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 Topic: Math & logic problems :-)

 (Read 13972 times)
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  • Re: Math & logic problems :-)
     Reply #30 - August 14, 2010, 05:25 AM
    hey, i think i said twice that it was the sum of the first n numbers Cheesy it was ignored for some reason.
  • Re: Math & logic problems :-)
     Reply #31 - August 14, 2010, 05:26 AM
    I'm not good with words Wink
  • Re: Math & logic problems :-)
     Reply #32 - August 14, 2010, 09:04 AM
    Quote from: serrated_colon on August 14, 2010, 01:59 AM
    Why did I not think of that? Nice solution, tlaloc.

    Well, it's not exactly my creation.

    I just noticed it's an arithmetic progression series and I remember the demonstration from school when i studied series and their convergence ^_^

    http://en.wikipedia.org/wiki/Arithmetic_progression
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #33 - August 14, 2010, 09:38 AM
    1x1, 3x1, 3x2, 5x2, 5x3, 7x3, 7x4....

      1,3,3,5,5,7,7,9,9,11,11,13,13,15,15...
      1,1,2,2,3,3,4,4,5,5,6,6,7,7,8
    X_______________________________

      n,           n+1,    n,       n+1,   
      (n+1)/2, n/2, (n+1)/2, n/2


    o_O
  • Re: Math & logic problems :-)
     Reply #34 - August 14, 2010, 01:07 PM
    Very nice Tlaloc, good use of Algebra  Afro

    burdenofbeing, well spotted pattern! Not seen this one before. Those are both equivalent and with a little tinkering will yield the same formula.

    Now I said not to google, but I did for purposes of seeing if there are more solutions. And there are two more, REALLY simple and neat ways to solve this problem if you think laterally (with creativity), rather than via tunnel vision (sheer brute force computational IQ).

    Hint: draw out the sequences, put them in triangles … or squares … see if you spot anything ...
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    Relativism | Cognitive dissonance | Cognitive biases | Memeplex | Consequentialism
  • Re: Math & logic problems :-)
     Reply #35 - August 14, 2010, 01:12 PM
    Quote from: Ephemeral on August 14, 2010, 02:39 AM
    OMFG IM GOING TO FAIL MATH!


    I like your idea in reverse psychology  Wink

    Solving problems, including maths, isn't just about computation - it's about creativity. An interesting video for you: http://www.ted.com/talks/dan_pink_on_motivation.html
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  • Re: Math & logic problems :-)
     Reply #36 - August 14, 2010, 05:47 PM
    well no more tinkering is needed, just one more step to be written down probably.
    the pattern goes like:

    [n x (n+1)/2 ], [(n+1) x n/2] which is n(n+1)/2 going forever.

    I tried extra hard to find something new Tongue

  • Re: Math & logic problems :-)
     Reply #37 - August 15, 2010, 11:46 PM
    I would have written it as (n2 + n)/2

    Obviously the same thing but that's how I see it in my head. If I were to do it on paper I would have used

    An2 + Bn+ C, where A is half the 2nd difference and C is the 0th term.
  • Re: Math & logic problems :-)
     Reply #38 - August 16, 2010, 12:57 AM
    There is a simpler way of explaining this.

    (This is a popular story, but is unverified and probably untrue).
    When Gauss (a famous mathematician) was a young boy and went to school, his math teacher wanted to keep the children occupied for a while, so he set a task. He told them to find the sum of 1 + 2 + 3 + ... + 100: the sum of all numbers from 1 up to 100.

    Confident that the children would spend a long time, he was rather surprised when the young Gauss rapidly found the correct answer. He asked how he did it, and the reasoning was something like this.

    He first listed the short version of the total sum, which we call S.
    S = 1 + 2 + 3 + ... + 98 + 99 + 100

    Rewriting slightly (which you can do since addition is commutative and associative  grin12).
    Think of this as 'folding' the numbers. Note that there are 50 terms in both rows.

    S =     1 +   2 +  3 + ... + 48 + 49 + 50
       + 100 + 99 + 98 + ... + 53 + 52 + 51

    Now we have 50 smaller additions which all yield the same number! 101!

    S = 101 + 101 + 101 + ... + 101 + 101 + 101

    When you add 101 fifty times, it's the same as 50 times 101.

    S = 50*101 = 5050.

    And that, my dear friends, is the correct answer! great

    This is a nice little application from the formula that has been mentioned.

    S = n(n+1)/2

    Here n is 100. We multiplied 50*101, which is n(n+1)/2 = (n/2)*(n+1) = 50*101.

    Hope this wasn't too confusing.
    Bukhari 62:142 - Narrated Anas bin Malik:
       The Prophet used to pass by (have sexual relation with) all his wives in one night, and at that time he had nine wives.
  • Re: Math & logic problems :-)
     Reply #39 - August 16, 2010, 04:57 AM
    give us more problems Tongue
    [13:36] <Fimbles> anything above 7 inches
    [13:37] <Fimbles> is wacko
    [13:37] <Fimbles> see
    [13:37] <Fimbles> you think i'd enjoy anything above 7 inches up my arse?
  • Re: Math & logic problems :-)
     Reply #40 - August 16, 2010, 12:34 PM
    Off topic;
    @ high octane; just asking, is it you on the avatar?
    "I'm standing here like an asshole holding my Charles Dickens"

    "No theory,No ready made system,no book that has ever been written to save the world. i cleave to no system.."-Bakunin
  • Re: Math & logic problems :-)
     Reply #41 - August 16, 2010, 01:19 PM
    There are 23 people in a room. What is the probability of two people sharing a birthday (ignore leap years, and assume births are uniformly spread throughout the year.) Show your working kids.
  • Re: Math & logic problems :-)
     Reply #42 - August 16, 2010, 01:21 PM
    Last post today, honest.

    @TheLastKnight.  Afro That's a bog standard graduate interview question now for many jobs!

    @Prince Spinoza: interesting .

    @Naija Infidel: Yes

    New problem:

    You are on the edge of a cliff. It is 100m high. You have a rope 75m long. You have a knife that can only be used to cut the rope. There are hooks attached to the cliff at the top and at half way (50m down). A big hungry bear is 2 min away from eating you for a snack. How do you get down safely, without risking a 25m fall?

    Put that in your pipe and smoke it! Grin
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  • Re: Math & logic problems :-)
     Reply #43 - August 16, 2010, 01:35 PM
    Strangle the bear with the rope.
  • Re: Math & logic problems :-)
     Reply #44 - August 16, 2010, 01:43 PM
    Attach the rope at the top. Climb to the second hook. Attach the end of the rope to the second hook. Let it hang down. Climb back up 20 metres. Cut the rope. Hold on tight. You'll drop to a height of 5 metres. Jump down.
  • Re: Math & logic problems :-)
     Reply #45 - August 16, 2010, 01:50 PM
    Or rather. Attach the rope at the top. Clib down 50 meteres. Attach the rope at the point. Climb up 25 metres. Cut the rope. You'll fall to a height of 25 metres. Climb back up to the mid point. Attach the other end of the rope to the hook (removing the first) so you have a 50m rope to climb down.
  • Re: Math & logic problems :-)
     Reply #46 - August 16, 2010, 02:21 PM
    Make a loop at one end of the rope, a fairly loose one, loop it over the first hook, climb down, tie the rope to the second hoop(make sure to leave enough slack), give the rope a big shake to take off the looped rope from the top hook (by creating a wave form), it will fall down and there you'll have 50m rope to safely climb down.
    "Between stimulus and response there is a space. In that space is our power to choose our response. In our response lies our growth and our freedom." - Viktor E. Frankl

    'Life is just the extreme expression of complex chemistry' - Neil deGrasse Tyson
  • Re: Math & logic problems :-)
     Reply #47 - August 16, 2010, 02:45 PM
    Quote from: Prince Spinoza on August 16, 2010, 01:19 PM
    There are 23 people in a room. What is the probability of two people sharing a birthday (ignore leap years, and assume births are uniformly spread throughout the year.) Show your working kids.


    take the 365^23 space to start off.

    for each pair of people, there are

    365 * (364 * 363 * 362 ....... 21 terms) cases in which only those 2 have identical birthdays and the rest have distinct birthdays.

    no of such pairs of people = 23C2 = 23*22/2.

    probablility =

    (23*22 /2 )* (365 * 364 * ........ 344 )/ 365 ^ 23.

    = 0.36


    the general formula is also easy to derive and for > 365 people the probability is one by pigeonhole principle.
  • Re: Math & logic problems :-)
     Reply #48 - August 16, 2010, 03:31 PM
    That is the famous "birthday paradox".
    So I already know the solution is close to 50% ;P

    You just have to calculate the probability of 23 people NOT sharing a birthday, and then subtract that from 1.

    Let's take one person alone. He shares bday with nobody. The chance of him not sharing a bday is 1.
    Let's add a 2nd person. The chance of him not sharing bday with 1st person is 364/365.
    Let's add a 3rd person. The chance of him not sharing bday with 1st and 2nd person is 363/365.
    Let's add the nth person. The chance of him not sharing bday with all the previous is (365+1-n)/365.

    So, the chance of nobody sharing a bday is found by multiplying all those:
    (365*364*...*(365-21)*(365-22)) / (365^23)

    Which can be written using factorials:
    365! / (365^23 * (365-23)!)

    Then take that result, and subtract it from 1.

    Use a calculator cause I cannot be bothered. But it's close to 0.5 ;P
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #49 - August 16, 2010, 03:50 PM
    yep, tlaloc has the correct answer. i've actually calculated the chance of there being only pairs who share a birthday. my probability discounts triplets or more.
  • Re: Math & logic problems :-)
     Reply #50 - August 16, 2010, 04:04 PM
    Quote from: Tlaloc on August 16, 2010, 03:31 PM
    That is the famous "birthday paradox".
    So I already know the solution is close to 50% ;P

    You just have to calculate the probability of 23 people NOT sharing a birthday, and then subtract that from 1.

    Let's take one person alone. He shares bday with nobody. The chance of him not sharing a bday is 1.
    Let's add a 2nd person. The chance of him not sharing bday with 1st person is 364/365.
    Let's add a 3rd person. The chance of him not sharing bday with 1st and 2nd person is 363/365.
    Let's add the nth person. The chance of him not sharing bday with all the previous is (365+1-n)/365.

    So, the chance of nobody sharing a bday is found by multiplying all those:
    (365*364*...*(365-21)*(365-22)) / (365^23)

    Which can be written using factorials:
    365! / (365^23 * (365-23)!)

    Then take that result, and subtract it from 1.

    Use a calculator cause I cannot be bothered. But it's close to 0.5 ;P


    Oh well done. Interestingly, the probability of 2 people sharing a birthday being 99% occurs with only 57 people. Odd when you first look at it, but it makes sense when you think about it.
  • Re: Math & logic problems :-)
     Reply #51 - August 16, 2010, 04:22 PM
    dang, i erred.
  • Re: Math & logic problems :-)
     Reply #52 - August 16, 2010, 04:24 PM
    Ooohhh the smart thread is still going  grin12
  • Re: Math & logic problems :-)
     Reply #53 - August 16, 2010, 05:30 PM
    So here is a "proof" that 2 = 1. Its simple enough. Can anyone tell me why this is and if there is a mistake ?

    Let
    a = b

    Note "a*a" means "a squared"
     
                  a*a = a*b
         a*a - b*b =  a*b - b*b
    (a + b)(a - b) = b(a - b)      (a-b) is the factored out
                a + b = b                (but a = b)
                a + a = a
                    2a = 1a              factor out a
                      2 = 1

    So whats going on here Huh?
    Nothing can be more contrary to religion and the clergy than reason and common sense. - Voltaire
  • Re: Math & logic problems :-)
     Reply #54 - August 16, 2010, 05:36 PM
    Quote from: canex on August 16, 2010, 05:30 PM
    So here is a "proof" that 2 = 1. Its simple enough. Can anyone tell me why this is and if there is a mistake ?

    Let
    a = b

    Note "a*a" means "a squared"
     
                  a*a = a*b
         a*a - b*b =  a*b - b*b
    (a + b)(a - b) = b(a - b)      (a-b) is the factored out
                a + b = b                (but a = b)              
                 a + a = a
                    2a = 1a              factor out a
                      2 = 1

    So whats going on here Huh?


    The mistake is the step in bold. You've divided by (a-b) which equals 0. You can't divide by 0.
  • Re: Math & logic problems :-)
     Reply #55 - August 16, 2010, 05:40 PM
    here's a classic, I'm sure many of you have seen it before, so if you have, let someone who hasn't figure it out!

    1
    11
    21
    1211
    111221
    312211
    13112221
    1113213211

    whats the next line...Huh?
  • Re: Math & logic problems :-)
     Reply #56 - August 16, 2010, 06:00 PM
    Quote from: stardust on August 16, 2010, 02:21 PM
    Make a loop at one end of the rope, a fairly loose one, loop it over the first hook, climb down, tie the rope to the second hoop(make sure to leave enough slack), give the rope a big shake to take off the looped rope from the top hook (by creating a wave form), it will fall down and there you'll have 50m rope to safely climb down.


    Feasible! Very close!  Smiley But there is a slightely more reliable approach where the rope is sufficiently tied to the hooks.

    Nice tries Prince Spinoza, I'll be honest, I didn't get this. I don't know anyone who has to be honest.
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  • Re: Math & logic problems :-)
     Reply #57 - August 16, 2010, 06:03 PM
    Another famous one -

    There are 3 doors. Behind 1 is a thousand dollars. Behind the other two is an empty bucket. You have to select one door. Once you do this, one of the other two doors will be opened to reveal one of the empty buckets. You now have the chance to stick with your original choice, or to switch to the other unopened door.

    What should you do, and why.
  • Re: Math & logic problems :-)
     Reply #58 - August 16, 2010, 06:09 PM
    Quote from: HighOctane on August 16, 2010, 06:00 PM
    Feasible! Very close!  Smiley But there is a slightely more reliable approach where the rope is sufficiently tied to the hooks.


    A 50m waveform is feasible? Tongue

    Quote
    Nice tries Prince Spinoza, I'll be honest, I didn't get this. I don't know anyone who has to be honest.


    Cut the rope into 2 pieces. One 25m, one 50m. Tie the 25m to the top hook. Make a loop in the other end. Loop the other piece through this. Tie the other end(s). Now you have a 50m rope. Climb down. Cut the knot. Tie oone end to the hook.  Pull the other out of the loop. You have another 5om of rope to climb down to the ground.
  • Re: Math & logic problems :-)
     Reply #59 - August 16, 2010, 06:10 PM
    Quote from: canex on August 16, 2010, 05:30 PM

    So whats going on here Huh?


    Nice one canex, I've seen this before but it's nice to solve again.  Afro
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