Printable

2003 AMC 10A Problem 15

Problem only Back to the problem

2003 AMC 10A Problem 15AMC 10 Step by Step

Name: Date: September 13, 2026

Divisibility & Factors · difficulty 2 of 5 · about 2 min · position P11-15

What is the probability that an integer in the set $\{1,2,3,...,100\}$ is divisible by $2$ and not divisible by $3$ ?

A)
$\frac{1}{6}$
B)
$\frac{33}{100}$
C)
$\frac{17}{50}$
D)
$\frac{1}{2}$
E)
$\frac{18}{25}$

Answer

Key insight

Count the 50 even numbers, then remove the 16 multiples of 6; 34 of 100 remain.

Solution

Multiples of $2$ up to $100$ : $\lfloor 100/2 \rfloor = 50$ .

Among these, the ones also divisible by $3$ are the multiples of $\operatorname{lcm}(2,3) = 6$ : $\lfloor 100/6 \rfloor = 16$ (namely $6, 12, \dots, 96$ ).

Even numbers that are not multiples of $3$ : $50 - 16 = 34$ . The probability is $\dfrac{34}{100} = \dfrac{17}{50}$ .

The answer is $\boxed{\textbf{(C)}\ \frac{17}{50}}$ .

Why this works

"Divisible by $2$ but not $3$ " is a set difference: (multiples of $2$ ) minus (multiples of $6$ ). Counting multiples uses floor division, and the floor matters: $100$ is not a multiple of $6$ , so the count is $16$ , not $16.67$ , which is why the answer is not the "expected" $\tfrac12 \cdot \tfrac23 = \tfrac13$ .

The trap

Assuming exactly one third of the evens are multiples of 3 and answering 1/3, when floor(100/6) = 16 is not 50/3.

Common mistakes

Assuming exactly one third of the evens are multiples of 3 and answering 1/3, when floor(100/6) = 16 is not 50/3.
Subtracting all $33$ multiples of $3$ from $50$ instead of only the even ones.

Techniques

Count the complement and subtract from the total