How many natural numbers can be selected from all the natural numbers from 1 to 1000, and the difference between any two natural numbers is not a multiple of 7 How many natural numbers are selected, and the sum of any two natural numbers is not a multiple of 7?

How many natural numbers can be selected from all the natural numbers from 1 to 1000, and the difference between any two natural numbers is not a multiple of 7 How many natural numbers are selected, and the sum of any two natural numbers is not a multiple of 7?

Divide all the natural numbers from 1 to 1000 into seven groups
They are (1) divisible by 7, (2) divisible by 7, remainder 1, (3) divisible by 7, remainder 2, (4) divisible by 7, remainder 3,
(5) It is divided by 7, the rest 4, (6) by 7, the rest 5, (7) by 7, the rest 6,
To meet the requirements, select at most one from each group,
The sum of any two natural numbers is not a multiple of 7

How many natural numbers are multiples of 5 or 6 in natural numbers from 1 to 80

80 / 5 = 16
80 / 6 = 13 two
30 and 60 are repeated
16 + 13-2 = 27

For a natural number less than 80, the sum of it and 3 is a multiple of 5, and the difference between it and 3 is a multiple of 6
Let the unknowns be determined by Diophantine equation

The sum of 3 is a multiple of 5
It means that the last position is 2 or 7
The difference from 3 is a multiple of 6
That means it's odd
So the last place can only be 7
And it can be explained that it is a multiple of 3
That is, the sum of each digit is a multiple of 3
So this natural number may be zero
27 57
Let this number be X
Then x + 3 = 5m
x-3=6n
0