take a bunch of unit cubes (1x1x1) and put them together to form one larger cube. (?x?x?)?

Question

I then take out my paint and paint 'some' (this term is suppose to be vague) of the faces/sides of the large cube. (when I paint a side, I paint the entire side)
Then the large cube is taken apart into its unit cubes and exactly 24 of the small cubes have no paint on them.

(A) How big was the large cube?

Answer 1

you had 24 in the middle= 2*3*4
and they didn't have paint on them so there should have been another cube on each direction which was painted, so adding 2 to each of these numbers.....the larger cube is 4*5*6

Answer 2

Suppose you have an n×n×n cube. First we find the number of pieces which could not possbily have been painted, which is necessarily no greater than the number which actually recieved no paint. The number which could not possibly have been painted is simply the number of pieces in a cube 2 units smaller, so (n-2)³≤24, thus n-2 ≤ ∛24, and since n is an integer this implies that n-2≤2, n≤4. However, if your cube is size three, then there are only 27 small cubes, and at least 9 of them have paint on them, leaving no more than 18 with no paint. But 24 have no paint. We therefore deduce that n≥4, and from this and the previous that n=4.

You didn't ask, but the cube was painted on four sides, and the two unpainted sides are opposite of each other.

Edit: actually, after reading antoine's answer, I realize that 3 sides with a common corner being painted also works.

Answer 3

let x be the dimension of one side of the large cube, then for having painted one side:
x^2(x-1) = 24
this tells you that you must have painted more than one side as the equation doesn't hold for x being an integer.
So consider 2 sides being painted. Either
(x-2)x^2 = 24 for opposite sides being painted or
x(x-1)^2 = 24 for perpendicular sides being painted
but again both of these equations don't hold.
We now consider 3 sides being painted:
one possibility is
(x-2)(x-1)x = 24
immediately this stands out as x = 4 is a solution
so the large cube was (4x4x4) with 3 sides painted

Answer 4

There's a reducing equation to solve this depending on sides painted. Since you're dealing only with whole numbers, you can try different combinations and see which works. I found this combination works:

x^3 - x^2 - x^2 - x(x-2) = 24

x = 4

The larger cube had 2 opposite sides and 1 adjoining side painted. Its volume was 64.

Three sides sharing 1 corner does not work as it would leave 27 smaller cubes without paint... sorry pascal...

Answer 5

4 faces of the massive cube have been painted and a couple of faces opposite one yet another weren't. the massive cube replaced into 5 x 5 x 5. That go away 40 5 small cubes in an oblong stable of three x 3 x 5 that have no components painted.

Answer 6

36