Friday Riddle

nudgie
nudgie Member Posts: 1,478 Member
edited March 2014 in Colorectal Cancer #1
Remember NO GOOGLING or searching the INTERNET allowed, and GOOD LUCK!!!!!!!!!!!

RIDDLE: You die and the devil says he'll let you go to heaven if you beat him in a game. The devil sits you down at a perfectly round table. He gives himself and you an infinite pile of quarters. He says "OK, we'll take turns putting one quarter down, no overlapping allowed, and the quarters must rest flat on the table surface. The first guy who can't put a quarter down loses." You guys are about to start playing, and the devil says that he'll go first. However, at this point you immediately interject, and ask if you can go first instead. You make this interjection because you are very smart and can place quarters perfectly, and you know that if you go first, you can guarantee victory. Explain how you can guarantee victory

Comments

  • robinvan
    robinvan Member Posts: 1,012
    #2
    Canadian Quarters?
    I realize at once that that 6 quarters will fit in a ring around the outside of the 1st quarter* laid and that each additional ring around will therefore approximate a hexagon with one additional quarter added to each side as it grows. Thus 1st ring - 6 quarters; 2nd ring - 12 quarters; 3rd ring - 18 quarters; 4th ring - 24 quarters; etc. etc. etc. Because of the single quarter in the middle the number of quarters to complete a ring will always be an odd number. ie 1,7,13,19,25,31... If I place the 1st quarter I will always be placing an odd number quarter.
    I will always complete the circle. I win!

    *I know this because the circumference of a circle is larger than its radius by a factor of 6.28 (C=2piR or C=piD) the 1st ring of quarters outside the single quarter can thus fit 6 "diameters" of quarters. Each successively larger ring will increase by a factor of 2pi or 6.28. But since we're dealing with whole quarters... approximately 6.

    With Canadian quarters, of course!

    I know this is pretty convoluted and am sure there is a more elegant way of saying it! Are you a Math teacher??

    Rob; in Vancouver
  • Patteee
    Patteee Member Posts: 945
    #3
    robinvan said:

    Canadian Quarters?
    I realize at once that that 6 quarters will fit in a ring around the outside of the 1st quarter* laid and that each additional ring around will therefore approximate a hexagon with one additional quarter added to each side as it grows. Thus 1st ring - 6 quarters; 2nd ring - 12 quarters; 3rd ring - 18 quarters; 4th ring - 24 quarters; etc. etc. etc. Because of the single quarter in the middle the number of quarters to complete a ring will always be an odd number. ie 1,7,13,19,25,31... If I place the 1st quarter I will always be placing an odd number quarter.
    I will always complete the circle. I win!

    *I know this because the circumference of a circle is larger than its radius by a factor of 6.28 (C=2piR or C=piD) the 1st ring of quarters outside the single quarter can thus fit 6 "diameters" of quarters. Each successively larger ring will increase by a factor of 2pi or 6.28. But since we're dealing with whole quarters... approximately 6.

    With Canadian quarters, of course!

    I know this is pretty convoluted and am sure there is a more elegant way of saying it! Are you a Math teacher??

    Rob; in Vancouver

    no idea if you are correct
    no idea if you are correct Rob- but I vote that YOU WIN!!
  • PGLGreg
    PGLGreg Member Posts: 731
    #4
    robinvan said:

    Canadian Quarters?
    I realize at once that that 6 quarters will fit in a ring around the outside of the 1st quarter* laid and that each additional ring around will therefore approximate a hexagon with one additional quarter added to each side as it grows. Thus 1st ring - 6 quarters; 2nd ring - 12 quarters; 3rd ring - 18 quarters; 4th ring - 24 quarters; etc. etc. etc. Because of the single quarter in the middle the number of quarters to complete a ring will always be an odd number. ie 1,7,13,19,25,31... If I place the 1st quarter I will always be placing an odd number quarter.
    I will always complete the circle. I win!

    *I know this because the circumference of a circle is larger than its radius by a factor of 6.28 (C=2piR or C=piD) the 1st ring of quarters outside the single quarter can thus fit 6 "diameters" of quarters. Each successively larger ring will increase by a factor of 2pi or 6.28. But since we're dealing with whole quarters... approximately 6.

    With Canadian quarters, of course!

    I know this is pretty convoluted and am sure there is a more elegant way of saying it! Are you a Math teacher??

    Rob; in Vancouver

    Quarters packed?
    I guess you're placing your quarters as close together as you can, but what if the Devil does not cooperate, and leaves a little extra space around the quarters he puts down? Then, a ring might have an odd number of quarters.

    --Greg
  • PGLGreg
    PGLGreg Member Posts: 731
    #5
    Symmetry.
    As Rob says, on your first turn, you put a quarter in the center. If the table is small enough, the game might be over, but if not, wherever the Devil places his quarter, on your next turn, you put a quarter opposite to where the Devil put his. That is, imagining a line drawn through the center of the table and the quarter the Devil just put down, you place your quarter on this line, and just as far from the center quarter as his quarter. Henceforth, after you take your turn, the pattern of quarters will always be symmetrical about the center quarter, because of the strategy you're following. If, after your turn, there is no room for the Devil to add a quarter, of course, you've won. If there is room for him to add a quarter, because of the symmetry of the pattern, there will always be room for you to add a quarter on the opposite side. So, eventually, you will place the last quarter that there is room for.

    --Greg
  • robinvan
    robinvan Member Posts: 1,012
    #6
    PGLGreg said:

    Quarters packed?
    I guess you're placing your quarters as close together as you can, but what if the Devil does not cooperate, and leaves a little extra space around the quarters he puts down? Then, a ring might have an odd number of quarters.

    --Greg

    Yikes...
    I guess I'm looking at a long time in a hot place!!

    There's got to be an easier answer to this!

    Rob; in Vancouver
  • robinvan
    robinvan Member Posts: 1,012
    #7
    PGLGreg said:

    Symmetry.
    As Rob says, on your first turn, you put a quarter in the center. If the table is small enough, the game might be over, but if not, wherever the Devil places his quarter, on your next turn, you put a quarter opposite to where the Devil put his. That is, imagining a line drawn through the center of the table and the quarter the Devil just put down, you place your quarter on this line, and just as far from the center quarter as his quarter. Henceforth, after you take your turn, the pattern of quarters will always be symmetrical about the center quarter, because of the strategy you're following. If, after your turn, there is no room for the Devil to add a quarter, of course, you've won. If there is room for him to add a quarter, because of the symmetry of the pattern, there will always be room for you to add a quarter on the opposite side. So, eventually, you will place the last quarter that there is room for.

    --Greg

    Wow!
    What a cool approach. I like the way you think. Sounds like this works!

    Rob; in Van

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