Question

The greatest positive integer x such that 23 to the power 6+x divides 2000 is

Answer 1

Answer:

The number 23 is prime and divides every 23rd number. In all, there are ⌊200023⌋ = 86 numbers from 1 to 2000 that are divisible by 23. Among those 86 numbers, three of them, namely 23,2⋅23 and 3⋅232 are divisible by 233. Hence 2389|2000! and x=89−6=83.

I really don't understand how 23,2⋅23 and 3⋅232 are divisible by 233 and I hope someone here can shed light on this step in particular, and the overall solution.