Question

If all the surfaces of a cube of \( 15 \, \text{cm} \) side are coloured black and then cut into smaller cubes of sides \( 3 \, \text{cm} \) each, then find how many cubes will have only one surface coloured in black?

• A. 18
• B. 24
• C. 36
• D. 54

Hint:

Remember the formulas for a cut painted cube:
- 3 faces painted = \( 8 \) (always the corners)
- 2 faces painted = \( 12(n - 2) \) (on the edges)
- 1 face painted = \( 6(n - 2)^2 \) (on the face centers)
- 0 faces painted = \( (n - 2)^3 \) (the inner core)

Answer 1

The Correct Option is D

Step 1: Find cuts per edge.
The big cube has side $15$ cm and the small cubes side $3$ cm, so each edge splits into $n=\dfrac{15}{3}=5$ pieces.
Step 2: Recall which cubes get exactly one painted face.
Only the cubes sitting at the centre of each face touch paint on a single side.
Step 3: Count centre cubes on one face.
Ignoring the outer ring leaves an inner block of $(n-2)\times(n-2)=(5-2)^2=9$ cubes per face.
Step 4: Use the standard formula.
One-face cubes $=6\times(n-2)^2$ since a cube has $6$ faces.
Step 5: Substitute the numbers.
\[ 6\times(5-2)^2 = 6\times 9 = 54 \]
Step 6: State the count.
So $54$ small cubes have exactly one black face.
\[ \boxed{54} \]