Today I find a very difficult problem
Can you solve this:

2*3+4=10

(I'm a genius)
Verificate it

...are we supposed to "prove" it using mathematics, or just offer some sort of explanation for it, because that would be too ambiguous.

Seek within your soul and the truth will become apparent.

Mods, can we have this spam locked and/or deleted?

Yeah, what is up with this anyway?  :huh:

Locked for being senseless. If the OP has a good reason for this topic, and can prove it to me, PM me with the explanation, else this stays locked.

sorry,but a (ex) friend of mine,who lives with me, is joking with my account.  i crash his head on the wall!! please,moderators,delete this topic  :blush:

[b]ex[/b]friend....crash his head on your wall...er, sorry man but I can't delete this thread...it might be used as evidence!

[quote="girino"]Can you solve this: $2\cdot3+4=10$
[/quote]
It is rather easy but is not very easy.
Here is a scetch only:

Let's start with Peano. Firstly we are to determine what we nean by addition $x+y$
We can define it by induction:
$x+0=x$
$x+y'=(x+y)'$
Now we have to prove that the operation occurs well-defined have to prove it's properties as well. Let's miss them and define multiplication:
$x\cdot 0 = 0$
$x\cdot y' = x\cdot y+x$
Again we have to prove ...  :D


A positive integer is written on each face of a cube. Each vertex is then assigned the product of the numbers written on the three faces intersecting the vertex. The sum of the numbers assigned to all the vertices is equal to 1001. Find the sum of the numbers written on the faces of the cube.

[hide="hint"] $1001=7*11*13$[/hide]
[hide="my_answer"] 31 [/hide]

[hide="Solution"] Let $a, b, c, a', b', c'$ be the numbers on the faces, where $a$ is opposite $a'$ and so forth.  Then the sum of the numbers assigned to the vertices is

$(a+a')(b+b')(c+c') = 1001$

Giving $a+a' = 7, b+b' = 11, c+c' = 13$.  Their sum is $\boxed{31}$. [/hide]


If $ a_1,a_2,...,a_n$ are positive integers , $ k \equal{} \max \{a_1,a_2,...,a_n\}$ and $ t \equal{} \min \{a_1,a_2,...,a_n\}$ , prove that :

  $ \left(\frac {a_1 ^2 \plus{} a_2 ^2 \plus{} \cdots \plus{} a_n ^2}{a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}\right)^{\frac {kn}{t}} \geq a_1 \cdot a_2 \cdots a_n$

Since $ \frac {k}{t}\geq 1$ and $ \frac {a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2}{a_1 \plus{} a_2 \plus{} ... \plus{} a_n} \geq 1$ it's enough to prove that :

$ \frac {a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2}{a_1\plus{}a_2\plus{}...\plus{}a_n} \geq \sqrt [n]{a_1a_2..a_n}$ But this it's obvious from

$ \frac {a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2}{n} \geq \left(\frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n}\right)^2$ ($ QM \geq AM$)
$ \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n} \geq \sqrt [n]{a_1a_2...a_n}$  ($ AM\geq GM$)

This is quite straightforward. We will show that
\[ \left(\frac {a_1 ^2 \plus{} a_2 ^2 \plus{} \cdots \plus{} a_n ^2}{a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}\right)^n \geq a_1a_2 \cdots a_n
\]
From weighted AM-GM inequality we have
\[ \frac {a_1 ^2 \plus{} a_2 ^2 \plus{} \cdots \plus{} a_n ^2}{a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}\ge (a_1^{a_1}a_2^{a_2}\cdots a_n^{a_n})^{\frac {a_1 \plus{} a_2 \plus{} \dots \plus{} a_n}{n}}
\]
Hence it remains to show that
\[ a_1^{a_1}a_2^{a_2}\cdots a_n^{a_n}\ge (a_1a_2\cdots a_n)^{\frac {a_1 \plus{} a_2 \plus{} \dots \plus{} a_n}{n}}
\]
Which is just Chebyshev's inequality in exponential form.


geometry
calculus
number theory
linear algebra
superior algebra
superior algebra theorems

I am a begginer in a commutative algebra.I am reading Atiyah-Mcdonald.But that book is too abstract & is not helping me anywhere.Can you plz suggest some good books(a little less abstact,some more motivation & a bit more examples).I equire 1 desperately.Please reply.

If you don't like abstract nonsense, you shouldn't be messing with commutative algebra- I doubt there's anything out there without a lot of abstraction.

I don't agree at all!

First of all, the book of Atyah-Macdonalds is excellent book and as far as I know,
the best book which gives introduction to commutative algebra.
Everyone learnes by this book.
Also, you surely must know commutative algebra in order to understand
algebraic geometry, algebraic number theory , representation theory,... ,
and other stuff and this book gives you an excellent introduction to an abstract
commutative algebra, which is the basis to study more advanced math.

Commutative algebra is definitely not an abstract nonsense - it is the tool
you have to know in order to understand modern math ( like calculus or linear algebra ).

Is there are any other good reference book?

Sorry, I am not an expert in algebra  :(  .
There is a book of S.Lang "Algebra", which contains a lot of information,
but for my opinion, it is more a "encyclopedia" but book to learn from
( but, may be, I am wrong ). There is also Van-der-Varden's " Algebra ",
but it is too old. Also there is 2 volumes of Samuel, Zarisski , " Commutative algebra ",
which contain A LOT of this stuff.
Hovever, I think , for the first time it is better to start with Atyah - Macdonalds,
which is an excellent book and contains a lot of nice excersizes.
After that you can choose exactly the field you are interested in ( as algebraic geometry,
representation theory, Lie algebras, ...) and try to find books in this field.

I too think Atiyah and Macdonald is a good book. It is not easy to read, but there are a lot of interesting exercisies in there that motivate you towards other topics in math.

I haven't quite got my nose in these by I've heard Zariski&Samuel - Commutative Algebra, Kunz - Introduction to Commutative Algebra and Algebraic Geometry, Matsumura - Commutative Algebra are THE books in Commutative Algebra.


Cancel out the $ \frac{1}{5}$ with $ 0.2$, the $ 5$ with $ \frac{1}{0.2}$, the $ \frac{1}{0.5}$ with $ 2$, and $ 0.5$ with $ \frac{1}{2}$ to obtain the answer of $ \boxed{0}$.


This is an example of an hypergeometric series, and in this case its sum is given by $S = \frac{1}{k \cdot k!}$. An hypergeometric series is one that verifies

$\frac{a_{n+1}}{a_{n}}= \frac{\alpha n+\beta}{\alpha n+\gamma}, \alpha \not= 0, \alpha+\beta-\gamma \not= 0$

and its sum is given by

$S = \frac{-\gamma a_{1}}{\alpha+\beta-\gamma}$.

[quote="kg2100"]Please tell me how to enter the math symbols easily, I know it can be done by use $..........$
But it is complex when I want to enter more symbols[/quote]
See here http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php  :wink:


ben hepsini yapmadığım için yorumda bulunmak istemiyorum.
ayrıca seçtiğiniz şıkla birlikte kısa bir yorum da yaparsanız sevinirim.
(sayısı fazla olduğundan dolayı ilk yıllardan bazılarını koyamadım)

büyük yarış bitti oylamaya katılan 4( :D ) kişinin verdiği oylar sonucu çok büyük bir farkla 2000 yılı şampiyon olmuştur bu ankette büyük emekleri geçen mathlinks.ro üyelerine minnetimi sunar çalışmalarınızın devamını dilerim. aralıkta hepinize başarılar.


[quote="0123456789"]How many pairs (x,y) positive integers satisfy the relation:

1!+2!+3!+...+x!=y^2

and is there a formula for finding the sum of the first n factorials?

I really need help, thanks.[/quote]

[hide=" Hint"]
Look at your equation modulus 5.
And find the two only solutions.
[/hide]

[quote="0123456789"]um...sorry. I don't really get it. Can you explain explain a little more? thanks.[/quote]

[hide=" Solution..."]
Take the equation $ 1!\plus{}2!\plus{}3!...\plus{}x!\equal{}y^2$ modulus $ 5$ :

If $ x\geq 4$, LHS$ \equal{}1!\plus{}2!\plus{}3!\plus{}4!\equal{}33\equal{}3\pmod{5}$ but RHS, as a square, can only be $ 0,1$ or $ 4\pmod{5}$

And so $ x\geq 4$ is impossible and we must have $ x<4$.

So we just have to check the three cases $ x\equal{}1$, $ x\equal{}2$ and $ x\equal{}3$ to find the two solutions $ (1,1)$ and $ (3,3)$
[/hide]


integration
limit
linear algebra
matrix
real analysis
real analysis unsolved

can someone help with problem 7b? thanks.

http://www.ams.jhu.edu/~eyink/Applied-Analysis/hw03.pdf

[/img]

Most of the people likely to help here are not necessarily likely to click on links or download documents. You'll get more response if you copy the problem here yourself.

[quote="Barnacle"]can someone help with problem 7b? thanks.
[/img][/quote]

Integrate $\dot{Y}=AY+YB$ by $t$ on $(0,\infty)$

If $X_{*}=-\int_{0}^\infty e^{tA}Ce^{tB}dt$ exist then $\lim_{t\to\infty}Y(t)=0$.

Hence
$0-C=A\int_{0}^\infty Y(t)dt+(\int_{0}^\infty Y(t)dt)B$.

$\int_{0}^\infty Y(t)dt=\int_{0}^\infty e^{tA}Ce^{tB}dt$.

Uniqueness...
Matrix equations $AX+BX=C$ is $n^{2}$ scalar equation fot $X_{ij}$.
Solutions this system exist for all C. Then solution unique.


function
trigonometry
real analysis
real analysis solved

[b]LET[/b] $\underline {\overline {\left|\ f:A\rightarrow R,\ f(x,y)=\sin x+\sin y-\sin x\cdot \sin y,\ \ A=\{\ (x,y)\ |\ 0<x,y\le \frac{\pi}{2}\le x+y<\pi\ \right| }}$[b].[/b]

[b]ASCERTAIN[/b] $Im\ f\ =\ \{\ f(x,y)\ |\ (x,y)\in A\ \} \ldots \ldots \ldots \ldots \ldots$[b]Answer:[/b] $Imf=\left[ \frac{24}{25},1 \right)$ [b](probable !).[/b]

:spider:  :first: Nobody ? :first:  :spider:

This didn't look very interesting- nobody bothered with it.

Note that $f'(x,y)=(\cos x(1-\sin y),\cos y(1-\sin x))$ is never zero in the interior of the domain. This means that both extrema will occur on the boundary.

The edge $x=\frac\pi2$ gives $f(x,y)=1$, which is clearly the overall maximum. The edge $y=\frac\pi2$ also gives $f(x,y)=1$.
The edge $x+y=\frac\pi2$ gives $f(x,y)=\sin x+\cos x-\sin x\cos x$. Defining $g(x)=f(x,\frac\pi2-x)$, we have $g'(x)=\cos x-\sin x-\cos^2 x+\sin^2 x$, which has zeros at $0,\frac\pi4,\frac\pi2$. At $\frac\pi4$, we have our minimum, with a value of $\sqrt2-\frac12$.

The range is $\left[\sqrt2-\frac12,1\right]$. Where did you get those wrong answers?


linear algebra
IMO
Sequence
algebra
maximization
Extremal combinatorics
IMO 1977

In a finite sequence of real numbers the sum of any seven successive terms is negative and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.

This is in my Top 10 of all IMO problems. Not only because I was a competitor then, but also because I was present to the "problem solving session" organized by the hosts and I wached the Czech student Martin Chadek presenting his solution, which made the jury K.O. Here it is:

Let $x_1,x_2,\ldots$ be the given sequence and let $s_n=x_1+x_2+\ldots+x_n$.
The conditions from the hypothesis can be now written as $s_{n+7}<s_n$ and $s_{n+11}>s_n$ for all $n\ge 1$.
We then have:
$0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6<s_{17}<s_{10}<s_3<s_{14}<s_7<0,$
a contradiction. Therefore, the sequence cannot have $17$ terms.
In order to show that $16$ is the answer, just take 16 real numbers satisfying
$s_{10}<s_3<s_{14}<s_7<0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6$.
We have $x_1=s_1$ and $x_n=s_n-s_{n-1}$ for $n\ge 2$. Thus we found all sequences with the given properties.

As far as I know, Martin only solved this problem and got a special prize for it :)

Other solutions are possible, including one using linear algebra. I'll post them if anyone is interested.

More generally, it can be shown that :
If $p,q$ are any two distinct positive integers such that none of them divides the other, and if a sequence has the property that the sum of any $p$ consecutive terms is always positive and the sum of any $q$ consecutive terms is negative then the maximal length of this sequence is $p+q-d-1$ where $d = \gcd (p,q).$

The proof I know is quite long, so that I will not write it here. It can be found in :
Quadrature, n°54 (Oct. Dec. 2004), ex. E-218, p.34-36.

Pierre.

Well, this was proved during the competition by GB7, John Rickard, a gold medalist in 1977!

[quote="enescu"]As far as I know, Martin only solved this problem and got a special prize for it :)[/quote]
No, he also fully solved Problem 4.

Does this work or have I deluded myself? Not fully detailed.
-------------------------------------------------
We claim that the answer is $16$. One can find a construction without too much difficulty.

Call an $n$-sum the sum of $n$ consecutive terms. If we suppose, for contradiction, that the sequence has $17$ terms, then every $4$-sum can be written as an $11$-sum with a $7$-sum subtracted, and must hence be positive. Similarly, every $3$-sum can be written as a $7$-sum with a $4$-sum subtracted, and must therefore be negative. And, finally, every term can be written as a $4$-sum with a $3$-sum subtracted, and must therefore be positive again. However, if every term is positive, then it's impossible for every $7$-sum to be negative, and we've reached a contradiction.

[quote=enescu][quote="Arne"] It is easy to show that 16 works :)[/quote]
This is the jury solution. They provided the example
\[ 5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5. \]
But how do you find such an example?
That's why I loved Martin's solution :)[/quote]

>16 years late, but I think you can motivate this solution by noticing that $\frac 27>\frac 3{11}$, meaning that if we can make it so that we get exactly two negative numbers of pretty large magnitude in each block of length $7$ and $3$ in each block of length $11$ we would be done by making the other numbers all equal to some positive number. It's not so obvious to pick the numbers $5$ and $-13$, but maybe this is easier?

\[20, 20, -51, 20, 20, 20, -51, 20, 20, -51, 20, 20, 20, -51, 20, 20.\]

Note that the block of $11$ reaches from the first $-51$ to right before the last $20$.


geometry
trapezoid
calculus
perimeter
area of a triangle
Heron\u0027s formula
cyclic quadrilateral

Could you guys help me with this problem?
 There is a farmer who has four fences , they are 4m,3m, 2m and 1m in length?

Fences:
_  _  _  _               / _  _  _            / _  _              / _

He wants to form a closed area to keep his sheep, (a trapezoid can be made from these)

What is the maxium area that can be Obtained?(any fence can be placed in any order, but all four are connected)

[quote="white_horse_king88"]Is a triangle allowed?[/quote]

Why not?

Well I don't think triangles are allowed. All four must be connected.

As far as I thought,
There are 4! or 24 possible cases in which 21 are repitative and only 3 trapezoids can be made out of these pieces....
 Just don't necessarily trust me , maybe I am wrong somwhere :lol:

[quote="jamshidi_f2004"]Well I don't think triangles are allowed. All four must be connected.[/quote]

I think the triangle referred to would be if, say, you put the 1 and 3 length fences in a row to make a fence of length 4, and then make a triangle with sides 2, 4, 4. That would be allowed.

Yeah, but I don't think that would give the maximun area...

There are 3 distinct orders for the edges.  One way you can approach this problem is to take one ordering and draw in a diagonal.  You can express the area of the quadrilateral easily in terms of the length of this diagonal, for example using Heron's formula.  If you know calculus, you can then find the maximum of this expression without a whole lot of difficulty, but I'm not sure you'll be able to do it without calculus.  I believe (although I won't promise) that the maximum will occur when the quadrilateral can be inscribed in a circle, which, interestingly enough, also occurs at a point when the different orderings all give the same area.  I don't really think this question can be solved by "Getting Started" - level methods, although there may be an elegant geometric proof that the cyclic quad is the largest.

Some comments -

As  JBL pointed out, the maximum area occurs when the quadrilateral is cyclic,  order of the sides does not matter, and the area is given by Brhamgupta's formula. [hide]that is if  $ s = (a+b+c+d)/2 $ then area is given by :
$ \sqrt {(s-a)(s-b)(s-c)(s-d) }$
(in this case this becomes $\sqrt{1.2.3.4} = \sqrt{24} $

The geometric proof ( Known to Brhamgupta or even earlier)   may not be very hard, but if you use calculus,  it is  easy to   see, specially  if you use Langrage multiplier  method:  (Google or Standard  Math Text can give more details)

[/hide]

BTW as this can be extended to more than 4 sides. As long as any one of the side is shorter than sum of the rest (so that one can  construct a polygon)   a cyclic polygon can always be made, at this will have the maximum area.

look for a cyclic quadrilateral.  check out mathworld for its definition and some area formulas with it.  I can't remember the name of the main formula based on sides and semi perimeter but it is pretty much the same as heron's formula for triangle.  A cyclic quadrilateral has maximum area for those length sides

Thanks a lot guys,
Brahmagupta's formula definitely works :)
That guy was a genious, when I saw the proof I was just amazed

By the way, once you have Brahmagupta's formula, Heron's formula can easily be derived. Can you see how?


I got my dad to explain how to prove this, but I don't really get it:

(n-1)(n+1)=(n^2)-1

Just use the FOIL method and you should be left with $n^{2}-1$.
If you don't know what that is, it's just a trick to multiply two binomials.

FOIL-
[hide]Using the distributive property, (n-1)(n+1)=n(n+1)+(-1)(n+1)
n(n+1)=n^2+n
-1(n+1)=-n-1
which also equals
n^2+n-n-1
which can again be simplified to n^2-1  :)[/hide]

[quote="7h3.D3m0n.117"][quote="Stargirl240"]FOIL-
[hide]Using the distributive property, (n-1)(n+1)=n(n+1)+(-1)(n+1)
n(n+1)=n^2+n
-1(n+1)=-n-1
which also equals
n^2+n-n-1
which can also be simplified to n^2-1  :)[/hide][/quote]

Actually that is the distributive property. FOIL is a little different.

FOIL stands for First, Outer, Inner, Last. For example, in $(x+y)(x+y)$

We multiply the first terms in each binomial (in this case it would be the x) so we get

$x^{2}$

Then, we do the outer terms, in this case it is

([b]x[/b]+y)(x+[b]y[/b])

so the next term in the expansion would be $xy$ so we have so far: $x^{2}+xy$.

Then, we do inner terms, which is

(x+[b]y[/b])([b]x[/b]+y)

Another xy.  so we have $x^{2}+2xy$

Finally, we do the last terms

(x+[b]y[/b])(x+[b]y[/b])

giving us the expansion $x^{2}+2xy+y^{2}$[/quote]
Oops that's sort of what I meant...  :blush: But isn't FOIL just the distributive property, basically?

FOIL is the same thing as the distributing process except that it only works on expressions in the form (a+b)(c+d)

7h3.D3m0n.117 is right. FOIL stands for first, outer, inner, last.  In plain English, this mean thats you multiply the first two terms (in the above example, a*c) then and the outer terms multiplied together (a*d) add the inner terms together (b*c) and finally add the last terms of each part multiplied together (b*d). This works out out be ab+bc+bd+ac.

"FOIL" is just an easier way to remember this process.

My recommendation would be to learn the distributive process down cold, then learn these tricks.


modular arithmetic
combinatorics proposed
combinatorics

you are allowed to take positive natural numbers, find a set such that it is not divisible by n
such that each consecutive sum is not divisible by n . Lets take n=100.

Consecutive sum would include individual terms too.

My answer is it doesn't exist. I couldn't find  a elegant proof but i had some idea and i tried with some numbers until n=6 on my computer and it failed to find any such set.

My reasoning is as follow. lets assume such a set exists for a particular n.

1) Now none of the elements should be divisible by n otherwise it would violate our assumption.
2) So all the elements modulo n lie in the interval [1,n-1]
For the 1st element we have n-1 choices (say we choose k)
for the 2nd element we have only n-2 choices ( we cannot choose n-k , say p)

3) For the 3rd element we have (n-2 choices if we consider 2nd element only say q).
Now q +p modulo n is not 0, , sum of 1st 2 elements is k+p,  now q + (k+p) modulo n is not 0 , this removes an additional number from the interval [1,n-1] because k+p is not 0. Now so 3rd element has only n-3 choices

 So the choices go on decreasing until the last element where we are left with 0 choices.

Does anyone have a better solution or actually a contradiction

Yes, your reasoning is true, and could be expressed more nicely:
Let you set be $ \{a_1,a_2,\dots,a_n\}$
then consider the sums $ a_1,a_1 \plus{} a_2,a_1 \plus{} a_2 \plus{} a_3,\dots, a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n$
there is $ n$ of them. So if we consider them $ \pmod{n}$ we have two cases:
1) either all of them are different. In which case all the remainders $ \{0,1,...,n \minus{} 1\}$ will appear. So $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_i\equiv 0\pmod{n}$ for some $ i$, contradiction!
2) some two of the sums are equal $ \pmod{n}$ so
$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_i\equiv a_1 \plus{} a_2 \plus{} \cdots a_j \pmod{n}$ for some $ i > j$
but then $ a_{j \plus{} 1} \plus{} \cdots \plus{} a_i\equiv 0\pmod{n}$ contradiction so our set can not satisfy that property.

[quote="Albanian Eagle"]Yes, your reasoning is true, and could be expressed more nicely:
Let you set be $ \{a_1,a_2,\dots,a_n\}$
then consider the sums $ a_1,a_1 \plus{} a_2,a_1 \plus{} a_2 \plus{} a_3,\dots, a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n$
there is $ n$ of them. So if we consider them $ \pmod{n}$ we have two cases:
1) either all of them are different. In which case all the remainders $ \{0,1,...,n \minus{} 1\}$
Cool ,
actually $ \{1,...,n \minus{} 1\}$ since individual numbers also can't be 0 , rest follows from pegion hole principle.


abstract algebra
advanced fields
advanced fields unsolved

prove the set of equivalence classes of ideals form an abelian group.
(already shown that for any ideal I there is an ideal J s.t. IJ is principal).

Multiplication is well defined, associativity/commutativity is easy, the neutral element is represented by $ (1)$ and inverses exist by the comment in the brackets. Nothing more is to show  :wink:


symmetry
geometry
geometric transformation
reflection
combinatorics proposed
combinatorics

[quote="Domi"]Aaron wins , he plays the middle cell and "symmetry" with Betty's play .

Domi

PS: sorry for my poor english[/quote]
What if Betty plays her O in the square next to the middle one? Then Aaron by symettry should play a X in a square next to another X, which can't happen!

And not only that, I'm almost sure that Betty is the ultimate winner. Take $ n\equal{}5$ for instance.

Eh, by the way, why is this topic in NT, and not in Combinatorics?

Yeah, my mistake, [size=150]mods please move it.[/size]

Domi means that Aaron plays symmetrically with the [i]same[/i] piece.

Unfortunately,
[quote="B_Gasperov"]Aaron always enters the symbol X and Betty always enters the symbol O. [/quote]

[quote="The QuattoMaster 6000"][quote="B_Gasperov"]Two players take turns entering a symbol in an empty cell of an 1 x n chessboard, where n is integer greater than 1. Aaron always enters the symbol X and Betty always enters the symbol O. Two identical symbols may not occupy adjacent cells. A player without a move loses the game. If Aaron goes first, which player has the winning strategy?

OK, for an even n its trivial because Betty's strategy of "symmetry" forces her win. But what about an odd n?[/quote]
I think that the reflection method still works, but it has to be modified a bit:
[hide="Solution"]
The case if $ n$ was even is described in your quote.  If $ n$ is odd, then we do the following: If Aaron places an X in a particular cell, then let Betsy put her O in the cell that is formed after "reflecting" Aaron's cell into the middle cell.  This works fine until Aaron puts his X in the middle square.  In this case, Betsy can put her Y into one of the squares adjacent to the middle square.  Assume at most one of them is open.  This means that none of them are open (Betsy would only have a Y in one of these adjacent squares if Aaron had an X in the other adjacent square.)  Without loss of generality, let the square to the left have a Y in it; then, the square to the right would have an X in it and two consecutive X's would be formed, which is a contradiction.  Also, notice that the squares adjacent to the adjacent squares are reflections over the middle square, so they cannot be both Y's.  Let Betsy put her Y into the square not adjacent to any Y's.  Now, notice that Aaron cannot do the reflection strategy back on Betsy or else two consecutive X's would be formed.  This forces Aaron to choose a new spot to place his X and Betsy continues the reflection strategy and wins.

Note: I realize that the wording of my proof is kind difficult to understand, especially because I didn't want to really add too much notation (I thought that would even make life more complicated) and I don't know how to add a diagram.  If you have any questions/counterexamples to my proof, then feel ask/comment.  [/hide][/quote]

What if Aaron places his X in the middle, and Betsy places her O directly to the left of it, then Aaron places an X two spaces to the right of his first X. Betsy cannot reflect that, because of her first O.

[code]-----
--X--
-OX--
-OX-X
OOX-X[/code]

See by your strategy, Betsy should place an O in that far left square, but she can't.

An easy solution  when n is odd ( except n=1 ) :D

Aaron plays anywhere and Betty plays on an extremity of the array . When Aaron plays again , appears X----X ( --- = empty ) and Betty can play beetween the "X" and wins .

Domi

Sorry for my very very bad english  :blush:


Prove that for any finite colouring of the 2-element suibsets of the positive integers, there is an infinite subset $S$ such that all its 2-elements subsets have the same color

I'll do it for two colors (red and white, say), because it's easier to discuss it this way; the proof for the general case is almost the same.

Take any natural $x_1$. There are infinitely many naturals $n>x_1$ s.t. all the $(x_1,n)$ have the same color (if this color is white, call $x_1$ white, and if it's red, call $x_1$ red; we do the same for all the other $x_k$'s). Now pick any $x_2>x_1$ among those $n$'s. There are infinitely many $n>x_2$ s.t. all the $(x_2,n)$ have the same color. Then pick any $x_3>x_2$ among these $n$'s and so on.

Among the naturals $x_1<x_2<x_3<\ldots$ there are infinitely many with the same color (I defined the color of an $x_k$ above). Suppose it's red, and $x_{i_1}<x_{i_2}<\ldots$ are all red. Then $\{x_{i_1},x_{i_2},\ldots\}$ is the set we're looking for.


inequalities
trigonometry
linear algebra
matrix

Prove, using algebra, the following inequality for $x_{i},y_{i}\in\mathbb{R}$:

$\left(\sum_{i=1}^{n}x_{i}y_{i}\right)^{2}=\left(\sum_{i=1}^{n}{x_{i}}^{2}\right)\left(\sum_{i=1}^{n}{y_{i}}^{2}\right)-\sum_{i<j}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}$

I was also curious about proving this with vectors.  Assume $\mathbf{x}=\sum_{i=1}^{n}x_{i}e_{i}$ and $\mathbf{y}=\sum_{i=1}^{n}y_{i}e_{i}$.  Then we would have:

$\left(\mathbf{x}\cdot\mathbf{y}\right)^{2}=\|\mathbf{x}\|^{2}\|\mathbf{y}\|^{2}-?$

Does anyone know how to represent that last term in terms of $\mathbf{x}$ and $\mathbf{y}$?  I'm drawing a blank...

The last part would be $\sin^2\theta||x||^2||y||^2$; this can be seen by the identity $\cos\theta=\frac{x\cdot y}{||x||\cdot||y||}$.

In three dimensions you have

$(\vec x\cdot\vec y)^2+|\vec x\times\vec y|^2=|\vec x|^2|\vec y|^2$, where

$\vec x\times\vec y=\left|\begin{matrix}\vec i &\vec j &\vec k\\ x_1 & x_2 & x_3\\ y_1 & y_2 & y_3\end{matrix}\right|$


1) How many 15-letter words can be produced from an alphabet which has three letter? Is it 3^15?
2) How many solutions are there of the equation x+y+z=15, where x,y,z are all non-negative integers?
I know that there should be a relationship between 1) and 2).
Please help me out.

#1: I'd say $ 3^{15}$ sounds right, as you have $ 3$ choices for each letter, and there are no restrictions.

[hide="2"]
This is a classic problem called balls and urns.
Basically, list out 15 as a series of ones:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

We want to split those ones up into 3 parts so that each part has some integer (greater than or equal to 0) number of ones in it. To do this, we can stick on two dividers.

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 | |

We want to pick two spots out of the 17 things above and move the dividers to those spots. For example,

|1 1 1 1 1 1 | 1 1 1 1 1 1 1 1 1

would represent the solution $ (x,y,z)\equal{}(0,6,9)$.

Obviously, there are $ \binom{17}{2}\equal{}136$ ways to do that.
[/hide]


Guess this woman's number:

(After going to the link, do not scroll it into the text.  That's
too far.)

The link...
http://www.birthdaycards.com/pickup?ID=A222-C8JJ-29S7-J4R

that's uhh pretty impossible to guess. Would would guess $\sqrt{53}$ (yay latex!)

You meant "Who would?"

You would!

By the way, this woman has more math under her belt (for future reference).

Wait, is she supposed to say something? I don't get it... :maybe:


search
number theory solved
number theory

Show that if $x,y,z$ are positive integers, then
$(xy+1)(yz+1)(zx+1)$ is a perfect square if and only if
$xy+1, yz+1, zx+1$ are all perfect squares.

I want to show $xy+1, yz+1, zx+1$ are pairwaise relative prime.
but.....

--------------
ah... I solved this problem right now...
without using $xy+1, yz+1, zx+1$ are pairwaise relative prime.....

Check Kedlaya's homepage. If I recall correctly, he has a solution in one of his expository papers.

could you give me the solution?
thank you

Search for the word "square" on this page:
[url]http://www-math.mit.edu/~kedlaya/papers/other.html[/url]


Diophantine equation
number theory proposed
number theory

Prove that there are infinitely many triples of positive integers $ (x; y; z)$ with $ (x,y,z)\equal{}1$ such that $ 14x^{2}\minus{}y^{2}\plus{}12\equal{}z^{2}$.

The triple $ (1; 1; 5)$ is a solution. I will prove there are infinitely many triples $ (x; y; 5)$.
Take $ x \equal{} 1 \minus{} a$, $ y \equal{} 1 \minus{} b$; plugging into the equation yields $ 14a^2 \minus{} 28a \minus{} b^2 \plus{} 2b \equal{} 0$. Take now $ b \equal{} \frac {p} {q} a$. This will give $ a \equal{} b \equal{} 0$, which we discard, and $ a \equal{} 2\frac {14q^2 \minus{} pq} {14q^2 \minus{} p^2}$, $ b \equal{} 2\frac {14pq \minus{} p^2} {14q^2 \minus{} p^2}$. Looking for a denominator $ \pm 1$ we fall upon a Pell equation $ p^2 \minus{} 14q^2 \equal{} 1$, with fundamental solution $ (15; 4)$.
Therefore there are infinitely many solutions with integer $ a,b$, hence integer $ x,y$.

I will prove there are infinitely many triples of positive integers $ (x;1;z)$. Obviously $ (x,y,z)\equal{}1$
If $ y\equal{}1$ then the equation becomes $ 14x^{2}\plus{}11\equal{}z^{2} \rightarrow z^{2}\minus{}14x^{2}\equal{}11$.
This last equation has a solution minimum $ (5;1)$ and therefore by theorem has infinite solutions.


This was the hardest problem, even though it was #2 of 45:

In the quaternion number system, there are special numbers $i$, $j$, and $k$ that satisfy \[ i^2=j^2=k^2=ijk=-1 \]. Find the result of the product $k\cdot i$. Note that unlike regular arithmetic, $i$, $j$, and $k$ do not commute.

(A) $-j$
(B) $j$
(C) $k$
(D) 1
(E) $-i$

Well, the idea is simple - you keep shifting over entries.  Is this pre-olympiad?  Questionable.

Anyways, start with

$ijk=-1$.  Right multiply by $i$ to get

$ijki=-i$.  Left multiply by $i$ to get

$iijki=-jki=-ii=1$.  So $-jki=1$.  Left multiply by $j$ to get

$-jjki=ki=j$

as desired.


[quote="nonie"]Find the smallest multiple of 3 with no odd digits.[/quote]

[hide]Uh...6?[/hide]

I think so. Otherwise you could say $-222\bar{0}$ which is infinitely small...

[quote="nonie"]Find the smallest multiple of 3 with no odd digits.[/quote]

OK, if it was 1 digit, that would be easy.  But let me change the question:

[b]Find the smallest 4-digit multiple of 3 with no odd digits and no 0's.[/b]

It shouldn't be too hard.  :D  :D  :D  :D  :D  :D  :D  :D

[hide]3-digit number: 204
4-digit number with no zeroes: 2226[/hide]

Do you have a method for solving these types of problems?

well for finding the smallest three digit multiple of three with all even digits you don't really have to use three digit multiples. you can do like 3 times 3 = 9 and 3 times 30= 90. After trying that out and you still don't have the answer, i think you have to use guess and check. by the way i got 204 for the answer to the problem.

 :furious:

Was this supposed to be hard..... the answer is 6. Its not 3 because 3 is odd and we dont want odd...........  ;)

3 consecutive even integers that addd up to 312

[quote="nonie"]whoops typo! I meant smallest [b]3 digit [/b]multiple of 3 with no odd digits...[/quote]

[hide]Way too easy!

It cannot be in the 100's, so we go to the 200's.  201 is a multiple of three, and then you go up to 204.  $\displaystyle{\boxed{204}}$[/hide]

No 3 consectutive Even Integers are going to all be multiples of 3!
Label the integers A, A+2, and A+4.

If A is a multiple of 3, then A+2 and A+4 won't be.  Good counter-example?

[quote="José"][quote="kishiman1"]try this ....

what are 3 consecutive even  integers that are multiples of 3 and add up to 312

NO CALCULATORS.[/quote]

The three are multiples of 3??? Can't be... If $3|x$, $3|x+2$ is not true (use mod to check it)[/quote]

There was nothing about multiples of three in kishiman's original post, or he edited it :ninja: But I'll find the numbers anyway.
[hide]
x-2, x, and x+2 are the numbers
3x=312
x=104
102, 104, 106 :D [/hide]

[quote="kishiman1"] that are multiples of 3 [/quote]

Here


May I ask a survey for the interview for Early Action/Decision this year? Has anyone got any notification so far? Or even have already interviewed? Please list your school.
Thank you.

PS I am worried about my interview because I haven't heard any info about it from Admission Office yet.

I was interviewed yesterday for Princeton ED.

Um, at this point I would actually suggest contacting the school concerning the interview if you're really gung-ho about it. They usually need the interview evaluations by the 'end of November' for EA/ED applications.

However, depending on your location, they may not have an alumnus in your area and therefore there won't be anyone to contact you. This will not be held against you in any way, so don't worry if you are not contacted.

But, like I said, if you really want an interview (or are afraid they have the wrong e-mail address/phone number for you) feel free to give the university a call (quickiest answer) or an e-mail and I'm sure they'll get back to ya as soon as they can.


inequalities
vector
geometry solved
geometry

in triangle $ABC$ find locus of $M$ wich that:
AM+BM<CM
AM+CM<BM
BM+CM<AM

Something is wrong here. If you add up the inequalities you get AM+BM+CM=0 which leads to A=B=C=M.

Misha

Even if you replace AM + BM < CM, AM + CM < BM and BM + CM < AM by AM + BM > CM, AM + CM > BM and BM + CM > AM, your locus becomes some strange figure enclosed by 3 quartics (= curves of 4th degree!). What exactly do you want to know about this locus?

  Darij

we can creat a triangle with $AM,BM,CM$ an we have $AM+BM+CM=0$
and the problem is true.

I think she mean AM,BM,CM are vector. But we can't compare two vector!!??


Bina and Andreas start at the same point on a circular track and begin running in opposite directions. After $ 35$ minutes they are both back at the starting point. Bina has run $ 12$ laps, while Andreas has run $ 17$ laps. How many times did they pass each other during their run, not including the times they were together at the start and finish?

I know the answer would be 28, but the answer would be hard to explain without a diagram. You basically draw a circle, with 29 ticks because they ran a total of 12+17=29 laps. Mark one of those as the starting point. The first time they meet, Bina ran 12/29 of the lap and Andreas ran 17/29 of the lap. Second time would be 24/29 of the lap and so on. Because 12 and 17 don't have an LCM that is less than 204, then they meet at all of the ticks once before they meet up again. Since we don't count the starting point, they meet 29-1=28 times.

I think the answer is 16, because 1 for every lap that Andreas runs, not including the ending point. It can't be 28, because how can Andreas pass Bina more than once in only one lap?

I know for a fact it is $ 28$, and they run at different speeds, so it is possible.

wut would u do if they had a lcm less than 204? y do u use 204 anyway? thx in advance!!!!!!!!! :D

well stupinerd, you gotta remember is 35 mins


[quote="GameBot"]A soccer jersey must be numbered with a one- or two-digit integer. Ellen has 8 each of the digits 1-8. What is the maximum number of jerseys she can number if no jersey number is repeated.[/quote]

Case 1: Ellen uses a one-digit number

There are 8 one-digit numbers.

Case 2: Ellen uses a two-digit number

There are $ 8 \times 8 \equal{} 64$ two-digit numbers.

This makes a total of $ 64 \plus{} 8 \equal{} \boxed{72}$.

She can make $ 8$ one digit, and $ 8\cdot 8$ two-digit jerseys, or 72.
EDIT: Beaten

This is incorrect. She only has 8 of each digit.

She makes the 8 1-digits numbers. Now she has 7 of each left, for 56 digits. These 56 digits can make 56/2=28 numbers. I leave it to the reader to verify that she can make all 28 of these numbers.

Our answer is 28+8=36.

I considered it as, she has 8 digits: 1-8

[quote="GameBot"]A soccer jersey must be numbered with a one- or two-digit integer. Ellen has 8[b] each [/b]of the digits 1-8. What is the maximum number of jerseys she can number if no jersey number is repeated.[/quote]

It clearly states that she has 8 of each number, that is 1111111122222222333333334444444455555555666666667777777788888888

The question actually asks for the number of jerseys she can make out of the 64 numbers, not how many choices for the first jersey there are. Thus, when you use all 8 numbers in a two-digit number, you don't have any left for a one-digit number. But since a one-digit number uses only one of the digits, more jerseys could be numbered by numbering the one-digit ones instead of those with repeated digits. Thus, the answer is $ 8$ one-digit numbers plus $ 8\minus{}1$ rows ($ \times$) of $ 8\minus{}1$ (subtracting the number in the row with a repeated digit), giving a total of $ \fbox{55}$ jersey numbers.

[quote="2000 Nats Solutions"]
5. To conserve digits, Ellen should number as many jerseys as possible with a one-digit number. She will
use 1 of each of the digits to create jerseys number 1 through 8. She will now have 7 each of the digits
1-8 remaining. This is a total of 56 digits, from which she can make 28 two-digit jersey numbers. She
will have no difficulty finding 28 unique jersey numbers. The total number of jerseys she can make is
8 + 28 = 36.[/quote]

So 36.

Oh yes, that's right, to consider each number used in a 10-row. Sorry!


nice prob!

[hide]
we have that $\forall k \in \{ 2,3...2006 \}$, $\log_kN=\frac{\log_NN}{\log_Nk} \implies \frac{1}{\log_kN}=\log_Nk$

so $H=\log_N2+\log_N3+\log_N4+...+log_N2006$. But the sum of the logs is the log of the product, then $H=\log_NN=1$[/hide]

Suppose that: $\ a^{log_bc}+b^{log_ca}=m$
Find the value of: $\ c^{log_ba}+a^{log_cb}$

 ;)  :huh:

hydro, $\log_kN$=$\frac{log_NN}{log_Nk}$
With N>0, N,k other 1

 :D

[hide]$\frac{1}{\log_a b}\Longleftrightarrow \log_b a$.

So rewrite equation as
$\log_N 2+\log_N 3+...+\log_N 2006=\log_{2006!}2006!=1$.[/hide]

since
$a^{log_bc}=c^{log_ba}$
and
$b^{log_ca}=a^{log_cb}$
so
$c^{log_ba}+a^{log_cb}=m$


geometry
circumcircle
inequalities
inequalities unsolved

\[ \begin{gathered} {\text{(x \plus{} y \plus{} z) (xR}}^2 _1 {\text{ \plus{} yR}}^2 _2 {\text{ \plus{} zR}}^2 _3 {\text{)}} \hfill \\
{\text{ }} \geqslant {\text{yza }}^{{\text{2 }}} {\text{ \plus{} zxb }}^{\text{2}} {\text{ \plus{} xyc }}^{\text{2}} {\text{ ,}} \hfill \\
{\text{where x, y}}{\text{. z are arbitrary real numbers and R}}_{\text{1}} \hfill \\
{\text{R }}_{\text{2}} {\text{ , R}}_{\text{3}} {\text{ are the distances from an arbitrary point}} \hfill \\
{\text{P to the vertices A, B, C, respectively}}{\text{. of the}} \hfill \\
{\text{triangle ABC}} \hfill \\
\end{gathered}
\]


[quote="sen"]Find all integral solutions of

$x^{2}-y^{2}=2xyz$[/quote]
[hide]
$x^{2}-y^{2}=2xyz$
$\iff x^{2}-x(2yz)-y^{2}=0$
$\iff x=\frac{2yz\pm\sqrt{4y^{2}z^{2}+4y^{2}}}{2}$
$\iff x=\frac{2yz\pm2y\sqrt{z^{2}+1}}{2}$
$\iff x=yz\pm y\sqrt{z^{2}+1}$
The stuff inside the radical, $z^{2}+1,$ can never be a perfect square since it is always 1 greater than a perfect square, unless $z=0.$
If $z=0,$ we get $x=\pm y$ which gives us the solutions $(a,a,0)$ and $(a,-a,0)$ for some integer $a.$

[/hide]


[quote="13375P34K43V312"]Prove that in any given base greater than 2, 11x11=121.[/quote]

[hide]Let the base be $a$.

$11_{a}=a+1$ and $121=a^{2}+2a+1$.

Also, it is true that $(a+1)(a+1)=a^{2}+2a+1$ so $11\times 11$ is always $121$.

Yay my 1000th post.[/hide]


1.  (somewhat well known).  For a set of two normal dice (i.e. labelled 1,2,3,4,5,6), the probability of rolling a sum of 2 is 1/36; the probability of rolling a sum of 3 is 2/36, and so on and so forth.  There is another pair of dice labelings that produce this probability distribution.  Find it.

2.  (variation)  Ana has three dice, one each of 4, 6, and 14 sides, each labelled as usual with 1, 2, ... , n.  Bob also has three dice, one each of 2 (i.e. a coin), 7, and 24 sides.  The 2- and 7-sided dice are labelled as usual.  The probability distributions for Ana and Bob are the same (i.e. they have the same chance of getting a sum of any given integer).  Find the numbers on Bob's 24-sided die, and the number of times each repeats.
For this one, ignore physical impossibilities such as the construction of regular 7-sided dice
Also, try to do it in some way other than brute force.  Warning: it's hard!

Martin Gardner talked a little about the first problem in a 1978 edition of [i]Scientific American[/i]; the second comes from the 'Pain' section of a sheet of problems given to the MN ARML team on the bus to Iowa.  Enjoy!

There are many other similar, simpler questions one can ask (if one wanted to).  The smallest:

Two non-standard (not necessarily identical), 2-sided dice, when rolled (flipped?) together, give the same distribution that two standard dice would.  Find all possible numberings for these two dice.
Same question, but with 3 dice.
Can you generalize to $n$ dice?

How about two 3-sided dice?

Finally:  Can you find four non-standard 2-sided dice which give the same distribution as two standard 4-sided dice?

1. I think Danbert means dice that have the same probabilities given that the dice each have six sides.
[hide="The solution"]
Let the number of ways each sum can be made be represented with the generating function...
$(x+x^2+x^3+x^4+x^5+x^6)(x+x^2+x^3+x^4+x^5+x^6)$
$(x+x^4)(x^0+x+x^2)(x^0+x)(x+x^3+x^5)$
$(x+x^4)(x+x^3+x^5)(x^0+x)(x^0+x+x^2)$
$(x^2+x^4+x^5+x^6+x^7+x^9)(x^0+2x+2x^2+x^3)$
So the new dice have the numbers (2,4,5,6,7,9), and (0,1,1,2,2,3)
[/hide]

[quote="gopherhole112"]1. I think Danbert means dice that have the same probabilities given that the dice each have six sides.[/quote]
Yes, that's right, but it's odd...  I got that the dice should be
[hide]1,2,2,3,3,4 and 1,3,4,5,6,8.  Oh, they're equivalent (add one on one die, subtract one on the other)  I suppose that my answer is the only one without zero's then.[/hide]
As for JBL's question,
[hide]I don't think so.  You obviously need exactly two 1's and two 4's on the coins, but then there must be exactly one 2; then it is impossible to get the required three ways to get a sum of 4.  If 0's are allowed, then there can be exactly one zero, no 1's, and one 2.  But then you can't have enough ways to get a sum of 4.  If I missed something here, let me know... It doesn't seem right.[/hide]

Danbert, you're right that it doesn't work if I restrict to positive integers, but your reasoning seems off -- I don't need any 4s, I have four dice to use.

All of my questions should be answered on the assumption that the dice are [i]not[/i] limited to the positive integers, and they all should have solutions when you expand only so far as to add 0.


trigonometry
geometry
incenter
inequalities
geometry proposed

Let $A', B' , C'$ be the projections of a point $M$ inside a triangle $ABC$ onto the sides $BC, CA, AB$, respectively. Deﬁne $p(M ) = \frac{MA'\cdot MB'\cdot MC'}{MA \cdot MB \cdot MC}$ . Find the position of point $M$ that maximizes $p(M )$.

Let $MA'=d_{a},MB'=d_{b},MC'=d_{c},\angle MAB=\alpha_{c},\angle MAC=\alpha_{b}$ we have
$MB\cdot MC=\frac{d_{b}d_{c}}{\sin\alpha_{c}\sin\alpha_{b}}=\frac{2d_{b}d_{c}}{\cos(\alpha_{c}-\alpha_{b})-\cos(\alpha_{c}+\alpha_{b})}\ge \frac{2d_{b}d_{c}}{1-\cos A}=\frac{d_{b}d_{c}}{\sin^{2}\frac{A}{2}}$ similarly we get
$MC\cdot MA\ge \frac{d_{c}d_{a}}{\sin^{2}\frac{B}{2}},MA\cdot MB\ge \frac{d_{a}d_{b}}{\sin^{2}\frac{C}{2}}$ multiply them and square root them we get
$\frac{d_{a}d_{b}d_{c}}{MA\cdot MB\cdot MC}\le \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$
Equality holds iff $M\equiv I$ incenter of $ABC$.

this is also in imo 01 shortlist... does the hungary-israel binational competition takes its problems from other sources?

let x = MÂC', y = MBA' and z = MCB'
so, sin(x) = MA'/MA sin(A-x) = MB'/MA,...
so (MA'.MB'.MC'/MA.MB.MC)²=sin(x)sin(A-x)sin(y)sin(B-y)sin(z)sin(C-z)
or sin(x)sin(A-x)<=(sin(x)+sin(A-x)/2)²   (AM-GM)
                         <=(sin(A/2))² by Jensen
so :
MA'.MB'.MC'/MA.MB.MC <=sin(A/2)sin(B/2)sin(C/2) with equality if M=I


probability
AMC
AIME
geometry
number theory

Hi! I' m trying to find a book about problem solving in number theory, but I can't find anything.
I'm looking for a book that starts from scratch in problem solving and gradually builds your skills.
Do you know any?

AOPS Volume 1 and Intro to Number theory (go to the bookstore on this site to buy them).

Can I find this book at bookstores or I have to buy it online?

I haven't seen any in the bookstore. You could probably find one in a used-book store but the probability of that is 0.0001%

Even on Amazon, not many people sell AoPS books.

The best place to buy those books is on this site and its the cheapest.

They do have the books on amazon but they are overpriced unless you buy the ones that are already used. So good luck with that.
 :)

Thank you all! I think I'll get the Introduction to number theory! :)

I have also heard that the "Number theory structures examples and problems" book by Titu Andreescu and Dorin Andrica is good!
Do you know anything about this book?

Does anyone know of a good intermediate number theory book? (AIME level) Also, does anyone know of a good intermediate geometry book?

Thanks

For Intermediate Geometry, try [url=http://www.amazon.com/exec/obidos/ASIN/0883856190/artofproblems-20]Geometry Revisited[/url].

Number theory structures examples and problems" book by Titu Andreescu and Dorin Andrica
This is a really wonderful book  which starts from the basics and gradually builds up the level.
Though it starts from basic divisibility the examples and additional problems are from IMO and national olympiads


Is there any pdf text for physics olympiad?

[quote="phucnv87"]http://www.jyu.fi/kastdk/olympiads/[/quote]
There are just the IPhO problems, but I want some introductory handouts with practices which are suitable for physics olympiad.

So this is what you need

http://hkpho.phys.ust.hk/frame_training.htm


assume $ P(x) \in Z[x]$ and we have $ deg(P)\geq 2$
let $ A$ = {$ p(i)|i\in Z$}
prove there exist $ a,b \in Z$ such that if we define:$ B$ = {$ an \plus{} b|n\in Z$}
we have $ A\cap B \equal{}\emptyset$

Solution...
[hide]Let $ k$ be a natural number such that $ |P(k \plus{} 1) \minus{} P(k)| \equal{} s \geq 2$.
You can easy to prove that this $ k$ is exist (because $ degP > 1$).
Look at the numbers:

$ P(k), P(k \plus{} 1), ..., P(k \plus{} s \minus{} 1)$.

There are $ s$ numbers, and $ P(k) \equal{} P(k \plus{} 1) [mod s]$, so there are not all rests of division by $ s$ in $ P(k), P(k \plus{} 1), ..., P(k \plus{} s \minus{} 1)$.

Let $ t$ - natural number such that $ P(k), P(k \plus{} 1), ..., P(k \plus{} s \minus{} 1) \neq t (mod s)$.

So, let $ a \equal{} s$, $ b \equal{} t$.

Obviously that $ P(ms \plus{} r) \equal{} P(r) \neq t (mod s)$ for all integer $ m$ and $ r \equal{} 0,1,2,...,s \minus{} 1$.

So, we find $ a \equal{} s$ and $ b \equal{} t$.[/hide]


trigonometry
inequalities
Euler
geometry
geometry proposed

Prove that

$ \frac{a^2}{bc}\plus{}\frac{b^2}{ca}\plus{}\frac{c^2}{ab}\geq4(\sin^2 \frac{A}{2}\plus{}\sin^2 \frac{B}{2}\plus{}\sin^2 \frac{C}{2})$

The inequality is equivalent to : $ \frac{a^3\plus{}b^3\plus{}c^3}{abc}\ge 4\cdot\sum\frac{1\minus{}\cos A}{2}\equal{}2\left (3\minus{}\sum\cos A \right)$ .

$ \Longleftrightarrow\ \frac{2s(s^2\minus{}6Rr\minus{}3r^2)}{4Rrs}\ge 2\left(3\minus{}1\minus{}\frac{r}{R}\right)\equal{}\frac{4R\minus{}2r}{R}$

$ \Longleftrightarrow\ 2s^3\minus{}12Rrs\minus{}6sr^2\ge 16Rrs\minus{}8r^2s\ \Longleftrightarrow\ 2s^3\plus{}2r^2s\ge 28Rrs$ $ \Longleftrightarrow\ s^2\plus{}r^2\ge 14Rr$

Since $ s^2\ge 16Rr\minus{}5r^2$ ([b]Gerretsen[/b]) it suffices to prove that : $ 16Rr\minus{}5r^2\ge 14Rr\minus{}r^2\ \Longleftrightarrow\ R\ge 2r$ ([b]Euler[/b]) .

Dear Ligouras,
your problem is so easy, because It equivalent with:
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{abc} \ge 4.(\frac {3}{2} \minus{} (a^2b \plus{} a^2c \plus{} b^2c \plus{} b^2a \plus{} c^2a \plus{} c^2b \minus{} a^3 \minus{} b^3 \minus{} c^3)/(abc))$
by Ravi substistution and Schur inequality, we completely our proof
dear constantin, your method in my country call GLA, it's too strong but do it will lost the beautiful of geometry, I think  :(
dear Ligouras: could you post all your inequality geometry in one box, so users can easy to follow and new posts will not under of your post. thanks :)

[quote="Ligouras"] [color=darkred]Prove that $ \frac {a^2}{bc} \plus{} \frac {b^2}{ca} \plus{} \frac {c^2}{ab}\ \ge\ 4\cdot\left(\sin^2 \frac {A}{2} \plus{} \sin^2 \frac {B}{2} \plus{} \sin^2 \frac {C}{2}\right)$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] $ 4abc\cdot\sum \sin^2\frac A2\equal{}$ $ \sum 2abc(1\minus{}\cos A)\equal{}$ $ \sum a\left[2bc\minus{}\left(b^2\plus{}c^2\minus{}a^2\right)\right]\equal{}$

$ \sum a\left[a^2\minus{}(b\minus{}c)^2\right]\equal{}$ $ \sum a^3\minus{}\sum a(b\minus{}c)^2\le \sum a^3$ $ \implies$ $ 4\cdot\sum \sin^2\frac A2\le \sum \frac {a^2}{bc}$ .

[b]Remark.[/b] I used only the relations $ \left\|\begin{array}{c}
2\sin^2x\equal{}1\minus{}\cos 2x\\\
2bc\cdot\cos A\equal{}b^2\plus{}c^2\minus{}a^2\end{array}\right\|$ .[/color]


Hey,
How do you prove that BF: FD = 2:1 using coordinate proof

 [img]http://img12.imagevenue.com/loc878/th_72867_1_122_878lo.jpg[/img]

I'd really appreciate any help thanks!
Steve

I'm assuming you mean that D and E are midpoints.

Alright, so WLOG set $ B=(0,0), C=(c,0), A=(a,b)$.

It follows that $ D=(\frac{a+c}{2}, \frac{b}{2})$ and $ E=(\frac{a}{2}, \frac{b}{2})$

Line BD can be represented as $ y=\frac{b}{a+c}\cdot x$, and EC can be represented as $ y=\frac{b}{a-2c}\cdot(x-c)$

Point F can be found by solving the following equation:

$ \frac{b}{a+c}\cdot x = \frac{b}{a-2c}\cdot(x-c)$

$ x(\frac{b}{a-2c}-\frac{b}{a+c})=\frac{bc}{a-2c}$

$ x[b(a+c)-b(a-2c)]=bc(a+c)$

$ x(3c)=c(a+c)$

$ x=\frac{a+c}{3}$

Plugging back in, $ y = \frac{b}{a+c}\cdot \frac{a+c}{3}= \frac{b}{3}$, meaning that $ F=(\frac{a+c}{3},\frac{b}{3})$.

Now, notice how...

$ B=(0\cdot\frac{a+c}{6},0\cdot\frac{b}{6}$)

$ F=(2\cdot\frac{a+c}{6},2\cdot\frac{b}{6}$)

$ D=(3\cdot\frac{a+c}{6},3\cdot\frac{b}{6}$)

Anything involving coordinates usually involves lots of tedious algebra.


Hi,

How many different values are taken by the expression $ \lfloor x\rfloor  \plus{} \lfloor 2x\rfloor  \plus{} \lfloor \frac{5x}{3}\rfloor \plus{} \lfloor 3x\rfloor \plus{} \lfloor 4x\rfloor$  for real $ x$ in the range $ 0 \leq x \leq 100$

[hide]Put $ y \equal{} {x\over 3}$ to get

$ E \equal{} [3y] \plus{} [5y] \plus{} [6y] \plus{} [9y] \plus{} [12y]$ for $ 0\leqslant y\leqslant{100\over 3}$

Write $ y \equal{} n \plus{} {k\over 180} \plus{} \beta$ where $ n \equal{} [y], k\in\mathbb{N}_0,0\leqslant k\leqslant 179, \beta\in\mathbb{R},0\leqslant\beta < {1\over 180}$

Then

$ E \equal{} 35n \plus{} \left[{k\over 15} \plus{} 12\beta\right] \plus{} \left[{k\over 20} \plus{} 9\beta\right] \plus{} \left[{k\over 30} \plus{} 6\beta\right] \plus{} \left[{k\over 36} \plus{} 5\beta\right] \plus{} \left[{k\over 60} \plus{} 3\beta\right]$

Within the interval $ [n,n \plus{} 1)$, the sum of the last five terms above will change at the values of $ k$ which are the multiples of any of the given denominators. Hence those "turning points" will be $ k\in\{0,15,20,30,36,40,45,60,72,75,80,90,100,105,108,120,135,140,144,150,160,165\}$, and there are $ 22$ of those. Hence for $ 0\leqslant y < 33$ we'll have $ 33$ unit intervals with $ 22$ values each, yielding $ 33\cdot 22 \equal{} 726$ values of $ E$.

For $ 33\leqslant y\leqslant{100\over 3}$, the range of $ k$ is clipped to $ 0\leqslant k\leqslant 60$, so we take only $ k\in\{0,15,20,30,36,40,45,60\}$, yielding $ 8$ additional values, hence the total number of values is $ N \equal{} 726 \plus{} 8 \equal{} 734$[/hide]


AMC
AIME
geometry
geometric transformation
reflection
complex numbers

Let $\alpha$ be a constant.  What curve in the complex plane represents the set {$\overline{z^2} | \arg{z} = \alpha$}?

(A) the ray $\arg{z} = 2\alpha$
(B) the ray $\arg{z} = -2\alpha$
(C) the ray $\arg{z} = -\alpha$
(D) None of (A), (B) and (C) is correct.

[quote="3X.lich"]Is this problem too hard or too technical to be here in the Intermediate Topics?  :D[/quote]

I think it's easier than most AIME probems and complex numbers aren't too advanced of a topic.  If anyone needs the vocabulary they can always look it up or ask.

I ll give it a shot... please tell me if this is wrong or not.

[hide]

We can express z as  .  So then .  It says the conjugate of  so I believe that the conjugate is just a reflection across the real axis:

Think of a complex number in the form a+bi.  The conjugate is a-bi.  If you plot these points they only differ in the sign of the b value which translates to a reflection across the real axis.

Back to the problem.  So since it is reflected across the real axis all the angles will just be negative.  Hence .  So that means that all complex numbers that have their angle equal to  will work.  This is the same   B[/hide]

Btw, is there a command for cis so that it isn't italicized like the commands for cos, sin...?  Or is cis meant to be typed like that?  I tried /cis and that obviously didn't work.

Nope, it doesn't like cis.  cis isn't really mathematician-notation, though -- it's equivalent to $e^{i\theta}$, and that's what most people would use.

Use {\rm whatever} to make 'whatever' unitalicized in math mode.

${\rm cis}\ \alpha$


probability
counting
derangement
factorial
combinatorics open
combinatorics

This is the problem:

Student A built a 6 digit string from the digits 1,2,3,4,5,6.
Each digit can be used only once.

Student B has to try and guess the string, he has only one guess.

I thing that the odds of matching the whole string are 1/6! = 1/720
and that a 5 digits match (out of the 6) is not possible.

I don't know the odds or how to calculate -
4 or 3 or 2 or 1 or 0 matches.

Can anyone please explain how to solve this problem and how to calculate the probability for each combination ?

Many thanks.

Tupbug.

The number of 6-digit strings (or better say permutations) that have exactly $ k$ matches with a fixed one is
\[ {6\choose k}\cdot !(6\minus{}k)\]
where $ !m$ is [url=http://en.wikipedia.org/wiki/Subfactorial]subfactorial[/url] of $ m$, giving the number of [url=http://en.wikipedia.org/wiki/Derangement]derangements[/url] of $ m$ elements.

Many thanks for your answer, but I still can't work it out.

According to the formula you sent the number is -

 ( 6 ) * (6-k)!
 ( k )

Isn't that formula just  equals 6!/k! ?

because  n!/{(n-k)!*k!}*(6-k)!= 6!/k!.

The other question is, when I use the formula anyway I get that
the number of permutations with k matches is:
6 matches - 1
5 matches - 6
4 - 30
3 - 120
2 - 360
1 - 720
0 - 720
Obviously I'm doing somthing wrong because the total amount of permutations should be 720 (at least I think so)/

Last question - I don't understand what m! is and how to use it.

Again many thanks for all the help.

Tupbug

Look carefully! It's not $ (6 \minus{} k)!$ but $ !(6 \minus{} k)$. The former is factorial while the latter is [b]subfactorial[/b]. Take a look at the links that I gave above.

Thank you very much  for the explanation it works perfectly.

Tupbug


Hi, I am deciding between the following four colleges: Georgia Tech, Carnegie-Mellon, Lehigh University, and the University of Florida.  I am interested in majoring in engineering, math, or physics (or maybe chemistry).  I was wondering how important your undergraduate school is if you plan to go to graduate school.  Also, one of the perks of the University of Florida is that I could graduate in three years (maybe less) and then go on to grad school.  I would appreciate any advice.  Thanks in advance.

By University of Florida, do you mean New College?

No, I think New College is Florida State University, but I'm not sure.

no, what school you go for your undergrads do not matter very much (just make sure that you don't go to some lamo community college...). You could still have a very good life without going to a famous university/college for your undergrads.

I'm not sure about the U of Florida, but I can give some info about Georgia Tech, Carnegie Mellon, and Lehigh.

Obviously they're all engineering schools, by far.  Out of those three, I would think Carnegie Mellon is the most prestigious, which would help you get into grad school; however, if you do go there, I hear that Pittsburgh is a horrible city.  Atlanta is also kind of bad, but Georgia Tech is a very good school.  You wouldn't go wrong there.  Lehigh is more rural; I think its in a valley.  I would think Lehigh is the third most prestigious out of the three.  Grad school is going to depend on how well you do at those schools, and doing well at those schools means that you have to like being there, usually.  I don't really see any kind of academic decisive factor between those three.

[quote="DPatrick"][quote="CheesyP18"]however, if you do go there, I hear that Pittsburgh is a horrible city.  [/quote]Where did you hear that?   I went to CMU and I really liked Pittsburgh.[/quote]

I am at CMU as I write this (for the CNA Summer School ([url=http://www.math.cmu.edu/cna/Summer06/index.html]link[/url]); this is my 9th day here), and ohgodohgodohgod I hate Pittsburgh with a passion already.  Everyone at the conference I have talked to agrees with me as well.

CMU has quite a nice campus, but things go downhill fast as soon as you leave the campus, and by that I mean venture one or two streets in the wrong direction.  The weather is also bad, it has gone from unbearably hot and humid to thunderstorms in the course of a day.

[quote="blahblahblah"]The weather is also bad, it has gone from unbearably hot and humid to thunderstorms in the course of a day.[/quote]

Well, that's true of the noncoastal areas of the United States in general in the summer. I greatly enjoyed my only visit to the city of Pittsburgh, in which I stayed in a downtown hotel back in the late 1980s, but I have no idea how well that experience matches the experience of someone who lives on campus at CMU. Because my son is interested in computer science, CMU is definitely a college on our radar screen here.

I freely admit that I am biased here; I have only lived in Vancouver and San Diego, which are some of the better climates around.

CMU is very good for (applied) math as well.


$ x,y,z\geq 0$,and $ a,b,c$ are positive constant.Denote
\[ F(x,y,z) \equal{} \frac {x}{ax \plus{} by \plus{} cz} \plus{} \frac {y}{ay \plus{} bz \plus{} cx} \plus{} \frac {z}{az \plus{} bx \plus{} cy}
\]
(1)if $ b^2\geq ca$ and $ c^2\geq ab$,then $ F(x,y,z)\geq \frac {3}{a \plus{} b \plus{} c}$

(2)if $ b^2\leq ca$ and $ c^2\leq ab$,then $ F(x,y,z)\leq \frac {3}{a \plus{} b \plus{} c}$

In fact ,I have solved the two problem ,but  I wonder how about the conditon if $ b^2\leq ca$ and $ c^2\geq ab$
I try to solved it last night ,but I failed.


Let given triangle$ ABC$. $ AA_{1}, BB_{1}, CC_{1}$ -- bisectors of $ \angle A, \angle B, \angle C$. Let
$ \angle C_{1}B_{1}A_{1}\equal{}90^{\circ}$.

Find $ \angle ABC$.

[quote="castigioni"]Angle B=120[/quote]

Can you prove it?

Let $ H\equal{}A_1C_1\cap BB_1, G\equal{}A_1C_1\cap AC$ then $ (A_1C_1HG)\equal{}\minus{}1$. But $ \angle A_1B_1C_1\equal{}90^o$ so $ C_1A_1$ is the bisector of angle $ BB_1G$. We get $ C_1$ is C-excenter of triangle $ BCB_1$. Therefore $ BC_1$ is the external bisector of angle $ B_1BC$, which follows that $ \angle ABC\equal{}120^o$


Eight toy camels and three toy pigs cost Gary $ \$85$. Twelve toy camels cost Larry $ \$96$. Assuming everyone bought their toys at the same store and there were no discounts, what is the cost of two toy pigs?

Since $ 12$ toy camels cost Larry $ \$96$, one toy camel cost him $ \$8$. So $ 3$ toy pigs cost $ 85\minus{}8*8\equal{}85\minus{}64\equal{}\$21$. This makes each toy pig $ \$7$, and $ 2$ toy pigs $ \boxed{\$14}$.

@above$,$ they have asked for $2$ toy pigs$,$ so the answer is $7*2=\fbox{14}.$


inequalities
geometry
incenter
geometry unsolved

Let $AD, BE, CF$ be the internal angle bisectors of $\triangle ABC$ with $D, E, F$ on the opposite side of the triangle to $A, B, C$ respectively. Let $X, Y, Z$ be points on the line segments $EF, FD, DE$ respectively. If $d(P, \ell)$ denotes the perpendicular distance of point $P$ from line $\ell$, show that
\[d(X, AB)+d(Y, BC)+d(Z, CA) \leq XY+YZ+ZX \]

[i]Alternative formulation.[/i] Let $I$ be the incenter of triangle $ABC$. Lines $AI,BI,CI$ meet the sides of triangle $ABC$ at $D,E,F$ respectively. Let $X,Y,Z$ be arbitrary points on segments $EF,FD,DE$ , respectively. Prove that $d(X,AB)+d(Y,BC)+d(Z,CA)\leq XY+YZ+ZX$, where $d(X,l)$ denotes the distance from a point $X$ to a line $l$.

R U sure is it correct.
I think it should be :
$d(X, BC) + d(Y, AC) + d(Z, AB) \leq XY + YZ + ZX$

I does seem strange that the perpendiculars are dropped to adjacent sides but that is the question that was set.

Does anyone have a solution because this is still unsolved from 2003.


MIT
college
Harvard
Princeton
topology
Galois Theory

I was wondering where I could obtain past exam papers for maths subjects from uni-s such as MIT, Harvard, Princeton, Cambridge, Caltech etc. I'm aware some MIT papers are available from MIT open courseware; but I couldn't find any papers from Harvard/Princeton on their websites; hopefully exams from those uni-s will contain challenging problems. I think I recall someone posting some links before in some thread in this forum, but I can't find it right now searching.
Specifically, I'm looking for exam papers in : Metric Spaces/ Topology (but not Algebraic Topology), and Abstract Algebra (Fields / Galois theory in particular), and Linear Algebra.
Thank you for your help.

For Harvard, you need to go to the course homepages.  They might be hard to find and some of them require a password but for certain classes, you can see old exams and problem sets.  With Princeton, they have a link with PDF's for old exams in many different classes.  Not sure how useful this is (under Practice Exams):
http://www.math.princeton.edu/undergraduate/
I believe Caltech also posts some old exams but I don't visit that site too often so I'm not too sure.  Cornell has a really long list of classes so you're bound to find some old exams there.  If you want Abstract Algebra, plenty of problems here:
http://www.math.jhu.edu/~mching/402/index.html
This is part 2 of the abstract algebra course.  Unfortunately, the page for part one seems to be gone.


The area of the floor in a square room is 225 square feet. The homeowners plan to cover the floor with rows of 6-inch by 6-inch tiles. How many tiles will be in each row?

225 square foot area $ \rightarrow$ 15 feet sides
6 inches = .5 feet, so 30 tiles can fit in a row.


algebra
polynomial
superior algebra
superior algebra solved

Prove that:

If $R$ is a commutative ring with multiplicative identity and $f=a_n x^n + \cdots + a_0$ is a zero divisor in $R[x]$, then there exists a nonzero $b \in R$ such that $ba_n = ba_{n-1} = \cdots = ba_0 = 0$

thanks.

I like this problem :).

Let $g(x)=b_0+b_1x+\ldots+b_mx^m$ be a polynomial of minimal degree $m\ge 1$ s.t. $fg=0$. Let $t\in R$ be such that $tb_m=0$. We then have $f(x)\cdot tg(x)=0$, and this means that $tg=0$, because of the minimality assumption.

We have $a_nb_m=0$, so $a_nb_i=0,\ \forall i$, which means that $(a_0+a_1x+\ldots+a_{n-1}x^{n-1})g=0$. We then get $a_{n-1}g=0$, and so on. In the end, we get $a_ib_m=0,\ \forall i$, so we have a contradiction (because $b_m$ is a polynomial of degree $0<m$, which multiplied with $f$ gives $0$). This means that there must be a polynomial of degree $0$ which multiplied with $f$ gives $0$, and this is our $b$.


I noticed that AoPS has a USAMTS page. Are they working together now or is AoPS just encouraging us to participate? But anyways, does anyone know when the year 16 problems are coming out?

Look at the USAMTS forum. AoPS is administering it now.

AOPS has now taken over running the competition. Complete details can be found by clicking the USAMTS link on the left sidebar. The first problem set will not be realized until the next school year.

This should be in the USAMTS forum. I am moving it.

Thanks for moving it.  All the answers given above are correct.  We are running the contest, and the questions for round 1 will be out in late summer.

This is really exciting... USAMTS is finally going to run efficiently! And, poking around the website, it looks like there will be lots of neat features I always wished could have been part of USAMTS (i.e. being plugged into the internet more).


induction
number theory unsolved
number theory

Let x be a real number such that $ x\plus{} \frac{1}{x}$ is an integer. Prove that $ x^n\plus{} \frac{1}{x^n}$
is also an integer for any positive number n.

Just use induction:
\[ \frac{1}{x^n} \plus{} x^n \equal{}(x \plus{} \frac 1{x})(x^{n \minus{} 1} \plus{} \frac 1{x^{n \minus{} 1}}) \minus{} x^{n \minus{} 2} \minus{} \frac {1}{x^{n \minus{} 2}}.
\]
Bye

[b]Generalization:[/b]  Let $ x, y, \alpha, \beta \in \mathbb{C}$ such that $ xy, x \plus{} y, \alpha \plus{} \beta, \alpha x \plus{} \beta y$ are integers.  Then $ \alpha x^n \plus{} \beta y^n$ is an integer for any positive integer $ n$.

Proof:  Let $ s_n \equal{} \alpha x^n \plus{} \beta y^n$.  Then $ s_0 \equal{} \alpha \plus{} \beta, s_1 \equal{} \alpha x \plus{} \beta y$ and $ s_{n\plus{}1} \equal{} (x \plus{} y) s_n \minus{} xy s_{n\minus{}1}$.

This problem is the special case $ y \equal{} \frac{1}{x}, \alpha \equal{} \beta \equal{} 1$.


AMC
AMC 10
algebra
polynomial
AMC 12
AIME
probability

This will be a good review for AMCs.

Here are the rules: you have to post an answer to the previous question and a question from AMC 10 or 12 (with answer choices), or some equally leveled question. Otherwise, you can aslo post any question you have regarding any Questions. Thanks! :lol:

I'll start, relatively hard AMC10 #25

In year N, the 300th say of the year is Tuesday. In the year N+1, the 200th day is also a Tuesday. On what day of the week did the 100th day of year N-1 occur?
A)Thursday
B)Friday
C)Saturday
D)Sunday
E)Monday

The first statement implies that year N or N+1 is a leap year.

If N was a leap year, then 566 days ago, was a Wendesday.

If N+1 is a leap year, then 565 days ago, was a Thursday.

We use the answer choices to determine Thursday.

I don't have a problem on me right now.

Just to keep this going.... I'll post a problem.

AMC 12 #23
How many perfect squares divide the product 1! * 2! * 3! ... 9! ?

A) 504
B) 672
C) 864
D) 936
E) 1008

I got the answer B.

Solution:

We can list out all the factorizations and count how many prime factors there are:

1
1 * 2
1 * 2 * 3
1 * 2 * 3 * 4
1 * 2 * 3 * 4 * 5
1 * 2 * 3 * 4 * 5 * 6
1 * 2 * 3 * 4 * 5 * 6 * 7
1 * 2 * 3 * 4 * 5 * 6 * 7 * 8
1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9

The prime factorization of this product will be:

$ 3^{13} * 2^{30} * 7^3 * 5^5$

The way we can count how many perfect squares there are is we can count the number of combinations of multiplying these numbers we can have, including the number to the zero power.
(i.e. 5^5 will have 3 possibilities, $ 5^0, 5^2$, and $ 5^4$)

Therefore, we have:

7 * 16 * 2 * 3 = [b]672[/b]

EDIT: Oh right, gotta add a problem. Here's one:

[b]AMC-12 #23[/b]

A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?

$ \textbf{(A)} \ \frac {1 \plus{} i \sqrt {11}}{2} \qquad \textbf{(B)} \ \frac {1 \plus{} i}{2} \qquad \textbf{(C)} \ \frac {1}{2} \plus{} i \qquad \textbf{(D)} \ 1 \plus{} \frac {i}{2} \qquad \textbf{(E)} \ \frac {1 \plus{} i \sqrt {13}}{2}$

Hmm, OK, I guess since no one else is posting, I'll reply to my own question.

First, we approach this problem with the understanding that if the given solution is a root, its complex conjugate must also be a root as well. Furthermore, the problem states that the coefficients of the polynomial are integers.

We can take a look at the constant term of the polynomial (the term without an x). We know that this is the product of all the roots.

Hmm...multiplying B, C, D, and E by their complex conjugates will all yield non-integer values.

This leaves answer choice A, and multiplying that number by its conjugate will give an integer value, 3.

Therefore, the answer is [b]A[/b]

Next Problem:

AMC12 #23

How many $ 15$-letter arrangements of  $ 5$ A's, $ 5$ B's, and $ 5$ C's have no A's in the first $ 5$ letters, no B's in the next $ 5$ letters, and no C's in the last $ 5$ letters?

$ \textbf{(A)}\ \sum_{k\equal{}0}^5\binom{5}{k}^3 \qquad
\textbf{(B)}\ 3^5\cdot 2^5 \qquad
\textbf{(C)}\ 2^{15} \qquad
\textbf{(D)}\ \frac{15!}{(5!)^3} \qquad
\textbf{(E)}\ 3^{15}$

Label the spaces like so:
_ _ _ _ _
_ _ _ _ _
_ _ _ _ _
Before accounting for permutations you can have either BCA or CAB in each column, in that order. If each column is BCA then there are $ \binom {5}{0}^3$ ways to arrange. If there are 4 BCA and 1 CAB then there are $ \binom {5}{1}^3$ ways. Proceed to 5 CAB and we get the answer A.

For each integer $ n \ge 4$, let $ a_n$ denote the base-$ n$ number $ 0.\overline{133}_n$. The product $ a_4a_5a_6...a_{99}$ can be expressed as $ \frac {m}{n!}$ where $ m$ and $ n$ are positive integers and $ n$ is as small as possible. Find $ m$. (Not putting answer choices because it makes it more fun  :P )

Heh. Whao, that last problem I posted was difficult, and I have no idea how I would've gotten that right if it was a short-answer question on the AIME. I just used elimination to get rid of some of the choices, and then I checked to find a way that answer choice A would work, and it does. This question could be a #8 on an AIME without the choices I think.

Hmm, the problem you just gave us crazypianist is extremely difficult. I'll admit, I looked at the solution. I think this could definitely be an AIME #12-14 without the answer choices. I'm not going to post the solution though, so someone else can solve it.

[quote="crazypianist1116"]Label the spaces like so:
For each integer $ n \ge 4$, let $ a_n$ denote the base-$ n$ number $ 0.\overline{133}_n$. The product $ a_4a_5a_6...a_{99}$ can be expressed as $ \frac {m}{n!}$ where $ m$ and $ n$ are positive integers and $ n$ is as small as possible. Find $ m$. (Not putting answer choices because it makes it more fun  :P )[/quote]

For $ a_x$, we just have $ \sum_{n\equal{}1}^\infty \frac{1}{x^{3n\minus{}2}}\plus{}\sum_{n\equal{}1}^\infty \frac{3}{x^{3n\minus{}1}}\plus{}\sum_{n\equal{}1}^\infty \frac{3}{x^{3n}}\equal{}\sum_{n\equal{}1}^\infty \frac{x^2\plus{}3x\plus{}3}{x^{3n}}$.  Using the formula for sum of an infinite geometric sequence, we find this sum to be $ \frac{\frac{x^2\plus{}3x\plus{}3}{x^3}}{1\minus{}\frac{1}{x^3}}\equal{}\frac{x^2\plus{}3x\plus{}3}{x^3\minus{}1}$.  Now, in order to cancel out terms, consider the denominator $ n^3\minus{}1\equal{}(n\minus{}1)(n^2\plus{}n\plus{}1)\equal{}(n\minus{}1)((n\minus{}1)^2\plus{}3(n\minus{}1)\plus{}3)$.  Therefore, each time $ x$ increases by 1, the second part of the denominator from the previous term cancels out and only $ n\minus{}1$ is left (of course, the last term will remain since there is nothing to make it cancel).  Therefore, the product is: $ \prod_{x\equal{}4}^{99} \left(\sum_{n\equal{}1}^\infty \frac{x^2\plus{}3x\plus{}3}{x^{3n}}\right)\equal{}\frac{99^2\plus{}3\cdot99\plus{}3}{\frac{98!}{2}\cdot21}\equal{}\frac{481}{\frac{98!}{2}}\equal{}\frac{962}{98!}$, so $ m\equal{}962$.

[quote="crazypianist1116"]It was a number 25 on an AMC12. I was keeping in line with the difficulty of the other questions. It actually only requires a knowledge of geometric series.

For response to the last question, are the trees distinguishable?[/quote]
I meant to say the probability. So I don't think it matters. (just assume they aren't)

[quote="Ihatepie"]Yeah that definitely isn't a AMC10/12 level problem.

New problem:

There are 4 maple and 6 oak trees. If they are placed randomly in a row, what is the probability that no two maple trees are next to each other?[/quote]

OK, I want to try and solve this using a bijection/counting approach, because I'm still experimenting with that. If there's anything wrong, please do tell.

The total number of ways to arrange the 4 maple and 6 oak trees is 10 C 4 = 210.

Now, we can count how many arrangements there are in which no two maple trees are next to each other. I want to start out with having all 4 maple trees together. Let M denote maple trees and O denote Oak trees.

MMMM

From here, just add oak trees between every maple tree:

MOMOMOM  <---this ensures us that no two maple trees are together.

Using casework, we can find how many ways there are to insert three oak trees in this arrangement.

Case 1: There is an oak tree at the beginning, but not at the end.

3C2 + 3 = 6  (there are a few mini-cases, but easy to figure out)

Case 2: There is an oak tree at the end, but not at the beginning.

3C2 + 3 = 6

Case 3: There are no oak trees at the beginning and the end.

1 + 6 + 3 = 10

Case 4: There are oak trees both at the beginning and the end.

3C1 = 3

Totalling these up, we get 6 + 6 + 10 + 3 = 25.

Therefore, the probability will be 25/210 or 5/42.

...I have a bad feeling about my answer. I won't post another question just yet because its probably wrong.

[quote="Ihatepie"]Yeah that definitely isn't a AMC10/12 level problem.

New problem:

There are 4 maple and 6 oak trees. If they are placed randomly in a row, what is the probability that no two maple trees are next to each other?[/quote]

[hide]There are $ \dbinom{10}{4} \equal{} 210$ distinct ways to arrange the trees.

Arrange the 4 maple trees in a row, and place an oak tree in between adjacent maple trees:

MOMOMOM

Now, we can use balls and urns to solve this:
We have 3 more oak trees left, and 4 maple trees that act as dividers between the 5 distinct locations for oak trees.
So, there are $ \dbinom{7}{3} \equal{} 35$ ways to place the remaining oak trees.

This gives a probability of $ \dfrac{35}{210} \equal{} \dfrac{1}{6}$.

@veezbo: I think you forgot the cases of: 3 oaks in the beginning/end and 2 oaks in the beginning and end.
[/hide]

If this is right, then here is the next problem: (Variation on 2003 AMC 12A #22)
Particles $ A$ and $ B$ move simultaneously in the coordinate plane via a sequence of steps each of length $ 1$.
Particle $ A$ starts at $ (0,0)$ and moves either right or up, both with equal probability.
Particle $ B$ starts at $ (7,5)$ and moves either left or down, both with equal probability.
What is the probability that both particles meet?

[quote="veezbo"]Ohhhhhhhhhhhh, you mean we had to find the union of the areas?![/quote]

No, it is the intersection. But nobody's solved the first extension yet... including me. :(

[quote="veezbo"][quote="isabella2296"]Bump.

New problem?

Two eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?[/quote]
...[long solution]...
[/quote]
I just wanted to point out that this can be solved another (shorter) way: ab>a+b=>(a-1)(b-1)>1. The only way this can't happen is when a=1, b=1, or a=b=2, for a total of 16 ways. The answer is then $ \frac{64\minus{}16}{64}\equal{}\frac{3}{4}$, since $ 64\equal{}8^2$ is the number of possibilities.

And as for the new problem, notice that $ P\equal{}(x^2 \plus{} 3x \plus{} 1)^2 \minus{}(3x \minus{} 30)$

[quote="alana.u.99"]lol wow you're smart i wood've never found dat  from this point on i'd just like look for a pattern! (:
i'm not too great in alegbra sooooooooooooooo
i took the amc10 this year and made it to aime and i found it really hard...i got a 4...
its my first year really doing math competitions and idk how to practice and stuffff[/quote]

Taking the AMC 10 for the first time ever and getting a 4 on the AIME the same year is extremely good.
I took the AMC 12 for the first time this year and failed miserably  :(

[quote="mathking123"][quote="veezbo"][quote="isabella2296"]Bump.

New problem?

Two eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?[/quote]
...[long solution]...
[/quote]
I just wanted to point out that this can be solved another (shorter) way: ab>a+b=>(a-1)(b-1)>1. The only way this can't happen is when a=1, b=1, or a=b=2, for a total of 16 ways. The answer is then $ \frac {64 \minus{} 16}{64} \equal{} \frac {3}{4}$, since $ 64 \equal{} 8^2$ is the number of possibilities.

And as for the new problem, notice that $ P \equal{} (x^2 \plus{} 3x \plus{} 1)^2 \minus{} (3x \minus{} 30)$[/quote]

Lol, yeah, my solution's tend to be long-winded and utterly useless. If a USAMO grader ever reads my solutions, they'll probably go mad. Half the stuff I do isn't even necessary, which is why I need more time on the AIME, and to the greatest extent, on the AMCs.

Hmm, I need to study samore number theory, but first, gotta get done with AOPS I stuff, like geometry that I thought I knew, but I guess I didn't.

Also, I'm quite sure that I've seen just about three problems in the last week that deal with trying to prove something to be a perfect square. I have no idea on how to do that. My first and only effort is to try and convert everything to mod 4, since perfect squares are only 0 or 1 mod 4. Other than that, I fail.

[quote="alana.u.99"]lol wow you're smart i wood've never found dat  from this point on i'd just like look for a pattern! (:
i'm not too great in alegbra sooooooooooooooo
i took the amc10 this year and made it to aime and i found it really hard...i got a 4...
its my first year really doing math competitions and idk how to practice and stuffff[/quote]
Welcome to the forums. :)

You just revived a topic from 2009, 9 years ago.

[quote=Stringintegral]Please anyone solve these problems too..
(1) Let a1 = 1
            a2 = 11
            a3 =  111
(Meaning whatever is the subscript of a, is that number of times 1 repeated in the number. (Kindly ignore the writing pattern as I really don't know LaTex programming)
Now find gcd (a45, a140)
Plz write some short trick for this.

(2) Find the summation of (n^2 + 1 ) n! Upto n

Plz provide a good solution[/quote]

We should be able to use division algorithm to find that (a45, a140) = (a45, a5) = a5.

[quote=song2sons][quote=mathchampion1]There's certainly nothing wrong with that provided that it is, indeed, a useful topic.[/quote]

Agreed. My question is how the heck he/she even arrived at this topic...[/quote]

By using the search function maybe?

Hey you all,
First of all understand this that I am new so I don't know where to post questions.  Now I have got where to post.

By the way I came here from someone's profile. But I doubt why you all are amazed.  Is there any restriction for entering the topic closed 9 years ago.

Thank you maths champion for taking my side. Please accept my friend request

Songstosons then tell me some other alternative to this topic where good questions are posted one by one


Given ABCD is a regulated quadrilateral inscribed in a circle (O) (which has $ AB\cdot CD \equal{} BC\cdot DA$).
Let K, H be the midpoints of AC and BD respectedly. Prove that AK+CK=BH+DH.

Are u sure? i think please chek your question again :wink:

Probably he means a cyclic [b][size=100]Harmonic quadrilateral[/size][/b], by the term [b][size=100]regulated[/size][/b] and the relation $ AH \plus{} CH \equal{} BK \plus{} DK$ instead of $ AK \plus{} CK \equal{} BH \plus{} DH.$ :maybe:

Kostas Vittas.


ok...I've been reading some posts on here and people are saying if you want to get into MIT (or a comparable school) you have to do all these olympiads and take many many AP courses and do research and win  a bunch of stuff and etc...

ok so what if I go to a school of class size 130 from which, on average, one person gets psat semifinalist every other year, and only 4 AP courses are offered

and there are pretty much no opportunities to do research here

and I had to start a math/science club myself because I found out about all these competitions and olympiads and wanted to do them and felt my school needed some sort of math/science club
---so chances are I won't do too outstandingly in these competitions because its my very first time (I'm postive I'll qualify for AIME though...hopefully)

does this mean I have no chance  of getting into a college like MIT??
---btw I have a 2300+ sat score, 230 psat, three 800s on subject tests (Math lvl 2, Physics, Molecular Bio), 5s on all APs so far, I'm rank 1 in my class and have been playing varsity tennis and soccer for my school since sophomore year (I'm a junior btw)....I know these stats are kind of standard but I can't exactly do any more in my school/area...I've pretty much exhausted every opportunity available here...so am I a hopelesss case for a college like MIT because I don't have alll the other academic credentials I first mentioned?

Colleges will look at your achievements in the context of the school you attend, and the opportunities that you have available to you. Just keep doing your best to learn and grow, and look for as many opportunities outside of your school as possible.

Also, you should keep in mind that most people on this forum are high school students who (unsurprisingly) don't actually know very much about the college admission process.

Instead of worrying about your chances, try to increase them.

Some applicants don't realize that two of the things that count the most in the application process are the essay(s) and the letters of recommendation. Your interview and overall presentation will also play a role. [color=blue][b]These things are all still within your control.[/b][/color]

[b]Essay:[/b] Find out essay questions from past years. MIT does not accept the Common Application; they have their own specific essay questions. On the other hand, the Common Application allows you to write your essay on [i]any[/i] topic of your choice!

Think carefully about what you're going to write about. [color=blue]Your choice of topic [i]itself[/i] will say a lot about you. [/color]Make sure it showcases who you are, some of your major interests, and some important personality traits (all three of which are closely related). Don't just re-write your activities in sentence form. Grades, scores, activities, and things all speak for you in an application. The essay is your only chance to directly speak for yourself (apart from the interview). So make it count!

You can look on [url=http://www.mitadmissions.org]the MIT Admissions[/url] website for some pointers as to what they're looking for.

[b]Letters of Recommendation:[/b] [color=blue]Pick your teachers carefully.[/color] Some schools will allow you to send more than the minimum number of letters (most won't) -- make sure that any additional letter you might want to send is at least as helpful to your case as the other letters; otherwise, it's clutter.

What your teachers say about you [color=blue]should affirm the personality traits, the strengths, and the interests[/color] that you bring out in your essay. Make sure you pick teachers that know enough about you to write convincingly about these.

Don't feel shy to [color=blue]ask your teachers what they will write[/color] about you, or to ]give them some background about what you do outside of their class that might help them write their letter. [color=blue]Tell them what you want![/color] It may be helpful to tell them [i]why[/i] you want a letter of recommendation from them, [i]what[/i] you want them to emphasize, and ask them if that's alright with them. And make sure you give them a friendly reminder to send it in on time.

The MIT Admissions website and the Common Application website have copies of their respective letter of recommendation forms. Take a look at these -- it might influence who you pick!

[b]Short Answer:[/b] There are going to be short answer questions as well. The Common Application will ask you to briefly discuss an activity that is important to you. MIT will ask you to list your top 5 most important activities, and give you 50 words each to explain them (or at least, they did two years ago). [color=blue]Pick every word carefully![/color]

[b]Resume:[/b] Some schools will allow you to submit a 'Resume'. [color=blue]Keep it short[/color], and make the most important points come out. Using space to [color=blue]explain [/color]one of your important projects ([color=blue]brief bullet points are good[/color]) is better than using it to list a fourth not-so-important math contest that you won (they'd probably get the point after seeing the more important ones). Of course, don't omit anything important.

If your resume is over [color=blue]one page long[/color] (size 10 font), it's time to start worrying that you're diluting your main points (for your reference, undergraduate students usually submit one page resumes to internship applications).

[b]Overall Presentation:[/b] There should be [color=blue]synergy[/color] between everything you present in your application -- you want to bring out your personality and best qualities.

Make sure everything is visually appealing and [color=blue]professional looking[/color]. It makes a positive impression.


Dear Mr.Kent,
                     I'm just a "NEW" and a beginner in math forum.
Though,i'm a "Mathematical Science Researcher,Enthusiast,"i
still unable to "CONVERT" my math ideas into simple mathematical equation,or say,notation.Can you,or anyone "HELP"
me to know on how will i used mathematical notation everytime
i exchange,discuss ideas(Mathematically) with other math forum member in this math forum website? Do i need a "Math Program"
to "TRANSLATE" my ideas,concepts into mathematical notation?

                        A Thousand Thanks.

                                            Sincerly,
                                           Mikhail

Postscript:
               Sometimes,i experienced problem with my "Password."
This makes my day difficult.If any reply related to my post message,you can used my very "ENCRIPTED" antonio65@Vfemail.net address. I adopt using encription after
experiencing problem with my password.

                                                   Same
                                                   ------

This mathematical notation is called Latex.

http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php

You need not download anything, just have dollar signs outside your latex code.
Like this:

$\$$ \sqrt{52}$\$$

to generate:

$\sqrt{52}$

Btw: Mr. K M you are very popular :D


I wonder what Americans call for AM-GM inequality, because I have got a problem about finding minimum for sum of areas which I used AM-GM inequality. However, I was received this statement from my professor "I really don't find your answer" :mad: :mad:

Anyone who is familiar with contest math, in the US or otherwise, would know the AM-GM inequality. That's a pretty standard abbreviation, although you could consider writing out the words it stands for.
College professors are not all familiar with contest math. You might want to find a reference and provide it to your professor.

Also, the professor may not be the one grading problem sets. They hire undergrads and grads to do such menial labor for them.

Very true; this applies even in graduate classes. I've been burned a few times myself.

It never hurts to show a few more steps and to write out in full words any special terms you use. Good mathematical writing is communicating, so it is important to be clear, not just brief.

As anyone who has taken a course with a truly bad marker knows, there is a big difference between 'good mathematical writing', and 'writing that will be understood by a marker with either a poor grasp of the english language or an insufficient grasp of mathematics to reasonably be a marker'.


logarithms
limit
LaTeX
real analysis
real analysis unsolved

Let $x_n \in \mathcal l R_+$ . If $x_{n+m}\leq x_nx_m$ prove that $\sqrt[n]{x_n}$ converges. I have a solution involving superior and inferior limits....

I also find a solution a solution involving superior and inferior limits in my book :P  :(

Take logarithms. Now we have a subadditive sequence $y_n=\log x_n$ ($y_{n+m}\le y_n+y_m$), and we are looking at the behavior of $z_n=\frac{y_n}n$.

We have $z_{n+m}=\frac{y_{n+m}}{n+m}\le\frac{y_n+y_m}{n+m}= \frac{n}{n+m}z_n+\frac{m}{n+m}z_m$.
It just works best here to invoke $\limsup$ and $\liminf$; the key lemma is that $\limsup_n z_n\le z_m$ for any $m$. In alternative language, we have that for all $n$ and all $\epsilon>0$, there exists $N$ such that $z_m<z_n+\epsilon$ for all $m>N$.
Now, either $z_n$ converges to some limit $L$ or it diverges to $-\infty$; then either the original sequence of roots converges to a finite limit or it diverges to zero.

After logarisming as jmerry did we get that $0<y_{n+m}<y_n+y_m$ and we have to prove that limit $\frac{y_n}{n}$ exists. That is well know problem and the limit is the inf of the $\frac{y_n}{n}$.

Proof:$\frac{y_n}{n}$ is bounded so the $\inf\frac{y_n}{n}$ exist and let it be $\alpha$. Then for any $\epsilon>0$  exist   $m: \alpha<\frac{y_n}{n}<\alpha+\frac{\epsilon}{2}$.  Then for any $n \in\mathbb{N}$  $n>m$ WE HAVE ${n=qm+r, r\in\{0,...,m-1} , y_n=y_{qm+r}\leq y_{qm}+y_r<qy_m+y_r$ SO  $\frac{y_n}{n}\leq \frac{q}{qm+r}y_m+\frac{y_r}{qm+r}=\frac{y_m}{m}\frac{qm}{qm+r}+\frac{y_r}{qm+r}$ SO  $\alpha<\frac{y_n}{n}<(\alpha+\frac{\epsilon}{2})\frac{qm}{qm+r}+\frac{y_r}{n}\leq \alpha+\frac{\epsilon}{2}+\frac{y_r}{n}$. But we can find some A s.t. for all n>A we get that $\frac{2y_r}{n}<\epsilon$
So finally we get that $\alpha<\frac{y_n}{n}<\alpha+\epsilon$  or $\lim_{n\to\infty}{\frac{y_n}{n}}$ and the problem is solved

Please tell me how to write the dot betwen the q and m in $LATEX$?

Just use "\cdot" :)
e.g. : $a \cdot (b+c) = a \cdot b + a \cdot c$

What if  for some k we will have that $x_k<1$? We cant then apply that problem ? :oops:

There's no problem with negative logarithms; as I said, the sequence $\frac{y_n}{n}$ either converges to its infimum or diverges to $-\infty$. If it diverges to $-\infty$, $x_n$ diverges to zero.