modular arithmetic
greatest common divisor

$ 2007$ integers are arranged on a circle such that for any five successive numbers, the sum of some three of them is twice the sum of the other two. Show that all the numbers are zero.

Let an arbitrary set of 5 numbers have $ a,b,c,d,e$ such that $ a \plus{} b \plus{} c \equal{} 2d \plus{} 2e$ satisfies the condition. Then $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 3d \plus{} 3e$ shows that the sum of any 5 consecutive numbers is divisible by 3.

Let the 2007 integers be $ a_1, a_2, \ldots a_{2007}$. Consider the first 6 numbers. Since $ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5$ and $ a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5 \plus{} a_6$ are both divisible by $ 3$, we get $ a_1 \equiv a_6 \pmod{3}$. This can be generalized to show that $ a_k \equiv a_{k \plus{} 5} \pmod{3}$. But since
$ (2007,5) \equal{} 1$, all numbers in the circle are equivalent $ \mod 3$.

Let $ x$ be such that all the numbers in the circle are equivalent to $ x \mod 3$. If we consider any 5 numbers, we have that the sum of 3 is equal to double the sum of 2. Written mod 3, this gives $ 3x \equiv 4x \pmod{3}$, so therefore $ x \equiv 0 \pmod{3}$ and all numbers in the circle are divisible by 3.

Suppose some numbers in the circle are nonzero and satisfy the conditions. Divide all numbers by the highest power of three dividing the greatest common divisor of all nonzero numbers. Such a number set should still satisfy the conditions, but now at least one number is not divisible by three. Contradiction


What is the Harpoon Mechanism? It has something to do with ionic bonding.


Is there a way to get some answers for the Aops book volume 1?  I cant find the solution book right now and i need to check some answers (not including proofs). i need answers for 360-400. thanks.

PS the font for everything on AoPS looks really small on my computer. looks like 6 font. how can i change this?

On my computer, you can change the font size by holding down "control" and spinning the little wheel on the mouse.


a is right. b and c do not appear to be correct. dunno about the others yet.

EDIT: d is correct

hint for b): $p(p-1)\binom{n}{p}=n(n-1)\binom{n-2}{p-2}$
hint for c): use above hint and $p^{2}=p(p-1)+p$

[hide="(b)"]Let $S$ be the desired sum.  We have

\begin{eqnarray*}S&=&\sum_{k=1}^{n-1}k(k+1)\binom{n}{k+1}\\ &=&2\sum_{k=1}^{n-1}\frac{k(k+1)}{2}\binom{n}{k+1}\\ &=&2\sum_{k=1}^{n-1}\binom{k+1}{2}\binom{n}{k+1}\end{eqnarray*}

The expression $\binom{k+1}{2}\binom{n}{k+1}$ is equivalent to forming a $k+1$-person committee from $n$ people, and then choosing two people already on the committee to be leaders.  But we could also form this committee by first choosing the two leaders, which can be done in $\binom{n}{2}$, and then choosing the remaining $k-1$ people to fill out the committee, which can be done in $\binom{n-2}{k-1}$ ways.  Therefore, $\binom{k+1}{2}\binom{n}{k+1}=\binom{n}{2}\binom{n-2}{k-1}$ (This is equivalent to the hint B23 gave).  Applying this identity, we obtain

\begin{eqnarray*}S&=&2\sum_{k=1}^{n-1}\binom{n}{2}\binom{n-2}{k-1}\\ &=&2\binom{n}{2}\sum_{k=1}^{n-1}\binom{n-2}{k-1}\\ &=&2\binom{n}{2}2^{n-2}\\ &=&2^{n-1}\binom{n}{2}\end{eqnarray*}[/hide]

[hide="part a"] Using Binomial Theorem, the sum is $(1-1)^{n}=0$[/hide]
[hide="part b"] Canceling terms in the numerator and denominator of each term, we get $n(n-1)+\frac{n(n-1)(n-2)}{1!}+\frac{n(n-1)(n-2)(n-3)}{2!}+...$
$=n(n-1)[1+{n-2\choose 1}+{n-2\choose 2}+...]=n(n-1)2^{n-2}={n\choose 2}2^{n-1}$.[/hide]

[hide=" :D $_{(C)}$"]
Notice it is just: \[(b)+\sum k{n \choose k}\] And we us $\sum k{n \choose k}=\sum (n-k){n \choose k}\implies \sum k{n \choose k}=n2^{n-1}$
So our answer is: \[\left( n+{n \choose 2}\right) 2^{n-1}={ n+1 \choose 2}2^{n-1}\] [/hide]

Another method is to write down the expansion of (1+x)^n and differentiating or integrating as required.

I got something different for f

[hide] We note that $\frac{2}{(j)(j+1)}\binom{n}{j-1}= \frac{2}{(n+1)(j+1)}\binom{n+1}{j}= \frac{2}{(n+1)(n+2)}\binom{n+2}{j+1}$

Substituting into $\sum_{j=1}^{n+1}(-1)^{j}\frac{\binom{n}{j-1}}{\frac{1}{2}(j)(j+1)}$, we obtain

$\frac{2}{(n+1)(n+2)}\sum_{j=1}^{n+1}(-1)^{j}\binom{n+2}{j+1}\\ \\ = \frac{2}{(n+1)(n+2)}[-(0-\binom{n+2}{1}+\binom{n+2}{0})] \\ \\ = \frac{2}{(n+1)(n+2)}[n+1] \\ \\ = \frac{2}{n+2}$.
[/hide]
EDIT: OH NOOOOO by trying a few small cases, I conclude that I am missing a negative sign somewhere. someone help me?

The negative shouldnt be inserted in line 3----> line 4

Hm lemme see...

 $\sum_{j=1}^{n+1}(-1)^{j}\binom{n+2}{j+1}+\binom{n+2}{1}-\binom{n+2}{0}= \sum_{j=0}^{n+2}(-1)^{j-1}\binom{n+2}{j}=-\sum_{j=0}^{n+2}(-1)^{j}\binom{n+2}{j}= 0$.

So in conclusion,  $\sum_{j=1}^{n+1}(-1)^{j}\binom{n+2}{j+1}=-[0-\binom{n+2}{1}+\binom{n+2}{0}]$

OH NO

EDIT: there were some silly errors that the brilliant Scrambled may or may not have helped me smooth out by shouting "wtf are you smoking."

Your formula doesn't work, farenjaht (just try some cases). In particular, why do you have $\frac{1}{k+1}$ in the denominator? Note that $\sum_{1 \leq k \leq j }k = \frac{j(j+1)}{2}$.

Phelpedo, I would like to turn your attention to the fact that I have solved problem [b](e)[/b], not problem (f). Thanks :)

Oh wow, they look so similar in summation form  :blush:


Three-digit powers of 2 and 5 are used in this cross-number
puzzle. What is the only possible digit for the outlined square?
\begin{align*}
&\text{ACROSS} &\text{DOWN}\\
&\textbf{2.}\quad 2^m &\textbf{1.}\quad 5^n\\
\end{align*}

[asy]/* AMC8 2000 #9 Problem */
draw((0,0)--(1,0)--(1,3)--(0,3)--cycle);
draw((0,2)--(3,2)--(3,1)--(0,1));
draw((2,2)--(3,2)--(3,1)--(2,1)--cycle, linewidth(2));
label("1", (0,3), SE);
label("2", (0,2), SE);[/asy]

[quote="FantasyLover"]1 would be 125, and 2 would be 256.

the number for outlined square is 6.

answer : 6[/quote]
Or 1 could be 625, but it doesn't matter since all (sufficiently large) powers of 5 end with 25.


If $\frac{AP}{PB}=\frac34$ and $\frac{AR}{RC}=\frac32$, prove that $C$ is the midpoint of $BZ$. (Refer to the attached diagram).

[hide="Mass Point Solution"]
Assign a weight of 4 to A and 3 to B. This means that P has a weight of 7.
Assign a weight of 6 to C. This means that R has a weight of 10.
[/hide]

Before I continue, I would like to know what that $1$ is doing above BC. Is $BC=1$?

i belive this is a direct application of menelaus

[quote="Hikaru79"][quote="chess64"]
Before I continue, I would like to know what that $1$ is doing above BC. Is $BC=1$?[/quote]
$1$ is the area of the trapezoid $PRBC$[/quote]

Well that could make this a little too tricky for mass points, especially considering that PRBC isn't a trapezoid


let P be a inner point in triangleA'B'C',let a,b,c be the sides of triangleABC,prove that
$\frac{PA'^{2}}{b^{2}+c^{2}}+\frac{PB'^{2}}{c^{2}+a^{2}}+\frac{PC'^{2}}{a^{2}+b^{2}}\leq \frac{1}{2}$

[quote="ityaa"]let P be a inner point in triangleA'B'C',let a,b,c be the sides of triangleABC,prove that
$\frac{PA'^{2}}{b^{2}+c^{2}}\frac{PB'^{2}}{c^{2}+a^{2}}\frac{PC'^{2}}{a^{2}+b^{2}}\leq \frac{1}{2}$[/quote]

Something must be wrong here. PA', PB', PC' can be arbitrarily small compared to a, b, c.

  Darij

[quote="ityaa"]something wrong,I have edited it.
another problem is:
let P be a inner point in triangleABC,let a,b,c be the sides of triangleABC,prove that
$\frac{PA^{2}}{b^{2}+c^{2}}+\frac{PB^{2}}{c^{2}+a^{2}}+\frac{PC^{2}}{a^{2}+b^{2}}\leq \frac{1}{2}$[/quote]
Let $P=B$ or tend.
it is obvious wrong...


Elegant solution, bruno.  Fermat-Euler, you shouldn't hide the problem...

[quote="boxedexe"][quote="Bruno Bonagura"][hide]
$A = \left(\frac{10^6 - 1}{9}\right)^2$[/hide][/quote]

You have a type here.  It should be $A = \left(\frac{10^6 - 1}{3}\right)^2$.  However, nice job. :)

Masoud Zargar[/quote]

oh... thx
I've fixed it!


Let $a,b,c$ are real number such that: $a^{2}+b^{2}+c^{2}=2(ab+bc+ca)$. Find the max and min of expression:
$H(a,b,c)=\frac{a^{3}+b^{3}+c^{3}}{(a+b+c)(a^{2}+b^{2}+c^{2}}$

From given condition we get $(b+c-a)^{2}=4bc \rightarrow bc$ is positive.Similarly $ac$ and $ab$ is positive.So $a,b,c$ are all positive or all negative. WLOG let all be positive and from simmetry let $a\geq b \geq c$.Then from given condition $a=(\sqrt{b}+\sqrt{c})^{2}$ and $a^{3}+b^{3}+c^{3}=\frac{(a+b+c)(a^{2}+b^{2}+c^{2}}{2}+3abc$.Then
$H(a,b,c)=\frac{a^{3}+b^{3}+c^{3}}{(a+b+c)(a^{2}+b^{2}+c^{2}}=\frac{1}{2}+\frac{3abc}{(a+b+c)(a^{2}+b^{2}+c^{2}}\geq \frac{1}{2}$.With equality $a=0,b=c$
Let's prove that $(a+b+c)(a^{2}+b^{2}+c^{2}) \geq 27abc$
$S=((\sqrt{b}+\sqrt{c})^{2}+b+c)((\sqrt{b}+\sqrt{c})^{4}+b^{2}+c^{2}) \geq 27(\sqrt{b}+\sqrt{c})^{2}bc$
From AM-GM inequality
$S=(\frac{(\sqrt{b}+\sqrt{c})^{2}}{4}+...+\frac{(\sqrt{b}+\sqrt{c})^{2}}{4}+b+c)(\frac{(\sqrt{b}+\sqrt{c})^{2}}{16}+...\frac{(\sqrt{b}+\sqrt{c})^{2}}{16}+b^{2}+c^{2}) \geq 6\sqrt[6]{\frac{(\sqrt{b}+\sqrt{c})^{8}}{4^{4}}ab}.18\sqrt[18]{\frac{(\sqrt{b}+\sqrt{c})^{64}}{4^{32}}a^{2}b^{2}}\geq 27(\sqrt{b}+\sqrt{c})^{2}bc$
$\Longrightarrow \frac{(\sqrt{b}+\sqrt{c})^{52}}{4^{26}}\geq a^{13}b^{13}$
So $H(a,b,c) \leq \frac{11}{18}$
Equality holds when $a=2b=2c$


[quote="Silverfalcon"]A trapezoid has three sides of length 2 and one side of length 1. Find its area.[/quote]

[hide]Since CD=2 and EF=1, DE=.5

therefore, $AE=\sqrt{2^2-.5^2}=\sqrt{3.75}$

Move BFC so side BC is on AD.

Now we have a rectangle

$1.5\sqrt{3.75}$ is our answer.[/hide]

you should probably put:

$\frac{3}{2} \cdot \frac{5\sqrt{15}}{2} = \frac{15\sqrt{15}}{4}$

just to be more accurate and simple looking.

[hide]The height would be :sqrt: (4-1/4)= :sqrt: (15/4)= :sqrt: (15)/2 as opposed to 5 :sqrt: (5)/2 as posted by xxreddevilzxx. Therefore, the area is 3 :sqrt: (15)/4.[/hide]


modular arithmetic
algebra
system of equations
number theory proposed
number theory

Determine all pairs $ (a,b)$ of positive integers, each integer having two decimal digits, such that $ 100a\plus{}b$ and $ 201a\plus{}b$ are both perfect squares.

[hide]
Let $ 100a\plus{}b\equal{}m^{2},201a\plus{}b\equal{}n^{2}$ so that
\[ n^{2}\minus{}m^{2}\equal{}101a\equiv 0\pmod{101}\Rightarrow (n\plus{}m)(n\minus{}m)\equiv 0\pmod{101}\]
However, we have $ 0<a,b\leq 99$ so $ m^{2}\leq 101*99<100^{2}\Rightarrow m\leq 99$ and $ n^{2}\leq 202\cdot 99<151^{2}\Rightarrow n\leq 150$. Also, $ n>m$.

If $ n\minus{}m$ were divisible by 101, then we would have $ n\minus{}m\equal{}101$ and $ a\equal{}n\plus{}m\equal{}101\plus{}2m$, which is not possible. So $ n\plus{}m$ is divisible by 101. We then have either $ n\plus{}m\equal{}101$ or $ n\plus{}m\equal{}202$. In the first case, $ a\equal{}n\minus{}m\equal{}101\minus{}2m$ and
\[ b\equal{}m^{2}\minus{}100a\equal{}m^{2}\plus{}200m\minus{}10100\equal{}(m\plus{}100)^{2}\minus{}20100\]
So $ (m\plus{}100)^{2}>20100>141^{2}$ and $ (m\plus{}100)^{2}<20200<143^{2}\Rightarrow m\equal{}42$, giving $ (17,64)$.

In the second case, $ 2a\equal{}n\minus{}m\equal{}202\minus{}2m$, so $ a\equal{}101\minus{}m$. Then
\[ b\equal{}m^{2}\minus{}100a\equal{}m^{2}\plus{}100m\minus{}10100\equal{}(m\plus{}50)^{2}\minus{}12600\]
So $ (m\plus{}50)^{2}>12600>112^{2}$ and $ (m\plus{}50)^{2}<12700<113^{2}$, so no solution.

The unique solution is therefore $ (17,64)$.
[/hide]

Sorry to revive but the problem is written wrong, it actually puts the restrinction that both $n^2$ and $m^2$ are four digit numbers. Here is the original paper (in Spanish). http://www.oei.es/oim/xixd2.PDF

Above: Yes, the problem does say that $n^2$ and $m^2$ must be four digit numbers. However, this information is unnecessary because, as established by the answers above, the only solution gives $n^2$ and $m^2$ with four digits. Indeed, $a = 17$ and $b = 64$ yields $100a + b = 42^2 = 1764$ and $201a + b = 59^2 = 3481$.

[quote=orl]Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=20109]Solar Plexsus:[/url]

Consider the Diophantine system of equations

$ (1) \;\; 100a \plus{} b \equal{} m^2$

$ (2) \;\; 201a \plus{} b \equal{} n^2$

where $ 10 \leq a,b < 100$ and $ 32 \leq m < n < 100$. By subtracting (1) from (2), we obtain

$ (3) \;\; 101a \equal{} n^2 \minus{} m^2 \equal{} (n \minus{} m)(n \plus{} m)$.

Since 101 is a prime and $ 0 < n \minus{} m < 101$, $ 101 | n \plus{} m$ by (3). This combined with the fact that $ 0 < n \plus{} m < 2*101$ implies that $ n \plus{} m \equal{} 101$, i,e $ n \equal{} 101 \minus{} m$. According to (3)

$ 101a \equal{} (101 \minus{} m)^2 \minus{} m^2 \equal{} 101^2 \minus{} 2 \cdot 101m$,

i.e.

$ a \equal{} 101 \minus{} 2m$.

From (2) we obtain

$ b \equal{} n^2 \minus{} 201a \equal{} (101 \minus{} m)^2 \minus{} 201(101 \minus{} 2m) \equal{} m^2 \plus{} 200m \minus{} 10100.$

Now $ 10 \leq b < 100$ implies that $ m \equal{} 42$, thus

$ 100a \plus{} b \equal{} 42^2 \equal{} 1764$.

by (1). In other words, the only solution of system of equations (1)-(2) is $ (a,b) \equal{} (17,64).$

Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=2975]jmerry:[/url]

The only such pair is $ a \equal{} 17,b \equal{} 64$, with $ 100a \plus{} b \equal{} 42^2 \equal{} 1764$ and $ 201a \plus{} b \equal{} 59^2 \equal{} 3781$

Let $ u^2 \equal{} 201a \plus{} b$ and $ v^2 \equal{} 100a \plus{} b$. Then by the four-digit restriction, $ 32\le u < v \le 99$. Now $ 101a \equal{} u^2 \minus{} v^2 \equal{} (u \minus{} v)(u \plus{} v)$. Since 101 is prime and $ u \minus{} v < 101$, $ 101|u \plus{} v$. Also, $ u \plus{} v < 202$, so $ u \plus{} v \equal{} 101$, leaving $ u \minus{} v \equal{} a$.
Now we have $ u \equal{} \frac {101 \plus{} a}{2}$, so
$ \left(\frac {101 \plus{} a}{2}\right)^2 \equal{} 201a \plus{} b$
$ 101^2 \plus{} 202a \plus{} a^2 \equal{} 804a \plus{} 4b$
$ 301^2 \minus{} 101^2 \plus{} 4b \equal{} a^2 \minus{} 602a \plus{} 301^2 \equal{} \left(301 \minus{} a\right)^2$
Therefore $ 200*402 \plus{} 4b \equal{} 4(100*201 \plus{} b)$ is a perfect square. Since squares greater than 10000 are at least 200 apart, there is at most one solution. We have $ \sqrt {100*201 \plus{} b} > 100\sqrt {2}$ by a small margin; guessing $ \sqrt {100*201 \plus{} b} \equal{} 142$ gives $ 142^2 \equal{} 20164 \equal{} 100*201 \plus{} 64$, so $ b \equal{} 64$ and $ 301 \minus{} a \equal{} 2*142 \equal{} 284$ so $ a \equal{} 17$.[/quote]

but if a = b = 99, then n > 100 ...


If,
$ \int \frac{dx}{(\sec x \plus{} \csc x \plus{}\tan x \plus{} \cot x)^2} \equal{} \frac{x}{a} \plus{} \frac{\sqrt2 \cos \left(x \plus{} \frac{\pi}{4}\right)}{b} \plus{} \frac{\cos 2x}{c} \plus{} d$,

then, $ a\plus{}b\plus{}c \equal{} ?$

hello
convert everything in the form of $ sin(x)$ and $ cos(x)$
ur question now will be::
$ \int \frac {dx}{(\frac {sinx \plus{} sin^2x \plus{} cosx \plus{} cos^2x}{sinxcosx})^2}$
=$ \int \frac {sin^2xcos^2xdx}{(sinx \plus{} cosx \plus{} 1)^2}$
=$ \frac {x}{2} \minus{} \frac {cos2x}{8} \plus{} \frac {cos(x \plus{} \frac {\pi}{4})}{\sqrt(2)}$+constant.

now u can find the value.
thank u

[quote="Conjurer"]I would just differentiate both sides, and try some values for x and solve the linear equations :D[/quote]

seems like a good idea

I couldn't find the Physics in the subject.

[quote="kabi"]
$ \int \frac {dx}{(\frac {sinx + sin^2x + cosx + cos^2x}{sinxcosx})^2}$

=$ \int \frac {sin^2xcos^2xdx}{(sinx + cosx + 1)^2}$

[/quote]

 :fool:
I forgot, I mean completely overlooked the square thing:

$ \int \frac{dx}{(\sec x + \csc x +\tan x + \cot x)^\color{red}2}$

rajiv,this forum is exclusively for physics problems. please post this in the iit jee thread.

[quote="aadil"]
rajiv,this forum is exclusively for physics problems.

[/quote]
I know! This is the consequence of having too many tabs open in one window!

[quote="aadil"]
please post this in the iit jee thread.

[/quote]
Right away, Sir!

But need not necessarily be the JEE thread. It'd better be in the Calculus section.


modular arithmetic
number theory proposed
number theory

(n+1)^k -1=n!
Find all solutions in positive integers to this equation.

[quote="sandu2508"][quote="Ulanbek_Kyzylorda KTL"]$ (n \plus{} 1)^k \minus{} 1 \equal{} n!$
Find all solutions in positive integers to this equation.[/quote]
Suppose that $ n \plus{} 1$ is not a prime. So it has a divisor smaller than $ n$ and different from $ 1$. Because $ n!$ is also divisible by that number we get contradiction. So $ n \plus{} 1$ is a prime, Let note it by $ p$. We get: $ p^k \equal{} (p \minus{} 1)! \minus{} 1$. But from Wilson theorem we get that $ p^k \equal{} (p \minus{} 1)! \minus{} 1\equiv \minus{} 2 \pmod p$. So 2 is divisible by $ p$. So $ n \equal{} 3$ and $ k \equal{} 1$.[/quote]
Unfortunately, there is an error here : we get $ p^k \equal{} (p \minus{} 1)! \plus{} 1$ and not $ p^k \equal{} (p \minus{} 1)! \minus{} 1$

$ p^{k}\equal{}(p\minus{}1)!\plus{}1$

$ \implies$  $ (p\minus{}2)!\equal{}p^{k\minus{}1}\plus{}...\plus{}1$

if $ p$ is prime $ p\minus{}1$is not prime for $ p>2$

$ \implies$  $ p\minus{}1|(p\minus{}2)!$  $ \implies$  $ p\minus{}1|p^{k\minus{}1}\plus{}..\plus{}1$  $ \implies$  $ p\minus{}1|k$

$ \implies$  $ k\geq p\minus{}1$  $ \implies$  $ p^{k}\geq p^{p\minus{}1}$

$ \implies$  $ (p\minus{}1)!\plus{}1\geq p^{p\minus{}1}$

We know $ (p\minus{}1)!<p^{p\minus{}1}$  $ \implies$  $ (p\minus{}1)!\plus{}1\equal{}p^{p\minus{}1}$

$ \implies$  $ (p\minus{}2)!\equal{}p^{p\minus{}2}\plus{}...\plus{}1$,$ (p\minus{}2)!>p^{p\minus{}2}$  Contradiction!

the rest is trivial :wink:

[quote="cnyd"]... $ p \minus{} 1$is not prime for $ p > 2$ $ \implies$  $ p \minus{} 1|(p \minus{} 2)!$  [/quote]

Hum, this seems wrong to me : Let $ p\equal{}5$, and $ 5\minus{}1\equal{}4$ does not divide $ (5\minus{}2)!\equal{}6$

sorry

Lemma: $ n\geq 5,n|(n \minus{} 1)!$  $ \iff$  $ n$ is coprime

if $ n$ is prime $ \implies$  $ n\nmid (n \minus{} 1)!$

if $ n$ is coprime

$ n \equal{} a.b$  $ \implies$

$ a,b\leq n \minus{} 1$  $ \implies$  if $ a\not \equal{} b$  $ \implies$  $ n|(n \minus{} 1)!$

if $ a \equal{} b$  $ \implies$  $ n \equal{} a^{2}$

$ n \equal{} p_{1}^{2a_{1}}....p_{k}^{2a_{k}}$

$ \implies$  if $ k\geq 2$  $ p_{i}^{2a_{i}}\leq n \minus{} 1$

and $ (p_{i}^{2a_{i}},p_{j}^{2a_{j}}) \equal{} 1$  for $ i\not \equal{} j$

$ \implies$  $ n|(n \minus{} 1)!$

if $ k < 2$  $ \implies$ $ k \equal{} 1$

$ n \equal{} p^{2a}$,$ [\frac {n \minus{} 1}{p}] \plus{} ... \plus{} [\frac {n \minus{} 1}{p^{2a \minus{} 1}}] \equal{} p^{2a \minus{} 1} \minus{} 1 \plus{} p^{2a \minus{} 2} \minus{} 1 \plus{} ... \plus{} p \minus{} 1 \equal{} p \plus{} p^{2} \plus{} .. \plus{} p^{2a \minus{} 1} \minus{} a$

if $ p \plus{} p^{2} \plus{} .. \plus{} p^{2a \minus{} 1} < 3a$ $ \implies$ $ p^{2a}\nmid (n \minus{} 1)!$

$ \implies$  $ p^{2a} < p.3a$

if $ p > 2$  $ \implies$  $ p^{2a \minus{} 1}\geq 3^{2a \minus{} 1}$

$ a\geq 3^{2a \minus{} 2}$  $ \implies$  $ a\leq 1$  $ \implies$  $ a \equal{} 1$

$ a \equal{} 1$  $ \implies$  $ n \equal{} p^{2}$  and $ p < 3$  $ \implies$  $ p \equal{} 2$   $ \implies$  $ n \equal{} 4$  isn't satisfies this condition

if $ p \equal{} 2$  $ \implies$  $ 2^{2a \minus{} 1} < 3a$  $ \implies$  $ a\leq 1$  $ \implies$  $ a \equal{} 1$  $ \implies$  $ n \equal{} 4$


Let's say we have a [i]nice[/i] representation of the positive integer $ n$ if we write it as a sum of powers of 2 in such a way that there are at most two equal powers in the sum (representations differing only in the order of their summands are considered to be the same).

a) Write down the 5 nice representations of 10.

b) Find all positive integers with an even number of nice representations.

[quote="basketball9"][hide="a"]$ 2^0 \plus{} 2^0 \plus{} 2^3$

   $ 2^2 \plus{} 2^1 \plus{} 2^1 \plus{} 2^0 \plus{} 2^0$

    $ 2^2 \plus{} 2^2 \plus{} 2^1$

   $ 2^ \minus{} 1 \plus{} 2^ \minus{} 1 \plus{} 2^2 \plus{} 2^2 \plus{} 2^0$

    $ 2^ \minus{} 1 \plus{} 2^ \minus{} 1 \plus{} 2^2 \plus{} 2^2 \plus{} 2^0$[/hide][/quote]
EDIT:
No, your last two are repeated.

I got five representations (between us) when the exponent
on a $ 2$ is limited to being nonnegative.
Use braces around the negative ones in the exponents.

You are missing:
[hide]
$ 2^1 \plus{} 2^3$

$ 2^0 \plus{} 2^0 \plus{} 2^2 \plus{} 2^2$

[/hide]

After looking at it further, I see no limit to the number of
"nice" representations for $ 10$ when the exponents can
be negative.

[hide]
$ 2^{ \minus{} 2} \plus{} 2^{ \minus{} 2} \plus{} 2^{ \minus{} 1} \plus{} 2^0 \plus{} 2^3$

$ 2^{ \minus{} 3} \plus{} 2^{ \minus{} 3} \plus{} 2^{ \minus{} 2} \plus{} 2^{ \minus{} 1} \plus{} 2^0 \plus{} 2^3$

And so on.

Keep splitting up the lead value of $ 1$ into sums of fractions
where the $ 2$ is raised to certain negative exponents.
[/hide]

So far for part B I got all odd powers of $2$ and all integers $n=2^{2k}-1$ for nonnegative integer $k$.

But I realizie there's a lot more...

I am going to take the question as meaning nonnegative powers, since otherwise part a makes no sense. Part a has been done, here is part b:

Let $k(n)$ denote the number of ways that n can be made permitted by the question, and let $f(n)$ be $k(n)$ reduced modulo 2. I will work only with $f$, although it is the same to use $k$.

Now, $f(1) = 1$, $f(2) = 0$. I will now generate two recurrences for $f$.

Suppose $n = 2m + 1$ is odd. Then there must be exactly one $2^0$ occurrence, by parity. Then, we have the rest being a representation of $2m$ using no $2^0$s. By dividing through by 2, we can see this is exactly equal to the number of representations of $m$. Therefore, $f(2m+1) = f(m)$.

Suppose $n = 2m$ is even. If there are no $2^0$ occurrences in a representation, then we have $f(m)$ ways as before. If there are two $2^0$ occurrences, subtract two, and we have the number of representations of $2m - 2$ using none of $2^0$, or rather we have $f(m-1)$ ways like this. Therefore, $f(2m) \equiv f(m) + f(m-1) 2$ mod 2.

Finally, I will show by strong induction that $f(n)$ is zero for $n \equiv 2$ mod 3. Note that $2m+1 \equiv 2$ iff $m \equiv 2$ mod 3, and $2m \equiv 2$ iff $m \equiv 1$ iff $f(m) = f(m-1)$. Therefore, it is obvious by strong induction. QED.


logarithms
real analysis
real analysis solved

I can help you with convergence:
The sequence a_n has the following properties:
a_(n+1)<a_n (not hard to prove) and it is all the terms are pozitive so a_n is inferior bounded! :D

About the limit.. I don't know how to do it.. perhaps it's 0! I don't know ....you should try to find a reccurence between terms of the sequence a_n...

lotza luck! :D

cheers! :D:D

looks better in this way
[img]http://www.sparetimegroup.net/~kuba/php4math/pb1.gif[/img]
 :D
i know about the convergence...
and an+1=(1-1/2n+1)an

Well, I really doubt that you can find exactly the limit. All I could do was to reduce that to finding the limit of sum (k>=1) 1/(k*(2^k-1)). If this one has limit s, then  I think that your limit is e^(-s).

Where did you find this? If you thought about it, it might not be soluble (meaning that maybe we can't find the limit, just like harazi said).

i just thought on it but i didn't know that it can't be solvable :(

harazi's equations can be derived the following way:
\[-\log \prod_{n=1}^{\infty} \left(1-(\frac 12)^n\right)=-\sum_{n=1}^{\infty} \log \left(1-(\frac 12)^n\right)=\sum_{n=1}^{\infty} \sum_{i=1}^{\infty} \frac{(\frac{1}{2})^{ni}}{i}=\sum_{n=1}^{\infty} \left(\sum_{d|n} \frac 1d\right) (\frac 12)^n\].
on the other hand
\[\sum_{k=1}^{\infty} \frac{1}{k(2^k-1)}=\sum_{k=1}^{\infty} \frac 1k \sum_{i=1}^{\infty} (\frac 12)^{ki}=\sum_{n=1}^{\infty} \left(\sum_{d|n} \frac 1d\right) (\frac 12)^n\].

the identity follows.

Peter


Show that if the sum of the elements that form the number n is divisible by 3 but not by 9, then n is not a sum of squares

I think it is: show that when the sum of digits is divisible by $3$ but not by $9$, then the number is not the sum of two quares.

[quote="srulikbd"][quote="pkerichang"][hide]
if a number is divisible by 3 but not by 9, it implies that it is either 3 or 6 mod 9.  However, quadratic residues of 9 are 0, 1, 4, 7.  and you can't get 3 or 6 from two of these.
[/hide][/quote]
why?[/quote]

Lets suppose $x$ and $y$ are the numbers which we are trying to find that $x^{2}$ + $y^{2}$ = $n$
Lets say $x\equiv a\mod9$ and $y\equiv b\mod 9$
$a^{2}$ and $b^{2}$ can only be $0,1,4,$or $7 \mod 9$
Then $x^{2}$ must either be $0, 1, 4$,or $7\mod 9$, same with $y^{2}$. There is no possible combination of $0,1,4,$ and $7$ that equals $3$ or $6\mod9$


Prove that for each natural number $s\geq3$ the equation:

$\frac{1}{x^{3}_{1}}+\frac{1}{x^{3}_{2}}+\cdots+\frac{1}{x^{3}_{s}}=\frac{1}{x^{3}_{s+1}}$

has $\infty$ solutions for positive integers $x_{1}, x_{2},\cdots,x_{s},x_{s+1}$.

Let's denote
$\frac{x_{s+1}}{x_{i}}=a_{i}$
We need to show that $\sum_{i=1}^{n}a_{i}^{3}=1$ has a solution in $\mathbb{Q}^{n}$ (in that way we can easily construct infinitely many solutions to the original equation. :wink: )
notice that $(\frac{1}{2},\frac{2}{3},\frac{5}{6})$ is a solution for $n=3$and that if $(a_{1},a_{2},...,a_{k})$ is a solution for $n=k$ then
$(\frac{a_{1}}{2},\frac{a_{2}}{2},...,\frac{a_{k}}{2},\frac{2}{3},\frac{5}{6})$ is a solution for $n=k+3$
so we only need solutions for $n=4$ and $n=5$
this is just a computational problem... :)


calculus
inequalities
LaTeX
linear algebra
matrix
absolute value
calculus computations

Hey,

[b]1. The problem statement, all variables and given/known data.[/b]
Let,
$ {{f(x)}, {{{f}^{\prime}}{(x)}}, {{{f}^{\prime\prime}}{(x)}},...,} \equal{} {{f}, {{f}^{\prime}}, {{f}^{\prime\prime}},...,}$

Given,
$ {{{f}^{2}} \plus{} {{{f}^{\prime}}^{2}}} \equal{} {0}$

Prove,
$ {f} \equal{} {0}$

In addition, what is $ {{{f}^{\prime}}{{}}}$?

[b]2. Relevant equations.[/b]
Calculus.

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
I understand that it's pretty intuitive that $ {{f(x)} \equal{} {0}}$ due to the nature of the sum of two squares, in other words both: $ {{\left({f(x)}\right)}^{2}}$ and $ {{\left({{{f}^{\prime}}{\left({x}\right)}}\right)}^{2}}$; must be positive. That being said, the only positive numbers when squared and summed up that equal zero, are zero and zero.

So,
$ {f(x)} \equal{} {0}$
$ {{{f}^{\prime}}{\left({x}\right)}} \equal{} {0}$
$ {\therefore}$
$ {f(x)} \equal{} {{{f}^{\prime}}{\left({x}\right)}}$

I get that, but explaining it like I did above seems well...not mathematically rigorous. I figure there must be a more mathematically rigorous way to show that $ {{f(x)} \equal{} {0}}$.

What is the more mathematically rigorous way to show that $ {{f(x)} \equal{} {0}}$ [b]?[/b]
[/hide]
Thanks,

-APSStudent

This has nothing to do with calculus, for one thing.

[b]Lemma:[/b]  A sum of non-negative real numbers $ a_1, a_2, ... a_k$ is zero if and only if $ a_1 \equal{} a_2 \equal{} ... \equal{} a_k \equal{} 0$.

[i]Proof.[/i]  Suppose that, without loss of generality, $ a_1 > 0$.  Since $ a_2, a_3, ... a_k \ge 0$, summing the inequalities gives $ a_1 \plus{} a_2 \plus{} ... \plus{} a_k > 0$; contradiction.

You're thinking too hard.  :)

Hey,
[quote="t0rajir0u"]
[b]Lemma:[/b]  A sum of non-negative real numbers $ a_1, a_2, ... a_k$ is zero if and only if $ a_1 \equal{} a_2 \equal{} ... \equal{} a_k \equal{} 0$.

[i]Proof.[/i]  Suppose that, without loss of generality, $ a_1 > 0$.  Since $ a_2, a_3, ... a_k \ge 0$, summing the inequalities gives $ a_1 \plus{} a_2 \plus{} ... \plus{} a_k > 0$; contradiction.
[/quote]
Thanks for the Lemma and Proof. However, while rewriting the proof I had a couple of questions.
[hide="Lemma and Solution"]
Let,
$ {{{\bigwedge_{i \equal{} m}^{n}}{{a}_{i}}} \equal{} {{a}_{m} \equal{} {a}_{m \plus{} 1} \equal{} {a}_{m \plus{} 2} \equal{} ... \equal{} {a}_{n \minus{} 2} \equal{} {a}_{n \minus{} 1} \equal{} {a}_{n}}}$

[b]Lemma:[/b] For non-negative real numbers,
$ {{a}_{1},{a}_{2},{a}_{3}{,}...,{a}_{k}} ,$

than the,
$ {{{\sum_{i \equal{} 1}^{k}}{{a}_{i}}} \equal{} {0}} , {\mbox{iff}} {\textcolor{white}{.}} {\textcolor{white}{.}} {{{\bigwedge_{i \equal{} 1}^{k}}{{a}_{i}}} \equal{} {0}}$

[b]Solution,[/b]
$ {\mbox{Let,}} {\textcolor{white}{.}} {\textcolor{white}{.}} {{a}_{1},{a}_{2},{a}_{3}{,}...,{a}_{k}} \in {|\mathbb{R}|}$

Prove,
$ {{{\sum_{i \equal{} 1}^{k}}{{a}_{i}}} \equal{} {0}} , {\mbox{iff}} {\textcolor{white}{.}} {\textcolor{white}{.}} {{{\bigwedge_{i \equal{} 1}^{k}}{{a}_{i}}} \equal{} {0}}$

Proof,
Suppose that, WLoG, $ {{{a}_{1}} > {0}}$

Since,
$ {{{a}_{2},{a}_{3},{a}_{4}{,}...,{a}_{k}} \geq {0}},$

At this step I do not undertsand why,
$ {{{a}_{2},{a}_{3},{a}_{4}{,}...,{a}_{k}} \geq {0}},$

Shouldn't,
$ {{{a}_{2},{a}_{3},{a}_{4}{,}...,{a}_{k}} > {0}},$

Since $ {{{a}_{1}} > {0}}$ and $ {{{\bigwedge_{i \equal{} 1}^{k}}{{a}_{i}}} \equal{} {0}}{\textcolor{white}{.}}{\textcolor{white}{.}}$[b]?[/b]

Also in the above proof I let $ {|\mathbb{R}|}$ represent set the non-negative real numbers, is this notation correct? I reasoned it was ok since technically $ {\mathbb{R}}$ represents the set of real numbers and the absolute value operation would reduce the set of real numbers to only the set of non-negative real numbers.
[/hide]
[quote="§outh§tar"]
Complex valued? :p
[/quote]
Would you please elaborate?

Thanks,

-APSStudent

[quote="APSStudent"]At this step I do not undertsand why,
$ {{{a}_{2},{a}_{3},{a}_{4}{,}...,{a}_{k}} \geq {0}},$

Shouldn't,
$ {{{a}_{2},{a}_{3},{a}_{4}{,}...,{a}_{k}} > {0}},$

Since $ {{{a}_{1}} > {0}}$ and $ {{{\bigwedge_{i = 1}^{k}}{{a}_{i}}} = {0}}{\textcolor{white}{.}}{\textcolor{white}{.}}$[b]?[/b][/quote]
You use a lot more $ \text{\LaTeX}$ than you need to; most of those brackets are unnecessary.  As for your question, $ a_2, ... a_k \ge 0$ follows from the definition of a non-negative real number.  You are also trying to prove that the $ a_i$ are equal, which means you can't assume it.

[quote="APSStudent"]Also in the above proof I let $ {|\mathbb{R}|}$ represent set the non-negative real numbers, is this notation correct? I reasoned it was ok since technically $ {\mathbb{R}}$ represents the set of real numbers and the absolute value operation would reduce the set of real numbers to only the set of non-negative real numbers.[/quote]
Notation is generally a matter of convention, not reasoning.  Common notations for the set of non-negative real numbers include $ \mathbb{R}_{\ge 0}$ and $ \mathbb{R}^{ + } \cup \{ 0 \}$.

[quote="APSStudent"]Would you please elaborate?[/quote]
The lemma and hence the conclusion fails if $ f$ is allowed to take complex values; in other words, the differential equation that inspires this problem has more solutions.

Hey,
[quote="t0rajir0u"]
You use a lot more $ \text{\LaTeX}$ than you need to; most of those brackets are unnecessary. As for your question, $ a_2, ... a_k \ge 0$ follows from the definition of a non-negative real number. You are also trying to prove that the $ a_i$ are equal, which means you can't assume it.
[/quote]
Ok, I understand that ${ {{a}_{2},{a}_{3},{a}_{4}},\dots,{a}_{k - 2},{a}_{k - 1},{a}_{k}} \geq {0}$ follows from the definition of a non-negative real number.
However, why is it that I can't assume the following?
${ {{a}_{1} = {a}_{2} = {a}_{3}} = \dots = {a}_{k - 2} = {a}_{k - 1} = {a}_{k}} = {0}$
I thought since it was stated in the Lemma,
[quote]
[b]Lemma[/b]: A sum of non-negative real numbers $ a_1, a_2, ... a_k$ is zero if and only if $ a_1 = a_2 = ... = a_k = 0$.
[/quote]
I could assume it. I know you pointed out that we are trying to prove it so we can't assume it, but I thought since it was stated in the Lemma we could assume it (kind of like a given). So, how can we tell it is not something we can assume?

[quote="t0rajir0u"]
Notation is generally a matter of convention, not reasoning.  Common notations for the set of non-negative real numbers include $ \mathbb{R}_{\ge 0}$ and $ \mathbb{R}^{ + } \cup \{ 0 \}$.
[/quote]
Ok, I see that the notation,
$ {{\mathbb{R}}_{\geq {0}}}$
is pretty intuitive implying all real numbers are greather than or equal to zero (ie. non-negative real numbers). However, the following notation,
$ {{\mathbb{R}}^{ + }} \cup {\{{0}\}}$
is a little unclear to me. Does it mean the following, "the set of all positive real numbers union with the set of zero represents the set of all positive real numbers and zero"?

[quote="t0rajir0u"]
The lemma and hence the conclusion fails if $ f$ is allowed to take complex values; in other words, the differential equation that inspires this problem has more solutions.
[/quote]
I see, thanks.

Thanks,

-APSStudent

[quote="APSStudent"]I know you pointed out that we are trying to prove it so we can't assume it, but I thought since it was stated in the Lemma we could assume it (kind of like a given). So, how can we tell it is not something we can assume?[/quote]
The statement is "P if and only if Q" which is really two statements:  P if Q and Q if P.  Q if P is trivial here (a sum of zeroes is zero), so the interesting direction is P if Q, and to prove this direction you need to assume P and prove Q, not the other way around.

[quote="APSStudent"]However, the following notation,
$ {{\mathbb{R}}^{ \plus{} }} \cup {\{{0}\}}$
is a little unclear to me. Does it mean the following, "the set of all positive real numbers union with the set of zero represents the set of all positive real numbers and zero"?[/quote]
Yes.

Hey,
[quote="t0rajir0u"]
The statement is "P if and only if Q" which is really two statements:  P if Q and Q if P.  Q if P is trivial here (a sum of zeroes is zero), so the interesting direction is P if Q, and to prove this direction you need to assume P and prove Q, not the other way around.
[/quote]
Thanks, I was reading a little bit more about the statement "P iff Q" and doesn't it mean a little bit more than the two statements: "P if Q and Q if P"?
Specifically, I believe it means the two statements: if P, then Q and if Q, then P. Also, when wrting a Proof given a statement, P iff Q; in order to prove that statement does it require that both statements: if P, then Q and if Q, then P; be proven?

Thanks,

-APSStudent

[quote="APSStudent"]Thanks, I was reading a little bit more about the statement "P iff Q" and doesn't it mean a little bit more than the two statements: "P if Q and Q if P"?
Specifically, I believe it means the two statements: if P, then Q and if Q, then P.[/quote]
"P if Q" and "if Q, then P" are the same statement.

[quote="APSStudent"]Also, when wrting a Proof given a statement, P iff Q; in order to prove that statement does it require that both statements: if P, then Q and if Q, then P; be proven?[/quote]
Yes.


Where can I get video tutorials/lectures on topics like Organic and Inorganic chemistry. Is there any portal like OCW. Please tell me if you know any.

I would just go with OCW.  It's MIT, and it's really comprehensive.  Other places that post lectures generallty require a student ID to get access.

I seem to remember that UC Berkeley had whole series of Organic Chemistry Lectures on Youtube. So try searching for that. I think they were quite decent.

BanishedTraitor would be correct.

http://youtube.com/user/ucberkeley

This is a link to Berkeley's archives so find what you need there.


limit
inequalities
function
real analysis
real analysis unsolved

Prove that if $a_{k}> 0, \sum a_{k}$ diverges, then $\sum \frac{a_{k}}{a_{k}+1}$ diverges.

I haven't been able to make any headway on this except that $\lim_{k \rightarrow \infty}a_{k}= 0$ since the case when it doesn't it is easy enough to prove the divergence of each series.

Hmmm so here's a pretty random way:

Suppose for the sake of contradiction that the second sum converges. Then

$\sum a_{i}-\sum \frac{a_{i}}{a_{i}+1}= \sum \frac{a_{i}^{2}}{a_{i}+1}$

diverges. As you established, we need only consider when $\lim_{n \to \infty}a_{n}= 0$. Assuming this, by limit comparison to $\sum \frac{a_{i}}{a_{i}+1}$, we know that $\sum \frac{a_{i}^{2}}{a_{i}+1}$ converges, which is a contradiction.

A more straightforward argument: the key observation is that for $a_{k}$ near zero, $\frac{a_{k}}{a_{k}+1}$ looks like $a_{k}.$ Let's turn this into an inequality, in the direction that we need:

For $0<x\le1,\ \frac{x}{x+1}\ge\frac{x}{2}.$

If $a_{k}>1$ for infinitely many $k,$ then the second series clearly diverges because the terms don't go to zero. Thus assume that there is some $N$ such that $a_{k}\le1$ for $k\ge N.$

Then $\sum_{k=N}^{\infty}\frac{a_{k}}{a_{k}+1}\ge \frac12\sum_{k=N}^{\infty}a_{k}=\infty.$

Now, try proving this:

Suppose $f: (0,\infty)\mapsto(0,\infty)$ is a nondecreasing function that has the property that $\lim_{x\to0^{+}}\frac{f(x)}{x}=C,$ where $C$ is a some fixed positive number.

Then for $a_{k}>0,$ $\sum_{k=1}^{\infty}a_{k}$ converges if and only if $\sum_{k=1}^{\infty}f(a_{k})$ converges.


natural number n is given and collection {n^2 +1, n^2 +2, ... , n^2 +2n}. For what n we can get numbers a,b,c for this collection satisfying c|(a^2 +b^2)  ?

 :wallbash_red:

I think we should assume that, but in problem there's nothing about that.

$ c\equal{}n^2\plus{}1$
$ a\equal{}n^2\plus{}n\plus{}1 \equiv n\bmod{n^2\plus{}1}$
$ b\equal{}n^\plus{}2 \equiv 1\bmod{n^\plus{}1}$
$ a^2\plus{}b^2\equiv n^2\plus{}1^2 \equiv 0\bmod n^2\plus{}1$

$ n>0$


limit
calculus
derivative
calculus computations

a) Let be $ f(x)\equal{}\frac{1}{1\plus{}e^{\frac{1}{x}}}$

 Find  $ \lim_{x \rightarrow 0}$  $ f'(x)$

b) If $ \lim_{h \rightarrow 0\plus{}}$ $ \frac{f(a\plus{}h)\minus{}f(a)}{h}$ and $ \lim_{h \rightarrow 0\minus{}}$ $ \frac{f(a\plus{}h)\minus{}f(a)}{h}$ both exist but have different values, show that $ f(x)$ is continuos at $ a$

part b should read "... $ f(x)$ is [b]dis[/b]continuous at $ a$."
as for part a, try it: take the derivative, and try to find the limit.

[quote="benorin"]part b should read "... $ f(x)$ is [b]dis[/b]continuous at $ a$."[/quote]
It's not necessarily true - think about $ y\equal{}|x|$

[b](a)[/b] Note that $ f$ satisfies $ f(x) \plus{} f( \minus{} x) \equal{} 1$.

Differentiate to get $ f'(x) \minus{} f'( \minus{} x) \equal{} 0 \implies f'(x) \equal{} f'( \minus{} x)$.

$ \therefore \lim_{x \to 0 ^ \plus{} } f'(x) \equal{} \lim_{x \to 0 ^ \plus{} } f'( \minus{} x) \equal{} \lim_{x \to 0 ^ \minus{} } f'(x)$

$ \therefore \lim_{x \to 0} f'(x)$ exists.

Now we can just calculate: $ f'(x) \equal{} \frac {1}{2x^2 (1 \plus{} \cosh x^{ \minus{} 1})}$

Note that $ \lim_{x \to 0} 2x^2 (1 \plus{} \cosh x^{ \minus{} 1}) > \lim_{x \to 0} 2x^2 \cosh x^{ \minus{} 1} \equal{} \lim_{z \to \infty} \frac {e^z \plus{} e^{ \minus{} z}}{z^2} \equal{} \infty$.

$ f'(x)$ is the reciprocal of that, and thus $ \lim_{x \to 0} f'(x) \equal{} 0$

[quote="The Zuton Force"]$ \therefore \lim_{x \to 0 ^ \plus{} } f'(x) \equal{} \lim_{x \to 0 ^ \plus{} } f'( \minus{} x) \equal{} \lim_{x \to 0 ^ \minus{} } f'(x)$

$ \therefore \lim_{x \to 0} f'(x)$ exists.[/quote]
Actually, no: you showed that if the limits from the left and right exist then they are equal, but you didn't show that those limits individually exist.  (Think of some highly occilatory even function.)

[quote="GioMott"][quote="benorin"]part b should read "... $ f(x)$ is [b]dis[/b]continuous at $ a$."[/quote]
It's not necessarily true - think about $ y \equal{} |x|$[/quote]

Duh! wasn't thinking...


一名热血青年对于现实中国的时代缴文
首先，我要声明的是，我非自围墙之内；因之，不晓得有什么繁文缛节，言辞略有唐突还望请见谅。
其次，我有三点企盼：
其一，请尊重他者之存在。
请不要再用激烈的言语来讽喻了，诸如“骑着自行车，想上月球？！”之云云，来对我们以伤害，这很刺耳，亦是无所用处。虽则，有来自民间人士的“无知”因素，是的，他们确实无知，最起码他们不懂的学术与技术的差别，何止在于纸面。可，这只能说明，我们的科技知识的传播是何等的匮乏。
当年，日本的《岩波书店》为了学术兴国起见，每本《岩波全书》仅售价七十钱啦！想想，我们现在高昂的书价，似乎知识真的就嬗变为财富了，可知识的本意却被严重曲解了，我不免惭愧难当。
但，您要知道。是中国人，谁都是希望自己的国家，能够用自己的运载火箭，载着我们国家的宇航员上月球的。民族的尊严，请不要随意地就此轻易的撕裂。我更加鄙视的是，我们所谓的“知识精英们”，总是欢喜于坐驾着白色的日本小汽车，之乎于欢娱——“经济的升腾”而到处讲演。是的，我们比之是更有勇气，最起码我们不惧怕现实中的繁难，我们敢于挺起自己坚实的脊梁。
其二，请学术的归于学术，尊重他者辛勤的劳动成果。就如同，农民，是不轻易舍弃任何一粒粮食的，面对自己数日的辛苦劳作，是谁都不愿意其无疾而终的，这点还请能够体谅作者的心情。如果，你还怜悯于我处境的尴尬，就请尽快恢复信件吧！我将翘首以盼。
其三，希望国家，能够给予当代年轻人，以合理的成长上升通道，构建一个科学高效的发展平台，能够指明方向，把脉时代。不应也不该仅局限于高考，不应也不该仅拘泥于天才，天才亦是世俗的产物。而目前，国家之急迫乃是兴学重教，革愚钝，启民智，比对日之二战前国家战略理念，我们差之不蒂半百。
最后，希望编辑同志，能够观之，察言之，再三思量后，取舍不敢强求，但求无憾。
程兄：你好！
怎樣的敬語，才會不致過火，亦無不妥，我已不知何措。只曉得能夠消除彼此的隔膜，稱兄道弟也未嘗不可。故國學大師王國維嘗有《人間詞話》，後有啟功先生之《論書絕句》，均載言學問之巍偉大哉，豈不積百思於一眸而信手拈來？！
今之天下，鬼馗之糗，畫符招魂者，有之；赤郎若貉，糙言謾文，諸如小剛者，不絕。
做人于斯國，難矣。慨以大言滔滔，直引饑兒之嗷嗷，各人之生活體態，相岔矣。君可高悠枕，小可尚須披星月，豈有紅木床？！君之幸，不謀為天下興；君之逸，豈不勞百姓役？！
  雖則文化悠遠，然而中華精神卻是已殘損，畸形的心態一直拖累著中國歷史之長長的倒影。為今之計，乃莫言。切勿言必曰無所知，無所為，無所寄予。有太多的過錯，為故人所歿，要我們這一代人來消磨。
  不願意，一點都不想聽，歷史老人們的佈告“人生雖難言成功與否，但業已盡力。。”
蕩然，推委之詞，交口銜語，我已木然。
岩波全書發刊之際瞻後言
岩波茂雄
時艱にして朝に諍臣なく野に義人なく舉世滔滔義をすて利に走りて恥づるを知らす輦轂の下薰化の重責を負へる者に縲絏の徒を出すが如さに至つては邦家の憂患之に過ぐるものはぃ。吾人は圖書に衣食する市井の一素町人に過ぎずと雖も先憂後樂君國に微力を捧げんとする奉公の至情に於ては敢て人後に落つるを潔とせざるもの、一の圖書一の雜誌を公にする場合と雖も常に出版の第一義に即し、未だ曾て學術と社會とを思はざるてとなかりしは自ら顧みて天地に恥ぢざる所である。創業二十年の紀念として

This is spam! :mad: Should be deleted!

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=135672[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=135673[/url]

Japanese Communities

kunny

Thanks for Kunny's attention with my topic.

you are always someone I am Engjoyble in Communitations.

Kunny, Kunny!

I am Jhahn Jsing-Jchi.

P.R.C's educational policy is not what you 've seen in the article ,I promise,and even I could not say the policy is harmful for the student of the mainland,because of some other reasons,the policy wll remain for considerable long time,but it always adjusts the fundmental situation of the educatiion.
That's all. :!:  :!:

但是我们是中国人，不是每个人都可以摆脱这种教育制度的。

[quote="chernfey"]但是我们是中国人，不是每个人都可以摆脱这种教育制度的。[/quote]
 what kind of Chinese Education System are you tolding about?
 Could you further explain to me and many another Chinese American Students?


colour unit cells of 2m+1 by 2n+1 greed using 2 colours in any way using 2 colours.a cell is called row dominant if in the corresponding row there are atleast n more cells of same colour.define column dominant similarly.show there are atleast m+n+1 cells which are column and roe dominantt.

[quote="who"]
[i]"Color unit cells in a $2m+1 \times 2n+1$ grid using two colors. A cell is called [b]row or column dominant[/b] if in that row or column respectively there are at least $n, m$ respectively more cells of the same color. Show that there are at least $m+n+1$ cells that are both column and row dominant"[/i][/quote]

hey whats the matter?most of the pro. posted by me in combinatorics proposed and own pro. remain unanswered.?

The reason no answer is posted is it too easy; there are m+1 dominants per column and n+1 dominants per row, easy to see there are m+n+1 squares that are both :wink:

There are $(2m+1)(2n+1)=4mn+2m+2n+1$ squares.  There are $(m+1)(2n+1)+(n+1)(2m+1)=4mn+3m+3n+2$ squares that are row or column dominant.  Thus $m+n+1$ squares must be row and column dominate.


Let $a,b,c$ be the side-lengths of a triangle. Prove that Schur's Inequality of fourth degree,
$\sum a^{4}+abc\sum a \geq \sum ab(a^{2}+b^{2})$,
is stronger than Schur's Inequality of third degree
$\sum a^{3}+3abc \geq \sum ab(a+b)$.

[quote="pvthuan"]Vasc, it is also a problem in chapter 7 of mine. Classical substitution works here. Let $a=x+y,$ ...  , where $x,y,z$ are positive real numbers, then the desired inequality takes the form
\[(x+y)(y+z)(z+x)(x+y+z)\geq x(y+z)^{3}+y(z+x)^{3}+z(x+y)^{3}. \]
Due to homogenuity, choose $x+y+z=1$. We simplify it to
\[x^{4}+y^{4}+z^{4}-3(x^{3}+y^{3}+z^{3})+3(x^{2}+y^{2}+z^{2})+xy+yz+zx\geq1+xyz. \]
Let $p=x+y+z$ ($=1$), $q=xy+yz+zx$, and $r=xyz$, then we transform the last into
\[q^{2}\geq3pr, \]
which is well-known.

 :lol:

We can solve a large class of inequalities by this way even those involving fractions, of course with some more difficulty.[/quote]

but you didnt prove that the fourth degree inequality is stronger than the third degree inequality . What you do is just prove Schur for degree 4 which is trivial.

From degree 4 we have

$\sum a^{4}+abc\sum a\geq\sum ab(a^{2}+b^{2})\geq 2\sum a^{2}b^{2}$

and
$\sum a^{4}+abc\sum a-2\sum a^{2}b^{2}\geq 0 \\\Longleftrightarrow (a+b+c)(\sum a^{3}+3abc-\sum ab(a+b))\geq 0$

by the way , why we need $a,b,c$ to be side of triangle ?

[quote="shyong"]
From degree 4 we have

$\sum a^{4}+abc\sum a\geq\sum ab(a^{2}+b^{2})\geq 2\sum a^{2}b^{2}$

and
$\sum a^{4}+abc\sum a-2\sum a^{2}b^{2}\geq 0 \\\Longleftrightarrow (a+b+c)(\sum a^{3}+3abc-\sum ab(a+b))\geq 0$

by the way , why we need $a,b,c$ to be side of triangle ?[/quote]

This is not a close argument to say that the first (1) is stronger than the second (2).
To find such a rigorous argument is the key to the problem.  :!:
Actually, (1) is stronger than (2) if and only if
$\sqrt a+\sqrt b+\sqrt c \geq 2 \max \{\sqrt a, \sqrt b, \sqrt c \}$.	 :lol:

[quote="Albanian Eagle"]Well, one definition that I know for "stronger inequality" is:
A>B is stronger than C>D if A+D>B+C
and in this case always 4-degree schur is "stronger" than 3-degree schur (guess why? :P )
So it's all a matter of definitions...[/quote]

Haha.. then you might as well define A>B stronger than C>D as AD<BC. At least it makes a bit more sense :wink: though not much more actually

In my statement above:
"stronger than"="stronger than or equivalent to". See the case $f(B,C)=g(B,C)$ for any a,b,c...

[quote="Nameless"][quote="Vasc"]In my opinion, the inequality
 $A \geq f(B,C)$
is stronger than
$A \geq g(B,C)$
if
$f(B,C)\geq g(B,C)$
for any a,b,c,....[/quote]
I don't think so. Example:
$\sum a^{2}\ge \sum bc$  (1)
$\sum a^{2}\ge \frac{(\sum a)^{2}}{3}$    (2)
$\frac{(\sum a)^{2}}{3}\ge \sum bc$
But (1) is equivalent to (2)[/quote]
[quote="Vasc"]Actually, (2) is strictly stronger than (1).
Sometimes, in order to prove (2), we say that "(2) is  equivalent to (1)", because we may bring (2) to the form (1).
But, in order to prove other inequalities, we can often do it using (2), and fail using (1).[/quote]
[quote="Albanian Eagle"]I think this is a very subjective issue. That's why i always thought of it as irrelevant. it not only depends on definitions but also on the objects you apply them...
I think for example that your definition Vasc is a special case of my definition above :wink:
and I said that $\sum a^{2}(a-b)(a-c)\geq \sum a(a-b)(a-c)$ :P[/quote]
I think Eagle understood that (1) is (S3) – Schur’s Inequality of third degree, and (2) is (S4) - Schur’s Inequality of fourth degree.  Anyway, $\sum a^{2}(a-b)(a-c)\geq \sum a(a-b)(a-c)$  is a joke.  :P

The statement below is valid in my opinion. Can someone to prove it?

(S4) is stronger than (S3) if
$\sqrt a+\sqrt b+\sqrt c \geq 2 \max \{\sqrt a, \sqrt b, \sqrt c \}$,
and (S3) is stronger than (S4) if
$\sqrt a+\sqrt b+\sqrt c \leq 2 \max \{\sqrt a, \sqrt b, \sqrt c \}$.

Using this, you can prove the inequality in
http://www.mathlinks.ro/Forum/viewtopic.php?t=123965

See:
[quote="silouan"]The last inequality mr Vasc wrote is equivalent to
$x^{4}+y^{4}+z^{4}+xyz(xy+zy+zx)\geq x^{2}y^{2}+z^{2}y^{2}+z^{2}x^{2}+x^{2}+y^{2}+z^{2}$[/quote]

FINALLY, I found out the way to prove Vasc's idea !
I must admitt it is a very nice and interesting result but also insist in that the formulation was far from rigorous. :wink:
About the last problem, I solved it by SOS. :)

[quote="Albanian Eagle"]I must admitt it is a very nice and interesting result but also insist in that the formulation was far from rigorous. :wink:
[/quote]
[quote="Vasc"]In my opinion, the inequality
 $A \geq f(B,C)$
is stronger than
$A \geq g(B,C)$
if
$f(B,C)\geq g(B,C)$ for any a,b,c,....[/quote]
What is far from rigorous here, Eagle ? :!:

Well i think there are many possibilities to apply that definition
but to arrive to your result I had to choose $A=abc, B=ab+bc+ca, C=a+b+c$ (btw is this also your proof? :) )
Bu we could get a different rezult taking different expressions for $A,B,C$

[quote="Albanian Eagle"] but to arrive to your result I had to choose $A=abc, B=ab+bc+ca, C=a+b+c$ (btw is this also your proof? :) )[/quote]
Yes, you are right.
[quote="Albanian Eagle"] Bu we could get a different rezult taking different expressions for $A,B,C$[/quote]
Can you give such an example?


superior algebra
superior algebra unsolved

Let $ \sigma: F \rightarrow K$ be a field embedding. Construct an extension field $ L$ of $ F$ such that $ L \minus{} F \equal{} K \minus{} \sigma(F)$.

[quote="ZetaX"]That equality makes no sense.[/quote]
May I know the reason?  :o
Should it be "$ \cong$" rather than "$ \equal{}$"?

the equality makes sense.

define $ L$ to be the disjoint union of $ F$ and $ K \minus{} \sigma(F)$. then there is an obivious bijection $ L \to K$, which defines a field structure on $ L$ such that $ F \subseteq L$ an extension of fields.


function
integration
calculus
parameterization
real analysis
real analysis solved

Let F be the set of all continuous functions f:[0, 1]->[0, 2] s.t.  \int {0 to 1} f(x)dx=1.

a) what are all the possible values of   \int {0 to 1} xf(x)dx?
b) if f and f^2 are in F then f is constant.

I'll post more later.

Okay... here is a nice solution:

a) We will prove that for f in F,  \int {0,1} xf(x) dx <=3/4.
Indeed we have that \int {0,1} xf(x) dx= \int {0,1/2} xf(x)dx+\int {1/2,1} xf(x)dx<=1/2 \int {0,1/2}f(x)dx+\int {1/2,1} xf(x) dx =1/2 (1-\int {1/2,1}f(x)dx) + \int {1/2,1} xf(x)dx=1/2+1/2 \int {1/2,1} (2x-1)dx<= 1/2+1/2 \int {1/2,1} (2x-1) 2dx<=1/2+(x2-x) (computed from 1/2 to 1)=3/4.

But we have more: \int {0,1} xf(x) dx<3/4. If \int {0,1} xf(x)dx=3/4 we would have \int {0,1/2}xf(x)dx=1/2 \int {0,1/2} f(x)dx and \int {1/2,1} xf(x)dx=2* \int {1/2,1} xdx, which means that \int {0,1/2}(2x-1)f(x)dx=0 and \int {1/2,1} x[f(x)-2]dx=0.
But we have that 2x-1<0 and x is in [0,1/2) which would imply f(x)=0, and for x>0, x in x(1/2,1] we would get that f(x)=2 so f is dicontinuous in x=1/2 which gives that f is not in F, false.

Also if f is in F then we have that: \int {0,1} xf(x)dx>1/4. Lets observe that f is in F if and only 2-f is in F si \int {0,1} x[2-f(x)]dx<3/4 so
\int {0,1} xf(x)dx>1/4. so the values that \int {0,1} xf(x)dx is in (1/4,3/4).

b) from \int {0,1} f(x)dx=1, \int {0,1} f^2(x)dx=1  and we get that \int {0,1} (f(x)-1)^2dx=0 and from this thing we get that f=1 constant.

cheers! :D :D

Good, Monsieur Lagrange , but the proof is not finished ... You proved only that the required set  is included in  ]1/4,3/4[ ....

I have also proved that the function is constant:

look at the end of the 1st post:

[b]b)[/b]
 from   \int {0,1} f(x)dx=1,  \int {0,1} f^2(x)dx=1 and we get that  \int {0,1} (f(x)-1)^2dx=0 and from this thing we get that f=1 constant.

cheers! :D :D

No, Maitre Lagrange .. Doesnt matter point b.  You haven't  finished the proof for the point a .
Let V be the required set of all the possible values etc .. You proved only that V  is included in ]1/4,3/4[ ..
But a complete proof needs to show also : ]1/4,3/4[ is included in V .. Do you follow me ?

Maitre Lagrange... hehe like it! :D :D thank you :D

I see what you mean.. sorry about that! I thought you haven't seen point b)... sorry my mistake!!!

Okay!!!

Now we will show that the integral takes any values in (1/4,3/4).
(this is the hard and long part) :D :D

Lets consider the function f:[0,1]-->[0,2].
f(x)=0, for x in [0,a]; 1 for x in [a+ \varepsilon; b- \varepsilon]; 2 for x in [b,1]; (x-a)/ \varepsilon for x in (a, a+ \varepsilon); and (x-b)/ \varepsilon +2 for x in (b- \varepsilon,b).
where 0<=a<b<=1, a+b=1; (b-a)/2>= \varepsilon >0. We have that f is continuous. So we have that:

 \int {0,1} f(x)dx=  \int {a,a+ \varepsilon} (x-a)/ \varepsilon dx+ \int {a+ \varepsilon; b- \varepsilon} 1dx+\int {b- \varepsilon,b} ((x-b)/ \varepsilon +2)dx+ \int {b,1} 2dx=(after computation)=1 which shows that f is in F. Then \int {0,1} xf(x)dx= \int {a, a+ \varepsilon} x(x-a)/ \varepsilon+dx + \int {a+ \varepsilon; b- \varepsilon} xdx+ \int {b- \varepsilon, b} ((x-b)/ \varepsilon +2) xdx+ \int {b;1} 2xdx=(after computations)= - \varepsilon/3 +(1-2a)/3* \varepsilon +3/4 -((1-2a)/2)^2.

Now lets consider the last result as a 2nd degree eqs which depends on y=(b-a)/2=(1-2a)/2 (parameter)
P_y( \varepsilon ) =-1/3* \varepsilon^2 + y* \varepsilon+3/4-y^2.

Because of the way we chose a,b, \varepsilon we have that a is in [0,1/2), y is in (0,1/2] and \varepsilon in (0,y]. On the interval (0,y] the function P_y is incresing, so P_y(0)<P_y( \varepsilon )<= P_y (y) which means that P_y( \varepsilon ) is in (3/4-y^2,3/4-y^2/3).

So, for y fixed  (a fixed), we consider all functions f with \varepsilon in (0,y], we get that P_y( \varepsilon )= \int {0,1} xf(x)dx is in  (3/4-y^2, 3/4-y^2/3).

Now lets consider y variable in (0,1/2] ( also a variable in [0,1/2]) we get that for all functions of the type we defined above, the integrals  \int {0,1} xf(x)dx are all in (3/4-y^2, 3/4-y^2/3) (where 0<y<=1/2)=(1/2,3/4).

Also if f(x)=1 (constant) for any x in [0,1] we have that f is in F and \int {0,1} xf(x)dx=1/2. So there is a subcrowd of F, F1 which is in F so that: { \int {0,1} xf(x)dx | f is in F1}= [1/2,3/4).
But we saw that if f is in F then so 2-f is in F and we also have that :
\int {0,1} x(2-f(x))dx=1- \int {0,1} xf(x)dx.
So we have that { \int {0,1} x(2-f(x))dx | f is in F1}=(1/4,1/2].

So this time I think I proved it both ways! :D :D

cheers! :D :D

[quote="Lagrangia"]Maitre Lagrange... hehe like it! :D :D thank you :D
[/quote]

Hi Lagrangia this for you

http://www-history.mcs.st-andrews.ac.uk/history/Mathematicians/Lagrange.html

oubinool


In rectangle $ ABCD$, we have $ AB\equal{}8$, $ BC\equal{}9$, $ H$ is on $ \overline{BC}$ with $ BH\equal{}6$, $ E$ is on $ \overline{AD}$ with $ DE\equal{}4$, line $ EC$ intersects line $ AH$ at $ G$, and $ F$ is on line $ AD$ with $ \overline{GF}\perp\overline{AF}$. Find the length $ GF$.
[asy]unitsize(3mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));

pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0);

draw(D--A--B--C--D--F--G--Ep);
draw(A--G);
label("$F$",F,W);
label("$G$",G,W);
label("$C$",C,WSW);
label("$H$",H,NNE);
label("$6$",(6,8),N);
label("$B$",B,NE);
label("$A$",A,SW);
label("$E$",Ep,S);
label("$4$",(2,0),S);
label("$D$",D,S);[/asy]$ \textbf{(A)}\ 16 \qquad
\textbf{(B)}\ 20 \qquad
\textbf{(C)}\ 24 \qquad
\textbf{(D)}\ 28 \qquad
\textbf{(E)}\ 30$

It is pretty obvious that we can solve this problem by using similar triangles. First we can extend line BC to line FG and call the point of intersection I. The length of FG is therefore FI + IG. FI = AB = 8, so we now need to find the length of IG. Let IG = x. This is also the height of triangle GCH. The base of this triangle is 3. This triangle is similar to triangle AGE. We can write the equation;

(GI/FG) = (CH/AE)
GI = x
FG = x + 8
CH = 3
AE = 5

So

x/(x+8) = 3/5
3 and 5 differ by 2, so if you multiply both of these numbers by four the numbers differ by eight. Just like the other fraction's numerator and denominator. 3/5 = 12/20, so x = 12, and FG = 20.

B is the answer.

Can somebody please help me with my solution?

I called $FD=x$ and $GF=y$. Through similarities of $\triangle GFE$ and $\triangle CDE$ we obtain that $GF=y=2x+8$.
After that, I know I want to find $x$ but I am not sure how

Are you talking about dropping one from $H$? I tried that, and it doesn't work because the altitude from H does not intersect FA at E

No,the one with HBA and something else. (I dont understand how you got that I was thinking that from my previous post !? )

$\frac {FG}{CD}=\frac {EG}{EC}=\frac {EG}{EG-GC}=$ $\frac {\frac{EG}{GC}}{\frac {EG}{GC}-1}=$ $\frac {\frac {AE}{CH}}{\frac {AE}{CH}-1}=$ $\frac {\frac 53}{\frac 53-1}=\frac 52$ .

 In conclusion $FG=\frac 52\cdot CD=\frac 52\cdot 8=20$ . The answer is $\boxed B$ .

[hide="Another Approach?"]
Once again $FD=x$. Since $\triangle GFE \tilde \triangle CDE$, $GF=2(x+4)=2x+8$. Now we also have \[(x+4)^2+(2x+8)^2=(x+4)\sqrt{5}\]
But solving for $x$ does not yield a positive answer? [/hide]

Don't you mean $(x+4)^2 + (2x+8)^2 = ((x+4)\sqrt{5})^2,$ which is always true?

[asy]unitsize(3mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));

pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0);

draw(D--A--B--C--D--F--G--Ep);
draw(A--G);
label("$F$",F,W);
label("$G$",G,W);
label("$C$",C,WSW);
label("$H$",H,NNE);
label("$6$",(6,8),N);
label("$B$",B,NE);
label("$A$",A,SW);
label("$E$",Ep,S);
label("$4$",(2,0),S);
label("$D$",D,S);[/asy]

Let $FD = x$. Because $GF \parallel CD$, it is easy to see that $\triangle{CDE} \sim \triangle{GFE}$, therefore $\frac{8}{4} = \frac{GF}{x+4}$, so $GF = 2x + 8$. Next note that $\triangle{BAH} \sim \triangle{FGA}$, so $\frac{6}{8} = \frac{x+9}{2x+8}$. Cross multiplying gives $8(x+9) = 6(2x+8) \rightarrow 8x + 72 = 12x + 48 \rightarrow 4x = 24 \rightarrow x = 6$. Plugging that into $2x + 8$, gives $\boxed{20}$.

[quote=Honestly]Ah the 7.5 year bump a sight to behold[/quote]

As I always stress, why extremise a bump? It's not a crime. What is wrong in bumping a thread?

[quote=HamstPan38825]:( stop bumping these and reminding me of how hard AMC 10 got over these years[/quote]

Why not?


Try $ a\equal{}b\equal{}1$ and $ c\equal{}0.$ :wink:

[quote="iptgut"][/quote]
So what???
$ 2<\frac{3}{\sqrt{2}}$

@iptgut:\[ \sum_{cyc}\sqrt {\frac {a}{b \plus{} a}}\leq \frac {3}{\sqrt {2}}\forall a,b,c > 0\]
can be proved. Is this what you meant?

@iptgut
The problem is also wrong for $ a,b,c>0$.Just try some number and you will discover that.
Substitute $ c\equal{}\frac{1\minus{}ab}{a\plus{}b}$ into the ineq,then try the values of $ (a,b)$ for which $ ab<1$,we will find some examples that don't satisfy the ineq.


does 27/25 make sense its probability right? shouldnt the answer be less than 1

He meant 27/35, and I'll go with that answer, even though I simply skimmed the thread.

Edited, although my question is why one method works and not the other. Shouldn't they both produce the same answer?

For your 27/35 you have to subtract the possibility that they both win.
There is a $\frac 15 \cdot \frac 27=\frac{4}{35}$ chance of that, leaving.
$\frac{27}{35}-\frac{4}{35}=\frac{23}{35}$

Comparing that to dice problems,
If the question was: A B and C are playing a game. The chance that A wins is 4:3 and the chance that B wins is 1:4. What is the probability that one of them wins?
You would use $\frac15+\frac47=\frac{27}{35}$

[quote="Phelpedo"]In a tennis tournament, the chance of player A winning is 4 to 3, and the chance of player B winning 1 to 4. What is the chance of A or B winnning?

My dilemma is whether the answer is 23/35 or 27/35. If you add the probability of A and B, 4/7 + 1/5, you get 27/35. If you do it another way, the probability of A  not winning is 3/7 and the probability of B not winning is 4/5, so if you multiply these you get 12/35 and subtracting from 1 gives 23/35. The question is, which one is right?

On a more general level, I compare this to dice problems. One is, what is the probability of rolling a 5 or a 6 on one six-sided die? In this case it would be 1/6+1/6 = 1/3. However, if the question is What is the probability of rolling at least one 6 on 2 different 6-sided die?, then the answer would but 1- (5/6)(5/6) = 11/36. So more generally, when do you use each method?[/quote]
In the case of the two dice problem, the outcome of each die is independent of the other.  So, both dice may come up with a 6, or neither die may come up with a 6, or one die may come up with a 6.

In the case of the tennis tournament, the two probabilities are not independent.  It isn't possible for A and B to both win.  That's why your second method of obtaining the probability isn't working.

$p(A)=\frac47\quad p(\overline{A})=\frac37$
$p(B)=\frac15\quad p(\overline{B})=\frac45$

$p(A\cap B)=0.$  This notation means the probability that A wins and B wins.

Remember that multiplying event probabilities only works if the events are independent of each other.  $p(A\cap B)\neq p(A)\times p(B)$

Contrast this to a two-dice problem.  (Event X is the probability that the Red die comes up with a 6.  Event Y is the probability that the Blue die comes up with a 6.)

$p(X)=\frac16\quad p(\overline{X})=\frac56$
$p(Y)=\frac16\quad p(\overline{Y})=\frac56$

\begin{eqnarray*} p(X\cap Y) &=& p(X)\times p(Y) = \frac{1}{36} \\ p(X\cap \overline{Y}) & = & p(X)\times p(\overline{Y})= \frac{5}{36} \\ p(\overline{X}\cap Y) & = & p(\overline{X})\times p(Y)= \frac{5}{36} \\ p(\overline{X}\cap \overline{Y}) & = & p(\overline{X})\times p(\overline{Y})= \frac{25}{36} \\ p(\hbox{total}) &=& \frac{1}{36} + \frac{5}{36} + \frac{5}{36} + \frac{25}{36} = 1 \end{eqnarray*}


vector
linear algebra
linear algebra unsolved

Give an example of a 2-dimensional subspace of
$ \mathbb{C}^3$ over Complex vector space.

thanks for any help

Pick two vectors from $ \mathbb{C}^3$ such that one is not a multiple of the other. For instance,

$ \langle (i,i,0),(i,0,i) \rangle$

is an example of such a subspace. However, observe that, for example, $ \langle (i,i,0),(\minus{}1,\minus{}1,0) \rangle$ is a two dimensional subspace on the [i]real[/i] space $ \mathbb{C}^3$, but a one dimensional subspace on the [i]complex[/i] space $ \mathbb{C}^3$.

thanks for that. How do i write a basis for this example?

The vectors I gave you do already form a basis, since they are linearly independent.


function
modular arithmetic
logarithms
number theory
totient function

1) find the first three and the last three digits of 2^6523.

2)given: 10 red points and 10 blue points in the plane, no three of them colinnear
   show that it is possible to draw 10 line segments joining a red point to a unique blue point in such a way that none of the line segments intersect.

have fun :)

1)
[hide]
We can first find the units digit with a pattern:
$2^1 = 2$
$2^2 = 4$
$2^3 = 8$
$2^4 = 16$
$2^5 = 32$
Since the pattern repeats after 4:
$6523\mod4 = 3$
So the units digit is 8

I'm not exactly sure how to find the rest, could someone maybe explain that for me?
[/hide]

For the last 3 digits:

[hide]
We can look at 2^6523 mod 1000.
But since we know both 2^6523 and 1000 are divisible by 8, we can look at 2^6523 in mod 125. So we have:

2^6523 mod 125.

By euler's totient function, $p^{\phi{(n)}} \equiv 1 \pmod n$ where $\phi{(n)}$ is the number of integers rel. prime to $n$, and $p$ and $n$ are also relatively prime. So we have:

2^100 = 1 mod 125

Therefore, (2^100)^65 = 2^6500  = 1 mod 125. So all we have to find now is 2^23 mod 125.  This can be done by noting that 2^7 = 128, so you have 2^23 = (2^7)^3 * (2^2)  = (3)^3 * (4) mod 125, or 108 mod 125.

So now we know that the last 3 digits of 2^6523 is 108 + 125m, where m is some integer from 0 to 7.  We know that these last 3 digits must be divisible by 8, so we find that m must equal 4, so the last digits are 108 + 125(4) = 608.

Hope that kinda helped... as far as the first 3 digits are concerned... I don't know how to do it unless I have a calculator....  :blush:
[/hide]

please look at problem 2. someone please solve this

2)given: 10 red points and 10 blue points in the plane, no three of them colinnear
show that it is possible to draw 10 line segments joining a red point to a unique blue point in such a way that none of the line segments intersect.

2 is easy...
Give one blue point and one red point a y-value from 1-10, so there is one of each color for each y-value.  Then give an arbitrary x value to each point, although make sure that none of them are collinear.  Then you have ten lines connecting the points, all parallel to each other and the x-axis.

[quote="Simo_the_Wolf"]2) consider the plane. Then order th red points by the y-value and the blue points by the x-value. then join the minimal red with the minimal blue and so on...[/quote]

I am confused as to how does this guarantee that there will be no intersections?
Also, the points cannot be moved around. Think of it as randomly putting 10 red and 10 blue points on a piece of paper, and then connecting a red with a unique blue.

[quote="alpha123"]1) find the first three and the last three digits of 2^6523.
[/quote]
im too lazy to do it, but maybe we could use log 1000?

1
[hide]Take logarithm of 2 multiply that by 6523 then subtract the whole number part (since that's just a bunch of 10's, and it helps to prevent ur calculator from overflowing), let that be x, and then calculate 10^x. First three digits is answer..[/hide]

[quote="alpha123"]
show that it is possible to draw ... in such a way that none of the line segments intersect.
[/quote]

[quote="alpha123"]
Also, the points cannot be moved around. Think of it as randomly putting 10 red and 10 blue points on a piece of paper, and then connecting a red with a unique blue.[/quote]

The way the originall problem is stated, I just had to show a way that it could be done.  Not that they are randomly placed.  And my solution works; it is one way that fits the criteria.

As for the first problem... I understand how to get the first several digits of a number, but don't know how to find the last few digits and would like to know.  So I guess this last half of my post is a bump.


$ x\plus{}y\plus{}z\equal{}n;x\plus{}y>z,x\plus{}z>y,y\plus{}z>x$ How many permutations of $ x,y,z$ in part $ n$?

I dont understand your question. Please be a little more clear. Are you asking for the number of non negative integer solutions of
x+y+z=n  provided
x+y>z
x+z>y and
y+z>x


Note that
$(a^3+abc)(b^3+abc)(c^3+abc)=abc(a^2+bc)(b^2+ca)(c^2+ab)$ and this can be rewritten both as
$\prod (a^2b+b^2c)$ or as $\prod (ca^2+bc^2)$. It follows by AM-GM that
\[ \min (2\sum a^2b,2\sum ca^2)\geq RHS  \]
Now if $a\geq c\geq b$ then $(a-b)(b-c)(c-a)\geq 0$ so $\sum ca^2\geq \sum a^2b$. Thus $LHS\geq 2\sum a^2b$
For $a\geq b\geq c$ we'll have instead  $LHS\geq 2\sum ca^2$ and we are done.

I hope it is ok


i never heard of any interesting science/math jobs except an opportunity at the university I will be attending.  there will be a lot of competition for it, so i don't think i'll get it.  i haven't got a summer job yet, i'm trying to get something for minimum wage.  if i don't get it, i may just study; there's a lot i would like to know

do you know what you'll be doing?

[quote="myc"]I'm a graduating senior and am planning my summer now.  What are some things you guys did over your pre-college summer?  Does anyone know any good science/math/or community service related internships/programs/jobs?

thanks![/quote]

Frankly, even if there was any available, it is probably too late to register for them anyway. So, don't bother with them.
The only exception is that jobs that college offers to students, but that is often only restricted to current students (i.e. not upcoming freshmen).

Hang out with friends, and strengthen the connections that you want to keep after college. (emphasis)

Make it a point to do something interesting:
- If you don't already have a job or internship already, this is a great chance to work on personal projects
- If you have any universities nearby, you can try e-mailing professors with research pursuits that appeal to you and ask if they need any help -- don't count on it, though.
- Learn something new
- You're much better off flipping burgers than wasting your time

You can also brace yourself for college life:
- Plan out some possible coursework paths for majors/minors you're interested in.
- Clubs and teams will not wait for the freshman class to feel settled down before getting started
- Start hunting for jobs for the semester. Be open minded and look for the benefit in everything -- some jobs are worth it because they will get you in the good books of some important person, otherwise achieve you campus connections, or because they are non-intensive and you can simultaneously study/do homework
- Look for scholarships, competitions, and other things you might want to invest your time into
- If you're not a flying genius, it helps to take a peek at coursework ahead of time
- Take the time to make a nice and informative personal website, and to upgrade your resume from college-application mode to real world mode

I would try to find a way to at least make some money. This might be money to help pay for books or perhaps pizza money for freshmen year. Oddly enough, I fall into both categories.


geometry
rectangle
ARML
geometric transformation
reflection

How many ways can one cover a 3 x 8 rectangle with 12 dominoes?

$ \textbf{(A)} 220\qquad\qquad\qquad\textbf{(B)}
153\qquad\qquad\qquad\textbf{(C)}
300\qquad\qquad\qquad\textbf{(D)}
452\qquad\qquad\textbf{(E)}
479$

See ARML Power Round #5

[hide]
Proceed Using Recursion. Let $ A_n$ designate the number of tilings of a 3 by n rectangle. There could be three

ABC
ABC

AS is a domino, BB is a domino, CC is a domino. There are $ A_{n\minus{}2}$ ways to tile the rest of the rectangle.

OR there could be

ABB
ACC

and its reflection. $ 2A_{n\minus{}2}$ ways to tile the rest (one for this and one for the reflection)

And there is

EFF
EDC
BDC
BAA

and it's reflection. 2$ A_{n\minus{}4}$;

Proceeding like this, we get $ A_n \equal{} 3A_{n\minus{}2} \plus{} 2A_{n\minus{}2} \plus{} 2A_{n\minus{}4}...2A_0;$

Since there is 1 way to tile a 3 x 0 rectangle (No tiles at all), we can use this recursion to get B.[/hide]


What is the smallest positive four-digit number that gives a quotient 432 with remainder 2 when divided by a positive one-digit number?
[hide]If the positve one-digit number was $1$, then the four-digit number would be $434$, but that's not an four-digit number. If the positve one-digit number was $2$ then the four-digit number would be $866$, which is not four-digits either. If the positve one-digit number was $3$, then the four-digit number would be $1298$. $1298$ would be the answer, I think[/hide]

[hide][color=indigo]If it is remainder 2 that means the one digit number is 3, 4, 5, 6, 7, 8, or 9. You want the smallest possible quotient, so you want the smallest of the numbers which is 3. 3*432+2 = 1298[/color] :D [/hide]

GOOD JOB EVERYONE

and ...

to Lucy, [hide="lucy only"]don't be afraid to say wrong answers..(you don't have to put the words "I think" at the end..)

regards <-copying Silverfalcon... :D [/hide]


How many different prime numbers are factors of $ N$ if
\[ \log_2 (\log_3 (\log_5 (\log_7 N))) \equal{} 11?
\]$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 7$

[quote="Silverfalcon"]How many different prime numbers are factors of $N$ if

\[ \log_2 (\log_3 (\log_5 (\log_7 N))) = 11?  \]

$\text{(A)} \ 1 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 4 \qquad \text{(E)} \ 7$[/quote]

[hide]5

Simplifying:
$N=7^{{{5^3}^2}^{11}}$
so the prime factors are 7,5,3,2,and 11 [/hide]

[quote="Silverfalcon"][quote="ch1n353ch3s54a1l"][quote="Silverfalcon"]How many different prime numbers are factors of $N$ if

\[ \log_2 (\log_3 (\log_5 (\log_7 N))) = 11?  \]

$\text{(A)} \ 1 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 4 \qquad \text{(E)} \ 7$[/quote]

[hide]5

Simplifying:
$N=7^(((5^3)^2)^{11})$
so the prime factors are 7,5,3,2,and 11 [/hide][/quote]

Are you sure? Because your answer is none of the given choices.  :)[/quote]

oops. forgot about those  :D o.0 I don't see how to get another answer...

[quote] 5

Simplifying:
N=7^{{{5^3}^2}^{11}}
so the prime factors are 7,5,3,2,and 11[/quote]
[hide]
N has only 1 prime factor, and that is 7.  5, 3, 2, and 11 are not factors of N because N is 7 raised to the power of a (large) number. [/hide]


calculus
AMC
AIME
geometry
search
MATHCOUNTS
AMC 10

It seems like everyone here is learning calculus from 8th grade and studying hardcore math from 9th grade. I just feel like I'm way behind.

Though I'm pretty skilled(enough to score top 5% among the competitors and win many good awards during 5-7 grade in korea)
but after I moved to this lame school in europe and I IMPROVED BY 0, LEARNED NOTHING for 3 years(4 considering that my skipping 8th grade counts).. NADA!!

only thing i did in 7th grade was place 1st in Math Count in Mediterranean and 7th in europe (which sucks coz top 5 went to state competition or something)

I just learned about those numerous math competitions after I came to the states in my junior year.

I have no experience in High School competition. Now I'm not even sure I'll be good at this.

Is being a Junior too late to prepare for AMC 12 and AIME and stuff???
I'm desperate to study for this because it will be a plus for college if i win something. and I have only 2 APs which leave a lot of free times.

How exactly do I prepare for these competitions?? I just don't know where to start.
I ordered Art and Craft of Problem Solving, but what else should I do.

Should I have enough time to improve till the competition starts? Will the competition be available on my senior year before college?

 It will be very generous of you to give me advice and telling me how to prepare for it.

Just wondering - in which European country did you live? And how old are you?

My opinion is - it is never too late to improve your problem solving skills. Even if it's a bit late to do really well in AMC/AIME etcetera (I don't know, maybe it's not in your case), it would still be useful during the rest of your life.

:) thx for replying..

I lived in Turkey... went to american school

I'm 15 now.. a year younger than my grade level coz i skipped 8th grade

it's never too late to prepare. just practice practice amc's and aime questions and you'll be fine. i'm sure there are plenty of high AMC scores that come from kids who are taking a regular grade level math class. (ex: friend took amc 10 last year as a freshman, got 137ish, took geometry then)

15? PLENTY of time!!

Haha, I started when I was 16 and I got quite far!

Just work hard, I'm sure you can get there.

lol great to hear that
so solution to success.. work hard

so can you tell me what kind of book i should buy?

when i solved the sample amc questions, i basically knew many of the algebra geometry stuffs except ones you have to know the exact formula or symbols

should i buy the AoPS book???
do you recommend buying the AoPS before starting ACoPS or just doing the ACoPS?
(I'm really short of money after buying the ACoPS)

I did everything without books, using internet resources only.

And in case you really want to buy books: there are LOTS of topics here in which the "which book should I buy?"-issue has been discussed. Have a look at those :)

Are there really resources on the internet that can substitute for A(C)oPS?? Like what?

There is Kalva, there is Mathworld, there are plenty of online lectures notes from programs like MOP and others, there is google (search for "mathematical olympiad" and see what you get). Oh, and the most important one: Mathlinks/AOPS.

I am convinced you don't need more than that to get very far.

sweet :)
that's amazing how you did everything from internet

[quote="yarakvar"]It seems like everyone here is learning calculus from 8th grade and studying hardcore math from 9th grade. I just feel like I'm way behind.
[/quote]

Hardcore math? :?
Anyways, I'm sure you can do well, I did Mathcounts as a recreational thing in 7th grade, and right after placing 11th in Chapter level in 8th grade, I thought, in this order, "Say, these math competitions are pretty fun, unlike school math", then "Gee, I want to do well in states", then "OH ****, I only have a month left to study". Well, I ordered the AoPS books, did as much as I could, did old Mathcounts competitions, and guess what? I improved from a measly 11th in chapters to 7th in states, in MA, which had I believe the second highest combined top 10 score in states. Now, that's an improvement! If only I had taken math seriously earlier...well, you have 24x as much time as I do, make good use of it. :D  :surf:

Yea, I know how you feel.  I wish I had discovered AoPS in 6th grade.  Granted, I've been a member of this site since 7th, but I never really took it seriously until recently :(

Same here; I didn't join until April of this last year. It would have been so nice to know about AoPS when doing MathCounts, which I didn't even start until 7th grade...

As for the original topic, I didn't start MathCounts until 7th grade, then went on to tie for first in the state (California) in 8th grade, and 42nd in the nation (kind of messed up on that test, but whatever). I took the AMC/AIME for the first time last year; I got a 134.5 on the AMC10, then got a 3 on the AIME, but after joining AoPS and doing a fair amount of practicing, I'm confident that I can get at least 10 on the AIME. I can almost always do 7 or 8 of the first 10 problems in under an hour now. :what?:

I didn't discover AoPS until my freshman year, and my AIME score went from 0 or 1 to a consistent 5-7 this summer :ninja: Anyway Im just in precalc, and calculus is pretty much useless for high school competition math except in some rare occasions in the AIME.
Read AoPS books and do lots and lots of tests and problems, you will notice all sorts of things in the problems that would have meant nothing before.

Kalva has lots of old AIMEs and USAMOs, although it's probably easier to just buy the AMC 12 bundle and AIMEs up from 1994.

I wasn't aware that we were doing 'hardcore' math.  :weightlift:

 :P

Fifteen is not late. The way I see it, the more interest a person has, the faster that person will progress. Don't worry too much about doing better than everyone else in some national competition; that's not the point of mathematics. You were top 7 in Europe??? Wow... You certainly beat me. I never got far in MathCounts, but the I look upon MC as a good bridge experience between the basics and serious competition math. It set me in the right direction for sure - I learned what I didn't know, and I realized what I needed to learn. If you have the interest, I'm sure you'll do fantastically.

The Art and Craft of Problem Solving is outstanding, and so are the AOPS books. I have the Salkind Algebra and Geometry books. I also ordered some of the old AMC and AIME problems. If you don't feel like spending the money, you can just go to Kalva, like somebody above me suggested.

BTW - I still don't know calculus. So sue me.  :cleaning:

[quote="white_horse_king88"]BTW - I still don't know calculus. So sue me. [/quote]

Be prepared to appear in court Wednesday, the nineteenth of October, 2005 at 9:00 AM.

Ooh, a court hearing. I sure wanna see this.

You don't need calculus on any of the AMC/AIME/USAMOs.  On the website, it is stated several times that the problems can be solved using Precalculus methods.

BTW, I don't know anyone that took Calculus in 8th grade, though I know a few people, including myself, who are taking/took it in 9th grade.

Wow, I know myself.

PS- 100th post.  Yeah!!!!!   :first:  :lol:  :jump:

I know someone who took calculus in 8th grade... personally, I self-studied Calc AB at the end of 9th. Not very important for problem solving, IMO. I was pretty in tune with competition math throughout middle school, placing in state competitions and such. But I bombed the AMC10 in eighth grade. Next year I switched districts and got myself in a school with a nice math club culture (which I proudly maintain now :D). That gave me the boost I needed to start AoPS and get the ball rolling. Since then, math has been my life (well, 90% anyway).

So yeah, a couple of years is plenty. Just focus and find the beauty in math. You can't miss it if you're looking hard enough.

[quote="Arne"]I did everything without books, using internet resources only.[/quote]Arne, you are a special case :D

If you are just starting up with the AMCs I stronly recommend you buy AoPS Volume 2. It's a great book for students just starting there. I will help you learn the necessary steps quickly and move on to the next level.

Of course a lot of information can be found on our site (including [url=http://www.mathlinks.ro/Forum/resources.php]resources[/url] from competitions, links, [url=http://www.mathlinks.ro/Classes/AoPS_C_About.php]online classes[/url], problems, discussions, tips from other people, etc.) but it's a good idea to have something that was structured and designed for students like you to begin with.

As you will advance in knowledge you will learn to use the internet resources better and better, but in the beginning it may be a bit overwhelming.

Actually, I'd recomend even buying AoPS volume 1 when you're starting out with the AMC's.  I was already an AIME qualifier last year, and I'm still learning a lot from volume 1.


For how many integers $n$ between 1 and 100 does $x^2+x-n$ factor into the product of two linear factors with integer coefficients?

$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 9 \qquad \text{(E)} \ 10$

[hide="hint"]Use the quadratic formula, and look at the $b^2-4ac$ part. You want to look closely at the c, which is n in this case. [/hide]

there's also a nice factoring way, as the problem suggests.

[hide="Answer"]Obviously, for integer coefficients the coefficient of the $x$'s in both terms must be $1$, so we have $(x-a)(x-b)=x^2-(a+b)x+ab=0$. $a+b=-1\Rightarrow a+1=b$, so we have $1<a(a+1)<100$, which has $9$ possible solutions. $\boxed{D}$[/hide]


How many rectangles of any size are on an 8 x 8 checker board?

sum of first 8 squares so

8(8+1)(16+1)/6=204

That's a formula. Why does it work? Can you figure the problem out without a formula?

[hide]for a 8x8 checkerboard, each number added to one dimension takes away from that dimension total. For example, 1x2 rectangles are the same as 1x1 rectangles on a 8x7 board.

thus, the number of rectangles is:
8x8 + 8x7 + 8x6 + 8x5 + ... + 8x1 + 7x8 + 7x7 + 7x6 + ... + 7x1 + ... + 1x1

which can be written as
(8 + 7 + 6 + 5 + 4 + 3 + 2 + 1)(8 + 7 + 6 + 5 + 4 + 3 + 2 + 1)

and 8 + 7 + 6 + 5  + 4 + 3 + 2 + 1 is....

(8 + 1)(8/2) = 36
so the total number of rectangles is ${36}^2$ = 1296 rectangles[/hide]

oh.  I looked at the fourth graders and middle schoolers and i thought they were the same.  So that's basically asking what's the sum of all the products of 1*1,1*2 to 8*7,8*8 so take the sum of the first 8 numbers and square it.

consider the chessboard.  it is 8x8 squares or alternatively you can look at it as 81 lines (9 vertical, 9 horizontal) that intersect in a plane.  any two horizontal and any two vertical lines together make a square (because a square is just two verticals and two horizontals).  Since there are 9 horizontals and 9 verticals, we only need to choose 2 of each.  the number of ways we can do that is the desired number of rectangles:

9c2 * 9c2 = 1296.


This is my 1st tourney. First 8 to sign up participates. Post your current rating. I am kind of active because I go on FTW about 3-5times a week.
1.Berzerk(1209)
2.
3.
4.
5.
6.
7.
8.

1.Berzerk(1209)
2.TwinDracula(1548)(hint: im overrated)
3.
4.
5.
6.
7.
8.
I JOINS!!!

1.Berzerk(1209)
2.TwinDracula(1548)(hint: im overrated)
3.Cogswell(1556)
4.Hexagonal Prism(1563)
5.
6.
7.
8.

I JOINS TOO!!

1ST TOURNEY!!

sorry berzerk.
You are going to have to play a load of 1500's.

(You can beat them)

1.Berzerk(1209)
2.Twindracula(1548)
3.Cogswell(1556)
4.Hexagonal Prism(1563)
5.Maybach(1499)
6.
7.
8.
Everyone is better than me so far.  :rotfl:
Signups will end on Febuary 5,2010.

Can I join???

I am 14_ _, though I got muted. I FAILED MY TESTING, I DIDN'T MAKE THE TEAM!!!  :mad:  :mad:  :mad:

[quote="yaofan"]haven't done these in a while...

1.Berzerk(1209)
2.Twindracula(1548)
3.Cogswell(1556)
4.Hexagonal Prism(1563)
5.Maybach(1499)
6. colinhy(14__)
7. abcak(1400\pm50)
8. yaofan (1574)[/quote]
Ok, so how about this?
Round1
Berzerk vs colinhy
Twindracula vs Cogswell
Hexagonal Prism vs Maybach
abcak vs yaofan
Try to get the results in 1 week.

[quote="TwinDracula"]is it a countdown or game???

cogswell never talks to me...so how do i get his attention to play me

(am i on my own???)[/quote]
Oops. It is a countdown round. You guys get to choose just ONE countdown or best 2 out of 3.

colinhy has not replied to me for 3 days. I'll give colinhy another 2 days.

[quote="TwinDracula"]well...

I think Cogswell beat me a week ago, but I didn't know if we were playing for fun of for your tournament.

hmm...
does he win?[/quote]
I think you and Cogswell can do another cd because there was no result from anyone else. Should I kick colinhy? colinhy had not login in 6 days.

Vame (1200)
EDIT: can't, nevermind, sorry.

Should I kick colinhy? The guy did not appear for about 10 days.

If you kick him, I would like to join. (1600+/-50)

I have quit the tourney. I have told that Lex1103 will host the tourney.


how do you get your avatar pic on to your account I have tried every thing posible and I cant get anything from my computer please hlp me  :what?:

to get an avatar:

choice #1
u can get the avatar pic from the internet
to do that go to google or yahoo or somewhere
and search in the pic you want
the pic has to be 80 by 100 pix. or 100 by 80 pix, something like that

*if the pic u want is larger than 80 by 100 pix, or cannot find a picture that is that small, copy and save the pic to ur computer. if you hav microsoft office picture manager or something similar to it you will be able to make the picture smaller. if u dont hav one, paste the pic to your microsoft word and crop it or make it smaller or whatever. make sure u save the smaller picture. look at choice #2 to use pictures from ur computer for ur avatar

anyway
after u get ur pic, copy the URL (the http://www.thing......)
then come to AoPs and log into ur account
click 'profile' (its in a blue box, beside 'private messages)
and click 'avatar control panel'
paste the URL where it asks u to paste it.
then click submit
if u did everything right, u should get an avatar

choice #2
for this one, ur getting a pic from ur computer.
go to ur account
click 'profile', then, 'avatar control panel'
where it says 'Upload Avatar from your machine', click browse
then find the pic you want and click 'open'
there should be something written in English on there after u click 'open'
click submit
u should hav an avatar after that

i hope that helps u

[quote="danb80"]An easy way to resize your picture for an avatar:

Go to [url]http://www.shrinkpictures.com/create-avatar/[/url] and follow directions--it's free![/quote]
Wow! That's nice!

Thanks! (Evn though I'm not supposed to be here :ninja: )

thank you Mr. B80 for the advice.  :clap2:  :clap2:

I am curious...

Who are you?(just give me a hint, since we can't post our real names..)

[quote="danb80"]DK:  Look under the "I don't know" topic.[/quote]

Oh.......well that was obvious...


For a math contest there is a shortlist with $ 46$ problems, of which $ 10$ are geometry problems. The difficulty of every two problems is different (so there are no two problems with the same difficulty). Let $ N$ be the number of ways the selection committee can select $ 3$ problems, such that
- Problem $ 1$ is easier than problem $ 2,$
- Problem $ 2$ is easier than problem $ 3$,
- There is at least one geometry problem in the test.
Calculate $ \frac{N}{4}$.

[hide="Solution"]Given an arbitrary selection of three problems, there is only one way to order them such that they are in ascending order of difficulty. Therefore, there are $ {46\choose 3} \equal{} 15180$ possible tests. However, we must compute the number of tests with no geometry problems. This is $ {36\choose 3} \equal{} 7140$. $ N \equal{} \frac{15180\minus{}7140}{4} \equal{} \boxed{2010}$.

[/hide]


abstract algebra
superior algebra
superior algebra unsolved

This problem was presented to me in my Abstract Algebra class.  The professor admitted he did not have an answer for the problem, but was hoping someone could come up with something.

Let [i]G[/i] be a finite abelian group.

For each [i]n[/i] (whcih is an element of the natural numbers), suppose the number of elements in
{ [i]x[/i] which is an element of [i]G[/i] | [i]x^n = e[/i] } is no larger than n.

Prove that [i]G[/i] is cyclic.


Give a sequence
$ A_k\equal{}\frac{1}{32}( (17\plus{}12\sqrt{2})^k\plus{}(17\minus{}12\sqrt{2})^k\minus{}2 )$

$ k$ is a natural number.
Prove that $ A_k$ is triangular number.

thanks!

Do not double post:

[url]http://www.mathlinks.ro/viewtopic.php?t=302438[/url]


algebra
polynomial
number theory solved
number theory

Let $P(x)$ a polynomial with integers coefficients such that $P(n) > n$ , for all positive integer $n$
Suppose that for each positive integer $m$  exists a term of the succession :

               $P(1),P(P(1)),P(P(P(1))),.........................$

that is multiple by $m$ .Prove that $P(x) = x+1$

 :juggle:  :play_ball:  :clap:

Look here: http://www.mathlinks.ro/Forum/viewtopic.php?t=75594  :D

Oops it had not given account me that already was resolute, thanks for the Link .................... :)  :roll:


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Hi.

Im stuck at the following problem and hope someone can shed some light.

Let A be a n*n matrix on field F and let T be linear operator (n*n) such that:

$T(X)= AX$

Prove the minimal polynomial of $T$ is the minimal polynomial of $A$.

note: The matrix representing T according to any base is $n^{2}*n^{2}$.

Thanks :)

[quote="jmerry"]First, show that $p(T)(X)=(p(A))X$ for any polynomial $p$.
Now, what does it mean for a polynomial $q$ to be the minimal polynomial of an operator?[/quote]

How do I actually show that?

Do I take a general polynomial like $Ax^{n}+Bx^{n-1}+....$

And a random n*n matrix and just do all the computations by substituting that matrix for x ?

Also im not familiar with the marking p(A)X ...

Does it mean if i have a polynomial: ax^2+bx+c , p(A)x would be aA^2X+bAX+cX ?

Thanks.

The definitions: If $p(x)=a_{m}x^{m}+a_{m-1}x^{m-1}+\cdots+a_{0}$, $p(T)=a_{m}T^{m}+a_{m-1}T^{m-1}+\cdots+a_{0}\cdot\text{Id}$, where $T^{k}=T\circ T\circ\cdots\circ T$ ($k$ times), and $\text{Id}$ is the identity operator $\text{Id}(X)=X$. Also, $p(A)$ is the matrix $a_{m}A^{m}+a_{m-1}A^{m-1}+\cdots+a_{0}I$.


1. Express $ \sum_{k \equal{} 0}^{n}{( \minus{} 1)^k{\binom{n}{k}}^2}$ in closed form.

2. Let $ u(i)$ denote the sum of the digits of the base-2 representation of $ i$. For $ k \le mn$, find $ \sum_{i \equal{} 0}^{2^k}{{( \minus{} 1)}^{u(i)}\binom{\binom{i}{n}}{m}}$.

3. Prove that $ \sum_{i \equal{} 0}^{2k}( \minus{} 1)^i\binom{n}{i}\binom{n}{2k \minus{} i} \equal{} ( \minus{} 1)^k\binom{n}{k}$

[hide]1.  Think about this as a product of two polynomials.  Then think about what coefficient you are trying to find.[/hide]


thanks this will really help :)

btw for the people who don't feel like opening the link(lol), it contains past amc tests and solutions

meh i used it and they are fine. But the book is good 2.

That Kansas University is pretty famous in forum. I've seen that site at least several YEARS ago. I'm not big fan of Problem "Set" for AMC though because working in order of difficulty helps you to predict types of questions you might encounter.

If you are looking for Problem Sets, I recommend ARML. That's an excellent source of problems for harder AMC and some easy-medium AIME.

[quote="rOadToAIME"][quote="MCTP"]I would be careful using those... Some of the problems have errors in them, which can really throw you off if you are taking the tests under timed conditions.[/quote]
Do you know which problems have errors? :huh:[/quote]
#4 on the 2001 AMC 12 is an example. I don't have an exhaustive list, of course, but I'm sure there are more.

[quote="AIME15"]I can't find ARML tests anywhere on the internet.  They don't allow people to publish their problems on the internet right?[/quote]

Only previous year is available (2007 for 2008). Older versions are in two books. I have the second book and it's very useful. I suggest you look at ARML Marathon or Problem Day (ARML problem per day) in ARML forum. That's the best source of online ARML problems.


The real numbers $a$, $b$, $x$, $y$ satisfy $ax + by = 3$, $ax^2 + by^2 = 7$, $ax^3 + by^3 = 16$, $ax^4 + by^4 = 42$. Find $ax^5 + by^5$.
Well the simple way is using coefficients, so I want a FULL-PROOF-NON-GUESSWORK-SOLUTION pls, can anyone do it? :?  :(

Whoa, this is one of the last problems on a previous AIME....it's at least Intermediate level.  If you want a solution, there should be one in one of the AIME practice series threads or whatever it's called.

Is there a formula for the sum of fifth powers? That won't help, but I can't see anything to do with coefficients.

where I saw the solution it has some trick to it, I was hoping someone could solve it without the trick. :?

What trick is it ? I first think of adding up all terms given and try to connect with the question by in vain .... maybe a hint from the trick ?  :D

well.... okay, look here,
[hide="Solution"]
 This needs a trick: $(ax^{n+1} + by^{n+1})(x+y) - (ax^n + by^n)xy = ax^{n+2} + by^{n+2}$. Hence $7(x+y) - 3xy = 16, 16(x+y) - 7xy = 42$. Solving $x+y = -14, xy = -38$. Applying again: $ax^5 + by^5 = 42(x+y) - 16xy = 20$.
[/hide]
this is not my solution but the solution from the test papers.

I see why that works, it you expand it, you'll see that everything cancels except $ax^{n+2}$ and $by^{n+2}$

But I didn't know that  :(

nor did I, that's why I asked if there was an alternate solution to the problem.

Another way to look at it is aka Fbonacci  that is the term is linear combination of previous two terms..
Obviously The terms go as: $T_n = 38 T_{n-2} - 14 T_{n-1} $

(For example check that $38 \cdot 3 - 14 \cdot 7 = 16$ and  $38 \cdot 7- 14 \cdot 16 = 42$
so next term would be $38 \cdot 16- 14 \cdot 42 = 20$
(And if you wanted the next term it would be:  $38 \cdot 42- 14 \cdot 20 =1316$

If you are wondering  from where did I pick up 38 and 14  and don't believe it that I ran  into them when I was on  my way to  visit my old pal Newton  [hide="click here"]okay if you  have to cross every $t$ and dot every $i$ you can do it by solving for p and q in
16 =  3p + 7q  and 42 = 7p + 16 q .
The method is general and will work every time. [/hide]

Added later: Sorry , did not see the solution, posted above.. yes it is similar and kown as Newton's sum method.

The problem was number 14 or 15 on an AIME.  You would be totally naive to expect the solution to be easy!  Those types of problems require a lot of investigation.

You might start investiagating by looking at how to get from $ax+by$ to $ax^2+by^2$.  Then, you would look for a generalization for $ax^n+by^n$ to $ax^{n+1}+by^{n+1}$.  So how do you get from $ax^n+by^n$ to $ax^{n+1}+by^{n+1}$?  Well, multiplying $ax^n+by^n$ by $x+y$ is a start.  Thus

\[(x+y)(ax^n+by^n)=ax^{n+1}+by^{n+1}+ax^ny+bxy^n\Rightarrow ax^{n+1}+by^{n+1}=(x+y)(ax^n+by^n)-xy(ax^{n-1}+by^{n-1}).\]

And we go from there.

Also, one could recognize that for example

$x^n+y^n = (x+y)(x^{n-1}+y^{n-1}) - xy(x^{n-2}+y^{n-2}) $

from Newton's sums, and try to integrate those $a$s and $b$s in there.

Cool solutions, Why didnt I think of them :wallbash_red:  :wallbash:  :wallbash_red:  :wallbash:  thx.

P.S I especially liked Gyan's solution even though it was hard to find the nth term, but it was a unique way of approaching the problem.


[quote="wilddeer"]:) http://www.imocn.com/UploadFile/2006-1/2006127102252763.gif[/quote]
It should be $f(x)=\frac{4^x}{4^x+2}$.

[quote="Fourier"]The answer is 1002,maybe i'm wrong.
it's an easy one.[/quote]
Oh, you're wrong. :(
I think the answer is $500$.

I think it's an easy question.F(x)+f(1-x)


this is REALLY easy, just givin for kicks

Solve for x,y,z
$z^x=x$
$z^y=y$
$y^y=x$

[quote="Melissa"][hide]Are they all equal to 1?[/hide][/quote]

I think they have to be? THis is sort of a dumb question.  :D

All equal to 1 is a trivial solution. There is another solution!

$z^x = x\\ z^y = y\\ y^y = x$

Then from first and third, $z^x = y^y$. Raise both sides to the $y$th power:

$z^(xy) = y^(y^2)\\ (z^y)^x = y^(y^2)$

But $z^y = y$ from second, so

$y^x = y^(y^2)$, then $x = y^2$. But from third equation $y^y = x$, so

$y^y = y^2\\ y = 2\\ x = 4$

From second $z^y = y$

$z^2 = 2\\ z = \pm \sqrt{2}$

Our solution set is then $(x,y,z)=(1,1,1),(4,2,\pm \sqrt{2})$

[color=green][size=75]($\LaTeX$ and formatting adjustments by chess64)[/size][/color]

[quote="MathMaven53"]Then from first and third, $z^x = y^y$. Raise both sides to the $y$th power:
[/quote]

or you can just substitute (2) into R.H.S... Same thing NEway. :P