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Difference between revisions of "2011 AMC 8 Problems/Problem 23"

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Therefore, the answer is <math>48+36=\boxed{\textbf{(D)}\ 84}</math>
 
Therefore, the answer is <math>48+36=\boxed{\textbf{(D)}\ 84}</math>
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==Unofficial Alternate Solution==
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We make four cases based off where the multiple of <math>5</math> is.
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Case 1: The first digit can't be <math>0</math>, so it must be <math>5</math>. There are <math>4\cdot3</math> to choose the other
  
 
==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}}

Revision as of 17:15, 7 August 2020

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?

$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108$

Solution

We can separate this into two cases. If an integer is a multiple of $5,$ the last digit must be either $0$ or $5.$

Case 1: The last digit is $5.$ The leading digit can be $1,2,3,$ or $4.$ Because the second digit can be $0$ but not the leading digit, there are also $4$ choices. The third digit cannot be the leading digit or the second digit, so there are $3$ choices. The number of integers is this case is $4\cdot4\cdot3\cdot1=48.$

Case 2: The last digit is $0.$ Because $5$ is the largest digit, one of the remaining three digits must be $5.$ There are $3$ ways to choose which digit should be $5.$ The remaining digits can be $1,2,3,$ or $4,$ but since they have to be different there are $4\cdot3$ ways to choose. The number of integers in this case is $1\cdot3\cdot4\cdot3=36.$

Therefore, the answer is $48+36=\boxed{\textbf{(D)}\ 84}$

Unofficial Alternate Solution

We make four cases based off where the multiple of $5$ is.

Case 1: The first digit can't be $0$, so it must be $5$. There are $4\cdot3$ to choose the other

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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