AMC 10 · 2005 · #12

Grade 7 probability
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
probability-basicprime-numbers identify-subproblemscasework ↑ Prerequisites: probability-basicprime-numbers
📏 Medium solution 💡 2 insights
Problem
Twelve fair six-sided dice are rolled at once. Multiply the twelve top numbers together. Find the probability that this product is a prime number.

Pick an answer.

(A)
$\left(\frac{1}{12}\right)^{12}$
(B)
$\left(\frac{1}{6}\right)^{12}$
(C)
$2\left(\frac{1}{6}\right)^{11}$
(D)
$\frac{5}{2}\left(\frac{1}{6}\right)^{11}$
(E)
$\left(\frac{1}{6}\right)^{10}$

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The word "prime" is the whole problem, so Tool #7 (Identify Subproblems) splits the work into three clean pieces: first decide exactly which rolls make the product prime, then count those winning rolls, then divide by the total number of rolls. The first piece is pure number theory — a prime has a single prime factor, which forces eleven of the dice to be 1 and the last to be a prime face. Tool #2 (Make a Systematic List) handles the counting: pick which die is the prime and which prime it shows. Tool #3 (Eliminate Possibilities) is the final match — the count collapses to a single power of 1/6, and only one answer choice has that clean form.

1STEP 1

When is the product prime?

A prime holds one prime factor used once, so eleven dice must show 1 and the last a prime face: 2, 3, or 5 — never 4 or 6.

product prime ⇔ eleven dice show 1 and one die shows 2, 3, or 5
2STEP 2

Count the winning rolls

Pick which die is the prime one (12 ways), then which prime it shows (3 ways); the rest are forced to 1, giving 36 winning rolls.

winning rolls = 12 × 3 = 36
3STEP 3

Count all possible rolls

Each of the twelve dice independently shows one of six faces, so all equally likely rolls number 6¹².

total rolls = 6×6×…×6₁₂ = 6¹²
4STEP 4

Divide and simplify

Divide 36 by 6¹²; since 36 = 6², two sixes cancel below and the probability left is (1/6)¹⁰ — choice (E).

P = 36/6¹² = 6²/6¹² = 1/6¹⁰ = (1/6)¹⁰ → (E)
Answer
(1/6)¹⁰
The result (1/6)¹⁰ is a positive number far below 1, which fits — a prime product is a rare event. It sits neatly between the other tiny choices: it is exactly 6(1/6)¹¹, larger than (C) 2(1/6)¹¹ and (D) 5/2(1/6)¹¹, which are the traps for a solver who counts only one or two prime faces instead of all three, or who forgets to place the prime on any of the twelve dice. Choice (B) (1/6)¹² is the trap for counting a single fixed winning roll and forgetting the 36 ways it can happen.
💡Key takeaway

A product is prime only when one die shows a prime (2,3, or 5) and the rest all show 1, so count 12×3=36 winning rolls out of 6¹² to get (1/6)¹⁰.

  • When is the product prime?
  • Count the winning rolls
  • Count all possible rolls
  • Divide and simplify

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