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12 Identical Balls
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Can you answer this riddle? You have 12 steel balls all exactly the same except for one. One of the balls is of ABNORMAL WEIGHT(not sure if it's heavier OR lighter). To determine... show more Update: I already solved it about 6 years ago, I am just curious to see if anyone else has a different answer. The answer would be really detailed, and is basically a proof. Update 2: GREENHORN---thats exactly what I was looking for!! When I solved it I quickly realized 2 things. 1. the second measurement is the most important and 2. the... show more Update 3: * 3 unknown balls I mean show more Update 4: Step 1 First weigh 4 balls on each side.. 1.A If the scale is even then your abnormal ball is in the 4 remaining balls Take 2 of those balls and measure... show more Update 5: one side and 2”C” balls,1”L”, and 1”H” on the other….The remaining balls will be 2”L” balls,1”H”, and 1”C” HHCL--------------CCLH ... show more 4 answers · Words & Wordplay
Best Answer Yes, this problem can be solved. I had found a solution for this problem soon after it was published many years ago, in the Journal of the American Mathematical Association, if I remember correctly. It requires some detailed explanation. Anyone interested in knowing the steps, please send me an e mail via the YA e mail system, when I would give the solution via the YA e mail system. If you have not registered your e mail with YA in your profile, then I would not be able to send you the solution. I pose this problem for people to think about, when I am in a group of strangers interested in jokes/chat/puzzles, and my experience is that very few persons come back with the right solution. It is not an easy problem, but not too difficult either.
EDIT: On second thoughts, I find that the solution would not fit in the limited space available in the YA e mail box. So I have posted the solution below. I request interested members to go through the solution and point out any errors they find, in the form of an e mail to me through the YA e mail system. Thanks.
Solution for the 12 Balls Problem 1. Place any 4 balls in the left pan, and any 4 balls in the right pan. Mark the remaining balls A, B, C and D lightly so that the markings don’t add to the weight. 2. Case 1: If the scale shows no variation, then the marked balls contain the defective ball. Now, retain only 3 balls in the left pan and remove the other 5 balls from both pans. Take any 3 of the marked balls and place them in the right pan. (a) If the scale is still even, this means the 4th marked ball is defective. To find out if this is heavier or lighter than the remaining 11 balls, weigh the 4th marked ball with any one of the remaining 11 equal balls. (b) If the right pan tilts down during the second weighing, this means that those 3 balls (say A,B,D) contain the defective ball which also would be heavier than the other 11 balls. To isolate the heavier ball, clear the scales of all balls, take any two of the balls A, B and D, place say A in the left pan and B in the right pan. Whichever pan tilts down contains the heavier ball. If however the scale is even during this third weighing, then ball D is the heavier ball. (c) During the second weighing, if the left pan tilts down, then it means that one of the balls A, B and D in the right pan is the defective ball which is also lighter than the other 11 balls. Follow the same procedure in 2(b) to isolate the lighter ball. 3. Case 2: If the scale during the first weighing shows say the right pan tilting down, this means one of these 8 balls is defective and balls A,B, C and D are of equal weight. Now lightly mark the balls in the left pan P, Q, R and S and those in the right pan T,U, V and W. Leave ball P in the left pan and ball T in the right pan, retain the balls U,V and W separately. Transfer balls Q, R and S from the left pan to the right pan (so that the right pan would have balls T, Q, R and S). Place balls A, B and C (which are of equal weight) in the left pan so that the left pan would now have P, A, B and C. So now we get the second weighing P, A, B and C in the left pan and T, Q, R and S in the right pan. (a): If the scale is now even, then it means that one of the balls removed from the right pan, viz. one of U,V and W is not only defective, but also is heavier than the other 11, since during the first weighing the right pan tilted down. The heavier among U,V and W can then be isolated in the 3rd weighing by weighing one of them against another of them in the manner mentioned in step 2 above. (b): If the right pan still tilts down, it means either P is lighter than the rest or T is heavier then the rest. Determination of which between P (lighter) and T (heavier)- only one of these is possible- is done by weighing either P or T against any of the other 10 balls which are of equal weight. (c): If now the left pan tilts down, it would mean that one of three balls Q, R or S transferred from the left pan to the right pan, is the lighter one. The lighter among Q, R and S can then be isolated in the third weighing by weighing two of them separately against each other, one in each pancontext