Question

Two adjacent faces of a cube are painted red and out of the remaining sides, two opposite faces are painted green. The cube is then cut into 125 equal smaller cubes. How many smaller cubes will have at most two sides painted?

• A. 121
• B. 123
• C. 117
• D. 113

Hint:

In painted-cube problems, count the 3-face corner cubes first. With \(n^3\) small cubes, “at most two faces painted” \(=\ n^3 - (\text{# of 3-face corners})\).

Answer 1

The Correct Option is B

The big cube is cut into \(5\times5\times5\) small cubes \(\Rightarrow\) total \(=125\).
“At most two sides painted” means we must \emph{exclude only} those small cubes that have 3 painted faces (corner cubes where three painted faces meet).
Painted faces: two adjacent Red faces and (from the remaining four) two \emph{opposite} Green faces.
Only the edge common to the two Red faces meets each Green face at its endpoints; hence exactly two corners have three painted faces (R–R–G).
Therefore, number with at most two faces painted \(=125-2=123\).
\(\boxed{123}\)