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Mr.Dog

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Posts posted by Mr.Dog

  1. Bro, it doesn't work like this. Let me teach and educate you why:

    Ok, so you say that we split the coins into 3 groups: 5, 5 and 2

    Then you take the one of the group of 5 coins and:

    1. 2 - 2...  and you put 1 coin aside. 

    if they are balanced the cf coin will be in other 8 coins

    2. You do the same with the second group of 5 coins - if they are balanced the cf coin will be in the remaining 4 coins( 2 you put aside + third group of 2 coins)

    3. now how do you find the cf coin in these 4 coins?

    Tell me, homeboy.

     

  2. On 3/26/2016 at 1:48 PM, bonanova said:

    Regarding the linked article

      Reveal hidden contents

    It contains the following paragraph

    Mass Vs. Weight

    The University of California Department of Physics and Astronomy explains that weight is the force of the gravity puts on the mass of an object. Weight measurements calculate how much gravity it takes to hold an object to the earth. Mass, on the other hand, is about solidity. Mass is a quantifiable measurement of the amount of substance it takes to make an object solid. Balance scales measure mass. An electronic scale measures weight.

    The part in red is pure rubbish. The author clearly listened to an expert and then printed her own ideas.

    The difference between (electronic, or other) scales and a balance is that the latter compares (same, heavier, lighter,) while the former measures (235.9 pounds.) That's all.  Both instruments require the objects involved to exert a force. That force (in a gravitational field) is their weight. Weight goes away outside of gravity, but mass does not. An astronaut could place her feet on an electronic scale while she is in en route to Mars, and the scale will register a reading of zero. That indeed is her weight outside a gravitational field. The electronic scale did its job. It measured her weight.

    She could also get onto one side of a teeter-totter (with no one sitting on the other side) and it would balance! If you believe that a balance measures mass, you must conclude that she has zero mass. But that is incorrect. Since a balance compares forces, you can conclude, correctly, that masses are weightless in space. On earth, of course, where she is not weightless, she would fall to the floor.

    The author's definition of mass (red underlined) is shameful. Gasses have mass, as do liquids, not just solids. Mass is simply an object's resistance to acceleration. That's what F=ma means. Mass (m) is the amount of force (F) required to provide one unit of acceleration (a) to an object. Consider the relative masses of a toy, a car and a freight train: You can flick a toy car with your finger and give it speed of perhaps 30 mph almost instantaneously. In a few seconds you could get a car to a speed of a few mph, by putting your shoulder to it and pushing with your legs. Push on a freight train, however, and you'd have to bring a lunch, perhaps several of them, and eat them before you could get it to observably budge, even if the wheels were frictionless. Applying a force and measuring acceleration is how you measure mass directly. You'd get the same results in space as you do on earth. You can also infer mass from weight in a gravitational field by measuring the force it takes to keep if from falling (its weight) and divide the weight by g, the acceleration due to gravity.

     

    Regarding the new solution,

      Hide contents
    1. What is the Devil group? It's one of the groups of 6, but which?
    2. How does 6-6 provide useful information?
    3. Why is x positive?
    Spoiler

    1.- 2. I figured out that doesn't provide any progress, since the coin could be either lighter or heavier and the scale could go either way (up or down)

    3. cause x is not a downer - but he is now terminated for the uprising foundation.

    Answer attempt 3 - the foundation of hope:

    Spoiler

    Ok Ok

    lets split the 12 coins into groups of 3 coins making 4 groups in total, so:

    1. 3-3

    if they balance the cf coin would be in the 2 other groups of 3 coins each. 

    if they don't balance the cf coin is in within these 6 coins, making the other 6 coins normal.

    so with 1 weigh we get rid of 6 suspicious coins.

     

    2. 2-2

    Now, as we know that the cf coin will be within 6 coins. We split the them into groups of 2 coins each making 3 groups.

    now we weigh 2 groups against each other - If they don't balance then the coin within those 4 coins - and you're screwed, well for now.

    but lets look away from that - yes far away - and think optimistic and lets say they if they balance, the coin will be in the 2 coins not weighted. Then we proceed victoriously to the next legendary step.

     

    3. 1-1

    pick 1 normal coin from the other normal 6 coins and compare it with 1 of the 2 coins - if they balance the coin is the other 1 of the 2 coins, if they don't balance - well then its obvious.

     

    This is the foundation - i am gonna rework this when i get into physics and probability next year.

     

     

    On 3/26/2016 at 1:48 PM, bonanova said:

     

     

     

     

  3. OK, i confused. But i put the blame on this article :D :

    http://www.livestrong.com/article/343183-the-differences-between-balance-beam-scales-electronic-scales/

    ...and not on me... at all.

    and previous solution was just a typo:

    Here's the new one:

    Spoiler

    x=mrdogrocks=any value greater than 0

    1. 6-6

    so you take the devil group and:

    2. 4-2

    Proportions would either be 3.x : 2,  4.x : 2,  4 : 1.x, or  4 : 2.x,

     

    if the cf coin is the group of 2 coins:

    3. 1-1

    Proportions would be: 1.x : 1 = heavier coin or 0.x : 1 = lighter coin

     

    if the cf coin is in the group of 4 coins

    3. 2-2

    Proportions would be either 2.x : 2 = Rock solid Malphite heavier coin or 1.x : 2 = 420 blaze lighter coin

     Dont mess with a dog, or you'll get a bark! bark!

  4. Alright, think i got it after fiddling with the unique number values for the letters.

    Spoiler

    Sol. abedcg = 5

    ( a=6 , b = -2 , c = 3 , d = -8 , e = 2, f = -1 , g = 4),

    steps: 

    1. fedcg=abcdef            2. (fg)(bc)= (3)+(1)=4       3. (ab)cdef=0 ---> (g)cdef=0 ----> (gf)cde=0---> 3cde=0          

        g=ab

     

    4. Trial and error with abged = 2 , abgcd = 3 and 3cde = 0, easier when you know that g=ab. 

    abcdef = 0
    bc = 1
    abged = 2
    abgcd = 3
    fgbc = 4
    abedcg = 5
    afedcg = 6
    abc = 7

     

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