Difference between revisions of "2018 AMC 10B Problems/Problem 5"
Benjamin2020 (talk | contribs) |
(→Solution) |
||
| Line 5: | Line 5: | ||
<math>\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}</math> | <math>\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Consider finding the number of subsets that do not contain any primes. There are four primes in the set: <math>2</math>, <math>3</math>, <math>5</math>, and <math>7</math>. This means that the number of subsets without any primes is the number of subsets of <math>\{4, 6, 8, 9\}</math>, which is just <math>2^4 = 16</math>. 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 <math>2^8 = 256</math>. Thus, the answer is <math>256 - 16 = 240</math>. <math>\boxed{D}</math> | Consider finding the number of subsets that do not contain any primes. There are four primes in the set: <math>2</math>, <math>3</math>, <math>5</math>, and <math>7</math>. This means that the number of subsets without any primes is the number of subsets of <math>\{4, 6, 8, 9\}</math>, which is just <math>2^4 = 16</math>. 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 <math>2^8 = 256</math>. Thus, the answer is <math>256 - 16 = 240</math>. <math>\boxed{D}</math> | ||
| + | |||
| + | ==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 permutations | ||
| + | |||
| + | <math>\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 15</math>. Using the answer choices, the only multiple of 15 is <math>\boxed{\textbf{(D) }240}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 18:18, 17 February 2018
Problem
How many subsets of 
contain at least one prime number?
Solution 1
Consider finding the number of subsets that do not contain any primes. There are four primes in the set: 
, 
, 
, and 
. This means that the number of subsets without any primes is the number of subsets of 
, which is just 
. 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 
. Thus, the answer is 
. 
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 permutations
. Using the answer choices, the only multiple of 15 is 
See Also
| 2018 AMC 10B (Problems • Answer Key • Resources) | ||
| 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. 
