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> (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 \boxed { (\textbf{D}) 240 }\indent</math> (Giraffefun) | ||
| + | |||
| + | ==Solution 2== | ||
| + | We can construct our subset by choosing which primes are included and which composites are included. There are <math>2^4-1</math> ways to select the primes (total subsets minus the empty set) and <math>2^4</math> ways to select which composites are included. Thus, there are <math>15*16</math> ways to choose a subset of the eight numbers, or <math>\boxed { (\textbf{D}) 240 }</math> | ||
==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 15:04, 16 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 
(Giraffefun)
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 which composites are included. Thus, there are 
ways to choose a subset of the eight numbers, or 
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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