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2016 AMC 10A Problem 12

2016 AMC 10A Problem 12AMC 10 Step by Step

Name: Date: October 28, 2026

Basic Probability · difficulty 2 of 5 · about 3 min · position P11-15

Three distinct integers are selected at random between $1$ and $2016$ , inclusive. Which of the following is a correct statement about the probability $p$ that the product of the three integers is odd?

A) $p<\dfrac{1}{8}$
B) $p=\dfrac{1}{8}$
C) $\dfrac{1}{8}<p<\dfrac{1}{3}$
D) $p=\dfrac{1}{3}$
E) $p>\dfrac{1}{3}$

Answer

Key insight

All three numbers must be odd. The first pick is odd half the time, each later pick slightly less, so p is below 1/8.

Solution

1. A product is odd only when every number in it is odd. One even number makes the whole product even. So we need all three picks to be odd.
2. From $1$ to $2016$ there are $1008$ odd and $1008$ even numbers.
3. Pick the numbers one at a time.
- First pick: $1008$ odd out of $2016$ , a chance of exactly $\tfrac12$ .
- Second pick: one odd number is used up, so $1007$ odd out of $2015$ left.
- Third pick: $1006$ odd out of $2014$ left.
4. So
$p = \frac12 \cdot \frac{1007}{2015} \cdot \frac{1006}{2014}.$
5. We do not need the exact value, only how it compares with $\tfrac18 = \tfrac12 \cdot \tfrac12 \cdot \tfrac12$ . Compare each fraction with $\tfrac12$ :
- Half of $2015$ is $1007.5$ , and $1007$ is less, so $\tfrac{1007}{2015} < \tfrac12$ .
- Half of $2014$ is $1007$ , and $1006$ is less, so $\tfrac{1006}{2014} < \tfrac12$ .
6. Two of the three factors are below $\tfrac12$ , so $p < \tfrac18$ .

The answer is $\boxed{\textbf{(A)}\ p<\dfrac{1}{8}}$ .

Why this works

When you pick without putting numbers back, each odd pick removes one odd number but the total shrinks by only one too, so the odd numbers become slightly less than half of what is left. The answer choices only ask which side of $\tfrac18$ the probability is on, so compare each factor with $\tfrac12$ instead of multiplying big numbers.

Alternative approach

Test the idea on a tiny case. From $1, 2, 3, 4$ pick three different numbers: there are $4$ ways, and none of them is all odd (there are only two odd numbers). So the chance is $0$ , far below $\tfrac18$ . With more numbers the effect shrinks, but it always pushes the probability below $\tfrac18$ .

The trap

Treating the three picks as if a number could be chosen twice, making each chance exactly one half and giving exactly 1/8.

Concise solution

A product of integers is odd exactly when every factor is odd. Among $1, 2, \dots, 2016$ there are $1008$ odd numbers and $1008$ even numbers.

Pick the three numbers one at a time without replacement. The first is odd with probability $\frac{1008}{2016} = \frac12$ . Given that, the second is odd with probability $\frac{1007}{2015}$ , and then the third with probability $\frac{1006}{2014}$ . So
$p = \frac12\cdot\frac{1007}{2015}\cdot\frac{1006}{2014}.$

Now compare each factor to $\frac12$ : $2\cdot 1007 = 2014 < 2015$ and $2\cdot 1006 = 2012 < 2014$ , so both later factors are strictly less than $\frac12$ . Hence
$p < \frac12\cdot\frac12\cdot\frac12 = \frac18.$

The answer is $\boxed{\textbf{(A)}\ p<\dfrac{1}{8}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques
Introduction to Algebra, Chapter 20: Special Functions (§20.3)

Techniques

Bound the quantity above/below or estimate to pin it down · Use an invariant, parity, or coloring argument