Required Mathematical Intelligence

Chapter 65 12 Table Tennis Problems

There are 12 ping-pong balls, one of which is substandard, but it is not known whether it is light or heavy.It is required to weigh three times with a balance to find out the bad ball.

[Answer: This is a difficult logical reasoning question.The difficulty of this problem lies in not knowing whether the unqualified ball is lighter or heavier than the qualified ball.To solve this problem, we must not only skillfully use various forms of reasoning, but also have a certain amount of ingenuity.

How many results will there be when weighing a table tennis ball with an unscaled balance?There are three different results, that is, the weight on the left side is heavier, lighter than or equal to the weight on the right side. In order to find out the unqualified table tennis ball by weighing three times, the balls must be divided into three groups (four balls each). ).Now, for the convenience of solving the problem, we number these three groups of table tennis balls as Group A, Group B and Group C respectively.

First, choose any two sets of balls and weigh them on the balance.For example, we put A and B groups on the balance and weigh them.There will be two situations:

In the first case, the balance is balanced on both sides.Then, the unqualified bad ball must be in group c.

Secondly, randomly take two balls (such as C1 and C2) from group c and place them on the left and right disks respectively, which is called the second time.At this time, two situations may occur:
1. Balance on both sides of the balance.Like this, bad ball must be in C3, C4.This is because, among the 12 table tennis balls, only one is a bad ball.Only when one of C1 and C2 is a bad ball, the two sides of the balance are not balanced.Now that both sides of the scale are balanced, it can be seen that C1 and C2 are both qualified strikes.

When weighing for the third time, one ball (such as C3) can be randomly taken out from C4 and C3, and placed on both sides of the balance with another qualified strike ball (such as C1), and the result can be deduced.At this time, there may be two results: if the two sides of the balance are balanced, then the bad ball must be C4; if the two sides of the balance are not balanced, then the bad ball must be C3.
2. The two sides of the balance are unbalanced.In this way, the bad ball must be in C1 and C2.This is because, only when one of C1 and C2 is a bad ball, the two sides of the balance cannot be balanced.This is the second time.

When weighing for the third time, one ball (such as C1) can be randomly taken from C2 and C1, and another qualified strike ball (such as C3) can be placed on both sides of the balance respectively, and the result can be deduced.The reason is the same as above.

The above is the analysis of the first situation after the first weighing.

In the second case, the two sides of the balance are unbalanced after the first weighing.This shows that Group C must be all qualified good balls, and unqualified bad balls must be in Group A or Group B.

We assume: Group A (with four balls A1, A2, A3, A4) is heavy, and Group B (with four balls B1, B2, B3, B4) is light.At this time, you need to take out A1 from the heavy tray and put it aside, take out A2 and A3 and put them in the light tray, and keep A4 in the heavy tray.At the same time, take out B1 and B4 from the light dish and put them aside, take out B2 and put them on the heavy dish, keep B3 in the light dish, and take another standard ball C1 and put it on the heavy dish.After such an exchange, there are three balls in each set: in the original heavy set, A4, B2, and C1 are now placed, and in the original light set, A2, A3, and B3 are now placed.
At this time, it can be called the second time.Three situations may arise after this weighing:
1. Balance on both sides of the balance.This shows that A4B2C1=A2A3B3, that is to say, these six are just strikes, and like this, the bad ball must be among A1 or B1 or B4 outside the plate.It is known that disk A is heavier than disk B.Therefore, A1 is either a strike, or heavier than a strike; while B1 and B4 are either strikes, or lighter than a strike.

At this time, you can put B1 and B4 at one end of the balance and weigh for the third time.At this time, three situations may also occur: (1) If both sides of the balance are balanced, it can be inferred that A12 is a bad ball that is not qualified. This is because there is only one bad ball among the 1 balls. The ball is a strike, and A4 is a strike; (1) B1 is lighter than B4, then B1 is a strike; (4) B1 is lighter than B4, then B1 is a strike, because B4 and B[-] are strikes, Or it is lighter than a good ball, so the third weighing is to compare which of the two light balls is lighter, and the lighter ball must be a bad ball.

2.放着A4、B2、C1的盘子（原来放A组）比放A2、A3、B3的盘子（原来放B组）重。在这种情况下，则坏球必在未经交换的A4或B3之中。这是因为已交换的B2、A2、A3个球并未影响轻重，可见这三只球都是好球。

以上说明A4或B3这其中有一个是坏球。这时候，只需要取A4或B3同标准球C1比较就行了。例如，取A4放在天平的一端，取C1放在天平的另一端。这时称第三次。如果天平两边平衡，那么B3是坏球；如果天平不平，那么A4就是坏球（这时A4重于C1）。

3.放A4、B2、C1的盘子（原来放A组）比放在A2、A3、B3的盘子（原来放B组）轻。在这种情况下，坏球必在刚才交换过的A2、A3、B2三球之中。这是因为，如果A2、A3、B2都是好球，那么坏球必在A4或B3之中，如果A4或B3是坏球，那么放A4、B2、C1的盘子一定重于放A2、A3、B3的盘子，现在的情况恰好相反，所以，并不是A2、A3、B2都是好球。

以上说明A2、A3、B2中有一个是坏球。这时候，只需将A2同A3相比，称第三次，即推出哪一个是坏球。把A2和A3各放在天平的一端称第三次，可能出现三种情况：（一）天平两边乎衡，这可推知B2是坏球；（二）A2重于A3，可推知A2是坏球；（三）A3重于A2，可推知A3是坏球。

According to the fact that Group A and Group B have different weights after the first time, we just assumed that Group A is heavier than Group B, and made the above analysis to explain how to deduce which ball is a bad ball in this case .If we now assume that the situation is that group A is lighter than group B, how can this be inferred?Please try to deduce it yourself. ]
(End of this chapter)