Difference between revisions of "2007 AMC 8 Problems/Problem 24"
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A bag contains four pieces of paper, each labeled with one of the digits <math>1</math>, <math>2</math>, <math>3</math> or <math>4</math>, 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 <math>3</math>? | A bag contains four pieces of paper, each labeled with one of the digits <math>1</math>, <math>2</math>, <math>3</math> or <math>4</math>, 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 <math>3</math>? | ||
<math> \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} </math> | <math> \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} </math> | ||
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==Solution 2== | ==Solution 2== |
Revision as of 13:47, 19 November 2023
Problem
A bag contains four pieces of paper, each labeled with one of the digits ,
,
or
, 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
?
Solution 2
The number of ways to form a 3-digit number is . 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
.
Therefore, the probability is
.
~abc2142
Video Solution
https://youtu.be/hwc11K02cEc - Happytwin
Video Solution 2
https://youtu.be/cUxFS9l-Pb4 - Soo, DRMS, NM
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.