AMC 10 · 2002 · #18
Grade 6 geometry-3dA 3x3x3 cube is made of 27 normal dice. Each die's opposite sides sum to 7. What is the smallest possible sum of all of the values visible on the 6 faces of the large cube?
Pick an answer.
AMC 10 2002 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 big cube is built from $27$ ordinary dice stacked $3\times 3\times 3$. On every die the numbers on opposite faces add to $7$. Orient each die so that the numbers showing on the six outer faces of the big cube add up to as little as possible, and find that smallest total.
Givens: $27$ dice form a $3\times 3\times 3$ cube; On each die, opposite faces sum to $7$, so the pairs are $(1,6)$, $(2,5)$, and $(3,4)$; Only faces on the outside of the big cube are counted; inside faces are hidden; Each die can be rotated freely before it is placed; Answer choices: (A) $60$, (B) $72$, (C) $84$, (D) $90$, (E) $96$
Unknowns: The smallest possible sum of all numbers visible on the outer surface of the big cube
Understand
Restated: A big cube is built from $27$ ordinary dice stacked $3\times 3\times 3$. On every die the numbers on opposite faces add to $7$. Orient each die so that the numbers showing on the six outer faces of the big cube add up to as little as possible, and find that smallest total.
Givens: $27$ dice form a $3\times 3\times 3$ cube; On each die, opposite faces sum to $7$, so the pairs are $(1,6)$, $(2,5)$, and $(3,4)$; Only faces on the outside of the big cube are counted; inside faces are hidden; Each die can be rotated freely before it is placed; Answer choices: (A) $60$, (B) $72$, (C) $84$, (D) $90$, (E) $96$
Plan
Primary tool: #14 Extreme Principle
Secondary: #7 Identify Subproblems, #17 Visualize Spatial Relationships
The question asks for a minimum, so Tool #14 (Extreme Principle) drives everything: push every visible face to its smallest legal value. Tool #7 (Identify Subproblems) makes this manageable by splitting the $27$ dice into groups that all behave the same way — corner dice show $3$ faces, edge dice show $2$, face-center dice show $1$, and the buried die shows none. Tool #17 (Visualize Spatial Relationships) supplies the key fact that faces meeting at a corner or along an edge are never opposite, so their small numbers can all show at once. Minimize each group, multiply by how many dice are in it, and add.
Execute — Answer: D
5.MD.C.4 Step 1 Sort the 27 dice by how many faces show
- Picture where each die sits.
- The $8$ dice at the corners of the big cube each stick out on $3$ faces.
- The $12$ dice in the middle of an edge each show $2$ faces.
- The $6$ dice in the center of a face each show $1$ face.
- The single die buried in the very center shows $0$ faces.
- Check the count: $8+12+6+1=27$, all dice accounted for.
💡 A die's position in the stack fixes how much of it pokes out, so grouping by position groups the dice by how many faces they contribute.
6.G.A.4 Step 2 Smallest total on a corner die
- A corner die shows $3$ faces that meet at one vertex.
- Faces meeting at a vertex are never opposite each other, so those three numbers come from three different pairs: one from $(1,6)$, one from $(2,5)$, one from $(3,4)$.
- To make the sum small, take the smaller number in each pair: $1$, $2$, and $3$.
- So the least a corner die can show is $1+2+3=6$.
💡 Because opposite faces sum to $7$, the three faces around a corner sit in three separate pairs, letting you show $1$, $2$, and $3$ together.
6.G.A.4 Step 3 Smallest totals on edge and face dice
- An edge die shows $2$ faces that share an edge; these are also never opposite, so they come from two different pairs.
- The smallest choices are $1$ and $2$, giving $1+2=3$.
- A face-center die shows just $1$ face, and the smallest number on any face is $1$.
- So an edge die can drop to $3$ and a face-center die to $1$.
💡 Adjacent faces are never opposite, so their smallest numbers can face out at the same time; a lone face just uses the $1$.
4.OA.A.3 Step 4 Add up the whole cube
- Multiply each smallest value by how many dice are in that group and add.
- Corners: $8\times 6=48$.
- Edges: $12\times 3=36$.
- Face centers: $6\times 1=6$.
- Buried die: $1\times 0=0$.
- Total $=48+36+6+0=90$.
- Because each die was minimized on its own and dice don't restrict one another, this really is the smallest possible sum, so the answer is (D).
💡 Minimizing every die separately and summing gives the true minimum, since one die's orientation never limits another's.
5.MD.C.4 Picture where each die sits. The $8$ dice at the corners of the big cube each st 6.G.A.4 A corner die shows $3$ faces that meet at one vertex. Faces meeting at a vertex 6.G.A.4 An edge die shows $2$ faces that share an edge; these are also never opposite, s 4.OA.A.3 Multiply each smallest value by how many dice are in that group and add. Corners Review
Reasonableness: The total $90$ sits right inside the answer range, and a rough estimate agrees: there are $8(3)+12(2)+6(1)=54$ visible faces, and the smallest numbers used average a little under $2$ per face, so a total near $90$ is expected. It must beat the largest choices ($96$) since we deliberately used the small faces, and it should exceed $60$ since corners are forced to include a $3$. Only $90$ fits both, confirming (D).
Alternative: Add the surfaces layer by layer instead of die by die: the top and bottom $3\times 3$ layers plus the four side walls each want their smallest numbers, and orienting every die to its minimal exposed set reproduces the same corner/edge/face contributions of $6$, $3$, and $1$, again totaling $90$.
CCSS standards used (min grade 6)
5.MD.C.4Measure volumes by counting unit cubes (Sorting the $27$ unit dice by position — $8$ corners, $12$ edges, $6$ face centers, $1$ core — and checking the count totals $27$.)6.G.A.4Represent three-dimensional figures using nets and find surface area (Reasoning about which faces of a die are exposed and using that faces meeting at a corner or edge are never opposite, so the small numbers $1,2,3$ can show together.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Multiplying each group's minimum by its count and adding: $8(6)+12(3)+6(1)=90$.)
⭐ Faces that meet at a corner or an edge are never opposite, so a die can show its smallest numbers ($1$, $2$, $3$) at once — minimize each die by its position, then add.
⭐ Faces that meet at a corner or an edge are never opposite, so a die can show its smallest numbers ($1$, $2$, $3$) at once — minimize each die by its position, then add.
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