A rectangular solid is built using $N$ cubes of a side length of 1cm. When viewed from such an angle such that only 3 of the sides of the rectangular solid are visible, there are 231 $cm^2$ of area which is not visible. What is the minimum number of $N$? So far, I got to this: If the solid has dimensions $(a, b, c)$, the total number of square faces for the solid$=6abc$. Hidden faces$=6abc-ab-bc-ac$. $231=6abc-ab-bc-ac$. I know that a sphere has the most volume for the least surface area, so I have to make $a$, $b$, and $c$ as close as possible to each other in order to have the most cubes completely hidden so that 231 is reached the fastest. However, I am stuck here. I know I could do trial and error but this is a timed competition problem and I don't even think you are allowed to use a calculator. Any help is appreciated, thanks!
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You can rearrange to $6abc-231=ab+ac+bc$ where $N=abc$. From there you try various $N$ with $6N>231$ until finding the least such for which $6N-231$ may be written as $ab+ac+bc$ where $abc=N$.
The first one I found (beginning at $N=39$) which worked was $N=45=3\cdot3\cdot5$ where $$6N-231=39=3\cdot3+3\cdot 5+3 \cdot 5.$$
So I think the minimal $N$ is $N=45$ where the dimensions of the rectangular solid are $3 \times 3 \times 5.$ I did use a calculator for this, but it would only require a simple one which might have been allowed in the competition...