, 2013

Since the edge length of each cube is 1 cm, then the length, width and height of the rectangular prism are 5 cm, 4 cm and 1 cm, respectively.

The top face of the prism has dimensions 5 cm by 4 cm, and so has area 20 cm$^2$.

Similarly, the bottom face on the opposite side of the prism has the same area, 20 cm$^2$.

The front and back faces of the prism each have dimensions 4 cm by 1 cm, and so have area 4 cm$^2$.

The right and left faces of the prism each have dimensions 5 cm by 1 cm, and so have area 5 cm$^2$.

Therefore, the surface area of the rectangular prism is $2\times(20+4+5)=2\times29=58$ cm$^2$.
Suppose that the rectangular prism in question is $l$ cm (cubes) in length and $w$ cm (cubes) wide, with $l\geq w$.

Then the top surface of the rectangular prism has area $(l\times w)$ cm$^2$, the front surface has area $(w\times 1)$ cm$^2$, and the right side has area $(l\times 1)$ cm$^2$.

Therefore, the surface area of the entire rectangular prism is $2\times(lw+w+l)$ cm$^2$.

Since the surface area is 180 cm$^2$, then $2\times(lw+w+l)=180$ or $lw+w+l=90$.

Adding 1 to both sides of this equation, we get $lw+w+l+1=91$, so $w(l+1)+1(l+1)=91$.

Factoring the left side of this equation one step further, we have $(w+1)(l+1)=91$.

Since both $l$ and $w$ are positive integers, then $(w+1)(l+1)$ is the product of two positive integers.

The right side, 91, can be written as the product of two positive integers in exactly two different ways, $1\times 91$ and $7\times13$.

Since both $l$ and $w$ are positive integers, then $2\leq(w+1)\leq(l+1)$ and so neither factor can be equal to 1.

Therefore $(w+1)=7$ and $(l+1)=13$ (since $l\geq w$), and so $w=6$ and $l=12$.

Since the rectangular prism is 6 cubes wide and 12 cubes in length, then it has $6\times 12=72$ cubes in total.
As shown in part (b), the surface area of the original prism without the removal of the internal prism, is $2\times(lw+w+l)$ cm$^2$.

To find the surface area of the frame, we must account for the area that is lost and that which is gained by removing the internal prism.

The area of the original prism that is lost is equal to the front and back rectangular faces of the internal prism.

Since the width of the original prism is $w$ cm and the internal prism is located $k$ cm from each side of the original prism, then the width of the internal prism is $(w-2k)$ cm.

Similarly, the length of the internal prism is $(l-2k)$ cm.

(Note that $w-2k>0$ and $l-2k>0$, so $w>2k$ and $l>2k$.)

Thus, the area from the original prism that is lost by removing the internal prism is $2\times(w-2k)\times (l-2k)$ cm$^2$.

The area that is gained by removing the internal prism is equal to the top, bottom, left, and right rectangular surfaces of the internal prism.

Since the width of the internal prism is $(w-2k)$ cm and its thickness is 1 cm, then the total area of the top and bottom faces is $2\times (w-2k)\times 1$ cm$^2$.

Similarly, since the length of the internal prism is $(l-2k)$ cm and its thickness is 1 cm, then the total area of the right and left faces is $2\times (l-2k)\times 1$ cm$^2$.

Summarizing, the surface area of the original prism without the removal of the internal prism, is $2\times(lw+w+l)$ cm$^2$.

The area of the original prism that is lost by removing the internal prism is, $2\times(w-2k)\times (l-2k)$ cm$^2$.

The area that is added to the area of the original prism by removing the internal prism is, $\text{(2 (w-2k) 1 + 2 (l-2k) 1) cm 2}$ or $2\times (w-2k+l-2k)=2\times (w+l-4k)$ cm$^2$.

That is, the surface area of the frame, in cm$^2$, is $$2\times(lw+w+l)-2\times(w-2k)\times (l-2k)+2\times (w+l-4k).$$ Since the surface area of the frame is 532 cm$^2$, then we equate this with the expression for the surface area and simplify the resulting equation. $$\begin{aligned} 2\times(lw+w+l)-2\times(w-2k)\times (l-2k)+2\times (w+l-4k)& =532\\ lw+w+l-(w-2k)\times (l-2k)+(w+l-4k)& =\tfrac{532}{2}\\ lw+w+l-(lw-2kw-2kl+4k^2)+(w+l-4k)& =266\\ 2w+2l+2kw+2kl-4k^2-4k& =266\\ w+l+kw+kl-2k^2-2k& =\tfrac{266}{2}\\ kw+kl-2k^2+w+l-2k& =133\\ k(w+l-2k)+1(w+l-2k)& =133\\ (w+l-2k)(k+1)& =133 \end{aligned}$$ Recall that $w>2k$ and $l>2k$, so then $(w+l-2k)$ is a positive integer since $w,l$ and $k$ are positive integers.

Therefore, $(w+l-2k)(k+1)$ is the product of two positive integers.

Expressed as the product of two positive integers, 133 can only be written as $1\times 133$ or $7\times 19$.

Note that $(k+1)>1$ , and $(w+l-2k)>2k+2k-2k=2k>1$, since $k$ is a positive integer.

That is, $(k+1)\neq1$ and $(w+l-2k)\neq1$, and so either $k+1=7$ or $k+1=19$.

If $k+1=7$, then $k=6$ and $w+l-2k=19$ or $w+l-12=19$, and so $w+l=31$.

Since $w>2k=12$, we require all possible values for $w$ and $l$ such that $w\ge 13$, $l\ge w$, and $w+l=31$.

Written as ordered pairs $(w,l)$ the only possibilities are $(13,18), (14,17)$ and $(15,16)$.

If $k+1=19$, then $k=18$ and $w+l-2k=7$ or $w+l-36=19$, and so $w+l=55$.

Since $w>2k=36$, then $l<55-36=29$.

This is not possible since $l\ge w$.

Therefore, the only possible values for $w$ and $l$ such that the frame has surface area 532 cm$^2$ are 13 and 18, or 14 and 17, or 15 and 16.