How many methods are there to divide 450 into the sum of several natural numbers How many kinds of methods are there to divide 450 into the sum of several natural numbers

How many methods are there to divide 450 into the sum of several natural numbers How many kinds of methods are there to divide 450 into the sum of several natural numbers

The number of your topic is too big. Let's use a "small even number with more factors" to illustrate this problem. Don't you want to explain it in detail? Then don't worry about my wordiness
Let there be a number: [12]. How many methods are there to divide it into the sum of several natural numbers?
We first point out that although the number "0" is a natural number, it is not within the scope of our textual research. In addition, for example, with 1 + 2, we will not consider 2 + 1 any more
There are two methods, one is the method of inserting board, the other is the method of exhausting
Let's start with the board inserting method
You put 12 small matchsticks on the table in a row, and there are 11 gaps between them. There are 11 ways to insert a board. For example, if it is inserted in the second gap, the left side of the board is 3; if it is inserted in the right side of the board, it is 9, so 12 = 3 + 9; if it is inserted in the third gap, it is 12 = 4 + 8, and so on, It's obvious that the first one is over calculated (because we have a rule before). 1 + 11,2 + 10,3 + 9,4 + 8,5 + 7,6 + 6,7 + 5,8 + 4,9 + 3,10 + 2,11 + 1. It can be seen that the second five should be over calculated, that is, we should add 1 to the answer 11 and divide by 2
Let's divide 12 into the sum of three numbers. For example, if we insert the third and the fifth slots, we will get 12 = 3 + 2 + 7. Obviously, there is a [repeat] here: 3 + 2 + 7 = 2 + 3 + 7 = 2 + 7 + 3 = 3 + 7 + 2 = 7 + 2 + 3 = 7 + 3 + 3 + 7 + 2. So, we have to divide "result" by 6, which is divided by "factorial of 3", How many kinds of inserting methods are there in [insert 2 boards]? There are so many kinds of inserting methods, that is, C (11,2) = 55. Note that these 55 methods include three states: one is that the number is different (such as 3 + 2 + 7), and the result is divided by 6; the other is that there are two same numbers (such as 5 + 5 + 2 = 12), Third, there are three numbers that are the same (4 + 4 + 4 = 12)
**In the same number, two 1's, two 2's, two 5's, two 2's, two 2's, two 2's, two 2's, two 5's, all of which need to be subtracted
Let's take a look at the [exhaustive method]. How many are the shapes of 1 +? How many are the shapes of 2 +? How many are the shapes of 3 +? How many are the shapes of 6 +? (no more)
1 + "one term", there is one;
1 + "binomial", the number 11 is divided into the sum of two, there are 1 + 10,2 + 9,3 + 8,4 + 7,5 + 6;
1 + "three, the number 11 divided into three sum, there are;
1 + "10 items", there is one
2 + ". 2 +". 2 + "
.
There is one of "6 +"
This is the exhaustive method. It's not easy. This is my idea for your reference
If you are interested, you can use [8] to split it up
How can 14 be divided into the sum of several natural numbers to make the greatest achievement?
Do you want to continue to disassemble a number?
a+b>ab?
It is equivalent to discussing the maximum of F = a + b-ab
The result is the maximum when a = B (a, b > = 2)
14 pairs to 7,7
And then it's divided into 3, 3, 4, 4
Compare 3,3,3,5 and 3,3,4,4
3,3,4,4 Max
Split into 7 + 7
Two three plus two four
The sum of number a, B and C is 80. The number a is twice that of number B, and the number B is three times that of number C. what are the numbers a, B and C?
Number b = 3 numbers C
A = 2 b = 6 C
C: 80 / (1 + 3 + 6) = 8
Number B: 8x3 = 24
Number A: 8x6 = 48
Let B be x, then a be 2x and C be (1 / 3) X
2X + X + (1 / 3) x = 80
The solution is: x = 80 * 3 / 10 = 24
Therefore, we can see that a is 2 * 24 = 48, B is 24, C is 1 / 3 * 24 = 8
There are 120 people in workshop a and workshop B. now 12 people are transferred from workshop a to workshop B. at this time, the number ratio of workshop a and workshop B is 7:5?
If the sum of a, B and C is 80, B is 4-11 times of a (i.e. 4 divided by 11), and C is one third of the sum of a and B, then the three numbers are
How to set up equations, please write it out
Let a x, b y, C Z
X+Y+Z=80
Y=4\11X
Z=1/3（X+Y）
X=44
Y=16
z=20
There are several workers in workshop a and workshop B. if 100 people are transferred from workshop B to workshop a, the number of people in workshop a is six times of the remaining number in workshop B; if 100 people are transferred from workshop a to workshop B, the number of people in the two workshops is equal, and the number of people in workshop a and workshop B is calculated
(100 + 100) / [(6 + 1) / - 2-1], = 200 ^ [7 ^ 2-1], = 200 ^ [3.5-1], = 200 ^ 2.5, = 80 (people), 80 + 100 = 180 (people), 180 + 100 × 2, = 180 + 200, = 380 (people). Answer: there are 380 people in workshop a and 180 people in workshop B
The average number of a, B and C is 80, and the ratio of a, B and C is 4:7; 1. What are the three numbers
80×3÷（4+7+2）=20
A = 20 × 4 = 80
B = 20 × 7 = 140
C = 20 × 1 = 20
Do not understand can ask, help please adopt, thank you!
The total number of a, B and C is 3 * 80 = 240
Number of a: 240 * 4 / (4 + 7 + 1) = 80
Number B: 240 * 7 / (4 + 7 + 1) = 140
C: 240-80-140 = 20
A: 80
B: 140
C: 20
There are several workers in workshop a and workshop B. if 100 people are transferred from workshop B to workshop a, the number of people in workshop a is six times of the remaining number in workshop B; if 100 people are transferred from workshop a to workshop B, the number of people in the two workshops is equal, and the number of people in workshop a and workshop B is calculated
(100 + 100) / [(6 + 1) / - 2-1], = 200 ^ [7 ^ 2-1], = 200 ^ [3.5-1], = 200 ^ 2.5, = 80 (people), 80 + 100 = 180 (people), 180 + 100 × 2, = 180 + 200, = 380 (people). Answer: there are 380 people in workshop a and 180 people in workshop B
The average number of a, B and C is 6, and their ratio is 1 / 2:2 / 3:5 / 6
Suppose a is 1 / 2x, then B is 2 / 3x and C is 5 / 6x
1/2X+2/3X+5/6X=3x6
3/6X+4/6X+5/6X=18
12/6X=18
2X=18
X=9
A: 1 / 2x = 1 / 2x9 = 9 / 2
B: 2 / 3x = 2 / 3x9 = 6
C: 5 / 6x = 5 / 6x9 = 15 / 2
The number of workers in workshop a is 25 of that in workshop B. later, the number of workers in workshop a increased by 20 and that in workshop B decreased by 35. In this way, the number of workers in workshop a is 79 of that in workshop B. now, how many workers are there in workshop a and workshop B?
Suppose the number of people in workshop B is x, then the number of people in workshop a is 79x. According to the meaning of the question, we can get: 79x-20 = (x + 35) × 25, 79x-20 = 25X + 35 × 25, 79x-20 = 25X + 14, & nbsp; 1745x = 34, & nbsp; & nbsp; & nbsp; X = 90, the number of people in workshop a: 79x = 90 × 79 = 70