AMC 10 · 2005 · #11
Grade 6 geometry-3dA wooden cube n units on a side is painted red on all six faces and then cut into n3 unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is n?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A wooden cube $n$ units on a side is painted red on all six outer faces, then sliced into $n^3$ unit cubes. Counting every little face of every unit cube, exactly one-fourth of those little faces are red. Find the value of $n$.
Givens: The big cube has side length $n$ and is painted red on all six faces; It is cut into $n^3$ unit cubes; Red little faces make up exactly $\tfrac14$ of all the little faces; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Unknowns: The side length $n$ of the original cube
Understand
Restated: A wooden cube $n$ units on a side is painted red on all six outer faces, then sliced into $n^3$ unit cubes. Counting every little face of every unit cube, exactly one-fourth of those little faces are red. Find the value of $n$.
Givens: The big cube has side length $n$ and is painted red on all six faces; It is cut into $n^3$ unit cubes; Red little faces make up exactly $\tfrac14$ of all the little faces; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #17 Visualize Spatial Relationships, #3 Eliminate Possibilities
Everything can be written in terms of the one unknown $n$, so Tool #4 (Introduce a Variable) drives the solution. Two counts are needed: the total number of little faces and the number of red little faces. The total is easy — $n^3$ cubes with $6$ faces each. The red count needs Tool #17 (Visualize Spatial Relationships): the red little faces are exactly the pieces of the big cube's painted skin, so they cover the original surface of $6$ faces of area $n^2$. Comparing red to total gives a clean fraction $\tfrac1n$; setting it equal to $\tfrac14$ pins down $n$. Tool #3 (Eliminate Possibilities) then matches the value to the answer list.
Execute — Answer: B
6.EE.A.1 Step 1 Count all the little faces
- Cutting the big cube gives $n^3$ unit cubes, and every unit cube has $6$ faces of its own.
- So the total number of little faces, painted or not, is $6$ times $n^3$.
- This is the whole pile of faces we are taking a fraction of.
💡 Count by cubes first, then multiply by the $6$ faces each cube carries.
6.G.A.4 Step 2 Count only the red faces
- Paint sits only on the outside skin of the big cube.
- When the cube is sliced, that red skin does not move — it just gets divided into unit squares.
- The skin is the big cube's $6$ faces, each an $n\times n$ square holding $n^2$ little squares.
- So the number of red little faces is $6$ times $n^2$, no matter how the red is spread among the small cubes.
💡 Red never appears on a fresh cut, so the red faces are just the original painted surface.
6.EE.A.3 Step 3 Compare red to total
- The fraction of little faces that are red is the red count over the total count.
- Since $n^3=n\cdot n^2$, the factor $6n^2$ in the top cancels with the same factor inside the bottom, leaving $\tfrac1n$.
- So the red share of all faces is simply $\tfrac1n$.
💡 Every unit cube surrenders $6$ faces but only the surface donates red, so the red share thins out as $\tfrac1n$.
6.RP.A.3 Step 4 Solve for n and pick the choice
- The problem says the red share equals one-fourth, so set $\tfrac1n=\tfrac14$.
- Two fractions with numerator $1$ are equal only when their denominators match, giving $n=4$.
- Scanning the answer list, $4$ is choice (B), and no other choice works, so the answer is (B).
💡 If two unit fractions are equal, their denominators are equal.
6.EE.A.1 Cutting the big cube gives $n^3$ unit cubes, and every unit cube has $6$ faces o 6.G.A.4 Paint sits only on the outside skin of the big cube. When the cube is sliced, th 6.EE.A.3 The fraction of little faces that are red is the red count over the total count. 6.RP.A.3 The problem says the red share equals one-fourth, so set $\tfrac1n=\tfrac14$. Tw Review
Reasonableness: Check $n=4$ directly. The total faces are $6\cdot 4^3=6\cdot 64=384$. The red faces are $6\cdot 4^2=6\cdot 16=96$. The ratio is $\tfrac{96}{384}=\tfrac14$, exactly the one-fourth the problem demands. It also makes sense that the red share drops as the cube grows: a bigger cube has proportionally more hidden interior, so a smaller slice of faces stays painted, matching the pattern $\tfrac1n$.
Alternative: Skip the shortcut fraction and solve the raw equation. Set red equal to one-fourth of total: $6n^2=\tfrac14\,(6n^3)$. Multiply both sides by $4$ to clear the fraction: $24n^2=6n^3$. Divide both sides by $6n^2$ (which is not zero): $4=n$. Same answer, $n=4$, choice (B).
CCSS standards used (min grade 6)
6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Writing the total number of little faces as $6n^3$ from $n^3$ cubes with $6$ faces each.)6.G.A.4Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area (Counting the red little faces as the big cube's painted surface: $6$ faces of $n^2$ unit squares, giving $6n^2$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Simplifying the ratio $\tfrac{6n^2}{6n^3}$ to $\tfrac{1}{n}$ by canceling the common factor $6n^2$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Setting the red share $\tfrac1n$ equal to $\tfrac14$ and reading off $n=4$.)
⭐ Cutting adds lots of bare inside faces but no new paint, so the red share is just $\tfrac1n$ — set that equal to $\tfrac14$ to get $n=4$.
⭐ Cutting adds lots of bare inside faces but no new paint, so the red share is just $\tfrac1n$ — set that equal to $\tfrac14$ to get $n=4$.
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