Can 2019 be split into several positive integers and added, and the reciprocal sum of positive integ

2019 can be split into the following 17 positive integers and the sum of the reciprocal of the positive integers is 1:

If a positive integer can be expressed as the sum of n positive integers. And the reciprocal sum of these n positive integers is 1Then we call this natural number a good number

1.Verification: If m is a good numberThen 2m+8 and 2m+9 are both [good numbers].

2.Try to determine whether the number of good numbers in a positive integer set is finite or infinite. and give reasons. Proof:

Let m=a1+a2+a3+......an

It can meet the ...... of 1 a1 + 1 a2 +1 an=1 then split 2m+8 into :

2a1+2a2+……2an+4+4

1/2a1+1/2a2+……1/2an+1/4+1/4(1/2)(1/a1+1/a2+……1 an)+1 22m+8 is a good number.

2m+9 split into:

2a1+2a2+……2an+3+6

1/2a1+1/2a2+……1/2an+1/3+1/6(1/2)(1/a1+1/a2+……1 an)+1 22m+9 is also a good number.

According to the conclusion of question 1, 2m+8 and 2m+9 are good numbers.

It's infinitely iterative.

There are infinite numbers.

Make it up slowly: start with 1=1 2+1 3+1 6.

In the above formula, the sum of the denominators of unit fractions = 8 4+(4+12) (2+3+6) 2+7+42

2017.It's close to 2019

We hope that those who are interested will explore together.

1. Use the ** watch to go to school, and bring it with you when school starts!

There are at least 9 even numbers, and the process is as follows.

Because the sum of the 52 natural numbers is 2019, and 2019 is an odd number, there must be an odd number of even numbers in these 52 numbers, and the odd number of odd numbers The question requires a minimum of how many even numbers, and we can consider the maximum number of odd numbers in reverse.

Because it is an odd number, the odd number is up to 51.

Since these 51 odd numbers are different from each other, the minimum value of their sum is 1+3+5+...101 = 51 = 2601>2019 so the condition is not met.

Since the number of odd numbers is odd, let's consider 49 odd numbers next, since these 49 odd numbers are different from each other, the minimum value of their sum is 1+3+5+...97 = 49 = 2401>2019 so the condition is not met.

Since the number of odd numbers is odd, let's consider 47 odd numbers next, and since these 47 odd numbers are different from each other, the minimum value of their sum is 1+3+5+...93 = 47 = 2209>2019 so the condition is not met.

Since the number of odd numbers is odd, let's consider 45 odd numbers next, and since these 45 odd numbers are different from each other, the minimum value of their sum is 1+3+5+.89 = 47 = 2025>2019 so the conditions are not met.

Since the number of odd numbers is odd, let's consider 43 odd numbers next, since these 43 odd numbers are different from each other, the minimum value of their sum is 1+3+5+...85 = 43 = 1849<2019 So the odd number is at most 43, then the even number is at least 52-43 = 9, and finally a set of 52 different natural numbers that meet the conditions is given to 43 odd numbers: 1, 3, 5, ,..

859 even numbers: 2, 4, 6, 8, 10, 12, 14, 16, 98

This depends on the smallest sum is 5+6=11, and the largest sum is 2018+2019=2227, and every number from 11 to 2227 can be sum, so there are a total of 2227-11+1=2217.

The question is how many different sums there are, then, in this sequence, each combination between the smallest sum 5 + 6 = 11 and the largest sum 2018 + 2019 = 4037 will produce a different sum, so just ask how many sums there are in 11 4037. Column formula: 4037-11+1=4027 kinds.

You can get two different sums.

[Right].

Because, 2019 (1 a) = 2019 a;

and a 1 natural numberThen, 2019a 2019

So, 2019 (1 a) 2019

This, as long as 55 is decomposed into prime factors, all the multiples of these prime factors are removed, and the remaining number is the number of the remaining 55 coprime.

In 1-2019, there are 403 multiples of 5, 183 multiples of 11, 36 multiples of 55, 403+183-36=550

So, in 1-2019, there are 1469 numbers coprime with 55

Perfect remainder: A positive integer divided by a positive integer n can only be ,n-1 。

Examine the remainders of 2019 natural numbers divided by 2018, and there are up to 2018 possible cases, so at least two of them must have equal remainders, then the difference between these two numbers with equal remainders is divisible by 2018.

The enumeration method is proved as long as there is one group, such as 2019-1=2018

Take advantage of the drawer principle.

The remainder divided by 2018 is regarded as 2018 drawers, and 2019 different natural numbers are put into these 2018 drawers, and the drawer principle knows that at least one drawer has at least 2 numbers, and two numbers are taken out in this drawer, and the remainder of these two numbers divided by 2018 is the same, and the difference between the two of them is divisible by 2018.

1) Let x=3a (a is a positive integer), y =2019-x =3 673-(3a) =3(673-3a), and y is a multiple of 3.

Let y=3b(b is a positive integer), and the honorable beam 3 673=2019=(3a) +3b) =9(a +b), 673=3(a +b), contradictory, unsolvable.

2) Let x=3a 1 (a is a positive integer) and y=3b 1 (b is a positive integer).

Answer: 3 673=2019=x +y =(3a 1) +3b 1) =3(3a +3b 2a 2b)+2,3[673-(3a +3b 2a 2b)]=2, contradiction, no solution.

x +y = 2019 without positive integer solution.