2015 AMC 10B Problems/Problem 25

Problem

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?

$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$

Solution

The surface area is $2(ab+bc+ca)$ , the volume is $abc$ , so $2(ab+bc+ca)=abc$ .

Divide both sides by $2abc$ , we get that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.$

First consider the bound of the variable $a$ . Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we have $a>2$ , or $a\ge 3$ .

Also note that $c\ge b\ge a>0$ , we have $\frac{1}{a}>\frac{1}{b}>\frac{1}{c}$ . Thus, $\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}$ , so $a\le 6$ .

So we have $a=3, 4, 5$ or $6$ .

Before the casework, let's consider the possible range for $b$ if $\frac{1}{b}+\frac{1}{c}=k>0$ .

From $\frac{1}{b}<k$ , we have $b>\frac{1}{k}$ . From $\frac{2}{b}\ge \frac{1}{b}+\frac{1}{c}=k$ , we have $b\le \frac{2}{k}$ . Thus $\frac{1}{k}<b\le \frac{2}{k}$

When $a=3$ , $\frac{1}{b}+\frac{1}{c}=\frac{1}{6}$ , so $b=7, 8, \cdots, 12$ . The solutions we find are $(a, b, c)=(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)$ , for a total of $5$ solutions.

When $a=4$ , $\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$ , so $b=5, 6, 7, 8$ . The solutions we find are $(a, b, c)=(4, 5, 20), (4, 6, 12), (4, 8, 8)$ , for a total of $3$ solutions.

When $a=5$ , $\frac{1}{b}+\frac{1}{c}=\frac{3}{10}$ , so $b=5, 6$ . The only solution in this case is $(a, b, c)=(5, 5, 10)$ .

When $a=6$ , $b$ is forced to be $6$ , and thus $(a, b, c)=(6, 6, 6)$ .

Thus, our answer is $\boxed{\textbf{(B)}\;10}$

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2015 AMC 10B Problems/Problem 25

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