Fun number one: front left, back right, left back, front right, left back, front right

Hello friend, Ni Xi Shi Boron

Don't be so complicated, the first time, put 1-6 on a plate, 7-12 on another plate, which side is light, the defective product is in that plate, the second time, take the light plate of 6, put three on each plate, still find out, the light plate out of three, the third time, put two plates, each plate put one, if the balance on both sides indicates that the other is not in the plate is defective, otherwise, which plate is light, the plate is defective.

(Super lucky): Left 5 right 5. If it is the same weight, take out one and weigh it with the remaining (defective). (I won't talk about it later).

For the first time, put 4 on the left and 4 on the right.

In this way, there will definitely be a group that will be selected to be different in weight.

The second time, put 2 on the left and 2 on the right.

In this way, there will definitely be a group that will be selected to be different in weight.

The third time, put 1 on the left and 1 on the right.

OK, I must have chosen.

I've done this"...Too lazy to write down the answers.

There are three groups: four in each group, the first group is numbered 1-4, the second group is 5-8, and the third group is 9-12

The first weighing: put the first group on the left side of the scale, and put the second group on the right.

a First possibility: balance. is different in the third group.

Next, you can put the number on the left and the number on the right three normal.

a.If balanced, the number 12 is different;

b.If the left is heavy and the right is light, it is different in the number, and it is heavier than the normal ball. Weigh it again:

9 to the left, 10 to the right, if balanced, 11 is different; If the left is heavy and the right is light, then the number 9 is different, and if the right is heavy and the left is light, then the number 10 is different.

c.If the left is light and the right is heavy, the principle is the same as B

b The second possibility: the left is heavier and the right is lighter, which is different in No. 1-8, but I don't know whether it is lighter or heavier than normal.

The second time: the left is numbered, and the right is numbered.

a.If balanced. is different in . It can be called the third time: put on the left and put on the right. If balanced, then 8 is different; If the left is heavy and the right is light, then 4 is different; If the left is light and the right is heavy, then 7 is different.

b.Still left heavy right light. is different in the position of unchanged.

It can be called the third time: put on the left and put on the right. If balanced, then 2 is different; If the left is heavy and the right is light, then 1 is different; If the left is light and the right is heavy, then 6 is different.

C: Light on the left and heavy on the right. is different in , because only they have changed their original position. It can be called the third time: 5,3 on the left, 9,10 on the right. If the left is light and the right is heavy, then 5 is different, and if the left is heavy and the right is light, then 3 is different.

c The third possibility: the left is light on the right, and the reason is the same as b

At this point, no matter what happens, you can find out the difference by weighing three times, and you can know whether it is lighter or heavier than normal.

Number the 12 balls as .12, and divided into three groups: group A; Group B ; Group C

The first time: put the two groups A and B on both sides of the scale, if they are the same weight, the abnormal ball is in group C, otherwise it is in groups A and B;

Discussed separately: 1) In the case of the anomaly in group C (i.e., the same weight of A and B), then.

The second time: three balls from group A are selected as standard balls to be placed on the left side of the scale, three balls from group C are placed on the right side of the scale, and if balanced, the abnormal ball is number 12; unbalanced, the anomalous ball is one of them, and it is known whether the anomalous ball is heavier or lighter than the standard ball;

The third ball is placed on the right side of the scale, and if it is balanced, the abnormal ball is 11; If there is an imbalance, the abnormal ball can be picked out according to the weight comparison of the above abnormal ball with the standard ball.

2) If the abnormal ball is in groups A and B (i.e., A and B are not the same weight), then group C is the standard ball, and A may be heavier than B.

The second time: the left side of the balance to put the ball, the right side of the ball, if the balance means that the abnormal ball must be numbered, and the abnormal ball must be lighter than the standard ball, the last time the weight of the ball can be picked out; If it is unbalanced (it must be on the left), it means that the abnormal ball is in group A, and the abnormal ball must be heavier than the standard ball, then any 2 balls of the last comparison ball can be picked out.

It can meet your needs. Here are a few examples:

Example 1: There are a few candles left on the table.

Topic: There were 12 lit candles on the table, but three were blown out by the wind, and soon two were blown out by another gust of wind, and finally there were a few candles left on the table.

Answer: 5 roots.

Example 2: How many lights are left?

Topic: There are 9 lights in the classroom, 3 are turned off, how many lights are left?

Answer: 9 lights.

Example 3: Soy sauce?

Topic: Xiao Ming's family has 16 catties of soy sauce, and 2 catties are taken away every month, will the soy sauce be used up in a few months?

Answer: After 7 months.

Example 4: Can the rim of the cup be turned face down?

Question: There are 14 cups on the table, 3 cups are facing up, now turn 4 cups at a time (turn the cups up to face down, and turn the cups down to face up). Q:

Can the cup be turned several times with the rim facing down? If not, then can you only do it with 6 flips per time? What about 7?

Answer None of them can be done, only 7 of them can do it.

The initial state is 3"+"，11"-", so multiplying 14 numbers makes the product -1, and turning 1 cup is to change +1 to -1 or -1 to +1, and when turning 1 cup, it is equivalent to multiplying the original state by -1.

When you turn the cup n times, it is equivalent to multiplying it by n"-1", so every time you turn an even number of cups, the initial state does not change"-1"of this result.

Therefore, turning 4 cups at a time and 6 cups at a time cannot change the product to yes"-1"of this result.

And every time you flip an odd number of cups, you can change the initial state"-1"of this result. So turning 7 cups at a time and turning an odd number of times will do it.

The specific operation is as follows: in the original state, 3 cups are facing up, and 11 cups are facing down;

Flip 2 cups with the rim facing up and 5 cups facing down, after flipping, 6 cups with the rim up and 8 cups with the rim facing down;

Flip 3 cups with the rim facing up, flip the 4 cups with the rim down, and after flipping, 7 cups with the rim facing up and 7 cups with the rim facing down;

Turn 7 cups facing up. After flipping, the 14 cups are all facing down, completing the task.

Finally, I would like to enclose the cover of "Fun Math Problems" for you!