2001 AMC 12 Problem 21

Adding $1$ to each equation factors the left sides: $$\begin{aligned} (a + 1)(b + 1) &= 525 = 3 \cdot 5^2 \cdot 7, \\ (b + 1)(c + 1) &= 147 = 3 \cdot 7^2, \\ (c + 1)(d + 1) &= 105 = 3 \cdot 5 \cdot 7. \end{aligned}$$ Since $525$ has a factor of $25$ while $147$ is not divisible by $5,$ the factor $a + 1$ must carry the $25.$ Among divisors of $525,$ only $a + 1 = 25$ makes $a = 24$ divide $8! = 40320.$ Then $b + 1 = 21,$ $c + 1 = 7,$ and $d + 1 = 15,$ giving $b = 20,$ $c = 6,$ $d = 14.$ (Indeed $24 \cdot 20 \cdot 6 \cdot 14 = 40320 = 8!.$) So $a - d = 24 - 14 = 10.$ Thus, the correct answer is D.