2015 AMC 12B Problem 23

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

The condition is

abc = 2(ab + bc + ca)

with

1 \le a \le b \le c

Show

a \le 6,

then for each fixed

a

factor the equation into a product equal to a constant

Numerically equal volume and surface area means

abc=2(ab+bc+ca).

Dividing by

abc

gives

1=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\le\frac{6}{a},

so

a\le6.

The cases

a=1

and

a=2

give no positive solutions. For

a\ne2,

set

\begin{gathered} u=(a-2)b-2a,\\ v=(a-2)c-2a. \end{gathered}

The equation then factors as

uv=4a^2.

For

a=3,

(b-6)(c-6)=36

gives

(b,c)=(7,42),

(8,24),

(9,18),

(10,15),

(12,12).

For

a=4,

(b-4)(c-4)=16

gives

(5,20),(6,12),(8,8).

For

a=5,

the congruence of the two factors modulo

3

leaves only the valid pair

(b,c)=(5,10),

and for

a=6,

(b-3)(c-3)=9

leaves only

(b,c)=(6,6).

Thus there are

5+3+1+1=10

triples.

Thus, the correct answer is B.

Problem 23 in other years