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

Revision as of 14:05, 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 1/3

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)

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.

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 ==Solution 1==
-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 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
 ==Solution 2==

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

Revision as of 14:05, 14 August 2021

Contents

Problem

Solution 1

Solution 2

See Also