2023/035) Prove that $4 . 29! + 5 ! \equiv 0 \pmod {31}$

We have as 31 prime  using Wilsons theorem

$30! + 1 \equiv 0 \pmod {31}$

or $30 * 29! + 1 \equiv 0 \pmod {31}$

as $30  \equiv - 1 \pmod {31}$

so we have  $-1 * 29! + 1 \equiv 0 \pmod {31}$

Or  $ 29! -1 \equiv 0 \pmod {31}$

Or $ 4 * 29! - 4 \equiv 0 \pmod {31}$

adding 124 which is multiple of 31 we get

$ 4 * 29! + 120 \equiv 0 \pmod {31}$

or $ 4 * 29! + 5! \equiv 0 \pmod {31}$

Proved