If \(p\) is a prime number then \((p - 1)! + 1\) is always divisible by \(p\). 

For example \(18! + 1\) is multiple of \(19, 22! + 1\) is multiple of 23 etc. For example if we have to find the remainder when \(28!\) Is divided by 29, 

We can apply this theorem, we know that \(28!+1\) is multiple of 29, hence when \(28!\) is divided by 29, remainder is -1 or 28.

\(\frac{{28!}}{{29}} = \frac{{(28!\, + 1) - 1}}{{29}} =  - 1\) or \(28\)

Example 27: Find the remainder when \(6.27!\) is divided by 31

Solution: \(6.27!\) can be written as \(1.2.3.27!\)

When \(1.2.3.(27!)\) is divided by 31, remainder is  

\((-30).(-29).(-28)(27!)\) that is equal to \((-30!)\), 

now the remainder is  \(\frac{{ - (30!)}}{{31}} = 1\)