2021/031) Given $(4-a)(4-b)(4-c)(4-d)(4-e)=12$ a,b,c,d,e are negative or positive integers which have different values Find the sum of a+b+c+d+e

First let us factor 12 we have 12 = 2 * 2 * 3.

The above has 3 factors and given expression has 5 factors so
12 = 1 * 1 * 2 * 2 * 3
In the above we ave 5 factors but not all 5 are different and to make them different we can have (knowing that -ve numbers are allowed
12 = 1 * (-1) * 2 * (-2) * 3
The (4-a,4-b,4-c,4-d,4-e) can be chosen as permutation of(1,-1,2,-2,3) giving a sum 3 and so a+b+c+d+e = 20 -3 or 17.