2016/016) Show that $4 * (29!) + 5! \equiv 0 (\,mod\, 31) $

because 31 is prime we have as per wilson's theorem
$30!  \equiv -1 (\,mod\, 31)\cdots(1) $
and also $30 * (-1) = -30  \equiv 1 (\,mod\, 31) =>30^{-1} = \equiv 1 (\,mod\, 31)\cdots(2)$
from (1) and (2)
$29!  \equiv 1 (\,mod\, 31)$
or $ 4 * 29!  \equiv 4 (\,mod\, 31) $
or $ 4 * 29! + 5!   \equiv 4 + 120 (\,mod\, 31) \equiv  124 (\,mod\, 31) \equiv 0 (\,mod\, 31)$