A math problem to find counterfeit and shoddy products, and a fun math problem to solve find defect

There is some discussion on this issue in my space, and I look forward to your opinions.

The answer is as follows: the first time is left and right

The second time is called left and right

The third time is called left and right

Judging principle: if the 1st left weight, the 2nd left weight, the 3rd right weight, then 1 weight on the contrary, if the 1st right weight, the 2nd right weight, the 3rd left weight, then 1 light for the layout clean, the following abbreviation, the record in turn of the results can not be indiscriminately noted: balance is opposite or balance).

If it is left and right, it is 2 heavy, and vice versa is light.

If the left and right sides are flat, it is 3 heavy, and vice versa.

If the left is flat and left, it is 4 heavy, and vice versa.

If it is left and right on the right, it is 5 heavy, and vice versa.

If the right side is flat, it is 6 heavy, and vice versa.

If the right side is flat, it is 7 heavy, and vice versa is light.

If the right is flat and the left is flat, it is 8 heavy, and vice versa.

If the left is flat, it is 9 heavy, and vice versa.

If it is flat on the right and left, it will be 10 heavy, and vice versa will be light.

If it is flat and left, it will be 11 heavy, and vice versa.

If it is flat on the right, it is 12 heavy, and vice versa.

Step 1: 4 balls on each side of the scale, with 4 balls on the outside. There are two cases in this way, balance balance (simple case) and balance imbalance (complex case).

Let's start with the simple case. The balance is balanced, then the remaining four balls have one bad ball and the other 8 balls are standard balls.

In the second step, take out 3 of the 4 balls and place them on the left and 3 from the standard ball on the right. If the balance is balanced, the remaining ball is bad, and the third step is to compare it with the standard ball to know the weight.

If it is not balanced, we might as well assume that the left is right, and we will know that it is biased. The third step is to take out one of the 3 balls on the left, one on the right, and one on the left. If Left = Right, what is left is a bad ball; Left >> Right, Left Bad; Left "Right, Right Bad." It's easy.

Let's talk about the complexities. The balance is unbalanced, so let's assume 4 balls on the left (numbers 1, 2, 3, 4) > 4 balls on the right (numbers 5, 6, 7, 8). There are two chances left and four more par balls to exploit.

In the second key step below, put 1, 2, 3, 8 on the left side of the scalePut 3 par balls on the right side of the scale +4;In other words, 1, 2, 3 are a group, the balance position is the same, 4, 8 is a group, they swap the position of the scale, 5, 6, 7 is a group, they take out of the balance.

Discussion, assuming that it is still left and right, then 4 and 8 are good balls, 1, 2, 3 have a bad ball and are heavy, and the problem is solved;

Suppose there is a bad ball in 4 and 8, but I don't know whether it is a light one or a heavy one, just compare either one with a standard ball;

Assuming Left = Right, there is a bad ball in 5,6,7 and it is on the light side, the problem is solved.

At this point, the problem is completely solved.

The classic topic of partition and subjugation of computers.

Check the divide and divide yourself.

This question is quite interesting, bookmark it.

I think 6 times. Weigh them one by one.

3 times, because you will know the answer after weighing 3.

Impossible! At least 3 times! 3 times the words are called so:

1. Weigh 5 cartons of milk first, if it is enough for 400*20*5, the fake box must be in the remaining 5 boxes, if it is not enough, that

2. Divide it again in the five fake boxes, and continue to weigh it!

The number of individuals in the sample is called the sample capacity. How to sample is a complex issue and there are no universal guidelines.

This depends on the specific situation, for example: if you are a multiple-choice question, you should compare others, and if you use the elimination method, you should exclude the remaining unreasonable, and you should choose this one overall.

On the other hand, because it is a random sample, the sampling is also universal and representative. It should be considered appropriate.

At present, there are many problems such as shoddy, insufficient area, and deceptive propaganda among real estate developers. Can developers who choose fake and shoddy products fill in the ** truthfully? Doubtful. It was inappropriate to select a number of developers as the targets of the investigation.

There are 32 provinces, cities and regions in the country, and if you choose 2 from each province and city, there will be 64 people, so it is almost appropriate.

Because points 1, 2, and 3 are in a straight line, point 3 may be in Yaogang Primary School in the northwest direction or Tamsui Town No. 2 Primary School in the southeast direction, and if there is an offset at point 2, point 3 may be in the Xingfu Kindergarten in the northwest direction.

The approximate location of 3 o'clock is Yaogang Primary School in the upper left and the Second Primary School in Tamsui Town in the lower right; This is followed by Zhangfu Kindergarten in the upper left.

No solution. "1 o'clock to 3 o'clock, and 1 o'clock to 2 o'clock, and 2 o'clock to 3 o'clock are all in a straight line", does not state that three points are collinear, which is nonsense.

It's possible in the circle.

There is a distance between the points, it is impossible to find the direction, although the 3 points are collinear, but I don't know where the 3 points are.

Solving with imaginary numbers is the same story as introducing the concept of imaginary numbers.

3 o'clock may be in the northwest direction of Yaogang Primary School or southeast direction of Tamsui Town No. 2 Primary School, if 2 o'clock is offset, 3 o'clock may be in the northwest direction of Xingfu Kindergarten.

Hello landlord

The first number of digits obtained is (n 1s) 2n-1 bits.

odd digits, then the middle number must be n

But after more than 10 years, I will give you two pictures to see, and you will probably understand a certain Schrödinger's cat.

10 1s = 1234567891098761421, the rule is how many 1s are multiplied and written from 1 to that number is written back to 1

123 is a set of numbers.

678 is a set of numbers.

The difference between the number tail and the number head of the latter group is 3

123,678,11,12,13,16,17,18

In groups of three, the latter group is 3 larger than the last one in the previous group

1,2,3,6,7,(8),(11)

This is 3 numbers 1, 2, 3 and 6, 7, 8 and then 11, 12, 13, and the difference is 5

It. I've tried, and the question must be found, but the answer is not necessarily the answer.

Type the title of this question on it, and then see if there is a similar question, click in to see if there should be an answer.

The original question is difficult to find, but it is easy to find a similar question, according to the gourd to draw a scoop "cottage" an answer to the answer.