Difference between revisions of "2007 AMC 8 Problems/Problem 24"
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==Solution 2== | ==Solution 2== | ||
Only (1,2,3) and (2,3,4) are possible combinations that sum to 3. The total ways to choose three numbers from the four numbers is 4 choose 3 which equals 4. The answer is <math>\frac{2}{4} = \boxed{\frac{1}{2}}</math>. | Only (1,2,3) and (2,3,4) are possible combinations that sum to 3. The total ways to choose three numbers from the four numbers is 4 choose 3 which equals 4. The answer is <math>\frac{2}{4} = \boxed{\frac{1}{2}}</math>. | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=23|num-a=25}} | {{AMC8 box|year=2007|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:06, 18 June 2024
Contents
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 1
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
Solution 2
Only (1,2,3) and (2,3,4) are possible combinations that sum to 3. The total ways to choose three numbers from the four numbers is 4 choose 3 which equals 4. The answer is 
.
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. 
