Why are there two multiples of 51 random numbers within 100? I'll kneel down for you, or I won't be able to hand in my homework tomorrow!

Why are there two multiples of 51 random numbers within 100? I'll kneel down for you, or I won't be able to hand in my homework tomorrow!

To the contrary:
Suppose there are 51 numbers between 1 and 100, and there is no multiple relationship between any two
Because it is 51, there must be numbers less than or equal to 50
Let the number of the 51 numbers less than or equal to 50 be n, and the set formed is a
Then the number of numbers greater than 50 is 51-n and the set is B
For any a belonging to a, there exists a positive integer k such that 50

Try to prove: take any 51 natural numbers from 1 to 100, and the difference between the two numbers is 50
It's a drawer problem,
I'm going to group up,

50 groups
1 51
2 52
3 53
.
50 100
There must be two numbers in a group, the difference is 50

From the 100 natural numbers from 1 to 100, arbitrarily take 51 numbers, of which there must be two numbers, the difference between them is 50, please say from 1 to 100 the 100 natural numbers% 1

1 to 100 were divided into 50 groups: (1,51) (2,52) (3,53) (4,54) （50,100）
Choose 51 numbers from these 50 groups. According to the drawer principle, one group must choose two numbers, and the difference between the two numbers is 50