- 追求自由Lv 51 decade agoFavorite Answer
Step 1. Number the 12 balls by 1, 2, ..... 12;
Step 2. Put 1-4, 5-8 onto the beam balance;
The remaining steps will depend on the outcome of the beam balance.
I give one example: 1-4 heavier than 5-8
Step 3. Put 1,2,5 on one side and 3,4,6 on the other side ;
Case 1: Balance
Step 4. If balance, then weight 7 & 8. The lighter one is the fake one.
Case 2: Unbalance
Step 4. If unbalance (say 1,2,5 heavier than 3,4,6), weight 1 & 2.
Step 5: If balance, then 6 is lighter.
Step 5: If unbalance, then the heavier is the fake one.
- WaiLv 61 decade ago
The 12 balls are to be distributed into 3 groups respectively into A, B, C, each 4 balls. Assume that the fake ball is in B.
A-0000 B-0000 C-0000
Let's weigh A & C for the 1st time; if they are equal; you will find that the fake ball is in B. If we weigh A & B for the 2nd time, you will find that either A or B is lighter or heavier. Then, distribute the 4 balls of Group B into 2 parts, and weigh for the 3rd time. You will find that one part is heavier and the other part is lighter. Let's assume that the lighter part is with problem. So, we weigh the 2 balls on the scale to see which one is lighter. Then you will find one of them is the fake ball. However, we have to weigh the balls for 4 times instead of 3 times.
2008-05-15 21:32:56 補充：
Another method: to distribute, A with 5 balls, B with 2 balls, C with 5 balls.
1st time of weighing: to weigh A & C, if they are equal.
2nd time of weighing: then try B, one is heavier, the other is lighter.
3rd time of weighing: to weigh one ball of Group B, and one ball either from A or C,
2008-05-15 21:36:08 補充：
Then you will find out the fake one.
In conclusion, if A-0000, B-0000, C-0000, it will make at least 4 times of weighing, or more.
if A-00000, B-00, C-00000, it will make at least 3 times of weighing, or more.
- 1 decade ago