# 2018 AMC 12B Problems/Problem 5

## Problem

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

$(\text{A}) \indent 128 \qquad (\text{B}) \indent 192 \qquad (\text{C}) \indent 224 \qquad (\text{D}) \indent 240 \qquad (\text{E}) \indent 256$

## Solution 1

Since an element of a subset is either in or out, the total number of subsets of the 8 element set is $2^8 = 256$. However, since we are only concerned about the subsets with at least 1 prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are 4 non-primes, there are $2^8 -2^4 = 240$ subsets with at least 1 prime so the answer is $\Rightarrow \boxed { (\textbf{D}) 240 }\indent$

## Solution 2

We can construct our subset by choosing which primes are included and which composites are included. There are $2^4-1$ ways to select the primes (total subsets minus the empty set) and $2^4$ ways to select the composites. Thus, there are $15*16$ ways to choose a subset of the eight numbers, or $\boxed { (\textbf{D}) 240 }$ .

## See Also

 2018 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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