Library / /62 of 740

, 2018

Combinatorics Difficulty 4.5 AIME Prove it United States

Problem:

Find the number of eight-digit positive integers that are multiples of $9$ and have all distinct digits.

Solution

Solution:

Note that $0+1+\cdots+9=45$ . Consider the two unused digits, which must then add up to $9$ . If it's $0$ and $9$ , there are $8 \cdot 7!$ ways to finish; otherwise, each of the other four pairs give $7 \cdot 7!$ ways to finish, since $0$ cannot be the first digit. This gives a total of $36 \cdot 7! = 181440$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.