Problem 276

Solution 1:

If the length, width and height of the rectangular prism are the
integers $\ell\text{ cm}$, $w\text{ cm}$ and $h\text{ cm}$ respectively, then the areas
of three of its faces are $\$\ell w\text{}$
cm}^2$,$$\ell h\text{}$ cm}^2$
and $wh\text{ cm}^2$.

Thus, each pair of areas shares a common factor, and the common factors
are the dimensions of the prism, $\ell$, $w$ and $h$.

The numbers $12$ and $15$ have $1$ and $3$ as common factors.

If one of $\ell$, $w$ or $h$ is equal to $1$, then the other two dimensions are
$12$ and $15$, but this is not possible since $12\times15$ is not equal to $20$.

If one of the dimensions is equal to $3$, then the other dimensions are $\dfrac{12}{3}=4$ and $\dfrac{15}{3}=5$, whose product gives the
required $20$.

Thus, $\ell$, $w$ and $h$ are equal to $3$, $4$
and $5$, in some order, and so the
volume of the rectangular prism is $\$V=3\text{ cm}\times4\text{ cm}\times5\text{}$ cm}$or$60 $\text{}$ cm}^3$.

Solution 2:

If the length, width and height of the rectangular prism are the
integers $\ell\text{ cm}$, $w\text{ cm}$ and $h\text{ cm}$ respectively, then the areas
of three of its faces are $\$\ell w\text{}$
cm}^2$,$$\ell h\text{}$ cm}^2$
and $wh\text{ cm}^2$.

Thus $\ell w$, $\ell h$ and $wh$ are equal to $12$, $20$ and $15$ in some order, and so $(\ell w)(\ell h)(wh)=12\times20\times15$
or $\ell ^2w^2h^2=3600$. Therefore,
$(\ell wh)^2=3600$, which gives $\ell wh=60$.

Measured in $\text{ cm}^3$, the
volume of the prism is $\$V=\ell$
wh=60$.