Difference between revisions of "2018 AMC 10B Problems/Problem 5"
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| − | Well, there are 4 composite numbers, and you can list them in a 1 number format, a 2 number, 3 number, and a 4 number format. Now, we can use | + | Well, there are 4 composite numbers, and you can list them in a 1 number format, a 2 number, 3 number, and a 4 number format. Now, we can use combinations. |
<math>\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 15</math>. Using the answer choices, the only multiple of 15 is <math>\boxed{\textbf{(D) }240}</math> | <math>\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 15</math>. Using the answer choices, the only multiple of 15 is <math>\boxed{\textbf{(D) }240}</math> | ||
Revision as of 23:03, 20 January 2019
Problem
How many subsets of 
contain at least one prime number?
Solution 1
Consider finding the number of subsets that do not contain any primes. There are four primes in the set: 
, 
, 
, and 
. This means that the number of subsets without any primes is the number of subsets of 
, which is just 
. The number of subsets with at least one prime is the number of subsets minus the number of subsets without any primes. The number of subsets is 
. Thus, the answer is 
Solution 2 (Using Answer Choices)
Well, there are 4 composite numbers, and you can list them in a 1 number format, a 2 number, 3 number, and a 4 number format. Now, we can use combinations.
. Using the answer choices, the only multiple of 15 is 
By: K6511
See Also
| 2018 AMC 10B (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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
