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

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==Solution 1==
 
==Solution 1==
  
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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 = 240</math>. <math>\boxed{D}</math>
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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>
  
 
==Solution 2 (Using Answer Choices)==
 
==Solution 2 (Using Answer Choices)==

Revision as of 19:48, 25 February 2018

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. There are four primes in the set: $2$, $3$, $5$, and $7$. This means that the number of subsets without any primes is the number of subsets of $\{4, 6, 8, 9\}$, which is just $2^4 = 16$. 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 $2^8 = 256$. Thus, the answer is $256 - 16 = \boxed{\textbf{(D) }240}$

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 permutations

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

By: K6511

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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