Problem
Counting painted unit cubes
Unit cubes are stacked into a staircase solid. Every face touching the air is painted, but not the faces on the table. There are 37 cubes in all. Count the cubes with exactly three painted faces.
Reasoning · Grade 6-1 Surface Area and Volume of Solids
Your answer
Watch out Counting every outer face overcounts: faces on the table stay unpainted.
How to solve
Strategy Make a Systematic List — Counting painted faces off the oblique drawing is hopeless, because most of the solid is hidden behind the front cubes. So I re-organise the picture into a top view with a number written on each of the 12 base squares — the height of that column. Once the solid is a 4-by-3 table of heights, every question about a single cube ('is the face on its left painted?') turns into a comparison of two numbers, and I can walk through the cubes in a fixed order without missing or repeating any. I check the finished list against a real stack of blocks at the end.
1STEP 1
Redraw the solid as a table of column heights
Redraw the solid as a table of heights.
16 + 13 + 8 = 37
This is the same move as counting volume by counting unit cubes: a stack of cubes is completely described by how tall each column is, and a 4-by-3 table of twelve numbers is far easier to work with than a drawing where half the solid is hidden.
5.MD.C.4Organize Information In More Ways2STEP 2
Work out which of a cube's six faces can be painted at all
Only faces with no neighbour get paint.
painted faces of a cube = #{up, left, right, front, back that have no neighbour}
Paint only reaches the outside of the solid, so 'is this face painted?' is really 'is this face part of the surface?'. Since no cube floats, the downward direction is settled for every cube at once and only five directions are left to check.
7.G.B.6Visualize Spatial Relationships3STEP 3
Turn 'is there a neighbour?' into a comparison of two heights
A neighbour is found by comparing two heights.
up painted ⇔ z = h, side painted ⇔ neighbour height ＜ z
Comparing two whole numbers is something a young solver can do instantly, and this step replaces every act of 3-D imagination with exactly that comparison.
5.MD.C.4Organize Information In More WaysAsking whether a side is painted becomes a comparison of two column heights, which needs no imagining at all.
Why?
A cube's side is hidden exactly when the neighbouring column reaches at least as high as that cube sits.
🧱Transitivity of orderIf the first is smaller than the second and the second smaller than the third, the first is smaller than the third, so pairwise comparisons chain into one full order.
Why?
Each cube has one definite neighbour in each direction, so one comparison settles one face and no face is judged twice.
🧱One-to-one correspondenceIf each thing here pairs with exactly one thing there, the two groups are the same size.
4STEP 4
Split the search: top cubes and buried-top cubes
A top cube needs two open sides.
top cube: 1 + (open sides) = 3 → 2 open sides; buried-top cube: 0 + (open sides) = 3 → 3 open sides
Splitting into two cases means you never have to hold five directions in your head at once — you only ask 'how many sides are open?' and compare with a target of 2 or 3.
4.OA.A.3Make A Systematic List5STEP 5
Check the 12 top cubes one by one
There are 6 such top cubes.
top cubes with exactly three painted faces: B1, B2, B3, B4, M2, F2 → 6
Twelve columns, one top cube each, four numbers to compare per cube — this is a list a fourth grader can write out completely, and writing it out is exactly what stops an off-by-one.
7.G.B.6Make A Systematic List6STEP 6
Hunt for buried-top cubes with three open sides
Just 1 buried-top cube has three open sides.
buried-top cubes with exactly three painted faces: M4 at level 3 → 1
This is the one cube that is easy to miss, because its top is covered — it only shows up if you remember that a cube can be painted on three sides without being painted on top. It sits in the notch of the staircase, where the drawing shows two of its side faces at once.
7.G.B.6Visualize Spatial Relationships7STEP 7
Add the two groups
Adding gives 7.
6 + 1 = 7
The two cases were chosen so that no cube can be in both, so the counts simply add.
4.OA.A.3Make A Systematic ListAnswer
7 cubes
6 + 1 = 7
The answer has to be a whole number between 0 and 37, and 7 is comfortably inside that range. A stronger check is to sort all 37 cubes by how many painted faces they have: the same scan gives 3 cubes with 0 painted faces (buried in the tall back-left block), 11 with 1, 12 with 2, 7 with 3 and 4 with 4, and 3 + 11 + 12 + 7 + 4 = 37, so every cube is accounted for exactly once. Those counts also pass an independent test: the total number of painted faces is 0(3) + 1(11) + 2(12) + 3(7) + 4(4) = 72. Counting the same paint a completely different way, the whole solid has 84 unit squares of surface, of which the 12 base squares rest on the table, giving 84 - 12 = 72 painted squares. The two 72s agree, so the face-by-face bookkeeping is right. Finally, a cube with 4 painted faces has to stick out at a step of the staircase, and there are exactly 4 such perches (the tops of Middle-1, Middle-4, Front-1 and Front-4), which matches.
Takeaway
Write the height of every stack on a top-view grid, and then 'is this face painted?' is just 'is the stack next door shorter than me?'
- Redraw the solid as a table of column heights
- Work out which of a cube's six faces can be painted at all
- Turn 'is there a neighbour?' into a comparison of two heights
- Split the search: top cubes and buried-top cubes
- Check the 12 top cubes one by one
- Hunt for buried-top cubes with three open sides
- Add the two groups