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

m (Solution 1)
(Solution 4)
Line 40: Line 40:
  
 
-goldenn
 
-goldenn
 +
 +
==Solution 5==
  
 
==Video Solution==
 
==Video Solution==

Revision as of 14:07, 10 August 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 - 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

Video Solution

https://youtu.be/6Bt4I23JL1s

~savannahsolver

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

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