Question

A rectangular box measures $a \times b \times c,$ where $a,$ $b,$ and $c$ are integers and $1\leq a \leq b \leq c.$ 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:

The condition is $abc=2(ab+ac+bc).$ Since $abc\le6bc$ , we have $a\le6$ . Also $a=1$ and $a=2$ give no positive solutions, so test $a=3,4,5,6$ .

For $a=3$ , $(b-6)(c-6)=36$ , giving $(b,c)=(7,42),(8,24),(9,18),(10,15),(12,12)$ . For $a=4$ , $(b-4)(c-4)=16$ , giving $(5,20),(6,12),(8,8)$ .

For $a=5$ , $(3b-10)(3c-10)=100$ , and the only solution with $a\le b\le c$ is $(b,c)=(5,10)$ . For $a=6$ , $(b-3)(c-3)=9$ , and the only solution with $a\le b\le c$ is $(6,6)$ .

The total number of triples is $5+3+1+1=10$ .

Thus, the correct answer is B.

2015 AMC 10B Problem 25

Solution:

Problem 25 in Other Years