Difference between revisions of "2018 AMC 12B Problems/Problem 5"
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Since an element of a subset is either in or out, the total number of subsets of the 8 element set is <math>2^8 = 256</math>. 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 <math>2^8 -2^4 = 240</math> subsets with at least 1 prime so the answer is <math>\Rightarrow (\text{D}) \indent</math> (Giraffefun) | Since an element of a subset is either in or out, the total number of subsets of the 8 element set is <math>2^8 = 256</math>. 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 <math>2^8 -2^4 = 240</math> subsets with at least 1 prime so the answer is <math>\Rightarrow (\text{D}) \indent</math> (Giraffefun) | ||
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{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}} | {{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}} | ||
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Revision as of 14:34, 16 February 2018
How many subsets of 
contain at least one prime number?
Solution 1
Since an element of a subset is either in or out, the total number of subsets of the 8 element set is 
. 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 
subsets with at least 1 prime so the answer is 
(Giraffefun)
See Also
| 2018 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 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 | |
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