30 balls, one ball is lighter than the others, ask for 3 times to weigh the ball, how to weigh it? T

The first time it is divided into 3 servings.

10 10 10 10.

Take 2 portions and place them on the scales.

1) If you balance it, take out another 10 pieces.

If it's not even, take out the light one.

2) Take out one of the 10 and divide the remaining 9 into 3 parts.

Take 2 servings, and if you balance it, take out 3 other servings.

If it's not even, take out the light one.

3) Take out 2 of the 3 out of one, take the other if it is balanced, and take out the light one if it is uneven.

There should be a scale, right!!

Divide the 30 balls into 3 groups of 10 balls for a total of 3 sets.

Take the balance to any two sets of scales (the first time).

Well, if the balance is not tilted, it means that the light ball is not in it (if it is tilted, the inclined group) take the group with the "light ball" and divide it into 4 groups, divide it into 3 3 3 1, put any two groups in it and weigh it, and the one with 1 is left, and it is useless to weigh it.

Or if there is no tilt, the light ball is not in the scale, what should I do? (The second time) is called a group with light balls.

There is still no inclination, that means that your character is too poor, it means that you can't weigh it 3 times, and it means that your question has a mistake should be 27 balls to be able to weigh 3 times.

Upstairs, you didn't notice that you still have a ball that is not weighed, if the balance is the light one?

Your answer is good, is the wrong question!!

The answer is D. There is at least one light ball in + than + heavier and , and there is at least one light ball in + than + lighter and at least one light ball in +, and the number of the two light balls is known to be as heavy as +.

Solution: Heavier than +, and at least one light ball, + lighter than +, and at least one light ball, + and + as heavy as the number of the two light balls is .

So the answer is:

Comment: This question examines the nature of the equation:Equation property 1, the result of adding (or subtracting) the same number (or subtract) on both sides of the equation is still equal;

Equation property 2, where both sides of the equation are multiplied (or divided) by the same number (the divisor is not 0) and the result is still equal.

Choose d because + is heavier than +, so there is at least one light ball in and , because + is lighter than +, so and at least one light ball, because + is as heavy as +, you can know that the number of the two light balls is

(1) From the previous two times, we can know that there is a light one, and there is also a light one.

2) From the third step, three of the four balls appear, and the two sides are equal, indicating that one is light on each side of the scale, so there is only a light sign on the right, and it is immediately known that the left side is light (because it is impossible to be light) and choose D.

4 and 5 reasons, 1 and 2 are heavier than 3 and 4, that is, one of the 3 and 4 balls is light, and again, a reasoning leads to the answer. Thank you for adopting!!

Summary. Finally, if the bad ball is in 4, divide the 4 into two groups, weigh the 2 balls in each group, if equal, the bad ball is in the other group, take one from the other group and the one of this group, if equal, the one that is not taken in the other group is the bad ball; If not, the ball from the other group is bad. Weigh 2 balls in each group, if unequal, then the bad ball is in the two, take any one from the other group and any one in this group, if equal, the other ball in this group is a bad ball, if not equal, then this one in this group is a bad ball.

30 balls, one of which I don't know how heavy, how to determine which ball is different, and how many times does it take?

Hello, I am Black Black Duck I am happy to serve you, I have received your question, I will reply in time after reading, please be patient

Please reply as soon as possible.

Hello, at least 4 times I can find a ball that I don't know how heavy.

First divided into 3 groups, each group of 10 years old model celebration balls to take two groups of AB weighing, if equal, then the bad ball cares about holding another group of C; If it is not equal, then take one of a A and the third group C and then weigh the tassel, if equal, the bad ball is in group B; If they are not equal, the bad ball is in Group A.

Secondly, Rana blind will have a group of 10 balls with bad balls divided into 3 groups of 3 groups of 3 groups, and the remaining one will take two groups of EF to weigh, if equal, the bad ball will be in the remaining 4; If it is not equal, then take one E and the third group G and re-weigh it, if it is equal, the bad ball is in group F; If not, the bad ball is in Group E. Weighing at least twice can determine which 3 or 4 balls are in the bad repulsion.

Finally, if the bad ball is in the 4, the branch sells the 4 into two groups, weighs the 2 balls in each group, if equal, the bad ball is in the other group, takes one from the other group and the one of this group, such as equal, and the one in the other group does not take the bad ball; If not, the ball from the other group is bad. Weigh the 2 balls in each group, if they are unequal, then the bad ball is in these two, take any one from the other group and any one in this group, if equal, the first to carry the other ball in this group is the bad ball, if not equal, then this one in this group is the bad ball.

If the bad ball is in the trace of 3, take any two weighing, if equal, the remaining one is the bad ball; If it is not equal, then take any one and the third one, if equal, then the one that is not taken is a bad ball, if it is not equal, then take the posture to know that this is a bad ball.

Hope it helps! Have a great day! If you find it helpful, please move your fingers and give it a thumbs up! Thank you

How many times? Correct answer.

4 times in total.

How to divide each time, can you break it down?

Look at the steps I posted above.

It was issued in four times.

One step at a time.

Step 1 4 to 4

First, the sky is flat.

The second step is to leave 4 on one side of 3 on the other side of the three scales good 1, flat The last remaining one and good scales can be seen light and heavy 2, uneven You can know that in these 3, it is light or heavy, take 2 of the 3 to the scale, flat, that is, the remainder, uneven, just now you can judge which one is now.

Second, the uneven sky remembers the severity.

The second step is to remove 3 on the left and about and turn one on the left to supplement 3 normal 4 to 4 and then scale 1, flat Take 2 pairs of scales for the 3 that are taken off, which is the rest, uneven, just now you can judge which one is now.

2, uneven a, the same as just now on the right of the 3 that did not move to take 2 to the scale flat, is the rest, uneven, just now you can judge which is now. B, and just the opposite is the two of the left and right are different, take one of them and compare it to know which one, and the relationship between the severity is also known.

Number the twelve balls as 1 12.

Place 1 4 on the left side of the scale and 5 8 on the right.

There are three outcomes:

One. Balance. Illustrating the problem is 9 12.

Put 1 3 on the left and 9 11 on the right.

There are three outcomes:

1.Balance. Explanation No. 12 is problematic.

Put number 1 on the left and number 12 on the right.

If the left is heavy, the number 12 is lighter, and the right weight is the number 12. It is impossible to balance.

2.Left weight. Illustration 9 11 has a ball that is light.

Put number 9 on the left and number 10 on the right.

The left weight is 10 lighter, the right weight is 9 lighter, and the balance is 11 lighter.

3.Right weight. Note 9 11 has a ball that is heavy.

Put number 9 on the left and number 10 on the right.

The left weight is 9 heavy, the right weight is 10 heavy, and the balance is 11 heavy.

Two. Left weight. The explanation of the problem is 1 8.

Put 1, 5 7 on the left and 8 11 on the right.

There are three outcomes:

1.Balance. Note 2 4 has a ball that is heavy.

Put number 2 on the left and number 3 on the right.

The left weight is 2 heavy, the right weight is 3 heavy, and the balance is 4 heavy.

2.Left weight. Indicate that No. 1 is heavy, or No. 8 is light.

Number 1 is placed on the left and number 2 is placed on the right.

The left weight is heavier than No. 1, and the balance is lighter than No. 8. It is impossible to weigh right.

3.Right weight. Description 5 No. 7 has a ball light.

Put number 5 on the left and number 6 on the right.

If the left weight is 6 lighter, the right weight is 5 lighter, and the balance is 7 lighter.

Three. Right weight. The explanation of the problem is 1 8.

Put 1, 5 7 on the left and 8 11 on the right.

There are three outcomes:

1.Balance. Illustration 2 4 has a ball that is light.

Put number 2 on the left and number 3 on the right.

If the left weight is lighter, the number 3 is lighter, the right weight is lighter than the number 1, and the balance is lighter than the number 4.

2.Right weight. Indicate that No. 1 is light, or No. 8 is heavy.

Number 1 is placed on the left and number 2 is placed on the right.

The left weight is lighter than No. 1, and the balance is heavier than No. 8. It is impossible to weigh right.

3.Left weight. Note 5 No. 7 has a ball that weighs.

Put number 5 on the left and number 6 on the right.

The left weight is 5 heavy, the right weight is 6 heavy, and the balance is 7 heavy.

In the first step, put five on the left and right to balance the scales. Getting a different ball must be in the remaining two. In the second part, I had already weighed one of the 10 balls as a standard to measure the weight of the other two.

It can make the balance of the exclusion, so that you can find the one with different weights just by weighing it twice.

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.

4 on the left, 4 on the right, 4 on the left

These are three groups. Weigh it once to determine which group it is in.

This group of 22 weighs was selected.

And then you know it all at once.

Compare the 12 balls three times to find an anomaly.

The 12 balls numbered 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, were compared for the first time.

1,2,3,4 and 5,6,7,8 were compared.

There are two scenarios:

1) If the weight is equal, it can be judged that the anomaly is in the four balls of 9, 10, 11, and 12.

A second weighing can be carried out at this point.

Comparing 9,10 and 1,2, there are two possibilities here.

1) The weight is equal, and the abnormality is in 11,12, at this time the third weighing can be carried out 1,11 comparison, weight.

If it is equal, it is anomaly No. 12, and if it is unequal in weight, it is anomaly No. 11.

2) The weight is not equal, the abnormal is in the 9,10 number, the same as above, compare the 1,9 and the weight is equal, it is.

Anomaly No. 10 and anomaly No. 9.

2) The weight is not equal, we have specified 1, 2, 3, 4 when numbering the weight is greater than 5, 6, 7, 8

At this point, a second weighing comparison is made.

Comparing 1,2,5,6 with 3,7,11,12, there are three possibilities.

1) The weight on both sides is equal, then you can judge that the weight of No. 4 is greater than No. 8, and then compare No. 4 with No. 1, the weight.

Equal is 8 good anomaly, unequal is 4 anomaly.

2) The weight remains the same, still 1, 2, 5, 6 side is greater than 3, 7, 11, 12, here you can judge 1 or 2.

The weight is greater than the weight of the 7, because the 5, 6, and 3 are placed on the opposite side, if there is an anomaly, that.

The weight on both sides will change.

At this time, the third weighing can be carried out, comparing 1 and 2, three possibilities can be obtained, the weight is equal, it is 7 different.

Often, if the weight of No. 1 is greater than that of No. 2, it means that No. 1 is heavier, and if the weight of No. 1 is less than that of No. 2, it is that No. 2 is heavier.

3) The weight change, 1, 2, 5, 6 side is less than 3, 7, 11, 12, here you can judge the weight of 5 or 6.

It is less than the weight of the 3.

As above, No. 5 and No. 6 can be compared, and the weight is equal to No. 3 anomaly, and No. 5 is greater than No. 6 is No. 6 anomaly, which is small.

So it was Anomaly 5.

1 Draw a sketch.

In the first step 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, the scales are equal, and the anomaly can be judged to be in 9, 10, 11, 12.

9,10 1, 2 Equals Anomalies at 11,12

The uneven anomaly is compared again at 9,10.

The anomaly can be detected.

uneven, and marked as 1,2,3,4,> 5,6,7,8

Step 2 Take 1,2,5,6 3,7,11,12

4>8 can be judged, and another 1,4 comparison can be made to determine the abnormality.

1 or 2 > 7 1,2 comparison = 7 anomalies; >1 abnormality; <2 Exceptions.

5 or 6< 3 5,6 comparison = 3 anomalies, >6 anomalies; <5 Exception.

(1) From the first and second balls, there must be a light ball in the ball and the ball, and there must be a light ball in the ball and the ball, so that it is concluded that both are standard balls;

2) From the situation of the third ball, both the number and the number are standard balls, assuming that the number is also a standard ball, from the "same weight" can be deduced: the number, the number is also a standard ball, which is inconsistent with the number and the number of balls must have a light ball, so it can be seen that the number ball is a light ball

So the ball is the standard ball, and then by the third "same weight", the ball is a light ball Answer: The number of the two light balls is and

So the answer is: