AMC 10 · 2003 · #15

Grade 7 probability
Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements
probability-basicmultiplescomplementary-counting complementary-countingidentify-subproblems ↑ Prerequisites: multiples
📏 Medium solution 💡 2 insights
Problem
You pick one integer at random from {1,2,3,…,100}, with every integer equally likely. Find the probability that the integer is divisible by 2 but not divisible by 3.

Pick an answer.

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

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

The condition has two parts glued together: "divisible by 2" AND "not divisible by 3." Counting both at once is awkward, so use tool #16 (Change Focus / Count the Complement): first count all the even numbers, then throw out the ones that are also multiples of 3. Tool #7 (Identify Subproblems) splits the work into three clean counts — how many multiples of 2, how many multiples of 6, and the difference. Tool #2 (Make a Systematic List) is the trustworthy way to count multiples in a range: they march in a steady sequence, so counting them never needs guesswork.

1STEP 1

Count the even numbers

Multiples of 2 run 2, 4, …, 100, so 100 split into groups of 2 gives 50 even numbers.

⌊ 100/2 ⌋ = 50
2STEP 2

Divisible by 2 and 3 means multiple of 6

Divisible by both 2 and 3 means divisible by their least common multiple 6, so the evens to drop are the multiples of 6.

lcm(2,3) = 6
3STEP 3

Count the multiples of 6

Multiples of 6 run 6, 12, …, 96, so 100 split into whole groups of 6 gives 16 of them.

⌊ 100/6 ⌋ = 16
4STEP 4

Subtract to get the favorable count

Take the 50 evens and remove the 16 that are also multiples of 3, leaving 34 integers that fit.

50 - 16 = 34
5STEP 5

Write and simplify the probability

All picks are equally likely, so the probability is 34/100, which reduces to 17/50 — choice (C).

P = 34/100 = 17/50 → (C)
Answer
17/50
The answer 17/50 = 0.34 sits sensibly between 0 and 1. Half of the numbers are even (0.5), and roughly a third of those even numbers get knocked out for being multiples of 3, leaving about two-thirds of a half — close to 0.33–0.34, which matches. It is also clearly bigger than 1/6 (too small, that's just the multiples of 6) and smaller than 1/2 (that would be all the evens), so choices (A) and (D) are ruled out.
💡Key takeaway

For "divisible by this but not that," count everything divisible by the first number, then subtract the ones also divisible by the second, and put that over the total.

  • Count the even numbers
  • Divisible by 2 and 3 means multiple of 6
  • Count the multiples of 6
  • Subtract to get the favorable count
  • Write and simplify the probability

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