Four positive integers $a$, $b$, $c$, and $d$ have a product of $8!$ and satisfy:
\[\begin{array}{rl} ab + a + b & = 524 \\  bc + b + c & = 146 \\  cd + c + d & = 104 \end{array}\]
What is $a-d$?
$\text{(A) }4 \qquad \text{(B) }6 \qquad \text{(C) }8 \qquad \text{(D) }10 \qquad \text{(E) }12$
Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows:
\begin{align*} (a+1)(b+1) & = 525 \\  (b+1)(c+1) & = 147 \\  (c+1)(d+1) & = 105 \end{align*}
Let $(e,f,g,h)=(a+1,b+1,c+1,d+1)$. We get:
\begin{align*} ef & = 3\cdot 5\cdot 5\cdot 7 \\  fg & = 3\cdot 7\cdot 7 \\  gh & = 3\cdot 5\cdot 7 \end{align*}
Clearly $7^2$ divides $fg$. On the other hand, $7^2$ can not divide $f$, as it then would divide $ef$. Similarly, $7^2$ can not divide $g$. Hence $7$ divides both $f$ and $g$. This leaves us with only two cases: $(f,g)=(7,21)$ and $(f,g)=(21,7)$.
The first case solves to $(e,f,g,h)=(75,7,21,5)$, which gives us $(a,b,c,d)=(74,6,20,4)$, but then $abcd \not= 8!$. We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by $7$. (Also, a - d equals $70$ in this case, which is way too large to fit the answer choices.)
The second case solves to $(e,f,g,h)=(25,21,7,15)$, which gives us a valid quadruple $(a,b,c,d)=(24,20,6,14)$, and we have $a-d=24-14 =\boxed{10}$.