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

(Solution 1)
(Solution 1)
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==Solution 1==
 
==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 total subsets, and 4 non-prime. Thus (2^8) - (2^4) = 240
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We use complementary counting, or Total- what we don't want = what we want. There are a total of <math>2^8</math> ways to create subsets (consider including or excluding each number) and there are a total of <math>2^4</math> subsets only containing composite numbers. Therefore, there are <math>2^8-2^4=240</math> total ways to have at least one prime in a subset.
  
 
==Solution 2 (Using Answer Choices)==
 
==Solution 2 (Using Answer Choices)==

Revision as of 12:06, 19 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

We use complementary counting, or 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. 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

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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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