Olympiad Maths Prep

Track / Stage 3 / 98 of 260 #98 of 2000

Problem 98

AMC 10/12, early questions
Combinatorics Difficulty 3.3 Find the answer

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

Official solution

We can separate this into two cases. If an integer is a multiple of 5,5, the last digit must be either 00 or 5.5.
Case 1: The last digit is 5.5. The leading digit can be 1,2,3,1,2,3, or 4.4. Because the second digit can be 00 but not the leading digit, there are also 44 choices. The third digit cannot be the leading digit or the second digit, so there are 33 choices. The number of integers is this case is 4⋅4⋅3⋅1=48.4\cdot4\cdot3\cdot1=48.
Case 2: The last digit is 0.0. Because 55 is the largest digit, one of the remaining three digits must be 5.5. There are 33 ways to choose which digit should be 5.5. The remaining digits can be 1,2,3,1,2,3, or 4,4, but since they have to be different there are 4⋅34\cdot3 ways to choose. The number of integers in this case is 1⋅3⋅4⋅3=36.1\cdot3\cdot4\cdot3=36.
Therefore, the answer is 48+36=(D) 8448+36=\boxed{\textbf{(D)}\ 84}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.