Difference between revisions of "2018 AMC 12B Problems/Problem 5"
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==Solution 1 == | ==Solution 1 == | ||
| − | 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> | + | 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> |
==Solution 2== | ==Solution 2== | ||
Revision as of 01:08, 19 February 2018
Contents
Problem
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 
Solution 2
We can construct our subset by choosing which primes are included and which composites are included. There are 
ways to select the primes (total subsets minus the empty set) and 
ways to select the composites. Thus, there are 
ways to choose a subset of the eight numbers, or 
(mathislife16).
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
