AMC 8 · 2003 · #12

Grade 7 probability

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

probability-basicdivisibility-rulesprime-factorizationcomplementary-countingsystematic-enumeration systematic-enumerationcaseworkcomplementary-counting ↑ Prerequisites: probability-basicdivisibility-rulesfraction-arithmetic

📏 Medium solution 💡 3 insights

Problem

When a fair six-sided dice is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the faces that can be seen is divisible by $6$ ?

Pick an answer.

(A)

1/3

(B)

1/2

(C)

2/3

(D)

5/6

(E)

AMC 8 2003 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.

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Toolkit + CCSS Solution

Understand

Restated: A fair standard six-sided die is tossed onto a table. One face — the bottom — is hidden, and the other five faces are visible. What is the probability that the product of the five visible numbers is divisible by $6$?

Givens: A standard die has faces $\{1, 2, 3, 4, 5, 6\}$; Exactly one face (the bottom) is hidden after a toss; The other five faces are visible; Each face is equally likely to be the bottom (die is fair); Answer choices: (A) $1/3$, (B) $1/2$, (C) $2/3$, (D) $5/6$, (E) $1$

Unknowns: The probability that the product of the five visible faces is divisible by $6$

Understand

Plan

Primary tool: #11 Find an Invariant

Secondary: #7 Identify Subproblems

There are only $6$ possible outcomes (one for each face being hidden), so we could check them one by one. But Tool #11 (Find an Invariant) gives a shorter path: ask whether the divisibility-by-$6$ status of the visible product is the same for every outcome. Tool #7 (Identify Subproblems) splits "divisible by $6$" into the two independent questions "divisible by $2$?" and "divisible by $3$?". If both answers stay "yes" no matter which face hides, the probability is forced to $1$.

Execute — Answer: E

#7 Identify Subproblems 4.OA.B.4 Step 1

Use Tool #7 to split the goal.
The product is divisible by $6$ exactly when it is divisible by $2$ and by $3$.
So replace one hard question with two easier ones.

$$\text{Product divisible by } 6 \iff \text{divisible by } 2 \text{ AND divisible by } 3$$

💡 Grade 4 factor-pair thinking: $6 = 2 \times 3$, and these two primes are independent factors.

#11 Find an Invariant 4.OA.B.4 Step 2

Subproblem A — is the product always divisible by $2$?
The even faces of the die are $\{2, 4, 6\}$ — three of them.
The hidden face is just one face, so at most one even number can be hidden.
That leaves at least two even numbers visible, so the product picks up a factor of $2$ no matter which face hides.

$$|\{2, 4, 6\}| = 3 \;\Rightarrow\; \text{at least } 3 - 1 = 2 \text{ evens visible}$$

💡 You cannot hide three things by removing only one — the count of evens is the invariant that protects the factor of $2$.

#11 Find an Invariant 4.OA.B.4 Step 3

Subproblem B — is the product always divisible by $3$?
The multiples of $3$ on the die are $\{3, 6\}$ — two of them.
Only one face hides, so at most one of $3$ or $6$ can be missing.
At least one stays visible, so the product always picks up a factor of $3$.

$$|\{3, 6\}| = 2 \;\Rightarrow\; \text{at least } 2 - 1 = 1 \text{ multiple of } 3 \text{ visible}$$

💡 Same idea: you cannot hide two faces with one bottom slot, so a multiple of $3$ always survives.

#11 Find an Invariant 7.SP.C.5 Step 4

Combine the subproblems.
Every outcome keeps a factor of $2$ and a factor of $3$, so every outcome's product is divisible by $6$.
The favourable count equals the total count, so the probability is $1$.

$$P = \dfrac{\text{favourable outcomes}}{\text{total outcomes}} = \dfrac{6}{6} = 1 \;\Rightarrow\; \textbf{(E)}$$

💡 Grade 7 probability: a certain event has probability $1$. Both subproblems are certain, so the combined event is certain too.

[1] #7 4.OA.B.4 Use Tool #7 to split the goal. The product is divisible by $6$ exactly when it i

[2] #11 4.OA.B.4 Subproblem A — is the product always divisible by $2$? The even faces of the die

[3] #11 4.OA.B.4 Subproblem B — is the product always divisible by $3$? The multiples of $3$ on t

[4] #11 7.SP.C.5 Combine the subproblems. Every outcome keeps a factor of $2$ and a factor of $3$

Review

Reasonableness: Spot-check the two riskiest outcomes — the ones where we lose a needed prime. Hide the $6$: visible faces are $\{1, 2, 3, 4, 5\}$, product $= 120 = 6 \times 20$. Hide the $3$: visible faces are $\{1, 2, 4, 5, 6\}$, product $= 240 = 6 \times 40$. Both pass. Every other hidden face leaves both $3$ and $6$ visible, which is even safer. All $6$ cases work, so $P = 1$ matches (E). The trap choices $1/3, 1/2, 2/3, 5/6$ are what you would compute if you forgot that hiding one face cannot remove all three evens or both multiples of $3$.

Alternative: Tool #2 (Make a List): enumerate the six outcomes directly. Hide $1$: product $720$. Hide $2$: $360$. Hide $3$: $240$. Hide $4$: $180$. Hide $5$: $144$. Hide $6$: $120$. Every product is a multiple of $6$, so $P = 6/6 = 1$. Same answer (E), reached by brute force instead of by the invariant argument.

CCSS standards used (min grade 7)

4.OA.B.4 Find all factor pairs for a whole number and identify factors and multiples (Recognizing $6 = 2 \times 3$ and identifying the even faces $\{2, 4, 6\}$ and the multiples of $3$ $\{3, 6\}$ on the die.)
7.SP.C.5 Understand that the probability of a chance event is a number between 0 and 1, with 1 indicating a certain event (Concluding that because every one of the $6$ outcomes yields a product divisible by $6$, the event is certain and its probability is $1$.)

⭐ Only one face hides, and the die has three evens and two multiples of $3$ — too many to wipe out with a single hidden face. So the product is always divisible by $6$, and the probability is $1$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 7)

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