function
modular arithmetic
number theory
totient function

[quote="Silverfalcon"]:D

What is the last digit of $2003^{2003}$

Hint: Pattern

Advance Hint: Mod[/quote]

[hide="answer"]
7
[/hide]

what do you mean in your hint, why do you need mod to do this?

[hide]because we are looking only for the last digit, we only need to focus on powers of 3, and their last digit. We can see that there is a pattern. 3,9,7,1. Therefore, we can conclude that the last digit of 2003^2003 is 7.[/hide]

plokoon51, thanks for your efforts.

[hide="extra for Lucy.."]let's see...
we could see the pattern like this..
2003^1 has the last digit of 3
2003^2 has the last digit of 9
2003^3 has the last digit of 7
2003^4 has the last digit of 1
2003^5 has the last digit of 3

now it's the pattern..every 5th powers has a number that has the last digit of 3
so we divide 2003 by 4 we get 500 and a remainder of 3..
the third part pattern is 7... so the answer is 7

and another tip,
you don't have to power 2003.. you could have just done this
3^1
3^2
3^3
etc.. because we are just focusing at the last digit and the last digit has nowhere to receive the upcoming ten's..
[/hide]

well, I know how to solve this problem, but i don't know what mod is.... so I don't reallly get how they relate...

This basically translates to the fact that you can mod out before or after you multiply.

[hide]What Hexahedron said is basically we can mod out before or after we multiply. We are looking for $2003^{2003} \equiv x \pmod{10} \implies (2 \pmod{10} \cdot 10 \pmod{10} \cdot 10 \pmod{10} \cdot 10 \pmod{10} + 3 \pmod {10})^{2003} \implies (2\cdot0\cdot0\cdot0+3)^{2003}=3^{2003}$.

Now, we are reduced to finding the last digit of $3^{2003}$. This post was a bit long and shows something kind of pointless, but I guess one aspect of mods is used here. I can't really think too well because I have to go and catch the bus for a math in a few minutes but I think the above is right.[/hide]

9, using the same method.

The pattern for 7 would be 7, 9, 3, and 1, and then 7 again.  This means that the last

digit 7 in $2003^{2003}$, when exposed to an exponent of 2002, would become a

3.  Once that is done, we use the pattern for 3 again, and since the pattern is 3, 9,

7, and 1, we conclude that the answer must be $\boxed{\boxed{\boxed{9}}}$

Well, in conclusion, is 9 or 7 the answer???

7 for the first question,
9 for the second

[quote="NightFlarer"]For an extra challenge, find the units digit of ((2003^2003)^2002)^2001[/quote]

I will find for you the last three.  ;)  (Note that the quantity is equal to $2003^{2003 \times 2002 \times 2001}$.

[hide]
We take $\bmod 1000$, so we take the exponent $\bmod \varphi(1000) = 400$ and find $\equiv 3^{6} \bmod 1000 = \boxed{729}$.
[/hide]

How about:  Find the units digit of $2003^{ ( 2002^{2001}) }$?  Or the last two, or three... or four ;)

[quote="Hokkage"][quote="Hexahedron"]You could also just forget about the 2000 part since it has no effect on the units digit.  So, this problem is equivalent to "what is the last digit of 3^2003"?[/quote]This uses Fermats Little Theorem, $2003^{2003} \equiv 3^{2003} \pmod{10}$.[/quote]

That's not Fermat's Little Theorem... that's just being able to take mods before multiplication (and by extension, exponentiation).

Fermat's Little Theorem isn't applicable to this situation because $10$ is composite.  You would use Euler's Totient Theorem, which is what I did.

[hide]$2003 \equiv 3(mod 4)$
${3^3} \equiv 7(mod 10) \implies \boxed{7}$[/hide]

Whoops  :roll:
I should have read your solution first

[quote="Silverfalcon"]:D

What is the last digit of $2003^{2003}$

Hint: Pattern

Advance Hint: Mod[/quote]
[hide]
$2003^{2003}$
We can ignore the "200" part of the base, since that doesn't contribute to the units digit. We have:
$3^{2003}$
$2003=3\pmod4$
$3^2003$ is equivilant to $3^4$ which has a last digit of 4.
[/hide]

[quote="t0rajir0u"]How about:  Find the units digit of $2003^{ ( 2002^{2001}) }$?  Or the last two, or three... or four ;)[/quote]

Is the last digit by any  [hide="chance"]$9$?[/hide]

Try these easy sums....

Find the last 3 digits of $9^{105}$

Find the last 2 digits of $7^{(7^7)}$

[quote="shadysaysurspammed"][quote="t0rajir0u"]How about:  Find the units digit of $2003^{ ( 2002^{2001}) }$?  Or the last two, or three... or four ;)[/quote]

Is the last digit by any  [hide="chance"]$9$?[/hide]

 [/quote]
FWIW: If you want to check your answer,  $2003^{2002^{2001}}$  ends in[hide] $\dots 234241$[/hide]
and $2003^{2003 \cdot 2002 \cdot 2001}$ ends in[hide] $\dots 3276729$ [/hide]

[quote="drunner2007"]Try finding the last three digits of $2003^{2003^{2002^{2001}}}$ if $a^{b^{c}}=a^{(b^{c})}$[/quote]
Last three digits of  are [hide="these"]403 [/hide], if you like to check.

[quote]Well, I figured out silverfalcon's problem easy enough, but what is Euler's Totient Theorem?
[/quote]
Euler's Totient function  for a positive integer $n$ is defined as number of positive integers less than $n$ which  have no common factors with $n$

and is written as $\phi (n)$
For example $\phi (10) = 4$ Because the only numbers whch have no common factors with 10 are $1,3,7,9$
and $\phi (7) = 6$    (Now the set of numbers are $1,2,3,4,5,6$
and $\phi (6) = 2$    ( The numbers now are $1,5$ only

Hope you get the idea.

The theorem says, for any $a$ which  has no common factors with $n$ ,   $a^{\phi (n) } \equiv 1 \bmod n$

Try it  for the case of n=10, here $\phi$ is 4, and you can see  , if you select any number (say a)  which has no common factor with 10, (that is - its not even and not divisible by 5)
you ould see $a^4 \equiv 1 \bmod 10$ (Or IOW the unit  digit would be 1)

In case when $n$ is prime,  (where you can see that phi would be nothing but (n-1), the theorem is also known as "Fermat's Little Theorem"

Hope that helps.

(You can also check out some standard math book or many internet sites such as [url=http://www.cut-the-knot.org/blue/Euler.shtml]Here[/url]

[quote="DanK"]Ok.  I already knew about Fermat's Little Theorem, so that makes the Totient Theorem easy enough...

One more thing, though.  What is $\varphi$ used for?[/quote]

I think $\varphi$ is $\phi$ except fancier?

[quote="weatherbottle"]It's easy to determine the last digit of large numbers using Euler's Totient Formula, but how do you do it for more digits?  Like for example, the last 3 digits of $3^{187}$.  :?[/quote]

In your question,we can write it as $(10-1)^{93}(9)$ and then apply binomial theorem.

I think you have to be really good at modular arithmetic and know how to manipulate stuff for finding more digits,its all practice,practice and *yawn* practice i guess....

[quote="DanK"]So people just use $\varphi$ because they think it looks cooler?[/quote]

Some people use it specifically to differentiate between the totient function and the constant $\phi = \frac{1 + \sqrt{5}}{2}$.  That way, it doesn't look like you're multiplying by $\phi$.  Usually, it's clear enough from context that you don't really have to make a distinction.

I've also seen people use $\varphi = \frac{1 - \sqrt{5}}{2}$.


induction
number theory proposed
number theory

For every positive integer $n$ we denote by $d(n)$ the sum of its digits in the decimal representation. Prove that for each positive integer $k$ there exists a positive integer $m$ such that the equation $x+d(x)=m$ has exactly $k$ solutions in the set of positive integers.

As I remmember ,   I had seen this problem before 2002,Maybe I am wrong.
I have a simple solution:
use induction:
if $x+d(x)=m_k$have exactly k solution,
then there exist $0\leq t\leq 8$so that $2+m_k+2t$is a multipy of 9
let $l=\frac{2+m_k+2t}{9}$
let $m_{k+1}=m_k+10^l+1$
then $x+d(x)=m_{k+1}$have exactly {k+1} solution.

Sorry,I didn't see how you apply induction and why there exist exactly $ k$ solutions but what about more than $ k$?
The solution below was rather long,because we should concentrate on [b]exactly[/b] $ k$ solutions...
By the way,soory for this bad-writing solution,because both the idea and construction are extraordinary complicated :blush:

This is the official solution...But I still missed the part for [b]exactly[/b] $ k$ solutions...
Is this part trivial to solve for I just missed some simple facts???

[quote="123eeee"]Hawk is right.[/quote]

Probably,but could anyone help me to explain?Thanks in advance :)

Thank you very much!I finally found the thing I missed,that is,if $ n<10^k,s(n)\equal{}9k\minus{}s(10^k\minus{}1\minus{}n)$ :lol:

Thank you so much for your explaination,I have rewritten it into English :)

First, let's make the observation that we are essentially tasked with finding $m$ so that the number of ways to represent it as a sum of at most $9$ $11$'s, at most $9$ $101$'s, at most $9$ $1001$'s, etc.

We will actually show the stronger statement that there exists such an $m$ which is $7$ (mod $9$). We'll show this by induction on $k$. When $k = 1$, $m = 88$ suffices.

Suppose it holds when $k = t.$ When $k = t+1$, let $9\ell - 2$ be the $m$ that works when $k = t.$ We claim that $8 \cdot (10^{\ell} + 1)$ works when $k = t+1.$ It's clear that we cannot use things $\ge 10^{\ell + 1}.$ If we use $8 \cdot (10^{\ell}+1)$, we cannot use anything else. Now, observe that we must otherwise use $7$ $(10^{\ell} + 1)$'s. In this case, observe that $10^{\ell} + 1 - 9(10 ^{\ell - 1} +1 + 10^{\ell - 2} + 1 + \cdot + 10 + 1 + 1 + 1) = 2 - 9 \ell.$ Hence, we must use $9$ of each of $10^{\ell - 1} + 1, 10^{\ell - 2} +1, \cdots, 10 + 1, 1+1$ except for some which sum to a total of $9 \ell -2.$ However, from our inductive hypothesis, there are exactly $t$ ways to exclude numbers with sum $9 \ell - 2.$ Hence, combining with the $1$ way from earlier ($8$ copies of $10^{\ell} + 1$), we have completed the induction.

$\square$


有好多关于北大的问题想咨询北大的学长。
MSN: golden_apple_is_mine@hotmail.com

还有被保送到北大的学友，或者决定上北大的学友。

我的高三生活基本上会是homeschooling,走很不一样的道路，因为各种原因，但绝不是回家修炼考试技能，呵呵。萍水相逢，尽是他乡之客……

看来你还是很心仪北大的呀？

[quote="Enya"]看来你还是很心仪北大的呀？[/quote]

 :rotfl:

毕竟对于很多出不去，不想出去的人来说，北大是最佳选择


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masalan [code]\$25\$[/code]
misheh:  $25$  (albateh niyazi be "\" nist)

thank you
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integration
logarithms
trigonometry
calculus
calculus computations

[hide="#2 Hint"]
$ \frac{\ln (x\plus{}1) \minus{} \ln x}{x(x\plus{}1)}$
$ \equal{} \frac{\ln (x\plus{}1) \minus{} \ln x}{x} \minus{} \frac{\ln (x\plus{}1) \minus{} \ln x}{x\plus{}1}$
$ \equal{}  \minus{} \frac{\ln x}{x} \plus{} \frac{\ln x}{x\plus{}1} \plus{} \frac{\ln (x\plus{}1)}{x} \minus{} \frac{\ln (x\plus{}1)}{x\plus{}1}$
[/hide]

#2) $ t \equal{} \displaystyle1 \plus{} {1\over{x}}$ , $ \minus{} dt \equal{} \displaystyle{1\over{x^2}}\,dx$, so
$ \displaystyle\int\frac {\ln(x \plus{} 1) \minus{} \ln(x)}{x(x \plus{} 1)}\,dx \equal{}
\int{\frac {\ln(1 \plus{} {1\over{x}})}{1 \plus{} {1\over{x}}}}{1\over{x^2}}\,dx \equal{}$
$ \minus{} \int{\frac {\ln(t)}{t}}\,dt \equal{}\minus{}1/2\, \left({\ln(t)}\right)^2 \plus{} c\equal{}\minus{} 1/2\, \left({\ln\left({1 \plus{} {1\over{x}}}\right)}\right)^2 \plus{} c$.

for number one..
use the substitution:: $ \tan\theta\equal{}\frac{2x\plus{}1}{\sqrt3}$
thx

[hide="Number 1"]Let's find the constants A, B, C, D such that

$ P \equal{} \int\frac {2x \minus{} 3}{(x^2 \plus{} x \plus{} 1)^2}\,dx \equal{} \frac {Ax \plus{} B}{x^2 \plus{} x \plus{} 1} \plus{} \int\frac {Cx \plus{} D}{x^2 \plus{} x \plus{} 1}\,dx$

Provided I have made no mistakes, one finds $ A \equal{} D \equal{} \minus{} \frac83$, $ B \equal{} \minus{} \frac73$, $ C \equal{} 0$. Therefore,

$ P \equal{} \minus{} \frac {8x \plus{} 7}{3(x^2 \plus{} x \plus{} 1)} \minus{} \frac83 \int \frac {1}{x^2 \plus{} x \plus{} 1}\,dx$

and

$ \int \frac {1}{x^2 \plus{} x \plus{} 1}\,dx \equal{} \int\frac {1}{\frac34 \plus{} \left(x \plus{} \frac12\right)^2}\,dx \equal{} \frac {2}{\sqrt {3}}\int\frac {\frac {2}{\sqrt {3}}}{1 \plus{} \left(\frac {2x \plus{} 1}{\sqrt {3}}\right)^2}\,dx$

$ \equal{} \frac {2}{\sqrt {3}} \arctan \left(\frac {2x \plus{} 1}{\sqrt {3}}\right) \plus{} C$.[/hide]

[hide="Number 2"]For x > 0, $ \frac{d}{dx}[\ln x \minus{} \ln(x\plus{}1)] \equal{} \frac{1}{x(x\plus{}1)}$, and $ \int[f(x)]^k \cdot f'(x)\,dx \equal{} \frac {[f(x)]^{k \plus{} 1}}{k \plus{} 1} \plus{} C$.[/hide]

For 2)
We have:
$ \int \frac {\ln(x \plus{} 1) \minus{} \ln x}{x(x \plus{} 1)}\ dx$
$ \equal{} \int [\ln x \ln(x \plus{} 1)]' \minus{} [\ln(x \plus{} 1)'\ln(x \plus{} 1) \plus{} \ln x'\ln x]\ dx$
$ \equal{} \ln(x \plus{} 1) \ln x \minus{} \frac {[\ln(x \plus{} 1)]^2}{2} \minus{} \frac {[\ln(x)]^2}{2} \plus{} C$.

[quote="Kouichi Nakagawa"][hide="#2 Hint"]
$ \frac {\ln (x \plus{} 1) \minus{} \ln x}{x(x \plus{} 1)}$
$ \equal{} \frac {\ln (x \plus{} 1) \minus{} \ln x}{x} \minus{} \frac {\ln (x \plus{} 1) \minus{} \ln x}{x \plus{} 1}$
$ \equal{} \minus{} \frac {\ln x}{x} \plus{} \frac {\ln x}{x \plus{} 1} \plus{} \frac {\ln (x \plus{} 1)}{x} \minus{} \frac {\ln (x \plus{} 1)}{x \plus{} 1}$
[/hide][/quote]
[hide="#2 Second Hint"]
$ \left(\frac{1}{2} \left\{\ln (x\plus{}a)\right\}^2\right) ' \equal{} \frac{\ln (x\plus{}a)}{x\plus{}a}$
$ \left(\ln (x\plus{}b) \ln (x\plus{}c)\right) ' \equal{} \frac{\ln (x\plus{}b)}{x\plus{}c} \plus{} \frac{\ln (x\plus{}c)}{x\plus{}b}$
[/hide]

#2)

$ \int \frac {\ln(x \plus{} 1) \minus{} \ln x}{x(x \plus{} 1)}\ dx\equal{}\minus{}\frac{\ln \left(1\plus{}\frac{1}{x}\right)}{1\plus{}\frac{1}{x}}\ \left(1\plus{}\frac{1}{x}\right)'\ dx$

$ \equal{}\minus{}\frac{1}{2}\left\{\ln \left(1\plus{}\frac{1}{x}\right)\right\}^2\plus{}C$.

For 2,

$ \int \frac {\ln(x \plus{} 1) \minus{} \ln x}{x(x \plus{} 1)}\ dx \equal{} \minus{}\int  \frac {\ln \left(1 \plus{} \frac {1}{x}\right)}{1 \plus{} \frac {1}{x}}\ \left(1 \plus{} \frac {1}{x}\right)'\ dx$

$ \equal{} \minus{} \frac {1}{2}\left\{\ln \left(1 \plus{} \frac {1}{x}\right)\right\}^2 \plus{} C$.


There are an infinite number of such polynomials, right?  Or is that a zero of the function?

I believe that's all we can do with the given info


If $ P(x) \equal{} x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$

and $ P(1) \equal{} 10$ , $ P(2) \equal{} 20$ and $ P(3) \equal{} 30$ find $ P(10) \plus{} P( \minus{} 6)$ = ?

I found the answer of this question in one long way, but I couldn't think of any shorter one.

[hide="Answer"]$ 8104$[/hide]

Oh, let me know if this is the right place to post this problem. ($ P(x)$ is a polynomial... :D)

It's just plug-and-chug...but this might be what you did.

[hide="Random Bash"]
$ P(1) \equal{} a \plus{} b \plus{} c \plus{} d \equal{} 9$ (1)

$ P(2) \equal{} 16 \plus{} 8a \plus{} 4b \plus{} 2c \plus{} d \equal{} 20\implies8a \plus{} 4b \plus{} 2c \plus{} d \equal{} 4$ (2)

$ P(3) \equal{} 81 \plus{} 27a \plus{} 9b \plus{} 3c \plus{} d \equal{} 30\implies27a \plus{} 9b \plus{} 3c \plus{} d \equal{} \minus{} 51$ (3)

$ P(10) \plus{} P( \minus{} 6) \equal{} 11296 \plus{} 784a \plus{} 136b \plus{} 4c \plus{} 2d \equal{} 11296 \plus{} 2(392a \plus{} 68b \plus{} 2c \plus{} d)$,

so we need to find $ 392a \plus{} 68b \plus{} 2c \plus{} d$, which we can obtain by adding and subtracting equations (1),(2), and (3) multiple times.

Let the number of equation (1)'s, (2)'s, and (3)'s be $ x,y,z$, respectively, so that we have

$ x \plus{} 8y \plus{} 27z \equal{} 392$ (n)
$ x \plus{} 4y \plus{} 9z \equal{} 68$ (a)
$ x \plus{} 2y \plus{} 3z \equal{} 2$ (b)
$ x \plus{} y \plus{} z \equal{} 1$ (c)

We solve with the last three equations, as it is easier.

$ y \plus{} 2z \equal{} 1$ (b)-(c)
$ 2y \plus{} 6z \equal{} 66\implies y \plus{} 3z \equal{} 33$ (a)-(b)

This gives us $ z \equal{} 32\implies y \equal{} \minus{} 63\implies x \equal{} 32$, which does indeed satisfy all four equations (a),(b),(c),(n).

Finally, we have that $ 392a \plus{} 68b \plus{} 2c \plus{} d \equal{} 32(a \plus{} b \plus{} c \plus{} d) \minus{} 63(8a \plus{} 4b \plus{} 2c \plus{} d) \plus{} 32(27a \plus{} 9b \plus{} 3c \plus{} d)$

$ \equal{} 32(9) \minus{} 63(4) \plus{} 32( \minus{} 51) \equal{} \minus{} 1596$, so

$ P(10) \plus{} P( \minus{} 6) \equal{} 11296 \plus{} 2(392a \plus{} 68b \plus{} 2c \plus{} d) \equal{} 11296 \plus{} 2( \minus{} 1596) \equal{} \boxed{8104}$.
[/hide]

Edit: Good catch. :)

The result is 8104, I'm sure. And as far as I know, a, b, c and d got infinite solutions and every single of the solutions can give you 8104 when tried. You probably did some mistake.

As for your solution, no it's not what I did... Mine is rather stupid but I found 3 equations about $ a, b, c, d$, which were;

$ c \minus{} 11a \equal{} 70$
$ 6a \plus{} b \equal{} \minus{} 25$
$ 6a \plus{} d \equal{} \minus{} 36$

Then, I don't know if this has any sense but I did this,

$ 4a \plus{} 2a \plus{} b \plus{} 25 \equal{} 0 \implies 4a \plus{} 2a \plus{} b \plus{} 38 \equal{} 13$ (I think I wrote it in 4n + 1 form, should be true but this is the part that I am suspicious)

Then I picked the closest number $ n$, in this case $ a$, which is 3. Then solved it for $ 2a \plus{} b \plus{} 38 \equal{} 1$ by choosing $ a$ as 3, gave me $ b \equal{} \minus{} 43$ Then found all of them etc. and put $ 10$ and $ \minus{} 6$ which gave me $ P(10) \equal{} 9676, P( \minus{} 6) \equal{} 1572$

I need correction though.

Oh and, I didn't get how you wrote the,

$ x \plus{} 8y \plus{} 27z \equal{} 392$ (n)
$ x \plus{} 4y \plus{} 9z \equal{} 68$ (a)
$ x \plus{} 2y \plus{} 3z \equal{} 2$ (b)
$ x \plus{} y \plus{} z \equal{} 1$ (c)

I saw the relationship between the coefficients $ a, b, c, d$ but... didn't get the logic you used.

My technique was to utilize the fact that through a linear combination (i.e. adding and subtracting multiple times) of $ a\plus{}b\plus{}c\plus{}d\equal{}10$, $ 8a\plus{}4b\plus{}2c\plus{}d\equal{}4$, and $ 27a\plus{}9b\plus{}3c\plus{}d\equal{}\minus{}51$, we could eventually find the value of $ 392a\plus{}68b\plus{}2c\plus{}d$.

In other words, $ x$ represented the number of $ a\plus{}b\plus{}c\plus{}d\equal{}10$ equations to add, $ y$ represented the number of $ 8a\plus{}4b\plus{}2c\plus{}d\equal{}4$ equations to add, and $ z$ represented the number of $ 27a\plus{}9b\plus{}3c\plus{}d\equal{}\minus{}51$ equations to add.

Then, I set up equations--I'll just explain the logic behind $ 1x\plus{}8y\plus{}27z\equal{}392$, as the rest are similar.

So for every $ a\plus{}b\plus{}c\plus{}d\equal{}1$, there is $ 1$ a added.

For every $ 8a\plus{}4b\plus{}2c\plus{}d\equal{}4$, there are $ 8$ a's added.

For every $ 27a\plus{}9b\plus{}3c\plus{}d\equal{}\minus{}51$, there are $ 27$ a's added.

Together, these prompt the equation $ 1x\plus{}8y\plus{}27z\equal{}392$, the coefficient of a in the desired expression $ 392a\plus{}68b\plus{}2c\plus{}d$.

Do you understand why this works?

Ahh now I see. I thought $ x, y, z$ represented the equations... Ok now I get it... For every $ x$ number of equation (1), there is an $ a$ added... and so on. Thanks a lot :)

Oh look what you did;

$ P(1) \equal{} a \plus{} b \plus{} c \plus{} d \equal{} 10$

the same mistake I did as I was solving it... You FORGOT the $ x^4$ at the equation, so $ P(1) \equal{} a \plus{} b \plus{} c \plus{} d \equal{} 9$

From $ P(1)\equal{}1*10,\ P(2)\equal{}2*10,\ P(3)\equal{}3*10$, we can set $ \frac{P(x)}{x}\equal{}(x\minus{}1)(x\minus{}2)(x\minus{}3)\plus{}10$, or $ P(x)\equal{}x(x\minus{}1)(x\minus{}2)(x\minus{}3)\plus{}10x$ because $ P(x)$ is monic with degree 4.

Well, that's how I got the 3 equations I listed above, divided with the multiplication of $ (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)$... I did divide it long and long to find the remainder, which is $ 10x$, which I used the equation of polynomials to write those equations.

But the way you say also allows us to write another equation, a fourth which will allow us to find the values of $ a, b, c, d$

But I would still love to hear a correction for my solution... I have a feeling that it is weird.

[quote="kunny"]From $ P(1) \equal{} 1*10,\ P(2) \equal{} 2*10,\ P(3) \equal{} 3*10$, we can
set $ \frac {P(x)}{x} \equal{} (x \minus{} 1)(x \minus{} 2)(x \minus{} 3) \plus{} 10$, or $ P(x) \equal{} x(x \minus{} 1)(x \minus{} 2)(x \minus{} 3) \plus{} 10x$ because $ P(x)$ is monic with degree 4.[/quote]

Actually, you can only claim that $ P(x)$ is of the form $ (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)(x \minus{} r) \plus{} 10x$.

Surprisingly, I also was guessing why he did put only "x" there... I decided to try it for like $ (x\plus{}1)$ or $ (x\minus{}1)$ etc. and the result was the same, 8104... That means these variables don't have something to do with the value of $ r$, but the $ P(10) \plus{} P(\minus{}6)$ Never checked that indepth.


hello, solving
$ 2x^2 \minus{} 3xy \minus{} 2y^2 \plus{} 2x \plus{} 11y \minus{} 12 \equal{} 0$  for $ y$ we get
$ y_1 \equal{} \minus{} 2x \plus{} 4$
$ y_2 \equal{} 1/2x \plus{} 3/2$
so we have
$ (y \plus{} 2x \minus{} 4)(y \minus{} 1/2x \minus{} 3/2)$
Sonnhard.

hello, solving
$ 6x^2\minus{}5xy\minus{}6y^2\minus{}5x\plus{}14y\minus{}4\equal{}0$
for $ y$ we get
$ y_1\equal{}\minus{}3/2x\plus{}2$
$ y_2\equal{}2/3x\plus{}1/3$
so we have
$ (y\plus{}3/2x\minus{}2)(y\minus{}2/3x\minus{}1/3)$
Sonnhard.

[quote="AndrewTom"]Factorise

(i) $ 2x^2 \minus{} 3xy \minus{} 2y^2 \plus{} 2x \plus{} 11y \minus{} 12$.

(ii) $ 6x^2 \minus{} 5xy \minus{} 6y^2 \minus{} 5x \plus{} 14y \minus{} 4$.

[hide=" Click to reveal hidden content"]

(i) $ (2x \plus{} y \minus{} 4)(x \minus{} 2y \plus{} 3)$

(ii) $ (3x \plus{} 2y \minus{} 4)(2x \minus{} 3y \plus{} 1)$

[/hide][/quote]

Dr Sonnhard Graubner,
you don't have the right answer for either problem.  A quick check
of the product of the constants you came up with in each problem
is different from the constant terms in each polynomial to be factored.
Please look at the solutions in the "reveal hidden content" given by
Andrew Tom.

From (i) you are missing a total factor of $ \minus{}2$, and from (ii) you are missing
a total factor of $ \minus{}6$.


calculus
integration
trigonometry
calculus computations

I've been trying to figure out how this integral is true...I got down to.... the integral of (sin^2(x)/cos^2(x)) or tan^2(x) which is just equal to tan(x)-x+C and obviously this uses trig identies....like sin^2(x)=1/2(1-cos(2x)) but I can't see where to use it, a little help would be great, thank in advance.

[img]http://i5.tinypic.com/121wt4h.jpg[/img]

the problem i have is similar to that but 4 is actually a^2 in my problem....but i got the a^2 to cancel out somewhere in the middle... look below for the original integral i need to work on....

[size=150][b]Int. ((x^2)/(((a^2)-(x^2))^(3/2)) dx[/b][/size]

thanks for anyone who can shed some light on this problem!

Post calculus questions in the College Playground section.

You have $\int\frac{x^{2}}{\left(a^{2}-x^{2}\right)^{\frac{3}{2}}}\,dx$.  One way is to make the substitution $x=a\sin\theta$ so that $dx=a\cos\theta\,d\theta$.  This reduces the integral to

\[ \int\frac{x^{2}}{\left(a^{2}-x^{2}\right)^{\frac{3}{2}}}\,dx=\int\tan^{2}\theta\,d\theta =\tan\theta -\theta =\frac{x}{\sqrt{a^{2}-x^{2}}}-\arcsin\frac{x}{a} \]

thank you for your help and sorry for posting in the wrong section...apparently i did it right and i just didnt see that last step. thanks again for clarifying that for me!! :)


[quote="nat mc"]B+M=27
B-M=7
2B=34
B=17, M=10[/quote]

next time nat mc, hide your answers using the "hide" option. when you reply or submit a message, there is a button that says "hide" next to the url. You most likely already knew this and I'm just saying this for nothing.  :|

btw can someone explain to me why the [hide]b-a=7[/hide] method is faster than [hide]b+(b+7)=27? (2b+7=27, 2b=20, b =10, so 10 and 10+7 (17))[/hide] or why you wouldnt do the second method? is it just preference?

i think it is just preference.  Whatever you feel more comfortable with.

I think the second method is bad because you can just think it.  If their total age was m and there difference was n, the younger one was m/2-n/2

and I just added 27 and 7 then div by two to get the older one.  Then figure for the younger.

[color=violet]   Mandy's age = m
                                            Billy's age = m+7

                                              m+ (m+7) = 27
                                                 2m+7= 27
                                                    2m=20
                                                     m=10

                                                  10+7=17

                                                 17+10=27

                                      Mandy is 10 and Billy is 17.[/color]

Yes, PLEASE keep it simplewith the black

[quote="Applesp Popo"]If Billy's and Mandy's ages put together make 27 years, and Billy is 7 years older than Mandy, how old are each of them?  :idea:[/quote]

The total is 27, and one age is $x$ and the other is $x+7$, so just subtract 7 and divide by 2 to get mandy is 10 and billy is 17.

[quote="Barnacle"]Assume their ages are 10 and 17.  Then the latter is 7 years greater than the former, and the sum of both ages is 27.  Therefore our assumption was correct and the problem is solved.

 :rotfl:[/quote]
:rotfl: good one

There's an easy way of doing this instead of using algebra linear equations..

Xander and Zoltaf's ages combined equals n. Zoltaf is m years older than Xander. How old is Xander and how old is Zoltaf?

[i]Zoltaf's age is (n/2)+(m/2)

Xander's age is (n/2)-(m/2)[/i]

Simple as that.

Drunken_Math, that fact is a result of doing the systems of equations.  If you know it off hand though, that's good.


[quote="AntonioMainenti"]I think a section to discuss math books of all kinds, and also a community/off-topic (most forums have this) section would be nice.[/quote]There are already such forums (for the record there are more than 300 subforums on this site :) ). Try the [url=http://www.mathlinks.ro/Forum/index.php?f=219]Other Topics[/url] forums, or the [url=http://www.mathlinks.ro/Forum/index.php?f=131]Other Problem Solving Topics[/url] forums.


geometry
combinatorics unsolved
combinatorics

Show that we cannot form more than $4096$ binary sequences of length $24$ so that any two differ in at least $8$ positions.

I haven't sat down with this problem properly yet, but this is the idea I got when I looked at it.

Consider one such set S and assume that it has n elements. Because any two sequences differ in at least 8 positions, by changing one bit (for example beginning with inverting the last bit) in every sequence, we generate n new sequences that are different. That's because if (b1,b2,...,b24), a sequence belonging to S, is inverted to become (b1,b2,...,b24') (the last bit inverted) then that sequence cannot belong to S because then the two sequences would differ in only one position. Now do the same by inverting the other bits. We will generate 24*n new sequences. That's because if you assume that (b1,b2,...,b24) and (c1,c2,...c24) are two sequences in S and for example (b1',b2,...b24) = (c1,c2',c3,...,c24), that would imply that the two sequences are equal in 22 values, a contradiction. So by inverting only one bit in all sequences in S and in all positions we get 24 new sets with n distinct sequences. That's also true if we invert any two bits and any three bits, giving us additional (24 2) + (24 3) new sets with n distinct elements, that is so far we have (1 + (24 1) + (24 2) + (24 3))*n distinct 24-length binary sequences.

Now if we invert 4 bits in sequences in S (observe, if it was unclear in the beginning, that when we invert bits in sequences in S we invert exactly the same positions in all sequences in S) and we have at least two that differ in exactly 8 positions, then by inverting (24 4) times, some sequences will repeat. Now what's left is to count how many repeats we at least must have and how many new elements it will generate. In our case, n = 4096, so we have to show that we will generate at least 1771 new sets with 4096 elements (when inverting 4 bits) because then we would have 4096*(1 + (24 1) + (24 2) + (24 3) + 1771) = 4096*4096 = 2^24 distinct 24-length sequences, and that's all we can have. So if we had 4097 or more elements, that would imply that we will get more than 2^24 binary sequences of length 24, a contradiction.

I hope this approach can finish it off.

Each element of $S$ will give $\binom{24}{4}$ sequences differing from it in exactly $4$ places. If $w$ is a sequence then the number of sequences of $S$ differing from $w$ in exactly $4$ places is at most $6$ since, each such sequence of $S$ corresponds to a $4$ element subset of $\{1,\ldots,24\}$ and all such subsets must be disjoint, because if they intersect then the corresponding sequences of $S$ differ in less than $8$ places.  Therefore number of such sequences is at least $4096*\binom{24}{4}/6 = 4096*1771$.

Actually such approach is quiet common in coding theory and gives one of the common bounds, Gilbert-Varshamov bound (though in these bounds typically only the order is given importance) and also in a construction (though not efficient) of codes known as 'asymptotically good codes'.

[quote="DK"]Yep, I also got the same stuff at the end.

Very nice problem actually. Interesting that it's connected to other scientific fileds.[/quote]

In fact, in Computer Science (my area!) it is very relatet to Hamming Error-Correcting Codes

Oh, I forgot to mention that there exists a code like that: it's the Golay code.


Are there any composite numbers of the form $10^{2^{n}}+1$, or are they all prime? Just wondering. :)

(I got bored, and was like, if $n$ has an odd prime divisor, they're not... but what about $2^{n}$)

Ravi B, why $10^{8}+1$ is composite follows from $10^{4}+1$is composite. (The two numbers are relatively prime after all).

I didn't mean that one implied the other.  I just meant that all three numbers are composite.  According to the computer, $10^{4}+1$ is a multiple of $73$, $10^{8}+1$ is a multiple of $17$, and $10^{16}+1$ is a multiple of $353$.

really? thanks. Did you just brute force it, or what? (or did you already know?)

I plugged in each number into a primality tester I found online.  I googled "primality tester", and chose the first link (some Java applet).


induction
function
floor function
logarithms
algebra proposed
algebra

prove that the application $ f: N$*--->$ N$* defined by:
$ f(n)=$  [$ 1+1/2+1/3+...+1/n$]
is surjective.

[hide="to help you, you can click hear"]define an other application $ g$ such that:
$ g(n)=1+1/2+1/3+...+1/n$
$ 1)$ prove that $ \forall n \in N$*: $ 1/2\leq g(2n)-g(n)<3/4$
$ 2)$ by induction (if you want) prove that :
$ (\forall p \in N$*) $ (\exists n \in N$*)  $ p \leq g(n)<p+1$
$ 3)$now you can deduce that $ f$ is surjective?
[hide="+exercice"]an other thing: Is $ f$ $ injective?$[/hide][/hide]

The English word is "function."  A similar problem was recently posted concerning the function $ f(n) \equal{} \lfloor \ln n \rfloor$.

t0rajir0u can you help me and write the solution or give me the link of it.


p.154 question 181:
Segments AD and BE are medians of right triangle ABC, and AB is its hypotenuse.  If a right triangle is constructed with legs AD and BE, what will be the length of its hypotenuse in terms of AB?
 :D

[hide]The idea is to relate the medians of the triangle to the sides. For example, $AD^2=AC^2+\frac{1}{4}BC^2$. [/hide]

I'll move this, just in case (it is a little hard, I think).

You posted this problem several times.

Plus this question has different text in the title and different emoticons, showing that it wasnt that you clicked the post topic button several times, but that you are spamming like a madman.


[url=http://www.alexchiu.com/affiliates/clickthru.cgi?id=saturn]Immortality site[/url]

A few weeks ago, I came upon a site whose webmaster claimed he had created immortality rings which could make you live forever. To achieve "eternal life" he says all that one should do is daily wear some rings he has made when you sleep. These are not just normal rings: they hold some magnets. These magnets are supposed to correct the magnetic flow through the body which has for effect that the body is healed and the aging process "reversed" (or countered). He explains exactly how this works on his site.

Obviously, it could take a while to see if he could be right about it... But on his site he has tons of testimonials that seem real to me. If the rings don't give eternal life, then they sure seem to do a lot of good for one's health. People seem to get in better shape starting within 24 hrs, and feel noticeably better in just 2 weeks. The rings seem to do all sorts of healing and improving one's health.

The best identified reason why the body ages is because the telomers of body cells shorten, and the telomerase (an enzyme that is supposed to counter the shortening of the telomers) gets used up over the years. One of the ways to counter this is by avoiding to drain the body from its enzymes (and consume enzyme supplements if necessary), and have a sufficient intake of minerals and vitamins. This means that one should avoid eating cooked and processed food (as when cooked or processed, food takes more enzymes from the body to be digested).

So if the rings really heal body tissues, like scars (as some have testified), then they might also heal the telomers, and then (maybe, as this is an unverified speculation) all of this could work!

Anyway, I haven't tried them myself yet, but I hope by recieving clicks, I will soon be able to order a free set of rings. Honestly, I think the guy doesn't seem to be a quack, considering the fact that on his site he even explains how to make the rings yourself, for those who don't want to get clicks for them.

So, please, click on my link leading to the site and look around it, as you may become interested in making an account yourself. After all, it wouldn't hurt to try, would it?

edit: I just ordered my first free pair of 1,000 gauss rings through the phone and should be receiving them in 2 to 4 weeks.

[quote="RobinHood3000"]Perhaps this is overly cynical of me, but I don't believe a word of it.[/quote]

Ya when ever i see stuff like this I don't believe it and think its a hoax or money-maker. I would only believe something like this if I had proof that it worked.

If this really worked, then it would be all over the news and stuff.

Well, it wouldn't hurt to try, I suppose.  Anyway, I still believe there is some truth to it no matter how ludicrous the claims may be.

I've seen that portrait before. It has appeared on at least one animutation, whose name is "We Drink Ritalin" (the song is a DDR track; I don't remember which mix). Yeah, and really, when you read the descriptions and such, do they [b]make sense[/b] to you at all? Certainly didn't for me.

Hope this message assuages your fears that such "immortality rings" could be genuine. :cool:

C'mon guys, I believe in the immortality rings! Sure, I've had some disappointments with the superpower wristband, the knowledge glasses, and the time-stop watch, but I really think this thing's for real. ;)

No offence, but how do you suppose he discovered this "amazing gadget" and if you check some 100 years later you think those people will still be alive. :maybe:

Isn't magnetism supposedly good for the body though? I've heard that Chinese people used to use magnets to do something to the "flux" of their bodies to cure certain illnesses. A lot of the people on the site's message board said there were some positive effects, too.

Magnetism? The website didn't really talk about how it worked. Sure maybe it could help, but I doubt it would really make you immortal. By the way I clicked that link like 20 times I hope that helps  ;) .

the magnetism part came from the patent abstract

If I made a huge immortality ring and put it around my head would it make me really smart?  :D

I'd take that over studying for classes.  :D

I'm inclined to believe some of the stuff. The immortality part sounds fake, although magnetotherapy has been used elsewhere, and they have been shown to work. After all, in the Far East and Hospitals, magnets have been used for quite a while.

Rings are too boring.  Maybe if it was an Immortality Group.

[quote="RobinHood3000"]Quite frankly, I think that the concept sounds just like something I'd get in my Inbox between the Generic Drug and Get a New Mortgage! e-mails.[/quote]

Dude, some of the mortgage emails are quality. You've really got to stop deleting them without reading them a little. They're good stuff.

Thanks for clicking my link guys, it's helped a lot! I just received enough for my first free set of rings!

Let us know if they help.  Just post here a hundred years from now.

I have to say the idea of the site is quite clever. This guy must get considerable publicity for the relatively low cost of sending out a few rings! Imagine how much he would have to spend on conventional advertising to achieve the same effect? If he was in turn to include commercial ads on his site, he could probably make a tidy profit - not to mention from anyone who actually buys his stuff.

dear dark_shadow:

what's with the pikachu???

Still, aging is caused by cells multiplying. You would think that cells live forever, but they don't. They have shorter DNA every time they multiply. That's probably why we age and even die.

I still say the Immortality Rings are codswallop.

darkshadow...really whats with the pikachu? its getting on my nerves and i don't know why

me? i don't know. pickachu just... never mind.

[quote="Athena"][quote="uglyman621"]hey i just had a project on cells  ;)[/quote]

who do u have for science?

and u actually (saturnlife) got free rings? do they deliver or sumthin. i thot he wasnt done manufacturing them yet.[/quote]
private discussions are bouncing back

and i believe (gameworld) that u r referring to dark_shadow and not darkshadow?

PIKACHU OWNS!!!!!

...yeah i'm referring to both darkshadows whatever...

FINAL FANTASY OWNS!!!

lol guys, i forgot i made this thread... btw, I received my rings on January 2nd or something like that. And I am honestly telling the truth. I don't experience anything unusual though, except when i wear them at night, it takes very long to go to sleep. But in the morning, I feel a little more awake. Bah... it's probably the placebo effect or something. Thanks to all who clicked! <3

rings that give you insomnia, lol, that's sure gonna sell


A line segment 40 cm long is to be divided into three segments whose lengths are integers. How many centimeters long is the longest piece if they for the sides of a right triangle?

This problem would be relatively easy if one knew their Pythagorean triples. We can just list some of them.
$ (3,4,5), (6,8,10), (5,12,13), (7,24,25), (10,24,26), \boxed{(8,15,17)}$ so the longest side is $ \boxed{17}$.


A number of triangles , all with a common orthocentre. are drawn with  their vertices on a fixed circle.

Prove that the feet of the altitudes of ALL the triangles lie on a circle.

Heh, this is just an immediate result of the properties of the nine-point circle.


function
induction
algebra unsolved
algebra

[quote="tfy"]Find all functions f(x) defined in nonnegative real numbers satisfying the following condtions:
(1) for all nonnegative real numbers, f(x) >=0 ;
(2) f(1) = 0.5 ;
(3) for all nonnegative numbers x and y, we have f(yf(x))f(x) = f(x + y).

I suppose the only solution to this problem is f(x) = 1 :help:  :help: /(x+1) ,but I can't prove it.I appreciate if anyone can help me.Could you do me a favour?[/quote]

[hide="My solution"]
Let $ P(x,y)$ the assertion $ f(y(f(x))f(x)\equal{}f(x\plus{}y)$

1) [u][b]Preliminary result[/b][/u] : $ f(x)>0$ $ \forall x$
Suppose we have $ f(b)\equal{}0$ for some $ b$
Then $ P(b,y)$ $ \implies$ $ f(y\plus{}b)\equal{}0$ and so $ f(x)\equal{}0$ $ \forall x\geq b$
Let then $ x\geq \max(0,b\minus{}1)$. we have $ x\plus{}1\geq b$ and so $ f(x\plus{}1)\equal{}0$. But then $ P(1,x)$ $ \implies$ $ f(\frac{x}{2})\equal{}2f(x\plus{}1)\equal{}0$
So $ f(x)\equal{}0$ $ \forall x\geq b$ $ \implies$ $ f(x)\equal{}0$ $ \forall x\geq \max(0,\frac{b\minus{}1}{2})$
And so, with a simple induction $ f(x)\equal{}0$ $ \forall x$, which is wrong since $ f(1)\neq 0$
So $ b$ does not exist and $ f(x)>0$ $ \forall x$
Q.E.D.

2) [u][b]we'll now show that $ f(x)$ is non increasing[/b][/u]
Suppose $ \exists a\geq 0$ such that $ f(a)>1$.
Then $ P(a,\frac{a}{f(a)\minus{}1})$ $ \implies$ $ f(\frac{af(a)}{f(a)\minus{}1})f(a)\equal{}f(\frac{af(a)}{f(a)\minus{}1})$ but this is impossible since $ f(a)>1$ and $ f(x)\neq 0$ $ \forall x$ so $ a$ does not exist and $ f(x)\leq 1$ $ \forall x$
So $ f(y(f(x))\leq 1$ and $ f(x\plus{}y)\leq f(x)$ and so $ f(x)$ is a non increasing function.
Q.E.D

3) [u][b]we'll now show that $ f(x)$ is injective[/b][/u]
Let $ a>b$ such that $ f(a)\equal{}f(b)$
$ P(b,a\minus{}b)$ $ \implies$ $ f((a\minus{}b)f(b))f(a)\equal{}f(b)$. We know that $ f(a)\equal{}f(b)\neq 0$ and so $ f((a\minus{}b)f(b))\equal{}1$. Let $ u\equal{}(a\minus{}b)f(b)$
$ P(u,x)$ $ \implies$ $ f(x)\equal{}f(x\plus{}u)$

So $ f(x)$ is periodic ($ u\neq 0$) and non increasing. So $ f(x)$ is constant. But the only constant function verifying $ P(x,y)$ are $ f(x)\equal{}0$ and $ f(x)\equal{}1$ and we know that $ f(1)\equal{}\frac{1}{2}$
So no such $ a$ and $ b$ exist and $ f(x)$ is injectice, so decreasing.
Q.E.D

4) [u][b]Final result[/b][/u] : $ f(x)\equal{}\frac{1}{x\plus{}1}$
$ P(1,2x)$ $ \implies$ $ \frac{1}{2}f(x)\equal{}f(2x\plus{}1)$
$ P(x,\frac{1}{f(x)})$ $ \implies$ $ \frac{1}{2}f(x)\equal{}f(x\plus{}\frac{1}{f(x)})$

So $ f(2x\plus{}1)\equal{}f(x\plus{}\frac{1}{f(x)})$ and, since $ f(x)$ is injective : $ 2x\plus{}1\equal{}x\plus{}\frac{1}{f(x)}$ and so $ f(x)\equal{}\frac{1}{x\plus{}1}$

And it's easy to check back that this solution indeed fits the initial requirements.
Q.E.D.
[/hide]


advanced fields
advanced fields unsolved

I want to find prime and jacobson radical radicals in Z[T]/(T^3), here Z = integers.

Is it true that Z[T]/(T^3) is a field, because T^3 is irreducibel over Z[T]. If it is true that
Z[T]/(T^3) is a field then 0 is the prime and jacobson radical radical.

Is it true please help.

You've already posted this and are receiving help at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=223018

Please don't double-post -- it's quite rude, both to the people who are already helping you and the rest of us.  I think you'll find that you'll have better luck if you take the time to actually think about the comments of other posters, rather than if you just sit and wait for them to do your work for you.


Prove that $ x(1\minus{}y)\plus{}y(1\minus{}z)\plus{}z(1\minus{}x)<\equal{} 1$ with x,y,z belonging to [0,1] . (sorry for my english :D )

If you subtract left side from right side, you get $ (1\minus{}x)(1\minus{}y)(1\minus{}z)\plus{}xyz$ which is nonnegative.

here's a gorgeus method in my opinion.I have a geometric solution.Lets build the equilateral triangle ABC with AB=AC=BC=1 . Lets take M in AB, N on AC and P on BC.Note that AM=x, NC=y, PB=z. We notice that  $ \sigma(AMN)\plus{} \sigma (NPC) \plus{}\sigma (BMP) \leq \sigma (ABC)$ This is equivalent to \[ x(1\minus{}y)\displaystyle\frac{sin60}{2}\plus{}y(1\minus{}z)\displaystyle\frac{sin60}{2}\plus{}z(1\minus{}x)\displaystyle\frac{sin60}{2} \leq 1 \cdot 1 \cdot \displaystyle\frac{sin60}{2}\] and that's it:D


Hi everyone!
I've seen some geometry proofs using directed angles/segments on the forum, but I'm unsure of the basic properties or the advantages of using them in problem solving. (eg, directed angles modulo :pi: ?)
Can anyone give me an introduction to the uses of those in problem solving?  Thanks! :lol:

Hmm... :(

The notions of directed angles and segments are really nice but correctly explaining them is such a hell of work...  :?

Anyway, if somebody can read German, there is an introduction into directed angles modulo 180 ([url=http://de.geocities.com/darij_grinberg/Dreigeom/Inhalt.html]"Orientierte Winkel modulo 180 und eine Lsung der WURZEL -Aufgabe Kappa 22 von Wilfried Haag"[/url]) on my website... but at first, it is not really axiomatic (I give some hints towards an axiomatic definition, but I don't really strictly prove all the basic properties), and then, don't forget that directed angles modulo 180 are only a part of the story... There are directed angles modulo 360 as well, they have a number of different properties (basically, directed angles modulo 180 are angles between lines, while directed angles modulo 360 are angles between directed lines or between vectors). And there are directed segments (best axiomatically defined using unit vectors). If somebody would take the burden of explaining this all in a clear way so that we can use it as a reference, I would be really glad...

  Darij


The equivalent form for this inequality:
let $a$ and $b$ are positive numbers such that $a^{2}+b^{2}=2.$ Prove that
$\frac{(a-1)^{3}}{a}+\frac{(b-1)^{3}}{b}\leq0.$

[hide]
Using the second form, WLOG let $a\geq b$. Since $a^{2}+b^{2}=2$, $a+b\leq 2$ by RMS-AM, so $(a-1)^{2}\leq (b-1)^{2}$. Also, $1-\frac{1}{a}\geq 1-\frac{1}{b}$. Hence, by Chebyshev,
\[(a-1)^{2}\left( 1-\frac{1}{a}\right)+(b-1)^{2}\left( 1-\frac{1}{b}\right) \leq \frac{(a-1)^{2}+(b-1)^{2}}{2}\left( 2-\frac{1}{a}-\frac{1}{b}\right)\quad (*)\]
Also, we have $\frac{1}{a}+\frac{1}{b}\geq 2$ which is evident from Holder:
\[(a^{2}+b^{2})\left( \frac{1}{a}+\frac{1}{b}\right)^{2}\geq (1+1)^{3}\]
So the right hand side of $(*)$ is nonpositive; the original expression is as well.
[/hide]

[quote="ElChapin"][quote="arqady"]Let $x$ and $y$ are positive numbers such that $x+y=1.$
Find the minimum value of $\frac{x+7y}{\sqrt{1-x}}+\frac{y+7x}{\sqrt{1-y}}.$[/quote]

$\frac{x+7y}{\sqrt{1-x}}+\frac{y+7x}{\sqrt{1-y}}=$
$\frac{1+6y}{\sqrt{y}}+\frac{1+6x}{\sqrt{x}}=$
$(\frac{1}{\sqrt{y}}+\frac{6y}{\sqrt{y}})+(\frac{1}{\sqrt{x}}+\frac{6x}{\sqrt{x}})\ge$      by AM-GM
$2\sqrt{6}+2\sqrt{6}=4\sqrt{6}$[/quote]
O.K. But what is the the minimum value of $\frac{x+7y}{\sqrt{1-x}}+\frac{y+7x}{\sqrt{1-y}}$ $?$ :wink:

[quote="arqady"]Let $x$ and $y$ are positive numbers such that $x+y=1.$
Find the minimum value of $\frac{x+7y}{\sqrt{1-x}}+\frac{y+7x}{\sqrt{1-y}}.$[/quote]

Let $x=\cos^{2}\theta ,\ y=\sin^{2}\theta \left(\theta \neq 0,\ \frac{\pi}{2}\right)$, we have

$P: =\frac{x+7y}{\sqrt{1-x}}+\frac{y+7x}{\sqrt{1-y}}$

$=\frac{1+6\sin^{2}\theta}{|\sin \theta |}+\frac{1+6\cos^{2}\theta }{|\cos \theta |}$

$=\frac{1}{|\sin \theta |}+\frac{1}{|\cos \theta|}+6(|\sin \theta |+|\cos \theta|)$. Set $|\sin \theta |+|\cos \theta |=t$, we have

$P: =f(t)=\frac{2t}{t^{2}-1}+6t\ (0<t\leq \sqrt{2})$.

$f'(t)=\frac{2}{(t^{2}-1)^{2}}(t+\sqrt{2})(\sqrt{3}t+1)(t-\sqrt{2})(\sqrt{3}t-1)$,

yielding the local minimum $f(\sqrt{2})=8\sqrt{2}$, which is the desired minimum value.

Another approach:

Let $\sqrt{1-x}=X,\ \sqrt{1-y}=Y$, by $x+y=1$, we have $X^{2}+Y^{2}=1$ and

$\frac{x+7y}{\sqrt{1-x}}+\frac{y+7x}{\sqrt{1-y}}$

$=6(X+Y)+\frac{X+Y}{XY}$ where $1<X+Y\leq \sqrt{2}$.

[quote="scorpius119"][hide]
Using the second form, WLOG let $a\geq b$. Since $a^{2}+b^{2}=2$, $a+b\leq 2$ by RMS-AM, so $(a-1)^{2}\leq (b-1)^{2}$. Also, $1-\frac{1}{a}\geq 1-\frac{1}{b}$. Hence, by Chebyshev,
\[(a-1)^{2}\left( 1-\frac{1}{a}\right)+(b-1)^{2}\left( 1-\frac{1}{b}\right) \leq \frac{(a-1)^{2}+(b-1)^{2}}{2}\left( 2-\frac{1}{a}-\frac{1}{b}\right)\quad (*) \]
Also, we have $\frac{1}{a}+\frac{1}{b}\geq 2$ which is evident from Holder:
\[(a^{2}+b^{2})\left( \frac{1}{a}+\frac{1}{b}\right)^{2}\geq (1+1)^{3}\]
So the right hand side of $(*)$ is nonpositive; the original expression is as well.
[/hide][/quote]
What is the simplest explanation of $(a-1)^{2}\leq (b-1)^{2}$?
The only way I can see is
\[\{a\geq b\implies 1-b\geq 1-a \qquad \text{ and }\qquad 2\geq a+b\implies 1-b\geq a-1 \}\]\[\implies 1-b=|1-b|\geq \pm (a-1)\geq |a-1|\implies (1-b)^{2}\geq (a-1)^{2}\]
where $b\leq \frac{a+b}{2}\leq 1$ but this is rather contrived...

[quote="me@home"][quote="scorpius119"][hide]
Using the second form, WLOG let $a\geq b$. Since $a^{2}+b^{2}=2$, $a+b\leq 2$ by RMS-AM, so $(a-1)^{2}\leq (b-1)^{2}$. Also, $1-\frac{1}{a}\geq 1-\frac{1}{b}$. Hence, by Chebyshev,
\[(a-1)^{2}\left( 1-\frac{1}{a}\right)+(b-1)^{2}\left( 1-\frac{1}{b}\right) \leq \frac{(a-1)^{2}+(b-1)^{2}}{2}\left( 2-\frac{1}{a}-\frac{1}{b}\right)\quad (*) \]
Also, we have $\frac{1}{a}+\frac{1}{b}\geq 2$ which is evident from Holder:
\[(a^{2}+b^{2})\left( \frac{1}{a}+\frac{1}{b}\right)^{2}\geq (1+1)^{3}\]
So the right hand side of $(*)$ is nonpositive; the original expression is as well.
[/hide][/quote]
What is the simplest explanation of $(a-1)^{2}\leq (b-1)^{2}$?
The only way I can see is
\[\{a\geq b\implies 1-b\geq 1-a \qquad \text{ and }\qquad 2\geq a+b\implies 1-b\geq a-1 \}\]

\[\implies 1-b=|1-b|\geq \pm (a-1)\geq |a-1|\implies (1-b)^{2}\geq (a-1)^{2}\]
where $b\leq \frac{a+b}{2}\leq 1$ but this is rather contrived...[/quote]
$(a-1)^{2}-(b-1)^{2}=(a-b)(a+b-2)\leq 0$. :wink:

you're right arqady $4\sqrt{6}$ cannot be the minimun
because the equalities:

$\frac{1}{\sqrt{y}}=\frac{6y}{\sqrt{y}}$and $\frac{1}{\sqrt{x}}=\frac{6x}{\sqrt{x}}$

cannot hold at same time.
You're right too kunny.

$\frac{x+7y}{\sqrt{1-x}}+\frac{y+7x}{\sqrt{1-y}}=$
$\frac{1+6y}{\sqrt{y}}+\frac{1+6x}{\sqrt{x}}=$
$(\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}})+(\frac{6y}{\sqrt{y}}+\frac{6x}{\sqrt{x}})$*

 by AM-GM we can proove the expresion takes the minimun when $x=y=1/2$
after evaluting * with these values we get

      $8\sqrt{2}$  :)

Thanks, Elchapin. :) Let me correct my answer in the last line partly.

[quote="kunny"][quote="arqady"]Let $x$ and $y$ are positive numbers such that $x+y=1.$
Find the minimum value of $\frac{x+7y}{\sqrt{1-x}}+\frac{y+7x}{\sqrt{1-y}}.$[/quote]

Let $x=\cos^{2}\theta ,\ y=\sin^{2}\theta \left(\theta \neq 0,\ \frac{\pi}{2}\right)$, we have

$P: =\frac{x+7y}{\sqrt{1-x}}+\frac{y+7x}{\sqrt{1-y}}$

$=\frac{1+6\sin^{2}\theta}{|\sin \theta |}+\frac{1+6\cos^{2}\theta }{|\cos \theta |}$

$=\frac{1}{|\sin \theta |}+\frac{1}{|\cos \theta|}+6(|\sin \theta |+|\cos \theta|)$. Set $|\sin \theta |+|\cos \theta |=t$, we have

$P: =f(t)=\frac{2t}{t^{2}-1}+6t\ (0<t\leq \sqrt{2})$.

$f'(t)=\frac{2}{(t^{2}-1)^{2}}(t+\sqrt{2})(\sqrt{3}t+1)(t-\sqrt{2})(\sqrt{3}t-1)$,

yielding the local minimum $f(\sqrt{2})=8\sqrt{2}$, which is the desired minimum value.[/quote]

[color=red]yielding the minimum value not local minimum[/color]

kunny

[b]Problem 2:[/b]Let $x$ and $y$ be positive numbers such that $x+y=1$.
Find the minimum value of
$P=\frac{2007x+2008y}{\sqrt{2007-x}}+\frac{2007y+2008x}{\sqrt{2007-y}}$

[quote="chien than"][b]Problem 2:[/b]Let $0<y<1$.
Find the minimum value of
$P=\frac{2007+y}{\sqrt{2006+y}}+\frac{2008-y}{\sqrt{2007-y}}$[/quote]If I didn't misunderstand... we just consider \[\frac{2007+y}{\sqrt{2006+y}}+\frac{2008-y}{\sqrt{2007-y}}=\frac{1}{\sqrt{2006+y}}+\sqrt{2006+y}+\sqrt{2007-y}+\frac{1}{\sqrt{2007-y}}\]
Or if $\frac{4013}{2}=c$ and $\frac12-y=b$ this is \[\frac{1}{\sqrt{c-b}}+\sqrt{c-b}+\sqrt{c+b}+\frac{1}{\sqrt{c+b}}\]
Now \[P'=\frac{-1}{2\sqrt[2/3]{c-b}}+\frac{-1}{2\sqrt[2/3]{c+b}}+\frac{1}{2\sqrt{c-b}}+\frac{1}{2\sqrt{c+b}}\]
\[=\frac12\left[\frac{1}{\sqrt{c-b}}\cdot \left(1-\frac{1}{c-b}\right)+\frac{1}{\sqrt{c+b}}\cdot \left(1-\frac{1}{c+b}\right)\right]\]
Where $b$ is solved for to find minimums...
 :blush: thought it would pan out


Three circles with radii $1,2,3$ respectively touch each other in a plane . In the region enclosed by the three circles , another smaller circle $O$ is drawn such that it touches the three circles . Find the radius of the smalle circle $O$ .

More generally if $a$, $b$ and $c$ are the radii of the given circles then the radius of the fourth circle enclosed by the given circles is given by

$\frac{abc}{ab+bc+ca+2\sqrt{abc(a+b+c)}}$


Do you have an IMO camp in your country ? Would you like to have one (if you don't ) ?  How long should it be ? Do you think it might be improved ? Is it fair to compare countries with varying IMO preparation camp's length, e.g. Germany, one week, and Viet Nam, 5 weeks ? Guess what's the reason why GDR/DDR and FRG were very strong in the past but when they were united a decade ago they could not keep up its high level ?

well, i have not attend any. (my math is not that good, so i have no chance)
but i still like to say something about the IMO Team training here.
The following text is from "introduction to math competitions", this book has only Chinese edition and its writer is a coach of Chinese IMO Team.

In Jan of every year, CMO is held to choose about twenty participants.
From March to the early May, they mainly have training. Every morning they have classes and every afternoon they discuss some problems, and they should do homework in the evening. Every week, they have at least one test, each test contains four problem and it lasts for three hours. The problems is not easier than IMO but the time limit is shorter.
After this, they will be two-day contest, called 'Chinese Team Selection Contest'. The result of this contest has a weight 0.5 and the results of tests at ordinary times also have a weight 0.5. Combining these, six members are chosen as Chinese IMO Team. Each person during this period have to do large quantities of problems.

Since the team has been determined in May, the training turns loose. The six members have classes only in the morning and they don't have ordinary tests. There are only two simulated contests mainly to train their psychological state. In this period, they don't do a lot of problem but spend more time on discussion. And they will have a holiday to relax about half a month before IMO.

[quote="liyi"]well, i have not attend any. (my math is not that good, so i have no chance)[/quote]

first of all thanks for shattering the "iron curtain" :D. Secondly I personally don't belive you don't stand a chance. Have you tried to get there? I qualified after 3 IMO preparation camps. Some people make it in their first. But you must definitely try. :)

math contest in china has three stages.
CMO is the second and the Team Selection Contest is the third.
I failed at the first stage(provincial contest), so i'm unable to attend CMO.
I'm definitely sure that I can do the problems, but maybe I was so nervous during the test that I even didn't work out some very easy problems. After I walk out the classroom, I immediately got the solution myself... But it is too late.
It seems that I'm not suitable to exams. I entered National Olympics in Informatics in August, and I didn't get a satisfactory prize. I'm sure I have the ability to get Silver Medal, but at last I only get Bronze. I made some stupid faults, for example, i missed an important condition in one problem, and i had the wrong filename of another problem...

I will enter a university next year, so I don't have any more chance to participate in Math Olympiad.

Do you have an IMO camp in your country? Yes - acutally two, one for possible candidates (in a couple of days) and one for the team (before IMO)
How long should it be ? a week plus a bit
Do you think it might be improved ? of course...  not many ppl do IMO things over here -- they should promote this

Yep.  In USA we have MOP, as some of you may have heard about.  Sadly, I haven't been there.  Yet, that is.  I hope to make it someday.

Strangely as it may seem, I've been to MOP (yey) in 2001 :D some of the people there still remember us (us= Andrei Jorza, Mihai Manea, Ana Caraiani, pliviu and me) :D

no I live in Romania (unfortunately :(). but since at that time the coach of the USA team was a Romanian (Titu Andreescu) we got around to a common trainning. That is when Ana, Liviu and me got really scared by the impressive Reid and Carroll duet. In the perspective of the IMO result that common training didn't help us too much, but rather the opposite :( as we got scared: "look this guy (ie Reid) went to 3 IMOs already and scored gold, also we went to IOI last year and also scored gold, do we have a chance? :?" or "look this guy (ie Carroll) went to IMO in 9th grade (ie Romania 1999) and also scored gold. In 2000 he decided not to foul around with IMO anylonger so he goes to putam and he becomes fellow .. damn we suck!" that's what we were thinking back then. Of course the results at the IMO after were desastrous. :(

[quote="Charlie"]Since Hong Kong is just a small city our training is rather different from all of you. We had a lecture class every weekend, two during the summer holiday and had our normal school life at the same time. Selection of team members are done in four tests, one is the Prelimary contest to select elgible students for training, and they would be trained for one year, with IMO team members chosen from three tests, Test 1, Hong Kong China Math Olympiad (aka Test 2  :) ) and APMO.[/quote]

we hv a hkmo camp.
However hkmo 's level is much lower than imo..  :(

[quote]he decided not to foul around with IMO anylonger so he goes to putam and he becomes fellow [/quote]

What is putam?

[quote="PiDeltaPhi"]What is putam?[/quote]

It is the major competition for undergraduate university students in the US.

Could you provide me a link,google doesnt seem to bring up some links...

Oh, it was typed incorrectly originally. Its "Putnam". I don't remember the original site, but google should work with the proper name.

What Charlie described sounds a bit like training in Singapore.  Since we're even smaller :D we manage to have training every Saturday for the team of 15.  During our one-week holiday in March we have training daily, and during our month-long June holidays the IMO team trains daily (and anybody else who has too much time on their hands and wants to pop in is welcome to do so).

Until last year we had a test in December to select the team of 15, but from this year onwards it will be selected directly from the SMO (which makes a bit more sense, because in the past SMO was never used to select anything).  The selection test for the IMO is done in April/May over 2 days (used to be 3, but it seems it doesn't make much difference, so now it's just plain IMO style).  We take the APMO too but our APMO results don't count in the selection.

haha... yes. and now we are training everyday. actually there are some odds days we don't train... but well... (=


modular arithmetic
search
AoPSwiki
articles

it's short for modulo. It's basically the remainder when divided by the modulus. For example,

7 is [i]congruent[/i] to 3 mod 4, since 7/4 has a remainder of 3.
The technical definition of a being congruent to b mod c is that ck+b=a for some integer k (for a<b).

ok........i didn't understand much of that.........
-jorian

When we say a number $n$ is congruent to $m$ mod $a$, we mean that $n-m$ is a multiple of $a$.
Basically, when you divide $n$ by $a$, it has a remainder $m$.
We write this as $n\equiv m\pmod{a}$.
For exmaple, when 15 is divided by 4, it has a remainder of 3. Thus, we say 15 is congruent to 3 mod 4, or $15\equiv 3\pmod{4}$

If $x \equiv y \pmod{z}$ it also means that $za+y=x$ for some integer z.
At my school there is an advanced math program. In it you learn mods, combinations, and about pascals triangle during pre alg which is in 6th grade. It helps the easing of being good at school math to being good at competition math.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=modulus&t=84613]This[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=modulus&t=90491]has been[/url]  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=modulus&t=28395]posted[/url] before. Try the [url=http://www.artofproblemsolving.com/Forum/search.php]Search[/url] button next time before you post.

Furthermore, the [url=http://www.artofproblemsolving.com/Wiki/index.php/Main_Page]AoPSWiki[/url] has a [url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction]pretty good article[/url] on modular arithmetic.

Locked.


John meant to divide a number by $ 2$, but he was careless and subtracted $ 2$ from the number instead.  He got an answer of $ 22$.  What would his answer have been had he actually divided by $ 2$ instead of subtracting $ 2$?

If we let the original number be $ x$, we get $ x\minus{}2\equal{}22$, so $ x\equal{}24$.  If we divide, we get $ \frac{24}{2}\equal{}\boxed{12}$


greatest common divisor
least common multiple

The greatest common divisor of two 3-digit numbers $m$ and $n$ is 6 and their least common multiple is a multiple of a six-digit number. Determine the algebraic sign of $tan(\frac{m}{350})$.

$tan(\frac{m}{350})$ is negative. this is equivalent to showing $m>\frac{350\pi}{2}$ which is the same as $m\geq 550$ since m is an integer. but since $lcm(n,m)=\frac{nm}{gcd(n,m)}=\frac{nm}{6}\geq 100000$, we have $m\geq \frac{600000}{999}>600$.


On my Latex Document, I get this annoying blue line that says "Create PDF with GO2PDF for free. If you wish to remove this line, click here to buy Virtual PDF."

How do I purge my document of this message?  :?

There is a link I posted to another freeware (much better) PDF printer. Uninstall GO2PDF and install that printer. It's much better, and in plus no nasty links ;)


superior algebra
superior algebra unsolved

Is it possible to product two groups together and take a quotient group of that product that you can't get by first taking quotients individually and then producting?

Notationally: Does there exist $ G$ and $ H$ groups and $ K\triangleleft G\times H$ such that $ \frac{G\times H}{K}\ncong \frac{G}{L}\times\frac{H}{J}$ for any $ L\triangleleft G$ and $ J\triangleleft H$?

Hmm...I think that is quantified properly.

Take $ G \equal{} H \equal{} Q_8 \equal{} \{\pm 1,\pm i,\pm j,\pm k\}\subseteq\mathbb{H}$
and $ K \equal{} \langle ( \minus{} 1, \minus{} 1)\rangle \unlhd Q_8\times Q_8$.
Then $ (Q_8\times Q_8)/K\cong Q_8/I\times Q_8/J$ with $ I,J\unlhd Q_8$ would give
$ |I|\cdot |J| \equal{} |K| \equal{} 2$, so WLOG $ |I| \equal{} 1,|J| \equal{} 2$, which gives $ I \equal{} 1,J \equal{} \langle \minus{} 1\rangle$.
So $ (Q_8\times Q_8)/K\cong Q_8\times C_2\times C_2$.
But the RHS has $ 24$ elements of order $ 4$, namly $ \{\pm i,\pm j,\pm k\}\times C_2\times C_2$
and the LHS has only $ 12$ such elements, namly $ (x,1)K,(1,x)K$ with $ x\in\{\pm i,\pm j,\pm k\}$.
Contradiction!

Thanks. I was working my way up there. I figured if it existed then it had to be kind of big, since my early attempts were giving me quotients of 1, $ \mathbb{Z}_2$, or $ \mathbb{Z}_3$, since I didn't have enough elements.


probability AMC AIME USA(J)MO USAMO function

Same rules as other marathons, AIME to USAMO questions.

Problem 1:
You have 8 different rings. Let n be the number of possible arrangements of 5 rings on the four fingers of one hand (each finger has zero or more rings, and the order matters). Find the three leftmost non-zero digits of n.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64995]376[/url]

[size=150]Problem 2[/size]

$m$ red socks and $n$ blue socks are in a drawer, where $m + n \le 1991$. If two socks are taken out at random, the chance that they have the same color is 1/2. What is the largest possible value of $m$? (AIME 1991)

[hide]
The probability that to pick 2 and get 1/2 is:
$\frac{m(m-1)+n(n-1)}{(m+n)(m+n-1)}=\frac{1}{2}\Rightarrow m^2-m-n+n=2mn\Rightarrow (m+n)^2=n+m$.
$\Rightarrow m-n=\sqrt{m+n}$. Notice that we have:
$m+n=x^2, m-n=x$. For maximum $x^2<1991$ is $x=43$.
Plugging $43$ back into the equation yields $m=\boxed{946}$. [/hide]

[b]Problem 3: [/b]Let $N$ denote number of 7 digit positive integers that have the property that their digits are in increasing order.  Determine the remainder of $N/1000$. (Repeated digits are allowed).

Once you pick out the digits, their order is determined since they must be increasing.
Breaking this up into cases depending on the number of different digits, the answer is
$9C7+9C6+...+9C1=501$

Edit: I forgot all the different cases for 6,5,..2 different digits

Yeah, that's what I was going to say.  :)

[quote="K81o7"][quote="chess64"][quote="jimhu"][b]Problem 3: [/b]Let $N$ denote number of 7 digit positive integers that have the property that their digits are in increasing order.  Determine the remainder of $N/1000$. (Repeated digits are allowed).[/quote]

Pick the 7 distinct digits and just put them in increasing order. First let's consider the case when there aren't any zeros.

Darn...I have to go. :([/quote]

Same idea, but repeated digits are allowed, right?  Otherwise it's just $\binom{9}{7}$.
[hide]
We may not pick any 0's because that would mean they would have to be at the front, which is impossible.  Thus, we may pick from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}.  Then, let $a_x$ be the number of the digit $x$ which we pick.  We want the number of nonnegative integer solutions to \[ a_1+a_2+...+a_9=7  \] which is $\binom{15}{7}$ (By an objects/dividers argument which I will give if necessary).  For any set of digits we choose, there is exactly one way to order them.  So $N=\binom{15}{7}=6435$, giving us a final answer of $\boxed{435}$[/hide]

Is that right?  Ah meh...I'll just post a problem...

[b] Problem 4: [/b] There are 5 divorced couples who are to be seated in a row of 10 chairs (one person to one chair).  How many ways are there to seat these five divorced couples so that no two ex-spouses are sitting next to each other?[/quote]
Couldn't you imagine this where the tow end chairs are connected., overocouting, and then take away the cases where the couples on the ends invalidates the contraints?

Nice Problem, Solution:
[hide]
Let there be $2$ new pairs formed for every $2$ chairs (1 and 2, 3 and 4, etc).
[b]Each pair needs these conditions:[/b]
1. Adjacent ppl have to be different. 2. The first person cant sit next to his ex in the adjacent pair.
[b]1st Pair: [/b]$10*9-10$ ways, since theres $5$ pairs can't go together.
[b]2nd Pair:[/b] $8*7-6-7$ ways, because theres $3$ couples left, and theres $7$ ways for the first guy to be next to the previous guy from the other pair.

And the same logic goes on for the next 3 pairs. We get:
$(10*9-10)(8*7-13)(6*5-7)(4*3-3)(2*1-1)=\boxed{721080}$.

Apparently this is out of $9!=3,628,800$ possible combinations without restrictions.

[/hide]

The first time you subtract 5, don't you have to subtract 10 instead (because there are 2 permutations for each couple)
And also, on the nth pair, there aren't necessarily 5-n couples left...
I solved this problem a while ago using inclusion exclusion...i'll solve it again tonight and check...

Let's make em harder.

A player starts with A blue cards, B red cards and C white cards. He scores points as he plays each card. If he plays a blue card, his score is the number of white cards remaining in his hand. If he plays a red card it is three times the number of blue cards remaining in his hand. If he plays a white card, it is twice the number of red cards remaining in his hand. What is the lowest possible score as a function of A, B and C and how many different ways can it be achieved?

[quote="jimhu"][quote="K81o7"]But jimhu, at a given time, you can't know how many couples remain...it depends on what you chose.

Here was my solution...
[hide]
Let $f_n(x)$ be the number of ways to arrange $x$ pairs so that exactly $n$ pairs of ex-spouses are adjacent.  Then we want to find $f_0(5)$, and we know \[ \sum\limits_{k-0}{5}{f_k(5)}=10!  \]
Now we will calculate each of $f_k(5)$.  Suppose we're counting the number of ways to have one pair of ex-spouses sitting adjacent.  Then, there are
$\binom{5}{1}$ possible couples to choose
$2^1$ ways to permute the couple.
$9$ possible positions for the couple.
And finally, there are $8!$ permutations for the remaining 8 people.  However, we note that we counted $f_1(5)$ once, but we counted $f_(2)$ twice (each possible pair of couples was counted twice), $f_3(5)$ thrice, $f_4(5)$ four times, and $f_5(5)$ five times.  Thus, we have \[ f_1(5)+2f_2(5)+3f_3(5)+4f_4(5)+5f_5(5)=8!*9*2*5=10!  \]
Thus, \[ f_0(5)=f_2(5)+2f_3(5)+3f_4(5)+4f_5(5)  \]

Going on, we find that if doing the same thing for two couples, there are
$\binom{5}{2}=10$ possible pairs of couples.
$2!=2$ ways to order those couples
$2^2=4$ possible permutations of those couples.
$7+6+5+4+3+2+1=28$ possible positions for the couples.
$6!=720$ permutations of the remaining people
\[ \implies \binom{2}{2}f_2(5)+\binom{3}{2}f_3(5)+\binom{4}{2}f_4(5)+\binom{5}{2}f_5(5)=720*28*4*10*2  \]

\[ \binom{3}{3}f_3(5)+\binom{4}{3}f_4(4)+\binom{5}{3}f_5(5)=\binom{5}{3}3!2^3([5+4+3+2+1]+[4+3+2+1]+[3+2+1]+[2+1]+[1])4!  \]

\[ \binom{4}{4}f_4(5)+\binom{5}{4}f_5(5)=\binom{5}{4}4!2^4([(3+2+1)+(2+1)+(1)]+[(2+1)+(1)]+[(1)])  \]

\[ f_5(5)=5!2^5=3840  \]

We could now just solve using inclusion-exclusion, but...
Plug and chug.  You get \[ f_4(5)=9600  \] \[ f_3(5)=326400  \] \[ f_2(5)=537600  \] \[ f_1(5)=1516800  \]
Then plugging that back into $f_0(5)=f_2(5)+2f_3(5)+3f_4(5)+4f_5(5)$, we have \[ f_0(5)=1234560  \]  Checking with $f_0(5)=10!-f_1(5)-f_2(5)-f_3(5)-f_4(5)-f_5(5)$, we get \[ f_0(5)=1234560  \]
so that is our answer.
Finally, checking with inclusion-exclusion, we have \[ f_0(5)=10!-10!+720*28*4*10*2-10*3!*2^3*35*4!+5*4!*2^4*15-5!2^5  \] \[ =1234560  \]
So our answer is $\boxed{1234560}$, triple-checked.
[/hide]Hmm[/quote]

I actually do think you know which couples remain.. After the first rearranged pair, there will be 3 couples remaining and 2 people who arent. After 2nd arranged pair, there is 1 couple remaining but 4 people who arent. And so on.. I don't think I have a mistake in that...[/quote]

Let's say $A_1$ cannot be adjacent to $B_1$, $A_2$ cannot be adjacent to $B_2$, and so on to $B_5$.  Then, in the first four people, you could have had, say, $A_1, B_1, C_1, D_1$ or a similar set in some order, or could have had $A_1, A_2, B_1, C_1$, or $A_1, A_2, B_1, B_2$....  So you could, by that point, have 3 couples left, 2 couples left, or 1 couple left...


AMC USA(J)MO USAMO inequalities geometry algebra polynomial

What kind of problems will be on this year's USAMO? Here are my predictions:

[hide="Definitely an inequality"]1) Geo
2) Combo
3) Inequality

4) Functional EQ
5) Geo
6) NT
[/hide]

I'm thinking the inequality will be a #2 or #4...definitely a functional EQ and probably heavy on GEO/NT.

[quote="pieterminate"]Hopefully no inequality, no chance I'll solve one :(

BTW, can we use pencils? I keep needing to erase  :oops:[/quote]

Read the manual. As far as I know, pencils are allowed, but pens are preferable (easier to read, especially after faxing). You can also cross things out...

And for questions, does it really matter... I hope there's an inequality, though.

[quote="paladin8"]
 You can also cross things out...[/quote]
Oh, that's good! :)

And of course it matters...I solved very few inequality ones unless they were super easy (2003 #5 :D )

I'm lucky that there should be at least 2 total geometry and inequality problems, at least one of which is likely a #3 or #6. I predict something like

1. geometry
2. inequality
3. combinatorics

4. weird game/invariant thingy
5. NT
6. geometry

I really think one of the #3/#6s will be a combinatorics.

Prediction:

#1 Discrete
#2 Geo
#3 Algebra

#4 Geo
#5 Inequality
#6 Discrete

Prediction Case II:

#1 Geo
#2 Inequality
#3 Discrete

#4 Algebra
#5 Discrete
#6 Geo

You know, it will seriously be

#1 Addition
#2 Subtraction
#3 Multiplication

#4 Division
#5 Linear Algebra
#6 European History

ermm...replace the linear algebra with european history because I dont think that they would go that high of a level this year.

If I am lucky enough to have it the way that beta posted, then I am so PWNING this!

[quote="beta"]1. NT
2.Geo
3. Combo.

4. Algebra(functions/polynomial)
5.Inequality
6. Geo[/quote]
Umm, wow, that's exactly what I was going to say.
Usually 5 is an inequality, 1 is nt, 6 is really hard geometry, ans so on. we have the same predictions

1. Number Theory
2. Classical Geometry
3. Discrete
4. Non-classical Geometry
5. Inequality
6. Discrete

[quote="bubala"]Hmmm... beta's prediction is wrong since 4 and 5 can't BOTH be algebra unless something crazy happens (like last year, when 7 out of the 6 problems were number theory).

But 6 is definitely going to be Romanian history, because that is the only country that matters, and as Valentin will be sure to tell you, Romania invented the USAMO.  Don't ask me how.[/quote]
Fine...

1. Algebra(functions/polynomial)
2.Geo
3. Combo.

4. NT
5.Inequality
6. Geo

Happy?

I disagree. This year there are going to be a huge number of participants; they have been trying to encourage new people to join, etc. So they will definitely not put a functional equation or an advanced polynomial as #1 since only a select few people know how to solve these. So I will take beta's prediction and modify it slightly:

1. Combo
2. Geo
3. Algebra (functions/polynomial)

4. NT
5. Inequality
6. Geo (I would REALLY prefer combo, but this is doubtful.)

2 things:

This year's USAMO was determined just under 1000 years ago (Mongol invasions triggered a reaction in math and science education).

Functional equations are hardly limited to a "select few." In fact, a fairly recent #4 functional equation was the second USAMO problem I ever solved.

[quote="Philip_Leszczynski"]So they will definitely not put a functional equation or an advanced polynomial as #1 since only a select few people know how to solve these. So I will take beta's prediction and modify it slightly:

1. Combo
2. Geo
3. Algebra (functions/polynomial)

4. NT
5. Inequality
6. Geo (I would REALLY prefer combo, but this is doubtful.)[/quote]

Umm... huh? A year ago, problems such as that (functions and polynomials) were the only USAMO problems I could solve. Those are still the easiest for me. I could say the same about a select few people knowing how to solve geometry problems. I like those predictions, though, since there appears no hope of just 1 geometry problem.

Does combo mean combinatorics?

If it were up to me, the test would have 3 Inequalities and 3 Number Theory. Too bad it's not.

Where do those [b]awsome[/b] fun & games things they always have fall under ?  :D

How about a [url=http://www.math.wisc.edu/%7Epropp/srat-Q]self referential[/url] USAMO...

Anyway I would guess:

1. Geometry
2. NT
3. Combinatorics
4. Geo
5. Inequalities
6. Algebra

But are we making educated guesses or are we just finding as many permutations of Algebra, Comb, NT, Geo, and Inequalities as we can think of?  :dry:

Eh 360 possibilities at any rate.

[quote="Sly Si"]In keeping with the spirit of randomdragon's post:

1. The Goldbach Conjecture (prove or disprove)
2. Romanian grammar
3. Presidential trivia

4. Music theory
5. Trivial inequality (disprove)
6. Structural engineering
7. Philosophy ("Prove that this question does not exist, and thus that the 2006 USAMO has only 6 questions.")[/quote]

Music theory,  romanian grammar??? lol :D  This is all math!!!!!!! :lol:

No, it's going to be:

1. NT:    prove fl(ittle)t
2. Algebra:   solve: x=x
3. Geometry:  Prove pythagorean theorem given that $a^2+b^2=c^2$ in a right triangle where c is the hypotnuse while a and b are the other sides.

4. Game thing:  You must contain 100 dogs in bags, and each bag must contain different amounts, what is the max number of bags they can use, given that you can place bags inside each other and each bag may countain anti-matter or dark matter. Source: Mathcounts
5. Inequality: prove or disprove:  $x^2+3xy+x^3+xy^4-xyz+xz+x^4z<yz$ for all $x=0, y=1, z=1$
6. Combinatorics:  How many ways are there to answer this question?

 :lol:

Hmm...I'll give a random guess...

1. NT
2. Geometry
3. Inequality

4. Geometry
5. Algebra
6. Combinatorics

i'm thinking a 4,6 geometry and a 2 or 5 inequality with the other one being a number theory.

[quote]2. Romanian grammar [/quote]

Woohoo! MOP here I come.

Too bad it's going to be Dutch grammar, because they're tricky like that.

No, the rules says that the graders won't grade if in different language so everyone gets 0s:roll:   But music theory certainly counts.  :D

[quote="MysticTerminator"][quote="filletwho"]While they might modify the AIME and AMC to be easier, I doubt they'd do that for the USAMO. It's not like people will feel bad if they can't solve a USAMO problem. Well, maybe...but the whole point of the USAMO is to select the best IMO team, not to make us rookies feel good.[/quote]

hmm, I remember zuming specifically saying at some point how #1 (usually #4 as well) on USAMO are just "gimme"s - useless for determining the team and mostly for MOP as well, so essentially just to make people feel good[/quote]

Ahh, now I feel stupid :(  ;)

hmmm.... what about predicting what someone else would predict?  :rotfl: nah, too late for that. Anyway,

[hide="at least this is hidden"]
1. Polynomial stuff
2. Geometry
3. NT
-----------------------
4. Geometry
5. Inequality
6. Algorithm/game[/hide]

[quote="Go Beyond"]Well, I pull this topic up again. After the test, does anyone feel this year's problems are so weird? In my opinion, #1,2,4,5 are all NT, and 3 is function, 6 is geo. Isn't that weird? (See you guys all predict very diversely) They have the same topic to propose again and again. Also, why there is only one geo this year, and that geo is the very last-hard one? I am quite good at geo, and I hope there is an easy one for 1245, but there isn't. (Most of you guys predict there are 2) :(
Anyway, my NT is not that bad. So I am moderate-lucky this year. :P , and hope my score ranks in at least 50%(about top 200, hope 120) overall.[/quote]

Hmm #1 was NT. #2 was not NT, it was I guess a mixture of Algebra and Combinatorics. #3 was a mix of NT and Algebra (Polynomials). #4 was Algebra, using a small amount of inequalities. #5...this was a strange Algebra/Game Theory/Number Theory kind of thing. And #6 was of course Geo  ;)

Yeah, I agree, I thought they were a bit weird too. It was fun though, I learned what it was like to think about USAMO problems for hours, actually figure out some complicated facts after long, careful thinking, and just break down a whole olympiad. I am very glad that I qualified for USAMO  :yup:

I think i got problems 1,2,4,5 maybe partial credit on 6.

I also think that I lost a few points for some little holes :P


Find the ordered pair of integers $ (x,\ y)$ such that $ 3x^2 \plus{} 4xy \minus{} 4y^2 \plus{} 4x \minus{} 16y \minus{} 28 \equal{} 0$.

[hide="solution"]
$ 3x^2 \plus{} 4xy \minus{} 4y^2 \plus{} 4x \minus{} 16y \minus{} 28 \equal{} 0$
clearly $ x\equal{}2k$, where $ k$ is an integer
$ \Rightarrow 3k^2\plus{}2ky\minus{}y^2\plus{}2k\minus{}4y\minus{}7\equal{}0$

for $ k\equal{}2$ we get $ y\equal{}\pm 3$ for higher positive $ k$ we will not get any solution in integer for y because the Discriminant cant be a perfect square, as the numbers increase the difference between squares increases.

same for negative k

so we get only two pairs of solution $ (4,3)$ and $ (4,\minus{}3)$

[/hide]

[hide]Write as $ 4x^2\minus{}(x\minus{}2y)^2\minus{}4x\plus{}8(x\minus{}2y)\minus{}28\equal{}0$

$ z\equal{}x\minus{}2y \implies 4x^2\minus{}z^2\minus{}4x\plus{}8z\minus{}28\equal{}0 \implies (2x\minus{}1)^2 \minus{} (z\minus{}4)^2\equal{}13$

We thus wish to solve $ u^2\minus{}v^2\equal{}13$ in the integers. We find that $ (u,v) \equal{} (\pm 7 , \; \pm 6)$

Thus, $ x \in \{ \minus{}3,4\}, \; z \in \{ \minus{}2, 10\}$.

Note that $ z$ is even, thus $ x\minus{}2y$ is even, thus $ x$ is even. So, we must have $ x\equal{}4 \implies y\equal{} \pm 3$.[/hide]

The given is equivalent to
$ (3x \minus{} 2y \minus{} 5)(x \plus{} 2y \plus{} 3) \equal{} 13$
One factor is $ \pm13$, the other $ \pm1$, giving $ (4,\pm3)$

Here was how I thought through the factorization:
I saw the first 3 terms could be combined as
$ (3x \minus{} 2y)(x \plus{} 2y)$
by inspection. Now I assumed a factorization in the form of
$ (3x \minus{} 2y \plus{} m)(x \plus{} 2y \plus{} n)$
So we want
$ m \plus{} 3n \equal{} 4$
$ 2m \minus{} 2n \equal{} \minus{} 16$
Solving, $ m \equal{} \minus{} 5$, $ n \equal{} 3$.

Nice. I should explain the motivation behind my factorization, which is the same as the above factorization. The idea was to separate the variables, which could be done by performing a substitution to get rid of the $ xy$ term.


What is the largest prime divisor of 855?


Given a quadrangle $ABCD$ with $H,K$ are the midpoint of $AC,BD$ respectively.
Suppose that $AC$ is the angle bisetor of $\angle BHD$ and $BD$ is the angle bisetor of $\angle AKC$.
Prove that $KA+KC=HB+HD$

Look at http://www.mathlinks.ro/Forum/viewtopic.php?t=23761 .

  darij

Thanks Darij !
The quadrangle in this problem is really a harmonic cyclic one  :D I saw it now

let $AC$ and $BD$ meets each other in point $Q$ so we will have:
\[\frac{AQ}{QC}=\frac{AK}{KC} \rightarrow
\frac{AH-QH}{AH+QH}=\frac{AK}{KC} \rightarrow
AK.AH+AK.QH=KC.AH-KC.QH \rightarrow AH(KC-AK)=QH(AK+KC)\]
and we have : \[ QH=\frac{|AK^2-KC^2|}{2AC} \]
$\rightarrow$ $ \frac{|AK^2-KC^2|}{2AC} \times (AK+KC)=AH(KC-AK) $ we suppose that $KC-AK$ is not $0$ $\rightarrow
 (AK+KC)^2=AC^2 \rightarrow AK+KC=AC $ but this is impossible because $AK+KC>AC$ $\rightarrow  AK=KC \rightarrow QH=0$
$\rightarrow$ $H$ is the midpoint of $AC$ and $BD$ and $AB=BC=CD=DA$ and $AB \parallel DC$ and $DA \parallel BC$
and now if we prove that one of this quadrilateral angles is $90$ my solution is complete. i think this quadrilateral is an inscribed quadrilateral and if it is an inscribed quadrilateral we r done.


I know the basic PIE formula such as finding the total number of elements in at least one of the sets. But, is there a formula to calculate how many things are in at least k (k>1) out of the n sets? How about the number of things that are in exactly k out of n sets?


inequalities
group theory
abstract algebra
linear algebra
matrix
vector
absolute value

Here is a lemma that appears in a proof of the famous theorem of Jordan-Schur:
  Any finite subgroup of $GL_n(C)$ has an abelian normal subgroup of index at most $ (\sqrt{8n}+1)^{2n^2}-(\sqrt{8n}-1)^{2n^2}$. (this is the theorem, but don't try to do this at home!!!; yet, in the following days I will post the steps that will allows us to have on the forum the proof of this magnificient theorem)
  The statement of the lemma is:
  If $ A,B$ are unitary matrices such that $||I-B||<2$  (here $||A||=\sqrt{\sum_{i,j}{|a_{ij}|^2}}$) and $A$ commutes with $ABA^{-1}B^{-1}$ then $A$ commutes with $B$.
  So, what do you say? Can we find a simple proof? The proof I have is not that simple.

[quote="harazi"] lemma :
  If $ A,B$ are unitary matrices such that $||I-B||<2$  and $A$ commutes with $ABA^{-1}B^{-1}$ then $A$ commutes with $B$.
  [/quote]

There is a similar one with conditions on the arguments of eigenvalues of B

Lemma: $A,B$ unitary matrices, $P_{B}(x)=(x-e^{ib_1})...(x-e^{ib_n})$
with $0\leq b_1\leq b_2\leq ...\leq b_n<2\pi$
suppose that $b_n-b_1<\pi$ and $A$ commute with $ABA^{-1}B^{-1}$

Then $AB=BA$

What we want to prove is (after some algebraic manipulation) that if $A,B$ are unitary, $\|I-B\|<2$, and $A$ commutes with $BAB^*$, then it commutes with $B$. To simplify matters a bit, notice that this ammounts to proving the assertion for an orthogonal projection $P$ instead of $A$: if $A=\sum_{i=1}^t\lambda_iP_i$ is the spectral decomposition of $A$, then $\sum_{i=1}^t\lambda_iBP_iB^*$ is the spectral decomposition of $BAB^*$, and a matrix commutes with $A$ iff it commutes with every $P_i$.

Since $P,BPB^*$ commute and they're projections of the same rank $r$, they can be simultaneously unitarily diagonalized, so we may as well assume that $P=\mbox{diag}(1,1,\ldots,0,0)$ ($r\ 1$'s), and $BPB^*$ also has $r$ entries equal to $1$ on its main diagonal, with the rest of its entries equal to $0$. We have to prove that if in addition to this, we also have $\|I-B\|<2\ (*)$, then these $r$ entries are the first $r$ entries (i.e. $BPB^*=P$).

Let $B_1$ be the submatrix formed by the first $r$ rows and $r$ columns of $B$, and $B_2$ the one formed by the last $n-r$ rows and $n-r$ columns (we disregard the trivial case $r=n$, when $P=I$). If $B=(b_{ij})$, then $(*)$ translates to $\mbox{Re}(\mbox{tr}\ B)>n-2$. Then at least one of the inequalities $\mbox{Re}(\mbox{tr}\ B_1)>r-1,\ \mbox{Re}(\mbox{tr}\ B_2)>(n-r)-1$ holds. Let's assume it's the first one (the other case is solved in a similar fashion). The non-zero row vectors of $B_1$ have norm $1$ (this follows from the fact that the matrices $P,BPB^*$ have the special forms described above). The absolute value of an entry of the form $(i,i)$ of $B_1$ is $\le 1$, and it's positive only if the corresponding row is non-zero. Since the sum of all the absolute values of entries $(i,i)$ is $>r-1$, all row vectors must be non-zero, so $B_1$ must be invertible. This means that $B_1B_1^*=I_r$, so the first $r$ entries of $BPB^*$ are $1$, as desired.

It's not exactly imple :).


Say a, b and c are the sides of a triangule. Prove:

$\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}<\frac{1}{16}$

The stronger inequality $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}<\frac{1}{20}$ holds and was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=22786 .

  darij


calculus
integration
logarithms
calculus computations

hello, alternatively you can set $ u\equal{}\sqrt[3]{\frac{x\minus{}1}{x}}$, from here we get $ x\equal{}\frac{1}{1\minus{}u^3}$ and
$ dx\equal{}\frac{3u^2du}{(1\minus{}u^3)^2}$ and we have to integrate $ 3\int\frac{u^2}{(1\minus{}u^3)(2\plus{}u)}\,du$.
Sonnhard.


Find all positive integers n such that:
sqrt(x-1)+sqrt(x-2)+...+sqrt(x-n)=sqrt(2003-x) has at least a real solution.

bye bye!!!

This is from Gazeta Matematica, isn't it? Anyway, it looks tough.

Yes that's right 3rd issue 2003!!!

I struggled on to this one and got nowhere.. perhaps you'll get somewhere!!!

cheers! :D :D

I think this was solved at the yahoo - group "imo - problems".

http://groups.yahoo.com/group/imo-problems/message/1654?expand=1

Is there an elementary solution?

cheers! :D :D :D


How many zeros are there in the product of $ 2^4$ and $ 5^5$?

This is equivalent to asking how many 10's we have in the factorization. Each 10 uses a 2 and a 5, so we can make 4 10's and thus the number will have $ \boxed{4}$ zeros at the end of it.

I still don't quite understand that..  :?:

Note that the only way we can "make" a $ 0$ is if we havew a factor of $ 2$ and $ 5$ (since $ 2(5)\equal{}10$). Thus, since there are $ 5$ factors of $ 5$, and only $ 4$ factors of $ 2$, $ \boxed{4}$ is the answer.


I know this can be a very touchy subject for many users on this forum but what is your belief on the status of Taiwan? Do you think it is a part of China and can be considered a special region like Hong Kong or Macau but radically different due to the difference of the beliefs of people there or do you think it should be its own country? Also, do you think it is right with Taiwan advertising a planned military demonstration on television and then doing it to show off their military prowess?

There has been a post discussing it in the Round Table. In that poll, the result seems to be that 50% say that Taiwan is independent and 40% say that Taiwan is a part of China while the rest say that they did not know about Taiwan.

You can try to do a search and discuss in that thread.

Communists actually do not believe in the concept of countries, but if I choose one, I choose that Taiwan belongs to China. It was separated just to keep old CKS in power with US aid.

now the result at  http://www.mathlinks.ro/Forum/viewtopic.php?t=20139 has changed.49% people think it is,and 41% think not,the others don't know about it.

i want to say :"certainly,it is."

Taiwan is a part of China since the old time.We chinese all know about it.Maybe you（non Taiwanese/Chinese） don't know about old time's,i will show you in the history book.maybe you can search it in Google!

here is the map about china.

Well I don't really mean what "being part of " means?  Is Belgium part of the Roman Empire?

The only thing that matters is who rules (centralized rule or not)

Therefore a couple of questions if I may

1) Can the Republic of China be seen as democracy?  Has Chang Kai Shek been elected for every time he was president?

2) How much of an influx was there of immigrants from the mainland when it became obvious the mainland was lost to the communists.  Once in a while I hear a critical voice deploring the suppression of locals by those from the mainland, and I would like to know more.

[quote="Ignite168"][quote="Totally Zealous"]Everyone non Taiwanese/Chinese says Taiwan isn't a part of China.
Everyone Taiwanese/Chinese says it is.

Since I'm Taiwanese/Chinese I'm saying that it is.

You don't need a passport to go to Taiwan or China if you Taiwanese/Chinese, so taiwan is a part of china.[/quote]

That's incorrect, no planes can go from China to Taiwan, (I believe a cargo plane did for the first time weeks ago) they have to go through Hong Kong first. You're going to need a visa or a special permit for leisure to go to Hong Kong and Taiwan but I hear you can take a boat from Xiamen to Taiwan with a special permit as well. Also, many Taiwanese people believe it isn't a part of China. Also, history isn't a good factor in determining borders, just look at the Middle East.[/quote]

Ignite168, I think that during the Chinese New Year, China allows people to travel to and from Taiwan. From Wikipedia, I have read that the ROC was not a democract until 1975 when Chiang Kai Shek died. From 1949 to 1975 it was a single-party rule.

I think that Taiwan was single party until 1987.

The official documents (told to us by the school Chinese teacher, whose ex-wife is Taiwanese) say that China is the land surrounded by mountains, rivers, and seas. Therefore it does not include Taiwan.

[quote="sapphyre571"]China is the land surrounded by mountains, rivers, and seas. Therefore it does not include Taiwan.[/quote]
If so, then China will lose all its islands, including Hai Nan.

[quote="andy_tok"]Taiwan is definitely a part of China, although it is separated mainly due to US's strategy in Asia-pacific.
Even Chiang Kai Shek didn't allow separating Taiwan from main land China.[/quote]

Now what was [i]THAT[/i]?
 :rotfl:

Even old CKS will protest against [i]this[/i] one!

Bubka please, that's not so crazy.  Both of them portrayed the other as a thug stealing land from the other.  CKS still claimed to be the ruler over mainland China and Taiwan together, even though he only controlled Taiwan.  For a long time (until 1971) the United Nations, who had to give a UN seat to China, UK, France, USSR and USA, went along with this and gave a UN Security Council seat to someone from Taiwan.

Another example is that CKS considered Nanjing the capital from 1928-2002, even though he didn't control that city after the Civil War.

Both sides still dream of "liberating" the other side and re-unification.  However the young generation of Taiwan has never been in China, feels rather Taiwanese, and therefore that might change.


symmetry
inequalities proposed
inequalities

Let $ x,\ y,\ z$ be non negative integers such that $ x\plus{}y\plus{}z\equal{}3k$ for some integer $ k\geq 1$.

Find the minimum value of $ x!y!z!$.

The minimum is $ (k!)^3$.

[i]Proof:[/i] Let $ (x,y,z)$ be a valid choice for which the minimum is attained. Suppose that we can find $ u,v$ among $ x,y,z$ with $ u>v\plus{}1$, then let $ w$ denote the remaining number and set $ u'\equal{}u\minus{}1,v'\equal{}v\plus{}1,w'\equal{}w$. This is a valid choice as well and we get $ u'!v'!w'!\equal{}u!v!w!\cdot\frac{v\plus{}1}{u}<u!v!w!$, contradicting the minimality. Hence, if we assume by symmetry that $ x$ is the smallest number, we must have $ y\minus{}x,z\minus{}x\in\{0,1\}$. This implies $ 0\equiv 3k\equiv x\plus{}y\plus{}z\equiv 3x\plus{}(y\minus{}x)\plus{}(z\minus{}x)\equiv (y\minus{}x)\plus{}(z\minus{}x)\bmod{3}$ and we get $ x\equal{}y\equal{}z$.


in my math team, a problem that our teacher posed to us was as follows,
find the first few lowest integers that are not the sum of 3 whole numbers
for example, 7 cannot be the sum of 3 perfect squares (knowing this by brute force)
the numbers do not need to be distinct, and they can be 0
is there a general way to do this? my teacher said that he did not know how to do this without brute force

A positive integer can be written as the sum of three perfect squares if and only if it is not of the form $4^k(8t + 7)$ where $k$ and $t$ are nonnegative integers.

care to explain why that is true?...you are better at math than our math team coach...sad for me

Well, first, if an integer is the sum of three squares, then it cannot be of that form.
Suppose that \[x^2 + y^2 + z^2 = 4^k(8t + 7).\] Looking mod 4, we get that $x$, $y$ and $z$ must be even, and hence we get a new equation to solve: \[x^2 + y^2 + z^2 = 4^{k - 1}(8t + 7).\] Etcetera, etcetera (descent) and we need to show that there do not exist $x$, $y$, $z$ such that \[x^2 + y^2 + z^2 \equiv 7\textrm{(mod 8)}\] and that is obvious since squares are 0, 1 or 4 mod 8.

The other direction is very hard. The proof is quite advanced, I don't remember who proved the theorem, but I won't write the proof here. You can find it an any good textbook on number theory.

Adding to Arne: Taking mod 8 one can  right away see that any number which is  $7 \bmod 8$ can not be written as sum of three squares.
Because
$n^2 \bmod 8$ can only be $0,1 \ or 4$ and you can not get 7 from 3 such numbers.

Going the other way :

It is also fairly easy to prove that any[b] prime [/b] not of the form $8k+7$  can be written as sum of three squares.
There are some other numbers which are not of the form $8k+7$ (eg $28$)  yet   can not be written as sum of  three squares.... as mentioned above all numbers not of the form $4^n(8k+7)$ can be written as sum of three (or less) squares.

For some more details please see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=triangular&t=5269]This AoPS thread.[/url]

i have to say, that is quite clever, but how do you figure in the 4^n, i know that trial and error can show you this, like with 28, but i don't see why, does it have something to do with looking at in mod 4?

The $4^n$ again has to do with the descent Arne was talking about.  Essentially you prove the base case, that is that the number cannot be of the form $8k+7$ and then multiplying both sides by $4$ and finding another set of three squares that don't work.  The $4^n$ simply accounts for all cases.

ok, then the next part of this problem that i got regards the [b]sum of 4 squares[/b], are there any numbers that cannot be the sum of 4 squares (not necessarily distinct), i think that every number can be the sum of 4 squares because you know the numbers that cannot be sum of 3 squares are spaced out and the number just below it can be the sum of 3 squares, so adding by 1^2 would fill in all the possibilities
like 7 cannot be sum of 3 squares, but 1^2 + 1^2 + 2^2 = 6 can so if it is the sum of 4 squares, it would just be
7 = 1^2 + 1^2 + 1^2 + 2^2, and similarly, all other cases would follow

So long as you allow 0 to be one of the squares when needed I believe it is possible.  In fact any number that has factors that can be expressed as the sum of four squares can itself be expressed as the sum of four squares using the identity
$(a^2+b^2+c^2+d^2)(e^2+f^2+g^2+h^2)=$
$(ae+bf+cg+dh)^2+(af-be+ch-dg)^2+(ag-bh-ce+df)^2+(ah-de+bg-cf)^2$