Find the largest prime factor that divides the number:

$ 3^{14}\plus{}3^{13}\minus{}12$

[quote="luiseduardo"]How do you discovered the number 4 ?[/quote]

$ 3^{14}\equal{}3\times 3^{13}$

So $ 3^{14}\plus{}3^{13}\equal{}3\times 3^{13}\plus{}3^{13}$

So $ 3^{14}\plus{}3^{13}\equal{}3\times 3^{13}\plus{}1\times 3^{13}$

So $ 3^{14}\plus{}3^{13}\equal{}(3\plus{}1)\times 3^{13}$

So $ 3^{14}\plus{}3^{13}\equal{}4\times 3^{13}$

And maybe you could kindly stop posting in olympiad fora.

[quote="luiseduardo"]Oh, Thanks. And I'm sorry. But this question is from a olympiad, what's the problem if I posted here ?[/quote]

If it is from an Olympiad, that's OK.

But it seems very very low level problem for an olympiad one.


How many ways are there to place 9 different flags on 5 different flagpoles if the order of the flags on the flagpoles is relevant and every flagpole flies at least one flag?

Not sure if this is exactly right, but...
This is similar to the balls-and-urns argument where there must be at least one ball in each urn. The solution to that is $\dbinom{8}{4}$. However, we want permutations, not combinations, so our solution is $\frac{8!}{4!}=1680$.

Actually, I'm not sure that's right, as we have distinguishable flags. Should I multiply my answer by $9!$ for the number of arrangements of the nine flags? Correct me, someone.

EDIT: On second thought, the answer should be $9!\dbinom{8}{4}$, as worthawholebean posted.

Yeah...on second thought, I think you're right about multiplying by $9!$; the turning combinations into permutations step is unnecessary when you do that, I think.

[quote="worthawholebean"][hide="I may be wrong..."]Label the flags 1-9. Then a sample arrangement would be 143|297|5|6|8, where pipes are divisions between flagpoles. Thus the answer is $_{8}C_{4}$ places to place the divisions (no two can be placed next to each other or on the ends) times $9!$, the number of ways to arrange the flags, which are distinguishable, or $25401600$.[/hide][/quote]

Yes I believe you are right, if the flagpoles are distinguishable.


Find all functions $f: \mathbb N \to \mathbb N$ such that for all positive integers $x$ and $y$,
\[f(x^2 + y^2) = f(x)^2 + f(y)^2.\]

[quote="mathmen"]find all function $ f;\mathbb{N}^* \to \mathbb{N}*$

such that $ \forall x;y \in \mathbb{N}^*$
$ f(x^2 \plus{} y^2) \equal{} f(x)^2 \plus{} f(y)^2$[/quote]
Once again, IMHO, a crazy problem. It seems that, since some days, we get a lot of abnormal problems that I thought impossible to find in a real contest / book / class. And I wonder if all these problems are maybe just invented by posters.

Here is a solution I gave in another forum :
The difficulty here is the fact that the function is not defined for $ 0$ :

Let $ f(1)\equal{}a$
$ f(2)\equal{}f(1^2\plus{}1^2)\equal{}f(1)^2\plus{}f(1)^2\equal{}2a^2$
$ f(5)\equal{}f(2^2\plus{}1^2)\equal{}f(2)^2\plus{}f(1)^2\equal{}4a^4\plus{}a^2$

Computation of $ f(7)$ :
$ f(50)\equal{}f(7^2\plus{}1^2)$
$ f(50)\equal{}f(5^2\plus{}5^2)$
So $ f(7)^2\plus{}a^2\equal{}2(4a^4\plus{}a^2)^2$
and $ f(7)\equal{}\sqrt{2(4a^4\plus{}a^2)^2\minus{}a^2}$

$ f(8)\equal{}f(2^2\plus{}2^2)\equal{}f(2)^2\plus{}f(2)^2\equal{}8a^4$

$ f(10)\equal{}f(3^2\plus{}1^2)\equal{}f(3)^2\plus{}a^2$

$ f(13)\equal{}f(3^2\plus{}2^2)\equal{}f(3)^2\plus{}4a^4$

Computation of $ f(3)$ :
$ f(125)\equal{}f(10^2\plus{}5^2)\equal{}f(10)^2\plus{}f(5)^2$
$ f(125)\equal{}f(11^2\plus{}2^2)\equal{}f(11)^2\plus{}f(2)^2$
and so $ f(11)^2\plus{}4a^4\equal{}(f(3)^2\plus{}a^2)^2\plus{}(4a^4\plus{}a^2)^2$
so $ f(11)^2\equal{}(f(3)^2\plus{}a^2)^2\plus{}(4a^4\plus{}a^2)^2\minus{}4a^4$

$ f(170)\equal{}f(11^2\plus{}7^2)$
$ f(170)\equal{}f(13^2\plus{}1^2)$
and so $ f(11)^2\plus{}2(4a^4\plus{}a^2)^2\minus{}a^2\equal{}(f(3)^2\plus{}4a^4)^2\plus{}a^2$
So $ f(11)^2\equal{}(f(3)^2\plus{}4a^4)^2\plus{}2a^2\minus{}2(4a^4\plus{}a^2)^2$

Comparing the equation  $ f(125)$ with the equation $ f(170)$, we get :
$ (f(3)^2\plus{}a^2)^2\plus{}(4a^4\plus{}a^2)^2\minus{}4a^4\equal{}(f(3)^2\plus{}4a^4)^2\plus{}2a^2\minus{}2(4a^4\plus{}a^2)^2$
$ (f(3)^2\plus{}a^2)^2\equal{}(f(3)^2\plus{}4a^4)^2\plus{}4a^4\plus{}2a^2\minus{}3(4a^4\plus{}a^2)^2$
$ (4a^2\minus{}1)f(3)^2\equal{}16a^6\plus{}12a^4\minus{}1$
$ f(3)\equal{}2a^2\plus{}1$

Synthesis up to now :

$ f(1)\equal{}a$
$ f(2)\equal{}2a^2$
$ f(3)\equal{}2a^2\plus{}1$
$ f(5)\equal{}4a^4\plus{}a^2$
$ f(7)\equal{}\sqrt{2(4a^4\plus{}a^2)^2\minus{}a^2}$
$ f(8)\equal{}8a^4$
$ f(10)\equal{}4a^4\plus{}5a^2\plus{}1$
$ f(13)\equal{}12a^2\plus{}1$

$ f(65)\equal{}f(7^2\plus{}4^2)$
$ f(65)\equal{}f(8^2\plus{}1^2)$
and so $ f(8)^2\plus{}f(1)^2\equal{}f(7)^2\plus{}f(4)^2$
so $ 64a^8\plus{}a^2\equal{}2(4a^4\plus{}a^2)^2\minus{}a^2\plus{}f(4)^2$
and $ f(4)^2\equal{}64a^8\plus{}a^2\minus{}2(4a^4\plus{}a^2)^2\plus{}a^2$
$ f(4)\equal{}\sqrt{32a^8\minus{}16a^6\minus{}2a^4\plus{}2a^2}$

$ f(185)\equal{}f(11^2\plus{}8^2)$
$ f(185)\equal{}f(13^2\plus{}4^2)$
and so $ f(11)^2\plus{}f(8 )^2\equal{}f(13)^2\plus{}f(4)^2$
$ f(11)\equal{}\sqrt{\minus{}32a^8\minus{}16a^6\plus{}142a^4\plus{}26a^2\plus{}1}$

$ f(130)\equal{}f(9^2\plus{}7^2)$
$ f(130)\equal{}f(11^2\plus{}3^2)$
and so $ f(9)^2\plus{}f(7)^2\equal{}f(11)^2\plus{}f(3)^2$
so $ f(9)^2\equal{}\minus{}32a^8\minus{}16a^6\plus{}142a^4\plus{}26a^2\plus{}1\plus{}(2a^2\plus{}1)^2\minus{}2(4a^4\plus{}a^2)^2\plus{}a^2$
$ f(9)\equal{}\sqrt{\minus{}64a^8\minus{}32a^6\plus{}144a^4\plus{}31a^2\plus{}2}$

$ f(85)\equal{}f(6^2\plus{}7^2)$
$ f(85)\equal{}f(9^2\plus{}2^2)$
and so $ f(6)^2\plus{}f(7)^2\equal{}f(9)^2\plus{}f(2)^2$
$ f(6)\equal{}\sqrt{\minus{}96a^8\minus{}48a^6\plus{}146a^4\plus{}32a^2\plus{}2}$

$ f(200)\equal{}f(10^2\plus{}10^2)$
$ f(200)\equal{}f(14^2\plus{}2^2)$
and so $ f(14)^2\plus{}f(2)^2\equal{}2f(10)^2$
$ f(14)\equal{}\sqrt{2(4a^4\plus{}5a^2\plus{}1)^2\minus{}4a^4}$

And finally, the equation allowing to find $ a\equal{}f(1)$
==================================================
$ f(205)\equal{}f(13^2\plus{}6^2)$
$ f(205)\equal{}f(14^2\plus{}3^2)$
and so $ f(13)^2\plus{}f(6)^2\equal{}f(14)^2\plus{}f(3)^2$
$ (12a^2\plus{}1)^2\minus{}96a^8\minus{}48a^6\plus{}146a^4\plus{}$ $ 32a^2\plus{}2\equal{}2(4a^4\plus{}5a^2\plus{}1)^2\minus{}4a^4\plus{}(2a^2\plus{}1)^2$
$ 144a^4\plus{}24a^2\plus{}1\minus{}96a^8\minus{}48a^6\plus{}146a^4$ $ \plus{}32a^2\plus{}2\equal{}32a^8\plus{}50a^4\plus{}2\plus{}80a^6\plus{}$ $ 16a^4\plus{}20a^2\minus{}4a^4\plus{}4a^4\plus{}4a^2\plus{}1$
$ 128a^8\plus{}128a^6\minus{}220a^4\minus{}36a^2\equal{}0$
$ 32a^6\plus{}32a^4\minus{}55a^2\minus{}9\equal{}0$
$ (a^2\minus{}1)(32a^4\plus{}64a^2\plus{}9)\equal{}0$
And so $ a\equal{}1$, unique positive integer root of this equation.

This implies $ f(n)\equal{}n$ $ \forall n\in\{1,2,3,4,5,6,7,8,9,10,11,13,14\}$

Now, we can use the equalities  :
$ (2n\minus{}1)^2\plus{}(n\minus{}3)^2\equal{}(2n\minus{}3)^2\plus{}(n\plus{}1)^2$
$ (2n)^2\plus{}(n\minus{}5)^2\equal{}(2n\minus{}4)^2\plus{}(n\plus{}3)^2$

Which allow to find $ f(n)$ for any $ n$ as soon as we know $ f(n)$ for $ n\in\{1,2,3,4,5,6\}$

And so $ \boxed{f(n)\equal{}n}$ $ \forall n$

[quote="nguyenvuthanhha"][quote="pco"] and so $ f(13)^2 \plus{} f(6)^2 \equal{} f(14)^2 \plus{} f(3)^2$
$ (12a^2 \plus{} 1)^2 \minus{} 96a^8 \minus{} 48a^6 \plus{} 146a^4 \plus{}$ $ 32a^2 \plus{} 2 \equal{} 2(4a^4 \plus{} 5a^2 \plus{} 1)^2 \minus{} 4a^4 \plus{} (2a^2 \plus{} 1)^2$

  [/quote]

[i]   Very brilliant idea  :lol:  but I think in that step , it should be :

   $ ( 8a^4 \plus{} 4 a^2 \plus{} 1)^2 \minus{} 96a^8 \minus{} 48a^6 \plus{} 146a^4 \plus{}$ $ 32a^2 \plus{} 2 \equal{} 2(4a^4 \plus{} 5a^2 \plus{} 1)^2 \minus{} 4a^4 \plus{} (2a^2 \plus{} 1)^2$

   Because if you can prove $ f(13) \equal{} 12a^2 \plus{} 1$ as you said in the previous step than :

    $ f(13) \equal{} 12a^2 \plus{} 1 \equal{} (f(3))^2 \plus{} 4a^4 \equal{} 8a^4 \plus{} 4a^2 \plus{} 1$ , so it's really trivial that $ a \equal{} 1$

    Am I right ?

 :lol:  [/i][/quote]

AAAaaaaaaaargh ! D**** !
I dont know from where I got $ f(13) \equal{} 12a^2 \plus{} 1$ and so the good value is  $ f(13) \equal{} (f(3))^2 \plus{} 4a^4 \equal{} 8a^4 \plus{} 4a^2 \plus{} 1$ and we dont have equation here.

But this error crashed down all my computations :

This modifies the computation of $ f(11)$ which becomes $ f(11) \equal{} \sqrt {32a^8 \plus{} 48a^6 \plus{} 30a^4 \plus{} 10a^2 \plus{} 1}$

Which modifies then the computation of $ f(9)$ which becomes $ f(9) \equal{} \sqrt {32a^6 \plus{} 32a^4 \plus{} 15a^2 \plus{} 2}$

Which modifies then the computation of $ f(6)$ which becomes $ f(6) \equal{} \sqrt { \minus{} 32a^8 \plus{} 16a^6 \plus{} 34a^4 \plus{} 16a^2 \plus{} 2}$

And then the final equation $ f(13)^2 \plus{} f(6)^2 \equal{} f(14)^2 \plus{} f(3)^2$ becomes ... an identity :(

And so I no longer got $ a$ :(

Thanks for your careful reading. I'll edit my post and try to find $ a$ in another way. [u][b]edited [/b][/u]: sorry, but my original post is too old to be edited.
What a pity :)

I got it !

We obtained $ f(6) \equal{} \sqrt { \minus{} 32a^8 \plus{} 16a^6 \plus{} 34a^4 \plus{} 16a^2 \plus{} 2}$

And it is very easy to see that  $ \minus{} 32a^8 \plus{} 16a^6 \plus{} 34a^4 \plus{} 16a^2 \plus{} 2 < 0$ $ \forall a\ge 2$ and so the only solution may be $ a \equal{} 1$

And so a correct (I hope) solution :
[quote="mathmen"]find all function $ f;\mathbb{N}^* \to \mathbb{N}*$

such that $ \forall x;y \in \mathbb{N}^*$
$ f(x^2 \plus{} y^2) \equal{} f(x)^2 \plus{} f(y)^2$[/quote]
Once again, IMHO, a crazy problem. It seems that, since some days, we get a lot of abnormal problems that I thought impossible to find in a real contest / book / class. And I wonder if all these problems are maybe just invented by posters.

Here is a solution I gave in another forum :
The difficulty here is the fact that the function is not defined for $ 0$ :

Let $ f(1) \equal{} a$
$ f(2) \equal{} f(1^2 \plus{} 1^2) \equal{} f(1)^2 \plus{} f(1)^2 \equal{} 2a^2$
$ f(5) \equal{} f(2^2 \plus{} 1^2) \equal{} f(2)^2 \plus{} f(1)^2 \equal{} 4a^4 \plus{} a^2$

Computation of $ f(7)$ :
$ f(50) \equal{} f(7^2 \plus{} 1^2)$
$ f(50) \equal{} f(5^2 \plus{} 5^2)$
So $ f(7)^2 \plus{} a^2 \equal{} 2(4a^4 \plus{} a^2)^2$
and $ f(7) \equal{} \sqrt {2(4a^4 \plus{} a^2)^2 \minus{} a^2}$

$ f(8) \equal{} f(2^2 \plus{} 2^2) \equal{} f(2)^2 \plus{} f(2)^2 \equal{} 8a^4$

$ f(10) \equal{} f(3^2 \plus{} 1^2) \equal{} f(3)^2 \plus{} a^2$

$ f(13) \equal{} f(3^2 \plus{} 2^2) \equal{} f(3)^2 \plus{} 4a^4$

Computation of $ f(3)$ :
$ f(125) \equal{} f(10^2 \plus{} 5^2) \equal{} f(10)^2 \plus{} f(5)^2$
$ f(125) \equal{} f(11^2 \plus{} 2^2) \equal{} f(11)^2 \plus{} f(2)^2$
and so $ f(11)^2 \plus{} 4a^4 \equal{} (f(3)^2 \plus{} a^2)^2 \plus{} (4a^4 \plus{} a^2)^2$
so $ f(11)^2 \equal{} (f(3)^2 \plus{} a^2)^2 \plus{} (4a^4 \plus{} a^2)^2 \minus{} 4a^4$

$ f(170) \equal{} f(11^2 \plus{} 7^2)$
$ f(170) \equal{} f(13^2 \plus{} 1^2)$
and so $ f(11)^2 \plus{} 2(4a^4 \plus{} a^2)^2 \minus{} a^2 \equal{} (f(3)^2 \plus{} 4a^4)^2 \plus{} a^2$
So $ f(11)^2 \equal{} (f(3)^2 \plus{} 4a^4)^2 \plus{} 2a^2 \minus{} 2(4a^4 \plus{} a^2)^2$

Comparing the equation  $ f(125)$ with the equation $ f(170)$, we get :
$ (f(3)^2 \plus{} a^2)^2 \plus{} (4a^4 \plus{} a^2)^2 \minus{} 4a^4 \equal{} (f(3)^2 \plus{} 4a^4)^2 \plus{} 2a^2 \minus{} 2(4a^4 \plus{} a^2)^2$
$ (f(3)^2 \plus{} a^2)^2 \equal{} (f(3)^2 \plus{} 4a^4)^2 \plus{} 4a^4 \plus{} 2a^2 \minus{} 3(4a^4 \plus{} a^2)^2$
$ (4a^2 \minus{} 1)f(3)^2 \equal{} 16a^6 \plus{} 12a^4 \minus{} 1$
$ f(3) \equal{} 2a^2 \plus{} 1$

Synthesis up to now :

$ f(1) \equal{} a$
$ f(2) \equal{} 2a^2$
$ f(3) \equal{} 2a^2 \plus{} 1$
$ f(5) \equal{} 4a^4 \plus{} a^2$
$ f(7) \equal{} \sqrt {2(4a^4 \plus{} a^2)^2 \minus{} a^2}$
$ f(8) \equal{} 8a^4$
$ f(10) \equal{} 4a^4 \plus{} 5a^2 \plus{} 1$
$ f(13) \equal{} (f(3))^2 \plus{} 4a^4 \equal{} 8a^4 \plus{} 4a^2 \plus{} 1$

$ f(65) \equal{} f(7^2 \plus{} 4^2)$
$ f(65) \equal{} f(8^2 \plus{} 1^2)$
and so $ f(8)^2 \plus{} f(1)^2 \equal{} f(7)^2 \plus{} f(4)^2$
so $ 64a^8 \plus{} a^2 \equal{} 2(4a^4 \plus{} a^2)^2 \minus{} a^2 \plus{} f(4)^2$
and $ f(4)^2 \equal{} 64a^8 \plus{} a^2 \minus{} 2(4a^4 \plus{} a^2)^2 \plus{} a^2$
$ f(4) \equal{} \sqrt {32a^8 \minus{} 16a^6 \minus{} 2a^4 \plus{} 2a^2}$

$ f(185) \equal{} f(11^2 \plus{} 8^2)$
$ f(185) \equal{} f(13^2 \plus{} 4^2)$
and so $ f(11)^2 \plus{} f(8 )^2 \equal{} f(13)^2 \plus{} f(4)^2$
$ f(11) \equal{} \sqrt {32a^8 \plus{} 48a^6 \plus{} 30a^4 \plus{} 10a^2 \plus{} 1}$

$ f(130) \equal{} f(9^2 \plus{} 7^2)$
$ f(130) \equal{} f(11^2 \plus{} 3^2)$
and so $ f(9)^2 \plus{} f(7)^2 \equal{} f(11)^2 \plus{} f(3)^2$
so $ f(9)^2 \equal{} \minus{} 32a^8 \minus{} 16a^6 \plus{} 142a^4 \plus{} 26a^2 \plus{} 1 \plus{} (2a^2 \plus{} 1)^2 \minus{} 2(4a^4 \plus{} a^2)^2 \plus{} a^2$
$ f(9) \equal{} \sqrt {32a^6 \plus{} 32a^4 \plus{} 15a^2 \plus{} 2}$

$ f(85) \equal{} f(6^2 \plus{} 7^2)$
$ f(85) \equal{} f(9^2 \plus{} 2^2)$
and so $ f(6)^2 \plus{} f(7)^2 \equal{} f(9)^2 \plus{} f(2)^2$
$ f(6) \equal{} \sqrt { \minus{} 32a^8 \plus{} 16a^6 \plus{} 34a^4 \plus{} 16a^2 \plus{} 2}$

And it is very easy to see that  $ \minus{} 32a^8 \plus{} 16a^6 \plus{} 34a^4 \plus{} 16a^2 \plus{} 2 < 0$ $ \forall a\ge 2$ and so the only solution may be $ a \equal{} 1$

This implies $ f(n) \equal{} n$ $ \forall n\in\{1,2,3,4,5,6,7,8,9,10,11,13\}$

Now, we can use the equalities  :
$ (2n \minus{} 1)^2 \plus{} (n \minus{} 3)^2 \equal{} (2n \minus{} 3)^2 \plus{} (n \plus{} 1)^2$
$ (2n)^2 \plus{} (n \minus{} 5)^2 \equal{} (2n \minus{} 4)^2 \plus{} (n \plus{} 3)^2$

Which allow to find $ f(n)$ for any $ n$ as soon as we know $ f(n)$ for $ n\in\{1,2,3,4,5,6\}$

And so $ \boxed{f(n) \equal{} n}$ $ \forall n$

Just a question:
What's the definition of $ \mathbb{N}^\ast$?

I've always thought that
\[ \mathbb{N} \equal{} \{ 1,2,3,\ldots \} \equal{} \mathbb{Z}_{>0}\] and
\[ \mathbb{N}^\ast \equal{} \{ 0,1,2,3,\ldots \} \equal{} \mathbb{Z}_{\geq 0}\]
But according to pco $ 0 \not\in \mathbb{N}^\ast$. Could you please explain this?

[quote="x164"]Just a question:
What's the definition of $ \mathbb{N}^\ast$?

I've always thought that
\[ \mathbb{N} \equal{} \{ 1,2,3,\ldots \} \equal{} \mathbb{Z}_{ > 0}\]
and
\[ \mathbb{N}^\ast \equal{} \{ 0,1,2,3,\ldots \} \equal{} \mathbb{Z}_{\geq 0}\]
But according to pco $ 0 \not\in \mathbb{N}^\ast$. Could you please explain this?[/quote]

As far as I know :

In some countries, (France, Morocco, ... for example), $ 0\in\mathbb N$ and $ \mathbb N^*\equal{}\mathbb N\backslash \{0\}$

In most other countries (and on Mathlink), $ 0\notin\mathbb N$ and $ \mathbb N_0\equal{}\mathbb N\cup\{0\}$. I did not think that $ \mathbb N^*$ was used in these countries.

I saw for the first time this problem in a forum in Morocco and this was the reason for which I supposed this was the first explanation.


1) Observe that 11|34!. This will give us one relationship between $a$, $b$, $c$ and $d$.

2) 34! must end with 7 zeros (Why?). Hence $b = 0$.

[hide]Assuming we didn't know that...

use divisibility rules for 9, 11, and realize that it must end in 7 zeroes.  That should help.[/hide]

It's a little easier to use the divisibility rule for 99.  Does everyone know that rule?

I solved this a while back in some Marathon. I'll try to find it.

EDIT:[quote="lotrgreengrapes7926"][quote="tim1234133"][size=150]Problem #?[/size]

Given that $34!=295 232 799 cd9 604 140 847 618 609 643 5ab 000 000$

determine the digits a,b,c,d[/quote]
[hide] $b=0$ obviously.
Umm...$4+a+c+d\equiv0\pmod9$.
$d-c-a-1\equiv0\pmod{11}$.
$a$ equals $8$ because it's the unit's digit of the product $1\cdot2\cdot3\cdot...\cdot9\cdot11\cdot12\cdot...\cdot19\cdot21\cdot...\cdot29\cdot31\cdot...\cdot34$, which is $8$.

$4+8+c+d\equiv0\pmod9\Longrightarrow c+d\equiv6\mod9$.
$d-c-8-1\equiv0\pmod{11}\Longrightarrow d-c\equiv-2\pmod{11}$.
$d=c-2$. $2c\equiv8\pmod9$. $c=4$. $d=2$.

$a=8, b=0, c=4, d=2$ [/hide] [/quote]

[quote="matt276eagles"][quote="lotrgreengrapes7926"]$a$ equals $8$ because it's the unit's digit of the product $1\cdot2\cdot3\cdot...\cdot9\cdot11\cdot12\cdot...\cdot19\cdot21\cdot...\cdot29\cdot31\cdot...\cdot34$, which is $8$[/quote]

This step is incorrect.  You want to evaluate $\frac{34!}{10^{7}}$.  Removing the multiples of 5 doesn't divide the original product by $10^{7}$.[/quote]

Oops.  :blush:

[hide] \begin{eqnarray*}\frac{34!}{10^{7}}&\equiv&3\cdot4\cdot6\cdot7\cdot8\cdot9\cdot11\cdot6\cdot13\cdot14\cdot3\cdot16\cdot17\cdot18\cdot19\cdot21\cdot11\cdot23\cdot24\cdot26\cdot27\cdot28\cdot29\cdot31\cdot16\cdot33\cdot34\\&\equiv&3\cdot4\cdot6\cdot7\cdot8\cdot9\cdot6\cdot3\cdot4\cdot3\cdot6\cdot7\cdot8\cdot9\cdot3\cdot4\cdot6\cdot7\cdot8\cdot9\cdot6\cdot3\cdot4\\&\equiv&2\pmod{10}\end{eqnarray*}

$b=0$
From divisiblity rules for 9 and 11:
$4+a+c+d\equiv0\pmod9$.
$d-c-a-1\equiv0\pmod{11}$.

$a=2$

$4+2+c+d\equiv0\pmod9\Longrightarrow c+d\equiv3\mod9$.
$d-c-2-1\equiv0\pmod{11}\Longrightarrow d-c\equiv3\pmod{11}$.

From the second equation, $d=c+3$ or $d=c-8$

Substituting into the first equation, $2c\equiv0,2\pmod9$.

Case 1: $d=c+3$ and $2c\equiv0\pmod9$. $(c,d)=(0,3)$ or $(c,d)=(9,12)$ (impossible). So, $(c,d)=(0,3)$.

Case 2: $d=c-8$ and $2c\equiv2\pmod9$. $(c,d)=(1,-7)$ or $(c,d)=(10,2)$, both are impossible.

$(a,b,c,d)=(2,0,0,3)$ [/hide]

Here's the approach based on 99.

[hide]We know that \[34! = 295 232 799 cd9 604 140 847 618 609 643 5ab 000 000\] is divisible by 99.

By the divisibility rule for 99, we get that \[2+95+23+27+99+cd+96+04+14+08+47+61+86+09+64+35+ab+00+00+00\] is also divisible by 99.

In other words, $cd+ab+670$ is divisible by 99.  lotrgreengrapes7926 already figured out that $ab = 20$.  Thus $cd+690$ is divisible by 99.  The multiples of 99 starting from 690 are 693 and 792.  Because 792 is too big, we have $cd+690 = 693$.  Thus $cd = 03$.

In summary, we have $a = 2$, $b = 0$, $c = 0$, and $d = 3$.[/hide]

I would venture to say that the answer is correct, nicely corresponding to the source (British MO 2003).  :lol:


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matrix
algebra theorems

Hello. Given a recurrence relation, is it possible to determine whether the sequence perodically recurs? If so , then is there a way to determine its period?
I'm novice to the theory of recurrence relations, so please anyone kindly recommend some articles or texts i should work on to develop my skills on the subject
Thankyou

There are some easy special cases.  If the recurrence is linear, this is possible if and only if the roots of the characteristic polynomial of the recurrence are roots of unity (edit: and either the roots are of multiplicity $ 1$ or the initial conditions are chosen appropriately).  For example, $ a_{n \plus{} 2} \equal{} a_{n \plus{} 1} \minus{} a_n$ has characteristic polynomial $ x^2 \minus{} x \plus{} 1$, whose roots are the primitive sixth roots of unity, so this recurrence has period $ 6$.  See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]here[/url] (especially Altheman's link at the bottom for a more systematic approach).

If the recurrence is of the form $ x \mapsto \frac {ax \plus{} b}{cx \plus{} d}$, this is possible if and only if some power of the matrix $ \left[ \begin{array}{cc} a & b \\
c & d \end{array} \right]$ is a multiple of the identity (i.e. has finite order in $ PSL_2(\mathbb{C})$).  This can also be determined by computing the characteristic polynomial.  See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=217361]here[/url], which applies these ideas to a different question.

Finally, there are some very specific tricks involving trigonometric identities.  Every periodic recurrence I have seen on a contest has been of one of these forms.


How do I change the width of the header so that it extends into the margin on both sides by an extra specified amount?

So far I have

[code]\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{array}
\usepackage{xy}
\usepackage{fancyhdr}
\usepackage{graphicx}
\usepackage{setspace}

\pagestyle{fancy}
\addtolength{\headwidth}{\marginparsep}
\addtolength{\headwidth}{\marginparwidth}

\fancyhf{}

\rhead{asdfasdfasdfasdf}
\chead{\Large \textbf{asdfasdfasdfasdfasdafsdf}}
\lhead{\includegraphics[width=30mm]{asdfasdfasdf.jpg}}
\rfoot{asdfasdfasdfasdfasdfsadfasdfasdfsadfasdf}
\cfoot{}
\lfoot{asdfasdfasdfasdfasdfasdfasdfasdf}

\fancyheadoffset{0 in}
\renewcommand{\headrulewidth}{2.5pt}

\pdfpagewidth 8.5in
\pdfpageheight 11in
\topmargin -0.6in
\headsep 0in
\textheight 8.3in
\textwidth 6.8in
\oddsidemargin -0.3in
\evensidemargin -0.3in
\headheight 84.6pt
\headsep 0.25in
\setlength{\parindent}{0in}[/code]

Use \fancyheadoffset. See [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=fancyhdr]fancyhdr manual[/url] page 12 section 12.


Ok quick question on the USNCO free response.  When, if ever, should you include states in reactions?  Are they required for question 5 (the reactions problem)?  Thanks.

i don't know if you are required to put states...
but how hard is it to do so?? ;)  ;)

just do it

Good point.  And also, is it at all better to put H30+ rather than H+, or use partial pressures (or even activities) in equilibrium expressions, or is that just overdoing it?

Activities are not necessary, certainly.  Partial pressures are necessary in $K_p$ expressions I think, and not $K_c$... I must admit I don't really understand the question - when do you have a choice?

$\text{H}_3\text{O}^+$ is way cooler than $\text{H}^+$.

Well what I meant is if they have a question such as "Write the equilibrium expression for this reaction" (not indicative of which form they mean), should you indicate partial pressures or just use molarity for everything?  It's really just style I suppose, and probably not even worth asking about.  Oops.

Hm... for "write the equillibrium expression," I would probably use concentrations, even if they're gas phase.  After all, it's just $(RT)^n$.

Ok thanks.  And I guess one other thing--what to do about signficant figures.  In the online answer keys, they are rather inconsistent with their usage.  Think those matter at all?

nope...
the choices in the MC are rarely "that" close, so that sig. fig. matters.

on free response it shouldnt be a problem. ;)


Let $ ABCD$ be a fixed quadrilateral such that $ AD//BC$, $ BC>AD$. Let $ I,J$ respectively be the midpoints of $ BC,AD$. Let $ M,N$ be the $ 2$ points moving on $ AB,DC$, respesctively such that $ MN//BC$. Denote $ X$ by the intersection of $ MC$ and $ IA$, denote $ Y$ by the intersection of $ CJ$ and $ AN$. Let $ P$ be the intersection of $ XY$ and $ IN$. Find the locus of $ P$.

It is easy to show by [b][size=100]Menelaos theorem[/size][/b], the concurrencies of $ P\equiv MJ\cap BD\cap IN$ and $ Q\equiv MI\cap AC\cap JN$ $ ($ consider for example the triangles $ \bigtriangleup ABD,\ \bigtriangleup CBD,$ with transversals the line segments $ MJ,\ IN$ respectively, for the first concurrency $ ).$

Because of now, the collinearity of the poins $ A\equiv XI\cap YN$ and $ C\equiv XM\cap YJ$ and $ Q\equiv MI\cap JN,$ based on the [b][size=100]Desarques theorem[/size][/b], we conclude that the triangles $ \bigtriangleup XIM,\ \bigtriangleup YNJ$ are perspective ans so, the line segments $ MJ,\ XY,\ IN,$ are concurrent at point $ P,$ as the point of intersection of $ MJ,\ IN.$

Hence, the locus of the point $ P\equiv XY\cap IN,$ is the line segment $ BD$ and the solution of the proposed problem is completed.

[b][size=100]REMARK.[/size][/b] - The problem is also true, for every pair of points $ I,\ J$ on $ BC,\ AD$ respectively, which are collinear with the point $ S\equiv AB\cap CD.$

Also, in the configuration of an arbitrary quadrilateral $ ABCD,$ instead of trapezium, the problem is true on condition of the concurrency $ R\equiv BC\cap MN\cap AD$ $ ($ in the trapezium here, we can say the the line segments $ BC\parallel MN\parallel AD,$ are concurrent at infinity $ ).$

Kostas Vittas.

[quote="vittasko"]It is easy to show by [b][size=100]Menelaos theorem[/size][/b], the concurrencies of $ P\equiv MJ\cap BD\cap IN$ and $ Q\equiv MI\cap AC\cap JN$ $ ($ consider for example the triangles $ \bigtriangleup ABD,\ \bigtriangleup CBD,$ with transversals the line segments $ MJ,\ IN$ respectively, for the first concurrency $ ).$

Because of now, the collinearity of the poins $ A\equiv XI\cap YN$ and $ C\equiv XM\cap YJ$ and $ Q\equiv MI\cap JN,$ based on the [b][size=100]Desarques theorem[/size][/b], we conclude that the triangles $ \bigtriangleup XIM,\ \bigtriangleup YNJ$ are perspective ans so, the line segments $ MJ,\ XY,\ IN,$ are concurrent at point $ P,$ as the point of intersection of $ MJ,\ IN.$

Hence, the locus of the point $ P\equiv XY\cap IN,$ is the line segment $ BD$ and the solution of the proposed problem is completed.

[b][size=100]REMARK.[/size][/b] - The problem is also true, for every pair of points $ I,\ J$ on $ BC,\ AD$ respectively, which are collinear with the point $ S\equiv AB\cap CD.$

Kostas Vittas.[/quote]
Dear [b]Mr. Kostas Vittas[/b], what a nice solution you offer to this problem. By chance, this is exactly the way I create this problem, thanks for your interest in it. Maybe you will make a connection to my other post: http://www.mathlinks.ro/viewtopic.php?t=276752


Fie $M$ o multime nevida cu $0\in M$ punct de acumulare si $f: M\to R$ o functie derivabila la dreapta in $0$ cu $f(0)=0$. Daca $(a_{n})_{n\geq1}$ este o progresie aritmetica de numere reale strict pozitive si $a_{n}\in M$ $\forall n\geq1$ calculati:
$\lim_{n\to\infty}\sum_{i=qn}^{pn}f(\frac{1}{a_{i}})$., unde $p>q$ sunt numere naturale nenule.

As putea sa scriu limita da e mult prea urata :)

Raspunsul e  $ \frac{f'(0) (\ln p-\ln q)}{r}$ unde r ratia progresiei.

Ma încumet, dar e urâtă rău.

[b]Pasul 1.[/b] Analizăm cazul $f(x) = x$.
Trebuie să calculăm $\lim_{n \to \infty}\sum_{i=qn}^{pn}\frac1{a_{0}+i \cdot r}= L$.
Dacă notăm $\theta = \frac{a_{0}}r$, atunci $L = \frac1{r}\cdot \lim_{n \to \infty}\sum_{i=qn}^{pn}\frac1{\theta+i}= \frac{M}{r}$.
Altfel zis,
\begin{eqnarray*}M &=& \lim_{n \to \infty}\sum_{i=0}^{(p-q)n}\frac1{\theta+qn+i}\\ \ &=& \lim_{n \to \infty}\frac1{(p-q)n}\sum_{i=0}^{(p-q)n}\frac1{\frac{\theta}{(p-q)n}+\frac{q}{p-q}+\frac{i}{(p-q)n}}.\end{eqnarray*}
Avem
\begin{eqnarray*}\lim_{n \to \infty}\frac1{(p-q)n}\sum_{i=0}^{(p-q)n}\frac1{\epsilon+\frac{q}{p-q}+\frac{i}{(p-q)n}}&=& \int_{0}^{1}\frac1{\epsilon+\frac{q}{p-q}+x}\, dx \\ \ &=& \ln \left( \frac{p+\epsilon (p-q)}{q+\epsilon (p-q)}\right) .\end{eqnarray*}
Când $\epsilon \to 0$, limita de mai sus tinde la $\ln \left( \frac{p}{q}\right)$, deci folosind o varianta a teoremei cleştelui obţinem $M = \ln \left( \frac{p}{q}\right)$.
Aşadar, $L = \frac1{r}\cdot \ln \left( \frac{p}{q}\right)$.

[b]Pasul 2.[/b] Cazul general.
Oricare ar fi $\epsilon > 0$, există $\delta > 0$ astfel încât $0 < x < \delta \, \, \Longrightarrow \, \, \left| \frac{f(x)}{x}-D \right| < \epsilon,$ unde $f^\prime (0) = D$.
Observăm că în suma $\sum_{i=qn}^{pn}f \left( \frac1{a_{i}}\right)$, argumentele lui $f$ sunt $\leq \frac1{qn}$, deci pentru $n$ mare ele sunt $\leq \delta$.
Rezultă că $\sum_{i=qn}^{pn}\frac{D-\epsilon}{a_{i}}< \sum_{i=qn}^{pn}f \left( \frac1{a_{i}}\right) < \sum_{i=qn}^{pn}\frac{D+\epsilon}{a_{i}}. \ \ \ \left( \star \right)$
Sumele din capete au limite $L_{1}= \frac{D-\epsilon}{r}\cdot \ln \left( \frac{p}{q}\right), \ L_{2}= \frac{D+\epsilon}{r}\cdot \ln \left( \frac{p}{q}\right)$.
Din prima inegalitate a relaţiei $\left( \star \right)$ obţinem $\frac{D-\epsilon}{r}\cdot \ln \left( \frac{p}{q}\right) \leq \liminf_{n \to \infty}\sum_{i=qn}^{pn}f \left( \frac1{a_{i}}\right)$. Luând $\epsilon \to 0$ obţinem
\[\frac{D}{r}\cdot \ln \left( \frac{p}{q}\right) \leq \liminf_{n \to \infty}\sum_{i=qn}^{pn}f \left( \frac1{a_{i}}\right) . \]
Analog,
\[\limsup_{n \to \infty}\sum_{i=qn}^{pn}f \left( \frac1{a_{i}}\right) \leq \frac{D}{r}\cdot \ln \left( \frac{p}{q}\right) . \]
Din astea două inegalităţi iese $\liminf = \limsup$, deci şirul e convergent şi are limita $\frac{D}{r}\cdot \ln \left( \frac{p}{q}\right)$.

se pare ca estetica uratului are si partile ei bune....Lol  :) mai e putin pana la bac:((((((((
BTW solutie f frumoasa perfect_radio :)

perfect_radio are un talent deosebit ( unic cred ) de a sari peste punctele esentiale si de a trece la concluzii. oricum, baiat destept  :idea:


geometry
incenter
circumcircle
geometry unsolved

Let $ ABC$ be a triangle and $ M,N$ points on $ AB$ respectively $ AC$ such that $ \frac{BM}{MA}\equal{}x$ and $ \frac{CN}{NA}\equal{}y$. Prove that $ MN$ passes through the incenter $ I$ of the triangle $ ABC$ if and only if $ a\equal{}bx\plus{}cy$, where $ a,b,c$ are the lenghts of the sides of the triangle.

PS: Also try to prove simillarly for $ G$(centroid) or $ O$(circumcenter) instead of $ I$ (the incenter).

[b]Cristea's theorem:[/b] $ D,E,F$ are three points on the sides $ CB,BA,AC$ of $ \triangle ABC$ and $ M \in AD.$ Then the transversal $ EF$ goes through $ M$ if and only if
\[ \overline{DC} \cdot \frac {\overline{EB}}{\overline{EA}} \plus{} \overline{BD} \cdot \frac {\overline{FC}}{\overline{FA}} \equal{} \overline{BC} \cdot \frac {\overline{MD}}{\overline{MA}} \]


Suppose that $f: R \rightarrow R$ is a function for which $f(2x-f(x))=x$ for all x, and let r be a fixed real number.

(a) Prove that if $f(x)=x+r$, then $f(x-nr)=(x-nr)+r$ for all positive integers n.
(b) Prove that if $f$ is a one-to-one function (i.e., $f(x)=f(y)$ implies $x=y$) then the property in (a) also hold for all integers n.

Euhm, I don't think I get it...just set $a=x-nr$. It's given that $f(a) = a+r$ for all $a\in\mathbb{R}$, so $f(x-nr) = (x-nr)+r$? :dry:

That's what I thought too at first. I think you have to interpret the $x$ in $f(x)=x+r$ to be a specific constant x, while the x in $f(2x-f(x))=x$ is arbitrary.

Don worry guys i'll get ur solution, that's if u don't have it yet, which i kinda doubt

Eh? I did get the solution, although I'm not sure of the importance that f(x) is a one-to-one function in part (b).

Myeah, the problem should say "If $f(a)=a+r$ holds for some real number $a$, prove that..." then. It's not thát hard: first part is quick induction, because it's true for $n=1$, i.e. $f(a-r) = a$, because it's given that $f(a-r) = f(2a-(a+r)) = f(2a-f(a)) = a$. Now suppose that, indeed, $f(a-nr) = (a-nr)+r$ for some positive integer $n$. Then \[\begin{aligned}(a-(n+1)r)+r & = a-nr = f(2(a-nr)-f(a-nr)) \\ & = f(2a-2nr-(a-nr+r)) = f(a-(n+1)r),\end{aligned}\] which completes the proof.
For the second one, we want to prove that $f(a+nr) = a+(n+1)r$ holds for all positive integers $n$; it's induction once again: note that $f(2(a+r)-f(a+r)) = a+r = f(a)$, so (one-to-one) $2a+2r-f(a+r) = a$, so $f(a+r) = a+2r$, which is our base case. Now suppose that, indeed, $f(a+nr) = a+(n+1)r$. Then clearly $f(2(a+(n+1)r)-f(a+(n+1)r)) = a+(n+1)r = f(a+nr)$, from which we get $2a+2(n+1)r-f(a+(n+1)r) = a+nr$, and $a+(n+2)r = f(a+(n+1)r)$, which completes the proof.

(I hope I didn't make any mistakes writing the proof. However, concept looks right to me.)

[quote="Phelpedo"]Why is it important that the function be one-to-one? The second argument looks like it works no matter what.[/quote]

I used it here: [quote="Kurt Gödel"]Note that $f(2(a+r)-f(a+r)) = a+r = f(a)$, so $2(a+r)-f(a+r) = a$.[/quote]We couldn't say that if it weren't given that $f$ is one-to-one  :wink:  (Later in the proof, it's used another time.)


inequalities
geometry
circumcircle
geometric transformation
reflection
inequalities proposed

Let $ABC$ be a triangle, the its circumcircle $C(O,R)$, the its incircle $C(I,r)$,
the orthocenter $H$ and the reflection $A'$ of the point $A$ w.r.t. the point $O$.

Prove that  $\underline {\overline {\left|\ \blacksquare\ \ AH\cdot AI +A'H\cdot A'I\ge 2R^2 .\ \ \blacksquare\ \right| }}$

i'm sorry... you wrote something like "the reflection $A'$ of the point  w.r.t. the point $O$... what is $A'$ and what does "w.r.t." means?

$A'$ is the reflection of $A$ [u]with respect to[/u] $O$.

So, $O$ is the midpoint of $AA'$.

Does that clear things up?


Could someone tell me what I should do to get a list of items like this?

a)

b)

c)

Because the $\verb+enumerate+$ command gives me a list like

1.

2.

3.

Thanks in advance for your help! :)

Use the line
\renewcommand{\labelenumi}{\alph{enumi})}
early on in your document, before you want to use enumeration. The extra ) there is to give a) rather than just a


geometry
geometric transformation
reflection
Support
calculus
integration
blogs

I was on some music forums and someone pointed out that there is a "White Nationalist" site with 88,000 members from aroudn the world, promoting "Racialist discussion board for pro-White activists and anyone else interested in "White survival". The membership is white only membership. LOL basically those people think that the white race is either superior and want non-white people out of the U.S. because supposedly they're the natives of the U.S. lol (But really the natives are American Indians).

What exactly do you guys think of this queer behavior and thought processes.

They even pointed out asians to lack intelligence and creativity in this thread called "Asians what do you think of them"

I find it quite stupid yet interesting at the same time that there's people who think like that. But I want to see your guy's opinion/comments on this type of shiz.

[i]This thread has been edited by a moderator to remove links to the mentioned website.[/i]

I agree with jimhu and Iversonfan2007.

Just because I am white, doesn't make me automatically superior to non-whites. The site is highly racist and should be shut down if possible.

[quote="TrueConservative"]Racially speaking, they [asians] are inferior to Europeans due to their lack of creativity and genius[/quote]

Not all asians are uncreative and stupid. An asian at my school won a creative writing contest. You have to be creative and intelligent for that.

yea, it isnt illegal but it is plain wrong

One thing I've noticed by reading through the forum is the weight given to the average IQ score of a particular race as a measure of that race's worth.  I find this a quite comical indicator of their ignorance.  :P

[quote="mysmartmouth"]The people discuss these issues, treating people like livestock.

However, a few positivish quotes:

I like asians they are equal to europeans as far as intelligence goes and I really like anime.

Only Asians can compete with Europeans as far as cultural acheivements. These people are worthy of respect.

...I practice a Chinese martial arts. Hating Asians would make me a hypocrite.

Yes, I agree, the Asians would be one of the last demographics I would deport. They seem to pose a drastically smaller problem than the other races here, and seem to in many circumstances actually contribute.

Even though these quotes are positive, the attitude still disgusts me.[/quote]

That is actually a (subtly) racist post in itself. Although it was probably unintentional,

"Only Asians can compete with Europeans as far as cultural acheivements. These people are worthy of respect."

Who said Europeans have the greatest cultural achievements? Why should any race be NOT worthy of (equal) respect?

"Yes, I agree, the Asians would be one of the last demographics I would deport. They seem to pose a drastically smaller problem than the other races here, and seem to in many circumstances actually contribute."

Why should anyone be given the "right" to deport anyone (I know its figuratively, but why are you even considering an option such as this)? Why should any whole race be deported in the first place? Why should the "whites" be deporting anyone? Any race is just as "American" as all of us. You are as "American" as you want to be. (Assuming your speaking of America.) I didn't realize that any whole race posed a problem for America.

[quote="ragnarok23"][quote="PenguinIntegral"][quote="Xantos C. Guin"]The site is highly racist and should be shut down if possible.[/quote]

Why should it be shut down? It's idiotic, yes. It's disgusting, yes. But the constitution protects people's freedom, and unless they are breaking the law, it should stay up as long as the site owner wishes.

Just because you do not agree with something does not entitle you to stop it or shut it down.[/quote]
Doesn't the constitution also protect people's religion and race...?[/quote]

The Constituion only allows you to practice whatever religion you want to.  There's nothing in the Constituion that can stop people from expressing their viewpoints on race and religion, even if they disagree with most of America (as far as i know).  Isn't that the whole point of freedom of speech?

[quote="surge"]
That is actually a (subtly) racist post in itself. Although it was probably unintentional....
[/quote]

No, no, no, no, no.  Sorry if that was taken that way.  Those were just quotes that I probably presented in the wrong way.  They were not my opinions, nor do I endorse them, believe they are justified, etc.

I just think that website is rude and wrong, but happily aops doesn't have that kind of forum.....That website must have been created in the slavery times or something. I don't know.

*PS I know computers weren't made then

[quote="ragnarok23"][quote="PenguinIntegral"][quote="Xantos C. Guin"]The site is highly racist and should be shut down if possible.[/quote]

Why should it be shut down? It's idiotic, yes. It's disgusting, yes. But the constitution protects people's freedom, and unless they are breaking the law, it should stay up as long as the site owner wishes.

Just because you do not agree with something does not entitle you to stop it or shut it down.[/quote]
Doesn't the constitution also protect people's religion and race...?[/quote]
Only from the government. Not from private people.

[quote="zeus_three"]PenguinIntegral, What?

The website should be shut down, full stop.  Its preposterous to claim that, just because of some idealistic, ephemeral, intangible concept or 'thing' called "freedom", then, we should go about allowing those things that [b]we know[/b] are intrinsically wrong.  It is absurd to use the law to override what we know is both morally right and correct.

This is not a matter of freedom for the sake of freedom, but freedom for the sake of our welfare.  If it were the former, then there'd be neither rules nor regulations.  It is a mistake, then,  to claim that we should allow these people to splash blatant racist remarks over the internet, because of some utopian concept called "freedom".[/quote]

Freedom exsists to prevet mob rule. If you were in power, then you would shut it down. If they were in power, they would deport blacks. In both cases, the law protects the minority from the majority, even when they "know" they are right. Just because you or the majority feels something is morally correct or right, does not make it so. The whole purpose of the first amendment is to give people the freedom to say whatever they want as long as it does not directly harm anyone, regardless of whether they are right or not. (The harm thing encompasses things like shouting "Fire!" in a crowded theater, or selling drugs falsely marked. In both cases, it isn't the use of speech itself, but that you are using it to directly harm another being).

Freedom exists for the welfare of the individual, not for the welfare of the community. That is the whole point of the constitution: that as long as you are not hurting anyone, you are free to do whatever you want.

People should largely be free, if they so desire, from the intervention of anther entity in their lives, whether it is you or the government.

One more thing to wobster: the other obvious reason that "white pride" is different from anyone elses pride is that we live in a society all of whose major institutions have been and continue to be white-dominated, and where the population is majority white.  White people don't [i]need[/i] pride groups -- they have the whole damn country (up to epsilonics).

[quote="deimos"]when one afroamerican meets other afroamerican (they don't know each other. they are complete strangers) why they call each other brother (it's clear that they say that just because they have the same colour of the skin)? isn't that some kind of racism?[/quote]

Is that a joke?  What exactly do you think the word "racism" [i]means[/i]?

[quote="JBL"][quote="wobster109"]how come something like "asian pride" is commonplace, while something like "white pride" gets called racism?[/quote]

Are you really so naive that you don't know the answer to this question?  "White pride" groups consistently assert the superiority of one racial group over another (in fact, typically, over all others).  "Black student associations" (or whatever your chosen equivalent is) consistently don't do that.[/quote]

I think that one of the major reasons that "white pride" groups consistently assert the supierority of one racial group over another is because anyone who wanted to set up a "white pride"-style group in a completely non-racist way would be deterred by the fact that it would be interpreted as racism. So the only "white pride" groups are those set up by racists- not that "white pride" has to be racist.

[quote="JBL"]

[quote="deimos"]when one afroamerican meets other afroamerican (they don't know each other. they are complete strangers) why they call each other brother (it's clear that they say that just because they have the same colour of the skin)? isn't that some kind of racism?[/quote]

Is that a joke?  What exactly do you think the word "racism" [i]means[/i]?[/quote]

I have pm you.

Morally:

If one deals with each person, not as a member of any race, religion, profession or nationality - but a a fellow human (made in the image of God, according to my religion) many of these issue not only aren't relevant, they can't arise.

Legally:

In the US the freedom of speech protects people (whether or not they are citizens) from most of the government's interference.  Governments can punish someone from maliciously crying, "Fire," in a crowded building and regulate demonstrations for things like only having one in a park at a time and the like.  There is no right to say what one wants on TV or in the newspaper and people don't have to listen.  It doesn't apply to a school controling student's statements in class or worker at work.  It doesn't protect a person's business (your customers can leave if they don't like what you have a right to say).  The government can't kick you out of school or deny you a license to work based on statements made that aren't otherwize a crime.  In these cases you may be barred from working about children if conviced of child prono - to protect the children and this isn't protected speech.

Perhaps too much, but new technology gives rise to new laws, but typically it takes quite some time.

Practically, if a site has that many followers it will be meaningless to close the site.  If it is based in the US and is closed, one can expect the added publicity of the closing to increase its draw among those so inclined and the site will migrate to a computer located in another country, with a reference at the orginal site, which probably can continue if it just has a reference to a non-US site.

It's usually better just to ignore this type of material rather than fighting to close it down.

Also, in the US we would and do allow those that would attack us to meet and plan things that do not constitute crimes.  If they are already wanted for having been convicted of something they will be arrested if found, but that has nothing to do with any legal meeing.

Why do we call certain issues "race issues?"  These are personal issues.

Enought from me.

[quote="tim1234133"][quote="JBL"]Are you really so naive that you don't know the answer to this question?  "White pride" groups consistently assert the superiority of one racial group over another (in fact, typically, over all others).  "Black student associations" (or whatever your chosen equivalent is) consistently don't do that.[/quote]

I think that one of the major reasons that "white pride" groups consistently assert the supierority of one racial group over another is because anyone who wanted to set up a "white pride"-style group in a completely non-racist way would be deterred by the fact that it would be interpreted as racism. So the only "white pride" groups are those set up by racists- not that "white pride" has to be racist.[/quote] "They're only white supremacists because everyone expects them to be white supremacists?"  That doesn't make any sense.  People expect a group calling itself a "white pride" group to be racist because every example of such a group that has arisen [i]has[/i] been white supremacist.  White people who want to go hang out only with white people and not be called racists can go join a country club.

[quote="wobster109"]JBL, that is holding whites to different standards than other ethnic groups.  whatever the underlying causes/implications may be, it is still racism that when another group does something, it isn't interpreted negatively.[/quote]
This (by itself) is an absurd objection to what I wrote: "white pride" groups [i]behave differently[/i] from nonwhite pride groups, so I am in fact applying the same standard to both.  Now, if you find me a racist Asian pride group and I refuse to call it racist, [i]that[/i] would be applying a double standard.

defaultvoid.com used to be seriously racist... there was this board dedicated for racism

[quote="zeus_three"]PenguinIntegral, What?

The website should be shut down, full stop.  Its preposterous to claim that, just because of some idealistic, ephemeral, intangible concept or 'thing' called "freedom", then, we should go about allowing those things that [b]we know[/b] are intrinsically wrong.  It is absurd to use the law to override what we know is both morally right and correct.

This is not a matter of freedom for the sake of freedom, but freedom for the sake of our welfare.  If it were the former, then there'd be neither rules nor regulations.  It is a mistake, then,  to claim that we should allow these people to splash blatant racist remarks over the internet, because of some utopian concept called "freedom".[/quote]

I agree with penguin integral-- no matter how much YOU believe it is wrong, you can't shut them down simply because you disagree with them. By doing so, you are endorsing the same sort of attitude as they are, the belief that what YOU think is correct and that everyone should conform to your belief. There is NO set definition of morality and you cannot just say that what they're talking about is "immoral"-- because to them, it's not.

Also, by raising their website to a standard of "instrinsic" morality you are only creating more problems. If you are entitled to shut down their website because it's "intrinsicly" wrong, then wouldn't people want to shut down websites supporting abortion, gay rights, or stem-cell research because these things are "instrinsicly wrong" as well?

These people can post what they believe. You can't stop them, but you can ignore them or try to reason with them. Simply claiming that they are immoral is not the solution.

I noticed a trend in these so-called "white supremacists." They tend to be the outcasts and often very stupid and weak. They are the remnants or members of the KKK and other organizations, which are still legal and from time to time try to intimidate people.

But unless they revert back to their old ways of violence, the government cannot ban or suppress them because they represent a point of view, no matter how ignorant it is.

By the way, I hope the white supremacists liked Survivor  :wink:

Race is a complicated issue in America. We like to think that in the 21st century racism is a thing of the past and that tolerance has replaced it. But, essentially we just fail to see racism under its new disguises: multiculturalism, political correctness, affirmative action, and my favorite, institutionalized racism.

To begin, there is a difference between being indecent and simply unacceptable. To be indecent is to hurt other's feelings. To be unacceptable is to hurt other's lives. When someone is just speaking their mind, it is not unacceptable but indecent. Likewise, if that person is taking a hand in murder, or advocating bodily and severe psychological harm to a person or group, that is unacceptable and criminal.

Unfortunately, in today's society, the line between indecency and unacceptability is skewed. We as a people really haven't concocted guidelines that establish standards of acceptability that is fair to all and realistic to institute. I as a human being have yet to figure out why we can have a Black Association of sorts but not a white, why we can have affirmative action for blacks and hispanics but marginalize whites and asians, why we can be a nation of great diversity but have most of our leading positions of power be occupied by whites. And, truthfully, those are tough questions to answer, if I consider the ramifications. But, those are the questions that must be answered.

I am familiar with the website that has been mentioned, and strangely I am at awe at the diverse backgrounds these people come from. A number of them do not come from backgrounds that were all about white nationalism, like kkk, skinheads, neonazis. A lot of them do, but not all. Rather, those that do not, become this way because of situations in their lives that make them this way, statistics about other ethnic populations that make them this way, the equal opportunity programs that they feel shortchanged by make them this way. What they percieve as the marginalization and eradication of the white race make them this way. And, sadly, this is what happens when we tend to create double-standards for people. Of course, it doesn't make it right to exercise hate, especially to the degree that white nationalists or any extreme terror groups assert. But, this serves as an example that when we as a people have double-standards we harbor bitterness within the minority that percieves themselves as the victim, and this can be said of any group.

Yet, as compelling as an issue that race is, it is a shame that open dialogue cannot be given to it. Because the line between decency and unacceptability is so precariously perched somewhere between two extremes, it's hard to have an honest conversation regarding race without being called racist. When my friends and I happen to be talking about race and ethnicity, it is very much a censored conversation with everyone tiptoeing around each other to sound "tolerant," or tasteless, when we should be aiming for truth and respect. I'm not just talking arguing for whites/white nationalists, either, though I cite them as an example, but everyone.

It's sad that humans have a petty habit to segregate themselves and degrade others, to exude prejudice and then cry foul when karma sets in. I am not targeting a specific group or taking any sides but iterating that prejudice towards any group is unjustified, whether the institutionalized racism against blacks or the affirmative action programs against whites or the attack on the Martians.

Just turn a deaf ear to hate websites and stuff.  They thrive off of people caring too much.  I'm sure you all have better things to worry about than this.

i'm assuming we're talking about de facto segregation and not de jure?

jcreinhardt is right; hate forums are crap and they're just there to attract attention. visit them all you want and spam if you wish, just don't make any comments that actually contributes to the site, unless you want a bunch of racists nagging you to join their cult [no offense intended to racists out there](i'm being serious)

Edited: I was wondering if people here think stereotyping is racism?

[quote="neelnal"]I was wondering if people here thing stereotyping is racism?[/quote]

ummm, clarification if possible about the question?

[quote="neelnal"]I was wondering if people here thing stereotyping is racism?[/quote]

imo stereotyping is human nature, however, racism is unnatural and generally hateful.


calculus
integration
calculus computations

\[ \int\limits_{ \minus{} 2}^2 {\left| {x(x \minus{} 1)} \right|dx}
\]

$ I \equal{} \int\limits_{ \minus{} 2}^2 {\left| {x\left( {x \minus{} 1} \right)} \right|}  \equal{} \int\limits_{ \minus{} 2}^0 {\left( {{x^2} \minus{} x} \right)dx}  \minus{} \int\limits_0^1 {\left( {{x^2} \minus{} x} \right)dx}  \plus{} \int\limits_1^2 {\left( {{x^2} \minus{} x} \right)dx}  \equal{} \frac{{17}}{3}$


Let p denote a prime number of the form $4k+3$ and let $x$ be any positive integer. Prove that the integer $x^2-xy+\frac{p+1}{4} y^2$  does not have any divisor of the form $kp-1$   :)  :)

Actually, I think $x,y$ had to be coprime, and the conslusion was supposed to be that there are no odd divisors $q$ of $x^2-xy+\frac{p+1}4y^2$ of the form $kp-1$. It's been discussed before (at least with prime $q$), but eh.. :)

Suppose there are such divisors $q$. Then $q|a^2+py^2$, where $a=2x-y$, and we have $\left(\frac{-p}q\right)=1\ (*)$. The LHS of $(*)$ is equal, because of the properties of the Jacobi symbol, to $\left(\frac{-1}q\right)\left(\frac pq\right)=(-1)^{\frac{q-1}2}(-1)^{\frac{p-1}2\cdot\frac{q-1}2}\left(\frac qp\right)$, which is $-1$, as can easily be seen. We thus have a contradiction.


find the 3rd term in the expansion of $ (2\minus{}x)^{\minus{}2}$

This is the same as $ \frac{1}{(2\minus{}x)^2}$.

[hide]$ (2\minus{}x)^2\equal{}4\minus{}2x\plus{}x^2$
$ (2\minus{}x)^{\minus{}2}\equal{}\frac{1}{4} \minus{} \frac{1}{2x} \plus{} \frac{1}{x^2}$

So I guess the third term is $ \frac{1}{x^2}$.[/hide]

nope problem in the fractions; you can't just split them up

[quote="grn_trtle"]$ (2 \minus{} x)^2 \equal{} 4 \minus{} 2x \plus{} x^2$
$ (2 \minus{} x)^{ \minus{} 2} \equal{} \frac {1}{4} \minus{} \frac {1}{2x} \plus{} \frac {1}{x^2}$[/quote]

Plug in $ x \equal{} 1$ to see why that's wrong. Generally, $ \frac {1}{a \plus{} b \plus{} c} \neq \frac 1 a \plus{} \frac 1 b \plus{} \frac 1 c$.

[hide="Hint"]Write it in the form $ k \left( \frac {1}{1 \minus{} r} \right)^2$, where $ r$ is some multiple of $ x$ and $ k$ is a constant.

Then note that $ \frac {1}{1 \minus{} r} \cdot \frac {1}{1 \minus{} r} \equal{} \frac {1}{1 \minus{} r} (1 \plus{} r \plus{} r^2 \plus{} \ldots)$

$ \equal{} \frac {1}{1 \minus{} r} \plus{} \frac {r}{1 \minus{} r} \plus{} \frac {r^2}{1 \minus{} r} \plus{} \frac {r^3}{1 \minus{} r} \plus{} \ldots$

$ \equal{} (1 \plus{} r \plus{} r^2 \plus{} r^3 \plus{} \ldots) \plus{} (r \plus{} r^2 \plus{} r^3 \plus{} \ldots) \plus{} (r^2 \plus{} r^3 \plus{} \ldots) \plus{} \ldots$

What happens when we collect like terms in this?

Now, what happens when we substitute in for $ r$ and multiply by $ k$?[/hide]

:rotfl:
I always get this mixed up

$ \frac {a \plus{} b}{c} \equal{} \frac {a}{c} \plus{} \frac {b}{c}$

$ \frac {c}{a \plus{} b}\ne\frac {c}{a} \plus{} \frac {c}{b}$

Thanks  :blush:

the answer is $ \frac{3x^2}{16}$ but i have no idea how u get that....

[hide]$ (2 \minus{} x)^2 \equal{} ( \minus{} x \plus{} 2)^2 \equal{} x^2 \minus{} 4x \plus{} 4$
therefore, $ (2 \minus{} x)^ \minus{} 2\equal{}\frac{1}{x^2\minus{}4x\plus{}4}$ then i can't type the latex from there on but someone else can...

and i repeat, the answer is 3x^2/16...

any takers?

[quote="Poincare"][hide]$ (2 \minus{} x)^2 \equal{} ( \minus{} x \plus{} 2)^2 \equal{} x^2 \minus{} 4x \plus{} 4$
therefore, $ (2 \minus{} x)^ \minus{} 2 \equal{} \frac {1}{x^2 \minus{} 4x \plus{} 4}$ [b]then i can't type the latex from there on but someone else can...[/b][/quote]
'nuff said [/b]

Long division is not the optimal way to go here. Observe that $ (2 \minus{} x)^{ \minus{} 2} \equal{} \frac {1}{4} \left( \frac {1}{1 \minus{} \frac {x}{2}} \right)^2$, and then use my hint. That way, you can get the whole series.

[hide="Answer"]
$ (2\minus{}x)^{\minus{}2} \equal{} \sum_{n\equal{}1}^{\infty}\frac{n}{2^{n\plus{}1}} x^{n\minus{}1}$
[/hide]

[b]@Steve:[/b] You're dividing in the wrong direction (ie, with decreasing exponents instead of increasing).

Your first term should be $ \frac {1}{4}$, like this:

$ \frac {1}{4 \minus{} 4x \plus{} x^2} \equal{} \frac {1 \minus{} x \plus{} x^2/4 \plus{} x \minus{} x^2/4}{4 \minus{} 4x \plus{} x^2} \equal{} \frac {1}{4} \plus{} \frac {x \minus{} x^2/4}{4 \minus{} 4x \plus{} x^2}$

Now make your next term have an exponent of $ 1$ on $ x$.


[quote="madness"]Question 1: How many integers between 1 and 1000000 have the sum of their digits equal to 18?
Answer : Consider a six digit number of the form $ \ce{x1x2x3x4x5x6}$ where x1 , x2 , x3 , x4 , x5 lie between 0 and 9 and x6 lies between 1 and 9 both limits included.
We have to find all numbers such that x1 + x2 + x3 + x4 + x5 + x6 = 18
The no. of nos. is the coeff of $ x^{18}$ in the expansion of :

           $ (1 \plus{} x \plus{} x^2 \plus{} ........... \plus{} x^9)^5 \cdot (x \plus{} x^2 \plus{} ............x^9)$

Solve  :)[/quote]

I don't quite follow....can u be more specific?

I meant let xi be the ith digit of the number.
Then sum of all xi = 18 (as per the problem)
To easily compute all xi's we use the multinomial theorem http://en.wikipedia.org/wiki/Multinomial_theorem
which is a part of generating functions http://en.wikipedia.org/wiki/Generating_functions
 :lol:

[quote="madness"]
The no. of nos. is the coeff of $ x^{18}$ in the expansion of :

           $ (1 \plus{} x \plus{} x^2 \plus{} ........... \plus{} x^9)^5 \cdot (x \plus{} x^2 \plus{} ............x^9)$

Solve  :)[/quote]

Sorry, but why is that???

kanna naan ezhudardhu laan solli solli ezhuduvein
puriyele na andha linkka paru
ennum puriyale na thiruppi thurippi padi
go in recursion
void read()
{
   int i=0,understood=0;
   see_above_explanations();
   cout<<"Have you understood"<<endl;
   cin>>understood;
   if(!understood)
   if(i<1000)
   read();
   else (if i==1000)
   cout<<"Memory overflow  :( "<<endl;
   else
   cout<<"Got it  :P "<<endl;
   return nothing;
}
 :wink:


In a triangle $ ABC$, let $ O$ be its orthocentre and $ AD,BE$ and $ CF$ be its altitudes. Then prove that $ \frac{OD^2}{a}\plus{}\frac{OE^2}{b}\plus{}\frac{OF^2}{c} \geq \frac{1}{abc}$ , where $ a,b$ and $ c$ are the sides of the triangle $ ABC$.

Manjil your inequality is clearly wrong . It is not even homogeneus , May you check it again?


algebra
polynomial
number theory proposed
number theory

Suppose that $p\in \mathbb Z[x]$ which $deg\ p>1$. Suppose that $A=\{p(n)|n\in\mathbb Z\}$. Prove that there exist $a,b\in \mathbb Z$ which if $B=\{an+b|n\in\mathbb Z\}$ then $A\cap B=\emptyset$

[quote="Omid Hatami"]Suppose that $p\in \mathbb Z[x]$ which $deg\ p>1$. Suppose that $A=\{p(n)|n\in\mathbb Z\}$. Prove that there exist $a,b\in \mathbb Z$ which if $A=\{an+b|n\in\mathbb Z\}$ then $A\cap B=\emptyset$[/quote]
I think $B=\{an+b|n\in Z\}$.

Nice, but very easy. Suppose the contrary, then for all $d>2$ and all $n$ the numbers $p(n),p(n+1),...,p(n+d-1)$ give all residues mod d. However, if $n$ is chosen such that $d=p(n+1)-p(n)>2$ (which can be achieved), the later cannot hold.


Probability with Combinations Part 1

We solve a basic probability problem with counting involving combinations.


I thought it was a sad story. In case you don't know what I am talking about- A mom saw that her baby was missing. She was a SUSPECT in the case. Then, she is invited to Nancy Grace's show and is completely attacked by words and is pressurized by that woman's pure arrogance. Accused of being a suspect and that she was involved, while the poor woman is just hurt by her baby's dissapearance.

The next day, she write's a statement including why she is about to commit suicide. Blaming Nancy Grace. Which is a perfectly good reason because she can't take it anymore.

What do you think of what happened? The victim's family has filed a lawsuit.

ROFL that's one of the funniest things I have ever read.

Of course she was a suspect, what's wrong with that? You need to check all possible bases and if there are no leads, then I would naturally be suspicious of the mother herself. She was invited to a show, she could've declined, but she apparently was greedy and wanted the money so she came on the show anyway. She had a choice and she chose money, Nancy Grace isn't arrogant at all, stop making ad hominem attacks.

She needs to learn how to solve her own problems and not take the easy way out of them. Her commiting suicide shows how much she's an idiot, she steps on a bit of misfortune, creates problems from that, and then has to take the easy way out.

It is not Nancy Grace's fault. The family deserves a dog's crap at most. I mean, look at the women, she was more upset over the Nancy Grace show than her baby's disappearence as shown by who she put the blame on in her suicide letter. And don't even make statements like "Which is a perfectly good reason because she can't take it anymore" as it isn't.

She's also going down for commiting suicide.

[quote="Treething"]So how would you treat your friend if he murdered his child?

I'd [b]definitely[/b] shower him with highly loving hugs, kisses, and other non-homosexual-yet-tender gestures of understanding.[/quote]

treething that is sort of scary but not really only if i didn't know you say things like that


A penny is thrown straight up off the top of the Empire State building at 30 feet/sec.  If the Empire State is 1440 feet high, and the equation for height of a falling projectile at time  t is  h = -16t2 + vt + d  , where v is the initial velocity and d the initial height, how long will the penny take before it hits 33rd street?  Use a calculator to evaluate to the nearest tenth of a second.

How long would the penny take to hit 33rd street if it were thrown straight down at 30 ft/sec?

Uhhhh....where is 33rd street?

Assuming 33rd street is the street where the penny will fall: Basically what you do is plug in the values for height (1440) and velocity (30), then set the height equal to zero (when the penny will be on the ground) and solve the quadratic equation.


Suppose $ P(x)\equal{}ax^2\plus{}bx\plus{}c$ be a polynomial such that $ a,b,c \in N$.we know that for all $ x \in R$ $ P(x) \geq 0$.Prove that:

$ (1)$ $ b$ is even.

$ (2)$  $ ac$ is a perfect square.

Nice,ha?

Is $ P(x) \equal{} 3x^{2} \plus{} x\plus{}1$ a counter-example? Or am I misinterpreting something?

[quote="TaiPan~SP!"]Is $ P(x) \equal{} 3x^{2} \plus{} x \plus{} 1$ a counter-example? Or am I misinterpreting something?[/quote]

[quote="irantst"]hi
how do you say that?[/quote]

Well his example is clearly a counterexample...the polynomial fulfils both your criteria: The coefficients $ a \equal{} 3$, $ b \equal{} c \equal{} 1$ are elements of $ \mathbb{N}$ and we have $ P(x)\geqslant0 \ \forall x\in\mathbb{R}$ because $ P(x) \equal{} \frac {11}{4}x^2 \plus{} (\frac {x}{2} \plus{} 1)^2> 0$.
But neither (1) nor (2) is true...


pigeonhole principle
quadratics
absolute value
number theory unsolved
number theory

Show that given any prime p, there exist integers x, y, z, w such that x^2+y^2+z^2-wp=0 and 0<w<p

Arne: Correct, but I want to see a proof of this fact not using any harder theorems (e.g. Dirichlets theorem on primes in arithmetic progressions) and fitting here  ;)

Another approach would simply be:
There are $x,y,z$, not all $ \equiv 0 \mod p$ that fulfill $x^2+y^2+z^2 \equiv 0$. This can be proven in various ways using known theorems like Chevalley-Warning, but more directly:
For $p=2$ check it manuelly, so let $p$ be odd.
Take $R:=\{ 0 , 1 , 2^2 , ... , \left( \frac{p-1}{2} \right)^2 \}$ and $S:=\{ -1-0 , -1-1 , -1-2^2 , ... , -1-\left( \frac{p-1}{2} \right)^2 \}$. Then by pigdeonhole-principle, there are some $r \in R, s \in S$ with $r \equiv s \mod p$, so there exists $x,y$ with $x^2 \equiv -1 - y^2 \mod p$ or $x^2+y^2+1^2 \equiv 0 \mod p$.

Now take from any solution the representatives $x,y,z$ of smallest possible absolute value (so $|x|,|y|,|z| < \frac {p}{2}$. Then one gets $0< x^2+y^2+z^2 < \frac{3}{4}p^2$, so one can also get $0<w<\frac{3}{4}p$ (and by $z=1$ even stronger results)

This is Korean MO 2000, Qn 1.
The official solution uses pigeon-hole principle, just like [i]ZetaX[/i]'s solution.
I remembered this problem because I encountered it in during training in 2003, and my solution was messy, using quadratic reciprocity (I split into cases... yuck, i got an ugly solution). And when I saw the official solution, I was startled by its elegance, which doesn't use 'harder' theorems. :)

[quote="Arne"]Well... any number which is not of the form $4^k(8t + 7)$ can be written as the sum of three squares so it's obvious.[/quote]
This claim seems useful, could you please post a proof?

Haha, nice joke. The proof is very hard and long. I am sure you can find it on the internet though. (By the way, you can prove for yourself that any number of the form $4^k(8t + 7)$ can NOT be written as the sum of three squares, that's easy, but the other direction is very hard.)

It follows indeed from the Hasse-Minkowski-global-to-local-principle (for $4$ variables) by some calculations. But to prove the principle, you will (at least as much as I know) have to use Dirichlets theorem again.

Sorry but what is the Chevalley-Warning teorem?

can someone answer?

Umm, sorry, didn't see that post for... four years  :blush:

See http://www.mathlinks.ro/Forum/viewtopic.php?p=328888 .


Which of the following sentences in each pair is correct or better?

(a) I usually do not go to school at 7.
(b) I do not usually go to school at 7.

(a) I always eat breakfast before I go to school. [edited]
(b) I have always eaten breakfast before I go to school.

(a) I usually do not go to school at 7.

(b) I do not usually go to school at 7.

these are both correct grammatically. subtle differences between them, depending on intonation, would only be heard in spoken english.

(a) I always eats breakfast before I go to school.

that should be "eat" (I eat).

(b) I have always eaten breakfast before I go to school.

normally this would be said "I have always eaten breakfast before going to school". the way you have it, the present "I go" doesn't correlate with the continual past action represented by "i have always eaten". so you change to "going", which is a noun and has no time dependency. you could also say it, "I always eat breakfast before going to school", which is more common.

anyway, these things are tough to explain, even if you know grammar, since there are always nuances of language beyond precise explanation. you just have to feel it

There is definitely a difference between (a) I usually don't go to school at 7, and (b) I don't usually go to school at 7. In (a), usually modifies don't. In (b), usually modifies go. The distinction may not be clear here, but consider "I always don't go to school at 7" and "I don't always go to school at 7". The former would never actually be heard - you would probably hear "I never go to school at 7". The latter, however, means "I sometimes go to school at 7". But since sometimes is not as common a word as never, I don't think the latter sounds weird at all. Basically, not-always=sometimes, but always-not=never. Then not-usually=occasionally, but usually-not=rarely (the former stressing "yes it does happen, but only occasionally" and the latter "it hardly ever happens"). A long explanation, but I hope it makes sense.

For your second question, there's not much difference between the two sentences. All I can think of is:

I always eat = habitual. This indicates a habit not likely to be broken.
I have always eaten = perfect habitual. This is used either to stress that it has always been a certain way, or to stress that the habit may be broken at some point.
I always ate = past. Usually, "always" doesn't take the past tense unless there is some modifier (last year, I always ate before going to school).

I hope my elaborate detail is not uncalled for. These aren't really grammatical truths about English, so don't think you always have to say it the right way. I think I'm just describing the way people interpret these sentences.

[quote="Hamster1800"]There is also a similar place where people often say what they don't mean. For example, suppose you make the statement, "Everyone knows what I'm talking about." Many people will respond negatively as, "No, everyone does not know what you're talking about," where they actually mean "No, not everyone knows what you're talking about." The difference is that the first means that nobody knows and the second means that some people don't know, but not necessarily everyone.[/quote]
[quote="Ravi B"]Even the Shakespearean line "all that glisters is not gold" would more logically (if less poetically) be written "not all that glisters is gold".[/quote]
$ \neg[\forall x (x\in P)] \Longleftrightarrow \exists x(x\notin P),$

$ \neg[\exists x (x\in P)] \Longleftrightarrow \forall x(x\notin P).$

[quote="Hamster1800"]There is also a similar place where people often say what they don't mean. For example, suppose you make the statement, "Everyone knows what I'm talking about." Many people will respond negatively as, "No, everyone does not know what you're talking about," where they actually mean "No, not everyone knows what you're talking about." The difference is that the first means that nobody knows and the second means that some people don't know, but not necessarily everyone.[/quote]
This is one of the blessings of the English language: we can distinguish between concepts such as $ \forall\,x\neg\,\mathrm{P}(x)$ and $ \neg\forall\,x\mathrm{P}(x)$ (where $ \mathrm{P}(x) \equal{}$ “$ x$ knows what you’re talking about”) in a grammatically precise way. However, there are languages in which both sentences are expressed in the same way. One example I know of is Finnish. In Finnish, “Everyone knows what I’m talking about,” is: [color=teal]Kaikki tietävät mistä puhun.[/color] But the negative sentence, [color=teal]Kaikki eivät tiedä mistä puhut,[/color] can mean either “Everyone doesn’t know what you’re talking about,” or “Not everyone knows what you’re talking about.” Context is needed to determine which the speaker means.


modular arithmetic
number theory open
number theory

hello guys.
i found this problem anywhere but i couldn't solve it...

you have any $k$ for which \[k\equiv 32768^{25}+1 \pmod{2^{376}}\] is true. you have now to show that \[k^{4^{13\cdot 37}}\equiv 1 \pmod{2^{1337}}\] is true...
i tried some numbers and always it was true...

please, please help me!

[quote="Naphthalin"]$k\equiv 32768^{25}+1 \pmod{2^{376}}$[/quote] Note that $32768^{25}=2^{375}$. Then check: is it -more generally- true for all numbers $k\equiv 1 \pmod{2^{375}}$?

are you really sure that this will work?
i tried... you're right, $32768^{25}=2^{375}$ is. but do anyone have any idea what to do next? because $1337$ is much bigger than $375$...

please, please help me!

Naphthalin

isn't there a way to compute $(1+2^{375}t)^{4^{13\cdot37}}\bmod 2^{1337}$? I haven't tried it out myself, but the classical ways (maybe including a little binomial properties) should work?

maybe that will work but i don't know anything about this...  :(

help!

Naphthalin

Do you know the binomial theorem?  The expression will look complicated, but you just need to notice some information about it.


Problem
(a) c(n,0)=c(n,n)=1 (n≥0)
(b) c(n+1,k)=2^k × c(n,k) + c(n,k-1) (n≥k≥1)

for all integer n that satisfy n≥k≥0, prove c(n,k)=c(n,n-k)

It is from IMO :) !(93->98)You can find!


What is the probability that a multiple of 864 is divisible by 1994.
Now i though the sample space is going to be infinite i am suprised how the book solved it

[hide="Solution"]We are looking for values of $ k$ such that $ 1994|864k \implies 997|432k$.

Hence, $ k$ is a multiple of 997.

Since there exists a multiple of 997 every 997 number starting from 1, the probability that a randomly selected integer will be multiple of 997 is $ \frac{1}{997}$.

Answer: $ \frac{1}{997}$.[/hide]


How many points, no three of which are collinear, determine 36 lines?

2 points determine a line, so nc2=36

n(n-1)=72

n=9,-8 so $ \boxed9$


pleaz state ur least favorite quotes here for discussion. and randomeness.

these r my least favorite quotes, from totally despise to slightly dislike, with my replys to them.

1. "The pen is mightier than the sword." let's just see what happens when i hire some samurais to assasinate whoever came up with that...

2. "practice makes perfect." yeah, and nobody's perfect. duh.

3. "with great victory, comes great sacrifice." then what happened all those times when i miraculously scored A's on my tests without studying a thing?

4. "you are what you eat." i haven't turned into a bowl of rice yet...

5. "revenge only brings more revenge." aw c'mon, NOBODY follows that... if they did, most of the major wars in the world would have never started.

sorry, but this was just REALLY random... just wanted to see what people thought... and i think 4 and 5 should flip, by the way. but i'm too lazy to do so.

Who made the quote about you being what you eat? That's ridiculous. That would also be pretty gross. :ewpu:

Quickster94 wrote: [quote]I hate ppl who say anything is possible, it's not really a quote, but there are so many things that r impossible... [/quote]

 Everything's possible with God on your side.  :wink:

PLEASE don't start a religious argument...

Or actually, please do.

According to the Bible, everything [i]is[/i] possible. The way my Bible phrases it, you don't even have to be Christian to do everything. You just need to use Jesus's name (or something like that).

Even with God, he couldn't say that 1=2 or 1=3, unless the language of numbers changed. Otherwise, it wuld never be true.

It's called redefining a word...

Anyway, I really dislike it when parents tell their children to eat their food because there are starving children elsewhere in the world.

So how does me eating the food help them?

I also dislike funerals. If I am dead, why would I want people to waste their time to mourn over me?

i got some!

"life is short" - name something in ur life longer than your life

"it's always in the last place you look" -someone plz name someone that would keep looking for something that you already found

o and this isn't a quote but when ppls say "can i ask u a question" -did i have a choice?!!?!?!?!?

i got a chainmail that said the exact same things as the person above me said.

going back to the topic, my least favorite quote is "there are 3 types of people in this world, those who can count and those who can't."

That quote is supposed to be funny, but I think it is kind of corny.

[quote="Nerd_of_the_Ages"]Dude, never take quotes literally.

That's like interpreting "A journey of a thousand miles begins with a step" as "not if you take an airplane or drive"

Never say never is kinda contradictory. They should change it to "Never say or write never except when stating this quote."

Actually, I don't have a ton of least favorite quotes, but there are a lot I like.

My favorite might be "Give a man a fish, and you feed him for a day. Teach a man to fish, and you feed him for a lifetime."[/quote]
Interesting.
I've heard a quote that was like "Give a man a fish and you feed him for a day. Give a fish a man, and you feed him for a week!"

[quote="crazychinesemaniac"]is eat your vegetables a quote...cus whoever came up with that one can die and also eat an apple a day keep the doctor away....does that really make any sense or was the person who came up with that one drunk at the time?[/quote]

True... I rarely get sick, and even if I do, I don't go the doctor/hospital. I have never been to a US hospital/doctor in the 6 years I've lived in the US.

I have 2 memories of going a hospital/doctor: Both in Taiwan, when my grandfather was certain (and was wrong) that I was sick and insisted on making me see a doctor.

I know I've been to a doctor another time before (when I was born), but that's all I remember.

Some really ridiculous qoutes:

"Don't break your neighbor's windows if your own are glass." - All the houses around mine are 1 story, and mine is a 2 story. We also have lots of redwood trees around our house, so we can throw stones at others windows from upstairs and they probably won't get past the trees. :D

"A penny earned is a penny saved." - Not that you can really do anything if you find a penny on the ground, but this quote is like saying a million dollars earned is a million dollars saved. If you got a million dollars, wouldn't you at least spend some of it? It should be "A penny earned is 0.9989773141592653589793238462... pennies earned." But, you can't spend 0.9989773141592653589793238462... pennies if you can't even spend a penny, so this quote contradicts itself.

"Nothing is certain but death and taxes." - Death is certain, but no one does nothing in the life time. If this person dies as a child, then this person doesn't have to pay taxes, because the parents would (or I should sa should).

All of these quotes are from Ben Franklin, and I did a biography report on him in 3rd grade. I'll try to look for some more quote I listed there and try to find a way to "embarrass" them. :D

"Genius must be born and never taught" -I forgot who said this. But it's definitely false.

"Not only does God play dice, he sometimes throws them where they cannot be found" - Stephen Hawking. God does not play dice. And Einstein disagrees as well.

"Sticks and stones may break my bones, but words will never hurt me" -I forgot. Self-explainable. (my friend's comeback: what if I throw a dictionary at you?)

"Sticks and stones may break my bones but words will never hurt me."

This isn't true because I heard somewhere someone tried to commit suicide because their friend sent a horrible email to them. Words can emmotionally hurt you.

"sticks and stones may break my bones but names will never hurt me."

*holds up sticks and stones* *grins evily* MUAHAHAHA :wallbash_red:

this should be called stupid quotes, not least favorite quotes.

"In ten years, YOU WILL LOOK EXACTLY LIKE YOOU DO NOW AND WILL OPEN A SCHOOL TEACHING CHILDREN TO PLAY CHILDREN'S CARD GAMES!"(dramatic music playing in the background)-cartoon show about childrens card games played by adults

I couldn't help but revive, but "God created the integers, the rest of the world is done by humanity" absolutely...DISGUSTS me.

"When life gives you lemons... freeze them and throw them at the people who are making your life difficult."
-Grumpy Cat.

[quote=anonomyous]Fall down seven times, get up 8.[/quote]
Oh yeah? WHAT ABOUT DEATH

[quote=mathblitz]hhehehe the Nerd of the ages:

teacher: follow the directions no "what ifs"

student: what if its a good question?

so funny!![/quote]

Right!  There is a book about What if.  Check out xkcd.com the writer is one of my heros.

Don't cry over spilt milk.
Yeah write say that to the babies WHO ONLY DRINK MILK!

Actually, why freeze them.  If you mean like liquid nitrogen freeze, thenthe lemons shatter.

What if no one made your life difficult.  What if you have a lot of self doubt.  What if you are feeling depressed for no reason?

[quote=dreambold22]pleaz state ur least favorite quotes here for discussion. and randomeness.

these r my least favorite quotes, from totally despise to slightly dislike, with my replys to them.

1. [b]"The pen is mightier than the sword." let's just see what happens when i hire some samurais to assasinate whoever came up with that...[/b]

2. "practice makes perfect." yeah, and nobody's perfect. duh.

3. "with great victory, comes great sacrifice." then what happened all those times when i miraculously scored A's on my tests without studying a thing?

4. "you are what you eat." i haven't turned into a bowl of rice yet...

5. "revenge only brings more revenge." aw c'mon, NOBODY follows that... if they did, most of the major wars in the world would have never started.

sorry, but this was just REALLY random... just wanted to see what people thought... and i think 4 and 5 should flip, by the way. but i'm too lazy to do so.[/quote]
And Then I use the pen and write a book about how samurais are bad before u even hire them.  Then riots break out and the locals trample over your samurai.  Have you seen the Mighty Med episode on netflix "The Pen is Mighty Med-ler than the sword"

Teach a dude how to fish, and he'll be set for life.
Give a man a fish, and he'll be set for a day.
GIve a man religion, and he'll die praying for a fish.


Is your question what is 0/0? Well if it is, then it is undefined.

Isn't the term indeterminant?

Getting a result of 0/0 can be useful.
For instance, you can have two functions, p(x) and n(x), where f(x)=p(x)/n(x). If for c, f(c)=p(c)/n(c)=0/0, then p(x) and n(x) have a common factor.

Excuse me, I should have said p(x) and n(x) need to be ploynomials. Because p(c)=0 and n(c) = 0,  (x-c) is a factor of p(x) and n(x).

[quote="LordoftheMorons"]$\frac{n}{0}, n \in \mathbb{R}$ is usually considered to be $\infty$.[/quote]

Well, $\frac{n}{n}, n \in \mathbb{R}$ is usually considered to be $1$ and $\frac{0}{n}, n \in \mathbb{R}$ is usually considered to be $0$.  :P

if you graph $y=x^0$, you see the limit as x approaches 0 is 1, but at 0 it is undefined (or at least that's what a calculator would tell you).  but if you graph $y=0^x$, the limit seems to be 0.  which is sort of strange, and that's probably why it is undefined.

edit: i thought we were talking about 0^0, when we are actually talking about 0/0

Since one of the previous posts mentioned limits (i'll try to be brief), $\frac{0}{0}$ is just an indeterminate form of a limit, and can be handled using something called L'Hopital's rule.  As for a specific value that the fraction is always without exception equal to, i've never heard of it, and it's  likely just undefined.

[quote="perfectnumber628"]if you graph $y=x^0$, you see the limit as x approaches 0 is 1, but at 0 it is undefined (or at least that's what a calculator would tell you).  but if you graph $y=0^x$, the limit seems to be 0.  which is sort of strange, and that's probably why it is undefined.[/quote]
This is an example of a double-limit changing depending on which limit is evaluated first.  You're implicitly looking at the solution to the limits

$\lim_{x \rightarrow 0} \lim_{y \rightarrow 0} x^y$ and $\lim_{y \rightarrow 0} \lim_{x \rightarrow 0} x^y$.  The first limit evaluates to 1 (since you take the inner limit first, and then the outer one), and the second one evaluates to 0.  This simply means that the function $f(x,y) = x^y$ isn't continuous at $(x,y) = (0,0)$, since the limit of the function as you approach the point changes depending on which path you take (the first path moves to the y-axis and then approaches zero along the x-axis, and the second path does the opposite).  So $\lim_{(x,y) \rightarrow (0,0)} f(x,y)$ is undefined.

Another easy example of this is the function $f(x,y) = \frac{x+y}{x-y}$.  Depending on how you approach $(x,y) = (0,0)$, the value of the limit changes.

Yes, it's said to be undefined because if it was defined, one could easily come to the conclusion that all numbers are equal...

Let's say
$\frac 0 0=$ Some number $z$
ok so...
$\frac 0 0=z$
$0z=0$ which is true for any real and\or complex number. This would imply that all numbers are equal.
That's why is just simply thought as undefined.

[quote="Jhonny Ruiz"]$0z=0$ which is true for any real and\or complex number. This would imply that all numbers are equal.
That's why is just simply thought as undefined.[/quote]
That step doesn't imply that all numbers are equal.  You've done an irreversible step: $0/0 = z \Rightarrow 0z = 0$, but not $0/0 = z \Leftrightarrow 0z = 0$.  So, you're not actually proving anything even if your manipulations with 0 were valid, since the steps are not reversible.

Divisibility by zero, strictly speaking, always produces an undefined quantity.

[quote="Jhonny Ruiz"][quote]Jhonny Ruiz wrote:
0z=0 which is true for any real and\or complex number. This would imply that all numbers are equal.
That's why is just simply thought as undefined.

That step doesn't imply that all numbers are equal. You've done an irreversible step: 0/0 = z \Rightarrow 0z = 0, but not 0/0 = z \Leftrightarrow 0z = 0. So, you're not actually proving anything even if your manipulations with 0 were valid, since the steps are not reversible.[/quote]

What do you mean those steps are irreversible...?
Generally, $(a)\frac{b}{a}=b$

So, why would something like...
$(0)\frac{0}{0}=(0)z$
$0=0z$ be irreversible?[/quote]

The function $f(x) = ax$ is only one-to-one iff $a \neq 0$.  When $a = 0$, $f(x) = 0 \forall x$, and so the function is no longer one-to-one and does not have an inverse.


here is a nice ineq:

Let a,b,c>0 be positive reals.
prove that:
(b+c)^1/2*( (a+b)^1/2 + (a+c)^1/2 ) >= (b+c)/2 +(ab)^1/2+(ac)^1/2 .

nighty-night!

i think an AM-GM and a Cauchy will do it...

[color=cyan]Don't you ever post a full solution?[/color]

Upon multiplying LHS by 1, we have 2 * sqrt((b+c)/2) * (sqrt((a+b)/2) + sqrt(a+c)/2)). By RMS/AM, sqrt((b+c)/2) >= (sqrt(b)+sqrt(c))/2; hence LHS >= 2 * (sqrt(b)+sqrt(c))/2 * (sqrt(a) + sqrt(b)/2 + sqrt(c)/2) = sqrt(ab)+sqrt(ac)+((sqrt(b)+sqrt(c))^2)/2. The last term is (b+c)/2 + sqrt(bc) > (b+c)/2 (because b,c > 0 by hypothesis). It is therefore a strict inequality.


Find all solutions to the functional equation $f(x)/f(y)=y/x$.

[hide]
$xf(x)=yf(y)$ for all $x$ and all $y$.  What can you conclude?  Is it possible that, say, $xf(x)$ is independent of $x$?  I think I've said too much.
[/hide]

[hide="hint"] let $af(a)=g(a)$ and see what you can get from that :) [/hide]


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[quote="7h3.D3m0n.117"]My reason against an infinite universe was because there must be infinite mass, which would be absurd.[/quote]
And why would it be absurd?
It's not like we're concerned about the gravitational force the universe exerts on a particle "outside" it, that would be infinite and absurd. Infinities in Physics are allowed as long as they don't ask for infinite energy somewhere that can be translated away from that place. If there were such a thing as a particle "outside" a given space of infinite mass, it would have infinite energy which we could have extracted.

The main objection to the infinitude of the universe comes from the Big Bang theory, which most people believe in, although it isn't completely established.

(I think before Einstein's relativity was available, the Newtonian cosmological model required an infinite universe for consistency)

[quote="7h3.D3m0n.117"]My reason against an infinite universe was because there must be infinite mass, which would be absurd.[/quote]

Actually, it's not as absurd as you might think. This is because though there is infinite mass, there is also an infinite amount of space where there is no mass.

An analogy for this--though I'm not sure how good it is--is the Dirichlet function in mathematics. This function is usually seen as being $ 1$ for any rational number and $ 0$ for any irrational number (those numbers can really be anything, but let's keep them $ 1$ and $ 0$ for convention's sake). If you were to plot this function, you'd see what appeared to be two lines, one at $ y\equal{}0$ and one at $ y\equal{}1$. After all, a line is simply an infinite amount of collinear points, right?

Except, in each of the "lines" there is also an infinite number of holes--all of the rational numbers for the points at $ 0$ and all of the irrational numbers for the points at $ 1$--so you could argue that nothing exists at all at $ y\equal{}0$ and $ y\equal{}1$ because "nothing" is an infinite amount of empty space. Yet the fact remains that there is something there, just not a line. Though there are infinite points, they don't take up the whole "space" of the line because of all the holes.

I guess what I'm trying to get at is that there can be an infinite amount of mass in the universe without every space in the universe being taken up by it, so the universe could look like it does to us today and still have an infinite amount of mass.

Did that make any sense?

it did make a lot of sense in fact i am making a thesis on functions corelating the infintisatioon of the universe .Well according to me the universe is at the infinity breiuyt.u see the breiuyt function co relates the dark matter along with the space time distortion in the universe .

well i might be wrong as i am just an XI grader !

Again, this topic is kind of too old to revive.  :wink:

before AstroPhys again points out the same thing, lets get the discussion going:
(I really don't understand why ppl get mad at revivals)

Here's what I think:
If for all line segments contained inside the universe, there exists another line segment greater than it contained in the universe, then universe is infinite.

What does "containing a line segment" even mean¿ The universe is not flat, it's bent by gravity; even more obstructive, the total structure may be even more complex.
Measuring volume in general is one way to define infinitude. Another one would be to check if the universe is compact (as a manifold).

I don't know this much. I'm sorry to have replied. That was "what I think"(=thought for now).

[quote="Akashnil"]
If for all line segments contained inside the universe, there exists another line segment greater than it contained in the universe, then universe is infinite.[/quote]

isnt it just a mathematical imagination?? :D and thats quite logical.
there is no need to say universe is bent due to gravity :P

but hypothetic research is a little different from reality :P  :P  :)

[quote="rituraj007"]isnt it just a mathematical imagination?? :D and thats quite logical.
there is no need to say universe is bent due to gravity :P[/quote]
But that intuition doesn't work that way: assume you draw a straight line here on earth (or any other sphere). After going around it once, you are at the same point again, but still, you can continue the line forever. So by that definition, earth would be infinite ;)
(And if necessary, by small adaptments, we may even avoid reaching any point again).

line segment is a part of a line and is not infinite
imagine earth be flat and all line segments on earth.they r clearly defined -so earth's not infinite :oops:

I don't understand your second sentence ("they are clearly defined"), but if we have an infinite line, we especially have segments of arbitrary length.

[quote="mathway"]Seriously?[/quote]
????
What?? do you mean?

@bdejean: Oh whoops, didn't realize i put an only there. Hmm... let's change that to for instance?

Do gluons and gravitons really travel at the speed of light?

If gravitons exist, then yes, gluons and gravitons travel at c.

LOL funny...
I know i read something about tis but i forgot where it was and what it said- if physicists confirm the existance of the graviton, what would it imply?

I haven't studied the Graviton in much detail, so I can't really tell you.

[quote="Kayli1999"]@bdjean: the graviton is still a theoretical particle, right?[/quote]

[quote="mathway"]Yes. Ha, i'm bdejean now.[/quote]

:)  As mathway confirmed, yes, it still is theoretical, but if gravity had a force carrier, it would be the Graviton. I think the presence of the Graviton may make it easier to incorporate a theory that includes both Gravity and the other fundamental forces, but that's just my guess.

I think it is not necessarily infinite because of time dilation.

[quote=n0vad3m0n]My friend raised a good question to whether the universe was infinite or not. Obviously, proof is not possible, but logic can be used. Opinions and support?

My reason against an infinite universe was because there must be infinite mass, which would be absurd.[/quote]

bruh that is common knowledge. There isn't infinite mass, the discovery that the universe is expanding was made by Edwin Hubble by using a red-shift blue-shift telescope and found all galaxies that weren't being pulled towards us(like Andromeda) had a blue-shift of some other gravitational pull. It is kind of like sewing some buttons on a cloth and then stretching it outwards.

[quote=n0vad3m0n]My friend raised a good question to whether the universe was infinite or not. Obviously, proof is no possible, but logic can be used. Opinions and support?

My reason against an infinite universe was because there must be infinite mass, which would be absurd.[/quote]

except, relativity uses $$1\over \sqrt{1-({v\over c})^2}$$ which tends to get larger and larger as $v$ approaches $c$ which makes mass nearly infinite near the speed of light.  due to length contraction density increases by the square of this factor.

you can modify newtons gravity with relativity, to define a surface that after passing halfway across particles of a given mass are accelerating at speed of light.

[quote=NumberX][quote=douglasubella]I don't think there is an infinity universe, because then by scientific research the universe expands, but then it can't already expand into itself.
Also, there can't be infinite mass, as mingxias said because of the reasons he said.

It's most likely that the infinite universe isn't there.[/quote]

umm you kind of misunderstand expansion. space doesnt expand into anything. the scale factor increase thats all. it makes perfect science.[/quote]

Oh, you're right, I made a mistake.


The measures of the three interior angles of a triangle are $ 50^\circ$, $ 55^\circ$ and $ x^\circ$. What is the degree measure of the largest interior angle of this triangle?

hello, it is $ 75^{\circ}\equal{}180^{\circ}\minus{}55^{\circ}\minus{}50^{\circ}$
Sonnhard.


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Hey, I got the prize today.
360 Problems .. and T-shirt !

They were really awesome! The book was hardcover too!

BTW, what is the level of difficulty of 360 Problems M C??

It ranges from medium AIME - hard USAMO, I think (probably with a few outliers). There are some old USAMO problems (a couple of 3/6's) but the beginning problems of each chapter are decently easy, too. Really good book in my opinion, though.

No wonder it looked hard...
Last time I read a post someone posted in this forum describing the USAMTS prizes, and it says the level of difficulty of this book is 9/10. I thought it meant that the book is for 9th and 10th graders... :D  Now I think it is based on the scale 1(the easiest) to 10(the hardest).

Anyway, I think it's really good book too!

[quote="d343seven"]I didn't get mine yet; did anyone else get/not get their books?[/quote]

I didn't receive it either. I was wondering to post thread with title "SHOULD I GET SCARED!?"  :P

I just got mine today. I live in Utah. I think if you live in the east you should get it later this week.

Hmmm...360 Problems book was REALLY hard. I thought I would use it to practice for AIME. Didn't work out too well. I thought the problems had a different feel than most contest problems.

[quote="d343seven"]I didn't get mine yet; did anyone else get/not get their books?[/quote]
me neither.  is it time to panic?  :D

I didn't get mine yet either.

Don't worry yet, guys.  They should come tomorrow or Friday.

JB

Don't get scared - many packages were mailed out at the end of last week, so it's too early to worry.

[quote="tongchen1226"]how were the book and t-shirt shipped? UPS like the mathematica?[/quote]

US mail.

I got mine today. YAY

Just wondering, mine didn't have certificate. I got shirt and 360 problem book but not the certificate.

 :?  :?

[quote="rrusczyk"]No certificates this year; we sent medals (as keychains) instead.[/quote]

Did everyone suppose to get it?

Should I be worried now? My stuff hasn't arrived. Were they shipped out in alphabetical order or something? :D

I just got my package today.

For any USAMTS graders and organizers, I just wanted to say thank you, again, for everything this year.  This competition has been absolutely amazing.  The problems were amazing, the grading was fair and efficient, and, of course, the prizes were great to receive.  I can only begin to understand all the work, time, and effort you all put in to making this competition run smoothly.  I really appreciate everything you all have done, and am really looking forward to next year.

JB

I didn't get the email from amazon yet, so I'm a little worried because i leave for india tomorrow, then as soon as i get back, I leave for PROMYS, so i'm worried i won't have it by then...

But yeah I got that book.  Sounds quite good.

[quote="mna851"]I didn't participate in the USAMTS this year.  I'm curious, is it possible to get a copy of 360 Problems through Amazon or somewhere other than as a prize for USAMTS?[/quote]Probably. Try searching on Amazon.

Arrived today, Eastern MA. So expect it to arrive soon if you haven't gotten it yet.

[quote="mna851"]I didn't participate in the USAMTS this year.  I'm curious, is it possible to get a copy of 360 Problems through Amazon or somewhere other than as a prize for USAMTS?[/quote]
Go to this web: http://www.gil.ro/en/index.html
This is the GIL publishing site... It says on the left column that for orders e-mail: sales@gil.ro

I got mine yesterday: t-shirt, silver medal, 360 problems, and the geometry book.

Got my t-shirt, key chain, and [i]The Elegant Universe[/i] yesterday :)  Thanks USAMTS!!!

I don't think admins answered my question.

Are all contestants receiving key chain? I'm asking this because I'm Honorable Mention..  :?

I think the key chain thing is the medals.  in other words, gold people get gold key chain, silver people get silver key chains, etc.

Has anyone else looked at the Diophantine EQ's book yet??  Cuz it's really good!

JB

Hm, so it seems like just about everyone's gotten their stuff already... I'm in NYC, and I haven't gotten it yet, should I be worried now?

[quote="joml88"]They would have sent it with all of your other stuff.[/quote]

They didn't so that's why I'm worrying. I would be OK if I didn't receive anything.. But I got all things except key chain (unless HM isn't suppose to get key chain?)

Indeed, the keychains only went to medalists.

Oh ok. Thanks.

Darn, I could've got keychain!  :ninja:


Let's say I have a buffer made of 0.1 M Hypochlorous Acid, HClO and 0.1 M Hypochlorite Ion, ClO–.  I react the acid with 0.05 M NaOH to form a buffer.  How do I calculate the buffer capacity?  In other words, how do I find how much H+ I could add without changing the pH by more than one unit, and how much OH- I could add with the changing the pH by more than one unit?

Thanks in advance!

Since there is always equilibrium between hypochlorous acid and its conjugate base we have for the pH of the solution

$ pH \equal{} pK_a \plus{} \log \frac{[ClO^\minus{}]}{[HClO]}$.

Now, you want pH - pKa = 1, and so

$ \log \frac{[ClO^\minus{}]}{[HClO]} \equal{} 1 <\equal{}> \frac{[ClO^\minus{}]}{[HClO]} \equal{} 10$.

For example, adding acid will decrease [ClO-] and increase [HClO]: now you just have to relate these concentrations with the quantity of acid or base added.