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Ask me a math problem:
I have answered both sides of this question, and it must be the correct answer.

There are two situations.

The first method: divide 12 balls into three groups, each group has four balls, two of which say that if the balance is flat, then the irregular balls are randomly selected from the remaining four groups and any three groups (standard balls) of the two weighing groups, and if the balance is flat, then the remaining one is weighed with any standard ball, which is the answer; If it is uneven, we know the weight of the irregular ball (assuming it is light), and then take out any two of these three and weigh them flat, then the remaining 1 and the standard ball are uneven, and the light side is irregular.

The second type; Divide 12 balls into three groups with four optional balls in each group. The two groups say that if the balance is uneven, the irregular ball is among the eight balls. At this time, we assume that the left end of the balance is light and the right end is heavy. Take the ball out of the balance, take out three suspects, put two at the right end of the balance, put 1 at the left end of the balance, then take out two suspects, and put 1 at the left end and the right end. Put a standard ball (any one of the remaining groups) at the left end (1). If it is flat, it is said that there are two suspected light balls; If it is still flat, the remaining one is irregular and heavy; If it is not uniform, the light ball at the left end is irregular and light; (2) If it is uneven, it is suspected that the heavy ball at the left end is irregularly heavy or the light ball at the right end is irregularly light, then take a standard ball and weigh it with one of the balls, and draw a conclusion. If the right end is heavy and the left end is light, it is suspected that the right end is heavy or the left end is light. If the two suspected heavy ones are leveled again, the left end is suspected to be irregular and light. If it is uneven, the heavy end is irregular and heavy.