Answer: Balance: Of the remaining four unweighted balls, take three and put two on one side of the balance, assuming one is on the left and the other is on the right. We can know that the ball balanced for the first time is normal. We call it a standard ball. Take a standard ball, put it on one side of the balance, and mark it with a pencil to indicate normality. The second result is as follows: Balance A 655.
A 1: obviously, in the only ball that is not weighed, it is impossible to know whether it is light or heavy; Here it is. Just weigh it twice.
A2: If it is unbalanced, we can mark the balls except the standard balls on the sinking side with+and the light ones with-,and the result is nothing more than
+,+,-or-,-,+three balls, take out one of them+,-and put it at one end of the balance. Take two of the first eight standard balls and put them at the other end. If the standard ball is heavy, it is obvious that we have deepened it. So that ball is what we are looking for. If the standard ball at that end is light, it means that our+ball is correct. We got what we wanted. * * * used three opportunities.
Next, solve the B imbalance:
B: unbalanced situation: (there are still two opportunities at this time)
We can assume that the sinking end may be heavy and the floating end may be light. We can use the same method here. Mark the ball with a pencil, indicate the possible heavy ball with a+sign, and indicate the possible light ball with a-sign. We assume that the left side sinks. Obviously, four unweighted balls are no problem. We can call it a standard ball. Then we take five balls that may be abnormal (that is, the balls marked with+or-are taken according to 3+2), suppose we take three balls with+and two balls with-(the same can be said for three balls with-and two balls with+), and then weigh them for the second time, and put the++combination at one end of the balance and-+at the other end. Note that we changed the original+ball to-ball, and the left side is still++ball, and the right side is-+ball and ordinary ball. The results are as follows: B2 1: If the balance result is unchanged, the problem ball is in the++on the left and the-on the right;
B22: If the exchange is unbalanced, it means that the ball is in the two balls we exchanged; B23: If the ball is balanced, the problem ball is not weighed in the balance;
The following+-of the three balls;
Next, we will deal with B 1, B2 and B3:
B2 1: If the balance result is unchanged, it means that the problem ball is in the++on the left and the-on the right. Next, there is a chance to find three balls, take+-and put two standard balls on the left side of the balance. If the left end sinks, we assume that the ball with+is correct. If the left end floats, the ball with-on the left end is correct.
B22: If the exchange is unbalanced, it means that this ball has a chance to determine two balls, a+ ball and a- ball, among the two balls we exchanged. It's simple. Put them on the left side of the balance and put two standard balls on the right side with standard balls.
If B23: If the ball is balanced, it means that the problem ball is among the+-three balls on the surface that did not participate in the balance weighing; The next thing to do is how to determine which of the three balls is a bad ball with only one chance. Smart words should know how to find it!