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

Revision as of 10:36, 15 July 2024

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 $48+36=\boxed{\textbf{(D)}\ 84}$ .

Video Solution

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

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

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

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.

@@ Line 13: / Line 13: @@
 Therefore, the answer is <math>48+36=\boxed{\textbf{(D)}\ 84}</math>.
-==Solution 2==
-We make four cases based off where the multiple of <math>5</math> digit (<math>0</math> or <math>5</math>) is. The number has to end with either <math>5</math> or <math>0</math> since it's a multiple of <math>5</math>. In all but the last case, the <math>5</math> and <math>0</math> are used at the end and in another spot which separates the cases.
-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 middle two digits. After that, the last digit has to be <math>0</math>, so there are a total of <math>1\cdot4\cdot3\cdot1=12</math> numbers.
-Case 2: The second digit can be <math>0</math> or <math>5</math>, leaving <math>2</math> choices. The first and third numbers can be chosen in <math>4\cdot3</math> ways, like last time. The last digit has to be <math>0</math> or <math>5</math>, but not the one we already used. There are a total of <math>4\cdot2\cdot3\cdot1=24</math> numbers.
-Case 3: There are the same choices, but the digits <math>0</math> and <math>5</math> are at the last and second-to-last spots. So there are <math>4\cdot3\cdot2\cdot1=24</math> numbers again.
-Case 4: There are <math>4\cdot3\cdot2</math> ways to choose the first three numbers. There has to be a <math>5</math> in the number because the largest digit is <math>5</math>. Coincidentally, there are <math>4\cdot3\cdot2\cdot1</math> numbers again.
-There are a total of <math>12 + 24 \cdot 3=\boxed{\textbf{(D)}\ 84}</math> numbers.
-<math>\footnotesize{\text{Alternate solution by Sotowa}}</math>
 ==Video Solution==

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

Revision as of 10:36, 15 July 2024

Contents

Problem

Solution 1

Video Solution

See Also