Congruence Modulo involving factorials

Question

How do I show that $23!\equiv 21! \pmod{101}$? I tried using a calculator but the numbers are so big that am finding it hard to prove. How can factorials be broken down so that they can be easily solved?

Answer 1

HINT: What is $\dfrac{23!}{21!}$? And what is that modulo $101$?

Answer 2

We have $$23!-21!=(23)(22)(21!)-21!=(21!)((23)(22)-1)=(21!)(505).$$ Note that $101$ divides $505$.