MATHCOUNTS
ARML
geometry
probability
3D geometry
number theory

Regarding team round:

I am the captain of my national state team, and even though MATHCOUNTS does not designate the captain to be much of absolute power other than recording official answers, my team pits me into a lot of responsibility. Not only do I have the call for writing the team answers, I also sometimes have trouble regarding whose answers should be trusted whenever there's a dispute (yes, I am the person in charge of resolving disputes).

Does anyone have any advice? I mean, in my case no one on our team seems to really dominate in ability but whenever there's a dispute, and time is running short, whose answer should we trust? (The first person, the person who corrected the problem, or me?) It's pretty clear that I have to come up with a final answer before time runs out, and every answer in nationals counts. We always strive to write something for every answer blank even if we're not sure, so we never leave anything blank.

A suggestion: try to get accustomed to your team members strengths - e.g. geometry, algebra, number theory, even calculator bashing/programming.

EDIT: Yay 2000th post

One questioin, do u know anyone in ur state team, as in, do u get to meet the people in practices?

If so, i would suggest that u look at how smart ur team members are. For example, if u know one of the team member is exceptionally good at geometry, and there is a geometry problem on the team round that you have 2 answers to, i would suggest going with the answer from the smarter person. Now i know that this wont work everytime, but the probability is higher with the person that gets those kinds of problems most often. However, the advice i gave you is only when you have little time left and need to answer quickly, otherwise, have a third person solve it.

I echo the sentiment that getting to know your team members is the best way possible.
Also, for my 3 years, I was never the captain, but we used a system in which we had a grid, 10 problems, four people, when you found an answer, you put it on the grid (paper in the center), once two or three people got the same answer, you can cross it out on the grid and write it down.
If two people disagree, perhaps getting a third person to tiebreak may be the best choice, or have the two check their work, if still resolution is not reached, explaining their work to each other?  I think the key thing is realizing that these discussions shouldn't take more than 1 minute, otherwise you're wasting time.  At some point, you're just going to have to pick one.
I don't think there's a need for the team captain to get involved in all of these discussions either, maybe only to make a decision at the very end... You want to resolve conflicts with as few people in the shortest amount of time possible

[quote="Quickster94"]I echo the sentiment that getting to know your team members is the best way possible.
Also, for my 3 years, I was never the captain, but we used a system in which we had a grid, 10 problems, four people, when you found an answer, you put it on the grid (paper in the center), once two or three people got the same answer, you can cross it out on the grid and write it down.
[/quote]

hey that's what our team does too!  :D

We had something like this, all of us were not sure on the answer for #8 on the team round this year at state, so we just put down one of the answers (there was little time left)

[quote="tinytim"]@Gauss1181:

Well, based on what you are saying, it seems to me that you are the strongest member of your team (you got first at states right?). If that is the case, then you should just solve the problem on your own if there is a dispute and trust your answer.[/quote]
No, I got second written (but best out of us 4 in countdown), but based on our practices, the people on my team thought I would be the best choice for team captain/resolving disputes. However, my question/problem is how I should resolve those disputes because no one on our team seems to show pure/clear dominance (like Bobby Shen/Darryl Wu style) in doing team round questions. We aren't sure who we should trust, whether it's the first person who did the question, the person who made the correction/thought the answer was wrong, or me. But our case specifically is when I have to resolve a dispute between 3 or 4 people [b]when time is running really short[/b].

the way our coach does it is the person with the best handwriting...well no one in our teams handwriting is gorgeous...so we kinda just picked a random person..(thank God its not me) and well thats how we went with it...

although u have the responsibility of writing down the answers at the end, if u dont write them  down, its not completely your fault but the entires team for arguing so much...

Source:my coach...

So in my chapter/state team, I'm the captain (even though I have the least MATHCOUNTS experience and the youngest), and divide problems up as first 5, odds, evens, last 5. (in order of ease, so in reverse order of mathematical capability). We somehow finish within the first 5 minutes of team, and then for 10 minutes double-check, where everyone does problems they didn't do, and finally for the last 5 minutes we all sit down and play Super Mario.  :roll: Go figure.

[quote="pythag011"]First of all, there are team captains?

Second of all, I suggest that you look at both their solutions and check them. Don't spend too much time arguing about who is correct though.. (I spent the whole team round at an arml practice once arguing with someone over answers (and 24 still doesn't equal 1...))

Also, how are team captains chosen?[/quote]
at least it wasn't at the real thing (argh power contest last year)

Hm. I was captain of my team by power of seniority, and we use the grid system. Both at chapter and state our team got the highest team round score, and I the system does help with efficieny and speed. Getting to know their strengths is a good thing, I would assign what problems people would do, but I don't know if other people do that. And most importantly, get to know your team mates! :)

I'm the captain for Puerto Rico. How would you program your calculator??

There are programmable calculators, like the TI83/84 brand.

Yeah, but how exactly do you program them? Do you have an example?

2009 Chapter Team #9
9. In the equation $ w^3 \plus{} x^3 \plus{} y^3 \equal{} z^3$, $ w^3$, $ x^3$, $ y^3$ and $ z^3$ are distinct, consecutive positive perfect cubes listed in ascending order. What is the smallest possible value of z?

A small trivial program:
:For(X,1,100)
:If x^2+(x+1)^2+(x+2)^2=(x+3)^2
:Disp x+3
:End

The best way to tiebreak is to always go with who you have a crush on. Failing that, trust someone else over your own answer. Or go with which sounds like a more likely answer. But this only applies if time is majorly short. :)

@ a lot of you guys: the point is not "how do I distribute the problems in the team round?" it's "how do I choose between multiple answers for one question?"

[quote="pythag011"]
Also, how are team captains chosen?[/quote]

Hmm, I guess I was wrong. I thought 1st place at state was captain (like on my team, that is me). I guess the team and coach just chooses then. That make sense though- rather than automatically having 1st place be captain, the team should choose someone who shows leadership, has good handwriting, listens to all opinions, and is respected by the other team members, not to forget to mention be one of the better, or the best, at mathCounts on the team (which isn't always necessarily the state winner- they could have been lucky, and also, nationals is a different level from state- some ppl may be better at these tougher problems).

I'm on the same team as pinkmuskrat.

Our coach chose pinkmuskrat as the team captain, and I am the scribe. I don't know why they didn't merge the two jobs, but  oh well. Our coach chose me as the scribe because I have the best handwriting  :winner_first: . In our team, the team captain is in charge of choosing which problems people do, based on our strengths and weaknesses. I think we should also employ the grid paper method. We don't really have disputes, though.

[quote="solafidefarms"]The best way to tiebreak is to always go with who you have a crush on. [/quote]

but what if you have an all guy team (and all the guy's are straight)? ;)

and i'm captain of my team, and i'm definitely not the smartest. it's just 'cause everyone else's handwriting sucks. but do captains have to be chosen on intelligence or handwriting?

[quote="james4l"]2009 Chapter Team #9
9. In the equation $ w^3 \plus{} x^3 \plus{} y^3 \equal{} z^3$, $ w^3$, $ x^3$, $ y^3$ and $ z^3$ are distinct, consecutive positive perfect cubes listed in ascending order. What is the smallest possible value of z?

[/quote]

On a TI-89 Titanium, you just type in (the home window):

solve((x-3)^3+(x-2)^3+(x-1)^3=x^3,x)

Output is the answer.

There is a problem when none of your team has a TI-89.

However in actuality TI-84 can do this, but I just wanted to demonstrate a basic "program".

Our team (includes pythag) hasn't even had a team meeting so I suppose that is when captains are chosen  :wink:

everyone on my team but me has a TI-84...

....mines not...when we put down the answers during the team round....everyone asked me 2wice what they were to make sure cause they couldnt read it...im kinda starting to think i missed a couple problems because of my handwriting...

@urc: :rotfl:

[quote="james4l"][quote="AIME15"]For choosing team captains: It's probably best to choose the one who works best with the rest as team captain, not necessarily the strongest in math.

@james4l: WHAT? NO TI-89 TITANIUM?

maybe i'll lend mine to you for CA team at nats[/quote]

I have a TI-nspire (which is low-quality calculator) and a TI-84 Plus (which is medium-quality calculator) but no TI-89 Titanium (which is high-quality calculator)[/quote]

I have a TI-89 Titanium but no mathematica.

[quote="pythag011"][quote="james4l"](quote)

I have a TI-nspire (which is low-quality calculator) and a TI-84 Plus (which is medium-quality calculator) but no TI-89 Titanium (which is high-quality calculator)[/quote]

I have a TI-89 Titanium but no mathematica.[/quote]

I have Mathematica but no TI-89 Titanium.

[quote="pinkmuskrat"][quote="Urc"][quote="solafidefarms"]The best way to tiebreak is to always go with who you have a crush on. [/quote]

but what if you have an all guy team (and all the guy's are straight)? ;)
[/quote]

Nice, Urc. LOL :D I really like that comment there. That's the same with my team. (I think?!)Regarding team captains, I think intelligence rules over handwriting, as long as handwriting is not absolutely terrible. I mean, it's just writing answers down, and they have to do it in sprint and target. Their handwriting MUST be okay if they were able to make nats.[/quote]

believe me, there are people on my team with HORRIBLE handwriting (sometimes their 4's look so bad the judge person thinks it's a 9). so they elected me as team captain because they don't wanna risk that, i guess.

This year especially, chapter and states don't mean a thing. They were more of a test to see who can make the least mistakes.

[quote="pinkmuskrat"][quote="Urc"][quote="solafidefarms"]The best way to tiebreak is to always go with who you have a crush on. [/quote]

but what if you have an all guy team (and all the guy's are straight)? ;)
[/quote]

Nice, Urc. LOL :D I really like that comment there. That's the same with my team. (I think?!)Regarding team captains, I think intelligence rules over handwriting, as long as handwriting is not absolutely terrible. I mean, it's just writing answers down, and they have to do it in sprint and target. Their handwriting MUST be okay if they were able to make nats.[/quote]

Yeah... btw, I'm straight. Unless the other two aren't (which I highly doubt!), we have that kind of team.
Oh, and you don't need good handwriting to get into nats. The cut off for Guam was 18. How hard is it to get 18 with bad handwriting...
And our #3 doesn't have the best handwriting...

But anyway, I agree with you. If you are smarter, then you should make the decisions!

pinkmuskrat wrote:
[quote]pythag011 wrote:

[quote]Also, how are team captains chosen? [/quote]

Hmm, I guess I was wrong. I thought 1st place at state was captain (like on my team, that is me). I guess the team and coach just chooses then. That make sense though- rather than automatically having 1st place be captain, the team should choose someone who shows leadership, has good handwriting, listens to all opinions, and is respected by the other team members, not to forget to mention be one of the better, or the best, at mathCounts on the team (which isn't always necessarily the state winner- they could have been lucky, and also, nationals is a different level from state- some ppl may be better at these tougher problems).[/quote]

Hmmm. Are the captains are decided by the coach only in certain states?  It looks like everybody has a different way of choosing the captain...

BTW: I'm straight.


algorithm
algebra
polynomial
modular arithmetic
function
number theory
greatest common divisor

I am having trouble using the division algorithm in $ \mathbb{F}_3$:

For $ x^6 \equal{} f(x)*g(x) \plus{} r(x)$ where $ f(x) \equal{} x^5 \plus{} 2x^4 \plus{} x^3 \plus{} x^2 \plus{} 2$, I want to find what $ g(x)$ is, and from that find the remainder $ r(x)$, for everything in $ \mathbb{F}_3$. I know the answer is: $ g(x) \equal{} x \plus{} 1$ and so $ 1 \plus{} x \plus{} 2x^2 \plus{} x^3$. This answer is easy to verify, but I don't know how you would get it. I guess you would use division, but I get $ g(x) \equal{} x \minus{} 2$ and $ r(x) \equal{} x^4 \minus{} x^2 \plus{} 2$ which is clearly different. What's the problem?

Thanks!

Your quotient isn't different. Your remainder is calculated incorrectly.
Long division is the right algorithm; it's the arithmetic you're getting wrong.

In the first division, I can see that there is an error with the remainder, because the coefficients are not in $ \mathbb{F}_3$ , but everything else looks ok. In the second one you still have $ \minus{} x^5 \plus{} x^4 \plus{} x^2 \minus{} 2x \plus{} 2$ when you subtract $ x^5 \plus{} 2x^4 \plus{} x^3 \plus{} x^2 \plus{} 2$ from $ \minus{} 2x^5 \minus{} x^4 \minus{} x^3 \minus{} 2x$, and the remainder here should be $ x^3\plus{}2x^2\plus{}x\plus{}1$! Where is the error? What should I do?

In the first, you didn't reduce mod $ 3$. In the second, you added instead of subtracting.

ah, i see it. Thanks for your answers :)

Well you could reduce the polynomials before using euclid to simplify things:

x^6 = x^3 (x^3-x) +x^4 = x^4 (by FLT) = x(x^3 - x) + x^2 = x^2 (again FLT)

and do the same for the other polynomial to make it easy.  It is useful to know that any polynomial in F_n can be reduced to a polynomial of degree at most n-1 by a similar fashion.

No, you can't. There is a difference between polynomials and polynomial functions. We might be thinking about the values on the field with $ 3^{60}$ elements for all you know, and those identities are false there.

In other words, all we are assuming is that the [url=http://mathworld.wolfram.com/FieldCharacteristic.html]characteristic[/url] is $ 3$.

well, a way to simplify would be nice. For example, to find that [b]241[/b]=10*[b]23[/b]+11, we [b]needn't [/b]use the division algorithm. It's just a by inspection sort of thing (choose the multiple of 23 to be 10, then the remainder is 11.). I wish there was an analogy to this for polynomials, but it appears that there is not, i.e. given

$ x^6\equal{}(x^5\plus{}2x^4\plus{}x^3\plus{}x^2\plus{}2)g(x)\plus{}r(x)$, come up with some g(x) and r(x), rather than computation with division algorithm:

 if $ g(x)\equal{}x$ we have $ x^6\equal{}(x^5\plus{}2x^4\plus{}x^3\plus{}x^2\plus{}2)\cdot x\plus{}r(x)\equal{}(x^6\plus{}2x^5\plus{}x^4\plus{}x^3\plus{}2x\plus{}r(x)$ and so $ r(x)\equal{}x^5\plus{}2x^4\plus{}2x^3\plus{}x$. But this is clearly wrong, since it is reduced $ \pmod{3}$ and does [b]not [/b]correspond to the previous (correct) answer $ r(x)\equal{}1\plus{}x\plus{}2x^2\plus{}x^3$...
Unless something here too is wrong?

The integer division you outlined is essentially a "guess" of the desired quotient.  This is easy in the integers because the integers are well-ordered; we have no such ordering in the polynomials.

You haven't fully divided out.  In your new computation, your remainder has degree $ 5$, but so does your divisor.  $ g(x) \equal{} x \plus{} 1$ is the correct choice.

for simplification you can take modulo $ p$ in $ \mathbb{Z}/p\mathbb{Z}$ after each subtraction.

A propos the last post: yes, it would be more time consuming that the division algorithm to guess a function $ g$, take the modulo of this and see if the degree of the (resulting) remainder is greater than $ g$, because polynomial multiplication is expensive $ (\mathcal{O}(n^2))$ steps, so the time would be multiple of this. Compare this with the few steps of division alg.

In fact, the full gcd algorithm for polynomials (over any field) is only as costly as a single polynomial multiplication. If you want to track more and find $ a,b$ such that $ ap\plus{}bq\equal{}\mathrm{gcd}(p,q)$, that's only a constant multiple of the effort.


geometry
perimeter
circumcircle
geometry unsolved

1Let ABC be a triangle ,H the orthercenter and M the midpoint of AC . Let l be the line

through M parallel to the bisector of angle AHC . Prove that l divides the triangle in two parts of equal

perimeter

2.Let M be the midpoint of the median AD of  ABC . The line BM intersects side AC at point N. Show that AB

is tangent to the circumcircle of NBC if and only if BM/BN=(BC/BC)^2

Problem 2: Please check your problem: (BC/BC)^2??
Problem 1:Construct the bisector $ BD$ of $ \angle ABC$. By angle chasing we can show that $ BD//HE$ so $ l//BD$
Let $ BC$ be extended to $ N$ so that $ BA \equal{} BN$ then $ l//BD//AN$.
But M is the midpoint of $ AC$ thus $ NL \equal{} LC$
$ \Rightarrow AB \plus{} BL \equal{} LC$
We end the proof  :)

[quote="poonprang"]Sorry BM/BN=(BC/BN)^2[/quote]
How? When $ \frac{BM}{BN}\equal{}\frac{3}{4}$ but $ (\frac{BC}{BN})^2\equal{}\frac{AC}{AN}\equal{}3$???

You meaned that M is the midpoint of median AD. So pls check this problem again !

Sorry I will check problem

Why  (BC/BN)^2=AC/AN


I am sorry for this request, but I would be very interested to know where I can find the problems of the [b]Belgium[/b] (not: Flanders) [b]Mathematical Olympiads[/b] from the last years (from 2000; most importantly, 2002-2004). All what I have found is a website, http://site.ifrance.com/omi-omb/SommaireENGL.htm , which doesn't seem to have any of the problems and which was last updated in 1999. I can read English, German, a bit of French, and quite often I understand Dutch (at least in geometry papers).

  Thanks!
  Darij

as it seems that this request has not been answered yet: this year's problems can be downloaded [url=http://dev.ulb.ac.be/urem/?Finale-de-la-32e-Olympiade]here[/url] (scroll to the bottom of the page) As far as I know, it's the first time the problems have ever been published on the net.


What's the next number in the sequence
1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, _

It's so difficult !! I've been lucky to find good answer !  :lol:
No explanation needed, it's just additions ...

[hide="solution"]Add the two previous terms up, and take the units digit  :D [/hide]
are you sure you're in middle school  :?

I'm sorry I've gone on this post by the page "unanswered post" and I've not seen the forum, further more I don't know really what's middle school, high school, ... compared with french system.

i don't know the french system but middle school usually means you're 11-13. high school means ages 14-18.

[quote="shyong"][hide="solution"]
Fibonanci Sequence taking only last digit [/hide][/quote]

Nice, i didn't notice that  :blush:
and it keeps following that rule.


geometry
3D geometry
linear algebra
matrix

Here's a problem I got from the Eli[b]as[/b] Saabs webpage. I've done it and got the answer, but I did it by listing, and that doesn't seem to be the way to do it. Does anyone know an easier way to do it? (besides listing)

Positive integers x,y,and n are each less than 20. If $x^2-y^3=n$, find the median of all possible values for the sum x+y+n

[hide="my answer"]
20
[/hide]

Sorry rcv, I typoed the y^2. It should be y^3

[quote="dwx314"]Does anyone know an easier way to do it?[/quote]

Making an organized list is a reasonable way to solve this problem.  But you must use your intellect to avoid listing unnecessary cases.

[hide]Remember $1\le x\le 19,\ 1\le y\le 19,\ 1\le n\le 19$.

What if $x=y+1$?  From the difference of consecutive squares, n will take odd values, beginning at 3.  Make a table:

$\begin{matrix}\\
x&y&n=x^2+y^2&x+y+n\\
2&1&3&6\\
3&2&5&10\\
4&3&7&14\\
5&4&9&18\\
6&5&11&22\\
7&6&13&26\\
8&7&15&30\\
9&8&17&34\\
10&9&19&38\\
11&10&21&--
\end{matrix}$

We don't have to go any farther, because n exceeds 19.

Now, what if $x=y+2$?

$\begin{matrix}\\
x&y&n=x^2+y^2&x+y+n\\
3&1&8&12\\
4&2&12&18\\
5&3&16&24\\
6&4&20&--
\end{matrix}$

Then, $x=y+3$:

$\begin{matrix}\\
x&y&n=x^2+y^2&x+y+n\\
4&1&15&20\\
5&2&21&--
\end{matrix}$

Finally, no higher values ($x=y+4$, $x=y+5$, etc.) produce a value of n less than 20.

There were only 13 items to be discovered, and a systematic method found them all quickly.  Now, we just need to find the median of these 13 values.  The "trick" is seeing the patterns and getting organized quickly.[/hide]


function
logarithms
ratio
parameterization
algebra
functional equation
real analysis

Put $f(0)=0$, and, as $y\rightarrow 0$, $\frac {x-y}{\ln x-\ln y}\rightarrow 0$, so $f(x)=0$, for all $x$, so only constant functions satisfy.

oh, so $f$ is defined on $\mathcal{R}_+^*$ ?

i should have known it's not that simple  :blush: .

Note first that $\frac{x-y}{\ln x-\ln y}$ is always strictly between $x$ and $y$. Using this, we have that either $f$ is constant or $f$ is strictly monotone. The functional equation is linear, so linear combinations of solutions are solutions.

[b]The only solutions are constants.[/b]
For a fixed constant $a$, we use the functional equation to compute $f(\frac xa)+2f(x)+f(ax)$ two different ways. If we merge the pairs $(\frac xa, x)$ and $(x,ax)$, we get $f(\frac xa)+2f(x)+f(ax)=4f(bx)$ where $b=\frac{(a-1)^2}{a\ln^2 a}$
If we first merge $(\frac xa, ax)$, then merge the result with $2f(x)$, we get $f(\frac xa)+2f(x)+f(ax)=4f(cx)$ where $c=\frac{a^2-2a\ln a-1}{2a\ln a\ln\left(\frac{a^2-1}{2a\ln a}\right)}$.

Of course, $b\neq c$. For $a=1000$, $\frac bc$ is approximately $1.2546$, and this ratio changes smoothly with $a$. We don't even need continuity, since we can show that $f(tx)=f(x)$ for all values of $t$ between 1 and 2; $\frac bc$ tends to 1 at $a=1$ and 2 at $a=\infty$.

[quote="harazi"]Find all continuous functions $f$ that verify for $x,y>0$ the relation $ f(x)+f(y)=2f(\frac{x-y}{lnx-lny})$. I think it is me the idiot, but I have no idea about their form.[/quote]

The N'th iterate of the transformation $(x,y) -> (x',y')$ of the recurrence
from Vess' thread, works for [b]real[/b] values of $N$ with the same formulas.
In other words, for any 2-parameter family $a(N),b(N)$ with

$a(N) - b(N) = 2^N p$     and

$a(N)/b(N) = q^{2^N} $   for constant $p,q > 0$,

then  $f(a(N)) + f(b(N)) = f(C)$  where $ C = p / log(q))$.
Such $a(t),b(t)$ parametrize the curves $ \frac{x-y}{lnx-lny} = C$.

[quote="harazi"]Find all continuous functions $f$ that verify for $x,y>0$ the relation $ f(x)+f(y)=2f(\frac{x-y}{lnx-lny})$. I think it is me the idiot, but I have no idea about their form.[/quote]

Harazi, see my and Jmerry postings in the original thread of Vess.


I already added you.  So all is good.  :-)

[quote="314159265358979323846"]Hi, can I join this group if I haven't already?
Am I supposed to be able to post?
I guess I'll find out.  Right now.[/quote]

[quote="tom11784"]Hi all.  For those of you don't know me I'm Tom and as you can see if you get here soon my work place can get very boring and allow me to think out the posts so that I'm last to respond to all 6 threads in this forum.

Dang I need to get a life.  :([/quote]

But if I reply, then you won't be last anymore! Does that mean that you'll have to reply again? If so, you can thank me for providing you with more entertainment at work! (And this is very silly of me, because I should be replying at work instead of during my lunch break.)

(And apologies to anyone who cares for making this entirely off topic. :-))

[quote="tom11784"][quote="intrepia"]Does that mean that you'll have to reply again?[/quote]Nah - I was last on all six for over 135 minutes - no need for more.[/quote]

It's a lot easier to do that here than at the RPI '07 forum of 'yore.  :-D.

yes - that was a truer sign of my boredom's dominance


Let {Z(k)} be a finite sequence of integers such that Sum{Z(k)}=0
Prove that Sum{(Z(k))37} is divisible by 399 .

sum(Z(i)^37-Z(i)) is divisible by 399 by Mathematical Induction.


Euler
number theory unsolved
number theory

[quote="rhbestrh"]Please help ![/quote]

$ \gcd\left(k,\phi(m)\right)\equal{}1$, where $ \phi(m)$ is the Euler function.


AMC
USA(J)MO
USAMO
AIME
Harvard
college
AMC 12

Now that the AIME II is over, we can discuss what we think the cutoff is. I'll start by saying 210. Thoughts?

By the way, when is the cutoff announced?

Yan Zhang

There just doesn't seem to be a limit to the number of math geniuses on this site...

In any case, were you referring to the 1st 160 cutoff Yan?

We got an insider from TJ now...

I don't think he is THE Yan Zhang, since that Yan has graduated high school and is at Harvard now.

I'm pretty sure he's the Yan Zhang in \mathbb{R}^- (i.e. he has also been to MOP).

I'll guess 215.

I believe Simon's guess to be slightly high.  I will go with 200-210.

a little off-topic, but do you get a certificate for participation in USAMO?  btw I am hoping the cutoff is at 200

Wait...are we talking about the cutoff for the first 160?

Yes, they do give certificates of participation for the USAMO. Actually, I don't think I ever got one for 2002 for some reason.

I keep on hearing about a 150 given to all sophomores and under who took the AMC 12 for the purposes of the USAMO index.  Is this true?

I am pretty sure that isn't true. What they may be refering to is that if you are a sophmore or less then getting a floor AIME gets you into USAMO (no matter how low AMC is, unless too many people qualify)

Right. But what that translates to is basically giving everyone in grade 10 and below an automatic 150 since there will almost certainly be someone who did well on the AMC12 but poorly on the AIME and still made USAMO.

[quote="Fierytycoon"]Geez..that fact that people always ask the same old questions about the AMC exams really shows how we need a good FAQ about this sort of thing.[/quote]

You're quite right. I am trying to work on part of the FAQ, describing the AMC contests in general, but I am mystified myself by the selection rules for the USAMO, so I will simply say here that anyone who can put a FAQ together about that will have my thanks.

[quote]I am mystified myself by the selection rules for the USAMO[/quote]

Hmm, what exactly are you confused about?

[quote="ComplexZeta"]Right. But what that translates to is basically giving everyone in grade 10 and below an automatic 150[/quote]

Yes! You've got it.

Marma103, you are right about what the rules say, but the rules themselves are apparently incorrectly stated.  We had a previous discussion of this topic in the following thread:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1831[/url]

The consensus view was that by "students who took AMC 10", the AMC rules meant "students in the 10th grade or below who took AMC 10 or 12".

Thank you Ravi, that's exactly what I meant... and thanks for the link, too.

They should've change the wording, though. I know kids who chose to take AMC-10 because of that wording.

[quote="texas137"]the first year that there were separate exams, there was a big advantage to taking the 10. Since then, the cutoffs have been sufficiently lower on the 12 than on the 10 that it is easier for students in grade 10 and below to qualify for the AIME by taking the 12. Once they take the AIME, they only have to make the AIME floor, not the USAMO index, so that removes any disadvantage they might have from a lower numerical score on the 12 than they would have had on the 10.[/quote]

This is correct, correct history, and correctly stated.

Several of us in the office go separately through several stages of checking, double-checking and cross-checking to make sure that no deserving student, and in particular no 7th, 8th, 9th or 10th grade student misses a valid invitation to USAMO.

Steve Dunbar
AMC Director

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamoqualstate.html

[quote]261 students from 190 schools
The USAMO index cutoff for participants was 210.
In addition, students in grades 7th through 10th achieving an AIME score of 7 or above
were also selected to write the USAMO.
[/quote]

Did it make a difference between AIME I and AIME II this year?

[quote="compucomp"]Wow, 210, I called it right!

I must be a prophet. :)

Yan P. Zhang[/quote]

I called it too, though I must admit to not being a prophet.

YES!

I squeezed past...barely...with a 7 on the AIME. (I'm in 8th grade.) I just got the email today (and I checked the website.)

Would it be possible for an AIME qualifier but non-USAMO qualifier to take the USAMO unofficially, but on the official day. Also, could a person take the AIME I unofficially and the AIME II officially?

Depends on what you mean by "unofficial". You can arrange for your school to put you in a closed room for the two days of 4.5 hrs that the USAMO takes. The problems become available on the webjust before the start, so you could certainly do this. Of course, you would have to wait for solutions before grading your own test. As for the AIME, same thing, just that if other students are officialy taking a test, you may not be able to sit in during it unofficialy.

When I was a sophomore, I was the only USAMO qualifier at my school, and my teacher called the AMC office to ask if someone else could take it since he had missed it by two points. They said that he could take it unofficially, but he had to take it later. But questions weren't on the web back then.

[quote="Pork_Chop8"]Also, could a person take the AIME I unofficially and the AIME II officially?[/quote]

I think this depends on whether you find a testing site with an extra exam that's willing to let you do this.  Usually the questions from the AIME-I are on AoPS and probably other online places a day after the actual exam, so if you wanted to take it for practice, you could lock yourself in your room for three hours =) .. I'm pretty sure people (such as MithsApprentice) have sat for both exams, but had their AIME II score count - you need to decide this before taking either, though.

official AMC rules state that the exam manager is only allowed to administer the AIME or USAMO to students who qualified for it. Now what you do on your own time, or what you work out informally with a teacher who is not administering the exam officially to anyone, is up to you.

[quote="JP"]Do the AIME I and the AIME II had the same cutoff?  This year's AIME II was much harder than AIME I.[/quote]

The cutoffs were the same.  While the AIME II did appear (at least statistically) to be harder than the AIME I, the deviation was not so large to expect more than a fraction of a problem difference for students between the exams.  For that reason I doubt they spent very long considering a different cutoff.

The floor for 10th graders and below was a 7 regardless of which exam was taken.


Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
1)\[ \sqrt [5]{\frac {a^4b \plus{} a^4c \plus{} b^4a \plus{} b^4c \plus{} c^4a \plus{} c^4b}{6}}\geq\sqrt [4]{\frac {a^3b \plus{} a^3c \plus{} b^3a \plus{} b^3c \plus{} c^3a \plus{} c^3b}{6}}\]
2)\[ \sqrt [5]{\frac {a^4b \plus{} a^4c \plus{} b^4a \plus{} b^4c \plus{} c^4a \plus{} c^4b}{6}}\geq\sqrt [3]{\frac {a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b\minus{}2abc}{4}}\]

[quote=arqady]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
1)\[ \sqrt [5]{\frac {a^4b \plus{} a^4c \plus{} b^4a \plus{} b^4c \plus{} c^4a \plus{} c^4b}{6}}\geq\sqrt [4]{\frac {a^3b \plus{} a^3c \plus{} b^3a \plus{} b^3c \plus{} c^3a \plus{} c^3b}{6}}\]
[/quote]

Use  AM-GM  and  Cauchy-Schwarz: $$\bullet 5\sqrt[5]{\sum\frac{a^4(b+c)}{6}}+3\sqrt[3]{\sum\frac{a^2(b+c)}{6}}\ge 8\sqrt[4]{\sum\frac{a^3(b+c)}{6}}$$
$$\bullet 4\sqrt[4]{\sum\frac{a^3(b+c)}{6}}+2\sqrt{\sum\frac{a(b+c)}{6}}\ge 6\sqrt[3]{\sum\frac{a^2(b+c)}{6}}$$
$$\bullet \sqrt[3]{\sum\frac{a^2(b+c)}{6}}\ge \sqrt{\sum\frac{a(b+c)}{6}}$$


calculus
integration
trigonometry
geometry
trapezoid
calculus computations

Help me to solve the double integral, thanks.
  $ I \equal{} \iint\limits_{R}{\cos \left( \frac {y \minus{} x}{y \plus{} x} \right).dx}$ where R is the trapezoidal region with vertices   $ A\left( 1,0 \right);\,\,B\left( 2,0 \right);\,\,C\left( 0,2 \right);\,\,D\left( 0,1 \right)$

One thing I'd try (without knowing yet how well it will work): make the substitution $ u \equal{} x \minus{} y,v \equal{} x \plus{} y,$ which is a linear map that converts that trapezoid to a different trapezoid. That doesn't rule out using another substitution later on.


modular arithmetic
geometry
3D geometry
arithmetic sequence
algebra
factorization
sum of cubes

Find an infinite arithmetic sequence such that no term in the sequence is either the sum of two squares or the sum of two cubes.

[hide="Two solutions"]No number which is $\equiv 3\pmod{4}$ can be sum of two squares since $n^{2}\equiv 0,1\pmod{4}$. No number which is $\equiv 3\pmod{7}$ can be sum of two cubes since $n^{3}\equiv 0,\pm 1\pmod{7}$. Therefore the sequence $a_{n}=28n+3$ would meet our demands.

Sum of two cubes is also $\not\equiv 4\pmod{7}$, hence $b_{n}=28n+11$ also works.[/hide]

[quote="boostage"][quote="matt276eagles"][hide]$f(n)=2(2n+1)=4n+2$ is a solution because every term is divisible by two but not four.

Generally, $g(n)=p(pn+1)=p^{2}n+p$ is a solution for prime $p$ because every term is divisible by $p$ but not $p^{2}$.[/hide][/quote]
I don't understand. Could you please explain why that makes the number inexpressible as the sum of squares or the sum of cubes?[/quote]

If $f(n) = 4n+2$, then $2 | f(n)$ but $4 \not| f(n)$. However, $\exists m\in\mathbb{Z}: f(n) = m^{2}, 2|f(n)\ \Rightarrow\ 4|f(n)$ and $\exists m\in\mathbb{Z}: f(n) = m^{3}, 2|f(n)\ \Rightarrow\ 8|f(n)$. Contradiction.

I had a perfectly good solution that I've later deleted due to a inexplicable massive electric discharge in my brain. Reposted in Post #2.

[quote="boostage"]But that only proves that no term in the sequence is itself a square or cube. But it can be the sum of [b]two[/b] squares or cubes. For example, $10$ is a number in the sequence $4n+2$, and $10=1^{2}+3^{2}$.[/quote]

Ah, yes, you're right.  I misread the problem.  Sorry for the confusion Farenhajt.


$\frac{1}{1+\frac{1}{2+\frac{2}{2+\frac{2}{4+\frac{4}{\cdots}}}}}$

Let the value of that be s.

$\frac{s}{s+\frac{s+1}{s+1+\frac{s+2}{\cdots}}}=v$

Compute

$\frac{1}{s+\frac{1}{v+\frac{1}{s+\cdots}}}$

Um, I'm not sure I understand what numbers are being used in the definitions of these numbers.

What is the source?  Do you have a solution?


Let be a non-negative integer. Determine a closed expression for the value of the sum:

S_n = \sum^{n}_{k=0} binom(n,k)/((k+1)*(k+3)).

Let fn(X)=(X+1)n
S_n= \int (X* \int fn(X)) taken for X=1

I am sorry, amfulger. But your expression is wrong. I guess you have to waste another sheet of paper.  :D

It's [tex]\frac{2^{n + 3} - n - 4}{2(n + 2)(n + 3)}[/tex].

Very similar to Canada MO, 1997, Problem 5: [url]http://www.kalva.demon.co.uk/canada/can97.html[/url]


What is the smallest number that can't be written as sum of five different perfect squares, but can be written as sum of both six and seven different perfect squares? I am looking for a strategical approach, mine is trial and error(Do not use 0). [hide="My answer"]224[/hide]

Every positive integer can be written as sumof five perfect squares. So what's the point¿

I am very sorry, squares must be different, of course. I have also corrected the question statement above.


Given the equation $(ax+1)^2=a(1-x^2)$ ($a>1$), prove that the positive root of the equation is less than $1$ while the negative one is greater than $-1$.

A grafical method works pretty well since the roots for LHS = $\frac{-1}{a}$ and the roots for RHS = $\pm 1$ so we see that $1>\frac{-1}{a}>-1$ and hence the intersection points must be within $\pm 1$ .

Analitically it's not difficult either. The equation is $(a^2+a)x^2+2ax+1-a=0$, and the solutions are $\displaystyle x_{1,2}={{-a\pm\sqrt{a^3+a^2-a}}\over{a^2+a}}$. Now it's really easy to show that (a) the discriminant is non-negative for $a>1$ and (b) the solutions lie in $(-1,1)$ for $a>1$.


[quote="communist"]#Prove that:
$ \frac {\sin^2 50 \minus{} \sin^2 40} {\cos^2 50.\cos^2 40} \equal{} 4\tan 10 (\tan 10 \plus{} \tan 40)$#[/quote]

I guess you mean degree.

Then $ L.H.S.\equal{}4 \cdot \frac {\sin^2 50 \minus{} \sin^2 40} {(2 \cos 50.\cos 40)^2}$
$ \equal{}4 \cdot \frac {\cos^2 40 \minus{} \sin^2 40} {(2 \sin 40.\cos 40)^2}$
$ \equal{}4 \cdot \frac{\cos 80}{\sin^2 80}$
$ \equal{}4 \cdot \frac{\sin 10}{\cos^2 10}$
$ \equal{}4 \cdot \frac{\tan 10}{\cos 10}$

So it remains to prove $ \frac{1}{\cos 10}\equal{}\tan 10 \plus{} \tan 40$
$ \Leftrightarrow \cos 40\equal{}\sin 10 \cos 40\plus{}\sin 40 \cos 10$  (by multiplying both sides by $ \cos 40 \cos 10$)

$ \Leftrightarrow \cos 40\equal{}\sin 50$ which is obviously true.


search
combinatorics unsolved
combinatorics

Between $2n-1$ positive integers, prove that we can choose n of them in which the sum of these $n$ integers is divisible by $n$.

Again?

Search for "Erdös Ginzburg Ziv", or "Erdos Ginzburg Zif", or combinations of these two.. :)

Can u give more detailed information ???
Or the most useful site explaining that theorem.
I just don't know .....  :maybe:

http://www.mathlinks.ro/Forum/viewtopic.php?t=28929
http://www.mathlinks.ro/Forum/viewtopic.php?t=57546 post nr. 10.
http://www.mathlinks.ro/Forum/viewtopic.php?t=19496 for the Combinatorial Nullstellensatz.


function
logarithms
algebra unsolved
algebra

[i][u]  Problem[/u][/i]

[i]Find all function $ f : R \to R$ such that

   $ f (x\plus{}y ) \ \equal{} \ f (x ) e^{f(y) \ \minus{} \ 1 }$  $ \forall x \ , \ y \in R$[/i]

[quote="nguyenvuthanhha"][i][u]  Problem[/u][/i]

[i]Find all function $ f : R \to R$ such that

   $ f (x \plus{} y ) \ \equal{} \ f (x ) e^{f(y) \ \minus{} \ 1 }$  $ \forall x \ , \ y \in R$[/i][/quote]
if $ y \equal{} 0$ then $ f(x) \equal{} f(x) e^{f(0) \minus{} 1} \to f(x) \equal{} 0$or $ f(0) \equal{} 1$
 With $ f(0) \equal{} 1$ then $ f(y) \equal{} e^{f(y) \minus{} 1}$
The equation is equivalent to:
$ f(x\plus{}y)\equal{}f(x)f(y)$
Let $ \ln f(x)\equal{} g(x)$ then $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$
Hence, $ g(x)\equal{}ax \to f(x)\equal{}e^{ax}\equal{}Ae^x$
But $ f(0)\equal{}1$ thus, $ f(x)\equal{}e^x\equal{}x\plus{}1$ ( contract)
Conclude that, $ f(x)\equal{}0$

A flaw in Hong Guy's argument is that $ g(x)\equal{}ax$ also works for $ a\equal{}0$. In this case, $ f(x)\equal{}1$ is another solution.

[quote="xxp2000"]A flaw in Hong Guy's argument is that $ g(x) \equal{} ax$ also works for $ a \equal{} 0$. In this case, $ f(x) \equal{} 1$ is another solution.[/quote]
You rae right, xxp2000.
I forgot that.
Thanks.


modular arithmetic
number theory
least common multiple
algebra theorems
algebra

So, what would be the definition of $x \equiv y \pmod{n}$?

I got an idea after searching for an explanation. What i understood is that $\frac{x-y}{n}\in \mathbb{Z}$

Well, let's give some examples:

$16 \equiv 8 \pmod{4}$
$4k \equiv 2k \pmod{2}$
$k \equiv 0 \pmod{k}, k \neq 0$ (is this one correct?)
$k \equiv k \pmod{k}, k \neq 0$
$k \equiv n \pmod{1}$

But how do we write several 'mods' for the same number? In fact, how do we write all the mods that the integer 'k' has?

$30 \equiv 15 \pmod{3}$ and $30 \equiv 15 \pmod{5}$

And this one?

$60 \equiv 30 \equiv 20 \equiv 15 \equiv 10 \equiv 5 \pmod{5}$

And i saw some time ago here on the forum something like:

Consider $f(x) \text{ modulo }2$

What does that mean?

Your examples are correct.
The notation $x\equiv y\pmod{n}$ means that $x$ and $y$ give the same remainder while divided by $n$, so the definition $x\equiv y\pmod{n}\iff \frac{x-y}{n}\in\mathbb{Z}$ is correct. The relation $\equiv$ is an equivalence relation: we have $x\equiv x\pmod{n},\; x\equiv y\pmod{n}\iff y\equiv x\pmod{n}$, $x\equiv y\pmod{n}\text{ and }y\equiv z\pmod{n}\implies x\equiv z\pmod{n}$.
To write the simultaneous conditions $x\equiv y\pmod{n}$ and $x\equiv y\pmod{m}$ in a closed form, you can just write $x\equiv y\pmod{\mathrm{lcm}(n,m)}$.
The notation $60 \equiv 30 \equiv 20 \equiv 15 \equiv 10 \equiv 5 \pmod{5}$ means that all numbers $60,\; 30,\; 20,\; 15,\; 10,\; 5$ give the same remainder while divided by $5$.
[i]Consider $f(x)$ modulo $2$[/i] means that we consider only remainders of $f(x)$ divided by $2$. Of course $\text{remainder while dividing by }n=\text{remainder modulo }n$.


Solve the equation: $\log_{x}(x+1)=\log\frac{3}{2}$

The rigth way is:
(Because the another don't have a algebraic solution)

${\log_{x}(x+1)=\log_{x}\left(\frac{3}{2}\right)}$

${x^{\log_{x}(x+1)}=x^{\log_{x}\left(\frac{3}{2}\right)}}$
${x+1=\frac{3}{2}}$
${x=\frac{1}{2}}$

for the another way, use newton method

http://mathworld.wolfram.com/NewtonsMethod.html

[quote="Thales418"]

for the another way, use newton method

http://mathworld.wolfram.com/NewtonsMethod.html[/quote]

by Newtons method is harder to solve

umm that answers wrong i think u simplified the question way too much by adding base x to the right side..cuz then its like a one step solution


Denote by $ X$ the left hand side.
by cauchy we have that $ 2abc(a+b+c)X\geq (a^{2}b+b^{2}c+c^{2}b)^{2}$ but $ (a^{2}b+b^{2}c+c^{2}b)^{2}\geq (abc)^{2}\frac{(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{4})}{(a+b+c)^{2}}\geq 3(abc)^{2}\frac{(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{3})}{(a+b+c)^{2}}$ using $ (\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{3}\geq \frac{(a+b+c)^{3}}{abc}$ which is well known we finish the problem  :)


Let $a_{1},a_{2},...,a_{n}$ non-negative reals with a sum of $1$. prove that $\sum_{i=1}^{n-1}a_{i}.a_{i+1}\leq \frac{1}{4}$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=30787 beginning with post #3.

  Darij


AMC USA(J)MO USAMO AIME geometry algebra polynomial

Let $N = .13605186556815063100136051\ldots$ be the decimal whose digits past the decimal point are the units digits of all triangular numbers (increasing from 1, 3, 6, 10, \ldots).  Determine with proof whether $N$ is rational or irrational.

[hide="RATIONAL"]
If we prove that the sequence $t_n = \frac{n(n+1)}{2}$ is perodic modulo 10 , then the number will be rational.
We'll prove that:
$t_{n+20}\equiv t_n \text{ (mod 10)}$
$\frac{(n+20)(n+21)}{2}=\frac{n^2+41n+420}{2}=\frac{n^2+n}{2}+\frac{40n+420}{2}=\frac{n(n+1)}{2}+(20n+210) \equiv \frac{n(n+1)}{2} \text{ (mod 10)}.$
So $t_n$ is periodic modulo 10, so the number is RATIONAL.
Exactly
$0.(13605186556815063100)$
[/hide]

krassi_holmz has the best way, in my opinion.  When I saw it, I was thinking "mod 10".

But, if the digits keep on repeating forever, doesn't that mean it's irrational?

[quote="chesspro"][quote="sapphyre571"]here's an easier way (though the proof isn't as great)
[hide]just continue the sequence and it repeats itself, therefore making itself rational. the actual exact fraction is
13605186556815063100/99999999999999999999 if i typed that correctly [/hide][/quote]

A decimal that is not found by division that repeats itself is not necessarily rational. For example, I could have the decimal .1212121212....1212123[non-repeating sequence]. Just because there are several {12} sequences does not mean that the decimal is rational.[/quote]so now you're asking me to prove that it repeats [b]forever[/b] in the same sequence
that should make it rational right?
great
i'm horrible with proofs
here's a shot
[hide]since the twentieth triangular number ends in a zero (.1360518655681506310[b]0[/b])
the sequence following that must always repeat istelf because the units digit of any number ending in zero
plus any number ending in one (the twentieth triangular number plus 21) will always end in a one. add 22 and it ends in a 3. and so on and so forth. sequence repeats itself infinitely. not my best proof. by far.[/hide]
i really need to get better at these proofs if i'm going to have a chance at any usamo problem next year
and my chances of qualifying next year as a freshman i feel are pretty high after a 190 index this year because i could have gotten a much higher score on my amc if i had not only taken it to qualify for aime and used my non-risk-taking test taking strategy. next year if i answer 24 or 25 questions and score an 9 on aime which i believe is possible because i had a 6 this year and didnt understand half the questions (i'm only in geometry) then i should be able to qualify for usamo next year.

I'm sorry to say this, but I don't think your proof is si preciese, because you assume the periodicity $t_n$ mod 10, but you haven't proved that it is.

Matt,

I am reading intro to number theory and I came across this problem.

It is conceivable to show that it is rational by demonstrating two things:

1) Repetition. I looked at an arithmetic series and determined that the decimal repeats every 20 digits by showing that $ 1/2*N(A_{1}\plus{}A_{n})$ has equivalent unit digits for a block of 20 numbers for some $ A_{n}$ that is a multiple of $ 20$ and for some $ A_{1}$ where $ A_{n}\minus{}19 \equal{} A_{1}$ . Beginning with $ A_{1}\equal{}1$ it is trivial to show that 20 is the first number that fits the requirements.

2) Having shown that it repeats, it is trivial to show that the decimal can be expressed as a geometric series in the form of $ 13605186556815063100/(10^{20}\minus{}1)$

Is there a better way to demonstrate that it repeats and where it first begins repeating?

The integers repeat modulo $ 10$. Thus, the triangular numbers repeat modulo $ 10$ since they are simply the partial sums.

(More detail: For any repeating sequence $ a_{1}, a_{2},\ldots, a_{m}$, the sum $ a_{1}\plus{}a_{2}\plus{}\cdots\plus{}a_{m}$ is constant, so we have $ p_{n\plus{}m}\equal{} p_{n}\plus{}s$ where $ s$ is the sum of the repeating parts, and $ p$ is the sequence of partial sums. It's quite clear that this sequence repeats modulo $ 10$.)

[quote="sapphyre571"]plus any number ending in one (the twentieth triangular number plus 21) will always end in a one. add 22 and it ends in a 3. and so on and so forth. sequence repeats itself infinitely. not my best proof. by far.[/quote]

You're getting there with this idea, but krassi_holmz's solution is a rigorous statement of ideas you're only getting at circumstantially.

Of course, the solution can be based on more general principles starting with this lemma:

[b]Lemma:[/b]  Let $ P(x)$ be a polynomial with integer coefficients.  Then for any two integers $ a, b$ we have $ a\minus{}b | P(a)\minus{}P(b)$.  (The proof is left as an exercise.)

[hide="Then..."] Recall that the triangular numbers take the form $ T_{n}\equal{}\frac{n(n\plus{}1)}{2}$.  We wish to determine whether this sequence is periodic $ \bmod 10$.  In other words, we want to determine whether $ n(n\plus{}1)$ is periodic $ \bmod 20$.

Well, let $ P(x) \equal{} x(x\plus{}1) \equal{} x^{2}\plus{}x$.  Then by the lemma,

$ (20\plus{}b)\minus{}b | P(20\plus{}b)\minus{}P(b)\Leftrightarrow$
$ P(20\plus{}b)\equiv P(b)\bmod 20$

Hence $ T_{20\plus{}n}\equiv T_{n}\bmod 10$ as krassi_holmz showed.  :) [/hide]


n -> IN
Find all the possible values for n where :
sqrt(1+5^n+6^n+11^n) -> IN

$ 1\plus{}5^n\plus{}6^n\plus{}11^n\equal{}k^2$ with $ k,n \in \mathbb N$
Taking mod $ 5$ notice that $ k^2 \equiv 0, 1$ or $ 4$ and
$ 1\plus{}5^n\plus{}6^n\plus{}11^n \equiv 1\plus{} 0^n\plus{}1^n\plus{}1^n \equiv 3$ Contradiction!
Then there is no solution...

Sorry I thought that by
[quote="mhdi"]n -> IN
[/quote]
You meant $ n \in \mathbb N$
it will be a lot easier if you use latex.

$ n \in \mathbb{N} \Longleftrightarrow n \in \mathbb{Z}^{\plus{}} \implies n > 0$.

[url=http://en.wikipedia.org/wiki/Natural_number]Wikipedia - Natural Numbers[/url].
It can mean both.  :wink:
$ \mathbb{Z}^{\plus{}}$ does not include zero, since zero is not positive or negative.

[quote]Mathematicians use N or \mathbb{N} (an N in blackboard bold, displayed as ℕ in Unicode) to refer to the set of all natural numbers. This set is countably infinite: it is infinite but countable by definition. This is also expressed by saying that the cardinal number of the set is aleph-nullhttp://en.wikipedia.org/wiki/Aleph_number#Aleph-null (\aleph_0).

To be unambiguous about whether zero is included or not, sometimes an index "0" is added in the former case, and a superscript "*" is added in the latter case:

    ℕ0 = { 0, 1, 2, ... } ; ℕ* = { 1, 2, ... }.

(Sometimes, an index or superscript "+" is added to signify "positive". However, this is often used for "nonnegative" in other cases, as R+ = [0,∞) and [b][size=200]Z+ = { 0, 1, 2,... }[/size][/b], at least in European literature. The notation "*", however, is standard for nonzero or rather invertible elements.)
[/quote]

Any way... how can people know what the other means by writting N?

Anyway, I prove that there is no solution for $ \{1,2,3,...\}$, the case for $ n\equal{}0$ is trivial...

How can you prove that there is no solution for {1,2,3...}? :)

my first reply:
[quote="mszew"]$ 1 \plus{} 5^n \plus{} 6^n \plus{} 11^n \equal{} k^2$ with $ k,n \in \mathbb N$
Taking mod $ 5$ notice that $ k^2 \equiv 0, 1$ or $ 4$ and
$ 1 \plus{} 5^n \plus{} 6^n \plus{} 11^n \equiv 1 \plus{} 0^n \plus{} 1^n \plus{} 1^n \equiv 3$ Contradiction!
Then there is no solution...[/quote]

Notice the following rule:
$ a | (a m\plus{}b)^n \minus{} b^n$ for $ a,m,b,n \in \mathbb N$ and $ 0 \le b < a$
for $ a\equal{}5$ and $ n\equal{}2$ implies $ k^2 \equiv 0, 1$ or $ 4(5)$
or for $ a\equal{}5,m\equal{}2,b\equal{}1$ implies $ 11^n \equiv 1^n(5)$

[quote="Wikipedia"]
In mathematics, a natural number can mean either an element of the set {1, 2, 3, ...} (i.e the positive integers or the counting numbers) or an element of the set {0, 1, 2, 3, ...} (i.e. the non-negative integers).
[/quote]

Not to seem quarrelsome. :wink:

I have seen $ \mathbb{N}$ defined as both $ \{0, 1, 2, ...\}$ and $ \{1, 2, ...\}$. I personally prefer the latter, but there's definitely no set in stone answer, I believe, or any good reason to pick one over the other (without a given context). That and if you [i]were[/i] going to find consistent answers to things, you wouldn't find them on Wikipedia. I suppose this is why we have $ \mathbb{Z}^{ \plus{} }$ to describe the second set. Which, incidentally, I think I've seen as $ \mathbb{Z}^{*}$ or something similar. Who said anything had to be simple?!

[b]EDIT:[/b] Said "first set" when I meant "second set".


Determine  the maximal  real  number  a  for  which  the  inequality
  x1^2+x2^2+x3^2+x4^2+x5^2>=a(x1*x2+x2*x3+x3*x4+x4*x5)
holds for  any  real  numbers  x1 , x2 , x3 , x4 , x5.
Yugoslavia  1996
thanks!

If the inequality holds for all real numbers a-1,...,a_5, then it holds when a_1 = ... = a_5 = a =/= 0. It follows that a \leq 1.
In another hand, if a = 1 the given inequality may be rewritten as
sum{ 1 \leq i \leq 5} (x_i - x_(i+1)) 2 /2  \geq 0 which holds trivially (where x_6 = x_1).

Thus, the maximal value is a = 1.

Pierre.

Doesn't x_5*x_1 appear in sum{ 1  i  5} (x_i - x_(i+1))2 ? We don't have x_5*x_1 in the initial inequality.

I  found  this  problem  at
http://imo.math.ntnu.edu.tw/English/past-exams/yogoslavia/yol.gif
It  is  the  same  i  posted....but  maybe  it  should appear  x5*x1  at  the  RHS....
Thanks!

sorry....tha  right  link  is
http://imo.math.ntnu.edu.tw/English/past-exams/yogoslavia/yo1.gif

AFAIK the answer is 2/ \sqrt 3.
A proof will be post soon.


A sign painter paints individual numerals for a row of 50 houses.
The houses are numbered with consecutive integers from 1 to 50.
How many 9s are painted?

Five 9's are painted: 9, 19, 29, 39, and 49.
Hope you find this useful!


function
search
real analysis
real analysis unsolved

Given any subset of $\mathbb{R}$, is there a function $f: \mathbb{R}\to \mathbb{R}$ continuous exactly on that set?

no! for example, it is quite clear that such a set must be measurable. There are nice description of such sets. This has already been discussed on the forum. Try to search ..

For $\epsilon>0$ let $U_{\epsilon}$ be the set of all $x\in\mathbb R$ for which $\exists\delta>0$ such that $|f(x')-f(x)|<\epsilon$ whenever $|x'-x|<\delta$. By construction $U_{\epsilon}$ is open. The set of continuity of $f$ is $\bigcap_{n=1}^{\infty}U_{1/n}$ which is a $G_{\delta}$-set.


A car is moveing with initial speed 20m/s . After 0.5 sec the car will stop. How shuld I pressent graphycally the relation between time and acceleration .

Acceleration is given by $ \textbf{a}\equal{}\frac{\Delta\textbf{v}}{t}\equal{}\frac{\minus{}20}{.5}\equal{}\minus{}40$.  I left out the units but that's easy to add in.  So on the interval in which the car is moving at constant velocity, acceleration is constant at 0.  Then it spikes down to -40 for .5 seconds, and then spikes back up to 0 when it stops.


let $ p_1 , p_2 , ...$ be a prime number sequance,prove that:
1) there exist infinitely many  $ n$ , such that $ p_n<\frac{p_{n\minus{}1}\plus{}p_{n\plus{}1}}{2}$

2) there exist infinitely many  $ n$ , such that $ p_n>\frac{p_{n\minus{}1}\plus{}p_{n\plus{}1}}{2}$

3) there exist infinitely many  $ n$ , such that $ {p_n}^2>p_{n\minus{}1}p_{n\plus{}1}$

4) there exist infinitely many  $ n$ , such that $ {p_n}^2<p_{n\minus{}1}p_{n\plus{}1}$

i will suppose that there are infinitely many p such that $ p,p\plus{}2$ are prime
we take $ (q_n)$ such that $ q_ {n\plus{}1}>q_n$ and $ p_{q_{n}},p_{q_n}\plus{}2$ are prime

(1) if it's finite then $ \exists m,\forall n\ge m: \ p_n\minus{}p_{n\minus{}1}<p_{n\plus{}1}\minus{}p_n$
so $ p_{n\plus{}1}\ge p_n\plus{}n$ so then are a finite $ p$ such that $ p,p\plus{}2$ both prim (impossible)

(2) if it's finite then $ \exists m,\forall n\ge m: \ p_n\minus{}p_{n\minus{}1}>p_{n\plus{}1}\minus{}p_n$
gives $ p_{n\plus{}1}\minus{}p_n\to \minus{}\infty$ (impossible)

[quote="aviateurpilot"]i will suppose that there are infinitely many p such that $ p,p \plus{} 2$ are prime
we take $ (q_n)$ such that $ q_ {n \plus{} 1} > q_n$ and $ p_{q_{n}},p_{q_n} \plus{} 2$ are prime[/quote]
we don't need this assumption really...

[quote](1) if it's finite then $ \exists m,\forall n\ge m: \ p_n \minus{} p_{n \minus{} 1} < p_{n \plus{} 1} \minus{} p_n$
so $ p_{n \plus{} 1}\ge p_n \plus{} n$ so then are a finite $ p$ such that $ p,p \plus{} 2$ both prim (impossible)[/quote]
I think this is a solution for this problem,not for (1)!
there exist infinitely many $ n$ such that $ p_n\ge\frac{p_{n\minus{}1}\plus{}p_{n\plus{}1}}{2}$

[quote](2) if it's finite then $ \exists m,\forall n\ge m: \ p_n \minus{} p_{n \minus{} 1} > p_{n \plus{} 1} \minus{} p_n$
gives $ p_{n \plus{} 1} \minus{} p_n\to \minus{} \infty$ (impossible)[/quote]
and this is a solution for this problem,not for (2)!
there exist infinitely many $ n$ such that $ p_n\le\frac{p_{n\minus{}1}\plus{}p_{n\plus{}1}}{2}$

[quote="ali666"][quote="aviateurpilot"] [quote] ( 1) if it's finite then $ \exists m,\forall n\ge m: \ p_n \minus{} p_{n \minus{} 1} < p_{n \plus{} 1} \minus{} p_n$
so $ p_{n \plus{} 1}\ge p_n \plus{} n$ so then are a finite $ p$ such that $ p,p \plus{} 2$ both prim (impossible)[/quote]
I think this is a solution for this problem,not for (1)!
there exist infinitely many $ n$ such that $ p_n\ge\frac {p_{n \minus{} 1} \plus{} p_{n \plus{} 1}}{2}$
 [/quote][/quote]
( 1) if it's finite then $ \exists m,\forall n\ge m: \ p_n \minus{} p_{n \minus{} 1} \le  p_{n \plus{} 1} \minus{} p_n$
$ \exists n: \ p_{n \plus{} 1}\minus{}p_n\ge 3$ then are a finite $ p$ such that $ p,p \plus{} 2$ both prime because if $ h>n$ we have $ p_{h\plus{}1}>p_h\plus{}2$(impossible)
now it's a solution for (1)

for (2) if it's finite then $ \exists m,\forall n\ge m: \ p_n \minus{} p_{n \minus{} 1} \ge  p_{n \plus{} 1} \minus{} p_n$
then $ \exists h>m,\forall n\ge h: \ p_{n\plus{}1}\minus{}p_n\equal{}A$ for some $ A \in \mathbb{N}$ (impossible)
now it's a solution for (2)
____________________
but havn't to use any assumption like "p ,p+2............"

Obviosly $ 2)\to 3)$. Let $ a_n\equal{}p_{n\plus{}1}\minus{}p_n$. If 2) is not true $ a_n\ge a_{n\minus{}1}$ for any $ n>N$. If $ a_n\equal{}a_{n\plus{}1}\equal{}...\equal{}a_{n\plus{}k\minus{}1}$, then for any prime $ p\le k$ must be $ p|a_n$. Let $ P_k\equal{}\prod_{i\le k} p_i$. Then if $ P_k\not |a_n$ and $ m\ge P_k$ we get $ a_{n\plus{}m}\ge a_n\plus{}P_k$. Let $ P_{k\minus{}1}\le a_N<P_k$, then $ a_{n\plus{}P_j}\ge a_N\plus{}P_j\to p_{N\plus{}2P_j}\ge p_N\plus{}2a_NP_j\plus{}P_j^2$ for any $ j\ge k$. Therefore for infinetely many n $ P_n>n^2\minus{}an\minus{}b$ (a,b -const). It give contradition with $ p_n>2nln n$ only for finite numbers n.
4) If 4) is not true, then $ (p_n\plus{}a_n)(p_n\minus{}a_{n\minus{}1})\le p_n^2$ for any $ n>N$ or $ p_n(a_n\minus{}a_{n\minus{}1})\le a_na_{n\minus{}1}$ or $ a_n\le \frac{p_na_{n\minus{}1}}{p_{n\minus{}1}}$ or $ p_n(a_{n\plus{}1}\minus{}a_n)\le a_n^2$. Therefore $ a_{n\plus{}1}\le a_n\plus{}[\frac{a_n^2}{p_n}]$ or
$ a_{n\plus{}N}\le a_N\plus{}\sum_{k\equal{}N}^{n\plus{}N}[\frac{a_k^2}{p_k}]$.
If conjecture for any big  x in interval $ (x,x\plus{}\sqrt x )$ exist prime is true, then $ a_{n\plus{}N}\le a_n\to \frac{p_n}{n}<const$ - contradition.

thanks for solutions :)
(1) is very easy problem...I know (2) has elementary solution...for (2) see also:[url]http://www.mathlinks.ro/viewtopic.php?t=150832[/url]
I am very interested see any solution for (2),(4)....
[quote="aviateurpilot"]
( 1) if it's finite then $ \exists m,\forall n\ge m: \ p_n \minus{} p_{n \minus{} 1} \le p_{n \plus{} 1} \minus{} p_n$
$ \exists n: \ p_{n \plus{} 1} \minus{} p_n\ge 3$ then are a finite $ p$ such that $ p,p \plus{} 2$ both prime because if $ h > n$ we have $ p_{h \plus{} 1} > p_h \plus{} 2$(impossible)
now it's a solution for (1)

for (2) if it's finite then $ \exists m,\forall n\ge m: \ p_n \minus{} p_{n \minus{} 1} \ge p_{n \plus{} 1} \minus{} p_n$
then $ \exists h > m,\forall n\ge h: \ p_{n \plus{} 1} \minus{} p_n \equal{} A$ for some $ A \in \mathbb{N}$ (impossible)
now it's a solution for (2)
____________________
but havn't to use any assumption like "p ,p+2............"[/quote]
you are right,but mixed (1),(2)!first solution,is a solution for (2),and second solution is for problem (1)!
[quote="shasha"]3) assuming the contrary gets $ a_n \equal{} p_n/p_{n \minus{} 1}$ is increasing which is
impossible because $ a_n\rightarrow1$ (it is a consequence of the prime numbe theorem).[/quote]
yes...but we didn't really need prime number theorem,this problem can be solved by more elementary way...
[quote="Rust"]Let $ P_k \equal{} \prod_{i\le k} p_i$. Then if $ P_k\not |a_n$ and $ m\ge P_k$ we get $ a_{n \plus{} m}\ge a_n \plus{} P_k$[/quote]
why?I dont get it... :oops:
[quote="Rust"]4) If 4) is not true, then $ (p_n \plus{} a_n)(p_n \minus{} a_{n \minus{} 1})\le p_n^2$ for any $ n > N$ or $ p_n(a_n \minus{} a_{n \minus{} 1})\le a_na_{n \minus{} 1}$ or $ a_n\le \frac {p_na_{n \minus{} 1}}{p_{n \minus{} 1}}$ or $ p_n(a_{n \plus{} 1} \minus{} a_n)\le a_n^2$. Therefore $ a_{n \plus{} 1}\le a_n \plus{} [\frac {a_n^2}{p_n}]$ or
$ a_{n \plus{} N}\le a_N \plus{} \sum_{k \equal{} N}^{n \plus{} N}[\frac {a_k^2}{p_k}]$.
If conjecture for any big  x in interval $ (x,x \plus{} \sqrt x )$ exist prime is true, then $ a_{n \plus{} N}\le a_n\to \frac {p_n}{n} < const$ - contradition.[/quote]
can you solve (4),without using this conjecture?

Yes. I prove, that from 4) we get $ a_{n\plus{}1}\le a_n\plus{}[\frac{a_n^2}{p_n}] \ \forall n>N.$
Because exist $ n>N$, suth that $ a_n^2<p_n$ (else in interval $ (p_n,2p_n)$ exist no more $ 2\sqrt{p_n}$ primes).
Therefore $ a_{n\plus{}1}\le a_n$ and $ a_{n\plus{}1}^2<p_n<p_{n\plus{}1}$. It give $ a_n\le const  \ \forall n>k>N$ - contradition.

[quote="Rust"]
Because exist $ n > N$, suth that $ a_n^2 < p_n$ (else in interval $ (p_n,2p_n)$ exist no more $ 2\sqrt {p_n}$ primes).
Therefore $ a_{n \plus{} 1}\le a_n$ and $ a_{n \plus{} 1}^2 < p_n < p_{n \plus{} 1}$. It give $ a_n\le const \ \forall n > k > N$ - contradition.[/quote]
more detail please,I don't understand:(

If $ a_n\equal{}p_{n\plus{}1}\minus{}p_n\ge \sqrt p_n$ for $ n>N$, then in interval $ (p_n,1p_n)$ number of primes no more $ \sqrt p_n$ contradition.


Να βρεθεί το min της συνάρτησης

$ f(x) \equal{} \dfrac{x^2}{8} \plus{} xcosx \plus{} cos2x$

[hide]1η Λύση

Διακρίνουσα ως προς x και έχουμε

D= $ \dfrac{1}{2}$
Γνωρίζουμε ότι $ f(x)\ge\minus{}\dfrac{D}{4a}\equal{}\minus{}1$
Άρα αρκεί να αποδείξουμε την ισότητα για $ x\equal{}\minus{}\dfrac{b}{2a}\equal{}\minus{}4cosx \longleftrightarrow cosx\equal{}\minus{}\dfrac{x}{4}$
Έστω Q(x)=x+4cosx , $ Q(0)>0$ και $ Q(\minus{}\dfrac{\pi}{2})<0$ . Έτσι σύμφωνα με το Θεώρημα του Bolzano ισχύει οτι υπάρχει λύση ώστε $ cosx\equal{}\minus{}\dfrac{x}{4}$
Άρα αποδείξαμε ότι min της f(x) είναι το - 1

2η Λύση

Παρηρούμε ότι $ \dfrac{x^2}{8}\plus{}xcosx\plus{}2cos^2x \minus{} 1 \equal{} {(\dfrac{x\plus{}4cosx}{2\sqrt{2}})}^2 \minus{} 1$ και πάλι θέλουμε $ x\equal{}\minus{}4cosx$
καθώς εύκολα φαίνεται ότι το ελάχιστο βγαίνει για $ x\plus{}4cosx\equal{}0$ και συνεχίζουμε όπως τη 1η λύση και καταλήγουμε σε min -1
[/hide]

Το θες μεσω ανάλυσης ή αλλιως ρε???? Για  να μην ψαχνω αδικα επειδη βαριεμαι....... :D
με προλαβες ρε!!!! :rotfl:

Δημήτρη, η πρώτη σου λύση μου φαίνεται λανθασμένη αφού δεν έχουμε τριώνυμο ως προς x, γιατί εμπλέκεται και το συνημίτονο.

Έχουμε ως προς χ γιατί στη συνέχεια βγαίνει η σχέση του χ με το συνημίτονο και μπορουμε να παρουμε και τις γραφικες παραστασεις $ \minus{} \dfrac{x}{4}$ και cosx .


PMWC Qualifying Test 2003 #15

At a carnival booth, you are allowed to throw a ball 3 times at a target. You must hit the target twice in a row to win. You must alternate between throwing with your right and left hand. You may start with whichever hand you wish. You hit the target 6/10 of the time with your right hand and 3/10 of the time with your left hand. What is the probability of winning if you make the best choice of starting hand?

does hitting three in a row count???

meaning, can we continue trying even after hitting the first two??

the problem is unclear.

[quote="NoSoupForYou"][hide="My solution"]$2(\frac{3}{10}*\frac{6}{10})=\frac{9}{25}$[/hide][/quote]

That's exactly what I did.

Even if you counted the possibility of hitting it three times in a row, the numbers come out the same.

$\frac 3{10} \cdot \frac 6{10} \cdot 1 = \frac {18}{100}$
$\frac 3{10} \cdot \frac 6{10} \cdot \frac 3{10} + \frac 3{10} \cdot \frac 6{10} \cdot \frac 7{10} = \frac {180}{1000}$

And seungsoo is correct  :D

I am confused!  Can you please simplify your answers? :)

Since people have already answered without using spoilers, I will explain this way also.
First we must see the probability of hitting all three times, which is
\[\frac{3*6*3}{1000}=\frac{18(3)}{1000}\]
Next, the probability of hitting only the first two times:
\[\frac{3*6*7}{1000}=\frac{18(7)}{1000}\]
This is the same as hitting only the last two times, so our final answer is:
\[\frac{18(3)+(18)(14)}{1000}=\frac{18(17)}{1000}=\frac{306}{1000}=\frac{153}{500}\]

BTW, I'm lefthanded, so it seems kind of odd that I would hit the target with my right hand twice as often.  ;)

Well, now you tell us that the game stops after you hit it twice...

[quote="spoudyal"]Nice signature bob.[/quote]

Thank you, :)


A two-pan balance scale comes with a collection of weights. Each weight weighs a whole number of grams. Weights can be put in either or both pans during a weighing. To ensure any whole number of grams up to 100 grams can be measured, what is the minimum number of weights needed in the collection?

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

I keep on getting a different answer from what the answer key, so help would be appreciated.

Thanks in advance.

I think the answer [hide="is"]

5. If you have the weights 1g, 3g, 9g, 27g, and 81g, you should be able to weigh anything from 1 through a 100.[/hide]

Do you know why they are all powers of three?

It would be powers of 2 if the weights were only allowed to be put on one side of the scale, but in this case, they are allowed on both sides.

But, I don't know why exactly it is powers of 3. Why?

Let's say we have weights $a_{1}, a_{2}, \cdots, a_{n}$. If one of them is on the left plate, we give it the number $1$. If one of them is on the right plate, we give it the number $-1$. If it's not on, we give it the number $0$. Having three different digits suggests using base 3. Now notice that in base 3, a $-1$ is a $2$ for the next higher number. This shows that all numbers are possible.

That's smart; why didn't I think of that?

Yeah, it is like something to do with a base three representation,

That's cool. I hope it comes up in nats this year

The school manual does not solely consist of school level problems.

[quote="rd5493"]a question like that would never be at nats.  Fanatic said his source was 04-05 warm-up 13 #8.  if he got that out of the school manual, then why would a school problem show up at the national cometition.  plz think b4 u post[/quote]
It could be Within the first ten probs. Actually there are some pretty hard questions in the manuals sometimes, while there some pretty easy ones in the first parts of nats.

yeah, i guess ur right, but wouldnt warm-up 13 be too easy of a worksheet to have a problem posted for the nat comp?  i just assumed it would be too easy since its in the school manual and everything... :blush:  :blush:


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

After using the series expansion $ \frac{\ln(1\plus{}z)}{z}\equal{}\sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k}{k\plus{}1} \, z^k$ the integral $ I$ becomes

$ \sum_{k\equal{}0}^\infty  \frac{(\minus{}1)^k}{k\plus{}1} \underbrace{\int_0^1 \cos^k (\pi x) \, dx}_{A_k}$ , whereas $ A_{2k}\equal{}\frac{1}{2^{2k}} {2k\choose k}$ and $ A_{2k\plus{}1}\equal{}0$ .

Eventually $ I\equal{}\sum_{k\equal{}0}^\infty  \frac{A_{2k}}{2k\plus{}1}\equal{}\arcsin 1$


superior algebra
superior algebra unsolved

let $ k$ be a field and $ G\equal{}GL_{m}(k)\times GL_{n}(k)$  acts on the set of $ m\times n$ matrices $ M_{m\times n}(k)$ as follow :$ (A,B). X\equal{}AXB^{\minus{}1}$ where $ (A,B)\in G$ and $ X\in M_{m\times n}(k)$ . Exhibit with justification a sub set $ S$ of $ M_{m\times n}(k)$ which contains exactly one and only one element of each orbit under this action .

Any idea will be highly appreciated .  thanks


trigonometry
modular arithmetic
algebra
function
domain
AMC

The first two terms of a sequence are $ a_1 \equal{} 1$ and $ a_2 \equal{} \frac {1}{\sqrt3}$. For $ n\ge1$,
\[ a_{n \plus{} 2} \equal{} \frac {a_n \plus{} a_{n \plus{} 1}}{1 \minus{} a_na_{n \plus{} 1}}.
\]What is $ |a_{2009}|$?

$ \textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 \minus{} \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 \plus{} \sqrt3$

[hide="Huh?"]Make the substitution $ a_n\equal{}\tan\theta_n\implies\theta_n\equal{}\arctan a_n$.
We have $ a_{n\plus{}2}\equal{}\tan\theta_{n\plus{}2}\equal{}\tan(\theta_{n\plus{}1}\plus{}\theta_{n})$.
Plugging in numbers, we get $ \tan\theta_3\equal{}\tan(45^\circ\plus{}30^\circ)\equal{}\tan75^\circ$.
$ \theta_4\equal{}\tan(75^\circ\plus{}30^\circ)\equal{}\tan(105^\circ)\equal{}\minus{}\tan75^\circ$.

And so on.
Working it out, we get:
$ a_1\equal{}1$.
$ a_2\equal{}\frac{1}{\sqrt{3}}$.
$ a_3\equal{}2\plus{}\sqrt{3}$.
$ a_4\equal{}\minus{}2\minus{}\sqrt{3}$.
$ a_5\equal{}\minus{}\frac{1}{\sqrt{3}}$.
$ a_6\equal{}2\plus{}\sqrt{3}$.
$ a_7\equal{}0$.

At this point, $ \tan\theta_7\equal{}\tan(255^\circ\plus{}360^\circ)$, and $ \tan(360^\circ\plus{}\alpha)\equal{}\tan\alpha$, so every term after this one should be $ \boxed{\textbf{(E)}\ 2\plus{}\sqrt{3}}$.

But apparently the answer is A...?
[/hide]

[hide="Click to reveal solution"]
The form of the recursion suggests tangents, so let $ a_i \equal{} \tan(\alpha_i)$.  Then $ \displaystyle \alpha_1 \equal{} \frac{\pi}{4}$ and $ \displaystyle \alpha_2 \equal{} \frac{\pi}{6}$, and $ \alpha_n \equal{} \alpha_{n\minus{}1} \plus{} \alpha_{n\minus{}2}$.  If we make everything have a denominator of $ 12$, then we have a Fibonacci-like sequence beginning with $ 3,2$, which after extensive table-making turns out to be periodic with period 24.  We have $ 2009 \equiv 17$ (mod $ 24$), and referring to the table which I'm too lazy to copy down here, we see that $ \alpha_{17} \equiv 0$ (mod $ \pi$), so the answer should be
$ \fbox{A}$.
[/hide]

[hide]
Consider another sequence $ \{\theta_1, \theta_2, \theta_3...\}$ such that $ a_n = \tan{\theta_n}$, and $ 0 \leq \theta_n < 180$.

The given recurrence becomes

\begin{align*} a_{n + 2} & = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\
\tan{\theta_{n + 2}} & = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\
\tan{\theta_{n + 2}} & = \tan{\theta_{n + 1} + \theta{n}} \end{align*}

It follows that $ \theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}$

Since $ \theta_1 = 45, \theta_2 = 30$, all terms in the sequence $ \{\theta_1, \theta_2, \theta_3...\}$ will be a multiple of $ 15$.

Now consider another sequence $ \{b_1, b_2, b_3...\}$ such that $ b_n = \theta_n/15$, and $ 0 \leq \b_n < 12$.

The sequence $ b_n$ satisfies $ b_1 = 3, b_2 = 2, b_{n + 2} \equiv b_{n + 1} + b_n \pmod{12}$.

We know that the sequence $ b_n$ is periodic. (To prove this, assume that it is not and you will run into a contradiction.)

Write out the first few terms of the sequence until it starts to repeat.

$ \{b_n\} = \{3,2,5,7,0,7,7,2,9,11,8,7,3,10,1,11,0,11,11,10,9,7,4,11,3,2,5,7,...\}$

Note that $ b_{25} = b_1 = 3$ and $ b_{26} = b_2 = 2$. Thus $ \{b_n\}$ has a period of $ 24$: $ b_{n + 24} = b_n$.

It follows that $ b_{2009} = b_{17} = 0$ and $ \theta_{2009} = 15 b_{2009} = 0$

Thus $ a_{2009} = \tan{\theta_{2009}} = \tan{0} = 0.$

Answer: $ |a_{2009}| = \boxed{\textbf{(A)}\ 0}$
[/hide]

Why is it is the sum of the angles $ (mod 180)$? I thought that the domain of tangent was from $ \pi/2$ to $ \minus{} \pi/2$. Why couldn't I do $ tan(45), tan(30), tan(75), tan( \minus{} 15)$?

You only need to consider the sum mod 180 because the period of the tangent function is 180 degrees.

It doesn't really matter what interval you restrict theta as long as its length is 180. I chose $ 0 \leq \theta < 180$ so I would only have to work with positive numbers. Your choice ($ \minus{} 90 \leq \theta < 90$) would work too.

(Actually, the domain of the tangent function is all real numbers except those in the form $ \frac {\pi}{2} \plus{} \pi k$, where $ k$ is an integer. The domain is restricted to $ \minus{} \frac {\pi}{2} < \theta < \frac {\pi}{2}$ only to define the arctangent function.)

When I do it on the interval from $ \minus{} 90$ to $ 90$ , i get :
$ tan(45), tan(30), tan(75), tan( \minus{} 15), tan(60), tan(45), tan( \minus{} 15), tan(30), tan(15), tan(45), tan(60), tan( \minus{} 15),$ which is a cycle since it begins with $ tan(45), tan(30)$again What am i doing wrong?
edit: ok I know what i did wrong. The opposite of $ 105$ is not $ \minus{}15$, It is $ \minus{}75$

Your fourth term should be $ \tan{(\minus{}75)}$ since $ 30 \plus{} 75 \equal{} 105 \equiv \minus{} 75 \pmod{180}$.

ok super newbie way (aka my way lol)

first term is 1, second is 1 over root 3, keep working it out until you get 0.

If a(n) is x, and a(n+1) is 0, then a(n+2) is x, and so on, so i got 0

=]

right answer, newbie way

[hide="alternate solution"]
The function points to tangent addition, so let $\tan{\alpha_n} = a_n$.  Then, $\alpha_1 = \tfrac{\pi}{4}$ and $\alpha_2 = \tfrac{\pi}{6}$.  We realize that $a_n$ can be written as
\[a_n = \tan{(F_{n - 2} \cdot \tfrac{\pi}{4} + F_{n - 1} \cdot \tfrac{\pi}{6})}\]  where $F_n$ is the fibonacci sequence such that $F_1 = 1$, $F_2 = 1$, and $F_n = F_{n - 1} + F_{n - 2}$.  So
\[a_{2009} = \tan{(F_{2007} \cdot \tfrac{\pi}{4} + F_{2008} \cdot \tfrac{\pi}{6})}\]
We try to establish some sort of cancellation between $F_{2007}$ and $4$, and $F_{2008}$ and $6$.  Taking the fibonacci sequence mod 4 results in 1, 1, 2, 3, 1, 0, 1, 1, etc.  Since $3|2007$, $F_{2007}$ is even, but not divisible by 4, so $F_{2007} \cdot \tfrac{\pi}{4}$ results in a fraction with denominator 2 when simplified.

Taking the fibonacci sequence mod 3 results in 1, 1, 2, 0, 2, 2, 1, 0, and it repeats from here.  Since $4|2008$, $F_{2008}$ is divisible by 3.  However, since $2008$ is not divisible by $3$, $F_{2008}$ is not even.  So the denominator is again 2 when simplified.

When adding, the result is $a_{2009} = \tan{(\text{something} \cdot \pi)}$ for a large integer "something".  Since $\tan{\pi k}$ is 0 for all integer $k$, the answer is $\boxed{\textbf{(A)}}$.
[/hide]

oof OR you can list out the cycle for the degrees (first notice it's $\tan(a+b)$) and find that it repeats every $24$ numbers
this took $15$ minutes of my time but at least i got it right XD

When seeing these types of problems and you don't see an immediate method, just list out a couple of terms and more often than not there will be a recurring pattern.

So misplaced I died.

Use tangent angle addition formula to obtain \[a_{n} = \tan{(F_{n-2} \cdot \tfrac{\pi}{4} + F_{n-1} \cdot \tfrac{\pi}{6})},\] plug in $n=2009$ and note the pisano periods of fibonacci numbers to finish


logarithms
calculus
calculus computations

Can someone show me how this series converges? Thanks.

The series is: 1/(ln(n)^(ln(n))

where n starts at 2 and goes to infinity.

Yes, it is convergent.
For $n$ large enough, it holds that
$\frac{1}{(\ln n)^{\ln n}} = \frac{1}{e^{\ln \ln n\cdot\ln n}} \leq \frac{1}{e^{2\ln n}} = \frac{1}{n^2}$


geometry
3D geometry
arithmetic sequence

Show that the cube roots of three distinct primes cannot be terms in an arithmetic progression (the primes do not have to appear consecutively in the arithmetic progression).

let the three primes be p<q<r. first of all, note that the cube root of a prime is irrational. now, note that their cube roots are terms in the same arithmetic sequence iff (q^(1/3)-p^(1/3))*k=r^1/3, where k is some rational number.
cube both sides, and we get:

k^3(q-p+3*p^(2/3)*q^(1/3)-3*p^(1/3)*q^(2/3))=r. this reduces to:

p^(1/3)*q^(1/3)*[p^(1/3)-q^(1/3)] = (pqr)^(1/3)/k=something rational, and because p, q, and r are distinct primes, (pqr) is not a perfect cube (fundamental theorem of arithmetic, i guess...), so (pqr)^1/3 is irrational. contradiction.

thus p^1/3, q^1/3, and r^1/3 cannot be members of the same arithmetic sequence.

sorry about that...
anyhow:
p^(1/3)+d=q^(1/3)
q^(1/3)+kd=r^(1/3), where k is rational (the idea is that the difference of every pair of terms is a rational multiple of the difference of the other pairs of terms)

multiplying the top eqn. by k and subtracting, we get:
k*p^(1/3)+r^(1/3)=(k-1)*q^(1/3)
cubing, we get:

k^3*p+r+3k(pr)^(1/3)[kp^(1/3)+r^(1/3)] = (k-1)^3*q

or, (pr)^(1/3)[kp^(1/3)+r^(1/3)]=something rational
(pqr)^1/3 = something rational

pqr cannot be a perfect cube because p, q, and r are each distinct primes. contradiction.


[quote][b]From the USAMO website: [/b]Selection will be based on the USAMO index which is defined as 10 times the student’s AIME score plus the student’s score on the AMC 12 or the AMC 10.[/quote]

This is what the cutoff is based on, so what did you guys get as your USAMO index?

My index is about the same as a lot of people's AMC scores  :(

i agree with treb. will someone adjust/make a new poll

256.5.......but my index doesn't matter since i'm a sophomore :D ...and what is this "trip to USAMO" that everyone keeps talking about....isn't the USAMO given at your school?

Is it really true that 228 won't make the USAMO this year? I was under the impression that an average cutoff was around 210.

I made USAMO in 10th grade with a 10 or 11 on the AIME, then screwed up last year. This year I made 2 or 3 stupid mistakes and got a 9 (as well as 2 stupid mistakes on the AMC-12 to get my 138).

Maybe we can try to get more of a concensus on what the actual cutoff score will be for AMC-12 takers. Thanks!

[quote="darktreb"]Those are some pretty poor ranges for the cutoff. The cutoff will definitely be higher than it was in 2003 when it was 228. But the odds of it being above 240 are probably pretty low. A better set of possible answers for the poll would be

220-230
230-234.5
235-239.5
240-250[/quote]

OK, I just read "USAMO Index," which says nothing about a cutoff value.  Isn't the poll asking what people's indices are?

yeah, but in order to be relevant in determining the cutoff (which will probably be between 220 and 250) it should break up that group.

[quote="Elemennop"]I did horrible.[/quote]

Yeah, I know how you feel.

Woo!  158!

Believe it or not, that's a school record!
I got a 3 on this (very easy) AIME.  Still the highest score in the school.
Oh well.

[quote=white_horse_king88]
I got a 209.5,

But I'm a freshman, so it doesn't matter. Hopefully a 9 one the AIME will do it, but maybe it won't. We'll see.
[/quote]

same here.... sophomore.... 118AMC12B, 9AIME... for a total of 208. hopefully a 9 will do for me, as it will for you. HOPEFULLY.

All the arguments I have seen on the boards here suggest something over 220 which is logical but for me quite disapointing (my index = 206   :mad: )


M is a point in the triangle ABC. find the smallest value of : MA/a + MB/b + MC/c

$ (\frac{MA}{a}+\frac{MB}{b}+\frac{MC}{c})^2\ge 3(\frac{MA.MB}{ab}+\frac{MB.MC}{bc}+\frac{MC.MA}{ca})\ge 3$
$ \Rightarrow \frac{MA}{a}+\frac{MB}{b}+\frac{MC}{c}\ge \sqrt{3}$


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a\plus{}b\plus{}c\equal{}3.$ Prove that:
\[ \frac{a^2\plus{}3a}{b^2\plus{}c^2}\plus{}\frac{b^2\plus{}3b}{a^2\plus{}c^2}\plus{}\frac{c^2\plus{}3c}{a^2\plus{}b^2}\geq6\]

The remain part is  :lol:
[hide]$ S_c  \equal{} \frac{{a^2  \plus{} b^2  \plus{} 3ab \minus{} c^2 }}{{(b^2  \plus{} c^2 )(a^2  \plus{} c^2 )}}$
$ S_a  \equal{} \frac{{c^2  \plus{} b^2  \plus{} 3cb \minus{} a^2 }}{{(b^2  \plus{} a^2 )(a^2  \plus{} c^2 )}}$
$ S_b  \equal{} \frac{{a^2  \plus{} c^2  \plus{} 3ac \minus{} b^2 }}{{(b^2  \plus{} a^2 )(b^2  \plus{} c^2 )}}$

Suppose that $ a \ge b \ge c$

We have $ S_b  \ge 0$ $ S_b  \plus{} S_c  \ge 0$ $ S_b  \plus{} S_a  \ge 0$[/hide]

Does anybody have another solution without using S.O.S?

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \frac {a^2 \plus{} 3a}{b^2 \plus{} c^2} \plus{} \frac {b^2 \plus{} 3b}{a^2 \plus{} c^2} \plus{} \frac {c^2 \plus{} 3c}{a^2 \plus{} b^2}\geq6
\]
[/quote]
This inequality follows from adding the following inequalities
\[ \sum \frac{2a^2\plus{}bc}{b^2\plus{}c^2} \ge \frac{9}{2}\]
and
\[ \sum \frac{ab\plus{}ac\minus{}bc}{b^2\plus{}c^2} \ge \frac{3}{2}\]
:)

can you post the solution for those inequalities?  :)


What is the remainder when $ 1999^{2008}$ IS DIVIDED BY 4?

[hide]1999 has a remainder of -1 when divided by 4, so it's just $ (\minus{}1)^{2008}$=1.[/hide]


The inner two terms and the outer two terms each sum to a second-order linear recurrence, so if we were solving for integers $x$ a quick calculation of their first few values would suffice, and I suspect it would lead us to a solution.

If we define $a=\sqrt{6}+\sqrt{5}$, $b=\sqrt{3}+\sqrt{2}$,
it's easy to check that $a^{-1}=\sqrt{6}-\sqrt{5}$, $b^{-1}=\sqrt{3}-\sqrt{2}$.
Therefore, $f(x)=g_{a}(x)+g_{b}(x)$, where $g_{c}(x)=c^{x}+c^{-x}$ (with $c>1$).

Properties of $g_{c}$:
- even function: $g_{c}(-x)=g_{c}(x)$;
- strictly increasing when $x>0$: $g_{c}'(x)=(c^{x}-c^{-x})\ln{c}>0$;
- $g(0)=2$.

So every equation $f(x)=k>4$ will have exactly two opposite solutions.

You can check that $f(2)=f(-2)=32$
($g_{a}(2)=2(6+5)=22$, $g_{b}(2)=2(3+2)=10$),
therefore $x=2$ and $x=-2$ are the only 2 solutions of $f(x)=32$.


algebra
polynomial
inequalities
triangle inequality
algebra unsolved

P(x) is a polynmial with real cofficients anf degree of P is n prove that
max{|$ a^i$ - P(i)|} $ \ge$ 1   with i in [0,n+1]

Hmm... according to a MOP handout, the statement should have $a\ge3$ a real number.

Assume for the sake of contradiction that $|a^i-P(i)|<1$ for $0\le i\le n+1$. We know that the $(n+1)^{th}$ forward difference of $P$ evaluated at $x=0$ is
\[0 = \Delta^{n+1}[P](0) = \sum_{i=0}^{n+1}\binom{n+1}{i}(-1)^{n+1-i}P(i).\]Thus by the triangle inequality, we have
\begin{align*}
(1+1)^{n+1} = \sum_{i=0}^{n+1}\binom{n+1}{i} &> \sum_{i=0}^{n+1}\binom{n+1}{i}|a^i-P(i)|\\
&= \sum_{i=0}^{n+1}\left|\binom{n+1}{i}(-1)^{n+1-i}a^i-\binom{n+1}{i}(-1)^{n+1-i}P(i)\right|\\
&\ge \left|\sum_{i=0}^{n+1}\binom{n+1}{i}(-1)^{n+1-i}a^i-\sum_{i=0}^{n+1}\binom{n+1}{i}(-1)^{n+1-i}P(i)\right|\\
&= |(a-1)^{n+1}-0|,
\end{align*}a clear contradiction.


For 11 sets $M_i(i=1..11)$ such that:$|M_i|=5$ and $M_i\cap M_j \neq \phi \forall 1\leq i,j \leq 11.m$ is maximal number so that:There exists $M{}_i{}_1,..M{}_i{}_m$ satisfying :$\bigcap\limits_{j=1}^{m}M{}_i{}_j\neq \phi$.Determine the least value of $m$

No.It's $M_{i_{j}}$,actually.Anybody had solution???

I think your meant was , find the minimum of $m$ s.t. at most $m$ of the $11$ sets have non-empty intersection  :?

Yes!!I do not know how write empty set .....Solution???


Can someone help me with this problem?
Prove that R x R ~ R ? I don't know how to construct a bijection between the two sets. Please help.

[quote="hadjimurad"]It might be easier to construct an injection $ f: \mathbb{R} \times \mathbb{R}\to\mathbb{R}$ and another injection $ g: \mathbb{R} \to \mathbb{R} \times \mathbb{R}$. If you can find two such injections, then the Cantor-Schroeder-Bernstein Theorem tells you that the two sets are equipollent. This is a neat trick that you can use in a lot of other places too.

http://en.wikipedia.org/wiki/Cantor-Schroeder-Bernstein_theorem[/quote]

Thanks for your help. Can you give me some ideas how to construct this injection?

Here's a possible way to construct an injection $ \mathbb{R}^2\rightarrow \mathbb{R}$: (With a bit more work, it can be transformed into an explicit bijection, but, as stated above, an injection is enough, since there are obvious injections $ \mathbb{R}\rightarrow \mathbb{R}^2$.)

First, $ f$ defined by $ f(x,y)\equal{}\left( \frac{1}{2}\plus{}\frac{1}{\pi}\arctan x,\frac{1}{2}\plus{}\frac{1}{\pi}\arctan y\right)$ is an injection $ \mathbb{R}^2\rightarrow [0,1)^2$ and the identity $ \iota$ is an injection $ [0,1)\rightarrow \mathbb{R}$. So if $ g$ is an injection $ [0,1)^2\rightarrow [0,1)$, then $ \iota \circ g\circ f$ is an injection $ \mathbb{R}^2\rightarrow \mathbb{R}$.

To construct $ g$, any element in $ [0,1)$ can be expressed uniquely as a binary expansion $ \sum_{k\equal{}1}^\infty \frac{a_k}{2^k}$ where each $ a_k$ is 0 or 1 and there are infinitely many 0's (to avoid duplicates like $ .1000\cdots \equal{}.0111\cdots$). In this case, define $ g$ by
\[ g\left( \sum_{k\equal{}1}^\infty \frac{a_k}{2^k},\sum_{k\equal{}1}^\infty \frac{b_k}{2^k}\right) \equal{}\sum_{k\equal{}1}^\infty \frac{a_k}{2^{2k\minus{}1}}\plus{}\frac{b_k}{2^{2k}}\]
That is, $ g$ is formed by interweaving the binary expansions of its two inputs. $ g$ isn't onto ($ .1010\cdots$ is not in the image of $ g$, for example) and is much harder to fix than finding a bijection $ \mathbb{R}\rightarrow [0,1)$ (which could be used to replace $ f,\iota$ with bijections), so it's a good thing we have Cantor-Schroeder-Bernstein.


What does the filldraw() command do? What are all its uses? I know it can be used to do stuff like:

[asy]filldraw(polygon(5),red,blue+linewidth(2));[/asy]

but what else?

I recommend that you read the Asymptote manual.  Go to [url]http://asymptote.sourceforge.net[/url] and click the Documentation or PDF links.


AMC USA(J)MO USAMO trigonometry function ratio induction

A calculator is broken so that the only keys that still work are the $ \sin$, $ \cos$, and $ \tan$ buttons, and their inverses (the $ \arcsin$, $ \arccos$, and $ \arctan$ buttons). The display initially shows $ 0$. Given any positive rational number $ q$, show that pressing some finite sequence of buttons will yield the number $ q$ on the display. Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians.

You start from 0 and have to try to get to any number.

I've seen this before (of course I never remember solutions). All I can remember was a couple of hints we got, we had to prove a couple of things like you could get from x to something related to x (might have been double, might have been something else).. that eliminates needing to think about sin, cos and tan, and then use those two facts to try and get from 0 to any number. So try playing around to see if you can get to double a number or half or something like that.

I don't remember exactly how to do this problem. However, I do know that if you set up a table of sorts with the given functions, you will be able to find a way to generate all integers, and also be able to create the ratios of any two integers. If I have some time I may play with this problem and find the old solution.

Funny problem. The best solution would be to buy a new calculator ...

Or :

Let's define :

f1 := sin(arcos(x)) = sqrt (1 - x^2);
f2 := sin(arctan(x)) = x / sqrt (1 + x^2);
f3 := cos(arctan(x)) = 1 / sqrt (1 + x^2);
f4 := tan(arccos(x)) = sqrt (x^2 - 1) / x;
f5 := tan(arcsin(x)) = x / sqrt (x^2 - 1);

Then

f4(f2(x)) = 1/x
g(x) = f4(f2(f3(x))) = sqrt(1 + x^2)

g(g(x) = sqrt (2 + x^2));

So :

g^n(x) = sqrt (n + x^2);
g^(n^2)(0) = n

[hide]pressing cos will give 1  (cos 0 = 1)

Now pressing   (arctan,  sin  )   again and again   will  give  ...
 sqrt(1/2),sqrt(1/3),sqrt(1/4) ....
So you can get   1/:sqrt:n for all positive integers n.

Above is because:
( sin (tarctan (1 / :sqrt: n))) = 1/( :sqrt: (n+1))   )

Now if I had  (1/x) key my work is done.  this is  because, now I can get any natural number (actually I do better than that, I can even get square-root  of any natural number) . also  if  I have   sqrt(1/n) I can always get   (sqrt (1+  1/n), sqrt(2+1/n),sqrt(3+1/n) or in general  sqrt(m+1/n) .. (press  arctan and then cos  and you get sqrt(n/n+1.)  Since any rational number can be expanded into finite terms of continued fraction, my work is done.

Now how do I get  (1/x).. this is not  difficult since, tan(pi/2-x) = 1/tan x)
 and you can get ( pi/2-x)  by  pressing  (arccos  and sin)

so pressing (arctan , sin , arccos, tan )  on  x we get 1/x.[/hide]

Actually by this method, you get all the numbers of the form sqrt(m/n) , that is, it also includes  any non-rational number whose square is rational.

My friend said me  many years ago that it is Landau problem. And Landau solution was the same as yours, Gyan.  :)

Myth - Is this the famous physicst  Landau? (Liquid helium)?

As far as I know it is famous russian mathematician.

You can start by using the cosine of 0 to get 1, and then work your way around from there.

I believe the correct approach is induction on the numerator or denominator of $ q$.

Edit:  Maybe not.  TripleM's approach seems viable.

[quote="t0rajir0u"]I believe the correct approach is induction on the numerator or denominator of $ q$.[/quote]
I think that actually, strong induction on the sum of the numerator and denominator of $ q$ is the most efficient strategy.  See what you think of my wiki write-up : [[1995 USAMO Problems/Problem 2]].

We define the functions
\begin{align*}
f(x) &= \sin\arctan{x} = \frac{x}{\sqrt{1+x^2}},\\
g(x) &= \cos\arctan{x} = \frac{1}{\sqrt{1+x^2}},\\
h(x) &= \tan\arccos{x} = \frac{\sqrt{1-x^2}}{x},
\end{align*}where $f,g$ are defined for all positive $x$ and $h$ for positive $x<1$. We first prove by strong induction on $n$ that all numbers of the form $\sqrt{m/n}$ for $0<m\le n$ can be achieved. The base case $n=1$ is trivial since we start with $0$ and $\cos0=1$. For the inductive step, note that (keep in mind that $m\le n$ here)
\[f\left(\sqrt{\frac{m}{n-m}}\right) = \frac{\sqrt{\frac{m}{n-m}}}{\sqrt{1+\frac{m}{n-m}}} = \sqrt{\frac{m}{n}},\quad
g\left(\sqrt{\frac{n-m}{m}}\right) = \frac{1}{\sqrt{1+\frac{n-m}{m}}} = \sqrt{\frac{m}{n}}.\]Since either $m\ge n-m$ or $n-m\ge m$, at least one of
\[\sqrt{\frac{m}{n-m}},\quad \sqrt{\frac{n-m}{m}}\]can be achieved by the inductive hypothesis, as desired.

Now we consider the case $m>n$. From the induction, we have that $\sqrt{n/(m+n)}$ is achievable, so
\[h\left(\sqrt{\frac{n}{m+n}}\right) = \frac{\sqrt{1-\frac{n}{m+n}}}{\sqrt{\frac{n}{m+n}}} = \sqrt{\frac{m}{n}}\]is as well, and we are done.

OK, this will be an annoying bump of an old topic, but since I'm a great fan of L. D. Landau... I have to fill in the details because Dau was mentioned here.

Moisei Isakovich Kaganov talks about it in his article [url=http://books.google.com/books?id=x9DyAAAAMAAJ&pg=RA2-PA48&lpg=RA2-PA48&dq=landau%27s+game+digits&source=bl&ots=TD9edo8piw&sig=LOi3q6uDxdfX6q3Knn7kXezXA2k&hl=en&ei=H4pPTrTdB8ecOrOVvcMC&sa=X&oi=book_result&ct=result&resnum=1&sqi=2&ved=0CBUQ6AEwAA#v=onepage&q&f=false]here[/url] (Quantum, march/april 1993). It was Landau's mental game - look at a license plate (back then, the plates had 4 numbers, in blocks by two)

[img]http://www.worldlicenseplates.com/jpglps/EU_USSR_GI1.jpg[/img]

Replace the "-" with "=" and make the equality correct by using mathematical symbols on both sides and digits on each side. Y. Gandel has shown that the solution always exists since $\sec \arctan \sqrt{N}=\sqrt{N+1}$. That kills the spirit in the game, doesn't it?

As Boris Gorobec notes [url=http://www.nkj.ru/archive/articles/5583/]here[/url] (Nauka i Zhisn, No. 1, 2000) that secant and cosecant are not in the Russian school curriculum anymore, and that regular students don't know what they are - hence, they couldn't use it in Landau's game, since if you use it as a cosine reciprocal - you need another 'one' for the numerator or exponent.

Since $\arccos\sin\theta = \frac{\pi}{2}-\theta$ and $\tan\left(\frac{\pi}{2}-\theta\right)=\frac{1}{\tan\theta}$ for $0<\theta <\frac{\pi}{2}$, we have for all $x>0$: \[\tan\arccos\sin\tan x = \tan\left(\frac{\pi}{2}-\arctan x\right)=\frac{1}{x} (*).\]

Also for $x\ge 0$, we have \[\cos\arctan\sqrt{x}=\frac{1}{\sqrt{x+1}} (**).\]

Hence, by $(*)$, \[\tan\arccos\sin\arctan\cos\arctan\sqrt{x}=\sqrt{x+1}.\]

We induct on the denominator of $r$, to prove that $\sqrt{r}$, for all non-negative rational $r$, can be obtained using the operations \[\sqrt{x}\mapsto\sqrt{x+1}\text{and} x\mapsto\frac{1}{x}.\]

If the denominator is $1$, we can obtain $\sqrt{r}$, for every non-negative integer $r$, by applying $\sqrt{x}\mapsto\sqrt{x+1}$ repeatedly. Now assume we can get $\sqrt{r}$ for all rational numbers $r$ with denominator up to $n$. In particular we can achieve \[\sqrt{\frac{n+1}{1}}, \sqrt{\frac{n+1}{2}}, \cdots , \sqrt{\frac{n+1}{n}}\]. Hence we can also get \[\sqrt{\frac{1}{n+1}}, \sqrt{\frac{2}{n+1}},\cdots , \sqrt{\frac{n}{n+1}},\] and $\sqrt{r}$, for any positive $r$ of denominator $n+1$ is obtainable by applying $\sqrt{x}\mapsto\sqrt{x+1}$.

Hence for any positive rational $r$, we can obtain $\sqrt{r}$, and in particuar we can obtain $\sqrt{q^2}=q$. $\Box$

[hide="Different"]
Recall the following trigonometric identities:

$f_1(x)=\tan(\sin^{-1}x)=\frac{x}{\sqrt{1-x^2}}$

$f_2(x)=\cos(\tan^{-1}x)=\frac{1}{\sqrt{1+x^2}}$

By simple algebra we can derive some new identities:

$f_3(x)=f_1(f_2(x))=\frac{1}{x}$

$f_4(x)=f_3(f_2(x))=\sqrt{1+x^2}$

Define $g_1(x)=f_4(x)$ and $g_{n+1}(x)=g_1(g_n(x))$ for $n \ge 1$. By trivial induction,

$g_n(x)=\sqrt{n+x^2}$

for $n \ge 1$. Define

$T=\{\sqrt{\frac{p}{q}} ; p,q \in \mathbb{N} ; \gcd (p, q)=1\}$

Note that $\mathbb{Q}\subset T$, so it suffices to show that all elements of $T$ are attainable. For any $t \in T$ define

$R(t)=\max\{p, q\} \mod \min \{p, q\}$

We show that all elements of $T$ are attainable by inducting on $R(t)$. The base case is $R(t)=0$. So let $t=\sqrt{\frac{p}{q}}$. We can assume that $p \ge q$ because the case $p \le q$ follows by taking $f_3(t)$. Then we have $R(t)= p \mod q =0$, so $q=1$. Then $t=\sqrt{\frac{p}{q}}=\sqrt{p}=g_p(0)$, which is attainable.

Now for the inductive step. Assume all $t$ with $0 \le R(t) < r$ are attainable. We show that all $t$ with $R(t)=r$ are attainable. Again, assume $p \ge q$ and write $p=qn+r$ with $0 \le r <q$. Since  $\gcd(p,q)=1$, $\gcd(r,q)=1$. So $\sqrt{\frac{r}{q}} \in T$. Now note that

$g_n(\sqrt{\frac{r}{q}})=\sqrt{n+(\sqrt{\frac{r}{q}})^2}=\sqrt{n+\frac{r}{q}}=\sqrt{\frac{qn+r}{q}}=\sqrt{\frac{p}{q}}$

Since $R(\sqrt{\frac{r}{q}})= q \mod r < r$, the hypothesis guarantees that $\sqrt{\frac{r}{q}}$ is attainable, so $\sqrt{\frac{p}{q}}$ is attainable. This completes the proof.
[/hide]


function
logarithms
floor function
AMC 12 B
algebra

Source: 2004 AMC 12 B

Given that $2^{2004}$ is a 604 digit number whose first digit is 1, how many elements of the set $S = \{2^0,2^1,2^2,\dotsc,2^{2003}\}$ have a first digit of 4?

A) 194
B) 195
C) 196
D) 197
E) 198

[hide="the answer is"]
B
[/hide]

But I have no clue how to get there or how to start.

[quote="t0rajir0u"]Multiply by $25$.  ;)[/quote]
That really doesnt help. Now we just want to see how many times $100\cdot2^{x-2}$ ends in $1$, and we're pretty much back at the original problem, because its pretty much how many times does $2^{x-2}$ end in $1$

Here's a hint, but it takes advantage of the calculator, i can't find a way to do it without one :( . Notice that the first digit of a power $a^b$, is $b \cdot \log{a}$, and then raising $10$ to the fractional part. So for instance, the first digit of $2^5$ is $5 \cdot \log{5}=3.49485...$, and $10^{.49485..}=3....$, just look at the first digit, i.e. greatest integer function. Also, notice that $10^x$ is $4$, where $.7>x>.6$. maybe this helps.

[quote="gopherhole112"][hide="maybe this will help..."]
If $2^a$ is in the set, then it has a first digit of 4 iff:
$\log(4\cdot10^\alpha)<\log(2^a)<\log(5\cdot10^\alpha)$, where $\alpha$ is the number of digits of $2^a$.
$\alpha=\left[a\cdot\log(2)\right]+1$

$\log(4)+\alpha<a\cdot\log(2)<\log(5)+\alpha$
$\log(4)+\left[a\cdot\log(2)\right]+1<a\cdot\log(2)<\log(5)+\left[a\cdot\log(2)\right]+1$
$\log(4)+1<${$a\cdot\log(2)$}$<\log(5)+1$

Note that [] denotes the greatest integer function, and {} denotes the fractional part function.

I'm not exactly sure where to go with this, but maybe it will help. I'm glad I took the AMC 12 A.
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uhh, I didnt really look at your work, but your last part has to be wrong because $\log(4)+1>1$ and $\{x\}<1$, so....

And anyways, like Shobhit said we want $\lfloor 10^{\{x\log(2)\}} \rfloor=4$, so calculator seems to be necessary, but AMC problems are supposed to be done without a caclulator, so there MUST be a non-calculator way.

[quote="Ravi B"][hide="Hint"]Look at the pattern of first digits.  For the powers of 2 with a particular number of digits, we have only the following patterns:
1, 2, 4, 8
1, 2, 4, 9
1, 2, 5
1, 3, 6
1, 3, 7

Note what's special when you have a 4.
[/hide][/quote]

[hide="so this means?"]
We are looking for the (number of powers of two that start with 8) + (the number of powers of 2 that start with 9).

It's worth pointing out that 8 & 9 are the only two digits that don't appear within 2 powers of two after a power of two that starts with a one, so I guess we could write 2004 - 3*(number of powers of 2 that start with 1), but we don't know how many start with 1 either...
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[hide]

[quote="nat mc"]...calculator seems to be necessary, but AMC problems are supposed to be done without a caclulator, so there MUST be a non-calculator way.[/quote]

huh isn't calculator allowed on AMC? :huh: :? :huuh:

[quote="Ravi B"][quote="calculuslover800"][quote="Ravi B"][hide="Hint"]Look at the pattern of first digits.  For the powers of 2 with a particular number of digits, we have only the following patterns:
1, 2, 4, 8
1, 2, 4, 9
1, 2, 5
1, 3, 6
1, 3, 7

Note what's special when you have a 4.
[/hide][/quote]

[hide="so this means?"]
We are looking for the (number of powers of two that start with 8) + (the number of powers of 2 that start with 9).

It's worth pointing out that 8 & 9 are the only two digits that don't appear within 2 powers of two after a power of two that starts with a one, so I guess we could write 2004 - 3*(number of powers of 2 that start with 1), but we don't know how many start with 1 either...
[/hide][/quote]You do know how many start with a 1.  They effectively told you that in the problem.[/quote]

[hide]

Ahh... that is a cool solution. Putting it together, we note that the only patterns of first digits are (thanks to Ravi B):

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

We see that if the first digit is 4, then the pattern will contain 4 different starting digits before going back to 1. So, in effect, if we subtract 3 times the number of these patterns we will be left over with exactly the number of times 8 and 9 appear as the first digit (since the other patterns of 3 will cancel out), which is equivalent to the number of times 4 appears as the first digit.

So we want $2004-3(\text{\# of times 1 is first digit})$ (thanks to calculuslover800).

But we see that for a number of any digits, there is always a power of $2$ that begins with $1$ (seeing the pattern, or verifying that if $2^n < 10^k \Rightarrow 2^{n+1} < 2 \cdot 10^k$), so the number of powers of $2$ that begin with $1$ is exactly the number of digits in the largest power of $2$ that begins with $1$. Since $2^{2004}$ has $604$ digits, the desired number is $603$.

Hence our answer is $2004-3(603) = 195$. [/hide]