Take any 11 numbers from 1 to 20 and prove that there must be two numbers, one of which is a multiple of the other

It is proved that: consider making drawers according to the principle that any two numbers in the same drawer have multiple relations. Divide these 20 numbers into the following ten groups according to odd numbers and their multiples, and regard them as 10 drawers (obviously, they have the above properties): {1, 2, 4, 8, 16}, {3, 6, 12}, {5, 10, 20}, {7, 14}, {9, 18}, {11}, {13}, {15}, {17}, {19} According to the principle of drawer, at least two numbers are taken from the same drawer. Because all the two numbers in the same drawer have multiple relationship, one of the two numbers must be the multiple of the other

There is a number, it is the square of a natural number, and its ten digits are 5, find its one digit number why?

First, let's set this number to 10A + B
Then, the square of 10A + B is 100A & # 178; + B & # 178; + 20ab
And because its tens are five, it's an odd number
That can only be more than 10. 30 such a number
So we can guess: B & # 178; = 16 or 36 [because the square of 16 is 256, the square of 36 is 1296]
So, her single digit is only 6
It's wrong to type 66 at the beginning. 66 is the square of root 66

The square of a natural number is a four digit number, the thousand digit number is 4, and the five digit number is 5______ ．

∵ the square of natural number is a four digit number, thousand digit number is 4, and ∵ 1002 = 10000, 92 = 81, ∵ the natural number can only be two digits, ∵ the one digit number is 5, ∵ 602 = 3600802 = 6400, ∵ its ten digit number may be 6 or 7, = 5625652 = 42

Take any 11 numbers from 1 to 20 and prove that there must be two numbers, one of which is a multiple of the other