Difference between revisions of "2018 AMC 10B Problems/Problem 5"
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<math>2^4</math> = 16 | <math>2^4</math> = 16 | ||
<math>256-16=240</math> | <math>256-16=240</math> | ||
| − | \boxed{\textbf{(D) }240} | + | <math>\boxed{\textbf{(D) }240}</math> |
-goldenn | -goldenn | ||
Revision as of 19:37, 25 January 2020
Contents
Problem
How many subsets of 
contain at least one prime number?
Solution 1
We use complementary counting, or Total- what we don't want = what we want. There are a total of 
ways to create subsets (consider including or excluding each number) and there are a total of 
subsets only containing composite numbers. Therefore, there are 
total ways to have at least one prime in a subset.
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
Solution 3
Subsets of include a single digit up to all eight numbers. Therefore, we must add the combinations of all possible subsets and subtract from each of the subsets formed by the composite numbers.
Hence:
By: pradyrajasai
Solution 4
Total subsets is 
Using complementary counting and finding the sets with composite numbers:
only 4,6,8 and 9 are composite. Each one can be either in the set or out:
= 16
-goldenn
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. 
