There are 12 balls that look exactly the same, one of them is a defective product, how to weigh it w

The 12 balls are divided into 3 groups of 4 each. They are denoted as Group 1, Group 2, and Group 3 respectively.

The process is as follows: take group 1 and group 2 and weigh them at both ends of the balance.

1: Balance balance.

then the defective product is in group 3. Group 3 has 4 balls, A, B, C, D. Take a, b at both ends of the balance weighed, if balanced, then in c, d in the two balls take either one and a weigh, if the c is taken, and the result and the same weight, then d is defective, otherwise, c is defective.

If the case is d, the analysis of c is also taken.

2. The balance is unbalanced.

Then take group 1 or group 2 and group 3.

Let's say you take group 1:

Balance: Explain that the defective product is in group 2, and the following analysis will determine the defective product in group 3 in the same as A1.

Unbalanced: Indicates that the defective product is in group 1, and the following analysis will determine the defective product in group 3 in the same way as in A1.

To sum up, it is understood.

the same weight, the remaining 2:2; Different, 2:2 after the same as.

Same weight, the remaining 1: the other 1;

Different, each group takes 1, with other groups 2:2

Quite simply, the defective product must be lighter or heavier than normal. Compared with the weight of 11, there must be a weight that is different. Which one is a defective product.

Divide the 27 balls into three piles, a pile of 9, and weigh two groups at random. If it is balanced, then the defective products are in the unselected pile; If it's not balanced, then the defective product is in the lighter pile on the scale.

Divide the selected 9 balls into three piles, a pile of three, and weigh two groups at random. If it is balanced, then the defective product is in the unselected pile, and if it is not balanced, then the defective product is in the lighter pile on the scale.

Repeat the above method, select two balls from the three balls for weighing, if balanced, the defective ball is the unselected ball, if not, the defective ball is the lighter one.

Expansion:

For this kind of ball weighing problem, it can be summarized as using the balance i times, and adopting a fixed weighing method (the difference between the i power of 3 and the difference between 3 and 2) can find a ball with different weights in each ball, and say whether it is lighter or heavier than other balls. If only the problem ball is required, you can instead identify 1 problem ball in the (3 to the i power and 2 to the 1) ball.

1. Divide these balls into 3 groups at random, that is, 9 balls per group, take two groups to weigh at random, if they are the same, the group that is not weighed has defective products, if not, the light group has defective products;

2 After the previous step, two groups have been excluded (a total of 18 balls), and the remaining 9 balls are randomly divided into 3 groups on average, and the steps are the same as above;

3 After the second part, two groups (6 balls) were excluded, and the remaining 3 balls were weighed at will, if they were the same weight, the unweighed was defective, and if not, the lighter one was defective.

1.Find 18 balls at random, 9 and 9 on both sides of the scale, if they are the same weight, 18 on one side, pick up the other 9; If one side is light, take these nine and set the others aside;

2.Take out six of the nine balls in your hand, weigh 3 and 3, if they weigh the same, put 6 aside and pick up the other 3; If one side is light, pick up these 3 and put the others aside;

3.Take two of the three in your hand, weigh 1, 1, if the same weight, the third is light; If one side is light, this is a light ball.

If it is only the condition in the question, and it is not told whether the ball is heavier or lighter than the other 7, it is impossible to distinguish the weight of the two times, and the question is not rigorous enough.

First of all, the idea of 332 is correct, so let's enumerate the possible steps:

Put the balance for the first time: take any 6 balls, 3 at each end of the scale;

1.Balance Balance:

In this case, you can make sure that the ball that is different is in the remaining two. Then you should take any one of the 6 balls used for the first time and put it on the balance section, and then take one of the remaining 2 balls and put it on the other end of the balance for a second comparison. (1) If it is the same weight, the unused one of the two is the ball to be found, and the problem is successfully solved; If it is different, then the one to be compared between the two is the ball to be found, and the problem is solved.

2.The balance is unbalanced:

At this point, you can't determine which of the 3, 3, and 2 balls you are looking for, so you won't be able to locate the ball you're looking for in the second weighing comparison, and the solution fails. Because the result of how to weigh for the second time does not have uniqueness, you choose the 3 on the left side of the scale, choose any 2 to weigh, if it is the same, then you have to take one from the balance, put the remaining one, and weigh it for the third time; If it's different, the first time you can be sure of finding the ball, it's 3 times; If it's the same, that is, all 3 balls are the same weight, sorry, you have to go to the right 3 and repeat the above steps; If you can't find it yet, you can compare the last 2 and repeat the comparison between 1 of the 6 and 2 in the first balance;

Therefore, to sum up, this question is not rigorous.

Put 3 scales on each side, if balanced, then light in the remaining 2, weigh again to compare.

If one side is light, then the light ball is in this set of 3.

Take this group of 3 light balls, take 2 of them and put them on both sides of the scale, if balanced, then the light is the remaining 1; If it is not balanced, the light one will come out.

The first time, put three balls on both sides, and if balanced, then put the remaining two on the second time.

If the first time the two sides are not equal, then the three on the side that weighs more.

Summary. Sorry to keep you waiting<>

<> this problem, first put 54 balls on both sides of the scale, a total of 108, if the two sides are balanced, it means that the remaining ball is defective, and only need to be used once. One side of the ball is heavier, indicating that the defective product is in it, and then the 54 balls containing the defective product are divided into two equal parts and placed on the balance, that is, 27 on one side

There are 109 balls, one of which is slightly heavier, and if you weigh it at least a few times with the balance, you will be sure to find a defective product.

Okay. Sorry to keep you waiting<>

<> this question first put 54 balls on both sides of the scale, a total of 108, if the two sides are balanced, the finger Mori rough means that the remaining spring plum ball is defective, at this time only need to be once. One side of it is heavier, indicating that the defective product is only in it, and then the 54 balls containing the defective product are divided into two equal parts and placed on the scale, that is, 27 on one side

No, no. At this time, 13 are placed on the left and right sides, and if the balance is balanced, the remaining one is defective, and it is weighed 3 times at this time. If one side is heavy, the defective product is in the 13 that are on the heavy side.

And so on, in the worst case, the defective product is found at the last time, so it takes 6 times.

The hall disturbance is 54 on one side, 27 on one side, 13 on one side, 6 on one side, 3 on one side, and 1 on one side, a total of 6 <>

Happy thick chain] because you need to be sure to find defective products, so you need to consider the worst-case scenario, so it takes at least six times.

At first, put 4 on each side of the scale, and keep 4 left.

Situation 1: If the two sides are flat, then the bad one must be in the 4 that are left. Number the 4 balls as 1, 2, 3, 4

Take out 1 and 2 first, and weigh it, if it is flat, then it means that the bad ones are in 3 and 4. Then since 1 and 2 are good, then 1 and 3 are called, and if 1 and 3 are flat, then 4 is bad. If 1 and 3 are not equal, then it must be 3.

Because 1 is intact, 1 and 2 are the same weight).If 1 and 2 are not even, then 3 and 4 must be intact, weigh 1 and 3 again, if 1 and 3 are tied, then it is 2, if 1 and 3 are not even, then it is 1

Scenario 2: If the two sides are uneven, then group the two sides. The heavier ones are divided into 1, 2, 3, and 4, and the lighter ones are divided into a, b, c, and dThen they exchanged the scales, weighing 1,2,a and 3,4,b.

If 1, 2, a and 3, 4, b are drawn, then that is, 1, 2, 3, 4 and.

A and B are equal weights, which means that there are no bad balls in 1, 2, 3, and 4, that is, the bad balls are on the light side. (Because the bad ball appears in the light ball group!) So that is to say, the light one in c and d is bad, and then it is called c and d to get the bad ball, and the light one is.

If 1,2,A and 3,4,b are not even, then it depends on which side is heavier. Let's say it's 1,2,a weight. (This is interchangeable with 3, 4, and b.) Then weigh 1 and 2.

If 1 and 2 are drawn, then it means that B is bad, because 1 and 2 are equal weight, that is, there is no bad ball in 1,2 (which is also a heavy ball), and A is from the light ball group, and A cannot be heavier than the other balls. So why is it 1,2,A heavy, the reason is obvious, 3,4,b has bad balls in it, and bad balls are light! But 3 and 4 come from the heavy ball group, that is, there can't be a light ball in 3 and 4, (otherwise 1, 2, 3, 4 will be light at the beginning!)

So B is a bad ball, and it's also a light ball.

If 1 and 2 are not drawn, then one of the 1,2 must be a bad ball, and since 1,2 is from the heavy ball group, the heavy one is bad.

In the same way, if 3, 4, and b are the heavy teasing side, then the process of pushing the mountain is the same as the above.