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2018 AMC 10B Problem 5

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2018 AMC 10B Problem 5AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 1 of 5 · about 2 min · position P1-10

How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?

A)
$128$
B)
$192$
C)
$224$
D)
$240$
E)
$256$

Answer

Key insight

Count all 2^8 subsets and subtract the 2^4 subsets built only from the four non-primes 4, 6, 8, 9.

Solution

The set has $8$ elements, so it has $2^8 = 256$ subsets in total.

The primes in the set are $2, 3, 5, 7$ ; the non-primes are $4, 6, 8, 9$ . A subset contains no prime exactly when it is a subset of $\{4, 6, 8, 9\}$ , and there are $2^4 = 16$ of those (including the empty set).

Every other subset contains at least one prime:
$256 - 16 = 240.$

The answer is $\boxed{\textbf{(D)}\ 240}$ .

Why this works

"At least one" is the signal for complementary counting: the complement "none" is a single clean case, here the subsets of the non-primes. The counts $2^n$ come from each element independently being in or out.

The trap

Miscounting the primes (there are exactly four: 2, 3, 5, 7) or forgetting that the empty set is one of the 16 prime-free subsets.

Common mistakes

Miscounting the primes (there are exactly four: 2, 3, 5, 7) or forgetting that the empty set is one of the 16 prime-free subsets.
Trying to count "at least one prime" directly by cases on the number of primes, which is slower and error-prone.

Techniques

Count the complement and subtract from the total