Difference between revisions of "2018 AMC 12B Problems/Problem 5"

Revision as of 17:54, 18 September 2021

Problem

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

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

Solution 1

Since an element of a subset is either in or out, the total number of subsets of the $8$ -element set is $2^8 = 256$ . However, since we are only concerned about the subsets with at least $1$ prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are $4$ non-primes, there are $2^8 -2^4 = \boxed{\textbf{(D) } 240}$ subsets with at least $1$ prime.

Solution 2

We can construct our subset by choosing which primes are included and which composites are included. There are $2^4-1$ ways to select the primes (total subsets minus the empty set) and $2^4$ ways to select the composites. Thus, there are $15\cdot16=\boxed{\textbf{(D) } 240}$ ways to choose a subset of the eight numbers.

Video Solution

https://youtu.be/8WrdYLw9_ns?t=253

~ pi_is_3.14

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 ==Problem==
-How many subsets of  <math>\{2,3,4,5,6,7,8,9\}</math>  contain at least one prime number?
+How many subsets of <math>\{2,3,4,5,6,7,8,9\}</math> contain at least one prime number?
-<math>(\text{A}) \indent 128 \qquad (\text{B}) \indent 192  \qquad (\text{C}) \indent 224  \qquad (\text{D}) \indent 240 \qquad (\text{E}) \indent 256  </math>
+<math>
+\textbf{(A) } 128 \qquad
+\textbf{(B) } 192 \qquad
+\textbf{(C) } 224 \qquad
+\textbf{(D) } 240 \qquad
+\textbf{(E) } 256
+</math>
 ==Solution 1 ==
-Since an element of a subset is either in or out, the total number of subsets of the 8 element set is <math>2^8 = 256</math>. However, since we are only concerned about the subsets with at least 1 prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are 4 non-primes, there are <math>2^8 -2^4 = 240</math> subsets with at least 1 prime so the answer is <math>\Rightarrow \boxed { (\textbf{D}) 240 }\indent</math>
+Since an element of a subset is either in or out, the total number of subsets of the <math>8</math>-element set is <math>2^8 = 256</math>. However, since we are only concerned about the subsets with at least <math>1</math> prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are <math>4</math> non-primes, there are <math>2^8 -2^4 = \boxed{\textbf{(D) } 240}</math> subsets with at least <math>1</math> prime.
 ==Solution 2==
-We can construct our subset by choosing which primes are included and which composites are included. There are <math>2^4-1</math> ways to select the primes (total subsets minus the empty set) and <math>2^4</math> ways to select the composites. Thus, there are <math>15*16</math> ways to choose a subset of the eight numbers, or <math>\boxed { (\textbf{D}) 240 }</math> .
+We can construct our subset by choosing which primes are included and which composites are included. There are <math>2^4-1</math> ways to select the primes (total subsets minus the empty set) and <math>2^4</math> ways to select the composites. Thus, there are <math>15\cdot16=\boxed{\textbf{(D) } 240}</math> ways to choose a subset of the eight numbers.
 == Video Solution ==

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