Use a scale to find out which number segments the defective product is 6 times

The problem of finding defective products is regular.

Generally, it is divided into three parts: a, a and b. b may be equal to or may be equal to a+1 or a-1, depending on the total.

Put two A at both ends of the balance, if the balance is balanced, the defective product is in the B, if the balance is unbalanced, then find out which part of the defective product is according to the difference between the defective product and **.

Once you've found it, continue to divide it into three parts.

This way you can eliminate two-thirds at a time, which is the fastest.

1 to 3 in one go.

4-9 pcs., twice needed.

10-27 pcs. It takes 3 times.

28-81 4 times.

82-243 5 times.

244-729 6 times.

The problem of finding defective products is regular.

Generally, it is divided into three parts: a, a and b. b may be equal to or may be equal to a+1 or a-1, depending on the total.

Put two A at both ends of the scale, if the balance is balanced, the defective product is in the B loss orange, if the balance is unbalanced, then according to the difference between the defective product and the ** carefully stare at the fault to find out which part of the defective product.

Once you've found it, continue to divide it into three parts.

This way you can eliminate two-thirds at a time, which is the fastest.

1 to 3 pieces, one time you can wide the air and get it done.

4-9 pcs., twice needed.

10-27 pcs. It takes 3 times.

28-81 4 times.

82-243 5 times.

244-729 6 times.

4 are divided into 1 group, a total of 3 groups and 1 other.

Take out 2 sets of scales.

1. Ping, then in the remaining 5, take 3 out of 5, and then take out 3 good scales.

a, flat, then in the remaining 2, and then just take a number and compare it with one of the 2 (not necessarily know the weight).

b. If it is uneven, it is in 3, and you know the weight (because the corresponding 3 are good), and then take 2 of the 3, and you can do it.

2. If it is uneven, then in the 8, remember the weight of the scale, switch one of each of the two groups, and then replace the remaining 3 in one group with a good scale.

C, flat, then in the replacement of the 3, then did not change the 3 is good, because of the memory of the weight, you know the weight of the defective product.

Then weigh the remaining 3 to know the defective product.

d. Uneven, remember the light and heavy sides.

e. If there is no change, the defective product is not replaced by 3 good ones, and it is the remaining 3, and you know the weight, and then weigh it.

F. If the weight of the side has changed, then look at the side of the 3 good ones, the opposite side is a defective product, and know the weight, and then weigh it.

Only the first one does not know the severity of the defective product, and the rest can be known.

From the meaning of the title, I think that the quality of defective products is not the same as that of **, assuming that the defective products are lower than the ** quality, then.

I weigh it on a scale for the first time, and I put 450 parts on the left and right sides respectively, and the defective product is on the light side.

For the second time, divide the 450 parts containing the defective product into two parts, and place 225 parts on the left and right sides of the balance.

Weigh the third time, and put 112 parts on the left and right sides of the balance.

Weigh the fourth time, and put 56 parts on the left and right sides.

Weigh the fifth time, and put 28 parts on the left and right sides.

Weigh the sixth time, and put 14 segments on the left and right sides respectively.

Weigh the seventh time, and put 7 parts on the left and right sides.

Weigh the eighth time, and put 3 parts on the left and right sides.

Weigh the ninth time, and put 1 grip sock part on the left and right sides.

Therefore, I think that by weighing at least nine times a fierce reed on a scale, this defective product will be found.

There is still a bit of a problem with the topic, you have to know whether the defective product is lighter or heavier than the first.

Let's assume that the defective product is light.

For the first time, divide into three portions of 4 4 and 5.

Put two 4s on the scale, if balanced, the defective product is in the 5, if it is not flat, then the light one has the defective product.

The second time if it's in 4. There are two on one side, and there is a defective product in the light one.

If it's in 5. Two on one side, when balanced, the remaining one is defective. There is a defective product in the unbalanced light.

The third time in two directly weighed once and it was solved.

Divided into two parts, defective products are generally light in this kind of question, so put 6 in each plate, 12, leave one next to it, good character, you will know it right away, bad character, just look at the lighter over there, and then divide the plate into 2 points, that is, 3 on one side, and then find the light one, and then take out one from the plate, one on each side, good character, and find out.

At least 3 times! 1 time: 6 on the left, 6 on the right, light and defective.

2 times: Then weigh 6 defective qualities, put 3 on the left, 3 on the right, and the light ones are defective. 3 times:

Then weigh the 3 defective products, put 1 on the left, put 1 on the right, and the light one is defective; If it is the same weight, the remaining one is defective. There is also a one by one.

Since the balance can provide three pieces of information each time, it is heavy on the left, heavy on the right, and the same weight.

Then the information that can be provided each time is 3. With reasonable use, the information provided m times is 3 m.

The information of each individual of n products is 1, that is, a total of n pieces of information are required. However, there is also information in the question raised by the subject that is whether the defective product is lighter or heavier than others, because this information actually brings the light and heavy attributes of each product, so it is equivalent to multiplying the entire product quantity by 2. The final amount of information is 2n.

Then the requirement is that the amount of information provided is at least higher than the amount of information required, i.e. 3 m 2n.

In this problem, m is 6, then n is an integer not more than 3 6 2, which is [729 2], that is, 364.

Let's take a simple example.

If we know that the defective product is light for 3 products, then we only need to weigh it once, put one on one side, and the light one is the one when it is heavy. Equal words are the rest.

But we don't know that the defective product is light, 1 time is not enough, because 1 side of one unless equal we know the rest is, one light and one heavy, we can't judge which side is defective, need a second time.

Then 2 times is definitely OK, and the continuation problem is that when you don't know the severity of the defective product for 2 times, you can have a few at most, according to our algorithm, it should not exceed 9 2 that is, 4.

But at this time, it is necessary to design the scale, specifically to put one (number) on the other side, and pick out one (such as 1) and the unweighed random one (such as 3) if it is not equal, and the defective product is still 1, and the equal is 2. If it is equal, it is still 13 scales, and at this time, the unequal defective product is 3, and the equal defective product is 4.

It can also be noted that 5 2 times is not possible, because the first time, if one on one side, the remaining 3 unknown information products cannot be weighed once. If there are 2 on one side, in the case of not equal, there are two tasks at this time, to judge which side of the defective product is and which one corresponds to the defective product, which is the amount of information of 2x2=4, and one weighing is not enough.

To sum up, this amount of information can be calculated at most, but the specific design of the name is more complicated, and if you are interested, you can find 13 designs that are 3 times.

Intuition is 3 6 (three to the sixth power). That is, 729 pcs.

Think for yourself.

To find defective products by the method of balance weighing, the measured object should be divided into (3) parts, of which at least (2) copies are the same quantity, and the difference between the number of two parts per l cannot exceed (1).

For example, if you find 1 of the 3 items, you will divide them into three parts, 1 for each part.

Find 1 of the 8 items, divide them into three parts, and put the two copies of the bright change on the balance to change the key spring and weigh it.