Select four numbers from 0, 1, 2, 5 and 8 to form the four digits that can be divisible by 2, 3, 5 and 8 at the same time, and the largest number among them

2X3X5=30
In other words, the last four digits are zero, and the first three digits can be divided by three,
The addition of numbers divisible by 3 can also be divisible by 3. 2 + 5 + 8 = 15 can be divisible by 3
That is to say, four numbers are 0.258, and then the maximum number is set
8520

Select three numbers from 0, 1, 2, 4 to form the maximum three digits that can be divisible by 2, 5 and 3 at the same time_ The minimum three digits are_ .

The characteristics of numbers divisible by 2,5,3 are as follows: the number of bits is 0, and the sum of numbers in each digit can be divisible by 3;
There are 120, 210, 240 and 420, of which the maximum three digit number is 420 and the minimum three digit number is 120
So the answer is: 420120

Choose 5 different numbers from the 10 numbers of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 to form a five digit number, so that it can be divisible by 3, 5, 7 and 13. What is the maximum number?

The five digits can be divisible by 3, 5, 7, 13 and, of course, by the least common multiple of 3, 5, 7, 13,
That is, the five digit number is a multiple of 3 × 5 × 7 × 13 = 1365,
Therefore, the maximum multiple of 1365 in five digits is 73 × 1365 = 99645,
However, two of the five digits in 99645 do not meet the requirements of the title
72 × 1364 = 98280 (two 8 repeats, unqualified)
71 × 1365 = 96915
70 × 1365 = 95550 (three 5 repeats, unqualified)
69 × 1365 = 94185
Therefore, the maximum number of five digits is 94185
A: the largest number is 94185

Choose five different numbers from the ten numbers 0-9 to form a 5-digit number so that it can be divisible by 3,5,7,13. The maximum number is () I need it urgently

ninety-four thousand one hundred and eighty-five
First, we multiply 3 5713 to get 1365, which is the least common multiple
And then set the equation
If the condition is divided by 1365, the infinite tends to 100000, and the five digit number cannot be repeated
Finally, a 69 fold 1365 digit number without repetition is calculated
And 94185

Five different numbers are selected from 10 numbers 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 to form a five digit number, so that it can be divisible by 3, 5, 7 and 13 Q: what's the maximum number?

3*5*7*13=1365
5 digits less than 98765
98765 / 1365 integral part = 72
1365*72＝98280
1365*71＝96915
1365*70＝95550
1365 * 69 = 94185

Choose five of the ten numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 to form a five digit number, so that the five digits are divisible by 3, 5, 7, 13______ .

The five digit number can be divisible by 3, 5, 7, 13, and of course, it can also be divisible by the least common multiple of 3, 5, 7 and 13, that is, the five digit number is a multiple of 3 × 5 × 7 × 13 = 1365, and the maximum multiple of 1365 in the five digit number is 73 × 1365 = 99645

0.25, 3 / 4, 8 and which number can form a proportion? What is the ratio

8/0.25=24/(3/4)
There can be a lot

Add another number to 345 so that the four numbers can form a proportional formula. The maximum number of groups that can be written should be written out

3:4=5:20/3
3:5=4:20/3
4:5=3:15/4
4:5=12/5:3

Use the four numbers 7.5, 3, 5, 2 to form different proportional formulas

7.5:5=3:2
7.5:3=5:2

Write two numbers that can be proportional to the three numbers 1, 3, 6, and list the proportion formula

18,2
1:3=6:18
3:1=6:2

Select four numbers from 0, 1, 2, 5 and 8 to form the four digits that can be divisible by 2, 3, 5 and 8 at the same time, and the largest number among them