Difference between revisions of "2018 AMC 10B Problems/Problem 5"
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By: K6511 | By: K6511 | ||
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+ | ==Solution 3== | ||
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+ | Subsets of <math>\{2,3,4,5,6,7,8,9\}</math> 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. | ||
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+ | Hence: | ||
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+ | <math>\binom{8}{1} - \binom{4}{1} + \binom{8}{2} - \binom{4}{2} + \binom{8}{3} - \binom{4}{3} + \binom{8}{4} - 1 + \binom{8}{5} + \binom{8}{6} + \binom{8}{7} + 1 = \boxed{\textbf{(D) }240}</math> | ||
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+ | By: pradyrajasai | ||
==See Also== | ==See Also== |
Revision as of 23:38, 6 June 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
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
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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