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Difference between revisions of "2018 AMC 12B Problems/Problem 5"

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==Solution 1 ==
 
==Solution 1 ==
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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)
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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 \boxed { (\textbf{D}) 240 }\indent</math> (Giraffefun)
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}}
 
{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:37, 16 February 2018

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 $\Rightarrow \boxed { (\textbf{D}) 240 }\indent$ (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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