Guest Report post Posted June 4, 2008 (edited) we can find it in atmost 3 weighings.(solution is already given by moderator) There is a formula to solve this kind of problems( i got this formula from combinatorics, and found suitable for these kind of problems) no. of uncertainties =x no. of outcomes for a weigh =y no. of weighings required =n then y^n >x ( the minimum value of n satisfying the equation is the answer) (this formula would be better for competitive exams) for this problem; no. of uncertainties = 2*12=24 ( each ball is heavier or lighter, we dont know) no. of possible outcomes of each weigh=3 ( each weigh says whether one half is heavy,light or equal to other) 3^n>24 ( 3 is the min. value of n satisfying the equation) therefore answer is 3. Edited June 4, 2008 by shobith Share this post Link to post Share on other sites

Guest Report post Posted September 7, 2008 To clarify: The solution of weighing 6 and 6, 3 and 3, and then finding the oddball doesn't work. When you get one group to be heavier and one group to be lighter, you still don't know whether the oddball is heavier or lighter-- if it's heavier, you want to start weighing the heavy group, if it's lighter, you want to start weighing the lighter 6. start with 5 balls on either side and go from there. 3 tries will do it. Share this post Link to post Share on other sites

Guest Report post Posted January 12, 2010 Could you do this same riddle with more balls, or is 12 the max? I think it is possible to determine the odd ball of 13 with still just 3 weighings. The problem would be determining if it was lighter or heavier... What would be the max number of balls you could have if you had 4 weighings? My first thought is more than 24 because if you weighed 6 v 6, worst case you would have 12 unknowns with 3 weighings left (same as this problem). Or you would have 6 possible heavys and 6 possible lights which is infomation you could use. So you could probably do 26 or something. Share this post Link to post Share on other sites

Guest Report post Posted March 24, 2010 Could you do this same riddle with more balls, or is 12 the max? Actually, it is possible to have up to 14 balls and determine which one is lighter/heavier in only three weighings. This makes the problem even more interesting; go ahead and try it Share this post Link to post Share on other sites

Guest Report post Posted March 24, 2010 What would be the max number of balls you could have if you had 4 weighings? My first thought is more than 24... The maximum number of balls for which you can solve the problem in 4 weighings is in fact 39. If you have an additional ball that you know is "normal", you can do up to 41 balls. My previous reply assumed such an "extra" ball. Without the extra ball, the maximum number of balls for which you can solve the problem in three weighings is actually 13, not 14. Share this post Link to post Share on other sites

Guest Report post Posted March 24, 2010 The maximum number of balls for which you can solve the problem in 4 weighings is in fact 39. Actually, you can do up to 40 balls in four weighings, not just 39. Share this post Link to post Share on other sites

Guest Report post Posted September 30, 2010 Didn't see it in the posts, i do miss things, but an alternative way to the solution for the situation where the 1234 = 5678 on the first weight, if you weighed 1,2 against 9,10 if equal then either 11 or 12 then weigh 11 against 1 if equal then 12 is odd and if unequal then 11 is odd but if 1,2 against 9,10 is unequal weigh 1 against 9 >> if equal 10 is odd if not 9 is odd, mark Share this post Link to post Share on other sites

Guest Report post Posted February 13, 2011 The least number of times is 4, according to me. First separate the balls into 3 groups of 4 balls each. 1st weighing - group 1 against group 2 2nd weighing - group 1 against group 3 By doing this, you know which two are equal. Consider the group which is heavier or lighter... Divide this group into two, X and Y. Also take two balls which u are certain are of normal weight. 3rd weighing - X against 2 normal balls If unequal, oddball is in X. If equal, oddball is in Y. Let's for now assume that the oddball is in Y. 4th weighing - one of the balls in Y against a normal ball. If unequal, that's your ball... if equal, the other ball in Y is the oddball Share this post Link to post Share on other sites

Guest Report post Posted May 16, 2011 I have a shorter solution. Put six balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of six. From these balls, put three balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of three. From the three balls, put a ball on each side of the scale. Case1: If the scales balance, the third ball is the lighter ball. Case2: If one is lighter, then that is the ball... tambay, i think you were finding out how to tell if that specific one is lighter or heavier, and even then, you'd have to have more steps to do that. the next step from case 1/2 is to take the other ball and see if it balances with one of the balls from case 1.if it does, then you have to try the other set of 3 to find out if that specific one is lighter or heaver by repeating the step that lead to case1/2. you would do the same for case 2, and if the two balls balanced (leaving one on the scale and replacing the other), then you know that the one you replaced is heavier my "case 3" would tell you if its lighter (take the step that lead to case 1/2 and replace one of the 2 balls, and if they balance, then the lighter one is the one you replaced, and to clarrify, NOT the one you replaced it with) Share this post Link to post Share on other sites

Guest Report post Posted June 7, 2011 I have a shorter solution. Put six balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of six. From these balls, put three balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of three. From the three balls, put a ball on each side of the scale. Case1: If the scales balance, the third ball is the lighter ball. Case2: If one is lighter, then that is the ball... Knowing what the site admin said, you could just rephrase this as I have a shorter solution. Put six balls on each side of the scale. Whichever is lighter or heavier, the lighter/heavier ball is in that group of six. From these balls, put three balls on each side of the scale. Whichever is lighter/heavier, the lighter/heavier ball is in that group of three. From the three balls, put a ball on each side of the scale. Case1: If the scales balance, the third ball is the lighter ball. Case2: If one is lighter, then that is the ball... Share this post Link to post Share on other sites

Guest Report post Posted June 8, 2011 I have a shorter solution. Put six balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of six. From these balls, put three balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of three. From the three balls, put a ball on each side of the scale. Case1: If the scales balance, the third ball is the lighter ball. Case2: If one is lighter, then that is the ball... The ball could possibly be heavier than the others so you can't determine it that way. Share this post Link to post Share on other sites