Answer the questions for $$

combatx
combatx Posts: 88 Arc User
edited December 2009 in Twilight Temple (West)
Windows 7 is cool, and what's cooler is its math input panel :D
So answer my high school/first year level math problems and you get money.

I havent played this game in like six months so i dont know the economy anymore but i'll base the rewards on a bit of what i remember:

Lets start off nice and easy:

For 250k:
Algebraically prove whether or not 4qlbph.png is true.

Hint: Set lna = ln(4^ln5) and lnb = ln(5^ln4)

For 250k:
Prove or disprove whether or not 5mw9s5.png is an odd function.

Hint: Remember that an odd function means f(-x) = -f(x)

For 500k: Solved! By Anirudh
Show that dg2sqo.png

Hint: Do some clever algebra and remember that mtmq1f.png.

For 500k:
Find all integers c with 0 <= c <= 30 for which there exist integers x and y such that 35x + 56y = c.

Hint: When is there a solution such that gcd(35, 56) = c?

For 1mil: Solved! By whiteshade
Prove that there is an infinite number of prime numbers, where a prime number 'p' is x|p if and only if x = 1 or p.

Hint: first prove that every integer x > 1 can be written as a product of primes, then base the infinite number of prime proof on a contradiction.

For 1mil:
Show that there is no perfect square whose last three digits are 341.

Hint: You should be able to insinuate that this question is simply: 16abrsk.png. Then remember that linear modular arithmetic problems can be written as linear diophantine equations.
I don't know if anything that I say is standard human nature. or if it matches up perfectly with some pathetic subculture. or that it fits the psychological profile of someone in need of this specific form of something.
Post edited by combatx on

Comments

  • Anirudh - Lost City
    Anirudh - Lost City Posts: 115 Arc User
    edited December 2009
    combatx wrote: »
    For 500k:
    Quote:
    Show that http://i45.tinypic.com/dg2sqo.png

    Hint: Do some clever algebra and remember that http://i49.tinypic.com/mtmq1f.png.

    4r705x.jpg

    Sorry took the pic with a 2 Megapixel mobile phone b:surrender
  • MagicHamsta - Lost City
    MagicHamsta - Lost City Posts: 10,466 Arc User
    edited December 2009
    Calcules confooses you too eh?

    b:chuckle
    darthpanda16: Firefox crashed on me. Aryannamage: I don't think I am a GM that would be new.
    Hawk:Do this. closing thread
    frankieraye: I'll see if we can replace the woman with a stick figure and the tiger fangs with marshmallows.//Issues like these need to get escalated quickly to minimize the damage.
    Kantorek: Yeah.. you should try it. It's awesome.
    Sihndra: Nope- not currently possible under any circumstances. Sorry.
    LokisDottir: I mean...not haunting the forums, nope nope..
    Konariraiden: You don't know what you are up against. You will lose.
    Waiting for...Hamster Packs!
    58% chance to get tokens
    41% chance to get an all class pet hamster....but they has already been freed by the magic hamster.
    1% chance to get ban hamstered with the message "Hamsters United!"
    [SIGPIC][/SIGPIC]
  • Anfisa - Lost City
    Anfisa - Lost City Posts: 274 Arc User
    edited December 2009
    Meh I could use some gold, why not?

    I'll edit this post once I'm done *edit, too tired to do more for now*

    The second 250k one:
    Ok I don't need to solve this completely out to tell that it's not an odd function. First off, -f(x) exists for all real numbers since x + sqrt(x^2+1) is always great than zero, and a of ln(a) has to be greater than zero it to be in the domain. Second f(-x) is not always in the domain. f(-x) equals ln(-x+sqrt(-x^2+1)) and -x^2 is always negative, so sqrt(-x^2+1) only exists on the interval -1 to 1. So clearly if -f(x) exists for all real numbers while f(-x) doesn't, they can't equal so the function cannot be odd.

    The first 500k one:
    If you go ahead and evaluate the limit, the you get 0/0. This means you can use L'Hopital's rule
    So thenI took the derivate of the top and bottom to go from limit as x goes to 0 of f(x)/g(x) to the limit as x goes to 0 of f'(x)/g'(x). This gives me the limit as x goes to 0 of sinx/1. Now evaluate it and I get 0. Proving that limit as x goes to zero of 1-cos(x)/x equals zero.
  • Whiteshade - Lost City
    Whiteshade - Lost City Posts: 36 Arc User
    edited December 2009
    for the first 1mil:

    lets say that the are finite numbers of primes, lets say they are P1, P2, P3,...,Pn

    assuming this are all the primes, they have to form all the possible numbers.

    the number who made of all the primes multiples +1 ( i.e [(P1*P2*P3*..*Pn)+1]) cannot be splitted to any of this listed primes, therefor there must be another prime, off that list who can split him, ( or he can be a prime himself) . we can now include that new prime to our list' call it P(n+1) but doing so will not solve anything cuz we can run this trick over and over again to add more and more primes.

    that was uclid's proof, a something more modern will involve the "zeta" function over the imaginary numbers (x axis is the real part, y axis is imaginary and z is the function value) , it was shown that all the points where this function meet the zero value represents a primary number, this function have infinite points who do so, therefor infinite primes ( this was riemanns work i think)
  • combatx
    combatx Posts: 88 Arc User
    edited December 2009
    Anirudh - Lost City wrote: »
    4r705x.jpg

    Sorry took the pic with a 2 Megapixel mobile phone b:surrender

    Woops i forgot to specify I didn't want people to use L'hopitals rule... oh well xD
    Much easier than i wanted it to be but you solved it :P
    MagicHamsta - Lost City wrote: »
    Calcules confooses you too eh?

    b:chuckle

    Calculus not so much... algebra and combinatorics... another story
    Anfisa - Lost City wrote: »
    Meh I could use some gold, why not?

    I'll edit this post once I'm done *edit, too tired to do more for now*

    The second 250k one:
    Ok I don't need to solve this completely out to tell that it's not an odd function. First off, -f(x) exists for all real numbers since x + sqrt(x^2+1) is always great than zero, and a of ln(a) has to be greater than zero it to be in the domain. Second f(-x) is not always in the domain. f(-x) equals ln(-x+sqrt(-x^2+1)) and -x^2 is always negative, so sqrt(-x^2+1) only exists on the interval -1 to 1. So clearly if -f(x) exists for all real numbers while f(-x) doesn't, they can't equal so the function cannot be odd.

    Sorry that's wrong, (-x)^2 is positive, not negative :p
    Whiteshade - Lost City wrote: »
    for the first 1mil:

    lets say that the are finite numbers of primes, lets say they are P1, P2, P3,...,Pn

    assuming this are all the primes, they have to form all the possible numbers.

    the number who made of all the primes multiples +1 ( i.e [(P1*P2*P3*..*Pn)+1]) cannot be splitted to any of this listed primes, therefor there must be another prime, off that list who can split him, ( or he can be a prime himself) . we can now include that new prime to our list' call it P(n+1) but doing so will not solve anything cuz we can run this trick over and over again to add more and more primes.

    that was uclid's proof, a something more modern will involve the "zeta" function over the imaginary numbers (x axis is the real part, y axis is imaginary and z is the function value) , it was shown that all the points where this function meet the zero value represents a primary number, this function have infinite points who do so, therefor infinite primes ( this was riemanns work i think)

    Yeah that was euclid's proof, i guess that description was okay even if you missed something (proving that all numbers can be written as a product of primes, your second sentence)
    I don't know if anything that I say is standard human nature. or if it matches up perfectly with some pathetic subculture. or that it fits the psychological profile of someone in need of this specific form of something.
  • Whiteshade - Lost City
    Whiteshade - Lost City Posts: 36 Arc User
    edited December 2009
    ill add one myself , for 250k ( since its easy) :
    show that:
    exp ( i*Pi) +1 =0

    where i = square root of (-1)
    Pi = 3.14159..

    also at the 1st 250k, its a simple answer so i might be wrong but u take lan ( Log e) from both sides of the equasion , than you get ln(5^ln4)=ln(4^ln5) ,
    the power of the 5 and 4 can be moved before the lans, and you get
    ln(4)ln(5)=ln(5)ln(4) wich is what were looking for.
    if this this proof is acceptable, you can pass the 250k reward to the one who will solve my question.

    i dun think ill solve more, dont wanna hug all the rewards.
  • Anfisa - Lost City
    Anfisa - Lost City Posts: 274 Arc User
    edited December 2009
    oh whoops, forgot the parentheses >.<
    Meh too tired to rework
  • combatx
    combatx Posts: 88 Arc User
    edited December 2009
    Whiteshade - Lost City wrote: »
    also at the 1st 250k, its a simple answer so i might be wrong but u take lan ( Log e) from both sides of the equasion , than you get ln(5^ln4)=ln(4^ln5) ,
    the power of the 5 and 4 can be moved before the lans, and you get
    ln(4)ln(5)=ln(5)ln(4) wich is what were looking for.
    if this this proof is acceptable, you can pass the 250k reward to the one who will solve my question.

    i dun think ill solve more, dont wanna hug all the rewards.

    That's pretty much exactly what i had :P
    http://i48.tinypic.com/a0v5ty.png
    Sure you can pass the reward on, i don't mind
    Anfisa - Lost City wrote: »
    oh whoops, forgot the parentheses >.<
    Meh too tired to rework

    hey don't give up, math is tricky
    it may be easier to prove that since an odd function is f(-x) = -f(x),
    it is also f(-x) + f(x) = 0 ;)
    I don't know if anything that I say is standard human nature. or if it matches up perfectly with some pathetic subculture. or that it fits the psychological profile of someone in need of this specific form of something.
  • threepointone
    threepointone Posts: 0 Arc User
    edited December 2009
    What is the mathematical name for this number:

    3.14159265358979323846264338327950288419716939937510

    Winner recieves 3 coins.
  • Column - Lost City
    Column - Lost City Posts: 201 Arc User
    edited December 2009
    I would assess the economy if you want some spam answers, 500k doesnt even get you a cheap "Honk" on Lost City anymore.
  • Devoted - Lost City
    Devoted - Lost City Posts: 3,634 Arc User
    edited December 2009
    threepointone wrote: »
    What is the mathematical name for this number:

    3.14159265358979323846264338327950288419716939937510

    Winner recieves 3 coins.

    mmmmmmmmm fffffff going to waterloo i should know this

    e?
  • Yourmom - Lost City
    Yourmom - Lost City Posts: 1,655 Arc User
    edited December 2009
    hmm something to do while i'm at work tomorrow, i'll have to come back to this.
    [SIGPIC][/SIGPIC]
  • combatx
    combatx Posts: 88 Arc User
    edited December 2009
    Column - Lost City wrote: »
    I would assess the economy if you want some spam answers, 500k doesnt even get you a cheap "Honk" on Lost City anymore.

    Column i miss you!!!
    I don't know if anything that I say is standard human nature. or if it matches up perfectly with some pathetic subculture. or that it fits the psychological profile of someone in need of this specific form of something.
  • Yourmom - Lost City
    Yourmom - Lost City Posts: 1,655 Arc User
    edited December 2009
    combatx wrote: »
    Column i miss you!!!

    No love for YM?

    I hate you so much right now.
    [SIGPIC][/SIGPIC]
  • Arganoth - Lost City
    Arganoth - Lost City Posts: 69 Arc User
    edited December 2009
    threepointone wrote: »
    What is the mathematical name for this number:

    3.14159265358979323846264338327950288419716939937510

    Winner recieves 3 coins.

    lul, Pi of course.
  • MagicHamsta - Lost City
    MagicHamsta - Lost City Posts: 10,466 Arc User
    edited December 2009
    Arganoth - Lost City wrote: »
    lul, Pi of course.

    technically you be incorrect.
    b:chuckle
    pi be infinite....that number stopped.
    b:sweat

    so the name of that number be.....
    3.141592653589793238462643383279502884197169399375 10
    b:chuckle
    me can has cheesecake now?
    darthpanda16: Firefox crashed on me. Aryannamage: I don't think I am a GM that would be new.
    Hawk:Do this. closing thread
    frankieraye: I'll see if we can replace the woman with a stick figure and the tiger fangs with marshmallows.//Issues like these need to get escalated quickly to minimize the damage.
    Kantorek: Yeah.. you should try it. It's awesome.
    Sihndra: Nope- not currently possible under any circumstances. Sorry.
    LokisDottir: I mean...not haunting the forums, nope nope..
    Konariraiden: You don't know what you are up against. You will lose.
    Waiting for...Hamster Packs!
    58% chance to get tokens
    41% chance to get an all class pet hamster....but they has already been freed by the magic hamster.
    1% chance to get ban hamstered with the message "Hamsters United!"
    [SIGPIC][/SIGPIC]
  • combatx
    combatx Posts: 88 Arc User
    edited December 2009
    Yourmom - Lost City wrote: »
    No love for YM?

    I hate you so much right now.

    Oh **** lol, sup ym totally didnt see u there
    havent slept and im malnurished and about to die etc etc
    but itll be over tomorrow morning b:sad
    I don't know if anything that I say is standard human nature. or if it matches up perfectly with some pathetic subculture. or that it fits the psychological profile of someone in need of this specific form of something.
  • angellicdeity
    angellicdeity Posts: 641 Arc User
    edited December 2009
    combatx wrote: »
    Windows 7 is cool, and what's cooler is its math input panel :D
    So answer my high school/first year level math problems and you get money.

    I havent played this game in like six months so i dont know the economy anymore but i'll base the rewards on a bit of what i remember:

    Lets start off nice and easy:

    For 250k:



    For 250k:



    For 500k: Solved! By Anirudh


    For 500k:


    For 1mil: Solved! By whiteshade



    For 1mil:

    Sounds like someone passed their MATH 135 course.
  • Anirudh - Lost City
    Anirudh - Lost City Posts: 115 Arc User
    edited December 2009
    combatx wrote: »
    Woops i forgot to specify I didn't want people to use L'hopitals rule... oh well xD
    Much easier than i wanted it to be but you solved it :P

    aww you should've mentioned that earlier :D I just wanted to keep it short and simple b:chuckle Thank you for the reward by the way! Never really did it for the money though b:victory
  • combatx
    combatx Posts: 88 Arc User
    edited December 2009
    angellicdeity wrote: »
    Sounds like someone passed their MATH 135 course.

    MATH 135 is one of the few courses i can say, although with not much practical use, is a very conceptual and mind-expanding course. I'd defend that course to death.

    Either way, only the last 3 questions require that level of algebra.
    Anirudh - Lost City wrote: »
    aww you should've mentioned that earlier :D I just wanted to keep it short and simple b:chuckle Thank you for the reward by the way! Never really did it for the money though b:victory

    Yeah my fault
    here's the solution without L'hopitals rule in case you were wondering:
    http://i46.tinypic.com/2zp32us.png
    I don't know if anything that I say is standard human nature. or if it matches up perfectly with some pathetic subculture. or that it fits the psychological profile of someone in need of this specific form of something.
  • Fishtaco - Lost City
    Fishtaco - Lost City Posts: 42 Arc User
    edited December 2009
    answer to question 1 for 250k

    5^(ln4)=4^(ln5)
    let x= 5^(ln4)
    thus
    x=4^(ln5)
    x/4=ln5
    e^(x/4)=5
    e^(5/4)(ln4)=5
    4^(5/4)=5
    5=5

    ps: im really poor so
    pls sir can i have moar
  • angellicdeity
    angellicdeity Posts: 641 Arc User
    edited December 2009
    Right from your math 135 notes:


    35x + 56y = c

    Then by the GCD Characterization theorem: if c is a common divisor of 35 and 56 there exist integer solutions - then c = n*gcd(35,56)


    Thus all integer solutions are given by n*7

    0 < 7n < 30 implies that:

    7, 14, and 21 are solutions to this Diophantine Equation














    Proof: Assume the number of primes is finite

    Then there exists primes p1 ... pk where k belongs to the set of positive integers


    Define N = p1*p2...pk + 1

    Since N is greater than any prime p1...pk and connot be prime:

    Assuming p_i > 1 (a prime number) divides N and pi divides the product p1 through pk

    Then p_i divides N + (-1)p1*....*pk (By Proposition 2.11 II of your MATH135 book)

    Therefore p_i divides 1


    Thus a contradiction, since p_i > 1 then p_i cannot divide 1

    Therefore N is prime

    Therefore there exist infinite many prime numbers
  • Yourmom - Lost City
    Yourmom - Lost City Posts: 1,655 Arc User
    edited December 2009
    angellicdeity wrote: »
    Sounds like someone passed their MATH 135 course.

    I'm confused. Does Canada use all the same course numbers across the country or something? AFAIK there was no MATH 135 at my school...
    [SIGPIC][/SIGPIC]
  • Devoted - Lost City
    Devoted - Lost City Posts: 3,634 Arc User
    edited December 2009
    i think it's just uw and maybe uoft
  • combatx
    combatx Posts: 88 Arc User
    edited December 2009
    Yourmom - Lost City wrote: »
    I'm confused. Does Canada use all the same course numbers across the country or something? AFAIK there was no MATH 135 at my school...

    yeah only UW and UofT
    I actually never took 135, i took 145, but theyre essentially the same with 145 being a bit more extended

    The reason i liked the course was because nothing was overly difficult, just really conceptual and a great intro to anything further down the math pathway

    The algebra isn't hard at all, it just deviates from the norm and for the first time you have to think critically and logically deduce conclusions by piecing together theorems and insinuating equations -- much different from the standard find x find y which we were sheeped to doing in high school.

    So yeah, i pulled questions out of the current 135 course; because the math isnt over the top of peoples heads for the most part, but it still requires a necessity to think critically.

    (Believe the only thing people haven't seen is modulo, which can be googled in 3 seconds and shown that you're just looking for the remainder between xy... hell, even one of the questions deals with gcd (greatest common divisor) which was taught in grade 3?)

    Of course I can't ask things related to combinatorics or vector spaces..
    I don't know if anything that I say is standard human nature. or if it matches up perfectly with some pathetic subculture. or that it fits the psychological profile of someone in need of this specific form of something.
  • angellicdeity
    angellicdeity Posts: 641 Arc User
    edited December 2009
    Yourmom - Lost City wrote: »
    I'm confused. Does Canada use all the same course numbers across the country or something? AFAIK there was no MATH 135 at my school...

    No the MATH 135 course code is associated with certain schools. Different schools have different course identification codes.


    This particular code belongs to the University of Waterloo - Waterloo Ontario, Canada.


    I recognized the content in this course with the questions - in particular the emphasis on Diophantine equations, modular arithmetic and prime numbers - 3 subjects covered in this particular course.
  • XXSayaXx - Lost City
    XXSayaXx - Lost City Posts: 57 Arc User
    edited December 2009
    mindbogling
    [SIGPIC][/SIGPIC]

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