There are 13 table tennis balls, one of which is defective, I don t know how heavy it is, and I use

Number the 13 balls ......13) Divided into three parts).

For the first time, the 1-5 ball is weighed with the 5-10 ball, and the two results (balance balance) (unbalanced).

i) If (balance balance) then the defective product is in (11-13), take 11 and 12 scales, if 11 is light, then take 13 and 11 names, if equal, then the defective product is 12, if the 11th ball is light, then the defective product is 11.

ii) If (unbalanced), then the defective product is in the 1-10 ball. If the weight of the 1-5 ball is light, take the 1-3 ball and the 11-13 ball, (a) if the 1-3 ball is light, the defective product is in the 1-3 ball and the defective product is relatively light, with the 1-2 ball weighing, the 1 ball is light, then the 1 ball is the defective product, if it is equal, the defective product is the 3 ball. (2) If the ball No. 1-3 is equal to the ball No. 11-13, it is said that the No. 10 ball is the same as the No. 10 ball.

If it is light, then No. 4 is defective, and if it is light, then No. 5 is a ball No. 5.

Weigh it out three times.

The first 5 5 scales, if it is heavy or light, it is on which side, or the same 2 situations The first way: find the heavy or light side and divide it into 3 parts: 2 2 1, two two scales, and there are two cases, the first one will find the remaining two phase scales, and the second is the remaining one.

The second is the one that remains!!

Please, give it to me!

The first time, put 6 table tennis balls on each side of the scale, if the balance does not react, the defective product is the one that is not weighed! If one of them is lighter, then the defective product is among the 6.

The second time, the 6 ping-pong balls mixed with the defective product are placed at each end of the scale (3 pieces each), and if one of them is lighter, the defective product is in the 3 pieces.

For the third time, put one of the three ping-pong balls mixed with defective products at each end of the scale, and leave one left! If the balance does not react, the defective product is the one that is not weighed! If one of them is lighter, then this is the defective product.

1, put 6 on one side first, if the same weight, then the rest of the unweighed one is defective.

2. If there is a light side, the defective product must be in those 6, and then divide these six into 2 parts, weigh 3 on one side, and the light side, the defective product must be there.

3, the remaining 3, only weigh two, if the two are the same weight, the one that is not weighed is defective, if one is light, the light one is defective.

I don't know the weight of the defective products The two methods upstairs are probably what my method is almost what is there Think about it again By the way, what is the balance with the game code It is easy to do without a trick.

There are 9 table tennis balls, one of which is defective (heavier or lighter than the others), there is a scale, but there is no weight, find out which defective product is the least how many times?

At least 3 times, set 9 balls for 1-9 balls. One. 1.

Weigh 123 V S456, if balanced, then 1-6 is **, defective in 789. 2.Weigh 7 vs 8, if the punch is balanced, then 9 is defective, if it is not balanced, then 3

Weigh 7 vs 1, if balanced, then 8 is defective, if unbalanced, then loose bucket 7 is defective. Two. 1.

Weigh 123 vs 456, if unbalanced. Assuming that 123 is heavier than 456, the defective product is in 123, and the defective product is heavier than **; Or the defective product is in 456, and the defective product is lighter than **. 7, 8, 9 for **.

2.Weigh 127 vs 345, if balanced, then 1, 2, 3, 4, 5, 7 are **, because the defective product is in 1-6, so 6 is defective, and it is lighter than the Zhengqi known product. If 127 is heavier than 345, the defective product is in 1,2 and heavier than **.

3.Weigh 1 vs 2, heavy as defective products. Back to II 2

If 127 is lighter than 345, then 3 is defective and heavier than **. Back to II 1If 123 vs 456 is unbalanced, and 456 is heavier than 123, the method is the same.

There are 9 table tennis balls, one of which is defective, heavier than the others, there is only one balance, it can only be weighed twice, how to find defective products?

Hello dear, I found out that there are 9 table tennis lifts for you, one of which is defective, and the defective product is lighter than **, but it can't be seen from the appearance. If there is only one balance, weigh it at least a few times to ensure that the defective products are found? Let's talk a little about how to weigh it.

Dial: If you weigh one by one, you may have to weigh it 9 times. In fact, both ends of the balance can be placed objects, if this slow balance, it means that the weight of both ends is equal, if not, the high end of the object is light.

According to this reason, first weigh out which part of a ping-pong ball is in Heng Heng Light, and in the part of the light, until you find out the defective product. This narrows the scope, that is, to achieve the purpose of reducing the number of weighings, if my answer is helpful to you, please give a like (rate in the lower left corner), look forward to your like, your efforts are very important to me, and your support is also the driving force for my progress. Finally, I wish you good health and a good mood!

Summary. 2) If there are 3 3 = 9 balls, you need to find the defective product at least 2 times: divide the 9, put one on each side of the scale, balance, and the remaining one has the defective product; Unbalanced, the defective product is on the side of the decline, and then the second measurement can be carried out according to the method of (1) above; If there is 1 more ball at this time, it will take at least 3 times to find the defective product;

Therefore, it is guaranteed to find defective products 3 times, so the number of balls may be 10

There is a box of table tennis balls, one of which is defective, and it is guaranteed to be weighed 3 times to find out this heavier table tennis ball. How many table tennis balls might be in this box.

Hello. There is a box of table tennis balls, one of which is defective, and the sail is weighed with a scale, which is guaranteed to be weighed 3 times to find out this heavier hole stuffy table tennis ball. This box of table tennis balls may have 10 najel bends.

According to the principle of balance balance, if there are 3 balls in the rubber grip (1), it takes at least 1 time to find out the defective product: divide the 3 into , put 1 on each side of the Zhongqing balance, balance, and the rest is defective, unbalanced, and the descending side is defective; If there is one more ball at this time, it will take at least 2 times to find the defective product;

2) If there are 3 3 = 9 balls, then the bad oak needs at least 2 times to find the defective product: divide the 9, put one on each side of the scale, balance, and the remaining part has the defective product; Unbalanced, the defective product is on the side of the decline, and then the second measurement can be carried out according to the method of (1) above; If there is 1 more ball at this time, it will take at least 3 times to find the second filial piety; Therefore, it is guaranteed to find the defective product 3 times, so the number of balls may be 10 missing.

Analysis] The balance is used to weigh the mass of the object, this question is not to weigh the mass of the object, but to use the balance to compare the size of the mass of the object, therefore, in the adjusted balance in the two plates respectively put the object, when the tray on which side rises, it means that the mass of the object in the tray is the same as the little brother oak

Summary. Six items, one of which is defective, that is, the weight is different, such a probability is one in six, when weighing, the first item to be placed on the scale, the remaining five in turn. The worst rule is to set the last time a defective product can be found, then you need to weigh it five times.

After subtracting one of the standard samples from the six items, there are five left, which is to be weighed five times. 6-1=5.

There are 6 table tennis balls, and there is a defective product with different weights in the middle of the period, and you will be sure to find the defective product if you weigh it at least a few times with the balance.

5 times. Steps.

Writing out the answer for you, please wait.

Steps are required. Use the worst-case principle.

Steps are required. Six items, one of which is defective, that is, the weight is different, such a probability is one in six, when weighing, the first item to be placed on the scale, the remaining five in turn. The worst rule is to set the last time a defective product can be found, then you need to weigh it five times.

After subtracting one of the standard samples from the six items, there are five left, which is to be weighed five times. 6-1=5.

This question is answered 3 times.

The first weighing: divide the balls into three groups and number them, the first group: 1, 2, 3, 4;Group 2:

5,6,7,8；Third group: 9, 10, 11, 12, place the first and second groups on either side of the scale. Two scenarios arise:

for balance or imbalance. Depending on the situation, start the second weighing.

1.If the balance is balanced, the second time: it means that the abnormal ball is in the third group, the first two groups of balls are normal, and then use 9, 10, 11 and 1, 2, 3 to weigh it, and the third time to weigh it:

If it is balanced, then use a ball and 12 to weigh it, and get that 12 is heavy or light, unbalanced, if it is heavy, take 9 and 10 and put it on the scale, you can 9 or 10 who is heavier; Balance is 11 weights.

2.If it is not balanced, it means that the four balls in the third group must be the standard weight, continue to test:

The second time: redivide the first group and the second group, take down the 3 balls of one group, take the other group of 3 balls, and then take the balls of the third group 3 to form a new group of balls, that is, form a group of groups, put them on the balance and weigh them, in this way, according to the situation that occurs, the third time:

If the balance is balanced, it means that the weight of the new two groups of balls is standard, the problem is that there is no balance inside, called, the unbalance is related to the first balance tilt situation to know whether it is light or heavy, the balance is 4 for the defective product, if it is not balanced, the direction of the imbalance has not changed, then the reversed balls are standard, weigh 1 and a standard ball, if the balance then 5 is defective,

If it is not balanced, the direction of the imbalance has changed, then there is a defective product among the 3 balls of the inversion, which is said to be defective if it is balanced, and if it is unbalanced, it is known that the defective product is known according to the result of the second tilt.

Up to 27.

9＝9 ……9 derives a heavy set of 9;

3＝3……3 derives a heavy set of 3;

1＝1……1 to find a defective product.

If there are less than 27, you can use this method to distinguish a defective product 3 times.