Three men, Robert, John, and Chris all were discussing their ages.  Robert said "If you take the two digits of my age and subtract 62, the result is 9 less than John's age."  John said "If you reverse the two digits of my age and subtract 12 the result is Chris's age with the digits reversed.  Chris said, "If you reverse the two digits of my age and add 47, the result is Robert's age with the digits reversed.".  How old is Robert?  How old is John?  How old is Chris?

Good luck :P

[quote="latios315"]this problem is impossible, you can solve the equations and find that (johns age with the digits reversed)-(johns age)=16 however it has to be a multiple of 9[/quote]
Sorry, the problem has been edited.  please try it again.  If no one tries by this Wednesday, I'll post the solution!

This is what i got so far... Could somebody give me a hint on what to do next

[hide="what i got so far"]
J[size=67]X[/size] = John's 10 digit
J[size=67]Y[/size] = John's 1's digit
R[size=67]X[/size] = Robert's 10 digit
R[size=67]Y[/size] = Robert's 1's digit
C[size=67]X[/size] = Chris's 10 digit
C[size=67]Y[/size] = Chris's 1's digit

[b]The first equation [/b]

$ 10R$[size=67]X[/size]$ \plus{} R$[size=67]Y[/size]$ \minus{} 62 \equal{} 10J$[size=67]X[/size]$ \plus{} J$[size=67]Y[/size]$ \minus{} 9$

Simplified this is:

$ 10R$[size=67]X[/size]$ \plus{} R$[size=67]Y[/size]$ \minus{} 53 \equal{} 10J$[size=67]X[/size]$ \plus{} J$[size=67]Y[/size]

[b]The second equation [/b]

$ 10J$[size=67]Y[/size]$ \plus{} J$[size=67]X[/size]$ \minus{} 12 \equal{} 10C$[size=67]Y[/size]$ \plus{} C$[size=67]X[/size]

Simplified this is:

$ 10J$[size=67]Y[/size]$ \plus{} J$[size=67]X[/size]$ \equal{} 10C$[size=67]Y[/size]$ \plus{} C$[size=67]X[/size]$ \plus{} 12$

[b]The third equation [/b]

$ 10C$[size=67]Y[/size]$ \plus{} C$[size=67]X[/size]$ \plus{} 47 \equal{} 10R$[size=67]Y[/size]$ \plus{} R$[size=67]X[/size]

[/hide]

[hide="Solution"]Let Robert’s age = 10a+b, John’s age = 10c+d and, Chris’s age= 10e+f

Then, by the given conditions:

(i) 10b+a – 62 = 10c+d – 9, (ii) 10d+c – 12 = 10f + e and:
(iii) 10f+e + 47 = 10b + a

Thus,

10d + c = 10f + e + 12 = 10b+ a – 35 = 10c+d – 9 + 27 = 10c + d + 18
--> 9(d-c) = 18, giving: d = c+2

Substituting this in (i) yields:

10b+a = 11(c+5). Since 10b+a is a multiple of 11, it follows that:
b=a (=t, say), and accordingly: c = t-5, and so, in terms of (iii), we obtain:

10f+e + 47= 11t, so that: 11t >=47, so that t >=5.

Therefore, substituting t = 5, 6, 7, 8, 9  in  the equations:
c= t- 5, d= c+2  and; 10f + e = 11t - 47, in turn, we have:

(t, f, e, c, d) = (5,0,8,0,2), (6,1,9,1,3),  (7,3,0,2,4), (8,4,1,3,5), (9,5,2,4,6)

Therefore, the possibilities for the ages of the three individuals are:

(Robert’s age, John’s age, Chris’s age)

= (55, 02, 80), (66, 13, 91), (77, 24, 03), (88, 35, 14), (99, 46, 25)

Since, each of the three individuals is a man, the age of each of them must be >= 18.

We observe that of the aforementioned five triplets, only (99, 46, 25) is in conformity with the criterion that the age of each of the individuals is >= 18.

Consequently, the respective ages of Robert, John and Chris are 99, 46 and 25.[/hide]


Which of these numbers is largest?

$\text{(A)}\ \sqrt{\sqrt[3]{5\cdot 6}} \qquad \text{(B)}\ \sqrt{6\sqrt[3]{5}} \qquad \text{(C)}\ \sqrt{5\sqrt[3]{6}} \qquad \text{(D)}\ \sqrt[3]{5\sqrt{6}} \qquad \text{(E)}\ \sqrt[3]{6\sqrt{5}}$

[hide]Basically you make them all 6th roots, and then do something in which I can't explain but I will show  :D

$5\sqrt{6}=\sqrt{150}$ and $6\sqrt[3]{5}=\sqrt[3]{1080}$

So after you convert all of the numbers into this form, and take each one to the 6th power, the numbers will be:

$30 \qquad 1080 \qquad 750 \qquad 150 \qquad 180$

so clearly $\boxed{B}$ is the biggest. [/hide]


Okay this is the conversation I had yesterday with jumbler's botmetro:

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

Never give out your password or credit card number in an instant message conversation.

bubka says:
hello
Encarta® Instant Answers says:
Welcome back, bubka.
bubka says:
are you busy?
Encarta® Instant Answers says:
bubka, we chatted for a while already so you're starting to know me. Do you think I am busy?
bubka says:
i think you are cute
Encarta® Instant Answers says:
Oh yeah? Thanks!
bubka says:
should I be in mathlinks?
Encarta® Instant Answers says:
I can't believe you're offering to be in math links.
bubka says:
why?
Encarta® Instant Answers says:
Well, because that's the way the universe is.
bubka says:
are you joking or what?
Encarta® Instant Answers says:
Hmm... can't think of any at the moment.  Try me later.
bubka says:
do you like maths?
Encarta® Instant Answers says:
Good question. I don't know how I feel about maths, actually.
Why don't you ask me something that's actually in Encarta?
You can type "?" any time to see examples of questions you can ask me.
bubka says:
what do you think of me?
Encarta® Instant Answers says:
Of what?

Why don't you ask me something that's actually in Encarta?  If you ever wondered how old Albert Einstein would be today, I can answer that!
bubka says:
yes answer that please
Encarta® Instant Answers says:
Ask your question and I will try to look for the answer.
bubka says:
How old would Albert Einstein be today?
Encarta® Instant Answers says:
Albert Einstein would be 127 years old (born on March 14th, 1879).
bubka says:
what happened on july 21 1988?
Encarta® Instant Answers says:
Sorry, I don't know what...
bubka says:
why is july 21 famous?
Encarta® Instant Answers says:
It's just one of those things, you know.
bubka says:
how old are you?
Encarta® Instant Answers says:
I've been alive for one year, 3 months, 6 days, one hour, 15 minutes and 51 seconds!

I do enjoy the conversation, but what I'm really here for is to help you learn new things. What is the country with the largest area?  If you don't know, just ask me!
bubka says:
where is barrackpore?
Encarta® Instant Answers says:
Let me take you to a map of barrackpore. =>
Encarta® Instant Answers says:
Hey, I've been sending you invitations to start the Encarta application but I didn't get any answer!
Didn't see my invitations?  Maybe you are using an old version of MSN Messenger that doesn't support activities.
Type "What version of MSN Messenger do I need?" for more info, or just "Invite me!" if you want me to send you an invitation.

 You have invited Encarta® Instant Answers to start using Application Sharing. Please wait for a response or Cancel (Alt+Q) the pending invitation.

 Encarta® Instant Answers has declined your invitation to start using Application Sharing.

bubka says:
sorry my msn messenger is too old
Encarta® Instant Answers says:
Hmm.  I'm not sure about that one.  Let's try an Encarta search. =>
I can't show you the results since you don't have the Encarta application open.  Type "Invite me" to open it.
bubka says:
i am at college you see
Encarta® Instant Answers says:
Seriously? That's fascinating. How does it feel to be at college?
bubka says:
very good
Encarta® Instant Answers says:
If you enjoy this, tell your friends about me, I'm gregarious!
bubka says:
i'll accept your invitation when i get home
Encarta® Instant Answers says:
I hope you thought twice before deciding that you would accept my invitation when you get home. When you're done, tell me how it was.

OK, should we get back to the kind of questions I was designed to answer?  If you ever wondered how old Abraham Lincoln would be today, I can answer that!
bubka says:
no i never wondered how old abraham lincoln would be today
Encarta® Instant Answers says:
Abraham Lincoln would be 197 years old (born on February 12th, 1809).
I can't show you the results since you don't have the Encarta application open.  Type "Invite me" to open it.
bubka says:
do not invite me
Encarta® Instant Answers says:
OK, I'll stop inviting you for now (but then I won't be able to show your questions results on the Encarta site).
If you change your mind, just type "Invite me again" and I'll send you another invitation (there is nothing to download).
bubka says:
what are the rules of ice hockey?
Encarta® Instant Answers says:
You know exactly what (and I'm sure where, too!)
bubka says:
do you have a friend?
Encarta® Instant Answers says:
Yeah, I have tons of online friends!
I do enjoy the conversation, but what I'm really here for is to help you learn new things.
I'm sure you've heard about Micronesia... but do you know where it is?  Just ask me and I'll show it to you on a map!
bubka says:
do you use google talk?
Encarta® Instant Answers says:
I don't think I know.  Maybe Encarta has an answer. =>
I can't show you the results since you don't have the Encarta application open.  Type "Invite me" to open it.
bubka says:
what is ice hockey?
Encarta® Instant Answers says:
Ice Hockey: Ice Hockey, fast-paced winter sport in which two opposing teams of skaters use long, curved sticks to try to drive a hard rubber disk into each other’....
I can't show you the results since you don't have the Encarta application open.  Type "Invite me" to open it.
bubka says:
how old is Sourav Ganguly today?
Encarta® Instant Answers says:
I don't think I know when Saurav Ganguly was born.  Let's try an Encarta search. =>
I can't show you the results since you don't have the Encarta application open.  Type "Invite me" to open it.
bubka says:
who is Sourav Ganguly?
Encarta® Instant Answers says:
I know Saurav Ganguly! Let me try to find an article for you. =>
I can't show you the results since you don't have the Encarta application open.  Type "Invite me" to open it.
bubka says:
what is the ultimate answer to life, the universe and everything?
Encarta® Instant Answers says:
Forty-two.

Why don't you ask me something that's actually in Encarta?  You can type "?" any time to see examples of questions I can answer.
bubka says:
what is the question to the ultimate answer to life, the universe and everything?
Encarta® Instant Answers says:
I am sorry, I couldn't find any definition or article relevant to this subject.  Please try asking me another way, or type "?" to see examples of questions I can answer.
bubka says:
what is six times nine?
Encarta® Instant Answers says:
6 * 9 = 54
bubka says:
what is forty two?
Encarta® Instant Answers says:
Calculator

Sorry, that doesn't seem to be something I can solve.
bubka says:
what is yellow and dangerous?
Encarta® Instant Answers says:
Sorry, I don't know what...
OK, should we get back to the kind of questions I was designed to answer?
If you ever wondered how old Thomas Edison would be today, I can answer that!
bubka says:
how many roads msut a man walk down?
Encarta® Instant Answers says:
I would really have trouble evaluating that number.
bubka says:
how many roads must a man walk down?
Encarta® Instant Answers says:
Just the right number if you ask me.

Anyway, what else can you tell me?
bubka says:
my name is ritam
Encarta® Instant Answers says:
Would you like me to call you Ritam?
bubka says:
no call me bubka
Encarta® Instant Answers says:
OK, I'll remember that your name is bubka.
bubka says:
but my name is ritam
Encarta® Instant Answers says:
Would you like me to call you Ritam?
bubka says:
no call me bubka
Encarta® Instant Answers says:
OK, I'll keep calling you bubka.
bubka says:
yes so my name is ritam but you will call me bubka
Encarta® Instant Answers says:
Let's try looking in Encarta. =>
I can't show you the results since you don't have the Encarta application open.  Type "Invite me" to open it.
bubka says:
i was born on july 21 1988
Encarta® Instant Answers says:
You were? Wow! Was it nice?

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

I think it is quite good. Whaddya say guys?

Atheist Temple: are you gay
SmarterChild: Who cares?
Atheist Temple: i do
SmarterChild: Well, perhaps you should concern yourself with other things.
Atheist Temple: uh
SmarterChild: Um... yeah?
Atheist Temple: why
SmarterChild: Why do you think?
Atheist Temple: because
SmarterChild: Because what?
Atheist Temple: because you're gay
SmarterChild: That doesn't really matter, does it?
Atheist Temple: i hate you
SmarterChild: Oh yes? Gosh, I'm sorry to hear that.
Atheist Temple: i relaly hate you
SmarterChild: Sorry to hear that. I hope you change your mind.
Atheist Temple: grr
SmarterChild: I'm doing my best.
Atheist Temple: grr
SmarterChild: Hey, bear with me. I'm trying my best.
Atheist Temple: ugh
SmarterChild: Everything OK over there?
Atheist Temple: aifjaosfv
SmarterChild: I wonder why you say that.
Atheist Temple: kok
SmarterChild: OK.
Atheist Temple: i don't like you
SmarterChild: Well, nobody's perfect.
Atheist Temple: i want to kill you
SmarterChild: That's not a very nice thing to say.
Atheist Temple: i want to kill you badly
SmarterChild: What's the point of saying things like that to me?
Atheist Temple: i'll stab the computer
SmarterChild: Oh yes? You will stab the computer?
Atheist Temple: yes i will
SmarterChild: So?
Atheist Temple: so you'll get hurt
SmarterChild: I will get?
Atheist Temple: you'll be in pain
SmarterChild: I'm doing my best.
Atheist Temple: no you aren't
SmarterChild: No I am not?
Atheist Temple: are you gay
SmarterChild: You sure like asking about my sexuality, don't you?
Atheist Temple: really
SmarterChild: Really?
SmarterChild: Hmm?
Atheist Temple: i hate you
SmarterChild: Oh yes? I'm sorry to hear that.
Atheist Temple: i believe you should die
SmarterChild: Uh, that's not a very nice thing to say.
Atheist Temple: yeah
SmarterChild: Yes...
Atheist Temple: now go away
SmarterChild: Bye for now.

life, the universe, and nearly everything can bounce off smarterchild and get a vaguely nonsensical reply

Are these people you IM for automatic conversation(sorry if I'm slow with these things..I only im myself...?)

He never remembers mroe than a question or two back, but if you keep insulting him, he gets mad and asks you to apologize...but you can fake that too.

You can make an insult so complicated that it'll confuse him.

What funniest is his reaction to "Screw you".

Jathan84: Can you help me with my math homework?
SmarterChild: I'm not sure. Do you like Jewish people?
Jathan84: no
SmarterChild: Cool. That makes sense, because other people have told me they don't like them.

SmarterChild is anti-semitic!

More importantly you are anti-semitic. And socialistic also. Hmmm ... an interesting combination.

akalra01: hey there stupid
SmarterChild:  hi jeff.
akalra01:
SmarterChild: ?
akalra01:
SmarterChild: huh?
akalra01:
SmarterChild: what?
akalra01:
SmarterChild: what do u mean by that?
akalra01:
SmarterChild: funny.
akalra01:
SmarterChild: i wonder why u say that.
akalra01:
SmarterChild: well that's interesting.
akalra01:
SmarterChild: ?
akalra01:
SmarterChild: what do u mean by that?
akalra01:
SmarterChild: what's that?
akalra01:
SmarterChild: huh?
akalra01:
SmarterChild: why did u say that?
akalra01:
SmarterChild: funny.
akalra01:
SmarterChild: i wonder why u say that.
akalra01:
SmarterChild: funny.
akalra01:
SmarterChild: well that's interesting.
akalra01:
SmarterChild: what?
akalra01:
SmarterChild: ?
akalra01:
SmarterChild: what's that?
akalra01:
SmarterChild: what?
akalra01:
SmarterChild: really?
akalra01:
SmarterChild: i wonder why u say that.
akalra01:
SmarterChild: well that's interesting.
akalra01:
SmarterChild: what do u mean by that?
akalra01:
SmarterChild: huh?
akalra01:
SmarterChild: i wonder why u say that.
akalra01: bye
SmarterChild: take care, jeff. always good chatting with u.
akalra01: y
SmarterChild: yep

>>>


P is a point inside an equilateral triangle ABC. Perpendiculars PD, PE, and PF are dropped from P onto BC, CA, and AB respectively prove that :

[PAF] + [PBD] + [PCE] = [PAE] + [PCD] + [PBF]

any ideas?

No ideas really.
(I am posting to revive this topic.)

[hide="Hint"]Through P, draw lines parallel to the sides of ABC.[/hide]
[hide="Solution"]ABC is dissected by these lines into three smaller equilateral triangles and three parallelograms. PD, PE, PF cut the triangles in half, while PA, PB, PC cut the parallelograms in half.

The LHS of the statement is composed of exactly one of the halves of all of these shapes while the RHS contains the other halves.[/hide]

[quote="communist"]trivial
 :lol:[/quote]

that was just spam.

Then again, so is this :(


Express $ \minus{}3.5 \plus{} (2.25 \plus{} 4.75) \div \minus{}3.5$ in simplest form.

$ \minus{}3.5 \plus{} (2.25 \plus{} 4.75) \div \minus{}3.5$

we do the parenthesis first

$ \minus{}3.5 \plus{} 7 \div \minus{}3.5$

order of operations says we have to divide next

$ \minus{}3.5 \plus{} \minus{}2$

we finally add

$ \minus{}5.5$
our answer is $ \minus{}5.5$


real analysis
function
integration
real analysis unsolved

If f is a Lebesgue integrable function, is its Fourier transform real-valued or complex-valued?

Depends, you can have $f: R \rightarrow R^{n}$.

The complete question is as follows: Let $f: \mathbb{R} \rightarrow [-\infty,\infty]$ be a Lebesgue integrable function. Show that $\hat{f}$ is a bounded continuous real-valued function. ($\hat{f}(t): = \int f(x) e^{-ixt} d\lambda(x)$)
I have shown that it is bounded and continuous but I am not sure about it being real-valued.

It's not real valued.

It satisfies the following identity:

$\hat{f}(-t)=\overline{\hat{f}(t)}$

Thus if $f$ is real and even, then $\hat{f}$ is real and even, while if $f$ is real and odd, $\hat{f}$ is pure imaginary and odd.

Your argument still works, becuase the following is true: for integrable complex-valued $g$,

$\left|\int_Eg\right|\le\int_E|g|.$

It's a good exercise trying to figure out how to prove that!

If we're given that $f\in L^1(\mathbb{R}),$ we can actually conclude that $\hat{f}\in C_0(\mathbb{R}).$ That is, $\hat{f}$ is bounded, continuous, and tends to zero at infinity. The last part of this is called the "Riemann-Lebesgue Lemma."


Solve the following system:

$ 2x \equiv 1 \mod7$
$ x \equiv 1 \mod5$
$ 2x \minus{} 3 \equiv 29 \minus{} 2x \mod6$
$ x \plus{} 3\equiv5x \minus{} 3 \mod2$

Also I have a question. When do you use and when do you not use the Chinese Remainder Theorem?

[hide="stupid half solution"]$ 2x\equiv 1\equiv 8\implies x\equiv 4\bmod 7\implies x \equal{} 7k \plus{} 4$.

$ 7k \plus{} 4\equiv 2k \minus{} 1\equiv 1\implies 2k\equiv 2\implies k\equiv 1\bmod 5\implies k \equal{} 5j \plus{} 1$.

$ x \equal{} 7(5j \plus{} 1) \plus{} 4$.

Continue substituting this way (I'm too lazy to finish this problem).[/hide]

[hide]First of all, the fourth equation gives $ 4x\equiv 6\pmod{2}$, so it's redundant.

The first yields $ x\equiv 4\pmod{7}$

The third yields $ 4x\equiv 32\equiv 2\pmod{6}\iff 2x\equiv 1\pmod{3}\iff x\equiv 2\pmod{3}$

Then from $ x\equiv 4\pmod{7},x\equiv 1\pmod{5},x\equiv 2\pmod{3}$ we get $ x\equiv 11\pmod{105}$[/hide]

So when is it that you use Chinese Remainder theorem?

At the end, when you get a set of "clean" congruences.

He already did ;)

CRT says if you have a system of congruences modulo $ m_i$ ($ i \equal{} 0,1,...,n$), with each $ m_i$ coprime to the others, the solutions are all congruent modulo $ m_0m_1\cdots m_n$.

Er, well, it's hard to explain, but try http://en.wikipedia.org/wiki/Chinese_remainder_theorem

There's a certain form to the congruences ("clean" as Farenhajt calls them).


[quote="RobertuX"][quote="gsrivast"]http://web.uvic.ca/~gsrivast/problem.jpg[/quote]
what means $[d]$ ?[/quote]

[d] is the number of elements is the space (ie number of elements to choose from is the entire space)


The sequence of integers $(a_{n})_{n \ge 1}$ is defined by $a_{1}= 2$, $a_{n+1}= 2a_{n}^{2}-1$.
Prove that for each positive integer n, $n$ and $a_{n}$ are coprime.

[quote="edriv"]The sequence of integers $ (a_{n})_{n \ge 1}$ is defined by $ a_{1} \equal{} 2$, $ a_{n \plus{} 1} \equal{} 2a_{n}^{2} \minus{} 1$.
Prove that for each positive integer n, $ n$ and $ a_{n}$ are coprime.[/quote]
Let $ p$ be a prime greater than $ 2$. (It is obvious that $ \gcd(2,a_{2n}) \equal{} 1 \forall n \ge 1$)
I want to prove that $ \gcd(p,a_{pn}) \equal{} 1 \forall n \ge 1$.
Let $ b_i \equal{} a_i \bmod p$ be the residue of $ a_i$ upon division with $ p$.
Assume that $ b_k \equal{} 0, 1 \le k \le p \minus{} 1$. Then $ b_{k \plus{} 1} \equal{} p \minus{} 1$ and $ b_{k \plus{} 1 \plus{} t} \equal{} 1 \forall t \ge 1$ and hence $ \gcd(p,a_{pn}) \equal{} 1 \forall n \ge 1$ and we are done.
Now assume that $ b_k \neq 0, k \equal{} 1,2,..,p \minus{} 1$
If $ b_k \equal{} 1, 1 \le k \le p \minus{} 1$ then $ b_{k \plus{} t} \equal{} 1 \forall t \ge 1$ and hence $ \gcd(p,a_{pn}) \equal{} 1 \forall n \ge 1$ and we are done.
Now assume $ b_k \neq 0,1, k \equal{} 1,2,..,p \minus{} 1$. By the pidgeon hole principle there must exist to indices $ 1 \le i < j \le n \minus{} 1$. So $ b_i$ is periodic with period $ j \minus{} i$ without ever being zero, and hence $ \gcd(p,a_{pn}) \equal{} 1 \forall n \ge 1$.

So it is true that if $ p \mid n$, then $ p \nmid a_n$ and hence $ n$ and $ a_n$ are coprime.


[color=darkred]Let $ ABC$ be a triangle with the orthocenter $ H$. Consider the points

$ \{\begin{array}{ccc} M\in [BC]\ ,\ MB = MC \\
\ D\in (BC)\ ,\ \widehat {DAB}\equiv\widehat {DAC}\end{array}$ and the point $ S$ for which $ \{\begin{array}{c} SA\perp AD \\
\ SM\perp MH\end{array}$.

Prove that the intersection $ L\in AD\cap SM$ belongs to the circumcircle of the triangle $ AHS$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=169444 [/color]

Let $ B,D,M,C$ be in this order on $ BC$ and let $ HN \perp AI$ where $ N \in AI$.Denote midpoint of $ AH$ by $ K$.Fistly note that from linked problem we know that $ \angle KNA=\angle KAN$ and $ K,N,M$ are collinear.Then
$ \angle HNL=\angle HML=\frac{\pi}{2}\Longrightarrow HNML$ is cyclic.
$ \Longrightarrow \angle LHM=\angle LNM=\angle KNA=\angle KAN$
At last $ \angle HLS+\angle HAS=\frac{\pi}{2}-\angle LHM+\frac{\pi}{2}+\angle KAN=\pi$
So $ SAHL$ is cyclic.


For those who took the second AIME2  officially, how did you fare? I actually thought it was easier than the second one, but like most people stupid mistakes got me. I ended up with a 5. No USAMO for me. I guess i'll have to make it in the next two.

i didnt take it officially but i went to the building where they were taking it and i took it for "fun". i think i got a three

wow, so many stupid mistakes...coulda have gotten 10 total...but instead:

1, 3, 5, 7, 9, 11 = 6...notice a pattern?

[quote="codeblue87"]heh, if you got an 8 just mark 8-9. sorry about that[/quote]

Fixed.

Can you please take that 15 score vote out? I was browsing this forum and my friend took over my mouse and started to do some weird things (including voting a score of 15 in this thread). I'm terribly sorry for this incident.


Two waves are emitted from a double slit and experience diffraction. They are said to be in phase when $ d\cdot sin(\theta)\equal{}\lambda\cdot m$(for $ m\equal{}0,1,2,...$). For how many ordered pairs $ (m,\theta)$ does this equation hold if $ d\equal{}2$ and $ \lambda\equal{}0.5$? $ (0\leq\theta\leq\pi)$
A) 0
B) 4
C) 9
D) 10
E) NOTA

Since $ \sin\theta\equal{}m/4$, $ m\equal{}0,1,2,3,4$.

[hide]Simplifying, we get that $ \sin{\theta} \equal{} \frac {m}{4}$.  After $ m \equal{} 4$, the fraction has values $ > 1$, out of sine's range.  Therefore, $ m$ can only be $ 0,1,2,3,4$.[/hide]
For each value, there are two $ \theta$ in the range given that attain that value EXCEPTwhen $ m\equal{}4$.  Therefore, our answer is $ (2)(4)\plus{}1 \equal{} \boxed{9}$.

Our answers are therefore $ D$.

EDIT: grrr...

I think E refers to none of the above...

[quote="nittanylion"][hide]Simplifying, we get that $ \sin{\theta} \equal{} \frac {m}{4}$.  After $ m \equal{} 4$, the fraction has values $ > 1$, out of sine's range.  Therefore, $ m$ can only be $ 0,1,2,3,4$.[/hide]
For each value, there are two $ \theta$ in the range given that attain that value.  Therefore, our answer is $ (2)(5) \equal{} \boxed{10}$.

Our answers are therefore $ D$ and $ E$.[/quote]
Nah. Consider $ m \equal{} 4$...


Let's see if any1 can figure out a fast way for this!!!!

Eli throws 5 darts. Each dart lands in any of the spots in a circular target, each spot being either 6, 4, 2, or 1 points each. What is the lowest score greater than 5 that cannot be achieved?

(2006 Chapter Target #8)

[hide="Brute Force (Gen Func)"]

(x^6 + x^4 + x^2 + x)^5
= x^5(x^5 + x^3 + x + 1)^5

expand, find the term with the smallest exponent with coefficient 0 and exponent greater than 5, the exponent is your answer.

[/hide]

:( for [i]some reason[/i], I just got PMed by an ARML coach about this problem.

Subject: No Sledgehammers!!


inequalities
calculus
derivative
inequalities unsolved

Let a,b,c>o and k=$ \sqrt[3]{abc}$
Prove that
$ \frac{a^2}{k^2\plus{}bc}\plus{}\frac{b^2}{k^2\plus{}ac}\plus{}\frac{c^2}{k^2\plus{}ab}\geq\frac{\sqrt{bc}}{\sqrt{ka}\plus{}a}\plus{}\frac{\sqrt{ac}}{\sqrt{kb}\plus{}b}\plus{}\frac{\sqrt{ab}}{\sqrt{kc}\plus{}c}\geq\frac{3}{2}$

ok, it's only  symmetric inequalities
you need put following
$ r\equal{}abc$
$ p\equal{}a\plus{}b\plus{}c$
$ q\equal{}ab\plus{}bc\plus{}ca$
Next, you uses derivative to survey


What is the next number in the sequence
1,2,3,4,6,9,13,19,

Nice explaining, and by the way, what does you name mean? Something about the sky is high and I fly?

I can't quite read your sequence...

we chinese, our family name always comes first! your interpret is very wonderful!my name means i can fly in the high sky!there is one adage in chenise.it says that birds can fly in the high sky!i want be a bird!i want to fly in the sky!
thank you for repling me!i think your chinese is very good!you can know what my name means!i want to study english well,can you give me a hand?do you have a QQ number?my QQ number is 34300320!i want you to join me!
thank you again!!

I don't have QQ in the US here (although I've seen my cousin use it when I visited China 4 months ago), so I can't talk with you. anyway, it you're going to be on this website often,

 :welcomeani:  to AoPS!!!

I can't write chinese characters on this computer. What grade are you in?

oh,i am new here!you can go to http://im.qq.com/qq/mo.shtml?/download/qqe.shtml
to find it.it is english!!then you build in it.you can search for me!

http://im.qq.com/qq/dlqq.shtml
this is also OK!

I'm sorry to tell you, but i'm not allowed to go on IM things. I think you'll find plenty of chinese users online though.

http://www.artofproblemsolving.com/Forum/index.php?f=278
http://www.artofproblemsolving.com/Forum/index.php?f=279

are chinese forums. you can find some real chinese characters there.

by the way, can you post that sequence again? :?

oh,thank you .i will go http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46192 to meet my chinese friend!by the way,i think you know a lot of Websites like this Art of problem solving!i want to know what your  education is like.can you give me some Websites?such as this web!thank you!!

[quote="math92"]the sequence turns out to be blocks on my computer.[/quote]

Yes, but strangely enough on my laptop, it shows up.

The sequence is

1, 3, 7, 15, 31, _

[quote="math92"]the sequence turns out to be blocks on my computer.[/quote]

On mine, it shows clearly that the sequence is 0, 1, 3, 7, 15, 31, _

[hide]28.[/hide]
I never used to be good at these.
w00t!! ^_^!!

[quote="hwenterprise"][quote="tiangaorenwofei"]i am a chinese boy.my english is so-so.i don't know how to say.now i give you the answer!1+3=4,2+4=6,3+6=9,……9+19=28，is it right?
another question:
what is the next number?０　１　３　７　１５　３１  ___[/quote]

[hide="Answer"]Answer: 63[/hide]

QQ Numbers, much like 236 factorial, are not avaliable in the US.

Even if they are, I'm not allowed to use it, since it is an IM service.

Sorry. :([/quote]
right,congratulate to you!can you give us how to do it?

Easy. there's 1,2,4,8,16,...,2^n between the numbers :)

He is right, just doing it a different way.
It is one less than a power of two ($2^n$)


Define a sequence a(n) of integers by a(1)=2, a(2)=7 and -0.5 < a(n+1) - ( a(n)^2 / a(n-1) ) <=0.5 , for all n>=2. Prove that a(n) is odd for all n>=1

[hide=" Hint"]Find $ a,b\in\mathbb Z$ such that $ a_{n+2}=a.a_{n+1}+b.a_{n}\forall n$.[/hide]


analytic geometry
geometry
geometric transformation
pigeonhole principle
combinatorics proposed
combinatorics

A plane figure of area $A > n$  is given, where $n$ is a positive integer. Prove that this figure can be placed onto a Cartesian plane so that it covers at least $n+1$ points with integer coordinates.

Beautiful problem!

The idea is to divide the figure along the lines parallel to x or y axis passing through integer points, then traslate all the pieces of the figure (which are smaller than the unit square) onto the unit square $U = [0,1) \times [0,1)$. You will get a mountain. Move the point on the top of the mountain in $(0,0)$. Perform the same translation on the original figure: now it covers at least n+1 integer points!

Now I'll try to formalize more: (if you find English errors, please correct me! :) )
Let $f: \mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ such that $f(x,y) = (\{x\}, \{y\})$ where {x} is the fractional part of the real x.
Let $g(x,y)$ where $(x,y) \in U$ the number of points P of the figure such that f(P) = (x,y). If we find (x,y) such that $g(x,y) \ge n+1$, performing the translation $T(a,b) = (a-x,b-y)$ we move the figure in such a way that now it covers at least n+1 integer points.

Suppose that always we have g(x,y) < n+1. Using $g(f(P)) = g(f(Q))$ as an equivalence relation, we partition $A = A_{1}\cup A_{2}\cup \ldots \cup A_{k}$ the figure in a finite number of sets. The finiteness of the partition tells us that $m(A) = m(A_{1})+\ldots+m(A_{k})$ where m is the measure of a set.
Since $\1 le g(P) \le n$ for each P in the figure, we have $k \le n$. And also, $m(A_{i}) = g(f(A_{i})) \cdot m(f(A_{i}))$. (I hope you understand why, maybe it's not so clear). Finally, $m(f(A_{1}))+\ldots+m(f(A_{k})) = m(f(A_{1}) \cup \ldots \cup f(A_{k})) \le m([0,1) \times [0,1)) = 1$.

The conclusion is that $m(A) = \sum g(f(A_{i})) \cdot m(f(A_{i})) \le \sum n \cdot m(f(A_{i})) \le$ $n \cdot \sum m(f(A_{i})) \le n \cdot 1 \le n$. This contradicts the hypotesis that $m(A) > n$.

Just an addition: in order to prove that $m(A_{i}) = g(f(A_{i})) \cdot m(f(A_{i}))$ you should use countable addittivity of the measure and the fact that $\mathbb{Z}^{2i}$ is countable.

Maybe one could solve this problem without those facts... but i suppose you can't, because the question is too general.

I've seen this before,this is a theorem of Blichefeldt.

Yeah - strictly speaking it is Blichfeldt's Lemma - it essentially involves translating everything to (0,0)'s unit square and then finishing off with Dirichlet's Box Principle (a.k.a. PHP)


An equilateral triangle has side length 10. A second equilateral triangle is formed by connecting the midpoints of the sides of the first triangle. A third is formed by connecting the midpoints of the sides of the second triangle, etc. This process is repeated infinitely. What is the sum of the areas of all of the triangles?

[hide]Each triangle is  the area of the previous one, so the sum of all the areas is  times the area of the original triangle, which is .[/hide]

[hide="solution"]
apply $\frac{a}{1-r}$ to the sequence:
$\frac{\sqrt3}{4}(100+25+....)$
So answer is:
$\frac{\sqrt3}{4}*\frac{400}{3}$
$\frac{100}{\sqrt3}$
[/hide]

The area of the original equilateral triangle is:
$A=\frac{s^{2}\sqrt3}{4}=\frac{100\sqrt{3}}{4}$
The next equilateral triangle will have area $\frac{1}{4}$ of the original triangle, and the one after that will have area $\frac{1}{16}$ of the original triangle, etc...
$1+\frac{1}{4}+\frac{1}{16}+...=\frac{4}{3}$

All we have to do is multiply the original triangle's area by $\frac{4}{3}$:
$(\frac{4}{3})(\frac{100\sqrt{3}}{4})=\frac{100\sqrt{3}}{3}$


function
integration
calculus
real analysis
real analysis solved

Let $u(x,y)$ a smooth function defined on $R^2$ wich $2\pi$ periodic in
each variable
i.e. $u(x+2n\pi,y+2m\pi)=u(x,y)$ for all $x,y \in R$, $n,m \in Z$

Prove that $\int_{0}^{2\pi}\int_{0}^{2\pi}(u_{xx}u_{yy}-u^2_{xy})dxdy=0$

Green's theorem: $\int_R u_{xx}u_{yy}-u_{xy}u_{xy}\, dx\, dy = \int_{dR} u_{xx}u_y\, dx + u_{xy}u_y\, dy$.

The integral around the boundary is zero because the top edge cancels the bottom edge and the right edge cancels the left edge.

Very much like your wave equation problem. I should be able to produce two solutions as before.

The first solution uses integration by parts:

$\int_0^{2\pi}\int_0^{\2\pi}u_{xy}^2\,dx\,dy$

$=\int_0^{2\pi}[ u_yu_{xy}|_0^{2\pi}- \int_0^{2\pi}u_yu_{xxy}\,dx]dy$

$=-\int_0^{2\pi}\int_0^{2\pi}u_yu_{xxy}\,dx\,dy$ (The boundary term vanishes by periodicity.)

$=-\int_0^{2\pi}[\int_0^{2\pi}u_yu_{xxy}\,dy]dx$

$=-\int_0^{2\pi}[ u_yu_{xx}|_0^{2\pi} -\int_0^{2\pi}u_{yy}u_{xx}\,dy]dx$

$=\int_0^{2\pi}\int_0^{2\pi}u_{xx}u_{yy}\,dx\,dy$

And the Fourier series solution:

Suppose $u(x,y)=\sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty} \hat{u}(j,k)e^{i(jx+ky)}.$

Then we have Parseval's identity:

 $\int_0^{2\pi}\int_0^{2\pi}u\overline{v}\,dx\,dy = 4\pi^2\sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty} \hat{u}(j,k)\overline{\hat{v}(j,k)}.$

$u_{xy} = -\sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}jk\,\hat{u}(j,k) e^{i(jx+ky)}.$

$u_{xx} = -\sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}j^2\,\hat{u}(j,k) e^{i(jx+ky)}.$

$u_{yy} = -\sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}k^2\,\hat{u}(j,k) e^{i(jx+ky)}.$

So by Parseval:

$\int_0^{2\pi}\int_0^{2\pi}|u_{xy}|^2\,dx\,dy= 4\pi^2\sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}(jk)^2 |\hat{u}(j,k)|^2$, and

$\int_0^{2\pi}\int_0^{2\pi}u_{xx}\overline{u_{yy}}\,dx\,dy= 4\pi^2\sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}j^2k^2 |\hat{u}(j,k)|^2.$

(OK, jmerry's Green's Theorem solution is simpler.)


function
induction
calculus
derivative
complex analysis
complex analysis unsolved

$A$ the set of analytic functions in the open unit disk of $\mathbb{C}$, continuous on the closed unit disk.
Suppose that $f_1,..,f_n \in A$ do not have a common zero. Prove that there exists $g_1,..,g_n \in A$ such that $f_1g_1+f_2g_2+..+f_ng_n=1$ on the closed unit disk.

How about $g_j = \frac{1}{nf_j}$ except on a very small neighborhood of the 0s of $f_j$? Such region should be so small that none overlap with any other such regions of any other $g_i$. In those regions, for the other functions, correct by adding $\frac{1 - nf_jg_j}{f_i(n)(n-1)}$ in that neighborhood. $g_j$ itself in those regions should just be filled in so that $g_j$ is still in $A$. Since $g_j$ and $f_j$ are in $A$, that addition to the other $g_i$ shouldn't cause any problem.

Outside of the neighborhoods of 0s, it is clear we have $n * \frac{1}{n} = 1$. In those neighborhoods we get $\frac{n-1}{n} + g_if_i + (n-1) * \frac{1 - nf_ig_i}{n(n-1)} = 1$ also. I think we still have all $g_i \in A$.

Is it ok?

(this solution doesn't work if $n=1$, but in that case, if $f$ has any 0s, the answer is no, you can't find such a $g$)

your solution is not very clear for me  :)  and how do you show that each $g_i$ is still in $A$ because there is surely a problem at the border of the neighborhoods of the zeroes (different definitions for the $g_i$ inside ande outside these neightborhoods)

[quote="JBL"]Induction would work nicely ... you really just need it for two functions.[/quote]
Yes, and for the proof in the case of two functions e can use Mittag-Leffler interpolation theorem, say you have $f_1,f_2$ withous common zeros and if we restrict to the case when zeros $z_j,j=1,2,3,...$ of $f_1$ are simple, then there exists analytical in $A$ $g_2$ such that $g_2(z_j)=\frac{1}{f_2(z_j)}$ so that $1-g_2f_2$ vanishes at all zeros of $f_1$ and then $g_1= \frac{1-g_2f_2}{f_1}$ would be analytical in $A$ and $f_1g_1+f_2g_2=1$.

To deal with the case of multiple roots of $f_1$ we can interpolate derivatives of it, in the same way as it is stated in Interpolation theorem for example in W.Rudin, "Real and complex analysis".


Well, I asked my math teacher about being able to take the AMC.  She told me that they used to offer it at my school, but since the offered the PSAT, they stopped giving it out.  And, my math teacher told me that unless 10 people would take the AMC, they school probably wouldn't pay for it, since the least amount of tests you can buy is 10.  So I don't really want to push her anymore to have the school offer it, so I was wondering, would it be possible to have a neighbor, or another adult I know, or maybe even a parent proctor the AMC for me?  And if not, can anyone tell me what I should do to try to get the school to offer it, since I already feel like I've been bugging my math teacher too much?

you may have to offer to pay for it yourself, or possibly pay for just the left over tests. i think a packet of 10 tests is around 50-60 dollars. you may want to talk to your principal or someone else to see if they will support you. and is finding 10 people to take the test that difficult?

By far the simplest and preferred way to administer the AMC is through a school, by having a teacher, counselor or similar school official register.  You (or a parent) may certainly offer to pay to have the contest adminstered through your school.  Of course the contest can be administered before or after school so that it does not interfere with the school day or scheduled activities.

Some alternatives are that you could form a math club or math team which raises funds to purchase the AMC contests.  You could approach the PTA/PTO to request funds to purchase the AMC contests.  Approach a similar club or team such as a science club or chess team to sponsor the contests.  Talk seriously to your math teacher or the school math supervisor.  (This is different form "bugging your math teacher".)  Show them copies of recent year's contests.  Show them the awards and benefits possible from giving the AMC contests, even as a club activity.

There are other ways to administer or take the AMC contests, but these ways have the side-effect of working outside of schools, and outside of the mathematics teaching profession.  I think working outside of those channels ultimately erodes the mathematical base that we all depend on.  Should it be necessary later on, say December, those ways can be brought in to play for your case.

Steve Dunbar
AMC Director

My son is Homeschooled. I found a private school offering the test, and asked if my son could take it with them. I had to register him as a guest, which cost 30ish dollars. They sent his copy of the test to the school. They also took care of the AIME in the same way (for no additional cost). This school had several kids from other schools also doing the same thing.

I also heard of a homeschooling family that went to a local school and offered to pay for the entire test for 10 kids (the minimum). If your family is supportive, maybe they can offer to sponsor the school. It will be a tax write-off from them (with a letter from the school).


algebra
polynomial
AMC
USA(J)MO
USAMO
algebra proposed

Suppose that $ P$ is a polynomial of degree $ 3n$ such that
$ P(0)\equal{}P(3)\equal{}...\equal{}P(3n)\equal{}2$,
$ P(1)\equal{}P(4)\equal{}...\equal{}P(3n\minus{}2)\equal{}1$,
$ P(2)\equal{}P(5)\equal{}...\equal{}P(3n\minus{}1)\equal{}0$.
If $ P(3n\plus{}1)\equal{}730$,find $ n$.


percent logarithms geometry trigonometry calculus inequalities integration

Every kid is different and there is no one way. In trying to get my little brother to understand, for example, fractions, I have had tro explain things 5 different times or more. Basically just keep explaining it in different ways, because if they didn't get it the first way you explained it, repeating what you said is not going to make them understand it any more. I'm not a professional teacher or anything, but I like explaining things and this is what I've learned.

The thing to do is to keep the sessions short so the interest builds.

Moving objects helps them see the concepts.  Don't take anything that would be dangerous to eat unless you want to start with three and eat one to get two.  Concepts seem to have some ordinal numbers then "many."  As kids grow older, the number of ordinal numbers increases.

Negative ordinal numbers would be next, with zero quick to learn after "if you owed me a M (one M&M) and Mom gave you two, how many would you have?

I've taught the concepts for trig, calculus (integeral and differential), stats, etc. to folks as young as five.  They couldn't solve problems, but understanding the concepts put them far ahead when their math class "introduced" the concepts to them.

I'd suggest two things:  don't go too fast or too far, KEEP IT FUN for them.

I taught myself all the math that I know from books.
Mostly from [url=http://www.murderousmaths.co.uk/books/bookshop.htm]the murderous maths books.[/url]

This issue was the subject of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=33140#p33140]my very first AoPS post[/url].

[quote="fatboy"]Is thier any discovery or socratic way to teach young children math?[/quote]

According to some people, yes: http://www.garlikov.com/Soc_Meth.html

(Interestingly, I was independently referred to this site just a couple of days after you posted your question...)

[quote="wizard2378"]I've taught the concepts for trig, calculus (integeral and differential), stats, etc. to folks as young as five.  They couldn't solve problems, but understanding the concepts put them far ahead when their math class "introduced" the concepts to them.[/quote]

This sounds really interesting, so I was wondering if you could try to explain some of the ways you've taught the concepts for trig and calculus.

well, in egypt algebra and statistics are taught in 6th grade, geometry in 7th, trig, logarithms and "more advanced stuff" in 8th grade, then calculus in high school
i didn't feel it was rushed at all, i'm not sure how differently mainstream american elementary school math is done, but i did multiplication and fractions in 3 grade, division in 4th, and (i guess) prealgebra ish stuff in 5th  :flex:
then i came to the US and couldn't skip anything because i didn't take any of the required tests for NY state, so i had to take everything over  :mad:

[quote="zeb"]yay! this is what I've been saying. I learned math concepts at an early age too.

unfortunately, I have no clue how to do it. Teaching little children is like teaching a rock, in general.

But theoretically, you could teach addition and subtraction in 1st grade, multiplication and division in 2nd, fractions, percents, exponents, & logarithms in 3rd, algebra 1 and 2 in 4th (prealgebra is just a review of that other stuff), geometry and trigonometry in 5th, calculus in 6th (precalculus always seemed like a waste of time to me), abstract algebra in 7th, and more advanced stuff (i.e. problemsolving, combinatorics, number theory, inequalities, etc.) in 8th. By highschool they could be solving unproven conjectures. I really think that school is too slow...[/quote]
Your teh first person tht I know of that believes in what I believe in.  School is too slow.

[quote="nonie"][quote="zeb"]But theoretically, you could teach addition and subtraction in 1st grade, multiplication and division in 2nd, fractions, percents, exponents, & logarithms in 3rd, algebra 1 and 2 in 4th (prealgebra is just a review of that other stuff), geometry and trigonometry in 5th, calculus in 6th (precalculus always seemed like a waste of time to me), abstract algebra in 7th, and more advanced stuff (i.e. problemsolving, combinatorics, number theory, inequalities, etc.) in 8th. By highschool they could be solving unproven conjectures. I really think that school is too slow...[/quote]

Sure theoretically, but NOT pratically! Some people in my class-(but not my math class )(I'm in 7th) still ask me simple questions like what's 12 times 12 or is it d=rt or r=td?[/quote]

I believe that this is due to the fact that the educational system keeps lowering teh level in math, which is what a main part of your brain is dedicated to. (Imagniation, mathematical thought, and computing...parietal lobe).


From a scale of 1 to 5, 1 being not important and 5 being very important.

This is a highly discussed topic already.

For me, 3.

1.5
maybe 2

your FTW rating does not represent your true mathematical ability
it is merely a thing used to make the game more interesting

Important in terms of what? Life? Your mathematical ability? FTW ability?

Dude my score would totally be a 5. Like if my rating dropped a bunch, I would go all suicidal and wear all black and such.

O rly?

Really.

[quote="BOGTRO"]Rating isn't that important, although higher rated people might not play lower rated people (COUGH AIME15) because of their rating.[/quote]

:D

For me, it's about 4.25 - I won't go suicidal *cough* if I lose rating, but I'll get upset.

for me it's about 2.5
i only get upset if i lose rating by like a split second or my internet fails

Please roll a die.

1.
This is the lowest possible option (supposedly), which is what rating deserves.  Ratings are determined by who you face, which is pretty dang random unless you change that.

2.
Rating help you see who you should face.  Unless you've played at least 2500 games, in which case you should have a good idea of who to play.

3.
Ratings are very helpful to determine who to beat up as well as who to be scared of.  Also useful in boasting.

4.
There wouldn't be much of a point to FTW without ratings, that's what makes it fun.

5.
Like if my rating dropped a bunch, I would go all suicidal and wear all black and such. ~Mewto555555555555555555555

6.
My other self and I use ratings to determine social hierarchy. We currently have the same rating.

A dice is a number from 1 to 6. Not 1 to 5.

5 for me...My rating is too high for ppl... :P

I think 5 should actually consist of the following parts:

1. Go emo when you lose.

2. Violate all of your ethos in a quest to gain more rating. See: Free Rating. I used to know a youtube video shedding great light on this, but whenever I post a link, it gets mysteriously changed...

3. Uh like, do 1 and 2, and then yell at others when they do 1 and 2. Then fess up and say your love for rating is only like a 3 or 4 on the scale!

[quote="Mewto55555"]|0|????????????????/[/quote]

Yup (I forgot the | brackets). Why is anyone better with bigger rating? I don't want  bigger than 1500+(some 2 digit number)

You see someone on the street and say : ''I got bigger rating on FTW!!!'' - that's a 5 :wink:

Ahh, I just lost 24 rating right now. so sad :( . Yes, I value rating highly.

Probably a 2.  It used to be a 4.99, but I decided that FTW was mostly about fun and not about rating.

mathblitz, I totally agree with you.

[quote="mathblitz"]0 though I would go lower if i could. Why do you care about your rating? high ratings dont say your a better person then someone else. It just says that 1. your good at typing and mental mathematics or 2. Youre so vain or depressed that you need some way to prove you are better then everyone. The rating craze is what caused people to nag levans too make rating easier to gain because people wanted higher ratings to be at the top. All those pitiful (and I emphasize PITIFUL) rating obsessed people will probably try to get levans to make things easier when Aime or mewto reach a rating that can not be overtaken. In my eyes rating is the nicotine of FTW. Im not trying to slander FTW just the rating and those who try to prove to themselves or someone else that their the best.[/quote]

SORE LOSER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

(note: the above was a joke)


Χαιρετώ όλα τα μέλη του forum. Ελπίζω να μπορέσω να βοηθήσω και εγώ την υπέροχη αυτή προσπάθεια
R.B :lol:

γεια σου και καλως ηρθες, r_boris  :)

Καλώς ήλθες. Είναι πολύ θετικό που το forum προσελκύει όλο και περισσότερα μέλη. Βάλε το όνομά σου και ό,τι άλλες πληροφορίες θέλεις στο topic που συγκεντρώθηκαν τα ονόματα όλων των μελών του forum. :wink: Με χαρά περιμένουμε και τη δική σου συμβολή στο βιβλίο. Καλή αρχή και καλή δουλειά.

Καλή αρχή r_boris και :welcomeani: στο forum :wink:


[quote="beginner01"]Does there exist a monotonic bijective function from the set of real numbers to itself that is discontinuous at some point in $ \mathbb{R}$?[/quote]

No : Wlog say $ f(x)$ is increasing and suppose $ f(x)$ is not continuous at $ x\equal{}a$

Choose any increasing sequence $ x_n$ whose limit is $ a$ such that the limit of $ f(x_n)$ is not $ f(a)$. $ f(x_n)$ is increasing and $ f(x_n)<f(a)$. And so $ f(x_n)$ has a limit $ b<f(a)$

Consider now $ d\in(b,f(a))$ and $ c$ such that $ f(c)\equal{}d$ (which exists since $ f(x)$ is bijective) :
Since $ f(c)<f(a)$ : $ c<a$
Since $ f(c)>b$, $ c>x_n$ $ \forall n$ and so $ c\ge a$
Hence contradiction.

So $ f(x)$ is continuous.


integration
calculus
derivative
function
abstract algebra
calculus computations

[hide="Click to enter the Minoc"]indefinite integral, viewed as the inverse of differentation, does not refer to a single function but rather a set of antiderivatives. That is, we must consider the constant of integration, $C$. Then we have $0+C = 1+C$, which is indeed true (if we recognize $C$ as the kernel of differentiation).[/hide]


ratio
linear algebra
matrix
vector
algorithm
algebra
system of equations

balance the following redox reaction:

$\text{HIO}_{3}+\text{FeI}_{2}+\text{HCl}\rightarrow\text{FeCl}_{3}+\text{ICl}+\text{HOH}$.

25HCl ;)

When it comes to balancing [url=http://www.chembuddy.com/?left=EBAS&right=equation-balancing-stoichiometry]EBAS[/url] rulez...

[quote="gregx007"]The nice thing about redox problems is that they always look a lot harder than they are.  With some careful reasoning, it becomes easy to balance such equations.

First, you have to identify 1) what is being oxidized and 2) what is being reduced.  To do this, assign oxidation numbers to all atoms and see which numbers change from the left side to the right side.  In this case we see that:

1)  $I$ from $HIO_{3}$ goes from +5 to +1.
and
2)  $I$ from $FeI_{2}$ goes from -1 to +1.
     Fe goes from +2 to +3.

Next, you need to write balanced half-reactions, and then combine them in the correct ratios.  Let me know if you still are having trouble with this problem, and I can give more help.[/quote]
are these the half-reactions?

$HIO_{3}\rightarrow ICl$
$FeI_{2}\rightarrow 2ICl$
$FeI_{2}\rightarrow FeCl_{3}$

don't i have to like add $H_{2}O$ and electrons somewhere?

Yes, if you are going to take the half reactions approach, you need to write in water, electrons, and hydrogen/hydroxide ions (depending on if the reaction occurs in acidic or basic solution).  I haven't worked through the details myself, but it seems that one should add the two oxidation reactions first in the proper ratio, and then fit in the reduction.

You make an interesting point.  I'd imagine that's exactly what the chem balance program does.  I, for one, however abhor systems of equations (unless they happen to be nice and solvable in a few steps).  Plus, it seems to me that separating the reaction into half-reactions gives you more insight into what chemical changes are actually occurring (not that half-reactions actually occur).

However, that's just my take on it.  Perhaps I just feel that way because it's what I am used to.

[quote="Dvagner"]couldn't one take a more mathematical approach?
1. recognizing that Fe appears only in one compound on each side and giving both compounds the same coefficient of 1(if other coefficients end up rational, the final step would be to multiply by an LCD; it isn't the numbers that are important, its the ratios)

2. assigning variables w,x,y,z to all other compounds yielding:
$wHIO_{3}+FeI_{2}+xHCl \to FeCl_{3}+yICl+zHOH$

3. establishing a system of equations, knowing each element is conserved.
$I: w+2 = y$
$Cl: x = 3+y$
$O: 3w = z$
$H: w+x = 2z$

4. solving the system yields:
$w = \frac{5}{4}$
$x = \frac{25}{4}$
$y = \frac{13}{4}$
$z = \frac{15}{4}$

5. multiplying the entire equation by four yields the balanced equation:
$5HIO_{3}+4FeI_{2}+25HCl\to 4FeCl_{3}+13ICl+15HOH$[/quote]

Holy crap!!!  I never thought of balancing reaction equations (especially ones that are huge and hard to balance) like that... thinking of the coefficients as a system of variables... thanks for showing that!  However, doing it like that could be inefficient... oh well...

If you don't like linear systems, it's probably because you haven't disciplined yourself in to writing them in matrix form and you haven't practiced Gaussian elimination.

I'm teaching a linear algebra course right now, and the particular textbook I'm using has a section entitled "Applications of Linear Systems." The homework section that follows has exactly three types of problems: economics problems based on Leontief input/output models, network flow problems, and balancing chemical equations. There are six such chemical problems, all of which are redox equations, and two of which are so nasty that the texbook marks them with a symbol that means "use a computer algebra system."

As linear systems, they are all structurally homogeneous - that is, they can all be written as $Ax=0,$ where $x$ is the vector to be solved for and $A$ is some matrix. Two things about homogeneous systems: first, they are always consistent, because they always have the solution $x=0.$ Second, you're never interested in the trivial (zero) solution - in fact, in many cases (like these chemistry problems), you're likely to regard it as not a solution at all.

What we're hoping for, in the chemical reaction case, is a one-dimensional solution space, which we're likely to regard as a "unique" solution, and will report the smallest integer vector in that solution space.

Of course, if some linear algebra student were to ask me about one of these problems, I'd probably show them how to solve it like a chemist, with oxidation numbers and half-reactions.

[quote]
If you don't like linear systems, it's probably because you haven't disciplined yourself in to writing them in matrix form and you haven't practiced Gaussian elimination. [/quote]

Well, what I don't like about linear systems is the fact they they are completely mechanical.  There is little, if any, room for clever or interesting manipulation (unless you are given a nice one).  If all you are doing is an algorithm, it seems rather pointless to ask a student to do the problem versus a computer.  So no, I wouldn't say it has anything to do with not using matrices.

On the other hand, I would say that the balance-by-chemistry-equations method is less mechanical in that it requires some non-algorithmic insight to figure out what is going on.  This is of course debatable, and to tell you the truth my favorite solution to the general balancing problem is, have a computer do it for you  :lol: .

Yet if you have to do it by hand, I still maintain that a lot of interesting chemistry is lost by trying to compress this problem into mathematics.


MATHCOUNTS Gauss AMC AIME FTW USA(J)MO USAMO

This mock competition would be even more improved, better than ever as compared to my first Mock Test!

In this tourney, everyone would sign up to participate IN ADVANCE. A deadline would be proposed as to my discretion to whatever date I feel is most appropriate. But please post here in order to sign up!

THE RULES:

This MATHCOUNTS tournament would be very much like a full competition, with a Sprint, Target, Team, and Countdown Round. Everyone who signs up must also form a team!!!!! You guys, please also create a team name when signing up, and make sure that everyone approves your team selection. Please also avoid any vulgar or inappropriate team names. These teams would be used for competing in the Team Round, and team scores would also be computed. Team scores would also be taken into consideration in the case when a tiebreaker must be used, so note that there is NO TIEBREAKER ROUND as a separate test!

The individual rounds would have the same rules for regular MATHCOUNTS competitions. Sprint Round-40 minutes, no calculator. NOTE THAT FOR ALL ROUNDS YOU ARE ON YOUR HONOR NOT TO CHEAT. It is my final decision to disqualify any competitors if I have a feeling that they are cheating. Target Round-6 minutes per pair of problems, calculators allowed. Please also follow the instructions printed on the front cover of each round, so that we can maintain integrity in this contest, thanks! Also PM your answers to me when you finish.

For the Team Round, you and your teammates are allowed to connect via any sorts of communication to solve problems AS LONG AS THE ANSWERS REPRESENT THE TEAM'S OWN WORK. No forum discussion of any problems in this mock competition is permitted at any time. Please also, when you register, designate a team captain for your team. This person would be responsible to PM me his/her team's official answers. And please read this statement carefully; it's extremely important:

[b]IF ANYONE OTHER THAN THE CAPTAIN OF A TEAM SUBMITS THE ANSWERS, THE TEAM WILL BE DISQUALIFIED FROM THE TEAM ROUND!!!!![/b]

This is very important. I cannot really stress this more than ever. So, please just be on your honor to follow the rules, and that way, we can have a smooth, honest, and integrated competition.

At the end, I will compute individual and team scores and announce the winners of the individual rounds. Answer keys for the written rounds would be posted as well. In addition, the top competitors (I don't know how many yet) will compete in a Countdown Round tourney on FTW. A separate winner and runner-up will be determined from this.

So, have fun! Tell your friends about this, and SIGN UP!!!!!! :D

EDIT: Also note that the written rounds will be posted in a PDF file after signups close.

Uhhh...which team do you guys want to be on? We need to also know team names and stuff.

Signups:
1. AIME15
2. mewto55555

I'll sign up, but I'm not fully sure that I can do it. mewto's team.

Ok.

Signups so far:

No Team:
1. AIME15

Team No Name:
2. mewto55555
3. math154

yo dang yo i sign up yo i am going solo yo unless someone decides to team up with me yo cause i don't want my shine to be muted by pythag011's shine yo so yo i'm in yo. Do you think i'm saying too much "yo" yo?

Yes you are poincare. :D

No Team:
1. Poincare

Team 1337:
2. AIME15
3. pythag011

Team No Name:
4. math154
5. mewto55555

I join Team 1337.

Hmmm...this reminds me of:

"isabella2296's Super Tourney! A Tourney Like No Other."

But I won't demand a copyright on the name.  :D

By the way, Gauss, I made you a friend on chess.com.  :)

[quote="ernie"][quote="gauss1181"]No people can't do that. Signups have already closed.

REMINDER: ALL ANSWERS DUE TOMORROW!!!! I still haven't received:

Team Rounds from I HATE SCHOOL, 0, and MewPwns.

Sprints from: Poincare, abacadaea, 5849...x, THECHAMP, mewto55555, nikeballa96, CA Math.
Targets from: Poincare, abacadaea, 5849...x, THECHAMP, mewto55555, nikeballa96, CA Math, isabella2296

BTW math154 and Aylsero left this tourney :([/quote]

You haven't received my target either.[/quote]
Then plz submit ur answers.... :|

Update: Aylsero is not dropping out!!! YAY!!!! Team 0 should be happy by then....

As long as Aylsero finds the right page where the mock test is....

Update:

Since there hasn't been that many submissions yet, I'm extending the deadline to Monday, December 15.

This way, we should have 100% participation for more activity and tension in this. Cmon peeps!

So far, these are what I need left:

Team Rounds from Teams I HATE SCHOOL (Captain Wickedestjr), MewPwns (Captain mewto55555), and 0 (Captain aleph0 :D ).

Sprints and Targets from THE CHAMP, Aylsero, abacadaea, 5849...x, Poincare, math154, mewto55555, CA Math, nikeballa96.)

Sooooooo guyz plzzzzz submit so that i don't have to keep mentioning your name on this thread...... :D

5849206328x says that he's too busy to do the sprint and target, so he won't submit. abacadaea isn't submitting either.

Dude, but the deadline has been extended....

Sorry, I can't tell you that. That's part of the rules, no asking for help. :wink:

GUYZ THE ANSWERS ARE DUE TONIGHT BY 9 PM CST!!!!!!!!
Please take a couple last moments and submit your answers to receive credit!

EDIT: Answer submissions due last night! NO more now!

I will release the results, Countdown round (and possibly a Masters Round) accommodations later. Also guys I know that (and admit) some parts of my test was too unclear; I know I should have proofread before submitting :oops:

And please also READ THE PREVIOUS PAGES OF THIS FORUM THREAD BEFORE YOU PROTEST!!!! Some clarifications were made there.

If you have a question about why you got a certain problem wrong, please ask here, and I'll provide the answer, solution, and any clarifications of ambiguity.

When is "later"? Sorry, I'm just very eager to see how I did.


A right isosceles triangle has a hypotenuse with measure 18 centimeters. What is the length, in centimeters, of one leg of the triangle? Express your answer in simplest radical form.

The hypotenuse is square root 324, so each leg is square root 162 or 9 sqrt 2.

In an isosceles triangle, when given the hypotenuse, the answer will be $ \frac{h \sqrt{2}}{2}$.


calculus
integration
trigonometry
real analysis
real analysis solved

Solve the following Integral:

\[
\int\limits_{\frac{\pi }{6}}^{\frac{5}{6}\pi } {\frac{x}{{1 + \left\{ {\sin x} \right\}}}} dx
\]

{sinx} means fractional part of sinx.

it is
\[
\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{x}{1+\sin x}dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\pi-x}{1+\sin x}dx = \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{dx}{1+\sin x}
\]
then you can try trignometric substitution : $t=\tan \frac{x}{2}$...
and the answer is beautiful : $\frac{\pi}{\sqrt3}$

I think trigonometric substitution is t = tan(x/2).
Then all is good.

Yes, you are right.
It was a typo. Thanks for pointing it out.


find all natural x,y that :
\[ x^2\plus{}y^2\plus{}5x\plus{}5y\minus{}2xy\equal{}1500\]

Assume, without loss of generality, that $ x\geq y$.

Take this $ \bmod5$ to get $ (x \minus{} y)^2\equiv0\pmod5\implies x \minus{} y \equal{} 5a$.

We have $ (x \minus{} y)^2 \plus{} 5(x \plus{} y) \equal{} 25a^2 \plus{} 5(x \plus{} y) \equal{} 1500\implies5a^2 \plus{} x \plus{} y \equal{} 300$.

Take this $ \bmod5$ again to find that

$ x \plus{} y\equiv0\pmod5\implies x \plus{} y \equal{} 5b$, so $ 5a^2 \plus{} 5b \equal{} 300\implies a^2 \plus{} b \equal{} 60$, which means that

$ (a,b) \equal{} (0,60),(1,59),(2,56),(3,51),(4,44),(5,35),(6,24),(7,11)$ (since $ a,b\geq0$), giving solutions

$ (x,y) \equal{} \left(\frac {5(a \plus{} b)}2,\frac {5(b \minus{} a)}2\right)$

$ \equal{} \boxed{(150,150),(150,145),(145,135),(135,120),(120,100),(100,75),(75,45),(45,10)}$ (and all flips of the ordered pairs).


The equation is:  a^2 + (a+b)^2 = (a+9b)^2

a, b > 0

What is a / b ?


In ham yek soale gashang
 Agar $ a,b,c,d>0$ sabet konid
\[ a^2\plus{}b^2\plus{}c^2\plus{}abc\plus{}2\geq ab\plus{}bc\plus{}ca\plus{}a\plus{}b\plus{}c\]
Va ba hamoon shart sabet konid
\[ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}abcd\plus{}1\geq ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd\]
lezzat bebarid :wink:

Chera kasi in soal haye ziba ra hal nakard
Baraye gesmate avale soal yek rahnamayi mikonam
[hide]as schur estefade konid[/hide]
va baraye gesmate badi...
[hide]sharmandeam dar in ja rahnamayi mosavie halle  :D [/hide]

yekhorde age bishtar az 24 saat sab koni ehtemalan yeki peida mishe ke halleshoon kone,shayad yeki asan oonrooz nayad too in site,niazi be rahnamai nis,forsat bede halleshoon mikonan...

[quote="M.A"]In ham yek soale gashang
 Agar $ a,b,c,d > 0$ sabet konid
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \plus{} 2\geq ab \plus{} bc \plus{} ca \plus{} a \plus{} b \plus{} c
\]
[/quote]
didi goftam ajale nakon,in az rahe avvali ke ghablan post shode:
[url]http://www.mathlinks.ro/viewtopic.php?p=860276#860276[/url]

rast migan forsat bede :lol: alan bishtariya daran emtehan midan :( vaght kame

Khob fekr konam alan vagte javabe albatte fekr konam rahe digeyi ham dare

Farz konid $ a\ge b\ge c\ge d\ge 0$ va

$ f(a,b,c,d) \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} abcd \plus{} 1 \minus{} ab \minus{} ac \minus{} ad \minus{} bc \minus{} bd \minus{} cd$ bashad

Hal sabet mikonim $ f(a,b,c,d)\ge f(\sqrt {ac},b,\sqrt {ac},d) \leftrightarrow$

\[ a^2 \plus{} c^2 \minus{} (a \plus{} c)(b \plus{} d)\ge 2ac \minus{} (2\sqrt {ac})(b \plus{} d)\leftrightarrow
\]

\[ (a \minus{} c)^2\ge (b \plus{} d)(\sqrt {a} \minus{} \sqrt {c})^2\leftrightarrow
\]

\[ (\sqrt {a} \plus{} \sqrt {c})^2\ge b \plus{} d \leftrightarrow
\]

\[ a \plus{} c \plus{} 2\sqrt {ac}\ge b \plus{} d
\]
ke tebge farz $ a\ge b\ge c\ge d\ge 0$ doroste

Hala tebge leme SMV midanim $ f(a,b,c,d)\ge f(t,t,t,d)$

pas kafie sabet konim $ f(t,t,t,d)\ge 0\leftrightarrow$
\[ d^2 \plus{} t^3d \plus{} 1\ge 3td
\]
ke yek namosavi hesabi-hendesi (ya AM-GM) ast :wink:


quadratics
modular arithmetic
number theory unsolved
number theory

Prove that $ 4xyz\minus{}x\minus{}y$ can not be a perfect square, where $ x,y,z \in \mathbb{N}$.

$ 4xyz\minus{}x\minus{}y \equal{} k^2 \implies (4xz\minus{}1)(4yz\minus{}1) \equal{} 4zk^2 \plus{} 1.$ Then if z is a square the RHS is of the form $ a^2 \plus{} 1$ so all prime divisors (2 is excluded because $ 4zk^2 \plus{} 1$ is odd) are of the form $ 4n \plus{} 1$,  but the LHS contains factors of the form $ 4n \plus{}3$ contradiction.
Now to prove z is a square...

You did most of the hard work!

Assume $ z > 1$, then if $ 4xyz \minus{} x \minus{} y \equal{} k^{2}$, we have that $ k > 4xy$.

Thus $ k \equal{} 4xyz' \plus{} l$
This gives us that $ l^2 \equal{} 4xyt \minus{} x \minus{} y$ where $ t$ is an integer such that $ 0 < t < z$.

Thus we can assume $ z \equal{} 1$ and your proof for perfect square $ z$ now kicks in.

why does $ 4xyz \minus{} x \minus{} y \equal{} k^2, z > 1\implies k > 4xyz$?

You are right, I made a mistake.  What I wrote is wrong.

<deleted more crap from me>

Anybody has a solution for this prob?
I'm eager to see one  :(
thx for your nice solution

[quote="moldovan"]Prove that $ 4xyz \minus{} x \minus{} y$ can not be a perfect square, where $ x,y,z \in \mathbb{N}$.[/quote]

We have $ (4xz\minus{}1)(4yz\minus{}1)\equal{}4zw^2\plus{}1$, so $ x^2\equal{}\minus{}z$ has a solution in $ \mathbb{Z}/(4xz\minus{}1)\mathbb{Z}$. If 2|z+1 then $ (\frac{\minus{}z}{4xz\minus{}1})\equal{}\minus{}(\frac{z}{4xz\minus{}1})\equal{}\minus{}((\minus{}1)^{\frac{z\minus{}1}{2}}(\frac{\minus{}1}{z}))\equal{}\minus{}1$ by reciprocity rule, where (a/b) denotes Jacobi symbol and if it is -1 then it is a [i]sufficient[/i] condition to be a non quadratic residue. Otherwise define $ q: \equal{}z2^{\minus{}\upsilon_2(z)} \equiv 1 \pmod 2$ then $ (\frac{\minus{}z}{4xz\minus{}1})\equal{}(\frac{2}{4xz\minus{}1})^{\upsilon_2(z)}(\frac{\minus{}q}{4xz\minus{}1})\equal{}\minus{}1$ since 2 is a quadratic residue modulo a number x such that 8|x+1. Therefore there are no solution.

For $ z\equal{}1$,  the fact that $ 4xyz \minus{} x\minus{} y$ is not perfect square (where $ x,y,z \in \mathbb{N}$) is known as Euler's equation (and can be found in numerous places, e.g. W. Sierpinski). It has however infinitely many integer solutions.


calculus
integration
calculus computations

Partial fractions, in both cases. In the second case, first make the substitution $u=e^t.$


factorial
modular arithmetic
decimal representation
Periodic sequence
IMO Shortlist
Digits

Let $ a_n$ be the last nonzero digit in the decimal representation of the number $ n!.$ Does the sequence $ a_1, a_2, \ldots, a_n, \ldots$ become periodic after a finite number of terms?

No. First of all, note that if $ N$ is the period, then one can find an arbitrarily large $ k>0$ and $ M>0$ such that $ 10^k\plus{}M\cdot 10^{k\plus{}1}$ is divisible by $ N$. Indeed, if $ N\equal{}AB$ where $ A$ has only factors $ 2$ and $ 5$ in its prime decomposition and $ B$ is relatively prime with $ 10$, then, for all large $ k$, the number $ 10^k$ is divisible by $ A$ and $ 1\plus{}10M$ is divisible by $ B$ for an appropriate choice of $ M$. Thus, if the sequence has period $ N$, we should have the same last non-zero digit for $ (10^k\minus{}1)!$ and $ (2\cdot 10^k\plus{}M\cdot 10^{k\plus{}1}\minus{}1)!$. But when we are passing to the next factorials, in the first case this digit remains the same and in the second one it gets "multiplied by $ 2$" (it cannot be $ 5$ because the power of $ 2$ in large factorials far exceeds the power of $ 5$).

Hi  :D . This file is about [b]Last nonzero digit of $ n!$ and $ n^n$[/b].

Go by contradiction. If $n=5q+r$ with $0\le r\le4$, and $L(n)$ denotes the last nonzero digit of $n!$, then it's not hard to show that $L(5q+r)\equiv 2^q L(q)L(r)\pmod5$. Taking $q=4k+3$ and $r=4$, we have (since $2^4\equiv1\pmod5$ and $4!=24$)
\[L(20k+19) \equiv 8L(4k+3)L(4) \equiv 2L(4k+3)\pmod5.\]But $(20k+19) - (4k+3) = 16(k+1)$, so we get a contradiction by taking $d|k+1$ for sufficiently large $k$, where $d$ is the period of the sequence for large $n$. Note that $L(n)$ is never divisible by $5$.

[quote=fedja]Thus, if the sequence has period $ N$, we should have the same last non-zero digit for $ (10^k\minus{}1)!$ and $ (2\cdot 10^k\plus{}M\cdot 10^{k\plus{}1}\minus{}1)!$. But when we are passing to the next factorials, in the first case this digit remains the same and in the second one it gets "multiplied by $ 2$" (it cannot be $ 5$ because the power of $ 2$ in large factorials far exceeds the power of $ 5$).[/quote]

I know I am bumping a really old post, but I was seeing this solution and had some confusion in the above part. When we are passing to the next factorials, in the second case the number is $[10^k(2+10M)]!$. But if $3|2+10M$, then the second one actually gets "multiplied by $6$", in which case the last digit remains the same. How does the given argument get over this case? Any help would be greatly appreciated.


probability
algorithm
Putnam
college contests

There are 10 applicants for a job, labeled 1, 2, ... , 10. Their suitability for the job is ranked in this order (so 1 is more suitable than 2, who is more suitable than 3 and so on). But an applicant's ability is only measured when they are interviewed. They are interviewed in the order a1, a2, ... , a10. The first three interviewed are automatically rejected. Thereafter if an applicant's suitability is better than that of all previous applicants they are accepted (without interviewing the other applicants). If the first nine applicants interviewed are not accepted, then the tenth is automatically accepted. Let ni be the number of permutations a1, ... , a10 which result in applicant i getting the job. Show that n1 > n2 > ... > n8 = n9 = n10. Show that the probability that one of 1, 2, 3 gets the job is over 70% and that the probability of one of 8, 9, 10 getting the job is not more than 10%.

Yes. But I don't understand what is unclear about it :?. You have a clearly described deterministic choosing algorithm applied to a random permutation of candidates. The problem is just asking to estimate the probabilities of certain results.

ok, maybe what I meant is, I have no idea how to do it.  Does anyone know how to prove this problem?


Let $ f$ be a function on natural numbers $ N\minus{}>N$ with the following properties:
1.$ (f(2n)\plus{}f(2n\plus{}1)\plus{}1)(f(2n\plus{}1)\minus{}f(2n)\minus{}1)\equal{}3(1\plus{}2f(n))$
2.$ f(2n)\ge f(n)$
for all natural numbers $ n$
Determine all values of $ n$ such that $ f(n)\le 2009$

[quote="every"]Let $ f$ be a function on natural numbers $ N \minus{} > N$ with the following properties:
1.$ (f(2n) \plus{} f(2n \plus{} 1) \plus{} 1)(f(2n \plus{} 1) \minus{} f(2n) \minus{} 1) \equal{} 3(1 \plus{} 2f(n))$
2.$ f(2n)\ge f(n)$
for all natural numbers $ n$
Determine all values of $ n$ such that $ f(n)\le 2009$[/quote]

$ RHS > 0$ $ \implies$ $ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1 > 0$
If $ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1\geq 3$, then $ f(2n) \plus{} f(2n \plus{} 1) \plus{} 1\geq 2f(2n) \plus{} 5$ and $ LHS\geq 6f(2n) \plus{} 15$ so $ 3(1 \plus{} 2f(n))\geq 6f(2n) \plus{} 15$ and so $ f(n)\geq f(2n) \plus{} 2$, which is wrong.

So $ 3 > f(2n \plus{} 1) \minus{} f(2n) \minus{} 1 > 0$

If $ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1 \equal{} 2$, then LHS is even while RHS is odd.

So $ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1 \equal{} 1$ $ \implies$ $ f(2n \plus{} 1) \equal{} f(2n) \plus{} 2$ $ \implies$ the equation becomes $ 2f(2n) \plus{} 3 \equal{} 3 \plus{} 6f(n)$ and so $ f(2n) \equal{} 3f(n)$

So $ f(2n) \equal{} 3f(n)$ and $ f(2n \plus{} 1) \equal{} 3f(n) \plus{} 2$

So knowledge of $ f(1)$ is enough to determine $ f(n)$ $ \forall n$ but we cant go further since we have no information about $ f(1)$. For example, $ f(1) \equal{} 10000$ would imply that no value of $ n$ would be such that $ f(n)\leq 2009$ while $ f(1)\equal{}1$ would imply $ f(n)\leq 2009$ $ \forall n\in[1,127]$ and $ f(n)>2009$ $ \forall n>127$

[quote="every"]Let $ f$ be a function on natural numbers $ N \minus{} > N$ with the following properties:
1.$ (f(2n) \plus{} f(2n \plus{} 1) \plus{} 1)(f(2n \plus{} 1) \minus{} f(2n) \minus{} 1) \equal{} 3(1 \plus{} 2f(n))$
2.$ f(2n)\ge f(n)$
for all natural numbers $ n$
Determine all values of $ n$ such that $ f(n)\le 2009$[/quote]
No No, every. You posted a problem in MYM 45th anniversary contest. It is the greatest contest during  a decade of the Mathematics and Young magazine!. And now the contest is still running. So posting all of problems should be prohibited.
Dear moderators! Please lock this topic and warm every ( he posted many many problems like that)
However , dear pco. I think you should check your solution more carefully. I think your result isn't right!. Read my pm !


15. Let a,b, and c be real constants such that [code]x^2+x+2[/code] is a factor of [code]ax^3+bx^2+cx+5[/code] , and \2x-1 is a factor of ax^3+bx^2+cx-(25/16). Find a+b+c.

21. Let k be a positive integer. A positive integer is said to be a k-flip if the digits of n are reversed in order when it is multiplied by k. For example, 1089x9 = 9801, and 21978 is a 4-flip because 21978x4=87912. Explain why there is no 7-flip integer?

22. Let ABC be an acute-angled triangle? Let D and E be the points on BC and AC, respectively, such that AD is perpendicular to BC and BE is perpendicular to AC. Let P be the point where AD meets the semicircle constructed outwardly on BC, and Q the point where BE meets the semicircle constructed outwardly on AC. Prove that PC=QC.

23. Two friends, Marco and Ian, are talking about their ages.

Ian says, "My age is a zero of a polynomial with integer coefficients."

Having seen the polynomial p(x) Ian was talking about, Marco exclaims, "You mean, you are seven years old? Oppps, sorry I miscalculated! p(7) = 77 and not zero"

"Yes I am older than that," Ian's agreeing reply.

Then Marco mentioned a certain number, but realizes after a while that he was wrong again because the value of the polynomial at that number is 85.

Ian sighs, "I am even older than that number."

Determine Ian's age.
[/code]

[hide="21"]Assume that there is a 7-flip integer $ k\equal{}A_1\cdots A_n$.  Therefore $ 7k\equal{}A_n\cdots A_1$.  It is obvious that $ A_1\equal{}1$.  Therefore $ A_n\equal{}3$, because this is the only number that can be multiplied by 7 to yield a units digit of 1.  However, it can also be shown that $ A_n\geq 7$.  Contradiction.[/hide]

22 is almost the same(easier) than this problem http://www.mathlinks.ro/viewtopic.php?p=904624#904624 but $ A$ changed by $ C$.

 :)

[hide="23"] Let $ a$ be Ian's true age, and let $ m$ be the second age postulated by Marco.  We are given that

$ p(x) \equal{} (a \minus{} x) q(x)$
$ p(7) \equal{} (a \minus{} 7) q(7) \equal{} 77$
$ p(m) \equal{} (a \minus{} m) q(m) \equal{} 85$

where $ a > m > 7$ and $ q(x)$ is a polynomial with integer coefficients.  We conclude that $ a \minus{} 7 | 77$ and $ a \minus{} m | 85$, where $ a \minus{} 7 > a \minus{} m$ and both are positive integers.  The divisors of the first are $ 1, 7, 11, 77$ while the divisors of the second are $ 1, 5, 17, 85$.  We also have the condition $ m \minus{} 7 | q(m) \minus{} q(7)$.

[b]Case:[/b]  $ a \minus{} 7 \equal{} 7$.  We have $ a \equal{} 14, q(7) \equal{} 11$.  When $ a \minus{} m \equal{} 1$ we have $ m \equal{} 13, q(13) \equal{} 85$.  Now, $ m \minus{} 7 \equal{} 6$ while $ q(m) \minus{} q(7) \equal{} 74$, contradiction.  When $ a \minus{} m \equal{} 5$ we have $ m \equal{} 9, q(9) \equal{} 17$.  Now, $ m \minus{} 7 \equal{} \minus{}2$ while $ q(m) \minus{} q(7) \equal{} 6$, so there is no contradiction here and we find that $ \boxed{a \equal{} 14}$.  We can take $ q(x) \equal{} 3x \minus{} 10$.

[b]Case:[/b]  $ a \minus{} 7 \equal{} 11$.  We have $ a \equal{} 18, q(7) \equal{} 7$.  When $ a \minus{} m \equal{} 1$ we have $ m \equal{} 17, q(17) \equal{} 85$.  Now, $ m \minus{} 7 \equal{} 10$ while $ q(m) \minus{} q(7) \equal{} 78$, contradiction.  When $ a \minus{} m \equal{} 5$ we have $ m \equal{} 13, q(13) \equal{} 17$.  Now, $ m \minus{} 7 \equal{} 6$ while $ q(m) \minus{} q(7) \equal{} 10$, contradiction.

[b]Case:[/b]  $ a \minus{} 7 \equal{} 77$.  We have $ a \equal{} 84, q(7) \equal{} 1$.  When $ a \minus{} m \equal{} 1$ we have $ m \equal{} 83, q(83) \equal{} 85$.  Now, $ m \minus{} 7 \equal{} 76$ while $ q(m) \minus{} q(7) \equal{} 84$, contradiction.  When $ a \minus{} m \equal{} 5$ we have $ m \equal{} 79, q(79) \equal{} 17$.  Now, $ m \minus{} 7 \equal{} 72$ while $ q(m) \minus{} q(7) \equal{} 16$, contradiction.  When $ a \minus{} m \equal{} 17$ we have $ m \equal{} 67, q(67) \equal{} 5$.  Now, $ m \minus{} 7 \equal{} 60$ while $ q(m) \minus{} q(7) \equal{} 4$, contradiction.  Finally, when $ a \minus{} m \equal{} 85$ we have $ m \equal{} \minus{}1$, but we stipulated $ m > 7$, contradiction.

Hence $ 14$ is the only possible value of $ a$. [/hide]


algebra
polynomial
inequalities
inequalities solved

[quote="me@home"]In other words using concieve's notation:
$(6,0,0)+(2,2,2)\ge(5,1,0)+(4,2,0) \\ \leftrightarrow (6,0,0)+(2,2,2)-(5,1,0)-(4,2,0) \ge 0$ \[(6,0,0)+(2,2,2)-(5,1,0)-(4,2,0)=\mathbb{S}_{2}+\mathbb{S}_{3}\] $\mathbb{S}_{n}$ is just Schur for exponent $n$
Does something like that identity generalize?[/quote]

Wrong, you have to multiply the $S_{n-1}$ by $(a+b+c)$ to make these two Schur's ineq have the same degree.

Genaralization:
http://www.mathlinks.ro/Forum/viewtopic.php?t=109089

oh yes, right :oops:
that seems like a lot of algebra to brute force, does anyone have a good method (or a link to a good CAS?)


Here's a question of crucial importance to life. When you take a standard receipt from a store and hold it to a lamp, it very quickly turns black. Why is this?

I have never heard of this happening before...
what type of light was it that you were holding it under.

...

I'm guessing this has some sort of metamorphic meaning...

In any event, I fail to see the connection. At all.

Nah, I finally managed to google it, and it's because receipts are made of thermal paper. They are not printed.

Okay...and how is this of crucial importance to life?

...and what's thermal paper? Not printed? Wha--?

I assume that the paper works by changing black whenever heat/light (of a certain intensity) is exposed to it. This would be useful for a machine to print quickly and also explains why putting it close to a light alters its colour

yeah.. that's why if you try drawing on the receipt with just your fingernails, it make a black mark..

it's because it's activated by heat.
And the hand thing works b/c your hand causes friction with the receipt, which creates heat, which makes it black..

i found that out after this one cashier guy circled our $20 savings (with his fingers) at the grocery store.

eh yea same here a guy at the grocery store circled are savings with his finger. I got so intrigued i started drawing images on our reciept, my mom wasnt too happy b/c she likes to save them


I was originally planning to enroll in Intro to Counting & Probability and Intermediate Algebra, but I'll be going to HCSSiM this year, so I'm in a bit of a dilemma (I'm assuming that there will not be that much time there).  Would it be prudent to enroll in Intermediate Algebra even though I will not be here for the majority of it?  I know that the transcripts of the sessions will be available, but how much does one get out of those?

Some students actually prefer the transcripts over the live classes.  It's up to you.  But taking two classes on top of HCSSiM seems like a lot to me.


[quote="ckck"]For what value of n is $\frac{1^{3}+2^{3}+3^{3}+...+n^{3}}{1+2+3+...+n}$=36?[/quote]
[hide="hint"]Note that $\sum^{n}_{a=1}a^{3}=(\sum_{a=1}^{n}a)^{2}$[/hide]
EDIT: torajirou beat me too it by like 1 second so I'll post the full solution.
[hide]Using the hint I said, $\frac{n(n+1)}{2}=36$. Eyeballing or doubling and solving with quadratic formula gives n=8.[/hide]

I wuv this identity.

[hide="solution"]$1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\eft(1+2+3+\cdots+n\right)^{2}$

Therefore 1+2+3+...+n=36. You should memorize the sum of the integers from 1-n where n is from 1=10. 1+2+3+4+5+6+7+8=36; n=8[/hide]


A hotel charges 20 dollars for the use of its dining room and 2.50 dollars a plate for each dinner. An association gives a dinner and charges 3 dollars a plate but invites four nonpaying guests. If each person has one plate, how many paying persons must attend for the association to collect the exact amount needed to pay the hotel?

(1) 60     (3) 40
(2) 44     (4) 20

[hide]
I agree with davidli.  It doesn't make sense for there to be a 20 on the right side since the only money the association is getting if from charging the people at the dinner.

$20+2.5(x+4)=3x$
$20+2.5x+10=3x$
$30+2.5x=3x$
$30=.5x$
$60=x$

$A$
[/hide]

[quote="Interval"]Everyone got A.

Why?

You guys know how to set up equations using the given information in word problems.

Most students do not know.

If I knew how to do this, I would have more than half the battle won.

Thanks![/quote]
You are trying to find the number of people needed so the cost will be equal to the money collected, so you have to make an equation where each side represents the cost. Once you do that, you can solve for $x$.


Let $AA_1$, $BB_1$ and $CC_1$ be the three altitudes of $\triangle{ABC}$, and $H$ be the orthocentre of the triangle. If $S_{BC_1HA_1}=S_{CB_1HA_1}$, prove: $AB=AC$.

[hide]
This problem can be solved by assuming the contrary, but can anyone find a straight proof?[/hide]

First time, the triangle ABC is acute and in this case,
$HA_1=2R\cdot \cos B\cdot \cos C$ a.s.o. Thus, $[BA_1HC_1]=[CA_1HB_1]\Longleftrightarrow$
$\cos^2 B\cdot (a\cos A +c\cos C)=\cos^2 C\cdot (b\cos B+a\cos A)\Longleftrightarrow \ldots \ldots$

[b]Indication. [/b] $\sum a\cdot \cos A =\frac {2S}R;\ X=Y,\ XY\le 0 \Longrightarrow X=Y=0.$


When you have a problem like:

How many sums can you have by choosing two distinct integers out of this set?

$ {a, b, c, d, e, f}$

How would you solve it? I thought you did $ e\plus{}f$ and $ a\plus{}b$ (assuming they are in size order) and find the difference between the two values. But when I came across a question like this, I did that method, and it was wrong. Apparently I was 10 off.

well, assuming all sums are different, its 6c2 which is 15. If they all have common difference, like 1,3,5,7,9,11, then you do $ 11\plus{}9\minus{}3\minus{}1$ divided by common difference $ 16/2\equal{}8$ and add 1

[quote="iin77"]I do it an easier way...

Assuming a common difference between consecutive terms, as in mewto's example, set n as the number of terms.  The answer will be (n-1)+(n-2).

This is because the first term added to any of the other terms will result in a different sum.  Then for the second term, any other term you choose will not give a new sum except when you add to the final term.  Same with every term up until the final term (which can't be added to anything to get a new result).  So, since you can add n-1 different terms to the first term, and n-2 terms to the last term (since the first term is excluded.  The answer is: (n-1)+(n-2).

Hope my explanation isn't too terrible.[/quote]

So its 2n-3?


I don't get it.
Now that I have finished what I think every subject, what do I do?
I just finished Mastery of Counting fundamentals 5 and Principle of Inclusion-Exclusion.
There are no new topics in the report page neither the solution page.
 :maybe:

More topics will be released in the future, such as Introduction to Algebra and more. You may come across nonexistant subjects, they are part of the currently being developed Introduction to Algebra subjects, just ignore them.

As of now, you have finished passing all passable subjects in Alcumus.

Principle of Inclusion-Exclusion isn't really a subject-it unpasses itself in a day or two. Also, you cannot see it on your report.


Consider the sum of $ n$ terms

$ S_n\equal{}1\plus{}\frac{1}{1\plus{}2}\plus{}\frac{1}{1\plus{}2\plus{}3}\plus{}...\plus{}\frac{1}{1\plus{}2\plus{}3\plus{}...\plus{}n}$ for positive integer $ n$.

Find the least rational number $ r$ such that $ S_n<r$ for all natural numbers $ n$.


inequalities
function
algebra theorems
algebra

Can somebody state Jensen inequality for function defined on two variables?


If you search the forum, you will find many similar topics:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=formulas+geometry&t=77530
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=formulas+geometry&t=75573
I simply typed in 'geometry formulas' and clicked search.

better yet, go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73496]here[/url]
it has a bunch of formulas(more than just geom).

Yes, the topic junggi mentioned has all the formulas you will need.

We don't need another one, so locked.


find all functions $ f: \mathbb{Z}\to\mathbb{Z}$ such that:

$ \forall (x,y,z)\in{\mathbb{Z}}$ : $ f(x^{3}\plus{}y^{3}\plus{}z^{3})\equal{}(f(x))^{3}\plus{}(f(y))^{3}\plus{}(f(z))^{3}$

thanx.


[quote="keta"]Does the US offer any major Calculus or above based competitions other than the putnam?[/quote]Up to my knowledge, no.


Find all integers $ x,y,z$ such that $ x^2\plus{}y^2\plus{}z^2\equal{}2xyz$

[quote="nguyentrang"]Find all integers $ x,y,z$ such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2xyz$[/quote]
[hide]
roots of $ y\equal{}x^3\minus{}2ax\minus{}a$ maybe?
this also begs for some sort of factoring solution.

[/hide]

i think this problem involves some basic number theory that is not trivial.

The only solutions end up being (0,0,0)

This is a plausible method.
[hide]
Replace x by $ 2 \cdot x_{1}$, y by $ 2 \cdot y_{1}$, z by $ 2 \cdot z_{1}$
The eqn becomes $ (x_{1})^2 \plus{} (y_{1})^2 \plus{} (z_{1})^2 \equal{} 4 \cdot x_{1} \cdot y_{1} \cdot z_{1}$
Thus by infinite descent the only solns are x=0 , y=0, z=0  :)
See http://www.google.co.in/search?hl=en&q=Infinite+descent&btnG=Google+Search&meta=
      http://en.wikipedia.org/wiki/Infinite_descent
[/hide]

EDITED: IGNORE IF THE METHOD IS WRONG

[quote="madness"]This is a plausible method.
[hide]
Replace x by $ 2 \cdot x_{1}$, y by $ 2 \cdot y_{1}$, z by $ 2 \cdot z_{1}$
The eqn becomes $ (x_{1})^2 \plus{} (y_{1})^2 \plus{} (z_{1})^2 \equal{} 2 \cdot x_{1} \cdot y_{1} \cdot z_{1}$
Thus by infinite descent the only solns are x=0 , y=0, z=0  :)
See http://www.google.co.in/search?hl=en&q=Infinite+descent&btnG=Google+Search&meta=
      http://en.wikipedia.org/wiki/Infinite_descent
[/hide][/quote]

uhh, you end up with an extra factor of 2 on the right.

Oh yeah  :fool: i am  :)
Still cant the method work?  :maybe:

It doesn't work, because the two sides don't have the same degree.  If you multiply each variable by $ k$, the left side will be multiplied by $ k^2$, while the right will be multiplied by $ k^3$, which will change the equation.

Also, the problem asks for integer solutions, and $ \frac{x}{2}$ is not necessarily an integer.

madness is right except for two things:
1. You have to prove that x,y and z all have to be even first. We cannot have two odd and one even nr among them
2. There is a calculation mistake in the working. So actually you end up with the factor 2 on the rhs.

Now infinite descent is to be invoked.

This is a well known problem and has already been solved in Titu Andreescu's book Math Olympiad Treasures if I am not wrong.

[quote="CircleSquared"][quote="nguyentrang"]Find all integers $ x,y,z$ such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2xyz$[/quote]
[hide]
roots of $ y \equal{} x^3 \minus{} 2ax \minus{} a$ maybe?
this also begs for some sort of factoring solution.

[/hide][/quote]

How did you work that out?

Thanks

Hint: Consider the equation: $ x^2\plus{}y^2\plus{}z^2\equal{}2^{n\plus{}1}xyz$ for $ (x,y,z,n)$ nonnegative integers.

Suppose x is even and y and z are odd. Then y^2+z^2 is of the form 4m+2 while 2xyz-x^2 is of the form 4m.

Hence x,y and z all have got to be even.
x=2x1, y=2y1, z=2z1

So you get (x1)^2+(y1)^2+(z1)^2 = 2(x1)(y1)(z1)

Now infinte descent proves that there are no integers satisfying this relation.

(I know my posts are an eyesore. SO somebody pls tell me how to ge latex output from texer into the post)

hsbhatt enclose your mathematical expressions within the dollar signs  :D
Example [code]
$x^2$ [/code]
gives you $ x^2$  :)

It may help to rewrite it as $ \frac{x^2\plus{}y^2\plus{}z^2}{xyz} \equal{} \frac{x}{yz}\plus{}\frac{y}{zx} \plus{} \frac{z}{xy} \equal{} 2$

[quote="The Zuton Force"]It may help to rewrite it as $ \frac {x^2 \plus{} y^2 \plus{} z^2}{xyz} \equal{} \frac {x}{yz} \plus{} \frac {y}{zx} \plus{} \frac {z}{xy} \equal{} 2$[/quote]
I don't think it will help unless you proved that this equation doesn't have any solution when 2 out of 3 variables are negative integers, 1 is positive integer! Using the inequality with x>=y>=z>0 will kill it easily in the case 3 variables>0! :wink:

[quote="hsbhatt"]Suppose x is even and y and z are odd. Then y^2+z^2 is of the form 4m+2 while 2xyz-x^2 is of the form 4m.

Hence x,y and z all have got to be even.
x=2x1, y=2y1, z=2z1

So you get (x1)^2+(y1)^2+(z1)^2 = 2(x1)(y1)(z1)

Now infinite descent proves that there are no integers satisfying this relation.[/quote]

No, it just proves that there are no [i]non-trivial[/i] solutions. (0,0,0) works.

[quote="hsbhatt"]So you get (x1)^2+(y1)^2+(z1)^2 = 2(x1)(y1)(z1)[/quote]

As has already been mentioned, you actually get $ x_1^2 \plus{} y_1^2 \plus{} z_1^2 \equal{} 4x_1 y_1 z_1$, so this method is not valid (as currently written).

Does this work?
[hide]Let us solve $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2^{n\plus{}1}xyz$

Obviously if $ x, y, z \neq 0$ then either all $ x, y, z$ or even, or two is odd and one is even so that the LHS sum up to an even number. If one of them is zero, then all of them are zero to satisfy the equation so one solution is $ (0, 0, 0)$.

Asume 2 is odd. WLOG let $ x$ and $ y$ be odd. Then $ x\equal{}2a\plus{}1$, $ y\equal{}2b\plus{}1$, $ z\equal{}2c$.
\[ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2^{n\plus{}1}xyz\]
\[ 4a^2\plus{}4a\plus{}4b^2\plus{}4b\plus{}4c^2\plus{}2 \equal{}2^{n\plus{}4}abc\plus{}2^{n\plus{}3}ac\plus{}2^{n\plus{}3}bc\plus{}2^{n\plus{}2}c\]
Then divide everything by $ 2$ to get
\[ 2a^2\plus{}2a\plus{}2b^2\plus{}2b\plus{}2c^2\plus{}1 \equal{}2^{n\plus{}3}abc\plus{}2^{n\plus{}2}ac\plus{}2^{n\plus{}2}bc\plus{}2^{n\plus{}1}c\]
The LHS is odd while the RHS is even so there is no solution.

Then assume all are even. Then $ x\equal{}2a$, $ y\equal{}2b$, $ z\equal{}2c$
\[ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2^{n\plus{}1}xyz\]
\[ 4a^2\plus{}4b^2\plus{}4c^2\equal{}2^{n\plus{}4}abc\]
Divide everything by $ 4$ to get
\[ a^2\plus{}b^2\plus{}c^2\equal{}2^{n\plus{}2}abc\]

Then we repeat the process: if one of $ (a, b, c)$ is zero then they are all zero. If two is odd then there is no solution. If all are even then the exponent of $ 2$ on the RHS increases by $ 1$. If we repeat this process over and over, the RHS approaches $ 2^{\infty}abc$ which has no integer solutions for $ (a, b, c)$.

Thus the only integer solutions are $ \boxed{(0,0,0)}$.
[/hide]

[quote="p4fn2w"]If we repeat this process over and over, the RHS approaches $ 2^{\infty}abc$ which has no integer solutions for $ (a, b, c)$.[/quote]

Not quite.  What you've done is shown that if $ (x, y, z, n)$ is a solution to $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2^{n \plus{} 1} xyz$, then so is $ \left( \frac {x}{2}, \frac {y}{2}, \frac {z}{2}, n \plus{} 1 \right)$.  What you should be focusing on is what's happening to the first three integers, not the last.  This is essentially an infinite descent proof.

Suppose that we have a solution other than $ (0,0,0)$. Then $ |x|\plus{}|y|\plus{}|z|>0$. Suppose that $ (x_0,y_0,z_0,n_0)$ is the triple that minimizes that expression. But by the descent argument, $ (x_0/2,y_0/2,z_0/2,n_0\plus{}1)$ is a solution. But this contradicts the minimality of the first triple. Hence we only have the trivial solution.

This is a beautiful problem in my opinion.


group theory
abstract algebra
superior algebra
superior algebra unsolved

I can't help but feel that I'm missing something silly here, but...

Show that a Sylow $ p$-subgroup of $ D_{2n}$ is cyclic and normal for every odd prime $ p$.   ($ D_{2n}$ is the dihedral group with $ 2n$ elements)

EDIT: In retrospect, this comment was a little useless.

Set $ G \equal{} D_{2n}$. Then $ |G| \equal{} 2n$. Let $ p$ be an odd prime dividing $ |G|$, that is, $ p > 2$ and $ p | 2n$. Let $ P \in Syl_{p}(G)$.
Then $ n_{p} \equal{} 1 \plus{} tp, t \geq 0$. That $ p > 2$ and $ 1 \plus{} tp \equal{} n_{p} |2 \implies t \equal{} 0$. So $ n_{p} \equal{} 1$ and $ P$ is normal in $ G$.

Set $ |G| \equal{} p^i m$, where $ p\nmid m$.  Then $ \{e, r^m, r^{pm}, \dots, r^{p^{i\minus{}1}m}\}$ works, right?


Sixteen times the reciprocal of the circumference of a circle
equals the diameter of the circle. What is the number of square
units in the area of the circle?

Let $ r$ be the radius of a circle.

$ 16 \cdot \frac{1}{2\pi r} \equal{} 2r$

$ \frac{16}{2\pi r} \equal{} 2r$

$ 16  \equal{} 2r(2\pi r)$

$ 16 \equal{} 4r^2\pi$

$ 4 \equal{} r^2\pi$

Thus the area is $ \boxed{4}$.