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

Revision as of 22:42, 15 October 2023

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

Video Solution 2

https://youtu.be/cUxFS9l-Pb4 - Soo, DRMS, NM

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.

@@ Line 16: / Line 16: @@
 https://youtu.be/hwc11K02cEc - Happytwin
+==Video Solution 2==
+https://youtu.be/cUxFS9l-Pb4 - Soo, DRMS, NM
 ==See Also==
 {{AMC8 box|year=2007|num-b=23|num-a=25}}
 {{MAA Notice}}

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