linear algebra
matrix
algebra
polynomial
vector
group theory
abstract algebra

Given a cyclic matrix (characteristic polynomial=minimal one) over a finite field K, with determinant 1, how many matrices with determinant 1 commute with this matrix? Suppose that the matrix is $2\times 2$. I have no idea how to solve it for larger dimensions.

Let $A$ be any such linear operator (acting on an $n$-dimensional vector space $V$ over $K$), and choose a basis $v_{0},\dots,v_{n-1}$ such that $Av_{i}=v_{i+1}$ for $0\le i\le n-2$. Let $p$ be the minimal/characteristic polynomial of $A$.
$A$ also acts on the vector space $W=K[x]/(p(x))$ by multiplication by $x$. We have a natural isomorphism from $V$ to $W$, given by $v_{i}\mapsto x^{i}$; this isomorphism commutes with the action of $A$.

Let $B$ be a linear map from $W$ to itself, and suppose $AB=BA$. Looking at the basis vectors, $B(x^{k})=B(A^{k}(1))=A^{k}(B(1))=x^{k}\cdot B(1)$, and the action of $B$ is multiplication by some polynomial $q=B(1)$ of degree less than $n$. This makes $B=q(A)$, and there are exactly $|K|^{n}$ such polynomials.

We didn't need to assume finiteness of the field; the same argument shows that the commutant is an $n$-dimensional vector space over any field.

Sorry, Jmerry, I don't understand a point: you have shown that the commutant of a cyclic matrix consists of the polynomials in that matrix. But I ask for the number of those elements of the commutant that have determinant 1.

I see. That's a different question, and it feels very strange.

Even for $2\times 2$ the problem is not completely obvious. For larger dimensions, there is nothing intelligent I can think of. :blush:

Here's what I've got:
Factor $p$ completely as $p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$. The polynomials in $A$ with determinant zero are exactly the multiples of $p_{i}$ for some $i$ (the zero divisors in the polynomial ring). For each $i$, there are $|K|^{n-\deg p_{i}}$ of these. Let $d_{i}=\deg p_{i}$. Since the $p_{i}$ are relatively prime, divisibility by the $p_{i}$ is independent and there are exactly $|K|^{n}\cdot \prod_{i=1}^{k}1-\frac1{|K|^{d_{i}}}$ polynomials with nonzero determinant.
Let $G$ be the group of invertible polynomials mod $p$; $G$ has $|K|^{n}\cdot \prod_{i=1}^{k}1-\frac1{|K|^{d_{i}}}$ members as above. The determinant is a homomorphism $\delta$ from $G$ to $K^{*}$. The number of elements with determinant 1 is precisely $\frac{|K|^{n}}{|\delta(G)|}\cdot \prod_{i=1}^{k}1-\frac1{|K|^{d_{i}}}$. Since $\delta$ is usually surjective, this is usually $\frac{|K|^{n}}{|K|-1}\cdot \prod_{i=1}^{k}1-\frac1{|K|^{d_{i}}}$.

An example (with $n=2$) in which $\delta$ is not surjective. Let $K$ be any finite field not of characteristic 2, and let $p(x)=(x-c)^{2}$. If $x\neq 0$, $\det(xI+yA)=x^{2}p(-\frac{y}{x})=(y-cx)^{2}$. The squares are a proper subgroup of $K^{*}$, and all nonzero determinants lie in this subgroup.
In all other cases with $n=2$, $\delta$ is surjective. Squares are always covered, so $\delta$ is surjective whenever $K$ has characteristic 2. If $p$ is irreducible, we cannot fit $|K|$ values onto $\frac{|K|-1}{2}$ quadratic residues without pilin up three and making $p$ constant. If $p(x)=(x-a)(x-b)$, we can rewrite this as $p(x)=(x-c)^{2}-d^{2}$. Given some nonsquare $t$, we have $t=t\cdot 1=\left(\frac{t+1}2\right)^{2}-\left(\frac{t-1}2\right)^{2}=r^{2}-s^{2}$, and $t=\left(\frac{s}{d}\right)^{2}\cdot p\left(\frac{dr}{s}+c\right)$; $p$ takes a nonsquare value.

There's nothing gained by forcing the determinant of $A$ to be 1; that just restricts the examples.


calculus
derivative
function
real analysis
real analysis unsolved

Suppose that $f$ is a function from $R^2$ to $R$ which has partial derivatives of any order at any variable. Is it continuous?  ;)

Is the intent of "partial derivatives of any order at any variable" is to restrict our attention to just $\frac{\partial^kf}{\partial x^k}$ and $\frac{\partial^kf}{\partial y^k}?$ If so we have the well-known example

$f(x,y)=\begin{cases}\frac{xy}{x^2+y^2}&(x,y)\ne(0,0)\\0&(x,y)=(0,0)\end{cases}.$

For each fixed $x$, this is $C^{\infty}$ as a function of $y,$ and vice versa.

On the other hand, if the partial derivatives you meant included the mixed partial derivatives, then I don't know the answer.

For the particular function above, $\frac{\partial^2f}{\partial x\partial y}$ does not exist at the origin.

We just need to apply this trick to something stronger- how about
$f(x,y)=e^{-\frac{x^2}{y^2}}\cdot e^{-\frac{y^2}{x^2}}$, extended to be 0 whenever $x=0$ or $y=0$. This isn't continuous at the origin, since along the line $x=y$ it has value $e^{-2}$.

It's obvious that $f$ is differentiable unless either $x=0$ or $y=0$. At the point $(0,y)$ (WLOG $x=0$), the term $e^{-\frac{x^2}{y^2}}$ is differentiable. The term $e^{-\frac{y^2}{x^2}}$ has partial derivatives $\frac{\partial^n}{\partial x^n}e^{-\frac{y^2}{x^2}}=r_n(x,y)e^{-\frac{y^2}{x^2}}$ where $r_n$ is rational, which tend to zero when $x=0$. Thus all of these derivatives are zero along that axis. Taking another partial derivative with respect to $y$ clearly makes the whole thing zero, and all partial derivatives of $e^{-\frac{y^2}{x^2}}$ exist and are zero. Multiplying by the other term and applying the product rule (as many times as necessary), all partial derivatives of $f$ are zero on the $y$-axis. In fact, $f$ is $C^\infty$ except at the origin.
At the origin, note that we've just calculated the partial derivatives on the axes and found that they're zero. This makes the partial derivatives at the origin zero.


If your salary is reduced by $ 25\%$, by what percent would it then have to be increased to bring it back to the original amount? Express your answer as a mixed number.

We have $ .75*x\equal{}1$, so $ x\equal{}1\frac{1}{3}$. Thus our answer is $ \boxed{33 \frac{1}{3}}$.

Be more clear. What is $ x$???

Let the original salary be $ a$ Then your salary after the decrease is $ \frac{3}{4}$ We need our salary to be increased to $ a$ or to be increased by $ \frac{4\minus{}3}{3}\equal{}\frac{1}{3}\equal{}33\frac{1}{3}\%$


i wanna understand maths more exact and better from pre-calc. if you know a source( i meant book: which explains a topic well and mostly easy to self-study) or problem book good for AIME and AMC.

noone knows a book (noone knows or know?)


logarithms
function
real analysis
real analysis unsolved

$ f(x)$ is continuous in $ [a,b]$ and twice differentiable in $ (a, b)$.
If $ f(a) \equal{} f(b), f'(a) \equal{} f'(b)$
prove that exist $ c \in (a, b)$ such that $ f''(c) \minus{} \lambda (f'(c))^{2} \equal{} 0$

[quote="Heebeen, Yang"]
If $ f(a) \equal{} f(b), f'(a) \equal{} f'(b)$
[/quote]

What does 'if' mean?

For the point $ c_1$ at which $ f'(c_1) \equal{} 0$  (this exists by mean value theorem, there is at least one such point), if $ f''(c_1) \equal{} 0$ as well, then we are done.
If that is not the case, then $ c_1$ is not our desired point, in the subsequent discussion we exclude the point that $ f'(x) \equal{} 0$.

Rewrite the equation as $ \frac {f''(c)}{f'(c)} \minus{} \lambda f'(c) \equal{} 0$, it is reminscent of the equation $ \Big.\left(\log f'(x) \minus{} \lambda f(x)\right)'\Big|_{x \equal{} c} \equal{} 0$.
Define $ g(x) \equal{} \log f'(x) \minus{} \lambda f(x)$, by the given condition we know that $ g(a) \equal{} g(b)$,
hence there exists, by mean value theorem, a $ c\in (a,b)$ such that $ g'(c) \equal{} \frac {g(b) \minus{} g(a)}{b \minus{} a} \equal{} 0$,
thus we are done.

But it is probably wrong as $ f'(x)$ can be negative, and... I guess there is such mean value theorem for complex function  :( .


algebra
polynomial
abstract algebra
function
Ring Theory
binomial coefficients
superior algebra

Let $ R\equal{}k[x_1,x_2,...x_n]$ be the graded polynomial ring over a field $ k$. Let $ f\in R$ be a nonzero   homogeneous polynomial of degree $ d>0$ and $ M$ be the graded $ R$ module $ R/(f)$. Compute the hilbert function of $ M$ and the corresponding hilbert polynomial $ P_{Q,M}$ where $ Q\equal{}(x_1,x_2...x_n)$.

So basically my idea right now is to just use the fact that we can expand the hilbert polynomial interms of a sum of binomial coefficients and then try to find enough values for the hilbert function to figure out what the coefficients are. Is this really how to do it or is there a faster way?

Try to consider the short exact sequence of graduated modules \[ 0\to R\stackrel{l_f}\to R(\minus{}d)\to M\to 0\] where $ l_f$ denotes multiplication with $ f$ and $ d\equal{}\deg(f)$. How do the corresponding Hilbert polynomials behave here? Note the shift in degrees to set up a relation between the Hilbert polynomials.

The basic idea is to find exact sequences with $ M$ and in which you know more about the Hilbert polynomials of the other modules involved.


Specifically in Illinois.  I only know about the AMC and its extensions through my school, and frankly I'm too lazy to actually look some up.  So what else is there that I can do, that I could just register for not through the school?

Search for similar threads on the forum or look in the Resources section of the website; your equation is a common one and has been answered before.

I have put together a fairly comprehensive [url=http://goodlymaths.0catch.com/Contests/directory.htm]list[/url] at my growing math contest resources site.  Hopefully this will be of use to you.

On another note, if anyone knows of any other math contests (it doesn't matter if they're state-wide, national, international, etc.), would you send them to me?  Thanks.


Prove that for every complex number $ z$ it happens that:

$ |z \plus{} 10| \plus{} |z \plus{} 11| \plus{} |z \plus{} 19| \le|z \plus{} 8| \plus{} |z \plus{} 12| \plus{} |z \plus{} 20|$

[quote="tchebytchev"]$ |z \plus{} 10| \plus{} |z \plus{} 11| \plus{} |z \plus{} 19| \minus{} |z \plus{} 8| \minus{} |z \plus{} 12| \minus{} |z \plus{} 20| \equal{} 10^2 \plus{} 11^2 \plus{} 19^2 \minus{} 8^2 \minus{} 12^2 \minus{} 20^2$ ,from $ |z \plus{} a| \equal{} |z|^2 \plus{} 2aRE(z) \plus{} a^2$.
[/quote]

Are you sure that $ |z \plus{} a| \equal{} |z|^2 \plus{} 2aRE(z) \plus{} a^2$?  If $ z \equal{} x \plus{} iy$, then
$ |z|^2 \plus{} 2aRE(z) \plus{} a^2 \equal{} x^2 \plus{} y^2 \plus{} 2ax \plus{} a^2 \equal{} (x\plus{}a)^2 \plus{} y^2 \equal{} |z \plus{} a|^2$, but not $ |z \plus{} a|$...

ok , bacause that i think that i see in your original problem an expression as $ |z\plus{}a|^2$.. :maybe:

Let $ z\equal{}a\plus{}bi$ then $ |z \plus{} 10| \plus{} |z \plus{} 11| \plus{} |z \plus{} 19| \le|z \plus{} 8| \plus{} |z \plus{} 12| \plus{} |z \plus{} 20|$ happens iff $ \sqrt{(a\plus{}10)^2\plus{}b^2}\plus{}\sqrt{(a\plus{}11)^2\plus{}b^2}\plus{}\sqrt{(a\plus{}19)^2\plus{}b^2} \leq \sqrt{(a\plus{}8)^2\plus{}b^2}\plus{}\sqrt{(a\plus{}12)^2\plus{}b^2}\plus{}\sqrt{(a\plus{}20)^2\plus{}b^2}$ consider the function $ f(x)\equal{}\sqrt{(a\plus{}x)^2\plus{}b^2}$, $ f''\equal{}\frac{b^2}{(b^2 \plus{} (a \plus{} x)^2)^{\frac{3}{2}}}$ this means $ f(x)$ is convex. Note that $ {19,11,10} < {20,12,8}$. Applying Karamata, we obtain the result.


geometry Asymptote trigonometry MATHCOUNTS

Isosceles triangle ABC with a vertex angle (<A) of 30 degrees is inscribed in square DEFG of side length 4. A coincides with D such that segments AB and AC trisect the corner. Square HIJK is then circumscribed around square DEFG such that angle the large square makes with the smaller square is 45 degrees. What percentage of the area of the larger square is in the triangle? Express your answer in simplest radical form as a common fraction.

[hide]Another Asymptote abuse!

In the figure below, we see that we can find the area of the shaded region by the formula $ A \equal{} \frac 12 ab\ \sin{C}$, and we already know that $ m\angle C \equal{} 30^\circ$, so $ \sin C \equal{} \frac 12$ by a well-known identity. Thus all we have left is to find $ ab$, and since $ a \equal{} b$, we just have to find $ a^2$. ($ a$ and $ b$ are the congruent sides of $ \triangle ABC$.) So the formula is then $ \frac 14 a^2$.

We immediately see that $ \triangle AGB$ is a 30-60-90 triangle, and we know that $ AG$, the longer leg, is $ 4$. Thus $ AB$ is $ \frac 4{\sqrt {3}} \cdot 2 \equal{} \frac 8{\sqrt {3}}$. We won't bother to simplify this because we're squaring this anyway. $ a^2 \equal{} \frac {64}3$. Thus $ \frac 14 a^2 \equal{} \frac {16}3$.

The area of the largest square in the picture below is twice the area of the square with side length 4, or $ 2 \cdot 16 \equal{} 32$.

Thus the fraction of the large square covered by the triangle is $ \frac {\frac {16}3}{32} \equal{} \boxed{\frac 16}$.

[asy]size(0, 150);
draw(scale(4) * unitsquare);
filldraw((0, 4) -- (4 * sqrt(3) / 3, 0) -- (4, 4 - 4 * sqrt(3) / 3) -- cycle, lightgrey);
draw((-2, 2) -- (2, 6) -- (6, 2) -- (2, -2) -- cycle);
label('A', (0, 4), NW);
label('B', (4 * sqrt(3) / 3, 0), S);
label('C', (4, 4 - 4 * sqrt(3) / 3), E);
label('G', (0, 0), SW);[/asy][/hide]

[asy]size(0, 150);
draw(scale(4) * unitsquare);
filldraw((0, 4) -- (4 * sqrt(3) / 3, 0) -- (4, 4 - 4 * sqrt(3) / 3) -- cycle, lightgrey);
draw((-2, 2) -- (2, 6) -- (6, 2) -- (2, -2) -- cycle);
label('A', (0, 4), NW);
label('B', (4 * sqrt(3) / 3, 0), S);
label('C', (4, 4 - 4 * sqrt(3) / 3), E);
label('G', (0, 0), SW);[/asy]
^ Thanks Yongyi for the picture :D I couldn't've made it myself. Oh yeah, your answer of 1.66666666...% doesn't make sense o.O Sorry.

How about mathcounts level?  :maybe:

Let's just assume :D the side length of the smaller square is 1. The area of the larger square is 2. Because angle BAG is 30 and GA is 1, BG is $ \frac {\sqrt3}{3}$. The area of triangle BAG is $ \frac {\sqrt3}{6}$. The smaller triangle inside the smaller square is $ \frac {1}{2}\cdot (1 \minus{} \frac {\sqrt3}{3})^2 \equal{} \frac {1}{6}$ The area of ABC is $ 1 \minus{} 2(\frac {\sqrt3}{6}) \minus{} \frac {1}{6} \equal{} \frac {5 \minus{} 2\sqrt3}{2}$ So the percentage is $ \frac {125 \minus{} 50\sqrt3}{3}$ %.

First of all, whoever told you that $ \frac 16$ was $ 1.6666666\%$ should explode. It's actually $ 16.66666\%$!

Next, you made an error here:

[quote="l Bob l"]
The smaller triangle inside the smaller square is $ \frac {1}{2}\cdot (1 \minus{} \frac {\sqrt3}{3})^2 \equal{} \frac {1}{6}$
[/quote]

Oops!  :blush: So the answer is $ \frac {50}{3}$%. (and not $ \frac {1}{6}$   :wink: )

This shows how bad I am at reading. At least I have an excuse. Do you? :P


limit
real analysis
real analysis unsolved

let {an} be increasing sequence of positive reals , suppose  \lim (n->00)an=00 show that  \sum arccos(an/a(n+1))=00

Are you sure about the problem? I think that there is something wrong.
Take an=(n!)2. Then \sum an/a(n+1)= \sum 1/(n+1)2 which is not 00

Heh, when a sequence diverges, we get some wacky stuff  :D

Just an application of  arccosx  \geq  1-x for x \in [0,1]


[quote="eLeM"][quote="AnLon"]Let a,b,c denote the lengths of the three sides oppisite to the three angles A,B,C of a triangle.
Show That:
$\frac{\pi}{3}\leq\frac{aA+bB+cC}{a+b+c}\leq\frac{\pi}{2}$

Any hints? :roll:[/quote]

Let $a \geq b \geq c$ so $A \geq B \geq C$. By Chebyshev inequality we have:

$\frac{aA+bB+cC}{a+b+c}\geq \frac{(\frac{a+b+c}{3})(A+B+C)}{a+b+c}=\frac{A+B+C}{3}=\frac{\pi}{3}$

Second inequality:

$2aA+2bB+2cC\leq\pi(a+b+c)$
$2aA+2bB+2cC\leq aA+bA+cA+aB+bB+cB+aC+bC+cC$
$A(b+c-a)+B(c+a-b)+C(a+b-c)\geq0$ which is true for $a,b,c$ sides of a triangle.[/quote]

i still got a question, is it only true for the condition$a \geq b \geq c$ and $A \geq B \geq C$ ?
and is it possible to solve the first inequality like the second inequality?
because i haven't learnt Chebyshev inequality yet :P

[quote="AnLon"]
i still got a question, is it only true for the condition$a \geq b \geq c$ and $A \geq B \geq C$ ?
and is it possible to solve the first inequality like the second inequality?
because i haven't learnt Chebyshev inequality yet :P[/quote]

Inequality holds for two sets arranged in the same way. If you have $a\geq b$ so also have $A \geq B$, because angle $A$ is opposite to side $a$. Hence we have $a,b,c$ and $A,B,C$ arranged in the same way and we can use Chebyshev.

First inequality can't be solved using $a+b>c$ because both sides of inequality can be equal when $a=b=c$ and $A=B=C=\frac{\pi}{3}$.


A locust, a grasshopper and a cricket are sitting in a long, straight ditch, the locust on the left and the cricket on the right side of the grasshopper. From time to time one of them leaps over one of its neighbours in the ditch. Is it possible that they will be sitting in their original order in the ditch after 1999 jumps?

[hide="My solution"]There are six possible arrangements for the locust, grasshopper, and cricket, as follows (L denotes the locust's position, C denotes the cricket's position, and G denotes the grasshopper's position):
(1) LGC
(2) GLC
(3) GCL
(4) LCG
(5) CLG
(6) CGL
Clearly, if it is possible to go to (a) from (b), it must be possible to go from (b) to (a).  From (1) it is possible to go to (2) and (4); from (2) it is possible to go to (1) and (3); from (3) it is possible to go to (2) and (6); from (4) it is possible to go to (1) and (5); from (5) it is possible to go to (4) and (6); and from (6) it is possible to go to (5) and (1).  This is hard to visualize; the best way to summarize this is with a graph:

(1) (2)
(3) (4)
(5) (6)

(don't know how to do this better; picture the two columns further apart, and lines between the following pairs of numbers: ((1),(2)), ((1),(4)), ((3),(2)), ((3),(6)), ((5),(4)), ((5),(6))).  Since any jump must take us from one side of the graph to the other, any sequence of jumps that ends at our starting point must have an even number of jumps, since it must end up on the same side that it starts out on.  Since we are taking 1999 jumps, and 1999 is an odd number, we cannot both start at (1) and end there after 1999 jumps.[/hide]

[hide]
0) LCG
1) LGC
2) LCG

he just said that that switch with one of their neighbors, the can keep switching with the same one over and over.  This still takes an even number of jumps, and if they jump in a braid fashion it will takes twelve jumps to return to the original position (even)  Thus they can never be in the same position in an odd number of jumps.[/hide]

[hide="Is this worth hiding?"]Such a jump is an odd permutation. The composition of an odd number of odd permutations is an odd permutation. The identity permutation, that is their position remains unchanged, is even. No composition of an odd number of odd permutations give the even identity permutation. Therefore, after any odd number of jumps, it's impossible for the insects to end up at the same place. Specifically for this problem, it's impossible for the insects to end up at their original positions after 1999 jumps.[/hide]


analytic geometry
modular arithmetic
symmetry

find integer numbers x,y such that:
      $ (x^3\plus{}1) \vdots (xy\minus{}1)$

It is easy to prove that $ xy\minus{}1$ divides $ x^2\plus{}y$ and $ x\plus{}y^2$. W.L.O.G., $ x>y$. We have $ x\plus{}y^2$<=$ xy\minus{}1$. Solve this with respect to $ x$ and the rest of the problem is easy. :)

why do you say WLOG x>y?
x and y are not symmetric in the problem I think.

I am not familiar with the notation $ \vdots$, but i remember that it means that $ a(xy \plus{} 1) \equal{} x^3 \plus{} 1$
[hide]
(nasty casework)
All coordinates $ (0,y)$ satisfy the condition.
If $ x \equal{} 1$, then $ a(y \plus{} 1) \equal{} 2 \implies y \plus{} 1 \equal{} \minus{} 2, \minus{} 1, 1, 2 \implies y \equal{} \minus{} 3, \minus{} 2, 0, 1$ for ordered pairs
$ (1, \minus{} 3), (1, \minus{} 2), (1,0), (1,1)$
If $ x \equal{} 2$, then $ a(2y \plus{} 1) \equal{} 9 \implies 2y \plus{} 1 \equal{} \minus{} 9, \minus{} 3, \minus{} 1, 1, 3, 9 \implies y \equal{} \minus{} 5, \minus{} 2, \minus{} 1, 0, 1, 4$ for ordered pairs
$ (2, \minus{} 5), (2, \minus{} 2), (2, \minus{} 1), (2,0), (2,1), (2,4)$
All ordered pairs $ ( \minus{} 1,y)$ satisfy the condition.
With really tedious casework, there are no new ordered pairs for $ x \equal{} \minus{} 2 , \minus{} 3$ other than the ones mentioned soon, and the ordered pairs for $ x \equal{} \minus{} 4$ are $ ( \minus{} 4,16), ( \minus{} 4,2), ( \minus{} 4,1), ( \minus{} 4,0), ( \minus{} 4, \minus{} 2), ( \minus{} 4, \minus{} 5)$

Otherwise, $ a(xy \plus{} 1) \equal{} (x \plus{} 1)\left(x^2 \minus{} x \plus{} 1\right)$
From this factorization, we see that 4 possible values of y are $ y \equal{} 0, 1, x \minus{} 1, x^2$

If there are other possible values of y, then there must be 2 factors of $ x^3 \plus{} 1$ that are 1 mod x. This is not possible if there are not 2 factors of $ x^2 \minus{} x \plus{} 1$ which are 1 mod x that are not 1, so assume that
$ mn \equal{} x^2 \minus{} x \plus{} 1 > 0; m, n \equiv 1 \pmod{x}$
because the product is positive, either m and n are both positive or they are both negative. Assume that they are positive but that they are either 1 or -1 mod x.
Therefore, the possible values of m or n, in strict ascending order, are $ |x| \minus{} 1, |x| \plus{} 1, 2|x| \minus{} 1, 2|x| \plus{} 1, 3|x| \minus{} 1, 3|x| \plus{} 1, \ldots$
By casework, $ m \equiv n \pmod{x}$
Suppose that x is positite. Then $ m \equiv n \equiv \minus{} 1 \pmod{x}$ because otherwise, $ x^2 \minus{} x \plus{} 1mn \ge (x \plus{} 1)(x \plus{} 1) \equal{} x^2 \plus{} 2x \plus{} 1 \implies 0 \ge x$ contradiction
Otherwise, $ n \equal{} \frac {x^2 \minus{} x \plus{} 1}{m} \le \frac {x^2 \minus{} x \plus{} 1}{x \minus{} 1} \equal{} x \plus{} \frac {1}{x \minus{} 1}$
So $ n \equal{} x \minus{} 1$ and by symmetry, $ m \equal{} x \minus{} 1$
But this does not satisfy $ mn \equal{} x^2 \minus{} x \plus{} 1$

So x is negative. Let $ z \equal{} \minus{} x > 1$
Then $ mn \equal{} z^2 \plus{} z \plus{} 1$ and the minimum possible values of m or n, in strict ascending order, are $ z \minus{} 1, z \plus{} 1, 2z \minus{} 1, 2z \plus{} 1, \ldots$
Again, $ m \equiv n \pmod{z}$
If $ m \equiv n \equiv 1 \pmod{z}$
Then $ z^2 \plus{} z \plus{} 1 \equal{} mn \ge (z \plus{} 1)(z \plus{} 1) \equal{} z^2 \plus{} 2z \plus{} 1 \implies 0 \ge z$ contradiction.
Otherwise, $ n \equal{} \frac {z^2 \plus{} z \plus{} 1}{m} \le \frac {z^2 \plus{} z \plus{} 1}{z \minus{} 1} \equal{} z \plus{} 2 \plus{} \frac {3}{z \minus{} 1} \implies n \le z \plus{} 2 \implies n \equal{} z \minus{} 1$
By symmetry, $ m \equal{} z \minus{} 1$
But this does not satisfy $ mn \equal{} z^2 \plus{} z \plus{} 1$

Answer: All the coordinates in the beginning of the solution plus $ (x,0)(x,1)(x,x \minus{} 1)(x,x^2)$ for all other x
[/hide]

my solution is probably too bashy


[quote="carlosbr"]Buenos dias ..
he estado bajando algunos libros via emule...
sin embargo.. como son muy pesadas..
no he podido adjuntarlos por aki...

estoy buscando la manera de compartir estoy archivos ..
y linkearlos desde aki para que los interesados puedan descargarlos sin problemas...

PD: Cualquier sugerencia para compartir archivos de gran tamaño enviar un PM (de 10 Mb hasta 25 Mb)

Carlos Bravo ;)
Lima -Peru[/quote]

¿Carlos no se pueden meter en Win Zip? No se si se pueden poner archivos en este formato aca... pero depronto funciona.

Bueno..
lo he probado ....

y no se ha logrado grandes avances .. [de 25 Mb a 20 o 23Mb]

....
mi pregunta iba por si alguien conoce algun lugar en la net ... donde .. uno puede colgar archivos
de estas dimensiones ....

De todas maneras gracias por el consejo...

Carlos Bravo ;)
Lima - Peru


quadratics
algebra
polynomial
calculus
integration
algebra solved

Does there exist a quadratic polynomial f(x) with integral coefficients with the property that, whenever x is a positve integer which consists only of 1's, f(x) is also a positive integer consisting only of 1's.

Just take f=9X^2+2X. The more interesting problem is to find all such polynomials of degree 2. I have a short and nice solution, but I prefer to let you think about this question.

:? I thought f(x) = 9x + 2x was of degree 2 ? :?

And I thought that quadratic means of degree 2. Ok, then I will present a polynomial of degree n: f=((1+9X)^n-1)/9. This one has integer coefficients. So, whatever quadratic means, you will have an example.

Ah, now I see - find [u]all[/u] such polynomials. Sorry :)


The book is correct. :lol:

P.S[hide]shouldn't this be in classroom mathematics :huh: [/hide]

[quote="nova"]Hi!

(using PEMDAS)
5(2-4+1) + 2/3(7-1) = ?

my book says the answer = [hide]-1[/hide] , but I seem to be dealing with the 2/3 fraction incorrectly, as I get the wrong answer.
So how should one correctly solve this expression?

thanks.[/quote]
You should evaluate it as follows:
$5(2-4+1)+\frac{2}{3}(7-1)$
$5(-1)+\frac{2}{3}(6)$
$-5+\frac{2\times 6}{3}$
$-5+\frac{12}{3}$
$-5+4$
$\boxed{-1}$


Why is 1 not a prime number. I've always learned that but I don't know why it's not. My AOPS book said that it wasn't but never gave an explanation because it said it was too complicated to get into. Can anyone tell me why it's not prime? God, I must look like an idiot now  :ewpu:  :fool:

i learned that if 1 were prime, it makes wacky things to prime factorizing, like you can make many primes 1 as you like because anything times 1 is the same number, like $3=3\cdot 1=3\cdot 1^2=\ldots=3\cdot 1^n$

Well, prime number is a number that is divisible by 1 and itself, so prime number has two divors.

But note that 1 is ONLY divisible by 1 and that's it! So, by the definition of the prime number, 1 is not a prime number.

[quote="Silverfalcon"]Well, prime number is a number that is divisible by 1 and itself, so prime number has two divors.

But note that 1 is divisible by 1 and that's it! So, by the definition of the prime number, 1 is not a prime number.[/quote]
Silverfalcon, 1 is divisible by itself! that's the definition you wrote above. perhaps you are contradicting your own words? it's funny thou becuz i don't like logic at all but it is essential in math... :roll:

It should've been ONLY by 1.

The definition says that divisible by 1 and itself such that itself $\neq 1$. But for 1, it doesn't work.

I was solving math problems when I posted this so I forgot to add word "ONLY." :)

What Palytoxin said is the main reason that $1$ is not considered prime.  In other terms, what he is saying is that if $1$ were considered prime then the fundamental theorem of arithmetic would be violated.  The fundamental theorem of arithmetic (also known as the unique factorization theorem) states that every positive number (with the exception of 1) can be represented by the product of some primes in exactly 1 way.

Additionally, if you accept $1$ as a prime, then there will be many instances in number theory that you will have to name $1$ as an exception.  Therefore, it is convenient to not consider $1$ as a prime.

Perhaps you would find this article of use: http://mathforum.org/library/drmath/view/57058.html

Well, definition of prime number is - number which has exactly 2 divisors, so..

There are no dumb questions, only dumb answers..

[quote="Valentin Vornicu"][quote="Silverfalcon"]The definition says that divisible by 1 and itself such that itself $\neq 1$. But for 1, it doesn't work. [/quote]Of course it doesn't work for 1. You say that a prime is a natural number [b]different from 1[/b] such that blah blah blah. Thus 1 is not a prime number. This is not an explanation :)[/quote]

Should've look at math world before said this.. Oh well...  :D

[quote="Xantos C. Guin"]Is $-1$ a prime?
$-1$ is divisible by $-1$ and $1$.[/quote]Primes are by definition natural numbers. $-1$ is not a natural number.

[quote="Xantos C. Guin"]Is $-1$ a prime?
$-1$ is divisible by $-1$ and $1$.[/quote]
No,  I think it's just if an integer is positive.


search
number theory unsolved
number theory

suppose that (xy+1)(yz+1)(zx+1) is a perfect square expression,how to prove that xy+1,yz+1,zx+1 are perfect squares respectively?

I'm sorry,
but there's another condition: x,y,z are positive integers.

See [url]http://www.mathlinks.ro/resources.php?c=200&cid=115&year=2007[/url]
By the way, it is still very hard to find the right problem using the search function. Please don't blame people if they post a problem again.


geometry
area of a triangle
Heron\u0027s formula

Triangle ABC has sides of length 13, 14, and 15. Triangle PQR is inside triangle ABC such that the corresponding sides are parallel, and 2 units away. Find the area of triangle PQR.

Don't know if I'm doing this right, but would each side just be $4$ less, so the sides being $9, 10, 11$, and from there just use Heron's Formula?

EDIT: Never mind, drew it wrong

[ABC] = 84
s = semiperimeter of ABC = (13+14+15)/2 = 21
r_1 = [ABC]/s = 84/21 = 4

r_2 = r_1 - 2 = 2

The areas of the incircles and the areas of the corresponding triangles are proportional.

(r_2/r_1)^2 = [PQR]/[ABC]

Plugging in our values and solving for [PQR], we get 21 as the area of PQR.


analytic geometry
graphing lines
slope
conics
hyperbola
calculus
derivative

finding the slope for simple lines is easy but how would we find the slope of a tangent like for something like this

x2+2xy-3y^2= 27  at a point say 2, 2
for some reason this seems like a hyperbola or something
and i have no clue as to how i would find the slope of this
does it make sense to seperate the first equation as in

x^2+2xy  and -3y^2-27=0
?
any hints would be appreciated
thanks  for your time

well i tried figuring it out
i took the derivative and ended up with

2(x'y +xy' (d/dx)) +2x-6y(d/dx)=0
so
2x'y+2xy'dy/dx +2x-6ydy/dx

then you get to solve for dy/dx=  -2x'y-2x/ 2xy'-6y
then i believe we can plug in our point 2, 2

-2(2) -2(2)/  2(2)-6(2)
-4-4/ 4-12

-8/-8

=the answer 1

is this the correct way of doing it..plz some1 verify

This is how i would do it $2x+2xy-3y^{2}=27$ first use implicit differentiation and get
$2+2x\acute{y}+2y-6y\acute{y}=0$
$\frac{dy}{dx}=\frac{2+2y}{-2x-6y}$
then just substitute
$\frac{dy}{dx}=\frac{6}{-12}$
$=\frac{3}{-8}$

Im pretty sure im right

RiemannZeta, that was almost right, but you copied the problem down wrong and also there was some arithmetic trouble when it came time to plug in $x=2$, $y=2$.

Seesaw, I agree with your work, except that there is also an arithmetic problem when you plug in for $x$ and $y$ at the end.

Tryingtolearnmath, I'm not sure what you mean when you write $\frac{d}{dx}$ by itself.

ok my bad, i wish people would use latex it makes life easyer


AMC
AMC 12
ratio
geometry
3D geometry
function
conics

AMC12 Mock Contest #1

(If you dont know what this is, see the Announcement or the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=55329#55329]Rules/Notes on Scoring[/url] for more information.)

1. The ratio (3^9)^3/(9^3)^9 is equal to
(A) 1/(9^8)
(B) 3^(-9)
(C) 6^27/(6^54*8^9)
(D) 9^16/3^59
(E) (1^3)^1/(3^1)^3

2.  Queen Rippledances palace features a big lawn.  The two gardeners, Tare and Mith, hate this lawn because it takes a long time to mow.  Tare can mow the lawn in 10 hours.  Mith can mow the lawn in 8 hours.  How long will it take the two to mow the lawn if they both work at the same time using two lawnmowers?
(A) 40/9 hours
(B) 9/2 hours
(C) 9 hours
(D) 79/18 hours
(E) none of these

3. The sum of 27 consecutive integers is 2646.  Find the sum of the digits in the lowest integer of these integers.
(A) 8
(B) 10
(C) 11
(D) 12
(E) 13

4. Find the sum of the squares of the roots of the equation x^4 - 3x:^2: + 2 = 0.
(A) 6
(B) 8
(C) 9
(D) 12
(E) 15

5. The volume of a cube is 216 cm^3.  Find the sum of the lengths of all the edges.
(A) 56
(B) 64
(C) 72
(D) 84
(E) 92

6. If f(x) = m^(5x) and m^3 = 16, then f(3) = ?
(A) 80
(B) 240
(C) 2^(20/9)
(D) 2^(20/3)
(E) 2^20

7. There are 40320 different arrangements of the letters CHINABOY.  In how many of these is B right after A?
(A) 4!
(B) 5!
(C) 6!
(D) 7!
(E) none of these

8. Find the range of the function f(x) = x + sqrt(1  x:^2:) defined on [-1, 1].  [Edited: Thanks, confuted!]
(A) [-:rt2:, :rt2:]
(B) [-1, :rt2:]
(C) [-1, 1]
(D) [-1/:rt2:, 1/:rt2:]
(E) [0, :rt2:]

9. Find the length of the major axis of the ellipse x:^2: + 5y:^2: = 5.
(A) 33
(B) 8
(C) 6
(D) 14
(E) none of these

10. Triangle ABC has vertices A(-1, 2), B(3,8), C(5,-2).  Find the area of triangle ABC.
(A) 12
(B) 26
(C) 42
(D) 56
(E) 66

11. MathFiend is making cookies.  She cuts circular cookies of radius 1 out from a 10 by 14 rectangular sheet of cookie dough so that each cookie occupies its own 2 by 2 square, and each cookie is tangent to its neighboring cookies or the edge of the sheet of cookie dough.  Find, to the nearest square unit, the area of the region that is between the outermost circles and the edges of the rectangle.  [Edited to clear up some confusion.]
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

12. THE BOB is throwing turtles at a dartboard with two regions.  If he hits the smaller region, he gets 15 points; if he hits the larger region, he gets 7 points.  What is the largest score that Bob cannot get?
(A) 67
(B) 79
(C) 83
(D) 89
(E) 97

13. For how many positive, integral values of n is n^3 - 8n:^2: + 20n  13 a prime number?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 or more

14. How many zeroes are at the end of (2003!)/(1003!)?
(A) 100
(B) 203
(C) 250
(D) 272
(E) none of these

15. Solve 2^(1-3x)  4 = a for x.  Then x =
(A) (5-log_2 a)/3
(B) (log_2 (a+2))/3
(C) (1 + log_2 (a+4))/3
(D) (1  log_2 (a+4))/3
(E) 1-log_2 (a+4):^3:

16.  For what values of p does the equation px:^2: - 4x + p = 0 have non-real solutions?
(A) -2 < p < 2
(B) p > 2 or p < -2
(C) p > 2
(D) p < 2
(E) all real values of p

17. There are four integers a, b, c, d such that d = 5a + 3b + 5c and d = 4a + 5b + 4c.  Suppose 131 :le: d :le: 150.  What is the value of a + b + c?  [Edited: Yikes!]
(A) 11
(B) 22
(C) 33
(D) 44
(E) 55

18. What is the coefficient of x^5*y^2*z^2 in the expansion of (x+y+2z)^9?
(A) 126
(B) 756
(C) 6048
(D) 2016
(E) 3024

19. Consider the triangular array
1
2 3
4 5 6
...........

What is the sum of the elements in the 100th row?
(A) 1000100
(B) 1000000
(C) 500000
(D) 500050
(E) 5000050

20. The quadratic polynomial f(x) satisfies f(0) = 3, f(1) = 5, f(2) = 8.  Find f(5).
(A) 22
(B) 23
(C) 24
(D) 25
(E) none of the above

21. What is the length of one edge of a regular tetrahedron inscribed in a sphere of unit radius?
(A) :rt3:
(B) sqrt(6)/2
(C) :rt3:/2
(D) 2*:rt2:/3
(E) 2*sqrt(6)/3

22. Fierytycoon has three coins, two of which are fair and one of which is double-headed.  He picks a coin at random and flips it 6 times.  It comes up heads each time.  What is the probability that [u]he[/u] picked the double-headed coin?
(A) 2/3
(B) 15/16
(C) 16/17
(D) 32/33
(E) 63/64

23. 223 unbendable sticks with lengths 1776, 1777, 1778, , 1998 are arranged to enclose the maximum possible area.  Which of the following is closest to the radius of the smallest circle that completely encloses the 223-gon?
(A) 25000
(B) 50000
(C) 75000
(D) 100000
(E) 125000

24. Two men (Rusczyk and Crawford) and two boys (Rep123max and Syntax Error) need to cross a lake.  Their boat will either carry one man or two boys.  What is the fewest number of one-way trips that it will take to get everybody to the other side of the lake?  [Assume that one person is required to row the boat.]
(A) 4
(B) 5
(C) 8
(D) 9
(E) 11

25. If arctan a + arctan b = 45 degrees, then a + b =
(A) 1/(ab)
(B) ab
(C) 1  ab
(D) 1 + ab
(E) none of the above

Enjoy!  Please PM your answers to me, mathfanatic, or e-mail to soccerdude9869 at netscape dot net.  I'll try to get back to you as soon as I can.

P.S. I tried to use as many AoPS names as I could...

It's up to you.  Submit solutions if you like.

on 11, should it be outside of all the circles, but inside the rectangle?

Oh.  Heh.  Yeah.  I'll change that.  Thanks, chinaboy!

some questions, spoiler just in case:

[hide]

11. I'm not getting any of the answers, sure they are right?

17. I'm getting multiple possibilities, maybe it's "distinct positive integers," or "what is b?"...

[/hide]

[hide]

17 should be "what is d" i think.[/hide]

[hide]but d is 131 to 150... none of those are possible answers. however, i believe b is one of them[/hide]

Whoops!  That was a BIG error on my part.  For #17 it should read "a + b + c", not "a".  Sorry.

I've also made an answer change for #8 (choice C read [-1, -1].  Heh) and reworded #11 after confuted suggested an answer based on a different interpretation of the problem statement.  Finally, I added a bit of information to #24 which I figured was implied but sort of isn't.  (No, the boat can't row itself back across the river...and yes, a single boy can row the boat by himself.  If he couldn't there would be no way to get everyone across!)

If there are any more errors, please mention them.  Thanks!  I am in the process of assembling an answer key.

I have assembled an answer key and have official answers to every question.  (At least, I think I do.)  So I'll be able to get back to everyone who submits answers fairly quickly.  Thanks to ComplexZeta for answering #18 - I didn't know the multinomial theorem but he cleared that up.

on 22, why does fierytycoon change genders on us?

cool questions

lool, yeah, me thinks that fierytycoon would not like that

[quote]why does fierytycoon change genders on us? [/quote]

I'm a skilled man, so I can be versatile like that.  :)

...

j/k

Whoops! Sorry, Fierytycoon.  Although I might say you are looking very attractive today.  :lol: </jk>  I changed it.

I know everyone is anxious to discuss the questions, so I've decided that [b]Tuesday, December 16th[/b] will be the last day to submit answers.  After that, I'll open up the problems to discussion on this board.  At that point, everyone is free to post their solutions, compare notes, etc.  This way, we'll be able to hear different people's methods of solving each problem.

Sooo...on wednesday, come and post your solutions!

Quick question: On 24 if the boat only carry "either one man or two boys" than how will Mr. Crawford and Mr. Rusczyk cross the river?

[hide=Sketchy solution for 25]
Note that $a=0,b=1$ works, so we have $a+b = 1 = \boxed{1+ab}$. Haven't looked at answers, so no way of determining if it is right; however, the tangent addition formula is similar to that, so it probably is the right answer.
[/hide]

This thread is over a year older than me

[quote=alphamom]
That symbol means "or," I believe.[/quote]

Please try not to bump threads up so much. That last post was made over 12 years ago!

Pretty sure 25 is wrong. Taking the tan of both sides, we obtain $\tan({\arctan{a}+\arctan{b}})=1$. Using the tan addition formula, $\frac{a+b}{1-ab}=1$. So $a+b=1-ab$, which #37's answer satizsfies.

Edit: google confirms my answer

[quote=Possible]For #3, you mean smallest instead of lowest right?[/quote]

Yeah probably.

[quote=smartguy888][quote=alphamom][quote=joml88]You can use \cap instead of what you used.  It loos like this  $P(A\cap B)$.  If what you used is just another convention that I have never seen then that's ok also.  Just wanted to let you know in case you didn't.[/quote]

That symbol means "or," I believe.[/quote]

No, it means the amount of items that are both in $A$ and $B$. Also don't bump 17-year old threads...[/quote]

Sorry, I just wanted to clarify the question.

$A\cap B$ is the [i]intersection[/i] of $A$ and $B$. $A\cup B$ is the [i]union[/i] of $A$ and $B$.

I am absolutely confused about this. ITS IN 2022 NOW

Let me just say the people in those problems and the person who made the problems probably have kids now

[quote name="bjump" url="/community/p26172617"]
25 seems a little too straightforward to be a 25
[/quote]

this test was made in 2003 so it is easier than normal


Mandelbrot Competition Round Four March 1991

4. Find the hypotenuse of a right triangle whose legs are 20806 and 408.

[hide="hint"]
no calculator
[/hide]

[hide]
$20806 = 2 \times 10403 = 2 \times 101 \times 103$
$408 = 2 \times 204 = 2^3 \times 17$

Next, we use the Pythagorean triple generating formulae $a = 2mn, b = m^2 - n^2, c = m^2 + n^2$.  Observe that if we set $a = 20806 = 2mn$, then the only positive value that $b = 408 = m^2 - n^2$ can take is $103^2 - 101^2 = 204 \times 2 = 408$, which satisfies the problem.  Therefore, our hypotenuse is $c = 103^2 + 101^2 = 10609 + 10201 = 20810$.  This can be verified by calculator.
[/hide]


1) Let a,b,c be three positive real numbers such that $ ab\plus{}bc\plus{}ca\equal{}1$. Prove that:
$ \frac{1\plus{}b^2c^2}{(b\plus{}c)^2} \plus{}\frac{1\plus{}c^2a^2}{(c\plus{}a)^2} \plus{}\frac{1\plus{}a^2b^2}{(a\plus{}b)^2} \geq  \frac{5}{2}$

[quote="mr.danh"]1) Let a,b,c be three positive real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that:
$ \frac {1 \plus{} b^2c^2}{(b \plus{} c)^2} \plus{} \frac {1 \plus{} c^2a^2}{(c \plus{} a)^2} \plus{} \frac {1 \plus{} a^2b^2}{(a \plus{} b)^2} \geq \frac {5}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57683


Support
geometry
MATHCOUNTS
trigonometry
LaTeX
geometric transformation
Gmaas

After a great deal of focus group testing, we are still unsure as to the list to include on the first AoPS T-shirt.  Let us know what you think of the following list:

10. You think math is a sport. -- anonymous
A You consider your primary language LaTeX. -- Ragingg
9. You've already figured out that this list is in base 11. -- MCrawford
8. You accidentally write your username instead of your name on a paper for school. -- MathFiend
7. Your idea of a perfect score is 42 points.
6. You have a favorite number and it's not a whole number. -- anonymous
5. You know at least 15 digits of pi.  You know someone who knows 1500. -- A+MATH
4. You try to write QED at the end of your English paper. -- jpark
3. The most entertaining math class you've ever taken was online. -- bleumoose
2. You know Simon's favorite factoring trick. -- DPatrick
1. Your homepage is http://www.artofproblemsolving.com. -- Ragingg

Are there any others that you think must absolutely be on the shirt?  Is this a shirt you would want?  Would you wear it to weddings and formal events?

hahaha, I think its great!!!  I would definitely want one!  Only I wouldn't wear it to formal occasions --> my mom would kill me.

[quote="Gyan"]Great .. Only think I will   enhance is the spelling of "if" to "iff" ..
Eg "You Know Yor're and AoPSer  iff "[/quote]

I would rather nix anything that would be factually incorrect in a mathematical sense.

that's awesome! I'm on there! I'd love to get one if they become available...then I can add another tally next to my "math shirts owned" column  :lol:

Another chance for me to get shirt to show the world my math pride! ^_^ :D
I want one!
Mummy would kill me if I wore it to anything fancy.
 :(
Meanie......

.....I such a geek...... :blush:

[quote]
I would rather nix anything that would be factually incorrect in a mathematical sense[/quote]
I agree, but IMO ,  a case can be made that "iff" may be factually  as correct as "if" . (that was my point) . but any way in either case shirt looks good. regards.

I'm first in line for a shirt. It looks very nice!

I would get one (given that my parents let me  :P ).  Do you have any idea what the rest of the shirt will look like yet or are you still developing that part?

I dunno. Okay here's my two cents:

The problem with putting authors is that it kind of alienates a lot of people who might want to buy it but don't exactly want a shirt with other people's usernames on it. I personally would never buy a shirt with other people's names on it and I know a lot of people who feel the same. (Someone would like me to point out that the MOP shirt was an exception. -.-) The thing is, identifying specific people makes it seem like that shirt and this community the result of efforts by 11 people, which, I would think you guys agree, is not true.

Furthermore, I think this shirt sends mixed messages in terms of personality. There will be people who want a nerdy shirt...a funny shirt...an AoPS-pride shirt...and stuff like that. Personally, I'd like a shirt that people who don't know about AoPS could see, laugh at, and then ask me exactly what this AoPS is. It's okay if they don't get all the jokes. The point of wearing one of these is to talk about it and then laugh at it. So far, the only ones I'm liking are A, 9, 7, 6 and 5. However, all of those statements are remarkable...it's just that some of them I can't picture myself wearing.

Also, as much as I agree with #1 and have put it to practice for a year now, I think actually putting a URL in the #1 spot is a level of commercialization that might not belong here. (Of course, I realize many people think differently.) It would be much more preferrable to have a person ask you for the URL (b/c if they're realli interested, they definitely will!) and attach a nice witty statement in the #1 spot. :-) It could/should even be AoPS related.

Again, there's definitely room for argumentation here as far as what should and shouldn't go on the list. :-) So far, the tone has been quite positive about the selections. Hopefully there's been no offense taken from this post. You guys all did a great job. :-D

Looking back, I think I feel that way too, perhaps more ideas would help.  :?

Uh...what Mith said. Yeah. Oh and Maybe put the list on the back and a picture in front. All my math shirts are t-shirts, mbe you could make a sweatshirt version.

I like it, although I'd make minor wording changes to some of them, esp. to make it "A[b].[/b] You consider ..."

We can't make word changes if we're going to quote people and if we quote people it seems reasonable to give proper attribution.

MithsApprentice, you don't think giving credit to the students who came up with the ones we plan to use helps foster a sense of community -- as if this is a shirt for AoPSers?  Do you think others will feel left out or simply dislike the fact that there are attributions to other students?

[quote]Personally, I'd like a shirt that people who don't know about AoPS could see, laugh at, and then ask me exactly what this AoPS is.[/quote]

I felt like that's part of what we were going for with some of these -- like Simon's favorite factoring trick and just the general set of references that somebody would have to ask about in order to understand.  You don't think any of these are going to be conversation starters?

[quote]So far, the only ones I'm liking are A, 9, 7, 6 and 5.[/quote]

Part of the problem is having something for everyone.  If the standard for a good shirt is that every buyer likes every one on the list, I doubt it will ever happen.  We also felt it best not to repeat some of the same themes even if more than one version was fairly funny.

Do you think we'd be better off scrapping the idea and just going with a proof-without-words kind of shirt or some neat mathematical art?  While I would personally prefer to wear that kind of shirt on most days, I just didn't think it stacked up as the kind of conversation starter that would bring new members to the community as well.  Do you think one of each would be preferable?

[quote]However, all of those statements are remarkable...it's just that some of them I can't picture myself wearing.[/quote]

Why is it that the others repel your overall interest?

[quote]Also, as much as I agree with #1 and have put it to practice for a year now, I think actually putting a URL in the #1 spot is a level of commercialization that might not belong here. (Of course, I realize many people think differently.) It would be much more preferrable to have a person ask you for the URL (b/c if they're realli interested, they definitely will!) and attach a nice witty statement in the #1 spot.  It could/should even be AoPS related. [/quote]

I think part of the idea is that it is fairly unique that a large number of people would have a math website as their homepage.  We can see the statistics of the number of people who visit our site who set it as their home page and the number is impressive enough to give us motivation.  To me it's easily one of the 10 most fascinating things about this community.  I also think it serves a purpose that is otherwise difficult to accomplish -- getting people to read the name of the site.  Most people are not going to stop you and have a conversation about your shirt.  They're going to see it and never say a word.  If the web address were not there, it would need to be somewhere else -- but why not there?

You are good about giving detailed explanations about ideas you have all over AoPS, so I'm interested in hearing your answers to these questions.

I rather enjoyed the "...it takes you less time to derive a formula than it does for your teacher to look it up", even if it does smack of arrogance and hubris :P .

Another note: I really enjoyed the '...you ask your school if the "Art of Problem Solving" counts as the required art class' (Not sure that quote is strictly true to the original).

Just if you guys decide to make another shirt.

Yeah, there were a few good ones that we didn't include either by vote of concern of appearance.

Hey, I made this shirt.  Cool.  I'm the calculator one.

[quote="MithsApprentice"]Since the shirt has already been completed and is now available for sale, I am going to make this a normal thread. It has been advised that moderators try to keep the top of the board as clean as possible.[/quote]

where can we find information on purchasing one?

[quote="aznness"]Hmm... I was one of the lucky few at my State that got one. However, all my friends thought it was kinda nerdy. I wore it with pride(not to be confused with nerdy pride).[/quote]

"do not less other people embarass you or discourage you (something like that) about math. math is cool."

-Greg Fishel, WRAL weatherman at local NC State Math Contest

[quote="MithsApprentice"]There a tab at the top of ur screen that says "t-shirts". Click it.[/quote]

now...why couldn't i find that? :)

There's a bunch of tabs on the top, right below 'Art of Problem Solving' on the top of the window.

About
Forum
MathJams
MyClasses
Blogs

and

T-Shirts!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[quote=Arze23]Woah this discussion is super old...
Now I made it go on the top of the fun factory forum[/quote]

You SHOULD have waited a year.

[quote=Danbert]I love it!   :D

I'd definitely wear it to all formal occasions possible![/quote]

funerals lol

another one: if you think about how you will have to prove everything


basically I am interested in how to write a proof such that when I give it to someone else (including myself), he/sher will completely convince about the correctness of my assertion (i.e. $\Sigma \vDash \tau$)

I was reading a book called "a mathematical introduction to logic" by Herbert B. Enderton. The author gives a good idea of what a proof should be, such as the idea of deduction of $\phi$ from $\Gamma$ as a sequence. And the idea of well-formed formula is absolutely fascinating. Unfortunately, he then starts to prove theorems using Polish notation which are incredibly hard to read.

So I am looking for some alternatives. I did go to amazon.com and found a book called Introduction to Mathematical Logic by Elliott Mendelson. But just one might not be enough.

I humbly await your reply :lol:

[quote="cheerful coffin"]basically I am interested in how to write a proof such that when I give it to someone else (including myself), he/sher will completely convince about the correctness of my assertion (i.e. $\Sigma \vDash \tau$)[/quote]

That sounds like a tough task. After all, when I assert one of my self-evidently true statements, not everone believes me.   :D

More seriously, take a look at

[url=http://www.amazon.com/gp/product/0521290384/]Proofs and Refutations: The Logic of Mathematical Discovery[/url]

and

[url=http://www.amazon.com/gp/product/0521675995/]How to Prove It: A Structured Approach[/url]

and other books you can find by following those Amazon.com links, and see what overall strategy you can develop to convince people your proofs are correct. A test you could apply to yourself in the real world is to convince someone that a theorem of non-Euclidean geometry is correct.

Good luck. This is the kind of thing I've enjoyed reading about for years.

I have read "How to prove it" as well.

While I find its example is quite helpful, I feel that it doesn't explain the reason of every step of the proof rigorously.

Same things go with the "practice makes perfect" and "read other people's proof" approach. I do know how to write a proof similar to those given by the profs, but sometimes I really wish there are guidelines on why the steps of the proof are valid.

Recently I came across a few propositional calculus deductive systems, including the Gentzen system and the Hilbert system. Those deductive systems are guidelines that I'm looking for in writing a proof I think. The rules of inference are useful for explaining how each part of the proof connects together. Unfortunately, nobody in my math. program seems to have done it before. So I am not sure if I am right about this.

On top of that, I still haven't decided which deductive system I should use. There are so many of them :o

[quote="drunner2007"]I've read the first half of "How to Prove It". It's definitely a good book: clear explanations, assumes just basic algebra, etc. It delves pretty deep into elementary set theory but then utilyzes the subject to teach you how to write proofs. It gives you a good foundation.

I still have a lot to learn though. In my eyes, the best way to get better is to practice writing proofs (accurate ones with peers that can evaluate them) and read other people's proofs. You get a feel for proof writing kind of like regular writing, at least that's what I've heard.[/quote]
I agree with Drunner2007, practise will help a lot. I don't really know many books, amazon search might help.
I do have a pdf though, its brief but it's a good start. :)

A VERY elementary book on logic is the MEGSSS book.  Does anyone know what I'm talking about?  It's the black and white spiral bound book...

what does MEGSSS stand for? Mathematics Education for Gifted Secondary School Students? :lol:

Yes.  The course it great, and the MEGSSS books are the BEST.  I think I remember something about ICS on the cover...

MEGSSS? That's something new to me...

And the MEGSSS book... it would be great if you can give me the title, so maybe I can search for it :roll:


algebra polynomial calculus integration function domain induction

$ p $ is a prime number . $ f(x_1,x_2,...,x_n) $ is the intergral coefficient polynomial . Its degrees from every variables are $ < p $ .For all $ x_1 , x_2 ,..., x_n \in Z $ , if $ f(x_1,x_2 ,..., x_n) \equiv 0 [ $ mod $ p] $ , then its coefficients are divisible by $ p $

The following holds in a unique factorization domain $A$: a polynomial of degree $n\ge 0$ from $A[X]$ has at most $n$ roots in $A$. This means that a polynomial of degree $n\ge 0$ which has at least $n$ roots is the null polynomial.

We first prove the assertion for one variable, and then proceed by induction on $n$: asssume we have proved it for $n-1$ variables. Then $\mathbb F_p[X_1,X_2,\ldots,X_{n-1}]$ is a unique factorization domain, and we can apply that sentence in the previous paragraph (here, $\mathbb F_p$ is the field with $p$ elements).


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hi,

how can i show that A5 (the group of sixty alternating permutations on
five symbols) cannot act primitive on a set of fifteen elements?

i understand that a random stabiliser would have four elements and perhaps
one needs to show that this isnt maximal?

plz let me know if this isn't the correct place for my problems
i have no intention to break the house rules  :blush:

fred

I'm a little upset that you interfere with my sustained efforts to keep one page of problems on this section (50 problems max), but this is the right place for your questions. I guess all I have to do is to solve your problems :)

i would start by looking at the
action on 2-element subsets.
The number of pairs is divisible by 7 which is
not a divisor of 60. Just a guess.

What about this:

Every subgroup of order 4 in $A_5$ is of the form $S=[(1),(ab)(cd),(ac)(bd),(ad)(bc)]$ with pairwaise different $a,b,c,d$. There are only 5 such (sylowian) subgroups and the intersection of two of them is $(1)$ or they are identical.
Let $M$ be a set with 15 elements.
Let $S_a$ denote the stabiliser (of order 4) of $a$ from $M$. Then you can divide $M$ into the subsets $U_a$ of that elements that have the same stabiliser as $a$. Its clear that $S_{xa}=x{S_a}x^{-1}$, so you have for every $b$ from $U_a$ that $S_{xb}=x{S_b}x^{-1}=x{S_a}x^{-1}=S_{xa}$. It follows that $xU_a$ and $U_a$ are identical or have no element in common. Because of the fact that not all elements can have the same stabiliser and there are only 5 stabilisers to 'choose' from, this implies that $A_5$ doesn't act primitve on $M$.

Please tell me if this is wrong...


algebra
function
domain
Ring Theory
calculus
integration
superior algebra

Let alpha : R ---> S be a ring homomorphism

(i) If R is a principal ideal domain (PID) and S is an integral domain. Show that alpha is EITHER injective OR Im(alpha) is a field.

(ii) Let S be a integral domain. By considering a ring homomorphism e_0: R = S[x]----> S by f(x) ---> f(0). Show that if R=S[x] is a PID, then S is a field.

Ok, in part (i), we have to prove that Im(alpha) is field if and only if Ker(alpha) is a maximal ideal of R. But I don't know how to prove it.

Part (ii) looks a bit hard to me, I dont know where to start

Any ideas would be greatful

Thank you so much

For (i), note that non-zero prime ideals in a PID are maximal.  For part (ii), use part (i).

Steve


geometry
circumcircle
inradius
incenter
geometry unsolved

I've corrected a typo!
The condition is equivalent to f(a,b,c) = 0 where
f(a,b,c) = 3abc-a^2(b+c-a)-b^2(c+a-b)-c^2(a+b-c)-(a+b+c)

First, let's the isosceles triangles
If g(a,c) = f(a,a,c) = c.a^2-2(c^2+1)a+c^3-c
As g(c-2,c) = c+4>0; g(c,c) = -3c<0; g(c+1,c) = -2c-2<0; g(c+4,c)=13c-8>0, we have
a=c-1 or a=c+2 or a = c+3
a=c-1 gives c=1, a=0 Impossible
a = c+2 gives c=4, a=6
a = c+3 gives c=1, a=4
We have only two isosceles triangles :
(1,4,4) and (4,6,6)

Now suppose a<b<c; suppose a and b are fixed and put h(x) = f(a,b,x) for x>b>a
h''(x) = 6x - 2(a+b) >0 for x>b
h'(b) = ab-a^2-1 >=0
Hence h is increasing for x>b
As h(b+1) = a(b-a-1)(b+2-a), if b>a+1 we have h(x)>0 for x>b;
hence b = a+1 and, as f(a,a+1,c) = c(c-a+1)(c-a-2) we have c=a+2
The other solutions are (n,n+1,n+2) with n>1

There is a little problem in your statement :
D lies on k means that A =60° or A = 120°
But, when A = 120, I cannot lie on k1 - because I is inside ABC - and OI and IH are not equal.
Fortunately, in any case, H lies on k1 and AP and MH are perpendicular

[quote="Beat"]:?  asdasdasd[/quote]

[b]This will be the last time you are doing it.  :mad:[/b]

  Darij

[quote="Beat"]Yes, the blame is mine. It will not happen again. I`m sure you find out why i did this.  :(  I`m guilty, sorry.[/quote]

Okay, but please do post the problems (I think there were two different of them) again!

Did anybody else get onto your account? In this case, I advise you to change your password!

Generally, remember that deleting one's problem after it has been solved is not only very unfriendly towards the forum, but also suggests that you are cheating on a mathematical contest. At least, there are numerous cheaters who do this. (Why do you think I usually quote or rewrite the problem when replying?)

  Darij

[quote="Beat"]I cheated, but not exactly! I was just wondering about this problem and post it. It was still in a contest, but i didn`t send the solution!
Forgive me.. Appologise to everyone. I didn`t wanted to be rood  :(
The problem:
Find all integer triangles with $R=2r+\frac{1}{2r}$, where $R$ is the circumradius and $r$ is the inradius[/quote]

Okay, this is nice. One of two problems ;) . There was also a second one, I used to believe...

Still you should not post problems from ongoing olympiads. Some other people who do participate might read your topics. Also, there is no proof that you didn't participate.

  Darij

Wait.. The problem wasn`t in olympiad it was in magazine- proposed problem.
And i didn`t send the solution, becouse i didn`t understand it
 :)
The second is quite stupid: In circle k with circumcentre O is inscribed an acute-angled triangle. The centre D of the circumcircle k1 of triangle BCO lies on k
Proove that the incentre and the orthocentre of triangle ABC lie on this circle k1,too.
And if H is the orthocentre of triangle ABC then find the angle between AP and MH, where AD and OH intersect BC at points M and P.

You don`t hate me right?
 :(
Everyone make mistakes, but not everyone becomes aware. I hope my english is clear. I mean i realized what i have done and i`m sorry. Hope you will forgive me

[quote="Beat"]The second is quite stupid: In circle k with circumcentre O is inscribed an acute-angled triangle. The centre D of the circumcircle k1 of triangle BCO lies on k
Proove that the incentre and the orthocentre of triangle ABC lie on this circle k1,too.
And if H is the orthocentre of triangle ABC then find the angle between AP and MH, where AD and OH intersect BC at points M and P.[/quote]

Okay, thanks.

[quote="Beat"]You don`t hate me right?
 :(
Everyone make mistakes, but not everyone becomes aware. I hope my english is clear. I mean i realized what i have done and i`m sorry. Hope you will forgive me[/quote]

No, I don't hate you. I just wanted you to explain the mistakes you did and to correct them by reposting the problems you had deleted and by telling where they were from. You have done it now and everything is okay.

  Darij


calculus
integration
limit
function
complex analysis

I've been reading the article and run into the following implication which i can't undertstand:
Fow $z=e^{it}$ the following holds $\displaystyle \int_{|u|=1} \frac{1}{(u-z)^2}(u\frac{w(u)}{w(z)}-2z+\frac{z^2w(z)}{uw(u)})du=\frac{2\pi i}{w(z)} \lim_{|z|>1, z\to e^{it}} res_{u=z} \frac{uw(u)}{(u-z)^2}$

As far as i understand they are considering our integral which depends on $z$ as when $|z|>1$ and $z\to e^{it}$. thus the integral $\int_{|u|=1} \frac{-2z}{(u-z)^2}du=0$ -it's ok and the integral $\int_{|u|=1} \frac{1}{(u-z)^2}\frac{z^2w(z)}{uw(u)}du =0$-that is what i can't understand ??? or maybe i've been wrond somewhere above???

Something is fishy here: the integral can be rewritten as $\int_{|u|=1}\frac1{uw(z)w(u)}\frac{(uw(u)-zw(z))^2}{(u-z)^2}\,du$. If $w$ is an analytic function (which, I hope, it is; otherwise there is no chance to apply the residue theorem), then there is no singularity at $u=z$ because the zero of the numerator cancels the zero of the denominator, so whatever residues may appear, they should be not at $u=z$. So, I should confess that I'm puzzled too. Maybe it'll help a bit if you tell what is written right before and right after this misterious formula. Also it may be helpful to know what exactly is assumed about $w$.


function
trigonometry
search
limit
calculus
calculus computations

$ f(x) = x\sin \frac{1}{x}$ and $ f(0)=0$

It is not on $ [0,1]$ but same idea.

Two things about sylow_theory's function:

(1) Its range is not contained in $ [0,1].$ That's the smaller objection. But ...

(2) The [i]only[/i] value that this function takes on infinitely many times is $ 0.$ It takes on every other value finitely many times, although the number of times it takes on the value $ y$ tends to infinity as $ |y|$ tends to $ 0.$

---

This problem has appeared on this forum before, in case someone wants to try using the search engine.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121848]This[/url] is a thread in which I sketched two constructions, and also asked if such a function can have bounded variation. No answer yet. :maybe:

[The link should work now]

[quote="mdk"]Does there exist a continuous real valued function f from [0,1] to [0,1] which takes on every value in [0,1] infinetely many times?[/quote]

you mean something like $ \textbf f(\theta)=\lim_{\alpha\to 0^{+}}\;\textbf{abs}\;\left(\sin\left( \frac{2\pi \theta}{\alpha}\right)\;\right)$, where $ \theta\in[0,1]\;$?

for example, if you put $ \alpha\;=\;2\times 10^{-10}$, then for $ \theta\;=\;\textbf k\;\cdot\;1\times 10^{-10}$,
( where $ \textbf k=1,2,3,....,10^{10}\;$), you will get the value $ 1$
[and similarly, other values $ \in[0,1)$ for different values of $ \theta$].

and so, as $ \alpha\to 0^{+}$ (i.e. $ \alpha$ becomes even smaller than above),
the [i]maximum [/i]value $ \textbf k$ can take will tend towards $ +\infty$!

or maybe $ \textbf f(\theta)=\lim_{\alpha\to 0^{+}}\;\textbf{abs}\left( \sin\left( \frac{1}{\alpha\cdot 2\pi \theta}\right)\;\right)$

( with the condition $ \textbf f(0)=\beta,\;\;\;where \;\;\;\beta\in[0,1]$ )

you can equally pick cosine instead of sine. :|

misan, for nearly every $ \theta,$ the limit you are using to define your function does not exist. Hence there is no example there, because the function has not been defined.

On a more general note, the pointwise limit of a sequence of continuous functions need not be continuous. You'll have to watch out for that if you're going to try any different limits.


We have (+) real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 1$.

Prove that: $ \frac{1}{{ab}} \plus{} \frac{1}{{bc}} \plus{} \frac{1}{{ca}} \ge 3 \plus{} \sqrt {\frac{1}{{a^2 }} \plus{} 1}  \plus{} \sqrt {\frac{1}{{b^2 }} \plus{} 1}  \plus{} \sqrt {\frac{1}{{c^2 }} \plus{} 1}$

I think that's very difficult!!!  : :|  :|  :|

[b]I'm edited my problem![/b]

By Engel Cauchy, $ \frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ac} \ge 9$. Each of the radicals is less than 1, so the ineq is true. Can't we change the 3 to a 6?

Nice problem!
Let $ 1/ab \equal{} x, 1/bc \equal{} y, 1/ac \equal{} z$. Then
$ xyz \equal{} xy \plus{} yz \plus{} xz$ and
$ x \plus{} y \plus{} z \ge 3 \plus{} \sqrt {xy/z \plus{} 1} \plus{} \sqrt {xz/y \plus{} 1} \plus{} \sqrt {yz/x \plus{} 1}$
But using the first condition, $ xy/z \equal{} (xyz \minus{} yz \minus{} xz)/z \equal{} xy \minus{} x \minus{} y$
So we have
$ x \plus{} y \plus{} z \ge 3 \plus{} \sqrt {xy \minus{} x \minus{} y \plus{} 1} \plus{} \sqrt {yz \minus{} y \minus{} z \plus{} 1} \plus{} \sqrt {xz \minus{} x \minus{} z \plus{} 1}$
$ x \plus{} y \plus{} z \ge 3 \plus{} \sqrt {(x \minus{} 1)(y \minus{} 1)} \plus{} \sqrt {(y \minus{} 1)(z \minus{} 1)} \plus{} \sqrt {(x \minus{} 1)(z \minus{} 1)}$
Now let $ m \equal{} \sqrt {x \minus{} 1}$ $ n \equal{} \sqrt {y \minus{} 1}$ $ p \equal{} \sqrt {z \minus{} 1}$. This simplifies to
$ m^2 \plus{} n^2 \plus{} p^2 \ge mn \plus{} mp \plus{} np$, clearly true.

An English tip: It's I've edited my problem (or I have edited my problem).

[quote="L_Euler"]We have (+) real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 1$.

Prove that: $ \frac {1}{{ab}} \plus{} \frac {1}{{bc}} \plus{} \frac {1}{{ca}} \ge 3 \plus{} \sqrt {\frac {1}{{a^2 }} \plus{} 1} \plus{} \sqrt {\frac {1}{{b^2 }} \plus{} 1} \plus{} \sqrt {\frac {1}{{c^2 }} \plus{} 1}$

I think that's very difficult!!!  : :|  :|  :|

[b]I'm edited my problem![/b][/quote]
The following stronger inequality is true too.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ ab\plus{}ac\plus{}bc\equal{}1.$ Prove that:

\[ \frac{1}{{ab}} \plus{} \frac{1}{{bc}} \plus{} \frac{1}{{ca}}\plus{} 3 \geq2\left(\sqrt {\frac{1}{{a^2 }} \plus{} 1} \plus{} \sqrt {\frac{1}{{b^2 }} \plus{} 1} \plus{} \sqrt {\frac{1}{{c^2 }} \plus{} 1}\right)\]


Who knows proof of this lemma
  (a^m-b^m; a^n-b^n)=a^(m;n)-b^(m;n)
where (a;b)=1

Let $ d \equal{} (a^m \minus{} b^m, a^n \minus{} b^n)$.  Then $ a^m \equiv b^m \bmod d, a^n \equiv b^n \bmod d \implies a^{mx \plus{} ny} \equiv b^{mx \plus{} ny} \bmod d$.  We can pick $ x, y$ so that $ mx \plus{} ny \equal{} (m, n)$, so $ d | a^{(m, n)} \minus{} b^{(m, n)}$.  On the other hand, $ a^{(m, n)} \minus{} b^{(m, n)} | d$.


group theory
abstract algebra
counting
distinguishability
geometry
geometric transformation
reflection

Hey!
I have N balls undistinguishable from one another.
I need to prove the following(no empty groups allowed):

The number of options to divide them to at most k groups
equal to:

The number of options to divide them to groups which each is with at most k balls.

Any idea?

Thanks!

Groups are indistinguishable, also.  Draw a diagram of a generic grouping (a "Ferrers diagram" or "Young diagram").  Reflect it across a line at a 45 degree angle with the axes of your diagram.  This is the bijection.


Sorry if It was posted.
Determine all the function $f: \mathbb{R} \to \mathbb{R}$ such that:
$f(x-f(y))=f(f(y))+xf(y)+f(x)-1, \forall x,y\in \mathbb{R}$

[quote="Altheman"]...let $x=0$...[/quote]

Don't forget you've constrained yourself with $x = f(y)$, so this step requires you to show that $f$ takes the value 0 at some point.


Consider 2004 binary strings of length 2004
We define for 2 strings A,B d(A,B) the number of positions they differ.
Prove one can find A,B,C s.t. d(A,B)=d(A,C)

yes...nice one. if it is wrong, then for every A there must be exactly one B such that d(A,B)=d for some given d=1,...,2004. for d=2004 that means we have exactly the contrast to A, call it Z ( :) ). now, taking d=1002 we have a string (call it N(in the middle of the alphabet :) )) which has distance 1002 to A, but thus distance 1002 to Z as well, QED.

Peter

You are a sharp beginner Peter... :D

Pierre.

Your proof is a bit wrong Peter.
you have 2004 numbers from 1 to 2004 and just 2003 string different from A so one of 1..2004 is missing.
Anyway it's easy


Assume that n^(1/2) is in lowest terms. How many possible values are there for n on the interval (1000, 1] if n is an integer?

Does n^(1/2) have anything to do with the problem?  The way I understand it, you're just asking for the number of integers on the interval [1,1000).

The answer to this interpretation of the problem is [hide]999[/hide].

Mabye you're asking for the number of values of n that n^(1/2) is an integer?

What restrictions apply to n if the square root of n is in lowest terms?

[quote]What restrictions apply to n if the square root of n is in lowest terms[/quote]
Huh ?? If  n is integer, square root of n is either an integer or irrational (or imaginary)

What ZennyK is saying (I think) is that, for all $n$, $n^{\frac{1}{2}}$ can be expressed in lowest terms as $a\sqrt{b}$.  How many values of $n$, from 1 to 999 inclusive, have that lowest terms expression with $a=1$?

So essentially, the problem is to find all values of n between 1 and 1000 that are not divisible by a perfect square, a.k.a. have no exponents (other than 1) in its prime factorization.  The way to solve this problem, then, is to find all products of distinct prime numbers that are less than or equal to 1000.

I have no idea where to go from here other than to brute-force it, but I thought of a creative way to do it that'll save a little time and make sure we get all the combinations.

We can represent each value for n as a binary number, where the nth digit represents whether the nth prime is included in the factorization: 1 means it is; 0 means it isn't:

1      2
01    3
11    6
001  5
101  10
and so on...

The problem with this, then, is efficiently calculating the actual values of n, and figuring out which (and, of course, how many) are below 1000.

It's 2^pi(1000), where pi(1000) signifies the number of primes less than or equal to 1000.


[quote="MoHaMeD"]At first we have to prove that $3/m$ by using $mod7$ and the fact that for any integer $k$ we have $k=0,1 or 6 mod7$ thus the equation becomes : $5^{3p}+7^n=k^3$ so $7^n=(k-5^p) (k^2+k5^p+5^{2p})$  at last we get $3k5^p=0 mod7$ which is untrue :).[/quote]

what about $m=0,n=1,k=2$  :)

Yeah you're right i forgot the case when $k-5^p=1$ regarding this case i found that $4/n$ by using $mod5$ this time if $p\neq 0$ .i'll try to look for a contradiction if possible


where can i get practice grissom tests?

(a link would be nice)

(yes, i know the tourny is in 4 days   :mad: )

http://www.mistacademy.com/grissom.htm

i wish i could go...

the link for the alg2 test doesnt work

it asks for a username and password    :huh:

i can't find anything else

try searching for it on forums or google


I was trying to log onto AoPS earlier, and a page came up where it said the "link is broken." Has this happened to anyone else today? Or is it just me?

I was messing around with something on the server.  It should be fine now.


[hide]
there are six possibilties for the result
M=metal, C=country, R=rap
1. R,R,R,C,M
2. C,C,C,R,M
3. M,M,M,C,R
4. R,R,C,C,M
5. R,R,M,M,C
6. C,C,M,M,R
for 1. there are $\binom{4}{3}\cdot\binom{5}{1}\cdot\binom{3}{1}$ ways or $60$
for 2. there are $\binom{5}{3}\cdot\binom{4}{1}\cdot\binom{3}{1}$ ways or $120$
for 3. there are $\binom{3}{3}\cdot\binom{5}{1}\cdot\binom{4}{1}$ ways or $20$
for 4. there are $\binom{4}{2}\cdot\binom{5}{2}\cdot\binom{3}{1}$ ways or $180$
for 5. there are $\binom{4}{2}\cdot\binom{3}{2}\cdot\binom{5}{1}$ ways or $90$
for 6. there are $\binom{5}{2}\cdot\binom{3}{2}\cdot\binom{4}{1}$ ways $120$
there are a total of $\binom{12}{5}$ ways or $792$
so the answer is:
$\frac{60+120+20+180+90+120}{792}$
simplified to:
$\frac{295}{396}$

[/hide]

it seems right, but i don't figure out how u figured which combinations to multiply

jorian

You are just choose so many CD's out of the number of CD's for each type of the CD's.

Like the first case is to choose 3 CD's out of 4 Rap, 1 CD out of 5 Country, and 1 CD out of 3 Heavy Metal.

And the last part is just choosing 5 CD's out of 12 which is all the possibilities.

[quote="pascal_1623"]Carlos has chosen 12 different compact discs (CDs) he would like to buy. Four are rap music, five are country, and three are heavy metal music. Carlos then randomly selects five of the 12 CDs to purchase. What is the probability that his purchase includes at least one CD from each of the three categories? Express your answer as a common fraction.[/quote]
[hide]
the probability that there are no country picked is $\frac{5^{5}}{12(11)10(9)8}$
the probability that there are no rap picked is $\frac{4^{5}}{12(11)10(9)8}$
the probability that there are no heavy picked is $\frac{3^{5}}{12(11)10(9)8}$
so the total chance is $\frac{12(11)10(9)8-4392}{12(11)10(9)8}$
simplified it is $\frac{1259}{1320}$
edit-oops...i forgot to count the different arrangements...
[/hide]

Sorry for reviving, but I would like to add a [url=http://www.artofproblemsolving.com/Wiki/index.php/PIE]PIE[/url] solution here (it works out quite nicely, and I did not find this solution anywhere else on the site when searching this problem).

[hide="Solution"]
We use complementary counting. If there are no country CDs, then we have $ \binom75$ ways to choose $ 5$ from the other CDs. Similarly, if there are no rap CDs, there are $ \binom85$ ways, and $ \binom95$ if there are no metal CDs. However, we are counting twice the cases in which two of the three groups have no CDs: we can find these to have $ \binom35$, $ \binom45$, and $ \binom55$ ways. But still, there is one last case to check: if all three have none. This case obviously has $ \binom05$ ways. Finally, we get $ \binom75\plus{}\binom85\plus{}\binom95\minus{}\binom35\minus{}\binom45\minus{}\binom55\plus{}\binom05\equal{}202$ (PIE), so the desired probability is $ 1\minus{}\frac{202}{\binom{12}5}\equal{}\boxed{\frac{295}{396}}$.
[/hide]

Nice math154! If the problem had said that all of the albums of the same group were identical, would there be an elegant balls and urns method?

It's alright. The good cases are just the number of solutions to $ a \plus{} b \plus{} c \equal{} 5$ for positive $ a,b,c$ which has $ \binom{5 \minus{} 1}{2 \minus{} 1} \equal{} 6$ solutions (and all solutions work since they must be between $ 1$ and $ 3$), but the total cases have a bit of casework: the number of solutions to $ a \plus{} b \plus{} c \equal{} 5$ for nonnegative $ a,b,c$ has $ \binom{5 \plus{} 3 \minus{} 1}{3 \minus{} 1} \equal{} 21$ solutions, but it is possible for some values to be out of their ranges (e.g., there are four metal CDs)--there happen to be $ 4$ of these cases, so the probability is $ \frac6{17}$. With some tweaking the problem might eliminate casework, though.

Sorry for the revive, but I need some help. This problem is from Intro to CP 15.26

What I did was: $$\frac{\binom{4}{1}\binom{5}{1}\binom{3}{1}\binom{9}{2}}{\binom{12}{5}}$$ since we choose 1 from each of rap, country, and heavy metal music, and this would leave us with 9 CD's left, after which it doesn't matter what types of CD's you choose, so you choose 2.

But clearly this is wrong, because the numerator is greater than the denominator. Could somebody help tell me what I did wrong (what was the flaw in my calculation/thinking)? Why doesn't my method work?

Thanks.

Say, for instance, that the CD's you choose to fulfill "1 from each category" are $R_1,C_1,H_1.$

This part is valid, but the part where you say
[quote=You]
and this would leave us with 9 CD's left, after which it doesn't matter what types of CD's you choose, so you choose 2.
[/quote]
is wrong. If the CDs chosen here are $R_2,C_2$ (this is just an example, it doesn't really matter which CDs get picked here,) then how is this different from initially choosing $R_2,C_2,H_1$ and later picking $R_1,C_1$? The order in which we pick the CDs here are irrelevant.


videos
calculus
geometry
summer program
MathPath
Duke
college

Does anyone know of a way I can take and finish Algebra II over the summer and have it credited? I have LOADS of time.

Jen

[quote="Jenn7"]Sorry if I bug any moderators by posting twice. I was actually tryin to edit the post, but .... Anyway, I found a way  :) . It's just going to cost me mucho dinero. I'm going to do it through ehighschool during summer. Yay![/quote]
huh? the georgia ehighschool? i was going to recommend aops with a supplement of video text algebra(for the accredited part.)

No, cobb county has its own ehighschool. Um, what does accredited mean? I was actually going to try to study on my own from a textbook, but it would be useless if my high school wouldn't credit it, and I'd have to take it again. 'Sides I get distracted really easily :lol: .

If you do the Cobb County eHighSchool, you need to understand a couple of things.

1. This is really intended to allow students to recover credit once they have failed the class once. Therefore it's taught at that level. You will definitely need to supplement from something a little more rigorous. (AOPS might take care of that.)

2. Do you really need to get that far ahead? As it stands now, you will be in Multi-variable Calculus as a Senior. Why don't you look for something that's NOT in the normal high school curriculum. More Geometry would be a great thing to do. I can suggest a couple of books if interested.

3. What about an AOPS class? I see that there is an Intro to Number Theory class this summer. You might enjoy that a lot.

4. What about a summer camp? There are lots of really good ones. Take a look at the list on this site.

Take care,
Fulton

1. Do you have any recommendations on what books I can buy?

2. It's not really a matter of getting ahead. I don't envision myself becoming an engineer or having a career in mathematics. If I remember correctly, Multivariable Calculus is Post-AP, and for medicine or law, that will probably be math beyond the requirements. It's more electives that I need. I've always wanted to be a lawyer or a doctor, but I'm still not sure, and I wanted to try the other electives(graphic communications, business, and accounting)...I would appreciate the suggestions for books though.

3. Maybe. That's something I hadn't thought about. I'll check it out later.

4. I'll check them out.

Thanks for the advice,
Jen.

I've come to my senses, lol, I'm not going to do Algebra II online.

About summer camps - are there any that's just right for someone taking Geometry that's challenging but not overwhelming? I've been looking at the ones on AOPS but there's so many, and most are for more talented mathletes.

Thanks, Jen.

Don't sell yourself short. You have the first attribute of a successful math student. You want to. Probably the best camp for middle school students is MathPath:

http://mathpath.org/

Try the qualifying quiz and see how you do. If you want to pass it by me and get some confidential feedback, I'll be happy to look at it.

One of the really neat things about MathPath is that it is for middle school students.

The other part is that it isn't cheap. But my understanding is that it's well worth the money.

Fulton

I attended MathPath last year and I thoroughly enjoyed it. There are several levels of difficulty so you won't feel overwhelmed. It was definitely worth the time and money. You may want to refer to [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=193]this link[/url] for any questions you have. There will be another Math Jam on May 1 if you have any more questions.

Thank you. I think I'll try the qualifying quiz. Just two questions: Around how many do we have to get right on the quiz, and are we allowed to look online to figure out how to solve it? Thanks, Jenn.

Another thing you could look into is perhaps the Duke TIP camp. It's probably less rigorous than MathPath or some other math camps, but it is pretty darn good.

Please try out for MathPath so we can share the unaccompanied minor fee or something. Please! Please!!!!! It's really not [i]that [/i]hard...except for one problem I despise....

Thanks. I just checked it out, but applications were due in February, so there's no way I can apply for it this year. Maybe next year. It's slightly more affordable than MathPath, and there's financial aid. I might even pick a course that's not math-related next year. Intro to Medicine sounds great :lol: .

[quote="gtf"]The test says that if you can do 3 out of the 6 (in whole or the sum of parts), then you should apply. I don't know exactly what it takes. Maybe knexpert can give us a clue.

As far as the Internet is concerned, I would say that it's OK if you want to understand a concept or find the definition of a term, but DON'T look for the solution on the Internet. Many times, what's more interesting to the people reading these applications is your thought process, not just whether or not you get the question right.[/quote]

Thanks. By the way, you said that if I ever saw this one teacher(Ms. McMullan?) at KMHS to tell him/her something. What  was it?

[quote="funcia"]Please try out for MathPath so we can share the unaccompanied minor fee or something. Please! Please!!!!! It's really not that hard...except for one problem I despise....[/quote]

Unaccompanied minor fee? Is that the one for traveling alone on an airplane? How can we share it?

Just that I said hi. She's a really neat lady and an awesome teacher. You will probably love her if you ever have her.

K. I'm sure I will. I won't know till Spring if I have her or not.

I'm so excited to be going to KMHS. I just signed up for Latin and Marching Band last Tuesday.

I thought that regular application was due in May(you're probably talking about something else..)
The flight instructions said that people from the same area taking that same airline(i think the one that sponsors mathpath) can share the unaccompanied minor fee.

We're talking about the same thing. If I get in, sure.

there are alot of colleges that have a summer session. u might be able to apply there as a joint enrollment student and take their algebra classes.

EPGY EPGY EPGY why does EPGY ring in my head like this


floor function
logarithms
calculus
integration
geometry
rectangle
function

Show that $ 999!$ has at least $ 1891$ digits.

Hint: [hide]1891 digits means that $ 999!$ is greater than $ 10^{1890}$[/hide]

Proof:

[hide]$ \displaystyle\frac {999!}{99!} \ge 10^{1800}$. This is proven by replacing the 900 numbers greater than $ 100$ with $ 100$.

Now, consider the numbers between $ 200$ and $ 299$, and the numbers between $ 500$ and $ 599$. If we divide each of those numbers by $ 100$, we get a string of 100 numbers greater than $ 2$, and a string of 100 numbers greater than $ 5$. Multiplying the $ n$th number in the first sequence with the $ n$th number in the second sequence yields 100 numbers that have products that are greater than or equal to 10. Therefore, there are at least $ 1900$ digits in $ 999!$, so there are at least $ 1891$ digits in $ 999!$.

Better proof: Replace all the numbers greater than $ 100$ with $ 100$, and replace all the numbers greater than $ 10$ but less than $ 100$ with $ 10$. You have $ 10^{1890}$, which is less than $ 999!$ and has $ 1891$ digits. Thus, $ 999!$ must have $ 1891$ digits.
$ \displaystyle\frac {999!}{99!} \ge 10^{1800}$. This is proven by replacing the 900 numbers greater than $ 100$ with $ 100$.

Now, consider the numbers between $ 200$ and $ 299$, and the numbers between $ 500$ and $ 599$. If we divide each of those numbers by $ 100$, we get a string of 100 numbers greater than $ 2$, and a string of 100 numbers greater than $ 5$. Multiplying the $ n$th number in the first sequence with the $ n$th number in the second sequence yields 100 numbers that have products that are greater than or equal to 10. Therefore, there are at least $ 1900$ digits in $ 999!$, so there are at least $ 1891$ digits in $ 999!$.

Better proof: Replace all the numbers greater than $ 100$ with $ 100$, and replace all the numbers greater than $ 10$ but less than $ 100$ with $ 10$. You have $ 10^{1890}$, which is less than $ 999!$ and has $ 1891$ digits. Thus, $ 999!$ must have $ 1891$ digits.[/hide]

On a futher note, would it be possible to find the exact number of digits?(without a ti-nspire from nats mc :P )

I don't think it would be too easy to calculate $ 1 \plus{} \left\lfloor\sum_{k \equal{} 1}^{999}\log k\right\rfloor$...

[quote="Darmani"]You can find pretty good upper and lower bounds using integrals, however.[/quote] Please elaborate.

He is most likely talking about interpreting the summation as a Riemann sum.

We can interpret the sum of logarithms from 1 to k as the total area of several rectangles each of width 1 and height log 1, log 2,...,log k. This can be plotted by looking at the graph of $ y \equal{} log x$, and drawing these rectangles with their right sides at the corresponding integral values of x. We thus see that this area lies entirely below the graph of $ y \equal{} log(x \plus{} 1)$, but is below the graph of $ y \equal{} log(x)$. Thus, $ \int_1^klog xdx < \sum_{i \equal{} 1}^klog x < \int_0^klog(x \plus{} 1)dx$. Note that the upper and lower bounds are always fairly close to each other, and are also asymptotic to each other. This is a good technique for finding asymptotic approximations to discrete sums; this approximation for the log function in particular is important in the derivation of Stirling's approximation.

See http://en.wikipedia.org/wiki/Summation#Approximation_by_definite_integrals

This may have already been said in the posts before this, yet:
[hide]
$ \log (999!) \equal{} \sum_{k \equal{} 1}^{999} \log k > \sum_{k \equal{} 10}^{99} 1 \plus{} \sum_{k \equal{} 100}^{999} 2 \equal{} 90 \plus{} 2 \times 990 \equal{} 1890$

Hence $ \log (999!)$ has more than 1891 digits.[/hide]


function
number theory open
number theory

two function $f,g$ such that  for each $n$ then $\phi(n)=2^{f(n)}.k$ (k odd) and
let $a$ is integer number ,$g(n)$ is greates integer number satisfying $2^{g(n)}$ divise $a^{f(n)}$ .find expression of $g(n)$.

two function $f,g$ such that  for each $n$ then $\phi(n)=2^{f(n)}.k$ (k odd) and
let $a$ is integer number ,$g(n)$ is greates integer number satisfying $2^{g(n)}$ divise $a^{f(n)}-1$ .find expression of $g(n)$.


Any one out there who has appeared for the JEE this year?

What do u feel about the level of the paper......expected cutoffs...Pl.reply

this years paper was a complete farce.I myself appeared for it and felt cheated. :rotfl: They nowhere appeared to test your mental calibre neither your stickability to any sort of problem . :mad: It was a banking clerical aptitude type of paper where the questions would be easy and the examiners wanted to eliminate those who would make silly mistakes.Among my own friends those who had done the hardest analysis of the questions in the previous two years are most depressed .A state is reached after deep thinking when you begin to ponder incessantly over simple questions.Rohit singh 0812 has also taken the exam and i suppose he will get a rank under 5 or 10. :ninja: I have studied with him.Coaching persons in my city are declaring a cut off of 190-200 with under 100 rank going in the range 360+.what is your expected score?

If FIITJEE's solutions are to be believed ,I am getting something in the range of 310 -325.
   And by the way ,I don't agree with your view that the questions were so easy that one had to ponder over them especially after solving the toughest possible problems.
   Basically one has to learn to think straight and thats what the IITs seem to want now.You may have different views but still...

(I don't intend to offend u in any way...just giving my opinion)

it seems that you have not understood my case.Had you made a habit of thinking over problems  for hours ,forming new ones and looking for special cases and beautiful ingenious approaches perhaps you would'nt be at sea with my words.Coming back to iit:
 :P Earlier good math students used to think to crack cpmt that do 100% paper in phy and chem in cpmt do little guesswork and use logic in bio with knowledge gained till 10th.All that will be reversed now.BIO STUDS WILL LOOK TO CRACK JEE! :wink:

That is the sad part of the story.....It'd really be irritating if someone interested in bones ,flesh and .....manages to clear the JEE.

This year's paper seemed relatively much easier than last year.The cut off should rather be atleast be 40%.
 And being interested in flesh and bones is in no way a factor in clearing JEE,more so in today's academic scenario with interdisciplinary work reaching new heights.Even physicists do a lot of Biology these days,especially the Molecular Biology part.It is just a huge misconception that Biology involves mugging up.In my opinion,Mathematics,Physics,Chemistry and Biology are greatly unified.
 After all,the IIT people are not that foolish!They have made a Biology course compulsory for all streams for atleast a semester.
Sathej

Well,opinions differ from person to person...Maybe it is because of my hatred towards the subject(Biology created nightmares for me during my 10th exams!) that has made me formulate such an opinion about biology and the people associated with it.

And the education system is to be blamed partly because no one told us at school about the pleasures of learning a subject like biology,emphasis was on mugging up and getting marks....the notion is that only those who want to become doctors would study biology(thats what atleast i think)Its not meant for engineers and mathematicians.

i think it was an easy paper but p'haps not as easy as all make it out to be, the match the following and assertion reasaon are sticky questions. top 50-ers would probably get 200 in paper 1 and around 160 in paper 2.  probably 350 odd woud finish
AIR 100.

Hello!
i m in X rite now...ya i saw this year's JEE paper..i currently don't know whether it is easy or not...my bro gave it and said...that iit don't need q's killers anymore..but want guys with brains and concepts...he himself suffered (got a mediocre score of 291) :rotfl: ....i knw a guy scoring 411..and yes he is very cool.. :blush:

[quote="kay-pee"]
 (got a mediocre score of 291) [/quote]
291 is not at all mediocre by any standards

well i was in the Indian system till 10th...and was planning to give the JEE...till i migrated....
i still had a passing interest in the exam so i decided to look up some of FIITJEE's model exams on the net....
the exam seemed to have a graduated system of Q's covering all ranges....but it wasn't too hard overall

PS: my views based on some model question paper from the net might not be too accurate

What makes the IIT JEE difficult is not the paper itself, but the time constraint, the examination stress and intense competition.
Sathej

Hi guys,

     I need a small favor of you. This might sound stupid. I would really appreciate it if any one of you could get hold of IIT-JEE 1998 Mathematics paper and post one of the 15 subjective problems which was on binomial co-efficients and primes. I tried searching on the net but in vain.

Thanks.

The problem reads as:
Let p be a prime and m a positive integer. By mathematical induction on m, or otherwise, prove that whenever r is an integer such that p does not divide r, p divides (mp)Cr.
 Here, nCr represents number of ways of choosing r things from n things.
Sathej

i know this is too late to ask but still why did many people get ranks with huge deviations(+/- :D )  from the expected ranks/predicted ranks by X,Y,Z...institutes?

This may primarily be attributed to the match the following questions the answers to which were hugely controversial........every institute gave its own set of answers....each different........since these questions carried a large chunk of the total marks alloted........the swing in marks (and hence the rank) may be due to this........

P.S....I myself feel cheated by over 150 ranks this year.....but under the assumption that the solns of FIITJEE are correct(in some way a hopeless assumption since they r defntly wrong....only IITppl know the correct answer and those ppl are not going to reveal it....so this wud remain a mystery forever.....)


Let $ n$ be a positive integer larger than $ 1$, and let $ a> 0$. Show that:

$ \frac{1\plus{} a \plus{} a^2 \plus{}... \plus{} a^n }{a\plus{}a^2\plus{}a^3\plus{}...\plus{}a^{n\minus{}1}}>\equal{}\frac{n\plus{}1}{n\minus{}1}$

$ A \equal{} 1 \plus{} a \plus{} ... \plus{} a^n \equal{} \frac {a^{n \plus{} 1} \minus{} 1}{a \minus{} 1}$
$ B \equal{} a \plus{} a^2 \plus{} a^3 \plus{} .. \plus{} a^{n \minus{} 1} \equal{} a(1 \plus{} a \plus{} a^2 \plus{} ... \plus{} a^{n \minus{} 2}) \equal{} a.\frac {a^{n \minus{} 1} \minus{} 1}{a \minus{} 1}$
=>$ \frac {A}{B} \equal{} \frac {a^{n \plus{} 1} \minus{} 1}{a^n \minus{} a}$
$ \frac {A}{B} \geq \frac {n \plus{} 1}{n \minus{} 1}$
<=>$ \frac {a^{n \plus{} 1} \minus{} 1}{a^n \minus{} a} \geq \frac {n \plus{} 1}{n \minus{} 1}$
<=>$ na^{n \plus{} 1} \minus{} a^{n \plus{} 1} \minus{} n \plus{} 1 \geq na^n \plus{} a^n \minus{} an \minus{} a$
<=>$ (n \minus{} 1)a^{n \plus{} 1} \plus{} a(n\plus{}1) \plus{} 1  \geq a^n(n \plus{} 1)\plus{}n$

I think is easy but not at all obvious how to finish from here. Can you post a conclusion?