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.
A number or a short expression. Spacing, $ signs and \frac vs / are all fine.
Official solution
If an integer is divisible by 36, it must also be divisible by 9 since 9 is a factor of 36. It is a well-known fact that, if N is divisible by 9, the sum of the digits of N is a multiple of 9. Hence, if N contains all the even digits, the sum of the digits would be 0+2+4+6+8=20, which is not divisible by 9 and thus 36. The next logical try would be 8640, which happens to be divisible by 36. Thus, N=8640≡640(mod1000).
Source: NuminaMath-1.5,
licensed Apache-2.0.
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