AMC 8 · 2000 · #8

Grade 3 arithmeticgeometry-3d

Archetype Pair-Sum Invariant └ Sub-archetype Universe Total Minus Known

set-partitionspatial-visualizationmulti-digit-arithmetic identify-subproblemspattern-recognition ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 2 insights 📊 Diagram

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Problem

Three dice with faces numbered $1$ through $6$ are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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.

View mode:

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

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

#9 Solve an Easier Problem 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.

$$1 + 2 + 3 + 4 + 5 + 6 = 21$$

💡 This is the invariant. Whichever face is up, down, front, back, left, or right, the six numbers still add to $21$.

#11 Find an Invariant 3.OA.A.1 Step 2

Scale up to three dice.
The total dots on all $18$ faces is $3$ copies of $21$.

$$3 \times 21 = 63$$

💡 Three identical groups of $21$ — a Grade 3 multiplication.

#9 Solve an Easier Problem 3.NBT.A.2 Step 3

Add the dots on the seven visible faces.
The problem shows $1, 1, 2, 3, 4, 5, 6$.

$$1 + 1 + 2 + 3 + 4 + 5 + 6 = 22$$

💡 A straight addition of seven small numbers.

#11 Find an Invariant 3.NBT.A.2 Step 4

Hidden dots $=$ total dots $-$ visible dots.
Subtract.

$$63 - 22 = 41 \;\Rightarrow\; \textbf{(D)}$$

💡 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.

[1] #9 3.OA.D.8 Easier problem first: find the total number of dots on one die. The six faces ar

[2] #11 3.OA.A.1 Scale up to three dice. The total dots on all $18$ faces is $3$ copies of $21$.

[3] #9 3.NBT.A.2 Add the dots on the seven visible faces. The problem shows $1, 1, 2, 3, 4, 5, 6$

[4] #11 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.1 Interpret 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.2 Fluently 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.8 Solve 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 3)

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