HI... if anyone could help me with these I'll be very thankful. I am lost as hell. Here's just a few of the problems I dont understand. I wont list them all because it would be too much for you.

Find x if the perimeter of the figure is 27.9 cm.
The measures are:
x/2, x/5, x + 0.8, x, 1.5x + 2, 2x - 2.8

Is there a way to make the following equation true?
6x/3 + 3= 2x

Is there a way to make the following equation false?
6x/3 + 3= 2x + 1

Which of the following equations could be associated as linear relations.
y= 3x + 10
y= 2x/3 - 5
y
Y1= 0.5x
y= 2x
F + 1= S- 2
y= 3 (x-4)
y= 2/x
A= (7.3)^2t

[quote="lingomaniac88"]Find $x$ if the perimeter of the figure is $27.9$ cm.
The measures are:
$\frac{x}2$, $\frac{x}5$, $x+0.8$, $x$, $1.5x+2$, $2x-2.8$
The sum of all these numbers is $6.2x$.  Set that equal to $27.9$.

Is there a way to make the following equation true?
$\frac{6x}3+3=2x$
This simplifies to $2x+3=2x$.

Is there a way to make the following equation false?
$\frac{6x}3+3=2x+1$
This simplifies to $2x+3=2x+1$.

Which of the following equations could be associated as linear relations.
Remember that a linear relation can be written in the form $ax+by=c$ or $y=mx+b$ or something of that nature.[/quote]

1) I'm assuming since 27.9/6.2= 4.5 that x= 4.5

2) My bad. What I meant was: Can you state the worth of x in order to make the following equation true?

3) Same as number 2 except a way to make the following equation false.

4) So I take the linear relation equations from this list are the following. (In bold)

[b]y=3x + 10 [/b]
[b]y= 2x/3 - 5 [/b]
Y1= 0.5x
y= 2x
F + 1= S- 2
[b]y= 3 (x-4)[/b]
y= 2/x
[b]A= (7.3)^2t[/b]

[quote="Dr. No"][quote="helpmeplz"]HI... if anyone could help me with these I'll be very thankful. I am lost as -. Here's just a few of the problems I dont understand. I wont list them all because it would be too much for you.

Find x if the perimeter of the figure is 27.9 cm.
The measures are:
x/2, x/5, x + 0.8, x, 1.5x + 2, 2x - 2.8

Is there a way to make the following equation true?
6x/3 + 3= 2x

Is there a way to make the following equation false?
6x/3 + 3= 2x + 1

Which of the following equations could be associated as linear relations.
y= 3x + 10
y= 2x/3 - 5
y
Y1= 0.5x
y= 2x
F + 1= S- 2
y= 3 (x-4)
y= 2/x
A= (7.3)^2t[/quote]
[hide]
[quote]Find x if the perimeter of the figure is 27.9 cm.
The measures are:
x/2, x/5, x + 0.8, x, 1.5x + 2, 2x - 2.8[/quote]
The perimeter is the distance around the outside of a figure.
Add the measures together then set the sum equal to 27.9 cm.
[quote]Is there a way to make the following equation true?
6x/3 + 3= 2x[/quote]
6x/3 + 3 = 2x + 3 ...
what do you think?
[quote]Is there a way to make the following equation false?
6x/3 + 3= 2x + 1[/quote]
Again, 6x/3 + 3 = 2x + 3 ... what do you think?
[quote]Which of the following equations could be associated as linear relations...[/quote]
A function is considered linear if and only if:

$f(\alpha\cdot x_{1}+\beta\cdot x_{2}) = \alpha\cdot f(x_{1})+\beta\cdot f(x_{2})$

which of the functions listed satisfy that property?[/hide][/quote]

Thanks for the help.. please read my answers above.

[quote="helpmeplz"]
Thanks for the help.. please read my answers above.[/quote]
1) I'm assuming since 27.9/6.2= 4.5 that x= 4.5

[color=blue]Correct[/color]

2) My bad. What I meant was: Can you state the worth of x in order to make the following equation true?

[color=blue]No, you can't. Ultimately you have: $0\cdot x+3 = 0$. No $x$ will make that true.[/color]

3) Same as number 2 except a way to make the following equation false.

[color=blue]It is false, so use whatever value of $x$ you like.[/color]

4) So I take the linear relation equations from this list are the following. (In bold)

[b]y=3x + 10 [/b]

[color=blue]Technically, no. No function of the form $y = mx+b$ is linear. I know that sounds weird, but it is the $b$ that messes things up. If $b = 0$ then everything works fine.[/color]

[b]y= 2x/3 - 5 [/b]

[color=blue]Again, no for the same reason.[/color]

Y1= 0.5x
y= 2x
F + 1= S- 2
[b]y= 3 (x-4)[/b]

[color=blue]No.[/color]
y= 2/x
[b]A= (7.3)^2t[/b]

[color=blue]Are you kidding?[/color]

[quote="helpmeplz"]2) My bad. What I meant was: Can you state the worth of x in order to make the following equation true?

3) Same as number 2 except a way to make the following equation false.[/quote]

I know what you meant.  I just felt that after what I did, the problem was trivial.


I once saw a Matlab demo where was given a picture of some face. Then Matlab plotted a 3D-function which approximates the face pretty well. Does anyone know what command gives that function or some other program to do such thing? It would be nice to make 3D-plots some of my own pictures.

It's impossible to get 3d data from 2d pictures
the only way this could happen is if they also got a relief map
perhaps you are looking for the matlab "mesh" function, which can draw any kind of surface
you could also use "fill3" to draw out polygons


Four points A,B,CD lie on a line in that order. Find the locus P such that $ \angle APB\equal{}\angle CPD$


Let $ A,B \in\mathcal{M}_n(\mathbb{C})$, such that $ AB\equal{}0$, show that $ A$ and $ B$ are simultaneously triangularizable.

P.S.: Not sure it is already on this forum.

[hide="Hint"]I posted a first step to solve this problem at the end of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=208827]Commutative transforms.[/url][/hide]
[hide="Second hint"]The next step is to induct over $ n.$[/hide]


find the smallest n ($ n\ge2$) /  $ \sqrt{\frac{\sum_{i\equal{}1}^{n}{i^{2}}}{n}}\in{IN}$
thanx.

[quote="mathematikos"]find the smallest n ($ n\ge2$) /  $ \sqrt {\frac {\sum_{i \equal{} 1}^{n}{i^{2}}}{n}}\in{IN}$
thanx.[/quote]

Hint: Note that
\[ \sum i^2 \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}
\]
.

yes,but if we use the formul,we get:
$ \sqrt{(n\plus{}1)(2n\plus{}1)}$ then the pb is to find $ min(n)$ !

Actual problem is to find minimal  $ n(n\geq2)$ for which  $ \frac {(n \plus{} 1)(2n \plus{} 1)}{6}$ is a perfect square, if such $ n$ exists of course.


geometry
trigonometry
angle bisector
perpendicular bisector
power of a point
geometry proposed

Let $ S$ be center of circle $ k$.
Let $ \mathcal{C}$ be circle with center at $ A$ and $ r\equal{}AB$.
Then $ C,D\in \mathcal{C}$. Let $ \angle CBD \equal{} \alpha$.
Then we have: $ \angle CAD \equal{} 2\alpha$ and $ \angle ABD \equal{} 60\minus{}\alpha$.
Beacuse triangles $ ABD$ and $ ACD$ are isoscale
we have $ \angle ADB \equal{} 60\minus{}\alpha$ and $ \angle ACD \equal{}\angle ADC \equal{}
90\minus{}\alpha$.
Then $ \angle BSE \equal{} 2\cdot \angle BDC \equal{} 60$.
So $ \overline{BE} \equal{}1$.

Now $ \angle BCE \equal{} 180\minus{} 60\minus{}(90\minus{}\alpha) \equal{} 30\plus{}\alpha$ and
$ \angle BEC \equal{} 120\minus{}2\alpha$ since $ ABED$ is cyclic.
Finaly $ \angle CBE \equal{}30\plus{}\alpha$ wich means $ \overline{CE} \equal{} \overline{BE} \equal{}1$.

Here is another solution.

Let $ S$ be the center of $ k$. Since $ ABDE$ is cyclic

$ \angle CBE\equal{}\angle ABE\minus{}60\equal{}180\minus{}\angle ADE\minus{}60\equal{}180\minus{}\angle ACD\minus{}60\equal{}\angle BCE$

so $ \triangle BCE$ is isosceles with $ EB\equal{}EC$. Extend $ AC$ to meet $ k$ at $ G$. Then

$ \angle CGE\equal{}180\minus{}\angle ABE\equal{}180\minus{}60\minus{}\angle CBE\equal{}180\minus{}60\minus{}\angle BCE\equal{}\angle GCE$.

so $ \triangle CGE$ is isosceles, with $ EC\equal{}EG$, so $ B,C,G$ lie on a circle centered at $ E$. Now $ \angle BSG\equal{}2\angle BAG\equal{}120$ and $ \angle BCG\equal{}180\minus{}\angle ACB\equal{}120$, so $ BSCG$ is cyclic, so $ S$ also lies on this circle centered at $ E$, so $ CE\equal{}SE\equal{}1$.

A simpler solution based upon Number1:
A is the excenter of the triangle BCD; since angle <BAC = 60, it follows that <BDC = 30 = <BDE = < BAE, hence BE = 1 ( radius of the excircle BADE ) and AE is the angle bisector of <BAC, thence the perpendicular bisector of BC and CE = BE.

Best regards,
sunken rock

ABED being cyclic and AB = AD, it follows that EA is the angle bisector of <BED. It is well known that the middle of the arc BC of the circle (BEC) is on EA and since that point and A lie on the perpendicular bisector of BC; it follows that AE is the perpendicu;ar bisector of BC, and CE = BE ( 1 ) and <BAE = 30, which gives BE = 1 ( the radius of circle BADE ), hence CE = 1.

Best regards,
sunken rock

Let $ O$ be the center of $ k$ and for any $ P,Q$ on $ k$ define the centerangle $ \angle POQ \equal{} \widehat{PQ}$.
Let $ A',B'$ be the intersection of $ AC,BC$ with $ k$ different from $ A,B$ respectively.
From the power of a point: $ CA \cdot CA' \equal{} CB \cdot CB' \Rightarrow CA' \equal{} CB'$
Let $ \widehat{BA} \equal{} \widehat{AD} \equal{} v$
$ \angle CAB \equal{} \frac {\widehat{BA} \plus{} \widehat{B'A'}}{2} \iff \widehat{B'A'} \equal{} 120^\circ \minus{} v$
$ \angle ABB' \equal{} \frac {\widehat{AD} \plus{} \widehat{DB'}}{2} \iff \widehat{DB'} \equal{} 120 \circ \minus{} v$
$ \widehat{DB'} \equal{} \widehat{B'A} \iff DB' \equal{} B'A \Rightarrow \triangle DB'C$ is isosceles.
$ \angle DB'C \equal{} \frac {\widehat{BA} \plus{} \widehat{AD}}{2} \equal{} v \Rightarrow \angle CDB' \equal{} 90 \minus{} \frac {v}{2} \equal{} \angle BCE$.
$ \angle CDB' \equal{} \angle EDB' \equal{} \frac {\widehat{B'A'} \plus{} \widehat{A'E}}{2} \iff \widehat{A'E} \equal{} 60 ^\circ$
$ \triangle OEA'$ equilateral $ \Rightarrow EA' \equal{} OE \equal{} 1$
$ \angle CEA \equal{} 120^\circ \minus{} v, \angle EA'C \equal{} \angle ECA' \equal{} 30 ^\circ \plus{} \frac {v}{2} \Rightarrow CE \equal{} A'E \equal{} 1$. QED

Let $\angle{ADC}=a$ and $O$ denote the center of circle $k$.

Since $AD=AC$, $\angle{BCE}=180-\angle{BCD}=180-(60+\angle{ACD})=120-a$. Note that $\angle{EBC}=180-a-\angle{ABC}=120-a$. Therefore $\triangle{ECB}$ is isosceles (1). By (1) and because $ABC$ is equilateral, $EA$ bisects $\angle{CAB}$ and therefore $\angle{EAB}=30$.

Hence $\angle{EOB}=60$ and $\triangle{EOB}$ is equilateral (2). By (1) and (2), it follows that $CE=EB=OE=1$.

[quote="Ahiles"]The vertices $ A$ and $ B$ of an equilateral triangle $ ABC$ lie on a circle $k$ of radius $1$, and the vertex $ C$ is in the interior of the circle $ k$. A point $ D$, different from $ B$, lies on $ k$ so that $ AD\equal{}AB$. The line $ DC$ intersects $ k$ for the second time at point $ E$. Find the length of the line segment $ CE$.[/quote]

By construction, both $D$ and $E$ are on $k$, the same as $A$ and $B$, duh.

Let <ABD=x. By angle chase, we have <CBE=90-x and <CEB=2x. Now by sine law, we have BC/sin(2x)=CE/cos(x) and AB/sin(x)=2. So BC/2sin(x)cos(x) =1/cos(x)=CE/cos(x) and so CE=1.

Very nice problem here! I used angle chasing for the solution. [quote=Konigsberg]How does everyone know that ABED is cyclic here?[/quote]

We are given that they lie on circle k.


algebra
polynomial
Euler
number theory unsolved
number theory

After multiplying out and simplifying polynomial $ (x \minus{} 1)(x^2 \minus{} 1)(x^3 \minus{} 1)\cdots(x^{2007} \minus{} 1),$ getting rid of all terms whose powers are greater than $ 2007,$ we acquire a new polynomial $ f(x).$ Find its degree and the coefficient of the term having the highest power. Find the degree of $ f(x) \equal{} (1 \minus{} x)(1 \minus{} x^{2})...(1 \minus{} x^{2007})$ $ (mod$ $ x^{2008}).$

The question is equivalent to finding such maximum $d<2008$ that [url=http://oeis.org/A010815]A010815[/url](d) is non-zero (follow the link for further references and formulas).

[quote="Rust"]Result is
\[f(x)=1+\sum_{k=1}^{36}(-1)^{k}(x^{k(3k-1)/2}+x^{k(3k+1)/2}). \]
Therefore $deg(f(x))=1962$.[/quote]

how did you get that expansion ? Thanks

In fact,it's also bacause the professor was born in [b][size=150]1962[/size][/b].
  :D

The known fact is called Eulers Pentagonal number theorem.

[quote="malinger"]I don't understand, why this theorem has to state the sum to infinity, not finite.[/quote]
I searched and I have http://en.wikipedia.org/wiki/Pentagonal_number_theorem That is all.

[quote="N.T.TUAN"]
I searched and I have http://en.wikipedia.org/wiki/Pentagonal_number_theorem That is all.[/quote]

That's all right, but just wondering, does anyone know the reason why to infinity?

I don't see the sense of your question...
Proofing it gives us this, so what's to discuss¿

One could form the problem with
$f(x)=\prod (1-x^{i})^{3}$ instead. This time the solution needs another nice identity.


Let f be a real-valued function such that f(m+n)=f(m)f(n). If f(4)=256 and f(k)=0.0625, find the value of k.

for both of you, i don't think that

$f(2)^2=256$ implies $f(2)=16$

doesn't it implie $f(2)= \pm 16$? But since we have not restriction on the iniectivity of the function, we can take one between $+16$ and $-16$ that help us to solve the equation, not?
And (I'm not sure of this), but can't we prove that the only function $\mathbb {R} \rightarrow \mathbb {R}$ that satisfy
$f(x+y)=f(x)f(y)$ $\forall x,y \in \mathbb {R}$ is the exponential one? ('cause I remeber something about the fact that the logaritm is the only function satisfying $f(xy)=f(x)+f(y)$)

my point was that you have to at least address the case where $f(2)=-16$


induction
number theory proposed
number theory

Let n be a natural number(n>3) and m a product of all primes that are smaller than n.
Solve equation m=2n+16.

Posted before:
http://www.mathlinks.ro/viewtopic.php?p=832717#832717

Anyway, I'd like to see a solution without using Bertrand's postulate

I don`t have much time but here are some steps:
1.  n is odd
2.  n=6k+1 (k is a natural number)
3.  7 is a solution
4.  if n is grater than 7 than there are no more solution
5.  4.   can be proved by induction(I think, it is enough to prove that m>4n)

Try this and correct me if I am wrong  :oops:


[quote="ashrafmod"]2)  a,b > 0

(a^2 +1)\b +  (b^2 +1)\a   >= 4[/quote]
By Cauchy in Engel form and $a+\frac1a\geqslant 2$:
$\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a}= \frac{a^{2}}{b}+\frac{b^{2}}{a}+\frac1a+\frac1b\geqslant$
$\frac{(a+b)^{2}}{a+b}+\frac1a+\frac1b= \left(a+\frac1a\right)+\left(b+\frac1b\right)\geqslant 4$.

[quote="ashrafmod"]1) if xyz(x+y+z) =1

prove
(x+ y)(z+ y) >=2
[/quote]
$xyz(x+y+z)=1\Rightarrow{y(x+y+z)=\frac{1}{zx}}\Rightarrow{yx+y^{2}+yz+xz=\frac{1}{xz}+xz}\Rightarrow(x+y)(z+y)=xz+\frac{1}{xz}\geq2$


for odd prime p and q  ,      $ q \mid a^p \minus{} 1$      prove that $ q \mid a \minus{} 1$ or $ q \equal{} 2kp \plus{} 1$

If $ (q,a \minus{} 1) \equal{} 1$, we have $ a^{q \minus{} 1} \minus{} 1\vdots q$.
$ \Rightarrow q \minus{} 1\vdots p$. Because $ q$ odd so $ q \minus{} 1\vdots 2p$.


Here is the answer to the question "what is 1+1?"
[hide]2[/hide]

 ;)

Why is this is the NH thread?

And HyperSet, I believe the whole of AoPS knows you like that song.

xD


If 6 computer programmers can write 8 computer programs in 3 days, how many programmers will be required to write a series of 20 similar programs in 15 days?

We have 2.5 times more programs, but 5 times more days, so we have $ 6\cdot2.5/5\equal{}3$ programmers

We see that the number of computer programmers is directly proportional to the number of computer programs written but inversely proportional to the number of days.

Hence:

Let $x$= the number of programmers

$\frac{6\cdot 3}{8}=\frac{15x}{20}$

$x=\boxed{3}$


I would be very grateful if you could list a couple of universities in Germany, Switzerland and Austria where you would recommend studying Mathematical Methods in Finance.

10x in advance


For a positive integer $ n$ there is a set $ S$ containing infinite positive integers such that the arithmetic mean of the integers in any nonempty subset of $ S$ is an integer. Prove or disprove.

(For finite, we can prove this easily be considering a set $ S$ consisting of $ a_k$ for $ k\equal{}1,2,3,..,n$ such that $ a_k \equal{} k n!$. )

What $ n$ has to do with that? :huh:

Anyway, such a set does not exist.
Suppose that $ S$ is such an infinite set. Let $ a$ and $ b$ be two distinct elements from $ S$. For each integer $ n > 1$ let's consider $ n \minus{} 1$ elements from $ S \minus{} \{a,b\}$, and let $ s$ be their sum.
Then $ a \plus{} s \equal{} 0 \mod[n]$ and $ b \plus{} s \equal{} 0 \mod[n]$, so that $ a \equal{} b \mod[n]$. It follows that $ a \minus{} b$ is a multiple of $ n$ for each $ n > 1$. Then $ a \equal{} b$, a contradiction.

Pierre.


function
inequalities
inequalities proposed

For three positive numbers $a,b,c$
$4(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b})^2 \ge 3(a^2+b^2+c^2)$

Let $a^2+b^2+c^2=3$
So now we want to find the minimum of $\sum \frac{a^2}{\frac{b+c}{2}} \geq \sum \sqrt{2} \frac{a^2}{\sqrt{b^2+c^2}} \geq \sum \sqrt{2} \frac{a^2}{\sqrt{3-a^2}}$.

But the function $f(x) = \frac{x}{\sqrt{3-x}}$ is convex for $x \in (0,3)$, so by Jensen, the minimum happens when $a=b=c$.


AMC
AIME
algebra
difference of squares
special factorizations
arithmetic series

Let $ N\equal{}100^2\plus{}99^2\minus{}98^2\minus{}97^2\plus{}96^2\plus{}\cdots\plus{}4^2\plus{}3^2\minus{}2^2\minus{}1^2$, where the additions and subtractions alternate in pairs. Find the remainder when $ N$ is divided by $ 1000$.

[hide]
We can get rid of the $ 100^2$ since it is divisible by 1000.

Using difference of squares repeatedly, we get

$ N\equal{}99^2\minus{}98^2\plus{}96^2\minus{}97^2\plus{}95^2\minus{}94^2\plus{}92^2\minus{}93^2\cdots\plus{}4^2\minus{}5^2\plus{}3^2\minus{}2^2\plus{}0^2\minus{}1^2$

$ N\equal{}(99\minus{}98)(99\plus{}98)\plus{}(96\minus{}97)(96\plus{}97)\plus{}(95\minus{}94)(95\plus{}94)\plus{}\cdots\plus{}(4\minus{}5)(4\plus{}5)\plus{}(3\minus{}2)(3\plus{}2)\minus{}1$

$ N\equal{}197\minus{}193\plus{}189\minus{}185\plus{}\cdots\minus{}9\plus{}5\minus{}1$

There are 25 pairs that each sum to 4, so the answer is $ 25\cdot4\equal{}\boxed{100}$[/hide]

[hide="another different way"]I noticed that each quadruple such as $ 100^2 \plus{} 99^2 \minus{} 98^2 \minus{} 97^2$ differed by 32. For example, $ 4^2\plus{}3^2\minus{}2^2\minus{}1^2\equal{}20$ and $ 8^2\plus{}7^2\minus{}6^2\minus{}5^2\equal{}52$ and 84.... then 126... so it forms an arithmetic series with 25 terms, and the 1st term is 20, and the last term is 788. Therefore, using the arithmetic series sum:

$ \frac{25(20\plus{}788)}{2}\equal{} 10100$ Divide 10100 by 1000, and get 100. [/hide]

[hide]
Our idea will be look at each block of 4 numbers and use difference of squares between the evens and the odds.

$ 100^2 \plus{} 99^2 \minus{} 98^2 \minus{} 97^2 \equal{} 2(100 \plus{} 98) \plus{} 2(99 \plus{} 97)$
$ 96^2 \plus{} 95^2 \minus{} 94^2 \minus{} 93^2 \equal{} 2(96 \plus{} 94) \plus{} 2(95 \plus{} 93)$
.
.
.
$ 4^2 \plus{} 3^2 \minus{} 2^2 \minus{} 1^2 \equal{} 2(4 \plus{} 2) \plus{} 2(3 \plus{} 1)$

and our sum is just:
$ 2(100 \plus{} 99 \plus{} 98 \plus{} \ldots 2 \plus{} 1) \equal{} 10100$

So our answer is $ \fbox{100}$
[/hide]

Same solution, I know, but it doesn't use any of the above ideas with arithmetic sequences etc, just good old triangular numbers.

Did anyone else just do it out? $ (4n)^2\minus{}(4n\minus{}2)^2\plus{}(4n\minus{}1)^2\minus{}(4n\minus{}3)^2\equal{}2(8n\minus{}2)\plus{}2(8n\minus{}4)\equal{}32n\minus{}12$

[quote="The Zuton Force"]Did anyone else just do it out? $ (4n)^2 \minus{} (4n \minus{} 2)^2 \plus{} (4n \minus{} 1)^2 \minus{} (4n \minus{} 3)^2 \equal{} 2(8n \minus{} 2) \plus{} 2(8n \minus{} 4) \equal{} 32n \minus{} 12$[/quote]That's essentially what I did, just not so rigorously.

[quote="The Zuton Force"]Did anyone else just do it out? $ (4n)^2 \minus{} (4n \minus{} 2)^2 \plus{} (4n \minus{} 1)^2 \minus{} (4n \minus{} 3)^2 \equal{} 2(8n \minus{} 2) \plus{} 2(8n \minus{} 4) \equal{} 32n \minus{} 12$[/quote]

Yeah... and then i lost my sigma somwhere and for some reason did $ 32n \minus{} 12$ instead of $ \sum^{25}_{k \equal{} 1} 32k \minus{} 12$...not noticing that the answer is way too small, and not greater than 1000... :wallbash_red:

[hide="Answer"]We group every four terms together. We can apply difference of squares on each. So, we have for the first four:

$ 100^2\plus{}99^2\minus{}98^2\minus{}97^2\equal{}(99\plus{}98)\plus{}(100\plus{}97)3\equal{}4(99\plus{}98)\equal{}2(100\plus{}99\plus{}98\plus{}97)$.

Likewise, we can see that this extends. So, $ N\equal{}2(100\plus{}99\plus{}98\plus{}\dots \plus{}1)\equal{}2(5050)\equal{}10100$. Answer is then $ \boxed{100}$.[/hide]

[quote="The Zuton Force"]Did anyone else just do it out? $ (4n)^2 \minus{} (4n \minus{} 2)^2 \plus{} (4n \minus{} 1)^2 \minus{} (4n \minus{} 3)^2 \equal{} 2(8n \minus{} 2) \plus{} 2(8n \minus{} 4) \equal{} 32n \minus{} 12$[/quote]

I did exactly that lol.  :)

[quote=HrishiP][quote=ishanvannadil2008]@above wow almost 13 year bump[/quote]

Tbh I think it isn't worth posting after every thread bump "@above you bumped a $x$ year old thread" or something along the lines of that since nobody really cares and it's just extra spam for people who look through this thread in the future (sorry if that sounded harsh).[/quote]

its only one post...


Supose I have a PDF in the directory "...\pdf" and some sounds in wav in the directory "...\pdf\sound"

How can I make a link in my pdf to play the sounds in some program?

The link is a offline one to outside the archive.
If someone have a good solutions, thanks.

[code]\usepackage{hyperref}
...
\href{sound/filename.wav}{Click to play sound}[/code]
Note:
1. You must use / not \ for paths as \ is seen as a latex command
2. Try to avoid spaces in file names but if you must have them then use quotes like "sound/file name.wav"
3. The sound will only play if the user's computer is set up to play files with whatever extension you are using.

After a long search on internet i found the solution

[code]\href{run:sound/audio.wav}{Play}[/code]

It need the command run

but thanks


calculus
integration
combinatorics unsolved
combinatorics

In how many ways could one ditribute cookies among 4 people $A,B,C and D$, if each person i to get at least one cookie?

[quote="boxedexe"]$a+b+c+d=n$, where $n$ is the number of cookies.  Then the number of positive integral solutions to the equation in terms of $n$ is $\binom{n-1}{3}$.  You can determine this by putting $3$ $)+($ signs in between the $n-1$ empty spaces of $1$_$1$_..._$1=n$.[/quote]

I sorry, but it should be 8 cookies in the problem text   :blush: .
Could you please explain your method? I would appreciate if you
could help me with both the general problem and $n=8$.

My try for $N=8$:
Since everybody gets a cookie $n$ is more or less reduced to 4.
So we have a distribution among $ABCD$ as follows (except the one cookie all of them get):
( 4 0 0 0 or 3 1 0 0 or 2 1 1 0 or 1 1 1 1 ) or  permutations of these.
So summing up the number of cases would give the answer, right?

The way you're doing it, you forgot 2 2 0 0. Yes, adding the permutations of each would result in the answer.

Question (to boxedexe): So in an equation of the form $a_{1}+a_{2}+a_{3}+\ldots+a_{n}=x$, the number of positive integral solutions is $\binom{x-1}{n-1}$?

If so, the answer to the problem would be $\boxed{35}$.

[quote="boxedexe"]$a+b+c+d=n$, where $n$ is the number of cookies.  Then the number of positive integral solutions to the equation in terms of $n$ is $\binom{n-1}{3}$.  You can determine this by putting $3$ $)+($ signs in between the $n-1$ empty spaces of $1$_$1$_..._$1=n$.[/quote]

Could you please explain how you get that formula again?
I mean \binomial{n-1}{x-1}. Please the easy way  :) .

Again thanks for your earlier answers Masoud.

How should one solve the problem when the cookies are
different?

[quote="boxedexe"]Multiply the above result by $n!$, that is $n!\binom{n-1}{3}$.[/quote]
Is it really correct to do so?
Another way I have seen is to consider #ways=#surjections([8]-->[4])=
\[\sum_{k=0}^{4}(-1)^{k}\binom{4}{k}(4-k)^{8}=40824\]

but $n!\binom{n-1}{3}$ would be 1411200.

[quote="JBL"][quote="color"]Is it really correct to do so?[/quote]

No, because A doesn't care which his "first" and "second" cookie are.[/quote]

Yes exactly, but do you have a solution, preferably a simple/easy one, to this problem?
Thanks for your reply.


Express the largest of the numbers below as a decimal.

\[ \sqrt{2\frac{1}{4}}\qquad (1.2)^2 \qquad \sqrt{2.56}\]

hello, we have $ \sqrt{2.56}\equal{}1.56>\sqrt{2\frac{1}{4}}\equal{}1.5>1.2^2\equal{}1.44$, so the largest number is $ \frac{16}{10}\equal{}\frac{8}{5}$.
Sonnhard.


ratio
LaTeX
algebra
polynomial
arithmetic sequence
geometric series

What is the name of sequences that have arithmetic and geometric qualities, but are not arithmetic or geometric? For example:
$ \sum_{n\equal{}1}^\infty \frac{n}{2^n} \equal{} \frac{1}{2} \plus{} \frac{2}{4} \plus{} \frac{3}{8} \plus{} \frac{4}{16}\ldots$
The numerators are in an arithmetic sequence, and the denominators are in a geometric sequence. Also, how would you find the sum of these types of sequences? I know that the infinite sum above is equal to 2, but I'm not sure why algebraically. Thanks in advance!

these are called arithmetico geometric series

let S be the sum
$ S \equal{} \frac12 \plus{} \frac24 \plus{} \frac38 \plus{} \frac {4}{16} \ldots$
$ \frac12S \equal{} \; \; \; \; \; \frac14 \plus{} \frac28 \plus{} \frac {3}{16} \ldots$
subtract and get
$ \frac12S \equal{} \frac12 \plus{} \frac14 \plus{} \frac18 \plus{} \frac {1}{16} \ldots$
which is geometric.

then solve for $ \frac12S$ and then find S
basically multiply an arithmeco geometric series by its 'geometric' ratio and subtract it from itself to form a geometric series...
hope you understand...
my LaTeX isnt that good :(

These kinds of sequences don't have a specific name; they're just one of the many many types with easy-to-describe mathematical patterns.

In general, for any polynomial $ P(n)$, you can use the method of finite differences to evaluate $ \sum_{n \equal{} 0}^{\infty}{P(n) r^n}$, where $ |r| < 1$ (so that $ \sum_{n \equal{} 0}^{\infty}{r^n}$ is itself a convergent series). The above problem is the case $ P(n) \equal{} n, r \equal{} \frac {1}{2}$. [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=1166263#1166263]This[/url] has some good information on finite differences, among other things.


Evaluate the infinity series

$ \sum^{\infty}_{k\equal{}0}\frac{2^k}{2^{2^k}\plus{}1}$

That series has already been posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125257]here[/url], although still without a solution.

$ \sum_{n}\zeta_n$ converges if and only if $ \sum_{n}2^n\zeta_{2^n}$ converges. I believe it's called Kummer's criteria .

Edit: Okey, then Cauchy Condensation Test  :wink: .

Actually that's Cauchy Condensation Test.


Let $ M$ be an arbitrary point inside triangle $ ABC$ and $ A_1B_1C_1$ be its cevian triangle. Call $ A_2$, $ B_2$, $ C_2$ be the midpoints of the segments $ B_1C_1$, $ C_1A_1$, $ A_1B_1$, respectively. Denote points $ A_3 \equal{} A_2M\cap BC$, $ B_3 \equal{} B_2M\cap CA$ and $ C_3 \equal{} C_2M\cap AB$. Prove that $ AA_3$, $ BB_3$, $ CC_3$ are concurrent at a point lies on the line connecting $ M$ and the centroid of triangle $ A_1B_1C_1$.

[b]EDIT.[/b] Thank you, dear Darij  :)

I am sorry to tell this, but this is a particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=5117 ...

  darij


Let $S_n$ and $T_n$ be the respective sums of the first $n$ terms of the two arithmetic series.  If $S_n : T_n = (7n+1) : (4n + 27)$ for all $n$, the ratio of the eleventh term of the first series to the eleventh term of the second series is:
\[ \text{(A)}\ 4 : 3\ \ \ \ \text{(B)}\ 3: 2\ \ \ \ \text{(C)}\ 7 : 4\ \ \ \ \text{(D)}\ 78 : 71\ \ \ \ \text{(E)}\ \text{undetermined} \]

[hide="Solution"]
Let $a_s$ be the first term of the first series and $d_s$ be the common difference of the first series, and similiarly define $a_t$ and $d_t$. We have
\[\frac{S_n}{T_n}=\frac{2a_s+(n-1)d_s}{2a_t+(n-1)d_t}=\frac{7n+1}{4n+27}.\]
We are looking for
\[\frac{a_s+10d_s}{a_t+10d_t}.\]
If we substitute $n=21$ into our equation, we can factor out 2 and be left with our desired ratio:
\[\frac{2a_s+20d_s}{2a_t+20d_t}=\frac{7(21)+1}{4(21)+27}\]
\[\frac{a_s+10d_s}{a_t+10d_t}=\frac{148}{111}=\frac{4}{3}.\]

Thus the answer is (A).[/hide]

Hmmm...

[hide]I did it differently and got a different answer.  Consider the sum of the first 11 terms as 7(11)+1=78.  So subtracting the sum of the first 10 terms one can get the value of the eleventh term.  So the sum of the first ten terms is 7(10)+1=71.  So the 11th term is 7.  Then the first 11 terms of the second sequence sum to 71, and the first 10 sum to 67.  Thus the eleventh term is equal to 4.  So shouldn't the ratio be 7:4.[/hide]

Sn/Tn are the ratios and not the actual figures.  That's where u went wrong.  Tricky tricky  :D


Let $ a,b,c$ be positive reals. Prove that
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \ge \frac{ab}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{bc}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{ca}{a^2\plus{}ab\plus{}b^2}.$

Not quite the nicest, but still a solution.

Dear canhang2007,

when you have any time, can you please post your solution on your beautiful blog.

Thank you very much. :)

[quote="can_hang2007"]Let $ a,b,c$ be positive reals. Prove that
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \ge \frac {ab}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {bc}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {ca}{a^2 \plus{} ab \plus{} b^2}.$[/quote]
This inequality is [b]Problem 2.24 - page 60[/b] in new book (where it is presented with a simple solution by Cauchy-Schwarz):

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://can-hang2007.blogspot.com/2009/12/new-book-inequalities-with-beautiful_20.html][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.


how far have u guys made it into mathcounts?

just curious

I made it to the school round by schooling all the other people in the preliminary school round.

Ignite and I went to nats last year. 56th and 114th, respectively.

[quote="worthawholebean"]Ignite went to nats last year. 56th[/quote]

I AM SUING YOU FOR LIBEL  :mad:  STOP LYING!

the yale center is starting math counts in all public school.. i may audition to do it.. seems like a very intresting thing.. :o


For each odd prime p, dtermine all integer i such that

                  1^(p-1)+2^(p-1)+...+i^(p-1)           is divisble by p.

Bomb.

IS BACK! :D

I think that $i=pk+1$

Proof : According to Fermat, if $(i,p)=1$, then $i^{p-1}=1 (p)$
This implies that $\sum_{i=1}^n  i^{p-1} = n (p)$ and like $p^{p-1}=0 (p)$, $n = pk+1$

euhh, my conclusion is false : the answer is $i=pk+k$ because you must consider al multiples of $p$.


algebra
polynomial
limit
algebra proposed

Find all real polynomials $ P$ so that we have $ P(x)P(2x^2\minus{}1)\equal{}P(x^2)P(2x\minus{}1)$ $ \forall x \in R$

$ degP \equal{} 0 \rightarrow P(x) \equal{} C$
$ degP > 1$

$ \Leftrightarrow \frac {P(2x^2 \minus{} 1)}{P(x^2)} \equal{} \frac { P(2x \minus{} 1)}{P(x)} \forall x \in R, P(x) \ne 0$
$ \Rightarrow \frac {P(2x^2 \minus{} 1)}{P(x^2)} \equal{} \lim_{n \rightarrow \plus{} \infty} \frac { P(2x^{\frac {1}{n}} \minus{} 1)}{ P( 2x^{\frac {1}{n}})} \equal{} k \ne 0$( $ P(1)$ can be equal to 0)
$ \Rightarrow P(2x \minus{} 1) \equal{} kP(x) \forall x$
$ \Rightarrow P(x) \equal{} 0$ has unique complex root  $ 1$

So , this equation has two solutions :$ P(x) \equal{} C , P(x) \equal{} k(x \minus{} 1)^n$( $ C,k$ are constants , $ n \in N^*$ )

Any constant polynomial satisfies our condition. Now consider $ P(x)$ is not a constant.

If $ P(1) = 0$ then put $ P(x) = (x-1)^r Q(x)$ where $ Q(1)\neq 0$. We have: $ (x-1)^r Q(x) (2x^2 - 2)^r Q(2x^2-1) = (x^2-1)^r(2x-2)^rQ(x^2)Q(2x-1)$, which means $ Q(x)Q(2x^2-1)=Q(x^2)Q(2x-1)$, so $ Q(x)$ also satisfies our condition and $ Q(x)\neq 0$. Thus we can assume that $ P(1)\neq 0$.

Since $ P(1)\neq 0$ then:
$ \frac{P(2x^2-1)}{P(x^2)}=\frac{P(2x-1)}{P(x)}=\lim_{n\to\infty} \frac{P(2x^{\frac{1}{2^n}}-1)}{P(x^{\frac{1}{2^n}})}=\frac{P(1)}{P(1)}=1$

Hence $ P(2x-1)=P(x),\forall x\in\mathbb{R}$. Consider $ x_0\neq 1$ then the sequence $ x_0,x_{n+1}=2x_n - 1$ approaches $ \infty$ when does $ n$. This means that $ P(x)$ is a constant.

Therefore that class of polynomial solutions for the problem is $ P(x) = C(x-1)^n$ where $ C$ is constant and $ n\in\mathbb{N}$.


modular arithmetic
number theory
number theory proposed

Let $m,n$ be positive intergers.
Prove that $4mn-m-n$ can never be a square

If $4mn - m - n = k^2$ is a perfect square, then we have $4k^2 + 1 = (4m - 1)(4n - 1)$.

Hence $(2k)^2 + 1$ must have a prime divisor congruent to $-1 \pmod{4}$, but we know that this cannot happen.

(Indeed, if $p$ is a prime, $n$ is a positive integer and $p$ divides $n^2 + 1$, then $p = 2$ or $p \equiv 1 \pmod{4}$.)

Can you prove that prime statement?

I'm not saying you're wrong, but I can't see an easy way to prove it. There probably is one though...

@kingof20penguins :
It's clear if $p = 2$. So consider $p$ odd prime.
If $p|n^{2}+1$, then $n^{2}\equiv-1(mod p)$ and $(n,p) = 1$.
By Fermat's Little Theorem, we have $n^{p-1}\equiv 1 (mod p)$.
But $n^{p-1}= (n^{2})^{\frac{p-1}{2}}\equiv (-1)^{\frac{p-1}{2}}(mod p)$.
Thus, $\frac{p-1}{2}$ must be even and $p \equiv 1 (mod 4)$.

This fact is very useful in NT..  :)


This is one of Maclaurin's inequalities, for the non-negative roots, via Viete's relations.
See [url]http://en.wikipedia.org/wiki/Maclaurin%27s_inequality[/url]

[quote="thuyanh158"]let $ P(x) \in R_{[x]}$ which $ P(x) \equal{} {x^n} \plus{} {a_1}{x^{n \minus{} 1}} \plus{} {a_2}{x^{n \minus{} 2}} \plus{} ... \plus{} {a_n}$
$ P(x)$ has $ n$ nonegative roots. Show that:
\[ {\left( {\frac {{{a_{n \minus{} 1}}}}{n}} \right)^n} \ge a_n^{n \minus{} 1}\]
[/quote]
this problem can be sloved by viete theorem and AM-GM inequality. easy to see :oops:


Has anyone managed to create a solution that combines both polar and nonpolar molecules?

If so, theoretically, could this be a much more effective solvent?

may be u would like tp have a look at [url=http://en.wikipedia.org/wiki/Phase_transfer_catalyst]this[/url]

[quote="Diophantient"]Has anyone managed to create a solution that combines both polar and nonpolar molecules?[/quote]

Nature already does that for millions of years at every second, when carbon dioxide dissolves in water to give its caracteristic acidic pH.


[quote="GameBot"]How many of the first 500 positive integers are divisible by 3, 4
and 5?[/quote]

$ LCM(3,4,5)\equal{}60$

$ \lfloor \frac{500}{60} \rfloor \equal{} 8$.


LaTeX
inequalities unsolved
inequalities

[quote="kihe_freety5"]prove that :
$ x,y,z >0$ we have :
\[ \sqrt {\frac{(x + y)}{z}}+\sqrt {\frac{(z + y)}{x}}+\sqrt {\frac{(x + z)}{y}}  \ge \sqrt{\frac{6(x + y + z)}{\sqrt[3]{xyz}}}\][/quote]
I helped you in latex :)

PS: The author of this problem is ductrung (pham huu duc)

[quote="manlio"]What page of the E-BOOK?  :maybe:

For many solutions look at

http://reflections.awesomemath.org/2007_2/MR_1(2007)solutions.pdf[/quote]
Oh, I am sorry, manlio. It is not the full of my ebook, I will post my full ebook soon. (By the way, my last ebook has some serious errors. Now, I have corrected them. :) )

[quote="shfdfzhjj"][quote="can_hang2007"][

Oh, I am sorry, manlio. It is not the full of my ebook, I will post my full ebook soon. (By the way, my last ebook has some serious errors. Now, I have corrected them. :) )[/quote]

But,where can I see your ebook?
Thank you very much. :)[/quote]
I posted it here http://www.mathlinks.ro/viewtopic.php?t=197674.
It had many mistakes, I will post my correction soon. :)

I have a solution!!
it is equavalent:
$ \frac {x \plus{} y}{z}$ + $ \frac {x \plus{} z}{y}$ + $ \frac {z \plus{} y}{x}$+2$ \sqrt {\frac {(x \plus{} y)(x \plus{} z)}{yz}}$ + 2$ \sqrt {\frac {(z \plus{} y)(x \plus{} y)}{xz}}$ + 2$ \sqrt {\frac {(z \plus{} y)(x \plus{} z)}{yx}}$ ≥ 6$ \frac {x \plus{} y \plus{} z}{\sqrt [3]{xyz}}$;
we prove that $ \frac{z}{x}$+$ \frac{z}{y}$+ 2$ \sqrt{\frac{(z\plus{}x)(z\plus{}y)}{yx}}$≥6$ \sqrt[3]{\frac{z^2}{xy}}$;
setting a=$ \frac {z}{x}$;b=$ \frac {z}{y}$
we need to prove that
a+b+$ \sqrt {(a \plus{} 1)(b \plus{} 1)}$≥6$ \sqrt [3]{ab}$
by AM-GM we have a+b+$ \sqrt {(a \plus{} 1)(b \plus{} 1)}$≥a+b+4$ \sqrt [4]{ab}$=a+b+$ \sqrt [4]{ab}$+$ \sqrt [4]{ab}$+$ \sqrt [4]{ab}$+$ \sqrt [4]{ab}$≥6$ \sqrt [3]{ab}$
so, we have ended!!!
thanks can_hang 2007


AMC
USA(J)MO
USAMO
Harvard
college
analytic geometry
Gauss

Can someone please explain what this is, and more importantly, how it is used? I gave a (past usamo qualifier and harvard soph) friend of mine this problem that i had that took me a while to do using functional remainders and such, but he did it in a very quick way using the chinese remainder theorem and of course, he was too busy to explain it.

The Chinese Remainder Theorem tells you that, given a system of modular equations
x :equiv: a1 (mod m1)
x :equiv: a2 (mod m2)
x :equiv: a3 (mod m3)
. . .
x :equiv: an (mod mn),
there is exactly one solution
x :equiv: z (mod m1m2m3. . .mn), provided that the pairwise greatest common divisor of the mi's is 1.  (That is, given any 2 of the mi's, they have no common factors.)

It doesn't go about telling you how to solve such a system, unfortunately.  However, that is the kind of thing you can best pick up with practice.  The most straight-forward method is just to solve the systems in pairs until you get the answer.  (Solving one pair isn't usually that hard.)

That's not quite true. The Chinese Remainder Theorem also provides a method for solving the  equations.

If you let M be the product of the mj's, and let tj = M/mj, and then find sj's such that tj*sj :equiv: 1 (mod mj);

Then x = a1*t1*s1 + a2*t2*s2 + ... + an*tn*sn is a solution to all of the original congruences. (Can you see why?)

An account of the Chinese Remainder Theorem can be found in most good number theory books - including a cheap one called [u]Number Theory[/u] by George Andrews.

Yeah, it can be used to find a solution as you said (or of course, you could just do it one equation at a time, building it up algebraically as you go). But usually it is more used for existence, quite often in an olympiad-type problem using this, you end up with, say, 2003 equations, the mods being all coprime, and then you can say there exists a solution to this.

Like that problem I posted ages ago here:

a lattice point is visible from the origin if there are no other lattice points on the line connecting the origin and that point. show there exists a 2003x2003 square of points all being invisible from the origin.

edit - actually, since you wanted to see how it worked, I'll put the answer here in spoiler anyway..

[hide]

Lets just let 2003=n for now

First point for noticing a point (x,y) is visible if gcd(x,y)=1.

Now we want to find a group of points all invisible, ie for each point the coordinates have a common factor. Lets choose n2 primes, p1 thru pn2. Then we decide that we want the first point in the square to have each coordinate divisible by p1, the second one p2 etc.

We end up with a lot of equations:

x :equiv: 0 mod p1

x :equiv: -1 mod p2

etc

x :equiv: 0 mod pn+1

etc etc

and same for the ys.

By chinese remainder theorem, exists an x and a y, and thus point (x,y) is the lower left corner of our huge invisible square.]

Note they didn't have to be primes, just all coprime to each other, primes makes it more obvious.

[/hide]

I know this is an old thread, but there is something I don't quite understand. As Joel stated after his series of congruences, "there is exactly one solution
x = z (mod m_1m_2m_3. . .m_n)" if gcd(m_1, m_2, .., m_n) = 1. But what is meant by "exactly one solution"? Isn't the general statement that there is exactly one solution to a = b (mod c) always true? For example, 7  :equiv: 1 (mod 3). Of course, 7 = 4 = 7 = 10 = 13 = ... (mod 3) too, but then again, x = z = z+m_1+m_2+...+m_n = z+2(m_1+m_2+...+m_n) (mod m_1m_2m_3. . .m_n).

It just means there is a single solution for x to all those equations. (of course this means there are infinite actual solutions because you could just keep adding the mod). In general there may be more than one solution that satisfies all of them..

Solve the system:
x :equiv: 3 (mod 4)
x :equiv: 8 (mod 9)

We can check with out too much trouble that any number x :equiv: 35 (mod 36) satisfies those equations.  The CRT says that [i]only[/i] numbers congruent to 35 (mod 36) satisfy it -- 35 (or -1, or 71, or . . .) is the unique solution.

Ah, silly me. Thx for the clarification.

I'm still unclear, can someone show me the process of solving this?  I don't understand, especially how to find the inverse.  Given that 3^302  :equiv: 4 (mod5), and 3^302  :equiv: 2 (mod 7) and 3^302  :equiv: 9 (mod 11), what is 3^302 (mod 385) ?  385 = 7*5*11.  How would I do this using the method gauss gave, rather than using substitutions?

Well, following gauss's notation, we have a1 = 4, a2 = 2, and a3 = 9.  Also, m1 = 5, m2 = 7, and m3 = 11.  Thus t1 = 7 * 11 = 77, t2 = 5 * 11 = 55, and t3 = 5 * 7 = 35.  Solving for the inverses, we get s1 = 3, s2 = 6, and s3 = 6.  Thus one solution to the equation is x = 4 * 77 * 3 + 2 * 55 * 6 + 9 * 35 * 6 = 924 + 660 + 1890 = 3474, which is 9 mod 385.

Well this is one case where examining the equations is a lot faster than using the Chinese Remainder Theorem. For example, noticing that 4 + 5 = 9 and 2 + 7 = 9 leads you to see that x = 9 (mod 385) fits the solution; and then the C.R.T. lets you know that this solution is unique.

Ravi correctly applied the C.R.T. formula to show how it works. The best way to understand though, is to see why it always works. Then you'll understand where all the terms come from, and why they are there.

thanks everyone.  But is there a methodical way for solving for the inverse, or is the best way to just plug in values until you get what you want?

There is a methodical way to solve for the inverses, but for small numbers it's a lot faster to just reduce and plug away. For example, to find the inverse of 77 mod 5, reduce it to 2 mod 5 and then notice that 2*3 = 6 = 1 mod 5. So 3 is the inverse.

The methodical way of finding the inverses in general is based on the Euclidean Algorithm. Recall that the EA takes two relatively prime integers a and b, and gives us integers x and y such that ax + by = 1. Well looking at this mod b shows us that ax = 1 mod b, so x is the inverse of a mod b. Try it out with some numbers to see how it works.

For small moduli, a sequential search for the inverse is fine; that's what I did for your problem.  For large moduli, it is more efficient to use the Euclidean algorithm for gcd; as a byproduct, that algorithm will give you the inverse solution.

(Oops.  I just noticed that gauss said the same thing as I was typing.)

To see why the C.R.T. formula works, think about the special case when a1 = 1, a2 = 0, and a3 = 0.  Then notice that a general vector (a1, a2, a3) is a linear combination of such unit vectors.


AMC
USA(J)MO
USAMO
search
algorithm
number theory solved
number theory

Prove that for each positive integer $n$ , theres exists an $n-$ digit number having only odd digitd and divisible by $5^{n}$

:(
Here is my solution though:

For $n=1$, it's true, since there is the $1$-digit number $x_1=5$, which is divisible by $5^1$ and has an odd digit.

Suppose that it is true for $n=k$, that is,
there is a $k$-digit number $x_k$, which is divisible by $5^k$, and has only odd digits

I'll prove that there is a $k+1$-digit number $x_{k+1}$ with odd digits and $x_{k+1}$ is divisible by $5^{k+1}$.

$*$ Notice that $10^{k-1} < x_k < 10^k$

$x_k$ is divisible by $5^k$, so we write:

$x_k = a \cdot 5^k$

For $t=0,1,2,3,4$ the numbers $t \cdot 2^{k+1}$ are all different in modulo $5$, since $2^{k+1} \neq 0 mod 5$

So, the 5 numbers $2^k + t \cdot 2^{k+1}$ are all different in modulo $5$, so $\exists t$ s.t.

$b = 2^k + t \cdot 2^{k+1} +a \equiv 0 mod 5$

I denote $x_{k+1} = b \cdot 5^k$

$x_{k+1} = $
$(2^k + t \cdot 2^{k+1} +a)5^k = $
$((1+2t)2^k+a)5^k = $
$(1+2t)10^k+a \cdot 5^k = $
$(1+2t)10^k+x_k$

This last number is the same as $x_k$, where we added on the left the odd digit $1+2t$. So, $x_{k+1}$ has $k+1$ odd digits.

On the other hand, it is divisible by $5^{k+1}$, since $x_{k+1} = b \cdot 5^k$ and $b$ is divisible by $5$.

Following this algorithm we generate the numbers
$5,\ 75,\ 375,\ 9375,\ 59375,\ 359375,\ 3359375,\ 93359375, ...$


geometry
3D geometry
sphere
calculus
calculus computations

My TA couldn't even do this problem for us...how annoying is that...anyways, here it is:

The circumference of a sphere was measured to be 73 cm with a possible error of 0.9cm.  Use differentials to estimate the maximum error in the calculated surface area.  BLANK

Estimate the relative error in the calculated surface area.  BLANK

Hint:  the circumference of a sphere of radius r is C=2pi r and its surface area is A=4pi rsquared

How do we calculate the surface area from the circumference? Probably by combining two given formulas into $A=4\pi(C/2\pi)^{2}=C^{2}/\pi$. The relevant differential is $dA=\frac{2C}{\pi}dC$. You have $C=73$ with $dC=0.9$.

Edit #1: TA=Teaching Assistant (in the USA)

Edit #2: There is an eatery called "Pie R square" on my campus.  :oops: Nerd alert level: Orange.


limit
real analysis
real analysis unsolved

Let $ a_{n}\equal{}\sum _{k\equal{}1}^n \frac{1}{3^{k^2}}$

Prove that:
a) $ a_{n}$ is convergent
b) if $ \lim_{n\rightarrow\infty }\equal{} \l$ then $ \l\in \mathbb R\minus{}\mathbb Q$

[quote="Carcul"][hide="Possible argument for (a)"]a_n is monotonically decreasing, because for $ x \geq 1,\, (3^{ \minus{} x^2})' < 0$, and is inferiorly limited (a_n > 0): so a_n converges.[/hide][/quote]
why a_n is monotonically decreasing,it's obvious increasing!for a_(n+1)-a_n=1/(3^((n+1)^2)))>0!

[quote="Carcul"]As far as I know, $ \frac {1}{3^{3^2}} < \frac {1}{3^{2^2}}$ for example.

[quote="Zhongym8903"]for a_(n+1)-a_n=1/(3^((n+1)^2)))[/quote]

Can I ask how did you arrive at that expression?[/quote]

for a_{n}=\sum_{k=1}^n\frac{1}{3{k^2}}!
did you see?it can easily give the expression!

[quote="Zhongym8903"]it can easily give the expression![/quote]

If it's easy than show us how.

$ a_{n\plus{}1}\minus{}a_{n}\equal{}\sum_{k\equal{}1}^{n\plus{}1}\frac{1}{3^{k^2}}\minus{}\sum_{k\equal{}1}^n\frac{1}{3^{k^2}}\equal{}\frac{1}{3^{(n\plus{}1)^2}}>0$. Thats what Svejk thought.

What Carcul thought (I guess) was $ a_{n}\equal{}\sum_{k\equal{}1}^{n}\frac{1}{3^{k^2}}\equal{}\sum_{k\equal{}1}^n b_n$, then $ b_n$ is decreasing.

Possible argument for b) http://www.mathlinks.ro/Forum/viewtopic.php?t=164347 , from a similar problem.

Ah yes, I made the confusion between $ a_n \equal{} \frac{1}{3^{n^2}}$ and $ a_n \equal{} \sum_{k\equal{}1}^n\frac{1}{3^{k^2}}$, and this last one is of course increasing, as Zhongym8903 showed. Sorry for that.

Analogous argument for $ b)$ in   :w00t: http://www.mathlinks.ro/viewtopic.php?t=112082   :wink:

An other one : $ l\equal{}\sum _{k\equal{}1}^{\infty} \frac{1}{3^{k^2}}  \notin Q$ because  that from any $ k$ you can't   find a periodically  sub-sequence of digits
 .:coool:

Suppose $ l\equal{}\frac{q}{p}$ is rational, so $ 3^{n^2}*p*\sum_{k\equal{}1}^{\infty}\frac{1}{3^{k^2}}$ is an integer. But you can easily figure out that if n is sufficiently large, then $ 0<3^{n^2}*p*\sum_{k\equal{}n\plus{}1}^{\infty}\frac{1}{3^{k^2}}<\frac{p}{3^{2n}}<1$. On the other hand, $ 3^{n^2}*p*\sum_{k\equal{}1}^{n}\frac{1}{3^{k^2}}$ is always an integer.


Let $ A$ be an $ n \times n$ matrix with complex entries.
Prove that if every eigenvalue of $ A$ is real, then $ A$ is
similar (over $ \bb{C}$) to a matrix $ B$ with real entries.

If any answer is shorter than jmerry's, it's wrong.  :P


Find a function $f(x)$ such that $f( f(x) ) = \exp(x)$.

[quote="Gib Z"]Find A function f(x) such that f( f(x) ) = exp(x)[/quote]

Rather classical problem and solution :

Let $ u < 0$
Let $ h(x)$ be any strictly increasing continuous bijection from $ ( - \infty,u]\rightarrow (u,0]$ such that : $ \lim_{x\rightarrow - \infty}h(x) = u$ and $ h(u) = 0$
Let $ \{a_n,n > 0\}$ the strictly increasing sequence defined by $ a_1 = u$, $ a_2 = 0$ and $ a_{n + 2} = e^{a_n}$ $ \forall n > 0$

Let then the sequence of functions $ \{g_k(x),k > 0\}$ defined as :
$ g_1(x)$ is the bijection from $ ( - \infty,a_1)\rightarrow (a_1,a_2)$ such that $ g_1(x) = h(x)$
$ \forall k > 1$ :
$ g_k(x)$ is the bijection from $ [a_{k - 1},a_k)\rightarrow [a_k,a_{k + 1})$ such that $ g_k(x) = e^{g_{k - 1}^{ - 1}(x)}$

Let then $ f(x)$ the strictly increasing continuous function defined as :

$ \forall x\in( - \infty,a_1)$ $ f(x) = g_1(x)$
$ \forall x\in[a_{k - 1},a_k)$, $ f(x) = g_k(x)$ $ \forall k > 1$

Then, if $ x < a_1$, $ f(x) = h(x)\in(a_1,a_2)$, so $ f(f(x)) = e^{g_1^{ - 1}(f(x))} = e^{^h^{ - 1}(h(x))} = e^x$
And, $ \forall k > 1$ :
$ \forall x\in[a_{k - 1},a_k)$, $ f(x) = g_k(x)\in[a_k,a_{k + 1})$. So $ f(f(x)) = e^{g_{k}^{ - 1}(f(x))} = e^{g_{k}^{ - 1}(g_k(x))} = e^x$

And so $ f(f(x)) = e^x$ $ \forall x\in\mathbb{R}$


:?  :?  :?  :?
let n,m be  given positive integers, n>1.
Find the condition of an integer a;(a,n) <>1; such that
 (x^m)-a is divisible by n^k and it is not divisible by n^(k+1),for every positive integer k.

i' ve just had the solution for n is a prime.
so sorry because I'm not very good at English. :D  :D  :D

Oh,I understand.Your idea is:Find the condition of a such that there is an integer x which satisfied the congurence.Well,it is a very difficult problems!!!


A parabola with equation $ y \equal{} x^2 \plus{} bx \plus{} c$ passes through the points $ (2,3)$ and $ (4,3)$.  What is $ c$?

$ \textbf{(A) } 2 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 11$

[hide]What I did was plug things in.

3 = 16+4b+c (4b+c=-11) and 3 = 4+2b+c are your equations. Multiply equation two by -2 to get -8 -4b -2c = -6, simplify this to -4b-2c=2. Add this equation to the first equation to get -c= -11. C is then 11.[/hide]

[hide]

since it passes through (2,3) and (4,3) we have

$y=(x-2)(x-4)+3$ $\implies y=x^2-6x+11$

so $c=11$
[/hide]

[hide="Answer"]$4+2b+c=3\Rightarrow 2b+c=-1$
$16+4b+c=3\Rightarrow 4b+c=-13$
$b=-6$, $c=11\Rightarrow \boxed{E}$[/hide]

[quote=Alex-131][quote=RedFireTruck]$b=-\frac{2+4}{2}=-3$

$3=2^2-3\cdot2+c$

$3=-2+c$

$c=\boxed{5}$[/quote]

Actually, the answer is 11?[/quote]

yeah, i should NOT have divided by 2. i fixed it now :)

[quote=RedFireTruck]$b=-(2+4)=-6$

$3=2^2-6\cdot2+c$

$3=-8+c$

$c=\boxed{11}$[/quote]

[quote=ishanvannadil2008][quote=kante314][quote=RedFireTruck]$b=-\frac{2+4}{2}=-3$

$3=2^2-3\cdot2+c$

$3=-2+c$

$c=\boxed{5}$[/quote]

This is crazy. You bumped so many threads. JUST WHYYYY[/quote]

Do you have any problem with someone wanting to be getting better at math?[/quote]

wdym how does bumping a 15 year old thread help you get better at math
plus there are already 4 sols

[quote=ishanvannadil2008][quote=kante314][quote=RedFireTruck]$b=-\frac{2+4}{2}=-3$

$3=2^2-3\cdot2+c$

$3=-2+c$

$c=\boxed{5}$[/quote]

This is crazy. You bumped so many threads. JUST WHYYYY[/quote]

Do you have any problem with someone wanting to be getting better at math?[/quote]

Posting on 15-year-old threads does nothing but clutter up the forums for other people who want to get better at math. Sharing a solution doesn't help you get better, especially when lots of them already exist.


This is from a Russian selection test (1999).

Show that each natural number can be written as the difference of two natural numbers having the same number of prime factors.

Can other problems from Russian sel tests be found? Do you know any more?

I have hundreds of problems from tests all around the world :) But try http://www.kalva.demon.co.uk, they have russian national olympiad problems there.

[quote="grobber"]I have a question:
Is it true that there are infinitely prime pairs p, p+2? I mean, is it a known fact? And if it is can we use it in competitions? It could be useful here for example.[/quote]

http://mathworld.wolfram.com/TwinPrimeConjecture.html

You gotta love these Russian problems! It took me a LOT of time to find a soln for this one! J

Here goes:

If n is even then we write n= 2n-n and it?s finished.
If n is odd consider this: let?s assume that n is divisible by p_1=2, p_2=3, .., p_k where p_k is the k?th prime nuber and n is not divisible by p_(k+1). Then we can write n as (p_(k+1)+1)*n-p_(k+1)*n. We can see that all the prime divisors of p_(k+1)+1 except for 2 are also divisors of  n, so we wrote n as a difference of 2 numbers for which the number of prime divisors is= (the number of prime divisors of n) +1.
 Hope it's correct! Bye!

Very nice. The answer looks real easy when you see it, but thinking of it can be pretty tough..

grobber

i am sorry for asking such a stupid question, but you said that "all the prime divisors of p_(k+1)+1 except for 2 are also divisors of n." what if n=7, then  p_(k+1)+1 is 12, therefore  (p_(k+1)+1)*n-p_(k+1)*n = 12*7- 11*7, but these two numbers dont have the same number of prime factors. could you please make this clear for me

thank you

Firstly, there was a tiny but obvious typo in Grobbers solution that I hadn't noticed before -
" lets assume that n is divisible by $p_1=2, p_2=3, .., p_k$" when that was of course not meant to include 2.

When n = 7, the first prime bigger than 2 than doesn't divide n is 3 (that is $p_{k+1}$, not 11). So we express n as 4n-3n.


geometry
circumcircle
incenter
geometry unsolved

Dear Jan and Mathlinkers,

let ABC be a triangle, O the circumcenter, I the incenter, A' the midpoint of BC, Fe the Feuerbach's point, P the intersection of OI and  BC, and P' the foot of the perpendicular on AO issue from P.
The result :  Fe, A' and P' are collinear .

Sincerely
Jean-Louis Ayme

Why do you own some problem which have been posted at "Geometry Unsolved Problems" ?

Dear thaithuan,
Sorry, I don't understand your message. My problem that I discovered to day and for which I have no solution presently, but perhaps to morrow, has therefore not been posted by myself. A possibility: this problem is not new...
Sincerely
Jean-Louis

Yes, problems like this belong into Unsolved Problems.

Really brief outline of the solution: Let B' and C' be the midpoints of CA and AB. During the proof of Theorem 13 in [url=http://www.cip.ifi.lmu.de/~grinberg/BalticFeuer.zip][i]From Baltic Way to Feuerbach - A Geometrical Excursion[/i][/url], I showed that < B'C'Fe = 90° - < (C'A'; IO). Similarly, < C'A'Fe = 90° - < (A'B'; IO). Now, you want to show that < FeA'O = < P'A'O. Using < FeA'O = < C'A'O - < C'A'Fe and (since the points A' and P' both lie on the circle with diameter OP) < P'A'O = < P'PO = 90° + < (AO; OI), this becomes a simple angle chase...

  darij

Dear Mathlinkers,
for more explanation, see for example
http://perso.orange.fr/jl.ayme   vol. 4  Symetriques de OI par rapport aux triangles de contact et médian p. 33
Sincerely
Jean-Louis


induction
inequalities
real analysis
real analysis solved

Study the convergence of $(u_{n})$, defined by :

$u_{0} \geq 0$ and, for all $n \in \mathbb{N}$, $u_{n+1} = \sqrt u_{n} + \frac{1}{n+1}$

Conjecture based on numerical evidence: $u_n\to 1.$ More specifically, as $n\to\infty,\,u_n=1+\frac2n-\frac1{n^2}+O(n^{-3}).$

[quote="Kent Merryfield"]Conjecture based on numerical evidence: $u_n\to 1.$ [/quote]
Yeah, ok. :)

[quote="Kent Merryfield"]More specifically, as $n\to\infty,\,u_n=1+\frac2n-\frac1{n^2}+O(n^{-3}).$[/quote]

Er.. could you explain this part?

Here is an elementary solution. First of all, try to prove by induction that $u_{n}\geq\epsilon >...0.$ After that try to prove that $u_{n}$ is either increasing either descreasing, so our sequence has limit finite or not. By passing to the limit in the recurence given, we obtain the result. ;)

Thanks cezar :)

Anyway, I'm still wanting an explanation from Kent Merryfield!

I said it was conjecture. It was based on trials using a spreadsheet; I haven't proved it yet.

If you're going to follow cezar lupu's suggestion, try showing that the sequence is eventually greater than 1 and eventually decreasing.

Alright, thanks for the advices :)

If you manage to prove it with your method, I'd be delighted to see it!  :)

We have $u_n \geq 1 + 2/n$ for $n > 5$ provable by induction and noticing that minimal value of sequence is for $u_0 = 0$.

Let $v_n = u_n - 1 - 1/n$ ($\geq 0$ for $n > 5$)

Using $\sqrt{1+x} \leq 1+x/2$

We get $v_{n+1} \leq v_n/2 + 1/n(n+1)$

We get $v_{n+1} \leq v_5/2^{n-3} + \sum 1/2^k * 1/k(k-1)$

$\sum 1/2^k * 1/k(k-1) \leq n/2 * (n/2*(n/2-1) + n/2 * 1/2^{n/2} \rightarrow 0$

So $v_n \rightarrow 0$ and $u_n \rightarrow 1$

I'd like to elaborate a bit upon tµtµ's and Kent's posts. This is all about perturbations of contractive mappings. Indeed, consider a more general problem. Let $F: (0.5,+\infty)\to (0.5,+\infty)$ be a contraction ($|F(x)-F(y)|<k|x-y|$, $k\in(0,1)$) with fixed point $1$. Create a sequence $x_{n_0}\in (0.5,+\infty)$, $x_{n+1}=F(x_n)+\frac1{n+1}$ for $n\ge n_0$. Suppose we can find another sequence $y_n$ ($n\ge n_0$) such that $y_{n+1}=F(y_n)+\frac 1{n+1}+O(n^{-m})$ where $m$ is some fixed number. Then $|x_n-y_n|=O(n^{-m})$. Indeed, if we denote $z_n=|x_n-y_n|$, we shall have $z_{n}\le k z_{n-1}+Cn^{-m}$. Writing the corresponding inequality for $z_n$, we get $z_n\le k^2 z_{n-2}+Ck (n-1)^{-m}+C{n}^{-m}$ and so on. Finally, when we go down to $n_0$, we get
\[ z_n\le z_{n_0}k^{n-n_0}+C\sum_{j=0}^{n-n_0+1}k^j (n-j)^{-m}  \]
(compare to tµtµ's post).
It is not hard to show that the right hand part is still $O(n^{-m})$, proving the claim. Now it is clear how to get Kent's result: all one needs is to find the coefficients $a,b$ such that
\[ 1+\frac a{n+1}+\frac b{(n+1)^2}=\sqrt{1+\frac an+\frac b{n^2}}+\frac 1{n+1}+O(n^{-3})\,.  \]
Expanding both sides in the inverse powers of $n$, we get
\[ 1+\frac an+\frac{b-a}{n^2}+O(n^{-3})=1+\frac{\frac a2+1}{n}+\frac{\frac b2-\frac{a^2}8-1}{n^2}+O(n^{-3})  \]
Thus $a=\frac a2+1$, whence $a=2$ (as Kent suggested) and $b-a=\frac b2-\frac{a^2}8-1$, whence $b=1$ (Here Kent, probably, just put a wrong sign; also compare this result with the beginning of tµtµ's post, it shows that $b$, if exists, must be non-negative).
In principle, you can find as many terms in the asymptotic series for $x_n$ as you wish but the farther you go, the more tedious the computation becomes and I do not see any clear pattern emerging.

Ah - I found my mistake. I did mean that negative sign on $b=-1$ when I wrote it, but I just went back and reexamined the spreadsheet I was playing with, and it turns out that I was actually tracking the recursion $u_{n+1}=\sqrt{u_n}+\frac1{n+2}.$

With that corrected, I now agree with tµtµ and fedja.

Alright, well, thanks to all of you! :)

Fedja, I really liked your approach!


Probability - Two Counting Problems in One!

We tackle a probability problem by breaking it into two counting problems: counting the number of successful outcomes and the number of possible outcomes.

It doesn't matter where the girl sits. The boy always has two spots to sit in according to the restriction, and there are a total of 5 seats left
so the probability is just $ \boxed{\frac{2}{5}}$.

Wait what? Gamebot didn't even post a question and you're answering the question!  :o

brightzhu is probably talking about the video :wink:

I agree, you could just see the two seat total out of 5 choices to get 2/5. It was still a big help!

I agree. When I saw what he was doing, I got kind of confused, but then I realized that his way worked as well

Oh I get it. The girl sits in a random position and 2 out of the remaining 5 work, which is $ \boxed{\frac{2}{5}} $.

Another way (not as simple)

B   G   B   G   B   G   B   G   B
1   2   3   4   5   6   7    8   9

There is a 1/5 chance the brother sits in seat 1, and if he does there is a 1/4 chance that his sister sits next to him in seat 2. 1/5 * 1/4 = 1/20 chance they are next to each other.

There is a 1/5 chance the brother sits in seat 3, and if he does there is a 2/4 chance that his sister sits next to him in seat 2 or seat 4. 1/5 * 2/4 = 2/20 chance they are next to each other.

There is a 1/5 chance the brother sits in seat 5, and if he does there is a 2/4 chance that his sister sits next to him in seat 4 or seat 6. 1/5 * 2/4 = 2/20 chance they are next to each other.

There is a 1/5 chance the brother sits in seat 7, and if he does there is a 2/4 chance that his sister sits next to him in seat 6 or seat 8. 1/5 * 2/4 = 2/20 chance they are next to each other.

There is a 1/5 chance the brother sits in seat 9, and if he does there is a 1/4 chance that his sister sits next to him in seat 8. 1/5 * 1/4 = 1/20 chance they are next to each other.

1/20 + 2/20 + 2/20 + 2/20 + 1/20 = 8/20 = 2/5 chance they sit next to each other.


uh

chinese

height: 5-6

weight: 140

typical chinese look

Second user: ernie!

Bah.

Orange skin, black hair, picture depicted in oyur avatar.

Me

Like [url=http://shotpolitics.com/wp-content/uploads/2008/03/mclovin.jpg]this[/url].

Next subject: $ _{\mathcal{T}} \mathcal{Z} _{\mathcal{F}}$

(Is it safe to assume that the subject will just be the last poster?)

[quote="tinytim"]Post what you think a particular AOPS user who you have never met or seen looks like in real life. (height, weight, hair color, etc.)

First user: Yongyi781

I know what he looks like so I will not guess.[/quote]

Hint: I got a perfect AMC 8.

Oh darn... uh-oh... :blush:

TZF: Hmm I believe I have met you at some point, or at least seen you, but I wouldn't know you by your appearance or name...Haha.

Anyway 5'7" Typical Asian look.

perfect628

Bah same as TZF rofl.

Me (no links please xD)

Chinese
5'3"
12 years old
has an isabella-ish face

rrusczyk (theres a picture of him on this website :D )

[url=http://www.artofproblemsolving.com/AboutUs/AoPS_A_Company.php]link to pic[/url]

Have fun trying to guess what I look like.

5'8" ish
non-asian
glasses
wears lots of math tshirts?
im totally guesing here.
i fail.
anyways

jeffhan1992

At above poster: I find issue with your choice of "jeffhan1992". I searched the forum for posts by him and it didn't reveal any results!

So I'm going to take the opportunity to respond to:

[quote="perfect628"]Anyway 5'7" Typical Asian look.[/quote]
[url=http://photos-h.ak.facebook.com/photos-ak-sf2p/v234/22/66/1227960671/n1227960671_30355047_1043.jpg]This is me =D[/url]

[b]Nominate:[/b] JBL

caucasian, 6ft, brown hair, 160 lbs

me(I [i]want[/i] you to get it wrong. Someone else on AoPS got my age horribly wrong. :rotfl: )

eh

2 yr old

3485 lbs

wears stretchy pants

wiry legs

puny head

etc.
[b]
Nominate: alexhhmun[/b]

Oh darn I know what he looks like. And I still remember.

:|

Nominate: tinytim :rotfl:
I actually found a picture of him on some Chinese website! (Obviously, it was about MATHCOUNTS). And I think I met him once at Nationals. He said something like,
"I think I got a 35."
And he was on the elevator with me once.

OK, stop. You are .4", green, have big teeth, triangles for eyes, and martian.

you're half correct :D

eggy looks like  :| because he's an egg who smells like a muffin man.

and aleph0, you rock like  :D

STOP Spamming. or I will report YOU (gauss and aleph) to the admins.

duh...we arent, we just love each other :D  :D  :D

I have a feeling eggy looks like the ninja smiley. He is 1 inch tall and weighs 1337 pounds.

[quote="gauss1181"]duh...we arent, we just love each other :D  :D  :D [/quote]

eggylv999 is an egg so I wanna eat him... :D

[quote="Poincare"]Since you're indian,
you'll be about 5 ft four inches and about 120 pounds.[/quote]

Almost correct.
5 ft 6, 100lbs.

BTW, who said all Indians are 5 ft 4?

Maybe you dont know Amitabh Bachchan 6ft5, Abhishek Bachchan 6ft6, Ishant Sharma (cricket) 6ft5, Khali....etc

1=2: I won't try and guess because I get an update every week. :P


calculus
derivative
calculus computations

edited...wrong question. thanks for telling me that

[quote="jackie-chan01"]Show me the path to solve it

Let $ S$ be the set of real values of parameter $ k$ for which the equation $ f(x) \equal{} 2x^{3} \minus{} 3(2 \plus{} k)x^{2} \plus{} 12kx$ has exactly one local minimum and exactly one local maximum.  $ S$ is a subset of
$ (a) ( \minus{} 4, \infty)$    $ (b) ( \minus{} 3,3)$       $ (c) (3,\infty)$   $ (d) ( \minus{} \infty,0)$[/quote]

take the first derivative, then determine for what $ k$ will $ f'(x)$ have two distinct roots.

[hide="not sure, BUT..."]I got $ S: \{k|k\neq 2\}.$  :( [/hide]

[hide="my solution"]$ f'(x) \equal{} 6x^{2} \minus{} (12 \plus{} 6k)x \plus{} 24k.$ $ f'$ has distinct roots for $ (k \plus{} 2)^{2} > 8k \Leftrightarrow k \neq2.$ :pilot:[/hide]

There's no appropriate answer among $ (a)$ through $ (d)$, but earldbest's answer is correct.

My solution:

$ f'(x)\equal{}6(x\minus{}2)(x\minus{}k)\Longrightarrow k\neq 2$.

A general rule of etiquette: [b]Do not ever remove content from a post that has been replied to.[/b] There's a reason you're not allowed to delete those posts. Threads don't make sense when the initial question disappears.

ok.. sorry for that.But as 2 users of the site pointed out that question was wrong,I decided to remove it. I promise it will not be repeated in future.


calculus
integration
inequalities
calculus computations

A possible solution.

Using the substitution $ x^{2}\equal{}y$ we get that

$ \int_{0}^{1}e^{\minus{}2x^{2}}dx\equal{}\frac{1}{2}\int_{0}^{1}\frac{e^{\minus{}2y}}{\sqrt{y}}dy$,

and hence we need to prove that

$ \pi\minus{}2<\int_{0}^{1}\frac{e^{\minus{}2x}}{\sqrt{x}}dx\equal{}\int_{0}^{\infty}\frac{e^{\minus{}2x}}{\sqrt{x}}dx\minus{}\int_{1}^{\infty}\frac{e^{\minus{}2x}}{\sqrt{x}}dx$.

It is not hard to check that the first integral equals $ \sqrt{\frac{\pi}{2}}$, and hence we need to show that

$ \int_{1}^{\infty}\frac{e^{\minus{}2x}}{\sqrt{x}}dx\leq\sqrt{\frac{\pi}{2}}\minus{}\pi\plus{}2$.

We have that

$ \int_{1}^{\infty}\frac{e^{\minus{}2x}}{\sqrt{x}}dx\leq\int_{1}^{\infty}e^{\minus{}2x}dx\equal{}\frac{1}{2e^{2}}<\sqrt{\frac{\pi}{2}}\minus{}\pi\plus{}2$,

since $ \frac{1}{2e^{2}}\equal{}0.0676$ and $ \sqrt{\pi/2}\minus{}\pi\plus{}2\equal{}0.11172$.

Of course if you are not allowed to use the calculators to check this inequality then perhaps a different approach is necessary for solving this problem.

Also, observe that we have an improvement of the lower bound of the integral $ \int_{0}^{1}e^{\minus{}2x^{2}}dx$, since

$ \frac{\pi}{2}\minus{}1<\frac{\sqrt{\pi/2}\minus{}\frac{1}{2e^{2}}}{2}$.


Let ABC be a triangle with <BAC = 60 . Let AP bisect<BAC
and let BQ bisect<ABC, with P on BC and Q on AC. If AB + BP = AQ + QB,
what are the angles of the triangle?

Is this problem difficult?
I heard somewhere the answer is 40, 60, 80 in degrees.

Thank you for the valuable link.
I'm sorry for double posting.
Do you know a correct solution clear for understanding to this one:
http://www.mathlinks.ro/Forum/viewtopic.php?t=212388 ?


Define $f$ on the positive integers by $f(n) = k^2 + k + 1$, where $n=2^k(2l+1)$ for some $k,l$ nonnegative integers.
Find the smallest $n$ such that $f(1) + f(2) + ... + f(n) \geq 123456$.

We are counting sums of powers of 2.
If n = 256 —> Sum of f(n<=256) = 132863 (just plug it in). Without the 2 to the power of 8 the amount will be too small.


Is any of your school has taken AP Physics C class now or before?
If you do, could you give me the e-mail of your AP Physics C teacher because next year, my physics teacher want to teach AP Physics C (and I really want to take this class, too), but he has never teach it before. So he wants ask teachers who have teach AP Physics C before give him some materials, advise, and so on, on the AP Physics C.
I need it soon, because course selection is this few weeks.
Thank you for your help!

The College Board's AP website has a list of resources to use, such as textbook recommendations.


Find the arithmetic mean of the first thousand positive odd numbers.

[quote="hockeyadi23"][hide]1000. Since the sum of the first n terms is n^2 and the mean will thus be n^2/n, which equals n, the answer is 1000.[/hide][/quote]

actually n 2  is the sum of n [b]ODD[/b] terms


Consider the matrix
\[ A \equal{} \begin{pmatrix} a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{nn}\end{pmatrix}
\]
where $ a_{ii} \equal{} 0$ for all $ i \equal{} \overline{1,n}$ and $ a_{ij} \equal{} 1$ or $ 2007$ for all $ i\neq j$, $ i, j \equal{} \overline{1,n}$. Prove that $ \det A\neq 0$ if $ n$ is an even number.

[quote="jmerry"]Why ask this in 2008? Can't you wait a few weeks so the year is valid again?

(That's actually a hint.)[/quote]
It's such a nice hint  :P

we see $ (\minus{}1)^ha_{1\alpha(1)}...a_{n\alpha(n)}$ is 0 iff $ \alpha(i)\equal{}i$ for some i, and it is odd for otherwise. the number of those which are equal to 0 is $ n!\minus{}\left(^n_2\right)(n\minus{}2)!\plus{}...\plus{}(\minus{}1)^{n\plus{}1}\left(^n_n\right)$ which is an odd number(n is even) so the determinant is equal to an odd number , it can't be equal to 0


There is a "proof" that 2=1 and it goes ,as I know it
Let's say that a=b

$ a\equal{}b$
$ a^2\equal{}ba$ -> multiply $ a$ on both sides.
$ a^2 \minus{} b^2\equal{}ba \minus{} b^2$ -> subtract $ b^2$ to both sides.
$ (a\plus{}b)(a\minus{}b)\equal{}b(a\minus{}b)$ -> factorization.
$ a\plus{}b\equal{}b$ -> divide $ (a\minus{}b)$ on both sides.
$ (b)\plus{}b\equal{}b$ -> Substituting $ a$ for $ b$ because $ a\equal{}b$
$ 2b\equal{}b$ -> adding
$ 2\equal{}1$ -> dividing by b on both sides.

I'll will post what is the flaw tomorrow unless someone get's it right today.

Um like, this is the most common proof when you divide by 0.
Because since $ a\equal{}b$, then $ (a\minus{}b)\equal{}(a\minus{}a)\equal{}0$
Meaning that the proof is incorrect because you can't divide by 0.

yep, that was quick.

Well here's a different question:

Find the value of
$ \sqrt{72\plus{}\sqrt{\plus{}72\sqrt{\plus{}72\cdots}}}$

$ \boxed{9}$.

Please don't turn this into a marathon.

EDIT: Let $ x$ be the value of the expression.  You eventually get $ (x\minus{}9)(x\plus{}8)\equal{}0$.

dividing by 0 is undefined...

first time i looked at this i really thought 2 could equal one but now i see carefully and u can't devide by 0

[quote="alanchou"]$ \boxed{9}$.

Please don't turn this into a marathon.

EDIT: Let $ x$ be the value of the expression.  You eventually get $ (x \minus{} 9)(x \plus{} 8) \equal{} 0$.[/quote]
How'd you solve it so fast?
is there a rule??? :( please tell me... Thanks

Let $ x\equal{}\sqrt{72\plus{}\sqrt{72\plus{}\sqrt{72\plus{}...}}}$
Then, $ x\equal{}\sqrt{72\plus{}x}$
Squaring both sides,
$ x^2\equal{}72\plus{}x$
$ x^2\minus{}x\minus{}72\equal{}0$
$ (x\minus{}9)(x\plus{}8)\equal{}0$

Hope that helps, woohoo :)

I'm sorry to say, but this problem has already been shown.

Look at my sig.  This problem has been on there for 36 days.


Let $ f: [0,1]\to\mathbb{R}$ be a continuous function on $ [0,1]$ such that
1)$ f(0)=0,f(1)=1$ and
2)$ 6f\left(\frac{2x+y}{3}\right)=5f(x)+f(y)\; \forall x\geq y\; \text{in}\; [0,1].$

Find $ f\left(\frac{8}{23}\right).$

You are true, i think so. Hmm, This problem is not GOOD!

[quote="Rust"]It posted before, and I prove, that if
$ f(ax+(1-a)y)=bf(x)+(1-b)f(y), a\not =b$
and f(x) continiosly, then $ f(x)\equiv const$.[/quote]

I suppose that $ a,b\not=0$, then it should be sth like this..

$ y=0$ gives $ f(ax)=bf(x)+c(1-b)$, with $ c=f(0)$.
Now suppose $ |a|\geq 1$.
$ x\rightarrow \frac{x}{a}$, and multiply it with $ b$, you get
$ bf(x)=b^{2}f(\frac{x}{a})+bc(1-b)$, do this over and over again, multiply it all, and get this:
$ f(ax)=b^{n+1}f(\frac{x}{a^{n}})+c(1-b^{n+1})$. Now letting $ n\to\infty$ gives $ f(ax)=c$.

If $ |a| \leq 1$ then instead of putting $ x\rightarrow \frac{x}{a}$, we put $ x\rightarrow ax$.

Exactly:
If $ f(ax+(1-a)y)=bf(x)+(1-b)f(y), \ a\not =b$
1) and exist $ x_{0}$, suth that $ f(x)$ continiosly in $ x_{0}$, then $ f(x) \equiv const$,
2) exist $ x_{0},C,\epsilon >0$,  suth that $ f(x)>C, \ |x-x_{0}|<\epsilon$ or $ f(x)<C, \ |x-x_{0}|<\epsilon$, then $ f(x)\equiv const$.

Final, There is not function satisfying that  condition, and pontios are true!


Numbers 1,2,3,...,n^2 are written in magic square n*n so that sums of each row and each column are equal (it's why is this square magic :)). For each 2 numbers in this magic square you make a vector beginning in center of smaller number and ending in center of bigger number. Prove that sum of these vectors is null vector. I hope It's written clearly, isn't it?

Place origin into the center of square.
Let number $k$ is written in point $A_k$, then we need to calculate $S=\sum_{i<j}\overrightarrow{A_iA_j}=\sum_{i<j}\overrightarrow{OA_j}-\overrightarrow{OA_i}$. Note that each $\overrightarrow{OA_k}$ is counted $k-1$ times with '+' and $n^2-k$ times with '-', so $S=\sum_{k=1}^{n^2}(2k-(n^2+1))\overrightarrow{OA_k}$.
We can state that $\sum_{k=1}^{n^2}(n^2+1)\overrightarrow{OA_k}=0$, since origin in the center of square. Thus $S=2\sum_{k=1}^{n^2}k\overrightarrow{OA_k}$   $(*)$.
Consider $y$-projection of $S$. Take two horizontal rows symmetrical wrt $O$. Then $y$-projections for corresponding vectors in different rows will have equal lengths and different signs. Due to $(*)$ $y$-projection of first row is $(a_1+a_2+...+a_n)\vec{h}$ and for second row is $-(b_1+b_2+...+b_n)\vec{h}$, where $a_i$, $b_i$ are numbers in these rows. By condition $a_1+...+a_n=b_1+...+b_n$, thus sum of $y$-projections for these two rows is $\vec{0}$. Therefore, for whole table $y$-projection of $S$ is $\vec{0}$. The same for $x$-projection.

That's all.

Thank you Myth, I really like your solution.

My friend also has found very nice solution making from these vectors and points an oriented graph and he proved, that there's Euler's path that comes back to point where you started it.

I also was thinking about half-combinatorical solution.
I don't see Euler's path here.

oh, it's not exactly as I write (my friend's solution). Vertices in his graph were columns and edges were vectors (after this part you can choose rows as vertices which is analogous). Than he poved that there's an Euler path :)

But maybe there's Euler's path also when you take numbers as vertices...

BTW: Myth, who's the man who is on your avatar? You? Or some other mathematicien? Or somebody imaginary?

[quote="ondrob"]BTW: Myth, who's the man who is on your avatar? You? Or some other mathematicien? Or somebody imaginary?[/quote]
Let it be a myth...


geometry
geometric transformation
reflection
perpendicular bisector
geometry proposed

Let three circles $ (O, R_a), (O, R_b), (O, R_c)$ are $ \omega_a, \omega_b, \omega_c$. The point $ A\in\omega_a, B\in\omega_b, C\in\omega_c$.
$ BC\cap \omega_b, \omega_c\equiv B_1, C_2$
$ CA\cap \omega_c, \omega_a\equiv C_1, A_2$
$ AB\cap \omega_a, \omega_b\equiv A_1, B_2$
Prove that $ \Delta(A_1A_2, B_1B_2, C_1C_2) , \Delta ABC$ are similar reverse.

Do you mean $ \Delta XYZ\sim \Delta ABC$?

No, But I discovered that it is very easy.

[b]Lemma:[/b]
 Given triangle $ ABC. (I,IA)\cap AB, AC\equal{}\{A_1,A_2\}$,$ (I,IB)\cap BC\equal{}\{B'\}, (I,IC)\cap BC\equal{}\{C'\}, C'A_1\cap B'A_2\equal{}\{A_3\}$. Then $ A,A_1,A_2,A_3$ are concyclic.
[b]Proof:[/b]
Denote $ K,J$ the intersections of a line through $ I$ and perpendicular to $ BC$ and $ AC,AB$, respectively. $ JB'\cap (I)\equal{}\{L,N\}$
Since $ IJ$ is the perpendicular bisector of $ [BB']$ we get $ AN//BB'$
Therefore $ \angle AA_2L\equal{}\angle ANL\equal{}\angle LB'C \Rightarrow B',C,L,A_2$ are concyclic.
$ \Rightarrow \angle B'A_2C\equal{}\angle B'LC\equal{}\angle C'A_1B$ (Because $ (A_1BC')$ is the reflection of $ (LB'C)$ across the line $ IJ$)
Or $ \angle AA_1A_3\equal{}\angle AA_2A_3 \Rightarrow$ QED.
[b]Return to our problem:[/b]
From the lemma above we get $ X\in \omega_a, Y\in \omega_b, Z\in \omega_c$ immediately.
$ \Rightarrow \angle ZXY\equal{}\angle A_1XA_2\equal{}\angle BAC$. Similarly we are done.

We only proff $ DG$ cut $ EF$ at $ c_3$.
$ H$ is symmetric of $ D$ so $ H\in c_3$.
$ \angle CFH\equal{}\angle DAG\equal{}180\minus{}\angle BAC\equal{}180\minus{}\angle BEH$
Then $ C,F,H,E\in$ one circle.
$ \angle CEF\equal{}\angle CHF\equal{}\angle ADG$
done.


Did the AMC page remove the links to purchasing the Contest Problem solving books 1-4? I really need these books for extra practice, yet I cannot find them on the page. Does anyone know other links that sell these books?

Try the MAA's web site: [url]http://www.maa.org[/url]

Click on "Bookstore", then choose "Problem Solving" from their category list.


In acute triangle $ABC$, $AB>AC$. $AD$ is the altitude. Point $E$ and $F$ lie in segment $AB$ and $AC$ respectively such that $\angle ADE=\angle ADF$.A line $l$ through $D$ and pararel with $EF$ intersect $BF$ and $AC$ in $P$ and $Q$ respectively.
Prove that $DP=DQ$

Let $L=EF\cap BC$, $R = AD\cap EF$. Then $DA$ and $DB$ are angle bisectors of $\angle FDE$ and hence $(L, R; E, F) =-1$. But the cross ratio is preserved under central projection hence $(D, R; X, A) =-1$ where $X = BE\cap AD$. Bearing this in mind the result follows easily after one applies Menelaus theorem to $\triangle QPE$ and line $AD$ and Thales theorem to $\triangle ADQ$ and $DQ\parallel RE$.

Thx xixas..

Is there any solution for this problem?

I need an euclid proof..

Thx..

Is there anyone can help me to solve this problem?

Thx very much..


calculus AMC AIME USA(J)MO USAMO HMMT Stanford

Calculus seems to open a new light on mathematics, so I was wondering do you think they will ever have Calculus problem solving contests like the AMC, AIME, USAMO?

I think they already have such contests for college students. I'm not sure, somebody else can elaborate.

Mu Alpha Theta has Calculus tests. Besides them, I dont see any. I dont see why AMC/AIME/USAMO would even consider having Calculus problems.

AMC/AIME/USAMO will never have Calculus problems. There are 7th graders (and maybe) younger who go to USAMO, and I highly doubt they will have taken Calculus by then. I knew two 7th graders who started Calculus by the end of the year, but that was after USAMO, for one thing, and they wouldn't have known it well enough anyway. Most people don't take Calculus until at least 11th grade, and they don't want to make it unfair for anyone.

Well, I don't know calculus, and I'm just dandy... though calculus sounds really cool and fun.  :idea:

It is most certainly not a requirement on most of any high school math competition tests.

IMO doesn't have calculus problems, so there would be no point in the AMC tests having calculus.

Other competitions, most notably HMMT, Rice, Stanford, and Mu Alpha Theta, are not limited by such restrictions. One question I've always had is...how to prepare for the calculus on those exams? I've taken my fair share of calculus based classes, and yet I never seem to find the ingenuity to solve the problems...

I was actually wondering if they would ever have problem solving contests LIKE AMC/AIME/USAMO that included calculus.

I was just wondering if there would be any, or am I wrong in assuming that calculus would have the breadth to have such contests? (like having single variable cal, multi-variable cal, differential equations...)

I mean with all the applications of calculus why isn't there a competition that incorporates calculus problem solving in an AMC/AIME/USAMO format for like college or perhaps even high school?

[quote="sr6622"]why isn't there a competition that incorporates calculus problem solving in an AMC/AIME/USAMO format for like college[/quote]

umm that is Putnam. its for College.

I would think most of those people good in math contests knew SOME calc.

I learned it in 9th grade myself.

[quote="sr6622"]Putnam has perliminary tests like AMC and AIME?[/quote]
No. The Putnam is a one-shot deal, on the first Saturday in December: 6 hours, 12 questions, all proofs. About 3,500 students take part, and the scores are not released until March.

Although Putnam questions can involve calculus, that is not usually the primary focus. The focus is on problem solving.

Here are two calculus-containing Putnam questions from years past that I like:

1. Is it possible for the arclength of the portion of a parabola contained within a circle of radius 1 to exceed 4?

2. Compute (with explanation):

$\int_0^{\infty}\left(x-\frac{x^3}2+\frac{x^5}{2\cdot4}-\frac{x^7}{2\cdot4\cdot6} +\cdots\right)\left(1+\frac{x^2}{2^2}+\frac{x^4}{2^24^2}+\frac{x^6}{2^24^26^2} +\cdots\right)\,dx$

why don't you start and IMO for college students then?

seems like an excellent opportunity.

[quote="sr6622"]Why isn't there an IMO for college students?[/quote]There is: [url]http://www.imc-math.org/[/url]

But there's not nearly as much need for one for college students as there is for high school students.  College students, unlike (most) high school students, generally find a sufficient intellectual challenge through their regular curriculum, so that they don't need the same degree of extracurricular contests to help motivate them.

Ohhh...Thank you for the website. I was just a bit curious because it seems like once people would get to college there would also be an IMO-type comeptition. (Although, I don't really know if this is like IMO, because the people represent their college/university not a certain country)

OH and about starting an IMO for college, I'm only a junior in high school and there is no way I could possibly do that.

the problem with calculus in high school contests is that coming up with new problems is hard
the difficulty of calculus questions cannot be too hard and thus many problems will have to be repeated
this is why imo doesnt have calculus problems

I get your point....

So, there is no IMO for college students (in which the students represent their country)?

No, there's no IMO equivalent for college students. Also, a lot of current Olympiad problems can be solved using calculus techniques, and sometimes they are significantly easier with calculus. But I think generally problem writers try to avoid this, for the reason that they don't want the advantage that one person has over another to be related to a topic that some high schools don't even offer.

I remember seeing a sequence problem by Titu, where if you recognized the similarity to the Taylor Series of cosh, then the direction to go in became quite clear quite fast. Titu said it was for that exact reason that he didn't use the problem, which is somewhat of a pity, since it was a great problem, the "calculus assistance" notwithstanding.

well Mu Alpha Theta is one contest that tests Calculus. A lot.

On our local contest (CSULB Math Day at the Beach), I did decide to allow a small amount of calculus into the contest. There was none in the individual round, but on the team round, two of the 8 questions used first-semester calculus. (I assumed that most high schools represented would have at least one member with AP calculus experience.) These two questions turned out to be the easiest (solved by 18 of the 22 schools) and the third easiest (solved by 12) of the 8 questions. That was no surprise to me, and the fact that the two combinatorics questions were the two hardest was also no surprise.

I agree with the last paragraph in Richard's post.

A short-answer Calculus exam is one of the five individual options at the Harvard-MIT Math Tournament. (See http://web.mit.edu/hmmt/ for details, archives.)

[quote="rrusczyk"][quote="gauss202"]There's a Calculus level in the [url=http://www.continentalmathleague.hostrack.com/]Continental Math League[/url] (not to be confused with [url=http://www.mathleague.com/contests.htm]Mathleague[/url]). The questions here are usually pretty easy (compared to AMC-12 or AIME).

There's also a Calculus contest here in the northwestern part of South Carolina called the [url=http://www.math.clemson.edu/events/CCC/2005/details.html]Clemson Calculus Challenge[/url]. I've never been, but I've heard the questions are pretty good. Some of them very challenging.

And the [url=http://math.cofc.edu/mathmeet/]College of Charleston Math Meet[/url] Level III contest has some calculus questions also.

As a Calculus teacher I would like to see more contests that involve Calculus. It invites students to deepen their understanding of the material, and to use their knowledge in new ways. Maybe down the line there will be more contests that involve Calculus as there are more and more high school students taking it now than there used to be.[/quote]

I'd hate to ever see calculus contests become the most important contests - it effectively locks out kids who haven't had the basics, and exacerbates the race to calculus that we see hindering kids developing thinking skills.  At this point, nearly all high school calculus contests I've seen have been straightforward computational events that test exposure far more than understanding.

Instead of having contests be twisted to fit the curriculum, I'd rather see the curriculum twisted to fit the better contests out there by including more number theory, combinatorics, real geometry, and -gasp- proofs, etc.  I'm not holding my breath.

I'd argue that a deeper understanding of the material *before* calculus will inspire a deeper understanding of calculus much more than calculus contests will - students who master MATHCOUNTS/AMC/AIME problem solving tend to find calculus a rather simple extension of ideas they already know.[/quote]

Amen to that!


geometry
3D geometry
tetrahedron
combinatorics proposed
combinatorics

We have to divide a cube onto k nonoverlapping tetrahedrons.
For what smallest k is it possible?

I think it's $5$. Since a tetrahedron can have at most $3$ faces lying on cube faces (and in this case the cube faces have a common vertex) and a face of a cube needs at leas two tetrahedrons to cover it, I think this already gives a lower bound of $4$ (draw a picture).

In this case (if we have $4$ terahedra), each face has to be covered by the faces of exactly two tetrahedra, so the $4$ tetrahedra have a total of $12$ faces which are isosceles right triangles lying on the faces of the cube. Since each tetrahedron has a face which is inside the cube, it means that each must have three faces which are isoscleles right triangles. This determines their shape and we see that there's still "room" inside the cube, so $4$ aren't enough.

The construction for $5$ is the one describes above: if we construct the four tetrahedrons properly, the shape of what's left inside is that of a fifth tetrahedron.


Let $ a,b,c$ be positive numbers such that $ abc\equal{}1$. Prove that:
$ \frac {1}{\sqrt {b\plus{}\frac {1}{a}\plus{}\frac {1}{2}}}\plus{}\frac {1}{\sqrt {c\plus{}\frac {1}{b}\plus{}\frac {1}{2}}}\plus{}\frac {1}{\sqrt {a\plus{}\frac {1}{c}\plus{}\frac {1}{2}}}\ge \sqrt {2}$

[i]Proposed by R.Stanojevíc[/i]


function
trigonometry
algebra
polynomial
calculus
calculus computations

I have no idea what you're asking. Can you give an example to illustrate what you mean by this question?

[quote="EUCLID 3.14"]If the function $f(t)$ is a function which contains only sin and cos terms to some power, then would the fourier series of $f(t)$ always have finite terms?[/quote]
If you mean something like "If $f(t)=P(\cos t,\sin t)$, where $P(x,y)\in\mathbb{R}[x,y]$, then Fourier series of function $f(t)$, denote $\frac{a_{0}}2+\sum_{k=1}^{\infty}a_{k}\cos kt+b_{k}\sin kt$, has only finite number of $k$, such that $|a_{k}|+|b_{k}|\ne0$", then it's true. This is particularly obvious, if you use "complex" Fourier series, i.e. $\sum_{k\in\mathbb{Z}}c_{k}e^{ikt}$. Just use connection between theese 2 kinds of Fourier series.

[quote="RIP"]If you mean something like "If $f(t)=P(\cos t,\sin t)$, where $P(x,y)\in\mathbb{R}[x,y]$, then Fourier series of function $f(t)$, denote $\frac{a_{0}}2+\sum_{k=1}^{\infty}a_{k}\cos kt+b_{k}\sin kt$, has only finite number of $k$, such that $|a_{k}|+|b_{k}|\ne0$", then it's true. This is particularly obvious, if you use "complex" Fourier series, i.e. $\sum_{k\in\mathbb{Z}}c_{k}e^{ikt}$. Just use connection between theese 2 kinds of Fourier series.[/quote]
That seems to be what I mean. Although, I am not sure what you mean when you say $P(x,y)\in\mathbb{R}[x,y]$. Still, I am just going to give an example anyway.

Say $f(t)= 5\sin^{2}(x)+\cos^{3}(3x)$ or $f(t) = 3\cos^{4}(5x)+5\sin^{8}(4x)$, then would its fourier series representation always have finite terms?

[quote="mlok"]RIP meant that $P(x,y)$ is a polynomial in two variables $x$ and $y$. And yes, in your two examples the Fourier series will be finite. The best way to see this is to use complex exponentials, as RIP suggested.[/quote]
More definitely, use Euler's formulae
\[e^{\pm iz}=\cos z\pm i\sin z \]

\[\cos z=\frac{e^{iz}+e^{-iz}}2 \]

\[\sin z=\frac{e^{iz}-e^{-iz}}{2i}\]
I forgot to mention that if we have $f(x)=\frac{a_{0}}2+\sum_{k=1}^{n}a_{k}\cos kx+b_{k}\sin kx$ then the last (finite) sum will be Fourier series for f(x) (the period is suggested to equal $2\pi$)

[quote="EUCLID 3.14"]Thanks guys. My original question is answered.
However, I still don't get what "$P(x,y)\in\mathbb{R}[x,y]$" means. I know that P(x,y) means that "$P(x,y)$ is a polynomial in two variables $x$ and $y$" but what does "$\mathbb{R}[x,y]$" mean...[/quote]
$\mathbb{R}[x,y]$ is notation for set (in fact, ring) of polynomials in 2 variables $x$ and $y$ with real coefficients (i.e. coefficients $\in\mathbb{R}$)

[quote="EUCLID 3.14"]I see.

I have one more question though. I was trying to expand $8\sin^{4}(5x)$ today using the Euler's formula but I got an answer which doesn't seem to make sense. I got: $\cos(20x)-4i\sin(10x)+3$. Can this be true? Is it possible to have a complex coefficient  (s) in your expansion?[/quote]
If in expression for $f$ there're no complex coefficients then its Fourier series ("real" one) has only real cofficients. In your example
\[8\sin^{4}5x=\cos20x-4\cos10x+3 \]
Check your calculation.


factorial
floor function
function
modular arithmetic

What is the rightmost digit of $ 1000000!$ (factorial of a million) ?

If that was too easy, then what is the rightmost non-zero digit of
$ 1000000!$ ? Also, how many zeros are there after this first non-zero digit from the right?

[hide]
1) The right most digit is $ 0$ obviously..

2)
 $ 5^8 < 10^6 < 5^9 \Rightarrow  \lfloor \frac {10^6}{5^1}  \rfloor +  \lfloor \frac {10^6}{5^2}  \rfloor +  \lfloor \frac {10^6}{5^3}  \rfloor  +  \lfloor \frac {10^6}{5^4}  \rfloor +  \lfloor \frac {10^6}{5^5}  \rfloor +  \lfloor \frac {10^6}{5^6}  \rfloor +  \lfloor \frac {10^6}{5^7}  \rfloor +  \lfloor \frac {10^6}{5^8}  \rfloor = 249998$

So there are $ 249998$ zeroes after the first non - zero digit from the right

[/hide]

What about the first non-zero right most digit?

Well you could attempt to evaluate $ 10^6 \mod 10^{249998}$, which may be possible if you know the carmichael function, but i dont so that would be futile...

Otherwise note that when we find the last digit of a product then we only need to consider the final digit of each term (i.e the lsat digit of $ 123 \times 321$ is the same as 3\times 1)

So consider the product $ 1 \times 2 \times 3 ...\times 1000000$. First we want to get rid of all the zeroes so take out all numbers that end in $ 5$ and $ 0$. We must also get rid of $ 249998$ factors of $ 2$
If we divide a number that end in $ 2$ by two then the last digit of the answer will either be $ 1$ or $ 6$
Similarly $ 4$ becomes $ 2$ or $ 7$...eventually we reallise the we need to evaluate

$ (1)^{10^5}\cdot (1)^{50000} \cdot (6)^{50000} \cdot (3)^{10^5} \cdot (2)^{50000}\cdot (7)^{50000} \cdot (6)^{50002}\cdot (3)^{24999} \cdot(8)^{24999} \cdot(7)^{10^5}\cdot(8)^{10^5}\cdot(9)^{10^5} \mod 10$

$ \Rightarrow (1)^{250000}(2)^{174999}(3)^{250000}(4)^{50002} \mod 10$
$ \Rightarrow (2)^{275003}(3)^{250000} \equiv 2^3 3 \equiv 4 \mod 10$

So the last non-zero digit is $ 4$!

Damn that's ugly...maybe you should just forget this post

i lost you after you said the 2 and 7 thing...

Haha yeah I couldn't expalian what i was doing because it was so long and ugly. Also I used a scientific calculator so the numbers are probably wrong. I'm sure there's a much shorter way!!

However, since there are more factors of $ 2$ than $ 5$ we can assume that the last non-zero digit will be even. So I have a $ \frac{1}{4}$ chance of being correct. :lol:

Wait...I have a question:

Since every factorial before $ 5!$ ends in 0, would the first non-zero right-most digit be simply $ 1! \times 2! \times 3! \times 4! \equiv 2 \times 6 \times 24 \equiv 288 \equiv 8 \pmod{10}$?

Can you explain you logic...

We are evaluating $ 10^6!$ not $ (10^6!)!$..

What I am saying is this:

1! ends in 1
2! ends in 2
3! ends in 6
4! ends in 4
5! ends n 0
6! ends in 0
7! ends in 0

Every number after 4! ends in 0.  So what I am saying is that the right most digit simply has to be 1*2*6*4 = 48 and 8 is the last digit.

EDIT: Nevermind...the simple 5! has 2 as it's right most non-zero digit whcih contradicts what I say...but 8 is still a possibility :D

As Ocha said , well i Ran it in a C++ program. If you Need to find the last non zero digit , first we get then no. of zeroes , now let a be 1000000! such that , a=10q+r , here r is the remainder and q is the quotient.
next time , take a as a-r/10 , that the quotient , now divide it again by 10...so keep on doing that for the no. of zeroes (let it be n ) we had attained + 1 . so that r would be that last non zero digit , because after you have taken mod10 n times , there are no zeroes left at the end.
 OR
Right 1000000! in the product of its propers divisors , Take Mod10^n+1 , and add the remainders.

[quote="Yongyi781"][quote="ocha"]Otherwise note that when we find the last digit of a product then we only need to consider the final digit of each term (i.e the lsat digit of $ 123 \times 321$ is the same as 3\times 1)
[/quote]

This won't work here because it includes multiples of 2 and 5, so the last digit will be 0. If you want the last non-zero digit, you certainly can't just take out all the multiples of 5!

The question itself is very tedious because you have to calculate all the primes below $ 10^6$.[/quote]

I know you can't [b]just[/b] take out all the multiples of $ 5$, that's why when I tried to take out multiples of two it became very very long...

That doesn't work either. After you do that, the last non-zero digit no longer depends only on the last digits of the numbers.

sure it does, if we take out the same number of factors of $ 2$ and $ 5$ then we are taking out factors of $ 10$ which effectively takes the zeroes of $ 10^6!$

For a small scale example, suppose we take the numbers $ 34\cdot 15 \equal{}510$.
Take out the $ 2$ from $ 34$ and the $ 5$ from $ 15$ and we have $ 17\cdot 3$.
The last numbers $ 7\cdot 3 \equal{} 21$
The last number of $ 21$ is $ 1$ and this is the last non-zero digit of 510.  :)

However this is irrelevant because RaviB has a much more elegant solution

Oh I thought you said take out completely the numbers that end in 5 or 0, not only the factors of 5.

....

Oh wait, you did say that.


Given a convex pentagon $ABCDE.$ Let $S,R$ be the intersection points of $AC$ or $AD$ and $BE$ respectively, and let the intersection points $T,P$ of $CA$ or $CE$ and $BD$ respectively.Let $Q$ be the intersection point of $CE$ and $AD.$ If all of $\triangle{ASR},\ \triangle{BTS},\ \triangle{CPT},\ \triangle{DQP},$ and $\triangle{ERQ}$ have the area of $1,$ then find the area of the following pentagons.

(1) The pentagon $PQRST.$
(2) The pentagon $ABCDE.$

Hint: Prove that BE || CD and CE || AB. The rest follows.

(1) The pentagon PQRST = 2.24
(2) The pentagon ABCDE = 15.33

the problem can be easily solved using the concept of areas
assume that the  $[PQRST]$ is $m$ . join $ET$ and assume that the $[ETP]$ is $t$([] denotes area)
then $[EST]$ is $= m+1-t$
now $\frac{[EBT]}{[ETP]}=\frac{[BTC]}{[CTP]}$
assume that $[BTC]=d$
then $\frac{m+1-t+1}{t}=\frac{d}{1}$.............$(1)$
similarly we can say that
$\frac{[ETC]}{[ETS]}=\frac{[BTC]}{[BTS]}$
which implies that
$\frac{t+1}{m+1-t}=\frac{d}{1}$.....................$(2)$
now from solving 1 and 2 we can say that
$t=\frac{m+1}{2}$
and $d=\frac{m+3}{m+1}$
and similarly we can say that
$[ARE]=[ASB]=\frac{m+3}{m+1}$
also observe that
$\frac{[ATE]}{[ETC]}=\frac{[ATB]}{[BTC]}$
therefore
$\frac{\frac{m+1}{2}+1+ \frac{m+3}{m+1}}{\frac{m+1}{2}+1}=\frac{\frac{m+3}{m+1}+1}{\frac{m+3}{m+1}}$
which by solving you get that
$m=\sqrt5$
area of the polygon $ABCDE$ can now be found out easily !!!!


search
number theory proposed
number theory

Find all positive integers that can be written in the form $\frac{(a+b+c)^{2}}{abc}$ where $a,b,c$ are positive integers.

It has been discussed before ,just search it.


The weather on Monday has a 50% chance of rain. For the following days of the week, the chance of rain on any given day decreases to 40% following a rainy day and increases to 60% if it does not rain the day before? What is the probability that it is raining on Wednesday?

I got a different answer.

[hide="My Answer"]

There are four cases to consider.

Case 1: It rains on Monday, doesn't rain on Tuesday, rains on Wednesday.

This is simply $ \frac{1}{2} \times \frac{3}{5} \times \frac{3}{5} \equal{} \frac{9}{50}$

Case 2: It rains on Monday, rains on Tuesday, rains on Wednesday.

This is $ \frac{1}{2} \times \frac{2}{5} \times \frac{3}{5} \equal{} \frac{3}{25}$

Case 3: It doesn't rain on Monday, doesn't rain on Tuesday, rains on Wednesday.

This is $ \frac{1}{2} \times \frac{3}{5} \times \frac{3}{5} \equal{} \frac{9}{60}$

Case 4: It doesn't rain on Monday, rains on Tuesday, rains on Wednesday.

This is $ \frac{1}{2} \times \frac{2}{5} \times \frac{3}{5} \equal{} \frac{3}{25}$

Adding it all together, we get $ \boxed{\frac{3}{5}}$.
[/hide]

[b]Rains on Monday[/b] $ 1/2$
* Rains on Tuesday $ 2/5$
** Rains on Wednesday $ 2/5$
*** Total: $ \frac 12 \cdot \frac 25 \cdot \frac 25 \equal{} \frac 2{25}$
* No rain on Tuesday $ 3/5$
** Rains on Wednesday $ 3/5$
*** Total: $ \frac 12 \cdot \frac 35 \cdot \frac 35 \equal{} \frac 9{50}$

[b]No rain on Monday[/b] $ 1/2$
* Rains on Tuesday $ 3/5$
** Rains on Wednesday $ 2/5$
*** Total: $ \frac 12 \cdot \frac 35 \cdot \frac 25 \equal{} \frac 3{25}$
* No rain on Tuesday $ 2/5$
** Rains on Wednesday $ 3/5$
*** Total: $ \frac 12 \cdot \frac 25 \cdot \frac 35 \equal{} \frac 3{25}$

Adding up: $ \frac {4 \plus{} 9 \plus{} 6 \plus{} 6}{50} \equal{} \frac {25}{50} \equal{} \frac 12$

@Isabella2296: P(No rain on Tuesday|No rain on Monday) = 2/5, not 3/5 (look at case 3) ;)
Also, in case 2: P(Rain on Wednesday|Rain on Tuesday) = 2/5, not 3/5 as well.

This can be done using marvok chains
Denote a day of rain r

$ Pr(r|r) \equal{} 0.4$
$ Pr(r'|r) \equal{} 0.6$
$ Pr(r|r') \equal{} 0.6$
$ Pr(r'|r') \equal{} 0.4$

$ T \equal{} \left[ \begin{array}{ c c } 0.4 & 0.6 \\
0.6 & 0.4 \end{array} \right]$

$ S_0 \equal{} \left[ \begin{array} {c} 0.5 \\
0.5 \end{array} \right]$

$ Pr(r_{wed}) \equal{} T^2 \times S_0$

$ Pr(r_{wed}) \equal{} \left[ \begin{array} {c} 0.5 \\
0.5 \end{array} \right]$

$ Pr(r_{wed}) \equal{} 0.5$

EDIT: for some reason I can see the matricies in Latex?

EDIT: Okay matricies are fixed, thanks yongyi781!

[quote="ocha"]
\[ T \equal{} \left[ \begin{array}{ c c } 0.4 & 0.6 \\
0.6 & 0.4 \end{array} \right]
\]
$ S_0 \equal{} \left[ \begin{array} {c} 0.5 \\
0.5 \end{array} \right]$[/quote]

You forgot the backslash in \begin{array} for both of them.