A new mind twister
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You have 12 balls - one of which is either lighter or heavier than the others. Using a balance scale determine the odd ball and whether it is heavy or light in 3 weighings. This is much more difficult than it first appears. Have fun. Dave Huff 'tis a silly place!
Ok, I got it now. The special step is to not compare the last 6 balls against either other once you've got down to 6! I won't post the full solution. The others can sweat it out. Roger Allen Sonork 100.10016 If I had a quote, it would be a very good one.
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You have 12 balls - one of which is either lighter or heavier than the others. Using a balance scale determine the odd ball and whether it is heavy or light in 3 weighings. This is much more difficult than it first appears. Have fun. Dave Huff 'tis a silly place!
Measure four against four. Possible results: 1) weigh the same - take three balls from a measured four and three of the remaining and weigh: if the same measure remaining ball is the one - measure it against known ball and you have whether it is heavy or light. if different you know whether it is heavy or light because you know the three balls already measured are the same. Measure one of the three unknows against another. If the same the last ball is the odd man. If not the same the heavier or lighter ball can be determined from previous step. This is the easy part. What to do if the 4 vs 4 do not weigh the same is where it gets hard. Dave Huff 'tis a silly place!
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But it will because the other eleven are all assumed to weigh the same. If each ball weighed a pound with the heavier one weighing two pounds, you'd be weighing 6 to 7. We know the heavier one is on the side that goes down. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country.What if this odd ball weighs half a pound? We do not know whether the odd ball is lighter or heavier. Regards Thomas Finally with Sonork id: 100.10453 Thömmi
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But it will because the other eleven are all assumed to weigh the same. If each ball weighed a pound with the heavier one weighing two pounds, you'd be weighing 6 to 7. We know the heavier one is on the side that goes down. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country. -
Won't work. What happens at stage two if the 3 & 3 balance? Paresh Solanki There is no substitute for genuine lack of preparation.
It can't because only one ball is a different weight. 6 and 6 will never balance because the heaviest is on one side. Taking the 6 with the different weighted ball and splitting it guarantees one will be a different weight; therefore, they won't balance. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country. -
But it will because the other eleven are all assumed to weigh the same. If each ball weighed a pound with the heavier one weighing two pounds, you'd be weighing 6 to 7. We know the heavier one is on the side that goes down. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country.But the heavier one could actually be lighter (read the question) so its doesn't help Roger Allen Sonork 100.10016 If I had a quote, it would be a very good one.
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Ok, I got it now. The special step is to not compare the last 6 balls against either other once you've got down to 6! I won't post the full solution. The others can sweat it out. Roger Allen Sonork 100.10016 If I had a quote, it would be a very good one.
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But the heavier one could actually be lighter (read the question) so its doesn't help Roger Allen Sonork 100.10016 If I had a quote, it would be a very good one.
Are we testing for either circumstance or one or the other at a time? If it's either, well I'm going back to the drawing board. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country. -
Are we testing for either circumstance or one or the other at a time? If it's either, well I'm going back to the drawing board. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country. -
It can't because only one ball is a different weight. 6 and 6 will never balance because the heaviest is on one side. Taking the 6 with the different weighted ball and splitting it guarantees one will be a different weight; therefore, they won't balance. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country.How? you do not know whether the different ball is lighter or heavier doing the 6:6 will tell you that something is different, but lets say you take the lighter group. If Odball is lighter, your plan will work, but if odball is heavier, then the 3:3 split will balance. you are stuck Paresh Solanki There is no substitute for genuine lack of preparation.
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You have 12 balls - one of which is either lighter or heavier than the others. Using a balance scale determine the odd ball and whether it is heavy or light in 3 weighings. This is much more difficult than it first appears. Have fun. Dave Huff 'tis a silly place!
- Put four balls in each scale to find a pile that contains normal balls. (If equal, then both weigthed piles, if not, the one not weigthted). 2) If eigth balls are found normal it is easy to find the odd one in two weighings. If not: Remove two ball from first scale, one from second scale and swap two of the remaining ones from scale to scale. Add one normal ball to have equal number of balls on each scale. Case 1: Scale is in balance - one of the three balls removed is the odd one. Case 2: Scale shifts - one of the two moved balls is odd. Case 3: Scale does not shift - one of the three unmoved balls are odd. 3.1) Place two of the odd balls from each scale on one of the scales. Balance with two normal balls on the other scale. Case 1: Scale in balance - The remaining ball is odd and it's status can be revealed from weighting 2). Case 2: The scale with the unknown balls are heavier/lighter - which one is odd can be determined from weighting 2) 3.2) Determine which is odd by comparing with a normal ball. 3.3) Do as 3.1) Is that the correct solution? /moliate
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Measure four against four. Possible results: 1) weigh the same - take three balls from a measured four and three of the remaining and weigh: if the same measure remaining ball is the one - measure it against known ball and you have whether it is heavy or light. if different you know whether it is heavy or light because you know the three balls already measured are the same. Measure one of the three unknows against another. If the same the last ball is the odd man. If not the same the heavier or lighter ball can be determined from previous step. This is the easy part. What to do if the 4 vs 4 do not weigh the same is where it gets hard. Dave Huff 'tis a silly place!
I already got that far:confused: Paresh Solanki There is no substitute for genuine lack of preparation.
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- Put four balls in each scale to find a pile that contains normal balls. (If equal, then both weigthed piles, if not, the one not weigthted). 2) If eigth balls are found normal it is easy to find the odd one in two weighings. If not: Remove two ball from first scale, one from second scale and swap two of the remaining ones from scale to scale. Add one normal ball to have equal number of balls on each scale. Case 1: Scale is in balance - one of the three balls removed is the odd one. Case 2: Scale shifts - one of the two moved balls is odd. Case 3: Scale does not shift - one of the three unmoved balls are odd. 3.1) Place two of the odd balls from each scale on one of the scales. Balance with two normal balls on the other scale. Case 1: Scale in balance - The remaining ball is odd and it's status can be revealed from weighting 2). Case 2: The scale with the unknown balls are heavier/lighter - which one is odd can be determined from weighting 2) 3.2) Determine which is odd by comparing with a normal ball. 3.3) Do as 3.1) Is that the correct solution? /moliate
moliate wrote: 2) If eigth balls are found normal it is easy to find the odd one in two weighings. If not: Remove two ball from first scale, one from second scale and swap two of the remaining ones from scale to scale. Add one normal ball to have equal number of balls on each scale. Case 1: Scale is in balance - one of the three balls removed is the odd one. At this point you have three balls and you still do not know whether they are heavier or lighter with only one weighing remaining. But its closer. Dave Huff 'tis a silly place!
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moliate wrote: 2) If eigth balls are found normal it is easy to find the odd one in two weighings. If not: Remove two ball from first scale, one from second scale and swap two of the remaining ones from scale to scale. Add one normal ball to have equal number of balls on each scale. Case 1: Scale is in balance - one of the three balls removed is the odd one. At this point you have three balls and you still do not know whether they are heavier or lighter with only one weighing remaining. But its closer. Dave Huff 'tis a silly place!
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Assuming you keep the two balls removed from the first scale separated from the one removed from the second one you could do it in one weighing... Or am I not thinking clearly here :confused: /moliate
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Narrowed down to max three balls in weighting 2) :~ - Unknown :) - Normal 1:~ 2:~ ----- 3:~ :) 1======= 2======= (scale 1 lighter) Take one :~ from each side: 2:~ 3:~ ----- :) :) 1======= 2======= Case 1: Scales in balance - 1:~ light Case 2: Scale 1 lighter - 2:~ light Case 3: Scale 2 lighter - 3:~ heavy Of course, replacing all light with heavy and the other way around will not change anything... The problem might be if the balls cannot be separated on the scale.. /moliate
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Narrowed down to max three balls in weighting 2) :~ - Unknown :) - Normal 1:~ 2:~ ----- 3:~ :) 1======= 2======= (scale 1 lighter) Take one :~ from each side: 2:~ 3:~ ----- :) :) 1======= 2======= Case 1: Scales in balance - 1:~ light Case 2: Scale 1 lighter - 2:~ light Case 3: Scale 2 lighter - 3:~ heavy Of course, replacing all light with heavy and the other way around will not change anything... The problem might be if the balls cannot be separated on the scale.. /moliate
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You have 12 balls - one of which is either lighter or heavier than the others. Using a balance scale determine the odd ball and whether it is heavy or light in 3 weighings. This is much more difficult than it first appears. Have fun. Dave Huff 'tis a silly place!
For convenience sake, name balls 1-12. * denotes normal balls. 1) Place balls 1-4 on one side of the scale, 5-8 on the other. If they weigh the same, 9-12 contains the odd ball (goto a.2). If they weigh different, 1-8 contain the odd ball (goto b.2). a.2) Place 9,10 on one scale, 11,1* on the other. If they weigh the same, 12 is the odd ball, weigh against a normal ball to determine heavy/light. If they weigh different, goto a.3 a.3) Place 9 on one, 10 on other. If they weigh same, 11 is the odd ball, and you know from a.2 whether it is heavy or light. If 9 is heavier, and 9,10 was heavier than 11,* then 9 is the heavy odd ball. If 9 is lighter, and 9,10 was heavier than 11,*, then 10 heavy odd ball. If 9 is heavier and 9,10 was lighter than 11,*, then 10 is the light odd ball. If 9 is lighter and 9,10 was lighter than 11,*, then 9 is the light odd ball. b.2) Place 1,5,6 on one side, 2,7,8 on the other. - If the scale is even, one of the following is odd: 3,4. Step (1) tells you if the odd ball is light or heavy, weigh them to figure out which is odd. - If the scale changes, the odd ball was moved, therefore, it is one of the following: 2,5,6. Goto c.3 - If the scale doesn't change, the odd ball wasn't moved, therefore, it is one of the following: 1,7,8. Goto d.3 c.3) The odd ball is 2,5,6. - If 1-4 was heavier than 5-8 (step 1), then 2 is heavy or 5 or 6 is light. Weigh 5,6. If they are even, 2 is the heavy odd ball. If 5 is light, 5 is the light odd ball. If 6 is light, 6 is the light odd ball. - If 1-4 was lighter than 5-8 (step 1), then 2 is light or 5 or 6 is heavy. Weigh 5,6. If they are even, 2 is the light odd ball. If 5 is heavy, 5 is the heavy odd ball. If 6 is heavy, 6 is the heavy odd ball. d.3) The odd ball is 1,7,8. Do something similar to c.3 to determine the answer.
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For convenience sake, name balls 1-12. * denotes normal balls. 1) Place balls 1-4 on one side of the scale, 5-8 on the other. If they weigh the same, 9-12 contains the odd ball (goto a.2). If they weigh different, 1-8 contain the odd ball (goto b.2). a.2) Place 9,10 on one scale, 11,1* on the other. If they weigh the same, 12 is the odd ball, weigh against a normal ball to determine heavy/light. If they weigh different, goto a.3 a.3) Place 9 on one, 10 on other. If they weigh same, 11 is the odd ball, and you know from a.2 whether it is heavy or light. If 9 is heavier, and 9,10 was heavier than 11,* then 9 is the heavy odd ball. If 9 is lighter, and 9,10 was heavier than 11,*, then 10 heavy odd ball. If 9 is heavier and 9,10 was lighter than 11,*, then 10 is the light odd ball. If 9 is lighter and 9,10 was lighter than 11,*, then 9 is the light odd ball. b.2) Place 1,5,6 on one side, 2,7,8 on the other. - If the scale is even, one of the following is odd: 3,4. Step (1) tells you if the odd ball is light or heavy, weigh them to figure out which is odd. - If the scale changes, the odd ball was moved, therefore, it is one of the following: 2,5,6. Goto c.3 - If the scale doesn't change, the odd ball wasn't moved, therefore, it is one of the following: 1,7,8. Goto d.3 c.3) The odd ball is 2,5,6. - If 1-4 was heavier than 5-8 (step 1), then 2 is heavy or 5 or 6 is light. Weigh 5,6. If they are even, 2 is the heavy odd ball. If 5 is light, 5 is the light odd ball. If 6 is light, 6 is the light odd ball. - If 1-4 was lighter than 5-8 (step 1), then 2 is light or 5 or 6 is heavy. Weigh 5,6. If they are even, 2 is the light odd ball. If 5 is heavy, 5 is the heavy odd ball. If 6 is heavy, 6 is the heavy odd ball. d.3) The odd ball is 1,7,8. Do something similar to c.3 to determine the answer.
You got it. One of the big tricks in figuring it out is to think of the balls not as known and unknown but as known, possibly lighter and possibly heavier. This allows you to cross reference and elimnate balls in the second weighing just as you did. Good job. Dave Huff 'tis a silly place!