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2021 AMC Fall 10B Problem 14

2021 AMC Fall 10B Problem 14AMC 10 Step by Step

Name: Date: October 28, 2026

Basic Probability · difficulty 3 of 5 · about 4 min · position P11-15

Una rolls $6$ standard $6$ -sided dice simultaneously and calculates the product of the $6{ }$ numbers obtained. What is the probability that the product is divisible by $4?$

A) $\frac34$
B) $\frac{57}{64}$
C) $\frac{59}{64}$
D) $\frac{187}{192}$
E) $\frac{63}{64}$

Answer

Key insight

Count failures: the product lacks a factor of 4 only if all dice are odd, or one die shows 2 or 6 and the rest are odd.

Solution

"Divisible by $4$ " happens in many ways, but failing happens in only a few, so count the failures and subtract.

1. Sort each die by how many 2s it gives. A product is divisible by $4$ when it has at least two factors of $2$ .
- $1, 3, 5$ : no factor of $2$ . Chance $\frac36 = \frac12$ .
- $2, 6$ : exactly one factor of $2$ . Chance $\frac26 = \frac13$ .
- $4 = 2 \times 2$ : two factors of $2$ . Chance $\frac16$ .
2. When does the product fail? When the six dice together give zero or one factor of $2$ .
3. Zero factors of 2. All six dice are odd:
$\left(\tfrac12\right)^6 = \tfrac{1}{64}.$
4. Exactly one factor of 2. One die shows $2$ or $6$ , and the other five are odd. The special die can be any of the $6$ dice:
$6 \times \tfrac13 \times \left(\tfrac12\right)^5 = \tfrac{6}{96} = \tfrac{1}{16} = \tfrac{4}{64}.$
5. Chance of failing: $\frac{1}{64} + \frac{4}{64} = \frac{5}{64}$ .
6. Chance of success:
$1 - \tfrac{5}{64} = \tfrac{59}{64}.$

The answer is $\boxed{\textbf{(C)}\ \frac{59}{64}}$ .

Why this works

Whether a product is divisible by $4$ depends only on how many factors of $2$ it has in total, so each die only needs to be sorted into "zero, one, or two 2s." When the event you want is big and messy, the opposite event is often small and easy to count. Count it and subtract from $1$ .

The trap

Treating a roll of 4 like a 2 or 6 (a 4 alone already makes the product divisible by 4), or forgetting that any of the 6 dice can be the lone even one.

Concise solution

Track only factors of $2$ . Each die contributes $0$ factors of $2$ (a roll of $1, 3, 5$ ; probability $\frac12$ ), exactly $1$ factor (a roll of $2$ or $6$ ; probability $\frac13$ ), or $2$ factors (a roll of $4$ ; probability $\frac16$ ).

The product fails to be divisible by $4$ when the total number of factors of $2$ is $0$ or $1$ .

Zero factors: all six dice odd, probability $\left(\frac12\right)^6 = \frac{1}{64}$ .
Exactly one factor: one die shows $2$ or $6$ and the other five are odd. Choose which die ( $6$ ways): $6 \cdot \frac13 \cdot \left(\frac12\right)^5 = \frac{6}{96} = \frac{1}{16} = \frac{4}{64}$ .

The failing probability is $\frac{1}{64} + \frac{4}{64} = \frac{5}{64}$ , so the product is divisible by $4$ with probability
$1 - \frac{5}{64} = \frac{59}{64}.$

The answer is $\boxed{\textbf{(C)}\ \frac{59}{64}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.3)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques

Techniques

Split into exhaustive cases and handle each · Count the complement and subtract from the total