Visible cubes

The outer shell (composed of unit cubes) is equal to the the volume of a 6x6x6 cube minus a 4x4x4 cube: $6^3-4^3=216-64=\boxed{152}$

The invisible cubes form a cube of dimensions $4\times 4\times 4$ . Thus the number of visible cubes is $6^3 - 4^3 = (6 - 4)(6^2 + 6\cdot 4 + 4^2) = 2\cdot 76 = \boxed{152}.$

General Case: Considering an $n\times n \times n$ cube, we can see $6n^2-12n+8$ unit cubes.

First proof: Let $w$ be the number of visible cubes. Using the already presented method of subtracting the "inner" cube, we have:

$w=n^3-(n-2)^3=(n-(n-2))(n^2+n(n-2)+(n-2)^2)=2(3n^2-6n+4)$ (using the formula for the difference of two cubes), which is the desired result.

Second proof: There are $8$ corner unit cubes, $n-2$ edge unit cubes on each edge(there are $12$ edges) and $6$ inner faces containing $(n-2)^2$ cubes. Thus,

$w=6(n-2)^2+8+12(n-2)=6n^2-12n+8$ .

All the small $1*1*1$ cubes on the faces, edges, or vertices of the big cube are visible. There are $6$ faces, $12$ edges, and $8$ vertices on a cube.

Each vertex of the big cube has $1$ small cube on it. $8$ vertices $*1$ cube per vertex $=8$ .

Each edge of the big cube has $4$ small cubes that are not already counted in the vertices. $12$ edges $*4$ cube per vertex $=48$ .

Each face of the big cube has $16$ small cubes in it that are not already counted in the edges or vertices. $6$ vertices $*16$ cube per vertex $=96$ .

Add them all up: $8$ for vertices $+48$ for edges $+96$ for faces $=152$ total cubes.

The answer is

There are 8 corners (sometimes also called "vertices") on the large cube. That's 8 small cubes

There are 12 edges to the cube. Each edge has 4 small cubes in the middle (we already counted the outer cubes of the edges as they are corners). So that's $4 \times 12 = 48$ small cubes.

Finally, each of the 6 faces of the cube has 16 small cubes that don't lie on any edge. That's a total of $6 \times 16 = 96$ small cubes to count for the faces.

So, the answer must be $8 + 48 + 96 = \large{\boxed{152}}$ small cubes being visible.

There are (6-2)^3=64 little cubes that are not visible (look at the inner 4×4×4 cube), so 216-64=152 little cubes are visible.

You can't just calculate $6 \times 6 \times 6$ because that gives you the number of faces not cubes. The cubes in the corners show 3 faces, and the ones in the edges show 2. Those extra faces need to be subtracted to get the correct solution.

Instead of counting the number of unit cubes on the surface, count the number of unit cubes that are not on the surface. The inside of the large cube is another cube with dimensions $4\times4\times4$ , so there are $64$ unit cubes inside the large cube. There are $216-64=\boxed{152}$ cubes which are on the outside of the large cube.

Side 1) 6 x 6= 36

Side 2) 5 x 6= 30

Sides 3+4) 2(5 x 5)= 50

Side 5) 4 x 5= 20

Side 6) 4 x 4= 16

Solution----- 36+30+50+20+16 = 152

2 whole faces: $6 \times 6 \times 2 = 72$

2 partial faces, with the opposite end rows being accounted for: $4 \times 6 \times 2 = 48$

2 partial faces with all 4 sides being accounted for: $4 \times 4 \times 2 = 32$

$72 + 48 + 32 = 152$

I didn't use the "inner cube method" (good method though: really fast!) because I didn't think about it; so I thought my solution could be useful to people who may want an alternative to that method. Consider one of the six surfaces of the cube - call it A. If you draw it (and please divide it into the 6x6 squares it is made of), you can see that: - There are $4$ cubes (at the vertices) that this surface has in common with two other surfaces; - There are $16$ cubes (at the corners) that this surface has in common with another surface; - There are $16$ cubes (in the middle) that this surface has only on itself.

And this happens for all the surfaces: at every vertice we have three surfaces that bump into each other and at every corner we have two of them. To count the total, we can simply add up all the cubes; but pay attention: this isn't simply multiplying every row of the table above by six. For example, think about the four cubes that appear on three surfaces. That means that if we count them on surface A and then consider surface B, which is consecutive to A, we don't need to count them again, because they are literally the same cubes; and this is the same for another surface, call it C, consecutive to A and to B. To be more general, if a cube is in common with $N$ surfaces, we have to multiply it for $\frac{6}{N}$ to get the total. So, for each row of the table above, we have:

Add up the numbers in the last column to get the total: $8 + 48 + 96 = 152$ .

I counted :) I counted the number of columns (6 6 would have repeated the corners): 20, multiplied 6 20, then accounted for the 16 squares visible on the top and the bottom. (16*2) + 120 = 152

Next level inner cube is 4x4x4. Subtract this sum from 6x6x6. 216 minus 64 is 152.

Outer cube - inner cube => (6x6x6)-(4x4x4)=216-64=152

I just counted them

露出3个面：8（8个角）；露出2个面： $12\times4=48$ （12条边）；露出一个面： $6\times4^2=96$ （6个面）。共计8+48+96=152个。

$\begin{aligned} \text{Number of outer cubes} &= 4n(n-1) + 2(n-2)^2 \\ &=4 \times 10 \times (10 -1) + 2 \times (10-2)^2 \\ &= 40 \times 9 + 2 \times 64 \\ & = 360 + 128 = 488 \end{aligned}$