From the nine numbers 1, 2, 3, and 9, take 5 different numbers to form a five digit progressive num

I don't know if you've ever learned permutations.

To be the smallest so the first two for.

Because the current 4 numbers are fixed, there are 5 ways to arrange the last number.

The first 3 are fixed and the last two numbers are arranged in 30 ways.

The first 2 numbers are fixed, and the last three numbers have 210 ways to arrange them" 71.

So fix the first two.

When the third number is 3, there are only 30 ways to arrange the last two numbers, and it is not enough to reach 71 numbers.

When the third number is 4, the last two numbers have 20 ways to arrange them, and the first one adds up to 50<71 .........At 5, there are 12 ways to arrange the last two numbers, and the first two are added to 72>71, so the first three numbers are 125, and because the 72nd number is the largest, it is 12589, so the 71st should be 12578

I don't know if I understand it.

Suppose the 10,000th digit of this number is 1

then there are 4c8 = 70.

So 71 is the smallest number with 2 as the 10,000th place.

So it's 23456

The best way to do this is to combine the most basic sorting with calculations.

First of all, the smallest number of the first class is the number where the 10,000 digit is 1, there are 4c8 70, then the 71st number is the smallest number with the 10,000 digit not 1, which is obviously 23456.

Kid, just use permutations and combinations.

You go and look at the textbook of the second year of high school, there is this content.

The method is the same as the one upstairs.

The answer is also true.

The 71st is 12578

This problem is essentially a combination problem, take any five out of nine numbers, because if it is a progressive number, there is only one arrangement, that is, from small to large. So how many combinations are there in total? 126 species. Then you can think for yourself, it's pretty simple.

Let the three numbers be a, b, c, and the six two-digit numbers are 10a+b, 10a+c, 10b+a, 10b+c, 10c+a, 10c+b, and the six numbers add up to 22a+22b+22c, so the answers are all 22

Regardless of which three numbers are chosen, the result is 22.

Because the sum of the six digits made up of every three numbers is an integer multiple of 22. The multiple also increases with the sum of the three digits, 6-24

The sum of the three numbers selected on the first day of junior high school.

3*5*7*11=1155

99999 (maximum five digits) 1155 86 (integer digits are 86) 1155 86 = 99330

The maximum five-digit number that can be divisible is 99330

But because there is a repetition of numbers, let's look at the second largest.

So the maximum number is 98175

Everyone has a scale in their hearts, but the difference is in the post--

This is a permutation problem, which is actually to choose 4 different numbers from 1-9 numbers, and then combine the 4 numbers into 5 digits.

There are possible combinations of c(9,4) to choose 4 different numbers, and then the 4 numbers in each case should be put on 5 digits is a permutation problem, first 3 positions indicate that there are no duplicate numbers, and two positions are left for the duplicate numbers, because the duplicate numbers in the two positions left are the same no matter how they are filled in and exchanged. There is a combination of c(4,3) unrepeating numbers in the first 4 digits, and then there are a(5,3) cases in the position of 5 digits that are not repeated, and the last 2 digits must be reserved for the remaining one with duplicate numbers, so the combination of situations is unique.

So in the end there is c(9,4)*c(4,3)*a(5,3)=30240.

There are a total of 30,240 such five-digit numbers.

It is not possible to list all the numbers. Here are some of the results and fortran** (using the enumeration algorithm).

abc is a natural number from 1-9.

6 2-digit sum royal stool = 20 (a + b + c) + 2 (a + b + c) = 22 (a + b + c) base split contains.

So after dividing by (a+b+c), it is equal to 22

3*5*7*11=1155

99999 (maximum five digits) 1155 86 (integer digits are 86) 1155 86 = 99330

The maximum five-digit number that can be divisible is 99330

But because there is a repetition of numbers, let's look at the second largest.

So the maximum number is 98175

The least common multiple of is 1155

This number must be a multiple of 1155.

The maximum is 98175 = 1155 x 85

Steps: 3 * 5 * 7 * 11 = 1155

1155*85=98175 (meet the requirements).

1155 * 86 = 99330 (does not meet the requirements, there is no 0 in 1-9) 1155 * 87 = 100485 (does not meet the requirements, not 5 digits).

(1):1.Choose 3 odd bai numbers and put them in the position of 1,3,5 a(3,5) 2The remaining number du is placed on the zhi2,4 bits a(2,6) 3The final DAO result is a(3,5)*a(2,6).

2): This question is actually an even number.

1.Choose two even numbers and place them in the position of 2,4 a(2,4) 2...The remaining numbers are placed in the 1,3,5 bits a(3,7) 3Final Result a(2,4)*a(3,7).

1. The first 3 odd numbers are selected from the 5 odd numbers, and the remaining positions are selected from the remaining 6 numbers 2 permutations: a(3,5)*a(2,6)=5*4*3*6*5=1800

2. The first 2 even positions are arranged from 2 of the 4 even numbers, and the remaining positions are arranged by 3 of the remaining 7 numbers: a(2,4)*a(3,7)=4*3*7*6*5=2520

Of these, there are 1,824 even numbers.

--Calculation process:

It is possible to make up five digits without repeating numbers:

c(4,3)*c(5,2)*p(5,5)-c(4,3)*c(4,1)*p(4,4)

4416.

Or: c(4,3)*c(4,2)*p(5,5)+c(4,3)*c(4,1)*(p(5,5)-p(4,4))).

4416.

Among them, even numbers have.

c(4,3)*c(4,2)*c(2,1)*p(4,4)+c(4,3)*c(4,1)*(2*p(4,4)-p(3,3))

1824 pcs.

Or: c(4,3)*c(5,2)*c(2,1)*p(4,4)-c(4,3)*c(4,1)*p(3,3).

1824 pcs.