The existing 1g, 2g, and 3g balance weights use 10 weights to weigh a 20g object

Let the number of weights of 1 gram, 2 gram and 3 gram be x, y and z respectively.

x+2y+3z=20

x+y+z=10

The above two equations form an indefinite system of equations.

The integer solution can be determined by assuming the value of an unknown quantity in turn, and it may be assumed that the value of x is in the order of , when x=0, solve the equation system composed of 2y+3z=20 and y+z=10 to obtain y=10, z=0

i.e. when x=0, y=10, z=0

When x=1, solve the equation system composed of 2y+3z=19 and y+z=9 to obtain y=8,z=1

i.e. when x=1, y=8, z=1

And so on. When x=2, y=6, z=2

When x=3, y=4, z=3

When x=4, y=2, z=4

When x=5, y=0, z=5

When x 6, y < 0, meaningless.

Therefore, all the methods have the above 6 sets of solutions.

1) It can be seen that when there are 3 weights of 1g, there should be 3 weights of 3g.

2) Let at least 1 weight be used, then there are 4 cases, that is, the situation that the unknown number in the above solution is excluded and 0 can be used).

2 2g weights, 4 3g and 1g weights.

4 2g weights, 3 3g and 1g weights.

6 2g weights, 2 3g and 1g weights.

8 2g weights, 1 3g and 1g weights.

1) All 2g weights, 2) 2g weights, 4 3g and 1g weights.

3) 4 2g weights, 3 3g and 1g weights.

4) 6 weights of 2g, 2 weights of 3g and 2 weights of 1g.

5) 8 weights of 2g, 1 weight of 3g and 1g each.

6) 5 weights of 3g and 1g.

Use a large scale, put 1 5g on one side, and 5 2g on one side, a total of 15 grams.

5+5x2=15, put 15 grams on the other side and you're good to go.

The working reference group and standard weights are usually divided into kilogram group (1-20kg), gram group (1-50g) and milligram group (1-500mg), and there can also be microgram group or other weight combinations (such as the weight increase group used on the bench scale) as needed. The combination of weights usually has and .

With a large balance, put 5 2G on one side, and 2 2G on the other 1 5G, because the balance is large, the 1G gap is completely negligible.

Or let the 5g weights rust and make them even.

It is also possible to cheat with a balance nut.

Summary. This is a classic math puzzle. Finding a 10-gram item with a weight of only 20 grams and a limit of 3 weighings requires some ingenuity.

Here's a workaround: The first weighing divides the weights into two parts, one containing 10 grams and 5 grams, and the other containing two 5 gram weights. Place the former on one side of the scale and the latter on the other, and if the balance is balanced, there must be an item weighing 10 grams in the first weighing.

Assuming that the left side is heavier, the weight on the left is 5 grams more than the weight on the right, so the side that contains an item weighing 10 grams in the first weighing must contain a weight of 10 grams. In order to find this item in the next two weighs, we need to divide it in half.

There is a balance that only has 20 grams of weights, and to weigh 10 grams of objects, you can only weigh it 3 times.

There is a balance that only has 20 grams of weights, and to weigh 10 grams of objects, you can only weigh it 3 times.

This 20g refers to a total of 20g of weights, or is there 5g, 10g, etc.

The weight is only 20 grams. One. Common.

This is a classic math puzzle. Finding a 10-gram item with a weight of only 20 grams and a limit of 3 weighings requires some ingenuity. Here's a workaround:

The first weighing is done by dividing the weights into two liquid halts, one containing 10 g and 5 g weights, and the other containing two 5 g weights. Place the former on one side of the scale and the latter on the other, and if the balance is balanced, there must be an item weighing 10 grams in the first weighing. Assuming that the left side is heavier, it means that the weight on the left is 5 grams more than the weight on the right, so the side that contains the buried lead of the object weighing 10 grams in the first weighing must contain 10 grams of weight.

In order to find this item in the next two weighs, we need to divide it in half.

The second weighing divided the 10-gram weight into two parts, one containing 5 grams and 2 grams of celery, and the other containing 3 grams of weights and one unused 5-gram weight.

1. Put 20 grams of weights on one side and objects on the other side, which is recorded as A, and A is 20 grams.

2. Put 20 grams of weights and A on one side for a total of 40 grams, and put objects on the other side as B, and B is 40 grams.

The sum of a and b is a 60-gram object.

There are a total of 7 masses, calculation process.

1 gram, 3 grams, 9 grams.

1+3=4 grams.

1+9=10 grams.

3 + 9 = 12 grams.

1 + 3 + 9 = 13 grams.

There are a total of 7 items of different quality.

1 gram, 3 grams, 9 grams, 4 grams, 10 grams, 12 grams, 13 grams.

For each weight combination, it's easy to figure it out, 1g 3g. 9g. 4g. 10g.12g of these qualities.

1g to 13g is fine.

The solution process is as follows:

1）1g=1

2）2g=3-1

3）3g=3

4）4g=3+1

5）5g=9-3-1

6）6g=9-3

7）7g=9+1-3

8）8g=9-1

9）9g=9

10）10g=9+1

11）11g=9+3-1

12）12g=9+3

13）13g=9+3+1

1g to 13g is fine.

1g=12g=3-1

3g=34g=3+1

5g=9-3-1

6g=9-3

7g=9+1-3

8g=9-1

9g=910g=9+1

11g=9+3-1

12g=9+3

13g=9+3+1

1g to 13g is fine.

Hello! 1g,3g,9g,27g.

1g can weigh 1g, 3g can weigh 1-4g, 9g can weigh 5-13g, and 27g can weigh 14 to 40 all.

There are 7 types in total:gramsgramsgrams+2 = 3 grams+5 = 6 grams+5 = 7 grams+2+5=8 gramsA moment forever 523 will answer for you, I wish you learning and progress If you agree with my answer, please click the [Accept as satisfactory answer] button in time The mobile phone questioner can comment "satisfied" on the client Your adoption is the driving force for me to move forward

If you have any new questions, please ask me for help, it is not easy to answer the questions, please understand

When the balance is not used, a total of 7 different object masses can be measured.

1. Use only 1 gram of weight. An object with a mass of 1 gram can be called 2 and only 2 grams of weights are used. An object with a mass of 2 grams can be called 3 and only 5 grams of weights are used.

An object with a mass of 5 grams can be weighed 4 and only 1 gram and 2 gram weights are used. An object with a mass of 3 grams5 can be weighed with only 1 gram and 5 gram weights. It can be said that an object with a mass of 6 grams 6 uses only 2 grams and 5 grams of weights.

Objects that can be said to have a mass of 7 grams7, all use the three weights of 1 gram, 2 grams, and 5 grams. An object with a mass of 8 grams can be weighed.

1:1g 2:1g,3g [1,4]3:

1g, 3g, 9g [1,13]4:1g,3g,9g,27g [1,40]5:1g,3g,9g,27g,81g [1,121]So with 5 weights (1g, 3g, 9g, 27g, 81g) you can weigh or skillfully weigh any item within 1 to 121 grams.

80 grams: Left 1 + item Right 8179 grams: Left 3 + item Right 81 + 1 78....