If $45=(6-a)(6-b)(6-c)(6-d)(6-e)$ then find $a+b+c+d=?$ Consider $(6-a)(6-b)(6-c)(6-d)(6-e)$ are five distinct factors of $45$. What is $a+b+c+d$ The problem I am facing is that I am supposing $b = 1$, $c = 5$, $d = 3$ The problem is coming in supposing the value of $a$ and $e$. So can you tell whether I am on the right path or not.

Any factor of $45$ can be written as followed : $$\pm 3^{r_i}5^{s_i}$$ where $0\leq r_i \leq 2$ and $0\leq s_i\leq 1$

So since they must be all different, we can write: $$(6-a)=3$$ $$(6-b)=5$$ $$(6-c)=-3$$ $$(6-d)=1$$ $$(6-e)=-1$$ From here it is easy to deduct that $$a=3$$ $$b=1$$ $$c=9$$ $$d=5$$ $$e=7$$

It follows $$a+b+c+d=3+1+9+5=18$$ Clearly, the solution is not unique.