1. 数論/NUMBER THEORY

について、ここに記述してください。

1.1. DIVISIBILITY 19

Problem 1.1.7. Find the greatest positive integer x such that 23^(6+x) divides 2000!

Solution.

The number 23 is prime and divides every 23rd number. Inall, there are 2000/23 = 86 numbers from 1 to 2000 that are divisible by 23.

Among those 86 numbers, three of them, namely 23^{2, 2*23}2 and 3*23^{2 are divisible by 23}2

Hence 23^89 |2000! and x= 89−6 = 83

1.1. Bertrand-Chebyshev Theorem

For all n (N, there exists a prime number p with n < p =<2n .

We will first prove the theorem for the case n<= 2047

Consider the following sequence of prime numbers: