Weigh 12 table tennis balls 3 times and find the lightest one

Let's number it first, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

1. The first time: weigh 4 on each side, left: 1, 2, 3, 4 right: 5, 6, 7, 8 and remaining: 9, 10, 11, 12

2. If it is the same weight, it means that there is a light one of the remaining 4;

2nd time: 2 more on one side, left: 9,10 right: 11,12 (this time it is called the remaining ball from last time, this time it is 1-8, 8 balls that are as light.) ）

Choose the lightest two, one on one side (there must be one lightest), plus four balls in 1-8 on one side (these 8 are the same weight, not in the way).

The third time: the lightest side is the lightest.

3. If it is not the same weight, the lightest of the 4 lightest ones, and the remaining 9-12 are the same heavy. For example, a group of 1-4 is lighter than a group of 5-8. The second time the left is 1,2+9,10 (or 11,12), the right is 3,4+11,12 (or 9,10), and this time it is 5-8

The lightest ball is the lightest on the lighter side, if it is 1,2

then the third side 1+5,6 (or 7,8); Side: 2+7,8 (or 5,6), the lightest one is the lightest ball.

Complete. It may be troublesome to look at it this way, but it is recommended that you take a pen and draw it to understand.

There is no solution to this question called 3 times, because no matter which side is heavier or lighter, it does not directly indicate whether the mass of the ball of different mass is lighter or heavier than that of the standard ball. actually quoted the mathematical model, and said that it was Microsoft's problem, which is really not ashamed.

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.

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.

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.

Number the 10 balls from 1 to 10, where the number is Group A, Group B is Group C.

For the first time, Group A and Group B are placed on the left and right sides of the scale, and the result will be two scenarios.

1. Balance. At this point, it can be said that there is no problem with all six balls. The problem is with the remaining four balls.

Take the right side of Group B and replace it with three balls from Group C. At this time, if it is balanced, there is no problem with the three balls, so it is certain that the number 10 ball is the different ball.

If it is unbalanced, it means that there are problem balls in Group C. The number 10 ball is a good ball. Make a note of whether Group C is heavy or light.

Leave the 7 on the right and the 8 on the other side of the scale, then if balanced, 9 is the problem ball, unbalanced, if the balance is the same as last time, then the 7th is the problem ball, if the weight is the opposite of the last time, the 8 is the problem ball.

2. Unbalanced, indicating that the balls are all good, note whether they are heavy or light over there.

Take the right group B and replace it with three good balls in group C, if balanced, the problem is in group B, and then put No. 4 and No. 5 on the left and right sides respectively, if balanced, No. 6 is the problem ball, if it is still unbalanced, look at the left side to determine who is the problem ball.

If it's not balanced, it's certain that there's something wrong with the sign. In the same way, you can determine that there is a problem with the ball.

Number 12 table tennis balls, 1-4 as a group, 5-8 as a group, and 9-12 as a group.

The first step: No. 1-4 and No. 5-8 are weighed, and there are two situations: first, the balance is balanced; Second, the balance is unbalanced.

If the balance is balanced, the special ball is in No. 9-12.

Step 2: Take the ball and weigh the ball, and there are two situations: the balance is balanced; The balance is unbalanced.

If the balance is balanced, the special ball is number 12.

Step 3: The 12th and the 1st are weighed together, and you can know whether the 12th is light or heavy.

If the balance is unbalanced, the special ball is in the middle of the sign and the weight of the special ball is known.

Step 3: Weigh No. 9 and No. 10, balance is No. 11 special, and unbalance is the ball in No. 9 and No. 10 that meets the severity of the situation.

If the balance is unbalanced, the special ball is in No. 1-8.

If the 1-4 is heavier.

Step 2: Take a group and weigh it as a group, and there are three situations: the balance is heavier and heavier.

If the balance is balanced, the special ball is the middle one and is heavier.

Step 3: Weigh 2 and 3, balance is special for No. 4, and heavy is special for unbalanced.

If it is heavier, the special ball is 1 or 5

Step 3: Weigh 1 and 12, the balance is 5 special and lighter, and the unbalanced is 1 special and heavier.

If it is heavier, the special ball is the middle one and is lighter.

Step 3: Weigh 6 and 7, balance is 8 special, unbalanced is light special.

If No. 1-4 is lighter. Same method as above.

The first step is to number the 12 balls from 1-12

The second step is to place balls 1-4 on the left side of the scale and balls 4-8 on the right side of the scale (this is the first time to weigh).

There are two scenarios at this time:

1。The balance balance, it means that the 1-8 balls are all standard balls, and the problematic balls are in the remaining 4 balls of 9-12, then take the ratio of the three balls of 9, 10, and 11 to any three of the 1-8 balls (the second time), and there are 3 situations:

1) If the balance is balanced, the remaining ball is different, and the weight of the remaining ball can be known by comparing it with a standard ball (the third scale).

2) 9, 10, 11 The three balls are heavier, then you can know that the balls are heavier, and among the three balls, you can take two of them to compare which of the three balls is different (the third time).

3) 9, 10, 11 The three balls are lighter, then you can know that the different balls are lighter, and among the three balls, you can compare two of them to know which of the three balls is different (the third ball).

2。The balance is unbalanced. This means that the remaining balls 9-12 are standard balls.

Assuming that 1-4 is heavier than 5-8 (anyway, there is one side heavier, as for which side does not matter), then take 2 balls from the heavier side of the scale, such as 1-2, then take 2 balls from the lighter side such as 5-6, and then take a standard ball such as 9 (a total of 5 balls), put them on the left side of the scale, and then put them on the right side of the scale: the remaining 3 standard balls are 10-12, 1 ball on the heavier side of the balance is like 3, and 1 ball on the lighter side of the balance is 7 (5 balls in total). This is the second time that there are two more cases:

1) The balance is balanced, then the ball is not the same in the 4th and 8th that is not weighed this time, then the 4th and 1 standard ball ratio (the third time), if balanced, then the 8 is different, and light; If it is not balanced, then 4 is not the same and heavy.

2) If the balance is unbalanced, there are 2 more cases:

The first is that the left side is heavy, and then there are two possibilities, either the left 1-2 ball is heavy, or the right 7 ball is light, then as long as the 1 ball and the 2 ball are weighed again (the third time), if balanced, the 7 ball is not the same ball, and it is light; If it is not balanced, the heavier ball in No. 1 and No. 2 is not the same ball and is heavier.

The first is that the left side is light, and then there are two possibilities, either the left 5-6 ball is light, or the right side of the 3 ball is heavy, then as long as the 5 ball and the 6 ball are weighed again (the third time), if balanced, the 3 ball is not the same ball, and the weight; If it is not balanced, the lighter ball in the number 5 and number 6 is not the same ball and is on the lighter side.

What is the difference between exact and subjective operations?

It's all hypothetical calculations!

Three bais are sufficient.

Divide into 4 du4 piles, 3 in each pile.

Take two piles and weigh it, zhi

1.Balance DAO

Then the false one is in the other 2 piles and one inner content, the two piles are called, the balance is the rule, the remaining one is false, and the unbalanced is the case.

There are false in the light, take two scales, balance the other, the remaining one is false, and the unbalanced light one is false;

2。Imbalance.

Then there is a false in the light.

Take any two scales, balance, the remaining one is false, and the unbalanced one is light.