Difference between revisions of "2011 AMC 8 Problems/Problem 23"
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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? | 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? | ||
<math> \textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108 </math> | <math> \textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108 </math> | ||
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==Solution== | ==Solution== | ||
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<math>\footnotesize{\text{Alternate solution by Sotowa}}</math> | <math>\footnotesize{\text{Alternate solution by Sotowa}}</math> | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/OOdK-nOzaII?t=48 | ||
==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 21:10, 17 April 2021
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
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
Unofficial Alternate Solution
We make four cases based off where the multiple of digit ( or ) is. The number has to end with either or since it's a multiple of . In all but the last case, the and are used at the end and in another spot which separates the cases.
Case 1: The first digit can't be , so it must be . There are to choose the middle two digits. After that, the last digit has to be , so there are a total of numbers.
Case 2: The second digit can be or , leaving choices. The first and third numbers can be chosen in ways, like last time. The last digit has to be or , but not the one we already used. There are a total of numbers.
Case 3: There are the same choices, but the digits and are at the last and second-to-last spots. So there are numbers again.
Case 4: There are ways to choose the first three numbers. There has to be a in the number because the largest digit is . Coincidentally, there are numbers again.
There are a total of numbers.
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.