_{2}$  are the traces of $P$ and $Q$ by the codition 6 points $A_{1}, A_{2}, B_{1}, B_{2}, C_{1},C_{2}$ lie on a conic.

I'll prove the generalization of this problem ( $ A_1, A_2, B_1, B_2, C_1, C_2 $ lie on a conic)

My solution:

[b]Lemma:[/b]

Let $ D, D' $ be two point on $ BC $ .
Let $ (w) $ be the circle pass through $ B $ and $ C $ .
Let $ (k) $ be the circle pass through $ D $ and $ D' $ .
Let $ M $ be the intersection of $ BC $ and the radical axis of $ \{ (w) , (k) \} $ .

Then $ \frac {MB}{MC}=\frac {BD}{DC} \cdot \frac{BD'}{D'C} $

Proof of the lemma:

Since $ MB \cdot (MB+BD+DC)=MB \cdot MC=MD \cdot MD'=(MB+BD) \cdot (MB+BD') $ ,

so we get $ MB=\frac{BD \cdot BD'}{DC-BD'} $ ,

hence

$ \frac {MB}{MC}=\frac {MB}{MB+BD+DC}=\frac {BD \cdot BD'}{BD \cdot BD'+(BD+DC)(DC-BD')} $

$ =\frac {BD \cdot BD'}{DC \cdot (BD+DC-BD')}=\frac {BD \cdot BD'}{DC \cdot (BC-BD')}=\frac {BD}{DC} \cdot \frac{BD'}{D'C} $ .

[b]Back to the main problem[/b]

Since  $ A_1, A_2, B_1, B_2, C_1, C_2 $ lie on a conic ,

so from Carnot theorem we get $ \frac{A_1C}{BA_1} \cdot \frac {A_2C}{BA_2} \cdot \frac {B_1A}{CB_1} \cdot \frac{B_2A}{CB_2} \cdot \frac{C_1B}{AC_1} \cdot \frac{C_2B}{AC_2}=1 $ . ... $ (1) $

From the lemma we get

$ \frac {XB}{XC}=\frac{BA_1}{A_1C} \cdot \frac{BA_2}{A_2C} $ ... $ (2) $

$ \frac {YC}{YA}=\frac{CB_1}{B_1A} \cdot \frac{CB_2}{B_2A} $ ... $ (3) $

$ \frac {ZA}{ZB}=\frac{AC_1}{C_1B} \cdot \frac{AC_2}{C_2B} $ ... $ (4) $

so combine $ (1),(2),(3),(4) $ we get $ X, Y, Z $ are collinear .

Q.E.D
____________________________________________________________
[b]Remark[/b]:

For the original problem ($ AA_1, BB_1, CC_1 $ are concurrent at $ P $, $ AA_2, BB_2, CC_2 $ are concurrent at $ Q $):
From the proof above we get $ X, Y, Z $ lie on the trilinear polar of the barycentric product of $ P \cdot Q $ :) .

[b]Another proof of the lemma :[/b]

Since $ M $ is the center of the involution determined by $ \{ (B,C),(D,D') \} $ ,

so we get $ (M, D; B, C)=( \infty , D'; C, B ) \Longrightarrow \frac{MB}{MC}=(\frac {BD}{DC})(\frac {BD'}{D'C}) $ .

[b]More general:[/b] A conic $\mathcal{C}$ cuts $BC,CA,AB$ at the pairs of points $\{A_1,A_2\},$ $\{B_1,B_2 \},$ $\{C_1,C_2 \}.$ $J,K$ are two arbitrary points on the plane and $\mathcal{K}$ is the conic through $A,B,C,J,K.$ The conic $\mathcal{K}_A$ through $A,A_1,A_2,J,K$ cuts $\mathcal{K}$ again at $U$ and $AU$ cuts $BC$ at $X.$ The points $Y \in CA$ and $Z \in AB$ are defined similarly. Then $X,Y,Z$ are collinear.

Let $JK$ cut $BC,CA,AB$ at $D,E,F.$ By Desargues theorem, the conics $\mathcal{K},$ $\mathcal{K}_A$ and $AU \cup JK$ of the pencil $\{A,J,K,U \}$ form an involution on the line $BC$ $\Longrightarrow$ $(D,A_1,B,C)=(X,A_2,C,B)$ $\Longrightarrow$

$\frac{DB}{DC} \cdot \frac{A_1C}{A_1B}=\frac{XC}{XB} \cdot \frac{A_2B}{A_2C}$ $\Longrightarrow \frac{XC}{XB}=\frac{DB}{DC} \cdot \frac{A_1C}{A_1B} \cdot \frac{A_2C}{A_2B}.$

Multiplying the cyclic expressions together gives the desired result.

P.S. Another proof, though not synthetic, is working on $\mathbb{C}^2$ and considering any projective transformation taking $J,K$ into the umbilics. This will produce the configuration of the original problem, then proceed as TelvCohl did.


geometry
function
quadratics
circumcircle
trigonometry
trig identities
Law of Sines

let  ABC be a regular triangle and P is a arbitary point in it the perpendicular of P meet the sides AB,AC,BC at
E,F,G repectively prove that
S(PAF)+S(PCG)+S(PBE) is not depended to P .

** S(X)=the area of X.

Let $AP'$ be a cevian containing $P$, then $S(PAF)+S(PCG)+S(PBE)$  depends lineary on position of $P$ on $AP'$. It is easy to see that the sum is $S(ABC)/2$ is $P$ coincides with $A$, more the sum is $S(ABC)/2$ for $P=P'$ either. Indeed, the sum for $P'$ also lineary depends on position $P'$ on $BC$, and it equal to $S(ABC)/2$ for both $B$ and $C$. But linear function take the same values iff it is constant.

We could do something like this:

Move $P$ s.t. the length $PG$ is fixed. If you denote by $x$ the distance $BG$, then $S[PEB]+S[PGC]+S[PFA]$ is a quadratic function of $x$. By eventually employing the same argument again, we can show that it equals half the area of $ABC$ in three distinct points (when $G=B,G=C$, and $G=$the midpoint of $BC$, for example), which means that it must be constant.

Edit: sorry, I hadn't seen Myth's post when I started typing mine :)

Anyway, my solution contains a mistake. Since, indeed, there is quadratic dependence, not linear. And we relly need to examine the sum at three points, not only in the end points.

There is also a more canonical solution. But first I rename the points to have symmetric notations:

[color=blue][b]Initial problem.[/b] Let P be a point inside an equilateral triangle ABC, and let X, Y, Z be the orthogonal projections of this point P on the sides BC, CA, AB of triangle ABC. Prove that $\left|PAY\right|+\left|PBZ\right|+\left|PCX\right|=\frac{\left|ABC\right|}{2}$, where the abbreviation $\left|P_1P_2...P_n\right|$ denotes the (non-directed) area of a polygon $P_1P_2...P_n$.[/color]

[i]Solution.[/i] Since < AYP = 90° and < AZP = 90°, the points Y and Z lie on the circle with diameter AP. Hence, the circumcircle of triangle YAZ has the segment AP as diameter. Thus, by the extended law of sines, $YZ=AP\cdot\sin\measuredangle YAZ$. But since the triangle ABC is equilateral, we have < CAB = 60°; in other words, < YAZ = 60°. Hence, $YZ=AP\cdot\sin 60^{\circ}=AP\cdot\frac{\sqrt3}{2}$. Thus, $YZ^2=\left(AP\cdot\frac{\sqrt3}{2}\right)^2=AP^2\cdot\frac34$, so that $AP^2=\frac43\cdot YZ^2$.

On the other hand, since the points Y and Z lie on the circle with diameter AP, the quadrilateral AYPZ is cyclic. Hence, < YPZ = 180° - < YAZ = 180° - 60° = 120°, so that $\cos\measuredangle YPZ=\cos 120^{\circ}=-\frac12$. By the cosine law in triangle YPZ, we have

$YZ^2=PY^2+PZ^2-2\cdot PY\cdot PZ\cdot\cos\measuredangle YPZ=PY^2+PZ^2-2\cdot PY\cdot PZ\cdot\left(-\frac12\right)$
$=PY^2+PZ^2+PY\cdot PZ$.

Hence,

$YZ^2\cdot\left(PY-PZ\right)=\left(PY^2+PZ^2+PY\cdot PZ\right)\cdot\left(PY-PZ\right)=PY^3-PZ^3$.

Thus,

$AP^2\cdot PY-AP^2\cdot PZ=AP^2\cdot\left(PY-PZ\right)=\frac43\cdot YZ^2\cdot\left(PY-PZ\right)$
$=\frac43\cdot\left(PY^3-PZ^3\right)$.

Similarly, $BP^2\cdot PZ-BP^2\cdot PX=\frac43\cdot\left(PZ^3-PX^3\right)$ and $CP^2\cdot PX-CP^2\cdot PY=\frac43\cdot\left(PX^3-PY^3\right)$.

Hence,

$\left(CP^2-BP^2\right)\cdot PX+\left(AP^2-CP^2\right)\cdot PY+\left(BP^2-AP^2\right)\cdot PZ$
$=\left(CP^2\cdot PX-BP^2\cdot PX\right)+\left(AP^2\cdot PY-CP^2\cdot PY\right)+\left(BP^2\cdot PZ-AP^2\cdot PZ\right)$
$=\left(AP^2\cdot PY-AP^2\cdot PZ\right)+\left(BP^2\cdot PZ-BP^2\cdot PX\right)+\left(CP^2\cdot PX-CP^2\cdot PY\right)$
$=\frac43\cdot\left(PY^3-PZ^3\right)+\frac43\cdot\left(PZ^3-PX^3\right)+\frac43\cdot\left(PX^3-PY^3\right)=0$.

But since $PX\perp BC$, the triangles CXP and BXP are right-angled, and thus, by Pythagoras, $CP^2=CX^2+PX^2$ and $BP^2=BX^2+PX^2$. Hence, $CP^2-BP^2=CX^2-BX^2$. But $CX^2-BX^2=\left(CX+BX\right)\cdot\left(CX-BX\right)=BC\cdot\left(CX-BX\right)=a\cdot\left(CX-BX\right)$, where a = BC = CA = AB is the sidelength of the equilateral triangle ABC. Thus, $CP^2-BP^2=a\cdot\left(CX-BX\right)$. Similarly, $AP^2-CP^2=a\cdot\left(AY-CY\right)$ and $BP^2-AP^2=a\cdot\left(BZ-AZ\right)$. Consequently, the equation $\left(CP^2-BP^2\right)\cdot PX+\left(AP^2-CP^2\right)\cdot PY+\left(BP^2-AP^2\right)\cdot PZ=0$ proven above becomes

$a\cdot\left(CX-BX\right)\cdot PX+a\cdot\left(AY-CY\right)\cdot PY+a\cdot\left(BZ-AZ\right)\cdot PZ=0$.

Dividing this by a and multiplying this by $\frac12$, we get

$\frac12\cdot\left(CX-BX\right)\cdot PX+\frac12\cdot\left(AY-CY\right)\cdot PY+\frac12\cdot\left(BZ-AZ\right)\cdot PZ=0$.

This is equivalent to

$\left(\frac12\cdot CX\cdot PX-\frac12\cdot BX\cdot PX\right)+\left(\frac12\cdot AY\cdot PY-\frac12\cdot CY\cdot PY\right)+\left(\frac12\cdot BZ\cdot PZ-\frac12\cdot AZ\cdot PZ\right)$
$=0$.

But since the triangle PCX is right-angled, its area equals one half of the product of the two catets: $\left|PCX\right|=\frac12\cdot CX\cdot PX$. Similarly, $\left|PBX\right|=\frac12\cdot BX\cdot PX$, $\left|PAY\right|=\frac12\cdot AY\cdot PY$, $\left|PCY\right|=\frac12\cdot CY\cdot PY$, $\left|PBZ\right|=\frac12\cdot BZ\cdot PZ$ and $\left|PAZ\right|=\frac12\cdot AZ\cdot PZ$. Thus, the above equation becomes

$\left(\left|PCX\right|-\left|PBX\right|\right)+\left(\left|PAY\right|-\left|PCY\right|\right)+\left(\left|PBZ\right|-\left|PAZ\right|\right)=0$.

Bringing all terms with minus signs onto the right hand side, we get

$\left|PAY\right|+\left|PBZ\right|+\left|PCX\right|=\left|PCY\right|+\left|PAZ\right|+\left|PBX\right|$.

On the other hand,

$\left(\left|PAY\right|+\left|PBZ\right|+\left|PCX\right|\right)+\left(\left|PCY\right|+\left|PAZ\right|+\left|PBX\right|\right)=\left|ABC\right|$.

Thus,

$\left|PAY\right|+\left|PBZ\right|+\left|PCX\right|=\frac{\left|ABC\right|}{2}$,

and the problem is solved.

  Darij

Am I missing something (It's quite late here...) or this is a proof without words:

One important thing: my solution immediately gives proof for the general case of regular $n$-gon (we need to consider oriented areas in this case).


modular arithmetic
quadratics
floor function
logarithms

How many decimal digits are there in the number 2005^2005? (ans: 6621)

and

How many integers between 1000 and 9999 are perfect squares and have sum of digits equal to 14? (ans: 0)

For the second problem, I don't think it's even necessary to put in between $ 1000$ and $ 9999$, but instead for all natural numbers, because none exist.

Here's why,

We know that a number, say $ x^2$, is congruent to the sum of it's digits in $ \pmod{9}$. So with this we can see that we are required to have $ x^2\equiv 14 \equiv 5\pmod{9}$, but this is impossible because the quadratic residues in $ \pmod{9}$ are $ \{0,1,4,7\}$ therefore none exist.

[quote="Mango 070"]Thanks, could you please explain why a number, in (mod 9), is congruent to the sum of the number's digits and what the quadratic residue in (mod 9) thing you're talking about is.[/quote]

Proof that a number is congruent to the sum of its digits mod 9:
$ n \equal{} \overline{a_na_{n \minus{} 1}\cdots a_0} \equal{} \sum_{k \equal{} 0}^{n}10^ka_k\equiv \sum_{k \equal{} 0}^{n}a_k\bmod 9$ because $ 10\equiv 1\bmod 9$.
(If the $ n \equal{} \overline{a_na_{n \minus{} 1}\cdots a_0}$ notation is unfamiliar: it means that $ n$ is composed of $ n$ digits $ a_i\in\{0,\ldots,9\}$).

Quadratic residues are the possible values a square can take while working in a particular residue class (all the possible values of $ x^2\bmod 9$, in this example)

Edit: Oh, and here's a [hide="lame solution to #1"]
A good thing to know is that the number of digits in an integer $ n$ is $ \lfloor\log_{10}(n)\plus{}1\rfloor$ (try to prove this if you want)

$ \lfloor\log 2005^{2005}\plus{}1\rfloor\equal{}\lfloor 2005\log 2005\plus{}1\rfloor$

$ \log 2005$ is clearly between $ 3$ and $ 4$, closer to the $ 3$ side. In any case, $ \log 2005\approx \log 2000\equal{}\log 2\plus{}3$

This is kind of stupid, but $ \log 2^{10}\equal{}10\log 2\equal{}\log 1024\approx \log 1000\equal{}3$, so $ \log 2$ is $ 0.3$ish, so $ \log 2005$ is $ 3.3$ish.

So our answer is slightly above $ 6617.5$. I guess that's as close as I can get without a calculator...
[/hide]

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=315899[/url]


AMC-12 2000, problem 23

Professor Gamble buys a lottery ticket, which requires that he pick six different integers from 1 through 46, inclusive. He chooses his numbers so that the sum of the base-ten logarithms of the six numbers is an integer. It so happens that the integers on the winning ticket have the same property - the sum of the base-ten logarithms is an integer. What is the probability that Professor Gamble holds the winning ticket?

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

[hide]
If the sum of the logs of the numbers is an integer, that means that the product of the integers is a power of 10. So the numbers must only be the products of powers of 2 and 5, and in the overall product there must be the same number of 2s and 5s. So here are the numbers to choose from:

1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40

or,

$1, 2^1, 2^2, 2^3, 2^4, 2^5, 5^1, 5^2, 2^15^1, 2^25^1, 2^35^1$

So now let's just try to equal powers of 10.

$10^1$: not possible
$10^2$: not possible
$10^3$: not possible
$10^4$: $(1*2*4*5*10*25)$
$10^5$: $(1*4*5*10*20*25)$, $(1*2*5*10*25*40)$
$10^6$: $(1*5*10*20*25*40)$

I count four cases, so the probability is $(B) \frac{1}{4}$
[/hide]

I've been busy that I was unable to post full problems of this year but this one is posted.

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Gambling+Professor&t=73513

Next time (this is to everyone), please use search button.


Solve for $ y$: $ 4\plus{}2.3y \equal{} 1.7y \minus{} 20$

hello, we have given
$ 4\plus{}2.3y\equal{}1.7y\minus{}20$
adding $ \minus{}1.7y$ on both sides we get
$ 4\plus{}0.6y\equal{}\minus{}20$
adding $ \minus{}4$
we get
$ 0.6y\equal{}\minus{}24$ therefore $ y\equal{}\minus{}40$ is the solution.
Sonnhard.


$ x^2 \plus{} x \minus{}1\equal{} y$
$ y^2 \plus{} y \minus{}1\equal{} z$
$ z^2 \plus{} z \minus{}1\equal{} x$

Some one knows where this problem was proposed?

It is proposed for IMO from Bulgaria in 70's I think.

First equation is $ x(x\plus{}1) \equal{} y\plus{}1$
Write the second equation as $ y(y\plus{}1) \equal{} z \plus{} 1$ and substitute: $ yx(x\plus{}1) \equal{} z\plus{}1$.
Write the third equation as $ z(z\plus{}1) \equal{} x \plus{} 1$ and substitute: $ zyx(x\plus{}1) \equal{} x \plus{} 1$.
Then either $ xyz \equal{} 1$ or $ x \equal{} \minus{}1$. If $ x \equal{} \minus{}1$, $ (x,y,z) \equal{} (\minus{}1,\minus{}1,\minus{}1)$ is easy to show.

So suppose $ xyz \equal{} 1$. Add all the equations together to get $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$. By AM-GM, $ x^2 \plus{} y^2 \plus{} z^2 \leq 3\sqrt[3]{x^2y^2z^2} \equal{} 3$, with equality iff $ x^2 \equal{} y^2 \equal{} z^2$. So each of $ x,y,z$ is $ \pm 1$. We already have a solution if any are $ \minus{}1$, so $ (x,y,z) \equal{} (1,1,1)$ is the other solution.


Let $ ABC$ be a triangle with centroid $ G$ and $ A_1,B_1,C_1$  midpoints of the sides $ BC,CA,AB$. A paralel through $ A_1$ to $ BB_1$ intersects $ B_1C_1$  at $ F$. Prove that triangles $ ABC$ and $ FA_1A$ are similar if and only if quadrilateral $ AB_1GC_1$ is cyclic.

1. AB1GC1 is cyclic => by angle chasing $ \angle ABB_1\equal{}\angle B_1A_1F$ and $ \angle CBB_1\equal{}\angle C_1B_1B\equal{}\angle C_1AG\equal{}\angle AA_1B_1$ hence result.
2. ABC=AA1F => again by angle chasing we prove that AB1GC1 is cyclic (prove is similar to 1 so I leave it to you).


Pascal\u0027s Triangle
geometric sequence

I've been looking at pascal's triangle and the binomial expansion theorem and all that kind of stuff, and I've been trying to figure out if something works.  But I got stuck with this expression-

y:^3:/x:^2: - y:^4:/x:^3: + y:^5:/x:^4: - y^6/x:^5: + ....

- because I don't think I know how to solve things like this.  Except I might, I got an answer, but it isn't the one I hoped I would get.  So if I"m right, then I'm wrong, but if I'm wrong then I might be right  :lol:  if that makes any sense.  So is this simplifiable to a non "...-ing" expression?

[quote="TripleM"]Think geometric progression.

a + ar + ar^2 + ar^3... = a/(1-r) as long as |r| < 1.[/quote]

It's indeed an Infinite geometric progression with a = y^3/x^2

and r= -y/x.

The series can be summed only if |-y/x| <1

i.e if |y| < |x| and x !=0

under the above restriction, the sum of the series , as stated by TripleM is

y^3/x^2/(1 + y/x) = y^3/x(x+y)


I think there are a lot of possibilities.

A possibility is

Jim:C
Jane:A
Lillian:B
Peter:D

[hide]If Jim has an A, then Jane does not have an A. By Jane's statement, this means that Jim doesn't have an A, which leads to contradiction. This means that Jim does not have an A.

Because of Lilian's second statement, Jim doesn't have a D.

Thus, Jim can have either a B or a C.

Let's assume that Jim has a C. This means that Peter has a D. Jane and Lilian got the A and B. Thus, two solutions are:
Jim: C
Lilian: A
Jane: B
Peter: D

Jim: C
Lilian: B
Jane: A
Peter: D

Let's assume that Jim has a B. This means Peter can have a C or a D. Let's assume that Peter has a C. This means that Lilian and Jane have the A and D. Thus, here are two more solutions.

Jim: B
Lilian: A
Jane: D
Peter: C

Jim: B
Lilian: D
Jane: A
Peter: C

Let's assume that Peter has a D. This means that Lilian and Jane have the A and C. If Jane has the C, then Jim has the D, which is a contradiction. So, the final solution is:

Jim: B
Lilian: A
Jane: C
Peter: D

There are 5 solutions altogether.[/hide]

OR...

[hide]They all are enemies and they lie about their grades. :rotfl: [/hide]

[quote="K Sengupta"]There are four grades A, B, C, D. A is the best and D is the worst.  Four students  Jim, Jane, Peter  and Lilian   appeared in an examination. After the results were published ,  they made  the following statements to a common acquaintance:

 Jim’s Statement:  : If I have  got A then Lilian  has  got D.

 Lilian’s Statement: : (i) If I got  C then Jim has got  D; (ii)  Jim’s  grade is better than Peter’s grade.

Jane’s Statement: If I didn’t  get A then Jim’s  grade is not  A.

Peter’s Statement :  If Jim’s grade is  A, then Jane’s grade is not B,  Lilian ’s grade is  C, but my grade is neither A nor B.

Can you help the acquaintance to determine the grades of each of the four students, given that all the above statements are true?[/quote] they can't all be true.
jim says that if he gets an A then lilian gets a d and peter says that if jim gets an a then lilian got a c they're contidicting eachother...


A slug climbs ten inches in ten minutes. It then rests two minutes. It continues climbing at a constant rate and rests for two minutes after climbing ten minutes. How many minutes will it take the slug to reach the top of a twenty-foot tower?

20 ft. = 240 inches. Therefore, it will take the slug 24 periods of climbing, which means 23 breaks in between, making it $ 240\plus{}46\equal{}\boxed{286}$


simplify:  ((sin x)^4+4((cos x)^2))^.5 +((cos x)^4 + 4((sin x)^2))^.5
(i dont have the answer)

Well it can be writen as $ \sin^{4}{x}\plus{}\cos^{4}{x}\plus{}4(\sin{x}\plus{}\cos{x})$.  I'm not sure how much farther though.

It might be helpful to also post the entire problem as well..

$ \sqrt{\sin^4 x \plus{} 4 \cos^2 x} \plus{} \sqrt{\cos^4 x \plus{} 4 \sin^2 x}$
$ \equal{} \sqrt{\sin^4 x \plus{} 4(1 \minus{} \sin^2 x)} \plus{} \sqrt{\cos^4 x \plus{} 4(1 \minus{} \cos^2 x)}$
$ \equal{} \sqrt{(\sin^2 x)^2 \minus{} 4 \sin^2 x \plus{} 4} \plus{} \sqrt{(\cos^2 x)^2 \minus{} 4 \cos^2 x \plus{} 4}$
$ \equal{} \sqrt{(\sin^2 x \minus{} 2)^2} \plus{} \sqrt{(\cos^2 x \minus{} 2)^2}$
$ \equal{} |\sin^2 x \minus{} 2| \plus{} |\cos^2 x \minus{} 2|$
$ \equal{} 2 \minus{} \sin^2 x \plus{} 2 \minus{} \cos^2 x$
$ \equal{} 4 \minus{} (\sin^2 x \plus{} \cos^2 x)$
$ \equal{} 4 \minus{} 1$
$ \equal{} 3$.

Note that for all real $ x$, $ |\sin x| \le 1$ and $ |\cos x| \le 1$, from which it follows that $ |\sin^2 x \minus{} 2| \equal{} 2 \minus{} \sin^2 x$, and so forth.


This is just a matter of plugging it into your calculator, I'm guessing.

[hide="answers"]
a) undefined, because $ 6.24\minus{}3.12\cdot2\equal{}0$
b)$ \minus{}.8901$
c) again, undefined
d) $ .7843$[/hide]

I thought it would be a good idea to bring up why A and C are undefined. You can [b]never[/b] divide a number by 0 because multiplication and divsion are interchangable meaning that the product of the divisor and quotient will equal the dividend, and in mutliplication of  2 numbers the product divided by one of te number will always equal the other number.

[quote="salahcool"]SOLUTION OF a is
6.19/0 =infinity                  :rotfl:[/quote]
Yes. The positive infinity.
See the endless ascending sequence $ \frac{1}{2}, \frac{1}{1}, \frac{1}{\frac{1}{2}}, \frac{1}{\frac{1}{3}}, \frac{1}{\frac{1}{4}}, \frac{1}{\frac{1}{10}}, \frac{1}{\frac{1}{100}}, \frac{1}{\frac{1}{1000}} ... \frac{1}{0}$
Then fewer is a discriminant then bigger is a fraction.


integration
trigonometry
calculus
calculus computations

Find the volume of the solid enclosed by the cylinder  $ {{x}^{2}}\plus{}{{y}^{2}}\equal{}9$ and the planes y + z = 5 and z = 1. Help me, please.

$ \begin{gathered}
  z \plus{} y \equal{} 5,{\text{ }}z \equal{} 1,{\text{ }}{x^2} \plus{} {y^2} \equal{} 9. \hfill \\
  D \equal{} \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ }} \minus{} \sqrt {9 \minus{} {x^2}}  \leqslant y \leqslant \sqrt {9 \minus{} {x^2}} ,{\text{ }} \minus{} 3 \leqslant x \leqslant 3} \right\}. \hfill \\
  V \equal{} \iint\limits_D {dxdy}\int\limits_1^{5 \minus{} y} {dz}  \equal{} \iint\limits_D {\left( {4 \minus{} y} \right)dxdy} \equal{} \int\limits_{ \minus{} 3}^3 {dx} \int\limits_{ \minus{} \sqrt {9 \minus{} {x^2}} }^{\sqrt {9 \minus{} {x^2}} } {\left( {4 \minus{} y} \right)dy}  \equal{}  \hfill \\
   \equal{} 2\int\limits_0^3 {\left. {\left( {4y \minus{} \frac{{{y^2}}}
{2}} \right)} \right|_{ \minus{} \sqrt {9 \minus{} {x^2}} }^{\sqrt {9 \minus{} {x^2}} }dx}  \equal{} 16\int\limits_0^3 {\sqrt {9 \minus{} {x^2}} dx}  \equal{} \left\{ \begin{gathered}
  x \equal{} 3\sin t,{\text{ }}dx \equal{} 3\cos tdt, \hfill \\
  0 \leqslant x \leqslant 3,{\text{ }}0 \leqslant t \leqslant \frac{\pi }
{2} \hfill \\
\end{gathered}  \right\} \equal{}  \hfill \\
   \equal{} 48\int\limits_0^{\pi /2} {\sqrt {9 \minus{} 9{{\sin }^2}t} \cos tdt}  \equal{} 144\int\limits_0^{\pi /2} {{{\cos }^2}tdt}  \equal{} 72\int\limits_0^{\pi /2} {\left( {1 \plus{} \cos 2t} \right)dt}  \equal{}  \hfill \\
   \equal{} \left. {72\left( {t \plus{} \frac{1}
{2}\sin 2t} \right)} \right|_0^{\pi /2} \equal{} 36\pi {\text{ }}\left( {{\text{cubic units}}} \right). \hfill \\
\end{gathered}$


limit
integration
inequalities
calculus
calculus computations

Find $ \lim_{n\to\infty} \frac{1}{n^2} \sum_{k\equal{}1}^n \sqrt[4]{k(k\plus{}1)(k\plus{}2)(k\plus{}3)}$.

[quote="shubin"]$ \lim_{n\to\infty} \frac{1}{n^2} \sum_{k\equal{}1}^n \sqrt[4]{k(k\plus{}1)(k\plus{}2)(k\plus{}3)}\equal{}\frac{1}{2}$[/quote]

How did you get it?

[quote="kunny"][quote="shubin"]$ \lim_{n\to\infty} \frac {1}{n^2} \sum_{k \equal{} 1}^n \sqrt [4]{k(k \plus{} 1)(k \plus{} 2)(k \plus{} 3)} \equal{} \frac {1}{2}$[/quote]

How did you get it?[/quote]
$ \int\limits_{0}^{1} \sqrt[4]{(x^2\plus{}3x\plus{}1)^2\minus{}1}dx \equal{}\frac{1}{2}$

[quote="Hong Quy"]$ \int\limits_{0}^{1} \sqrt [4]{(x^2 \plus{} 3x \plus{} 1)^2 \minus{} 1}dx \equal{} \frac {1}{2}$[/quote]

$ \int_{0}^{1} x \, dx \equal{} \frac {1}{2}$.

One may prove like this:

$ \frac {1}{n^2} \sum_{k \equal{} 1}^n \sqrt [4]{k(k \plus{} 1)(k \plus{} 2)(k \plus{} 3)}$ $ \equal{} \frac {1}{16} \sum_{k \equal{} 1}^n \sqrt [4]{\left(\frac {4k}{n}\right)\left(\frac {4k \plus{} 1}{n}\right)\left(\frac {4k \plus{} 2}{n}\right)\left(\frac {4k \plus{} 3}{n}\right)} \cdot \frac {4}{n}$ $ \equal{} \frac {1}{16} \sum_{k \equal{} 1}^n \frac {4k \plus{} c_k}{n} \cdot \frac {4}{n}$

for some $ 0 < c_k < 3$, by intermediate value theorem applied to the inequality

$ \frac {4k}{n} < \sqrt [4]{\left(\frac {4k}{n}\right)\left(\frac {4k \plus{} 1}{n}\right)\left(\frac {4k \plus{} 2}{n}\right)\left(\frac {4k \plus{} 3}{n}\right)} < \frac {4k \plus{} 3}{n}.$

Then it converges to  $ \frac {1}{16}\int_{0}^{4} x \, dx \equal{} \int_{0}^{1} t \, dt$.


induction
graph theory
combinatorics proposed
combinatorics

There is a complete Graph $K_n$ and all its edges are colored in two colors, prove there exists a Hamilton Cycle such that going arround it one only change of color once.

Ok, I hope it's ok.....
So, just do it by induction...I mean, consider a bicolored $K_{n+1}$. Take some $K_n$ of it, apply induction, and the vertex left is $A$. If the cycle is of the form $A-X_1-B-X_2-C$, where all the edges in the $A$ and $C$ chains are of the same color, and $X_i$ are "intermediate" vertices. Now just consider the color of $A-X_i$ and look to the left or right of it.....
I hope I didn't miss any cases..... :oops:


geometry
parallelogram
circumcircle
analytic geometry
function

Does anyone know any general theorems that give you information as to how a line can bisect the area of a polygon? I think I have seen such a thm. before, but I thought it was too difficult to learn at the time.

I found one. Can anyone find the proof?
If a line bisects the area of a parallelogram, then the line passes through the intersection of the diagonals of the parallelogram.

if it is a regular polygon, a line through the circumcenter suffices, one exercise related exercise that you should work on is: prove that such a line exists
[hide="hint"]continuity[/hide]

I'll followed your hint Altheman. Tell me if you see anything incorrect.
Thm: There exists atleast one line that bisects the area of a concave polygon.
Proof: Consider a concave polygon $A$ and an arbitrary line $l_k$ situated in the coordinate plane. Furthermore, let $l_k$ be defined by the equation $y=mx+k$ where $m$ is a constant. Finally, note that $A$ and $l_k$ are infinite sets of pts.
Now define $A(k)$ to equal the lower area bounded by $A$ and $l_k$. We see that, for different values of k, $A(k)$ ranges from $0$ to $A$'s entire area. It is also easy to see that $A(k)$ is a continuous function of $k$ since $A$ is concave.
Thus, if we let $k_0$ be the value of $k$ that makes $A(k)=0$, and $k_t$ be the value of $k$ that makes $A(k)$ equal the total area of $A$, then there exists some value $k'$ such that $k_0<k'<k_t$ and $A(k')$ is equal to half the area of $A$ by the intermediate value thm. of differential calculus.

I don't know how to rigorously prove that the function $A(k)$ is continuous since $A$ is concave. Could someone help?


I saw this inequality 1 years ago. It was a 4th problem of fKMO (final KMO).
Try to expand left side. :D  It is so amazing!

It is, indeed, Little Gauss.

Actually do you have solutions to the problems (from the same competition) I posted at
http://www.mathlinks.ro/Forum/viewtopic.php?t=138536
http://www.mathlinks.ro/Forum/viewtopic.php?t=121381

I would be glad.  :)


[quote="Rep123max"]Can someone please help me on this science stuff because my teacher doesnt teach us anything and finals are next week!

The first part I am pretty sure I am right I just want to double check:
Classify each of the following mixtures as homogeneous or heterogeneous:
a. lemonade with pulp [u]heterogeneuous[/u]
b. coffee with cream and sugar [u]homogeneous[/u]
c. vinegar [u]homogeneous[/u]
d. dessert gelatin with pears in it [/u]heterogeneous[u]

Then, they ask this question:
Explain why the air we breathe and solids, such as steel and bronze are solutions or homogeneous mixtures, just like salt water is. I have NO idea!!

Then, this has everyone stumped because it says nowhere in our book about this:
After boiling a chicken in a stock pot, you let the chicken stock cool down and place it in the refrigerator overnight. The next day you notice that a solid layer of fat has formed on the top of a gelatinous mixture.
a. Explain why the chicken fat rose to the top.
b. Explain what the gelatinous mixture is underneath the layer of fat.
c. Dexcribe how you could separate the fat from the gelatinous mixture.

[b]ANY HELP IS APPRECIATED!!!!!!![b][/b][/quote]
Air: air is a mixture of N2, O2, CO2, Ar, and other trace elements. Anyway, that stuff is all solutions because if you heat/cool it enough (depending on what it is), it will be a liquid with a bunch of components. Well, that's a nice way of thinking of it at least.

chicken: a) density b) "broth" that's cooled down enough to congeal c) The fat and broth will have different melting points. Figure it out from there. ...or use a spoon.

The first part looks right.  For the second part, just use what you know about homogeneous mixtures -- in air, is the nitrogren one big sea with little blobs of oxygen floating in it, or is it all mixed in together?

For the third part, a you should figure out yourself, b I'm not sure what they want, and c you should be able to do by thinking about liquids and solids.

Very standard questions of science...intro to chemistry...

For your first question regarding why the air we breathe and steel is homogeneous, think about the definition of homogeneous.

I'll show ou an example:

If you took a salad, I bet you could pick out the lettuce and cabbage for me. This is a heterogeneous mixture because you could easily seperate the parts by physical means. Now, suppose you put that salad into a blender. (Not that you would want to) When it comes out, all the parts will look the same. At this point, I bet you couldn't diffrentiate between lettuce and cabbage (even though before they entered the blender, they were totally different). This is a homogeneous mixture.

The same thing applies to air and steel like your teacher put it. If I gave you a sample of our atmosphere (and you could see its parts) you would not be able to reach out and grab the carbon dioxide or the oxygen. It's mixed so well that you'd have to be close to the atomic level to see different parts. This is why it's a homogeneous mixture. Same thing with the steel. It is so well mixture that you could not tell what is what.

The only difference between a homogeneous mixture and a solution is that in a solution, there is a solvent and a solute. And one or more substances must be completely dissolved into another. Solutions usually refer to liquids.

For your other problem,

a) Pretty much common sense. The fat rose to the top b/c it is the least dense of the substances originally in the chicken soup (aka. the mixture).

b) The gelatinous mixture is basically the rest of the mixture that the fat seperated out of. There may even be some bits of chicken in it still.

c) This is also common sense. Just take a knife (or a fork even) and cut off the fat at the top and remove it. It is basically already seperated. All you have to do is seperate the layers and remove the fat. The more professional way would be to take a tiny tube and suck the fat into it. This is how most chemists remove certain substances from a mixture. There are other ways which you will learn in high school chemistry

But it's not really seperable -- you couldn't get the sugar out.  (I don't remember exactly the definitions of homogeneous and heterogeneous . . . I think I just went on the basis of "if it looks homogeneous, it probably is.")

Well, what do [i]you[/i] think will happen?

[quote="Tare"]Question: For b in the first part if the coffee is cold enough then you couldn't you argue it's heterogeneous because the sugar content near the bottom is higher than the top and thus the sugar is not spread throughout the coffee equally?[/quote]

well sure...you could argue it that way...but ppl would still call it homogeneous...

You could also argue that if you were to raise the temperature of the mixture enough, then the coffee would evaporate and you could collect the gas and cool is again. This way you will definitely have seperated coffee and sugar since the boiling point of sugar is much much higher than that of water.

However, I think we all know what he meant by coffee, cream, and sugar.

Maybe you could do a little experimenting with sugar and water.

hint- when you make coffee, when you make tea, and you put sugar into it, the coffee and the tea... they are hot. so you can put lots and lots of sugar into it. wow that was a really obvious hint. sry.

[quote="Rep123max"]it starts to turn into a solid (im scientifically challenged)[/quote]
Mmm I was going to say the same thing until I realized that in order to put the maximum amount of sugar in water and melt it it has to be hot/warm so it's probably just going to become cold and thus some of the sugar will not dissolve anymore and sink to the bottom.

Tare, sugar does not melt in hot water. It dissolves.

Speaking of which, I should try the supersaturation experiment with sugar, since it didn't work with salt (fairly predictable, it turns out)

:oops: just keep in mind all of this I just know and that I didn't learn any of this...

You actually conducted an experiment? how?

Mmmm, take some water and dissolve a cystalline solid of some sort in it. Salt is a bad choice. Keep adding and stirring until the solid will not dissolve anymore and stays on the bottom [i]in solid form[/i]. Heat the mixture while stirring until all the crystals are completely dissolved. If you add too much in the first step, you won't be able to get it to dissolve. If you use some exotic substance, solubility may not increase with heat, which would also be a problem. Anyway, once it's all dissolved, allow the solution to cool slowly. In theory, the solute should stay dissolved, creating a supersaturated solution. It doesn't work with salt because the solubility of salt in water doesn't increase enough with temperature. Once you have the supersaturated solution and it's cooled back down to room temperature, if you add a [i]single grain[/i] of the solid crystal to the mixture, all of the excess crystals that were originally laying on the bottom should drop out of solution. I've seen pictures in the chemistry book, but haven't gotten it to work yet. I need to try it with sugar, but first I should make sure we have enough that we won't run out.

Ooooh...I did something like that for my fourth or third grade science report for Japanese school over at San Jose and I did solutability testing of sugar and water within different temperatures of water but I've never tried that...*looks in pantry* mmm, must ransack Cooking Class :twisted:

homogeneous -> cannot be separated by physical methods, i.e. filtering.

*nods rippledance*...thats a very textbooky definition though

Basically, the solubility of liquids increases as the temperature increases.  That's a basic textbook principle.  See what you can do with that.

BTW, does IPC stand for "introduction to physical chemistry"?  I can't imagine what else it stands for.

Hmm, this reminds me of the first few weeks of Earth Science class when we were defining basic stuff like solution.
Mixture: Can be separated by physical means.
Solution: Can not be separated by physical means.

So Mixture=heterogeneous and solution=homogeneous, si?


Hello everyone,
When can a regular polygon have n vertices,i.e., for which n there exists a regular polyhedra of n vertices.
This question came up due to a question posed by the ALBANIAN EAGLE in reference to the question 'What are the number of ways of arranging n objects on a polyhedra with n vertices',which I posted in combinatorics open problem section.
Thank You
Have a nice time. :?:  :)

[quote="darij grinberg"]Let me just refer you to http://en.wikipedia.org/wiki/Regular_polyhedra . From an elementary viewpoint, your question is pretty hard to formalize - what exactly is a regular polyhedron? What is a polyhedron? The simplest definition that comes into my mind uses some basic linear optimization theory.

  darij[/quote]

Well, got the answer for the number of symmetries of a regular polygon from http://mathworld.wolfram.com/RegularPolyhedron.html

Anyway Thanks for refering to the http://en.wikipedia.org/wiki/Regular_polyhedra .
 :)  :)


I am taking a class from AoPS and need 5 things in a header, but I only know how to make 3 (rhead, chead, lhead).  How do you add multiple lines in it :?:
Thanks,
minicon

Just put \\ in your text eg \chead{this is on \\ 2 lines}. You may need to increase the headheight using the geometry package.


modular arithmetic
number theory
number theory unsolved

Show that there exists a positive interger n such as all it's digit are 1 and it is divisible by 2003!

Yes definitely.. there are infinitely many such numbers, the number $n = 2003$ can be replaced with any positive integer coprime with $10$ and $9$, and if you do not restrict yourself to the decimal system, the number can just be replaced with any positive integer coprime to the base, and another suitable number. (E.g. in base $5$, to get a number consisting of a string of ones that is divisible by $n$, $n$ must be coprime with $4$ as well.) In fact, he 'suitable number' condition can also be dropped, as we can always construct a number that is also divisible by that 'suitable number' as well. All these can be illustrated with the case of $n=2003$.

For the case of $n=2003$, since $n$ is prime, by Fermat's Little Theorem, $10^{2002} \equiv 1 \pmod{2003}$. The integer $\frac{10^{2002}-1}{9}$ then is our desired number. In particular, the sequence $\frac{10^{2002k}-1}{9}$ for $k = 1, 2, 3, \ldots$ generates an infinite number of integers divisible by $2003$.

Indeed you don't need it to be coprime with something other than the base (like the $9$ here).
And also indeed, any number coprime to the base works, but all these stuff was posted before.


Γενίκευση μιας απο παλιο διαγωνισμο

Aν $ a_{1}^{2}\plus{}a_{2}^{2}\plus{}....\plus{}a_{n}^2\equal{}c_1$ και $ a_{1}\plus{}a_{2}\plus{}....\plus{}a_{n}\equal{}c_2$  με  $ a_i>0, n>2$ να βρεθεί το $ (a_1)_{max}$ και $ (a_1)_{min}$ (συνθήκη υπαρξης για min)

Efarmozwntas tin anisotita Cauchy-Scharz gia tis $ n \minus{} ades$ $ (1,1,...,1),(a_1,a_2,...,a_n)$ pernoume oti:

                          $ n(a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2)\geq(a_1 \plus{} a_2 \plus{} ... \plus{} a_n)^2$
  or
                          $ nc_1 \geq c_2^2$ $ (1)$

     Twra,efarmozwntas tin Cauchy-Scharw gia tis $ (n \minus{} 1) \minus{} ades$ $ (1,1,...,1),(a_2,a_3,...,a_n)$ pernoume:

                          $ (n \minus{} 1)(a_2^2 \plus{} ... \plus{} a_n^2)\geq(a_2 \plus{} ... \plus{} a_n)^2$ or
                          $ (n \minus{} 1)(c_1 \minus{} a_1^2)\geq(c_2 \minus{} a_1)^2$   or
                          $ na_1^2 \minus{} 2c_2a_1 \plus{} (c_2^2 \minus{} nc_1 \plus{} c_1)\leq0$

Theoroume to aristero melos tis anisotitas polyonymo ws pros $ a_1$.

I diakrinousa tou polyonymou aftou einai:

                          $ D \equal{} 4c_2^2 \minus{} 4n(c_2^2 \minus{} nc_1 \plus{} c_1)$ or
                          $ D \equal{} 4(1 \minus{} n)(c_2^2 \minus{} nc_1) > 0$ logw tis $ (1)$ kai epeidi $ n > 2$

Epeidi $ D > 0$ to polyonymo exei dyo anises pragmatikes rizes oi opoies kai antistoixoun me tin elaxisti kai megisti timi tou $ a_1$.

                          Oi rizes tou polyonymou einai:

                          $ \frac {c_2 \plus{} \minus{} \sqrt {(1 \minus{} n)(c_2^2 \minus{} nc_1)}}{n}$,  oi opoies kai
einai i megisti kai elaxisti timi antoistoixws tou $ a_1$.

~Kostas

Όμορφα, αλλά θυμίσου ότι $ a_i > 0$...

Για τα $ c_1, c_2$ θα πρέπει λοιπόν να έχουμε $ c_1 > 0, c_2 > 0$ και επίσης $ \frac {c_2^2}{n} \leq c_1 < c_2^2$ (όχι μόνο $ nc_1\geq c_2^2$).

Γεωμετρικά, η πρώτη συνθήκη είναι μία σφαίρα με ακτίνα $ \sqrt {c_1}$ ενώ η δεύτερη αντιπροσωπεύει ένα επίπεδο κάθετο στο $ (1,1\ldots 1)$ και που συναντάει τους άξονες στο $ c_2$. Η τομή τους άρα στο θετικό $ 2^n$-μόριο θα είναι ένας κύκλος (ok, σφαίρα :P) που οι ακραίες αποστάσεις του από τον άξονα $ x_1$ υπολογίζονται εύκολα αν θέσουμε τις υπόλοιπες μεταβλητές ίσες μεταξύ τους (λόγω συμμετρίας). Τότε καταλήγουμε στην εξίσωση που είχες γράψει κι εσύ πριν: $ (n \minus{} 1)x^2 \plus{} (c_2 \minus{} x)^2 \equal{} (n \minus{} 1)c_1$.

Όχι ότι η ουσία αυτού που κάνεις είναι διαφορετική, απλώς θα ήταν κρίμα να μην τονιστεί η γεωμετρική ερμηνεία εδώ :)

Cheerio,

Durandal 1707

Hey,i geometriki lysi einai i "swsti" lysi gia to provlima kata tin gnwmi mou aplws den mporousa na to kataferw
geometrika gia afto kai anatreksa se algevrikes methodous.
Dystyxws den mporw na dw/katalavw giati isxyei i afstiri anisotita $ \frac {c_2^2}{n}<c_1$. To deksi melos tis anisotitas pou
egrapses $ c_1 < c_2^2$ isxyei profanws i afstiri anisotita.eimai entaksei me afto.

Gia paradeigma,i isotita panw stin Cauch-Schwarz isxyei otan ola ta $ a_1,a_2,...,a_n$ einai isa metaksi tous dld:
$ a_1 \equal{} a_2 \equal{} ...a_n$. giati kati tetoio den mporei na isxyei?   :huh:

Thanks( Isws na einai profanes giati einai afstiri i anisotita aplws na min mporw na to dw twra.who knows?)

[quote="new_member"](Isws na einai profanes giati einai afstiri i anisotita aplws na min mporw na to dw twra.who knows?)[/quote]

Φυσικά και είναι προφανές: I mucked it up :D

Ήθελα να τονίσω ότι υπάρχει και η αυστηρή ανισότητα στα άκρα και πήρα κατά λάθος και το σημείο επαφής μαζί, sorry :) Edited now...

Cheerio,

Durandal 1707