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morgant
11-05-2003, 05:38 PM
You have 8 identical balls. One is a hair heavier than the other 7. You cannot tell the heavier ball by sight or touch. But you have a scale to weigh the balls against eachother. You may only use the scale twice, how do you definitively find the one heavier ball??

docidiot
11-05-2003, 05:46 PM
put three balls on either side of the scale. if they balance out, then weigh the other two to determine which is heavier. if they do not balance out, take the three balls that are heavier and weigh two of them. if one is heavier, you will be able to tell. if they balanace, the third ball is hte heaviest.

BruceZ
11-05-2003, 05:50 PM
I think you mean a balance not a scale.

Here is a similar problem (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=158052&page=&view=&sb =5&o=).

pokerlover
11-05-2003, 05:50 PM
You split the balls into 3 groups. 3/3/2. You then weigh the 2 groups of 3 on the scale. If one side is heavier then that group must have the heavier ball. Then take that group with the heavier ball and weigh any 2. If the ball is identical than the ball that you did not way must be the heavier ball. If one is heavier then the answer is obvious.

If you weight the 2 sets of 3 balls and they are identical then the heavier ball must be in the set of two balls that you didn't weight and by weighing those 2 you will determine the heavier ball.

Boris
11-05-2003, 05:58 PM
[ QUOTE ]
You have 8 identical balls.

[/ QUOTE ]

How the hell can you walk?

Utah
11-05-2003, 06:08 PM
12 balls. 1 ball is heavier OR lighter than the other balls (but you dont know which). You can use the balance 3 times.

One of my favorite all time puzzles.

pokerlover
11-05-2003, 06:14 PM
3 groups of 4. Weigh the first 2 groups. If one is heavier break it down and weigh 2 and 2. then weight one and one.

If the 1st 2 are the same weigh the 3rd group as 2 and 2. Then break that down to 1 and 1.

docidiot
11-05-2003, 06:17 PM
that doesnt work because the ball could be lighter. for instance, when you go for the 1 and 1, you cant distinguish which one is the normal one.

Utah
11-05-2003, 07:35 PM
docidiot is correct - that is what makes the puzzle so devious.

When you weigh 4v4 and it tilts to one side it could be because either the one side is heavier and the other side is lighter. Therefore, you haven't eliminated any balls.

I assure you though that the problem is solvable (and without using any gimmicky tricks or "outside the box" thinking)

pokerlover
11-05-2003, 08:39 PM
My mistake I just assumed it was heavier....I will think about this one

Boris
11-05-2003, 10:12 PM
OK. I think I have an answer.

Split the balls into 3 groups of 4. For simplicity lets number the balls 1-12 with the first 4 being in group A, the next 4 in group B, etc...

If group 1 = group 2, then obviously the oddball is in group 3.
CASE #1

In this case our next measurement shall be balls
9 and 10 against balls 12 and 1 (or any ball from groups A or B). If this group balances then it must 11 that is the oddball. The third measurement is balls 1 and 11 v. 2,3.

If balls 9 and 10 are heavier than 12 and 1, then it must be that either 9 or 10 is heavy, or 12 is light.

the third measurement is 1 and 9 against 2 and 10.

CASE #2: group A is heavier than group B.

measurement number 2 is balls 1, 5, 9, 10 vs. 3, 7, 8, 11.
If this measurement balances then the oddball is either 2, 4 or 6 and either 2 or 4 is heavy, or 6 is light. The third measurement is then 2, 10 vs. 4, 11.

if balls 1,5,9,10 are heavier than 3,7,8,11 then either 1 is heavy, or 7 or 8 is light. The third measurment is balls 7, 10, vs. 8, 11.

If balls 1, 5, 9, 10 are lighter than 3, 7, 8, 11 then 5 is the oddball and it is light.

Utah
11-05-2003, 10:59 PM
Very good. This is a fairly tricky puzzle as you need to mix in standard balls and you need to cross mix potentially Heavy and Light balls on the same side of the scale - as you did. Very nice work.

In case 2, you used HLX v HLL. You can also use HHL v HHX.

You have some extra balls added that are not needed but they dont affect the results. e.g., in the case of 1 and 9 versus 2 and 10. 1 and 2 are indentical weight so you can just weigh 9 v 10. You did this in another spot as well.

Boris
11-06-2003, 12:06 AM
good puzzle. not sure why i added the extra balls in the last tests. i did this at work and kind of got busted by the boss. oh well.

pokerlover
11-06-2003, 10:22 AM
Well done. I thought about it last night and I gave up when I started tilting /images/graemlins/wink.gif. Good job and thanks for the answer. My father and wife also thank you as none of us could think of the answer. WTG