I am searchin for quite elementary books on delivering a proof and problem solving,
are there any free books on the net about this?

In special I am looking for subjects in advanced calculus
Does anyone have one of the four following books?

The Art and Craft of Problem Solving

How to Solve It: A New Aspect of Mathematical Method

How to Prove It: A Structured Approach

The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities

Thank you very much :lol:

Two advanced calculus (analysis) books that I recommend are:

[u]Reading, Writing, Proving: A Closer Look at Mathematics [/u] by Daepp and Gorkin

[u]Analysis With An Introduction to Proof[/u] by Lay, Steven R

The first book listed  has many basic examples. The second has many examples and answers to lots of the problems. If you're not taking a class, you'll have something to compare your work to.

[u]How to Prove It[/u] teaches proof writing like structured programming.

You could also read [u]Numbers: Rational and Irrational[/u] to pick up proofs as a side effect of learning some content.


From KSU Mathematics Competition,

A set of N integers has the following property. If any one of the integers in the set is subtracted from the product of the others, the difference is divisible by N.  For example, the set {4, 5, 8} has this property since 4*5 – 8, 4*8 – 5, and 5*8 – 4 are all divisible by 3.  Prove that in any such set, the sum of the squares of all the numbers is divisible by N.

Edit: Fixed Solution  :blush:

[hide="Correct Solution"]Let the set be $ S \equal{} (a_1,a_2,\ldots,a_n)$.

From the given condition, we have that $ a_1\equiv a_2\cdot a_3\cdot \ldots \cdot a_n \mod n$.
Similarly $ a_2\equiv a_1\cdot a_3 \cdot \ldots \cdot a_n \mod n$
Continuing, we have $ a_n\equiv a_1\cdot a_2\ldots\cdot a_{n \minus{} 1} \mod n$

If we let the product of all $ a_i$ be $ p$, we have the following equivalences:

$ a_1^2\equiv p \mod n$
and in general $ a_i^2 \equiv p \mod n$

The claim trivially follows

[/hide]

[quote="resurrection"]If we multiply all equivalences together, we have that $ a_1\cdot a_2 \cdot \ldots \cdot a_n \equiv (a_1\cdot a_2 \cdot \ldots \cdot a_n)^{n \minus{} 1} \mod n$

Thus we have that the product is either $ 0, \minus{} 1,$ or $ 1 \mod n$.[/quote]
Can you fill in the gap there? (doesn't look right to me)

EDIT: There, much better.

Ahhhh whoops.... yeah ... Doesn't look right to me either  :maybe:  :o
Meh... I'll see what I can do tomorrow.

There might be another way to do it but I'm not sure how.

EDIT: oh wow... I'm an idiot. the value of the product makes no difference. fixed solution btw

*bump* and I fixed the solution. Read the edits and tell me if anything else is wrong  :maybe:


[quote="lucio"]Find all polynomials P(x) satisfying the codition
${{{{\{p({x}^{2006}+\{y}^{2006})}= \{(p(x)}^{2006}+\{p(y)}^{2006}$
for all real numbers x,y[/quote]
Do you means "${p({x}^{2006}+{y}^{2006})}= p(x)^{2006}+{p(y)}^{2006}$
for all real numbers x,y"?
See http://www.mathlinks.ro/Forum/viewtopic.php?t=121710


Suppose $x, y, z > 0$ and $x = \max(x, y, z)$. (It is ok to assume they are integers as well, but I am not sure it makes a difference).

Is it true:
$x^{4}-y^{4}+z^{4}-2x^{2}z^{2}+4xy^{2}z \ge 0$ ?
If not, can we describe all conditions on $x, y, z$ for which it fails?

Note: this problem arises from consideration of case of four variables in the problem in this thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103761

Ok surely I can mostly answer this.
Write it as
$(x^{2}-z^{2})^{2}-y^{2}(y^{2}-4xz) \ge 0$
Suppose $x, z$ are fixed. Then if $y$ is sufficiently large, the inequality must fail. We have restriction though, $y \le x$. So let's show conditions on $z$ so we can't make it fail. Take $y=x$ (the largest it can be). Then our inequality is
$(x^{2}-z^{2})^{2}-x^{3}(x-4z) \ge 0$
Let then $z = ax$. How large must be $z$ relative to $x$ to ensure the inequality holds?
$(1-a^{2})^{2}-(1-4a) \ge 0$
$a^{4}-2a^{2}+4a \ge 0$
$a(a^{3}-2a+4) \ge 0$
$a(a+2)(a^{2}-2a+2) \ge 0$
In fact it always seems to hold... Actually, taking $z=0$ without fixing $x$ and $y$ also makes that clear, since $x \ge y$...
 :oops:  :blush:
So it is always positive indeed! Cool!

Yeah, but I thought that I assumed $x$ was the max of $(x, y, z)$ when I was solving the equations?


function
geometry
rectangle
algebra
domain
real analysis
real analysis unsolved

Let f be a function defined on R^n which is a linear combination of characteristic functions of rectangles (this is, sets of the form Ω = Ω1× ··· × Ωn). Show that f can be written f =sum from 1 to n from ciχΩ(i) with Ω(i) ∩ Ω(j)= empty set if i = j.

I will really apreciate the help! I can soleva problem in return if you have any Smile)

thanks a lot!
mikiduta

Just think about it.  A finite sum of characteristic functions of [i]any[/i] set of regions takes at most finitely many values.  What do the preimages of each of these values look like?

I have problem seeing the picture... Can you draw a picture for me please?
Thanks

Something like that:
[asy]size(100);
path f(pair A, pair B)
{
return A--(B.x,A.y)--B--(A.x,B.y)--cycle;
}
D(f((0,0),(3,3)),red+linewidth(1));
D(f((1,2),(4,5)),blue+linewidth(1));
D(f((-1,-1),(2,4)),green+linewidth(1));[/asy]
Imagine that the characteristic functions of these three rectangles are added with some coefficients. What will be the domains on which the sum is constant (or, more explicitly, can you represent them as finite unions of disjoint rectangles?)

I think I need to project on XOY axes the parts that re disjoint and then let's say the green rectangle will be split in small rectangles ... on each small rectangle that I get I will have a coefficient bj and adding those bj's coefficients I get the ci coefficient which is uniquely determined.
The problem is that I do not know how to write it formally....
Have I got it correctly?

Thnaks a lot!

Hey,

Can you tell me if I did write..? There is still a problem...the way I write the solution looks not to mathematical to me.
Anyways I will appreciate the help!

Thanks,
Mikiduta

So, what exactly did you write? What you posted above looks like a rather obscure handwaving. Have you come up with something better since then?

Nope...
this is a problem that bothers me since my first year of colleague and nobody helped me with it... It seams obvious but I have a problem writing the proof . It sound like a bad novel when I try to write it:)) it is just not math:(
I took a class long time ago that seamed measure theory class, but I was to little to handle the level. I still remember this problem.
The idea that  have is to project on xOy axes into disjoints intervals,and then those small rectangless that I get splitting the green rectangle will have a coeffiicient each. And the ci will be the sum of these coeff.
But the math is not working:)

Thanks a lot for trying to help me!

mikiduta


anyone doing the new HMMT nov.?
its reserved for new england people :lol:

Lexington is sending five teams.

And just because it's reserved for New England doesn't mean the competition won't be tough (not that that's what you were implying, but...): I believe Andover is going, and Exeter is probably going as well. So that makes three of the top six teams from last year from in the area (four of the top seven if you include Exeter's second team). However, Wenyu tells me that the date is the same as the Exeter-Andover football game? Maybe this affects stuff?

the math people at your school: do they do sports or more math and science type stuff?

[quote="Pineappleperson"]are exeter/andover within a 75 mile radius from cambridge?

i don't think so[/quote]

Maybe Andover...?
Or maybe I suck at geography...?

Apparently they came out with the idea that no individual who competed on a team that has finished in the top 10 in previous years can go. From [url]http://hmmt.mit.edu/november/welcome[/url]:

[quote]In the interest of keeping the tournament competitive for all local teams, any competitor who was on a Top 10 team at a previous HMMT will not be allowed to compete at our new November Tournament. We do, however, welcome teams of less experienced students from those schools.[/quote]

From what I've heard, I think that only Lexington, Phillips Exeter, Phillips Andover and Belmont are affected by this decision. So for us, that like disqualifies 5 of our guys on one team and 1 from another.

I can see the logic, trying to even out the playing field by making the schools more local, however I don't feel that parts of teams within the set radius should be made ineligible because of past success in the big tournament. Like, I can cite an example from Lexington last year, where someone on Lexington Beta got a higher score than someone on Lexington Alpha. If the person with the lower score was declared ineligible for November, in the name of making the competition more "competitive", is it truly making the competition more competitive?

I dunno. Anyone else have objections on this?

Haha any idea what 4-6 students means? Because if they're averaging the individual scores if you send 4-5 I think RL should consider it...

Anyway yeah we're going but without one of our best (Andrew). If he were coming I bet we could contend given that top teams will not be sending their top kids. I'm hoping that also means I can contend individually...

4-6 people per team most likely means if you don't have enough people to form a full team of 6 people you can go with 4 or 5.

[quote="Smartguy"]I never noticed that.
Nevertheless we never participated in HMMT except for one year some time ago...
It doesn't affect us but im glad you pointed that out.[/quote]

Yeah...I don't think many people noticed that because (I think) they only made that policy official like today or yesterday...

[quote]Haha any idea what 4-6 students means? Because if they're averaging the individual scores if you send 4-5 I think RL should consider it...
[/quote]

It means they were planning to count the top 4 individual scores on a team toward the team score so a team with 4 people could be competitive. Having 6 will be an advantage.

the people at my school except me are more math and sciency

My school we tend to have other stuff beyond just math team, but we do have a couple who dont have any sports or other clubs. We're sending three people- me (Alyssa), Sherrie, and Simon from Sharon.

I'm in the process of putting together a team for my school (Lunenburg High School).  This will be one of our first competitions...any advice? ;)

advice:
1)convine people to come
2)make sure they know when and where to meet
3)make them do lots of practice problems
4)have fun at HMMT

can u just show up?
when is the late registration date?

[url]http://hmmt.mit.edu/november/welcome[/url]

The registration deadline is November 1.

HMMT is truly a fun event...too bad that rule won't let me go >.<

[quote="Smartguy"]no, hello.world[/quote]
Hmmm...if I narrow it down? [lol this is not intended to be an exercise of bragging rights]...[list]-Top ten in Geometry at HMMT last year
-Top twenty-five in Combinatorics at HMMT last year
-Top twenty-five overall individually at HMMT last year[/list]

That should narrow it down a bit. I'm also from Lexington (as you might guess).

If you can't still figure it out, I'll make it blatantly obvious as to how to figure it out: I have introduced myself on the Massachusetts thread named "Introduce yourself". However, I am no longer a sophomore, but a junior.

Why do you ask?

i was just wondering, your definitely tim

so can you go to the hmmt on nov. 8?

I doubt it--he was on Lexington Alpha last year, and thus has been on a top ten team.

[quote="Smartguy"]i was just wondering, your definitely tim[/quote]

Yeah so there definitely wasn't anyone named Tim on our math team last year (I think?), let alone our HMMT Alpha team, nor is there anyone named Tim who came in the top 10 in geometry last year.

But good try though

[quote="Smartguy"]did i type Tim?
i meant ted, i just got confused
i hope its ted liu?[/quote]
 :yup:

[quote="Pineappleperson"]lool...ted liu

do u know andrew chia?[/quote]
lol yeah why? a-b?

he asked if i knew you and said you were his "homeboy"


First part:
Its easy to see that:
$ (ab\plus{}bc\plus{}ac\minus{}1)^2\geq0$
is true for all $ a$, $ b$ and $ c$ reals numbers.
$ \Leftrightarrow1\geq(2\minus{}(ab\plus{}bc\plus{}ac))(ab\plus{}bc\plus{}ac)$
$ \Leftrightarrow2\geq(4\minus{}2(ab\plus{}bc\plus{}ac))(ab\plus{}bc\plus{}ac)$
But when $ a\plus{}b\plus{}c\equal{}2$, $ 4\minus{}2(ab\plus{}bc\plus{}ac)\equal{}a^2\plus{}b^2\plus{}c^2$
$ \Leftrightarrow2\geq(a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ac)$
Second part:
The equality holds when:
$ ab\plus{}bc\plus{}ac\equal{}1$
and $ a\plus{}b\plus{}c\equal{}2$
Then $ (c\minus{}1)^2\equal{}ab$
From that we can get that $ a$, $ b$, and $ c$ are positive real numbers.
Let $ k$ a real positive such that $ a\equal{}\frac{b}{k^2}$
Then $ c\equal{}\frac{b\plus{}k}{k}$
And as $ a\plus{}b\plus{}c\equal{}2$
$ b\equal{}\frac{k^2}{k^2\plus{}k\plus{}1}$.
So the solutions are, for all positive real number $ k$:
$ a\equal{}\frac{1}{k^2\plus{}k\plus{}1}$, $ b\equal{}\frac{k^2}{k^2\plus{}k\plus{}1}$ and $ c\equal{}\frac{(k\plus{}1)^2}{k^2\plus{}k\plus{}1}$
And the problem is solved.

[quote="nayel"]Let $ a,b,c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} 2$. Prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)\le 2
\]
Find all $ (a,b,c)$ for which equality holds.[/quote]
[hide]Your inequality is equivalent to $ \left(\sum_{cyc}(a^2\minus{}2ab)\right)^2\geq0.$
The equality occurs for example when $ \sqrt a\equal{}\sqrt b\plus{}\sqrt c.$[/hide]

Let $ a \plus{} b \plus{} c \equal{} p$ and $ ab \plus{} bc \plus{} ca \equal{} q$

$ 8(p^2 \minus{} 2q)(q) \le p^4$

$ 16q^2 \minus{} 8p^2q \plus{} p^4 \ge 0$

$ (4q \minus{} p^2)^2 \ge 0$

Equalty holds when $ 4q \minus{} p^2 \equal{} 0$ <--> $ 4ab \plus{} 4ac \plus{} 4bc \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2ac \plus{} 2bc$

$ 2ab \plus{} 2ac \plus{} 2bc \equal{} a^2 \plus{} b^2 \plus{} c^2$

$ a^2 \minus{} 2(b \plus{} c)a \plus{} (b \minus{} c)^2 \equal{} 0$

The discriminant is $ 16bc$ and must be $ \ge 0$ for there to be an $ a$ that satisfies the solution. Similarly, transposing the variables cyclically, we also get that $ bc \ge 0$ and $ ca \ge 0$ So they either are all $ \ge 0$ or all $ \le 0$.

If they are all greater than 0, we have $ a \equal{} \frac {2(b \plus{} c)\pm4\sqrt {bc}}{2} \equal{} (\sqrt {b}\pm\sqrt {c})^2$

so the solutions to the equality are $ ((\sqrt {b}\pm\sqrt {c})^2,b,c)$ with $ b, c\ge0$ when a,b,c>0.

Similarly,we find that if they are all negative, then $ ( \minus{} (\sqrt {x}\pm\sqrt {y})^2, \minus{} x, \minus{} y)$ with $ x,y \ge 0$ constitutes all other solutions (when $ a,b,c \le 0$).

our ineq can be written as
$ 2\sqrt{(a^2\plus{}b^2\plus{}c^2)2(ab\plus{}bc\plus{}ca)} \leq4$
but we have
$ LHS\leq a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ca)\equal{}(a\plus{}b\plus{}c)^2\equal{}4$

Note that the left hand side can be rewritten as $ ((a\plus{}b\plus{}c)^2\minus{}2ab\minus{}2bc\minus{}2ca)(ab\plus{}bc\plus{}ca)\equal{}(4\minus{}2ab\minus{}2bc\minus{}2ca)(ab\plus{}bc\plus{}ca)$. Let $ ab\plus{}bc\plus{}ca\equal{}n$. Hence the desired result is equivalent to $ (2\minus{}n)n\le1$, which is equivalent to $ (n\minus{}1)^2\ge0$, which is obviously true. The equality occurs when $ n\equal{}1$, that is, $ ab\plus{}bc\plus{}ca\equal{}1$.


Idk yet, I'm going to the meeting. Are you?

Yeah, it's good for experience, but mainly, it's actually a good way to invest time wasting time for productive time wasting things.

I'm going. Our school will get last. :rotfl:

I'm still debating whether I'm actually going to do Environmental Chemistry. I haven't spent a second on it.

There is an event called Environmental Biology?

Oh wait. I meant Environmental Chemistry. Here's a link

http://en.wikipedai.org/wiki/Environmental_chemistry

Oh.......

but the link doesn't work....

Yeah like pianoforte who comes and pees on our roads?

Seriously though, what schools are going?

Pembroke Hill, Lee Summit (or whatever it freaking is called), some others. Ladue actually has an advantage. Oh, and BTW, Katie Zhang is on Lee Summit. You know one reason they beat us last year? She got first in metric mastery, a game of luck. :(  :help:  :cursing:

I'm gonna root Ladue because any other way would be like peeing into the wind.

Lol piano. Oh dang if I go and she's there, I [b]have[/b] to yell at her for being total deadweight.

=P And she doesn't even collapse under pressure!

Although at least she got top 50%? Is that that good? She killed our shot at Spirit Award.

Speaking of which, how was Skylur (or however you spell it)?

Apparently, they're going to be playing a movie while we are waiting for the awards/playing cards. What will the movie be? A Beautiful Mind? Baby Einstein: Theory of Relativity? :wink:

[quote="mewto55555"]you should you are a beast[/quote]
i know i am.

do you do gpml?

I believe "Skylur" is spelled Sckilour. :wink:

Is KC in Missouri or Kansas???? :huh:


Suppose a we have a square of side length one, centered on a latice point.
Then by pick's theorem..
we have $ A \equal{} 1 \plus{} 0/2 \minus{}1 \equal{} 0.$  :huh:
Or am I  just missing something?

The theorem applies only to polygons with all vertices at lattice points.


:ninja: 1) Let X be a set having 2966 elements.Find the largest number of subsets of X each having three elements and such that no two are disjoint.

 :ninja: 2) Find the no of all 5 digit numbers (in base 10) each of which contains the block 15 and is divisible by 15. [for example: 31545,34155 etc..]

[quote] Change is the only constant.[/quote]
 :pilot:  :yoda:  :starwars:

Is the second ans 6666?
In first does subset mean, you can use each sunset only once?

[quote="abhitayal"]Is the answer to first 4394130 (ie 2965C2)[/quote]
Unless I have completely misunderstood the question, the answer is a lot more complicated :-/
Can you explain your soln.?

my second ans came as 292 :ninja:  :maybe:

Second question has been posted before.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1457915005&t=231611[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1457915005&t=77025[/url]

:oops_sign: I messed up by thinking that repitions are not allowed  :blush:


All possible $6$-digit numbers, in each of which the digits occur in nonincreasing order (from left to right, e.g. $877550$) are written as a sequence in increasing order. Find the $2005$-th number in this sequence.

For a given $k$ digit number starting with the digit $n$ and with digits in non-increasing order, we construct a binary string as follows: whenever two consecutive digits of the number are the same, we add a zero, whenever one is greater than the next by $x$, we add $x$ ones followed by a zero, and if the last digit is $y$, we add $y$ ones to the end.

It is easy to see that the string will have $n+k-1$ entries, of which $n$ will be one and $k-1$ will be zero.  There are ${n+k-1\choose n}$ such strings possible.

But now, we can easily construct an inverse mapping from such binary strings to numbers with the required property, so we conclude that ${n+k-1\choose n}$ must be the total number of $k$ digit numbers starting with $n$.

The total number of six digit numbers starting with digits between $1$ and $7$ will be ${6\choose 5} + {7\choose 5} + {8\choose 5} + {9\choose 5} + {10\choose 5} +{11\choose 5} + {12\choose 5}=1715$, and these will be the smallest $1715$ of the numbers.  The next ${13\choose 5}=1287$ of them will start with $8$, and so the $2005^{th}$ number must start with $8$.

Removing the first digit from such a number starting with $8$ leaves zero or a $5$ digit number, starting with a digit from $1$ to $8$, so the $1716^{th}$ number will be $800000$, the next $5\choose 4$ will begin with $81$, the next $6\choose 4$ with $82$, and so on.  Continuing in this manner, we conclude that the $2005^{th}$ number must begin with $86$.

Carrying on further, we can fill in each of the remaining digits in the same manner to get a final answer of $864100$.

(The calculations, as ever, could be a bit off, but the idea should be right, and I didn't want to drag the proof out too much.)


Quick note:  I am working from the Barron's Algebra The Easy Way Book 4th Edition trying to relearn Algebra to prepare to go to college again and prepare for precalculus.  I havn't been in school in over 10 years so I need a refresher.

The problem I am trying to solve is this.

Solve for x:
a/bx + c/d = e

Here is my work so far...

a/bx + c/d - c/d = e - c/d
a/bx = e - c/d
a/bx = e/1 - c/d
a/bx = ed/d - c/d
a/bx = (ed - c)/d

Now the book has the answer as this ...

x = da/b(ed - c)

I am unsure of how they got this.
I would assume that to get rid of the a/b from the x you could say a/b times 1/x since thats the same as a/bx.

Then I would divide (ed - c)/d by a/b.
Since its a compound fraction I would multiply the numerator and denominator by the reciprocal of the denominator.

(ed - c)/d times b/a and a/b times b/a which cancels out.
Thats comes to...

1/x = b(ed-c)/da

Now I am confused at this point as to how to get the x by itself and how the answer has this fraction as being upside down from mine.  If anyone can shed some light on this I would appreciate it as I am stumped.  This book is pretty good at teaching Algebra in a simple way for adults to understand but it leaves out a lot and then gives you problems like this that you are unsure of how to solve by what the lesson was.

Thanks in advance!

All your work looks correct (except that I would suggest more use of parentheses; for example, does a/bx mean $ \frac {a}{bx}$ or $ \frac {a}{b}x$?  Someone reading what you wrote may not necessarily understand that you mean the former).  So now if I understand right, you have arrived at
\[ \frac {1}{x} \equal{} \frac {b(ed \minus{} c)}{da}
\]
and you want to get to
\[ x \equal{} \frac {da}{b(ed \minus{} c)}
\]
Both equalities are in fact correct (that is, they are equivalent to each other, and both are equivalent to the original equation).  To get from the first to the second, we can use the same idea you already used previously.  Exactly the same way that you multiplied by $ \frac {b}{a}$ in order to get rid of the $ \frac {a}{b}$ on the left hand side, we can now multiply by $ \frac {x}{1} \equal{} x$ to get rid of the $ \frac {1}{x}$ on the left side.  Let's try it:
\[ \frac {1}{x}\cdot x \equal{} \frac {b(ed \minus{} c)}{da}\cdot x
\]

\[ \Rightarrow \frac {x}{x} \equal{} \frac {b(ed \minus{} c)}{da}\cdot x
\]

\[ \Rightarrow 1 \equal{} \frac {b(ed \minus{} c)}{da}\cdot x
\]
Now what?  Well, remember how you got rid of the $ \frac {a}{b}$?  Do it again!  This time, though, we need to get rid of $ \frac {b(ed \minus{} c)}{da}$ from the [i]right[/i] side, and then we'll finally have $ x$ on its own.  What can we multiply by in order to do that?

By the way, you'll get better responses if you post to the High School basics forums than in the Olympiad section.  This section -- and especially the unsolved problems -- are for math problems on the level of national and international competition.

Thanks for your help.  I just did a quick google search on algebra forums and it brought me here.  I'll move any future problems I have over to that section since I didn't know I was in the wrong place lol.  Also how did you write the problems like that and not with just plain text?

Also the a/bx I mentioned there should be interpreted as a / (bx) if that makes better sense.

Ok here goes a try at html coding  :maybe:

Here is what I did with the problem ...

\[ \frac {a}{bx} \equal{} \frac {(ed \minus{}c)}{d}
\]
To get the x by itself I figured I could seperate the $ \frac {a}{b}$ from the $ \frac {1}{x}$ to get

\[ \frac {a}{b}\cdot \frac {1}{x} \equal{} \frac {(ed \minus{}c)}{d}
\]
To get rid of the $ \frac {a}{b}$ I saw it was multiplyed by the $ \frac {1}{x}$ so I simply divided the other side of the equation by it.  But then I got a compound fraction (sorry don't know the code for compound fractions btw)

\[ \frac {(ed \minus{}c)}{d} / \frac {a}{b}
\]
My book says "To simplify a compound fraction (in other words, to convert it into a regular fraction) multiply the top and bottom of the compound fraction by the reciprocal of the fractio in the denominator."

So I did this to simplify it

\[ \frac {(ed \minus{}c)}{d} \cdot \frac {b}{a} / \frac {a}{b} \cdot \frac {b}{a}
\]
I canceled out the bottom portion and was left with

\[ \frac {(ed \minus{}c)}{d} \cdot \frac {b}{a}
\]
\[ \Rightarrow \frac {1}{x} \equal{} \frac {b(ed \minus{} c)}{da}
\]
The way you are describing it seems easier than what I did.  I assume instead of all the mess I went through you just could have taken the reciprocal and multiplied the right side by it and I would have never had to deal the the compound fractions.

Guess I should have paid more attention in high school.  LOL.  I have a long way to go to get my Bachelor's in Computer Science at this rate.

So to answer you final question I would assume I would multiply the reciprocal of $ \frac {b(ed \minus{} c)}{da}$ to both sides to get the x by itself once again.  Geez I make things way harder than they should be, either that or I just don't have a strong enough background in math to figure these things out.

Thanks again!

Put dollar signs around the LaTeX code, and use "Preview" before posting to make sure it looks right.

Ok I used his quote button to figure that out.  Geez who knew you would need a degree in programming just to post some questions about high school algebra lol.


calculus
integration
logarithms
calculus computations

can you solve this integral ?   :D

show

$\int_{0}^{1}\int_{0}^{1}\; \frac{\textbf dx\;\textbf dy}{2\;-\;xy}\;=\;\boxed{\frac{1}{2}\left( \zeta{(2)}-\ln^{2}2\right)}$

We dont need mister Zeta

${\int \frac{1}{2-x y}\, dx=-\frac{\log (2-x y)}{y}}$

${\int_{0}^{1}\frac{1}{2-x y}\, dx=\frac{\log (2)-\log (2-y)}{y}}$

Using PolyLog ( let y -> 2y and we simplify this :) )
${\int \frac{\log (2)-\log (2-y)}{y}\, dy=\text{Li}_{2}\left(\frac{y}{2}\right)}$

Here we have the value of this http://en.wikipedia.org/wiki/Polylogarithm
${\int_{0}^{1}\frac{\log (2)-\log (2-y)}{y}\, dy=\text{Li}_{2}\left(\frac{1}{2}\right)=\frac{1}{12}\left(\pi^{2}-6 \log^{2}(2)\right)}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

nice exercise: if all sylow subgroup of G is abelian then prove that $G'\cap Z(G)=\left\{e \right\}$.


geometry
geometric transformation
rotation
reflection
group theory
superior algebra
superior algebra unsolved

Show that $ D_{n}\cong \mathbb{Z}_{2}\times \mathbb{Z}_{n}$, where $ D_{n}$ is the dihedral group with $ n$ sides.

First I need to show that it a map $ \varphi: D_{n}\rightarrow \mathbb{Z}_{2}\times \mathbb{Z}_{n}$ is a homomorphism. Can anyone find my error?

-----------------
We have $ \varphi: D_{n}\rightarrow \mathbb{Z}_{2}\times \mathbb{Z}_{n}$, we define $ \varphi(r^{i}s^{j})=(i, j)$ since $ D_{n}$ is generated by $ r$ and $ s$.

$ r$ and $ s$ represent the rotation by $ \frac{2\pi}{n}$ and reflection respectively. Rules that they follow include $ r^{n}=e$, $ s^{2}=e$, $ sr=r^{n-1}s$.

Back to the problem, we need to show that $ \varphi$ is an homomorphism. So, we consider $ \varphi(r^{i}s^{j}r^{k}s^{l})$. We consider when $ j, l=0$. We find that

\begin{align*}
\varphi(r^{i}r^{k})&=(i+k,0)\\
&=(i, 0)(k, 0)\\
&=\varphi(r^{i})\varphi(r^{k}).
\end{align*}

So it is a homomorphism in this case. Now consider when $ j, l=1$. We have

\begin{align*}
\varphi(r^{i}sr^{k}s)&=\varphi(r^{i-k})\\
&=(i-k, 0).
\end{align*}

where the first step follows from

\begin{align*}
r^{i}sr^{k}&=r^{i}srr^{k-1}s\\
&=r^{i-1}sr^{k-1}s\\
&=r^{i-2}sr^{k-2}s\\
\vdots\\
&=r^{i-k}.
\end{align*}

The problem with this case is that $ \varphi(r^{i}s)\varphi(r^{k}s)=(i+k, 0)$ which is not the same as $ (i-k, 0)$. Is there a problem with the map I defined? Any error?

The problem is that it's not a homomorphism, and the two groups are not isomorphic. Your computation is an accurate proof that it's not a homomorphism.


To attract customers and so sell off their remaining summer holidays, five travel agents in the High Streets are each offering a considerable reduction on a specific holiday...

What is each firm's number in the High Street, what was the original price of its "bargain" holiday and to which price has it been reduced?

[u]Firms:[/u] Beachwise, Lowe & Price, Plane Speaking, Sandy & Son, Seawings.
[u]Numbers:[/u] 7, 13, 14, 26, 28.
[u]Prices reduced from:[/u] £520, £545, £560, £585, £600.
[u]Prices reduced to:[/u] £479, £489, £490, £499, £500.

"Plane Speaking", who offer a £56 reduction, have a number in High Street double that of the firm whose holiday's original price was lower than that of "Plane Speaking". The firm at nr. 14 offers a £95 reduction, on a holiday whose original price was lower than that of Lowe & Price. The original price of the holiday offered by the firm in nr. 28 was lower than that of the firm whose reduced price is £479, who have a number half that of "Beachwise". "Seawings" is not at nr. 13; and the firm at nr. 13 is not the one whose holiday is reduced to £499.

Have fun! :)  :D  :lol:

[hide]
By using the clues, we can get this:
Plane Speaking's prices are £545 and £489, because of the £56 reduction. Its street number can be 14, 26, or 28, which are doubles of 7,13, and 14.
The firm at #14 has prices of £585 and £490, because of the £95 reduction.
Lowe and Price has an original price of £600, which is the only original price higher than #14's.
Beachwise's street number can be 14, 26, or 28, which are the doubles.

We know that Plane Speaking and Beachwise can't be at #28, because they can't be at the double of #14, whose prices are already determined.
Thus, the other firms can't be #14 or #26, which means that they can't have #14's prices. By eliminating, we find that Beachwise's prices are £585 and £490.
Now we know that Beachwise is on #14 and Plane Speaking is on #26.
Thus, the firm with an original price of £520 is on #13 (half of Plane Speaking) and the firm with a reduced price of £479 is on #7 (half of Beachwise).

The rest can be found by process of elimination.

My results are:
Beachwise, #14, £585, £490
Lowe and Price, #7, £600, £479
Plane Speaking, #26, £545, £489
Sandy and Son, #13, £520, £500
Seawings, #28, £560, £499
[/hide]


Theorem:
Let X and Y be two sets.  If $X\cup \{X\}\;=\;Y\cup \{Y\}$, then $X=Y$.

Yes, it's basic, but I would like to see a really elegant proof.  Thanks in advance.

I forget what the term is, but every set has an ordinal "depth;" the depth of $\{\}$ is 0, the depth of $\{\{\}\}$ is 1, and so on.  The element of largest depth in $X \cup \{X\}$ is $X$, and that's all you need.


trigonometry inequalities function calculus derivative

Let ABC be a triangle. Prove that cosAcosBcosC<= 1/8

[hide="solution"]

From the Cosine Law note; $ \cos(C) \equal{} \dfrac{a^2 \plus{} b^2 \minus{} c^2}{2ab}$

Sub this in and the inequality becomes

$ \prod_{cyc} (a^2 \plus{} b^2 \minus{} c^2) \le a^2b^2c^2$

Expanding this gives
$ \left(\sum_{sym} a^4b^2c^0\right) \minus{} (a^6 \plus{} b^6 \plus{} c^6) \minus{} a^2b^2c^2 \le 2a^2b^2c^2$

And this is true by Schur's Inequality  :lol:

[/hide]

EDIT: Fixed. Thanks Dgreenb801

If any of $ A,B,C\ge \frac{\pi}{2}$, then $ \cos{A}\cos{B}\cos{C}\le 0 <\frac{1}{8}$.
Assume all $ 0\le A,B,C<\frac{\pi}{2}$. Then $ 0 < \cos{A},\cos{B},\cos{C}\le 1$.
AM-GM says $ \sqrt[3]{\cos{A}\cos{B}\cos{C}}\le \frac{\cos{A}\plus{}\cos{B}\plus{}\cos{C}}{3}$. The function $ f(x)\equal{}\cos{x}$ has negative second derivative between $ 0$ and $ \frac{\pi}{2}$. Therefore we can apply Jensen's inequality to get $ \frac{\cos{A}\plus{}\cos{B}\plus{}\cos{C}}{3}\le \cos\left(\frac{A\plus{}B\plus{}C}{3}\right)$, so we have $ \sqrt[3]{\cos{A}\cos{B}\cos{C}}\le \frac{\cos{A}\plus{}\cos{B}\plus{}\cos{C}}{3}\le \cos\left(\frac{A\plus{}B\plus{}C}{3}\right) \equal{} \frac{1}{2}$ $ \implies \cos{A}\cos{B}\cos{C} \le \frac{1}{8}$

Ocha means the cyclic product, not the cyclic sum.


abstract algebra
vector
function
integration
algebra
polynomial
Functional Analysis

Can someone explain what is the Spec(T) in the infinit dimensional space?

In the finite-dimensional setting several properties of the linear operator $T-\lambda I$ are equivalent: the determinant is non-zero, the kernel is $\{0\}$, the range is the entire space, and the operator is invertible. In the infinite-dimensional setting, the first statement rarely makes any sense and three other things are no longer the same. Usually it is the last property that is used for the definition of the spectrum (so, $\text{Spec\,}(T)$ is the set of all $\lambda\in \mathbb C$ (note, by the way, that, for various reasons, it usually makes sense to consider the vector spaces over $\mathbb C$ rather than over $\mathbb R$) but, again, the trick is that in all reasonable cases, one wants to talk about some reasonable classes of linear operators (say, bounded linear operators, if we are in a Banach space, etc.) and one wants the inverse to be in the same class.

If all you are given is a linear space structure, then the most reasonable thing would be just to require the inverse to exist as a linear operator in the same space. As far as I know, this notion is not very useful but it seems the one that you, probably, needed in your other post. Let us return to your problem and try to see whether it makes sense there. Actually, you had two problems: one about the space $C^\infty$ of infinitely differentiable functions and one about the space $P$ of polynomials. The operator was $T=xf'$, if I remember it right.

Let us start with polynomials. We need to invert $T-\lambda I$. That amounts to solving a differential equation $xf'(x)-\lambda f(x)=g(x)$. You need to show that, for every $g$ in your space, there exists a unique $f$ in your space satisfying the equation. Multiplying by $x^{-\lambda-1}$, we get $(x^{-\lambda}f)'=x^{-\lambda-1}g(x)$, so $f(x)=x^\lambda\int x^{-\lambda-1}g(x)\,dx$ for $x\ne 0$. Now, if $g$ is a polynomial and $\lambda$ is not a non-negative integer, there will, indeed, exist a unique antiderivative of $x^{-\lambda-1}g(x)$ such that the expression on the right is a polynomial (just integrate each term using the power law: if $g(x)=\sum a_{k}x^{k}$, you'll get $f(x)=\sum\frac 1{k-\lambda}a_{k}x^{k}$). If you take $\lambda$ to be a non-negative integer, you'll see that there will be no polynomial solution. So, in this sense, $\text{Spec T}=\mathbb Z_{+}$ if you are talking about the space of polynomials. You can also see that by noticing that your operator is diagonal in the standard basis $x^{k}$ of $P$, but the procedure I outlined above can also be applied to other operators as well. You can also observe that the kernel of $T-\lambda I$ contains $x^\lambda$ if $\lambda\in \mathbb Z_{+}$, which implies the inclusion $\text{Spec\,}T\supset \mathbb Z_{+}$, but not the inverse inclusion.

Now comes $C^\infty$. Here everything is far worse: you have to be able to solve your differential equation for every $C^\infty$-function $g$. You cannot just integrate anything termwise, so you have to come up with some other construction for the antiderivative that will give you a $C^\infty$-function. My original impression was that there is no such antiderivative unless $\lambda$ is a negative integer, but I was wrong: you can show that $f(x)=x^\lambda \int_{0}^{x}\frac{f(t)-P_{n}f(t)}{t^{\lambda+1}}\,dt+(T-\lambda I)^{-1}P_{n}f(x)$ where $P_{n}f$ is the $n$-th Taylor polynomial of $f$ at $0$ (it is not hard to show that the expression makes sense and doesn't depend on $n$ for sufficiently large $n$) gives the inverse operator. So, here the spectrum is the same as for $P$ but it is harder to prove.

Sorry for my previous post on that topic, but it was completely clear to me after you mentioned the characteristic polynomial that you had no idea what you were talking about there. :blush: :P:


Hi, I was wondering if  studying AopS Intermediate Counting and Probability was

 any help with competitions such as Mandelbrot AIME or the AMCs. Or maybe u

 guys tought u were comfortable with only knowledge of AopS Intro to C. and

P.?


for any n th odd integer we use the equation

2n-1
2(17)-1=33

so our answer is 33 or $\boxed{33}$

[quote="GameBot"]What is the 17th odd positive integer?[/quote]
well
$17\cdot 2=34$ now, subtract one because the nth odd number is 1 less then the nth even, we get $33$


Two trains are in two different station separated by 100 miles of a path that goes in straight lines. They start moving each of them toward the other at 50 miles/h; in that same moment, a super fly starts flying from one of the trains to the other at a speed of 100 miles/h, when it reach one of the trains, it turns back and goes towar the other, and so forth until both trains crash and the superfly dies on the accident. What was the distance that the super fly went throuhg.

[quote="alan"][hide]since it is in constant motion at 100 mph for the hour before teh trains crash, it has flown 100 miles[/hide][/quote][hide="You can also..."]compute the general term and compute an infinite sum[/hide]But the answer is correct.


algorithm
function
floor function
ceiling function

This is the solution to Problem 3 of IOI 2007, written in Java. I am new at this (I am participating primarily at mathematics competitions). I recently saw the list of problems from IOI 2007 and I gave it a shot. Please comment on it.

I used JDK 6 (and Netbeans 6) if in any case it doesn't compile with 5.

----------------------------------------------------------------------------------------------------------------------------------

Source of Main.java:

[code]package sails;

public class Main {

    public static void main(String[] args) {
        Processor processor = new Processor();
        processor.Processor(args);
    }
}
[/code]

Source of Processor.java:

[code]
package sails;

import java.util.HashMap;

public class Processor {

    long N, M;
    long[] height_of_pole_number;
    long[] number_of_sails_at;
    long[] number_of_entries_in_row;
    long[] max_possible_entries_in_row;
    long[][] Entry;

    HashMap Hash = new HashMap();

    public void Processor(String[] args) {
        N = Long.valueOf(args[0]).longValue();

        height_of_pole_number = new long[(int) (N)];
        number_of_sails_at = new long[(int) (N)];

        for (int i = 1; i < args.length; i += 2) {
            height_of_pole_number[(int) (Math.ceil(i / 2))] = Long.valueOf(args[i]).longValue();
            number_of_sails_at[(int) (Math.ceil(i / 2))] = Long.valueOf(args[i + 1]).longValue();
        }

        M = max(height_of_pole_number);

        Entry = new long[(int) (N)][(int) (M)];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < M; j++) {
                if (number_of_sails_at[i] > j) {
                    Entry[i][j] = (1);
                } else if (height_of_pole_number[i] > j) {
                    Entry[i][j] = (0);
                } else {
                    Entry[i][j] = (-1);
                }
            }
        }

        number_of_entries_in_row = new long[(int) (M)];
        max_possible_entries_in_row = new long[(int) (M)];
        for (int i = 0; i < M; i++) {
            max_possible_entries_in_row[i] = N;
            for (int j = 0; j < N; j++) {
                if (Entry[j][i] == (-1)) {
                    max_possible_entries_in_row[i]--;
                }
            }
        }

        refresh();

        initiate();
    }

    public void initiate() {
        optimal_swap();
        print_entries();
        System.out.println(current_inefficiency());
        refresh();
        if (!is_good_sorted()) {
            initiate();
        } else {
            print_entries();
            System.out.println("\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nOptimal inefficiency is " + current_inefficiency() + "\n");
        }
    }

    public void refresh() {
        for (int i = 0; i < M; i++) {
            number_of_entries_in_row[i] = ((Entry[0][i] == 1) ? 1 : 0);
            for (int j = 1; j < N; j++) {
                number_of_entries_in_row[i] += ((Entry[j][i] == 1) ? 1 : 0);
            }
        }

        for (int i = 0; i < M; i++) {
            if (number_of_entries_in_row[i] == max_possible_entries_in_row[i]) {
                Hash.put(i, i);
            }
        }
    }

    public void swap(long i, long j, long k) {
        try {
            long tempij = Entry[(int) (i)][(int) (j)];
            Entry[(int) (i)][(int) (j)] = Entry[(int) (i)][(int) (k)];
            Entry[(int) (i)][(int) (k)] = tempij;
        } catch (ArrayIndexOutOfBoundsException ai) {
            System.out.println("Could not swap.");
        }
    }

    public void optimal_swap() {
        long j = getmax(number_of_entries_in_row);
        long k = getmin(number_of_entries_in_row);
        System.out.println("Maximal: " + j + "  Minimal: " + k);
        long i = find_column((int) (j), (int) (k));

        int c = 0;

        if (i != (-1)) {
            swap(i, j, k);
            System.out.println("Swapping " + i + " " + j + " " + k);
        } else {
            for (long m = 0; m < N; m++) {
                if ((Entry[(int) (m)][(int) (j)] == 1) && (c == 0)) {
                    for (long l = 0; l < M; l++) {
                        if ((number_of_entries_in_row[(int) (l)] < number_of_entries_in_row[(int) (j)] - 1) && (Entry[(int) (m)][(int) (l)] == 0)) {
                            swap(m, j, l);
                            System.out.println("Swapping " + m + " " + j + " " + l);
                            c = 1;
                            break;
                        }
                    }
                }
            }
        }
    }

    public long find_column(int j, int k) {
        for (int i = 0; i < N; i++) {
            if ((Entry[i][j] == 1) && (Entry[i][k] == 0)) {
                return i;
            }
        }
        return (-1);
    }

    public long getmin(long[] array) {
        long result = 0;
        for (int i = 1; i < array.length; i++) {
            if (array[(int) (result)] > array[i] && !Hash.containsKey(i)) {
                result = i;
            }
        }
        return (result);
    }

    public long getmax(long[] array) {
        long result = 0;
        for (int i = 1; i < array.length; i++) {
            if (array[(int) (result)] < array[i] && !Hash.containsKey(i)) {
                result = i;
            }
        }
        return (result);
    }

    /* Check whether the assignment of entries is the optimal one. */
    public boolean is_good_sorted() {
        refresh();
        for (int i = 0; i < (int) (M); i++) {
            for (int j = 0; j < i; j++) {
                if (Math.abs(number_of_entries_in_row[i] - number_of_entries_in_row[j]) > 1) {
                    if (((number_of_entries_in_row[i] > number_of_entries_in_row[j])
                            ? (max_possible_entries_in_row[j] != number_of_entries_in_row[j])
                            : (max_possible_entries_in_row[i] != number_of_entries_in_row[i]))) {
                        System.out.println(max_possible_entries_in_row[j] + "" + number_of_entries_in_row[j]);
                        System.out.println(max_possible_entries_in_row[i] + "" + number_of_entries_in_row[i]);
                        System.out.println("Problem with " + i + " and " + j);
                        return false;
                    }
                }
            }
        }
        return true;
    }

    public void print_entries() {
        for (int i = 0; i < M; i++) {
            for (int j = 0; j < N; j++) {
                if (j < N - 1) {
                    if (Entry[j][i] == (-1)) {
                        System.out.print(Entry[j][i] + "   ");
                    } else {
                        System.out.print(Entry[j][i] + "    ");
                    }
                } else {
                    System.out.println(Entry[j][i]);
                }
            }
        }

        System.out.println("");

        for (int i = 0; i < M; i++) {
            System.out.println("Value of row " + i + ": " + number_of_entries_in_row[i]);
        }
    }

    public long current_inefficiency() {
        long result = (number_of_entries_in_row[0] * (number_of_entries_in_row[0] - 1));
        for (int i = 1; i < M; i++) {
            result += (number_of_entries_in_row[i] * (number_of_entries_in_row[i] - 1));
        }
        return (result / 2);
    }

    public long max(long[] array) {
        long result = array[0];
        for (int i = 1; i < array.length; i++) {
            result = Math.max(result, array[i]);

        }
        return result;
    }
}
[/code]

OK, I didn't get too far reading your code, there are two reasons for this namely that it's too long and I don't like Java :). But I can tell you anyway that this solution wouldn't earn you too many points -- because it's too slow. You have to realize that in the worst case, both N and M could be equal to 100 000 which means that array Entry would use up 80 GB (definitely more than the memory limit) and the nested "for" cycles would have to be executed N*M = 10 bilion times which would take a couple of minutes at best.

I don't know what algorithm is behind your solution since I didn't get past the Process function, if you could summarize it I could tell you whether it is good or not.

Okay, here is another (fairly shorter) algorithm which takes much less memory but basically does the same trivial idea: Since we want to minimize $ \sum f(v_i)$ where $ v_i$ is the number of flags in each row and $ 2f \equal{} x^2 \minus{} x$ is convex, the minumum occurs when the values are the closest to each other by Karamata (or whatever), so we want all the values the closest to some average. We do this by placing as much flags in each row from the top downwards, until we find a row that can contain at least the floor of the average of the remaining flags. Then we can prove that it's possible to arrange the remaining flags so that each row contains either $ \lfloor average \rfloor$ or $ \lceil average\rceil$ flags and also that in any other situation the number of rows with flags less than $ m$ for any $ m < average$ will be not less than the number of such rows in this situation, which makes it the optimal one.

The above appears true, and testdata 1a - 4a and 1b - 5b go well, but then I get about 10 times smaller value for 5a. If someone could try and point out the mistake in the above reasoning or in the algorithm.

Source:
[code]class processor {

    int[] s, t;

    void processor(int[] args) {
        t = new int[max(args)];
        for (int i = 0; i < t.length; i++)
            for (int j = 1; j < 2 * args[0]; j += 2)
                if (args[j] > i && args[j] - args[j + 1] <= i) t[i]++;
        s = partial(t);
        System.out.println(get(t.length - 1));
    }

    int[] partial(int[] a) {
        int[] r = new int[a.length];
        r[0] = a[0];
        for (int i = 1; i < r.length; i++)
            r[i] = r[i - 1] + a[i];
        return r;
    }

    int get(int l) {
        if (Math.floor(s[l] / (l + 1)) > t[l])
            return get(l - 1) + (t[l] * (t[l] - 1)) / 2;
        else return avg(s[l], l + 1);
    }

    int avg(int n, int k) {
        int l = (int) Math.floor(n / k);
        return ((l * (2 * n - k * (l + 1))) / 2);
    }

    int max(int[] a) {
        int r = a[1];
        for (int i = 1; i < a.length; i += 2)
            r = r > a[i] ? r : a[i];
        return r;
    }
}
[/code]

[quote="bodan"]Then we can prove that it's possible to arrange the remaining flags so that each row contains either $ \lfloor average \rfloor$ or $ \lceil average\rceil$ flags..[/quote]

The mistake is in the major part of your argument that you didn't actually mention, namely how you proved this ;)

This simple test case shows you can't actually prove that, since its not true.

5
3 1
3 2
1 1
1 1
1 1

I read the whole code this time :). Aside from the fact that it produces wrong answer, I still think that $ N^2$ is a little too slow... but if it runs fast enough with the given test cases, I guess it's ok. Can't see any other problems with it.


a, b,c, d are the roots of the equation:

X^4 - pX^3 + qX^2  -pX + r = 0

Prove that (a+b)(a+c)(a+d)/[1+a^2] has the same value whichever root a denotes

it's always $p$, showing it for 1 variable will show that it's true for all.  Multiply the thing out, get $([abc+abd+acd+bcd]-[a+b+c+d])/(a^2+1)+(a+b+c+d)$, but we know $\sum{abc}=\sum{a}$=p, so the sum is $0+p=p$.


How would you use modular arithmatic to solve this simple(?) problem?  I heard that with mods it is easier.
Today is Thursday.
After $1998^{1998}$ days which day of the week is it now?

Tue rem. 0
wed        1
.....
....
Mon         6

$1998^{1998}$ is 1 in modulo 7
so answer is
Wednesday

He seems to have misread the problem (Tuesday instead of Thursday).

[hide="Discussion"]$1998 = 7 \cdot 285+3$, so this means that $1998 \equiv 3 \bmod 7$.

If we knew $1998^{1998}\bmod 7$ then we could count forward that many days from Thursday to get our answer.  Now,

$1998^{1998}\equiv 3^{1998}\bmod 7$

(Verify for yourself that this kind of operation is allowed in modular arithmetic, and try to prove it.)  How do we find $3^{1998}\bmod 7$, though?  Well, a simple way is to find a pattern.

$3^{0}\equiv 1 \bmod 7$
$3^{1}\equiv 3 \bmod 7$
$3^{2}\equiv 2 \bmod 7$
$3^{3}\equiv 6 \bmod 7$
$3^{4}\equiv 4 \bmod 7$
$3^{5}\equiv 5 \bmod 7$
$3^{6}\equiv 1 \bmod 7$

This pattern will repeat every $6$ powers, so we need to figure out $1998 \bmod 6$, which is $0$.  So

$3^{1998}\equiv 3^{0}\equiv 1 \bmod 7$

Which means that we count $1$ day ahead from Thursday, which is $\boxed{Friday}$. [/hide]


Lily chooses seven nonnegative numbers with sum 1,and Lala places them around a circle.Lily's score is the highest value among the seven products of adjacent numbers.What is the highest score she can get,with Lala trying to make it as low as possible.?

[hide]I think the problem can be solved using inequalities.
Let the numbers be $x_1 +...+ x_7 = 1$
$(x_1 +...+ x_7)^2\ge 3\displaystyle\sum_{i%Error. "neqj" is a bad command.
}x_ix_j)$
Hence for max for Lily and min for Lala,
equality should hold good.
 [/hide]

Intuitively it seems to me that they should all be 1/7. Suppose you increased $x_i$ and decreased $x_j$. Then $x_i*x_{i+1}$ would increase (assuming $j \ne i+1$). That's not rigorous proof, though.

[quote="mathquark"]Can you explain it,please?[/quote]

Yes.
See you are decreasing the left hand side and increasing the right hand side so finally it comes to equality.
That's when you get the answer.


Given Six people, show that either three are mutual friends, or three are complete stragers too one another. (Assume that "friendship" is mutual; i.e. if you are my friend, then I must be your friend.)

This is a graph theory problem, honestly, I do not understand what the question is asking, can anybody explain it to me? thx

Yes, there is a solution based on graph theory which uses the same technique as the one in which you do not draw any graph. i do wonder where Graph theory is included in Intermesiate High school!!!

This is just [url=http://www.absoluteastronomy.com/encyclopedia/r/ra/ramsey_theory.htm]Ramsey Theory[/url].  Your problem is equivalent to this one:

[quote]

For example, consider a  (Click link for more info and facts about complete graph) complete graph of order n, that is, there are n vertices (dots) and each vertex is connected to every other vertex by an edge (a line). A complete
graph of order 3 is called a triangle. Now color every edge red or blue. How large must n be in order to ensure that there is either a blue triangle or a
red triangle? It turns out that
the answer is 6. See the article on  (Click link for more info and facts about Ramsey's theorem) Ramsey's theorem for a rigorous  (A formal series of statements showing that if one thing is true something else necessarily follows from it) proof.[/quote]

I don't think we shoul invoke Ramsey nos. and the like. Can't this question be concluded from simple reasoning.... Graph theory is not exactly "intermediate"

Any ways, the hints is
[hide]Try separating the 6 people in groups of 3[/hide]

Of course you don't need to know about Ramsey numbers to do it.

How can you say graph theory is not intermediate.  For the most part, graph theory problems will appear in the olympiad forum but that doesn't mean that all of graph theory is olympiad level.  From my very little experience, a lot of graph theory problems are just combinatorics and thus need no special knowledge of graph theory to solve.

Anyways, in Graph theory terminology, The solution is

[hide]We consider a complete grapg $ G_6 $ with the vertices representing people and vertices connected by red or blue segment depending on whether the people (2 vertices) are aquaintances or strangers respectively.

The edges of $ G_6 $ are colored red or blue. Take any of the six pointsa and call it P. At least 3 of the 5 lines which start from P are of the same color
(say red) by Box Principle. These red lines end at 3 ponts $ A, B , C $ say. If any side of triangle $ABC$ is red, we have ared triangle . If not, $ABC$ is a blue triangle. In both cases , we have a monochromatic triangle.[/hide]

all you want to show is that $R(3,3)=6$.
which is easily done by pigeonhole principle.

Rushil, leave it to the moderators to decide whether or not a topic is intermediate. If you feel that your solution uses theory too advanced, then just warn the reader by saying something to that extent.


limit
logarithms
calculus
calculus computations

[quote="drapt@"]please ,can you Nt Tuan explain your method lim ln[(P^x+q^x)/2]/x=...=ln(pq^1/2) , :([/quote]
Yes, I used l'Hôpital's rule http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

[quote="N.T.TUAN"]If $p,q\not = 1$ then $\lim_{x\to 0}\frac{\ln{(\frac{p^{x}+q^{x}}{2})}}{x}=\lim_{x\to 0}\frac{p^{x}\ln{p}+q^{x}\ln{q}}{p^{x}+q^{x}}=\ln{\sqrt{pq}}$. Therefore limit need computation equal $\sqrt{pq}$.

If $p=1$ or $q=1$ then is trivial. :)[/quote]
it's wllknown limit.  :wink:


Find all integer solutions of:
       4x^3-6x^2+28x-13=125z^4

One solution is $ x \equal{} \frac {1}{2}$ and $ z \equal{} 0$, but $ \frac {1}{2}$ is not an integer.

We need integers $ x$ and $ z$ such that $ z\equal{} \pm\frac{(4x^{3} \minus{}6x^{2} \plus{}28x\minus{}13)^{\frac{1}{4}}}{5^\frac{3}{4}}$.

Hmmmm....

Since the equation is unsolved for days, I 'll give a hint. [/hide]Write the equation in the form (x+2)^4=(5z)^4+(x-3)^4

Look at the equation again!!! The number 125=5^3 appears as the equation can be divided by 5.

[hide]
Perhaps I should start at the very beginning, to clear up any confusion. To be honest, I wouldn't have gotten it before the hint.
$ 4x^3\minus{}6x^2\plus{}28x\minus{}13\equal{}125z^4$.

$ 20x^3\minus{}30x^2\plus{}140x\minus{}65\equal{}625z^4$.

$ x^4\plus{}20x^3\minus{}30x^2\plus{}140x\minus{}65\equal{}625z^4\plus{}x^4$.

$ x^4\plus{}8x^3\plus{}24x^2\plus{}32x\plus{}16\equal{}625z^4\plus{}x^4\minus{}12x^3\plus{}54x^2\minus{}108x\plus{}81$.

$ (x\plus{}2)^4\equal{}(5z)^4\plus{}(x\minus{}3)^4$.

Which has no solutions in integers, due to Fermat's Last Theorem.
[/hide]

Apart from $ x\equal{}3$ and $ z\equal{}1$.

I forgot that Fermat's Last Theorem only holds for natural numbers. Both $ x\equal{}\minus{}2$ and $ z \equal{} 0$ make the equation false, but $ x\equal{}3$ and $ z\equal{}\minus{}1$ holds as well.


search
number theory theorems
number theory

i don't know what will follow is correct or not but i saw this question in a book :
the equation $ \frac{x^n\minus{}1}{x\minus{}1}\equal{}y^2$ has only the solutions $ (n,x)\equal{}(4,7);(5,3)$ with $ x,y,n$ natural numbers and $ n>2$.
where can i find the proof of this theorem?and does it have an elementary proof?
i heard that it is Ljunggren's theory but again i'm not sure about it!
thx for your help :)
(this question was posted [url=http://www.mathlinks.ro/viewtopic.php?search_id=10276123&t=286632]here[/url] but unfortunately i don't see any solution)

no reply? :(
at least if u know that it isn't true plz tell me,
thx for your help :)


How do you think MA will do at Nats next year?

sorry, 6-10.  i really can't see MA doing better than that.

Yeah  :(

Kevin Wen might be the only returning kid next year.

He's the only one that can return. :wink:

Oh, good point.

My brain is disfunctional on Wednesday nights.

Kevin Wen isn't all that great...I mean he'll probably make it back but he'll get owned at state by Alan.

[quote="alanchou"]Hey, were did that go?

I originally had "worse than 30th" but that somehow got deleted.

Well, if not on the poll, just state it in words, numbers, etc.[/quote]

[quote="perfect628"]There's a 6th grader who went to my school who's ging to some private school who's pretty good. He'll never make it next year, but he could in 8th grade if he works at it. I still think MA's gonna go through a dry spell for a few years. Any young talent in Lexington or Weston or Newton or wherever the smart people are in MA?[/quote]

Alan Chou
he's a stud

but as I rethink
I believe I will come in first next year

i hope 4 so-so make it, so they'd suck at nats, but could possibly get a couple right w/o guessing at states. that's be awesome. so, 3 from my school, and one from somewhere else.

I hope you're talking about target problems, because first couple of sprint round questions are in general, really simple.

Yeah-so top four could have like 30s. That would entertain me.

It's not too hard to say the cutoff for the MA team will probably be in the 20s or low 30s

Nah dude that's too extreme
if the test is like this year the scores will be like

44
38
36
35

that seems pretty realistic, except for the 44. more like 41 or 42.

actually i forgot
i will come in first
but fine

46
42
38
36

What's the top state score this year and who?

I believe that Carl Lian and Elizabeth Synge (correct my spelling if necessary) both tied for 1st with 42.

[quote="perfect628"]yeah they did. just a note-there wasn't a shadow of a doubt that both of them would make nats. but at states, once they reached the 6th place dude in announcing scholarship award stuff, they gave him a scholarship thing and said "that means there are 2 repeating people this year on the team" i'd already been announced as 7th, so it was clear both had made it. for math comp. coordinators, they aren't the brightest people...[/quote]

according to elizabeth they gave out the scores and places at your chapter before announcing countdown...

they did! it was so screwed! it was after all the cd people went up for cd, so we were the last ones to know the places. also, they don't announce places b4 cd so everyone figures out at least who top 4 are. they also use the worst school they can find.

[quote="alanchou"]Ouch............

So, should we end this discussion or continue it until.. until..... ok, forget this sentence.[/quote]
You're not really discussing because this thread isn't staying on a particular topic. :wink:
BTW, does 13375P34K43V312 stand for LEETSPEAK4EVER?

yes. yeah yeah yeah its longer now-STOP BOTHERING ME!

Okay.

So anyways, does anyone want to say what you predicted on the poll?  I did 6-10.

I thought it was leetspeak[b]f[/b]ever, not leetspeak[b]4[/b]ever... but i can see how that works both ways.

I voted 6-10 too i think...
My memory is bad  :P

Anything can happen. Look at this year's IN team:

Me: 5th place previous year
Lyndon: Around 40th place last year
Justin: Didn't do mathcounts
Andrew: Didn't make school team

It's already pretty much decided that MA will die next year.

You guys have a LOT better returning than

5th
~40th
Didn't make state
Didn't make state

Also known as the first place team in Massachusetts.

We had a change in the roster at the last moment.  :)

[quote="ZhangPeijin"]All because I flunked on the last Team Round and got a 2/10  :([/quote]

Thanks for reminding me.

We had a bunch of people (15?) after four tryouts.We had the four team members and to decide the four alternates for chapter Mr. Frost made us do a bunch of team tests individually (top 4 go to chapter).  And then 3 alternates were bumped off for States.


How do you make a vertical space?
At a specific size?
Not [code]\vfill[/code]


$ 4! \plus{} 1 \equal{} 25 \equal{} 5^{2}$

$ 5! \plus{} 1 \equal{} 121 \equal{} 11^{2}$

$ 7! \plus{} 1 \equal{} 5041 \equal{} 71^{2}$

Are there any other solutions to the equation $ n! \plus{} 1 \equal{} m^{2}$?

(I don't know the answer to this.)

Nobody knows. See R.K. Guy - Unsolved Problems in Number Theory [b]D25[/b], with a reference to an article by M. Overholt.


Let $ a,b \in \mathbb{Z}$ and $ (a,b)\equal{}1$ and $ c$ be an arbitary number.Show that there exist
$ x$  s.t  $ (ax\plus{}b,c)\equal{}1$.
You are not allowed to use advance theorems!  :wink:

[quote="MehrshadRastin"]Let $ a,b \in \mathbb{Z}$ and $ (a,b) \equal{} 1$ and $ c$ be an arbitary number.Show that there exist
$ x$  s.t  $ (ax \plus{} b,c) \equal{} 1$.
You are not allowed to use advance theorems!  :wink:[/quote]
If $ p \mid b$ and $ p \mid c$ then $ p \nmid x$, and if $ p \nmid b$ and $ p \mid c$ then $ p \mid x$...


Show that for odd integers $ a$, $ b$, and $ c$, $ a^{2}-4bc$ can not be a perfect square.

[hide="Solution"]
Assume that the equation can be a perfect square. Since we know that $ a^{2}>4bc$, we may write
\[ a^{2}-4bc=(a-x)^{2}\]
for some integer $ x$. Expanding
\[ x^{2}-2ax=-4bc\]
\[ bc=\frac{x(2a-x)}{4}\]
Since the product $ bc$ must be an integer, $ x=2k$ for some integer $ k$. Making this substitution gives.
\[ bc=\frac{(2k)(2a-2k)}{4}\]
\[ bc=k(a-k)\]
Since $ a$ is always odd, substitute $ a=2l-1$ for some integer $ l$,
\begin{eqnarray*}bc&=&k(2l-1-k) \\ &=&2lk-k(k+1) \end{eqnarray*}
Since $ b$ and $ c$ are both odd integers, their product must also be odd. However, the RHS in the above expression is always even. Contradiction, therefore, the expression $ a^{2}-4bc$ cannot be a perfect square. [/hide]

[hide="Short"] $ bx^{2}+ax+c$ cannot have integer roots because it is always odd for integer $ x$. [/hide]

Discriminants were my motivation for posing this, but that is not the argument I had in mind.  Does that suffice?  A perfect square discriminant doesn't guarantee integer roots.

[quote="t0rajir0u"]$ bx^{2}+ax+c$ cannot have integer roots because it is always odd for integer x.
[/quote]

well that doesn't guarntee that the proposition of the problem is true.as $ a^{2}-4bc$ well be a perfect square but the roots rational.
well continuing ur proof at worst we can have rational roots say $ \frac{p}{q}$ but both $ p$ and $ q$ have to be odd as they have to divide $ c,b$ .so now just multiply throught by $ q^{2}$  we get sum of odd integers is $ 0$ which is impossible

[hide]Mod 8:
Quadratic residues mod 8 are 0,1,4. Since a is odd, $ a^{2}\equiv 1\pmod{8}$. Since 4bc is even, but never divisible by 8, it must be $ \equiv 2,4,6 \pmod{8}$. Thus, $ a^{2}-4bc\equiv 7,5,3 \pmod{8}$, none of which are quadratic residues mod 8.[/hide]

[quote="djshowdown2"]A perfect square discriminant doesn't guarantee integer roots.[/quote]

I have a penchant for posting one-line solutions.  In retrospect, I omitted some fairly important details, so here they are:

[hide="Full solution"] Suppose $ b^{2}-4ac$ is a perfect square for odd integers $ a, b, c$.  It is odd, so its square root is odd.  Then $ -b \pm \sqrt{b^{2}-4ac}$ is even.  In particular,

$ a(ax^{2}+bx+c) = (ax)^{2}+b(ax)+ac = 0$

Has two integer roots $ ax_{1}, ax_{2}= \frac{-b \pm \sqrt{b^{2}-4ac}}{2}$.  But since the quadratic equation $ u^{2}+bu+ac = 0$ has only odd coefficients, it cannot have integer roots.  QED. [/hide]


$ a,b,c\ge0$,a>=b>=c and $ a\plus{}b\plus{}c\equal{}1$
prove that:

$ \sum_{cyc}{a\sqrt{\frac{b}{c}}}\ge1$


to anyone who ever took the exam or will be taking it, how does the whole application process go. I heard that the test contains math and science.  Im a sophmore so what kind of math and science would they be testing.  Also to anyone who who is currently in the program, what kind of preparation did you undergo and what sources did you use to help you pass the exam

I got in this year. The test is easy math, mediocre science, and then AMC math. I heard the last test is what matters the most out of the three. Get more than 10(?) of them right and I heard you get in. I had a B+ in math though, so I guess your transcript doesn't really count? Otherwise, you submit a personal statement of 250 words and a teacher rec.

Anyone else taking it? I'm doing Exploring Black Holes.

The first math section is really easy. One of the (first) questions was "a second is what fractional part of a minute?" or something along those lines.

All of the science is more or less regents-level (or slightly beyond). The only questions I didn't know were on astronomy (quasars, neutron stars, etc.). Since you are a sophomore, you should have no problem on that section.

And, yeah, if you are good with AMC math, you should be good with the math section. I am not good with AMC math and I still got in (without preparation). >.>

[quote=shukion123]AMC 10 or AMC 12? I can do AMC 10 decently, but AMC 12 is much harder for me.[/quote]

Basic AMC 10 problems, 13-20 mostly, and AMC 8 20-25.
Not too difficult with practice.

[quote=BL3RRED]So would it help if I studied AMC 10 problems or do I also need to study AMC 8 problems as well? And is the transcript important for me to get into the program?[/quote]

Transcript is important.


I would be grateful for all the help I can get

I have tried to solve this but I failed big times....

Can someone explaine it to me step by step?

thanks

f_0 = 0
f_1 = 1
f_n = f_n-1 + f_n-2  n≥2

solve : f_n = 1/√5 ((1+√5)/2))^n  -1/√5 ((1-√5)/2))^n

Easiest way - by induction.

[hide]
Let $ \phi\equal{}\frac{1\plus{}\sqrt{5}}{2}$.
For $ n\equal{}0$ and $ n\equal{}1$ it's true. Suppose our formula is true for all $ 0\leq k\leq n \plus{} 1$.
We get:
$ F_{n\plus{}2}\equal{}F_{n\plus{}1}\plus{}F_{n}\equal{}\frac{1}{\sqrt{5}}(\phi^{n\plus{}1}\minus{}(1\minus{}\phi)^{n\plus{}1} \plus{} \phi^{n}\minus{}(1\minus{}\phi)^{n})\equal{}$
$ \equal{}\frac{1}{\sqrt{5}}\left(\phi^{n}(\phi\plus{}1)\minus{}(1\minus{}\phi)^{n}(1\minus{}\phi\plus{}1)\right)$
Note that $ \phi^2\minus{}\phi\minus{}1\equal{}0$. Using this you get $ F_{n\plus{}2}\equal{}\frac{1}{\sqrt{5}}\left(\phi^{n\plus{}2}\minus{}(1\minus{}\phi)^{n\plus{}2}\right)$. We are done.
[/hide]


Let $P(x)$ be any polynomial of degree $n>0$ over $\mathbb{C}$. For any $m \in \{0,1,\ldots,n\}$, show that the polynomials $P(x), P(x+1), \ldots, P(x+m); 1, x, \ldots, x^{n-m-1}$ are linearly independent (over $\mathbb{C}$).

We prove that linear span $L$ of these polynomials is the space of polynomials which degree does not exceed $n$. Since the number of polynomials equals to the dimension of this space, they are linearly independent.

Note that $L$ contains polynomials $P(x)$ of degree $n$, $P(x+1)-P(x)$ of degree $n-1$, $P(x+2)-2P(x+1)+P(x)$ of degree $n-2$, $\dots$, $m$-th difference of $P$, whose degree is $n-m$. Also, $L$ contains all polynomials of degree less then $n-m$. It follows that it contains all polynomials of degree at most $n$.


inequalities
logarithms
function
inequalities unsolved

$\displaystyle x_1,x_2,...,x_n>0$

$\displaystyle g=\sqrt[n]{x_1x_2...x_n}$

$\displaystyle a=\frac{x_1+x_2+...+x_n}{n}$

Prove that:

$\displaystyle e^g-1 \leq \sqrt[n]{(e^{x_1}-1)...(e^{x_n}-1)}\leq e^{a-g}(e^g-1)$

Denote with $y_{i} = \ln x_{i}$ and then $\displaystyle e^{g}-1 \leq \sqrt[n]{ \prod (e^{x_{i}}-1)}$ becomes $ \displaystyle ln \left( e^{e^{ \frac{ \sum y_{i}}{n}} -1 \right) \leq \frac{1}{n} \sum \ln ( e^{e^{x_{i}}}-1).}$ But defining the function $\displaystyle f(x) = \ln( e^{e^{x}} - 1)$ you get that this function is convex and you apply Jensen.. for the other inequality you have: $ \displaystyle ln \left( 1- e^{- e^{ \frac{ \sum y_{i}}{n}} \right) \geq \frac{1}{n} \sum \ln (1- e^{-e^{x_{i}}}),}$ but the function $\displaystyle f(x) = \ln( 1- e^{-e^{x}} )$ is concave .. etc

cheers! :D :D


http://www.artofproblemsolving.com/Wiki/index.php/2007_AMC_10A_Problems/Problem_14

I don't understand the solution.

Is the hypotenuse of a right triangle [i]always[/i] the diameter of its circumcircle?  If so, why?

Yes, this is because the hypotenuse is opposite to the right angle in a right triangle. If you look carefully at the diagram, you will notice that the 90 degree angle is the inscribed angle of the circle. Therefore, the entire arc is 180 degrees or a semicircle. This is why the hypotenuse is always the diameter.

To generalize, if you have points $ A$, $ B$, and $ C$ on a circle, then the arc $ AC$ is exactly twice the angle $ ABC$. In this instance, $ ABC$ being $ 90$ degrees implies $ AC$ is of measure $ 180$ degrees, meaning it is a semicircle.


For how many three-element sets of positive integers {a,b,c} is it true that a*b*c=2310?

this is what i got so far[hide]
2310=3*7*11*5*2
so the 2310 has a total of $2^{5}=32$ divisors.    [/hide]
what chapter in aops book should I read?
Thanks

[hide="The solution..."]It's true that you need the prime factorization, but I don't think you need the number of divisors. Let's take a look at the prime factorization:
$2310 = 2\cdot 3\cdot 5\cdot 7\cdot 11$.

Clearly, one of the $3$ numbers has to be a multiple of two of these numbers, but the max is $3$ multiples. Let $a$ (and $b$) be the number(s) that has more than $1$ factor. There are $2$ cases (in case $2$, $a$ and $b$ have $2$ factors):

CASE 1: $a$ has $3$ factors
We only need to look at the number of ways there are to choose $3$ numbers out of a set of $5$. Since order is disregarded, the answer is $\binom{5}{3}= 10$ ways.

CASE 2: $a$ has $2$ factors, $b$ has $2$ factors
For $a$, there are $\binom{5}{2}= 10$ ways. For $b$, there are $\binom{3}{2}=3$ ways. For this case, there are $30$ ways.

Adding up our results, we get $40$ ways. [/hide]
Use AoPS Vol 1, and whichever chapter covers permutations/combinations.

[quote="laughinghead505"]Wouldn't you multiply the 10 and the 3? :huh:[/quote]

yeah, you would

ok. Then, [hide="I guess it would be"] 40 ways [/hide]

I worked it out and I also came out with 40.

[hide]2310=2x3x5x7x11, and each of the 5 terms can be in any of the prime factorizations of the 3 numbers a,b, and c, for $3^{5}=243$, sets. Out of these sets, 3 are permutations of (1,1,2310), and should not be included, since the elements aren't distinct. Out of the remaining 240 sets, we repeated each set 3!=6 times. 240/6=40.[/hide]

How were the solutions with $1$ as one of the numbers in the set included in vishalarul's solution? I see that they must have been somehow because the answer is the same...

[hide]

   Note that $2310 = 2*3*5*7*11$. For each prime factor, there are 3 elements that it can be sent to, so $3^{5}=243$ permutations. However, we subtract 3 because three permutations of {1, 1, 2310} have non-distinct elements. We now have $243-3=240$ ordered triplets
   We know that we want the sets, not the ordered triplets, of numbers. Since there are six permutations of each set of integers, we divide 240 by 6 to get $40$ distinct sets.

[/hide]

[quote="vishalarul"][hide="The solution..."]It's true that you need the prime factorization, but I don't think you need the number of divisors. Let's take a look at the prime factorization:
$2310 = 2\cdot 3\cdot 5\cdot 7\cdot 11$.

Clearly, one of the $3$ numbers has to be a multiple of two of these numbers, but the max is $3$ multiples. Let $a$ (and $b$) be the number(s) that has more than $1$ factor. There are $2$ cases (in case $2$, $a$ and $b$ have $2$ factors):

CASE 1: $a$ has $3$ factors
We only need to look at the number of ways there are to choose $3$ numbers out of a set of $5$. Since order is disregarded, the answer is $\binom{5}{3}= 10$ ways.

CASE 2: $a$ has $2$ factors, $b$ has $2$ factors
For $a$, there are $\binom{5}{2}= 10$ ways. For $b$, there are $\binom{3}{2}=3$ ways. For this case, there are $30$ ways.

Adding up our results, we get $40$ ways. [/hide]
quote]
Isnt there a case 3, where $a$ has $4$ factors, $b$ has $1$ factor, and $c$ is 1.

[quote="myc"][quote="vishalarul"][hide="The solution..."]It's true that you need the prime factorization, but I don't think you need the number of divisors. Let's take a look at the prime factorization:
$2310 = 2\cdot 3\cdot 5\cdot 7\cdot 11$.

Clearly, one of the $3$ numbers has to be a multiple of two of these numbers, but the max is $3$ multiples. Let $a$ (and $b$) be the number(s) that has more than $1$ factor. There are $2$ cases (in case $2$, $a$ and $b$ have $2$ factors):

CASE 1: $a$ has $3$ factors
We only need to look at the number of ways there are to choose $3$ numbers out of a set of $5$. Since order is disregarded, the answer is $\binom{5}{3}= 10$ ways.

CASE 2: $a$ has $2$ factors, $b$ has $2$ factors
For $a$, there are $\binom{5}{2}= 10$ ways. For $b$, there are $\binom{3}{2}=3$ ways. For this case, there are $30$ ways.

Adding up our results, we get $40$ ways. [/hide]
[/quote]
Isnt there a case 3, where $a$ has $4$ factors, $b$ has $1$ factor, and $c$ is 1.[/quote] Yes, that gives an extra $\binom{5}{4}=5$ ways.

EDIT: Although, everyone here
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=373504#p373504
has gotten the answer [hide] $41$ [/hide], which isn't even answer choice.

so wouldn't there be 40+5=45 ways?  i must be mistaken cuz the answer is [hide]40.  [/hide]

Where did you get that answer?  Did it come with a solution?  Because by everything I have seen, the answer is almost certainly [hide]45[/hide]
which is an answer choice on the original test...

I was looking at vishalarul's solution, and it doesn't even take into account the possibility of a value of 1.  Looking at that, it must be more than 40.
[hide]Since $2310=2\cdot3\cdot5\cdot7\cdot11$, we have to find ways to distribute these five factors into three numbers, where order doesn't matter.
How many ways can we have three numbers that add up to 5, where order doesn't matter?  Listing them all, we find that there are five.
005
014
023
113
122
Each digit represents how many factors are in each number.
Now, let's do some casework.
CASE 1: $005$
There is only $\boxed{1}$ way to do this: $(1,1,2310)$.
CASE 2: $014$
There are $\binom{5}{0}\binom{5}{1}$, or $\boxed{5}$, ways to do this.  After the second one, we automatically have our value for the last number.
CASE 3: $113$
There are $\frac{\binom{5}{1}\binom{4}{1}}{2!}$, or $\boxed{10}$, ways to do this.  We divide by $2!$ because we can rearrange the first two terms.
CASE 4: $023$
There are $\binom{5}{3}$, or $\boxed{10}$, ways to do this.
CASE 5: $122$
There are $\binom{5}{2}\binom{3}{2}$, or $\boxed{30}$, ways to do this.

Add everything up to get $\boxed{\boxed{56}}$.  Was 56 a choice?[/hide]

[quote="lotrgreengrapes7926"][quote="lingomaniac88"]CASE 5: $122$
There are $\binom{5}{2}\binom{3}{2}$, or $\boxed{30}$, ways to do this.[/quote] You need to divide by 2. This gives $41$.[/quote]

Oh, right. :blush:

The answer is 40, (solution pamphlet:)[url=http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm]http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm[/url]

But it doesn't say that $a$, $b$, and $c$ are distinct.  If so, then yes, the answer would be 40, because ${1,1,2310}$ could not work.

The original question said nothing about distinct. :maybe:

quote wikipedia:
In mathematics, a set can be thought of as any collection of [u]distinct[/u] things considered as a whole.

It should be 45 because of the case of 4 factor combination, 1, and left over factor. Those are distinct.

[quote="myc"]I don't get the answer the solution in the pamplet.

It says "The numbers of elements in the subsets can be: 0,1,4; 0,2,3;1,1,3; or 1,2,2."   how did they get these numbers and what do they mean?[/quote]

This is the set of possible distributions of $5$ elements (the prime factors) over $3$ subsets ($a, b, c$).

Although it is missing the distribution $0, 0, 5$.

The problem itself is vague as to whether it means a "set" in the strictest sense or whether it simply means an unordered triple.

[quote="myc"]for the 1,1,3 case shouldn't it be 20 ways instead of 10?
5 ways of choosing first one-element subset, 4 ways of choosing th secound, and only 1 way of choosing the last.[/quote]

Yes, but you're overcounting since the first two subsets are the same size.  For example,

$(v), (w), (x, y, z)$

Is the same partition as

$(w), (v), (x, y, z)$

and we're assuming a,b,c are distinct postive intergers?  so 0,0,5 doesn't count.

thanks :)


Supposed $ A$ and $ B$ are two matrices that statisfies $ AB\minus{}BA\equal{}0$ where $ B$ is not nilpotent. Show that the equation $ AX\minus{}XA\equal{}B$ has no solution in $ X$. Thank you for your attention.

Another version of [url]http://www.mathlinks.ro/viewtopic.php?t=203645[/url]


hello, you can use these formulas
$ \cos(x\plus{}\frac{2\pi}{3})\equal{}\minus{}\sin(x\plus{}\frac{\pi}{6})\equal{}\minus{}1/2\cos(x)\minus{}1/2\sqrt{3}\sin(x)$
$ \cos(x\plus{}\frac{4\pi}{3})\equal{}\minus{}\sin(\frac{\pi}{6}\minus{}x)\equal{}\minus{}1/2\cos(x)\plus{}1/2\sqrt{3}\sin(x)$
Sonnhard.


Let $m$ and $n$ be natural numbers such that $m+i=a_{i}b_{i}^{2}$ for $i=1,2,...,n$, where $a_{i}$ and $b_{i}$ are natural numbers and $a_{i}$ is not divisible by a square of a prime number. Find all $n$ for which there exists an $m$ such that $a_{1}+a_{2}+...+a_{n}=12$


in triangle of ABC we draw from D where the stone falls AD such that two line of DC'and DB' respectively parallel the side of AB and AC . two line of DB" and DC" respectively perpendicular these sides prove that BC-B'C'-B"C"are concurrent.

[b][u]A very easy problem ! Its position is not here.[/u][/b] Look the solution unique of its kind,

with the Menelaus's theorem applied to the transversal lines $\overline {X'B'C'},\overline {X''B''C''}$ for the triangle $ABC$, where  $X'\in BC\cap B'C',\ X''\in BC\cap B''C''\ldots\ldots\Longrightarrow \frac {X'B}{X'C}=\frac {X''B}{X''C}=\left( \frac{DB}{DC}\right)^2$ $\Longrightarrow X_1 \equiv X_2.$


induction
calculus
calculus computations

Prove that for a positive integer n and any real numbners a and b.

$\frac{1}{(x-a)^n(x-b)^n}$ =
$\frac{(-1)^n} {(a-b)^{2n}}\sum_{0}^{n-1}(C_r)((\frac{b-a} {x-a})^{n-r}+(\frac{a-b} {x-b})^{n-r})$

where $C_r$ is the coefficient of $h^r$ in the expansion of

$(1-h)^{-n}$

A proof by induction is quite long
Can anyone suggest a short proof?

An idea is to set $\boxed{x= \frac{b-a}{2} \cdot y +\frac{b+a}{2}}$ but it is just an idea, i don't have a solution.

Start with the partial fraction decomposition

$ \frac{1}{x-\alpha}\cdot\frac{1}{x-\beta}=-\frac{1}{\beta-\alpha}\cdot\frac{1}{x-\alpha}+\frac{1}{\beta-\alpha}\cdot\frac{1}{x-\beta} $ .

Then apply $ \left(\frac{d}{d\alpha}\right)^n $ and $ \left(\frac{d}{d\beta}\right)^m $ on it.

Finally you get

$ \frac{1}{(x-\alpha)^{n+1} \, (x-\beta)^{m+1}} $

$ =\sum_{k=0}^n \frac{(-1)^k}{(\alpha-\beta)^{m+1+k}}\, \frac{(m+k)!}{m!\, k!}\, \frac{1}{(x-\alpha)^{n+1-k}} $

$ +\sum_{k=0}^m \frac{(-1)^k}{(\beta-\alpha)^{n+1+k}}\, \frac{(n+k)!}{n!\, k!}\, \frac{1}{(x-\beta)^{m+1-k}} $ .

See this [url=http://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Identit%C3%A4ten:_Partialbruchzerlegungen#2]link[/url].


At a women's doubles tennis tournament, there were three teams of
two women. After the tournament, each woman shook hands once
with each of the other players except her partner. What is the
number of handshakes that occurred?

6 woman, handshakes: $ \frac{6\cdot 5}{2}\minus{}3\equal{}27$

Teams. A1,B2,C3

Hnadshakes:

AB
A2
Ac
A3
1b
12
1c
13
bc
b3
2c
23

12 total

From a different viewpoint (as I now understand this problem), of Team AB, A shakes hands $ 4$ times and so does B.  In fact, every team member shakes hands $ 4$ times.  But this will result in 2 handshakes between each of the two players, so we must divide by $ 2$ to get the desired answer of $ \boxed{12}$.

I got correct, just I thought 6*5/2=30... Clumsy me!

I understand Bugis method. Handshake combos= 6c2- 3 (partener combos)

ill answer this in a series of steps

[hide]1. we know the following facts

 I.there are 3 teams with 2 woman

 II.there are 6 woman all together

 III.1 woman shakes 1 person and doesn't shake with her again

2.   1 woman can shake with 6 people and that woman can shake with another 5 people so we get the equation 5*6=30

3.but we're overcounting so we need to subtract by 3 30-3=27

4.[hide]27[/hide][/hide]


[quote="Valentin Vornicu"]let's count a little bit :)

find the number of numbers with 5 digits which can be formed with the digits 1,2,4,7, 9 and 0.

what about the number of numbers with 5 digits that can be formed with 1,2,4,7,9 and 0, but no digit occurs twice in the same number?[/quote]

5*6*6*6*6?

5*5*4*3*2?

[quote="Valentin Vornicu"]I thought that you are beyond beginner :)[/quote]

The answers are wrong?! :?

no they are not, but they are supposed to be for more beginner people :) let's encourage them to post here :)

Hm there seems to have been quite some "market spoiling" in the Beginner's Corner...in future can the not-so-beginners try to give hints at most, instead of giving solutions?

that would be recommended. I was expecting more beginners to ask questions and post problems. By now only one has done that :)

Actually, Valentin, this question really is a bit too easy...my school classmates learn it (and more complicated counting stuff) in grade 9, so anybody who is doing olympiads should be familiar with it.

we don't do much of combinatorics here in romania, and the question is for beginner's. it does not matter too much that it's too easy. there are a lot of easy questions that people can learn alot from :)

well eum... that's like one of the questions we got at school... on day 1... to show what combinatorics are... and most of us just saw that on first sight

I think - beginners section or not! - "problems" should be things to "think" about... so a little harder would not mind.

well let me just say this: this is the first question that was addressed to my colleagues majoring in MATH, at Univ. Bucharest. The answers were as follows: 5, 25, 10, 6^5, etc. After quite a while I decided to shout the answer. So if people majoring in maths here had no idea about this question I supposed that if is of somewhat difficulty :) that's why I put it here :D.


What are some ways of solving x^x=c? (I only know of one, using Newton's root-approximation method.)

There is no exact method -- the function x^x doesn't have a nice inverse.  (It's particularly poorly behaved over the negative reals.)  The best you can do is various approximations.

Well, you can solve this with the Lambert W function. W is defined as the inverse function to xe^x, in the same way as log(x) is the inverse of e^x ... so if

te^t = x

then

t = W(x)

W(x) is nicely behaved and single valued if x is positive - there are some technicalities to handle branches if x is negative. We can avoid these if we assume c is greater than or equal to 1, in which case

x^x =c

has solution

x = e^(W(log(c))) = log(c) / W(log(c))

[see [url]http://en.wikipedia.org/wiki/Lambert's_W_function[/url] for more details]

Of course, you then need to be able to find the value of W(log(c)) - but W(x) is really no stranger than functions that we all know and love, like log(x) or sin(x) ... except that you don't find W(x) on many calculators !

And that it's very difficult to find specific values for ...

It's not [i]that[/i] difficult to find values for W(x). Conceptually, it is no more complicated than log(x). For x>0 it has derivatives and a Taylor series. Algebra packages like Maple and Mathematica implement W(x) as just another function.

If you wanted to solve x^x=e, for example, you could either leave the solution symbolically as

x = 1/W(1)

or you could look up a value for W(1), which is approximately 0.567143, and so get an approximate numeric value

x  :cong: 1.763224


modular arithmetic
Euler
function
number theory unsolved
number theory

Find the last three digits of the number $2003^{{2002}^{2001}}$.

how did you know that $2003^{n} \pmod {1000}$ had period 400? and how did you know that $2002^{n} \pmod{25}$ had period 20.

$\varphi(1000)=400, \varphi(25)=20$, so we don't need to know the order, just Euler-Fermat suffices (note that period and order must not be the same, for example the function $\sin(x)$ has the period $42\pi$).

[hide="Solution"]
We desire to find $2003^{2002^{2001}}\pmod{1000}$.  To do so, we find it $\pmod 8$ and $\pmod{125}$ and then solve the equations as we did before.  Note that this $\pmod 8$ is going to be $3^{2002^{2001}}$ or $(3^2)^{2002^{2000}*1001}\equiv 1\pmod 8$ and $\pmod{125}$ gives us $3^{2002^{2001}}\pmod{125}$.  Since $\phi(125)=100$, we desire $2002^{2001}\pmod{100}\equiv 2^{2001}\pmod{100}$.  Take this $\pmod 25$ and $\pmod 4$ to give us $x\equiv 0\pmod 4$ and since $\phi(25)=20$ we get $2^{2001}\equiv 2\pmod{25}$ so $x\equiv 0\pmod 4, x\equiv 2\pmod{25}$ therefore $x=4k=25m+2$ or $m+2\equiv 0\pmod 4$ therefore $m\equiv 2\pmod 4=4m'+2$ giving us $25(4m'+2)+2=100m'+52$ therefore $x\equiv 52\pmod{100}$.  Therefore we desire $3^{52}\pmod{125}$.  Since $3^6=729 \equiv -21 \pmod{125}$, we get $3^{52}\equiv (-21)^8*3^4 \pmod{125}$.  This is the same thing as $(73)^4 \pmod{125}$ or $(79)^2\pmod{125}\equiv 116\pmod{125}$  Finally, $2003^{2002^{2001}}=x\equiv 1\pmod 8$ and $x\equiv 116\pmod{125}$.  Therefore $8a+1=125b+116$.  Take $\pmod 8$ to give $5b+4\equiv 1\pmod{8}$ or $b\equiv 1\pmod 8=8b'+1$.  Therefore $125(8b'+1)+116=1000b'+125+116=1000b'+241$.  I conclude the answer to be $\boxed{241}$.  [/hide]

Time for $\lambda$!

Since $2003^{100} \equiv 1 \pmod {1000}$, we wish to find $2002^{2001} \pmod {100}$. This reduces to finding $2^{1999} \pmod {25}$ or $1999 \pmod {20}$ which is just $-1$. So $2002^{2001} \equiv 4 \cdot 2^{-1} (\pmod {25}) \pmod {100} \equiv 52 \pmod {100}$. So our original problem reduces to $2003^{52}$ and then we proceed as above.


inequalities
probability
vector
trigonometry
algebra
polynomial
linear algebra

use \displaystyle{} around your [tex]\TeX[/tex] to make it look much nicer.

Let [tex]a_1,\dots,a_n, b_1\dots,b_n[/tex] be real numbers such that
[tex]\displaystyle{\sum_{k=1}^na_i^2=1}[/tex],  [tex]\displaystyle{\sum_{k=1}^nb_i^2=1}[/tex] and [tex]\displaystyle{\sum_{k=1}^na_ib_i=0}[/tex]. Show that [tex]a_i^2+b_i^2\le1[/tex] for all [tex]i=1,\dots,n[/tex].

WLOG we will show that this is true for [tex]a_1, b_1[/tex].

By Cauchy,

[tex]\displaystyle \Big(\sum_{i=1}^na_i^2 - a_1^2\Big)\Big(\sum_{j=1}^nb_j^2 - b_1^2\Big) \ge \Big(\sum_{k=1}^na_kb_k - a_1b_1\Big)^2[/tex]

Plugging in our known values for the summations, we have

[tex](1-a_1^2)(1-b_1^2) \ge (0-a_1b_1)^2 = a_1^2b_1^2[/tex]

Distributing and adding [tex]a_1^2+b_1^2-a_1^2b_1^2[/tex] to both sides yields our desired inequality.

[tex]a_1^2+b_1^2 \le 1[/tex], which generalizes for all [tex]a_i, b_i[/tex].

[quote="Osiris"][quote="probability1.01"]By Cauchy,

[tex]\displaystyle \Big(\sum_{i=1}^na_i^2 - a_1^2\Big)\Big(\sum_{j=1}^nb_j^2 - b_1^2\Big) \ge \Big(\sum_{k=1}^na_kb_k - a_1b_1\Big)^2[/tex]
[/quote]

Doesn't Cauchy-Schwarz only accept positive terms? How can you justify that the inequality will hold when you artificially insert -a12 and -b12?[/quote]

Do you know a proof for Cauchy?  If not, I would suggest looking one up.  Once you know a derivation you should be able to answer the question of whether or not Cauchy is true when member(s) of either sequence are negative or nonpositive.

Osiris: first, Cauchy accepts negative terms -- you can prove this easily just by looking at Cauchy for positive terms and noting that switching some of the terms to negative can only decrease the absolute value of the smaller side while the larger side will remain unchanged.  However, looking at any derivation of Cauchy should also make it clear.  In this case, however, it wouldn't even matter since the expression he gave can also be written as

[tex]\displaystyle \Big(\sum_{i=2}^na_i^2\Big)\Big(\sum_{j=2}^nb_j^2\Big) \ge \Big(\sum_{k=2}^na_kb_k\Big)^2[/tex]

from which the stated result still follows.

Edit: Oh, I just understood what you mean!  Yes, the larger side does have to be a sum of squares, not a sum or difference.  But if you re-write it as I did, it doesn't matter at all.  I don't know why he chose to use the notation he did instead of mine, but I hope my explanation clears it up.

Er, yeah. What I meant was that in any of the brackets, all of the terms (a12 through to an2, and likewise for b and aibi) are supposed to have + signs in front of them.

I derived Cauchy in its vector form, though I suspected it of being loopy logic. Cauchy's inequality is ridiculously easy to remember in its vector form:

|[b]a[/b] . [b]b[/b]| :le: |[b]a[/b]||[b]b[/b]|, where . denotes the dot product, and [b]a[/b] and [b]b[/b] are two vectors. (So Cauchy's algebraic form specifies two vectors in [b]R[/b]n.)

Proof:

[tex]|\mathbf{a} \cdot \mathbf{b}|=|\mathbf{a}||\mathbf{b}|\cos\theta[/tex],

where [tex]\theta[/tex] is the angle between vectors [tex]\mathbf{a}[/tex] and [tex]\mathbf{b}[/tex]. [tex]\cos\theta[/tex] attains maximal value at 1, therefore

[tex]|\mathbf{a}||\mathbf{b}|\cos\theta \leq |\mathbf{a}||\mathbf{b}|[/tex]

and the result follows.

Clearly from this proof (and definition of dot product), every term inside each bracket must have a + sign.

Here is a proof for Cauchy Schwarz:

The polynomial f(x) = :Sigma: (ax-b)^2 is clearly non-negative. We expand this polynomial into ( :Sigma: a^2)x^2 - ( :Sigma: (2ab)) x + :Sigma: b^2. Since the polynomial is non-negative, the discriminant must be non-positive, i.e. ( :Sigma: (2ab))^2 - 4( :Sigma: a^2 :Sigma: b^2) :le: 0. Dividing by 4 and shifting terms around, we obtain ( :Sigma: ab)^2 :le: ( :Sigma: a^2 :Sigma: b^2), which clearly still holds for negative a and b.

Sorry, dunno how to TEX.

Quote:Let [tex]a_1,\dots,a_n, b_1\dots,b_n[/tex] be real numbers such that
[tex]\sum_{k=1}^na_i^2=1[/tex],  [tex]\sum_{k=1}^nb_i^2=1[/tex] and [tex]\sum_{k=1}^na_ib_i=0[/tex]. Show that [tex]a_i^2+b_i^2\le1[/tex] for all [tex]i=1,\dots,n[/tex].

Since this has been proved by probability1.01 above, I am posting a different solution here using Linear Algebra.

[hide]Let [tex]{\bf v_1} = (a_1,\dots,a_n)[/tex] and [tex]{\bf v_2} =(b_1\dots,b_n)[/tex] be vectors in [tex]\mathbb R^n}[/tex]. Extend [tex]{\bf v_1}, {\bf v_2}[/tex] to an orthonormal basis [tex]{\bf v_1}, {\bf v_2},\ldots,{\bf v_n}[/tex]  of [tex]\mathbb R^n}[/tex]. Make an [tex]n\times n[/tex] orthogonal matrix [tex]U[/tex] using [tex]{\bf v_1}, {\bf v_2},\ldots,{\bf v_n}[/tex]  as rows of [tex]U[/tex]. Then the transpose [tex]U^T[/tex] is an orthogonal matrix too. In particular, each row vector of [tex]U^T[/tex] has length 1. This implies that [tex]a_i^2+b_i^2\le1[/tex] for all [tex]i=1,\dots,n[/tex]. Q.E.D.[/hide]


integration
real analysis
real analysis solved

Compare numbers(clarify which one is greater or equal to) $\sqrt{e-1}$ or $\int_0^1 \sqrt{x}e^xdx$

Fast,  with Cauchy-Schwarz : $\int_0^1 \sqrt{x}e^x.dx \leq  \sqrt{\int_0^1x.dx}\sqrt{\int_0^1e^{2x}.dx} = \frac{\sqrt{e^2-1}}{2} < \sqrt{e-1} $
:cool:

i used  Cauchy-Schwarz  but you got sharper estimate:
 $\int_0^1 \sqrt{x}e^x.dx =\int_0^1 (\sqrt{x}e^{\frac{x}{2}})e^{\frac{x}{2}}.dx  \leq  \sqrt{\int_0^1xe^x.dx}\sqrt{\int_0^1e^x.dx} = \sqrt{e-1} $


show that the maximale dimension of a vectoriel space who don't containe a inversible martix is $ n(n\minus{}1)$?

it's easy.
hint :le noyau d une forme lineare est un hyperplan et les seuls formes linears de Mn(K) sont de la forme f(A)=ktrace(A)


Results for the 2006 University of Maryland Math Competition are here: http://www.math.umd.edu/highschool/mathcomp/2006winner.html

First Prize
  Brian Lawrence                 Montgomery Blair High School
Second Prize
  Richard McCutchen             Montgomery Blair High School
Third Prize
  Qinxuan Pan              Thomas S. Wootton High School

Wow ... who would have guessed Brian Lawrence would win!  :rotfl: And Richard McCutchen is no surprise either.

I got an honorable mention, which is fine, but chess64 beat me for first in Baltimore County. :mad:  Just kidding, nice job. :)

Not yet :(

EDIT: But I just found out that I got 18th place in MD.

I got 20th place.  We probably got the same score on Part II and his extra 2 points on Part I won it for him.

Apparantly both of us did well on Part II since chess64 jumped 16 places and I jumped 21 places from Part I to Part II.

[quote="E^(pi*i)=-1"]I got 20th place.  We probably got the same score on Part II and his extra 2 points on Part I won it for him.[/quote]

Lol, were the scores that close?

I dunno... Dr. Asaro is probably going to tell us our scores next Thursday though.

[quote="mathgeniuse^ln(x)"]I am sorry, but what was the score of Brian Lawrence?[/quote]

Perfect.

Do you know this for sure, or is this just a good guess?

For sure. He's won first place every year he's taken it, except for last year when he got second. And he got a perfect 2 years ago, I think.

EDIT: typo

Well, one person got a perfect score on both Parts, and Brian Lawrence won, so it must have been him.  :-)

Part I had a lot of ties, so Part II probably broke a lot of those.

Students qualifying for Part II were about the top 10%.  Students ranked in the top 100 were about the top 4%.  Students with Honorable Mention were about the top 1%.  Students winning prizes, overall or county, were the top fraction of a percent.

And yes, indeed, I can give you the scores on Thursday.

Good job!

Best,
Dr. Catherine Asaro


logarithms
function
real analysis
real analysis unsolved

show, for $ |z|\,<\,1$ :D

$ \ln\,\left(\Gamma\,(z)\right)\; \equal{}\;\minus{}\ln z\minus{}\gamma\cdot z\plus{}\frac{1}{2}\zeta(2)\cdot z^{2}\minus{}\frac{1}{3}\zeta(3)\cdot z^{3}\plus{}\frac{1}{4}\zeta(4)\cdot z^{4}\minus{}\frac{1}{5}\zeta(5)\cdot z^{5}\plus{}\frac{1}{6}\zeta(6)\cdot z^{6}\minus{}\;\;\;\ldots\;\;\;\ldots\;\;\;\plus{}\minus{}\;\;\;\ldots$

( where $ \gamma$ is Euler's constant )

many of your problems is particular case of the properties wellknown function.
http://dlmf.nist.gov/Contents/GA/7/


function
combinatorics proposed
combinatorics

1) Compute the number of integer sequences $ (a_1, \ldots, a_n)$ such that $ a_1 \equal{} 1$ and $ 1 \leq a_{i \plus{} 1} \leq a_i \plus{} 1$ for $ 1 \leq i \leq n \minus{} 1$.

2) Compute the number of integer sequences $ (a_1, \ldots, a_n)$ such that $ a_1 \equal{} 1$ and $ 1 \leq a_{i \plus{} 1} \leq 1 \plus{} \max\{a_1, \ldots, a_i\}$ for $ 1 \leq i \leq n \minus{} 1$.

For 2, the best form I can get is the following: if $ \Delta^k P(x)$ is the $ k$th forward difference of $ P$, and $ P(x) \equal{} x^n$, then the number of sequences is
$ \sum_{k\equal{}1}^n \left( \frac{\Delta^k P(0)}{k!} \right)$
Do you have a simpler closed form in mind? If not I can post the derivation.

It's actually a well-known sequence, though not one that has what I would be inclined to call a closed formula.  Does have a nice (exponential) generating function, though.  Anyhow, I'd be happy to see the derivation.

Casework on what the maximum of the whole sequence is; call that max $ k$. Use PIE to figure out the number of $ n$-term integer sequences with elements from a set of $ k$ elements that uses each element once; this simplifies to $ \Delta^k P(0)$. There is only one way to assign the numbers $ \{1,2,\ldots k\}$ to the elements of the set we're picking from to form a valid sequence from this, so we divide by $ k!$. Now just sum over all $ k$.

The identity

$ x^n \equal{} \sum_{k \equal{} 0}^{n} S(n, k) k! {x \choose k}$

gives $ \Delta^r P(0) \equal{} \sum_{k \equal{} 0}^{n} S(n, k) k! {0 \choose k \minus{} r} \equal{} S(n, r) r!$, which gives the answer as $ \sum_{k \equal{} 1}^{n} S(n, k) \equal{} B_n$, the [url=http://en.wikipedia.org/wiki/Bell_numbers]Bell numbers[/url].

With that knowledge, there should be a nice explicit bijection to partitions of $ [n]$.

Edit:  Here it is.  The value of $ a_k$ describes the index of the partition to which $ k$ belongs.  At each step, we either add $ k$ to an existing partition or create a new partition and place $ k$ into it.

Edit #2, 3:  Cute.  1) has a bijection to Dyck paths: whenever $ a_{i \plus{} 1} \equal{} a_i \plus{} 1$, we go to the right, and whenever $ a_{i \plus{} 1} < a_i \plus{} 1$ we need to go up $ a_i \plus{} 1 \minus{} a_{i \plus{} 1}$ times first and then go to the right.  Finally, we go up $ a_n$ times.  For a solution in the spirit of 2), we can biject to plane trees by setting $ a_0 \equal{} 0$ as the root and connecting $ a_i$ to the largest $ j < i$ such that $ a_i \equal{} a_j \plus{} 1$.  Either way, the answer is the [url=http://en.wikipedia.org/wiki/Catalan_number]Catalan numbers[/url].

Edit #4:  Are you using these characterizations to find a short way of proving that $ C_n \sim 4^n$ on the one hand but $ B_n$ is superexponential?


Given $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}3$.Prove that:
$ \frac{a}{\sqrt{4b\plus{}3c^2}}\plus{}\frac{b}{\sqrt{4c\plus{}3a^2}}\plus{}\frac{c}{\sqrt{4a\plus{}3b^2}} \ge\ \frac{3}{\sqrt{7}}$

[quote="nguoivn"]It's obviously trues because: $ 4\sum\ ab \plus{} 3 \sum\ a^2b \leq\ 21$
Now, try with my old and harder problem (with the same condition)
$ \frac {a}{\sqrt {4b \plus{} 4c^2 \plus{} 1}} \plus{} \frac {b}{\sqrt {4c \plus{} 4a^2 \plus{} 1}} \plus{} \frac {c}{\sqrt {4a \plus{} 4b^2 \plus{} 1}} \geq\ 1$
 :)[/quote]
I can solve this inequality with $ c \ge b \ge a$ :(  :blush:


calculus derivative integration function

Hi, I'm trying to study some pchem, but I can hardly understand the math. If I'm trying to be as efficient as possible and learn only the most important math stuff I need, what math topics should I read up on?

sorry if this is overly vague, but I really can't give any more details other than I plan to start cracking on "Physical Chemistry: A Molecular Approach" --McQuarrie

My math is limited so far to only BC Calc. :(

I would start here: http://en.wikipedia.org/wiki/Partial_derivatives because I have see partial derivatives often in pchem books.

Though, frankly, I am also having difficulty with pchem math.

@adchia
You should mostly learn basic differntiation and integration techniques .It will help you in 90% of the cases as most derivations require calculus only.

The math you need is:

- differential calculus of functions of a single variable and of several variables (concept of partial derivatives and relations between them, total differential);

- integral calculus (primitives, ordinary integrals, and line integrals);

- Taylor series;

- basic facts about logarythmic and exponential functions;

- a good ability to interpret graphs/diagrams (such as Clayperon (or P-V) diagrams, T-S diagrams, and so on).

ok thanks! off i go to find an MV calc book.