Difference between revisions of "2018 AMC 10B Problems/Problem 5"

Revision as of 19:15, 1 January 2020

Problem

How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?

$\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}$

Solution 1

Consider finding the number of subsets that do not contain any primes and minus that number from the total number of subsets.There are 2^8

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.

$\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 15$ . Using the answer choices, the only multiple of 15 is $\boxed{\textbf{(D) }240}$

By: K6511

Solution 3

Subsets of $\{2,3,4,5,6,7,8,9\}$ 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:

$\binom{8}{1} - \binom{4}{1} + \binom{8}{2} - \binom{4}{2} + \binom{8}{3} - \binom{4}{3} + \binom{8}{4} - 1 + \binom{8}{5} + \binom{8}{6} + \binom{8}{7} + 1 = \boxed{\textbf{(D) }240}$

By: pradyrajasai

2018 AMC 10B (Problems • Answer Key • Resources)
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@@ Line 7: / Line 7: @@
 ==Solution 1==
-Consider finding the number of subsets that do not contain any primes. There are four primes in the set: <math>2</math>, <math>3</math>, <math>5</math>, and <math>7</math>. This means that the number of subsets without any primes is the number of subsets of <math>\{4, 6, 8, 9\}</math>, which is just <math>2^4 = 16</math>. 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 <math>2^8 = 256</math>. Thus, the answer is <math>256 - 16 = \boxed{\textbf{(D) }240}</math>
+Consider finding the number of subsets that do not contain any primes and minus that number from the total number of subsets.There are 2^8
 ==Solution 2 (Using Answer Choices)==

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