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2011 AMC 8 Problems/Problem 23

Revision as of 16:06, 3 November 2023 by Iceskate1 (talk | contribs) (→‎Solution 1)

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?

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

Solution 1

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 $5$ digit ($0$ or $5$) is. The number has to end with either $5$ or $0$ since it's a multiple of $5$. In all but the last case, the $5$ and $0$ are used at the end and in another spot which separates the cases.

Case 1: The first digit can't be $0$, so it must be $5$. There are $4\cdot3$ to choose the middle two digits. After that, the last digit has to be $0$, so there are a total of $1\cdot4\cdot3\cdot1=12$ numbers.

Case 2: The second digit can be $0$ or $5$, leaving $2$ choices. The first and third numbers can be chosen in $4\cdot3$ ways, like last time. The last digit has to be $0$ or $5$, but not the one we already used. There are a total of $4\cdot2\cdot3\cdot1=24$ numbers.

Case 3: There are the same choices, but the digits $0$ and $5$ are at the last and second-to-last spots. So there are $4\cdot3\cdot2\cdot1=24$ numbers again.

Case 4: There are $4\cdot3\cdot2$ ways to choose the first three numbers. There has to be a $5$ in the number because the largest digit is $5$. Coincidentally, there are $4\cdot3\cdot2\cdot1$ numbers again.

There are a total of $12 + 24 \cdot 3=\boxed{\textbf{(D)}\ 84}$ numbers.

$\footnotesize{\text{Alternate solution by Sotowa}}$

Video Solution

https://youtu.be/OOdK-nOzaII?t=48

https://youtu.be/QWJ-zWaLtzY ~David

https://youtu.be/4Teb7HbmARc ~savannahsolver

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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