How to do these type of solid object questions in general?

Question

Assume that n copies of unit cubes are glued together side by side to form a rectangular solid block. If the number of unit cubes that are completely invisible is 30, then what is the minimum possible value of n?

First of all, it's only given this much in the question so, I don't clearly know how the cubes the glued, only side by side or randomly on any side of the cube, the question itself is unclear to me, please help...

Answer 1

The question means that the cubes are glued together to form a rectangular parallelepiped. Suppose they form a $l\times w\times h$ brick, so that $lwh$ small cubes are used.

Then the number of invisible cubes is $lwh$ minus the number of cubes we can see. We see $2$ faces with $lw$ cubes, $2$ with $lh$ cubes, and $2$ with $wh$ cubes, giving $$lwh-2lw-2lh-2wh,$$ but the cubes along the edges have been subtracted twice, so we have to add $2(l+w+h)$ back in. Now the cubes at the corners have been added in $3$ times and subtracted out $3$ times, so we must subtract them.

Thus the number of invisible cubes is $$ lwh-2lw-2lh-2wh+2(l+w+h)-8=30\tag1$$ and we must find positive integers $l,w,h$ that satisfy $(1)$ and minimize $lwh$.

I won't take this any further, since your question only asks for an explanation of the problem. I suggest you work on it, and ask another question if you run into difficulty.

EDIT

This isn't a very good way of looking at it, on second thought. The invisible cubes themselves form a cuboid of dimension $(l-2)\times(w-2)\times(h-2)$.

Answer 2

The question is not worded well. Literally, I would read "glued together side by side" to mean the cubes are glued together in a long straight row, so that if you look at the row of cubes from certain directions you will see each cube (except the ones at the ends) has a cube glued to its "left" side and one glued to its "right" side, as seen from your point of view.

But no blocks would be invisible in such a configuration, and the problem also says "solid block", which suggests more than just a single row.

So imagine you took a Soma cube whose pieces are made by gluing together small wooden cubes, and you solved it by putting all the pieces into one large cubic region with three small wooden cubes along each edge. And then imagine you glued all the pieces to each other in this configuration so they cannot come apart again. That is the kind of "solid block" that I think the problem statement was meant to describe.

In the Soma cube there are $27$ small cubes but only one is invisible. So you will need more cubes.

Answer 3

The different decompositions of $30$ as a product of $3$ integers are :

$$30=\begin{cases}5 \times 3 \times 2\\ 6 \times 5 \times 1\\ 10\times 3 \times 1\\ 15\times 2 \times 1\\ 30 \times 1 \times 1\\\end{cases}$$

giving in each case the following total number of cubes (we have added $2$ to each factor in order to account for "bordering") :

$$n=\begin{cases}7 \times 5 \times 4 &=& 140\\ 8 \times 7 \times 3&=&168\\ 12\times 5 \times 3&=&180\\ 17\times 4 \times 3&=&204\\ 32 \times 3 \times 3 &=& 288\\\end{cases}$$

The least cube-consuming solution is ... the first one with $n=7 \times 5 \times 4 = 140$ cubes.