Brain Teaser!! Need some Help!

[AnimeKnight]

You have 12 balls, one of which is different. The ball is indistinguishable from the rest except that it is either heavier or lighter (but you don't know which). How can you determine which all is the different one in 3 weighings on a balance scale?

 

[Platypus]

4x3=12. Measure them in sets of 4 on the scale. Therefore you can find their weight difference. Does that make sense?

 

[MrCodeDude]

When you take out the balls, unless it's slightly heavier or lighter, you should be able to feel the difference when you take the balls out If not, I have no clue.
-- mrcodedude

 

[MrCodeDude]

<< 4x3=12. Measure them in sets of 4 on the scale. Therefore you can find their weight difference. Does that make sense? >>


That would only tell you which cluster of 4 the group is in.
-- mrcodedude

 

[Platypus]

Yeah... and then you find which one is heavier or lighter.... You have narrowed it down to 4 balls.

 

[IcemanJer]

<< That would only tell you which cluster of 4 the group is in. >>

then you can narrow it down even more, and do 2x2.

 

[viewton]

1st: put 6 balls on each side of the balance, whichever side is heavier, remove those balls.
2nd: take the 6 &quot;light&quot; balls, and place 3 each, on each side of the balance = repeat above.
3rd: take 2 balls, leave last ball out = if two balls balance, then the remaining is the diff ball. if the 2 balls don't balance, then one of them....hmmmm...this would only work if you knew if the diff ball was heavier or lighter....I give up

 

[viewton]

ACK! :Q here's the answer...this might be where corporaterecreation was heading...too late for me to read the whole thing:

This is long and complicated, please bear with me. First of all, for the sake of answering this, we will label the balls a, b, c, d, e, f, g, h, i, j, k, and l. The letter x represents any ball that has been identified as not the odd ball. ok. here goes.
Step 1: weigh abcd against efgh. if they balance, move to step two. If they are unbalanced move to step 3.
Step 2: wiegh ijk-xxx (x can be any letter from a to h, since they all balanced out, indicating the odd ball is not included in a-h). If they balance, weigh l to x. If l rises, its light, if it falls, it is heavy and we have one ball identifiable! if they are not balanced, move to step 4.
Step 3: weigh abef against cgxx.If they balance, go to step 5. if they do not balance go to step 6.
Step 4: weigh i against j. Which ever of these two reacts the same as the weighing in step 2 (i.e. if ijk rose while the xxx sunk) it is the odd one and weighs as its movement indicates, (i.e. if it rises then its the light one, but i am sure you could figure that one out.) If this weighing
balances, then it is k that is the odd ball and it wieghs as step two's weighing indicates.
Step 5: weigh d against x. if they balance, h is the odd ball and weighs as step 1 indicated. (Keep in mind that in step one, when the two sides where unequal, either one side contained a heavy ball or the other contained a light ball. This way you can tell how the last ball to be eliminated weighs without weighing it individually to an x) If they are
unbalanced, d is as its reaction indicates.
Step 6: this part gets a little confusing, so please bare with me. because of step three, three balls were eliminated. for instance, if, in step one, abcd was the light side and efgh was the heavy side and you get to step 3
and abef is the light side, you know that it is not e or f because if it was, abef would have sank. you also know that it is not c. If it were c, the side containing cgxx would have risen. The results can be switched around, but the key is that once a ball has been identified as a possible light ball, it cannot be found to be heavy. once step 1 has identified four as possible light and four as possible heavy balls, then they cannot be the other. With that said, move on to step 7.
Step 7: Ok, now that you have eliminated three balls, you must take, of the remaining balls, one from abcd and one from efgh, which ever of those are left, and place them on the same side of the scale. Weigh them against xx. if the side of the balance with the two non-eliminated balls reacts in the same manner as abcd did in step one, it is the ball from that group that is on the scale. if it reacts like efgh did in step one, it is the ball from that group that is on the scale. if the balls balance, it is the one ball left that is it and it weighs as step 1's weighing indicates.

 

[AnimeKnight]

<< << That would only tell you which cluster of 4 the group is in. >>

then you can narrow it down even more, and do 2x2.
>>


remember you only have 3 tries

 

[WJB]

i say, take the balls into a room filled with valuable glass objects and throw them around. which ever ball causes more or less damage is the lighter or heavier 1.

 

[Poncherelli]

its called the 12 coin problem, do a search on yahoo and you'll find a bunch of different sites with different solutions

 

[AdamDuritz99]

lol i saw this on paypal a few months ago. I figured it out after like 30 mins. i would post it but it's kinda long and confusing. And someone already posted one way to do it.


peace
sean

 

[AdamDuritz99]

*EDIT* sorry double post

 

[Qman71]

Well, I cheated and searched and found the answer here
This is a tough one!

 

[yiwonder]

BTW, this is a repost. A few months ago, there was a thread about this.

Anyway, I've got the answer from that thread.

 

[MrCodeDude]

<< 1st: put 6 balls on each side of the balance, whichever side is heavier, remove those balls.
2nd: take the 6 &quot;light&quot; balls, and place 3 each, on each side of the balance = repeat above.
3rd: take 2 balls, leave last ball out = if two balls balance, then the remaining is the diff ball. if the 2 balls don't balance, then one of them....hmmmm...this would only work if you knew if the diff ball was heavier or lighter....I give up >>


Yeah, I was on that train of thought too.. But then, you'd have to know whether the ball was lighter or heavier. Because, Step 1 is flawed because you think the X ball is lighter.

 

[Poncherelli]

organize things as follows. The first weighing would represent the nines place, the second weighing the threes place, and the final weighing the units place. I chose &quot;the left pan goes down&quot; to represent the digit 1 and &quot;the right pan goes down&quot; represents the digit -1. The digit 0 represents the pans balancing. Then I placed the 12 coins on the pans in such a way that, if any one of them turned out to be counterfeit and heavy, the outcome of the weighing would be a positive number in balanced ternary notation. Here is how the numbers are placed on the pans according to these rules.

left pan right pan
(9) first weighing 5,6,7,8,9,10,11,12
(3) second weighing 2,3,4,11,12 5,6,7
(1) third weighing 1,4,7,10 2,5,8,11

Obviously this assignment of coins to pans cannot work because the number of coins per pan in the first two weighings are not equal. However, if we interpret the sign of the outcome for certain coins oppositely, we can insure that the coin assignment to the pans is four coins per pan. Accordingly, I chose to place the coins 7, 9, 11, and 12 to be placed so that the outcome weighing would have a sign opposite to the correct sign. This produced the assignment

left pan right pan
(9) first weighing 5,6,8,10 7,9,11,12
(3) second weighing 2,3,4,7 5,6,11,12
(1) third weighing 1,4,10,11 2,5,7,8

For example, if the counterfeit coin is 6 and heavier than the other coins, then, in the first weighing, the left pan goes down, in the second weighing the right pan goes down, and in the last weighing the pans balance. This produces the base three number (1,-1,0)=9-3+0=6, indicating that the counterfeit coin is the sixth coin and it is heavier than the other coins because 6 is positive. If the counterfeit coin is 5 and lighter than the other coins, then, in the first weighing, the right pan goes down, in the second weighing the left pan goes down, and in the last weighing the left side goes down. This produces the base three number (-1,1,1)=-9+3+1=-5, indicating that the counterfeit coin is the fifth coin and it is lighter than the other coins. The coins 7, 9, 11, and 12 are treated oppositely. If the counterfeit coin is 7 and is heavier than the other coins, then, in the first weighing, the right pan goes down, in the second weighing the left pan goes down, and in the last weighing the right pan goes down. This produces the base three number (-1,1,-1)=-9+3-1=-7, indicating, for this special case, that the counterfeit coin is the seventh coin and it is heavy because we reverse the sign for the specially treated coins.

You cannot fail to notice that this method should be able to handle thirteen coins. However, to maintain the same number of coins per pan, we need a coin guaranteed to be good. The table below takes care of this situation with G representing the known good coin.

left pan right pan
(9) first weighing 5,6,8,10,13 7,9,11,12,G
(3) second weighing 2,3,4,7,13 5,6,11,12,G
(1) third weighing 1,4,10,11,13 2,5,7,8,G

 

[chiwawa626]

Realy AdamDuritz99? j/k

edit: i was gonna harass u bout the double post! darn u fixed it..

 

[AdamDuritz99]

<<Realy AdamDuritz99? j/k

edit: i was gonna harass u bout the double post! darn u fixed it.. >>

lol, yah, stupid mistake on my part, but i fixed it quickly.


peace
sean