Difference between revisions of "2009 AMC 8 Problems/Problem 13"
(→Solution 1) |
MRENTHUSIASM (talk | contribs) (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 | + | 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 16:32, 14 August 2021
Contents
Problem
A three-digit integer contains one of each of the digits , , and . What is the probability that the integer is divisible by ?
Solution 1
The three digit numbers are . The numbers that end in are divisible are , and the probability of choosing those numbers is .
Solution 2
The number is divisible by 5 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.