2009 AMC 8 Problems/Problem 13: Difference between revisions
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A three-digit integer contains one of each of the digits <math> 1</math>, <math> 3</math>, and <math> 5</math>. What is the probability that the integer is divisible by <math> 5</math>? | A three-digit integer contains one of each of the digits <math> 1</math>, <math> 3</math>, and <math> 5</math>. What is the probability that the integer is divisible by <math> 5</math>? | ||
<math> \textbf{(A)}\ \frac{1}{6} \qquad | <math> \textbf{(A)}\ \frac{1}{6} \qquad | ||
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\textbf{(D)}\ \frac{2}{3} \qquad | \textbf{(D)}\ \frac{2}{3} \qquad | ||
\textbf{(E)}\ \frac{5}{6}</math> | \textbf{(E)}\ \frac{5}{6}</math> | ||
==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>. | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=12|num-a=14}} | {{AMC8 box|year=2009|num-b=12|num-a=14}} | ||
Revision as of 15:58, 25 December 2012
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 
.
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 | ||
