2010 AIME II 第 1 题

Since $36 = 4 \cdot 9,$ the digit sum must be divisible by $9$ and the last two digits must form a multiple of $4$

Distinct even digits come from $\{0, 2, 4, 6, 8\},$ so the digit sum is even and hence must be $18.$ Use the largest such digits.

Since $36 = 4 \cdot 9,$ the number $N$ must be divisible by both $4$ and $9.$ Its digits are distinct members of $\{0, 2, 4, 6, 8\},$ whose total is $20,$ so $N$ cannot use all five. The digit sum must be a multiple of $9,$ and being even it must be $18;$ the only such digit sets are $\{4, 6, 8\}$ and $\{0, 4, 6, 8\}.$

The largest number formed from $\{0, 4, 6, 8\}$ is $8640,$ which ends in $40,$ a multiple of $4.$ So $N = 8640,$ and the remainder upon division by $1000$ is $640.$