From 1, 2, 3, 4 Take any 26 numbers out of the 50 numbers, 49 and 50, then at least two of the 26 numbers are coprime. Q: why?

From 1, 2, 3, 4 Take any 26 numbers out of the 50 numbers, 49 and 50, then at least two of the 26 numbers are coprime. Q: why?

Construct 25 drawers: two adjacent numbers are one, namely (1, 2) (3, 4) (5, 6) (7, 8) (49, 50); choose 26 numbers, at least one group of numbers is in them, because two adjacent natural numbers are coprime, so there are at least two coprime numbers in the 26 numbers

From 1, 2, 3, 4 Take any 26 numbers out of the 50 numbers, 49 and 50, then at least two of the 26 numbers are coprime. Q: why?

Construct 25 drawers: two adjacent numbers are one, namely (1, 2) (3, 4) (5, 6) (7, 8) (49, 50); choose 26 numbers, at least one group of numbers is in them, because two adjacent natural numbers are coprime, so there are at least two coprime numbers in the 26 numbers

Taking any number () from 1 to 10 can ensure that at least two of them are mutually prime. Why?

1、 Take any 6 numbers to ensure that at least two numbers are coprime
reason:
They were divided into 5 groups
(1,2)
(3,4)
(5,6)
(7,8)
(9,10)
When any 6 numbers are selected,
There must be two numbers in one of the above five groups (that is, there must be two numbers in the same group),
Because the two numbers in each group above are coprime, so
Taking (6) numbers from 1 to 10 can ensure that at least two of them are coprime
2、 If you take any 2-5 numbers, you cannot guarantee that at least two numbers are coprime. (counterexample: 2,4,6,8,10)
3、 According to (1) and (2) above,

It is proved that if n + 1 numbers are taken from 1,2,2n, there must be two coprime numbers
2 N, this problem has nothing to do with the value of N, it has been proved

Counter proof: suppose that N + 1 is not coprime, then we can find a number that is a factor of other n numbers at the same time. We all know that 1 does not belong to the scope of consideration, so we start from 2, Suppose that the minimum number is 2, then because the other n numbers are all integral multiples of 2, we know that the number of integral multiples of 2 in the range of 2n will not exceed n, which also includes 2. From the hypothesis, we can know that there are n + 1 multiples of 2 in 2n. The root of the contradiction between the two is that the original proposition is true, and it is even more impossible if it is a number larger than 2