= 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

== Bertrand-Chebyshev Theorem ==

https://proofwiki.org/wiki/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:
  2,3,5,7,13,23.43....