2016 AMC 10A Problem 5

P1-10~1 minPrint

Number Properties Solid Geometry Use the answer choices (mod checks, size, form) to eliminate or select

A rectangular box has integer side lengths in the ratio $1: 3: 4$ . Which of the following could be the volume of the box?

Show solution without answering

Key insight

Call the sides k, 3k and 4k. The volume is 12 times k cubed, so it must be 12 times a perfect cube. Only 96 = 12 times 8 works.

Solution

1. The sides are in the ratio $1 : 3 : 4$ and are whole numbers, so they are $k$ , $3k$ and $4k$ for some positive whole number $k$ .
2. The volume is
$k \cdot 3k \cdot 4k = 12 \cdot k \cdot k \cdot k = 12k^3.$
3. List the possible volumes by trying small $k$ :

$k$	box	volume
$1$	$1 \times 3 \times 4$	$12$
$2$	$2 \times 6 \times 8$	$96$
$3$	$3 \times 9 \times 12$	$324$

The volumes jump from $96$ straight to $324$ , so the only choice on the list is $96$ .

The answer is $\boxed{\textbf{(D)}\ 96}$ .

Why this works

A ratio of whole numbers can be written with one scaling number $k$ . Since volume multiplies three lengths, it grows like $k^3$ , so only a few volumes are possible. Being a multiple of $12$ is not enough; what is left after dividing by $12$ has to be a perfect cube.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Write the side lengths as $k$ , $3k$ , $4k$ for some positive integer $k$ . The volume is
$k \cdot 3k \cdot 4k = 12k^3.$

So the volume must be $12$ times a perfect cube: $12 \cdot 1 = 12$ , $12 \cdot 8 = 96$ , $12 \cdot 27 = 324$ , and so on. Among the choices, only $96 = 12 \cdot 2^3$ has this form (a $2 \times 6 \times 8$ box).

The answer is $\boxed{\textbf{(D)}\ 96}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Algebra, Chapter 20: Special Functions (§20.3)
Introduction to Geometry, Chapter 15: Curved Surfaces