How many 3-letter words can we make from the letters A, B, C, and D, if we are allowed to repeat letters, and we must use the letter A at least once?

This is $ 4^3\minus{}3^3\equal{}\boxed{37}$ (complementary counting).

I believe this solution is shorter and thus easier to comprehend that the solution posted.

Answer = (number of 3 lettered words) - (number of words containing no As)

(number of 3 lettered words) = 4*4*4 (since we have 4 letter to pick from)
(number of words containing no As) = 3*3*3 (we exclude A from our options)

Answer = 4^3 - 3^3

Answer = 37

But I'm pretty sure "AAA" or "AAC" are words. Just saying.

I got confused because AAD and ACA are not words but they count. :(

I attempted casework on this and I was off by a margin..

Can someone solve this using casework?

Edit: I see where I went wrong with casework, I didn't count that the A's could be in different positions oops :P

this question is easy. No one should get it wrong


function
advanced fields
advanced fields unsolved

Find the formula for $ \sum_{i \equal{} 1}^{n} i^a$ where $ a$ is a natural number.

I know there is a formula, I just don't know what it is.

See this: [url]http://www.xula.edu/math/Research/Colloquium/Past/Colloquium20030320-Paper-SahooP.pdf[/url]

The formula for kth power of first n positive integers, in terms of Bernoulli numbers is:

    $ S_k(n) = \frac {1}{k + 1}\sum_{j = 0}^{k}\binom{k + 1}{j}B_j( n + 1 ) ^{k + 1 - j}$

I knew I read it someplace... So I googled a bit, and voila :lol:

My favorite form of this:
$ \sum_{j\equal{}1}^n j^k\equal{}\frac{n^{k\plus{}1}}{k\plus{}1}\minus{} n^k\sum_{m\equal{}0}^{k\minus{}1}\binom{k}{m}\zeta(\minus{}m)n^{\minus{}m}$

This puts it in terms of powers of $ n$ rather than $ n\plus{}1$. The reference to the zeta function is unnecessary, but I like it anyway.


binomial coefficients
real analysis
real analysis unsolved

$ \sum_{n \equal{} 1}^{\infty}\frac {4^n}{n^3\binom {2n}{n}} \equal{} ?$

[color=green][Edited by moderator. The right code is \binom. And also \infty rather than \propto][/color]

A useful bit of notation: \binom{m}{k} bets you $ \binom{m}{k}$, without the bar.

$ \sum_n \binom{2n}{n}x^n$ is a nice series, and multiplying by $ n^3$ isn't too bad. Taking reciprocals of the terms- that gets hairy. Binomial coefficients in the denominator don't usually behave nicely.

hello, we get
$ \sum_{n\equal{}1}^{\infty} \frac{4^n}{n^3\binom{2n}{n}}\equal{}\pi^2\log(2)\minus{}\frac{7}{2}\zeta(3)$.
Sonnhard.

I suppose starting with this
\[ \sum _{n\equal{}1}^{\infty }{\frac {{x}^{n}}{{2\,n\choose n}}}\equal{}{\frac {\sqrt
{x} \left( 4\,\sqrt {x}\minus{}{x}^{3/2}\plus{}4\,\sqrt {4\minus{}x}\arcsin \left( 1/2\,
\sqrt {x} \right)  \right) }{ \left( \minus{}4\plus{}x \right) ^{2}}}\]
You can get there by doing three successive integrals.  But it doesn't look pretty.


All fixed. =D

Thanks!

-Jay, Pres. Radnor Math Club

[i]Moderator note: removed 'please read AMC Director' as the problem has been fixed.  -mathfanatic[/i]

Wow... that is a position that no coach/student/AMC Director wants to be in.  And wasn't Radnor the AHSIMC (sp?) winner?

The deadline was today for late registration, actually. You should've just sent it in. And well...it's kind of a week before the test, and since AMC  needs to actually send it to you several days ahead of time, which requires that you send them the forms and stuff which also takes several days, I'm not sure there would be enough time to let you register. Good luck though.

Wow, Jay, I'm sorry.  Something similar happened to me--only I'm the only one at my school who actually cares, so I might just take it as a Visitor (IF I can figure out how.)  You may just want to call their office (there's an -800 number somewhere, or): $\footnotesize \mbox{\text 402-472-2257}$ or e-mail your director.  I've seen one of the other directors--mine over the mountain west--post, but I doubt if he'll post anything right away for how much stuff they must be getting.  Just make sure you make any calls [b]before 5:00 [/b]P.S.T.

Eddy

Yay!  Registration approved.

Thanks guys!


Let $A$ be a nonempty set of positive integers, and let $N(x)$ denote the number of elements of $A$ not exceeding $x$. Let $B$ denote the set of positive integers $b$ that can be written in the form $b=a-a^{\prime}$ with $a\in A$ and $a^{\prime}\in A$. Let $b_1<b_2<\cdots$ be the members of $B$, listed in increasing order. Show that if the sequence $b_{i+1}-b_i$ is unbounded, then $\lim_{x\to \infty}\frac{N(x)}{x}=0$.

I'll try to reconstruct my solution on this one...

We argue by contradiction. Suppose that $b_{i+1}-b_i$ is unbounded, and $\limsup_{x\to\infty}\frac{N(x)}{x}=a>\frac1M$

Let $c_0=0$. Let $c_1$ be an arbitrary integer not in $B$. For $1< k <M$, let $c_k=b_j-1$ for some $j$ such that $b_j-b_{j-1}>c_{k-1}+1$. Then, for any $k>i$, $c_k-c_i\not\in B$. The $M$ sets $A_k=A+c_k$ are therefore disjoint.

By the density condition, we can find some $t>0$ such that the set of $x$ for which $\frac{N(x)}{x}\ge \frac1M+t$ is unbounded. Find such an integer $x$ with the additional property that $x>\frac{c_{M-1}}{Mt}$.
Now, we count the elements of $A_0\cup A_1\cup\cdots\cup A_{M-1}$ in $\{1,2,\dots,x+c_{M-1}\}$. There are at least $x\left(\frac1M+t\right)$ elements of each $A_k$ in this range, so in total there are at least $x+tMx$ elements in the set because the $A_k$ are disjoint. On the other hand, there are at most $x+c_{M-1}$ elements in this union, so $tMx\le c_{M-1}$, contradicting our choice of $x$.

The only way out of this contradiction is to say that the sets were not disjoint, and $b_{j-1}-b_j$ was in fact bounded.


calculus
integration
algebra
polynomial
calculus computations

If $ I_{n}$ denotes $ \frac{1}{n!}\int_{0}^{x} t^{n}e^{\minus{}t}\ dt$, find a relation connecting $ I_{n}$ and $ I_{n\minus{}1}$. Deduce that the value of $ e^{x}$ exceeds its Maclaurin polynomial of degree $ n$ by $ \frac{e^{x}}{n!}\int_{0}^{x} t^{n}e^{\minus{}t}\ dt$.

$ I_n \equal{} \frac{1}{n!} \int_0^xt^ne^{\minus{}t}\,dt \equal{} \frac{1}{n!} \left[\minus{}x^ne^{\minus{}x} \plus{} \int_0^x nt^{n\minus{}1}e^{\minus{}t}\,dt\right]$

$ \equal{} \minus{} \frac{x^n}{n!}e^{\minus{}x} \plus{} \frac{1}{(n\minus{}1)!}\int_0^x nt^{n\minus{}1}e^{\minus{}t}\,dt$

$ \equal{}> I_n \equal{} I_{n\minus{}1} \minus{}  \frac{x^n}{n!}e^{\minus{}x}$.

Now we have

$ I_1 \equal{} I_0 \minus{} x e^{\minus{}x} \equal{} 1 \minus{} e^{\minus{}x} \minus{} xe^{\minus{}x}$

$ I_2 \equal{} I_1 \minus{} \frac{x^2}{2!} e^{\minus{}x} \equal{} 1 \minus{} e^{\minus{}x} \minus{} xe^{\minus{}x} \minus{} \frac{x^2}{2!} e^{\minus{}x}$

$ I_3 \equal{} I_2 \minus{} \frac{x^3}{3!} e^{\minus{}x} \equal{} 1 \minus{} e^{\minus{}x} \minus{} xe^{\minus{}x} \minus{} \frac{x^2}{2!} e^{\minus{}x} \minus{} \frac{x^3}{3!} e^{\minus{}x}$

$ ...$

$ I_n \equal{} 1 \minus{} e^{\minus{}x} \minus{} xe^{\minus{}x} \minus{} \frac{x^2}{2!} e^{\minus{}x} \minus{} ... \minus{} \frac{x^{n\minus{}1}}{(n\minus{}1)!} e^{\minus{}x} \minus{} \frac{x^n}{n!} e^{\minus{}x}$

and denoting the Maclaurin polynomial of $ e^x$ of order n by $ P_n$ this is

$ I_n \equal{} 1\minus{}P_ne^{\minus{}x} <\equal{}> e^x \equal{} P_n \plus{} e^x I_n \equal{} P_n \plus{} \frac{e^x}{n!} \int_0^xt^ne^{\minus{}t}\,dt$.


topology
vector
function
Functional Analysis
absolute value
advanced fields
advanced fields open

Hi!

First of all, a very simple problem.
1. Suppose $\{A_n\}$ is a family of subsets of metric space $X$. It is known that for any increasing sequence $\{n_k\}$ there is a subsequence $\{n_{k_l}\}$ s.t. it is possible to select $a_{n_{k_l}}\in A_{n_{k_l}}$ and $a_{n_{k_l}}\to a$.
Prove that it is possible to select $a_n\in A_n$ in such a way that $a_n\to a$.

And now the problem on which I got stuck.
The same condition as above, but we replace the metric space $X$ with reflexive separable banach space and the strong convergence to the weak convergense. Is the satement still true?
Now formally.
2. Suppose $\{A_n\}$ is a family of subsets of reflexive separable banach space $X$. It is known that for any increasing sequence $\{n_k\}$ there is a subsequence $\{n_{k_l}\}$ s.t. it is possible to select $a_{n_{k_l}}\in A_{n_{k_l}}$ and $a_{n_{k_l}}\rightharpoonup a$.
Is it true that we can select $a_n\in A_n$ in such a way that $a_n\rightharpoonup a$?

Fortunately, I avoided to prove it, since all my sets $A_n$ consists of single points. But the general question seems to be very interesting. I have a feeling that I am missing something simple here. :?

Ops...  :blush:
It was really easy. The weak topology of separable banach space is metrizable. So [b]2[/b] comes immediately from [b]1[/b].

BTW, these are interesting problems themself to show that $X^*$ is separable and then to show how to metrize the weak topology of $X$.

[quote="Myth"]
It was really easy. The weak topology of separable banach space is metrizable. So [b]2[/b] comes immediately from [b]1[/b].[/quote]
To be honest, I thought so for the first ten minutes after seeing your question too and almost posted it :blush: but then I realized that it is not that easy. You can metrize the unit ball (or a ball of any finite radius), but not the entire space. So, if all $A_n$ are all contained in some fixed ball, then the metrization argument is perfect. But if we do not assume such a uniform bound a priori, there are counterexamples. To describe one of them, let us assume that $X$ is just a separable Hilbert space. Let $a=0$. Let $\{e_j\}_{j\geq 0}$ be an orthonormal basis in $X$ and let $p(x)=\sum_{j\geq 0}2^{-j}|(x,e_j)|$ be the seminorm that is used for metrization of balls. Then $x_n$ weakly converges to $0$ if and only if the sequence $\|x_n\|$ is bounded and the sequence $p(x_n)$ converges to $0$. For each pair of positive integers $k,\ell$ choose some vector $v_{k\ell}$ such that $\|v_{k,\ell}\|=k$, $p(v_{k\ell})=\frac1\ell$. Choose also some bijection $\psi:\mathbb N\to \mathbb Q$. The sets $A_n$ will consist only of vectors $v_{k\ell}$ and are defined according to the following rule: $v_{k,\ell}\in A_n$ if and only if $\psi(n)\notin[k,\ell)$.

Note that the sequence $v_{k_n\ell_n}$ weakly converges to $0$ if and only if $k_n$ are bounded and $\ell_n\to\infty$. So, if we have a sequence $v_{k_n\ell_n}\in A_n$ weakly converging to $0$, we would have some $K$ such that all $A_n$ except finitely many contain a vector $v_{k,\ell}$ with $k\leq K$, $\ell\geq K+1$. But it cannot happen if $\psi(n)\in [K,K+1)$ and the set of such $n$ is infinite. On the other hand, if $M\subset\mathbb N$ is an infinite subset, then either $\psi(M)\cap [\ell,+\infty)$ is non-empty for all $\ell$, and then we can choose vectors $v_{1,r}$ in some $A_{n_r}$, $n_r\in M$, which will give us a subsequence weakly converging to $0$, or $\psi(M)\subset (-\infty,K)$ for some $K$ and then $v_{K,n}\in A_n$ for all $n\in M$.

Yes, I forgot to say about this aspect of metrization :(
To be honest, by some speculations in my mind I came to conclusion that $A_n$ are necessarily bounded. :(
It happens very often to me, since I very rare write anything to a piece of paper and all my solutions live only in my head, and it leads to a number of errors.

Your construction is very complicated. How did you imagine it?

Initially, in my paper I defined $A_n$ as sets lying in $\varepsilon$-neighbour of some point $u\in W_q^2(\Omega)$, so clearly their were bounded. Further, I slightly changed my approach and got that $A_n$ are just single points.

Anyway, the problem I formulated is very nice (in my opinion, of course).

[quote="Myth"]Your construction is very complicated. How did you imagine it?
 [/quote] I was lucky to start thinking in the right way. Since weak convergence is all about boundedness of norms and $p(x_n)$ tending to $0$, I defined the sets $N(R,\delta)=\{n\in \mathbb N: A_n\cap B(R)\cap\{p<\delta\}=\varnothing\}$ and tried to express the given conditions in terms of these sets. After some time I managed to do it and the problem reduced to the following: does there exist a collection of sets $N(R,\delta)$ that is decreasing in both $R$ and $\delta$ and such that a) for every $R$ there exists $\delta$ such that $N(R,\delta)$ is infinite and b) For every function $\delta(R)$ there exists $\rho$ such that $\bigcap_{R<\rho}N(R,\delta(R))$ is finite. This reminded me of the well-known problem about construction of a family of subsets of $\mathbb N$ of cardinality continuum with the property that every 2 sets have finite intersection. The rest was fairly straightforward, but it took me almost the entire evening to finish the solution.

[quote="Myth"]Anyway, the problem I formulated is very nice  [/quote]
Yes, it was beautiful. The same can be said about many other problems you proposed (in my humble opinion, of course).

[quote="Myth"] I very rare write anything to a piece of paper and all my solutions live only in my head[/quote] I envy you here. I often need to write the statement of the problem down before I can start solving it. As a result, my office is usually a complete mess with paper lying (and flying) everywhere :)

I will include both cases (bounded and unbounded) in my future functional analysis course.

I know reviving old threads isn't always welcome, but while leafing through Paul Halmos's wonderful "A Hilbert Space Problem Book" I came across the following problem: The weak topology of an infinite-dimensional Hilbert space is not metrizable.

So fedja's counterexample is the rule, not the exception :). I hope the following is a correct proof of the theorem.

If the weak topology of an infinite-dimensional hilbert space $H$ were metrizable, the space would be first-countable under this topology. It thus suffices to prove that $0$ does not have a countable basis of weakly open neighborhoods. This is equivalent to proving the following:

We cannot find a sequence $x_1,x_2,\ldots$ of vectors in $H$ s.t. for every vector $x$ and every $\varepsilon>0$ there is a $k$ and an $n$ s.t. $|(y,x_i)|<\frac 1n,\ \forall i\in\overline{1,k}\Rightarrow |(y,x)|<\varepsilon$.

Suppose the contrary. Extract a linearly independent subset from $(x_i)_i$ and perform the Gram-Schmidt orthonormalization on it to get an orthonormal system $(e_i)_i$ s.t. the span of $e_1,\ldots,e_k$ equals the span of $x_1,\ldots,x_{m_k}$ for some $m_k\ge k$, for all $k$. Complete this orthonormal system to a maximal one in $H$. Choose $\varepsilon>0,n$ and $x_1,\ldots,x_k$ as described above. Now choose $x$ s.t. it has only non-zero Fourier coefficients when represented in the orthonormal basis constructed above, and consider a vector $y$ orthogonal to all the $e_i$ 's which appear in the decompositions of $x_1,\ldots,x_k$ (there are finitely many such $e_i$'s), but not orthogonal to $x$. For a complex number $\alpha$ with large enough absolute value, we will have $|(\alpha y,x_i)|=0<\frac 1n,\ \forall i\in\overline{1,k}$, but $|(\alpha y,x)|>\varepsilon$, contradicting the existence of the countable basis.

Is it correct? :?


[hide] Let $ x$ be the number of $ 10$ dollar bills. Then, $ x\plus{}(x\plus{}1)\plus{}(x\plus{}3)\equal{}10 \implies x\equal{}2$. Thus, he has $ 2$ $ \$10$ dollar bills, $ 3$ $ \$5$ dolllar bills, and $ 5$ $ \$1$ dollar bills (this also checks with the value of the money that we were given).[/hide]

[quote="sheeprule"]If someone has 10 bills in their wallet and in all they equal 40 dollars, if they have one more 5 dollar bill than 10 dollar bill, and two more 1 dollar bills than 5 dollar bills...

how many of each do they have?[/quote]

This question gives too much information.  You don't need to be told the relationship
of the number of 1-dollar bills to the number of 5-dollar bills.

 It could certainly could have been:

"Suppose someone has 10 bills in their wallet, and in all they equal 40 dollars.
If they have one more 5-dollar bill than 10-dollar bill and at least one 1-dollar bill,
then how many of each type of bill do they have?"

Candidates:

A) 1 ten & 2 fives
B) 2 tens & 3 fives

Because there is a total of 10 bills (given), then scenario A) would
mean there are 7 ones, but that totals to only 27 dollars.

But in scenario B), there would have to be 5 ones using the same reasoning
because there is a total of 5 bills between the tens and fives.  That equals to
(tens) 2 bills + (fives) 3 bills + (ones) 5 bills = 40 dollars.


solve for x

\[ \minus{}1\plus{}x\equal{}\sum_{k\equal{}1}^{2000} \binom{2000}{k}\]

hello, $ x\equal{}\sum_{k\equal{}1}^{2000}\binom{2000}{k}\plus{}1$.
Sonnhard.

EDIT: Wow, epic failed.

[hide="Solution"]
We have $ x \equal{} \sum_{k \equal{} 1}^{2000}\binom{2000}{k} \plus{} 1$.

It is known that $ \sum_{k \equal{} 0}^{n}\binom{n}{k} \equal{} 2^n$.

Thus, $ x \equal{} 2^{2000} \plus{} 1\minus{}\binom{2000}{0}\equal{}2^{2000}$.

Answer: $ 2^{2000}$[/hide]

Yep, it's 2^2000 as tonypr said because k starts from 0, not 1.

Hint: how many subsets exist of a set with n elements? Count it two different ways.

[quote="ckck"]Time to be stupid here and ask...how does that work?? D:[/quote]

Using binomial theorem, we have $ (x\plus{}1)^n\equal{}\binom{n}{0}x^n\plus{}\binom{n}{1}x^{n\minus{}1}\plus{}\cdots\plus{}\binom{n}{n\minus{}1}x^{1}\plus{}\binom{n}{n}x^0$

If $ x\equal{}1$, we will have $ 2^n\equal{}\binom{n}{0}\plus{}\binom{n}{1}\plus{}\cdots\plus{}\binom{n}{n\minus{}1}\plus{}\binom{n}{n}.$

[quote="ckck"]Time to be stupid here and ask...how does that work?? D:[/quote]
Another way would be to look at Pascal's triangle. Knowing that each number on the triangle is the number of ways you can get to it from the top moving either left or right, try summing the numbers in the rows of the triangle. If you're not seeing it,
[hide="Hint"]Count the number of ways you can get to any number on a certain row.
[hide="Another"]Now count it in a different way.[/hide][/hide]
The idea is really similar to the one behind ThinkFlow's method.

Once you know the identity, the problem is really straightforward. Just plug in and isolate x.


geometry
geometric transformation
rotation
reflection
floor function
search
group theory

"In how many ways may we color the edges of a (regular) (2n + 1)-gon
free to move in three-dimensional space so that n edges are colored
red and n + 1 edges are colored blue? Your answer may depend on
whether n is even or odd."

This is a question from a textbook of mine and I was shocked that they would include such a hard problem. Do you guys think this is olympiad level or is it more routine than that?

Well, it depends what the content of the textbook is.  This is a pretty standard problem from Polya theory and we can hit it directly with the lemma that isn't Burnside's.  Our set $ X$ is the set of $ n$-subsets of $ [2n \plus{} 1]$ (the red vertices), and our group $ G$ is the dihedral group acting in the obvious way.  The identity element fixes all $ \binom{2n \plus{} 1}{n}$ elements of $ X$.  Since $ (n, 2n \plus{} 1) \equal{} 1$, the non-identity rotations don't fix any element of $ X$, and we only have to consider the reflections.  If $ n$ is even then each reflection fixes $ \binom{n}{n/2}$ subsets, while if $ n$ is odd then each reflection fixes $ \binom{n}{(n \minus{} 1)/2}$ subsets.  Thus, the number of orbits (which is what we're after) is given by
\[ \frac {\binom{2n \plus{} 1}{n} \plus{} (2n \plus{} 1) \cdot \binom{n}{\lfloor n/2 \rfloor}}{2(2n \plus{} 1)}.\]
This is sequence [url=http://www.research.att.com/~njas/sequences/A007123]A007123[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

Here are some other threads that use a similar technique:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=319975
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=314174
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=288514
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=14125
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=259872
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=252879
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=228658
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=225043
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=194846
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=192829
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=179833
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=106466
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106467
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=325429882&t=109090


[quote="Interval"]The student radio station has a broadcasting range of 24 miles. Describe the locus of points which represents the outer edge of the broadcasting range.[/quote]
[hide]I think it is a circle with radius of 24 miles. [/hide]

hard extention,

if the locus was graphed on a cartesion coordinate plane and the center at the origin, where each unit square was a house, how many houses would the locus pass through?


The prime factorization of 2160 is $ 2^4 \times 3^3 \times 5$. How many of its positive integer factors are perfect squares?

Squares include even exponent of 2,3 and 5.
For 2: 0,2,4 can be the exponents
3: 0,2
5: 0

That's 3*2*1=6 possibilities


inequalities algebra function domain induction

How  do you prove Bernoulli's Inequality, which says that for x>-1, x not equal to 0, and n a positive integer greater than 1, then:

$(1+x)^{n}>1+nx$.

[hide="Proof"]Let $f\left(x\right)=\left(1+x\right)^{n}-1-nx$.  Then you have \[f'\left(x\right)=n\left(1+x\right)^{n-1}-n.\] Given the domain, $f'\left(x\right)>0$ for $x>-1$ and $x\neq 0$.  Also, we have $f\left(0\right)=0$ so the function is greater than $0$ since it is increasing.  This proves the inequality.

Alternatively, you could consider the binomial series which would give you the same result.[/hide]

You can also prove it by induction in $n$ (more simply than the other proof above):

Base case: $n = 2$:
$(1+x)^{2}> 1+2x$
$1+2x+x^{2}> 1+2x$
$x^{2}> 0$ which is true for all real $x \not= 0$

Inductive Step: assume $(1+x)^{n}> 1+nx$
Now because $x >-1$, we can multiply both sides of the inequality by $1+x$, so:
$(1+x)^{n+1}> 1+(n+1)x+nx^{2}> 1+(n+1)x$, so
$(1+x)^{n+1}> 1+(n+1)x$, which completes the inductive proof.

[quote="maokid7"]yes but the induction proof only proves that it is true for integers and not all real values.[/quote]

n[i] is [/i]a positive integer

oops sorry.
normally when i see this inequality it stats that $n$ is a real number such that
$n\geq 1$

[quote="amcavoy"]Let $f\left(x\right)=\left(1+x\right)^{n}-1-nx$.  Then you have \[f'\left(x\right)=n\left(1+x\right)^{n-1}-n.\] Given the domain, $f'\left(x\right)>0$ for $x>-1$ and $x\neq 0$.  Also, we have $f\left(0\right)=0$ so the function is greater than $0$ since it is increasing.
[/quote]
I would rather think $f'(x)<0$ if $-1<x<0$ and $f'(x)>0$ if $x>0$, hence it has a  (nonnegative) minimum at x=0.


[quote="Cezar Lupu"]Exista inel cu 2006 elemente si exact 10 elemente inversabile?[/quote]

Fie cele zece elemente inversabile $x_{1},x_{2},...,x_{10}$. Desigur unul dintre elemente este $1$ deci din cele 9 (inpar) elemente ramase una trebuie sa fie egal cu inversa sa. Fie asta $x\ne 1$ Deci $x=x^{-1}$ deci $x^{2}=x$ rezulta $x(x-1)=0\Rightarrow x=1$ imposibil

da :) si in plus nu te-ai folosit deloc de ipoteza ca inelul are 2006 elemente, sau o fi doar o capcana :P ma mai gandesc :roll:

[quote="maky"]hm...
nu cred ca daca $x=x^{-1}$ rezulta ca $x^{2}=x$... :blush:[/quote]

Da am observat dupa ce am trimis:) Rusinica:((

cred ca ce a vrut cckek sa spuna este ca daca din 9 elemente exista
sigur 2 elemente care au acelasi invers(deoarece 9 impar) astfel ar rezulta
ca daca au acelasi invers sunt identice deci multimea n-ar mai fi de 9
elemente ci defapt doar de 8 si impreuna cu elementul unitate n-ar mai
face 10 ci doar 9.


inequalities
LaTeX
inequalities proposed

Hi !
Below you can find an inequality invented by me and I know it's really easy, but the problem [imaginary anyway] is to solve it in the most original way you can. Interesting ? I hope so  :)  Enjoy  :D

a,b,c are non-negative reals:

$a^3 + b^3 + c^3 \geq \sqrt{a^2c^2(b+a)(b+c)+b^2c^2(a+b)(a+c)}$

.

is it SO EASY that nobody wanna try ?  :maybe:

[quote="Megus"]a,b,c are non-negative reals:

$a^3 + b^3 + c^3 \geq \sqrt{a^2c^2(b+a)(b+c)+b^2c^2(a+b)(a+c)}$[/quote]

If this is really easy, please post your solution. I don't find it easy; it took me a week until I found ANY solution. Well, here is my solution. No doubt it is very original. Enjoy.

We have to prove that

$a^{3}+b^{3}+c^{3}\geq \sqrt{a^{2}c^{2}\left( b+a\right) \left( b+c\right) +b^{2}c^{2}\left( a+b\right) \left( a+c\right) }$.

In other words, we must prove

$\left( a^{3}+b^{3}+c^{3}\right) ^{2}\geq a^{2}c^{2}\left( b+a\right) \left( b+c\right) +b^{2}c^{2}\left( a+b\right) \left( a+c\right) $.

We will solve a more symmetric formulation of this inequality:

[i]If we denote $x=b^{2}c^{2}\left( a+b\right) \left( a+c\right)$, $y=c^{2}a^{2}\left( b+c\right) \left( b+a\right)$, $z=a^{2}b^{2}\left( a+b\right) \left( a+c\right)$, then prove that all three inequalities $\left( a^{3}+b^{3}+c^{3}\right) ^{2}\geq y+z$, $\left( a^{3}+b^{3}+c^{3}\right) ^{2}\geq z+x$ and $\left( a^{3}+b^{3}+c^{3}\right) ^{2}\geq x+y$ hold.[/i]

This problem is symmetric in a, b and c; hence, we can WLOG assume $a\leq b\leq c$. Then, we can denote u = b - a and v = c - b, and get $u\geq 0$ and $v\geq 0$. On the other hand, since u = b - a and v = c - b, we have b = a + u and c = b + v; hence we can express b, c, x, y and z through a, u and v, and thus, after a short calculation,

$\left( a^{3}+b^{3}+c^{3}\right) ^{2}-\left( x+y\right) $
$=10a^{4}v^{2}+68au^{3}v^{2}+62au^{2}v^{3}+60a^{2}uv^{3}+91a^{2}u^{2}v^{2}$
$+40a^{3}u^{2}v+52a^{3}uv^{2}+6a^{4}uv+40au^{4}v+6uv^{5}+15u^{2}v^{4}$
$+6av^{5}+v^{6}+30auv^{4}+a^{4}\left( a-v\right) ^{2}+15a^{2}v^{4}+21u^{3}v^{3}$
$+18u^{4}v^{2}+9u^{5}v+16au^{5}+32a^{2}u^{4}+13a^{4}u^{2}+30a^{3}u^{3}$
$+20a^{3}v^{3}+3u^{6}+2a^{5}u+63a^{2}u^{3}v$,

what is clearly $\geq 0$. Hence, $\left( a^{3}+b^{3}+c^{3}\right) ^{2}\geq x+y$. Now, since $a\leq b\leq c$, we have $c\geq a$, so that $c^{2}\left( b+c\right) \geq a^{2}\left( b+a\right) $, thus also $c^{2}a^{2}\left( b+c\right) \left( b+a\right) \geq a^{2}b^{2}\left( a+b\right) \left( a+c\right) $, or, in other words, $y\geq z$. Similarly, $x\geq z$. Hence, the inequality $\left( a^{3}+b^{3}+c^{3}\right) ^{2}\geq x+y$ implies both $\left( a^{3}+b^{3}+c^{3}\right) ^{2}\geq z+x$ and $\left( a^{3}+b^{3}+c^{3}\right) ^{2}\geq y+z$. And the proof is complete.

This solution shows that the inequality is very crude. Equality is only achieved for a = b = c = 0. Probably there are some sharper versions of this inequality. Most likely these sharper versions are much easier to prove.

I am running out of time. Please post your elegant solutions.

  Darij

First case a+b < 2c then

a^3+b^3+c^3 ≥ c√(a+b)√(c(a^2+b^2)+ 2abc) ≥ c√(a+b)√(c(a^2+b^2)+ab(a+b))

so a^3 +b^3+c^3 ≥ c(a+b)√(c(a+b)^2) = c√c *(a+b)√(a+b)

Using a^3 +b^3 ≥ [(a+b)^3]/4

a^3 +b^3+c^3 ≥ [(a+b)^3]/4 +c^3 ≥ c√c *(a+b)√(a+b)
(equality when a = b = c = 0 ).

Second case a+b>2c then inequality is equalent to :

(a+b)^3 +c^3 ≥ 3abc(a+b)+ c√(a+b)*√(c(a^2+b^2)+ab(a+b)) or

(a+b)^3 +c^3 ≥ 3abc(a+b)+ c√(a+b)*√(c(a+b)^2+ab(a+b-2c))

We can consider mixing variables or Sturm metod - pair (a,b) change with
((a+b)/2,(a+b)/2)
Right part will remain constant, left part increase, because a*b increases.
So we can say a=b and using our second case a>c.

So 2a^3 +c^3  ≥ 2ac√a(a+c).

Squaring : 4a^6 +c^6 ≥ 4(a^4)*(c^2) remembering that a > c - obliviously.
(equality when a = b = c = 0 ).

Well, this problem is not too difficult. We can use mixing variables.
Let $ f(a,b,c)=a^3+b^3+c^3-\sqrt{a^2c^2(a+b)(b+c)+b^2c^2(a+b)(a+c)}$
               $ =a^3+b^3+c^3-c\sqrt{(a+b)[ab(a+b)+c(a^2+b^2)]} $
And it is easy to see that:
$ f(a,b,c) \geq f(\sqrt{\frac{a^2+b^2}{2}},\sqrt{\frac{a^2+b^2}{2}},c) $.
Now,we just need prove for $ a=b $.That means:
$ 2a^3+c^3 \geq 2\sqrt{a^3c^2(a+c)} $
But this is clear by Cauchy:
$ 2a^3+c^3 \geq 2\sqrt{a^3c^2 \cdot 2c} \geq2\sqrt{a^3c^2(a+c)}$ (if $a \leq c $)
or
$ a^3+(a+c)(a^2-ac+c^2) \geq 2\sqrt{a^3(a+c)(a^2-ac+c^2)} \geq 2\sqrt{a^3c^2(a+c)} $.(if $ a \geq c $). :)

Ok, here is my solution :
$2(a^3 + b^3 + c^3) \geq a^2(b+c) + b^2(a+c) + c^2(a+b)$
because {3,0,0} majorizes {2,1,0} this inequality is true due to the Muirhead's Theorem, so we have [using Cauchy]:
$a^3 + b^3 + c^3 \geq \frac{a^2(b+c) + b^2(a+c) + c^2(a+b)}{2} \geq \sqrt{(a^2(b+c)+b^2(a+c))c^2(a+b)} = \sqrt{a^2c^2(a+b)(b+c) + b^2c^2(a+c)(a+b)}$

Darij: What do you think of my solution ?  :D
prowler: use LateX - it's not so hard....
Anh Cuong: nice solution  :)

Muuuh. I knew you would have something like this... And, when one reads the solution, it looks really simple...  :D  :D  :D

Just a remark: Here -

[quote="Megus"]$2(a^3 + b^3 + c^3) \geq a^2(b+c) + b^2(a+c) + c^2(a+b)$
because {3,0,0} majorizes {2,1,0} this inequality is true due to the Muirhead's Theorem,[/quote]

you needn't even to apply Muirhead. Just remember that $b^3+c^3\geq b^2c+c^2b$ (this is clear by rearrangement, or by a simple application of $b^3+c^3=\left(b+c\right)\left(b^2-bc+c^2\right)$), and similarly $c^3+a^3\geq c^2a+a^2c$ and $a^3+b^3\geq a^2b+b^2a$; adding these three inequalities together yields

$\left(b^3+c^3\right)+\left(c^3+a^3\right)+\left(a^3+b^3\right)\geq \left(b^2c+c^2b\right)+\left(c^2a+a^2c\right)+\left(a^2b+b^2a\right)$,

or, in other words,

$2\left(a^3+b^3+c^3\right)\geq a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)$,

qed..

  Darij

Yeah, I needn't but it's really making life simpler  :) I love majorization  :P - ok, I'll try to invent a new problem [inequality of course  :D ], today or tomorrow.

Ji Chen, do you have a nice proof for it?


when can you take the next PMWC qualifying test?

Mostly likely Nov, 2008. you may email them to ask, see this:

http://www.txstate.edu/mathworks/contact.html

Erm... what is PMWC?

Sorry if this is considered as spam.

I don't really know the full form (and i'm too lazy to google it) but its an international mathematics competition for middle schoolers a wee bit harder than mathcounts

It's the Primary World Mathematics Competition.

Ernie, look here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=215511


Jasmine wishes to purchase some trading cards. She has $ \$7.50$ and the cards each cost $ \$0.85$, tax included. What is the most number of cards she can buy?

$ \lfloor\frac{7.50}{0.85}\rfloor\equal{}8$

answer : 8


Ben performed the following incorrect operations on a number. First he added -5 instead of subtraction -5. Then he multiplied his result by $ \frac{1}{4}$ instead of dividing by $ \frac{1}{4}$. Finally he squared the last result instead of taking the square root. Ben's final result was $ \frac{225}{16}$. If Ben had performed the correct operations, what would the result have been?

We can work backwards to solve this problem. To get $ \frac{225}{16}$, the last step Ben performed was to square. Therefore, we take the square root to get $ \frac{15}{4}$. Since he multiplied by $ \frac{1}{4}$, we'll divide $ \frac{15}{4}$ by $ \frac{1}{4}$ to get $ 15$. He added $ \minus{}5$, so we'll subtract $ \minus{}5$ to get $ 15 \minus{} (\minus{}5) \equal{} 20$. The original number was then $ 20$.

However, we want to know what the correct result should have been. First we subtract $ 5$ to get $ 15$. Then, we divide by $ \frac{1}{4}$ to get $ 60$. Finally, we take the square root to get $ 2\sqrt{15}$.

[quote="isabella2296"]
However, we want to know what the correct result should have been. First we subtract $ 5$ to get $ 15$. Then, we divide by $ \frac {1}{4}$ to get $ 60$. Finally, we take the square root to get $ 2\sqrt {15}$.[/quote]

Actually, I think that we are supposed to subtract $ \minus{}5$ to get $ 25$ instead.  Then, we divide by $ \frac {1}{4}$ to get $ 100$.  Taking the square root of $ 100$ we get $ \boxed {10}$


Find 2005 positive integers, not necessarily distinct, whose sum is equal to their product

   :starwars:  i just think the starwors thing is cool :gleam:

very nice now try my challange problem in the olympiad section.
The title is very hard problem. :D

[hide="A description of all such sets of integers"]
I worked on this problem before.  Any such set can be generated by taking an arbitrary set of numbers, $a_n$, and adding to this set a number of $1$s equal to $\prod_{k=1}^{n} a_k - \sum_{k=1}^{n} a_k$.
[/hide]

[hide="Does it help that"]They are the roots of $x^{2005}-kx^{2004}+k=0$? :?[/hide]

[quote="t0rajir0u"]$\prod_{k=1}^{\infty} a_k - \sum_{k=1}^{\infty} a_k$[/quote]
Typo Error in $\infty$ :D


limit
logarithms
calculus
calculus computations

$ \lim_{x\to\infty}(1\plus{}\dfrac{1}{x})^{2x}$

good luck!

Do you regard $ \lim_{x\to\infty}\left(1\plus{}\frac1x\right)^x$ as famous enough that you should just know it at a glance?

This $ \lim_{x \to \infty} x^{\frac {1}{x}} \equal{} 1$ can easily be shown by taking the log and then by taking the exponentiation of the result.

We know that the log is differentiable at 1, which gives us:

$ \lim_{h \to 0} \log \left((1 \plus{} h)^{\frac {1}{h}} \right) \equal{} \lim_{h \to 0} \frac {\log \left(1 \plus{} h\right)}{h} \equal{} \lim_{h \to 0} \frac {\log \left(1 \plus{} h\right) \minus{} \log 1}{h} \equal{} 1$

Taking the exponential gives: $ \lim_{h \to 0} \left(1 \plus{} h \right)^{\frac {1}{h}} \equal{} e$. If you set $ h \equal{} \frac {1}{x}$, you get $ \lim_{x \to \infty} \left(1 \plus{} \frac {1}{x} \right)^{x} \equal{} e$.

Now you have $ \left(1 \plus{} \frac {1}{x} \right)^{2x} \equal{} \left(1 \plus{} \frac {1}{x} \right)^{x} \cdot \left(1 \plus{} \frac {1}{x} \right)^{x}$

is how to do the same way that other:

$ \lim_{x\to\infty}(1\plus{}\dfrac{1}{x})^{x\plus{}2}$

tha answer is $ e^2$

$ \lim_{x\to\infty}(1+\frac{1}{x})^{x+2}$ for this answer is e because it is $ \lim_{x\to\infty}(1+\frac{1}{x})^{x}$ $ \lim_{x\to\infty}(1+\frac{1}{x})^{2}
= e . 1 = e$

$ \lim_{x\to\infty}(1 + \frac {1}{x})^{x + 2}$
$ = \lim_{x\to\infty}(1 + \frac {1}{x})^{\frac {x(x + 2)}{x}}$
$ = \lim_{x\to\infty} e^{\frac {x+2}{x}}$

$ = e$

(Just for another way of doing it)


I wish to disguise myself for a few days with the use of possibly makeup, wigs, clothing, etc.  Instead of a wig I might get a haircut, but I wish to be completely indistinguishable.  I know this is random, but if there are any tips, it would be much appreciated.  Thanks!

[quote="JBL"][quote="CheesyP18"]I wish to be completely indistinguishable.[/quote]

Indistinuishable from what?  Perhaps "unrecognizable," instead?[/quote]

I apologize.  Unrecognizable was the word I should have used.

Try new insoles in your shoes.  It will give you a totally different gait, if you get something with insanely high or low arches (opposite of what you're used to).  Have fun!

i dont know whether you are a girl or a guy, but girls will pay close attention to facial details (i.e eyelashes, eyebrows, makeup, hairline, as in pulled back or not)  try a different body spray.  if you are known for wearing abercrombie, wear american eagle.  people, esp girls pay attention to that kind of thing.

are colored contacts an option?

nails? painted, long, bitten

glasses?

ears (earrings or not, how many piercings)

birthmarks? (cover/ make using makeup)

this sounds fun.  good luck!

[quote="eryaman"]Hmm well, I think they say that the triangular region with vertices at the eyes and nostrils are the "defining" characteristics of a face, so you might want to do something with those areas.[/quote]

Congratulations, you turned makeup into a geometry problem.   Weird. . .

To answer star 99:

I'm a guy who usually wears khaki pants or shorts, polo shirts or just random t-shirts that I find at Goodwill.

I wear glasses, but I have another set of glasses that I'm probably going to use.

I bite my nails, so I guess I should stop that.

No piercings, and all of my birthmarks are covered by clothes.

I like the insoles idea, and I think it'll increase my height just a little.

Put a paper bag over your head.  Nobody will recognize you.  Or try the cardboard box trick.

If you can get your eyebrows going a different way, that would help.
Of course, a haircut is mandatory.
The clothes you wear sound somewhat like mine...you could try jeans with plad, and that would drmatically change your appearance  :D

I heard somewhere that concentrated sulphuric acid, if allowed to work upon your face, makes it nicely unrecognizable. But do not try that, you won't enjoy the experience.

Try wearing a burkha. It should be effective.

Be a cross dresser!

The army got fooled with that trick.

Make yourself look older. Like, say, 63.

Make sure to have a different voice. Like Mrs. Doubtfire!

[quote="star99"]long hair <--> short hair
flat hair <--> spiked hair

you could dress like a girl...that would be much easier...so many more things to do...[/quote]

Not necessarily...
There are some pretty weird outfits you could wear out there...

[quote="CheesyP18"]Why did the moderator move this to the Games & Fun Factory forum?  I want serious replies, and I had more serious replies over at the Round Table.  Could someone move this thread back over there?[/quote]

The moderator that forum must have thought G & FF was a better fit for this topic.

Relax cheesy. If you want serious replies, give us a low-down on the situation. We must know the circumstances under which you want to go around incognito before offerign any real help. And if you hide facts from your friends, it means you don't consider them your friends. Two kinds of people can give helpful advice on such things - professional experts and friends. We are not the former, you know that very well. We are willing to be the latter (I am, at least) but only if you are open enough. Oterwise, you are free to seek out experts. We won't mind.

Come on! Buy makeup, plaster, call your Uncle Fred and Aunt Jack, and be Mrs. Doubtfire! That's as good as it can get!

Of course I'm trying to hide a fact from a friend, but it has no malicious intent.  I only need to be unnoticed for a few days, after which I can safely reveal myself.  I was thinking insoles and a full wig/beard/mustache setup.  I know where to get the insoles, but I'm having a problem locating a wig/beard/mustache mask.

[quote="star99"]long hair <--> short hair
flat hair <--> spiked hair

you could dress like a girl...that would be much easier...so many more things to do...[/quote]

cough girlslockerroom cough

ewwwwww.

lets not go there.  sorry for suggesting that.

but schools out in most places.

ewwwwwwww.

there's still the bathroom problem.

wow - cheesy has 42 posts exactly! What a lucky boy!
But you can dress like a girl (unless you have huge beards or bushy moustaches - in which case you have to shave these off). As I suggested in the first post, burkha isgenerally the champion's choice in these cases.


If $\alpha$ and $\beta$ are two irrational numbers and $A =${$[n \alpha ] : n \in N$}, $B =${$[n \beta ] : n \in N$}. Find the condition that $A$ and $B$ form a partition of $N$.
$[x]$ represents the greatest integer less than or equal to $x$

It's $\frac 1\alpha+\frac 1\beta=1$. Try searching the forum (and the web, in general) for "Beatty sequence". It's been discussed here a number of times.

P.S.

Clearly, this is not an "open question", and it's hard for me to believe you thought it was unsolved :). If you don't have a solution to such a problem, you should post it in the "Unsolved Problems" section, and if you do have a solution and you're posting it as a challenge, it should go into the "Proposed and Own Problems" section.


Euler limit logarithms algebra polynomial

Well-known.  In fact, Euler showed that it diverges.

[hide="Demonstration"] We make use of the Euler product

$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{k}= \lim_{n \to \infty}\prod_{p \in \mathbb{P}, p \le n}\frac{1}{1-\frac{1}{p}}\propto \ln n$

We know that the LHS diverges.  Taking natural logs,

$\lim_{n \to \infty}\sum_{p \in \mathbb{P}, p \le n}-\ln (1-\frac{1}{p}) \propto \ln \ln n$

Note that the Taylor polynomial for $-\ln (1-\frac{1}{p})$ begins with a term $\frac{1}{p}$, and that subsequent terms are negligible because their sum converges.  Hence

$\lim_{n \to \infty}\sum_{p \in \mathbb{P}, p \le n}\frac{1}{p}\propto \ln \ln n = \infty$ [/hide]


Gauss
number theory unsolved
number theory

Sequence $(a_n)$  is given such that
                   $2^n=\sum_{d|n}{a_d}$ for all $n\geq 1$.
Prove that $a_n$ divises by $n$. :lol:

I think  if $m$ divise $n$  then $a_m$ divise $a_n$ :P

By Mobius inversion formula we have \[ a_n=\sum_{d|n}2^d\mu\left(\frac{n}{d}\right) \] and the result follows from Gauss' theorem.


I know that if two proportions are equal the following equality holds:

$ \frac{a}{b}\equal{}\frac{c}{d}\equal{}\frac{a\plus{}kc}{b\plus{}kd}$

Can you do the same if you are trying to compare two expressions?

ex)    $ \frac{a\plus{}c}{b\plus{}d}?\frac{n\plus{}c}{m\plus{}d}?\frac{a\minus{}n}{b\minus{}m}$

Thank you for your help.

im not really sure what u mean. could u give a numerical example?

Yes, in the following sense.

[b]Lemma:[/b]  For [b]positive[/b] $ a, b, c, d$, if $ \frac{a}{b}>\frac{c}{d}$ then $ \frac{a}{b}>\frac{a\plus{}c}{b\plus{}d}>\frac{c}{d}$

Note the strict inequalities.


I've heard Waterloo's Problems Problems Problems series only contain problems and solutions but no actual texts. Can you still learn some good techniques just from reading the solutions? Or would it be better to get some texts like the Art of Problem Solving?

I have a copy of one of the books and I think the solution is understandable.  Although I have no copy of Art of Problem Solving but I know that it is one of the best books ever written when it comes to problem solving.  I wish I could have a copy of it.

It's a good thing to use to study for FGH, PCF or Euclid.


What are frustums, and can they vary in size and shape?

I posted this to have a post to look for to learn about frustums and their formulas. If anyone knows any frustum formulas, hopefully you will list them.  :)

Start from

[url]http://en.wikipedia.org/wiki/Frustum[/url]
[url]http://www.mathwords.com/f/frustum.htm[/url]

and for advanced reading go to

[url]http://mathworld.wolfram.com/PyramidalFrustum.html[/url]
[url]http://mathworld.wolfram.com/ConicalFrustum.html[/url]

frustums are just cones with the top cut off. you can calculate the volume or surface area by just subtracting the big cone volume by the smaller cone volume and etc.

yep and u just subtract yea i looked on wiki, wiki rocks!

lol frustums ./.... I just think of them as yoplait yogurt containers  :roll:


For $a,b,c$ sides of a triangle, prove $\sum\sqrt{ab+(a-c)(b-c)}\ge \sum\sqrt{ab}\ge\sum \sqrt{ab-(a-c)(b-c)}$ :D

Nobody? [hide="Hint"]
After the standard substitution $a=x+y$ etc. and renaming $x,y,z$ by $a,b,c$ the inequality chain writes with the same notation as in http://www.mathlinks.ro/Forum/viewtopic.php?t=67457
simply as the $n=\frac12$ case of $\sum S^n\ge\sum\left(\frac{S+T}2\right)^n\ge\sum T^n$. (See now how I came up with this Vornicu-Schur variation :) ? And the door is wide open for composing more variations :) :) )
From here it should be clear how to proceed, at least for the right inequality which can in fact easily be shown to be true for all $n\ge0$. Note that the left one is false for $0<n< 0.20967346$ with $a=99.4, b=1, c=0$ as a counter-example :D. It is not too hard to prove it though for $n=\frac12$, nor to show that it holds for all $n\ge\frac12$.
On the contrary, I guess it must be quite tough to show the left one e.g. for $n=\frac14$. A case for Vasc ;) ? [/hide]


Since it's over, it's no use to talk about who should have won...
We should all start working.

[quote="Meghrabi"][quote="mihail911"]Now that it's over, what do u guys think of the IMO results?  I personally am surprised that Russia got first place, I expected China to.[/quote]

you act like they have ended 10th or smthng.[/quote]

no, i knew russia would get in the top 5..just seeing as china got first last year...whatever, it was a stupid prediction.


I have a very interesting inequality.
a,b,c>0 then
3(a/b+b/c+c/a) \geq 9+ \sum ab(a-b) 2 /(a+c)(b+c)
sorry about my bad English..

i don't think that the inequality is good
take a=2b=2c and b large
you will have
3(2+1+1/2)\geq 9+2b^2/3 which is not true for b very large


let $ x$ , $ y$ , $ z$ $ \in ]0,\plus{}\infty[$ , prove that :

$ \frac{x\plus{}y}{x^2\plus{}xy\plus{}y^2}$ $ \plus{}$ $ \frac{y\plus{}z}{y^2\plus{}yz\plus{}z^2}$ $ \plus{}$ $ \frac{z\plus{}x}{z^2\plus{}zx\plus{}x^2}$ $ \geq \frac{1}{x\plus{}y\plus{}z}$

created by yassine

thank's Marius Mainea

nice prof even your result is the best


Souce: USSR Olympiad
165. Factor the following expression:
$a^{10}+a^5+1$.

[hide]Let $w$ be a cubic root of unity. Then $w^3 = 1$ and $w^2 + w + 1 = 0$. So we have $w^3 \cdot w^3 \cdot w + w^3 \cdot w^2 + 1 = w^2 + w + 1 = 0$. Hence $w$ and $w^2$ are roots of the polynomial. So $(a - w)(a - w^2) = a^2 + a + 1$ is a factor. Then use division of polynomials to get the others.[/hide]


When fitting "ten-speed" gears to a bicycle, you choose two front sprockets, which must have different numbers of teeth between 40 and 62 (inclusive) and five back sprockets, which must have different numbers of teeth between 11 and 35 (inclusive). The gear ratios available to you are then obtained by dividing the number of front sprocket teeth by the number of back sprocket teeth. For example. if you choose 44 and 60 tooth front sprockets and 14, 18, 22, 26, and 30 tooth back sprockets, your ratios are

44/14
44/18
44/22
44/26
44/30
60/14
60/18
60/22
60/26
60/30

If you choose badly, some of the ratiuos will be the same (for example, 44/22 = 60/30 above) and you will not obtain ten different ratios. Supposing you choose as badly as possible, what is the smallest number of ratios you wil end up with?


out of 16 people who book a flight 1 do not turn up. What is the probability that in a 48-seat plane, if 50 people book for that flight, that there will not be enough seats?

[hide]
Let $p$ be the probability that a person does not turn up. $p = \frac{1}{16}$.

So if $49$ people show up, the probability of that is $\binom{50}{49}(1-p)^{49}p$.
And if $50$ people do, the probability is $(1-p)^{50}$.

Then add them up, and I don't want to do that.[/hide]


I'm doing test A, but I don't think I'm going to apply for early decision. Should I still include the 20 bucks grading fee when I send my solutions in? or do I include that later with the application?

I would also like to know the answer to this question.

Just send it with the rest of the application (to avoid having to write multiple checks.) We usually don't review your application until it's mostly complete, anyway.

Good question!

[quote="roadnottaken"]I took test a but plan on also taking test b (deadline problem, see my post in "Acceptance" thread).  Should I pay a grading fee for both tests, or just for one?

Ironically, this forum is a much better place to go for information pertaining to awesomemath than is its own forum (especially since that forum is now locked).[/quote]
I think it's 20 bucks if you do multiple admin tests.

[quote="roadnottaken"]Do you mean in addition to the $\$50$?[/quote]
I think so.
[quote]There is an application fee of \$50 (\$30 for processing the application materials, and \$20 for grading the admission test). If a student wants to take a second test (if time allows and spots are still available), an additional\$20 will be charged.[/quote]
It's on the bottom of [url]http://www.awesomemath.org/application.shtml[/url]


1. how can we construct a sequence whose set of limit points is exactly the set of integers?

2. can there exist a sequence whose set of limit points is exactly 1, 1/2, 1/3, ...?

The set of limit points must be closed (why?), so the answer to (2) is no.

Suppose $(a_{n})$ is a sequence converging to $0$. Then the sequence $a_{1},a_{1}+1,a_{2},a_{2}+1,a_{3},a_{3}+1,\dots$ has limit points at $0$ and $1$. This generalizes to any finite set of limit points, but you want an infinite one. For this, consider the sequence of sequences
$a_{1},a_{2},a_{3},\dots$
$a_{1}+1,a_{2}+1,a_{3}+1,\dots$
$a_{1}+2,a_{2}+2,a_{3}+2,\dots$
$\dots$
and arrange all of these numbers into a single sequence.


modular arithmetic
quadratics
number theory
prime factorization

$ 2^{x} \plus{} 3^{y} \equal{} Z^{2}$

Solve for positive integers X,Y,Z.

[hide="Hint"]
Treat the two cases x = 2k, x=2k+1 separately.
[/hide]

Case $ x \equal{} 2k \plus{} 1$
[hide]
We have $ x^{2k \plus{} 1} \plus{} 3^y \equal{} z^2 \implies 2 \cdot 4^k \plus{} 3^y \equal{} z^2$. Examining this mod 6, we have $ 2 \cdot 4^k \equiv 2 \pmod 6$ and $ 3^y \equiv 3 \pmod 6$, whic means that $ z^2 \equiv 5 \pmod 6$. However, 5 is not a quadratic residue mod 6, so there is no solution for $ x \equal{} 2k \plus{} 1$.
[/hide]

Case $ x \equal{} 2k$:
[hide="Progress"]
We have $ 2^{2k} \plus{} 3^y \equal{} z^2 \implies z^2 \minus{} (2^k)^2 \equal{} 3^y \implies (z \minus{} 2^k)(z \plus{} 2^k) \equal{} 3^y$

Thus since $ 3^y$ is always odd and positive if $ y$ is an integer, this implies that $ z$ must be odd and greater than $ 2^k$ (so that the $ (z \minus{} 2^k)$ term does not become negative). Also, since the product of $ (z \minus{} 2^k)$ and $ (z \plus{} 2^k)$ yields a number with only 3's in its prime factorization, the terms also must have only 3's in its prime factorization.

Quick brute-forcing yields $ 2^4 \plus{} 3^2 \equal{} 5^2$ is a solution. Working...
[/hide]

[quote="Yongyi781"]
We have $ x^{2k \plus{} 1} \plus{} 3^y \equal{} z^2 \implies 2 \cdot 4^k \plus{} 3^y \equal{} z^2$. Examining this mod 6, we have $ 2 \cdot 4^k \equiv 2 \pmod 6$ and $ 3^y \equiv 3 \pmod 6$, whic means that $ z^2 \equiv 5 \pmod 6$. However, 5 is not a quadratic residue mod 6, so there is no solution for $ x \equal{} 2k \plus{} 1$.
[/quote]

alternatively, for case $ x \equal{} 2k \plus{} 1$

$ 2^x \equiv 2\mod3$

$ 3^y \equiv 0\mod3$

$ 2\mod3$ isn't a square residue (:

Bah, I have no idea why I chose mod 6 :P

To continue from where Yongyi left off...

[hide="Hint"]Basically you have reduced the problem to $ 3^{a}\plus{}2^{k\plus{}1}\equal{}3^{b}$.  But then $ a\equal{}0$, or else we have two terms divisible by 3 and one term that isn't.  Now we have $ 3^{b}\minus{}2^{k\plus{}1}\equal{}1$.[/hide]

[hide]
Firstly, since $ z^2 \not \equiv 2 \pmod{3}$, we have $ 2^x\equiv 1\pmod{3}$, so $ x\equiv 0\pmod{2}$.
Now, since $ 2^x$ is at least 4, and $ z^2 \equiv 1 \pmod{4}$, we have $ 3^y\equiv 1\pmod{4}$, so $ y\equiv0\pmod{2}$ as well.
Now, we can write $ x \equal{} 2x_1$ and $ y \equal{} 2y_1$ for some positive integers $ x_1$,$ y_1$.
Thus, the original equation becomes $ (2^{x_1})^2 \plus{} (3^{y_1})^2 \equal{} z^2$.
It is well known that the solutions to $ a^2 \plus{} b^2 \equal{} c^2$ are $ a \equal{} 2rs$, $ b \equal{} r^2 \minus{} s^2$, $ c \equal{} r^2 \plus{} s^2$, for arbitrary integers $ r, s$.
Since $ 3^{y_1}$ is always odd, $ a \equal{} 2^{x_1}$ and $ b \equal{} 3^{y_1}$, so we have:
$ 2^{x_1} \equal{} 2rs$, $ 3^{y_1} \equal{} r^2 \minus{} s^2$, $ z \equal{} r^2 \plus{} s^2$.
From the first equation, $ r, s$ are both powers of 2. Then, since $ 3^{y_1}$ is odd, $ s \equal{} 1$.
So $ 2^{x_1} \equal{} 2r$ and $ 3^{y_1} \equal{} r^2 \minus{} 1 \equal{} (r \plus{} 1)(r \minus{} 1)$, which means that both $ r \plus{} 1, r \minus{} 1$ are powers of 3, so $ r \equal{} 2$, since r is also a power of 2.
From this, we see that the only solution is $ 2^{x_1} \equal{} 4$,$ 3^{y_1} \equal{} 3$, $ z \equal{} 2^2 \plus{} 1 \equal{} 5$, so $ (x,y,z) \equal{} \boxed{(4,2,5)}$.[/hide]


Determine all the integers $n >1$ such that $\sum_{i=1}^{n}x_i^2\ge x_n\cdot \sum_{i=1}^{n-1}x_i$ for all real numbers $x_1,x_2,\cdot \cdot \cdot ,x_n$.

Terribly sorry! this inequality is so stupid. (I have just solve it  :mad: )


When looking at boundary behaviour of analytic functions defined on $ \mathbb{D} \equal{} \{z \in \mathbb{C} : |z| < 1\}$, it turns out that it's very giving to look at so called non-tangential limits at the boundary. We say that a sequence $ (z_n)$ in $ \mathbb{D}$ converges nontangentially to $ 1$ if there is an angle with vertex at $ 1$, symmetric with respect to $ [0,1)$ and with opening less than $ \pi$ such that $ z_n$ belongs to this angle for all $ n$ large. (Of course, we also require the sequence to converge in the normal fashion).

An elementary characterization is the following. $ z_n \equal{} r_ne^{i\theta_n}$ converges nontangentially to $ 1$ if and only if $ \limsup_{n\to\infty} \frac {|\theta_n|}{1 \minus{} r_n} < \infty$. This notation simply means that the sequence is bounded.

Now, I have shown this with an extremely messy brute-force calculation. However, it is easy to see that the condition is equivalent to the alternative condition that $ \limsup_{n\to\infty} \frac {|1 \minus{} z_n|}{1 \minus{} |z_n|} < \infty$. During the lecture my professor commented that this condition is equivalent to the following statement, which should follow from elementary geometry:
(*) In an angle of the above type, with a fixed opening, there is a $ C > 0$ such that $ |z \minus{} 1| \leq C(1 \minus{} |z|)$ for all $ z$ in the angle.

Now I'm certainly not familiar with elementary geometry, so I don't see why this follows from geometric considerations, but I know you guys are. So I ask if you can tell me why (*) holds? Basically, I can show that it holds by writing $ z \equal{} re^{it}$ as $ z \equal{} 1 \plus{} \rho e^{i\theta}$ for suitable $ \rho$ and $ \theta$ so that $ z$ is in the angle, and then just brute force it to show the inequality. What I am looking for is a more succinct argument.

To mods: I thought this topic was fun to put here, since I really might get elementary solutions. If you feel that it's inappropriate, move it to complex analysis.

[quote="Kalle"]analytic functions... converges nontangentially... During the lecture my professor... If you feel that it's inappropriate, move it to complex analysis.[/quote]Yea, you might get better responses if this is moved to Complex Analysis forum.

I guess I might have added too much background to the problem, making it look scary. It's really not an advanced problem, and I think you could solve it. Anyway, I realized earlier that basically the thing I ask for is more or less obvious if we look at an angle against a straight line instead of a circle. So if we just approximate the circle with its tangent it's quite clear.


I'm taking it next year.  What review book or resource do you think was/would be the best for Physics B?  Just want to know before I take the class.

[quote="hello"]Did anyone here take the physics B exam. How would you rate it? I think the free response was ok and the multiple choice was pretty hard.[/quote]
I took it last year ('03), after learning the curriculum from a review book over the course of a month. My algebra is strong, but I'd assume that if you're here, so is yours. I got a 5 without much difficulty, and I didn't know a few of the topics.

I felt the same about the multiple choice and the free response. BTW-I already had a thread on this...


logarithms
floor function
number theory proposed
number theory

To do this by hand...

I remember, from all of those years ago when I used a base-10 log table to compute with, that $ \log 2\approx.30103.$ (It's an easy number to remember.)

Hence, $ \log(2^{81})\approx 81\cdot.30103\approx24.38$ (that much arithmetic is easy to do by hand).  But if $ \log2\approx.3,$ then $ \log2.5\equal{}\log10\minus{}2\log 2\approx .4.$ Hence, $ 2\times 10^{24}<2^{81}<2.5\times 10^{24}.$

Bulgarian TST for IMO 2008.
For every natural number $ n$ we mark the first digit of the number $ 2^{n}$ with $ a_{n}$. Is the number $ 0,a_{1}a_{2}a_{3}\cdots$ rational?

[quote="Johan Gunardi"]Cool. I didn't know that the first digit of $ 2^n$ is periodic. However, you still need to prove it, right?[/quote]
It simply isn't.

[quote="Johan Gunardi"]Find the first digit of $ 2^{81}$.[/quote]

this the simple way of solving this kind of question :

we have to find $ 0 < k < 10$  knowing that $ 2^{81} \equal{} k[10]$.

$ 2^3 \equal{} 8 \equal{} \minus{} 2[10]$==> $ 2^9 \equal{} 2[10]$ so $ 2^{81} \equal{} 2[10]$  :mad:

[quote="Ibn Rochde"]
this the simple way of solving this kind of question :

we have to find $ 0 < k < 10$  knowing that $ 2^{81} \equal{} k[10]$.

$ 2^3 \equal{} 8 \equal{} \minus{} 2[10]$==> $ 2^9 \equal{} 2[10]$ so $ 2^{81} \equal{} 2[10]$  :mad:[/quote]

Can someone explain this solution?

That solution is wrong: it finds the _last_ digit instead of the _first_ digit.

Could you explain what he was doing? I don't understand the $ []$ at all, unless he was implying that meant modular? :S

Also, the left-most digits are not periodic for $ 2^n$ ? Mhmm, is there a proof for this? They look it up to whatever exponent I used. >_<

The $ [n]$ are some (rather unused and old) way to express $ \mod n$.

Assume the leading digit of $ 2^n$ is periodic with period $ m$. This means that all the elements of the sequence $ a_k \equal{} \left( (k \cdot m \cdot \log_{10}(2)) \mod 1 \right)$ are in the interval $ I \equal{} [0,\log_{10}(2)[$ (here I used $ \mod 1$ to express fractional parts).
But for any irrational $ \alpha$, the numbers $ (\alpha \cdot k) \mod 1$ are dense in $ [0,1]$, thus not restricted to $ I$. As $ m \cdot log_{10}(2)$ is irrational, we reached a contradiction.

On a more intuitive level, you'll notice that the second digit is increasing between "periods" (compare $ 64$ and $ 65536$).  You can verify directly by calculator that if you repeat this calculation the second digit is going to "overflow."

[quote="ZetaX"]

Assume the leading digit of $ 2^n$ is periodic with period $ m$. This means that all the elements of the sequence $ a_k \equal{} \left( (k \cdot m \cdot \log_{10}(2)) \mod 1 \right)$ are in the interval $ I \equal{} [0,\log_{10}(2)[$ (here I used $ \mod 1$ to express fractional parts).
But for any irrational $ \alpha$, the numbers $ (\alpha \cdot k) \mod 1$ are dense in $ [0,1]$, thus not restricted to $ I$. As $ m \cdot log_{10}(2)$ is irrational, we reached a contradiction.[/quote]
Could someone explain this in more detail? I do not doubt the proof at all, I tried looking it up and finding various methods, all awfully similar but I don't understand why the whole $ \log_{10}{x}$ and what does $ k$ mean? And why is $ m$ multiplying with the $ \log$?
etc...  THanks a plenty :)

The leading digit of a number $ n$ is determined by the fractional part of $ \log_{10} n$.  The integer part tells you how many digits there are; once you ignore that, you're left with information about the rest of the number.  If the leading digit of $ 2^n$ has period $ m$, then the sequence of numbers

$ A_k \equal{} 2^{km}$

all have the same leading digit (which, by plugging in $ k \equal{} 0$, is $ 1$), which means that the sequence of fractional parts

$ a_k \equal{} \left( \log_{10} a_k \right) \bmod 1 \equal{} (km \log_{10} 2) \bmod 1$

all tell you the same leading digit.  Specifically, the relevant calculation is that this sequence is in the interval ZetaX gave.  ZetaX then cites a lemma that tells us this is impossible.

Mhmm sorry to bring this up again :P but could you explain this some more?

I grabbed a calculator and tried $ \log_{10}{49985275}$. This has 8 digits. The number on the calculator reads $ 7.698824086...$. I tried a few other examples and it seems the integer part gives me the number of digits $ \minus{} 1$. To relate it to my example, I got a $ 7$ for the integer part but the number really has $ 8$ digits. Do they differ by one for all numbers? I tried 5 and they all seemed to work this way.

Now the fractional part, $ 0.698824086...$ and the leading digit $ 4$... I can't see the relation. Even with the few other cases I tried. :(.
The rest of your explanation helped significantly though. Thanks very much, to ZetaX as well :)

P.S. -> Does the lemma that he cited have a specific name? So I may study these sort of questions more.

Again, thanks :)

You are correct. It is the number of digits minus one. If a number $ a$ has $ n$ digits, then

$ 10^{n \minus{} 1}\le a < 10^{n}$

(Because $ 10^{n \minus{} 1}$ is the first $ n$ digit number and $ 10^{n}$ is the first $ n \plus{} 1$ digit number). Taking the log,

$ \log(10^{n \minus{} 1})\le \log a < \log(10^{n})$
$ n \minus{} 1\le \log a < n$

For your second question, you say the fractional part is $ .698824086$. Well $ \log 4\approx .6002$ and $ \log 5\approx .6989$, so

$ \log 4\le 0.698824086 < \log 5$.

As for the name of that lemma, I don't know.

The lemma goes by the name of the [url=http://en.wikipedia.org/wiki/Equidistribution_theorem]equidistribution theorem[/url].

Mhmm, my issue with that $ \log$ stuff is unless one has them memorized, a calculator is needed >_<.

But aside ffrom thank you guys very much :) I am such an amateur but I shan't give up xD

Also, for the fractional part stuff. Is it safe to say... :

let $ n \equal{} n_1n_2...n_n.0$ (0s just for simplicity)
Then $ n_1 \equal{} x$ where $ \log_{10}{x} \leq \log_{10}{n } [\mod 1]$. Here I am borrowing ZetaX's notation for the fractional part (ie. $ \mod 1$) and $ x$ is the greatest such integer. That is, the greatest integer $ x$ such that $ \log_{10}{x} \leq \log_{10}{n} [\mod 1]$

Hopefully I didn't confuse myself and get screwed up. My apologies if I did. :P

This problem has got a large number of posts for such a specific, and non-generalised problem.

Instead...anyone got a solution to the following generalisation?

What is the first digit of $ a^b$ where $ a$ and $ b$ are natural numbers base $ n$? Giving your answer base $ n$.

This is an example of a question where a sufficiently detailed specific solution already contains all of the ingredients of the general solution; that is, ZetaX's discussion tells you immediately that the leading digit is the unique positive integer $ d$ such that $ \log_n d < \{ b \log_n a \} < \log_n (d \plus{} 1)$, where $ \{ x \}$ denotes the fractional part of $ x$.  In other words,

$ d \equal{} \lfloor n^{\{ b \log_n a \} } \rfloor$.

(Are you expecting a "simpler" expression?)  A similar expression is readily obtained for the first $ k$ digits, from which it follows that the first $ k$ digits of an exponential sequence such as $ 2^n$ are equidistributed; in other words, one can find a power of two beginning with any prescribed digit sequence.


Show that $ 1-(cosn\theta+isinn\theta)$=$ -2isin\frac{n\theta}{2}(cos\frac{n\theta}{2}+isin\frac{n\theta}{2})$

Find the sum of the series $ z+z^{2}+z^{3}+...+z^{n}$ for $ z\neq1$

(iii)If $ z=cos\theta+isin\theta$, show that for $ sin\frac{\theta}{2}\neq0$,

$ cos\theta+cos2\theta+cos3\theta+...+cosn\theta$=[url=http://img95.imageshack.us/my.php?image=39986391td8.png][img]http://img95.imageshack.us/img95/1276/39986391td8.th.png[/img][/url]

The first question is a trivial use of a double angle formula for cosine.

Treat your second problem as a geometric sum.

$ \sum_{k=1}^{n}\cos k\theta=\Re\sum_{k=1}^{n}e^{ik\theta}=\frac{1}{2}\sum_{k=1}^{n}(e^{ik\theta}+e^{-ik\theta})$.

$ \sin{n\theta}= 2 \sin{\frac{n \theta}{2}}\cos{\frac{n \theta}{2}}$ and

$ \cos{n \theta}= 1-2 \sin^{2}{\frac{n \theta}{2}}$. Plug these identities in and the first part is easily solved.

$ z+z^{2}+z^{3}+.........+z^{n}= \frac{z^{n+1}-z}{z-1}$, is a geometric series summation.

Now $ \cos{\theta}+\cos{2 \theta}+..........+\cos{n \theta}= Re(z+z^{2}+.......+z^{n}) = Re(\frac{z^{n+1}-z}{z-1})$ and the relationship can be obtained in this way.

Edit: Oops, boxedexe beat me to it.


2n+1 segments are marked on a line .Each of the segments intersects at least n other segments.prove that one of the segments intersects all the other segments

assume the segment $ AB$ (from left to right) we call $ A,B$ the start and end point of segment $ AB$ respectively...
take the first segment that ended...
name it $ a$ now there are $ n$ other segments $ x_{1},...,x_{n}$ which intersect $ a$,now between $ x_{i}$s choose the segment that ended last and name it $ x$,we claim that $ x$ is the desired segment...
now if we name the end point of $ a$ as $ A$ then $ A \in x_{1},...,x_{n}$ so $ x_{i}$s must intersect each other,so $ x$ intersects $ a,x_{i}$s now note that other than $ a,x_{i}$s there are $ n$ segments left,name them $ y_{1},...,y_{n}$ now we know that every $ y_{i}$ must intersect at least on of $ a,x_{i}$s so it mst intersect $ x$...
so $ x$ intersects $ a,x_{i}$s,$ y_{i}$s and we are done...
[hide="another similar problem"]
in a graph with $ 2n+1$ vertices,every $ n$ vertices have a common neighbour,prove that there exists a vertex which is connedcted to every other vertices...[/hide]


modular arithmetic
pigeonhole principle
combinatorics proposed
combinatorics

Let $ A$ be a set containing $ 2001$ positive integers. Prove that $ A$ contains a subset $ B$, such that

1. $ B$ has at least $ 668$ elements.

2. For any $ a,b \in B$, $ a\plus{}b$ doesn't belong to $ B$.($ a$ and $ b$ are not necessary distinct).

By Dirichlet's Theorem, there exists a prime $ p\equiv2\pmod3$ which is greater than any element of $ A$. Call a number [i]good[/i] if and only if its least positive residue modulo $ p$ lies in $ \left(\frac{p}3,\frac{2p}3\right)$, i.e., $ \left[\frac{p\plus{}1}3,\frac{2p\minus{}1}3\right]$.

Note that for all $ a_i\in A$, $ a_i,2a_i,\ldots,(p\minus{}1)a_i$ are different residues modulo $ p$. So there are $ \frac{p\plus{}1}3$ integers $ k\in[1,p\minus{}1]$ for which $ ka_i$ is good. Hence there are $ 2001\cdot\frac{p\plus{}1}3\equal{}667(p\plus{}1)$ pairs of $ (k,a_i)$ for which $ ka_i$ is good. By Pigeonhole Principle, there exists $ k$ such that the set $ B\equal{}\{a_i|ka_i\text{ is good}\}$ contains 668 elements.

We shall prove that $ B$ is sum-free. It is enough to prove that $ k$ times the sum of any pair in $ B$ is not good. But $ k$ times the sum of any pair is in the interval $ \left[\frac{2p\plus{}2}3,\frac{4p\minus{}2}3\right]$, i.e., $ \left[0,\frac{p\minus{}2}3\right]\cup\left[\frac{2p\plus{}2}3,p\minus{}1\right]$ which is not good. Therefore the set $ B$ satisfies the desired conditions.

Thank you, but this is actually the official solution, replicated word for word, well, even if it is not - I've been already aware of it. Any other solutions?


DOes anyone know if 117 (the cutoff for amc10a) is going to be the cutoff for aime on amc10b?

im pretty sure the cutoff for amc10a doesnt carry over to amc10b...

Yes, the cutoff for the 10A does not carry over to the 10B. The cutoff is determined by the lowest score of the top 5% of the people who took the test, I believe.

i thought amc10 cutoff was determined by the top 1%..? not 5? o.O

Yep, the AMC 10 cutoff is the top 1%. Though I think that should be even further below 120 for the 10B than it was for the 10A.

really..why should it be lower...o.O ah well..i'm hoping its lower =X all i want to get into is aime...=( i don't care if i get a 0 on it..lol

Yeah, me too, i got 117  :ninja:
four stupid mistakes, two blank

Well, you might still be able to make it with 114 or 117. They might lower the cutoff again.

Why must all the B tests that have been administrated be harder than the A tests?!? They canceled the A test because it was cold outside.. because i didnt see any snow or ice.. i am sad

The difficulty of the tests don't usually matter, it's how well you did in comparison to other people in the US (or taiwan, ...). So if you think it was hard, a lot of other people will too.

[quote="shen02"]is it sad if i scored 114 both times?
first attempt: 4 unanswered, and 18 correct.
second attempt: 19 correct, and 6 wrong...

if the cutoff is lowered to 115.5 for the B test, i swear i'm going to just break down in tears... :([/quote]

THAT HAPPENED TO ME TOO! I GOT 114 ON A AND B!!!

and same here! for attempts

8th grader here.

ah that sucks. yea my score differential is pretty enormous. 109.5 on amc A and 132 on amc B. but i thought the B was harder for some reason...

I thought B was EXTREMELy difficult compared to the A.

138 on the B anyways though [officially]. I took A officially and had 144

I didnt take the amc10a, so i dont know about difficulties. but from what ive heard so far, i think the cutoff may be ~114.

Well, theres a kid at my school who got a 115 on the amc 10a(missed by 1.5, fewest you can miss by) and a 99 on the amc 12b(so if the cutoff isnt lowered, thats super unlucky for him).. sad thing is i dont want him to qualify because hes annoying as hell

lol maybe because it was my first time taking AMC10 on A, but I thought B was much easier... and I actually studied for A. I got 112.5 on A and 133.5 on B... but then again, i did miss number 1 on A =]


[quote="thanhnam2902"]Let $ ABC$ is a triangle. Prove that: $ \boxed{p^2\le5R^2\minus{}Rr\plus{}9r^2}$[/quote]
After Ravi's substitution we need to prove that
$\sum_{cyc}(5x^4y^2+5x^4z^2+10x^3y^3-6x^4yz-22x^3y^2z-22x^3z^2y+30x^2y^2z^2)\geq0$,
which is true by easy SOS.


[quote="Bovy11"]Let $ n$ be a positive integer such that all its digits are the same (for example $ 222$, $ 555555$),and it is dividible by $ 1998$.

Determine the minimum $ n$.[/quote]

[hide=" This may be true"]

Min value of n $ \equal{}\underbrace{222\cdots222}_{27}\equal{}1998\times111222333444555666777889$
Reference:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=236702[/url][/hide]

uh wat lol

1998 = 999 * 2

obv the smallest value is a lot of 2's, cuz 2...2|{4...4, 6...6, 8...8} given each thing has the same amount of digits

so now if 1998 | 2...2, it must be 9k digits long

222,222,222/1998 = 111,111,111/999 = 1001001/9

now that doesn't work

so let's try 18 digits

and we get that turns out to be like 1001001001001001/9

which doesn't work either

so we try 27 digits

which is like 1001001001001001001001001/9 which is an integer so we good, smallest thing is 2...2 with 27 digits

[quote]obv the smallest value is a lot of 2's, cuz 2...2|{4...4, 6...6, 8...8} given each thing has the same amount of digits [/quote]

Completely flawed. It is entirely possible that 1998 does not divide 222,222,222, but it does divide 3 times that.

And it does.[b] 666,666,666[/b] is smallest.


Now it was made in 2006 so it's 2 years and 6 months old.

Seriously who keeps rating my posts spam when they aren't?

Lol, rhetorial question? Or sarcastical?

I thought math154's occupation was n00b,

Seriously, yes. All the people here are nonsense.

Ha you posted here so you are nonsense too.

I just realized that I have posted about 500 posts in October. *gulp* :o

EDIT: Checking, I actually DID post about 500 posts in October. My guesstimate was correct!!!

I just looked at my posts and the October posts go back to the 26th page, and I've only posted 67 pages.

$ \frac{26}{67}\cdot1333\approx517$. Yay.

Now I did.
So shut up.

[hide="and lay off"] :D  :D  :D [/hide]

Spam isn't cool. That's why people not from Missouri are invading your happy little forum. So it isn't happy anymore.

Except this isnt spam. This is the most intelligent people in Missouri posting the most intelligent things they know. Missouri is mostly rural you know.

Well, the rural people made up Framer's Rule, so they can't be that stupid.

Lol like Kramers rule with farmers. Thats pretty punny.

Can you post the expanded version of Cramer's Rule for 1337 variables again? I forgot it. :maybe:

I know! It's the only one I can't remember!


geometry
3D geometry
geometric transformation
reflection
vector
AMC
AIME

Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12.$  A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P,$ which is 7 units from $\overline{BG}$ and 5 units from $\overline{BC}.$  The beam continues to be reflected off the faces of the cube.  The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\sqrt{n},$ where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime.  Find $m+n.$

[hide="Answer"]It moves $5$ to the side and $7$ up, and we must end up at another vertex. Every time it reflects, it will move $5$ units to the side and $7$ units either up or down. If it hits a face partway through a reflection, it will continue the remaining number of units in the opposite direction. We first have that both $5x$ and $7x$ are multiples of $12$ for $x=12$. The length $AP$ is $\sqrt{5^2+7^2+12^2}=\sqrt{218}$, so we multiply that by $12$, and we have $12\sqrt{218}\Rightarrow 12+218=\boxed{230}$.[/hide]

Your answer is correct but I think you should explain more.  Unless someone knows how to do the problem already, your solution doesn't shed much light on the essence of the problem.

[hide="Hint"]
Reflect the cube the face that the beam hits.  Keep reflecting each time the beam hits another side.[/hide]

I'll post a solution if no one else does and I have time later...

Well, just ignore this if you haven't done the problem yet.

[hide]

Split vector AP up into its components. You have a vertical vector of 7 and a horizontal vector of 5, if we just look at what happens on face BCPG. Of course, since this is 3D, there is a depth vector, but it's not necessary since it is simply the side length of the cube. Each reflection of the vector AP travels 12 units deep. So it's negligible. Well, now we want after a certain number of reflections, which are always at the original angle (so the same vector is preserved throughout), to have the components to both be divisible by 12, as this will guarantee a corner.

The reason for this is because the vertical vector of magnitude 7 units moves up and down the BCPG face if we look at it in 2 dimensions. And after moving up and down and up and down, we want it to land on either side BG or side CP (from the 2D perspective). We want the horizontal vector of magnitude 5 units to move left and right, bouncing boppity bop on the walls, until it lands on PG or BC in the 2D perspective. Thus, for some minimum value of $x$, $5x$ and $7x$ are divisible by $12$. Well, clearly $x=12$.

So the actual vector reflects exactly 12 times, so the length is 12 times the length of the original vector, which by the Pythagorean theorem is: $\sqrt{218}$. So $12\sqrt{218}$ is the answer. And in AIME, it's 230 I guess, but that doesn't matter that much.  :P

Haha... I was high on sugar when I wrote this so it's pretty overkill.

[/hide]

[url]http://www.artofproblemsolving.com/Wiki/index.php/2002_AIME_I_Problems/Problem_11[/url]

For the solution on this page, I don't understand how the photons are in exactly the same position relative to the cubes they are at any moment after the space if divided. I understand the rest, however, by similar triangles. Can somebody help me out?

bump please
i am really struggling with this problem. is my question too confusing?

What it's saying is that since you've divided the entire space into cubes, at any moment the 2nd photon is in a cube, and you can take that cube and rotate/flip it or whatever and make it look like the cube the 1st photon is in.

Sorry for the bump, but I just wanted to share this with everyone because I find it super cool :-D

I coded a 3D simulation of this in Python. It requires both Pygame (so you'll need to copy into your own editor to run it, you can't just run it on AoPS) and some patience as it takes a bit of time before the laser actually hits the corner again, but it's cool xD.

If you want to see the laser at a position just press the space bar to pause the simulation and press the space bar a second time to resume it.

Enjoy haha
[pywindow]
import pygame
import random
pygame.init()

def dot_product(vector1, vector2):
    value = vector1[0]*vector2[0]+vector1[1]*vector2[1]+vector1[2]*vector2[2]
    return value

def add_vector(shape, vector):
    # shape is a list of vectors of the same length as the given vector

    new_shape = []
    for vertex in shape:
        new_vertex = []
        for i in range(len(vertex)):
            new_vertex.append(vertex[i] + vector[i])
        new_shape.append(tuple(new_vertex))

    return new_shape

def project(vector):
    plane = (0, 0, 1)
    direction = (1, 1, -3)

    t = -(dot_product(plane, vector)/dot_product(plane, direction))

    projected_vector = (vector[0]+t*direction[0],
                        vector[1]+t*direction[1])
                        # vector[2]+t*direction[2])

    return projected_vector

side_length = 500
cube = [
    (0, 0, 0),
    (side_length, 0, 0),
    (side_length, side_length, 0),
    (0, side_length, 0),
    (0, 0, side_length),
    (side_length, 0, side_length),
    (side_length, side_length, side_length),
    (0, side_length, side_length)]

start_vector = (100, 75, 0)
cube = add_vector(cube, start_vector)

projected_cube = []
for vertex in cube:
    projected_cube.append(project(vertex))

window = pygame.display.set_mode((1000, 800))
pygame.display.set_caption('Cube')

print(projected_cube)
color_of_cube = (255, 255, 255)
line_width = 3

for i in range(7):
    if i != 3:
        pygame.draw.line(window, color_of_cube, projected_cube[i], projected_cube[i+1], line_width)

pygame.draw.line(window, color_of_cube, projected_cube[0], projected_cube[3], line_width)
pygame.draw.line(window, color_of_cube, projected_cube[4], projected_cube[7], line_width)

for i in range(4):
    pygame.draw.line(window, color_of_cube, projected_cube[i], projected_cube[i+4], line_width)

pygame.display.flip()

speed = 1/72

laser_dir = [speed*12, -speed*5, -speed*7]
laser_location = [0, 0, 0]

paused = False
color = [255, 0, 0]

while True:
    if paused == False:
        laser_location_projected = project(tuple(laser_location))
        laser_location_projected_rounded = (int(laser_location_projected[0]+0.5), int(laser_location_projected[1]+0.5))
        laser_location_projected_shifted = list(add_vector([laser_location_projected_rounded], start_vector)[0])

        """
        for i in range(3):
            if color[i] >= 253:
                color[i] = 0
            else:
                color[i] += (i+1)
        """

        pygame.draw.circle(window, tuple(color), laser_location_projected_shifted, 1)

        triple_bounce = [False, False,False]
        for i in range(len(laser_location)):
            if laser_location[i] < 0 or laser_location[i] > side_length:
                triple_bounce[i] = True
                laser_dir[i] *= -1
        if triple_bounce == [True, True, True]:
            pass
            # color = (random.randint(0, 255), random.randint(0, 255), random.randint(0, 255))

        for i in range(len(laser_location)):
            laser_location[i] += laser_dir[i]

        pygame.display.flip()

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            pygame.quit()
            break
        elif event.type == pygame.KEYDOWN:
            if event.key == pygame.K_SPACE:
                paused = not paused
[/pywindow]

@above the python code has an error in it
[quote=Python]ImportError: No module named pygame on line 1[/quote]
@below I'll try running it on my computer I guess

you probably have to run it in repl.it or another python runner

[quote=peace09][quote=HumanCalculator9]My computer still gave me that error...[/quote]

Same here[/quote]

Same here

@above2 no need to say Same here thats what the upvote button is for :|

Yeah I know

AoPS doesn’t support the module Pygame, so you’ll need to copy the code into your own editor that has Pygame and run it there.

Sorry, if I could make it run here I would but I can’t :/

I think I know why my computer gave me an error...
Maybe my python version isn't new enough

No, you probably just don’t have Pygame installed onto your computer...

If you want to somehow download it just search up Pygame and it’ll give you their website with instructions how to download it.

I’ll post a screenshot of the sim here later if you’re too lazy to download it, I can’t do it right now because I’m on mobile.


function
modular arithmetic
number theory
greatest common divisor
group theory
number theory open

Maybe this is boring, but I asked myself this question a few days ago and couldn't find a very satisfying answer:

Let $ m$ be a positive integer that[b] has no primitive roots[/b] and let $ d$ be a positive integer. How many integers $ x$ with $ 1\leq x \leq m$ and $ gcd(x,m)=1$ exist, so that $ ord_{m}(x)=d$?.

Personally, I didn't get much further than this (and these results are trivial  :oops: ):
[hide]
It is obvious and well known, that there is some residue $ x$ with order $ d$, if and only if $ d\mid \lambda(m)$ where $ \lambda$ is the carmichael function; in particular, there is some $ x$ with $ ord_{m}(x)=\lambda(m)$. The obvious lower boundary for the number of such residues is $ \prod_{p^{k}\parallel m}\varphi(\varphi(p^{k}))$ (number of primitive roots modulo $ p^{k}$, respectively number of residues $ x$ modulo $ 2^{v_{2}(m)}$ with $ ord_{2^{v_{2}(m)}}(x)=2^{v_{2}(m)-2}$ if $ v_{2}(m)\geq 3$). However, this boundary is very rudimentary since such a residue must not neccessarily be congruent to a primitive root modulo every $ p^{k}$. Also, for every $ d\mid \lambda(m)$ there are at least $ \varphi(d)$ residues with order $ d$, but this is not very sharp as well since altogether, we would only cover $ \lambda(m)<\varphi(m)$ residues.

Maybe this can be killed by some theorem in group theory, but I must admit that I'm not very familiar with group theory (yes, I know, how embarrassing  :blush: ).
[/hide]

First let's find how many are the solutions to:
$ x^{d}\equiv 1 \pmod m$
so, the integers whose order is a divisor of d.
Let's write $ m = p_{1}^{e_{1}}\cdots p_{k}^{e_{k}}$
It is a solution (mod m) if and only if it is a solution mod every $ p_{i}^{e_{i}}$.
And modulo $ p_{i}^{e_{i}}$, as I think you know, there are exactly $ \mbox{gcd}(\phi(p_{i}^{e_{i}}), d)$ solutions.
So, the solutions to $ x^{d}\equiv 1 \pmod m$ are:
$ \prod_{1 \le i \le k}\mbox{gcd}(\phi(p_{i}^{e_{i}}), d)$.

If we define $ \psi(d)$ as the number of residues with order d, what we got is:
$ \sum_{a \mid d}\psi(a) =$ the formula described above.
To get a closed formula for $ \psi$, we use Moebius' inversion formula:
$ \psi(d) = \sum_{a \mid d}\mu(\frac{d}{a}) \prod_{1 \le i \le k}\mbox{gcd}(\phi(p_{i}^{e_{i}}), a)$

This is what I got... It's not a very good formula, I see, but I don't think you can find something very simple.

Ouch... :blush:
This was staring just right into my face, how could I fail to think of the Möbius' inversion formula? :wallbash:

But, however...
[quote="edriv"]And modulo $ p_{i}^{e_{i}}$, as I think you know, there are exactly $ \frac{\phi(p_{i}^{e_{i}})}{\mbox{gcd}(\phi(p_{i}^{e_{i}}), d)}$ solutions.[/quote]
Euhm...ok, it's already a bit late, so maybe my brain has already gone asleep, but shouldn't it be ${ gcd(\varphi(p_{i}^{e_{i}}), d)}$ rather than $ \frac{\phi(p_{i}^{e_{i}})}{gcd(\phi(p_{i}^{e_{i}}), d)}$?

Anyway...thanks...  :oops:

Oh, you're absolutely right, now I'll edit my last post...  :blush:
Your brain is certainly more awake than mine!

P.S. See you next week in Vietnam!  :)


can we calculate sin20 as a real number ?

i've already known this equation but my question is what sin20 is...
do you know its value?

$0.342020143325669...$ or what do you wan't to hear¿
You can always express $\sin\left( r \cdot \pi \right)$ in terms of roots (radicals) if $r$ is rational.

As Zetax said:
Let $\alpha$ a "well chosen" cubic root of $-4\sqrt3+4i$.
Then $sin(\frac{\pi}{9})=\frac{\alpha}{4}+\frac{1}{\alpha}$.

thank you loup blanc this is what i want to hear¿


I don't really know anything at all about astronomy, but I'm going into 9th grade and they have a full year course, Advanced Astronomy. It said that "there is a strong mathematical approach in this subject." I was wondering what kind of math is used in astronomy, and how useful it is. Thanks for any help.

if u look at most college course catalogs, astronomy is listed with physics and has a very heavy math requirement cuz it's a math-based science. in cosmology and more advanced astronomy it's almost entirely math-based, but if you're taking a 9th grade class that emphasizes that there will be little math, it probably just means that you'll be doing fairly straightforward stuff like calculating magnitude (that would be logs) and distance using parallax (trig), distance modulus (more logs), etc.
probably won't be a problem.
best of luck! :lol:


integration
function
inequalities unsolved
inequalities

let $f(x)=a_1.x+a_2.x^2+\dots +a_nx^n$ then we have $\int _{0}^{1} \frac{f(x)^2}{x} dx= LHS\geq 0$

-Another similarly problem:
For any $a_1,a_2,...,a_{n}$ we have:
$\sum_{i=1}^n\sum_{j=1}^n\frac{a_{i}.a_{j}}{1+|i-j|} \geq 0$
Try it! :D

[quote="hxtung"]Let $a_1,a_2,...,a_n \in R$. Prove that:
$\displaystyle \sum_{i=1}^n \sum_{j=1}^n \frac {a_ia_j}{i+j} \geq 0$.[/quote]

This is very similar to problem 4 of the Baltic Way 1990. That problem 4 demands to prove that

$\displaystyle \sum_{i=1}^n \sum_{j=1}^n \frac {a_ia_j}{i+j-1} \geq 0$.

The solution is almost the same as Sam-n gave to Hxtung's problem; the only thing different is that you have to integrate the function $\displaystyle \frac{f^2\left(x\right)}{x^2}$ instead of $\displaystyle \frac{f^2\left(x\right)}{x}$.

  Darij


[hide="Answer"]144. It's the Fibonacci sequence.[/hide]

[hide]fibonacci (every term is equal to the sum of the 2 preceding it, except the first two which are defined). 144. [/hide][/hide]

its 144 from the famous Fibanocci sequence

[quote="4everwise"][hide]144, fibonnaci sequence. I just found the sum of the first 10 terms and subtracted by 1.[/hide][/quote]

I think you added one.
by the way, nice observation, 4everwise. nth term in this sequence is sum of the first n-2 numbers plus one.

[quote="math92"]What is the twelfth number (the 2nd missing number) in the following pattern?
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ____, ____[/quote]

The next number is the sum of the previous two numbers.

So the next two numbers are $34+55=89$

And the next is $55+89=144$

[quote="math92"]What is the twelfth number (the 2nd missing number) in the following pattern?
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ____, ____[/quote]

This is the fibianacci sequence.

[hide]
34+55=79
55+79=134
[/hide]

[hide]The Fibonacci Sequence

$F_n=F_{(n-2)}+F_{(n-1)}$

Therefore, the two Missing terms are $34+55=89$, and $55+79=144$[/hide]


A database of famous mathematical proofs.  If I knew any html I would make this myself.  It could work just by having viewers submit proofs, and having viewers also be able to bring attention to false proofs.  So one of you guys should make it, because right now all I have is google.

Google is a powerfull tool for the one that knows how to use it ;)

There is already such a site: http://planetmath.org ...


quadratics
number theory
Diophantine equation
number theory unsolved

Find all integer solutions of the equation $ y^2 \minus{} 5x^2 \equal{} 44.$

Four solutions are  $ (\pm 1,\pm 7)$, so there exists solutions. If $ y^2\leq 5x^2$ or $ y^2< 44$, then there are no solutions.
If $ (x_1,y_1)$ are solutions, then $ y_1^2\minus{}44\equiv 0\: (\mod 5)$ i.e. $ y_1^2 \plus{}1\equiv 0 (\mod 5)$.
By brute force, the quadratic residues are $ \overline{2}\equal{}5\mathbb{Z}\plus{}2$ and $ \overline{3}\equal{}5\mathbb{Z}\plus{}3$.
So the solution set of the diophantine equation is $ \{(\pm\sqrt{\frac{y^2\minus{}44}{5}},\pm y)\: |\: y\equiv 2(\mod 5)\: \: \text{or}\: \: y\equiv 3(\mod 5), y^2\geq 49\}$.

$ y \equal{} 12 \equal{} 2\mod 5\to x \equal{} \sqrt {20}$ not $ \in \mathbb{Z}.$ In what cases we should take $ y \equal{} 5k \plus{} 2$ and in what $ y \equal{} 5k \plus{} 3$ ;) ? Equation has infinitely many solutions. I think there exist a recurence formula for them. Ok, I found a solution: [url]http://mathworld.wolfram.com/PellEquation.html[/url]. Use this technique to get all pairs of solutions.

I dont think this problem has infinitely many solutions, i think there are only 3 (+-5,+-13), (+-1,+-7) and (+-2.+-8).
First we can easily see $ 5(x\minus{}1)(x\plus{}1) \equal{} (y\plus{}7)(y\minus{}7)$ and from here on there is just som divisibility theory to get the result. Maybe i made a mistake.

If you clearly read my link you understand, that it has infinitely many solutions.

Interesting theory, i didnt read your link, because i tried to solve it alone. It really was to hard for me.

Pell's equations always have infinitely many solutions if they have one.  For example, the next solution is $ (7, 17)$.  The theory is worth learning.

I think it's not so worth  :D

First solve "unit" form of this equation $ r^2 \minus{} 5s^2 \equal{} 1$. First solution is $ (9, 4)$, then use recurence formula $ r_{i \plus{} 1} \equal{} r_1 r_i \plus{} 5s_1 s_i$ and $ s_{i \plus{} 1} \equal{} r_1 s_i \plus{} s _1 r_i$ to get all other solutions.

Then build set of soutions of $ t^2 \minus{} 5m^2 \equal{} 44$ using it's set of fundamental solutions:$ A \equal{} {(t, m)} \equal{} {(7, 1), (8, 2), (13, 5), (17, 7), (32, 14), (43, 19)}$ and using the identity $ t^2 \minus{} 5m^2 \equal{} (t^2 \minus{} 5m^2)(r^2 \minus{} 5s^2) \equal{} (tr \plus{} 5ms)^2 \minus{} 5(ts \plus{} mr)^2 \equal{} 44$.

This gives $ tr \plus{} 5ms$ for the new value of $ t$ and $ ts \plus{} mr$ for $ s$.

So for every start point from $ A$ we get: $ t_{i,\;k \plus{} 1} \equal{} t_{i,\;1} r_{k} \plus{} 5m_{i,\;1} s_{k}$ and $ m_{i,\;k \plus{} 1} \equal{} t_{i,\;1} s_{k} \plus{} m_{i,\;1} r_{k}$ for $ 1\leq i\leq 6.$ Then merge solutions for all $ i$.


x is a real numbers , then show that

\[ x^{6}\minus{}x^{5}\plus{}x^{4}\minus{}x^{3}\plus{}x^{2}\minus{}x\plus{}1\geq \frac{1}{2}\]


modular arithmetic
combinatorics proposed
combinatorics

An even number of politicians are sitting at a round table. After a break, they come back and sit down again in arbitrary places. Show that there must be two people with the same number of people sitting between them as before the break..

[b]Additional problem:[/b]
Solve the problem when the number of people is in a form $6k+3$.

For the first one, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34679]this[/url].

1. It comes from polish final round in 1989.

The second one can be solved by using the same idea (but it requires a bit more work). It all comes down to this:

If $n=6k+3$ and $\sigma$ is a permutation of the residues modulo $n$, then we can't have both $\{\sigma(i)-i\ |\ i\in\overline{0,n-1}\}=\overline{0,n-1}$ and $\{\sigma(i)+i\ |\ i\in\overline{0,n-1}\}=\overline{0,n-1}$ (everything takes place in $\mathbb Z_n$).

Assume the contrary (and remember, all the equalities take place in $\mathbb Z_n$; it's because I don't want to write $\pmod n$ all the time). We then have $\sum i^2=\sum(\sigma(i)+i)^2=\sum(\sigma(i)-i)^2\Rightarrow 4\sum\sigma(i)i=0\Rightarrow\sum\sigma(i)i=0$. This means that $\sum i^2=\sum\sigma(i)^2+\sum i^2=2\sum i^2\Rightarrow\sum i^2=0$. This yields a contradiction, since $\sum i^2=\frac{n(n+1)(2n+1)}6$ is easily seen to be different from $0$ modulo $n$.

I hope it's correct.

And it looks like for all $n=6k+1,6k+5$ we can find a counterexample.

With the notations above, for all $i\in\mathbb Z_n$, take $\sigma(i)=2i$. We get $\sigma(i)-i=i,\ \forall i$, and $\sigma(i)+i=3i,\ \forall i$, and the requirement that both sets $\{\sigma(i)-i\}$ and $\{\sigma(i)+i\}$ are equal to $\mathbb Z_n$ is met.


john, who is 5 foot tall, sees the top of a mountain at 34  degrees.
if he was 50 miles away, how high is the mountain?

Wrong forum.  ;)

[hide="wrong solution"]$\tan34=\frac{x}{50}$
$x=33.7$
about 178075 feet. [/hide]

it isnt thaat hard tho...
i learned this in my school a year ago... :huh:

this belongs in high school, despite easiness  ;)

[quote="aznmonkey1992"]( in a childish voice) okay, mommy :lol:[/quote]
lol.  :rotfl:
i agree that this does not belong here

I thought I did but I don't remember clearly so I won't say anything in case i'm wrong.*zips lips*

[quote="B4k4ka"]dont you need a calculator to solve problems like these?

Does anyone know how to figure out Tan, Sin, Cos without a calculator?[/quote]

Only certain values, otherwise either a table or calculator is necessary.

Trigonometry is incredibly difficult for this forum...unless it's easy.

Sine, Cosine and Tangent require a calculator, but how do you figure it out with nothing but addition? (calculators do that)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47804]do not click this[/url]

This is getting off topic. May a mod please move this to a higher forum?

[quote="236factorial"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47804]do not click this[/url]

[/quote]i didn't click it yet.  Will something bad happen if i click it? :stink:

[quote="mathmanman"]Haha no, it's just a 'nasty' demonstration about trigo :)
(i.e. : calculate sin 3° without a calculator..)[/quote]
oh, i'll click it now...


Find: $ lim\displaystyle \left ( \frac{cos^{2}1}{1.2}\plus{} \frac{cos^{2}2}{2.3}\plus{}...\plus{} \frac{cos^{2}n}{n(n\plus{}1)}\right)$

[quote="thanhnam2902"]Find: $ lim\displaystyle \left ( \frac {cos^{2}1}{1.2} \plus{} \frac {cos^{2}2}{2.3} \plus{} ... \plus{} \frac {cos^{2}n}{n(n \plus{} 1)}\right)$[/quote]

Here it is!  :lol:  (specify your limit boundary / point)

Denote
\[ f(z)\equal{}\sum_{n\equal{}1}^\infty\frac{z^n}{n(n\plus{}1)}\equal{}\frac{\ln(1\minus{}z)\plus{}z}{z}\minus{}\ln(1\minus{}z),\]
then the answer is
\[ \frac12(f(1)\plus{}\mathop{\mathrm{Re}}f(e^{2i}))\equal{}1\minus{}\sin^21\cdot\ln(2\sin1)\plus{}\left(\frac12\minus{}\frac\pi4\right)\sin2.\]


does anyone know a website with old math team tests/practice math team tests. Not AMC or Mathcounts. it would be best if the tests were like the 25 question + 3 tiebrakers at the end test.
(BTW, this is for ALABAMA geometry math team [i.e. Vestavia tourny, grissom tourny, hoover tourny, state tourny, etc.) thus, any help would be greatly appreciated  :lol:

http://www.mistacademy.com/contests.htm#alcontests

Just click around on the links.

http://www.quia.com/pages/pizitztourn.html

for pizitz.

http://www.hoover.k12.al.us/hhs/Math/PKustos/tournament/tests.html

for hoover.

I think I have more somewhere but i kant find them. hope this helps.

OHHHH YEA, that helps (as in all 3 sites help)

are there anymore sites?

are you being sarcastic??? (I can never tell on the internet...) You could, you know, google for tests yourself...

i tried googleing them, but i didnt get "math team" tests. i got tests with like 5 questions and tests that are NOT what i am looking for.

thus, i came to ask the people of the AoPS forum if they knew any...

so, i nEEEEEEDDD MOREEEE!!!!