geometry
3D geometry
number theory unsolved
number theory

Show that 26 is the only number between a square and a cube.


calculus
integration
logarithms
Putnam
calculus computations

The integral equals: $2-\frac{\pi^{2}}{6}-2\ln 2+(\ln 2)^{2}$.
To see this use the following expansions:

$\ln(1-x)=-\sum_{m=1}^{\infty}\frac{x^{m}}{m}$, $-1\leq x<1$,

$\ln(1+x)=-\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{n}}{n}$, $-1<x\leq 1$
and
$\frac{1}{2}(\ln(1-z))^{2}=\sum_{n=1}^{\infty}\frac{H_{n}}{n+1}z^{n+1}$, $|z|<1$, wher $H_{n}$ is the $n^{th}$ harmonic number

and

$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{2}}=\frac{\pi^{2}}{12}$
Let $I$ be the value of your integral, then

$I=\sum_{n,m=1}^{\infty}\frac{(-1)^{n}}{nm(n+m+1)}=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n}H_{n+1}}{n}-\frac{(-1)^{n}H_{n+1}}{n+1}\right)$.
From now on, using some simple *rearrangements*, the calculations are straightforward.

[quote="didilica"]
From now on, using some simple *rearrangements*, the calculations are straightforward.[/quote]

i dont see it :wink:
please elaborate.

haha, this problem is suddenly popping up in so many places.

i was given the exact problem in my putnam meetings the other night.
and the same problem was used in our university's challenge problem.

my solution is almost identical to the didilica's, so i agree with the answer.

i wonder if there's a shorter and/or more elegant way to do this!


1.	How many positive integers are divisors of 444,444?
[hide]96[/hide]

[quote="helga7032Z"]i believe the answer to this problem is:

[hide]96.

[/hide]hi roland[/quote]
eh.. next time can you explain?

I believe prime factorization of $ 444444$ is $ 2^2\cdot3\cdot7\cdot11\cdot13\cdot37$

So (exponent+1)$ \rightarrow 3\cdot2\cdot2\cdot2\cdot2\cdot2$ and multiply to get $ \boxed{96}$

[quote="helga7032Z"]i believe the answer to this problem is:

[hide]96.

[/hide]hi roland[/quote]
I believe that you should post a solution with your answers?
Because he told us the answers, and wanted a solution.


Prove that the following equation has exactly one root in $ (\frac{1}{2},1)$:
$ \frac{e^x}{(x\plus{}1)^2}\equal{}x$

Let $ f(x) \equal{} e^x \minus{} (x)(x \plus{} 1)^2$.

We can use crude approximations to show that:

$ f(1/2) > 0 > f(1)$

Hence there is at least one root due to the intermediate value theorem. (since obviously f is continuous).

Then $ f'(x) \equal{} e^x \minus{} (x \plus{} 1)(3x \plus{} 1)$

again, we can use crude approximations to show that

$ f'(x) < 0$ for $ x\in[0.5,1]$ [that is a very weak inequality...]

In particular, $ e^x\le e^{1} < 3$

and $ \minus{} (x \plus{} 1)(3x \plus{} 1) \le \minus{} 3.75 < \minus{} 3$

Summing those gives us the result...

Hence $ f(x)$ is strictly decreasing in that interval and it has only one root.

This is not a formal proof but,
Consider the graphs of $ y \equal{} e^x$ and $ y \equal{} f(x) \equal{} x(x \plus{} 1)^2$in the interval $ (\frac {1}{2},1)$ (at this point it is almost evident)
Note that $ f$ and $ e^x$ both are increasing continuous functions, for $ x > 0$
Since $ f(\frac {1}{2}) < \sqrt {e}$ and $ f(1) > e^1$, So(Here it comes the key of this problem):
"The line joining $ f(\frac {1}{2})$ and $ f(1)$ must intercept the graph $ y \equal{} e^x$ somewhere, when $ x\in(\frac {1}{2},1)$, it cannot intercept it twice since (reread the above conditions) for $ x \equal{} 1$ $ f$ is still inside the region bounded by $ x \equal{} 0$ and $ y \equal{} e^x.$"

 :)


[b]1.2[/b] [i] A coin is flipped 1001 times.  What is the probablity that at least 3 heads will appear consecutively in the set of heads and tials?[/i]

[i]Masoud Zargar[/i]

[quote="draperp"]There are $1001-2=999$ ways that the three flips could be consecutive.  Therefore, the probability is $\frac{999}{2^{3}}=\frac{999}{8}$.[/quote]

Umm, probability is never greater than 1, and what if I flipped an alternating pattern of tails and heads? That means that the probability of this is lower than 1.

Your avatar is a game of Go.  It looks like the players are not very aggressive.  :)

Edit:  A recursion is possible, but solving it would not be very nice as it would include the number of ways that at least two consecutive heads can be flipped.  Two recursions, really.   :|

If you find the recursion, you can relate it to a simpler sequence.

[hide="this is what I got"]
let $p_{n}$ be the total number of $n$ term sequences that satisfy the given condition.  then we wish to calculate $p_{n+1}$.
If we stick a head or a tail to the end of one of the $p_{n}$ sequence, it'll still work.  so $2p_{n}$ sequences work.
Otherwise, if we stick a head to the end of a sequence, and it contains 2 consecutive heads at the end, and no consecutive 3 heads block before, we got a new satisfying sequence.
In the latter case, the last two must be heads, and the third one from the last must be tails.  All the terms before can be anything as long as it does not have a consecutive 3 block.
So we get:

$p_{n+1}= 2p_{n}+(2^{n-3}-p_{n-3})$.

With $p_{0}= p_{1}= p_{2}= 0$, and $p_{3}= 1$.  And to get the probability, divide $p_{n}$ by $2^{n}$.
[/hide]

Someone can figure out the close form from now on...


[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=188613[/url]

I guess this is a reminder....


function
superior algebra
superior algebra unsolved

I just want to make sure I did this right.

Suppose the group G acts on the set X.  Let x be in X.  Find a bijection from the set of right cosets of the stabilizer to the G-orbit of x.  Prove that the function is indeed a bijection.

We need to prove a bijection between the right cosets of the stabilizer and the orbit.

Defining the function as f: H--> Orb(x)
H=gG_x

One-to-one
f(g_1H)=f(g_2H)
g_1x=g_2x
x=(g_1)(g_2)^{-1}x  This stabilizes x, through moving stuff around we know that g_1 and g_2 are in H.  This satisfies 1-1.

Onto

Given some element b in Orb(x)
b=g*x=f(gh)

A cleaner proof I suppose:

f(gGx)=Orb(x)

1-1

f(g1Gx)=f(g2Gx)

I'm going to use ' to denote inverse

g2'g1x=x, this shows that g2'g1 is in the left coset.  g2'g1x=x shows that g1x=g2z.  So we are done

Onto

Given any gx=y with y  in Orb(x) we know that f(gGx)=y,this shows that it is onto.


calculus
LaTeX
symmetry
calculus computations

1) Post this in the right section (Computations and tutorials).

2) Learn how to use LaTeX.

3) We're not going to do your homework for you.

[hide="#1"]$f(x) = g(x)\sqrt{x} \Rightarrow f'(x) = g'(x)\sqrt{x}+\frac{g(x)}{2\sqrt{x}} \Rightarrow f'(4) = 16$[/hide]

[hide="#2"]For the first three, use $[f(x)+g(x)]' = f'(x) + g'(x)$, the product rule, and the quotient rule, and for the forth:

$\left[\frac{f(x)}{f(x)-g(x)}\right]' = \frac{f'(x)\left[f(x)-g(x)\right]-f(x)\left[f'(x)-g'(x)\right]}{\left[f(x)-g(x)\right]^2}$[/hide]

Jmerry has a good point, if you don't understand a concept we'd be glad to help, but don't post a list of homework problems.

For the millionth time, I am NOT, NOT, NOT, posting homework.
I am preparing for a state test that I need to pass.  I am NOT, NOT, NOT in high school.

To Jimmy:

Thank you for your help.  I can now go back into the test prep book and solve similar questions.

I know, that's why I didn't post that, but at least try to learn LaTeX and post in the right section! You just keep ignoring that!

[quote="symmetry"]I need help with the following two questions.

1)  If f(x) = sqrt{x} times g(x), where g(4) = 8 and g ' (4) = 7, find f '(4).

2)  If f(3) = 4, g(3) = 2, f '(3) = -6 and g '(3) = 5, find the following numbers a through d:

a) (f + g) ' (3)

b) (fg) ' (3)

c) (f/g) ' (3)

d) (f/(f - g)) ' (3)[/quote]symmetry,

Key to both problems: Differentiate the general expression first, then plug in specific values of x.

For example, 1) $ f(x)g(x) \Rightarrow \frac{\mathrm{d}fg}{\mathrm{d}x} = \frac{\mathrm{d}f}{\mathrm{d}x}g(x) + f(x)\frac{\mathrm{d}g}{\mathrm{d}x} = \frac{1}{2\sqrt{x}}g(x) + \sqrt{x}\cdot g'(x) $


Reading a proof of "R\Q is not a countable union of closed subsets of R, (Q is such a union as [b]Q= union {q}[/b])

Suppose by contradiction, that [b]R\Q [/b]=[b] countable union {Fn}[/b], Fn is equals to it's closure. Then for every n, Fn is nowhere dense,
since
[b] interior(Fn closure)[/b] =[b] int(Fn)[/b] is contained in [b]int(R\Q) [/b]= [b]empty set[/b]. (Since Q is dense in R). Hence,
[b]R[/b]=[b](R/Q) union Q[/b] =[b] [countable union Fn] union [union{q}] [/b]
is a countable union of nowhere dense sets.  :?
But by Baire's Category Theorem, [b]int {[countable union Fn] union [union{q}] } [/b]= empty set does not equals to R = int R...so a contradiction...

on the sentence marked with  :? , why it can be said that =[b] [countable union Fn] union [union{q}] [/b] is a countable union ofnowhere dense sets??  [b]Q= union {q}[/b] is nowhere dense?? why?

Use the single element sets as your other nowhere dense sets, not their union.

so does it means every singleton {q} is nowhere dense.....so is it's union?


I did it with long way.Is there any short solutions? :blush:
[hide][img]http://img180.imageshack.us/img180/9038/17688677se5.gif[/img][/hide]

What's the long way?

[hide="maybe"]
Find the second symmetry $ab+ac+bc$. Then use $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-ac-bc)$ to find $abc$. Then use vieta's, create the polynomial with roots a, b, c, then find $s_{4}$
[/hide]

That's monstrous.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131617

i don't understand vieta's theory completely. :blush:


function
limit
integration
calculus
complex analysis
complex analysis unsolved

Suppose $ \sum_{0}^{\infty}|a_n|^2$ converges.  Show that $ f(z)\equal{}\sum_{0}^{\infty}a_nz^n$ is anlytic for $ |z|<1$.

More over, prove $ \lim_{r\rightarrow 1}\int_{0}^{2\pi}|f(re^{it})|^2dt$ exists and find its value.

First question is definition of radius of convergence. Second is Parseval's formula.

i understand the first question, but i stilldont know how to get the second one!
any help :maybe: ?

Do you know Parseval's formula? Here is a derivation

$ |f(re^{it})|^2 \equal{} \left(\sum_{n \equal{} 0}^\infty a_n r^n e^{int} \right)\left(\sum_{n \equal{} 0}^\infty \overline{a_n} r^n e^{ \minus{} int} \right) \equal{} \sum_{m,n \equal{} 0}^\infty a_n\overline{a_m}r^{m \plus{} n}e^{i(n \minus{} m)t}$
It's trivial to justify interchanging the sum and integral so we get
$ \int_0^{2\pi}|f(re^{it})|^2 \frac {dt}{2\pi} \equal{} \sum_{m,n \equal{} 0}^\infty a_n\overline{a_m}r^{m \plus{} n}\int_0^{2\pi}e^{i(n \minus{} m)t} \frac {dt}{2\pi} \equal{} \sum_{n \equal{} 0}^\infty |a_n|^2r^{2n}$


For a world record, George Adrian picked 15,832 pounds of
apples in 8 hours. Assuming he maintained a constant rate of
picking, how many pounds of apples did he pick in 3 hours?

To figure this out, let's find out how many apples he picked per hour, and then multiply by 3.

$ \frac{15832}{8} \equal{} 1979$

$ 1979 \cdot 3 \equal{} 5937$

So our answer is $ \boxed{5937}$.

Hmmmmmmmm. I have 8 hours of free time today! Why not go pick over 7 tons of apples!!! :rotfl:


A bookstore sells used books at a 40$ \%$ discount off the regular price of the book when it was new. The price of a used book is $ \$18$. What was the regular price of that book when it was new?

since there is $ 40\%$, that means the book costs $ 60\%$ of it's original price, so we just divide
$ 60\%$ is equal to $ 0.6$, so we get...

$ \frac {18} {0.6} \equal{} 30$


Let $ a$ and $ b$ be natural numbers such that
\[ \frac{a}{b}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{1318}+\frac{1}{1319}. \]
Prove that $ a$ is divisible by $ 1979$.

I will assume paragraphs 1. and 2. of http://www.mathlinks.ro/Forum/viewtopic.php?t=150391 (or http://www.problem-solving.be/pen/viewtopic.php?t=25 ) post #4 as known.

Note that $ 1979$ is a prime (I am wondering whether the IMO contestants had to check that by hand or they were supposed to know that). We are going to prove $ v_{1979}\left(\frac {a}{b}\right)\geq 1$, because this will yield $ v_{1979}\left(a\right) \equal{} v_{1979}\left(\frac {a}{b}\cdot b\right) \equal{} \underbrace{v_{1979}\left(\frac {a}{b}\right)}_{\geq 1} \plus{} \underbrace{v_{1979}\left(b\right)}_{\geq 0\text{, since }b\text{ is an integer}}\geq 1$, so that $ 1979\mid a$, and thus the problem will be solved.

In order to prove that $ v_{1979}\left(\frac {a}{b}\right)\geq 1$, we note that $ \frac {a}{b} \equal{} 1 \minus{} \frac {1}{2} \plus{} \frac {1}{3} \minus{} \frac {1}{4} \plus{} \cdots \minus{} \frac {1}{1318} \plus{} \frac {1}{1319} \equal{} \sum_{i \equal{} 1}^{\left\lfloor 2\cdot 1979/3\right\rfloor}\frac {\left( \minus{} 1\right)^{i \minus{} 1}}{i}$, so we have to show that $ v_{1979}\left(\sum_{i \equal{} 1}^{\left\lfloor 2\cdot 1979/3\right\rfloor}\frac {\left( \minus{} 1\right)^{i \minus{} 1}}{i}\right)\geq 1$. This is a particular case ($ p \equal{} 1979$) of the following theorem:

[color=blue][b]Theorem 1.[/b] Let $ p$ be a prime such that $ p > 3$. Then, $ v_{p}\left(\sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {\left( \minus{} 1\right)^{i \minus{} 1}}{i}\right)\geq 1$.[/color]

[i]Proof of Theorem 1.[/i] First we prove that

[color=green][b](1)[/b][/color] $ p \minus{} \left\lfloor \left\lfloor 2p/3\right\rfloor /2\right\rfloor \equal{} \left\lfloor 2p/3\right\rfloor \plus{} 1$.

In fact, since $ p$ is a prime and $ p > 3$, we have $ 3\nmid p$. Thus, either $ p\equiv 1\mod 3$ or $ p\equiv 2\mod 3$.

If $ p\equiv 1\mod 3$, then $ \frac {p \minus{} 1}{3}\in\mathbb{Z}$, so that $ \left\lfloor\frac {2p}{3}\right\rfloor \equal{} 2\cdot\frac {p \minus{} 1}{3}$ (since $ 2\cdot\frac {p \minus{} 1}{3}\in\mathbb{Z}$ (because $ \frac {p \minus{} 1}{3}\in\mathbb{Z}$) and $ 2\cdot\frac {p \minus{} 1}{3}\leq\frac {2p}{3} < 2\cdot\frac {p \minus{} 1}{3} \plus{} 1$) and therefore $ \left\lfloor\left\lfloor\frac {2p}{3}\right\rfloor /2\right\rfloor \equal{} \frac {p \minus{} 1}{3}$ (since $ \left\lfloor\frac {2p}{3}\right\rfloor /2 \equal{} \left(2\cdot\frac {p \minus{} 1}{3}\right)/2 \equal{} \frac {p \minus{} 1}{3}$ and $ \frac {p \minus{} 1}{3}\in\mathbb{Z}$), so that

$ p \minus{} \left\lfloor\left\lfloor\frac {2p}{3}\right\rfloor /2\right\rfloor \equal{} p \minus{} \frac {p \minus{} 1}{3} \equal{} 2\cdot\frac {p \minus{} 1}{3} \plus{} 1 \equal{} \left\lfloor\frac {2p}{3}\right\rfloor \plus{} 1$,

and therefore [color=green][b](1)[/b][/color] is proven for the case $ p\equiv 1\mod 3$.

If $ p\equiv 2\mod 3$, then $ \frac {p \minus{} 2}{3}\in\mathbb{Z}$, so that $ \left\lfloor\frac {2p}{3}\right\rfloor \equal{} 2\cdot\frac {p \minus{} 2}{3} \plus{} 1$ (since $ 2\cdot\frac {p \minus{} 2}{3} \plus{} 1\in\mathbb{Z}$ (because $ \frac {p \minus{} 2}{3}\in\mathbb{Z}$) and $ 2\cdot\frac {p \minus{} 2}{3} \plus{} 1\leq\frac {2p}{3} < \left(2\cdot\frac {p \minus{} 2}{3} \plus{} 1\right) \plus{} 1$) and therefore $ \left\lfloor\left\lfloor\frac {2p}{3}\right\rfloor /2\right\rfloor \equal{} \frac {p \minus{} 2}{3}$ (since $ \left\lfloor\frac {2p}{3}\right\rfloor /2 \equal{} \left(2\cdot\frac {p \minus{} 2}{3} \plus{} 1\right)/2 \equal{} \frac {p \minus{} 2}{3} \plus{} \frac {1}{2}$ and $ \frac {p \minus{} 2}{3}\in\mathbb{Z}$), so that

$ p \minus{} \left\lfloor\left\lfloor\frac {2p}{3}\right\rfloor /2\right\rfloor \equal{} p \minus{} \frac {p \minus{} 2}{3} \equal{} \left(2\cdot\frac {p \minus{} 2}{3} \plus{} 1\right) \plus{} 1 \equal{} \left\lfloor\frac {2p}{3}\right\rfloor \plus{} 1$,

and therefore [color=green][b](1)[/b][/color] is proven for the case $ p\equiv 2\mod 3$.

Hence, [color=green][b](1)[/b][/color] is proven in both possible cases, and we can step to the actual proof of Theorem 1.

We work with fractions modulo $ p$, noting that we can divide by every $ i\in\left\{1;\;2;\;...;\;p \minus{} 1\right\}$ modulo $ p$. Obviously,

$ \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {\left( \minus{} 1\right)^{i \minus{} 1}}{i} \equal{} \sum_{\substack{1\leq i\leq \left\lfloor 2p/3\right\rfloor ; \\
i\text{ is odd}}}\frac {\left( \minus{} 1\right)^{i \minus{} 1}}{i} \plus{} \sum_{\substack{1\leq i\leq \left\lfloor 2p/3\right\rfloor ; \\
i\text{ is even}}}\frac {\left( \minus{} 1\right)^{i \minus{} 1}}{i}$
$ \equal{} \sum_{\substack{1\leq i\leq \left\lfloor 2p/3\right\rfloor ; \\
i\text{ is odd}}}\frac {1}{i} \plus{} \sum_{\substack{1\leq i\leq \left\lfloor 2p/3\right\rfloor ; \\
i\text{ is even}}}\frac { \minus{} 1}{i} \equal{} \sum_{\substack{1\leq i\leq \left\lfloor 2p/3\right\rfloor ; \\
i\text{ is odd}}}\frac {1}{i} \minus{} \sum_{\substack{1\leq i\leq \left\lfloor 2p/3\right\rfloor ; \\
i\text{ is even}}}\frac {1}{i}$
$ \equal{} \left(\sum_{\substack{1\leq i\leq \left\lfloor 2p/3\right\rfloor ; \\
i\text{ is odd}}}\frac {1}{i} \plus{} \sum_{\substack{1\leq i\leq \left\lfloor 2p/3\right\rfloor ; \\
i\text{ is even}}}\frac {1}{i}\right) \minus{} 2\sum_{\substack{1\leq i\leq \left\lfloor 2p/3\right\rfloor ; \\
i\text{ is even}}}\frac {1}{i} \equal{} \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i} \minus{} 2\sum_{\substack{1\leq i\leq \left\lfloor 2p/3\right\rfloor ; \\
i\text{ is even}}}\frac {1}{i}$
$ \equal{} \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i} \minus{} 2\sum_{j \equal{} 1}^{\left\lfloor \left\lfloor 2p/3\right\rfloor /2\right\rfloor}\frac {1}{2j} \equal{} \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i} \plus{} \sum_{j \equal{} 1}^{\left\lfloor \left\lfloor 2p/3\right\rfloor /2\right\rfloor}\frac {1}{ \minus{} j}$.
$ \equiv\sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i} \plus{} \sum_{j \equal{} 1}^{\left\lfloor \left\lfloor 2p/3\right\rfloor /2\right\rfloor}\frac {1}{p \minus{} j}$
$ \equal{} \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i} \plus{} \sum_{i \equal{} p \minus{} \left\lfloor \left\lfloor 2p/3\right\rfloor /2\right\rfloor}^{p \minus{} 1}\frac {1}{i}$      (here we substituted $ j$ by $ p \minus{} i$ in the second sum)
$ \equal{} \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {1}{i} \plus{} \sum_{i \equal{} \left\lfloor 2p/3\right\rfloor \plus{} 1}^{p \minus{} 1}\frac {1}{i}$      (by [color=green][b](1)[/b][/color])
$ \equal{} \sum_{i \equal{} 1}^{p \minus{} 1}\frac {1}{i}\mod p$.

Now, $ \sum_{i \equal{} 1}^{p \minus{} 1}\frac {1}{i}\equiv 0\mod p$, as can be shown in different ways - e. g. as follows:

$ \sum_{i \equal{} 1}^{p \minus{} 1}\frac {1}{i} \equal{} \frac12\cdot\left(\sum_{i \equal{} 1}^{p \minus{} 1}\frac {1}{i} \plus{} \sum_{i \equal{} 1}^{p \minus{} 1}\frac {1}{i}\right)$
$ \equal{} \frac12\cdot\left(\sum_{i \equal{} 1}^{p \minus{} 1}\frac {1}{i} \plus{} \sum_{i \equal{} 1}^{p \minus{} 1}\frac {1}{p \minus{} i}\right)$             (here we substituted $ i$ by $ p \minus{} i$ in the second sum)
$ \equal{} \frac12\cdot\sum_{i \equal{} 1}^{p \minus{} 1}\left(\frac {1}{i} \plus{} \frac {1}{p \minus{} i}\right)\equiv\frac12\cdot\sum_{i \equal{} 1}^{p \minus{} 1}\left(\frac {1}{i} \plus{} \frac {1}{ \minus{} i}\right) \equal{} \frac12\cdot\sum_{i \equal{} 1}^{p \minus{} 1}0 \equal{} 0\mod p$;

hereby, we were allowed to divide by $ 2$ because $ 2\not\equiv 0\mod p$ (since $ p > 3$). Thus,

$ \sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {\left( \minus{} 1\right)^{i \minus{} 1}}{i}\equiv \sum_{i \equal{} 1}^{p \minus{} 1}\frac {1}{i}\equiv 0\mod p$,

so that $ v_{p}\left(\sum_{i \equal{} 1}^{\left\lfloor 2p/3\right\rfloor}\frac {\left( \minus{} 1\right)^{i \minus{} 1}}{i}\right)\geq 1$, and Theorem 1 is proven.

  Darij

The generalisation of the above is : Let $ p$ be a prime greater than $ 3$, and $ q\equal{}[\frac{2p}{3}]$ where $ [x]$ is the greatest integer function.Let $ m$ and $ n$ be positive integers such that$ \frac{m}{n}\equal{}1\minus{}\frac{1}{2}\plus{}\frac{1}{3}\minus{}\frac{1}{4}.....\plus{}(\minus{}1)^{q\minus{}1}\frac{1}{q}$. then $ m$ is divisible by $ p$.

Just an observation: if I'm not mistaken, this problem (and manjil's generalization) is a direct consequence [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150392]A 24[/url]. Why?

First, I claim that when $ 0 \leq k < p$, $ {p \minus{} 1 \choose k} \equiv (\minus{}1)^k \mbox{ mod p}$. If $ k \equal{} 0$, this is obvious. Otherwise, this is a consequence of the fact that $ \binom{p \minus{} 1}{k} \equal{} \frac{(p\minus{}1) \cdot (p\minus{}2) \cdot \ldots \cdot (p\minus{}(k\plus{}1)}{k \cdot (k\minus{}1) \cdot \ldots \cdot 1}$; taking the numerator mod $ p$ and cancelling wtih the denominator, our result follows easily.

Next, observe that $ p \binom{p \minus{} 1}{k\minus{}1} \equal{} k\binom{p}{k}$; this is a trivial consequence of the definition of $ \binom{n}{k}$. Rearranging, we see that $ \frac{\binom{p\minus{}1}{k\minus{}1}}{k} \equal{} \frac{\binom{p}{k}}{p}$.

Now, let $ p$ be a prime and $ q \equal{} \lfloor \frac{2p}{3} \rfloor$. We wish to show that $ 1 \minus{} \frac{1}{2} \plus{} \frac{1}{3} \ldots \plus{} (\minus{}1)^{q\minus{}1} \frac{1}{q}$ is 0 mod $ p$, where $ \frac{1}{n}$ is taken to be the inverse of $ n$ modulo $ p$. Our sum is

$ \sum_{k \equal{} 0}^q (\minus{}1)^k \frac{1}{k} \equiv \sum_{k \equal{} 0}^q \frac{\binom{p\minus{}1}{k\minus{}1}}{k} \equiv \sum_{k\equal{}0}^q \frac{{n \choose k}}{p} \mbox{ mod } p$. However, by problem A 24, $ \sum_{k\equal{}0}^q {n \choose k} \equiv 0 \mbox{ mod } p^2$, so $ \sum_{k\equal{}0}^q \frac{{n \choose k}}{p} \equiv 0 \mbox{ mod } p$, as desired.

[Clearly, we may also go in reverse and use this generalization to solve problem A 24.]

We don't need generalizations :D

[url]https://artofproblemsolving.com/community/c6h61006p367332[/url]


If would appreciate it if you could tell me how to solve these 2 problems

1st Problem

If Bob can decorate a room for a party in 8 hours and Jim can do it in 12 hours. If Jen helps, all three can do the job in 3 hours working together. How long would it take Jen if she works alone?

2nd Problem

2/3 of men are married to 3/4 of the women (assume each woman is married to exactly one man and conversely). What is the ratio of number of men to the number of women

Here's the first one:

[hide]
If Bob can decorate the room in 8 hours, he can do 1/8 of a room per hour. So Jim can do 1/12 per hour. If Jen helps, they can do 1/3 per hour. 1/8 + 1/12 = 5/24, and 1/3 = 8/24. So Jen can do 8/24 - 5/24 = 3/24 = 1/8 per hour, or one room in eight hours.[/hide]

[hide="hint for 2"]

Let x be the total number of couples.

Then there are $\frac{3}{2}x$ men and $\frac{4}{3}x$ women.

[/hide]

THANKS FOR 1ST one ... now so im guessing for the 2nd one i would do 2/3(x) / 3/4(x) and get 8:9 ?

[quote="xxxxx"][quote="fero4u123"]If would appreciate it if you could tell me how to solve these 2 problems

1st Problem

If Bob can decorate a room for a party in 8 hours and Jim can do it in 12 hours. If Jen helps, all three can do the job in 3 hours working together. How long would it take lois if she works alone?

2nd Problem

2/3 of men are married to 3/4 of the women (assume each woman is married to exactly one man and conversely). What is the ratio of number of men to the number of women[/quote]

[hide]
$\frac{1}{8}+\frac{1}{12}+\frac{1}{x}=\frac{1}{3}$

$3+2+\frac{24}{x}=8$

$5x+24=8x$

$x=\boxed{8}$
[/hide]

what's it mean by conversely?[/quote]
Basically meaning that also assume that one man is married to a woman as well.

conversly means "also reverse in order" so that would mean to say each woman is married to exactly one man / each man is married to exactly one woman

ok thx.

number 2: (is this right?)
[hide]

$\frac{2}{3}m=\frac{3}{4}w$

$m=\frac{9}{8}w$

$\boxed{8: 9}$[/hide]

yeah it makes no sense if lois just appears

[hide="xxxxxhowevermany xs"]
9:8
is what i get
so there were 9 men
6=3/4x
8=x
x is number of women
9:8[/hide]

Well, since 2/3 of the men is equal to 3/4 of the women, there must be more men than women.

[quote="fero4u123"]*im sorry that lois was suppose to say Jen * i will change and 9:8 does seem right because it makes the 1:1 ratio but so does (8/9):1 ? im not sure of exactly how u are suppose to work the problem[/quote]

well i mean it is sort of gross if you are going to start cutting women in ninths....go ahead if you want to


induction
number theory solved
number theory

Let p,q be natural numbers, and n=2^q(2^p+1).Prove that the binomial number Cn,k( combinations) is odd if and only if k is in the set {0,n,2^q,2^(p+q)}

The sufficience:

If k is 0 or n the corresponding binomial coefficient is 1, so it's done. For the other 2 we write the binomial coefficient:

[2^q+2^(p+q)]!/[2^q]!*[2^(p+q)]!. To prove that it's odd all we have to do is prove that the exponent of 2 in the numerator=the exponent of 2 in the denominator:

 \sum {k=1->00} [2^q(2^p+1)/2^k] =
 \sum {k=1->00} ([2^q/2^k]+[2^(p+q)/2^k]), but it's easy to prove that for each k>=1 we have [2^q(2^p+1)/2^k]=[2^q/2^k]+[2^(p+q)/2^k], where [x]=integer part of x.

I think the necessity has a similar idea, but I haven't given it any thought yet.

This a special case of a more general (and very nice :D)  result :

Write n = sum 2^ai (n in base 2)

C(n,k) is odd iff k = sum 2^bj where bj  {ai}, i.e. the bj are chosen among the binary digits of n

It's fun to show using modular polynomials.

. (1+a)^2 = 1+a (mod 2)
. by induction (1+a)^(2^n) = 1 + a^(2^n) (mod 2) : it means C(2^n,k) are even for k != 0,2^n
. (1+a)^(2^n + 2^p) = 1 + a^(2^n) + a^(2^p) + a^(2^(n+p)) (mod 2)
. .....
. by induction :
(1+a)^(sum 2^ai) = 1 + sum a^(something) (mod 2) where something is a sum of some of the ai.
Hence the result.

So :
. C(2^n, k) is odd for k != 0,2^n
. C(2^m+2^n, k) is odd for k != 0,2^m,2^n,2^(m+n)
. C(2^l+2^m2^n,k) is odd for k != 0,2^l,2^m,2^n,2^(l+n), .., 2^(l+m+n).
. .....

Nice isn't it  :)


Find the local mimimum of $f(x)=x^4+2x^3+3x^2+5x+4$.

Masoud Zargar

Do you mean local maximum?
Your function is unbounded above.
Besides, your function has no local maximums.
$f'(x) = 4x^3+6x^2+6x+5=0$ if and only if x = -1.143955(approx.)
$f''(x)=12x^2+12x+6=7.976135>0=$ when x =-1.143955.
Therefore, f(x) has only one local minimum but it has no maximums.

lol, what a foolish problem...very well then; find the local minimum.

So in other words, find the root of $4x^3+6x^2+6x+5=0$. (Just to get rid of the calculus in it.)
Can this root even be found exactly (without ugly cubic formula)?

absolutely...I made the problem, so yes. :D

are you saying that is supposed to factor nicely?

Yes, it is a foolish problem, for beginners ! [color=darkred]But we must make it.[/color]
$\left(\forall\right) x\in R\ ,\ f''(x)>0$ $\Longrightarrow$  $f'\nearrow$ $\Longrightarrow$ $\left(\dot {\exists}\right)\ \alpha\in (-2,-1)$ so that
$f'(x)\ .s.s.\ (x-\alpha )$ $\Longleftrightarrow$ $\min_{x\in R} f(x)=f(\alpha)=\frac 38(2\alpha^2+8\alpha +9)\ .$
[u]Notations.[/u] $x\ .s.s.\ y\Longleftrightarrow x=y=0\ \vee\ xy>0\ ;$
$\left(\dot {\exists }\right) \Longleftrightarrow$ exists and is uniquely .

[quote="Virgil Nicula"]
$\left(\dot {\exists }\right) \Longleftrightarrow$ exists and is uniquely .[/quote]
Isn't there an international symbol for that? Namely : $\exists !$?

Yes, but I use often this symbol which is more suggestive, in my opinion !

Well, my solution goes as follows; I am currently busy with other math.

1. Take the derivative.
2. Let $x=y-\frac{b}{3}$, where $b$ is in the second coifficient.
3. Simplify and set $y=z+\frac{p}{3z}$, where $p$ is the second coifficient of the new equation.
4. Solve the quadratic equation and solve  for $x$.

Masoud Zargar


Inegalitatea $\ (1)$
Folosind spargerile cu medii obtinem inegalitatea $\frac{\frac{4}{7}}{3a+b}+\frac{\frac{1}{7}}{3b+c}+\frac{\frac{2}{7}}{3c+a}\geq\frac{1}{2b+c+a}$.Adunand si analoagele, obtinem $\ (1)$.
Inegalitatea $\ (2)$
Tot cu spargerile cu medii avem inegalitatea $\frac{\frac{a}{3}}{3}+\frac{\frac{8b}{9}}{4}\geq\frac{9}{a+2b}$ si iese inegalitatea $\ (2)$.
Inegalitatea $\ (3)$
..nu mai scriu ca oricum e la fel, tot cu spargeri cu medii. :)

$(2)$ Cautam o spargere convenabila:
$\frac{x}{a}+\frac{y}{b}\geq \frac{9}{a+2b}$
$\frac{x^{2}}{ax}+\frac{y^{2}}{by}\geq \frac{(x+y)^{2}}{ax+by}\Rightarrow\ x=1,\ y=2$
Rezulta:
$\frac{1}{a}+\frac{2}{b}\geq \frac{9}{a+2b}$
si
$\frac{1}{c}+\frac{2}{a}\geq \frac{9}{a+2b}$
Ramane sa demonstram ca
$\frac{2}{b}+\frac{4}{c}\geq \frac{18}{b+2c}\Leftrightarrow \frac{1}{b}+\frac{2}{c}\geq \frac{9}{a+2b}$, care este evident.


Solve the equation:

\[ 2^{x\plus{}1}\equal{}(x\minus{}1)(x\minus{}2)\]

[quote="moldovan"]Solve the equation:
\[ 2^{x \plus{} 1} \equal{} (x \minus{} 1)(x \minus{} 2)\]
[/quote]

The function $ 2^{x \plus{} 1} \minus{} (x \minus{} 1)(x \minus{} 2)$ is strictly increasing. We have just one solution $ x\equal{}0$


ARML MATHCOUNTS geometry AMC USA(J)MO USAMO trigonometry

P.K,tell me the thoughts crossing your minds,even the randomest.

why do people seem to get confused with what they feel about themselves??!!!

gah you know what I really need to go to the bathroom right now and why did mandelbrot send me a shirt that was so small that's sort of weird heh when I was thinking of it I accidentally thought of mathcounts but we all know that's not true as THEY EXIST ONLY IN THE MINDS OF HORRIBLE PEOPLE THAT oh man I think I'll go to the bathroom now it'll be pretty awesome maybe there'll be like an orchestra y'know like in heaven with the angels and they're all like whoa if you mistype angels it looks like angles angles angles like geometry yea geometry is awesome except for solid geometry because that's like oh no I haven't learned that yet I should though that sounds sort of like fun but then again arrgh

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GET IT IN YOUR HEAD PEOPLE

does that mean i can submit it tomorrow and not be late or do i have to submit before tomorrow?

my thoughts...
[size=200]$7\frac{1}{2}$ more days of school!![/size]

why do we drive on a parkway...and park on a driveway??

Would you stone me if I dropped out of MIMC?  I totally forgot about it, and at this point, I have too much work...

Edit: 10 more posts...

well...schools almost over!!!!!
12 more days!

[quote="fireemblem13"][size=200]$7\frac{1}{2}$ more days of school!![/size][/quote]
[quote="math92"]12 more days![/quote]

I have like around a month left.[img]http://smilies.sofrayt.com/fsc/sobbing.gif[/img]

a bear came to me bearing a bear...i can bear a bear but not a bear's bear.........get it???!!!

I went shopping for me yesterday and the bill was like over 5000 rupees,so i thiought the bags would be goddarn heavy and when i carried them,they were so light,i'm so disappointed....

ou know some typos which can be really misunderstood,like when you write 'busy',you accidentally type the 't' next to y... :o

acidpop,did you mean you can't be the mother of the bear's child bear?That's totally possible ...

I have no chance of qualifying for IMO. Got 13 out of 40 in Practcie Test I and wrote a horrible selection test I some hours ago. Dash it!

i just saw my chem teacher from last year- and in his classroom, there's this clock the has element symbols instead of numbers- like where 1 should be, it says H, then He, according to atomic number.

[quote="MysticTerminator"]man MIMC that makes no sense at all oh man oh man should I do it no I think not HAHA LIKE DESCARTES EH GET IT GET IT this thread is such a good stress reliever wait no it isn't I wasn't even stressed in the first place YOU LIE[/quote]

YOU BETTER DO IT

IT'S DUE TODAY

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I'm so sad i didn't make usamo this year...
I can solve ~3 problems per practice, sometimes some partial stuff on other ones...
I might have done well...

[quote="Iversonfan2005"][quote="perfectnumber628"]wondering if that is all you think about lately

(by the way, i just sent in my mimc answers)[/quote]

this is the only way I can get people to remember...by spamming about this. We only have around 30 people that sent in their answers out of 120 that signed up! :o , so either people forgot or don't care. *hoping that it was the former*, I'm trying to remind people.[/quote]

maybe it's because most of them (like me?) didn't sign up  :o

whoops,didn't submit my answers,i shiuld sttill get a 50 though,according to the rules,right?

currently thinking about writing "the greatest english paper you've ever written" because that's what my teacher said to do... :(  :oops:

When was the last time i had a bath......

today i realized i don't have school this friday!  i'll probably forget, and then remember on thursday, and it'll make my whole day  :roll:

i wish aops didn't have censors

don't eat the fish

You should put that on the wish list...I'll back it

Oh mann..,..a huge chemistry test and REALLY HUGE physics test this week...arrrrrrrghhhhhh

[quote="fireemblem13"]these last 3 or 4 days of school are gonna be awesome. :lol:[/quote]

i need to edit that. yesterday(monday), we had our annual class trip to schlitterbahn and i lost my glasses. :blush:

[quote="perfectnumber628"]i should probably be working on my english homework right now.  odd how that works out.[/quote]
:huh:

it was 2:48 when you wrote that. are you already out of school?

EDIT: oh you're in NY. that means it's 4:48. it's still kind of early to start on homework. don't you think?

[quote="perfectnumber628"]actually, school ends at 2:13 ish, but if i stay after (like today) i get home at like, 3:15.

and it is ALWAYS too early to do homework.[/quote]

do u mean

it is NEVER too early to do homework.

cuz if it's ALWAYS too early then if i procrastinate, it's still early?

procrastination is an important skill to have.

even more important: when in doubt, use law of cosines!  :lol:

i'm out of school before all of you. i got out last thursday!
 :rotfl:      :rotfl:   :rotfl:
i'm already enjoying the benefits of sleeping in and being lazy and getting more stuff done than when i was in school and stuff!

[size=200][color=red]I'M FREE!!!!!!!!!!!!!!![/color][/size]

I may not even be free for a month ;) .

BIRTHDAY TOMORROW!

random comment: no school tomorrow!  actually, we have a 4-day weekend: tomorrow is off because we didn't have snow days, and monday is memorial day.

[size=200]SCHOOL IS OUT!!!!!!![/size] (well, yesterday, but i couldn't get on to the internet) [size=200]WOOOOOHOOOOOO!!!!!!!!!![/size]

I'm so exited! My birthday was yesterday!!!!! .......wait. D'oh!

arml in some smallish number of days!!!  :clap:  :trampoline:

All math competitions gone! NOOOOOOOOOOOOOOOOOoooooooooooooooooooooooooooo!!!!!!!!!!!!!!!!!!


[quote="Denbox"]$x,y,z \in R+$.Prove that:
$(x+y)\sqrt \frac{xy}{(x+z)(y+z)}+(y+z)\sqrt \frac{yz}{(x+y)(x+z)}+(z+x)\sqrt \frac{xz}{(y+z)(y+x)}$
$\geq x+y+z$ :?:[/quote]
Take a look at
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=462855#p462855]Moldova TST 2006 Test II problem 3[/url] :wink:

I have a nice solution.
Let $a= y+z, b= z+x, c= x+y$ then $a, b, c$ are side lenghts of a triangle $ABC$.
And now, we have to prove that
$\sum_{}asin\frac{A}{2}\geq  p$
First, note that $cos\frac{B-C}{2}= \frac{b+c}{a}sin\frac{A}{2}$ therefor $\sum_{}asin\frac{A}{2}\geq  \sum_{}(b+c)sin^{2}\frac{A}{2}= \frac{1}{2}\sum_{}(b+c)(1-cosA)= 2p-\frac{1}{2}\sum_{}(b+c)cosA$
On the other hand, $a= bcosC+ccosB$. Applying this equality, the inequality is clearly true. The equality holds iff $A= B= C$ or $x= y= z$.


I m given these rules for division:

a)$ t|t$
b)$ 1|t$
c)$ t|0$
d)$ t|s\wedge s|r \rightarrow t|r$
e)$ s \neq 0 \wedge t|s \rightarrow t \le s$
f)$ t|s \wedge s|t \rightarrow s\equal{}t$
g)$ t|s\rightarrow t|(r\cdot s)$
h)$ t|s \wedge t|r \rightarrow t|(s\plus{}r)$

note: $ \wedge$ = "and" , $ \rightarrow$ = implies

how would I show $ (t|s\wedge t|s')\rightarrow t\equal{}1$?

Basically $ s'$ is the successor of $ s$ like 4 is the successor of 3. Thus the statement above is saying if $ t$ divides $ s$and $ t$ divides the successor of $ s$, then $ t$ is $ 1$.

What would be a rigorous way to show this?

So are we assuming $ t$ is a positive integer then?  If these divisibility rules are allowed to extend to the integers, we use h.
You have $ t|s \implies t|\minus{}s$.  So because $ t|s'$, we have $ t|s'\minus{}s$, so $ t|1$.  If $ t$ is given to be a positive integer, then $ 1|t, t|1 \implies t\equal{}1$.

these rules are only for the natural numbers. Though, I wish we were in the integers.

There are many arithmetic progressions of the form ks-1 (k>1), ks-2 (k>2)... ks-n (k>n).
try to use that, even though I ain't exactly getting anywhere

I don't think it's possible just from those rules.  If you had the rule $ \left(t | s \right)\wedge\left( t | (r \plus{} s)\right) \to t | r$ then you would be all set.

The reason it's not possible:
Rule (e) is clearly irrelevant.  For rules (d), (g) and (h), we always end up with $ t$ dividing something larger than what we started with.  However, our conclusion can only come from rule (f), which requires that we prove $ t | 1$.  Thus, it's only possible if one of $ s, s'$ is smaller than or equal to 1, which is not true in general.

what about, call s'=s+1 and s=s then for the statement to be correct we must have t|(1+s), but t|s, hence by h) t|1, or t=1

It is not possible to conclude from the facts $ t | s$, $ t | s \plus{} 1$ and (h) that $ t | 1$.  This was exactly the point of my comment.

Another way to see that this cannot work:
Instead of using the normal definition of $ a|b$, we write $ a|b$ iff $ b\equal{}0$ or $ 0<a \leq b$ (thus we see $ 0$ as some infinite number and $ |$ as standard order relation)
It's easy to check that all the mentioned statements are fulfilled, but the conclusion is not.

What if we are also given that $ a|b \Leftrightarrow \exists c, a\cdot c\equal{}b$? What conditions do we need to show the requested result?

Then it works as this is the standard definition of $ |$, thus all the "known" properties hold.

Hmm..I was asking which of the properties listed above are necessary to prove the result? Some of them are given by the definition, but not all of them. For example, what if these weren't integers, and therefore not given e? Is e necessary?

yes Hamster1800. It's stupid of me not to include that we're given the definition $ a|b \Leftrightarrow \exists c, a\cdot c\equal{}b$. I just thought it was assumed. my fault.

[quote="mstrip"]yes Hamster1800. It's stupid of me not to include that we're given the definition $ a|b \Leftrightarrow \exists c, a\cdot c \equal{} b$. I just thought it was assumed. my fault.[/quote]

If we have this, we can reason by contradiction.

Prove$ (t|s\wedge t|s')\rightarrow t\equal{}$

Need to show for two cases $ s\equal{}0$ and $ s\neq0$

$ s\equal{}0$

we simply have the case $ t|1\rightarrow t\equal{}1$

by hypothesis, we have $ t|1$
by proposition b) we have $ 1|t$
then by prop. f) we have $ t|1 \wedge 1|t \rightarrow t\equal{}1$

$ s\neq0$

let $ s'\equal{}s\plus{}1$
we'll try to prove a contradiction
by hypothesis we have$ t|s$ and $ t|s'$ and assume $ t\neq1$
then by prop e)$ s \neq 0 \wedge t|s \rightarrow t \le s$
thus we have two cases: i) $ t\equal{}s$ and ii)$ t<s$
case i)
if $ t\equal{}s$ then we know that $ t|s\equal{}1$ but $ t\nmid s'\equal{}s\plus{}1$.
how do we know this? proceed by induction.
base case:( since $ t\neq 1$) we start with t=2
clearly, $ 2|2$ and for $ n \in\mathbb{N} , 2|2n$ but it's clear that $ 2\nmid 2n\plus{}1$
step:Assume true for $ k$ where $ k\equal{}mn$ for $ m,n\in\mathbb{N}$

this is where I'm stuck. I just dont see how i can show definitively for $ k\plus{}1$

the idea of this proof is to show that for $ t\equal{}s$, we can't have $ t\equal{}s$ since $ t\nmid s'$ and thus doesn't satisfy our hypothesis. and for $ t<s$, show that becuase $ t<s$ and $ t<s'$ for some arbitrary $ s$, then work backwards inductively ( method of infinte descent?) such that we get $ t<1$ which contradicts the fact that $ t\in\mathbb{N}$. So I sorry that I m so bad at both cases.

any help is very much appreciated.

Thanks

I'd like to point out that in no way is property (e) given by the standard definition of divisibility. It happens to be true in the naturals, but it is not true in general. My question is: Is property (e) necessary?

well this is a proof in the naturals ( at least attempting a proof). I don't know if it's "necessary" but if you have a different idea of how to start,  it's probably better than my approach.

Let $ (a,b)$ denote the greatest common divisor of $ a$ and $ b$. This is defined as being the number $ g$ such that $ g|a$, $ g|b$, and $ d|a, d|b \implies d|g$. Now consider $ g' \equal{} (b\minus{}a,b)$. Suppose $ d|a$ and $ d|b$. Then $ d|b\minus{}a$ and $ d|b$. Thus, $ g' | g$. Then suppose $ d|b\minus{}a$, $ d|b$. Then $ d|b\minus{}(b\minus{}a) \equal{} a$, so $ g | g'$. Therefore, $ (a,b) \equal{} (b\minus{}a,b)$, and the result follows.

Notice that there was no way in which the ordering was required.

I hate to be a jerk, but the reason I didn't use subtraction was because I m not given subtraction. If i was, as you showed, this problem would be a little bit better.

am I at least on the right track( in terms of my proof), or can you suggest improvements?

thanks

You are actually given subtraction, and here's why:

$ t|a, t|b \implies t|a, t|\minus{}1\cdot b \implies t|a\plus{}\minus{}1\cdot b \implies t|a\minus{}b$

As far as I'm aware, Peano arithmetic deals with $ \mathbb{N}_0$, so the use of $ \minus{}1$ is not valid.

That is correct.  Hamster's proof is bogus for Peano arithmetic.

Unrelatedly, a notational point: the problem with the notation $ \bf N$ is that it is unenlightening -- there's no way to know whether that's meant to include 0 or not, unless you've already established context.  The problem with the notation $ \mathbb{N}_0$ is that it's equally unenlightening.  I strongly advocate the use of notations that no one can possibly be confused by, especially things like $ {\bf Z}_{\geq 0}$.  (No one uses $ \bf Z$ for anything other than the integers, so not great likelyhood of confusion there.)

I haven't worked with peano arithmetic directly before, so I'll trust you that my proof doesn't work. How would you go about the proof then?

don't be shy, someone please help.

Also, I was wondering if my attempt to prove the question in my previous post was a good start, or should I have attempted another direction?

thanks

Let's say we're given induction for this problem $ (t|s\wedge t|s')\rightarrow t\equal{}1$

if we applied induction:

base: $ s\equal{}1$

$ t|1$ and $ t|2$. Since we're working in $ \mathbb{N}$ with peano arithmetic, then $ t\equal{}1$

step: assume true for $ s \equal{}k$, then $ (t|k\wedge t|k')\rightarrow t\equal{}1$
we will note that proposition (h) $ \rightarrow t|2k\plus{}1$, in other words t divides odd numbers by our induction step
assume $ k' \equal{} k\plus{}1$

attempt to show for $ s\equal{}k\plus{}1$ in which we have $ t|k\plus{}1 \wedge t|k\plus{}2 \rightarrow t|2k \plus{}3$ , in other words t divides an odd number.

Does this prove our assumption then that $ t|k\plus{}1 \wedge t|k\plus{}2 \rightarrow t\equal{}1$?

Also, what would be wrong if I took $ k\plus{}1 \equal{} m$ and then showed that $ t|m \wedge t|m\plus{}1$ is in the form of our induction step, thus proving $ t\equal{}1$ for $ s\equal{}k\plus{}1$?

thanks for the help

[quote="mstrip"]$ s \equal{} k \plus{} 1$ in which we have $ t|k \plus{} 1 \wedge t|k \plus{} 2 \rightarrow t|2k \plus{} 3$ , in other words t divides an odd number.

Does this prove our assumption then that $ t|k \plus{} 1 \wedge t|k \plus{} 2 \rightarrow t \equal{} 1$? [/quote]

Er, no.  This doesn't do anything to show that $ t \equal{} 1$ in this case.

[quote="mstrip"]Also, what would be wrong if I took $ k \plus{} 1 \equal{} m$ and then showed that $ t|m \wedge t|m \plus{} 1$ is in the form of our induction step[/quote]

You seem to be misunderstanding induction.  What we are trying to prove is the statement

$ P(n) \equal{} \text{If t divides n and t divides n \plus{} 1 then t \equal{} 1}$

for all positive integers $ n$.  The inductive hypothesis is that we assume $ P(k)$ for some [b]fixed[/b] $ k$ and look at

$ P(k \plus{} 1) \equal{} \text{If t divides k \plus{} 1 and t divides k \plus{} 2 then t \equal{} 1}$

Unfortunately, the set of $ t$ such that $ t | k, t | k \plus{} 1$ is by no means necessarily related to the set of $ t$ such that $ t | k \plus{} 1, t | k \plus{} 2$, so in fact induction is a very poor choice of strategy here.

$ (s|s\wedge s|s')\rightarrow s \equal{} 1$
You can prove that by induction.

Also $ t \leq s$ but $ s\equal{}1$ so $ t\equal{}1$.

[quote="Fraenkel"]$ (s|s\wedge s|s')\rightarrow s \equal{} 1$
You can prove that by induction.[/quote]

$ s | s$ is given.  Moreover, [b]there is nothing on which to induct.[/b]  This is not a statement about all natural numbers $ n$.

fraenkel, so then how do you do it by induction? it seems doable, but then t0rajir0u says induction wouldn't be a good idea. So who's right? and how would you prove it( as in I need someone to walk me through it) please?

thanks!

[b]Given subtraction:[/b]

$ t | k, t | k\plus{}1 \implies t | (k\plus{}1) \minus{} k \implies t | 1 \implies t \equal{} 1$

[b]Given the standard definition of divisibility:[/b]

$ t | k \Leftrightarrow \exists a : k \equal{} at$
$ t | k\plus{}1 \Leftrightarrow \exists b : k\plus{}1 \equal{} bt$
$ (k\plus{}1) \minus{} k \equal{} 1 \equal{} bt \minus{} at \equal{} (b \minus{} a)t \implies$
$ t | 1 \implies$
$ t \equal{} 1$

if we did not have subtraction then we can't do it?

If we had neither subtraction nor the standard definition of divisibility, then [b]no.[/b]

in my previous post i said that in order for the statement to be true t|1, now by b) and f) t|1 "and" 1|t implies that 1=t. If there's anything wrong with that please pint it out coz i don't get it :!:

[quote="ggbg90"]in my previous post i said that in order for the statement to be true t|1[/quote]

You can't conclude this without either subtraction or the standard definition of divisibility.

... or the variation on rule (h) that I suggested back in my first post on the thread.

[quote="Hamster1800"]Your proof with the standard definition of divisibility relies on the fact that multiplication distributes over subtraction, which is sufficient to fix my proof from before. Why is this true?[/quote]

[b]Definition:[/b]  We say that $ a \minus{} b \equal{} c$ when $ a \equal{} b \plus{} c$.

[b]Given distributivity over addition[/b] (which is not an axiom in Peano arithmetic):

$ a \minus{} b \equal{} c \Leftrightarrow$
$ a \equal{} b \plus{} c \implies$
$ da \equal{} d(b \plus{} c) \equal{} db \plus{} dc \Leftrightarrow$
$ da \minus{} db \equal{} dc \equal{} d(a \minus{} b)$

Alternately, prove that $ a \minus{} b \equal{} a \plus{} (\minus{}1)b$.

So, I know I'm just making a bother of myself now, but I would like to suggest that any solution using subtraction is "inferior" from the point of view of Peano arithmetic to any solution not using subtraction (i.e., solutions either using my proposed additional axiom or solutions that use a quantified definition of divides) because Peano arithmetic doesn't have subtraction and negative numbers -- you can sort-of define subtraction, but you have to decide what to do when you subtract a larger number from a smaller number.  So, there is a lot more hidden behind any proof that has a "$ \minus{}$" in it than there is in proofs that can avoid it.

but we are assuming that t|s and t|s+1, so in h) we can put t|s "and" t|1 implies t|s+1. We know that t|s and t|s+1 by our assumption, so if t doesn't divide 1 then condition h) fails

[quote="ggbg90"]but we are assuming that t|s and t|s+1, so in h) we can put t|s "and" t|1 implies t|s+1. [/quote]

h) tells us that if $ t | s$ and $ t | 1$ then $ t | s \plus{} 1$.  It gives us no information about what happens if $ t | s$ and $ t | s \plus{} 1$.

Here's another example:  Say that h) was the statement "if I light a match and it is not raining then I will start a fire."  You are attempting to conclude from this that "if I have started a fire and it is not raining then I lit a match."  But there are many other ways to start a fire (for example, a flamethrower).

Put another way:  you are assuming that $ t | 1$, and from the premise $ t | s$ you can conclude that $ t | s \plus{} 1$.  All this tells you is that it's [b]possible[/b] that $ t | 1$, not that it [b]must be true.[/b]


[hide]1)     x = y
2)     x^2 = xy      (multiply both sides by x)
3)     x^2 - y^2 = xy - y^2  (subtract y^2 from each side)
4)     (x + y)(x - y) = y(x - y)     (factor)
5)     x + y = y     (divide out (x - y))
6)     2y = y      (substitute y for x from #1)
7)     2 = 1      (What do you think of that?!)
[/hide]

 :D  I copied this of a site!....Cool, huh?

[color=white]BTW:we cannot divide by (x-y) so don't annoy me too much[/color]

Actually, one step doesn't work.  You can't divide out (x-y) since that's equal to 0 and you can't divide by 0.

It's a perfectly legal answer. Since that's not true, we know $x\not=y$.
Either that, or it's an inequality. :P

[quote="fredbel6"]If I'd get a Euro every time I see that proof  :( ... Let me say it this way:
[img]http://users.telenet.be/frederic_bel6/Fun/everytimeyouprovethat2improved.jpg[/img][/quote]

 :lol: nice response

Yes, nice reply! (reminds me of another forum, when someone was spamming and then some other person gave a link to a pic with a 5yo kid praying: "Please God, make it stop!")

But what about those who don't like kittens? You know, the kind of people that have been terorized in their childhood by kittens and want to get even with them...

For an explanation of the picture, please see this Wikipedia article

[url]http://en.wikipedia.org/wiki/Every_time_you_masturbate%E2%80%A6_God_kills_a_kitten[/url]

yea, I've definitely heard the phrase several times

so I was reading the wiki article and then linked to dysphemism and then to Gresham's Law where I found

"In democracy, a less intelligent majority may be able to out-vote an exceptionally intelligent minority, effectively driving out the most intelligent voters by guaranteeing that their votes will be defeated. See also: IQ"

[quote="fredbel6"]If I'd get a Euro every time I see that proof  :( ... Let me say it this way:
[img]http://users.telenet.be/frederic_bel6/Fun/everytimeyouprovethat2improved.jpg[/img][/quote]

I hate kittens!! :diablo:

[color=white]and mystic is right,"less-inteligent people rock"[/color]


limit
function
calculus
Asymptote
calculus computations

$ N$ is a positive integer. Show that the equation $ \sqrt{2x\plus{}3} \equal{} \frac{N}{x^{3}}$ has exactly one real root.

Let $ f(x) \equal{} \sqrt {2x \plus{} 3} \minus{} N/x^3\colon [ \minus{} 3/2, \plus{} \infty)\to {\bf R}.$  [b]1)[/b] $ \lim_{x\to( \minus{} 3/2)^ \plus{} } f(x) \equal{} 8N/27 > 0,$  [b]2)[/b]
$ \lim_{x\to0^ \minus{} } f(x) \equal{} \plus{} \infty,$ [b]3)[/b] $ \lim_{x\to0^ \plus{} } f(x) \equal{} \minus{} \infty,$  [b]4)[/b] $ f'(x) \equal{} {1\over \sqrt {2x \plus{} 3}} \plus{} {3N\over x^4} > 0$ for all $ x > \minus{} 3/2.$ [b]1)--4)[/b] togheter with Weierstrass theorem on continuous functions imply the existence of a single
real numbers such that $ f(x) \equal{} 0$

[b] Edit [/b] Add the fifth condition $ \lim_{x\to\plus{}\infty} f(x)\equal{}\plus{}\infty$

P.S. may be you thought of a simpler proof (differential calculus free?)[/u]

Thanks, Arrigo-Sacchi. By the way, what does Weierstrass's theorem state exactly?

Sketching the graphs. But perhaps there's a better way.

Let $ K$ be a compact subset of $ {\bf R}^n$ and let $ f\colon K\to{\bf R}$ be a continuous function on $ K.$ For instance $ n \equal{} 1$ and $ K$ a closed and bounded interval $ [a,b]$ . Then $ f$ is bounded in $ [a,b],$ $ m \le f(x)\le M,$ and moreover there exist $ x_1,$ $ x_2$ such that $ f(x_1) \equal{} m,$ $ f(x_2) \equal{} M.$

In your problem it sufficies to find two real positive values  $ y_1$ and $ y_2$ such that $ f(y_1) < 0$ and $ f(y_2) > 0.$ This is not difficult as the behaviors at $ x\to0^ \plus{}$ and $ x\to\ \plus{} \infty$ show (by the way in my post I forgot to write $ \lim_{x\to \plus{} \infty} f(x) \equal{} \plus{} \infty$ becaquse otherwise the function could be negative having $ y \equal{} 0$ as horizontal aymptote


When I see some users' mini-profiles on the left of their post, I see two national flags below their rating.  Is this new?  How can I turn it off or on?
EDIT:  Ok, I see this can be edited in the profile options under Contact information.  The first flag represents the country the user was originally from, and the second represents te country the user is currently residing in.

Why does mine come above my rating?

[quote="SimonM"]Out of interest, can we request new flags (my country isn't there :()?[/quote]What country would that be? We're using the ISO 3661 standards btw.

I always thought the flags were a nice little touch, and I'm glad they're back. And having two flags allows for a little more nuance in the personal preferences.

I'd like to see more individual users turn this preference on.

[quote="JBL"]Guernsey (one of the Channel Islands between England and France): https://www.cia.gov/library/publications/the-world-factbook/geos/gk.html

Total area, 78 square km, or about half the size of Washington, D.C.[/quote]

Yeah, JBL's got it :) You're not the only site which doesn't have it. I had to request it on ProjectEuler (and  on a few other sites)

Sarnia Cherie!

BTW, that CIA page could do with some improvements though.

Our suffrage is 16. I have the right to vote too!

[quote="Carcul"]One more time: how can we make a flag appear? Why hasn't some administrator explained this? Is that a secret?[/quote]

More specifically, go to your profile, click "profile", then click "personal". You will see two drop down menus where you can choose your flags.

[quote="jmerry"]It's in the first post- you can indicate your country in your profile. It's also an old feature reenabled; I haven't done anything, since I was already set up for it.[/quote]

Why was it disabled before?

[quote="BanishedTraitor"]Why was it disabled before?[/quote]We made some upgrades to the site (major ones) a while ago, and I didn't have time to upgrade the flag display up until recently.

Thank you very much Jmerry and Buzzer11.


When will the WOOT message boards and classroom be open, and how would one navigate to those pages?

Thanks.  :lol:

--200th post!--

We'll likely activate 2008-09 WOOT Message Boards and classroom towards the end of the month.  There won't be any classes until September, but we'll start some message board discussion when we open the board, by posting occasional problems to challenge enrolled students.

[quote="cognos599"]i have a question that is somewhat related to this topic and i figured it would be useless to make another topic, but just for conformation, students can type latex in the WOOT classroom? because i remember attending some math jams and only the instructor could use latex[/quote]

Students can use LaTeX in the classroom, even in the Math Jams (we sometimes don't mention that at the start of Math Jams, because, particularly in large Math Jams, it leads to some confusion, as some students think we mean that they have to know LaTeX, etc).

I (my parents) purchased the WOOT class for me today, but I still cannot find the corresponding message board. Is it under the Online School forum? I looked there, but I can't find it.
edit: Yay, thanks!


Using a balance scale, it was determined that an algebra book
and a geometry book together balance a trigonometry book,
and two trigonometry books balance three geometry books.
How many geometry books are needed to balance four algebra
books?

Let $ a$= the weight of an algebra book, $ g$= the weight of a geometry book, and $ t$= the weight of a trigonmetry book. We have the following:

$ a \plus{} g \equal{} t$
$ 3g \equal{} 2t$

Therefore, $ 2a \plus{} 2g \equal{} 2t \implies 2a \plus{} 2g \equal{} 3g \implies 2a \equal{} g \implies 4a \equal{} 2g$.

So, $ \boxed{2}$ geometry books balance $ 4$ algebra books.


Can the number obtained by writing numbers from 1 to n be a palindrome?

[hide="this is what i have done"]
we have to see if the number - $ 12345....(n\minus{}2)(n\minus{}1)(n)$ where n is replaced by its digits form a palindrome.
We see that $ n$ has to end with $ 321$ to make the number palindrome.
so $ (n\minus{}1)$ ends in 0.
consider - $ 123456789101112131415161718192021222324252627282930313234.....$
for the whole number to be a palindrome we chose n such that the digit after n in this sequence is 0 and n in the sequence starts from the left hand most and keeps on going until when we choose.
so $ n$ is of $ 1987654321$ or of the form $ X9(X\minus{}1)8(X\minus{}1)7(X\minus{}1)6(X\minus{}1).....$ where $ (X\minus{}1)$ is replaced by its digits.
this means that $ (n\minus{}1)$ ends $ 7X6X5X4X3X2X1X0$. BUT $ (n\minus{}1)$'s last digits are $ 54320$ so even if we set X=2 this is not possible so the number cant be a palindrome . However i dont think we are done as we also have to consider the case when n has digits $ 3201$.[/hide]
 i think it is a beautiful question!

c'mon mathlinkers! no one interested in this problem?

1) This is not a personal help line.  There is no reason to go bumping posts a mere 12 hours after they were posted while they still remain at the top of the forum.

2) This question is very easy to search for: putting "Russia 1996" or "Russia palindrome" into the search function leads to threads like
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19872
with very little effort.


What is the value of the expression $ 5^3 \minus{} 3 \times 5^2 \plus{} 3 \times 5 \minus{} 1$?

Whenever you see a question in this form ($ ax^3 \minus{} 3x^2 \plus{} 3x \minus{} 1$), the answer is simply $ (x \minus{} 1)^3$, so in this case, the answer is $ \boxed{64}$.

hello, it is $ 125\minus{}75\plus{}15\minus{}1\equal{}64$.
Sonnhard.

The order of operations goes as follows:

P: Parentheses
E: Exponents
M/D: Multiplication/Division(in the order from left to right)
A/S: Addition/Subtraction(in the order from left to right)

There aren't any parentheses, so we do exponents first.

$ 5^3 \equal{} 125$

and

$ 5^2 \equal{} 25$

$ 125 \minus{} 3 \times 25 \plus{} 3 \times 5 \minus{} 1$

Now multiplication.

$ 3 \times 25 \equal{} 75$

$ 3 \times 5 \equal{} 15$

$ 125 \minus{} 75 \plus{} 15 \minus{} 1$

Finally addition and subtraction.

$ 125 \minus{} 75 \equal{} 50$

$ 50 \plus{} 15 \equal{} 65$

$ 65 \minus{} 1 \equal{} \boxed{64}$

EDIT: Beaten to it.


在[url]http://en.wikipedia.org/wiki/The_High_School_Affiliated_to_Renmin_University_of_China[/url] 有一句：2 Gold Medalists in the International Mathematics Olympiad

Out of the 6 Chinese Competitors, who is from 人大附中？   （对不起。我的汉语不好....）

我在韩国的大元外语高中上学，我的学校和人大附中 are sister schools。上个暑假我到北京去访问。Just curious....[/url]

According to the listing on most Chinese websites, there was only one team member from 人大附中 in 2008, Ruixiang Zhang (张瑞祥), who earned a gold medal.


function
logarithms
algebra unsolved
algebra

Let $ h: \mathbb{R} \minus{} > \mathbb{R}$ a function that satisfies:
i) $ h(xy) \equal{} h(x) \plus{} h(y)$
ii)$ h(max(\frac {m}{2^p3^q},\frac {n}{2^p}) \plus{} min(\frac {n}{2^p3^q},\frac {m}{2^p}))\equal{}\log_2 (q\plus{}1)  \forall m, n \in \mathbb{Z} , \forall p,q \in \mathbb{N}$
Show that if $ x \in \mathbb{Q}$ then $ h(x) \equal{} 0$.

[quote="MellowMelon"]You haven't written the second condition well.

- What you've said is basically the equivalent of saying $ h(z)$, which means nothing. What's the other side of the equation?
- What are $ m,n,p,q$? Integers? Positive integers?[/quote]

Now I guess it's ok.

From the first condition: $ h(x) \equal{} c\log_a x , a > 0$,  $ a \neq 1, c \neq 0$ or $ h(x) \equal{} 0$. I was trying to prove that $ h(x) \equal{} c\log_a x$ cannot be satisfied. But it's not working.

[quote="danilorj"]Let $ h: \mathbb{R} \minus{} > \mathbb{R}$ a function that satisfies:
i) $ h(xy) \equal{} h(x) \plus{} h(y)$
ii)$ h(max(\frac {m}{2^p3^q},\frac {n}{2^p}) \plus{} min(\frac {n}{2^p3^q},\frac {m}{2^p})) \equal{} \log_2 (q \plus{} 1) \forall m, n \in \mathbb{Z} , \forall p,q \in \mathbb{N}$
Show that if $ x \in \mathbb{Q}$ then $ h(x) \equal{} 0$.[/quote]

It seems obviously wrong.

Take property ii with $ p\equal{}q\equal{}m\equal{}n\equal{}1$ and you have : $ h(max(\frac {1}{6},\frac {1}{2}) \plus{} min(\frac {1}{6},\frac {1}{2})) \equal{} \log_2 (2)$ And so :

$ h(\frac{2}{3})\equal{}1\neq 0$

ii) $ h( \max ( \frac{m}{2^{p}(q\pi \plus{} 1 )}, \frac{n}{2^{p}}) \plus{}  min (\frac{m}{2^{p}}, \frac{n}{2^{p}(q\pi \plus{} 1)})) \equal{} log _{2} (q \plus{} 1)$

Now the second condition is correct.

Still doesn't look right. You need to clarify the first condition too (sorry for not noticing in first reply). If it's for all reals $ x,y$, then $ x \equal{} 0$ gives $ h(y) \equal{} 0$ for all reals $ y$, trivializing the problem.

Perhaps the function is $ \mathbb{R}^\plus{} \rightarrow \mathbb{R}$?

[quote="danilorj"]ii) $ h( \max ( \frac {m}{2^{p}(q\pi \plus{} 1 )}, \frac {n}{2^{p}}) \plus{} min (\frac {m}{2^{p}}, \frac {n}{2^{p}(q\pi \plus{} 1)})) \equal{} log _{2} (q \plus{} 1)$

Now the second condition is correct.[/quote]
Let $ n>0,m<0$, then we get $ h(\frac{n\plus{}m}{2^p})\equal{}log_2(q\plus{}1)\not \equal{}0.$

I received an e-mail that this question is totally incorrect. So forget about it.


calculus
integration
algebra
function
domain
abstract algebra
analytic geometry

@ integral domain:

[quote="Jacobson"]Must $ A$ be an integral domain if $ A_p$ is an integral domain for every prime ideal $ p$?[/quote]

What about $ A \equal{} \mathbb{Z}\slash 6\mathbb{Z}$ ? This is hardly an integral domain, but the localizations at the prime ideals $ \left(2\right)$ and $ \left(3\right)$ are (if I am not mistaken, they are $ \mathbb{Z}\slash 2\mathbb{Z}$ and $ \mathbb{Z}\slash 3\mathbb{Z}$, up to isomorphism).

  darij

@Noetherian:

[quote="Jacobson"]in same spirit:

Let $ A$ be a ring such that each local ring $ A_p$ is Noetherian. Must $ A$ itself be Noetherian?[/quote]

http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2005;task=show_msg;msg=2110.0001 gives a nice counterexample, which would be even more nice if I would understand it. Here is a modification which I can prove: Let $ A \equal{} \oplus_{i\in I}F$, where $ F$ is a field and $ I$ is an infinite index set. Then, for every prime ideal $ \mathfrak{p}$ of $ A$, the ring $ A_{\mathfrak{p}}$ is a field and therefore Noetherian, but $ A$ itself is not Noetherian. How to show that $ A_{\mathfrak{p}}$ is a field? For every $ i\in I$, if $ \mathfrak{p}$ contains an element with nonzero $ i$-th coordinate (say $ \left(x_1,x_2,...,x_{i \minus{} 1},x_i,x_{i \plus{} 1},...\right)$ with $ x_i\neq 0$), then $ \mathfrak{p}$ contains the element $ \left(0,0,...,0,1,0,0,...\right)$, where the $ 1$ stands on the $ i$-th place (because $ \left(0,0,...,0,1,0,0,...\right) \equal{} \left(0,0,...,0,\frac {1}{x_i},0,0,...\right)\cdot\left(x_1,x_2,...,x_{i \minus{} 1},x_i,x_{i \plus{} 1},...\right)$). Conclude from this that there exists a subset $ J\subseteq I$ such that $ \mathfrak{p} \equal{} \oplus_{i\in I}\left\{\begin{array}{c}F,\text{ if }i\in J; \\
0,\text{ if }i\notin J\end{array}\right.$; in other words, $ \mathfrak{p}$ is the set of all elements of $ I$ whose $ i$-th coordinates are $ 0$ for $ i\notin J$, but all other coordinates can be arbitrary. Now show that $ \left|I\setminus J\right| \equal{} 1$ because $ \mathfrak{p}$ is a prime ideal (in fact, if $ I\setminus J$ would contain two distinct elements $ u$ and $ v$, then the product of $ \left(0,0,...,0,1,0,...\right)$ with $ 1$ on the $ u$-th place and $ \left(0,0,...,0,1,0,...\right)$ with $ 1$ on the $ v$-th place would be $ 0$ and thus lie in $ \mathfrak{p}$, but none of the two factors lie in $ \mathfrak{p}$). Conclude from this that $ A_{\mathfrak{p}}$ is a field, namely $ F$.

I hope I didn't make any stupid mistakes here.

  darij

thanks darij

Can someone expand my list?

[b]Local property[/b]
-flatness
-injectivity
-surjectivity

[b]Non-local property[/b]
-being an integral domain
-being noetherian

a geometric motivation for darij's example: due to stone's representation theorem, the question, if noetherian is a local property, for boolean rings is if there is a totally disconnected hausdorff space, which is not artinian for closed subsets. well, $ \mathbb{N}$ is an example.

further local properties for rings: reduced, normal (given integral domains), regular
there are more for schemes.

[quote="Jacobson"]Must $ A$ be an integral domain if $ A_p$ is an integral domain for every prime ideal $ p$?[/quote]

If I recall correctly, for $ A$ to be an integral domain, $ Spec(A)$ must be connected. If that's the case, it's quite easy to construct such an example: let $ F$ be a field and $ A \equal{} F \times F$. Then $ A$ has two prime ideals $ 0 \times F$ and $ F \times 0$. Their corresponding localisations are isomorphic to F and hence integral domains - on the other hand $ A$ clearly is not.

Geometrically, we're just taking the union of two points.

another local property: being torsion-free


Here is a well used inequality and I think it needs a new topic to put nice solutions generalisations and maybe some stronger ones.
Here is the problem
$a,b,c\geq 0$ . Prove that
$(a+b+c)^6\geq 64(a^2+bc)(b^2+ca)(c^2+ab)$

[quote="silouan"]
$a,b,c\geq 0$ . Prove that
$(a+b+c)^6\geq 64(a^2+bc)(b^2+ca)(c^2+ab)$[/quote]
silouan,it follows from a wonderful hungkhtn's inequality:
$a+b+c\geq\frac{2}{3}\left(\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+ab}\right).$ ;)

[quote="libra_gold"]It is equivalent to

$360(3,2,1)+3(6,0,0)+36(5,1,0)+90(4,2,0)\geq 102(4,1,1)+132(3,3,0)$

Then you can use Muirhead to solve it anyway.[/quote]

How you arrive at such expression ? And is this the expansion of first or second inequality ? (I have checked and the expansion of first inequality doesnt look such way,and I doubt the second inequality can be " expanded" ? :?: )

[quote]Here is a well used inequality and I think it needs a new topic to put nice solutions generalisations and maybe some stronger ones.
Here is the problem

Prove that
\[ (a+b+c)^6\ge 64(a^2+bc)(b^2+ca)(c^2+ab) \][/quote]

My direct solution for it is using $(a-b)^2(b-c)^2(c-a)^2 \ge 0$ and SOS method, but it's truly hard,

Also, Manlio has asked me this question one time, in topic [b]Nice[/b].

I have a nice solution to this inequality in Algebraic Inequality.
I finished the last correction of the book.  :(


[quote="LoboSolitario"]Let $A(x)$ and $B(x)$ polynomials with degree greater than 1 and assume that exists polynomials $C(x)$ and $D(x)$ such that:
\[ A(x)\cdot C(x)+B(x)\cdot D(x)=1,~\forall\, x\,\in\,\mathbb{R}.  \]
Prove that $A(x)$ isn't divisible by $B(x)$.[/quote]

Assume for the sake of contradiction that $B(x)|A(x)$. Then $A(x) = Q(x)B(x)$ so $A(x)C(x)+B(x)D(x) = B(x)[Q(x)C(x)+D(x)] = 1 \Rightarrow B(x)|1$. But since $B$ has degree greater than $1$, it obviously cannot divide something of degree $0$. Contradiction.


geometry
trigonometry
similar triangles
power of a point
trig identities
Law of Cosines

Prove the Power of a Pt. Thm. for the case where the pt. of intersection lies within the circle. There is a simple proof, different from the more complete one in AOPS vol. II.

Couldn't get a good picture up. I'm still learning how to do that kind of thing  :roll:

Well, while we're on the subject, I know two good geometry programs :
[url=http://www.matf.bg.ac.yu/~janicic/gclc/]GCLC[/url] (Ugh, I don't like that new thing about proving theorems at all, though)
[url=http://www.mit.edu/~ibaran/kseg.html]KSEG[/url]

Do you mean AoPS I?!  I'm sure they don't do Power of a Point in Volume II—it seems too elementary.  Someone else is borrowing my Vol. I right now, so how do they prove it there?  [hide]Do they use similar triangles?[/hide]

Oh, I'm sorry. I accidentally claimed that they proved the pwr. of a pt. thm. in AOPS II. Instead, I was thinking of the proof of the Law of Cosines, where the power of a pt. thm. was used throughout most of the proof.

Similar triangles is the right track. Someone should give a clear solution though.


if 3^1993 = 100[i]k[/i]+[i]n[/i], where [i]n[/i]<100 is a nonnegative integer, what is [i]n[/i]?

[hide="um"]
23?

The solution is a little above mathcounts
[/hide]

[quote="unimpossible"]no it is not.

notice 9^10 ends in 01[/quote]

okay.

Thats exactly the point, so if 9^10 is 1 mod 100, then 3^20 is and so is 3^1980.

So we want to find just 3^13 mod 100, which is 23.

[quote="andrewliu"]if 3^1993 =
 100[i]k[/i]+[i]n[/i]
, where [i]n[/i]<100 is a nonnegative integer, what is [i]n[/i]?[/quote]

That part gives away that it is (mod 100).

[quote="501316"]My idea is that u can find a pattern in the 1's digits. 3^1=3; 3^2=9; 3^3=27; 3^4=81; 3^5=243 and so on. If u look at the ones digit, it always is 3, 9, 7, 1, 3, 9, ... To find the units digit of 3^1933, just divide 1933 by 4 and find the remainder. Since every 4th power's units digit is 1, and u get a remainder of 1 which tells u to go 1 further than the 4th possible digit, then the units digit is 3. Since anything multiplyed by 100 has a units digit of 0 the answer is 3. Hope it helps  :D  :lol:  :)  :oops:  :wink:[/quote]

Not asking for units digit, asking for modular residue in modulo 100, or in simpler words, the first 2 digits. Although there indeed is a pattern for the hundred's digit, this solution is not very "elegant" and takes a while.

You first want to write it as $ 3^{1993}\equiv{x}\pmod{100}$. Based on unimpossible's observation (you can confirm it by entering it in to your calculator), we can rewrite is as $ 3^{20} * 3^{20}... 3^{13}$. Since $ 3^{20}\equiv{1}\pmod{100}$, they cancel out, leaving out $ 3^{13}\equiv{x}\pmod{100}$. This number is small enough for your calculator to handle; you get

$ \boxed{n \equal{} 23}$


Let $M$ be the midpoint of side $BC$ of a triangle $ABC$. Let the median $AM$ intersect the incircle of $ABC$ at $K$ and $L$, $K$ being nearer to $A$ than $L$. If $AK=KL=LM$, prove that the sides of triangle $ABC$ are in the ratio $5: 10: 13$ in some order.
 :wink:

You will find a solution in this forum very easy , because it is posted again recently. I'm not at home this moment , so I can not search.I hope , somebody will send you the link. Ah! I think Nicula has recently sent a nice solution.
   I had proposed the same problem in the past too !!!
[u]Babis[/u]

Well , see

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=incircle&t=49419


This has appeared a ton of times on mathcounts cd rounds when they say like leaps = some lops = some loops and then they want you to find the leaps for something. I always get intimidated by this and screw up, does anyone have any tips for this?

Lets say that $ \frac{a}{b}logs\equal{}\frac{c}{d}fraps$ and $ \frac{e}{f}fraps\equal{}\frac{g}{h}$ slaps.

If we put in numbers for the variables and think it through, we can find a pattern. Lets say we want to go from slaps to fraps. If we have $ x$ slaps, we have $ x*\frac{h}{g}*\frac{e}{f}$ fraps. Can you see a pattern of multiplying certain reciprocals by certain fractions?

Now if you wanted to from slaps to logs, just convert slaps to fraps and then convert fraps to logs using similar logic.


function
limit
logarithms
integration
calculus
derivative
calculus computations

If $f(x)$ be a differentaible function such that $ \lim_{x \longrightarrow \infty} f(x) = a \in \mathbb{R} $. Suppose that $ \lim_{x \longrightarrow \infty} x f'(x) $ exists. Find this limit.

The limit $\lim_{x\to\infty} xf'(x)$ must be zero. If not, WLOG, suppose it is a positive real. Then there exists $X>1$ such that $xf'(x)>a>0$ for $x>X$. $f'(x)>a/x$ for $X>0$. Choose $b>X$.

Let us prove that $f(x)>f(b)+a\ln x-a\ln b$ for $x>b$.
To see this, let $g(x) = f(x)-f(b)-a\ln x+a\ln b$. In fact, $g'(x) = f'(x) - \frac{a}{x} > 0$ for $x>b$ and $g(b)=0$. Thus $g(x)>0$ for $x>b$.

$f(x)>f(b)+a\ln x-a\ln b$ implies that $f$ is unbounded. Contradiction.

A small comment: A more natural thought is use $\int_b^x f'(x)dx > \int_b^x \frac{a}{x}dx$ to get the contradiction. But is $f'(x)$ integrable? We need some words to argue. So I use derivatives instead of integration in the proof.


Hello.

Does anybody know if there's a way to get two lines of conditions below a sum in LaTeX? i.e. sum over d<x and Y_1 < T < Y_2, with the "d<x" and " Y_1 < T < Y_2 " on two separate lines under the sum? (because it looks ugly with one rediculously long line)  I tried using double subscripts and putting in a \\ but to no avail.

Thanks,

ian

$\sum_{\stackrel{d<x}{Y_{1}<T<Y_{2}}}$

I don't know how to make them equally sized (the second line has bigger chars). :(

\sum_{\stackrel{d<x}{Y_{1}<T<Y_{2}}}

You can view the code of a $\LaTeX$ picture by putting your mouse over it or viewing the picture's properties.

Use the \atop command as follows: $\sum_{ d<x \atop Y_{1}<T<Y_{2}}$.

The problem with \stackrel and \atop is that the size of the text may be too small. So amsmath has \substack which also allows as many lines as you like. To compare:
\substack: $\sum_{\substack{d<x\\Y_{1}<T<Y_{2}}}$ \stackrel: $\sum_{\stackrel{d<x}{Y_{1}<T<Y_{2}}}$  \atop: $\sum_{d<x \atop Y_{1}<T<Y_{2}}$
The substack version looks nice and uses \sum_{\substack{d<x\\Y_{1}<T<Y_{2}}. It can be extended to say for example: $\sum_{\substack{d<x\\Y_{1}<T<Y_{2}\\a>2\\b\neq0}}$ with each line being separated by \\.


if $ {S_a}\plus{}{S_b}\plus{}{S_c}\geq 0$ and $ {S_a}{S_b}\plus{}{S_b}{S_c}\plus{}{S_c}{S_a}\geq 0$
can we prove that:$ {S_a}{(b\minus{}c)^2}\plus{}{S_b}{(a\minus{}c)^2}\plus{}{S_c}{(a\minus{}b)^2}\geq 0$
thank you for your help~

[quote="xiaohui"]if $ {S_a} \plus{} {S_b} \plus{} {S_c}\geq 0$ and $ {S_a}{S_b} \plus{} {S_b}{S_c} \plus{} {S_c}{S_a}\geq 0$
can we prove that:$ {S_a}{(b \minus{} c)^2} \plus{} {S_b}{(a \minus{} c)^2} \plus{} {S_c}{(a \minus{} b)^2}\geq 0$
thank you for your help~[/quote]
Let $ S_a\plus{}S_b\geq0.$
Hence, $ {S_a}{(b \minus{} c)^2} \plus{} {S_b}{(a \minus{} c)^2} \plus{} {S_c}{(a \minus{} b)^2}\equal{}$
$ \equal{}{S_a}{(b \minus{} c)^2} \plus{} {S_b}{(a \minus{} b\plus{}b\minus{}c)^2} \plus{} {S_c}{(a \minus{} b)^2}\equal{}$
$ \equal{}(S_a\plus{}S_b)(b \minus{} c)^2 \plus{} 2S_b(a \minus{} b)(b\minus{}c) \plus{} (S_c\plus{}S_b)(a \minus{} b)^2\geq0$
since, $ S_b^2\minus{}(S_a\plus{}S_b)(S_c\plus{}S_b)\leq0.$ :wink:

$ S_{b}^{2}\minus{}(S_{a}\plus{}S_{b})(S_{c}\plus{}S_{b})\leq0.$
yes,it is the key to the problem,thank you~


Let $ a,b,c\geq0$.Prove that:
$ (a^2\minus{}bc)\sqrt{b\plus{}c}\plus{}(b^2\minus{}ca)\sqrt{c\plus{}a}\plus{}(c^2\minus{}ab)\sqrt{a\plus{}b}\geq0$

Let $ x \equal{} \sqrt {a \plus{} b}, y \equal{} \sqrt {b \plus{} c}, z \equal{} \sqrt {c \plus{} a}$

then this inequality is equivalent to $ \sum_{cyc} x(y^4 \plus{} z^4 \minus{} x^2y^2 \minus{} x^2z^2) \geq 0$,

which is true by Muirhead's inequality. ($ \sum_{sym} x^4y \geq \sum_{sym} x^3y^2$)

[quote="M.A"]I think SOS might be useful and this: [b]2(a^2 -bc)= (a-b)(a+c)+(a-c)(a+b)[/b][/quote]
I think you're right , M.A . This inequality is equivalent to :
$ \sum (a \minus{} b)^2.[\frac {1}{2}\frac {\sqrt {a \plus{} c}\sqrt{b \plus{} c}}{\sqrt {a \plus{} c} \plus{} \sqrt {b \plus{} c}} ]\geq 0$

The following inequality is also true
[img]http://s3.sinaimg.cn/middle/0018zOAxgy70KPcMrfA92&690[/img]


[quote="13375P34K43V312"]For what real number value of $n$ does the system of equations

$nx+y=1$
$ny+z=1$
$nz+x=1$

have no solution?[/quote]
[hide]Adding them we get $x(n+1)+y(n+1)+z(n+1)=3$
$(n+1)(x+y+z)=3$
So, if n=-1, then there is no solution...[/hide]

[quote="13375P34K43V312"]Yes, nice solution.[/quote]
This is a great problem...
This and that one other one that I said was great...


inequalities
geometry
circumcircle
number theory
prime numbers
national olympiad

5. http://www.mathlinks.ro/Forum/viewtopic.php?p=910666#910666
For all positive reals $ a$, $ b$, and $ c$, what is the value of positive constant $ k$ satisfies the following inequality?
$ \frac{a}{c\plus{}kb}\plus{}\frac{b}{a\plus{}kc}\plus{}\frac{c}{b\plus{}ka}\geq\frac{1}{2007}$

6. http://www.mathlinks.ro/Forum/viewtopic.php?p=910667#910667
$ ABC$ is a triangle which is not isosceles. Let the circumcenter and orthocenter of $ ABC$ be $ O$, $ H$, respectively,
and the altitudes of $ ABC$ be $ AD$, $ BC$, $ CF$. Let $ K\neq A$ be the intersection of $ AD$ and circumcircle of triangle $ ABC$, $ L$ be the intersection of $ OK$ and $ BC$, $ M$ be the midpoint of $ BC$, $ P$ be the intersection of $ AM$ and the line that passes $ L$ and perpendicular to $ BC$, $ Q$ be the intersection of $ AD$ and the line that passes $ P$ and parallel to $ MH$, $ R$ be the intersection of line $ EQ$ and $ AB$, $ S$ be the intersection of $ FD$ and $ BE$.
If $ OL \equal{} KL$, then prove that two lines $ OH$ and $ RS$ are orthogonal.

7. http://www.mathlinks.ro/Forum/viewtopic.php?p=910669#910669
In each $ 2007^{2}$ unit squares on chess board whose size is $ 2007\times 2007$,
there lies one coin each square such that their "heads" face upward.
Consider the process that flips four consecutive coins on the same row, or flips four consecutive coins on the same column.
Doing this process finite times, we want to make the "tails" of all of coins face upward, except one that lies in the $ i$th row and $ j$th column.
Show that this is possible if and only if both of $ i$ and $ j$ are divisible by $ 4$.

8. http://www.mathlinks.ro/Forum/viewtopic.php?p=910671#910671
For all positive integer $ n\geq 2$, prove that product of all prime numbers less or equal than $ n$ is smaller than $ 4^{n}$.

Thank you very much!  :wink: I posted help you all link.  :)

[quote="modal_soul"]8. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=910671#910671
For all positive integer $ n\geq 2$, prove that product of all prime numbers less or equal than $ n$ is smaller than $ 4^{n}$.[/quote]Anyone know why this one was deleted? I reposted it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=364002]here[/url].


Let $ K$ denote the intersection of the two diagonals of a quadrilateral $ ABCD$ where $ \angle ABC\equal{}\angle ADC\equal{}90^o$ and $ AC\equal{}AB\plus{}AD$.Prove that the radius of inscribed of triangle $ ABK$ and $ ADK$ are equal

[quote="woofunky"]Let D'=>triangle ADC and triangle D'BC is congruent.
AB+AD=AB+AD'=AD'=AC
so, triangle AD'C is isoceles triangle. Let, angle AD'C =a
angle DAC=a, and angle DCA=angle BCD'=90-a
angle D'CA=a, so, angle BCA=2a-90 ,and angle BAC is 2a
Look at triangle AD'C, you can fing 4a=180, a=45..
so, quadrilateral ABCD is square. so, triangle ABK, ADK is congruent.
so, the radius is same.[/quote]
Sorry，but you are wrong

That \[ \Leftrightarrow \frac{{S_{\vartriangle ABK} }}
{{S_{\vartriangle ADK} }} \equal{} \frac{{BK}}
{{KD}} \equal{} \frac{{AB \plus{} BK \plus{} KA}}
{{AD \plus{} DK \plus{} KA}} \equal{} \frac{{AB \plus{} KA}}
{{AD \plus{} KA}}\]

Let $ \angle BAK \equal{} \theta$ ,$ \angle DAK \equal{} \phi$ and $ \angle BKA \equal{} \psi$.Then use the Law of Sine: $ BK: KA: AB \equal{} \sin \theta : \cos \phi : \sin \psi$ ,$ AK: KD: DA \equal{} \cos \theta : \sin \phi : \sin \psi$.Then $ \frac{{BK}}{{KD}} \equal{} \frac{{BK}}{{KA}} \cdot \frac{{KA}}{{KD}} \equal{} \frac{{\sin \theta \cos \theta }}{{\sin \phi \cos \phi }}$ ,$ \frac{{AB \plus{} AK}}{{AD \plus{} AK}} \equal{} \frac{{1 \plus{} {\textstyle{{AB} \over {AK}}}}}{{1 \plus{} {\textstyle{{AD} \over {AK}}}}} \equal{} \frac{{\cos \theta \left( {\cos \phi  \plus{} \sin \psi } \right)}}{{\cos \phi \left( {\cos \theta  \plus{} \sin \psi } \right)}}$.
So $ \sin \theta \cos \theta  \plus{} \sin \theta \sin \psi  \equal{} \sin \phi \cos \phi  \plus{} \sin \phi \sin \psi$.
As $ \frac{\pi }{2} \equal{} \theta  \plus{} \psi  \minus{} \phi$ ,everything left is so easy.

Stop this topic!
This problem is in a problem of a continous contest in "Math and Youth Magazine", so I think the topic should be locked in order to ensure the justice for this contest.
Last note: In fact, this problem has many synthetic proofs and excellent extensions. However, we will refer to them after the contest of M&Y finishes.
[hide="To Mathlinkers of Vietnam"]Mong các bạn đến từ Việt Nam đừng nên post những bài từ THTT số mới nhất, đó ko phải là bài tập bắt buộc nên hãy tự suy nghĩ để giải quyết chúng.[/hide]


Arrange these Phosphorous Trihalides in the order of Bond Angle

$ PF3, PCl3, PBr3, PI3$.

Not meant for chemrock, he has already been told the ans

[hide="Answer"]Bond angle increases from left to right.[/hide]

the answer i think is

[hide]$ {PCl3} < {PBr3} < {PF3} < {PI3}$[/hide]

Valerium reconsider ur answer a little. :D

forget about whatever the answer is if the answer by Carcul(the logical one) is not right then how are we supposed to know them without any data or experiment....not to sound bad what's the use of such a problem :(

well u all the posted ur answers but has anyone given an explanation for it? :huh:

@pardesi: Once u reason out then obviously u will find the answer :)

[quote="pardesi"]i guess it will be impossible to answer the reason without reasoning the answer...and that too insuffcient[/quote]

Ho Ho Mokka Da!

@Carcul: One small thing I think you are missing.

I am not missing anything. The halogen size increases, that's a fact. I have said anything else.

Varun you are right!! Its illogical to compare PCl3 with PI3.. But definitely you can logically compare all adjacent groups???

[quote="pardesi"]the bond angle is PI3>PBr3>PCl3. but i wont say anything about where pf3 should be placed, as i think there is an anomaly in case of pf3 as the hybridisation of pf3 is sp3, but in case of the others, as the size of halogen is great, the overlapping of orbitals is reduced ,and therefore greater deviation of bond angle from 109 deg. [/quote]

EDIT: never mind.

[quote="VARUNARASU"]the bond angle is PI3>PBr3>PCl3. but i wont say anything about where pf3 should be placed, as i think there is an anomaly in case of pf3 as the hybridisation of pf3 is sp3, but in case of the others, as the size of halogen is great, the overlapping of  orbitals is reduced ,and therefore greater deviation of bond angle from 109 deg.
personally i agree with pardesi, without experimental evidence we are in no position to answer the question, but of course we mite get such questions in our exams. :wink:[/quote]
ur right. PF3 cannot be placed without factual evidence but for the rest it is correct. :)

[quote="VARUNARASU"][quote="pardesi"]see i can put all the things i know into pen and paper but can never predict until i am given the data or ofc the answer what's the order
1)the electronegativity difference is decreasing down the group and that of iodine and phosphorous are nearly the same so logically the bond angle shud be decreasing taking this into account as the bond pairs are attracte dless towards the central atom as we move down so repulsion decreases

2)the sizes as Carcul said is increasing so in that way the bond pair-bond pair repulsion is increasing so looking at it this way the bond angle shud increase as we move down...
so the thing happening depends on what factor outweighs what....which is IMPOSSIBLE to predict unless u r given the data or u know the answer :)[/quote]
considering your first point pardesi, as the electroneg diff, decreases down the group, bond angle shud increase not decrease.[/quote]

why as the elevtronegativity of the central atom decreases so the eletrons tend to be relatively more away from the central atom hence they are more away than they would have been if they were near the angular points(wher the repulsion is max)  :)

Answer

$ PI_3 > PBr_3 > PCl_3 < PF_3$

Care to explain??

[quote="pardesi"]
why as the elevtronegativity of the central atom decreases so the eletrons tend to be relatively more away from the central atom hence they are more away than they would have been if they were near the angular points(wher the repulsion is max)  :)[/quote]

Dei only the surrounding atom changes and not the central atom :)

[quote="Carcul"][b]Cyber-Shreyas[/b], your answer is wrong.[/quote]

I verified values. Wiki says http://en.wikipedia.org/wiki/Phosphorus_chlorides I am wrong.

However a book of mine says PF3 > PCl3 because there is back bonding which makes the BP - BP repulsion slightly significant increasing the P-X-P bond. :(


For how many prime numbers between 10 and 100 is the last digit also a prime number?

the prime numbers are 2, 3, 5, and 7

however, prime numbers ending in 2 and 5 are not prime, so the only digits that the prime numbers can end in are 3 and 7

let's list the numbers out
13
17
23
37
43
47
53
67
73
83
97

we have 11 numbers

can anyone confirm?
i feel like i made a mistake somewhere....

since 2 and 5 are prime number we can conclude that 15 12 25 22 are prime # hence as they say that you can include any number as long as they have ANY PRIME #
SO THE LIST WILL BE WHAT VALLON22 WROTE

If this is a straight line [showing his audience a straight line drawn by a ruler], then it necessarily ensues that the sum of the angles of the triangle is equal to two right angles, and conversely, if the sum is not equal to two right angles, then neither is the triangle rectilinear.
Aristotle (384-322 B. C. E)


Given are triangle $ABC$ with any point $M$ on plane prove that
$\sqrt{bc}MA^{2}+\sqrt{ca}MB^{2}+\sqrt{ab}MC^{2}\ge abc$
similar like
$aMA^{2}+bMB^{2}+cMC^{2}\ge abc$
 :D

We have:
$\left(\sum\sqrt{bc}.\overrightarrow{MA}\right)^{2}\ge 0\Longrightarrow\sum bc.MA^{2}+2\sum a\sqrt{bc}.\overrightarrow{MB}.\overrightarrow{MC}\ge 0\Longrightarrow\sum bc.MA^{2}+\sum a\sqrt{bc}.(MB^{2}+MC^{2}-a^{2})\ge 0\Longrightarrow\left(\sum\sqrt{bc}\right)\cdot\left(\sum{\sqrt{bc}.MA^{2}}\right)\ge\sum a^{3}\sqrt{bc}\Longrightarrow\sum\sqrt{bc}.MA^{2}\ge\frac{\sum a^{3}\sqrt{bc}}{\sum\sqrt{bc}}$
We have to prove that:
$\frac{\sum a^{3}\sqrt{bc}}{\sum\sqrt{bc}}\ge\prod a\Longleftrightarrow\sum\frac{a^{3}\sqrt{bc}}{\prod a}\ge\sum\sqrt{bc}\Longleftrightarrow\sum\frac{a^{2}}{\sqrt{bc}}\ge\sum\sqrt{bc}$
But we easy see that:
 $\sum\frac{a^{2}}{\sqrt{bc}}\ge\frac{(\sum a)^{2}}{\sum\sqrt{bc}}\ge\frac{\left(\sum\sqrt{bc}\right)^{2}}{\sum\sqrt{bc}}=\sum\sqrt{bc}\; \mathbb{GED}$


Let be given a natural number $n\ge 2$.Find all real numbers $\alpha$ such that for all sets of positive real numbers $a_1\ge a_2\ge a_3\ge ...\ge a_n$ we have:
$\frac{a_1}{a_n}+\frac{a_2}{a_{n-1}}+....+\frac{a_{n-1}}{a_2}+\frac{a_n}{a_1}\le n+\frac{(a_1-a_n)^2+(a_2-a_n)^2+...+(a_{n-1}-a_n)^2+(a_n-a_n)^2}{a_n^\alpha}$

If $\alpha>2:$
For $a_2=a_3=\ldots=a_n=r>2^{\frac{1}{\alpha-2}}$
$a_1=2r$
The term of the left is equal to $n+\frac{1}{2}$
and the term of the right is equal to $n+\frac{r^2}{r^\alpha}$
But $r^{\alpha-2}>2 \Rightarrow \frac{1}{2}>r^{2-\alpha}$ (contradiction)
If $\alpha<2:$
analogously
If  $\alpha=2:$
adding extrems It prove the inequality


i think this should be moved to high school... it gave me a headache... T-T

This problem isn't hard... but I'll move it to mathcounts

[hide="hint"]think about how many digits are possible for each place. [/hide]

[hide][quote="4everwise"]How many odd numbers with third digit 5 are there between 20000 and 69999 inclusive?[/quote]

5*10*1*10*5=2500 numbers[/hide]


For a triple $ (m,n,r)$ of integers with $ 0 \leq r \leq n \leq m \minus{} 2$, define
$ p(m,n,r) \equal{} \sum_{k \equal{} 0}^{r}{m \plus{} n \minus{} 2(k \plus{} 1) \choose n}{r \choose k}$
Prove that $ p(m,n,r)$ is positive and that
$ \sum_{r \equal{} 0}^{n}{p(m,n,r} \equal{} {m \plus{} n \choose n}$

It looks like the factor $ (\minus{}1)^k$ is missing in the definition of $ p(m,n,r)$.
If so, first notice that
\[ \sum_{r \equal{} 0}^{n} p(m,n,r) \equal{} \sum_{r \equal{} 0}^{n} \sum_{k \equal{} 0}^{r}(\minus{}1)^k {m \plus{} n \minus{} 2(k \plus{} 1) \choose n}{r \choose k} \equal{} \sum_{k \equal{} 0}^{n} (\minus{}1)^k {m \plus{} n \minus{} 2(k \plus{} 1) \choose n} \sum_{r \equal{} k}^{n} {r \choose k} \equal{} \sum_{k \equal{} 0}^{n} (\minus{}1)^k {m \plus{} n \minus{} 2(k \plus{} 1) \choose n} {n\plus{}1\choose k\plus{}1} \equal{} {m\plus{}n\choose n} \minus{} \sum_{k \equal{} \minus{}1}^{n}  {m \plus{} n \minus{} 2(k \plus{} 1) \choose n} \cdot (\minus{}1)^{k\plus{}1} {n\plus{}1\choose k\plus{}1}\]
The first factor of the summand in the last sum is the coefficient of $ x^{m \plus{} n \minus{} 2(k \plus{} 1)}$ in $ x^n (1\plus{}x)^{\minus{}(n\plus{}1)}$ while the second factor is the coefficient of $ x^{k\plus{}1}$ in $ (1\minus{}x)^{n\plus{}1}$, so that the sum equals the coefficient of $ x^{m \plus{} n}$ in
\[ x^n (1\plus{}x)^{\minus{}(n\plus{}1)} (1\minus{}x^2)^{n\plus{}1} \equal{} x^n (1\minus{}x)^{n\plus{}1}\]
i.e., the coefficient of $ x^m$ in $ (1\minus{}x)^{n\plus{}1}$ which is zero since $ m>n\plus{}1$.
Therefore,
\[ \sum_{r \equal{} 0}^{n} p(m,n,r) \equal{} {m\plus{}n\choose n}.\]


calculus
integration
real analysis
real analysis unsolved

Find: $ \int \frac{xarctgx}{\sqrt{x^{2}\minus{}1}} \,dx$

P.S. Do not use integration software :-)

Integrate by parts and you're left with $ \int \frac{\sqrt{x^2 \minus{} 1}}{x^2 \plus{} 1} \, dx$.

Write that as $ \int \frac1{\sqrt{x^2 \minus{} 1}} \cdot \frac{x^2 \minus{} 1}{x^2 \plus{} 1} \, dx$.

Hence, it suffices to compute $ \int \frac1{\left( x^2 \plus{} 1 \right) \sqrt{x^2 \minus{} 1}} \, dx$.

Make the substitution $ t \equal{} \frac1{x^2} \, \Longrightarrow \, \int \frac1{\left( 1 \plus{} t \right) \sqrt{1\minus{}t}} \, dt$.

Final substitution $ u^2 \equal{} 1\minus{}t \, \Longrightarrow \, \int \frac1{u^2 \minus{} 2} \, du$.

For final answer use integration software.

What do you think? Was this integral interesting and/or difficult?

[color=darkblue][b][u]The Abel's substitution.[/u][/b] $ \boxed {t \equal{} \frac {2ax \plus{} b}{\sqrt {ax^2 \plus{} bx \plus{} c}}}$

for $ \int\frac {1}{\left(ax^2 \plus{} bx \plus{} c\right)\cdot\sqrt [n] {ax^2 \plus{} bx \plus{} c}}\ \mathrm {dx}\ .$

Although $ a\ne b$, you can try similarly and here :

$ I(a,b)\equiv\int\frac {1}{(x^2 \plus{} a)\sqrt {x^2 \plus{} b}}\ \mathrm {dx} \equal{} F\left(\frac {x}{\sqrt {x^2 \plus{} b}}\right) \plus{} \mathcal C$,

where $ F\equiv\int\frac {1}{(b \minus{} a)t^2 \plus{} a}\ \mathrm {dt}\ .$[/color]


Let a finite sequence consist of $N$ positive integers. There are $n$ distinct positive integers in this sequence with the condition that $N\geq 2^{n}$.
Show that we can find a block of consecutive integers whose product is equal to a perfect square.

So far I considered the following cases
-------------------------------------
1)Let $S=\{a_{1}<a_{2}<...<a_{n}\}$ be these integers. And let the holes be the proper subsets of $S$ and therefore a total of $2^{n}-1$. And we have $N>2^{n}-1$ but that doesnt seem to work.

2)Let the pigeons be the different possible consecutive blocks. Which is $(N-1)+(N-2)+...+2+1 = \frac{N(N-1)}{2}\geq 2^{n-1}(2^{n}-1)$. But I have no idea what the holes I have to take.

I got a seperate question. Do the Pigeonhole skill eventually get good and fast?

It is certainly not easy. Pigeonhole is really ingenuity; you can memorize many of the constructions, but hard work can only get you so far.

[hide="Solution"]

Consider the sequences
$a_{1}\\ a_{1}a_{2}\\ ... \\ a_{1}a_{2}...a_{n}$

Each of the integers can be written as $a_{1}^{k_{1}}a_{2}^{k_{2}}...a_{n}^{k_{n}}$. If any of the above have $k_{1}...k_{n}$ all even, we are done. Otherwise, there are $2^{n}-1$ different possibilities for the parities of the $k_{i}$. By pigeonhole, some two are equal. Their difference is a string of the form $a_{i}...a_{j}$ and all the $k_{i}$ are even. [/hide]

Note: this is a rather brutal estimate.

[quote="Phelpedo"]It is certainly not easy. Pigeonhole is really ingenuity; you can memorize many of the constructions, but hard work can only get you so far.

[hide="Solution"]

Consider the sequences
$a_{1}\\ a_{1}a_{2}\\ ... \\ a_{1}a_{2}...a_{n}$

Each of the integers can be written as $a_{1}^{k_{1}}a_{2}^{k_{2}}...a_{n}^{k_{n}}$. If any of the above have $k_{1}...k_{n}$ all even, we are done. Otherwise, there are $2^{n}-1$ different possibilities for the parities of the $k_{i}$. By pigeonhole, some two are equal. Their difference is a string of the form $a_{i}...a_{j}$ and all the $k_{i}$ are even. [/hide]
[/quote]
Can you be more explanatory. I am not good with this pigeons as you are. Thanks.

Let me give a similar problem that may seem more natural:

Prove that any 16-digit number contains a subsequence of digits whose product is a perfect square

[hide]
Let $a_{i}$ be the $i$th digit. Consider the products
$a_{1}$

$a_{1}\cdot a_{2}$

$a_{1}\cdot a_{2}... a_{16}$

Note that each of these numbers can be factorized as $2^{w}3^{x}5^{y}7^{z}$. If any of these $16$ products has $w,x,y,z$ all even, then we have a perfect square. Otherwise, there are $2^{4}-1=15$ possible parity combination. By pigeonhole, some two are equal. Divide them to get a product $a_{i}...a_{j}$ with all even exponents. [/hide]

[quote="Phelpedo"]

Consider the sequences
$a_{1}\\ a_{1}a_{2}\\ ... \\ a_{1}a_{2}...a_{n}$

Each of the integers can be written as $a_{1}^{k_{1}}a_{2}^{k_{2}}...a_{n}^{k_{n}}$. If any of the above have $k_{1}...k_{n}$ all even, we are done. Otherwise, there are $2^{n}-1$ different possibilities for the parities of the $k_{i}$. By pigeonhole, some two are equal. Their difference is a string of the form $a_{i}...a_{j}$ and all the $k_{i}$ are even. [/quote]

I don't see the pigeonhole though -- there are $2^{n}-1$ different parities, but only $n$ different sequences.

Ok, I can't edit the post but the $n$ in the sequences should be a capital $N$, which by problem statement is greater than $2^{n}-1$.

Thank you very much, Phelpedo! That really is a tough pigeonhole argument and tought to explain but you did a good job :). I only have one complaint I think you should have used $b_{1}, b_{1}b_{2},...,b_{1}b_{2}...b_{N}$ because the way you wrote it it was confusing it seems the $a_{i}$ where distint but with these $b_{j}$ you can say they arent necessarily distinct.


geometry
geometric transformation
homothety
rotation
circumcircle
rectangle
geometry proposed

ABC is a triangle and M, N, P are pts on (AC), (AB), (BC) respectively s.t. MP, PN, NM are perpendicular to BC, AB, CA respectively. Show that if the centroid of MNP is the Lemoine point of ABC then ABC is equilateral.

This is interesting. If I remember correctly (I am writing from memory now), triangle MNP is the pedal triangle of one of the two Brocard points of triangle ABC, and triangle MNP is the image of triangle ABC in a spiral homothety with the center of rotation at a Brocard point. Maybe, this helps?

  Darij

I don't think MNP is the pedal triangle of a Brocard point, but the second part (the one about the similarity or spiral homothety) is, indeed, true. The triangles ABC and MNP share one of the Brocard points of ABC.

It didn't help me, though :). I'd love to see a solution which uses that.

[quote="grobber"]I don't think MNP is the pedal triangle of a Brocard point,[/quote]

Yes, sorry, I was wrong. As I've said, I was writing from memory. Actually, the points M, N, P are the intersections of the lines CA, AB, BC with the antiparallels to AB, BC, CA through the symmedian point of triangle ABC. Hence, these points M, N, P lie on the second Lemoine circle of triangle ABC. Therefore, the circumcenter of triangle MNP is the center of this second Lemoine circle, i. e. the symmedian point of triangle ABC. Now, since we assumed that this symmedian point coincides with the centroid of triangle MNP, we conclude that the circumcenter of triangle MNP coincides with the centroid of triangle MNP. Hence, triangle MNP is equilateral, and triangle ABC, being similar to triangle MNP, must be equilateral, too. Theorem proven.

Ah yes, and how I showed that the points M, N, P are the intersections of the lines CA, AB, BC with the antiparallels to AB, BC, CA through the symmedian point of triangle ABC ?

Well, let K' be the circumcenter of triangle MNP, and let L be the point diametrically opposite to M on the circle MNP. Then, < MPL = 90, but we know that < MPB = 90, hence L lies on the line BC. It is very easy to show that the segment LM is antiparallel to AB; now, the point K' is the midpoint of this segment (since L and M are diametrically opposite points on the circle MNP, and K' is the center of the circle), and hence, K' lies on the C-symmedian of triangle ABC. Similarly, K' lies on the A-symmedian and on the B-symmedian, and consequently, K' is the symmedian point of triangle ABC. This not only entails that the point M is the intersection of the line CA with the antiparallel to AB through the symmedian point of triangle ABC (and similarly for N and P), but this also immediately implies that the circumcenter of triangle MNP is the symmedian point of triangle ABC, the only fact we really needed in our proof.

  Darij

Wow! It's so cool! :). I had a completely different approach:

I tok the point X to be the common point (X=K(ABC)=G(MNP)). X is projected in M', P' on NP, NM respectively, and it's projected in B', C' on AC, AB respectively. Triangles ABC and NPM are directly similar, X is K(ABC) and X=G(MNP)=the isogonal conjugate of K(MNP), so XP'/XM'=XC'/XB'=>XP'*XB'=XM'*XC', and since XB'MP' and XC'NM' are rectangles, it means that S(XMP')=S(XNM') (*). Remember that X is the centroid of MNP and P', M' are its projections on the sides. If we write all the relations analogous to (*) it's easy to derive the fact that MNP is equilateral.