Question

A rectangular box measures $a \times b \times c,$ where $a,$ $b,$ and $c$ are integers and $1 \le a \le b \le c.$ The volume and the surface area of the box are numerically equal. How many ordered triples $(a, b, c)$ are possible?

$4$

$10$

$12$

$21$

$26$

Solution:

Numerically equal volume and surface area means $abc = 2(ab + bc + ca).$ Rearranging shows $a \le 6,$ and $a = 1$ or $a = 2$ give no solutions.

For each remaining $a,$ the equation factors: $a = 3$ gives $(b - 6)(c - 6) = 36$ with $5$ solutions; $a = 4$ gives $(b - 4)(c - 4) = 16$ with $3$ solutions; $a = 5$ gives $(3b - 10)(3c - 10) = 100$ with $1$ valid solution; and $a = 6$ gives $(b - 3)(c - 3) = 9$ with $1$ valid solution. That is $5 + 3 + 1 + 1 = 10$ triples.

Thus, the correct answer is B.

2015 AMC 12B Problem 23

Solution:

Problem 23 in Other Years