1)Let the real $ 2nx2n$ matrix $ X$ have the form :$ X \equal{} \left[\begin{array}{cc}A & B \\
C & D\end{array}\right]$
Where $ A,B,C,D$ are $ nxn$ matrices that commute with one another.Prove that $ X$ is invertible if and only if $ (AD \minus{} BC)$ is invertible.
2)Let $ A,B,C \in M_n{(\mathbb{C})}$ such that : $ AB \equal{} BA,BC \equal{} CB,AC \equal{} CA$ prove that :
 $ M \equal{} \left[\begin{array}{ccc}A & B & C \\
C & A & B \\
B & C & A\end{array}\right]$ is invertible if and only if $ A^3 \plus{} B^3 \plus{} C^3 \minus{} 3ABC$ is invertible.
------------1+1-------------

[quote="zengzeng"][quote="Carcul"]1) $ \det(X) \equal{} \det(AD \minus{} BC)$, under the given conditions.[/quote]
Do You can to explain more problem ?  :blush:[/quote]
[hide="solution1"]Take $ \lambda \notin SP_c(D)$ then we have $ (D \minus{} \lambda I_n) \in GL_n({\mathbb{C}})$
We have : $ \left[\begin{array}{cc}D \minus{} \lambda I_n & \minus{} B \\
O & I_n\end{array}\right]\left[\begin{array}{cc}A & B \\
C & D \minus{} \lambda I_n\end{array}\right] \equal{} \left[\begin{array}{cc}(D \minus{} \lambda I_n)A \minus{} BC & O \\
C & D \minus{} \lambda I_n\end{array}\right]$
then $ \det\left[\begin{array}{cc}A & B \\
C & D \minus{} \lambda I_n\end{array}\right] \equal{} \det((D \minus{} \lambda I_n)A \minus{} BC)$ (by $ (D \minus{} \lambda I_n) \in GL_n({\mathbb{C}}))$
Put $ P(\lambda) \equal{} \det\left[\begin{array}{cc}A & B \\
C & D \minus{} \lambda I_n\end{array}\right] \minus{} \det((D \minus{} \lambda I_n)A \minus{} BC)$
 $ P(\lambda) \equal{} O ,\forall \lambda \notin SP_c(D)$ ,$ deg(P) \leq n$ then $ P(\lambda) \equal{} O$ for all $ \lambda \in C$
and $ P(O) \equal{} O$ (use continoun of $ P(\lambda)$) [b]as Carcul wrote[/b]:$ \det(X) \equal{} \det(AD \minus{} BC)$
------------ :P -------------- :P -------------[/hide]

[color=blue] O ke ! I use above problem [/color]
 [hide="solution2"]Take $ \lambda \notin SP_c(A)$ then $ A_1 \equal{} A \minus{} \lambda I_n \in GL_n(\mathbb{C})$
We have :
$ \det\left[\begin{array}{ccc}A_1 & B & C \\
C & A_1 & B \\
B & C & A_1\end{array}\right] \equal{} \det\left[\begin{array}{ccc}I_n & O & O \\
\minus{}A_1^{\minus{}1}C & I_n & O \\
\minus{}A^{\minus{}1}_1B & O & I_n\end{array}\right]\det\left[\begin{array}{ccc}A_1 & B & C \\
C & A_1 & B \\
B & C & A_1\end{array}\right]\equal{}\det\left[\begin{array}{ccc}A_1 & B & C \\
O & A_1 \minus{} BCA_1^{ \minus{} 1} & B \minus{} C^2A^{ \minus{} 1}_1 \\
O & C \minus{} B^2A_1^{ \minus{} 1} & A_1 \minus{} CBA_1^{ \minus{} 1}\end{array}\right] \equal{} \det(A_1)\det((A_1 \minus{} BCA_1^{ \minus{} 1})(A_1 \minus{} CBA_1^{ \minus{} 1}) \minus{} (B \minus{} C^2A_1^{ \minus{} 1})(C \minus{} B^2A_1^{ \minus{} 1})) \equal{} \det(A_1)(A_1^2 \minus{} 3BC \plus{} B^3A_1^{ \minus{} 1} \plus{} C^3A_1^{ \minus{} 1}) \equal{} \det(A_1^3 \plus{} B^3 \plus{} C^3 \minus{} 3A_1BC)$
Put $ G(\lambda) \equal{} \det\left[\begin{array}{ccc}A_1 & B & C \\
C & A_1 & B \\
B & C & A_1\end{array}\right ] \minus{} \det(A_1^3 \plus{} B^3 \plus{} C^3 \minus{} 3A_1BC)$
then $ G(\lambda) \equal{} O,\forall \lambda \notin SP_c(\mathbb{C}),\deg(G) \leq 3n$ by continoun of $ G(\lambda)$ we have $ G(\lambda) \equal{} O$ for all $ \lambda \in C$ and implies $ G(O) \equal{} O$
Or :$ \det\left[\begin{array}{ccc}A & B & C \\
C & A & B \\
B & C & A\end{array}\right] \equal{} \det(A^3 \plus{} B^3 \plus{} C^3 \minus{} 3ABC)$[/hide]

2: See the discussion in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=268054]this thread[/url] on diagonalizing (block) circulant matrices, or triangularizing them over rings in which we can't divide by $ n$. In other words, $ \mathbb{C}$ is completely irrelevant, and the result is the same even in characteristic 3. Commutativity only comes in when you try to multiply the three diagonal entries together.


inequalities
function
ceiling function
floor function
inequalities open

I had posted this problem here http://www.mathlinks.ro/Forum/viewtopic.php?t=221022 but no one was able to solve it fully. So I am posting it here once again!

For a natural number $ n$ and $ n>1$ find the minimum $ k$ for which this inequality is always true!
   $ \frac{1}{n}\plus{}\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}....\plus{}\frac{1}{k}>1$.

For relevant discussion see the above URL.

$ H(k) \minus{} H(n \minus{} 1) > 1\Leftrightarrow H(k) > 1 \plus{} H(n \minus{} 1)$, where $ H(m)$ is the $ m$th [url=http://mathworld.wolfram.com/HarmonicNumber.html]harmonic number[/url].  Let $ \psi: \mathbb{R}^ \plus{} \to \mathbb{R}^ \plus{}$ be any strictly increasing function satisfying $ \psi (m) \equal{} H(m)$ for each $ m\in \mathbb{N}$, then $ \psi$ is invertible and its inverse is a strictly increasing function.  Therefore, the original inequality is satisfied if and only if $ k > \psi^{ \minus{} 1}\left(1 \plus{} \psi(n \minus{} 1)\right)$, and the least such integer is
\[ \boxed{k \equal{} \left\lceil \psi^{ \minus{} 1}\left(1 \plus{} \psi(n \minus{} 1)\right) \right\rceil}
\]
A convenient choice of $ \psi$ is $ \psi(x) \equal{} \psi_0(x \plus{} 1) \plus{} \gamma$ where $ \psi_0$ is the [url=http://mathworld.wolfram.com/DigammaFunction.html]digamma function[/url].  In this case $ \psi^{ \minus{} 1}\left(1 \plus{} \psi(x \minus{} 1)\right)$ is an analytic function, and after some tedious series manipulation you can find that $ en \minus{} \psi^{ \minus{} 1}\left(1 \plus{} \psi(n \minus{} 1)\right)$ is a positive strictly increasing function of $ n$, and that it approaches the constant $ \frac {e \plus{} 1}{2}$.

This is sufficient to prove $ \left \lfloor en \right \rfloor \minus{} 1 \leq k\leq \left \lfloor en \right \rfloor$, which is essentially what was already done in the previous thread.  Determining which of these values $ k$ actually is for any $ n$ would involve determining how the fractional remainder behaves as a function of $ n$, which seems like it would require more sophisticated tools.  One rather crude result I did get using only series is this: If $ [en] > \frac {e \minus{} 1}{2}$ (where $ [\cdot ]$ denotes the fractional part) then $ k \equal{} \left \lfloor en\right \rfloor$.  Similarly, if $ [en]<\frac{17}{20}$ then $ k\equal{}\left\lfloor en\right \rfloor \minus{}1$.  However, this method doesn't help if $ \frac{17}{20}\leq [en] \leq \frac{e\minus{}1}{2}$.

You may get some better ideas if you have this moved to the analysis forum.


15 lines drawn in a plane, with no 3 concurrent and no 2 parallel, divide the plane into 121 non-overlapping regions. Some of these regions are completely bounded by line segments and some are not. The number which are completely bounded by the line segments is?

[hide="just an attempt..."]

The first 3 lines form 1 region completely bounded by segments. Each line before the third, must have the maximum possible number of intersection points with the other lines, because of the initial conditions. If we imagine to draw the line from a starting point, the first intersection doesn't form any region completely bounded, but each intersection before the first will form a region bounded by segments. So the fourth line has 3 intersection points and will form $3-1=2$ other region more then the first 3 lines. We have 15 lines, so the number of completely bounded regions is
$\frac{13*14}{2}=91$ [/hide]


Find all positive integer $n$ such that $n|2^{n}+2$. :)

I think this one is hard. As far as i know, the original problem is: Prove that if $n$ satisfy the condition then  $n$ must be even, which is not too hard.

[quote="t_c_lhp"] As far as i know, the original problem is: Prove that if $n$ satisfy the condition then  $n$ must be even, which is not too hard.[/quote]
Are you sure that this was the original problem because I don't think that $2^{n-1}$ has any odd divisors :rotfl: . And if $n$ has an odd divisor this divisor must devides $2^{n-1}$ ;)

Yes it is really hard to find all possible values .

[quote="t_c_lhp"][quote="Davron"]Sorry but i think this problem is from APMO 1997, and the answer must be 946...

[color=red][posted before] [/color] :)

Davron[/quote]

I  saw the problem and maybe it is impossible to find all such number[/quote]
Yes, it's difficult and maybe unsolved problem. But there is infinitely many solutions of it.
Even numbers $n_1=2, n_{k+1}=2^{n_k}+2$ for $k\geq 1$ satisfy even to 2 conditions:
1. $n_i|2^{n_i}+2$ and
2. $n_i-1|2^{n_i}+1$.
It's easily checked by induction.

[quote="ZetaX"][quote="pavel kozlov"][quote="Rust"]$n=2^k,k\in Z_+ (0,1,2,...)$[/quote]
It isn't true.[/quote]
But it was true![/quote]
Why? $2^n+2$ isn't divisible even by $4$.

it used to be divisible by 4 :D

 the problem has been edited now :D

[quote="t_c_lhp"]Sorry for my stupidness, i've edited it  :blush: . Now it turns to untrivial problem. :P[/quote]
Look at this post and the ones before ;)

[quote="pavel kozlov"][/quote][quote="ZetaX"][quote="pavel kozlov"][quote="Rust"]$n=2^k,k\in Z_+ (0,1,2,...)$[/quote]
It isn't true.[/quote]
But it was true![/quote][quote="pavel kozlov"]
Why? $2^n+2$ isn't divisible even by $4$.[/quote]

It is not divisible by four, because it has only one two in it's prime factorisation.

[quote="Davron"][/quote][quote="pavel kozlov"][/quote][quote="ZetaX"][quote="pavel kozlov"][quote="Rust"]$n=2^k,k\in Z_+ (0,1,2,...)$[/quote]
It isn't true.[/quote]
But it was true![/quote][quote="pavel kozlov"]
Why? $2^n+2$ isn't divisible even by $4$.[/quote][quote="Davron"]

It is not divisible by four, because it has only one two in it's prime factorisation.[/quote]

 :oops: it's my fault at all.

[quote="Davron"][/quote][quote="pavel kozlov"][/quote][quote="ZetaX"][quote="pavel kozlov"][quote="Rust"]$n=2^k,k\in Z_+ (0,1,2,...)$[/quote]
It isn't true.[/quote]
But it was true![/quote][quote="pavel kozlov"]
Why? $2^n+2$ isn't divisible even by $4$.[/quote][quote="Davron"]

It is not divisible by four, because it has only one two in it's prime factorisation.[/quote]
I didn't ask why $2^n+2$ isn't divisible by 4. I asked ZetaX why 'it was true'?


In this post I will examine what effect each of the four rounds has on the overall score. I won't analyze all 117 teams but will rather confine this analysis to the 30 North American teams who scored 112 or higher. This includes all of the teams who earned national honors in Division B but omits the two teams from Taiwan (whose profile is a little different, anyway.)

First, the high, low, and mean scores on these four rounds for these 30 teams:

$\begin{array}{cccc}\text{Round}&\text{High}&\text{Low}&\text{Mean}\\ \text{Team}&32&12&24.7\\ \text{Power}&39&16&27.4\\ \text{Indiv}&86&53&73.5\\ \text{Relay}&28&2&15.6\end{array}$

Now to do something a little more sophisticated - but only a little. I'm not a statistician. I'm not looking for statistical significance here but rather for some insights into what wins at ARML. First up is the covariance matrix among the four components:

$\begin{array}{ccccc}&\text{Team}&\text{Power}&\text{Indiv}&\text{Relay}\\ \text{Team}&30.0&12.7&29.6&3.2\\ \text{Power}&12.7&28.1&19.7&3.3\\ \text{Indiv}&29.6&19.7&74.8&11..3\\ \text{Relay}&3.2&3.3&11.3&33.7\end{array}$

Now I'll process this matrix. I'll take the square root of the terms on the main diagonal, so that those terms will now be the standard deviation of the scores. (Note: I'm not using sampling theory here - all divisions are by $n,$ not $n-1.$) The off-diagonal entries will be the correlation coefficients between those respective components.

$\begin{array}{ccccc}&\text{Team}&\text{Power}&\text{Indiv}&\text{Relay}\\ \text{Team}&5.47&0.44&0.62&0.10\\ \text{Power}&0.44&5.30&0.43&0.11\\ \text{Indiv}&0.62&0.43&8.65&0.22\\ \text{Relay}&0.10&0.11&0.22&5.81\end{array}$

What conclusions can I draw? None of the correlations is all that high, so the rounds are measuring somewhat different things. The strongest correlation is between the team round and the individual round. That's not too surprising, as both rounds have the same type of question. The correlations involving the power round are weaker; proofs are different. But, at least among this group of 30 competitive teams, the relays are nearly uncorrelated with any other round - they're either completely different or heavily random. But the relays [i]matter[/i]. Note that the range between top and bottom is larger than the range on either the team or power rounds by themselves, and nearly as large as the range on the individual round. Note also that the standard deviation for the relay round is a little larger than the standard deviation of either the team or power rounds. The standings of the top 5 Division A teams at Las Vegas turned nearly completely upside down on the relay round.

These results might look different in different years. I'd imagine you could find years in which the standard deviation of the power round is smaller than this but the standard deviation of the individual round is larger.

I tried a few years back to do some analysis of this nature, but I don't know enough statistics to do it well.  Instead, I looked at a very simple measure: If you threw out an event in any given year, would the same team still win?  For example, this year, if you get rid of the team event, Chicago ties with Exeter.  Toss power, and Georgia wins.  Toss individual, and Lehigh wins.  Toss relay, and Lehigh and Exeter tie.  So this year, any of the four rounds could have been the "deciding round".  Incidentally, this is the first time ever that four teams have come "within one event" of winning or tying for the championship.

Historically, in 33 years of ARML, the team event has been "deciding" 6 times, power 5 times, individual 14 times, and relay 9 times.  (Note that these don't need to add up.)

I also looked at the number of times the winning team had the top score on each event.  For the team event, this happened 23 times (you would expect this number to be higher because there are few possible scores on team), 17 for power, 22 for individual, and 13 for relay.

By either of my measures, it looks like the individual has the most influence on who wins ARML.  But that's to be expected, given that it's worth half your score.  And this doesn't say anything about what rounds are correlated to what, just what rounds are correlated with who finishes first.

Let me repeat the same analysis, but for 60 teams rather than 30. Now the team scores go down to 78, which is just above the level at which we would start to include alternate teams.

First, the high, low, and mean scores on these four rounds for these 60 teams:

$\begin{array}{cccc}\text{Round}&\text{High}&\text{Low}&\text{Mean}\\ \text{Team}&32&8&20.8\\ \text{Power}&39&12&23.8\\ \text{Indiv}&86&34&59.9\\ \text{Relay}&28&0&10.9\end{array}$

Now the covariance matrix:

$\begin{array}{ccccc}&\text{Team}&\text{Power}&\text{Indiv}&\text{Relay}\\ \text{Team}&42.6&18.5&65.9&19.4\\ \text{Power}&18.5&34.1&59.0&18.2\\ \text{Indiv}&65.9&59.0&236.8&66.9\\ \text{Relay}&19.4&18.2&66.9&43.0\end{array}$

Now I'll process this matrix. I'll take the square root of the terms on the main diagonal, so that those terms will now be the standard deviation of the scores. (Note: I'm not using sampling theory here - all divisions are by $n,$ not $n-1.$) The off-diagonal entries will be the correlation coefficients between those respective components.

$\begin{array}{ccccc}&\text{Team}&\text{Power}&\text{Indiv}&\text{Relay}\\ \text{Team}&6.52&0.48&0.66&0.45\\ \text{Power}&0.48&5.84&0.66&0.47\\ \text{Indiv}&0.66&0.66&15.39&0.66\\ \text{Relay}&0.45&0.47&0.66&6.56\end{array}$

Compared to what we had with the smaller sample of teams, we now have that everything - including the relay round - correlates pretty well to the individual round, and that the individual round makes much more of the difference between the strong teams and the weak teams. The relay round is as good a marker for the difference between a stronger and a weaker team as the team or power rounds.

There are probably some other observations worth making, but I'll let other eyes look this over.

Sly Si, could you explain what you mean by the 'deciding' event?  Is an event a 'deciding' event in a particular year if omission of that event would result in a different team winning?  Because in that case, your numbers seem a little low - for example, under this interpretation, this year there were four deciding events, so since your numbers sum to 34 that means that there were 3 years in which there were no deciding events - highly unlikely, given how close the competition usually is.

Yes, "deciding" event means what you take it as--an event that, if omitted, would change the winner (or change from a win to a tie or vice versa).  And actually, there were several years when the competition was not close enough to have one.  Assuming the data I have is correct, those years and the margins of victory were:

1977 (Massachusetts by 5)
1981 (NYC by 37)
1982 (NYC by 30)
1983 (NYC by 19)
1987 (NYC by 10)
1989 (Chicago by 16)
1994 (NYC by 14)
1996 (SFBA by 12)
1997 (Minnesota by 9)
2002 (TJ by 18)
2004 (TJ by 16)

So the competition is not always as close as recent years would suggest.  Also note that if no rounds were "deciding", what it really means is that the winning team built up its lead little by little over all the rounds, rather than shooting out ahead in one round.  When you think of it that way, it doesn't seem too implausible.


Solve the system in positive integers
$x^2 = 2(y+z)$
and
$x^6= y^6+z^6+31(y^2+z^2)$

[quote="M4RI0"]Is $(x;y;z)=(2;1;1)$ the only solution?[/quote]

I guess so, but maybe I'm wrong  :)


Euler
FTW
modular arithmetic
number theory

What is the remainder when $ 19^{2001}$ is divided by $ 23$?

You use mod to solve this problem. You try to find $ 1\equiv 23 (\text{mod}23)$ or $ \minus{}1\equiv 23 (\text{mod}23)$

But I couldn't find a pair yet.

I think there's a formula for that.
Alan said it somewhere...
In another problem like this.

a lot of these problems just require playing around with it.

maybe plug in -4 for 19?

[quote="the future"]Yea I know the formla, you keep subtracting 22 from it but then you get

19^{21} mod 23 which is still hard to do.[/quote]

cant you subtract again, multiply both sides by 19, and you are good, or can you not do that.

i have just a few questions
what does "|" mean in 23|2001?
why do you have 2070-69=2001?

$ a|b$ means that $ b\equal{}ak$ for some integer $ k$.

And I forgot I'm supposed to use 22.  So yeah, we have $ 2001\equal{}1980\plus{}21\equiv \minus{}1\mbox{ (mod 22)}$.  From there, we find $ 19^{\minus{}1}$.  Of course, since this is modular arithmetic, we have some other stuff...

Hmm, let's see, we have $ 19k\equiv 1$ (mod 23).  and then $ 19k\minus{}1\equal{}23n$ for some integer $ n$.  Taking mod 19, we get $ \minus{}1\equiv 4n$ (mod 19).  Then, $ 4n\plus{}1\equal{}19r$, for some integer $ r$.  Taking mod 4, we get $ 3r\equiv 1$ (mod 4), in which we get $ r\equiv 3$ (mod 4).  Therefore, let $ r\equal{}\minus{}1$.  Then, we get $ n\equal{}\minus{}5$ and $ k\equal{}\minus{}6$, or, since this is mod 23, $ \boxed{17}$.  Was that right?

[quote="alanchou"]$ a|b$ means that $ b \equal{} ak$ for some integer $ k$.

And I forgot I'm supposed to use 22.  So yeah, we have $ 2001 \equal{} 1980 \plus{} 21\equiv \minus{} 1\mbox{ (mod 22)}$.  From there, we find $ 19^{ \minus{} 1}$.  Of course, since this is modular arithmetic, we have some other stuff...

Hmm, let's see, we have $ 19k\equiv 1$ (mod 23).  and then $ 19k \minus{} 1 \equal{} 23n$ for some integer $ n$.  Taking mod 19, we get $ \minus{} 1\equiv 4n$ (mod 19).  Then, $ 4n \plus{} 1 \equal{} 19r$, for some integer $ r$.  Taking mod 4, we get $ 3r\equiv 1$ (mod 4), in which we get $ r\equiv 3$ (mod 4).  Therefore, let $ r \equal{} \minus{} 1$.  Then, we get $ n \equal{} \minus{} 5$ and $ k \equal{} \minus{} 6$, or, since this is mod 23, $ \boxed{17}$.  Was that right?[/quote]
I don't really understand your solution... was the first part that you deleted necessary to understanding this?

The correct part of the deleted stuff was that $ 19^{22}\equiv 1$ (mod 23).  Obviously I didn't use trial and error.

using eulers theorem. i could get it down to $ 19^{22}mod{23}$

Well, I suppose you could call it Euler's theorem, since Fermat's is the case where you have mod $ p$.

The fact that the description said "FTW" makes it seem unnecessary to do all this work

except that you need parentheses, that's okay

$ a^{b^c}\neq (a^b)^c$

[hide="The worst way to get the wrong answer"]well u can see that $ 19^3$ $ \equiv$ $ 5(\mbox{mod} 23)$ so we know that  in mod 23, $ 19^{2001}$ = $ 5^{667}$. Keep going

$ 5^2$ $ \equiv$ $ 5(\mbox{mod} 23$,
so in mod 23, $ 5^{667}$ $ \Longrightarrow$ $ 2^333 \times 5$
since $ 2^9$ $ \equiv$ $ 6(\mbox{mod} 23)$
so we now have $ 6^{27}$ $ \times$ $ 5$
$ \Longrightarrow$ $ 9^9$ $ \times$ $ 5$ $ \Longrightarrow$ $ 16^3$ $ \times$ $ 5$
$ \Longrightarrow$ $ 2\times{5}$ $ \Longrightarrow$ $ \fbox{10}$

the answers probly wrong but this is a good way of doing such probs
[/hide]

[quote="abacadaea"]im pretty sure if
$ a\equiv b \pmod{m}$
then
$ ac\equiv bc \pmod{m}$ only holds true if c is a multiple of m[/quote]

...
That makes both of them $ \equiv 0 \pmod{m}$...

[quote="cognos599"][hide="The worst way to get the wrong answer"]well u can see that $ 19^3$ $ \equiv$ $ 5(\mbox{mod} 23)$ so we know that  in mod 23, $ 19^{2001}$ = $ 5^{667}$. Keep going

$ 5^2$ $ \equiv$ $ 5(\mbox{mod} 23$, <==here
so in mod 23, $ 5^{667}$ $ \Longrightarrow$ $ 2^333 \times 5$
since $ 2^9$ $ \equiv$ $ 6(\mbox{mod} 23)$
so we now have $ 6^{27}$ $ \times$ $ 5$
$ \Longrightarrow$ $ 9^9$ $ \times$ $ 5$ $ \Longrightarrow$ $ 16^3$ $ \times$ $ 5$
$ \Longrightarrow$ $ 2\times{5}$ $ \Longrightarrow$ $ \fbox{10}$

the answers probly wrong but this is a good way of doing such probs
[/hide][/quote]
some of your latex's messed up, but I can see at least one mistake,if thats what I assume it is (see arrow), and the rest I can't really tell.

[quote="abacadaea"]im pretty sure if
$ a\equiv b \pmod{m}$
then
$ ac\equiv bc \pmod{m}$ only holds true if c is a multiple of m[/quote]

No, it works for the general case as well.

In your solution, you said that 5[u]=[/u]5^2(mod 23).  Obviously, this is incorrect.  I believe you mean that 2[u]=[/u]5^2(mod 23).  Also, you divided wrong.  333/9≠27, rather it equals 39.  This would've made it congruent to 6^39(mod 23)*5(mod 23)
[u]=[/u]9^13*5 (mod 23).  Unless you have the heart to multiply out 9^13*5 and then mod it out, this solution is virtually useless.  However, it looks very useful if the number 2001 had more factors =).


Pluto has now officially been demoted to an icy dwarf-- so now, there are only 8 official planets in the solar system. Any thoughts on this?

From the new definition:

[quote]1) A planet is a celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit.[/quote]

[source: http://www.iau2006.org/mirror/www.iau.org/iau0602/index.html ]

I just don't understand the hype about Pluto being a planet... it was known for long that the only reason could be the historical one.

[quote]1) A planet is a celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit.[/quote]

Which point, a, b and/or c, does Pluto do not fulfill?

Didn't Segna, Charon (pluto's moon), and two other bodies get accepted as planets a couple weeks ago?

No, where did you get that nonsense from¿ (especially calling charon a planet is just absolutely weird)

No It was Ceres, Charon and Xena. Some experts were going to debate on whether to consider them planets, but I don't think they were accepted.

they had a bunch of scientists debate the definition of "planet," and charon satisfied one of the proposed, but ultimately voted down definitions.  about 4 years ago, i had a teacher who taught that pluto was a kuiper belt object (small rocky asteroid things outside of neptune with elliptical orbits).  pluto does not have enough gravity to have cleared the area around it of other objects (condition c).

i agree with boy soprano. hows the lassiter tournament going by the way?

I dunno, it still seems like pretty big news to me, as it blows, you know, 100 years of elementary textbooks out of the water.

I don't get it.  :maybe:  Pluto is obviously a planet. "Intersects Neptune's orbit" yeah, well Neptune "intersects pluto's orbit".  I orbits around the sun and even has a moon so it;s obviously a planet! Besides, Hades would be demoted from Greek methology then!  :D

Well, it's better to "kick-out" Pluto, than to add 3 or 4 new planets (that was the other option on the Congress), 'cos that would become endless. Every year or so, a new asteroid would become a planet etc.
I think that this is the right choice.

[quote="Boy Soprano II"]It is my opinion that this concept is being blown out of proportion.  Debating whether Pluto is a planet or not is like debating over which is the best way to construct the reals from the rationals.  It is purely a matter of definition among scientists.  I can't understand how classifying Pluto as a "planet" or not would affect any astronomical research.  As for elementary-school textbooks, they should simply state that sometimes Pluto is considered a planet, and sometimes it is not.

If I am badly mistaken, though, please correct me.  I have practically no experience in astronomy, but I cannot see why this is particularly relevant.[/quote]
Well, I think that's the problem : it DOES affect astronomical research, as research costs money.
When they were preparing the[url=http://en.wikipedia.org/wiki/New_Horizons]New Horizon[/url]mission by NASA to fly by Pluto and the Kuiper Belt, the debate started again.  A common argument among proponents was : Pluto is the only planet never visited by a spacecraft, therefore this mission is of historic importance and must now be explored by New Horizon.  If Pluto were downgraded then, it might have cut their funds.  However, that particular mission is off and doing well anyway  :wink:

[quote="Treething"]Does it really matter? Sure, it's not gonna officially be considered a planet, but will textbook writers actually change what it says in the textbooks? No, at least not for a while - that's kind of like redefining $\frac{\pi}{2}$ to be 45 degrees.[/quote]

Maybe it looks weird to our young naive minds, but when my grandmother was born, no books talked about Pluto, the ninth planet, either.

[quote="nutz_for2.718281828"]You've GOT to be kidding. It's STILL a planet! It's at least big enough![/quote]

Just for the record, Pluto is the 17th largest thing in the solar system (not counting the sun).  Our moon, 4 of Jupiter's moons, one each of Saturn and Neptune's moons and the object 2003 U313 (which was going to be one of the new planets if they had used the earlier proposed definition, with potential proposed name "Xena") are all larger.  So, size along is a very poor grounds for including Pluto.

[quote="JBL"]
Just for the record, Pluto is the 17th largest thing in the solar system (not counting the sun).  Our moon, 4 of Jupiter's moons, one each of Saturn and Neptune's moons and the object 2003 U313 (which was going to be one of the new planets if they had used the earlier proposed definition, with potential proposed name "Xena") are all larger.  So, size along is a very poor grounds for including Pluto.[/quote]
Minor question : do you speak of a ranking by mass or by volume?

[quote="fredbel6"]Minor question : do you speak of a ranking by mass or by volume?[/quote]

Volume.  (Radius, actually.)  The information can be found at wikipedia or at [url=http://www.nineplanets.org/datamax.html]this website[/url], whose name seems suddenly out of date.  (They don't include 2003 UB313, which is just slightly larger than Pluto.)  Switching the metric to mass doesn't change the ranking, except that the size of 2003 UB313 becomes more speculative.  Ranking by mass has the nice property that the first 9 objects on the list are one star and 8 planets.  In a ranking by radius, Jupiter's moon Ganymede and Saturn's moon Titan just pass Mercury, but Mercury is considerably more dense and massive.

[quote="Boy Soprano II"]As for elementary-school textbooks, they should simply state that sometimes Pluto is considered a planet, and sometimes it is not.[/quote]

Kinda like "a, e, i, o, u, and sometimes y" :lol:

Calling Charon a planet wouldn't exactly have been "nonsense" under the proposed (failed) definition. It's almost as big as Pluto and could be considered a binary planet.

[quote="nutz_for2.718281828"]You've GOT to be kidding. It's STILL a planet! It's at least big enough![/quote]

But Xena is bigger, so it must be a planet too. Which opens the door to lots of Kuiper Belt objects, some of which are potentially bigger than Pluto.

(a) is in orbit around the Sun
our Sun?
Well, Pluto officially... died :rotfl:
It is just a matter of definition, there is no point arguing

Another proposed definition was that a planet has more mass than the sum of all other matter in its orbit.

whatever happened to the term planetoid?

I think "planetoid" was a considered replacement for the originally proposed (and much mocked) term "pluton" for those planets that are kind of sketchy in their planet-ness, like Pluto.  With the official term "dwarf planet", it's no longer needed.

And I think "Pluto is not officially a planet, but has traditionally been considered to be the ninth planet" will look just fine in textbooks.  Until the new editions go out, I guess teachers will just have to make the corrections in their lessons.

it'll be just like chemistry, where nomenclature has changed repeatedly and you have to understand historical details

i think pluto shouldn't be a planet, because then all kinds of other asteroids would also have to be planets (they're bigger, etc.) and that's too many for the average person to remember.  and it's kind of funny how people are like "oh poor pluto  :( " because i don't think pluto cares one way or another about this.  :D

I'll bet people born under Pluto do care....anyway, Pluto is like a 1/2 mile in diameter or something like that. I guess our history's been altered forever!

1/2 mile¿!? More like 1423 ;)
And well... people believing this nonsense with "born under Pluto" should just change to "born under spam" ;)

Neptune needs 164.88, Pluto 248.09 years to go around the sun one time. This is very near to $\frac{2}{3}$ in relation, thus collison will not happen in next time.
Additionally, the orbit of Pluto is a bit 'skew' in relation to Neptunes orbit, which makes it even less realistic (don't want to look up if it is even possible).

Look at all this other stuff circulating around here... crashes are very rare (remember the last big one on Jupiter), and really big ones (not those small ones some people fear to happen on earth) are a very special event. One between planets is near to be excluded since all big objects that could collide have already collided.

[quote]Look at all this other stuff circulating around here... crashes are very rare (remember the last big one on Jupiter), and really big ones (not those small ones some people fear to happen on earth) are a very special event. One between planets is near to be excluded since all big objects that could collide have already collided.[/quote]
Well, there has to be a reason why Uranus has such an unbelievably tilted axis?

[quote="ZetaX"]since all big objects that could collide have already collided.[/quote]
Except the Milky Way and the Galaxy Andromeda are expected to collide in a few billion years... :roll:

What about the rumour that 2002 NT7 will hit the earth on 1 Feb. 2019 ?  :maybe:
See [url]http://en.wikipedia.org/wiki/(89959)_2002_NT7[/url]
[url]http://en.wikipedia.org/wiki/Palermo_Technical_Impact_Hazard_Scale[/url]

I said [b]BIG[/b] (look at my post in the collisions, I said there that your's not big). This small thingy is not even near to anything big. Big starts about the size of Pluto or so.

And as you can read there, there is (as good as) no chance that it hit's earth in nearer future.

[quote="futuremontrealian"]I'm sorry if this is a stupid question, but it just hit me.  Is pluto going to hit neptune?  Obviously their orbits crossed paths, but I always assumed that they'd never actually touch each other.  I never bothered to consider the possibility that in a few million/billion years they could have a "smashing" party.  Sorry for the pun.  That would be awesome though.  I'd love to study that as an impact.  I better go wiki it.

And I always thought pluto should be out of the planetary league.  It never really fit in very well.  I bet it'll confuse third graders around the world though.  "My Very Educated Mother Just served us... nachos?"[/quote]

No, Pluto and Neptune orbit the sun on different planes and therefore will not hit each other.

At least, that's what I've always been taught.

But it would be fun to watch if they did. :rotfl:

[quote="fredbel6"]Well, there has to be a reason why Uranus has such an unbelievably tilted axis?[/quote]

*titters*

pluto isn't the planet. it was "the planet" just because it was the ONLY "planet" discovered by an american astronomer

Guys this is probably said somewhere here but where is the proof? I don't stay with news..

Proof of what¿  :huh:
That this decision was made¿ Look at my previous posts, I gave one.


In the prime factorization of $ 100!*99!*...*3!*2!*1!$, one factor is $ 3^x$.  Find x.

[hide] We have that this number is $ 2^{99}*3^{98}*4^{97}...*100$.  Okay, so take all of the multiples of 3.  The sum of the exponents of all of them is $ 98 \plus{} 95 \plus{} 92... \plus{} 2$.  Okay, now take all of the multiples of 9.  The sum of the exponents of these is $ 92 \plus{} 83 \plus{} 74... \plus{} 2$.  Now of 27, we get $ 74 \plus{} 47 \plus{} 20$.  Finally, of 81, we get $ 20$.  Okay, so we  have that the sum for 27 and 81 together is $ 161$.  Now we can find the sum of the threes using the sum formula for an arithmetic sequence.  So, we get $ \frac {(33)(2 \plus{} 98)}{2} \equal{} 50*33 \equal{} 1650$.  We can also do the same thing for the nines to obtain $ \frac {(11)(94)}{2} \equal{} 47*11 \equal{} 517$.  We add $ 161 \plus{} 517 \plus{} 1650 \equal{} \boxed{2328}$.  Okay... I could have made a mistake...
[/hide]
EDIT: Oh, whoops... yeah, you're right...
Okay, solution completely changed

Sorry, that's the answer for $ 100!$. This is asked for $ 100! \cdot 99! \cdots 1!$.


Έστω οι διακεκριμένοι αριθμοί l1,l2,...,ln e R-{-1,-2,...,-n}
Να δείξετε ότι ο πίνακας με γραμμές
ri=[$ \frac{1}{li\plus{}1}$,$ \frac{1}{li\plus{}2}$,...,$ \frac{1}{li\plus{}n}$], i e {1,2,...,n} είναι αντιστρέψιμος.

[quote="mast"]Έστω οι διακεκριμένοι αριθμοί $ a_1,a_2,...,a_n \in \mathbb{R}\minus{}\lbrace\minus{}1,\minus{}2,...,\minus{}n\rbrace$

Να δείξετε ότι ο πίνακας

$ \left[\begin{array}{c c c c}
\frac {1}{a_1 \plus{} 1} & \frac {1}{a_1 \plus{} 2} & \hdots & \frac {1}{a_1 \plus{} n} \\
 &  &  &  \\
\frac {1}{a_2 \plus{} 1} & \frac {1}{a_2 \plus{} 2} & \hdots & \frac {1}{a_2 \plus{} n} \\
 &  &  &  \\
\vdots & \vdots & \ddots & \vdots \\
 &  &  &  \\
\frac {1}{a_n \plus{} 1} & \frac {1}{a_n \plus{} 2} & \hdots & \frac {1}{a_n \plus{} n}
\end{array}
\right]$

είναι αντιστρέψιμος.[/quote]

Παραθέτω τη λύση που σκέφτηκα
[hide]Θέτουμε $ D_k \equal{} det \left[\begin{array}{c c c c} \frac {1}{a_k \plus{} k} & \frac {1}{a_k \plus{} 2} & \cdots & \frac {1}{a_k \plus{} n} \\
& & & \\
\frac {1}{a_{k \plus{} 1} \plus{} k} & \frac {1}{a_{k \plus{} 1} \plus{} k \plus{} 1} & \cdots & \frac {1}{a_{k \plus{} 1} \plus{} n} \\
& & & \\
\vdots & \vdots & \ddots & \vdots \\
& & & \\
\frac {1}{a_n \plus{} k} & \frac {1}{a_n \plus{} k \plus{} 1} & \cdots & \frac {1}{a_n \plus{} n} \end{array} \right]$ όπου $ k\in N\cap [1,n \minus{} 2]$.

Καλούμε $ k_{ij}$ το στοιχείο που βρίσκεται στην $ (i \minus{} k \plus{} 1) \minus{}$γραμμή και $ (j \minus{} k \plus{} 1) \minus{}$στήλη της $ D_k$ όπου $ i,j\in N \cap [k,n]$.Αφαιρώντας από την $ i \minus{}$γραμμή $ \frac {a_k \plus{} k}{a_i \plus{} k}$ φορές την $ k \minus{}$γραμμή  $ \forall i \in N \cap [k \plus{} 1,n]$ παίρνουμε μια ίση ορίζουσα. Ας ονομάσουμε τώρα $ k'_{ij}$ το στοιχείο που βρίσκεται στην $ (i \minus{} k \plus{} 1) \minus{}$γραμμή και $ (j \minus{} k \plus{} 1)$στήλη της όπου $ i,j\in N \cap [k,n]$.Τότε προφανώς $ k'_{ik} \equal{} 0, \forall i\in N\cap [k \plus{} 1,n]$ και

$ {k'_{ij} \equal{} \frac {1}{a_i \plus{} j} \minus{} (\frac {a_k \plus{} k}{a_i \plus{} k})(\frac {1}{a_k \plus{} j}) \equal{} \frac {a_ia_k \plus{} ja_i \plus{} ka_k \plus{} kj \minus{} a_ia_k \minus{} ka_i \minus{} ja_k \minus{} jk}{(a_i \plus{} j)(a_i \plus{} k)(a_k \plus{} j)} \equal{} \frac {(j \minus{} k)(a_i \minus{} a_k)}{(a_i \plus{} j)(a_i \plus{} k)(a_k \plus{} j)}}$.

Επομένως θα ισχύει:

$ {D_k \equal{} det \left[\begin{array}{c c c c} \frac {1}{a_k \plus{} k} & \frac {1}{a_k \plus{} 2} & \cdots & \frac {1}{a_k \plus{} n} \\
& & & \\
0 & \frac {a_{k \plus{} 1} \minus{} a_k}{(a_{k \plus{} 1} \plus{} k \plus{} 1)(a_{k \plus{} 1} \plus{} k)(a_k \plus{} k \plus{} 1)} & \cdots & \frac {(n \minus{} k)(a_{k \plus{} 1} \minus{} a_k)}{(a_{k \plus{} 1} \plus{} n)(a_{k \plus{} 1} \plus{} k)(a_k \plus{} n)} \\
& & & \\
\vdots & \vdots & \ddots & \vdots \\
& & & \\
0 & \frac {a_n \minus{} a_k}{(a_n \plus{} k \plus{} 1)(a_n \plus{} k)(a_k \plus{} k \plus{} 1)} & \cdots & \frac {(n \minus{} k)(a_n \minus{} a_k)}{(a_n \plus{} n)(a_n \plus{} k)(a_k \plus{} n)} \end{array} \right]}$

Φαίνεται τώρα οπτικά καλύτερα το γεγονός ότι ο παράγοντας $ \frac {a_i \minus{} a_k}{a_i \plus{} k}$ είναι κοινός στην $ i \minus{}$γραμμή.Αυτό εξηγείται και γιατί τα $ i$ παραμένουν αμετάβλητα σε μια γραμμή άρα για μία γραμμή όταν παίρνουμε τα στοιχεία από τις στήλες της αυτός ο παράγοντας είναι σταθερός.Άρα:

$ D_k \equal{} det \left[\begin{array}{c c c c} \frac {1}{a_k \plus{} k} & \frac {1}{a_k \plus{} 2} & \cdots & \frac {1}{a_k \plus{} n} \\
& & & \\
0 & \frac {a_{k \plus{} 1} \minus{} a_k}{(a_{k \plus{} 1} \plus{} k \plus{} 1)(a_{k \plus{} 1} \plus{} k)(a_k \plus{} k \plus{} 1)} & \cdots & \frac {(n \minus{} k)(a_{k \plus{} 1} \minus{} a_k)}{(a_{k \plus{} 1} \plus{} n)(a_{k \plus{} 1} \plus{} k)(a_k \plus{} n)} \\
& & & \\
\vdots & \vdots & \ddots & \vdots \\
& & & \\
0 & \frac {a_n \minus{} a_k}{(a_n \plus{} k \plus{} 1)(a_n \plus{} k)(a_k \plus{} k \plus{} 1)} & \cdots & \frac {(n \minus{} k)(a_n \minus{} a_k)}{(a_n \plus{} n)(a_n \plus{} k)(a_k \plus{} n)} \end{array} \right] \equal{} \frac {1}{a_k \plus{} k}\prod_{i \equal{} k \plus{} 1}^n {\frac {a_i \minus{} a_k}{a_i \plus{} k}} det \left[\begin{array}{c c c c} \frac {1}{(a_{k \plus{} 1} \plus{} k \plus{} 1)(a_k \plus{} k \plus{} 1)} & \cdots & \frac {n \minus{} k}{(a_{k \plus{} 1} \plus{} n)(a_k \plus{} n)} \\
& & & \\
\vdots & \ddots & \vdots \\
& & & \\
\frac {1}{(a_n \plus{} k \plus{} 1)(a_k \plus{} k \plus{} 1)} & \cdots & \frac {n \minus{} k}{(a_n \plus{} n)(a_k \plus{} n)} \end{array} \right]$

Εύκολα διαπιστώνουμε με τον ίδιο τρόπο ότι ο παράγοντας $ \frac {j \minus{} k}{a_k \plus{} j}$ είναι σταθερός σε κάθε στήλη $ \forall j\in N \cap [k \plus{} 1,n]$.Άρα και σύμφωνα με τα παραπάνω:

$ D_k \equal{} \frac {1}{a_k \plus{} k}\prod_{i \equal{} k \plus{} 1}^n {\frac {a_i \minus{} a_k}{a_i \plus{} k}} \prod_{j \equal{} k \plus{} 1}^n {\frac {j \minus{} k}{a_k \plus{} j}} det \left[\begin{array}{c c c c} \frac {1}{a_{k \plus{} 1} \plus{} k \plus{} 1} & \cdots & \frac {1}{a_{k \plus{} 1} \plus{} n} \\
& & & \\
\vdots & \ddots & \vdots \\
& & & \\
\frac {1}{a_n \plus{} k \plus{} 1} & \cdots & \frac {1}{a_n \plus{} n} \end{array} \right] \equal{} {\frac {1}{a_k \plus{} k}\prod_{i \equal{} k \plus{} 1}^n {\frac {a_i \minus{} a_k}{a_i \plus{} k}} \prod_{j \equal{} k \plus{} 1}^n {\frac {j \minus{} k}{a_k \plus{} j}} D_{k \plus{} 1}}, \forall k\in N \cap [1,n \minus{} 2]$.

Επίσης

$ det \left[\begin{array} {c c c c} \frac {1}{a_{n \minus{} 1} \plus{} n \minus{} 1} & \frac {1}{a_{n \minus{} 1} \plus{} n} \\
& & & \\
\frac {1}{a_n \plus{} n \minus{} 1} & \frac {1}{a_n \plus{} n} \end{array} \right] \equal{} {\frac {1}{a_{n \minus{} 1} \plus{} n \minus{} 1} \frac {1}{a_n \plus{} n} \minus{} \frac {1}{a_n \plus{} n \minus{} 1} \frac {1}{a_{n \minus{} 1} \plus{} n}} \equal{} \frac {a_n \minus{} a_{n \minus{} 1}}{(a_{n \minus{} 1} \plus{} n \minus{} 1)(a_n \plus{} n)(a_n \plus{} n \minus{} 1)(a_{n \minus{} 1} \plus{} n)}\neq 0$.

αφού οι αριθμοί είναι διακεκριμένοι.Από τα παραπάνω και αυτό συνάγουμε επαγωγικά ότι $ D_1 \neq 0$ άρα ο πίνακάς μας είναι αντιστρέψιμος.[/hide]
Ελπίζω να μη έκανα κάποιο λάθος με τόσους δείκτες  :maybe: Πάντως κάτι μου λέει ότι υπάρχει και πιο απλή λύση  :roll:

Ωραία η λύση σου Παύλο.
Το πρόβλημα θα μπορούσε να λυθεί και αποδεικνύοντας ότι οι στήλες του πίνακα είναι γραμμικώς ανεξάρτητες. Αφήνω την απόδειξη σε όποιν θέλει να την προσπαθήσει.

mast, για γραψε την λυση σου να την δουμε.  :roll:

Λοιπόν, ονομάζω την ι-οστή στήλη $ c_i$, και έστω $ l_1,l_2,...,l_n$ πραγματικοί αριθμοί με $ \Sigma l_ic_i\equal{}0$. Τότε, αν πάρω τις συνιστώσες του αθροίσματος  $ \Sigma l_ic_i$, τότε έχω ότι για κάθε j στο {1,2,...,n}, ισχύει ότι $ \Sigma \frac{l_i}{a_j\plus{}i}$, όπου η άθροιση γίνεται στο i. Οπότε, αν θεωρήσω την συνάρτηση
f(x)=$ \Sigma \frac{l_i}{x\plus{}i}$, τότε η συνάρτηση έχει ν ρίζες. Όμως, η συνάρτηση μπορεί να γραφτεί ως ρητή συνάρτηση με f(x)=$ \frac{P(x)}{\Pi (x\plus{}a_i)}$ , όπου το P(x) είναι πολυώνυμο n-1 βαθμού. Οπότε το P είναι το μηδενικό πολυώνυμο και η f είναι η μηδενική συνάρτηση. Από αυτό προκύπτει ότι τα $ l_i$ είναι μηδέν και κατα συνέπεια οι στήλες είναι γραμμικώς ανεξάρτητα διανύσματα. Άρα, ο πίνακας είναι αντιστρέψιμος.

Αν μπορεί κάποιος mod ας διορθώσει τη λύση μου στην πρώτη ορίζουσα στο στοιχείο (1,2) αντί $ a_k\plus{}2$ να βάλει $ a_k\plus{}k\plus{}1$ γιατί μου βγάζει ένα μήνυμα ότι δεν μπορώ.Ευχαριστώ.


pigeonhole principle
modular arithmetic
inequalities
number theory
prime numbers
Diophantine equation
number theory proposed

Let $ a$ be a fix natural number . Prove that the set of prime divisors of $ 2^{2^{n}} \plus{} a$ for $ n \equal{} 1,2,\cdots$ is infinite

[quote="khashi70"]Let $ a$ be a fix natural number . Prove that the set of prime divisors of $ 2^{2^{n}} + a$ for $ n = 1,2,\cdots$ is infinite[/quote]
$ x_n=2^{2^n}+a$
We denote $ S$ be the set of primes p such that exist an positive integer M which for all $ n\geq M$ then $ p|x_n$ .We will so that $ S$ is a finite set . This result may be easy because $ \gcd(x+a,x^2+a)$ is a divisor of $ a(a+1)$ , which means that if $ p\in S$ then  $ p|a(a+1)$ . Other hand ,by the condition $ p|x_n$ we have  $ p=2$ or $ p$ is divisor of $ a+1$ .
Lemma 1 :Consider $ k$ is a positive greater than $ v_2(a)$ then $ x_k$ is not a power of 2 .For each pair of  natural numbers $ (m,n)$ such that  $ n\geq v_2(\varphi(x_k))$ and $ m\equiv n (\mod \varphi(x_k))$we have   $ x_m\equiv x_n (\mod x_k)$ .
Proof : Let $ x_k=2^{v}.u$ ,by the condition we have $ n\geq (v+\varphi(u)-1)$ and $ m\equiv n (\mod \varphi(x_k))$ . We have
\[ x_m-x_n=2^{2^n}(2^{2^m-2^n}-1)
\]
By Euler's theorem we have $ \varphi(u)|2^m-2^n$ therefore $ u|2^{2^m-2^n}-1$. Other hand $ 2^n\geq v$ so $ x_k|x_m-x_n$ .
Lemma 2 :Let $ n,t$ be  fixed positive integers,$ t$ is even , consider sequence $ y_k=2^{2^{n+kt}}+a$ then exist a positive integer $ k$ such that $ y_k$ have a prime divisor not in $ S$ .
Let $ a+1=2^{u}.v$ .We will show that exist an integer $ k$ which satisfy
\[ y_k\equiv a+1 (\mod v^2)
\]
This can be done by similar trick above , hence we can write :$ y_k=2^uv+t.v^2=v(2^u+tv)$ which have a prime divisor not divide $ v$ and different from 2 . Lemma 2 has been prove .
Now let consider our problem . Take $ a$ number in the sequence ,which has a prime divisor not in S , by the definition of $ S$ exist a numbers $ (m,n)$ satisfy lemma 1 and then by lemma 2 ,exist $ m$ such that it has a different prime divisor ,and not in S ... So on ,we can  construct a infinite set of prime numbers

But ... where do we know that there exists such $ r$ from? Even there is something more incomprehensible - if $ a\equal{}2$, then $ a^{2^k}\plus{}a$ and $ 2^{2^k}\plus{}a$ will coincide and since there exists big enough $ n$ such that $ p_i^r|2^{2^n}\plus{}a$, then $ p_i^r|a^{2^n}\plus{}a$ :maybe:

but $ k$ is a fix number and each of $ a^{2}+a$ , $ a^{4}+a$ , $ \cdots$ , ${ a^{2^{k}}+a}$ has a factorization by prime numbers ... so we can choose this $ r$ .

oh , yes... I thought you take an infinite set $ a^{2^k}\plus{}a$.

I have two questions regarding your solution and I hope you will explain it to me

1. why can't s=k+1

2. how do you get the last statement I thought that a is given and fix so how did you deduce that
$ p_{1}^{r}|2^{2^{u}} \plus{} a$
$ p_{1}^{r}|2^{2^{u \plus{} s}} \plus{} a ( 1\leq s \leq k )$
Thus
$ p_{1}^{r}|a^{2^{s}} \plus{} a$

I'm sorry I just can't get it.

1-Our $ k \plus{} 1$ consecutive numbers are $ u$, $ u \plus{} 1$, $ \cdots ,u \plus{} k$

2-$ p_{1}^{r}|a^{2^{u}} \plus{} a$ and
$ a^{2^{u \plus{} s}} \equal{} (a^{2^{u}})^{s}$ so :
$ a^{2^{u \plus{} s}}\equiv ( \minus{} a)^{2^{s}}$ (mod $ p_{1}^{r}$)
I hope u get ur answers .

A very useful

$ \textsc{Kobayashi}$ Theorem. [color=blue]If the positive integer sequence $ (a_n)_{n\geq 1}$ is not bounded, and has a finite set of prime divisors, then the translated sequence $ (a_n \plus{} a)_{n\geq 1}$ has an infinite set of prime divisors, for any not-null integer $ a$.[/color]

I have a short proof of this, using $ \textsc{Thue}$'s Theorem that the Diophantine equation $ Ax^3 \plus{} By^3 \equal{} C$ has finitely many integer solutions.

Now, it's an overkill, since $ (2^{2^n})_{n\geq 1}$ has only $ 2$ as prime divisor.

Dear mavropnevma
Can you show us your nice proof of this
Kobayashi Theorem?
Thanks in advance.

No reference, sorry (I found about it in some  note in the Monthly, but can't remember where...). As for the proof, I gave enough hints, I think (look it over again, it's not difficult  :) ).

[quote="mavropnevma"]A very useful

$ \textsc{Kobayashi}$ Theorem. [color=blue]If the positive integer sequence $ (a_n)_{n\geq 1}$ is not bounded, and has a finite set of prime divisors, then the translated sequence $ (a_n \plus{} a)_{n\geq 1}$ has an infinite set of prime divisors, for any not-null integer $ a$.[/color]
[/quote]

Here is a reference to Kobayashi's paper:

  Kobayashi, Hiroshi
  On existence of infinitely many prime divisors in a given set.
  Tokyo J. Math.  4  (1981), no. 2, 379-380.

However, this is far above Olympiad level.

here is my solution. is it correct??

suppose that the set of prime divisors of this sequence is finite, namely $\{p_1,...,p_k\}$. let $n_1$ be a natural number such that $n_1>v_2((p_1-1)(p_2-1)....(p_k-1)a)$. suppose that ${v_{p_i}(2^{2^{n_1}}}+a)=\alpha_i$. we want to prove there exists $n_2$ and $t$ such that $2^{n_1}+t\phi(p_1^{\alpha_1+1})\phi(p_2^{\alpha_2+1})....\phi(p_k^{\alpha_k+1})=2^{n_2}$ (I) or in other words
$p_1^{\alpha_1}.....p_k^{\alpha_k}(p_1-1).....(p_k-1)|2^{n_1}(2^{n_2-n_1}-1)$.
note that because of the choice of $n_1$, the exponent of $2$ on the right hand side is more than the one in the left hand side, so we can omit them. now the left hand side is equal to an odd number $m$, and there exists a number $s$ such that $2^s\equiv 1$ $(mod$ $m)$. now let $n_2=n_1+s$, so such $n_2$ exists that (I) holds.
now back into the main problem, we get that for each $i$, we have $v_{p_i}(2^{2^{n_2}}+a)=v_{p_i}(2^{2^{n_1}}+a)$, but $2^{2^{n_2}}+a>2^{2^{n_1}}+a$, a contradiction. so the set of prime divisors of the sequence is infinite.

[quote="mavropnevma"]A very useful
using $ \textsc{Thue}$'s Theorem that the Diophantine equation $ Ax^3 \plus{} By^3 \equal{} C$ has finitely many integer solutions. [/quote]

Someone can post reference to THUE 's Theorem on Diophante Equation ??
Thanks!

Let $\{x_n\}=\{2^{2^n}+a\}$ , first suppose $a$ is odd. Now let set of prime divisors of $\{x_n\}$ is finite. Now consider the sub sequences $\{a_i\}$ for finite number of $i's$. Now construct those such as in each of $\{a_i\}$ set of prime divisors of all element is same. Now certainly size at least one of those $\{a_i\}$ must be infinite. Consider the $\{a_k\}$ , contains maximum number of prime divisors and infinitely many terms. Now so maximum power of at least one of the prime dividing terms of $\{a_k\}$ must not be bounded. Now suppose ${\{a_k}\}=\{a'_1,a'_2,a'_3,......\}$. Now suppose we’ve like $v_p(x_m),v_p(x_n)\alpha>3$ then it’s easy to see $v_p(|m-n|)\geq (\alpha-2)$. Now suppose in the sequence $\{a_k\}$ a very large number is $n$ which is nothing but maximum power of a prime $p$ dividing terms of $\{a_k\}$ and whose power is unbounded. Now suppose $v_p(a_i)=n$ and first term greater than $a_i$ is $a_m$ such as $v_p(a_m)\geq n$. Now as we took $n$ to be a very large number so the gap between $m$ and $i$ must be greater than $p^{n-2}$ and that is also very large. Now by our assumption in that large gap, powers of $p's$ is less than $n$. So there must be always another prime with unbounded power, if not then there would enter some terms of another sequences other than $\{a_k\}$. Then so again we would get another sub sequence of infinite length. And similarly going on we would get all of the sub sequences are of infinite length because of large $n$, it’s not difficult show infinitely many $x_n$ can’t be power of a prime. Now so lets back to case where in $\{a_k\}$ there are at least two primes with unbounded power, now basically similarly we get all prime dividing terms of $\{a_k\}$ are unbounded. Now just consider two consecutive very large terms of $\{a_k\}$ and notice similarly we’ll get another sub sequence with infinite length. So going on as previous we’re getting contradiction, now if $a$ would be even then similarly we would do, and so we’re now done to show $\{x_n\}$ contains infinitely many prime divisors.

The solution may be pre-posted but still I am reposting. :D

Let  primes dividing $f(n)$ as $n$ varies over all the positive integers be $p_1,p_2,....,p_k$.

Then let $M=\sum_{i=1}^{k}v_2(p_i-1)$.

And let $f(M)=\prod_{i=1}^{k}p_i^{a_i}$.

Now we show that there exists a $k$ such that $n=\prod_{i=1}^kp_i^{a_i+1}$ and $n|f(M+k)-f(M)$ where $f(n)=2^{2^n}+a$.

Clearly, $f(x+y)-f(x)=2^{2^x}(2^{2^x(2^y-1)}-1).$  $\forall$ $x$ and $y$

Now put $x=M$ and $y=\phi(\prod_{i=1}^{k}p_i^{a_i}r_i)$ where $p_i-1=2^{v_2(p_i-1)}r_i $

We now claim that $\phi(n)|2^M(2^y-1)$.

Clearly $\phi(n)=\prod_{i=1}^{k}p_i^{a_i}(p_i-1)$.Now $M=\sum_{i=1}^kv_2(p_i-1)=v_2(2^{M}(2^y-1))=v_2(\phi(n))$.

Now $y=\phi(\prod_{i=1}^{k}p_i^{a_i}r_i)=\phi(\frac{\phi(n)}{\prod_{i=1}^k2^{v_2(p_i-1)}})$ and hence

$\frac{\phi(n)}{\prod_{i=1}^k(2^v_2(p_i-1)}|2^y-1$.

This indicates $\phi(n)|2^M(2^y-1).$

Now we know that $f(M+y)-f(M)=2^{2^m}(2^{2^M(2^y-1)}-1)$ and hence $n|f(M+y)-f(M)$.

Now take any of the primes $p_i$.

If $v_{p_i}(f(M))<v_{p_i}(f(M+y))$ then contradiction arises because then $ v_{p_i}(f(M+y)-f(M))=v_{p_i}(f(M)) < v_{p_i}(n)$.

Similarly the other case may also be contradicted.

So only case is that $v_{p_i}(f(M))=v_{p_i}(f(M+y))$.Now this happens for all odd primes $p_i$.

Notice that it indicates that the odd parts of both $f(M+y)$ and $f(M)$ are equal and hence it indicates that $f(M+y)=2^{l_1}f(M)$ for

some.Now this indicates after some easy analysis that $v_2(a)=2^M+l_1$

Notice that we can once again carry on the same arguement this time $M$ replaced by $M+y$.In that case we have to suitably

choose a number $s$ in place of $y$ which will depend on the exponents of the primes in $f(M+y)$.We will again get

$f(M+y+s)=2^{l_2}f(M+y)$ and thus $v_2(a)=2^{M+y}+l_2$.

We can continue the process continously then we notice that $l_i$ is a strictly decreasing sequence.Again as all the $l_i$s ar positive

 integers so eventually we will get one $t$ such that $l_t=0$.And hence we will get two $x,y$ with $x \neq y$ such that $f(x)=f(y)$

 which is clearly a contradiction. :D

consider the sequence $a_{n}=2^{2^n} $ . Its an unbounded sequence and each term has only 1 prime divisor , viz , $2$ . hence , by Kobayashi's theorem , we get that the set of prime divisors of  $(2^{2^n}+a)$,$ n=1,2,3 , .....$  is infinite (as $a>0$ ) :)

[hide="solution"]
Assume for contradiction that the only primes that divide $2^{2^{n}} + a$ where $n$ ranges over all positive integers are $P_1, ..., P_N$, where $N$ is some positive integer. Let $r$ be any positive integer. Then there exists an $n_0$ which depends on $r$ so that for all positive integers $n \geq n_0$, we have $2^{2^{n}} + a > (P_1 P_2 \cdots P_N)^r$. This means that among the numbers $2^{2^{n_0}} + a, 2^{2^{n_0} + 1} + a, \cdots 2^{2^{n_0} + N} + a$, we can find two of these numbers and some prime $P_i$ so that $P_i ^ r$ divides both of these numbers. So take $0 \leq \ell < m \leq N$, and some prime $P_i$ so that $2^{2^{n_0} + \ell} \equiv -a \pmod {P_i ^ r}$ and so that $2^{2^{n_0} + m} \equiv -a \pmod {P_i ^ r}$. Then we easily get that $a^{2^{m - \ell}} \equiv -a \pmod {P_i^r}$. This means that $a^{2^{m - \ell}} + a \geq P_i ^r$ But note that $m -\ell$ is at most $N$, so the previous inequality can not hold for large enough $r$, which is a contradiction, so we are done.
[/hide]

[quote="FelixD"]@mavropnevma: have you got any references to Kobayashi's theroem?  :)[/quote]
See this, it should help you.
http://leonettipaolo.wordpress.com/2013/01/20/kobayashi-theorem/
And the original paper is attached too.

Insta-kill by Kobayashi's Theorem but let's be a bit more modest. For $a=1$ we know that Fermat numbers are pairwise coprime, so we are done and we are free to assume $a>1$. Suppose this set is finite and the set of all primes dividing at least one term of this sequence is $P=\{p_1,\dots,p_k\}$. Write each term as $p_1^{e_1}\cdot \dots \cdot p_k^{e_k}$ and choose the index $1 \le i \le k$ such that $p_i^{e_i}$ is maximal. Since we are assigning to each term one out of $k$ indices, we have by the Pigeon-hole Principle, $x_n=2^{2^n}+a$ and $x_m=2^{2^m}+a$ assigned the same index $i$ for some $n+k \ge m>n>N$ where $N>0$ is a sufficiently large constant. Now, let us observe that \begin{align*} p_i^{a_i}= p_i^{v_{p_i}(\gcd(x_m,x_n))} \ge \min (\sqrt[k]{x_n},\sqrt[k]{x_m})=O(\exp(\exp(n))) \end{align*}however, we have \begin{align*} p_i^{a_i} \mid 2^{2^n}\cdot ((2^{2^n})^{2^{m-n}-1}-1) \end{align*}yielding that \begin{align*} 1 \equiv ((2^{2^n})^{2^{m-n}-1} \equiv (-a)^{2^{m-n}-1} \bmod p_i^{a_i} \end{align*}and we see that $p_i^{a_i} \le |(-a)^{2^{m-n}-1}-1|=O(1)$ since $m-n$ is bounded above by $k$. This is a contradiction and the result follows from our deduction.

[b]Kobayashi's theorem[/b] kills it since $a_n=2^{2^{n}}$ is not bounded and $2^{2^{n}}$ has only $2$ as prime divisor.

[quote=khashi70]Assume that this set be finite i.e there exist $ p_{1}, p_{2}, \cdots p_{k}$ which they are all the prime divisor of $ 2^{2^{n}} \plus{} a$
now consider these numbers :
$ a^{2} \plus{} a$, $ a^{4} \plus{} a$ , $ \cdots a^{2^{k}} \plus{} a$ and assume $ r$ is a number such that none of this number is divisible by $ p_{i}^{r}$ for all $ i \equal{} 1,2,\cdots,k$
now we know if $ n$ be sufficient big then if factorize $ 2^{2^{n}} \plus{} a$ by $ p_{1}, p_{2}, \cdots p_{k}$ one of this prime number has a power bigger than $ r$ .
so if we consider $ k \plus{} 1$ consecutive number $ n$ which are sufficient big then by Pigeonhole theorem there exist two of this number which one of these prime number (for example) $ p_{1}$ has a power bigger than $ r$ in both which means :
$ p_{1}^{r}|2^{2^{u}} \plus{} a$
$ p_{1}^{r}|2^{2^{u \plus{} s}} \plus{} a$ ( $ 1\leq s \leq k$ )
Thus $ p_{1}^{r}|a^{2^{s}} \plus{} a$
which is contradiction ...
So the problem is solved .[/quote]

NAILED DAT PROB  :)

[quote=mavropnevma]
$ \textsc{Kobayashi}$ Theorem. [color=blue]If the positive integer sequence $ (a_n)_{n\geq 1}$ is not bounded, and has a finite set of prime divisors, then the translated sequence $ (a_n \plus{} a)_{n\geq 1}$ has an infinite set of prime divisors, for any not-null integer $ a$.[/color]

I have a short proof of this, using $ \textsc{Thue}$'s Theorem that the Diophantine equation $ Ax^3 \plus{} By^3 \equal{} C$ has finitely many integer solutions.
.[/quote]
To use Thue's Theorem you must have irreducibility of $f(x,y)=Ax^3 \plus{} By^3$. So what are your conditions for $A,B$ since for $A=B\neq 0$ this polynomial is reducible: $f(x,y)=A(x+y)(x^2-xy+y^2)$

[hide=One liner]The only prime that divides powers of two is two. So by Kobayashi, adding $a$ will give infinitely many primes.[/hide]

[quote=fukano_2]The only prime that divides powers of two is two. So by Kobayashi, adding $a$ will give infinitely many primes[/quote]
I would really like to see a one-liner for Kobayashi :)

[quote=SenatorPauline] I would really like to see a one-liner for Kobayashi :)[/quote]

$A=\left \{2 ^{2^n}: n \in \mathbb{N} \right \}$ has finite prime divisors, so $A'=\left \{2 ^{2^n}+a: n \in \mathbb{N} \right \}$ doesn't by Kobayashi's.

[b]Iran TST 2009 P2.[/b] Show that \[\forall a\in\mathbb{N}: \not\exists k\in\mathbb{N}: k=\left|\{p\in\mathbb{P}: \exists n\in\mathbb{N}: p\mid 2^{2^n}+a\}\right|.\]

[b]Solution 1.[/b] Assume on the contrary that \[\exists a\in\mathbb{N}: \exists k\in\mathbb{N}: k=\left|\{p\in\mathbb{P}: \exists n\in\mathbb{N}: p\mid 2^{2^n}+a\}\right|.\] And assume that \[\{p\in\mathbb{P}: \exists n\in\mathbb{N}: p\mid 2^{2^n}+a\}=\{p_1,\dots,p_k\}.\] Choose \[r=\max_{1\le i\le k}\left\{\left\lfloor \log_{p_i}{(a^{2^k}+a)}\right\rfloor+1\right\}\] such that \[\forall i\in\{1,\dots,k\}: p_i^r>a^{2^k}+a.\] Choose \[n=\left\lfloor\log_2{\left(\log_2{\left(\prod_{i=1}^k p_i^r-a\right)}\right)}\right\rfloor+1\] such that \[2^{2^n}+a>\prod_{i=1}^k p_i^r.\] Then \[\forall m\ge n: \prod_{i=1}^k p_i^{\nu_{p_i}(2^{2^m}+a)}=2^{2^m}+a>\prod_{i=1}^k p_i^r.\] This means that \[\forall m\ge n:\exists i\in\{1,\dots,k\}:\nu_{p_i}(2^{2^m}+a)>r.\] Now by PP for some $j\in\{1,\dots,k\}$ \[\exists n+s>n+t\in\{n,n+1,\dots,n+k\}: \nu_{p_j}(2^{2^{n+s}}+a),\nu_{p_j}(2^{2^{n+t}}+a)>r.\] Accordingly \[-a\equiv 2^{2^{n+s}}=(2^{2^{n+t}})^{2^{s-t}}\equiv (-a)^{2^{s-t}}\equiv a^{2^{s-t}}\pmod {p_j^r},\] which means \[p_j^r\mid a^{2^{s-t}}+a\le a^{2^k}+a,\] contradicting the prior assumption that \[\forall i\in\{1,\dots,k\}: p_i^r>a^{2^k}+a,\] which establishes our proof.

[b]Solution 2.[/b] Since \[1=\left|\{p\in\mathbb{P}: \exists n\in\mathbb{N}: p\mid 2^{2^n}\}\right|,\] the result follows from Kobayashi's Theorem.

The set of prime divisors of $2^{2^n}$ is infinite (since $\{2^n\mid n=1,2,\ldots\}$ is a subset of this). By Kobayashi's Theorem the result follows.


There are 10 teams in a school district competition. Given that one team plays other only once. What is the total numbers of games played in district competition?

Explain your answer.

[hide]Assume there are teams A-J.  Team A would play B, then C, then D, and on to J.  Thats 9 games.  Team B would then play C, D, and on to J, making 8 games (hes already played with A).  Team C would then play 7 games.  So D 6 games.  E 5, and on and on.

9+8+7...+1 = 45 total games[/hide]

[hide="HINT"] How many teams are we CHOOSING out of 10? [/hide]

[hide]The number of games is equal to C(10,2)=45[/hide]


Let $ n$ be a fixed natural number. Find all natural numbers $ m$ such that

$ \frac{1}{a^n}\plus{}\frac{1}{b^n}\geq a^m\plus{}b^m$

for all positive $ a,b$ such that $ a\plus{}b\equal{}2$

none: try $ a\equal{}\minus{}1$, $ b\equal{}3$. This was on a MOP team contest X.X

[quote="not_trig"]none: try $ a \equal{} \minus{} 1$, $ b \equal{} 3$. This was on a MOP team contest X.X[/quote]

I said "positive" a,b.

I think it's from Bulgaria, but I don't know the solution.

It must be from Bulgaria since I know its solution.
I assume the author is Oleg Mushkarov. He has worked over this problem and its generalization - for $ n$ variables. :)

Let $ b\ge 1\ge a$. Since $ a \plus{} b \equal{} 2$ $ \Rightarrow 1 \minus{} b \equal{} \minus{} (1 \minus{} a)$.

$ \star$ $ \frac {1}{a^n} \plus{} \frac {1}{b^n} \minus{} a^{n \plus{} 1} \minus{} b^{n \plus{} 1} \equal{} \frac {(1 \minus{} a)(1 \plus{} a \plus{} ... \plus{} a^{2n})}{a^n} \plus{} \frac {(1 \minus{} b)(1 \plus{} b \plus{} ... \plus{} b^{2n})}{b^n} \equal{}$ $ \frac {(1 \minus{} a)}{a^nb^n}\sum_{i \equal{} 0}^{2n}(b^na^i \minus{} a^nb^i) \equal{}$ $ \frac {(1 \minus{} a)}{a^nb^n}\left(\sum_{i \equal{} 0}^{n \minus{} 1}(b^na^i \minus{} a^nb^{2n \minus{} i}) \plus{} (a^nb^n \minus{} a^nb^n) \plus{} \sum_{i \equal{} n \plus{} 1}^{2n}(b^na^i \minus{} a^nb^{2n \minus{} i})\right) \equal{}$ $ \frac {(1 \minus{} a)}{a^nb^n}\left(\sum_{i \equal{} 0}^{n \minus{} 1}b^na^i(1 \minus{} a^{n \minus{} i}b^{n \minus{} i}) \minus{} \sum_{i \equal{} n \plus{} 1}^{2n}a^nb^{2n \minus{} i}(1 \minus{} a^{i \minus{} n}b^{i \minus{} n})\right) \equal{}$ $ \frac {(1 \minus{} a)}{a^nb^n}\left(\sum_{i \equal{} 0}^{n \minus{} 1}b^na^{n \minus{} i}(1 \minus{} a^{i \plus{} 1}b^{i \plus{} 1}) \minus{} \sum_{i \equal{} 0}^{n \minus{} 1}a^nb^{n \minus{} i}(1 \minus{} a^{i \plus{} 1}b^{i \plus{} 1})\right) \equal{}$ $ \frac {(1 \minus{} a)}{a^nb^n}\left(\sum_{i \equal{} 0}^{n \minus{} 1}(1 \minus{} a^{i \plus{} 1}b^{i \plus{} 1})(b^na^{n \minus{} i} \minus{} a^nb^{n \minus{} i})\right)$

$ \Rightarrow \frac {1}{a^n} \plus{} \frac {1}{b^n} \minus{} a^{n \plus{} 1} \minus{} b^{n \plus{} 1} \equal{} \frac {(1 \minus{} a)}{a^nb^n}\left(\sum_{i \equal{} 0}^{n \minus{} 1}(1 \minus{} a^{i \plus{} 1}b^{i \plus{} 1})b^{n \minus{} i}a^{n \minus{} i}(b^i \minus{} a^i)\right)\ge 0$ $ \Leftrightarrow \frac {1}{a^n} \plus{} \frac {1}{b^n}\ge a^{n \plus{} 1} \plus{} b^{n \plus{} 1}$.

$ \star$ $ \frac {1}{a^n} \plus{} \frac {1}{b^n} \minus{} a^{n \plus{} 2} \minus{} b^{n \plus{} 2} \equal{}$ $ \frac {(1 \minus{} a)(1 \plus{} a \plus{} ... \plus{} a^{2n \plus{} 1})}{a^n} \plus{} \frac {(1 \minus{} b)(1 \plus{} b \plus{} ... \plus{} b^{2n \plus{} 1})}{b^n} \equal{}$ $ \frac {(1 \minus{} a)}{a^nb^n}\left(\sum_{i \equal{} 0}^{2n \plus{} 1}a^nb^n(a^{i \minus{} n} \minus{} b^{i \minus{} n})\right) \equal{}$ $ (1 \minus{} a)\left(\sum_{i \equal{} 0}^{n}(\frac {1}{a^{n \minus{} i}} \minus{} \frac {1}{b^{n \minus{} i}} ) \plus{} \sum_{i \equal{} n \plus{} 1}^{2n \plus{} 1}(a^{i \minus{} n} \minus{} b^{i \minus{} n})\right) \equal{}$ $ (1 \minus{} a)\left(\sum_{i \equal{} 0}^{n}\frac {b^{n \minus{} i} \minus{} a^{n \minus{} i}}{a^{n \minus{} i}b^{n \minus{} i}} \plus{} \sum_{i \equal{} 0}^{n}(a^{n \plus{} 1 \minus{} i} \minus{} b^{n \plus{} 1 \minus{} i})\right)$ $ \equal{} (1 \minus{} a)\left((a^{n \plus{} 1} \minus{} b^{n \plus{} 1}) \plus{} \sum_{i \equal{} 0}^{n \minus{} 1}(a^{n \minus{} i} \minus{} b^{n \minus{} i})(1 \minus{} \frac {1}{a^{n \minus{} i}b^{n \minus{} i}})\right)$

$ (a^{n \plus{} 1} \minus{} b^{n \plus{} 1}) \plus{} \sum_{i \equal{} 0}^{n \minus{} 1}(a^{n \minus{} i} \minus{} b^{n \minus{} i})(1 \minus{} \frac {1}{a^{n \minus{} i}b^{n \minus{} i}}) \equal{}$ $ (a \minus{} b)(b^n \plus{} b^{n \minus{} 1}a \plus{} ... \plus{} a^n) \plus{} \sum_{i \equal{} 0}^{n \minus{} 1}(a \minus{} b)(b^{n \minus{} i \minus{} 1} \plus{} b^{n \minus{} i \minus{} 1}a \plus{}$ $ \cdots \plus{} a^{n \minus{} i \minus{} 1})(1 \minus{} \frac {1}{a^{n \minus{} i}b^{n \minus{} i}}) \equal{}$ $ \minus{} (b \minus{} a)\left( (b^n \plus{} b^{n \minus{} 1}a \plus{} ... \plus{} a^n) \minus{} \sum_{i \equal{} 0}^{n \minus{} 1}(b^{n \minus{} i \minus{} 1} \plus{} b^{n \minus{} i \minus{} 1}a \plus{} ... \plus{} a^{n \minus{} i \minus{} 1})(\frac {1}{a^{n \minus{} i}b^{n \minus{} i}} \minus{} 1)\right)$ $ < 0$ for some $ (a_0,b_0)$.

$ \Rightarrow \frac {1}{a^n} \plus{} \frac {1}{b^n}\ge a^{n \plus{} 2} \plus{} b^{n \plus{} 2}$ doesn't hold every time.

Now I'll finish with the following obvious statement - $ a^t \plus{} b^t\ge a^{t \minus{} 1} \plus{} b^{t \minus{} 1} \equal{} \frac {a \plus{} b}{2}(a^{t \minus{} 1} \plus{} b^{t \minus{} 1})$ $ \Leftrightarrow a^t \plus{} b^t\ge ab^{t \minus{} 1} \plus{} ba^{t \minus{} 1}$.

$ \Rightarrow m$ is every nature in $ [1;n \plus{} 1]$ .

[quote="BG Yoda"] $ \minus{} (b \minus{} a)\left( (b^n \plus{} b^{n \minus{} 1}a \plus{} ... \plus{} a^n) \minus{} \sum_{i \equal{} 0}^{n \minus{} 1}(b^{n \minus{} i \minus{} 1} \plus{} b^{n \minus{} i \minus{} 1}a \plus{} ... \plus{} a^{n \minus{} i})(\frac {1}{a^{n \minus{} i}b^{n \minus{} i}} \minus{} 1)\right)$ $ < 0$
[/quote]

Why is this?

[quote="jbmorgan"][quote="BG Yoda"] $ \minus{} (b \minus{} a)\left( (b^n \plus{} b^{n \minus{} 1}a \plus{} ... \plus{} a^n) \minus{} \sum_{i \equal{} 0}^{n \minus{} 1}(b^{n \minus{} i \minus{} 1} \plus{} b^{n \minus{} i \minus{} 1}a \plus{} ... \plus{} a^{n \minus{} i \minus{} 1})(\frac {1}{a^{n \minus{} i}b^{n \minus{} i}} \minus{} 1)\right)$ $ < 0$
[/quote]

Why is this?[/quote]

Let $ S \equal{} (b^n \plus{} b^{n \minus{} 1}a \plus{} ... \plus{} a^n) \minus{} \sum_{i \equal{} 0}^{n \minus{} 1}(b^{i} \plus{} b^{i \minus{} 1}a \plus{} ... \plus{} a^{i})(\frac {1}{a^{i}b^{i}} \minus{} 1)$.
Notice that for $ ab \equal{} 1$ then $ S > 0$ !!!. Notice now that if $ ab\rightarrow 0$ then $ S < 0$  :wink: $ \Rightarrow S(ab)$ change its sign in $ (0;1)$.

So the inequality doesn't hold for all $ (a;b)$  :)

However, the mistake is in me indeed, because I missed to add I didn't mean $ \frac {1}{a^n} \plus{} \frac {1}{b^n}\le a^{n \plus{} 2} \plus{} b^{n \plus{} 2}$ holds every time, but it is understood exactly like that. I wanted to say there exist some couples $ (a_0;b_0)$, for which $ \frac {1}{a_0^n} \plus{} \frac {1}{b_0^n}\le a_0^{n \plus{} 2} \plus{} b_0^{n \plus{} 2}$, therefore $ m$ can't be $ n \plus{} 2$.

PS I've fixed it  :D


In the sequence 1,1,2,3,5,8,3,1,4,... each term starting with the third is the sum of the two preceding terms.Addition is done mod 10. Prove that the sequence is purely periodic. What is the maximum possible legth of the period?

once again, you can take a look at [url=http://www.mathlinks.ro/Forum/weblog_entry.php?t=156030]here[/url]. Again, this has something to do with my signature. :wink:

And since the period of the Fibonacci sequence $ \mod 2$ is $ 3$ and $ \mod 5$ is $ 20$, we conclude that the length of the period $ \mod 10$ is $ 3\times 20=\boxed{60}$.

There are an infinite number of ordered pairs $ (a_{n}, a_{n+1})$, but only a finite number of ordered pairs $ (m,n), \; 0 \leq m,n < 10$.

By pigeonhole, one such ordered pair must repeat. Since two specified numbers always have the same sum in a given modulo, it must be periodic.


Romanian selection tests, proposed by Gheorghe Eckstein

Let $1\ge x_1\ge x_2\ge \ldots \ge x_n \ge 0$ be real numbers such that $\sum _{i=1} ^{n} x_i^2=1$.
Prove that the least integer greater than or equal to $\sum _{i=1} ^n x_i$ is $m$ then $\sum _{i=1} ^m x_i \ge 1$.

I think a more general problem was given in Leningrad mathematical olympiad and something tells me this more general problem was already discussed on the forum of inequalities.

Even if it was discussed... Who can find it now?

Unfortunately, I don't know the nice approach :(

Suppose the contrary.
Let $s=\sum_{i=1}^mx_i<1$. First of all, it is clear $m^2<n$, otherwise consider $1>s^2\geq \sum x_i^2=1$ -- contradiction.
Suppose $x_{m+1}=x$, then $1=\sum x_i^2\leq (s-(m-1)x)^2+(m-1)x^2+(n-m)x^2$, where $m\geq s+(n-m-1)x$ and $x\leq s/m$. It gives some estimate on $s$, which is very close to 1.


For $a,b,c$ positive reals such that $abc=1$ prove that

$2\sum a^{2}+12 \geq 3\sum a+3\sum ab$

This is my solution. I was posted it in VI Forum (Vietnam Inequality Forum - [url]http://www.bdt.ch06k.com/[/url]  :)
[quote]I am thankful to you because you take part in VietNam's inequality forum. I hope this forum is useful to you for solve nice and difficult Inequalities ;) For above inequality, it is not hard problem so I can solve it :D
Using classical substitution, we put:
$p=a+b+c,q=ab+bc+ca,r=abc$. The inequality becomes of type:
$2(p^{2}-2q)+12\ge 3p+3q$

we have a well-known inequality ... this is Schur's inequality of second degree: $q\le \frac{p^{3}+9}{4p}$
After that, we will prove that:
$2p^{2}-3p-\frac{7(p^{3}+9)}{4p}+12\ge 0\\ \iff \frac{(p-3)(p^{2}-9p+21)}{4p}\ge 0$
The inequality is true because $p\ge 3$ (by Am-GM), and $p^{2}-9p+21=(p-\frac{9}{4})^{2}+\frac{3}{4}\ge 0$
we are done...Equality occurs for $a=b=c=1$[/quote]


We are given an infinite sequence of $ 1$'s and $ 2$'s with the following properties:

(1) The first element of the sequence is $ 1$.

(2) There are no two consecutive $ 2$'s or three consecutive $ 1$'s.

(3) If we replace consecutive $ 1$'s by a single $ 2$, leave the single $ 1$'s alone, and delete the original $ 2$'s, then we recover the original sequence.

How many $ 2$'s are there among the first $ n$ elements of the sequence?

[i]P. P. Palfy[/i]

This is an interesting question, especially back in the year 1981.

However nowadays it is too easy to find the relevant information [url=http://oeis.org/A189572]on OEIS[/url] (except that here it's using $1, 2$ instead of $0, 1$).
The number of $2$'s among the first $n$ elements is just the partial sum of the above linked sequence.

To get a closed form result, I copy here the most useful information from the above OEIS page:
$a(n) = \lfloor \alpha n + \beta\rfloor - \lfloor \alpha (n-1)+\beta\rfloor$, with $\alpha = \sqrt 2 -1$, $\beta = 1 - \frac{\sqrt 2}2$.

It follows that the number of $2$'s among the first $n$ elements is given by $\lfloor \alpha n + \beta \rfloor$ with $\alpha, \beta$ as above.

This sequence is related to the so called [url=https://en.wikipedia.org/wiki/Hofstadter_sequence]Hofstadter sequences[/url].


algebra
polynomial
number theory unsolved
number theory

Show that the polynomial $ 343x^{5}\plus{}539x^{4}\plus{}336x^{3}\plus{}56x^{2}\plus{}16x\plus{}1$ can never be a power of $ 7$.

$ 343x^{5}\plus{}539x^{4}\plus{}336x^{3}\plus{}56x^{2}\plus{}16x\plus{}1 \equal{} 16x\plus{}1\mod7$ which means $ x \equal{} 3\mod7$ since we want the polynomial to be devisible by $ 7$ in order to ba a power of $ 7$.

assuming $ 343x^{5}\plus{}539x^{4}\plus{}336x^{3}\plus{}56x^{2}\plus{}16x \equal{} 7^{n}\minus{}1$ and noticing that the right side of equasion is always devisable by $ 3$ we can conclude that $ x(343x^{4}\plus{}539x^{3}\plus{}336x^{2}\plus{}56x\plus{}16)$ must also be devisable by $ 3$. we can check by $ mod3$ that $ (343x^{4}\plus{}539x^{3}\plus{}336x^{2}\plus{}56x\plus{}16)$ cannot be devisable by $ 3$ therefore $ x$ must be.

now knowing $ x \equal{} 3\mod7$ and $ x \equal{} 0\mod3$ we can conclude that $ x \equal{} 3,24or45\ mod49$. now we can check by modulo $ 49$ that if $ x \equal{} 3,24or45\ mod49$ than $ 343x^{5}\plus{}539x^{4}\plus{}336x^{3}\plus{}56x^{2}\plus{}16x\plus{}1$ isn't devisable by $ 49$.

therefore the only power of $ 7$ theoretically still posible is $ 7$. it's easy seeing that this also is imposible since $ p(0) \equal{} 1$ and $ p(1) \equal{} 1291$ and polynomial strictly rising after $ x \equal{} 0$

$ 343x^{5}\plus{}539x^{4}\plus{}336x^{3}\plus{}56x^{2}\plus{}16x\plus{}1$ cannot be a power of $ 7$


Find all positive integers N such that the product 2005 x N has exactly 6 divisors.

:) Hi, there.
It seems to be not a very hard problem.

We have 2005 / 5 = 401.

I don't really know, but It's look like 401 is prime.
(Because of my current situation I can't check it, Sorry :-( )
But, It does not matter. Because idea of solution will work
even 401 is not prime.

So, suppose 401 is prime.

2005 has 401, 5 and 1 as divisors.

Hence we have 3 divisors of 2005.

If we choose N as an any prime number except of 1,5,401
then
2005*N = (1*401*5 )*N
has as divisors
 1 , 401 , 5 , 401*5, 401*N, 5*N
a total number exactly 6.

But, You have to check whether 401 is prime.
I hope so :oops:

errr...
[hide]if i'm not mistaken $2005$ divides $2005$ as well :P
so we have: $2005=5*401$
which gives us $(1+1)(1+1)=4$ divisors...
so if we want six, we can't multiply with a random prime; 'cause then we'ld get $(1+1)(1+1)(1+1)=8$ divisors
so we should multiply with either $5$ or $401$ to make $(2+1)(1+1)=6$ divisors...

in case you didn't know: $x=m^an^bp^c$ (with $m,n$ and $p$ prime) has $(a+1)(b+1)(c+1)$ divisors... you can still expand that of course ;)[/hide]

I'd go about it like this:
[hide] 6 = 1*6, 2*3

So, the number must be of the form a^5 or a*b^2

Since we already have 2005, it can't be a^5

2005= 5*401, so it must be 5*401^2 and 401 * 5^2, which gives 804,005 and 10,025[/hide]

:oops:

N = 5 , seems to be unique answer.

The divisors of 2005*5 are
 1, 5, 25, 401, 2005, 2005*5.


\[
\begin{array}{l}
 Avem(x_n )_{n \ge 0} sir,a \in R. \\
 Demonstati:\mathop {\lim }\limits_{n \to \infty } (x_n ) = a \Leftrightarrow \mathop {\lim }\limits_{n \to \infty } (4x_{n + 2}  - 4x_{n + 1}  + x_n ) = a \\
 \end{array}
\]

Incearcă să nu mai bagi tot textul in formatul $\LaTeX$. Uite postul tău editat:

[quote="silviu"]
 Avem $(x_n )_{n \ge 0}$ un şir, $a \in \mathbb{R}$.
 Demonstraţi că: \[ \lim_{n \to \infty } x_n  = a \Leftrightarrow \lim_{n \to \infty } (4x_{n + 2}  - 4x_{n + 1}  + x_n ) = a \] [/quote]

Excuse, I don't know Romanian.
Consider $y_n=2x_{n+1}-x_n$. Then $\lim_{n\to\infty}(2y_{n+1}-y_n)=a$. It folows that
\[\lim_{n\to\infty}\left((2y_{n+1}-y_n)+\frac{1}{2}(2y_{n}-y_{n-1})+\frac{1}{4}(2y_{n-1}-y_{n-2})+...\right)=2a,\]
i.e. $\lim 2y_n=2a$, so $\lim (2x_{n+1}-x_{n})=\lim y_n=a$. Apply the same trick to obtain result.


Let
$ a_{11} x_1 \plus{} \cdots \plus{} a_{1n}x_n \equal{}0$
$ \cdots$
$ a_{r1} x_1 \plus{} \cdots \plus{} a_{rn}x_n \equal{} 0$

be $ r$ equations in $ n$ unknowns with $ n>r$ and with $ a_{ij}$ integers and

$ |a_{ij} | \le A$ for all $ i,j$ Then there are integers $ x_1 , \cdots , x_n$

not all zero, such that $ |x_j | \le 2(2nA)^{r/(n\minus{}r)}$ $ (1 \le i \le n )$

Prove this one using Pigeon hole

Hmm..I think the bound can easily be reduced to $ (2An)^{\frac {r}{n-r}}$ :wink:

Let $ k>0$ be an integer.
Let's consider all the $ n$-uples $ (x_1,\cdots,x_n)$ where $ x_i \in \{1,\cdots,k\}$ for each $ i$.
There are $ k^n$ such $ n$-uples.

Each of them, when substituing in the given system, leads to a $ r$-uple $ (b_1,\cdots,b_r)$, where $ b_i = \sum_i a_{i,j}x_j$.
Note that $ |b_i| \leq \sum_i |a_{i,j}\parallel{}x_j| \leq nAk$. (1)
Moreover, if for some $ n$-uple $ (x_1,\cdots,x_n)$ we may obtain the value $ b_i = nAk$ then $ a_{i,j}=A$ for each $ j$. Since the $ x_j$ are positive, it means that no $ n$-uples will give the value $ b_i = -nAk$. Using (1) and the last remark, it follows that there are at most $ 2nAk$ possible values for each $ b_i$ so that there are at most $ (2nAk)^r$ possible $ r$-uples.

If $ (2nAk)^r < k^n$ (*) then, from the pigeon-hole principle, we may find two $ n$-uples, say $ (x_1,\cdots,x_n)$ and $ (y_1,\cdots,y_n)$, which gives the same $ r$-uple. Using the linearity of the LHS of each of the equations of the given system, it follows that $ (z_1,\cdots,z_n)$ is a non-trivial solution of the system, where $ z_i = x_i - y_i$ for each $ i$.
Note that, for each $ i$, since $ x_i,y_i \in \{1,\cdots, k\}$ it follows that $ z_i$ is an integer with $ |z_i| < k$.

Now, for $ k=[(2nA)^{\frac {r}{n-r}}]+1$, we have $ k>(2nA)^{\frac {r}{n-r}}$, which easily leads to (*).
And $ |z_i| \leq k-1 \leq (2nA)^{\frac {r}{n-r}}$, as claimed.

Pierre.


Yea....
I was thinking that if people like llamel, mewto, fantasy, and all the other people who are above 1450 should get like felt like first round byes.
If there were, say 10 others that joined, we could have a quick first round of single elimination. There would be around 8 people left. Then, the tourney would continue as regularly, single elimination until the final match, which in my opinion should be 3 rounds, but I would appreciate feedback for this rating grabbing scheme.

[=P]eace.

WOW you managed to double post a thread.
I will join with 1402.

superllamel is on vacation.

i think he wont be back until like the 20th.  :P  :P  :P

Ernie is right.

I can't join because I'm on vacation and have no internet access.  :D

idk if i should
i'm gonna lose a lotta rating if i join this...
o well
i'll join with highest rating = 1464

u know...
u really should ad this tourny
like i'm advertising mine in urs
LOOK AT MY SIG EVERYONE!!!


[hide]
Simplifying gives:
$\frac{1}{6} - \frac{1}{20} - \frac{1}{42} - ... - \frac{1}{9702}$
Dont know what else to do though.
I'll try again later when I have more time.
[/hide]
[/hide]

[quote="AntonioMainenti"]I think we would need to find the general $n\text{th}$ partial sum. For this problem we just need the $99\text{th}$ partial sum, which can also be written as \[\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{98}\right)-\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\ldots+\frac{1}{99}\right).\]Maybe there is no explicit formula for the partial sums? Is that possible? I don't really know much about the theory of series..[/quote]

From what I know, there is no good formula for sum of a harmonic series.
But I would think that some stuff should cancel out in here.
I'll post if i get anything worthwhile

There is no nice formula for the partial sums. There are, however, good estimates.

Doesn't this equal $\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+...+\frac{1}{9,702}$ whatever that is?

As jmerry says, there is no nice formula for the partial sums.  We can approximate them, however, by $1 - \ln 2$.  (That's the limit if you continue the series forever.)  Is that good enough for the original poster?


A piggy bank has 1.05 dollars in coin change. Figure out which coins are in the bank, if you know that the bank

cannot make change for a nickel

cannot make change for a dime

cannot make change for a quarter

cannot make change for a half dollar

cannot make change for a dollar

and none of the coins is a dollar

[hide] Since of none of the coins is a dollar coin, we need some quarters and half dollars to fall into place (if we use nickels or dimes, we will be able to make change eventually). Since $ 2$ half dollars make a dollar, we must have $ 1$ half dollar. Since $ 2$ quarters make a half dollar, we must have $ 1$ quarter. This is a total of $ 75$ cents, so we need $ 30$ cents more. This can be accomplished with $ 3$ dimes, which satisfies both requirements (the amount of money and the no change requirement) So the bank has $ 1$ half dollar, $ 1$ quarter, and $ 3$ dimes.[/hide]


Find the sum of all integers between 1 and 9999 that are divisible by 5 but not by 6 nor 7.

This might be intermediate...

We use the inclusion-exclusion principle here. First we find all the multiples of 5, subtract all the multiples of 30, subtract multiples of 35, then add multiples of 210.

1999 - 333 - 245 + 47 = 1468.

He asked for the sum of all such numbers.

Ah. Then you would just do the same thing...

Add the sum of the first 1999 multiples of 5,
Subtract the sum of the first 333 multiples of 30,
Subtract the sum of the first 245 multiples of 35,
and add the sum of the first 47 multiples of 210.

5(1999)(2000)/2 - 30(333)(334)/2 - 35(245)(246)/2 + 210(47)(48)/2
9995000 - 1668330 - 1054725 + 236880
= 7508825.


Let $ x;y;z$ be the sides of a triangle.Prove that:
$ \frac{x}{\sqrt[3]{y^3\plus{}z^3}}\plus{}\frac{y}{\sqrt[3]{z^3\plus{}x^3}} \plus{}\frac{z}{\sqrt[3]{x^3\plus{}y^3}} < 2\sqrt[3]{4}$

$ \sum{\frac {x}{\sqrt [3]{y^3 \plus{} z^3}}} \equal{} \frac {1}{\sqrt [3]{2}}\sum{x\sqrt [3]{\frac {2}{y^3 \plus{} z^3}}} \leq \frac {1}{\sqrt [3]{2}}\sum{2\frac {x}{y \plus{} z}} \leq \frac {3}{\sqrt [3]{2}} \equal{} \frac {3\sqrt [3]{4}}{2} < 2\sqrt [3]{4}$

Where the first inequality is by the power mean, the second inequality by Nesbit's, Cauchy...

I think this is true for all positives...

 :)

Your solution is wrong.  $ \sum_{cyc}\frac {x}{y \plus{} z} \equal{} \frac {1}{2}\sum_{cyc}(x \plus{} y)\sum_{cyc}\frac {1}{x \plus{} y} \minus{} 3\geq \frac {3}{2}$.

[quote="chien than"]Let $ x;y;z$ be the sides of a triangle.Prove that:
$ \frac {x}{\sqrt [3]{y^3 \plus{} z^3}} \plus{} \frac {y}{\sqrt [3]{z^3 \plus{} x^3}} \plus{} \frac {z}{\sqrt [3]{x^3 \plus{} y^3}} < 2\sqrt [3]{4}$[/quote]
By Holder inequality, we have:
$ \sqrt[3]{4(y^3\plus{}z^3)} \equal{}\sqrt[3]{(1^3\plus{}1^3)(1^3\plus{}1^3)(y^3\plus{}z^3)}\ge y\plus{}z$
So:
$ \frac{1}{\sqrt[3]{4}}(\frac {x}{\sqrt [3]{y^3 \plus{} z^3}} \plus{} \frac {y}{\sqrt [3]{z^3 \plus{} x^3}} \plus{} \frac {z}{\sqrt [3]{x^3 \plus{} y^3}}) \le \sum \frac{x}{y\plus{}z}$
Because $ x<y\plus{}z,y<z\plus{}x,z<x\plus{}y$:
$ \frac{x}{y\plus{}z}<\frac{2x}{x\plus{}y\plus{}z} (\Leftrightarrow x<y\plus{}z)$
Similarly, we have:
$ \sum \frac{x}{y\plus{}z} <\frac{2(x\plus{}y\plus{}z)}{x\plus{}y\plus{}z}\equal{}2$
We are done!


probability function calculus geometry integration

This problem is fairly well-known, so don't post a solution if you have seen it please.

Suppose you have parallel lines drawn on a table, and the distance between two consecutive lines is d. Now suppose you have a needle of length l<d. If you drop the needle, what is the probability that it will land on one of the lines?

I was thinking along the lines of expressing probability as a function, most likely a circle, then... use calculus to find the average probability in terms of the area of the circle (semicircle, maybe)

hm .... ok i havent' seen this before...

I guess i have a way to do it using calculus, as mentioned above, as for how to do it without calculus is beyond me....

[hide]

so, needle of length l falling between lines with spacing d.

:theta: is the angle that the needle falls at, with respect to the parallel lines, so theta can be anything, and there's an equal probability of each theta occuring...

for a given theta i want to find how close to the lines the center of the needle can be, in that case.. one of the endpoints of the needle will be on one of the lines...., so the center of the needle would be (l sin :theta: )/2 away from the line....

picture 2 parallel lines going horizontally.... the center has to be at least (l sin  :theta: )/2 above the bottom line.. and (l sin  :theta: ) / 2 below the top line to not hit the lines at all...

the center of the needle has an equal chance to be anywhere between the bottom or top lines..... so the chance that it lands on one of the lines is l sin  :theta:

theta can be anything, but it can be assumed to be between 0 and  :pi: /2 to make things easier....

so p ( :theta: ) = l/d * sin  :theta:  and theta goes from 0 to  :pi: /2

so the average value of anything in calculus land is...

1  / (b-a ) * integral from a to b of f(x) dx

in this case.... the average probability is...

1 / ( :pi: /2 - 0) * integral from 0 to  :pi: /2 of l/d * sin  :theta: , d :theta:

so that's 2 /  :pi:  * l/d * [-cos  :theta: ]from 0 to (:pi: / 2)

2 /  :pi: * l / d * -1 * (cos  :pi: /2 - cos 0)

2 /  :pi: * l / d * -1 * ( 0 - 1)

= (2 l /  :pi: d)

[/hide]

hm... anyone know if that's right?

yay!

care to explain the non-calculus way now?

I don't know of a non-calculus way. That's why I posted it in advanced problems. As far as I'm concerned, good problem-solving ith calculus is perfectly legitimate, but calculus problems probably don't belong on the other boards.

[quote="ComplexZeta"]I don't know of a non-calculus way. [/quote]

There is a nice one in "Proofs from The Book" (very nice book, highly recommended). It uses the linearity of mean (in probability) the show that it's possible to use a circle instead of a straight needle.

Ooh. That is a clever proof. For anyone who wants an additional challenge, what happens if l>d? The answer is unfortunately not very nice.

hmm ok.. so L > d
....
that would mean that for some  :theta: (angle of the needle with respect to the lines), there are no solutions...

picture the needle having both endpoints on parallel lines that are d apart, with angle  :theta:

so  sin  :theta:  =  d / L
 :theta:  = arcsin (d / L)
this is the max  :theta:  for given values of L and d

so P( :theta:  ) is still ( L sin  :theta: ) / d when  :theta:  is in an acceptable range, that is... 0 <  :theta:  < arcsin (d / L)

but when arcsin (d / L) <  :theta:  <  :pi: /2 ..... P( :theta: ) = 0....

we still want to find the average value of P( :theta: ) between 0 and  :pi: /2 .... so to do that.... we integrate P( :theta: ) from 0 to  :pi: /2 , and multiply by 1/ ( :pi: /2 - 0 )

this is the same as integral of...

(L sin  :theta: ) / d from 0 to arcsin (d/L) *upper limit of theta for given L and d*

turns into....

L / d * (- cos  :theta: ) [from 0 to arcsin d/L)

L / d * (- cos (arcsin d/L) + cos 0)

L/d * (- :sqrt: ((L^2 - d^2) / d)) + 1)

( L -  :sqrt: (L^2 - d^2) ) / d .... hm....

oh yeah... multiply by 2/  :pi:

and so i get an answer of...

[b] p = 2 * :sqrt: (L^2 - d^2) / (  :pi: * d) [/b]

hm.. that's pretty ugly....

some simple evalutations of it...

if L = d = 1....

p = 2 * 1 / ( :pi: * 1) = 2 /  :pi:  (as expected from my earlier post)...

if L = 2, d = 1... u get....

p = (4 - 2  :rt3: ) /  :pi:  which is....roughly p = .17 or so, i guess the formula makes sense.. anyone know if its right?[/b]

How can the probability be less than 2pi if l>d? Somehow your answer is very close to the probability that it doesn't hit a line though. It doesn't make senes to me that you ended up with a d that isn't in the form l/d since if we multiply everything by a constant, the probability should remain the same.

Simon, all of the d's are in l/d form -- :sqrt:(l2 - d2)/d = :sqrt:((l/d)2 - 1)

Ha. You're right. For some reason I thought the d in the denominator would stay there. Nice catch.


Euler
complex numbers
advanced fields
advanced fields unsolved

My previous question too dumb to be real.  Found answer.

This question:  Using three dimensions  y;  x-real;  x-imaginary;  draw graph of y=(-1)^x.  What dors it look like?

Largely irrelevant question:  What does graph of  e^ix  look like in same three dimensions?

Ähmm... your graph will have four, not three dimensions (or you must look at $\Re (y)$ and $ \Im (y)$ ), and you will first have to define powers to complex numbers accuratly (see your other tread on that topic)

Indeed.  Even my question containeth mistake.  Question One from double-question post should be:

Fop x real,  y complex,  graph y=(-1)^x

Second question should be framed similarly.

Your "largely irrelevant" question is actually the same question. We would graph $(-1)^z$ by first choosing $t$ such that $(-1)=e^t.$ This $t$ could be $\pi i,$ and would certainly be a pure imaginary number. Then we would define $(-1)^z=e^{tz}.$

You would still need four dimensions for the graph, and I don't believe there are mant people who can visualize a four-dimensional object.

The slice of the graph for $z$ real would be a helix.

Thanx for very clear answer.  But wait.  You say 3D slice for x-real is helix.  Is SEVERELY discontinuous helix??  Like y=(-1)^(1/10)  has ten roots and very *next* value of x real  ((which is WHAT??))  has ???  roots?

The helix I was referring to is a perfectly ordinary smooth helix, but it's only one of the branches. If you superimpose all of the branches, you get a bizarre tube of "fibers", each wrapped around the cylinder at different rates, and collectively dense in the cylinder.

The helix is given when using jmerry's definition as all tripletts $(t,\cos(t),\sin(t))$, since the formula simply reduces to eulers formula!


superior algebra
superior algebra unsolved

Let H be a subgroub of G.

Prove that the number fo left cosets = the number of right cosets.

Consider the bijection $ g\mapsto g^{\minus{}1}$.

[b]Hi there, mlok,

Shouldn't it be something like[/b] $ gH \mapsto Hg^{ \minus{} 1}$ [b]?[/b]

[b]All the Best Regards,[/b]

it's the same (variables are not fixed! and "$ (gH)^{\minus{}1}\equal{}H g^{\minus{}1}$").


So, i was thinking about the problems of math olympiads. I'll elaborate. The math olympiads concentrate on the creative part of mathematics. The olympiad problems require a "flash" of the mind, an outstanding idea coming from nowhere, an insight to combine this and that and this way make it work!! But, the thing is that through the reasoning at these olympiad problems you can only solve olympiad problem and olympiad problems are "made up". The hard part of them is identifying the right idea or the right trick. From the time you do this you have finished, because the problem has been set up in such way that the trick will work. But if one changes some things provided by the problem many times the problem gets exponentially harder. Instead of speaking abstractly i'll give an example. Consider the following olympiad problem:

[i]There are $ 10001$ students at an university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There is a total of $ k$ societies. Suppose that the following conditions hold:

i.) Each pair of students are in exactly one club.

ii.) For each student and each society, the student is in exactly one club of the society.

iii.) Each club has an odd number of students. In addition, a club with $ 2m\plus{}1$ students ($ m$ is a positive integer) is
in exactly $ m$ societies.

Find all possible values of $ k$.[/i]

This problem is not that difficult. But if we change the sentence [i]"In addition, a club with $ 2m\plus{}1$ students ($ m$ is a positive integer) is in exactly $ m$ societies"[/i] with [i]"In addition, a club with $ 2m\plus{}7$ students ($ m$ is a positive integer) is in exactly $ m$ societies"[/i] the problem gets much-much harder(can you solve it?). So, what i am trying to say is that mathematics is a jungle. And in every jungle most parts are cruel and unapproachable, but there are also some beautiful places(with nice ponds etc). Olympiads of mathematics present only these places but fail to present the greatest parts of the jungle.

I'd like to hear everyone's opinion...

Well, yes, but Olympiads don't have time to present such questions and that's not the purpose of Olympiads.  I don't think anyone ever claimed that Olympiad math represents all of math.

On perusing your post, I get the impression that you're suggesting how to approach or solve IMO problems (or whatever), it would be much better if you delineate your views. What would like us to do? Listen to your views on -"How to solve Math problems at the Olympiad" or debate (I don't see any topic to debate upon  :lol: ) or provide our views?

I also can't understand on which aspect should we present our views/opinions upon, is it on- How to approach  IMO problems, or on corroborating your proposition that "Math is a jungle with cruel and nice parts" ( I feel this a weak analogy).

But, I agree that most IMO problems require "insight" but these can be developed with a great amount of practice.

Olympiad math aims to get people interested in math. I think it succeeds admirably. You don't get people interested by showing the ugly parts; you show the beautiful parts.

There are some olympiad problems that touch on more complicated ideas, and in which the details don't work out so nicely. Vietnam seems to be particularly fond of them.

I think that you must also remember that if you call math olympiad "jungle," it often is a "jungle" with more than one opening for sunlight. The true beauty of olympiad problems isn't solely on the insightful solution. It is the presence of more than one good solutions in many cases (there are, of course, some problems in which only one solution exists).

OK, let me explain my thoughts further.

In the olympiad problems the path to a solution usually is:
1) Identify the basic insight and understand the problem
2) Select an approach to use (ex. induction, contradiction etc.)
3) Start solving
From the time you do so most likely your technique will work, because as "silverfacon" said there are many solutions to olympiad problems. Olympiad problems are made up in such a way that identifying the insight is the hard part and they promote creativity. But in most problems(not olympiad) identifying an insight does not always give a solution many times you should use brute force or design an algorithm and enter it to a computer to help you solve the problem and other times the problem appears unsolvable.

I mean that such problems as the the new version of the former olympiad (with $ 2m\plus{}7$) i stated in my previous post can not be solved with olympiad methods. However, i do agree that olympiads achieve to get students interested in math.

Really, if anyone solve the (with the $ 2m\plus{}7$) version of the problem in my previous post i'd like to see the solution (maybe i should move it to the combinatorics section :?: )

[quote="filosofimenos"]Olympiad problems are made up in such a way that identifying the insight is the hard part and they promote creativity. But in most problems(not olympiad) identifying an insight does not always give a solution many times you should use brute force or design an algorithm and enter it to a computer to help you solve the problem and other times the problem appears unsolvable. [/quote]
I think this is widely recognized.  What's your point?  (My own pet example of this phenomenon is [url=http://mathworld.wolfram.com/Self-AvoidingWalk.html]self-avoiding walks[/url].)

It would probably be appropriate to post the part of your post that is actually a math problem in the relevant part of the forum.  What follows is some rambling about broad similarities between olympiad and research mathematics.

Mathematical olympiads ask you to solve problems in a fixed amount of time with a fixed set of resources, and so they go out of their way to actually provide problems that are tractable within these limitations.  One natural consequence of this is that problems on olympiads tend to be non-open and have relatively short solutions.  Normal mathematicians also restrict themselves to problems tractable with a given collection of resources, but it happens that the constraints here are much broader (many person-months or -years, as much computing power is available and so on) so that the class of problems solved is correspondingly larger.  But it is still the case that much new mathematics conforms to some rather tight boundaries -- for example, papers longer than 200 pages are quite rare, as are papers with more than a handful of authors.

I'm relatively early in the portion of my life in which I'm comfortable calling myself a mathematician as well as a student of mathematics, but already I have had the opportunity to consider solving a decent variety of problems (within my own little corner of combinatorics).  From among these, I am always looking for those that combine tractability with interest (where both are relative to me -- my abilities and my tastes, respectively).  There are a great many problems on which I've [i]not[/i] spent my time, either because I believed them not very interesting (even if solvable) or because I didn't believe that I could solve them (even if I thought that knowing the solution would be very interesting).*  I don't think this is terribly different from olympiad problem-solving, really, except that I take a larger role in choosing the problems I attempt than olympiad competitors do.

* Of course, to eventually be "successful" as a research mathematician, you must solve problems that [i]other[/i] people think are interesting, and to be "very successful" you must solve problems that other people think are both interesting and difficult.  I put some terms in quotation marks because their definitions are far from universal.


vector
abstract algebra
superior algebra
superior algebra unsolved

[quote="withoutwarning"]Find a Vector space Y such that the map f: (V_1/W_1)X(V_2/W_2)->(V_1 tensor V_2)/Y (where V_1, V_2 are arbitrary vector spaces, and W_1, W_2 are subspaces of V_1 and V_2, respectively) is a bilinear, well defined map.

What I'm having a lot of trouble with is composing this vector space Y such that the map f is well defined.

Any help would be greatly appreciated[/quote]

Take Y = V1 ⊗ V2.

[quote="withoutwarning"]If you do that though, then (V_1 tensor V_2)/Y =0.[/quote]

Exactly, and any map to 0 is well-defined, since you send everything to 0.

Oh sorry I guess I didn't phrase my question correctly. It's supposed to be find a vector space Y contained in (V_1 tensor V_2) such that
f:(V_1/W_1)X(V_1/W_1)->(V_1 tensor V_2)/Y is a tensor product of (V_1/W_1)X(V_2/W_2). Sorry about that.

You mean, you want to show that V1/W1 ⊗ V2/W2 is isomorphic to (V1⊗V2)/Y for some subspace Y≤V1⊗V2?

Define a homomorphism F from V1⊗V2 to V1/W1 ⊗ V2/W2 by defining a bilinear map f from V1×V2 to V1/W1 ⊗ V2/W2 by f(v1,v2) = (v1+W1) ⊗ (v2+W2).  Since the image of f contains a generating set of V1/W1 ⊗ V2/W2, the homomorphism F is surjective.  Just take Y=ker(F) and use the first isomorphism theorem.