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. 
 A: Hint: $45=1\cdot (-1)\cdot 3\cdot (-3)\cdot 5$
A: 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.
A: You can suppose a as "-3" and b as "-9" 
So this will solve your problem 
(6-(-3))=9
(6-(-9))=15
Both 9 and 15 are factors of 45
A: Notice that $45=3\times3\times5$. Now it is given to us that there are 5 factors and all must be distinct. So let's try to arrange it so that condition is satisfied.
$45=3\times3\times5\times1\times-1$. Oops, $3=3$ (Not distinct)
This false example made me realise that I have only three factors and I have to make $5$ from them. Then I remembered The dialogue of the movie Stand And Deliver that Negative times Negative Equals Positive.
So, $45=3\times3\times5=1\times3\times3\times5\times1=1\times3\times-3\times5\times-1$.
Is this enough.....??
