Three cube sides are painted black. The cube is now cut to exactly 64 cubes that are the same. What is the largest number of cubes that can have a wall that is painted black?

Only the small cubes cut from the edges of the large cube will have at least two faces painted red. All the rest will have either 1 or 0 faces painted red. There are 8 small cubes which have exactly three sides painted. These are the 8 small cubes located at the 8 vertices (corners) of the big cube. There are 12 edges of the big cube.  Each of these 12 edges of

the big cube consists of the edges of 10 small cubes.  But 2 of

these are at a corner of the big cube, and we have already

conted these.  So we only need to count the other 8 cubes along

each of the 12 edges, which have exactly 2 faces painted red.    

So that's an additional 12x8 or 96.

So the total is 8 small cubes with exactly three faces painted red

plus 96 cubes with exactly two faces painted red. That's 8+96 = 104.

Another way to count them is to start with the 1000 cubes and  

start subtracting.

There are the inner 8x8x8 or 512 small cubes cut from the inside

of the big cube which have no faces painted.

On each of the 6 faces of the big cube there are 8x8 or 64 faces

of small cubes which only have 1 face painted.  So that's

64x6 or 384 small cubes which have 1 face painted.  So subtracting

1000 small cubes  

- 512 small cubes which have no faces painted red

 488 small cubes which have at least one face painted red.

- 384 small cubes which have exactly one face painted red.  

 104 small cubes with at least two faces painted red.

Edwin

Step-by-step explanation:

the correct answer is x/3+10≥32

hope this !

x equals 1

step-by-step explanation:

f(4)=f(×)

x=1