calculate the sum of 5 numbers, knowing the product of their values minus 4

Question

$a,b,c,d$ and $e$ are five different integers. If $(4-a)(4-b)(4-c)(4-d)(4-e)=12$, then calculate $a+b+c+d+e$.

Answer 1

I found the solution: $12$ can be expressed as a product of five integers only as $1\times(-1)\times2\times(-2)\times3 =12$, the $a-4=1, b-4=-1, c-4=2, d-4=-2$ and $e-4=3$. we deduce from that $a+b+c+d+e=23$.