geometry
geometric transformation
ratio
incenter
linear algebra
matrix
rectangle

View all the problems: http://www.mathlinks.ro/Forum/resources.php?c=143&cid=61&year=2007

sorry for bad translation
Be free to ask me

8.1.(O. Podlipsky)
Given reals numbers $a$, $b$, $c$. Prove that
at least one of three euqations
$x^{2}+(a-b)x+(b-c)=0$, $x^{2}+(b-c)x+(c-a)=0$, $x^{2}+(c-a)x+(a-b)=0$
has a real root.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832699#832699

8.2. (N. Agakhanov)
The numbers $1,\,2,\dots,100$ are written in cells of a table
$10\times 10$, each number is written once. By one move, Nazar may
interchange numbers in any two cells. Prove that he may get a table,
in which sum of numbers in every two adjacent (by side) cells is composite
after at most 35 such moves.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832703#832703

8.3.(S. Berlov, F. Petrov, A. Akopyan)
Given a rombus $ABCD$. A point $M$ is chosen on its side $BC$.
The lines, which pass through $M$ and are perpendicular to $BD$ and $AC$,
meet line $AD$ in points $P$ and $Q$ respectively. It appears that
lines $PB,QC,AM$ have a common point. Find all possible values of a ratio
$BM/MC$.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832705#832705

8.4.(A. Akopyan, A. Akopyan, A. Akopyan, I. Bogdanov)

A conjurer Arutyun and his assistant Amayak are going to show
following super-trick. A circle is drawn on the board in the room.
Spectators mark 2007 points on this circle, after that Amayak
removes one of them. Then Arutyun comes to the room and
shows a semicircle, to which the removed point belonged. Explain,
how Arutyun and Amayak may show this super-trick.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832707#832707

8.5.(folclore) A distance between Maykop and Belorechensk is 24 km.
Two of three friends need to reach Belorechensk from Maykop, and another
friend want to reach Maykop from Belorechensk. They have ony bike, which is
initially in Maykop. Each guy may go on foot (with velocity
at most 6 kmph) or on a bike (with velocity at most 18 kmph). It is
forbidden to leave a bike on a road. Prove that all of them may achieve
their goals after 2 hours 40 minutes. (Only one guy may seat on the bike
simultaneously).
http://www.mathlinks.ro/Forum/viewtopic.php?p=832709#832709

8.6.(S. Berlov) A line, which passes through the incentre $I$ of a triangle
$ABC$, meets its sides $AB$ and $BC$ in points $M$ and
$N$ respectively. Triangle $BMN$ is acute. Points $K,L$ are chosen
on the side $AC$ so that $\angle ILA=\angle IMB,\ \angle IKC=\angle INB$. Prove that $AM+KL+CN=AC$.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832713#832713

8.7.(V. Senderov)  For an integer $n>3$ denote by $n?$ the product
of all primes, which are less then $n$. Solve an equation $n?=2n+16$.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832717#832717

8.8.(A. Badzyan)
Given a matrix $\{a_{ij}\}_{i,j=0}^{9}$, $a_{ij}=10i+j+1$. Andrey is going
to cover its entries by 50 rectangles $1\times 2$ (each such rectangle contains
two adjacent entries) so that the sum of 50 products in these rectangles
is minimal possible. Help him.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832718#832718

9.1.(S. Berlov) Unitary quadratic trinomials $f(x)$ and $g(x)$ satisfy
the following interesting condition: $f(g(x))=0$ and $g(f(x))=0$
do not have real roots. prove that at least one of equations
$f(f(x))=0$ and $g(g(x))=0$ does not have roots too.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832720#832720

9.2.(N. Agakhanov, I. Bogdanov)
100 fractions re written on a board, their numerators
are numbers from 1 to 100 (each once) and denominators
are also numbers from 1 to 100 (also each once). It appears
that the sum of these fractions equals to $a/2$ for some odd $a$.
Prove that it is possible to interchange numerators of two
fractions so that sum becomes a fraction with odd denominator.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832723#832723

9.3.(K. Sukhov) Two players by turns draw diagonals in a regular
$(2n+1)$-gon ($n>1$). It is forbidden to
draw a diagonal, which was already drawn, or inersects an odd number of already
drawn diagonals. The player, who has no legal move, loses.
Who has a winning strategy?
http://www.mathlinks.ro/Forum/viewtopic.php?p=832730#832730

9.4.(V. Astakhov) $BB_{1}$ is a bisector of an acute triangle
$ABC$. A perpendicular from $B_{1}$ to $BC$ meets
a smaller arc $BC$ of a circumcircle of $ABC$ in a point $K$.
A perpendicular from $B$ to $AK$ meets $AC$ in a point $L$.
$BB_{1}$ meets arc $AC$ in $T$. Prove that $K$, $L$, $T$
are collinear.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832732#832732

9.5.(F. Petrov) Two numbers are written in each vertice of a convex 100-gon.
Prove that it is possible to remove a number from each vertice so that
remaining numbers in any two adjacent vertices would be different.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832736#832736

9.6.(V. Filimonov)
Given acute triangle $ABC$. Points $M$ and $N$ are midpoints
of $AB$ and $BC$ respectively, point $BH$ is an altitude of $ABC$.
Circumcircles of $AHN$ and $CHM$ meet in $P$~($P\ne H$).
Prove that $PH$ contains a midpoint of $MN$.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832738#832738

9.7-8.8

9.8.(A. Golovanov)
Dima has written number $1/80!,\,1/81!,\,\dots,1/99!$ on 20
infinite pieces of papers as decimal fractions
(the following is written on the last piece:
$\frac{1}{99!}=0{,}{00\dots 00}10715\dots$, 155 0-s before 1).
Sasha wants to cut a fragment of $N$ consecutive digits from one
of pieces without the comma. For which maximal $N$ he may
do it so that Dima may not guess, from which piece Sasha has cut his fragment?
http://www.mathlinks.ro/Forum/viewtopic.php?p=832741#832741

10.1.(A. Poliansky) Faces of a cube $9\times 9\times 9$ are partitioned
onto unit squares. The surface of a cube is pasted over
by 1458 strips $2\times 1$ without overlapping.
Prove that the number of bent strips is odd.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832743#832743

10.2.(A. khrabrov) Given polynomial
$P(x) = a_{0}x^{n}+a_{1}x^{n-1}+\dots+a_{n-1}x+a_{n}$.
Put
$m=\min \{ a_{0}, a_{0}+a_{1}, \dots, a_{0}+a_{1}+\dots+a_{n}\}$. Prove that
$P(x) \ge mx^{n}$ for $x \ge 1$.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832745#832745

10.3=9.4

10.4.(K. Knop, O. Leontieva)
Arutyun and Amayak show another effective trick.
A spectator writes down on a board a sequence of $N$ (decimal) digits.
Amayak closes two adjacent digits by a black disc. Then Arutyun comes
and says both closed digits (and their order). For which minimal $N$
they may show such a trick?
http://www.mathlinks.ro/Forum/viewtopic.php?p=832747#832747

10.5.(N. Agakhanov) Given a set of $n>2$ planar
vectors. A vector from this set is called {\it long},
if its length is not less than the length of the sum of other
vectors in this set. Prove that if each vector is long,
than th sum of all vectors equals to zero.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832749#832749

10.6.(S. Berlov) Two circles $\omega_{1}$ and $\omega_{2}$
intersect in points $A$ and $B$. Let $PQ$ and $RS$
be segments of common tangents to these circles
(points $P$ and $R$ lie on $\omega_{1}$, points $Q$ and $S$ lie on $\omega_{2}$).
It appears that $RB\parallel PQ$. Ray $RB$ intersects $\omega_{2}$
in a point $W\ne B$. Find $RB/BW$.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832750#832750

10.7.(D. Karpov)
Given a convex polyhedron $F$. Its vertice $A$ has degree 5,
other vertices have degree 3. A colouring of edges of $F$ is called nice,
if for any vertex except $A$ all three edges from it have different colours.
It appears that the number of nice colourings is not divisible by 5.
Prove that there is a nice colouring, in which some three consecutive
edges from $A$ are coloured the same way.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832752#832752

10.8=9.8

11.1.(N. Agakhanov)
Prove that for $k>10$ Nazar may replace in the following
product some one $cos$ by $sin$ so that new function $f_{1}(x)$
would satisfy inequality $|f_{1}(x)|\le 3\cdot 2^{-1-k}$ for all real $x$.
\[f(x) = \cos x \cos 2x \cos 3x \dots \cos 2^{k}x \]
http://www.mathlinks.ro/Forum/viewtopic.php?p=832754#832754

11.2 (A. Polyansky)

An incircle of triangle $ABC$ touches its sides
$BC$, $AC$, $AB$ in points $A_{1}$, $B_{1}$, $C_{1}$ respectively.
A segment $AA_{1}$ intersects an incircle in a point
$Q\ne A_{1}$. A line $\ell$ through $A$ is parallel to
$BC$. Lines $A_{1}C_{1}$ and $A_{1}B_{1}$ intersect $\ell$ in points $P$ and $R$
respectively. prove that $\angle PQR=\angle B_{1}QC_{1}$.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832755#832755

11.3=10.3

11.4.(A. Golovanov)
An infinite sequence $(x_{n})$ is defined by its first
term $x_{1}>1$, which is a rational number and relation
$x_{n+1}=x_{n}+1/[x_{n}]$ for all positive integers $n$
($[x]$ is the maximal integer, which does not exceed $x$).
Prove that this sequence contains an integer.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832757#832757

11.5=9.5

11.6.(N. Agakhanov, I. Bogdanov)
Do there exist non-zero reals $a$, $b$, $c$ such that
for any $n>3$ there exist a polynomial
$P_{n}(x) = x^{n}+\dots+a x^{2}+bx+c$, which have
exactly $n$ (not necessary distinct) integral roots.
http://www.mathlinks.ro/Forum/viewtopic.php?p=832759#832759

11.7. (A. Zaslavsky)
Given a tetrahedron $T$. Aleksei wants
to find two its edges $a,b$ with no common vertices so
that $T$ is covered by balls with diameters $a,b$. Can he
always find such a pair?
http://www.mathlinks.ro/Forum/viewtopic.php?p=832762#832762

11.8.(I. Bogdanov, G. Chelnokov)
Given an undirected graph with $N$ vertices. For any set of $k$ vertices ($1\le k\le N$)
there are at most $2k-2$ edges, which join vertices of this set.
prove that the edges may be coloured in two colours so that each cyle
contains edges of both colors. (Graph may contain multiple edges).
http://www.mathlinks.ro/Forum/viewtopic.php?p=832765#832765

Please post all these problems in sub-boxes of Olympiad section and create links here :) ( See  topics in this box as examples )

Sorry, my internet is too slow for this routine job. I think, if anybody wants to discuss some problem, he may make a post with it in corresponding section and leave a comment here with a link. For som problems, I doubt that they deserve special topics.

[quote="Fedor Petrov"]I think, if anybody wants to discuss some problem, he may make a post with it in corresponding section and leave a comment here with a link. [/quote]
But if so, this topic is very long bacause anyone post one link :D

After some links are posted, I may copy them to my initial post. But make about 30 posts in different sections is hard for me.

Ok, I will help you  :wink:

http://www.mathlinks.ro/Forum/resources.php?c=143&cid=61&year=2007

[quote="Fedor Petrov"]Some corrections

1) 11.3=10.4

2) In 10.1, there are 243 rectangles, not 1458[/quote]Done. Thanks :)


function
parameterization
geometry open
geometry

For  $x >$ 0 let us consider the family of functions  $f(x)=f(p,q,r;x): = px^{q}+(1-p)x^{r}$ where $p \in [0,1]$  and $q, r$ are real numbers . Let  ${\mathcal T}$  be the subset of all $(a,b,c)$ from  $(0,\infty)\times (0,\infty)\times (0,\infty)$  having the property that $a, b, c$  are the lenght-sides of a triangle.
[b]Find all parameters  $p, q, r$  for which the implication  $(a,b,c) \in{\mathcal T}\; \; \; \Longrightarrow \; \; \left(f(a),f(b),f(c)\right) \in{\mathcal T}$ is true .[/b]

Triangle preserving means $f(x+y) + f(x+z) > f(y+z)$.  For continuous increasing $f$ this is equivalent to superlinear, $f(y)+f(z) \geq f(y+z)$.  For $f(0)=0$ this is the same as convexity of $(-f)$.   So the condition is $q,r \leq 1$.


Find all polynomials $ P(x)\in\mathbb R[x]$ such that \[ P(x)P(x+1)=P(x^2)\] for all $x \in \mathbb R$.

[quote="MJ GEO"]P(X)P(X+1)=P(X^2)[/quote]

Let $ A(x)$ be the assertion $ P(x)P(x\plus{}1)\equal{}P(x^2)$
$ P(x)\equal{}0$ is a solution. So consider from now non-zero polynomials.

Let $ z\in\mathbb C$ any root of $ P(x)$

$ A(z)$ $ \implies$ $ P(z^2)\equal{}0$. So $ z$ root $ \implies$ $ z^2$ root $ \implies$ $ |z|\equal{}0$ or $ |z|\equal{}1$ (else $ P(x)$ would have infinitely many roots).

$ A(z\minus{}1)$ $ \implies$ $ P((z\minus{}1)^2)\equal{}0$. So $ z$ root $ \implies$ $ (z\minus{}1)^2$ root and so $ |z\minus{}1|\equal{}0$ or $ |z\minus{}1|\equal{}1$

So the only roots may be $ 0,1,e^{i\frac{\pi}3}$ and $ e^{\minus{}i\frac{\pi}3}$

So $ P(x)\equal{}x^m(x\minus{}1)^n(x^2\minus{}x\plus{}1)^p$

Plugging back in the original equation, we get $ x^m(x\minus{}1)^n(x^2\minus{}x\plus{}1)^p(x\plus{}1)^mx^n(x^2\plus{}x\plus{}1)^p$ $ \equal{}x^{2m}(x^2\minus{}1)^{n}(x^4\minus{}x^2\plus{}1)^{p}$

and so $ m\equal{}n$ and $ p\equal{}0$

Hence the result :
$ P(x)\equal{}0$
$ P(x)\equal{}x^n(x\minus{}1)^n$


How can I demonstrate that
1) if f:R->R, f(x) is even then f'(x) is odd.
2) if f:R->R is a derivable function and it's even then
f'(0)=0?
3) if g:R->R, g(x)=((x^3)/3 +1)f(x) then g'(0)=1, knowing that f(x) is even and derivable.

All help appreciated.

First, what's the definition of an even function? Or more specifically, what identity must it satisfy?

If you can answer this, then you should be able to answer the question from here.

Also, if $ g(x) \equal{} \left(\frac {x^3}{3} \plus{} 1\right)f(x)$, and if $ f$ is specified as written, the given value of $ g'(0)$ isn't correct.

for f to be an even function f(-x)=f(x). After this am I just supposed to write
f'(-x) ?
and about 3) I have to agree there might be a mistake with g'(0)=1.


So the harmonic series diverges.

But does the sequence 1/1, 1/2, 1/3, ... 1/9, 1/11, 1/12, ...
converge or diverge, where every 1/n is removed if n has a 0 in its base ten representation?

Actually not true, neelnal.  You'll find that this converges.  ;)

[hide="Hint"] Take the sum from $n = 1$ to $n = 10^{k}$ and find a way to bound it in terms of $k$.  Er, either that or it was from $n = 10^{k}$ to $n = 10^{k+1}$.  [/hide]

[quote="neelnal"]Still diverges because you can still infinitely group terms that will add up to greater than 1.[/quote]

Um, no. If you're going to claim something, at least try proving it first.
[hide]
There are $9^{n}$ $n$-digit numbers. Each is at least $10^{n-1}$, so the reciprocal is at most $\frac{1}{10^{n-1}}$. Then the sum of all $n$-digit numbers is at most

$\frac{9^{n}}{10^{n-1}}$.

So if we sum over all $n$, we have our desired summation is at most

$\sum_{n=1}^{\infty}\frac{9^{n}}{10^{n-1}}= 9 \sum_{n=0}^{\infty}\left(\frac{9}{10}\right)^{n}= 90$,

so it is finite (converges).
[/hide]

EDIT: Beat by a couple of seconds :o.

There's an interesting concept to be toyed with here - how thickly spread within $\mathbb{N}$ is some subset of $\mathbb{N}?$

Try out this definition: Suppose $S\subset\mathbb{N}.$

Call $S$ [i]abundant[/i] or [i]thick[/i] iff $\sum_{n\in S}\frac1n=\infty.$

Call $S$ [i]sparse[/i] or [i]thin[/i] iff $\sum_{n\in S}\frac1n$ converges.

If $S$ is all of $\mathbb{N},$ then it's abundant.

Any finite set is sparse.

The even numbers are abundant, as are the odd numbers.

The perfect squares are sparse.

Any infinite arithmetic progression is abundant.

Any infinite geometric progression (of integers!) is sparse.

By the proof that t0rajir0u and paladin8 have provided, the set of integers whose decimal representations contain no zeros is sparse.

What about the primes? Sparse or abundant?

And what other interesting sets do you want to ask about in this way?

And $S=\{p\in \mathbb{P}: p\equiv a\mod{b}\}$ where $\gcd{(a,b)}=1$ is abundent

[quote="Altheman"][quote="seamusoboyle"]Primes are abundent[/quote]

why?

The message contains only quote and code tags. Please add some text of your own.[/quote]
See http://en.wikipedia.org/wiki/Proof_that_the_sum_of_the_reciprocals_of_the_primes_diverges .


ratio
quadratics
algebra
quadratic formula

try this one out even though i got it but try it : the ratio of the length to its width is the same as that of the diagonal to the length. if the width is 2, how many units are in the length of the diagonal.

not that hard

[hide]call the diagonal x, the width is 2, so the length is  :sqrt: (x 2 -4)

So (:sqrt: (x 2 -4))/2=x/(:sqrt: (x 2 -4))

So 2x=x 2 -4, x 2 -2x-4=0.  Use the quadratic formula so x=1+ :sqrt: 5[/hide]


quadratics
modular arithmetic
geometry
3D geometry
arithmetic sequence
number theory proposed
number theory

Let $A$ be the set of primes $p$ such that $p \mid 2^{n}-3$ for some $n$; and let $\mathbb{P}$ denote the set of all primes. Show that the sets $A$ and $\mathbb{P}\setminus A$ are both infinite.

[hide="Second part."]Let $p$ be an arbitrary prime of the form $24k+17$. We will show that there isn't $n$ such that $p|2^{n}-3$ and the proof will be finished. So, suppose that such $n$ exists and it is even. But by the Quadratic Reciprocity theorem it can be easily checked that $3$ is not a quadratic residue $\mod{p}$. Analogically, if $n$ would be odd then $6$ would be a quadratic residue, which is also impossible as one can verify.[/hide]

Good!

And how about a proof not resorting to Dirichlet's theorem? (and. even more so, to any of its generalizations, e.g. Chebotarev...)

I remember an old olympiad problem which takes care of the first part; it states that there is an infinite sequence $(a_{n})_{n}$ of mutually coprime positive integers of the form $2^{n}-3$. Such a sequence can be constructed recursively: if we already have $a_{1},\ldots,a_{n}$, then choose $k$ such that $a_{i}|2^{k}-1,\ i=\overline{1,n}$, and set $a_{n+1}=2^{k}-3$.

[hide="First part."]I hope I'm not wrong here. Suppose that $A$ is finite and let $p_{1}, p_{2}, p_{m}$ be its all elements. Then, taking $n=(p_{1}-1)(p_{2}-1)...(p_{m}-1)$ implies that $p_{i}|2^{n}-1$ for $i=1,2,...,m$, so of course $p_{i}\not | 2^{n}-3$ for $i=1,2,...,n$, contradiction (since $2^{n}-3$ is greater than $1$). One should also add that $A$ is not empty, but this is obvious.[/hide]

EDIT: It's the same as grobber's post.

:)

Yes, the first part is much easier. The challenge now remains to prove the second part without resorting to Dirichlet's theorem.

And the first (little) generalization:

Let $a,b,c,d$ be integers, and denote by $A$ the set of primes $p$ for which $p \mid ca^{n}+db^{n}$ for some $n$. Suppose one of $A$ and $\mathbb{P}\setminus A$ is finite. Then, show that $abcd(a^{2}-b^{2})(c+d) = 0$.

Hint for a proof without any Dirichlet-type theorems:

[hide]Observe that, for $p \in A$ and for any $n$, $p \mid 2^{n}-1$ implies $p \mid 3^{n}-1$. Thus, the observation that [edit!] $\mathbb{P}\setminus A$ is finite would imply that, for some integer $B > 0$ and for all $n$, the set of prime factors of $2^{n}-1$ is a subset of that of $B(3^{n}-1)$, so for every $M$ there is an integer $N = N(M)$ such that $B^{N}(3^{n}-1)^{N}/(2^{n}-1)$ is an integer for $1 \leq n \leq M$. Moreover, $N = o(M)$. Obtain a contradiction somehow, by constructing appropriate identities involving linear combination of those integers, and for $M$ sufficiently large (relative to $B$). Maybe this last sentense is not too much of a hint... :) [/hide]

But, of course, there may be other, simpler [and elementary! that's the whole point!] solutions.

[quote="Megus"][quote="vess"]Thus, the observation that $A$ is finite[/quote]

Should have been $\mathbb{P}\backslash A$ I suppose  :)[/quote]

 :D Of course...  :blush: Thanks! I've edited.

By the way, here is how to eliminate the easy argument with Dirichlet's theorem:

Prove that there are infinitely many primes $p \notin A$ with $p \equiv 1 \pmod{12}$ (so that $(3/p) = 1$, and... you can't handle the even $n$ so easily).

More generally: there are infinitely many $p \notin A$ inside any given arithmetic progression (of course, with coprime difference and first term). This follows easily from Chebotarev's theorem. But the existence of infinitely many primes in $12k+1$ is, of course, elementary, and I propose the above problem: show that there are infinitely many $p$ of the form $12k+1$ outside of $A$. Dirichlet's theorem IS allowed to this end (Chebotarev is NOT), but it is not likely to be of much use since $(3/p) = 1$ (implied by $p=12k+1$) invalidates Tomci0's simple argument.

By Cebotarev, one may still give a similar proof by finding infinitely many primes $p = 12k+1$ for which $3, 6$, and $12$ are NOT cubes modulo $p$ and solve similarly the problem by considering $n=3k, 3k+1, 3k+2$ separately; but how about an elementary proof?

So, my new problem is: show, elementarily (without Chebotarev) that there are infinitely many primes $p =12k+1$ that do not divide any number of the form $2^{n}-3$.


yeah, sorry about the blank space in the attachment though

go to http://mnspe.org/mathcounts/comppdfs/test.pdf . I think you missed some of the horizontally symetrical ones.

Did you only do one kind of symmetry? That's what I did on the problem, because it said it had to have at least one line of symmetry, and i didn't specify vertical or horizontal.


AMC
AIME
algebra
binomial theorem
number theory

I remember seeing a problem from AIME like this one:

Find the remainder when 6^83 + 8^83 is divided by 49.

I tried to use the pattern (hahah) and it didn't work. Plus, I couldn't use calculator so it got pretty nasty. I know this problem was solvd using binomial theorem but I wonder if I can do this using modular arithmetic. I'm not very good at it so I'm willing to learn it. I saw this kind of problem many times and always, I skipped it. :(

Thanks for the great help!

Why not use  (7-1)^83 + (7+1)^83.

(Notice that  all terms except  83*7 (+-) 1 would be zero mod 49.)
You would be left with something like 2*7*83  ...

If you are lazy, and not that   83 =( 7*something -1 ) the answer would be  (-14) or 35.

I believe that that is exactly the method he was refering to when he mentioned the binomial theorem.

A result related to what you are looking for (but which is not so useful in this context) is Fermat's Little Theorem, a very useful result from number theory.  I don't know if anyone has posted an explanation or proof of it here, but it's probably worth a search.  You might also try Euler's theorem which is a generalization of Fermat's, and which you should be able to find in the same places.

[quote] ...but which is not so useful in this context[/quote]
FWIW actually this is  what made the answer almost jump out for me. Noticing that phi(49) = 42 and  and 83=84-1  ( = 2*42-1)
The expression was (All mod 49)

    (6)^(-1)+(8)^-1
 == -8 - 6  ==  -14 = 35
(Of course  since 6*8 = -1 "reciprocals" of 6 and 8 are -8 and -6)

*******
(If my first post was not very clear  about how expanding via  Bionomial theorem  make  the probelm easy ..)

([hide]7+1)^83 =  (Terms containg Higher_than_two  powers of 7   hence divisible by 49) + 83*7 + 1
and
(7-1)^83 = (Same as above)  +83*7 -1

Hope that helps. [/hide]

Also as JBL said, Euler's theorem which is a generalization of Fermat's,  is  helpful,  You may like to look at  in any good reference but basically it is:

if p is prime , and    a and p has no common factors (that is a is not a multiple of p) then:

a^(p-1) == 1 mod p

if  p is not prime,  replace (p-1) by phi(p) ,( that is number of  positive integers less than p which  has no factors common with p)

and you have  a^(phi(p)) = = 1 mod p.


I'm going to probably have to make a choice. Let's see what you think.

in early February of next year, there's probably going to be a conflict between swimming and math. I can either join the Mathcounts Chapter Competition, or join my school swimming championships.

First, let me say what would go wrong if I picked swimming over Mathcounts:

- I wouldn't be able to go to states or anything above that, and I have a great math ability (I love math too). My Mathcounts coach wouldn't be too happy either.

Then, let my say about if I pick Mathcounts over swimming:
- I have a local record in an event, and I want to drop it again, but if I don't do the swimming champs, I can't improve that record. I love swimming too.

- The coach and most of the school would wonder why I didn't go for such an important meet for the school!!!

Both of them are important to school. If you were faced with this decision and you had conditions like the one I mentioned above, which one would you choose?
Teh answers might be biased, I know.

When I get into high school, it can get worse than this (3 choices?)

Try the Round Table forum for these questions. Not that it's bad to ask, but it's not a math problem. No offense whatsoever.

EDIT: Of course, that was edited by the suggestion of the next person. :blush:

I disagree.  The best place to put this would be in: "The Round Table", for serious discussion.  However, I can understand that he is putting it here because he is looking for a "mathcounts competitor" point of view.  As for your dilemma 236!, I dont wanna sound corny, but this is one of those "follow your heart" (jeez I said it) kinda situations.  Which one are you more passionate about?  Which will be more helpful to you in the future?  If this is your last year to compete for mathcounts, then mabye you should go.  However, if you've competed before and you think that you've had enough experience and your no longer interested, mabye its time to take the plunge (pun intended).  You've also noted that both desicions would help your school, I've however always felt you can only improve your school by improving yourself, and then the results will follow.  I've tried not to be too biased, but I've always preffered mathematics over swimming, good luck, and make the[i][b] right [/i][/b] desicion. (pun intended)

                                            Swimming vs Mathcounts

choose the one that is more fun for you and the one that will help you in the future.  THis is a really hard choice for you.  If i were you i would choose Mathcounts because you can always join another swim team or go to your local pool (i understand that you want to hold the record). And also this is your last year of Mathcounts if you are going into 8th grade.  That's my opinion.

Ok, here's what I think.
I'll try to be impartial.  It seems like you have a great ability at math and swimming.  But, if you are as good as you say at Mathcounts and by what you've shown us, you probably could be in the top 4 in your state and in the top 50 or 100 in the nation!  Can you truthfully say that about swimming? I'm sure there are many swimmers that are better, no offense, but too many people are involved in sports to make you stand out.  think about it

Uh... But I'm too stupid in math to be any better that the top four in my [i]chapter[/i]!

[quote="236factorial"][quote="GoBraves"]Do mathcounts...obviously

I used to swim and i hated swim practice...[/quote]

Bias, bias, bias,
was it because you couldn't float?  :D[/quote]

no...it was more towards... boring

[quote="236factorial"]really? I think swimming's fun. Just read my blog  :)
Well, math is fun in that other sensation.[/quote]

Swimming is fun

SWIM PRACTICE is not..

I sorta get your point there   :D

One day, one of the swimmers who were faster than me pushed it a little too hard, and I still feel sore after two days.

We're not talking practice here. Gobraves, go you like swim competitions?

i like playing in the pool not swimming laps.

[quote="236factorial"]We're not talking practice here[/quote]

Maybe we should be. Since you can't seem to decide between the competitions, let's try this: Which do you like more, math practice or swim practice?

[quote]Uh... But I'm too stupid in math to be any better that the top four in my chapter![/quote]

In 7th grade I got 5th in my school but 3rd in state. You never know....

[quote="nat mc"]In high school, you can do swimming but not MathCounts. This is your last chance[/quote]

That's right, but it's also the last chance to set a record for the middle school swim team.
the last chance factor makes the decision harder. Also, the school would sort of.... you know....

well it just all depends on which one u like better... we can't really tell you which one to choose because 1. this is a math site and most ppl'll choose math 2. if u ever regret what u choose, ull be blaming us... lol hope u don't... just choose which one u like better... like which one ur friends are in, and stuff like that

I think you should choose without caring what people in your school will think about it. Your math teacher may be upset for a while and so might the swim team or whatever, but this is your choice so don't let other people change your mind just so they won't be upset with you.


calculus
integration
function
trigonometry
geometry
geometric transformation
calculus computations

I just introduced myself to integrals, and it seemed easy at first. Unfortunately...[hide="Its not"]Pfft easy circle, right? No, not right.  Can someone lead me through integrating with radicals? I just didn't know how to treat it algebraically. I didn't know if I should apply the formula to the variables inside the root, or the root as a whole. I was applying to the entire square root because the variables cant be treated independently, its like saying $ (a\plus{}b)^{2}\equal{}a^{2}\plus{}b^{2}$. And if I WERE to square it, when do I reverse that to retain validity?

Only doing the positive form of the square root, it should be $ 2\pi$.
[/hide]Find $ \int^{2}_{\minus{}2}f(x)\,dx$ where $ f(x)\equal{}\sqrt{4\minus{}x^{2}}$.

[hide="Opinion"]Let I be your integral. As f(x) is an even function, then $ I \equal{} 2\int_{0}^{2}f(x)\,dx$. This type of function, with a term $ \sqrt{a^{2}\minus{}b^{2}x^{2}}$, is more easily handed by making the trigonometric substitution $ x \equal{}\frac{a}{b}\sin t$ or $ x \equal{}\frac{a}{b}\cos t$. Try it.[/hide]

exactly, substitute $ x\equal{}\sin w$ to get $ 2\pi$ as a result for the integral

That should be $ x\equal{}2\sin w,$ from which $ dx\equal{}2\cos w\,dw$ and $ \sqrt{4\minus{}x^{2}}\equal{}2\cos w.$ The integral becomes:

$ \int_{\minus{}\frac{\pi}{2}}^{\frac{\pi}{2}}4\cos^{2}w\,dw$

Now, you need to know how to integrate $ \cos^{2}w.$

But, in my opinion: why are we doing this? Our geometric knowledge should not be discarded. We do already know the value of this integral, so we don't need any additional manipulations.

We could even do $ \int_{\minus{}1}^{1}\sqrt{4\minus{}x^{2}}\,dx$ by a purely geometric argument.

OK, so polar translation (after a sort). It makes sense. Thought my algebra was wrong, I was flabbergasted!

$ y\equal{}\sqrt{4\minus{}x^{2}}$, so $ y^{2}\plus{}x^{2}\equal{}4$. We have the top half of a circle of radius $ 2$. What is the area under this curve?

Yes Hilbert, but if I understood right, Igiul already know that. IHis purpose was really to work the integral without the geometric knowledge.


geometry
geometric transformation
reflection
trigonometry
calculus
percent
AMC

Ughh...English pulls me back because I am a really bad writer

Oh well at least I'm unique on this forum  :wink:

Same here I stink at writing but excel at everything else

[color=red]8th but I'm not too smart...[/color]

I love my math SAT score, I hate *ma* language scores...

Grading systems vary greatly from place to place, so this poll may not accurately reflect the quality of students here. In some countries, it is customary for grades to be much more normally distributed than is typical in the United States. Also, some top universities (like mine and Chicago, for example) don't like it to be easy for someone to be a "straight-A student", so it's pretty hard to consistently get perfect grades (but most people still have pretty high averages).

For example, at my school, we have the following (fairly standard) weighting system:

A = 4.0
A- = 3.7
B+ = 3.3
B = 3.0
B- = 2.7
lower follows the same pattern, but is hardly relevant.

I know that in my year, the 20th percentile is around 3.65 or so. So, you can make an A- average and still be in the top quintile. However, the 80th percentile is still above 3.0. So you can make above a B average and still be in the bottom quintile. The average grades are heavily concentrated between B+ and A-.

Personally, I get mostly A, A-, and B+. So I marked the 2nd choice on the poll. However, according to the percentile of my GPA, I should probably be in the first spot (assuming the poll is referring to American high school grading standards).

As and Bs during interim time (usually high Bs). Then straight As by quarter.

[quote="Temperal"]I fail every class, all the time. On purpose. Then I write long papers on how the American education system has failed me.[/quote]

That's funny.

I fail in elective classes  :oops: otherwise I get A's

As and Bs.  :roll: I don't really care about grades.

Mostly A+'s, except of course for good ol' Chinese where I got C+

[quote="dbkarp"]If u r on AoPS how can u be stupid? [/quote]

try spammers :wink:

We have to take Algebra I at highest in 8th grade, and I'm sure I already know all the stuff in their textbook, so yeah, that's how bored I get.

I get As.

Someone at school came up with this grading system:
F=fantastic
D=something, I forgot what it is
C=common
B=bad
A=atrocious

[quote="kimmystar94"]and bowling?[/quote]

The class is seniors only actually.

 :D

I have straight As, but then again, I'm only in 7th grade, which is going to be extremely easy compared to the other grades.

I still haven't gotten any b's on my report card ever. I guess that would be smart. But then again, I'm only in 6th, which is not a very long time to get straight a's, and I always get 92-94 in some class (usually english), but never below 92.

From quarter to quarter, my grades fluctuate between A- and A+; through the year, though, they usually end in the A range.

To my chagrin, A-'s bring down the GPA.

[quote="D3m0n Shad0w"]I'm a fantastic writer, such as J.K. Rowling.[/quote]

 :huh: Does that mean that you're as good of a writer as her, or does that mean that you know that J.K. Rowling is a writer?  Hmm...

It's a general forum poll. Revival is fine.

Err...if i didnt have to take english id have straight A's!!!!Well english and
!!!Ceramics!!! thats not even a class???well at my school you have to take it for a quarter!!!And im awefull at it...i didnt even make one pot!!!so in the grade book thats a * id rather not discuss that!!!

Hmm...

Orchestra - 101% (I play every single instrument so the teacher forces me to spots where we don't have enough people for extra credit)

Science - 98% (Easy.)

Language Arts - 95% (Our teacher is incredibly hard...he teaches us things called "real grammar.")

Social Studies - 89% (Only this trimester, since I moved and I lost the folder where I had kept all the SS Homework and we usually do HW grades in one bunch)

PF - 105% (Easy.)

Geometry - 97% (Our teacher is bad, and we have hard tests. But I have AoPS)

[quote="D3m0n Shad0w"]I get straight A's. [b]But[/b] then again, 5th grade is pretty easy. It's challenging for me[b],[/b] to keep up with the social stuff[b],[/b] and the important facts such as education.

I've never gotten a B or lower on my interim/report card. I guess you would call that pretty smart :D

The thing that worries me is math. I'm a fantastic writer,[b] such as J.K. Rowling[/b], and I got a scholarship [b]to a writing camp[/b], so that's not weighing me down. [b]But[/b] Math...I got a 91% on my last test...Decimals and Algebra. I would've had a [b]lower one[/b] but the teacher said that he forgot to mark it (it what?) [b](missing comma)[/b] and he won't lower my grade.[/quote]

I don't mean to be condescending, but one should live up to their statements...

dude, happyme, it's the internet. capitalization, punctuation, and intelligence in general aren't really mandatory.

social studies ? 91%
PE/health = 98%
French I = 100%
Language Arts = 97%
Science = 95%
Art = 92% (i slack off too much)
Geometry = 9 something percent idk what.

i capitalized just for Lazarus. :D

This quarter:
4.00 unweighted, 4.17 weighted.
Nuff said.

please take ur AIME score off of ur sig.


For every $ 0 \le i \le 17$, $ a_i \equal{} \{ \minus{} 1, 0, 1\}$.
How many $ (a_0,a_1, \dots , a_{17})$ $ 18 \minus{}$tuples are there satisfying :
$ a_0 \plus{} 2a_1 \plus{} 2^2a_2 \plus{} \cdots \plus{} 2^{17}a_{17} \equal{} 2^{10}$

$\textbf{(A)}\ 9  \qquad\textbf{(B)}\ 8  \qquad\textbf{(C)}\ 7  \qquad\textbf{(D)}\ 4  \qquad\textbf{(E)}\ 1$

The set $ \{ \minus{}1,0,1\}$ is the same as $ \mod3$ residues. Putting the equation into mod 3 gives us:

$ a_0 \minus{}a_1\plus{}a_2\minus{}a_3\plus{}....\minus{}a_{17} \equiv 1 \mod 3$

Notice how much alike it looks like the divisibility rule of 11.... Now I'm stuck, I got nothing.. But I think we should look at numbers $ 1 \mod 33$. The only problem is that the numbers would have 18 digits....

[hide]$ 2^{10}$ can be written as $ 2^{10},2^{11}\minus{}2^{10},2^{12}\minus{}2^{11}\minus{}2^{10}$ etc. all the way to $ 2^{17}\minus{}2^{16}\minus{}\dots\minus{}2^{11}\minus{}2^{10}$, hence there are $ 8$ such sets and the answer is $ B$.[/hide]


Find the value : $\large \sum\limits_{i=1}^{n} \frac{1}{4^icos^2\frac{a}{2^i}}$

I tried working on the problem. I almost had it, but I'm not sure now. Can you help me out by telling me if [hide]$\frac{1}{3}\cos{\frac{a}{2}}$[/hide] is correct?

drunner, the sum is not infinite.  It should be a function of $n$ as well.

Why do people use i as a variable? It is so confusing!
$\sum_{k=1}^{n} \frac{1}{4^kcos\frac{a}{2^k}}$

But anyone tell me the solution of this problem ?

it seems quite hard, maybe you can try posting it in more advanced forums?

yeah your right t0. I kept interpreting it as infinite because I'm so accustomed to infinite sums.


Let $ a,b,c\geq0$. Prove that:
$ 3(1\minus{}a\plus{}a^2)(1\minus{}b\plus{}b^2)(1\minus{}c\plus{}c^2)\geq 1\plus{}abc\plus{}a^2b^2c^2$

[quote="Inequalities Master"]Let $ a,b,c\geq0$. Prove that:
$ 3(1 \minus{} a \plus{} a^2)(1 \minus{} b \plus{} b^2)(1 \minus{} c \plus{} c^2)\geq 1 \plus{} abc \plus{} a^2b^2c^2$[/quote]
Use this inequality
\[ (1\minus{}x\plus{}x^2)^3\ge\frac{(x^6\plus{}x^3\plus{}1)}{3}\]


[b]Consider the continued fraction below:

1+1/1+1/1+1/1+1/...

We can consider this "infinite fraction" as a sequence of terms, u_n, where

u_1 = 1+1
u_2 = 1+1/1+1
u_3 = 1+1/1+1/1+1
...

Determine a generalized formula (recursion formula) for u_n+1 in terms of u_n.
[/b]

the closest i've come to figuring this out is

u_n+1 = u_n/u_n-1, u_1 = 1+1, n>2

but then, --->  u_3 = 1+1/(1+1)/1+1 ?

urgh, help !

sorry i dont know how to use that latex stuff.

I assume that "1+1/1+1/1+1/1+1/... " means

$ 1\plus{}\frac{1}{1\plus{}\frac{1}{1\plus{}\frac{1}{1\plus{}\cdots}}}$

and that you're definitions for the $ u_n$ were

$ u_1 \equal{} 1\plus{} 1$

$ u_2 \equal{} 1\plus{} \frac{1}{1\plus{}1}$

$ u_3 \equal{} 1\plus{} \frac{1}{1\plus{}\frac{1}{1\plus{}1}}$

$ u_4 \equal{} 1\plus{} \frac{1}{1\plus{}\frac{1}{1\plus{}\frac{1}{1\plus{}1}}}$

Now, if you  look closely at these, you will see that each one is the denominator of the fraction in the next one. So, now how can you express $ u_{n\plus{}1}$ in terms of $ u_n$?

Note that you can click on the latex and see what I typed in to make it.

no it wouldnt be u_1... id have to break that term down right?

u_n+1 = 1+1/u_n

lol dammit that wasnt too hard, why did it take me so long?

thanks for your help!

A bit of an extension:
Using the recursive formula we now have, we can easily find a way to determine the limit (if it exists) of this continued fraction: since at it reaches its limit, one term will nearly equal the next, we can put in $ u_{n\plus{}1}\equal{}u_n\equal{}x$, giving us the equation $ x\equal{}1\plus{}\frac{1}{x}$.
Solving this will give 2 possible limits, one of which will be clearly absurd. (I'll leave it to you to find it for yourself :P)
Of course, this doesn't prove what the limit is; rather, it only shows what the limit must be if it exists. However, once the limit is found, one can use a method of choice (whee, delta-epsilon arguments) to prove that the limit must exist.

Actually, I think I'll go ahead and say that the ratio it approaches is the golden ratio. In fact, each term (known as a convergent), will be of the form $ \frac{F_{n\plus{}1}}{F_n}$, which is easily provable by induction. This either proves the limit of the continued fraction or of the ratio of fibonacci numbers, whichever you prefer. :)


some observation about the biology of mosquitos
http://www.e-monsite.com/aply123/rubrique-1134044.html


Enathalpy Change equals Internal energy change if
a)all reactants and products are in solution
b)if reaction is carried out in a closed Vessel
c)number of moles of gaseous products equal the number of moles of gaseous reactants
d)reaction is carried out at constant pressure

[hide="Answer"]By exclusion of parts, c) is the correct answer. Let's see why. In (a), the fact that all reacting species are in solution does not have nothing to do with $ \Delta H$ or $ \Delta U$. For a general process, $ \Delta H \equal{} \Delta U \plus{} \Delta(PV)$. For an isochoric process, $ \Delta H \equal{} \Delta U \plus{} V \Delta(P)$, and for an isobaric process, $ \Delta H \equal{} \Delta U \plus{} P \Delta(V)$. Therefore, neither (b) nor (d) are correct. However, (c) is not completely correct, since that is only true for standard $ \Delta H$ and $ \Delta U$.[/hide]


algebra
polynomial
modular arithmetic
LaTeX

I have to prove that if two integers x,y does not divided with 3 then (x^2-y^2) always divided with 3. (my question:Can we make the last part like that ->  (x^2-y^2)=0mod3 ?)

What i have done since now.

I said that since x,y are always divided with 3 with have four cases.
1. x=3i+1 and y=3j+1
2. x=3i+1 and y=3j+2
3. x=3i+2 and y=3j+1
4. x=3i+2 and y=3j+2

where i,j are integers.

case 1: If i prove that x^2=(3i+1)^2 = 1mod3 and y^2=(3j+1)^2 = 1mod3 then we have 1-1 = 0 mod 3 that is correct!
How to prove that (3i+1)^2=1mod3?

same thing for the other cases.

In general, if $ P$ is a polynomial and $ x \equiv y \pmod{n}$, then $ P(x) \equiv P(y) \pmod{n}$.
Try to prove this.

A Hint:
[hide]
Show $ x \minus{} y \mid P(x) \minus{} P(y)$.
[/hide]

Then it is clear that $ (3i\plus{}1)^2 \equiv 1^2 \equiv 1 \pmod{3}$.

The point is that you don't need to expand $ (3j\plus{}2)^2$, you can just do

$ (3j\plus{}2)^2 \equiv 2^2 \equal{} 4 \equiv 1 \pmod{3}.$

probably not optimal, but here it goes: x^2-y^2=(x+y)(x-y). since x,y are not divisible by 3, you can get a few cases: x=1,y=1(mod3) x=2,y=2(mod2), x=1,y=2(mod3), and x=2,y=1(mod3). plugging in, you see that for (1-1)(1+1), (2-2)(2+2), (1-2)(1+2), and (2-1)(2+1), all are congruent to 0 (mod 3).

The squares are congruent to 1 or 0 mod 3.

To directly answer your question [quote="a.katerina"](3j+2)^2=9j^2+12j+4=3(3j^2+4j)+4=....????[/quote] it may help to write $ 4 \equal{} 1 \cdot 3 \plus{} 1$ ;).  (What hjbrasch and Nukular have said is also correct, of course.)

Since you're already using some [[LaTeX]] codes, you might as well enclose them in dollar signs to make everything look nicer -- see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 and also the first link for more about LaTeX and using it on the forum.


linear algebra
matrix
geometry
geometric transformation
reflection
linear algebra open

What are the matrices in $M_n(C), M_n(Q), M_n(Z)$ which can be represented as a product of matrices of finite multiplicative order belonging to each of the thrre $M_n(C), M_n(Q), M_n(Z)$?

Let E be  the matrices in Mn(R) which can be represented as a product of matrices of finite multiplicative order belonging to Mn(R).
If A is in E then det(A)=+-1.Conversely if det(A)=+-1 then A is in E.
PROOF: we may suppose that A is in SLn(R) (det(A)=1).The permutation matrices are in E; thus it is enough to show that any matrix U=I+tMij (where t is a real, i<>j, Mij is the matrix s.t. its elements are zero except that of indices i,j which is 1) is in E.
Fact: U is the product of the 2 reflections:
V=diag(1,..,1,-1,1,..1) where -1 is in position j and
V+tMij.
Remark: we use only reflections!

I reconsider my last post: the question about Mn(R) was not raised by Harazi but my proof holds, with the same result, in Mn(Q) and more generally in Mn(K) where K is a field where x^k=1 implies x=+-1.

1°) The proof in my first post holds also in Mn(Z) with the result GLn(Z) (use elementary operations and euclidean algorithm).
2°)Let F be the matrices in Mn(C) which can be represented as a product of matrices of finite multiplicative order belonging to Mn(C).
If A is in F then det(A) is in V, the set of the roots of 1.
We show a very partial result:
Let G be the matrices B in Mn(C) such that det(B) is in V.
Then F is allwhere dense in G.
PROOF: Let B in G; by polar decomposition B=HN: H is a hermitian >0 matrix s.t. det(H)=1 and H  is similar to a real matrix in E (see first post) thus H is in F. N is an unitary matrix which is similar to a diagonal matrix s.t. each entry is in U (elements of modulus 1 in C).
V is allwhere dense in U and N is in the closure of F.


Let $ f: [1,\plus{}\infty) \rightarrow \mathbb{R}$ be a strictly increasing function, which satisfies the relation $ (f \circ f \circ f)(x)\equal{}x^2\minus{}2x\plus{}2, \forall x \in [1,\infty)$. Determine the value of $ f(1)$.

[quote="moldovan"]Let $ f: [1, \plus{} \infty) \rightarrow \mathbb{R}$ be a strictly increasing function, which satisfies the relation $ (f \circ f \circ f)(x) \equal{} x^2 \minus{} 2x \plus{} 2, \forall x \in [1,\infty)$. Determine the value of $ f(1)$.[/quote]
Clearly $ (f \circ f \circ f)(1) \equal{} 1$,and we have that $ (f \circ f \circ f \circ f)(x) \equal{} f(x)^2\minus{}2f(x)\plus{}2$.It follows that $ f(1) \equal{} (f \circ f \circ f \circ f)(1) \equal{} 1$.


Prove that for $ x,y$ real positive numbers:
$ (x^6\minus{}y^6)^2\plus{}27x^4y^4(x^2\minus{}y^2)^2\geq16x^3y^3(x^3\minus{}y^3)^2$

[quote="gilcu3"]Prove that for $ x,y$ real positive numbers:
$ (x^6 \minus{} y^6)^2 \plus{} 27x^4y^4(x^2 \minus{} y^2)^2\geq16x^3y^3(x^3 \minus{} y^3)^2$[/quote]

$ L\minus{}R \equal{}$$ (x^8\minus{}2x^4y^4\plus{}y^8\plus{}4yx^7\plus{}10y^2x^6\plus{}4y^3x^5\plus{}4y^5x^3\plus{}10y^6x^2\plus{}4y^7x)(x\minus{}y)^4$


Find the number of degrees in the measure of one of the base angles of an isosceles triangle, if the measure of the vertex angle is $ 36^{\circ}$ less than the sum of the measures of the base angles.

Call each of the base angles $ x$. Then, the vertex angle would be $ 2x \minus{} 36$, since the base angles are congruent in an isosceles triangle. Because the sum of all angles in a triangle is $ 180^{\circ}$, we call set up and solve an equation.
$ x \plus{} x \plus{} (2x \minus{} 36) \equal{} 180\implies2x \plus{} 2x \minus{} 36 \equal{} 180\implies4x \minus{} 36 \equal{} 180\implies4x \equal{} 216 \implies x \equal{} \fbox{54}$.


Well, it's easy to calculate the probability of having a certain hand in poker. Is that what you want to know?

i think the probablity of getting a pair in your hand is:
$\frac{3}{676}$

let K be the first card you receive.

out of remaining 51 cards, there are three remaining Ks.

so the probability of having a pair is $\frac{1}{17}$

What do you think about this website?
http://www.pokibot.com/

[quote="kotsba"]hello you math fans
i love to play poker,mostly texas holdem
i play it on the net mostly cuz in my country there are no casinos
now i know that in [url=http://www.pacificpoker-online.com]poker[/url] you can calculate odds and statistics and that can help me win more hands
since im not good at math i would be more than gald to hear some tips
thanks[/quote]

what country is that?

[quote="pohoatza"]pikachu :)[/quote]
isn't that a pokemon? :?

[quote="lingomaniac88"]Finding the odds of getting a pair?  Here's what I would do:
[hide]Let the cards be $A$, $B$, $C$, $D$, and $E$.  Cards $A$ and $B$ will be the pair.
$A$ can be anything.  Let it have a value of $x$.
$B$ must also be $x$.  There would be 3 cards out of the 51 left that give a value of $x$.  So that gives us $\frac{1}{17}$.
Card $C$ can't have a value of $x$, or else we'll have a three-of-a-kind.  So that leaves 48 out of 50 cards for $C$, or $\frac{24}{25}$.
Now, $D$ can't be $x$, either.  It can't be the same value as $C$, because that will form a two pair.  So we're left with $49-2-3$ cards, or 44 possible cards out of 49.  That means $\frac{44}{49}$.
Finally, $E$.  It can't have a value of $x$, nor can it be the same value as $C$ or $D$.  So that leaves us with $48-2-3-3$ cards, or 40 out of 48, or $\frac{5}{6}$.

So we multiply: $1\cdot\frac{1}{17}\cdot\frac{24}{25}\cdot\frac{44}{49}\cdot\frac{5}{6}$, or $\frac{176}{4165}$.

But we just considered one case.  There should be a total of $_5C_2$, or 10.  So we multiply that fraction by 10 to get $\boxed{\frac{352}{833}}$.
At least, that's what I think...[/hide][/quote]

We're talking about having a pocket pair.

i tend to think that the whole category of Poker game is Overplayed.
i even think it's less mathematical. partly because Blackjack was like already the most strategic Poker game in casinos, and partly because Poker is a very specific and therefore narrow form of math model. Some rules are unbelievably complicated though, compared to the IQ/degree of people who play them.
I thought this guy was a spammer promoting his website. At least he asked a math question. Then another replier just posted his website and asked "how do you think of it". When the real spammer appears(well, why would a real spammer appear only in a reply) i don't think it's a spamming topic anymore.
then here we go calculating odds of "pairs". 3/676 seems ridiculous cos I get pairs all the time, and i don't even play much Poker.

Then "Pikachu" was inserted non-sensely. It's a Pokemon character and Pokemon isn't Poker Monster.
after one of our member's comprehensive calculation, another Poker player stunned us with the term "pocket pair". Poker, Pocket, PocketMonster...... What on earth is a pocket pair.....
I don't expect answers because when it comes to Poker i just don't know any math. When hitting a Blackjack, i just follow instincts, really bad instincts.

anyway Poker is supposed to be fun for avg-IQ folks. That's why many rules reduce the effect of good strategies. It's the Vegas thing that brings us math without our analyzing it or even realizing it.

Pocket pair means getting a pair in your hand in texas hold'em. A hand consists of 2 cards.

Also, your post didn't help much with the spamming eh? You talk about people posting spam and you come post here doing exactly what those other spammers did. Completely unnesessary.

Oh, in Texas Hold'em.  I didn't realize that.  Yeah, then it would be $\frac{1}{17}$.

I have a site on poker calculators that also has videos for some simple math calculations needed to help you be profitable at the game. There are 3 distinct types of poker calculators - Mathematical, Empirical, and Operative - these are also described in detail.

[url]http://www.PokerCalculatorReport.com[/url]

[quote]
after one of our member's comprehensive calculation, another Poker player stunned us with the term "pocket pair". Poker, Pocket, PocketMonster...... What on earth is a pocket pair..... [/quote]

Stunned [b]us?[/b] don't include us into your lack of knowledge.


floor function
factorial
number theory
prime factorization

The number 10!, when written in the base 12 system, ends with exactly how many zeros? :?:

[hide="intuitive hint"]we know how to deal with such problems in base 10, since 10=2*5. it's not much different with 12=2^2*3. to add one zero in base 12, you must have one factor of 12.
just making sure, is this the correct reasoning?
if so, it's pretty straightforward from here[/hide]

You're on the right track but you are not quite there yet. There is a simple trick used to solve problems like this.

The general problem is not actually straightforward in base $ 12$.  That is, in base $ 10$ it's easy to show that only the $ 5$s need to be counted, but in base $ 12$ it's not clear whether to count the $ 2$s or the $ 3$s.

[hide="my answer"]there are 8 factors of 2 in 10! and 4 factors of 3. to make a product of 12 you need to have 2^2*3. thus there are 4 zeros when 10! is written in base 12?[/hide]

*edited: thanks to correction by xpmath

I agree.

I assume you meant 4 factors of 3?

mihail911's solution already takes that into consideration, although it is not easy to see if you are not familiar with the basic principle.

[b]Lemma:[/b]  The number of times a prime $ p$ can divide into $ n!$ is given by

$ \left\lfloor \frac{n}{p} \right\rfloor \plus{} \left\lfloor \frac{n}{p^2} \right\rfloor \plus{} \left\lfloor \frac{n}{p^3} \right\rfloor \plus{} ...$

Explanation:  Every multiple of $ p$ between $ 1$ and $ n$ contributes one power of $ p$; that is the first term.  But the multiples of $ p^2$ contribute two powers of $ p$, so we have to count them again; that is the second term.  And the multiples of $ p^3$ contribute yet another power of $ p$...

mihail911 has therefore calculated that there are $ 5$ multiples of $ 2$, $ 2$ multiples of $ 4$, and $ 1$ multiple of $ 8$ between $ 1$ and $ 10$, so there are $ 5 \plus{} 2 \plus{} 1 \equal{} 8$ total times $ 2$ divides into $ 10!$.  He then calculated similarly that there are $ 3$ multiples of $ 3$ and $ 1$ multiple of $ 9$, so there are $ 3 \plus{} 1 \equal{} 4$ total times $ 3$ divides into $ 10!$.  In other words, he has calculated that the prime factorization of $ 10!$ begins

$ 10! \equal{} 2^8 3^4 ...$

Oh, wow. My method of solving this is rather crude; I'm not sure if it's just a coincidence that I got 4. Please correct me if this is wrong.

[hide="my method"]I listed 10!:

1*2*3*4*5*6*7*8*9*10
Next I grouped the values that make 12 multiple times

(2*6)(3*4)(8*9*10)(1)(5)(7)
(1, 5, and 7 are useless)

We can see that 2 and 6 are a pair; 3 and 4 are also a pair. That's two so far.
Now, the prime factorization of 8*9*10 is
(2*2*2)(3*3)(2*5)
I regrouped that into 5(2*2)(3)(2*2)(3)=5(4*3)(4*3). That's two other 12s. Another way I thought of
this is that 10*9*8=720 = 5 * $ 12^2$. And we know that the value $ n^k$ has k zeros when counting in base $ n$. (E.g.10^2 has 2 zeros in base 10.) [/hide]

Is this logic correct?

It works, but for larger numbers your method will get difficult.  Can you do the same for, say, $ 100!$?

But for 100!, I just know that there are 50 multiples of 2, 25 multiples of 4, 12 multiples of 8, 3 multiples of 32, 1 multiple of 64...  :lol:

That's what I'm saying.  (It's also not obvious that you should count powers of $ 2$ and not powers of $ 3$.)

the answer is:

[hide]4.

you take the prime factorization of the base number
          12 = 2 x 2 x 3
Then divide the factorial by each of those.
          10/2 = 5
          5/2 = 2.5 (you drop any decimals you get)
          2/2 = 1
Then you add all your answers together
          5 + 2 + 1 = 8
There are two 2's so you end up with 16 2's
Then you do the same procedure with the 3's
          10/3 = 3.33333 (drop the decimal)
          3/3 = 1
So you get a total of four 3's

Then you pair up the 3's and 2's

You get an answer of 4.

This is a simple and fun trick that is very useful.

[/hide]


A set of four consecutive integers has a sum of 22. If each integer of the set is increased by 2 and then multiplied by 20, what is the sum of the new set of integers?

We know that $ n \minus{} 2 \plus{} n \minus{} 1 \plus{} n \plus{} n\plus{}1 \equal{} 22$. Add 2 to each term to get $ n \plus{} n \plus{} 1 \plus{} n \plus{} 2 \plus{} n\plus{}3 \equal{} 30$, and then multiply each side by 20, to give us an answer of $ \boxed{600}$.


Joe's French poodle, FooFoo, is tied to the corner of the barn which measures $ 40$' x $ 30$'. FooFoo's rope is $ 5$0' long. In terms of $ \pi$, over how many square feet can FooFoo wander?
[asy]size(150);
draw((40,0)--(40,30)--(0,30)--(0,0)--(40,0)--(70,40));
draw(Circle((72,40),2));
label("30'",(0,15),E);
label("40'",(20,0),N);
label("50'",(55, 20),E);[/asy]

The distinct figures, in clockwise order, are a quarter-circle with radius $ 50 \minus{} 30 \equal{} 20$, a three-quarter-circle with radius $ 50$, and a quarter circle with radius $ 10$, for a total area of

$ \frac14\pi\cdot20^2 \plus{} \frac34\pi\cdot50^2 \plus{} \frac14\pi\cdot10^2 \equal{} 100\pi \plus{} 1875\pi \plus{} 25\pi \equal{} \boxed{2000\pi}$.


geometry
geometric transformation
dilation

when u treat an allyl chloride with highly concentrated solution of sodium ethoxide in ethanol then u get $ S_{N}2$ products but when u do that in a dilute solution u get otherwise.why is it so

SN2 is a bimolecular reaction, so its rate depends on the concentration of both substracte and nucleophile. If the nucleophile concentration is low, then SN1 (whose rate depends only on alkyl halide concentration) is favoured.

but that doesn't ensure that if the concn. is high then $ S_{N}2$ is favoured

Well, from a certain point of view you are right, because sodium ethoxide is also quite a strong base, so E2 would perhaps dominate in an allyl chloride (specially because the product have an extend $ \pi$ system).
However, speaking only in SN2, why doesn't a high conc. favour SN2? Can you give a counter example?

my basis of argument would be ethoxide is very well solvated in ethanol(see u have bot the organic ad polar requirements) but then in order for $ S_{N}2$ to happen at a substasntial rate the nu must be free to move .also we normally have aprotic polar solvents for $ S_{N}2$ here the solvent is protic thus reduces th rate for above mentioned reason....


Hello I understand this problem but i want someone to show me a different way how to solve these problems. Thanks

$3x^3-2x^2-7x-2$ one way i did  this was to find the factors of -2 and than plug them for x but thats when x is by its self. Now i have a 3 infront of x so please show me how to solve this thanks

Ryan

By solve, i hope you mean factor.

$f(x) = 3x^3 - 2x^2 - 7x - 2$
$f(2) = 0$
which means
$(x - 2)$ is a factor.
Divide f(x) by $(x-2)$ to get $3x^2 + 4x + 1$.
This means, $f(x) = (x-2)(3x^2 + 4x + 1)$. Can you see now that when we evaluated f(2) = 0, we know it's a factor, try it with the factored f(x). We can easily factor the quadratic:
$f(x) = (3x+1)(x-2)(x+1)$

Yes I mean factor! Thanks So the quadratic formula $-b+- sqrtb^2-4ac/2a$ comes in handy thanks again!

The factor theorem applies to polynomials with leading coefficients other than one, also.  The rule is, any rational root must be of the form $\displaystyle\frac{a}{b}$, where $a$ is divisible by the constant term and $b$ is divisible by the leading coefficient.  In this case, the possible roots are $\pm \left(\frac{1}{3},\frac{2}{3},1,2\right)$.


I made this problem up, and I can't tell how hard it is to solve (I'm using a simpler case of the original problem). Let's see what people come up with.

Four infinite geometric sequences have distinct first terms of the form 1/n, where n is an odd positive integer other than 1. The common ratio of each sequence is equal to the first term of that sequence. The sum of the sum of these four sequences equals 1. If the first terms of these four sequences are 1/a, 1/b, 1/c, and 1/d, find the lowest possible value of a+b+c+d.

I hope that's comprehensible. By "sum of the sum of these four sequences," I mean take the sum of each of these four infinite geometric sequences, and then add these four resulting quantities.

[quote="0714446459923"][hide]
(1/n)/(1-(1/n))=(n-1)^-1

5,5,5,5 is the smallest solution so the answer is 20
if one of the numbers was 3, then another of the numbers would have to be higher. Since it is odd it can't be 6, so if the one of the numbers is 3, another number has to be at least 7 and so there are no solutions wiht a sum less than 20.

[/hide][/quote]

oh, ok, got it now

The sum of the sequences gives $\frac 1 {a-1} + \frac 1 {b-1} + \frac 1 {c-1} + \frac 1 {d-1} = 1$
I don't know how to prove it, but inspection/intuition gives the minimum being when each $x - 1 = 4$, so the sum $a+b+c+d = 20$

[hide]The sum of the sequences can be expressed as $\sum_{n=1}^\infty \frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}+\frac{1}{d^n}=1$.

From Power Mean Inequality, $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}+\frac{1}{d^n}\geq 4\left( \frac{\frac 1a +\frac 1b +\frac 1c +\frac 1d}{4}\right)^n$.

Summing up the inequalities, $1\geq \frac{4r}{1-r}\Rightarrow 5\geq \frac{4}{1-r}$ where $r=\frac{\frac 1a +\frac 1b +\frac 1c +\frac 1d}{4}$.

Isolate $r$ to get $r\leq \frac 15$ so the sum of the reciprocals is less than or equal to $\frac 45$.

Finally, $(a+b+c+d)\left( \frac 1a +\frac 1b +\frac 1c +\frac 1d \right)\geq 16$ from Cauchy, so $a+b+c+d\geq \boxed{20}$, where the minimum occurs when $a=b=c=d=5$.[/hide]

Does everybody have poor vision? The problem says "distinct first terms", hence $a=b=c=d=5$ is out of question.

[quote="deej21"]:(
Well...
$5-\epsilon, 5+\epsilon, 5 - 2\epsilon, 5 + 2\epsilon$

 :D[/quote]

And what about "where n is an odd positive integer" part?  :roll:

[hide]I got 28.

Basically, we need to find four distinct euclidean fractions with even denominators that have a sum of 1, and have to find the least possible value of this. I then remembered $\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}=\frac{3}{4}$, leaving euclidean fraction $\frac{1}{4}$ (as 1-3/4 is 1/4). Thus the answer is (2+1)+(4+1)+(6+1)+(12+1), which should come out to 28.

EDIT: I can't add: I meant 28.

Q'ed I hope.[/hide]

[quote="10000th User"][quote="hockeyadi23"][hide]I got 26.

Basically, we need to find four distinct euclidean fractions with even denominators that have a sum of 1, and have to find the least possible value of this. I then remembered $\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}=\frac{3}{4}$, leaving euclidean fraction $\frac{1}{4}$ (as 1-3/4 is 1/4). Thus the answer is (2+1)+(4+1)+(6+1)+(12+1), which should come out to 26.

Q'ed I hope.[/hide][/quote]How do you know that's the minimum? Proof please.[/quote]

I'm working on it right now, but first I have to get my sister from the bus stop!

[quote="hockeyadi23"](2+1)+(4+1)+(6+1)+(12+1), which should come out to 26.[/quote]

This should come out to 28.

And the proof isn't difficult: Put $a'=a-1, b'=b-1, c'=c-1, d'=d-1$. WLOG we can assume $a'<b'<c'<d'$, which gives ${1\over a'}>{1\over b'}>{1\over c'}>{1\over d'}$, and that gives $1={1\over a'}+{1\over b'}+{1\over c'}+{1\over d'}<{4\over a'}$, giving $a'<4$. Since $a'$ is even, we get $a'=2$.

This gives ${1\over b'}+{1\over c'}+{1\over d'}={1\over 2}$, and the similar reasoning gives ${1\over 2}<{3\over b'}$ or $b'<6.$ Since $b'$ is also even and different from $a'=2$, we get $b'=4.$

Then ${1\over c'}+{1\over d'}={1\over 4}$, which implies ${1\over 4}<{2\over c'}$, and that gives $c'<8$ or $c'=6$.

Finally this gives ${1\over d'}={1\over 12}$, or $d'=12$

Hence the desired numbers are $a=2+1=3, b=5, c=7, d=13$ and the sum is $a+b+c+d=28.$

Lol my b I can't add... kk my proof will be a bit informal but basically, there are only two sets of four denominators which will sum up to less than 24 (I'm using 24 because I'm subtracting 4 from 28, the result which I got): (denominators only): 2, 4, 6, 10 and 2, 4, 6, 8. (Since 2, 4, 8, 10 already adds up to 24, we can leave it out).
Then, we test each set to see if their reciprocals will give us a sum of 1. The first set (1/2+1/4+1/6+1/10) gives us 61/60, and the second (1/2+1/4+1/6+1/8) gives us 25/24.

Thus, as all the possible sets of four distinct even denominators either sum up to greater than 24, or have reciprocals don't add up to one, 28 is the smallest answer. I know this is pretty informal but I figured for a problem like this, an informal proof should be sufficient.

[quote="0714446459923"][hide]first assume
a<b<c<d

obviously a has to be 3 or else b=a
b has to be 5 so
c has to be 7 so
d has to be 13 so
you have your answer - 28[/hide][/quote]

To hide your text, you need to highlight the text to be hidden. I have an example above, where I hid your text.

Question. How did you assume a=3? Did you use the solution if all the numbers were not distinct, to assume that a<5?

28 is right. It seems legitimate to say that 3 is the lowest denominator, then 5 and 7 are the next lowest (keeping in mind that they have to be distinct), then 13 is the denominator that works to sum to 1. How difficult did you guys find the problem?

well we already know that a>1 from the prolem.
if a<b<c<d then 1/(a-1)>1/(b-1)>1/(c-1)>1/(d-1)
and 1/(a-1)+1/(b-1)+1/(c-1)+1/(d-1)=1
therefore
1/(a-1)>1/4
a<5

a similar argument forces b to be 5.


The name of your link is correct but the actual url is missing the ending parenthesis, here is the [url=http://en.wikipedia.org/wiki/E_(mathematical_constant)]link[/url]

I didn't get that at all! What's a tangent?

There's a link to the Wikipedia article on tangent lines in the [i]first sentence.[/i]

A tangent should be where the line only touches a curve once.


Mathnerd314 River has a bridge over it in the shape of a circular arc, and the highest point of the bridge is 10 meters above the water. Mathnerd314, who is 9 meters tall (not really), gets on a boat and rows 27 meters from the bank. When he stands up, he finds that he can just fit under the bridge, with his feet at water level. (Then he finds out--the hard way--that standing up in a boat is a very dangerous thing to do!!!) How wide is Mathnerd314 River?

Due to lack of response for six days, I will post a hint:

[hide="hint"]
Draw the rest of the circular sector and draw right triangles. Remembering that radii of the same circle are always equal, try to set two expressions equal by the Pythagorean Theorem.
[/hide]

[hide]Can't you use $x^2+y^2=100$ as the equation of the circle, see that $x=\sqrt{19}$ when y=9, and use that to find the width? [/hide]

EDIT: Nah, this doesn't look like a circle.

[hide="Solution"]Denote by $x$ the distance between mathnerd's position and the symmetrial axis of the bridge, and by $r$ the radius of the bridge. Then we have

$x^2=r^2-(r-1)^2\iff x^2=2r-1$
$(27+x)^2=r^2-(r-10)^2\iff 729+54x+x^2=20r-100$

From the first equation $x^2=2r-1.$ Substituting that into the second equation, we obtain $x=\frac{r-46}{3}.$ Plugging that back into the first equation, we get $r^2-110r+2125=0$, and the solutions are $r_1=85$ and $r_2=25.$ From $x=\frac{r-46}{3}$ we get $x_1=13$ and $x_2=-7$ (positive and negative solutions mean that mathnerd might be standing either nearer the bank he embarked from or nearer the opposite bank), and the width of the river is given by $d=2(27+x)$, which yields either $d_1=80$ or $d_2=40.$[/hide]


geometry calculus derivative probability

hello everyone,

i am suneet and am studying in class 11. the kvpy scholarship has always been a dream for me like thousands of my age fellows.i want to know the approx. cutoff for the written exam and is the written exam marks also counted in final result or only the interview matters, what level of questions are asked in interview, and lastly, those who are receiving the scholarship,what marks they got in the written test(if written test marks are counted?)
hoping for generous answers
suneet :D  :D

hello and welcome to aops :D

first of all  we r not told our written marks so cant tell the exact cutoff
but the marks r counted in final result

and the interview is not too diff--u shud bjust b able to convince the interviewers :D

Can you point me anywhere in this wide wide web (www) where I can atleast get some previous years qurestions. I know they are not disclosed bt maybe someone somewhere has posted them???? :)  :roll:

ofc on kvpy's site
some quesn r available:P:

but u r in 12th ,no?

I need the interview questions ASAP. I've asked a few qualifiers from past years but more info would be helpful. I guess it won't be as easy as the NTSE interview was :(

i feel the ques wud b in no way helpful,coz well they can ask nething... :D

i never prepared or knew abt ques while going 4 interview :)

Basically I was worried whether questions will be based on a lot of hard factual stuff too? The application part is good as I am told though that isn't the major prob. Thing is I'm too lazy to study too much and want to know if there is a very serious need too.  :oops:

there is no serious need of nething..

just application based :D

I'M PANICCING! I asked a senior and he told me all those wierd class 12 questions they asked!!
Can other KYPYers please please post the questions they were asked? Including stuff like aim of life. And how important are the certificates you carry?
I was thinking of just revising my coaching notes....
I'll prolly have to try and mug up all those NCERT books in 24 hours?  :ninja:
I'm DEAD!!!

i never did ne of these things..i didnt even carry a certificate(dont remember though)
there was no question that i cudn't answer then and can answer now ,ie to say there is no need of class 12th knowledge!!

and no need 2 panic at all!!!

{Mind u ,this is strictly applicable if u r appearing at Delhi centre,i dont really kno abt other centres :P  }

I WAS already DEAD when I learnt They asked wierd questions such as brahmgupta theorem etc.
One of my seniors who qualified for the same 4-5 yrs ago even told me that they had to solve specific problems on the blackboard in front of the interviers. :noo:

btw best of luck to all the candidates especially arnav( His interview would have probably started by now)  :rotfl:

I'm at Allahabad centre on 12th at 1:30 pm
Brahmagupta theorem!
I've only HEARD of it!!!! Something to do with geometry I guess(And hope)
Rituraj, man you are a genius.... 1st rank in NSTSE and all. But there are some rather under blessed guys like me who don't know as much :( I don't even have full 11th knowledge  :maybe:

[quote="nuclear_alchemist"]Brahmagupta theorem![/quote]
We had it discussed by a teacher in a IX or X class.

Just checked wiki on it. Pretty elementary theorem. And I'm sure we must have had it as a proving question in 8-9. But how many theorems is one supposed to mug up?  :o

they must just expect u to hav an idea abt wat u kno...
mayb things were diff 4-5 yrs ago :oops:

Maybe they were a lot diffrent, since he was asked to choose any 2 subjects from PCM on which questions were asked.

I agree that Brahnagupta theorem is an elementry one, but still it is greek to a person who has never heard it(I also didn't knew it though i have studied Vedic maths)  :oops:

My senior was asked bio even though he had PCM.  :o

Guess than I will have to prepare class 10 bio in around 2 hrs (Dont know that even that would be enough)

Do you people have your interviews tomorrow?
All the best to the future KVPY Scholars{All those taking the interview from Aops}.

Mine is day after tomorrow. I have never done anything like this before and I know it is risky! But I'll take the risk.  :maybe:  Here's my no.- 9415005333. If any Aops has his/her interview tomorrow please call me and tell me the questions you were asked. Especially if you have your centre in Allahabad. Thanks!

what do u think is the risk?? :rotfl:

questions r asked in All 4 subjects,just the proportion may vary :)

Shouldnt post personal no.s in public forum.

i am ruined.................

just try to b calm nuclear..........(evn though i have now found,that it is actually difficult........)
i suffrd only coz ov dat........... :(

just scared to even sit on the chair...........
i was asked a ques on blackboard which i was explainig them ............
explanation of 1 part reqd me to say..........
"acc is d 2nd derivative of disp"

which i said:-
"disp is the 2nd derivative of acc" :|

also/...........wen i was asked "where r u staying in deli"

i had to answer inspection quarter............
insted i said headquarter............... :o

i am totally ruind................

just dont be scared anyone.................

also previous kvpy scholars..........
pl tell ........

if i am not able to answer some ov there ques then is there any scope ov selection..........????????
some questions were ov 12th class of electricity and magnetism.................

they didnt evn ask dat hav u done these chapters. or not............
direct ques......................

btw truly........ritu..........rupa maam was a superb interviewer..........

i forgot to say the most imp thing..........

all the best to all of u giving interviews...........especially pulkit............. :rotfl:

......................

well i said,try ur best to smile :rotfl:  :P  :rotfl:

[quote="Arnav...!!!"]rupa maam was a superb interviewer..........[/quote]
Kindly elucidate........ Share your experience.... :)

And I thought that NTSE interview would be the would be the worst interview of my life.
kvpy interview for me started in a bad way and ended in the worst possible manner. :(

Yeps i was able to handle the physics part well( thanks to Pr R. Ghosh, really felt confortable answering her). But maths and chemistry ruined me. I told him(chemistry interviewer) that i love orgo.He than asked me 3 questions on organic ,couldn't answer any one of them(even the simplest one- phenols are acidic or basic). Maths mam asked me about probablity( I did my first probability question of my life in the interview itself  :ninja:  ).

Arnav you do have a chance since i know i am out of the competition.

P.S I doubt that a drank the water meant for Maths mam in my nervousness) :oops:

hey man y do u hav 2 b nervous???yyy not stay calm???

When you have 8 eyes glaring upon you and you know that these 20 minutes are the most important minutes in your schooling life, it is natural to get nervous. I couldn't answer the phenol question even though i told them that  i loved orgo. That really ignited nervousness in me. :maybe: .

btw how long was your interview (to past as well as present candidates)

my chances are tending to nil........... :(
btw wat was the sequence of ur being asked questions........subjectwise?????

mine was...............-
Physics
Biology
Mathematics
Chemistry............

i was so nervous i couldnt make out wether they are asking me maths or bio..........
maths maam was speaking digits and i thougth it is bio ques.............
wen i realized it is maths i said,...........o it is a maths ques....... :rotfl:

all the teachers started laughing at that tym........... :(

btw pulkit do tell how many ques from each subject did they ask...???and which were ur "subjects you like"????????

mine were -P & M
P-3 ques
B-1 ques
M-2 ques
C-2 ques

while others of my grp were interviewed 4 abt 25 min or so,i was done off in abt 15min....the case wasn't too different in nTSE as well.. :maybe:

seq
P..too many..interlinked
C..2??
M..1..+1 by physics teacher :oops:
B 1/2(half)(i dont think they listened the answer)

not at all asked wat  subj i liked...or i dont rem.if i was asked,i had replied "science" and that's all :P
though later i did say..i wud give pref to bio over maths.... :rotfl:

i said that i loved the sciences equally though i preferd physics.
They asked me(in the same order)
5-6 questions from physics(many were interlinked)
2 question from maths(both were of probability)
3 question from chem
0 question from bio :) (The only fact that made me feel good)

probability would hav been easy i suppose :maybe:  :huh:

one ov my frends ws also asked prob...........he said they were too easy??????

As i told you in one of my previous post that it was the very first time i attempted a probability ,   :mad:  thus i was totaly new to the concept of probability it got coupled with my nervousness and ...... well you can understand.
What more can you expect from the guy who forgot that opposite poles repel each other,- during the interview. :(

o ya i forgt to tell..............
btw if u consider interlinkd ques............
i was asked many acc to dat.......... :huh:

P-4
B-2
M-2
C-3

i think they were tryin to see if the boy can anser any of the ques in any one of the subjects.............. :(
which actually i didnt do..........
i could not anser 2 from physics............those were of electromag.........
2 from bio..........i gav answer to his first ques which on giving his 2nd interlinked ques he prooved wrong.............. :rotfl:

M-i answered 1 correctly........but in the second ques.......i gave anser correct but a littl concept mistake...............

C-i ansered 2 and not the 3rd............

well ya ...........

if the intervew takes abt 30 min for 1 intervew is it gud or baaaaaaaddddd

the bio teacr was telling me dat xylm tissue is mad ov dead cells..........
but me foolish first said yes and then said no..........phloem contains dead cells.......??? :oops_sign:  :fool:

both hav dead tissues..but genarallytalking,most cells of xylem r dead while phloem living :P

and well cant really say longer  or shorter interviews r better,but i my case certainlythe shorter ones were much better :rotfl:

Lol, I can't edit my post now. It says only a mod can do it. Also I think anyone with brains(i.e. nearly everyone here) can easily extract the info I gave coz of iisc's carelessness!
And now about my interview. MOVED- http://www.mathlinks.ro/viewtopic.php?t=251255


The vertices of right triangle RAG with hypotenuse AG are
R(-2, 4), A(7, 4) and G(x, y).  If the measure of angle RAG = 45 degrees, what are the possible coordinates of point G?

This is a very specific triangle, 45-45-90 which makes it easier to work with.  There are two candidates for G, and each must have an $ x$-coordinate of -2.  Otherwise, there can't be a right angle.  So now we just need to find the two $ y$-coordinates.  We know that $ \overline{RA}$ has length 9, so the hypotenuse must have length $ 9\sqrt{2}$.  Now use the distance formula:
$ 9\sqrt{2}\equal{}\sqrt{(7\minus{}\minus{}2)^{2}\plus{}(4\minus{}y)^{2}}$
Square both sides and you will end up with a quadratic, easily solved by the quadratic formula.  Your two roots are the two possible values of $ y$.

[quote="JRav"]This is a very specific triangle, 45-45-90 which makes it easier to work with.  There are two candidates for G, and each must have an $ x$-coordinate of -2.  Otherwise, there can't be a right angle.  So now we just need to find the two $ y$-coordinates.  We know that $ \overline{RA}$ has length 9, so the hypotenuse must have length $ 9\sqrt {2}$.  Now use the distance formula:
$ 9\sqrt {2} \equal{} \sqrt {(7 \minus{} \minus{} 2)^{2} \plus{} (4 \minus{} y)^{2}}$
Square both sides and you will end up with a quadratic, easily solved by the quadratic formula.  Your two roots are the two possible values of $ y$.[/quote]

it's easier:
$ X_G\equal{}\minus{}2$
$ Y_G\equal{}4\pm(7\minus{}(\minus{}2))$

Well you should probably explain how you got that but I'm assuming you did this:
$ \overline{RA}\equal{}\overline{RG}\equal{}9$.  Then consider $ \overline{RA}$ to be a radius of length 9 so it sweeps out a circle or radius 9, centered at $ (\minus{}2,4)$.

Thank you JRav!
I'm not very able in speaking english, so explain something could be very difficult for me! :)

We could also include the coordinates of $ G$ when $ RA$ [i]is[/i] the hypotenuse, namely $ (2.5, 8.5)$ and $ (2.5, \minus{}0.5)$, since the problem allows us to do so.

The problem specifies that $ \overline{AG}$ is the hypotenuse so we can't use that point.


algebra
polynomial
linear algebra
matrix
algorithm

[quote="Third Edition"]Hint: Apply the division algorithm to divide $ X^{2009}$ by the characteristic polynomial $ \chi_A(X)$ of $ A$. Then plug $ A$ into the relation and remember the result of $ \chi_A(A)$ ...[/quote]

i really haven't known what you mean...
what is characteristic polynomial $ \chi_A(X)$ of $ A$?
what i have done is try computing A^1, A^2, A^3, ...
but i cant see the pattern :huh:

[quote="GTK X Hunter"]what is characteristic polynomial $ \chi_A(X)$ of $ A$?[/quote]Oouups, the other day you seemed to know the Cayley-Hamilton theorem. So I assumed that you are familiar with the basics of linear algebra.  :o
[quote]what i have done is try computing A^1, A^2, A^3, ...
but i cant see the pattern :huh:[/quote]You cannot very easily. But all your homework - I suspect your postings are your homework ;) - are about the beginning of eigen value theory, ring theory, polynomials and, in particular, Euclidean rings.

If you don't know, this is a good opportunity to learn some theory in order to avoid tedious calculations - if this can be of any use to find a solution at all. Good luck !!!  :)

characterstic polynomial is $ \minus{}X^3\plus{}6X^2\minus{}10X\plus{}5$
then what to do?

[quote="coquitao"]Why don't you try to diagonalize $ A$?[/quote]

how? teach me please^_^

[quote="Third Edition"][quote="GTK X Hunter"]what is characteristic polynomial $ \chi_A(X)$ of $ A$?[/quote]Oouups, the other day you seemed to know the Cayley-Hamilton theorem. So I assumed that you are familiar with the basics of linear algebra.  :o [/quote]

don't be so shocked master, because when you answer using Cayley-Hamilton theorem indeed i haven't known about it, hehehe... i'm so sorry... >.<

[quote="Third Edition"][quote="GTK X Hunter"]what i have done is try computing A^1, A^2, A^3, ...
but i cant see the pattern :huh:[/quote]You cannot very easily. [/quote]

can't this way be used??

[quote="Third Edition"]But all your homework - I suspect your postings are your homework ;) - are about the beginning of eigen value theory, ring theory, polynomials and, in particular, Euclidean rings.

If you don't know, this is a good opportunity to learn some theory in order to avoid tedious calculations - if this can be of any use to find a solution at all. Good luck !!!  :)[/quote]

actually this is not my homework, and even i haven't got linear algebra in my college :noo:
so i will be happy if you'd like to teach me...^^
i will study about some theories more, thanks you so much...^^

[quote="JAS"]characterstic polynomial is $ \minus{} X^3 \plus{} 6X^2 \minus{} 10X \plus{} 5$
then what to do?[/quote]

how did it come?? let me know please^^

i have calculated
$ |A\minus{} \lambda I|\equal{}0$
$ \minus{}\lambda^3\plus{}6\lambda^2\minus{}10\lambda\plus{}5\equal{}0$

I found the three eigen values, they are $ \lambda \equal{} 1, \lambda \equal{} \frac{\minus{}5\plus{}\sqrt5}{2}, \lambda \equal{} \frac{\minus{}5\minus{}\sqrt5}{2}$
then what do i have to do to find $ A^{2009}$?

Let $ \lambda_1, \lambda_2, \lambda_3$ denote the eigenvalues of $ A$. Let $ E_j$ be the matrix $ \prod_{i\equal{}1, i \ne j}^{n} \frac{A\minus{}\lambda_i}{\lambda_j \minus{}\lambda_i}$, i.e. the projection onto the eigenspace associated with $ \lambda_j$. Then $ A$ decomposes as $ \sum_{i\equal{}1}^{3}\lambda_i E_i$ and $ A^{2009} \equal{} \sum_{i\equal{}1}^{3} \lambda_i^{2009} E_i$.


geometry
inequalities
symmetry
calculus
calculus computations

We want to construct a paralelipipedic box (without a cover) with $ 12\,\,m^2$ of hard paper in such a way that the volume of box be a maximum. Find the required dimensions of the box.

Call the dimensions $ b$, $ h$, $ w$.  Then $ SA\equal{}2bh\plus{}2wh\plus{}bw\equal{}12$ and $ V\equal{}bwh$.  Once we substitute in, it's just a matter of differentiating.

I know how to solve it. I just post it because it could look interesting. By the way, why don't you show your reasoning?

This is a case in which the technique of Lagrange multipliers is reasonably efficient:

[hide="Details hidden for the sake of those who would like to try it themselves."]Take $ \frac {\partial}{\partial b}: \ hw \equal{} \lambda(2h \plus{} w)$

Take $ \frac {\partial}{\partial h}: \ bw \equal{} \lambda(2b \plus{} 2w)$

Take $ \frac {\partial}{\partial w}: \ bh \equal{} \lambda(2h \plus{} b)$

We know that any solutions with any of $ b, h,$ or $ w$ equal to zero cannot represent a maximum of volume; hence we ignore those.

Subtract the third equation from the first. We get:

$ h(w \minus{} b) \equal{} \lambda(w \minus{} b).$

Thus either $ h \equal{} \lambda$ or $ w \equal{} b.$

Let's try $ h \equal{} \lambda:$ In the first equation, dividing by $ h$ leads to $ w \equal{} 2h \plus{} w$ or $ h \equal{} 0.$ Since this cannot be, we conclude that $ w \equal{} b.$

With $ w \equal{} b,$ the second equation becomes $ b^2 \equal{} \lambda(4b),$ hence $ \lambda \equal{} \frac {b}4.$ Put that in the third equation to get

$ bh \equal{} \frac {b}4(2h \plus{} b)$

$ bh \equal{} \frac12bh \plus{} \frac {b^2}4.$ Divide by $ 4b$ to get

$ h \equal{} \frac {h}2 \plus{} \frac {b}4.$ So $ h \equal{} \frac {b}2 \equal{} \frac {w}2.$

The box must have a square base and be half as tall as it is wide.

(Lagrange multipliers usually brings us to this point - the "shape" of the correct answer.)

Based on this shape, if we express everything in terms of the height $ h,$ we have a volume of $ V \equal{} 4h^3$ and a surface area of $ A \equal{} 12h^2.$ With $ A \equal{} 12\text{m}^2,$ we get a $ 2\times 2\times 1$ box of volume $ V \equal{} 4\text{m}^3.$

Asking for such a box with a given volume and the least surface area would give us the same answer - or at least the same shape of the answer.[/hide]

-----

We could phrase this as an inequality problem: for positive real numbers $ x,y,z,$ prove that

$ xyz\le \frac1{6\sqrt {3}}(2xy \plus{} 2xz \plus{} yz)^{3/2}$

[hide="Hint."]And it should be possible to find a way to use AM-GM to prove that.[/hide]

Yes Professor Merryfield, this is a typical example of application of Lagrange multipliers. Regarding your inequality:

[hide]For positive real numbers x, y, z, the numbers $ 2xy, 2xz, yz$ are also positive, and so the AM-GM inequality tells us that

$ \sqrt [3]{2xy \cdot 2xz \cdot yz} \leq \frac {2xy \plus{} 2xz \plus{} yz}{3}$

$ \equal{} > 4x^2y^2z^2 \leq \frac {(2xy \plus{} 2xz \plus{} yz)^3}{27}$

$ \equal{} > (xyz)^2 \leq \frac {(2xy \plus{} 2xz \plus{} yz)^3}{108}$

$ \equal{} > xyz \leq \frac {(2xy \plus{} 2xz \plus{} yz)^{3/2}}{6 \sqrt {3}}$

QED

This inequality shows that, for an open, paralelipipedic box, whose volume is V and lateral surface area is S, we have

$ V \leq \frac {S^{3/2}}{6 \sqrt {3}}$.

In this particular case, S = 12, and so $ V \leq \frac {(12)^{3/2}}{6 \sqrt {3}} \equal{} 4$, and the maximum volume is 4 (units). However, how does this help in finding the dimensions of the box?[/hide]

Carcul: the equality case for AM-GM is that all of the numbers being averaged are the same. So we have equality in the inequality if $ 2xy\equal{}2xz\equal{}yz.$ Hence, $ y\equal{}z\equal{}2x.$

Yet another argument:

Build two copies of the open-topped box. Invert the second over the first to make a closed box. This closed box has twice the surface area and twice the volume of a single box. We maximize the volume for a given surface area of the half-box precisely when we maximize the volume for a given surface are for the closed box. Symmetry considerations strongly suggest that that happens  when the closed box is a perfect cube. Working back from that, we see that the volume of the closed box is maximized when it has a square base and is half as tall as it is wide - that is, it is half of a cube.

The same argument suggests that if we allow the container to have any shape whatsoever, we get maximum volume for a given surface area if the open container has the shape of a hemisphere.

[quote="Kent Merryfield"]the equality case for AM-GM is that all of the numbers being averaged are the same.[/quote]

Of course, I was missing that. Thanks.


This was on page 504 of the AOPS Introduction to Geometry book

Since you're studying geometry and trigonometry,I will give you a problem. A ship sails the ocean. It left Boston with a cargo of wool. It grosses 200 tons. It is bound for Le Havre. The mainmast is broken, the cabin boy is on deck, the are 12 passengers aboard, the wind is blowing east-north-east, the clock points to a quarter past 3 in the afternoon.It is the month of may. How old is the captain?

-Gustave Flaubert

Does this havean answer?

What does this question have to do with trigonometry?

[quote="MrBob"]
-Gustave Flaubert
[/quote]
Flaubert was a novellist, so I think it is a joke...

He's 85. Quite simple really.

No, you see, this quote was used as an example of how people automatically assume that just because a word problem has a set of numerical conditions, that it can actually be solved. You fell into this very common trap, I'll give you a different problem to work on:

A marathon runner can run 42 yards in 1 minute. How far can he run in 1 hour and 32 minutes?

marathon runner?

is it 26 miles 385 yards?

EDIT : @rofler

oh, and why is the captain 85 years old??
a captain of a ship must be 85 years old??

Not only did you somehow manage to multiply 42 by 92, but 123456789 is right.

And you clearly missed the point of what I was saying if you're asking how I got him to be 85 years old.


[quote="kingkong"]Let ABCD be a tetrahedron and M be midpoint of BC. Prove that:
If d(M; (ADB)) = d(M; (ADC)) then BC $\perp$ (ADM).
(d(M;(ADB)) : distance from M to plane (ADB)).
[img]http://pic19.picturetrail.com/VOL1079/4259104/8941288/137802830.jpg[/img][/quote]

Right or wrong?

[quote="Shinta"]It is wrong.
Draw line Bx // AD, line Cy // AD. Get B' then get C' on Cy satifies that M is the midpoint of B'C'. If BC $\perp$ AD then you have B'C' $\not\perp$ AD. So if B $\equiv$ B' and C $\equiv$ C' then M be midpoint of BC and d(M; (ADB)) = d(M; (ADC)) but BC $\not\perp$ AD. Right? :D[/quote]

Thanks a lots!


geometry
3D geometry
dodecahedron
probability
expected value

I don't know if this was posted before.  :maybe:

You roll a dodecahedron dice with a different number on each face, with each face being equally likely to show up. What's the expected value of the number of times you have to roll the dice to roll all the different numbers?

[hide]I would say it is the inverse of the probability of rolling all the numbers:
$\frac{1}{(\frac{1}{12})^{12}}$.[/hide]

[quote="Pedro Ponte"][hide]I would say it is the inverse of the probability of rolling all the numbers:
$\frac{1}{\frac{1}{12}^{12}}$.[/hide][/quote]

Even if what you say is true, that's not the correct number.

It was just a guess. That should be wrong.

But that's a very interesting problem. If you are a "little" lucky you just have to roll it $12$ times. If you have bad luck you may pass your entire life rolling the dice.  :lol:

[hide="I had"]
(12^12)/(12!)
but that's over 18000 so it seems way too high
but I used the crappy calculator on my computer
so maybe I just punched it in wrong
[/hide]

I think it is more complicated than that. Let $E_{n}$ be the expected number of rolls it takes to get $n$ different numbers. Clearly $E_{1}= 1$. So then

$E_{2}= \frac{11}{12}(E_{1}+1)+\frac{1}{12}(E_{2}+1)$

because there is a $\frac{11}{12}$ chance that it will take just one roll more than it takes to get $1$ number and a $\frac{1}{12}$ chance that we will make a failed roll and start over again. This recursion can generalize to

$E_{n+1}= \frac{12-n}{12}(E_{n}+1)+\frac{n}{12}(E_{n+1}+1)$

or

$E_{n+1}= E_{n}+\frac{12}{12-n}$.

This makes sense because $\frac{12}{12-n}$ is the inverse of the probability of rolling a new number (i.e. the expected number of rolls to get a new number). So basically we can solve this out to get

$E_{12}= 12\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{12}\right) \approx 37.24$.

[quote="mustafa"]For these kind of problems, isn't the expected value infinity? Because it is possible that you just never roll a 6 or something?[/quote]

The probability that you never roll a 6 is also infinitely small...

[quote="mustafa"]For these kind of problems, isn't the expected value infinity? Because it is possible that you just never roll a 6 or something?[/quote]

It's possible. Just like it's possible you can flip a coin forever without getting tails, but that's not real life. And that's only one case when expected value is the average.

By the way, paladin8 has the right answer.


Connie spent the weekend making cookies. She made 60 sugar cookies, 80 chocolate
chip cookies and 100 peanut butter cookies. She plans to make packages
of cookies that each contain an identical assortment of whole cookies.
How many cookies are in a package, assuming that she makes as
many packages as possible and uses all of the cookies she made?

gcf(60,80,100)=20, so 20 packages.

Then there are 3+4+5=12 cookies per package.


Se defineste un sir de numere reale in modul urmator: u0=a , un=bun-1+c pentru n>=1.
a)Sa se determine coeficientii b si c pentru ca acest sir sa fie o progresie aritmetica si sa se calculeze in acest caz suma s1=u0+............+un.
b)Considerand sirul numeric dat initial sa se determine un numar d, in functie de b si c, astfel incat sirul cu termenul general un-d sa fie o progresie geometrica . In acest caz sa se exprime un in functie de n si de coeficientii a,b,c si sa se calculeze suma s2=u0+...............+un

Tiberiu Roman , Bucuresti

Daca $a_{n+1}= x a_{n}+y$, atunci $a_{n}= \left( a_{1}+\frac{y}{x-1}\right) x^{n-1}-\frac{y}{x-1}$. Putem sa demonstram chestia asta prin inductie, ori sa consideram progresia geometrica $a_{n}+\frac{y}{x-1}$ de ratie $x$.


Find $ n\in Z^ \plus{}$ and $ n > 1$ such that $ \frac {2^n \plus{} 1}{n^2}\in Z$

That is indeed a very hard problem. It was a problem on the IMO.
For solution look here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=55436
:wink:


In a circle of radius $ 5$ units, $ CD$ and $ AB$ are perpendicular diameters. A chord $ CH$ cutting $ AB$ at $ K$ is $ 8$ units long. The diameter $ AB$ is divided into two segments whose dimensions are:

$ \textbf{(A)}\ 1.25, 8.75 \qquad\textbf{(B)}\ 2.75,7.25 \qquad\textbf{(C)}\ 2,8 \qquad\textbf{(D)}\ 4,6$
$ \textbf{(E)}\ \text{none of these}$

[hide]

Let the center of the circle be $ O$. Since $ \triangle{CHD}$ has $ CD$ as the diameter, $ \angle{CHD} \equal{} 90^\circ$. Also, by Pythagorean Theorem, we have $ HD\equal{}6$. So, we have $ \triangle{CHD}\sim\triangle{COK}$. Hence, $ \frac {8}{5} \equal{} \frac {6}{OK}$ and $ OK \equal{} 15/4$. So, $ AK \equal{} 5/4$ and $ BK \equal{} 35/4$, or $ \textbf{(A)}$.

[/hide]


20 numbers are written in a row such that the sum of any three succesive numbers is negative.
Can the sum of all numbers be positive.

Answer without full solution:

[hide]Yes.  First consider the 5 number sequence 1000, -501, -501, 500, 0.  Their sum is positive!  Consider the features of this sequence (they are purposely apparent :)).  You should be able to generalize (in fact there is a specific sequence you can find that will work well).  [/hide]

Here's the next question:
Find the least value of $k$ such that:
We can put 20 numbers in a  row such that the sum of any 3 of them is negative and the sum of any $k$ of them is positive
Here's the solution:
[hide]Note that $k=18$ doesn't work, indeed $(a_1+a_2+a_3)+...+(a_{16}+a_{17}+a_{18})<0$ So $k\ge 19$.
Now here's example for $k=19$, $(a, a, -b,a, a, -b,..., a, -b, a)$ Put $a=1000,b=2001$.[/hide]


suppose {an} is a sequence of complex numbers such that ∑an diverges. Then what is the radius of convergence R of the power series ∑ an(z-1)^n /2^n

or

summation an into (z-1) raise to power n divided by 2 raise to the power n[/code]

bn=an/ 2^n
bn+1= an+1/ 2^n+1
by ratio test
bn/bn+1=  2* an/an+1
since series an diverges so an/an+1<1
so bn/bn+1<2
is it right?


A deck of 32 cards has 3 suits and 2 jokers.Each suit has cards numbered from 1 -> 10,and each joker is numbered 0.One chooses a hand of cards from the deck.A card numbered k has value 2^k ,and the value of a hand is the sum of the values of the cards in the hand. Determine the number of hands of value 2004

I think this problem is not hard,but i'm a chicken  :oops:
Who can help me?

[hide="hint"]the possible values of a hand are:
$ a_0 \times 2^0\plus{}a_1 \times 2^1\plus{} \cdots \plus{} a_{10} \times 2^{10}$
with $ a_0 \in \{0,1,2\}$ at most two jokers
and $ a_i \in \{0,1,2,3\}$ at most three different suits

On the other hand:
$ 2004\equal{}4\plus{}16\plus{}64\plus{}128\plus{}256\plus{}512\plus{}1024$[/hide]

[quote="mszew"][hide="hint"]the possible values of a hand are:
$ a_0 \times 2^0 \plus{} a_1 \times 2^1 \plus{} \cdots \plus{} a_{10} \times 2^{10}$
with $ a_0 \in \{0,1,2\}$ at most two jokers
and $ a_i \in \{0,1,2,3\}$ at most three different suits

On the other hand:
$ 2004 \equal{} 4 \plus{} 16 \plus{} 64 \plus{} 128 \plus{} 256 \plus{} 512 \plus{} 1024$[/hide][/quote]

I think it's just a hint,but i knew it,too.I want a certain solution,and result.  :(


ratio
geometry
trigonometry
analytic geometry

Take a $ \triangle ABC$.
Choose point $ D$ on $ AB$ such that $ \frac{AD}{AB}\equal{} 0.25$
Choose point $ E$ on $ BC$ such that $ \frac{BD}{BC}\equal{} 0.25$
Choose point F on CA such that $ \frac{CF}{CA}\equal{} 0.25$

Now draw the lines $ AE$, $ BF$ and $ CD$. Let their points of intersection make the $ \triangle PQR$.

What is the ratio of the areas of $ \triangle ABC$ and $ \triangle PQR$?

Do not use trigonometry or coordinate geometry..

See the attached diagram for additional notation.

$ PX, FU\perp BC$ and $ PY, FV\perp AB$

$ {x\over h_{a}/4}\equal{}{BP\over BF}\equal{} k$

$ {y\over 3h_{c}/4}\equal{}{BP\over BF}\equal{} k$

$ [BPE]\plus{}[BPA] \equal{}{[ABC]\over 4}\implies{kh_{a}\over 4}\cdot{a\over 8}\plus{}{3kh_{c}\over 4}\cdot{c\over 2}\equal{}{[ABC]\over 4}$

and from there

$ {k\over 16}\plus{}{3k\over 4}\equal{}{1\over 4}\implies k \equal{}{4\over 13}$

Now $ y \equal{}{3h_{c}\over 13}\implies [APB]\equal{}{3\over 13}[ABC]$

In the similar fashion, we find $ [BQC] \equal{} [CRA] \equal{}{3\over 13}[ABC]$, which gives

$ {[PQR]\over[ABC]}\equal{}{4\over 13}$


Find the equation of a circle that passes throu points (2,7) and (-6,1) and has a radius of 5sqrt2

[hide="Hint"]Consider the circles centered at (2,7) and (-6,1), respectively, each with radius $5\sqrt{2}$

What do their intersections have to do with the circle you are looking for?
[/hide]

one of there points of intersection should be The center of the circle we are looking for but i dont know how to find the intersection of the two circles. I was absent from school that whole week so i realy dont know how to get the intersection of a cricle and how to find the equation using several points on the circle

Write out their equations. You then have a system of two equations in two variables. You can then try subtracting them from each other, substituting, etc.


find the value of [1/(sin 10)]-4sin70[/list]

[quote="Kouichi Nakagawa"][Solution]
$= \frac{1-4\sin 70^\circ \sin 10^\circ}{\sin 10^\circ}$
$= \frac{1+2 (\cos (70^\circ+10^\circ)-\cos (70^\circ-10^\circ))}{\sin 10^\circ}$
[/quote]

how to get this? :/

From the sum of angles for cosine $\cos \left( \alpha \pm \beta \right) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta$ .


Prove Pascal's Identity: $ \binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$

[hide="hint"]
$ {n\choose k}=\frac{n!}{k!(n-k)!}$[/hide]

There is a classic combinatorial argument for this identity.

n-1 choose k-1 ways to choose k items with some designated item, n-1 choose k ways to choose k items without this designated item.

[hide]So first, you can transform $ \binom{n-1}{k-1}+\binom{n-1}{k}$ into $ \frac{(n-1)!}{(k-1)!(n-k)!}+\frac{(n-1)!}{k!(n-1-k)!}$. Next, I added both fractions together. This gave me $ \frac{(n-1)!k!(n-1-k)!+(n-1)!(k-1)!(n-k)!}{(k-1)!(n-k)!k!(n-k-1)!}$. I distributed $ (n-1)!(k-1)!(n-k-1)!$ from both parts of the numerator and got $ \frac{(n-1)!(k-1)!(n-k-1)!(k+n-k)}{(k-1)!(n-k)!k!(n-k-1)!}$, which simplifies to $ \frac{n(n-1)!}{(n-k)!k!}$, which simplifies to $ \frac{n!}{k!(n-k)!}$. This equals $ \binom{n}{k}$.[/hide]