Question

Let $N$ be the greatest integer multiple of $36$ all of whose digits are even and no two of whose digits are the same. Find the remainder when $N$ is divided by $1000.$

Answer

Solution:

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 $640.$

2010 AIME II Problem 1

Solution:

Problem 1 in Other Years