2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 5 Easy

How many subsets of \left\{2,3,4,5,6,7,8,9\right\} contain at least one prime number? (2018 AMC 10B Problem, Question#5)

  • A.

    128

  • B.

    192

  • C.

    224

  • D.

    240

  • E.

    256

Answer:D

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 \left\{4, 6, 8,9\right\},  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=(\rm D)~240.

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.  \left( \begin{array}{l} {4}\\{1} \end{array}\right)+\left( \begin{array}{l} {4}\\{2} \end{array}\right)+\left( \begin{array}{l} {4}\\{3} \end{array}\right)+\left( \begin{array}{l} {4}\\{4} \end{array}\right)=15. Using the answer choices, the only multiple of 15 is (\rm D)~240.

Subsets of \left\{2,3,4,5,6,7,8,9\right\} indude a singe digit up to all eight numbers.Therefore,we must add the combinations of all possible subsets and subtract from each of the subsets fomed by the composite numbers.

Hence:

\left( \begin{array}{l} {8}\\{1} \end{array}\right)-\left( \begin{array}{l} {4}\\{1} \end{array}\right)+\left( \begin{array}{l} {8}\\{2} \end{array}\right)-\left( \begin{array}{l} {4}\\{2} \end{array}\right)+\left( \begin{array}{l} {8}\\{3} \end{array}\right)-\left( \begin{array}{l} {4}\\{3} \end{array}\right)+\left( \begin{array}{l} {8}\\{4} \end{array}\right)-1+\left( \begin{array}{l} {8}\\{5} \end{array}\right)+\left( \begin{array}{l} {8}\\{6} \end{array}\right)+\left( \begin{array}{l} {8}\\{7} \end{array}\right)+1=(\rm D)~240.