Crosswords1 min ago
a riddle
4 Answers
There are twelve balls, marbles whatever, balls and they all look alike except that one is EITHER lighter OR heavier and you have a scale and you can only use it three times, and so how would you find what ball it is and whether the ball is heavier or lighter? If you can answer me I would love it, I can't figure it out, he says to get the answer you have to switch your paradigm, or your normal way of thinking.
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The riddle should specify that one ball is heavier or lighter than the rest. It doesn't matter which BUT the riddle has to tell you and the key is that the scales are balance scales. Let us for this example say one ball is heavier. You weigh six balls on one side of the (balance) scales and six on the other ... discard the lightest set. Next weigh three on one side and three on the other .. again discard the lightest set. Last weigh you put one ball on one side of the scales and one on the other : if they weigh the same then the third ball is the heavy one .... if not then its the heaviest on the scales.
The question doesn't need to tell you whether the odd ball is lighter or heavier, just that it is not the same weight.
With twelve balls (knowing that one & only one is 'fake') you have 24 possibilities, ball 1 is too heavy, ball 2 is too heavy . . . etc. or ball 1 is too light, ball 2 is too light . . . etc.
Each of 12 balls can be either heavy or light (but only one is fake) so 2*12 =24 possibilities
Using a twin pan balance, for each weighing you can get one of three results, right pan is heavy, left pan Is heavy, or both balance. So with 3 weighings you can get 3*3*3 =27 possible outcomes.
27 outcomes is more than enough to distinguish between 24 possibilities.
The rest is left for the reader!
With twelve balls (knowing that one & only one is 'fake') you have 24 possibilities, ball 1 is too heavy, ball 2 is too heavy . . . etc. or ball 1 is too light, ball 2 is too light . . . etc.
Each of 12 balls can be either heavy or light (but only one is fake) so 2*12 =24 possibilities
Using a twin pan balance, for each weighing you can get one of three results, right pan is heavy, left pan Is heavy, or both balance. So with 3 weighings you can get 3*3*3 =27 possible outcomes.
27 outcomes is more than enough to distinguish between 24 possibilities.
The rest is left for the reader!
As this is still showing as unanswered I will expand on my earlier answer. The usual method requires you to make one weighing, and tailor the later weighings based on that. (you have a branching flow-chart)
1, Put 4 balls in each pan, and note the result. There are 3 possible results:
a, pans balance
b, left pan heavy
c, right pan heavy
b&c can be treated the same if we just refer to 'heavy pan' and 'light pan' as they are mirror images.
2, a (pans balanced) the false coin is among the remaining 4 balls, and you have identified the first 8 balls as 'good'. Weigh 3 'good' balls against 3 of the untested balls.
If it balances, the only unweighed ball is fake, and you can tell whether it is heavy or light by weighing against any other ball.
If the 3 against 3 doesn't balance then you know that the group of 3 you have just used which were previously unweighed contains the fake. Take two of those three and weigh against each other. The one which replicates the result of the 3*3 weighing is the fake, and you know whether it is heavy or light. If this balances then again it identifies the unweighed ball.
b, If the pans don't balance in weighing 1 you have 4 balls which haven't been weighed, and these must be genuine, mark them 'G' for good. The 4 balls from the pan which went down mark 'H' for (potentially) heavy. The other four mark 'L' for (potentially Light).
Now for the tricky part! Weigh H1, H2, H3, & L1 against H4, G2, G3 & G4. (so 3 heavy & 1 good balls against 1 heavy & 3 good balls)
If the pan with 3 heavy balls goes down then one of those 3 is false, As in 2a you can identify which by weighing 1 against 1 (either the pans balance meaning the unweighed ball of the three is false, or one pan goes down, and this contains the false ball)
If the pan with one heavy ball goes down then either that one heavy ball is false, or the one light ball in the other pan is light. Weigh either of these against a known good ball to decide which.
If the two pans balance then the false ball is one of the three untested 'light' balls. weigh one against one to find your result as above.
1, Put 4 balls in each pan, and note the result. There are 3 possible results:
a, pans balance
b, left pan heavy
c, right pan heavy
b&c can be treated the same if we just refer to 'heavy pan' and 'light pan' as they are mirror images.
2, a (pans balanced) the false coin is among the remaining 4 balls, and you have identified the first 8 balls as 'good'. Weigh 3 'good' balls against 3 of the untested balls.
If it balances, the only unweighed ball is fake, and you can tell whether it is heavy or light by weighing against any other ball.
If the 3 against 3 doesn't balance then you know that the group of 3 you have just used which were previously unweighed contains the fake. Take two of those three and weigh against each other. The one which replicates the result of the 3*3 weighing is the fake, and you know whether it is heavy or light. If this balances then again it identifies the unweighed ball.
b, If the pans don't balance in weighing 1 you have 4 balls which haven't been weighed, and these must be genuine, mark them 'G' for good. The 4 balls from the pan which went down mark 'H' for (potentially) heavy. The other four mark 'L' for (potentially Light).
Now for the tricky part! Weigh H1, H2, H3, & L1 against H4, G2, G3 & G4. (so 3 heavy & 1 good balls against 1 heavy & 3 good balls)
If the pan with 3 heavy balls goes down then one of those 3 is false, As in 2a you can identify which by weighing 1 against 1 (either the pans balance meaning the unweighed ball of the three is false, or one pan goes down, and this contains the false ball)
If the pan with one heavy ball goes down then either that one heavy ball is false, or the one light ball in the other pan is light. Weigh either of these against a known good ball to decide which.
If the two pans balance then the false ball is one of the three untested 'light' balls. weigh one against one to find your result as above.
Or, for those who don't like the idea of a branching diagram to answer the question, as long as you can label, or keep track of, all 12 balls.
Label the 1 to 12 and make 3 weighings:
1,2,3,4, against 5,6,7,8
2,3,8,9 against 1,4,10,11
2,5,10,12 against 4,6,8,11
Tabulating the results of these three weighings will allow you to identify the fake ball, and whether it is light, or heavy.
But it does take the fun out of it ;)
Label the 1 to 12 and make 3 weighings:
1,2,3,4, against 5,6,7,8
2,3,8,9 against 1,4,10,11
2,5,10,12 against 4,6,8,11
Tabulating the results of these three weighings will allow you to identify the fake ball, and whether it is light, or heavy.
But it does take the fun out of it ;)
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