I've a question.... If I have 12 balls, 11 of those balls weights exactly the same... 1 is either lighter OR heavier, you don't know... You have a weighing scale. (not a machine that shows the weight, but a device where you put mass on both sides and you can see which one is heavier) You can only weight three times to find out which ball does not belong to the group and if its heavier OR lighter than the other balls....
Are you sure that you can only weight it 3 times only and you don't know if that one ball is lighter or heavier?
weight 6 against 6. it would show you which side is heavier or lighter. then weight those 6 balls 2 against 2. you should then know what to do
i don't know if my answer is correct -- but here my guess: divide the balls into 2 groups of 6. weigh it the first time -- (1st weigh is gone) that will tell you which side is heavier or lighter then divide the "heavier/lighter" side of 6 into 2 groups of 3. weigh it the second time -- (2nd weigh is gone) that will tell you again which side is heavier or lighter then (here's the part which is a bit unorthodox) use your hands and test which balls is possibly the heavier or the lighter one. << a valid method since it allows a step of hypothesis and a step-by-step testing phase. then use the scale for the last time -- divide the balls 2 vs. 1 -- and test if your hypothesis is correct. (3rd weigh is gone) however, if your hypothesis is wrong -- then you will have to make a guess between the last two balls.
this only work if you know if the ball is heavier or lighter, but you dont know if its lighter or heavier.... you need to find that out yet. so if you weight 6 on 6, which one do you need to choose? There is a chance that the ball is lighter than the normal one (then you pick the light side) BUT what if the ball is heavier than the other ones, then I need to pick the heavy side, but I dont know if that ball is light or heavy.... this is basically what wind says And we are not allowed to use hands to weight
okay number all the balls 1- 12 step 1.take 1234 and 5678 and weigh them if they weigh the same then 9,10,11,12 has the odd ball or if 1234-5678 weighs diff. step 2 then u found the odd ball batch lets say its in 1234 step 3 then u weigh 1,2-3,4 then see which one is heavier lets say its 1 and 2 then wala you got your answer
hey, I haven't though about this earlier... Thanks, but there is still one thing you haven't take it into consideration... You don't know if the odd ball is light or heavy... so lets say of 1234-5678 weight diff, then we know its in there, but which group? Since the ball can be lighter than the other OR heavier than the others....
Done this awhile ago, can't really remember.. but i think this is the right way to do it. First i would label the balls from 1-12 and divide them into 3 sets. Step 1 - weigh 1,2,3 and 4 against 5,6,7 and 8. If they weigh the same then 9,10,11,12 is obviously going to have the odd ball. So the next step would be Step 2 - weigh 9 against 10. If they weigh the same, then the odd ball is either going to be 11 or 12. So the final step would be Step 3 - weigh any ball apart from 12 against 11. If 11 is heavier/lighter then 11 is the odd ball. If it aint then we could assume 12 is the odd ball. However if step 2 doesn't weigh the same (9 against 10), then step 3 would be Step 3 - Weigh any ball apart from 10 against 9, if 9 is heaver/lighter then 9 is the odd ball. If it aint then 10 is the odd ball. If 1,2,3,4 and 5,6,7,8 doesn't weigh the same then the odd ball has to be in 1,2,3,4,5,6,7,8 and 9,10,11,12 are standard balls. Now we label the balls heavy, light and normal. So say 1,2,3,4 weighed lighter than 5,6,7,8 we would label them L1 L2 L3 and L4. Then we would label 5,6,7,8 H1 H2 H3 H4 and 9,10,11,12 N1 N2 N3 N4. The nxt step would then be Step 2. Weigh L1 L2 H1 against H2 L3 N1. Now we could have three possible results which are: 1. If L1 L2 H1 weighs less than H2 L3 N1 then either L1 / L2 is light or H2 is heavy. 2. If L1 L2 H1 weighs more than H2 L3 N1 then either H1 is heavy or L3 is light. 3. If they are equal then the remaining balls contain the odd one. So either H3 / H4 is heavy or L4 is light. So now here's step 3 to the three possible results from step 2. Step 3 1. Weigh L1 against L2, if they are equal then H2 is the odd ball. If they aren't equal then the lighter one is the odd ball. 2. Weigh H1 against any ball apart from L3. If they are equal then L3 is the odd ball. If they aren't equal then H1 would be the odd ball (H1 should weigh more than the other ball). 3. Weigh H3 against H4. If they are equal then L4 is the odd ball. If they aren't equal then the heavier ball is the odd one. Actually i think this should be the right way... man i've typed alot lol
lol its correct... I've found the solution yesterday thanks to the idea of nycko thanks though... Now you can help me to think of an assigment that the idiots have to do that opposed me and said that I would lose this bet .
wtf ...too complicated... this is the solution. 1. weight 6 vs 6. Result: keep the one on that is lighter. 2. From the 6 balls, weight the 2 against 2. - if it is equal, then the odd ball is one of the two in your hands, all u do is weight the two and see which one is lighter- > u are done, if not then go to 3. 3. take the lighter of the two sets, and then weight them, the lighter one is it.. done. if ur phsyics teacher think this is wrong, then pull ur pants off and show him the REAL balls ask him which one is heavy, urs or his. LOL
with this you just assume that the odd ball is lighter.... but there is a chance that the odd ball heavier... so this won't work and is incorrect.... so not only will my physics teacher say its wrong... but I also agree that this is wrong lol... So does that mean I still have to pull my pants off?
it don't matter which one is heavier or lighter...the odd one would show up anyway.... yea. pull it down and show him the heavier balls...make sure he knows the physics of what goes up must come down...don't let those dumb idiots touch try the soln again if he don like it ...lol
no the odd one won't show up. If you take the light side as you said, then you weight 2 on 2, if they are equal, then weight the other 2 left of the group of 6. what if they are equal again? Cuz the odd ball is in the other group of 6? well, I will pull my pants off the next tuesday when I see him lol
Lol I remember hearing a riddle like this once and it pissed me off cause I actually had to use a scrap paper to help me answer.
