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

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?

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.

How to solve

Strategy Find an Invariant

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.

1STEP 1

Split the goal with Tool #7: the product is divisible by 6 exactly when it is divisible by both 2 and 3.

Product divisible by 6 ⇔ divisible by 2 AND divisible by 3

2STEP 2

Subproblem A: three faces are even {2, 4, 6}; hiding one leaves at least two evens, so a factor of 2 always survives.

|{2, 4, 6}| = 3 → at least 3 - 1 = 2 evens visible

3STEP 3

Subproblem B: {3, 6} are the only multiples of 3; one hidden face removes at most one, so a multiple of 3 always stays visible.

|{3, 6}| = 2 → at least 2 - 1 = 1 multiple of 3 visible

4STEP 4

Both subproblems always hold, so every outcome's product is divisible by 6 — the probability is 1, choice (E).

P =

\frac{\text{favourable outcomes}}{\text{total outcomes}}

\frac{6}{6}

= 1 → (E)

Answer

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 × 20. Hide the 3: visible faces are {1, 2, 4, 5, 6}, product = 240 = 6 × 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

\frac{1}{3}

\frac{1}{2}

\frac{2}{3}

\frac{5}{6}

are what you would compute if you forgot that hiding one face cannot remove all three evens or both multiples of 3.

💡Key takeaway

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.

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