I don't know how hard this problem is.
Prove that for all k such that $ k \equiv 32 \mod 128$ and k has exactly 12 divisors, k is the expressible as the sum of two squares.

[quote="K81o7"][hide]$ k$ is divisible by $ 2$ exactly five times.  Then then follows that $ k$ is of the form $ 2^{5}p$ where $ p$ is some prime.  We then  notive $ p \equiv 1\mod{4}$ because of the given congruence, yielding the desired result.[/hide][/quote]
[hide]How does having p=prime and p being the sum of two squares give the desired result for k? (I'm probably missing something obvious...)[/hide]

K81o7 is using the fact that $ n$ can be written as the sum of two squares if and only if each odd prime $ p$ which divides $ n$ an odd number of times is congruent to 1 mod 4.

So to elaborate (I suppose no reason to hide it now...), we can prove that any prime $ 1 \mod 4$ can be written as a sum of two squares.  Then $ p=(a+bi)(a-bi)$ for some $ a, b$.  Then notice $ 2=(1+i)^{5}(1-i)^{5}$.  Therefore, $ 2^{5}p=\left[(a+bi)(1+i)\right]^{5}\left[(a-bi)(1-i)\right]^{5}$.  These two are conjugates, and so if they're $ c+di$ and $ c-di$, then $ c^{2}+d^{2}$ is our number.  (what I used generalizes to what JBL says, and then can also be generalized even further: first using the Gaussian integers, and then using other rings for other equations.)

[quote="Esteban8231"][quote]we can prove that any prime 1 \mod 4 can be written as a sum of two squares[/quote]

How do you prove it?[/quote]

You use Gaussian Integers.

[quote="Esteban8231"][quote]we can prove that any prime 1 \mod 4 can be written as a sum of two squares[/quote]

How do you prove it?[/quote]

[b]Theorem:[/b] If $ p\equiv 1 (\bmod 4)$ then $ p \equal{} a^{2}\plus{}b^{2}$. (Fermat's theorem).

[b]Proof:[/b] Given any finite field $ F$ we know that $ F^{*}$ (the group of non-zero elements under multiplication) is cyclic of order of $ p\minus{}1$. Now since $ 4 |(p\minus{}1)$ it means that there is an element $ a\in F^{*}$ of order $ 4$ (you can use Sylow theorem here or the fact its abelian). It must mean that $ a^{2}\equal{}\minus{}1$ in other words $ a^{2}\equiv\minus{}1 (\bmod p)$ and so we see that $ p|(n^{2}\plus{}1)$. We switch to the Gaussian integers $ \mathbb{Z}( i )$. Now $ n^{2}\plus{}1 \equal{} (n\plus{}i)(n\minus{}i)$. Now if $ p$ is an irreducible element then it means $ p$ is a prime since the Gaussian integers form a Principal Ideal Domain. Hence, if $ p$ is irreducible then $ p|(n\plus{}i)$ or $ p|(n\minus{}i)$. So $ n\pm i \equal{} p(a\plus{}bi)$ for some $ a,b\in\mathbb{Z}$, which is easily seen to lead to a contradiction. Since $ p$ aint irreducible it means $ p \equal{} (x\plus{}iy)(z\plus{}wi)$ where neither $ x\plus{}iy, z\plus{}wi$ are units in $ \mathbb{Z}( i )$. Take the norm of both sides to get $ p^{2}\equal{} (x^{2}\plus{}y^{2})(z^{2}\plus{}w^{2})$. Since neither $ x^{2}\plus{}y^{2},z^{2}\plus{}w^{2}$ are $ 1$ it means $ p \equal{} x^{2}\plus{}y^{2}$. Q.E.D.

[b]Bonus Question:[/b] So that $ a,b$ are [i]unique[/i].  :wink:


trigonometry algebra polynomial logarithms

yeah, he wanted me to post some problems for him that are USC level (#20-25, no geometry). here they are:

1. Find the smallest positive number from this list:

$a)$ $10-3\sqrt{11}$
$b)$ $3\sqrt{11}-10$
$c)$ $18-5\sqrt{13}$
$d)$ $51-10\sqrt{26}$
$e)$ $10\sqrt{26}-51$

2. If $\tan{a}$ and $\tan{b}$ are the roots of $x^2-nx+m$, then $\cot{a}$ and $\cot{b}$ are the roots of $x^2-rx+s$. $rs$ is necessarily:
$a)$ $nm$
$b)$ $\frac{1}{mn}$
$c)$ $\frac{n}{m^2}$
$d)$ $\frac{m}{n^2}$
$e)$ $\frac{n}{m}$

3. What is the smallest integer $q$ such that $\frac{q-13}{5q+6}$ is an integer?
$a)$ $45$
$b)$ $68$
$c)$ $155$
$d)$ $226$
$e)$ none of the above

4.$\frac{\sin{10^\cir}+\sin{20^\circ}}{\cos{10^\circ}+\cos{20^\circ}}$ equals:
$a)$ $\tan{10^\circ}+\tan{20^\circ}$
$b)$ $\tan{30^\circ}$
$c)$ $\tan{15^\circ}$
$d)$ $\frac{\tan{60^\crc}}{4}$
$e)$ $1$

5. The number of integers $(a,b)$ that satisfy $0<a<b$ and $\sqrt{1984}=\sqrt{a}+\sqrt{b}$ is:
$a)$ $0$
$b)$ $1$
$c)$ $2$
$d)$ $3$
$e)$ $4$

[hide="Problem 1"]
The answer choices can be expressed as:

A) $\sqrt{100}-\sqrt{99}$
B) $\sqrt{99}-\sqrt{100}$
C) $\sqrt{324}-\sqrt{325}$
D) $\sqrt{2601}-\sqrt{2600}$
E) $\sqrt{2600}-\sqrt{2601}$

We can eliminate B, C, and E, since they are all negative.
This leaves A and D as potential answers.  We know that D will obviously be smaller than C, since the numbers are larger.  So our answer is $\boxed{D}$.
[/hide]

[hide="Problem 2"]
We know that for any polynomial, the polynomial with inverse roots is the polynomial with inverse coefficients.  So $mx^2-nx+1 = x^2-rx+s$.  Now, we can multiply a polynomial by anything, and still have the same roots.  So we can multiply the RHS by m, giving us $mx^2-nx+1 = mx^2-mrx+ms$.  To make these polynomials equivalent, $n=mr$ and $ms=1$.  Multiply these equations together to get $m^2rs=n$.  So $rs=\frac{n}{m^2}$, so our answer is $\boxed{C}$.
[/hide]

[hide="Problem 3"]Just plug in 13 and get 0 as an output.  Therefore, the answer is $\boxed{E}$[/hide]

[hide="Alternate Problem 3"]
We know that $5q+6$ divides $q-13$.  So $5q+6$ also divides $5q-65$.  So it also divides $5q-65-(5q+6)$.  So $5q+6$ divides $-71$, which means that it divides $71$.  Set $5q+6=71$, and get $q=13$, so the answer is $\boxed{E}$.[/hide]

[hide="Problem 4"]
Using the identity:
$\frac{\sin x + \sin 2x}{\cos x + \cos 2x} = \tan 3x/2$, $\frac{\sin{10^\cir}+\sin{20^\circ}}{\cos{10^\circ}+\cos{20^\circ}} = \tan 30/2 = \tan 15^\circ$.  So the answer is clearly $\boxed{B}$.
[/hide]

[hide="Problem 5"]
Simplify the radical, and we get $8\sqrt{31}$.  Now, we simply find the ways to add two distinct positive integers and get 8.  There are 3 ways: (1,7), (2,6), and (3,5).  So the answer is 3, which is $\boxed{D}$.[/hide]

Can I have some more problems, oh and my 5 bucks?  :D

lol, ok. i'll post some more problems by noon today.

1. $n$ and $n+1$ are two consecutive positive integers and both have four positive divisors and have the same sum of divisors as well. What is the number of divisors of their product?

$a)$ $6$
$b)$ $12$
$c)$ $16$
$d)$ $20$
$e)$ $22$

2. If $a$ divides $b$, and $\log{(b/a)^{b/2}}+\log(\sqrt{a/b} )^{9a}=1$, then find $b^2-a^2$.

$a)$ $360$
$b)$ $396$
$c)$ $462$
$d)$ $480$
$e)$ $500$

3. Four positive integers $p,q,r,s$ satisfy $pqrs=10!$. Find the minimum value of $p+q+r+s$.

$a)$ $170$
$b)$ $175$
$c)$ $180$
$d)$ $185$
$e)$ $190$

4. What is the smallest value of a positive integer $n$ if $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+....+\frac{1}{n(n+1)}\geq 1$?

$a)$ $10$
$b)$ $100$
$c)$ $1000$
$d)$ $2002$
$e)$ no value of $n$ possible

5. How many positive integers $k\leq 1000$ can be written in the form $a^2-b^2$?

$a)$ $100$
$b)$ $250$
$c)$ $700$
$d)$ $750$
$e)$ $751$

Problem 5 seems inherently flawed...

EDIT: Never mind, he fixed it up.

[quote="mysmartmouth"][hide="Problem 4"]
If we write out the first few sums, we get a sequence: $\frac{1}{2}$, $\frac{2}{3}$, $\frac{3}{4}$, $\frac{4}{5}$, etc.  We see that the sum will never exceed 1, therefore, our answer is $\boxed{E}$.
[/hide]

In problem 2, your parenthesis don't make sense.[/quote]

Fixed another mistake in the problem as well :D

Also, for problem 4, can you generalize WHY it never exceeds $1$. Or in other words, how do you KNOW that as $n \longrightarrow \infty$, the sum will still be $<1$ besides using patterns of course.

[hide="hint"]telescope[/hide]

Problem 4:

If you want to telescope it, then write it as:
$(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...(\frac{1}{n}-\frac{1}{n+1})$
We see that all of it will telescope out, leaving us with $1-\frac{1}{n+1}$.  Since $n$ is a positive number, $1-\frac{1}{n+1}$ will always be less than 1.  $\boxed{QED}$

[hide="Problem 1"]
For a number to have four divisors, it must be in the form of $x^3$ or $ab$.
We can prove two things, which I will call my lemmas:
A) Any two consecutive numbers with four divisors will BOTH be of the form $ab$.
B) The two consecutive integers $ab$ and $cd$ will not share any factors (other than 1).

It is trivially easy to prove B.  But to prove A requires a bit more thought:

We know that $n$ is either of the form $p^3$ or $pq$.  If it's of the form $p^3$ then $n+1 = p^3+1$ which factors as $(p+1)(p^2+p+1)$.  But $p+1$ is even so we can write it as $2 * ((p+1)/2) * (p^2+p+1)$.  Additionally, $p^2+p+1$ is an integer greater than 1.  But $n+1$ must be of the form $p^3$ or $pq$ as well and we already have the factors 2 and $p^2 + p +1$.  Thus, $(p+1)/2$ can only equal 1 making $p=1$ which obviously doesn't work. $\boxed{QEFD}$.

Therefore, both of our lemmas A and B are true.  Therefore, the product of $ab$ and $cd$ will be $abcd$, where all factors are prime.  Therefore, $abcd$ will have 16 factors.  Our answer is $\boxed{C}$.

The approach I used in this problem disregarded another bit of information given in the problem.  Turns out the fact that n and n+1 have the same sum of divisors is extra info not needed to solve the problem.
[/hide]

[hide="Problem 2"]
$\log{(b/a)^{b/2}}+\log(\sqrt{a/b} )^{9a}=1$

$\frac{b}{2}\log{(b/a)-\frac{9a}{2}\log(b/a)=1}$

$(\frac{b-9a}{2})(\log(b/a))=1$

$(b-9a)(\log(b/a))=2$

$\log((\frac{b}{a})^{b-9a})=2$

$(\frac{b}{a})^{b-9a}=10^2$

We know that a divides b, so $b/a$ is an integer.  Let's try letting it equal 10, so the exponent will be two.  Set up the simple equations, and get $a=20$ and $b=2$.  So $a^2-b^2=396$.  Thus, our answer is $\boxed{B}$.
[/hide]

[hide="Problem 3"]AM-GM gives an answer of 175, so the answer is $\boxed{B}$.[/hide]

[hide="Problem 4"]
See above.
[/hide]

[hide="Problem 5"]
We know that $a^2-b^2=(a+b)(a-b)$.  So we break it down into two cases: when a+b is odd, and when a+b is even:

Case 1:  If a+b is odd, then we can write it like this: $(a+b)(a-b)=x$, where x is an odd integer. Then, we could set $a+b=x$ and $a-b=1$, which makes $2a$ an even integer, which makes $a$ an integer.  Since this works for any odd $a+b$, all odd values of $k$ satisfy the condition.

Case 2:  If $a+b$ is even, then $a-b$ is even as well.  Therefore, $k$ is divisible by 4.  So all multiples of 4 will satisfy the condition.

Now we add 500 odd integers from case 1, and the 250 multiples of 4 from case 2, and get $500+250=750$, so the answer is $\boxed{D}$.

[/hide]

[quote="mysmartmouth"]
I'm still working on problem 3.  I used AM-GM to get an answer of B (I think that's what I got...) but AM-GM doesn't require integers, so I'm gonna work the problem some more before I post my final answer...[/quote]

You are correct though on it. Although it doesn't require integers, the closest answer is of course B. And since it's multiple choice, you have to go with the closest answer.

Is there an alternate solution to problem 3, either not using AM-GM, or finding a case of 175 and proving (by AM-GM) that you can't go lower?


let  a,b,c natural                                           $ \sqrt{\frac {2005}{a \plus{} b}} \plus{} \sqrt{\frac {2005}{a \plus{} c}} \plus{} \sqrt{\frac {2005}{c \plus{} b}}$ integer find    a,b,c

[quote="XAKKER"]let  a,b,c natural                                           $ \sqrt {\frac {2005}{a \plus{} b}} \plus{} \sqrt {\frac {2005}{a \plus{} c}} \plus{} \sqrt {\frac {2005}{c \plus{} b}}$ integer find    a,b,c[/quote]

http://www.mathlinks.ro/viewtopic.php?p=233034#233034


function
number theory theorems
number theory

Is there a formula for such a function:
for a natural number $ n$ , this function give the sum of all the natural coprimes to $ n$ and less than it .
excepting : $ \frac{n(n\plus{}1)}{2}\minus{}\sigma(n)$

$ \sigma(n)$ is the sum of all the natural divisors of $ n$
and $ \sigma(n)\equal{}\prod_{i\equal{}1}^{r}\frac{p_i^{s_i\plus{}1}\minus{}1}{p_i\minus{}1}$
where $ n\equal{}p_1^{s_1}p_2^{s_2}...p_r^{s_r}$ is the canonical expansion of n

What properties has this function ???

If $ (k, n)\equal{}1$ then $ (n\minus{}k, n)\equal{}1$ as well, so this sum is equal to $ n\frac{\varphi(n)}{2}$.

so this function is a multiplicative one :D nice

thank you , TomciO


algebra
polynomial
limit
real analysis
real analysis unsolved

P $\in C[X]$
F a close  set (adherence(F)=F)
prove that $P(F)$ is a close set  ( adherence(P(F))=P(F) )

Let $L$ be a point of the adherence of $P(F)$ and $z_n$ be a sequence in $F$ such that $\exists \lim P(z_n)=L$. Hence $z_n$ is bounded. Since $P$ is continuous we have that $L=\lim_{n\to \infty}P(z_n)=P(\lim_{n \to \infty} z_{k_n})\in P(F)$ where $z_{k_n}$ is a convergent subsequence of $z_n$. We have that $z_n$ is bounded so it must have a convergent subsequence.
I just proved that for every $L\in$ adherence of  $P(F)$ we have $L\in P(F)$. Hence,  $adherenceP(F)\subset P(F)$. Obviously we get that $P(F)\subset adherenceP(F)$. So the equality follows.  Am I wrong? I may used something not provided but I don't realize what.

xirti : but you have 2 cas : if P is a constant ,it's trivial ,in the second cas we can apply ur proof... :D( just to be sure that your sequense must be bounded)

Thank you for for completing the proof ;)


Hello, I'm fairly new at this forum, but a very good friend send me here, because he said people here would help me with this problem:

Party

The president of a company is organising a party for his employees. The company has a hierarchical structure; that is, the supervisor relation forms a tree rooted at the president. In order to make the party fun for all attendees, the president does not want both an employee and his or her supervisors to sit at the same table.

The problem is to compute the minimal number of tables necessary to make a seating arrangement that satisfies the above requirement. The input of the program is the immediate supervisor relation of the company. The supervisor relation is defined in terms of the immediate supervisor relation as follows. A person P is supervisor of a person Q if P is the immediate supervisor of Q, or there are persons P1, ..., Pk (1<k) such that P=P1, Q=Pk and Pi is the immediate supervisor of Pi+1 (i=1, ..., k-1).

The number of attendees is N (1<=N<=200), and there are T (2<=T<=10) seats at each table.

Input data

The employees of the company are identified by the first N (1<=N<=200) natural numbers. The president of the company is identified by number 1, and has no supervisor. The first line of the text file DAY2B.DAT contains T, the number of seats per table (2<=T<=10). The second line contains the number N (1<=N<=200) of the attendees, each employee identified by numbers from 1 to N will attend the party. In the third line you find N numbers. The i-th number in the line is the immediate supervisor of the i-th person. The first number in the line is 0, indicating that the president has no supervisor.

Output Data

Write the minimal number of tables needed to seat all attendees in the specified way in the text file DAY2B.SOL.

Example input

4
13
0 1 9 9 9 2 2 1 1 7 8 8 10

Example output

5

This problem was part of the problem set of the CEOI 95. I will be waiting for your help.

So the min number of tables is the maximum depth of the tree: we must have at least this many, since the people in the maxdepth branch must sit at different tables, and we can use exactly this many, since we can have everyone of the same supervisor level sitting at the same table. It's easy to implement a fast tree-merge or just flood-fill to find the max depth.

EDIT: whups, missed the maximum people at each table. Never mind. Makes it a decently hard problem :).

No matter you've never mind the maximum people at each table. Could you explain me your idea with more details, specially when you talk about "a tree-merge or just flood-fill"?

It's not actually the basis of a correct solution, but here goes in pseudo-C++

Floodfill:
int N;
int supervisorOf[N][N];
for(i=0 to N)
 for(j=0 to N)
  supervisorOf[i][j]=0;
read(supervisorNumber);//get rid of the presidents' supervisor
for(i=1 to N)
 read(supervisorNumber)
 supervisorNumber--;//silly 1-based indexing.
 supervisorOf[supervisorNumber][i]=1
if(N<2){
 return N;
}
int visited[N]
for(i=0 to N)
 visited[i]=0
visited[0]=1;
int cDepth=1;
vector<int> curr;
curr.push_back(0);//start at the presidents level
while(curr.size()>0){
 vector<int> newPoss;
 for possibleSup in curr{
  for(i=0 to N){
   if !visited[i] and supervisorOf[possibleSup][i]{
    newPoss.push_back(i)
    visited[i]=1
   }
  }
 }
 curr=newPoss;
 cDepth++;
}
return cDepth

Tree merging might work too, but I need go do other stuff right now. Will try to post it later.


function
complex analysis
complex analysis unsolved

Are there three non-constant entire functions f,g,h such that $ e^f+e^g+e^h=1$? What about two such functions?

[quote="harazi"]Are there three non-constant entire functions f,g,h such that $e^f+e^g+e^h=1$? What about two such functions?[/quote]

 There are no such two functions. If there are, namely

 $e^f + e^g = 1$ , since $e^g$ omits the value $0$ thus $e^f$ must omit the value 1.
 Then $e^f$ omits the two value $0$ , $1$ , so it must be constant according to little Picard theorem.

Modulo Picard's theorem, that wasn't difficult. But what about three functions?

I do not know. But Pierre must know... :)

Pierre was here today. And where is his comments?  :)  :)  :)

I do not know, probably he did not see this post. Or probably he does not know either.

Let

  $g(x) = (e^x + 1)/ (e^{-x} + 1)$

 Then it can be easily seen that $g$ is an entire function without zero.
Therefore there exists an entire functions $h$ such that $g = e^h$

 So what we have now is

  $e^h(e^{-x} + 1) = e^x + 1$  thus

  $e^{h-x} + e^h - e^x = 1$ thus

  $e^{h-x} + e^h + e^{/pi.ix} = 1 $ holds for all x.

 Is is correct?

Let

  $g(x) = (e^x + 1)/ (e^{-x} + 1)$

 Then it can be easily seen that $g$ is an entire function without zero since  the denominator and the nominator have all the common roots ( which are all simple )
Therefore there exists an entire functions $h$ such that $g = e^h$

 So what we have now is

  $e^h(e^{-x} + 1) = e^x + 1$    thus

  $e^{h-x} + e^h - e^x = 1$    thus

  $e^{h-x} + e^h + e^{{\pi}ix} = 1 $ holds for all x.

 Is is correct?

Next time simply edit your post instead of send it second time.
See my reply above.

So sorry & embarrasing. Thanks for being cool!! This one drives me nuts!!

[quote="pbornsztein"]What do I have to know? :? [/quote]
Was that problem solved in AMM?

[quote="pbornsztein"]I don't know  :D

Pierre.[/quote]

Pierre is just too lazy to search...  ;)

[quote="orl"][quote="pbornsztein"]I don't know  :D

Pierre.[/quote]

Pierre is just too lazy too search...  ;)[/quote]
I agree  :lol:  :lol:  :lol:

I will defend Pierre. Those who show how strong they are in analysis could probably find a solution. Of course, in more than 5 minutes... :)


inequalities
Euler
inequalities proposed

Prove that

$ \frac{27}{2}Rr\leq m_a^2\plus{}m_b^2\plus{}m_c^2\leq\frac{27}{4}R^2$

.

[quote="kuing"][quote="Ligouras"]Prove that

$ \frac {27}{2}Rr\leq m_a^2 \plus{} m_b^2 \plus{} m_c^2\leq\frac {27}{4}R^2$

.[/quote]
\[ \sum {m_a^2 } \equal{} \frac {3}{4}\left( {a^2 \plus{} b^2 \plus{} c^2 } \right) \equal{} \frac {3}{2}\left( {s^2 \minus{} 4Rr \minus{} r^2 } \right)\]
then Gerretsen Inequality, Euler Inequality, can prove it easy.[/quote]

Thanks  :wink:


There is a Set S, which has 3 rules.
I. 2 is in S
II. If x is in S, then 3x, and x+7 is also in set S.
III. Numbers that do not follow these rules do not belong in set S.
Given these set of rules, find the smallest possible integer greater then 2007 that does not belong in set S.
(reworded a BIT)

Also an old mathcounts, I think

[hide="hint"]
Consider the numbers mod 7
[/hide]

[hide="Another hint"] Apply the $ 3x$ rule over and over again, and keep looking $ \bmod 7$. [/hide]

[hide="hint"]
apply 3x part to '2 is in set' part and since x+7 is in, apply to mod 7, and you will find your answer quickly[/hide]

[hide]2008? I mean, it's not divisible by 3, and 2006 isn't divisible by 7. [/hide]

Unfortunately, $ 2008 \equal{} 2 \cdot 3 \plus{} 7(286)$, which is in the set.  The key to this problem is to use [b]both[/b] the $ 3x$ rule and the $ x \plus{} 7$ rule.


Let ABC be a triangle and M be a point in it. Prove that for every M we have   MA +MB+MC>= 2*( \sum a* ma)/(a+b+c).
    Here ma, mb, mc are the lenghts of ABC 's medians.

I have been trying this inequality for quite sum time now.
and i'd be very grateful if u could post ur proof ,Treegoner
best regards

My solution uses Ptolemee theorem.
  Let A1 be a point such that ABA1C is a parallelogram.
  Thanks to Ptolemee theorem in BA1CM,we have
      MA1*a <= MB*c + MC*b.
   Furthermore, we have MA1>=2m_a-MA.
   Thus, MA*a+MB*c+MC*b >=2m_a *a.
    Similarly, we have the result. :D


Αυτό το πρόβλημα το έδωσε ο Σιλουανός στις 2-11-07 με τον τίτλο "ΘΕΩΡΙΑ ΑΡΙΘΜΩΝ" αλλά το είδα μόλις τώρα:
[b]Έστω $ p$ ένας πώτος αριθμός και $ m,n$ ακέραιοι μεγαλύτεροι του 1 . Υποθέτουμε ότι ο $ n$ διαιρεί τον
$ m^{p(n \minus{} 1)} \minus{} 1$ . Δείξτε ότι ο $ n$ και ο $ m^{n \minus{} 1} \minus{} 1$ έχουν μέγιστο  κοινό διαρέτη μεγαλύτερο της μονάδας.[/b]

Με το "κινέζικο θεώρημα υπολοίπου" (δεν ξέρω αν έτσι λέγεται, πάντως έτσι είναι η μετάφρασή του) βγαίνει μια πολύ σύντομη και απλή λύση.
Αρχικά θα δώσω μια διατύπωση του θεωρήματος:
-αν $ a \equiv b (mod n)$ τότε υπάρχει c|n με c>1 τέτοιο ώστε $ a \equiv b (mod c)$

Τώρα θα επισημάνω ένα κενό στην εκφώνηση που παρατήρησα:
-το n είναι ο μέγιστος κοινός διαιρέτης του n και του $ m^{(n \minus{} 1)} \minus{} 1$, αφού
και τα δύο διαιρούνται με το n γιατί
$ m^{p(n \minus{} 1)} \minus{} 1 \equiv 0 (mod n) < \equal{} > m^{p(n \minus{} 1)} \equiv 1^{p} (mod n) < \equal{} >$
(από το θεώρημα του Fermat)  $ m^{n \minus{} 1} \equiv 1 (mod n) < \equal{} > m^{n \minus{} 1} \minus{} 1\equiv 0 (mod n)$
Προφανώς, δεν ήταν σκοπός η άσκησης να συμπεριληφθεί το n, γιατί τότε η λύση θα ήταν προφανής. Γι' αυτό θεώρησα ότι ο μέγιστος κοινός διαιρέτης έπρεπε να είναι μικρότερος του n.

Αν θέλαμε να μη ληφθεί το n υπόψιν στην εύρεση κοινού διαιρέτη, τότε το n πρέπει να δοθεί σύνθετος για να υπάρχει λύση στην άσκηση (αν το n είναι πρώτος, τότε το c που χρησιμοποιείται παρακάτω είναι ίσο με n και έχουμε την περίπτωση που παρατήρησα παραπάνω) και να η λύση:

Αρκεί να δειχτεί ότι οι αριθμοί έχουν τουλάχιστον ένα κοινό διαιρέτη >1.
Από το θεώρημα που προαναφέρθηκε συνάγεται ότι:
(  επειδή $ m^{p(n \minus{} 1)} \minus{} 1 \equiv 0 (mod n) < \equal{} > m^{p(n \minus{} 1)} \equiv 1^{p} (mod n)$
   $ < \equal{} >$  (από το θεώρημα του Fermat)  $ m^{n \minus{} 1} \equiv 1 (mod n) < \equal{} >$
   $ m^{n \minus{} 1} \minus{} 1\equiv 0 (mod n)$  )
υπάρχει c>1 τέτοιο ώστε c|n και $ m^{n \minus{} 1} \minus{} 1 \equiv 0 (mod c)$   )
άρα το $ m^{n \minus{} 1} \minus{} 1$ και το n έχουν τουλάχιστον ένα κοινό διαιρέτη c μεγαλύτερο του ένα.

Διωρθώστε με αν κάνω κάπου λάθος ή αν έχει η λύση κάποιο κενό. Πάντως αυτός ο τρόπος είναι πιο σύντομος και εύκολος (Η λύση του Ηλία δεν είναι λάθος, αλλά στην ουσία αποτελεί απόδειξη του θεωρήματος η διαδικασία).

Η εκφώνηση μια χαρἀ είναι
Το παρακάτω είναι λάθος αφού  στα modula δεν ισχύοθν οι ιδίοτητες των πραγματικών αριθμών απαραίτητα . Δεν μπορούμε να πάρουμε p-osti ρίζα εκεί. Και μην ξεχνάς ότι το θεώρημα Fermat ισχύει για n  πρώτο ...

[quote="maniopas"]  του n και του $ m^{(n \minus{} 1)} \minus{} 1$, αφού
και τα δύο διαιρούνται με το n γιατί
$ m^{p(n \minus{} 1)} \minus{} 1 \equiv 0 (mod n) < \equal{} > m^{p(n \minus{} 1)} \equiv 1^{p} (mod n) < \equal{} >$
(από το θεώρημα του Fermat)  $ m^{n \minus{} 1} \equiv 1 (mod n) < \equal{} > m^{n \minus{} 1} \minus{} 1\equiv 0 (mod n)$
[/quote]

Ναι, δίκιο έχεις.


Prove that $ a \plus{} b \plus{} c \equal{} 0 \implies a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc.$

Factorize $ a^3\plus{}b^3\plus{}c^3\minus{}3abc$ or substutite $ c\equal{}\minus{}(a\plus{}b)$ to $ a^3\plus{}b^3\plus{}c^3\minus{}3abc$.

If the condition $ a \neq b$ is imposed, the left direction also holds by factorization.  mathwizarddude, you have an uncanny knack for posting problems that have already been posted at least twice on the forum  :)  Try a search next time.

You can also show this with Newton sums if you let a, b, and c be the roots of a cubic polynomial.


[hide="solution"]
If $ n^{2}$ is divisible by 6 that means that the prime factors of $ n^{2}$ include 2 and 3. As $ n$ is an integer, and $ n^{2}$ is $ n\cdot n$ that means that $ n^{2}$'s prime factors include $ 2^{2}$ and $ 3^{2}$. The square root of $ n^{2}$ is $ n$, so, if we treat $ m^{2}$ as all other factors of $ n^{2}$,and $ m$ as an integer, then $ n^{2}\equal{}2^{2}\cdot3^{2}\cdot m^{2}$ which means that $ n\equal{}2\cdot3\cdot m$ or $ \boxed{\frac{n}{6}\equal{}m}$
It seemed so harmless, but it was hard-ish to explain. :lol:
[/hide]

assume that $ n\equal{}6k\plus{}r$ then $ n^2\equal{}36k^2\plus{}12kr\plus{}r^2\equal{}6m\plus{}r^2$ where m is a natural number
since $ n^2$ is divisible by 6 so $ r^2\equal{}0$ which means $ r\equal{}0$ , in other words ,n is di visible by 6

[quote="SoNYDvDPro"]Prove that if n is a natural number and n^2 is divisible by 6, then n is divisible by 6.[/quote]

[hide="Proof without word"]$ 6\mid n^2\implies 3\mid n^2\wedge 2\mid n^2$.

$ \begin{array}{c}3\mid n^2\implies 3\mid n \\
2\mid n^2\implies 2\mid n\end{array}\}\implies 6\mid n$.[/hide]

Standard generalisation:

An integer $ a$ is called squarefree, if it's not divisble by any perfect square other than $ 1$. Let $ k \geq 2$, show that:
$ a$ is squarefree iff $ a|n^k$ implies $ a|n$.

$ 2,3$ must appear in the prime factorization of $ n^2$. In the case that either $ 2$ or $ 3$ only had a multiplicity of one in said factorization, then $ n$ would not be an integer.  Thus, all forms of $ n$ who are integers must be divisible by $ 6$ given that $ 6\mid{n^2}$.


calculus
greatest common divisor
modular arithmetic
integration
number theory
algebra
difference of squares

What's the greatest integer that divides $ p^{4}-1$ for every prime p greater than 5?

12, 30, 48, 120, 240

gosh i'm useless in number theory...

explanations too please

Number theory doesn't belong in the calculus forum.  The solution is simply to consider the different primes, i.e. what power of 2 must divide that expression, what power of 3 must divide that expression, etc.

[hide="Hint"]
Consider $ p=6k+1$ and $ p=6k-1$ separately and take the $ \gcd$.[/hide]

A few nice facts:

All primes can be expressed in the form $ 6k \pm 1$
$ p^{4}-1$ can be expressed as as difference of squares:
$ (p^{2}-1)(p^{2}+1)$ which itself can have a difference of squares.
$ (p+1)(p-1)(p^{2}+1)$

[quote="DiscreetFourierTransform"]All primes can be expressed in the form $ 6k \pm 1$[/quote]

Don't be sloppy!

umm you guys should treat me as if I don't know anything about number theory, in fact, i really don't. Please explain.

[hide]First, note that this number must be a factor of $ 7^{4}-1 = 2400 = 2^{5}\times 3 \times 5^{2}$.
It must also be a factor of $ 11^{4}-1 = 14641-1 = 14640= 2^{4}\times 3 \times 5 \times 61$
This can serve as a good guide for us, since we already know the GCF is a divisor of $ 2^{4}\times 3 \times 5 = 240$

[hide="2s"]Note that all numbers can be written in the form $ 4k+1$ or $ 4k-1$

$ (4k+1)^{4}-1 = (16k^{2}+8k+1)^{2}-1 \equiv (8k+1)^{2}-1$ $ \equiv 64k^{2}+16k+1-1 \equiv 0 \pmod{16}$
and
$ (4k-1)^{4}-1 = (16k^{2}-8k+1)^{2}-1 \equiv (8k-1)^{2}-1$ $ \equiv 64k^{2}-16k+1-1 \equiv 0 \pmod{16}$

So, it's necessarily divisible by $ 16$.[/hide]
[hide="3s"]
Also note that primes greater than $ 5$ are of the form $ 6k \pm 1$.
$ (6k \pm 1)^{4}-1 \equiv (\pm 1)^{4}-1 \equiv 0 \pmod{3}$
So $ 3$ is a necessary factor.[/hide]
[hide="5s"]
$ (6k \pm 1)^{4}-1 \equiv (k \pm 1)^{4}-1 \equiv k^{4}\pm 4k^{3}+6k^{2}\pm 4k \equiv k^{4}\pm k^{3}+k^{2}\pm k$.

Indeed, this is equivalent to $ 0 \pmod{5}$ unless $ p=6k+1,k \equiv-1 \pmod{5}$ or $ p=6k-1,k\equiv 1 \pmod{5}$.

However, in the first case, we get $ p=6(5j-1)+1 = 30j-5$, which is not a prime, and in the second case, we get $ 6(5j+1)-1=30j+5$, also not prime.

So for all prime numbers, we have one factor of $ 5$.[/hide]

Compiling our results, we have $ 2^{4}$, $ 3$, and $ 5$, which are coprime, so the greatest common factor is $ 2^{4}(3)(5) = \boxed{\text{E. }240}$

[i]Why was this question marked as spam?[/i][/hide]

Some side explanations:

[hide]
All primes greater than $ 5$ are of the form $ 4k \pm 1$ because they are all odd. A number is either $ 4k-1$, $ 4k$, $ 4k+1$, or $ 4k+2$, and the other two categories are even only.

All primes greater than $ 5$ are also of the form $ 6k \pm 1$.
$ 6k$ is divisible by 6, $ 6k \pm 2 = 2(3k \pm 1)$, which is even, and $ 6k+3 = 3(2k+1)$, which is divisible by $ 3$. This means all primes must fall into the remaining two categories, $ 6k \pm 1$.

$ \equiv \pmod{n}$ (elementarily) indicates that two numbers differ by a multiple of $ n$. If a number is divisible by $ n$, then it mus be $ \equiv 0 \pmod{n}$, since $ 0$ and the number differ by the number itself, which would then be an integral multiple of $ n$.
[/hide]

[quote="DiscreetFourierTransform"]Ok, fair enough.

All except 2. (is that what you meant?)[/quote]
That is what I meant, but you're only half-way to fixing it :wink:

[quote="A M Y"][i]Why was this question marked as spam?[/i][/quote]
Because only one person (me) had rated it, and I rated it when it was in the Calculus forum, where it clearly doesn't belong.  The beauty of the rating system is that, rather than complain about the fact that I have done grievous damage to some other poster's status, or whatever, you can simply rate the post up again (which you and others presumably have done).


inequalities
algebra
function
domain
calculus
derivative
inequalities unsolved

i can't understand what you mean .please write completely

sorry ,i think when you let a>b>c then 1/1+a<1/1+b<1/1+c
it tells that  using this inequalities is the wrong way
i also use this way with null results

ah sorry i was stupid, i will try it later.

[quote="cuong"]let        0 <= a,b,c<=1 prove that:

(a/1+a)^3+(b/1+b)^3+(c/1+c)^3>=abc(1/(1+a)^3+1/(1+b)^3
+1/(1+c)^3[/quote]

A helpful idea for solving the problem would be to use LaTeX. ;)

[i][b]Problem.[/b]

If a, b, c are three reals lying in the interval [0; 1], then prove that

$(\frac{a}{1+a})^3+(\frac{b}{1+b})^3+(\frac{c}{1+c})^3 \geq abc((\frac{1}{1+a})^3+(\frac{1}{1+b})^3+(\frac{1}{1+c})^3)$.[/i]

All I have is a messy algebraic proof of this inequality (please excuse me - my head is aching of inequalities, I really must rest for a while until I can find any nice new ideas):

$((\frac{a}{1+a})^3+(\frac{b}{1+b})^3+(\frac{c}{1+c})^3) - abc((\frac{1}{1+a})^3+(\frac{1}{1+b})^3+(\frac{1}{1+c})^3)$
$=((a+1)(b+1)(c+1))^{-3}$
$\cdot(a(1-b^2c^2)+b(1-c^2a^2)+c(1-a^2b^2)+2bc(1-a^2)+2ca(1-b^2)+2ab(1-c^2))$
$\cdot((\frac12a^2+a+\frac12)(b-c)^2+(\frac12b^2+b+\frac12)(c-a)^2+(\frac12c^2+c+\frac12)(a-b)^2)$,

what is $\geq 0$ because the numbers a, b, c lie in the interval [0; 1]. This proves the inequality

$(\frac{a}{1+a})^3+(\frac{b}{1+b})^3+(\frac{c}{1+c})^3 \geq abc((\frac{1}{1+a})^3+(\frac{1}{1+b})^3+(\frac{1}{1+c})^3)$.

[quote="ondrob"]Didn't you mean (1/a+a)^3+(1/b+b)^3+(1/c+c)^3 instead of (a/1+a)^3+(b/1+b)^3+(c/1+c)^3 ?
Because as you write it, it is very easy:
8(a^3+b^3+c^3) >= 3abc >= abc(1/(1+a)^3+1/(1+b)^3
+1/(1+c)^3[/quote]

Ondrob, I fear I don't understand you. If you mean the inequality

$(\frac{1}{a}+a)^3+(\frac{1}{b}+b)^3+(\frac{1}{c}+c)^3 \geq abc((\frac{1}{1+a})^3+(\frac{1}{1+b})^3+(\frac{1}{1+c})^3)$,

then it is actually much weaker than the original inequality

$(\frac{a}{1+a})^3+(\frac{b}{1+b})^3+(\frac{c}{1+c})^3 \geq abc((\frac{1}{1+a})^3+(\frac{1}{1+b})^3+(\frac{1}{1+c})^3)$,

since for any positive number x, we have $\frac{1}{x}+x\geq\frac{x}{1+x}$ (in fact, $(\frac{1}{x}+x)-\frac{x}{1+x}=\frac{x^3+x+1}{x^2+x}\geq 0$).

  Darij

Left side of your inequality begin with (a/1+a). It means $\frac{a}{1}+a$ or $\frac{a}{1+a}$?

Oohoh, now I see (Darij's solution). I understand now :)

[quote="ondrob"]Left side of your inequality begin with (a/1+a). It means $\frac{a}{1}+a$ or $\frac{a}{1+a}$?[/quote]

Ah, that clears up the confusion! Of course, with a/1+a, he meant $\frac{a}{1+a}$, and not $\frac{a}{1}+a=2a$. This is very bad style, but people use it...

  Darij

It is clear that the domain is closed and that $f$ is bounded here.

Calculating second derivatives, we get that $\frac {d^2f}{dx^2}= \frac {6(1-x)(x+yz)}{(1+x)^5}\geq 0$ and similarly for the others so that $f$ is convex on $[0,1]^3$.  Unfortunately, the inequality sign is the wrong way to use this.

i can say darij's solution is so beautiful, i also find the way to prove it but such a terrible solution , i write this in two pages by:
let x=a/1+a , y=b/1+b ,z=c/1+c
then x,y,z<=1/2
after i used a lot of inequallities to solve it
it was too long

Very nice and hard inequality.
Using Chebyshev’s inequality, we get:
$RHS \geq  [\frac{a}{(a+1)^2}+\frac{b}{(b+1)^2}+\frac{c}{(c+1)^2}][\frac{a^2}{a+1}+\frac{b^2}{b+1}+\frac{c^2}{c+1}]$
and
$LHS \leq  [\frac{a}{(a+1)^2}+\frac{b}{(b+1)^2}+\frac{c}{(c+1)^2}][\frac{bc}{a+1}+\frac{ca}{b+1}+\frac{ab}{c+1}]$.
Thus, it remains to show that
$\frac{a^2}{a+1}+\frac{b^2}{b+1}+\frac{c^2}{c+1} \geq  \frac{bc}{a+1}+\frac{ca}{b+1}+\frac{ab}{c+1}$,
which is equivalent to
$(1-a^2)(b-c)^2+(1-b^2)(c-a)^2+(1-c^2)(a-b)^2 \geq  0$. ;)


calculus
geometry
integration
rectangle
Gauss
3D geometry
sphere

I am not sure what exams my son should take for college admission, SATII or AP. I don't want him spend time and energy on these exams unless they are absolutely necessary.
1) I noticed that most universities ask for 3 SAT II. However, I was told by many that AP's carry more weight (because they are harder exams). So, what would be the better strategy: taking SATII or AP? If YOU have to choose between SAT II and AP exams (instead of taking both), which would you choose (or which would add weight to your application)?
2) My child didn't do that well on verbal (570). He is not fond of writing either. He took math IIC (800). What would be good SATII subjects to take other than Writing? "Good" subjects to me are those that a student could study on his own without taking any courses. For example, I don't consider foreign languages be a good subject in this sense.
3) what are the differences between AP physics B and C (other than C uses calculus)? I looked at the discriptions; they seem to cover similiar topics. In fact, B seems to cover a few more areas than C. If a student studies AP physics B, will he sufficiently prepare himself for SAT II physics as well?
Thanks!

I really don't know that much, but here's what I know -

1.  SAT II, for the top colleges, are usually mandatory.  If you don't take them, it's very likely that you won't get admitted.  Aps on the other hand, are different.  Colleges don't require you to take them, but since they're much harder, they have more weight, and might even get you credits if you get a good enough score on the test.

2. I'm pretty sure that SAT II writing is mandatory for most colleges.

3.  AP physics C probably goes more in depth thatn Physics B, which is why Physics B can cover more areas.

ditto on 1 and 2. Take all of the tests you can - SAT II and AP serve two very different and distinct purposes. If money is a concern, there are fee-reduction options; talk to someone at the collegeboard or a guidance counselor at your school.

3. Physics C is much more in depth than Physics B. Also, do not just write it off as "it uses Calculus while B uses algebra" - it uses calculus because most of physics can not be expressed using algebra beyond a very basic and simplified level. For example, Physics C: Electricity and Magnetism requires a fairly indepth knowledge of Maxwell's equations, which govern ALL phenomena of electricity and magnetism down to the quantum scale (at which point they cease to hold), while Physics B requires naught but the knowledge of a few equations with three variables each, which only hold for DC circuits that are not in a changing magnetic field, etc etc. (Okay, a few of them also hold for AC circuits, with a few modifications, but the point remains.)

Yes, the Physics B test will prepare one for the SAT II Physics test, with the possible exception of a few minor (and easy) topics. However, topics are weighted differently between the two tests, so while one thing may be very important for passing one test, it may not be necessary for the other.

I agree with Matt's response to 1 and 2; take as many as you can.  However, from what I've heard, colleges don't care that much about SATIIs (that is to say, they can't make you stand out that much; every competitive applicant takes at least three, including the writing) compared to APs.  So if your son is looking to strengthen his college app, then he should try to take lots of APs rather than SAT IIs.

And Physics B will easily prepare a student for the SAT II Physics (you might need to review a few things, but most of it is common to both exams).  But in general, taking an SAT II in the same subject that you've already taken an AP in does little to help your app, unless you're still looking for the three SAT IIs.

Although it does add up, fee is not a big issue. But time is. I would like to give my son enough time to explore and "play" (computer games, math competitions, Legos, etc.) To Prep for exams does not have much of value other than for college application.

[quote="h_s_potter2002"]
2. I'm pretty sure that SAT II writing is mandatory for most colleges.
[/quote]

I think that SAT II Writing will be eliminated, since writing will be part of the SAT I. Same is true for SAT II Math I. Some colleges already ask applicants to take Math IIc and two others. I thought with a low verbal score, my son probably should take subjects that give his test report some balance. I would love to hear suggestions from you, especially from someone who may not be as strong in lib. arts as in math/science.

It sounded to me that Physics C might be a good choice. A student may gain a deeper understanding of physics as well as prepare for AP B/C and SAT II.

[quote="confuted"]
topics are weighted differently between the two tests, so while one thing may be very important for passing one test, it may not be necessary for the other.[/quote]
That's very interesting. In what ways do they weight topics differently, and what is the main focus of each test in your view?

And how much and to what extent does an EM course use calculus?

Thank you all for your thoughts and insights.

[quote="istudio"]It sounded to me that Physics C might be a good choice. A student may gain a deeper understanding of physics as well as prepare for AP B/C and SAT II.
[/quote]
Providing that he has some grounding in physics going in, or is planning on teaching himself (and is prepared to learn the basics before jumping into the "meat" of physics C), that would be a wonderful idea. Physics is certainly a lot of fun.
[quote="istudio"]
That's very interesting. In what ways do they weight topics differently, and what is the main focus of each test in your view?

And how much and to what extent does an EM course use calculus?
[/quote]
I'll answer the second question first. I wouldn't say that the EM course [i]uses[/i] Calculus so much as it [b]abuses[/b] it. (Mechanics does a much better job of actually using it) I should explain the abuse comment. In a calculus course, students will learn how to do things the "right way," like finding $\displaystyle\int_0^3 x^2\,dx$. There's plenty to learn like that. EM, however, will throw in the $\displaystyle \oint$ from multivariable calculus (Calc III), but the student won't have to actually know the  [i]Calculus[/i] behind that symbol; it will be more important to understand the physical concept. If the student understands that the $\oint$ is taking the contributions from a bunch of small "surface patches" rather than a bunch of small rectangles, that's pretty much all that's necessary, because the differentials in the integrals don't use the same variable as the rest of the terms. Also, there will be no bounds on any of the integrals. Assuming you have a basic knowledge of Calculus, I'll demonstrate:

Gauss' law for electricity is $\displaystyle \oint{\vec{E}\cdot d\vec{A}}=\frac{q_{enc}}{\epsilon_0}$. $q_{enc}$ is the amount of charge "inside" the closed surface that you pick in order to find the electric field $E$ from the charge $q_{enc}$. $\epsilon_0$ is a constant. The explanation is, of course, not adequate, but if you want more, I suppose you can read about it on any of the thousands of websites, or in a book, or even in my paper [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicExam.php]here[/url].  But suppose we have a positive charge $Q$, and for the closed surface around it, we choose a sphere of radius R, to make our math easy. Physically, we're allowed to choose any closed surface we'd like, and it will come out the same, but there's no sense in choosing something which would make the math hard.
$\displaystyle \begin{eqnarray*} \oint{\vec{E}\cdot d\vec{A}&=&\frac{Q}{\epsilon_0}\\
(\vec{E})(4\pi R^2)&=&\frac{Q}{\epsilon_0}\\
\vec{E}&=&\frac{Q}{4\epsilon_0\pi R^2}\end{eqnarray*}$
(In the second step, I just substituted $4\pi R^2$ for $\oint d\vec{A}$ because that's the surface area of my sphere. I used no real calculus).

I don't know how well I just made my point, but essentially every time calculus appears in E&M, it's just a "cheat" to get rid of it, and the equations are expressed using Calculus to show that they hold [i]in general[/i] and not just in special cases like the algebra equivalents. Mechanics uses Calculus "correctly" a wee bit more, but on the AP test, the student is often expected to do little more than set up the integral (write down the equation and plug in the variables) and perhaps perform a few simple manipulations.

On a different note, however: You said that you wanted your student to have time for fun, and not spend all his time studying for the SAT IIs and AP tests. The remedy for this is rather simple and much more honest: do not study for standardized tests meant to test one's existing abilities. If one wishes to learn something new, then learn it for the sake of knowing it, and then take a test to prove that you now know it, if need be. The test should never be the motivation to learn; studying for those tests cheats yourself. If you've no passion for the subject, but cram facts into your head for a few days before the test, ace the test, and promptly forget the material as you never cared in the first place - [i]what have you shown?[/i] My answer is "nothing of value."

I just realized that I answered the second question first, and did so in great depth, and did not answer the first question. To avoid making the post overly long with an edit, I'll just post again. You asked how they are weighted differently.
[b]SAT II[/b]
Mechanics - 34 to 38%
Electricity and Magnetism - 22 to 26%
Waves - 15 to 19%
Heat, Kinetic Theory, and Thermodynamics - 8 to 12%
Modern Physics (Quantum, Atomic, Nuclear, Special Relativity, etc) - 8 to 12%
Miscellaneous (measurement, math skills, lab skills, history, etc) - 2 to 4%

[b]AP Physics B[/b]
Mechanics - 35%
Thermal Physics - 10%
Electricity and Magnetism - 25%
Waves - 5%
Optics - 10%
Atomic and Nuclear - 10%
(It's a bit hard to see the difference between Physics B and the SAT II when breaking it down into such broad categories; the differences become more apparent on individual topics, but I don't have such a detailed breakdown of the SAT II)

[b]AP Physics C[/b]
Mechanics - 50%
Electricity and Magnetism - 50%
Wave motion, and Thermal, Optical, Atomic, Nuclear, Quantum, Relative, and historical physics are not covered.

I can only speak for my own university here - an "urban comprehensive university" that awards masters degrees but not (for the most part) doctorates. From my perspective, I see SAT II's and AP's as serving very different purposes.

The SAT II's are admissions tests, used (among many other criteria) in the decisions about whether to admit a student and forgotten about afterwards. We don't even ask for them, although if a student chooses to supply them, there's a very slight chance they might exempt that student from our own Math and/or English placement tests (tests to determine whether remedial courses are required.) Whether to take them, and which ones to take, depend on the admissions expectations of the colleges you intend to apply to.

AP exams are taken not as admissions tests, but to actually get college credit for the subjects in question. Note that on average high school students take the largest number of AP exams in their senior years. They take the tests in May but the results aren't known until mid-July. That's far, far too late to be used in an admissions decision, and late enough that it requires us to use some guesswork with summer freshman orientation and registration. Yes, an admissions committee could make use of AP exams taken in the junior year or earlier, but they aren't going to expect that of everyone they see.

How these exams are accepted for course credit varies widely - not just university to university, but often department to department within the same university.

Some comments on specific AP exams:

AP English is always a good thing - preferrably the Composition and Literature exam. If you're taking any AP exams, and you're sufficiently far advanced in English to be allowed to take this in your senior year, do so. In our case, passing this exempts you from taking our standard freshman English course. Other colleges may vary widely, but assume that this credit is always a good thing.

AP Physics B versus AP Physics C: In our case (and this probably holds true in a reasonable number of other places), Physics B earns credit for a specific sequence of Physics courses that counts for general education science and counts for Biology majors, but does [i]not[/i] count for engineering or physical science majors. (Even though this meets the major requirements for Biology majors, many medical schools are reluctant to accept AP exams for meeting their admissions requirements.) In our case, we do accept each of the halves of Physics C as meeting a semester each of the Physics requirements for Engineering, Physics, Chemistry, Geology, and Math majors. Around the country, Physics departments may be among the most reluctant to accept AP credit as credit - check the details for where you're going.

Yes, there's a lot of calculus in an E & M course - in fact, it should be vector calculus, with divergences, curls, line integrals, surface integrals, and so on. Since that lies beyond the syllabus of AP Calculus BC, I don't really know what the expectations of AP Physics C E & M are.

If you are a Biology (or Biochemistry) major, the single AP exam that would do you the most good (as far as we are concerned) is AP Chemistry. We're serious about this mattering more than AP Biology.

There are several AP exams that bear little relation to any specific course credit, especially for someone planning to major in that subject or something related. Examples of this include AP Environmental Science and AP Statistics. (AP Statistics may do a social science major some good, but the path to a statistics major goes through calculus). Physics B more or less falls into this category, since it's not for Physics or Engineering majors. I wouldn't encourage you to spend too much effort on these.

The fees for AP exams, high as they are, are much, much lower than the tuition cost for taking that many units of college courses. For the most part, they don't lead to students graduating in less than four years, (so if you go to an expensive college they will get you to pay that much) but they do lower the chances of taking longer than that.

[quote="Kent Merryfield"]Yes, there's a lot of calculus in an E & M course - in fact, it should be vector calculus, with divergences, curls, line integrals, surface integrals, and so on. Since that lies beyond the syllabus of AP Calculus BC, I don't really know what the expectations of AP Physics C E & M are.[/quote]
The AP Physics C E&M exam uses the integral forms of Maxwell's equations, which are mathematically much simpler than the differential forms - which is how they expect students to handle it.

[quote="Kent Merryfield"]
The fees for AP exams, high as they are, are much, much lower than the tuition cost for taking that many units of college courses.[/quote]
Indeed; taking (and passing with 4s and 5s all around) 13 tests, I was awarded 56 (fifty-six!) credits from MTU; the current cost for a credit hour is 235 dollars, meaning I got over 13,000 dollars worth of credits, which doesn't even take into account the time it would have taken or the cost of room and board. Of course, I wouldn't have taken all of those exact credits at the university, but I still saved many thousands of dollars on classes I would have taken. After some fee reductions, I spent a total of $296 on AP exams. Surely, that's a good return on the money. Take as many AP exams as you can (and more!)

Thank you all! I now began to see the picture more clearly. I was a bit surprised by the topic breakdown for different physics exams.
I didn't know one could ACTUALLY use APs for college units (not merely adding some weights to the application).  And I never thought taking AP would save money and time in a long run. I like that a lot!
Here are some more questions (Sorry ;) )
1) Do all or most universities accept APs for credits as MTU?
2) How do universities view community college transcript in comparison to an AP report? I recognize that AP is standardized while the qulities of community college courses vary from campus to campus.
3) Matt had 56 AP units, which was very inpressive and was as many units as what was asked of a transfer student. Again, if a student already took college courses (say, calc and physics) at a community college, is there any point to take AP in the same subjects? Vice versa?

[quote="Kent Merryfield"]
How these exams are accepted for course credit varies widely - not just university to university, but often department to department within the same university.[/quote]

That's what I was afraid of. What does WTU stand for, Matt?

[quote="Kent Merryfield"]AP English is always a good thing - preferrably the Composition and Literature exam. If you're taking any AP exams, and you're sufficiently far advanced in English to be allowed to take this in your senior year, do so. In our case, passing this exempts you from taking our standard freshman English course. [/quote]

It would be great if one doesn't have to take freshman English or other required lib. arts courses.

[quote="istudio"]1) Do all or most universities accept APs for credits as MTU?[/quote]

Practices vary widely. Each university sets its own policies for dealing with AP credits. The College Board has a [url=http://www.collegeboard.com/ap/creditpolicy/]Web page on AP credit policies[/url] with sublinks to lots of different colleges' policy pages.

To give one example, Harvard University doesn't accept credits from ANY other college as credits toward Harvard graduation, but will give a student advanced standing (such that the student can graduate in three years rather than four) if the student has a specified set of AP scores (all 5s in the required tests) or IB scores. Harvard nonetheless advises [url=http://www.fas.harvard.edu/~fdo/publications/0304/as/general.htm#5]it may not be a good idea to take advanced standing[/url], and not many Harvard students do.

By the way, [url=http://www.mtu.edu/]MTU[/url] is Michigan Technological University of Houghton, Michigan on the Upper Peninsula north of Wisconsin. It hosts a Minnesota Public Radio affiliate station.

[quote="istudio"]2) How do universities view community college transcript in comparison to an AP report? I recognize that AP is standardized while the qulities of community college courses vary from campus to campus.[/quote]

This also varies immensely. A fairly commonly encountered situation is that each state will regulate the transferability of credits from community colleges in that state to that state's flagship university. Anything goes as to out-of-state universities, public or private, taking transfer credits. It's always most prudent to ask in advance.

Again, Harvard's answer will be that credits from other schools do NOT count toward Harvard graduation. However, they can fulfill course prerequisites for more advanced Harvard courses. Note that incoming math majors at Harvard are advised NEVER to go beyond the [url=http://www.math.harvard.edu/undergrad/Pamphlets/freshmenguide.html]legendary Math 55 course[/url] in their first year, no matter how many previous math courses they have taken.

[quote="istudio"]3) Matt had 56 AP units, which was very inpressive and was as many units as what was asked of a transfer student. Again, if a student already took college courses (say, calc and physics) at a community college, is there any point to take AP in the same subjects? Vice versa?[/quote]

I expect to have my son take some AP tests simply to ratify that he learned something in for-credit courses he will take on some of the AP subjects. Some courses he takes are noncredit courses organized by our local homeschooling group, some are in the [url=http://www.itcep.umn.edu/umtymp/]University of Minnesota Talented Youth Mathematics Program[/url], and some will eventually be through the [url=http://www.mrs.umn.edu/academic/advising/pseo%20students.htm]Post-Secondary Enrollment Options Program[/url] at the U of MN. He will take AP courses in selected subjects as he learns the material.

Hope this helps! Good luck in your planning, and feel free to ask follow-up questions as needed.

[quote="istudio"]1) Do all or most universities accept APs for credits as MTU?
2) How do universities view community college transcript in comparison to an AP report? I recognize that AP is standardized while the qulities of community college courses vary from campus to campus.
3) Matt had 56 AP units, which was very inpressive and was as many units as what was asked of a transfer student. Again, if a student already took college courses (say, calc and physics) at a community college, is there any point to take AP in the same subjects? Vice versa?[/quote]
1) Nearly all; criteria and credits awarded differ by school. MTU is pretty generous with the credits; MIT, on the other hand, accepts nothing or nearly nothing, simply because the institute does not feel that AP is up to its standards. Some colleges/universities also place an upper cap on the number of credits you can recieve.
2) I've no idea. It probably differs between colleges.
3) I wouldn't think so, assuming a) the credits will transfer and b) no further credits could be gained by taking the AP test.

[quote="istudio"]It would be great if one doesn't have to take freshman English or other required lib. arts courses.[/quote]

I'm not sure what you were thinking when you wrote this, but I'd like to say something. As an academic advisor (which is what I am), I don't particularly like to see students having schedules with nothing but math and science/engineering courses on them in any particular semester. It's healthier and more balanced to also go visit the other side of the campus. The real advantage of having AP credit in English and in US History is that it may free you up to take deeper and more interesting humanities courses earlier in your college career. I also like to see those with AP credit in a second language either take the next course or two in that language (language departments tend to be experienced at figuring out the appropriate placement) or start a third language.

[quote]How do universities view community college transcript in comparison to an AP report? I recognize that AP is standardized while the qualities of community college courses vary from campus to campus.[/quote]

Now I'm back to speaking only for myself (and for CSULB). We do accept community college courses - after all, something like half of our graduates transfer to us after taking 60-70 units at a community college. Yes, the quality varies - but the quality can just as easily vary from instructor to instructor at the same institution. We would regard an AP exam and a community college credit in the same subject as redundant - we'll only give you credit for it once. However, there are courses like Calculus III, Linear Algebra, and Organic Chemistry for which no AP exam exists because they lie beyond the existing tests. The only way to transfer credit for such subjects is to transfer them from a college.

[quote="confuted"]Of course, I wouldn't have taken all of those exact credits at the university, but I still saved...[/quote]

Every year, we have a couple of dozen freshmen with 40-55 units of AP credit. It's quite rare for any of them to try to graduate in three years - mostly they take at least four. They are significantly less likely to take 5, 6, or 7 years to graduate. I wouldn't emphasize the money saved as much as I would point out that for the same money and time, you get to pursue your degree of choice with less time pressure, less postponing of vital requirements until your senior year, and more time to explore those parts of your subject that lie beyond the bare minimum for graduation.

Some state universities - I know this happens in the UC system - have rules against continuing at the unversity after you have already accumulated a certain number of units over the minimum number required for graduation. Generally, AP units - and possibly other units earned while still in high school - will not count against this limitation.

[quote="Kent Merryfield"][quote="istudio"]It would be great if one doesn't have to take freshman English or other required lib. arts courses.[/quote]

I'm not sure what you were thinking when you wrote this, but I'd like to say something. As an academic advisor (which is what I am), I don't particularly like to see students having schedules with nothing but math and science/engineering courses on them in any particular semester. [/quote]

I was thinking about "REQUIRED" freshman English and lib. arts [b]courses.[/b] I agree with you whole heartedly on having a well-rounded education. However, I would much prefer [b]required units[/b] for different subject areas instead of [b]required courses[/b]. "Required units" in various areas will encourage students to develop a balanced curriculum and at same times give students some freedom to choose based on their interests. Matt made an excellent point on learning for passion in his earlier post.

There are also couple of other reasons: first, I compared the scope of AP English with that of freshman English. I have to say that AP Eng seems harder, more objective (graded by collegeboard not an instructor/prof.), more substantial, and more focused on lit. and lang. Secondly, I had my child's age in mind. A student may be able to understand Diff. Eq or quantum physics, but she or he might not mature enough to discuss Coming to Age or Catcher in the Rye. Lib. Arts involve different kind of mental abilities, critical thinking skills, or experience.

[quote="Kent Merryfield"]Some state universities - I know this happens in the UC system - have rules against continuing at the unversity after you have already accumulated a certain number of units over the minimum number required for graduation. Generally, AP units - and possibly other units earned while still in high school - will not count against this limitation.[/quote]
Thank you for pointing this out!!! I think I have to be very careful about units earned prior to transfer - this is another thing that I would have never thought of or known about, had you not mentioned.

tokenadult, you are so very resourceful. I really enjoyed your website. Like you, I am homeschooling my kids. Your website is a treasure to me. Thanks.

[quote="istudio"]I forgot to ask this:
Some universities (e.g. Berkeley and Stanford) require at least two of the three foreign languages: Russian, German, and French for a Ph.d in Math. Why? And why not other languages such as Spanish, Chinese, etc.?[/quote]

You can add Princeton to that list. I just came across that requirement on a Web search the other day. Why those languages and not others? Mostly because those are the languages in which the best non-English-language mathematical literature was published during most of the twentieth century. Today, I would definitely add Chinese to that list, as it is surely even more useful for CURRENT literature than is German (although so many classic papers in mathematics, notably Riemann's paper that posed the Riemann hypothesis, were written in German that it would be a shame not to know some German). Spanish, alas, has a lot of speakers but not nearly so much high-level scholarship published in that language as French, a language with many fewer speakers. Most of the Russian or German journals I have perused at the U of MN Math Library are publishing most of their articles in English, as are most Japanese journals. French journals seem to have the strongest tendency to publish articles in the language of the country of publication rather than in English.

Hope this helps!

physics B supposedly has everything you need to learn for SAT II, but of the people i know, only the physics C students can get 800 on sat II, just because if you know more in depth about physics in general, Sat II questions become common sense


Let $ f: X\to Y$ be a function. Prove that there exists a function $ g: Y\to X$ such that $ f\circ g \equal{} I_{y}$ if and only if $ f$ is a surjection.

So we use the axiom of choice to construct the function $ g: Y\to X$. We construct $ g$ by picking a point in the preimage of $ y$ which is an element $ x\in X$. $ f\circ g \equal{} I_{y}\implies f(g(y)) \equal{} f(x) \equal{} y,\ \forall x\in X$ and so $ f$ is a surjection.

Given that $ f$ is a surjection then $ \forall x\in X, y \equal{} f(x)$. This implies that $ f\circ g \equal{} I_{y}$.

Is this correct?

Please edit that to state the problem in a way that doesn't stretch the page.

This does require the Axiom of Choice in general. Construct $ g$ by picking a point in the preimage of $ y$ for each $ y$; these are nonempty by hypothesis.

Your logic doesn't really make sense. Separate the two directions. One is a proof, the other is a construction.

Why is there a need to construct the function $ g: Y\to X$? Isn't it already implicitly given?

You have two things to prove:

1) If $ f$ is a surjection, there exists a function $ g$ such that $ f\circ g \equal{} I_{Y}$.
2) If there exists a function $ g$ such that $ f\circ g \equal{} I_{Y}$, then $ f$ is a surjection.

You've muddled these two parts up with each other.  In particular, your first argument begins, "So we use the axiom of choice...," which is the beginning of a construction to solve (1).  However, that argument ends, "... and so $ f$ is a surjection," which should be the conclusion of a proof of (2).  You should clearly state your premises for each part of the argument.

I want to prove that $ f\circ g \equal{} I_{y}\Leftrightarrow f$ is a surjection.

So I first did the $ \Rightarrow$ direction: Construct $ g$, and given $ f\circ g \equal{} I_{Y}$ prove that $ f$ is surjective.

Then I did the $ \Leftarrow$ direction: Given $ f$ is a surjection prove $ f\circ g \equal{} I_{Y}$.

[quote="ragingbullrocky"]So I first did the $ \Rightarrow$ direction: Construct $ g$, and given $ f\circ g \equal{} I_{Y}$ prove that $ f$ is surjective.[/quote]

No.  You've muddled the two together.  The $ \Rightarrow$ direction is: [i]Given[/i] a function $ g$ such that $ f\circ g \equal{} I_{Y}$, prove $ f$ is surjective.  Similarly, the $ \Leftarrow$ direction is: given a surjection $ f$, [i]prove there exists[/i] (= construct, in this case) a function $ g$ such that $ f\circ g \equal{} I_{Y}$.


inequalities
function
inequalities proposed

Let $a, b, c>0$ such that $abc=1.$ Prove the following inequality:

    $\frac{a^{2}+bc}{a^{2}(b+c)}+\frac{b^{2}+ca}{b^{2}(c+a)}+\frac{c^{2}+ab}{c^{2}(a+b)}\geq ab+bc+ca.$ ;)

Easy...

Notice that $ab=\frac{1}{c}, ac=\frac{1}{b}, bc=\frac{1}{a}$

Then write the given inequality in the form

$\frac{(a-b)(a-c)}{a^2(b+c)} + \frac{(b-a)(b-c)}{b^2(a+c)} + \frac{(c-a)(c-b)}{c^2(a+b)} \geq 0$

WLOG $a \geq b \geq c$

Factor $(b-c)$ from the second and third term to obtain

$\frac{(a-b)(a-c)}{a^2(b+c)} + \frac{a^2(b-c)(b^2-c^2)}{b^2c^2(a+b)(a+c)} \geq 0$

Which is trivially true.

Or you may deduce the last inequality by Sander's Inequality.

(Actually Sander's Inequality is perfectly the same as Vornicu-Schur's Inequality. Sander proved it before Vornicu. Sander is an undergraduate student at the University of Illinois at Urbana-Champaign. In ML usually one refer Sander's Inequality to Vornicu-Schur's Inequality.)

[quote="Detective"](Actually Sander's Inequality is perfectly the same as Vornicu-Schur's Inequality. Sander proved it before Vornicu. Sander is an undergraduate student at the University of Illinois at Urbana-Champaign. In ML usually one refer Sander's Inequality to Vornicu-Schur's Inequality.)[/quote]What year is did he find it? A friend of mine has an old course (1992 I believe) in which it states that convex functions are schur functions.

[quote="Detective"](Actually Sander's Inequality is perfectly the same as Vornicu-Schur's Inequality. Sander proved it before Vornicu. Sander is an undergraduate student at the University of Illinois at Urbana-Champaign. In ML usually one refer Sander's Inequality to Vornicu-Schur's Inequality.)[/quote]

I'm going to make the wild guess here that you are Sander.

[quote="cezar lupu"]Come on, it's not Sander's inequality. It's just the extension of Schur. ;)[/quote]
Agree. Come on, this must have been discovered by many many people here before. But here, in Mathlinks, we must show some respect to our beloved webmaster.  :D  :D

With all respect that I have for Valentin, this isn't his inequality. :|

Why we need to repeat the same things? Or  the ineq is Vornicu or Sander or generalised Shur it is the same thing because this is the same ineq. :!: So don't stick at names. Only we use names , because we want the others  understand what ineq we use. So in mathlinks we name this ineq as Vornicu-Shur, so this is its name.


This is a thread for you to post information about you, your avatar, your signature, and anything else you would like the rest of the Art of Problem Solving - Mathlinks community to know about yourself.

Remember that all information posted here is available to anyone who visits this website. Thus, I advise you to use discretion in what personal information you choose to disclose. Parents of children under the age of 13 can choose to have their child's information removed. Also, you may remove your post or ask a moderator to do so for you at any time.

Keep in mind that this thread could and probably will get very long over time as more and more users post. So, please just make one post on this thread. In the future, should you like to add or change information, please just edit your post.

Also, this is not a thread for discussion. If you would like to discuss what someone writes here, please quote them and start another topic.

I'm a 12 year old homeschooler living in Minnesota. My user name is what I like to envision myself as, and my signature tagline is an antischool quote. My avatar goes hand in hand with my username.

anybody here have a xanga? mines is [url]http://www.xanga.com/poisoned_cherries[/url]

My avatar is a photoshopped gopher with a skull on his head and a chicken bone. Its made to look like a cubone (pokemon).

my siggy says where i got it and the amount of people who were on at once

I am 13, homeschooled forever, and like gardening and web pages. My avatar is a daylily that I sort of like, but I like the picture more.  We have 16 sheep, and I have 40 chickens. The sheep's relatives are rather tasty.

Billy Dorminy

hi, my name is David and I'm a mathoholic.
i go to public school (i know, it's amazing since this a "smart forum") and i feel left out and lonely. i have no friends. my parents take advantage of me. i have no money and live in a trailer. i drive a crappy neon and don't have a job. and i live in south carolina, redneck capital of the world.

Hey! I'm Kristy. I'm 11, about to be 12 in two days. i'm in sixth grade. i feel bad sometimes not knowing a lot of the stuff, but i live. i have a bunch of questions about how do u get those avatars and all that, and also i'm hidden somehow and i don't know how to get it to go back. anyways,I live in Vestavia Hills, Alabama.  :)

and i have a question, are homeschooled ppl usually smarter than public school ppl?

Hey, my name is what my username is (not literally)
My avatar is some character from my favorite game.
My sig, I had no idea what i was going to put for my sig so i just put something.
Oh yah, and i'm twelve soon to be 13 on jan. 10.:D

I almost forgot I love HALO, it's awesome! I like doing math and love playing games!  :D

My name's Lynnelle and I'm going to be thirteen on Jan. 15. Besides the obvious (math) I also love to read, write, and draw. My avatar is a random lightbulb I made on MS Paint. I forget why I made it but it seems appropriate for this forum. My siggy is pretty much my personal motto.

[quote="Jeffreywang"]Oh yah, and i'm twelve soon to be 13 on jan. 10.:D[/quote]
You're exactly five days older than me!

Hi! I'm Emily, but I usually go by Emma. I absolutely LOVE maths! I also enjoy soccer, surfing the net, chattin on IM (feel free to say hello sometime), reading, and singing. I'm currently a freshman (9th grade) and will be 15 on January 5. hoorah! My avatar is Josh Groban, who is my absolute FAVORITE singer ever! He's AMAZING!!!  :lol:

Hi! I'm Jarey - Yeah. It's just pronounced Jerry. I like to be unique!  :D . BTW: These smileys are way too cool. I am a freshman, and I'll turn 15 August 8th. Yeah - A long way. My signature is expressive of my pessimism - yes I'm a down to earth platonic type of guy. Very pragmatic. My Avatar is my old favorite game - Starcraft. My username, is well, random. Yes. I love math, and I love HALO. And I love to debate. Yes. That about sums it up. And I play saxophone, the best instrument ever.

Hey guys, I'm Yifan. To me, math is pretty cool and fun. I like playing Halo, as well as playing baseball, listening to music, sleeping and some other stuff.

My avvy is the Red Sox logo. As you can see I'm a huge Red Sox fan. It used to be a "Yankees suck" logo, but the Sox lost 5 straight games after I changed it to that, so I took it away. This works good though, since they won the World Series.

Hi my real name is Katie.  I live in the Fort Worth Dallas area of Texas and I absolutly love math.  I go to public school which is way too slow for me especially in math.  I also like to work with horses of which I have 2 of my own.  I am 14 and will be 15 on November 2nd.  I am an 8th grader but I am taking some high school classes for credits.

btw I also like theses smilies  :lol: they are so holiday like.

My avatar is SN standing for Super Naterul. I live in Washington. I homeschooled.  My signature is just a saying that my brother heard somewhere. I like the smileys and especially these: :alien:  :rotfl:  :ninja: .

I'm MAcky/Mause/Vito/Marianne (I honestly don't care which.), and I don't understand hte purpose of avatar explanations, but mine is a photo of myself.

I'm a modeler and graphics artist at Buchholz High. I'm not insanely sociable. XD I'm also obsessive and infatuated. I live in Gainesville,FL [huge stupid mistake... obnoxious fool town in which it is impossible to get a 6000sq ft brick house built] now, but I used to live in Silicon Valley and Tampa, as well as a few palces inbetween, including Italy.

I'm very into programming, statistics, and economics. I love Macs. And to risk sounding even more strange, I drive a Mercedes SLK convertible...

Please don't post here desriptions anymore. Refer to this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=15747 from now on :)


function
logarithms
algebra proposed
algebra

For any positive integer $n$, we define
\[f( n) = n +\max \left\{ m \in \mathbb N \cup \{0\}: 2 ^{2 ^{m}} \leq  n\cdot 2^{n}\right\}.\]
Find the image of function $f$.

[quote="diegu"]for any positive integer,we have:
$ f( n) \equal{} n$ +$ max \{ m \in N : 2 ^{2 ^{m} } \leq n\cdot 2^{n}\}$

find the image of function f[/quote]

[u]Remark[/u]: just notice that in order $ f(1)$ be defined, the $ \mathbb N$ inside the definition must include $ 0$, else $ max \{ m \in N : 2 ^{2 ^{m} } \leq 1\cdot 2^{1}\}$ would not be defined :(
So I consider the problem rewritten as :
"Let $ \mathbb N$ be the set of positive integers and $ f(n) : \mathbb N\to\mathbb N$ such that $ \forall n\in\mathbb N$ :  $ f( n) \equal{} n$ +$ max \{ m \in \mathbb N\cup\{0\} : 2 ^{2 ^{m} } \leq n\cdot 2^{n}\}$. Find $ f(\mathbb N)$"

Then :
$ f(n) \equal{} n \plus{} [\log_2(n \plus{} \log_2(n))]$  and so :

$ \forall n\in[2^k,2^{k \plus{} 1} \minus{} k)$ : $ f(n) \equal{} n \plus{} k$
$ \forall n\in[2^{k \plus{} 1} \minus{} k,2^{k \plus{} 1})$ : $ f(n) \equal{} n \plus{} k \plus{} 1$

So $ f([2^k,2^{k \plus{} 1}) \equal{} [2^k \plus{} k,2^{k \plus{} 1} \minus{} 1]\cup[2^{k \plus{} 1} \plus{} 1,2^{k \plus{} 1} \plus{} k]$

So $ f(\mathbb N) \equal{} \mathbb N\backslash\{2^k,\forall k\in\mathbb N\}$


Congratulations for the brilliant idea of having a section of open question. It's absolutely wonderful!

thanks.

if you have any other ideas, put them on the wish list, you never know when they are going to become true :D :D

Even if latex is not up yet,i think that we should familiarize the users with latex and point them to speicific Latex guides,Latex command references.We could suggest Latex editors etc. etc.

Windows Latex Editors:

Commercial Ware:

WinEdt [url]http://www.winedt.com/[/url]
MathType [url]http://www.dessci.com/[/url]

Freeware:

TexShell [url]http://www.projectory.de/texshell/[/url]
WinShell [url]http://www.winshell.de/[/url]
TexnicCenter [url]http://www.toolscenter.org/products/texniccenter/[/url]

Linux Editors:

LyX [url]http://www.lyx.org/[/url]
Klyx [url]http://www.lyx.org/about/klyx.php3[/url]
Kile [url]http://kile.sourceforge.net/[/url]
Texmaker [url]http://perso.club-internet.fr/pascal.brachet/texmaker/[/url]
Emacs [url]http://www.gnu.org/software/emacs/emacs.html[/url]

Latex Reference:

[url]http://www.ctan.org/[/url]
[url]http://www.latex-project.org/[/url]

Edit:Added Mindspa's suggestions[/url]

And [url=http://www.mathlinks.ro/viewtopic.php?t=38]here[/url] are some instructions that Valentin once gave to those people who don't have LaTeX and would like to get a free (31-day trial, and if you can stand the recurring message, longer I suppose) version of LaTeX online: (I edited them for clarity.)

To install a Windows-based version of LaTeX editor:

1) Install MikTex (http://www.miktex.org/), a LaTex/Tex environment which will compile the document you write and create an .dvi file which can be viewed/printed with a viewer included in the MikTex package.

2) Install WinEdt (http://www.winedt.com/).  Don't bother to read through the many help files, because if you don't have time to read them throughly they will confuse you even more.

3) After installing WinEdt, it should detect the presence of the MikTex environment - if it doesn't then you should try to reinstall MiTtex again, and if this doesn't solve the problem then start reading those docs ;)

4) Download an example file (http://www.mathlinks.go.ro/style/example.tex) to get the necessary headers to cut and paste.

i think a poll for the survey would be better.

In my opinion Texmaker and emacs are together much better for Linux.

I've heard of them before but I never use Linux., so I don't know the real effect...

[quote="liyi"]how many users here use LaTeX?[/quote]

Me,me,me  :D


This is '08 National Target, #8.

To win a game with a fair coin, you must flip $ n \plus{} 1$ heads in a row, where $ n$ is the total number of tails flipped so far. So if your first flip is heads, you win and the game is over; if your first flip is tails but your next two in a row are heads, you win and the game is over; and so on. What is the probability that you win by flipping exactly four heads in a row? Express your answer as a common fraction.

I know the [hide="answer,"] $ \frac{21}{1024}$ [/hide] but I do not know how to get that answer. Please help!

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1459137#1459137]this[/url] for a very badly written post of mine that shows almost no steps at all.

Alternatively, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1961107126&t=274892]this[/url] for a slightly better written post. Or you could search "2008 national target" and hopefully find some better threads.

Incase you're wondering, only five people solved this problem at the actual competition in 2008.

This involves some casework, but don't overthink it like I did... :blush:


Hi

New to the site, looks good!

I'm trying to calculate the Nth digit of Pi.

David Bailey, Peter Borwein, and Simon Plouffe in 1995 came up with a formula for this that calculated the nth digit and returns the result in base 16.

[code]Pi = SUMk=0 to infinity 16-k  [ 4/(8k+1) - 2/(8k+4) - 1/(8k+5) - 1/(8k+6) ].[/code]

I am trying to interpret this equation into computer code.

[code]
    Dim lngDigitToFind
    Dim lngAnswer
    Dim intLooper
    Dim lngTotalRunning

    intLooper = 0

    lngDigitToFind = round(Request.form("digit"))

    Do until intLooper = lngDigitToFind
        intLooper = intLooper + 1
        lngTotalRunning = lngTotalRunning + intLooper
    Loop

    If lngDigitToFind > 0 then

        lngAnswer = round(lngTotalRunning * (((4/((8*lngDigitToFind)+1)) - (2/((8*lngDigitToFind)+4)) - (1/((8*lngDigitToFind)+5)) - (1/((8*lngDigitToFind)+6))))*((1/16)^lngDigitToFind),10)

    end if[/code]

Is my interpretation correct?  And what am I meant to get from the result?  I don't think I fully understand the equation, if someone could step through it in an algortihmic fashion it would help a lot.

Thanks,

Tom[/quote]

An image of the formula if it helps

[img]http://crd.lbl.gov/~dhbailey/bbp-formula.bmp[/img]

Found a C version of it!

http://crd.lbl.gov/~dhbailey/pi/piqp.c

Is that an approximation formula? Or is it exact?

I can hardly believe it is a correct formula... however base 16 there are many incredible tricks :)

[ I think however a proof of this theorem may come in pretty handy when evaluating it... do you have any? ]

The formula (it's exact) and the idea of the proof can be found here:
http://mathworld.wolfram.com/PiFormulas.html , number (30)


Let $ \dfrac{1}{3}\leq a,b,c\leq 3$.Prove that:
$ \dfrac{a}{a\plus{}b}\plus{}\dfrac{b}{b\plus{}c}\plus{}\dfrac{c}{c\plus{}a}\geq\dfrac{7}{5}$.

[quote="Inequalities Master"]Let $ \dfrac{1}{3}\leq a,b,c\leq 3$.Prove that:
$ \dfrac{a}{a \plus{} b} \plus{} \dfrac{b}{b \plus{} c} \plus{} \dfrac{c}{c \plus{} a}\geq\dfrac{7}{5}$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175423


Let $a,b,c$ be positive integers such that $1<a<b<c$. Suppose that $(ab-1)(bc-1)(ca-1)$ is divisible by $abc$. Find the values of $a,b,c$

[hide]By opening parenthises we obtain $abc\mid ab+bc+ca-1$. So if $a\geq 3$ then $abc\geq 3bc>ab+bc+ca-1$. Contradiction. Therefore, $a=2$. Then $c\mid 2b-1$ (since we know $c\mid ab-1$). It follows that $c=2b-1$ (since $c>b$). Then $b\mid 2(2b-1)-1$ (since $b\mid ac-1$), i.e. $b=3$. And finally $c=5$.
Answer: $a=2,b=3,c=5$.[/hide]


Here is a place where you can sing some creative songs/parodies about the muffin man. It's fun!

Here are some of mine:

Do you know the mitten man the mitten man the mitten man
Do you know the mitten man who lives on Snowy Lane!

And some funny ones, thanks to others:

By mz94:

im a little muffin short and stout
here is my frosting, here is the grout
when i get all cooked up here me shout
if you put me in your stomach i would really glout (frown)

By 1=2:

Do you know the mutton man the mutton man the mutton man
Do you know the mutton man who lives on Bloody Lane!

What a shie muffin man!

Mie muffin man!
Mie muffin man!
Mie muffin man!

By shaggy75:

Bie, Muffin-Man!
Bie, Muffin-Man!
Bie, Muffin-Man!

By BanishedTraitor:

Die, Muffin-Man!
Die, Muffin-Man!
Die, Muffin-Man!

By lingomaniac88:

Hie, gauss1181!
Hie, gauss1181!
Hie, gauss1181!

I'm the Muffin-Man!

Hie, everybody!
Hie, everybody!
Hie, everybody!

By DonkeyKong:

Who is the Muffin Man? Do you want me to say hie to him as well? GO MUFFINS! GO MUFFINS! GO MUFFINS!!!!!

Is there a reason to be obsessed about the Muffin Man? :P

I just am. And I live on Drury Lane! My favorite quote is:

Hie, Muffin-Man!

[quote="xpmath"]I remember when you said the tug-of-war games won't help you at all :roll:[/quote]

Don't speak nonsense. Everyone knows that a random thread about a character from an old children's song will help everyone unspeakably!

[quote="miyomiyo"]To me, "hie" continues to sound like an exclamation you might make while being punched in the stomach.[/quote]

From a medical perspective hie= Hyper-IgE Syndrome which is caused by an inherited abnormality in the gene. Hence explains the absurd circumstance we are currently in.

According to dictionary.com, "Hie, Muffin Man" is telling Muffin Man to hasten; speed; go in haste.

Can [i]you[/i] run faster than a speeding muffin?

No. Cuz the muffin man runs wwwaaaayyy faster than me in order to deliver all those muffins.

p.s. people, don't just think I am the muffin man even though i said i live on drury lane. I"M NOT THE MUFFIN MAN!

NO!!!!!!!

MIYOMIYO IS TOTALLY WRONG!!!! :furious:  :starwars:

[quote="miyomiyo"]You also don't know how to use Caps Lock.  Or make a thread.

Somebody lock this.[/quote]

I am in favour.


[hide] It is not that hard to brute force, since we merely need to multiply the previous factorial's residue by the extra term and reduce.

$1!\equiv 1$
$2!\equiv 2$
$3!\equiv 6$
$4!\equiv 11$
$5!\equiv 3$
$6!\equiv 5$
$7!\equiv 9$
$8!\equiv 7$
$9!\equiv 11$
$10!\equiv 6$
$11!\equiv 1$
$12!\equiv 12$

$1+2+6+11+3+5+9+7+11+6+6+12=79\equiv 1 \pmod {13}$

[/hide]

I'm sure there is a much more elegant method.


[hide="Solution"]Taking weights yields that the ratio of 20% solution to 30% solution is 3:2.
Thus we have that there are $ \frac{3}{5}*45\equal{}27$ oz of the 20% solution.[/hide]

[quote="pascal_1623"]I forget what year it was...

How many ounces of 20% acid solution must be mixed with a 30% acid solution to obtain 45 ounces of 24% acid solution?[/quote]
[hide]Can't this also be solved with a system of equations?
Let x be the amount of 20% solution and y the 30% solution.
x+y=45
0.2x+0.3y=0.24(45)=10.8

so 2x+3y=108, 3x+3y=135, so x=27.[/hide]
That works too. I think it's more comprehensive (just to me, no offense mathcrazed)

If you're familiar with weighted averages, mathcrazed's answer is much faster and easier.

Let's say we have a set of number $ S\equal{}\lbrace x_0,x_1,\hdots,x_n\rbrace$ and the mean is given as $ m$.

If there are $ a$ values less than $ m$ which have an average of $ n$ and there are $ b$ values greater than $ m$ which have an average of $ c$.  The ratio $ a: b\equal{}(c\minus{}m): (m\minus{}n)$.  I might've screwed up considering I'm watching some stupid chinese show =P.


conics
parabola
hyperbola
geometry unsolved
geometry

[quote="derezin"][quote="misan"]hi, what is a nice [b][u]continuous [/u][/b] mechanical construction  method
for a parabola, and also for a hyperbola?  :maybe:[/quote]

Lol  :P[/quote]

huh ?  :maybe:

 :D


What is the cut off scores in different states who advanced to nats last year?

In North Dakota, it was 24.. a 4 way tie for 3rd and 4th...

24?
Wish i lived in North Dakota

Kentucky was a 35 (yeah i know. its still a low score compared to other states)

32 (7th ( :P ) through 3rd written) in KS

[quote="13375P34K43V312"]Search "The Ultimate National Information Thread" or something, it was started by Ignite168.[/quote]

Link:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84737

Thank you! Soon, we will see the cut off of this year. :-)

In Arkansas, 4th place got about 26 or 27

i have no clue, my teacher didnt tell me, but it was probably around 30-35 for wv

Ugh, I wish I was you people...
In texas this year, the cutoff was like a 40...
I got like 38-ish and got 7th...

[quote="Art of Owna"][quote="indianamath"]39 in IN last year.[/quote]

and i got more than half of that..so im happy.. :P[/quote]
barely lol.

[quote="Art of Owna"][quote="indianamath"]39 in IN last year.[/quote]

and i got more than half of that..so im happy.. :P[/quote]

what did u get? it's not like saying ur score will let ppl narrow down the list anyways...

[quote="indianamath"]last year he got 21.[/quote]

Im not that stupid anymore  :D

i just found out, and i was right.  in WV it went like this

1st-35
2nd-34
3rd-32 (me)
4th-31

So the cutoff for WV was 31.

it's okay. last year, i got a 24... what rank were you last year? i was 64th...

lol don't rub it in.

he did this year's test for practice as a freshman and he got 25/14.

Hey..they also graded my wrong because of an eraser (seriously) but a 21- i say 80th?  not bad eh...?

o yea, and if i lived in certain states or territories i wouldve still mad nats.

maryland's cutoff was 37
1st: 45
2nd:37
3rd:37
4th:37

Kyyuan, this is for last year. In the 'Cutoff Scores' thread, he was asking about cutoffs for this year.

In VA, 5 people tied for 4th with a 37.
I was one of them, but tiebreakers brought me down to 6th

[quote="jhredsox"]about the high cutoff scores...if you get like constand mid 20's on state sprints, shouldn't you skip the middle 10 or so and go to the last 10 so that you get those right?

jorian[/quote]
uhh, no because the questions aren't weighted and the middle questions are easier.


Let $ o$ be an odd whole number and let $ n$ be any whole number.  Which of the following statements about the whole number $ (o^2\plus{}no)$ is always true?

\[ \textbf{(A)}\ \text{it is always odd} \\
\textbf{(B)}\ \text{it is always even} \\
\textbf{(C)}\ \text{it is even only if n is even} \\
\textbf{(D)}\ \text{it is odd only if n is odd} \\
\textbf{(E)}\ \text{it is odd only if n is even}
\]

[hide="Click for solution"]
The number $ o^2$ is always odd.  Now $ no$ is odd if $ n$ is odd and even if $ n$ is even.  Thus the sum $ o^2\plus{}no$ is even if $ n$ is odd, since the sum of two odd numbers is even; the sum is odd if $ n$ is even, since the sum of an odd number and an even number is odd.  Thus, the answer is $ \boxed{\textbf{(E)}}$.
[/hide]


four points, are midpoints of sides of two different quadrilaterals at the same time.  show that the two quadrilaterals have the same area.

[hide="Solution"]
Vectors:

\[ [ABCD] \equal{} |\frac{1}{2} (D \minus{} B) \times (C \minus{} A)|\]

\[ \equal{} |[ \frac{A\plus{}D}{2} \minus{} \frac{A\plus{}B}{2} ) ]\times [\frac{C\plus{}D}{2} \minus{} \frac{A\plus{}D}{2}]|\]

[/hide]

i dont understand your solution
what are A, B, C and D? the four given points?

how do you write the area of any quadrilateral in terms of vectors?!

JavaMan is saying that the area of a quadrillateral is dependant solely on the location of its midpoints of its sides

Here A,B,C,D are vectors OA, OB, OC, OD but he assumed O as an origin of plane, so from here it follows.

The midpoints form a parallelogram with area half of the original quadrilateral...


How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$?

A. 768

B. 801

C. 934

D. 1067

E. 1167

Or maybe you are just too smart, that is the correct answer.

[quote="adalton"][quote="akkaren"][hide="Solution"] (For each division, take the greatest integer less than or equal to.) 2001/3+2001/4-2001/12-2001/12-2001/20+2001/60=801, B

...and I probably did that wrong. Seems too easy. [/hide][/quote]
Shouldn't it be:
[hide]
$\lfloor\frac{2001}{3}\rfloor+\lfloor\frac{2001}{4}\rfloor-\lfloor\frac{2001}{12}\rfloor-\lfloor\frac{2001}{15}\rfloor-\lfloor\frac{2001}{20}\rfloor+\lfloor\frac{2001}{60}\rfloor=801$
[/hide][/quote] [quote="my lack of clarity"](For each division, take the greatest integer less than or equal to.)[/quote] That's what I meant. I've been too lazy to figure out how to do the latex thing.

Not the latex, you subtracted 2001/12 twice.

Oh, okay. (Actually, I reread mine, checking it against yours, and didn't see the error. My apologies, but please don't assume I'm jumping to conclusions about my infallibility.)

It's still what I meant. Sorry. I didn't notice.

No worries, I was just pointing it out so others don't get confused.


algorithm
number theory
least common multiple
Euclidean algorithm
number theory unsolved

Find all integers n which are divisible by all integers less than square root of n.
Prove that this set of integers is the whole set

I've found the numbers to be 1?, 2, 3, 4, 6, 8 , 12, 24
But how to [b]prove [/b] that they are the ones, and that there are no more.
Help me, you're my only hope[/list]

Take $k^2$ to be the greatest square that is at most $n$, so that $n=k^2+r$ with $r$ between $0$ and $2k$. Now if $n$ is a nonsquare, $k \mid k^2+r$ so $r=k$ or $r=2k$.

In the first case we have also $k-1 \mid k(k+1)$, and since the only factor $k-1$ and $k(k+1)$ can share is $2$, we have $k-1 \mid 2$, then $k=2$ or $k=3$. These give $6, 12$ respectively. If $r=2k$, $k-1 \mid k(k+2)$ so $k-1 \mid 3$, then $k=2, k=4$ which give $8, 24$ respectively.

The only case left is $n=k^2$. Here $k-1 \mid k^2$; since $k-1 \mid k^2-1$, $k-1 \mid 1$, $k=2, n=4$. We have forgotten the case where $k-1$ is 0, which would give $1, 2, 3$.

[quote="spx2"]i got lost at $k|k^2+r$
why is this ?[/quote]
$\sqrt{n} > \sqrt{k^2} = k$

i don't see how that may justify what he states...

[quote="Jenny11235"]Find all integers n which are divisible by all integers less than square root of n.
[/quote]

$k$ is smaller than the square root of $n$, so this has to be satisfied by the question

Excuse my stupid questions....

1)   k - 1  divides k(k+1), you say the only factor they can share is 2. I know they are consecutive so the product is even. But how do we know this is the highest factor?

2)  k-1 divides k(k+2) so k-1 divides 3. I don't get this. How do you know this?

Also the question is.....
Find all integers n which are divisible by all integers [b]less than [/b]square root of n.

if k^2 is the greatest square of most n,   n = k^2 + r, where r is 0 to 2k is what was proposed,

is it not that r goes from 1 to 2k + 1.
For example numbers that are divisible by 1,2,3 and 4  potentially, are in the bound 17-25.
Can't have 16 as sqrt is 4. All numbers less than this are 1, 2,3,   and 4 is not less than?

What difference does this make to the answer?

Also what about this way to prove there are no more....
numbers above 25 need a factor 5 (As well as 1,2,3,4). But it already needs a factor 12   (coming from 3 and 4), So it needs a factor 60. So n is at least 60. But 7<sqrt(60), so it also needs a factor 7. So n is at least 420. And so on.
Is there a way to prove it from this angle. Perhaps by a sequence or limit or something?

Thanks a lot.

For questions 1) and 2) use the eucidean algorithm

For the limit question:
There is a solution using Bertrand's postulate, but it's a lot harder this way (proving the postulat)

Both of your questions (1) and (2) can be answered with the following:

Let k, j be positive integers.  By the Euclidean Algorithm, $gcd(k,k+j)=gcd(k,k+j-k)=gcd(k,j)|j$.
By the above lemma, $gcd(k,k+1)|1$ and thus any two consecutive integers are relatively prime.  So in this case, if $k-1|k(k+1)$, then $k-1|k+1$ since $gcd(k-1,k)=1$.  Since $gcd(k-1,k+1)|2$ by the lemma, we know $k-1|2$, for otherwise $k-1$ would not divide $k+1$.

Similarly, if $k-1|k(k+2)$ then $k-1|k+2$.  Since $gcd(k-1,k+2)|3$ as well, $k-1|3$.

As to your other question, reindexing r between 1 and 2k+1 will not change the results, for $k+2k+1=(k+1)^2$, and it is only a matter of whether you would prefer to consider the perfect squares written as $(k+1)^2$ with $k\geq0$ or $k^2$ with $kgeq1$. It makes no difference.

I believe your "limit" idea would work if you used it in combination with the above lemma.  You are basically trying to prove that for sufficiently large k, the lcm of the first $k-1$ integers is always greater than $k^2$, which is equivalent to your problem.  You could prove this by noting that for $k>5$, $k-3$ does not divide $(k-1)(k-2)$, so the lcm of the first $k-1$ integers is at least $2(k-1)(k-2)$, which is greater than $k^2$ for all $k>7$ by simple algebra, leaving a finite number of possibilities for k that you can verify.


inequalities
trigonometry
function
triangle inequality

[quote="April"]Let $a,\,b,\,c$ be the side lengths of a right triangle. Find the least value of $P=\frac{(a+b)(b+c)(c+a)}{abc}$[/quote]

[hide]
$1$ ?  (Just using the triangle inequality.)
[/hide]

Obviously that's not correct. Using the triangle inequality, you could be tempted to call $1$ the minimum, but this value can never be achieved.

You have to find a value that is always smaller or equal to the expression, and then show that it can be achieved, hence the minimum value. Here, it is $8$, because $a+b \geq 2 \sqrt{ab}$ and multiply for all constants. Of course this can be reached if and only if the triangle is equilateral.

[quote="Jan"]
Of course this can be reached if and only if the triangle is equilateral.[/quote]

But the problem says "of a [b]right[/b] triangle".

[hide="Here is my messy solution...I hope it works"]

wlog let $c$ be the hypotenuse, and again wlog let $c=1$.

Then $\frac{(a+b)(b+c)(a+c)}{abc}=\frac{a+b}{a}+\frac{b+c}{b}+\frac{a+c}{c}=(1+\tan \theta)(1+\frac{1}{\sin \theta})(1+\cos \theta)$, where $\theta= \angle CBA$.

After some multiplying out, this becomes

$1+1+(\tan\theta+\cot \theta)+(\sin \theta+\frac{1}{\sin \theta})+(\cos \theta+\sec \theta)$.

Note that $\tan \theta+\cot \theta \geq 2$ by simple AM-GM.

Here's where my solution gets ugly.

I differentiated $(\sin \theta+\frac{1}{\sin \theta})+(\cos \theta+\sec \theta)$ w.r.t $\theta$, rearranged and got that the there is a turning point iff
$\sin \theta(\frac{1}{\cos^{2}\theta}-1)=\cos \theta(\frac{1}{\sin^{2}\theta}-1)$.

Clearly if $\sin \theta > \cos \theta$ then LHS > RHS, and vice versa. Hence the only turning point is when $\cos \theta=\sin \theta$ and $\theta = \frac{\pi}{4}$. This is clearly a minimum (as $\theta \rightarrow 0$ and $\rightarrow \frac{\pi}{2}$, the expression becomes large).

At this point $\sin \theta+\frac{1}{\sin \theta}+\cos \theta+\sec \theta= 3 \sqrt{2}$, it also coincides with the minimum of $\tan \theta+\cot \theta$ at $2$. Hence the minimum of the whole expressionis $4+3\sqrt{2}$.

I don't have much time, so apologies for that being a bit rushed. If someone would like to check my solution (particularly the ugly bit  :) ) then I'd really appreciate it.

Incidentally, does anyone know why \cosec \theta doesn't show up?

[/hide]

Because the engine uses American notation: only \csc and not \cosec, only \tan and not \tg, only \cot and not \ctg, only \arctan and not \arctg, and there's neither \arccot nor \arcctg. (Though the missing functions can be "simulated" via \text{} function.)

My mistake, I didn't read the solution careful enough.

Anyway, here's a solution if the triangle is right-angled:

[hide]
It is clear that the minimum $4+3\sqrt{2}$ can be achieved by taking $a=b$.

Now, I will show that it is the minimum value:
substituting $c=\sqrt{a^{2}+b^{2}}$:
\[(a+b)(ab+(a+b)\sqrt{a^{2}+b^{2}}+a^{2}+b^{2}) \geq (4+3\sqrt{2})ab\sqrt{a^{2}+b^{2}}\]
We will prove this statement in the following way:
\[(a+b)(ab+(a+b)\sqrt{a^{2}+b^{2}}+a^{2}+b^{2}) \geq 2\sqrt{ab}(ab+2\sqrt{ab}\sqrt{a^{2}+b^{2}}+a^{2}+b^{2}) \]
So now we only have to prove that
\[a^{2}+b^{2}+ab \geq 3 \sqrt{\frac{ab(a^{2}+b^{2})}{2}}\]
This is obviously true, since
\[\left(\frac{a^{2}+b^{2}}{2}+ab\right)+\frac{a^{2}+b^{2}}{2}\geq 2\sqrt{\frac{ab(a^{2}+b^{2})}{2}}+\sqrt{\frac{ab(a^{2}+b^{2})}{2}}\]
By AM-GM and QM-GM. [/hide]

[quote="April"][color=darkred]Let $a,b,c$ be the side lengths of a [b][u]right[/u][/b] triangle.
Find the least value of $P=\frac{(a+b)(b+c)(c+a)}{abc}\ .$[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Suppose w.l.o.g. that $a^{2}=b^{2}+c^{2}\ .$ Therefore,

$\{\begin{array}{ccc}4bc\le (b+c)^{2}\le 2(b^{2}+c^{2}) & \Longrightarrow & \{\begin{array}{c}\frac{2a}{b+c}\ge \sqrt 2\\\\ \frac{(b+c)^{2}}{bc}\ge 4\\\\ \frac{a(b+c)}{bc}\ge \frac{4a}{b+c}\end{array}\\\\ \mathrm{A.M.\--\ G.M.}& \Longrightarrow & \frac{2a}{b+c}+\frac{b+c}{a}\ge 2\sqrt 2\end{array}\|\ \bigoplus\ \mathrm{(sum)}$ $\Longrightarrow$

$\frac{(b+c)^{2}}{bc}+\frac{a(b+c)}{bc}+\frac{b+c}{a}\ge 4+3\sqrt 2$ $\Longrightarrow$ $\frac{(b+c)(c+a)(a+b)}{abc}\ge 4+3\sqrt 2\ .$

[b][u]Remark.[/u][/b] Generally,

$\frac{(a+b)(b+c)(c+a)}{abc}\ge 4+2\sqrt 2+\sqrt 2\cdot\max\{\frac{a}{\sqrt{b^{2}+c^{2}}}\ ,\ \frac{b}{\sqrt{c^{2}+a^{2}}}\ ,\ \frac{c}{\sqrt{a^{2}+b^{2}}}\}\ .$
You can use the inequality $\frac{2a}{b+c}\ge \sqrt 2\cdot\frac{a}{\sqrt{b^{2}+c^{2}}}$ as the first inequality in the above sum a.s.o.

[b][u]Particular case.[/u][/b]  If the triangle $ABC$ isn't acute, then

$\frac{(a+b)(b+c)(c+a)}{abc}\ge 4+3\sqrt 2$ because $\max\{\ldots \ ,\ \ldots\ ,\ \ldots \}\ge 1\ .\blacktriangleleft$[/color]

[hide]
WLOG let $c$ be the hypotenuse. Observe that
\begin{eqnarray*}\frac{(a+b)(b+c)(c+a)}{abc}&=&2+\frac{a}{b}+\frac{b}{a}+\frac{a+b}{c}+c\cdot \frac{a+b}{ab}\\ &=&2+\frac{a}{b}+\frac{b}{a}+\frac{2(a^{2}+ab+b^{2})}{c(a+b)}\\ \end{eqnarray*}
By AM-GM, $\frac{a}{b}+\frac{b}{a}\geq 2$. We now show
\[\frac{2(a^{2}+ab+b^{2})}{c(a+b)}\geq 3\sqrt{2}\]
equivalent to
\[\sqrt{2a^{2}+2b^{2}}\leq \frac{(a+b)(a^{2}+ab+b^{2})}{3ab}\]
But $\sqrt{2a^{2}+2b^{2}}=a+b+\frac{(a-b)^{2}}{a+b+\sqrt{2a^{2}+2b^{2}}}$ and $\frac{(a+b)(a^{2}+ab+b^{2})}{3ab}=a+b+\frac{(a+b)(a-b)^{2}}{3ab}$. So it only remains to prove that
\[\frac{1}{a+b+\sqrt{2a^{2}+2b^{2}}}\leq \frac{a+b}{3ab}\]
After clearing denominators, this becomes
\[(a+b)(a+b+\sqrt{2a^{2}+2b^{2}})\geq 3ab\]
which is obvious by RMS-AM and AM-GM:
\[(a+b)(a+b+\sqrt{2a^{2}+2b^{2}})\geq 2(a+b)^{2}\geq 8ab>3ab\]
So we find the minimum to be $\boxed{4+3\sqrt{2}}$, attained for isosceles right triangles.
[/hide]