12 glass balls, one of which weighs differently than the others, is distinguished by a balance in th

Question 1: Is it lighter or heavier than the others?

I'm going to assume light: divide 12 glass balls into 2 parts (5 parts 1) and weigh them on a scale. If the balance is balanced, then the light glass ball is in the unweighed 2, and then the unweighed 2, and the one with the height of the balance 2 is the one on the side.

If it is not balanced, divide the 5 sides of the balance 2 into 2 parts (2 1 parts)) and weigh them with the balance. If the scales are balanced, then the light glass ball is in the unweighed one. If it is not balanced, weigh the two sides of the balance that are high, and the one that is higher is lighter.

If it's heavy, then it's the low one.

Please forgive me for the inappropriate language.

Take out any 6 first, divide into 2 teams, 3 on the left side of the scale, 3 on the right, and see the inclination:

If it is equal, then take another 3 weighings, that is, 1 left, 1 right, if equal, then the weight of the last one is different.

If it is not drawn, then arbitrarily take out 1 team and continue to weigh as before.

Divide the 12 balls into 3 groups of 4 each.

For the first time; Let's start with two groups. to find out which one is in the group.

Second; Divide the balls in the identified group into 2 groups of 2 each. Again, you can determine which group the ball is in.

the third time; I weigh two of those sets separately and find the different balls.

You figured out whether the special ball is lighter than the normal ball or a rewrite. Eldest brother!

Personally, I don't think there is a solution without knowing the severity, see if there is a high energy that can be solved!

A: It takes at least 2 times.

Method: 1st time: Divide into 3 portions, 3 portions each, put 2 portions on both sides of the scale, if the same, then in the third part, if not, in the lighter part.

The second time: Divide into 3 parts, 1 for each part, put 2 parts on both sides of the scale, if the same, the third, if not, the lighter one.

Summary. Divide 26 into three parts, which are

Weigh 9 and 9 respectively, if equal, then the weight is in 8, so that the 8 are divided into three parts, weigh with the balance, if equal, then the remaining 2 weigh again, you can find the ball you are looking for; If you don't wait, divide the heavy 3 into 2 groups, and judge it.

If it is 9 and 9, then divide the heavy part into any two groups, you can judge which group the heavy ball is in, and then divide it, take any two of the three groups, and judge it.

A: Weigh at least 3 times to ensure that you find the orb.

There are 26 glass balls, one of which is a bit heavier than the other 25. Weigh at least a few times with a balance to ensure that you can find the orb.

Dear, in. 3 times.

Process. I'm writing.

Well, good. Trouble can you hurry up.

Right away. Because I'm in a bit of a hurry.

Divide 26 into three parts, which areWeigh 9 and 9 respectively, if equal, then the weight is in 8, so that the 8 are divided into three parts, weigh with the balance, if the branch is equal, then the remaining 2 weigh again, you can find the ball you are looking for; If you don't wait, divide the heavy 3 into 2 groups, and judge it. If it is 9 and 9 are not equal, then divide the heavy part into any two of the fierce positive call group, you can judge which group the heavy ball is in, and then divide the book into two, take two of the three groups, and judge the answer:

Weigh it at least 3 times to ensure that the orb is found.

Look at it. It's not something that can be made clear in a sentence or two.

Dear, no, look down.

First, the 12 balls are divided into three groups: 4a, 4b, and 4c, with four in each group

Step 1: Weigh 4a and 4b first, and there will be two situations:

In the first case: equal, then it can be judged that the ball you are looking for is in 4c, and 4a and 4b are normal balls;

Step 2: Divide 4c into four 1c, weigh any two 1c, and you can get two results:

1. Equal, then the third step here is: take off the 1c on either side and put the third 1c, you will get two answers:

1. If it is equal, then the fourth 1c is the ball to be found;

2. If it does not wait, the third 1c is the ball you are looking for.

1. If it is equal, then the 1c taken is the ball to be found;

2. If it does not wait, then the remaining 1c on the balance is the one found.

The second case: it is not equal, and it is assumed that 4a is light and 4b is heavy, and 4c is a normal ball. Now.

4a is divided into two 2a; Divide 4b into 3b and 1b;

Step 2: Put 4C 1B on the left side of the balance and 3B 2A on the right side of the balance, you can get the following two situations:

1. Equal, then the ball found is in the remaining 2A and is a light ball, the third step here is to divide 2A into two 1A, and then place it on both sides of the scale, the light is the ball found.

2. There are two situations:

1. When the left is light and the right is heavy, the ball found is in 3b and is a heavy ball, and the next third step here is: divide 3b into three 1b, take any two of the 1b to weigh, you can get:

1. If it is equal, then the remaining 1b is the ball to be found;

2. If you don't wait, the heavy 1b is the ball you are looking for.

2. When the left is heavy and the right is light, the ball you are looking for is in 2A and is a light ball or 1B and is a heavy ball, the next third step is: divide 2A into two 1A, put 1A and 1B on the left side of the scale, and put 2C on the right side, then you can get:

1. If it is equal, then the remaining 1a is the ball found;

2. If it is not equal, there are two situations:

1. When the left is light and the right is heavy, 1a is the ball found;

2. When the left is heavy and the right is light, 1b is the ball you are looking for.

If the ball is lighter than the others, you can't find the one you weigh on the light side.

Divide the 12 balls into three sets of ABC, each group of 4, and put the two groups of AB on the scale for the first time, either flat or uneven

If it is drawn, the ball to be found is in group C, and then group C is divided into C1C2 and C3C4. Compare the two balls in group A with C1C2, if the ball is drawn, the ball is in C3C4, and then use the ball in group A and C3, if it is drawn, the ball to be found is C4, and if it is not drawn, the ball to be found is C3

If it is uneven, it is in C1C2. Comparing the ball in group A with C1, if the ball is looking for C2, if it is not even, it is C1 (the weight is directly seen).

If it's not a draw, the ball is in group AB. Then compare the ball in group A with group C, if it is not drawn, it will be in group A, and if it is drawn, it will be in group B. Then you can find out by following the method of .

This question.

I think it should be six six points first.

If it's heavy on the left, it's three three.

If the left side is still sinking, put one aside (a ball, a scale, a ball), if it's the same, it's the ball, and if one side is toppling, it's the ball! There is no need to divide ABCD

I came up with it myself in 3 minutes.

Number the 12 balls as .12, and divided into three groups: group A; Group B ; Group C

The first time: put the two groups A and B on both sides of the scale, if they are the same weight, the abnormal ball is in group C, otherwise it is in groups A and B;

Discuss separately: (1) the case of the abnormal ball in group C (i.e., A and B are the same weight), then.

The second time: three balls from group A are selected as standard balls to be placed on the left side of the scale, three balls from group C are placed on the right side of the scale, and if balanced, the abnormal ball is number 12; unbalanced, the anomalous ball is one of them, and it is known whether the anomalous ball is heavier or lighter than the standard ball;

The third ball is placed on the right side of the scale, and if it is balanced, the abnormal ball is 11; If there is an imbalance, the abnormal ball can be picked out according to the weight comparison of the above abnormal ball with the standard ball.

2) If the abnormal ball is in groups A and B (i.e., A and B are not the same weight), then group C is the standard ball, and A may be heavier than B.

The second time: the left side of the balance to put the ball, the right side of the ball, if the balance means that the abnormal ball must be numbered, and the abnormal ball must be lighter than the standard ball, the last time the weight of the ball can be picked out; If it is unbalanced (it must be on the left), it means that the abnormal ball is in group A, and the abnormal ball must be heavier than the standard ball, then any 2 balls of the last comparison ball can be picked out.

First, divide the 12 balls into three equal portions of four.

Take out two of them and place them on both sides of the scale (for the first time).

Scenario 1: Balance balance.

Then the eight balls that are weighed are normal, and the special ones are in the four.

Take out three of the remaining four balls and set them aside and put three normal balls on the other side (2nd time) Situation 1-1: Balance balance.

The special one is the one that remains. Take out any one and the special one from the normal one and put them on both sides of the scale, that is, you know whether the special ball is light or heavy. the third time;

Scenario 1-2: The balance is unbalanced.

The special balls are in the three above the scale, and you know whether they are heavy or light.

Take two out of the remaining three and weigh them. the third time;

Case 1-2-1 balance.

The special ball is the one that is left, and I know how heavy it is.

Situation 1-2-2 The balance is unbalanced.

Based on the light and heavy characteristics of the special ball known above, you will know which one is the special ball.

Scenario 2: The balance is out of balance.

The special balls are inside the eight that are placed on the scale.

The four balls on the heavier side are counted as a1a2a3a4 and the light ones are counted as b1b2b3b4.

The remainder is determined to be four normal, denoted as C.

Set a1b2b3b4 aside, b1 and the three normal c-balls aside. (Second) Scenario 2-1: The balance is balanced.

The special ball is in a2a3a4, and you know that the special ball is heavier.

Weigh a2a3 to know which of the three is special, and you know the severity. the third time;

Scenario 2-2: The balance is unbalanced, and the A1 side is heavier.

Special balls are between A1 and B1.

Just take a normal scale, and you will know which one is special, and you will know the severity. (3rd) Situation 2-3: The balance is unbalanced, and the B1 side is heavier.

The special ball is in the middle of b2b3b4, and you know that the special ball is lighter.

Weigh B2B3 to know which one is special, and you know how heavy it is. the third time;

First divide into three piles of 4 4 4 4, and choose any two piles to weigh.

If it is equal, take the remaining pile, divide it into 112, and take 11 weighings; Then take the remaining 2 and take one and the 1 just now, equal is the rest of the difference, unequal is the new one.

Is the quality different whether it is lighter or heavier?

That's right.

Solution: Divide the 9 glass balls into three groups, the first time: weigh two of them, if the balance is balanced, the lighter one is in the remaining group, and the lighter glass ball can be found again;

If the balance is unbalanced, the lighter ball is at the rising end of the balance tray;

The second time: divide the lighter group into three groups, weigh two of them, that is, the lighter glass ball can be found;

So it only takes 2 times to find that lighter orb.

Find two balls and place them at opposite ends of the scale, and if the mass is not equal, congratulations on your luck. If the two balls are of equal mass, take out any of them.

Divide the remaining 8 balls into two equal groups, place them in the balance, and take out the light group.

Divide the 4 balls into two groups and take out the light ones.

If the mass is the same, then the other is light, and the mass is not the same, then this is the lighter.

Supply and demand can be found out in 4 steps! I don't know if it's the right answer.

Common sense understanding:

The so-called weighing once refers to putting the items that need to be weighed and then observing the equilibrium state of the balance, so as to draw conclusions. This is a one-time weighing.

You put the ball on both sides of the scale, and you have to observe the result every time you put it, so it's actually a weighing.

This kind of question tests the student's logical thinking and problem-solving ability, and there is no need for the landlord to dwell on these places.

Two times is enough, the key is the process of analysis, and the method of releasing the ball.

Let's say these 9 different balls are heavier than the others.

First: five on each side of the scale. The balls on both sides are denoted as sets A, BCount the five balls in the sinking cavity as set B (because different balls are heavier than the others, then there are different balls in B).

The second time: put the five balls of A on both sides of the scale, and mark them as C and D. At this time, the C side sinks.

Then take a ball from B (there are 4 balls in B) and put it on the D side, if the D side sinks, the ball placed in the D side is a different ball; If the balance is over, there is no difference in the ball, and the ball is released. First take a ball in B (there are 3 balls left in B) and put it on the C side, then take a ball in B (there are 2 balls in B) and put it on the D side, if the C side sinks, the balls placed on the C side are different balls; If the D side sinks, the ball placed on the D side is a different ball; If the two sides of the CD are balanced, there is no difference in the ball, and the ball is released. At this time, there are two balls, which are placed on both sides of the balance CD, and the analysis method is the same as in the previous step, if the C side sinks, the one placed on the C side is a different ball; If the D side sinks, the ball is placed on the D side.

In the same way, when different balls are lighter than other balls, it is the same as this measurement method.

It's just that the results of the analysis are reversed. The sinking side is no different ball.