A cube of volume 1000 cm3 is divided into small cubes of 1 cm3 each, and all…

Concept: When a cube built from unit cubes has every outer face painted, a unit cube stays completely unpainted only if it touches NO outer face. Along each edge of length n unit cubes, the cubes lying strictly inside — one full layer in from every face — form a smaller cube of edge (n − 2), so unpainted cubes = (n − 2)3. Every other cube touches at least one painted face, so painted cubes = n3 − (n − 2)3.

Applying this to the given cube:

1. The volume is 1000 cm3, split into 1 cm3 unit cubes, so the edge length n satisfies n3 = 1000, giving n = 10 (since 103 = 1000).

2. Total unit cubes = n3 = 103 = 1000.

3. Cubes with no side painted = (n − 2)3 = (10 − 2)3 = 83 = 512.

4. Cubes with at least one side painted = total − unpainted = 1000 − 512 = 488.

Cross-check (classifying by number of painted faces):

• Corner cubes (3 faces painted): always 8, for any edge length.

• Edge cubes excluding corners (2 faces painted): 12(n − 2) = 12 × 8 = 96.

• Face-only cubes excluding edges (1 face painted): 6(n − 2)2 = 6 × 64 = 384.

Adding these independently: 8 + 96 + 384 = 488, matching the subtraction result and confirming the split is consistent.

So, 488 of the 1000 small cubes have at least one side painted, and 512 have no side painted at all.