2018 AMC 10B Problem 5

Below is the professionally curated solution for Problem 5 of the 2018 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 10B solutions, or check the answer key.

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Concepts:subsets complementary counting prime

Difficulty rating: 1200

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

$128$

$192$

$224$

$240$

$256$

Solution:

Count the complement. The set has $2^8 = 256$ subsets total. A subset avoids every prime exactly when it sticks to the non-primes $\{4, 6, 8, 9\},$ and there are $2^4 = 16$ of those. So $256 - 16 = 240$ subsets contain at least one prime. Thus, D is the correct answer.

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