Mobius function
induction
number theory
prime numbers
relatively prime
prime factorization
greatest common divisor

Let $ (a_{n})_{n\ge 1}$ be a sequence of positive integers satisfying $ (a_{m},a_{n}) = a_{(m,n)}$ (for all $ m,n\in N^ +$). Prove that for any $ n\in N^ + ,\prod_{d|n}{a_{d}^{\mu (\frac {n}{d})}}$ is an integer. where $ d|n$ denotes $ d$ take all positive divisors of $ n.$ Function $ \mu (n)$ is defined as follows: if $ n$ can be divided by square of certain prime number, then $ \mu (1) = 1;\mu (n) = 0$; if $ n$ can be expressed as product of $ k$ different prime numbers, then $ \mu (n) = ( - 1)^k.$

I think this problem is a consequence of an Iran TST's problem : exist an integer sequence $ {b_d}$ such that : $ a_n \equal{} \prod_{d|n}b_d$ and this problem follows from Mobius inverse function . More than ,we can find $ b_n$ in closed form as follow :
Let $ E_n$ be product of $ a_{\frac {n}{d}}$ with d is a square free divisor of n and has an even numbers of prime factors and $ O_n$ be the product of $ a_{\frac {n}{d}}$ with d is a square free divisor of n and has an odd numbers of prime factors then $ b_n \equal{} \frac {E_n}{O_n}$

here is my solution ,i think it is more simple than the offical one.

Dear sholly,
You prove that RHS of (*) $ \geq$0.
And since we are done.
I don't understand.
You can explain for me. Thank you.

We first prove that the sequence $ a_{n} $ is of the following form:
$i)$every $ a_{n} $ is of the form $ kb_{n} $ for all positive integers $ n $ and a certain positive integer $ k $. (this is obvious, because k is the greatest common divisor of all the terms of the sequence a)
$ii)$ $ b_{n} $ has the same set of exponents in its prime factorizations the number $n$, but the primes may be different. Moreover, like the sequence of positive integers is generated form the (infinite) set of prime numbers, the sequence $ b_{n} $ should also be generated from a certain set of prime numbers, all of which are relatively prime to the previously arbitrarily chosen positive integer $ k $.

Hope this part made sense. Now we prove this claim.

[hide]If $ d $ divides $ n $ then $ a_{d}=a_{(n,d)}=(a_{n},a_{d}) $, similarly if $ a_{d} $ divides $ a_{n} $ then $ d $ divides $ n $.
Therefore the number of divisors of $ n $ from the sequence of positive integers and number of divisors of $ a_{n} $ from the sequence $ a_{1}, a_{2},... $ have to be equal.

Let $ k=(a_{1}, a_{2},...)$ be the greatest common divisor of all terms of the sequence $ a_{1}, a_{2},... $ and let $ b_{1}, b_{2},... $ be the sequence obtained by $ kb_{n}= a_{n} $.  Each divisor of $ a_{n} $ from the set $ a_{1}, a_{2},... $ corresponds to a divisor of $ b_{n} $. Then the previous paragraph implies that the number of divisors of $ n $ and $ b_{n} $ have to be equal. This means that the sets of exponents of $ n $ which is $ x_{1}, x_{2},... $ and the set of exponents of $ b_{n} $ which is $ y_{1}, y_{2},... $ satisfy the equality:
$ (x_{1}+1) (x_{2}+1)...(x_{l}+1)=(y_{1}+1) (y_{2}+1)...(y_{j}+1) $
Moreover, the analogous equality has to be satisfied for all $ d $ and $ b_{d} $ where $ d $ divides $ n $ (in fact all terms of the sequence and their corresponding indices have this property). Therefore terms of the sequence with a prime number as an index are prime numbers themselves and $b_{1}=1 $.

We want to prove that the sets $ x_{1}, x_{2},... $ and $ y_{1}, y_{2},... $ are equal. We prove first that the biggest elements (WLOG $x_{1} $and $y_{1}$)  of both sets are equal. If they are not we can WLOG assume that $x_{1}$ is bigger than $y_{1}$ and consider the divisor which has $ x_{1}-1, x_{2},... $ in the exponents. This divisor has number of divisors $ (x_{1}) (x_{2}+1)...(x_{l}+1) $. The corresponding divisor (which is a divisor of the number with exponents in the prime factorization:  $ y_{1}, y_{2},... $) cannot have number of divisors equal to $ (x_{1}) (x_{2}+1)...(x_{l}+1) $. Because all those product are lowered to at most this number $ (y_{1}) (y_{2}+1)...(y_{l}+1) $ which is smaller than $ (x_{1}) (x_{2}+1)...(x_{l}+1) $. We could not increase any of the numbers in the product because that way we would not look at the exponents of a divisor. Therefore the biggest exponents in the prime factorization are always equal.

Consider $ b_{p^{m}}=\prod{p_{i}}^{z_{i}} $ where $p$ is prime. Then this number has $m+1 $ divisors. If $(z_{1}+1)(z_{2}+1)... =m+1$ and the biggest prime powers have to be equal we have that $ b_{p^{m}} ={p_{i}}^{m} $for a certain prime $p_{i}$.

Now suppose $ n $ is a composite number with its prime factorization with exponents $ x_{1}, x_{2},... $. Clearly for all divisors of d, the number $ b_{d} $ has to divide $ b_{n} $. If we take just all the prime power divisors we conclude that $ b_{n} $ is divisible by $ \prod {b_{d}} $
where the product runs over all the prime powers $p^{m}$ such that $m$ is the largest positive integer such that $p^{m}$ divides $n$ . if $ b_{n} $ is bigger than that product its number of divisors would be to large, therefore $ b_{n} =\prod {b_{d}} $.

ii) is proven.[/hide]

We need to prove that $\prod_{d|n}{a_{d}^{\mu (\frac{n}{d})} }$ is an integer. The case $n=1$ is obvious therefore we will prove the statement for $n>1$.
$\prod_{d|n}{a_{d}^{\mu (\frac{n}{d})} = \prod_{d|n}{k^{\mu (\frac{n}{d})}\prod_{d|n}{b_{d}^{\mu (\frac{n}{d})}={k^{\sum_{d|n}{\mu (\frac{n}{d}})}\prod_{d|n}{b_{d}^{\mu (\frac{n}{d})}=\prod_{d|n}{b_{d}^{\mu (\frac{n}{d})}}}}}}}$
because of the well-known identity $\sum_{d|n}{\mu (\frac{n}{d}})=0$ for $n>1$.

We can replace $b_{d}$ with $d$ in the product $\prod_{d|n}{b_{d}^{\mu (\frac{n}{d})}}$ without changing its value. This is so, because the product depends only on the exponents of the prime factorization of the numbers $b_{d}$ which is precisely of the same nature as that of the number $d$. This is that we proved above.
Therefore we have: $\prod_{d|n}{b_{d}^{\mu (\frac{n}{d})}=\prod_{d|n}{d^{\mu (\frac{n}{d})}}}$
We can switch the roles of $d$ and $\frac{n}{d}$ to obtain: $\prod_{d|n}{d^{\mu (\frac{n}{d})}=\prod_{d|n}{{(\frac{n}{d})}^{\mu (d)}=\prod_{d|n}{{n}^{\mu (d)}\prod_{d|n}{{d}^{-\mu (d)}=\prod_{d|n}{{d}^{-\mu (d)}}}}}}$ where the last equality holds because of the previously mentioned identity (the roles of $d$ and  $\frac{n}{d}$  are switched again here).

Clearly, it is enough to consider the square-free divisors of $n$. We will prove that this product is an integer with induction on the number of prime factors of $n$. The base of induction is when $n=p$ is prime. Then we have:
$\prod_{d|n}{{d}^{-\mu (d)}={p}^{-\mu (p)}{1}^{-\mu (1)}=p}$
Now suppose $n'=pn$ where $p$ and $n$ are relatively prime and $p$ is a prime number. The induction hypothesis states that $\prod_{d|n}{{d}^{-\mu (d)}=R}$ is an integer. We will prove that $\prod_{d|n'}{{d}^{-\mu (d)}}$ is an integer. As suggested above, we will only consider square-free divisors.

Indeed, we can partition the set of divisors of $n'$ into two sets of divisors, one set contains the divisors which are relatively prime to $p$ and the other set contains those divisible by $p$. The former ones are red, the latter ones are blue. Note that the product of the induction hypothesis runs over all red divisors of $n'$ and is equal to the positive integer $R$.

So, we have:  $\prod_{d|n'}{{d}^{-\mu (d)}=\prod_{blue}{{d}^{-\mu (d)}\prod_{red}{{d}^{-\mu (d)}=\prod_{red}{{(pd)}^{-\mu (pd)}\prod_{red}{{d}^{-\mu (d)}=\prod_{red}{{(pd)}^{-\mu (p)\mu (d)}\prod_{red}{{d}^{-\mu (d)}=\prod_{red}{{(pd)}^{\mu (d)}\prod_{red}{{d}^{-\mu (d)}}}}}}}}}}$
The last expression is equal to $\prod_{red}{p}^{\mu (d)}=\prod_{d|n}{p}^{\mu (d)}={p}^{\sum_{d|n}{\mu (d)}}=1$

Therefore our product is always an integer. QED

We prove that for every prime number $p$,
$$\nu_p(\prod\limits_{d|n}a_d^{\mu(\frac{n}{d})})\geq 0 \qquad(\star)$$

Let $\alpha(n)=\nu_p(a_n)$ for $n\geq 1$, then $(\star)$ is equivalent to
$$f(n):=\sum\limits_{d|n}\alpha(d)\mu(\frac{n}{d})\geq 0$$
Due to Mobius inversion formula, we know that
$$\alpha(n)=\sum\limits_{d|n}f(d)$$

For positive integers $d$ and $x$, define
$$S(d,x)=\sum\limits_{k|dx,k \nmid d}f(k)$$
[b]Claim[/b]: For any positive integer $d$ and coprime positive integers $x,y$, we have
$$\min\{S(d,x),S(d,y)\}=0$$
[i]Proof of the claim:[/i]
Apply $\gcd(a_{dx},a_{dy})=a_d$.$\blacksquare$

Now we have the following observation.
[b]Lemma:[/b]
If $S(1,l)=0$ for all $l\mid n$, then $f(n)=0$.
[i]Proof of the lemma:[/i]
Note that if $f(n)\neq 0$, then $f(h)\neq 0$ for some $h\mid n$, $1<h<n$. Apply induction to finish.$\blacksquare$

If $S(1,l)=0$ for all positive integer $l$, $f(n)=0$ would hold for any positive integer $n>1$. Hence we might assume that $S(1,l)$ is not always $0$. Suppose that $u$ is the smallest positive integer such that $S(1,u)>0$. Obviously $u>1$.
From the claim we know that $S(1,x)=0$ for all positive integer coprime to $u$, because $S(1,u)>0$. In addition, $S(1,l)=0$ for all positive divisor $l$ of $u$ other than $u$ itself due to the minimality of $u$, indicating that $f(l)=0$ for $l|n,1<l<n$.
Observe that for any positive divisor $h$ of $u$ smaller than $u$, $S(h,\frac{u}{h})=f(u)>0$, therefore for any integer $y$ coprime to $u$, $S(h,y)=0$.
We wish to show that $f(h\cdot y)=0$ for $h|u,h<u$ and $\gcd(y,u)=1$. Note that above already holds for $h=1$. Then just apply induction on $h$ and use the fact that $S(h,y)=0$.
We have so far shown that $$f(n)\neq 0 \quad \text{only if} \quad u\mid n$$
Apply the method above for $u\mathbb{Z}$, consider the behaviour of $S(u,x)$ and eliminate the terms $f(k)$ where $u\nmid k$. This would show that there exist a finite or infinite (possibly none) sequence of positive integers $j_0<j_1<\dots$ such that $j_{i-1}|j_i$ for $i>0$, $f(j_i)>0$ and $f$ vanishes at all other positive integer values. Hence $f$ is always nonnegative.


geometry
rectangle
circumcircle
geometry proposed

On sides $AC$ and $BC$ of acute-angled triangle $ABC$ rectangles with equal areas $ACPQ$ and $BKLC$ were built exterior. Prove that midpoint of $PL$, point $C$ and center of circumcircle are collinear.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=28928 .

  darij

Take such a point $X$ that quadrilateral $CLXP$ parallelogram. Then $\angle XLC=180^\circ-\angle PCL=\angle ACB$. Additionally equal areas $ACPQ$ and $BKLC$ imply $\frac{XL}{CL}=\frac{PC}{CL}=\frac{BC}{AC}$. So triangles $XCL,\ BCA$ are similar hence $\angle LCX=\angle CAB$. That's $\angle XCL+\angle OCB=90^\circ$ so $\angle XCO=180^\circ$

Let OC meets LP at N then it suffices to prove that $\frac{CL}{CP}$=$\frac{sin(PCN)}{sin(LCN)}$ but $\frac{CL}{CP}$=$\frac{AC}{BC}$ , LCN=A and PCN=B  so the pervious relation is true .


[quote="mai quoc thang"]a,b,c>0;abc=1.Prove that:
\[ \frac {a^{3}}{b^{3}} \plus{} \frac {b^3}{c^3} \plus{} \frac {c^3}{a^3} \plus{} \frac {(ab^{2} \plus{} bc^{2} \plus{} ca^{2})^{3}}{9} \geq \frac {1}{4}[(\frac {a}{b} \plus{} \frac {b}{c})^{3} \plus{} (\frac {b}{c} \plus{} \frac {c}{a})^{3} \plus{} (\frac {c}{a} \plus{} \frac {a}{b})^{3}]
\]
.
     :D:D:D:D:D[/quote]
Let $ x\equal{}\frac{a}{b},$ $ y\equal{}\frac{b}{c}$ and $ z\equal{}\frac{c}{a}.$
Thus, your inequality is equivalent to $ \sum_{sym}(11x^3\minus{}15x^2y\plus{}4xyz)\geq0,$ which true by Schur and Muirhead.

[quote="Sudoku"]
Assume $ x\geq y\geq z > 0$[/quote]

Can we do that?  :wink:

[quote="mai quoc thang"]$ a,b,c > 0;abc \equal{} 1$.Prove that:
\[ \frac {a^{3}}{b^{3}} \plus{} \frac {b^3}{c^3} \plus{} \frac {c^3}{a^3} \plus{} \frac {(ab^{2} \plus{} bc^{2} \plus{} ca^{2})^{3}}{9} \geq \frac {1}{4}[(\frac {a}{b} \plus{} \frac {b}{c})^{3} \plus{} (\frac {b}{c} \plus{} \frac {c}{a})^{3} \plus{} (\frac {c}{a} \plus{} \frac {a}{b})^{3}]
\][/quote]
Setting $ a/b\equal{}x;b/c\equal{}y;c/a\equal{}z$
The above inequality 's equivalent to
$ x^3\plus{}y^3\plus{}z^3\plus{}(1/9)(x\plus{}y\plus{}z)^3\geq (1/4)((x\plus{}y)^3\plus{}(y\plus{}z)^3\plus{}(z\plus{}x)^3)$ (symmetric)
We can assume $ x\geq y\geq z$ :)


Hold it there, this problem is not middle school level BASICS. Perhaps the mods move this to appropriate section as the poster seems to be a new user. By the way mitix,  :welcomeani:

:welcome:  Limits is too hard for the basics forum but dont worry new members do it all the time. A moderator will move it.

Did someone say moderater? Not to fear it's SUPERMOD!  :D

Yeah, like they said this is way too hard for this forum.

I'm moving it to [url=http://www.artofproblemsolving.com/Forum/index.php?f=150]Intermediate Forum[/url].

[quote="paladin8"][hide]
We have $\frac{1}{a_ka_{k+1}} = \frac{1}{(a_1+(k-1)r)(a_1+kr)} = \frac{1}{ra_k}-\frac{1}{ra_{k+1}}$ by the partial fraction decomposition. So we know it telescopes.

Thus $b_n = \frac{1}{ra_1}-\frac{1}{ra_{n+1}}$

So $\lim_{n \rightarrow \infty}{\frac{b_n}{a_1}} = \frac{\frac{1}{ra_1}}{a_1} = \frac{1}{ra_1^2}$? Is that right...[/hide][/quote]

Yes, u r right !!!


[quote="MCrawford"]The consecutive angles of a trapezoid form an arithmetic sequence.  If the smallest angle is $75^\circ$, then the largest angle is
\[ \text{(A)} 95^\circ\ \ \ \text{(B)} 100^\circ\ \ \ \text{(C)} 105^\circ\ \ \ \text{(D)} 110^\circ\ \ \ \text{(E)} 115^\circ \][/quote]

[hide]the angles added is 360
the average of the angles should be 90
90-75=15       15+90=[b]105[/b]
C[/hide]

[hide]
The average size of the angle would be 90 degrees.
So the two angles that are in the middle of the consecutive numbers add up to 180 degrees, and so do the outer ones. Which gives you this equation:
$x+75=180$
$x=105$
So, the answer is $C$[/hide]


If $x_{k+1}=\sqrt{x_k+\sqrt{x_k+\sqrt{x_k+\sqrt{x_k...}}}}$
for natural number $k$
What is $x_{\infty}$

[hide]x of infinity = x of infinity + 1

so x = square root x + square root of x +...

x^2 - x = x

x = 0 or 2

You need to specify what the original value of x is (like, what x1 is) [/hide]

PS: How do I make it look fancy like the original post?

For yours it would look like
[code]x_{\infty}=x_{\infty+1}

so x=\sqrt{x+\sqrt{x+\sqrt{x+...}}}

x^2-x=x

x=0 or x=2[/code]

Then you put $---$s around the equations
Which would look like:
$x_{\infty}=x_{\infty+1}$
so $x=\sqrt{x+\sqrt{x+\sqrt{x+...}}}$
$x^2-x=x$
$x=0$ or $x=2$


4. In the game of primes, players roll a fair cube whose faces are labeled 2, 3, 5, 7, 11, and 13. What is the smallest integer sum greater than 2 that cannot be made by rolling the die once, twice or three times?

[hide="Strategy and Answer"]30
We know that everything under, say 20, is likely doable. Now note that all primes are odd except for 2. So let's look for an even number that is just too high to be written with 2 odd primes and a 2. That number is the first even number greater than 13+13+2=28, or [b]30[/b].[/hide]


Let $ a$, $ b$, $ c$ three positive reals. Prove that

\[ 12(\sum{\frac{a^{3}+abc}{b+c})\geq\left(\sum_{cyc}\left(\frac{a^{2}}{b+c}+\frac{3a}{2}\right)\right)^{2}}\]

I think that the inequality is false, if you put a=25, b=c=1, the inequality is on the reverse side.


Let $p_{1},\ldots,p_{k}$ be distinct primes. Prove that if $x^{2}\equiv 0 \pmod{p_{1}\ldots p_{k}}$ then $x^{2}\equiv 0 \pmod{(p_{1}\ldots p_{k})^{2}}$

if p|x^2, then p|x [blah consider prime factorization...common, trivial lemma], so p^2|x^2...then crt over all primes

Generally, if $p | ab$ then $p | a \vee p | b$.  In other words, $\mathbb{Z}_{p}$ is an integral domain.


If $ a,b,c>0$ and $ abc\equal{}1$, then
$ \prod(a\plus{}\sqrt{b}\plus{}\sqrt[3]{c})\geq\frac{27}{8}$

[quote="lipako"]If $ a,b,c > 0$ and $ abc \equal{} 1$, then
$ \prod(a \plus{} \sqrt {b} \plus{} \sqrt [3]{c})\geq\frac {27}{8}$[/quote]

$ \prod(a \plus{} \sqrt {b} \plus{} \sqrt [3]{c}) \ge 27$- by holder inequality  :)


superior algebra
superior algebra unsolved

If p is a prime number, show that in S_p there are (p-1)!+1 elements x satisfying x^p = e.
----

I know that in S_p there are p! elements, and by Lagrange, x^(p!)=e

Now where do I go?

it is obvious that we have $ (p\minus{}1)!$ $ p$-cycles, each of one has order $ p$, but $ (e)^{p} \equal{} e$ so it depends if your cuestion is about $ (p\minus{}1)!\plus{}1$ elements of order $ p$ or only the condition for such numbers is that given $ a$, $ a^{p} \equal{} e$.

[quote="alejnaser"]it is obvious that we have $ (p \minus{} 1)!$ $ p$-cycles, each of one has order $ p$, but $ (e)^{p} \equal{} e$ so it depends if your cuestion is about $ (p \minus{} 1)! \plus{} 1$ elements of order $ p$ or only the condition for such numbers is that given $ a$, $ a^{p} \equal{} e$.[/quote]

Yes it is for the element x such that x^p=e.

But then we have $ (p \minus{} 1)! \plus{} 1$ of such elements, the $ (p \minus{} 1)!$ $ p$-cycles and $ e$, I mean, it is clear that $ e$ is not one of that $ p$-cycles. Am I missing something? I never used that $ p$ is prime. I think we have to find elements of order $ p$.

You did use that $ p$ is prime--if $ p\equal{}ab$ is not prime, then a product of an $ a$-cycle with a disjoint $ b$-cycle will satisfy $ x^p\equal{}e$.

Oh! Sure....Thanks, and then we are done  :D

[quote="LydianRain"]You did use that $ p$ is prime--if $ p \equal{} ab$ is not prime, then a product of an $ a$-cycle with a disjoint $ b$-cycle will satisfy $ x^p \equal{} e$.[/quote]

Can you explain this reasoning more?

Let $ p\equal{}ab$, $ \sigma$ be an $ a$-cycle and $ \tau$ a $ b$-cycle (disjoints), then $ (\sigma\tau)^{p} \equal{} \sigma^{p}\tau^{p}$ since disjoint cycles commutes, but then $ \sigma^{p}\tau^{p} \equal{} (\sigma^{a})^{b}(\tau^{b})^{a}$. But $ \sigma^{a} \equal{} \tau^{b} \equal{} e$ and then we have $ (\sigma\tau)^{p} \equal{} e$.

[quote="alejnaser"]Let $ p \equal{} ab$, $ \sigma$ be an $ a$-cycle and $ \tau$ a $ b$-cycle (disjoints), then $ (\sigma\tau)^{p} \equal{} \sigma^{p}\tau^{p}$ since disjoint cycles commutes, but then $ \sigma^{p}\tau^{p} \equal{} (\sigma^{a})^{b}(\tau^{b})^{a}$. But $ \sigma^{a} \equal{} \tau^{b} \equal{} e$ and then we have $ (\sigma\tau)^{p} \equal{} e$.[/quote]

Oh i understand the proof now.  But why does p have to be a prime anyway?  You just showed that x^p=e even for some p=ab with p not prime.

[quote="alejnaser"]it is obvious that we have $ (p \minus{} 1)!$ $ p$-cycles, each of one has order $ p$, but $ (e)^{p} \equal{} e$ so it depends if your cuestion is about $ (p \minus{} 1)! \plus{} 1$ elements of order $ p$ or only the condition for such numbers is that given $ a$, $ a^{p} \equal{} e$.[/quote]

Why do we have (p-1)! p-cycles, and not p! p-cycles?

For example, $ (123) \equal{} (231) \equal{} (312)$, so we have to divide by $ p$ to eliminate equals $ p$-ciclos.

How do you know you divide by p?

Also, why does p have to be a prime anyway? Your proof seems to just show that x^p=e even for some p=ab with p not prime.

Sorry, I am new to all this stuff...

$ p$ has to be prime because otherwise, the answer is [i]greater[/i] than $ (p\minus{}1)! \plus{} 1$.  What we're saying is that you can get that many elements, no matter what, but only when $ p$ is prime are we guaranteed that we've got all of them.  Is it clear?

As for why we divide by $ p$, alejnaser has shown the general situation pretty well.  When $ p\equal{}2$, the only $ 2$-cycles are $ (12)$ and $ (21)$.  But these are the same--they're both just the permutation that swaps $ 1$ and $ 2$.  Similarly, with $ p\equal{}3$, it looks like we have $ 6\equal{}3!$ different cycles, but each permutation can be written in $ 3$ different ways, so we have to divide by $ 3$.

This is a good general hint for problems like this: if you are having trouble understanding them in the abstract, work out small cases in detail until they make sense.  Make [i]sure[/i] you understand why when $ p\equal{}2$, there are $ (2\minus{}1)!\plus{}1\equal{}2$ solutions to $ a^p\equal{}e$, and then move on to $ p\equal{}3$, convincing yourself that there are $ (3\minus{}1)!\plus{}1\equal{}3$ solutions.  Then try $ p\equal{}4$, and see why the right answer is different than $ (4\minus{}1)!\plus{}1$. ($ 4$ is not prime)  By the time you get to $ p\equal{}5$, hopefully you will understand what we are saying...


Determine if there exists a positive integer n such that n has exactly 2000 prime divisors and 2n + 1
is divisible by n.

I think he means 2^n + 1 instead of 2n+1. If I'm true, then your problems turns over a IMO 2000 problem. See here:http://www.mathlinks.ro/viewtopic.php?p=354115#354115


algorithm
articles
number theory
Euclidean algorithm

How do you solve this kind of congruence aside from brute force?
$ 34x \equiv 60$ (mod m).

This is equivalent to

$ 34x \plus{} my \equal{} 60$.

Use [url=http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm]the Extended Euclidean algorithm[/url].

What part of the wiki article is unclear?  The computations are straightforward, although they can look intimidating if you don't understand their motivation, which I'll clarify:

What we're really after right now is a solution to the linear Diophantine $ 17x \plus{} 49y \equal{} 30$.  We want to take this solution $ \bmod 49$ to figure out the answer to the original question.  Instead, let's take it $ \bmod 17$:

$ \minus{} 2y \equiv \minus{} 4 \bmod 17 \Leftrightarrow y \equiv 2 \bmod 17$.

Let's substitute $ y \equal{} 17y' \plus{} 2$.  Then

$ 17x \plus{} 49 \cdot 17y' \plus{} 98 \equal{} 30 \Leftrightarrow$
$ 17(x \plus{} 49y') \equal{} \minus{} 68 \Leftrightarrow$
$ x \plus{} 49y' \equal{} \minus{} 4$.

To find a particular solution, simply set $ y' \equal{} \minus{} 1$ and we obtain $ x \equal{} \boxed{45}$.

The Extended Euclidean Algorithm generalizes this idea; we repeatedly rewrite a modular equation in terms of a linear Diophantine, reduce via a modulus to obtain another modular equation which we rewrite as another linear Diophantine, and so forth until we obtain a solution that we can plug back in to calculate a solution to the original linear Diophantine (or modular equation).


geometry
3D geometry
quadratics
ratio
trigonometry
algebra
quadratic formula

Let $ r$ be a positive real number such that \[ \sqrt[4]{r}\minus{}\frac{1}{\sqrt[4]{r}}\equal{}14\] Prove that \[ \sqrt[6]{r}\plus{}\frac{1}{\sqrt[6]{r}}\equal{}6\]

Square both sides of the equation that is given to arrive at
$ r^{1/2} \plus{} r^{ \minus{} 1/2} \equal{} 198$.
Set $ r^{1/6} \plus{} r^{ \minus{} 1/6}$ equal to $ x$. We will prove that if $ r$ is a positive real number, then $ x$ must be $ 6$. Cube the equation $ r^{1/6} \plus{} r^{ \minus{} 1/6} \equal{} x$ to get
$ r^{1/2} \plus{} r^{ \minus{} 1/2} \plus{} 3r^{1/6}r^{ \minus{} 1/6}(r^{1/6} \plus{} r^{ \minus{} 1/6}) \equal{} x^3$
The expression $ r^{1/6}r^{ \minus{} 1/6}$ is equal to $ 1$. $ r^{1/6} \plus{} r^{ \minus{} 1/6}$ is equal to $ x$. Our first result was $ r^{1/2} \plus{} r^{ \minus{} 1/2} \equal{} 198$. So
$ 198 \plus{} 3x \equal{} x^3$
$ 0 \equal{} x^3 \minus{} 3x \minus{} 198$
It is easy to verify that the only rational root of this cubic, which must be an integer divisor of $ 198$, is $ 6$. All other roots of the cubic must also be roots to the quadratic
$ \frac {x^3 \minus{} 3x \minus{} 198}{x \minus{} 6} \equal{} x^2 \plus{} 6x \plus{} 33$
However the discriminant of this quadratic is $ \minus{} 96$, thus both roots of the quadratic are imaginary. However, it was given that $ r$ is real, so $ r^{1/6} \plus{} r^{ \minus{} 1/6}$ must be real also. Thus $ x$ cannot be imaginary, and the only value of $ x$ that satisfies the constraints is $ 6$.

Help I'm stuck

Let $ x\equal{}\sqrt[12]{r}$. Then $ x^3\minus{}\frac{1}{x^3} \equal{} 14$. Multiply everything by $ x^3$ to make a "quadratic":
\[ x^6 \minus{} 14x^3 \minus{} 1 \equal{} 0\]

Using the quadratic formula, we see that $ x^3\equal{}\frac{14\pm10\sqrt{2}}{2}\equal{}7\pm5\sqrt{2}$

Now what? Is there a way to take roots of that radical?

I don't think you're getting anywhere by defining $ x$. Although, since only the positive root of the quadratic is possible, you could try proving

$ (7 \plus{} 5\sqrt {2})^{2/3} \plus{} \frac {1}{(7 \plus{} 5\sqrt {2})^{2/3}} \equal{} 6$, but the cubic root involved looks very hard to deal with...

Edit: What I really mean is that you could probably solve the quadratic for $ \sqrt[4]{r}$, and eliminate the definition. But I'm still not sure if your way leads anywhere.

Note that we need $ x > 1$ for the first condition to be satisfied, so we must take the positive root.

Let $ a_n \equal{} x^n \minus{} \frac {1}{x^n}$.  We are given that $ a_3 \equal{} 14$ and we want to prove that $ \frac {a_4}{a_2} \equal{} 6$.  But we know that $ a_0 \equal{} 0$ and that

$ a_{n \plus{} 1} \equal{} \left( x \plus{} \frac {1}{x} \right) a_n \minus{} a_{n \minus{} 1} \equal{} \frac {a_2}{a_1} a_n \minus{} a_{n \minus{} 1}$.

This gives

$ a_3 \equal{} \frac {a_2}{a_1} a_2 \minus{} a_1 \equal{} 14 \implies$
$ a_4 \equal{} \frac {a_2}{a_1} a_3 \minus{} a_2 \implies$
$ \frac {a_4}{a_2} \equal{} \frac {a_3}{a_1} \minus{} 1 \equal{} \frac {14}{a_1} \minus{} 1$.

It therefore suffices to prove that $ a_1 \equal{} 2$.  Note that $ x \minus{} \frac {1}{x} \equal{} a_1 \implies x \plus{} \frac {1}{x} \equal{} \sqrt {a_1^2 \plus{} 4} \implies a_2 \equal{} a_1 \sqrt {a_1^2 \plus{} 4}$.  The condition therefore gives

$ a_3 \equal{} \frac {a_2}{a_1} a_2 \minus{} a_1 \equal{} a_1 (a_1^2 \plus{} 4) \minus{} a_1 \equal{} a_1^3 \plus{} 3a_1 \equal{} 14$.  The only positive root of this cubic is $ a_1 \equal{} 2$.  (This gives $ x \equal{} 1 \plus{} \sqrt {2}$, which is indeed the cube root of $ 7 \plus{} 5 \sqrt {2}$.)  The key point here is that we don't need to calculate $ x$ to calculate values of $ a_n$.

@distracted523: I doubt that can be simplified...If you do the cubic formula or whatever you get something like that anyway.

@t0rajir0u: nice...I was trying to find some way of condensing the problem into that type of recurrence, but I couldn't figure how to deal with the + vs -. Considering the ratio $ \frac{a_{2n}}{a_n}$ where $ a_n\equal{}x^n\minus{}x^{\minus{}n}$ is smart.

This has a nice interpretation in terms of trig identities... $ \frac{\sin 2x}{\sin x}\equal{}2\cos x$, etc.

Indeed.  The identity $ x \plus{} \frac {1}{x} \equal{} \sqrt {\left( x \minus{} \frac {1}{x} \right)^2 \plus{} 4}$ is essentially a hyperbolic trigonometric identity.  (On further inspection, the identity $ a_3 \equal{} a_1^3 \plus{} 3a_1$ requires much less work than I put in, but I was working for generality.)


Prove that for a,b,c>0 we have sum a(2a+3b+3c)/(4a^2+3(b+c)^2)<=3/2.

Is this so ugly? Of course! But maybe it has an interesting solution. Or maybe not!

A marvellous and very hard inequality.
I would like to see your  solution, Harazi.

Thank you very much.

All we need is a little care with computations and some luck to apply Cauchy and Schur:
     First of all, the inequality can be written sum [(b+c)^2-2a(b+c)]/[3(b+c)^2+4a^2]>=0, or sum (3b+3c-2a)^2/[3(b+c)^2+4a^2]>=3.
     But the last sum is by Cauchy at least (3*sum (b+c)/a-6)^2/[12+3*sum((c+b)/a)^2]. Thus, it is enough to prove that the last quantity is at least 3. After expanding everything, it becomes sum (b+c)*(c+a)/ab>=2 sum (b+c)/a, or sum a^3+3abc>=sum ab(a+b) , which is true by Schur.


Prove that for each positive integer n, the number of ordered pairs (x;y) of integers satisfying: $\large x^2-xy+y^2=n$ is finite and divisible by 6

another solution
we have,if (x,y) is solution then:
$n=x^2-xy+y^2=\frac{1}{2}(x+y)^2+\frac{1}{2}(x^2+y^2)\geq \frac{1}{2}(x^2+y^2)$
There only finite many ordered pairs (x,y)do so.Hence,there are finitely many solutions
Otherwise,if(x,y) is solution then${(x,y),(y,y-x),(y-x,-x),(-x,-y),(-y,x-y),(x-y,x)}$is solution.Hence,the number of ordered pairs(x,y)divisible by 6


probability
counting
distinguishability
AMC
AIME
floor function
function

[b]Problem[/b] A fair die is rolled four times. Find the probability that each number is no smaller than the preceding number.

Why is this problem equivalent to counting the ways to put four balls into six distinguishable boxes?
Why can we think of the counting problem as counting the ways to arrange 4 balls and 5 dividers to separate the six boxes?

How would you phrase the solution in a balls and urns fashion?

See [[2001 AIME I Problems/Problem 6]].

Note that given any set of four numbers $ \{a,b,c,d\}$, there is exactly one way to permute them into a non-decreasing order. Thus, the number of ways to roll four dice in a non-decreasing order is just the number of sets $ \{a,b,c,d\}$ with $ 1 \le a,b,c,d \le 6$.

How many such sets are there? Consider 6 containers, which we will label,
\[ \lfloor 1 \rfloor\ \lfloor 2 \rfloor\ \lfloor 3 \rfloor\ \lfloor 4 \rfloor\ \lfloor 5 \rfloor\ \lfloor 6 \rfloor
\]With each dice roll, we drop a "ball" into the respective container. Eg, with the rolls $ \{1,3,3,6\}$, we would have
\[ \lfloor \circ \rfloor\ \lfloor\ \rfloor\ \lfloor \circ\,\circ \rfloor\ \lfloor\ \rfloor\ \lfloor\ \rfloor\ \lfloor \circ \rfloor
\]We can now replace the containers with just their boundaries, like
\[ \circ\, |\,|\circ\circ\, |\, |\, |\, \circ
\]There are $ 5$ "dividers" and $ 4$ "balls", and so $ \frac {(5 \plus{} 4)!}{5!4!} \equal{} 126$ ways to do this.

The probability of course is $ \frac {126}{6^4} \equal{} \frac {7}{72}$.

Whoa! Cool diagram but unfortunately, I'm pretty dumb and don't understand how to replace the containers with the boundaries.

$ \circ\, |\,|\circ\circ\, |\, |\, |\, \circ$
Are the first two boundaries to the right of the first circle on the left from the the two gaps between the three urns to the right? Does a gap between urns (that is, each urn consisting of both the left and right "L's" of the floor function)?

How can we use the idea of "dropping a ball into containers" for other counting problems?

Ah, sorry, I should've been clearer about the boundaries thing. I've boxed the nine items in particular that we are shuffling around; since we cannot place a ball before the first container or the last container, we can just ignore the outermost boundaries.

\[ \lfloor \boxed{\circ} \boxed{\rfloor\ \lfloor}\ \boxed{\rfloor\ \lfloor} \boxed{\circ}\,\boxed{\circ} \boxed{\rfloor\ \lfloor}\ \boxed{\rfloor\ \lfloor}\ \boxed{\rfloor\ \lfloor}\boxed{\circ} \rfloor \Longrightarrow \circ\, |\,|\circ\circ\, |\, |\, |\, \circ\]

"Balls and urns" can be used for many counting problems, though it often takes a bit of time to recognize how to use the argument. I would suggest searching for such either on the wiki or the forums and taking a look at a couple of problems.

[quote="aufha"]How can we use the idea of "dropping a ball into containers" for other counting problems?[/quote]
The canonical example is this kind of problem: How many solutions are there to $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 10$ where $ a,b,c,d,e$ are nonnegative integers?

[quote="LightEntropy"]Trying to understand this, this is a first time explanation for me.
So you can choose any example that would fit under the given conditions in order to find all such examples that fit the given conditions? If so why does it work that way?[/quote]
What you said doesn't sound right, and I am not sure what you're referring to that needs clarification.  :(  Can you elaborate?


im in AP chemistry, and i feel listless and unchallenged, what can I learn independantly? is organic really the only "good" path or are there other introductory college level subjects that can arouse interest?

I would recommend trying out some USNCO problems (just google USNCO if you don't know where to find them).  The local are written ~ AP level, while I believe the national is much harder than AP.  If you find some challenges here, perhaps you could take the material you don't know from those exams and study that.

I'm sure there are plenty of other things you could do as well.  There's nothing wrong with organic--a lot of people love it.  Really though, you should be able to study out of a text of any of the main branches of chemistry (minus biochem) without having a real orgo experience.

Whatever you ultimately decide, good luck!


search
complex numbers
algebra unsolved
algebra

Consider an integer $n > 1$ and a disc $D$ in the complex plane.
Prove that for any complex numbers $z_1, z_2, \cdots , z_n \in D$, there exists $z \in D$ such that

$z^n = z_1z_2 \cdots z_n$.

BOmb

Edited by pbornsztein to make it readable (come on, it's time to learn tex basis!)

It is from Romania TST 2004 :first selection test. isn't it? :o

Armanf is right, it is Romanian selection tst 2004. Here is the link :
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5450[/url]
Try to search for the problems berfore posting them  ;)


number theory
relatively prime
number theory unsolved

Prove that if $ n>1$ is a integer such that $ n|3^n\plus{}4^n$ then $ 7|n$.

[hide]
Let $ p$ be the smallest prime divisor of $ n$. Note that $ p\ne 2$, since $ 3^n\plus{}4^n$ is odd. Therefore $ p$ is an odd prime.

$ 3^n\equiv \minus{}4^n\mod p$.

Let $ g$ be a generator mod $ p$, and say $ g^x\equiv 3$ and $ g^y\equiv 4$. Also note that $ g^{\frac{p\minus{}1}{2}}\equiv \minus{}1\mod p$. Therefore

$ g^{xn}\equiv g^{\frac{p\minus{}1}{2}\plus{}yn}\mod p$

$ xn\equiv \frac{p\minus{}1}{2}\plus{}yn\mod p\minus{}1$

Now because $ p$ is the smallest prime divisor of $ n$, we must have $ \gcd(p\minus{}1,n)\equal{}1$, since any prime dividing $ p\minus{}1$ must be less than $ p$. Thus, we can divide the $ n$ out and get

$ x\equiv \frac{p\minus{}1}{2}n^{\minus{}1}\plus{}y\mod p\minus{}1$.
$ x\equiv \frac{p\minus{}1}{2}\plus{}y \mod p\minus{}1$

Therefore we must have

$ g^x\equiv \minus{}g^{y}\mod p$
$ 3\equiv \minus{}4\mod p$
$ 7\equiv 0\mod p$

Hence $ p\equal{}7$, so $ 7|n$.
[/hide]

[quote="ishfaq420haque"]what is generator mod???please help[/quote]
It is a primitive root mod n, a number that if you multiply it by itself enough times, you can get any element modulo n (more precisely: any element of the multiplicative group modulo n).
g is called a primitive root of $ Z/pZ$ iff $ \displaystyle (\mathbb{Z}/p\mathbb{Z})^\times \equal{}\{1,2,3,\ldots,p\minus{}1\}\equal{} \{ 1, g, g^2,\ldots,g^{p\minus{}2}\}.$
I copied the definition from [url]http://planetmath.org/encyclopedia/EveryPrimeHasAPrimitiveRoot.html[/url] - a great article, and there's more information on wikipedia.

@tjhance: beautiful solution.

i need some help tjhance
1. How can the existence of a generator mod p be proved?
2. I dont understand why you assumed the inverse of n modulo p-1 to be 1

1. bambaman posted a link to such a proof.

2. $ n$ and $ n^{ \minus{} 1}$ are both relatively prime to $ p \minus{} 1$, so they are both odd. If $ n^{ \minus{} 1} \equal{} 2k \plus{} 1$, then $ \frac {p \minus{} 1}{2}n^{ \minus{} 1}\equiv \frac {p \minus{} 1}{2}(2k \plus{} 1)\equiv p \minus{} 1 \plus{} \frac {p \minus{} 1}{2}\equiv \frac {p \minus{} 1}{2}\mod p \minus{} 1$.

[quote="jbmorgan"]Prove that if $ n > 1$ is a integer such that $ n|3^n + 4^n$ then $ 7|n$.[/quote]
Let $ p$ be the smallest prime divisor of $ n$
$ p|n|3^n + 4^n$ so $ (p,3) = (p,4) = 1$
$ \exists a$ such that $ 4a\equiv 1 (mod\, p)$
$ p|3^n + 4^n|(3a)^n + (4a)^n$ so $ p|(3a)^n + 1|(3a)^{2n} - 1$
Let $ h = ord_{3a}(p)$ so $ h|(2n,p - 1) = 2$
1.$ h = 1$ then $ 3a\equiv 1(mod\, p)$ so $ 3.(4a)\equiv 4(mod\, p)$ so $ 4\equiv 3(mod\, p)$
so $ p = 1$ not satisfy
2.$ h = 2$ then $ (3a)^2\equiv 1(mod\, p)$ so $ 9.(4a)^2\equiv 16(mod\, p)$ so $ 7\equiv 0(mod\, p)$
so $ p = 7$ as desire!


Solve the rational inequality without a graphics calculator.
$\frac{x-2}{2(x-3)}>\frac{x}{x+3}$

[hide="Answer"]$\frac{x-2}{2(x-3)}>\frac{x}{x+3} \Rightarrow (x-2)(x+3)>2x(x-3)$
$x^2+x-6>2x^2-6x \Rightarrow 0>x^2-7x+6 \Rightarrow 0>(x-1)(x-6)$

From here, we see that both numbers will be negative in the interval $(-\infty,1)$, resulting in an answer greater than $0$, so the answer is $1<x<6$.[/hide]

[quote="JesusFreak197"]the answer is $1<x<6$.[/quote]

The interval you gave has a serious flaw to the problem. Can you explain why????

[hide="hint"]
look at the fractions
[/hide]

Whoa... that's odd. What's wrong with my solution, other than that the answer doesn't work for the whole interval?

[quote="peeta"][hide="answer"]$3> x>6$
peeta[/hide][/quote] :rotfl:  :rotfl:  :rotfl: keke peeta, I like your answer  :rotfl:  :rotfl:  :rotfl:

You must remember that when you multiply by a negative number on both sides, the inequality sign flips.  Thus, you cannot multiply like you did, unless you broke it down into cases where one was positive and the other is negative, or both negative, and stuff like that.

Ohh, right. It didn't look negative, but it would turn out that way for $1<x<3$, which was the part of the interval that didn't work, and when $x=3$ it's obviously undefined. :P


In any triangle  ABC  show that
[tex] 2p(\Sigma\sqrt(p-a)) \leq 3(\Sigma\sqrt(bc(p-a)) [/tex]

[quote=nickolas]In any triangle  ABC  show that
$$2p\sum_{cyc}\sqrt{p-a}\leq 3\sum_{cyc}\sqrt{bc(p-a)} $$[/quote]
Let $a=y^2+z^2$, $b=x^2+z^2$ and $c=x^2+y^2,$ where $x$, $y$ and $z$ are positives.
Thus, we need to prove that
$$3\sum_{cyc}x\sqrt{(x^2+y^2)(x^2+z^2)}\geq2(x^2+y^2+z^2)(x+y+z)$$ or after squaring of the both sides $$\sum_{cyc}(5x^6-8x^5y-8x^5z-3x^4y^2-3x^4z^2-16x^3y^3-8x^4yz-16x^3y^2-16x^3z^2y+x^2y^2z^2)+$$
$$+18\sum_{cyc}xy(x^2+y^2)\sqrt{(x^2+z^2)(y^2+z^2)}\geq0$$ and after using C-S it's enough to prove that

 $$\sum_{cyc}(5x^6-8x^5y-8x^5z-3x^4y^2-3x^4z^2-16x^3y^3-8x^4yz-16x^3y^2-16x^3z^2y+x^2y^2z^2)+$$
$$+18\sum_{cyc}xy(x^2+y^2)(xy+z^2)\geq0$$ or
$$\sum_{cyc}(5x^6-8x^5y-8x^5z+15x^4y^2+15x^4z^2-16x^3y^3-8x^4yz+2x^3y^2z+2x^3z^2y+x^2y^2z^2)\geq0,$$
which is $f(w^3)\geq0,$ where $f$ is a concave function.
Now, $y=z=1$ gives $$(x-1)^4(5x^2+4x+8)\geq0,$$ which says that $f(w^3)\geq0$ is true even for all reals $x$, $y$ and $z$.
Done!

In triangle,prove

\[\sqrt { \left( s-b \right)  \left( s-c \right) a}+\sqrt { \left( s-c
 \right)  \left( s-a \right) b}+\sqrt { \left( s-a \right)  \left( s-b
 \right) c}\geq 3\,{3}^{3/4}\sqrt {2}{r}^{3/2}\]

[quote=arqady][quote=xzlbq][quote=Dr Sonnhard Graubner]hello, this here
$2p(\Sigma\sqrt{(p-a)}) \leq 3(\Sigma\sqrt{bc(p-a)}) $?
Sonnhard.[/quote]

this is sqing' inequality

see:https://artofproblemsolving.com/community/c6t243f6h1738187_simple_inequality
2#[/quote]
Are you sure? See please the starting post in this topic.  ;)[/quote]

I am bad

[quote=xzlbq]In triangle,prove

\[\frac{1}{3}\,\sqrt {2}s \left( \sqrt {s-a}+\sqrt {s-b}+\sqrt {s-c} \right) \geq
\sqrt { \left( s-b \right)  \left( s-c \right) a}+\sqrt { \left( s-c
 \right)  \left( s-a \right) b}+\sqrt { \left( s-a \right)  \left( s-b
 \right) c}\][/quote]

as use arqady subs:

Let $a=y^2+z^2$, $b=x^2+z^2$ and $c=x^2+y^2,$ where $x$, $y$ and $z$ are positives.
Thus, we need to prove that

\[\frac{1}{3}\,\sqrt {2} \left( {x}^{2}+{y}^{2}+{z}^{2} \right)  \left( x+y+z
 \right) \geq yz\sqrt {{y}^{2}+{z}^{2}}+zx\sqrt {{x}^{2}+{z}^{2}}+xy\sqrt
{{x}^{2}+{y}^{2}}\]


vector
LaTeX
combinatorics unsolved
combinatorics

Let (c_1,c_2, ..., c_n) be in [b]Z+[/b]^n, with all c_j < k+1.  Let S be a set such that no more than m elements (c_1, ..., c_n) can have the same c_1, ..., c_j-1, [not c_j], c_j+1, ..., c_n; for some m < k.  What is the maximum number of unique elements of S?  Prove it.

Alex


Does anyone know of any good, safe sites that have applications for the TI-89 Titanium(other than the TI site)?

Well, the TI website archive is the best by far, but the next best one that I've found is [url]http://www.calcgames.org/[/url].  Contrary to it's name, it doesn't just have games, but, unfortunately, it doesn't have many math-related programs.

Actually, ticalc.org is the "ti site archive" that I referred to.  Sorry.


My physics class divided the topics of electric field/magnetic field and electromagnetic waves into two disjoint topics (for obvious reasons), but we never tried to connect the two. Can anyone help provide some sort of explaination about how the two are connected?

My postulate was that the wave creates the fields, but I can't see how this could be correct. Say you have a source creating a E/B-field. The wave emanating off of this source doesn't decrease in magnitude (according to the wave equation; this may be where my misunderstanding comes in), but the fields decrease in magnitude of $\frac{1}{r^{2}}$ as you get a distance $r$ away.

This may be more advanced now, but we also seem to talk about the direction of the wave in the sense that it has a single direction, but if a source is creating an electromagnetic wave it seems like it shouldn't just have 1-direction, or even two in the sense of radiating out in a circle, but it should have three dimensions radiating spherically outward.

Final exam in one week. Nothing of this sort will probably be asked, but I would like to know for curiosity sake and just in case some sort of conceptual question is asked about it.

Thanks.

Electromagnetic waves don't behave exactly like ripples on water or sound in the air. The really are oscillating, sinusoidal, perpendicular electric and magnetic fields: they are not oscillating circles or spheres. When a source emits many of them in all directions, we experience an intensity decrease with distance because there are fewer individual waves per unit area.


[color=darkblue][u]Here is the first my topic. Please, give me a little  indulgence ![/u]

Let $A,X,C,Y,B$ be five collinear points in this order so that

the divisions $(A,C,X,B)$ and $(B,C,Y,A)$ are harmonically, i.e.

$\frac{XA}{XC}=\frac{BA}{BC}$ and $\frac{YB}{YC}=\frac{AB}{AC}$. Prove that $XY\le \frac{1}{3}\cdot AB$.

We'll have the equality if and only if $CA=CB$.

[b]P.S.[/b] Thank you very much, [i]Mr. Virgil Nicula[/i] ![/color]

nice!

write $\frac{XA}{XC}=\frac{BA}{BC}\Longleftrightarrow \frac{XA}{AB}=\frac{XC}{BC}=\frac{AC}{AB+BC}\Rightarrow XC=\frac{AC \cdot BC}{AB+BC}$

similary we have $YC=\frac{AC\cdot BC}{AB+AC}$, thus $XY=(AC \cdot BC) (\frac{1}{AC+2BC}+\frac{1}{2AC+BC})$ , so by clearing the denominator it remais us to prove that : $(AC+2BC)(BC+2AC)\geq 9AC \cdot BC$, which is AM-GM, so the equality holds if $AC=BC$.


function
Ramsey Theory
combinatorics unsolved
combinatorics

Prove there is a function  $f$ from N to N, such that among $f(n)$ points in the plain, no 3 collinear, there exist a convex n-agon.

This is the famous Erdös-Szekeres problem (the happy-ending one).
It is conjectured that $f(n)=2^{n-2} + 1.$

Pierre.

Pierre, can you prove for every $n$, there exist a integer $f(n)$, i am not locking for sharp bounds, but only for the idea, i am sure it has to do something with Ramsey, but how can I aply it?

[quote="Pascual2005"]i am sure it has to do something with Ramsey, but how can I aply it?[/quote]

Sure it has to do with it. While solving this problem, G.Szekeres found a new proof of Ramsey theorem (he didn't know about it and used it as a crucial lemma).

We will use two easy lemmas, that I will not prove :

Lemma 1. (Esther Klein-Szekeres theorem)
Among any set of $5$ points in the plane, no three collinear, we may find $4$ which are the vertices of a convex quadrilateral.

Lemma 2.
Let $n \geq 4$ be an integer. If $n$ points in the plane, no three collinear, are such that any $4$ of them are the vertices of a convex quadrilateral, then these $n$ points are the vertices of a convex $n$-gon.

Theorem (Erdös-Szekeres, 1935)
Let $n \geq 3$ be an integer. There exists an integer $n_0 > 0$ such that in each set of at least $n_0$ points in the plane, no three collinear, we may find $n$ which are the vertices of a convex $n$-gon.

Proof.
Let's choose $n_0 = R_4(5,n)$, where $R_4(5,n)$ denotes the Ramsey number such that  for any two-coloring in red and blue of the $4$-sets of $\{1, \cdots, R_4(5,n) \}$ there exist either a $5$-set with all its $4$-subsets blue or a red $n$-set with all its $4$-subsets red.
Now, let's consider a set of at least $n_0$ points in the plane, no three collinear. Color in blue each group of $4$ points which are not the vertices of a convex quadrilateral, and in red each group of $4$ points which are the vertices of a convex quadrilateral.
From lemma 1, there is no $5$-set with all its $4$-subsets blue. Thus, from Ramsey theorem, there exists a $n$-set with all its $4$-subsets red. According to lemma 2, these $n$ points are the vertices of a convex $n$-gon.

Pierre.

Pierre I think you posted a site with an article of this problem. Do you remember this site? thanks!

Try this :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=szekeres&t=21981

Pierre.


Find the range of $ f(x) \equal{} 8 \cos x \minus{} 15 \sin x$.

$ (8\cos x\minus{}15\sin x)^2\plus{}(15\cos x\plus{}8\sin x)^2\equal{}17^2$, yielding $ |8\cos x\minus{}15\sin x|\leq 17.$

[quote="kunny"]$ (8\cos x \minus{} 15\sin x)^2 \plus{} (15\cos x \plus{} 8\sin x)^2 \equal{} 17^2$, yielding $ |8\cos x \minus{} 15\sin x|\leq 17.$[/quote]

But you need to prove $ 15\cos x \plus{} 8\sin x$ can be any number between $ 17$ and $ \minus{}17$.

Solution:

Let $ cos k\equal{}\frac{8}{17}$, then $ sin k\equal{}\frac{15}{17}$

$ f(x)\equal{}8 \cos x \minus{} 15 \sin x\equal{}17 cos (x\plus{}k)$

So $ \minus{}17 \le f(x) \le 17$

[quote="kunny"][quote="stephencheng"][quote="kunny"]$ (8\cos x \minus{} 15\sin x)^2 \plus{} (15\cos x \plus{} 8\sin x)^2 \equal{} 17^2$, yielding $ |8\cos x \minus{} 15\sin x|\leq 17.$[/quote]

But you need to prove $ 15\cos x \plus{} 8\sin x$ can be any number between $ 17$ and $ \minus{} 17$.

[/quote]

That's obvious, as $ (15\cos x \plus{} 8\sin x)^2 \geq 0.$ :wink:[/quote]

You showed that $ |8\cos x \minus{} 15\sin x|\leq 17$

But you didn't show that for every $ k$ between $ 17$ and $ \minus{}17$ we can always find $ x$ satisfying $ 8\cos x \minus{} 15\sin x\equal{}k$.

For example, $ |cos^2 x| \le 1$ but $ \minus{}1 \le cos^2 x \le 1$ is not the range for $ cos^2 x$.


Using mathematical induction, prove that if
0 (zero) is less than "a" which is less than "b", then 0 is less than "a" to the nth power which is less than "b" to the nth power, where n is a natural number.

Usually 'proof'-type of problems aren't HSB level, but this one might be OK since it's a basic application of induction.

[hide="Step 1"]Prove that the statement is true for $ n\equal{}1$, ie $ 0<a^1<b^1$. I hope you don't have trouble with this part  :roll: [/hide]

[hide="Step 2"]Assume that the statement is true for $ n\equal{}k$, ie $ 0<a^k<b^k$. Using this assumption and the $ n\equal{}1$ part, you can deduce within a few steps that the statement is also true for $ n\equal{}k\plus{}1$.[/hide]

And, by induction hypothesis, $ 0<a^n<b^n$ for all naturals $ n$.

See if you can do it.

Well, I don't know about most high schools...but my high school teaches basic induction in the Algebra 2/Trigonometry class, which is usually a freshmen or sophomore class...so it seems okay, I guess...

[quote]

Step 2
Assume that the statement is true for , ie . Using this assumption and the  part, you can deduce within a few steps that the statement is also true for .

And, by induction hypothesis,  for all naturals .

See if you can do it.[/quote]

Is it something like this?

a<b (Given)

a^k<b^k (Given)
So multiplying these two inequalities, we get

a^k * a < b^k * b
a^(k+1) < b^(k+1)
Which should complete the proof
Please tell me if I did somthing wrong...

[quote]Quote:

Step 2
Assume that the statement is true for , ie . Using this assumption and the part, you can deduce within a few steps that the statement is also true for .

And, by induction hypothesis, for all naturals .

See if you can do it.

Is it something like this?

a<b (Given)

a^k<b^k (Given)
So multiplying these two inequalities, we get

a^k * a < b^k * b
a^(k+1) < b^(k+1)
Which should complete the proof
Please tell me if I did somthing wrong...[/quote]

I don't see why not. What do you guys think?

There's room for clarification; it would be nice to mention the signs $ a$ and $ b$; the argument does not hold for all real numbers.

For example $ \minus{}5 < \minus{}3$ and $ \minus{}1 < 1$ does not imply that $ (\minus{}5)(\minus{}1) < (\minus{}3)(1) \iff 5 < \minus{}3$.

Where possible, my suggestion is to use inequality addition over inequality multiplication.

For example, we have $ b^k < a^k$, so multiplying by $ b \in \mathbb{R}^\plus{}$ gives $ b^{k\plus{}1} < a \cdot b^k$

Also, multiplying the inequality $ 0 < b^k$ by the quantity $ b\minus{}a$, which is positive, gives $ 0 < b^{k\plus{}1} \minus{} ab^k$, hence $ a b^k < b^{k\plus{}1}$

Transitively, $ a^{k\plus{}1} < b^{k\plus{}1}$

I was thinking about this: $ a<b\text{ and }a^k<b^k\Rightarrow a^{k\plus{}1}\equal{}aa^k<ab^k<bb^k\equal{}b^{k\plus{}1}$


factorial
function
ceiling function
algebra
floor function
puzzles

It's interesting.  I wonder if there exist cards that don't have solutions.

[quote="D3m0n Shad0w"]I play the 24 math game and theres this one card that stumped all the teachers, my brother, and everyone. The numbers are:

7,3,3,5

All prime numbers. Its hard...If one of you could post that would be great![/quote]
Can you do factorials?
[hide="If so..."]$7+3!+3!+5=7+6+6+5=24$[/hide]

This would also work iff factorials are allowed:
[hide]$\frac{5!}{3!}+7-3=20+7-3=\boxed{24}$[/hide]

I'm reasonably certain the classic game only allows addition, multiplication, subtraction, and division.
[hide="Another silly one, with derangements..."] $(5-!3) \times 7+3 = 24$ [/hide]
Edit:  Also, you do realize this game becomes rather trivial with factorials, [hide="Since the following type of silliness is also possible..."] $(3+3+5-7)! = 24$ [/hide]

hmm yeah my sister asked me and i didn't get it...is there a solution? (the original 24 game only allows you to use addition, subratraction, multiplication, and division.)

[quote="7h3.D3m0n.117"]is there a solution? (the original 24 game only allows you to use addition, subratraction, multiplication, and division.)[/quote]

No.

nope

$(5\cdot3-7)\cdot3$ lol thanks chess64  :lol:

Didn't think so.  I did some quick mental casework on every possibility for the last step, and prospects don't look good.

[quote="D3m0n Shad0w"]I play the 24 math game and theres this one card that stumped all the teachers, my brother, and everyone. The numbers are:

7,3,3,5

All prime numbers. Its hard...If one of you could post that would be great![/quote]

no, u can use powers

$(7-5)^{3}*3$

I remember this one from 6th grade

-jorian

lol you can't

its the STANDARD 24 CARD GAME...if you check the rules powers don't apply

[quote="chess64"][quote="7h3.D3m0n.117"]nope

$5\cdot3-7\cdot3$ lol thanks chess64  :lol:[/quote]

Wow, I'm the only one who saw this? It only took me like 3 seconds, and I'm not even good at this game.

(BTW, Henry, you mean $(5\cdot 3-7)\cdot 3$ though ;))[/quote]
OH, YES, I SERIOUSLY DO REMEMBER THIS ONE!
-jorian

yeah 7h3.D3m0n.

remember order of operations in first grade?

[quote="nutz_for2.718281828"]I got one using ceiling functions:

ceiling{5/3} +7*3 =24[/quote]
What is a ceiling fuction?

[quote="7h3.D3m0n.117"]umm like

$\lceil 2.3 \rceil$

ceiling functions mean you would round up no matter what decimal  is

guess what a floor function is  :D

and this is getting off topic[/quote]
Some of the other posts are off topic...
But I don't think you're allowed to use floor/ceiling functions in the 24 game.

EDIT: then $\lceil 5/3 \rceil+7*3 =2+21=23\ne24$


invariant
Ring Theory
superior algebra
superior algebra unsolved

Let $ A$ be a commutative ring and $ G$ a finite group of automorphisms of $ A$. Let $ S\subset A$ be a multiplicatively closed set.
Denote by $ A^G \equal{} \{a\in A: \sigma(a)\equal{}a,\forall  \sigma\in G\}$ and $ S^G\equal{}S\cap A^G$.
The action of  $ G$ on $ A$ extends naturally to an action on $ S^{\minus{}1}A$.
Prove that $ (S^G)^{\minus{}1}A^G\simeq (S^{\minus{}1}A)^G$.

I additionally assume that $ S$ is invariant under $ G$: $ \sigma(S)\equal{}S$ for all $ \sigma\in G$.

[b]Fact 1:[/b] It is easy to see that the operation of $ G$ on $ A$ naturally extends to $ S^{\minus{}1}A$ by setting $ \sigma\left(\frac{a}{s}\right)\equal{}\frac{\sigma(a)}{\sigma(s)}$.

[b]Fact 2:[/b] The natural homomorphism $ (S^G)^{\minus{}1}A\rightarrow S^{\minus{}1}A$ is an isomorphism. [hide="Hint"]Use $ \frac{1}{s}\equal{}\frac{s'}{s^G}$ with $ s^G\equal{}\prod_{\sigma\in G}\sigma(s)$ and $ s'\equal{}\prod_{\sigma\in G,\sigma\neq e_G}\sigma(s)$ for every $ s\in S$.[/hide]

[b]Fact 3:[/b] For every ring $ A$ with an $ G$-operation the subring $ A^G$ is the equalizer of two natural ring homomorphisms; more precisely the sequence
\[ 0\rightarrow A^G\stackrel{\subseteq}{\rightarrow} A\rightrightarrows \oplus_{\sigma\in G}A\,\,\, (*)\]
where each of the double arrow is defined by $ a\mapsto (\sigma\mapsto a)_\sigma$  and $ a\mapsto(\sigma\mapsto\sigma(a))_\sigma$, respectively, is exact.

This is all we need for the proof: Localizing the exact sequence (*) with respect to $ S^G$ gives the exact sequence
\[ 0\rightarrow (S^G)^{\minus{}1}A^G\stackrel{\subseteq}{\rightarrow} (S^G)^{\minus{}1}A\rightrightarrows \oplus_{\sigma\in G}(S^G)^{\minus{}1}A\,\,\,(**)\] Thus we have natural isomorphisms
\[ (S^G)^{\minus{}1}A^G\cong ((S^G)^{\minus{}1}A)^G\cong (S^{\minus{}1}A)^G\]
where we used (**) for the first isomorphism and Fact 2 for the second.


Let $ a$ , $ b$ and $ c$ be positive real numbers. Prove that :  $ \sqrt{\frac{a^{2}\plus{}bc}{b\plus{}c}}\plus{}\sqrt{\frac{b^{2}\plus{}ca}{c\plus{}a}}\plus{}\sqrt{\frac{c^{2}\plus{}ab}{a\plus{}b}} \geq \sqrt{3(a\plus{}b\plus{}c)}$

[quote="canada"]Let $ a$ , $ b$ and $ c$ be positive real numbers. Prove that :  $ \sqrt {\frac {a^{2} \plus{} bc}{b \plus{} c}} \plus{} \sqrt {\frac {b^{2} \plus{} ca}{c \plus{} a}} \plus{} \sqrt {\frac {c^{2} \plus{} ab}{a \plus{} b}} \geq \sqrt {3(a \plus{} b \plus{} c)}$[/quote]
Easy Holder:
$ \left(\sum_{cyc}\sqrt {\frac {a^{2} \plus{} bc}{b \plus{} c}}\right)^2\sum_{cyc}(b \plus{} c)(a^2 \plus{} bc)^2\geq\left(\sum_{cyc}(a^2 \plus{} bc)\right)^3$ and
$ \left(\sum_{cyc}(a^2 \plus{} bc)\right)^3\geq3(a \plus{} b \plus{} c)\sum_{cyc}(b \plus{} c)(a^2 \plus{} bc)^2$,
which is equivalent to the obvious: $ \sum_{cyc}(a^6 \plus{} a^3b^3 \plus{} 3a^4bc \minus{} 5a^2b^2c^2)\geq0.$
The following inequality seems true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0.$ Prove that:
\[ \sqrt {\frac {a^{2} \plus{}2bc}{b \plus{} c}} \plus{} \sqrt {\frac {b^{2} \plus{} 2ca}{c \plus{} a}} \plus{} \sqrt {\frac {c^{2} \plus{} 2ab}{a \plus{} b}} \geq 3\sqrt {\frac{a \plus{} b \plus{} c}{2}}\]


A right angle has its vertex at the centroid of an equilateral triangle of side $1$ unit.  Find the maximum possible area interior to both the angle and the triangle.

we can assume that a maximum either occurs at an endpoint or when the sides are equal, the two rays can touch one side, this gives 1/(2root3)^2, intersects two sides, 2(1/root3)^2*1/2*sin 30 sin 45/sin 75=(root 3 -1)/6, which is greater than the other one, we also have at a vertex, which would be (1/2)(1/3)(1/root 3), which is also less

(root 3 -1)/6


A contest began at noon one day and ended $ 1000$ minutes later.  At what time did the contest end?

\[ \textbf{(A)}\ 10: 00 \text{ p.m.} \qquad
\textbf{(B)}\ \text{midnight} \qquad
\textbf{(C)}\ 2: 30 \text{ a.m.} \\
\textbf{(D)}\ 4: 40 \text{ a.m.} \qquad
\textbf{(E)}\ 6: 40 \text{ a.m.}
\]

[hide="Click for solution"]
Since $ 1000$ minutes $ \equal{} \frac{1000}{60}$ hours $ \equal{} 16\frac{2}{3}$ hours $ \equal{}16$ hours, $ 40$ minutes, the contest ended $ 16$ hours $ 40$ minutes past noon or at $ 4: 40 \text{ a.m.}$, or answer $ \boxed{\textbf{(D)}}$.
[/hide]

Wow AIME15, it looks like a clear scandal to get more posts.  At least post the solution in the same post...

That is a blatant, false accusation.  The [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182]Resources Section[/url] lacks AMC 8/AJHSME problems, so I am trying to add some.  As for the posting solution in the same post, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191609]this[/url].

Couldn't you just post your problems on the AoPS wiki?  :roll:


function
algebra
polynomial
complex analysis
complex analysis unsolved

let $ f(z)$ be an entire function , Prove the family $ \{f_k(z)\}_{k\in \mathbb{Z}}$ of functions $ f_k(z)\equal{}f(kz)\ ,k\equal{}1,2,...$ , in annulus $ r_1<|z|<r_2$ are normal iff $ f(z)$ be a polynomial.

HomeWork

Either I don't get something or this is not true. Consider $ f(z) \equal{} z$; then the family $ \{ f_{k}(z)\}$ is not normal since it is not pre-compact; indeed, if it were, the sequence of functions $ \{ f_k(z)\}_{k\in\mathbb N}$ would have to have a convergent subsequence. But it obviously doesn't, as any neighborhood of any $ f_k(z)$ contains only finitely many other $ f_n(z)$. Can you clarify?

Xixas, we allow $ \infty$ as a limit.

If $ f_{k}$ converges normally to infinity, then it converges uniformly to infinity on some smaller annulus $ r_1' < |z| < r_2'$. This means that we can choose $ N$ so that $ f$ is big on all annuli $ \{kr_1' < |z| < kr_2'\}$ for $ k \geq N$. Hence $ f$ has a pole at infinity, and so is a non-constant polynomial.

The other case is that $ f_{k}$ has a subsequence $ f_{k_n}$ converging to some analytic function $ g$. Then $ f_{k_n}$ is uniformly bounded in some smaller annulus $ r_1' < |z| < r_2'$, and hence $ f$ is uniformly bounded on $ \bigcup \{k_nr_1' < |z| < k_nr_2'\}$. By the proof of Liouville's theorem, $ f$ is hence constant.

If it is not a polynomial, then it has an essential singularity at infinity, so use Picard or something?


[quote="yookon88"]p+q+r=1 (p,q,r>0)

7(pq+qr+rp)<=2+9pqr

can anyone help me solve this?[/quote]

hi there

homogenize, do calculations and schur saves you.........

Yes, get everything to the same degree, then it should come easily.

[hide]

The inequality becomes

Expanding and rearranging, we get

Which holds by Muirhead, because . Schurs works fine here, too, I believe.

edit> Nevermind Schurs, I don't believe that works, but Rearrangement works perfectly here on the three pairs (p,q), (p,r), and (q,r), and Muirhead just shortens it a bit.

[/hide]

By the way, this problem is from 1999 UK.

[quote="yookon88"]Thank you socrates and Elemennop for your help! It worked.
By the way what does homogenize mean?[/quote]

this form (of same degree)  $7(pq+qr+pr)(p+q+r)\leq2(p+q+r)^3 + 9pqr$
 :)

Elemennop wrote:Expanding and rearranging, we get

Which holds by Muirhead, because . Schurs works fine here, too, I believe.

edit> Nevermind Schurs, I don't believe that works, but Rearrangement works perfectly here on the three pairs (p,q), (p,r), and (q,r), and Muirhead just shortens it a bit.

[/hide]

Good.

But don't you see using Rearrangerment and Muirhead's is quite complicated, while, at least, you're not allowed to use them in some high school contests. Let we try for another proof, I think there is a simpler solution for this problem.

[quote="yookon88"]Thank you socrates and Elemennop for your help! It worked.
By the way what does homogenize mean?[/quote]

Yes, Homongenize means to get every term to the same total degree, which lets you use alot more theorems/tricks

and while Muirhead is usually viewed as being too advanced for high school contests, Rearrangement is perfectly valid. However, anything solved with Muirhead should be solvable by AMGM, so there's probably a solution of that sort here.


floor function
number theory proposed
number theory

the answer, i think, is 224.
notice that $ \frac {a_{n \minus{} 1} \plus{} a_{n \plus{} 1}}{a_n}$ is constant and you're asking the value of $ \lfloor\frac {a_{2006} \plus{} a_{2008}}{a_{2007}} \minus{} \frac {2007}{a_{2006}a_{2007}}\rfloor$


linear algebra
matrix
geometry
rectangle
rotation
absolute value
combinatorics unsolved

In a matrix $2n \times 2n$, $n \in N$, are $4n^2$ real numbers with a sum equal zero. The absolute value of each of these numbers is not greater than $1$. Prove that the absolute value of a sum of all the numbers from one column or a row doesn't exceed $n$.

:huh: it seems wrong...
Doesn't the matrix with $1$'s on the first line, $-1$'s in the second and $0$'s in the rest of the matrix give a counter example?

(total sum is $0$, all absolute values are at most $1$, and the sum of the firt line is bigger than $n$.)

No, we have to show that there's some line with sum not greater than $n$.

This problem is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41119]C4 from the ISL 2004[/url].

I remember doing this several months ago, so I'll try to reconstruct my solution... I think it went as follows:

Let $R_1, R_2, \ldots, R_{2n}$ be the rows and $C_1, \ldots, C_{2n}$ the columns. Obviously we can permute rows and columns without altering the sum of the board or of each row and column, so assume that $R_1, \ldots, R_k$ have non-negative sum, $C_1, \ldots, C_m$ have non-negative sum and the rest of the rows and columns have negative sum. There are four kinds of squares: type 1 (on positive row and column), type 2 (on positive row, negative column), type 3 (negative row, positive column) and type 4 (both negative). Essentially, we want to show that we need only consider the case when all squares of each type are equal.

Note that the four types of squares form 4 rectangles on the board. Let $(i, j)$ be the square on $R_i, C_j$. Replace each number by the arithmetic mean of the numbers on its row and of the same type (including itself). That is, replace the numbers on $(1, 1), (1, 2), \ldots, (1, m)$ by the mean of them all, and so on. This doesn't change the sum of each row (or of the whole board), doesn't make any absolute value go over $1$, and doesn't lower the minimal sum of a line (in absolute value).

The first two claims are obvious. For the third: the point is that all positive columns end up equal and all negative rows end up equal; it is clear that, since the sum is unchanged, this doesn't lower the minimal sum.

Now, rotate the board and perform the same operation again. Analogously, the minimal sum is not smaller afterwards, and now all numbers of each type are equal. Let the type 1s be $a$, the type 2s $b$, the type 3s $c$, and the type 4s $d$. We know that $kma+(2n-m)kb+m(2n-k)c+(2n-m)(2n-k)d=0$ and $-1 \leq a, b, c, d \leq 1$. Besides, $ka+(2n-k)c>0, kb+(2n-k)d<0, ma+(2n-m)b>0, mc+(2n-m)d<0$ (we needn't consider equality cases since they would solve the problem at once).

Now it boils down to lots of cases and calculation. I may complete this later, but for now...

Proof by contradiction.  Let $c_1,c_2,...c_{2n}$ be the column sums.  Then permute the columns so that $c_1\ge c_2...\ge c_{2n}$.  We can assume that $c_n \ge 0$ (otherwise multiply each element of the matrix by $-1$ and rearrange to get what we want).  Let $a_1,a_2,...a_{2n}$ be the sum of the elements of the first $n$ elements in each row.  Now we have assumed $c_1,c_2,...c_n > n$, but ${a_i}\leq n$, and $n^2 <\sum_{i=1}^{n}c_i=\sum_{i=1}^{2n}a_i$, so there must be at least $n+1$ positive $a_i$.

If we define $b_1,b_2,...b_{2n}$ to be the sum of the last $n$ elements of each row, since $\sum_{i=n+1}^{2n}c_i=-\sum_{i=1}^{n}c_i<- n^2$, so there must be at least $n+1$ negative $b_i$.  Hence there is a $k$ such that $a_k > 0$ and $b_k <0$.  The absolute value of the sum of the elements of this row is $|a_k+b_k|$, which has absolute value at most the maximum of $|a_k|$ and $|b_k|$ or $n$.

Similar to the difficult ISL 1988 #14.

Assume the contrary, for contradiction. Let $x, y$ be numbers such that there are $x$ rows with sum greater than $n$ and $y$ columns with sum greater than $n$. Then, there are $2n-x$ rows with sum less than $-n$ and $2n-y$ columns with sum less than $-n.$

Note that we can reorder rows without altering row or column sums. Similarly, we can reorder columns, and so WLOG assume that the rows are ordered in nondecreasing order by sum from top to bottom and the columns are ordered in nondecreasing order by sum from left to right.

Then, let $A$ be the sum of the numbers in the top-left $x \times y$ corner of the matrix, $B$ be the sum of the numbers in the top-right $x \times (2n-y)$ corner of the matrix, $C$ be the sum of the numbers in the bottom-left $(2n-x) \times y$ corner of the matrix, and $D$ be the sum of the numbers in the $(2n-x) \times (2n-y)$ corner of the matrix.

From our assumption, we have:

$$A+B > xn, \qquad (1)$$

$$C+D < -(2n-x)n, \qquad (2)$$

$$A+D > yn, \qquad (3)$$

$$B+C < -(2n-y)n. \qquad (4)$$

By $(1) + (3) - (2) - (4),$ we have:

$$A - C > 2n^2.$$

This implies by the Triangle Inequality that there must be more than $2n^2$ numbers in $A$ and $C$, i.e. $xy + (2n-x)(2n-y) > 2n^2 \Leftrightarrow (x-n)(y-n) > 0.$

WLOG that $x > n$ and $y>n$, as otherwise we can just negate all the numbers in the array.

Observe that $A \le xy$, and so together with $(1) + (3)$, we have:

$$A+B+D = (A+B) + (A+D) - A > xn + yn - xy. \qquad (5)$$

It's also clear that

$$C \ge -(2n-x)(2n-y), \qquad (6)$$

and so we've

$$0 = A+B+C+D > xn+yn - xy -(2n-x)(2n-y).$$

This is easily seen to be equivalent to:

$$(4n-3x)(4n-3y) > xy.$$

However, we have that $xy > 0$, $|x| \ge |4n-3x|$, and $|y| \ge |4n-3y|$ from $x, y > n$, and so equality must hold in both cases, i.e. $|x| = |4n-3x|, |y| = |4n-3y|$. However, this implies that $x = y = 2n$, i.e. all the rows have positive sum, clearly absurd.

Hence, we've attained the desired contradiction, and we're done.

$\square$


limit
integration
trigonometry
calculus
geometry
real analysis
real analysis unsolved

$ 1\blacktriangleright$ $ \lim_{n\to\infty }\left(\int_a^b(x\minus{}a)^n(b\minus{}x)^n\ \mathrm {dx}\right)^{\frac 1n}\equal{}\left(\frac {a\minus{}b}{2}\right)^2$, where $ a<b\ .$

$ 2\blacktriangleright$ $ \lim_{n\to\infty }\frac {1}{n^2}\cdot \sum_{k\equal{}0}^{n\minus{}1}\left[k\int_k^{k\plus{}1}\sqrt {(x\minus{}k)(k\plus{}1\minus{}x)}\ \mathrm {dx}\right]\equal{}\frac {\pi}{16}\ .$

For the first one, you may use this:

$ \int_{a}^{b} (x\minus{}a)^m (b\minus{}x)^n \, dx \ \equal{} \ \frac{m!n!}{(m\plus{}n\plus{}1)!}(b\minus{}a)^{m\plus{}n\plus{}1}$

and for $ r > 0$,

$ \lim_{n\to\infty} \frac{\sqrt[n]{(rn)!}}{n^{r}} \ \equal{} \ \left( \frac{r}{e} \right)^r$

Or $ \lim_{n \to \infty} \sqrt [n] {\int_a^b f^n (x) \, dx} \equal{} \max_{x \in [a,b]} f(x)$.

[quote="Virgil Nicula"]$ 1\blacktriangleright$ $ \lim_{n\to\infty }\left(\int_a^b(x \minus{} a)^n(b \minus{} x)^n\ \mathrm {dx}\right)^{\frac 1n} \equal{} \left(\frac {a \minus{} b}{2}\right)^2$, where $ a < b\ .$

$ 2\blacktriangleright$ $ \lim_{n\to\infty }\frac {1}{n^2}\cdot \sum_{k \equal{} 0}^{n \minus{} 1}\left[k\int_k^{k \plus{} 1}\sqrt {(x \minus{} k)(k \plus{} 1 \minus{} x)}\ \mathrm {dx}\right] \equal{} \frac {\pi}{16}\ .$[/quote]
$ 2\blacktriangleright$
let $ I\equal{}\int_k^{k \plus{} 1}\sqrt {(x \minus{} k)(k \plus{} 1 \minus{} x)}\ \mathrm {dx}$

put $ x\equal{}k\plus{}\cos^2\,\theta,\;\;\;\quad\;\;\;dx\equal{}\minus{}\,\sin\,2\theta\;d\theta$

so, $ I\equal{}\int_{\frac{\pi}{2}}^0\;\sqrt{\;\cos^2\,\theta\cdot\sin^2\,\theta\;}\;(\minus{}\,\sin\,2\theta)\;d\theta$
$ \equal{}\frac{1}{2}\int_0^{\frac{\pi}{2}}\;\sin^2\,2\theta\;d\theta$
$ \equal{}\frac{1}{4}\int_0^{\frac{\pi}{2}}\;(1\minus{}\cos\,4\theta)\;d\theta$
$ \equal{}\frac{\pi}{8}$

then, $ L\equal{}\lim_{n\to\infty }\frac {1}{n^2}\cdot \sum_{k \equal{} 0}^{n \minus{} 1}\;k\cdot\frac{\pi}{8}$
$ \equal{}\frac{\pi}{8}\lim_{n\to\infty }\frac {1}{n^2}\cdot \sum_{k \equal{} 0}^{n \minus{} 1}\;k$
$ \equal{}\frac{\pi}{8}\lim_{n\to\infty }\frac {1}{n^2}\cdot \dfrac{n(n\minus{}1)}{2}$
$ \equal{}\frac{\pi}{16}$

[ PS: the first limit is also easy to show after you make a similar trigonometric substitution as was done above ]

For 2,

The definite integral is the area of the semi circle with radius $ \frac{1}{2}$ centered on $ \left(k\plus{}\frac{1}{2},\ 0\right)$,yielding the value is $ \frac{\pi}{8}.$

$ \therefore \lim_{n\to\infty }\frac {1}{n^2}\cdot \sum_{k \equal{} 0}^{n \minus{} 1}k\int_k^{k \plus{} 1}\sqrt {(x \minus{} k)(k \plus{} 1 \minus{} x)}\ \mathrm {dx}\equal{}\frac{\pi}{8} \lim_{n\to\infty }\frac {1}{n^2}\cdot \sum_{k \equal{} 0}^{n \minus{} 1} k$

$ \equal{}\frac{\pi}{8}  \lim_{n\to\infty }\frac {1}{n}\cdot \sum_{k \equal{} 0}^{n \minus{} 1} \frac{k}{n}\equal{}\frac{\pi }{8}\int_0^1 x\ dx\equal{}\frac{\pi}{16}.$


[quote="Together"]$a,b,c,d>0$ .
 $p_1=\frac{a+b+c+d}{4} , p_2=\frac{ab+bc+cd+da+bd+ac}{6} , p_3=\frac{abc+bcd+cda+dab}{4} , p_4 =abcd$

Prove that: if $p_1=4$ then
a, $\frac{3}{4}p_1 \ge 31.p_4 +1$
b,$7.p_1 + 216p_3 \ge 108p_3 + 256p_4.$[/quote]

Excuse me,but what do you want to prove,I think maybe there is something wrong with it.

Your inequalities are weird since they include $p_1$ which is said to be $4$ and second one have $p_3$ on both side  :o
Probably you mixed some indices, didn't you ?


Is it known if there are infinetely many primes in the fibonacci sequence ?
If yes, is it know any elementary approach ?
Generally, what do you know about this topic ?

I feel that this should be an open question, or if itisn't then there must be a very difficlut approach... but who knows... ?

Have a look here :
http://mathworld.wolfram.com/FibonacciPrime.html

Pierre.


inequalities
analytic geometry
graphing lines
slope
algebra unsolved
algebra

Question # 01

Solve the inequality and sketch the solution on a coordinate line.
	2x-1>11x+9

Question # 02

Show that (5,-2), (6,5) and (2,2) are the vertices of right triangle.
By using

1)	Slope formula
2)	Distance formula

Question # 03

Determine whether the equation represents a circle, if the equation represents a circle, find the center and radius.

x2+y2-2x-4y-11=0

This is not olympiad worthy, and shouldn't be answered here.


Given regular pentagon $ABCDE$, a circle can be drawn that is tangent to $\overline{DC}$ at $D$ and to $\overline{AB}$ at $A$. The number of degrees in minor arc $AD$ is

$\textbf{(A)}\ 72 \qquad
\textbf{(B)}\ 108 \qquad
\textbf{(C)}\ 120 \qquad
\textbf{(D)}\ 135 \qquad
\textbf{(E)}\ 144$

[asy]
size(100); defaultpen(linewidth(0.7));
draw(rotate(18)*polygon(5));
real x=0.6180339887;
draw(Circle((-x,0), 1));
int i;
for(i=0; i<5; i=i+1) {
dot(origin+1*dir(36+72*i));
}
label("$B$", origin+1*dir(36+72*0), dir(origin--origin+1*dir(36+72*0)));
label("$A$", origin+1*dir(36+72*1), dir(origin--origin+1*dir(36+72)));
label("$E$", origin+1*dir(36+72*2), dir(origin--origin+1*dir(36+144)));
label("$D$", origin+1*dir(36+72*3), dir(origin--origin+1*dir(36+72*3)));
label("$C$", origin+1*dir(36+72*4), dir(origin--origin+1*dir(36+72*4)));[/asy]

[hide]Let $O$ be the center of the circle. The intercepted arc $AD$ is equal to $\angle AOD$. $\angle ODC$ and $\angle OAB$ are both right angles. Since the measures of the angles of the pentagon $ABCDO$ add up to 540, the measure of $\angle AOD$, in degrees, is 540-2*108-2*90=540-2*198=144. [b](E)[/b][/hide]

[hide]Center of circle is O

Interior angle of Pentagon is 108 degrees.

angle OAB and angle ODC are 90 degrees

Angle EAO and angle EDO are both 18 degrees- $108-90=18$

From Quadrilateral AEDO-  $316-(108+18+18)=216$ (360- (angle AED + angle EAO + angle EDO)) gives major arc AD

360-216=144 degrees gives Minor arc AD[/hide]

Just wondering, can you just do $ (\frac{2}{5})(360) \equal{} 144$?
Because the pentagon is regular...

Clarification/help would be appreciated.

Also, again, is it ok to revive old posts like these ahsme problems?

Yes, it is ok to revive old contest problem threads. It is better to do this than having many users make a ton of threads.

And, 2/5(360) is acceptable, but to some, it may not be obvious.

How would one get the solution of 2/5*360?

I got two upvotes on my post for no apparent reason (lawl), but I still don't have an answer...


calculus
derivative
inequalities
topology
calculus computations

Question: Suppose $ f$ is differentiable on $ [a,b]$, $ f(a)\equal{}0$ and there is a fixed real number $ A$ such that $ |f'(x)|\leq A|f(x)|\text{ }\forall x \in [a,b]$. Prove that $ f: [a,b]\mapsto 0$

Answer: By the mean value theorem it is true that on any closed interval $ [a,b]$ we have that $ \frac{f(b)\minus{}f(a)}{b\minus{}a}\leq \sup_{x\in(a,b)}|f'(x)|$. So in our case $ \frac{f(b)}{b\minus{}a}\leq |f'(\xi)|{\color{red}(*)}$ where $ \xi$ is the value that $ |f'(\xi)|\equal{}\sup |f'(x)|$. So then stringing the given inequality together with $ \color{red}(*)$ we get that $ \frac{f(b)}{b\minus{}a}\leq |f'(\xi)|\leq A|f(b)|\implies \frac{f(b)}{b\minus{}a}\leq A|f(b)|$. From here it is clear that since $ b,a$ were arbitrary and $ A$ fixed, the only way this is satisfied is if $ f(b)\equal{}0$. Finally noting again that $ b$ was arbitrary we can conclude that $ f\equal{}0$

Is that correct?

You want to show that the roots of f form an open set in the subspace topology.

You can, for example, show that the derivative is zero in a neighborhood of any root.

Done.

[quote="Mathstud28"]
Is that correct?[/quote]
Nope, you've made quite a mess out of it. The first incorrect (or, at least, unjustified) step is
[quote]
$ \frac{f(b)}{b\minus{}a}\leq |f'(\xi)|\leq A|f(b)|\dots$
[/quote]
I see only that $ |f'(\xi)|\le A|f(\xi)|$, which, in general, may be much larger than $ |f(b)|$.

Here's what I'd recommend: Don't start with the initial condition $ f(a)\equal{}0$. Show that $ |f(x)|\le e^{A(x\minus{}a)}|f(a)|$ when you start somewhere else, and then push $ f(a)$ to zero.

If you can do this, the general uniqueness theorem for differential equations is in reach.

Show that this is true if $ |a \minus{} b| < \frac {1}{A}$ (your original argument works here if you take $ b \equal{} \xi$) and "induct" on the length of the interval.

The structure of the MVT doesn't let you choose that way; $ \xi$ is determined by the choice of $ b$, and isn't equal.

Sorry, that was worded poorly.  We'll show the statement is true if the interval has length less than $ \frac {1}{A}$.  Since $ f$ is differentiable, it is continuous, hence $ |f(x)|$ assumes a maximum at some $ \xi \in [a, b]$.  Suppose $ |f(\xi)| > 0$.  By the MVT, there is some $ z \in [a, \xi]$ such that $ \frac {f(\xi) \minus{} f(a)}{\xi \minus{} a} \equal{} \frac {f(\xi)}{\xi \minus{} a} \equal{} f'(z)$.  On the other hand, since $ |f(\xi)| \ge |f(z)|$ it follows that

$ A |f(z)| < \left| \frac {f(\xi)}{b \minus{} a} \right| \le A |f(z)|$;

contradiction.  Now just note that an interval $ [a, b]$ of arbitrary length contains an interval $ [c, d]$ of length less than $ \frac {1}{A}$ such that $ f(c) \equal{} 0$.


Euler
inequalities
trigonometry
probability
number theory
prime numbers
relatively prime

Here's my question:

[b]Does the sum of the reciprocals of all the members of the set of prime numbers converge or diverge?[/b]

I think that it diverges, but I'm not totally sure.
Here's my reasoning:

Since we know that

\[\sum_{n=1}^\infty \stackrel{1}{\overline{n}}\]

diverges, and we also know, by the prime number theorem, that the proportion of prime numbers in a given set of positive integers approaches a limit ( it doesn't matter what limit) as the set size approaches $\infty$, we can conclude that the proportion relating the sum of the reciprocals of the all the members of the set of prime numbers to the sum of the reciprocals of all the members of the set of positive integers also approaches a limit.

Thus, since the limit of
\[\sum_{n=1}^\infty \stackrel{1}{\overline{n}}\]
is $= \infty$, and the proportion relating the sum of the reciprocals of the all the members of the set of prime numbers to the sum of the reciprocals of all the members of the set of positive integers approaches a limit, and $\infty$ multiplied by any limit (except 0)$= \infty$, then it can be concluded that the limit of the sum of the reciprocals of all the members of the set of prime numbers does not exist. ( $= \infty$  and thus diverges)

Is my reasoning correct here?  :) Or am I missing something? :maybe:

[quote]... and we also know, by the prime number theorem, that the proportion of prime numbers in a given set of positive integers approaches a limit ( it doesn't matter what limit) as the set size approaches $\infty,$ ...[/quote]

No, we don't know that. That's not what the prime number theorem says. The proportion of primes in a set of integers tends to zero. Now we can use the prime number theorem anyway, by using the correct estimate of the proportion of primes, to prove that this series diverges.

But the prime number theorem is a very big gun - a major theorem, a conjecture that was still unsettled a century after Euler gave the elegant proof of the result you want here, that the sum of the reciprocals of the primes diverges.

Euler's hint: consider $\prod_{p\text{ prime}}\left(1-\frac1p\right)^{-1}.$

All right, thanks for setting me straight and pointing out my error.
    Hmmm...I wonder if all series, that are made up of the sums of the reciprocals of an aleph-null set of positive numbers, diverge.

Of course not. Just consider the reciprocals of the perfect squares. This - the convergence or divergence of the sum of reciprocals - is a very interesting way to look at the "thickness" of a subset of $\mathbb{N},$ but it has nothing to do with cardinality. And the result says that in some sense, primes are plentiful.

Erdos offered $\$3000$ for a proof of the following conjecture.

[quote]Let $A\subset\mathbb N$ be a set such that $\sum_{n\in A}\frac{1}{n}=\infty$. Then $A$ contains arbitrarily long arithmetic progressions.[/quote]

The special case $A=\{\textrm{primes}\}$ [url=http://arxiv.org/abs/math.NT/0404188]has been proved recently[/url].

[quote="mlok"]Erdos offered $\$3000$ for a proof of the following conjecture.

[quote]Let $A\subset\mathbb N$ be a set such that $\sum_{n\in A}\frac{1}{n}=\infty$. Then $A$ contains [u][b]arbitrarily long arithmetic progressions[/b][/u].[/quote]

The special case $A=\{\textrm{primes}\}$ [url=http://arxiv.org/abs/math.NT/0404188]has been proved recently[/url].[/quote]

what does [u][b]arbitrarily long arithmetic progressions[/b][/u] means?

[quote="color"]There is a beutiful proof of the divergence of the harmonic prime series, try some of these sites

http://planetmath.org/encyclopedia/PrimeHarmonicSeries.html

http://en.wikipedia.org/wiki/Proof_that_the_sum_of_the_reciprocals_of_the_primes_diverges[/quote]

the 4th proof seems to be simple. But how to prove the Dusart's inequality?

I just received a poster "The Year of Euler" published by [url=http://www.maa.org]MAA[/url]. It highlights "some of Euler's most famous results":
\[e^{i\theta}=\cos\theta+i\sin\theta\]
\[\sum_{k=1}^{\infty}\frac{1}{k}=\prod_{p}\frac{1}{1-1/p}\]
\[\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6}\]
\[V-E+F=2\]

The way #2 is stated does not do it justice: one could just put $\infty$ on the left. The main point of the product-sum formula is $\sum_{k=1}^{\infty}\frac{1}{k^{s}}=\prod_{p}\frac{1}{1-1/p^{s}}$, where $\mathrm{Re}\,s>1$.

I also wonder why #3 is so important. Does anyone know?

[quote="mlok"]
I also wonder why #3 is so important. Does anyone know?[/quote]

Historic value. Apparently, Euler established his reputation as mathematician by computing that sum. It had been an open problem for some time.

I just came across a curious proof of $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$: [url=http://arxiv.org/abs/math.GM/0701039]How to compute $\sum 1/n^{2}$ by solving triangles[/url]

Application of #3:

[b]Theorem.[/b] The probability that two random integers are relatively prime is equal to $\frac{6}{\pi^{2}}$.

I'm not saying that #3 is useless. I just think it's not nearly as important as the other three.


How would one go about writing a program that outputs itself? (C++ or otherwise)

write a program to open and print a file "Prog.txt", then create this text file in notepad, copy the program listing from the compiler window and paste it in this file. Running the program outputs itself. (Don't ask me how you print the file - I am always weak in file handling and can go only about as far as #include <fstream.h> before starting to wonder "Now what?")


In an automobile commercial that is currently running, a car is shown suspended from a helicopter above a straight level road. The voice-over explains that the car is dropped from "over 4000 feet" while another car on the road accelerates from a standing start at the same moment "from the same distance." The car on the road is shown as passing through the target spot before the falling car crashes into it. No parachute is shown.

The filmed commercial clearly uses special effects techniques, but the question is this: is the scenario plausible? Calculate some pertinent numbers.

Well, the time it takes for the suspended car to hit ground
could be determined by $t=\sqrt{\frac{2h}{g}}$ (free fall)
Here $h=4000$ feet. In the same time (roughly 16-17 seconds)
the car has to cover 4000 feet, which means it has to
keep an average speed of approximately 300 km/h which is too high
for most everyday usual cars. And also one has to take into account
that it has to accelerate to this speed during a part of the time.
Actually this will make it almost impossible for the car to make it.
Unless it is a very special car  :lol:, for a common sports
car data look [url=http://en.wikipedia.org/wiki/BMW_Z8]here[/url] for example.

One could also argue easier, namely by only considering the acceleration.
A car that could accelerate with acceleration $g$ during this time is rare :) .
So it is not plausible i think.

Another relevant number to look at would be the friction coefficient of tire rubber against asphalt. If it is less than $1$, there is no way for a normal car (i.e., not equipped with "anti-wings" and similar things) to develop a horizontal acceleration exceeding that of the free fall no matter how powerful the engine is. According to [url=http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm#coef]this site[/url] it is at best $0.8$, end of story.

The sport of drag racing involves an automobile accelerating from a standing start to the end of a quarter-mile track. In publishing the results, both the elapsed time and the final velocity are given. A "top fuel" dragster might occasionally turn in times around 5 seconds and top speeds of around 300 miles per hour. This suggests that the horizontal acceleration of such a machine might approach $3g$ for a few seconds. Such machines do come equipped with elaborate means, mostly aerodynamic, for increasing the "down force" on the drive wheels. (As fedja put it, "anti-wings.) They're also not meant to be driven on streets and highways.

The car in the commercial is a street-legal sedan, available at enough dealerships and in sufficient quantity to justify television advertising (mostly during sporting events.) Color's very simple estimate of speed already rules out any mass-produced sedan I know of.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Let $ R$ be a finite ring with $ 1$. Prove that $ R\equal{} \mathbb{F}_1 \times \mathbb{F}_2 \times ...\times \mathbb{F}_n$ , where $ \mathbb{F}_i$ are finite fields.

So, what should the original statement say to preserve that answer? No nilpotent elements, at least. Is that enough?

every finite commutative(!) ring $ R$ with 1 is a finite direct product of local rings, namely $ R/p^n$ where $ p$ runs through the prime ideals of $ R$ and $ n$ is a large integer.  thus if the ring is reduced (no nilpotent elements besides $ 0$), we must have $ n \equal{} 1$ and $ R/p$ is a finite integral domain, thus a field. conversely, every finite product of finite fields is a commutative reduced ring with $ 1$.

Do you have a noncommutative counterexample? That case is why I didn't say more myself.

somehow gap lied ;-), look at olorin's post here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=46392


3y+a=a(y-2)-6

how can you solve this in terms of y and a?

pls&&thx


can $ \infty$ be used as a number?
prove or disprove it.

if it's a real number, is it odd or even?

5.8 isn't odd or even, is it?

Infinity is not a number.  It is a concept.   :D

Anyways, alan tells me that
$ \lim_{x\rightarrow 0}{\frac{x}{x}}\neq 1$
?
isn't that true?

...Because it's not a number...

Real numbers are defined.  But $ \infty\equal{}\frac{1}{0}$ which is undefined.   :P

I'd suggest this topic stops. It's by no means MC level to discuss infinity.

[quote="ra5249"]this should not be under mathcounts, as proofs and infinity have nothing to do with mathcounts >.<
and zhang, that limit is equal to one
idk why alan tells you otherwise[/quote]

maybe because he wasn't really paying attention to what I said...


ALEXANDRA ONLY GOT 3RD?! BSHEN GOT 1ST
i got 5th epic phail

dude fifth at your chaps is pretty beastly
yury had a forty-five, alex had a forty-four
blah we messed up in team round =(

this chapter was easy, so the ranking doesn't actually show how good you are going to do at states, with the exception of bobby shen.

and there was score differences of only 1 or 2 in the top 10, and couple of ties.

so be happy, be healthy, be glad.

dingzhou's avatar is true iff the reader is me. IFF. So don't say that Lawrence.

[quote="ra5249"]What were the results at your chapters?
AUSTIN
1. Kevin Tian
2. Yury Aglyamov
3. Alex Ilic
4. Allan Sadun
5. Austin Fields
6. Steven Chen[/quote]

WOW, WHO IS KEVIN TIAN, HE MUST BE REALLY SMART?!?!?!?!?!?!!?!?!?

obviously your not from texas, so you shouldn't be in this forum

obviously anyone who thought i was not kidding, is most likely from texas, OWNAGE!!!!!!!!!
sorry guys.

1. Given collinear points A, B, and C, AB=12, BC=35, and AC=23. Which point is between the other two?
2. Given collinear points A, B, and C, AB=3, BC=7, and AC=10. Which point is between the other two?
3. Given collinear points A, B, and C, AB=132, BC=101, and AC=31. Which point is between the other two?
4. Given collinear points A, B, and C, AB=7, BC=4, and AC=11. Is point B between the other two?
5. Given collinear points A, B, and C, AB=5, BC=9, and AC=4. Is point C between the other two?
6. Given collinear points A, B, and C, AB=12, BC=1, and AC=11. Is point A between the other two?
7. If B is between A and C, and AB=7, and AC:BC=3:2, find AC and BC.
8. A line is _____ times longer than that of a ray.
9. If B is between A and C, and AB=4, and AC:BC=9:4, find AC and BC.
10. If B is between A and C, and AB=1/3, and AC:BC=4:3 1/3, find AC and BC.

Texas people are smart. I am from illionios and i came here just to look at what Texas people talk about. I think the best score in my chapter was 39.

lol 39 was like the 7th highest score in our chapter and it would've been like 20th at some chapters.

What about u what did u get at ur chapter?
I think I got like 30 :blush:

why are people still posting in this???? lol


function
integration
calculus
calculus computations

Sum the series $ \sum_{n\equal{}0}^\infty{2n \choose n}^{\minus{}1}$.

hello, we have $ \sum_{n\equal{}0}^{\infty}\binom{2n}{n}^{\minus{}1}\equal{}\frac{2}{27}(18\plus{}\sqrt{3}\pi)$
Sonnhard.

May we set an ODE in which the function $ f(x) \equal{} \sum_{n \ge 0}\binom{2n}{n}^{ \minus{} 1}x^n$ is the only root?

\begin{eqnarray*}
   \sum\limits_{n=0}^{\infty }{\frac{(n!)^{2}}{(2n)!}}&=&\sum\limits_{n=0}^{\infty }{n\cdot \frac{\Gamma (n)\Gamma (n+1)}{\Gamma (2n+1)}} \\
 & =&\sum\limits_{n=0}^{\infty }{\left\{ n\int_{0}^{1}{x^{n-1}(1-x)^{n}\,dx} \right\}} \\
 & =&\int_{0}^{1}{\left\{ \frac{1}{x}\sum\limits_{n=0}^{\infty }{n\left( x-x^{2} \right)^{n}} \right\}\,dx} \\
 & =&\int_{0}^{1}{\frac{1-x}{\left( 1-x+x^{2} \right)^{2}}\,dx}.
\end{eqnarray*}

Sonnhard, please don't post if you have no solution.

[quote="Bipshula"]\begin{eqnarray*} \sum\limits_{n = 0}^{\infty }{\frac {(n!)^{2}}{(2n)!}} & = & \sum\limits_{n = 0}^{\infty }{n\cdot \frac {\Gamma (n)\Gamma (n + 1)}{\Gamma (2n + 1)}} \\
& = & \sum\limits_{n = 0}^{\infty }{\left\{ n\int_{0}^{1}{x^{n - 1}(1 - x)^{n}\,dx} \right\}} \\
& = & \int_{0}^{1}{\left\{\frac {1}{x}\sum\limits_{n = 0}^{\infty }{n\left( x - x^{2} \right)^{n}} \right\}\,dx} \\
& = & \int_{0}^{1}{\frac {1 - x}{\left( 1 - x + x^{2} \right)^{2}}\,dx}. \end{eqnarray*}[/quote]

I think, $ \sum_{n={\color{red}{1}}}^\infty \frac {(n!)^2}{(2n)!} = \int_0^1\frac{1-x}{(1-x+x^2)^2}dx = \frac{1}{27}(9+2\sqrt{3}\pi)$.

So, $ \sum_{n=0}^\infty \frac {(n!)^2}{(2n)!} = 1+\sum_{n=1}^\infty \frac {(n!)^2}{(2n)!} = 1 + \frac{1}{27}(9+2\sqrt{3}\pi) = \frac{2}{27}(18+\sqrt{3}\pi)$.

[quote="Bipshula"]Sonnhard, please don't post if you have no solution.[/quote]
I always think of the same thing, too.
Anyone can give a calculation result by CAS.
I think that it is important that I show a process.

[quote="Kouichi Nakagawa"]
I think, $ \sum_{n \equal{} {\color{red}{1}}}^\infty \frac {(n!)^2}{(2n)!} \equal{} \int_0^1\frac {1 \minus{} x}{(1 \minus{} x \plus{} x^2)^2}dx \equal{} \frac {1}{27}(9 \plus{} 2\sqrt {3}\pi)$.[/quote]
Yes, thanks for pointing that out.


Billy Bob says hi a lot. He says hi to 2/3 of the people at the mall. There are 672 people at the mall. How many people did he say hi to?

[hide="Is there a trick to this question?!?"]2/3*672=448[/hide]

[quote="alex__ib"]The problem is impossible since you said that he says "high" to 2/3 people and we are asked for "hi" 's  :P
If you edit your post, then i will agree with funcia![/quote]

You're right, tricky question!

[quote="Arvind_sn"]Billy Bob says hi a lot. He says high to 2/3 of the people at the mall. There are 672 people at the mall. How many people did he say hi to?[/quote]\

0  :rotfl:

he said 'high' to 2/3, not 'hi'
 :rotfl:

[quote="Arvind_sn"]Billy Bob says hi a lot. He says high to 2/3 of the people at the mall. There are 672 people at the mall. How many people did he say hi to?[/quote]

2x/3=672
x=672*3/2

tricky tricky!  :wink:
[hide]$\boxed{0}$[/hide]

Incorrect. Please, before posting, see the post before

[hide]448/672[/hide] and no the post doesn't say High anywhere

[quote="Arvind_sn"]Guys I just made a typo... :oops: anyway, moogra you might want to check your answer.[/quote]

[hide]In that case, it is obviously 448[/hide]


Euler
algebra
polynomial
modular arithmetic

In 1772 Euler discovered the curious fact that $n^2 +n+41$ is prime when $n$ is any of $0,1,2, \cdots, 39$. Show that there exist $40$ consecutive integer values of $n$ for which this polynomial is not prime.

Let $p_{1},p_{2},\dots,p_{40}$ be distinct odd primes such that -163 is a square modulo every $p_{i}.$

Our goal is to find such $m$ that $(m+i)^{2}+(m+i)+41$ is divisible by $p_{i}$ for every $i=1,2,\dots,40.$ We have
\[4((m+i)^{2}+(m+i)+41)\equiv (2(m+i)+1)^{2}+163\pmod{p_{i}}\]
Since -163 is a square modulo every $p_{i}$ and $p_{i}$ is odd, the congruence $(2(m+i)+1)^{2}\equiv-163\pmod{p_{i}}$ have a solution $m\equiv a_{i}\pmod{p_{i}}.$
Using CRT, we can combine these solutions into
\[m\equiv m_{0}\pmod{\prod_{i=1}^{40}p_{i}}.\]
Taking large enough $m$ satisfying this congruence completes the proof.

we'll solve this in another way by using the fact mentioned:
Let $ p_0,p_1,...,p_{39}$ be the primes that we get when we put $ 0,1,...,39$ in the polynomial $ f(n) \equal{} n^2 \plus{} n \plus{} 41$,respectively(I mean that $ p_0 \equal{} 41,...,p_{39} \equal{} 1601$).By using CRT we can find an integer $ k$ that satisfies these congruences:
$ k \equiv \minus{} (2j \plus{} 1) \mod{p_{j}},0\leq{j}\leq39$. then we'll have $ k(k \plus{} 2j \plus{} 1) \equiv 0 \mod{p_{j}}$ but it's known that
$ j^2 \plus{} j \plus{} 41 \equal{} p_{j}$,or $ j^2 \plus{} j \plus{} 41 \equiv 0 \mod{p_{j}}$ by adding up these two congruences we get
$ (k \plus{} j)^2 \plus{} (k \plus{} j) \plus{} 41 \equiv 0 \mod{p_{j}}$.It means that when we put the numbers $ k,k \plus{} 1,...,k \plus{} 39$ in $ f(n)$ we get composite numbers.DONE.