we khnow we can just to weight the coins 3 time and we donot khow that this coin has lowest weight or the highest
122166
This problem was fun and took me a few minutes to realise the solution.  First I will restate the problem just for clarification.\n\nWe have 12 coins, one of which is fake (lighter or heavier).  We can only make 3 weight comparisons to discover which of the coins is the fake one.  My solution will tell you which coin is fake and if it is heavy or light in exactly 3 comparisons.\n\nI number the coins 1 - 12 and make the following 3 comparisons on a scale\n\nleft    vs   right\n-------------------\n1 2 3 10  vs  4 5 6 11\n1 2 3 11  vs  7 8 9 10\n1 4 7 10  vs  2 5 8 12\n\nnow I know which coin is false.  How do I do this? well I created a table which has 12 unique outcomes of 3 weighing comparisons on which side of the scale went down. for example left left left indicates that all 3 comparisons had the left side dropped.  Here is my table:\n\nfake coin  | correspondin weight comparisons\n--------------------------------------------\n1          |  left left left (heavy) or right right right (light)\n2          |  left left right (heavy) or right right left (light)\n3          |  left left balanced (heavy) or right right balanced (light)\n4          |  right balanced left (heavy) or left balanced right (light)\n5          |  right balanced right (heavy) or left balanced left (light)\n6          |  right balanced balanced (heavy) or left balanced balanced (light)\n7          |  balanced right left (heavy) or balanced left right (light)\n8          |  balanced right right (heavy) or balanced left left (light)\n9          |  balanced right balanced (heavy) or balanced left balanced (light)\n10         |  left right left (heavy) or right left right (light)\n11         |  right left balanced (heavy) or left right balanced (light)\n12         |  balanced balanced right (heavy) or balanced balanced left (light)\n\nnow I can find exactly which coin was the fake one by comparing the results of my 3 wieghs against my table. For example if I get two balances and then a right, I know the 12 coin is the fake and is heavy. Note that if I get balanced balanced left the the 12 coin is light.\n\nNotice that I first created this table of possible outcomes and then generated from it which coins would be compared against which. \n\nHeavy or Light:  I created this table in which no result on any row is the reverse of any result on any other row (as the reverse indicates whether it is light or heavy)\n\nThis is the simplest solution to this problem.  I hope this made sense.  Was really fun.
20761
well its not the most important but its the first imprestion we get but its not the most important its 3rd mabe 4th
379846
theoretically and very rare, yes i'd say it is possible. but due to gravity, if it does land on its edge after throwing, it will fall again and land on the sides. gravity pulls the object mainly from the center, which is in the middle of the curcumference of the coin which is located on either sides but not on the edges. so even if it did land on its edge, which is a very rare possibility, 1/10000000, gravity would pull on ur coin causing it to fall down on its side.
178794
IF we know the coin is heavy and we have 5 scale usage , we can find out the heavy coin amist 243 coins.\n\nFor the first question, with 3 scale usage and when u dont know is the coin is heavier or liter\nNumber the coins A-L. \n\nBegin by weighing four against four. If they balance, then weigh any of the remaining three against any three of the good coins. \n\nIf they balance then the odd one is the remaining coin and can be \nidentified (whether it's lighter or heavier) in the final weighing. \n\nIf three against three do not balance, take the three containing the odd \ncoin and weigh any one against another. \n\nThus: \n\nIf the first weighing of four against four does not produce a balance, then \nthe second weighing involves three against three with the coins switched \nbetween the two pans and a good coin introduced. \n\nSo: \n\nIf A+B+C+D > E+F+G+H \n\nWe try A+B+E against C+F+J \n\nIf A+B+E + C+F+J then either D is heavier or G or H is lighter, so weigh \nG against H. \n\nIf A+B+E > C+F+J then either F is lighter or A or B is heavier, so we \nweigh A against B. \n\nIf A+B+E < C+F+J then either E is lighter or C is heavier, so weigh \neither against a good coin, eg. K against E. \n\nAn alternative second weighing is A+B+E against C+D+F, which follows \nsimilar lines to the above.
645642
it jibbs chain hang low
413516
The 1st time was ok, the 2nd time was ok, the 3rd 4th and 5th were a little much.
485693
I think it seems that way because of the saying.  When 1 and then the second thing happen all in a short period of time we notice and when we get to the 3rd thing we especially take notice.  Sometimes a 4th or even 5th thing happens but we stop counting OR wait for the 6th - 3 and 3 again?\nPersonally, 3 is my lucky number so when 3 seemingly bad things happen, I look for the silver lining.
134910
Answer: 1,3,9 and 27 kg\n\nWay of solving:\n\n1) 4 pieces are very small number to weigh such number (40 of different weights. So I just tried to understand a most efficient way of weighing different weights using a small number of pieces.\n\n2) Firstly, I understand the most efficient set of pieces when we put some thing we are weighing on one scale and pieces on another. There are 15 (2^4-1) ways of taking different non-empty subset from set of 4 pieces. So, in most effective set all those subsets must have different weights. Eady to see, that pieces which are powers of 2: 1,2,4,8 kg make such a set (15 different weighings from 1 to 15 kg). \n\n3) Next I guessed that it is possible to put pieces not only on another but also at the same scale as we put some thing we are weighing. So, the weight of little piece can be not only added to the big piece, but also subtract from it, giving more possible combinations of weights.\nThen I guessed that some subset1 without smallest piece (ss1), ss1 + smallest piece (ss1+), another subset without smallest piece (ss2), ss2 with smallest piece on another scale (ss2-) must be consecutive numbers. So, ss1 = ss2 + 3. So, if smallest piece is x kg, the next piece must be 3 * 1st=3x kg.\nThe same situation is with other pieces: 3rd = 3 * 2nd = 9x kg, 4th = 3* 3rd = 27x kg.\nThis set of pieces makes possible 40 different weights.\nAlso, we know that sum of weights of those pieces is 40 kg.\nSo, x + 3x + 9x + 27x = 40kg\nx = 1kg\n\nAnd the final answer is: 1 kg, 3 kg, 9 kg and 27 kg.
293159
they quit making it, but it is still considered legal tender.  too many folks were confusing it for quarters, and there were issues with vending machines
536822
Rarely does a team with 3+ turnovers win, unless teh other team turns it over as often.
29930
This is a classic scale puzzle.  The best solution I've seen so far is here:\nhttp://www.mazes.com/scales/Scale301.html\nYou just need to substitute in twelve identical weights for twelve identical coins in the solution.
395462
Another coin in the slot.
201073
Nope.  We don't want people who ultimately are in control of the game to be able to gamble on the outcome.
564870
there are different reasons. 3 is a very powerful number in a lot of cultures. also, in fairy tales they like to isolate people, things, etc. having thins come in 3s allow them to create a patten, and then have the third person or thing break this pattern.
605228
They can only count to 4
216295
we need more information.  We don't need your average--we need to know how many total points you have and how many possible points there are, so we know how much weight to give the 100 points.
35254
listen the origine of this is difficult but if want to know contact me\n\nI can tell how you can get them:\nAny year that can satisfy 2 of this 3 conditions (1&2) or\n(1&3) and fail in the third\n1) the year must be divisible by 4\n2) the year must NOT be divisible by 100\n3) the year must be divisible by 400\n\nExample: 1996 this satisfy condtion 1 and 2 >> 1996 is leap\nExample: 2000 this satisfy condtion 1 and 3 >> 2000 is leap\nExample: 1900 this satisfy condtion 1 only >> 1900 is not leap [[[1900 is divisible by 4 and  it is not leap because it is divisible by 100 and not by ]]]\n\nWhy? you can contact me rarebonheur@yahoo.com