2016/17 AHSMC Part 1 Problem 8

Since $7$, $8$ and $9$ are pairwise relatively prime, the six-digit number must be a multiple of $7 \times 8 \times 9 = 504$. When $523999$ is divided by $504$, the remainder is $343$. To get a multiple of $504$, we subtract $343$ from $999$ to obtain $656$. This is one of the answers, and we have $523656 = 7 \times 8 \times 9 \times 1039$. We can get another answer by subtracting $504$ from $656$ to obtain $152$, and we have $523152 = 7 \times 8 \times 9 \times 1038$. This subtraction cannot be repeated without reducing the difference below $523000$. Hence $656$ and $152$ are the only possible answers.

Alternative Solution: The six digit number should have the form $504n$ where $n$ is a positive integer. The conditions of the problem lead to the inequality $523100 \le 504n \le 523999$ or equivalently $1038 \le n \le 1039$, and hence $n = 1038$ or $n = 1039$. With these two values of $n$, one obtains two three-digit numbers having the requested properties. The answer is (C).