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==
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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>)
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If we randomly arrange the three digits, the probability of the last digit being <math>5</math> is <math>\boxed{\textbf{(B)}\ \frac13}</math>.
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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 $1$, $3$, and $5$. What is the probability that the integer is divisible by $5$?

$\textbf{(A)}\  \frac{1}{6}  \qquad \textbf{(B)}\   \frac{1}{3}  \qquad \textbf{(C)}\   \frac{1}{2}  \qquad \textbf{(D)}\  \frac{2}{3}   \qquad \textbf{(E)}\   \frac{5}{6}$

Solution

The three digit numbers are $135,153,351,315,513,531$. The numbers that end in $5$ are divisible are $5$, and the probability of choosing those numbers is $\boxed{\textbf{(B)}\ \frac13}$.

Alternate Solution

The number is odd if and only if the number ends in $5$ (also $0$, but that case can be ignored, as none of the digits are $0$) If we randomly arrange the three digits, the probability of the last digit being $5$ is $\boxed{\textbf{(B)}\ \frac13}$.

Note: The last sentence is true because there are $3$ randomly-arrangeable numbers)

See Also

2009 AMC 8 (ProblemsAnswer KeyResources)
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

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