Difference between revisions of "2009 AMC 8 Problems/Problem 13"
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==Solution== | ==Solution== | ||
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>. | ||
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+ | ==Alternate Solution== | ||
+ | The number is odd 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>. | ||
+ | |||
+ | Note: The last sentence is true because there are <math>3</math> randomly-arrangeable numbers) | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=12|num-a=14}} | {{AMC8 box|year=2009|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:09, 8 August 2018
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 .
Alternate Solution
The number is odd 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.