Analysis and solution: For example, seven consecutive integers, 842, 843, 844, 845, 846, 847 and 848, can be divisible by 2, 3, 4, 5, 6, 7 and 8 respectively, that is to say, they are not prime numbers.
Comments: Some students may say how this was discovered, through the quality table or ... we noticed it (n+ 1)! +2，(n+ 1)！ +3，(n+ 1)！ +4，…，(n+ 1)！ The n numbers +(n+ 1) are divisible by 2, 3, 4, …, (n+ 1) respectively, and are continuous n composite numbers.
Where n! Represents the product of 1 to n, that is,1× 2× 3× …× n. 。
2. Write five prime numbers from small to large, so that the number behind is larger than the previous ones 12.
Analysis and solution We know that 12 is a multiple of 2 and 3. If the prime number at the beginning is 2 or 3, then the sum of 12 must also be a multiple of 2 or 3, which will be a composite number, so try from 5.
Five, 17, 29, 4 1, 53 are five prime numbers that meet the conditions.
3.9 consecutive natural numbers, all greater than 80, so how many prime numbers are there at most?
Natural numbers greater than 80 are not prime numbers as long as they are even numbers, so the more odd numbers, the better. Nine consecutive natural numbers have at most five odd numbers, and their digits should be 1, 3, 5, 7, 9. But numbers with five digits greater than 80 must not be prime numbers, so there are only four digits at most.
Verification 10 1, 102, 103, 104, 105, 106, 107,/.
That is, nine consecutive natural numbers greater than 80, of which there can be at most four prime numbers.