Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2007 AMC 8 Problems/Problem 24"

(→‎Solution)
(→‎Solution 2)
Line 10: Line 10:
 
The number of ways to form a 3-digit number is <math>4 \cdot 3 \cdot 2 = 24</math>. The combination of digits that give us multiples of 3 are (1,2,3) and (2,3,4), as the integers in the subsets have a sum which is divisible by 3. The number of 3-digit numbers that contain these numbers is <math>3! + 3! = 12</math>.  
 
The number of ways to form a 3-digit number is <math>4 \cdot 3 \cdot 2 = 24</math>. The combination of digits that give us multiples of 3 are (1,2,3) and (2,3,4), as the integers in the subsets have a sum which is divisible by 3. The number of 3-digit numbers that contain these numbers is <math>3! + 3! = 12</math>.  
 
Therefore, the probability is <math>\frac{12}{24} = \boxed{\frac{1}{2}}</math>.
 
Therefore, the probability is <math>\frac{12}{24} = \boxed{\frac{1}{2}}</math>.
 +
 +
~abc2142
  
 
==Video Solution==
 
==Video Solution==

Revision as of 22:14, 20 August 2021

Problem

A bag contains four pieces of paper, each labeled with one of the digits $1$, $2$, $3$ or $4$, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3$? $\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4}$

Solution

The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4). The number of ways to choose three digits out of four is 4. Therefore, the probability is $\boxed{\textbf{(C)}\ \frac{1}{2}}$.

Solution 2

The number of ways to form a 3-digit number is $4 \cdot 3 \cdot 2 = 24$. The combination of digits that give us multiples of 3 are (1,2,3) and (2,3,4), as the integers in the subsets have a sum which is divisible by 3. The number of 3-digit numbers that contain these numbers is $3! + 3! = 12$. Therefore, the probability is $\frac{12}{24} = \boxed{\frac{1}{2}}$.

~abc2142

Video Solution

https://youtu.be/hwc11K02cEc - Happytwin

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_24&oldid=160673"