Difference between revisions of "2009 AMC 8 Problems/Problem 13"

(Solution 1)
(Undo revision 160260 by Raina0708 (talk) LaTeX makes the solution look neat and professional. I PM'ed Raina0708 and asked the user NOT to remove the LaTeX. I will undo this change.)
(Tag: Undo)
 
Line 10: Line 10:
  
 
==Solution 1==
 
==Solution 1==
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 1/3
+
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==
 
==Solution 2==

Latest revision as of 15:32, 14 August 2021

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 1

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 $\boxed{\textbf{(B)}\ \frac13}$.

Solution 2

The number is divisible by 5 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png