There were a few problems on this sprint that bugged me.. and please show methods that DON'T involve a calculator...

28. After collecting for the candy drive, Sally had eight $1 bils, five $5 bills, and 3 $10 bills. She randomly selected three bills without replcement. What is the probability that she chose one of each type of bill? Express your answer as a common fraction.

29. In rectangle ABCD, H is the midpoint of BC, E lies on AD, and F lies on AB. In rectangle CEFG, H lies on FG and HG = 3 in. m<DEC=45 :^o: What is the number os square inches in the difference between the areas of the two rectangles? Note: There was no drawing to accompany this problem.

[hide]I just solved 28. First, I segmented this problem. The number of ways to choose three different bills is 8C1*5C1*3C1, which is 8*5*3. The number of ways to choose 3 bills out of 16 is 16C3, which is 16*15*14/3*2, which simplifies to 8*5*14. Putting these together, you get $\frac{8*5*3}{8*5*14}$. That equals, by canceling and NOT multiplying out, 3/14, the answer.[/hide]

coins? I thought you said dollars at first

Coins, dollars, all the same thing ;)  it really was dollars... sry y'all

without handy paper and pencil, I'm not bothering to do the problem right now. It is a good one for me to think about...

[quote="hockeyadi23"]There were a few problems on this sprint that bugged me.. and please show methods that DON'T involve a calculator...

29. In rectangle ABCD, H is the midpoint of BC, E lies on AD, and F lies on AB. In rectangle CEFG, H lies on FG and HG = 3 in. m<DEC=45 :^o: What is the number os square inches in the difference between the areas of the two rectangles? Note: There was no drawing to accompany this problem.[/quote]

[hide]In the picture, all angles are either 90 or 45, so there are a lot of isoceles right triangles which have hypotenuses = legs times root2.  The area of the big rectangle is $(\frac{9}{\sqrt{2}})( 6\sqrt{2}) = 54$ While the area of the small rectangle is $(3)(9)=27$ so the difference of their two areas is $54-27=27$[/hide]

I solved number 29, and here is my attempt at "drawing" the picture on here.
[hide] The segments are AB, BC, CD, AD, FG, GC, CE, and EF. They are whole segments, though it probably does not look like it on here. Point F in on AB, point H is on BC and FG, and point E is on AD. It would be a good idea to rewrite this on paper.  ;)
[code]
         A____F_____B
         |    /'\   |
         |   /   \  |
         |  /      \|H
         | /        |\
        E|/         | \G
         |\         | /
         |  \       |/
         |    \     |/
         |      \   |/
         |       \  |/
         |_________\|
        D           C         [/code] [/hide] That was fun. :lol: Once you have the picture, solving it is not hard at all.
[hide] Since it tells you that measure of angle DEC =45, then the measures of angles DCE, HCG, CHG, FHB, BFH, AFE, and AEF are all 45.....The final answer is 27. I don't have time to finish the problem right now but I'll try to later.[/hide]

Use the [c-o-d-e][/c-o-d-e] tags (minus the hyphens) to evenly space the things so you don't have to use asterisks.

For example:
[code]
 ( )
 \|/
 / \
[/code]

[quote]Use the [c-o-d-e][/c-o-d-e] tags (minus the hyphens) to evenly space the things so you don't have to use asterisks.
[/quote] Thanks, the asterisks were a pain. Is there a program on here that lets you draw pictures directly?

[quote="frost13"][quote="NEEOMstudent7"]I think I have #28.
[hide]If you take 8*5*3 divided by 16*15*14 (the number of each bill possible to be picked divided by the total number of bills without replacement)  And cross out, you get 1/2*14, or 1/28.[/hide][/quote]

[hide]NEEOM, that is the probability that you would pick a ten, five, and one dollar bill in that order.  You need to take your answer and multiply it by 3! sinec that is how many ways you can rearrrange the different dollar values.(10,5,1);(10,1,5);(5,1,10);(5,10,1);(1,5,10);(1,10,5)

So $\frac{1}{28}$ x $6$ = $\frac{6}{28} = \frac{3}{14}$
[/hide][/quote]

[hide]Thanks, I see what you mean now.  I put them in a certain order, but you can change the order they are in.  [/hide]

That combination thing is what screws me up in probability questions, and I am getting much better at probability problems. You will be lucky not to see one probability question on a competition, because it has not happened before, at least not since 2000.


$(\sum_{i=1}^{n}{a_i}{b_i}{c_i})^2 \leq  (\sum_{i=1}^{n}{a_i}^2)(\sum_{i=1}^{n}{b_i}^2)(\sum_{i=1}^{n}{c_i}^2)$.

Is the above inequality true in general?

Besides, in what other directions can we modify the Cauchy-Schwarz Inequality?

Best Regards.

I think that this topic shouldn`t be here. I think it should be in and advanced forum (maybe intermediate topics?).

If that's middle school, then I'm seriously behind  :D

Let's move it to intermediate topics?

Just moved it to Olympiad :lol: It might be too far, but it's better than before in a forum too low...


Suppose that $ \omega$ is a primitive 2007th root of unity. Find $ (2^{2007} \minus{} 1) \cdot \sum_{j\equal{}1}^{2006} \frac{1}{2 \minus{} \omega^j}$.

Let $ f(x) \equal{} x ^{2007} \minus{} 1 \equal{} \prod_{j \equal{} 0}^{2006}(x \minus{} \omega^j)$,

then $ P'(x) \equal{} 2007x^{2006} \equal{} P(x)\sum_{j \equal{} 0}^{2006}\frac {1}{x \minus{} \omega^j}$

and we will get the value if let $ x \equal{} 2$.


There are $n!$ permutations of ${1,2,...,n}$. How many of them satisfy $s_k \geq k-2$ for all $k = 1,2,...,n$?

The only $s_k$ I know is $x^k+y^k$. What is $s_k$ in this case?

[quote="DPopov"]$s_k$ denotes the $k$th number in the permutation, at least that is how I understood the question.[/quote]
Yes, you understood it right

[hide]
For $s_k=1$ we have $1 \geq k-2 \Rightarrow k \leq 3$.  Thus we have three options,  $k=1,2, or 3.$
For $s_k=2$, we find $k \leq 4$, so we have four possible places, but one of these spaces is already taken by 1, thus we only have three options.  Similarly we have three options for $2,3,...,(n-2)$.  Once we get to $n-1$, we have $k \leq n+1 $, but there are already $n-2$ numbers laid down, so there are really only two possible choices.  And then when we get to $n$ there will only be one space where it can go.  Thus the anwer is $\boxed{2 \cdot 3^{n-2}}$
[/hide]


$ A, B, C$ and $ D$ are distinct positive integers such that the product $ AB \equal{} 60$,  the product $ CD \equal{} 60$ and $ A \minus{} B \equal{} C \plus{} D$ .  What is the value of $ A$?

Fastest way for this one is definately guess and check. There aren't too many factors of 60 so you should find it quite easily.

Just look at the factors that are closest to each other and easily enough we find

$ 20 \minus{} 3 \equal{} 12 \plus{} 5$

Therefore, $ \boxed{A \equal{} 20}$.

EDIT: Thanks jxl, typo.

Though $ 20$ is the correct answer, $ 20\minus{}3\ne 15\plus{}4$.  It's actually $ 20\minus{}3\equal{}12\plus{}5$.

[quote="modularmarc101"]Fastest way for this one is definately guess and check. There aren't too many factors of 60 so you should find it quite easily.
[/quote]
I thought there was some kind of formula or an even faster way of solving it. (I read it somewhere)


Hello,
Last time I a posted a similar thread, which turned out to be unhelpful (I hate to repost things).  Here I am doing it again and really need help right away this time.

I participate on a community forum, which is a non Aops. We have a subsection for mathematics, but the inability to type math symbols is a huge drawback. Can anyone really help explain how to install latex on a discussion forum. The adminstrators of that forum do not even know what a latex is so I think they would be happy if I bring back to them some instruction that is very fundamental. Thank you for your help.

It very much depends what your system allows. Can LaTeX be installed on the server (root access is required or help from the host)? If not, can [url=http://www.forkosh.com/mimetex.html]MimeTeX[/url] be installed as a CGI program? Is it a phpBB forum (like AoPS) or another type? Is it even a forum powered by PHP?

Full instructions, particularly for phpBB forums, are given in [url=http://www.mayer.dial.pipex.com/tex.htm]LatexRender[/url] for both LaTeX and MimeTeX as well as an email address for help.

So I take it Invisionfree forums don't allow LaTeX installation?  :(

It [i]could[/i] host LaTeX but you would have to persuade them to do so and, to be honest, you're unlikely to be sucessful. Unfortunately, mathematics in general, and LaTeX in particular, is very much a minority activity.

Hello friends,
Thanks for your answers. I asked the question you raised about who powers the forum I was referring to and here I paste the response I got from the admin of the forum in question.

"Akech,
Yes it's supported by PHP. The only reason why the math characters aren't showing isn't because it doesn't support it. It's just that I'm having hard time trying to get it working. I've uploaded files that should make it work on the server yet still nothing seems to be functioning as needed.

Anyway, help me or anyone who has an idea on how it could easily be installed."

Can anyone step in to help now?
Thanks,
Akech

I can help if you get the forum admin to email me with details of what errors he is seeing - the more detail, the easier it is to solve. The email address is in the readme.txt file of the LatexRender download.

Stevem,
Thank you very much. I sent your note to the admin of the forum in question.
Akech


Hm... ok

i dont have PM so i cant reply your PM, so i need a topic to do this.

 um pe no saco isso

I never send Spam on PM,  its ilogic block it. becuase i never send spam on PM, so why block this!?

and a dozen of people etc... it is too exaggerate it

if i found 17 persons, i can block someone in this forum!?

hm... this sound good.

i willl make some plans using this ILOOGIC MATH though

it's wonderfull, a new kind of math with a perfect structure.

amazing

it's possible to "unblock' my PM?

because i want to talk with brambila about math in portuguese, and i dont want others people post on it saying stupid things like "What is this, please, talk in english" etc....

its a lost of time

bye

hey buda

i dont have private massagens to post back

i'm sorry for the jokes

and PS: It's really idiot ignore me on my true posts, because all solutions on my true post is true.

so apologize too


Show that the area of a first-quadrant triangle bounded by the axes and the line y - 5 = m(x - 2) is at least 20.

$ S \equal{} \frac {1}{2}\left(2 \minus{} \frac {5}{m}\right)(5 \minus{} 2m)$

$ \equal{} 10 \plus{} \frac {1}{2}\left( \minus{} 4m \minus{} \frac {25}{m}\right)\ (m < 0)$

$ \geq 10 \plus{} \sqrt {( \minus{} 4m)\cdot \left( \minus{} \frac {25}{m}\right)}$

$ \equal{} \boxed{20}$ the equality holds $ \minus{} 4m \equal{} \minus{} \frac {25}{m}\ (m < 0)\Longleftrightarrow m \equal{} \minus{} \frac {5}{2}$.

hello, your equation is $ y \equal{} mx \minus{} 2m \plus{} 5$ and the intersection point with the $ y$-axes is
$ P_1(0,5 \minus{} 2m)$
and with the $ x$-axes
$ P_2(\frac {2m \minus{} 5}{m},0)$ if $ m$ is not zero.
We have the both conditions
$ m\le 5/2$ and $ m\geq5/2$ because the triangle should be in the first quadrant, hence we get $ m \equal{} 5/2$ so the area is equal
$ A \equal{} 1/2\frac {(2m \minus{} 5)^2}{|m|} \equal{} 0$ if $ m \equal{} 5/2$.
Oh i have forgotten the case $ m<0$ is possible so we have $ 1/2\frac{(2m\minus{}5)^2}{\minus{}m}\geq20\Leftrightarrow (m\plus{}5/2)^2\geq0$
Sonnhard.

[quote="kunny"]$ (1) S \equal{} \frac {1}{2}\left(2 \minus{} \frac {5}{m}\right)(5 \minus{} 2m)$

$ (2) \equal{} 10 \plus{} \frac {1}{2}\left( \minus{} 4m \minus{} \frac {25}{m}\right)\ (m < 0)$

$ (3)\geq 10 \plus{} 2\sqrt {( \minus{} 4m)\cdot \left( \minus{} \frac {25}{m}\right)}$[/quote]

1.  In (2), I see the justification for the constraint m < 0, but where do you use that fact?

2.  In (3), what's the justification for the inequality?

Thanks

I used $ Arithmetic\ Mean\geq Geometric\ Mean$, or $ \boxed{\frac{a\plus{}b}{2}\geq \sqrt{ab}}$ for $ a>0,\ b>0$.

hello, for all possible $ m$ is the area not at least $ 20$.
Sonnhard.

[quote]$ \geq 10\plus{}2\sqrt{(\minus{}4m)\cdot\left(\minus{}\frac{25}{m}\right)}$[/quote]

Should be:

$ \geq 10\plus{}\sqrt{(\minus{}4m)\cdot\left(\minus{}\frac{25}{m}\right)}$


quadratics
number theory proposed
number theory

This is a really really cute problem:

Suppose that some positive integer can be written as the sum of 4 odd distinct squares. Prove that it can be written as the sum of 4 even distinct perfect squares!

If there is no distinct in the question then it's quite simple, but Arne does the 'distinct' make it much hrader?

Bomb

Well, not really, if you look at the solution...

Hint
[hide]
\[(4a + 1)^2 + (4b + 1)^2 + (4c + 1)^2 + (4d + 1)^2 = 4 \left[\dots\right]\][/hide]

So does the following solution work
Since n is sum of 4 odd squares its divisible by 4, so n/a is integer, so by Lagranges' theorem , we are done...

Bomb

Yes, there is, there is a hint... (It's not a link, just some hidden information)

Plz look, I just fixed my reply....

Bomb

Well, Lagrange's theorem doesn't tell you that these squares will be distinct :)
So, that's not sufficient.

[quote="borz"]$4 (a+\frac{1}{2})^2+4 (b+\frac{1}{2})^2+4 (c+\frac{1}{2})^2+4 (d+\frac{1}{2})^2$
Is that what you mean?[/quote]

No, of course not. The expressions in the brackets might not be integers in your case.

[quote="bomb"]Nuce soln, Arne!!

Bomb[/quote]

So, you managed to finish the identity?

$(4a + 1)^2 + (4b + 1)^2 + (4c + 1)^2 + (4d + 1)^2$ equals \[4\left[(a + b + c + d + 1)^2 + (a + b - c - d)^2 + (a - b - c + d)^2 + (a - b + c - d)^2\right]\] and it's easy to verify that the squares are distinct. Obviously you can take $a$, $b$, $c$ and/or $d$ negative to get numbers which are congruent to 3 (mod 4). Pretty nice. :)

Is this result strengthen-able? Suppose by Dirichlet theorem...

Bomb

[Can someone write the case where the squares are 3modulo4? Thanks

But I did mention that already...

Suppose you want to use $(4n - 1)^2$ in some way...
Then you just write it as $\left(4 \cdot (-n) + 1\right)^2$, do you see why that works?

I began solving the problem by checking in the following example:
1+9+25+49=84
but 4+16+36+64=120
And i gave up as i told to myself that maybe some conditions are missing.

A similar problem, inspired by Arne...

A positive integer n is expressible as sum of 5 odd perfect squares.
Prove that n is expressible as sum of three distinct prefect squares.

Bomb

Clearly $n \equiv 5 \mod 8$, so it can be written as sum of three squares. But all quadratic residues $\mod 8$ are $\{0,1,4\}$, and the only possibilty to get the residue $5$ is having all of them once, so all numbers are different.

what results are known about numbers which can be written as a sum of $ n $ perfect squares? (like the one ZetaX used in the previous post)

a link would be appreciated

[b]1)[/b] Sum of one square: trivial ;)

[b]2)[/b] Sum of two squares:
A positive integer $n$ can be represented as sum of two perfect squares iff all prime factors $p \equiv 3 \mod 4$ of $n$ occur an even number of times in the factorisation of $n$. Then there are exactly
\[ \mathrm{A}_2 (n) : = 4 \cdot \sum\limits_{\substack{ d \in \mathbb{N} \\ d|n \\ d \equiv 1 \mod 2}} (-1)^{\dfrac{d-1}{2}} = 4 \cdot \prod_{\substack{p \in \mathbb{P} \\ p\equiv 1 \mod 4}} (v_p(n)+1) \]
different solutions $(a,b) \in \mathbb{Z}^2$ to $n= a^2+b^2$.

[b]3)[/b] Sum of three squares:
Write $n$ as $n=4^k u$, $k,u \in \mathbb{N}_0$ with $4 \nmid u$ (but $u$ can be even). Then $n$ can be written as sum of three squares iff $u \not\equiv 7 \mod 8$.

[b]4)[/b] Sum of four squares:
Every positive interger can be written as sum of four squares, and there are
\[ \mathrm{A}_4 (n) = 8 \cdot \sum\limits_{ \substack{ d \in \mathbb{N} \\ 4 \nmid d \mid n } } d \]
different solutions $(a,b,c,d) \in \mathbb{Z}^4$ to $n=a^2+b^2+c^2+d^2$.

[b]5)[/b] Sum of five squares:
As corollary to 4) every integer can be written as sum of five squares, but there is one thing to say: except of some small numbers (all $<100$ or so), every positive integer can be written as sum of five nonzero perfect squares.

...

[b]8)[/b] Sum of eight squares:
 (just mentioned here because of it's beautyness)
There are
\[ \mathrm{A}_8 (n) = 16 \cdot \sum\limits_{ \substack{ d \in \mathbb{N} \\ d \mid n } } (-1)^{n-d} d^3 \]
different solutions $(a,b,c,d,e,f,g,h) \in \mathbb{Z}^8$ to $n=a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2$.

[b]rest)[/b] Corollarys and further results:
From 3) one easily gets that every positive integer can be written as sum of three triangular numbers. Further, every positive integer can be written as sum of $k$ $k$-gonal numbers.
There are also formulas like the ones given in 2), 4) and 8) for representations as sum of $k$ squares, but they are not that nice as the ones mentioned.
Also for every $k$ there is a number $n_k$ so that every integer can be written as sum of $n_k$ nonnegative $k$-th powers of integers (Hilbert-Waring-Theorem).

ZetaX how do you derive sum of three triagular numbers from 3).
Thanks.

Bomb

Consider $n$ as the positive integer we want to write as sum of three triangular numbers.
Then $8n+3$ can be written as sum of three squares by 3) and looking $\mod 8$ gives that all of them are odd. So $8n+3 = (2a+1)^2+(2b+1)^2+(2c+1)^2 = 4(a(a+1)+b(b+1)+c(c+1))+3$ which gives $n=\dfrac{a(a+1)}{2}+\dfrac{b(b+1)}{2}+\dfrac{b(b+1)}{2}$, so $n$ is sum of three triangular numbers.


(1) In how many ways can a number be written as the sum of consecutive integers?

(2) 9 can be expressed as the sum of three consecutive integers: $ 9\equal{}2\plus{}3\plus{}4$ and it can be expressed in three different ways as the sum of consecutive integers: $ 9\equal{}4\plus{}5\equal{}2\plus{}3\plus{}4$. Is there a number which can be expressed as the sum of 99 consecutive integers and in 99 different ways as the sum of consecutive integers?

(Sums are to be counted if they are made out of one single number.)

Do you mean positive consecutive integers only?

 :)

[quote="bertram"](1) In how many ways can a number be written as the sum of consecutive integers?

(Sums are to be counted if they are made out of one single number.)[/quote]
Hint: The sum of k+1 consecutive positive integers, starting from h is
N = h + (h+1) + (h+2) + …. + (h+k) = (2h+k)(k+1)/2
with $ h \ge 1$ and $ k \ge 0$
(2h+k)(k+1) = 2N is the product of two factors, one odd and the other even, with
k+1 < 2h + k
Solution:[hide]There are as many different sums as odd factors for N.
For all integers, k=0 is a solution (a “sum” of one single number). If N is a power of 2, it is the only solution.
Each odd factor of N (other  than 1) generates a different sum.
If N = (p)(2m+1), p>0, m>0, the solution is either
(h, k) = (p-m, 2m) for  m<p
(h, k) = (m-p+1, 2p-1) for $ m \ge p$
If p is another (not the same) odd factor, it generates a different (h,k).[/hide]
[quote="bertram"](2) 9 can be expressed as the sum of three consecutive integers: $ 9 \equal{} 2 \plus{} 3 \plus{} 4$ and it can be expressed in three different ways as the sum of consecutive integers: $ 9 \equal{} 4 \plus{} 5 \equal{} 2 \plus{} 3 \plus{} 4$. Is there a number which can be expressed as the sum of 99 consecutive integers and in 99 different ways as the sum of consecutive integers?[/quote]
Yes, infinitely many.
Hint:
From above, for k=98, N = 99(h+49) is a sum of 99 consecutive integers
$ 99 \equal{} 3^2 * 11$
Solution:
[hide]$ N \equal{} 3^a * t^b * 11^c$ has 99 odd factors iff (a+1)(b+1)(c+1) = 99
$ 3^{10} * 5^2 * 11^2 \equal{} 178623225 \equal{} 99 * 1804275$ is the smallest solution
178623225 = 1804226+1804227+ .... +1804323+1804324[/hide]


function
induction
algebra unsolved
algebra

$LHS - RHS$ is an increasing function, I'm too lazy to prove, so $n = 1$ gives only solution.  Can someone else find a neat way to prove it?

[quote="luimichael"]Find all possible solutions to the following equation:
$3^n+4^n+...+(n+2)^n=(n+3)^n$ , $n \in \mathbb{N}$.[/quote]

       We have : $\large{3^n+4^n+...+(n+2)^n=(n+3)^n\,=\,S_n}$

        By induction , we can prove that $\large{\,\,\frac{(n+2)(n+3)}{2}\,|\,[S_n\,-\,(2^n+1)]}$

     Does it help ???

[quote="t0rajir0u"][quote="luimichael"]Note that $n=2$ also gives a solution as $3^2+4^2=5^2$.
Besides, $n=3$ also works: $3^3+4^3+5^3=216=6^3$.[/quote]

Oops, I meant $n = 2$.  And I thought I'd checked $n = 3$...

molina, what is the statement you proved by induction?[/quote]

      Hi , i used induction , see :

   Calculate

    $\large{\sum_{i=1}^{n}k^i}$ using binomial teorem , then use induction.

    Sorry , but I am so lazy to do it...


[quote="rnwang2"]If $f(x)=2\cdot f(x-1)$ for all integers $x$, and $f(n)=3$ for some integer $n$, find the value of $f(n-5)\cdot f(n+5)$.[/quote]
[hide]
$f(n-5)=32*f(n)$
$f(n+5)=\frac{f(n)}{32}$
$f(n)^{2}=9$[/hide]
To ckck, read the question again.

[quote="ckck"][quote="rnwang2"]If $f(x)=2\cdot f(x-1)$ for all integers $x$, and $f(n)=3$ for some integer $n$, find the value of $f(n-5)\cdot f(n+5)$.[/quote]
[hide]So plug in n to have $2(n-1)=3$. Solving for this we get $n=\frac{5}{2}$. So using this get $\frac{-75}{4}$.[/hide][/quote]
Umm...no. Check your first step.

[quote="ckck"]Ok i know I did the problem wrong. But bpms, can you explain your solution? I don't the first step.[/quote]
[hide]
$f(x)=2\cdot f(x-1)\implies \frac{f(x)}{f(x-1)}=2$.

This condition is satisfied by the exponential equation $f(x)=2^{x}$.

$\frac{f(n-5)}{f(n)}=2^{-5}\implies f(n-5)=\frac{f(n)}{32}$.

$\frac{f(n+5)}{f(n)}=2^{5}\implies f(n+5)=32\cdot f(n)$.

The answer is $[f(n)]^{2}=3^{2}=\boxed{9}$.[/hide]


probability
percent
symmetry
ratio
Pascal\u0027s Triangle

9 fair coins are flipped. What is the probability that at least 4 are heads?

The probability should be greater than $ \frac{1}{2}$.  Try again :)

lol whoops.

Too lazy to type up the fractions in Google again.

 :oops:

[hide]hint: binomial distribution [/hide]

[hide]USing complimentary counting, we find it's $ 1\minus{}\frac{130}{512}$, so $ 382/512\equal{}191/256.$Notice we found the numerator by choosing 3 coins to be heads, then 2, then 1, and then 0.[/hide]

i counted the ways to get 4 or more heads using combinations. then i divided by 512

[hide="WAY #1"]
We can easily do this by computing 2^9 (the number of total possibilities of flipping 9 coins), and then find the number of ways by computing (9 choose 4)+(9 choose 5)...+(9 choose 8=9 choose 1=9)+1.  This works because for each choice you are counting a different way of getting 4 heads, 5 heads, 6 heads...This equals 382/512=191/256[/hide]
[hide="WAY#2"]
Similar to the first way but less tedious, we can compute the number of ways 3 heads, 2 heads, 1 head, and 0 heads can occur-this probability subtracted from 1 will similarly yield the answer because the probability that at most 3 heads will occur+the probability that at least 4 heads will occur will ultimately equal 1 (duh!).  Thus, we have the answer is:
1-(([9 choose 3]+[9 choose 2]+[9 choose 1]+[9 choose 0])/512)=
1-(130/512)=382/512=191/256.....again[/hide]

[hide="Note"] The probability that at least $ 5$ coins are heads is exactly $ \frac{1}{2}$ (can you figure out why? (use symmetry!)).  So all you have to do is add the probability that [i]exactly[/i] $ 4$ of the coins are heads.  :) [/hide]

9th line of pascal's triangle:
1 9 36 84 126 126 84 36 9 1
Now, that means that there's 1 way of getting 9 heads, 9 ways of getting 8 heads, 36 ways of getting 7 heads, etc.
So there's 126 ways of getting 4 heads, 126 ways of getting 5 heads, 84 ways of getting 6 heads, etc.
Now add em up and put that ratio over $ 2^9$
$ \frac{(1\plus{}9\plus{}36\plus{}84\plus{}126\plus{}126)}{2^9}\equal{}\frac{382}{512}\equal{}\frac{191}{256}$


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $S$ be a commutative subalgebra of $M_n(K)$ and $A_{11},...,A_{pp}$ be some matrices which belong to $S$. Is it true that we can find $X$ in $S$ having the same determinant as the matrix having as entries $A_{ij}$ (which is a $npxnp$ matrix)?

In fact, one can prove (try it, it's really nice!) that if $A_{11},...,A_{pp}$ are $nxn$ matrices which commute then the determinant of the big matrix $(np)x(np)$ with entries $A_{ij}$ as blocks is $det(\sum_{\sigma\in S_p}{\epsilon(\sigma)\cdot A_{1\sigma(1)}...A_{p\sigma(p)})}$. And here is a nice application of this result: if $A$ is a real matrix whose characteristic polynomial has no real roots then for any matrices $A_{ij}$ lying in the algebra generated by $A$ over $R$ we have $det(A_{ij})\geq 0$. This generalizes a problem posted long time ago by Moubi and solved by Amfulger.

Does S  contains the diagonal matrices  ?

Why? Is this important? I really don't see how this could help. :?

If it does $X=diag(1,..,1,det(\sum_{\sigma\in S_p}{\epsilon(\sigma)\cdot A_{1\sigma(1)}...A_{p\sigma(p)})}$

I see you like very much jokes, Moubi, today. Ok, I will be bad: no, it's arbitrary subalgebra, it does not necessarily contain all diagonal matrices. :P  :P  :P


geometry
3D geometry
pyramid
sphere
geometry proposed

Let $n \geq 2$ be a natural number. A pyramid $P$ has base $A_1A_2...A_{2n}$ and apex $O$. The polygon $A_1A_2...A_{2n}$  is regular and the point $C$ is its centre. The line $OC$ is perpendicular to the plane of the base of $P$. A sphere passes through $O$ and meets each of the line segments $OA_i$ internally. For each $i = 1, 2,..., 2n$, let $X_i$ be the point (other than $O$) where the sphere meets $OA_i$. Prove that $OX_1+OX_3 + ... + OX_{2n-1} =  OX_2+OX_4 + ... + OX_{2n}$

We can actually project all the points and segments to the plane of the base and thus get back a plane figure. Now it should be easy to handle.

Let $U$ be the center of sphere and $V$ it's projection on a plane of poligon.

$A_iX.A_iO = A_iO(A_iO - OX_i)=A_iU^2 - R^2 = A_iV^2 +UV^2 -R^2$

Summing we just need to prove that for any point $V$ in plane $A_1A_2..A_n$ that

$A_1V^2+A_3V^2+...+A_{2n-1}V^2 = A_2V^2+A_4V^2+...+A_{2n}V^2$

But this is corect from analitic geometry and complex numbers.

[quote="cadge_nottosh"]Use intersecting chords theorem to make the condition equivalent to some sums of squares being equal, then use the gpat (I missed this last step, but it is still 7-, probably, as it is an easy application, and I proved it for 2 divides n)[/quote]

interesecting chords thm? where's the cyclic quad?

Well, it is easy to see that it suffices to prove $\Sigma A_{2i+1}X_{2i+1} = \Sigma A_{2i}X_{2i}$.
But if $K$ is the centre of the sphere, $R$ its radius, then $A_{i}X_{i}.A_{i}O = A_{i}K^{2} - R^{2}$.
Then GPAT finishes it off.

Well, if $(X_{i},M_{i}) , (Y_{i}, N_{i})$ are two sets of points with masses associated with them, with the total mass of $X$ being $M$, of $Y$ being $N$, and the mean square distance of the two sets of points is defined to be \[ \frac{1}{MN}\sum{M_{i}N_{i}|X_{i}Y_{i}|^2 }\], with $\sigma(X)$ being the mean square distance of the points $X_i$ to their centroid $X^{\prime}$, then the Generalised Parallel Axis Theorem states that the mean square distance of the $X_{i}$ and $Y_{i}$ is\[ \sigma(X) + |X^{\prime}Y^{\prime}|^2 + \sigma(Y) \].

Jack


geometry
incenter
inradius
trigonometry
trig identities
Law of Cosines

Triangle ABC has AB equal to 8, BC equal 10, and CA equal 12. Point D is on BC such that AD bisects angle BAC. Point E is on CA such that BE bisects angle ABC. Lines AD and BE intersect at F. What is the distance from F to A?

Written by: Me  :)

[hide]F is the incenter of the triangle. The area is $ 15 \sqrt {7}$ by Heron's, so since $ A \equal{} rs$, the inradius is $ \sqrt {7}$. By the law of cosines, $ cos A \equal{} \frac {9}{16}$. If the incircle is tangent to AC at X, then FAX is right, so $ FA \equal{} r/sin(\frac {\angle BAC}{2}) \equal{} \sqrt {7}/(\sqrt {\frac {1 \minus{} \frac {9}{16}}{2}}) \equal{} 4\sqrt {2}$.[/hide]


Define a sequence $ a_n$ as $ a_1\equal{}1,\ a_2\equal{}a,\ a_{n\plus{}2}\equal{}a_{n\plus{}1}\minus{}a_n\ (n\equal{}1,\ 2,\ \cdots)$.
Find the value of $ a\in\mathbb{R}$ such that $ \sum_{k\equal{}1}^{2008} a_k\equal{}0.$

$ a_n\equal{}Re[(1\minus{}a)\minus{}i\frac{1\plus{}a}{\sqrt 3}]exp\frac{\pi ni}{3}$ and
$ \sum_{k\equal{}1}^n a_k\equal{}a_2\plus{}a_{n\minus{}1}$. It give solution.

[quote="nayel"]Easily verify that $ a_1 \equal{} 1,a_2 \equal{} a,a_3 \equal{} a \minus{} 1,a_4 \equal{} \minus{} 1,a_5 \equal{} \minus{} a,a_6 \equal{} \minus{} a \plus{} 1,a_7 \equal{} 1,a_8 \equal{} a$. Thus $ a_n \equal{} a_{n \plus{} 6}$. Also note that $ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5 \plus{} a_6 \equal{} 0$ and $ 2008 \equal{} 6\cdot 334 \plus{} 4$. Thus $ 0 \equal{} \sum_{n \equal{} 1}^{2008}a_n \equal{} \sum_{n \equal{} 1}^{2004}a_n \plus{} 1 \plus{} a \plus{} (a \minus{} 1) \plus{} ( \minus{} 1) \equal{} 0 \plus{} 2a \minus{} 1$
$ \implies \boxed{a \equal{} \frac 12}$.[/quote]
This is a standard solution. :)

$ a_n \equal{} (1 \minus{} a \minus{} \omega )\omega ^{n \minus{} 1} \plus{} (\omega \minus{} a)(1 \minus{} \omega )^{n \minus{} 1},$ where $ \omega \equal{} \frac {1\pm \sqrt {3}i}{2}.$


calculus
integration
trigonometry
function
algebra
rational function
real analysis

problem31: find integral(-2pi to 2pi)(1/((e^sinx) +1))(1/(e-cosx))dx.

We have:

$I=\int_{-2\pi}^{2\pi}\frac{dx}{(e^{\sin x}+1)(e-\cos x)}=\left(\int_{-2\pi}^{0}+\int_{0}^{2\pi}\right)\frac{dx}{(e^{\sin x}+1)(e-\cos x)}$

For the first integral, use the substitution $x=-t$ and you get that:

$I=\int_{0}^{2\pi}\frac{dx}{e-\cos x}$, and this integral can be calculated by the standard techniques.

sorry i can't integrate integral(0 to 2pi)dx/(e-cosx)

Last integral was posted in this forum few time ago. you can find it

It has an elementary antiderivative. Smash it with a substitution to make it a rational function.

"antiderivative"?
well i can't think of any suitable substitution:)

It's a classic trick: substituting $t=\tan\frac{x}{2}$ converts any rational function of $\sin x$ and $\cos x$ into a rational function.


Let $ \delta ABC$
PROVE THAT: $ \sum \frac{1\plus{}2cos^2 A}{sin B} \geq 3$

[quote="nguyensilva"]Let $ \delta ABC$
PROVE THAT: $ \sum \frac {1 \plus{} 2cos^2 A}{sin B} \geq 3$[/quote]

$ \sum \frac {\sqrt{1 \plus{} 2cos^2 A}}{sin B} \geq 3$
I'm sorry


There are six English words containing six letters that follow a certain property.
This property is that the first letter of the word alone is an English word, the first two letters alone form an English word, the first three letters, and so on.

Can you find all six?

I know the first letter must be either "A" or "I", but I can't get any further.

[quote="hm29168"]I know the first letter must be either "A" or "I"[/quote]Brilliant.

[quote="alkjash"]If you search for 2-letter words, there are like 20 different possibilities just for the first 2 letters...[/quote]Your point?  The fact that the first 3, first 4, first 5, and all 6 have to be words limits the possibilities down to six.

so we have 2*26^5 possibilities = 23762752. Ouch.

2 letters
[hide="a- words"]
aa -- hawaiian term for lava
ab -- ?, not sure if it's a word
ac -- ?, doubt it
ad -- advertisement
ae -- Nope.
af -- Nope.
ag -- Agricultural or "Silver"
ah -- vocalic sound, does that count?
ai -- Nope.
aj -- no
ak -- no
al -- hmmm...
am -- uh, yeah that's a word
an -- it's a word
ao -- no
ap -- only in an acronym
aq -- no
ar -- no
as -- word
at -- word
au -- only "gold"
av -- no
aw -- interjection
ax -- cutting tool
ay -- interjection
az -- no[/hide]

Definitely a word: ad, am, an, as, at, ax (6)
Maybe a word (or too technical): aa, ab, ac, ag (4)
Interjectional: ah, aw, ay (3)
Total=13
everything else a- is definitely not a word

...

id = a word
if = a word
im = "instant messaging"
in = a word
io = Jupiter moon
ip = "internet protocol"
iq = "intelligence quotient"
is = a word
it = a word

definitely a word: id, if, in, is, it (5)
maybe a word: io (1)
common acronyms: im, ip, iq (3)

Ok!

ad, am, an, as, at, ax, id, if, in, is, it (11) are our possibilities for the first two letters!

23762752 possibilities =>  5026736 possibilities left

well I helped eliminate 18736016 possibilities!

New stuff: look at the scrabble 3-letter list... [removed all uncommon words]
add ado ads
amp
and ant any
ash ask
ate
axe

ids
iff ifs
ink inn ins
ism
its

Total: 19 3-letter words = 333944 possibilities left

antics
That works right?
OH I NEED TO FIND 6.
dang.
Guys, I found another one, but it includes a censored word...
[hide]
assess[/hide]
#3.
Intoes

what's 'anti' and 'asse'?
the four letter word is giving the pain (as four letter words are so often wont to).

Anti is a permitted word, unfortunately asse is not.  Unfortunately, int, intoe, and intoes are all not words.

Current words: ANTICS

Are these words fairly common and not obscure like "kiva?" (an Anasazi religious area, go to Mesa Verde.) And yes, I know that's not a likely word because it doesn't start with A or I. I just needed an example.

Search asse online.
http://dictionary.reference.com/search?q=asse

Hm, I guess there are seven.

Current words: ANTICS, ASSESS

Search Int online.
http://dictionary.reference.com/search?q=int&x=0&y=0
Search Intoe online.
http://medical-dictionary.thefreedictionary.com/metatarsus+varus
Note the part that says "also called intoe"

Int is an abbreviation for integer, intoe is an obscure medical term that has never been used.  Ever.

But medical terms are still english.
Anyways, no abbreviations?
That sucks.
found another.
annals.

Ann and Anna are people's names. Do they count?

They're other things too.
Search them on http://www.dictionary.com

What, can't be proper nouns?
Anyways, Ann is a cape in MA.


We start with a line of a poem, like

I went to the town one day to get a can of noodles,

and someone adds on to the poem and displays the whole poem, like this:

I went to the town one day to get a can of noodles,
When I saw someone walking down with four very cute poodles.

And we do that again and again, until none can find a suitable line for the poem.

I went to the town one day to get a can of beans,

I went to the town one day to get a can of beans,
I saw somebody wearing bright purple jeans,
And was trying, by any means,

I went to the town to get a can of beans,
I saw somebody wearing bright purple jeans,
And was trying, by any means,
To turn this into a movie scene,

[b]I went to the town to get a can of beans,
I saw somebody wearing bright purple jeans,
And was trying, by any means,
To turn this into a movie scene.

And then I saw a big giant bubble,
And I knew that we were headed for trouble.[/b]

Why don't we make this into four rhyming lines per stanza?  or atleast two...just for easierness and better readability and all that

sorry for adding an extra line

I went to the town to get a can of beans,
I saw somebody wearing bright purple jeans,
And was trying, by any means,
To turn this into a movie scene.

And then I saw a big giant bubble,
And I knew that we were headed for trouble,
It popped and everything turned to rubble,
And then something happened having to do with flubber.

Then Robin Williams came passing by,
Followed by a stupid fly,
And a large man who wanted to die,
Being all alone under the sky.

(Who else thinks the rhyme sucks? Oh and so does the tempo)

Nice try, miyomiyo.

I went to the town to get a can of beans,
I saw somebody wearing bright purple jeans,
And was trying, by any means,
To turn this into a movie scene.

And then I saw a big giant bubble,
And I knew that we were headed for trouble,
It popped and everything turned to rubble,
And then something happened having to do with flubber.

Then Robin Williams came passing by,
Followed by a stupid fly,
And a large man who wanted to die,
Being all alone under the sky.

Then the world turned to orange,
Everybody climbed up Mt. Blorange, (an actual mountain in Wales, U.K.)

Trying to make impossible rhymes, huh?

I went to the town to get a can of beans,
I saw somebody wearing bright purple jeans,
And was trying, by any means,
To turn this into a movie scene.

And then I saw a big giant bubble,
And I knew that we were headed for trouble,
It popped and everything turned to rubble,
And then something happened having to do with flubber.

Then Robin Williams came passing by,
Followed by a stupid fly,
And a large man who wanted to die,
Being all alone under the sky.

Then the world turned to orange,
Everybody climbed up Mt. Blorange
Everyone there was chewing a lozenge,
And then they murdered the cast of Entourage.

They turned their heads to see something silver,
And then da da da da da Bob the Builder,
Talked to his wife with some jewelry, trying to guilt her,
For hiring some clowns and some stilters.

Their heads stayed like that for a month
And they talked about the function $y=2^{n+1}$

I went to the town to get a can of beans,
I saw somebody wearing bright purple jeans,
And was trying, by any means,
To turn this into a movie scene.

And then I saw a big giant bubble,
And I knew that we were headed for trouble,
It popped and everything turned to rubble,
And then something happened having to do with flubber.

Then Robin Williams came passing by,
Followed by a stupid fly,
And a large man who wanted to die,
Being all alone under the sky.

Then the world turned to orange,
Everybody climbed up Mt. Blorange
Everyone there was chewing a lozenge,
And then they murdered the cast of Entourage.

They turned their heads to see something silver,
And then da da da da da Bob the Builder,
Talked to his wife with some jewelry, trying to guilt her,
For hiring some clowns and some stilters.

Their heads stayed like that for a month
And they talked about the function $y = 2^{n+1}$
And then they decided to stop rhyming

I went to the town to get a can of beans,
I saw somebody wearing bright purple jeans,
And was trying, by any means,
To turn this into a movie scene.

And then I saw a big giant bubble,
And I knew that we were headed for trouble,
It popped and everything turned to rubble,
And then something happened having to do with flubber.

Then Robin Williams came passing by,
Followed by a stupid fly,
And a large man who wanted to die,
Being all alone under the sky.

Then the world turned to orange,
Everybody climbed up Mt. Blorange
Everyone there was chewing a lozenge,
And then they murdered the cast of Entourage.

They turned their heads to see something silver,
And then da da da da da Bob the Builder,
Talked to his wife with some jewelry, trying to guilt her,
For hiring some clowns and some stilters.

Their heads stayed like that for a month
And they talked about the function $y=2^{n+1}$
And then they decided to stop rhyming
No they didn't!

There once was a man from Nantucket

I went to the town to get a can of beans,
I saw somebody wearing bright purple jeans,
And was trying, by any means,
To turn this into a movie scene.

And then I saw a big giant bubble,
And I knew that we were headed for trouble,
It popped and everything turned to rubble,
And then something happened having to do with flubber.

Then Robin Williams came passing by,
Followed by a stupid fly,
And a large man who wanted to die,
Being all alone under the sky.

Then the world turned to orange,
Everybody climbed up Mt. Blorange
Everyone there was chewing a lozenge,
And then they murdered the cast of Entourage.

They turned their heads to see something silver,
And then da da da da da Bob the Builder,
Talked to his wife with some jewelry, trying to guilt her,
For hiring some clowns and some stilters.

Their heads stayed like that for a month
And they talked about the function
And then they decided to stop rhyming
No they didn't!

There once was a man from Nantucket
And in his hand was a great big bucket
A lady named Nan ran away with that man
It seemed quite evil to Stan.


Can somebody please explain to me how exactely a voltaic cell works? There's a whole chapter about it in my chemistry textbook but it simply says what happens when it works, not why and how it works.

Essentially, when an active metal and a less active metal ion are in contact, the more active metal loses its electrons, which are captured by the metal ion, which becomes a metal.  Thus, we have a transfer of electrons.  In a voltaic cell, this transfer occurs through the movement of electrons through a wire - producing an electric current.

Does that help?

yeah as inteluser said, the cell basically works basically because of difference of Electron potential of the metals ($E^0$).  and the transfer of electrons creates the voltage in the cell.


Salut, :)
Cette section de mathlinks est devenue ma premiere place d'aide avec mon francais. Alors je vous remercie, mais j'ai une autre chose a vous demander... Chaque annee dans mon ecole il y a un "French Speech Contest," ou on doit parler pour 3 à 5 minutes sur un sujet interessant. J'ai fait ce concours l'année passée mais je n'ai pas gagné. La personne qui a gagnée avait ecrit du sujet "un interpretation des reves" ou en anglais "an interpretation of dreams." Alors, a-t-on quelque idée que vous pouvez m'offrir?
Merci encore! ;)

P.S. Si vous pouvez parler en anglais, pouvez vous mettre votre reponse dans cette langue. Ca sera plus facile pour moi à comprendre... :D

C'est évident, non?

Les mathématiques sont le sujet le plus intéressant!  :)

Je suis désolé, mais j'écrit mon réponse en français, parce que j'ai besoin du exercice. :oops:

C'est pas besoin d'exercise? (No 'pun' intended,...mais je me demandais cela, c'est tout simple comme ça).


Hello yet again.

Can someone please explain to me what does 4a and 4b mean?  I actually got (nC2) as the answer in a)...using a different explanation involving Fibonnaci numbers.  However...I have no idea what does 4b mean.

http://www.cms.math.ca/Competitions/COMC/COMC00/solutions2000.pdf

Thanks!

We have a set $T$, consisting of n terms.  For example, if $n=5$, then $T = \{1,4,5,9,14\}$.  The first part of the question asks you to find the [b]number[/b] of positive integers that can be expressed as the sum of two distinct (ie., different) elements of $T$.  For example, $5$ is one of the expressable integers, because $1 + 4 = 5$ and $1, 4$ are distinct terms in $T$.

In part b, the question is the same except we are asking for the number of positive integers that can be expressed as the sum of [b]three[/b] distinct elements of T.  But, say in the case $n=7$, we have $T = \{1,4,5,9,14,23,37\}$, and we dont count $37+4+5=9+14+23=46$ twice, even though we can make the sum in two ways.  We are counting the possible sums (answers), and not counting different arrangements of terms that yield the same sum twice (or more).

ps.  I don't know what your math ability is, but I would like to give some unsolicited advice that may or may not apply to you:  Since the COMC is quite close at hand, be more focused on being able to solve the other questions of the COMC than solving the last questions (70/80 not attempting B4 is better than 64/80 attempting B4 but messing up B1)  Read very carefully, and if it helps, circle the keyword which indicates what they are asking for in the question.  Do quick mental checks when writing so that you don't have to rewrite stuff.  Pace yourself during number crunching so you don't screw up [on part A's.]  Try mock tests where you time yourself and write it up.   (Also:  include rough work for all part A questions, so if you get it wrong accidentally, you dont lose all 5 marks.)

Thanks for the bit of wisdom.

I understood the question completely, but what I don't understand is the official solution to the specific COMC problem.  Understandably, I surely would like to know at least how 4b or 4a was derived so that I could maybe learn some new techniques or something...I dunno.

As for mock practise tests, my school has a very nice boardroom where I'll be taking the COMC, and I've already done several timed practise tests in there.  My (self-given) scores range between a 65-75 (the 75 being that easier COMC last year).  However, I guess you could say I'm really concerned that this year's COMC will be a big step up from last year's.

You advice does have some point to it, I sometimes miss the question because I don't answer what the Q's are asking for (for example, asking for the number of terms in a sequence; I put the last term).  All the more to worry...haha.

what steps of the solution dont you understand?  I think its very clear.

For part a, how does the fact that t_c + t_d > t_a + t_d complete the proof?  We know that t_a, t_b, and t_c are all different, and t_a + t_b is at most t_{d-1} + t_{d-2} but I can't quite place why right after, we say "and there are no two pairs that add to the same integer."

Can we also necessarily say that t_c + t_d vs. t_a + t_b???  What if it's t_d + t_b etc.???

For b, how is it 2 < c < f-2???  I get the 2 < c part...

Also, how are there f-5 choices as follows?

Finally, how did we get those "maximal values" at the end of the proof?

Thanks!

[quote="lowest_reynolds"]For part a, how does the fact that t_c + t_d > t_a + t_d complete the proof?  We know that t_a, t_b, and t_c are all different, and t_a + t_b is at most t_{d-1} + t_{d-2} but I can't quite place why right after, we say "and there are no two pairs that add to the same integer."
[/quote]
We supposed $t_a + t_b = t_c + t_d$, justified that they must be distinct, and that wlog $t_d$ was the largest.  Then $t_d \ge t_b + t_a$, and so $t_d + t_c > t_a + t_b$, contradiction.

[quote]
For b, how is it 2 < c < f-2???  I get the 2 < c part...
[/quote]
$t_f$ is the largest.  Then choosing $t_f$ automatically chooses $t_a, t_b$ as $t_{f-1}, t_{f-2}$.  So we need $c < f-2$ because $t_c < t_f$ and it cant be equal to $t_a$ nor $t_b$.
[quote]
Also, how are there f-5 choices as follows?
[/quote]
We can choose c = 1...f, except for c = 1,2,f,f-1,f-2.  So f choices minus 5 choices.
[quote]
Finally, how did we get those "maximal values" at the end of the proof?
[/quote]
read carefully you should get it


[quote="arqady"][quote="gemath"]Given $ x,y,z>0,x+y+z=1$ prove that
\[ (2+\frac{\sqrt{3}x}{y^{2}+yz+z^{2}})(2+\frac{\sqrt{3}y}{z^{2}+zx+x^{2}})(2+\frac{\sqrt{3}z}{x^{2}+xy+y^{2}})\ge 27 \]
It is really true, but I haven't solution!  :)[/quote]
$ \left(2+\frac{x}{y^{2}+yz+z^{2}}\right)\left(2+\frac{y}{z^{2}+zx+x^{2}}\right)\left(2+\frac{z}{x^{2}+xy+y^{2}}\right)\ge27$
true too with same conditions and nice enough.[/quote]

ok you are right arqady,and yours is very nice but I wrote wrong my inequality must be
\[ (2+\frac{\sqrt{3}x}{\sqrt{y^{2}+yz+z^{2}}})(2+\frac{\sqrt{3}y}{\sqrt{z^{2}+zx+x^{2}}})(2+\frac{\sqrt{3}z}{\sqrt{x^{2}+xy+y^{2}}})\ge 27 \]
I think it look not nice with yours but maybe harder, I haven't solved it. :)


LaTeX
real analysis
real analysis unsolved

Hello,

Can anyone do this question please ?

Let X and Y be sets, and let f : X → Y be a function. If A c ("Contained in and equal to") X, write f[A] for the set {f(x) | x ∈ A}. Prove that f is a surjection if and only if Y − f[A] c ("Contained in and equal to") f[X − A] for all A c ("Contained in and equal to") X.

Note: c means "contained in and equal to where there is a horizontal line below the symbol "c" ", I am sorry because I don't know how to type this symbol in and "c" is the closest representation to it.

I totally can't do it, can someone please please please do it, I would really appreciate it.

(->) if f is surjection and there exists an A st Y-f[A] not contained in f[X-A]
then (Y-f[A] union f[A]) is not contained in (f[X-A] union f[A]) since f[A] and Y-f[A] are disjoint
but the former is Y, and the latter is f[X]. here is our contradiction.

(<-) forevery A, Y-f[A] c f[X-A]. let A=empty set.
then Y-f[A]=Y, and f[X-A]=f[X]. so we have Ycf[X]. this is the definition of surjective.

sorry i dont know latex yet, but ill learn it soon.


inequalities
induction
inequalities unsolved

Posted before :

[hide="hint"]$ \frac{1}{n^{2}}<\frac{1}{n(n\minus{}1)}\equal{}\frac{1}{n\minus{}1}\minus{}\frac{1}{n}$ then $ \sum_{i\equal{}1}^{n}\frac{1}{n^{2}}<2\minus{}\frac{1}{n}$[/hide]

[quote="quangpbc"]Posted before :

[hide="hint"]$ \frac{1}{n^{2}}<\frac{1}{n(n\minus{}1)}\equal{}\frac{1}{n\minus{}1}\minus{}\frac{1}{n}$ then $ \sum_{i \equal{} 1}^{n}\frac{1}{n^{2}}< 2\minus{}\frac{1}{n}$[/hide][/quote]

Yes, exactly,
The idea is to prove first:
$ \sum_{i \equal{} 1}^{n}\frac{1}{i^{2}}< 2\minus{}\frac{1}{n}$

Another way to do it:

$ \sum_{i \equal{} 1}^{n}\frac{1}{i^{2}}\equal{}1\plus{}\sum_{i \equal{} 2}^{n}\frac{1}{i^{2}}<1\plus{}\sum_{i \equal{} 2}^{n}\frac{1}{i(i\minus{}1)}\equal{}2\minus{}\frac{1}{n}$

[quote="quangpbc"]Posted before :

[hide="hint"]$ \frac{1}{n^{2}}<\frac{1}{n(n\minus{}1)}\equal{}\frac{1}{n\minus{}1}\minus{}\frac{1}{n}$ then $ \sum_{i \equal{} 1}^{n}\frac{1}{n^{2}}< 2\minus{}\frac{1}{n}$[/hide][/quote]

I mean that: If you try to use induction on $ \sum_{i \equal{} 1}^{n}\frac{1}{i^{2}}< 2$ it will not work

same if u use CS etc

But if u use the idea to prove first $ \sum_{i \equal{} 1}^{n}\frac{1}{i^{2}}< 2\minus{}\frac{1}{n}$ it becomes a peace of cake  :lol: .


Let $k=2^{2^{n}}+1$ for some $n\in\mathbb{N}$. Show that $k$ is prime iff $k|3^{\frac{k-1}{2}}+1$.

HERE MY SOLUTION(I don't read Rust's solution) :lol:
If  $2^{2^{n}}+1$ prime then $p=-1 mod 3$  therefor $(3|p)(p|3)=1$
  Hence $3^{\frac{p-1}{2}}+1=0 mod p$
 If $3^{\frac{k-1}{2}}+1 =0 mod k$ then k is prime .
Use lemma :
If $(x,y)=1$ then $x^{2^{n}}+y^{2^{n}}$ has only prime factor form
$2^{n+1}m+1 : lol:$HAHAHAHA

Cuopbientcd.

[hide=Solution]
If $3^{(k-1)/2}+1\equiv 0\pmod k$, then $\text{ord}_k(3)|k-1.$ But notice that $\text{ord}_k(3) \nmid (k-1)/2.$

Thus, $\text{ord}_k(3)=k-1.$ Therefore, $\phi(k)=k-1$ which implies $k$ is prime.

Assume that $k$ is prime. Now, our objective is to show that $3^{(k-1)/2)}\equiv -1\pmod k$ which by Euler's Criterion, simplifies to proving that $\left(\dfrac{3}{k}\right)=-1.$

By Quadratic Reciprocity, $\left(\dfrac{3}{k}\right)\left(\dfrac{k}{3}\right)=-1^{(k-1)/2}=-1^{\left(2^{2^{n-1}}\right)}=1.$

Note that $k\equiv 2\pmod 3.$ Thus, $k$ is not a quadratic residue mod $3.$ Therefore, $\left(\dfrac{3}{k}\right)=-1$ and we can finish up with Euler's Criterion. $\qquad$ $\qquad$ $\blacksquare$ [/hide]

Solution with Aahan Catterjee, Aditya Khurmi, Anshul, Anushka Aggarwal, Derek Liu, Grant Yu, Max Lu, William Wang, Eric Shen, Paul Hamrick, and others:

First, suppose $k$ is prime. We know $k \equiv 1 \pmod 4$, and $k \equiv 2 \pmod 3$. Then \begin{align*} 	3^{\frac{k-1}{2}} &= \left( \frac 3k \right) \pmod k \\ 	\left( \frac k3 \right) &= \left( \frac 3k \right) 		\qquad \text{by quadratic reciprocity} \\ 	\left( \frac{k}{3} \right) 	&= \left( \frac{2}{3} \right) = -1. \end{align*} So this completes one direction.

Conversely, assume \begin{align*} 	3^{\frac{k-1}{2}} &\equiv -1 \pmod k \\ 	3^{k-1} &\equiv 1 \pmod k \end{align*} The order of $3 \pmod k$ should divide $k-1 = 2^{2^n}$, but it shouldn't divide $\frac{1}{2}(k-1) = 2^{2^n-1}$. So the order is $k-1=2^{2^n}$ exactly.

In particular, by Euler's little theorem, we get $k-1 = 2^{2^n} \mid \varphi(k)$ and in particular $\varphi(k) \ge k-1$. Obviously this can only occur when $k$ is prime.

[quote=v_Enhance]
First, suppose $k$ is prime. We know $k \equiv 1 \pmod 4$, and $k \equiv 2 \pmod 3$. Then \begin{align*} 	3^{\frac{k-1}{2}} &= \left( \frac 3k \right) \pmod k \\ 	\left( \frac k3 \right) &= \left( \frac 3k \right) 		\qquad \text{by quadratic reciprocity} \\ 	\left( \frac{k}{3} \right) 	&= \left( \frac{2}{3} \right) = -1. \end{align*} So this completes one direction.
[/quote]
-----
Can you give me concepts or a explain for this?

Read the article https://en.wikipedia.org/wiki/Legendre_symbol

Everything you need is in the first and third section.

Solved with [b]fukano_2, E_6. [/b]

If $k\mid 3^{\frac{k-1}2}$, then \[-1\equiv 3^{\frac{k-1}2}\equiv 3^{\frac{2^{2^n}}2}\equiv 3^{2^{2^n-1}}\pmod k\implies 3^{2^{2^n}}\equiv 1\pmod k. \] From this we have that the order of $3$ mod $k$ is just $2^{2^n}=k-1$, so $k$ must be prime.

Now suppose that $k$ is prime. Note that $k\equiv 2^{2^n}+1\equiv 1\pmod 4$ and $k\equiv 2^{2^n}+1\equiv (-1)^{2^n}+1\equiv 2\pmod 3$, so \[3^{\frac{k-1}2}\equiv \left(\frac3k\right)=(-1)^{\frac{(3-1)(k-1)}4}\left(\frac k3\right)=\left(\frac k3\right)=\left(\frac 23\right)=-1\pmod k,\] as needed.

Start with if.We have that $k$ is congruent to $1$ mod $4$,but $2$ mod $3$.Thus,by quadratic reciprocity,$3$ is a non-quadratic residue mod $p$,which proves the if part by Euler's criterion.
Now,the only if part.We have that $3^{2^{2^n}}$ is $1$ mod $k$.This gives that order of $3$ mod $k$ is ${2^{2^n}}$.Thus,$k-1 | \varphi(k)$,which is sufficient to give $k$ prime.

For one direction, suppose that $k \mid 3^{(k-1)/2} + 1$. Then $3^{k-1} \equiv 1 \pmod k$ is the precise order by definition of $k$, so $k-1 \mid \phi(k) \leq k-1$, and $k$ must be prime.

On the other hand, if $k$ is prime, Euler's criterion suffices as $\left(\frac 3k\right) = \left(\frac k3\right) = \left(\frac 23\right) = -1$.


The median of a trapezoid cuts the trapezoid into two regions whose areas are in the ratio 1:2. Find the ratio of the smaller base of the trapezoid to its longer base.

[hide]Let the shorter base be $a$ and the longer base be $b$.

$2\cdot\frac{a+\frac{a+b}2}2=\frac{b+\frac{a+b}2}2$

$2a+a+b=b+\frac{a+b}2$

$6a+2b=2b+a+b$

$\frac{a}{b}=\boxed{\frac15}$[/hide]


Find the sum of the values of $x$ which satisfy the following equation:${\left({x}^{2}-5x+5\right)^{\left({x}^{2}-7x+12\right)}=1}$

[hide]if the exponent is 0, the value is 1. If the base is 1, the value is 1. If the base is -1 and the exponent is even, the value is 1. The values of x that satisfy the equation are $1, 2, 3$, and $4$. The sum is $10$.[/hide]

[quote="Ubemaya"][hide]if the exponent is 0, the value is 1. If the base is 1, the value is 1. If the base is -1 and the exponent is 0, the value is 1. The values of x that satisfy the equation are $1, 2, 3$, and $4$. The sum is $10$.[/hide][/quote]

[hide]I think you mean if the base is -1 and the exponent is even.[/hide]

If the base is -1 there are only 2 solutions...

[quote="alfalfa"]Note that the exponent is even for all integer values of x.  If the base is -1 there are infinite solutions.[/quote]
There are only two ways the base can be -1, and the exponent is dependant on the base.

The answer is actually 4, because x can also have a value of -6, because, if you remember your textbooks, a minus sign comes after you multiply exponentially with the exception of when a negative number is inside parentheses and the exponent is outside the parentheses.

[quote="myspacesucks"]NO, the ^2 is not outside parentheses, it is inside, so you do the number^2 and then you add the negative.[/quote]
ya but when you are substituting a number it is always going to be in parentheses.

when substituting its a cardinal rule that parenethesis are present.

This is getting absurd. Consider the most primitave definition of $x^{2}=x \cdot x$. So you are claiming that $-6 \cdot-6$

Also, try factoring ${x}^{2}-5x+6$. You get $(x-2)(x-3).$Clearly, $-6$ isn't a solution.

[quote="mysmartmouth"]$-6^{2}= 36$
$(-6)^{2}= 36$

Order of operations.

If you don't believe me, type in

-6^2

in Google.  Then type in (-6)^2[/quote]
I suggest that you read the argument before you make claims. We're saying that
$x^{2}$ isn't -36 if x=-6.

[quote="myspacesucks"]NO!!!! It isn't!!!  Go ask your math teacher what -6^2 is!!!!![/quote]
I did. My math teacher agrees with me.

there was a long and funny discussion (ok, not very long :P ) in the classroom forum. someone thoroughly explained it. it was a guy who was discussing with his wife which it was. (yay, the wife won :D ). look it up.

jorian


[quote="ashrafmod"]Find the sum of the series
$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+..$

where the denominators consist of all natural numbers that have no prime factors other than 2[/quote]

Presumably, "other than 2 and 3."  Also, 1 is such a number (it has no prime factors, and so certainly none other than 2 or 3).

Going off of that idea, what is the sum of the reciprocals of the natural numbers with no prime factors other than $2, 3, 5, 7, ...$?

[quote="mathgeniuse^ln(x)"][quote="cincodemayo5590"][hide="Solution"]
The sum is equal to:
\[\left( 1+\frac{1}{2}+\frac{1}{4}+\ldots \right) \left( 1+\frac{1}{3}+\frac{1}{9}+\ldots \right) = (2)\left( \frac{3}{2}\right) = \boxed{3}\]
[/hide][/quote]

That would be correct if 1 was not in it.[/quote]
I believe you meant to write that that would be correct if 1 was in it.

[quote="mathgeniuse^ln(x)"][quote="cincodemayo5590"][hide="Solution"]
The sum is equal to:
\[\left( 1+\frac{1}{2}+\frac{1}{4}+\ldots \right) \left( 1+\frac{1}{3}+\frac{1}{9}+\ldots \right) = (2)\left( \frac{3}{2}\right) = \boxed{3}\]
[/hide][/quote]

That would be correct if 1 was not in it.[/quote]

1 is a natural number with no prime factors other than 2 or 3. So $\frac{1}{1}$ should be in the sum. If for some odd reason its not, then just subtract it. I like the form of the sum in my solution, though.

[quote="mathgeniuse^ln(x)"]Yeah, I meant 1 in it.

And yeah, cincodemayo55 good solution[/quote]

 :)  Thanks. But I think I've seen a problem like this before, somewhere...

woa this problem is old...WOOO...i have seen it on an amc once...

This is from James Stewart's calculus book. chapter 11

[quote="aahmadi"]If t0rajir0u is asking "what is the sum of the reciprocals of the natural numbers with no prime factors other than 2, 3, 5, 7?" then I figure we would have [hide]$\left( 1+\frac{1}{2}+\frac{1}{4}+\ldots \right) \left( 1+\frac{1}{3}+\frac{1}{9}+\ldots \right) \left( 1+\frac{1}{5}+\frac{1}{25}+\ldots \right) \left( 1+\frac{1}{7}+\frac{1}{49}+\ldots \right) = \left(\frac{2}{1}\right) \left(\frac{3}{2}\right) \left(\frac{5}{4}\right) \left(\frac{7}{6}\right) =\frac{35}{8}$[/hide][/quote]

Good.  Now extend that pattern to all of the primes.

In other words, what is the sum of the reciprocals of the natural numbers?

[hide="And..."] How can you use this result to show that the sum of the reciprocals of the [b]primes[/b] diverges? [/hide]

[quote="aahmadi"]If you pay attention to the written series, then you wouldn't include 1, I don't know why you would pay attention to what he wrote, since it doesn't really make any sense. That's why I presumed that the 1 was not included. Either way, both our solutions get the point across, but cincodemayo5590's solution is prettier.

And I also don't know what boxedexe meant by "I believe you meant to write that that would be correct if 1 was in it." Since he did include 1 in his solution... :huh:

If t0rajir0u is asking "what is the sum of the reciprocals of the natural numbers with no prime factors other than 2, 3, 5, 7?" then I figure we would have [hide]$\left( 1+\frac{1}{2}+\frac{1}{4}+\ldots \right) \left( 1+\frac{1}{3}+\frac{1}{9}+\ldots \right) \left( 1+\frac{1}{5}+\frac{1}{25}+\ldots \right) \left( 1+\frac{1}{7}+\frac{1}{49}+\ldots \right) = \left(\frac{2}{1}\right) \left(\frac{3}{2}\right) \left(\frac{5}{4}\right) \left(\frac{7}{6}\right) =\frac{35}{8}$[/hide][/quote]
"it" referring to the original problem.  :wink:

Oh, my bad boxedexe........and yeah t0rajir0u, thanks for the enlightening advice on the harmonic series, it was just a couple of hours ago when I was thinking, "Why the hell does the harmonic series diverge?", but yeah, obviously you are always multiplying by a number a little bit bigger than 1 therefore it diverges.

[quote="aahmadi"]but yeah, obviously you are always multiplying by a number a little bit bigger than 1 therefore it diverges.[/quote]

This is not, in fact, a valid reason to conclude that the series diverges.  For example, if $a_{n}= \frac{n^{2}+2n+1}{n^{2}+2n}> 1$ for $n \geq 1$, the product $a_{1}\cdot a_{2}\cdot a_{3}\cdots\cdots$ converges to 2.

You can build as many such counter-examples as you like: take any positive increasing bounded sequence $0 < b_{1}< b_{2}< \ldots < B$.  Then the sequence $a_{n}= \frac{b_{n+1}}{b_{n}}$ has the property that $a_{n}> 1$ and the infinite product $a_{1}\cdot a_{2}\cdot \cdots$ converges.

its more easy like that.....
 :D

[hide]${\sum_{\alpha =0}^{\infty }\sum_{\beta =0}^{\infty }\frac{1}{2^{\alpha }3^{\beta }}=\sum_{\alpha =0}^{\infty }\frac{\sum_{\beta =0}^{\infty }\frac{1}{3^{\beta }}}{2^{\alpha }}}$

${\sum_{\beta =0}^{\infty }\frac{1}{3^{\beta }}=\frac{3}{2}}$

${\sum_{\alpha =0}^{\infty }\sum_{\beta =0}^{\infty }\frac{1}{2^{\alpha }3^{\beta }}=\frac{3}{2}\sum_{\alpha =0}^{\infty }\frac{1}{2^{\alpha }}}$

${\sum_{\alpha =0}^{\infty }\frac{1}{2^{\alpha }}=2}$
${\sum_{\alpha =0}^{\infty }\sum_{\beta =0}^{\infty }\frac{1}{2^{\alpha }3^{\beta }}=\frac{3 .2}{2}=3}$[/hide]

since 1 is not in the first equation

just subtract this

[quote="t0rajir0u"][hide="The result to which I hint goes something along these lines..."] Let $H_{n}= 1+\frac{1}{2}+...+\frac{1}{n}$.  As $n$ becomes large we know that $H_{n}\approx \ln n$.

As $n$ becomes large we can write

$H_{n}\approx \left( 1+\frac{1}{2}+\frac{1}{4}+... \right) \left( 1+\frac{1}{3}+\frac{1}{9}+... \right) \left( 1+\frac{1}{5}+\frac{1}{25}+... \right) ... = \prod_{p \in \mathbb{P}}\left( \frac{p}{p-1}\right)$

We can therefore write

$\ln H_{n}\approx \sum_{p \in \mathbb{P}}-\ln (1-\frac{1}{p}) \approx \sum_{p \in \mathbb{P}}\frac{1}{p}$

Which is a [b]very unrigorous[/b] demonstration that the sum of the reciprocals of the primes up to $n$ is proportional to $\ln \ln n$.

We conclude that the sum of the reciprocals of the [b]primes[/b] diverges. [/hide]
aahmadi, [hide="the actual lemma concerning your conjecture..."] $\prod_{k = 1}^{n}(1+a_{n})$ converges [b]iff[/b] $\sum_{k=1}^{n}a_{n}$ converges. [/hide][/quote]

Not a hope that I (or any of you) will give a proof of this, but it'll be good for you to see.

If $(a,b)=1$

Then

$\sum_{p\equiv a (mod{b})}\frac{1}{p}$

Diverges. Personally i think this is amazing, it basically says that the sum of the recipicols of the primes in nearly every arithmetric progression diverges.


probability
expected value
pifinity is ze best

Two standard dice are rolled. What is the expected number of 1's obtained? Express your answer as a common fraction.

The probability of getting a one, is $ \frac {1}{6}$. Multiplying this by $ 2$ gives $ \boxed{\frac {1}{3}}$.
I think.........

The solution here is by far the most straightforward solution, and the "right" way for this problem, particularly on a timed test.
But there are other nuggets that you can get from it that may be helpful in other problems.

For example, you can calculate:
$P($neither die is a 1$) = (\frac{5}{6})^2 = \frac{25}{36}$

$P($exactly one die is a 1$) = (\frac{5}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{5}{6}) = \frac{10}{36}$

$P($both dice are 1s$) = (\frac{1}{6})^2 = \frac{1}{36}$

First, note that the sum of all those probabilities is $1$.  That's very reassuring.
Second, you can compute the EV with this info.
In the first case, you get no rolls of $1$, so your expected number of 1s for that part is $0 \cdot \frac{25}{36} = 0$
In the second case, you get one $1$ to add to your total, so it contributes $1 \cdot \frac{10}{36} = \frac{10}{36}$
In the third case, you get $2$ ones, so it contributes $2 \cdot \frac{1}{36} = \frac{2}{36}$
The total EV is thus $0 + \frac{10}{36} +  \frac{2}{36} = \frac{12}{36} = \frac{1}{3}$

Again, this calculation is probably unnecessarily complicated for this problem.   It's easier to look at each die individually, and realise that each die will give a one $\frac{1}{6}$ of the time, so when you roll two of them you'll get a one $\frac{2}{6}$ of the time.  But I'm sure there are other situations where you have to use the formula:

$EV = \sum P($event$) \cdot V($event$)$

Let $X_i$ be the number of 1's rolled on die $i$.  $E(X_1+X_2) = E(X_1) + E(X_2) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$.  It doesn't even matter if the dice rolls are i.i.d.  The solution is much more involved than it needs to be.

The long explanation... or $2 \cdot 1/6 = \boxed{1/3}$

or its like you expect to get one 1 In 6 rolls. do proportions and get 1/3

[quote=ZanTwilight]Yeah, the solution to this problem is way more confusing than it needs to be.  There is a 1/6 chance of rolling a one on each dice, so add the probabilities and you get 1/3.  I don't even know what the solution was talking about half the time, but it seems like I didn't need to.[/quote]

The question asks for the expected number of 1's coming up. The solution is doing it in the proper way expected values are calculated.
The fun thing is, the expected number of any number showing up after two rolls is also 1/3. 6*1/3=2. While you roll two dice,  you actually get two numbers. I wonder if you can even calculate this way: 2/6=1/3 (since 6 is the sum of the expected number for all 6 numbers).


The operation $ \oplus$ is defined as $ m \oplus n \equal{} m^2 \minus{} mn \minus{} n^2$ and the operation $ \bowtie$ is defined as $ m \bowtie n \equal{} 2(m\minus{}n)$. Compute $ (3 \oplus 4) \bowtie (4 \oplus 3)$.


Square $ ABCD$ has midpoints $ E$, $ F$, $ G$, and $ H$. $ AB \equal{} 15$ centimeters. Find the area of the shaded interior square in square centimeters.

[asy]pair A = (0,50);
pair B = (50,50);
pair C = (50,0);
pair D = (0,0);
pair E = (50,25);
pair F = (25,0);
pair G = (0,25);
pair H = (25,50);

draw(A--B--C--D--cycle);
draw(A--E);
draw(B--F);
draw(C--G);
draw(D--H);
fill((10,20)--(20,40)--(40,30)--(30,10)--cycle);

label("$A$", A, NW);
label("$B$", B, NE);
label("$C$", C, SE);
label("$D$", D, SW);
label("$E$", E, ENE);
label("$F$", F, S);
label("$G$", G, W);
label("$H$", H, N);[/asy]

Of course, you must prove that your new triangles are congruent to the old small ones, but that's brilliant. The way I solved this was by similar triangle bashing...

It is a very elegant solution indeed. :)  To similar triangle bash, we let the congruent smaller triangles be of area $ x$ so that the triangle of the next size is $ 4x$ and of the next size, $ 5x$ such that $ 5x \equal{} \frac12\cdot\frac {15}2\cdot15 \equal{} \frac {15^2}4$.

Then the desired area is $ 15^2 \minus{} 4(4x) \equal{} 225 \minus{} 4\left(\frac45\cdot\frac {15^2}4\right) \equal{} 15^2 \minus{} \frac45\cdot15^2 \equal{} \frac15\cdot15\cdot15 \equal{} \boxed{45}$.


The curves $ y\equal{}6 \times 5^{x}$ and $ y\equal{}9^{x}$ intersect at the point $ Q$. Find the coordinates of $ Q$.

Dividing by 5^x yields
(9/5)^x=6, from which x=A=log6/log(9/5).  Therefore, the point is (log6/log(9/5),9^A=5*6^A).

We can make the $ x$-value a bit cleaner if you write as $ x\equal{}\frac{\log6}{\log9\minus{}\log5}$. I'm not sure how to make $ y$ look cleaner than $ y\equal{}9^{\frac{\log6}{\log9\minus{}\log5}}$. :(

Hi.

For the x-coordinate, I took logs of both sides to get

$ \log_{3} 6 \plus{}x \log_{3} 5 \equal{} x \log_{3} 9$

$ \log_{3} 2 \plus{} 1 \plus{} x \log_{3} 5 \equal{} 2x$

$ x(2\minus{} \log_{3} 5) \equal{} 1 \plus{} \log_{3} 2$

$ x \equal{} \frac{1\plus{} \log_{3} 2}{2 \minus{} \log_{3} 5}$


calculus
integration
calculus computations

hello, do you mean with $ \cot^{\minus{}1}(x)\equal{}arccot(x)?$
Sonnhard.


Define the sequence $\left\{u_n\right\}_{n \geq 1}$ by \[ u_1 = 1,\ \ u_n = (n - 1)u_{n - 1} + 1.  \] Find all positive integers $n$ such that $n$ divides $u_n$.


function
topology
advanced fields
advanced fields unsolved

Let M be a metric space. If f: M $ \rightarrow$ M is a continuous function in which d(f(x),f(y)) $ \ge$ d(x,y) for every x,y $ \in$ M, then f is a homeomorphism of M on itself.

That's just wrong of $ f$ is not surjective.


On the Fourth of July, Billy Bob goes to see the Space Shuttle Discovery take off. He is 10 miles from the launch pad of the Space Shuttle. If sound travels at 761 miles per hour, how long is the interval between the rocket's launch and when Billy Bob hears the rocket? Round to the nearest tenth of a second.  :D

[hide="If he is that far away, how can he see it?"]
$(\frac{10 miles}{761 miles})(3600 seconds)= 47.3 seconds$[/hide]

:fool: I don't know, I just picked a distance. And Billy Bob is afraid of loud noises.  :(

Oh yes, the distance is very unreasonable, 47 seconds!!!!!


Let $x,y,z$ be real numbers such that $x+y+z=1, \; x, y, z>0$. Prove that \[(1+x)(1+y)(1+z) \geq (1-x^{2})^{2}+(1-y^{2})^{2}+(1-z^{2})^{2}\] [hide]The given inequality is equivalent to \[xy+yz+zx \geq 2(xy+yz+zx)^{2}+3xyz \quad \Longleftrightarrow\]  \[(x+y+z)^{2}(xy+yz+zx) \geq 2(xy+yz+zx)^{2}+3xyz(x+y+z) \quad \Longleftrightarrow\]  \[\left(x^{2}+y^{2}+z^{2}\right)(xy+yz+zx) \geq 3xyz(x+y+z)\] which is true because \[\left(x^{2}+y^{2}+z^{2}\right)(xy+yz+zx)-3xyz(x+y+z)\geq\]  \[\geq (xy+yz+zx)^{2}-3xyz(x+y+z)=\]  \[= \frac{1}{2}\left[x^{2}(y-z)^{2}+y^{2}(z-x)^{2}+z^{2}(x-y)^{2}\right] \geq 0\] and the proof is complete. $\blacksquare$ [/hide]

Homogenizing gives \[(x+y+z)(2x+y+z)(x+2y+z)(x+y+2z)\ge \sum_{cyc}((y+z)(2x+y+z))^{2}\] By using the ravi-substitution, we obtain \[(a+b+c)(a+b)(b+c)(c+a)\ge 2\sum_{cyc}(a(b+c))^{2}\] which is equivalent to \[\sum_{sym}a^{3}b\ge\sum_{sym}a^{2}b^{2}\] which is true. $\Box$
(not the most beautiful way....but straightforward)


I have a question
\[ \text{Let} \; x,y,z \; \text{be nonegative real numbers such that} \; xy+yz+zx=1,\;\text{prove that} \; x+y+z\geq\sqrt3,\; \text{and}\; (x+yz)(y+zx)(z+xy)\leq1.  \]
The first one is easy of course. I have not got a solution for the second one.

i think the second must be wrong. if any of $x,y,z=0$, we have the problem is obvious, so we might suppose $x,y,z>0$

then, after some algebraic manipulations and a division by $xyz$ our original ineq turns into

$(x+y+z-1)^2+xyz\leq 2$,

but this is false since we may take $y=z, x>\sqrt 2+1$ and prove that this is false!


limit
trigonometry
calculus
calculus computations


In general, how can I find the inverse (assuming it exists) of a m by n matrix, nonsquare? For example, I want to find the right inverse using a method rather than by guess and check. How can I find the right inverse of a 2 by 3 matrix where the first row is $ 1   1    0$ and the second row is $ 0   1   1$?

You're not looking for [i]the[/i] right/left inverse; they're not unique.

You could adapt the standard square algorithm: augment $ \begin{bmatrix}A&I_m\end{bmatrix}$ and perform Gaussian elimination, with back-substitution. If $ A$ is $ m\times n$ with full rank and $ m\ge n$, this leads to $ \begin{bmatrix}I_n&B\\ 0_{m\minus{}n,n}&C\end{bmatrix}$. Now, $ BA\equal{}I_n$.

That gives left inverses. For right inverses, use columns instead of rows.


Let $a_1 , a_2, ... a_{2000}$ be  integers in interval $[-1000, 1000]$ such that $\sum_{i=1} ^{2000}a_i=1$ .show that
sum of some of the $a_i$s is equal to zero.

First we show we can relabel them in such a way that partial sums always lie in the interval $[-999,1000]$ Next there are $2000$ partial sums, none of which is $0$, so there are 2 equal ones, their difference gives the desired set.


quadratics
algebra
binomial theorem
number theory unsolved
number theory

if $ p$ divides $ 2^{2^n} \plus{} 1$ then prove that $ 2^{n\plus{}2}$ divides $ p\minus{}1$

[hide]What i have so far

Since $ 2^{2^n}$ is congruent to $ \minus{}1$ mod $ p$

We have obviously $ 2^{n\plus{}1}$ is the order.

Coz if d is the order then d divides  $ 2^{n\plus{}1}$

So d is a power of 2.

But then if d is less than $ 2^{n\plus{}1}$  then obviously $ 2^{2^n}$ is congruent to $ \minus{}1$ mod $ p$ becomes false.

So d is $ 2^{n\plus{}1}$.

We also have $ 2^{p\minus{}1}$ is congruent to $ 1$ mod $ p$

So $ 2^{n\plus{}1}$. divides $ p\minus{}1$
But sadly the question asked us to prove that $ 2^{n\plus{}2}$ divides $ p\minus{}1$

I am just a 2 away :rotfl:

Any help would be appreciated.

[/hide]

This doesn't seem to be true for $ n \equal{} 0, 1$.

It holds true for $ n\ge 2$, I believe.

[hide="Solution"]
Let the order of $ 2\bmod p$ be $ k$.  Then, we have that $ 2^{2^n} \plus{} 1\equiv 0\bmod p$, so $ 2^{2^n}\equiv \minus{} 1\bmod p$, so $ 2^{2^{n \plus{} 1}} \equiv 1\bmod p$.  Thus, $ k|{2^{n \plus{} 1}}$, but $ k$ is not a factor of $ 2^{n}$, so $ k \equal{} 2^{n \plus{} 1}$.  Thus, $ k|(p \minus{} 1)\implies 2^{n \plus{} 1}|(p \minus{} 1)$, meaning that $ 8|(p \minus{} 1)$.  Now, we use quadratic reciprocity to show that $ 2$ is a quadratic residue $ \bmod p$.  Notice that $ (\frac {2}{p}) \equal{} ( \minus{} 1)^{\frac {p^2 \minus{} 1}{8}} \equal{} 1$, so $ 2$ is a quadratic residue $ \bmod p$.  Hence, there is an $ x$ so that $ x^2 \equiv 2\bmod p$, meaning that $ x^{2^{n \plus{} 1}}\equiv \minus{} 1\bmod p$, but $ 2^{2^{n \plus{} 2}}\equiv 1\bmod p$.  Letting the order of $ x\bmod p$ be $ l$, we have that $ l|2^{n \plus{} 2}$, but not a factor of $ 2^{n \plus{} 1}$.  Hence, $ l \equal{} 2^{n \plus{} 2}$, so $ l|(p\minus{}1)\implies 2^{n \plus{} 2}|(p \minus{} 1)$. [/hide]

Another solution by assuming $ n\ge 2$.
Since $ n > 1$, $ f_{n \minus{} 1} \equal{} 2^{2^{n \minus{} 1}} \plus{} 1$ is defined.
Also we observe that
$ (f_{n \minus{} 1})^{2^{n \plus{} 1}} \equal{} (f_n \plus{} 2^{2^{n \minus{} 1} \plus{} 1} )^{2^n}$.
By Binomial Theorem, we can easily deduce that $ (f_{n \minus{} 1})^{2^{n \plus{} 1}} \equiv\ \minus{} 1 \bmod{f_n}$.
This implies that $ f_n$ divides $ (f_{n \minus{} 1})^{2^{n \plus{} 1}} \plus{} 1$.
Also , since $ p$ divides $ f_n$, $ p$ divides $ (f_{n \minus{} 1})^{2^{n \plus{} 1}} \plus{} 1$.
Now I use a very well known result: If $ t$ is any postive integer greater than $ 1$,  and let $ n$ be a postive integer. If $ p$ is a prime divisor of $ t^{2^n} \plus{} 1$, then $ p \minus{} 1$  is divible by $ 2^{n \plus{} 1}$. The lemma is very simple to prove using order of modulos.
So setting $ t \equal{} f_{n \minus{} 1}$ in the lemma,we are done.


is chromate really $Cr(O_{2})_{2}^{-}$ aka diperoxochromate?
what is dichromate(by that I mean what's the coordinate structure)? i really don't know for this one

(Edited: my apologies  :oops: )  Dichromate is $Cr_{2}O_{7}^{2-}$.

For your first question--no!  Peroxide bonds are very weak, and you will very rarely find them.  Chromate, $CrO_{4}^{2-}$ has a Lewis structure of the following, where the periods are just placeholders:
. . . O
. . . |
O = Cr = O
. . . |
. . . O

Note that this is stabilized by lots of resonance, which is good.  Make sure you see where the various lone pairs reside, and why $Cr$ has a formal charge of 0.

Dichromate, as t0rajir0u said, is $Cr_{2}O_{7}^{4-}$.  See the following for a hint as to its structure: [hide]Its structure is similar to that of chromate; $Cr$ is single bonded to $Cr$.  Click the following for the structure:
[hide]
. . . O . . O
. . . || . .||
O - Cr - Cr - O
. . . || . ||
. . . O . .O[/hide][/hide]

I stand corrected.  Thank you for pointing that out.

There's no Cr-Cr bond. There's a "bridge" oxygen: Cr-O-Cr. Dichromate, $(\text{Cr}_{2}\text{O}_{7})^{2-}$ is really just a partial anhydride of $(\text{CrO}_{4})^{2-}.$ In either case, this is "hexavalent" chromium in which the Cr atoms have a +6 oxidation state. It's an oxidizing agent, with the chromium likely to return to a +3 oxidation state by pulling electrons off from some other molecule.

Physical chemists used to love "chromic acid", which was potassium dichromate dissolved in concentrated sulfuric acid, as a cleaning fluid. Nothing else was quite as effective at removing all organic residues from glassware. These days, their safety/environmental officers mostly won't let them use the stuff.

Does anyone remember the movie [i]Erin Brockovitch[/i]? The toxic substance at the core of the legal case was hexavalent chromium.

i knew the formula, i was just asking for the structure, thanks Kent; so there's no coordination in this polyatomic ion?

Another error on my part.  I must have written this post rather hastily.

In Dvagner's first post, he asks if dichromate involves a peroxide. Emphatically not! The use of "chromic acid" to clean glassware may have fallen out of favor because of environmental toxicity and difficulty of disposal, but it would never have been in favor in the first place had it involved a peroxide. We want to chew up and dissolve organic residue on glassware, not cause it to explode!


Can someone please compile a list of the AIME II problems?
just the problems?

Thanks.


[quote="boukharfane"]solve for x,y and z reels the system:
x^3+y^3+z^3=x+y+z
x^2+y^2+z^2=xyz[/quote]

[hide]use identity (x^3+y^3+z^3 -3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)[/hide]


Agar f(x)=x+(x betavane 1/3)
Angah makuse f(x) ra biabid?

salam
tabe makose in tabe ra mitavan ba estefade az formul kardano bedast avard.
formul kardano:
yeki az rishehaye moadele daraje sevom $t^{3}+pt+q=0$ ebarat ast az:
$t=\sqrt[3]{\frac{-q+\sqrt{q^{2}+\frac{4}{27}p^{3}}}{2}}+\sqrt[3]{\frac{-q-\sqrt{q^{2}+\frac{4}{27}p^{3}}}{2}}$
hal agar gharar dahim $t=x^{\frac{1}{3}}$ va $p=1$ va $q=-y$
mitavanim rishie yektaye moadeleh $x+x^{\frac{1}{3}}-y=0$ ra bedast biavarim va az anja makos tabe bedast miayad:
$x^{\frac{1}{3}}=\sqrt[3]{\frac{y+\sqrt{y^{2}+\frac{4}{27}}}{2}}+\sqrt[3]{\frac{y-\sqrt{y^{2}+\frac{4}{27}}}{2}}\\ \Rightarrow{x=(\sqrt[3]{\frac{y+\sqrt{y^{2}+\frac{4}{27}}}{2}}+\sqrt[3]{\frac{y-\sqrt{y^{2}+\frac{4}{27}}}{2}})^{3}}\\ \Rightarrow{f^{-1}(x)=(\sqrt[3]{\frac{x+\sqrt{x^{2}+\frac{4}{27}}}{2}}+\sqrt[3]{\frac{x-\sqrt{x^{2}+\frac{4}{27}}}{2}})^{3}}$


[hide]5,because 21+4 = 25 and 5*5=25.You could also use guess and check. I got it on my first try.[/hide]

be careful when you see quadratics, and also, math92, can you tell us if there are any restrictions as to what type of number is $x$ allowed to be?

[quote="math92"]Solve this by using any method:

$x^2-4=21$[/quote]

[hide]x^2-25=0

x={5,-5}[/hide]

it's a second degree equation so there's two answers. which are $\pm5$

[hide]
Remember that $5^2=(-5)^2=25$. Therefore, $x=\pm 5$[/hide]

yup yup yup =] thats exactly what I did Drunken_Math. It's almost always easier.
a^2 - b^2 = (a - b)(a + b)


Is your hand writing beautiful? My question is: how to have patient to write neatly if it takes so much time?
If you are thinking this guy... Yes, that's one more stupid question.

Example:
[img]http://spiff.rit.edu/classes/phys311.old/lectures/proj/proj_14.jpeg[/img]

Yeah, it has to be a girl. My handwriting is horrible...

Does this really belong in Round Table?

I have awful handwriting unless I write really slowly, which takes too long. It's really annoying particularly for things like maths and physics, when there are loads of symbols all over the place- the solution is to use a computer.  :)

And I think that it must be a girl too.

it doesn't have to be girl. i'm male and /in elementary school/ my chemistry teacher (female) said that i have a nicer handwritting than most of the girls in the class. and then she started to talk than that's the female handwritting so she forgot why she took my notebook-i was to be questioned.  :P  :lol:

I have terrible handwriting. I actually got 10% off on my ap english timed writing for handwriting.

I have nice handwriting but only utilize it when I have little time constraints, like homework due in a week. But if I am in math or physics class or doing a timed writing, I don't care how I write, just as long as it is readable.

I usually type all my English assignments, unless it's writing two sentences or something.

For math, on the other hand, I usually write quickly and sometimes am unable to read my own handwriting. Like when you have a negative sign right in front of a fraction, and you miss seeing it because it looks just like a continuation of the vinculum, and you only realize it after you've filled up a page with calculations... Yeah.

I only write pretty when I have homework due in a week or something.

that handwriting honestly looks like a girls. velocity equals orginal velocity plus acceleration times time. I am in 8th grade, but I did some physics over summer.

My HANDWRITING is okay. I write a plain alphabet manuscript that is straight and just looks like times new roman. Basic, and not pretty. I think it is just neat and normal.

If i am taking notes in class, I just scribble cursive down

[quote="now a ranger"]cursive just never worked for me. I tried, but it just wouldn't work because my hands didn't cooperate. IT works well in manuscript fast.  :oops:[/quote]

I have the same problem with printing. It is way too slow.

Usually when I'm frustrated or really want to get something done I just scribble, but otherwise, if I really want something to look good, I can write pretty well.

My USAMTS papers...ugh those were like completely illegible.

Why were our USAMTS papers messy? Didn't you have over a month to turn them in?

My teachers tell me that i am going to fail exams because of my handwriting.....enough said.  :wink:

My handwriting is bad, but legible. However, the person who sat next to me in bio last year...

Teacher: "You can handwrite or type your lab reports, except for YOU. YOU have to type it."

[quote="mathnerd314"]For math, on the other hand, I usually write quickly and sometimes am unable to read my own handwriting. Like when you have a negative sign right in front of a fraction, and you miss seeing it because it looks just like a continuation of the vinculum, and you only realize it after you've filled up a page with calculations... Yeah.[/quote]

Good to hear that other people have that problem too...  :blush:

I'm a girl, and my handwriting starts out all nice at the beginning of an assignment, but it gets really sloppy throughout it.  I can always read it, and when I take notes I just write as fast as I can, then clean it up and rewrite them that night.

My math calculations are awful though - nothing is in the same direction, or same size, and equations go all over the paper.  Assignments are nice and clean, but on the standardized tests where they give you a piece of scrap paper and then ask you to turn it in so they can see your calculations - I don't know how anyone will be able to get anything other than a whole bunch of numbers and arrows and various circled and boxed numbers out of mine.  I pity those people - hope that they get payed a lot.

[quote="mustafa"]The trick to having good handwriting, at least for me, is to do one of two things.

1. TRY! I never do :rotfl: . Because of this, my handwriting is HORRIBLE!

2. Write big and loopy. My biggest problem is that when I draw loops, like in the letter "e", it looks more like a straight line. If I write big and loopy, my handwriting is pretty legible. If you look at the best handwriting that doesn't take concentration, it is probably loopier than yours.[/quote]
Hate to break it to you but your handwriting is really not all that nice :D

[quote="mustafa"]my handwriting is HORRIBLE!
[/quote]

I know :wink:

Why do girls spend so much time to make their handwriting pretty?

when i read what i have writen, i can't tell the difference between "n" and "u", "m" and "nu" / or "un", "2" and small "z" (so i cross my little and big "z")

a few days ago, my german teeacher asked me to read my homework, and i read "bauk" insted "bank" (in croatian, word bauk has a completely different meaning :blush: )

sometimes I'll start an assignement or smething in print, and it'll gradually evolve into cursive.
i also write two different ways for math: one when I'm doing homework/tests, another when I know I'll never need to see it again, I just have to write it to get to the answer.

I am probably one of those girls who doesn't care about her handwriting. When people were signing yearbooks last year, everyone took care to write in pretty letters with hearts over the 'i's, but mine was just scribbles. No teacher has ever had trouble reading it though.

my handwriting is not really neat but it's legible

[quote="i_like_pie"]I'd say my handwriting is pretty neat. Some of the girls in my classes think its beautiful.  :D :love:[/quote]
dude, that's not normal. :o

I'd say my handwriting is pretty bad, but my teachers seem to have no problems with it. The funny thing is that the teachers I had at Korea were like "OMFG WHAT IS THIS" when reading my handwriting. :huh:
When I'm writing fast, it gets pretty illegible, which is why I have trouble reading my bio notes...r=v,P=D,a=e=o,g=y,l=i=j,...bleh

My handwriting has been described as "messy beautiful"

well, my handwriting has been described as indecipherable  :D

I'm too lazy to try to write nicely, I always end up connecting my letters so it's a cursive-print mix.

I just don't see that point of writing nicely. If I can read it, that's all I care about :wink:

[quote="junggi"][quote="i_like_pie"]I'd say my handwriting is pretty neat. Some of the girls in my classes think its beautiful.  :D :love:[/quote]
dude, that's not normal. :o [/quote]
Even if I write really fast, it's still pretty legible. :P

My handwriting has gradually gotten better over the years somehow.

I ahve always had the worst handwriting in my clasas. Every year, my teachers like yell at me for it, but I've never done a thing.

[quote="Max300"][quote="Ignite168"][quote="lyra"]Although I know that mathematics is a field dominated by boys (look around the math team room, 1, 2,3 a record number of girls), as a girl my self I find it rather insulting that it should be assumed that females have better hand writting than males.  While this may not be the intent, it implies that the person writting such a comment feels that girls have nothing more important to do than perfect their hand writting.

~lyra[/quote]

They don't.[/quote]

dude stfu we're all sick and tired of you turning every forum into a females are inferior thread[/quote]

Please point out one place I've done that and one place where I've said females are inferior. They just have a different place in society.

[quote="Ignite168"][/quote][quote="Max300"][quote="Ignite168"][quote="lyra"]Although I know that mathematics is a field dominated by boys (look around the math team room, 1, 2,3 a record number of girls), as a girl my self I find it rather insulting that it should be assumed that females have better hand writting than males.  While this may not be the intent, it implies that the person writting such a comment feels that girls have nothing more important to do than perfect their hand writting.

~lyra[/quote]

They don't.[/quote]

dude stfu we're all sick and tired of you turning every forum into a females are inferior thread[/quote][quote="Ignite168"]

Please point out one place I've done that and one place where I've said females are inferior. They just have a different place in society.[/quote]
Those posts were deleted, so I can't.


search
number theory
least common multiple
factorial
puzzles

[hide="progress?"]Well, all of the primes greater than 5 are on there, and the rest of the numbers are either of the form $ x^y$ or else $ 6x$[/hide]

i dont see a patern unless mufasta is right and that seems awfully complicated!

It looks like either differences or all the numbers that satisfy some rule.
differences:1,1,1,1,2,2,3,1,2,4,1,3,2,2,1,5,4,2,4
That looks pretty random so I think it's the second type.
How about the numbers that aren't there?
1,2,3,4,10,12,14,15,18,20,21,22,25,26,28,30,33,34,35,36,38,39,40,42,44,45,46
Hmm. I'm not sure. It could be something to do with the words for the numbers:
five, six, seven, eight, nine, eleven, thirteen, sixteen, seventeen, nineteen, twenty-three, twenty-four, twenty-seven, twenty-nine, thirty-one, thirty-two, thirty-seven, forty-one, forty-three, forty-seven
Or it could be how you write the digits for the numbers, maybe in some other base or in roman numerals? (I'm so not making seventeen more lists. Three is enough.)

if you assume that the 24 is supposed to be a 25, then its pretty easy
the first term is 5 and each other number is only in the sequence if it isn't a factor of the LCM of the previous terms

[quote="i_like_pie"][quote="Jean-Louis"]could u determine the next few terms of the sequence

5, 6, 7, 8, 9, 11, 13, 16, 17, 19, 23, 24, 27, 29, 31, 32, 37, 41, 43, 47,......[/quote]
Except for the 6 and the 24, those are all powers of primes.

It's not an easily-recognizable sequence, so it could be anything, or it could even be two sequences put together, which makes it much too hard.

One possibility: Powers of primes > 4 combined with factorials that have n > 2. This would make the next term 49, and then 53.[/quote]

Yeah, but 25 isn't on there...


If $(a^{2}+b^{2})(c^{2}+d^{2})-(ac+bd)^{2}=4$, determine all possible values of $ad-bc$.

[quote="babbomammo"](a² + b²)(c² + d²) = (ad - bc)² +(ac + bd)²

  so (ad - bc)² = 4
  so  ad - bc = + 2 or -2[/quote]
I don't understand how the equation $ (a^{2}+b^{2})(c^{2}+d^{2})-(ac+bd)^{2}=4 $ can be rearranged to $(a^2 + b^2)(c^2 + d^2) = (ad - bc)^2 +(ac + bd)^2$.  Can someone please explain?  Thanks.

We have

$(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + 4$

From babbomammo's equation $(a^2 + b^2)(c^2 + d^2) = (ad - bc)^2 + (ac + bd)^2$ which is basically a given if you simplify it out, you can just match the terms up.

$(ad - bc)^2 + (ac + bd)^2 = (ac + bd)^2 + 4$
$(ad - bc)^2 = 4$

The result follows.

$ (a^{2}+b^{2})(c^{2}+d^{2}) = (ad-bc)^{2}+(ac+bd)^{2} $ is an identity (that is, it holds for all values of $a$,$b$,$c$,$d$. In fact, it is a special (two-variable) case of Lagrange's Identity.


group theory
abstract algebra
superior algebra
superior algebra solved

Hi! I've just finished eleventh grade so I'm not so good at 12th grade algebra (groups,rings etc.) so I could use some help with this problem:

We have a group with 2002 elements, which has only 1 subgroup with 26 elmts and only 1 subgroup with 77 elmts, and they are both commutative.
Show that the group is commutative.

I found that that the 2 subgroups cyclic (I hope I didn't make a mistake). Can I actually get the fact that the group is cyclic? Please help.

Let G, H, and K be the mentioned groups with |G|=2002, |H|=26 and |K|=77. Then show that

(i) use Lagrange's theorem to show that the intersection of H and K is trivial, i.e. consists only of the identity.
(ii) Since H is unique of order 26 and for any g in G, g^{-1}Hg is also a subgroup of G of the same order, they must coincide. The same idea applies for K. (this means that H and K are normal subgroups) Thus if h is H and k is K, then h^{-1}Kh=K and k^{-1}Hk=H, and so

h^{-}k^{-1}hk belongs to both H and K (why?) and so it is equal to ....?

(iii) by (i) and counting elements |G|=|HK| and so G=HK, which means that for any x,y in G, there exist h,h' in H and k,k' in K such that x=hk and y=h'k'.

Use (iii) and (ii) to conclude that xy=yx; i.e. G is commutative.

Comment: It is true that H and K are cyclic. Also, since gcd(26,77)=1, it is true that G is also cyclic. This goes as follows: If h is a generator of H and k is a generator of K, then it is easy to prove from above that hk has order 2002 (using the fact that 26 and 77 are coprime), and so hk generates G.

Tahnks a lot! You really helped. I'll try to learn more about this subject. I like it but I don't know that much, do I?    :). For example today is the first time I,ve heard of normal subgroups(but I got the idea).

Thanks again!
Cheers!


graph theory
combinatorics unsolved
combinatorics

Let k be a natural number. You are given k colours which are used to colour all edges of a complete graph with n vertices. Determine the minimum value of n such that there necessarily exists a triangle with its edges all the same colour.

Bomb

I think that the answer to this problem is given by the Ramsey theorem.


analytic geometry
trigonometry
geometry
similar triangles

hint:
[hide]
You could brute force with trigonometry but using similar triangles would be much quicker.
[/hide]

EDIT: Your value for C is wrong in your post above, what did you do to obtain it?

[hide="Hint:"]
What you need to do is find the total distance from Bob to Alice, from the right triangle. Then, Bob will have traveled a [i]fraction[/i] of that distance. Do you see how to figure that fraction out?

After finding that fraction, you can multiply it by the change in position for horizontal and vertical components (x and y coordinates) from Bob to Alice, and add the two to the x and y coordinates, respectively.

[hide="The Answer is:"]
\[\left( 1+\frac{15}{ \sqrt{41}}, 1+\frac{12}{ \sqrt{41}}\right) \]
[/hide] [/hide]

[quote="brookr"]Hello everyone,

little problem im stuck on  :)

[img]http://img262.imageshack.us/img262/8163/problem13483ow2.jpg[/img]
(scale is meters)

bob (1,1) wishes to visit to alice (6,5) taking the [b]diagonal[/b] path straight to her. He only manages to travel 3 meters, what are bobs coordinates after he has travelled these 3 meters?

[hide]
This is as far as i have got, i think the solution involves calculating the length of C and then the angle, then using the angle to establish the affect of the 3 meters on the coordinates?

A = 5
B = 4
C = 4.2

[img]http://img262.imageshack.us/img262/2686/problem134832rf5.jpg[/img]
[/hide][/quote]

just to let you know, if a=5 and b=4, then c=$\sqrt{41}$ which definatly isnt 4.2

oh yeah sorry, C is wrong, i must of done:

$\sqrt{5}$ + $\sqrt{4}$

instead of to the power of two then square rooting the result  :blush:

-- just looking through some of the replies now

Ok i think i've solved it with your help cincodemayo  :)

The distance between bob and alice is sqrt{41} which is 6.

The distance bob travelled is 2 meters, so i did the total distance divided by the distance travelled (6/2) to get 3.

I then multipled this by the diffrence between the coordinates and got 15 for x (5x3) and 12 for the y (4x3).

For x i then divided by the total distance to travel (15/6) and added 1 to get 3.5 for the new x coordinate, did the same for y to get (3.5,3)

thanks alot  :)

[quote="brookr"]
The distance between bob and alice is sqrt{41} which is 6.
[/quote]

Not true.
$\sqrt{41}$ is roughly $6.40312424$


[quote="vinskman"]$ P$ is an interior point of a triangle.The distances from $ P$ to each side are $ x,y,z$ corresponding to $ a,b,c$ respectively. Find the maximum value of $ xyz$.[/quote]
IT very easy :)
$ xyz\equal{}\frac{8S(PBC)S(PCA)(PAB)}{abc}\le\ \frac{8(S(PBC)\plus{}S(PCA)\plus{}S(PAB))^3}{27abc}\equal{}\frac{8S(ABC)^3}{27abc}$


the amino acid valine becomes succinic acid in the krebs cycle, but valine has 5 carbons, and succinic acid only has 4 carbons.. will someone please explain what happens to that carbon?


modular arithmetic
function
number theory

Find the last three digits of the number .

2003 power 2002 power 2001

ie: [2003]^{2002}^2001

thanks alot

hint

[hide]Try writing out the powers and notice a pattern[/hide]

Also don't forget $ x^{y^{z}}$  is read as $ x^{(y^{z})}$

[hide="Assuming the above is true and with a lot of scratch work. But Not Sure"]
We know that since $ 352\equiv 12\bmod{20}$, the last two digits must be $ 41$.

Not sure if I did the modular arithmetic work correctly, but for the scratch work for the 3rd digit:
[list]$ 3^{352} \equal{} 9^{176} \equal{} 81^{88}\equiv 561^{44}\bmod{1000}$
$ 561^{44}\equiv 721^{22}\equiv 841^{11}\bmod{1000}$
$ 841^{11} \equal{} 841^{10}841^{1}$
$ 841^{10}\equiv 281^{5} \bmod{1000}$
$ 281^{5}\equiv 401 \bmod{1000} \implies 401(841)\equiv \boxed{241}\bmod{1000}$[/list][/hide]

It's been posted several times, but I can't find the other threads.

[hide="Solution"] $ \lambda(1000) \equal{} 100$, where $ \lambda(n)$ is the [url=http://mathworld.wolfram.com/CarmichaelFunction.html]Carmichael Function[/url].  Hence

$ 2003^{2002^{2001}} \equiv 3^{2^{2001}} \bmod 1000$

We seek to reduce $ 2^{2001} \bmod 100$.  Well, $ 2^{2001} \equiv 0 \bmod 4$ and $ 2^{2001} \equiv 2 \bmod 25$, so $ 2^{2001} \equiv 52 \bmod 100$ and

$ 2003^{2002^{2001}} \equiv 3^{52} \bmod 1000$

For the rest, we work $ \bmod 8, \bmod 125$.  Now,

$ 3^{52} \equiv 9^{26} \equiv 1 \bmod 8$

and

$ 3^{52} \equiv 81 \cdot 729^{8} \equiv ( \minus{} 44) \cdot ( \minus{} 21)^{8} \equiv ( \minus{} 44) \cdot 441^4 \equiv ( \minus{} 44) \cdot ( \minus{} 59)^{4} \equiv ( \minus{} 44) \cdot 3481^2 \equiv ( \minus{} 44) \cdot ( \minus{} 19)^2 \equiv ( \minus{} 44) \cdot 361 \equiv ( \minus{} 44) \cdot ( \minus{} 14) \equiv 616 \equiv 116 \bmod 125$

hence

$ 2003^{2002^{2001}} \equiv \boxed{241} \bmod 1000$. [/hide]


inequalities
search
inequalities unsolved

$ a,b,c$ are non negative reals .prove that $ \sum_cyc{\frac{1}{(a\plus{}b)^2}} >\equal{}\frac{9}{4(ab\plus{}bc\plus{}ca)}$

This is the famous Iran 96 inequality.
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=814239985&t=32470
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=814239985&t=3547
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