AMC 10 · 2007 · #11
Grade 3 geometry-3dThe numbers from 1 to 8 are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same. What is this common sum?
Pick an answer.
AMC 10 2007 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: The numbers $1$ through $8$ are placed on the eight corners of a cube, one number per corner. The arrangement is chosen so that the four corner numbers around every face add up to the same total. Find that common total.
Givens: The numbers $1, 2, 3, \dots, 8$ are placed on the eight vertices of a cube, one number per vertex; Each of the cube's six faces has the same sum of its four corner numbers; Answer choices: (A) $14$, (B) $16$, (C) $18$, (D) $20$, (E) $24$
Unknowns: The common sum shared by all six faces
Understand
Restated: The numbers $1$ through $8$ are placed on the eight corners of a cube, one number per corner. The arrangement is chosen so that the four corner numbers around every face add up to the same total. Find that common total.
Givens: The numbers $1, 2, 3, \dots, 8$ are placed on the eight vertices of a cube, one number per vertex; Each of the cube's six faces has the same sum of its four corner numbers; Answer choices: (A) $14$, (B) $16$, (C) $18$, (D) $20$, (E) $24$
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #16 Change Focus / Count the Complement, #1 Draw a Diagram
Trying to actually place $1$ through $8$ so every face matches is a maze of guessing. The winning move is Tool #17 (Visualize Spatial Relationships): picture the cube and notice that a face and the face directly across from it never share a corner, yet between them they use up all eight corners. That single spatial fact lets Tool #16 (Change Focus) stop worrying about individual placements and instead add up totals, because the two opposite faces together must hold all eight numbers. Tool #1 (Draw a Diagram) keeps the top-face and bottom-face corners straight while doing it.
Execute — Answer: C
2.NBT.B.5 Step 1 Add up all eight numbers
- No matter how the numbers are arranged, the eight corners always hold the numbers $1$ through $8$.
- Add them once: $1+2+3+4+5+6+7+8 = 36$.
- This is the grand total of everything written on the cube.
💡 The corners always carry the same eight numbers, so their overall total is fixed before any arranging begins.
K.G.B.4 Step 2 Look at a face and its opposite
- Picture the cube's top face and its bottom face.
- The top face rests on four corners and the bottom face rests on the other four corners, and no corner belongs to both.
- So the top and bottom faces together cover all eight corners of the cube exactly once.
💡 Opposite faces of a cube face away from each other, so they can never share a corner and between them they sweep up every vertex.
3.OA.C.7 Step 3 Split the grand total in half
- Let the common face sum be $S$.
- The top face adds to $S$ and the bottom face also adds to $S$.
- Since those two faces together contain all eight numbers, their combined total is the grand total: $S + S = 36$.
- So $2S = 36$, which means $S = 36 \div 2 = 18$.
- The common sum is $18$, which is choice (C).
💡 If two equal groups together make $36$, each group is just half of $36$.
2.NBT.B.5 No matter how the numbers are arranged, the eight corners always hold the number K.G.B.4 Picture the cube's top face and its bottom face. The top face rests on four corn 3.OA.C.7 Let the common face sum be $S$. The top face adds to $S$ and the bottom face als Review
Reasonableness: The result $18$ is one of the listed choices, and it should land in the middle: the smallest four numbers $1,2,3,4$ sum to $10$ and the largest four $5,6,7,8$ sum to $26$, so any face total has to sit between $10$ and $26$ — and $18$ is comfortably inside that range. A different count confirms it: each vertex touches $3$ faces, so adding all six face totals counts every number three times, giving $3 \times 36 = 108$; sharing that across the six faces gives $108 \div 6 = 18$, the same answer.
Alternative: Count each vertex's contribution instead of pairing faces. Every corner number is used by exactly $3$ faces, so if you add up the sums of all six faces you are really adding every number three times over: the six face-sums together equal $3 \times (1+2+\dots+8) = 3 \times 36 = 108$. Because all six faces share one common sum $S$, that same grand tally is $6S$. Setting $6S = 108$ gives $S = 18$.
CCSS standards used (min grade 3)
2.NBT.B.5Fluently add and subtract within 100 (Adding the eight vertex numbers $1+2+\dots+8$ to get the grand total $36$.)K.G.B.4Analyze and compare two- and three-dimensional shapes (Recognizing that a cube's two opposite faces share no vertex and together cover all eight vertices.)3.OA.C.7Fluently multiply and divide within 100 (Halving the grand total ($2S = 36$, so $S = 18$) to find the common face sum.)
⭐ Two opposite faces of a cube use all eight corners once, so each face just carries half of the total $1+2+\dots+8 = 36$, which is $18$.
⭐ Two opposite faces of a cube use all eight corners once, so each face just carries half of the total $1+2+\dots+8 = 36$, which is $18$.
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