Difference between revisions of "2015 AMC 10B Problems/Problem 11"

Revision as of 22:30, 16 June 2022

Problem

Among the positive integers less than $100$ , each of whose digits is a prime number, one is selected at random. What is the probability that the selected number is prime?

$\textbf{(A)} \dfrac{8}{99}\qquad \textbf{(B)} \dfrac{2}{5}\qquad \textbf{(C)} \dfrac{9}{20}\qquad \textbf{(D)} \dfrac{1}{2}\qquad \textbf{(E)} \dfrac{9}{16}$

Solution 1

The one digit prime numbers are $2$ , $3$ , $5$ , and $7$ . So there are a total of $4\cdot4=16$ ways to choose a two digit number with both digits as primes and $4$ ways to choose a one digit prime, for a total of $4+16=20$ ways. Out of these $2$ , $3$ , $5$ , $7$ , $23$ , $37$ , $53$ , and $73$ are prime. Thus the probability is $\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}$ .

Solution 2 (Listing)

Since the only primes digits are $2$ , $3$ , $5$ , and $7$ , 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 $20$ cases and $8$ of these are prime. So the answer is $\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}$ . ~JH. L

Video Solution

https://youtu.be/cL9wo9kcOGg

~savannahsolver

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.

@@ Line 9: / Line 9: @@
 ==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.
-<math>2</math>- Prime;
-<math>3</math>- Prime;
+*2- Prime;
-<math>5</math>- Prime;
+*3- Prime;
-<math>7</math>- Prime;
+*5- Prime;
-<math>22</math>- Composite;
+*7- Prime;
-<math>23</math>- Prime;
+*22- Composite;
-<math>25</math>- Composite;
+*23- Prime;
-<math>27</math>- Composite;
+*25- Composite;
-<math>32</math>- Composite;
+*27- Composite;
-<math>33</math>- Composite;
+*32- Composite;
-<math>35</math>- Composite;
+*33- Composite;
-<math>37</math>- Prime;
+*35- Composite;
-<math>52</math>- Composite;
+*37- Prime;
-<math>53</math>- Prime;
+*52- Composite;
-<math>55</math>- Composite;
+*53- Prime;
-<math>57</math>- Composite;
+*55- Composite;
-<math>72</math>- Composite;
+*57- Composite;
-<math>73</math>- Prime;
+*72- Composite;
-<math>75</math>- Composite;
+*73- Prime;
-<math>77</math>- Composite.
+*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

Page

Toolbox

Search