You have 12 balls. Eleven weigh exactly the same. One weighs slightly more or less than the other 11. You have a double-beam balance scale that you may use 3 times. Identify the ball that is different, and tell me if it weighs more or less than the other 11.
For the heaviest: Place six balls on each side of the scale. From the side that is heavier, split the six into two groups of three, on on each side of the scale. From the side that is heavier take all three and place any two on opposite sides of the scale. If they balance, the heaviest is in your hand. If they do not, the answer is obvious! Please Register or Log in to view the hidden image! For the lightest, replace "heavier" with "lighter", and "heaviest" with "lightest" in the above. :m: Peace.
Goofyfish is right almost. But after the first weigh we still would not know whether the odd ball is heavier or lighter. If it is lighter then the heavier group does not show any difference on the second weighing. we will be left out with only one more weighing. http://www.math.umd.edu/undergraduate/pme/puzzles/Internet10_ans.html
1 - weigh two sets of four - if any difference then take the heavier set. if not then take the set you did not weigh. 2 - split the set you grabbed into two sets of two. weigh them and determine if there is a difference. there will be. 3 - take the heavier side of the two sets of two and split them. weigh each ball individually. result - you will have determined which ball is heavier.
no... I think dribbler has it. I count three times. And if the four he weighs originally on each side are equal in weight, then he has the two groups of two which only require two more weighings... I think dribbler has it.
No, count again jokerz...tiz three. you do not need to weigh one becuase you know if it is heavier if the first weighing is even. Peace Goofy, but it looks like Nader is right, as much as you do not want me to be, i am correct.
it is 4 joker. you're taking only the heavier group for weighing. what if the odd ball is lighter than the others..? you 3rd weighing will show both the balls balance and leave you blinking. you need 4 tries to find the odd ball then. the link given above gives a good solution.
jagass! review my method again and you will note that there are only three actual weighings. my method incorporates the same formula that is noted on the reference site.
I hate puzzles, I would just weigh all 12 till I found the one that wasn't the same and smash it with a rock so it would never bother me again.
if the odd ball happens to be lighter than other balls then your method, ofcourse wieghs only three, and does find your own ball is odd. it does not work. as i said your method works only if the odd ball is heavier than other balls. the solution from the site is complete and is significantly different. your method is basically Goofyfish's method and the difference is - he knows its limitations whereas you don't know wtf you are talking about.
I say take all of the balls, and begin throwing them at the instructor. Threaten with more pummeling unless the answer is relevealed.
