sorry i dunno if this is allowed here..but is anyone up for studying SATs this summer?
and i also have questions..when studying for SATs, is it really worth it to take classes?
n do universities look at both of ur SAT I AND SAT II or just I?
does it make a difference to take some SAT IIs?
i'm currently in gr.10 n i kinda wanna make some SAT buddies so if anyone is interested can u pls PM me :D

lol i'm in toronto..n i'd love to go to MIT if they accept me which i doubt.. :roll:

Don't take classes, they are a huge rip off. My cousin took ACT prep classes and they told her that if in doubt, she should measure the angles with her protractor (on SAT geometry questions). Their "inside" strategies include eliminating wrong answers and picking from the rest. Save your money.

Instead, if you really wanna prepare, buy/borrow:

http://www.amazon.com/Official-SAT-Study-Guide/dp/0874477182

and maybe this

http://www.amazon.com/Real-SATs-Third-College-Board/dp/0874477050/ref=sr_1_5?ie=UTF8&s=books&qid=1214788458&sr=1-5

If your vocab is bad, you can find either an SAT list, or you can study from:

http://www.amazon.com/Verbal-Advantage-Steps-Powerful-Vocabulary/dp/0375709320

I highly recommend the last one as a way to study vocab. You can probably find the audiobooks online. Finally you can read.

To answer your question, they look at both, SAT I and II, or else they would not ask for them. Do yourself a favour and score triple 800s on your subjects, cause the tests aren't hard. People tend to overestimate the importance of SATs. They are important, as in, you don't wanna screw them up, but they are unlikely to make you stand out of the crowd. The most useful part about the SAT I, is that it allows places like MIT to reject people with 4.3 GPAs who have 1400 SAT scores (ie. people whose schools just inflate their marks).

Hope this helps

Dimitri

wow THX Dimitri! that really helps
but the thing is i'm really worried about the essay part n i just want to have ppl look it over for me
is there any book that u'd recommend that helps to improve this?
thx XDDD


Let $P = \{4, 8, 9, \ldots\}$ be the set of perfect powers, or the set of positive integers of the form $a^b$, where $a$ and $b$ are integers greater than $1$. Prove that

$\sum_{j \in P} \frac{1}{j-1} = 1$

[quote="fanaticsm"]Let $P = \{4, 8, 9,...\}$ be the set of perfect powers, or the set of positive integers of the form $a^b$, where $a$ and $b$ are integers greater than $1$. Prove that

$\sum_{j \in P} \frac{1}{j-1} = 1$[/quote]

You defined P wrong.  You made it out to be a finite set.  I fixed it in my quote above.

So indeed you want us to show that $\underset{i=2}{\overset{+\infty }{\sum }}\frac{1}{i^{2}-1}=1$
Is that right ??  :?

That would be a very foolish thing to do, amirthlusa.

Here's a useful [hide="hint"]Expand each term as a geometric series, and rearrange to $\sum_{m,n>1}\frac1{n^m}$. That sum can be dealt with by normal methods.[/hide]

This is in the section of Zeitz that assumes college material.

Maybe I'm crazy, but this seems to involve primes, so maybe it'll end up being the Zeta function.

The zeta function could appear, but that's a sign you did the sums in the wrong order.

Also, it's not really about primes; there's no term for $2^33^4.$

[quote="blahblahblah"]No.  Read the problem.[/quote]
Ok I'm sorry  :blush:  :blush:
I think he wanted us to show the following :
$\underset{i=2}{\overset{+\infty }{\sum }}\frac{1}{i^{2}-1}+\underset{i=2}{\overset{+\infty }{\sum }}\frac{1}{i^{3}-1}+\underset{i=2}{\overset{+\infty }{\sum }}\frac{1}{i^{4}-1}+...=\underset{k=2}{\overset{+\infty }{\sum }}\left( \underset{i=2}{\overset{+\infty }{\sum }}\frac{1}{i^{k}-1}\right)=1 $
Is that right ??  :?

[quote="jmerry"]rearrange to $\sum_{m,n>1}\frac1{n^m}$[/quote]
isn't it $\sum_{m,n>1, r>0}\frac1{n^{m^r}}$.

[quote="erdos"]$\underset{i=2}{\overset{+\infty }{\sum }}\frac{1}{i^{2}-1}+\underset{i=2}{\overset{+\infty }{\sum }}\frac{1}{i^{3}-1}+\underset{i=2}{\overset{+\infty }{\sum }}\frac{1}{i^{4}-1}+...=\underset{k=2}{\overset{+\infty }{\sum }}\left( \underset{i=2}{\overset{+\infty }{\sum }}\frac{1}{i^{k}-1}\right)=1$ [/quote]
No. That equation is false, because it's double-counting the numbers that are perfect powers in more than one way. There is only one term for $\frac1{64-1}$, and your sum has three copies: $\frac1{2^6-1}$, $\frac1{4^3-1}$, and $\frac1{8^2-1}$.

Here's a full explanation of my [hide="method"]
Expand each term $\frac{\frac1n}{1-\frac1n}$ as a geometric series, so we have $\sum_{n\in P}\sum_{k\ge1}\frac1{n^k}$. Each perfect power can be written in exactly one way as a perfect power raised to the 1, and in exactly one way as an integer other than a perfect power raised to a power greater than one. If we change the first type to the second in the above sum, and then arrange our terms by base and exponent, the sum becomes $\sum_{m,n>1}\frac1{n^m}$; this covers each perfect power in the same number of ways.

Now, we need to evaluate $\sum_{n=2}^\infty\sum_{m=2}^\infty\frac1{n^m}$.
The inner sum is a geometric series, with sum $\frac1{n(n-1)}$. The outer sum is then $\sum_{n=2}^\infty\frac1{n(n-1)}=\sum_{n=2}^\infty\left(\frac1{n-1}-\frac1n \right)=1$ and we are done.[/hide]


can some one will send me the notes of iit of maths or tell m sites where i can get it.

sites yes [url=http://iit-redefined.zzl.org/index.php]this[/url] one


USA1 Brian Lawrence 6 7 0 7 7 1 |  28
USA2 Tedrick Leung   7 1 0 7 3 1 |  19
USA3 Arnav Tripathy  7 7 0 7 1 0 |  22
USA4 Alex (Lin) Zhai  7 7 3 7 7 1 |  32
USA5 Eric Larson       7 7 0 7 1 0 |  22
USA6 Sherry Gong     7 7 3 7 7 1 |  32

Medal cut-offs: Gold 29, Silver 21 and Bronze 14.

It's pretty sad that Brian and Arnav both got silver,
well, Iurie Boreico got silver too.

[quote="xxazurewrathxx"]Hmm, wow the people who got silvers before (Sherry and Alex) got golds this time, and the people who got golds (Brian and Arnav) got silvers.[/quote]

Coincidence? I THINK NOT!


probability Mafia AMC AIME Support function geometry

Okay, so everyone has a score, and you use spells and attacks and crap to raise or lower your own or other people's scores.

So here's the deal. You make everything up. There aren't really any rules, except:
**No single move shall change a player's score by more than +/- 1000 points (negative scores are allowed)
**No effect shall last more than 100 posts

The object of the game isn't really to win, it's just to see how messed up this game can get (sort of like Yu-Gi-Oh on TV, but that's another story)

Also, declare and adjust for any rules violations that may be encountered (what rules? lol)

Add yourself to the point list as you post.

Since there's no one to attack yet...

------

Early Bird Special +300, also makes user immune to breakfast-type attacks for 50 turns

[b]Score[/b]
randomdragoon: 300

I exist: + 500
Dude I made MOP last year: +41
Dude why did you not make MOP last year!?!?: -299 to randomdragoon

[b]Score[/b]
randomdragoon: 1
probability1.01: 541

cheesecakes: +42
jelly!: +10 to probability

[b]Score[/b]
randomdragoon: 1
probability1.01: 551
junggi:42

Complement is -0.01: -0.01

Score
randomdragoon: 1
probability1.01: 550.99
junggi:42

moderator of game 25: +25
messing up on probability assignments left right and center: -0.89
ToW3 went to 3414 pages: +3414/3=+1138
asian: +10 to me and junggi

Score
randomdragoon: 1
probability1.01: 550.99
junggi: 52
bubka: 1172.11

I pwn!

Adding more than 1000: -138 to bubka
Everyone else being asian too: +10 to me and randomdragoon
1% bonus: +0.1 to me

Score
randomdragoon: 11
probability1.01: 561.09
junggi: 52
bubka: 1034.11

me making the analogy to calvinball: +50
me not being asian: +100
me lying about not being asian: +100
probability1.01 and bubka having decimal scores: round down, -100

randomdragoon: 11
probability1.01: 461
junggi: 52
bubka: 934
iamagenius: 250

JELLY!!! *no effect until 3140 A.D.*

BREAD!!! -400 bubka

TEA!!!!! +21 randomdragoon

Hurtz Doughnut -40 iamagenius

BUMP +200 points 1=2

randomdragoon: 32
probability1.01: 461
junggi: 52
bubka: 534
iamagenius: 210
1=2: 200

At this moment, I have 467 posts. [b]+467[/b]

Summons a giant Pac-man to eat 1=2!  But it only eats his fingernail... [b]-2[/b]

Throws a chicken at randomdragoon!  Pecks randomdragoon to death. [b]-31.999[/b] *like Stewie Griffin* BLAST!

randomdragoon: 0.001
probability1.01: 461
junggi: 52
bubka: 534
iamagenius: 210
1=2: 198
lingomaniac88: 467

Pebble of No Real Significance: Has no effect
The Forty-nine Spell Effect: Has no effect
The Evil Wrath of Bob the Edible Jellybean: Has no effect
Right, so everything is still the same as before. Carry on.

pikachu is in the cool club, which automatically gives me +800 points because daermon>pikachu

randomdragoon: 0.001
probability1.01: 461
junggi: 52
bubka: 534
iamagenius: 210
1=2: 198
lingomaniac88: 467
daermon:  800

cheburashka atomatically gives me +500 and after +50 per tern for as long as someone is playing

Also I cast russian cold

causes 200 damage to all non russians per attempt at reducing a russians health

finally I round randomdragoons heath up to 1 using reverse pyschology
randomdragoon: 1
probability1.01: 461
junggi: 52
bubka: 534
iamagenius: 210
1=2: 198
lingomaniac88: 467
daermon: 800
ars:500

I use the power of ketchup to move everybody's points too 100.
(random dragoon however avoids this asn stays at 1)
AZN POWER!!!!!!!!!!!!!!!!!!!!!!!!!

All the azns get +(100 times # of posts before them)
That means i have 1300 and everybody else has

randomdragoon: 1
probability1.01: 100+
junggi: 100+
bubka: 100+
iamagenius: 100 (non asian, skrew u. jk)
1=2: 100+
lingomaniac88: 100+
daermon: 100+
ars:100 (Edit- i guess ars is russian. SKREW U!!! jk)
To tell the truth, this game is like a game where everybody's God

i summon hitler. hitler reduces all jews and liberals to 1.

randomdragoon: 1
probability1.01: 1
junggi: 1
bubka: 1
iamagenius: 1
1=2: 1
lingomaniac88: 1
daermon: 1
ars:1

unfortuently, i activates the anti hitler attack which removes the effects of hitler

ashpotter:1300
randomdragoon: 1+
probability1.01: 100+
junggi: 100+
bubka: 100+
iamagenius: 100
1=2: 100+
lingomaniac88: 100+
daermon: 100+
ars:100

you can't really give somebody an ability... but oh well, thanks! =D

(btw, did you know that nobody is paying attention to HP anymore and everybody's just using as currency for these awesomelicioustastic items. LOL)

Uses my Lazer Staff of Satan to Obliterate Prime Factorization.  Now I wield all my stuff.  Including my Vacuum Cleaners and Magical Microwaves.

Throws a Vacuum Cleaner at markkram.  Doing so causes him/her to get brainwashed and lose all knowledge of their Genesis Ability.  HP are gone too.

mz94:Ghost + THIEF ability + casper the ghost ice cream bar + Magical TV (1) + Halo of Angelicy (1) + Amulet of Shizzatzzcka (Unwielded) + poison blister bombs(1) + invisibility cloak(unremovable) + Paper Mario Wormholes (Unassembled) + Lazer Staff of Satan (1) + Rubber Chicken (1) + Lazer Staff of Satan (1) +Amulet of Shizzatzzcka (Unwielded)

infinity4ever: 554025000 + poison blister bombs (2 Wielded) + Vacuum Cleaner (9 Wielded) + THIEF ability (2) + random junk + Energy Drain Ability + mega energy drain ability + Magical TV (3 Wielded) + Magical Microwaves (11 Wielded) + Halo of Angelicy (Wielded) + Amulet of Shizzatzzcka (Wielded) + invisbility cloak (unremovable) + Paper Mario Wormhole (Assembled Wielded) +  shield of honor (Wielded)

Nerd_of_the_Ages: 5000 + SuperBomb (10) + THIEF ability + Halo of Angelicy (1) + Amulet of Shizzatzzcka (Unwielded) + Lazer Staff of Satan (1) +560000 + poison blister bombs (1)

Shaggy75: Nothing

Prime Factorization: -52000 + resident evil 4 (2)+ THIEF ability + Genesis ability + obi wan's jedi belt (unwielded) + poison blister bombs (1) + honey coated nobel peace prize (1) + authentic invisibility cloak(1) + monkey-loaded bazooka(1)

12markkram34: 0

Farrell: 2048585 + obi wan's jedi belt (unwielded) + entire ratchet and clank arsenal (and clank) + Godric Gryffindor's sword + Elder Wand + Ressurection Stone + Real Invisibility Cloak(2)(unremovable) + Paper Mario Wormhole (Unassembled) + Moldy Apple (1) + invisbility cloak(unremovable) + Paper Mario Wormholes (Unassembled)

Asiodea: 0.0000001 + Halo of Angelicy (1) + Amulet of Shizzatzzcka (Unwielded) + Paper Mario Wormhole (Unassembled) + Lazer Staff of Satan (1)

My authentic invisibility cloak reveals that you destroyed a clone image of me. Your futile attack is rendered back at you, costing you 3/4 of your points which I absorb. I then use my monkey-loaded bazooka on Farell, which does 300 000 damage + public embarassment. Finally, I reward myself a bottle of soap.

mz94:Ghost + THIEF ability + casper the ghost ice cream bar + Magical TV (1) + Halo of Angelicy (1) + Amulet of Shizzatzzcka (Unwielded) + poison blister bombs(1) + invisibility cloak(unremovable) + Paper Mario Wormholes (Unassembled) + Lazer Staff of Satan (1) + Rubber Chicken (1) + Lazer Staff of Satan (1) +Amulet of Shizzatzzcka (Unwielded)

infinity4ever: 138506250 + poison blister bombs (2 Wielded) + Vacuum Cleaner (9 Wielded) + THIEF ability (2) + random junk + Energy Drain Ability + mega energy drain ability + Magical TV (3 Wielded) + Magical Microwaves (11 Wielded) + Halo of Angelicy (Wielded) + Amulet of Shizzatzzcka (Wielded) + invisbility cloak (unremovable) + Paper Mario Wormhole (Assembled Wielded) + shield of honor (Wielded)

Nerd_of_the_Ages: 5000 + SuperBomb (10) + THIEF ability + Halo of Angelicy (1) + Amulet of Shizzatzzcka (Unwielded) + Lazer Staff of Satan (1) +560000 + poison blister bombs (1)

Shaggy75: Nothing

Prime Factorization: 415518750 + resident evil 4 (2)+ THIEF ability + Genesis ability + obi wan's jedi belt (unwielded) + poison blister bombs (1) + honey coated nobel peace prize (1) + authentic invisibility cloak(1) + bottle of soap(1)

12markkram34: 0

Farrell: 1748585 + obi wan's jedi belt (unwielded) + entire ratchet and clank arsenal (and clank) + Godric Gryffindor's sword + Elder Wand + Ressurection Stone + Real Invisibility Cloak(2)(unremovable) + Paper Mario Wormhole (Unassembled) + Moldy Apple (1) + invisbility cloak(unremovable) + Paper Mario Wormholes (Unassembled)

Asiodea: 0.0000001 + Halo of Angelicy (1) + Amulet of Shizzatzzcka (Unwielded) + Paper Mario Wormhole (Unassembled) + Lazer Staff of Satan (1)

i use may ghostly underworld powers to reveal yr true form, which i kill, disable yr invisible cloak, and give farrell and infinity each 1/2 of yr hp

i also award shaggy a sword of the living dead, which he can use to revive himself and/or drain someone's hp

mz94:Ghost + THIEF ability + casper the ghost ice cream bar + Magical TV (1) + Halo of Angelicy (1) + Amulet of Shizzatzzcka (Unwielded) + poison blister bombs(1) + invisibility cloak(unremovable) + Paper Mario Wormholes (Unassembled) + Lazer Staff of Satan (1) + Rubber Chicken (1) + Lazer Staff of Satan (1) +Amulet of Shizzatzzcka (Unwielded)

infinity4ever: 346265625 + poison blister bombs (2 Wielded) + Vacuum Cleaner (9 Wielded) + THIEF ability (2) + random junk + Energy Drain Ability + mega energy drain ability + Magical TV (3 Wielded) + Magical Microwaves (11 Wielded) + Halo of Angelicy (Wielded) + Amulet of Shizzatzzcka (Wielded) + invisbility cloak (unremovable) + Paper Mario Wormhole (Assembled Wielded) + shield of honor (Wielded)

Nerd_of_the_Ages: 5000 + SuperBomb (10) + THIEF ability + Halo of Angelicy (1) + Amulet of Shizzatzzcka (Unwielded) + Lazer Staff of Satan (1) +560000 + poison blister bombs (1)

Shaggy75: Nothing + sword of the living dead

Prime Factorization: 0 + resident evil 4 (2)+ THIEF ability + Genesis ability + obi wan's jedi belt (unwielded) + poison blister bombs (1) + honey coated nobel peace prize (1) + authentic invisibility cloak(1)(disabled) + bottle of soap(1)

12markkram34: 0

Farrell: 209507960+ obi wan's jedi belt (unwielded) + entire ratchet and clank arsenal (and clank) + Godric Gryffindor's sword + Elder Wand + Ressurection Stone + Real Invisibility Cloak(2)(unremovable) + Paper Mario Wormhole (Unassembled) + Moldy Apple (1) + invisbility cloak(unremovable) + Paper Mario Wormholes (Unassembled)

Asiodea: 0.0000001 + Halo of Angelicy (1) + Amulet of Shizzatzzcka (Unwielded) + Paper Mario Wormhole (Unassembled) + Lazer Staff of Satan (1)

Getting sick and tired of this game, I quit and hand down all my items to Aisodea.

mz94:Ghost + THIEF ability + casper the ghost ice cream bar + Magical TV (1) + Halo of Angelicy (1) + Amulet of Shizzatzzcka (Unwielded) + poison blister bombs(1) + invisibility cloak(unremovable) + Paper Mario Wormholes (Unassembled) + Lazer Staff of Satan (1) + Rubber Chicken (1) + Lazer Staff of Satan (1) +Amulet of Shizzatzzcka (Unwielded)

infinity4ever: 346265625 + poison blister bombs (2 Wielded) + Vacuum Cleaner (9 Wielded) + THIEF ability (2) + random junk + Energy Drain Ability + mega energy drain ability + Magical TV (3 Wielded) + Magical Microwaves (11 Wielded) + Halo of Angelicy (Wielded) + Amulet of Shizzatzzcka (Wielded) + invisbility cloak (unremovable) + Paper Mario Wormhole (Assembled Wielded) + shield of honor (Wielded)

Nerd_of_the_Ages: 5000 + SuperBomb (10) + THIEF ability + Halo of Angelicy (1) + Amulet of Shizzatzzcka (Unwielded) + Lazer Staff of Satan (1) +560000 + poison blister bombs (1)

Shaggy75: Nothing + sword of the living dead

12markkram34: 0

Farrell: 209507960+ obi wan's jedi belt (unwielded) + entire ratchet and clank arsenal (and clank) + Godric Gryffindor's sword + Elder Wand + Ressurection Stone + Real Invisibility Cloak(2)(unremovable) + Paper Mario Wormhole (Unassembled) + Moldy Apple (1) + invisbility cloak(unremovable) + Paper Mario Wormholes (Unassembled)

Asiodea: 0.0000001 + Halo of Angelicy (1) + Amulet of Shizzatzzcka (Unwielded) + Paper Mario Wormhole (Unassembled) + Lazer Staff of Satan (1) +
resident evil 4 (2)+ THIEF ability + Genesis ability + obi wan's jedi belt (unwielded) + poison blister bombs (1) + honey coated nobel peace prize (1) + authentic invisibility cloak(1)(disabled) + bottle of soap(1)

I declare that all these items are idiotic, so they're all gone.

mz94:0
infinity4ever: 346265625
Nerd_of_the_Ages: 565000
Shaggy75: 0
12markkram34: 0
Farrell: 209507960
Asiodea: 0.0000001

im a ghost! i dont get any hp!

i give my ghost powers to infinity

infinity4ever: 346265625 + ghost powers
Nerd_of_the_Ages: 565000
Shaggy75: 0
12markkram34: 0
Farrell: 209507960
Asiodea: 0.0000001

Sayanoras evry1!
im off to destroy Canada!!

taking mz's lead, i commit suicide and become a ghost to haunt another thread.

i hand my ghost powers over to nerd

Nerd_of_the_Ages: 565000 + ghost powers
Shaggy75: 0
12markkram34: 0
Farrell: 209507960
Asiodea: 0.0000001

taking infinity's lead, i also leave this post

Nerd_of_the_Ages: 565000 + ghost powers
12markkram34: 0
Farrell: 209507960
Asiodea: 0.0000001

Wow. This thread is falling faster than a bad nosejob.

too bad nobody's really playing anymore...  :|

:starwars: isn't this cool; i now use my superlasergun to touch all my friends and they now have it too.

It can do 11000000 damage.  :juggle: this is my jugglerballattack it does 50000000 damage now all my friends have it. :flex:

:spider:  :gathering: my evil spider venompoisonbomb kills all the enemies of my friends. The happy gathering is for friends only.  :play_ball: they are cool. :trampoline: my trampolinebazookamachinegunchainsaw kills anyone ten feet away, i give it to all my friends~!@#%^&*~!!!Charge! 40000 damage done to enemies.hahahahaha!!!!!

alexluh: please, though this is the Games and fun Factory, if you don't have anything to say that makes ANY sense at all, just don't post.

Could a mod lock this thread?


If the pattern continues, what is the next term in the sequence 12, 7, 15, 10, 18, $ \_\_\_\_$?

The pattern is (-5),(+8),(-5),(+8), etc. making the next number 18-5=13.


1. $ a^2\plus{}b^2\plus{}(ab)^2 \equal{} 4567654.$
Find all integer values of $ a$ and $ b$ so this equation is true.

2. $ a^2\plus{}b^2\plus{}c^2\plus{}(ab)^2\plus{}(bc)^2\plus{}(ac)^2\plus{}(abc^2) \equal{} 456787654$
Find all integer values of $ a$, $ b$, and $ c$ so this equation is true.

3. $ (ab)^2 \equal{} a^2\plus{}b^2\plus{}c^2$.
Find all integer values of $ a$, $ b$, and $ c$ so this equation is true.

[hide="solution to 1"]Note that $ 4567654\equiv2\pmod4$ and that $ p^2\in\{0,1\}\pmod4$ for any $ p\in\mathbb{Z}$.

By checking all possible sums of 3 squares $ \pmod4$, we see that the three squares must be $ 1\plus{}1\plus{}0\pmod4$, or two of the squares must be odd and one square must be even.

Assume for the sake of contradiction that exactly one of $ a$, $ b$, and $ ab$, is even.
If $ a$ is even, then so is $ ab$, which is a contradiction.
If $ b$ is even, then so is $ ab$, which is also a contradiction.
If $ ab$ is even, then at least one of $ a$ and $ b$ is even, which is a contradiction.
Thus, the original assumption is false and there are no solutions to the equation.[/hide]

[quote="Fraenkel"][hide="Hint"]
Simon's fav factoring trick.
[/hide][/quote]

That would involve factoring a large number, which is not fun (note that I say not fun, not that it can't be done, or even that it is difficult).

It only reduces problem 1 to:
[hide]$ (a^2 \plus{} 1)(b^2 \plus{} 1) \equal{} 4,567,655$[/hide]

reduces to 4567655. One factor is 5. so let $ a\equal{}2$. all you have to do now is check the other number.
$ 913531 \minus{}1\equal{}913531$ is not a square. No solutions.

Try No 3.

[quote="Fraenkel"]reduces to 4567655. One factor is 5. so let $ a \equal{} 2$. all you have to do now is check the other number.
$ 913531 \minus{} 1 \equal{} 913531$ is not a square. No solutions.

Try No 3.[/quote]

That's a minor technicality.  Anyway, your solution would only be valid if you knew that 913,531 was prime.  You have only shown that there would not be a solution IFF $ a\equal{}2$.  However, say $ 913,531\equal{}pq$. Then, $ p\minus{}1$ could be a perfect square and so could $ 5q\minus{}1$.

Also, $ 913,531\minus{}1\neq913,531$  :wink:

I think I have most of 3, but I've got to go now (I'll post it if I finish)

[hide="3"]
Simple casework with mod 3 quadratic residues gives no solution.

[/hide]


Euler
LaTeX
number theory
number theory proposed

$n$ divides $fi(a^n-1)$  i know it has trivial solution with primitive roots because n is the order of $a^n-1$ and thats it but i tried this problem a year ago when i knew nothing of primitive roots now it is trivial to me but before it was so hard but anyway i never found a solution without 'elementary' number theory  could there be one?? i m very interested

At first, "n is the order of $a^n-1$" is nonsense; what is correct is that n is the order of a modulo $a^n-1$. Secondly, this proof has nothing to do with primitive roots, and can be easily elementarized as follows:

We have to show that for any two natural numbers n and a, we have $n\mid \phi\left(a^n-1\right)$.

In fact, dividing $\phi\left(a^n-1\right)$ by n with remainder, we get $\phi\left(a^n-1\right)=qn+r$, where q and r are integers and $0\leq r<n$. We want to show that r = 0, since this will imply $\phi\left(a^n-1\right)=qn$ and thus $n\mid \phi\left(a^n-1\right)$, and this will solve the problem.

By the Euler-Fermat theorem, since the numbers a and $a^n-1$ are coprime, we have $a^{\phi\left(a^n-1\right)}\equiv 1\mod a^n-1$. In other words, $a^{qn+r}\equiv 1\mod a^n-1$, or, equivalently, $a^{qn}a^r\equiv 1\mod a^n-1$. Now, obviously $a^n\equiv 1\mod a^n-1$, and thus $a^{qn}=\left(a^n\right)^q\equiv 1^q=1\mod a^n-1$. Hence, $a^{qn}a^r\equiv 1\mod a^n-1$ becomes $a^{qn}a^r\equiv a^{qn}\mod a^n-1$. In other words, $a^n-1\mid a^{qn}a^r-a^{qn}$, or, equivalently, $a^n-1\mid a^{qn}\left(a^r-1\right)$. Now, since the numbers $a^n-1$ and $a^{qn}$ are coprime (since $a^n-1$ and a are coprime), this yields $a^n-1\mid a^r-1$. But since r < n, this is only possible if $a^r-1=0$, i. e. if $a^r=1$. Thus, r = 0, and we are done.

  darij

very nice darij!! anyway for those who dont know the other solution here it goes:

lemma 1.
if n is the order of a modulo m and x is such that $a^x=1(mod m)$ then n divides x this is proved in any number theory book and i really dont know all the  latex keys to post it correctly.

corolary1 n divides $fi(m)$

now it is clear that n is the order of a modulo $a^n-1$ so n divides $fi(a^n-1)$

and were done


function
real analysis
real analysis unsolved

This isn't hard at all, but I thought it was rather interesting:

Let $(X,d)$ be a complete separable metrix space, and let $f:X\to\mathbb R$ be a function. Show that the set of accumulation points in $X$ where $f$ has limit $\infty$ is at most countable.

Let $S$ be the set of all these points. Consider $ x \in S $. Then let $ p \in Q $ such that $f(x) < p$. Moreover there is $q \in Q$ such that $ t\in \{y\in X: d(x,y)<q\} $ implies $f(t) >p$. So for every $ x\in S $ there is at least an order pair $(p,q)$ for which the above conditions hold. Now let $S(p,q)$ be the set of all points of $S$ which can be corralated with the pair $(p,q)$. Let L be a countable base of X. Then every open ball with center at a point of $S(p,q)$ and radious $q$ equals to the union of countable open sets of L. But since $d(x,y)\geq q$ for all $x,y \in S(p,q)$ these balls are disjoint. So $S(p,q)$ is at most countable and we are done.
But, do we really need the fact that X is complete ? don't beat me if  i wrote nonsences cause i'm sleeppy right now...

This isn't a know theorem ? that the set of points, of such a function on X, in which f is has discontinuity of the first kind, is at most countable ?

what about you proof,grobber? it seems Anto's solution doesn't really need completeness

I saw the problem in a book and I used the completeness, so it's rather pointless to post the solution. As a consequence, I will not :).


algebra
polynomial
vector
Ring Theory
advanced fields
advanced fields unsolved

Let $ I\equal{}(f_1, ... , f_m)$ be an ideal of the polynomial ring $ \mathbb{C}[x_1,...,x_n].$ Let $ R\equal{}\frac{\mathbb{C}[x_1,...,x_n]}{I},$ and suppose $ R$ is finite dimensional as a vector space over $ \mathbb{C}.$ Is it true that we always have $ \dim_{\mathbb{C}}R \leq \prod_{j\equal{}1}^m \deg{f_j}$? Here $ \deg{f_j}$ is the total degree of $ f_j.$

I am guessing no, but my guess doesn't necessarily make it so.
For a counter example one should look for a non-trivial ideal $ I$ with a Grobner basis wrt the lex order that has a polynomial in only the last of the variables and that (the basis) is larger than the given system of generators. Google did not help me find one.


In case you don't know, mole day is on October 23 (This Saturday!) and starts at 6:02 am. Go to http://www.moleday.org/ for more information.

So, what is everyone (anyone) doing to celebrate?

Woah! Their site is hilarious. I thought you meant the animal mole like groundhog day or something. Now I have to tell everyone at school about Mole Day, since we are starting chemistry! :D  :D  :D

Thankfully, my teachers didn't make me do any projects for mole day.  Last year my chemistry teacher made everyone do projects.  She still does it s just i don't have that class   :) ...actually I really liked chemistry just not the little projects  :( .


If ABC is a self-conjugate triangle. D, E, F are the mid-points of AC, CB, AB respectively. Does CF cut DE into 2-halves?

False (the solution is the next exercise)


You have just made it to the final level of the MAO Gameshow! There are three
doors in front of you. Two of them contain a goat, and one contains a million
dollars. After you pick a door, the announcer will show you a goat behind one
of the unchosen doors. He then asks you if you want to stay with your original
choice, or change to the next door. You decided long before the show started
that you would always stay if you got this far. What is the probability of winning
the million dollars with this strategy?

Not changing cases, the probability is $ 1/3$ while switching yields a $ 2/3$.  This is a famous problem called the Monty Hall Paradox.

LOL i remember reading about this in "The Curious Incident of the Dog in the Night-time" a looooooooooooooooooooooooooooooooong time ago.

I...don't quite get it. :blush:

What? I think I half get it.Could someone please explain it...I think I'm having a brain blank.

ok

here it is:(restate)

you have 3 doors; 1 has the grand prize, the other 2 have a one-way trip to Loserville :D

They are A, B, and C

you pick C

lovely choice

well, I will tell you A has a one-way trip to Loser-ville

do you choose B now or stick with C

the trick here is switching gives you a higher probability to win

this is why:

if you stick with the box you picked, you only have a 1/3 probability to win; if you picked the right box first

if you exchange boxes, you will win if you pick a wrong box, because then its not in the box he shows you and its not in the box you picked(if you picked the wrong box).

the probability of picking a wrong box first is 2/3; so this is the MONTY HALL PARADOX

Don't believe me?

Suppose there are 647236 boxes. 647235 are losers
1 has a Grand Prize

you pick one

i declare 647234 of them losers, leaving you with one you picked and one other one.

Would you still keep your choice or would you pick the one i conspicuously "just happened" to skip?


inequalities
function
number theory unsolved
number theory

Do you have any idea how to show the product of all primes less than some x (a primordial) is less than e^x ?

There is an elementary proof of the fact that this product it is less than $4^n$, i think that this proof is due to Erdos.
But as far i know, it is an open problem to find an elementary proof of the fact that this product is less than $3^n$ , so the same holds for $e^n$. (I'll post you the book in which i read this, but not know cause i'm not home. )
So if you want a proof of this, you should expect a non-elementary proof.

I read this is not quite true!
It is true for all "small" $n$, but it seems not to hold for all $n$, even such an integer was not found?!?
It should be something like the inequality between prime counting function and $Li$?!?
Rosser and Schoenfeld proved (1975) that:
$ 0.998684p_n < ln(p_1 p_2 ... p_n) < 1.001102p_n $
The lower inequality holds for $p>1319007$ and the upper inequality for $p>2$.
I assume the product of all primes less than $n$ is smaller than $3^n$.
Any further reference is welcomed.

Yes , for the value $3^n$ the inequality holds for all n. This fact has been proved.
But for the value $e^n$ i don't know any results. My previous post was about the case $3^n$, for which i know that it doesn't exist any elementary proof.


induction
inequalities proposed
inequalities

Prove that , for all $ n \in N*$ and for all $ x$ , we have:

$ 1 \plus{} cosx \plus{} \frac {cos2x}{2} \plus{} \frac {cos3x}{3} \plus{} ... \plus{} \frac {cosnx}{n} \ge 0$

who have some ideas?  :oops:

to Obel1x: i think induction is useful method for this. but this Problem can be sloved by Rolle theorem


real analysis
calculus
integration
real analysis unsolved

When we construct a non Lebesgue-measurable subset $E$ of the reals (with the traditional classes of equivalence modulo the rational numbers), we can insure that $\mu^{*}(E)<\epsilon$, where $\epsilon>0$ is as small as desired. Can we also make sure to have $\mu^{*}(E)>\eta$ for some $\eta>0$?

Forgot about [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=vitali&t=52701]this[/url] already? :)

Of course, that problem asks for a proof that we can have $\mu^{*}(E)$ in any subinterval of $[0,1]$, but it's clear that this implies the positive answer to your question.

Well, I completely forgot about it since I had never read the solutions (it looked too complicated), but today, as I was reading some integration stuff again, I realized that the information we have on the Vitali sets is very meager, and this question came to my mind. I'm going to go through your proof now ...


calculus
integration
trigonometry
calculus computations

Let $ g(x)\equal{}4\sin x\plus{}3\cos x\plus{}2,$ the denominator of this expression.

$ g(0)\equal{}5$ and $ g(\pi)\equal{}\minus{}1.$ By the Intermediate Value Theorem, there is a point $ a$ such that $ g(a)\equal{}0.$ Therefore, this integral is improper.

It should be possible to show that the integral is divergent.

But the principal value still makes sense :) (OK, OK, I remember our discussion about Bictor717's post a couple of years ago :P)


AMC AIME USA(J)MO USAMO USAMTS geometry Princeton

It has started.  AoPS Kings.  Please sign-up here.  This will have 3 levels, one for 6-8 graders, the second for people with AIME scores less than 8, and the third for the rest.  We really want at least 100 signed up for each level, so [size=200]PLEASE SIGN-UP NOW.[/size]There will be a detailed description in a few weeks.  Also, please put this in your signature.  It helps.  This will be bigger than the MIMC, so please sign-up.

I'll join, can you sign up for more than one level?

No, just one level.  Choose the one you think you fit best in.  When, you sign-up, please include your level.

Is the 6-8th grade supposed to be easier than the less than 8 on AIME?

I don't really know which one to sign up for...  :maybe:

Also, are you talking about average practice AIME score right now, or what you got on the real thing?

[quote="mad_skillz_aops"][quote="Iversonfan2007"]dude u spelled south carolina as "south caralina" on ur sig, wtf is with that, if you're gonna post some kind of contest test thing (aops kings...?) on this forum, get the name right.[/quote]
I did get the name right.   And Sean is running it too, so you might stop insulting it.[/quote]

no u just changed ur sig after reading my post :wink: , and i wasn't insulting the test, since it might bring this forum back to life.

Well, you said "...some kind of contest test thing..."  You called it some kind, contest, test, and thing.  That's a bunch of names, but that doesn't matter.

I'll join on level 2.

What are the prizes going to be?  :lol:

[quote="mad_skillz_aops"]I think I'll keep the prizes in suspense.  But they will be worth the time.[/quote]

are they objects or something intangible?

Probably not.

When are you aiming for this? Like next 1 week or so would be great. :)

Sign me up for level 3(even though I never wrote AIME :D ).

Sign me up for Level 3, please.
Would you message everyone who signed up once you get the release date?

We will message everyone when we post the problems, but not when we find out the release date.

[quote="SplashD"][quote="mad_skillz_aops"]We already have our prizes.[/quote]

are they the same for each level?[/quote]

The prizes should probably be commensurate with the level to discourage "playing down" as a form of cheating.

[quote="mad_skillz_aops"]It should be out around October 15th-20th.[/quote]

Is it just me or is it October 15th already?

Um... Hello everyone :blush:
Just before the deadline I would like to participate in your contest
It seems fun :)
I decided to be in Level 3
 :)

Hi, could I sign up for level 3. Thanks.

Signup closed, thread locked.  Look for the test on November 1st, in a new thread.  We will link to it from here.

I wanna sign up for the one for 6-8 graders i am in 8ty grade


How many of the numbers in the following list can be simplified? \[ \sqrt2,\sqrt3,\sqrt4,\ldots,\sqrt{100}\] For example, $ \sqrt{45}\equal{}3\sqrt5$ can be simplified, but $ \sqrt{30}$ cannot.

Heres a quick solution:

There are $25$ multiples of $4$, $11$ multiples of $9$, and $2$ multiples of both. This is a total of $34$ numbers that can be simplified throuigh factoring out a $2$ or $3$. The only two other optionsare to factor out a $5$ or $7$. Counting these up, we see there are $5$ additional cases, giving as a total of $\boxed{39}$.

[quote=jbatterson]Ugh... missed 49x2[/quote]

is that jbatterson himself????

[hide=sol] first we find the squares of the prime numbers that are less than ten:
$2^2=49$,$3^2=9$,$5^2=25$,$7^2=49$ From here we see that any multiples of these numbers will be able to be simplified. We also realize that because we dont want to count the multiples of 9 that are also multiples of four we subtract 2 from the multiples of nine. we can do the same and not count the fourth multiple of 5^2 .So there are $25+9-2+3-1+1=39$ multiples[/hide]


inequalities
search
function
calculus
derivative
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Consider the following inequality
\[ \frac{1}{(a\plus{}b)(a\plus{}kb)}\plus{}\frac{1}{(b\plus{}c)(b\plus{}kc)}\plus{}\frac{1}{(c\plus{}a)(c\plus{}ka)} \ge \frac{9}{2(k\plus{}1)(ab\plus{}bc\plus{}ca)}.\]
a) Prove that the inequality holds for $ 0 \le k \le \frac{4}{11}$, $ k\equal{}1$ and $ k \ge \frac{11}{4}.$

b) Prove that the inequality holds for all $ k \ge 0$ (I have no nice solution for this case :().

:)

See
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1813350011&t=147606

A partial solution is based on the inequality
$ \sum \sqrt\frac a{kb+c} \ge \frac 3{\sqrt{p+1}}$,
which holds for ${ p\ge \frac {49+9\sqrt{17}}{32}}$.

[quote="Vasc"]See
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1813350011&t=147606

A partial solution is based on the inequality
$ \sum \sqrt\frac a{kb + c} \ge \frac 3{\sqrt {p + 1}}$,
which holds for ${ p\ge \frac {49 + 9\sqrt {17}}{32}}$.[/quote]
Yes, VasC. I proved it by $ 2(ab+bc+ca) =a(b+c)+b(c+a)+c(a+b)$ and Cauchy Schwarz. :)

Hi dear can_hang2007, I see

\[ \sum\frac{1}{b^2\plus{}kbc\plus{}c^2}\ge\frac{9}{(k\plus{}2)(ab\plus{}bc\plus{}ca)}\]

holds for all $ k\ge 2$, is it new?

[quote="encyclopedia"]Hi dear can_hang2007, I see
\[ \sum\frac {1}{b^2 \plus{} kbc \plus{} c^2}\ge\frac {9}{(k \plus{} 2)(ab \plus{} bc \plus{} ca)}
\]
holds for all $ k\ge 2$, is it new?[/quote]
Actually, this result is not new, encyclopedia. And it comes from Iran 96 Inequality. ;) :)
The function $ f(k) \equal{}\frac{k\plus{}2}{b^2\plus{}kbc\plus{}c^2}$ has the derivative
\[ f'(k)\equal{}\frac{(b\minus{}c)^2}{(b^2\plus{}kbc\plus{}c^2)^2} \ge 0\]
So $ f(k)$ is increasing and it suffices for us to show that the inequality holds for $ k\equal{}2$, which is just equivalent to Iran 96 Inequality. :)


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ P(x)$ be a polynomial of degree $ n$. Let
\[ P(k) \equal{} \frac{k}{k\plus{}1}, \; \; k \equal{} 0, 1,\ldots, n.\]
Find explicitly an expression for $ P(m)$ for $ m > n$.

By Lagrange Interpolation Formula:
$ P(x)\equal{}\sum_{i\equal{}0}^{n}P(x_i)\prod_{j\neq i}\frac{x\minus{}x_j}{x_i\minus{}x_j}$.In our case $ x_i\equal{}i$ and $ P(x_i)\equal{}\frac{i}{i\plus{}1}$,where $ i\equal{}0,1,\dots,n$.

Er, we can do a little better than that. :)

First, consider the polynomial $ Q(x) \equal{} (x\plus{}1)P(x) \minus{} x$.  The condition tells us that $ Q(x)$ is zero at the points $ k\equal{}0,1,\ldots,n$.  So $ Q(x) \equal{} Cx(x\minus{}1)\ldots(x\minus{}n)$, since $ Q$ has degree $ n\plus{}1$.

What is $ C$?  Well, we can plug in $ x\equal{}\minus{}1$ to get $ 1 \equal{} C(\minus{}1)(\minus{}2)\ldots(\minus{}n\minus{}1) \equal{} C (\minus{}1)^{n\plus{}1} (n\plus{}1)!$.  Solving for $ C$ and plugging in, we get:

$ Q(x) \equal{} x \cdot  \frac{1\minus{}x}{2} \cdot \frac{2\minus{}x}{3} \ldots \frac{n\minus{}x}{n\plus{}1}$.  So we have $ P(x) \equal{} \frac{x \plus{} x \cdot  \frac{1\minus{}x}{2} \cdot \frac{2\minus{}x}{3} \ldots \frac{n\minus{}x}{n\plus{}1}}{1\plus{}x}$.

We can clean up a little when $ m > n$ is an integer.  Then $ P(m) \equal{} \frac{m \plus{} (\minus{}1)^{n\plus{}1}{m\choose{n\plus{}1}}}{m\plus{}1}$.

Whoops, typo here.  I meant to conclude that: $ P(x) \equal{} \frac {x \minus{} x \cdot \frac {1 \minus{} x}{2} \cdot \frac {2 \minus{} x}{3} \ldots \frac {n \minus{} x}{n \plus{} 1}}{1 \plus{} x}$.

My value for $ P(m)$ is still correct.

i have to say your solution makes perfect sense to me but how would one use Lagrange's Interpolation Formula to solve this?


calculus
induction
number theory unsolved
number theory

[quote="moldovan"]Find the number of ten-digit natural numbers (which do not start with zero) containing no block $ 1991$.[/quote]

I hope somebody will give a non-ugly solution :

[hide="Ugly direct calculus"]
We have $ T\equal{}9\times 10^9$ ten digits numbers not starting with $ 0$

Let $ N_1$ the quantity of such numbers containing exactly once the block $ 1991$
Let $ N_2$ the quantity of such numbers containing exactly twice the block $ 1991$
Let $ N_3$ the quantity of such numbers containing exactly thrice the block $ 1991$
No ten digit number contains the block $ 1991$ more than thrice

We are looking for $ T\minus{}N_1\minus{}N_2\minus{}N_3$

1) Obviously $ N_3\equal{}1$ (number $ 1991991991$)

2) $ N_2\equal{}4267$
We have $ 10^3\minus{}1$ numbers containing exactly twice blocks $ 1991$ in positions $ 1$ and $ 4$ (the $ \minus{}1$ is for eliminating $ 1991991991$)
We have $ 10^2$ numbers containing exactly twice blocks $ 1991$ in positions $ 1$ and $ 5$
We have $ 10^2$ numbers containing exactly twice blocks $ 1991$ in positions $ 1$ and $ 6$
We have $ 10^2\minus{}1$ numbers containing exactly twice blocks $ 1991$ in positions $ 1$ and $ 7$ (the $ \minus{}1$ is for eliminating $ 1991991991$)
We have $ 9\times 10^2$ numbers containing exactly twice blocks $ 1991$ in positions $ 2$ and $ 5$
We have $ 9\times 10$ numbers containing exactly twice blocks $ 1991$ in positions $ 2$ and $ 6$
We have $ 9\times 10$ numbers containing exactly twice blocks $ 1991$ in positions $ 2$ and $ 7$
We have $ 9\times 10^2$ numbers containing exactly twice blocks $ 1991$ in positions $ 3$ and $ 6$
We have $ 9\times 10$ numbers containing exactly twice blocks $ 1991$ in positions $ 3$ and $ 7$
We have $ 9\times 10^2\minus{}1$ numbers containing exactly twice blocks $ 1991$ in positions $ 4$ and $ 7$ (the $ \minus{}1$ is for eliminating $ 1991991991$)
So $ N_2\equal{}10^3\plus{}3\times 10^2\plus{}3\times 9\times 10^2\plus{}3\times 9\times 10\minus{}3\equal{}4267$

3) $ N_1\equal{}6,391,463$
We have $ 10^6$ numbers starting with $ 1991$
We have $ 9\times 10^5$ numbers containing $ 1991$ starting at position $ 2$
We have $ 9\times 10^5$ numbers containing $ 1991$ starting at position $ 3$
We have $ 9\times 10^5$ numbers containing $ 1991$ starting at position $ 4$
We have $ 9\times 10^5$ numbers containing $ 1991$ starting at position $ 5$
We have $ 9\times 10^5$ numbers containing $ 1991$ starting at position $ 6$
We have $ 9\times 10^5$ numbers containing $ 1991$ starting at position $ 7$
We have none containing $ 1991$ starting beyond position $ 7$
So $ N_1\plus{}2N_2\plus{}3N_3\equal{}64\times 10^5$
So $ N_1\equal{}6400000\minus{}2\times 4267\minus{}3\equal{}6391463$

And the required number is $ 9\times 10^9\minus{}6391463\minus{}4267\minus{}1$ $ \equal{}\boxed{8,993,604,269}$
[/hide]

[hide="A rather ugly induction"]
Let $ a_n$ the quantity of $ n\minus{}$digits numbers not starting with $ 0$, containing no $ 1991$ blocks and ending with $ 199$
Let $ S_n$ the total quantity of $ n\minus{}$digits numbers not starting with $ 0$ and containing no $ 1991$ blocks

$ S_n\equal{}10S_{n\minus{}1}\minus{}a_{n\minus{}1}$
$ a_n\equal{}S_{n\minus{}3}\minus{}a_{n\minus{}3}$

We can start with $ S_1\equal{}9$, $ a_1\equal{}a_2\equal{}0$, $ a_3\equal{}1$ and $ a_4\equal{}9$ and so :
$ a_1\equal{}0$ and $ S_1\equal{}9$
$ a_2\equal{}0$ and $ S_2\equal{}90$
$ a_3\equal{}1$ and $ S_3\equal{}900$
$ a_4\equal{}9$ and $ S_4\equal{}8,999$
$ a_5\equal{}90$ and $ S_5\equal{}89,981$
$ a_6\equal{}899$ and $ S_6\equal{}899,720$
$ a_7\equal{}8990$ and $ S_7\equal{}8,996,301$
$ a_8\equal{}89891$ and $ S_8\equal{}89,954,020$
$ a_9\equal{}898821$ and $ S_9\equal{}899,450,309$
$ a_{10}\equal{}8987311$ and $ S_{10}\equal{}\boxed{8,993,604,269}$
[/hide]

[hide="Simplifying my previous induction"]
Using the induction given in my previous post, we can simplify and :

Let $ S_n$ the number of $ n\minus{}$digits numbers not starting with $ 0$ and containing no $ 1991$ blocks. Then :

$ S_1\equal{}9$
$ S_2\equal{}90$
$ S_3\equal{}900$
$ S_4\equal{}8999$
$ S_{n\plus{}4}\equal{}10S_{n\plus{}3}\minus{}S_{n\plus{}1}\plus{}9S_{n}$ $ \forall n\geq 1$

[/hide]


Which post-count ranks do you like better, AoPS or Mathlinks?

Definitely AoPS. I prefer Riemann Hypothesis than "Hyper Active"

Can you please show me the old AoPS ranks? I have never seen them.

  dg

Looks like a crushing majority likes the AoPS ranks better :).

I think that the AoPS ranks go somewhat like this:

P vs. NP,  Hodge Conjecture, Poincare Conjecture, Riemann Hypothesis, Yang Mills Thoery,  Navier Stokes Equations, and Birch and Swinnerton Dyer

The AoPS ranks are named after the 7 millennium problems.  [url=http://www.claymath.org/millennium]More about them here[/url][/url]

Also, instead of the number of stars filled in showing your rank, you have a number of question marks. It's curiosity!

It looks like they returned the old ranks... for the AoPSers at least.  yah!!!  I think they mispelled Riemann however.

O and by the way I like the stars better.

Many of us now have the millennium problems with stars now!  That looks good!

when i replied, his rank is Poincare Conjecture and mine was still Hyper Active.

[quote="Rushil"]Just to know, since I joined quite a time after the merger, what were the original Mathlinks ranks????[/quote]Not very meaningfull. Something like "Junior Member, Member, Old Member, Active Member, Hyper Active" or something like that  :blush: In the end the best deal was to keep the AoPS old rank names, and the stars (there were question marks before).

[quote="Rushil"]No offence, but really funny ranks, Valentin!!!!  :lol:[/quote]None taken  :D However the topic is so old, that it begs to be closed :D


Hi!!!! I was looking for an e-book about non-euclidian geometry... If anyone can help me, please PM me or just quote this topic. Thanks!!!

You can read about it here:
http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Non-Euclidean_geometry.html

Thanks a lot GG!!!! But any e-book with proofs or logic steps or explanations about non-euclidian geometry???

[quote="Davron"]Thats for you ...[/quote]

Davron, this pdf is corrupted. Can you send it again?

I will try to send it later, becasue I couldn't find it (now) on my computer, sorry.

davron


I am trying to understand why I got the following problems wrong...

find all scalars c such that the statement is true:

1.)
i + cj -3k is a linear combination of i + j, and j + 3k.

the answer for this is 0... but I can't understand really what it's askign for.
My book is really of no use.  Can someone help explain?

2.)
the vector 3i - 2j + ck is in the span of i + 2j -k, and j + 3k
the answer is -27

I have the answers, but I am trying to understand why the answers are the way they are.

You will find the value of $ c$ such that $ i\plus{}cj\minus{}3k\equal{}s(i\plus{}j)\plus{}t(j\plus{}3k)$ for some real numbers $ s,\ t$.

Can you explain a bit more?  I mean, I am trying to find out why the answer to the first one is 0...
I should note, that I am a newbie so I may be missign somethgin obvious here.

For 1, you have $ i \plus{} cj \minus{} 3k \equal{} s(i\plus{}j) \plus{} t(j\plus{}3k).$  Since the unit vectors don't combine, you can just equate their corresponding coefficients to get $ i \equal{} si, cj \equal{} sj \plus{} tj, \minus{}3k \equal{} 3tk$ and then solve for $ c.$

Do the similar setup for 2.

Ah OK - now I get it.

One last somewhat related question.  What's the difference of somethign being a linear combination of something, and being in the span of something?

Nothing: the span of some vectors is defined to be the set of their linear combinations.


#1: What are all odered triples of real numbers $(a,b,c)$ which satisfy:

$c^a=b^{2a},$ $2^c=2(4^a),$ and $a+b+c=10$

Source: Florida Mathematics League, Constest 2, Problem 6 December 7, 1993

--Removed Number 2 Because Everyone did this...--

lol... I got a 6 on it with no difficulty.

I would appreciate it if you could please show your solutions!  :P   Try to hide your answers aswell *hint hint*  :roll:

[quote="4everwise"]Number 2 is the same as one of the problems in the California Math League problem we had TODAY! How uncreative is that??!!!??  ;)  :)
[hide]
The answers are 4 and 7/2.[/hide][/quote]

I know. But you do realize that all the state math leagues are the same...So I took that math league too. I got a 5 :( I accidentally got 601 for #2 :blush: But all of them were simple, even #6. I still can't believe I missed #2 ...

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=355356#p355356

lol, I had a contest today, Math League. and this same question was on it...now, different contests having same questions on the same say is uncreative.

EDIT: Don't worry, I had a simple problem...I read number 2 wrong and instead of using $10x=2005$, I used $9x=2005$ :(.  Now I realize that I need more math practice (I always did and will do, but now it'll be intense independent maths studying).

Masoud Zargar

Okay everyone: If you took a "math league" contest today, then you took the contest offered by http://www.mathleague.com which had 6 problems, about Pythagorean triples, Halloween, Equilateral triangles, differences of square roots, squares inside other squares, and four numbers which are summed two at a time.

Discuss solutions [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=355356#p355356]here[/url]!

Is anyone going to even attempt this problem... or do you just want a solution?

I got, (4, -3, 9) and (20/9, 7/3, 49/9).

From first equation we get $c=b^2$ second one gives us c=2a+1.
Plug into the one with 10, and solve


$ ABCD$ is a square. Let $ E$ be a point on the segment $ BC$ and $ F$ be a point on the segment $ ED$. If $ DF \equal{} BF$ and $ EF \equal{} BE$, then $ \angle DFA$ is

$\textbf{(A)}\  45^\circ \qquad\textbf{(B)}\  60^\circ \qquad\textbf{(C)}\  75^\circ \qquad\textbf{(D)}\  80^\circ \qquad\textbf{(E)}\  85^\circ$


geometry
trapezoid
trigonometry
analytic geometry
geometry unsolved

Let $ ABC$ be an acute triangle. Circle $ \omega_{BC}$ has segment $ BC$ as its diameter. Circle $ \omega_A$ is tangent to lines $ AB$ and $ AC$ and is tangent externally to $ \omega_{BC}$ at $ A_1$. Points $ B_1$ and $ C_1$ are defined analogously. Prove that lines $ AA_1$, $ BB_1$, and $ CC_1$ are concurrent.

I posted this problem in High school section long time ago, and received a hard solution by [b]Virgil Nicula[/b]. But I just want to see a synthetic solution, so come on!!!  :D

$ (I)$ is incircle. $ A$ is external similarity center of $ \omega_A, (I)$ and $ A_1$ is internal similarity center of $ \omega_A, \omega_{BC}$ $ \Longrightarrow$ internal similarity center $ J$ of $ (I), \omega_{BC}$ is on $ AA_1.$ Let $ A'$ be midpoint of $ BC,$ let $ U, D, P \in BC$ be orthogonal projections of $ A, I, J$ and let $ AA_1J$ cut $ BC$ at $ X.$ $ J \in A'I$ and $ A'I$ cuts $ AD$ at its midpoint $ M$ (well known). $ \overline{A'D} \equal{} \frac {c \minus{} b}{2}$ and $ \overline{JP} \equal{} \frac {a}{a \plus{} 2r} \cdot \overline{ID} \equal{} \frac {ar}{a \plus{} 2r}.$ By Menelaus theorem for $ \triangle ADX$ cut by the line $ A'JM,$

$ \frac {\overline{A'D}}{\overline{A'X}} \equal{} \frac {\overline{MD}}{\overline{MA}} \cdot \frac {\overline{JA}}{\overline{JX}} \equal{} \frac {\overline{AU} \minus{} \overline{JP}}{\overline{JP}} \equal{}$ $ \frac {h_a(a \plus{} 2r) \minus{} ar}{ar} \equal{} \frac {2pr \plus{} 2h_ar \minus{} ar}{ar} \equal{} \frac {b \plus{} c \plus{} 2h_a}{a}$

$ \overline{BX} \equal{} \overline{BA'} \plus{} \overline{A'X} \equal{} \frac {a}{2} \plus{} \frac {c \minus{} b}{2} \cdot \frac {a}{b \plus{} c \plus{} 2h_a} \equal{} \plus{} \frac {a(c \plus{} h_a)}{b \plus{} c \plus{} 2h_a},$ [color=white]_[/color] $ \overline{CX} \equal{} \minus{} \frac {a(b \plus{} h_a)}{b \plus{} c \plus{} 2h_a}$

$ \Longrightarrow\ \frac {\overline{BX}}{\overline{CX}} \equal{} \minus{} \frac {c \plus{} h_a}{b \plus{} h_a} \equal{} \minus{} \frac {ac \plus{} 2pr}{ab \plus{} 2pr}.$

Similarly, if $ K, L$ are internal similarity centers of $ (I), \omega_{CA}$ and $ (I), \omega_{AB}$ and if $ BB_1K, CC_1L$ cut $ CA, AB$ at $ Y, Z,$ then

$ \frac {\overline{BX}}{\overline{CX}} \cdot \frac {\overline{CY}}{\overline{AY}} \cdot \frac {\overline{AZ}}{\overline{BZ}} \equal{} \minus{} \frac {ac \plus{} 2pr}{ab \plus{} 2pr} \cdot \frac {ba \plus{} 2pr}{bc \plus{} 2pr} \cdot \frac {cb \plus{} 2pr}{ca \plus{} 2pr} \equal{} \minus{} 1$

Label $ (K) \equiv \omega_{A}.$ $ X,Y$ are the projections of the midpoint $ M$ of $ BC$ on $ AB,AC.$ $ P,Q$ are the tangency points  of $ (K)$ with $ AB,AC$ and $ B',C'$ are the projections of $ A_1$ on $ AB,AC.$ In the trapezoid $ XMKP,$ we get

$ A_1B' \equal{} \frac {XM \cdot KA_1 \plus{} KP \cdot MA_1}{MK} \equal{} \frac {KA_1(XM \plus{} MA_1)}{MK} \equal{} \frac {KA_1(a \plus{} h_c)}{2MK}$

Analogously, in the trapezoid $ YMKQ,$ we have $ A_1C' \equal{} \frac {KA_1(a \plus{} h_b)}{2MK}$

$ \frac {A_1B'}{A_1C'} \equal{} \frac {a \plus{} h_c}{a \plus{} h_b} \equal{} \frac {1 \plus{} \sin B}{1 \plus{} \sin C } \equal{} \frac {\sin \angle BAA_1}{\sin \angle CAA_1}$

Cyclically, we have the expressions

$ \frac {\sin \angle CBB_1}{\sin \angle ABB_1} \equal{} \frac {1 \plus{} \sin C}{1 \plus{} \sin A} \ , \ \frac {\sin \angle ACC_1}{\sin \angle BCC_1} \equal{} \frac {1 \plus{} \sin A}{1 \plus{} \sin B}$

$ \frac {\sin \angle BAA_1}{\sin \angle CAA_1} \cdot \frac {\sin \angle CBB_1}{\sin \angle ABB_1} \cdot \frac {\sin \angle ACC_1}{\sin \angle BCC_1} \equal{} \frac {1 \plus{} \sin B}{1 \plus{} \sin C}\cdot \frac {1 \plus{} \sin C}{1 \plus{} \sin A}\cdot \frac {1 \plus{} \sin A}{1 \plus{} \sin B} \equal{} 1$

$ \Longrightarrow$ Lines $AA_1,BB_1,CC_1$  concur due to Ceva's theorem.

[b]Remark:[/b] Let $ D$ denote the intersection of the ray $ AA_1$ with $ BC.$ Then we have

$ \frac {BD}{CD} \equal{} \frac{AB}{CA} \cdot \frac {\sin \angle BAA_1}{\sin \angle CAA_1} \equal{} \frac{AB}{CA} \cdot \frac {(1 \plus{} \sin B)}{(1 \plus{} \sin C)}$

Then, using barycentric coordinates at this point, we obtain

$ D  \equiv \left (0: \frac {b}{1 \plus{} \sin B}: \frac {c}{1 \plus{} \sin C} \right ) \equiv \left (0: \frac {1}{ca \plus{} S}: \frac {1}{ab \plus{} S} \right)$

By cyclic reasoning, it follows that $ AA_1,BB_1,CC_1$ concur at Paasche point of $\triangle ABC$

$ X_{1123} \left (\frac {1}{bc \plus{} S}: \frac {1}{ca \plus{} S}: \frac {1}{ab \plus{} S} \right)$


A geometric solid can be formed by gluing together some of the faces of 10 cubes, each having 3-cm edges. Of all such possible solids that use all 10 cubes, there is one with the least surface area. What is the number of square centimeters in the total surface area of this solid?

This will be a 5x2x1 prism. As there are 3 cm per 1 cube attached, it is really a 15x6x3.

Our answer is then $ 2(15\cdot6\plus{}15\cdot3\plus{}6\cdot3)\equal{}2(90\plus{}45\plus{}18)\equal{}270\plus{}36\equal{}306$

The answer is 270 not 306 your not supposed to add on the 36.


calculus
integration
function
geometry
calculus computations

A $ 10^{th}$ degree function $ y \equal{} ax^{10} \plus{} bx^9 \plus{} cx^8 \plus{} dx^7 \plus{} ex^6 \plus{} fx^5 \plus{} gx^4 \plus{} hx^3 \plus{} ix^2 \plus{} jx \plus{} k \ (a\neq 0)$ touches the line $ y \equal{} px \plus{} q$ at $ x \equal{} \alpha ,\ \beta ,\ \gamma ,\ \delta ,\ \epsilon \ (\;\text{where}\;\;\;\quad \alpha < \beta < \gamma < \delta < \epsilon)$

Find the area of the region bounded by these graphs in terms of $ a,\ \alpha ,\ \beta ,\ \gamma ,\ \delta,\ \epsilon$

 :rotfl:

The area is
$ F(a,\ \alpha ,\ \beta ,\ \gamma ,\ \delta,\ \epsilon)$
where $ F$ is the third Haldeman function.


algorithm
number theory
Euclidean algorithm

i made up this question myself

wat are the last 2 digits of 17^19^23^29....^97

i tried solving and i got the answer through brute force as 53

when i tried a logical approach this is how i did it

17^40 = 1 mod 100
trying to express the power's (19^23^29...) in terms of 40

19^2 = 1 mod 40
the powers of 19 ie 23^29...^97 can be written as 2x+1
hence (19^23^29.....) = 19 mod 40

hence the powers (19^23^29.....) can be written as 40k+19

hence 17^(40k+19) = 17^19 mod 100
i m not able to solve 17 ^19 mod 100 n hence stuck
plz help..

separate it into mod 4 and mod 25.  Mod 4, 17^19 is just 1, and mod 25, the exponents are taken mod 20, you take the multiplicative inverse of 17, which is 13 (if you use -3, it's easier).  So then you want something 13 mod 25 and 1 mod 4...and 13 does the job.

hey krishanu could u plz explain in more detail the theorem's etc u used..it will really help me as i have just started number theory and my concepts are not very strong.. :(

thanks

[quote="K81o7"]Well, by the Chinese Remainder Theorem, $17^{19}\equiv x\mod{100}\iff 17^{19}\equiv x \mod{4}, \mod{25}$.  So it suffices to determine $17^{19}$ in those two moduli.  Mod 4, it's clearly just 1^19=1.  Mod 25, the exponent can be reduced mod $\phi(25)=20$, by Euler's Totient Theorem.  So then $17^{19}\equiv 17^{-1}\mod{25}$.  Sorry for the mistake I made, I did this mod 20 instead.  What it should be is $3$, because $17^{-1}\equiv (-8)^{-1}$, and $-8*3 \equiv 1\mod{25}$.  So we have $17^{19}\equiv 3\mod{25}, 1\mod{4}$.  Again using the CRT to combined these, we have $17^{19}\equiv 53\mod{100}$...[/quote]

thanks boss,
got a couple of questions more
u said [b](17^-1) = (-8)^-1 mod 25[/b]
does the above mentioned statement arise from the fact that
17 = (-8) mod 25
hence (17^-1) = (-8)^-1 mod 25
my doubt is (-8)^-1 is equal to -(1/8) then how did u proceed from here on..

no, he's using $17^{-1}$ to mean the inverse of 17 modulo 25, not $\frac{1}{\17}$.  however, the concepts are closely related: for example, one can show that for $p \ge 5$ a prime, we have $2^{-1}+3^{-1}+6^{-1}\approx\equiv \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\approx\equiv 1 \mod{p}$ - this was the main part to the solution to an old IMO problem. (i use the approximate notation because this doesn't really hold under strict definitions of modular arithmetic.)

the inverse of 17 modulo 25 is the integer $a$ such that $17a \equiv 1\mod{25}$.  The integer is unique if we work modulo some prime, up to multiples of that prime.  i don't remember if it's unique otherwise - maybe so.

thanks teki-teki

this is how i have done,plz tell me if there is anything wrong..

m trying to solve
17x = 1 mod 100
using euclidean algorithm this is wat i get

100 = 17*5+15----> eqn 1
17 = 15*1+2---->eqn 2
15 = 7*2 + 1
2 = 2*1

now working bacwards
1 = 15-(7*2)
1 = 15 - (17-15*1)*7
1 = 100-(17*5) - 17*7 + 15*7
1 = 100 - 17 *12 - 15*7
1= 100 - 17*12 + (100-17*5)*7
1 = 100(8) - 17*47
which is of the form
100(8)+17(-47)=1
17* (-47) = 1 mod 100
since we are looking for a positive no hence x = (-47)+100 = 53
m not too sure abt the reasoning of my last step..
plz tell me

thanks

FWIW the answer is quite easy, May be  it's already given (have not looked at all the posts).. any way peek if still interested.

[hide]Obviously $19^{odd \ number }\equiv-1 \equiv 19 \bmod 20$
So we want to find $17^{20k+19}$
Since  for any $n$ which is coprime to  $100$ we have $n^{20}\equiv 1 \bmod 100$
[hide="because"]For a number n which is coprime to 100 (or 2 and 5) $n^{2}\equiv 1\bmod 4$ and $n^{\phi(25)}= n^{20}\equiv 1\bmod 25$ [/hide]

 So $17^{20k}\equiv 1 \bmod 100$ and  all we have to do is to find  the last 2 digits of $17^{19}$  which is $53$

(If  we  notice 17*3 = 51  (which is 1 mod 25) so we  don't even need pen or paper and can  easily see the above number is  is 3 mod 25, and 1 mod 4   (because 17 is 1 mod 4) so you get the answer as   53) [/hide]


Prove that there is no bijective function $f : \left\{1,2,3,\ldots \right\}\rightarrow \left\{0,1,2,3,\ldots \right\}$ such that $f(mn)=f(m)+f(n)+3f(m)f(n)$.

Let $g(n)=3f(n)+1$, then $g(mn)=g(m)g(n)$ (g is strongly multilikative).
If f is bijective, then $g: N\to 3N+1$ too bijective.
Let $q_{1},q_{2},q_{3},q_{4}$ primes form $q_{i}=2(mod \ 3).$
Then exist distinct primes $p_{1},p_{2},p_{3},p_{4}$, suth that
$g(p_{1})=q_{1}q_{2},g(p_{2})=q_{3}q_{4},g(p_{3})=q_{1}q_{3},g(p_{4})=q_{2}q_{4}.$
It give contradition with bijective, because $g(p_{1}p_{2})=g(p_{3})g(p_{4}).$


inequalities
number theory proposed
number theory

I found some results and my proof is false .Can you help me?
Consider $M=n(n-1).......(n-k+1)$ with $n \geq k \geq 1$.
1, Prove that : If $k\geq 2$ then $M\neq a^m$ with $a$, $m$ are positive integers and $m \geq 2$.
2, Prove that : If $n \geq 2k$ then $M$ has a prime factor $>k$.
3, If $k$ is prime then $M$ has a prime factor $>k$.

[i]Edited by Myth[/i]

As I recall, the first one was proved by Erdos, and it is definitely no piece of cake. I don't know if you'll receive a solution here.. :? Maybe I'm mistaking it for some other problem thoug.

[quote="K09"]1, Prove that : If k >= 2 then M#a^m with a,m are positive integers and m >= 2.[/quote]
It was proven for small $k$ ($k=2,3,4,5$) here.

[quote]2,Pro ve that : If n >= 2k then M has a prime factor >k[/quote]
Bertran's postulat?

[quote="K09"]3, If $k$ is prime then $M$ has a prime factor $>k$.[/quote]
What if $k=p$ and $n=p+1$?  :? For example, $n=4$ and $k=3$...

2) Is a result found by Chebyshev, and rediscovered by Schur, and then by Erdös when he was young.
The proof is quite long and there is many computations to check the small values of $k$.
Sorry, I can't give references until some days...

Pierre.


Hi, is it possible to figure the sum of the factors of a number from its prime factorization, without figuring out the original number? I would like to know if there's such a convenient formula.

Thanks.

The formula for the sum of the divisors of a number is

$ \prod_{k \equal{} 1}^{n}{\sum_{j \equal{} 0}^{e_k}{p_k^j}}$

where $ N \equal{} p_1^{e_1}\cdot p_2^{e_2} \cdots p_n^{e_n}$ is the prime factorization of $ N$.

Yes in fact.
[img]http://mathworld.wolfram.com/images/equations/DivisorFunction/NumberedEquation4.gif[/img]
Basically its saying that for a number $ N$, the sum of divisors is the sum of the divisors of each prime to the highest power times each other.

Take the example 24, with the prime factorization $ 2^3 \cdot 3$.  The sum of the divisors of $ 24$ is the same as the sum of divisors of $ 8$
times the sum of the divisors of $ 3.$  The sum of the divisors of $ 8$ is $ 1 \plus{} 2 \plus{} 4 \plus{} 8 \equal{} 15$ and the sum of the divisors of $ 3$ is just $ 1 \plus{} 3 \equal{} 4$.

Therefore, the formula is saying that the sum of divisors of $ 24$ is $ 15 \cdot 4 \equal{} \fbox{60}$.

Let's check to see if this holds - $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 6 \plus{} 8 \plus{} 12 \plus{} 24 \equal{} \fbox{60}$.

Sure enough, we get the right answer.

Let's get a formal formula for this:
For a prime power $ p_i^{a_i}$, the divisors are $ 1, p_i, p_i^2, \cdots p_i^{a_i}$, so
[img]http://mathworld.wolfram.com/images/equations/DivisorFunction/NumberedEquation5.gif[/img]

Therefore, for a number $ n$, the sum of divisors is:
[img]http://mathworld.wolfram.com/images/equations/DivisorFunction/NumberedEquation6.gif[/img]

I hope I was clear about this - it is a bit difficult to explain.

I took all pictures from here since I don't know how to do that kind of LaTeX.

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

EDIT: Beaten.  :l

Thanks. I find Adithya's example very helpful.  :)

[quote="Arrange your tan"][quote="AdithyaGanesh"]  The sum of the divisors of $ 24$ is the same as the sum of the divisors of $ 8$ times $ 3$.  [/quote]

This should instead be:

"The sum of the divisors of $ 24$ is the same as the product of the sum
of divisors of $ 8$ and the sum of the divisors of $ 3$."

Or, it should be:

"The sum of the divisors of $ 24$ is the same as the sum of divisors of $ 8$
times the sum of the divisors of $ 3$."

What is meant is $ (1 \plus{} 2 \plus{} 4 \plus{} 8)(1 \plus{} 3)$.  It's order of operations, where
the multiplication is done after the respective sums.

Yours means add up the divisors of $ 8$, and then multiply that result by $ 3$.[/quote]

Good point, that sure makes it a lot clearer.


calculus
geometry
topology
integration
number theory
graph theory
projective geometry

This is a thread for discussion of your favorite math books. These would be books that had a great impact on your understanding of a particular aspect of mathematics. Perhaps after a while these can be sorted and categorized for easier reference. The format should be -- Branch of Mathematics (Recommended level) - Book, Author

Number Theory (High School level) - Continued Fractions by C.D. Olds
Number Theory (High School level) - Number Theory by George Andrews
Graph Theory (High School level) - Introduction to Graph Theory by Richard Trudeau
Recreational Mathematics (High School level) - The Enjoyment of Mathematics by Rademacher and Toeplitz
General Interest (High School level) - Journey through Genius: The Great Theorems of Mathematics by William Dunham
General Interest (High School level) - The Music of the Primes by Marcus du Sautoy

History of Mathematics (Early Undergrad level) - Men of Mathematics by E.T. Bell
Calculus Techniques (Early Undergrad level) - Schaum's Outline of Advanced Calculus by Wrede and Spiegel
Advanced Calculus Theory (Early Undergrad level) - Calculus I & II by Tom Apostol
Non-Euclidean Geometry (Early Undergrad level) - Affine and Projective Geometry by M.K. Bennett
Riemannian Geometry (Early Undergrad level) - Riemannian Geometry: A Beginner's Guide by Frank Morgan
Topology (Early Undergrad level) - Elementary Topology by Michael Gemignani

Galois Theory (Upper Undergraduate level) - Galois Theory by Ian Stewart
Fourier Analysis (Upper Undergraduate level) - Fourier Series and Integrals by Dym and McKean
Mathematical Physics (Upper Undergraduate level) - Mathematics of Classical and Quantum Physics by Byron and Fuller

A great mathematics bibliography - http://www.ericweisstein.com/encyclopedias/books/Mathematics.html

General Interest (High School level) - Art of Problem Solving, Volume 2 by Richard Rusczyk and Sandor Lehoczky

Best. Book. Ever.

General textbook for high school (upper high school)

Further Mathematics RI Porter


I know that they are taking extra people in the MOP this year. Could someone explain this year's selection process?

And I'm not quite in favor of the 9th grade system. Why not 9th and under?

 :(

You'll get your chance in a bit - I'm fairly sure, based on something Chuck [Straley] said, that the MOP administration wishes to repeat the extended MOP next year, so they probably plan to make this regular (unless there are any huge stock market crashes in the next year or so).

Anyway, DNAS looks well positioned to get in "the hard way" sometime in the next couple of years, whether or not there is a ninth-grade MOP. With AoPS raising the level of math development in middle-school-age people, in a few years all this qualifying for MOP will get to be much more competitive. (I can already see that high school class of 2010 has some outstanding young people coming up, who will have to compete with one another.) But sooner or later, I think most of the younger regular participants here who maintain the desire will be able to attend MOP.

Um, what exactly is MOP?

I tried googleing it, but I just come up with all these chinese and arabic websites about stuff non-relate ot math.

[quote="h_s_potter2002"]Um, what exactly is MOP?
[/quote]

Math Olympiad(Summer) Program, it's a camp for top-USAMO scorers and to select the IMO team.

[quote="filletwho"]If Dnas makes it into the top twelve next year as an eighth grader, will they let him into MOP?[/quote]

Yes.  In fact, he can get there even if he's not in the top 12 - the automatic invites reach down further than just 12.


topology
advanced fields
advanced fields unsolved

Let $A=\{\#(X) | X$is a compact metric space$\}$.Find elements of set $A$.

Judging by the title, I assume that by $(X)$ you mean the cardinality of $X$. You should mention that though. I don't think it's standard notation.

Anyway, the answer is, I believe, that a compact metric space can either be at most countable, or of cardinality $c$ (continuum cardinality), and these are the only possibilities.

There are plenty of examples: for finite cardinals any finite susbpace of the real line, for $\aleph_0$, a convergent real sequence together with its limit, and for $c$, the circle. We must now prove that there are no other possibilities. First of all, a compact metric space has a countable dense subset, so all its points are limits of sequences formed with elements from a countable set. This means that it has cardinality $\le c$.

Now let $X$ be an uncountable compact metric space. We want to show that it has continuum cardinality. Let $\mathcal B$ be a countable basis of $X$. For any point $x\in X$ which has a neighborhood containing at most countably many points of $X$, remove from $X$ the points contained in an element of $\mathcal B$ with this property. We're left with a compact metric space $\widetilde X$, with the same cardinality as $X$, which has the additional property of being perfect (in fact, all its points have uncountably many other points in every neighborhood). Take two points $x_0,x_1\in\widetilde X$, and let $F_0,F_1$ be the disjoint closures of two open sets, one around $x_0$, and one around $x_1$. Now repeat the procedure for each $F_0,F_1$. We find disjoint compact sets $F_{00},F_{01}\subset F_0,\ F_{10},F_{11}\subset F_1$. Then repeat for the four new sets, and so on. Let $t$ be a binary sequence, and let $t_1,t_2,\ldots$ be its initial sequences (for instance, $t=(1,0,0,0,1,\ldots),\ t_1=(1),\ t_2=(1,0),\ t_3=(1,0,0)$, et.). Since $F_{t_n}$ are compact, non-empty and form a descending sequence, they have non-empty intersection. We can thus find a point $x_t\in\bigcap_nF_{t_n}$. For different $t$, the points $x_t$ are clearly different, so we have a one-to-one mapping from the set of binary sequences to the set of points of $\widetilde X$, which means that $\widetilde X$ (and thus $X$) has cardinality $\ge c$. Since we already know that it has cardinality $\le c$, we're done.

If you look at the tex in his original post, you see that he wrote #(X); for the tex to render correctly he needed to write

$A=\{\#(X): X \ \text{is a compact metric space}\}$

[color=green][And a moderator can make that happen.] [/color]


If $ 0<a<b$ then: $ \frac{a\plus{}b}{2}\minus{} \sqrt{ab} \le \frac{(b\minus{}a)^2}{8a}$.

[quote="soruz"]If $ 0 < a < b$ then: $ \frac {a \plus{} b}{2} \minus{} \sqrt {ab} \le \frac {(b \minus{} a)^2}{8a}$.[/quote]

Ineq
$ <\equal{}>(\sqrt{a}\plus{}\sqrt{b})^2 \ge 4a$
It's true  :)


[quote="AndrewTom"]The points $ Q, R$ lie on the circle $ \gamma$, and $ P$ is a point such that $ PQ, PR$ are tangents to $ \gamma$. $ A$ is a point on the extension of $ PQ$ and $ \gamma '$ is the circumcircle of triangle $ PAR$. The circle $ \gamma '$ cuts $ \gamma$ again at $ B$ and $ AR$ cuts $ \gamma$ at the point $ C$. Prove that $ \angle PAR \equal{} \angle ABC$.[/quote]

The points $ A,P,R$ and $ B$ are cocyclic Hence $ \angle BAP \equal{} \angle BRL$ where $ L$ is a point on the extention of $ (PR)$ and because $ (PR)$ is tangent to $ \gamma$ we have $ \angle BRL \equal{} \angle RCB$ and Thus $ \angle BAP \equal{} \angle RCB$ . Let $ I \equal{} (BC) \cap (PA)$ we have $ \angle ICA \equal{} \angle PAB$ and Thus it follows that $ \angle IBA \equal{} \angle PAR$ Hence the result ... :P

[u][b]Figure:[/b][/u]:

[img]http://img202.imageshack.us/img202/7499/andrew.png[/img]


Two mixtures of 28% acid [72% water] and 33% acid [67% water] are combined along with 2 gallons of water to form a mixture of 30% acid [70% water]; ten more gallons of the 33% mixture than the 28% mixture were used.  How many gallons of acid should be added to the 30% mixture to obtain a 40% acid mixture?


Hmmmm, maybe this should be in intermediate, but anyway.

If $n=a^2+1$

Prove that $n$ has no prime factors in the form $4n+3$

(nice, isn't it!?!)

Nice result.

Suppose $p=4k+3$ divides $a^2+1$. Then $(a, p)=1$, so $p \mid a^{p-1}-1$. Now since $p \mid a^2+1$, $p \mid a^{p-1}+a^{p-3}$. So $p \mid a^{p-1}+a^{p-3}-a^{p-1}+1=a^{p-3}+1$. Also $p \mid a^{p-3}+a^{p-5}$, so $p \mid a^{p-5}-1$ (note that we can't do this for $p=3$ because it's too small, but it's a case we can check manually).

Now we can go from $p-1$ to $p-5, p-9, ...,$ to $p-4m-1$. If $p=4k+1$ this causes no problems, but if $p=4k+3$ then $4k+3-4k-1=2$, so $p \mid a^2-1$. Then as $p \mid a^2+1, p \mid 2$, contradiction.

note that this is a result of Euler's criterion, which states that, for an odd prime $ p $, $ r $ is a quadratic residue $ \pmod p $ if and only if $ r^{\frac {p-1}{2}} \equiv 1 \pmod p $.  in this case, if $ p = 4k+3 $, then $ (-1)^\frac {p-1}{2} = (-1)^{2k+1} \equiv -1 \pmod p $, so -1 isn't a quadratic residue $ \pmod p $.


geometry
search
parameterization
number theory
number theory open

Is it possible to find a point in the unit square that lies at a rational distance from each of the four vertices of the square?

This [i]is[/i] number theory, not geometry, at its core.

[hide="Comment."]I had a student obsessing about this problem several years ago. I followed his stuff far enough to do a cheap computer search that confirmed that there no answers with denominators less than 10,000. I heard someone say that this is actually a notorious unsolved problem, but I'm too lazy to confirm that.[/hide]

Let the square be $ {(0,0),(0,1),(1,1),(1,0)}$ , the point $ (x,y)$ and $ a,b,c,d,p,q,r,s$ integer. Then we have
$ x^2 \plus{} y^2 \equal{} a, x^2 \plus{} (y \minus{} 1)^2 \equal{} b, (x \minus{} 1)^2 \plus{} (y \minus{} 1)^2 \equal{} c, (x \minus{} 1)^2 \plus{} y^2 \equal{} d$. Hence
$ y \equal{} (1 \plus{} a \minus{} b)/2 \equal{} (1 \plus{} d \minus{} c)/2,x \equal{} (1 \plus{} a \minus{} d)/2 \equal{} (1 \plus{} b \minus{} c)/2$. Then $ (1 \plus{} a \minus{} d)^2 \plus{} (1 \plus{} a \minus{} b)^2 \equal{} 4a,(1 \plus{} a \minus{} d)^2 \plus{} (1 \minus{} a \plus{} b)^2 \equal{} 4b, (1 \plus{} d \minus{} a)^2 \plus{} (1 \plus{} a \minus{} b)^2 \equal{} 4c, (1 \plus{} d \minus{} a)^2 \plus{} (1 \plus{} b \minus{} a)^2 \equal{} 4d$.
As a result, $ 2 \plus{} 2a^2 \plus{} b^2 \plus{} d^2 \equal{} 2a(b \plus{} d \plus{} 2),2 \plus{} 2a^2 \plus{} b^2 \plus{} d^2 \equal{} 2a(b \plus{} d) \plus{} 4b,2 \plus{} 2a^2 \plus{} b^2 \plus{} d^2 \equal{} 2a(b \plus{} d) \plus{} 4c,2 \plus{} 2a^2 \plus{} b^2 \plus{} d^2 \equal{} 2a(b \plus{} d) \plus{} 4d$. Hence $ a \equal{} b \equal{} c \equal{} d$. Then $ x \equal{} y \equal{} a \equal{} 1/2$. condictration.

It's not that simple. For one thing, you cannot assume that $ a,b,c,d$ are integers - they merely need to be the squares of rationals.

It is possible to prove that $ x$ and $ y$ must be rational numbers. Given that, we may multiply everything by the common denominator of $ x$ and $ y$ to get this: can we find integer $ x$ and $ y$ and integer $ N$ such that the distance from $ (x,y)$ to each of $ (0,0),(N,0),(0,N),$ and $ (N,N)$ is an integer?

This is about patching together Pythagorean triples.

I am sorry for it should be rational in the proof above.

We can consider multiplying a real factor $ k$. Then $ x\equal{}y\equal{}k/2, a\equal{}k^2/2$.

I am sorry for my carelessness.

It is too bad to do so. Sorry.

I guess this is an open question, or at least I haven't heard anyone to have the proof. I think the solution requires algebraic geometry but I have no idea how to prove it.

[b]Idea[/b]:1, Verify the equivalence between that all parameters are rational and that all parameter are integer.  Then prove the integer case.
2,Let the point be $ (x,y)$. Since Pythagorean triples, we have $ 4|y, (x,y)\equal{}1$.
3,Consider $ \mathbb{Z}\{d^{1/2},d\in\mathbb{N}\}$. To prove $ 2|x^{1/2}$ yields a contradiction.

Perhaps it would help.

The square:$ \{(a,a),(a,\minus{}a),(\minus{}a,\minus{}a),(\minus{}a,a)\}$ and we can let $ 4|(a,y)$.


Fie $P(x)=a_{0}x^{n}+a_{1}x^{n-1}+...+a_{n}$ cu coeficienti numere intregi .Cate triplete de numere intregi (a,b,c) $a\neq b\neq c \neq a$ exista, astfel incat $P(a)=b$, $P(b)=c$ si $P(c)=a$


inequalities
number theory unsolved
number theory

Let $a_{0}$ be positive integer. We define :
$a_{i+1}=a_{i}/2$, when $a_{i}$ is even
$a_{i+1}=3a_{i}-1$ when $a_{i}$ is odd
Prove that if there exists $a_{n}$ such that $a_{0}=a_{n}$, than $2^{n}>a_{0}$

Tiks: you're getting $2^na_0 \geq a_0$ so how do you conclude that $a_0 <2^n$  :). BTW I think it is considered the second hardest problem from first round of this year Polish Olympiad (the hardest is surely stereometry problem - I hate stereometry  :D) - my proof is 4 pages long brutal solution :P

[quote="Tiks"]You can easy see that $a_{n+1}\geq a_n/2$for evry $n\in{N}$=>
$a_n\geq a_{0}/2^n$=>if $a_n=a_{0}$=>$a_{0}<2^n$ :rotfl:  :rotfl:[/quote]

$a_n\geq a_{0}/2^n \leftrightarrow 2^na_n\geq a_0$

But is it obvious $a_n\geq 1$  ???

For equality to hold, we have $a_n=1 \rightarrow a_0=1$ and this is periodic with lenght 2,  so n=3 is the smallest solution to this, and indeed $8 \geq 1$, so the inequality is strict

Here's an idea. Maybe one could write a_n like:

a_n = ((a_0)*3^(t) - 3^(t-1) - 2^(l_1)*3^(t-2) - ... - 2^(l_s)*3^(t - s - 1))/2^k

where t + k = n and l_1, l_2,...,l_s are exponents of the denominator when we multiplied by 3. Setting a_n = a_0 we get:

(3^t - 2^k)*a_0 = 3^(t-1) + 2^(l_1)*3^(t-2) + ... + 2^(l_s)*3^(t - s - 1)

Maybe from here one can derive the desired inequality?

Yeah... it's the right way to the solution....but you also have to do something more with this equality... :lol:

Can this sequences a_n be bounded or not?

Bomb


integration
geometric series
real analysis
real analysis unsolved

$ \sum\limits_{m,n \equal{} 1}^{\infty} \frac {1}{(n \plus{} 1)^{m \plus{} 1}}$

[hide="solution"] $ \sum_{n,m = 1}^\infty{\frac {1}{(n + 1)^{m + 1}} = \sum_{n = 1}^\infty{\frac {1}{n(n + 1)}}}$ which telescopes to $ 1$.  It can also be written nicely as $ \sum_{m = 2}^\infty{\left( \zeta (m) - 1\right) } = 1$.[/hide]

Thanks JoeBlow,i have the problem,which  i don't solve. Can u help me  kill it?
$ \int\limits_{1}^{3} t^t dt \geq 7$
 and $ \sum\limits_{n\equal{}1}^{\infty} \frac{n!}{2^{n!}} <1.1$

The word is "calculate".

For the first, see techniques [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34003]here[/url]. For the second, sum a few terms and overestimate the tail with a geometric series


Suppose that p is a prime congruent to 2 mod 3.Show that there exists$\frac{p(p+1)}{3}$ subsets with p elements of a$\frac{p(p+1)}{3}$-element set, such that the intersection of any two of the subsets has at most 3 elements.

Sorry, but there is clearly a mistake in it, but I don't see a canonical correction.


geometry
rectangle
parallelogram
rhombus
geometric transformation
reflection
symmetry

Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:
(a) a square
(b) a rectangle
(c) a parallelogram with unequal sides
(d) a rhombus
(e) a quadrilateral with no special properties

A diagram would help, but I don't mind if the proof is without diagram.

[quote="worthawholebean"][hide="Hint"]Think symmetry. Do some angle chasing.[/hide][/quote]
I did and found that 2 angles are of 90. One is of 120 and one is of 60. But that doesn't match with the answer mentioned in the book. The answer mentioned is Rhombus.

No, I don't think so. Opposite angles of a rhombus are congruent, and 60-90-90-120 doesn't have two pairs of congruent angles.

My guess is that there is a wording issue in the question, because I got the same thing you got (60-90-90-120). The only rhombus I can see is if you take four trisectors, one from each vertex, with the trisectors reflections of each other over the axes of symmetry of the rectangle.

[quote="Potla"]A rhombus is definitely not the answer, as a rhombus with any of its angles 90 is a rectangle... :maybe:[/quote]
It's particularly a square not only a rectangle.