function
complex analysis
complex analysis unsolved

We know that an analytic function cannot have an accumulation point of zeros in its domain. It is possible to construct analytic functions having an accumulation point on the boundary of its domain. The examples I have seen cannot be extended continuously to the boundary of the domain. Does there exist a function f, analytic on the unit disk, which can be extended continuously to the unit circle, such that there is an accumulation point of zeros which lies on the unit circle?

Yes. Take an infinite Blashke product in the unit disk with zeroes on $ [0,1)$ and multiply by $ z\minus{}1$ to kill the only discontinuity at $ 1$.

I don't know much about Blaschke products. Why is the only discontinuity at $ z \equal{} 1$?

What can, in general, be said about continuity of Blaschke products on the unit circle?

They are analytically extendable throughout arcs free of accumulation points of their zeroes. (Just show that the product converges perfectly well as long as the distance from the point to the zero set is positive)


limit
calculus
real analysis
real analysis unsolved

I need help finding where i made a mistake finding this limit:

$\lim_{n \to \infty} \left(1-\frac{2}{n}\right)^{n\sqrt{2}}$

What I did was say $u=-2n$ so the limit becomes
$\left(\lim_{u \to -\infty} \left(1+\frac{1}{u}\right)^{-u}\right)^{\frac{\sqrt{2}}{2}}$
Now, $-u$ would approach $\infty$ and $\lim_{u \to -\infty} 1+\frac{1}{u}=\lim_{u \to +\infty} 1+\frac{1}{u}=1$, so i set the limit in the parentheses to $e$ and since that's raised to the $\frac{\sqrt{2}}{2}$ power, said the whole limit is $e^{\frac{\sqrt{2}}{2}}$.  The thing is, that's not one of the answer choices (this is from some previous calculus competition).  Where did I mess up?  :huh:

How the heck do you raise something between $0$ and $1$ to a large power and end up with something $>1$?

Think about that; it's not hard to fix your solution.

[quote="RaduSorici"]maybe u want $n = -2u$[/quote]
[hide]
with $n=2u$
$\lim_{n \to \infty} \left(1-\frac{2}{n}\right)^{n\sqrt{2}}= \lim_{u \to \infty} \left(1+\frac{1}{-u}\right)^{2u\sqrt{2}}= \left(\lim_{u \to \infty} \left(1+\frac{1}{-u}\right)^{-u}\right)^{-2\sqrt{2}}=$
                           $e^{-2\sqrt{2}}$ [hide] props to tong and radusorici for the help [/hide][/hide]
[quote="blahblahblah"]How the heck do you raise something between 0 and 1 to a large power and end up with something >1? [/quote]
that's how you know it was wrong :roll:

 :D my first post with a hidden message!


search
inequalities proposed
inequalities

Here is another problem from my friend Sayar. It tells, let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \sqrt {12abc} \leq 1$
 :)

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \sqrt {12abc} \equal{} (a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ca) \plus{} \sqrt {12abc}\le 1 \minus{} 2\sqrt {3abc(a \plus{} b \plus{} c)} \plus{} \sqrt {12abc} \equal{} 1$

 because $ (x \plus{} y \plus{} z)^2\ge 3(xy \plus{} yz \plus{} zx)$

[quote="Sunkern_sunflora"]Here is another problem from my friend Sayar. It tells, let $ a,b,c > 0$ and $ a + b + c = 1$. Prove that
$ a^2 + b^2 + c^2 + \sqrt {12abc} \leq 1$
 :)[/quote]

$ a^2 + b^2 + c^2 + \sqrt {12abc} \leq 1 \Leftrightarrow
a^2 + b^2 + c^2 + \sqrt {12abc} \leq (a + b + c)^2 \Leftrightarrow
a^2 + b^2 + c^2 + \sqrt {12abc} \leq a^2 + b^2 + c^2 +2(ab + bc + ca) \Leftrightarrow
(ab + bc + ca) \geq \sqrt{3abc} \Leftrightarrow
(ab + bc + ca)^2 \geq 3abc \Leftrightarrow
a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c) = (ab)^2 + (bc)^2 + (ca)^2 + 2abc \geq 3abc \Leftrightarrow
(ab)^2 + (bc)^2 + (ca)^2 \geq abc = abc(a + b + c) = abac + bcab + acbc$

Which is true :)

My proof is same as that of Marius. But I think it can be solved in other techniques as well.
I am in search of an SOS solution, can anybody help me??? :wink:

[quote="Sunkern_sunflora"]I am in search of an SOS solution, can anybody help me??? :wink:[/quote]
Well Marius' solution it is easy to put in S.o.S. form
$ \iff \sqrt {12abc}\le 1 \minus{} (a^2 \plus{} b^2 \plus{} c^2) \equal{} 2(ab \plus{} bc \plus{} ca)$
$ \iff 12abc\le 4(ab \plus{} bc \plus{} ca)^2$
$ \iff 3abc(\sum a)\le \sum (a^2b^2) \plus{} 2\sum (ab\cdot bc) \equal{} \sum a^2b^2 \plus{} 2abc\sum a$
$ \iff 0\le 2(\sum (a^2b^2) \minus{} abc\sum a) \equal{} \sum ((ab)^2 \plus{} (ac)^2 \minus{} 2\cdot ab\cdot ac)$
$ \iff 0\le \sum a^2(b \minus{} c)^2$

Problem ta sayar dada r agei ar keu baniechilo. Austrian Math Olympd. ([url=http://www.mathlinks.ro/viewtopic.php?t=214641]ei link ta dekho[/url])


videos
logarithms
function
algebra unsolved
algebra

I've been out of school for several years. I've forgotten a lot of math and now I'm paying for it.

I made this equation for a video game raid. Basically a monster has 2 attacks. If people are close he hits them with a blast that's equal to 50 thousand damage divided equally among the people he hits. If people are far away he hits with a chaining attack that starts with 1000 damage and doubles for each person it hits.

Basically I came up with a little thought problem. I'm curious where exactly the numbers meet. I'm using this to formulate a better strategy. i.e. how many people should be up close and how many far away.

Anyway enough about games, here is the equation I made to model the situation

1000 * 2^x = 50000/x

Now I'm trying to find for x here and having signifigant difficulty, maybe someone can tell me where I'm going wrong or fill in the missing link?

I start with

1000 * 2^x = 50000/x

=

2^x = 50/x

=

x = log2(50/x)

=

x = log2(50)-log2(x)

=

x +log2(x) = log2(50)

And thats where I get stuck. I'm not sure hot to simplify the problem further short of graphing it out on a calculator. Can anyone give me some pointers? At this point, haven't struggled with it all day, I care less about if the equation accurately models the situation and more about just how do I find X.

Thank you

$ 1000\cdot 2^x \equal{} \frac {50000}{x}\Leftrightarrow x \equal{} \frac {W(50\log 2)}{\log 2}\approx 3.74059$ where $ W$ is the [url=http://en.wikipedia.org/wiki/Lambert_W_function]Lambert W[/url] function.

Once you start doing math out of the sheltered context of a textbook you'll find that most of the equations you want to solve can't be solved algebraically.  Instead, what you want to do is to find numerical approximations, since that's all you need for practical purposes anyway.  There are a variety of methods for doing this - a basic one that's easy to implement is [url=http://mathworld.wolfram.com/NewtonsMethod.html]Newton's method[/url].


induction
number theory solved
combinatorics
number theory
prime numbers
combinatorics solved
Arrangements

[hide]1. First of all it is clear that it is impossible for $n=pq$.

2. Let's prove that all remaining $n$ are good. By induction.
Base:
1) $n=p^k$, $k>1$,
2) $n=p^kq$, $k>1$,
3) $n=pqr$.

Consider an arbitrary $n\neq pq$, which is distinct from 1), 2) and 3). Then $n=s^{k}\cdot m$, where $m\neq pq$.
Suppose the sequence on circle for $m$ is $m_1=m,m_2,....,m_t$. Then consider a new sequence $m,sd_1,sd_2,...,sd_l,m_2,...,m_t$, where $sd_1,...,sd_l$ all divisors of $m$ which are not coprime with $s$ and $d_l=m_2$.[/hide]

Here is my candidate for the solution with the most overkill.  We will use the following results from graph theory.

Dirac's Condition: Let $G$ be a graph with $v \ge 3$ vertices.  If every vertex has degree at least $\frac{v}{2}$, then $G$ has a Hamiltonian cycle.

Chvatal's Condition: Let $G$ be a graph with $v$ vertices ($v$ odd).  If every vertex has degree at least $\frac{v-1}{2}$, and at most $\frac{v-1}{2}$ vertices have degree exactly $\frac{v-1}{2}$, then $G$ has a Hamiltonian cycle.

We can check that the divisor graph obeys one of these two conditions, provided that $n$ is not the product of two distinct primes.

Can you find a solution with more overkill? :)

He-he. In another thread rrusczyk wrote that convex hull and discrete continuity are too complicated ideas...
Your idea with Hamiltonian graph is absolurely right and clear. It was my first idea too. I always have in my mind this nice theorem.

an easy one
we proof by induction
we only need to proof that if n is an int and p is a prime and it's true for n it's true for N*p

some thing else how can I get some of those stars !!!!!!!

I claim that all composites except for numbers of the form $pq$ work, where $p$ and $q$ are distinct primes. It is easy to see that $pq$ fails, and that numbers of the form $p^k$ work.

I will now show that all numbers of the form $p^aq^b$ work for all pairs of $a$ and $b$ except for when they are both one. We first isolate the factors of the form $p^k$, and place them such that $p$ and $p^a$ are adjacent. Now, "split" open this circle and place $pq$ next to $p$ and $p^aq$ next to $p^a$. Since the numbers left all have some factor of $q$, they are all not pairwise prime, so "closing" the circle with these numbers yields a valid construction.

I now show that numbers with $\geq 3$ primes work. For the three primes, take the previous construction, and split the circle open at $p$ and $pq$. Note that all the numbers we have left to place have some factor of the third prime $r$, so we can simply place $pqr$ next to $pq$ and $pr$ next to $p$ to achieve a valid construction. Now we can see that we can achieve four primes in a similar fashion: split at $p$ and $pr$, and suppose the fourth prime is $s$. We can place $ps$ next to $p$ and $prs$ next to $pr$ and place all factors that are divisible by $s$. This can clearly be extended to an indefinite number of primes, so we are done.

We claim that such an arrangement is possible iff $n$ is not the product of two distinct primes. First, we show that such an arrangement is impossible in the latter case; let $n=pq$, then the divisors of $n$ greater than $1$ are $p,q,pq$, but $p$ must be next to $q$ in any such circle, contradiction.

Now, we show that such an ordering is possible for any other $n$. First, in the case that $n=p^k$ for some prime $p$ and integer $k\ge 2$, note that any ordering works. Additionally, remark that in the case $n=pqr$ for some primes $p,q,r$, we can take the ordering $\{p,pq,q,qr,r,rp,pqr\}$.

In the case that $n$ is square-free, either we can use the construction for $n=pqr$, or $n$ has at least four prime divisors and we can take some prime divisor $p\mid n$, then take the construction for $n/p$, choose two adjacent elements $a,b$ in that construction, then insert the multiples of $p$ in some consecutive order with $ap$ next to $a$ and $bp$ next to $b$. Similarly, in the case that $n$ is not square-free, choose a prime $p\mid n$ such that $n/p$ is not square-free and insert multiples of $p$ similarly to before, or $n=p^k$ and we have already found the construction thereof.

At least one of the criteria $\exists p\mid n \mbox{ s.t. } p^2\mid n$, $\exists p\ne q\ne r\ne p \mbox{ s.t. } pqr\mid n$ must hold as long as $n$ is not the product of exactly two distinct primes, hence we are done.

[b]Claim: [/b] We can create such a circle for all composite numbers other than numbers of the form $n = pq$ where $p,q$ are distinct primes.

If $n = pq$ with $p,q$ distinct primes, the three divisors that are greater than $1$ are $p, q, pq$. Obviously they do not meet the condition as $p,q$ are forced to be next to each other, and they are relatively prime to each other.

There are a few cases we have to look at. The first case is if $n = p^k$ where $k \geq 2$. Obviously, all numbers of this form must satisfy the condition as all divisors $>1$ are multiples of $p$.

The next case we have to look at is $p^aq^b$ with $a,b \geq 2$. Notice that all of the terms with the exception of the powers of $p$ and $q$ must share a common divisor as they contain powers of both $p$ and $q$. So, start with a circle of the divisors of $p^aq^b$ other than 1 and the powers of $p,q$. All of these divisors have both powers of $p$ and $q$. If you place all of the powers of $p$ between one pair of terms and all of the powers of $q$ between another pair of terms, you have a circle that satisfies the conditions. So, you can do it for this case as well.

Another case to consider is the case where we have $pq^n$ with $n \geq 2$. Note that all of the divisors other than $1$ and $p$ have a power of $q$ in them. So, you can place all of the divisors greater than $1$ except for $p$ into a circle in such a way that there is at least one pair of adjacent numbers that are multiples of $p$ and place the $p$ in between that pair of adjacent numbers. So, it is possible for this case as well.

There is one last case to consider, which is $n = p_1^ap_2^bp_3^c\ldots$ where $a,b,c,\ldots \geq 1$. First, we can start by creating a circle that contains the integers $p_1p_2,p_2,p_3,\ldots,p_np_1$. All multiples of $p_1$ go in between $p_np_1$ and $p_1p_2$ and so on with multiples of $p_k$ going between $p_{k-1}p_k$ and $p_kp_{k+1}$. So, it is also possible to create such a circle for this case.

As all of the cases have been addressed, it has been proven that you can create such a circle for all composite numbers other than $n=pq$ with $p,q$ distinct primes.

First we shall show that for all numbers $n=p_1p_2p_3\dots p_k$ there is a possible arrangment, where $k \geq 3$.

We define a $p_i$-[b]sector[/b] to be a circle section of angle $\alpha$ where all the numbers that are on the circumference of the circle are divisible by $p_i$.
We firstly throw on the numbers $p_i$ and $p_ip_{i+1}$ onto the circle for all $i$, where $1 \leq i \leq k$ and where all the indexes are taken $\text{mod}\; k$.

Now we devise the following algorithm:
[color=#f00][b]The algorithm:[/b][/color]
We start by taking the $p_1$ sector, and we place all the numbers divisible by $p_1$ and who divide $n$, into the $p_1$ sector and we place this between $p_1$ and $p_1p_2$, now we do this for the $p_2$ sector, then for $p_3$, etc...
When this is done we take a look at the repeating numbers and delete one of that number until we have just one of that number on the circle.

By this way every single divisor is on the circle and the neighboring condition is satisfied.Because we have that in the $p_i$-sector the $GCD$ is equal to at least $p_i$.

Now we shall show that it is possible for $n=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}$, where $k \geq 3$
We still use our algorithm but this time when we encounter a number of the form $p_1^{\beta_1}p_2^{\beta_2}\dots p_k^{\beta_k}$ we take a look at the exponents.
Now if we have let's say number $d$, where $d \mid n$, depending on the exponent we place $d$ in the $p_i$ sector, if we have that $v_{p_i}(d) \geq v_{p_j}(d)$ for all $j$, where $1 \leq i,j \leq k$.
Now again we delete until we are left with just one of $d$.

Now we take a look at trivial $n=p^{\alpha}$, where $\alpha > 1$.This is an easy construction. We just place $p^i$, in ascending order on the circle, where $k \geq i > 0$.

Now we look at $n=p^{\alpha}q^{\beta}$.
Obviously when $n=pq$ this is impossible.
So let's assume $WLOG$ that $\alpha \geq \beta$.
We start off with the construction for $p^2q$.We place the numbers in the following order: $p,p^2,p^2q,q,pq$ onto the circle.
Again we divide into the $p$ sector and $q$ sector. We now use our algorithm with the divisor types $p^{\gamma}q^{\delta}$ to generalize for $p^{\alpha}q^{\beta}$.

Thus giving us the answer of $n \in \mathbb{N} \backslash (\mathbb{P} \cup \{1\} \cup \{pq \mid p\neq q;p,q\in \mathbb{P}\})$,where $\mathbb{P}$ is the set of primes.
In other words the only solutions are all composite numbers except of the form $n=pq$, where $p,q$ are distinct primes

We claim that all composite numbers not in the form of $pq$ work, where $p$ and $q$ are primes. \\

We have several cases.
It is easy to see that numbers in the form $p^a$ work, where $p$ is a prime. \\

For numbers in the form $n=p^aq^b$ where $a\geq2$ and $b\geq1$, we can take all of the factors of the form $p^k$, and put them in the order $p, p^2, \dots , p^a$. We then put $pq$ next to $p$, and $p^aq$ next to $p$ and $p^a$, respectively. Then, the rest of the factors have $q$ as a divisor, so none of them are relatively prime.

Let $n=p_1^{n_1}p_2^{n_2}\dots p_k^{n_k}$, where $n_i\geq 1$ and $k\geq 3$. We start by putting the integers $p_1p_2,p_2,p_3,\ldots,p_np_1$ on the circle. We then put all multiples of $p_1$ in between $p_np_1$ and $p_1p_2$ and so on with multiples of $p_k$ going between $p_{k-1}p_k$ and $p_kp_{k+1}$. Thus, it is also possible to create such a circle for this case.\\

Thus, we have proved that all cases work other than the case $pq$, as desired.

Extremely clunky solution:
We claim that it works if the sum of the exponents of the prime factorization is $\ge 3$ or it is a prime power with exponent $\ge 2$. Another way of saying this is not $p_1 \cdot p_2$ where $p_1$ and $p_2$ are primes. (and of course also not $1$ or any prime, as the problem statement asks).

It is obvious that $p_1 \cdot p_2$ is impossible as we must place $p_1,p_2,p_1p_2$, and since there are only 3, inevitably, $p_1$ and $p_2$ must be adjacent.

The case $p^k$ works as they all share a common factor of $p$.

The case $p_1^{e_1}p_2^{e_2}$ works as long as at least one of the exponents is $\ge 2$. For this we can place it around the circle with $n$ at the very top, then circling around with all the powers of $p_1$ on one side and $p_2$ on the other and then spamming the multiples of $p_1p_2$ that aren't $n$ at the bottom.

If there are three or more primes, then the circle can be constructed as follows:
$$p_1,(\text{all } d \mid n, p_1 \mid d, d \ne p_1p_2,p_1p_k),p_1p_2, p_2, (\text{all } d \mid n, p_1 \nmid d, p_2 \mid d, d\ne p_2p_3), p_2p_3,\ldots, p_k, p_k^2,\ldots , p_k^{e_k},p_kp_1.$$
This works for more than 3 primes since from there, $p_kp_1 \ne p_1p_2$.

We define $p_1,p_2,p_3,\ldots$ as distinct arbitrary primes.
The answer is any $n$ that cannot be expressed in the form $p_1p_2$.
First we prove that no numbers of the form $p_1p_2$ work. The factors greater than $1$ here are $p_1,p_2,p_1p_2$. We can easily see by placing them in a circle (there is actually only one way to do this, since reflections and rotations don't change the result) that this arrangement doesn't work, since we have $p_1$ next to $p_2$.
Now we prove that everything else works.
First, consider a number of the form $p_1^k$ for some integer $k>1$. Clearly, all factors greater than $1$ of this number are divisible by $p_1$, so this works.
The remainder of the solution assumes that $n$ is divisible by at least $2$ distinct primes.
Next, consider a number of the form $p_1p_2p_3\ldots p_n$, with $n>2$. For convenience, we define $p_{n+1}=p_1$.
We first place $p_1,p_2,p_3,\ldots,p_n$ a circle in that order. Then, we put $p_i p_{i+1}$ between $p_i$ and $p_{i+1}$. Afterwards, we may place any other divisors in the circle. We may do this by placing all divisors divisible by $p_1$ in between $p_1$ and $p_1p_2$. Then place all remaining divisors divisible by $p_2$ in between $p_2$ and $p_2p_3$, and so on. That is, we place all remaining divisors divisible by $p_i$ in between $p_i$ and $p_i p_{i+1}$. This clearly works, since all the numbers between $p_i$ and $p_i p_{i+1}$ share a factor of $p_i$, and $p_i p_{i+1}$ shares a factor of $p_{i+1}$ with $p_{i+1}$.
Now suppose a number is of the form $p_1^2p_2$. We give the following explicit construction, which can easily be verified to work. Starting from the top and going clockwise, we place: $p_1,p_1^2,p_1p_2,p_2,p_1^2p_2$.
Finally, we can induct to every other number. Given that $p_1^{k_1}p_2^{k_2}p_3^{k_3}\ldots p_n^{k_n}$ works, for positive integers $k_1,k_2,\ldots,k_n$, $p_1^{k_1+1}p_2^{k_2}p_3^{k_3}\ldots p_n^{k_n}$ will work as well. All factors of $p_1^{k_1+1}p_2^{k_2}p_3^{k_3}\ldots p_n^{k_n}$ are factors of $p_1^{k_1}p_2^{k_2}p_3^{k_3}\ldots p_n^{k_n}$, except for those divisble by $p_1^{k_1+1}$. Thus we can put all these "new factors" in between $p_1$ and $p_1 p_2$. Then all the numbers between $p_1$ and $p_1 p_2$ share a factor of $p_1$. Nothing else is changed, so the arrangement works. Thus we have finished the inductive step.
Now, since $p_1$ is arbitrary, we can start from either $p_1^2p_2$ or $p_1p_2p_3\ldots p_n$ with $n>2$, and repeatedly apply the "inductive step" to reach any other composite number. Thus, we have proven that everything else works. We are done. $\blacksquare$

The answer is all $n$ which is not the product of two primes. Clearly, products of two primes fail, because in a circle of three elements, every two elements are adjacent. Yet the two distinct prime factors will be adjacent to each other, violating the problem condition. To show that all other numbers work, we proceed by induction on the number of distinct prime factors in $n$.

[b]Base Case[/b]. We will show that $n=p^k$ for $k \geq 2$ satisfy the conditions. Consider arranging the $k$ numbers in a circle in the order $p, p^2, p^3, p^4, \cdots, p^k$. This clearly works. (In fact, any placement of the numbers around the circle works!)

We also need to show that $n=pqr$ satisfies the conditions, for primes $p, q, r$. This is not hard; arrange the numbers in the order $1, p, pq, q, qr, r, pr, pqr$. Clearly no two divisors are relatively prime.

[b]Inductive Step[/b]. Assume that $n= p_1^{k_1} p_2^{k_2}\cdots p_i^{k_i}$ satisfies the condition. We will show that $$n' = p_1^{k_1} p_2^{k_2}\cdots p_i^{k_i}p_{i+1}^{k_{i+1}}$$ satisfies the conditions. Arrange the divisors of $n$ in a valid circle, which we know exists. Then, insert all divisors of $n'$ that contain $p_{i+1}$ (aka all divisors of $n'$ that are not divisors of $n$) consectutively into the circle, between $n$ and some number $j$. We can arrange the numbers such that the number adjacent to $n$ is $n'$, while the number adjacent to $j$ is $p_{i+1} \cdot j$. Since all the numbers between $n$ and $j$ satisfy the conditions, and so do the numbers in the consecutive ``block" (since they all contain the factor $p_{i+1}$) and the numbers between the two blocks, we have a valid placement for $n'$, as desired.

Thus the induction is complete, and the assertion holds for all $n$ not the product of two primes.

We claim the answer is all $n$ not of the form $pq$ where $p\neq q$. It is clear that such $n$ fail. Otherwise, we do casework on the number of distinct prime divisors of $n$.

Case 1: $n=p^k$. This clearly works because all divisors $>1$ have common factor $p$.

Case 2: $n=p^a q^b$ where wlog $a\geq 2, b\geq 1$. Then, we may construct:
\[\text{multiples of $p$}, pq, \text{multiples of $q$} , p^2q\]

Case 3: $n$ has $p_1,\ldots,p_m$ as $m$ distinct prime divisors. Then, we may place
\[p_1p_2,\text{multiples of $p_2$},p_2p_3, \text{multiples of $p_3$},p_3p_4,\ldots\]
Thus, all $n\neq pq$ work, and we are done. $\blacksquare$.

The answer is all composite numbers not of the form $pq$. It's easy to see that this fails. We provide the construction for all other composites :
First, $p^k$ works, so ignore that
It's easy to get the construction for $p^a \times q^b, \max{(a,b)} \geq 2$
For all others -
Let $n=p_1^{a_1} \times \dots \times p_k^{a_k}$. Start by placing $p_1$ anywhere on the circle. Then add $p_1^1, \dots, p_1^k$ in any arbitrary order, clockwise (all numbers are added adjacent to the previous one - figure is attached for reference). Then add all numbers divisible by $p_1p_2$. Then add all powers of $p_2$. Keep doing that with $p_3$ and so on. At the end, add all remaining numbers divisible by neither $p_1$ or $p_k$, but divisible by say $p_i$ to the group of numbers not coprime to $p_i$ (for example, by adding them between $p_i$ and $p_ip_{i+1}$). Also add all remaining numbers not divisible by $p_1p_k$ but divisible by $p_1$  anticlockwise to $p_1$, and do the same with $p_k$ except add them anticlockwise. In the middle of the two groups, add numbers divisible by $p_1p_k$ (at least $p_1p_k$ must have been unused ;) ). This satisfies the required conditions $\blacksquare$

The answer is all composites excluding semi-primes. It's easy to see semi-primes, i.e. $n = p_1 \cdot p_2$, must fail, as $p_1$ and $p_2$ are adjacent in all possible arrangements. We now proceed with induction on the number of distinct primes dividing $n$. (In this solution, we assume the first and last terms of our arrangement on paper are actually adjacent in the circle.)

[b]Base Cases (Prime Powers and $K = 2$ Primes, Excluding Semi-Primes):[/b] If $n = p_1^{e_1}$ where $e_1 \ge 2$, then the construction $$p_1, p_1^2, \ldots, p_1^{e_1}$$ obviously works.

If $n = p_1^{e_1} \cdot p_2^{e_2}$ such that $e_1 = e_2 = 1$ doesn't hold, then the construction $$p_1, p_1^2, \ldots, p_1^{e_1}, p_1 \cdot p_2, p_2, p_2^2, \ldots, p_2^{e_2}, \{p_1^{a} \cdot p_2^{b}\}$$ clearly works, where $a$ ranges from $1$ to $e_1$ inclusive, and $b$ ranges from $1$ to $e_2$ inclusive, such that $a = b = 1$ doesn't occur.

[b]Inductive Hypothesis:[/b] Assume the desired construction is possible for all $n$ with $k = m$ distinct prime divisors. We will show it is also possible for all $n$ with $k = m+1$ distinct prime divisors.

[b]Induction Step:[/b] Suppose $$d_1, d_2, \ldots, d_i$$ is a satisfactory arrangement of all the divisors, other than $1$, of some arbitrary $n$ with $k = m$ distinct prime divisors. If $p_{m+1}$ denotes our $m+1$th prime divisor, then the construction
$$p_{m+1}, p_{m+1}^2, \ldots, p_{m+1}^{e_{m+1}}, d_1 \cdot p_{m+1}^{e_{m+1}}, d_1 \cdot p_{m+1}^{e_{m+1} - 1}, \ldots, d_1, \{d_2 \cdot p_{m+1}^{t} \}, \{d_3 \cdot p_{m+1}^{t} \},$$ $$ \ldots, \{d_{i-1} \cdot p_{m+1}^t \}, d_i, d_i \cdot p_{m+1}, \ldots, d_i \cdot p_{m+1}^{e_{m+1}}$$  clearly works, where $t$ ranges from $0$ to $e_{m+1}$ inclusive. $\square$

Because all composites are divisible by at least $2$ distinct prime divisors, we're done. $\blacksquare$

Clearly numbers of the form \(pq\) where \(p,q\) are two primes don't work. We show that numbers of the form \[p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdots p_k^{\alpha_k}\] where at least two of the \(\alpha_i\) are positive. If we show this, we can conclude that this set of numbers are the final answer, because
the set of all numbers of the form \(pq\) and the other set are two disjoint sets whose union is the set of all composite natural numbers. For the numbers of the form we wanted to show work, start with \(p_1\) and go all the way till \(p_1^{\alpha_1}\), then we place \(p_1^{\alpha_1}p_2\) and then we place \(p_2\) and then place \(p_1^{\alpha_1}p_2^2\) and go all the way up till \(p_1^{\alpha_1}p_2^{\alpha_2}\), and then place \(p_1^{\alpha_1}p_2^{\alpha_2}p_3\) and then place \(p_3\) and keep following this pattern, in the end you would reach \(p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdots p_k^{\alpha_k}\). You can easily see that this way, no two adjacent numbers are relatively prime, and thus our desired conclusion follows.

Outline (indices taken mod k):
Numbers of the form $pq$ don't work, then for powers of primes it works. For anything else we have $p_1^{e_1}p_2^{e_2}\dots p_k^{e_k}$. First place $p_1p_2,p_2p_3,\dots,p_kp_{k+1}$ on the circle in that order. Then for all other numbers we can just place them between the numbers $p_{i-1}p_i$ and $p_ip_{i+1}$ if $p_i$ is a divisor of that number.

The claim is that all the composite numbers $n\neq pq,$ where $p,q$ are primes, work.

First observe that $n=pq,$ $p,q$ primes, does not work. Indeed, in this case divisors of $n$ are $1,p,q,pq$, so only way we can arrange them in a circle is $p$ then $pq$ and then $q$, but clearly $p,q$ are also adjacent here, so this does not work.

Now assume $n=p^{\alpha}q^{\beta}, \alpha\geq 2, \beta\geq 1.$ To see that this works fine, isolate $p, p^{2},\dots,p^{\alpha}$ then put $p^{\alpha}q$ adjacent to $p^{\alpha}$. Now the remaining factors are $p^{\alpha}q^2,\dots, p^{\alpha}q^{\beta}.$ So we simply put all these factors after $p^{\alpha}q$. So last adjacent pair will be $(p^{\alpha}q^{\beta},p)$ which have gcd $p$. So this construction clearly works.

At last, assume $n=p_{1}^{k_{1}}p_{2}^{k_{2}}\cdots p_{r}^{k_{r}}, r\geq 3, k_{i}\geq 1, i=1,2,3,\dots,r.$ Observe that when $k_{i}>1$ we can just arrange all the divisors like previous arrangement. So now we only have to check the case when $k_{i}=1$, that is $n=p_{1}p_{2}\cdots p_{r},p_{1}\geq 3.$ For this, put all $p_{1}p_{2},\dots p_{r}p_{1}$. Now put the divisors containing $p_{1}$ between $p_{r}p_{1}, p_{1}p_{2}$ then put all divisors containing $p_{2}$ between $p_{1}p_{2},p_{2}p_{3}$ and we continue this way, put $p_{i}$ between $p_{i-1}p_{i},p_{i}p_{i+1}.$ So we see this clearly works. Hence this completes the proof.$\boxed{}$

The answer is all not of the form $n \equiv pq$,clearly any number of this form doesn't work,to show number of the form \[p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdots p_k^{\alpha_k}\] use the above(no below) construction.

We claim the answer is all $n$ with $1$ or at least $3$ prime divisors.

[color=#ff9a00]$n$ has one prime divisor[/color]
[b]$n=p^k$[/b]
This clearly works if we place it in the form of $p, p^2, p^3, ... , p^k.$

[color=#ff0]$n$ has at least $3$ prime divisors[/color]

We partition the set of divisors of $n$ into $f(n)+2$ sets, where $f(n)$ is the number of prime divisors of $n$ as follows, where $n = p_1^{e_1}p_2^{e_2}...p_k^{e_k}.$

[list]
[*]  [b]$S_i \text{with}1 \leq i \leq k$[/b] where all $x \in S_i$ is in the form of $p_i^{m}, 1 \leq m \leq e_i, 1 \leq i \leq k$
[*] [b]$S_{i,j}, 1 \leq i < j \leq k$[/b]where all $x \in S_{i, j}$ is in the form of $p_1^{x_1},p_2^{x_2}, ... ,p_k^{x_k}, \text{where at exactly 2 of the} x_i, x_j \geq 1$
[*] [b]$S_{i_1, i_2, ..., i_m}$[/b]where all $x \in S_{i_1, i_2, ..., i_m}$ is in the form of $p_1^{x_1},p_2^{x_2}, ... ,p_k^{x_k}, \text{where at least 3 of the} x_{i_1}, x_{i_2}, ... , x_{i_m} \geq 1$
[/list]

Now, we develop our algorithm as follows:
[list]
[*]Place all the $x \in S_i$ all adjacent
[*]Place all the $x \in S_{i,j}$ adjacently and then stick them between the $S_i$ and $S_j$ block
[*]For all $x \in S_{i_1, i_2, ... , i_m}, $ first stick them all together, and then place it in between $S_{i_1}$ and $S_{i_2}$
[\list]

This algorithm clearly works, so we are done.

We will prove that the only time that $n$ doesn't work is when $n=pq$ where $p$, $q$ are distinct primes.

It is easy to see that if $n=pq$ where $p$ and $q$ are distinct primes, it is impossible to place $p, q, pq$ in a circle such that $p$ and $q$ are not bordering. If $n=p^k$ it is easy to see that it will work since any divisor of $n$ that is greater than 1 will be a multiple of $p$. Now we will prove that other $n$ work as well.

Let $n=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k}$  where $p_1, p_2, \dots, p_k$ are distinct primes and is in strictly increasing order. That is, $p_1<p_2<\dots<p_k$.

Now we define $A_i$ from $1\le i\le k$ to be the [b]strictly increasing[/b] set containing the divisors of $n$ such that the divisor is a multiple of $p_i$ but is not a multiple of $p_j$ with $j<i$. We will now perform two changes on $A_1$.
[list]
    [*] Take the last element of $A_1$ and move it to the front of the set without changing the order of the other elements.
    [*] Take the element $p_1\cdot p_2$ and move it to the end of the set without changing the order of the other elements.
[/list]

For the second operation, it is quite obvious that the number $p_1\cdot p_2$ does appear in $A_1$ since it is a multiple of $p_1$. Now imagine if we strung together all of these sets into a circle so that the last element of some $A_j$ and the first element of $A_{j+1}$ are neighboring. If $j=k$ then the last element of $A_k$ and the first element of $A_1$ are neighboring. We will claim that this construction works. It is easy to see that the first element of $A_1$ and the last element of $A_k$ are not coprime because the first element of $A_1$ is now $n$. Notice that the last element of $A_1$ is now $p_1p_2$ and the first element of $A_2$ is $p_2$ so the connection between $A_1$ and $A_2$ works. Furthermore, it is easy to see that if $1<j<k$ then the last element of $A_j$, $p_jp_{j+1}\dots p_k$ and the first element of $A_{j+1}$ which is $p_{j+1}$ are obviously not coprime so we are done. $\blacksquare$

Every number that isn't equal to pq with p and q being prime works. To prove this we will show that n works if it is one of the following:
-$p^2q$
-$pqr$
-$p^k$
where $p, q,$ and $r$ are prime and $k$ is a positive integer. Then we will prove that if $n$ works then $xn$ works where $x$ is any positive integer.

Note that putting the divisors greater than 1 on a circle is equivalent to putting the proper divisors greater than 1 on a line. We will now prove it is possible for the three cases mentioned earlier. Just do the following:
-$p^2q$: $p^2$  $p$  $pq$  $q$
-$pqr$: $p$  $pq$  $q$  $qr$  $r$  $pr$
-$p^k$: $p$  $p^2$ $\ldots$  $p^k$
Now we will prove that if you multiply $n$ by any integer you will still be able to put the divisors on a circle. It suffices to show that you can multiply $n$ by a prime $p$ and still be able to put the divisors on a circle. If it is possible to put the divisors of $n$ on a circle satisfying the given property, for each divisor $d$, insert $pd$ after $d$. Then, before $n$, insert $p$ and $pn$.
Now we will prove it is impossible to put the divisors of $pq$ on a circle so that neighbors are not relatively prime. We are putting $p, q,$ and $pq$ on a circle. Then, $p$ and $q$ are next to each other, a contradiction.

The answer is all $n$ not equal to the product of two distinct primes. We first show all such $n$ work.

Case 1: $n=p^k$, where $p$ is prime and $k\ge 2$.
Then all divisors of $n$ greater than $1$ are divisible by $p$, so any arrangement works.

Case 2: $n=p^aq^b$, where $p,q$ are distinct primes and $a\ge 2$.
First, arrange some divisors as follows: $p, pq, q, p^2q$.
Then for each remaining divisor, place it next to $p$ if it's divisible by $p$, or next to $q$ if it's divisible by $q$.

Case 3: $n=p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k}$, where the $p_i$ are distinct primes and $k\ge 3$.
Arrange some divisors as follows: $p_1, p_1p_2, p_2, p_2p_3, p_3, \ldots p_kp_1$. For each remaining divisor, if it's divisible by $p_i$, put it next to $p_i$ on the circle. This ensures no two adjacent divisors are relatively prime.

Now, we show $n=pq$ does not work. The three divisors that must be arranged are $p$, $q$, and $pq$, and no matter the arrangement, $p$ and $q$ are adjacent, a contradiction.

Let $n$ be [i]juxtaposable[/i] if we can arrange $n$ in such a way, and let $\tau(n)$ be the number of divisors of $n$, $\omega(n)$ of which are prime.

[b]Claim:[/b] if $n$ is juxtaposable, $\omega(n)\ge1$ and $q\nmid n$ is a prime, then $q^kn$ is juxtaposable for any integer $k\ge2$. Moreover, if $\omega(n)\ge2$ then $q^kn$ is juxtaposable for any integer $k\ge1$.
First, we show that $q^kn$ is juxtaposable for $k\ge2$.
Suppose we know that $n=\prod_{i=1}^{\omega(n)}p_i^{a_i}$ is juxtaposable with $d_1,d_2,\ldots,d_{\tau(n)-1}$ arranged clockwise in the circle, where $d_{\tau(n)-1}=n$.
Insert $qd_1$ to the left of $d_1$, and $q^2d_i$ in between $qd_1$ and $d_1$, and each $q^ad_i$ and $q^a$ in between $qd_1$ and $q^2d_1$.
We end up with:
$$qd_1,q,q^2,\ldots,q^k,q^3d_1,q^3d_2,\ldots,q^3d_{\tau(n)-1},q^4d_1,q^4d_2,\ldots,q^4d_{\tau(n)-1},\ldots,q^kd_1,q^kd_2,\ldots,q^kd_{\tau(n)-1},q^2d_1,d_2,d_3,\ldots,d_{\tau(n)-1}$$
written in the circle. Then $\gcd(qd_1,d_{\tau(n)-1}\ge\gcd(d_1,d_{\tau(n)-1})>1$, $\gcd(q^ad_i,q^bd_j)\ge q^{\min\{a,b\}}>1$ for all $3\le a,b\le k$ and $1\le i,j\le\tau(n)-1$, and $\gcd(q^2d_1,d_2)\ge\gcd(d_1,d_2)>1$, so this construction works.
Then, suppose $\omega(n)\ge2$ and $k=1$. We have $\tau(n)\ge4$. If $i=1$ or $i\ge3$, insert $qd_i$ to the right of $d_i$. Then, insert $qd_2$ to the left of $d_2$ and $q$ between $qd_1$ and $qd_2$. We end up with:
$$d_1,qd_1,q,qd_2,d_2,d_3,qd_3,\ldots,d_{\tau(n)-1},qd_{\tau(n)-1},$$
which, similarly to before, shows that $n$ is juxtaposable.

We claim that the only $n$ that is not juxtaposable are the product of two primes. Now we induct on $\omega(n)$, with the following two base cases.
If $\omega(n)=1$, $n=p^a$, so we can simply write the divisors:
$$p,p^2,\ldots,p^a.$$
If $\omega(n)=2$, $n=p^aq^b$ and $(a,b)\ne(1,1)$. Note that $p^a$ and $q^b$ are juxtaposable, so if either $a$ or $b$ is at least two, by the claim, $n$ is juxtaposable.

Then the induction step is just the claim, so we're done.

[b]USAMO 2005.1.[/b] Determine all composite positive integers $n$ for which it is possible to arrange all divisors of $n$ that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.

[b]Solution.[/b] We claim that only all numbers that are products of two distinct primes do not satisfy the conditions. It is clear that we cannot, for a number of the form $n=pq$ where $p,q$ are distinct primes, arrange the divisors $p,q,pq$ in a circle so that no two adjacent divisors are relatively prime, because $p$ and $q$ must be adjacent regardless of the arrangement.

For the other direction, note that the case when $n=1$ and $n=p$ for some prime is vacuous. If $n$ is a perfect prime power $p^k$, we can arrange the divisors arbitrarily. If $n=p^aq^b$ for distinct primes $p,q$ and $a>1$, we can place $pq$ and $p^2q$ and place the multiples of $p$ in one arc and the multiples of $q$ in the other arc. This obviously satisfy the problem condition.

It suffices to consider the case when $n=\prod\limits_{j=1}^kp_j^{e_j}$ for $k\ge 3$ and pairwise distinct primes $p_1,\dots,p_k$. Then we can place $p_1p_2,p_2p_3,\dots,p_{k-1}p_k,p_kp_1$ in that order and place the multiples of $p_j$ in the arc $p_{j-1}p_j$ and $p_jp_{j+1}$ that does not contain the number $p_{j+1}p_{j+2}$, indices taken modulo $k$. It is easily seen that this arrangement satisfies the problem condition, and hence completing the proof.

Denote the multiples of $n$ as $m(n).$ Clearly, $pq$ doesn't work and $p^k$ works for all primes $p$ and $q.$ Also, $p^k$ for $k>1$ works. We claim all other positive integers work. For $n=p^kq^{\ell}$ with $\ell\ge 1$ and $k>1$ the construction $$\dots,pq,m(p),p^2q,m(q),\dots$$ suffices, where the multiples of $pq$ are counted in multiples of $p.$ When $n$ has at least three prime factors $p_1,p_2,\dots p_i,$ the construction $$\dots,m(p_1),m(p_1p_2),m(p_2),\dots,m(p_{i-1}p_i),m(p_i),m(p_ip_1),\dots$$ works. $\square$

We claim that all numbers not of the form $p_1p_2$ for distinct primes $p_1$ and $p_2$ work.

Clearly $p^k$ always works and $p_1p_2$ does not.
Let $n=p_1p_2…p_m$ for $m\geq 3$. Arrange the primes $p_1$ thru $p_m$ clockwise in a circle, and put $p_1p_m$ between $p_1$ and $p_m$. Now, between $p_i$ and $p_{i+1}$ insert all divisors whose smallest prime factor is $p_i$, such that $p_ip_{i+1}$ is adjacent to $p_{i+1}$. This satisfies the requirements.

Now suppose $n=p_1^{k_1}p_2^{k_2}…p_m^{k_m}$. We proceed by induction. Clearly our condition can be satisfied for $k_1=k_2=…=k_m=1$. Now suppose one of the exponents is greater than $1$, WLOG let it be $k_1$. Then if there exists a valid drawing for $p_1^{k_1-1}p_2^{k_2}…p_m^{k_m}$, we just need to insert all divisors of $n$ that are divisible by $p_1^{k_1}$, and so if we place $p_1^{k_1}q$ clockwise adjacent to $p_1^{k_1-1}q$, then we have a valid configuration for $n$, completing the inductive step. Thus, any number with three or more prime factors satisfies the problem statement.

For $n=p_1^2p_2$, we can arrange the numbers consecutively as follows: $p_1,p_1^2,p_1^2p_2,p_2,p_1p_2$, and using the inductive statement above, any number with two prime factors and an exponent greater than one in its prime factorization works, and we are done.

The answer is all prime powers and numbers with at least 4 divisors, none of which is equal to 1. (This is equivalent to saying that it works for positive integers not of the form $n=pq$, where $p$ and $q$ are primes).  For prime powers, every arrangement works. So assume $n$ is not the power of a prime.
		[hide=Claim]
			All numbers with at least four divisors work.
		[/hide]
		[hide=Proof of the claim]
			If $n$ consists of exactly two primes, then one of them must appear with power at least 2. So say $n=p^aq^b$ where $a\ge 2$. Put $pq$ opposite of $n$ on the circle. For any other divisor of $n$, put on arc $pq,n$ if it has $p$ as a prime factor and on the other arc otherwise. If it contains both, we may simply choose one arc, and it is easy to check that this construction works. If $n$ has at least three prime factors, then let $n=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$. To the right of of $n$, place $s_1=p_2p_3\ldots p_{k}$, then $s_2=p_1p_3\ldots p_{k}$ etc. until $s_{k}=p_1p_2\ldots p_{k-1}$. If any divisor has a $p_1$ in it, put it on the arc $ns_1$, if it has a $p_2$ in it, place it on arc $s_1s_2$ etc.. If any divisor has multiple prime factors, simply put it on any of the arc corresponding to one of these prime factors.
		[/hide]
		Now assume that $n$ is has less than four divisors and is not a prime power, i.e. it is of the form $pq$ and thus has divisors $pq,p,q$. This forces $p$ and $q$ besides each other, so this case does not work.

I believe the following solution is much cleaner than a lot of the solutions on the thread, so long as it's not a fakesolve (if anyone could check for me).
Call all $n$ that satisfy the condition expressible. The key claim is the following:
[b]Claim: [/b] Suppose $n$ is expressible and let $p$ be a prime such that $\gcd(n,p)=1$. Then, for every integer $k\geq 1$, $np^k$ is also expressible.
[i]Proof.[/i] Suppose the divisors of $n$ are $d_1, \dots, d_m$ arranged in that order as well. Then, we will arrange the divisors of $np^k$ as such:
\begin{align*}
\dots , d_2, d_1, pd_1, (\text{all other divisors that are a multiple of } p), pd_2, \dots, pd_m, d_m.
\end{align*}
Since $p^2$ where $p$ is a prime is expressible, all  nonsquarefree $n$ are expressible. Now, we look at squarefree $n$. Clearly, $pq$ is not expressible but $pqr$ is, as the following arrangement works:
\begin{align*}
p, pq, pqr, q, qr, r, pr.
\end{align*}
Hence, all squarefree $n$ that have more than $3$ distinct prime divisors are expressible. Therefore, all composite $n$ are expressible except for when $n=pq$ where $p,q$ are distinct primes.

My first combinatorics post :-D. Hope I didn't fakesolve
The answer is all $n$ that are not represented as $n=p_1p_2$ for distinct primes $p_1,p_2$.

There are three cases for $n$:

[b]1)[/b]$n=p^a$. This is always possible since all divisors of $n$ greater than 1 are divisible by $p$ and thus are not coprime with each other.

[b]2)[/b]$n=p_1p_2$ for distinct primes $p_1,p_2$. This is impossible since there are only three divisors of $n$ greater than 1 and $p_1,p_2$ are always adjacent in a circle and are relatively prime.

[b]3)[/b]$n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ for distinct primes $p_1,p_2, \cdots p_k$. We set the construction of this case like this: Define $b_i$ as a group of divisors of $n$ greater than 1 that are divisible by $p_i$, not divisible by $p_j$ and the last term being divisible by $p_{i+1}$ for $1 \leq j \leq k-1, 1 \leq i \leq k, j<i$ (To avoid confusion we define $k+1 \equiv 1$). Note that in this configuration, the last term of $b_k$ will be divisible by $p_1$. For example terms of $b_1$ will include all divisors that are divisible by $p_1$ except the last term of $b_k$, terms of $b_2$ will not be divisible by $p_1$, terms of $b_3$ will not be divisible by $p_1, p_2$ and so on. Now when these groups are in order in a circle, all the divisors adjacent to each other will not be coprime and we're done.

There are three cases to consider, when $n = p_1^{e_1}$, $n = p_1p_2$ and $n = p_1^{e_1} p_2^{e_2}p_3^{e_3} \cdots p_k^{e_k}$. For the first case $n = p_1^{e_1}$ it is obvious on how to put the factors in the circle. For the second case $n = p_1p_2$ we claim that it is impossible to put them on a circle and this obvious as to why. Now we look at the case when $n = p_1^{e_1} p_2^{e_2}p_3^{e_3} \cdots p_k^{e_k}$. Here is the construction:

We claim that all $n$ will work except for those in the form $n=p_{1}*p_{2}$, where $p_{1}$ and $p_{2}$ are primes.

First we will show that $n=p_{1}*p_{2}$ does not work (trivial)

[b]Construction:[/b]

We start with an empty circle. Let $n=p_{1}^{a_{1}}*p_{2}^{a_{2}}*...$. Add $n$ to this circle. Then, pick a prime from the prime factorization of $n$ (WLOG $p_{1}$) and repeatedly add multiples of it to the circle, ensuring that the last added factor contains another factor.

For example: consider $n=2^{2}*3^{2}=36$

Once such configuration would be 36 - 2 - 4 - 18 - 12 - 6 - 3 - 9 - 36

This construction only doesn't work when a "transition" that is already taken: meaning that the transition (which is the product of 2 primes) is $n$, and since we have already proved that, we are done. $\blacksquare{}$

Another solution cuz why not

We claim that $n$ can be all compositive positive integers except for those in the form $p_{1}*p_{2}$.

We begin by adding $n$ to the circle. We can split the problem up into 3 cases as follows:

$\textbf{Case 1}$

$n=p_{1}^{q_{1}}$, where $p_{1}$ is prime. It is trivial that such a configuration must occur.

$\textbf{Case 2}$

$n=p_{1}*p_{2}$, where $p_{1}$ and $p_{2}$ are both prime. It is also trivial that there are no configurations that satisfy the problem conditions.

$\textbf{Case 3}$

$n=p_{1}^{q_{1}}*p_{2}^{q_{2}}*...*p_{k}^{q_{k}}$, where $p_{i}$ is prime for all $1 \le i \le k$. For simplicity, WLOG $p_{1} < p_{2} < ... < p_{k}$. Define a "transition" as $p_{a}*p_{a+1}$ for $1 \le a \le k-1$. Add all "transitions" in order $p_{1}, p_{2}, ..., p_{k}$ to the circle in a clockwise manner. Then, for the remaining factors, insert them immediately after the second occurrence of the least prime factor, reading from $n$, EXCEPT for $p_{1}$, where you can insert it after $n$.

This completes the proof. $\blacksquare{}$

[hide=sol]
Firstly, note that it is impossible for all numbers of the form $n = pq.$ Now, note that it is possible for all numbers of the form $p^k.$ If it is possible for a number $n = p_1k^1p_2k^2\cdots p_t^{k_t},$ then we claim that it is possible for $np_{t+1}^{k_{t+1}}.$ Suppose the arrangement for $n$ is $a_1,a_2,\dots,a_{b}.$ Then, we arrange factors of $np_{t+1}^{t+1}$ in the following manner:
$\begin{matrix}
a_1&a_2&\cdots&a_b\\
p_{t+1}a_1&p_{t+1}a_2&\cdots&p_{t+1}a_b\\
p_{t+1}^2a_1&p_{t+1}^2a_2&\cdots&p_{t+1}^2a_b\\
\vdots&\vdots&\cdots&\vdots\\
p_{t+1}^{k_{t+1}}a_1&p_{t+1}^{k_{t+1}}a_2&\cdots&p_{t+1}^{k_{t+1}}a_b\\
\end{matrix}$
We take the first row in order, then add the reverse of the second row, add the third row in order, and so on.[/hide]

[b][color=#00f]Claim:[/color][/b] We claim that only composite $n$ that are not of the form $n = pq$ where $p, q$ are distinct primes work.

To quickly show $n = pq$ is not feasible, observe that $p$ and $q$ must be adjacent in the circle, which is absurd. To finish our claim, we aim to demonstrate a construction for every other possible prime factorization, which we will do in cases.

[b][color=#ff9a00]Case 1: [/color][/b]$n = p^a$ for some prime $p$ and some integer $a \ge 2$.

Every divisor besides $1$ of $n$ is divisible by $p$, so any arrangement works.

[b][color=#ff9a00]Case 2: [/color][/b]$n = p^a q^b$ work for distinct primes $p, q$ and $a, b$ are positive integers with $(a, b) \ne (1, 1)$.

We demonstrate a construction. Observe that for numbers of the form $n = p^a q^b$, we can construct the cycle (arrangement of the divisors in the circle)
$$p \to p^2 \to p^3 \dots \to p^a \to$$
$$p^2q \to p^3q \to \dots \to p^aq \to$$
$$pq^2 \to p^2q^2 \to p^3q^2 \dots \to p^aq^2 \to$$
$$pq^3 \to p^2q^3 \to p^3q^3 \dots \to p^aq^3 \to$$
$$\dots$$
$$pq^b \to p^2q^b \to p^3q^b \dots \to p^aq^b \to$$
$$q \to q^2 \to q^3 \dots \to q^b \to pq$$

[b][color=#ff9a00]Case 3: [/color][/b]$n$ has at least $3$ prime factors.

Let $n = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}$ be the prime factorization of $n$, where $p_i$ are distinct prime factors of $n$, and $p_1 < p_2 \dots < p_k$. Observe that we can classify any divisor of $d~|~n$ with a $k$-sized binary tuple, where a $1$ in the $i$-th position means $p_i~|~d$, and a $0$ denotes otherwise. For example, if a divisor generated the tuple $(0, 1, 1, 0, 1)$, it has $p_2, p_3, p_5$ as factors.

With this new definition in mind, observe that it suffices to construct a cycle of all binary tuples of size $k$ except the all-$0$ tuple, such that any adjacent tuples have a $1$ in the same position (This is sufficient as we can group terms with the same generated tuple together.) We proceed with induction for $k \ge 3$, here is a construction for $k = 3$:
$$(1, 0, 0) \to (1, 1, 0) \to (0, 1, 0) \to (0, 1, 1) \to (0, 0, 1) \to (1, 0, 1) \to (1, 1, 1)$$
Now assume that for $k = t$, there exists some cycle $X_1 \to X_2 \dots \to X_{2^t - 1}$ that works. To finish the induction, we construct a valid cycle for $k = t + 1$. Without loss of generality, assume that $X_{2^t - 1}$ is the all-$1$ tuple of size $t$. Suppose $A_i$ and $B_i$ are the tuple $X_i$ with an extra $0$ and $1$ appended to the rightmost end respectively for integers $1 \le i \le 2^t - 1$. Then
$$A_1 \to A_2 \dots \to A_{2^t - 1} \to B_1 \to (0, 0, \dots, 1) \to B_2 \dots \to B_{2^t - 1}$$
is a valid sequence, where the tuple between $B_1$ and $B_2$ contains $t$ zeroes. Therefore, the induction is complete, and all positive integers $n$ with at least $3$ prime factors work.

Collectively, all of our steps show that only composite positive integers $n$ that satisfy the condition cannot be expressed as the product of two distinct primes. Our proof is complete. $\blacksquare$

The answer is all integers not of the form $pq$, where $p$ and $q$ are prime numbers. We will show that all other integers work. It's easy to see that numbers of the form $p^k$, $k\ge 1$ work because $p$ divides each one of the divisors of $p^k$.

Next, consider the case $n=p^{k_1}\cdot q^{k_2}$ where $k_1$, $k_2$ are not both equal to $1$. Also, WLOG $k_1\ge k_2$. Place $p$ and $q$ in the circle first. We can then find two numbers that are able to "fill" the gap between $p$ and $q$, say $pq$ and $p^2q$. It's easy to see then that we can place the remaining divisors of $p^{k_1}\cdot q^{k_2}$ to make a good circle, as shown in the following construction:
[center][img width=25]https://i.ibb.co/py2JVsc/IMG-0217.jpg[/img][/center]

Finally, consider the case $p_1^{k_1}p_2^{k_2}\cdots p_m^{k_m}$. Like before, place $p_1$, $p_2$, $\ldots$, $p_m$ in the circle first. Then, use $p_1p_2$, $p_2p_3$, $\ldots$, $p_{m-1}p_m$ to "fill" the gaps between adjacent primes. It's to easy to see then that we can place the remaining divisors of $n$ to make a good circle, shown in the construction below.
[center][img width=25]https://i.ibb.co/MCyySc3/IMG-0218.jpg[/img][/center]
It's trivial to see that all $n=pq$ fail, so we're done. $\blacksquare$

All positive integers work except $\left\{1,pq\right\}$ where $p$ and $q$ are two distinct primes.

Divide into cases where the prime factorization has $\ge 3$ primes in it, $2$ or $1$.

$p^k$ clearly works for $k\ge 1$ and $pq$ clearly doesn't work for distinct $p$, $q$.

For the base case for $k=3$, we have the image below.

[center][img]https://i.imgur.com/IWR9PAq.png[/img][/center]

Now for an increase in the value of $k$ to $k+1$, note that the divisors that get introduced are $p_{k+1} \times \left\{\text{set of divisors of } p_1\cdot p_2 \cdots p_k\right\}$. For this we can select a divisor from the configuration for $k$ from induction and its neighboring divisor and then insert the new divisors as shown below.

[center][img]https://i.imgur.com/nUhzWDR.png[/img][/center]

For increasing the power of a prime (WLOG $p_1$) in the factorization from $e$ to $e+1$, we can see that the new divisors that get introduced are those who have $\nu_{p_1}(\text{divisor}) = e+1$. Now note in the configuration of the inductive hypothesis, that the divisor adjacent to $p_1$ must also have $p_1$ in it. Thus we can insert all the new divisors in between the two selected divisors as shown below.

[center][img]https://i.imgur.com/yLGWlQQ.png[/img][/center]

And we are done. :yoda:


Let k be an integer,k>1.
    S be the set of all polynominal P with integer coefficients satisfy
P(1)=P(k)=0 and P#0
    For every P in S ,f(P) is the minimum coefficients of P
    Find the maximum of f(P) when P in S

I think he means that P(x) is not identically equal to zero.


In year 1984 I read the fantastic book of Hofstadter : GODEL,ESCHER,BACH.

Now, after 20 year, I am reading it again for the second time.

It is really one of the most beautiful books I have ever read.

I would like to know if you have read it and your opinion about it.

You're dating yourself.

Many people of my generation credit that book with giving them an interest in <fill in the math/logic/AI-related field of your choice here>.

If you like Escher, Bach, or Lewis Carroll, give this one a shot.  It's pretty long, so you may want to wait until you have a stretch of time.


Please help me figure this word problem out (set up and solutions)

The organizer for a church picnic ordered coleslaw, potato salad, and beans amounting to 50 pounds. There was to be three times as much potato salad as coleslaw. The number of pounds of beans was to be 6 less than the number of pounds of potato salad. Find the number of pounds of each.

Thanks! :!:

Set up 3 equations:
c=coleslaw, p=potato salad, b=beans (all in pounds)
$ c\plus{}p\plus{}b\equal{}50$
$ 3c\equal{}p$
$ b\equal{}p\minus{}6$
The topic says solve by elimination, but I would suggest substitution.
First, substitute $ 3c$ for $ p$ in $ b\equal{}p\minus{}6$, which yields $ b\equal{}(3c)\minus{}6$
Then, substitute these equations into the first equation:
$ c\plus{}(3c)\plus{}(3c\minus{}6)\equal{}50$
Combine terms and solve for c: $ 7c\equal{}56, c\equal{}8$
Thus, $ c\equal{}8, p\equal{}24, b\equal{}18$

ak he kinda said help with elimination so he prolly meant elimination *ur solution was still good*
basically elimination is subtracting or adding equations to eachother in an effort to ELIMINATE one or more variables
ex:
$ b\equal{}p\minus{}6$ basically means $ b\minus{}p\equal{}\minus{}6$
$ c\plus{}b\plus{}p\equal{}50$
$ \underline{\minus{} (b\minus{}p\equal{}\minus{}6)}$
$ c\plus{}0b\plus{}2p\equal{}56$
b was eliminated and you can eliminate c in the same way using $ 3c\equal{}p$
srry cant align in LaTeX yet

Elimination Method:

Same variables as the second post in this topic:

$ c\plus{}p\plus{}b\equal{}50$
$ 3c\minus{}p\equal{}0$
$ p\minus{}b\equal{}6$

Let us eliminate p first.  So we solve the first two equations first.

$ c\plus{}p\plus{}b\equal{}50$
$ 3c\minus{}p\equal{}0$
--------------
$ 4c\plus{}b\equal{}50$

Now we eliminate p from the first and last equation.  To do this, we need to multiply the first equation by -1.:

$ \minus{}1(c\plus{}p\plus{}b\equal{}50)$
$ p\minus{}b\equal{}6$
$ \downarrow$
$ \minus{}c\plus{}\minus{}p\plus{}\minus{}b\equal{}\minus{}50$
$ p\minus{}b\equal{}6$
------------------
$ \minus{}c\plus{}\minus{}2b\equal{}\minus{}44$

Now we find the value of either c or b....let us do c first.  To do so, we need to multiply the first equation by 2.:

$ 2(4c\plus{}b\equal{}50)$
$ \minus{}c\plus{}\minus{}2b\equal{}\minus{}44$
$ \downarrow$
$ 8c\plus{}2b\equal{}100$
$ \minus{}c\plus{}\minus{}2b\equal{}\minus{}44$
--------------
$ 7c\equal{}56$
$ c\equal{}8$

We know c is equal to 8.  Now we plug it into one of the equations we used in the previous step:

$ 4(8)\plus{}b\equal{}50$
$ 32\plus{}b\equal{}50$
$ b\equal{}18$

We know know that c is 8 and b is 18.  Now we plug these in to our first equations and get p.

$ 8\plus{}p\plus{}18\equal{}50$
$ p\equal{}24$

So we conclude:
$ c\equal{}8$      $ b\equal{}18$    $ p\equal{}24$


integration
trigonometry
function
calculus
derivative
calculus computations

Please help me with this problem about fourier series:
$a_n = \frac{2}{\pi} \left ( \int_{0}^{\pi} (x \pi - x^2) \cos nx \, dx \right )$
$b_n = 0$
$f(x) = x(\pi -x)$ .
 I'm asking about the expansion of $f(x)$ thanks. :)

What is the actual function you are talking about? From your integral formulas, it appears that you are talking about this function:

$f(x)=\{\begin{array}{cc}x(\pi-x),&x\in[0,\pi]\\-x(\pi+x),&x\in[-\pi,0]\end{array},$ continued to be periodic of period $2\pi.$

We could also write this as $f(x)=\pi|x|-x^2$ on $[-\pi,\pi].$

On the other hand, that might not be the function you mean.

Any time I see someone say they are interested in the the Fourier series of "$f(x)=$ (some formula)," the statement is nearly always incomplete or misleading. We really need to know what interval that formula is valid for, and what formulas are valid on other intervals.

If that really is the function you intend, we note the following properties: it is an even function, so its Fourier series in real form will have only cosine terms, or its complex Fourier coefficients will be real and even. It is a continuous function but not a $C^1$ function. In particular, its derivative has jump discontinuities. From that I predict that the Fourier coefficients will be $O(n^{-2})$ as $n\to\infty.$

Here's what I would actually do to calculate this:

I'm going to find $f'$ and $f''$. $f'(x)=\pi-2x$ on $(0,\pi)$ and $-\pi-2x$ on $(-\pi,0)$. Note the jump discontinuities at $0$ and $\pi.$ There's an upward jump of size $2\pi$ at $0$ and another upward jump of size $2\pi$ at $\pi.$ Thus (and now I'm talking about the "weak" or "distributional" derivative) $f''(x)=-2+2\pi\delta(x)+2\pi\delta(x-\pi).$

If $f(x)=\frac{a_0}2+\sum_{n=1}^{\infty}a_n\cos nx,$ then $f''(x)=-\sum_{n=1}^{\infty} n^2a_n\cos nx.$

So, for $n\ge 1,$ we have

$-n^2a_n=\frac1{\pi}\int_{-\pi}^{\pi}(-2+2\pi\delta(x)+2\pi\delta(x-\pi))\cos nx\,dx$

$=2+2\cos(\pi n)$ which equals $4$ if $n$ is even and $0$ if $n$ is odd.

Thus $a_n=0$ for odd $n$ and for positive even $n,\,a_n=-\frac4{n^2}$

We can't ever see the constant term from the derivative, so that's a separate calculation:

$a_0=\frac2{\pi}\int_0^{\pi}(\pi x-x^2)\,dx=\frac2{\pi}(\frac{\pi^3}2-\frac{\pi^3} 3)=\frac{\pi^2}3.$

Thus $f(x)=\frac{\pi^2}6-\sum_{k=1}^{\infty}\frac1{k^2}\cos(2kx).$

Go ahead and put $x=0$ into that to see a famous result.

You could just integrate by parts in your original integral. What I've done by considering the series of the derivatives is to capture the same information that would come from that integration by parts. And in dealing with the derivatives, you don't have to take it all the way to Dirac delta functions - it's just that the maximum computational advantage happens when you go that far.

That is the function I mean, sorry not including the interval that formula is valid for. I suppose to write down the interval that you mention. It's the same with my question.
thank's.  :lol:


This is from the 1979 USAMO.

Find all sets of 14 or less fourth powers that sum to 1599.

[hide]mod 16, we have all fourth powers congruent to 1 or 0. so the sum of 14 or less fourth powers will be congruent to anything from 0 to 14. it cannot be congruent to -1 mod 16, which is what 1599 is.[/hide]

My official guess is essentially that it is coincidental.

It's also "kind of" nice if you look at powers of 2 -- for example, n^4 mod 2^4, n^3 mod 2^3, which I can at least explain some of the niceness of -- all of the even numbers always go to 0.

[quote="s0mp"]has anyone noticed this pattern? squares look good in mod 4 (2^2), cubes look good mod 9 (3^2), fourth powers look good mod 16 (4^2), and does this hold for kth powers (k^2)? mod 25 the fifth powers are either 0,1,-1,7,or-7, so i dont know if thats really "nice"

this takes some experimenting.[/quote]
That's very interesting, methinks.


Suppose that 7 boys and 13 girls line up in a row. Let S be the number of
places in the row where a boy and a girl are standing next to each other. For
example, for the row
GBBGGGBGBGGGBGBGGBGG
we have S = 12. If all possible orders of these 20 people are considered, what is
the average value of S? Generalize this result to a group of m boys and n girls. :maybe:  :maybe:  :maybe:  :maybe:

Associate to every ordering of boys and girls a sequence that describes the way the sequence switches between boys and girls as you move from left to right (after giving the first gender0:  say $ A$ means gender changes and $ B$ means gender stays the same.  For your sequence we have

$ GABABBAAAABBAAAABAB$.

What is the value of $ S$ in terms of this sequence?  What kinds of sequences are permissible?


geometry
function
calculus
calculus computations

The rate at which the volume of a snowball evaporates from solid to gas is proportional to the surface area of the snowball. Suppose the snowball has been observed at one time to have a radius of 0.5 ft and six months later a radius of 0.25 ft.

(a) Express the radiuis of the snowball as a function of time.

So $\frac{dV}{dt}= k(4\pi r^{2})$

$4\pi r^{2}\frac{dr}{dt}= k(4\pi r^{2})$

$\frac{dr}{dt}= k$

$r = kt+C$

How would I use the initial conditions to find $k$ and $C$?

Thanks

[quote="ragingbullrocky"]How would I use the initial conditions to find $k$ and $C$?[/quote]

Here is the key:

[quote="ragingbullrocky"]Suppose the snowball has been observed at one time to have a radius of 0.5 ft and six months later a radius of 0.25 ft.[/quote]

Let the time at which the snowball had a radius of 0.5 ft be $t=0$, where $t$ is in years, so that the later observation (when the radius was 0.25 ft)  corresponds to $t=0.5$; the points are then $r(0)=.5,\quad r(.5)=.25$ and we had that $r(t)=kt+C$... you solve.


Euler
real analysis
real analysis unsolved

1. Prove that if $ (x{}_n{}_k)$ is a sub-sequence of $ (x_n)$ then $ \forall k\in\mathbb{N},\ \ n_k\geq k$

2. Prove that $ (x_n)$ is bounded if and if has a sub-sequence $ (x{}_n{}_k)$ safty$ \forall k\in\mathbb{N}, \ \ x{}_n{}_k \geq k$

It boils down to writing out the definitions. What are you having troubles with?

[quote="euler_vn"]
2. Prove that $ (x_n)$ is bounded if and if has a sub-sequence $ (x{}_n{}_k)$ safty$ \forall k\in\mathbb{N}, \ \ x{}_n{}_k \geq k$[/quote]
What do you mean here dear Mr euler_vn?

I think you should mean [quote]Prove that $ (x_n)$ is [b]not[/b] bounded if and only if it has a sub-sequence $ (x{}_n{}_k)$ satisfy$ \forall k\in\mathbb{N}, \ \ x{}_n{}_k \geq k$ [/quote]


algebra
polynomial
superior algebra
superior algebra unsolved

find the degree of the splitting field of $ x^6 \plus{} 1$ over $ \mathbb{F}_2$.

Do you know how to factor this polynomial into irreducibles?


parameterization
advanced fields
advanced fields unsolved

Hello to everyone,
I am new to this forum and I hope that this is the appropriate section to post this question.
Let say we have to predict the unknown parameters A and theta at the following system

y = A*sin(w*t+theta) where w is known and A,theta unknown

This must be done with an online method such as sliding mode and Lyapunov

Any idea how to begin with?
Is it possible to convert this sytem from nonlinear to linear (parametrization)
something like y = a*u ???

Whatever comes to your mind would be appreciable.
Thank you in advance,
Nikolaos


1. Fie $ A \in \mathcal{M}_{m,n}$ cu elemente numere reale pozitive.Numim "transformare" inlocuirea tuturor elementelor de pe o linie sau de pe o coloana cu inversele lor.Sa se arate ca putem efectua o succesiune de "transformari" care modifica matricea $ A$ in matricea $ B$ cu proprietatea ca produsul tuturor elementelor de pe fiecare linie si de pe fiecare coloana este cel putin $ 1$.

2.Fie $ A \in \mathcal{M}_n(\mathbb{C})$ o matrice cu proprietatea ca:

\[ S(A)\equal{}S(A^2)\equal{}\cdots S(A^n)\equal{}0\],
unde $ S(A^k)$ este suma tuturor elementelor matricei $ A^k$, $ k\equal{}\overline{1,n}$.Sa se arate ca:

$ a)$ $ \det A \equal{}0$.

$ b)$ $ S(A^k)\equal{}0$, $ \forall k\in\mathbb{N^{*}}$.

$ c)$Sa se dea exemplu de matrice nenula $ A$ cu proprietatea din enunt.

3. Pentru $ a>0$ definim sirul $ (x_n)_{n\geq 1 }$ prin relatia de recurenta:

\[ x_{n\plus{}1}\equal{}a\plus{}ax_1x_2x_3 \cdots x_{n\minus{}1}\plus{}(x_1x_2 \cdots x_{n\minus{}1})^2 \text{ ,} \forall n\geq 2\]

cu $ x_1\equal{}1$ si $ x_2\equal{}1\plus{}a$.

$ a)$Sa se arate ca $ x_{n\plus{}1}\equal{}x_n^2\minus{}ax_n\plus{}a$ $ \forall n\geq 2$.

$ b)$ Sa se arate ca \[ \lim_{n\to \infty} \sum_{k\equal{}1}^{n} \frac{a^k}{x_k}\equal{}a(a\plus{}1)\] .

4.Fie $ a \in (0,1)$ un numar real si $ a\equal{}0,a_1a_1 ...a_n...$, cu $ a_1,a_2..,a_n.. \in \{0,1,2,..9\}$ reprezentarea sa zecimala.

$ a)$ Sa se arate ca pentru orice $ x \in (0,1)$ exista si este finita limita:
\[ \lim_{n \to\infty} (a_1x\plus{}a_2x^2\plus{}\cdots \plus{}a_nx^n)\].

$ b)$Daca notam cu $ f_a(x)\equal{}\lim_{n \to\infty} (a_1x\plus{}a_2x^2\plus{}\cdots \plus{}a_nx^n)$, $ x\in (0,1)$, sa se arate ca functia $ f_a: (0,1) \to  \mathbb{R}$ este o functie rationala daca si numai daca numarul $ a$ este numar rational.
(Functia $ f_a$ este rationala daca $ f_a(x) \equal{}\frac{P(x)}{Q(x)}$, cu $ P,Q \in \mathbb{Z}[X]$).

[u][b]Subiectele de la clasa a XI-a [/b][/u]
http://www.mathlinks.ro/viewtopic.php?t=238240
http://www.mathlinks.ro/viewtopic.php?t=238236
http://www.mathlinks.ro/viewtopic.php?t=238235
http://www.mathlinks.ro/viewtopic.php?t=238232


calculus
integration
calculus computations

These integral ones are tough to me:

$1/(sqrt(x)-1) dx$
$cos(sqrt(x)) dx$
$x{}^3e{}^x{}^2 dx$

hints:
 [hide="first"]change $\sqrt{x}-1=t$[/hide]
  [hide="second"] change $x=t^2$ and then integrate by parts[/hide]
[hide="third"]change $x^2=t$ and then integrate by parts twice[/hide]


group theory
abstract algebra
invariant
superior algebra
superior algebra solved

By the standard techniques, the group is a semidirect product of its Sylow subgroups. The only possible variation is in the choice of the automorphism we map a generator of the smaller subgroup to. The choice of generator is arbitrary- varying this, we see that all of the nontrivial choices are isomorphic.

[quote="Alope"]How can you show that up to isomorphism, there are at most two groups of order pq where p,q are primes (not necessarily distinct)?  So that if p=q how can you prove that there are two groups of order pq and if p<q there is one group if p does not divide q-1 and there are two groups if p does divide q-1?[/quote]

Let the group be $G$ and let $a\in G$ be a nonzero element.  There will then be a minimum $m>0$ such that $ma=0$.  Let $A=\{m_{1}a: 0\le m_{1}\le (m-1)\}$.  Consider the sets $A+x$ for $x\in G$.  It is easy to see that any two such sets must be disjoint or equal.  Hence $m$, the number of elements in $A$, must divide $pq$, the number of elements in $G$.  If $m=pq$, then $a$ generates a cyclic group of order $pq$ and we are done.  Otherwise $m>1$ so $m=p$ or $m=q$.  Assume without loss of generality that $m=q$.  There will therefore be $p$ distinct sets $A+x$:  let us refer to them as $A+x_{0}$, $A+x_{1}$, ..., $A+x_{p-1}$.  Every element $g\in G$ can then be written uniquely as $g=m_{1}a+x_{i}$ where $0\le m_{1}\le (q-1)$ and $0\le i\le (p-1)$.  We can then write:

$x_{0}+a=m_{0}a+x_{f(0)}$
$x_{1}+a=m_{1}a+x_{f(1)}$
...
$x_{p-1}+a=m_{p-1}a+x_{f(p-1)}$

or

$x_{i}=x_{i}+qa=m_{i}a+x_{f(i)}+(q-1)a$
$x_{i}=m_{i}a+m_{f(i)}a+x_{f(f(i))}+(q-2)a$
...
$\displaystyle x_{i}=\sum_{j=0}^{q-1}m_{f^{j}(i)}a+x_{f^{q}(i)}$

From this it follows that $f^{q}(i)=i$.  This establishes an equivalence class on the numbers $0\le i\le p-1$ where each set consists of $\{ i,f(i),f(f(i)),...,f^{q-1}(i)\}$.  Each set must have either $1$ or $q$ members.  If we assume $p\ne q$ (we will deal with the $p=q$ case later), then $q$ does not divide $p$ and so there must at least one set with $1$ member so that $i=f(i)$:

$x_{i}+a=m_{i}a+x_{i}$

To simplify notation we let $x=x_{i}$ and $k=m_{i}$ to get:

$x+a=ka+x$ (Eq 1)

What is the order of $x$ (the minimum $m$ such that $mx=0$)?  From Eq 1 it follows that $a,x$ generate a subgroup of order $qm$, from which we can conclude that $m=p$.  We next ask:  what is the order of $x+a$?  Let this order be $t$.  We can write:

$ka+x=x+a=(t+1)(x+a)$

Repeated application of Eq 1 gives us:

$\displaystyle ka+x=\sum_{j=1}^{t+1}k^{j}a+(t+1)x$

Given that representation in this form is unique, we can conclude that $p|t$.  Unless $x+a$ generates a cyclic group of order $pq$, in which case we are done, we can assume $t=p$:

$\displaystyle ka+x=\sum_{j=1}^{p+1}k^{j}a+x$

From this we can conclude that:

$\displaystyle k\equiv \sum_{j=1}^{p+1}k^{j} \mod q$
or $k^{2}\equiv k^{p+2} \mod q$
or $k^{p}\equiv 1 \mod q$

since we can easily show $k$ is not divisible by $q$.  From this it follows that $p|(q-1)$.  This pretty much defines the group up to isomorphism.  If another group satisfies the same but a different $p$th root of unity appears in Eq 1, $k^{\prime}$, then there will exist an $r$ such that $k^{\prime r}\equiv k \mod q$ and we can write:

$(rx)+a=ka+(rx)$

showing it is an isomorphic group.  This leaves the $p=q$ case.  In this case, we can show, without going through as many details, that unless $x$ generates a cyclic group of order $p^{2}$, that $x$ has order $p$, so it easily follows that each group element can be written in the form $ia+jx$.  From this point, we can skip the counting argument and skip to Eq 1.  In this case we will end up concluding that $k=1$, meaning the group is abelian;  however, unlike in the $p\ne q$ case where the group being abelian implies it must be cyclic, there is another abelian group of order $p^{2}$ consisting of all elements of the form $ia+jx$. QED

[quote="NoYoo-hooForMe"]...[/quote]

Actually I believe there is a fallacy in the above proof in that the invariant $f(i)=i$ may just be the trivial one where $x_{i}=0$ or $x_{i}$ is a multiple of $a$--in which case the remainder of the argument doesn't hold.  I was attempting to show that there must be subgroups of order $p$ and $q$.  The Sylow theorems will buy us this.  Once we have generators for these two subgroups, I believe that the remainder of my argument which constructs the structure of the group of order $pq$ still holds.

What does this expression mean ?
" up to isomorphism"
or "up to an isomorphism" ?!??!

well there are of courses thousands of different groups of order 15 for instance
but a group theorist will only be interested, if two different groups are also non isomorphic

so when someone says there are only two groups order four up to isomorphism, they mean : there are two nonisomorphic groups, and all the rest is isomorphic with one of them

thank you fredbel...i encountered allot this expression lately...
i really didnt knew what it meant.

[url]http://en.wikipedia.org/wiki/Up_to_isomorphism[/url]
whenever you have a question like that, google, but usually wikipedia suffices, has the answer


If your hot coffee loses 50 kJ of energy as it ocols and the volume shrinks because of thermal contraction, how large is the change in the internal energy of the coffee?  How large is the change in the enthalpy?  Explain.

a) +50 kJ
b) -50 kJ
c) >50 kJ
d)>-50kJ
e)<50 kJ
f)<-50 kJ

Considering the hot coffee as our system, then, as it transfer heat to the surroundings we have $ Q \equal{} \minus{} 50\,\,kJ$, and as its volume decreases the surroundings do work on the system and so $ W > 0$. Therefore, denoting by $ U$ the internal energy of the coffee, and assuming that 1) there is no mass transfer, 2) the system is at rest, and 3) there is transfer of energy only under the form of heat and work, then

$ \Delta U \equal{} Q \plus{} W \equal{} \minus{} 50 \plus{} W$,

and because $ W > 0$ we must have $ \Delta U > \minus{} 50$. From the definition of enthalpy $ H$,

$ H \equal{} U \plus{} PV$,

we have

$ \Delta H \equal{} \Delta U \plus{} \Delta (PV)$.

Assuming constant pressure (which is valid if the coffee is "open" to the atmosphere), we have

$ \Delta H \equal{} \Delta U \plus{} P \Delta V$.

For liquids and solids, in most of the cases of interest we have $ \Delta V \approx 0$, and so we can write

$ \Delta H \approx \Delta U$.


Suppose that for any real x with |x| <> 1, a function f(x) satisfies f((x-3)/(x+1)) + f((x+3)/(x-1)) = x. Find all possible f(x).

This should be classical. Denote y=(x-3)/(x+1). So, f(y)+f((-3+y)/(1+y))=(3+y)/(1-y).Similarly, put z=(3+x)/(1-x)  and find that f(z)+f((3+z)/(1-z))=(z-3)/(z+1). Put here z=y and add this to the previous one. Use also the relation in the problem and you will find that f(y)=(7y+y^3)/(2-2y^2). Again, I don't want to check that this is a solution.

I Think the problem is not correct
Problem is :f((x-3)/(x+1))+f((3+x)/(1-x))=x
[img]http://www.geocities.com/duyduca1/Math2.gif[/img]

Yeah! But the solution is correct, since I solved your problem!


The goodFood catering company finds that competitors cater lunch for a group of 100 people for $5 each. The manager of GoodFood calculates that for each 25 cent discount per lunch, it is possible to sell an additional 10 lunches. If each lunch costs GoodFood$2 to prepare, how many lunches should be prepared to maximize the profit?

Don't even know where to start. :(

Your post has been rendered incomprehensible by the dollar signs in it: this forum interprets dollar signs as $\LaTeX$. Either disable BBCode (which also removes smilies) or replace dollar signs by $\$$ (dollar backslash dollar dollar).


Suppose that $M$ is an arbitrary point in the equilateral triangle $ABC$.$MH \perp BC(H \in BC);MK \perp CA( K \in CA) ;ML \perp AB(L \in AB)$.Prove that:$BH+CK+AL=DK+CH+BL$

Draw a figure:
Let angle MAB = x, angle MBC = y, angle MCA = z.
Let k stand for sqrt 3(AL + BH + CK)
2 AK = 2 AM Cos (60 - x) =  AM (Cos x + k Sin x) = AL + k.ML
2 BL = BH + k. MH, 2 CH = CK + k. MK
Adding: 2(AK + BL + CH) = (AL + BH + CK) + k(ML + MH + MK)
                                      = (AL + BH + CK) + k.h
                                      =(AL + BH + CK) + 2S/a
where h is the altitude, a is the length of each side and S is the area of the equilateral triangle.
Note S = k a^2/4
Denoting (AK + BL + CH) by S' and (AL + BH + CK ) by S''
2S' - S'' = 2kS/a = 2 k^2 a /4 = 3a /2
Observe S' + S'' = perimeter = 3a
Solving S' = 3a /2 = S'' as required

vprasad_nalluri

[quote="chien than"][color=brown]Suppose that $M$ is an arbitrary point in the equilateral triangle $ABC.\,MH \perp BC\,(H \in BC);\,MK \perp CA\,(K \in CA);\,ML \perp AB\,(L \in AB)$.
Prove that: $BH+CK+AL=DK+CH+BL$[/color][/quote]
[color=indigo][b]Proof.[/b] Assume that: $BC=CA=AB=a$

We have:
$\left \{ \begin{array}{c}&MB^{2}-MC^{2}=HB^{2}-HC^{2}=a(HB-HC)\\ &MC^{2}-MA^{2}=KC^{2}-KA^{2}=a(KC-KA)\\ &MA^{2}-MB^{2}=LA^{2}-LB^{2}=a(LA-LB)\end{array}\right\|\\ \Longrightarrow a(HB+KC+LA-HC-KA-LB)=0\\ \Longrightarrow BH+CK+AL=CH+AK+BL$[/color]

I have other problem:
Let $ABC$ be an equilateral triangle.A circle $(O)$ cuts $AB;BC;CA$ at $M;N;P;Q;R;S$.Prove that:
$AR+BM+CP=AQ+BS+CN$

[quote="chien than"]
[color=darkred]Let $ABC$ be an equilateral triangle. A circle $(O)$ cuts $AB,\,BC,\,CA$ at $M,\,N,\,P,\,Q,\,R,\,S$.
Prove that: $AR+BM+CP=AQ+BS+CN$[/color][/quote]
[color=indigo][b]Proof.[/b]
Let $H,\,K,\,L$ be the projections of $O$ on $BC,\,CA,\,AB$ respectively.
We have:
$BH+CK+AL=CH+AK+BL\\ \Longrightarrow\left\{\begin{array}{c}&BM+MH+CP+KP+AR+RL=CN+NH+AQ+KQ+BS+SL\\ &MH=HN,\,KP=KQ,\,RL=LS\end{array}\right\|\\ \Longrightarrow AR+BM+CP=AQ+BS+CN$[/color]


number theory
least common multiple
function
MATHCOUNTS
greatest common divisor

first meet 1994
1. Which number is larger $\sqrt{3\sqrt{6\sqrt{2}}}$ or $\sqrt{6\sqrt{2\sqrt{3}}}$

3. Sam taps on the table at a steady rate of 40 taps per minute, hile Sandor taps at a rate of only 32 times per minute. If I hear the taps coincide, how many seconds must I wait to hear them coincide again?

1: [hide]Just set them equal to each other:
$\sqrt{3\sqrt{6\sqrt{2}}} ?= \sqrt{6\sqrt{2\sqrt{3}}}$
$\sqrt{6\sqrt{2}} ?= 2\sqrt{2\sqrt{3}}$
$6\sqrt{2} ?= 8\sqrt{3}$
obviously:
$\3sqrt{2} < 4\sqrt{3}$
EDIT: I meant $3\sqrt{2} < 4\sqrt{3}$; typo
[/hide]

[hide]For the other problem just find the least common multiple, which, since 30 = 2^5 and 40 = 2^3*5, is 160 since 2^3*5 = 160...[/hide]

I think this is the answer to number 2

[hide]40 taps per minute translates to $\displaystyle\frac{40}{60}$ taps per second, and similarly 32 taps per minute becomes $\displaystyle\frac{32}{60}$. $lcm(32,40) = 160$, so the next time the taps coincide is after $\boxed{160}$ seconds.[/hide]

Those solutions to #2 are not correct because it's not that Sandor taps every 40 seconds and Sam taps every 32 seconds but that Sandor taps 40 times every minute and Sam taps 32 times every minute.

[hide]40/60 is 2/3 taps per 60 minutes, and 32/60 is 8/15 taps every minute, so we find common multiple of 10/15 and 8/15, which is 160/15, so every 10 2/3 seconds their taps will coincide?[/hide]

2) [hide] Sam taps once every 60/40 = 3/2 seconds and Sandor taps once every 60/32 = 15/8 seconds. LCM of 12/8 and 15/8 is 60/8, so they will coincide every [b]7.5 seconds[/b][/hide]

I think that's right

[quote="Apprentice"]are the taps equally spaced?
Um, nevermind, stupid question[/quote]

The question says that the rate for taps are [b]steady[/b]!

reformulate problem #2
is
we split 60 into 40 with this function wich outputs every second some of those guys tap
$f(n)=n*\frac{40}{60}$ with $n\in \(0,+\infty \)$
so we have associated to every tap number $n$ a moment in time $f(n)$.
now for the other its the same and we get
$g(n)=n*\frac{32}{60}$ with $n\in \(0,+\infty \)$

we see that $0\leq f(n),g(n) \leq 60$

in the first second 0 they coincide.
the question is to find all the other coincidence moments reformulated as
find numer of pairs $(i,j)$  with $i\neq j$ so that $f(i)=g(j)$
thus numer of pairs of tap numbers different so that the moment of the tap is the same
for both those dudes

so for that we can do the following equation
$i%Error. "neqj" is a bad command.
$ and also $f(i)=g(j)\Rightarrow i*\frac{40}{60}=j*\frac{32}{60}$
$\Rightarrow 40i=32j$ [b]*[/b] now for w00t i,j does this happen ?
we can first do $gcd(40,32)=8$ no ?
ok so $5i=4j$ so if we talk mod 4
$i=0$ and mod 5 $j=0$
so i is multiple of 4 and j is multiple of 5.But also because $5i=4j$ it happens
in the other way around i mean i is multiple of 5 and also j is multiple of 4.
But also both are multiples of 8 because we divided equation [b]*[/b]
by 8.
so we find that they are both multiples of $4*5*8=160$.
Is this right ?

Can't it be
[hide]
7.5 seconds? If both nx(40/60) and nx(32/60) are whole numbers than the taps occur at the same time. Since the greatest common factor of 32 and 40 is eight we have 60/8 or 7.5. If they ask for a whole number of seconds, it think it's 15.[/hide]

always mistakes in interpretation are saw if you try to explain why you do those numbers you do...
so xplain why you are getting 7.5

here's what I got for this mathcounts looking problem:
[hide="solution"]We see one taps 5 times in 7.5 seconds, the other 4 times in 7.5 seconds. Thus this is the first time they coincide. (Very simle problem; it works because this is the lowest simple subdivision which remains integral).
[/hide]
Too many hard problems make you look at all problems like that  :P I prescibe you to less AIME.. use as a controlled substnace

It is definitely $7.5$ seconds.  One is tapping every $1.5$ seconds, the other every $1.875$ seconds ($\frac{3}{2}$ and $\frac{15}{8}$).  $1.875$ goes into $7.5$ 4 times, and $1.5$ goes into $7.5$ 5 times, and this is the lowest time they coincide by multiplying them both by an integer, hence, the lowest interval the taps actually coincide.

I am looking for the most creative/ingenious solution to a problem, PM me with the question/answer if you think your problem is 'game'.


geometry
circumcircle
parallelogram
geometry unsolved

Two chords $ AC$ and $ BD$ of a circle with center $ M$ intersect at $ P$. The circumcenters of the triangles $ PBC$ and $ PDA$ have their centers in $ E$ and $ F$, respectively, and intersect for a second time in $ Q$. Prove that we have $ MF \equal{} QE$.

Easy to show that PE is perp to AD, PF is perp to BC, so MEPF is a parallelogram, thus MF=EP=QE.

ME⊥BC  ∠BPX=∠FPD=90-∠PAD=90-∠PBC
=> PF⊥BC    =>  FP∥ME
Analogously  FM∥EP
=>  MFPE is a parallelogram
=> MF=PE=QE


1. "62xy427" is a seven-digit number divisible by 99.
Find 950x + 24y + 1

2. "abc", "def" are 3-digit numbers. abc + def is divisible by 37.
Is the 6-digit number "abcdef" divisible by 37? Explain why or why not.

3. N is a 1990-digit number divisible by 9.The sum of its digits is N_1. The sum of the digits in N_1 is N_2. The sum of the digits in N_2 is N_3. Find N_3.

quotes surround one number... for example

"abc" where a = 2, b = 4, c = 6  is 246.

I hope this is more understandable.

[quote="Chinaboy"]2. abc, def are 3-digit numbers. abc + def is divisible by 37.
Is the 6-digit number abcdef divisible by 37? Explain why or why not.[/quote]
I'm not answering the question... but by "the 6-digit number" do you mean abcdef?

[hide]1. using rule of divisiblility of 11 and 9 we get only possibily for x and y is x=4 and y=4 so the answer is 3800+96+1=3897\

2. abcdef= abc*1000+def=abc+def +999*abc, both def+abc and 999*abc are divisible by 37 so abcdef is too.

3. there are quite a few answers, N could be 9....999 or 8......888 in one case N_3 is 9 in the other 8.[/hide]

On #1 you missolved for one variable, and on you've got #3 wrong because I mistated the question... hehehe... Sorry. I'll edit it.

1)[hide]x=2 y=4 so answer is 1997[/hide]

yup, that's right... it would be nice if you added solutions so we can see different ways.

3) [hide]well let us say N is 999....999, then the sum is 1990*9, the sum of those is 18 and the sum of that is 9. that can be proven because the biggest numbe N could be is 1990*9 which equals 17910, so the sum of digit of N_1 is less than 91, which means the sum of digit of N_2 is 9, because any number divisble by 9 is only divisble by 9 if the sum of digits is and that cycle can keep giong until it reaches 9. since our N_2 is less than 91 and is divisble by 9, the sum of digits is 9. so N_3 is 9.[/hide]

you're close... but not exactly right... ur on the right track on finding the maximum for each N


If $x$ and $y$ are integers and $xy+7x+5y=55$, find all pairs of $x$ and $y$ that work.

Generalize: If $x$ and $y$ are integers and $xy+ax+by=c$, find all pairs of $x$ and $y$ that work.

[hide="Hint"] $(x+n)(y+m) = xy+ny+mx+nm$ [/hide]

[hide]$xy+7x+5y=55$
$(x+5)(y+7)=90=2\cdot3^{2}\cdot5$
$x+5=\pm1,\pm2,\pm3,\pm5,\pm6,\pm9,\pm10,\pm15,\pm18,\pm30,\pm45,\pm90$
$y+7=\pm90,\pm45,\pm30,\pm18,\pm15,\pm10,\pm9,\pm6,\pm5,\pm3,\pm2,\pm1$
$x=\text{something something...}$
$y=\text{something something...}$
We can solve $ax+by+dxy=c$ in a similar way using SFFT.
I don't know if this is the best way...it would be pretty bruteforce-like if we get a lot of divisors...
(edit:added negative sol. too....I can't believe I actually went back and added the \pm's... :ninja: )[/hide]

You omitted the negative solutions ;)

The general solution depends on the factors of the final RHS, so it doesn't exactly have a closed form.


[quote="maxal"]Your answer is not integer when $ n$ is not a power of $ 2$[/quote]
 Oke you can give for me a answer ! Thanks you very much !
 I will edit answer  it last  ....  :blush:


am auzit`o azi si mi s`a parut draguta
fie $ABC$ un triunghi, $D$ punctul de tangenta al cercului inscris $\omega$ cu $BC$ si $S$ a doua intersectie cu cercul a dreptei $AB$. dreptele $BS$ si $CS$ intersecteaza cercul a doua oara in punctele $K$ si $L$. sa se arate ca dreptele $CK$ si $BL$ se intersecteaza pe $AD$.

s-a dat prin iran parca, vezi aici [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=2619[/url], problemuta se mai poate extinde de exemplu : fie $M,N$ intersectiile lui $AK$ si $AL$ a doua oara cu cercu. atunci dreptele $EM$ si $FN$ se taie pe $AD$. unde $E,F$ sunt punctele de tangenta ale inscrisului cu $AB,AC$.

[quote][color=darkblue]Fie $ABC$ un triunghi, $D$ punctul de tangenta al cercului inscris $\omega$ cu $BC$ si $S$ a doua intersectie cu cercul a dreptei $AD$. Dreptele $BS$ si $CS$ intersecteaza cercul a doua oara in punctele $K$ si $L$. Sa se arate ca
1. Dreptele $CK$, $BL$ si $AD$ sunt concurente.[/color][/quote]
[color=darkred][u][b]NOU[/b][/u] : [b]2. Patrulaterul $SKDL$ este armonic, adica $SK\cdot DL=SL\cdot DK$,
adica dreptele $SS$, $BC$, $KL$ sunt concurente, adica dreptele $AD$, $KK$, $LL$ sunt concurente.[/b][/color]

[quote="Virgil Nicula"][color=darkred][u][b]NOU[/b][/u] : [b]2. Patrulaterul $SKDL$ este armonic, adica $SK\cdot DL=SL\cdot DK$.[/b][/color][/quote]

Se foloseste acest fapt in doua din solutiile din link.

mai mult, daca numim $E,F$ punctele de tangenta cu $AB,AC$, atunci hexagonul $SELDKF$ e un hexagon din acela "minune", inscriptibil si cu diag.concurente deci are loc relatia $SE \cdot LD \cdot KF = EL \cdot DK \cdot FS$, deci se putea da si asta ca un punct $3$.

Si acum, in urma ultimei observatii a lui Pohoatza, aceasta "dulce" problema "s-a topit", adica a fost "distrusa". Intr-adevar, ramane o problema draguta !


function
topology
advanced fields
advanced fields unsolved

Let $X$ be a $\sigma$-compact metric space. Prove that there is no separately continuous function $f: X\times X\to I=[0,1]$ such that for all continuous $g: X\to I$, we can find $y\in X$ with $g(x)=f(x,y),\ \forall x\in X$.

The problem it generalizes is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=others+amm&t=46361]this one[/url] (both parts, after a slight modification of the first question in that topic).

Another possible generalization: show that for [b]any[/b] metric space there is no such [b]continuous[/b] function $f$. The difference with grobber's version is that the restriction on $f$ is now much stronger but the $\sigma$-compactness condition is no longer there. I still wonder whether we can relax continuity to separate continuity and remove the $\sigma$-compactness condition simultaneously :maybe:.


In how many ways can seven stuffed animals are distributed among four children.

Write a generating function using power series for this problem.
Write the power series as a negative power of a binomial
Use the formula for C(-n,k) to find the coefficient of the power seires that solves the problem

[hide="Solution"]Each child contributes $ (1\plus{}x\plus{}\cdots\plus{}x^{7})$, so the generating function is $ (1\plus{}x\plus{}\cdots\plus{}x^{7})^4$

Not sure what you mean by part 2.

We want the coefficient of $ x^7$ in the expansion of $ (1\plus{}x\plus{}\cdots\plus{}x^{7})^4$. This is pretty annoying to evaluate... I guess sum over the partitions of $ 7$ into 3 integers

Or you can be clever and use balls and urns, which gives $ \binom{7\plus{}3}{3}\equal{}\boxed{120}$[/hide]

[hide]First note that $ \binom{ \minus{} n}{k} \equal{} \frac {( \minus{} n)( \minus{} n \minus{} 1)( \minus{} n \minus{} 2)\cdots( \minus{} n \minus{} k \plus{} 1)}{k!}$
$ \equal{} \frac {( \minus{} 1)^k(n)(n \plus{} 1)(n \plus{} 2)\cdots(n \plus{} k \minus{} 1)}{k!} \equal{} ( \minus{} 1)^k \binom{n \plus{} k \minus{} 1}{k}$

Then notice the relationship of this formula to the original problem. The original problem (assuming the stuffed animals are identical) is a classic "ball-in-boxes" problem. If there are $ k$ (identical) stuffed animals and $ n$ children, then the number of ways to distribute the stuffed animals among the $ n$ children is $ \binom{n \plus{} k \minus{} 1}{k}$.

[hide]So the generating function for this counting problem is $ \sum_{k \equal{} 0}^\infty \binom{n \plus{} k \minus{} 1}{k} x^k \equal{} \sum_{k \equal{} 0}^\infty ( \minus{} 1)^k \binom{n \plus{} k \minus{} 1}{k} ( \minus{} x)^k \equal{} (1 \minus{} x)^{ \minus{} n}$

Note how this idea is related to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1487045#1487045]this problem[/url].[/hide][/hide]

But how can you solve the problem like:

Find the coefficient of x^n in each of the following generating funcitons
F = 2(1-x)^-4 + 3(1-x)^-5 term n=6

With the [url=http://en.wikipedia.org/wiki/Binomial_Theorem#Newton.27s_generalized_binomial_theorem]Extended Binomial Theorem[/url]


calculus
derivative
real analysis
real analysis unsolved

Hello, I have an exercise I find it is false ... I don't know where I'm wrong

I have to show that $|th(x)-th(y)|$ and $|x-y|$ are equivalent distance, i.e. it exists $C_1, C_2$ such that
\[C_1 |x-y|\leq |th(x)-th(y)|\leq C_2 |x-y|\]
I find that we can find such a $C_2$ ($C_2=1$) but not $C_1$ ...

Maybe you can tell us what $th$ means first? :?

I think it's wrong too. The existance $C_1 > 0$ such that for all $x, y$ we have $\frac{|th x - th y|}{|x-y|} \geq C_1$ contradicts with the fact that the derivative of $th$ tends to $0$ when $x$ tends to infinity.


geometry
geometric transformation
rotation

Spinners prefer the old ball bcos they can grip it better  :huh:   :P   :rotfl:

[quote="Cyber-Shreyas"]Good yes, then can't fastbowlers grip the old ball better as well?  :P[/quote]

They want to swing the ball while spinners want to spin  :huh:   :P
Im a cricketer before a physicist  :D

Give a proper explanation using the basic concepts of physics.

Seamers want air drift so they bowl with the new ball seam while spinners want to spin the ball n hence they require better grip for rotation so they like an older n rougher ball  :maybe:

1) Why does an old ball offer better grip?
2) Whats the problem if fast bowlers have better grip as well?

Your answer is missing one key point on the significance of the old ball.

1. this is because the surface of the ball gets rough and then it  offers a better grip./....
2. For fast bowlewrs , their main aim is to bowl as fast as possible and they want easy release  of the ball which offers them spped as well as swing..

You have posted the same thing as rohit??? Your argument is wrong. Looking at the angle of fast bowlers yes they do want to bowl fast but they want to swing as well.... And from the same argument then spinners want to bowl slow???

The point is that an old ball with a rough surface has a high FRICTION coefficient. ( None of you used friction). When it lands on the ground, high frictional torque acts on it providing more spin.


geometry circumcircle trigonometry india

Let $ABC$ be an acute-angled triangle and $\Gamma$ be a circle with $AB$ as diameter intersecting $BC$ and $CA$ at $F ( \not= B)$ and $E (\not= A)$ respectively. Tangents are drawn at $E$ and $F$ to $\Gamma$ intersect at $P$. Show that the  ratio of the circumcentre of triangle $ABC$ to that if $EFP$ is a rational number.

I have tried to search for this problem, since I am sure it already appeared on this forum, but it turned out easier to solve it again.

The core of the problem is the following "theorem":

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle and X, Y, Z the feet of its altitudes from the vertices A, B, C, respectively. Also, let H be the point of intersection of these altitudes, i. e. the orthocenter of triangle ABC. Since < AXB = 90° and < AYB = 90°, the points X and Y lie on the circle with diameter AB. Then, the tangents to this circle at the points X and Y both pass through the midpoint R of the segment CH.[/color]

[i]Proof of Theorem 1.[/i] The center of the circle with diameter AB is the midpoint M of the segment AB.

It is well-known that the feet X, Y, Z of the altitudes of triangle ABC, the midpoint M of its side AB and the midpoint R of the segment CH all lie on the nine-point circle of triangle ABC. Since < RZM = 90°, the segment RM is a diameter of this nine-point circle. Hence, we also have < RYM = 90°, so that $RY\perp MY$. Now, since M is the center of the circle with diameter AB, and Y is a point on this circle, the relation $RY\perp MY$ shows that the line RY is the tangent to the circle with diameter AB at the point Y (since the tangent to a circle at a point is perpendicular to the radius through this point). Hence, the point R lies on the tangent to the circle with diameter AB at the point Y. Similarly, the point R lies on the tangent to the circle with diameter AB at the point X. Hence, Theorem 1 is proven.

Now, your points E and F are the points Y and X from Theorem 1, and your point P is the point of intersection of the tangents to the circle with diameter AB at the points X and Y, i. e. the midpoint R of the segment CH (after Theorem 1). Hence, the problem requires to prove that the ratio between the circumradii of triangles ABC and YXR is a rational number. But this is trivial: Since the points Y, X and R lie on the nine-point circle of triangle ABC, the circumcircle of triangle YXR is the nine-point circle of triangle ABC, and thus its radius is $\dfrac12$ of the circumradius of triangle ABC (this is a trivial property of the nine-point circle); in other words, the ratio between the circumradii of triangles ABC and YXR is 2, and since 2 is a rational number, the problem is solved.

  darij

Since $\angle OFP=\angle OEP=90^o$ the quadrilateral $EOFP$ is cyclic. And it's circumcenter have the diameter $PO$. It's easy to see by angle chasing, and using the cyclic quadrilaterals $OEFP,BCEF$, (Where $O$ is the midpoint of $BC$) that $\angle OPE=\angle A$. Hence, $2R\sin A= BC=2 OE=\frac 1 2 OP \sin A=r\sin A$ ,where $R$is the circumradius of $ABC$, and $r$ the one of the $EFP$. Hence the ratio of the two radius is $\frac 1 2$.

Lol, this is probably the trolliest problem ever posed in IMOTC (yeah, I know, I just made that word up).

[hide]
Let $M$ be the midpoint of $AB.$ Observe that $M$ is the center of $(BCEF),$ so $ME \perp CP$ and $MF \perp BP,$ so $MEFP$ is cyclic. Let $R$ be the circumradius of $\triangle ABC.$ By sine rule on $\triangle DEF,$ where $D$ is the foot of perpendicular on $A,$ $\dfrac{EF}{\sin (\pi - 2C)} = R \implies EF = R \sin 2C$ (since the radius of the nine point circle is half the circumradius). By sine rule on $\triangle MEF,$ the circumradius of $\triangle MEF$ is
\[\dfrac{R \sin 2C}{2\angle MEF}  = \dfrac{R}{2} \dfrac{\sin 2C}{\sin 2 \angle EBC} = \dfrac{R}{2} \dfrac{\sin 2C}{2 \sin \angle EBC \cos \angle BCE} = \dfrac{R}{2},\]
so the ratio is $\dfrac{1}{2},$ which is clearly a rational number.
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Dear Mathlinkers,
with harmonic division, we have the nature of the point of intersection of the tangent and the C-altitude... the C-Euler's point...
Very nice problem which permit us to consider the Euler's circle of ABC..
Sincerely
Jean-Louis


Nice problem :) . I think you mean:

If $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=1$, then prove that \[ \sum\frac{1}{a^{3}+b^{3}+abc}\leq\frac{1}{a+b+c}.\]

From a well-known inequality, we have \[\sum\frac{1}{a^{3}+b^{3}+abc}\leq\frac{1}{abc}\], therefore we only need to prove that \[\frac{1}{a+b+c}\geq\frac{1}{abc}\] \[\iff abc\geq a+b+c\] \[\iff abc(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})\geq a+b+c\] \[\iff a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\geq abc(a+b+c).\]But the last one is an aplication of the famous $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$, for the numbers $ab, bc$ and $ca$.
Hence the problem is solved!

What happened? Did I have a mistake in my proof, or what? :?:


graph theory
combinatorics proposed
combinatorics

At a party there are eight guests, and each participant can't talk with at most three persons. Prove that we can group the persons in four pairs such that in every pair a conversation can take place.

[quote="pbornsztein"]Hmm.. I think there is a straightforward proof but is it not a direct consequence of Hall's theorem?

Pierre.[/quote]

Hall's theorem only applies to bipartite graphs.

Groumpf.. :blush:

Well, the graph whose vertices are the persons and any edge connect two of them if and only if they can have a conversation has an hamiltonian cycle, because each vertex has degree at least $4$, so that it satisfies Dirac's condition.
To find the desired pairing, you only have to cut the hamiltonian cycle into appropriate pairs.

Pierre.

Here's another (clumsier) way:

Suppose that no such grouping exists; that is, every grouping has at least one pair of people who can't talk to each other. We show that at least one person can't talk to more than 3 people. Call a pair of people "forbidden" if they can't talk to each other. There are $\frac{{{8}\choose{2}}{{6}\choose{2}}{{4}\choose{2}}}{4!}= 105$ possible groupings. Label the people from 1 to 8. WLOG assume that 1 and 2 can't talk to each other. There are $\frac{{{6}\choose{2}}{{4}\choose{2}}}{3!}=15$ groupings that contain the pair (1 2). Similarly, there 15 - 3 = 12 additional groupings that contain the pair (3 4), 10 additional groupings containing the pair (5 6), and 8 additional groupings that contain the pair (7 8). We continue the process of constructing "forbidden" pairs until every person can't talk with 3 other people. Eventually we find that there are 15+12+10+8+10+7+5+3+5+5+5+5+5+5 = 100 groupings containing these pairs (just some counting). But there are 105 groupings in total, so there is at least one more "forbidden" pair. Contradiction.


how to show that any numbers of the form n^4+4^n are not prime?

[quote="itazzirun"]how to show that any numbers of the form $n^{4}+4^{n}$ are not prime?[/quote]

[hide]For n=even, the number is even and is not prime.
if n=odd, let $n=2k+1$.
$n^{4}+4\cdot4^{n-1}=n^{4}+4\cdot4^{2k}=n^{4}+4\cdot{2^{k}}^{4}$
let $2^{k}=a$
then
$n^{4}+4a^{4}=n^{4}+4a^{4}+4n^{2}a^{2}-4n^{2}a^{2}=(n^{2}+2a^{2})^{2}-(2an)^{2}$
$=(n^{2}+2a^{2}+2an)(n^{2}+2a^{2}-2an)$
and the right term is equal to 1 for n=1 and is prime.
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