2011 AMC 8 Problems/Problem 23
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 |
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