Find $ f : R--->R $ satisfying :
 $ f(\frac{x+y}{x-y}) = \frac{f(x)+f(y)}{f(x)-f(y)} $

from observation, we have if f(x)=x
f((x+y)/(x-y))=(f(x)+f(y))/(f(x)-f(y))

Can you prove that f(x)=x is the only solution?

Here is one thing you know about f:
Let x=x and y=-x, then $f(0)=\frac{f(x)+f(-x)}{f(x)-f(-x)}$
Then, by manipulating the algebra
$\begin{eqnarray*} f(0)*f(x)-f(0)*f(-x)=f(x)+f(-x) \\ f(0)*f(x)-f(x)=f(0)*f(-x)+f(-x) \\ f(x)*(f(0)-1)=f(-x)*(f(0)+1) \\ \frac{f(x)}{f(-x)}=\frac{f(0)+1}{f(0)-1}$
This is also true when x=0, which gives you an equation for f(0).  It turns out f(0)=0.
Then $\frac{f(x)}{f(-x)}=-1$ Therefore f is odd, with f(0)=0.

I don't know if f(x)=x is the only solution, but I have a hunch it is.
For f(x)=x^n, f(x)=x^1 is the only solution, for sure, since all the others give you $\frac{(x+y)^n}{(x-y)^n}=\frac{x^n+y^n}{x^n-y^n}$ which is only true for n=1.

Here is another function question:
Find all differentiable functions for which $f(x)+f(y)=f(\frac{x+y}{1-xy})$ for real x, y and xy!=1.  Trig sub?

[quote="Wumbate"]\[\frac{f(x)}{f(-x)}=\frac{f(0)+1}{f(0)-1}\]
[b]This is also true when x=0, which gives you an equation for f(0).  It turns out f(0)=0.[/b][/quote]
Hmm, are you sure? Because if $x=0$, \[\frac{f(0)}{f(0)}=1=\frac{f(0)+1}{f(0)-1}\] which implies that $1=-1$  :!: :?: , an obvious contradiction.
What went wrong? The idea :idea: is that $y\neq x$ otherwise wacky things will come up ;)
Yes, $f(x)=x$ is one solution but this is not the way to obtain this function.

[quote="Wumbate"]OK but can you prove it is the only solution?[/quote]
I'll wait until you guys are hungry and thirsty for a proof  :D

Isn't this a little iffy... don't you have to say explicitly that f is injective in the problem statement, as if f(x) is a constant function this statement is meaningless for all x,y pairs as we get division by zero on the RHS.

[quote="mathnovice"]Does anyone know where functional equations are studied, i.e. which branch of math?[/quote]

Functional Analysis... maybe...  :D

[quote="Nukular"]if f(x) is a constant function this statement is meaningless for all x,y pairs as we get division by zero on the RHS.[/quote]

All that shows is that constant functions don't work.  It's clear that a non-injective function won't work because then you get a contradiction, as you said -- they can't possible satisfy the condition.  So injection is a property that follows from the givens, not a property that needs to be stated in the question.


Let $O$ be the excenter opposite angle $B$ of triangle $ABC$ and $D$ be the center of the circle passing through $A,B$ and $O$. Prove that $A,B,C,D$ lie on a circle.

Hi
OBC=(180-B)/2
OCB=(180-C)/2
BOC=180-(OBC+OCB)=(B+C)/2
As BDC=2*BOC so:
BDC=(B+C)=180-A
B,D,C,A are on a circle
Good Luck


function
real analysis
real analysis unsolved

Suppose that $f_1,f_2,...,f_n: \mathbb{R}\rightarrow \mathbb{R}$ are the periodic functions and $f=f_1+f_2+...+f_n , f:\mathbb{R} \rightarrow \mathbb{R}$ has an finite limit in $+\infty$.
Prove that $f$ is the constant function.

Let $T_1,\ldots,T_n>0$ be periods for $f_1,\ldots,f_n$ respectively. Now let's look at what happens to $f(x+p_1T_1+\ldots+p_nT_n)$ as some of the natural numbers $p_i$ go to $\infty$.

Fix an $i$ and make just $p_i\to\infty$ while keeping all the other $p_j$'s constant. This will keep $f_i$ constant, and will show that $\sum_{j\ne i}f_j(x+\sum p_kT_k)$ has a limit when $p_i\to\infty$. Now increase some $p_j,\ j\ne i$ (for a single fixed $j\ne i$) with $1$, and then perform the same operation as above. $\sum_{j\ne i}f_j(x+\sum p_kT_k)$ will have the same limit that it had the last time, and this shows that $f(x+p_1T_1+\ldots+p_jT_j+\ldots+p_nT_n)=f(x+p_1T_1+\ldots+(p_j+1)T_j+\ldots+p_nT_n)$, which, in turn, means that $f_i$ has $T_j$ as a period. This means that all the $T_k$'s are periods for all the $f_k$'s, and from here the conclusion follows immediately.

Sorry if it's a bit unclear, ehsan. Try it with $n=2$ to see what's going on.

A little bit clearer. By induction. For $n=1$ it is very simple. Assume it is true for $n-1$ and consider the function $ f(x+T_1)-f(x)$ that we know it has a limit 0. On the other hand, this function is the sum of $n-1$ periodic funtions $ f_i(x+T_1)-f_i(x)$ with $i>1$. Thus it must be constant and since it tends to 0, it must be 0. Thus $T_1$ is period for $f$ apply now the case $n=1$.

what a beautiful solutions :w00t:
thanks all of you  :blush:


LaTeX
linear algebra
matrix
induction
binomial coefficients

Prove that
F_1*(n/0)+F_2*(n/1)+...+F_(n-1)*(n/n-1)+F_n*(n/n)=F_(2n)
Where F_n denotes the nth Fibbonacci number (F_1=1,F_2=1,F_n=F_n-2+F_n-1), and (a/b) is a choose b=(a!/(b!*(a-b)!)).  a/b is not a fraction.  There are no fractions in this problem.

I think the indices are shifted by 1 for the first two terms. It should read as

F_1*(n/1)+F_2*(n/2)+...+F_(n-1)*(n/n-1)+F_n*(n/n)=F_(2n)

Quote: Here is the [tex]\LaTeX[/tex] version of the problem. Prove that

[tex]\displaystyle \sum^n_{i=1}{n \choose i}F_i=F_{2n}
[/tex] Where [tex]F_n[/tex] denotes the nth Fibbonacci number.

It's not a hint, I wrote too much.

[hide]

Binomial theroem works for matrices if AB=BA.

Let A=(a_ij) be the 2x2 matrix where a_11=1, a_12=a_21=1, a_22=0.

Let I=(b_ij) be the 2x2 identity matrix where b_11=1, b_12=b_21=0, b_22=1.

Then A^2 = I+A

Define X_n=(x_ij) to be the 2x2 matrix where x_11=F_{n+1} , x_12=x_21=F_n, x_22=F_{n-1} for n :ge: 1, define X_0=I

Then X_n = A^n by induction.

So X_{2n} = A^{2n} = (I+A)^n =  :Sigma: (n/i) A^i =  :Sigma: (n/i)X_i

Comparing the x_12 elements of the matrices on both sides yields the desired result.

[/hide]

Magnara wrote:Wow.  That was pretty cool.  However, this was from a n.t. book that doesn't even touch on matrices.  Anyone have any other ideas?

[hide]

It can also be proved using the explicit formula for Fibbonacci numbers F_n.

Let r and s be the roots of x^2-x-1=0, then F_k=a r^k + b s^k

So  :Sigma: (n/i) F_i = a  :Sigma: (n/i) r^i + b :Sigma: (n/i) s^i = a[(1+r)^n - 1] + b [(1+s)^n -1] = a r^{2n} + b s^{2n} - (a+b) = F_{2n} since 1+r = r^2, 1+ s = s^2 and a+b = 0.

[/hide]

You can almost certainly do this with induction, although you may have to do 2 seperate inductive steps.  In particular, the additive formulae for the Fibonacci sequence and binomial coefficients will probably come into play.

I bet you can also think of a clever model that this identity represents.  I'll go think about that.  Sounds like fun.

Here's a model for it, with explanation.  Feel free to ask if you want me to clarify better.

Consider a block-walking problem as follows: how many ways are there to get from the dot marked 1" in the diagram to the dot marked 2n, given that I always walk diagonally to the right (either up or down) or straight to the right?  We can see immediately that this obeys the Fibonacci recursion, since you can arrive at each point either from the point directly before it or the point 2 spaces before it.  We also see that the number of ways to get to 1" is 1 and the number of ways to get to 2" is 1, making this match the Fibonacci numbers exactly.  Thus, for the specific question posed (number of routes from 1" to 2n,) the answer is precisely [tex]F_{2n}[/tex].

Now, let us count it another way.  It will take at least n steps to make this trip.  Thus, we can assign an order to the first n steps and follow it up by assigning an order to whatever happens to remain.  For example, our trip might consist of 3 horizontal motions, n - 3 diagonal motions, and then we would have n - 3 spaces left to cover.  One can see this since each horizontal motion moves us forward 2 while each diagonal motion moves us forward 1.  There are [tex]\displaystyle {n \choose 3}[/tex] ways to choose which three motions will be horizontal and, as shown above, exactly [tex]F_{n - 3}[/tex] ways to make the final n - 3 moves.

Summing this over all i, we see that we do not have double counting, since each grouping has a distinct collection for its first n moves while within a group, the Fibonacci product guarantees that each member will be distinct, and we do not have undercounting since every path is composed of n moves followed by possible some other moves.

Therefore [tex]\displaystyle \sum_{i=1}^{n} {n \choose i} F_{i} = F_{2n}[/tex]


geometry
circumcircle
trigonometry
trig identities
Law of Sines
geometry solved

let O be interior to a triangle ABC . Let A'.B',C' be the point of intersection of the line perpentcular to BC,CA,AB  and this lines passing through O .
    prove that AO* CB = OB*CA =OC *AB when  and only when equilateral triangleA'B'C'

To be honest, I must say the problem is very easy, it uses a trick which can be applied to a dozen of past IMO problems.

[color=blue][b]Problem.[/b] Let O be a point in the plane of a triangle ABC. Let A', B', C' be the feet of the perpendiculars from the point O to the lines BC, CA, AB.

Prove that $AO\cdot BC=BO\cdot CA=CO\cdot AB$ holds if and only if the triangle A'B'C' is equilateral.[/color]

[i]Solution.[/i] Since < AB'O = 90° and < AC'O = 90°, the points B' and C' lie on the circle with diameter AO. In other words, the circumcircle of triangle AB'C' is the circle with diameter AO. Or, equivalently, we can say that AO is a diameter of the circumcircle of triangle AB'C'. Now, the extended law of sines states that in every triangle, the length of a side equals the diameter of the circumcircle, multiplied with the sine of the opposite angle. Hence, in triangle AB'C', we have $B^{\prime}C^{\prime}=AO\cdot\sin\measuredangle B^{\prime}AC^{\prime}$. In other words, $B^{\prime}C^{\prime}=AO\cdot\sin A$. Similarly, $C^{\prime}A^{\prime}=BO\cdot\sin B$ and $A^{\prime}B^{\prime}=CO\cdot\sin C$. Thus, the triangle A'B'C' is equilateral if and only if B'C' = C'A' = A'B' if and only if $AO\cdot\sin A=BO\cdot\sin B=CO\cdot\sin C$ if and only if $AO\cdot BC=BO\cdot CA=CO\cdot AB$ (this is because $\sin A : \sin B : \sin C = BC : CA : AB$ by the sine law, applied to triangle ABC). Hence, the problem is solved.

Note that there are (in general) two points O in the plane of triangle ABC which satisfy $AO\cdot BC=BO\cdot CA=CO\cdot AB$. These are called the two [b]isodynamic points[/b] of triangle ABC. They are the isogonal conjugates of the two Fermat points of triangle ABC, and one of them always lies outside of triangle ABC, while the other one lies inside the triangle ABC if and only if all angles of triangle ABC are < 120°.

  Darij

[b]It is known that [/b]$\underline {\overline {\left| B'C'=AO\cdot \sin A\right| }}.$[b]Therefore [/b]$B'C'\cdot 2R=AO\cdot BC$[b], a.s.o. Thus,[/b]

$AO\cdot BC=BO\cdot CA=CO\cdot AB\Longleftrightarrow B'C'\cdot 2R=C'A'\cdot 2R=A'B'\cdot 2R\Longleftrightarrow$

$B'C'=C'A'=A'B'.$


If $\log_{2n} 1994 = \log_n \left(486 \sqrt{2}\right)$, compute $n^6$.

In original test, the problem never told you this but to give your answer in exact form, write your answer in form $p_1^{e_1} \cdot p_2^{e_2} \cdot p_3^{e_3} \cdots$ where $p_i$ is the prime and $e_i$ is the corresponding exponent with $i = 1,2,3,...$

This is actually a simple problem, but it is impossible to have an integer value of $ n$ with the given values. $ 1944$ was the number used in the original problem and it turns out that it works out very nicely.

[hide="My Solution"]
Factorizing the two integers in the problem, we obtain $ 486 \equal{} 2\cdot 3^5$ and $ 1944 \equal{} 2^3\cdot 3^5$.

Let $ a \equal{} \log_{2n}{1944} \equal{} \log_n{486\sqrt{2}}$. We have

$ 2^a\cdot n^a \equal{} 2^3 \cdot 3^5 (1)$ and $ n^a \equal{} 2\cdot 3^5\cdot 2^{\frac{1}{2}}\implies n^a \equal{} 2^{\frac{3}{2}}\cdot 3^5 (2)$.

Substituting $ (2)$ into $ (1)$, we have

$ 2^a \cdot 2^{\frac{3}{2}}\cdot 3^5 \equal{} 2^3 \cdot 3^5\implies 2^a \equal{} 2^{\frac{3}{2}}\implies a \equal{} \frac{3}{2}$.

Finally, using $ (2)$, we have
$ n^{\frac{3}{2}} \equal{} 2^{\frac{3}{2}}\cdot 3^5\implies n^6 \equal{} \boxed{2^6\cdot 3^{20}}$.
[/hide]


We have u(n+1)= \sqrt(u(n)2+u(n)), so {u(n)} is increasing, so it has a limit, which can only be infinity.
We calculate lim(u(n+1)-u(n))=u(n)/[\sqrt(u(n)2+u(n))+u(n)], which tends to 1/2, so, by Cesaro-Stolz, the limit of u(n)/n is 1/2.


MATHCOUNTS
quadratics
algebra
quadratic formula

From now on I'll randomly post problems from the last year's MAML Olympiad questions that I liked (and appears also in MATHCOUNTS)

25.) Find the sum of all solutions to (x:^2:+3x+1)^(x:^2:+2x-8)=1

All you need to know is the Quadratic Formula and the basic rules of exponent and you're all set, which I think is learned by many in 8th grade or so.

Ah, you fell right into the trap my friend, I fell down that hole the first time too.

I'll give you a hint, but think about the problem hard for a while and it may dawn on you, so just don't look at the hint *audience: Then why write a hint!?*

Hint: [hide]additive inverse[/hide]

So close, yet you somehow got the signs mixed up somewhere...

Thats still not what I get, maybe you should show your solution :P

Although both didn't give their solution I see where they went wrong, just compare the two answers, see what they could have potentially tripped up on, then figure out what they did.

WARNING: The below is a message towards TripleM or whoever solved this already but is too old/experienced to post the answers themselves. All people who's still working on this, please don't read any further.

I believe what DaSweepa did was [hide]he got the solutions for x2+3x+1=1 and x2+2x-8=0, then added them together to get -5[/hide]

Judging from the fact that he forgot [hide]to solve for x2+3x+1=-1[/hide] I think ml2 did solve for that but instead of [hide]x=-1, -2[/hide] I think he got [hide]x=-1, 2[/hide]

Reply to Tare:

[hide]

You mean x=-1 and -2 right? Though x=-1 doesn't give an even value for the power so you discount it..

[/hide]

i did some snipping and editting to get us back on track


This is a NYCIML problem, I was wondering if anyone could help me with it?

Find all ordered pairs (x,y) of real numbers for which x^2+xy+x=14 and y^2+xy+y=28

Smiles,
lyrA

An alternative [hide="hint"]Try adding the two equations.[/hide]

Ooh, thank you, that makes sense now.  I am on the Blair math team, but my teacher didn't explain it really.  I think  understand now, thanks!

Smiles,
lyra

Ooh, thank you, that makes sense now.  I am on the Blair (w/brian lawrence) math team, but my teacher didn't explain it really.  I think  understand now, thanks!

Smiles,
lyra

[quote="Pakman2012"]I like JBL's method more, because it seems more... obvious to me at least.

$(x^2+y^2+2xy)+(x+y)=42$
$(x+y)^2+(x+y)=42$
$(x+y)(x+y+1)=42$
$6*7$ or $-7*-6$

$y=6-x$
$x^2+x(6-x)+x=14$
$x=2$
$y=4$

or

$y=-7-x$
$x^2+x(-7-x)+x=14$
$-6x=14$
$x=-7/3$
$y=-14/3$

EDIT: Looking back at the first equations... one is half the other.. hence y=2x[/quote]

Nice solution, but I think that you are forgeting two cases.

$y=7-x$
$x^2+x(7-x)+x=14$
$8x=14$
$x=7/4$
$y=21/4$
and
$y=-6-x$
$x^2+x(-6-x)+x=14$
$x=-14/5$
$y=-16/5$

Smiles,
lyra

Watch out--in the equation you're going off of, $(x+y)(x+y+1)=42$, the second factor has to be one more than the first.  That's why you can't have $7*6$ or $-6*-7$.  And if you plug in those last two answers you got, you'll find that they don't work in the original equations.

And as your edit suggests, it might be easier to factor the original equations right off the bat and notice that $y=2x$.  It then becomes a simple quadratic in one of the variables.  That's how I did it.


Hi Valentin,

First of all, congratulations on your site! It has been of great help in my mathematical skill development. Since I am familiar with the LaTeX language and I also have a lot  of documents written in MS Word before I learned this language,  I was wondering if you, or any other user of this site know of an efficient Word to LaTeX conversion tool. It does not have to be for free, but it would be neat if it had a reasonable price. However, this is not mandatory. Have you ever used such a tool?

It seems that no tool works perfectly.
You can have a try of Word2TeX...
Mathtype can convert the formula (maybe only formula, not the whole document) to TeX too.


The sequence of positive integers (1,1,1,1,2) has the property that the sum of any three numbers in the sequence is divisible by the sum of the other two. Is there a sequence of five distinct positive integers with this property?

Nope! :) To be more accurate, if you name the numbers $a_1\le a_2\le\ldots,\le a_5$, then we must have $a_3=a_4$.

We have $\displaystyle a_1+a_2+a_3\le \frac 32 (a_4+a_5)\Rightarrow a_1+a_2+a_3=a_4+a_5\ (*)$. Now $a_3+a_4|a_1+a_2+a_5=2(a_1+a_2)+a_3-a_4$, and if $a_3<a_4$ then that last number is $<2(a_1+a_2)\le 2(a_3+a_4)$, so $a_1+a_2+a_5=a_3+a_4$. Subtracting $(*)$ from the last relation gives $a_3-a_5=a_5-a_3$, which is absurd.


algebra
system of equations
algebra proposed

Find all real solutions $ (x,y,z)$ of the following system of equations:

$ xy\equal{}z\minus{}x\minus{}y$
$ xz\equal{}y\minus{}x\minus{}z$
$ yz\equal{}x\minus{}y\minus{}z$

hello, the following solutions i have found
$ x\equal{}\minus{}2,y\equal{}\minus{}2,z\equal{}0$
$ x\equal{}\minus{}2,y\equal{}0,z\equal{}\minus{}2$
$ x\equal{}\minus{}1,y\equal{}\minus{}1,z\equal{}\minus{}1$
$ x\equal{}0,y\equal{}\minus{}2,z\equal{}\minus{}2$
$ x\equal{}0,y\equal{}0,z\equal{}0$
Sonnhard.

Here my solution to get Dr Sonnhard's solutions:

[hide]
From the first equation we express $ z$ as $ x$ and $ y$: $ z \equal{} xy \plus{} x \plus{} y$.

Substituting in the second one we get: $ x(xy \plus{} x \plus{} y) \equal{} y \minus{} xy \minus{} x \minus{} y$ or $ x^2y \plus{} x^2 \plus{} 2xy \plus{} 2x \equal{} 0$. Factorising we get: $ x(y \plus{} 1)(x \plus{} 2) \equal{} 0$

So we have $ 3$ cases. Fist case: $ x \equal{} 0$.

From the first equation we get $ z \equal{} y$. From the last one we get: $ y^2 \plus{} 2y \equal{} 0$. The solution for this case are $ x \equal{} 0, y \equal{} 0, z \equal{} 0$ and $ x \equal{} 0, y \equal{} \minus{} 2, z \equal{} \minus{} 2$

Second case: $ x \equal{} \minus{} 2$

From the first equation we get: $ y \plus{} z \plus{} 2 \equal{} 0$. From the last one we get: $ yz \equal{} \minus{} 2 \minus{} y \minus{} z \equal{} 0$ So or $ y \equal{} 0$ or $ z \equal{} 0$. This will give us the permutation of the solution of first case: $ x \equal{} \minus{} 2, y \equal{} \minus{} 2, z \equal{} 0$ and $ x \equal{} \minus{} 2, y \equal{} 0, z \equal{} \minus{} 2$.

The last case: $ y \equal{} \minus{} 1$.

From the first equation we get $ z \equal{} \minus{} 1$ and form the last one $ x \equal{} \minus{} 1$. So the solution are $ x \equal{} 0, y \equal{} \minus{} 2, z \equal{} \minus{} 2$ and its permutations, $ x \equal{} y \equal{} z \equal{} 0$ and $ x \equal{} y \equal{} z \equal{} \minus{} 1$[/hide]

[hide]
$ xy \equal{} z \minus{} x \minus{} y \Longleftrightarrow xy \plus{} x \plus{} y \plus{} 1 \equal{} z \plus{} 1 \Longleftrightarrow (x\plus{}1)(y\plus{}1) \equal{} (z\plus{}1)$
$ xz \equal{} y \minus{} x \minus{} z \Longleftrightarrow xz \plus{} x \plus{} z \plus{} 1 \equal{} y \plus{} 1 \Longleftrightarrow (x\plus{}1)(z\plus{}1) \equal{} (y\plus{}1)$
$ yz \equal{} x \minus{} y \minus{} z \Longleftrightarrow yz \plus{} y \plus{} z \plus{} 1 \equal{} x \plus{} 1 \Longleftrightarrow (y\plus{}1)(z\plus{}1) \equal{} (x\plus{}1)$

Multiply all three equations to get $ [(x\plus{}1)(y\plus{}1)(z\plus{}1)]^2 \equal{} (x\plus{}1)(y\plus{}1)(z\plus{}1)$.

Therefore, $ (x\plus{}1)(y\plus{}1)(z\plus{}1) \equal{} 0$ OR $ (x\plus{}1)(y\plus{}1)(z\plus{}1) \equal{} 1$

If $ (x\plus{}1)(y\plus{}1)(z\plus{}1) \equal{} 0$:
One of $ x,y,z$ is $ \minus{}1$. WLOG, let $ z \equal{} \minus{}1$.
Then, equations 2 and 3 become: $ y\plus{}1 \equal{} 0$ and $ x\plus{}1 \equal{} 0$.
Thus, $ (x,y,z) \equal{} (\minus{}1,\minus{}1,\minus{}1)$ is the only solution in this case.

If $ (x\plus{}1)(y\plus{}1)(z\plus{}1) \equal{} 1$:
Then, $ (x\plus{}1)(y\plus{}1) \equal{} \dfrac{1}{(z\plus{}1)} \equal{} (z\plus{}1)$, so $ z\plus{}1 \equal{} \pm 1$. Similarly, $ x\plus{}1 \equal{} \pm1$ and $ y\plus{}1 \equal{} \pm1$

Clearly an even number of $ x\plus{}1,y\plus{}1,z\plus{}1$ must equal $ \minus{}1$ otherwise $ (x\plus{}1)(y\plus{}1)(z\plus{}1) \equal{} \minus{}1$.

So, the only possible solutions in this case are:
$ (x\plus{}1,y\plus{}1,z\plus{}1) \equal{} (1,1,1); (1,\minus{}1,\minus{}1); (\minus{}1,1,\minus{}1); (\minus{}1,\minus{}1,1)$,
all of which satisfy the system of equations.

Thus, the solutions are:
$ (\minus{}1,\minus{}1,\minus{}1)$
$ (0,0,0)$
$ (0,\minus{}2,\minus{}2)$
$ (\minus{}2,0,\minus{}2)$
$ (\minus{}2,\minus{}2,0)$
[/hide]


Fie $x_0>0,p>1$, p numar natural si definim sirul $x_{n+1}=\sqrt[p]{{px_n  + 1}} - 1,\forall n \in N.$
Calculati   $\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\limits_{k = 0}^n {x_k } }}{{\ln n}},\mathop {\lim }\limits_{n \to \infty } n^{x_n } ,\mathop {\lim }\limits_{n \to \infty } x_n ^{x_n } $

(Am reusit sa dem. ca $x_n \to 0$ )

It seems this one was posted by Lagrangia or Grobber

$\frac{1}{x_n^2}-\frac{1}{x_{n-1}^2}$ tends to $\frac{1}{p-1}$

With Cesaro theorem $x_n\sim \frac{1}{\sqrt{(p-1)n}}$

...


inequalities induction calculus factorial

Show that

    (1/sqrt(4n+1))<(1.3.5...2n-1)/(2.4.6...2n)<=(1/sqrt(3n+1))
Please do not use induction method.

I don't think calculus is better than induction, certainly not in the beginner's corner :)

a way of brute forcing these problems involving factorials is Stirlings formula:
n! is approximately equal to n^(n+1/2)*e^(-n)*sqrt (2*pi) (and within 1% error i think).
that expression is just equal to
(2n-1)! / 2^2n * (n!)^2
which should be nicer after stirlings ... then it should be trivial (tho this isnt really appropriate for beginners corner  :D ).

how you are going to proof an inequality with stirling's formula??? it is something that says how things behave near infinity, but there may be a finite number of exceptions...if you want to proof it with stirling, you need an error term bound, and at that time it is really :? . (and the 1% error bound is not always, only from a certain point on, try n=1 in stirling, the error is greater, above 7%)

Peter


The function $ f$ on the non-negative integers takes non-negative integer values and satisfies $ f(4n) \equal{} f(2n) \plus{} f(n), f(4n\plus{}2) \equal{} f(4n) \plus{} 1, f(2n\plus{}1) \equal{} f(2n) \plus{} 1$ for all $ n$. Show that the number of non-negative integers $ n$ such that $ f(4n) \equal{} f(3n)$ and $ n < 2m$ is $ f(2m\plus{}1)$.

tdl could you please post the solution to it?


Let $ a_1,b_1,c_1$ are positive integers such that $ a_1^2 \plus{} b_1^2 \equal{} c_1^2$ ,and define three sequences  $ \{a_n\},\{b_n\},\{c_n\}$as:

$ a_{n \plus{} 1} \equal{} | \minus{} a_n \minus{} 2b_n \plus{} 2c_n|$,

$ b_{n \plus{} 1} \equal{} | \minus{} 2a_n \minus{} b_n \plus{} 2c_n|$,

$ c_{n \plus{} 1} \equal{} \minus{} 2a_n \minus{} 2b_n \plus{} 3c_n$.　$ (n \equal{} 1,2,3,...)$

(1)Show that $ a_n^2 \plus{} b_n^2 \equal{} c_n^2$ and $ c_n\ge c_{n \plus{} 1} > 0$ .

(2)If there exists $ m\in \mathbb N$ such that $ c_m > c_{m \plus{} 1} \equal{} c_{m \plus{} 2}$,then determine the ratio $ a_m: b_m: c_m$.

[hide="hints for 1"]Pythagorean theorem and Triangle Inequality[/hide]

This is do-able.

Well I know that in a^2 + b^2 = c^2
a = v^2 - u^2
b = 2uv
c = u^2 + v^2

for any u, v that are integers
(does this help?)


topology
advanced fields
advanced fields unsolved

I am stuck in the following problem.

Prove that a point $ x \in X$ not belonging to a closed set $ A$ has a non-zero distance from $ A$. That is, we have $ \inf_{y \in A} \| x\minus{}y \| > 0$.

Any help please?Thanks

Assume opposite, that $ \inf_{y\in A}\| x \minus{} y\| \equal{} 0$. That means for every $ \epsilon > 0$ exists $ y\in A$ so that $ \|x \minus{} y\| < \epsilon.$
Form a sequence $ \epsilon_n : \equal{} \frac {1}{n}$ and $ n\in\mathbb{N}.$ Now we have that exist sequence $ y_n\in A$ which converge to $ x$ $ (y_n\to x).$ Because $ A$ is closed it follows that $ x\in A,$ but we assumed that $ x\notin A.$

assume for a contradiction that $ \inf_{y\in A} \|x\minus{}y\|\equal{}0$. this means there are points of $ A$ arbitrarily close to $ x$, and thus there is a sequence of points in $ A$ that converges to $ x$. but since $ A$ is closed, we get $ x\in A$, a contradiction.

x is in "the complement of A", which is open.

There is a ball B(x, a), centered at x, with radius a > 0, such that B(x, a) is a subset of the complement of A.

This means that d(x, A) is at least equal to a > 0.


This should be easy. ;)

Prove
$ (a^7 \plus{} b^7 \plus{} c^7)(a^2 \plus{} b^2 \plus{} c^2) >\equal{} (a^4 \plus{} b^4 \plus{} c^4)(a^5 \plus{} b^5 \plus{} c^5)$

using [b]ONLY Cauchy Schwarz[/b] Inequality

Lets suppose that $ a,b,c > 0$...I can do it with holder...Let $ s_n \equal{} a^n \plus{} b^n \plus{} c^n$. Then
\[ (s_7)(s_7)(s_2)(s_2)(s_2)\ge (s_{\frac {7 \plus{} 7 \plus{} 2 \plus{} 2 \plus{} 2}{5}})^5 \equal{} (s_4)^5
\]
Likewise,
\[ (s_2)(s_2)(s_7)(s_7)(s_7)\ge (s_{\frac {2 \plus{} 2 \plus{} 7 \plus{} 7 \plus{} 7}{5}})^5 \equal{} (s_5)^5
\]
Multiplying these together gives
\[ (s_7)(s_7)(s_2)(s_2)(s_2)(s_2)(s_2)(s_7)(s_7)(s_7) \equal{} (s_7)^5(s_2)^5\ge (s_4)^5 (s_5)^5
\]
Then taking the $ 5$th root of both sides finishes the problem. Equality occurs when $ a \equal{} b \equal{} c$

I feel like there should be an obvious cauchy solution but I don't see it at the moment.

Let $ s_n$ be defined as in the post above. By BCS, we have $ s_{n\plus{}1}\cdot s_{n\minus{}1}\ge (s_n)^2$. Multiplying these inequalities for $ n\equal{}3,4,5,6$ gives us the desired result.


Let $ABC$ is an rightangled triangle with $AB$ as hypotenuse. Let $D$ is a point on $AB$. Points $O_{1},  O_{2}$ are the circumcenters of the triangles $ACD, BCD$ respectively. Let $E=AO_{2}\cap{BO_{1}}$. Prove that   $\angle{BCE}=\angle{ACD}$.

If I remember well it's the same as this one:
http://www.mathlinks.ro/Forum/viewtopic.php?t=18550

Here it follows a sketch of a different proof.

If J=AO_1^B_O2, let K be the antipode of J wrt c(ABC). Then if F=AB^CK and E'=AO_2^CK, as HB//AK and O_2 is the midpoint of BH, where H is the antipode of B on c(O_2) and lies also on AC, it holds that {C,F;E',K} is an harmonic quadruple.
Similarly if I is the antipode of A wrt c(O_1) and lies also on BC, {BC,BF;BO_1,BK} is an harmonic quadruple.
Therefore, by the unicity of the fourthy harmonic, CK,AO_2 and BO_1 are concurrent.
From here, by easy angle chasing one can derive the thesis.


algebra
polynomial
abstract algebra
group theory
superior algebra
superior algebra unsolved

Given a prime $p$, show that there is an irreducible polynomial $P$ of $\mathbb{Q}[X]$ with degree $p$ and such that for any root $x_{0}$ of $P$, all the other roots of $P$ lie in $\mathbb{Q}[x_{0}]$.

Let $L$ be an abelian extension of $\mathbb Q$ whose degree is divisible by $p$ (for instance, we could take it to be the splitting field of $x^{p^2}-1$). $G=\mbox{Gal}(L/\mathbb Q)$ has a subgroup $H$ of order $\frac{[L: \mathbb Q]}p$, so the subfield $K$ of $L$ fixed by $H$ is a Galois extension of $\mathbb Q$ of degree $p$.

Because $K/\mathbb Q$ is finite and separable, it's simple: $K=\mathbb Q(\alpha)$ for some $\alpha$. The degree of the extension being $p$, this must be the degree of the minimal polynomial of $\alpha$. Since the extension $K/\mathbb Q$ is normal, all conjugates of $\alpha$ lie in $K$, and if $\beta$ is such a conjugate, then $\mathbb Q(\beta)$ is a subfield of $K$ different from $\mathbb Q$. Since $p=[K: \mathbb Q]=[K: \mathbb Q(\beta)][\mathbb Q(\beta): \mathbb Q]$ is a prime and $[\mathbb Q(\beta): \mathbb Q]>1$, we get $\mathbb Q(\beta)=K$, and we're done.

My solution is the same modulo a few details:

[quote="grobber"]Let $L$ be an abelian extension of $\mathbb Q$ whose degree is divisible by $p$ (for instance, we could take it to be the splitting field of $x^{p^2}-1$).[/quote]

What do you mean exactly by "abelian" ? Do you mean Galois and abelian ? I took the splitting field of $X^{p}+2$ (whose degree is a power of $p$ since this polynomial is irreducible). Is that also correct ?

[quote="grobber"] Since the extension $K/\mathbb Q$ is normal, all conjugates of $\alpha$ lie in $K$ [/quote]

What do you call a "conjugate" in this context ?

Thanks


[hide="my solution"]$ f(n) = \sum_{k = 0}^{\lfloor \frac {n}{2} \rfloor}_{n - k}C_{k}$

[b]Q1[/b]$ f(6) = 1 + 5 + 6 + 1 = 13$
[b]Q2[/b]$ f(f(6)) = f(13) = 1 + 12 + 55 + 120 + 126 + 56 + 7 = 377$
[b]Q3[/b]When $ n > 3$, $ \lfloor \frac {n}{2} \rfloor > 1$, so $ f(n) > 1 + (n - 1) = n$, so it's easy to check that $ n = 1,2,3$ satisfies the condition. The answer for Q3 is $ 3$
[b]Q4[/b]Obviously $ n > 4$, so $ f(n) \ge 1 + (n - 1) + \frac {(n - 2)(n - 3)}{2!}$, so we must have $ (n - 2)(n - 3) = 2$, i.e. $ n = 3$, contradiction. So the only solution is $ n=4$ after checking, answer $ =1$[/hide]

[quote="eureka-123"]Ans2 377
Ans3 2 :(  :(  :([/quote]

For Ans 1 and Ans 4 I made some mistakes , now corrected.

For Ans 3 I don't think you are correct as $ 1\equal{}1, 2\equal{}2\equal{}1\plus{}1, 3\equal{}2\plus{}1\equal{}1\plus{}2\equal{}1\plus{}1\plus{}1$

[quote="eureka-123"]let f(n) denote number of ways the positive integer n can be expressed as sum of 1's and 2's
for eg: f(4)=5  => 4=1+1+1+1=1+1+2=1+2+1=2+1+1=2+2

Q1 Find value of f(6)
Q2 Find value of f(f(6))
Q3 Find the number of solutions to equation f(n)=n  where n belongs to set of natural numbers
Q4 Find the number of solutions of equation f(n)=n+1 where n belongs to set of natural numbers[/quote]
$ 1\plus{}1\plus{}2$ is the same as $ 2\plus{}1\plus{}1$ .Why you caunt this as two different cases.

@ enndb0x
because it is permutation of cases :lol:

@ stephencheng
btw can u guide me how to attempt these type of questions........i am weak in algebra a bit........ :(  :(  :(


A dealer packages marbles in two different size box sizes.  One size holds 5 marbles and the other holds 12 marbles.  If the dealer packaged 99 marbles and used more than 10 boxes, how many boxes of each size did he use?

[hide]The greatest multiple of 12 less than 99 is 96.  But this is not a multiple of 5 less than 99. 84 is a multiple of 5 less than 99, but this will use only 7+3=10 boxes.  The other multiple that will work is 24.  So he used [b]2[/b] boxes of 12 and  [b]15[/b] boxes of 5.[/hide]


geometry
calculus
integration
calculus computations

Sketch and find the area bounded by the region.

x=y^2

x=8-2y

First find intersection points
$y^{2}=8-2y$
$y=-4,2$
Now compute the integral
$\int_{-4}^{2}\left(\left(8-2y\right)-\left(y^{2}\right)\right)dy=\left[8y-y^{2}-\frac{y^{3}}{3}\right]_{-4}^{2}=\boxed{36}$
(I may have made some mistake, so double-check what I wrote)


limit
integration
logarithms
real analysis
real analysis unsolved

Find $\lim_{n\to\infty}n\cdot\int_{0}^{1}\ln (1+x^{n}) dx$.
[hide="What I got"] $\frac{\pi^{2}}{12}$ [/hide]

Let $x_{n}=n\int_{0}^{1}\ln(1+x^{n})dx$.

Calculate:

$\lim_{n\rightarrow \infty}n\left(x_{n}-\frac{\pi^{2}}{12}\right)$.

Here is my solution.

Using the substitution $x^{n}=y$ we get that

$x_{n}=\int_{0}^{1}\frac{\ln(1+y)}{y}y^{1/n}dy\rightarrow \int_{0}^{1}\frac{\ln(1+y)}{y}dy=\frac{\pi^{2}}{12}$. The convergence follows based on the Bounded Convergence theorem.

Now,

$n(x_{n}-\frac{\pi^{2}}{12})=\int_{0}^{1}\frac{\ln(1+y)}{y}\frac{y^{1/n}-1}{\frac{1}{n}}dy\rightarrow \int_{0}^{1}\frac{\ln(1+y)\ln y}{y}dy=-\frac{3}{4}\zeta(3)$, where the convergence follows based on Lebesque convergence theorem.


Let $ n$ be a positive integer satisfying the following property: If $ n$ dominoes are placed on a $ 6 *6$ chessboard, with each domino covering exactly 2 unit squares,  then one can place another domino on the board without moving the $ n$ dominoes already there.
Determine the maximum value of $ n$.

11.
If the dominoes are arranged to allow the maximum number of 1*1 gaps, 12 separate 1*1 gaps are possible. In this configuration, 12 dominoes are used. Therefore, if 11 dominoes are used, there must be a way to add another dominoe.

it's easy to see that we can place 12 dominoes so that there's no room for a 13th: in each of columns 1, 3, 5, put 2 vertical dominoes covering pairs (2,3) and (5,6), respectively (2, for example, referring to the second square in the column counted from the top), and in each of columns 2, 4, 6, cover (1,2) and (4,5).

so $ n\leq 11$.

now to show that 11 works. if we place 11 dominoes on the board, 14 squares are uncovered, 7 of each color (assuming standard checkerboard coloring). we wish to show that two of these uncovered squares are adjacent. assume the contrary. then each of the white squares that's next to at least one of the 7 uncovered black squares must be covered. it's pretty clear that these 7 black squares will border a minimum number of white squares when they're along the boundary of the board. but even in that case, the minimum number of white squares they border is 12 (when the black squares are consecutive around the boundary of the board), leaving at most 6 uncovered white squares - a contradiction.


$N$ dwarfs of heights $1,2,...,N$ are arranged in a circle. For each pair of neighbouring dwarfs the positive difference between the heights is calculated: the sum of these $N$ differences is called the "total variance" $V$ of the arrangement.
Find (with proof) the maximum and minimum possible values of $V$.

[hide="Minimum"]
Let $V_{m}$ be the minimum $V$ when $N=m$. Clearly $V_{2}=2$. Now, let's examine what happens if we have $N$ dwarves in optimal order and add another:
Let's say we add the new dwarf in between dwarves of heights $a$ and $a+b$. Let the new dwarf have height $a+b+c$.
$V_{n+1}= V_{n}+(a+b+c-a)+(a+b+c-a+b)-b=V_{n}+2c$. So we need to minimize $c$, which is the difference between the new dwarf's height and that of his greatest neighbour. If we put the new dwarf beside the second tallest dwarf, this will be $1$. Hence, $V_{n+1}=V_{n}+2$. From here we can see that $V_{n}=2(n-1)$.
[/hide]


geometry
perimeter
trigonometry
ratio
AMC
AIME
Pythagorean Theorem

Triangle $ABC$ is isosceles, with $AB=AC$ and altitude $AM=11.$  Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$  Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers.  Find $a+b.$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=450[/img]

[hide]
We start by setting $\arctan{x}=3*\arctan{x/11}$
Seeing this by angle addition formulas we get:
$tan(\arctan{x}-\arctan{x/11})=10x/(11+x^2)$
$tan(2\arctan{x/11})=22x/(121-x^2)$
We apply triangle addition formula with triangles and we get:
$10x(121-x^2)-22x(11+x^2)=0$
Solving we get $x=11/2$
Therefore the perimeter is $11+\sqrt{605}$
We get $a+b=616$.
[/hide]

[quote="shobber"]Who can find a geometrical solution for this problem?[/quote]

I couldn't, I think it would be A LOT tougher to do it that way.
[hide]
Call the angle $\theta$, so we know that $\tan{\theta}=\frac{x}{11}$, and $\tan{3\theta}=x \Longrightarrow \frac{3\tan{\theta}-\tan^3{\theta}}{1-3\tan^2{\theta}}=x$, after substituting in $\frac{x}{11}$ for $\tan{\theta}$, we have $x=\frac{11}{2}$, so the perimeter is $\boxed{\sqrt{605}+11}$[/hide]

I have a geometric way.  But, I don't get all the credit for it...I did it exactly like Shobhit at first, but then when I checked my answer I caught a glimpse of one of the alternate (geometric) solutions ;)

[hide="Pure Geometry"]
Let $\angle CAM=\theta$ so that $\angle CDM=3\theta.$  Now, we can see pretty easily that $\angle ACD=2\theta.$  So drawing in its bisector, $\overline{CP},$ is natural since it gives us an isosceles triangle.  Let $x=AP=PC.$  So then by the Pythagorean theorem we have

\begin{eqnarray*} CM &=& \sqrt{x^2-(11-x)^2}\\ &=& \sqrt{22x-121}\\ DC &=& \sqrt{22x-120}\\ AC &=& \sqrt{22x} \end{eqnarray*}

The angle bisector theorem then gives

\begin{eqnarray*} AP\cdot CD &=& DP\cdot AC\\ x\sqrt{22x-120} &=& (10-x)\sqrt{22x}\\ \Leftrightarrow 11x^3-60x^2 &=& 11x^3-220x^2+1100x\\ \Leftrightarrow 20x(8x-55) &=& 0\\ &\Rightarrow& x=55/8 \end{eqnarray*}

So then $AC=\frac{\sqrt{605}}2,$ $CM=\frac{11}2$, and the perimeter is $11+\sqrt{605}.$  Our answer is 616.[/hide]

[hide="One minute solution... surprising that nobody saw it!"]
Let $BM=MC=x$ and $AB=AC=y$. By Stewart's Theorem, $2y^2x=11^2(2x)+11x^2$, and since $x\ne 0$, we get \[ y^2=121+\frac{11}{2}x.  \] We also have \[ y^2=11^2+x^2  \] by the Pythagorean Theorem, so $x=11/2$ and $y=\sqrt{605}/2$. Hence, the perimeter is $2x+2y=11+\sqrt{605}$, so the answer is $\boxed {616}$.

Note: the relation between the angles was not required!
[/hide]

[quote="chess64"][hide="One minute solution... surprising that nobody saw it!"]
Let $BM=MC=x$ and $AB=AC=y$. By Stewart's Theorem, $2y^{2}x=11^{2}(2x)+11x^{2}$, and since $x\ne 0$, we get \[y^{2}=121+\frac{11}{2}x.\] We also have \[y^{2}=11^{2}+x^{2}\] by the Pythagorean Theorem, so $x=11/2$ and $y=\sqrt{605}/2$. Hence, the perimeter is $2x+2y=11+\sqrt{605}$, so the answer is $\boxed{616}$.

Note: the relation between the angles was not required!
[/hide][/quote]
lol your solution just makes mine look soo dumb... I extended D to P. Used Menelaus twice, worked out some ratios and eventually used Stewart's theorem. I should've used Stewarts theorem from the start  :rotfl:

Something missing here... I didn't know stewarts theorem..so i looked it up...now... if BM=MC = x then

2x(y^2) = (11)^2(2x) + 2(x^3)

check this out...[url]http://mathworld.wolfram.com/StewartsTheorem.html[/url]

Am i missing out something??? :huh:

[quote="Leao"]I don't understand what you are trying to say...These AIME problems always get done by new and different people; you may have done this problem before but this problem is new to me and many others...and isn't this the thread for the discussion of this problem?[/quote]
Ok, maybe you're right...

[quote="chess64"][hide="One minute solution... surprising that nobody saw it!"]
Let $BM=MC=x$ and $AB=AC=y$. By Stewart's Theorem, $2y^{2}x=11^{2}(2x)+11x^{2}$, and since $x\ne 0$, we get \[y^{2}=121+\frac{11}{2}x.\] We also have \[y^{2}=11^{2}+x^{2}\] by the Pythagorean Theorem, so $x=11/2$ and $y=\sqrt{605}/2$. Hence, the perimeter is $2x+2y=11+\sqrt{605}$, so the answer is $\boxed{616}$.

Note: the relation between the angles was not required!
[/hide][/quote]

how did you get 11x^2 on the right side according to stewart's theorem? does $11x^{2}= mn^{2}+nm^{2}$??? am i missing something?

$m(\widehat{BAM})=\phi$ $\Longrightarrow$ $BM=\boxed{\tan 3\phi =11\cdot\tan\phi}$ $\Longrightarrow$
$\tan \phi=\frac{1}{2}$ $\Longrightarrow$ $BC=AM=11\ ,\ AB=AC=\frac{11\sqrt 5}{2}$ $\Longrightarrow$
the perimeter of the triangle $ABC$ is $P=(1+\sqrt 5)\cdot 11$ a.s.o.

[quote="chess64"][hide="One minute solution... surprising that nobody saw it!"]
Let $BM=MC=x$ and $AB=AC=y$. By Stewart's Theorem, $2y^{2}x=11^{2}(2x)+11x^{2}$, and since $x\ne 0$, we get \[y^{2}=121+\frac{11}{2}x.\] We also have \[y^{2}=11^{2}+x^{2}\] by the Pythagorean Theorem, so $x=11/2$ and $y=\sqrt{605}/2$. Hence, the perimeter is $2x+2y=11+\sqrt{605}$, so the answer is $\boxed{616}$.

Note: the relation between the angles was not required!
[/hide][/quote]

uhhh....you assumed that $BC=11$

[quote="Virgil Nicula"]No ! $\tan\phi=\frac{1}{2}$ $\Longrightarrow$ $AM=2\cdot BM$, i.e. $11=AM=BC$.[/quote]

i know that $\tan\phi=\frac{1}{2}$ is true, but his solution does not say [i]why[/i] it is true, furthermore, once you have that piece of information, the problem is solved already, it directly follows that $MB=\frac{11}{2}$, $AB=...$, his solution with stewart's theorem is just complicating things

[quote="chess64"]Note: the relation between the angles was not required![/quote]This is definately not true.  :wink:

Is there a way to do this with  trig formulas? I don't always remember stewart's or menalaus's theorem.  :(

Oh snap I screwed up the equation. But I still got the right answer :D


Do, Re, Mi, Fa, So, La, Ti, want to have their pictures taken with at least one of the other six. However, Do and Ti as well as Mi and Fa do not go along well with each other,and because of this, they do not want to be seen in any picture with the other. How many pictures are possible?

there are total of $ 7x6/2 \equal{} 21$ possiblities.
subtract the two that are incompatible, is 19 possible pictures

EDIT: ignore

[quote="RobRoobiks"]there are total of $ 7x6/2 \equal{} 21$ possiblities.
subtract the two that are incompatible, is 19 possible pictures[/quote]

It's [i]at least[/i] one.  Meaning that you are going to have to complementary count or do casework.

2 people photo 2_C_7 =21  Out: (1 7)  1 (3 4) 1
3 people photo 3_C_7 =35  Out: (1 7 x) 5 (3 4 x) 5
4 people photo 4_C_7 =35  Out: (1 7 x x) 10 (3 4 x x) 10, add 1 since (1 7 3 4) repeates
5 people photo : The two not taking has to be 1 from (1 7) and one from (3 4), a total of 4 combinations
6 people photo : None eithre 17 or 34 will always be together
7 people photo : None

Summing 21+35+35+4+1-10-10-5-5= 66 pictures

[hide]Hmm... This is what I'm getting.

There are a total of
\[ \binom{7}{2} \plus{} \binom{7}{3} \plus{} \dots \plus{} \binom{7}{7} \equal{} 2^7 \minus{} 7 \minus{} 1 \equal{} 128 \minus{} 8 \equal{} 120\]
ways to have a picture with at least two of them, without any restrictions.

Now we count the number of ways Do and Ti are together:

We must have
\[ \binom{5}{0} \plus{} \binom{5}{1} \plus{} \dots \plus{} \binom{5}{5} \equal{} 2^5 \equal{} 32\]
pictures, because once they are chosen, we have 5 people left, out of whom we can form committees that can accompany Do and Ti.

Similarly, the number of arrangements with Mi and Fa is also $ 2^5 \equal{} 32.$

Now we must subtract the ones with both Do and Ti, and Mi and Fa in it, because they are getting counted twice.

Once all the four are in the picture, there are 3 others left, with whom we can form committees to accompany the four in the picture in
\[ \binom{3}{0} \plus{} \binom{3}{1} \plus{} \binom{3}{2} \plus{} \binom{3}{3} \equal{} 2^3 \equal{} 8\]
ways.

Adding the initial two cases, we have 32+32=64. And subtracting this, we are left with 64-8=56.

We must subtract this count from the total of 120, which gives us 120-56=64.
[/hide]
@adithya, Principle of Inclusion-Exclusion is also a good candidate for "at least" problems.

I found my error: I forgot to subtract the initial two from
2 people photo 2_C_7 =21 Out: (1 7) [b]1[/b] (3 4)[b] 1 [/b]

Possible groups(7,0),(2,5),(3,4)(3,2,2)
=7C7+7C2+7C4 +(7C3X4C2x2C2)=7+21+35+35x6=7x(1+3+5+5x6)=7x(39)=2 73

Possibilities to be subtracted-
1.(7,0)=1
2.(2,5)= 1(when both in the group of two) +5C3(when both are in group of 5)
=1 +10=11
3.(3,4)=5C1(when both are in group of 3) +5C2(when both are in group of 4)=15
4.(3,2,2)=5C1(when both are in group of 3) +1(when both in group of 2)=6
total=1+11+15+6=33
required cases=273-33=240

For simplicity, I am reframing the question as [b]there are seven persons A, B, C, D, E, F and G who want to have picture with atleast one of the other six. However A and B as well as C and D do not go along well with each other and hence cannot be photographed together. How many pictures are possible?[/b]
Here is the beauty solution:

From A and B , selection can be made in 3 ways (A or B or none of them)

From C and D, selection can be made in 3 ways ( C or D or none of them)

For E, F, G selection can be made in 23 ways ( 2 ways for each; selected or not selected)

Now out of total {3^2}{2^3} = 72 cases I need to subtract the cases when only one person is selected ( i.e. 7 ways; any one of the seven) and when no person is selected ( i.e. 1 way only).

So, in all, total pictures possible = 72 - 7 - 1 = 64.  :)
It can be found here also:
[url]http://totalgadha.com/mod/forum/discuss.php?d=7405#74200[/url]

[hide="please see this"] can anybody comment on my answer :) [/hide]


geometry
3D geometry
sphere
calculus
calculus computations

Given a sphere $ S: x^2 \plus{} y^2 \plus{} (z \minus{} 1)^2 \equal{} 1$ in $ xyz$ space and a line $ L: 2x \plus{} 3y \equal{} 1$ in $ x\minus{}y$ plane.Let $ \pi$ be the plane which is tangent to the sphere $ S$ and intersects the line $ L$ with $ x \minus{} y$ plane. Find the intersection point of the plane $ \pi$ and the sphere $ S$.

The equation $ (2x\plus{}3y\minus{}1)\plus{}kz\equal{}0$ ($ k$ is an arbitrary real number) denotes the plane-family that intersects $ L$, except $ z\equal{}0$. The plane $ \pi_1: {z\equal{}0}$ is tangent to $ S$,and the tangent point is $ (0,0,0)$.

 Now, let's think about $ \pi$ is not $ z\equal{}0$.
The condition of tangency : $ \frac{\left|k\minus{}1\right|}{\sqrt{2^2\plus{}3^2\plus{}k^2}}\equal{}1$. Here, we get $ k\equal{}\minus{}6$.

 We have to find the intersection point of $ \pi_2: {2x\plus{}3y\minus{}6z\equal{}1}$ and $ S$. We know the vecter $ (0,0,1)$ to the intersection point, it is the normal vecter of the plane $ \pi_2$. So, the point we want to find is $ (0,0,1)\minus{}(2t,3t,\minus{}6t)$. Because this point is on the sphere $ S$, we get $ t\equal{}\frac{1}{7}$. :  finally, $ \left(\frac{2}{7}, \frac{3}{7}, \frac{1}{7}\right)$ is the point.

 Conclusion : the answer is $ (0,0,0)$ and $ \left(\frac{2}{7}, \frac{3}{7}, \frac{1}{7}\right)$.

That's almost correct. :) $ (0,\ 0,\ 0)$ wasn't the answer.

My Solution:

Let $ T(a,\ b,\ c)$ be the tangent point and $ C(0,\ 0,\ 1)$ be the center of the Sphere $ S$. The equation of the tangent plane is $ ax \plus{} by \plus{} (c \minus{} 1)z \equal{} 1$, so the line of intersection is $ ax \plus{} by \equal{} c$, which coincides with $ L: 2x \plus{} 3y \equal{} 1$. Thus we have $ a \equal{} 2k,\ b \equal{} 3k,\ c \equal{} k\ (k\neq 0)$.Since $ T$ is on $ S$, we have $ a^2 \plus{} b^2 \plus{} (c \minus{} 1)^2 \equal{} 1\Longleftrightarrow 4k^2 \plus{} 9k^2 \plus{} (k \minus{} 1)^2 \equal{} 1\ \therefore k \equal{} \frac17\ \because k\neq 0$, yielding $ T\left(\frac 27,\ \frac 37,\ \frac 1 7\right)$

$ (0,0,0)$ is the trivial answer. The plane $ z\equal{}0$ intersects line $ L$ and tangent to sphere $ S$.

 Your tangent plane $ ax\plus{}by\plus{}(c\minus{}1)z\equal{}1$ cannot denote the plane $ z\equal{}0$.


calculus
derivative
analytic geometry
graphing lines
slope
quadratics
calculus computations

find the equations of the two vertical tangents to x^2+y^3=3xy

[hide]
Deriving implicitly, we get
$ 2x \plus{} 3y^2\frac{dy}{dx} \equal{} 3x\frac{dy}{dx} \plus{} 3y$

$ \frac{dy}{dx} \equal{} \frac{3y\minus{}2x}{3y^2 \minus{} 3x}$

We wish to find the place where $ 3y^2 \minus{} 3x \equal{} 0$, since that is where the slope is infinite, or vertical.

So, we have $ y^2 \minus{} x \equal{} 0$, or $ y^2 \equal{} x$ Replacing $ y^2$ for x in the original equation, we have
$ y^4 \plus{} y^3 \equal{} 3y^3$
So, we have $ y \equal{} 2$

Finding x by replacing y with 3 in the original equation, we obtain the quadratic
$ x^2 \minus{} 6x \plus{} 8 \equal{} 0$
$ (x\minus{}4)(x\minus{}2) \equal{} 0$

So, the points are

$ \boxed{(2,2);(4,2)}$
[/hide]


When expanding the expression $ (x+y)^n$ for a non-integer n, the series converges iff $ \abs{x} > \ab{y}$, But how does this make sense, since x+y = y+x, but switching the two values around and expanding can change the convergence?  :huh:

Is this just another odd behavior of infinity?

Unlike the standard Binomial Theorem, the Extended version is no longer symmetric in x and y. As your index of summation goes to infinity, the exponent on one variable goes to infinity and the exponent on the other variable goes to -infinity.

If you choose x and y so that the exponent on the smaller one is increasing and the exponent on the larger one is decreasing, then the sum converges to the desired value. If you switch them, then your successive terms are getting bigger and bigger, and there's no convergence.

Yeah, I just find it weird that it's not symmetric. For example, if one wanted to use the extended binomial theorem to estimate $ (1/4.127)^{4.5}$, $ (.127 \plus{} 4)^{\minus{}4.5}$ should equal $ (4 \plus{} .127)^{ \minus{} 4.5}$, yet one converges and the other doesn't...I guess for this one, you would choose the divergent one, but still... it's weird.

I mean the two expressions that you listed are equal. It's just the summation that's not symmetric -- you have to pick one variable to give rising exponents and the other variable to give falling exponents.

$ (1+x_*)^r \& =\sum_{k=0}^\infty {r \choose k} x_*^{r-k}$ converges for $ |x_*| < 1$. That makes sense, right? Otherwise it would blow up to infinity. Now, you can write any general $ (x+y)^r$ in this form by pulling out a factor so that $ |x_*| < 1$.


Most of the time sums like that can be solved simply by finding a way to telescope it.


algebra
function
domain
geometry
3D geometry
Ring Theory
number theory

find all x,y in integer such that :
       x^2+4=y^3

[quote="nhat"]i don't understand what you say[/quote]
In my opinion, he advise you to work in the euclidean domain $\mathbb{Z}[i]$.
_______________
blang

$x^2+4=(x+2i)(x-2i)$ let an irreducible $p$ in $Zi$ divide both factors, then it divides $4i$, so its norm divides 4, so we get that $2$ divides $y^3$, and mod 4 it is impossible. Now both factors are cubes in $Zi$, finish it this way:

$(a+bi)^3=a^3-3b^2a+3a^2bi-b^3i$ so we have: $2=3a^2b-b^3$ so $b=2$, or $b=+-1$. iff $b=+-2$, now, finish triying this values, and then finding $x$ by comparing the real aprt of both numbers.

[quote="Pascual2005"]$x^2+4=(x+2i)(x-2i)$ let an irreducible $p$ in $Zi$ divide both factors, then it divides $4i$, so its norm divides 4, so we get that $2$ divides $y^3$, and mod 4 it is impossible. .[/quote]

why is it impossible mod 4? 2^2 + 4 = 2^3... and 2 + 2i,  2 - 2i are definately not coprime.

I think what you should do for the case x is even, just divide through by 4 and you get another diophantine equation which is not too hard to attack.

do you have the another solution without use Z[i]? :(  :(  :(  :(

I think the only solutions are (2,2), (-2,2), because, if $Y^3$ is a sum of squares, then $Y^3$ is $1 \ mod\ 4$, so $Y^3=1+4d$ and hence $Y-1=4E$ because always $Y^2+Y+1$ is odd, then by a divisivility critery $Y^2+Y+1=1 \ mod \ 4$ WHICH is IMPOSSIBLE, because $Y$ and $Y+1$ are not multiples of $4$.

Best regards.

RX the Crazy Math

ok zschool you are right, from there we get the solutions so this is solved, and i dont have  a solution without $Zi$, sorry!


Here's the other one, but i'm not sure of my solution so just look at my answer first. if you have other ways of doing this, can you show it to me and i'll show mine later...thanks!

Solve:

$ |4x \plus{} 3| \minus{} |x \plus{} 5| \leq 8$

[hide]My answer for this is $ \frac { \minus{} 16}{5} \leq x \leq \frac {10}{3}$
[/hide]

btw, im sorry, i dont know how to hide this, you might as well teach me when you answer this...
(edit: now i know...thanks daan!:))

To hide something, type "hide" between [] in front of it, and "/hide" between [] after it.

I can only type it here using a mix of $ \text{\LaTeX}$ and regular text:
$ \text{[hide]}$Here comes your hidden stuff$ \text{[/}$hide].

When typing a message, you can also use the buttons (where it says "message body"). Hover over it with your cursor and you'll get some info.

BTW, I don't know the solution to your other problem.

Your answer is true. And the best way for solving this type of problems is ,of course, examination of cases (you can allways put them into a table and then take interval by interval).

[quote="Jakey"]Here's the other one, but i'm not sure of my solution so just look at my answer first. if you have other ways of doing this, can you show it to me and i'll show mine later...thanks!

Solve:

$ |4x \plus{} 3| \minus{} |x \plus{} 5| \leq 8$

{hide=My answer for this is $ \frac { \minus{} 16}{5} \leq x \leq \frac {10}{3}$}
This is my answer
{/hide}

btw, im sorry, i dont know how to hide this, you might as well teach me when you answer this...[/quote]

Your answer is correct.

@kunny...i didn't graph it, i used the properties of absolute value:

but this needs verification though.
$ |4x \plus{} 3| \equal{} \begin{cases}4x \plus{} 3 & \text{if }x\ge \minus{} \frac {3}{4} \\
\minus{} (4x \plus{} 3) & \text{if }x < \minus{} \frac {3}{4}\end{cases}$

and $ |x \plus{} 5| \equal{} \begin{cases}x \plus{} 5 & \text{if }x\ge \minus{} 5 \\
\minus{} (x \plus{} 5) & \text{if }x < \minus{} 5\end{cases}$.

So, when $ x\geq \minus{} \frac {3}{4}$ and $ x\geq \minus{} 5$, $ 4x \plus{} 3 \minus{} x \minus{} 5\leq8, x\leq\frac {10}{3}$
     when $ x\geq \minus{} \frac {3}{4}$ and $ x < \minus{} 5$, $ 4x \plus{} 3 \plus{} x \plus{} 5\leq8, x\leq0$
     when $ x < \minus{} \frac {3}{4}$ and $ x\geq \minus{} 5$, $ \minus{} 4x \minus{} 3 \minus{} x \minus{} 5\leq8, x\geq \minus{} \frac {16}{5}$
     when $ x\leq \minus{} \frac {3}{4}$ and $ x < \minus{} 5$, $ \minus{} 4x \minus{} 3 \plus{} x \plus{} 5\leq8, x\geq \minus{} 2$

then my answer was originally $ \minus{} 2\leq x\leq0$ since it's the intersection of the 4 but i accidentally tried a number outside $ \minus{} 2\leq x\leq0$ but inside $ \minus{} \frac {16}{5}\leq x\leq\frac {10}{3}$...so, is anything wrong with my solution?

O.K. but before answering your question, let me ask you the following question.

What do you think that for what we should solve inequality, say, $ x^2>x\plus{}2$?

[quote="kunny"]O.K. but before answering your question, let me ask you the following question.

What do you think that for what we should solve inequality, say, $ x^2 > x \plus{} 2$?[/quote]

it's $ x<\minus{}1,x>2$

Your answer is correct.

That means that which graph is located over or below. In this case, solving the inequality means that we should find the range of $ x$ such that the graph of $ y \equal{} x^2$ is above than the line of $ y \equal{} x$.

[quote="kunny"]Here is a standard solution.

[quote="Jakey"]

Solve:

$ |4x \plus{} 3| \minus{} |x \plus{} 5| \leq 8$[/quote]

We are to classify the 3 cases;

$ 1^\circ x\leq \minus{} 5$

$ 2^\circ \minus{} 5 < x\leq \minus{} \frac {3}{4}$

$ 3^\circ x > \minus{} \frac {3}{4}$[/quote]

thanks! i think it's better to graph...


[i][color=red]First I think that you must correct the problem(typos). Do you mean $x^2 + 2mx - 1$ ?[/color][/i]

This function in continuous and changes the monotony at $x_0 = - m$. If  $-m <0$ , then the function is never an injection. So we must have $m \leq 0$.  The answer is $m<0$ , because for $m=0$ the second branch becomes a line parallel to  X'X  and the injectivity  does not hold.

 [b]Ah! An other thought :[/b]

 The least value of the function is $- m^2 - 1$, so we mast have $- m^2 - 1 \geq -1 = f(0)$. Finally we must have $m<0$ , since for $m = 0$ we have problem with the other part($f(x) = -1 , x>0$ )

 [u]Babis[/u]

[quote="nemesis"]$f: R\to R$
$fx)= 2^2+2mx-1$, for x<=0
        $mx-1$, for x>0
If the function is injective, what is the $m$'s interval ?[/quote]

Yes, typo....  :blush:

The function was :
$fx)= x^2+2mx-1$, for x<=0
        $mx-1$, for x>0


X1,...,Xn are independant random variable with integrable distribution,
 evalute E(X1|X1+X2+...Xn)  ( conditional expectation )
I Think that if they have the same distibution so the E(Xi|X1+...+Xn) is the same for all i , so = E/n with E=E(X1)
 can someone help me understanding how to solve the general case ?

if they have the same distributiton then
$ E[X_i|X_1\plus{}X_2\plus{} \ldots\plus{}X_n]$
hence
$ X_1\plus{}\ldots\plus{}X_n \equal{} \sum_{i\equal{}1}^n E[X_i|X_1\plus{}X_2\plus{} \ldots\plus{}X_n]$
so
$ E[X_i|X_1\plus{}X_2\plus{} \ldots\plus{}X_n] \equal{} \frac{X_1\plus{}X_2\plus{} \ldots \plus{} X_n}{n}$
Rq: there is a classical proof of the law of large numbers that begins with this result.

Oh yes you'r right , I made a mistake, it shold be X1+X2+...+Xn/n , But what i'm looking for first is when they dont have the same distribution , in that case what simplfication can we find for it ?
Thank's Aleck , alaways there for help !


inequalities
three variable inequality
trigonometry

a new problem.....own~ :) need a easy proof...

the other  aside ....and i find the equalities is very uesful~ :)

The problem is equivalent to the following inequality
$ (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b) \geq \sqrt{(a^2\plus{}b^2\minus{}c^2)(a^2\plus{}c^2\minus{}b^2)(c^2\plus{}b^2\minus{}a^2)}$
All we need to do is to prove the trivial inequalities  ( notice that it can only be used when a, b,c are the sides of an acute triangle that satisfies the conditions given)
$ [(a\plus{}b\minus{}c)(a\plus{}c\minus{}b)]^2\geq (a^2\plus{}b^2\minus{}c^2)(a^2\plus{}c^2\minus{}b^2)$
Eventually, we get what we want! :)

[quote="great math"]The problem is equivalent to the following inequality
$ (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \geq \sqrt {(a^2 \plus{} b^2 \minus{} c^2)(a^2 \plus{} c^2 \minus{} b^2)(c^2 \plus{} b^2 \minus{} a^2)}$
All we need to do is to prove the trivial inequalities  ( notice that it can only be used when a, b,c are the sides of an acute triangle that satisfies the conditions given)
$ [(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)]^2\geq (a^2 \plus{} b^2 \minus{} c^2)(a^2 \plus{} c^2 \minus{} b^2)$
Eventually, we get what we want! :)[/quote]
Yes, it is. See also my book: [b][i]Old And New Inequalities 2[/i][/b], Vo Quoc Ba Can - Cosmin Pohoata, GIL publishing house 2008 for another nice solution. :)

[quote="thegod277"]How can I read your book? :D[/quote]
Buy it, my friend! :)

Ai cho em muon quyen "Old and new inequalities 2" duoc khong (Vietnamese)

[quote="great math"]The problem is equivalent to the following inequality
$ (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \geq \sqrt {(a^2 \plus{} b^2 \minus{} c^2)(a^2 \plus{} c^2 \minus{} b^2)(c^2 \plus{} b^2 \minus{} a^2)}$
All we need to do is to prove the trivial inequalities  ( notice that it can only be used when a, b,c are the sides of an acute triangle that satisfies the conditions given)
$ [(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)]^2\geq (a^2 \plus{} b^2 \minus{} c^2)(a^2 \plus{} c^2 \minus{} b^2)$
Eventually, we get what we want! :)[/quote]
thanks my friend,it's so easy.i have only proof it by a complex method. :(

so what is the simple solution for these inequalities????
 :D

any body please help me to solve this problem.thanks
I don't understand these solutions

[quote="ye109"]For acute triangle $ ABC$, a new problem.....own~ :)

$ \cos{A} \plus{} \cos{B} \plus{} \cos{C}\leq\sqrt {2\cos{A}\cos{B}\cos{C} \plus{} 2},$

need an easy proof...[/quote]$ 2 \plus{} 2\cos{A}\cos{B}\cos{C} \minus{} (\cos{A} \plus{} \cos{B} \plus{} \cos{C})^2 \equal{} \frac {s^2 \minus{} 2R^2 \minus{} 8Rr \minus{} 3r^2}{2R^2}\geq 0,$

which is the Walker inequality $ s^2\geq 2R^2 \plus{} 8Rr \plus{} 3r^2$ for non-obtuse triangle, see :

A. W. Walker, Problem E 2388, [url=http://www.maa.org/pubs/monthly.html]Amer. Math. Monthly[/url], 79 (1972), 1135.

[quote="ye109"]$ \cos A \plus{} \cos B \plus{} \cos C \geq 1 \plus{} \sqrt {2\cos A\cos B\cos C}.$[/quote]See : http://www.mathlinks.ro/viewtopic.php?t=228364

note that the first one is equivalent to

$ \sum (\cos B\plus{}\cos C)^2\leq 3$...

this is equivalent to [url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=176396577&t=219301[/url]...


induction
number theory solved
number theory

It's mine problem and I hope the following formulation is correct: let 1+1/2+1/3+...+1/n=A(n)/B(n), where A(n) and B(n) are coprime integers. Prove that there exist infinitely many n, for which A(n) is not the power of a prime.

I guess, A(n) is called numerator. For B(n) it is obvious: take n=p, p is prime. Then B(n) is divisible by p and aswell divisible by 2^k, where 2^k<p<2^(k+1), so is not a power of prime.

Harazi wrote:

>Prove that there exist infinitely many n such that the denominator of the >irreducible fraction 1+1/2+...+1/n is not a prime power. I don't know >how to solve this one. I've tried to take n a prime, so that the >denominator should be divisible by p, but I don't know how to continue.

Well, I have to apologise for my poor english. We I said denominator, I was refering to A(n), not to B(n). Sorry.

I thought you did harazi. Indeed A(n) is caled the numerator.
In harazi's first post of this problem, he was suggesting taking n=p-1 where p is a prime. It is easy to see that p|A(n) and p does not divide B(n)), so if A(n) is a power of a prime, A(n) must be a power of p.
Don't know how to continue either.

I asked a question here. Nobody care to answer?

I do not know the answer. But it does not look natural. Too many powers of 3 in denominator must be substracted.

Dear Fedor, please post your solution to this problem. It is too difficult for me, but I like it very much and I would like to see its idea. Thank you very much.

Well, I do not remmeber, but try.

Assume a contrary.
Denote 1+1/2+...+1/n=f(n)

1. B(n)>n/2. Indeed, consider k such that 2^k<=n<2^{k+1}. Then B(n) is divisible by 2^k (because there exists exactly one summand 1/a with a divisible by 2^k).

2. Take n=p-1, where p is large prime. Then A(p-1) is divisible by p and so A(p-1) is a power of p, but it is more than p (otherwise f(p-1)<=(p-1)/((p-1)/2)<2 - a contradiction). So, A(p-1) is divisble by p^2.

3. I claim that A(p^k-1) is a power of p, certainly more than p itself. Indeed, we have it for k=1, then consider A(p^{k+1}-1). The sum of all summands 1/(p*r+i), i=1,2,...,p-1 is divisible by p. (the sum of p-1 consecutive such summands is equal 1+1/2+...+1/(p-1) modulo p, so is divisible by p). And what about the sum of other summands, like 1/(p*r)?
Indeed, their sum equals x=f(p^k-1)/p. Since f(p^k-1) is divisible by p^2 (induction proposition), x is divisible by p. So,
A(p^{k+1}-1) is divisible by p and therefore is a power of p.

4. Well, then consider f(p^k-p). We have f(p^k-p)=f(p^k)+1/(1-p^k)+1/(2-p^k)+...+1/((p-1)-p^k). So, f (p^k-p) is divisible by p. Then A(p^k-p) is divisble by p and is therefore a power of p.

5. So, f(p^k-1) and f (p^k-p) are divisible both by p^n for any n and large enough k. Their differnce, 1/(1-p^k)+1/(2-p^k)+...+1/((p-1)-p^k) is also divisble by p^n. Take n such that 1+1/2+...+1/(p-1) is not divisible by p^n and large k>n. Then
1/(1-p^k)+1/(2-p^k)+...+1/((p-1)-p^k)=1+1/2+...+1/(p-1) modulo p^n - a contradiction.

This problem is really extremely difficult, but extraordinary beautiful. So many ideas in a single problem. Wow!
    Thank you Fedor for this masterpiece.

At fact, it was included to the problem set in the last moment, to replace other problem (harder). It was even solved by Andrei Khaliavin (that says a lot about Andrei, more than 42 points in Glasgow, I think). There was another problem (a graph with 1014 colours), which was not solved by anybody, and yet another, solved also by only one guy, Almaz Mirgasimov. In general, that olympiad was the hardest ever.

Fedor, are you sure your last post ended up where it has supposed to? because I don't find the connection with those written above it ..  :?

Ah, yes. It was proposed on All-Russian Olympiad in 2002, and I've just commented this event.

[quote="Fedor Petrov"]At fact, it was included to the problem set in the last moment, to replace other problem (harder).[/quote]

Which was the problem that was exchanged due to its difficulty ? Where do you know that from, Fedor ?

[quote="Fedor Petrov"] In general, that olympiad was the hardest ever.[/quote]

What do you mean ? The hardest All-Russian olympiad or the hardest olympiad in general ?

You have one of the problems of the 2002 russian olympiad in the combinatorics section.
Seems very difficult.

Pierre.

[quote="orl"]
What do you mean ? The hardest All-Russian olympiad or the hardest olympiad in general ?[/quote]

The hardest All-Russian for sure and the hardest national olympiad ever, I suppose (but may not be sure).

Hi Fedor,

which was the replaced harder problem ?

Oh, my God, harder than this one? What do they want to do to those participants?

It still may be used, so I must hold my tongue.

@above Fedor Petrov
Can you say the harder problem now please ?  :)

@below
Thanks. that's would be great

Good question. I do not remember it but may try to reconstruct using colleagues' help.

[quote=Fedor Petrov]Well, now I (again) know the harder problem. But it remains unclear whether it was used on some competition , and if not, I certainly should not share it here.[/quote]

U joined October 2003 and ur still active :0 :O :o

[quote=azerluigi][quote=Fedor Petrov]Well, now I (again) know the harder problem. But it remains unclear whether it was used on some competition , and if not, I certainly should not share it here.[/quote]

We’ll make sure to remind you again in 17 years then. :-)

P.S.: I am quite surprised you managed to see post 19, AoPS gets flooded with so much stuff. Maybe it was because it was close to the IMO.[/quote]

I got a link in PM)


$If x,y,z>0$    and	      $x^{2}+y^{2}+z^{2}=3$

proove that
$\frac{xy}{xy+x+y}+\frac{yz}{yz+y+z}+\frac{zx}{zx+z+x}\leq1$

 :blush:

At first use the inequality $x+y\geq 2\sqt {xy}$. then we put $\sqrt{xy}=a$ and the

inequality reduses to show that $\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\leq 1$

which is equivalent to $2abc+2(ab+bc+ca)\leq 8$.

but this is true due to AM-GM and using the fact that $a^2+b^2+c^2\leq 3$

opening, we have to prove that $2x^2y^2z^2 + \sum{x^2y^2z} \leq 2xyz + \sum{x^2y}$.

by $MA-MG$, we have $(\frac{x^2+y^2+z^2}{3})^3 \ge xyz \Longrightarrow xyz \leq 1$

$\Longrightarrow 2x^2y^2z^2 \leq 2xyz$.

now, we have to prove that $(x^2+y^2+z^2)\sum{x^2y} \ge 3\sum{x^2y^2z}$.

this is the same that prove that $\sum{x^4y} + \sum{x^3y} \ge 2\sum{x^2y^2z}$.

by bunching we have $\sum{x^4y} \ge \sum{x^2y^2z}$ and $\sum{x^3y} \ge \sum{x^2y^2z}$.

so, the problem is solved.

E.L

Since $x^2+y^2+z^2=3$, we have $\sqrt{\frac{x^2+y^2+z^2}{3}}=\sqrt{\frac33}=1$. By the AM-QM inequality, we have $\sqrt{\frac{x^2+y^2+z^2}{3}} \geq\frac{x+y+z}{3}$. Thus, $1\geq\frac{x+y+z}{3}$. Hence,

$yz+y+z=yz+1\cdot\left(y+z\right)\geq yz+\frac{x+y+z}{3}\cdot\left(y+z\right)=yz+\frac{zx+xy+\left(y+z\right)^2}{3}$.

By the AM-GM inequality, $\left(y+z\right)^2\geq 4yz$, so this becomes

$yz+y+z\geq yz+\frac{zx+xy+4yz}{3}=\frac{7yz+zx+xy}{3}$.

Hence, $\frac{yz}{yz+y+z}\leq\frac{yz}{\left(\frac{7yz+zx+xy}{3}\right)}=\frac{3yz}{7yz+zx+xy}=\frac{3a}{7a+b+c}$, where we use the abbreviations a = yz, b = zx, c = xy.

Now, using the sign $\sum$ for cyclic sums, we get

$\sum\frac{yz}{yz+y+z}\leq\sum\frac{3a}{7a+b+c}$.

But we have to prove that $\sum\frac{yz}{yz+y+z}\leq 1$. Hence, it remains to show that $\sum\frac{3a}{7a+b+c}\leq 1$. But

$1-\sum\frac{3a}{7a+b+c}=\sum\frac13-\sum\frac{3a}{7a+b+c}=\sum\left(\frac13-\frac{3a}{7a+b+c}\right)$
$=\sum\frac{\left(c-a\right)-\left(a-b\right)}{3\left(7a+b+c\right)}=\frac13\left(\sum\frac{c-a}{7a+b+c}-\sum\frac{a-b}{7a+b+c}\right)$
$=\frac13\left(\sum\frac{b-c}{7c+a+b}-\sum\frac{b-c}{7b+c+a}\right)=\frac13\sum\left(b-c\right)\left(\frac{1}{7c+a+b}-\frac{1}{7b+c+a}\right)$.

Now, if $b\geq c$, then $7c+a+b\leq 7b+c+a$ (since $\left(7c+a+b\right)-\left(7b+c+a\right)=-6\left(b-c\right)$), so that $\frac{1}{7c+a+b}\geq\frac{1}{7b+c+a}$, and thus $\frac{1}{7c+a+b}-\frac{1}{7b+c+a}\geq 0$; on the other hand, $b\geq c$ yields $b-c\geq 0$. Thus, in the case $b\geq c$, we get $\left(b-c\right)\left(\frac{1}{7c+a+b}-\frac{1}{7b+c+a}\right)\geq 0$. On the other hand, if $b\leq c$, then we similarly obtain $\left(b-c\right)\left(\frac{1}{7c+a+b}-\frac{1}{7b+c+a}\right)\geq 0$. Hence, in every case, we get $\left(b-c\right)\left(\frac{1}{7c+a+b}-\frac{1}{7b+c+a}\right)\geq 0$, and thus

$1-\sum\frac{3a}{7a+b+c}=\frac13\sum\left(b-c\right)\left(\frac{1}{7c+a+b}-\frac{1}{7b+c+a}\right)\geq 0$,

so that $\sum\frac{3a}{7a+b+c}\leq 1$, and the problem is solved.

  Darij


1) Let $ a_1,a_2,..,a_n$ are $ n$ positive real numbers and $ A \equal{} [a_{ij}]_{1 \leq i,j \leq n}$ such that :
    $ a_{ij} \equal{} \frac {e^{\frac { \minus{} (a_i \minus{} a_j)^2}{4(a_i \plus{} a_j)}} \plus{} e^{\frac { \minus{} (a_i \plus{} a_j)}{4}}}{\sqrt {a_i \plus{} a_j}}$
Prove that $ \det(A) \geq 0$
 2)Let $ b_1,b_2,..,b_n$ are $ n$ positive real numbers and $ B \equal{} [b_{ij}]_{1 \leq i,j \leq n}$ such that :
   $ b_{ij} \equal{} \frac {b_ib_j(b_i^2 \plus{} b_j^2)}{(b_i \plus{} b_j)(b_i^3 \plus{} b_j^3)(b_i^4 \plus{} b_j^4)}$
Prove that $ \det(B) \geq 0$

[hide="solution"]
First, we note that
$ \int\limits_{0}^{ \plus{} \infty}{e}^{u^2t}.cos(au)\ du \equal{} \sqrt {\pi}\frac {{e}^{ \minus{} \frac {a^2}{t}}}{\sqrt {t}}$
Then we know that $ \sqrt {\pi}(\frac {{e}^{ \minus{} \frac {(a_i \minus{} a_j)^2}{(a_i \plus{} a_j)}} \minus{} {e}^{ \minus{} \frac {a_i \plus{} a_j}{4}}}{\sqrt {a_i \plus{} a_j}}) \equal{} \int\limits_{0}^{ \plus{} \infty}{e}^{u^2(a_i \plus{} a_j)}.cos((a_i \minus{} a_j)u)\ du \minus{} \int\limits_{0}^{ \plus{} \infty}{e}^{u^2(a_i \plus{} a_j)}.cos((a_i \plus{} a_j)u)\ du \equal{} \int\limits_{0}^{ \plus{} \infty}{e}^{u^2(a_i \plus{} a_j)}.2(sin(a_iu)).sin(a_j)u)\ du \equal{} \int\limits_{0}^{ \plus{} \infty} f_i(u).f_j(u)\ du$ by taking $ f_i(u) \equal{} {e}^{a_i.u^2}.\sqrt {2}.sin(a_i.u)$. Now, construct a matrix with entries $ \int\limits_{0}^{ \plus{} \infty} f_i(u).f_j(u)\ du$, with $ f_i(u)$ is as defined above, then the matrix is semidefinite positife (By noting that $ < f_i,f_j > \equal{} \int\limits_{0}^{ \plus{} \infty} f_i(u).f_j(u)\ du$ is an inner product) and the determinant of the matrix is greater than or equal to $ 0$. Moreover, if we divide each row by $ \sqrt {\pi}$ (In other words we do the first elementary row opeation),we obtain a matrix as stated in the problem given and the determinant remains nonnegative.  And the proof is finished. I am sorry if I am wrong.
[/hide]


probability
number theory
relatively prime

A deck of forty cards consists of four 1's, four 2's,..., and four 10's.  A matching pair (two cards with the same number) is removed from the deck.  Given that these cards are not returned to the deck, let $m/n$ be the probability that two randomly selected cards also form a pair, where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

[hide="Answer"]First case: the first card picked is one of the two remaining of the same number as the first pair. Chance is $\frac{2}{38}\cdot \frac{1}{37}=\frac{1}{703}$.

Second case: the first card picked is not one of the two remaining of the same number as the first pair. Chance is $\frac{4}{38}\cdot \frac{3}{37}=\frac{6}{703}$.

Therefore, the total chance is $\frac{6+1}{703}=\frac{7}{703}\Rightarrow 7+703=\boxed{710}$.[/hide]

i got a different answer...

[hide]let's say that our matching pair were 2 ones. (matching pair doesn't really matter. our probability of picking the other 2 ones would be $\frac{2}{38}*\frac{1}{37}$ If we want to choose any pair of the cards 2-10, the probability to pick one pair is $\frac{4}{38}*\frac{3}{37}$ Remember that this value is for one certain number, like a pair of 2's. To find the value for the rest of the cards, we'd multiply it by 9. Therefore, our probability is $\frac{2}{38}*\frac{1}{37}$ $+$ $9$$(\frac{4}{38}*\frac{3}{37})$. Our probability is $\frac{55}{703}$. Adding 55 and 703, we get 758 as our answer.[/hide]

Oh, whoops, I think you're right, I should have multiplied the chance by $9$ because there are $9$ ways to do it. :oops:

[hide]This problem is best approached with simple casework.

The probability that my next pair is of the same card type is $\frac{1}{38}\cdot \frac{1}{37}$

The probability that my next pair is of a different card type is $\frac{4}{38}\cdot \frac{3}{37}$. However, there are 9 ways this can happen, so this case contributes $9\left(\frac{4}{38}\cdot \frac{3}{37} \right)$.

In all, our probability is $\frac{1}{38}\cdot \frac{1}{37}+9\left(\frac{4}{38}\cdot \frac{3}{37} \right)=\frac{55}{703}$, so the answer is $55+703=\boxed{758}$.[/hide]

This is trivial

The answer is just $\frac{10 \cdot \binom{4}{2} - 5}{\binom{38}{2}} = \boxed{758}.$


LaTeX
calculus
number theory
number theory open

Why don't we try a giant group effort to solve the Goldbach conjecture (as far as I know, no one has been able to do it by themselves :lol: )?

In case that's too tough for y'all, you can try this one, by me (unfortunately I have no Latex skeelz and I'm too lazy too develop some): let P be a set of odd primes, and suppose we had a simultaneous system of congruences like this:

x != a11, a12 (mod p1)
x != a21, a22 (mod p2)
etc.

For all the primes in P. Let g(P, a11, a12, etc.) be the minimum positive value of x that satisfies this equation, and f(P) be the maximum value of g(P, a11, a12, etc.) for all choices of a11, a12, etc. What is f(P)? What can we say about f(P)? (It would be really nice if f(P) < max(P)^2, and it isn't too hard to see that f(P) <= p1*p2*... - 2^|P|).

I'd rather like to see here a massive try /group effort in order to attack ;)  [b]ABC conjecture[/b]. :weightlift:
This conjecture is in the very core of number theory and more important than Goldbach. :)

esti ,i dont like abc conjecture
it seems much more calculus related than nt-related.
it is also not at all natural,and doesnt make too much sense,
i mean it makes sense but not in a natural inteligible way.
abc conjecture is a pretty weird conjecture

oooh... :o  sorry...i see it would imply a more elegant solution to FLT and catalan...
indeed it does have some big importance now  :blush: sorry i didnt check out
enough material before...

Well, and Goldbach is as good as for sure also something that wants to be solved by asymptotic reasons ;)

yes zetax...abc conjecture presents some kind of higher bound


Let $ a,b,c$ be real numbers such that:
$ a\plus{}b\plus{}c\equal{}0, a^2\plus{}b^2\plus{}c^2\equal{}6$
Find the min, max value of $ P\equal{}a^2b\plus{}b^2c\plus{}c^2a$

[quote="mr.danh"]Let $ a,b,c$ be real numbers such that:
$ a \plus{} b \plus{} c \equal{} 0, a^2 \plus{} b^2 \plus{} c^2 \equal{} 6$
Find the min, max value of $ P \equal{} a^2b \plus{} b^2c \plus{} c^2a$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162227


Hey people ,
If you were at
1 ) A maths quiz , and
2 ) A computer related quiz ,
What team names would you give provided there was a prize for the best team name ??

for mathcounts we wanted to call ourselves "The Flaming Cabbage" just for fun, but our coach wouldn't let us.....we had a mascot picture and everything!  :D

I've been The Knights Who Until Recently Said Ni, Los Gauchos, and The Unblondes.

Electric Pig and CBASS. Two ultimate club team names

[quote="furious"]math1337z[/quote]

very nice. very nice ;)

[quote="furious"]math1337z[/quote]

 :rotfl:  :rotfl:  :rotfl:

[quote="math92"]Admin edit: don't do this anymore![/quote]

 What did you do?

the Super stinky slimy salamanders... sigh I still remember that from 4th grade, kinda sad  :blush:


Prove that

$ \frac{cot x}{1\minus{}tan x}\plus{} \frac{tanx}{1\plus{} cot x} \equal{} cscsecx \plus{}  1$

hello, i have got
$ \frac{\cot(x)}{1\minus{}\tan(x)}\plus{}\frac{\tan(x)}{1\plus{}\cot(x)}\equal{}2\csc(4x)(2\cos(2x)\plus{}\sin(2x))$
Sonnhard.


induction
number theory unsolved
number theory

[quote="Tiks"][quote="kcbu"]Let ${x_{n}}$ and
$\{{\begin{array}{*{20}c}{x_{1}= 1}\\{x_{n+1}= ({1+\frac{3}{n}})x_{n}+2-\frac{3}{n}{\text{ }}n \geqslant 1}\\ \end{array}}$
Prove that $\boxed{x_{n}\in \mathbb{N}}$[/quote]
Prove by induction that
\[\boxed{x_{n+1}=\frac{n(n+1)(n+5)}{6}+1}\]
:wink:[/quote]
I knew. But i don't understand.
How do u get one ???

[quote="kcbu"]
I knew. But i don't understand.
How do u get one ???[/quote]

If the question comes to that,there are two ways to work it out:to guess(with some strategy,of cours :) ) or to calculate.

So,about guessing,firstly you should substitute $x_{n}=a_{n}+1$ to make the condition more understandable,that is you will get that
\[a_{n+1}=(1+\frac{3}{n})a_{n}+2\]

Now it is not hard to suggest that $a_{n}$ should have the following form:$a_{n}=\frac{nb_{n}}{3}$ in case to keep $a_{n+1}$  beeing a positive integer number.
Now we see that $b_{n}=\frac{3a_{n}}{n}$, and let's look at some of the first members of this sequence.We can easily find them:
$b_{1}=0,b_{2}=3,b_{3}=7,b_{4}=12,b_{5}=18,b_{6}=25,b_{7}=33$
,don't you see how they grow?I believe you do,because it is not hard to notice that $b_{n+1}=b_{n}+n+2$, and therefore $b_{n+1}=\frac{n(n+5)}{2}$. :wink:

About the second way,it is more directe but also more complicated.
So,we have-as before-that $a_{n+1}=(1+\frac{3}{n})a_{n}+2$.
Setting $b_{n}=1+\frac{3}{n}$ we see that
\[a_{n+1}=b_{n}a_{n}+2=a_{n-1}b_{n}b_{n-1}+2(b_{n}+1)=...=b_{n}b_{n-1}...b_{1}a_{1}+2(\sum_{m=0}^{n-2}b_{n}...b_{n-m}+1)=2(\sum_{m=0}^{n-2}b_{n}...b_{n-m}+1)\]
Then
\[b_{n}b_{n-1}...b_{n-m}=\frac{(n+3)(n+2)(n+1)}{(n+2-m)(n+1-m)(n-m)}\]
and one more thing to help is that
\[\frac{1}{(n+2-m)(n+1-m)(n-m)}=\frac{1}{2}(\frac{1}{n-m}+\frac{1}{n+2-m}-\frac{2}{n+1-m})\]
Afterwards,substituting the two realtions above in the relation with $a_{n+1}$ and $b$s you will come to the desired expression. :wink:


Ring Theory
quadratics
superior algebra
superior algebra unsolved

Does a ring $\mathbb{Z}[c]$ (where $c \in \mathbb{C}$ is an algebraic number) with property that every rational prime of the form $4k+1$ in $\mathbb{Z}[1]$ is also a prime in $\mathbb{Z}[c]$ exist ? Give an example of such ring (should be UFD, but I will be satisfied with any answer at first  :) )

I don't have many ideas, but I think the answer is no. Note that prime integers of the form $p = 4k + 1$ have a very special property: They can be decomposed into the sum of two squares $p = m^2 + n^2 (m < n)$ in one and only one way. For example, $5 = 1^2 + 2^2, 13 = 2^2 + 3^2, 17 = 1^2 + 4^2, 29 = 2^2 + 5^2$. Therefore, if $\mathbb Z[i] \subseteq \mathbb Z[c]$ then the answer is obviusly no. If $\mathbb Z[c]$ contains any non-integer rational, then, the answer is no. But I don't have quite a lot of ideas in general.

A=Z[\sqrt{3}]  this ring satisfies our condition.

If <p> not prime ideal, then there are (a+\sqrt{3}b), (c+\sqrt{3}d) in A\<p>  which these product in <p>.

So ac+3bc=0(mod p) and ad+bc=0 (mod p) .
a^2=3b^2 (mod p) is impossible, because p=4k+1.

Counterexample to the above: $(5+2\sqrt{3})(5-2\sqrt{3})=13$

There's clearly no hope of doing it with a quadratic irrational.

I forgot to mention that $c$ is algebraic integer of degree at least 2 (otherwise ring $\mathbb{Z}[\frac{1}{2}]$ suffices the conditions).

Anybody thinking of this problem ? :?

[quote="Megus"]I forgot to mention that $c$ is algebraic integer of degree at least 2 (otherwise ring $\mathbb{Z}[\frac{1}{2}]$ suffices the conditions).[/quote]No, it does not! Note that there are no primes in $\mathbb{Z}\left[\frac{1}{2}\right]$. Note that if $p\neq2$, then $p=2\cdot\frac{p}{2}$, and if $p=2=4\cdot\frac12$, hence your ring does not satisfy the conditions!

Any way, as I already said, if $i\in\mathbb{Z}[c]$ or $\frac{p}{q}\in\mathbb{Z}[c]$ for $(p,q)=1$ and $|q|\neq1$, then the answer is NO, the ring does not satisfy your conditions!


$ F(s)\equal{}\frac{e^{\sqrt{s^2\minus{}1}x}}{(s\minus{}1)(1\plus{}e^{\sqrt{s^2\minus{}1}\pi})}$. Help me.

The inverse Laplace transform is a function of $ s$ as well as $ x$ ? or is it a typo?


My favorite number is a positive four-digit integer. My second favorite number is also a four-digit integer, and used the same four digits. When I subtract the two i get a third four digit number, three of whose digits are 2,3, and 9. What is the fourth digit of the difference?

the answer is 4 here is the solution...

let the first number be A and the second B and let C be their difference

since both numbers have the same digits  A ≡ B (mod 9) => C ≡ 0 mod(9)

if 9|C then the sum of C's digets must also be divisiable by 9 so

9+2+3 = 14 ≡ 5 (mod 9)

9-5=4

[quote="Wichking"]since both numbers have the same digits  A ≡ B (mod 9) [/quote]Can someone explain why?

Let $ x$ be represented by the digits $ x_kx_{k \minus{} 1}\cdots x_2x_1$. Then, $ x\equiv x_k\cdot10^{k \minus{} 1} \plus{} x_{k \minus{} 1}\cdot10^{k \minus{} 2} \plus{} \cdots \plus{} x_210 \plus{} x_1\equiv x_k \plus{} x_{k \minus{} 1} \plus{} \cdots \plus{} x_2 \plus{} x_1\bmod9$. So, $ x\bmod9$ is just the sum of the digits of $ x$ modded by 9.

because

(10^n)-1=99999..999≡0 (mod 9)

sice every number in base 10 can be written like that :

(10^n)*x1+(10^(n-1))*x2+....+10*x(n-1)+xn

(those numbers 1,2...,n-1,n are indexes)

the sum x1+x2+...+x(n-1)+x(n)≡(10^n)*x1+(10^(n-1))*x2+....+10*x(n-1)+xn (mod 9)

you can use the same trick to check if a number is divisiable by 3...


Let S denote the set of all (positive) divisors of 60^5. The product of all the numbers
in S equals 60^e for some integer e. What is the value of e?

[hide]
$ 60^5 \equal{} 2^{10}3^55^5$; there are $ (5 \plus{} 1)(5 \plus{} 1) \equal{} 36$ factors in $ S$ with exponent of $ 2^m$ for each $ m$, so taking the product of all of the powers of $ 2$ we get a exponent of $ 2^{36(1 \plus{} 2 \plus{} \cdots \plus{} 10)} \equal{} 2^{55 \cdot 36}$. Thus the answer is $ 1980/2 \equal{} \boxed{990}$.
[/hide]

?  isn't the answer 198 because $ 60^5 \equal{} 2^{10}*5^5*3^5$ which has $ 11*6*6 \equal{} 396$ divisors, of which are all in pairs which multiply to 60 which means $ e \equal{} 396/2 \equal{} 198$?

edit: nevermind... all are in pairs which multiply to $ 60^5$ which means $ e\equal{}396*5/2 \equal{} 990$.
By the way, this is probably easier (and a better proof) than just considering the 2's.

[quote="facis"]?  isn't the answer 198 because $ 60^5 \equal{} 2^{10}*5^5*3^5$ which has $ 11*6*6 \equal{} 396$ divisors, of which are all in pairs which multiply to 60 which means $ e \equal{} 396/2 \equal{} 198$?

edit: nevermind... all are in pairs which multiply to $ 60^5$ which means $ e \equal{} 396*5/2 \equal{} 990$.
By the way, this is probably easier (and a better proof) than just considering the 2's.[/quote]

It is.  :wink: , and could be applied to all types of problems like these.

yes to generalize for a number k
the product of its factors is K ( # of  factors) / 2

[b]mathemonster[/b] — Actually, it would be that for any number $ n^{k}$, the product of its factors is $ \sqrt {n^{k\phi}}$, where $ \phi$ is the number of distinct factors.

whoops it the product of the divisors of x is x ^ number of divisors/2


The price of a stereo is $ \$200$ on Monday. On Tuesday the price from Monday is reduced by $ 10\%$. On Wednesday, Tuesday's price is increased by $ 10\%$. What is the price of the stereo on Wednesday?

If $ P$ is the decimal form of the percentage, to reduce N by that much percent, you compute $ N \times (1-P)$, to increase, you compute $ N \times (1+P)$.

In this case, the price undergoes a decrease by $ .10$ and an increase by $ .10$  So, we need to compute $ 200 \times (.9) \times (1.1)$.  My mental math goes as follows:

$ 200 \times (.9) \times (1.1) = 2 \cdot 9 \cdot 11= 18 \cdot 11 = 198$

[size=75][color=red]Added $ \text{\LaTeX}$ to make clearer. -Isabella2296[/color][/size]

Or you can pretty much just find 90% of 200, which is 180, then add 10% of 180, which is 18, so 180+18=198


[size=150][b][color=red]T-Man Conjecture[/color][/b][/size]

[b][color=darkblue]maybe make a new field of math: "Macro-Mathematics" --- calculate not by data of figure, but by number of unit.[/color][/b]

[color=red]1. Embryonic Proposition:[/color]

For each n belong N, nE[1,9], let S(n) be the subset of the set S={1,2,...,9}. Question: which is the minimum of "n", to make the each number of S(n) calculates the sum "24" surely all be in, just through "+" "-" "*" "/" the four arithmetic without the order of number position. (no method of exhaustion)

[color=olive]Like a classical "Sum 24 of Card Arithmetic Game" .... and apparently the "9" seems suitable number, but not minimum.
Example. For n=9 only 1 subset {1,2,3,4,5,6,7,8,9}, find (9+8+7+6+5-4-3-2)*1=24. [/color]

.

[color=red]2. Expanding Proposition:[/color]

For each n&m belong N, n>=4 , mE[2,n], let S(m) be the subset of the set S={2,3,...,n}, and a arbitrary number X be the one of the set S={2,3,...,n}. Question: if m being right to make the each number of S(m) calculates the sum X surely all be in, just through "+" "-" "*" "/" the four arithmetic without the order of number position; then is ANY number M being, M E [m,n], M belong N, to get the same result ?

.

[color=red]3. T-Man Conjecture & T-Man Number:[/color]

Assume that the "Expanding Proposition" be established, apparently each natural number X between 2 to n can correspond with a minimum of m, and thinking generally those numbers of m will not be equal --- some will be bigger, and some smaller.

Here I conjecture that those X which maximum of m corresponding to are "Prime Number", and minimum of m corresponding to are "Composite Number". Further more, this kind "Prime Number" is different from usual "Prime Number" --- [color=darkblue]just filtered by "Multiplication"[/color] --- they all be filtered by "Four Arithmetic" within limit and necessity even limitless.

So this kind "Prime Mumber" can be called "Further Prime Number", same as "Further Composite Number". Those two kinds number also can be named "T-Man Number" according to this "[color=darkblue]T-Man Conjecture[/color]" me found.
[color=olive](Why "T-Man" named ? see the attached words)[/color]

In essential nature "T-Man Conjecture" expanded the definition about "Prime Number" and "Composite Number" of classical mathematical theory. Original "Composite Number" definition: a integer can be factored as two positive integers other than itself or 1 by "Multiplication" (more factors can be seen as two factors). For the particularity of single "Multiplication", there is just one time assembling requested by two factors. So it is "Composite Number". Contrarily zero time assembling make "Prime Number".

Here this Number's Judging Differentiation is like "T-Man Conjecture" me found --- "a minimum of m" shows the most choices assembling, so the corresponding number X is the "Further Composite Number". Contrarily the least choices assembling make the "Further Prime Number". In this kind judgement the more(most) choices make "Composite Number", the less(least) choices make "Prime Number".

Also say the classical mathematical definition is "one-dimensional", and me given here the definition is "two-dimensional". Of course the "three-dimensional" or more dimensions is also being. I think that kind definition mode is similar to this.

[color=red]4. (Further Conjecture)[/color] [color=olive]A good prospect of application:[/color]

Seen from another angle for understanding, "[color=blue][b]Goldbach Conjecture[/b][/color]" also can be thought as "T-Man Conjecture" extending --- make 1 time "Addition" into original "Multiplication" of 2 numbers case about "Prime Number" and "Composite Number", with specific limit of that 2 numbers.

The best benefit from "T-Man Conjecture" is calculating not by data of figure, just by number of unit and the type of mathematics operations. Throught two or three even more equality or equation we can narrow the result area greatly using result intersections, then calculating result exactly faster by old mathematical way. I think it is better to used on rocket technology etc real-time processing, and other more fuzzy processing of sociological field.

Follow is the simple operation axiom from "T-Man Conjecture", further reseach let the professional to do.

...... ......

more info welcome to my blog:
http://tanweekly.blogspot.com/

P.S.
[color=red][b]Why "T-Man Conjecture" named ? [/b][/color]
Just to disseminate widely of my Special Paranormal Phenomenon with own experience, as a warning to others and to the world.

some more relevant info, please visit the follow pages:
(English) http://tanweekly.blogspot.com/2006/05/be-transparent-man-what-can-i-do-what.html
(Chinese) http://www.bailingren.com/special/index.htm

"T-Man", the character "T" has three meaning: [1] "T" the first character of my last name. [2] "T" the first character of Chinese Spelling Forms of my Chinese code name "Tuo Mingrum". [3] "T" the first character of "Transparent Man".

It seems like "X-Man" --- from a movie "X-Man" --- another word for psychic man or magic man .... Few days ago I had watched the "The X-Files" of FOX. "X-Files" mean the files about unexplainable phenomenon, I think maybe the "X-Man" called from the "X-Files".

[color=darkblue]"The Truth Is Out There"[/color]

====================================
[b][color=olive]"They can see around faces which I see sharing my eyes,
they can listen around talks which I listen sharing my ears,
they can know all my acts which I think about sharing my head! "[/color][/b]

[quote=Idefix][quote=NoDealsHere]Dead dead[/quote]

Why are you reviving all these dead threads?[/quote]

fun fun

Well, I guess this is called the [b]fun[/b] factory, so...