Difference between revisions of "2009 AMC 8 Problems/Problem 13"
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The three digit numbers are <math>135,153,351,315,513,531</math>. The numbers that end in <math>5</math> are divisible are <math>5</math>, and the probability of choosing those numbers is <math>\boxed{\textbf{(B)}\ \frac13}</math>. | The three digit numbers are <math>135,153,351,315,513,531</math>. The numbers that end in <math>5</math> are divisible are <math>5</math>, and the probability of choosing those numbers is <math>\boxed{\textbf{(B)}\ \frac13}</math>. | ||
| − | == | + | ==Solution 2== |
The number is divisible by 5 if and only if the number ends in <math>5</math> (also <math>0</math>, but that case can be ignored, as none of the digits are <math>0</math>) | The number is divisible by 5 if and only if the number ends in <math>5</math> (also <math>0</math>, but that case can be ignored, as none of the digits are <math>0</math>) | ||
If we randomly arrange the three digits, the probability of the last digit being <math>5</math> is <math>\boxed{\textbf{(B)}\ \frac13}</math>. | If we randomly arrange the three digits, the probability of the last digit being <math>5</math> is <math>\boxed{\textbf{(B)}\ \frac13}</math>. | ||
Revision as of 21:25, 3 September 2020
Contents
Problem
A three-digit integer contains one of each of the digits 
, 
, and 
. What is the probability that the integer is divisible by ?
Solution
The three digit numbers are 
. The numbers that end in are divisible are , and the probability of choosing those numbers is 
.
Solution 2
The number is divisible by 5 if and only if the number ends in (also 
, but that case can be ignored, as none of the digits are )
If we randomly arrange the three digits, the probability of the last digit being is .
Note: The last sentence is true because there are randomly-arrangeable numbers)
See Also
| 2009 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
