Difference between revisions of "2015 AMC 10B Problems/Problem 11"
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<math>\textbf{(A)} \dfrac{8}{99}\qquad \textbf{(B)} \dfrac{2}{5}\qquad \textbf{(C)} \dfrac{9}{20}\qquad \textbf{(D)} \dfrac{1}{2}\qquad \textbf{(E)} \dfrac{9}{16}</math> | <math>\textbf{(A)} \dfrac{8}{99}\qquad \textbf{(B)} \dfrac{2}{5}\qquad \textbf{(C)} \dfrac{9}{20}\qquad \textbf{(D)} \dfrac{1}{2}\qquad \textbf{(E)} \dfrac{9}{16}</math> | ||
− | ==Solution== | + | ==Solution 1== |
The one digit prime numbers are <math>2</math>, <math>3</math>, <math>5</math>, and <math>7</math>. So there are a total of <math>4\cdot4=16</math> ways to choose a two digit number with both digits as primes and <math>4</math> ways to choose a one digit prime, for a total of <math>4+16=20</math> ways. Out of these <math>2</math>, <math>3</math>, <math>5</math>, <math>7</math>, <math>23</math>, <math>37</math>, <math>53</math>, and <math>73</math> are prime. Thus the probability is <math>\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}</math>. | The one digit prime numbers are <math>2</math>, <math>3</math>, <math>5</math>, and <math>7</math>. So there are a total of <math>4\cdot4=16</math> ways to choose a two digit number with both digits as primes and <math>4</math> ways to choose a one digit prime, for a total of <math>4+16=20</math> ways. Out of these <math>2</math>, <math>3</math>, <math>5</math>, <math>7</math>, <math>23</math>, <math>37</math>, <math>53</math>, and <math>73</math> are prime. Thus the probability is <math>\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}</math>. | ||
− | ==Solution 2== | + | ==Solution 2 (Listing)== |
− | + | Since the only primes digits are <math>2</math>, <math>3</math>, <math>5</math>, and <math>7</math>, it doesn't seem too hard to list all of the numbers out. | |
+ | |||
+ | *2- Prime; | ||
+ | *3- Prime; | ||
+ | *5- Prime; | ||
+ | *7- Prime; | ||
+ | *22- Composite; | ||
+ | *23- Prime; | ||
+ | *25- Composite; | ||
+ | *27- Composite; | ||
+ | *32- Composite; | ||
+ | *33- Composite; | ||
+ | *35- Composite; | ||
+ | *37- Prime; | ||
+ | *52- Composite; | ||
+ | *53- Prime; | ||
+ | *55- Composite; | ||
+ | *57- Composite; | ||
+ | *72- Composite; | ||
+ | *73- Prime; | ||
+ | *75- Composite; | ||
+ | *77- Composite. | ||
+ | |||
+ | Counting it out, there are <math>20</math> cases and <math>8</math> of these are prime. So the answer is <math>\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}</math>. | ||
+ | ~JH. L | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | https://youtu.be/RZDFs3qrw7Y | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 16:13, 2 August 2022
Contents
Problem
Among the positive integers less than , each of whose digits is a prime number, one is selected at random. What is the probability that the selected number is prime?
Solution 1
The one digit prime numbers are , , , and . So there are a total of ways to choose a two digit number with both digits as primes and ways to choose a one digit prime, for a total of ways. Out of these , , , , , , , and are prime. Thus the probability is .
Solution 2 (Listing)
Since the only primes digits are , , , and , it doesn't seem too hard to list all of the numbers out.
- 2- Prime;
- 3- Prime;
- 5- Prime;
- 7- Prime;
- 22- Composite;
- 23- Prime;
- 25- Composite;
- 27- Composite;
- 32- Composite;
- 33- Composite;
- 35- Composite;
- 37- Prime;
- 52- Composite;
- 53- Prime;
- 55- Composite;
- 57- Composite;
- 72- Composite;
- 73- Prime;
- 75- Composite;
- 77- Composite.
Counting it out, there are cases and of these are prime. So the answer is . ~JH. L
Video Solution 1
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.