Fie $\ a\geq\ 2$ un numar natural.Sa se arate ca multimea {$\ a^2+a-1,a^3+a^2-1,...,a^{n+1}+a^n-1...$} contine o submultime infinita de numere relativ prime doua cate doua.
  $\text{ Mircea Becheanu, Bucuresti}$

frumoasa problema.
sper sa fie buna solutia asta...
[hide="solutie"]presupun ca am construit o multime $A=\left\{n_{1},n_{2},\ldots,n_{k}\right\}$, astfel incat numerele $a^{n_{i}+1}+a^{n_{i}}-1$ sa fie doua cate doua prime intre ele pentru $i=\overline{1,k}$.
notam cu $P_{i}=\left\{p\mbox{ prim }\Big| ~ p|(a^{n_{i}+1}+a^{n_{i}}-1)\right\}$, si cu
$Q=\bigcup_{i=\overline{1,k}}P_{i}$. multimea $Q$ contine doar numere prime.
pentru inceput observam ca $P_{i}\cap P_{j}=\emptyset$ pt $i\neq j$.
notam
$s=\prod_{q\in Q} (q-1)$.
Presupunem ca exista $p\in Q$ astfel incat $a^{s+1}+a^{s}-1 \equiv 0 (\bmod ~ p)$. Cum $(p-1)|s$, rezulta ca $a^{s}\equiv 1 (\bmod ~ p)$, deci $a^{s+1}\equiv 0(\bmod ~ p)$, adica $a\equiv 0(\bmod ~ p)$. deoarece $p\in Q$, rezulta ca $p\in P_{r}$, cu $1\leq r\leq k$. atunci, rezulta ca $a^{r+1}+a^{r}-1\equiv 0(\bmod ~ p)$, dar cum
$a\equiv 0(\bmod ~ p)$, rezulta $-1\equiv 0(\bmod ~ p)$, contradictie.
alegem $n_{k+1}=s$, si pentru $j=\overline{1,k+1}$ notam $X_{j}=a^{n_{j}+1}+a^{n_{j}}-1$.
Initial numerele $X_{1},X_{2},\ldots,X_{k}$ erau relativ prime (din ipoteza de inductie), iar $X_{k+1}$ este relativ cu $X_{j}$ pentru orice $j=\overline{1,k}$, deoarece $X_{k+1}$ nu are factori primi ce se afla in $Q$, iar toti factorii primi ai lui $X_{j}$ sunt in $Q$.
deci $X_{1},X_{2},\ldots,X_{k+1}$ sunt prime intre ele. acest fapt ne spune ca putem gasi o infinitate de numere, aplicand algoritmul de mai sus.[/hide]


logarithms
limit
function
calculus
derivative
algebra
binomial theorem

The tricky thing about this problem comes as a result of its purpose. In particular, you can't differentiate the exponential while solving this, because this limit needs to be calculated before the exponential can be differentiated.

Say $ n^x \equal{} e^{\ln x}$, then the problem becomes $ \lim_{x \to 0} \frac {e^{x \ln n}}{x \ln n} \equal{} 1$, or equivalently that $ \lim_{x \to 0} \frac {e^x \minus{} 1}{x} \equal{} 1$

From here, it depends on how you've defined the function $ e^x$.

- If you've [i]defined[/i] it in terms of its power series, then you can just substitute directly.
- If you've [i]defined[/i] it as the inverse of the natural logarithm, then you can express this statement equivalently as $ \text{exp} '(0) \equal{} 1$, and use the derivative of the logarithm to help you out*.
- If you've [i]defined[/i] it as $ e^x \equal{} \lim_{n \to \infty} \left(1 \plus{} \frac {x}{n} \right)^n$, then you can use the binomial theorem and assert that the terms you want to drop out will drop out.

Have I missed any definitions? I guess the point is, for some of these functions, there are many interesting relations that they satisfy. We have to pick one way to define them, and then show that all the other forms are equivalent.

* - This obviously implies that you've defined the natural logarithm first somehow, and works depending on how you have done so.

oh.my english is't good.i can't translate your text well.but if you let u=$ n^x \minus{} 1$ then  using $ \lim_{x\rightarrow 0}ln(u \plus{} 1)^{\frac {1}{u}}$=1.is it easier ,athough liking your answer?


How many multiples of 11 are there between 100 and 1000?

basically there are 900 numbers between 100 and 1000.
So to find how many numbers that are divisible by 11, you do 900/11, which
is 81.81818181.....

You round down since its asking numbers between 100 and 1000.

So your answer is $ \mathfrak{81}$.

$ 110, 121, ...990$

Divide by 11.

$ 10, 11, ..., 90$

Clearly, there are $ 90\minus{}9 \equal{} \boxed{81}$ multiples.

OR

$ \lfloor{\frac{1000}{11}\rfloor} \minus{} \lfloor{\frac{100}{11}\rfloor}$

$ 90 \minus{} 9 \equal{} \boxed{81}$

The smallest # multiple of 11 is 110 in the 3digit  the largest 990

990-110/11   +1 =81

[hide="MULTIPLES ARE AWSOME"]81 IS THE ANSWER YOU SEEK : )[/hide][/quote]

[size=200][color=#FF80BF][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH IT IS GREATLY APPRECIATED : )[/u][/i][/b][/color][/size]

[size=150][color=#8000FF][b][i][u]I HAVE 99.9898989898989899898989898% CONFIDENCE THAT IM CORRECT [/u][/i][/b][/color][/size]

[size=50]I KNOW THIS IS SO SMALL BUT WILL YOU POST ME IF IM WRONG : ) THANKS!![/size]

 :starwars:  :starwars:  :starwars:  :starwars:


Twelve students are to be divided among Mr. Mirus's and Ms. Batty's classes. No teacher is to have more than 8 students. How many different groups of students could be in Mr. Mirus's class?

We can't have any more than 8 students in each class, so these are our possible ways:

8,4
7,5
6,6
5,7
4,8

There are $ \boxed{5}$ different ways.

Yes, izzy, this is what I thought.  But you can have a different 8 people or 7 people.  For example, a class could be like (random names):

Abigail
Brendan
Connie
Don
Julia
Sarah

or

Avery
Brianna
Connie
Don
Johnny
Sarah

so how many different ways are possible?

But when I typed in 5, it was counted as correct...hmm...

[quote="GameBot"]Twelve students are to be divided among Mr. Mirus's and Ms. Batty's classes. No teacher is to have more than 8 students. How many different [b]groups[/b] of students could be in Mr. Mirus's class?[/quote]

read the question ernie--groups means amount not certain ppl

The most a class can have is 8 students, so we find that these are the possible answers...(the first number is one class, the second number is the second class)

8,4
7,5
6,6
5,7
4,8

There are $ \boxed{5}$ different ways.

Oh, duh...

And mz, i almost never read.  :P

lol.
try reading once in a while; you'll like it :P

now it's 4 posts!  :P
ernie, ernie, ernie.....

:P  :P  :P

lol ernie no problem i do that a lot

--i messed up my cd round like that--i read 8000miles as 8000mph

[quote="isabella2296"]It [b]is[/b] a bit helpful - but Ernie's a fan of only reading the last line, and skimming it, too.  :P

(haha, i just realized that in three consecutive posts, we all used to  :P  guy.)[/quote]

Izzy is right.  :P  (just to keep it going)

this has seriously disintegrated to spam stop posting here for no reason pplz

 :P

[quote="myyellowducky82"][quote="isabella2296"]Heed mz's advice. We shall all stop spamming.

This forum should get a mod...hmm...I wouldn't mind being a mod!!!  :P[/quote]
wait so regular users can be mods???[/quote]

NICE GOING YOU BROKE THE CHAIN!!!! :P

yes they can: admins i think appoint you or somethin like that. All mods, i think started as regular users

Just curious: do the admins just choose people to be mods, or do you have to ask or something?


Solve this equation:
 $\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{3}{2}\sqrt{\frac{x}{x+\sqrt{x}}}$

$x>0$ and then $x\geq \sqrt{x}$ gives $x\geq 1$.
Multiplying with $\sqrt{x+\sqrt{x}}$ gives $x+\sqrt{x}-\sqrt{x^{2}-x}={3\over 2}\sqrt{x}$.
Dividing with $\sqrt{x}$ gives $\sqrt{x}-\sqrt{x-1}={1\over 2}$
and then $\sqrt{x}+\sqrt{x-1}={1\over \sqrt{x}-\sqrt{x-1}}=2$.
The LHS is strictly increasing with unique solution $x={25\over 16}$.


Let $ABC$ ($\angle{ACB}=90$) be right angle triangle. Consider a variable point $D$ on side $AC$. Circle with center $O$ through $D$ and tangents $AB$ at $A$.Prove that circumcircle of $\triangle{AOD}$ through a fixed point (not $A$) when $D$ is vary.

probelm is wrong.

what are B and C for?
notice that all $ \triangle AOD$ are homothec with center $ A$ so their circumcircles are tangent to each other.hence they only pass through $ A$.


In each cell of $ 2007 x 2007$ table there is an odd integer. Denote by $ Z_{i}$ the sum of the numbers in $ i\minus{}$th row and by $ S_{j}$ the sum of the numbers in $ j\minus{}$ th column.
Let $ A\equal{}\prod_{i\equal{}1}^{2007}Z_{i}$ and $ B\equal{}\prod_{j\equal{}1}^{2007}S_{j}$.
Show that $ A\plus{}B$ can not be equal to zero.

See here http://www.mathlinks.ro/Forum/viewtopic.php?t=149440 .


If $ f(x)\equal{}\frac{x(x\minus{}1)}{2}$, then $ f(x\plus{}2)$ equals:
$ \textbf{(A)}\ f(x)\plus{}f(2) \qquad\textbf{(B)}\ (x\plus{}2)f(x) \qquad\textbf{(C)}\ x(x\plus{}2)f(x) \qquad\textbf{(D)}\ \frac{xf(x)}{x\plus{}2}\\
\textbf{(E)}\ \frac{(x\plus{}2)f(x\plus{}1)}{x}$

[hide]
Looking at each of the options individually and very quickly testing each out,

$ \boxed{\text{E}}$
[/hide]

I think there would definitely be a direct method to find the answer, but I can't figure out what that method is either.  :P

[hide="Click for solution"]
$ f(x\plus{}2)\equal{}\frac{(x\plus{}2)(x\plus{}2\minus{}1)}{2}$ and $ f(x\plus{}1)\equal{}\frac{(x\plus{}1)(x\plus{}1\minus{}1)}{2}$.  So $ \frac{x\plus{}1)}{2}\equal{}\frac{f(x\plus{}1)}{x} \implies f(x\plus{}2)\equal{}\frac{(x\plus{}2)f(x\plus{}1)}{x}$, or $ \boxed{\textbf{(E)}}$.
[/hide]


Let $ p(n)$ denote the number of unrestricted partitions of $ n$. Show that $ p(n)\geq 2^{\lfloor \sqrt{n} \rfloor }$ for all $ n\geq 2$.

[hide="Solution"]
Clearly, it works for $ n \equal{} 2$ as there are $ 2$ partitions of $ 2$ (namely $ 2$ and $ 1 \plus{} 1$).  Then, it suffices to prove that $ p(n)\ge 2^{\sqrt {n}}$ for when $ n$ is a perfect square.  This is because $ p(n) > p(n \minus{} 1)$, but unless $ n$ is a perfect square, the floor of $ \sqrt {n}$ is equal to that of $ \sqrt {n \minus{} 1}$.  Now, let $ n \equal{} k^2$.  Consider the set $ \{ 1, 2, 3, \cdots , k \}$.  There are $ 2^k \equal{} 2^{\sqrt {n}}$ such subsets of these.  Notice that the maximal such sum of any subset is $ \frac {k^2 \plus{} k}{2}$.  To each subset, we will add one more number.  Let this number be $ l_1$ for subset $ S_1$.  Say that the sum of the elements of $ S_1$ is $ a_1$.  Then, we define $ l_1$ to be $ n \minus{} a_1 \equal{} k^2 \minus{} a_1\ge k^2 \minus{} (\frac {k^2 \plus{} k}{2}) \equal{} \frac {k^2 \minus{} k}{2} > k$ when $ k > 3$.  We quickly test the possibilities for $ k \equal{} 2, 3$.  These yield that $ n \equal{} 4, 9$.  For $ n \equal{} 4$, there are $ 5$ partions (namely $ 1 \plus{} 1 \plus{} 1 \plus{} 1$, $ 1 \plus{} 2 \plus{} 1$, $ 2 \plus{} 2$, $ 4$, and $ 1 \plus{} 3$.)  For $ n \equal{} 9$, there are at least $ 8$ possibilities (such as $ 9$, $ 8 \plus{} 1$, $ 7 \plus{} 2$, $ 6 \plus{} 3$, $ 5 \plus{} 4$, $ 1 \plus{} 1 \plus{} 7$, $ 1 \plus{} 1 \plus{} 1 \plus{} 6$, and $ 1 \plus{} 2 \plus{} 6$).  Now, we only need to consider $ k > 3$, so $ l > k$ for all subsets $ S_i$.  Notice as there are $ 2^k$ subsets, there are $ 2^k$ such partions created (notice that when we add $ l$, a partition is made because of the definition of $ l$.)  We now show that these subsets are unique.  Notice that every two distinct subsets differ in an element, say $ e \le k$.  Let subset $ S_i$ have $ e$ and subset $ S_j$ not.  Then, we add $ l_i$ to $ S_i$ and $ l_j$ to $ S_j$.  Since $ l_i, l_j > k$, $ S_i$ still will have $ e$, but $ S_j$ still will not.  Then, these two subsets are still distinct after adding the specified element.  We have thus created $ 2^k$ subsets (that can be turned into partitions of $ n$), so $ 2^k\le p(n)$.
[/hide]


Which of the following is equal to $ 1\plus{}\frac{1}{1\plus{}\frac{1}{1\plus{}1}}$?

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

We have $ 1 \plus{} \frac {1}{1 \plus{} \frac {1}{2}} \equal{} 1 \plus{} \frac {1}{\frac {3}{2}} \equal{} 1 \plus{} \frac {2}{3} \equal{} \frac {5}{3}$.
So the answer is $ \text{C}$.

Also stuff in that form always have consecutive Fibonacci numbers on the top and bottom.
A,D,E are eliminated.
B is easily wrong and 1+1/2 is not 2.

hello, why don't you use $ \text{\LaTeX}$?
Sonnhard.

[quote=Complex_Ninja]And not revive the thread for no apparent reason.. But he is a new member though  :|[/quote]

but i like reviving threads for no apparent reason! :)

[hide=Solution]
$\frac{1}{1+1}=\frac12$

$1+\frac12=\frac32$

$\frac{1}{\frac32}=\frac23$

$1+\frac23=\boxed{\frac53}$
[/hide]

[quote=RedFireTruck][quote=Complex_Ninja]And not revive the thread for no apparent reason.. But he is a new member though  :|[/quote]

but i like reviving threads for no apparent reason! :)
[/quote]
Go to fun factory then


The increasing sequence $ 1,4,5,16,17...$ consists of all positive integers which are powers of $ 4$ or sums of distinct powers of $ 4$(cannot use a power of $ 4$more than once-so you cant have $ 4^{1} \plus{} 4^{1} \equal{} 8$on this list).  Find the 50th term of this sequence.

[hide="Solution"]
In base 4, for each digit, we may have either a 0 or a 1.

$ a_n$ strictly increases with $ n$, hence if we write $ n$ in binary and $ a_n$ in base 4, the digits will be the same.

$ 50 \equal{} 110010_2 \implies \boxed{a_{50} \equal{} 110010_4 \iff a_{50} \equal{} 1284}$
[/hide]

So this would work with any rational number greater than 2 right?

If this was a sequence of 3s, 5s, or even something like 4.5, you could apply t hat right?

But if it were like 1.5 it wouldnt work because 1.5<2 and it would not be strictly increasing?

(because in order to do that thing with base 2, woundnt you assume $ 4^{n} > 4^{n \minus{} 1} \plus{} 4^{n \minus{} 2} \plus{} .... \plus{} 4 \plus{} 1$ for integer n>0)

The important part of the solution is a bijection between the indices of the sequence written in base $ 2$ and the sequence itself written in base $ 4$ (in particular, the digits used are exactly the same).  The reason this bijection works is because ordering both sets gives the same sequence of binary digits.  A similar idea holds for any [b]real[/b] number greater than $ 2$ (why would you restrict to rationals?).

Edit:  You need to be more specific when you say "it would not be strictly increasing."  The problem with numbers less than $ 2$ is that the ordering of the corresponding sequence is different:  in particular, $ 2$ is the smallest real number $ a$ that satisfies

$ a^n > a^{n\minus{}1} \plus{} a^{n\minus{}2} \plus{} ... \plus{} a \plus{} 1$

for all $ n$.


In a chess tournament each player plays every other player once. A player gets 1 point for a win, 0.5 point for a draw and 0 for a loss. Both men and women played in the tournament and each player scored the same total of points against women as against men. Show that the total number of players must be a square.

Let $k$ be the number of men and $m$ the number of women, with $n=k+m$. The sum of the scores of man-against-man matches is $\binom{k}{2}$, so that is also the total points that men scored against women. Analogously, women score $\binom{m}{2}$ in total against men. But $km$ opposite-sex matches are played, so
\[ \binom{k}{2}+\binom{m}{2}=km \\ k^2-k+m^2-m=2km \\ (k-m)^2=k+m=n, \]
which is what we wanted to prove.


Very coll, very funny problem. I got very surprised and a bit angry when I saw the solution of this problem:
Show that $ \prod_{1\le i < j\le 2009}(i^\dfrac{1}{i} - j^\dfrac{1}{j})\in \mathbb{Q}$.

Look at $ i \equal{} 2$ and $ j \equal{} 4$. :)

Yes, correct :!:  I think this problem is very funny, because can make people loose hours in it, without concluding anything relevant.

This reminds me of a joke problem I encountered in middle school:  compute the product

$ (n \minus{} a)(n \minus{} b)(n \minus{} c) ... (n \minus{} z)$.

:P


inequalities
geometry
circumcircle
angle bisector
geometry proposed

In a triangle $ABC$ denote the length $l_a$ of the $A$-angle-bisector
and the length $m_a$ of the A-median a.s.o. Prove that:

$1.\blacktriangleright \boxed {a^2+b^2+c^2=1\Longrightarrow a^2m_a+b^2m_b+c^2m_c\le \frac 12}\ .$

$2.\blacktriangleright \boxed {\ a^2l_a+b^2l_b+c^2l_c\le 2p^2\sqrt{2Rr}\ }\ .$

I added a new geometrical inequality at the previous message !

Nice problem !

$\displaystyle (a^2m_a+b^2m_b+c^2m_c)^2\leq (a^2+b^2+c^2)(a^2m_{a}^2+b^2m_{b}^2+c^2m_{c}^2)\leq \frac{1}{4}$
It is enough to  prove right inequality
$\displaystyle (a^2m_{a}^2+b^2m_{b}^2+c^2m_{c}^2)\leq \frac{1}{4}$ or $\sum a^2(\frac{b^2+c^2}{2}-\frac{a^2}{4})\leq \frac{1}{4}$

Equalent to $4\sum a^2b^2-\sum a^4\leq 1=(a^2+b^2+c^2)^2$ and  this lead us to

$2\sum a^2b^2 \leq 2\sum a^4$ q.e.d

P.S I understood how you composed second one:

I think you used that $\displaystyle \frac{h_a}{l_a}\geq \sqrt{\frac{2r}{R}}$
look http://www.mathlinks.ro/Forum/viewtopic.php?t=18137

From here $\displaystyle p\sqrt{2Rr}\geq a\cdot l_a$ sum them and we are done.

I think we can also change our problem to

$\displaystyle a^2l_a+b^2l_b+c^2l_c \leq min(2p^2\sqrt{2Rr},\frac{4}{3\sqrt{3}}p^3)$

[color=darkred][b]Lemma.[/b] Let $ABC$ be a triangle. Denote the points $D\in (BC)$ and the second intersection $E$ of the line $AD$ with the circumcircle $w$ of the triangle $ABC$. Then for any point $M\in (AE)$ there is the following inequality: $\boxed {a\cdot AD\le \frac{c^2\cdot DC+b^2\cdot DB}{2\sqrt{p(M)}}}$, where $p(M)$ is the power of the point $M$ w.r.t. the circle $w$.[/color]
[color=darkblue][b]Proof.[/b] Apply the Stewart's theorem to the cevian-line $AD$ in the triangle $ABC$: $a\cdot AD^2+a\cdot DB\cdot DC=$ $c^2\cdot DC+b^2\cdot DB$.But $p(D)=$ $DB\cdot DC=$ $DA\cdot DE$. Thus, $AE=AD+DE=$ $AD+\frac{DB\cdot DC}{AD}=$ $\frac{AD^2+DB\cdot DC}{AD}=$ $\frac{a\cdot AD^2+a\cdot DB\cdot DC}{a\cdot AD}=$ $\frac{c^2\cdot DC+b^2\cdot DB}{a\cdot AD}$ and $AE=AM+ME\ge 2\sqrt{MA\cdot ME}=$ $2\sqrt{p(M)}$. Therefore, $\frac{c^2\cdot DC+b^2\cdot DB}{a\cdot AD}\ge 2\sqrt{p(M)}$, i.e. $a\cdot AD\le \frac{c^2\cdot DC+b^2\cdot DB}{2\sqrt{p(M)}}\ .$

[u]Two particular cases.[/u]

$1.\blacktriangleright M: =I\Longrightarrow \frac{DB}{c}=\frac{DC}{b}=\frac{a}{b+c}\Longrightarrow$ $al_a\le \frac{a(c^2b+b^2c)}{2(b+c)\sqrt{2Rr}}$ $\Longrightarrow$ $l_a\le \frac{bc}{2\sqrt{2Rr}}$ $\Longrightarrow$ $l_a\le \frac{2Rh_a}{2\sqrt{2Rr}}$ $\Longrightarrow$ $\boxed {\frac{h_a}{l_a}\ge \sqrt{\frac{2r}{R}}}$ (a well-known inequality). Therefore, $\sum a^2l_a\le \frac{abc}{2\sqrt{2Rr}}\cdot \sum a=2p^2\sqrt{2Rr}\ .$

$2.\blacktriangleright M: =G$ and $a^2+b^2+c^2=1\Longrightarrow DB=DC=\frac a2\ ,\ p(G)=\frac{a^2+b^2+c^2}{9}\Longrightarrow$
$am_a\le \frac 34 a(b^2+c^2)\Longrightarrow \sum a^2m_a\le \frac 34\sum a^2(b^2+c^2)=$$\frac 32\sum (ab)^2\le \frac 32\cdot \frac 13\left(\sum a^2\right)^2=\frac 12\ .$[/color]


In how many different ways can 12 dimes be divided into three
piles with an odd number of dimes in each pile?

No matter how many combinations you try, you will not get $ 3$ piles, with odd numbers of coins, to combine to $ 12$ total coins.  An odd number plus an odd number plus an odd number ALWAYS equals another odd number.  Since $ 12$ is even, the answer is $ \boxed{0}$, since it is impossible to make an even number with the restrictions given.

[hide="what I thought"]This is similar to what Alcumus said, but I know that an odd number plus an odd number is an even number, and an even number plus an odd number is an odd number.  Therefore, three odd numbers add up to make an odd number.  12 is an even number, so the answer is 0.[/hide]

[hide="My Solution"]Note that any odd number can be written in the form $2p+1$.

Let the $1st$ pile contain $2a+1$ dimes, the $2nd$ pile contain $2b+1$ dimes, and the $3rd$ pile contain $2c+1$ dimes.

We know that the number of dimes in the three piles must add up to 12:

$2a+1+2b+1+2c+1=12$

$2a+2b+2c+1+1+1=12$

$2(a+b+c)+3=12$

$2(a+b+c)=9$.

$a+b+c=\frac{9}{2}$.

Because $a, b,$ and $c$ must be whole numbers, this equation is impossible.

Therefore, there is no way that 12 dimes can be divided into three piles with an odd numer of dimes.

[hide="answer"]There are $\boxed{0}$ ways for 12 dimes to be divided into three piles, each containing an odd number of dimes.[/hide][/hide]

We use generating functions:[hide = Solution]$\left (\frac{x}{1-x^2}\right )^3$ is the product of the three generating functions, and it doesn't contain a $x^{12}$ term. So, the answer is $0$[/hide]

[hide=Sol]Odd+Odd+Odd = Odd.
12 ≠ Odd.
Ans = 0[/hide]

troll question lol

[hide=sol]3 odd numbers are always odd.[/hide]

[hide]Three odd numbers can never equal an even number.

Example:
$3+5+7=15$[/hide]


[quote="flyover"]Can anyone please prove this identity, its been bothering me for a while now.

tan(x-y) + tan(y) / 1 - tan(x-y) * tan(y)  = tan(x)[/quote]Uh, just let $ x\equal{}y\plus{}z$ and you'll get the angle addition formula for tan... unless you want to prove $ \tan(\alpha\plus{}\beta)\equal{}\frac{\tan\alpha\plus{}\tan\beta}{1\minus{}\tan\alpha\tan\beta}$...  :roll:

[quote="flyover"]Can anyone please prove this identity, its been bothering me for a while now.

tan(x-y) + tan(y) / 1 - tan(x-y) * tan(y)  = tan(x)[/quote]
[hide]\[ \frac{\tan(x\minus{}y)\plus{}\tan(y)}{1\minus{}\tan(x\minus{}y)\cdot\tan(y)}\equal{}\tan((x\minus{}y)\plus{}(y))\equal{}\tan(x)\][/hide]


[quote="amir2"]$a,b,c \ge 0$ prove:
$\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}\ge \sum_{cyc}a\sqrt{2a^2+bc}$[/quote]

Already discussed, i think....

$a^4+a^2b^2+b^4\geq \frac{3}{4}(a^2+b^2)^2$


Uitati aici o problema de Mortici, care eu consider ca e gresita, intrucat am gasit un contraexemplu.

Fiind date cinci puncte P1, P2, P3, P4, P5 in planul xOy, avand coordonatele intregi, demonstrati ca exista 1 <= i <j <= 5 astfel incat segmentul (Pi, Pj) contine un punct cu ambele coordonate intregi.

Incercati si voi si vedeti ce parere aveti daca P1(6, 5) P2(7, 5) P3(7, 6) P4(6, 6) P5(0, 4)

Problema se gaseste in cartea lui Cristinel Mortici, dar nu este facuta de el, ci este o bucata dintr-o problema propusa la Putnam prin anii 70-80. Se foloseste principiul lui Dirichlet si paritatea coordonatelor (ie unul din mijloacele segmentelor formate de cele 5 varfuri va avea ambele coordonate numere intregi).

Mersi,am inteles. Legat de contraexemplul meu, am fost neatent.


topology
geometry
rectangle
inequalities
analytic geometry
advanced fields
advanced fields unsolved

I wrote a proof to the following problem the other day. Is it rigorous, or are there some flaws?
I'm sure there is an easier way to prove this; does anyone have any suggestions?

[b]Theorem[/b]: The rectangle $R=[a_{1},b_{1}]$x$\ldots$x$[a_{n},b_{n}]$ in $R^{n}$ is closed.

[b]Proof[/b]: We will show that the compliment $R^{n}-R=\{x=(x_{1},\ldots,x_{n}) \in R^{n}: x_{i}\notin [a_{i},b_{i}]$ for some $i \in \{1,\ldots,n\}\}$ is open.
Let $x \in R^{n}-R$. Then there exists an $i \in \{1,\ldots,n\}$ such that $x_{i}\notin [a_{i},b_{i}]$. Now define $\delta =$min$(|x_{i}-b_{i}|,|a_{i}-x_{i}|)$. We claim that $B(x,\delta) \subset R^{n}-R$, proving $R^{n}-R$ is open.
Let $y=(y_{1},\ldots,y_{n}) \in B(x,\delta)$, so that $\delta > \parallel y-x\parallel ={\sqrt{\sum_{t=1}^{n}(y_{t}-x_{t})^{2}}}\geq \sqrt{(y_{i}-x_{i})^{2}}=|y_{i}-x_{i}|$.
Now suppose that $y_{i}\in [a_{i},b_{i}]$. We proceed in cases.
In the first case, assume $x_{i}<a_{i}$, so that $\delta=|a_{i}-x_{i}|=a_{i}-x_{i}$. We know $\delta > |y_{i}-x_{i}|=y_{i}-x_{i}$, so $a_{i}-x_{i}>y_{i}-x_{i}$, or $y_{i}<a_{i}$. But this contradicts the fact that $y_{i}\in [a_{i},b_{i}]$.
In the second case assume $x_{i}>b_{i}$. Then $\delta=|x_{i}-b_{i}|=x_{i}-b_{i}$, and $\delta>|y_{i}-x_{i}|=x_{i}-y_{i}$. So $x_{i}-b_{i}>x_{i}-y_{i}$, or $y_{i}>b_{i}$. Once again, contradiction.
Thus $y_{i}\notin [a_{i},b_{i}]$, from which we deduce $y \in R^{n}-R$, and in turn $B(x,\delta) \subset R^{n}-R$. $x$ was arbitrary, so this proves that $R^{n}-R$ is open.
Therefore $R$ is closed. [b]QED[/b]

It's fine and neat except the $\le$ sign should be $\ge$. I do not think there is a shorter way of proving the statement if one has to write down every detail as you did.

the inequality sign is fixed  :)

so when you're proving that a set is closed (in an undergraduate level topology course), then you almost always resort to proving that the compliment is open?

Sometimes one may prefer to prove that it contains all its boundary points instead but, if you look at how it is done and require each step to be brought to the very basic level, you usually find that it is still a proof of the openness of the complement in some disguised form, which is not surprising because that is how the closed sets are defined. So, the answer to your last question is  "Yes".  :)

compl[b]e[/b]ment

You could also use the fact that unions of open sets are open. Since $\mathbb R^{n}\setminus R=\bigcup_{i=1}^{n}\{x\in\mathbb R^{n}\colon x_{i}<a_{i}\}\cup \bigcup_{i=1}^{n}\{x\in\mathbb R^{n}\colon x_{i}>b_{i}\}$, this reduces the problem to showing that an "open half-space" is actually open.  :) Not much of a simplification, actually.

Couldn't you use the fact that if x is in the complement of the interval then the coordinates of x belong to the respective complements of the components of the interval (which we know to be open) and therefore x belongs to an open set?

But the point $(1,5)$ lies outside of the square $[0,2]\times [0,2]$ even though the first coordinate, $1$, is within $[0,2]$.

[quote="Victor96"]I think the most important thing is what is open according your assumption.  Open balls?  Open retangles?[/quote]

The definition I'm using:
A set $S$ is open if for every $a \in S$ there exists a $\delta>0$ such that the ball $B(a,\delta)$ is entirely contained in $S$.

Oh, then I think mlok's decomposition has offered a very good way of simplifying the problem because you need only to bound your radius using the information of one coordinate.


integration
algebra
polynomial
calculus
real analysis
real analysis unsolved

For natural number n, consider: $ H_k \equal{} \frac{(\minus{}1)^{n\minus{}k}}{n\cdot k!(n\minus{}k)!}\int_0^n\frac{q(q\minus{}1)\cdots(q\minus{}n)}{q\minus{}k}\ ,\ \ k\equal{}0,1,\ldots,n.$

Prove that $ H_k$ is positive and satisfies: $ H_k\equal{}H_{n\minus{}k},\ H_0\plus{}H_1\plus{}\cdots\plus{}H_n\equal{}1.$

Just consider Lagrange's interpolation polynomial for $ f(x)\equal{}\frac{1}{n}$ on segment $ [0,n]$ with centers 0, 1, 2, ..., $ n$. Then $ H_{0}\plus{}H_{1}\plus{}\cdots\plus{}H_{n}\equal{}1$ is a fact that $ \int_0^nf(x)dx\equal{}1$.

Don't you use Newton-Cotes formula?
So we have to consider the formula of error estimates afterwards. But the original problem precedes this theorem in my textbook. Any hints for a natural solution without numerical analysis?

What "error estimates"? We get the exact identity for the case $ f(x)\equal{}1/n$ , since $ 1/n$ is a zero-degree polynomial.

[quote]Prove that $ H_k$ is positive and satisfies: $ H_k \equal{} H_{n \minus{} k}$[/quote]

Hello, I'm new here.
Any ideas in proving this side? I tried expanding the product but failed.

It was an obvious part of the problem. You just need to change $ q\to n\minus{}q$ in the integral (dunno how to say it in english properly).


What product will be obtained when 2-(3-cyclohexenyl)acetic acid is treated with $ I_2/NaHCO_3$?

Is it alpha iodonation by any chance?? :huh:

look out for methyl carbonyl chemrock....

i think because...
it is similar top iodoform i.e. iodine + a base..

Excuse me it is acetic acid :huh:  :huh:  :huh:

sorry carcul I didn't mean acetic acid!
I told mr.valerium that it is acetic acid and not acetone.

you get a lactone. the double bond goes and O attaches to one of the carbons(Halo lactonisation)

Good that might be right I never thought of that :o  :o

That's right: iodo-lactonization is the answer.


When a right triangle is rotated about one leg, the volume of the cone produced is $800 \pi$ $\text{cm}^3$.  When the triangle is rotated about the other leg, the volume of the cone produced is $1920 \pi$ $\text{cm}^3$.  What is the length (in cm) of the hypotenuse of the triangle?

[hide="Solution"]Let the legs of the right triangle be $x$ and $y$. We're trying to find $\sqrt{x^2+y^2}$. Applying the formula for a cone's volume, $\frac 13 xy^2=800$ and $\frac 13 x^2y=1920$. To find $x$ and $y$, we multiply the equations together to obtain $(\frac 13 xy^2)(\frac 13 x^2y)=\frac 19 x^3y^3=(800)(1920)$, so that $xy=240$. Therefore, $x=24$ and $y=10$, and the hypotenuse is $\boxed{026}$. [/hide]


The frequency distribution shows the number of CD's that students in a math class listened to in their leisure time last week. Find the mode of the data.

none here either. check over their behind that tree please.

There should be a distribution with 1 mark for 0, 2 for 1, 1 for 2, continuing...the answer is 3.


[quote="Andreas"]The non-negative real numbers $a, b, c, d$ add up to $1$. Prove the inequality

$\left| ab - cd \right| \leq \frac{1}{4}$.[/quote]
you mean $a,b,c,d<1$ or $a+b+c+d<1$ ?

I think this works:

Assume the opposite. Let's say $ab-cd > \frac{1}{4}$.

Clearly, $ab > \frac{1}{4}$, which implies $\sqrt{ab} > \sqrt{\frac{1}{4}}$.

By AM-GM, $\frac{a+b}{2} \ge \sqrt{ab}$. The maximum of $a+b$ is $1$.

So we have $\frac{1}{2} \ge \sqrt{ab} > \sqrt{\frac{1}{4}}$.

So, $\frac{1}{2} > \frac{1}{2}$. Contradiction.

If we remove the absolute value signs, we have also $cd - ab > \frac{1}{4}$, which leads back to the argument above.

The case for equality is when $a = .5$, $b=.5$, $c=0$, $d=0$, or $a=0$, $b=0$, $c=.5$, $d=.5$.


What is the coefficient of the term containing $x^{6}$ in the expansion of $(4x^{2}-\frac{y}{2})^{9}$?

[quote="xpmath"][hide]To get an $x^{6}$ from an $x^{2}$, you need it multiplied three times. There fore, we need to simplify
$\binom{9}{3}(4x^{2})^{3}(-\frac{y}{2})^{6}$ which is
$84*64x^{6}*(\frac{y^{6}}{64})$ so I think the answer is 5376?[/hide]

I don't know, it could be wrong.[/quote]
$(4x^{2})^{3}=4096x^{6}$. I think you typed it in wrong or something because you got the right answer, but the wrong number.

The coefficient is just $84$.  $y$ is considered a variable the same as $x$, so it is not part of the coefficient.

[quote="xpmath"][quote="diophantient"][hide]
Each term is of the form $4^{n}\cdot x^{2n}\cdot y^{9-n}\cdot .5^{9-n}\cdot (-1)^{9-n}\cdot \binom{9}{n}$.
n=3 is what we want, this turns into $\binom{9}{3}=84$[/hide][/quote]
But don't forget about the 4, as it will still be a coeifficient[/quote]
The $4^{3}$ is cancelled out by the $.5^{9-3}$


Find the next number in the sequence

1,2,2,3,2,4,2,4,3,4,2,6,2,4,4

Ive thought about this for a while and I couldn't get it. Could you post a hint that doesn't give away too much?


OK, try this one - I like it and I have a solution.

A word of length n that consists only of the digits "0" and "1" is called a bit-string of length n. (For example, 01101 is a bit-string of length 5.) Consider the sequence s(1), s(2), ... of bit-strings of length n > 1 which is obtained as follows :
(1) s(1) is the bit-string 00...01, consisting of n-1 zeros and a "one" ;
(2) s(k+1) is obtained as follows :
(a) Remove the digit on the left of s(k). This gives a bit-string t of length n-1.
(b) Examine whether the bit-string t1 (length n) is already in the set
{s(1), s(2), ..., s(k)}.
If this is the case, then s(k+1) = t1. If this is not the case then s(k+1) = t0.

For example, if n = 3 we get :
s(1) = 001 -> s(2) = 011 -> s(3) = 111 -> s(4) = 110 -> s(5) = 101 ->
s(6) = 010 - > s(7) = 100 -> s(8) = 000 -> s(9) = 000 -> ...

Suppose N = 2^n.
Prove that the bit-strings s(1), s(2), ..., s(N) of length n are all different.

I have a nice and simple proof that 00...0 is the only one that can repeat, but I'm not sure how to go about showing that it must occur first at step s(2^n).  I guess I haven't given it enough thought yet.

BTW, you might want to edit your statement of the procedure for determining s(k+1).  Its sort of the opposite of what you want  (ie if t1 is in the set then s(k+1) should be t0, not t1).  It rather confused me to begin with.

A soln for this can be found on Mathlinks  :lol: I couldn't solve it on my own completely, so Arne helped me (he gave me a BIG hint).

The official solution can be found at

http://www.mathlinks.ro/phpBB/viewtopic.php?t=836.


Arqady, Vasc, have you got solution for it?

The following estimate is nice and easy
\[\sum_{\rm cyclic}\frac a{\sqrt{b^{2}+c^{2}+d^{2}}}\geq2. \]

But doesn't that mean that equality case is when $a=\sqrt{b^{2}+c^{2}+d^{2}}$ and so on?

if so

$\sum{\frac{a}{\sqrt{b^{2}+c^{2}+d^{2}}}}=4$

what am I doing wrong...?

[quote="Allnames"][quote="shfdfzhjj"][quote="pvthuan"]Prove that if $ a,b,c,d\geq0$, then
\[ \frac a{\sqrt {a^{2} \plus{} b^{2} \plus{} c^{2}}} \plus{} \frac b{\sqrt {b^{2} \plus{} c^{2} \plus{} d^{2}}} \plus{} \frac c{\sqrt {c^{2} \plus{} d^{2} \plus{} a^{2}}} \plus{} \frac d{\sqrt {d^{2} \plus{} a^{2} \plus{} b^{2}}}\leq\frac {4\sqrt3}3.
\]
[/quote]
I have an ugly proof. :([/quote]
Yes,I have a proof,But i think it is quite nice(I know it from Can hang 2007,but i find the solution independly)[/quote]

Can you show it? :maybe:


[i]Let $ p$ be a fix odd prime .The sequence $ \{ x_n \}$ was defined by : $ x_{n \plus{} 2} \ \equal{} \ 4x_{n \plus{} 1} \ \minus{} \ x_n \ \forall \ n \ \in \ \mathbb{N}$

 Choose $ x_0 \ ; \ x_1$  such that :

   For all positive integer $ k$ , there exists an index $ i$ such that : $ 4p^2 \ \minus{} \ 8p \plus{} 1 | x_i \ \minus{} \ (2p)^k$

( $ \mathbb{N}$ is the set of all nonegative integer )[/i]

[hide]
First, lets say that $ 4p^{2}\ \minus{}\ 8p\plus{}1 | x_{k\minus{}2}\ \minus{}\ (2p)^{k\minus{}2}$ and $ 4p^{2}\ \minus{}\ 8p\plus{}1 | x_{k\minus{}1}\ \minus{}\ (2p)^{k\minus{}1}$ and let $ 4p^{2}\ \minus{}\ 8p\plus{}1\equal{}A$.

We have that $ x_{k\minus{}2}\equiv(2p)^{k\minus{}2} (mod A)$ and $ x_{k\minus{}1}\equiv(2p)^{k\minus{}1} (mod A)$.

Then,
 $ x_{k}\equiv 4x_{k\minus{}1}\minus{}x_{k\minus{}2}\equiv 4(2p)^{k\minus{}1}\minus{}(2p)^{k\minus{}2}\equiv(2)^{k\plus{}1}(p)^{k\minus{}1}\minus{}(2)^{k\minus{}2}(p)^{k\minus{}2}
\equiv(2p)^{k\minus{}2}(8p\minus{}1)\equiv(2p)^{k\minus{}2}(4p^2\minus{}8p\plus{}1\plus{}8p\minus{}1)\equiv(2p)^{k\minus{}2}(4p^2)\equiv(2p)^{k} (mod A)$

Thus if we choose $ x_{0}, x_{1}$ such that $ 4p^{2}\ \minus{}\ 8p\plus{}1 | x_{0}\ \minus{}\ 1$ and $ 4p^{2}\ \minus{}\ 8p\plus{}1 | x_{1}\ \minus{}\ (2p)$ then we will have a sequence that satisfies the requirements. Letting $ x_{0}\equal{}1$ and $ x_{1}\equal{}2p$ works and we are done.
[/hide]

[quote="Bacteria"]
First, lets say that $ 4p^{2}\ \minus{} \ 8p \plus{} 1 | x_{k \minus{} 2}\ \minus{} \ (2p)^{k \minus{} 2}$ and $ 4p^{2}\ \minus{} \ 8p \plus{} 1 | x_{k \minus{} 1}\ \minus{} \ (2p)^{k \minus{} 1}$ and let $ 4p^{2}\ \minus{} \ 8p \plus{} 1 \equal{} A$.

We have that $ x_{k \minus{} 2}\equiv(2p)^{k \minus{} 2} (mod A)$ and $ x_{k \minus{} 1}\equiv(2p)^{k \minus{} 1} (mod A)$.

Then,
 $ x_{k}\equiv 4x_{k \minus{} 1} \minus{} x_{k \minus{} 2}\equiv 4(2p)^{k \minus{} 1} \minus{} (2p)^{k \minus{} 2}\equiv(2)^{k \plus{} 1}(p)^{k \minus{} 1} \minus{} (2)^{k \minus{} 2}(p)^{k \minus{} 2} \equiv(2p)^{k \minus{} 2}(8p \minus{} 1)\equiv(2p)^{k \minus{} 2}(4p^2 \minus{} 8p \plus{} 1 \plus{} 8p \minus{} 1)\equiv(2p)^{k \minus{} 2}(4p^2)\equiv(2p)^{k} (mod A)$

Thus if we choose $ x_{0}, x_{1}$ such that $ 4p^{2}\ \minus{} \ 8p \plus{} 1 | x_{0}\ \minus{} \ 1$ and $ 4p^{2}\ \minus{} \ 8p \plus{} 1 | x_{1}\ \minus{} \ (2p)$ then we will have a sequence that satisfies the requirements. Letting $ x_{0} \equal{} 1$ and $ x_{1} \equal{} 2p$ works and we are done.
[/quote]

  Quite nice :lol:


How many different combinations of 5 dollar bills and 2 dollar bills can be used to make a total of 17 dollars? Order doesn't matter in this problem.

a) 2
b) 3
c) 4
d) 5
e) 6

[hide]
You nedd an even number of dollars left after you subtract the five dollar bills. Thus you can only subtract an odd number of five dollar bills. It must be <17, so the number of 5 dollar bills can only be 1,3.
The number of corresponging 2 dollar bills is 6 and 2.
Thus the answer is
$(1,6)$
$(3,1)$
[/hide]

[hide]2,answer A. I won't bother with a solution as biffandoc's already got one and it's kinda obvious.[/hide]

[hide]Two. Three fives and one two, and six twos and one five.[/hide]

[hide]The combinations are
3 5s 1 2

1 5   6 2s

So there are 2 combinations

$A$[/hide]

[hide]There are two ways[/hide]  what if order matters?


We have
$ P\leq \frac{(a\plus{}1)(b\plus{}1)(c\plus{}1)\plus{}(a\minus{}1)(b\minus{}1)(c\minus{}1)}{abc\plus{}1}$
$ \equal{}\frac{2(abc\plus{}a\plus{}b\plus{}c)}{abc\plus{}1}\equal{}2\plus{}\frac{2(a\plus{}b\plus{}c\minus{}1)}{abc\plus{}1}$
Because $ x,y\geq 1\Rightarrow (x\minus{}1)(y\minus{}1)\geq 0\Rightarrow xy\plus{}1\geq x\plus{}y$ then $ abc\plus{}1\geq ab\plus{}c\geq a\plus{}b\plus{}c\minus{}1$
Then $ P\leq 4$

Let $ a,b,c$ be positive numbers satisfying $ a,b,c \geq 1$ .Prove that
      $$(a+1)(b+1)(c+1)\leq 4(abc+1)$$

Let a, b, and c be positive real numbers. [url=https://artofproblemsolving.com/community/c6h1179230p5701994]Prove that [/url]
$$(a+1)(a+b)(b+c)(c+16)\ge81abc$$
[url=https://artofproblemsolving.com/community/c6h866p2820]h[/url]

[quote=hoangclub]Let $ a,b,c$ be positive numbers satisfying $ a,b,c \geq 1$ .Find the maximum of express
      $ P\equal{} \frac{(1\plus{}a)(1\plus{}b)(1\plus{}c)}{abc\plus{}1}$.[/quote]

We will prove that $P\leq 4$, or $(\frac{a+1}{2})(\frac{b+1}{2})(\frac{c+1}{2})\leq \frac{abc+1}{2}$
Firstly, we have $(\frac{x+1}{2})(\frac{y+1}{2})=\frac{xy+1}{2}-\frac{(x-1)(y-1)}{4}\leq \frac{xy+1}{2}\ \ \forall x,y\geq 1$
$\Rightarrow LHS\leq (\frac{ab+1}{2})(\frac{c+1}{2})\leq \frac{abc+1}{2}$ ($Q.E.D$)
$P_{max}=4$ at $a=b=c=1$. $\square$

$(ab-1)(c-1) \geq 0 \implies abc+1 \geq ab+c \implies a+b+c+ab+bc+ca \leq 3abc+3 \implies \frac{(a+1)(b+1)(c+1)}{abc+1} \leq 4$


(a)Find a number divisible by $ 2$ and $ 9$ which has exactly $ 14$ divisors .
(b)Find a number divisible by $ 2$ and $ 9$ which has exactly $ 15$ divisors .
(c)Find a number divisible by $ 2$ and $ 9$ which has exactly $ 17$ divisors .

(a) The only option is $ 2\cdot 3^6 \equal{} 1458$.
(b) The two options are $ 2^2 \cdot 3^4 \equal{} 324$ and $ 2^4 \cdot 3^2 \equal{} 144$.
(c) Such a number does not exist.  :(


MATHCOUNTS induction calculus function algebra polynomial Gauss

This is a challenge based on the 2000 National Mathcounts Masters Round:

The sum of the first n positive integers is given by the formula n(n+1)/2.
The sum of the first n squares is given by n(n+1)(2n+1)/6.

1) Find a formula for the sum of the first n cubes.  How did you derive this?
2)          "                                first n 4th powers.  How did you derive this?
3) Give a method of finding a formula for the sum of the first n xth powers.

I am too lazy to do it myself right now, but I think you can prove all three with induction (well, at least the first two definitely).

Fierytycoon

Sure, you can prove that a formula works using induction (in fact, the n(n+1)/2 example is the classic example of induction), but can you make your own formula?  Can you come up with some method?

Since this isn't particularly easy, I should point out that there's a nice section in [u]Concrete Mathematics[/u] by Ronald Graham, Donald Knuth, and Oren Patashnik on this topic. It has to do with Bernoulli numbers if you want to do that, but you can solve individual cases with finite calculus, generating functions, or various other methods. By the way, if you're looking for a terrific (but extremely challenging) math book, check out Graham, Knuth, Patashnik. For some reason, a friend of mine had this book and didn't like it, and so he gave it to me. I wonder what other math books he has that he doesn't like...

I've actually worked out a method long ago for extracting the explicit formula of the sums. But it's _heavy_ in calculation. Want me to demonstrate for some simple cases?

Sure, Mindspa.  And I have a method that's pretty easy to understand (I think it falls under the "generating functions" category.  If anyone wants, I'll demonstrate also.

Yes. there is a pretty simple recursive way to find the formulas for the sum of nth powers. But to jump to a particular formala for n non-recursively, I think you need a generating function that involves Bernoulli numbers.

I say it again, it's heavy on calculation really brute force, no ingeniuity, I remember doing this method a long time ago when I searched for a general formula, when I didn't even know calculus.

Let's make a factored table first for the first eightterms in the sum S^4_n = 1^4 + ... + n^4,

{1,1},{2,17},{3,2*7^2},{4,2*3*59},{5,11*89},{6,5^2*7*13},{7,7*2^2*167},{8,2^2*3*17*43}

We know from formulas for 1,2,3 that they contained a factor of n. So in order to be able to get a factor n in this formula (wishful thinking) we need to have S^4_n = n*q(n)/30 (S^4_n needs to be multiplied by 30 for it to be able to be divisible by n as seen in the table) for some polynomial q(n), we now build a similar table for q(n) = 30*S^4_n/n with seven terms,

{1,2*3*5},{2,3*5*17},{3,2^2*5*7^2},{4,3^2*5*59},{5,2*3*11*89},{6,5^3*7*13},{7,2^3*3*5*167}.

Now again we do some wishful thinking and assume that S^4_n is divisible by (n+1) thus q(n) = p(n)(n+1) for some polynomial p(n). A table for p(n) = q(n)/(n+1) (the job now is easier because we only delete factors).

{1,3*5},{2,5*17},{3,5*7^2},{4,3^2*59},{5,11*89},{6,5^3*13},{7,3*5*167}.

Some wishful thinking yet again gives us that it must contain a factor of (2n + 1) because we can clearly spot the "sequence" 3,5,7,9,11,13,15 in the factors. Thus p(n) = r(n)(2n+1) for some polynomial r(n) = p(n)/(2n+1).

{1,5},{2,17},{3,5*7}={3,35},{4,59},{5,89},{6,5^3}={6,125},{7,167}.

We now see that the factors look somewhat an arithmetic sequence. Let's "guess" the difference r(n) - r(n-1) = (n-1)*6 + 6 = 6n thus r(n) = r(n-1) + 6n, wishful thinking again makes us want to be able to divide r(n) by 6, because the and it looks from the table as we can succed by adding 1 to each of the factors. We get (r(n)+1)/6 = m(n) for some polynomial m(n) without writing out the table for m(n) we see after writing it out that it is the sequence 1,3,6,10,... which is the n(n+1)/2 thus m(n) = n(n+1)/2.

Using all the polynomials we get m(n) = n(n+1)/2, thus r(n) = 6n(n+1)/2 - 1 why p(n) = (3n(n+1)-1)(2n+1) and q(n) = (3n(n+1)-1)(2n+1)(n+1) and thus S^4_n = n(n+1)(3n(n+1)-1)(2n+1)/30 for the nth term. I said it was heavy on calculation :(.

Yo G. Dig the avatar man..... Ramanujan is definitely my hero when it comes to contemporary mathematicians... Some of the stuff he's come up with, you can only come up with when you are on pot, crack, or both. For example,

1/:pi: =  ((:sqrt: 8 )/9801):Sigma: ((4n)!(1103+26390n))/(((n!)^4)396^(4n)) which is infinite sum.

And in the famous words of my little bro, "Damn dude, thas messed up..."

Peace,
IW.OAV

[quote="Idiot_without_a_village"]Some of the stuff he's come up with, you can only come up with when you are on pot, crack, or both.[/quote]

High on mathematics...

[color=cyan]I believe this is the method that gauss was talking about, and it doesn't rely on the amount of wishful thinking required by MindSpa's solution.  You need the formulas for the sums of the powers of the positive integers up to (m - 2) in order to get (m - 1).

Let S(n,m) = :Sigma: i^m = 1^m + 2^m + ... + n^m
Write each i^m as ((i-1) + 1)^m and expand using the binomial theorem.  Then, regroup terms according to their powers.  Subtract all the mth powers, which will leave you with an n^m on the left side.  Substitute in for all the sums of smaller powers.  Subtract.  Divide.  And you will have your answer.  It's pretty nice up until about the 4th powers.  After that, I wouldn't want to touch it without a computer-algebra system or something that will keep me from writing everything out.

You can also do it a different way:  you can calculate any of the base values easily.  You also "know" that the sum of the mth powers will give you an (m+1)st degree polynomial.  So the first (m+2) values uniquely determine it.  So solve a system of linear equations for the coefficients.

The first way I like a lot better, and I think in general involves less calculation.[/color]

This is the way I work out sums of powers easily (though not any specific one, this is recursive), its probably the same as JBLs one here but it doesn't look exactly the same..
All we do is expand i^n - (i-1)^n. This gives us a whole bunch of terms..
Now just add that up from i=0 to whatever. This gives i^n on the LHS, and a constant multiplied by each of the sums of smaller powers on the RHS, along with the one we are looking for.
ie, to find the sum of squares we get i^3 - (i-1)^3 = 3i^2 -3i + 1

So i^3 = 3(sum of squares) - 3i(i+1)/2 + i
And then rearrange that to get the sum of squares.

[color=cyan]Yeah, your way is basically the same as the first way I did it, but it looks a lot nicer if you sum it up afterwards (like you did) instead of before-hand (like I did).[/color]

Here's an even faster way to produce them recursively. Take the polynomial f_k(n) for the sum of the first n kth powers. To get f_(k+1)(n) for the sum of the first n (k+1)th powers, multiply the previous formula by k and then integrate with respect to n. Then take that and add an n term that makes the sum of the coefficients equal to 1.

For example, to find the sum of the first n cubes from the forumula for the sum of the first n squares:

Multiply f_2(n) = n^2/2 + n/2 by 2. Then integrate with respect to n to get n^3/3 + n^2/2. Now add n/6 to make the sum of the coefficients (i.e. 1/3 + 1/2 + 1/6) = 1. So f_3(n) = n^3/3 + n^2/2 + n/6.

Similarly, multipling this by 3 and integrating you get n^4/4 + n^3/2 + n^2/4. The sum of it's coefficients is already 1, so you don't need to add anything. Therefore f_4(n) = n^4/4 + n^3/2 + n^2/4.

You can keep doing this recursively to get the higher and higher Sum of Powers formulas.

Yeah I've heard of that way as well. Never thought about why it works though, the other method is probably quite a lot easier to remember, though the integral method is quite a bit faster. Probably best to remember both..


trigonometry
inequalities
function
geometry
circumcircle
cyclic quadrilateral
geometry proposed

In a cyclic quadrilateral $ABCD$, $AB=a$, $BC=b$, $CD=c$, $\angle ABC = 120^\circ$ and $\angle ABD = 30^\circ$. Prove that

(1) $c \ge a + b$;

(2) $|\sqrt{c + a} - \sqrt{c + b} | = \sqrt{c - a - b}$.

Since $ABCD$ is a cyclic quadrilateral, we have that $\angle ACB=30$ and $\angle CDA=180-\angle ABC=60$, then we have that $AC=\frac{c\sqrt3}{2}$.

Now, by the cosine law in $\triangle ABC$ we have that
$a^2+ab+b^2=\frac{3c^2}{4} \ \ \ (1)$.

Then, $a)$ follows from $(1)$ and the inequality $a^2+ab+b^2\geq\frac{3(a+b)^2}{4}$.

And we have $b)$ follows from $|\sqrt{c+a}-\sqrt{c+b}|=\sqrt{c-a-b}\Leftrightarrow (\sqrt{c+a}-\sqrt{c+b})^2=c-a-b \\\\\Leftrightarrow 2a+2b+c=2\sqrt{(c+a)(c+b)} \Leftrightarrow 4a^2+4ab+4b^2=3c^2$
which is clearly equivalent to $(1)$.

Hello,
    The first part of the problem can be done differently using Jensen's inequality. The problem reduces to showing that sin x + siny <=1 if x + y =60. Here, sinx is a concave function. Hence, Jensen's inequality for concave functions gives the result.
Sanjith

[hide="A elementary proof (similarly rather with the Campos-solution) for this nice and easy problem."]I suppose w.l.o.g. the circumradius $R=1$. Thus, $AD\equiv d=1$, $CD=c=2$, $AC\equiv e=\sqrt 3$, $AB=a$, $BC=b$, $BD\equiv f=\sqrt{4-b^2}\ .$
From the [i]Ptolemeu's theorem[/i] $ac+bd=ef$ results $\boxed {\ a^2+ab+b^2=3\ }\ .$
I remark $3=a^2+ab+b^2\ge3ab\Longrightarrow \boxed {\ ab\le 1\ }\ .$
Therefore, the proposed problem is equivalently with the following conditioned inequality: $a>0$, $b>0$, $a+b<\sqrt 3$, $a^2+ab+b^2=3\Longrightarrow$
$a+b\le 2\ \ \wedge\ \ \left|\sqrt{2+a}-\sqrt{2+b}\right|=\sqrt{2-a-b}\ \ (1)\ .$
$1^{\circ}\blacktriangleright ab\le 1\ \wedge\ a^2+ab+b^2=3\Longrightarrow a^2+2ab+b^2\le 4\Longrightarrow a+b\le 2\ .$
$2^{\circ}\blacktriangleright(1)\Longleftrightarrow4+a+b-2\sqrt{4+2(a+b)+ab}=2-a-b\Longleftrightarrow$
$a+b+1=\sqrt{4+2(a+b)+ab}\Longleftrightarrow(a+b)^2+2(a+b)+1=4+2(a+b)+ab\Longleftrightarrow$
$a^2+2ab+b^2+1=4+ab\Longleftrightarrow a^2+ab+b^2=3$, what is true.[/hide]

Exactly!!  Ptolemys theorem gives the shortest , simplest, direct and best proof!! No need for dirty trigno!! I did this in the paper!!!

Exactly I did the same thing though I did not assume R=1. :P


Suppose you are living in a city where all the
residents have been allotted a three digit number between $ 100$ and
$ 999$ by a numbering machine which generates three digit numbers
randomly.

Now, on the centinery 'foundation day' of the city, there was a huge
gathering at the city esplanade where all residents including you
are to be present and where the given three digit numbers of the
residents were checked at the entrance gate. it was announced by the
city municipality that the first person who will be having a number
which has been recorded earlier with another resident while
entering, that is the first person who has a repeat number will be
the winner of a luxury car.

When do you think your chance of winning better ? if you're the
$ 13^{th}$, $ 31^{st}$, $ 251^{st}$, $ 501^{st}$ visitor ?

If the $ 501^{st}$ visitor goes, does this mean that no one else won before him?

[quote="Brut3Forc3"]I think it means he is the 501st in line, i.e. in calculating the probability, you must account for the possibility that someone before him won.[/quote]

yes, thats correct.

[hide]In order for the $ 13^{th}$ person to win, the first twelve must get different numbers.  The first person gets a number.  The second person now has a $ \frac{899}{900}$ chance of not winning.  Continue for eleven iterations to get $ \frac{899\cdot898\cdot...889}{900^{11}}$.  Let that be $ a$ (for referencing).  The $ 13^{th}$ person now must choose from twelve numbers, so we multiply to get $ \frac{a}{75}$ as the probability.

For the $ 31^{st}$ visitor to win, the first thirty cannot win.  We continue our expression of $ a$ by multiplying by $ a\cdot \frac{888\cdot887\cdot...871}{900^{18}}$.  For the next number, there is a $ \frac{1}{30}$ chance that the person will win.  This is a greater probability than $ \frac{a}{75}$, so we will let this be represented as $ \frac{b}{30}$.

For the $ 251^{st}$ visitor, we have $ b\cdot \frac{250}{900}\cdot \frac{888\cdot887\cdot...651}{900^{238}}$.  I think this is MUCH smaller than the probability for the $ 31^{st}$ person to win.

And that logic holds for the $ 501^{st}$ person as well.

Therefore, the $ 31^{st}$ person will most likely win of these four.

I don't know how to do it, but there is a way to maximize the chances of winning.  Earlier is better, but there needs to be enough numbers in the pool to choose from.

[/hide]


Έστω $ n$ θετικός ακέραιος μεγαλύτερος του $ 2$ κια έστω $ f$ ένα πολυώνυμο βαθμού όχι μεγαλύτερου του $ n\minus{}2$. Εάν $ a_{1},a_{2},...,a_{n}$ είναι μιγαδικοί αριθμοί και $ p(t)\equal{}\prod^{n}_{i\equal{}1}(t\minus{}a_{i})$ αποδείξτε ότι $ \sum^{n}_{i\equal{}1}\frac{f(a_{i})}{p^{'}(a_{i})}\equal{}0$.

Από το Lagrange Interpolation formula έχουμε ότι $ f(x) \equal{} \displaystyle\sum_{i \equal{} 1}^{n}f(a_i)\displaystyle\prod_{j\neq i}\frac {x \minus{} a_j}{a_i \minus{} a_j}$ και επειδή $ deg(f)\le n \minus{} 2$ έχουμε ότι ο συντελεστής του $ x^{n \minus{} 1}$του $ f(x)$ που είναι $ \sum^{n}_{i \equal{} 1}\frac {f(a_{i})}{p^{'}(a_{i})}$ θα είναι 0.


Not sure where this should go, but what do you guys make of this?

http://mathworld.wolfram.com/DiagonalParadox.html

No mater what the value of n is the line will move 1 unit north total and 1 unit east total.
Therefore the line will always be 2 units long.

Yes, but as the sizes of the up and downs approaches 0, the line approaches that of the diagonal. However, the distance traveled for the up and down route is always 2 while the diagonal is always $\sqrt{2}$.

I'm thinking that as the up and down approaches 0, the line would become the diagonal, but they have different lengths. Can anyone expand for me why my reasoning is incorrect, because obviously, it is, and it has not created a huge impact on mathematics that is felt everywhere (even Russels Paradox at least inspired some people to create a whole new set theory)

I know that the lines have to move up and down only, but at such small sizes, wouldn't it be like a limit?

Upon thinking about this a bit more, no matter how small the steps get, there is always a diagonal that is $\sqrt{2}$ the size of the step--which was at first confusing because of my idea of a limit.

However, if my liit idea really is correct, the steps become infinitely small, but then, the diagonal really won't go any distance because the steps are so small--and I'm rambling now so I'll go to sleep.

Hope some of you can help me shed some light on this :D

Here's what I make of it.
A limit of something as it approaches a value is not necessarily that something evaluated at that value.
\[\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1\] does not imply \[\frac{\sin 0}{0}=1\]
The value of the limit is merely an indication of the function's behavior as the function approaches a value not in the function's domain. In the case of the diagonal, taking the limit of the steps are a bad indication of how long the diagonal because they will always be perpendicular while the diagonal is a straight line. As the length of the individual steps approaches 0, the steps still do not behave like the diagonal. Therefore, the limit of the length of the staircase as the length of the individual steps approaches 0 is not the same as the length of the diagonal.

It actually doesn't approach the diagonal of a square. It just looks that way since it gets really small, but no matter how many steps you take you'll never look closely and see anything that resembles a straight line. That's the intuitive answer, and the mathematical one is obvious: it's always a total of twice as long.

I think it's like this:

square with length a, diagonal d

d_n  is the n-th length of the "step" diagonal.

d_n = sum(1 to n) of (a_n) where a_n = a/n

d_n = a * sum(1 to n) of (1/n)

so to get the length of the diagonal we take the limit for n -> infinity.
this gives

d_step = limit(n->infinity) a * sum (1 to n) of (1/n)

but sum(1 to infinity) of 1/n diverges!!

But, no matter how many subdivisions you pick, the total length of the stepwise diagonal must be 2. How can this be reconciled with the fact that the sum (2a/n from n=1 to infinity) diverges as n->infinity?

Or, taking a finite case, let's say n=1000; then the sum from n=1 to 1000 of 1/n is 7.485..., not 2. Is the statement that the total length of the stepwise diagonal must be 2 incorrect? How can the line be any longer than 2 units while only travelling up and right through the square, without reversing direction or exceeding the bounds of the square?

edit: The reason for this is that sum is wrong, the sum is not from n=1 to infinity at all, but is simply a/n where n=constant added to itself n times! So the sum is 2a/n times n which obviously equals 2a or the length of the stepwise diagonal.


for how many integer pairs (x,y) exist where $ \sqrt{1984}\equal{}\sqrt{x}\plus{}\sqrt{y}$ and $ 0<x<y?$


AMC USA(J)MO USAMO geometry AIME inequalities Harvard

I never thought calling about 45 people would take so long... maybe I will never be getting a call? :( :( :(

[quote="alc"]Ninth graders going to MOP post here!

I (9th Grade) got my call at about 4:30 central time today, so I'll bet a lot of you have been called or will be getting called soon.[/quote]

congrats!!! Are you sure you are in Red Group?

I still haven't get called.. I'm losing hope :(

Don't mind beta.  His "losing hope" means he "only" has a 99% chance, not a 99.9% chance.

Only 99%? Start getting worried, people :P.

meh.. im as good as a freshman.. though i am actually a soph.. ugh no phone call here.

Still no call!!!!

Could it be "green card(which I wrote on my S-III that I will receive it on June 6th, 2005)"?

Could it be that I didn't score high enough?

Could it be that they haven't called yet?

Could it be faxing error?

Could it be reaching the wrong number

Could it be....

It's freshman that wouldn't have qualified for MOP the "regular way."

I'm young enough to be a freshman, should've stayed back a grade :D

[quote="alc"]Is the red group the group of ninth graders not winners or HMs?[/quote]

yes

Yes, and maybe some sophomores this year, depending on what Mr. Feng decides to do.

I'm young enough to be an 8th grader.  I should stay back next year. :D

Well you should be able to qualify for MOP the "regular way" next year. But I probably won't make it this year, so I have more incentive to stay back ;)

I got called, but maybe it's because I'm in Lincoln.  Anyway, what was the HM cutoff this year?

[quote="darktreb"]The USA has the potential to win IMO, but certain steps have to be taken, I think, and certainly this would be one of them. (I'm aware the US has won IMO once, though just once)[/quote]
US started to participate in IMO since 1974 and won IMO 4 times: 1977, 1981, 1986 and 1994.

[quote]I'm sure he meant that the most notable year was 1994.[/quote]

Yes, sorry for the confusion.

[quote="geoffreynham"]Math, unfortunately, cannot be represented in a short T.V. clip - for what is there to show?  A head bent down over a piece of paper?  A long and unintelligble proof?  A close zoom-in on a sheet of undecipherable gibberish?  Hardly exciting stuff.  [/quote]
I wasn't clear and will clarify what I meant.  I am not at all talking about televising the competition(s).  Many people do not even know about the AMCs.  There may be talent out there that has not been identified.  If parents can see higher limits for their children and the opportunities available, it will be very different.  I am myself an example of such a case.  Despite both of us being professionals we had no idea about enrichment except grade skipping and doing higher level courses.

How I approach schools now is very different from what I did 2 years back.  My younger one has benefited more from my awareness than my older one.

What if there were a panel discussion on Math education, competitions, etc. and comparisions made with the rest of the world! ... Or 60 minutes did something on this... I am positive it will make a difference because I know.  It has made a huge difference in our small community.  Luckily people have kids in different schools within my circle of friends which means there is going to be a change in other schools besides ours.

This discussion has certainly gone away from the topic.

[quote="rrusczyk"]I'm not optimistic about the mainstream press being helpful in this regard.  Unfortunately, I think the only way to make progress in this battle is through widespread intense local efforts.  Note, that this how MATHCOUNTS succeeds - they have a tremendous network of volunteers.  How to organize and mobilize such a network is unclear.[/quote]

While the mainstream press might not be helpful, finding a few people within the press willing to play up the image isn't impossible.  At least, seeing Olson's book gives some hope.

What might pique the interest of the press is for somebody to put together a study on what Olympiad students do after the contests are over.  Given the number of Wall Street millionaires, technology millionaires, notable professors, and entrepreneurs that have gone through the MOP program, a picture of dramatic success shouldn't be hard to paint.

Are you serious? A lot of the people that go to MOP become millionaires later on?

[quote="MCrawford"]Given the number of Wall Street millionaires, technology millionaires, notable professors, and entrepreneurs that have gone through the MOP program, a picture of dramatic success shouldn't be hard to paint.[/quote]

O.O Hmm..where is this information available?

Cool, about what percentage of them become millionaires?

No idea about the percentage, and certainly the information isn't publicly available.  MCrawford and I are mostly speaking anecdotally - we know several MOPers who are doing quite well, and others who are early in financial careers (i.e. they'll be doing quite well in a few years).

For the sake of mention, I went to MOP with at least 3 Google researchers who are doing quite well.  That's just one in a reasonably long list of anecdotes.

Makes me kind of wonder why somebody hasn't written a book already.

Funny how the word "millionaire" automatically and dramatically increases the appeal of just about anything. If such a book were to be written I would hope it would include endeavors of other successful former MOPpers (who may not necessarily have chosen as lucrative a path as others)

[quote="darktreb"] If such a book were to be written I would hope it would include endeavors of other successful former MOPpers (who may not necessarily have chosen as lucrative a path as others)[/quote]

Like, say, AoPS instructors :)

Actually, Steve Olson's book usefully cites some studies, which I looked up at the University of Minnesota Library, on past IMO winners from various countries. Yes, getting involved in competition culture is definitely better for young people's futures than the kind of mind-rotting stuff I did in junior high and high school. The people at [url=http://www.jhu.edu/gifted/set/index.html]SET[/url] appear to gather some of these data, and many SET members are also math competitors.


complex analysis
function
complex analysis unsolved

Prove that there is an unique holomorphic function $f$ such that

        \[ f(\frac{1}{n})=\frac{1}{n}\cdot e^{\frac{n}{n-1}}.  \]

$n$ from positive integers¿ And holomorphic on the whole plane¿
Then (with some problem, see below)
The value $f(0)=0$ exists by continuity. Then we have a set $P=\{0\} \cup \left\{ \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, ...\right\}$ that has a accumulation point, namely $0$.
Since two holomorphic functions are identical when they are the same at some set with an accumulation point, we get that $f(x) = x \cdot e^{\frac{1}{1-x}}$.
But that has a singularity at $x=1$...  :?


[b](a)[/b] Does there exist a positive integer $n$ such that the decimal representation of $n!$ ends with the string $2004$, followed by a number of digits from the set $\left\{0;\;4\right\}$ ?

[b](b)[/b] Does there exist a positive integer $n$ such that the decimal representation of $n!$ starts with the string $2004$ ?

The (b)-part is difficult (very technical) but really well known. Haven't thought about (a) yet.

For b), have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=langlois&t=23417

Pierre.

But the problem does not imply that the last four non-zero digits are $2,0,0,4$...


geometry
projective geometry
geometry proposed

Let $a$, $b$ and $l$ be (independent) lines in the projective plane and let $P$, $Q$ and $R$ be fixed points on $l$. Let $\Delta ABC$ be a triangle such that $A \in a$, $B \in b$, $P \in BC$, $Q \in CA$ and $R \in AB$. Show that all possible points $C$ lie on a fixed line. Projective geometry again.

Consider a central proection which put line l to infty.Then angles $(A'B',a')$ , $(A'B',b')$ , $(A'B' , A'C')$ , $(A'C' , B'C')$-constant. The result follows from gomothety in point $D'=a' \cap b'$.Also locus of C is a line passing throug a point D.


The first is in the OEIS, http://www.research.att.com/~njas/sequences/A001566 .  In order to figure out what the sequence is, I suggest comparing the first few values to the [hide]Lucas numbers.[/hide]

Well, that rather defeated the purpose of my hide tags  :maybe:

The Lucas numbers are a Fibonacci-type sequence that begins $ L_{0}= 2, L_{1}= 1$.  So the first few terms are $ 2, 1, 3, 4, 7, 11, \ldots$.


AMC
AIME
USAMTS
ceiling function
modular arithmetic
AIME I

The sequence 3, 15, 24, 48, ... is those multiples of 3 which are one less than a square. Find the remainder when the 1994th term is divided by 1000.

I looked at the Kalva's solution but I have different idea.. I wonder if there are any other ways solving it.

Note: I just wrote what I had in my paper.. It's not edited so it's not like nice USAMTS solution I try to write. :D

[hide="My answer"]
I decided to write the question in math term. So, we are looking for multiples of 3 that is in form of $k^2-1$.

Note that $k^2-1 = (k+1)(k-1)$. So I decided to plug in some numbers that make it multiple of 3.

\[a_1 = (2+1)(2-1)\\
a_2 = (4+1)(4-1)\\
a_3 = (5+1)(5-1)\\
a_4 = (7+1)(7-1)\\
a_5 = (8+1)(8-1)\\
a_6 = (10+1)(10-1)\\
a_7 = (11+1)(11-1)\\
a_8 = (13+1)(13-1)\\
a_9 = (14+1)(14-1)\\
a_{10} = (16+1)(16-1)\\
a_{11} = (17+1)(17-1)\\
a_{12} = (19+1)(19-1)\]

(I just wrote out some terms to make sure I'm seeing the pattern right).

So, $a_n = (m+1)(m-1)$ is how it's expressed. From this, I developed formula with $a_8$ (I picked that one.. Seem pretty confident number).

Let $Q$ to be positive integer then the value $m$ and the value of $n$ in $a_n$ can be written as $13+3Q$ and $8+2Q$ respectively.

In our problem, $8+2Q = 1994$ so $Q = 993$. Putting in $13+3Q=m$ tells us that $m=2992$.

So.. $a_{1994} = (2992+1)(2992-1)$.. You can multiply it.. I won't..

Since $2993 \equiv -7 \mod 1000$ and $2991 \equiv -9 \mod 1000$, the product will be $63 \mod 1000$. Therefore, the answer is $\fbox{063}$. [/hide]

$n = k^2 - 1 = (k-1)(k+1)$

If $3|(k-1)(k+1)$, then $3|k-1$ or $3|k+1$.

The values of $k$ which work are in the form of $3n-1$ and $3n+1$ where $n$ is a positive integer.

For the nth term, the corresponding value for $k$ is given by $\frac{3\lceil\frac{n}{2}\rceil}{2} - 1$ when $n$ is odd and $\frac{3\lceil\frac{n}{2}\rceil}{2} + 1$ when $n$ is even.

$k=3(1994/2) + 1 = 2992$.
$2992^2 - 1 = 992^2 - 1 = (-8)^2 - 1 = 63\pmod{1000}$.

I don't know what Kalva did, but I found the 1995th positive integer that is not a multiple of 3 (not 1994th because 1 doesn't count), squared it, and subtracted 1, since every integer that isn't a multiple of 3 will have a square that is 1 mod 3. Of course on the actual AIME I probably wouldn't calculate the entire square, just the last three digits, but I had a calculator handy. Your way of finding the residue mod 1000 is much smarter, of course.

I basically did what WarpedKlown1335 did at the start...
You know that 3 divides k-1 or k+1.  So then the sequence of possible k goes
2 4 5 7 8 10 11 ...
So then the 1994th k must be
$4+3(996)$
So the 1994th term is
$2992^2-1=(3000-8)^2-1=3000^2-16*3000+64-1=63(mod1000)$


1. What are the dimensions of the shaded square inside the 3-4-5 right angle triangle?

2. Find the position of the square that has the maximum area.

3. Can you generalise for an x-y-z right angle triangle?

4. What would be the dimensions of the square in terms of x, y, and z?

1, 4
[hide]The square splits the entire triangle into a bunch of similar triangles. If the side of the square is $s$, then the little right triangle at the top has legs of $s$ and $3 - s$. Since it is similar to the largest triangle, we form the proportion $\frac {3-s}{s} = \frac {3}{4}$ and solving for $s$ we get $\frac {12}{7}$. If we have an $x, y, z$ right triangle, we just substitute $x$ and $y$ in for $3$ and $4$ in the equations and we get that the side of the square is $\frac {xy}{x+y}$.[/hide]

[hide="2"]
as it is drawn below
[/hide]

[hide="3"]
call z the hypotenuse, x the shorter side, and s the side of the square.

$\displaystyle \frac{x-a}{a} = \frac{a}{y-a}$
$\displaystyle a^2 = xy - a(x+y) + a^2$
$\displaystyle a(x+y) = xy$
$\displaystyle a = \frac{xy}{x+y}$
[/hide]

EDIT: Ignore number 3. I don't know if what i did for number 3 is generalization or not....(realized that after looking at tarquin's number [b]4[/b]


Noah has to fit 8 species of animals into 4 cages of the Arc.  He planes to put two species of animal in each cage.  It turns out that, for each species of animal, there are at most 3 other species with which it cannot share a cage.  Prove that there is a way to assign the animals to the cages so that each species shares a cage with a compatible species.

[quote="Singular"]

 Each stage, we are guaranteed to find a pair, because the string necessarily contains all pairs formed of unused letters.  The 4 pairs are our solution.

[/quote]

Why is that? I don't understand. Can anybody explain?

Sorry for the reply to an old topic.  :oops:

Isn't the problem a direct conseguence of Dilworth's Theorem?

I think we can generalize it into $2n$ and $n$ ccases,where $3$ is replaced with $n-1$.

[quote="Singular"]It is obvious we can remove the "at most" condition and suppose that there are 4 other compatible species for each specie.

Consider the graph of $K_8$ with 3 edges at each node removed so that in total we remove 12 edges.  The degree of each node is even, so we can find an euler circuit through this graph.  Label the nodes A..H.  The circuit looks something like ABCB...  where we write each letter 2 times. Taking every two adjacent letters (where the first and last are considered adjacent) , we find we have 16 'adjacencies'.

Every adjacent pair of letters is a legal pair.  Since we never use the same edge twice in a circuit, we have all 16 legal pairs.  Since we use all edges in a circuit, the following method is guaranteed to work:

we take our 'circuit string' and pick any pair in it, and erase all the letters that the pair uses.  We repeat this 3 more times.  Each stage, we are guaranteed to find a pair, because the string necessarily contains all pairs formed of unused letters.  The 4 pairs are our solution.

Example:

AEBFCGDHCEDFAGBH; pick AE
BFCGDHC DF GBH; pick GB
FC DHC DF H; pick DF
C HC H; pick HC and done.[/quote]

I also find this a bit too optimistic

Take the following circuit for eg:

4-5-1-2-5-3-1-6-7-8-2-7-6-4-3-8

we can remove 1-2 first, then 3-4, then we are left with

5      5     6-7-8     7-6      8

where 5 can never be paired with anything

It appears this algorithm is not at all intuitive and will need at least a 1-ply look-ahead to work properly,

actually 6 and 7 doesn't satisfy the problem requirement in my previous example, a correct example would be:

4-5-1-2-5-3-1-6-7-8-2-7-3-4-6-8

again remove 12, then 34, and 5 is left hanging


group theory
abstract algebra
algebra
polynomial
superior algebra
superior algebra unsolved

Hi!

Can anyone give a hint for this problem:

Let K be a field, g an element of Aut(K)(=group of ring automorphisms of K) of infinite order.
If F is the fixed field of the cyclic subgroup (of Aut(K)) generated by g, and K/F is an algebraic extension,
show that K/F is a normal extension.

Take $x \in K$, and let $P$ be its minimal polynomial (which exists since $K/F$ is algebraic). If $\alpha$ is a root of $P$, then $P(\alpha)=0 \Rightarrow P(g^{k}(\alpha))=0$ for every integer $k$ (since $g^{k}$ fixes the coefficients of $P$ by hypothesis). Since $P$ has only finitely many roots and since $<g>$ has infinite order, we have $g^{k}(\alpha)=g^{m}(\alpha)$ for some $n \neq m$. But then $\alpha=g^{k-m}(\alpha) \in K$, and the problem is solved.


function
algebra
domain
polynomial
calculus
integration
functional equation

Can anyone of you help me how to prove that
1) a function is injective, surjective and bijective?
2) a function is constant, continuous and monotonous?
by giving me an examples?

Another thing what is the differences between solving the functional equation that has real numbers and quotient number? I don't quite understand the notes our teacher has given us.

For 2,

A constant function is a function in which the y value is a constant for whatever x value you "plug" into it.
Ex: y=3

A continuous function is a function which you can "draw" without lifting up your pencil, such as y=x, y=x^2, y=2^x, y=lnx, y=sinx, etc....

I know that a monotonically increasing function is a function in which for all pairs x_1<x_2 in its domain, f(x_1)<f(x_2), and f(x_1)>f(x_2) for a monotonically decreasing function.

I am not sure if I answered the last one correctly.

-interesting_move

Constant means f(x) = C for a constant C, for all x (or put another way, f(x) = f(y) for all x, y).
The formal definition is continuous is that for any a, the limit of f(x) as x goes to a is f(a). If you don't need to do it formally, interesting_move has it right.
Monotonous means either f(x) >= f(y) for all x>=y, or f(x)<=f(y) for all x<=y.

Injective means one to one; ie, if f(x) = f(y) then x=y.
Surjective means onto; ie, for all y in the range of the function, you can find an x with f(x)=y.
Bijective means injective and surjective.

I'm not quite sure what you mean by quotient number, but I'll take a guess that you mean Q, ie the rational numbers. A function is something which takes a value, and outputs another value. If you have a function from the reals, that means f(any real number) exists. If you restrict things to the rationals, there is no such thing as f(sqrt(3)) for example - f is only defined for rational numbers. Often you can prove something is true for rational numbers then use something like continuity to prove it for all reals - like the famous Cauchy equations.

[quote="ken_himura1984"][quote="TripleM"]Injective means one to one; ie, if f(x) = f(y) then x=y.
[/quote]

But I found out that it has another explanation which confuses with this one.

It states that,
                        f : A -> B
            if  both a1 and a2 are elements of A, then a1 is not equal to a2,
            which implies that f(a1) is not equal to f(a2).

Your say that f(x) = f(y) and this makes me confused.[/quote]

Take the contrapositive of what he said, e.g, p => q is logically equivalent to ~q => ~p.

So, MMM defined injective as: f(x)=f(y) implies x=y.

The contrapositive is x != y implies f(x) != f(y), which is essentially what you wrote.

[quote="ken_himura1984"][quote="TripleM"]The formal definition is continuous is that for any a, the limit of f(x) as x goes to a is f(a). If you don't need to do it formally, interesting_move has it right.[/quote]

I don't understand what you mean by this sentences. Can you simplify it?

And also, can U please help me hoe to prove the function below?

 how to prove

1) f(an) = nf(a)

2) f(a/n) = f(a)/n

3) f(a(m/n)) = f(a)(m/n)
    where m and n are elements of N[/quote]

Hi,

First understand that f(x) is a general representation of any relation between the variable x and another variable y. The former is called an independent variable and the latter, the dependent  variable. The set of all possible values ,the independent variable assumes ,is said to be the domain while that of the dependent variable is called the range of the function

examples of function :

y = sin(x)
y = ax^2  + bx +c

etc.

Note, that in the above examples y has been explicitly defined in terms of x.

We can also have analytically defined functions, that is functions that are not defined by a single relationship over the entire domain of definition .

Examples

f(x) = x-1 for    x <= 1

      = sin(x)  for    1 < x <2

      = log(x)  for 2<= x

With this background ,let's try to answer ur queries.

[quote]
How to prove

1) f(an) = nf(a)

2) f(a/n) = f(a)/n

3) f(a(m/n)) = f(a)(m/n)
    where m and n are elements of N[/quote]

U can prove the above only if the function f(x) is defined in a PARTICULAR way.

For example if f(x) = x (the Identity function), u can prove all of the above identities.

If f(x) = x for all real values of x,

f(an) = an = na= nf(a)  { since f(a) = a}

similarly f(a/n) = a/n = f(a)/n

I leave the proof of the third to u .

--------------------------------------------------------------------------------------------

Since u r new to functions , it's almost certain that u haven't been introduced to the concept of limits of functions. Once you understand the concept of limits, u 'll appreciate the more formal and analytical definition of continuous functions based on Limits. As of now, just visualise any function whose graph can be plotted without any breaks as a continuous function,as defined in some of the earlier replies to your message.

Some examples of continuous functions  could be y=x,y=2,y=sin(x),y=some polynomial in x etc.

To give u an idea of Discontinous functions,consider the greatest integer function

y= [x] where [a] denotes the largest integer not exceeding a

Ex. [3.5] = 3

      [2.3]=2

      [-1.3]= -2

       [ 3] = 3

So for any integer [x] = x

Now let's try to construct the graph of [x].

let's take the case of  0<= x <1

here [x] = 0 ,i.e., it's constant over the entire range 0 to 1

similarly [x] =1  for  1<= x <2

and        [x] =2  for  2<= x <3

and so on and so forth.

If you plot these points on a graph sheet, u 'll get a series of steps with breaks at integral values of x. In other words, the function [x] is discontinuous at integers.

[quote="ken_himura1984"]santosh_blr thanks for your explanation. Now at least what function is all about.

We all know surjective function means "onto" but there is also a "into" function too. What does into means ?[/quote]

Consider the association

 f : A-> B  ( A= domain, B= Co-domain}

Where A = {1,2,3} and B= {4,5}

   Also let the function be a constant one with value 5

Then f(1) = f(2)=f(3) = 5

So the range of f  ,denoted by ran(f ), = {5}  a SUBSET of B = {4,5}

[b]Definition [/b]: When the range of a function (In this case {5} )is a PROPER subset of its Co-domain ( in this case, it's B), we say the function is injective

I got confused the way they solve this question.

[b]Let R be the real numbers. Find all functions f:R -> R such that for all real numbers x and y,
                              f(xf(y)+x) = xy+f(x)[/b]

The solution should be:
Taking x = 1, y = -1-f(1) and letting a = f(y)+1, we get

           f(a) = f(f(y)+1) = y+f(1) = -1

Putting y = a, and letting b = f(0), we get

           b = f(xf(a)+x) = ax+f(x),

so, f(x) = -ax+b. Putting this into the equation, we have

           a^2xy-abx-ax+b = xy-ax+b,

Equating coefficients, we get a = +/-1 and b = 0, so f(x) = x or f(x) = x. We can easily check both are solutions.

My question is
1)  from the 1st step, why (what is the purpose) we let  y = -1-f(1) and       a = f(y)+1

2) cAN U help me to get the equation   " a^2xy-abx-ax+b = xy-ax+b,"
    by substitution. I can't seem to get that equation for a long period.

I'll try to explain the answer in a way that you are likely to actually come up with as you solve it - remember, when writing down the solution, you don't always write it down in the same order as you tried when solving it.

[quote="ken_himura1984"]I got confused the way they solve this question.

My question is
1)  from the 1st step, why (what is the purpose) we let  y = -1-f(1) and       a = f(y)+1

2) cAN U help me to get the equation   " a^2xy-abx-ax+b = xy-ax+b,"
    by substitution. I can't seem to get that equation for a long period.[/quote]

The question says that equation f(xf(y)+x) = xy+f(x) is true for *any* x and y. You can substitute any x and y you want, and that equation will be true. Usually when trying to solve these equations, you find a particular substitution that will make things a lot easier - which often involves making part of the equation 0 to cancel things out.

In this case you could think: It would be nice if we could have most of the left hand side cancel out. If we can find some y with f(y) = -1, then we have almost solved the problem - since then for that fixed y we have f(0) = xy + f(x), f(x) = f(0) - xy and that gives us the only possible solution.

Now that we know we should look to see if f(y) = -1 for some y (which may not be possible, but its a good start). Now we start playing around with some values for x or y. Lets put x=1.
We know that f(f(y) + 1) = y + f(1) for all y. Hey - now we can choose y = 1-f(1) and we have f(something) = -1, which is exactly what we want! Now all you have to do is put all this information together.

We call a that value with f(a) = -1, and we call f(0) = b. From the first step we did, we have f(x) = b - ax.

f(xf(y)+x) = xy+f(x)
b - a(xf(y) + x) = xy + (b-ax)
b - a(x(b-ay) + x) = xy + b - ax
b - abx + xya^2 - ax = xy + b - ax
-abx + xya^2 = xy

Remember this is true for all x and y - so the coefficient of y must be the same on each side. So xa^2 = x, so a^2 = 1. Then we have -abx = 0, so ab = 0 (we can divide by x since this is true for EVERY x), so b = 0.

So f(x) = x or f(x) = -x. They both work.


linear algebra
matrix
number theory
greatest common divisor
calculus
integration
linear algebra unsolved

Are you sure about the wording ? Try the diagonal matrix with $x$ on its main diagonal such that $x^3=x+1$ ?

Yes, why? You are trying to say that the problem is wrong? :huh:

Consider $a$ a real number such that $a^3=a+1$  with a calculator you have

$a\sim 1.3247...$

Take $A=diag(a,a,a,a)$ diagonal matrix we have $A^3=A+I$

But $det(A^4-I)=(a^4-1)^4\sim ((1.3247)^4-1)^4\sim 18.697 \neq 1$

Sorry, I have made a stupid mistake. It's not $M_{4}(\mathbb{R})$ it is $M_{4}(\mathbb{Z})$.
I must have been drunk when I typed this problem. :oops:

P.S. I have edited the initial problem. :)

A is invertble

gcd of $x^3-x-1$ and $x^4-1$ is 1

$A^4-I$ is invertible since its entries are in Z

$det(A^4-I)=\pm1$

This is a Romanian forum (ie Romanian speakers!). Moved to the College Playground.

[quote="alekk"]you know, there are a lot of invertible matrices, with integral entries, such that $|det(A)| \neq 1$ ..[/quote]Not invertible in $\mathcal{M}_n(\mathbb{Z})$ :P

Well, that didn't work very well. I was trying to merge two topics. For what it's worth, Moubinool's post #7 should be at the top - that's the problem statement.


quadratics
number theory unsolved
number theory

Problem)
Prove that for any arbitrary positive integer $ n$ which is not the square of an integer,
there exists a prime number $ p$ such that $ n$ is a quadratic nonresidue modulo p.

Comment) It's just my quess, so I don't know if it is true.
Also I guess that it will need Dirichlet's theorem. If you can, please don't use Dirichlet's theorem

i think that there is a generalisation,
for $ m>1$
$ \forall p\in\mathbb{P},\exists k\in\{0,1,...,p\minus{}1\}: \ n\equiv k^{m}(mod\ p) \Longrightarrow \exists A\in\mathbb{N}: \ n\equal{}A^m$

how can we prove it?

It must be generalization of IMO shortlist 2007 N2, maybe?

for your exemple $ m\equal{}2$
i think that so for $ n\in\mathbb{N}$ $ \exists h: \ h^2x\plus{}1$ have a primp divisor p of that form $ p\equal{}4k\plus{}3$ (i am not sur)
so $ h^2x$ is not a quadratic nonresidue modulo $ p$ ==> $ x$ is not a quadratic nonresidue modulo $ p$

See http://www.mathlinks.ro/viewtopic.php?t=64322 for a proof for the case $ m \equal{} 2$ without Dirichlet's theorem.

The generalisation is wrong, e.g. $ 16$ is a $ 8$-th power modulo all primes.
But with some adaption, it is still correct:
Let $ m$ be a natural number and $ a$ some integer. Assume that $ a$ is a $ m$-th modulo all but finitely many primes.
a) if $ 8 \nmid m$, then $ a$ is the $ m$-th power of an integer.
b) if $ 8|m$, then $ a$ is the $ \frac m2$-th power of an integer.

This is sometimes called Grunwalds theorem and there is some funny story around this: Grunwald even "proved" (in 1933) what aviateurpilot conjectured: if something is a $ m$-th power modulo all primes, then it's a $ m$-th power. This was "reproven" 1942 by Whaples, and it needed six more years untill Wang found the easy counterexample I gave above and has shown the correct version.

See some texts on Class Field Theory to see some proofs.


Source: AHSME 1973 #14

Each valve A, B, and C, when open, releases water into a tank at its own constant rate.  With all three valves open the tank fills in 1 hour, with only valves A and C open it takes 1.5 hours, and with only valves B and C open it takes 2 hours.  The number of hours required with only valves A and B open is
$\text{(A)} \ 1.1 \qquad \text{(B)} \ 1.15 \qquad \text{(C)} \ 1.2 \qquad \text{(D)} \ 1.25 \qquad \text{(E)} \ 1.75$

[hide="Solution"]
$(C)$
Let $A$, $B$, and $C$ denote each respective valve's rate.
Thus,
\[ A+B+C = 1  \]
\[ (A+C)1.5 = 1 \implies A+C = 2/3  \]
\[ (B+C)2 = 1 \implies B+C = 1/2  \]
We can quickly substitute each of the second and third equations into the first to find that $A = 1/2$ and $B = 1/3$, so together they will take $1/(1/2 + 1/3) = 6/5 = 1.2$ hours.
[/hide]

Classic question...

[hide]Let the amount of time it takes to fill a tank be $a$, $b$, and $c$ for valves A, B, and C respectively. Let x be the amount of time it takes for valves A and B to fill the tank. We then have

1/a+1/c=1/1.5=2/3
1/b+1/c=1/2
1/a+1/b=1/x

Add: 2(1/a+1/b+1/c)=7/6+1/x

since 1/a+1/b+1/c=1/1=1,

1/x+7/6=2, and x=6/5, or 1.2 hours.  [/hide]


logarithms
combinatorics unsolved
combinatorics

We have n coins, and at least one of them is bad. The bad coins weigh the same, and so do the good coins. The bad coins, however, are lighter.

Find the number of bad coins in O((logn) 2 ) using a balance scale

how much times should we use the balancE

As stated, $ C \cdot \log_{2}(n)$ times at most, $ C$ constant.


Καλησπέρα σε όλους!!
Μία βοήθεια για την άσκηση....

[img]http://img142.imageshack.us/img142/9162/92231888xt2.jpg[/img]

Ευχαριστώ πολύ

Αν γίνεται, να είναι σε Latex για να διαβάζονται πιο εύκολα.

1.
Για $ n \equal{} 2$ έχουμε δυνατά υποσύνολα πλήθους $ 1 \equal{} \frac {2(2 \minus{} 1)}{2}$
Έστω ότι για $ n \equal{} k$ έχουμε $ \frac {k(k \minus{} 1)}{2}$  υποσύνολα
Για $ n \equal{} k \plus{} 1$ έχουμε k περισσότερα υποσύνολα απο $ n \equal{} k$ (όλα τα υποσύνολα που σχηματίζονται με έναν από τους προηγούμενους k αριθμούς που και τον καινούριο αριθμό) άρα το πλήθος των υποσυνόλων είναι:
$ \frac {k(k \minus{} 1)}{2} \plus{} k \equal{} \frac {k(k \plus{} 1)}{2}$

2.Α.
Εδώ θα εφαρμοσθεί ο αρισμός για 2 τύπους και μετά θα γενικευτεί για n
(Εφαρμόζουμε εφαρμογή στους φυσικούς και όχι στη δομή των τύπων, γιατί η τελευταία δε δίδεται από την εκφώνηση)

2.Β.
Είναι απλή εφαρμογή του 2.Α.

Σε ευχαριστώ πολύ για την απάντηση και
συγνώμη για αυτή τη μορφή!
Αλλά αν μπορείς να μου πεις και τα άλλα
2 ερωτήματα λίγο πιο αναλυτικά όπως το πρώτο :lol: .
Αν πάλι όχι δεν πειράζει.Σε ευχαριστώ και πάλι.

[quote="maniopas"]Δεν ξέρω πώς να εισάγω το συγκεκριμένο συμβολισμό.[/quote]
Μήπως όμως μπορείς να μας δώσεις περισσότερη βοήθεια;

[quote="maniopas"]2.Α.
Ελλείψη συμβολισμού θα γράψω το ζητούμενο για απόδειξη ως εξής:
$ (Y_1 and ... and Y_n) \equal{} not(notY_1 or ... or notY_n)$

Απόδειξη:
Για $ n \equal{} 1$ είναι $ Y_1 \equal{} not(notY_1)$, που ισχύει
Έστω ότι ισχύει για $ n \equal{} k$ τότε
για $ n \equal{} k \plus{} 1$ είναι $ (Y_1 and ... and Y_{k \plus{} 1}) \equal{} not(notY_1 or ... or notY_{k \plus{} 1}) < \equal{} >$
$ (Y_1 and ... and Y_k and Y_{k \plus{} 1}) \equal{} not(notY_1 or ... or notY_k ornotY_{k \plus{} 1}) < \equal{} >$
$ (Y_1 and ... and Y_k) and (Y_{k \plus{} 1}) \equal{} not(notY_1 or ... or notY_k ornotY_{k \plus{} 1}) < \equal{} >$
$ not(notY_1 or ... or notY_n) and not(notY_{k \plus{} 1}) \equal{} not(notY_1 or ... or notY_k or notY_{k \plus{} 1}) < \equal{} >$
$ not[ not(notY_1 or ... or notY_n) and not(notY_{k \plus{} 1}) ] \equal{} not[not(notY_1 or ... or notY_k or notY_{k \plus{} 1})] < \equal{} >$
$ (notY_1 or ... or notY_n) or (notY_{k \plus{} 1}) \equal{} (notY_1 or ... or notY_k or notY_{k \plus{} 1})$, που ισχύει
άρα αν η παράσταση ισχύει για $ n \equal{} k$,θα ισχύει και για $ n \equal{} k \plus{} 1$

Σημείωση:
Για τη λύση θεώρησα ως δεδομένη την πρόταση ότι $ not(X and Y) \equal{} notX or notY$ και την αντιμεταθετικότητα της noτ.

Το 2.Β είναι απλή εφαρμογή του 2.Α (με αντικατάσταση)
Δεν το γράφω γιατί μου "βγαίνει ο πάτος" με αυτό το συμβολιασμό.[/quote]
Σ' ευχαριστώ πολύ για την βοήθειά σου


Bill has a bucket of [b]R[/b]ed, [b]W[/b]hite and [b]B[/b]lue marbles.

[b]R[/b]ed marbles = 2 more than [b]B[/b]lue marbles

[b]W[/b]hite marbles = twice the [b]R[/b]ed marbles

Which could be the total # of marbles? It's multiple choice: 24, 25, 26, 27, 28

The answer is 26

There are 7 Red, 5 Blue and 14 White

R = 2 + B

W = 2R

B = R - 2

26 = (R-2) + (2R) + 2 + B

26 = (7 - 2) + (2*7) + 2 + 5

26 = 5 + 14 + 2 + 5

[quote="stevenmeow"]what are you asking us?
i have verified your answer by the way...[/quote]

I guess that I'm asking what the equation(s) is/are.

I got the answer by plugging in the possibilities and trial and error.

I'd like to know how to setup the problem in case I don't have multiple choice in the future.

[quote="stevenmeow"]ok ill start similarly to you

red=r, blue=b, white=w
r=2+b
w=2r=2(2+b)=4+2b
add all of them together

total=b+r+w
=b+(2+b)+(4+2b)
=4b+6
the total must be in the form 4b+6
where b is a whole number
in other words, the number b must be in the series
6, 10, 14, 18, 22, 26, 30...
that adds 4 each term[/quote]

Thank you

:)


geometry
parallelogram
geometry unsolved

Consider 5 points A, B, C, D, and E are drawn so that ABCD and BCED are parallelogram and cyclic quadrilateral. Let l be line passing through A. Suppose l intersect segment CD and line BC at F and G so that EF = EG = EC. Prove that l bisects anlge DAB.


Two Points $C$ and $D$ with constant distance move on a semicircle with diameter $AB$. Lines $AC$ and $BD$ intersect at $E$, lines $AD$ and $BC$ intersect at $F$.
Prove that the quadrilateral $AEBF$ has constant area!

Ok, I'll post my solution:

[hide]

Obviously $E$ is the orthocenter of the triangle $ABF$ so $EF\perp AB$. Hence the claim that $AEBF$ has constant area is equivalent to $EF$ being constant.
Now obviously the quadrilateral $CEDF$ is cyclic and the diameter of its circumcircle is $EF$. But $CD$ is fixed, hence $\angle CAD$ is fixed and hence $\angle CFD$ and $\angle CED$ is fixed. Thus $EF$ is fixed.
qed

[/hide]


[quote="malinger"]Let $X$ be a fixed point on bisector $AA'$ of $\triangle ABC$. $B'=BX\cap AC,C'=CX\cap AB, P=A'B'\cap CC',Q=A'C'\cap BB'$. Prove that $\angle PAC=\angle QAB$[/quote]

One direction of http://www.mathlinks.ro/Forum/viewtopic.php?t=21297 .

  darij


calculus
integration
trigonometry
logarithms
calculus computations

Calculate:
1. $ I_1 \equal{} \int_{\frac { \minus{} \pi}{2}}^{\frac {\pi}{2}}\frac {1 \plus{} sinx}{1 \plus{} cosx}e^{x} dx$

2. $ I_2 \equal{} \int_{0}^{\frac {\pi}{3}}ln(1 \plus{} \sqrt {3} tgx) dx$

hello, a simple calculation shows that
$ e^x\frac {1 \plus{} \sin(x)}{1 \plus{} \cos(x)} \equal{} \frac {1}{2}e^x\left(\tan\left(\frac {1}{2}x\right) \plus{} 1\right)^2$.It follows
$ \int \frac {1}{2}e^x\left(\tan\left(\frac {1}{2}x\right) \plus{} 1\right)^2dx \equal{} e^x\tan\left(\frac {1}{2}x\right) \plus{} C$. Using this, so your integral gives
$ e^{\frac {\pi}{2}} \plus{} e^{ \minus{} \frac {\pi}{2}}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, a simple calculation shows that
$ e^x\frac {1 \plus{} \sin(x)}{1 \plus{} \cos(x)} \equal{} \frac {1}{2}\left(\tan\left(\frac {1}{2}x\right) \plus{} 1\right)^2$.It follows
$ \int \frac {1}{2}\left(\tan\left(\frac {1}{2}x\right) \plus{} 1\right)^2dx \equal{} e^x\tan\left(\frac {1}{2}x\right) \plus{} C$. Using this, so your integral gives
$ e^{\frac {\pi}{2}} \plus{} e^{ \minus{} \frac {\pi}{2}}$.
Sonnhard.[/quote]

I don't see how $ e^x\frac {1 \plus{} \sin(x)}{1 \plus{} \cos(x)} \equal{} \frac {1}{2}\left(\tan\left(\frac {1}{2}x\right) \plus{} 1\right)^2$ and $ \int \frac {1}{2}\left(\tan\left(\frac {1}{2}x\right) \plus{} 1\right)^2dx \equal{} e^x\tan\left(\frac {1}{2}x\right) \plus{} C$.

Hello, sorry, it was a typo it must be:
$ e^x\frac{1\plus{}\sin(x)}{1\plus{}\cos(x)}\equal{}\frac{1}{2}e^x\left(\tan\left(\frac{1}{2}x\right)\plus{}1\right)^2$
and
$ \int\frac{1}{2}e^x\left(\tan\left(\frac{1}{2}x\right)\plus{}1\right)^2dx\equal{}e^x\tan\left(\frac{1}{2}x\right)\plus{}C$.
Sonnhard.

OK, but may I know how you come up with the identity $ e^x\frac{1\plus{}\sin(x)}{1\plus{}\cos(x)}\equal{}\frac{1}{2}e^x\left(\tan\left(\frac{1}{2}x\right)\plus{}1\right)^2$ (from sketch)?  Thanks.

hello, this is easy,
$ \frac{1}{2}\left(\tan\left(\frac{1}{2}x\right)\plus{}1\right)^2\equal{}
\frac{1}{2}\left(\frac{\sin\left(\frac{1}{2}x\right)\plus{}\cos\left(\frac{1}{2}x\right)}{\cos\left(\frac{1}{2}x\right)}\right)^2$.
Sonnhard.

Kirill's answer seems to be correct:

$ I_2\equal{}\int_{0}^{\frac {\pi}{3}}\log(1 \plus{} \sqrt {3} \tan x) dx\equal{}$

$ \equal{}\int_{0}^{\frac {\pi}{3}}\log(1 \plus{} \sqrt {3} \tan \left(\frac{\pi}{3}\minus{}x\right) dx\equal{}$

$ \equal{}\int_{0}^{\frac{\pi}{3}}\log\left[1\plus{}\frac{3 \cos x\minus{}\sqrt{3} \sin x}{cos x\plus{}\sqrt{3} \text sin x}\right]dx\equal{}$

$ \equal{}\int_{0}^{\frac{\pi}{3}}\log\left[\frac{4 \cos x}{\cos x\plus{}\sqrt{3} \sin x}\right]\equal{}$

$ \equal{}\minus{}I_2\plus{}\int_{0}^{\frac{\pi}{3}}2\log 2 dx$

Hence, $ I_2\equal{}\frac{\pi}{3}\log 2$.

($ \log\equal{}\ln$)


integration
real analysis
real analysis unsolved

Prove that $ \boxed {\int_0^{\infty}\frac {1}{\left(x^4 \plus{} 1^2\right)\left(x^4 \plus{} 2^2\right)\ldots\left(x^4 \plus{} n^2\right)}\ \mathrm {dx} \equal{} \frac {\pi}{2}\cdot\sum_{k \equal{} 1}^n\frac {( \minus{} 1)^{k \minus{} 1}\cdot\sqrt {2k}}{(n \minus{} k)!\cdot(n \plus{} k)!}\ }$ .
See http://www.mathlinks.ro/viewtopic.php?t=189621

The residue theorem yields the result almost immediately but it would be interesting to see if someone could come up with a real variable proof.


algebra
polynomial
Vieta
algebra unsolved

I'm looking for the most beautiful proof of identity (if there exist any beautiful):

$ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}ac\minus{}bc)$

Thanks in advance!  :)

Compute the determinant of

$ \left[ \begin{array}{ccc} a & b & c \\
c & a & b \\
b & c & a \end{array} \right],$

first by expansion (LHS), and then by computing its eigenvalues (RHS).

First proof: Expand the RHS?

Second proof:

Look at the LHS as a polynomial of $ a$. Take $ a\equal{}\minus{}(b\plus{}c)$, then the LHS is 0, so $ a\plus{}b\plus{}c$ divides $ P$. As every term on the LHS is of degree 3, we have that $ a^3\plus{}b^3\plus{}c^3\equal{}(a\plus{}b\plus{}c)(xa^2\plus{}xb^2\plus{}xc^2\plus{}yab\plus{}ybc\plus{}yca)$, the coefficients of the squares are the same because of symmetry.

Now we have $ a^3\equal{}xa^3$ if we look just at the $ a^3$ term, so $ x\equal{}1$. Similarly, $ y\equal{}\minus{}1$.

Bugi, I've asked this because I don't like expansion proof of this...
Thanks for nice proofs.
 :)

I have answered the proof before, but I can't find it.
There were at least 4 proofs.
For example, one of them is using
\[ a^3 \plus{} b^3 \equal{} (a \plus{} b)^3 \minus{} 3ab(a \plus{} b)\]
or Vieta or Complex number.

$ \boxed{(a \plus{} b \plus{} c)(a \plus{} b\omega \plus{} c\omega ^ 2)(a \plus{} b\omega ^ 2 \plus{} c\omega) \equal{} a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc}$ for $ \omega \equal{} \frac { \minus{} 1\pm \sqrt {3}i}{2}$

You are welcome. :)

I have just found the solution [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22252]here[/url]
Please take a look in #8 & #9 post.


What is the sum of outer angles of any concave quadrilateral?360?Any proof?

[hide]It is 360 degrees. If the quadrilateral is $ABCD$ and $\angle BCD>180^\circ$, then draw in $\overline{AC}$ to split the quadrilateral into two triangles.

The sum of the angles of the quadrilateral is the sum of the angles of the triangles, which adds to 360 degrees.[/hide]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=397070&highlight=#397070]Once is enough[/url] :!:


One more than the reciprocal of a particular number is $ \frac{7}{3}$. What is the original number expressed as a common fraction?

so $ \frac73$ is one more, so $ \frac43$ is the number, reciprocol of a number is $ \frac43$ so the real number is $ \frac34$

So its just 7/3-3/3=4/3 then we flip it to get 3/4.


$ \forall m,n \in \mathbb Z^\plus{}$ , prove that :

$ \frac{(3n)!\cdot(3m)!}{(m!)\cdot (n!)\cdot (m\plus{}n)!} \in \mathbb Z$.

 :wink:

[quote="Hidden Scofield"]$ \forall m,n \in \mathbb Z^ \plus{}$ , prove that :

$ \frac {(3n)!\cdot(3m)!}{(m!)\cdot (n!)\cdot (m \plus{} n)!} \in \mathbb Z$.

 :wink:[/quote]

I will instead show that $ \frac {(2n)! \cdot (2m)!}{(m!) \cdot (n!) \cdot (m \plus{} n)!} \in \mathbb Z$.

We just have to show that $ e_p(2n!) \plus{} e_p(2m!) \ge e_p(m!) \plus{} e_p(n!) \plus{} e_p((m \plus{} n)!)$

Which is the same as $ \left[\frac{2n}{p^r}\right] \plus{} \left[\frac{2m}{p^r}\right] \ge \left[\frac{m}{p^r}\right] \plus{} \left[\frac{n}{p^r}\right] \plus{} \left[\frac{m \plus{}n}{p^r}\right]$ For every prime p and factor r.

We'll set $ a \equal{} \frac{n}{p^r}$ and $ b \equal{} \frac{m}{p^r}$ and then we get:

$ [2a] \plus{} [2b] \ge [a] \plus{} [b] \plus{} [a\plus{}b]$

It is easy to see that $ [2a] \equal{} 2[a] \plus{} [\{2a\}], [2b] \equal{} 2[b] \plus{} [\{2b\}], [a\plus{}b] \equal{} [a] \plus{} [b] \plus{} [\{a\plus{}b\}]$
And then we just have to show that:
$ [\{2a\}] \plus{} [\{2b\}] \ge [\{a\}\plus{}\{b\}]$
$ RHS < 2$.. If $ RHS \equal{} 0$ it is obviously correct. But if $ RHS \equal{} 1$ then we have $ {a} \ge \frac{1}{2}$ or $ {b} \ge \frac{1}{2}$ which means that $ LHS \ge 1$.

Thus $ \frac {(2n)! \cdot (2m)!}{(m!) \cdot (n!) \cdot (m \plus{} n)!} \in \mathbb Z$.

[quote="Hidden Scofield"]What is e_p(2n)! ?? :huh:[/quote]
$ e_p(x)$ is the largest integer so $ p^{e_p(x)}|x$, where $ p$ is a prime and $ x$ an integer.
And we know that $ e_p(n!) \equal{} \left[\frac{n}{p}\right] \plus{} \left[\frac{n}{p^2}\right] \plus{} \left[\frac{n}{p^3}\right] \plus{} \dots$ were $ [x]$ denotes the largest integer $ \le x$