I competed in the Silver division, and once again failed to have a perfect score due to a character error.

My code in outofhay read:  weight[y][z] = z;
When it should have read: weight[y][x] = z;

So I'm guessing around a 700 score (Perfect on 1 and 2, and 4/11 test cases on 3).

Feel free to ask about any problems.  I've solved all silver, and I think all of the gold but I haven't put them to the test yet.

how do you have access to both silver and gold problems?

Since the contest is over, the grading program is put into Analysis Mode.  Anybody with a username and password can log in at ace.delos.com/ioigate and look at gold, silver, and bronze problem sets.

I'm silver... perfect on #2 and #3 (in fact i used almost the same algorithm on both) but for some reason timed out in 3 test cases in #1... dang, close to perfect.

Well, I participated in Gold. The 2nd and the 3rd were easy. I had no idea how to do the 1st. Can someone give me some direction?

got 7/11 on outofhay, perfect for yogurt and alibi.

which was the most difficult prob for all, i think yogurt was easiest , but have no clue why then they included it in Gold too.

alibi was dijkstra's and outofhay was prim's, right paladin8?

I would say the most difficult problem was definitly the first one in Gold.  All the others were much simpler algorithms.

And yes I used Dijkstra and Prim's on 2 and 3 for silver.  You probably timed out on yogurt because you did a O(n 2 ) algorithm which is too slow for the input size.  Actually I believe the correct algorithm is still considered O(n 2 ), but it has a key optimization.  I'll just tell you that you don't necissarily need to check every previous week every time.

Why would you do O(N^2) when O(N) is almost as easy to think of and equally easy to code and you don't even have to worry about optimisation  :)

All my cases for yogurt ran <= 0.02s

Here's my code for yogurt:
-----------------
TotalCost = 0
Get NumWeeks and CostPerWeek
Minimum = MAXINT - 1 - CostPerWeek
For Each Input do
{
    Minimum = Minimum + CostPerWeek
    Get DailyCost , Requirement
    If(DailyCost < Minimum){

                          Minimum = DailyCost
    }
    TotalCost += Requirement * Minimum;
}
-------------------------

Is this solution O(N)?

That is very similar to my program, but is a little faster (my slowest was .05 s).  The reason I chose O(N 2 ) was because it was the first thing that came to my mind and I knew it would satisfy the time limits.

Yeah I realized that I could make it a lot faster after the contest. Probably could've gotten a perfect score... dang.


modular arithmetic
number theory proposed
number theory

Show that the equation
\[ a^4\plus{}2b^4\equal{}c^4\]
has no solutions in positive integers.

It is even true that the equation $ x^4\plus{}2y^2\equal{}z^4$ has no solutions in positive integer numbers. But I can't prove it now. I know it could be proved using the method of infinite descent.

Could anyone please post the proof of $ a^4 \plus{} 2b^2 \equal{} c^4$ having no non-trivial integer solution using infinite descent? I have been trying to find it for a long time but all I found is the following proof.

Let us consider the solutions where $ (a,c) \equal{} 1$. This implies $ (a,b,c)$ is primitive. First we have the following lemma: all primitive integer solutions to $ a^2 \plus{} 2b^2 \equal{} c^2$ are given by $ a \equal{} \pm (n^2 \minus{} 2m^2),b \equal{} 2nm,c^2 \equal{} n^2 \plus{} 2m^2$, where $ (n,m) \equal{} 1$. (For a proof see [url=http://www.mathlinks.ro/weblog_entry.php?t=201834]here[/url].)

Therefore in our problem we conclude that $ a^2 \equal{} \pm (n^2 \minus{} 2m^2), b \equal{} 2nm, c^2 \equal{} n^2 \plus{} 2m^2$, where $ (n,m) \equal{} 1$. Since $ (a,b) \equal{} 1$, $ n$ must be odd. Hence $ c$ is odd and thus $ c^2 \equal{} n^2 \plus{} 2m^2\equiv 1\pmod 8$ implies $ 2\mid m$. Now $ a^2 \equal{} 2m^2 \minus{} n^2$ implies $ a^2\equiv \minus{} 1\pmod 4$, contradiction. Hence $ a^2 \equal{} n^2 \minus{} 2m^2$.

From the last equations we get $ c^2 \plus{} a^2 \equal{} 2n^2$, $ c^2 \minus{} a^2 \equal{} 4m^2$. Hence $ (c \plus{} a)(c \minus{} a) \equal{} 4m^2$ implies $ c \plus{} a \equal{} 2x^2,c \minus{} a \equal{} 2y^2$, as $ (c \plus{} a,c \minus{} a) \equal{} (c \plus{} a,2c) \equal{} 2$. Then $ (2n)^2 \equal{} 2(c^2 \plus{} a^2) \equal{} (c \plus{} a)^2 \plus{} (c \minus{} a)^2 \equal{} 4x^4 \plus{} 4y^4$ which implies $ x^4 \plus{} y^4 \equal{} n^2$, which is well known to have no non-trivial solutions in integers.

Nice solution!
Thank you for the lemma about the equation $ a^{2}\plus{}2b^{2}\equal{}c^{2}$!


Prove, using basic geometry that if a triangle has two equal angle bisectors than the triangle is necessarily isoceles.

well this the well-known steiner-lehmus theorem,
this is a known proof-
Construct G such that AEGD is a parallelogram.

∴AD=BE, ∠CAD=∠DAB, ∠CBE=∠EBA.

But AD=EG ∴EG=EB ∴ΔBEG is isosceles ∴ ∠EGB=∠EBG.

Suppose AC≠BC, w.l.o.g., AC < BC.

∴∠CAB > ∠CBA ∴∠DAB > ∠EBA.

∴ DB > EA (*)

But ∠EGD=∠EAD=∠DAB > ∠EBA=∠DBE.

So ∠DGB < ∠DBG ∴ DB < GD=EA contradicting (*)

Likewise the assumption AC > BC leads to a contradiction.

∴ AC=BC ∴ΔABC is isosceles.

I don't follow how $ AD\equal{}EG$ if $ ADEG$ is a parellogram. Did you make it a square?
Also, how would you construct the paralleogram, internal or external or is it fixed by your criteria?

Use contradiction--prove that the angle bisectors can't be equal if the triangle is not isosceles.

I've tried that, it didn't go anywhere.
Could you post a list of checkpoints I should follow to prove this problem?

Well, I believe [url=http://forumgeom.fau.edu/FG2005volume5/FG200525.pdf]this[/url] is the proof geniusbliss mentioned.

yes math154, thanks,

[hide]i saw the mistake in the proof i gave so i tried to edit my post and as i finished everything
suddenly my pc got switched off and moreover i was having trouble with geogebra[/hide]


In a triangle $ ABC$, $ M$ is the midpoint of $ AB$ and $ P$ an arbitrary point on side $ AC$. Using only a straight edge, construct point $ Q$ on $ BC$ such that $ P$ and $ Q$ are at equal distance from $ CM$.

$ BP\cap MC \equal{} \{I\}, AI\cap BC \equal{} \{Q\}$.
[hide="Note"]Use lemma about trapezoid and [b]Thales[/b]'s theorem[/hide]


vector
abstract algebra
group theory
superior algebra
superior algebra unsolved

$A=\{m2^{n}|m, n\in Z\}$  is  not  finitely generated group

[quote="dhdung"]$A=\{m2^{n}|m, n\in Z\}$  is  not  [color=blue]finitely generated[/color][/quote]

What do you mean? as ideals, moduls, vector spaces, etc ?

Suppose it were ideals. Because $A$ is trivially finitely generated as an ideal in $A$, the ring would be bigger than $A$. But since $A$ contains 1, this is not possible. As for modules, the only reasonable ground rings would be $A$ and $\mathbb Z$, and $A$ has been ruled out before, leaving us with $\mathbb Z$. $A$ is not a $\mathbb Q$-vector space, but obviously no vector space over a finite field, hence no vector space at all. Finally, "finitely generated" could mean "as an algebra", but since $A$ is obviously f.g. as a $\mathbb Z$-algebra, this is also impossible. Hence, the question asks for being finitely generated as a $\mathbb Z$-module, or, equivalently, as an abelian group. Q.E.D. owk

@owk: LOL ... you forgot the monoids ;-)
@dhdung: every element of $A=\mathbb{Z}[1/2]$ can be uniquely written in the form $a/2^{n}$, where $a \in \mathbb{Z}$ odd and $n \in \mathbb{N}$. check that $1/2^{n}$ does not lie in the subgroup generated by a finite number of elements $a/2^{i}$ where $i<n$.

Thanks you very much , I understand your idea. Thanks alot.

how about problem 1 in the following link? Prove G is generated by $x$ and $1$
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=136063[/url]


Given that $ 12a\plus{}10b \equal{}1020$, what is the value of $ \frac{a}{5}\plus{}\frac{b}{6}$?

How do you get the answer? Apparently, it's 17

Noting that $ 12\cdot 5\equal{}10\cdot 6\equal{}60$, dividing both sides of the equation by $ 60$ gives us \[ \frac{a}{5}\plus{}\frac{b}{6}\equal{}\frac{1020}{60}\equal{}\boxed{17}\]

An easier way is that if you noticed that 1020 was divisible by 12, then you'd get 85. And 85/5 = 17. :wallbash:


For a,b be non- negative real numbers that satisfy $ a^2 \minus{} 2ab \plus{} a \minus{} 2b \geq 0$
Find the greatest value of M=$ a^2 \minus{} 5b^2 \plus{} 3a$

The problem can be solved with the little bit change in condition as follows

\[ a^2\minus{}2ab\plus{}a\minus{}2b \ge 0\]

\[ \Leftrightarrow (a\plus{}1)(a\minus{}2b) \ge 0\]

Hence, $ a \ge 2b \ \forall a,b \in R$.

Rearrange the $ b^2$ in terms of $ a$ and $ M$ , we have

\[ a^2\plus{}3a\minus{}M \equal{} 5b^2\]

Compare with the obtained condition, we get

\[ a^2\minus{}12a\plus{}4M \ge 0 \ \forall a \in R\]

Thus, $ \Delta \le 0 \Rightarrow  M \ge 9$

[u][i][b]Remark[/b][/i][/u]. There is no largest value since if a comes to infinity then the condition is satisfied but not maximum value of M


It is wrong for $ x\equal{}y\equal{}100$,  $ a\equal{}1000000$.

[quote="Myth"]It is wrong for $ x \equal{} y \equal{} 100$,  $ a \equal{} 1000000$.[/quote]
I'm sorry for it, a = 8

[quote="frankvista"]Can anyone prove it[/quote]

http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=48&Id=12617


calculus
integration
LaTeX
function
logarithms
calculus computations

Hello
Well I m not getting the question hope this post would help u in someway.
 [hide="May be the answer if I have got ur question correctly"]
Is ur question:
$ \int\frac {dx}{(x \minus{} 2)\sqrt {x^2 \minus{} 1})}$?
[/hide][hide="answer"]
just put $ x \minus{} 2$=$ \frac {1}{t}$
well my answer is coming
$ \frac {log\frac {x \minus{} 2}{2x \plus{} \sqrt3\sqrt {x^2 \minus{} 1} \minus{} 1}}{\sqrt3}$

[/hide]
Thank u.

how did u get that answer? answer is correct, but i couldn't reach. i got

-1/sqrt(3t^2+4t+1)

kan: some of what you write is already $ \text{\LaTeX}$ code. Just enclose it in dollar signs to have it show up properly on this site.

ok

$ \int{\frac{\minus{}1}{3t^2\plus{}4t\plus{}1}}dt$

Hello
[hide="solution"]
what is left now?
$ \int\frac{\minus{}dt}{\sqrt{3t^2\plus{}4t\plus{}1}}$
=$ \frac{\minus{}1}{\sqrt3}\int{dt}{\sqrt{t^2\plus{}\frac{4t}{3}\plus{}\frac{1}{3}}}$
=$ \frac{\minus{}1}{\sqrt3}\int\frac{dt}{\sqrt{(t\plus{}\frac{2}{3})^{2}\minus{}\frac{1}{9}}}$
=$ \frac{\minus{}1}{\sqrt3}log{(t\plus{}\frac{2}{3} \plus{}\sqrt{t^2\plus{}\frac{4t}{3}\plus{}\frac{1}{3}})}$
now again put $ t\equal{}\frac{1}{x\minus{}2}$[/hide]

[quote="kabi"]Hello
[hide="solution"]
what is left now?
$ \int\frac { \minus{} dt}{\sqrt {3t^2 \plus{} 4t \plus{} 1}}$
=$ \frac { \minus{} 1}{\sqrt3}\int{dt}{\sqrt {t^2 \plus{} \frac {4t}{3} \plus{} \frac {1}{3}}}$
=$ \frac { \minus{} 1}{\sqrt3}\int\frac {dt}{\sqrt {(t \plus{} \frac {2}{3})^{2} \minus{} \frac {1}{9}}}$
=$ \frac { \minus{} 1}{\sqrt3}log{(t \plus{} \frac {2}{3} \plus{} \sqrt {t^2 \plus{} \frac {4t}{3} \plus{} \frac {1}{3}})}$
now again put $ t \equal{} \frac {1}{x \minus{} 2}$[/hide][/quote]

Haha :D
Hello, Kabi

Have you learnt how to use Latex to write the equations/formulae here?
Great..

 :clap2:

[quote="vinskman"]
Haha :D
Hello, Kabi

Have you learnt how to use Latex to write the equations/formulae here?
Great..

 :clap2:[/quote]
Hello
Well I learnt from others post.I m glad that I have solved that problem .
Thanks.

[quote="kan"]How do we integrate this,

1/{(x-2)sqrt(x^2-1)} with restpect to x, when x is not eqaul to 2 and x less than -1.[/quote]

Hello, we can find not difficultly this integral, using hyperbolical functions:

$ \begin{gathered}
  \int {\frac{{dx}}
{{\left( {x \minus{} 2} \right)\sqrt {x^2  \minus{} 1} }} \equal{} } \left\{ \begin{gathered}
  x \equal{} \cosh \left( t \right), \hfill \\
  dx \equal{} \sinh \left( t \right)dt \hfill \\
\end{gathered}  \right\} \equal{} \int {\frac{{\sinh \left( t \right)dt}}
{{\left( {\cosh \left( t \right) \minus{} 2} \right)\sqrt {\cosh ^2 \left( t \right) \minus{} 1} }} \equal{} } \int {\frac{{dt}}
{{\cosh \left( t \right) \minus{} 2}}} . \hfill \\
  \cosh \left( t \right) \minus{} 2 \equal{} \cosh ^2 \left( {\frac{t}
{2}} \right) \plus{} \sinh ^2 \left( {\frac{t}
{2}} \right) \minus{} 2\left( {\cosh ^2 \left( {\frac{t}
{2}} \right) \minus{} \sinh ^2 \left( {\frac{t}
{2}} \right)} \right) \equal{} 3\sinh ^2 \left( {\frac{t}
{2}} \right) \minus{} \cosh ^2 \left( {\frac{t}
{2}} \right). \hfill \\
  \int {\frac{{dt}}
{{\cosh \left( t \right) \minus{} 2}}}  \equal{} \int {\frac{{dt}}
{{3\sinh ^2 \left( {\frac{t}
{2}} \right) \minus{} \cosh ^2 \left( {\frac{t}
{2}} \right)}}}  \equal{} \int {\frac{{\left( {\cosh \left( {\frac{t}
{2}} \right)} \right)^{ \minus{} 2} }}
{{\left( {\sqrt 3 \tanh \left( {\frac{t}
{2}} \right)} \right)^2  \minus{} 1}}dt}  \equal{}  \hfill \\
  \frac{{2\sqrt 3 }}
{3}\int {\frac{{d\left( {\sqrt 3 \tanh \left( {\frac{t}
{2}} \right)} \right)}}
{{\left( {\sqrt 3 \tanh \left( {\frac{t}
{2}} \right)} \right)^2  \minus{} 1}}}  \equal{}  \minus{} \frac{{2\sqrt 3 }}
{3}{\text{arctanh}}\left( {\sqrt 3 \tanh \left( {\frac{t}
{2}} \right)} \right) \plus{} C \equal{}  \hfill \\
  \left[ {\tanh \left( {\frac{t}
{2}} \right) \equal{} \frac{{\sinh \left( t \right)}}
{{1 \plus{} \cosh \left( t \right)}} \equal{} \frac{{\sinh \left( {{\text{arccosh}}\left( x \right)} \right)}}
{{1 \plus{} \cosh \left( {{\text{arccosh}}\left( x \right)} \right)}} \equal{} \frac{{\sqrt {x^2  \minus{} 1} }}
{{1 \plus{} x}}} \right] \hfill \\
   \equal{}  \minus{} \frac{{2\sqrt 3 }}
{3}{\text{arctanh}}\left( {\frac{{\sqrt {3x^2  \minus{} 3} }}
{{1 \plus{} x}}} \right) \plus{} C \equal{} \frac{{\sqrt 3 }}
{3}\ln \left( {\frac{{x \minus{} \sqrt {3x^2  \minus{} 3}  \plus{} 1}}
{{x \plus{} \sqrt {3x^2  \minus{} 3}  \plus{} 1}}} \right) \plus{} C. \hfill \\
\end{gathered}$
Formula $ {\text{arctanh}}\left( x \right) \equal{} \frac{1}{2}\ln \left( {\frac{{1 \plus{} x}}{{1 \minus{} x}}} \right)$
So, we have

$ \boxed{\int {\frac{{dx}}
{{\left( {x \minus{} 2} \right)\sqrt {x^2  \minus{} 1} }} \equal{} \frac{{\sqrt 3 }}
{3}\ln \left( {\frac{{x \minus{} \sqrt {3x^2  \minus{} 3}  \plus{} 1}}
{{x \plus{} \sqrt {3x^2  \minus{} 3}  \plus{} 1}}} \right) \plus{} C} }$


consider the number A=3^0+3^1+3^2...+3^95. show that A is an even number that is divisible by 13.

[hide="hidden"]
Since A is the sum of an even number of odd integers, A must be odd.  Odd + odd = even.

$3^0 \equiv 1 \pmod{13}$

$3^1 \equiv 3 \pmod{13}$

$3^2 \equiv 9 \pmod{13}$

$3^3 \equiv 1 \pmod{13}$

$3^4 \equiv 3 \pmod{13}$

$3^5 \equiv 9 \pmod{13}$

And so on...

$3^{95} \equiv 9 \pmod{13}$

Since each group of 3 adds up to 13, then A is divisible by 13.[/hide]


geometry
circumcircle
geometric transformation
homothety
Pythagorean Theorem

[quote="vidyamanohar"]How to find the radius of a circle touching three given circles which  touch each other pairwise externally ?[/quote]

Do you mean how to find the radius of the circle that is tangent to three other circles? Because then wouldn't the radius of the other circles be needed?

[quote="Hexaditidom"]This kind of problem is called a [url=http://mathworld.wolfram.com/SoddyCircles.html]soddy circle[/url]. :)

According to the page, the radius of an inner soddy circle ($r_4$) based on the radii of the other three ($r_1,r_2,r_3$) can be found by the formula:

$ r_4 = \frac{r_1r_2r_3}{r_1r_2 + r_2r_3 + r_1r_3 + 2\sqrt{r_1r_2r_3(r_1+r_2+r_3)}} $

You can even find the radius of a circle that surrounds all three circles by a similar formula:

$ r_4 = \frac{r_1r_2r_3}{r_1r_2 + r_2r_3 + r_1r_3 - 2\sqrt{r_1r_2r_3(r_1+r_2+r_3)}} $[/quote]

Thanks.  It could have been nice if proof is available for this.

[quote="Hexaditidom"]This kind of problem is called a [url=http://mathworld.wolfram.com/SoddyCircles.html]soddy circle[/url]. :)

According to the page, the radius of an inner soddy circle ($r_4$) based on the radii of the other three ($r_1,r_2,r_3$) can be found by the formula:

$ r_4 = \frac{r_1r_2r_3}{r_1r_2 + r_2r_3 + r_1r_3 + 2\sqrt{r_1r_2r_3(r_1+r_2+r_3)}} $

You can even find the radius of a circle that surrounds all three circles by a similar formula:

$ r_4 = \frac{r_1r_2r_3}{r_1r_2 + r_2r_3 + r_1r_3 - 2\sqrt{r_1r_2r_3(r_1+r_2+r_3)}} $[/quote]

Oh, I see you wanted a formula, I thought this was a problem or something.. ;)

[quote="santosh_blr"][quote="WindSlicer"]Is there a nice geometrical proof for this?[/quote]

Yes of course there's a nice proof to that..i just worked it out..but can't post the same as explnation needs to be augmented by good diagrams which I'm very uncomfortable at producing  :D ..

However it would be a good exercise for u if u take a shot at proving the result...u have to have ur basics on relationships between triangles and their incircles and  circumcircles and common  tangency of circles right...

happy proving  :lol:[/quote]

Though not very 'nice', I have worked out the solution using properties of triangles, conditional identities in a triangle and few big calculations.  It takes a little time to type the steps.  As soon as I type the solution I will upload it.

Let (A), (B), (C) be 3 externally tangent circles with radii a, b, c and WLOG, assume  $a \le b \le c.$ Let (D) be a circle with (unknown) radius d tangent to (A), (B), (C). Put the inversion center at the tangency point O of (A), (B) and let (A), (B) meet their center line AB at points P, Q different from O. Inversion with center O carries the circles (A), (B) into lines a', b' perpendicular to AB and the circles (C), (D) into 2 congruent tangent circles (C'), (D'), both tangent to the parallel lines $a' \parallel b'.$ The points P, Q are carried into the points $P' \equiv a' \cap AB,\ Q' \equiv b' \cap AB$ and by the basic properties of inversion,

$OP' = \frac{r^{2}}{2a},\ \ OQ' = \frac{r^{2}}{2b}$

where r is radius of the inversion circle (O). The solution is greatly simplified, if we choose $r = 2\sqrt{ab}$ ($\sqrt{2ab}$ would also serve the purpose, but we stick with the 1st choice). Then OP' = 2b and OQ' = 2a, so that radii of the inverted circles (C'), (D') are c' = d' = a + b. The inversion center O is the homothety center of the circles (C'), (C) and their homothety coefficient is equal to the power of the inversion center O to the inverted circle (C') divided by the power of inversion $r^{2}.$ Therefore,

$\frac{c'}{c}= \frac{OC'^{2}-c'^{2}}{r^{2}}$

Calculating $OC'^{2}$ and substituting $c' = a+b,\ r^{2}= 4ab,$

$OC'^{2}= \frac{4ab(a+b)}{c}+(a+b)^{2}$

Let R' be the midpoint of P'Q', so that $OR' = \frac{OP'-OQ'}{2}= b-a.$ The centers C', D' of the inverted circles (C'), (D') are on a normal to AB through R'. Using Pythagorean theorem for the right triangle $\triangle OR'C',$

$OC'^{2}= OR'^{2}+R'C'^{2}= (b-a)^{2}+R'C'^{2}= \frac{4ab(a+b)}{c}+(a+b)^{2}$

$R'C' = \sqrt{\frac{4ab(a+b+c)}{c}}$

If the circle (D) is tangent to the circles (A), (B), (C) externally, its center D is closer to the inversion center O than the center C of the circle (C). Consequently, the center D' of the inverted circle (D') is farther form the inversion center O than the center C' of the inverted circle (C'). Thus $R'D' = R'C'+c'+d' = R'C'+2(a+b).$ On the other hand, if the circle (D) is tangent to the circles (A), (B), (C) internally, $R'D' = R'C'+c'+d' = R'C'-2(a+b).$ In exactly the same way as for the circles (C), (C'), we can find

$R'D' = \sqrt{\frac{4ab(a+b+d)}{d}}$

Combining,

$\sqrt{\frac{4ab(a+b+d)}{d}}= \sqrt{\frac{4ab(a+b+c)}{c}}\pm 2(a+b)$

which is an equation for d in terms of a, b, c. Reducing by 2 and squaring the equation,

$\frac{ab(a+b)}{d}+ab = \frac{ab(a+b)}{c}+ab \pm 2(a+b) \sqrt{\frac{ab(a+b+c)}{c}}+(a+b)^{2}$

Canceling equal terms $ab$ on both sides and then reducing by (a + b),

$\frac{ab}{d}= \frac{ab}{c}\pm \sqrt{\frac{4ab(a+b+c)}{c}}+(a+b)$

Calculating d,

$d = \frac{abc}{ab+bc+ca \pm 2 \sqrt{abc(a+b+c)}}$

which is the desired result, symmetrical with respect to a, b, c, as expected. The plus sign is for (D) touching (A), (B), (C) externally, while the minus sign is for (D) touching (A), (B), (C) internally. The expression can be easily converted into the Soddy circles equation.


I was just wondering what age group the Alcumus problems were for. Like what level of math does it test (ex. middle school, high school, college)?

It complements material found in the Art of Problem Solving Introduction to Counting and Probability book. We typically suggest this for high-level middle school students. However it can be used at any age.


In the set of equations $ z^x \equal{} y^{2x}, 2^z \equal{} 2\cdot4^x, x \plus{} y \plus{} z \equal{} 16$, the integral roots in the order $ x,y,z$ are:

$ \textbf{(A)}\ 3,4,9 \qquad\textbf{(B)}\ 9, \minus{} 5 \minus{} ,12 \qquad\textbf{(C)}\ 12, \minus{} 5,9 \qquad\textbf{(D)}\ 4,3,9 \qquad\textbf{(E)}\ 4,9,3$

[hide]D. Being cheap, from the second equation, we see that $ z \equal{} 2x \plus{} 1$. D is the only solution that satisfies this.[/hide]

Or, being cheap from the first equation (which is what I did), $ z\equal{}y^2$ and it's D.

[hide="If you are dumb enough to not look at the answer choices..."]
Like Zhero and dgreenb801 said, $ z\equal{}y^2$ and $ z\equal{}2x\plus{}1$.  Then $ x\equal{}\frac{z\minus{}1}{2}\equal{}\frac{y^2\minus{}1}{2}$.  From the third equation $ \frac{y^2\minus{}1}{2}\plus{}y\plus{}y^2\equal{}16$.  This equation has one integral root $ y\equal{}3 \implies x\equal{}4, y\equal{}3, z\equal{}9$ or $ \boxed{\textbf{(D)}}$.
[/hide]


inequalities
number theory unsolved
number theory

This, also like the ABC inequality I posted was given in 2001 for the 9th grade in the county Arges, in a math contest.

Any ideas?

This is exactly the kind of problems I hate. Someone( who has nothing else to do) combines a lot of stupid techniques and creates a monster with no intelligent idea. I' m not against theorems and lemmas and all this stuff, but they must be used to create interesing, difficult and beautiful problems, not something like this( which cannot be called a problem).


AMC
AMC 10
AIME
geometry
trigonometry
complex numbers

[quote="seohyun Kim"]I came here first time and I have many problems in AMC :|
I think many users are really good at math and plz help me
what do i have to study(part) for amc 10 and 12?
I tried to solve amc 12 and I have got almost 100 but mostly I get 10 corrects (OTL) lol
I've got 94 in 2001, 97.5 in 2000 and 97.5 in 2006( :blush: )
I can not see 3 problems at 2006 so I could get more than 100 but ... after number 15 (sometimes 13 or 14) I even cant try to solve problem ....
also my score is close and is it better to take AMC10 instead of 12? and is it true that blank will count as 1.5 point?
plz help me also I am asian so I'm really bad at English plz understand this....[/quote]

Try to take AMC 12. Don't take AMC 10 because it is harder to make AIME that way. You have to make 120 on AMC 10 to make AIME, but only 100 to make AMC 12, and it is harder to make a 120 because of the 1.5 reduction.

[quote="mdk"][quote="Mathlete9366"]The best advice I have is practice. If you think that you just walk into the test with only a vague idea of what will be on the test, practice will really help your scores (because there are some patterns to the types of problems that the AMC asks).

The best way to actually learn the material (in my opinion) is through the purchase of the Art of Problem Solving by Richard Rusczyk (and others...) located in this site's bookstore (there is a link in the upper-left corner of this webpage). Unfortunately this book is in english, and I don't think it is available in any other languages (somebody correct me on that if I am wrong).

Good Luck!

EDIT: Art of Problem Solving (AoPS) Volume I should be enough to get a 100 on the AMC 12. I am not absolutely positive about that though, and of course it does not hurt to eventually work through Volume II.[/quote]

I know that I can get at least a 100 on the AMC 12 and I am planning to do well on the AIME. Would you recommend Art of Problem Solving (AoPS) Volume I or (AoPS) Volume 2?[/quote]

Get Volume I to reinforce skills, especially geometry, if you need it. But get Volume II for sure.

[quote="NeverOddOrEven"]is it really more difficult to make aime through the amc 10 than the amc 12? It seems much easier to me.[/quote]

It may have been last year. But this year, do yourself a favor and take the AMC 12.

[quote="NeverOddOrEven"]is it really more difficult to make aime through the amc 10 than the amc 12? It seems much easier to me.[/quote]
It depends on 2 things. First, If you are good with trig and logs and other things like that that aren't on the 10 but are on the 12 then the 12 is better. Second, if you make a lot of careless mistakes, but can answer almost all the questions bar 1 or 2 then the 12 is better. If not, then go for the 10...

[quote="seohyun Kim"]thanks for help and now I 'm worring about taking 10 or 12 ..
if i can i will take both of them and
anybody know what part(of math) is in the 12 but not in 10?
also what books will help a lot?

I totally agree with this!
If Timmy has 3 apples and Johnny has 2 apples, why don't they just shut up and eat?[/quote]
The 12 has logs, trig, complex numbers, and some other stuff (I think) that I can't think of at the moment.

[quote="seohyun Kim"]
I totally agree with this!
If Timmy has 3 apples and Johnny has 2 apples, why don't they just shut up and eat?[/quote]

 :D

[quote="bpms"]
The 12 has logs, trig, complex numbers, and some other stuff (I think) that I can't think of at the moment.[/quote]

The 10 and 12 effectively cover the same material (including some of the same questions) but the 12 gets a bit harder towards the end.

I always take both tests (provided that you can) so you have double chance to make the AIME.

It used to be you got 2.5 points for a blank, and because the AMC 10 problems was enough easier, it was easier if you are good at general problem solving.

But, like people have said, a lot of the same problems are used on both tests.

Also, as long as you don't guess, and just make sure of the answers you have, you will be better off on the 12.
(just spend your time making sure you have the first 15 (or how ever many you can do) correct and you should do fine (you get an extra 15 points for the blank problems).

[quote="seohyun Kim"]I came here first time and I have many problems in AMC :|
I think many users are really good at math and plz help me
what do i have to study(part) for amc 10 and 12?
I tried to solve amc 12 and I have got almost 100 but mostly I get 10 corrects (OTL) lol
I've got 94 in 2001, 97.5 in 2000 and 97.5 in 2006( :blush: )
I can not see 3 problems at 2006 so I could get more than 100 but ... after number 15 (sometimes 13 or 14) I even cant try to solve problem ....
also my score is close and is it better to take AMC10 instead of 12? and is it true that blank will count as 1.5 point?
plz help me also I am asian so I'm really bad at English plz understand this....[/quote]
What grade are you in?


Um... it seems few care about their rating, but anyways...

A lot of the time I'll accidentally click on the little cubes above someone's post. How do I "un-rate" it so that it would be as if I hadn't rated it?

You can't. You can, however, click again to replace the rating you accidentally gave with an accurate one.

so when u rate the first time, think carefully before u rate.


The circumfence of a circle is 12.56cm. What is the area of the circle? Let pi be 3.14.

yA, LOL the only time the area and the circumference are equal, when r=2

Ya, that point has the same area as its circumference

Excuse the decimal fractions  ;)
$\frac{12.56}{3.14}=2r$
$2r=4$
$r=2$
$2^2*\pi=a$
$a=12.56$

Or, you could immediate realize that $\frac{12.56}{3.14}=4$  and that the circumference is the same as the area.

[hide="clik dis cause its da answer"]
C=2*3.14*r
A=pi*r^2
r=2
pi*4=[b]12.56[/b]

[/hide]

try learning $\LaTeX$, its much pretttier

i do know $\text{\LaTeX}$
[hide="clik dis"]
$\displaystyle \frac{12.56}{3.14}=2r
2r=4
r=2
r^2*\pi=A
12.56=A
12.56$[/hide]

[quote="math92"]i do know $\LaTeX$[/quote]Then use it :) Also please stop using chatspeak. Thanks.

[quote="Barnacle"][hide]
A=pi*r^2
C=2*pi*r

A=pi*(C/2pi)^2=C^2/4pi = 12.56^2 / (4*3.14) = 12.56 cm^2

... nice choice of numbers  ;)
[/hide][/quote]

I never realized that.
2*2 is 4, and 2+2 is 4, so it explains why the numbers are the same.


calculus
integration
trigonometry
calculus computations

Solve the integral:

$ \sqrt {\frac { \cos3x}{\cos^3x\minus{}1}} \cdot dx$

hello, do you meant $ \int \sqrt{\frac{\cos(3x)}{\cos^3(x)\minus{}1}}\,dx$?
Sonnhard.

Yes it is:
$ \int \sqrt {\frac {\cos(3x)}{\cos^3(x) \minus{} 1}}\,dx$
Sorry! :D


Er, you might try to want to express it in LaTeX: put dollar signs before and after the equation you want to TEX (dollar sign ............... dollar sign), and use the proper syntax ; I learned how to do this like only two or three days ago... :D

$(1) \frac{3x}{x+y} - \frac{2y}{x}$

$(2) c^{-1} + d^{-4}$

$(3) \frac{6x^{2xy^{3}}-12w^{3x^{2y}}}{2wxy}$

$(4) \frac{\frac{7a}{a-b}}{\frac{4b}{a+b}}$

[hide]Answers:

(1) By simple addition and simplification, you should get:

$\frac{3x^{2} - 2y(x+y)}{x(x+y)} =$
$\frac{3x^{2}-2xy-2y^{2}}{x(x+y)}$

(2) By simple addition and simplification, you should get:

$c^{-1}+d^{-4} =$
$\frac{1}{c} + \frac{1}{d^{4}}$
$\frac{c+d^{4}}{cd^{4}}$[/hide]

Er... I'll work on the rest later... too lazy for now  :)

man i know just i got 56k and its takes ages to download


Let a an integer $ m$ be an $ i$th pharlgebus of another integer $ n$ if $ m$ divides exactly $ i$ of the factors of $ n$ (including $ n$ itself). How many numbers are a 3rd pharlgebus of 2008?

you made up that pharlgebus just for this problems its not a real mathematical term right??? kinda like those the operationis defined by a#b=$ \frac {a + b}{a\times b}$.

[quote="zolojetto"]you made up that pharlgebus just for this problems its not a real mathematical term right???[/quote]

Yes, it's just a term defining a set mathematical operation  :wink: .

hmm...ok thats what i though...i dont exactly want to do this how i would do it if i had too :maybe:  :( uh does any one know an easier way to do this than writing all of the factors and counting the number of pharlgebus's!!!

[quote="ZhangPeijin"]Hint: Basically the 3rd number has to be a factor of exactly 3 of the factors of 2008.[/quote]

NO REALLY


calculus
integration
quadratics
algebra unsolved
algebra

Let m and n be positive integers such that the equation x2 - mx + n = 0 has real roots a and b. Prove that a and b are integers if and only if [ma] + [mb] is the square of an integer. Here,[x] is the greatest integer not exceeding x.

If a anb b are integers the conclusion is obvious. If [ma]+[mb] is the square of an integer it follows easily that it can be onle $m^{2}$. therefore both a and b are rational numbers. However, a and b are are the roots of a monic 2-degree polynomial. Therefore, a and b must be integers.

 :) Hi
m=a+b
If a & b r integers then [ma]=ma & [mb]=mb
thus [ma]+[mb]=m(a+b)=m^2
Which is a perfect square. Hence the first part is true.
if [ma]+[mb]=k^2 for some integer k
then m([a]+[b])=k^2
i.e. m divides k &[a]+[b] also divides k
But m=a+b
So this is only possible when [a]=a & [b]=b
i.e. both a & b r integers.
Plz tell me whether i m correct or not. I gave this sol. in INMO 2007.

[quote="Asmita"]
if [ma]+[mb]=k^2 for some integer k
then m([a]+[b])=k^2
[/quote]
Why? You are wrong! I think  :wink:

i think asmita's solution is wrong...

a + b = m
so a = m - b
so[ma] + [mb] = m^2 +[-mb] + [mb] = k^2
its not difficult to see that
[-x] + [x] = 0(if and only if x is integer)
                 -1(iff x is not integer)
suppose k not equal to m
k^2 - m^2 = -1,only non- negative integral solution is k = 0 and m = 1
but if m= 1 then the equation does not have real roots.
so a contradiction!!
hence k = m
[-mb] + [mb] =0
so mb is an integer
b is rational
but the leading coefficient of the quadratic is 1.
so b has to be integer and then obviously a is an integer

[hide="Les Solutiones"]This can also be very easily done using a simple double bound inequality. Assume that say, a and b are not integers.
   $[ma]< ma < [ma]+1$
   $[mb] < mb < [mb]+1$
Adding the two we get,
    $[ma]+[mb] < ma+mb < [ma]+[mb]+2$
Now, very clearly we have two consecutive perfect squares $x^{2}$ and $m^{2}$. This is possible iff $x= 0$. So $m=1$. But this implies that $ma = a$ and $mb = b$ So we get $a,b<1$. But this implies that $ab < 1$ and $ab$ is a positive integer$(ab=n)$ This cannot happen. Hence the contradiction.

  AR[/hide]


Sean is four years younger than Mike. In five years, Mike will be twice as old as Sean is now. What is the number of years in Mike's current age?

how do u do this? wouldn't the answer be 3?

What I have noticed in competitions and sorts is that people get awestruck(?) by these kinds of problems. What only requires simple algebra is overthought and missed. Clearly, these problems should be practiced more often.

EDIT: By "these" problems I mean easy problems that are hard :P

[size=95]Last edited by [url=http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=136993]TrueshotBarrage[/url] on Jan 25, 2014, 9:48 pm, edited 1 time in total.[/size]

[list] [/list]


$ (a_n) :  a_{n+1}= \frac{1}{4-3a_n}  , n \geq 1 $ .
 Find $ a_1 $ such that : $ (a_n) $ converges and find $ \lim_{n \to \infty}a_n $ ?

The answer is

$\lim_{n\to\infty} a_n=\frac{1}{3}\ \left(a_1\neq 1,\frac{4}{3}\right)$

$\lim_{n\to\infty} a_n=1\ (a_1=1)$

Solving for the given recursive formua, we obtain $a_n=\frac{(a_1-1)3^n-3(3a_1-1)}{(a_1-1)3^{n+1}-3(3a_1-1)}$


Just search for it  :wink:
http://en.wikipedia.org/wiki/Butterfly_theorem
http://www.cut-the-knot.org/pythagoras/Butterfly.shtml

I know. I got it from Wikipedia.

But I was asking for YOU to prove it.


abstract algebra
algebra
polynomial
linear algebra
linear algebra unsolved

define $f:M_n(C) \to C^n$ by $f(M)=(tr(M),tr(M^2),..,Tr(M^n))$.
Find the rank of $df(M)$, the differential of $f$ at the point $M$.

The differential of $f$ at $M_{0}$ is a linear map defined by

$d_{M}f(H)=\left ( tr(H),2tr(M_{0}H),3tr(M_{0}^2H),...,ntr(M_{0}^{n-1}H) \right)$

Study the dimension of the kernel and use rank theorem

$H$ is in $Ker (d_{M_0}f)$

$tr(H)=tr(M_{0}H)=...=tr(M_{0}^{n-1}H)=0$

Let's post a solution for this one. I will admit Moubinool's formulae since they are correct. Define $ g_k(A)=tr(M^k A)$ and observe that we must actually find the dimension of the space generated by these $g_k$ (use here duality in finite dimensional spaces). Of course, if $P$ is the minimal polynomial of $M$ then if $ p=degP$, all $ g_i$ with $i\geq k$ are in the subspace generated by $g_0,g_1,...,g_{k-1}$ and so the rank is at most $k$. On the other hand, the family $ g_0,...,g_{k-1}$ is free, because of the fact that $ tr(AX)=0$ for all $X$ implies $A=0$. Thus the rank is the degree of the minimal polynomial of $M$.


For whatever images I need/want to upload...


geometry
complex numbers
number theory
prime numbers
number theory proposed

Prove that $2005^2$ can be written in at least $4$ ways as the sum of 2 perfect (non-zero) squares.

consider $2005=|(1 \pm 2i)(1 \pm 20i)|$, so you have
$2005^2=|(1 \pm 2i)(1 \pm 20i)(1 \pm 2i)(1 \pm 20i)|$ where you can choose all $\pm$ independent of each other.
This gives the solutions $2005^2=2005^2+0^2=1037^2+1716^2=1203^2+1604^2=1995^2+200^2=1357^2+1476^2$

You can even prove more in this way:

The number of integer solutions $n=x^2+y^2$ is $4 \left( a(n)-b(n) \right)$ where $a(n)$ is the number of divisors of $n$ that are congruent to $1 \mod 4$ and $b(n)$ the number of divisors congruent to $3 \mod 4$.

The proof is more or less constructive, simply consider the prime decomposition in gaussian integers

Hehe... very smart workaround with complex numbers! :D

I did note the numbers had something special these but I could not establish this link :)

Anyone up for a more basic proof? (the simplified version was asked to all grades, so even to the 14yo pupils...)

[edit: problem proposed by me]

a bit more basic, but it's more or less the same as ZetaX did: write 2005 as a sum of two squares (it can be done by checking for a few, or not a few but not so many, values) . It can be done in two diferent ways. Then use the very-well-known identity $(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2$ , and you have two possibilites for chosing a,b and two for c,d, hence there will be (not so sure, but almost) 4 ways to write 2005.2005 as a sum of two positive squares.

I'm not sure if RaMlaF's idea works for four really different solutions, but it surely works if you do the following (which is exactly the same as I did in my post before, but without complex numbers):

Write all prime factors ($5,401$) of $2005$ as sum of two squares and then apply the formula in RaMlaF's post

[quote]ZetaX[/quote]

How do u did that...the complex thing...and really dont get it...can someone explain please

[quote="Peter VDD"]Indeed, that was my solution too. However I didn't know the "well-known" formula before.
But apparently the jury found it too hard since they gave this product rule as a hint. :)[/quote]

Correction: that was part of a very similar question proposed by me :D So I guess they picked half of my question, and half of yours.

What did you propose then exactly? (as this is exactly the question I submitted :))

Or is it still secret?

[did it use the wording [i]'good/better numbers'[/i]? As that was something I did not write.]

Hm... I had a lot of alternative versions for this question. Here are some of them.

Let's call an integer lucky if it can be written as the sum of two squares.

a) Prove that the product of two lucky integers is again lucky.
b) Prove that there does not exist a lucky integer a such that 7a is lucky.
c) Find all prime numbers p such that if a is lucky, then pa is "unlucky", for all a.
d) Find all prime numbers p such that if a is lucky, then pa is lucky too, for all a.

As you can see they picked part (a) (which is way too easy but anyway).
For the (b)-part of the question they picked your proposal.

Ah I see. But your (a) part was also in my question apparently. :)

The only thing I wonder is: why they didn't ask for 4 right-angled triangles with integer sidelenghts and hypothenusa 2005? I think it would have been a much better wording... :)

PS: why don't you post them in pre-olympiad? They may be nice challenges to people.

Well, it's a number theoretic problem (or rather [i]exercise[/i]) so why would you try to "cover" it with geometry? I mean, it doesn't change much...

[quote="Peter VDD"]why don't you post them in pre-olympiad? They may be nice challenges to people.[/quote]

Done :)


a_1, a_2, ... , a_n are real numbers such that 1/2 ≤ a_i ≤ 5 for each i. The real numbers x_1, x_2, ... , x_n satisfy 4x_i2 - 4a_ix_i + (a_i - 1)2 = 0. Let S = (x_1 + x_2 + ... + x_n)/n, S2 = (x_12 + x_22 + ... + x_n2)/n. Show that √S2 ≤ S + 1.

elmist is wrong! Problem is :

$a_{1},a_{2},...,a_{n}$ are reals such that $\frac{1}{2}\leq a_{i}\leq 5\forall i$. The reals $x_{1},x_{2},...,x_{n}$ satisfy $4x_{i}^{2}-4a_{i}x_{i}+(a_{i}-1)^{2}\leq 0\forall i$. Prove that $\sqrt{\frac{\sum_{i=1}^{n}x_{i}^{2}}{n}}\leq \frac{\sum_{i=1}^{n}x_{i}}{n}+1$.

[quote]elmist is wrong[/quote]
He copied from kalva  :lol:


Find the volume, in cubic centimeters, of a cube whose diagonal is 4 centimeters longer than its edge.

Is the diagonal a space diagonal? Or is it between two vertices on the same face?

[hide]The diagonal of a cube refers to the space diagonal. Therefore we have that $ s\sqrt{3}\equal{}s\plus{}4$
$ s\sqrt{3}\minus{}s\equal{}4$
$ s(\sqrt{3}\minus{}1)\equal{}4$
$ s\equal{}\frac{4}{\sqrt{3}\minus{}1}$
$ s\equal{}\frac{4\cdot(\sqrt{3}\plus{}1)}{2}$
$ s\equal{}2\sqrt{3}\plus{}2$[/hide]


calculus
integration
function
limit
inequalities
real analysis
real analysis solved

Let $f:[0,\infty)\to(0,\infty)$ a continous function such that $\lim_{n\to\infty} \int^x_0 f(t)dt$ exists and it is finite. Prove that
\[ \lim_{x\to\infty} \frac 1{\sqrt x} \int^x_0 \sqrt {f(t)} dt = 0. \]

[i]Radu Miculescu[/i]

There is a mistake in the proposed problem, please correct it!

[quote="Moubinool"]$f:[0,+\infty[\rightarrow ]0,+\infty[$ continous

Suppose $\lim_{x\rightarrow +\infty}\int_{0}^{x}f(t)dt=L$ where $L$ is a real number

Prove that $\lim_{x\rightarrow +\infty}\frac{1}{\sqrt x}\int_{0}^{x}f(t)dt=0$

[/quote]
Isn't it obvious(denominator goes to infty and numerator is bounded)? :?

Deleted all the stupid stuff I had written here (see below) :).

Let's consider the function $f$ s.t. $f(n)=0,f(n+\frac{1}{2^{n}})=0,f(n+\frac{1}{2^{n+1}})=n$,$f$ is linear on the intervals $[n,n+\frac{1}{2^{n+1}}]$ and $[n+\frac{1}{2^{n+1}},n+\frac{1}{2^{n}}]$, for all $n{\in}{\mathbb{N}}$ and $f$ is zero elsewhere.$f$ is continious and the integral $\int_{0}^{+\infty}{f(t)dt}$ converges because $\int_{n}^{+\infty}{f(t)dt}=\sum_{k=n}^{+\infty}{\frac{k}{2^{k+1}}} {\rightarrow} 0$.But even $f(x)$ doesn't tend to $0$.So dear [b]grobber[/b] we don't have that $xf(x) {\rightarrow} 0$.I hope everything is ok.

:) Terribly sorry about that. The problem I was talking about also had $f$ decreasing, I think. I'm not sure anymore. Sorry again :blush:.

Ok, one more try:

We can find $x_0$ s.t. $\int_{x_0}^x f(t)dt$ is as small as we want, for all $x\ge x_0$. Once more, write $\int_0^x \sqrt{f(t)}dt=\int_0^{x_0}\sqrt{f(t)}dt+\int_{x_0}^x\sqrt{f(t)}dt$. When we divide this by $\sqrt x$ and make $x$ large, the first summand becomes as small as we want, and the second summand is, from Cauchy's inequality, $\le \sqrt{\int_{x_0}^xf(t)dt}$, which we can make as small as we want, as noted above.

Is it Ok now? :?

Moubi, didn't it take less than 4 minutes to solve this problem? It is hyper-classical oral Ecole Polytechnique problem.


Will you folks [b]PLEASE[/b] stop spamming this forum!

Today, I have deleted about 70 absolutely useless posts.

If you do not have anything mathematically relevant to add to a topic, do not say anything!  Do not say "I agree".  Do not say "I don't know".  Do not say "twin".  Do not say "the answer is 3, but I'm too lazy to explain."

This forum is for MATHCOUNTS-level discussions of mathematical topics.

If you do not have anything mathematically relevant to add to a topic, do not say anything!

Be warned that if you continue spamming as I have seen now some of these replies, you might take a small undesired vacation until you cool off ;)

Whew!! Thank you for saying that! I can't stand all these strange posts when I'm trying to look for good solutions.

Again, Thanks very much.

Vacation? Like you're banned or something?

Meaning you cant post for a period of time.

WHOOHOO!!!! :D
An announcement to stop spamming!  :jump:
Thank You!

As for being suspended...Hey, if it works!

And that twin thing was definately getting annoying

Also,  could people hide answers?

O and spamming also constitutes of not editing your previous posts when you can (thus adding one to your post count)... I run other forums and find that one of the most annoying things. Idk what do you moderators think?

Thanks for this post by the way.

As a moderator and also a student who made many posts in this forum, I suggest people not to spam because it is very annoying.

Like someone mentioned, if you want to add something to what you said, please use "edit" button. I'm not a moderator of this forum so I won't comment on this issue much (since rcv is a great moderator who is doing awesome job with this forum. Nice job by the way :) ) but like people say "do not spam."

im sorry i think i was spamming. forgive a newbie for being unfamiliar with forum protocol  :)

Does using the edit button mean we can only have one post in any thread? Or are we allowed to make a new post as long as someone's posted after us and we're replying to their post?

[quote="Valentin Vornicu"][quote="sirorange"]It's just a way to prevent double posting, and such. You can definetly post a reply to someone's prior post.[/quote]Exactly. Having [b]two consecutive[/b] posts for no reason is what we don't want.[/quote]

What I was referring to. I also am not a moderator so I won't comment too much, but there are times where an edit is apt, and other times when it isn't. Certainly two consecutive posts is spamming, but say in this case, this material is continuing the conversaion in a way that I couldn't add to the previous post really... else it wouldn't make sense.

O and one/two/three word (or letter) posts usually don't cut it either.

I like spam <.< . So yummy...

While on the topic of moderators, you said my moderator privileges have been suspended. For how long? It's been around six months now...

How can you become a mod? How can your priveliges become suspended?  :?

Mr. Rusczyk or another admin may make you an admin. But I can't say why exactly you are made an admin.

To have mod privleges revoked/suspended you must have abused your power or have posted something inappropriate.

How many people are Admins and what does it take to become one?

yay i didn't realize that our admin, Valentin, has changed his avatar that looks almost the same as gr8sk8r's avatar! perfect deception! :D
ps: btw, yoda is cool  :lol:
ps ps: i just checked and we have 5 admins currently and plenty of mods. but i think 5 admins are good enough just like the united nations with only 5 permanent countries, too many gets chaotic and maybe too much disagreements...

yeah Valentin's and gr8k8r have similar avatars...cool

[quote="Treething"][b]I like spam[/b] <.< . So yummy...

While on the topic of moderators, you said my moderator privileges have been suspended. For how long? It's been around six months now...[/quote]As long as you keep doing the same mistakes over and over again they will remain suspended.

Math92 you are spamming in this very thread. Be warned!

PS Don't bother thinking about Admin positions. It won't happen :)

[quote="nat mc"]How did you become an Admin?[/quote]I see you guys don't seem to get this: [b]PLEASE STOP SPAMMING[/b] including this thread. If one more offtopic post appears I will take measures!

Spam in my post was intended as the canned meat/MSG mixture. Sorry for the weak pun :P .

And yes, I have noticed that math92 has gone on an immense spamming spree. Yesterday he had 1175 posts, now he has 1205.

No, you don't become an admin by reaching 1,000,000 posts.

Sorry if someone's mentioned this, but I think that it's alright if you spam in the test forum. However, if you're only trying to increase your post count, the test forum won't help. Although, posts have laid in there for up to three months. Also, the fun and games factory doesn't seem to care as much. (My Apologies to those mods.)

PS. Valentin's avatar is cool.

[quote="RC-7th"]Sorry if someone's mentioned this, but I think that it's alright if you spam in the test forum. However, if you're only trying to increase your post count, the test forum won't help. Although, posts have laid in there for up to three months. Also, the fun and games factory doesn't seem to care as much. (My Apologies to those mods.)
[/quote]

There are three moderators for the fun and games factory, but you really can't "spam" in that forum. Then, do the moderators just watch to see if there are arguments/inappropriate posts floating around?

Note: this "double-post phenomenon" has happened to me occasionally, where you submit a message, and the computer takes forever to load, and you press the submit button again, and all of a sudden you have two posts in a row. Or maybe the message doesn't appear right away the first time, so you write a new one and then both appear.

Although I don't think statements such as "that's a great solution" or "I agree with ____'s answer" are that bad in terms of being annoying. Maybe if that's the only content in the message....

I think it has been mentioned above by rcv, that posting something along the lines of " I agree" is unacceptable and considered spam.  However, when someone has posted a problem and another user has posted the answer and [b]solution [/b], one's curiousity tends to wander and he/ or she ends up reading the solution, regardless it being hidden or not.  And not lets say that the user who posted the solution wants to know whether the solution is correct, but the originator of the problem has not been offline for some time.  So imagine yourself in this position, you want to know whether your correct are not........  So why is it so offensive to write " I agree", simply as a confrimation that you have solved problem correctly , yes I know you are not including a mathematical explanation but suppose your mathematical explanation is identical or very similar to the one posted above, it would seem strange and even suspicious to post the same explanation twice.  I know that this " I agree",   priveledge is being abused, and we can't always tell when its being used or misused, but in my opinion I think it would be best to ease up on this rule.................

On the topic of F&G... its spam depending on the topic. I think spam is just a post which doesnt contribute to the thread or help anyone in any way. Saying 'I Agree' usually is spam. And except for those cases where someone wants to know if their solution is correct, and there it is overrused - you see 5 posts one after the other of just 'I agree.' -Just as a note i dont think math92 is spamming all that much.. atleast half  :)  or more of his posts do contribute.

To lighten everyone's curiosity, check out the definition of spam
[url]http://dictionary.reference.com/search?q=spam[/url]

[quote="G-UNIT"]...So why is it so offensive to write " I agree", simply as a confrimation that you have solved problem correctly...[/quote]

If you want to know whether your solution is correct or not, you have a few options.

- simply wait for the poster of the problem to come back and see
- say "Is that correct? That's what I got also, and I did the same thing," or something to that effect. I think that's less likely to be considered spamming.
- post your solution, even if it is similar to the other person's solution

I'm not a moderator, but those are my ideas.

[quote="G-UNIT"]...So why is it so offensive to write " I agree"...[/quote]
When a single thread contains one solution (or one answer), followed by a string of "I agree" responses, a good part of that is unnecessary -- thus spam.

When a single user posts a response to every thread, saying "I agree", a good part of that is unnecessary -- thus spam.

If the original poster of the question knows the solution, and posted the question as a challenge, they have an obligation to monitor the thread and provide feedback when necessary.  Such feedback may include confirming a correct solution.  If your question and/or solution were from a book or other reference, it is a good idea to list your reference.

If you solve the problem yourself, and you want to take the time to write up your solution, that is OK.  [This is why the use of "hide" tags is encouraged -- Each user can solve a problem without influence of previous solutions.]  If your original solution happens to be similar to a previous solution, that is OK.  In practice, most original solutions will offer something unique -- different approach, different terminology, different notation -- all of which are beneficial to other students learning to solve problems effectively.  [A sincere and original [b]solution[/b] is not spam.  If you are too lazy to write up a solution, and merely respond with the same answer or the words "I agree", that may be considered spam.]

If you are the first person to note that a solution is incorrect, you should feel free to explain the error in the previous solution -- better yet, write up what you believe is a correct solution.

And, if there is a controversy over which of two solutions is correct, you should feel free to say "I agree with solution X because ..." or "I disgree with solution Y because..."  If the author of the incorrect solution says "Oh.  I see my mistake", the controversy is over.

Use common sense.  These are general guidelines.

[quote="Palytoxin"]let's face it, here are the statistics for high posting rates (as off Tuesday midnight)

math92                   1364 posts           17.49 posts/day
amirhtlusa               1169 posts           16.24 posts/day
BornForMath            795 posts             14.45 posts/day
Barnacle                  333 posts             22.20 posts/day
Palytoxin                  314 posts            14.95 posts/day

and yes i got a high rate  :roll:[/quote]

Just because you've got a high posting rate doesn't mean you spam a lot. Some are one the internet more than others. Adminastrators and Moderators can have high posting rates, but they almost never spam.

Like you, Palytoxin, you almost never spam, and it seems like you post a lot every single day. To find if a person spams consistently, a more detailed analysis of his of her posts is needed.

And everyone spams sometime, except maybe for rcv  :)

[quote="236factorial"]Like you, Palytoxin, you almost never spam, and it seems like you post a lot every single day...[/quote]

I don't think amirhtlusa and BornForMath don't spam very often either. :)

I'm on the high posting list???
I post 17 times a day???
whoa.
Anyways i think that i have been spamming less everyday.  At least i think i have been.

[quote="hockeyadi23"]O and spamming also constitutes of not editing your previous posts when you can (thus adding one to your post count)... I run other forums and find that one of the most annoying things....[/quote]

What do you mean?  [i]Not[/i] editing is spamming? :?:

Also, I've seen some text beneath the username of the posters.  Mine is "New Member",but I have seen "Yang-Mills THeory", "Riemann Hypothesis", etc.  What does this mean?  And what are the stars?  (I'm sorry but I'm a new AoPSer and haven't a clue.)

The mathematical names are the ranks on AoPS, and the stars correspond. The ranks depend on how many posts you have, and naturally some people will spam just to get a high post count and thus lots of stars. But it's funny that one goes from zero to a half to one star, then jumps to two, then two and a half, then three, four, and five, without intermediate stops at one and a half, three and a half, or four and a half. Administrators have six stars to distinguish them, regardless of how many posts they have.

Hope that helps.

What irks me is that some people will post random things just to get their post count up.  If all you care about is your post count, maybe you need to get out a little bit more...


Ring Theory
superior algebra
superior algebra unsolved

Let $ p$ be a prime number in $ \mathbb{Z}$, consider the ring: $ \mathbb{Z}_{(p)} \equal{} \{\frac{m}{n} \in \mathbb{Q}: m, n \in \mathbb{Z}$, and $ \gcd(p, n) \equal{}1 \}$.

Then for any integer $ n \geq 1$, identify the ring $ \mathbb{Z}_{(p)}/p^n \mathbb{Z}_{(p)}$, where $ p^n \mathbb{Z}_{(p)}$ is the ideal of $ \mathbb{Z}_{(p)}$ generated by $ p^n$.

Localization is exact, so this ring is the same as $ \left(\mathbb{Z}/p^n\right)_{(p)}$; however any value not divisible by $ p$ in $ \mathbb{Z}/p^n$ is already invertible, so localizing does nothing; this ring is just $ \mathbb{Z}/p^n$.

$ \mathbb{Z}_{(p)}$ is just a localization of the ring $ R \equal{} \mathbb{Z}$ at a non-prime ideal $ (p)$. This makes it a lot easier but the context from which I saw this problem, the term localization was not one of the vocabularies there [or perhaps it was assumed to be one of the things a reader must know] and so is there a different argument that does not use the properties of localization?

HINT what's the image of $ \mathbb{Z}$ in $ \mathbb{Z}_{(p)}/p^{n}\mathbb{Z}_{(p)}$ ?


group theory
abstract algebra
superior algebra
superior algebra theorems

If $ F$ is a free group and $ K$ and $ R$ are two normal subgroups, is it true that if $ F/K$ and $ F/R$ are isomorphic then $ R\equal{}K$?

A related question: If $ G$ is isomorphic to $ G'$ can there be an onto homomorphism that is not an isomorphism? The answer is probably yes, but can some sort of freeness or universal property save it?

1) no .. this would mean that all isomorphic groups generated by a specific number of elements are equal.
2) look at $ G \equal{} G'$. for example $ G \equal{} S^1$ and $ x \mapsto x^n$.

No, of course... what I meant to ask is with the condition that $ R \subset K$.

How about this: let  $ G \equal{} \mathbf{Q} / \mathbf{Z}$ be an additive group.

Pick a set of generators of $ G$ - the set of relations satisfied by the generators induces an isomorphism $ G \equiv F/R$, where $ F$ is a free group on this set of generators.

Now consider the additional relation $ 1/2 \equal{} 0$. This generates the normal subgroup $ K \subset F$, and $ F/K \equal{} \mathbf{Q} / \frac 1 2 \mathbf{Z}$ which is also isomorphic to $ G$.

[quote="darij grinberg"][quote="lzw75"]How about this: let  $ G \equal{} \mathbf{Q} / \mathbf{Z}$ be an additive group.

[...][/quote]

The original poster is talking about free groups.
I am wondering, do free groups belong into "Superior Algebra - Groups, Fields, Rings, Ideals" at all? It seems that the only reasonable approach to them is via topology...

  darij[/quote]

$ G$ is not free, but $ F$ is. :)

The example I can think of: Let $ F$ be the free group on infinitely many generators $ x_1,x_2,\dots$, $ R\equal{}\{e\}$, and $ K$ is the smallest normal subgroup containing $ x_1$. The quotient $ F/K$ is the group with one relation $ x_1\equal{}e$, which is isomorphic to the free group on $ x_2,x_3,\dots$, which is isomorphic to $ F$.

Actually, any pair with a nontrivial quotient of a group isomorphic to the original can be adapted to this, since every group is a quotient of a free group.


Thank you people for pointing out the problem herein. See this code? Looks good and efficient. (Note that this for loop is the fast way to calculate nCr in [i]any[/i] language: at the beginning of each iteration of for, res contains nCi, and you *know* nCi is integral; thus the multiplication results in another integer, then the division results in yet another integer.)
[code]
int chooose(int n,int r)
{
 int res=1;
 for(int i=0; i<r; ++i)
 {
  res=res*(n-i)/(i+1);
 }
return res;
}
label(string(chooose(3,2)),(2,0));
[/code]

However, in Asymptote, unlike C, a/b for int a,b returns a [i]real[/i]. Uh-oh, you have a syntax error. So you want quotient(a,b) here, which does return an integer. :)

[asy]int chooose(int n,int r)
{
 int res=1;
 for(int i=0; i<r; ++i)
 {
  res=quotient(res*(n-i),i+1);
 }
 return res;
}
label(string(chooose(3,2)),(2,0));[/asy]

Nope. It's because you try to assign a fractional value to an integer variable ;).

It's the extra semicolon in the for condition. I think the division should be okay, although I'm not convinced it would work as you would expect.

fedja was right. Also, 'postfix expressions are not allowed' :).

If you strongly prefer the C++ conventions, you can just put

int operator/(int a,int b){return floor((a+0.0)/b);}

to the beginning of your code. Then the division of 2 integers will give you the integer part of the quotient like in C++. Having the same operator / to give you the integer value when possible and the real value when impossible seems rather problematic from the programming standpoint.


Congratulations to the Maryland MathCounts team-- 5th place at nats! :lol:

2009 maryland team:

Sam Zbarsky
Jason Hyun
Charlie Pasternak
Diwakaran Ilangovan

Does anyone know the individual results??

I'm not from MD but I at least know the following (places in top 58):

Sam Zbarsky-14th
Diwakaran Ilangovan-16th
Jason Hyun-47th ( :o )


The remainder of three positive intergers when divided by 5 are 1,2, and 3. What is the remainder when the product of these integers are divided by 5?
I did this:
$ \ n\cdot m\cdot p\equiv 1\cdot 2\cdot 3\pmod 5$
Then I did $ \ 1\cdot 2\cdot 3 \equal{} 6$
then 6/5=1R1
So i got $ \boxed 1$
anything wrong with this?
(my mom took the answer book)

That looks fine to me.  :) You can also check it by trying an example, such as 6, 7, and 8. In that case, you can see that $ 6 \times 7 \times 8 \equiv 1 \pmod 5$.


integration
calculus
calculus computations

[b]Problem.[/b] Find $\int\frac{dx}{x-1}$.

We have $\int\frac{dx}{x-1}=\ln|x-1|+C$, obviously, but $\int\frac{dx}{x-1}=-\int\frac{dx}{1-x}=-\ln|1-x|+C$.
Huh, what's the matter with me?

You just forgot the $-$ sign coming from the chain rule when you differentiate $-x$, so the second antiderivative is actually $-(-\ln|1-x|)+C$ :)


Looks good to me, except in the third last line you have:

$x^{3}+y^{3}+z^{3}+3n+6xyz = 7+3 \cdot 336 = 343$ when I think you mean
$x^{3}+y^{3}+z^{3}+3n+6xyz = 7+3 \cdot 336+6xyz =343$

but that doesn't affect the actual solution


Find:
$ \int_{0}^{\infty}e^{\minus{}px} (\sin{qx} \plus{} \cos{qx})$ $ dx$

Just write out the sines and cosines using complex notation, and then you only have to integrate exponential functions.. the answer will be after short calculation:

\[ \int_0^{\infty}e^{\minus{}px}(\cos qx\plus{}\sin qx) dx \equal{} \frac{p\plus{}q}{p^2\plus{}q^2}\]

(Of course if q is real and p has a positive real part...)

Alternatively, integrating by parts twice, we get

$ I \equal{} \int_{0}^\infty\, e^{ \minus{} px}(\sin{qx} \plus{} cos{qx})\, dx$

$ \equal{} \frac {e^{ \minus{} px}}{q}(\sin qx \minus{} \cos qx) \plus{} \frac {p}{q}\int\, e^{ \minus{} px}(\sin qx \minus{} \cos qx)\, dx$, (putting $ u \equal{} e^{ \minus{} px}$ and $ dv \equal{} (\sin qx \plus{} \cos qx)\,dx$)

$ \equal{} \frac {1}{q} \plus{} \frac {p}{q}(\frac { \minus{} 1}{q}e^{ \minus{} px}(\sin qx \plus{} \cos qx) \minus{} \frac {p}{q}I)$

(on putting in the limits and with $ u \equal{} e^{ \minus{} px}\,$ and $ dv \equal{} ( \sin qx \minus{} \cos qx)\,dx$)

$ \Rightarrow\, I \equal{} \frac {1}{q} \plus{} \frac {p}{q^{2}} \minus{} \frac {p^{2}}{q^{2}}I$

$ \Rightarrow\, \frac {p^{2} \plus{} q^{2}}{q^{2}}I \equal{} \frac {p \plus{} q}{q^{2}}$

$ \Rightarrow\, I \equal{} \frac {p \plus{} q}{p^{2} \plus{} q^{2}}$.


Find the value of A if ((-x-11)/((x-4)(x+1))) equals ((A/x-4)+(B/x+1))

2
-1
-3
-2
none of these

[hide="solution"]
standard partial fractions method.
simplifying the RHS, we get the numerator to be:
$ (A \plus{} B)x \minus{}(4B \minus{} A)$
comparing the coefficients of LHS and RHS, we get
$ A \plus{} B \equal{} \minus{}1$ and $ 4B \minus{} A \equal{} 11$
solving, we get $ A\equal{} \minus{}3$
:)

[/hide]


A blank page, how to draw a  line?
Suppose that the width, height , and position  were  appointed .

For  example,  how to write  the  code  to  get  the  line  in  the  attachment?

Many thanks!

Like this?
\[ \text{
\begin{tabular}{rl}
\vline&
\begin{tabular}{l}
\vspace{0.5cm}
\\
\tiny{Labradore}
\\
\vspace{1cm}
\\
Labradore
\vspace{3cm}
\end{tabular}
\end{tabular}
}\]
You do not need dollar signs and \text{} in your document and you can adjust spacing to make it look nicer, but the general idea should be clear. There are many other ways, of course.

Thank you very much!

I only pay attention to   the   line,    not  the    words.

I want  to  draw  a  line  on  a  [color=red]blank [/color]page,  if  the width, height , and position of  the line  were appointed(especially   transverse line  or vertical line)

How to write the code with \LaTex?

Is   it  probably  to  draw the line with asymptote?

Try
[code]
\documentstyle{article}
\begin{document}
ttt
\newpage
\vspace{1cm}

\begin{tabular}{crlc}
\hspace{0.5cm}
&\vline&
\begin{tabular}{l}
\vspace{0.5cm}
\\
\tiny{Labradore}
\\
\vspace{1cm}
\\
Labradore
\vspace{3cm}
\end{tabular}
&\hfill
\end{tabular}
\vfill
\newpage
ttt
\end{document}
[/code]
and see if it does what you want. If not, tell what exactly is not to your liking.

How  to  change   the   length(for example, I want  the  length  is  the   textheight),  the  thickness,  and the position(for example, the  leftmost  or  the  middle   of   the   page )of  the  line?

And  is   it  probably  to  write  [color=red]anything(or insert  picture)[/color]  on   the  right   blank   of   the   page?

Thank  you!


Two vertical poles of different heights stand on ground level, as shown but not to scale. Straight wires from the top of each pole to the base of the other pole cross at a point 24 feet above ground. If the shorter pole is 40 feet, then what is the height of the taller pole?

I would appreciate it if you include an explanation along with your solution.

[hide=Hint]
Use similar triangles[/hint]

In general, if $ a$ and $ b$ are the two outer poles, then the middle pole of length $ x$ satisfies $ \frac1x\equal{}\frac1a\plus{}\frac1b$.

Draw the $ 24$ foot pole. Then, use similar triangles to prove it.

lol this is from the Forest Park Math comp in STL.

Yea... use similar triangles.


I'll start first.

Everyone after me says a word/phrase that pops up in their head when they see the word/phrase of the person above them, and then they say another word that the person below them has to associate with in their head.

Example: Person A says ninjas.
Person B says turtles. Person B says his own word: macaroni
Person C: cheese.......etc. and so on.

OK? Got it? Let's begin!

Me start: Donkey Kong

donkey kong??

uhhh....

stupid ape... (that's what i think it translates to in some language..)

aren't you guys supposed to say your own word (look at person B in the model)?

anyway...

midnight (because of Charlie Bone series)

dragonforce

curly fries ( jack in the box makes good curly fries)

ciabatta

I have no clue what that means...

Hie, Muffin-Man!

DK! DONKEY KONG! DONKEY KONG IS HERE!

p.s. DonkeyKong why did you change your avatar??? i likeed the one where you portrayed yourself as the 800 lb ape, plz keep it  im beggin ya i loved the old one so much :maybe:


modular arithmetic
quadratics
real analysis

Using the in/out theorem, we get: LHS = 2*(p-1)!*(2p-1)C(p-1) = 2(p+1)(p+2).....(p+(p-1)).

Now, calculating the LHS gives (p-1)! + (1+2+..+p-1)p + ([1+2+....+p-1)^2-[1^2+2^2+....+(p-1)^2]p^2

The (1+2+..+p-1)p term is equal to p^2(p-1)/2

The last term in there is equal to: ([p(p-1)/2]^2 - [(p-1)p(2p-1)/6]^2)p^2.  It's pretty obvious that all of those are equal to 0 mod p^3

So we have 2*(p-1)!*(2p-1)C(p-1) = 2[(p-1)! + p^2(p-1)/2] mod p^3

I'm not sure if that means anything.

[quote="mathfanatic"]Let $p$ be a prime greater than 3.  Prove that
\[\binom{2p}{p} \equiv 2 \pmod{p^3}.\]

Any help would be appreciated.[/quote]

You could try to prove that

\[\binom{a}{b}\equiv \binom{ap}{bp} \pmod{p^3}\]

where $p$ is a prime greater than 3 and $a,b$ are nonzero natural numbers.
As a particular case you'll obtain the solution to your problem.

Well, I can prove it $\pmod{p^2}$, but I'm not sure how to do it for $p^3$...

Anyways, using Vandermonde's Identity,

\[\sum_{k=0}^r \binom{m}{r-k}\binom{n}{k} = \binom{m+n}{r},\]

setting $p=m=n=r$, and simplifying a bit, we get

\[\binom{2p}{p} \equiv \binom{p}{0}^2+\binom{p}{1}^2+\cdots + \binom{p}{p}^2 \equiv \binom{p}{0}^2+\binom{p}{p}^2 \equiv 2 \pmod{p^2},\]

since every term except the first and the last have a factor of $p^2$ in them, we're done...

First I prove a result that I will use later.

I will use QR to abbreviate Quadratic Residue, that is, a perfect square.

Lemma: if p > 3 is a prime, then the sum of the QR's mod p is 0.

Proof:
 Consider the sum $\sum_{i=1}^{p-1} i^2 = \frac{(p-1)(p)(2p-1)}{6}$.

First note the fraction at the right is divisible by p as p > 3.

Note this sum adds in each quadratic residue exactly twice, but it doesn't matter since $p \ne 2$.

This proves the lemma.
---- end proof ----

Now to the main claim:

$\text{Claim: }{2p \choose p} \equiv 2 \pmod{p^3}$
ok, be prepared for some algebra.

First, we use vandermonde's convolution to rewrite

${2p \choose p} = \sum_{i=0}^{p} {p \choose i}^2
 = 2 + \sum_{i=1}^{p-1} {p \choose i}^2$,

so we want to show $\sum_{i=1}^{p-1}{p \choose i}^2 \equiv 0 \pmod{p^3}$

Again, we can rewrite

$\sum_{i=1}^{p-1}{p \choose i}^2 = \sum_{i=1}^{p-1}\frac{p^2}{i^2}{p-1 \choose i-1}^2$

so pulling out the factor of $p^2$, we must show

$\sum_{i=1}^{p-1}\frac{{p-1 \choose i-1}^2}{i^2}
= \sum_{i=1}^{p-1}\frac{1}{(i!(p-i)!)^2} \equiv 0 \pmod{p}$

(NOTE: $1/x \pmod{p}$ just means the inverse of $x$, modulo $p$. It is guaranteed to exist for all things we will use because they are all nonzero and $\mathbb{Z}_p$ is a field)

However, if we look at $i!(p-i)!$, we see it is equal to
$(-1)^{p-i}i(p-1)!$, so $(i(p-i)!)^2 \equiv i^2 \pmod{p}$, so

$\sum_{i=1}^{p-1}\frac{1}{(i!(p-i)!)^2} \equiv \sum_{i=1}^{p-1}\frac{1}{i^2} \equiv 0 \pmod{p}$,

By the lemma above, since this is twice the sum of the QRs $\pmod{p}$.


Can't quite get this. Please show work! Thanks!

Determine whether each ordered pair is a solution of the given equation.

-6x+5y=-6  (1,0);(2,6/5)

It says that their both yes. I'm having the most problem with the fraction.  :idea:

Just plug in each $ x$ and $ y$ values in the equation.  If the equation is true, then the ordered pair is a solution.

$ \minus{} 6x \plus{} 5y \equal{} \minus{} 6$

$ \minus{} 6 \cdot 1 \plus{} 5 \cdot 0 \equal{} \minus{} 6$

$ \minus{} 6 \plus{} 0 \equal{} \minus{} 6$

$ \minus{} 6 \equal{} \minus{} 6$

True.

$ \minus{} 6x \plus{} 5y \equal{} \minus{} 6$

$ \minus{} 6 \cdot 2 \plus{} 5 \cdot \frac {6}{5} \equal{} \minus{} 6$

$ \minus{} 12 \plus{} 6 \equal{} \minus{} 6$

$ \minus{} 6 \equal{} \minus{} 6$

True.

Therefore, $ (1,0)$ and $ (2,\frac{6}{5})$ are solutions to $ \minus{}6x \plus{} 5y \equal{} \minus{}6$.

Oh, I get it now. I was doing that totally wrong. Thanks

Now I don't get these. Same type of problem...but squared

y=2xsquared  (1,2);(3,18)     answer:yes;yes

I don't know what to do when I get to the y and I have to 2=2xsquared.

How do I unsquare the x or do I need to do that?

Please show work.

PS does the same rule apply when the x is tripled or quadruple times?

To "unsquare" x you would take the square root, but this is unnecessary. Just plug in values.

Example: (1,2) is a valid solution to $ y \equal{} 2x^2$ because when you plug it in, you get $ 2 \equal{} 2(1^2)$ which simplifies to $ 2 \equal{} 2$, which is obviously true.

In case you're curious, here's how to do it with a sqraure root:
\[ y\equal{}2x^2\\
\sqrt y\equal{}\sqrt{2x^2}\\
\sqrt y\equal{}x\sqrt 2\\
\sqrt 2\equal{}1\sqrt 2\\
\sqrt 2\equal{}\sqrt 2
\]
But if that's too complicated, don't worry about it.

Perhaps you'd like to take a look at your thread in Classroom Math?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=227383


algebra
polynomial
calculus
integration
complex analysis
complex analysis unsolved

The problem is this:

Let M>0 and a>0, a not necessarily an integer. Suppose f is entire and

|f(z)|[u]<[/u]M|z|^(a)

for all z in the complex plane C. Show that f is a polynomial.

--
At first I thought this was the same as proving Extended Liouville's Theorem, but it really isn't, huh. How do I deal with the condition that a is not necesarily an integer, and that |f(z)|[u]<[/u]M|z|^(a) instead of |f(z)|<M|z|^(a), and still apply Extended Liouville's Theorem?

The difference between $ \le$ and $ <$ is insignificant. So is the difference between integer and non-integer $ a$.

The proof I like here: Let $ C$ be a circle of radius $ R>1$, centered at $ z_0$. Let $ n>a$. By the Cauchy integral formula, $ f^{(n)}(z_0)\equal{}\int_C \frac{f(z)}{(z\minus{}z_0)^{n\plus{}1}}\,dz$, so $ |f^{(n)}(z_0)|\le \int_C \frac{M'|z\minus{}z_0|^a}{|z\minus{}z_0|^{n\plus{}1}}\,|dz|\le 2\pi CM'R^{a\minus{}n}$. This goes to zero as $ R\to\infty$, so $ f^{(n)}$ is identically zero, and $ f$ is a polynomial with degree $ \le a$.

(When we shift the center, we get a new constant $ M'$ in the bound, but not a new exponent.)

$ f(z)\equal{}\sum_{n\geq 1}a_{n}z^n$

$ a_{n}r^n\equal{}\frac{1}{2\pi}\int_{0}^{2\pi}f(r.e^{ix})e^{\minus{}ixn}dx$

$ |a_{n}r^n|\leq M.r^{a}$

$ |a_{n}|\leq \frac{M}{r^{n\minus{}a}}$  fixed $ n$ such that it is the smallest integer with $ n\minus{}a>0$

tends $ r$ to +oo

the coeficient $ a_{n}$ are zero for $ n>a$

$ f$f is a polynomial


Let $ 21 $ points be on one circle . Prove that : There is at least 100 arc which are defined by these 21 points form angles $ \leq 120 $ degree .

Suppose there are none... :P
draw an edge between $2$ points if their central angle is bigger than $120$. Use Turans Theorem to see there is a $K_3$ in our graph (Turans theorem is in the theorems and formulas sections). Prove a $K_3$ cannot exist.

I can't understand your solution , Pascual .
  I don't have a solution for this problem but I think we can solve this problem without Graph ?


How many ways are there to put 4 balls in 3 boxes if the balls are distinguishable but the boxes are not?

One can proceed with casework:

Case 1: boxes hold 4, 0, 0 balls

There is $ 1$ way to do this.

Case 2: boxed hold 3, 1, 0 balls

Choose which ball is separate in $ 4$ ways.

Case 3: boxes hold 2, 2, 0 balls

This symetric, so put one of the balls in a box. We can choose which of the other three pairs up with it in $ 3$ ways. Alternatively, we can do $ \binom{4}{2}$ and then divide by $ 2$ for overcounting.

Case 4: boxes hold 2, 1, 1 ball

Choose which two are in the box which holds two balls; the other balls have one remaining choice. This is done in $ \binom{4}{2} \equal{} 6$ ways.

These cases total $ 1 \plus{} 4 \plus{} 3 \plus{} 6 \equal{} \boxed{14}$.

MRGH I GOT 23
indistinguishable boxes mrph

[hide][quote=Alcumus solution]
(2,2,0): There are $\binom{4}{2} = 6$ ways to choose the balls for the first box, and the remaining go in the second box. However, the two pairs of balls are interchangeable, so we must divide by 2 to get $6 / 2 = 3$ arrangements.
[/hide]

Why are the two pairs of balls interangeable? I thought they were distinguishable.

Huh. So far all of the problems in this topic have been the exact same:
4 distinguishable things, 3 indistinguishable containers. The first one was cousins in rooms. The second was foods in bags. This one is balls in boxes.

AHHHHHH, i thought 4C2 was 12

HAHHAHAHAHAH

Is $4$C$2$ basically $\frac{4 \cdot 3}{2}?$

[quote=TheReaper10]Why I keep get these balls and boxes...[/quote]

Distinguishability, it's fine you can change the numbers, what is distinguishable or not so they are easy to get a lot of, and the AoPS staff have decided to put them there.

[quote=Wood2009][quote=TheReaper10]Why I keep get these balls and boxes...[/quote]

Stop postfarming and clogging up gamebot threads please.[/quote]

This itself is postfarming, some people are gaining hundreds of posts by a week...

Pretend for a sec that the boxes are distinguishable: Then because the balls are all distinguishable, everything is overcounted by a factor of 6x  EXCEPT for the case when two boxes are empty, which is only overcounted by 3x. Hence we can count where both balls and boxes are distinguishable, take out the 3x case tied to the 4/0/0 ball box split, then divide by 6x (as everything else is overcounted by 6x), and finally add back the 3x case, which only needs to be counted once:  $$ \frac{3^4 - 3}{6} + 1 = \boxed{14}$$

[quote=ilovepizza2020]It's $\frac{4!}{2!\cdot 2!}$[/quote]

[quote=Wood2009][quote=TheReaper10]Why I keep get these balls and boxes...[/quote]

Stop postfarming and clogging up gamebot threads please.[/quote]

[quote=peelybonehead]Why can’t I use stars and bars here?[/quote]

stars and bars only works with indistinguishable balls and distinguishable boxes


The unit hypercube in 8 dimensions can be defined as the set of points $ (x_1, x_2, \ldots, x_8)$ such that $ 0 \leq x_i \leq 1$ where $ i$ is one of $ \{1, 2, \ldots, 8\}$. Suppose the hypercube is cut into smaller hyper-rectangular boxes with side lengths of $ \frac {1}{3}$ or $ \frac {2}{3}$ by hyperplanes whose equations are $ x_i \equal{} \frac {1}{3}$ where $ i$ is an element of $ \{1, 2, \ldots, 8\}$. Color the hyper-rectangular box with vertex on the origin red, color those adjacent to red hyper-rectangular boxes blue, and those adjacent to blue hyper-rectangular boxes red where "adjacent" means bordering along one of the hyperplanes $ x_i \equal{} \frac {1}{3}$. If $ R$ equals the total volume of the red hyper-rectangular boxes and $ B$ equals the total volume of the blue hyper-rectangular boxes, compute $ R \minus{} B$.

[hide="Solution without answer"]Let $ R_n$ and $ B_n$ be the total volume of the red and blue boxes respectively if we did the process for a hypercube of $ n$ dimensions. If we know $ R_n$ and $ B_n$, observe that when going to $ n\plus{}1$ dimensions, the following happens:
- The additional dimension of the boxes whose volumes we know will measure 1/3.
- The rest of the hypercube will be a copy of the boxes we already know, except that the colors are flipped and the additional dimension measures 2/3.
This gives the following recurrence.
$ R_{n\plus{}1} \equal{} R_n/3 \plus{} 2B_n/3$
$ B_{n\plus{}1} \equal{} B_n/3 \plus{} 2R_n/3$
We have $ R_1 \equal{} 1/3$, $ B_1 \equal{} 2/3$ easily, so it's matter of using the recurrence to compute $ R_8 \minus{} B_8$. Someone else can do the computation if they feel like it.[/hide]

[hide][quote="MellowMelon"]
This gives the following recurrence.
$ R_{n \plus{} 1} \equal{} R_n/3 \plus{} 2B_n/3$
$ B_{n \plus{} 1} \equal{} B_n/3 \plus{} 2R_n/3$
We have $ R_1 \equal{} 1/3$, $ B_1 \equal{} 2/3$ easily, so it's matter of using the recurrence to compute $ R_8 \minus{} B_8$. Someone else can do the computation if they feel like it.
[/quote]
Now it's easy:
\[ R_{n \plus{} 1} \minus{} B_{n \plus{} 1} \equal{} \minus{}\frac{R_n \minus{} B_n}{3}.\]
Since $ R_1 \minus{} B_1 \equal{} \minus{}1/3$, it follows that $ R_n \minus{} B_n \equal{} (\minus{}1/3)^n$.  In particular, $ R_n \minus{} B_n \equal{} 1/3^8$.
[/hide]


Θα ήθελα να ρωτίσω σε τι αναφέρονται τα ακόλουθα προβλήματα

[color=red]Yang-Mills Theory
Poincare Conjecture
Riemann Hypothesis
Navier-Stokes equations[/color]

Gianni mporeis na koitakseis edw gia perissoteres plirofories gia ta provlimata pou theleis

http://www.claymath.org/millennium/

Dipla apo kathe provlima exei to "Official Problem Description" pou eksigei ti akrivws leei to kathe provlima.

H eikasia tou Poincare einai pleon Thewrima, afou APODEIXTHKE perisi ton Aprilio (oute xrono den exei kleisei akomi) apo enan Rwsso onomati Grigori (Grisha) Perelman.

Alexandros

Τελικά τον βρήκαν για να του δώσουν το 1 εκατομμύριο δολλάρια;  :P

Ela nte!! Meta tin dialeksi pou ekane sto MIT, tin kopanise! Den kserw an ton vrikan. :)

Alexandros

[quote="thatin"]den einai provlimata einai vathmides ierarxisis twn melwn tou forum  :rotfl:

diladi twra tha vgalloun tin Poincare Conjecture?[/quote]

 :rotfl: Χαχα , σωστό .... :lol:

thatin
[hide]Απο την Έδεσσα είσαι?[/hide]

[hide="@giannis"]/\/4!
eisai sta iwannina?
[/hide]


Fie ABCD un tetraedru , M si N mijloacele respectiv pentru [AB] si [CD].Demonstrati ca:
a) 2MN<BC+AD
b) 2(AN+BN)<AC+AD+BC+BD
c) 4MN<AC+AD+BC+BD

   :!: Dati doua rezolvari acestei probleme.

Mihnea,vezi ca e un forum special pentru asta, http://www.mathlinks.ro/Forum/index.php?f=263
Si in legatura cu punctul a,se face o constructie ajutatoare($P$ mijl $AC$) si de aici inegalitatea triunghiului si e imediat.Pt celelalte subpuncte,tot inegalitatea triunghiului daca nu ma insel..

Si cum ramane cu cele doua solutii?Nu ziceai asa, Mihnea?

eu si cu andy ciupan suntem de la aceeasi scoala si de fapt problema asta mi-a dat-o profesorul lui de la clasa!!Eu i-am mai adus cateva imbunatatiri!!!

Mihnea, tu n-ai fost la Pitagora?Am vazut ca a luat excelent Andrei Ciupan(locul I la amandoua).


Let $ {0 < a,b,c < 1}$ and $ {ab \plus{} bc \plus{} ac \equal{} 1}$ then find mininum value of $ {\frac {1}{1 \minus{} a} \plus{} \frac {1}{1 \minus{} b} \plus{} \frac {1}{1 \minus{} c}}$.
Looking forward to your nice solutions , dear mathlinkers! :)

umm... :maybe:

by Cauchy-Schwarz and AM-GM, $ \sum \frac {1}{1 \minus{} a} \ge \frac {9}{3 \minus{} \sum a} \ge \frac {9}{3 \minus{} \sqrt {3}} \equal{} 9 \plus{} 3\sqrt {3}$.

Equality holds for $ a\equal{}b\equal{}c\equal{} \frac{1}{\sqrt{3}}$.


again a problem i can solve  :D
we have $f(x+1)+94\leq f(x+95)=f(x+5\cdot 19)\leq f(x)+95$, so $f(x+1)\leq f(x)+1$. but since $f(x+94)\geq f(x)+94$, we must always have inequality.

Peter

[quote="N.T.TUAN"]General problem

Suppose $ f$ is a real functions satisfying $ f(x+a) \leq f(x)+a$ and $ f(x+b) \geq f(x)+b$ for all real $ x$, here $ a,b\in\mathbb N$ and $ \gcd (a,b)=1$. Prove that $ f(x+1) = f(x)+1$ for all real $ x$.[/quote]

we use $ f(n+ax)\le f(x)+ax$ and $ f(n)\ge f(n-b)+b\ge f(n-2b)+2b\ge ...\ge f(n-yb)+yb$
we have $ gcd(a,b)=1$ so:
$ \exists (x,y)\in N^{*}^{2}: \ ax-by=1$
so, $ f(n+1)=f(n+ax-by)\le f(n-by)+ax\le f(n)-by+ax\le f(n)+1$
and $ \exists (x,y)\in N^{*}^{2}: \-ax+by=1$
so, $ f(n+1)=f(n+ax-by)\ge f(n-by)+ax\ge f(n)-by+ax\ge f(n)+1$
then $ f(n+1)=f(n)+1$

[quote="Thamdu"][quote="Arne"]Suppose $f$ is a real functions satisfying $f(x + 19) \leq f(x) + 19$ and $f(x + 94) \geq f(x) + 94$ for all real $x$. Prove that $f(x + 1) = f(x) + 1$ for all real $x$.[/quote]
This is easy and old problem.[/quote]

You really should avoid such posts :
- claiming that a problem is easy is useless : if the solution is given and easy, everybody knows it. If the solution is given and difficult, it seems to be false. If the solution is not given and you claim it's easy without giving the solution, you look like a liar.
- claiming that the problem is old is useless : everybody can see that the original post was more than six years old ...

[quote="Thamdu"][quote="Arne"]Suppose $f$ is a real functions satisfying $f(x + 19) \leq f(x) + 19$ and $f(x + 94) \geq f(x) + 94$ for all real $x$. Prove that $f(x + 1) = f(x) + 1$ for all real $x$.[/quote]
This is easy and old problem.[/quote]

What was that? I agree totally Patrick.


[u][b]Inequality:[/b][/u]

[i]Let $a,b,c$ three positive real numbers. Prove that the following inequality holds true:[/i]

\[ \frac{yz}{x(y+z)}+\frac{zx}{y(z+x)}+\frac{xy}{z(x+y)}\geq 2\cdot\frac{x^2(y+z)+y^2(x+z)+z^2(x+y)}{(x+y)(y+z)(z+x)}  \]

Yassin your inequality is in fact my inequality after making the substitutions $x=\frac{1}{a}, y=\frac{1}{b}$ and $z=\frac{1}{c}$. Now this problem becomes [color=blue] Inequality *56 [/color]. ;)


probability
LaTeX
number theory
prime numbers
set theory

My teacher gave us a problem today, out of the blue. We haven't studied probability in about a year now, and I'm extremely rusty, since I was bad at it even when we were studying the subject.

Given a set of numbers consisting of $ \{1,2,3,4,5,6,7,8,9,10\}$, what is the probability of picking 2 or more prime numbers at random? Picked numbers aren't replaced.

Well...my reasoning went like this, and I still don't honestly see why it is wrong.
There are 4 possible prime numbers, 2,3,5,7.
In the first possibility, there are only 2 primes...so...
$ \frac{4}{10}*\frac{3}{9}*\frac{6}{8}*\frac{5}{7}\equal{}\frac{1}{14}$
3 primes...
$ \frac{4}{10}*\frac{3}{9}*\frac{2}{8}*\frac{5}{7}\equal{}\frac{1}{42}$
and 4...
$ \frac{4}{10}*\frac{3}{9}*\frac{2}{8}*\frac{1}{7}\equal{}\frac{1}{210}$

Hmm...now I'm seeing a problem I didn't notice before...How to apply these numbers to each other...This is wrong, but why?

How would you do this with combinations? I'm almost positive they're involved...I would guess something like...
$ 4c2*6c2\plus{}4c3*6c1\plus{}4c4*6c0$?

P.S. How do you type out the combinations the other way, within parentheses? And how do you represent the dot for multiplication? I don't like the asterisk.

Er... how are you picking numbers?  Are you picking a random number of numbers, or are you considering a random subset?  The problem is very unclear.

$ \text{\LaTeX}$ answers:  {n \choose k} and \cdot.

I'm sorry, I forgot that set means a lot of things. I meant, basically, you're pulling those 10 numbers out of a hat, one at a time, without replacing them. No set theory or anything of the sort.

[quote="Icy"]you're pulling those 10 numbers out of a hat, one at a time, without replacing them.[/quote]

Are you asking for the probability of picking two or more prime numbers [i]on the first two tries[/i]?  Please try to be specific.  Not reading the question correctly is the #1 reason people get probability questions wrong, in my mind.

You want \mathbb or possibly \mathcal.  These show up as $ \mathbb{Z}$ and $ \mathcal{S}$, respectively.

Haha, I didn't realize that I didn't say how many are pulled out. :D You're pulling out exactly 4 numbers once. They want the probabilty of 2 or more of these numbers being prime. So...in this one instance where you're pulling out 4 numbers, what is the probability that 2, 3, or 4 of these 4 numbers are prime? If I still haven't worded it well enough, I'll give up for now and go dig up the problem to copy word for word.

[quote="Icy"]you're pulling out exactly 4 numbers once. [/quote]

Thank you.  This was the missing information I wanted.

[hide="Solution"] We can pick either $ 2$ primes and $ 2$ composites, $ 3$ primes and $ 1$ composite, or $ 4$ primes.  The probability is therefore

$ \boxed{\frac{{4\choose 2}{6\choose 2}\plus{}{4\choose 3}{6\choose 1}\plus{}{4\choose 4}{6\choose 0}}{{10\choose 4}}}$ [/hide]

Makes sense now that I look at it. What was so illogical about my first attempt though? I'm not seeing it, aside from the fact that it does not return the right answer.

The last factor in the second expression should be $ \frac{6}{7}$ (there are six non-primes) rather than $ \frac{5}{7}$.

My way returns $ \frac{11}{105}$ and the method with combinations returns $ \frac{23}{42}$ Clearly there is something wrong with my method, but I don't know what. It looks logically sound to me.

Oh, I thought I was wrong before, but my initial thought was correct.

The issue with your solution is that it doesn't matter what order you draw the cards in as long as you get the right collection of cards in the end.  The first probability you've got needs to be divided by $ {4\choose 2}$ and the second by $ {4\choose 3}$.

Ah, I get it. I think I was looking at permutations, not combinations. Thanks.


Mr. Okokok announced to his class that the class's average was 82% and after removing an unusual score, the new average was 84%. What was an unusual score that Mr. Okokok removed if there are 20 people in the class?

[hide="An odd way of doing it"]
The total percentage must be 1640 with the odd score.
The total percentage must be 1596 without the odd score.

The odd score must be 44%.

Stupid people =o.
[/hide]

[hide]the total score is 1640

the score without the score is 1596

the score is then 44%

that [b]is[/b] kind of an unusual score :P [/hide]

[hide]
The total with the score is 1640
The total without the score is 1596
1640-1596=[b]44[/b][/hide]

Har har, been peeking at my English test scores, Silverfalcon?

Haha...

Could've been :lol:

Although I didn't know Mr.Okokok was your teacher! :D


[quote="1/(ln x)"]An ideal gas system expands from 1.0 m^3 to 3.0 m^3 at atmospheric pressure while absorbing $ 5.0 \times 10^{4}$ J of heat in the process. What is the change in internal energy of the system. Since the system absorbs heat, (1) the temperature of the system must increase, (2) the internal energy of the system must increase or (3) none of the preceding.

Please check my work:
$ \Delta U \equal{} Q \minus{} W$
= $ 5.0\times 10^{4} J \minus{} (101325\frac {N}{m^{2}})(3.0 \minus{} 1.0 m^{3})$
= $ 5.0\times 10^{4} J \minus{} 202650 J$
= -152650 J

So, the change in internal energy of the system is $ \minus{} 152650 J$, and the choice (1), (2), (3), is (3).[/quote]

Your calculation is correct. Thus internal energy is lost, and it follows that (2) is false. Nonetheless, for an ideal gas : $ PV \equal{} nRT \rightarrow P\Delta V \equal{} nR\Delta T$ if pressure is constant. Therefore, since the volume increases so does the temperature; (1) is true.

[quote="Carcul"]No, that is wrong. The internal energy of an ideal gas is only a function of temperature: therefore, if U decreases then also the temperature must decrease. The correct answer is (3).[/quote]

That's an astute observation, however, the data given in the problem is inconsistent. If an ideal gas at constant pressure undergoes a volume expansion, the temprature must increase (per my original answer). The internal energy: $ U\equal{}C_V\cdot T \plus{} const.$, to a linear approximation.

$ C_V \equal{} \frac{R}{\gamma \minus{}1}$.

For an ideal gas, $ \gamma > 1 \therefore C_V > 0$.

Hence, $ \Delta U \equal{} C_V \Delta T > 0$

Since $ \Delta U > 0$ then the heat absorbed must be $ Q > P\Delta V \equal{} 202 kJ$.

[quote="Dr. No"]$ C_V \equal{} \frac{R}{\gamma \minus{} 1}$[/quote]

The correct expression is $ C_V \equal{} \frac{nR}{\gamma \minus{}1}$

[quote="Dr. No"]For an ideal gas, $ \gamma > 1 \therefore C_V > 0$[/quote]

First, why one thing imply the other? Second, $ \gamma > 1$ and $ C_V > 0$ holds for any substance.

[quote="Dr. No"]Hence, $ \Delta U \equal{} C_V \Delta T$[/quote]

We have $ dU \equal{} C_V dT$, and considering that Cv does not vary with temperature then we can write $ \Delta U \equal{} C_V \Delta T$. This implyes that, if $ \Delta U < 0$ then also $ \Delta T < 0$ since $ C_V$ is always positive.

[quote="Dr. No"]$ C_V \equal{} \frac {R}{\gamma \minus{} 1}$[/quote]

[quote="Carcul"]The correct expression is $ C_V \equal{} \frac {nR}{\gamma \minus{} 1}$[/quote]
True, I was working on a specific basis and forgot to include the factor of $ n$. However, having done so does not affect the conclusion.

[quote="Dr. No"]For an ideal gas, $ \gamma > 1 \therefore C_V > 0$[/quote]

[quote="Carcul"]First, why one thing imply the other? Second, $ \gamma > 1$ and $ C_V > 0$ holds for any substance.[/quote]
It is true that $ C_V$ is positive definite for any substance, yet I had defined it in terms of $ R$ and $ \gamma$ for an ideal gas in particular. It was already demonstrated that the temperature must increase in the circumstances in question, and thus the only way to get $ \Delta U < 0$ is to have $ C_V < 0$. What I was pointing out is that with $ \gamma > 1$ this could never be the case. $ C_V$ is always positive.

[quote="Dr. No"]Hence, $ \Delta U \equal{} C_V \Delta T$[/quote]

[quote="Carcul"]We have $ dU \equal{} C_V dT$, and considering that Cv does not vary with temperature then we can write $ \Delta U \equal{} C_V \Delta T$. This implyes that, if $ \Delta U < 0$ then also $ \Delta T < 0$ since $ C_V$ is always positive.[/quote]
As I said, the reason for my proof that $ C_V$ is always positive, above, is because we already knew that $ \Delta T > 0$ in this case and therefore, calculating a loss of internal energy, i.e. $ \Delta U < 0$, is a mistake. Not that the original calculation was made incorrectly, but that the heat absorbed has to be more than was indicated in the problem statement.

[quote="Carcul"]Yes Dr. No, there is an inconsistency, that's why the correct answer is (3), right? Did you think for a moment that finding the inconsistency could be the all purpose of the problem?

Also, we already know that Cv is always positive, then why bother to express it in terms of R and gamma?[/quote]

No, the correct answer is that the problem is inconsistently stated. As long as this is the case a correct answer to the problem does not exist.

An analogy would be, "What is the minimum number of daily calories needed to sustain a purple zebra with orange stripes?"

You might be tempted to infer that it would be the same as the minimum number of daily calories required to sustain an ordinary black & white zebra -- but that would be an error. Since for all we know purple zebras with orange stripes don't exist, the question has no answer.

Likewise, the question of the given problem asks about the thermodynamic properties of an ideal gas subject to contraints that are contrary to the known behavior of ideal gases.  Ergo, the question has no answer, not even option (3).

What I thought about why the question was asked is immaterial.

The reason I found it necessary to express $ C_V$ in terms of $ R$ and $ \gamma$ was because I didn't introduce the fact that $ C_V$ is always positive [i]a priori[/i].  I deduced that $ \Delta T$ was positive from the ideal gas law. $ \Delta U$ was found to be negative by direct calculation from the data given in the problem. From which I determined if $ \Delta U < 0$ and $ \Delta T > 0$ then $ C_V < 0$. (Even though it is a known fact that $ C_V$ is always positive, it does no harm to do what I did and show by calculation that $ C_V > 0$.)  Whether one invokes it as a theorem, "$ C_V$ is always postive," or shows it by calculation: $ C_V \equal{} \frac{nR}{\gamma \minus{} 1} > 0$ makes no difference.

One still has $ \left(\Delta T > 0 \wedge \Delta U < 0 \Rightarrow C_V < 0\right)$, but $ C_V > 0$. By [i]modus tollens[/i], we conclude $ \neg\left(\Delta T >0 \wedge \Delta U < 0 \right)$.  This contradicts the problem statement.

[quote="Dr. No"]Whether one invokes it as a theorem, " is always postive," or shows it by calculation[/quote]

But you haven't show anything. If you want to be complete, then you must also show that $ \gamma > 1$. Otherwise, saying that Cv > 0 because gamma > 1 is as good as Cv > 0. Also, that's not a theorem, since there are some processes and systems for which the heat capacities are negative. But in "ordinary" thermodynamics, we can safely consider them to be positive.

How about a related problem? Solve the exercise posted by 1/(ln x) but ignoring the "at atmospheric pressure" part.

[quote="Carcul"]How about a related problem? Solve the exercise posted by 1/(ln x) but ignoring the "at atmospheric pressure" part.[/quote]

Okay.

$ Q = C_P \Delta T$

$ C_P = \frac{\gamma nR}{\gamma - 1}$

$ \Rightarrow \Delta T = \frac{Q}{C_P} =\frac{\gamma-1}{\gamma}\frac{Q}{nR}$

$ \Delta U = C_V \Delta T$

$ \because C_V=\frac{nR}{\gamma - 1} \Rightarrow \Delta U =\frac{Q}{\gamma}$

$ Q = 5\times 10^4\,\, \text{J}$

$ \gamma = \frac{7}{5}$ for a diatomic ideal gas.

$ \Delta U = \frac{5}{7} \cdot (5\times 10^4\,\, \text{J}) = 4\times 10^4\,\, \text{J}$ (to one significant digit.)

$ -P\Delta V = \Delta U - Q = (4\times 10^4 - 5\times 10^4)\,\, \text{J} = -1\times 10^4\,\, \text{J}$.

$ \Delta V = (3 m^3 - 1 m^3) = 2 m^3$.

$ P = \frac{1\times 10^4}{2}\,\frac{\text{J}}{\text{m}^3} = 5\,\text{kPa}$.

$ nR\Delta T = P\Delta V = 1\times 10^4 \,\text{J}$

$ \Rightarrow n\Delta T = \frac{1\times 10^4}{8.3145}\, \frac{\text{J}}{\text{J/mol K}} = 2000 \,\,\text{mol K}$ (to one significant digit.)

Since $ n > 0 \Rightarrow \Delta T > 0$.

[quote="Carcul"]Your answer is wrong. Who told you that the pressure remains constant during the process? And in any case your first two derivations are totally wrong, because you mixed up temperature variations at constant pressure and at constant volume.[/quote]

I solved the problem you asked me to solve, that is the same one given with the exception that the process was not carried out at atmospheric pressure. The calculations I made were all based upon the ideal gas law and the data given in the problem.

As we've already discussed the original problem is flawed. If you assume that the pressure is not constant, you will obtain a result that once again contradicts the problem statement even as modified.

[quote="Carcul"]Solving a problem in a correct way to get a contradictory result (because of the given data) is not the same as solving erroneously a problem and getting a contradictory result (or a "correct" result). I was referring to your reasoning, which is wrong independently of the data being contradictory or not.[/quote]

What?!?


projective geometry
conics
geometry proposed
geometry

Let a cute angle $xOy$ .Let three points namely $C_{1},B_{1},A_{1}$ putted in this way  on $Ox$  and three other points   $A,B,C$ putted in this way  on $Oy$ .
We have $C_{1}B\cap B_{1}A=M$ , $B_{1}A\cap A_{1}B=D$ ,
$C_{1}C\cap B_{1}A=E$ , $A_{1}A\cap C_{1}C=K$ , $B_{1}C\cap A_{1}B=L$ ,$C_{1}C\cap A_{1}B=Z$ .Prove that $M,K,L$  are on the same line

For easiness, at the places of the points A, B, C, we change the labels of A, C, with the labels C, A, respectively.

     So by Pappus theorem we have that the points K, L, M, are collinear.

     The Pappus theorem states that if three points A1, B1, C1, are on a given line (e), in arbitrary order and other three points A, B, C, are on another given line (e'), in arbitrary also order, then the point K, as the intersection point of AC1, A1C, the point L, as the intersection point of AB1, A1B and the point M, as the intersection point of BC1, B1C, are collinear.

     Kostas Vittas.

This theorem is still true , if the six points are on a circle or a conic  and is well known as ,,Pascal Theorem,, with many usefull aplications.

  Babis


quadratics
number theory proposed
number theory

Find all integers $ n$ such that $ 2n+1$, $ 3n+1$ and $ 6n+1$ are simultaneously perfect squares.

Remark: this is a stronger problem than http://www.mathlinks.ro/Forum/viewtopic.php?t=23649, but lacks the same kind of short solution (in fact, this one here seems to be more ugly ;) ).

In American Mathematical Monthly has following problem: [i]Find all $ n\in\mathbb N$ such that $ 2n+1$ and $ 3n+1$ are perfect squares.[/i]

And its answer is $ n=y_{k}^{2}-x_{k}^{2}(k\geq 0),$ here $ x_{m}=u_{m}+2v_{m},y_{m}=u_{m}+3v_{m}(m\geq 0)$ and $ u_{0}=5,v_{0}=2$.

It seems to be equivalent to the equation:

$ (2+\sqrt{3})^{x}+(2-\sqrt{3})^{x}=y^{2}$

(where $ 4(3n+1)=y^{2}$)

which i think it's quite well-known, though I could neither solve it nor find any reference...

i found the following result:
$Lemma:$ if $2n+1$ & $3n+1$ are both perfect squares,then $40|n$,its almost trivial,now set $n=60m$,then say $6n+1=240m+1=a^2$, $a$ odd let $a=2b+1$,then
$b(b+1)=60m$,taking cases (since $gcd(b,b+1)=1$), $b$ or $b+1$$=1,2,3,4,5,6,10,12,15,20,30,60,m$, we get valid values $m=0,4,7,61$ but $4,7,61$ leads to contradiction for $2n+1$,hence i found the only solution $n=0$

Hi, thanks
but i dont get it.. m is not prime, so the values of b and b+1 to be among the numbers of the list u wrote
as far as i understood u've tried to make a list of the divisors of 60m but i disagree that the list u wrote is that list

If we denote $2n+1=x^2$, $3n+1=y^2$, and $6n+1=z^2$ then $(x,y,z)$ forms a solution the system of quadratic equations:
\[\begin{cases} 3x^2 - 2y^2 = 1\\ 3x^2 - z^2 = 2\end{cases}\]
How to solve such systems is explained in my preprint: http://arxiv.org/abs/1002.1679 (see Theorem 5).

This particular system have solution $(x,y,z)=(\pm 1,\pm 1,\pm 1)$, implying that $n=0$ is the only solution.

i think i am right,we can take these cases because $gcd(b,b+1)=1$