The minimum multiple of natural number a is 51, and its factors are______ ．

The minimum multiple of natural number a is 51, and its factors are______ ．

The minimum multiple of natural number a is 51. The number is 51. Its factors are 1, 3, 17 and 51. So the answer is 1, 3, 17 and 51

Choose 51 numbers from 1 ~ 100, among which there must be 8 numbers whose common factor is greater than 1. Who can prove it!

There are 25 prime numbers in 100, which is similar to the eratosthene sieve method. The classification of 1 ~ 100 is as follows: Class 1: 1 (take out separately, a total of 1) class 2: multiples of 2 (2,4,6,...) Class 3: multiples of 3 (3,9,15,...) Class 4: multiples of 5 (5,25,35,55,65,85,95, 7 in total)

Choose 51 numbers from 1,2,3.100, and prove that there must be 8 numbers among them, and their greatest common divisor is greater than 1

There are 49 multiples of 2, 33-16-1 = 16 multiples that are multiples of 3 but not multiples of 2 (multiples of 3 to multiples of 6), and so on. There are 6 multiples of 5 that are not multiples of 2 or 3, 3 multiples that are multiples of 7, 3 multiples that are not multiples of 2, 25 prime numbers, and 1, The prime number group itself is also coprime. In order to choose a number whose greatest common divisor is not greater than 1, that is, to be coprime, each group can take up to 7 numbers in addition to the prime array. Therefore, take 7 multiples of 2, 7 multiples of 3, 6 multiples of 5, 3 multiples of 7, 25 prime numbers and 1. There are 49 numbers in total. Any 8 of these 49 numbers are coprime, so if you take 2 more, you must meet the requirements of the topic

What is the least common multiple of Coprime number
If both numbers are coprime, what is the least common multiple of the two numbers?

If two numbers have only the common divisor 1, then they are coprime numbers
So the least common multiple of Coprime number is the product of two numbers