# What is the minimum number of blocks to build this?

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 say this is a timed competition problem - what competition is it from? – Chris Apr 14 '13 at 23:38
I'm pretty sure I got this from the AIME. – Ovi Apr 14 '13 at 23:48
I think the equation you get is $231=ab+ac+bc$, and you are trying to minimize $abc$ subject to this equation (and to all variables being positive integers). – Gerry Myerson Apr 15 '13 at 12:56
One solution is $a,b,c=4,9,15$ giving $N=540$, but I don't know if that's minimal. – Gerry Myerson Apr 15 '13 at 13:02
@Gerry Myerson: Under your interpretation the minimal number of cubes, namely $N=115$, would result when $a=b=1$, $c=115$. Maybe the "invisible area" incorporates the hidden surfaces of the little cubes. – Christian Blatter Apr 15 '13 at 13:19

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...

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I think you and OP have misunderstood the question. The total area of your solid is only 78, so it's impossible for the hidden area to be 231. – Gerry Myerson Apr 15 '13 at 12:52
The hidden area does not only refer to the 3 sides you are not viewing, but also to the area of the cubes which are completely within the solid. Those 1cm cubes have all 6 of their sides hidden. The total number of blocks is abc, so the total area of all the blocks is 6*abc. To find the hidden sides of the cubes, we subtract the visible area from the total area, hence the formula $231=6abc−ab−bc−ac$ – Ovi Apr 15 '13 at 13:32
OK, evidently I'm the one who misunderstood the question. – Gerry Myerson Apr 15 '13 at 23:53