geometry
3D geometry
number theory
prime factorization

1. Today all of farmer Bill's chickens, ducks and geese laid one egg. Bill has twice as many chickens as ducks, 25 more chickens than geese and 3 more ducks than geese. The weight of a goose egg is twice as much as the weight of a chicken egg. It's also triple as much as the weight of a duck egg. The weight of three chicken eggs is 75g more than the weight of one goose egg. What's the weight of the eggs Bill's chickens, ducks and geese laid all together today?

2. 8 people wish to form pairs and play a card game called Bridge. How many games are needed so that all possible teams play each other?

3. I rolled a regular dice (which is a cube that has the numbers from 1 to 6 on its sides, so that the sum of the numbers on the facing sides is always 7) 10 times. The product of the 10 numbers I rolled is 7776. What is the highest possible sum of these 10 numbers.

[hide="da answer to 2"]
2.  6 games[/hide]

that is incorrect math92

[hide="hint"]  The problem asks for "[b] all possible teams[/b] [/hide]

For problem # 2 [hide]Since there will be four teams and each of these is an independant event, 4 x 4 = 16.

6 games must be played  :lol: [/hide]

For problem # 3 [hide] Did you mean to type "of these five numbers?"
Here's why:

First, you get the prime factorization of 7776, which is 3^5 * 2^5  This means that the only possible sum of the 10 numbers would be 15+10 or 25.

However, if you had typed, "of these five numbers," it would be different because you get 5(3*2) or 30

[/hide]

I didn't feel like doing # 1, because of the eggs  :P

let me try # 3

[hide]since we now know the prime factorization of 7776 is 3^5 * 2^5, let's pair the 3's and 2's up into 6's.
5 "6" rolls contribute to the sum of 30, and then we can roll 5 ones. That's a sum of 35.[/hide]

[hide="number 2"]

there are 8 players so there are 8 choose 2 teams which is 28 different teams. However, one person is on 7 different teams. This means that a team cannot play against any of the twelve other teams that have one of their players, leaving 16 teams for every team to play against. This means that the number of games to play is 28*16/2 because each of the 28 teams will play against 16 teams but each game is counted twice. This gives us $\boxed{224 \text{ games}}$

oops there's only 15 other teams so the answer is

$\frac{28\cdot15}2 = \boxed{210 \text{ games}}$

[/hide]

[hide="Problem 1"]
$c=2d$, $c=g+25$, $d=3+g$. So $(c,g,d)=(44,22,19)$.

I'll use the same variables for the eggs  :blush:
$g=2c$, $g=3d$, and $3c=g+75$, so $(c,g,d)=(75,150,50)$.

All the chickens, geese, and ducks each laid 1 egg. $44 \cdot 75 + 22 \cdot 150 + 19 \cdot 50 = 7550 \text { I hope that's right.}$  :winner_first:  :agent: Oh yeah  :lol:
[/hide]

[quote="Hamster1800"][hide="Problem 1"]
[quote="chess64"]
$c=2d$, $c=g+25$, $d=3+g$. So $(c,g,d)=(44,22,19)$.

I'll use the same variables for the eggs  :blush:
$g=2c$, $g=3d$, and $3c=g+75$, so $(c,g,d)=(75,150,50)$.

All the chickens, geese, and ducks each laid 1 egg. $44 \cdot 75 + 22 \cdot 150 + 19 \cdot 50 = 7550 \text { I hope that's right.}$  :winner_first:  :agent: Oh yeah  :lol:
[/quote]

It should be for the number of birds $(c,g,d)=(44,19,22)$.

however, I agree on the weights. This gives:
$44\cdot75 + 19\cdot150 + 22\cdot50 = \boxed{7250 \text{ grams}}$

[/hide][/quote]
Oops, wrong order  :blush:


inequalities
trigonometry
inequalities unsolved

For any angles A, B and C, show that
$ (cos\frac{A}{2}\plus{}cos\frac{B}{2}\plus{}cos\frac{C}{2})(tan\frac{A}{2}\plus{}tan\frac{B}{2}\plus{}tan\frac{C}{2}\minus{}tan\frac{A}{2}tan\frac{B}{2}tan\frac{C}{2})\ge4$

[quote="miwalin"]For any angles A, B and C, show that
$ (cos\frac {A}{2} \plus{} cos\frac {B}{2} \plus{} cos\frac {C}{2})(tan\frac {A}{2} \plus{} tan\frac {B}{2} \plus{} tan\frac {C}{2} \minus{} tan\frac {A}{2}tan\frac {B}{2}tan\frac {C}{2})\ge4$[/quote]
Id est, we need to prove that:
$ \sum_{cyc}\frac{1}{\cos{\frac{A}{2}}\cos{\frac{B}{2}}}\geq4,$ which is obvious.

A,B and C should be angles of a triangle.

[quote="miwalin"]For any angles A, B and C, show that
$ (cos\frac {A}{2} \plus{} cos\frac {B}{2} \plus{} cos\frac {C}{2})(tan\frac {A}{2} \plus{} tan\frac {B}{2} \plus{} tan\frac {C}{2} \minus{} tan\frac {A}{2}tan\frac {B}{2}tan\frac {C}{2})\ge4$[/quote]

 $ tan\frac {A}{2} \plus{} tan\frac {B}{2} \plus{} tan\frac {C}{2}\equal{} \frac{4R\plus{}r}{s}$

 $ tan\frac {A}{2}tan\frac {B}{2}tan\frac {C}{2}\equal{} \frac{r}{s}$

 $ cos\frac {A}{2} \plus{} cos\frac {B}{2} \plus{} cos\frac {C}{2} \ge 3 \sqrt [3] {cos\frac {A}{2} cos\frac {B}{2} cos\frac {C}{2}} \equal{} 3 \sqrt [3] {\frac{s}{4R}}$

 $ \Leftrightarrow 4 \sqrt [3] {\frac{27R^2}{4s^2}} \ge 4$

[quote="Stephen"][quote="Sergic Primazon"][quote="miwalin"]For any angles A, B and C, show that
$ (cos\frac {A}{2} \plus{} cos\frac {B}{2} \plus{} cos\frac {C}{2})(tan\frac {A}{2} \plus{} tan\frac {B}{2} \plus{} tan\frac {C}{2} \minus{} tan\frac {A}{2}tan\frac {B}{2}tan\frac {C}{2})\ge4$[/quote]

 $ tan\frac {A}{2} \plus{} tan\frac {B}{2} \plus{} tan\frac {C}{2} \equal{} \frac {4R \plus{} r}{s}$

 $ tan\frac {A}{2}tan\frac {B}{2}tan\frac {C}{2} \equal{} \frac {r}{s}$

 $ cos\frac {A}{2} \plus{} cos\frac {B}{2} \plus{} cos\frac {C}{2} \ge 3 \sqrt [3] {cos\frac {A}{2} cos\frac {B}{2} cos\frac {C}{2}} \equal{} 3 \sqrt [3] {\frac {s}{4R}}$

 $ \Leftrightarrow 4 \sqrt [3] {\frac {27R^2}{4s^2}} \ge 4$[/quote]

$ 4 \sqrt [3] {\frac {27R^2}{4s^2}} \ge 4$

Why does this works?[/quote]

Since $ \frac {R}{s} \equal{} \frac {1}{sinA \plus{} sinB \plus{} sinC}\ge\frac {2}{\sqrt{27}}$


linear algebra
matrix
combinatorics unsolved
combinatorics

To solve the problem of arranging n different elements in circles we can use this recursion G(n) = nG(n-1)
How to explain it combinatorically?
And how to show the solution n! directly?

Thanx!

$G(n) = nG(n-1)$ can be interpreted like this...  for every arrangement into circles of $1, 2, ..., n-1$, we have $n$ places to place a new object $n$: to the right of wherever $1$ is, to the right of wherever $2$ is, ..., to the right of wherever $n-1$ is, or to the right of no one -- i.e., at a table by himself.

as for proving the formula directly... if you've taken abstract algebra you've probably seen the group $S_n$ of permutations of $1, ..., n$, and how each permutation of $1, ..., n$ can be written as a product of cycles.  for instance

$\left( \begin{matrix} 1&2&3&4&5&6&7\\4&7&3&2&6&5&1 \end{matrix} \right) = (1,4,2,7)(3)(5,6)$

this gives an obvious 1-1 correspondence between the permutations of $1,...,n$ and the arrangements of $1,...,n$ into circles.


[size=150][color=red]Rules[/color][/size]

[quote="The Rules"]This game is called Friend or Foe, it was a TV show once, but I think this game is simpler.

I will be looking for 6-8 signups (maybe more).
Everyone will play against everyone else, 2 people per game, and a game will be very quick. You either PM me "friend" or "foe".

If both players pick friend, each gets 1 point. If one picks foe and the other picks friend, the one picking foe gets 2 points. If both pick foe, no one gets any points.

After everyone has played everyone else, the top 3 scorers will go into the final round, and each will play the two other players twice. The winner is declared champion. I'll figure out how to break ties if we ever need to.

I'll be setting this up in a table, maybe LaTeX, maybe not.

Who wants to play? Go ahead and ask questions if you don't get something.

-PMing others is allowed.
-There is no giving points to others.[/quote]

[size=150][color=violet]Specifics:[/color][/size]
[quote="The Specifics"]You are PMing me 1 word, 'friend', or 'foe'.
The PM should have this style header: FoF, Round x, opponent, (friend)/(foe).
Where FoF is the title (never changes), round x is the round number (should be correct), opponent is the name of your opponent, and friend or foe is the word you are choosing.
An example:
FoF, Round 3, PeterVDD, foe

The body of the PM should be
I am choosing the word 'foe' against 'PeterVDD' for round '3'.[/quote]

[size=125][color=green]Players:[/color][/size]
[quote="The Players"]1) funcia
2) mustafa
3) pianoforte
4) jhredsox
5) i_like_pie
6) ignite168 [color=white]ProtestanT[/color]
7) perfect628
8) fireemblem 13
9) Go Around the Tree
10) Walk Around the River

Replacements:
[/quote]

[color=blue][b]Status: It is currently round 2[/b][/color]
Matchups:
Funcia vs. Pianoforte
Mustafa vs. jhredsox
i_like_pie vs perfect628
ignite 168 vs Go Around the Tree
fireemblem13 vs Walk Around the River

[hide="ROUND 1 RESULTS"]Funcia FRIEND vs. Mustafa FOE
Pianoforte FOE vs. jhredsox FOE
i_like_pie FOE vs ignite168 FOE
perfect628 FOE vs Fireemblem13 FOE
Go Around the Tree FRIEND vs Walk Around the River FOE

Mustafa gets 2, WAtR gets 2[/hide]

[hide="TABLE"]
$\begin{tabular}[t]{|c|c|c|c|c|c|c|c|c|c|c|c|}\multicolumn{7}{l}{{\bf{\large Preliminary Scorebaord}}}\\\hline{\bf Players}&{\bf funcia}&{\bf must}&{\bf piano}&{\bf jh}&{\bf i-like}&{\bf ign}&{\bf perfect}&{\bf fire}&{\bf GAtT}&{\bf WAtR}&{\bf SCORE}\\\hline funcia& X &0&-&-&-&-&-&-&-&-& 0\\\hline mustafa &2& X &-&-&-&-&-&-&-&-& 2\\\hline pianoforte&-&-& X &0&-&-&-&-&-&-& 0\\\hline jhredsox &-&-&0& X &-&-&-&-&-&-& 0\\\hline i-like-pie&-&-&-&-& X &0&-&-&-&-& 0\\\hline ignite168 &-&-&-&-&0& X &-&-&-&-& 0\\\hline perfect628&-&-&-&-&-&-& X &0&-&-& 0\\\hline fireemblem13 &-&-&-&-&-&-&0& X &-&-& 0\\\hline Go Around the Tree&-&-&-&-&-&-&-&-& X &0& 0\\\hline Walk Around the River&-&-&-&-&-&-&-&-&2& X & 2\\\hline \end{tabular}$
Players going down, Opponents to the right. [/hide]

I'm in. This was a pretty good show. Who was the host? Now I can'y remember. Some woman with glasses, right?

will the results be revealed round by round, or will the first matches be revealed as a block

Round by round.

Ignite is in to replace. I'm gonna be editing the 1st post shortly, and then the game will begin.

i wanna play.

I used to watch that game show back in the day(2003) when i was in 3rd grade.

It was hosted by that lady with glasses and freakish lipstick lol. I think named kennedy

yay! it was that lady! i always thought it would be cule if they did that on dog eat dog
too bad

It is now round 1

Matchups:
Funcia vs. Mustafa
Pianoforte vs. jhredsox
i_like_pie vs ignite168
perfect628 vs Fireemblem13
Go Around the Tree vs Walk Around the River

Please read the first post again. I added a new section, called Specifics, telling you how to send in a PM.[b] Please send your PM like stated in the Specifics.
[/b]
You have 48 hours (or so) to PM me your orders. I will update the round as soon as all orders are in, so please send them in when you see this post. It takes only a minute to do so.

oh wow this is scary :lol:

EDIT:
LALALA

Round 2 has begun! Check the first and second posts for changes.

Round 2 matchups:
Funcia vs. Pianoforte
Mustafa vs. jhredsox
i_like_pie vs perfect628
ignite 168 vs Go Around the Tree
fireemblem13 vs Walk Around the River
(5/10 submissions received)

Round 3 matchups: (in advance)
Funcia vs. perfect628
mustafa vs. i-like-pie
pianoforte vs. Go Around the Tree
ignite168 vs. Walk Around the River
fireemblem13 vs. jhredsox

Oh, and, I'm going on vacation tomorrow, and though I'll have internet access, it may be sporadic. Still try to meet the 48 hour time limits if you can, though. (if you are going to be gone, PM me your actions in advance)

mustafa so it was you and the little blue men plotting against me the whole time! I am so outraged! :mad:

Wtf my post was deleted.

Anyway, I said I'm just going to reveal my choices publicy:

Round 1 Klebian just messaged me on AIM and told me I was replacing. I had no idea who I was going against or anything, so I just went "Foe" because I had no time to talk. It turns out this was a good idea as my opponent was mean spirited.  :mad:

Round 2 I am going "Friend" because Go Around the Tree is obviously a very nice person like me for going Friend the 1st round.

Round 3 I am going "Foe" because Walk Around the River is obviously a very mean person for going Foe 1st round and he has already confirmed he's going Foe against me.

im sry bout bein away, on christmas vacation (startin 12/22), ill be able to be on more. then ill join next game

[quote="bubka"]Has anyone watched 'Nice guys finish first', a documentary by richard dawkins? It was shown to us in the KVPY camp last year.

In it there was a game like this:
Two players play the game.
Each chooses to either cooperate or defect independently.
If both defect both get one point.
If both cooperate both get three points.
If one cooperates and the other defects, the defecter alone gets four points.
It was a kind of a psychology test.

Klebian, I would like you to compare this scoring system to your own and find out what went wrong with yours. I believe you are smart enough to find it.[/quote]
yep. the thing is, the reward of both players cooperating added together has to be Greater than the defecter alone, or else there is no point in both players being friend (they can alternate friend and foe)

[quote="bubka"]In it there was a game like this:
Two players play the game.
Each chooses to either cooperate or defect independently.
If both defect both get one point.
If both cooperate both get three points.
If one cooperates and the other defects, the defecter alone gets four points.
It was a kind of a psychology test.[/quote]

Prisoner's Dilemma?

Yes, with the current rules, everyone should go foe. It's incredibly stupid not to. On the show, people went friend because there was money involoved, and people felt bad for each other, and other stuff. Now, foe is the sole smart decision.

ya, this is kind dumb

everyone goes foe:no points

everyone goes foe, u go friend once:that person gets 2 points

do u get what im trying to say?

(really, no offense to klebian)

-jorian

Prisoner's Dilemma is bang on target.

The situation is like two prisoners in the same cell (strangers to each other). They are both suspicious. Now they make a plan to escape. If they both decide to cooperate, they may escape. If they both decide to defect, then both will remain locked. But if one sticks to cooperation, and the other defects, he will get away while the other maybe caught. That is the dilemma.

In probabilistic terms, the expected total score for cooperation (5 in the original example) should be greater than the expected total score for defection (4 in the original example). (And, yes, in hte experiment money [i]was[/i] involved, but that is a moot point :) )

[quote="mustafa"]Nash's Equilibrium? Is that what it's called?[/quote]
that is where we all go friend because it is for the greater good of all of us i think

http://en.wikipedia.org/wiki/Nash_equilibrium

I have nothing to comment on the game.

I [i]think[/i] I have all the submissions, but I don't think I'll get them in today. Sorry about this.

Really, I don't mean to offend you Klebian

-jorian

[quote="perfect628"]This is dumb(no offense) so i quit. Someone can replace me if they want to.[/quote]

dont be mad because your a crappy player with 0 points

[quote="now a ranger"]just an idea

Maybe we can have a chatroom on AIM where klebian comes online and invites two people who are competing against each other to the chat room. Then, the two competitors can say why each other should vote friend, etc. Then, you message Klebian on AIM in private messaging(not chatroom), and then your vote will be revealed.

This would largely decrease the amount of foes most likely, and no doubt make the game much more interesting.

I think this could be taken into consideration[/quote]

nah

personally i think it's a great idea

just to see who wants it, put [color=green]yay[/color] in ur post if u want it

[color=green]yay[/color]

-jorian

Hmm... I would have to make an AIM account, but I don't really mind that. At any rate, I think that it would be better than what's going on now, although I have a secret plan to win :diablo:

YAY

My opinion is that you need 1 on 1 communication to produce a vote that is not FOE, unless you already know the opposing person in real life.

6 votes and i will tell klebian (majority)
-jorian

woppty tee do

YIPEEE

HOORAY! lol

Actually, I think Klebian is on vacation. I am at my hotel with a labtop, and I'm about go to shopping in Atlanta lol.

counting mutafa as a yay, we have 3 yays. 3 more to go!

-jorian


[quote="dhvdochilinh"]let $ a,b,c$ are three positive real numbers,prove that
\[ \frac {{{{(a \plus{} b \minus{} c)}^2}}}{{{c^2} \plus{} {{(b \plus{} a)}^2}}} \plus{} \frac {{{{(b \plus{} c \minus{} a)}^2}}}{{{a^2} \plus{} {{(b \plus{} c)}^2}}} \plus{} \frac {{{{(c \plus{} a \minus{} b)}^2}}}{{b \plus{} {{(c \plus{} a)}^2}}} \ge \frac {3}{5}
\]
i have three keys for this ineq :D[/quote]
Maybe $ \frac {{{{(a \plus{} b \minus{} c)}^2}}}{{{c^2} \plus{} {{(b \plus{} a)}^2}}} \plus{} \frac {{{{(b \plus{} c \minus{} a)}^2}}}{{{a^2} \plus{} {{(b \plus{} c)}^2}}} \plus{} \frac {{{{(c \plus{} a \minus{} b)}^2}}}{{b^2 \plus{} {{(c \plus{} a)}^2}}} \ge \frac {3}{5}$ $ ?$
If so it's true for all reals $ a,$ $ b$ and $ c$ such that
$ ((a\plus{}b)^2\plus{}c^2)((a\plus{}c)^2\plus{}b^2)((b\plus{}c)^2\plus{}a^2)\neq0.$ :wink:

$ A \equal{} \frac {{{{(a \plus{} b \minus{} c)}^2}}}{{{c^2} \plus{} {{(b \plus{} a)}^2}}} \plus{} \frac {{{{(b \plus{} c \minus{} a)}^2}}}{{{a^2} \plus{} {{(b \plus{} c)}^2}}} \plus{} \frac {{{{(c \plus{} a \minus{} b)}^2}}}{{b^2 \plus{} {{(c \plus{} a)}^2}}} \ge \frac {3}{5}$

From Andreescu;

$ A\ge \frac {(a \plus{} b \plus{} c)^2}{c^2 \plus{} (a \plus{} b)^2 \plus{} a^2 \plus{} (b \plus{} c)^2 \plus{} b^2 \plus{} (c \plus{} a)^2}\ge\frac {3}{5}$ $ \iff$

$ 5(a \plus{} b \plus{} c)^2\ge3[a^2 \plus{} b^2 \plus{} c^2 \plus{} (a \plus{} b)^2 \plus{} (b \plus{} c)^2 \plus{} (c \plus{} a)^2]$

$ 5(a \plus{} b \plus{} c)^2 \minus{} 3[a^2 \plus{} b^2 \plus{} c^2 (a \plus{} b)^2 \plus{} (b \plus{} c)^2 \plus{} (c \plus{} a)^2]\ge0$

Expanding it leads to;

$ \minus{}4a^2 \minus{}4b^2 \minus{} 4c^2 \plus{} 4ab \plus{} 4bc \plus{} 4ca\ge0$ $ \iff$

$ a^2 \plus{} b^2 \plus{} c^2\leq ab \plus{} bc \plus{} ca$

I edited my solution and found this which is obviously wrong.. so i don't see how the given holds :blush:  :blush: .

[quote="dhvdochilinh"]let $ a,b,c$ are three positive real numbers,prove that
\[ \frac {{{{(a \plus{} b \minus{} c)}^2}}}{{{c^2} \plus{} {{(b \plus{} a)}^2}}} \plus{} \frac {{{{(b \plus{} c \minus{} a)}^2}}}{{{a^2} \plus{} {{(b \plus{} c)}^2}}} \plus{} \frac {{{{(c \plus{} a \minus{} b)}^2}}}{{b \plus{} {{(c \plus{} a)}^2}}} \ge \frac {3}{5}
\]
i have three keys for this ineq :D[/quote]

because the ineq is homogenuous, we can assume that $ a \plus{} b \plus{} c \equal{} 1$
ineq \[ \Leftrightarrow \]\[ \sum {(\frac {{{{(1 \minus{} 2a)}^2}}}{{{a^2} \plus{} {{(1 \minus{} a)}^2}}}} \minus{} \frac {1}{5} \ge 0 \Leftrightarrow \sum {\frac {{(3a \minus{} 1)(3a \minus{} 2)}}{{{a^2} \plus{} {{(1 \minus{} a)}^2}}}} \ge 0
\]
we can apply $ Chebyshev$ inequality ,we have
$ LHS \geq$
\[ \frac {1}{3}(3a \plus{} 3b \plus{} 3c \minus{} 3)(\sum {\frac {{3a \minus{} 2}}{{{a^2} \plus{} {{(1 \minus{} a)}^2}}}} ) \ge 0
\]
($ a \plus{} b \plus{} c \equal{} 1$)
(Q.E.D) :D

[quote="dhvdochilinh"][quote="dhvdochilinh"]let $ a,b,c$ are three positive real numbers,prove that
\[ \frac {{{{(a \plus{} b \minus{} c)}^2}}}{{{c^2} \plus{} {{(b \plus{} a)}^2}}} \plus{} \frac {{{{(b \plus{} c \minus{} a)}^2}}}{{{a^2} \plus{} {{(b \plus{} c)}^2}}} \plus{} \frac {{{{(c \plus{} a \minus{} b)}^2}}}{{b \plus{} {{(c \plus{} a)}^2}}} \ge \frac {3}{5}
\]
i have three keys for this ineq :D[/quote]

because the ineq is homogenuous, we can assume that $ a \plus{} b \plus{} c \equal{} 1$
ineq
\[ \Leftrightarrow
\]

\[ \sum {(\frac {{{{(1 \minus{} 2a)}^2}}}{{{a^2} \plus{} {{(1 \minus{} a)}^2}}}} \minus{} \frac {1}{5} \ge 0 \Leftrightarrow \sum {\frac {{(3a \minus{} 1)(3a \minus{} 2)}}{{{a^2} \plus{} {{(1 \minus{} a)}^2}}}} \ge 0
\]
we can apply $ Chebyshev$ inequality ,we have
$ LHS \geq$
\[ \frac {1}{3}(3a \plus{} 3b \plus{} 3c \minus{} 3)(\sum {\frac {{3a \minus{} 2}}{{{a^2} \plus{} {{(1 \minus{} a)}^2}}}} ) \ge 0
\]
($ a \plus{} b \plus{} c \equal{} 1$)
(Q.E.D) :D[/quote]
You must explain that,why do you use chebyshev :D,have fun,my friend

This is 1997 Japan mathematical olympiad finals, Problem 2.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=258462&search_id=2048105115[/url]

I think this forum should be locked, because the inequality has been appeared many times.

Japanese communities Moderator

kunny


superior algebra
superior algebra unsolved

I'm confused by the composition series.

a) Find a composition series of S3 X S3. Is S3 X S3 solvable?
b) Proof that a finite direct product of solvable groups is solvable.

Essential property: If $ N$ is normal in $ G$, $ N\times H$ is normal in $ G\times H$. We can deal with the components of the product one at a time.


Not sure where to post this:

$ 1, 2, 4, 7, 8, 11, 14, 16, 17, 19, 22, 26, 28, 29, 41, 44,$;

what is the next number in this serquence?

In general, such things go best in the "Puzzles and Brainteasers" forum.  [url=http://www.research.att.com/~njas/sequences/index.html]OEIS[/url] doesn't like this sequence, which is not terribly surprising, although it does start out looking an awful lot like [url=http://www.research.att.com/~njas/sequences/A000069]A000069[/url].

Thanks. I can't find the puzzles and brainteasers forum.

[hide=" click to reveal hidden content"]

Looking at the numbers that are missing, we get multiples of $ 3$ and multiples of $ 5$ and numbers which contain the digit $ 3$ and numbers which contain the digit $ 5$, so the next term must be $ 46$.

[/hide]

Another brainteaser I've just been asked is:  what are the next two letters in the sequence $ A, E, F, H, I, K, L, M, ...?$

[hide=" click to reveal hidden content"]

It looks like we're missing the letters with round bits, so the next two letters in the sequence must be $ N$ and $ T$.

[/hide]

Can someone post me a link to "puzzles and brainteasers"? Thanks.

Thanks JBL. I'll use this for puzzles in future.


On Monday, Millie puts a quart of seeds, $ 25\%$ of which are millet, into a bird feeder.  On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left.  Each day the birds eat only $ 25\%$ of the millet in the feeder, but they eat all of the other seeds.  On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?

$ \textbf{(A)}\ \text{Tuesday}\qquad \textbf{(B)}\ \text{Wednesday}\qquad \textbf{(C)}\ \text{Thursday} \qquad \textbf{(D)}\ \text{Friday}\qquad \textbf{(E)}\ \text{Saturday}$

If you want to know the answer, click here:
[hide="click here for answer"]D[/hide]

Similar to the AMC 8 this year, wouldn't E be an answer as well? More than half the seeds will be millet on that day as well. The question didn't specify the first day the birds will find that there will be more than half, just on which day. Granted I put D because I assumed they were asking for first.

I thought she was refilling what the birds had eaten, not adding another quart. I got Wednesday.

this problem was relatively easy. i found the general method, got excited and then messed up

i will kill some birds to make myself feel better

also, E is correct under the description

That problem was easy but I messed up on it :(. For some reason I thought the millet level only had to pass .5

heh, i messed up my arithmetic and this was complete arithmetic. but e is still technically correct :D

Technically, all the answers are correct. The next week has Monday-Sunday too :P

for some reason I thought that the question asked for when there were more millet seeds than other seeds  :noo:  :wallbash:

looks like I'm joining the bird killing club

I hated that question, just started with 1 quart millet and 3 quart other. Then just pure computation with ugly fractions...

This question does not satisfy the description #2 on the cover, which said "This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D, and E. [b]Only one of these is correct.[/b]"

Both D and E is correct, because nothing stated that the birds stop eating the seeds after finding that more than half the seeds in the feeder are millet, and it makes sense that the birds continue eating the seeds even after finding that more than half the seeds in the feeder are millet. Because of both answers being correct, and the first page saying there's only one correct answer, my friend got really confused...
(Basically, he just thought too much into the problem)

But then...it's possible that birds only eat food on weekdays...and rest on weekends :)

Yes, both of those answers are certainly correct.  The other ones are too if you think about it.  But I'd appreciate it if they accepted D or E as correct.

While E may [i]technically[/i] satisfy the problem, the problem's intent was obvious. I don't think that any answer besides D should be accepted. How many people got the question wrong because it was ambiguous, and how many because they made a mistake? I think the latter greatly outnumbers the former.

I agree that the protest should not be carried out. It is clear what the problem wants.

Can anyone tell me how this question was tricky, and how they got wednesday and saturday?

I seriously don't get it. When I did it, I thought this question was easy enough to be #1. However after comparing my answer (which turned out to be the right one) with all the other people I know who took the test (some better than me, some worse) and my dad, they ALL got wednesday. Then I got scared.

I still don't get how you end up with wednesday or saturday.

A lot of people thought there needed to be over a half quart of millet seeds, instead of 3/4 quart.

I got Wednesday by calculating when the number of millet seeds would be more than one half the number of other seeds.  Epic fail.

Have the results of that protest been announced?

If not, can someone please pm me the results when they are?

I filed a protest, I assume the only one, and I have not gotten any sort of response other than, what was I assume, a standard, "We will forward this to the right people" type of response.  I'm sure they'll post the result on that thread or on a new thread.  I hope it comes soon.

Just a general note, there was a topic about a person protesting this and the debate got pretty heater (Personally, seeing the protest angered me a LOT, although i don't know why). Please keep this thread to solving the question when it asks for the first day, and don't turn this into the other thread. Thanks.

I'm not going to let this devolve into another protest thread. I will delete any such posts. If you have a compelling reason to post something in the protest thread, PM me and I will unlock it.

If we protest and the answer is changed to include Saturday, is it changed for everyone that put Saturday?  Or when one protests and it is accepted, only the protestor's score is changed?  Or has the announcement already been made and Im missing something?

And besides, im not trying to start an argument again, but does anyone remmeber this infamous AIME problem from three years ago?
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1054291590&t=81167
I think that if they were willing to accept an answer (and for people that write 5 as $ \frac{5}{1}$, then they would be willing to accept anything)..and remember that was an AIME

[quote="RunpengFAILS"][hide="Solution"]Assume a quart equals 100 seeds. The number of millet seeds left after the birds are finished eating on the $ n$th day is $ \frac {\frac {75}{4}(1 - \left(\frac {3}{4}\right)^n)}{1 - \frac {3}{4}}$ (From $ 25\times{\frac {3}{4}} + 25\times{\left(\frac {3}{4}\right)^2} + \cdots$). This value PLUS the extra 25 millet seeds that will be added the next day must be greater than 75 (that's how many other seeds will be added). Therefore we have $ \frac {\frac {75}{4}(1 - \left(\frac {3}{4}\right)^n)}{1 - \frac {3}{4}} > 50\implies{\left(\frac{3}{4}\right)^n<\frac{1}{3}\implies{n=4}}$. But remember we have to add a day to this so the answer is FRIDAY (D).[/hide][/quote]
I'm confused, why would you multiply it by (3/4)^n? Why isn't it just 25*3/4+25*3/4... You're only taking away 1/4 of what's there, so why would it be (3/4)^n...


$x\equiv{1}\pmod{2}$
$x\equiv{2}\pmod{3}$
$x\equiv{3}\pmod{4}$
.                        .
.                        .
.                        .
$x\equiv{n}\pmod{n+1}$

Express $x$ in terms of $n$.

You mean you want to express $x \bmod lcm(2, 3, 4, ... n+1)$ in terms of $n$, don't you?

[hide="Hint"] What is $x+1$?  :) [/hide]


Suppose $ f(a \plus{} b) \equal{} f(a) \plus{} f(b)$. If $ f(1) \equal{} 1$, then evaluate $ f(n)$.

[hide="comments"]
This problem was from a book from China. (挑战奥数阶梯训练营)

I'm glad I'm Chinese (so I can read these types of books easily) as China is great at math olympiads for some weird reason. (I am NOT from China!)
[/hide]

Another nice problem from the same book (translated then edited for sheer fun):

If real numbers $ x,y$ satisfy $ f(x \times y) \equal{} f(x) \times f(y)$, and $ f(0) \neq 0$, find $ f(2008)$.

[hide="hint"]
Try substituting $ y \equal{} 0$.
[hide="solution"]
If $ y \equal{} 0$, $ f(0) \equal{} f(x) \times f(0)$. Since $ f(0) \neq 0$, we can divide by $ f(0)$ to get $ f(x) \equal{} 1$. Thus $ f(2008) \equal{} 1$.
[/hide]
[/hide]

[hide="Solution"]
It's easy to see that $ f(0)\equal{}0$ and that $ f(x\plus{}1)\equal{}f(x)\plus{}1$
So, the answer is $ f(x)\equal{}x$
[/hide]

I'm really having a problem understanding...

wouldn't the solution f(X)=1 have to make the initial equation also valid?

I just don't see it..

It does. Perhaps you're confusing the first problem and the second problem :wink:

The original equation was $ f(xy) \equal{} f(x) f(y)$. If $ f\equal{}1$, then $ f(xy)\equal{}f(x)f(y) \iff 1 \equal{} 1 \cdot 1$.

My solution was for the first problem.
Solution to the second problem is very easy and already written by ThetaPi.

Is this in chinese?
What book?
traditional or mordern?

It was published in 2006, and yes, it is in Chinese.

Thanks TZF, I looked over the solutions again, and I can see now... So if f(0) did = 0, I guess the equation would be satisfied as f(x)=x.

I'm just a new problem solver, and I'm just beginning to delve into the subject of (relatively) more advanced mathematics. :)

The first equation is known as the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url], and all you can conclude is that $ f(n) \equal{} n$ for [i]integers[/i] $ n$.  With no other restrictions, there are pathologically discontinuous solutions.

I'm looking into functional equations right now, and it is a very interesting (albeit difficult) subject.  Certainly something which will take a long while of contemplation to grasp.

[quote="t0rajir0u"]The first equation is known as the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url], and all you can conclude is that $ f(n) \equal{} n$ for [i]integers[/i] $ n$.[/quote]

well, since $ f(1) \equal{} qf(1/q)$ for all integers $ q\not\equal{}0$, we get that $ f(1/q)\equal{}1/q$ for all   nonzero integers.  then $ f(p/q) \equal{} pf(1/q) \equal{} p/q$ as long as $ p,q$ are integers with $ q$ nonzero, so $ f$ fixes all rationals.

I should've been more specific.  I mean that the reasoning given in Bugi's solution, as it stands, only allows us to conclude that $ f(n) \equal{} n$ for integers.


First game [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=293724]here[/url]

I will start:

[b]25[/b]

That is unfair, as girls are outnumbered 3 to 1 on this forum.

OMG GIRLLSSS WIN!!!!!!!!!! yes!!!

good try boys

congrats everyone!!! :D  :P  :rotfl:

plz don't make another one right now... i'm eating lunch now!

gg...lolz...sry jjfun...trying to post and help

soory boyz, I failed,
the net connection went haywire :(

hmm good job everyone who participated!!!

let's make round 8 in an hour..... or something


Prove that the following equation does not have an integer solution(y is not 0)

$ x^2 \plus{} x$ = $ y^3 \plus{} y^2 \plus{} y$

[hide="Hint"]$ x^2\plus{}x\equal{}y^3\plus{}y^2\plus{}y \iff (x\minus{}y)(x\plus{}y\plus{}1)\equal{}y^3$[/hide]

oh... due to my computer.

now what should I do?


Hello,

I saw this handy feature on other fora like :
[url]http://www.skyscrapercity.com[/url]

When you click the quote button there, you get this :
[quote][QUOTE=slby;11119668]I will be going to Brunei for a visit after Chinese New Year, anyone out there has any good recommendation? (Interesting places to visit?)[/quote][/quote]
"slby" indicates that slby has written the message, 11119668 is the number of the post.
When you submit this, the quoted text will provide a link to the original post.

This is extremely handy, it allows you to see in what kind of context someone wrote that.
Think of what it could mean in Mafia Games. :P

Those fora do allow nested quoting but the quote button itself gives you a textbox that has automatically erased the higher degree quotes.  This might be good or bad, but fact is that that link makes nested quotes kind of superfluent.

Is this worth considering, and is it possible?
Thank you.

[quote="fredbel6"][quote="anirudh"][quote="[url=http://google.com]Here[/url], Someone"]hello[/quote]

The code is:

[code][quote="[url=http://google.com]Here[/url], Someone"]hello[/quote][/code]

Generalized code:

[code][quote="[url=URL GOES HERE]Text that is clicked to go to the URL[/url] ADDITIONAL TEXT (OPTIONAL)"]QUOTED TEXT[/quote][/code]

Like this? :maybe:[/quote]
That's cool and I hadn't thought of that  :)  but the difference is that those fora do it for you, by default. :D   I don't think anyone will ever bother to do this.[/quote]

Aah. I see.

That'll require some patching up inside includes/bbcode.php for an admin.

Yes. Its possible.


Let $a\in(0,4)$ and x, y, z > 0 such that:\[xyz+x^2+y^2+z^2=a\]
Prove that:\[x+y+z<3\]

If $x^2+y^2+z^2+xyz < 4$, suitably increase the value of one variable until it equals 4. When $x^2+y^2+z^2+xyz=4$, put $x=2cosA$ and so on, then $\sum cosA \le \frac{3}{2}$ is trivial.


So, I want to find and add all the multiples of 3 and 5 under 1000. So, when i tried this in java, Here is some of my code:

public class Euler {

     public static void main (String args[]) {
           int multiples5[];
           int multiples3[];
           int counter = 0;
           int counter2 = 0;
           while(counter < 200) {
                  counter++;
                  multiples5[counter2]= 5 * counter;
           }       counter2++;
       }
}

But , when i put in the multiples5 array , it gives me a error and says "Local Variable may not have been intialized" plz help

thank you
poincare

Well the error says exactly what's wrong. You haven't initialized multiples5. You probably want something like "int multiples5[] = new int[200]".

oops, didn't notice that!
haven't used arrays for a while now!


algebra
polynomial
number theory unsolved
number theory

Was wondering whether there has been any investigation into the properties/distribution of these for a given cyclotomic field of prime degree Q(z_p) (where z_p^p = 1)?
For example, the imaginary sextic subfield of Q(z_163) admits two non-trivial principal prime cofactors 787 and 919. The real octic subfield of Q(z_257) has 17^2 as the only non-trivial cofactor of the field discriminant. Apart from needing to obey obvious rules mod p to fit in the right residue class little seems to be known about these.
Does anyone know anything a little more scientific about the ramification behaviour of such primes in these subfields - even for small (p-1)/d = 6 or 8 say?

In answer to my own thread after some investigations - I am confusing the polynomial discriminant with the field discriminant. In fact 787*919 is known as the index of the power order in the rings of integers (assuming we take the mimimal polynomial defining the sextic subfield of Q(z_163)). The field discriminant stays constant at -163^5 though the index increases where a minimal polynomial is not used.

So is it the case that q1=787 and q2=919 do not ramify in the above field despite having a repeated root of the sextic
x^6 + x^5 + 14x^4 - 178x^3 - 552x^2 + 1854x +5023
factored mod q1 & q2? This is only a specific example of course - what's the correct statement regarding q1, q2,...in Q(z_m) generally?


MATHCOUNTS
Gauss
number theory
relatively prime

OK, well...the MathCounts training session with Mr. Reiter was awesome!  The last problem on the last team round, however, was a problem written by Dr. Reiter himself, a couple of weeks prior to the session.  The answer key was wrong, as well as every single student (team) answer.  Dr. Reiter worked the problem on the board (along with the students help), and eventually, a consensus was reached.  Here is the problem.  Please post solutions along with correct answers.  (Shobhit, before you blow this off as MathCounts, note that many of these problems were above MathCounts level).

2006 MathCounts Training, Team Round 7
Problem Number 10
Dr. Harold Reiter

[size=125][b]How many three digit numbers can be written as a sum of a 3-digit number and its (one-, two-, or three-digit) reversal?[/b][/size]

[hide]$(100a + 10b + c) + (100c + 10b + a) = 101(a + c) + 20b$

Since 101 and 20 are relatively prime and the choices for $a + c$ and $b$ are all less than 20, each choice of $a + c$ and $b$ gives a unique outcome for $101(a + c) + 20b$.

We also want $101(a + c) + 20b$ to be a 3 digit number. $100 \leq 101(a + c) + 20b \leq 999$ implies that either $1 \leq a + c \leq 8$ (8 outcomes) and $0 \leq b \leq 9$ (10 outcomes), or that $a + c = 9$ (1 outcome) and $0 \leq b \leq 4$ (5 outcomes).

Hence there are (8)(10) + (1)(5) = [b]85[/b] possible outcomes.[/hide]

Yes, Gauss got the answer.  That's the answer Dr. Reiter got as well.


Let $ a,b,c>0$ such that $ : 4(ab\plus{}bc\plus{}ca)\equal{}(a\plus{}b\plus{}c)^2$
Find max :$ P\equal{}\frac{(a^2b\plus{}b^2c\plus{}c^2a)^2}{(ab\plus{}bc\plus{}ca)^3}$

[quote="DCTPKTCSPHN"]Let $ a,b,c > 0$ such that $ : 4(ab \plus{} bc \plus{} ca) \equal{} (a \plus{} b \plus{} c)^2$
Find max :$ P \equal{} \frac {(a^2b \plus{} b^2c \plus{} c^2a)^2}{(ab \plus{} bc \plus{} ca)^3}$[/quote]
It's easy.
Observe that the ineq is homogeneous so we can always assume $ a\plus{}b\plus{}c\equal{}1\implies ab\plus{}bc\plus{}ca\equal{}\frac{1}{4}   (*)$. From Cauchy schwartz;
$ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\geq (\sum a^2b)^2\implies (a^2b \plus{} b^2c \plus{} c^2a)^2\leq\frac{1}{4}(**)$
Using (*) & (**), we have:
$ P\leq \frac{(\frac{1}{4})^2}{(\frac{1}{4})^3}\implies P\leq 4$

[quote="Potla"][quote="DCTPKTCSPHN"]Let $ a,b,c > 0$ such that $ : 4(ab \plus{} bc \plus{} ca) \equal{} (a \plus{} b \plus{} c)^2$
Find max :$ P \equal{} \frac {(a^2b \plus{} b^2c \plus{} c^2a)^2}{(ab \plus{} bc \plus{} ca)^3}$[/quote]
It's easy.
Observe that the ineq is homogeneous so we can always assume $ a \plus{} b \plus{} c \equal{} 1\implies ab \plus{} bc \plus{} ca \equal{} \frac {1}{4} (*)$. From Cauchy schwartz;
$ (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)\geq (\sum a^2b)^2\implies (a^2b \plus{} b^2c \plus{} c^2a)^2\leq\frac {1}{4}(**)$
Using (*) & (**), we have:
$ P\leq \frac {(\frac {1}{4})^2}{(\frac {1}{4})^3}\implies P\leq 4$[/quote]
hello! polta! I'm sorry but your using Cauchy is wrong.

[quote="Potla"][quote="DCTPKTCSPHN"]Let $ a,b,c > 0$ such that $ : 4(ab \plus{} bc \plus{} ca) \equal{} (a \plus{} b \plus{} c)^2$
Find max :$ P \equal{} \frac {(a^2b \plus{} b^2c \plus{} c^2a)^2}{(ab \plus{} bc \plus{} ca)^3}$[/quote]
It's easy.
Observe that the ineq is homogeneous so we can always assume $ a \plus{} b \plus{} c \equal{} 1\implies ab \plus{} bc \plus{} ca \equal{} \frac {1}{4} (*)$. From Cauchy schwartz;
$ (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)\geq (\sum a^2b)^2\implies (a^2b \plus{} b^2c \plus{} c^2a)^2\leq\frac {1}{4}(**)$
Using (*) & (**), we have:
$ P\leq \frac {(\frac {1}{4})^2}{(\frac {1}{4})^3}\implies P\leq 4$[/quote]
I'm sorry,but I don't agree with you :huh:

[quote="Potla"] From Cauchy schwartz;
$ (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)\geq (\sum a^2b)^2$
[/quote]
Maybe $ (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)\geq (\sum \sqrt{a^2b})^2$ $ ?$ :wink:


Given the expression
$P = \frac{8-7-6-5}{4-3-2-1}$
we can place as many brackets as we want anywhere on the numerator, but we must place at the same places on the denominator.

[color=blue](a)[/color] Find the least possible value of $P$
[color=blue](b)[/color] Find all integral values of $P$

Just try every possible way of putting brackets?  :D

You do it then  :D
Or is there any way to avoid that?


does the product of the diagonals of a trapezoid divided by two equal the area?
 :?:  :stink:

I don't think so.

The area of a trapezoid, if you don't know, is $ \frac{h(b_1\plus{}b_2)}{2}$.

In a quadrilateral, the area equals $ \frac{d_1d_2}{2}$ iff it is orthogonal.  Since not all trapezoids are orthogonal, the answer to your question is no.

oh sorry, i was thinking about a rhombus
 :oops:

and btw, is there anyway to find the area of an isoceles trapezoid using the diagonals?
i didn't know what orthogonal meant..

I think you should see [url=http://en.wikipedia.org/wiki/Isosceles_trapezoid]this[/url]


Let p be an odd prime number .
f(x) and g(x) are polynomials over $F_{p}$
f(x)=(p+1)$x^{p}$+p$x^{p-1}$+....+2x+1
g(x)=$x^{p^{p}}$-x;
prove or disprove that f(x) divides g(x) ???

Field with $p$ elements.

You have to show that $f$ is irreducible,:

$f(x)|x^{p^{p}}-x$ for $deg(f)=p$ iff $f$ has no double roots and $f$ either splits completely or is irreducible.

Yes,it is true.
in fact I want to prove that f(x) is irreducible over $F_{p}$

The product of all monic irreducible polynomials with degree a divisor of $d$ in $F_{p}$ is $X^{p^{d}}-X$

$f(X)$ is irreducible of degree d iff $f(X)|X^{p^{d}}-X$
and $(f(X),X^{p^{\frac{d}{k}}}-X)_{k|d}=1$


Find all quadruples $ (a, b, c, d)$ of distinct positive integers satisfying $ a \plus{} b \equal{} cd$ and $ c \plus{} d \equal{}ab$.

[hide="I had 1 of 10 on this problem in a important competition last year..."]
Add and subtract one from another to get $ ab\minus{}a\minus{}b\plus{}cd\minus{}c\minus{}d\equal{}0$ or $ (a\minus{}1)(b\minus{}1)\plus{}(c\minus{}1)(d\minus{}1)\equal{}2$
1) $ (a\minus{}1)(b\minus{}1)\equal{}2$
a) $ a\equal{}3,b\equal{}2$
b) $ a\equal{}2,b\equal{}3$
2) $ (a\minus{}1)(b\minus{}1)\equal{}1$
$ a\equal{}2\equal{}b$
3) $ (c\minus{}1)(d\minus{}1)\equal{}2$
Similar to 1)
It has 9 solutions[/hide]


The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle.  The circles are tangent to one another and to the sides of the rectangle as shown in the diagram.  The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\frac 12\left(\sqrt{p}-q\right),$ where $p$ and $q$ are positive integers.  Find $p+q.$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=405[/img]

[hide="Answer"]For convenience, we set $r=1$. Obviously, the longer dimension of the rectangle is $14$, so we simply need to find the smaller. Note that, in the row with only $6$ circles, if we draw the lines connecting the centers of two circles, one directly above the other, in the first and third rows, that line will be tangent to at least one circle in the second row, so the centers of the circles in the second row are directly above or below the points of tangency between circles in the first and third rows. Therefore, if we take two tangent circles, one from the first row and one from the second, and draw two radii that touch at the point of tangency, then draw the tangent line from the center of the circle in the first row to the circle in the second, and finally draw the radius to that point of tangency on the second circle, we have a right triangle. The shorter leg is half the length of the second, so we have a 30-60-90 triangle, and the longer leg has length $\sqrt{3}$. We double this and add $2$ to find the height of the rectangle, so the ratio is $\frac{14}{2+2\sqrt{3}}=\frac{7(1-\sqrt{3})}{-2}=\frac{7}{2}(\sqrt{3}-1)=\frac{1}{2}(\sqrt{147}-7)$, and our answer is $147+7=\boxed{154}$.[/hide]

[hide]connect the center of the top left circle, the bottom left circle, and the center of the circle to the right of the bottom left circle, to get a right triangle. assume WLOG that the radius is 1. then the base of the triangle is 2, and hypotenuse is 4, and the height is then $\sqrt{4^2-2^2}=2\sqrt{3}$. So the height of the rectangle is $2+2\sqrt{3}$ and the width is obviously 14. divide the latter by the former to get the answer $\frac{1}{2}(\sqrt{147}-7)$, so $154$.[/hide]

This problem is fun.

[hide="Answer"]

We can say $r$ as the length of the radius of the circles but since our problem asks for ratio, it won't matter what number we'll say $r$ is. So, let $r = 1$. Then the long side is $2(7) = 14$. Now, the real problem is the shorter dimension. Considering four circles that are tangent (one on the first row, two on the second, one on the third), we can form a 30-60-90 triangle by connecting the center of the circle in the first row, either of the circle in the second row and the tangent point of two circles in the second row. In this triangle, $\sqrt{3}$ part is the part of the height. Since there are two of them, the shorter dimension is 2 (two radii that weren't included in the right triangle calculation) + $2 \sqrt{3}$. Hence, the ratio is $\frac{14}{2 \sqrt{3} + 2} = \frac{1}{2} (7 \sqrt{3} - 7)$. But the question has $7 \sqrt{3}$ under one square root so that is $\sqrt{147}$. Hence, our final answer is $147+7 = \fbox{154}$. [/hide]


real analysis
probability
real analysis unsolved

Consider the sequence $\{2^{n}x\}$.If $x$ is an irrational number. Prove:$\{2^{n}x\}$  is dense in $[0,1]$

$\{x\}$ is the fractial part of x.
Thanks for help

what about $x$? Is it irrational? Or just real?

The assumption is that $x$ is not a [b]dyadic[/b] rational. (Edit: I was wrong -- the assumption is stronger.)

Thank you two.
$x$ is a fixed real number such the condition.Maybe it called $x$ is not a dyadic rational.
Can somebody prove it or disprove it? :?:

Oh,sorry :blush:
I find it will be wrong if $x$ is an rational number.
What about when $x$ is an irrational number?

Diogene seems to have killed the original problem.

How about this:

Let $A$ be the set of all $x\in[0,1]$ for which the set $\{\{2^{n}x\}: n\in\mathbb{N}\}$ is dense in $[0,1].$

What can we say about the Lebesgue measure of $A?$

[quote="Diogene"]For $x=\sum_{k=1}^{\infty}\frac 1{2^{k^{2}}}$ I obtain $\{2^{n}x\}< x$, so non dense in $[0,1]$.
:cool:[/quote]
Thank you.Here is my proof for $x=\sum_{k=1}^{\infty}\frac 1{2^{k^{2}}}$ is irrational,If not let $x=\frac{m}{n}$,There is a $N$ such that $2^{(N+1)^{2}-2}>n\cdot 2^{N^{2}}$
So $0<\frac{m}{n}-\sum_{k=1}^{N}\frac 1{2^{k^{2}}}=\sum_{k={N+1}}^{\infty}\frac 1{2^{k^{2}}}\le \frac{1}{2^{(N+1)^{2}-1}}\le \frac{1}{n2^{N^{2}+1}}$
That a contradiction with $\frac{m}{n}-\sum_{k=1}^{N}\frac 1{2^{k^{2}}}=\frac{t}{2^{N^{2}}n},t \in \mathbb{N}$

Kent's problem maybe hard,and Can somebody prove or disprove:
$\{\{2^{n}\pi\}: n \in \mathbb{N}\}$ is dense in $[0,1]$?

[quote="Kent Merryfield"]
How about this:

Let $A$ be the set of all $x\in[0,1]$ for which the set $\{\{2^{n}x\}: n\in\mathbb{N}\}$ is dense in $[0,1].$

What can we say about the Lebesgue measure of $A?$[/quote]

I believe it can be shown that if $(a_{n})_{n\ge 1}$ is a sequence of positive reals increasing towards $\infty$, then for almost all $x\in[0,1]$ the set $\{\{a_{n}x\}\ |\ n\ge 1\}$ is dense in $[0,1]$.

I didn't work it out in detail, but what I was trying to do seemed to work.

[quote="Kent Merryfield"]
Let $A$ be the set of all $x\in[0,1]$ for which the set $\{\{2^{n}x\}: n\in\mathbb{N}\}$ is dense in $[0,1]$.

What can we say about the Lebesgue measure of $A$?
[/quote]

[b]Hint 1 (weak form of the Borel-Cantelli lemma).[/b] Let $x_{1}, x_{2}, \dots$ be an i.i.d sequence of random variables: $x_{i}= 0$ w.p. $1/2$, $x_{i}= 1$ w.p. $1/2$. Prove that with probability $1$ every finite sequence of zeros and ones appears in $(x_{i})_{i}$ as a subsequence.
[hide]
Consider a sequence $d_{1},\dots, d_{k}$. The probability that $(x_{1},\dots,x_{k}) \neq (d_{1},\dots,d_{k})$, and $(x_{k+1},\dots,x_{2k}) \neq (d_{1},\dots,d_{k})$, etc is less than $\left(1-\frac{1}{2^{k}}\right)^{m}$ for every $m$. So $(d_{1},\dots,d_{k})$ occurs in $(x_{i})_{i}$ w.p. $1$. The number of finite sequences of zeros and ones is countable. So all of them appear in $(x_{i})_{i}$ with probability $1$.
[/hide]

[b]Hint 2.[/b] Consider a sequence $(x_{i})_{i}$ s.t. every finite sequence of zeros and ones appears in it. Let $r=\sum_{i=1}^{\infty}x_{i}/2^{i}$. Then the set $\{\{2^{n}r\}: n\in\mathbb{N}\}$ is dense in $[0,1]$.

Conclude that the measure of $A$ is $1$.


In the triangle $ABC$, $\angle{C}=90^{\circ},\angle {A}=30^{\circ}$ and $BC=1$. Find the minimum value of the longest side of all inscribed triangles (i.e. triangles with vertices on each of three sides) of the triangle $ABC$.

let $D\in BC,E\in AC,\F\in AC$
you can prove easily,that minimum area of $DEF$ will occur when $DEF$ be a equalateral triangle.
now let $BD=x$ hence $DE^{2}=\frac{7}{4}x^{2}-2x+1$ and minimum of $DE$ is when $x=\frac{4}{7}\rightarrow DE=\sqrt{\frac{3}{7}}$

[quote="Iso92"]How do you know that $CE=\frac{\sqrt{3}x}{2}$ and $BF=1-\frac{x}{2}$?[/quote]

let $H$ be feet of perpendicular from $D$ to $AB$ rotate $AB$ around $D$ by angle $60^{\circ}$ this line intersect $AC$ at $E$,let $H'$ be rotation of point $H$ .

it's clear that $DH'\perp BC\Longrightarrow DH'=CA=\frac{\sqrt{3}}{2}x\ \ , \ H'E=HF=1-x\Longrightarrow \ BF=1-\frac{x}{2}$ .

[quote="Vo Duc Dien"]"you can prove easily that minimum area of DEF will occur when be a equalateral triangle. "

That's not right. If F -> B, E -> C and no matter where D is on BC, the area of DEF -> 0.[/quote]

I think the claim of Amir.S is correct.Suppose $DEF$ is a triangle with $DF>DE>EF$ and $DF$ is the minimum of the longest side of all possible inscribable triangles.Then by assumption there exists a point $E'$ on $(AC)$ such that $DF=DE'$.Consider the triangle $DE'F$.Now slide $DE'F$ by moving $E'$ and $F$ along $AC$ and $AB$ such that $DF,DE'$ become shorter at the same rate and $E'F$ becomes longer gradually.At a point of time $DE'F$ will become equilateral with sides shorter than $DF$,a contradiction.The same reasoning goes for isosceles triangles.


calculus
integration
trigonometry
calculus computations

Find the indefinite integral
$ \int xe^{x}\cos x dx$.

Begin by integrating by parts with $ u\equal{} x$ and $ dv\equal{} e^{x} \cos x$, so that $ du \equal{} dx$ and $ v\equal{} \frac{1}{2} e^{x} (\sin x \plus{} \cos x)$.

This gives $ I\equal{}\frac{1}{2}xe^{x}(\sin x \plus{} \cos x) \minus{} \frac{1}{2}\int e^{x}(\sin x \plus{} \cos x) \ dx$ which seems easy to integrate by parts.

Hope this helps.

[quote="AndrewTom"]Begin by integrating by parts with $ u \equal{} x$ and $ dv \equal{} e^{x} \cos x$, so that $ du \equal{} dx$ and $ v \equal{} \frac {1}{2} e^{x} (\sin x \plus{} \cos x)$.

This gives $ I \equal{} \frac {1}{2}xe^{x}(\sin x \plus{} \cos x) \minus{} \frac {1}{2}\int e^{x}(\sin x \plus{} \cos x) \ dx$ which seems easy to integrate by parts.

Hope this helps.[/quote]

Thanks. AndrewTom, I didn't see this type of integration parts. I tried $ u \equal{} xe^{x},dv\equal{}\cos x$ and $ u\equal{}x \cos x, dv\equal{}e^{x}$, and it became very complicated. Thank you very much, i hope you will solve the other problems I posted.

Or this:

$ \int xe^x\cos x\,dx\equal{}\int\text{Re}\,(xe^{(1\plus{}i)x})\,dx\equal{}\text{Re}\left(\int xe^{(1\plus{}i)x}\,dx\right)$

$ \equal{}\text{Re}\left(\frac1{1\plus{}i}xe^{(1\plus{}i)x}\minus{}\frac1{1\plus{}i}\int e^{(1\plus{}i)x}\,dx\right)$

$ \equal{}\text{Re}\left(\frac{1\plus{}i}{(1\plus{}i)^2}xe^{(1\plus{}i)x}\minus{}\frac1{(1\plus{}i)^2}e^{(1\plus{}i)x}\right)\plus{}C$

$ \equal{}\text{Re}\left(\frac{e^x(\cos x\plus{}i\sin x)((1\plus{}i)x\minus{}1)}{(1\plus{}i)^2}\cdot\frac{(1\minus{}i)^2}{(1\minus{}i)^2}\right)\plus{}C$

$ \equal{}\frac14\text{Re}\left(e^x(\cos x\plus{}i\sin x)((x\minus{}1)\plus{}ix))(\minus{}2i)\right)\plus{}C$

$ \equal{}\frac{e^x}4\text{Re}\left(((x\minus{}1)\cos x\minus{}x\sin x)\plus{}i((x\minus{}1)\sin x\plus{}x\cos x)(\minus{}2i)\right)\plus{}C$

$ \equal{}\frac{e^x}2((x\minus{}1)\sin x\plus{}x\cos x)\plus{}C$

As a side benefit, we have also computed that

$ \int e^x\sin x\,dx \equal{} \frac{e^x}2(x\sin x\minus{}(x\minus{}1)\cos x)\plus{}C.$

I would like this method a little better for a definite integral, say $ \int_{\minus{}\infty}^0e^x\cos x\,dx.$ But once we get started, we don't need any new clever ideas. And it would work just as well for $ \int xe^{ax}\cos bx\,dx$ for any $ a$ and $ b.$


A square is inscribed in a semicircle, and a second square is inscribed in the whole circle, as shown below. What is the ratio of the area of the smaller square to the area of the larger square? Express your answer as a common fraction.

Hm.. the picture is slanted. Dont use it much  :rotfl:

[hide="Answer"]The center of the circle will be in the middle of a side of the smaller square. Therefore, if we say that the center of the circle splits the side of the square into two segments of length $x$, and we draw the radius to one of the two corners of the square that touch the outside of the circle, then we have a right triangle with legs $x$ and $2x$ and hypotenuse $r$. Therefore, $r^2=5x^2$, and the area of the square is $4x^2$, so the area of the square in terms of $r$ is $\frac{4}{5}\cdot r^2$. With the larger square, drawing the diagonals gives two diameters of the circle. We can proceed from there to see that each side of the square has length $r\sqrt{2}$, so the total area of the big square is $2r^2$. Therefore, $\frac{A_{small}}{A_{large}}=\frac{\frac{4}{5}\cdot r^2}{2r^2}=\boxed{\frac{2}{5}}$.[/hide]

Good, that's what I want (that's what I had), but your final step is messed up in latex terms (and I can't see what's wrong with the code).

how is his final step messed up in latex? it looks fine to me. :?

I think he just changed it  :)
it had 25 at the end, but his steps were correct, so I assumed a latex error.

If you edit it quickly enough, it doesn't register the edit. You must have seen it just a few seconds after I posted it, because I think I had it corrected within about 30 seconds. ;)

[quote="math92"]it doesn't say he edited it though.
oh, well. i don't really care.[/quote]

I think if you edit it within 5 minutes, it doesn't count, something like that.
Let's stop this, because it has no revelance  :D


number theory
greatest common divisor
modular arithmetic

The sequence is 0, 30, 24 .... None of the answers is small enough to be a factor of those terms, except for 1. The answer is (A).

EDIT: I just realized that and I edited that. Ignore Ihatepie's post because you can't see it anymore.

@Ihatepie: That is annoying and makes your message ugly.

@pi 2^5-2=30.

The message is too small. Please make the message longer before submitting.

[quote="sunehra"]The largest of the following integers which divides each of the numbers of the sequence $ 1^5 \minus{} 1,\, 2^5 \minus{} 2,\, 3^5 \minus{} 3,\, \cdots, n^5 \minus{} n, \cdots$ is:

 $ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 120\qquad \textbf{(E)}\ 30$[/quote]

$ 2^5\minus{}2\equal{}30$, $ x(x^4\minus{}1)\equal{}0 \pmod{3}, \equal{}0(\pmod{5}), \equal{}0(\pmod{2}$ by Fermat's little theorem, so the answer is $ E$

Err, I thought, for the third term:

$ 3^5 \minus{} 3 \equal{} 240$

EDIT: Refering to PowersOfPI's post.

[hide="Solution"]
$ n^5\equiv n\pmod{5}\implies n^5 \minus{} n \equiv 0\pmod{5}$, so it's a multiple of $ 5$.

$ n\equiv n^k\pmod{2}\implies n\equiv n^5\pmod{5}\implies n^5 \minus{} n\equiv 0\pmod{2}$ so it's a multiple of $ 2$ also.

Similarly $ n^5 \minus{} n\equiv 0\pmod{3}$ so it's also a multiple of $ 3$.

Since $ 2,3$ and $ 5$ all divide $ n^5 \minus{} n$, then $ n^5 \minus{} n\equiv 0\pmod{30}$. This is the maximum because neither $ 60$ nor $ 120$ divide $ 30$(second term).

So the answer is $ \textbf{E}$

Edit: Sorry didn't see stephencheng's post.
[/hide]

Ignore my post.

Just by inspection, we can limit the choices to (C) and (E).


geometry
circumcircle
incenter
geometry proposed

Let triangle $ ABC$ with circumcircle $ (O)$. A circle $ (J)$ intangent $ (O)$ at $ T$ and tangent $ AB,AC$. Denote $ K,L$ be the incenter of $ ABT,ACT$ respectively. Prove that: $ (KLT)$ tangent $ (O)$

Do you mean circumcircle of $ KLT$ is tangent to $ O$?

No, it mean two circle: circumcircle of triangle $ KLT$ and circumcircle of triangle $ ABC$ is tangent.

What is $ K,L$?If $ K,L$ are circumcenter's of $ \triangle{ABT}$ and $ \triangle{ACT}$ then $ K \equal{} L$,since $ A,B,C,T$ are concyclic?Maybe misunderstand something :maybe: ?

Sorry erken! I have fix this problem! $ K,L$ be the incenter of $ 2$ this triangle!  :blush:
[img]http://i225.photobucket.com/albums/dd18/nktp/2circle.png[/img]

This problem is very easy.We only need to prove that $ \frac{TG}{TR}\equal{}\frac{TK}{TP}$.Which is obviously true.(Look at my picture)

[url=http://imageshack.us][img]http://img516.imageshack.us/img516/2882/nktpjf0.png[/img][/url]

Yes, [b]Erken[/b], I think the property $ KL\parallel DH$ is the main idea of this problem. To prove this property, let us consider $ AK$, $ AL$ as the angle bisectors of two angles $ \angle ADT$ and $ \angle AHT$ (I think the result $ TD$ passes through the midpoint of arc $ AB$ of the circumcircle of triangle $ ABC$ is well-known).
We have
\[ \frac {KT}{KD} \equal{} \frac {AT}{AD} \equal{} \frac {AT}{AH} \equal{} \frac {GT}{GH}
\]
Therefore $ KL\parallel DH$, and now we only have to consider the common tangent at $ T$ of two circle $ (O)$ and $ (J)$, we will get the result the circumcircle of triangle $ KLT$ is tangent to $ (O)$ immediately.


here is a nice sequence for beginers in sequences:
if a_1=2 find a_n if
a_(n+1)=-a_n+2*(-1)^n.

cheers! :D


Let be $ a_1,a_2,...,a_n\in \mathbb{N}^*$ , not two equals . Prove that :

$ a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2\ge \frac {(a_1 \plus{} a_2 \plus{} ... \plus{} a_n)^2}{n} \plus{} \frac {n^3 \minus{} n}{12}$

Put $ n\equal{}2,a_1\equal{}2,a_2\equal{}1$ :wink:

I guess that Lagrange identity kills it.
.
.
.
Yes, it does and I guess this was the expected proof...
[hide]Lagrange identity and the formula for $ \sum^{n}_{i\equal{}1}{i^2\plus{}i^3}$ prove that
$ \sum_{i\equal{}1}^n{a_i^2}\minus{}\frac{1}{n}\left(\sum_{i\equal{}1}^n{a_i}\right)^2\equal{}\frac{1}{2n}\sum_{i,j\equal{}1}^n{(a_i\minus{}a_j)^2}\ge\frac{1}{2n}\sum_{i,j\equal{}1}^n{(i\minus{}j)^2}\equal{}\dfrac{n^3\minus{}n}{12}$

Equality happens when the $ a_i$ are consecutive positive integers.[/hide]

:)


Albatross has a 8 by 8 by 8 cube made out of 1 by 1 by 1 cubes and he paints the outer surface of the cube orange. After that he takes the cube apart back into the small cubes. What is the probabilty that a randomly selected cube has exactly one orange face?

Find a general formula for a N by N by N cube.

MUST MAKE MORE ALBATROSS PROBLEMS! HELP THE CAUSE WRITE YOUR OWN!

To have one side, it must have been on a face not touching the edge,

That gives us 6 sets of 6x6x1 cubes

216/512 = 27/64

Formulas:

no faces:
(n-2)3

one face:
6(n-2)2

two faces:
12(n-2)

three faces:
8

6 sides. Each side has (8-2)^2 sides painted orange.

6 * 36 = 216

Total number of cubes: 8*8*8= 512

Probability: 216/512 = 59/128

I hope that was clear enough.  :)


Could someone please explain to me the characteristics of balanced and unbalanced forces?

Also, in which of the following situatins do unbalanced forces act, and why?
(a) A car is traveling at 20 metres per second, and the driver steps on the icfalbrake.
(b) A chair is dragged across a floor at constant velocity.
(c) A ploane makes a turn at a constant speed.
(d) An object is dropped vertically toward the ground.

Thanks in advance!

If the acceleration is 0, then there is a balanced force.  If not, then it is not balanced.  So the constant speed examples involved balanced forces.

If an object moves at a constant speed but changes direction, is there a balanced or unbalanced force?

Thanks!

Unbalanced. The plane in (c) is acted on by a combination of aerodynamic and gravitational forces whose net vector sum points toward the center of the circle. Note that the plane is being accelerated in that direction. That the speed is constant is simply an indication that the net force is perpendicular to the direction of motion - but there is a nonzero net force.

The phrase JRav should have used rather than "constant speed examples" is "constant velocity examples." The chair in (b) is a constant velocity example. The plane in (c) does not have constant velocity.

Incidentally, if the object in (d) is dropped from a high altitude and we only ask about it very late in its fall, then the forces acting on it are approximately balanced, balanced between the downward gravitational force and the upward air resistance. This approximate balance is the practical meaning of "terminal velocity." Of course, for the purpose of this problem, I would assume we are asking about it early in its fall and I would answer "unbalanced."


inequalities
function
induction
algebra unsolved
algebra

Find all functions $f: \mathbb N \to \mathbb N$ such that
\[ 2n \le f(f(n)) \plus{} f(n) \le 2n \plus{} 1\]
for any $n \in \mathbb N$.

[quote="moriarti"]Find all functions $ f: N \minus{} > N$ such that

$ 2n \le f(f(n)) \plus{} f(n) \le 2n \plus{} 1$

for any n.[/quote]

First, we can see that $ f(n)$ is injective :
Let $ n_1<n_2$ such that $ f(n_1)\equal{}f(n_2)$. Then $ f(f(n_1))\plus{}f(n_1) \le 2n_1\plus{}1$ $ < 2n_2\le f(f(n_2))\plus{}f(n_2)$ which is impossible since $ f(f(n_1))\plus{}f(n_1) \equal{} f(f(n_2))\plus{}f(n_2)$. Hence $ f(n)$ is injective.

With $ n\equal{}1$, we get $ 2\le f(f(1))\plus{}f(1) \le 3$
$ f(f(1))>0$ and $ f(1)>0$, so $ 1\le f(1)\le 2$
If $ f(1)\equal{}2$, then $ 2\le f(2)\plus{}2 \le 3$ and so $ f(2)\equal{}1$

But then, setting $ n\equal{}2$ in the original inequation, we get $ 4\le f(f(2))\plus{}f(2)\le 5$ $ \iff$ $ 4\le f(1)\plus{}1\equal{}3 \le 5$, which is impossible, so $ f(1)\equal{}1$

We can extend this demo thru induction :
Suppose $ f(p)\equal{}p$ $ \forall p<n$. Then :
Since $ f(n)$ is injective, $ f(x)>n\minus{}1$ $ \forall x\ge n$ (and so $ f(n)\ge n$ and $ f(n\plus{}1)\ge n$)
We know that $ 2n\le f(f(n))\plus{}f(n)\le 2n\plus{}1$
If $ f(n)>n\plus{}1$, we get $ f(f(n))<n$, which is impossible since $ f(n)>n\plus{}1$ implies $ f(f(n))>n\minus{}1$
If $ f(n)\equal{}n\plus{}1$, then $ 2n\le f(f(n))\plus{}f(n)\le 2n\plus{}1$ implies $ n\minus{}1\le f(n\plus{}1)\le n$ and so $ f(n\plus{}1)\equal{}n$

But then, since $ 2n\plus{}2\le f(f(n\plus{}1))\plus{}f(n\plus{}1)\le 2n\plus{}3$, we get $ 2n\plus{}2\le (n\plus{}1)\plus{}n\le 2n\plus{}3$, which is wrong.
So $ f(n)\equal{}n$

And, since $ f(1)\equal{}1$, the induction is terminated.

Hence the unique solution $ \boxed{f(n)\equal{}n}$ $ \forall n$


inequalities
function
inequalities unsolved

prove that :
[img]http://img361.imageshack.us/img361/4547/my1zw9.png[/img]

hello, what is $ E\left(\frac{n}{10}\right)$?
Sonnhard.

Not hard at all. Let $ n \equal{} 10m \plus{} k$

$ 2^{69m \plus{} 6k \plus{} 2} < 5^{30m \plus{} 3k \plus{} 1}$

Which is true since:

$ 2^{6k \plus{} 2} \equal{} 4^{3k \plus{} 1} < 5^{3k \plus{} 1}$ (*)

and $ 2^{69} < 5^{30}$

Since $ 2^{23}\equal{}8388608<9765625\equal{}5^{10}$

So therefore: $ 2^{69m} < 5^{30m}$ (**)

Multiplying * and ** we get our desired inequality.


All right guys,

Mr. Anderson is ignoring my PM's.  So it's time to teach me a lesson by BOMBARDING ME with private messages.  They don't need to contain any particularly useful substance, except the words "Mr. Anderson," hopefully coupled with "I'm going to enjoy watching you die."

And do it quickly, before he sees this thread and shamelessly deletes it and I cannot edit my posts anymore(!)

Love,
DBR

[hide="EDIT:"]
01001001 00100000 01000001 01001101 00100000 01000111 01001111 01001001 01001110 01000111 00100000 01010100 01001111 00100000 01000101 01001110 01001010 01001111 01011001 00100000 01010111 01000001 01010100 01000011 01001000 01001001 01001110 01000111 00100000 01011001 01001111 01010101 00100000 01000100 01001001 01000101 00100000 01001101 01010010 00100000 01010010 01010101 01010011 01001000[/hide] MR. ANDERSON. THIS IS THE LAST TIME. STOP REPLACING "BOMBARDING HIM" WITH "BOMBARDING ME." I AM BEING BOMBARDED ME.

[quote="infinity4ever"]I think that's a result of Mr. Anderson's [i]mod powers[/i].  :P[/quote]

Wow he's pretty good at modular arithmetic... :P

[quote="gighiuhui"][quote="1=2"]Why are we bombarding YOU with private messages???[/quote]

What?

I am not Mr. Anderson.  [/quote]

Check your first post.  :wink:

[quote="gighiuhui"]All right guys,

Mr. Anderson is ignoring my PM's.  So it's time to teach him a lesson by BOMBARDING ME with private messages.  They don't need to contain any particularly useful substance, except the words "Mr. Anderson," hopefully coupled with "I'm going to enjoy watching you die."

And do it quickly, before he sees this thread and shamelessly deletes it(!)

Love,
DBR[/quote]

I HAVE to quote this so it will be saved forever, and can't be edited.  :rotfl:

[quote="gighiuhui"][quote="1=2"]Why are we bombarding YOU with private messages???[/quote]

What?

I am not Mr. Anderson.

Altheman==Mr. Anderson.

See, he is living two lives.  In one life, he works as a take-home-test-writer for a respectable math program.  But in the other life, he goes by the alias Altheman and is guilty of virtually every AoPS crime we have a law for.[/quote]
LOL YAYAYAYAYAY DBR

[quote="le00327146"]So...who are we bombarding?[/quote]

GAH MR. ANDERSON

CHANGE MY POSTS SO I STOP GETTING BOMBARDED

That's what he changed.  He cut the hard line.  We're trapped.

Be calm.  Give him your PMs.  He'll be able to track it.  We have no choice.

lol DBR... Alex actually isn't responding to any of my PMs either  :P So I'm all up for any annoying things you are planning.

[quote="gighiuhui"][quote="le00327146"]So...who are we bombarding?[/quote]

ME of course![/quote]

 :rotfl:

Although the first post says "HIM" now...

[quote="junggi"]ROFL I THINK ALEX PWNS DBR[/quote]

He is the gatekeeper.  He is holding all the keys.  He is guarding all the doors.  So someday, one of us is going to have to fight him.

But, my dear friends, I say to you tonight that we must face the simple truth: tomorrow is today.  Right here, in this very room, at this very moment we are confronted with the fierce urgency of now.

Take him.

> zephyredx used Zap Cannon!
> zephyredx's attack missed!
> Altheman used MASTER BALL!
*struggle animation*
> All right! zephyredx was caught!

Shoot! Now I'm stuck in a pokeball. Maybe I shouldn't have spammed Altheman.

Hmmm... I wonder what Alex will do when he sees this...  :cursing: Lol

Huh thats odd.  He replied to our PM at 1 AM during camp... so why not all of these?  :maybe:

I have an idea: if we whine so much, the whole world will be filled with whining, and he will hear, and voila!

Awww, I missed all the hilarity. I demand more god power abuse! MORE!

Dude... David that is not cool. You need to leave people whose last names are "Anderson" alone about the fact.

We PMed him one night when we were still at Awesomemath at like one thirty because we saw he was online and he actually replied...

Lol. I remember that :) And yet then he replied.

Life on the Mississippi is a different thing to Mr. Anderson from what it was to Mark Twain. The living writer sends a wife-weary, word-worn, middle-aged reporter vagabonding away from life's respectable responsibilities along the banks of the Mississippi and the Ohio.
-------------
Nickysam

[url=http://www.drivenwide.com]buzz[/url]


Argh. Never mind, I caught my mistake.  :blush:

Yes, you messed up.

(First of all, learn LATEX ... it's easier!)

So...

$ \frac{(m*79) \plus{} (n*87)} {m\plus{}n} \equal{} 84$

then...

$ 79m \plus{} 87n \equal{} 84m \plus{} 84n$

which simplifies to...

$ 3n \equal{} 5m$

You said that $ m/n \equal{} 5/3$

BUT... do it out!

$ \frac{3n} {n} \equal{} \frac{5m} {n}$

which is...

$ 3 \equal{} \frac{5m}{n}$

which is...

$ \frac{3}{5} \equal{} \frac{m}{n}$

which is the answer. I bet someone will have a more clear answer, but w/e.

[quote="AIME15"]Out of interest, what was the problem?[/quote]

Thanks for bring this up.

In the future, please don't edit your previous posts if you've made an error. Just leave it be so other users may learn  :wink:

[quote="nick42"]i'll assume he was doing problems out of a book and then looked at the answers (:noo:) Then he saw that his answer was different from the book's and quickly posted it here to make sure that he wasn't insane or something.  :D

and apparently, he messed up and the book is right.[/quote]

 Now that you say it like thaaat...  :( . And yeah, you're right.  :P


Prove that $ x^{2}+y^{2}=x^{2}y^{2}$ has apart from $ x=y=0$ no integer solutions.


geometry trapezoid rectangle trigonometry

Calculate the area of the isosceles trapezium:

[img]http://img301.imageshack.us/img301/9803/cals.jpg[/img]

Partition the trapezoid into the two triangles and a middle rectangle.  The area of the middle rectangle is easy.  For the  triangles, the base has length 4 and then you can use the fact that $ \tan{60}\equal{}\frac{x}{4}$, where $ x$ is the height.

it almost looks like the solution was on the paper but someone erased it...

hmmm


[i]*Please excuse me if this is in the wrong forum.[/i]

What score do you need on USAMTS to qualify for AIME

And how would the index work then?

[quote="Wiki"]
Further, after the third round, given a high enough score, a student may qualify to take the AIME exam instead of qualifying through the AMC 10 or 12 competitions.[/quote]
As for your second question, it counts as getting a 100 on the AMC 12, or the lowest score needed to qualify that year.

I'm also taking AMC 10/12 so if I get a higher index would that count instead?

They take whichever score is the best one for your index.

[quote="AwesomeToad"]What score do you need on USAMTS to qualify for AIME[/quote]
The USAMTS website says:
[quote="the USAMTS website"]Participants who score well after 3 rounds will be invited to participate in the American Invitational Mathematics Examination (AIME), the second step in the process of selecting the USA Mathematical Olympiad team.[/quote]
So I would guess it is on a case by case basis.

[quote="PowerOfPi"][quote="AwesomeToad"]What score do you need on USAMTS to qualify for AIME[/quote]
[quote="USAMTS"]Participants who score well after 3 rounds will be invited to participate in the American Invitational Mathematics Examination (AIME), the second step in the process of selecting the USA Mathematical Olympiad team.[/url]
So I would guess it is on a case by case basis.[/quote][/quote]

What do you mean by "case by case" basis?

[quote="AIME15USAMO"]I'm assuming it means how hard the problems are each year, and how well people do on them.[/quote]

So sort of like a 'Top x' sort of thing (assuming x to be an integer :lol:)

The minimum score required to qualify for AIME from USAMTS is a score of 69 out of 75 possible possible points after 3 rounds of USAMTS.
In computing the USAMO index, we use a score of 100 on the AMC 12.

Steve Dunbar
MAA Director, American Mathematics Competitions

Sorry to revive, but is 69/75 strictly set for each year? Or does it also depend slightly on the difficulty of the USAMTS problems, in a similar way to how the amc's allow slightly less than 100 or 120 to qualify?


Ring Theory
Galois Theory
advanced fields
advanced fields unsolved

Let $ K$ be a Galois extension of $ \mathbb{Q}$.

For each prime ideal $ \mathcal{P} \lhd \mathcal{O}_K$, let $ I_\mathcal{P}$ be the inertia group.

Show that the groups $ I_\mathcal{P}$ (for all possible $ \mathcal{P}$) generate $ \Gamma(K : \mathbb{Q})$.

Let $ G^\prime$ denote the group generated by all the inertia groups. Can you show that the field of elements fixed under $ G^\prime$ is $ \mathbb Q$? If so this implies that $ G^\prime$ is the Galois group of $ K|\mathbb Q$.  :P


I moved it to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=129662]here[/url]
where it will more likely be answered

$a\geq 1$ $\Rightarrow $ $b+c\leq-1$ $\Rightarrow $ $(A(m)\geq G(m))$ $(b+c)(b^{2}-bc+c^{2})\leq (b+c)|bc|\leq 0\leq b^{4}+c^{4}$ $\Rightarrow $ $a^{4}+b^{4}+c^{4}\geq a^{3}+b^{3}+c^{3}$
$\Rightarrow $ $\sum{a^{4}}-3abc\geq \sum{a^{3}}-3abc=(\sum{a})(\sum{a^{2}}-\sum{bc})=0$ $\Rightarrow $ $min\geq 0$.

It's too late, I'll write whole solution tomorrow.  :wink:

Thank you mods for moving this :) (I think it was in the IMO discussion forum before)


algebra
polynomial
complex analysis
algebra unsolved

Let $P (x)$ be polynomial degree $n$ with distinct real zeros $x_1, x_2, \dots, x_n$. For $0 \leq k \leq n - 2$ prove:
\[ \sum_{i = 1}^n {\frac{{x_i}^k}{P' (x_i)}} = 0. \]

Greetings!

See http://www.mathlinks.ro/Forum/viewtopic.php?t=65189 for a special case.
But here similar solutions apply, like using the residue theorem for $\frac{x^k}{P(x)}$ instead.

What is residue theorem? Can you explain a little bit?

See my posts (#52, #56 ) in http://www.mathlinks.ro/Forum/viewtopic.php?p=317935#p317935
Well, I think it's hard to handle when you aren't used to it by knowing it from complex analysis, but when being able to handle it; it is the ultimative killer for a lot of problems of that type.

It is very easy.
$Let \ f(x) \ is \ polinom \ (degree\le (n-1)), \ then \ f(x)=\sum_{i=1}^n \frac{f(x_i)(x-x_1) \dots(x-x_{i-1})(x-x_{i+1}) \dots (x-x_n)}{P^{'}(x_i)}.$ $\ Therefore \ (n-1)'th \ coefficient \ f(x) \ is \ \sum_{i=1}^n \frac{f(x_i)}{P^{'}(x_i)}.$
$For \ polinoms \ x^k, \ k=0,1, \dots,n-2 \ this \ coefficient \ equal \ 0. \ Therefore$ \[ S_k=\sum_{i=1}^n \frac{x_i^k}{P^{'}(x_i)} =0, \ k=0,1, \dots n-2.  \]
$S_{n-1}=1, \ S_n=x_1+\dots+x_n, S_k=(n-1)'th \ \ coefficient \ for \ polinom \ x^k (mod \ P(x)).$


This is a math olympiad problem and was given as an assignment to me. Could someone please solve this?

[b]GIVEN ANY SET OF TEN DISTINCT NUMBERS IN THE RANGE 10, 11, ... 99, PROVE THAT WE CAN ALWAYS FIND TWO DISJOINT SUBSETS WITH THE SAME SUM.

[/b]

i think this is an old imo problem - maybe not, maybe a national olympiad, but i've seen it before...

edit: yay, 200 posts.

[quote="boxedexe"]1) Find the number of nonempty subsets.
2) Find the number of possible sums.
3) Use Dirichlet's Box Principle (aka Pigeonhole principle) to finish the problem.[/quote]

But there's something lacking. The problem is asking for the disjoint subsets. How do you do that?

There's an easy way to satisfy this condition.


[quote="{x}"]Find all polynomial $f \in \mathbb{Z} \left[ X \right]$ such that for all positive integer $m, n$:
\[ ( f ( m ), f ( n ) ) = 1 \Longleftrightarrow ( m, n ) = 1  \][/quote]
Let p|f(x^q) and p is not divide f(x) then p is not divide x (if p|x p|(f(x^q)-f(x)) contradition). for any k p|f(x^q-kp), if (k,x)=1 then (x^q-kp,x^q)=1  (f(x^q-kp),f(x^q))>1 contradition.
Let p and q is prime and q|f(p). If q is not equal p then (p, p+kq)=1 (k is not divide p) (f(p),f(p+kq)>1 contradition.
Therefore, we prove  if x=p(1)^a(1)*...p(k)^a(k), then f(x)=p(1)^b(1)*...*p(k)^b(k). This condition satisfied only f(x)=x^m (m -natural const).

Shortly prove. Let p and q primes q|f(p), if(p,q)=1 we have (p,p+q)=1 (f(p),f(p+q))>1 contradition. Let m=deg(f(x)). Then exist N for any primes p>N f(p)=p^m.  Use Lejander approccimation with (p1,...,pk) ,k>m, and pi>N we have f(x)=p^m.

Used only f(x) - polinomial in Z[x], and (m,n)=1 ---> (f(m),f(n))=1. Not used (f(m),f(n))=1 --->(m,n)=1.

Very nice in #3 !
I'm impatient for the reverse side :)


algebra
polynomial
linear algebra
linear algebra unsolved

For any $A\in\mathcal l M_n(\Re)$ we denote $m_A$ the minimal polynomial of $A$.Let $U,V\in \mathcal l M_n(\Re)$.Prove that $\frac{m_{UV}}{m_{VU}} \in\{ 1,X,\frac{1}{X} \}$.

Notice that $X*m_{uv}(vu)=0_n$ and $X*m_{vu}(uv)=0_n$  thus $m_{vu}/X*m_{uv}$ and $m_{uv}/X*m_{vu}$  so $\exists P,Q \in R[X]$ such that $m_{vu}/m_{uv}=P/X$ and $m_{uv}/m_{vu}=Q/X$ thus $PQ=X^2$then the answer to your question is directly concluded :)


http://mathprob.com/_private/MC/2005Sprint17.gif

You need to click on it.

[quote="Internet Explorer"]You are not authorized to view this page
You might not have permission to view this directory or page using the credentials you supplied.

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

If you believe you should be able to view this directory or page, please try to contact the Web site by using any e-mail address or phone number that may be listed on the mathprob.com home page.

You can click  Search to look for information on the Internet.

HTTP Error 403 - Forbidden
Internet Explorer [/quote] Others may have this same problem :wink:

Forbidden...

Use [img][/img] instead of [url][/url].

[quote="#H34N1"]You click it. Then you refresh the page. :wink:

Its a security measure I put in to prevent hotlinking but have to change everything...[/quote]

it still doesnt work

[quote="ckck"]Now it works.[/quote]

not for me.  but im safari on a macbook, so that could be why

Forbidden
You don't have permission to access /_private/MC/2005Sprint17.gif on this server.

No, I'm not on Mac

Okay. Then Copy and paste it into your web browser.
It will say forbidden. Then click the go button again.

Click the link, then click the Go button, not Refresh.

I think Quevvy's answer and solution are correct.

[quote="davidli"][quote="Quevvy"]Since I'm the only one that can see it....  :P
[hide]$w=6$ so then $x=22,34$
$x=22\rightarrow y=14,58$ or $x=34\rightarrow y=2,70$

$y=2\rightarrow z=78,74$
$y=14\rightarrow z=100,58$
$y=58\rightarrow z=134,18$
$y=70\rightarrow z=146,6$

$z=6,18,58,74,\boxed{78,100,134,146}$

$146+134+100+78=\boxed{\boxed{458}}$[/hide][/quote]

[hide]For y=14, z=90,62. So instead the answer should be 78+90+134+146=448.[/hide][/quote]
Oh yeah, sorry.


The information in the Online School page about selecting regular classes in Independent Study seems to be contradictory.
Do we get to choose two classes or just one?

On the "About" section:
[quote]Enrollment in [b]one[/b] of our regular online classes. Students enroll in one of our regular online classes. Students can choose any of the classes we offer. Students do not need to select the class at the time of enrollment in the independent study. Click here for more details on which classes students can choose.[/quote]
And on the "Classes" section:
[quote]Students will also see other subjects in the [b]two[/b] subject courses they select.[/quote]

Thanks for pointing that out.  Independent Study students choose [b]one[/b] companion class.


AMC
AIME
pigeonhole principle
AoPSwiki
number theory
least common multiple
greatest common divisor

As a starter, I'll post a solution to number 7.

Question 7:
Let $N = \sum_{k = 1}^{1000}{k(\lceil{\log_{\sqrt{2}}k}\rceil-\lfloor{\log_{\sqrt{2}}k\rfloor)}}$

Find the remainder when $N$ is divided by 1000.  ($\lfloor{k}\rfloor$ is the greatest integer less than or equal to $k$, and $\lceil{k}\rceil$ is the least integer greater than or equal to $k$.)

[hide]
Answer: $\boxed{477}$

Notice that $k(\lceil{\log_{\sqrt{2}}k}\rceil-\lfloor{\log_{\sqrt{2}}k\rfloor)}$ is equal to $k(\lceil{2\log_{2}k}\rceil-\lfloor{2\log_{2}k\rfloor)}$

Hence, we have:

$\sum_{k = 1}^{1000}{k(\lceil{2\log_{2}k}\rceil-\lfloor{2\log_{2}k\rfloor)}}$

It is clear that the difference will be $0$ when $k = 2^{n}$, and the rest of the time $\lceil{2\log_{2}k}\rceil-\lfloor{2\log_{2}k\rfloor}$ will equal 1, meaning

$\sum_{k = 1}^{1000}{k(\lceil{2\log_{2}k}\rceil-\lfloor{2\log_{2}k\rfloor)}}$ will equal $k$ when $k \neq 2^{n}$ and will equal $0$ when $k = 2^{n}$.

Since $2^{10}= 1024$, there are exactly 10 powers of 2 $(0,1,2,...,9)$ less than 1000.  We also know that $2^{0}+2^{1}+2^{2}+...+2^{n}= 2^{n+1}-1$

We then take the sum of the first 1000 naturals $\frac{(a)(a+1)}{2}= \frac{(1000)(1001)}{2}= 500500$

We must take into account the zeros generated by our summation, leading us to $500500-1023= 499477$.

Take $499477\mod{1000}$ (or just take the last three digits) to get $\boxed{477}$[/hide]

Thanks for this! I just put up the answers for people to determine their scores.

I particularly liked #8 (well okay it wasn't that hard) :

 We can see that they must have a root in common for them to  both be factors of the same cubic.

Let this root be $a$.

We then know that $a$ is a root of \[Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0\], so $x = \frac{-k}{5}$.

We then know that $\frac{-k}{5}$ is a root of $Q_{1}$ so we get:
\[\frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k\] or $k^{2}=30k$, so $k=30$ is the highest.

We can trivially check into the original equations to find that $k=30$ produces a root in common, so the answer is $\boxed{030}$.

#3  The complex number $z$ is equal to $9+bi$, where b is a positive real number and $i^{2}=-1$.  Given that the imaginary parts of $z^{2}$ and $z^{3}$ are the same, what is b equal to?

[hide]

Answer: $\boxed{15}$

The easiest way to solve this is by expansion.  $z^{2}= 81+18bi-b^{2}$

Multiplying this by z, we get $(9+bi)(81+18bi-b^{2}= 9*81+9*18bi-9b^{2}+81bi-18*81b^{2}-b^{3}i$

Throwing away all real parts of $z^{3}$, we get $imag(z^{2})= 18bi$ and $imag(z^{3})=243bi-b^{3}i$, where $imag(x)$ is the imaginary part of x.

Since the $imag(z^{2}) = imag(z^{3})$, we can set their imaginary parts equal to each other to get

$18bi = 243bi-b^{3}i \Leftrightarrow b^{3}i = 225bi \Leftrightarrow b^{2}= 225$

The problem states that $b > 0$, so

$b = \boxed{15}$
[/hide]

Would someone please work #6 (about the frogs)?

Frogs? I forget. It was on the list of problems I got, though.

If you write out all the multiples of three and 13 for problem number six, and note how many possible ways there are to get to that number it isn't hard to work up to 39.  there is only one way to get to 3 , 6, 9 , 12.  To get to thirteen you can go from either o or 3,6,9,12 for a total of 5 ways.  To get to fifteen you can go from 13 or 12 for a total of six ways.  The rest follows the same.

[quote="SuperThread"]But you don't have to go to fifteen...

Do you have an actual answer?

PS: I'm going to kill whoever wrote #2.[/quote]

Agreed with the #2 comment, I completely missed the "next to the walkway" part...

And the mentioned strategy for #6 works, you can count the number of ways to 13, then go to 26, then to 39, and it [i]should[/i] all work out if you're careful (and include stuff like the path that skips all multiples of 13)...of course, I counted it up wrong and got an answer that was 20 too low...oh well

It was awful. I got #5 wrong because I thought that the Farenheit was ranging from 32 - 100 instead of to 1000.
I found the patterns of 9 with 5 that works in the pattern. Anyone else did that problem differently?
1000 / 9 = 111 (2 left over) pattern starts at 36, 111 - 4 = 107
107 * 5 = 535 (because 5 work, 4 don't), from 32 - 36 2 work, from 999 - 1000 they both work. 535 + 2 + 2 = 539

If everyone who has a solution posts a complete one, rather than an outline, it will benefit everyone much more.

Thanks!

[quote="calc rulz"]Thanks for this! I just put up the answers for people to determine their scores.

I particularly liked #8 (well okay it wasn't that hard) :

 We can see that they must have a root in common for them to  both be factors of the same cubic.

Let this root be $a$.

We then know that $a$ is a root of
\[Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0 \]
, so $x = \frac{-k}{5}$.

We then know that $\frac{-k}{5}$ is a root of $Q_{1}$ so we get:
\[\frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k \]
or $k^{2}=30k$, so $k=30$ is the highest.

We can trivially check into the original equations to find that $k=30$ produces a root in common, so the answer is $\boxed{030}$.[/quote]

This is EXACTLY what I did, except when I plugged it in, I said $(2x+5)(x+6)$ doesn't have a $(x-5)$ as a factor like $(x-5)(x+6)$, therefore my answer is wrong and I will guess 000.

#5: A Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest integer.

How many Fahrenheit temperatures between 32 and 1000 inclusive have this property?

[hide="Solution"]Let $T=9a+b+32$ be the Fahrenheit temperature.

We have:

$T=\left[\frac{9}{5}\left[ \frac{5}{9}(T-32) \right]+32\right]$
$\implies 9a+b=\left[\frac{9}{5}\left[ \frac{5}{9}(9a+b) \right]\right]$
$\implies 9a+b=\left[\frac{9}{5}\left[ 5a+\frac{5}{9}b \right]\right]$
$\implies 9a+b=\left[9a+\frac{9}{5}\left[\frac{5}{9}b \right]\right]$
$\implies b=\left[\frac{9}{5}\left[\frac{5}{9}b \right]\right]$

Checking values of $b$ from $0$ through $8$ gives that:
$b=0,2,4,5,7$

We must find all integers $k,32 \leq k \leq 1000$ such that $k=9n+32+{0,2,4,5,7}$, of which there are $\lfloor \frac{1000-32}{9}\rfloor (5)+4=\boxed{539}$[/hide]

#14: A sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$, $a_{n+1}a_{n-1}=a_{n}^{2}+2007$.

Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}$

[hide="Solution"]
Ok, so I don't understand why this is a number 14, and moreover, how this is a particularly good problem, since it is so easy to go about by estimation.

Note that the end behavior of this sequence is roughly that of a geometric series... that is, for high values of $a$, we have
$a_{n+1}=\frac{a_{n}^{2}+2007}{a_{n-1}}\approx \frac{a_{n}}{a_{n-1}}a_{n}=Ra_{n}$

We want to approximate $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=R+\frac{1}{R}$

If $R$ is sufficiently large, we may use $\lfloor R+\frac{1}{R}\rfloor = \lfloor R \rfloor$

Additionally, because of the multiplicative nature of the sequence, we may divide out a factor of 3 from all $a$:
$a_{0}=a_{1}=1$ and $a_{n+1}a_{n-1}=a_{n}^{2}+223$

Calculating out the first few terms, we have:
$a_{2}=224$
$a_{3}=224^{2}+223 < (225)224$

The ratio between terms is already under $225$.
By the next 2 terms, the effect of the $+223$ is already negligible. Clearly, the $R$ value cannot be less than $224$. Our $R$ value is converging on some number between $224$ and $225$, so the answer is $\lfloor R \rfloor = \boxed{224}$.
[/hide]

[quote="crazy_insanity"]#4

Three planets orbit a star circularly in the same plane.  Each moves in the same direction and moves at constant speed.  Their periods are $60$,$84$, and $140$.  The three planets and the star are currently collinear.  What is the fewest number of years from now that they will all be collinear again?

[hide]
Answer: $\boxed{105}$

We essentially must find the $\frac{LCM}{GCF}$

This is done by taking the prime factorization of each period,

$2^{2}*3*5$, $2^{2}*3*7$, and $2^{2}*5*7$

and multiplying the shared primes of the least degree

$2^{2}*3*5*7 = 420$

However, we must then divide the LCM by the GCF in order to eliminate overlooked times of collinearity.

$\frac{420}{2^{2}}$

This equals $\boxed{105}$
[/hide][/quote]
Where did the formula $\frac{LCM}{GCF}$ come from?  It's not even right.  For example, if two planets have periods of 1 year and 2 years, respectively, then they will next be collinear with the star in 1 year, whereas your formula says 2 years.

#6-- The Frog Problem

Solution:

[hide]
So I fail to solve this problem, I go home, I think to myself "if I want to finally make GA ARML, I should crack open AOPS Volume II." So I do. I open to the "Counting in the Twighlight Zone" chapter (for no particular reason) and subsequently learn how to do this problem.

All this is made clear in the image below.

Draw a graph (edges and vertices) with directional arrows that show where the frog is allowed to go. (Some arrows have been omitted for comprehensibility.) The 13s are on the top arch and the 3s are on the bottom.

The whole idea is to work backward, from end to beginning.

Start from the end, #39, and at 36 note how many ways you can get from 36 to 39 (there is 1). Then at 33 note the number of ways to 39, which is 2 (either straight there, or through 36). A couple steps later, note that from 27 there are 4 ways.

Then note that at 26 there are 5 ways; either straight there, or any of the 4 ways from 27 (that is, 1+4=5). As you go back along the 3s, continue adding the number of paths from two points you could go next (a multiple of 3 and a multiple of 13).

For instance, from 24, you could get to 39 by any of the 4 ways from 27 or any of the 5 ways from 26. None of these paths overlap because going through 26 or not makes each path unique.

Thus there are 4+5=9 ways from 24, 9+5=14 ways from 21... there are 24 ways from 15, 24+5=29 ways from 13, 24+29=53 ways from 12... 24+5(29)= $\makebox{169}$ ways from 0, which is the answer.[/hide]

Yay for DP!

You can alternately do case analysis on which multiples of 13 are used.

Oh and btw, for number 10, why does it follow that the "mirror" of any shading must also exist? It doesn't seem immediately obvious to me.

Guys come tomorrow for the AIME mathjam so you can see my neat one-liner solutions for #14 and #15 ;)

The frog one was actually very similar to the block walking ones, where you fill in the numbers.  I did that with a number line.

[quote="futuremontrealian"]I need to see #13.

I got 20 and I was very sure.   I had the height as sqrt(10) and base as 2sqrt(2)

multiple gives sqrt(80), divide by two and gives you sqrt(20)

???[/quote]

It's not a triangle. It's a pentagon.

and there lies the problem :)

...stupid diagram drawing...

[quote="JSteinhardt"]Oh and btw, for number 10, why does it follow that the "mirror" of any shading must also exist? It doesn't seem immediately obvious to me.[/quote]

Basically, pigeonhole principle. If we have USUS, for example, and each of the other 5 rows has at least one of the second or fourth boxes shaded, then there will be one column with four shaded boxes. So at least one can't have either the second or fourth shaded, and therefore is SUSU.

Okay, for the people who asked for problems, I typed them up (or in some cases copied from what had already been typed up in this thread) on the AoPSWiki page.  You can access them at

[url]http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_I_Problems[/url].

I did number 5 in a slightly different way than that previously mentioned, unfortunately I made a error on the exam, giving me 537 instead of 538

The correct version of my solution:
[hide]

For every value of C, through this conversion, the same value of C is returned...so convert the F range to a C range
32F=0C
1000F=537.7 -> 538 (this was my mistake[rounding down, not up] :ewpu: )

Range of 0-538, 539 possible values
[/hide]

Lets get our solutions onto the AoPSWiki ([url]http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_I_Problems[/url], the solutions links).

Its easy there to post alternative solutions, edit each others proofs, post brilliant one-liner solutions, discuss the problems on talk pages, etc. Saves us some scrolling time too  :lol:

I did the frog problem in the precise way you described. My tree was beautiful.  :P

I'm particularly proud of this one. Others of you guys have probably come up with much better solutions, but I like mine. I also discovered a way to prove Ptolomey's theorem, heh. This was my favorite problem of all.
[hide="9"]Let the point where CB's extension hits the circle be D, and the point where the hypotenuse hits that circle be E. Clearly EB=DB. Let EB=x. Draw the two perpendicular radii to D and E. Now we have a cyclic quadrilateral. Let the radius be length r. We see that since the cosine of angle ABC is $\frac{15}{17}$ the cosine of angle EBD is $-\frac{15}{17}$. Since the measure of the angle opposite to EBD is the complement of this one, its cosine is $\frac{15}{17}$. Using the law of cosines, we see that $x^{2}+x^{2}+\frac{30x^{2}}{17}=r^{2}+r^{2}-\frac{30r^{2}}{17}$
This tells us that r=4x.

Now look at the other end of the hypotenuse. Call the point where CA hits the circle F and the point where the hypotenuse hits the circle G. Draw the radii to F and G and we have cyclic quadrilaterals once more.
Using the law of cosines again, we find that the length of our tangents is 2.4x. Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, 8x+2.4x+x=34. Solving we find that $4x=\frac{680}{57}$ so our answer is 737.[/hide]


(a^3+1)(a^3+1)(b^3+1)\ge (a^2b+1)^3 by holder. Multiply this cyclically to get the result.


Hey people can you recommend any good scholarships to apply for ?
or how can i get money for mit? I applied for fin aid, and I need a big help but i have an unusual situation and my fasfa form do not show that i need money..
I did the fastweb.com search but dont know if those are the best scolarships there are...
so can anybody suggest something? :?

another question: do you think  that i got in early that will help in the fin aid descision?

Manuel - Best resource is MIT financial aid office, call/email them,  MIT from my experience (compared to some others) is one of the most sraight-forward school as finaid is concerned... talk about your "unusual situaion" to them.. they  DO pay attention to that.

External Scholarships in practical sense may not help as much as you think,  if the amount of external scholarship is less than the MIT financial help. (MIT scholarships are not considered merit scholarship so unless the external scholarship is more than the financial aid, it will just get absorbed in financial help)

Intel, Elk, Coca-cola are few scholarships which go above 10K.   There are of course, many others...and some schools (Eg, Caltech, Duke)  do give Merit scholarships where one need not  have to worry about filling out fafsa etc...

Good luck. (PM me if you think you could use my pespective  or - I have one son in MIT)

i'm pretty sure that gettin i early helps- that means they really really want you here and would be willing to give you as much as it may take to make that happen... you should really call and ask