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2005 AMC 10A Problem 11

2005 AMC 10A Problem 11AMC 10 Step by Step

Name: Date: September 13, 2026

Solid Geometry · difficulty 2 of 5 · about 3 min · position P11-15

A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^3$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n$ ?

A)
$3$
B)
$4$
C)
$5$
D)
$6$
E)
$7$

Answer

Key insight

Red faces are the original surface, 6n^2 unit squares, out of 6n^3 unit-cube faces in all, so the fraction is exactly 1/n.

Solution

Count red faces and total faces separately.

Red faces: paint only reaches the outside of the big cube. Each of its $6$ faces is an $n \times n$ grid of unit squares, so there are $6n^2$ red unit faces.

Total faces: there are $n^3$ unit cubes with $6$ faces each, so $6n^3$ faces.

The condition says
$\frac{6n^2}{6n^3} = \frac{1}{n} = \frac{1}{4},$
so $n = 4$ .

The answer is $\boxed{\textbf{(B)}\ 4}$ .

Why this works

Cutting does not create or destroy paint: the red unit faces tile the original surface exactly. Comparing surface (grows like $n^2$ ) to total face count (grows like $n^3$ ) always produces the ratio $1/n$ , so the answer is forced without ever classifying cubes as corner, edge, or interior.

Alternative approach

Test $n = 4$ : the $4 \times 4 \times 4$ cube has $64$ small cubes with $384$ faces, and its surface is $6 \cdot 16 = 96$ squares; $96/384 = \frac{1}{4}$ .

The trap

Counting red cubes (cubes with at least one painted face) instead of red faces.

Common mistakes

Counting red cubes (cubes with at least one painted face) instead of red faces.
Using the surface area $6n^2$ but comparing it to $n^3$ (the number of cubes) rather than $6n^3$ faces, which gives $n = 24$ , not among the choices.

Techniques

Set up the equation/formula and compute; no special trick needed