Solve an arcane problem in the sixth grade of primary school, please explain it in detail, thank you

Solution: Knowing from the meaning of the question:

The number of turns of wheel B is 4 5 of wheel A

Then the ratio of the circumference of A to the circumference of B is: 4:5

B wheel circumference: cm).

B wheel diameter: 157 cm).

A wheel diameter: 50 4 5 = 40 (cm).

The number of turns of wheel B is 4 5 of wheel A

Then the circumference of A is the circumference ratio of B: 4 5 1 = 4 5 Circumference of B's wheel: cm.

Diameter: 157 cm.

A diameter: 50 4 5 = 40 cm.

Let the diameter of the wheel of A be R1 and the diameter of Car B R2

50 * 50 circles of A, multiplied by circumference of A = 40 circles of B, multiplied by circumference of B) r1 * (circumference of A + circumference of B).

Solve the equation. r1=40cm

r2=50cm

The original water storage of a reservoir is certain, and the river water is evenly deposited every day. 5 pumping machines can be drained for 20 consecutive days; Six of the same pumping machines can be drained for 15 days. If 6 days of draining is required, how many of the same pumps are needed?

How many pumping machines can pump the original water and 20-day inflow water in the reservoir for 1 day? 20 5 = 100 (sets).

How many pumps can pump the original water from the reservoir and the water that flows in 15 days for 1 day? 6 15 = 90 (sets).

How many pumps can pump the water flowing in each day for 1 day?

100-90) (20-15) = 2 (sets).

How many pumps can pump the original water for 1 day?

100-20 2=60 (sets).

If the pumping is done in 6 days, how many pumps are needed?

60 6 2 = 12 (sets).

A: If the pumping is completed in 6 days, a total of 12 pumping machines are required.

A is 50 circles, and B is 50 4 5 = 40 circles. The formula for perimeter c= d). Let the diameter of A be d A, and B's be d B, then D B D A

D A D B

Simultaneous solution: D A = 40 cm D B = 50 cm.

A: The diameter of A is 40 cm and the diameter of B is 50 cm.

Let the circumference of A be x and the circumference of B be x+

50x=50*4/5（x+

x=nail diameter=

B diameter = (12 56 31 4) 3 14 14

The equiquantity relation is the distance at which wheel A rolls = the distance at which wheel B rolls.

Solution: Let the circumference of wheel B be xcm, then the circumference of wheel A is (40x=50(, x=157.)A wheel circumference =

The diameter of B = 157 pi, and the diameter of A = pi.

Let the diameter of the first wheel be x, and the diameter of the wheel will be y. Then x+ , 50 x=50*4 5 y

Solve the equation x+ , 5x=4y

Result: x=40

y=50

Long meters = 121 decimeters.

Width meter = 26 decimeters.

High meter = 21 decimeters.

then the maximum can be accommodated.

500 (only).

500 is correct and cannot be calculated by volume because cardboard boxes cannot be compressed to fill the space.

Long meters = 121 decimeters.

Width meter = 26 decimeters.

High meter = 21 decimeters.

Only if: 121 6 = 20 ......1，26÷5=5……1，21÷4=5……1, the remainder is the smallest, that is, it can accommodate the most.

Therefore: 20 5 5 = 500 (only).

Therefore, a container is long, wide and high, and can accommodate up to 500 cardboard boxes with a length of 6 decimeters, a width of 5 decimeters, and a height of 4 decimeters.

I wish you progress in your studies and go to the next level! (*

Container volume: cubic meters).

Cardboard box volume: 6x5x4 = 120 (cubic decimeter) unit: 120 1000 = cubic meter).

A: Up to 550 pieces.

I don't know if the landlord has learned the volume. by volume).

Is it , take the whole number 121*26*21= 66066 round up to 66000

6*5*4=120 divide by , resulting in 550.

1. The smallest of these divisors is:

That is, find the least common multiple of these numbers.

Only the least common multiple of is required, i.e. 7 8 9 5 = 2520n The minimum is 2520

143 n The remainder is 0

It is necessary to find the lesser of the total number of occurrences of the factor from 1 to 2003.

It is easy to know that if the number of occurrences of factor 13 is less than that of factor 11, it is only necessary to require the number of occurrences of factor 13 from 1 to 2003.

13*13*13 = 2197 > 2003, and none of these numbers contains three or more factors of 13.

2003 13 = 154 remainder 1, a total of 154 numbers divisible by 13.

2003 (13*13) = 11 remainder 144, a total of 11 numbers divisible by 169.

Therefore, the number of factors 1 to 13 in 2003 is:

154 - 11)*1 + 11*2 = 165.

Therefore, the maximum power of 143 to the power of 165 is divisible by 2003!

3. Which scores you wrote are not clear.

n=2*3*2*5*7*2*3=2520

143^n=11^n*13^n

Because 13 11, then only look at 2003! How many prime factors are there13

Of the 2003 numbers from 1 to 2003, multiples of 13 are 154, multiples of 13 2 are 11, then 2003! It contains 154 + 11 = 165 prime factors 13

The maximum value of n is 165