Difference between revisions of "2018 AMC 10B Problems/Problem 5"
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<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 1== | + | ==Solution 1 - Complementary Counting== |
− | We use complementary counting, or | + | We use complementary counting, or |
+ | |||
+ | <math>\text{total - what we don't want = what we want}</math>. | ||
+ | |||
+ | 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 (the composite numbers are <math>4, 6, 8, 9</math>). 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 12:11, 4 August 2020
Contents
[hide]Problem
How many subsets of contain at least one prime number?
Solution 1 - Complementary Counting
We use complementary counting, or
.
There are a total of ways to create subsets (consider including or excluding each number) and there are a total of subsets only containing composite numbers (the composite numbers are ). Therefore, there are 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.
. Using the answer choices, the only multiple of 15 is
By: K6511
Solution 3
Subsets of 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:
By: pradyrajasai
Solution 4
Total subsets is 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: = 16
-goldenn
Video Solution
~savannahsolver
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 |
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