Difference between revisions of "2011 AMC 8 Problems/Problem 23"
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=22|num-a=24}} | {{AMC8 box|year=2011|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:38, 15 July 2024
Contents
Problem
How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?
Solution 1
We can separate this into two cases. If an integer is a multiple of 
the last digit must be either 
or 
Case 1: The last digit is The leading digit can be 
or 
Because the second digit can be but not the leading digit, there are also 
choices. The third digit cannot be the leading digit or the second digit, so there are 
choices. The number of integers is this case is 
Case 2: The last digit is 
Because 
is the largest digit, one of the remaining three digits must be There are ways to choose which digit should be The remaining digits can be or 
but since they have to be different there are 
ways to choose. The number of integers in this case is 
Therefore, the answer is 
.
Video Solution
https://youtu.be/OOdK-nOzaII?t=48
See Also
| 2011 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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. 
