9 table tennis balls tell the abnormal ball is lighter or heavier, how to weigh it 2 times to find o

Here's how: The premise is to know that the ball is lighter or heavier to judge.

1. Grouping: Divided into three groups, 3 in each group.

2. Select the abnormal group: take any two groups of upper scales, if the same weight, then the abnormal ball is in the third group, if the weight is different, select the lighter or heavier group.

3. Select the abnormal ball: take any two balls on the scale, if the same weight, then the abnormality is the third ball, if the weight is different, select the lighter or heavier ball, complete.

First of all, you need to know whether the anomalous ball is lighter or heavier before you can proceed, assuming that the anomalous ball is heavier, number the ball by 1-9:

The first time to take the 6 balls of 1-6 and weigh them, if 1+2+3 = 4+5+6, then the abnormal balls are inside, and the 1-6 are all qualified balls;

If 1+2+3 > 4+5+6, then the anomalous ball is inside;

If 1+2+3 < 4+5+6, then the anomaly ball is in the number;

After finding out the three balls where the abnormal balls are located, the three balls are weighed for the second time, numbered by X, Y, and Z:

Weigh x and y, if x=y then z is an anomaly ball;

If x>y then x is an anomaly;

If x< y is an anomalous ball.

Divide it into three piles, a, b, c, take any two piles and compare, you can definitely find out with the above method.

Summary. Hello dear, the answer is as follows.

35 are divided into 3 groups, respectively (12, 12, 11), put 2 of them first, that is, 12 on each side of the balance, if it is not balanced, the defective product is on the light side;

Divide the 12 on the light side of the last time into 3 groups, respectively (4, 4, 4), take 2 of them arbitrarily, put 4 on each side of the scale, if it is not balanced, the defective product is on the light side;

Then divide the 4 on the light side of the last time into 3 groups, respectively, (1, 1, 2), take the 3rd group, put 1 on each side of the scale, if it is not balanced, the defective product is on the light side;

Then divide the 2 pieces of the 3rd group into (1,1), put 1 on each side of the scale, if it is not balanced, the defective product is on the light side.

In this way, it takes 4 times to find the defective product.

There are 35 table tennis balls, one of which is unqualified, and I don't know how many times to weigh it at least to ensure that this unqualified one is found.

Dear, hello, the answer is as follows 35 divided into 3 groups, respectively, (12,12,11), first put 2 of them, that is, 12 on each side of the balance, if it is not balanced, the defective product is on the light side; Divide the 12 on the light side of the last time into 3 groups, respectively, (4,4,4), arbitrarily take 2 of them, put 4 on each side of the scale, Song You Heng if not wild to do balance, defective products on the light side; Then divide the 4 on the light side of the last time into 3 groups, respectively, (1, 1, 2), take the 3rd group, put 1 on each side of the scale, if it is not balanced, the defective product is on the light side; Then divide the 2 pieces of the 3rd group into (1,1), put 1 on each side of the scale, if it is not balanced, the defective product is on the light side. In this way, it takes 4 times to grind the potatoes and find the defective product.

The school buys a number of books of science and technology, literature and art books, story books, and Youyin, and each student borrows 4 books from them, so at least several students need to borrow books to ensure that there are two people borrowing the same type of books.

At a minimum, a minimum of 4 students. Because there are three kinds of books such as Song, two books per person, there are three different choices: (pants, cherry blossoms, science and technology, literature and art), (literature and art stories) (science and technology stories).

Divide a number by 2 and leave 1 divided by 3 and divide by 4 and 3, what is the minimum number of numbers.

Adding 1 is a multiple of 2,3,4, so minimum = [2,3,4]-1 = 12-1 = 11

So the minimum is equal to 11

One, one side of six is called to exclude the light six, two, the remaining six are three on one side of the three to exclude the light three, three and the remaining three pick out two of them to weigh, if the two picked out have the same quality, then the unselected is the quality abnormal, if one of the two picked out is heavier, it is the quality abnormal.

Put the twelve ping-pong balls in half on both sides of the weight, put the heavy side in half and put it in half, then take out the three on the heavy side, and put the other two on it, if it is the same weight, it is the remaining one.

Put 6 on one side and weigh one side, divide the heavy side into 3 and then weigh it, and then take out two of the heavy side to weigh it, which side is heavy is which ball is abnormal. If it's the same weight, it's the ball left.

If the weight is abnormally heavy or light, I will do it, such a condition I don't have enough brains - -

Let's not talk about the process, remind me, don't compare with each other, take the right comparison, the process is a bit complicated.

Six-six symmetry, three-three symmetry, one-to-one symmetry.

Divide the table tennis balls into three groups of four.

Take out two of the sets and put them on the scale. The first category is the balance of the scales, and the anomalous balls are in the third group. Replace the third set of three balls, and the first set of three balls, if balanced, the remaining ball in the third group is abnormal.

If it is unbalanced, it is clear whether the abnormal ball is light or heavy, and you can find it by drawing two arbitrarily and weighing it for the third time.

The second category is the first time that the balance is weighed, then it is recoded, the first group is 1234 on the light side, and the second group is 5678 on the heavy side. Any one of the third group that is determined to be normal is a standard ball of 0.

12 and 0 are on the left, and 456 is on the right. Weigh the second time. That is, 12 on the left do not move, and 56 on the right do not move. If it is balanced, it proves that 12456 is normal, 378 is abnormal, and 78 can be compared.

If the second weighing is unbalanced. Proof that the anomalous orb is in 12456.

Since 124 may be light, 56 may be heavy. If it is 120 heavy and 456 light, then 4 is abnormal and is a light ball. If 120 is light and 456 is heavy, it means that 4 is normal.

Either 56 heavy, or 12 light. Take 01 and 25 to compare, and if it is balanced, it will be 6 heavy. If the imbalance is 01 heavy, it will be 2 light, and 01 will be 5 light.

Codes 1-12

First: 1 2 3 4-5 6 7 8

1.There is a problem with balance - 9 10 11 12.

The second time: 9 10 11-1 2 3

There is a problem with balance-12.

Unbalanced - left-handed.

3rd 9-10

There is a problem with Balance-11.

There is a problem with the left weight -9.

There is a problem with the right weight -10.

2.Unbalanced - left-handed.

The second time: 1 2 5-3 6 10

Balance the third time: 7-8

Left heavy - 8 light.

Balance - 4 weight.

Right heavy - 7 light.

Unbalanced - left-handed.

Third time: 1-2

Balance - 6 light.

Left weight - 1 weight.

Right weight - 2 weight.

Unbalanced - right-weighted.

Third time: 3-9

Balance - 5 light.

Left weight - 3 weight.

The first step is six on one side The second step is three on one side The third step is one on the side Got it?

This question tests a person's divergent thinking, and the premise of this answer is to use a balance, and it takes at least 3 times to weigh a slightly heavier unqualified product.

If you don't use a scale, this question doesn't make much sense!

2 times 26 points 13+13

Take out one of the heavier ones, and it becomes 12+1

12 is divided into 6+6

If two 6s weigh equally, the one taken out is on the heavier side.

Divide the 12 balls into three groups of four each.

Then weigh the two groups. (1st), two cases, flat, uneven.

Case 1 (relatively simple): Flat: The ball with abnormal weight is in 4 that are not weighed. Take out 3 of them and weigh any 3 of them in the previous two groups. (2) Two situations: flat and uneven.

Situation 1-1: Ping: Then the one that is not weighed is abnormal in weight, and the result will be known by weighing it one by one (3).

Situation 1-2: Uneven: You already know whether the weight abnormality is light or heavy (one side is normal, the other side is abnormal, you can know the abnormality side is heavy or light), weigh two of the 3 abnormal weights (3rd), whether flat or uneven, you know the result.

Situation 2 (more complicated, it is recommended to write on paper): uneven: divide the heavy side into group A1 A2 A3 A4.

The light side is divided into group B, and the normal side is divided into group C. Weigh a1, a2, b1, and a3 b2 c1. (2) Three cases:

A1 side is heavy, A1 side is light, flat.

Situation 2-1: A1 side weight: It means that one of A1 A2 is heavy or B2 is light, and B1A3 is normal. Weigh a1 and a2 to know the result (p. 3). If it is flat, it means that B2 is light; If it's not flat, whichever is heavier is which.

Situation 2-2: A1 side is light: A3 is heavy or B1 is light, A1A2B2 is normal. Weigh A3 with C1 to know the result (3rd), if flat, B1 is light. If it is uneven, A3 is heavy.

Situation 2-3: Ping: A4 is heavy or B3 and B4 is light. B3 and B4 are called (the 3rd) three cases: B3 is heavy, B3 is light, and flat.

Case 2-3-1: B3 side is heavy, indicating that B4 is light.

Situation 2-3-2: B3 is light, indicating that B3 is light.

Case 2-3-3: Flat, indicating that A4 is heavy.

At this point, all the balls have been judged.

For the first time, divide 12 balls into 3 equal parts, 4 each, weigh them once, and determine which pile the unqualified ones are in.

The second time, take out 4 balls, take out two at random and weigh them once, if they are not balanced, you need to change the ball again to see which ball is defective.

If it is balanced, you also need to change the ball scale again to see which ball is defective.

Using a scale, weigh the 12 balls in half, i.e., divide the balls into two parts, each with the same amount. The first time you remove 6 balls, the second time you remove 3 balls, and the third time you can find it.

For the first time, divide the 12 balls into 3 equal parts, 4 pieces each, weigh them once, and determine which pile the unqualified ones are in.

The second time, take out 4 balls, take out two random scales once, if it is not balanced, you need to change the ball scale again to see which ball is defective.

Three times, the first time five on one side, the light five out, just take four and put two on one side, take out the light one and then one on the other, if the second balance is the one that is not weighed, of course this is too coincidental%....So at least three times.

1: The 11 balls are numbered as 1 5 on the left handicap, 6 10 on the right handicap, 11 do not put on the balance If the left is light, then the unqualified ball is on the 1 5.

If the right side is light, then the unqualified ball is on the 6-10.

If the left and right sides are the same, the non-qualifying ball is 11.

2: If the left side is light, then take the number and put the left plate number on the right disk, and the 5th number does not repeat the process in 1.

3: If the left side is still light, then the number is placed on both sides, and the unqualified ball can be determined.

First weighing: Divide them into groups of four. Take out two sets first, if the two groups have the same weight, then the bad table tennis ball must be in the remaining four, if the first two sets of weights are different, then you need to weigh the remaining group, in short, find out the four that are different.

The second weighing: first take out the two balls and weigh them separately, if the weight is the same, then the bad balls are in the remaining two.

The third time, weigh the remaining group with bad balls, one of the two balls, and the weight is different.