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A Math Question for you Racers

Discussion in 'Advanced Tech Section' started by Boost Engineer, Mar 28, 2009.

  1. Boost Engineer

    Joined:
    May 19, 2004
    Things have been kinda slow around here the last couple of days.

    I posted this topic on another website, can you figure it out?

    Most racers have been around gram scales used to measure pistons, bearings, connecting rods, etc for weight.

    So here is your chance to see if you WOULD HAVE won that Million Dollar prize or if you were actually looking at the Cute Teacher's legs during class.

    What we know:

    We have a Gram Scale that can accurately read 1 gram of weight change.

    We have 12 stacks of Silver Dollars.

    There are 12 Silver Dollars in each stack.

    Each stack is given a number.

    A Counterfeit coin weighs 1 gram more than a legal coin.

    One Stack is Counterfeit coins.

    You would get ONE PICK to win the Million Dollars, if you guessed wrong you would give up your race car.

    Which stack has the counterfeit coins? How would you find out the answer? Post your ideas. (You cannot get 11 of your buddies together, each one picks a number, and then you share the money!)

    Have Fun!

    Tom Vaught
     
  2. Chevalade

    Joined:
    Sep 27, 2007
    Just start adding stacks of coins to the scale and when it jumps by 24 grams instead of 12, pick that stack.
     
  3. Boost Engineer

    Joined:
    May 19, 2004
    OK, another clue:

    You can move one coin, you can move a stack of coins or you can move any or all of the coins BUT the gram scale will be off when you do this. When you are ready with your "guess" the gram scale will be turned on and you will be required to make your choice. You get only one guess.

    Good Luck.

    Tom Vaught
     
  4. BottleFed70

    Joined:
    Sep 7, 2006
    Remove the number of coins from each stack that corresponds to the stack number/label -1 . (Stack 1 remove 0 coin, stack 2 remove 1 coins, stack 3 remove 2 coins, etc)
    Weigh all of the stacks (minus coins removed).
    Take the total weight and divide by 12. Keep note of the remainder.


    If the remainder is 7, the stack that has 7 coins remaining is the cointerfit stack.
    If the remainder is 2, the stack that has 2 coins remaining is the counterfit stack.. etc.
     
  5. Chevalade

    Joined:
    Sep 27, 2007


    Shoulda stated that in the first post.....lol.



    :agree:
     
  6. BottleFed70

    Joined:
    Sep 7, 2006
    I had to edit for the math to work right, but I think that's the correct way to approach this puzzle.
     
  7. Chevalade

    Joined:
    Sep 27, 2007
    That is the only way given the set parameters. If the scale was left on the whole time, and you could only add to the scale, then my first way would have worked. Needed the part about the scale on/off. :bang:



    The other option would be to physically pick up all the stacks till you find the heaviest one.....
     
  8. Chevalade

    Joined:
    Sep 27, 2007


    Found the flaw to this.....you have the right idea, but you need to weigh the coins you pulled off.....not the left over stacks. Subtract total number of coins from weight on scale......the number left over is your stack.
     
  9. BottleFed70

    Joined:
    Sep 7, 2006
    Wouldn't it work either way?
     
  10. Chevalade

    Joined:
    Sep 27, 2007
    re edited it....now it works.
     
  11. Chevalade

    Joined:
    Sep 27, 2007
    Nope, because if you take 12 + 11 + 10 + 9 + 8 + 7 etc.....and add 2 for say stack 2 you get 80 total.....divided by 12 gives 6.666666

    Now if you do the same thing.....12 + 11 ....etc........and end up with 80 grams, you subtract total coins (78) and that leaves you with 2 grams/stack 2
     
  12. Boost Engineer

    Joined:
    May 19, 2004
    You would only weigh one stack of coins, but it is comprised of coins from each stack and does not number 12 coins.

    Each stack is labeled 0, 1, 2, 3 and so on. You take 0 coins from stack 0, 1 coin from stack 1, 2 coins from stack 2, etc. When you are finished creating your stack, you'll have a total number that should equal 66 grams (assuming 1 real coin weighs one gram and a counterfeit coin weighs two grams).

    If the scale reads higher than 66 grams, subtract that number of grams and you have named the stack that is counterfeit (3 grams more = stack 3)

    Tom Vaught
     
  13. gotboost89gt

    Joined:
    Feb 9, 2008
    have any more ??
     
  14. Boost Engineer

    Joined:
    May 19, 2004
    I will see what else is in "the book" lol

    Tom Vaught
     
  15. gotboost89gt

    Joined:
    Feb 9, 2008
  16. Boost Engineer

    Joined:
    May 19, 2004
    Another question for sharp thinkers:

    gotboost89gt, chevalade, & bottlefed70 each attend different schools: A, B, &C.

    They each have a test tomorrow


    gotboost89gt is not in school A. chevalade is not in school B. The student at school A does not have a test in mathematics. The one at school B has a test in English.
    Chevalade does not have a test in Science.

    In what subject is Bottlefed70 having a test in and at what school?
     
  17. whiteghost

    Joined:
    May 31, 2007
    chevalade is at school C test in math
    gotboost school B english
    bottlefed school A science
     
  18. 302tt

    Joined:
    Aug 4, 2004
    I agree. Turns out they all failed though.
     
  19. mustangcobra438

    Joined:
    Mar 18, 2008
    Bottlefed70=class A/Science test
    Gb=Class B/English test
    Chev=Class C/Math test
     
  20. whiteghost

    Joined:
    May 31, 2007
    here's an old one for you guys that takes a bit of creative thinking. Please don't reply if you've heard it before.

    A cunning innocent man is imprisoned in by a sadistic king for a crime he did not commit. On the day of his planned execution he is brought before the king and given a chance to save his life by playing a game. The game is this- The king has two pieces of folded paper, written on one, the king says is "death". If the prisoner chooses this one he will be put to death within the hour. Written on the other, the king says, is "life". If the prisoner chooses this on he will live and be set free. The prisoner knows that 50-50 odds are far too generous for this King to be offering. He knows that both papers say "death" and he cannot win the king's game playing by his rules, yet the prisoner calmly chooses a piece of paper and moments later is set free. How did he do it?
     
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