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

Latest revision as of 12:30, 28 May 2023

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 - Complementary Counting

We use complementary counting, or

$\text{total - what we don't want = what we want}$ .

There are a total of $2^8$ ways to create subsets (consider including or excluding each number) and there are a total of $2^4$ subsets only containing composite numbers (the composite numbers are $4, 6, 8, 9$ ). Therefore, there are $2^8-2^4=240$ 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.

$\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 15$ . 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

Solution 4

Total subsets is $(2^8) = 256$ 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: $2^4$ = 16 $256-16=240$ $\boxed{\textbf{(D) }240}$

-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 $2^4-1=15$ (As there is one solution not containing any primes)

The possibilities of the set containing composites are $2^4=16$ (There can be a set with no composites)

Multiplying this we get $15 \cdot 16 = \boxed{\textbf{(D) }240}$ .

-middletonkids

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/RkQPY-_Qieo

~Education, the Study of Everything

Video Solution 1

https://youtu.be/6Bt4I23JL1s

~savannahsolver

Video Solution 2

https://youtu.be/5UojVH4Cqqs?t=1609

~ pi_is_3.14

@@ Line 46: / Line 46: @@
 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 countaining composites are <math>2^4=16</math> (There can be a set with no composites)
+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 = \text{\bold{(D)}240}</math>.
+Multiplying this we get <math>15 \cdot 16 = \boxed{\textbf{(D) }240}</math>.
-==Video Solution==
+-middletonkids
+==Video Solution (HOW TO THINK CRITICALLY!!!)==
+https://youtu.be/RkQPY-_Qieo
+~Education, the Study of Everything
+==Video Solution 1==
 https://youtu.be/6Bt4I23JL1s
 ~savannahsolver
+== Video Solution 2==
+https://youtu.be/5UojVH4Cqqs?t=1609
+~ pi_is_3.14
 ==See Also==

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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