Difference between revisions of "2018 AMC 10B Problems/Problem 5"
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The possibilities of primes are <math>2^4-1=15</math> (As there is one solution not containing any primes) | The possibilities of primes are <math>2^4-1=15</math> (As there is one solution not containing any primes) | ||
− | The possibilities of the set | + | The possibilities of the set containing composites are <math>2^4=16</math> (There can be a set with no composites) |
− | Multiplying this we get <math>15 \cdot 16 = \ | + | Multiplying this we get <math>15 \cdot 16 = \boxed{\textbf{(D) }240}</math>. |
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+ | -middletonkids | ||
==Video Solution== | ==Video Solution== |
Revision as of 14:12, 10 August 2020
Contents
[hide]Problem
How many subsets of contain at least one prime number?
Solution 1 - Complementary Counting
We use complementary counting, or
.
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 (the composite numbers are ). 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
Solution 5
We multiply the number of possibilities of the set having prime numbers and the set having composites.
The possibilities of primes are (As there is one solution not containing any primes)
The possibilities of the set containing composites are (There can be a set with no composites)
Multiplying this we get .
-middletonkids
Video Solution
~savannahsolver
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 |
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