Question

Four positive integers $a,$ $b,$ $c,$ and $d$ have a product of $8!$ and satisfy

$ab + a + b = 524,$ $bc + b + c = 146,$ $cd + c + d = 104.$

What is $a - d?$

$4$

$6$

$8$

$10$

$12$

Solution:

Adding $1$ to each equation factors the left sides: $(a + 1)(b + 1) = 525 = 3 \cdot 5^2 \cdot 7, \quad (b + 1)(c + 1) = 147 = 3 \cdot 7^2, \quad (c + 1)(d + 1) = 105 = 3 \cdot 5 \cdot 7.$

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.

2001 AMC 12 Problem 21

Solution:

Problem 21 in Other Years