AMC 8 · 2000 · #8
Easy mode Grade 3Problem
Picture three dice stacked on top of each other. Each die has faces numbered 1 through 6.
Three dice together have 18 faces in total. In this stacked view, 7 of those faces are visible. The other 11 faces are hidden — some on the back, some on the bottom, some squeezed between dice.
Add up the dots on all the hidden faces. What is that total?
Pick an answer.
AMC 8 2000 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: Three standard dice (faces $1$ through $6$) are stacked. Of the $18$ total faces, $7$ are visible and $11$ are hidden (the backs, bottoms, and the faces pressed between adjacent dice). The visible faces show $1, 1, 2, 3, 4, 5, 6$. Find the total number of dots on the hidden faces.
Givens: Each die has faces numbered $1, 2, 3, 4, 5, 6$; There are $3$ dice, giving $18$ faces in all; $7$ faces are visible, showing $1, 1, 2, 3, 4, 5, 6$; $11$ faces are hidden; Answer choices: (A) $21$, (B) $22$, (C) $31$, (D) $41$, (E) $53$
Unknowns: The total number of dots on the $11$ hidden faces
Understand
Restated: Three standard dice (faces $1$ through $6$) are stacked. Of the $18$ total faces, $7$ are visible and $11$ are hidden (the backs, bottoms, and the faces pressed between adjacent dice). The visible faces show $1, 1, 2, 3, 4, 5, 6$. Find the total number of dots on the hidden faces.
Givens: Each die has faces numbered $1, 2, 3, 4, 5, 6$; There are $3$ dice, giving $18$ faces in all; $7$ faces are visible, showing $1, 1, 2, 3, 4, 5, 6$; $11$ faces are hidden; Answer choices: (A) $21$, (B) $22$, (C) $31$, (D) $41$, (E) $53$
Plan
Primary tool: #11 Find an Invariant
Secondary: #9 Solve an Easier Problem
We are never told which face of each die points where, and we cannot see eleven of them — so trying to identify each hidden face is the wrong path. The invariant (Tool #11) is the sum of the six faces on a single die: $1+2+3+4+5+6 = 21$ no matter how the die is turned. Tool #9 (Solve an Easier Problem) tells us to first solve for one die, then scale to three. Once we know the grand total of dots on all three dice, the hidden total is just total minus visible — a single subtraction.
Execute — Answer: D
3.OA.D.8 Step 1 - Easier problem first: find the total number of dots on one die.
- The six faces are $1, 2, 3, 4, 5, 6$ and their sum does not depend on which way the die is turned.
💡 This is the invariant. Whichever face is up, down, front, back, left, or right, the six numbers still add to $21$.
3.OA.A.1 Step 2 - Scale up to three dice.
- The total dots on all $18$ faces is $3$ copies of $21$.
💡 Three identical groups of $21$ — a Grade 3 multiplication.
3.NBT.A.2 Step 3 - Add the dots on the seven visible faces.
- The problem shows $1, 1, 2, 3, 4, 5, 6$.
💡 A straight addition of seven small numbers.
3.NBT.A.2 Step 4 - Hidden dots $=$ total dots $-$ visible dots.
- Subtract.
💡 Everything on the dice is either seen or hidden, so the two parts must add back to the whole. Subtracting the seen part leaves the hidden part.
3.OA.D.8 Easier problem first: find the total number of dots on one die. The six faces ar 3.OA.A.1 Scale up to three dice. The total dots on all $18$ faces is $3$ copies of $21$. 3.NBT.A.2 Add the dots on the seven visible faces. The problem shows $1, 1, 2, 3, 4, 5, 6$ 3.NBT.A.2 Hidden dots $=$ total dots $-$ visible dots. Subtract. Review
Reasonableness: Check the size of the answer. There are $11$ hidden faces, each showing a number between $1$ and $6$, so the hidden total must lie between $11 \times 1 = 11$ and $11 \times 6 = 66$. Our answer $41$ sits comfortably in that range. The average hidden face value is $41 / 11 \approx 3.7$, very close to a die's average face value of $21/6 = 3.5$ — exactly what we should expect when the hidden faces are a mostly random mix. The arithmetic also checks: $22 + 41 = 63 = 3 \times 21$.
Alternative: Tool #2 (Make a List): for each die separately, list its six faces, cross off the ones that are visible on that die, and add up what is left. This works but forces us to figure out which visible face belongs to which die — extra bookkeeping. Tool #11 (Find an Invariant) skips all of that by using one fact about a die's total, which is why we preferred it.
CCSS standards used (min grade 3)
3.OA.A.1Interpret products of whole numbers as a number of equal groups (Computing the dots on all three dice as $3$ groups of $21$, that is $3 \times 21 = 63$.)3.NBT.A.2Fluently add and subtract within $1000$ (Adding the seven visible face values to get $22$, then subtracting $63 - 22 = 41$ to find the hidden total.)3.OA.D.8Solve two-step word problems using the four operations, including assessing the reasonableness of answers (Chaining "one die totals $21$" $\to$ "three dice total $63$" $\to$ "$63 - 22 = 41$ hidden," and sanity-checking that $41$ lies in the plausible range $[11, 66]$.)
⭐ The six faces of a die always add to $21$, no matter how it is turned. Three dice hold $63$ dots in all, so the hidden dots are just $63 - 22 = 41$ — one subtraction once you spot the invariant.
⭐ The six faces of a die always add to $21$, no matter how it is turned. Three dice hold $63$ dots in all, so the hidden dots are just $63 - 22 = 41$ — one subtraction once you spot the invariant.
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