Can you use a weightless balance to find this fake pearl in just three scales? 5

Take out 3 pearls from each of the two boxes A and B and put them on both sides of the scale, if balanced, then the false pearls are in the remaining two in box B, take out one of them and put one of the A boxes on both sides of the scale, if balanced, the rest is false, otherwise the balance is false; If the first measurement is unbalanced, then the false pearl in the first 3, take one from each of the scales, if the balance is balanced, then the balance is taken is false, if it is not balanced, then continue to take one, the balance is taken is false, and the unbalanced is the rest is false.

First, put 2 boxes of A pearls on the left disk of the adjusted balance, and 2 boxes of B pearls on the right disk - B1 and B2 (weighing at one time), if balanced, B1 and B2 are true;

Replace B3 and B4 into the right disc to test the balance (secondary weighing), if balanced, B3 and B4 are true; Thus no weighing b5 is false;

Finally, by comparing B5 with one of A (three weighing), you can determine whether B5 is lighter or heavier than a real pearl.

Replace B3 and B4 into the right disk to test the balance (secondary weighing), if it is not balanced, one of B3 and B4 is false.

By comparing a fake B1 or B2 with a B or B (three weighing), you can tell whether a fake B1 or B2 is lighter or heavier than a real pearl.

First put 2 boxes of A pearls on the left disk of the adjusted balance, and 2 boxes of B pearls on the right disk - B1 and B2 (one weighing), if it is not balanced, one of B1 and B2 is true; The remaining 3 are true.

Then put only 1 box of pearls on the left disk of the balance, and put B1 into the right plate to test the balance (secondary weighing), if the balance, B1 is true and B2 is false; If it is unbalanced, b2 is true and b1 is false.

Finally, by comparing a fake B1 or B2 with a pearl (three weighings), it is possible to determine whether the fake B1 or B2 is lighter or heavier than the real pearl.

Summary. Because the last one must have been two.

Then multiply by 2 in that order.

In the end, it is 32, but we can also come up with a 16 alone, because for example, the 32 of the first time is the same, and we can use the extra 16.

How many pearls can you weigh five times on a balance and find only one pearl that is too light to be qualified?

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Good. Up to 48 can find out which one is not qualified. Equation 2 2 2 2 2 2 = 32 + 16 = 48

Thank you. Because the last one must have been two. Then multiply by 2 in that order. In the end, it is 32, but we can also come up with a 16 alone, because for example, the 32 of the first time is the same, and we can use the extra 16.

Okay thank you. You're welcome, dear. <>

The first three are weighed, that is, three are placed on the left, and three are also placed on the right (1) If the balance is balanced, then the false one is among the three that are not weighed Take two of the remaining three to weigh, if the balance is balanced, then the remaining one is false, if it is not balanced, then the light one is false

2) If the balance is unbalanced, then the false one is on the light side Take two on the light side to weigh it, and then the first step is the same as if the balance is balanced, then the remaining one is false, if it is not balanced, then the light one is false

A: You need to weigh at least 2 times to find out

So the answer is: 2

Answer 2, at least twice to find the fake pearl.

243 pcs. First time:

Divide 243 into three portions of 81 capsules.

Two of them will be weighed. If it is not the same, there is a substandard one in the light one; If it is the same, then there is a disqualification in the third part.

The second time: Weigh the light part of the first 81 pieces, and then divide it into three parts, 27 pieces each, and find out which one of them is light.

The principle is the same as above. The third time:

Divide the second light part of 27 pieces, divide it into three parts, 9 pieces each, and then according to the above principle, find out the light one of the congratulatory letters.

Fourth: Divide the third light pat spine into 9 pieces, divide them into three parts, 3 pieces each, and then find the lightest part according to the above principle.

Fifth: Divide the light part of the fourth time into three parts, 1 piece each, and then according to the above principle, find out the light one, which is the unqualified one. Thank you.

Divide the pearl into 3 parts equally, weigh two of them, if they are equal, then the third part of Xun's god contains the one that is too light, if it is not equal, the light part contains the loss of mu, and so on, the last time is 3.

So you can find a light one from up to 243.

Divided into 3 groups, a group of 3 pieces, respectively numbered as A, B, C, group A and group B compared, 1) assuming that group A and group B are equal in weight, then the false beads must be in group C, take out two of the remaining 3 to compare, if the weight is not equal,The light one is the false bead, and if the weight is equal, the last one left is the false bead.

2) Assuming that the weight of group A and group B is not equal, then the light group must have false beads, take out two from the group to compare, the weight is equal, then the third is a false column, and the weight is not equal, then the light one is a false bead.

Answer: Answer: At least 2 times can find it, the first time: take any 4 of the 5 pearls, divide them into two equal parts, 2 pieces each, and put them separately.

At both ends of the scale, if the balance scale is balanced, the one that is not taken is a false pearl, if the balance scale is not balanced; The second time: put the 2 pearls on the higher end of the scale at both ends of the scale, and the higher end is a fake pearl

Hello The process is as follows:

1.Method 1 (step-by-step method):

243 divided into 3 equal parts The first balance is divided into 81 on each side, and the balance is found in the remaining 81; It is also divided into 3 points equally, 27 on each side of the scale, and the balance is found in the remaining 27; Divide it into 3 points equally, put 9 on each side of the scale, and find the balance in the remaining 9; And so on, and the rest is.

The imbalance is found at the slightly heavier end, divided into 3 points equally, 27 on each side of the scale, and the unbalance is found at the slightly heavier end, 9 on each side of the balance, and so on until the last one on both sides of the scale, the heavier is.

2.Rule of thirds: 3 5 = 243 5 times.

Hope it helps you.

At least once, put 121 pieces on each end of the balance to weigh, if the two sides are equal, the remaining one is slightly heavier.

Number the 8 beads, 1-8, for the first time, take out the 1-6 from the 8 Chunhui good beads, and put 3 at each end of the Biyou scale, 1, 2, 3 on the left;4, 5, 6 on the right; 7,8 left.

Situation 1: If the balance is balanced, it means that the 6 beads are true, and the remaining 2 beads have problems, then select 1 from the 6 normal, and select 1 of the remaining 2 to put on the balance for comparison, (the second time) if the balance is balanced, it means that this one is true and the other is false, if the balance is not balanced, it means that this one is false. (This situation can be identified 2 times).

Scenario 2 If there are 3 parties on both sides and the balance is unbalanced, then one of the numbers 1-6 is false, and numbers 7 and 8 are true.

The second time, take out No. 1 and No. 4, No. 3 and No. 5 switch positions and weigh again, that is, put 26 on the left side of the balance, put 35 on the right side, if balanced, it means that there is a problem with No. 1 or No. 4, and then take any one true and No. 1 once to distinguish. If the balance is unbalanced and the tilt direction remains unchanged, it means that No. 2 or No. 5 is false, and it can be identified by weighing the lead once, and if the tilt direction of the balance changes, it means that No. 3 or No. 5 is false, and it can be identified by weighing it again.

Unfortunately, there are no bounty points!!

4 weighed, which pile is light, which cover Lachang leakage pile.

2 weighs things smoothly, which pile is light, which pile is weighed which pile.

One is one scale, which one is lighter, and that one is fake.

Hello! The 9 pearls are divided into 3 groups of 3 each.

For the first time, put a set on each side of the scale.

If balanced, then the false ones are in the remaining group.

If it is not balanced, then the false one is in the light group.

By weighing it once in this way, it is possible to determine which group (3) the fake ones are in.

The second time, take 2 of the group containing the false and place them on either side of the scale.

If it is not balanced, then the light is fake.

If balanced, the one that remains is fake.

If it helps you, hope.